Automorphic Forms on GL(2)^{†}

H.Jacquet and R. P. Langlands

† Appeared as vol. 114 of Springer Lecture Notes in Math., Springer–Verlag, Berlin–Heidelberg, New York, 1970, pp. 1–548

Table of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii

Chapter I: Local Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 § 1. Weil representations . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 § 2. Representations of GL(2, F) in the non-archimedean case . . . . . . . . . . 13 § 3. The principal series for non-archimedean fields . . . . . . . . . . . . . . . 48 § 4. Examples of absolutely cuspidal representations . . . . . . . . . . . . . . 65 § 5. Representations of GL(2, ℝ) . . . . . . . . . . . . . . . . . . . . . . . 80 § 6. Representation of GL(2, ℂ) . . . . . . . . . . . . . . . . . . . . . . . 115 § 7. Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 § 8. Odds and ends . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

Chapter II: Global Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 157 § 9. The global Hecke algebra . . . . . . . . . . . . . . . . . . . . . . . . 157 §10. Automorphic forms . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 §11. Hecke theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 §12. Some extraordinary representations . . . . . . . . . . . . . . . . . . . . 209

Chapter III: Quaternion Algebras . . . . . . . . . . . . . . . . . . . . . . . 222 §13. Zeta-functions for M(2, F) . . . . . . . . . . . . . . . . . . . . . . . 222 §14. Automorphic forms and quaternion algebras . . . . . . . . . . . . . . . . 245 §15. Some orthogonality relations . . . . . . . . . . . . . . . . . . . . . . . 253 §16. An application of the Selberg trace formula . . . . . . . . . . . . . . . . 267

Introduction

Two of the best known of Hecke’s achievements are his theory of L-functions with grössencharakter, which are Dirichlet series which can be represented by Euler products, and his theory of the Euler products, associated to automorphic forms on GL(2). Since a grössencharakter is an automorphic form on GL(1) one is tempted to ask if the Euler products associated to automorphic forms on GL(2) play a role in the theory of numbers similar to that played by the L-functions with grössencharakter. In particular do they bear the same relation to the Artin L-functions associated to two-dimensional representations of a Galois group as the Hecke L-functions bear to the Artin Lfunctions associated to one-dimensional representations? Although we cannot answer the question definitively one of the principal purposes of these notes is to provide some evidence that the answer is affirmative.

The evidence is presented in §12. It come from reexamining, along lines suggested by a recent paper of Weil, the original work of Hecke. Anything novel in our reexamination comes from our point of view which is the theory of group representations. Unfortunately the facts which we need from the representation theory of GL(2) do not seem to be in the literature so we have to review, in Chapter I, the representation theory of GL(2, F) when F is a local field. §7 is an exceptional paragraph. It is not used in the Hecke theory but in the chapter on automorphic forms and quaternion algebras.

Chapter I is long and tedious but there is nothing hard in it. Nonetheless it is necessary and anyone who really wants to understand L-functions should take at least the results seriously for they are very suggestive.

§9 and §10 are preparatory to the Hecke theory which is finally taken up in §11. We would like to stress, since it may not be apparent, that our method is that of Hecke. In particular the principal tool is the Mellin transform. The success of this method for GL(2) is related to the equality of the dimensions of a Cartan subgroup and the unipotent radical of a Borel subgroup of P GL(2). The implication is that our methods do not generalize. The results, with the exception of the converse theorem in the Hecke theory, may.

The right way to establish the functional equation for the Dirichlet series associated to the automorphic forms is probably that of Tate. In §13 we verify, essentially, that this method leads to the same local factors as that of Hecke and in §14 we use the method of Tate to prove the functional equation for the L-functions associated to automorphic forms on the multiplicative group of a quaternion algebra. The results of §13 suggest a relation between the characters of representations of GL(2) and the characters of representations of the multiplicative group of a quaternion algebra which is verified, using the results of §13, in §15. This relation was well-known for archimedean fields but its significance had not been stressed. Although our proof leaves something to be desired the result itself seems to us to be one of the more striking facts brought out in these notes.

Both §15 and §16 are after thoughts; we did not discover the results in them until the rest of the notes were almost complete. The arguments of §16 are only sketched and we ourselves have not verified all the details. However the theorem of §16 is important and its proof is such a beautiful illustration of the power and ultimate simplicity of the Selberg trace formula and the theory of harmonic analysis on semi-simple groups that we could not resist adding it. Although we are very dissatisfied with the methods of the first fifteen paragraphs we see no way to improve on those of §16. They are perhaps the methods with which to attack the question left unsettled in §12.

We hope to publish a sequel to these notes which will include, among other things, a detailed proof of the theorem of §16 as well as a discussion of its implications for number theory. The theorem has, as these things go, a fairly long history. As far as we know the first forms of it were assertions about the representability of automorphic forms by theta series associated to quaternary quadratic forms.

As we said before nothing in these notes is really new. We have, in the list of references at the end of each chapter, tried to indicate our indebtedness to other authors. We could not however acknowledge completely our indebtedness to R. Godement since many of his ideas were communicated orally to one of us as a student. We hope that he does not object to the company they are forced to keep.

The notes^{∗} were typed by the secretaries of Leet Oliver Hall. The bulk of the work was done by Miss Mary Ellen Peters and to her we would like to extend our special thanks. Only time can tell if the mathematics justifies her great efforts.

New York, N.Y. August, 1969 New Haven, Conn.

∗ that appeared in the SLM volume

Chapter I: Local Theory

§1 Weil representations. Before beginning the study of automorphic forms we must review the representation theory of the general linear group in two variables over a local field. In particular we have to prove the existence of various series of representations. One of the quickest methods of doing this is to make use of the representations constructed by Weil in [1]. We begin by reviewing his construction adding, at appropriate places, some remarks which will be needed later.

In this paragraph F will be a local field and K will be an algebra over F of one of the following types:

(i) The direct sum F ⊕ F.

(ii) A separable quadratic extension of F. (iii) The unique quaternion algebra over F. K is then a division algebra with centre F. (iv) The algebra M(2, F) of 2 × 2 matrices over F.

In all cases we identify F with the subfield of K consisting of scalar multiples of the identity. In particular if K = F ⊕ F we identify F with the set of elements of the form (x, x). We can introduce an involution ι of K, which will send x to x^{ι}, with the following properties:

(i) It satisfies the identities (x + y)^{ι} = x^{ι} + y^{ι} and (xy)^{ι} = y^{ι}x^{ι}.

(ii) If x belongs to F then x = x^{ι}. (iii) For any x in K both τ(x) = x + x^{ι} and ν(x) = xx^{ι} = x^{ι}x belong to F.

If K = F ⊕ F and x = (a, b) we set x^{ι} = (b, a). If K is a separable quadratic extension of F the involution ι is the unique non-trivial automorphism of K over F. In this case τ(x) is the trace of x and ν(x) is the norm of x. If K is a quaternion algebra a unique ι with the required properties is known to exist. τ and ν are the reduced trace and reduced norm respectively. If K is M(2, F) we take ι to be the involution sending

x = ( \begin{matrix}a & b \\ c & d \end{matrix} )

to

x = ( \begin{matrix}d & −b \\ −c & a \end{matrix} )

Then τ(x) and ν(x) are the trace and determinant of x.

If ψ = ψ_{F} is a given non-trivial additive character of F then ψ_{K} = ψ_{F} ∘ τ is a non-trivial additive character of K. By means of the pairing

⟨x, y⟩ = ψ_{K}(xy)

we can identify K with its Pontrjagin dual. The function ν is of course a quadratic form on K which is a vector space over F and f = ψ_{F} ∘ ν is a character of second order in the sense of [1]. Since

ν(x + y) − ν(x) − ν(y) = τ(xy^{ι})

and

f(x + y)f^{−1}(x)f^{−1}(y) = ⟨x, y^{ι}⟩

the isomorphism of K with itself associated to f is just ι. In particular ν and f are nondegenerate.

Let 𝒮(K) be the space of Schwartz-Bruhat functions on K. There is a unique Haar measure dx on K such that if Φ belongs to 𝒮(K) and

Φ′(x) = ∫_{K} Φ(y) ψ_{K}(xy) dy

then

Φ(0) = ∫_{K} Φ′(x) dx.

The measure dx, which is the measure on K that we shall use, is said to be self-dual with respect to ψ_{K}.

Since the involution ι is measure preserving the corollary to Weil’s Theorem 2 can in the present case be formulated as follows.

Lemma 1.1. There is a constant γ which depends on the ψ_{F} and K, such that for every function Φ in 𝒮(K)

∫_{K}(Φ ∗ f)(y) ψ_{K}(yx) dy = γf^{−1}(x^{ι}) Φ′(x)

Φ ∗ f is the convolution of Φ and f. The values of γ are listed in the next lemma.

Lemma 1.2 (i) If K = F ⊕ F or M(2, F) then γ = 1.

(ii) If K is the quaternion algebra over F then γ = −1. (iii) If F = ℝ, K = ℂ, and

ψ_{F} (x) = e^{2πiax},

then

γ = \frac{a}{|a|} i

(iv) If F is non-archimedean and K is a separable quadratic extension of F let ω be the quadratic character of F^{∗} associated to K by local class-field theory. If U_{F} is the group of units of F^{∗} let m = m(ω) be the smallest non-negative integer such that ω is trivial on

U_{F}^{m} = {a ∈ U_{F} | α ≡ 1 (mod 𝔭_{F}^{m})}

and let n = n(ψ_{F}) be the largest integer such that ψ_{F} is trivial on the ideal 𝔭_{F}^{−n}. If a is any generator on the ideal 𝔭_{F}^{m+n} then

γ = ω(a)\frac{∫_{U_{F}} ω^{−1}(α) ψ_{F} (αa^{−1}) dα}{|∫_{U_{F}} ω^{−1}(α) ψ_{F} (αa^{−1}) dα|}.

The first two assertions are proved by Weil. To obtain the third apply the previous lemma to the function

Φ(z) = e^{−2πzz^{ι}}.

We prove the last. It is shown by Weil that |γ| = 1 and that if ℓ is sufficiently large γ differs from

∫_{𝔭_{K}^{−ℓ}} ψ_{F} (xx^{ι}) dx

by a positive factor. This equals

∫_{𝔭_{K}^{−ℓ}} ψ_{F} (xx^{ι}) |x|_{K} d^{×}x = ∫_{𝔭_{K}^{−ℓ}} ψ_{F} (xx^{ι})|xx^{ι}|_{F} d^{×}x

if d^{×}x is a suitable multiplicative Haar measure. Since the kernel of the homomorphism ν is compact the integral on the right is a positive multiple of

∫_{ν(𝔭_{K}^{−ℓ})} ψ_{F} (x) |x|_{F} d^{×}x.

Set k = 2ℓ if K/F is unramified and set k = ℓ if K/F is ramified. Then ν(𝔭_{K}^{−ℓ}) = 𝔭_{F}^{−k} ∩ν(K). Since 1 + ω is twice the characteristic function of ν(K^{×}) the factor γ is the positive multiple of

∫_{𝔭_{F}^{−k}} ψ_{F} (x) dx + ∫_{𝔭_{F}^{−k}} ψ_{F} (x) ω(x) dx.

For ℓ and therefore k sufficiently large the first integral is 0. If K/F is ramified well-known properties of Gaussian sums allow us to infer that the second integral is equal to

∫_{U_{F}} ψ_{F} (\frac{α}{a}) ω (\frac{α}{a}) dα.

Since ω = ω^{−1} we obtain the desired expression for γ by dividing this integral by its absolute value. If K/F is unramified we write the second integral as

∑_{j=0}^{∞}(−1)^{j−k} {∫_{𝔭_{F}^{−k+j}} ψ_{F} (x) dx − ∫_{𝔭_{F}^{−k+j+1}} ψ_{F} (x) dx}

In this case m = 0 and

∫_{𝔭_{F}^{−k+j}} ψ_{F} (x) dx

is 0 if k − j > n but equals q^{k−j} if k − j ≤ n, where q is the number of elements in the residue class field. Since ω(a) = (−1)^{n} the sum equals

ω(a) \{q^{m} + ∑_{j=0}^{∞}(−1)^{j}q^{m−j} (1 − \frac{1}{q})\}

A little algebra shows that this equals \frac{2ω(a)q^{m+1}}{q+1} so that γ = ω(a), which upon careful inspection is seen to equal the expression given in the lemma.

In the notation of [19] the third and fourth assertions could be formulated as an equality

γ = λ(K/F, ψ_{F}).

It is probably best at the moment to take this as the definition of λ(K/F, ψ_{F}).

If K is not a separable quadratic extension of F we take ω to be the trivial character.

Proposition 1.3 There is a unique representation r of SL(2, F) on 𝒮(K) such that

(i) r( ( \begin{matrix}α & 0 \\ 0 & α^{−1} \end{matrix} ) )Φ(x) = ω(α) |α|_{K}^{1/2}Φ(αx)

(ii) r( ( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ) )Φ(x) = ψ_{F} (zν(x))Φ(x)

(iii) r( ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) )Φ(x) = γΦ′(x^{ι}).

If 𝒮(K) is given its usual topology, r is continuous. It can be extended to a unitary representation of SL(2, F) on L^{2}(K), the space of square integrable functions on K. If F is archimedean and Φ belongs to 𝒮(K) then the function r(g)Φ is an indefinitely differentiable function on SL(2, F) with values in 𝒮_{(}K).

This may be deduced from the results of Weil. We sketch a proof. SL(2, F) is the group

generated by the elements ( \begin{matrix}α & 0 \\ 0 & α^{−1} \end{matrix} ), ( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ), and w = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) with α in F^{×} and z in F

subject to the relations

(a)  w ( \begin{matrix}α & 0 \\ 0 & α^{−1} \end{matrix} ) = ( \begin{matrix}α^{−1} & 0 \\ 0 & α \end{matrix} ) w

(b)  w^{2} = ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} )

(c)  w ( \begin{matrix}1 & a \\ 0 & 1 \end{matrix} ) w = ( \begin{matrix}−a^{−1} & 0 \\ 0 & −a \end{matrix} ) ( \begin{matrix}1 & −a \\ 0 & 1 \end{matrix} ) w ( \begin{matrix}1 & −a^{−1} \\ 0 & 1 \end{matrix} )

together with the obvious relations among the elements of the form ( \begin{matrix}α & 0 \\ 0 & α^{−1} \end{matrix} ) and ( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ).

Thus the uniqueness of r is clear. To prove the existence one has to verify that the mapping specified by (i), (ii), (iii) preserves all relations between the generators. For all relations except (a), (b), and (c) this can be seen by inspection. (a) translates into an easily verifiable property of the Fourier transform. (b) translates into the equality γ^{2} = ω(−1) which follows readily from Lemma 1.2.

If a = 1 the relation (c) becomes

∫_{K} Φ′(y^{ι}) ψ_{F} (ν(y))⟨y, x^{ι}⟩ dy = γψ_{F} (−ν(x)) ∫_{K} Φ(y)ψ_{F} (−ν(y))⟨y, −x^{ι}⟩ dy  (1.3.1)

which can be obtained from the formula of Lemma 1.1 by replacing Φ(y) by Φ′(−y^{ι}) and taking the inverse Fourier transform of the right side. If a is not 1 the relation (c) can again be reduced to (1.3.1) provided ψ_{F} is replaced by the character x → ψ_{F} (ax) and γ and dx are modifed accordingly. We refer to Weil’s paper for the proof that r is continuous and may be extended to a unitary representation of SL(2, F) in L^{2}(K).

Now take F archimedean. It is enough to show that all of the functions r(g)Φ are indefinitely differentiable in some neighborhood of the identity. Let

N_{F} = \{( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) | x ∈ F \}

and let

A_{F} = \{( \begin{matrix}α & 0 \\ 0 & α^{−1} \end{matrix} ) | α ∈ F^{×}\}

Then N_{F} wA_{F} N_{F} is a neighborhood of the identity which is diffeomorphic to N_{F} × A_{F} × N_{F}. It is enough to show that

ϕ(n, a, n_{1}) = r(nwan)Φ

is infinitely differentiable as a function of n, as a function of a, and as a function of n_{1} and that the derivations are continuous on the product space. For this it is enough to show that for all Φ all derivatives of r(n)Φ and r(a)Φ are continuous as functions of n and Φ or a and Φ. This is easily done.

The representation r depends on the choice of ψ_{F}. If a belongs to F^{×} and ψ′_{F} (x) = ψ_{F} (ax) let r′ be the corresponding representation. The constant γ′ = ω(a)γ.

Lemma 1.4 (i) The representation r′ is given by

r′(g) = r (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g ( \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} ))

(ii) If b belongs to K^{∗} let λ(b)Φ(x) = Φ(b^{−1}x) and let ρ(b)Φ(x) = Φ(xb). If a = ν(b) then

r′(g)λ(b^{−1}) = λ(b^{−1})r(g)

and

r′(g)ρ(b) = ρ(b)r(g).

In particular if ν(b) = 1 both λ(b) and ρ(b) commute with r.

We leave the verification of this lemma to the reader. Take K to be a separable quadratic extension of F or a quaternion algebra of centre F. In the first case ν(K^{×}) is of index 2 in F^{×}. In the second case ν(K^{×}) is F^{×} if F is non-archimedean and ν(K^{×}) has index 2 in F^{×} if F is ℝ.

Let K′ be the compact subgroup of K^{×} consisting of all x with ν(x) = xx^{ι} = 1 and let G_{+} be the subgroup of GL(2, F) consisting of all g with determinant in ν(K^{×}). G_{+} has index 2 or 1 in GL(2, F). Using the lemma we shall decompose r with respect to K′ and extend r to a representation of G_{+}.

Let Ω be a finite-dimensional irreducible representation of K^{×} in a vector space U over ℂ. Taking the tensor product of r with the trivial representation of SL(2, F) on U we obtain a representation on

𝒮(K) ⊗_{ℂ} U = 𝒮(K, U)

which we still call r and which will now be the centre of attention.

Proposition 1.5 (i) If 𝒮(K, Ω) is the space of functions Φ in 𝒮(K, U) satisfying

Φ(xh) = Ω^{−1}(h)Φ(x)

for all h in K′ then 𝒮(K, Ω) is invariant under r(g) for all g in SL(2, F).

(ii) The representation r of SL(2, F) on 𝒮(K, Ω) can be extended to a representation r_{Ω} of G_{+} satisfying

r_{Ω}( ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) )Φ(x) = |h|_{K}^{1/2}Ω(h)Φ(xh)

if a = ν(h) belongs to ν(K^{×}). (iii) If η is the quasi-character of F^{×} such that

Ω(a) = η(a)I

for a in F^{×} then

r_{Ω}( ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) ) = ω(a) η(a)I

(iv) The representation r_{Ω} is continuous and if F is archimedean all factors in 𝒮(K, Ω) are

infinitely differentiable.

(v) If U is a Hilbert space and Ω is unitary let L^{2}(K, U) be the space of square integrable functions from K to U with the norm

∥Φ∥^{2} = ∫ ∥Φ(x)∥^{2} dx

If L^{2}(K, Ω) is the closure of 𝒮(K, Ω) in L^{2}(K, U) then r_{Ω} can be extended to a unitary representation of G_{+} in L^{2}(K, Ω).

The first part of the proposition is a consequence of the previous lemma. Let H be the group of matrices of the form

( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} )

with a in ν(K^{×}). It is clear that the formula of part (ii) defines a continuous representation of H on 𝒮(K, Ω). Moreover G_{+} is the semi-direct of H and SL(2, F) so that to prove (ii) we have only to show that

r_{Ω} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g ( \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} )) = r_{Ω} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} )) r_{Ω}(g) r_{Ω} (( \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} ))

Let a = ν(h) and let r′ be the representation associated ψ′_{F} (x) = ψ_{F} (ax). By the first part of the previous lemma this relation reduces to

r′_{Ω}(g) = ρ(h) r_{Ω}(g) ρ^{−1}(h),

which is a consequence of the last part of the previous lemma.

To prove (iii) observe that

( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) = ( \begin{matrix}a^{2} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} )

Chapter 1 7 and that a^{2} = ν(a) belongs to ν(K^{×}). The last two assertions are easily proved.

We now insert some remarks whose significance will not be clear until we begin to discuss the local functional equations. We associate to every Φ in 𝒮(K, Ω) a function

W_{Φ}(g) = r_{Ω}(g) Φ(1)  (1.5.1)

on G_{+} and a function

φ_{Φ}(a) = W_{Φ} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ))  (1.5.2)

on ν(K^{×}). The both take values in U.

It is easily verified that

W_{Φ} (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = ψ_{F} (x)W_{Φ}(g)

If g ∈ G_{+} and F is a function on G_{+} let ρ(g)F be the function h → F (hg). Then

ρ(g) W_{Φ} = W_{r_{Ω}}(g)Φ

Let B_{+} be the group of matrices of the form

( \begin{matrix}a & x \\ 0 & 1 \end{matrix} )

with a in ν(K^{×}). Let ξ be the representation of B_{+} on the space of functions on ν(K^{×}) with values in U defined by

ξ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} )) φ(b) = φ(ba)

and

ξ (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) φ(b) = ψ_{F} (bx) φ(b).

Then for all b in B_{+}

ξ(b)φ_{Φ} = φ_{r_{Ω}}(b)Φ.  (1.5.3)

The application Φ → φ_{Φ}, and therefore the application Φ → W_{Φ}, is injective because

φ_{Φ}(ν(h)) = |h|_{K}^{1/2} Ω(h) Φ(h).  (1.5.4)

Thus we may regard r_{Ω} as acting on the space V of functions φ_{Φ}, Φ ∈ 𝒮(K, Ω). The effect of a

matrix in B_{+} is given by (1.5.3). The matrix ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) corresponds to the operator ω(a) η(a)I.

Since G_{+} is generated by B_{+}, the set of scalar matrices, and w = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) the representation r_{Ω}

on V is determined by the action of w. To specify this we introduce, formally at first, the Mellin transform of φ = φ_{Φ}.

If μ is a quasi-character of F^{×} let

φ̂(μ) = ∫_{ν(K^{×})} φ(α) μ(α) d^{×}α.  (1.5.5)

Appealing to (1.5.4) we may write this as

φ̂_{Φ}(μ) = φ̂(μ) = ∫_{K^{×}} |h|_{K}^{1/2}μ(ν(h)) Ω(h) Φ(h) d^{×}h.  (1.5.6)

If λ is a quasi-character of F^{×} we sometimes write λ for the associated quasi-character λ ∘ ν of K^{×}. The tensor product λ ⊗ Ω of λ and Ω is defined by

(λ ⊗ Ω)(h) = λ(h) Ω(h).

If α_{K} : h → |h|_{K} is the module of K then

α_{K}^{1/2}μ ⊗ Ω(h) = |h|_{K}^{1/2} μ(ν(h)) Ω(h).

We also introduce, again in a purely formal manner, the integrals

Z(Ω, Φ) = ∫_{K^{×}} Ω(h) Φ(h) d^{×}h

and

Z(Ω^{−1}, Φ) = ∫_{K^{×}} Ω^{−1}(h) Φ(h) d^{×}h

so that

φ̂(μ) = Z(μα_{K}^{1/2} ⊗ Ω, Φ).  (1.5.7)

Now let φ′ = φ_{r_{Ω}(w)Φ} and let Φ′ be the Fourier transform of Φ so that r_{Ω}(w) Φ(x) = γΦ′(x^{ι}). If μ_{0} = ωη

φ′ (μ^{−1}μ_{0}^{−1}) = Z (μ^{−1}μ_{0}^{−1}α_{K}^{1/2} ⊗ Ω, r_{Ω}(w)Φ)

which equals

γ ∫_{K} μ^{−1}μ_{0}^{−1}(ν(h))Ω(h) Φ′(h^{ι}) d^{×}h.

Since μ_{0}(ν(h)) = η(ν(h)) = Ω(h^{ι}h) = Ω(h^{ι})Ω(h) this expression equals

γ ∫_{K} μ^{−1}(ν(h))Ω^{−1}(h^{ι}) Φ′(h^{ι}) d^{×}h = γ ∫_{K} μ^{−1}(ν(h))Ω^{−1}(h) Φ′(h) d^{×}h

so that

φ′(μ^{−1}μ_{0}^{−1}) = γZ (μ^{−1}α_{K}^{1/2} ⊗ Ω^{−1}, Φ′) .  (1.5.8)

Take μ = μ_{1}α_{F}^{s} where μ_{1} is a fixed quasi-character and s is complex number. If K is a separable quadratic extension of F the representation Ω is one-dimensional and therefore a quasicharacter. The integral defining the function

Z(μα_{K}^{1/2} ⊗ Ω, Φ)

is known to converge for Re s sufficiently large and the function itself is essentially a local zetafunction in the sense of Tate. The integral defining

Z(μ^{−1}α_{K}^{1/2} ⊗ Ω^{−1}, Φ′)

converges for Re s sufficiently small, that is, large and negative. Both functions can be analytically continued to the whole s-plane as meromorphic functions. There is a scalar C(μ) which depends analytically on s such that

Z(μα_{K}^{1/2} ⊗ Ω, Φ) = C(μ)Z(μ^{−1}α_{K}^{1/2} ⊗ Ω^{−1}, Φ′).

All these assertions are also known to be valid for quaternion algebras. We shall return to the verification later. The relation

φ̂(μ) = γ^{−1}C(μ)φ′(μ^{−1}μ_{0}^{−1})

determines φ′ in terms of φ.

If λ is a quasi-character of F^{×} and Ω_{1} = λ ⊗ Ω then 𝒮(K, Ω_{1}) = 𝒮(K, Ω) and

r_{Ω_{1}}(g) = λ(detg)r_{Ω}(g)

so that we may write

r_{Ω_{1}} = λ ⊗ r_{Ω}

However the space V_{1} of functions on ν(K^{×}) associated to r_{Ω_{1}} is not necessarily V. In fact

V_{1} = {λ_{φ} | φ ∈ V }

and r_{Ω_{1}}(g) applied to λ_{φ} is the product of λ(detg) with the function λ · r_{Ω}(g)_{φ}. Given Ω one can always find a λ such that λ ⊗ Ω is equivalent to a unitary representation.

If Ω is unitary the map Φ → φ_{Φ} is an isometry because

∫_{ν(K^{×})} ∥φ_{Φ}(a)∥^{2} d^{×}a = ∫_{K^{×}} ∥Ω(h) Φ(h)∥^{2}|h|_{K} d^{×}h = ∫_{K} ∥Φ(h)∥^{2} dh

if the measures are suitably normalized.

We want to extend some of these results to the case K = F ⊕ F. We regard the element of K as defining a row vector so that K becomes a right module for M(2, F). If Φ belongs to 𝒮(K) and g belongs to GL(2, F), we set

ρ(g) Φ(x) = Φ(xg).

Proposition 1.6 (i) If K = F ⊕ F then r can be extended to a representation r of GL(2, F) such that

r (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} )) Φ = ρ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} )) Φ

for a in F^{×}.

(ii) If Φ̃ is the partial Fourier transform

Φ̃(a, b) = ∫_{F} Φ(a, y) ψ_{F} (by) dy

and the Haar measure dy is self-dual with repsect to ψ_{F} then

[r(g)Φ̃] = ρ(g)Φ̃

for all Φ in 𝒮(K) and all g in G_{F}.

It is easy to prove part (ii) for g in SL(2, F). In fact one has just to check it for the standard generators and for these it is a consequence of the definitions of Proposition 1.3. The formula of part (ii) therefore defines an extension of r to GL(2, F) which is easily seen to satisfy the condition of part (i).

Let Ω be a quasi-character of K^{×}. Since K^{×} = F^{×} × F^{×} we may identify Ω with a pair (ω_{1}, ω_{2}) of quasi-characters of F^{×}. Then r_{Ω} will be the representation defined by

r_{Ω}(g) = |detg|_{F}^{1/2}ω_{1}(detg)r(g).

If x belongs to K^{×} and ν(x) = 1 then x is of the form (t, t^{−1}) with t in F^{×}. If Φ belongs to 𝒮(K) set

θ(Ω, Φ) = ∫_{F^{×}} Ω((t, t^{−1})) Φ((t, t^{−1})) d^{×}t.

Since the integrand has compact support on F^{×} the integral converges. We now associate to Φ the function

W_{Φ}(g) = θ(Ω, r_{Ω}(g)Φ)  (1.6.1)

on GL(2, F) and the function

φ_{Φ}(a) = W_{Φ} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ))  (1.6.2)

on F^{×}. We still have

ρ(g)W_{Φ} = W_{r_{Ω}}(g)Φ.

If

B_{F} = \{( \begin{matrix}a & x \\ 0 & 1 \end{matrix} ) | a ∈ F^{×}, x ∈ F \}

and if the representations ξ of B_{F} on the space of functions on F^{×} is defined in the same manner as the representation ξ of B_{+} then

ξ(b)φ_{Φ} = φ_{r_{Ω}(b)Φ}

for b in B_{F}. The applications Φ → W_{Φ} and Φ → φ_{Φ} are no longer injective.

If μ_{0} is the quasi-character defined by

μ_{0}(a) = Ω((a, a)) = ω_{1}(a) ω_{2}(a)

then

W_{Φ} (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = μ_{0}(a) W_{Φ}(g).

It is enough to verify this for g = e.

W_{Φ} (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} )) = θ (Ω, r_{Ω} (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} )) Φ)

Chapter 1 11 and

( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) = ( \begin{matrix}a^{2} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a^{−1} & 0 \\ 0 & a \end{matrix} )

so that

r_{Ω} (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} )) Φ(x, y) = |a^{2}|_{F}^{1/2}ω_{1}(a^{2})|a|_{K}^{−1/2} Φ(ax, a^{−1}y).

Consequently

\begin{aligned}W_{Φ} (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} )) & = ∫_{F^{×}} ω_{1}(a^{2})ω_{1}(x)ω_{2}^{−1}(x)Φ(ax, a^{−1}x^{−1}) d^{×}x \\ & = ω_{1}(a)ω_{2}(a) ∫_{F^{×}} ω_{1}(x)ω_{2}^{−1}(x)Φ(x, x^{−1}) d^{×}x \end{aligned}

which is the required result.

Again we introduce in a purely formal manner the distribution

Z(Ω, Φ) = Z(ω_{1}, ω_{2}Φ) = ∫ Φ(x_{1}, x_{2}) ω_{1}(x_{2}) ω_{2}(x_{2}) d^{×}x_{2} d^{×}x_{2}.

If μ is a quasi-character of F^{×} and φ = φ_{Φ} we set

φ̂(μ) = ∫_{F^{×}} φ(α) μ(α) d^{×}α.

The integral is

\begin{matrix}∫_{F^{×}} μ(α)θ (Ω, r_{Ω} (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} )) Φ) d^{×}α \\ = ∫_{F^{×}} μ(α) \{∫_{F^{×}} r_{Ω} (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} )) Φ(x, x^{−1})ω_{1}(x)ω_{2}^{−1}(x) d^{×}x\} d^{×}α \end{matrix}

which in turn equals

∫_{F^{×}} μ(α)ω_{1}(α)|α|_{F}^{1/2} {∫_{F^{×}} Φ(αx, x^{−1})ω_{1}(x)ω_{2}^{−1}(x) d^{×}x} d^{×}α.

Writing this as a double integral and then changing variables we obtain

∫_{F^{×}} ∫_{F^{×}} Φ(α, x)μω_{1}(α)μω_{2}(x)|αx|_{F}^{1/2} d^{×}α d^{×}x

so that

φ̂(μ) = Z (μω_{1}α_{F}^{1/2}, μω_{2}α_{F}^{1/2}, Φ) .  (1.6.3)
Let φ′ = φ_{r_{Ω}(w)Φ}. Then

φ′(μ^{−1}μ_{0}^{−1}) = Z (μ^{−1}ω_{2}^{−1}α_{F}^{1/2}, μ^{−1}ω_{1}^{−1}α_{F}^{1/2}, r_{Ω}(w)Φ)

which equals

∫ ∫ Φ′(y, x)μ^{−1}ω_{2}^{−1}(x)μ^{−1}ω_{1}^{−1}(y)|xy|_{F}^{1/2} d^{×}x d^{×}y

so that

φ′(μ^{−1}μ_{0}^{−1}) = Z(μ^{−1}ω_{1}^{−1}α_{F}^{1/2}, μ^{−1}ω_{2}^{−1}α_{F}^{1/2}, Φ′).  (1.6.4)

Suppose μ = μ_{1}α_{F}^{s} where μ_{1} is a fixed quasi-character and s is a complex number. We shall see that the integral defining the right side of (1.6.3) converges for Re s sufficiently large and that the integral defining the right side of (1.6.4) converges for Re s sufficiently small. Both can be analytically continued to the whole complex plane as meromorphic functions and there is a meromorphic function C(μ) which is independent of Φ such that

Z(μω_{1}α_{F}^{1/2}, μω_{2}α_{F}^{1/2}) = C(μ)Z(μ^{−1}ω_{1}^{−1}α_{F}^{1/2}, μ^{−1}ω_{2}^{−1}α_{F}^{1/2}, Φ′).

Thus

φ̂(μ) = C(μ)φ′(μ^{−1}μ_{0}^{−1})

The analogy with the earlier results is quite clear.

Chapter 1 13 §2 Representations of GL(2, F) in the non-archimedean case. In this and the next two paragraphs the ground field F is a non- archimedean local field. We shall be interested in representations π of G_{F} = GL(2, F) on a vector space V over ℂ which satisfy the following condition.

(2.1) For every vector v in V the stabilizer of v in G_{F} is an open subgroup of G_{F}.

Those who are familiar with such things can verify that this is tantamount to demanding that the map (g, v) → π(g)v of G_{F} × V into V is continuous if V is given the trivial locally convex topology in which every semi-norm is continuous. A representation of G_{F} satisfying (2.1) will be called admissible if it also satisfies the following condition

(2.2) For every open subgroup G′ of GL(2, O_{F}) the space of vectors v in V stablizied by G′ is finite-dimensional.

O_{F} is the ring of integers of F.

Let ℋ_{F} be the space of functions on G_{F} which are locally constant and compactly supported. Let dg be that Haar measure on G_{F} which assigns the measure 1 to GL(2, O_{F}). Every f in ℋ_{F} may be identified with the measure f(g) dg. The convolution product

f_{1} ∗ f_{2}(h) = ∫_{G_{F}} f_{1}(g) f_{2}(g^{−1}h) dg

turns ℋ_{F} into an algebra which we refer to as the Hecke algebra. Any locally constant function on GL(2, O_{F}) may be extended to G_{F} by being set equal to 0 outside of GL(2, O_{F}) and therefore may be regarded as an element of ℋ_{F}. In particular if π_{i}, 1 ≤ i ≤ r, is a family of inequivalent finite-dimensional irreducible representations of GL(2, O_{F}) and

ξ_{i}(g) = dim(π_{i}) tr π_{i}(g^{−1})

for g in GL(2, O_{F}) we regard ξ_{i} as an element of ℋ_{F}. The function

ξ = ∑_{i=1}^{r} ξ_{i}

is an idempotent of ℋ_{F}. Such an idempotent will be called elementary.

Let π be a representation satisfying (2.1). If f belongs to ℋ_{F} and v belongs to V then f(g) π(g)v takes on only finitely many values and the integral

∫_{G_{F}} f(g) π(g)v dg = π(f)v

may be defined as a finite sum. Alternatively we may give V the trivial locally convex topology and use some abstract definition of the integral. The result will be the same and f → π(f) is the representation of ℋ_{F} on V. If g belongs to G_{F} then λ(g)f is the function whose value at h is f(g^{−1}h). It is clear that

π(λ(g)f) = π(g) π(f).

Moreover Chapter 1 14 (2.3) For every v in V there is an f in ℋ_{F} such that πf(v) = v.

In fact f can be taken to be a multiple of the characteristic function of some open and closed neighborhood of the identity. If π is admissible the associated representation of ℋ_{F} satisfies

(2.4) For every elementary idempotent ξ of ℋ_{F} the operator π(ξ) has a finite-dimensional range.

We now verify that from a representation π of ℋ_{F} satisfying (2.3) we can construct a representation π of G_{F} satisfying (2.1) such that

π(f) = ∫_{G_{F}} f(g) π(g) dg.

By (2.3) every vector v in V is of the form

v = ∑_{i=1}^{r} π(f_{i}) v_{i}

with v_{i} in V and f_{i} in ℋ_{F}. If we can show that

∑_{i=1}^{r} π(f_{i}) v_{i} = 0  (2.3.1)

implies that

w = ∑_{i=1}^{r} π(λ(g)f_{i})v_{i}

is 0 we can define π(g)v to be

∑_{i=1}^{r} π(λ(g)f_{i})v_{i}

π will clearly be a representation of G_{F} satisfying (2.1).

Suppose that (2.3.1) is satisifed and choose f in ℋ_{F} so that π(f)w = w. Then

w = ∑_{i=1}^{r} π(f ∗ λ(g)f_{i})v_{i}.

If ρ(g)f(h) = f(hg) then

f ∗ λ(g)f_{i} = ρ(g^{−1})f ∗ f_{i}

so that

w = ∑_{i=1}^{r} π(ρ(g^{−1})f ∗ f_{i})v_{i} = π(ρ(g^{−1})f) {∑_{i=1}^{r} π(f_{i})v_{i}} = 0.

It is easy to see that the representation of G_{F} satisfies (2.2) if the representation of ℋ_{F} satisfies (2.4). A representation of ℋ_{F} satisfying (2.3) and (2.4) will be called admissible. There is a complete correspondence between admissible representations of G_{F} and of ℋ_{F}. For example a subspace is invariant under G_{F} if and only if it is invariant under H_{F} and an operator commutes with the action of G_{F} if and only if it commutes with the action of ℋ_{F}.

Chapter 1 15

¿From now on, unless the contrary is explicitly stated, an irreducible representation of G_{F} or ℋ_{F} is to be assumed admissible. If π is irreducible and acts on the space V then any linear transformation of V commuting with ℋ_{F} is a scalar. In fact if V is assumed, as it always will be, to be different from 0 there is an elementary idempotent ξ such that π(ξ) ≠ 0. Its range is a finite-dimensional space invariant under A. Thus A has at least one eigenvector and is consequently a scalar. In particular there is a homomorphism ω of F^{×} into ℂ such that

π (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} )) = ω(a)I

for all a in F^{×}. By (2.1) the function ω is 1 near the identity and is therefore continuous. We shall refer to a continuous homomorphism of a topological group into the multiplicative group of complex numbers as a quasi-character.

If χ is a quasi-character of F^{×} then g → χ(detg) is a quasi-character of G_{F}. It determines a one-dimensional representation of G_{F} which is admissible. It will be convenient to use the letter χ to denote this associated representation. If π is an admissible reprentation of G_{F} on V then χ ⊗ π will be the reprenentation of G_{F} on V defined by

(χ ⊗ π)(g) = χ(detg)π(g).

It is admissible and irreducible if π is.

Let π be an admissible representation of G_{F} on V and let V^{∗} be the space of all linear forms on V. We define a representation π^{∗} of ℋ_{F} on V^{∗} by the relation

⟨v, π^{∗}(f)v^{∗}⟩ = ⟨π(f̌)v, v^{∗}⟩

where f̌ν(g) = f(g^{−1}). Since π^{∗} will not usually be admissible, we replace V^{∗} by Ṽ = π^{∗}(ℋ_{F})V^{∗}. The space Ṽ is invariant under ℋ_{F}. For each f in ℋ_{F} there is an elementary idempotent ξ such that ξ ∗ f = f and therefore the restriction π̃ of π^{∗} to Ṽ satisfies (2.3). It is easily seen that if ξ is an elementary idempotent so is ξ̌. To show that π̃ is admissible we have to verify that

Ṽ(ξ) = π̃(ξ)Ṽ = π^{∗}(ξ)V^{∗}

is finite-dimensional. Let V (ξ̌) = π(ξ̌)V and let V_{c} = (1 − π(ξ̌))V. V is clearly the direct sum of V (ξ̌), which is finite-dimensional, and V_{c}. Moreover Ṽ(ξ) is orthogonal to V_{c} because

⟨v − π(ξ̌)v, π̃(ξ)ṽ⟩ = ⟨π(ξ̌)v − π(ξ̌)v, ṽ⟩ = 0.

It follows immediately that Ṽ(ξ) is isomorphic to a subspace of the dual of V (ξ̌) and is therefore finite-dimensional. It is in fact isomorphic to the dual of V (ξ̌) because if v^{∗} annihilates V_{c} then, for all v in V,

⟨v, π^{∗}(ξ)v^{∗}⟩ − ⟨v, v^{∗}⟩ = −⟨v − π(ξ̌)v, v^{∗}⟩ = 0

so that π^{∗}(ξ)v^{∗} = v^{∗}.

π̃ will be called the representation contragradient to π. It is easily seen that the natural map of V into Ṽ^{∗} is an isomorphism and that the image of this map is π^{∗}(ℋ_{F})Ṽ^{∗} so that π may be identified with the contragredient of π̃.

If V_{1} is an invariant subspace of V and V_{2} = V_{1} \ V we may associate to π representations π_{1} and π_{2} on V_{1} and V_{2}. They are easily seen to be admissible. It is also clear that there is a natural embedding of Ṽ_{2} in Ṽ. Moreover any element ṽ_{1} of Ṽ_{1} lies in Ṽ_{1}(ξ) for some ξ and therefore is determined by its effect on V_{1}(ξ̌). It annihilates (I − π(ξ̌))V_{1}. There is certainly a linear function ṽ on V which annihilates (I − π(ξ̌))V and agrees with Ṽ_{1} on V_{1}(ξ̌). ṽ is necessarily in Ṽ so that Ṽ_{1} may be identified with Ṽ_{2} \Ṽ. Since every representation is the contragredient of its contragredient we easily deduce the following lemma.

Chapter 1 16 Lemma 2.5 (a). Suppose V_{1} is an invariant subspace of V. If Ṽ_{2} is the annihilator of V_{1} in Ṽ then V_{1} is the annihilator of Ṽ_{2} in V.

(b) π is irreducible if and only if π̃ is.

Observe that for all g in G_{F}

⟨π(g)v, ṽ⟩ = ⟨v, π̃(g^{−1})ṽ⟩.

If π is the one-dimensional representation associated to the quasi-character χ then π̃ = χ^{−1}. Moreover if χ is a quasi-character and π any admissible representation then the contragredient of χ ⊗ π is χ^{−1} ⊗ π̃.

Let V be a separable complete locally convex space and π a continuous representation of G_{F} on V. The space V_{0} = π(ℋ_{F})V is invariant under G_{F} and the restriction π_{0} of π to V_{0} satisfies (2.1). Suppose that it also satisfies (2.2). Then if π is irreducible in the topological sense π_{0} is algebraically irreducible. To see this take any two vectors v and w in V_{0} and choose an elementary idempotent ξ so that π(ξ)v = v. v is in the closure of π(ℋ_{F})w and therefore in the closure of π(ℋ_{F})w∩π(ξ)V. Since, by assumption, π(ξ)V is finite dimensional, v must actually lie in π(ℋ_{F})w.

The equivalence class of π is not in general determined by that of π_{0}. It is, however, when π is unitary. To see this one has only to show that, up to a scalar factor, an irreducible admissible representation admits at most one invariant hermitian form.

Lemma 2.6 Suppose π_{1} and π_{2} are irreducible admissible representations of G_{F} on V_{1} and V_{2} respectively. Suppose A(v_{1}, v_{2}) and B(v_{1}, v_{2}) are non-degenerate forms on V_{1} × V_{2} which are linear in the first variable and either both linear or both conjugate linear in the second variable. Suppose moreover that, for all g in G_{F}

A(π_{1}(g)v_{1}, π_{2}(g)v_{2}) = A(v_{1}, v_{2})

and

B(π_{1}(g)v_{1}, π_{2}(g)v_{2}) = B(v_{1}, v_{2})

Then there is a complex scalar λ such that

B(v_{1}, v_{2}) = λA(v_{1}, v_{2})

Define two mappings S and T of V_{2} into Ṽ_{1} by the relations

A(v_{1}, v_{2}) = ⟨v_{1}, Sv_{2}⟩

and

B(v_{1}, v_{2}) = ⟨v_{1}, T v_{2}⟩,

Since S and T are both linear or conjugate linear with kernel 0 they are both embeddings. Both take V_{2} onto an invariant subspace of Ṽ_{1}. Since Ṽ_{1} has no non-trivial invariant subspaces they are both isomorphisms. Thus S^{−1}T is a linear map of V_{2} which commutes with G_{F} and is therefore a scalar λI. The lemma follows.

An admissible representation will be called unitary if it admits an invariant positive definite hermitian form.

We now begin in earnest the study of irreducible admissible representations of G_{F}. The basic ideas are due to Kirillov.

Chapter 1 17 Proposition 2.7. Let π be an irreducible admissible representation of G_{F} on the vector space V.

(a) If V is finite-dimensional then V is one-dimensional and there is a quasi-character χ of F^{×}

such that

π(g) = χ(detg)

(b) If V is infinite dimensional there is no nonzero vector invariant by all the matrices ({1 \atop 0} {x \atop 1}), x ∈ F.

If π is finite-dimensional its kernel H is an open subgroup. In particular there is a positive number ϵ such that

( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )

belongs to H if |x| < ϵ. If x is any element of F there is an a in F^{×} such that |ax| < ϵ. Since

( \begin{matrix}a^{−1} & 0 \\ 0 & a \end{matrix} ) ( \begin{matrix}1 & ax \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )

the matrix

( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )

belongs to H for all x in F. For similar reasons the matrices

( \begin{matrix}1 & 0 \\ y & 1 \end{matrix} )

do also. Since the matrices generate SL(2, F) the group H contains SL(2, F). Thus π(g_{1})π(g_{2}) = π(g_{2})π(g_{1}) for all g_{1} and g_{2} in G_{F}. Consequently each π(g) is a scalar matrix and π(g) is onedimensional. In fact

π(g) = χ(detg)I

where χ is a homorphism of F^{×} into ℂ^{×}. To see that χ is continuous we need only observe that

π (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} )) = χ(a)I.

Suppose V contains a nonzero vector v fixed by all the operators

π (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) .

Let H be the stabilizer of the space ℂv. To prove the second part of the proposition we need only verify that H is of finite index in G_{F}. Since it contains the scalar matrices and an open subgroup of G_{F} it will be enough to show that it contains SL(2, F). In fact we shall show that H_{0}, the stabilizer of v, contains SL(2, F). H_{0} is open and therefore contains a matrix

( \begin{matrix}a & b \\ c & d \end{matrix} )

with c ≠ 0. It also contains

( \begin{matrix}1 & −ac^{−1} \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}1 & −dc^{−1} \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}0 & b_{0} \\ c & 0 \end{matrix} ) = w_{0}.

If x = \frac{b_{0}}{c} y then

( \begin{matrix}1 & 0 \\ y & 1 \end{matrix} ) = w_{0} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) w_{0}^{−1}

also belongs to H_{0}. As before we see that H_{0} contains SL(2, F).

Because of this lemma we can confine our attention to infinite-dimensional representations. Let ψ = ψ_{F} be a nontrivial additive character of F. Let B_{F} be the group of matrices of the form

b = ( \begin{matrix}a & x \\ 0 & 1 \end{matrix} )

with a in F^{×} and x in F. If X is a complex vector space we define a representation ξ_{ψ} of B_{F} on the space of all functions of F^{×} with values in X by setting

(ξ_{ψ}(b)φ)(α) = ψ(αx)φ(αa).

ξ_{ψ} leaves the invariant space 𝒮(F^{×}, X) of locally constant compactly supported functions. ξ_{ψ} is continuous with respect to the trivial topology on 𝒮(F^{×}, X).

Proposition 2.8. Let π be an infinite dimensional irreducible representation of G_{F} on the space V. Let 𝔭 = 𝔭_{F} be the maximal ideal in the ring of integers of F, and let V′ be the set of all vectors v in V such that

∫_{𝔭^{−n}} ψ_{F} (−x)π (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) v dx = 0

for some integer n. Then (i) The set V′ is a subspace of V. (ii) Let X = V′ \ V and let A be the natural map of V onto X. If v belongs to V let φ_{v} be the

function defined by

φ_{v}(a) = A (π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v) .

The map v → φ_{v} is an injection of V into the space of locally constant functions on F^{×} with value in X.

(iii) If b belongs to B_{F} and v belongs to V then

φ_{π(b)v} = ξ_{ψ}(b)φ_{v}.

If m ≥ n so that 𝔭^{−m} contains 𝔭^{−n} then

∫_{𝔭^{−m}} ψ(−x)π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v dx

is equal to

∑_{y∈𝔭^{−m}/𝔭^{−n}} ψ(−y)π \begin{matrix}( \\ ( & \begin{matrix}1 & y \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix} ∫_{𝔭^{−n}} ψ(−x)π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v dx.

Thus if the integral of the lemma vanishes for some integer n it vanishes for all larger integers. The first assertion of the proposition follows immediately.

To prove the second we shall use the following lemma.

Chapter 1 19 Lemma 2.8.1 Let 𝔭^{−m} be the largest ideal on which ψ is trivial and let f be a locally constant function on 𝔭^{−ℓ} with values in some finite dimensional complex vector space. For any integer n ≤ ℓ the following two conditions are equivalent

(i) f is constant on the cosets of 𝔭^{−n} in 𝔭^{−ℓ} (ii) The integral

∫_{𝔭^{−ℓ}} ψ(−ax) f(x) dx

is zero for all a outside of 𝔭^{−m+n}.

Assume (i) and let a be an element of F^{×} which is not in 𝔭^{−m+n}. Then x → ψ(−ax) is a non-trivial character of 𝔭^{−n} and

∫_{𝔭^{−ℓ}} ψ(−ax) f(x) dx = ∑_{y∈𝔭^{−ℓ}/𝔭^{−n}} ψ(−ay) {∫_{𝔭^{−n}} ψ(−ax) dx} f(y) = 0.

f may be regarded as a locally constant function on F with support in 𝔭^{−ℓ}. Assuming (ii) is tantamount to assuming that the Fourier transform F′ of f has its support in 𝔭^{−m+n}. By the Fourier inversion formula

f(x) = ∫_{𝔭^{−m+n}} ψ(−xy) f′(y) dy.

If y belongs to 𝔭^{−m+n} the function x → ψ(−xy) is constant on cosets of 𝔭^{−n}. It follows immediately that the second condition of the lemma implies the first.

To prove the second assertion of the proposition we show that if φ_{v} vanishes identically then v is fixed by the operator π(({1 \atop 0} {x \atop 1})) for all x in F and then appeal to Proposition 2.7.

Take

f(x) = π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v.

The restriction of f to an ideal in F takes values in a finite-dimensional subspace of V. To show that f is constant on the cosets of some ideal 𝔭^{−n} it is enough to show that its restriction to some ideal 𝔭^{−ℓ} containing 𝔭^{−n} has this property.

By assumption there exists an n_{0} such that f is constant on the cosets of 𝔭^{−n_{0}}. We shall now show that if f is constant on the cosets of 𝔭^{−n+1} it is also constant on the cosets of 𝔭^{−n}. Take any ideal 𝔭^{−ℓ} containing 𝔭^{−n}. By the previous lemma

∫_{𝔭^{−ℓ}} ψ(−ax) f(x) dx = 0

if a is not in 𝔭^{−m+n−1}. We have to show that the integral on the left vanishes if a is a generator of 𝔭^{−m+n−1}.

If U_{F} is the group of units of O_{F} the ring of integers of F there is an open subgroup U_{1} of U_{F} such that

π \begin{aligned}( \\ ( & \begin{matrix}b & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v = v

for b in U_{1}. For such b

π \begin{aligned}( \\ ( & \begin{matrix}b & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}_{𝔭^{−ℓ}} ψ(−ax) f(x) dx = ∫_{𝔭^{−ℓ}} ψ(−ax)π \begin{aligned}( \\ ( & \begin{matrix}b & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v dx

is equal to

∫_{𝔭^{−ℓ}} ψ(−ax)π \begin{aligned}( \\ ( & \begin{matrix}1 & bx \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π \begin{aligned}( \\ ( & \begin{matrix}b & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v dx = ∫_{𝔭^{−ℓ}} ψ (−\frac{a}{b}x) f(x) dx.

Thus it will be enough to show that for some sufficiently large ℓ the integral vanishes when a is taken to be one of a fixed set of representatives of the cosets of U_{1} in the set of generators of 𝔭^{−m+n−1}. Since there are only finitely many such cosets it is enough to show that for each a there is at least one ℓ for which the integral vanishes.

By assumption there is an ideal 𝔞(a) such that

∫_{𝔞(a)} ψ(−x)π (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} )) v dx = 0

But this integral equals

|a|π \begin{matrix}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix} ∫_{a^{−1}𝔞(a)} ψ(−ax)π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v dx

so that ℓ = ℓ(a) could be chosen to make

𝔭^{−ℓ} = a^{−1}𝔞(a).

To prove the third assertion we verify that

A (π \begin{aligned}( \\ ( & \begin{matrix}1 & y \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v) = ψ(y) A(v)  (2.8.2)

for all v in V and all y in F. The third assertion follows from this by inspection. We have to show that

π \begin{aligned}( \\ ( & \begin{matrix}1 & y \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v − ψ(y)v

is in V′ or that, for some n,

∫_{𝔭^{−n}} ψ(−x)π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π \begin{aligned}( \\ ( & \begin{matrix}1 & y \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v dx − ∫_{𝔭^{−n}} ψ(−x) ψ(y)π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v dx

is zero. The expression equals

∫_{𝔭^{−n}} ψ(−x)π \begin{aligned}( \\ ( & \begin{matrix}1 & x + y \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v dx − ∫_{𝔭^{−n}} ψ(−x + y)π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v dx.

If 𝔭^{−n} contains y we may change the variables in the first integral to see that it equals the second.

It will be convenient now to identify v with φ_{v} so that V becomes a space of functions on F^{×} with values in X. The map A is replaced by the map φ → φ(1). The representation π now satisfies

π(b)φ = ξ_{ψ}(b)φ

if b is in B_{F}. There is a quasi-character ω_{0} of F^{×} such that

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω_{0}(a) I.

If

w = \begin{aligned}( \\ ( & \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} & ) \\ & & ) \end{aligned}

the representation is determined by ω_{0} and π(w).

Proposition 2.9 (i) The space V contains

V_{0} = 𝒮(F^{×}, X)

(ii) The space V is spanned by V_{0} and π(w)V_{0}.

For every φ in V there is a positive integer n such that

π \begin{aligned}( \\ ( & \begin{matrix}a & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ = φ

if x and a − 1 belong to 𝔭^{n}. In particular φ(αa) = φ(a) if α belongs to F^{×} and a − 1 belongs to 𝔭^{n}. The relation

ψ(αx)φ(α) = φ(α)

for all x in 𝔭^{n} implies that φ(α) = 0 if the restriction of ψ to α𝔭^{n} is not trivial. Let 𝔭^{−m} be the largest ideal on which ψ is trivial. Then φ(α) = 0 unless |α| ≤ |ϖ|^{−m−n} if ϖ is a generator of 𝔭.

Let V_{0} be the space of all ϖ in V such that, for some integer ℓ depending on φ, φ(α) = 0 unless |α| > |ϖ|^{ℓ}. To prove (i) we have to show that V_{0} = 𝒮(F^{×}, X). It is at least clear that 𝒮(F^{×}, X) contains V_{0}. Moreover for every φ in V and every x in F the difference

φ′ = φ − π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ

is in V_{0}. To see this observe that

φ′(α) = (1 − ψ(αx))φ(α)

is identically zero for x = 0 and otherwise vanishes at least on x^{−1}𝔭^{−m}. Since there is no function in V invariant under all the operators

π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

the space V_{0} is not 0.

Before continuing with the proof of the proposition we verify a lemma we shall need.

Lemma 2.9.1 The representation ξ_{ψ} of B_{F} in the space 𝒮(F^{×}) of locally constant, compactly supported, complex-valued functions on F^{×} is irreducible.

For every character μ of U_{F} let φ_{μ} be the function on F^{×} which equals μ on U_{F} and vanishes off U_{F}. Since these functions and their translates span 𝒮(F^{×}) it will be enough to show that any non-trivial invariant subspace contains all of them. Such a space must certainly contain some non-zero function φ which satisfies, for some character ν of U_{F}, the relation

φ(aϵ) = ν(ϵ) φ(a)

for all a in F^{×} and all ϵ in U_{F}. Replacing φ by a translate if necessary we may assume that φ(1) ≠ 0. We are going to show that the space contains φ_{μ} if μ is different from ν. Since U_{F} has at least two characters we can then replace φ by some φ_{μ} with μ different from ν, and replace ν by μ and μ by ν to see it also contains φ_{ν}.

Chapter 1 22

Set

φ′ = ∫_{U_{F}} μ^{−1}(ϵ)ξ_{ψ} \begin{aligned}( \\ ( & \begin{matrix}ϵ & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} ξ_{ψ} \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ dϵ

where x is still to be determined. μ is to be different form ν. φ′ belongs to the invariant subspace and

φ′(aϵ) = μ(ϵ)φ′(a)

for all a in F^{×} and all ϵ in U_{F}. We have

φ′(a) = φ(a) ∫_{U_{F}} μ^{−1}(ϵ)ν(ϵ)ψ(axϵ) dϵ

The character μ^{−1}ν has a conductor 𝔭^{n} with n positive. Take x to be of order −n − m. The integral, which can be rewritten as a Gaussian sum, is then, as is well-known, zero if a is not in U_{F} but different from zero if a is in U_{F}. Since φ(1) is not zero φ′ must be a nonzero multiple of φ_{μ}.

To prove the first assertion of the proposition we need only verify that if u belongs to X then V_{0} contains all functions of the form α → η(α)u with η in 𝒮(F^{×}). There is a φ in V such that φ(1) = u. Take x such that ψ(x) ≠ 1. Then

φ′ = φ − π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ

is in V_{0} and φ′(1) = (1 − ψ(x))u. Consequently every u is of the form φ(1) for some φ in V_{0}. If μ is a character of U_{F} let V_{0}(μ) be the space of functions φ in V_{0} satisfying

φ(aϵ) = μ(ϵ)φ(a)

for all a in F^{×} and all ϵ in U_{F}. V_{0} is clearly the direct sum of the space V_{0}(μ). In particular every vector u in X can be written as a finite sum

u = ∑ φ_{i}(1)

where φ_{i} belongs to some V_{0}(μ_{i}).

If we make use of the lemma we need only show that if u can be written as u = φ(1) where φ is in V_{0}(ν) for some ν then there is at least one function in V_{0} of the form α → η(α)u where η is a nonzero function in 𝒮(F^{×}). Choose μ different from ν and let 𝔭^{n} be the conductor of μ^{−1}ν. We again consider

φ′ = ∫_{U_{F}} μ^{−1}(ϵ)ξ_{ψ} (( \begin{matrix}ϵ & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) φ dϵ

where x is of order −n − m. Then

φ′(a) = φ(a) ∫_{U_{F}} μ^{−1}(ϵ)ν(ϵ)ψ_{F} (axϵ) dϵ

The properties of Gaussian sums used before show that φ′ is a function of the required kind.

Chapter 1 23

The second part of the proposition is easier to verify. Let P_{F} be the group of uppertriangular matrices in G_{F}. Since V_{0} is invariant under P_{F} and V is irreducible under G_{F} the space V is spanned by V_{0} and the vectors

φ′ = π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π(w)φ

with φ in V_{0}. But

φ′ = {φ′ − π(w)φ} + π(w)φ

and as we saw, φ′ − π(w)φ is in V_{0}. The proposition is proved.

To study the effect of w we introduce a formal Mellin transform. Let ϖ be a generator of 𝔭. If φ is a locally constant function on F^{×} with values in X then for every integer n the function ϵ → φ(ϵϖ^{n}) on U_{F} takes its values in a finite-dimensional subspace of X so that the integral

∫_{U_{F}} φ(ϵϖ^{n})ν(ϵ) = φ̂_{n}(ν)

is defined. In this integral we take the total measure of U_{F} to be 1. It is a vector in X. φ̂(ν, t) will be the Formal Laurent series

∑_{t} t^{n}φ̂_{n}(ν)

If φ is in V the series has only a finite number of terms with negative exponent. Moreover the series φ̂(ν, t) is different from zero for only finitely many ν. If φ belongs to V_{0} these series have only finitely many terms. It is clear that if φ is locally constant and all the formal series φ̂(ν, t) vanish then φ = 0.

Suppose φ takes values in a finite-dimensional subspace of X, ω is a quasi-character of F^{×}, and the integral

∫_{F^{×}} ω(a)φ(a) d^{×}a  (2.10.1)

is absolutely convergent. If ω′ is the restriction of ω to U_{F} this integral equals

∑_{n} z^{n} ∫_{U_{F}} φ(ϖ^{n}ϵ) ω′(ϵ) dϵ = ∑_{n} z^{n} φ̂_{n}(ω′)

if z = ω(ϖ). Consequently the formal series φ̂(ω′, t) converges absolutely for t = z and the sum is equal to (2.10.1). We shall see that X is one dimensional and that there is a constant c_{0} = c_{0}(φ) such that if |ω(ϖ)| = |ϖ|^{c} with c > c_{0} then the integral (2.10.1) is absolutely convergent. Consequently all the series φ̂(ν, t) have positive radii of convergence.

If ψ = ψ_{F} is a given non-trivial additive character of F, μ any character of U_{F}, and x any element of F we set

η(μ, x) = ∫_{U_{F}} μ(ϵ) ψ(ϵx) dϵ

The integral is taken with respect to the normalized Haar measure on U_{F}. If g belongs to G_{F}, φ belongs to V, and φ′ = π(g)φ we shall set

π(g) φ̂(ν, t) = φ′(ν, t).

Proposition 2.10 (i) If δ belongs to U_{F} and ℓ belongs to ℤ then

π \begin{aligned}( \\ ( & \begin{matrix}δϖ^{ℓ} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ̂(ν, t) = t^{−ℓ}ν^{−1}(δ) φ̂(ν, t)

(ii) If x belongs to F then

π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ̂(ν, t) = ∑_{n} t^{n} \{∑_{μ} η(μ^{−1}ν, ϖ^{n}x)φ̂_{n}(μ)\}

where the inner sum is taken over all characters of U_{F} (iii) Let ω_{0} be the quasi-character defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω_{0}(a) I

for a in F^{×}. Let ν_{0} be its restriction to U_{F} and let z_{0} = ω_{0}(ϖ). For each character ν of U_{F} there is a formal series C(ν, t) with coefficients in the space of linear operators on X such that for every φ in V_{0}

π \begin{aligned}( \\ ( & \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} & ) \\ & & ) \end{aligned} φ̂(ν, t) = C(ν, t) φ̂(ν^{−1}ν_{0}^{−1}, t^{−1}z_{0}^{−1}).

Set

φ′ = π \begin{aligned}( \\ ( & \begin{matrix}δϖ^{ℓ} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ.

Then

φ′(ν, t) = ∑_{n} t^{n} ∫_{U_{F}} ν(ϵ) φ(ϖ^{n+ℓ} δϵ) dϵ.

Changing variables in the integration and in the summation we obtain the first formula of the proposition.

Now set

φ′ = π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ.

Then

φ′(ν, t) = ∑_{n} t^{n} ∫_{U_{F}} ψ(ϖ^{n}ϵx) ν(ϵ) φ(ϖ^{n}ϵ) dϵ.

By Fourier inversion

φ(ϖ^{n}ϵ) = ∑_{μ} φ̂_{n}(μ) μ^{−1}(ϵ).

The sum on the right is in reality finite. Substituting we obtain

φ′(ν, t) = ∑_{n} t^{n} {∑_{μ} ∫_{U_{F}} μ^{−1}ν(ϵ) ψ(ϵϖ^{n}x) dϵ φ̂_{n}(μ)}

Chapter 1 25 as asserted.

Suppose ν is a character of U_{F} and φ in V_{0} is such that φ̂(μ, t) = 0 unless μ = ν^{−1}ν_{0}^{−1}. This means that

φ(aϵ) ≡ νν_{0}(ϵ) φ(a)

or that

π \begin{aligned}( \\ ( & \begin{matrix}ϵ & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ = νν_{0}(ϵ)φ

for all ϵ in U_{F}. If φ′ = π(w)φ then

π \begin{aligned}( \\ ( & \begin{matrix}ϵ & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ′ = π \begin{aligned}( \\ ( & \begin{matrix}ϵ & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π(w)φ = π(w)π \begin{aligned}( \\ ( & \begin{matrix}1 & 0 \\ 0 & ϵ \end{matrix} & ) \\ & & ) \end{aligned} φ.

Since

\begin{aligned}( \\ ( & \begin{matrix}1 & 0 \\ 0 & ϵ \end{matrix} & ) \\ & & ) \end{aligned} = \begin{aligned}( \\ ( & \begin{matrix}ϵ & 0 \\ 0 & ϵ \end{matrix} & ) \\ & & ) \end{aligned} \begin{aligned}( \\ ( & \begin{matrix}ϵ^{−1} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

the expression on the right is equal to

ν^{−1}(ϵ)π(w)φ = ν^{−1}(ϵ)φ′,

so that φ′(μ, t) = 0 unless μ = ν.

Now take a vector u in X and a character ν of U_{F} and let φ be the function in V_{0} which is zero outside of U_{F} and on U_{F} is given by

φ(ϵ) = ν(ϵ) ν_{0}(ϵ)u.  (2.10.2)

If φ′ = π(w)φ then φ′_{n} is a function of n, ν, and u which depends linearly on u and we may write

φ′_{n}(ν) = C_{n}(ν)u

where C_{n}(ν) is a linear operator on X.

We introduce the formal series

C(ν, t) = ∑ t^{n}C_{n}(ν).

We have now to verify the third formula of the proposition. Since φ is in V_{0} the product on the right is defined. Since both sides are linear in φ we need only verify it for a set of generators of V_{0}. This set can be taken to be the functions defined by (2.10.2) together with their translates of power ϖ. For functions of the form (2.10.2) the formula is valid because of the way the various series C(ν, t) were defined. Thus all we have to do is show that if the formula is valid for a given function φ it remains valid when φ is replaced by

π \begin{aligned}( \\ ( & \begin{matrix}ϖ^{ℓ} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ.

By part (i) the right side is replaced by

z_{0}^{ℓ}t^{ℓ}C(ν, t) φ̃(ν^{−1}ν_{0}^{−1}, t^{−1}z_{0}^{−1}).

Chapter 1 26 Since

π(w)π \begin{aligned}( \\ ( & \begin{matrix}ϖ^{ℓ} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ = π \begin{aligned}( \\ ( & \begin{matrix}1 & 0 \\ 0 & ϖ^{ℓ} \end{matrix} & ) \\ & & ) \end{aligned} π(w)φ

and π(w)φ̂(ν, t) is known we can use part (i) and the relation

( \begin{matrix}1 & 0 \\ 0 & ϖ^{ℓ} \end{matrix} ) = ( \begin{matrix}ϖ^{ℓ} & 0 \\ 0 & ϖ^{ℓ} \end{matrix} ) ( \begin{matrix}ϖ^{−ℓ} & 0 \\ 0 & 1 \end{matrix} )

to see that the left side is replaced by

z_{0}^{ℓ}t^{ℓ}π(w)φ̂(ν, t) = z_{0}^{ℓ}t^{ℓ}C(ν, t)φ̂(ν^{−1}ν_{0}^{−1}, t^{−1}z_{0}^{−1}).

For a given u in X and a given character ν of U_{F} there must exist a φ in V such that

φ̂(ν, t) = ∑ t^{n}C_{n}(ν)u

Consequently there is an n_{0} such that C_{n}(ν)u = 0 for n < n_{0}. Of course n_{0} may depend on u and ν. This observation together with standard properties of Gaussian sums shows that the infinite sums occurring in the following proposition are meaningful, for when each term is multiplied on the right by a fixed vector in X all but finitely many disappear.

Proposition 2.11 Let 𝔭^{−ℓ} be the largest ideal on which ψ is trivial.

(i) Let ν and ρ be two characters of U_{F} such that νρν_{0} is not 1. Let 𝔭^{m} be its conductor. Then

∑_{σ} η(σ^{−1}ν, ϖ^{n})η(σ^{−1}ρ, ϖ^{p})C_{p+n}(σ)

is equal to

η(ν^{−1}ρ^{−1}ν_{0}^{−1}, ϖ^{−m−ℓ})z_{0}^{m+ℓ}νρν_{0}(−1)C_{n−m−ℓ}(ν)C_{p−m−ℓ}(ρ)

for all integers n and p. (ii) Let ν be any character of U_{F} and let ν̃ = ν^{−1}ν_{0}^{−1}. Then

∑_{σ} η(σ^{−1}ν, ϖ^{n})η(σ^{−1}ν̃, ϖp)C_{p+n}(σ)

is equal to

z_{0}^{p}ν_{0}(−1)δ_{n,p} + (|ϖ| − 1)^{−1}z_{0}^{ℓ+1}C_{n−1−ℓ}(ν)C_{p−1−ℓ}(ν̃) − ∑_{−2−ℓ}^{−∞} z^{−r}C_{n+r}(ν)C_{p+r}(ν̃)

for all integers n and p.

The left hand sums are taken over all characters σ of U_{F} and δ_{n,p} is Kronecker’s delta. The relation

( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) = − ( \begin{matrix}1 & −1 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}1 & −1 \\ 0 & 1 \end{matrix} )

implies that

π(w)π \begin{aligned}( \\ ( & 1 1 0 1 & ) \\ & & ) \end{aligned} π(w)φ = ν_{0}(−1)π \begin{aligned}( \\ ( & \begin{matrix}1 & −1 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π(w)π \begin{aligned}( \\ ( & \begin{matrix}1 & −1 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ

for all φ in V_{0}. Since π(w)φ is not necessarily in V_{0} we write this relation as

π(w) \{π \begin{aligned}( \\ ( & 1 1 0 1 & ) \\ & & ) \end{aligned} π(w)φ − π(w)φ\} + π^{2}(w)φ = ν_{0}(−1)π \begin{aligned}( \\ ( & \begin{matrix}1 & −1 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π(w)π \begin{aligned}( \\ ( & \begin{matrix}1 & −1 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ.

The term π^{2}(w)φ is equal to ν_{0}(−1)φ.

We compute the Mellin transforms of both sides

π \begin{aligned}( \\ ( & \begin{matrix}1 & −1 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ̂(ν, t) = ∑_{n} t^{n} \{∑_{ρ} η(ρ^{−1}ν, −ϖ^{n})φ̂_{n}(ρ)\}

and

π(w)π \begin{aligned}( \\ ( & \begin{matrix}1 & −1 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ̂(ν, t) = ∑_{n} t^{n} ∑_{p,ρ} η(ρ^{−1}ν^{−1}ν_{0}^{−1}, −ϖ^{p})z_{0}^{−P} C_{p+n}(ν)φ̂_{p}(ρ)

so that the Mellin transform of the right side is

ν_{0}(−1) ∑_{n} t^{n} ∑_{p,ρ,σ} η(σ^{−1}ν, −ϖ^{n})η(ρ^{−1}σ^{−1}ν_{0}^{−1}, −ϖ^{p})z_{0}^{−p}C_{p+n}(σ)φ̂_{p}(ρ).  (2.11.1)

On the other hand

π(w)φ̂(ν, t) = ∑_{n} t^{n} ∑_{p} z_{0}^{−p}C_{p+n}(ν)φ̂_{p}(ν^{−1}ν_{0}^{−1})

and

π \begin{aligned}( \\ ( & 1 1 0 1 & ) \\ & & ) \end{aligned} π(w)φ̂(ν, t) = ∑_{n} t^{n} ∑_{p,ρ} z_{0}^{−p}η(ρ^{−1}ν, ϖ^{n})C_{p+n}(ρ)φ̂_{p}(ρ^{−1}ν_{0}^{−1})

so that

π \begin{aligned}( \\ ( & 1 1 0 1 & ) \\ & & ) \end{aligned} π(w)φ̂(ν, t) − π(w)φ̂(ν, t)

is equal to

∑_{n} t^{n} ∑_{p,ρ} z_{0}^{−p}[η(ρνν_{0}, ϖ^{n}) − δ(ρνν_{0})]C_{p+n}(ρ^{−1}ν_{0}^{−1})φ̂_{p}(ρ).

Here δ(ρνν_{0}) is 1 if ρνν_{0} is the trivial character and 0 otherwise. The Mellin transform of the left hand side is therefore

∑ t^{n} ∑_{p,r,ρ} z_{0}^{−p−r}[η(ρν^{−1}, ϖ^{r})−δ(ρν^{−1})]C_{n+r}(ν)C_{p+r}(ρ^{−1}ν_{0}^{−1})φ̂_{p}(ρ)+ν_{0}(−1) ∑ t^{n}φ̂_{n}(ν). (2.11.2)

The coefficient of t^{n}φ̂_{p}(ρ) in (2.11.1) is
ν_{0}(−1) ∑_{σ} η(σ^{−1}ν, −ϖ^{n}) η(ρ^{−1}σ^{−1}ν_{0}^{−1}, −ϖ^{p})z_{0}^{−1}C_{p+n}(σ)  (2.11.3)

and in (2.11.2) it is

∑_{r}[η(ρν^{−1}, ϖ^{r}) − δ(ρν^{−1})]z_{0}^{−p−r}C_{n+r}(ν)C_{p+r}(ρ^{−1}ν_{0}^{−1}) + ν_{0}(−1)δ_{n,ρ}δ(ρν^{−1})I  (2.11.4)

These two expressions are equal for all choice of n, p, ρ, and ν.

If ρ ≠ ν and the conductor of νρ^{−1} is 𝔭^{m} the gaussian sum η(ρν^{−1}, ϖ^{r}) is zero unless r = −m − ℓ. Thus (2.11.4) reduces to

η(ρν^{−1}, ϖ^{−m−ℓ})z_{0}^{−p−m−ℓ}C_{n−m−ℓ}(ν)C_{p−m−ℓ}(ρ^{−1}ν_{0}^{−1}).

Since

η(μ, −x) = μ(−1) η(μ, x)

the expression (2.11.3) is equal to

ρ^{−1}ν(−1) ∑_{σ} η(σ^{−1}ν, ϖ^{n}) η(ρ^{−1}σ^{−1}ν_{0}^{−1}ϖ^{p})z_{0}^{−p}C_{p+n}(σ).

Replacing ρ by ρ^{−1}ν_{0}^{−1} we obtain the first part of the proposition.

If ρ = ν then δ(ρν^{−1}) = 1. Moreover, as is well-known and easily verified, η(ρν^{−1}, ϖ^{r}) = 1 if r ≥ −ℓ,

η(ρν^{−1}, ϖ^{−ℓ−1}) = |ϖ|(|ϖ| − 1)^{−1}

and η(ρν^{−1}, ϖ^{r}) = 0 if r ≤ −ℓ − 2. Thus (2.11.4) is equal to

ν_{0}(−1)δ_{n,p}I+(|ϖ|−1)^{−1}z_{0}^{−p+ℓ+1}C_{n−ℓ−1}(ν)C_{n−ℓ−1}(ν^{−1}ν_{0}^{−1})− ∑_{r=−ℓ−2}^{−∞} z_{0}^{−p−r}C_{n+r}(ν)C_{n+r}(ν^{−1}ν_{0}^{−1}).

The second part of the proposition follows.

Proposition(2.12) (i) For every n, p, ν and ρ

C_{n}(ν)C_{p}(ρ) = C_{p}(ρ)C_{n}(ν)

(ii) There is no non-trivial subspace of X invariant under all the operators C_{n}(ν). (iii) The space X is one-dimensional.

Suppose ρνν_{0} ≠ 1. The left side of the first identity in the previous proposition is symmetric in the two pairs (n, ν) and (p, ρ). Since (η^{−1}ρ^{−1}ν_{0}^{−1}, ϖ^{−m−ℓ}) is not zero we conclude that

C_{n−m−ℓ}(ν) C_{p−m−ℓ}(ρ) = C_{p−m−ℓ}(ρ) C_{n−m−ℓ}(ν)

for all choices of n and p. The first part of the proposition is therefore valid in ρ ≠ ν̃.

Now suppose ρ = ν̃. We are going to that if (p, n) is a given pair of integers and u belongs to X then

C_{n+r}(ν)C_{p+r}(ν̃)u = C_{p+r}(ν̃)C_{n+r}(ν)u

for all r in ℤ. If r ≪ 0 both sides are 0 and the relation is valid so the proof can proceed by induction on r. For the induction one uses the second relation of Proposition 2.11 in the same way as the first was used above.

Chapter 1 29

Suppose X_{1} is a non-trivial subspace of X invariant under all the operators C_{n}(ν). Let V_{1} be the space of all functions in V_{0} which take values in X_{1} and let V′_{1} be the invariant subspace generated by V_{1}. We shall show that all functions in V′_{1} take values in X_{1} so that V′_{1} is a non-trivial invariant subspace of V. This will be a contradiction. If φ in V takes value in X_{1} and g belongs to P_{F} then π(g)φ also takes values in X_{1}. Therefore all we need to do is show that if φ is in V_{1} then π(w)φ takes values in X_{1}. This follows immediately from the assumption and Proposition 2.10.

To prove (iii) we show that the operators C_{n}(ν) are all scalar multiples of the identity. Because of (i) we need only show that every linear transformation of X which commutes with all the operators C_{n}(ν) is a scalar. Suppose T is such an operator. If φ belongs to V let T_{φ} be the function from F^{×} to X defined by

T φ(a) = T (φ(a)).

Observe that T φ is still in V. This is clear if φ belongs to V_{0} and if φ = π(w)φ_{0} we see on examining the Mellin transforms of both sides that

T φ = π(w)T φ_{0}.

Since V = V_{0} + π(w)V_{0} the observation follows. T therefore defines a linear transformation of V which clearly commutes with the action of any g in P_{F}. If we can show that it commutes with the action of w it will follow that it and, therefore, the original operator on X are scalars. We have to verify that

π(w)T φ = T π(w)φ

at least for φ on V_{0} and for φ = π(w)φ_{0} with φ_{0} in V_{0}. We have already seen that the identity holds for φ in V_{0}. Thus if φ = π(w)φ_{0} the left side is

π(w)T π(w)φ_{0} = π^{2}(w)T φ_{0} = ν_{0}(−1)T φ_{0}

and the right side is

T π^{2}(w)φ_{0} = ν_{0}(−1)T φ_{0}.

Because of this proposition we can identify X with ℂ and regard the operators C_{n}(ν) as complex numbers. For each r the formal Laurent series C(ν, t) has only finitely many negative terms. We now want to show that the realization of π on a space of functions on F^{×} is, when certain simple conditions are imposed, unique so that the series C(ν, t) are determined by the class of π and that conversely the series C(ν, t) determine the class of π.

Theorem 2.13 Suppose an equivalence class of infinite-dimensional irreducible admissible representations of G_{F} is given. Then there exists exactly one space V of complex-valued functions on F^{×} and exactly one representation π of G_{F} on V which is in this class and which is such that

π(b)φ = ξ_{ψ}(b)φ

if b is in B_{F} and φ is in V.

We have proved the existence of one such V and π. Suppose V′ is another such space of functions and π′ a representation of G_{F} on V′ which is equivalent to π. We suppose of course that

π′(b)φ = ξ_{ψ}(b)φ

Chapter 1 30 if b is in B_{F} and φ is in V′. Let A be an isomorphism of V with V′ such that Aπ(g) = π′(g)A for all g. Let L be the linear functional

L(φ) = Aφ(1)

on V. Then

L (π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ) = Aφ(a)

so that A is determined by L. If we could prove the existence of a scalar λ such that L(φ) = λφ(1) it would follow that

Aφ(a) = λφ(a)

for all a such that Aφ = λφ. This equality of course implies the theorem.

Observe that

L (π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ) = π′ \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} Aφ(1) = ψ(x)L(φ).  (2.13.1)

Thus we need the following lemma.

Lemma 2.13.2 If L is a linear functional on V satisfying (2.13.1) there is a scalar λ such that

L(φ) = λφ(1).

This is a consequence of a slightly different lemma.

Lemma 2.13.3 Suppose L is a linear functional on the space 𝒮(F^{×}) of locally constant compactly supported functions on F^{×} such that

L (ξ_{ψ} \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ) = ψ(x) L(φ)

for all φ in 𝒮(F^{×}) and all x in F. Then there is a scalar λ such that L(φ) = λφ(1).

Suppose for a moment that the second lemma is true. Then given a linear functional L on V satisfying (2.13.1) there is a λ such that L(φ) = λφ(1) for all φ in V_{0} = 𝒮(F^{×}). Take x in F such that ψ(x) ≠ 1 and φ in V. Then

L(φ) = L (φ − π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ) + L (π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ) .

Since

φ − π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ

is in V_{0} the right side is equal to

λφ(1) − λψ(x)φ(1) + ψ(x)L(φ)

so that

(1 − ψ(x)) L(φ) = λ(1 − ψ(x)) φ(1)

which implies that L(φ) = λφ(1).

Chapter 1 31

To prove the second lemma we have only to show that φ(1) = 0 implies L(φ) = 0. If we set φ(0) = 0 then φ becomes a locally constant function with compact support in F. Let φ′ be its Fourier transform so that

φ(a) = ∫_{F} ψ(ba) φ′(−b) db.

Let Ω be an open compact subset of F^{×} containing 1 and the support of φ. There is an ideal 𝔞 in F so that for all a in Ω the function φ′(−b)ψ(ba) is constant on the cosets of 𝔞 in F. Choose an ideal 𝔟 containing 𝔞 and the support of φ′. If S is a set of representatives of 𝔟/𝔞 and if c is the measure of 𝔞 then

φ(a) = c ∑_{b∈S} ψ(ba)φ′(−b).

If φ_{0} is the characteristic function of Ω this relation may be written

φ = ∑_{b∈S} λ_{b}ξ_{ψ} \begin{aligned}( \\ ( & \begin{matrix}1 & b \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ_{0}

with λ_{b} = cφ′(−b). If φ(1) = 0 then

∑_{b∈S} λ_{b}ψ(b) = 0

so that

φ = ∑ λ_{b} \{ξ_{ψ} \begin{aligned}( \\ ( & \begin{matrix}1 & b \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ_{0} − ψ(b)φ_{0}\}

It is clear that L(φ) = 0.

The representation of the theorem will be called the Kirillov model. There is another model which will be used extensively. It is called the Whittaker model. Its properties are described in the next theorem.

Theorem 2.14 (i) For any φ in V set

W_{φ}(g) = (π(g)φ)(1)

so that W_{φ} is a function in G_{F}. Let W (π, ψ) be the space of such functions. The map φ → W_{φ} is an isomorphism of V with W (π, ψ). Moreover

W_{π(g)φ} = ρ(g)W_{φ}

(ii) Let W (ψ) be the space of all functions W on G_{F} such that

W \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} g = ψ(x)W (g)

for all x in F and g in G. Then W (π, ψ) is contained in W (ψ) and is the only invariant subspace which transforms according to π under right translations.

Since

W_{φ} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = (π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ) (1) = φ(a)

the function W_{φ} is 0 only if φ is. Since

ρ(g)W (h) = W (hg)

the relation

W_{π(g)φ} = ρ(g)W_{φ}

is clear. Then W (π, ψ) is invariant under right translation and transforms according to π.

Since

W_{φ} (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = (π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π(g)φ) (1) = ψ(x)\{π(g)φ(1)\}

the space W (π, ψ) is contained in W (ψ). Suppose W is an invariant subspace of W (ψ) which transforms according to π. There is an isomorphism A of V with W such that

A(π(g)φ) = ρ(g)(Aφ).

Let

L(φ) = Aφ(1).

Since

L(π(g)φ) = Aπ(g)φ(1) = ρ(g)Aφ(1) = Aφ(g)

the map A is determined by L. Also

L (π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ) = Aφ \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = ψ(x)Aφ(1) = ψ(x)L(φ)

so that by Lemma 2.13.2 there is a scalar λ such that

L(φ) = λφ(1).

Consequently Aφ = λW_{φ} and W = W (π, ψ).

The realization of π on W (π, ψ) will be called the Whittaker model. Observe that the representation of G_{F} on W (ψ) contains no irreducible finite-dimensional representations. In fact any such representation is of the form

π(g) = χ(detg).

If π were contained in the representation on W (ψ) there would be a nonzero function W on G_{F} such that

W (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = ψ(x)χ(detg)W (e)

In particular taking g = e we find that

W \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = ψ(x)W (e)

However it is also clear that

W \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = χ (det ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) W (e) = W (e)

so that ψ(x) = 1 for all x. This is a contradiction. We shall see however that π is a constituent of the representation on W (ψ). That is, there are two invariant subspaces W_{1} and W_{2} of W (ψ) such that W_{1} contains W_{2} and the representation of the quotient space W_{1}/W_{2} is equivalent to π.

Chapter 1 33 Proposition 2.15 Let π and π′ be two infinite-dimensional irreducible representations of G_{F} realized in the Kirillov form on spaces V and V′. Assume that the two quasi-characters defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I π′ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω′(a)I

are the same. Let {C(ν, t)} and {C′(ν, t)} be the families of formal series associated to the two representations. If

C(ν, t) = C′(ν, t)

for all ν then π = π′.

If φ belongs to 𝒮(F^{×}) then, by hypothesis,

π(w)φ̂(ν, t) = π′(w)φ̂(ν, t)

so that π(w)φ = π′(w)φ. Since V is spanned by 𝒮(F^{×}) and π(w)𝒮(F^{×}) and V′ is spanned by 𝒮(F^{×}) and π′(w)𝒮(F^{×}) the spaces V and V′ are the same. We have to show that

π(g)φ = π′(g)φ

for all φ in V and all g in G_{F}. This is clear if g is in P_{F} so it is enough to verify it for g = w. We have already observed that π(w)φ_{0} = π′(w)φ_{0} if φ_{0} is in 𝒮(F^{×}) so we need only show that π(w)φ = π′(w)φ if φ is of the form π(w)φ_{0} with φ_{0} in 𝒮(F^{×}). But π(w)φ = π^{2}(w)φ_{0} = ω(−1)φ_{0} and, since π(w)φ_{0} = π′(w)φ_{0}, π′(w)φ = ω′(−1)φ_{0}.

Let N_{F} be the group of matrices of the form

( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )

with x in F and let ℬ be the space of functions on G_{F} invariant under left translations by elements of N_{F}. ℬ is invariant under right translations and the question of whether or not a given irreducible representation π is contained in ℬ arises. The answer is obviously positive when π = χ is onedimensional for then the function g → χ(detg) is itself contained in ℬ.

Assume that the representation π which is given in the Kirillov form acts on ℬ. Then there is a map A of V into ℬ such that

Aπ(g)φ = ρ(g)Aφ

If L(φ) = Aφ(1) then

L (ξ_{ψ} \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ) = L(φ)  (2.15.1)

for all φ in V and all x in F. Conversely given such a linear form the map φ → Aφ defined by

Aφ(g) = L(π(g)φ)

satisfies the relation Aπ(g) = ρ(g)A and takes V into ℬ. Thus π is contained in ℬ if an only if there is a non-trivial linear form L on V which satisfies (2.15.1).

Chapter 1 34 Lemma 2.15.2 If L is a linear form on 𝒮(F^{×}) which satisfies (2.15.1) for all x in F and for all φ in 𝒮(F^{×}) then L is zero.

We are assuming that L annihilates all functions of the form

ξ_{ψ} \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ − φ

so it will be enough to show that they span 𝒮(F^{×}). If φ belongs to 𝒮(F^{×}) we may set φ(0) = 0 and regard φ as an element of 𝒮(F). Let φ′ be its Fourier transform so that

φ(x) = ∫_{F} φ′(−b)ψ(bx) db.

Let Ω be an open compact subset of F^{×} containing the support of φ and let 𝔭^{−n} be an ideal containing Ω. There is an ideal 𝔞 of F such that φ′(−b)ψ(bx) is, as a function of b, constant on cosets of 𝔞 for all x in 𝔭^{−n}. Let 𝔟 be an ideal containing both 𝔞 and the support of φ′. If S is a set of representatives for the cosets of 𝔞 in 𝔟, if c is the measure of 𝔞, and if φ_{0} is the characteristic function of Ω then

φ(x) = ∑_{b∈S} λ_{b}ψ(bx)φ_{0}(x)

if λ_{b} = cφ′(−b). Thus

φ = ∑_{b} λ_{b}ξ_{ψ} \begin{aligned}( \\ ( & \begin{matrix}1 & b \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ_{0}.

Since φ(0) = 0 we have

∑_{b} λ_{b} = 0

so that

φ = ∑_{b} λ_{b} \{ξ_{ψ} \begin{aligned}( \\ ( & \begin{matrix}1 & b \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ_{0} − φ_{0}\}

as required.

Thus any linear form on V verifying (2.15.1) annihilates 𝒮(F^{×}). Conversely any linear form on V annihilating 𝒮(F^{×}) satisfies (2.15.1) because

ξ_{ψ} \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ − φ

is in 𝒮(F^{×}) if φ is in V. We have therefore proved

Proposition 2.16 For any infinite-dimensional irreducible representation π the following two properties are equivalent:

(i) π is not contained in ℬ.

(ii) The Kirillov model of π is realized in the space 𝒮(F^{×}).

A representation satisfying these two conditions will be called absolutely cuspidal.

Chapter 1 35 Lemma 2.16.1 Let π be an infinite-dimensional irreducible representation realized in the Kirillov form on the space V. Then V_{0} = 𝒮(F^{×}) is of finite codimension in V.

We recall that V = V_{0} + π(w)V_{0}. Let V_{1} be the space of all φ in V_{0} with support in U_{F}. An element of π(w)V_{0} may always be written as a linear combination of functions of the form

π \begin{aligned}( \\ ( & \begin{matrix}ϖ^{p} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π(w)φ

with φ in V_{1} and p in ℤ. For each character μ of U_{F} let φ_{μ} be the function in V_{1} such that φ_{μ}(ϵ) = μ(ϵ)ν_{0}(ϵ) for ϵ in U_{F}. Then

φ̂_{μ}(ν, t) = δ(νμν_{0})

and

π(w)φ̂_{μ}(ν, t) = δ(νμ^{−1})C(ν, t).

Let η_{μ} = π(w)φ_{μ}. The space V is spanned by V_{0} and the functions

π \begin{aligned}( \\ ( & \begin{matrix}ϖ^{p} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} η_{μ}

For the moment we take the following two lemmas for granted.

Lemma 2.16.2 For any character μ of Û_{F} there is an integer n_{0} and a family of constants λ_{i}, 1 ≤ i ≤ p, such that

C_{n}(μ) = ∑_{i=1}^{p} λ_{i}C_{n−i}(μ)

for n ≥ n_{0}.

Lemma 2.16.3 There is a finite set S of characters of U_{F} such that for ν not in S the numbers C_{n}(ν) are 0 for all but finitely many n.

If μ is not in S the function η_{μ} is in V_{0}. Choose μ in S and let V_{μ} be the space spanned by the functions

π \begin{aligned}( \\ ( & \begin{matrix}ϖ^{p} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} η_{μ}

and the functions φ in V_{0} satisfying φ(aϵ) = φ(a)μ^{−1}(ϵ) for all a in F^{×} and all ϵ in U_{F}. It will be enough to show that V_{μ}/V_{μ} ∩ V_{0} is finite-dimensional.

If φ is in V_{μ} then φ̂(ν, t) = 0 unless ν = μ and we may identify φ with the sequence {φ̂_{n}(μ)}. The elements of V_{μ} ∩ V_{0} are the elements corresponding to sequences with only finitely many nonzero terms. Referring to Proposition 2.10 we see that all of the sequences satisfying the recursion relation

φ̂_{n}(μ) = ∑_{i=1}^{p} λ_{a}φ̂_{n−i}(μ)

for n ≥ n_{1}. The integer n_{1} depends on φ.

Lemma 2.16.1 is therefore a consequence of the following elementary lemma whose proof we postpone to Paragraph 8.

Lemma 2.16.4 Let λ_{i}, 1 ≤ i ≤ p, be p complex numbers. Let A be the space of all sequences {a_{n}}, n ∈ ℤ for which there exist two integers n_{1} and n_{2} such that

a_{n} = ∑_{1≤i≤p} λ_{i}a_{n−i}

for n ≥ n_{1} and such that a_{n} = 0 for n ≤ n_{2}. Let A_{0} be the space of all sequences with only a finite number of nonzero terms. Then A/A_{0} is finite-dimensional.

We now prove Lemma 2.16.2. According to Proposition 2.11

∑_{σ} η(σ^{−1}ν, ϖ^{n})η(σ^{−1}ν̃, ϖ^{p})C_{p+n}(σ)

is equal to

z_{0}^{p}ν_{0}(−1)δ_{n,p} + (|ϖ| − 1)^{−1}z_{0}^{ℓ+1}C_{n−1−ℓ}(ν)C_{p−1−ℓ}(ν̃) − ∑_{−2−ℓ}^{−∞} z_{0}^{−r}C_{n+r}(ν)C_{p+r}(ν̃).

Remember that 𝔭^{−ℓ} is the largest ideal on which ψ is trivial. Suppose first that ν̃ = ν. Take p = −ℓ and n > −ℓ. Then δ(n − p) = 0 and

η(σ^{−1}ν, ϖ^{n})η(σ^{−1}ν, ϖ^{p}) = 0

unless σ = ν. Hence

C_{n−ℓ}(ν) = (|ϖ| − 1)^{−1}z_{0}^{ℓ+1}C_{n−1−ℓ}(ν)C_{−2ℓ−1}(ν) − ∑_{−2−ℓ}^{−∞} z_{0}^{−r}C_{n+r}(ν)C_{−ℓ+r}(ν)

which, since almost all of the coefficients C_{−ℓ+r}(ν) in the sum are zero, is the relation required.

If ν ≠ ν̃ take p ≥ −ℓ and n > p. Then η(σ^{−1}ν, ϖ^{n}) = 0 unless σ = ν and η(σ^{−1}ν, ϖ^{p}) = 0 unless σ = ν̃. Thus

(|ϖ| − 1)^{−1}z_{0}^{ℓ+1}C_{n−1−ℓ}(ν)C_{p−1−ℓ}(ν̃) − ∑_{2−ℓ}^{−∞} z_{0}^{−r}C_{n+r}(ν)C_{p+r}(ν̃) = 0.  (2.16.5)

There is certainly at least one i for which C_{i}(ν̃) ≠ 0. Take p − 1 − ℓ ≥ i. Then from (2.16.5) we deduce a relation of the form

C_{n+r}(ν) = ∑_{i=1}^{q} λ_{i}C_{n+r−i}(ν)

where r is a fixed integer and n is any integer greater than p.

Lemma 2.16.3 is a consequence of the following more precise lemma. If 𝔭^{m} is the conductor of a character ρ we refer to m as the order of ρ.

Chapter 1 37 Lemma 2.16.6 Let m_{0} be of the order ν_{0} and let m_{1} be an integer greater than m_{0}. Write ν_{0} in any manner in the form ν_{0} = ν_{1}^{−1}ν_{2}^{−1} where the orders of ν_{1} and ν_{2} are strictly less than m_{1}. If the order m of ρ is large enough

C_{−2m−2ℓ}(ρ) = ν_{2}^{−1}ρ(−1)z_{0}^{−m−ℓ} \frac{η(ν_{1}^{−1}ρ, ϖ^{−m−ℓ})}{η(ν_{2}ρ^{−1}, ϖ^{−m−ℓ})}

and C_{p}(ρ) = 0 if p ≠ −2m − 2ℓ.

Suppose the order of ρ is at least m_{1}. Then ρν_{1}ν_{0} = ρν_{2}^{−1} is still of order m. Applying Proposition 2.11 we see that

∑_{σ} η(σ^{−1}ν_{1}, ϖ^{n+m+ℓ})η(σ^{−1}ρ, ϖ^{p+m+ℓ})C_{p+n+2m+2ℓ}(σ)

is equal to

η(ν_{1}^{−1}ρ^{−1}ν_{0}^{−1}, ϖ^{−m−ℓ})z_{0}^{m+ℓ}ν_{1}ρν_{0}(−1)C_{n−m−ℓ}(ν)C_{p−m−ℓ}(ρ)

for all integers n and p. Choose n such that C_{n}(ν_{1}) ≠ 0. Assume also that m + n + ℓ ≥ −ℓ or that m ≥ −2ℓ − n. Then η(σ^{−1}ν_{1}, ϖ^{n+m+ℓ}) = 0 unless σ = ν_{1} so that

η(ν_{1}^{−1}ρ, ϖ^{p+m+ℓ})C_{p+n+2m+2ℓ}(ν_{1}) = η(ν_{2}ρ^{−1}, ϖ^{−m−ℓ}z_{0}^{m+ℓ}ν_{1}ρν_{0}(−1)C_{n}(ν_{1})C_{p}(ρ).

Since ν_{1}^{−1}ρ is still of order m the left side is zero unless p = −2m − 2ℓ. The only term on the right side that can vanish is C_{p}(ρ). On the other hand if p = −2m − 2ℓ we can cancel the terms C_{n}(ν_{1}) from both side to obtain the relation of the lemma.

Apart from Lemma 2.16.4 the proof of Lemma 2.16.1 is complete. We have now to discuss its consequences. If ω_{1} and ω_{2} are two quasi-characters of F^{×} let ℬ(ω_{1}, ω_{2}) be the space of all functions φ on G_{F} which satisfy

(i) For all g in G_{F}, a_{1}, a_{2} in F^{×}, and x in F

φ (( \begin{matrix}a_{1} & x \\ 0 & a \end{matrix} ) g) = ω_{1}(a_{1})ω_{2}(a_{2}) |\frac{a_{1}}{a_{2}} |^{1/2} φ(g).

(ii) There is an open subgroup U of GL(2, O_{F}) such that φ(gu) ≡ φ(g) for all u in U.

Since

G_{F} = N_{F} A_{F} GL(2, O_{F})

where A_{F} is the group of diagonal matrices the elements of ℬ(ω_{1}, ω_{2}) are determined by their restrictions to GL(2, O_{F}) and the second condition is tantamount to the condition that φ be locally constant. ℬ(ω_{1}, ω_{2}) is invariant under right translations by elements of G_{F} so that we have a representation ρ(ω_{1}, ω_{2}) of G_{F} on ℬ(ω_{1}, ω_{2}). It is admissible.

Chapter 1 38 Proposition 2.17 If π is an infinite-dimensional irreducible representation of G_{F} which is not absolutely cuspidal then for some choice of μ_{1} and μ_{2} it is contained in ρ(μ_{1}, μ_{2}).

We take π in the Kirillov form. Since V_{0} is invariant under the group P_{F} the representation π defines a representation σ of P_{F} on the finite-dimensional space V/V_{0}. It is clear that σ is trivial on N_{F} and that the kernel of σ is open. The contragredient representation has the same properties. Since P_{F} /N_{F} is abelian there is a nonzero linear form L on V/V_{0} such that

σ̃ \begin{aligned}( \\ ( & \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} & ) \\ & & ) \end{aligned} L = μ_{1}^{−1}(a_{1})μ_{1}^{−1}(a_{2})L

for all a_{1}, a_{2}, and x. μ_{1} and μ_{2} are homomorphisms of F^{×} into ℂ^{×} which are necessarily continuous. L may be regarded as a linear form on V. Then

L (π \begin{aligned}( \\ ( & \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} & ) \\ & & ) \end{aligned} φ) = μ_{1}(a_{1})μ_{2}(a_{2})L(φ).

If φ is in V let Aφ be the function

Aφ(g) = L(π(g)φ)

on G_{F}. A is clearly an injection of V into ℬ(μ_{1}, μ_{2}) which commutes with the action of G_{F}.

Before passing to the next theorem we make a few simple remarks. Suppose π is an infinitedimensional irreducible representation of G_{F} and ω is a quasi-character of F^{×}. It is clear that W (ω ⊗ π, ψ) consists of the functions

g → W (g)ω(detg)

with W on W (π, ψ). If V is the space of the Kirillov model of π the space of the Kirillov model of ω ⊗ π consists of the functions a → φ(a)ω(a) with φ in V. To see this take π in the Kirillov form and observe first of all that the map A : φ → φω is an isomorphism of V with another space V′ on which G_{F} acts by means of the representation π′ = A(ω ⊗ π)A^{−1}. If

b ( \begin{matrix}α & x \\ 0 & 1 \end{matrix} )

belongs to B_{F} and φ′ = φω then

π′(b)φ′(a) = ω(a){ω(α)ψ(ax)φ(αa)} = ψ(ax)φ′(αa)

so that π′(b)φ′ = ξ_{ψ}(b)φ′. By definition then π′ is the Kirillov model of ω ⊗ π. Let ω_{1} be the restriction of ω to U_{F} and let z_{1} = ω(ϖ). If φ′ = φω then

φ′(ν, t) = φ̂(νω_{1}, z_{1}t).

Thus

π′(w)φ′(ν, t) = π(w)φ̂(νω_{1}, z_{1}t) = C(νω_{1}, z_{1}t)φ̂(v^{−1}ω_{1}^{−1}ν_{0}^{−1}, z_{0}^{−1}z_{1}^{−1}t^{−1}).

The right side is equal to

C(νω_{1}, z_{1}t)φ′(ν^{−1}ν_{0}^{−1}ω_{1}^{−1}, z_{0}^{−1}z_{1}^{−2}t^{−1})

so that when we replace π by ω ⊗ π we replace C(ν, t) by C(νω_{1}, z_{1}t).

Suppose ψ′(x) = ψ(bx) with b in F^{×} is another non-trivial additive character. Then W (π, ψ′) consists of the functions

W′(g) = W (( \begin{matrix}b & 0 \\ 0 & 1 \end{matrix} ) g)

with W in W (π, ψ).

The last identity of the following theorem is referred to as the local functional equation. It is the starting point of our approach to the Hecke theory.

Theorem 2.18 Let π be an irreducible infinite-dimensional admissible representation of G_{F}. (i) If ω is the quasi-character of G_{F} defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

then the contragredient representation π̃ is equivalent to ω^{−1} ⊗ π. (ii) There is a real number s_{0} such that for all g in G_{F} and all W in W (π, ψ) the integrals

\begin{aligned}∫_{F^{×}} W (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s−1/2} d^{×}a & = Ψ(g, s, W) \\ ∫_{F^{×}} W (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s−1/2}ω^{−1}(a) d^{×}a & = Ψ̃(g, s, W) \end{aligned}

converge absolutely for Re s > s_{0}. (iii) There is a unique Euler factor L(s, π) with the following property: if

Ψ(g, s, W) = L(s, π)Φ(g, s, W)

then Φ(g, s, W) is a holomorphic function of s for all g and all W and there is at least one W in W (π, ψ) such that

Φ(e, s, W) = a^{s}

where a is a positive constant. (iv) If

Ψ̃(g, s, W) = L(s, π̃)Φ̃(g, s, W)

there is a unique factor ϵ(s, π, ψ) which, as a function of s, is an exponential such that

Φ̃ (( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, 1 − s, W ) = ϵ(s, π, ψ)Φ(g, s, W)

for all g in G_{F} and all W in W (π, ψ).

To say that L(s, π) is an Euler product is to say that L(s, π) = P^{−1}(q^{−s}) where P is a polynomial with constant term 1 and q = |ϖ|^{−1} is the number of elements in the residue field. If L(s, π) and L′(s, π) were two Euler factors satisfying the conditions of the lemma their quotient would be an entire function with no zero. This clearly implies uniqueness.

If ψ is replaced by ψ′ where ψ′(x) = ψ(bx) the functions W are replaced by the functions W′ with

W′(g) = W (( \begin{matrix}b & 0 \\ 0 & 1 \end{matrix} ) g)

and

Ψ(g, s, W′) = |b|^{1/2−s}Ψ(g, s, W)

while

Ψ(g, s, W′) = |b|^{1/2−s}ω(b)Ψ̃(g, s, W).

Thus L(s, π) will not depend on ψ. However

ϵ(s, π, ψ′) = ω(b) |b|^{2s−1}ϵ(s, π, ψ).

According to the first part of the theorem if W belongs to W (π, ψ) the function

W̃(g) = W (g)ω^{−1}(detg)

is in W (π̃, ψ). It is clear that

Ψ̃(g, s, W) = ω(detg)Ψ(g, s, W̃)

so that if the third part of the theorem is valid when π is replaced by π̃ the function Φ̃(g, s, W) is a holomorphic function of s. Combining the functional equation for π and for π̃ one sees that

ϵ(s, π, ψ)ϵ(1 − s, π̃, ψ) = ω(−1).

Let V be the space on which the Kirillov model of π acts. For every W in W (π, ψ) there is a unique φ in V such that

W \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = φ(a).

If π is itself the canonical model

π(w)φ(a) = W (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) w)

where

w = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) .

If χ is any quasi-character of F^{×} we set

φ̂(χ) = ∫_{F^{×}} φ(a)χ(a) d^{×}a

if the integral converges. If χ_{0} is the restriction of χ to U_{F} then

φ̂(χ) = φ̂(χ_{0}, χ(ϖ)).

Thus if α_{F} is the quasi-character α_{F} (x) = |x| and the appropriate integrals converge

\begin{aligned}Ψ(e, s, W) & = φ̂(α_{F}^{s−1/2}) = φ̂(1, q^{1/2−s}) \\ Ψ(e, s, W) & = φ̂(α_{F}^{s−1/2}ω^{−1}) = φ̂(ν_{0}^{−1}, z_{0}^{−1}q^{1/2−s}) \end{aligned}

if ν_{0} is the restriction of ω to U_{F} and z_{0} = ω(φ). Thus if the theorem is valid the series φ̂(1, t) and φ̂(ν_{0}^{−1}, t) have positive radii of convergence and define functions which are meromorphic in the whole t-plane.

It is also clear that

Ψ(w, 1 − s, W) = π(w)φ̂(ν_{0}^{−1}, z_{0}^{−1}q^{s−1/2}).

If φ belongs to V_{0} then

π(w)φ̂(ν_{0}^{−1}, z_{0}^{−1}q^{−1/2}t) = C(ν_{0}^{−1}, z_{0}^{−1/2}q^{−1/2}t)φ̂(1, q^{1/2}t^{−1}).

Choosing φ in V_{0} such that φ̂(1, t) ≡ 1 we see that C(ν_{0}^{−1}, t) is convergent in some disc and has an analytic continuation to a function meromorphic in the whole plane.

Comparing the relation

π(w)φ̂(ν_{0}^{−1}z_{0}^{−1}q^{−1/2}q^{s}) = C(ν_{0}^{−1}, z_{0}^{−1/2}q^{−1/2}q^{s})φ̂(1, q^{1/2}q^{−s})

with the functional equation we see that

C(ν_{0}^{−1}, z_{0}^{−1}q^{−1/2}q^{s}) = \frac{L(1 − s, π̃)ϵ(s, π, ψ)}{L(s, π)}.  (2.18.1)

Replacing π by χ ⊗ π we obtain the formula

C(ν_{0}^{−1}χ_{0}^{−1}, z_{0}^{−1}z_{1}^{−1}q^{−1/2}q^{s}) = \frac{L(1 − s, χ^{−1} ⊗ π̃)ϵ(s, χ ⊗ π, ψ)}{L(s, χ ⊗ π)}.

Appealing to Proposition 2.15 we obtain the following corollary.

Corollary 2.19 Let π and π′ be two irreducible infinite-dimensional representations of G_{F}. Assume that the quasi-characters ω and ω′ defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I π′ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω′(a)I

are equal. Then π and π′ are equivalent if and only if

\frac{L(1 − s, χ^{−1} ⊗ π̃)ϵ(s, χ ⊗ π, ψ)}{L(s, χ ⊗ π)} = \frac{L(1 − s, χ^{−1} ⊗ π′)ϵ(s, χ ⊗ π′, ψ)}{L(s, χ ⊗ π′)}

for all quasi-characters.

We begin the proof of the first part of the theorem. If φ_{1} and φ_{2} are numerical functions on F^{×} we set

⟨φ_{1}, φ_{2}⟩ = ∫ φ_{1}(a)φ_{2}(−a) d^{×}a.

The Haar measure is the one which assigns the measure 1 to U_{F}. If one of the functions is in 𝒮(F^{×}) and the other is locally constant the integral is certainly defined. By the Plancherel theorem for U_{F}

⟨φ, φ′⟩ = ∑_{n} ∑_{ν} ν(−1)φ̂_{n}(ν)φ′_{n}(ν^{−1}).

The sum is in reality finite. It is easy to se that if b belongs to B

⟨ξ_{ψ}(b)φ, ξ_{ψ}(b)φ′)⟩ = ⟨φ, φ′⟩.

Suppose π is given in the Kirillov form and acts on V. Let π′, the Kirillov model of ω^{−1} ⊗ π, act on V′. To prove part (i) we have only to construct an invariant non-degenerate bilinear form β on V × V′. If φ belongs to V_{0} and φ′ belongs to V′ or if φ belongs to V and φ′ belongs to V′_{0} we set

β(φ, φ′) = ⟨φ, φ′⟩.

If φ and φ′ are arbitrary vectors in V and V′ we may write φ = φ_{1} +π(w)φ_{2} and φ′ = φ_{1} +π′(w)φ′_{2} with φ, φ_{2} in V_{0} and φ′_{1}, φ′_{2} in V′_{0}. We want to set

β(φ, φ′) = ⟨φ_{1}, φ′_{1}⟩ + ⟨φ_{1}, π′(w)φ′_{2}⟩ + ⟨π(w)φ_{2}, φ′_{1}⟩ + ⟨φ_{2}, φ′_{2}⟩.

The second part of the next lemma shows that β is well defined.

Lemma 2.19.1 Let φ and φ′ belong to V_{0} and V′_{0} respectively. Then

(i)

⟨π(w)φ, φ′⟩ = ν_{0}(−1)⟨φ, π′(w)φ′⟩

(ii) If either π(w)φ belongs to V_{0} or π′(w)φ′ belongs to V′_{0} then

⟨π(w)φ, π′(w)φ′⟩ = ⟨φ, φ′⟩.

The relation

π(w)φ̂(ν, t) = ∑_{n} t^{n} ∑_{p} C_{n+p}(ν)φ̂_{p}(ν^{−1}ν_{0}^{−1})z_{0}^{−p}

implies that

⟨π(w)φ, φ′⟩ = ∑_{n,p,ν} ν(−1)C_{n+p}(ν)φ̂_{p}(ν^{−1}ν_{0}^{−1})z_{0}^{−p}φ′_{n}(ν^{−1}).  (2.19.2)

Replacing π by π′ replaces ω by ω^{−1}, ν_{0} by ν_{0}^{−1}, z_{0} by z_{0}^{−1}, and C(ν, t) by C(νν_{0}^{−1}, z_{0}^{−1}t). Thus

⟨φ, π(w)φ′⟩ = ∑_{n,p,ν} ν(−1)C_{n+p}(νν_{0}^{−1})z_{0}^{−n}φ′_{p}(ν^{−1}ν_{0})φ̂_{n}(ν^{−1}).  (2.19.3)

Replacing ν by νν_{0} in (2.19.3) and comparing with (2.19.2) we obtain the first part of the lemma.

Because of the symmetry it will be enough to consider the second part when π(w)φ belongs to V_{0}. By the first part

⟨π(w)φ, π′(w)φ′⟩ = ν_{0}(−1)⟨π^{2}(w)φ, φ′⟩ = ⟨φ, φ′⟩.

It follows immediately from the lemma that

β(π(w)φ, π′(w)φ′) = β(φ, φ′)

so that to establish the invariance of β we need only show that

β(π(p)φ, π′(p)φ′) = β(φ, φ′)

Chapter 1 43 for all triangular matrices p. If φ is in V_{0} or φ′ is in V′_{0} this is clear. We need only verify it for φ in π(w)V_{0} and φ′ in π′(w)V′_{0}.

If φ is in V_{0}, φ′ is in V′_{0} and p is diagonal then

β(π(p)π(w)φ, π′(p)π′(w)φ′) = β(π(w)π(p_{1})φ, π′(w)π′(p_{1})φ′)

where p_{1} = w^{−1}pw is also diagonal. The right side is equal to

β(π(p_{1})φ, π′(p_{1})φ′) = β(φ, φ′) = β(π(w)φ, π′(w)p′).

Finally we have to show that*

β (π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ, π′ \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ′) = β(φ, φ′)  (2.19.2)

for all x in F and all φ and φ′. Let φ_{i}, 1 < i < r, generate V modulo V_{0} and let φ′_{j}, 1 ≤ j ≤ r′, generate V′ modulo V′_{0}. There certainly is an ideal 𝔞 of F such that

π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ_{i} = φ_{i}

and

π′ \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ′_{j} = φ′_{j}

for all i and j if x belongs to 𝔞. Then

β (π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ_{i}, π′ \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ′_{j}) = β(φ_{i}, φ_{j}).

Since 2.19.2 is valid if x is in 𝔞 and φ is in V_{0} or φ′ is in V′_{0} it is valid for all φ and φ′ provided that x is in 𝔞. Any y in F may be written as ax with a in F^{×} and x in 𝔞. Then

( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} )

and it follows readily that

β (π \begin{aligned}( \\ ( & \begin{matrix}1 & y \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} ϖ, π′ \begin{aligned}( \\ ( & \begin{matrix}1 & y \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ′) = β(φ, φ′).

Since β is invariant and not identically zero it is non-degenerate. The rest of the theorem will now be proved for absolutely cuspidal representations. The remaining representations will be considered in the next chapter. We observe that since W (π, ψ) is invariant under right translations the assertions need only be established when g is the identity matrix e.

If π is absolutely cuspidal then V = V_{0} = 𝒮(F^{×}) and W (({a \atop 0} {0 \atop 1})) = φ(a) is locally constant with compact support. Therefore the integrals defining Ψ(e, s, W) and Ψ̃(e, s, W) are absolutely

* The tags on Equations 2.19.2 and 2.19.3 have inadvertently been repeated.

Chapter 1 44 convergent for all values of s and the two functions are entire. We may take L(s, π) = 1. If φ is taken to be the characteristic function of U_{F} then Φ(e, s, W) = 1.

Referring to the discussion preceding Corollary 2.19 we see that if we take

ϵ(s, π, ψ) = C(ν_{0}^{−1}, z_{0}^{−1}q^{−1/2}q^{s})

the local functional equation of part (iv) will be satisfied. It remains to show that ϵ(s, π, ψ) is an exponential function or, what is at least ast strong, to show that, for all ν, C(ν, t) is a multiple of a power of t. It is a finite linear combination of powers of t and if it is not of the form indicated it has a zero at some point different form 0. C(νν_{0}^{−1}, z_{0}^{−1}t^{−1}) is also a linear combination of powers of t and so cannot have a pole except at zero. To show that C(ν, t) has the required form we have only to show that

C(ν, t)C(ν^{−1}ν_{0}^{−1}, z_{0}^{−1}t^{−1}) = ν_{0}(−1).  (2.19.3)

Choose φ in V_{0} and set φ′ = π(w)φ. We may suppose that φ′(ν, t) ≠ 0. The identity is obtained by combining the two relations

φ′(ν, t) = C(ν, t)φ̂(ν^{−1}ν_{0}^{−1}, z_{0}^{−1}t^{−1})

and

ν_{0}(−1)φ̂(ν^{−1}ν_{0}^{−1}, t) = C(ν^{−1}ν_{0}^{−1}, t)φ′(ν, z_{0}^{−1}t^{−1}).

We close this paragraph with a number of facts about absolutely cuspidal representations which will be useful later.

Proposition 2.20 Let π be an absolutely cuspidal representation of G_{F}. If the quasi-character ω defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

is actually a character then π is unitary.

As usual we take π and π̃ in the Kirillov form. We have to establish the existence of a positive-definite invariant hermitian form on V. We show first that if φ belongs to V and φ̃ belongs to Ṽ then there is a compact set Ω in G_{F} such that if

Z_{F} = \{( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) | a ∈ F \}

the support of ⟨π(g)φ, φ̃⟩, a function of g, is contained in Z_{F} Ω. If A_{F} is the group of diagonal matrices G_{F} = GL(2, O_{F}) A_{F} GL(2, O_{F}). Since φ and φ̃ are both invariant under subgroups of finite index in GL(2, O_{F}) it is enough to show that the function ⟨π(b)φ, φ̃⟩ on A_{F} has support in a set Z_{F} Ω with Ω compact. Since

⟨π (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) b) φ, φ̃⟩ = ω(a)⟨π(b)φ, φ̃⟩

it is enough to show that the function

⟨π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ, φ̃⟩

has compact support in F^{×}. This matrix element is equal to

∫_{F^{×}} φ(ax)φ̃(−x) d^{×}x.

Since φ and φ̃ are functions with compact support the result is clear.

Choose φ̃_{0} ≠ 0 in Ṽ and set

(φ_{1}, φ_{2}) = ∫_{Z_{F} \G_{F}} ⟨π(g)φ_{1}, φ̃_{0}⟩\overline{⟨π(g)φ_{2}, φ̃_{0}⟩} dg.

This is a positive invariant hermitian form on V.

We have incidentally shown that π is square-integrable. Observe that even if the absolutely cuspidal representation π is not unitary one can choose a quasi-character χ such that χ ⊗ π is unitary.

If π is unitary there is a conjugate linear map A : V → Ṽ defined by

(φ_{1}, φ_{2}) = ⟨φ_{1}, Aφ_{2}⟩.

Clearly Aξ_{ψ}(b) = ξ_{ψ}(b)A for all b in B_{F}. The map A_{0} defined by

A_{0}φ(a) = \overline{φ}(−a)

has the same properties. We claim that

A = λA_{0}

with λ in ℂ^{×}. To see this we have only to apply the following lemma to A_{0}^{−1}A.

Lemma 2.21.1. Let T be a linear operator on 𝒮(F^{×}) which commutes with ξ_{ψ}(b) for all b in B_{F}. Then T is a scalar.

Since ξ_{ψ} is irreducible it will be enough to show that T has an eigenvector. Let 𝔭^{−ℓ} be the largest ideal on which ψ is trivial. Let μ be a non-trivial charcter of U_{F} and let 𝔭^{n} be its conductor. T commutes with the operator

S = ∫_{U_{F}} μ^{−1}(ϵ)ξ_{ψ} (( \begin{matrix}ϵ & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & ϖ^{−ℓ−n} \\ 0 & 1 \end{matrix} )) dϵ

and it leaves the range of the restriction of S to the functions invariant under U_{F} invariant. If φ is such a function

Sφ(a) = φ(a) ∫_{U_{F}} μ^{−1}(ϵ)ψ(aϵϖ^{−ℓ−n}) dϵ.

The Gaussian sum is 0 unless a lies in U_{F}. Therefore Sφ is equal to φ(1) times the function which is zero outside of U_{F} and equals μ on U_{F}. Since T leaves a one-dimensional space invariant it has an eigenvector.

Since A = λA_{0} the hermitian form (φ_{1}, φ_{2}) is equal to

λ ∫_{F^{×}} φ_{1}(a)\overline{φ}_{2}(a) d^{×}a.

Chapter 1 46 Proposition 2.21.2. Let π be an absolutely cuspidal representation of G_{F} for which the quasicharacter ω defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

is a character.

(i) If π is in the Kirillov form the hermitian form

∫_{F^{×}} φ_{1}(a)\overline{φ}_{2}(a) d^{×}a

is invariant. (ii) If |z| = 1 then |C(ν, z)| = 1 and if Res = 1/2

|ϵ(s, π, ψ)| = 1.

Since |z_{0}| = 1 the second relation of part (ii) follows from the first and the relation

ϵ(s, π, ψ) = C(ν_{0}^{−1}, q^{s−1/2}z_{0}^{−1}).

If n is in ℤ and ν is a character of U_{F} let

φ(ϵϖ^{m}) = δ_{n,m}ν(ϵ)ν_{0}(ϵ)

for m in ℤ and ϵ in U_{F}. Then

∫_{F^{×}} |φ(a)|^{2} da = 1.

If φ′ = π(w)φ and C(ν, t) = C_{ℓ}(ν)t^{ℓ} then

φ′(ϵϖ^{m}) = δ_{ℓ−n,m}C_{ℓ}(ν)z_{0}^{−n}ν^{−1}(ϵ).

Since |z_{0}| = 1

∫_{F^{×}} |φ′(a)|^{2} da = |C_{ℓ}(ν)|^{2}.

Applying the first part of the lemma we see that, if |z| = 1, both |C_{ℓ}(ν)|^{2} and |C(ν, z)|^{2} = |C_{ℓ}(ν)|^{2} |z|^{2ℓ} are 1.

Proposition 2.22. Let π be an irreducible representation of G_{F}. It is absolutely cuspidal if and only if for every vector v there is an ideal 𝔞 in F such that

∫_{𝔞} π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v dx = 0.

It is clear that the condition cannot be satisfied by a finite dimensional representation. Suppose that π is infinite dimensional and in the Kirillov form. If φ is in V then

∫_{𝔞} π \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ dx = 0

if and only if

φ(a) ∫_{𝔞} ψ(ax) dx = 0

for all a. If this is so the character x → ψ(ax) must be non-trivial on 𝔞 for all a in the support of φ. This happens if and only if φ is in 𝒮(F^{×}). The proposition follows.

Chapter 1 47 Proposition 2.23. Let π be an absolutely cuspidal representation and assume the largest ideal on which ψ is trivial is O_{F}. Then, for all characters ν, C_{n}(ν) = 0 if n ≥ −1.

Take a character ν and choose n_{1} such that C_{n_{1}}(ν) ≠ 0. Then C_{n}(ν) = 0 for n ≠ n_{1}. If ν̃ = ν^{−1}ν_{0}^{−1} then, as we have seen,

C(ν, t)C(ν̃, t^{−1}z_{0}^{−1}) = ν_{0}(−1)

so that

C_{n}(ν̃) = 0

for n ≠ n_{1} and

C_{n_{1}}(ν)C_{n_{1}}(ν̃) = ν_{0}(−1)z_{0}^{n_{1}}.

In the second part of Proposition 2.11 take n = p = n_{1} + 1 to obtain

∑_{σ} η(σ^{−1}ν, ϖ^{n_{1}+1})η(σ^{−1}ν̃, ϖ^{n_{1}+1})C_{2n_{1}+2}(σ) = z_{0}^{n_{1}+1}ν_{0}(−1) + (|ϖ| − 1)^{−1}z_{0}C_{n_{1}}(ν)C_{n_{1}}(ν̃).

The right side is equal to

z_{0}^{n_{1}+1}ν_{0}(−1) · \frac{|ϖ|}{|ϖ| − 1}.

Assume n_{1} ≥ −1. Then η(σ^{−1}ν, ϖ^{n_{1}+1}) is 0 unless σ = ν and η(σ^{−1}ν̃, ϖ^{n_{1}+1}) is 0 unless σ = ν̃. Thus the left side is 0 unless ν = ν̃. However if ν = ν̃ the left side equals C_{2n_{1}+2}(ν). Since this cannot be zero 2n_{1} + 2 must be equal n_{1} so that n_{1} = −2. This is a contradiction.

Chapter 1 48 §3. The principal series for non-archimedean fields. In order to complete the discussion of the previous pragraph we have to consider representations which are not absolutely cuspidal. This we shall now do. We recall that if μ_{1}, μ_{2} is a pair of quasi-characters of F^{×} the space ℬ(μ_{1}, μ_{2}) consists of all locally constant functions f on G_{F} which satisfy

f (( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} ) g) = μ_{1}(a_{1})μ_{2}(a_{2}) |\frac{a_{1}}{a_{2}} |^{1/2} f(g)  (3.1)

for all g in G_{F}, a_{1}, a_{2}, in F^{×}, and x in F. ρ(μ_{1}, μ_{2}) is the representation of G_{F} on ℬ(μ_{a}, μ_{2}).

Because of the Iwasawa decomposition G_{F} = P_{F} GL(2, O_{F}) the functions in ℬ(μ_{1}, μ_{2}) are determined by their restrictions to GL(2, O_{F}). The restriction can be any locally constant function on GL(2, O_{F}) satisfying

f (( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} ) g) = μ_{1}(a_{1})μ_{2}(a_{2})f(g)

for all g in GL(2, O_{F}), a_{1}, a_{2} in U_{F}, and x in O_{F}. If U is an open subgroup of GL(2, O_{F}) the restriction of any function invariant under U is a function on GL(2, O_{F})/U which is a finite set. Thus the space of all such functions is finite dimensional and as observed before ρ(μ_{1}, μ_{2}) is admissible.

Let ℱ be the space of continuous functions f on G_{F} which satisfy

f (( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} ) g) = |\frac{a_{1}}{a_{2}} | f(g)

for all g in G_{F}, a_{1}, a_{2} in F^{×}, and x in F. We observe that ℱ contains ℬ(α_{F}^{1/2}, α_{F}^{−1/2}). G_{F} acts on ℱ. The Haar measure on G_{F} if suitably normalized satisfies

∫_{G_{F}} f(g) dg = ∫_{N_{F}} ∫_{A_{F}} ∫_{GL(2,O_{F})} |\frac{a_{1}}{a_{2}} |^{−1} f(nak) dn da dk

if

a = ( \begin{matrix}a_{1} & 0 \\ 0 & a_{2} \end{matrix} ) .

It follows easily from this that

∫_{GL(2,O_{F})} f(k) dk

is a G_{F} -invariant linear form on ℱ. There is also a positive constant c such that

∫_{G_{F}} f(g) dg = c ∫_{N_{F}} ∫_{A_{F}} ∫_{N_{F}} |\frac{a_{1}}{a_{2}} |^{−1} f (na ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) n_{1}) dn da dn_{1}.

Consequently

∫_{GL(2,O_{F})} f(k) dk = c ∫_{F} f (( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) dx.

If φ_{1} belongs to ℬ(μ_{1}, μ_{2}) and φ_{2} belongs to ℬ(μ_{1}^{−1}, μ_{2}^{−1}) then φ_{1}φ_{2} belongs to ℱ and we set

⟨φ_{1}, φ_{2}⟩ = ∫_{GL(2,O_{F})} φ_{1}(k) φ_{2}(k) dk.

Clearly

⟨ρ(g)φ_{1}, ρ(g)φ_{2}⟩ = ⟨φ_{1}, φ_{2}⟩

so that this bilinear form is invariant. Since both φ_{1} and φ_{2} are determined by their restrictions to GL(2, O_{F}) it is also non-degenerate. Thus ρ(μ_{1}^{−1}, μ_{2}^{−1}) is equivalent to the contragredient of ρ(μ_{1}, μ_{2}).

In Proposition 1.6 we introduced a representation r of G_{F} and then we introduced a representation r_{Ω} = r_{μ_{1},μ_{2}}. Both representations acted on 𝒮(F^{2}). If

Φ (a, b) = ∫_{F} Φ(a, y)ψ(by) dy

is the partial Fourier transform

[r(g)Φ] = ρ(g)Φ  (3.1.1)

and

r_{μ_{1},μ_{2}}(g) = μ_{1}(detg) |detg|^{1/2}r(g).  (3.1.2)

We also introduced the integral

θ(μ_{1}, μ_{2}; Φ) = ∫ μ_{1}(t)μ_{2}^{−1}(t)Φ(t, t^{−1}) d^{×}t

and we set

W_{Φ}(g) = θ(μ_{1}, μ_{2}; r_{μ_{1},μ_{2}}(g)Φ).  (3.1.3)

The space of functions W_{Φ} is denoted W (μ_{1}, μ_{2}; ψ).

If ω is a quasi-character of F^{×} and if |ω(ϖ)| = |ϖ|^{s} with s > 0 the integral

z(ω, Φ) = ∫_{F^{×}} Φ(0, t)ω(t) d^{×}t

is defined for all Φ in 𝒮(F^{2}). In particular if |μ_{1}(ϖ)μ_{1}^{−1}(ϖ)| = |ϖ|^{s} with s > −1 we can consider the function

f_{Φ}(g) = μ_{1}(detg) |detg|^{1/2}z(α_{F} μ_{1}μ_{2}^{−1}, ρ(g)Φ)

on G_{F}. Recall that α_{F} (a) = |a|. Clearly

ρ(h)f_{Φ} = f_{Ψ}  (3.1.4)

if

Ψ = μ_{1}(deth) |deth|^{1/2}ρ(h)Φ.

We claim that f_{Φ} belongs to ℬ(μ_{1}, μ_{2}). Since the stabilizer of every Φ under the representation g → μ_{1}(detg) |detg|^{1/2}ρ(g) is an open subgroup of G_{F} the functions f_{Φ} are locally constant.

Chapter 1 50 Since the space of functions f_{Φ} is invariant under right translations we need verify (3.1) only for g = e.

f_{Φ} \begin{aligned}( \\ ( & \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} & ) \\ & & ) \end{aligned} = z (μ_{1}μ_{2}^{−1}α_{F} , ρ \begin{aligned}( \\ ( & \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} & ) \\ & & ) \end{aligned} Φ) μ(a_{1}a_{2}) |a_{1}a_{2}|^{1/2}.

By definition the right side is equal to

μ_{1}(a_{1}a_{2})|a_{1}a_{2}|^{1/2} ∫ μ_{1}(t)μ_{2}^{−1}(t) |t| Φ(0, a_{2}t) d^{×}t.

Changing variables we obtain

μ_{1}(a_{1})μ_{2}(a_{2}) |\frac{a_{1}}{a_{2}} |^{1/2} ∫ μ_{1}(t)μ_{2}^{−1}(t) |t| Φ(0, t) d^{×}t.

The integral is equal to f_{Φ}(e). Hence our assertion.

Proposition 3.2. Assume |μ_{1}(ϖ)μ_{2}^{−1}(ϖ)| = |ϖ|^{s} with s > −1.

(i) There is a linear transformation A of W (μ_{1}, μ_{2}; ψ) into ℬ(μ_{1}, μ_{2}) which for all Φ in 𝒮(F^{2}), sends W_{Φ} to f_{Φ}.

(ii) A is bijective and commutes with right translations.

To establish the first part of the proposition we have to show that f_{Φ^{∼}} is 0 if W_{Φ} is. Since

N_{F} A_{F} ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) N_{F} is a dense subset of G_{F} this will be a consequence of the following lemma.

Lemma 3.2.1. If the assumptions of the proposition are satisfied then, for all Φ in 𝒮(F^{2}), the function

a ⟶ μ_{2}^{−1}(a) |a|^{−1/2}W_{Φ} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

is integrable with respect to the additive Haar measure on F and

∫ W_{Φ} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} μ_{2}^{−1}(a) |a|^{−1/2}ψ(ax) da = f_{Φ^{∼}} (( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) .

By definition

f_{Φ^{∼}} (( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) = ∫ Φ^{∼}(t, tx)μ_{1}(t)μ_{2}^{−1}(t) |t| d^{×}t

while

W_{Φ} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} μ_{2}^{−1}(a) |a|^{−1/2} = μ_{1}(a)μ_{2}^{−1}(a) ∫ Φ(at, t^{−1}) μ_{1}(t) μ_{2}^{−1}(t) d^{×}t.  (3.2.2)

After a change of variable the right side becomes

∫ Φ(t, at^{−1})μ_{1}(t)μ_{2}^{−1}(t) d^{×}t.

Computing formally we see that

∫ W_{Φ} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} μ_{2}^{−1}(a) |a|^{−1/2}ψ(ax) da

is equal to

∫_{F} ψ(ax) {∫_{F^{×}} Φ(t, at^{−1})μ_{1}(t)μ_{2}^{−1}(t) d^{×}t} da = ∫_{F^{×}} μ_{1}(t)μ_{2}^{−1}(t) {∫_{F} Φ(t, at^{−1})ψ(ax) da} d^{×}t

which in turn equals

∫_{F^{×}} μ_{1}(t)μ_{2}^{−1}(t) |t| {∫_{F} Φ(t, a)ψ(axt) da} d^{×}t = ∫_{F^{×}} Φ (t, xt)μ_{1}(t)μ_{2}^{−1}(t) |t| d^{×}t.

Our computation will be justified and the lemma proved if we show that the integral

∫_{F^{×}} ∫_{F} |Φ(t, at^{−1})μ_{1}(t)| d^{×}t da

is convergent. It equals

∫_{F^{×}} ∫_{F} |Φ(t, a)| |t|^{s+1} d^{×}t da

which is finite because s is greater than −1.

To show that A is surjective we show that every function f in ℬ(μ_{1}, μ_{2}) is of the form f_{Φ} for some Φ in 𝒮(F^{2}). Given f let Φ(x, y) be 0 if (x, y) is not of the form (0, 1)g for some g in GL(2, O_{F}) but if (x, y) is of this form let Φ(x, y) = μ_{1}^{−1}(detg)f(g). It is easy to see that Φ is well-defined and belongs to 𝒮(F^{2}). To show that f = f_{Φ} we need only show that f(g) = f_{Φ}(g) for all g in GL(2, O_{F}). If g belongs to GL(2, O_{F}) then Φ((0, t)g) = 0 unless t belongs to U_{F} so that

f_{Φ}(g) = μ_{1}(detg) ∫_{U_{F}} Φ((0, t)g)μ_{1}(t)μ_{2}^{−1}(t) dt.

Since

Φ((0, t)g) = μ_{1}^{−1}(t)μ_{2}^{−1}(detg)f (( \begin{matrix}1 & 0 \\ 0 & t \end{matrix} ) g) = μ_{1}^{−1}(t)μ_{2}(t)μ_{2}^{−1}(detg)f(g)

the required equality follows.

Formulae (3.1.2) to (3.1.4) show that A commutes with right translations. Thus to show that A is injective we have to show that W_{Φ}(e) = 0 if f_{Φ} is 0. It follows from the previous lemma that

W_{Φ} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

is zero for almost all a if f_{Φ} is 0. Since W_{Φ} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} is a locally constant function on F^{×} it

must vanish everywhere.

We have incidentally proved the following lemma.

Lemma 3.2.3 Suppose |μ_{1}(ϖ)μ_{2}^{−1}(ϖ)| = |ϖ|^{s} with s > −1 and W belongs to W (μ_{1}, μ_{2}; ψ). If

W \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = 0

for all a in F^{×} then W is 0.

Theorem 3.3 Let μ_{1} and μ_{2} be two quasi-characters of F^{×}.

(i) If neither μ_{1}μ_{2}^{−1} nor μ_{1}^{−1}μ_{2} is α_{F} the representations ρ(μ_{1}, μ_{2}) and ρ(μ_{2}, μ_{2}) are equivalent and irreducible.

(ii) If μ_{1}μ_{2}^{−1} = α_{F} write μ_{1} = χα_{F}^{1/2}, μ_{2} = χα_{F}^{−1/2}. Then ℬ(μ_{1}, μ_{2}) contains a unique proper invariant subspace ℬ_{s}(μ_{1}, μ_{2}) which is irreducible. ℬ(μ_{2}, μ_{1}) also contains a unique proper invariant subspace ℬ_{f}(μ_{2}, μ_{1}). It is one-dimensional and contains the function χ(detg). Moreover the G_{F} -modules ℬ_{s}(μ_{1}, μ_{2}) and ℬ(μ_{2}, μ_{1})/ℬ_{f}(μ_{2}, μ_{1}) are equivalent as are the modules ℬ(μ_{1}, μ_{2})/ℬ_{s}(μ_{1}, μ_{2}) and ℬ_{f}(μ_{2}, μ_{1}).

We start with a simple lemma.

Lemma 3.3.1 Suppose there is a non-zero function f in ℬ(μ_{1}, μ_{2}) invariant under right translations by elements of N_{F}. Then there is a quasi-character χ such that μ_{1} = χα_{F}^{−1/2} and μ_{2} = χα_{F}^{1/2} and f is a multiple of χ.

Since N_{F} A_{F} ({0 \atop 1} {−1 \atop 0})N_{F} is an open subset of G_{F} the function f is determined by its value at ({0 \atop 1} {−1 \atop 0}). Thus if μ_{1} and μ_{2} have the indicated form it must be a multiple of χ.

If c belongs to F^{×} then

( \begin{matrix}1 & 0 \\ c & 1 \end{matrix} ) = ( \begin{matrix}c^{−1} & 1 \\ 0 & c \end{matrix} ) ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ( \begin{matrix}1 & c^{−1} \\ 0 & 1 \end{matrix} ) .

Thus if f exists and ω = μ_{2}μ_{1}^{−1}α_{F}^{−1}

f \begin{aligned}( \\ ( & \begin{matrix}1 & 0 \\ c & 1 \end{matrix} & ) \\ & & ) \end{aligned} = ω(c)f \begin{aligned}( \\ ( & \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} & ) \\ & & ) \end{aligned} .

Since f is locally constant there is an ideal 𝔞 in F such that ω is constant on 𝔞 − {0}. It follows immediately that ω is identically 1 and that μ_{1} and μ_{2} have the desired form.

The next lemma is the key to the theorem.

Lemma 3.3.2. If |μ_{1}μ_{2}(ϖ)| = |ϖ|^{s} with s > −1 there is a minimal non-zero invariant subspace X of ℬ(μ_{1}, μ_{2}). For all f in ℬ(μ_{1}, μ_{2}) and all n in N_{F} the difference f − ρ(n)f belongs to X.

By Proposition 3.2 it is enough to prove the lemma when ℬ(μ_{1}, μ_{2}) is replaced by W (μ_{1}, μ_{2}; ψ). Associate to each function W in W (μ_{1}, μ_{2}; ψ) a function

φ(a) = W \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

on F^{×}. If φ is 0 so is W. We may regard π = ρ(μ_{1}, μ_{2}) as acting on the space V of such functions. If b is in B_{F}

π(b)φ = ξ_{ψ}(b)φ.

Appealing to (3.2.2) we see that every function φ in V has its support in a set of the form

{a ∈ F^{×} | |a| ≤ c}

where c = c(φ) is a constant. As in the second paragraph the difference φ − π(n)φ = φ − ξ_{ψ}(n)φ is in 𝒮(F^{×}) for all n in N_{F}. Thus V ∩ 𝒮(F^{×}) is not 0. Since the representation ξ_{ψ} of B_{F} on 𝒮(F^{×}) is irreducible, V and every non-trivial invariant subspace of V contains 𝒮(F^{×}). Taking the intersection of all such spaces we obtain the subspace of the lemma.

We first prove the theorem assuming that |μ_{1}(ϖ)μ_{2}^{−1}(ϖ)| = |ϖ|^{s} with s > −1. We have defined a non-degenerate pairing between ℬ(μ_{1}, μ_{2}) and ℬ(μ_{1}^{−1}, μ_{2}^{−1}). All elements of the orthogonal complement of X are invariant under N_{F}. Thus if μ_{1}μ_{2}^{−1} is not α_{F} the orthogonal complement is 0 and X is ℬ(μ_{1}, μ_{2}) so that the representation is irreducible. The contragredient representation ρ(μ_{1}^{−1}, μ_{2}^{−1}) is also irreducible.

If μ_{1}μ_{2}^{−1} = α_{F} we write μ_{1} = χα_{F}^{1/2}, μ_{2} = χα_{F}^{−1/2}. In this case X is the space of the functions orthogonal to the function χ^{−1} in ℬ(μ_{1}^{−1}, μ_{2}^{−1}). We set ℬ_{s}(μ_{1}, μ_{2}) = X and we let ℬ_{f}(μ_{1}^{−1}, μ_{2}^{−1}) be the space of scalar multiples of χ^{−1}. The representation of G_{F} on ℬ_{s}(μ_{1}, μ_{2}) is irreducible. Since ℬ_{s}(μ_{1}, μ_{2}) is of codimension one it is the only proper invariant subspace of ℬ(μ_{1}, μ_{2}). Therefore ℬ_{f}(μ_{1}^{−1}, μ_{2}^{−1}) is the only proper invariant subspace of ℬ(μ_{1}^{−1}, μ_{2}^{−1}).

If |μ_{1}(ϖ)μ_{2}^{−1}(ϖ)| = |ϖ|^{s} then |μ_{1}^{−1}(ϖ)μ_{2}(ϖ)| = |ϖ|^{−s} and either s > −1 or −s > −1. Thus if μ_{1}^{−1}μ_{2} is neither α_{F} nor α_{F}^{−1} the representation π = ρ(μ_{1}, μ_{1}) is irreducible. If ω = μ_{1}μ_{2} then

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

so that π is equivalent to ω ⊗ π̃ or to ω ⊗ ρ(μ_{1}^{−1}, μ_{2}^{−1}). It is easily seen that ω ⊗ ρ(μ_{1}^{−1}, μ_{2}^{−1}) is equivalent to ρ(ωμ_{1}^{−1}, ωμ_{1}^{−1}) = ρ(μ_{2}, μ_{1}).

If μ_{1}μ_{2}^{−1} = α_{F} and π is the restriction of ρ to ℬ_{s}(μ_{1}, μ_{2}) then π̃ is the representation on ℬ(μ_{1}^{−1}, μ_{2}^{−1})/ℬ_{f}(μ_{1}^{−1}, μ_{2}^{−1}) defined by ρ(μ_{1}^{−1}, μ_{2}^{−1}). Thus π is equivalent to the tensor product of ω = μ_{1}μ_{2} and this representation. The tensor product is of course equivalent to the representation on ℬ(μ_{2}, μ_{1})/ℬ_{f}(μ_{2}, μ_{1}). If μ_{1} = χα_{F}^{1/2} and μ_{2} = χα_{F}^{−1/2} the representations on ℬ(μ_{1}, μ_{2})/ℬ_{s}(μ_{1}, μ_{2}) and ℬ_{f}(μ_{2}, μ_{1}) are both equivalent to the representations g → χ(detg).

The space W (μ_{1}, μ_{2}; ψ) has been defined for all pairs μ_{1}, μ_{2}.

Proposition 3.4 (i) For all pairs μ_{1}, μ_{2}

W (μ_{1}, μ_{2}; ψ) = W (μ_{2}, μ_{1}; ψ)

(ii) In particular if μ_{1}μ_{2}^{−1} ≠ α_{F}^{−1} the representation of G_{F} on W (μ_{1}, μ_{2}; ψ) is equivalent to ρ(μ_{1}, μ_{2}).

If Φ is a function on F^{2} define Φ^{ι} by

Φ^{ι}(x, y) = Φ(y, x).

To prove the proposition we show that, if Φ is in 𝒮(F^{2}),

μ_{1}(detg) |detg|^{1/2}θ(μ_{1}, μ_{2}; r(g)Φ^{ι}) = μ_{2}(detg) |detg|^{1/2}θ(μ_{2}, μ_{1}; r(g)Φ).

Chapter 1 54 If g is the identity this relation follows upon inspection of the definition of θ(μ_{1}, μ_{2}; Φ^{ι}). It is also easily seen that

r(g)Φ^{ι} = [r(g)Φ]^{ι}

if g is in SL(2, F) so that it is enough to prove the identity for

g = ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) .

It reduces to

μ_{1}(a) ∫ Φ^{ι}(at, t^{−1})μ_{1}(t)μ_{2}^{−1}(t) d^{×}t = μ_{2}(a) ∫ Φ(at, t^{−1})μ_{2}(t)μ_{2}^{−1}(t) d^{×}t.

The left side equals

μ_{1}(a) ∫ Φ(t^{−1}, at)μ_{1}(t)μ_{2}^{−1}(t) d^{×}t

which, after changing the variable of integration, one sees is equal to the right side.

If μ_{1}μ_{2}^{−1} is not α_{F} or α_{F}^{−1} so that ρ(μ_{1}, μ_{2}) is irreducible we let π(μ_{1}, μ_{2}) be any representation in the class of ρ(μ_{1}, μ_{2}). If ρ(μ_{1}, μ_{2}) is reducible it has two constituents one finite dimensional and one infinite dimensional. A representation in the class of the first will be called π(μ_{1}, μ_{2}). A representation in the class of the second will be called σ(μ_{1}, μ_{2}). Any irreducible representation which is not absolutely cuspidal is either a π(μ_{1}, μ_{2}) or a σ(μ_{1}, μ_{2}). The representations σ(μ_{1}, μ_{2}) which are defined only for certain values of μ_{1} and μ_{2} are called special representations.

Before proceeding to the proof of Theorem 2.18 for representations which are not absolutely cuspidal we introduce some notation. If ω is an unramified quasi-character of F^{×} the associated L-function is

L(s, ω) = \frac{1}{1 − ω(ϖ) |ϖ|^{s}} .

It is independent of the choice of the generator ϖ of 𝔭. If ω is ramified L(s, ω) = 1. If φ belongs to 𝒮(F) the integral

Z(ωα_{F}^{s} , φ) = ∫_{F^{×}} φ(α)ω(α) |α|^{s} d^{×}α

is absolutely convergent in some half-plane Re s > s_{0} and the quotient

\frac{Z(ωα_{F}^{s} , φ)}{L(s, ω)}

can be analytically continued to a function holomorphic in the whole complex plane. Moreover for a suitable choice of φ the quotient is 1. If ω is unramified and

∫_{U_{F}} d^{×}α = 0

one could take the characteristic function of O_{F}. There is a factor ε(s, ω, ψ) which, for a given ω and ψ, is of the form ab^{s} so that if φ̂ is the Fourier transform of φ

\frac{Z(ω^{−1}α_{F}^{1−s}, φ̂)}{L(1 − s, ω^{−1})} = ε(s, ω, ψ)\frac{Z(ωα_{F}^{s} , φ)}{L(s, ω)}.

If ω is unramified and O_{F} is the largest ideal on which ψ is trivial ε(s, ω, ψ) = 1.

Proposition 3.5 Suppose μ_{1} and μ_{2} are two quasi-characters of F^{×} such that neither μ_{1}^{−1}μ_{2} nor μ_{1}μ_{2}^{−1} is α_{F} and π is π(μ_{1}, μ_{2}). Then

W (π, ψ) = W (μ_{1}, μ_{2}; ψ)

and if

L(s, π) = L(s, μ_{1}) L(s, μ_{2})
L(s, π̃) = L(s, μ_{1}^{−1}) L(s, μ_{2}^{−1})
ε(s, π, ψ) = ϵ(s, μ_{1}, ψ) ε(s, μ_{2}, ψ)

all assertions of Theorem 2.18 are valid. In particular if |μ_{1}(ϖ)| = |ϖ|^{−s_{1}} and |μ_{2}(ϖ)| = |ϖ|^{−s_{2}} the integrals defining Ψ(g, s, W) are absolutely convergent if Re s > max{s_{1}, s_{2}}. If μ_{1} and μ_{2} are unramifed and O_{F} is the largest ideal of F on which ψ is trivial there is a unique function W_{0} in W (π, ψ) which is invariant under GL(2, O_{F}) and assumes the value 1 at the identity. If

∫_{U_{F}} d^{×}α = 1

then Φ(e, s, W_{0}) = 1.

That W (π, ψ) = W (μ_{1}, μ_{2}; ψ) is of course a consequence of the previous proposition. As we observed the various assertions need be established only for g = e. Take Φ in 𝒮(F^{2}) and let W = W_{Φ} be the corresponding element of W (π, ψ). Then

φ(a) = W \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

belongs to the space of the Kirillov model of π. As we saw in the closing pages of the first paragraph

Ψ(e, s, W) = ∫_{F^{×}} W \begin{matrix}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix} |a|^{s−1/2} d^{×}a = φ̂(α_{F}^{s−1/2})

is equal to

Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , Φ)

if the last and therefore all of the integrals are defined.

Also

Ψ̃(e, s, W) = Z(μ_{2}^{−1}α_{F}^{s} , μ_{1}^{−1}α_{F}^{s} , Φ).

Any function in 𝒮(F^{2}) is a linear combination of functions of the form

Φ(x, y) = φ_{1}(x)φ_{2}(y).

Since the assertions to be proved are all linear we need only consider the functions Φ which are given as products. Then

Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , Φ) = Z(μ_{1}α_{F}^{s} , φ_{1})Z(μ_{2}α_{F}^{s} , φ_{2})

so that the integral does converge in the indicated region. Moreover

Z(μ_{2}^{−1}α_{F}^{s} , μ_{1}^{−1}α_{F}^{s} , Φ) = Z(μ_{2}^{−1}α_{F}^{s} , φ_{1})Z(μ_{1}^{−1}α_{F}^{s} , φ_{2})

also converges for Re s sufficiently large. Φ(e, s, W) is equal to

\frac{Z(μ_{1}α_{F}^{s} , φ_{1})}{L(s, μ_{1})} \frac{Z(μ_{2}α_{F}^{s} , φ_{2})}{L(s, μ_{2})}

and is holomorphic in the whole complex plane. We can choose φ_{1} and φ_{2} so that both factors are 1.

It follows from the Iwasawa decomposition G_{F} = P_{F} GL(2, O_{F}) that if both μ_{1} and μ_{2} are unramified there is a non-zero function on ℬ(μ_{1}, μ_{2}) which is invariant under GL(2, O_{F}) and that it is unique up to a scalar factor. If the largest ideal on which ψ is trivial is O_{F}, if Φ_{0} is the characteristic function of O_{F}^{2} , and if Φ_{0}^{∼} is the partial Fourier transform introduced in Proposition 1.6 then Φ_{0}^{∼} = Φ_{0}. Consequently

r_{μ_{1},μ_{2}}(g)Φ_{0} = Φ_{0}

for all g in GL(2, O_{F}). If W_{0} = W_{Φ_{0}} then, since Φ_{0} is the product of the characteristic function of O_{F} with itself, Φ(e, s, W_{0}) = 1 if

∫_{U_{F}} d^{×}α = 1.

The only thing left to prove is the local functional equation. Observe that

Φ̃(w, s, W) = Φ̃(e, s, ρ(w)W ),

that if W = W_{Φ} then ρ(w)W = W_{r(w)Φ}, and that r(w)Φ(x, y) = Φ′(y, x) if Φ′ is the Fourier transform of Φ. Thus if Φ(x, y) is a product φ_{1}(x)φ_{2}(y)

Φ̃(w, s, W) = \frac{Z(μ_{1}^{−1}α_{F}^{s} , φ̂_{1})}{L(s, μ_{1}^{−1})} \frac{Z(μ_{2}^{−1}α_{F}^{s} , φ̂_{2})}{L(s, μ_{2}^{−1})}.

The functional equation follows immediately.

If μ_{1}μ_{2}^{−1} is α_{F} or α_{F}^{−1} and π = π(μ_{1}, μ_{2}) we still set

L(s, π) = L(s, μ_{1}) L(s, μ_{2})

and

ε(s, π, ψ) = ε(s, μ_{1}, ψ) ε(s, μ_{2}, ψ).

Since π̃ is equivalent to π(μ_{1}^{−1}, μ_{2}^{−1})

L(s, π̃) = L(s, μ_{1}^{−1})L(s, μ_{2}^{−1}).

Theorem 2.18 is not applicable in this case. It has however yet to be proved for the special representations. Any special representation σ is of the form σ(μ_{1}, μ_{2}) with μ_{1} = χα_{F}^{1/2} and μ_{2} = χα_{F}^{−1/2}. The contragredient representation of σ̃ is σ(μ_{2}^{−1}, μ_{1}^{−1}). This choice of μ_{1} and μ_{2} is implicit in the following proposition.

Proposition 3.6 W (σ, ψ) is the space of functions W = W_{Φ} in W (μ_{1}, μ_{2}; ψ) for which

∫_{F} Φ(x, 0) dx = 0.

Theorem 2.18 will be valid if we set L(s, σ) = L(s, σ̃) = 1 and ε(s, σ, ψ) = ε(s, μ_{1}, ψ) ε(s, μ_{2}, ψ) when χ is ramified and we set L(s, σ) = L(s, μ_{1}), L(s, σ̃) = L(s, μ_{2}^{−1}), and

ε(s, σ, ψ) = ε(s, μ_{1}, ψ) ε(s, μ_{2}, ψ) \frac{L(1 − s, μ_{1}^{−1})}{L(s, μ_{2})}

when χ is unramified.

W (σ, ψ) is of course the subspace of W (μ_{1}, μ_{2}; ψ) corresponding to the space ℬ_{s}(μ_{1}, μ_{2}) under the transformation A of Proposition 3.2. If W = W_{Φ} then A takes W to the function f = f_{Φ^{∼}} defined by

f(g) = z(μ_{1}μ_{2}^{−1}α_{F} , ρ(g)Φ^{∼})μ_{1}(detg) |detg|^{1/2}.

f belongs to ℬ_{s}(μ_{1}, μ_{2}) if and only if

∫_{GL(2,O_{F})} χ^{−1}(g)f(g) dg = 0.

As we observed this integral is equal to a constant times

∫_{F} χ^{−1} (w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) f (w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) dx = ∫_{F} f (w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) dx

which equals

∫ z (α_{F}^{2} , ρ(w) ρ \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} Φ^{∼}) dx = ∫ \{∫ Φ^{∼}(−t, −tx) |t|^{2} d^{×}t\} dx.

The double integral does converge and equals, apart from a constant factor,

∫∫ Φ^{∼}(t, tx) |t| dt dx = ∫∫ Φ^{∼}(t, x) dt dx

which in turn equals

∫ Φ(t, 0) dt.

We now verify not only the remainder of the theorem but also the following corollary.

Corollary 3.7 (i) If π = π(μ_{1}, μ_{2}) then

ε(s, σ, ψ) \frac{L(1 − s, σ̃)}{L(s, σ)} = ε(s, π, ψ) \frac{L(1 − s, π̃)}{L(s, π)}

(ii) The quotient

\frac{L(s, π)}{L(s, σ)}

is holomorphic (iii) For all Φ such that

∫ Φ(x, 0) dx = 0

the quotient

\frac{Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , Φ)}{L(s, σ)}

is holomorphic and there exists such a Φ for which the quotient is one.

The first and second assertions of the corollary are little more than matters of definition. Although W (μ_{1}, μ_{2}ψ) is not irreducible we may still, for all W in this space, define the integrals

\begin{aligned}Ψ(g, s, W) & = ∫ W (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s−1/2} d^{×}a \\ Ψ̃(g, s, W) & = ∫ W (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s−1/2}ω^{−1}(a) d^{×}a. \end{aligned}

They may be treated in the same way as the integrals appearing in the proof of Proposition 3.5. In particular they converge to the right of some vertical line and if W = W_{Φ}

\begin{aligned}Ψ(e, s, W) & = Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , Φ) \\ Ψ̃(e, s, W) & = Z(μ_{2}^{−1}α_{F}^{s} , μ_{1}^{−1}α_{F}^{s} , Φ). \end{aligned}

Moreover

\frac{Ψ(g, s, W)}{L(s, π)}

is a holomorphic function of s and

\frac{Ψ̃(g, 1 − s, W)}{L(1 − s, π̃)} = ε(s, π, ψ)\frac{Ψ(g, s, W)}{L(s, π)}.

Therefore

Φ(g, s, W) = \frac{Ψ(g, s, W)}{L(s, σ)}

and

Φ̃(g, s, W) = \frac{Ψ̃(g, s, W)}{L(s, σ̃)}

are meromorphic functions of s and satisfy the local functional equation

Φ̃(wg, 1 − s, W) = ε(s, σ, ψ) Φ(g, s, W).

To compete the proof of the theorem we have to show that ε(s, σ, ψ) is an exponential function of s and we have to verify the third part of the corollary. The first point is taken care of by the observation that μ_{1}^{−1}(ϖ) |ϖ| = μ_{2}^{−1}(ϖ) so that

\frac{L(1 − s, μ_{1}^{−1})}{L(s, μ_{2})} = \frac{1 − μ_{2}(ϖ) |ϖ|^{s}}{1 − μ_{1}^{−1}(ϖ) |ϖ|^{1−s}} = −μ_{1}(ϖ) |ϖ|^{s−1}.

If χ is ramified so that L(s, σ) = L(s, π) the quotient part (iii) of the corollary is holomorphic. Moreover a Φ in 𝒮(F^{2}) for which

Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , Φ) = L(s, σ) = 1

can be so chosen that

Φ(εx, ηy) = χ(εη)Φ(x, y)

for ε and η in U_{F}. Then

∫_{F} Φ(x, 0) dx = 0.

Now take χ unramified so that χ(a) = |a|^{r} for some complex number r. We have to show that if

∫_{F} Φ(x, 0) dx = 0

then

\frac{Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , Φ)}{L(s, μ_{1})}

is a holomorphic function of s. Replacing s by s − r + 1/2 we see that it is enough to show that

(1 − |ϖ|^{s+1})∫∫ Φ(x, y) |x|^{s+1} |y|^{s} d^{×}x d^{×}y

is a holomorphic function of s. Without any hypothesis on Φ the integral converges for Re s > 0 and the product has an analytic continuation whose only poles are at the roots of |ϖ|^{s} = 1. To see that these poles do not occur we have only to check that there is no pole at s = 0. For a given Φ in 𝒮(F^{2}) there is an ideal 𝔞 such that

Φ(x, y) = Φ(x, 0)

for y in 𝔞. If 𝔞′ is the complement of 𝔞

∫∫ Φ(x, y) |x|^{s+1} |y|^{s} d^{×}x d^{×}y

is equal to the sum of

∫_{F} ∫_{𝔞′} Φ(x, y) |x|^{s+1} |y|^{s} d^{×}x d^{×}y

which has no pole at s = 0 and a constant times

{∫_{F} Φ(x, 0) |x|^{s} dx} {∫_{𝔞} |y|^{s} d^{×}y}

If 𝔞 = 𝔭^{n} the second integral is equal to

|ϖ|^{ns}(1 − |ϖ|^{s})^{−1}

If

∫_{F} Φ(x, 0) dx = 0

the first term, which defines a holomorphic function of s, vanishes at s = 0 and the product has no pole there.

If φ_{0} is the characteristic function of O_{F} set

Φ(x, y) = {φ_{0}(x) − |ϖ|^{−1}φ_{0}(ϖ^{−1}x)} φ_{0}(y).

Then

∫_{F} Φ(x, 0) dx = 0

and

Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , Φ)

is equal to

{∫ (φ_{0}(x) − |ϖ|^{−1}φ_{0}(ϖ^{−1}x)) μ_{1}(x) |x|^{s} d^{×}x} {∫ φ_{0}(y) μ_{2}(y) |y|^{s} d^{×}y}

The second integral equals L(s, μ_{2}) and the first equals

(1 − μ_{1}(ϖ) |ϖ|^{s−1}) L(s, μ_{1})

so their product is L(s, μ_{1}) = L(s, σ).

Theorem 2.18 is now completely proved. The properties of the local L-functions L(s, π) and the factors ϵ(s, π, ψ) described in the next proposition will not be used until the paragraph on extraordinary representations.

Proposition 3.8 (i) If π is an irreducible representation there is an integer m such that if the order of χ is greater than m both L(s, χ ⊗ π) and L(s, χ ⊗ π̃) are 1.

(ii) Suppose π_{1} and π_{2} are two irreducible representations of G_{F} and that there is a quasicharacter ω such that

π_{1} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I π_{2} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

Then there is an integer m such that if the order of χ is greater than m

ε(s, χ ⊗ π_{1}, ψ) = ε(s, χ ⊗ π_{2}, ψ)

(iii) Let π be an irreducible representation and let ω be the quasi-character defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

Write ω in any manner as ω = μ_{1}μ_{2}. Then if the order of χ is sufficiently large in comparison to the orders of μ_{1} and μ_{2}

ε(s, χ ⊗ π, ψ) = ε(s, χμ_{1}, ψ) ε(s, χμ_{2}, ψ).

It is enough to treat infinite-dimensional representations because if σ = σ(μ_{1}, μ_{2}) and π = π(μ_{1}, μ_{2}) are both defined L(s, χ ⊗ σ) = L(s, χ ⊗ π), L(s, χ ⊗ σ̃) = L(s, χ ⊗ π̃), and ε(s, χ ⊗ σ, ψ) = ε(s, χ ⊗ π, ψ) if the order of χ is sufficiently large.

If π is not absolutely cuspidal the first part of the proposition is a matter of definition. If π is absolutely cuspidal we have shown that L(s, χ ⊗ π) = L(s, χ ⊗ π̃) = 1 for all π.

According to the relation (2.18.1)

ε(s, χ ⊗ π, ψ) = C(ν_{0}^{−1}ν_{1}^{−1}, z_{0}^{−1}z_{1}^{−1}q^{−1/2}z^{−1})

if the order of χ is so large that L(s, χ ⊗ π) = L(s, χ^{−1} ⊗ π̃) = 1. Thus to prove the second part we have only to show that if {C_{1}(ν, t)} and {C_{2}(ν, t)} are the series associated to π_{1} and π_{2} then

C_{1}(ν, t) = C_{2}(ν, t)

if the order of ν is sufficiently large. This was proved in Lemma 2.16.6. The third part is also a consequence of that lemma but we can obtain it by applying the second part to π_{1} = π and to π_{2} = π(μ_{1}, μ_{2}).

We finish up this paragraph with some results which will be used in the Hecke theory to be developed in the second chapter.

Lemma 3.9 The restriction of the irreducible representation π to GL(2, O_{F}) contains the trivial representation if and only if there are two unramified characters μ_{1} and μ_{2} such that π = π(μ_{1}, μ_{2}).

This is clear if π is one-dimensional so we may as well suppose that π is infinite dimensional. If π = π(μ_{1}, μ_{2}) we may let π = ρ(μ_{1}, μ_{2}). It is clear that there is a non-zero vector in ℬ(μ_{1}, μ_{2}) invariant under GL(2, O_{F}) if and only if μ_{1} and μ_{2} are unramified and that if there is such a vector it is determined up to a scalar factor. If π = σ(μ_{1}, μ_{2}) and μ_{1}μ_{2}^{−1} = α_{F} we can suppose that π is the restriction of ρ(μ_{1}, μ_{2}) to ℬ_{S}(μ_{1}, μ_{2}). The vectors in ℬ(μ_{1}, μ_{2}) invariant under GL(2, O_{F}) clearly do not lie in ℬ_{s}(μ_{1}, μ_{2}) so that the restriction of π to GL(2, O_{F}) does not contain the trivial representation. All that we have left to do is to show that the restiction of an absolutely cuspidal representation to GL(2, O_{F}) does not contain the trivial representation.

Suppose the infinite-dimensional irreducible representation π is given in the Kirillov form with respect to an additive character ψ such that O_{F} is the largest ideal on which ψ is trivial. Suppose the non-zero vector φ is invariant under GL(2, O_{F}). It is clear that if

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

Chapter 1 62 then ω is unramified, that φ(ν, t) = 0 unless ν = 1 is the trivial character, and that φ(ν, t) has no pole at t = 0. Suppose π is absolutely cuspidal so that φ belongs to 𝒮(F^{×}). Since π(w)φ = φ and the restriction of ω to U_{F} is trivial

φ̂(1, t) = C(1, t) φ̂(1, z_{0}^{−1}t^{−1})

if z_{0} = ω(ϖ). Since C(1, t) is a constant times a negative power of t the series on the left involves no negative powers of t and that on the right involves only negative powers. This is a contradiction.

Let ℋ_{0} be the subalgebra of the Hecke algebra formed by the functions which are invariant under left and right translations by elements of GL(2, O_{F}). Suppose the irreducible representation π acts on the space X and there is a non-zero vector x in X invariant under GL(2, O_{F}). If f is in ℋ_{0} the vector π(f)x has the same property and is therefore a multiple λ(f)x of x. The map f → λ(f) is a non-trivial homomorphism of ℋ_{0} into the complex numbers.

Lemma 3.10 Suppose π = π(μ_{1}, μ_{2}) where μ_{1} and μ_{2} are unramified and λ is the associated homomorphism of ℋ_{0} into ℂ. There is a constant c such that

|λ(f)| ≤ c ∫_{G_{F}} |f(g)| dg  (3.10.1)

for all f in ℋ if and only if μ_{1}μ_{2} is a character and |μ_{1}(ϖ)μ_{2}^{−1}(ϖ)| = |ϖ|^{s} with −1 ≤ s ≤ 1.

Let π̃ act on X̃ and let x̃ in X̃ be such that ⟨x, x̃⟩ ≠ 0. Replacing x̃ by

∫_{GL(2,O_{F})} π̃(g)x̃ dg

if necessary we may suppose that x̃ is invariant under GL(2, O_{F}). We may also assume that ⟨x, x̃⟩ = 1. If η(g) = ⟨π(g)x, x̃⟩ then

λ(f) η(g) = ∫_{G_{F}} η(gh) f(h) dh

for all f in ℋ_{0}. In particular

λ(f) = ∫_{G_{F}} η(h) f(h) dh.

If |η(h)| ≤ c for all h the inequality (3.10.1) is certainly valid. Conversely, since η is invariant under left and right translations by GL(2, O_{F}) we can, if the inequality holds, apply it to the characteristic functions of double cosets of this group to see that |η(h)| ≤ c for all h. Since

η (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) h) = μ_{1}(a) μ_{2}(a) η(h)

the function η is bounded only if μ_{1}μ_{2} is a character as we now assume it to be. The finite dimensional representations take care of themselves so we now assume π is infinite-dimensional.

Since π and π̃ are irreducible the function ⟨π(g)x, x̃⟩ is bounded for a given pair of non-zero vectors if and only if it is bounded for all pairs. Since G_{F} = GL(2, O_{F}) A_{F} GL(2, O_{F}) and μ_{1}μ_{2} is a character these functions are bounded if and only if the functions

⟨π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} x, x̃⟩

are bounded on F^{×}. Take π and π̃ in the Kirillov form. If φ is in V and φ̃ is in V then

⟨π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ, π̃(w)φ̃⟩

is equal to

⟨π^{−1}(w) π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ, φ̃⟩ = μ_{1}(a) μ_{2}(a) < π \begin{aligned}( \\ ( & \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π^{−1}(w)φ, φ̃ >

Thus η(g) is bounded if and only if the functions

⟨π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ, φ̃⟩

are bounded for all φ in V and all φ̃ in 𝒮(F^{×}).

It is not necessary to consider all φ̃ in 𝒮(F^{×}) but only a set which together with its translates by the diagonal matrices spans 𝒮(F^{×}). If μ is a character of U_{F} let φ_{μ} be the function on F^{×} which is 0 outside of U_{F} and equals μ on U_{F}. It will be sufficient to consider the functions φ̃ = φ_{μ} and all we need show is that

⟨π \begin{aligned}( \\ ( & \begin{matrix}ϖ^{n} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ, φ_{μ}⟩  (3.10.2)

is a bounded function of n for all μ and all φ. The expression (3.10.2) is equal to φ̂_{n}(μ). If φ belongs to 𝒮(F^{×}) the sequence {φ̂_{n}(μ)} has only finitely many non-zero terms and there is no problem. If φ = π(w)φ_{0} then

∑_{n} φ̂_{n}(μ)t^{n} = C(μ, t) η(t)

where η(t) depends on φ_{0} and is an arbitrary finite Laurent series. We conclude that (3.10.1) is valid if and only if μ_{1}μ_{2} is a character and the coefficients of the Laurent series C(μ, t) are bounded for every choice of μ.

It follows from Proposition 3.5 and formula (2.18.1) that, in the present case, the series has only one term if μ is ramified but that if μ is trivial

C(μ, |ϖ|^{1/2} μ_{1}^{−1}(ϖ) μ_{2}^{−1}(ϖ)t) = \frac{(1 − μ_{1}(ϖ)t^{−1})(1 − μ_{2}(ϖ)t^{−1})}{(1 − μ_{1}^{−1}(ϖ) |ϖ| t)(1 − μ_{2}^{−1}(ϖ) |ϖ| t)}.

The function on the right has zeros at t = μ_{1}(ϖ) and t = μ_{2}(ϖ) and poles at t = 0, t = |ϖ|^{−1}μ_{1}(ϖ), and t = |ϖ|^{−1}μ_{2}(ϖ). A zero can cancel a pole only if μ_{2}(ϖ) = |ϖ|^{−1}μ_{1}(ϖ) or μ_{1}(ϖ) = |ϖ|^{−1}μ_{2}(ϖ). Since μ_{1} and μ_{2} are unramified this would mean that μ_{1}^{−1}μ_{2} equals α_{F} or α_{F}^{−1} which is impossible when π = π(μ_{1}, μ_{2}) is infinite dimensional.

If C(μ, t) has bounded coefficients and μ_{1}μ_{2} is a character the function on the right has no poles for |t| < |ϖ|^{−1/2} and therfore |μ_{1}(ϖ)| ≥ |ϖ|^{1/2} and |μ_{2}(ϖ)| ≥ |ϖ|^{1/2}. Since

|μ_{1}(ϖ)μ_{2}^{−1}(ϖ)| = |μ_{1}(ϖ)|^{2} = |μ_{2}^{−1}(ϖ)|^{2}

where μ_{1}μ_{2} is a character these two inequalities are equivalent to that of the lemma. Conversely if these two inequalities are satisifed the rational function on the right has no pole except that at 0 inside the circle |t| = |ϖ|^{−1/2} and at most simple poles on the circle itself. Applying, for example, partial fractions to find its Laurent series expansion about 0 one finds that the coefficients of C(μ, t) are bounded.

Chapter 1 64 Lemma 3.11 Suppose μ_{1} and μ_{2} are unramified, μ_{1}μ_{2} is a character, and π = π(μ_{1}, μ_{2}) is infinite dimensional. Let |μ_{1}(ϖ)| = |ϖ|^{r} where r is real so that |μ_{2}(ϖ)| = |ϖ|^{−r}. Assume O_{F} is the largest ideal on which ψ is trivial and let W_{0} be that element of W (π, ψ) which is invariant under GL(2, O_{F}) and takes the value 1 at the identity. If s > |r| then

∫_{F^{×}} |W_{0} \begin{matrix}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix}| |a|^{s−1/2} d^{×}a ≤ \frac{1}{(1 − |ϖ|^{s+r})(1 − |ϖ|^{s−r})}

if the Haar measure is so normalized that the measure of U_{F} is one.

If Φ is the characteristic function of O_{F}^{2} then

W_{0} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = μ_{1}(a) |a|^{1/2} ∫_{F^{×}} Φ(at, t^{−1}) μ_{1}(t) μ_{2}^{−1} d^{×}t

and

∫_{F^{×}} |W_{0} \begin{matrix}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix}| |a|^{s−1/2} d^{×}a ≤ ∫∫ Φ(at, t^{−1}) |a|^{s+r}|t|^{2r} d^{×}a d^{×}t.

Changing variables in the left-hand side we obtain

∫_{O_{F}} ∫_{O_{F}} |a|^{s+r}|b|^{s−r} d^{×}a d^{×}b = \frac{1}{(1 − |ϖ|^{s+r})(1 − |ϖ|^{s−r})}.

Chapter 1 65 §4. Examples of absolutely cuspidal representations In this paragraph we will use the results of the first paragraph to construct some examples of absolutely cuspidal representations.

First of all let K be a quaternion algebra over F. K is of course unique up to isomorphism. As in the first paragraph Ω will denote a continuous finite-dimensional representation of K^{×} the multiplicative group of K. If χ is a quasi-character of F^{×} and ν is the reduced norm on K we denote the one-dimensional representation g → χ(ν(g)) of K^{×} by χ also. If Ω is any representation χ ⊗ Ω is the representation g → χ(g) Ω(g). If Ω is irreducible all operators commuting with the action of K^{×} are scalars. In particular there is a quasi-character Ω of F^{×} such that

Ω(a) = ω(a)I

for all a in F^{×} which is of course a subgroup of K^{×}. If Ω is replaced by χ ⊗ Ω then ω is replaced by χ^{2}ω. Ω̃ will denote the representation contragredient to Ω.

Suppose Ω is irreducible, acts on V, and the quasi-character ω is a character. Since K^{×}/F^{×} is compact there is a positive definite hermitian form on V invariant under K^{×}. When this is so we call Ω unitary.

It is a consequence of the following lemma that any one-dimensional representation of K^{×} is the representation associated to a quasi-character of F^{×}.

Lemma 4.1 Let K_{1} be the subgroup of K^{×} consisting of those x for which ν(x) = 1. Then K_{1} is the commutator subgroup, in the sense of group theory, of K^{×}.

K_{1} certainly contains the commutator subgroup. Suppose x belongs to K_{1}. If x = x^{ι} then x^{2} = xx^{ι} = 1 so that x = ±1. Otherwise x determines a separable quadratic extension of F. Thus, in all cases, if xx^{ι} = 1 there is a subfield L of K which contains x and is quadratic and separable over L. By Hilbert’s Theorem 90 there is a y in L such that x = yy^{−ι}. Moreover there is an element σ in K such that σzσ^{−1} = z^{ι} for all z in L. Thus x = yσy^{−1}σ^{−1} is in the commutator subgroup.

In the first paragraph we associated to Ω a representation r_{Ω} of a group G_{+} on the space 𝒮(K, Ω). Since F is now non-archimedean the group G_{+} is now G_{F} = GL(2, F).

Theorem 4.2 (i) The representation r_{Ω} is admissible.

(ii) Let d = degree Ω. Then r_{Ω} is equivalent to the direct sum of d copies of an irreducible

representation π(Ω). (iii) If Ω is the representation associated to a quasi-character χ of F^{×} then

π(Ω) = σ(χα_{F}^{1/2}, χα_{F}^{−1/2}). (iv) If d > 1 the representation π(Ω) is absolutely cuspidal.

If n is a natural number we set

G_{n} = {g ∈ GL(2, O_{F}) | g = I(mod 𝔭^{n})}

We have first to show that if Φ is in 𝒮(K, Ω) there is an n such that r_{Ω}(g)Φ = Φ if g is in G_{n} and that for a given n the space of Φ in 𝒮(K, Ω) for which r_{Ω}(g)Φ = Φ for all g in G_{n} is finite dimensional.

Any

g = ( \begin{matrix}a & b \\ c & d \end{matrix} )

Chapter 1 66 in G_{n} may be written as

g = ( \begin{matrix}1 & 0 \\ ca^{−1} & 1 \end{matrix} ) ( \begin{matrix}a & b′ \\ 0 & d′ \end{matrix} )

and both the matrices on the right are in G_{n}. Thus G_{n} is generated by the matrices of the forms

( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} )  ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )  w ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) w^{−1}  w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) w^{−1}

with a ≡ 1 (mod 𝔭^{n}) and x ≡ 0 (mod 𝔭^{n}). It will therefore be enough to verify the following three assertions.

(4.2.1) Given Φ there is an n > 0 such that

r_{Ω} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} Φ = Φ

if a ≡ 1 (mod 𝔭^{n})

(4.2.2) Given Φ there is an n > 0 such that

r_{Ω} \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} Φ = Φ

if x ≡ 0 (mod 𝔭^{n}).

(4.2.3) Given n > 0 the space of Φ in 𝒮(K, Ω) such that

r_{Ω} \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} Φ = Φ

and

r_{Ω}(w^{−1}) r_{Ω} \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} r_{Ω}(w)Φ = Φ

for all x in 𝔭^{n} is finite-dimensional.

If a = ν(h) then

r_{Ω} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} Φ = |h|_{K}^{1/2} Ω(h) Φ(xh).

Since Φ has compact suport in K and is locally constant there is a neighborhood U of 1 in K^{×} such that

Ω(h) Φ(xh) |h|_{K}^{1/2} = Φ(x)

for all h in U and all x in K. The assertion (4.2.1) now follows from the observation that ν is an open mapping of K^{×} onto F^{×}.

We recall that

r_{Ω} \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} Φ(z) = ψ(xν(z)) Φ(z)

Let 𝔭^{−ℓ} be the largest ideal on which ψ is trivial and let 𝔭_{K} be the prime ideal of K. Since ν(𝔭_{K}^{m}) = 𝔭_{F}^{m}

r_{Ω} \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} Φ = Φ

for all x in 𝔭^{n} if and only if the support of Φ is contained in 𝔭_{K}^{−n−ℓ}. With this (4.2.2) is established.

Φ satisfies the two conditions of (4.2.3) if and only if both Φ and r(w)Φ have support in 𝔭_{K}^{−n−ℓ} or, since r(w)Φ = −Φ′, if and only if Φ and Φ′, its Fourier transform, have support in this set. There is certainly a natural number k such that ψ(τ(y)) = 1 for all y in 𝔭_{K}^{k}. Assertion (4.2.3) is therefore a consequence of the following simple lemma.

Chapter 1 67 Lemma 4.2.4 If the support of Φ is contained in 𝔭_{K}^{−n} and ψ(τ(y)) = 1 for all y in 𝔭_{K}^{k} the Fourier transform of Φ is constant on cosets of 𝔭_{K}^{k+n}.

Since

Φ′(x) = ∫_{𝔭_{K}^{−n}} Φ(y) ψ(τ(x, y)) dy

the lemma is clear.

We prove the second part of the theorem for one-dimensional Ω first. Let Ω be the representation associated to χ. 𝒮(K, Ω) is the space of Φ in 𝒮(K) such that Φ(xh) = Φ(x) for all h in K_{1}. Thus to every Φ in 𝒮(K, Ω) we may associate the function φ_{Φ} on F^{×} defined by

φ_{Φ}(a) = |h|_{K}^{1/2} Ω(h) Φ(h)

if a = ν(h). The map Φ → φ_{Φ} is clearly injective. If φ belongs to 𝒮(F^{×}) the function Φ defined by

Φ(h) = |h|_{K}^{−1/2} Ω^{−1}(h) φ(ν(h))

if h ≠ 0 and by Φ(0) = 0

belongs to 𝒮(K, Ω) and φ = φ_{Φ}. Let 𝒮_{0}(K, Ω) be the space of functions obtained in this way. It is the space of functions in 𝒮(K, Ω) which vanish at 0 and therefore is of codimension one. If Φ belongs to 𝒮_{0}(K, Ω), is non-negative, does not vanish identically and Φ′ is its Fourier transform then

Φ′(0) = ∫ Φ(x) dx ≠ 0.

Thus r_{Ω}(w)Φ does not belong to 𝒮_{0}(K, Ω) and 𝒮_{0}(K, Ω) is not invariant. Since it is of codimension one there is no proper invariant subspace containing it.

Let V be the image of 𝒮(K, ω) under the map Φ → φ_{Φ}. We may regard r_{Ω} as acting in V. ¿From the original definitions we see that

r_{Ω}(b)φ = ξ_{ψ}(b)φ

if b is in B_{F}. If V_{1} is a non-trivial invariant subspace of V the difference

φ − r_{Ω} \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ

is in V_{0} ∩ V_{1} for all φ in V_{1} and all x in F. If φ is not zero we can certainly find an x for which the difference is not zero. Consequently V_{0} ∩ V_{1} is not 0 so that V_{1} contains V_{0} and hence all of V.

r_{Ω} is therefore irreducible and when considered as acting on V it is in the Kirillov form. Since V_{0} is not V it is not absolutely cuspidal. It is thus a π(μ_{1}, μ_{2}) or a σ(μ_{1}, μ_{2}). To see which we have to find a linear form on V which is trivial on V_{0}. The obvious choice is

L(φ) = Φ(0)

if φ = φ_{Φ}. Then

L (r_{Ω} (( \begin{matrix}a_{1} & 0 \\ 0 & a_{2} \end{matrix} ) φ)) = χ(a_{1}a_{2}) |\frac{a_{1}}{a_{2}} | L(φ).

To see this we have only to recall that

r_{Ω} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = Ω(a)I = χ^{2}(a)I

and that

r_{Ω} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} Φ(0) = |h|_{K}^{1/2} Ω(h) Φ(0)

where a = ν(h) so that |h|_{K}^{1/2} = |a|_{F}^{2} and Ω(h) = χ(a)I. Thus if

Aφ(g) = L(r_{Ω}(g)φ)

A is an injection of V into an irreducible invariant subspace of ℬ(χα_{F}^{1/2}, χα_{F}^{−1/2}). The only such subspace is ℬ_{s}(χα_{F}^{1/2}, χα_{F}^{−1/2}) and r_{Ω} is therefore σ(χα_{F}^{1/2}, χα_{F}^{−1/2}).

Suppose now that Ω is not one-dimensional. Let Ω act on U. Since K_{1} is normal and K/K_{1} is abelian there is no non-zero vector in U fixed by every element of K_{1}. If Φ is in 𝒮(K, Ω) then

Φ(xh) = Ω^{−1}(h) Φ(x)

for all h in K_{1}. In particular Φ(0) is fixed by every element in K_{1} and is therefore 0. Thus all functions in 𝒮(K, Ω) have compact supports in K^{×} and if we associate to every Φ in 𝒮(K, Ω) the function

φ_{Φ}(a) = |h|_{K}^{1/2} Ω(h) Φ(h)

where a = ν(h) we obtain a bijection from 𝒮(K, Ω) to 𝒮(F^{×}, U). It is again clear that

φ_{Φ_{1}} = ξ_{ψ}(b)φ_{Φ}

if b is in B_{F} and Φ_{1} = r_{Ω}(b)Φ.

Lemma 4.2.5 Let Ω be an irreducible representation of K^{×} in the complex vector space U. Assume that U has dimension greater than one.

(i) For any Φ in 𝒮(K, U) the integrals

Z(α_{F}^{s} ⊗ Ω, Φ) = ∫_{K^{×}} |a|_{K}^{s/2} Ω(a) Φ(a) d^{×}a
Z(α_{F}^{s} ⊗ Ω^{−1}, Φ) = ∫_{K^{×}} |a|_{K}^{s/2} Ω^{−1}(a) Φ(a) d^{×}a

are absolutely convergent in some half-plane Re s > s_{0}. (ii) The functions Z(α_{F}^{s} ⊗ Ω, Φ) and Z(α_{F}^{s} ⊗ Ω^{−1}, Φ) can be analytically continued to functions

meromorphic in the whole complex plane. (iii) Given u in U there is a Φ in 𝒮(K, U) such that

Z(α_{F}^{s} ⊗ Ω, Φ) ≡ u.

(iv) There is a scalar function ε(s, Ω, ψ) such that for all Φ in 𝒮(K, U)

Z(α_{F}^{3/2−s} ⊗ Ω^{−1}, Φ′) = −ε(s, Ω, ψ) Z(α_{F}^{s+1/2} ⊗ Ω, Φ)

Chapter 1 69 if Φ′ is the Fourier transform of Φ. Moreover, as a function of s, ε(s, Ω, ψ) is a constant times an exponential.

There is no need to verify the first part of the lemma. Observe that α_{F} (ν(x)) = |ν(x)|_{F} = |x|_{K}^{1/2} so that

(α_{F}^{s} ⊗ Ω)(x) = |x|_{K}^{s/2} Ω(x).

If Φ belongs to 𝒮(K, U) set

Φ_{1}(x) = ∫_{K_{1}} Ω(h) Φ(xh).

The integration is taken with respect to the normalized Haar measure on the compact group K_{1}. Φ_{1} clearly belongs to 𝒮(K, U) and

Z(α_{F}^{s} ⊗ Ω, Φ) = Z(α_{F}^{s} ⊗ Ω, Φ_{1})  (4.2.6)

and the Fourier transform Φ′_{1} of Φ_{1} is given by

Φ′_{1}(x) = ∫_{K_{1}} Ω(h^{−1}) Φ′(hx)

The function Φ′_{1}(x^{ι}) belongs to 𝒮(K, Ω) and

Z(α_{F}^{s} ⊗ Ω^{−1}, Φ′) = Z(α_{F}^{s} ⊗ Ω^{−1}, Φ′_{1}).  (4.2.7)

Since Φ_{1} and Φ′_{1} both have compact support in K^{×} the second assertion is clear.

If u is in U and we let Φ_{u} be the function which is O outside of U_{K}, the group of units of O_{K}, and on U_{K} is given by Φ_{u}(x) = Ω^{−1}(x)u then

Z(α_{F}^{s} ⊗ Ω, Φ_{u}) = cu

if

c = ∫_{U_{K}} d^{×}a.

If φ belongs to 𝒮(K^{×}) let A(φ) and B(φ) be the linear transformations of U defined by

\begin{aligned}A(φ)U & = Z(α_{F}^{s+1/2} ⊗ Ω, φ^{u}) \\ B(φ)u & = Z(α_{F}^{−s+3/2} ⊗ Ω^{−1}, φ′u) \end{aligned}

where φ′ is the Fourier transform of φ. If λ(h) φ(h) = φ(h^{−1}x) and ρ(h) φ(x) = φ(xh) then

A(λ(h)φ) = |h|_{K}^{s/2+1/4}Ω(h) A(φ)

and

A(ρ(h)φ) = |h|_{K}^{−s/2−1/4}A(φ)Ω^{−1}(h).

Since the Fourier transform of λ(h)φ is |h|_{K}ρ(h)φ′ and the Fourier transform of ρ(h)φ is |h|_{K}^{−1}λ(h)φ′, the map φ → B(φ) has the same two properties. Since the kernel of Ω Chapter 1 70 is open it is easily seen that A(φ) and B(φ) are obtained by integrating φ against locally constant functions α and β. They will of course take values in the space of linear transformations of U. We will have

α(ha) = |h|_{K}^{s/2+1/4}Ω(h) α(a)

and

α(ah^{−1}) = |h|_{K}^{−s/2−1/4}α(a) Ω^{−1}(h)

β will satisfy similar identities. Thus

\begin{matrix}α(h) = |h|_{K}^{s/2+1/4} Ω(h) α(1) \\ β(h) = |h|_{K}^{s/2+1/4} Ω(h) β(1) \end{matrix}

α(1) is of course the identity. However β(1) must commute with Ω(h) for all h in K^{×} and therefore it is a scalar multiple of the identity. Take this scalar to be −ε(s, Ω, ψ).

The identity of part (iv) is therefore valid for Φ in 𝒮(K^{×}, U) and in particular for Φ in 𝒮(K, Ω). The general case follows from (4.2.6) and (4.2.7). Since

ε(s, Ω, ψ) = −\frac{1}{c} Z(α_{F}^{3/2−s} ⊗ Ω^{−1}, Φ′_{u})

the function ε(s, Ω, ψ) is a finite linear combination of powers |ϖ|^{s} if ϖ is a generator of 𝔭_{F}. Exchanging the roles of Φ_{u} and Φ′_{u} we see that ε^{−1}(s, Ω, ψ) has the same property. ε(s, Ω, ψ) is therefore a multiple of some power of |ϖ|^{s}.

We have yet to complete the proof of the theorem. Suppose φ = φ_{Φ} belongs to 𝒮(F^{×}, U) and φ′ = φ_{r_{Ω}(w)Φ}. We saw in the first paragraph that if χ is a quasi-character of F^{×} then

φ̂(χ) = Z(α_{F} χ ⊗ Ω, Φ)  (4.2.8)

and, if Ω(a) = ω(a)I for a in F^{×},

φ′(χ^{−1}ω^{−1}) = −Z(α_{F} χ^{−1} ⊗ Ω^{−1}, Φ′).  (4.2.9)

Suppose U_{0} is a subspace of U and φ takes its values in U_{0}. Then, by the previous lemma, φ̂(χ) and φ′(χ^{−1}ω^{−1}) also lie in U_{0} for all choices of χ. Since φ′ lies in 𝒮(F^{×}, U) we may apply Fourier inversion to the multiplicative group to see that φ′ takes values in U_{0}.

We may regard r_{Ω} as acting on 𝒮(F^{×}, U). Then 𝒮(F^{×}, U_{0}) is invariant under r_{Ω}(w). Since r_{Ω}(b)φ = ξ_{ψ}(b)φ for b in B_{F} it is also invariant under the action of B_{F}. Finally r_{Ω}(({a \atop 0} {0 \atop a}))φ = ω(a)φ so that 𝒮(F^{×}, U_{0}) is invariant under the action of G_{F} itself. If we take U_{0} to have dimension one then 𝒮(F^{×}, U_{0}) may be identified with 𝒮(F^{×}) and the representation r_{Ω} restricted to 𝒮(F^{×}, U_{0}) is irreducible. From (4.2.8) and (4.2.9) we obtain

\begin{aligned}φ̂(α_{F}^{s−1/2}χ) & = Z(α_{F}^{s+1/2}χ ⊗ Ω, Φ) \\ φ′(α_{F}^{−s+1/2}χ^{−1}ω^{−1}) & = −Z(α_{F}^{−s+3/2}χ^{−1} ⊗ Ω^{−1}, Φ′) \end{aligned}

so that

φ′(α_{F}^{−s+1/2}χ^{−1}ω^{−1} = ε(s, χ ⊗ Ω, ψ) φ̂(α_{F}^{s−1/2}χ).

Thus if π_{0} is the restriction of r_{Ω} to 𝒮(F^{×}, U_{0})

ε(s, χ ⊗ π_{0}, ψ) = ε(s, χ ⊗ Ω, ψ)

so that π_{0} = π(Ω) is, apart from equivalence, independent of U_{0}. The theorem follows.

Let Ω be any irreducible finite-dimensional representation of K^{×} and let Ω act on U. The contragredient representation Ω̃ acts on the dual space Ũ of U. If u belongs to U and ũ belongs to Ũ

⟨u, Ω̃(h)ũ⟩ = ⟨Ω^{−1}(h)u, ũ⟩.

If Φ belongs to 𝒮(K) set

Z(α_{F}^{s} ⊗ Ω̃, Φ; u, ũ) = ∫_{K^{×}} |ν(h)|^{s} Φ(h) ⟨Ω(h)u, ũ⟩ d^{×}h

and set

Z(α_{F}^{s} ⊗ Ω̃, Φ; u, ũ) = ∫_{K^{×}} |ν(h)|^{s} Φ(h) ⟨u, Ω̃(h)ũ⟩ d^{×}h.

Theorem 4.3 Let Ω be an irreducible representation of K^{×} in the space U.

(i) For any quasi-character χ of F^{×}

π(χ ⊗ Ω) = χ ⊗ π(Ω).

(ii) There is a real number s_{0} such that for all u, ũ and Φ and all s with Re s > s_{0} the integral

defining Z(α_{F}^{s} ⊗ Ω, Φ; u, ũ) is absolutely convergent. (iii) There is a unique Euler factor L(s, Ω) such that the quotient

\frac{Z(α_{F}^{s+1/2} ⊗ Ω, Φ, u, ũ)}{L(s, Ω)}

is holomorphic for all u, ũ, Φ and for some choice of these variables is a non-zero constant. (iv) There is a functional equation

\frac{Z(α_{F}^{3/2−s} ⊗ Ω̃, Φ′, u, ũ)}{L(1 − s, Ω̃)} = −ε(s, Ω, ψ) \frac{Z(α_{F}^{s+1/2} ⊗ Ω, Φ, u, ũ)}{L(s, Ω)}

where ε(s, Ω, ψ) is, as a function of s, an exponential.

(v) If Ω(a) = ω(a)I for a in F^{×} and if π = π(Ω) then

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I.

Moreover L(s, π) = L(s, Ω), L(s, π̃) = L(s, Ω̃) and ε(s, π, ψ) = ε(s, Ω, ψ).

The first assertion is a consequence of the definitions. We have just proved all the others when Ω has a degree greater than one. Suppose then that Ω(h) = χ(ν(h)) where χ is a quasi-character of F^{×}. Then π(Ω) = π(χα_{F}^{1/2}, χα_{F}^{−1/2}) and if the last part of the theorem is to hold L(s, Ω), which Chapter 1 72 is of course uniquely determined by the conditions of part (iii), must equal L(s, π) = L(s, χα_{F}^{1/2}). Also L(s, Ω̃) must equal L(s, π̃) = L(s, χ^{−1}α_{F}^{1/2}).

In the case under consideration U = ℂ and we need only consider

Z(α_{F}^{s} ⊗ Ω, Φ; 1, 1) = Z(α_{F}^{s} ⊗ Ω, Φ).

As before the second part is trivial and

Z(α_{F}^{s} ⊗ Ω, Φ) = Z(α_{F}^{s} ⊗ Ω, Φ_{1})

if

Φ_{1}(x) = ∫_{K_{1}} Φ(xh).

The Fourier transform of Φ_{1} is

Φ′_{1}(x) = ∫_{K_{1}} Φ′(hx) = ∫_{K_{1}} Φ′(xh)

and

Z(α_{F}^{s} ⊗ Ω̃, Φ′) = Z(α_{F}^{s} ⊗ Ω̃, Φ′_{1}).

It is therefore enough to consider the functions in 𝒮(K, Ω).

If φ = φ_{Φ} is defined as before then φ lies in the space on which the Kirillov model of π acts and

φ̂(α_{F}^{s−1/2}) = A(α_{F}^{s+1/2} ⊗ Ω, Φ).

The third assertion follows from the properties of L(s, π). The fourth follows from the relation

φ′(α_{F}^{1/2−s}ω^{−1}) = −Z(α_{F}^{3/2−s} ⊗ Ω^{−1}, Φ′),

which was proved in the first paragraph, and the relation

\frac{φ′(α_{F}^{1/2−s}ω^{−1})}{L(1 − s, π̃)} = ε(s, π, ψ)\frac{φ̂(α_{F}^{s−1/2})}{L(s, π)},

which was proved in the second, if we observe that Ω̃(h) = Ω^{−1}(h). φ′ is of course π(w)φ.

Corollary 4.4 If π = π(Ω) then π̃ = π(Ω̃).

This is clear if Ω if of degree one so suppose it is of degree greater than one. Combining the identity of part (iv) with that obtained upon interchanging the roles of Ω and Ω̃ and of Φ and Φ′ we find that

ε(s, Ω, ψ) ε(1 − s, Ω̃, ψ) = ω(−1).

The same considerations show that

ε(s, π, ψ) ε(1 − s, π̃, ψ) = ω(−1).

Consequently

ε(s, π̃, ψ) = ε(s, Ω̃, ψ).

Replacing Ω by χ ⊗ Ω we see that

ε(s, χ^{−1} ⊗ π̃, ψ) = ε(s, χ^{−1} ⊗ Ω̃, ψ) = ε(s, χ^{−1}π(Ω̃), ψ)

for all quasi-characters χ. Since π̃ and π(Ω̃) are both absolutely cuspidal they are equivalent. There is a consequence of the theorem whose significance we do not completely understand.

Chapter 1 73 Proposition 4.5 Let Ω be an irreducible representation of K^{×} on the space U and suppose that the dimension of U is greater than one. Let Ũ be the dual space of U. Let π be the Kirillov model of π(Ω), let φ lie in 𝒮(F^{×}), and let φ′ = π(w)φ. If u belongs to U and ũ belong to Ũ the function Φ on K which vanishes at 0 and on K^{×} is defined by

Φ(x) = φ(ν(x)) |ν(x)|^{−1}⟨u, Ω̃(x)ũ⟩

is in 𝒮(K) and its Fourier transform Φ′ vanishes at 0 and on K^{×} is given by

Φ′(x) = −φ′(ν(x)) |ν(x)|^{−1}ω^{−1}(ν(x)) ⟨Ω(x)u, ũ⟩

if Ω(a) = ω(a)I for a in F^{×}.

It is clear that Φ belongs not merely to 𝒮(K) but in fact to 𝒮(K^{×}). So does the function Φ_{1} which we are claiming is equal to Φ′. The Schur orthogonality relations for the group K_{1} show that Φ′(0) = 0 so that Φ′ also belongs to 𝒮(K^{×}).

We are going to show that for every irreducible representation of Ω′ of K^{×}

∫ \frac{Φ_{1}(x), ⟨u′, Ω̃′(x)u′⟩ |ν(x)|^{3/2−s} d^{×}x}{L(1 − s, Ω̃′)} = − ∫ \frac{ε(s, Ω′, ψ) Φ(x) ⟨Ω′(x)u′, u′⟩ |ν(x)|^{s+1/2} d^{×}x}{L(s, Ω′)}

for all choices of u′ and u′. Applying the theorem we see that

∫ {Φ_{1}(x) − Φ′(x)} ⟨u′, Ω̃′(x)u′⟩ |ν(x)|^{3/2−s} d^{×}x = 0

for all choices of Ω′, u′, u′, and all s. An obvious and easy generalization of the Peter-Weyl theorem, which we do not even bother to state, shows that Φ_{1} = Φ′.

If

Ψ(x) = ∫_{K_{1}} ⟨u, Ω̃(hx)ũ⟩ ⟨Ω′(hx)u′, u′⟩ dh

then

∫_{K^{×}} Φ(x)⟨Ω′(x)u′, u′⟩ |ν(x)|^{s+1/2} d^{×}x = ∫_{K^{×}/K_{1}} φ(ν(x)) |ν(x)|^{s−1/2} Ψ(x) d^{×}x

while

∫_{K^{×}} Φ_{1}(x) ⟨u′Ω̃′(x), u′⟩ |ν(x)|^{3/2−s} d^{×}x = − ∫_{K^{×}/K_{1}} φ′(ν(x)) ω^{−1}(ν(x)) |ν(x)|^{1/2−s} Ψ(x^{−1}) d^{×}x

If Ψ is 0 for all choice of u′ and u′ the required identity is certainly true. Suppose then Ψ is different from 0 for some choice u′ and u′.

Let U be the intersection of the kernels of Ω′ and Ω. It is an open normal subgroup of K^{×} and H = U K_{1} F^{×} is open, normal, and of finite index in K^{×}. Suppose that Ω′(a) = ω′(a)I for a in F^{×}. If h belongs to H

Ψ(xh) = χ_{0}(h) Ψ(x)

where χ_{0} is a quasi-character of H trivial on U and K_{1} and equal to ω′ω^{−1} on F^{×}. Moreover χ_{0} extends to a quasi-character χ of K^{×} so that

∫_{K^{×}/H} Ψ(x) χ^{−1}(x) = ∫_{K^{×}/F^{×}} ψ(x) χ^{−1}(x) ≠ 0

χ may of course be identified with a quasi-character of F^{×}.

Lemma 4.5.1 If

∫_{K^{×}/F^{×}} Ψ(x) χ^{−1}(x) ≠ 0

then Ω′ is equivalent to χ ⊗ Ω.

Ω′ and χ ⊗ Ω agree on F^{×} and

∫_{K^{×}/F^{×}}⟨u, { \atop χ ⊗ Ω}(x)ũ⟩ ⟨Ω′(x)u′, u′⟩ ≠ 0.

The lemma follows from the Schur orthogonality relations.

We have therefore only to prove the identity for Ω′ = χ ⊗ Ω. Set

F (x) = ∫_{K_{1}}⟨u, Ω̃(hx)ũ⟩ ⟨Ω(hx)u′, u′⟩ dh.

u′ and u′ now belong to the spaces U and Ũ. There is a function f on F^{×} such that

F (x) = f(ν(x))

The identity we are trying to prove may be written as

\frac{∫ φ′(a) χ^{−1}(a) ω^{−1}(a) f(a^{−1}) |a|^{1/2−s} d^{×}a}{L(1 − s, χ^{−1} ⊗ π̃)} = ε(s, χ ⊗ π, ψ)\frac{∫ φ(a) χ(a) f(a) |a|^{s−1/2} d^{×}a}{L(s, χ ⊗ π)}. (4.5.2)

Let H be the group constructed as before with U taken as the kernel of Ω. The image F′ of H under ν is a subgroup of finite index in F^{×} and f, which is a function on F^{×}/F′, may be written as a sum

f(a) = ∑_{i=1}^{p} λ_{k} χ_{i}(a)

where {χ_{1}, ⋯, χ_{p}} are the characters of F^{×}/F′ which are not orthogonal to f. By the lemma Ω is equivalent to χ_{i} ⊗ Ω for 1 ≤ i ≤ p and therefore π is equivalent to χ_{i} ⊗ π. Consequently

ε(s, χ ⊗ π, ψ) = ε(s, χχ_{i} ⊗ π, ψ)

and

\frac{∫_{F^{×}} φ′(a) χ^{−1}(a) χ_{1}^{−1}(a) ω^{−1}(a) |a|^{1/2−s} d^{×}a}{L(1 − s, χ^{−1} ⊗ π̃)} = ε(s, χ ⊗ π, ψ)\frac{∫_{F^{×}} φ(a) χ(a) χ_{i}(a) |a|^{s−1/2} d^{×}a}{L(s, χ ⊗ π)}.

The identity (4.5.2) follows.

Now let K be a separable quadratic extension of F. We are going to associate to each quasi-character ω of K^{×} an irreducible representation π(ω) of G_{F}. If G_{+} is the set of all g in G_{F} whose determinants belong to ν(K^{×}) we have already, in the first paragraph, associated to ω a representation r_{ω} of G_{+}. To emphasize the possible dependence of r_{ω} on ψ we now denote it by π(ω, ψ). G_{+} is of index 2 in G_{F}. Let π(ω) be the representation of G_{F} induced from π(ω, ψ).

Chapter 1 75 Theorem 4.6 (i) The representation π(ω, ψ) is irreducible.

(ii) The representation π(ω) is admissible and irreducible and its class does not depend on the

choice of ψ. (iii) If there is no quasi-character χ of F^{×} such that ω = χ_{0}ν the representation π(ω) is absolutely

cuspidal. (iv) If ω = χ_{0}ν and η is the character of F^{×} associated to K by local class field theory then π(ω)

is π(χ, χ_{η}).

It is clear what the notion of admissibility for a representation of G_{+} should be. The proof that π(ω, ψ) is admissible proceeds like the proof of the first part of Theorem 4.2 and there is little point in presenting it.

To every Φ in 𝒮(K, ω) we associate the function φ_{Φ} on F_{+} = ν(K^{×}) defined by

φ_{Φ}(a) = ω(h) |h|_{K}^{1/2} Φ(h)

if a = ν(h). Clearly φ_{Φ} = 0 if and only if Φ = 0. Let V_{+} be the space of functions on F_{+} obtained in this manner. V_{+} clearly contains the space 𝒮(F_{+}) of locally constant compactly supported functions on F_{+}. In fact if φ belongs to 𝒮(F_{+}) and

Φ(h) = ω^{−1}(h) |h|_{K}^{−1/2} φ(ν(h))

then φ = φ_{Φ}. If the restriction of ω to the group K_{1} of elements of norm 1 in K^{×} is not trivial so that every element of 𝒮(K, ω) vanishes at 0 then V_{+} = 𝒮(F_{+}). Otherwise 𝒮(F_{+}) is of codimension one in V_{+}.

Let B_{+} be the group of matrices of the form

( \begin{matrix}a & x \\ 0 & 1 \end{matrix} )

with a in F_{+} and x in F. In the first paragraph we introduced a representation ξ = ξ_{ψ} of B_{+} on the space of functions on F_{+}. It was defined by

ξ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ(b) = φ(ba)

and

ξ \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ(b) = ψ(bx) φ(b).

We may regard π(ω, ψ) as acting on V_{+} and if we do the restriction of π(ω, ψ) to B_{+} is ξ_{ψ}.

Lemma 4.6.1 The representation of B_{F} induced from the representation ξ_{ψ} of B_{+} on 𝒮(F_{+}) is the representation ξ_{ψ} of B_{F}. In particular the representation ξ_{ψ} of B_{+} is irreducible.

The induced representation is of course obtained by letting B_{F} act by right translations on the space of all functions φ̃ on B_{F} with values in 𝒮(F_{+}) which satisfy

φ̃(b_{1}b) = ξ_{ψ}(b_{1}) φ̃(b)

for all b_{1} in B_{+}. Let L be the linear functional in 𝒮(F_{+}) which associates to a function its value at 1. Associate to φ̃ the function

φ(a) = L (φ̃ \begin{matrix}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix}) = L (ρ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ̃(e))

The value of φ̃(({a \atop 0} {x \atop 1})) at α in F_{+} is

\begin{aligned}L (φ̃ \begin{matrix}( \\ ( & \begin{matrix}αa & αx \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix}) & = L (ξ_{ψ} \begin{aligned}( \\ ( & \begin{matrix}1 & αx \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ̃ \begin{matrix}( \\ ( & \begin{matrix}αa & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix}) \\ & = ψ(ax)L (φ̃ \begin{matrix}( \\ ( & \begin{matrix}αa & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix}) = ψ(αx)φ(αa). \end{aligned}

Since F^{×}/F_{+} is finite it follows immediately that φ is in 𝒮(F^{×}) and that φ̃ is 0 if φ is. It also shows that φ can be any function in 𝒮(F^{×}) and that if φ′ = ρ(b) φ̃ then φ′ = ξ(b) φ for all b in B_{F}. Since a representation obtained by induction cannot be irreducible unless the original representation is, the second assertion follows from Lemma 2.9.1.

If the restriction of ω to K_{1} is not trivial the first assertion of the theorem follows immediately. If it is then, by an argument used a number of times previously, any non-zero invariant subspace of V_{+} contains 𝒮(F_{+}) so that to prove the assertion we have only to show that 𝒮(F_{+}) is not invariant.

As before we observe that if Φ in 𝒮(K, ω) = 𝒮(K) is taken to vanish at 0 but to be nonnegative and not identically 0 then

r_{ω}(w) Φ(0) = γ ∫_{K} Φ(x) dx ≠ 0

so that φ_{Φ} is in 𝒮(F_{+}) but φ_{r_{ω}(w)Φ} is not.

The representation π(ω) is the representation obtained by letting G_{+} act to the right on the space of functions φ̃ on G_{+} with values in V_{+} which satisfy

φ̃(hg) = π(ω, ψ)(h) φ̃(g)

for h in G_{+}. Replacing the functions φ̃ by the functions

φ′(g) = φ̃ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g)

we obtain an equivalent representation, that induced from the representation

g → π(ω, ψ) (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g ( \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} ))

of G_{+}. It follows from Lemma 1.4 that this representation is equivalent to π(ω, ψ′) if ψ′(x) = ψ(ax). Thus π(ω) is, apart from equivalence, independent of ψ.

Since

G_{F} = \{g ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) | g ∈ G_{+}, a ∈ F^{×}\}

Chapter 1 77 φ̃ is determined by its restrictions to B_{F}. This restriction, which we again call φ̃, is any one of the functions considered in Lemma 4.6.1. Thus, by the construction used in the proof of that lemma, we can associate to any φ̃ a function φ on F^{×}. Let V be the space of functions so obtained. We can regard π = π(ω) as acting on V. It is clear that, for all φ in V,

π(b)φ = ξ_{ψ}(b)φ

if b is in B_{F}. Every function on F_{+} can, by setting it equal to 0 outside of F_{+}, be regarded as a function F^{×}. Since

φ̃ \begin{matrix}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix} (α) = φ(αa)

V is the space generated by the translates of the functions in V_{+}. Thus if V_{+} = 𝒮(F_{+}) then V = 𝒮(F^{×}) and if 𝒮(F_{+}) is of codimension one in V_{+} then 𝒮(F^{×}) is of codimension two in V.

It follows immediately that π(ω) is irreducible and absolutely cuspidal if the restriction of ω to K_{1} is not trivial.

The function φ in V_{+} corresponds to the function φ̃ which is 0 outside of G_{+} and on G_{+} is givne by

φ̃(g) = π(ω, ψ)(g)φ.

It is clear that

π(ω)(g)φ = π(ω, ψ)(g)φ

if g is in G_{+}. Any non-trivial invariant subspace of V will have to contain 𝒮(F^{×}) and therefore 𝒮(F_{+}). Since π(ω, ψ) is irreducible it will have to contain V_{+} and therefore will be V itself. Thus π(ω) is irreducible for all ω.

If the restriction of ω to K_{1} is trivial there is a quasi-character χ of F^{×} such that ω = χ ∘ ν. To establish the last assertion of the lemma all we have to do is construct a non-zero linear form L on V which annihilates 𝒮(F^{×}) and satisfies

L (π \begin{aligned}( \\ ( & \begin{matrix}a_{1} & 0 \\ 0 & a_{2} \end{matrix} & ) \\ & & ) \end{aligned} φ) = χ(a_{1}a_{2}) η(a_{2}) |\frac{a_{1}}{a_{2}} |^{1/2} L(φ)

if π = π(ω). We saw in Proposition 1.5 that

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} φ = χ^{2}(a) η(a)φ

so will only have to verify that

L (π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} φ) = χ(a)|a|^{1/2}L(φ)

If φ = φ_{Φ} is in V_{+} we set

L(φ) = Φ(0)

so that if a is in F_{+}

L (π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ) = r_{ω} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} Φ(0) = χ(a) |a|^{1/2} L(φ).

If ε is in F^{×} but not in F_{+} any function φ in V can be written uniquely as

φ = φ_{1} + π \begin{aligned}( \\ ( & \begin{matrix}ε & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ_{2}

with φ_{1} and φ_{2} in V_{+}. We set

L(φ) = L(φ_{1}) + χ(ε) L(φ_{2}).

Theorem 4.7 (i) If π = π(ω) then π = π(ω^{ι}) if ω^{ι}(a) = ω(a^{ι}), π̃ = π(ω^{−1}) and χ ⊗ π = π(ωχ′) if χ is a quasi-character of F^{×} and χ′ = χ ∘ ν.

(ii) If a is in F^{×} then

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a) η(a)I.

(iii) L(s, π) = L(s, ω) and L(s, π̃) = L(s, ω^{−1}). Moreover if ψ_{K}(x) = ψ_{F} (ξ(x)) for x in K and if λ(K/F, ψ_{F}) is the factor introduced in the first paragraph then

ε(s, π, ψ_{F}) = ε(s, ω, ψ_{K}) λ(K/F, ψ_{F})

It is clear that χ ⊗ π(ω, ψ) of G_{+}. However by its very construction χ ⊗ π(ω, ψ) = π(ωχ′, ψ). The relation

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a) η(a)I

is a consequence of part (iii) of Proposition 1.5 and has been used before. Since η′ = η ∘ ν is trivial and ω(ν(a)) = ω(a) ω^{ι}(a)

π̃ = ω^{−1}η^{−1} ⊗ π = π(ω^{−ι})

To complete the proof of the first part of the theorem we have to show that π(ω) = π(ω^{ι}). It is enough to verify that π(ω, ψ) = π(ω^{ι}, ψ). If Φ belongs to 𝒮(K) let Φ^{ι}(x) = Φ(x^{ι}). Φ → Φ^{ι} is a bijection of 𝒮(K, ω) with 𝒮(K, ω^{ι}) which changes π(ω, ψ) into π(ω^{ι}, ψ). Observe that here as elsewhere we have written an equality when we really mean an equivalence.

We saw in the first paragraph that if φ = φ_{Φ} is in V_{+} then

φ̂(α_{F}^{s−1/2}) = Z(α_{K}^{s}ω, Φ)

and that if φ′ = π(w)φ and Φ′ is the Fourier transform of Φ then, if ω_{0}(a) = ω(a) η(a) for a in F^{×},

φ′(ω_{0}^{−1}α_{F}^{s−1/2}) = γZ(α_{K}^{1−s}ω^{−1}, Φ′)

if γ = λ(K/F, ψ_{F}). Thus for all φ in V_{+} the quotient

\frac{φ̂(α_{F}^{s−1/2})}{L(s, ω)}

has an analytic continuation as a holomorphic function of s and for some φ it is a non-zero constant. Also

\frac{φ′(w_{0}^{−1}α_{F}^{1/2−s})}{L(1 − s, ω^{−1})} = λ(K/F, ψ_{F}) ε(s, ω, ψ_{K}) \frac{φ̂(α_{F}^{s−1/2})}{L(s, ω)}.

To prove the theorem we have merely to check that these assertions remain valid when φ is allowed to vary in V. In fact we need only consider functions of the form

φ = π \begin{aligned}( \\ ( & \begin{matrix}ϵ & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ_{0}

where φ_{0} is in V_{+} and ε is not in F_{+}. Since

φ̂(α_{F}^{s−1/2}) = |ε|^{1/2−s} φ̂_{0}(α_{F}^{s−1/2})

the quotient

\frac{φ̂(α_{F}^{s−1/2})}{L(s, ω)}

is certainly holomorphic in the whole plane. Since

φ′(ω_{0}^{−1}α_{F}^{1/2−s}) = ω_{0}(ε) ω_{0}^{−1}(ε) |ε|^{1/2−s}φ′_{0}(ω_{0}^{−1}α_{F}^{1/2−s}) = |ε|^{\frac{1}{2} −s}φ′_{0}(ω_{0}^{−1}α_{F}^{1/2−s})

the functional equation is also satisfied.

Observe that if ω = χ ∘ ν then π(ω) = π(χ, χ_{η}) so that

L(s, ω) = L(s, χ) L(s, χ_{η})

and

ε(s, ω, ψ_{K}) λ(K/F, ψ_{F}) = ε(s, χ, ψ_{F}) ε(s, χ_{η}, ψ_{F})

These are special cases of the identities of [19].

Chapter 1 80 §5. Representations of GL(2, ℝ). We must also prove a local functional equation for the real and complex fields. In this paragraph we consider the field ℝ of real numbers. The standard maximal compact subgroup of GL(2, ℝ) is the orthogonal group O(2, ℝ). Neither GL(2, ℝ) nor O(2, ℝ) is connected.

Let ℋ_{1} be the space of infinitely differentiable compactly supported functions on GL(2, ℝ) which are O(2, ℝ) finite on both sides. Once a Haar measure on G_{ℝ} = GL(2, ℝ) has been chosen we may regard the elements of ℋ_{1} as measures and it is then an algebra under convolution.

f_{1} × f_{2}(g) = ∫_{G_{ℝ}} f_{1}(gh^{−1}) f_{2}(h) dh.

On O(2, ℝ) we choose the normalized Haar measure. Then every function ξ on O(2, ℝ) which is a finite sum of matrix elements of irreducible representations of O(2, ℝ) may be identified with a measure on O(2, ℝ) and therefore on GL(2, ℝ). Under convolution these measures form an algebra ℋ_{2}. ℋ_{ℝ} will be the sum of ℋ_{1} and ℋ_{2}. It is also an algebra under convolution of measures. In particular if ξ belongs to ℋ_{2} and f belongs to ℋ_{1}

ξ ∗ f(g) = ∫_{O(2,ℝ)} ξ(u) f(u^{−1}g) du

and

f ∗ ξ(g) = ∫_{O(2,ℝ)} f(gu^{−1}) ξ(u) du.

If σ_{i}, 1 ≤ i ≤ p, is a family of inequivalent irreducible representations of O(2, ℝ) and

ξ_{i}(u) = dim σ_{i} trace σ_{i}(u^{−1})

then

ξ = ∑_{i=1}^{p} ξ_{i}

is an idempotent of ℋ_{ℝ}. Such an idempotent is called elementary.

It is a consequence of the definitions that for any f in ℋ_{1} there is an elementary idempotent ξ such that

ξ ∗ f = f ∗ ξ = f.

Moreover for any elementary idempotent ξ

ξ ∗ ℋ_{1} ∗ ξ = ξ ∗ C_{c}^{∞}(G_{ℝ}) ∗ ξ

is a closed subspace of C_{c}^{∞}(G_{ℝ}), in the Schwartz topology. We give it the induced topology.

A representation π of the algebra ℋ_{ℝ} on the complex vector space V is said to be admissible if the following conditions are satisfied.

(5.1) Every vector v in V is of the form

v = ∑_{i=1}^{r} π(f_{i})v_{i}

Chapter 1 81 with f_{i} in ℋ_{1} and v_{i} in V.

(5.2) For every elementary idempotent ξ the range of π(ξ) is finite dimensional.

(5.3) For every elementary idempotent ξ and every vector v in π(ξ)V the map f → π(f)v of ξℋ_{1}ξ into the finite dimensional space π(ξ)V is continuous.

If v = ∑_{i=1}^{r} π(f_{i})v_{i} we can choose an elementary idempotent ξ so that ξf_{i} = f_{i}ξ = f_{i} for 1 ≤ i ≤ r. Then π(ξ)v = v. Let {φ} be a sequence in C_{c}^{∞}(G_{ℝ}) which converges, in the space of distributions, towards the Dirac distribution at the origin. Set φ′_{n} = ξ ∗ φ_{n} ∗ ξ. For each i the sequence {φ′_{n} ∗ f_{i}} converges to f_{i} in the space ξℋ_{1}ξ. Thus by (5.3) the sequence {π(φ′_{n})v} converges to v in the finite dimensional space π(ξ)v. Thus v is in the closure of the subspace π(ξℋ_{1}ξ)v and therefore belongs to it.

As in the second paragraph the conditions (5.1) and (5.2) enable us to define the representation π̃ contragredient to π. Up to equivalence it is characterized by demanding that it satisfy (5.1) and (5.2) and that there be a non-degenerate bilinear form on V × Ṽ satisfying

⟨π(f)v, ṽ⟩ = ⟨v, π(f̌)ṽ⟩

for all f in ℋ_{ℝ}. Ṽ is the space on which π̃ acts and f̌ is the image of the measure f under the map g → g^{−1}. Notice that we allow ourselves to use the symbol f for all elements of ℋ_{ℝ}. The condition (5.3) means that for every v in V and every ṽ in Ṽ the linear form

f → ⟨π(f)v, ṽ⟩

is continuous on each of the spaces ξℋ_{1}ξ. Therefore π̃ is also admissible.

Choose ξ so that π(ξ)v = v and π̃(ξ̌)ṽ = ṽ. Then for any f in ℋ_{1}

⟨π(f)v, ṽ⟩ = ⟨π(ξfξ)v, ṽ⟩.

There is therefore a unique distribution μ on G_{ℝ} such that

μ(f) = ⟨π(f)v, ṽ⟩

for f in ℋ_{1}. Choose φ in ξℋ_{1}ξ so that π(φ)v = v. Then

μ(fφ) = μ(ξfφξ) = μ(ξfξφ) = ⟨π(ξfξφ)v, ṽ⟩ = ⟨π(ξfξ)v, ṽ⟩

so that μ(fφ) = μ(f). Consequently the distribution μ is actually a function and it is not unreasonable to write it as g → ⟨π(g)v, ṽ⟩ even though π is not a representation of G_{ℝ}. For a fixed g, ⟨π(g)v, ṽ⟩ depends linearly on v and ṽ. If the roles of π and π̃ are reversed we obtain a function ⟨v, π̃(g)ṽ⟩. It is clear from the definition that

⟨π(g)v, ṽ⟩ = ⟨v, π̃(g^{−1})ṽ⟩.

Let 𝔤 be the Lie algebra of G_{ℝ} and let 𝔤_{ℂ} = 𝔤 ⨂_{ℝ} ℂ. Let 𝔄 be the universal enveloping algebra of 𝔤_{ℂ}. If we regard the elements of 𝔄 as distributions on G_{ℝ} with support at the identity we can take their convolution product with the elements of C_{c}^{∞}(G_{ℝ}). More precisely if X belongs to 𝔤

X ∗ f(g) = \frac{d}{dt} f( exp(−tX))|_{t=0}

Chapter 1 82 and

f ∗ X(g) = \frac{d}{dt} f(g exp(−tX))|_{t=0}

If f belongs to ℋ_{1} so do f ∗ X and X ∗ f.

We want to associate to the representation π of ℋ_{ℝ} on V a representation π of 𝔄 on V such that

π(X) π(f) = π(X ∗ f)

and

π(f) π(X) = π(f ∗ X)

for all X in 𝔄 and all f in ℋ_{1}. If v = ∑ π(f_{i}) v_{i} we will set

π(X)v = ∑_{i} π(X ∗ f_{i})v_{i}

and the first condition will be satisfied. However we must first verify that if

∑_{i} π(f_{i}) v_{i} = 0

then

w = ∑_{i} π(X ∗ f_{i})v_{i}

is also 0. Choose f so that w = π(f)w. Then

w = ∑_{i} π(f) π(X ∗ f_{i})v_{i} = ∑_{i} π(f ∗ X ∗ f_{i})v_{i} = π(f ∗ X){∑ π(f_{i})v_{i}} = 0.

¿From the same calculation we extract the relation

π(f) {∑_{i} π(X ∗ f_{i})v_{i}} = π(f ∗ X) {∑ π(f_{i})v_{i}}

for all f so that π(f)π(X) = π(f ∗ X).

If g is in G_{ℝ} then λ(g) f = δ_{g} ∗ f if δ_{g} is the Dirac function at g. If g is in O(2, ℝ) or in Z_{ℝ}, the groups of scalar matrices, δ_{g} ∗ f is in ℋ_{1} if f is, so that the same considerations allow us to associate to π a representation π of O(2, ℝ) and a representation π of Z_{ℝ}. It is easy to see that if h is in either of these groups then

π(AdhX) = π(h) π(X) π(h^{−1}).

To dispel any doubts about possible ambiguities of notation there is a remark we should make. For any f in ℋ_{1}

⟨π(f)v, ṽ⟩ = ∫_{G_{ℝ}} f(g)⟨π(g)v, ṽ⟩ dg.

Thus if h is in O(2, ℝ) or Z_{ℝ}

⟨π(f ∗ δ_{h})v, ṽ⟩ = ∫_{G_{ℝ}} f(g) ⟨π(gh)v, ṽ⟩ dg

and

⟨π(f)π(h)v, ṽ⟩ = ∫_{G_{ℝ}} f(g)⟨π(g)π(h)v, ṽ⟩ dg

so that

⟨π(gh)v, ṽ⟩ = ⟨π(g) π(h)v, ṽ⟩.

A similar argument shows that

⟨π(hg)v, ṽ⟩ = ⟨π(g)v, π̃(h^{−1}ṽ⟩.

It is easily seen that the function ⟨π(g)v, ṽ⟩ takes the value of ⟨v, ṽ⟩ at g = e. Thus if h belongs to O(2, ℝ) or Z_{ℝ} the two possible interpretations of ⟨π(h)v, ṽ⟩ give the same result.

It is not possible to construct a representation of G_{ℝ} on V and the representation of 𝔄 is supposed to be a substitute. Since G_{ℝ} is not connected, it is not adequate and we introduce instead the notion of a representation π_{1} of the system {𝔄, ε} where

ε = ( \begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix} ) .

It is a representation π_{1} of 𝔄 and an operator π_{1}(ε) which satisfy the relations

π_{1}^{2}(ε) = I

and

π_{1}(AdεX) = π_{1}(ε)π_{1}(X) π_{1}(ε^{−1}).

Combining the representation π with 𝔄 with the operator π(ε) we obtain a representation of the system {𝔄, ε}.

There is also a representation π̃ of 𝔄 associated to π̃ and it is not difficult to see that

⟨π(X)v, ṽ⟩ = ⟨v, π̃(X̌)ṽ⟩

if X → X̌ is the automorphism of 𝔄 which sends X in 𝔤 to −X.

Let

φ(g) = ⟨π(g)v, ṽ⟩.

φ is certainly infinitely differentiable. Integrating by parts we see that

∫_{G_{ℝ}} f(g)φ ∗ X(g) dg = ∫_{G_{ℝ}} f ∗ X̌(g) φ(g) dg

The right side is

⟨π(f) π(X̌)v, ṽ⟩ = ∫_{G_{ℝ}} f(g)⟨π(g) π(X̌)v, ṽ⟩

so that

φ ∗ X̌(g) = ⟨π(g)π(X̌)v, ṽ⟩.

Assume now that the operators π(X) are scalar if X is in the centre ℨ of 𝔄. Then the standard proof, which uses the theory of elliptic operators, shows that the functions φ are analytic on G_{ℝ}. Since

\begin{aligned}φ ∗ X̌(e) & = ⟨π(X̌)v, ṽ⟩ \\ φ ∗ X̌(ε) & = ⟨π(ε) π(X̌)v, ṽ⟩ \end{aligned}

and G_{ℝ} has only two components, one containing e and the other containing ε. The function φ vanishes identically if ⟨π(X̌)v, ṽ⟩ and ⟨π(ε) π(X̌)v, ṽ⟩ are 0 for all X in 𝔄. Any subspace V_{1} of V invariant under 𝔄 and ε is certainly invariant under O(2, ℝ) and therefore is determined by its annihilator in Ṽ. If v is in V_{1} and ṽ annihilates V_{1} the function ⟨π(g)v, ṽ⟩ is 0 so that

⟨π(f)v, ṽ⟩ = 0

for all f in ℋ_{1}. Thus π(f)v is also in V_{1}. Since ℋ_{2} clearly leaves V_{1} invariant this space is left invariant by all of ℋ_{ℝ}.

By the very construction any subspace of V invariant under ℋ_{ℝ} is invariant under 𝔄 and ε so that we have almost proved the following lemma.

Lemma 5.4 The representation π of ℋ_{ℝ} is irreducible if and only if the associated representation π of {𝔄, ε} is.

To prove it completely we have to show that if the representation of {𝔄, ε} is irreducible the operator π(X) is a scalar for all X in ℨ. As π(X) has to have a non-zero eigenfunction we have only to check that π(X) commutes with π(Y) for Y in 𝔄 with π(ε). It certainly commutes with π(Y). X is invariant under the adjoint action not only of the connected component of G_{ℝ} but also of the connected component of GL(2, ℂ). Since GL(2, ℂ) is conected and contains ε

π(ε) π(X) π^{−1}(ε) = π(Adε(X)) = π(X).

Slight modifications, which we do not describe, of the proof of Lemma 5.4 lead to the following lemma.

Lemma 5.5 Suppose π and π′ are two irreducible admissible representations of ℋ_{ℝ}. π and π′ are equivalent if and only if the associated representations of {𝔄, ε} are.

We comment briefly on the relation between representations of G_{ℝ} and representations of ℋ_{ℝ}. Let V be a complete separable locally convex topological space and π a continuous representation of G_{ℝ} on V. Thus the map (g, v) → π(g)v of G_{ℝ} × V to V is continuous and for f in C_{c}^{∞}(G_{ℝ}) the operator

π(f) = ∫_{G_{ℝ}} f(x) π(x) dx

is defined. So is π(f) for f in ℋ_{2}. Thus we have a representation of ℋ_{ℝ} on V. Let V_{0} be the space of O(2, ℝ)-finite vectors in V. It is the union of the space π(ξ)V as ξ ranges over the elementary idempotents and is invariant under ℋ_{ℝ}. Assume, as is often the case, that the representation π_{0} of ℋ_{ℝ} on V_{0} is admissible. Then π_{0} is irreducible if and only if π is irreducible in the topological sense.

Suppose π′ is another continuous representation of G_{ℝ} in a space V′ and there is a continuous non-degenerate bilinear form on V × V′ such that

⟨π(g)v, v′⟩ = ⟨v, π′(g^{−1})v′⟩.

Then the restriction of this form to V_{0} × V′_{0} is non-degenerate and

⟨π(f)v, v′⟩ = ⟨v, π′(f̌)v′⟩

for all f in ℋ_{ℝ}, v in V_{0}, and v′ in V′_{0}. Thus π′_{0} is the contragredient of π_{0}. Since

⟨π_{0}(f)v, v′⟩ = ∫_{G_{ℝ}} f(g) ⟨π(g)v, v′⟩

we have

⟨π_{0}(g)v, v′⟩ = ⟨π(g)v, v′⟩.

The special orthogonal group SO(2, ℝ) is abelian and so is its Lie algebra. The onedimensional representation

( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) → e^{inθ}

of SO(2, ℝ) and the associated representation of its Lie algebra will be both denoted by κ_{n}. A representation π of 𝔄 or of {𝔄, ε} will be called admissible if its restrictions to the Lie algebra of SO(2, ℝ) decomposes into a direct sum of the representations κ_{n} each occurring with finite multiplicity. If π is an admissible representation of ℋ_{ℝ} the corresponding representation of {𝔄, ε} is also admissible. We begin the classification of the irreducible admissible representations of ℋ_{ℝ} and of {𝔄, ε} with the introduction of some particular representations.

Let μ_{1} and μ_{2} be two quasi-characters of F^{×}. Let ℬ(μ_{1}, μ_{2}) be the space of functions f on G_{ℝ} which satisfy the following two conditions.

(i)

f (( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} ) g) = μ_{1}(a_{1}) μ_{2}(a_{2}) |\frac{a_{1}}{a_{2}} |^{1/2} f(g)

for all g in G_{ℝ}, a_{1}, a_{2} in ℝ^{×}, and x in ℝ.

(ii) f is SO(2, ℝ) finite on the right. Because of the Iwasawa decomposition

G_{ℝ} = N_{ℝ} A_{ℝ} SO(2, ℝ)

these functions are complete determined by their restrictions to SO(2, ℝ) and in particular are infinitely differentiable. Write

μ_{i}(g) = |t|^{s_{i}} (\frac{t}{|t|})^{m_{i}}

where s_{i} is a complex number and m_{i} is 0 or 1. Set s = s_{1} − s_{2} and m = |m_{1} − m_{2}| so that μ_{1} μ_{2}^{−1}(t) = |t|^{s} (\frac{t}{|t|})^{m}. If n has the same parity as m let φ_{n} be the function in ℬ(μ_{1}, μ_{2}) defined by

φ_{n} (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a_{1} & 0 \\ 0 & a_{2} \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} )) = μ_{1}(a_{1}) μ_{2}(a_{2}) |\frac{a_{1}}{a_{2}} |^{1/2} e^{inθ}.

Chapter 1 86 The collection {φ_{n}} is a basis of ℬ(μ_{1}, μ_{2}).

For any infinitely differentiable function f on G_{ℝ} and any compactly supported distribution μ we defined λ(μ)f by

λ(μ)f(g) = μ̌(ρ(g)f)

and ρ(μ)f by

ρ(μ)f(g) = μ(λ(g^{−1})f).

If, for example, μ is a measure

λ(μ)f(g) = ∫_{G_{ℝ}} f(h^{−1}g) dμ(h)

and

ρ(μ)f(g) = ∫_{G_{ℝ}} f(gh) dμ(h).

In all cases λ(μ)f and ρ(μ)f are again infinitely differentiable. For all f in ℋ_{ℝ} the space ℬ(μ_{1}, μ_{2}) is invariant under ρ(f) so that we have a representation ρ(μ_{1}, μ_{2}) of ℋ_{ℝ} on ℬ(μ_{1}, μ_{2}). It is clearly admissible and the associated representation ρ(μ_{1}, μ_{2}) of {𝔄, ε} is also defined by right convolution.

We introduce the following elements of 𝔤 which is identified with the Lie algebra of 2 × 2 matrices.

U = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ,  J = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) ,  V_{+} = ( \begin{matrix}1 & i \\ i & −1 \end{matrix} ) ,  V_{−} = ( \begin{matrix}1 & −i \\ −i & −1 \end{matrix} ) ,

X_{+} = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) ,  X_{−} = ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) ,  Z = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) ,

as well as

D = X_{+}X_{−} + X_{−}X_{+} + \frac{Z^{2}}{2} ,

which belongs to 𝔄.

Lemma 5.6 The following relations are valid

(i) ρ(U)φ_{n} = inφ_{n}  (ii) ρ(ε)φ_{n} = (−1)^{m_{1}}φ_{−n}
(iii) ρ(V_{+})φ_{n} = (s + 1 + n)φ_{n+2}  (iv) ρ(V_{−})φ_{n} = (s + 1 − n)φ_{n−2}
(v) ρ(D)φ_{n} = \frac{s^{2}−1}{2}φ_{n}  (vi) ρ(J)φ_{n} = (s_{1} + s_{2})φ_{n}

The relations (i), (ii), and (vi) are easily proved. It is also clear that for all φ in ℬ(μ_{1}, μ_{2})

ρ(Z) φ(e) = (s + 1) φ(e)

and

ρ(X_{+}) φ(e) = 0.

Chapter 1 87 The relations

Ad ( ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) ) V_{+} = e^{2iθ}V_{+}

and

Ad ( ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) ) V_{−} = e^{−2iθ}V_{−}

show that ρ(V_{+})φ_{n} is a multiple of φ_{n+2} and that ρ(V_{−})φ_{n} is a multiple of φ_{n−2}. Since

V_{+} = Z − iU + 2iX_{+}

and

V_{−} = Z + iU − 2iX_{+}

the value of ρ(V_{+})φ_{n} at the identity e is s + 1 + n and that of ρ(V_{−})φ_{n} = s + 1 − n. Relations (iii) and (iv) follow.

It is not difficult to see that D belongs to ℨ the centre of 𝔄. Therefore ρ(D)φ = λ(Ď)φ = λ(D)φ since D = Ď. If we write D as

2X_{−} X_{+} + Z + \frac{Z^{2}}{2}

and observe that λ(X_{+})φ = 0 and λ(Z)φ = −(s + 1)φ if φ is in ℬ(μ_{1}, μ_{2}) we see that

ρ(D)φ_{n} = \{−(s + 1) + \frac{(s + 1)^{2}}{2} \} φ = \frac{s^{2} − 1}{2}φ_{n}.

Lemma 5.7 (i) If s − m is not an odd integer ℬ(μ_{1}, μ_{2}) is irreducible under the action of 𝔤.

(ii) If s − m is an odd integer and s ≥ 0 the only proper subspaces of ℬ(μ_{1}, μ_{2}) invariant under 𝔤 are

\begin{aligned}ℬ_{1}(μ_{1}, μ_{2}) & = ∑_{{n≥s+1 \atop n=s+1(mod 2)}} ℂφ_{n} \\ ℬ_{2}(μ_{1}, μ_{2}) & = ∑_{{n≤−s−1 \atop n=s+1(mod 2)}} ℂφ_{n} \end{aligned}

and, when it is different from ℬ(μ_{1}, μ_{2}),

ℬ_{s}(μ_{1}, μ_{2}) = ℬ_{1}(μ_{1}, μ_{2}) + ℬ_{2}(μ_{1}, μ_{2}).

(iii) If s − m is an odd integer and s < 0 the only proper subspaces of ℬ(μ_{1}, μ_{2}) invariant under 𝔤 are

\begin{aligned}ℬ_{1}(μ_{1}, μ_{2}) & = ∑_{{n≥s+1 \atop n=s+1(mod 2)}} ℂφ_{n} \\ ℬ_{2}(μ_{1}, μ_{2}) & = ∑_{{n≤−s−1 \atop n=(s+1)(mod 2)}} ℂφ_{n} \end{aligned}

and

ℬ_{f}(μ_{1}, μ_{2}) = ℬ_{1}(μ_{1}, μ_{2}) ∩ ℬ_{2}(μ_{1}, μ_{2}).

Since a subspace of ℬ(μ_{1}, μ_{2}) invariant under 𝔤 is spanned by those of the vectors φ_{n} that it contains this lemma is an easy consequence of the relations of Lemma 5.6.

Before stating the corresponding results for {𝔄, ε} we state some simple lemmas.

Chapter 1 88 Lemma 5.8 If π is an irreducible admissible representation of {𝔄, ε} there are two possibilities:

(i) The restriction of π to 𝔄 is irreducible and the representations X → π(X) and X → π(Adε(X)) are equivalent.

(ii) The space V on which π acts decomposes into a direct sum V_{1} ⊕ V_{2} where V_{1} and V_{2} are both invariant and irreducible under 𝔄. The representations π_{1} and π_{2} of 𝔄 on V_{1} and V_{2} are not equivalent but π_{2} is equivalent to the representation X → π(Adε(X)).

If the restriction of π to 𝔄 is irreducible the representations X → π(X) and X → π(Adε(X)) are certainly equivalent. If it is not irreducible let V_{1} be a proper subspace invariant under 𝔄. If V_{2} = π(ε)V_{1} then V_{1} ∩ V_{2} and V_{1} + V_{2} are all invariant under {𝔄, ε}. Thus V_{1} ∩ V_{2} = {0} and V = V_{1} ⊕V_{2}. If V_{1} had a proper subspace V′_{1} invariant under 𝔄 the same considerations would show that V = V′_{1} ⊕ V′_{2} with V′_{2} = π(ε)V′_{1}. Since this is impossible V_{1} and V_{2} are irreducible under 𝔄.

If v_{1} is in V_{1}

π_{2}(X) π(ε)v_{1} = π(ε) π_{1}( ad ε(X))v_{1}

so that the representations X → π_{2}(X) and X → π_{1}(Adε(X)) are equivalent. If π_{1} and π_{2} were equivalent there would be an invertible linear transformation A from V_{1} to V_{2} so that Aπ_{1}(X) = π_{2}(X)A. If v_{1} is in V_{1}

A^{−1}π(ε) π_{1}(X)v_{1} = A^{−1} π_{2}( ad ε(X)) π(ε) v_{1} = π_{1}(Adε(X)) A^{−1} π(ε) v_{1}

Consequently {A^{−1}π(ε)}^{2} regarded as a linear transformation of V_{1} commutes with 𝔄 and is therefore a scalar. There is no harm in supposing that it is the identity. The linear transformation

v_{1} + v_{2} → A^{−1}v_{2} + Av_{1}

then commutes with the action of {𝔄, ε}. This is a contradiction.

Let χ be a quasi-character of ℝ^{×} and let χ(t) = t^{c} for t positive. For any admissible representation π of 𝔄 and therefore of 𝔤 we define a representation χ ⊗ π of 𝔤 and therefore 𝔄 by setting

χ ⊗ π(X) = \frac{c}{2} trace X + π(X)

if X is in 𝔤. If π is a representation of {𝔄, ε} we extend χ ⊗ π to {𝔄, ε} by setting

χ ⊗ π(ε) = χ(−1) π(ε)

If π is associated to a representation π of ℋ_{ℝ} then χ ⊗ π is associated to the representation of ℋ_{ℝ} defined by

χ ⊗ π(f) = π(χf)

if χf is the product of the functions χ and f.

Chapter 1 89 Lemma 5.9 Let π_{0} be an irreducible admissible representation of 𝔄. Assume that π_{0} is equivalent to the representation X → π_{0}(Adε(X)). Then there is an irreducible representation π of {𝔄, ε} whose restriction to 𝔄 is π_{0}. If η is the non-trivial quadratic character of ℝ^{×} the representations π and η ⊗ π are not equivalent but any representatin of {𝔄, ε} whose restriction to 𝔄 is equivalent to π_{0} is equivalent to one of them.

Let π_{0} act on V. There is an invertible linear transformation A of V such that Aπ_{0}(X) = π_{0}(Adε(X))A for all X in 𝔄. Then A^{2} commutes with all π_{0}(X) and is therefore a scalar. We may suppose that A^{2} = I. If we set π(ε) = A and π(X) = π_{0}(X) for X in 𝔄 we obtain the required representation. If we replace A by −A we obtain the representation η ⊗ π. π and η ⊗ π are not equivalent because any operator giving the equivalence would have to commute with all of the π(X) and would therefore be a scalar. Any representation π′ of {𝔄, ε} whose restriction to 𝔄 is equivalent to π_{0} can be realized on V_{0} in such a way that π′(X) = π_{0}(X) for all X. Then π′(ε) = ±A.

Lemma 5.10 Let π_{1} be an irreducible admissible representation of 𝔄. Assume that π_{1} and π_{2}, with π_{2}(X) = π_{1}(Adε(X)), are not equivalent. Then there is an irreducible representation π of {𝔄, ε} whose restriction to 𝔄 is the direct sum of π_{1} and π_{2}. Every irreducible admissible representation of {𝔄, ε} whose restriction to 𝔄 contains π_{1} is equivalent to π. In particular η ⊗ π is equivalent to π.

Let π_{1} act on V_{1}. To construct π we set V = V_{1} ⊕ V_{2} and we set
π(X) (v_{1} ⊕ v_{2}) = π_{1}(X)v_{1} ⊕ π_{2}(X)v_{2}

and

π(ε) (v_{1} ⊕ v_{2}) = v_{2} ⊕ v_{1}.

The last assertion of the lemma is little more than a restatement of the second half of Lemma 5.8.

Theorem 5.11 Let μ_{1} and μ_{2} be two quasi-characters of F^{×}.

(i) If μ_{1}μ_{2}^{−1} is not of the form t → t^{p} sgn t with p a non-zero integer the space ℬ(μ_{1}, μ_{2}) is irreducible under the action of {𝔄, ε} or ℋ_{ℝ}. π(μ_{1}, μ_{2}) is any representation equivalent to ρ(μ_{1}, μ_{2}).

(ii) If μ_{1}μ_{2}^{−1}(t) = t^{p} sgn t, where p is a positive integer, the space ℬ(μ_{1}, μ_{2}) contains exactly one proper subspace ℬ_{s}(μ_{1}, μ_{2}) invariant under {𝔄, ε}. It is infinite dimensional and any representation of {𝔄, ε} equivalent to the restriction of ρ(μ_{1}, μ_{2}) to ℬ_{s}(μ_{1}, μ_{2}) will be denoted by σ(μ_{1}, μ_{2}). The quotient space

ℬ_{f}(μ_{1}, μ_{2}) = ℬ(μ_{1}, μ_{2})/ℬ_{s}(μ_{1}, μ_{2})

is finite-dimensional and π(μ_{1}, μ_{2}) will be any representation equivalent to the representation of {𝔄, ε} on this quotient space.

(iii) If μ_{1}μ_{2}^{−1}(t) = t^{p} sgn t, where p is a negative integer, the space ℬ(μ_{1}, μ_{2}) contains exactly one proper subspace ℬ_{f}(μ_{1}, μ_{2}) invariant under {𝔄, ε}. It is finite-dimensional and π(μ_{1}, μ_{2}) will be any representation equivalent to the restriction of ρ(μ_{1}, μ_{2}) to ℬ_{f}(μ_{1}, μ_{2}). σ(μ_{1}, μ_{2}) will be any representation equivalent to the representation on the quotient space

ℬ_{s}(μ_{1}, μ_{2}) = ℬ(μ_{1}, μ_{2})/ℬ_{f}(μ_{1}, μ_{2}).

Chapter 1 90 (iv) A representation π(μ_{1}, μ_{2}) is never equivalent to a representation σ(μ′_{1}, μ′_{2}).

(v) The representations π(μ_{1}, μ_{2}) and π(μ′_{1}, μ′_{2}) are equivalent if and only if either (μ_{1}, μ_{2}) = (μ′_{1}, μ′_{2}) or (μ_{1}, μ_{2}) = (μ′_{2}, μ′_{1}).

(vi) The representations σ(μ_{1}, μ_{2}) and σ(μ′_{1}, μ′_{2}) are equivalent if and only if (μ_{1}, μ_{2}) is one of

the four pairs (μ′_{1}, μ′_{2}), (μ′_{2}, μ′_{1}), (μ′_{1}η, μ′_{2}η), or (μ′_{2}η, μ′_{1}η). (vii) Every irreducible admissible representation of {𝔄, ε} is either a π(μ_{1}, μ_{2}) or a σ(μ_{1}, μ_{2}).

Let μ_{1}μ_{2}^{−1}(t) = |t|^{s}(\frac{t}{|t|})^{m}. s − m is an odd integer if and only if s is an integer p and μ_{1}μ_{2}^{−1}(t) = t^{p} sgn t. Thus the first three parts of the lemma are consequences of Lemma 5.6 and 5.7. The fourth follows from the observation that π(μ_{1}, μ_{2}) and σ(μ′_{1}, μ′_{2}) cannot contain the same representations of the Lie algebra of SO(2, ℝ).

We suppose first that s − m is not an odd integer and construct an invertible transformation T from ℬ(μ_{1}, μ_{2}) to ℬ(μ_{2}, μ_{1}) which commutes with the action of {𝔄, ε}. We have introduced a basis {φ_{n}} of ℬ(μ_{1}, μ_{2}). Let {φ′_{n}} be the analogous basis of ℬ(μ_{2}, μ_{1}). T will have to take φ_{n} to a multiple a_{n}φ′_{n} of φ′_{n}. Appealing to Lemma 5.6 we see that it commutes with the action of {𝔄, ε} if and only if

\begin{aligned}(s + 1 + n)a_{n+2} & = (−s + 1 + n)a_{n} \\ (s + 1 − n)a_{n−2} & = (−s + 1 − n)a_{n} \end{aligned}

and

a_{n} = (−1)^{m}a_{−n}.

These relations will be satisfied if we set

a_{n} = a_{n}(s) = \frac{Γ\frac{(−s+1+n)}{2}}{Γ\frac{(s+1+n)}{2}}

Since n = m(mod 2) and s−m−1 is not an even integer all these numbers are defined and different from 0.

If s ≤ 0 and s − m is an odd integer we set

a_{n}(s) = lim_{z→s} a_{n}(z)

The numbers a_{n}(s) are still defined although some of them may be 0. The associated operator T maps ℬ(μ_{1}, μ_{2}) into ℬ(μ_{2}, μ_{1}) and commutes with the action of {𝔄, ε}. If s = 0 the operator T is non-singular. If s < 0 its kernel is ℬ_{f}(μ_{1}, μ_{2}) and it defines an invertible linear transformation from ℬ_{s}(μ_{1}, μ_{2}) to ℬ_{s}(μ_{2}, μ_{1}). If s > 0 and s − m is an odd integer the functions a_{n}(z) have at most simple poles at s. Let

b_{n}(s) = lim_{z→s}(z − s) a_{n}(z)

The operator T associated to the family {b_{n}(s)} maps ℬ(μ_{1}, μ_{2}) into ℬ(μ_{2}, μ_{1}) and commutes with the action of {𝔄, ε}. It kernel is ℬ_{s}(μ_{1}, μ_{2}) so that it defines an invertible linear transformation from ℬ_{f}(μ_{1}, μ_{2}) to ℬ_{f}(μ_{2}, μ_{1}). These considerations together with Lemma 5.10 give us the equivalences of parts (v) and (vi).

Now we assume that π = π(μ_{1}, μ_{2}) and π′ = π(μ′_{1}, μ′_{2}) or π = σ(μ_{1}, μ_{2}) and π′ = σ(μ′_{1}, μ′_{2}) are equivalent. Let μ_{i}(T) = |t|^{s_{i}}(\frac{t}{|t|})^{m_{i}} and let μ′_{i}(t) = |t|^{s′_{i}}(\frac{t}{|t|})^{m′_{i}}. Let s = s_{1} −s_{2}, m = |m_{1} −m_{2}|, s′ = s′_{1} − s′_{2}, m′ = |m′_{1} − m′_{2}|. Since the two representations must contain the same representations of the Lie algebra of SO(2, ℝ) the numbers m and m′ are equal. Since π(D) and π′(D) must be Chapter 1 91 the same scalar Lemma 5.6 shows that s′ = ±s. π(J) and π′(J) must also be the same scalar so s′_{1} + s′_{2} = s_{1} + s_{2}. Thus if η(t) = sgn t the pair (μ_{1}, μ_{2}) must be one of the four pairs (μ′_{1}, μ′_{2}), (μ′_{2}, μ′_{1}), (ημ′_{1}, ημ′_{2}), (ημ′_{2}, ημ′_{1}). Lemma 5.9 shows that π(μ′_{1}μ′_{2}) and π(ημ′_{1}, ημ′_{2}) are not equivalent. Parts (v) and (vi) of the theorem follow immediately.

Lemmas 5.8, 5.9, and 5.10 show that to prove the last part of the theorem we need only show that any irreducible admissible representation π of 𝔄 is, for a suitable choice of μ_{1} and μ_{2}, a constituent of ρ(μ_{1}, μ_{2}). That is there should be two subspace ℬ_{1} and ℬ_{2} of ℬ(μ_{1}, μ_{2}) invariant under 𝔄 so that ℬ_{1} contains ℬ_{2} and π is equivalent to the representation of 𝔄 on the quotient ℬ_{1}/ℬ_{2}. If χ is a quasi-character of F^{×} then π is a constituent of ρ(μ_{1}, μ_{2}) if and only if χ ⊗ π is a constituent of ρ(χμ_{1}, χμ_{2}). Thus we may suppose that π(J) is 0 so that π is actually a representation of 𝔄_{0}, the universal enveloping algebra of the Lie algebra of Z_{ℝ}\G_{ℝ}. Since this group is semi-simple the desired result is a consequence of the general theorem of Harish–Chandra [6].

It is an immediate consequence of the last part of the theorem that every irreducible admissible representations of {𝔄, ε} is the representation associated to an irreducible admissible representation of ℋ_{ℝ}. Thus we have classifed the irreducible admissible representations of {𝔄, ε} and of ℋ_{ℝ}. We can write such a representation of ℋ_{ℝ} as π(μ_{1}, μ_{2}) or σ(μ_{1}, μ_{2}).

In the first paragraph we associated to every quasi-character ω of ℂ^{×} a representation of r_{ω} of G_{+} the group of matrices with positive determinant. r_{ω} acts on the space of functions Φ in 𝒮(C) which satisfy

Φ(xh) = ω^{−1}(h) Φ(x)

for all h such that hh̄ = 1. All elements of 𝒮(C, ω) are infinitely differentiable vectors for r_{ω} so that r_{ω} also determines a representation, again called r_{ω}, of 𝔄. r_{ω} depended on the choice of a character of ℝ. If that character is

ψ(x) = e^{2πuxi}

then

r_{ω}(X_{+}) Φ(z) = (2πuzz̄i)Φ(z).

Lemma 5.12 Let 𝒮_{0}(ℂ, ω) be the space of functions Φ in 𝒮(C, ω) of the form

Φ(z) = e^{−2π|u|zz̄}P (z, z̄)

where P (z, z̄) is a polynomial in z in z̄. Then 𝒮_{0}(ℂ, ω) is invariant under 𝔄 and the restriction of r_{ω} to 𝒮_{0}(ℂ, ω) is admissible and irreducible.

It is well known and easily verified that the function e^{−2π|u|zz̄} is its own Fourier transform provided of course that the transform is taken with respect to the character

ψ_{ℂ}(z) = ψ(z + z̄)

and the self-dual measure for that character. From the elementary properties of the Fourier transform one deduces that the Fourier transform of a function

Φ(z) = e^{−2π|u|zz̄} P (z, z̄)

where P is a polynomial in z and z̄ is of the same form. Thus r_{ω}(w) leaves 𝒮_{0}(ℂ, ω) invariant. Recall that

w = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) .

Chapter 1 92 𝒮_{0}(ℂ, ω) is clearly invariant under r_{ω}(X_{+}). Since X_{−} = Adw(X_{+}) it is also invariant under X. But X_{+} X_{−} − X_{−} X_{+} = Z so that it is also invariant under Z. We saw in the first paragraph that if ω_{0} is the restriction of ω to ℝ^{×} then

r_{ω} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = (sgn a) ω_{0}(a)I

thus r_{ω}(J) = cI if ω_{0}(a) = a^{c} for a positive a. In conclusion 𝒮_{0}(ℂ, ω) is invariant under 𝔤 and therefore under 𝔄.

If

ω(z) = (zz̄)^{r} \frac{z^{m}z̄^{n}}{(zz̄)^{\frac{m+n}{2}}}

where r is a complex number and m and n are two integers, one 0 and the other non-negative, the functions

Φ_{p}(z) = e^{−2π|u|zz̄} z^{n+p} z̄^{m+p},

with p a non-negative integer, form a basis of 𝒮_{0}(ℂ, ω). Suppose as usual that \frac{∂}{∂z} = \frac{1}{2} \frac{∂}{∂x} + \frac{1}{2i} \frac{∂}{∂y} and that \frac{∂}{∂z̄} = \frac{1}{2} \frac{∂}{∂x} − \frac{1}{2i} \frac{∂}{∂y}. Then the Fourier transform Φ′_{p} of Φ_{p} is given by

Φ′_{p}(z) = \frac{1}{(2πiu)^{m+n+2p}} \frac{∂^{n+p}}{∂z^{n+p}} \frac{∂^{m+p}}{∂z̄^{m+p}} e^{−2π|u|zz̄}

which is a function of the form

(i sgn u)^{m+n+2p}e^{−2π|u|zz̄} z̄^{n+p} z^{m+p} + ∑_{q=0}^{p−1} a_{q}e^{−2π|u|zz̄} z̄^{n+q} z^{m+q}.

Only the coefficient a_{p−1} interests us. It equals

\frac{(i sgn u)^{m+n+2p−1}}{2πiu} \{p(n + m + 1 + p − 1)\}.

Since

r_{ω}(w) Φ(z) = (i sgn u) Φ′(z̄)

and

r_{ω}(X_{−}) = (−1)^{m+n} r_{ω}(w) r_{ω}(X_{+}) r(w)

while

r_{ω}(X_{+})Φ_{p} = (2πui)Φ_{p+1}

we see that

r_{ω}(X_{−})Φ_{p} = (2πui)Φ_{p+1} − (i sgn u)(n + m + 2p + 1)Φ_{p} + ∑_{q=0}^{p−1} b_{q}Φ_{q}.

Since U = X_{+} − X_{−} we have

r_{ω}(U)Φ_{p} = (i sgn u)(n + m + 2p + 1)Φ_{p} − ∑_{1=0}^{p−1} b_{q}Φ_{q}

and we can find the functions Ψ_{p}, p = 0, 1, ⋯, such that

Ψ_{p} = Φ_{p} + ∑_{q=0}^{p−1} a_{pq} Φ_{q}

while

r_{ω}(U)Ψ_{p} = (i sgn u)(n + m + 2p + 1)Ψ_{p}.

These functions form a basis of 𝒮_{0}(ℂ, ω). Consequently r_{ω} is admissible.

If it were not irreducible there would be a proper invariant subspace which may or not contain Φ_{0}. In any case if 𝒮_{1} is the intersection of all invariant subspaces containing Φ_{0} and 𝒮_{2} is the sum of all invariant subspaces which do not contain Φ_{0} both 𝒮_{1} and 𝒮_{2} are invariant and the representation π_{1} of 𝔄 on 𝒮_{1}/𝒮_{2} ∩ 𝒮_{1} is irreducible. If the restriction of π_{1} to the Lie algebra of SO(2, ℝ) contains κ_{p} it does not contain κ_{−p}. Thus π_{1} is not equivalent to the representation X → π_{1}(Adε(X)). Consequently the irreducible representation π of {𝔄, ε} whose restriction to 𝔄 is π_{1} must be one of the special representations σ(μ_{1}, μ_{2}) or a representation π(μ_{1}, μ_{2}η). Examining these we see that since π contains κ_{q} with q = sgn u(n + m + 1) it contains all the representations κ_{q} with q = sgn u(n + m + 2p + 1), p = 0, 1, 2, · · ·. Thus 𝒮_{1} contains all the functions Ψ_{p} and 𝒮_{2} contains none of them. Since this contradicts the assumption that 𝒮_{0}(ℂ, ω) contains a proper invariant subspace the representation r_{ω} is irreducible.

For the reasons just given the representation π of {𝔄, ε} whose restriction to 𝔄 contains r_{ω} is either a σ(μ_{1}, μ_{2}) or a π(μ_{1}, μ_{1}η). It is a π(μ_{1}, μ_{1}η) if and only if n + m = 0. Since

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a) sgn aI = ω(a) η(a)I,

we must have μ_{1}μ_{2} = ω_{0}η in the first case and μ_{1}^{2} = ω_{0} in the second. ω_{0} is the restriction of ω to ℝ^{×}. Since the two solutions μ_{1}^{2} = ω_{0} differ by η they lead to the same representation. If n + m = 0 then μ_{1}^{2} = ω_{0} if and only if ω(z) = μ_{1}(ν(z)) for all z in ℂ^{×}. Of course ν(z) = zz̄.

Suppose n+m > 0 so that π is a σ(μ_{1}, μ_{2}). Let μ_{i}(t) = |t|^{s_{i}}(\frac{t}{|t|})^{m_{i}}. Because of Theorem 5.11 we can suppose that m_{1} = 0. Let s = s_{1} − s_{2}. We can also suppose that s is non-negative. If m = |m_{1} − m_{2}| then s − m is an odd integer so m and m_{2} are determined by s. We know what representations of the Lie algebra of SO(2, ℝ) are contained in π. Appealing to Lemma 5.7 we see that s = n + m. Since μ_{1}μ_{2} = ηω_{0} we have s_{1} + s_{2} = 2r. Thus s_{1} = r + \frac{m+n}{2} and s_{2} = r − \frac{n+m}{2}. In all cases the representation π is determined by ω alone and does not depend on ψ. We refer to it as π(ω). Every special representation σ(μ_{1}, μ_{2}) is a π(ω) and π(ω) is equivalent to π(ω′) if and only if ω = ω′ or ω′(z) = ω(z̄).

We can now take the first step in the proof of the local functional equation.

Theorem 5.13 Let π be an infinite-dimensional irreducible admissible representation of ℋ_{ℝ}. If ψ is a non-trivial additive character of ℝ there exists exactly one space W (π, ψ) of functions W on G_{ℝ} with the following properties

(i) If W is in W (π, ψ) then

W (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = ψ(x) W (g)

for all x in F.

Chapter 1 94

(ii) The functions W are continuous and W (π, ψ) is invariant under ρ(f) for all f in ℋ_{ℝ}.

Moreover the representation of ℋ_{ℝ} on W (π, ψ) is equivalent to π. (iii) If W is in W (π, ψ) there is a positive number N such that

W \begin{aligned}( \\ ( & \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = O(|t|^{N})

as |t| → ∞.

We prove first the existence of such a space. Suppose π = π(ω) is the representation associated to some quasi-character ω of ℂ^{×}. An additive character ψ being given the restriction of π to 𝔄 contains the representation r_{ω} determined by ω and ψ. For any Φ in 𝒮(ℂ, ω) define a function W_{Φ} on G_{+} by

W_{Φ}(g) = r_{ω}(g) Φ(1)

Since ρ(g) W_{Φ} = W_{r_{ω}(g)Φ} the space of such functions is invariant under right translations. Moreover

W_{Φ} (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = ψ(x) W_{Φ}(g)

Every vector in 𝒮(ℂ, ω) is infinitely differentiable for the representation r_{ω}. Therefore the functions W_{Φ} are all infinitely differentiable and, if X is in 𝔄,

ρ(X)W_{Φ} = W_{r_{ω}(X)Φ}.

In particular the space W_{1}(π, ψ) of those W_{Φ} for which Φ is in 𝒮_{0}(ℂ, ω) is invariant under 𝔄. We set W_{Φ} equal to 0 outside of G_{+} and regard it as a function on G_{ℝ}.

We want to take W (π, ψ) to be the sum of W_{1}(π, ψ) and its right translate by ε. If we do it will be invariant under {𝔄, ε} and transform according to the representation π of {𝔄, ε}. To verify the second condition we have to show that it is invariant under ℋ_{ℝ}. For this it is enough to show that 𝒮_{0}(ℂ, ω) is invariant under the elements of ℋ_{ℝ} with support in G_{+}. The elements certainly leave the space of functions in 𝒮(ℂ, ω) spanned by the functions transforming according to a onedimensional representation of SO(2, ℝ) invariant. Any function in 𝒮(ℂ, ω) can be approximated uniformly on compact sets by a function in 𝒮_{0}(ℂ, ω). If in addition it transforms according to the representation κ_{n} of SO(2, ℝ) it can be approximated by functions in 𝒮_{0}(ℂ, ω) transforming according to the same representation. In other words it can be approximated by multiples of a single function in 𝒮_{0}(ℂ, ω) and therefore is already in 𝒮_{0}(ℂ, ω).

The growth condition need only be checked for the functions W_{Φ} in W_{1}(π, ψ). If a is negative

W_{Φ} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = 0

but if a is positive and

Φ(z) = e^{−2π|u|zz̄}P (z, z̄)

it is equal to

e^{−2π|u|a}P (a^{1/2}, a^{1/2}) ω(a) |a|^{1/2},

and certainly satisfies the required condition.

We have still to prove the existence of W (π, ψ) when π = π(μ_{1}, μ_{2}) and is infinite dimensional. As in the first paragraph we set

θ(μ_{1}, μ_{2}, Φ) = ∫_{ℝ^{×}} μ_{1}(t) μ_{2}^{−1}(t) Φ(t, t^{−1}) d^{×}t

for Φ in 𝒮(ℝ^{s}) and we set

W_{Φ}(g) = μ_{1}(detg) |detg|^{1/2} θ(μ_{1}, μ_{2}, r(g)Φ)
= θ(μ_{1}, μ_{2}, r_{μ_{1},μ_{2}}(g)Φ).

r_{μ_{1},μ_{2}} is the representation associated to the quasi-character (a, b) → μ_{1}(a) μ_{2}(b) of ℝ^{×} × ℝ^{×}. If X is in 𝔄

ρ(X) W_{Φ}(g) = W_{r_{μ_{1},μ_{2}} (X)Φ}(g)

Let W (μ_{1}, μ_{2}; ψ) be the space of those W_{Φ} which are associated to O(2, ℝ)-finite functions Φ. W (μ_{1}, μ_{2}; ψ) is invariant under {𝔄, ε} and under ℋ_{ℝ}.

Lemma 5.13.1 Assume μ_{1}(x) μ_{2}^{−1}(x) = |x|^{s} (\frac{x}{|x|})^{m} with Re s > −1 and m equal to 0 or 1. Then there exists a bijection A of W (μ_{1}, μ_{2}; ψ) with ℬ(μ_{1}, μ_{2}) which commutes with the action of {𝔄, ε}.

We have already proved a lemma like this in the non-archimedean case. If Φ is in 𝒮(ℝ^{2}) and ω is a quasi-character of ℝ^{×} set

z(ω, Φ) = ∫ Φ(0, t) ω(t) d^{×}(t)

The integral converges if ω(t) = |t|^{r}(sgn t)^{n} with r > 0. In particular under the circumstances of the lemma

f_{Φ}(g) = μ_{1}(detg) |detg|^{1/2}z(μ_{1}μ_{2}^{−1}α_{ℝ}, ρ(g)Φ)

is defined. As usual α_{ℝ}(x) = |x|. A simple calculation shows that

f_{Φ} (( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} ) g) = μ_{1}(a_{1}) μ_{2}(a_{2}) |\frac{a_{1}}{a_{2}} |^{1/2} f_{Φ}(g).

If Φ^{∼} is the partial Fourier transform of Φ introduced in the first paragraph then

ρ(g) f_{Φ^{∼}} = f_{Φ_{1}^{∼}}

if Φ_{1} = r_{μ_{1},μ_{2}}(f)Φ. A similar relation will be valid for a function f in ℋ_{ℝ}, that is

ρ(f)f_{Φ^{∼}} = f_{Φ_{1}^{∼}}

if Φ_{1} = r_{μ_{1},μ_{2}}(f)Φ. In particular if f_{Φ^{∼}} is O(2, ℝ)-finite there is an elementary idempotent ξ such that ρ(ξ)f_{Φ^{∼}} = f_{Φ^{∼}}. Thus, if Φ_{1} = r_{μ_{1},μ_{2}}(ξ)Φ, f_{Φ^{∼}} = f_{Φ_{1}^{∼}} and Φ_{1}^{∼} is O(2, ℝ) finite. Of course f_{Φ^{∼}} is O(2, ℝ)-finite if and only if it belongs to ℬ(μ_{1}, μ_{2}).

We next show that given any f in ℬ(μ_{1}, μ_{2}) there is an O(2, ℝ)-finite function Φ in 𝒮(ℝ^{2}) such that f = f_{Φ^{∼}}. According to the preceding observation together with the self-duality of 𝒮(ℝ^{2}) under Fourier transforms it will be enough to show that for some Φ in 𝒮(ℝ^{2}), f = f_{Φ}. In fact, by linearity, it is sufficient to consider the functions φ_{n} in ℬ(μ_{1}, μ_{2}) defined earlier by demanding that

φ_{n} ( ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) ) = e^{inθ}

n must be of the same parity as m. If δ = sgn n set

Φ(x, y) = e^{−π(x^{2}+y^{2})}(x + iδy)^{|n|}

Then

ρ ( ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) ) Φ = e^{inθ}Φ

Since ρ(g)f_{Φ} = f_{ρ(g)Φ} when detg = 1 the function f_{Φ} is a multiple of φ_{n}. Since

\begin{aligned}f_{Φ}(e) & = (i)^{|n|} ∫_{−∞}^{∞} e^{−πt^{2}}t^{|n|+s+1} d^{×}t \\ & = (i)^{n} \frac{π^{\frac{−(|n|+s+1)}{2}}}{2} Γ\frac{(|n| + s + 1)}{2} \end{aligned}

which is not 0, the function f_{Φ} is not 0.

The map A will transform the function W_{Φ} to f_{Φ^{∼}}. It will certainly commute with the action of {𝔄, ε}. That A exists and is injective follows from a lemma which, together with its proof, is almost identical to the statement and proof of Lemma 3.2.1.

The same proof as that used in the non-archimedian case also shows that W (μ_{1}, μ_{2}; ψ) = W (μ_{2}, μ_{1}; ψ) for all ψ. To prove the existence of W (π, ψ) when π = π(μ_{1}, μ_{2}) and is infinitedimensional we need only show that when μ_{1} and μ_{2} satisfy the condition the previous lemma the functions W in W (μ_{1}, μ_{2}; ψ) satisfy the growth condition of the theorem. We have seen that we can take W = W_{Φ} with

Φ^{∼}(x, y) = e^{−π(x^{2}+y^{2})}P (x, y)

where P (x, y) is a polynomial in x and y. Then

Φ(x, y) = e^{−π(x^{2}+u^{2}y^{2})}Q(x, y)

where Q(x, y) is another polynomial. Recall that ψ(x) = e^{2πiux}. Then

W_{Φ} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = μ_{1}(a) |a|^{1/2} ∫_{−∞}^{∞} e^{−π(a^{2}t^{2}+u^{2}t^{−2})}Q(at, ut^{−1}) |t|^{s}(sgn t)^{m} d^{×}t

The factor in front certainly causes no harm. If δ > 0 the integrals from −∞ to −δ and from δ to ∞ decrease rapidly as |a| → ∞ and we need only consider integrals of the form

∫_{0}^{δ} e^{−π(a^{2}t^{2}+u^{2}t^{−2})}t^{r} dt

Chapter 1 97 where r is any real number and u is fixed and positive. If v = \frac{u}{2} then u^{2} = v^{2} + \frac{3u^{2}}{4} and e^{− \frac{3}{4} πu^{2}t^{−2}}t^{r} is bounded in the interval [0, δ] so we can replace u by v and suppose r is 0. We may also suppose that a and v are positive and write the integral as

e^{−2πav} ∫_{0}^{δ} e^{−π(at+vt^{−1})^{2}} dt.

The integrand is bounded by 1 so that the integral is O(1). In any case the growth condition is more than satisfied.

We have still to prove uniqueness. Suppose W_{1}(π, ψ) is a space of functions satisfying the first two conditions of the lemma. Let κ_{n} be a representation of the Lie algebra of SO(2, ℝ) occurring in π and let W_{1} be a function in W_{1}(π, ψ) satisfying

W_{1} (g ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} )) = e^{inθ}W_{1}(g).

If

φ_{1}(t) = W_{1} \begin{aligned}( \\ ( & \begin{matrix}\frac{t}{|t|^{1/2}} & 0 \\ 0 & \frac{1}{|t|^{1/2}} \end{matrix} & ) \\ & & ) \end{aligned}

the function W_{1} is completely determined by φ_{1}. It is easily seen that

\begin{aligned}ρ(U)W_{1} \begin{aligned}( \\ ( & \begin{matrix}\frac{t}{|t|^{1/2}} & 0 \\ 0 & \frac{1}{|t|^{1/2}} \end{matrix} & ) \\ & & ) \end{aligned} & = i n φ_{1}(t) \\ ρ(Z)W_{1} \begin{aligned}( \\ ( & \begin{matrix}\frac{t}{|t|^{1/2}} & 0 \\ 0 & \frac{1}{|t|^{1/2}} \end{matrix} & ) \\ & & ) \end{aligned} & = 2t \frac{dφ_{1}}{dt} \\ ρ(X_{+})W_{1} \begin{aligned}( \\ ( & \begin{matrix}\frac{t}{|t|^{1/2}} & 0 \\ 0 & \frac{1}{|t|^{1/2}} \end{matrix} & ) \\ & & ) \end{aligned} & = i u t φ_{1}(t). \end{aligned}

Thus if φ_{1}^{+} and φ_{1}^{−} correspond to ρ(V_{+})W_{1} and ρ(V_{−})W_{1}

φ_{1}^{+}(t) = 2t\frac{dφ_{1}}{dt} − (2ut − n) φ_{1}

and

φ_{1}^{−}(t) = 2t\frac{dφ_{1}}{dt} + (2ut − n) φ_{1}(t).

Since

D = \frac{1}{2}V_{−} V_{+} − iU − \frac{U^{2}}{2}

ρ(D)W_{1} corresponds to

2t \frac{d}{dt} (t \frac{dφ_{1}}{dt} − 2t \frac{dφ_{2}}{dt} ) + (2nut − 2u^{2}t^{2})φ_{1}.

Finally ρ(ε)W_{1} corresponds to φ_{1}(−t).

Chapter 1 98

Suppose that π is either π(μ_{1}, μ_{2}) or σ(μ_{1}, μ_{2}). Let μ_{1}μ_{2}^{−1}(t) = |t|^{s}(sgn t)^{m}. If s − m is an odd integer we can take n = |s| + 1. From Lemma 5.6 we have ρ(V_{−})W_{1} = 0 so that φ_{1} satisfies the equation

2t \frac{dφ_{1}}{dt} + (2ut − n)φ_{1} = 0.

If the growth condition is to be satisfied φ_{1} must be 0 for ut < 0 and a multiple of |t|^{n/2}e^{−ut} for ut > 0. Thus W_{1} is determined up to a scalar factor and the space W (π, ψ) is unique.

Suppose s − m is not an odd integer. Since ρ(D)W_{1} = \frac{s^{2}−1}{2}W_{1} the function φ_{1} satisfies the equation

\frac{d^{2}φ_{1}}{dt^{2}} + \{−u^{2} + \frac{nu}{t} + \frac{(1 − s^{2})}{4t^{2}}\} φ_{1} = 0

We have already constructed a candidate for the space W (π, ψ). Let’s call this candidate W_{2}(π, ψ). There will be a non-zero function φ_{2} in it satisfying the same equation as φ_{1}. Now φ_{1} and all of its derivatives go to infinity no faster than some power of |t| as t → ∞ while as we saw φ_{2} and its derivations go to 0 at least exponentially as |t| → ∞. Thus the Wronskian

φ_{1} \frac{dφ_{2}}{dt} − φ_{2} \frac{dφ_{1}}{dt}

goes to 0 as |t| → ∞. By the form of the equation the Wronskian is constant. Therefore it is identically 0 and φ_{1}(t) = α φ_{2}(t) for t > 0 and φ_{1}(t) = β φ_{2}(t) for t < 0 where α and β are two constants. The uniqueness will follow if we can show that for suitable choice of n we have α = β. If m = 0 we can take n = 0. If μ_{1}(t) = |t|^{s_{1}}(sgn t)^{m_{1}} then π(ε)W_{1} = (−1)^{m_{2}}W_{1} so that φ_{1}(−t) = (−1)^{m_{1}}φ_{1}(t) and φ_{2}(−t) = (−1)^{m_{2}}φ_{2}(t). Thus α = β. If m = 1 we can take n = 1. From Lemma 5.6

π(V_{−1})W_{1} = (−1)^{m_{1}}sπ(ε)W_{1}

so that

2t \frac{dφ_{1}}{dt} + (2ut − 1)φ_{1}(t) = (−1)^{m_{1}}sφ_{1}(−t).

Since φ_{2} satisfies the same equation α = β.

If μ is a quasi-character of ℝ^{×} and ω is the character of ℂ^{×} defined by ω(z) = μ(zz̄) then π(ω) = π(μ, μη). We have defined W (π(ω), ψ) in terms of ω and also as W (μ_{1}, μ_{2}; ψ). Because of the uniqueness the two resulting spaces must be equal.

Corollary 5.14 Let m and n be two integers, one positive and the other 0. Let ω be a quasi-character of ℂ^{×} of the form

ω(z) = (zz̄)^{r− \frac{m+n}{2}} z^{m} z̄^{n}

and let μ_{1} and μ_{2} be two quasi-characters of ℝ^{×} satisfying μ_{1}μ_{2}(x) = |x|^{2r}(sgn x)^{m+n+1} and μ_{1}μ_{2}^{−1}(x) = x^{m+n} sgn x so that π(ω) = σ(μ_{1}, μ_{2}). Then the subspace ℬ_{s}(μ_{1}, μ_{2}) of ℬ(μ_{1}, μ_{2}) is defined and there is an isomorphism of ℬ(μ_{1}, μ_{2}) with W (μ_{1}, μ_{2}; ψ) which commutes with the action of {𝔄, ε}. The image W_{s}(μ_{1}, μ_{2}; ψ) of ℬ_{s}(μ_{1}, μ_{2}) is W (π(ω), ψ). If Φ belongs to 𝒮(ℝ^{2}) and W_{Φ} belongs to W (μ_{1}, μ_{2}; ψ) then W_{Φ} belongs to W_{s}(μ_{1}, μ_{2}; ψ) if and only if

∫_{∞}^{∞} x^{i} \frac{∂^{j}}{∂y^{j}} Φ(x, 0) dx = 0

Chapter 1 99 for any two non-negative integers i and j with i + j = m + n − 1.

Only the last assertion is not a restatement of previously verified facts. To prove it we have to show that f_{Φ^{∼}} belongs to ℬ_{s}(μ_{1}, μ_{2}) if and only if Φ satisfies the given relations. Let f = f_{Φ^{∼}}. It is in ℬ_{s}(μ_{1}, μ_{2}) if and only if it is orthogonal to the functions in ℬ_{f}(μ_{1}^{−1}, μ_{2}^{−1}). Since ℬ_{f}(μ_{1}^{−1}, μ_{2}^{−1}) is finite-dimensional there is a non-zero vector f_{0} in it such that ρ(X_{+})f_{0} = 0. Then

f_{0} (w ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} )) = f_{0}(w)

and f is orthogonal to f_{0} if and only if

∫_{ℝ} f (w ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} )) dy = 0.  (5.14.1)

The dimension of ℬ_{f}(μ_{1}^{−1}, μ_{2}^{−2}) is m + n. It follows easily from Lemmas 5.6 and 5.7 that the vectors ρ(X_{+}^{p}) ρ(w) f_{0}, 0 ≤ p ≤ m + n − 1 span it. Thus f is in ℬ_{s}(μ_{1}, μ_{2}) if and only if each of the functions ρ(X_{+}^{p}) ρ(w) f satisfy (5.14.1). For f itself the left side of (5.14.1) is equal to

∫ \{∫ Φ^{∼} ((0, t) w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) μ_{1}(t) μ_{2}^{−1}(t) |t| d^{×}t\} dx.

Apart from a positive constant which relates the additive and multiplicative Haar measure this equals

∫∫ Φ^{∼}(−t, −tx)t^{m+n} sgn t dt dx

which is

(−1)^{m+n−1}∫∫ Φ^{∼}(t, x)t^{m+n−1} dt dx

or, in terms of Φ,

(−1)^{m+n−1} ∫ Φ(t, 0) t^{m+n−1} dt.  (5.14.2)

By definition

r_{μ_{1},μ_{2}}(w) Φ(x, y) = Φ′(y, x)

and an easy calculation based on the definition shows that

r_{μ_{1},μ_{2}}(X_{+}^{p}) Φ(x, y) = (2iπuxy)^{p} Φ(x, y).

Thus r_{μ_{1},μ_{2}}(X_{+}^{p}) r_{μ_{1},μ_{2}}(w) Φ is a non-zero scalar times

\frac{∂^{2p}}{∂x^{p} ∂y^{p}} Φ′(y, x)

For this function (5.14.2) is the product of a non-zero scalar and

∫∫ \frac{∂^{2p}}{∂x^{p} ∂y^{p}} Φ′(0, x) x^{m+n−1} dx.

Integrating by parts we obtain

∫ \frac{∂^{p}}{∂y^{p}} Φ′(0, x) x^{m+n−p−1} dx

except perhaps for sign. If we again ignore a non-zero scalar this can be expressed in terms of Φ as

∫ \frac{∂^{m+n−p−1}}{∂y^{m+n−p−1}} Φ(x, 0) x^{p} dx.

The proof of the corollary is now complete.

Before stating the local functional equation we recall a few facts from the theory of local zeta-functions. If F is ℝ or ℂ and if Φ belongs to 𝒮(F) we set

Z(ωα_{F}^{s} , Φ) = ∫ Φ(a) ω(a) |a|_{F}^{s} d^{×}a.

ω is a quasi-character. The integral converges in a right half-plane. One defines functions L(s, ω) and ε(s, ω, ψ_{F}) with the following properties:

(a) For every Φ the quotient

\frac{Z(ωα_{F}^{s} , Φ)}{L(s, ω)}

has an analytic continuation to the whole complex plane as a holomorphic function. Moreover for a suitable choice of Φ it is an exponential function and in fact a constant.

(b) If Φ′ is the Fourier transform of Φ with respect to the character ψ_{F} then

\frac{Z(ω^{−1}α_{F}^{1−s}, Φ′)}{L(1 − s, ω^{−1})} = ε(s, ω, ψ_{F})\frac{Z(ωα_{F}^{s} , Φ)}{L(s, ω)}.

If F = ℝ and ω(x) = |x|_{ℝ}^{r}(sgn x)^{m} with m equal to 0 or 1 then

L(s, ω) = π^{− \frac{1}{2} (s+r+m)} Γ(\frac{s + r + m}{2})

and if ψ_{F} (x) = e^{2πiux}

ε(s, ω, ψ_{F}) = (i sgn u)^{m} |u|_{ℝ}^{s+r− \frac{1}{2}} .

If F = ℂ and

ω(x) = |x|_{ℂ}^{r} x^{m}x̄^{n}

where m and n are non-negative integers, one of which is 0, then

L(s, ω) = 2(2π)^{−(s+r+m+n)}Γ(s + r + m + n).

Recall that |x|_{ℂ} = xx̄. If ψ_{F} (x) = e^{4πi Re(wz)}

ε(s, ω, ψ_{F}) = i^{m+n}ω(w) |w|_{ℂ}^{s−1/2}.

Chapter 1 101

These facts recalled, let π be an irreducible admissible representation of ℋ_{ℝ}. If π = π(μ_{1}, μ_{2}) we set

L(s, π) = L(s, μ_{1}) L(s, μ_{2})

and

ε(s, π, ψ_{ℝ}) = ε(s, μ_{1}, ψ_{ℝ}) ε(s, μ_{2}, ψ_{ℝ})

and if π = π(ω) where ω is a character of ℂ^{∗} we set

L(s, π) = L(s, ω)

and

ε(s, π, ψ_{ℝ}) = λ(ℂ/ℝ, ψ_{ℝ}) ε(s, ω, ψ_{ℂ/ℝ})

if ψ_{ℂ/ℝ}(z) = ψ_{ℝ}(z + z̄). The factor λ(ℂ/ℝ, ψ_{ℝ}) was defined in the first paragraph. It is of course necessary to check that the two definitions coincide if π(ω) = π(μ_{1}, μ_{2}). This is an immediate consequence of the duplication formula.

Theorem 5.15 Let π be an infinite-dimensional irreducible admissible representation of ℋ_{ℝ}. Let ω be the quasi-character of ℝ^{×} defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

If W is in W (π, ψ) set

\begin{aligned}Ψ(g, s, W) & = ∫_{ℝ^{×}} W (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s−1/2} d^{×}a \\ Ψ̃(g, s, W) & = ∫_{ℝ^{×}} W (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) ω^{−1}(a) |a|^{s−1/2} d^{×}a \end{aligned}

and let

\begin{aligned}Ψ(g, s, W) & = L(s, π) Φ(g, s, W) \\ Ψ̃(g, s, W) & = L(s, π̃) Φ̃(g, s, W). \end{aligned}

(i) The integrals defined Ψ(g, s, W) and Ψ̃(g, s, W) are absolutely convergent in some right halfplane.

(ii) The functions Φ(g, s, W) and Φ̃(g, s, W) can be analytically continued to the whole complex plane as meromorphic functions. Moreover there exists a W for which Φ(e, s, W) is an exponential function of s.

(iii) The functional equation

Φ̃(wg, 1 − s, W) = ε(s, π, ψ) Φ(g, s, W)

is satisifed. (iv) If W is fixed Ψ(g, s, W) remains bounded as g varies in a compact set and s varies in the

region obtained by removing discs centred at the poles of L(s, π) from a vertical strip of finite width.

We suppose first that π = π(μ_{1}, μ_{2}). Then W (π, ψ) = W (μ_{1}, μ_{2}; ψ). Each W in W (μ_{1}, μ_{2}; ψ) is of the form W = W_{Φ} where

Φ(x, y) = e^{−π(x^{2}+u^{2}y^{2})}P (x, y)

Chapter 1 102 with P (x, y) a polynomial. However we shall verify the assertions of the theorem not merely for W in W (π, ψ) but for any function W = W_{Φ} with Φ in 𝒮(ℝ^{2}). Since this class of functions is invariant under right translations most of the assertions need then be verified only for g = e.

A computation already performed in the non-archimedean case shows that

Ψ(e, s, W) = Z(μ_{1}α_{ℝ}^{s}, μ_{2}α_{ℝ}^{s}, Φ)

the integrals defining these functions both being absolutely convergent in a right half-plane. Also for s in some left half-plane

Ψ̃(w, 1 − s, W) = Z(μ_{1}^{−1}α_{ℝ}^{1−s}, μ_{2}^{−1}α_{ℝ}^{1−s}, Φ′)

if Φ′ is the Fourier transform of Φ.

Since Φ can always be taken to be a function of the form Φ(x, y) = Φ_{1}(x) Φ_{2}(y) the last assertion of part (ii) is clear. All other assertions of the theorem except the last are consequence of the following lemma.

Lemma 5.15.1 For every Φ in 𝒮(ℝ^{2}) the quotient

\frac{Z(μ_{1}α_{ℝ}^{s_{1}}, μ_{2}α_{ℝ}^{s_{2}}, Φ)}{L(s, μ_{1}) L(s, μ_{2})}

is a holomorphic function of (s_{1}, s_{2}) and

\frac{Z(μ_{1}^{−1}α_{ℝ}^{1−s_{1}}, μ_{2}^{−1}α_{ℝ}^{1−s_{2}}, Φ′)}{L(1 − s_{1}, μ_{1}^{−1}) L(1 − s_{2}, μ_{2}^{−1})}

is equal to

ε(s_{1}, μ_{1}, ψ) ε(s_{2}, μ_{2}, ψ) \frac{Z(μ_{1}α_{ℝ}^{s_{1}}, μ_{2}α_{ℝ}^{s_{2}}, Φ)}{L(s_{1}, μ_{1}) L(s_{2}, μ_{2})}.

We may as well assume that μ_{1} and μ_{2} are characters so that the integrals converge for Re s_{1} > 0 and Re s_{2} > 0. We shall show that when 0 < Re s_{1} < 1 and 0 < Re s_{2} < 1

Z(μ_{1}α_{ℝ}^{s_{1}}, μ_{2}α_{ℝ}^{s_{2}}, Φ) Z(μ_{1}^{−1}α_{ℝ}^{1−s_{1}}, μ_{2}^{−1}α_{ℝ}^{1−s_{2}}Ψ′)

is equal to

Z(μ_{1}^{−1}α_{ℝ}^{1−s_{1}}, μ_{2}^{−1}α_{ℝ}^{1−s_{2}}, Φ′) Z(μ_{1}α_{ℝ}^{s_{1}}, μ_{2}α_{ℝ}^{s_{2}}, Ψ)

if Φ and Ψ belong to 𝒮(ℝ^{2}).

The first of these expressions is equal to

∫ Φ(x, y)Ψ′(u, v)μ_{1}(\frac{x}{u}) μ_{2}(\frac{y}{v} ) |\frac{x}{u}|^{s_{1}} |\frac{y}{v} |^{s_{2}} d^{×}x d^{×}y du dv

if we assume, as we may, that d^{×}x = |x|^{−1} dx. Changing variables we obtain

∫ μ_{1}(x) μ_{2}(y) |x|^{s_{1}} |y|^{s_{2}} {∫ Φ(xu, yv) Ψ′(u, v) du dv} d^{×}x d^{×}y

The second expression is equal to

∫ μ_{1}^{−1}(x) μ_{2}^{−1}(y) |x|^{1−s_{1}} |y|^{1−s_{1}} {∫ Φ′(xu, yv) Ψ(u, v) du dv} d^{×}x d^{×}y

which equals

∫ μ_{1}(x) μ_{2}(y) |x|^{s_{1}} |y|^{s_{2}} {∫ |xy|^{−1} Φ′(x^{−1}u, y^{−1}v) Ψ(u, v) du dv} d^{×}x d^{×}y.

Since the Fourier transform of the function (u, v) → Φ(xu, yv) is the function |xy|^{−1}Φ′(x^{−1}u, y^{−1}v) the Plancherel theorem implies that

∫ Φ(xu, yv) Ψ′(u, v) du dv = ∫ |xy|^{−1}Φ′(x^{−1}u, y^{−1}v) Ψ(u, v) du dv.

The desired equality follows.

Choose Φ_{1} and Φ_{2} in 𝒮(ℝ) such that

L(s, μ_{i}) = Z(μ_{i}α_{ℝ}^{s}, Φ_{i})

and take Ψ(x, y) = Φ_{1}(x) Φ_{2}(y). The functional equation of the lemma follows immediately if 0 < s_{1} < 1 and 0 < s_{2} < 1. The expression on one side of the equation is holomorphic for 0 < Re s_{1} and 0 < Re s_{2}. The expression on the other side is holomorphic for Re s_{1} < 1 and Re s_{2} < 1. Standard and easily proved theorems in the theory of functions of several complex variables show that the function they define is actually an entire function of s_{1} and s_{2}. The lemma is completely proved.

For π = π(μ_{1}, μ_{2}) the final assertion of the theorem is a consequence of the following lemma.

Lemma 5.15.2 Let Ω be a compact subset of 𝒮(ℝ^{2}) and C a domain in ℂ^{2} obtained by removing balls about the poles of L(s_{1}, μ_{1}) L(s_{2}, μ_{2}) from a tube a_{1} ≤ Re s_{1} ≤ b_{1}, a_{2} ≤ Re s_{2} ≤ b_{2}. Then

Z(μ_{1}α_{ℝ}^{s_{1}}, μ_{2}α_{ℝ}^{s_{2}}, Φ)

remains bounded as Φ varies in Ω and (s_{1}, s_{2}) varies in C.

The theorems in the theory of functions alluded to earlier show that it is enough to prove this when either both a_{1} and a_{2} are greater than 0 or both b_{1} and b_{2} are less than 1. On a region of the first type the functions Z(μ_{1}α_{ℝ}^{s}, μ_{2}α_{ℝ}^{s}, Φ) is defined by a definite integral. Integrating by parts as in the theory of Fourier transforms one finds that

Z(μ_{1}α_{ℝ}^{σ_{1}+iτ_{1}}, μ_{2}α_{ℝ}^{σ_{2}+iτ_{2}}, Φ) = O(τ_{1}^{2} + τ_{2}^{2})^{−n}

as τ_{1}^{2} + τ_{2}^{2} → ∞ uniformly for Φ in Ω and a_{1} ≤ σ_{1} ≤ b_{1}, a_{2} ≤ σ_{2} ≤ b_{2} which is a much stronger estimate than required. For a region of the second type one combines the estimates just obtained with the functional equation and the known asymptotic behavior of the Γ-function.

Now let ω be a quasi-character of ℂ^{×} which is not of the form ω(z) = χ(zz̄) with χ a quasicharacter of ℝ^{×} and let π = π(ω). W (π, ψ) is the sum of W_{1}(π, ψ) and its right translate by ε. It is easily seen that

Φ(g, s, ρ(ε)W) = ω(−1) Φ(ε^{−1}gε, s, W)

and that

Φ̃(wg, s, ρ(ε)W) = ω(−1) Φ̃(wε^{−1}gε, s, W)

Thus it will be enough to prove the theorem for W in W_{1}(π, ψ). Since

Φ(εg, s, W) = Φ(g, s, W)

and

Φ̃(wεg, s, W) = Φ̃(wg, s, W)

we can also take g in G_{+}. W_{1}(π, ψ) consists of the functions W_{Φ} with Φ in 𝒮_{0}(ℂ, ω). We prove the assertions for functions W_{Φ} with Φ in 𝒮(ℂ, ω). Since this class of functions is invariant under right translations by elements of G_{+} we may take g = e.

As we observed in the first paragraph we will have

\begin{aligned}Ψ(e, s, W) & = Z(ωα_{ℂ}^{s}, Φ) \\ Ψ̃(w, 1 − s, W) & = λ(ℂ/ℝ, ψ) Z(ω^{−1}α_{ℂ}^{1−s}, Φ′) \end{aligned}

in some right half plane and the proof proceeds as before. If ω(z) = (zz̄)^{r}z^{m}z̄^{n} and p − q = n − m the function

Φ(z) = e^{−2π|u|zz̄}z^{p}z̄^{q}

belongs to 𝒮_{0}(ℂ, ω) and

\begin{aligned}Z(ωα_{ℂ}^{s}, Φ) & = 2π ∫_{0}^{∞} e^{−2π|u|t^{2}}t^{2}(s + r + p + m) dt \\ & = π(2π|u|)^{−(s+r+p+m)} Γ(s + r + p + m) \end{aligned}

Taking p = n we obtain an exponential times L(s, ω). The last part of the theorem follows from an analogue of Lemma 5.15.2.

The local functional equation which we have just proved is central to the Hecke theory. We complete the paragraph with some results which will be used in the paragraph on extraordinary representations and the chapter on quaternion algebras.

Lemma 5.16 Suppose μ_{1} and μ_{2} are two quasi-characters for which both π = π(μ_{1}, μ_{2}) and σ = σ(μ_{1}, μ_{2}) are defined. Then

\frac{L(1 − s, σ̃) ε(s, σ, ψ)}{L(s, σ)} = \frac{L(1 − s, π̃) ε(s, π, ψ)}{L(s, π)}

and the quotient

\frac{L(s, σ)}{L(s, π)}

is an exponential times a polynomial.

Interchanging μ_{1} and μ_{2} if necessary we may suppose that μ_{1}μ_{2}^{−1}(x) = |x|^{s}(sgn x)^{m} with s > 0. According to Corollary 5.14, W (σ, ψ) is a subspace of W (μ_{1}, μ_{2}, ψ). Although W (μ_{1}, μ_{2}, ψ) is not irreducible it is still possible to define Ψ(g, s, W) and Ψ̃(g, s, W) when W lies in W (μ_{1}, μ_{2}, ψ) and to use the method used to prove Theorem 5.15 to show that

\frac{Ψ̃(wg, 1 − s, W)}{L(1 − s, π̃)}

is equal to

ε(s, πψ)\frac{Ψ(g, s, W)}{L(s, π)}

Applying the equality to an element of W (σ, ψ) we obtain the first assertion of the lemma. The second is most easily obtained by calculation. Replacing μ_{1} and μ_{2} by μ_{1}α_{ℝ}^{t} and μ_{2}α_{ℝ}^{t} is equivalent to a translation in s so we may assume μ_{2} is of the form μ_{s}(x) = (sgn x)^{m_{2}}. There is a quasi-character ω of ℂ^{×} such that σ = π(ω). If ω(z) = (zz̄)^{r}z^{m}z̄^{n} then μ_{1}(x) = |x|^{2r+m+n}(sgn x)^{m+n+m_{2}+1}, μ_{1}(x) = x^{m+n}(sgn x)^{m_{2}+1} so that r = 0. Apart from an exponential factor L(s, σ) is equal to Γ(s + m + n) while L(s, π) is, again apart from an exponential factor,

Γ (\frac{s + m + n + m_{1}}{2} ) Γ (\frac{s + m_{2}}{2} )  (5.16.1)

where m_{1} = m + n + m_{2} + 1 (mod 2). Since m + n > 0 the number

k = \frac{1}{2}(m + n + 1 + m_{1} − m_{2}) − 1

is a non-negative integer and m_{2} + 2k = m + n + m_{1} − 1. Thus

Γ (\frac{s + m_{2}}{2} ) = \{\frac{1}{2^{k+1}} ∏_{j=0}^{k}(s + m_{2} + 2j)\}^{−1} Γ (\frac{s + m + n + m_{1} + 1}{2}) .

By the duplication formula the product (5.16.1) is a constant times an exponential times

\frac{Γ(s + m + n + m_{1})}{∏_{j=0}^{k}(s + m_{2} + 2j)}.

If m_{1} = 0 the lemma follows immediately. If m_{1} = 1

Γ(s + m + n + m_{1}) = (s + m + n) Γ(s + m + n)

and m_{2} + 2k = m + n. The lemma again follows.

Chapter 1 106 Lemma 5.17 Suppose ω(z) = (zz̄)^{r}z^{m}z̄^{n} is a quasi-charaacter of ℂ^{×} with mn = 0 and m + n > 0. Suppose μ_{1} and μ_{2} are two quasi-characters of F^{×} with μ_{1}μ_{2}(x) = |x|^{2r}x^{m+n} sgn x and μ_{1}μ_{2}^{−1}(x) = x^{m+n} sgn x. Then for every Φ in 𝒮(ℝ^{2}) such that

∫ x^{i} \frac{∂^{j}Φ}{∂yj}(x, 0) dx = 0

for i > 0, j ≥ 0, and i + j + 1 = m + n the quotient

\frac{Z(μ_{1}α_{ℝ}^{s}, μ_{2}α_{ℝ}^{s}, Φ)}{L(s, π(w))}

is a holomorphic function of s and for some Φ it is an exponential.

If W_{Φ} belongs to W (μ_{1}, μ_{2}, ψ) this is a consequence of Corollary 5.14 and Theorem 5.15. Unfortunately we need the result for all Φ. The observations made during the proof of Lemma 5.16 show that if π = π(μ_{1}, μ_{2}) the quotient

\frac{Z(μ_{1}α_{ℝ}^{s}, μ_{2}α_{ℝ}^{s}, Φ)}{L(s, π)}

is holomorphic. Since L(s, π) and L(s, σ) have no zeros we have only to show that the extra poles of L(s, π) are not really needed to cancel poles of Z(μ_{1}α_{ℝ}^{s}, μ_{2}α_{ℝ}^{s}, Φ). As in the proof of Lemma 5.16 we may take r = 0. We have to show that Z(μ_{1}α_{ℝ}^{s}, μ_{2}α_{ℝ}^{s}, Φ) is holomorphic at s = −m_{2} − 2j, 0 ≤ j ≤ k if m_{1} = 0 and at s = −m_{2} − 2j, 0 ≤ j ≤ k if m_{1} = 1. We remark first that if μ_{1} and μ_{2} are two quasi-characters of ℝ^{×}, Φ belongs to 𝒮(ℝ^{2}), and Re s is sufficiently large then, by a partial integration,

∫ μ_{1}(x) μ_{2}(y) |x|^{s} |y|^{s} Φ(x, y) d^{×}x d^{×}y = −\frac{1}{s} ∫ μ_{1}(x) μ_{2}(y) η(y) |x|^{s} |y|^{s+1} \frac{∂Φ}{∂y}(x, y) d^{×}x d^{×}y

if η(y) = sgn y. Integrating by parts again we obtain

∫ μ_{1}(x)μ_{2}(y) |x|^{s} |y|^{s} Φ(x, y) d^{×}x d^{×}y = \frac{1}{s(s + 1)} ∫ μ_{1}(x) μ_{2}(y) |x|^{s} |y|^{s+2} \frac{∂^{2}Φ}{∂y^{2}}(x, y) d^{×}x d^{×}y.

If Φ belongs to 𝒮(ℝ^{2}) the function defined by

∫ Φ(x, y) |x|^{s+1} |y|^{s} d^{×}x d^{y}y  (5.17.1)

is certainly holomorphic for Re s > 0. We have to show that if

∫ Φ(x, 0) dx = 0

it is holomorphic for Re s > −1. Suppose first that Φ(x, 0) ≡ 0. Since

Φ(x, y) = y \frac{∂Φ}{∂y}(x, 0) + ∫_{0}^{y}(y − u)\frac{∂^{2}Φ}{∂y^{2}}(x, u) du

Chapter 1 107 the function

Ψ(x, y) = \frac{1}{y} Φ(x, y)

is dominated by the inverse of any polynomial. Thus (5.17.1) which equals

∫ Ψ(x, y) |x|^{s+1} |y|^{s+1} η(y) d^{×}x d^{×}y

is absolutely convergent for Re s > −1. In the general case we set

\begin{aligned}Φ(x, y) & = \{Φ(x, y) − Φ(x, 0)e^{−y^{2}}\} + Φ(x, 0)e^{−y^{2}} \\ & = Φ_{1}(x, y) + Φ_{2}(x, y). \end{aligned}

Since Φ_{1}(x, 0) = 0 we need only consider

∫ Φ_{2}(x, 0)e^{−y^{2}} |x|^{s+1} |y|^{s} d^{×}x d^{×}y

which is the product of a constant and

Γ(\frac{s}{2}) ∫ Φ_{2}(x, 0) |x|^{s} dx.

The integral defines a function which is holomorphic for Re s > −1 and, when the assumptions are satisfied, vanishes at s = 0.

We have to show that if 0 ≤ j ≤ m + n − 1 and j − m_{2} is even then Z(μ_{1}α_{ℝ}^{s}, μ_{2}α_{ℝ}^{s}, Φ) is holomorphic at −j. Under these circumstances the function Z(μ_{1}α_{ℝ}^{s}, μ_{2}α_{ℝ}^{s}, Φ) is equal to

∫ η(x)^{m_{1}} η(y)^{m_{2}} |x|^{m+n} |x|^{s}|y|^{s} Φ(x, y) d^{×}x d^{×}y

which equals

\frac{(−1)^{j}}{∏_{i=0}^{j−1}(s_{i})} ∫ η(x)^{m_{1}} |x|^{s+m+n} |y|^{s+j} \frac{∂^{j}Φ}{∂y^{j}}(x, y) d^{×}x d^{×}y.

The factor in front is holomorphic at s = −j. If

Ψ(x, y) = x^{m+n−j−1} \frac{∂^{j}Φ}{∂y^{j}}(x, y)

the integral itself is equal to

∫ |x|^{s+j+1}|y|^{s+j} Ψ(x, y) d^{×}x d^{×}y.

Since, by assumption,

∫ Ψ(x, 0) dx = 0,

it is holomorphic at s = −j.

We observe that if m + n is even

Φ(x, y) = e^{−π(x^{2}+y^{2})}xy^{m+n}

satisfies the conditions of the lemma and, if r = 0 and m_{2} = 0, Z(μ_{1}α_{ℝ}^{s}, μ_{2}α_{ℝ}^{s}, Φ) is equal to

∫ e^{−π(x^{2}+y^{2})} |x|^{m+n+s+1} |y|^{m+n+s} d^{×}x d^{×}y

which differs by an exponential from Γ(s + m + n) and L(s, π(ω)). If m_{2} = 1 we take Φ(x, y) = e^{−π(x^{2}+y^{2})} y^{m+n+1} to obtain the same result. If m + n is odd and m_{2} = 0 the polynomial factor will be y^{m+n+1} but if m + n is odd and m_{2} = 1 it will again be xy^{m+n}.

Proposition 5.18 Suppose π and π′ are two infinite-dimensional irreducible admissible representations of ℋ_{ℝ} such that, for some quasi-character ω of F^{×},

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I π′ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I.

If

\frac{L(1 − s, χ^{−1} ⊗ π̃)}{L(s, χ ⊗ π)} ε(s, χ ⊗ π, ψ) = \frac{L(1 − s, χ^{−1} ⊗ π̃′)}{L(s, χ ⊗ π′)} ε(s, χ ⊗ π′, ψ)

for all quasi-characters χ and π and π′ are equivalent.

Suppose π = π(μ_{1}, μ_{2}) or σ(μ_{1}, μ_{2}). From Lemma 5.16 and the definitions the expression on the left is equal to

(i sgn u)^{m_{1}+m_{2}} |u|^{2s+s_{1}+s_{2}−1}π^{2s+s_{1}+s_{2}−1} \frac{Γ (\frac{1−s−r_{1}+m_{1}}{2} ) Γ (\frac{1−s−r_{2}+m_{2}}{2} )}{Γ (\frac{s+r_{1}+m_{1}}{2} ) Γ (\frac{s+r_{2}+m_{2}}{2} )}

if χ is trivial and μ_{i}(x) = |x|^{r_{i}}(sgn x)^{m_{1}}. If χ(x) = sgn x and n_{i} is 0 or 1 while m_{i} +n_{1} = 1 (mod 2) the quotient is

(i sgn u)^{m_{1}+m_{2}} |u|^{2s+s_{1}+s_{2}−1}π^{2s+s_{1}+s_{2}−1} \frac{Γ (\frac{1−s−r_{1}−n_{1}}{2} ) Γ (\frac{1−s−r_{2}+n_{2}}{2} )}{Γ (\frac{s+r_{1}+n_{1}}{2} ) Γ (\frac{s+r_{2}+n_{2}}{2} )}.

If we let π′ be π(μ′_{1}, μ′_{2}) or σ(μ′_{1}, μ′_{2}) we obtain similar formulae with r_{i} replaced by r′_{i} and m_{i} by m′_{i}.

Consider first the quotients for π. The first has an infinite number of zeros of the form −r_{1} − m_{1} − 2p where p is a non-negative integer and an infinite number of the form −r_{2} − m_{2} − 2p where p is a non-negative integer, but no other zeros. Similarly the zeros of the second are at points −r_{1} − n_{1} − 2p or −r_{2} − n_{2} − 2p. Thus if the quotients are equal r_{1} + m_{1} ≡ r_{2} + n_{2} ≡ r_{2} + m_{2} + 1 (mod 2). Moreover if r_{1} + m_{1} = r_{2} + m_{2} + 1 (mod 2) then π = σ(μ_{1}, μ_{2}) and, as we saw in Theorem 5.11, σ(μ_{1}η, μ_{2}η) = σ(μ_{1}, μ_{2}) so that the two quotients are equal. As a result either r_{1} + m_{1} = r_{2} + m_{2} + 1 (mod 2) and r′_{1} + m′_{1} = r′_{2} + m′_{2} + 1 (mod 2) or neither of these congruences hold.

Suppose first that π = π(μ_{1}, μ_{2}) and π′(μ′_{1}, μ′_{2}). Then the first quotient for π has zeros at the points −r_{1} − m_{1}, −r_{1} − m_{1} − 2, ⋯ and −r_{2} − m_{2}, −r_{2} − m_{2} − 2, ⋯ while that for π′ has zeros at −r′_{1} − m′_{1}, −r′_{1} − m′_{1} − 2, ⋯ and −r′_{2} − m′_{2}, −r′_{2} − m′_{2} − 2, · · ·. Thus either r_{1} + m_{1} = r′_{1} + m′_{1} or r_{1} + m_{1} = r′_{2} + m′_{2}. Interchanging μ′_{1} and μ′_{2} if necessary we may assume that the first of these two alternatives hold. Then r_{2} +m_{2} = r′_{2} +m′_{2}. Moreover r_{1} +r_{2} = r′_{1} +r′_{2} and |m_{1} −m_{2}| = |m′_{1} −m′_{2}|. If m_{1} = m′_{1} it follows immediately that μ_{1} = μ′_{1} and μ_{2} = μ′_{2}. Suppose that m_{1} ≠ m′_{1}. Examining the second quotient we see that either r_{1} + n_{1} = r′_{1} + n′_{1} or r_{1} + n_{1} = r′_{2} + n′_{2}. The first equality is incompatible with the relations r_{1} + m_{1} = r′_{1} + m′_{1} and m_{1} ≠ m′_{1}. Thus r_{1} + n_{1} = r′_{2} + n′_{2}. For the same reason r_{2} + n_{2} = r′_{1} + n′_{1}. Interchanging the roles of μ_{1}, μ_{2} and μ′_{1}, μ′_{2} if necessary we may suppose that m_{1} = 0 and m′_{1} = 1. Then r_{1} = r′_{1} + 1. Since r_{1} + r_{2} = r′_{1} + r′_{2} we have r_{2} = r′_{2} − 1 so that m_{2} = 1, m′_{2} = 0. Thus n_{1} = n′_{2} = 1 and r_{1} = r′_{2} so that r_{2} = r′_{1}. It follows that μ_{1} = μ′_{2} and μ_{2} = μ′_{1}.

Finally we suppose that π = σ(μ_{1}, μ_{2}) and π′ = σ(μ′_{1}, μ′_{2}). Then there are quasi-characters ω_{1} and ω′_{1} of ℂ^{×} such that π = π(ω_{1}) and π′ = π(ω′_{1}). Replacing ω_{1} by the quasi-character z → ω_{1}(z̄) does not change π(ω_{1}) so we may suppose that ω_{1}(z) = (zz̄)^{r}z^{m} while ω′_{1}(z) = (zz̄)^{r′}z^{m′}. Since ω_{1} and ω′_{1} must have the same restriction to ℝ^{×} the numbers 2r + m and 2r′ + m′ are equal while m ≡ m′ (mod 2). Apart from a constant and an exponential factor the quotient

ε(s, π, ψ) \frac{L(1 − s, π̃)}{L(s, π)}

is given by

\frac{Γ(1 − s − r)}{Γ(s + r + m)}

whose pole furthest to the left is at 1 − r. Consequently r = r′ and m = m′.

Corollary 5.19 Suppose π and π′ are two irreducible admissible representations of ℋ_{ℝ}. Suppose there is a quasi-character ω of ℝ^{×} such that

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I π′ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

If for all quasi-characters χ, L(s, χ ⊗ π) = L(s, χ ⊗ π′), L(s, χ^{−1} ⊗ π) = Ł(s, χ^{−1} ⊗ π′), and ε(s, χ ⊗ π, ψ) = ε(s, χ ⊗ π′, ψ) then π and π′ are equivalent.

Combining Lemma 5.16 with the previous propositon we infer that there is a pair of quasicharacters μ_{1} and μ_{2} such that both π and π′ are one of the representations π(μ_{1}, μ_{2}) or σ(μ_{1}, μ_{2}). However the computations made during the proof of Lemma 5.16 show that L(s, χ ⊗ π(μ_{1}, μ_{2})) differs from L(s, χ ⊗ σ(μ_{1}, μ_{2})) for a suitable choice of χ.

Let K be the quaternion algebra of ℝ. We could proceed along the lies of the fourth paragraph and associate to every finite-dimensional irreducible representation Ω of K^{×} a representation π(Ω) of G_{ℝ}. Since we have just classified the representations of G_{ℝ} we can actually proceed in a more direct manner.

We identify K with the algebra of 2 × 2 complex matrices of the form

z = ( \begin{matrix}a & b \\ −b̄ & ā \end{matrix} )

Then

z^{ι} = ( \begin{matrix}ā & −b \\ b̄ & a \end{matrix} )

and ν(z) = z^{ι} is the scalar matrix (|a|^{2} + |b|^{2})I while τ(z) is the scalar matrix (a + ā + b + b̄)I. Let ρ_{1} be the two dimensional representation of K^{×} associated to this identification and let ρ_{n} be the nth symmetric power of ρ_{1}. Any irreducible representation is equivalent to a representation of the form χ ⊗ ρ_{n} where χ is a quasi- character of ℝ^{×}. Thus

(χ ⊗ ρ_{n})(h) = χ(ν(h)) ρ_{n}(h)

Since ν(h) is always positive we may suppose that χ is of the form χ(x) = |x|^{r}.

Let Ω be a finite dimensional representation and let Ω act on U. In the first paragraph we introduced the space 𝒮(K, U). It is clear that if Φ is in 𝒮(K, U) the integrals

Z(α_{ℝ}^{s} ⊗ Ω, Φ) = ∫_{K^{×}} Ω(h) |ν(h)|^{s} Φ(h) d^{×}h

and

Z(α_{ℝ}^{s} ⊗ Ω^{−1}, Φ) = ∫_{K^{×}} Ω^{−1}(h) |ν(h)|^{s} Φ(h) d^{×}h

converge absolutely in some right half-plane.

Proposition 5.20 Suppose χ(x) = |x|^{r} and Ω = χ ⊗ ρ_{n}. Let ω be the character of ℂ^{×} defined by ω(z) = (zz̄)^{r−1/2}z^{n+1}. Set L(s, Ω) = L(s, ω) and

ε(s, Ω, ψ_{ℝ}) = λ(ℂ/ℝ, ψ_{ℝ}) ε(s, ω, ψ_{ℂ/ℝ})

The quotient

\frac{Z(α_{ℝ}^{s+1/2} ⊗ Ω, Φ)}{L(s, Ω)}

can be analytically contained to the whole complex plane as a holomorphic function. Given u in U there exists a Φ in 𝒮(K, U) such that

\frac{Z(α_{ℝ}^{s+1/2}, Φ)}{L(s, Ω)} = a^{s}u.

For all Φ the two functions

\frac{Z(α_{ℝ}^{3/2−s} ⊗ Ω^{−1}, Φ′)}{L(1 − s, Ω̃)}

and

−ε(s, Ω, ψ_{ℝ}) \frac{Z(α_{ℝ}^{s+1/2} ⊗ Ω, Φ)}{L(s, Ω)}

are equal. Finally Z(α_{ℝ}^{s+1/2} ⊗ Ω, Φ) is bounded in any region obtained by removing discs about the poles L(s, Ω) from a vertical strip of finite width.

Suppose K_{1} is the subgroup of K^{×} formed by the elements of reduced norm one. Let Φ_{1} be the function on ℝ defined by

Φ_{1}(t) = ∫_{K_{1}} Ω(h) Φ(th) dh

Φ_{1} belongs to 𝒮(ℝ) and if ω_{0} is the quasi-character of ℝ^{×} defined by Ω(t) = ω_{0}(t)I the function ω_{0}(t)Φ_{1}(t) is even. Moreover if the multiplcative Haar measures are suitably normalized

Z(α_{ℝ}^{s+1/2} ⊗ Ω, Φ) = Z(α_{ℝ}^{2s+1}ω_{0}, Φ_{1}).

Since ω_{0}(t) = |t|^{2r}t^{n} we can integrate by parts as in the proof of Lemma 5.17 to see that for any non-negative integer m

Z(α_{ℝ}^{2s+1}ω_{0}, Φ_{1}) = \frac{(−1)^{m}}{∏_{j=0}^{m−1}(2s + 2r + n + j + 1)} ∫ η(t)^{m+n} |t|^{2s+2r+m+n+1} \frac{∂^{m}Φ_{1}}{∂t^{m}} d^{×}t.

The integral is holomorphic for Re(2s + 2r + m + n) > −1 and, if \frac{∂^{m}Φ_{1}}{∂t^{m}} vanishes at t = 0, for Re(2s + 2r + m + n) > −2. Thus the function on the left has an analytic continuation to the whole complex plane as a meromorphic function with simple poles. Since

L(s, Ω) = 2(2π)^{−(s+r+n+1/2)}Γ(s + r + n + \frac{1}{2})

Chapter 1 111 we have to show that its poles occur at the points s + r + n + \frac{1}{2} + j = 0 with j = 0, 1, 2, · · ·. Since \frac{∂^{m}Φ_{1}}{∂t^{m}} vanishes at 0 if m + n is odd its only poles are at the points 2s + 2r + 2n + 2j + 1 = 0 with n + 2j ≥ 0. To exclude the remaining unwanted poles we have to show that \frac{∂^{m}Φ_{1}}{∂t^{m}} = 0 at 0 if m < n. If we expand Φ in a Taylor’s series about 0 we see that \frac{∂^{m}Φ_{1}}{∂t^{m}} = 0 at 0 unless the restriction of ρ_{n} to K_{1} is contained in the representation on the polynomials of degree m on K. This can happen only if m ≥ n.

Since Ω̃ is equivalent to the representation h → Ω^{−1}(h^{ι}) the quotient

\frac{Z(α_{ℝ}^{3/2−s} ⊗ Ω^{−1}, Φ′)}{L(1 − s, Ω̃)}

is also holomorphic. The argument used to prove Lemma 5.15.1 shows that there is a scalar λ(s) such that, for all Φ,

\frac{Z(α_{ℝ}^{3/2−s} ⊗ Ω^{−1}, Φ′)}{L(1 − s, Ω̃)} = λ(s) \frac{Z(α_{ℝ}^{s+1/2} ⊗ Ω, Φ)}{L(s, Ω)}.

We shall used the following lemma to evaluate λ(s).

Lemma 5.20.1 Let φ be a function in 𝒮(ℂ) of the form

φ(x) = e^{−2πxx̄}P (x, x̄)

where P is a polynomial in x and x̄. Suppose φ(xu) = φ(x) ω^{−1}(u) if uū = 1. Define the function Φ in K^{×} by

Φ(z) = φ(α) ω(α) (αᾱ)^{− \frac{1}{2}} ⟨u, Ω̃(z)ũ⟩

if ν(z) = αᾱ. Then Φ extends to a function in 𝒮(K) and its Fourier transform is given by

Φ′(z) = −λ(ℂ/ℝ, ψ_{ℝ}) φ′(α) ω^{−1}(α) (αᾱ)^{− \frac{1}{2}} ⟨Ω(z)u, ũ⟩

if φ′ is the Fourier transform of φ.

By linearity we may assume that φ is of the form

φ(x) = e^{−2πxx̄}(xx̄)^{p}x̄^{n+1}

where p is a non-negative integer. We may suppose that the restriction of ρ_{n} to the elements of norm one is orthogonal and identify the space U on which its acts with its dual Ũ. Then Ω̃ = α_{ℝ}^{−r−n} ⊗ ρ_{n}. Thus if

z = ( \begin{matrix}a & b \\ −b̄ & ā \end{matrix} )

the value of Φ at z is

e^{−2π(aā+bb̄)}(aā + bb̄)^{r+n+p}⟨u, Ω̃(z)ũ⟩ = e^{−2π(aā+bb̄)}(aā + bb̄)^{p}⟨u, ρ_{n}(z)ũ⟩

The expression on the right certainly defines a function in 𝒮(K).

We are trying to show that if

F (z) = φ(α) ω(α) (αᾱ)^{− \frac{1}{2}} Ω^{−1}(z) when z = αᾱ then the Fourier transform of F is given by

F′(z) = −λ(ℂ/ℝ, ψ_{ℝ}) φ′(α) ω^{−1}(α) (αᾱ)^{− \frac{1}{2}} Ω(z).  (5.20.2)

If h_{1} and h_{2} have norm one

F (h_{1}zh_{2}) = Ω(h_{2}^{−1}) F (z) Ω(h_{1}^{−1})

and therefore

F′(h_{1}zh_{2}) = Ω(h_{1}) F′(z) Ω(h_{2})

In particular if z is a scalar in K the operator F′(z) commutes with the elements of norm one and is therefore a scalar operator. The expression F_{1}(z) on the right of (5.20.2) has the same properties so that all we need do is show that for some pair of vectors u and ũ which are not orthogonal

⟨F′(z)u, ũ⟩ = ⟨F_{1}(z)u, ũ⟩

for all positive scalars z.

If we only wanted to show that F′(z) = c F_{1}(z) where c is a positive constant it would be enough to show that

⟨F′(z)u, ũ⟩ = c⟨F_{1}(z)u, ũ⟩.  (5.20.3)

Once this was done we could interchange the roles of φ and φ′ and Φ and Φ′ to show that c^{2} = 1. To obviate any fuss with Haar measures we prove (5.20.3).

Recall that if

a(θ) = ( \begin{matrix}e^{iθ} & 0 \\ θ & e^{−iθ} \end{matrix} )

then, apart from a positive constant,

∫_{K_{1}}⟨u, Ω̃(k)ũ⟩ f(k) dk

is equal to

⟨u, ũ⟩ ∫_{0}^{π} sin(n + 1)θ sin θ f(a(θ)) dθ

if f is a class function on K_{1}, the group of elements of norm one. The equality is of course a consequence of the Weyl character formula and the Schur orthogonality relations.

If x is a positive scalar in K then, apart from a positive constant, Φ′(x) is given by

∫_{K^{×}} Φ(z) ψ_{ℝ}(τ(xz)) |ν(z)|^{2} d^{×}z

which is a positive multiple of

∫_{0}^{∞} t^{3}φ(t) {∫_{K_{1}}⟨u, Ω̃(k)ũ⟩ ψ_{ℝ}(xt τ(k)) dk} d^{×}t.

Since τ(k) is a class function this expression is a positive multiple of

⟨u, ũ⟩ ∫_{0}^{∞} t^{3}φ(t) {∫_{0}^{π} sin(n + 1)θ sin θ ψ_{ℝ}(2xt cos θ) dθ} d^{×}t

Integrating the inner integral by parts we obtain

⟨u, ũ⟩\frac{n + 1}{4πiux} ∫_{0}^{∞} t^{2}φ(t) \{∫_{0}^{π} cos(n + 1)θ ψ_{ℝ}(2xt cos θ) dθ\} d^{×}t.

On the other hand if x, which is a positive real number, is regarded as an element of ℂ then φ′(x) is a positive multiple of

∫_{C^{×}} φ(z) ψ_{ℝ}(τ(xz))zz̄d^{×}z

or of

∫_{0}^{∞} t^{2}φ(t) {∫_{0}^{2π} e^{−i(n+1)θ}ψ_{ℝ}(xt cos θ) dθ)} d^{×}t.

Since

∫_{0}^{2π} e^{−i(n+1)θ}ψ_{ℝ}(xt cos θ) dθ = 2 ∫_{0}^{π} cos(n + 1)θ ψ_{ℝ}(xt cos θ) dθ

and λ(ℂ/ℝ) = i sgn u the identity (5.20.3) follows for any choice of u and ũ.

To evaluate λ(s) we choose Φ as in the lemma and compute

⟨Z(α_{ℝ}^{s+ \frac{1}{2}} ⊗ Ω, Φv), ṽ⟩ = ∫ Φ(z) |ν(z)|^{s+ \frac{1}{2}} ⟨Ω(z)v, ṽ⟩ d^{×}z

and

⟨Z(α_{ℝ}^{\frac{3}{2} −s} ⊗ Ω^{−1}, Φv), ṽ⟩ = ∫ Φ(z) |ν(z)|^{\frac{3}{2} −s}⟨v, Ω̃(z)ṽ⟩ d^{×}z.

The first is equal to

∫_{K^{×}/K_{1}} |ν(z)|^{s+ \frac{1}{2}} \{∫_{K_{1}} Φ(zk)⟨Ω(zk)v, ṽ⟩ dk\} d^{×}z.

Since

∫_{K_{1}}⟨Ω(zk)v, ṽ⟩⟨u, Ω̃(zk)ũ⟩ dk

is, by the Schur orthogonality relations, equal to

\frac{1}{deg Ω}⟨v, ũ⟩⟨u, ṽ⟩

the double integral is equal to

\frac{1}{deg Ω}⟨v, ũ⟩⟨u, ṽ⟩ ∫_{K^{×}} φ(α) ω(α) (αᾱ)^{s} d^{×}z

where αᾱ = ν(z). If the Haar measure on ℂ^{×} is suitably chosen the integral here is equal to Z(ωα_{ℂ}^{s}, φ). The same choice of Haar measures lead to the relation

⟨Z(α_{ℝ}^{\frac{3}{2} −s} ⊗ Ω^{−1}, Φv), ṽ⟩ = \frac{−λ(ℂ/ℝ, ψ_{ℝ})}{deg Ω}⟨v, ũ⟩⟨u, ṽ⟩Z(ω^{−1}α_{ℂ}^{1−s}, φ′).

Since L(s, Ω) = L(s, ω) and L(s, Ω̃) = L(s, ω^{−1}) we can compare the functional equation for Z(ωα_{ℂ}^{s}, φ) with that for Z(α_{ℝ}^{s+1/2} ⊗ Ω, Φv) to see that

λ(s) = −λ(ℂ/ℝ, ψ_{ℝ}) ε(s, ω, ψ_{ℂ/ℝ})

as asserted.

If

φ(x) = e^{−2πxx̄}

then Z(α_{ℂ}^{s}ω, φ) is an exponential times L(s, ω) so that Z(α_{ℝ}^{s+1/2} ⊗ Ω, Φv) is, with a suitable choice of v and ũ, a non-zero scalar times an exponential times L(s, ω)u. The last assertion of the proposition is proved in the same way as Lemma 5.15.2.

We end this paragraph with the observation that the space W (π, ψ) of Theorem 5.13 cannot exist when π is finite-dimensional. If W = W (π, ψ) did exist the contragredient representation π̃ on the dual space W̃ would also be finite dimensional and π̃(X_{+}) would be nilpotent. However if λ is the linear functional φ → φ(e) then π̃(X_{+})λ = −2iπa if ψ(x) = e^{2iπax}.

Chapter 1 115 §6. Representations of GL(2, ℂ). In this paragraph we have to review the representation theory of G_{ℂ} = GL(2, ℂ) and prove the local functional equation for the complex field. Many of the definitions and results of the previous paragraph are applicable, after simple modifications which we do not always make explicit, to the present situation.

The standard maximal compact subgroup of GL(2, ℂ) is the group U(2, ℂ) of unitary matrices. ℋ_{1} will be the space of infinitely differentiable compactly supported functions on G_{ℂ}. ℋ_{2} will be the space of functions on U(2, ℂ) which are finite linear combinations of the matrix elements of finite dimensional representations. ℋ_{ℂ} = ℋ_{1} ⊕ ℋ_{2} can be regarded as a space of measures. Under convolution it forms an algebra called the Hecke algebra. The notion of an elementary idempotent and the notion of an admissible representation of ℋ_{ℂ} are defined more or less as before.

Let 𝔤 be the Lie algebra of the real Lie group of GL(2, ℂ) and let 𝔤_{ℂ} = 𝔤 ⊗_{ℝ} ℂ. 𝔄 will be the universal enveloping algebra of 𝔤_{ℂ}. A representation of 𝔄 will be said to be admissible if its restriction to the Lie algebra of U(2, ℂ) decomposes into a direct sum of irreducible finite dimensional representations each occurring with finite multiplicity. There is a one-to-one correspondence between classes of irreducible admissible representations of ℋ_{ℂ} and those of 𝔄. We do not usually distinguish between the two. The representation π̃ contragredient to π and the tensor product of π with a quasi-character of ℂ^{×} are defined as before.

If μ_{1} and μ_{2} are two quasi-characters of ℂ^{×} we can introduce the space ℬ(μ_{1}, μ_{2}) and the representation ρ(μ_{1}, μ_{2}) of ℋ_{ℂ} or of 𝔄 on ℬ(μ_{1}, μ_{2}). In order to study this representation we identify 𝔤_{ℂ} with 𝔤_{ℓ}(2, ℂ) ⊕ 𝔤_{ℓ}(2, ℂ) in such a way that 𝔤 corresponds to the elements of X ⊕ X̄. If 𝔄_{1} is the universal enveloping algebra of 𝔤_{ℓ}(2, ℂ) we may then identify 𝔄 with 𝔄_{1} ⊗ 𝔄_{1}.

In the previous paragraph we introduced the elements D and J of 𝔄_{1}. Set D_{1} = D ⊗ 1, D_{2} = 1⊗D, J_{1} = J ⊗1, and J_{2} = 1⊗J. These four elements lie in the centre of 𝔄. A representation of 𝔄 is admissible if its restriction to the Lie algebra of the group SU(2, ℂ) of unitary matrices of determinant one decomposes into the direct sum of irreducible finite dimensional representations each occurring with finite multiplicity.

The first part of the next lemma is verified by calculations like those used in the proof of Lemma 5.6. The second is a consequence of the Frobenius reciprocity law applied to the pair SU(2, ℂ) and its subgroup of diagonal matrices.

Lemma 6.1. Let

μ_{1}(z) = (zz̄)^{s_{i}− \frac{1}{2} (a_{i}+b_{i})}z^{a_{i}}z̄^{b_{i}}

and

μ_{1}μ_{2}^{−1}(z) = μ(z) = (zz̄)^{s− \frac{1}{2} (a+b)}z^{a}z̄^{b}

where a_{i}, b_{i}, a, and b are non-negative integers and a_{i}b_{i} = ab = 0.

(i) On ℬ(μ_{1}, μ_{2}) we have the following four relations

\begin{aligned}ρ(D_{1}) & = \frac{1}{2}\{(s + \frac{a − b}{2})^{2} − 1\} I \\ ρ(D_{2}) & = \frac{1}{2}\{(s + \frac{b − a}{2})^{2} − 1\} I \\ ρ(J_{1}) & = \{(s_{1} + s_{2}) + \frac{a_{1} − b_{1} + a_{2} − b_{2}}{2} \} I \\ ρ(J_{2}) & = \{(s_{1} + s_{2}) + \frac{b_{1} − a_{1} + b_{2} − a_{2}}{2} \} I \end{aligned}

Chapter 1 116

(ii) ρ(μ_{1}, μ_{2}) is admissible and contains the representation ρ_{n} of the Lie algebra of SU(2, ℂ) if and only if n ≥ a + b and n ≡ a + b (mod 2) and then it contains it just once.

ρ_{n} is the unique irreducible representation of SU(2, ℂ) of degree n + 1. Let ℬ(μ_{1}, μ_{2}, ρ_{n}) be the space of functions in ℬ(μ_{1}, μ_{2}) transforming according to ρ_{n}.

Theorem 6.2. (i) If μ is not of the form z → z^{p}z̄^{q} or z → z^{−p}z̄^{−q} with p ≥ 1 and q ≥ 1 then ρ(μ_{2}, μ_{2}) is irreducible. A representation equivalent to ρ(μ_{1}, μ_{2}) will be denoted by π(μ_{1}, μ_{2}),

(ii) If μ(z) = z^{p}z̄^{q} with p ≥ 1, q ≥ 1 then

ℬ_{s}(μ_{1}, μ_{2}) = ∑_{{n≥p+q \atop n≡p+q (mod 2)}} ℬ(μ_{1}, μ_{2}, ρ_{n})

is the only proper invariant subspace of ℬ(μ_{1}, μ_{2}). σ(μ_{1}, μ_{2}) will be any representation equivalent to the representation on ℬ_{s}(μ_{1}, μ_{2}) and π(μ_{1}, μ_{2}) will be any representation equivalent to the representation on the quotient space

ℬ_{f}(μ_{1}, μ_{2}) = ℬ(μ_{1}, μ_{2})/ℬ_{s}(μ_{1}, μ_{2})

(iii) If μ(z) = z^{−p}z̄^{−q} with p ≥ 1, q ≥ 1 then

ℬ_{f}(μ_{1}, μ_{2}) = ∑_{{|p−q|≤n<p+q \atop n≡p+q (mod 2)}} ℬ(μ_{1}, μ_{2}, ρ_{n})

is the only proper invariant subspace of ℬ(μ_{1}, μ_{2}). π(μ_{1}, μ_{2}) will be any representation equivalent to the representation ℬ_{f}(μ_{1}, μ_{2}) and σ(μ_{1}, μ_{2}) will be any representation equivalent to the representation on the quotient space

ℬ_{s}(μ_{1}, μ_{2}) = ℬ(μ_{1}, μ_{2})/ℬ_{f}(μ_{1}, μ_{2}).

(iv) π(μ_{1}, μ_{2}) is equivalent to π(μ′_{1}, μ′_{2}) if and only if (μ_{1}, μ_{2}) = (μ′_{1}, μ′_{2}) or (μ_{1}, μ_{2}) = (μ′_{2}, μ′_{1}).

(v) If σ(μ_{1}, μ_{2}) and σ(μ′_{1}, μ′_{2}) are defined they are equivalent if and only if (μ_{1}, μ_{2}) = (μ′_{1}, μ′_{2}) or (μ_{1}, μ_{2}) = (μ′_{2}, μ′_{1}).

(vi) If μ(z) = z^{p}z̄^{q} with p ≥ 1, q ≥ 1 there is a pair of characters ν_{1}, ν_{2} such that μ_{1}μ_{2} = ν_{1}ν_{2}

and ν_{1}ν_{2}^{−1} = z^{p}z̄^{−q} and σ(μ_{1}, μ_{2}) is equivalent to π(ν_{1}, ν_{2}). (vii) Every irreducible admissible representation of ℋ_{ℂ} or 𝔄 is a π(μ_{1}, μ_{2}) for some choice of μ_{1}

and μ_{2}.

The proofs of the first three assertions will be based on two lemmas.

Lemma 6.2.1. If there exists a proper invariant subspace V of ℬ(μ_{1}, μ_{2}) which is finite dimensional then μ_{1}μ_{2}^{−1}(z) = z^{−p}z̄^{−q} with p ≥ 1, q ≥ 1 and V = ℬ_{f}(μ_{1}, μ_{2}).

Lemma 6.2.2. Let V be a proper invariant subspace of ℬ(μ_{1}, μ_{2}) and let n_{0} be the smallest integer such that some subspace of V transforms according to the representation ρ_{n_{0}} of the Lie algebra of SU(2, ℂ). Either

V = ∑_{n≥n_{0}} ℬ(μ_{1}, μ_{2}, ρ_{n})

Chapter 1 117 or V contains a finite-dimensional invariant subspace.

Grant these lemmas for a moment and let V be a proper invariant subspae of ℬ(μ_{1}, μ_{2}). As in the case of the non-archimedean and real fields there is an invariant non-degenerate bilinear form on ℬ(μ_{1}, μ_{2}) × ℬ(μ_{1}^{−1}, μ_{2}^{−1}). The orthogonal complement V^{⊥} of V in ℬ(μ_{1}^{−1}, μ_{2}^{−1}) is a proper invariant subspace. By Lemma 6.2.1 they cannot both contain an invariant finite dimensional subspace. Therefore by Lemma 6.2.2 one of them is of finite codimension. The other must be of finite dimension. If V is finite-dimensional then μ_{1}μ_{2}^{−1}(z) = z^{−p}z̄^{−q} and V = ℬ_{f}(μ_{1}, μ_{2}). If V^{⊥} is finite dimensional then μ_{1}μ_{2}^{−1}(z) = z^{p}z̄^{q}. Since the orthogonal complement of ℬ_{f}(μ_{1}, μ_{2}) is ℬ_{s}(μ_{1}, μ_{2}) we must have V = ℬ_{s}(μ_{1}, μ_{2}).

We shall now show that ℬ_{f}(μ_{1}, μ_{2}) is invariant when μ_{1}μ_{2}^{−1}(z) = z^{−p}z̄^{−1}. It will follow from duality that ℬ_{s}(μ_{1}, μ_{2}) is invariant when μ_{1}μ_{2}^{−1}(z) = z^{p}z̄^{q}. Every irreducible finite-dimensional representation π of 𝔄 determines a representation π of G_{ℂ}. If π acts on X there is a nonzero vector v_{0} in X such that

π \begin{aligned}( \\ ( & \begin{matrix}z & x \\ 0 & z^{−1} \end{matrix} & ) \\ & & ) \end{aligned} v_{0} = z^{m}z̄^{n}v_{0}

for all z in ℂ^{×} and all x in ℂ. v_{0} is determined up to a scalar factor and m and n are non-negative integers. Moreover there is a quasi-character ω_{0} of ℂ^{×} such that

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω_{0}(a)I

Thus

π \begin{aligned}( \\ ( & \begin{matrix}z_{1} & x \\ 0 & z_{2} \end{matrix} & ) \\ & & ) \end{aligned} v_{0} = ω_{1}(z_{1}) ω_{2}(z_{2}) v_{0}

where ω_{1}ω_{2}^{−1}(z) = z^{m}z̄^{n}. π is determined up to equivalence by ω_{1} and ω_{2} so we write π = κ(ω_{1}, ω_{2}). As long as ω_{1}ω_{2}^{−1}(z) = z^{m}z̄^{n} with non-negative integers m and n the representation κ(ω_{1}, ω_{2}) exists. By the Clebsch–Gordan formula the restriction of κ(ω_{1}, ω_{2}) or its contragredient to SU(2, ℂ) breaks up into the direct sum of the representations ρ_{i} with |m − n| ≤ i ≤ m + n and 1 ≡ m + n (mod 2). Let π be κ(ω_{1}, ω_{2}) and let π̃, the contragredient representation, act on X̃. To each vector ṽ in X̃ we associate the function

φ(g) = ⟨v_{0}, π̃(g)ṽ⟩

on G_{ℂ}. The map ṽ → φ is linear and injective. Moreover π̃(g)ṽ → ρ(g)φ while

φ (( \begin{matrix}z_{1} & x \\ 0 & z_{2} \end{matrix} ) g) = ω_{1}^{−1}(z_{1}) ω_{2}^{−1}(z_{2}) φ(g)

so that if μ_{1} = ω_{1}^{−1} α_{ℂ}^{−1/2} and μ_{2} = ω_{2}^{−1} α_{ℂ}^{1/2} the function φ belongs to ℬ(μ_{1}, μ_{2}). As we vary ω_{1} and ω_{2} the quasi-characters μ_{1} and μ_{2} vary over all pairs such that μ_{1}μ_{2}^{−1}(z) = z^{−p}z̄^{−q} with p ≥ 1 and q ≥ 1.

We have still to prove the two lemmas. Suppose V is a proper finite-dimensional subspace of ℬ(μ_{1}, μ_{2}). The representation of 𝔄 on V is certainly a direct sum of irreducible representations each occurring with multiplicity one. Let V′ be an irreducible subspace of V and let Ṽ′ be the dual space of V′. Let λ be the linear functional λ : φ → φ(e) on V′. If π is the representation of 𝔄 or of G_{ℂ} on V′ then

π̃ \begin{aligned}( \\ ( & \begin{matrix}z_{1} & x \\ 0 & z_{2} \end{matrix} & ) \\ & & ) \end{aligned} λ = μ_{1}^{−1}(z_{1}) μ_{2}^{−1}(z_{2}) (z_{1}z̄_{1}z_{2}^{−1}z_{2}^{−1})^{− \frac{1}{2}} λ

Thus if ω_{1} = μ_{1}^{−1} α_{ℂ}^{−1/2} and ω_{2} = μ_{2}^{−1} α_{ℂ}^{1/2} the representation π̃ is κ(ω_{1}, ω_{2}). It follows immediately that μ_{1}μ_{2}^{−1} is of the form μ_{1}μ_{2}^{−1}(z) = z^{−p}z̄^{−q} with p ≥ 1 and q ≥ 1 and that V′ and therefore V is ℬ_{f}(μ_{1}, μ_{2}).

To prove the second lemma we regard 𝔤 as the real Lie algebra of 2 × 2 complex matrices. Then

𝔞 = \{( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) | a ∈ ℂ\}

is the centre of 𝔤 and

𝔲 = \{( \begin{matrix}ia & b \\ −b̄ & −ia \end{matrix} ) | a ∈ ℝ, b ∈ ℂ\}

is the Lie algebra of SU(2, ℂ). If

𝔤 = \{( \begin{matrix}a & b \\ b̄ & −a \end{matrix} ) | a ∈ ℝ, b ∈ ℂ\}

then 𝔲 ⊕ 𝔤 is the Cartan decomposition of the Lie algebra of the special linear group. The space 𝔤_{ℂ} = 𝔤 ⊗_{ℝ} ℂ is invariant under the adjoint action of 𝔲 on 𝔤_{ℂ}. Moreover 𝔲 acts on 𝔤_{ℂ} according to the representation ρ_{2}. One knows that ρ_{2} ⊗ ρ_{n} is equivalent to ρ_{n+2} ⊕ ρ_{n} ⊕ ρ_{n−2} if n ≥ 2, that ρ_{2} ⊗ ρ_{1} is equivalent to ρ_{3} ⊕ ρ_{1} and, of course, that ρ_{2} ⊗ ρ_{0} is equivalent to ρ_{2}. The map of 𝔤_{ℂ} ⊗ ℬ(μ_{1}, μ_{2}, ρ_{n}) into ℬ(μ_{1}, μ_{2}) which sends X ⊗ f to ρ(X)f commutes with the action of 𝔲. Thus ρ(X)f is contained in

ℬ(μ_{1}, μ_{2}, ρ_{n+2}) ⊕ ℬ(μ_{1}, μ_{2}, ρ_{n}) ⊕ ℬ(μ_{1}, μ_{2}, ρ_{n−2}).

It is understood that ℬ(μ_{1}, μ_{2}, ρ_{ℓ}) = 0 if ℓ < 0.

Now let V be a proper invariant subspace of ℬ(μ_{1}, μ_{2}). Let n_{0} be the smallest non-negative integer n for which V contains ℬ(μ_{1}, μ_{2}, ρ_{n}). If n ≥ n_{0} set

V (n) = ∑_{{n≥k≥n_{0} \atop k≡n_{0} (mod 2)}} ℬ(μ_{1}, μ_{2}, ρ_{k})

If V contains every V (n) there is nothing to prove so assume that there is a largest integer n_{1} for which V contains V (n_{1}). All we need do is show that V (n_{1}) is invariant under 𝔤. It is invariant under 𝔞 and 𝔲 by construction so we need only verify that if X lies in 𝔤_{ℂ} then ρ(X) takes V (n_{1}) into itself. It is clear that ρ(X) takes V (n_{1} − 2) into V (n_{1}) so we have only to show that it takes ℬ(μ_{1}, μ_{2}, ρ_{n_{1}}) into V (n_{1}). Take f in ℬ(μ_{1}, μ_{2}, ρ_{n_{1}}) and let ρ(X)f = f_{1} + f_{2} with f_{1} in V (n_{1}) and f_{2} in ℬ(μ_{1}, μ_{2}, ρ_{n_{1}+2}). Certainly f_{2} lies in V. Since

V ∩ ℬ(μ_{1}, μ_{2}, ρ_{n_{1}+2})

Chapter 1 119 is either 0 or ℬ(μ_{1}, μ_{2}, ρ_{n_{1}+2}) and since, by construction, it is not ℬ(μ_{1}, μ_{2}, ρ_{n_{1}+2}) the function f_{2} is 0.

The first three assertions of the theorem are now proved and we consider the remaining ones. We make use of the fact that D_{1}, D_{2}, J_{1} and J_{2} generate the centre of 𝔄 as well as a result of Harish-Chandra to be quoted later. Suppose π and π′ are two irreducible representations of 𝔄 which are constituents of ρ(μ_{1}, μ_{2}) and ρ(μ′_{1}, μ′_{2}) respectively. Assume π and π′ contain the same representations of the Lie algebra of SU(2, ℂ) and are associated to the same homomorphism of the centre of 𝔄 into ℂ. Comparing the scalars π(J_{1}) and π(J_{1}) with π′(J_{1}) and π′(J_{2}) we find that μ_{1}μ_{2} = μ′_{1}μ′_{2}. Let μ_{1}μ_{2}^{−1}(z) = (zz̄)^{s− \frac{a+b}{2}} z^{a}z̄^{b} and let μ′_{1}μ′_{2}^{−1}(z) = (zz̄)^{s′− \frac{a′+b′}{2}} z^{a′}z̄^{b′}. Comparing π(D_{1}) and π(ν_{2}) with π′(D_{1}) and π′(D_{2}) we see that

(s + \frac{a − b}{2})^{2} = (s′ + \frac{a′ − b′}{2} )^{2}

and

(s + \frac{b − a}{2})^{2} = (s′ + \frac{b′ − a′}{2} )^{2}.

These relations will hold if μ_{1}μ_{2}^{−1} = μ′_{1}μ′_{2}^{−1} or μ_{1}^{−1}μ_{2} = μ′_{1}μ′_{2}^{−1} and therefore, when μ_{1}μ_{2} = μ′_{1}μ′_{2}, (μ_{1}, μ_{2}) = (μ′_{1}, μ′_{2}) or (μ_{1}, μ_{2}) = (μ′_{2}, μ′_{1}). If neither of these alternatives hold we must have

s = \frac{a′ − b′}{2} , s′ = \frac{a − b}{2},

or

s = \frac{b′ − a′}{2} , s′ = \frac{b − a}{2}.

Since μ_{1}μ_{2} = μ′_{1}μ′_{2} the integers a + b and a′ + b′ must have the same parity. Let μ = μ_{1}μ_{2}^{−1} and μ′ = μ′_{1}μ′_{2}^{−1}. In the first case μμ′ is of the form μμ′(z) = z^{2p} and μμ′^{−1} is of the form z̄^{2q} and in the second μμ′(z) = z̄^{2p} while μμ′^{−1}(z) = z^{2q}. Since {μ_{1}, μ_{2}} is not {μ′_{1}, μ′_{2}} neither p nor q is 0. In the first case μ = z^{p}z̄^{q} and μ′ = z^{p}z̄^{−q} and in the second μ = z^{q}z̄^{p} while μ′ = z^{−q}z̄^{p}.

In conclusion we see that π and π′ contain the same representations of the Lie algebra of SU(2, ℂ) and are associated to the same homomorphism of the centre of 𝔄 into ℂ if and only if one of the following alternatives holds.

(i) For some pair of quasi-characters ν_{1} and ν_{2} we have {π, π′} = {π(ν_{1}, ν_{2}), π(ν_{1}, ν_{2})} or {π, π′} = {π(ν_{1}, ν_{2}), π(ν_{2}, ν_{1})}.

(ii) For some pair of quasi-characters ν_{1} and ν_{2} we have {π, π′} = {σ(ν_{1}, ν_{2}), σ(ν_{1}, ν_{2})} or

{π, π′} = {σ(ν_{1}, ν_{2}), σ(ν_{2}, ν_{1})}. (iii) For some pair of quasi-characters ν_{1} and ν_{2} with ν_{1}ν_{2}^{−1}(z) = z^{p}z̄^{q} where p ≥ 1, q ≥ 1 we have

{π, π′} = {σ(ν_{1}, ν_{2}), π(ν′_{1}, ν′_{2})} where ν_{1}ν_{2} = ν′_{1}ν′_{2} and ν′ν′_{2}^{−1}(z) is either z^{p}z̄^{−1} or z^{−p}z̄^{q}. (iv) For some pair of quasi-characters ν_{1} and ν_{2} with ν_{1}ν_{2}^{−1}(z) = z^{−p}z̄^{−q} where p ≥ 1, q ≥ 1

we have {π, π′} = {σ(ν_{1}, ν_{2}), π(ν′_{1}, ν′_{2})} where ν_{1}ν_{2} = ν′_{1}ν′_{2} and ν′_{1}ν′_{2}^{−1}(z) is either z^{p}z̄^{−q} or z^{−p}z̄^{q}. The remaining assertions are now all consequences of a theorem of Harish-Chandra which,

in the special case of interest to us, we may state in the following manner.

Chapter 1 120 Lemma 6.2.3. If π is an irreducible admissible representation of 𝔄 there exists a pair of quasicharacters μ_{1} and μ_{2} such that ρ(μ_{1}, μ_{2}) and π contain at least one irreducible representation of the Lie algebra of SU(2, ℂ) in common and are associated to the same homomorphism of the centre of 𝔄 into ℂ. When this is so π is a constituent of ρ(μ_{1}, μ_{2}).

As before χ ⊗ π(μ_{1}, μ_{2}) is π(χμ_{1}, χμ_{2}) and χ ⊗ σ(μ_{1}, μ_{2}) is σ(χμ_{1}, χμ_{2}). If

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω_{0}(a)I

then π̃ = ω_{0}^{−1} ⊗ π.

Theorem 6.3. Let π be an infinite dimensional irreducible admissible representation of ℋ_{ℂ} and let ψ be a non-trivial additive character of ℂ. There is exactly one space W (π, ψ) of functions on G_{ℂ} which satisfies the following three conditions.

(i) Every function W in W (π, ψ) satisifies

W (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = ψ(x) W (g).

(ii) The functions in W (π, ψ) are continuous and W (π, ψ) is invariant under the operators ρ(f)

for f in ℋ_{ℂ}. Moreover the representation of ℋ_{ℂ} on W (π, ψ) is equivalent to π. (iii) If W is in W (π, ψ) there is a positive number N such that

W \begin{aligned}( \\ ( & \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = O(|t|^{N})

as |t| → ∞.

Since every π is of the form π = π(μ_{1}, μ_{2}) the existence is rather easy to prove. If Φ is in 𝒮(ℂ^{2}) let

θ(μ_{1}, μ_{2}, Φ) = ∫_{ℂ^{×}} Φ(t, t^{−1}) μ_{1}(t) μ_{2}^{−1}(t) d^{×}t

We let W (μ_{1}, μ_{2}, ψ) be the space of functions on G_{ℂ} of the form

W (g) = W_{Φ}(g) = μ_{1}(detg) |detg|_{ℂ}^{1/2} θ(μ_{1}, μ_{2}, r(g)Φ)

where Φ in 𝒮(ℂ^{2}) is SU(2, ℂ)-finite under the action defined by r. It is clear that W (μ_{1}, μ_{2}, ψ) = W (μ_{2}, μ_{1}, ψ) and that W (μ_{1}, μ_{2}, ψ) is invariant under right translations by elements of ℋ_{ℂ} and of 𝔄.

The existence of W (π, ψ) will, as before, be a consequence of the following analogue of Lemma 5.13.1.

Chapter 1 121 Lemma 6.3.1 Suppose μ_{1}μ_{2}^{−1}(t) = (tt̄)^{s− \frac{a+b}{2}} t^{a}t̄^{b} with Re s > −1. Then there is a bijection A of W (μ_{1}, μ_{2}, ψ) with ℬ(μ_{1}, μ_{2}) which commutes with the action of ℋ_{ℂ}.

As before A associates to W_{Φ} the function

f_{Φ^{∼}}(g) = μ_{1}(detg) |detg|_{ℂ}^{1/2} z(μ_{1}μ_{2}^{−1}α_{ℂ}, ρ(g)Φ^{∼})

The proof of course proceeds as before. However we should check that A is surjective. Theorem 6.2 shows that, under the present circumstances, there is no proper invariant subspace of ℬ(μ_{1}, μ_{2}) containing ℬ(μ_{1}, μ_{2}, ρ_{a+b}) so that we need only show that at least one nonzero function in ℬ(μ_{1}, μ_{2}, ρ_{a+b}) is of the form f_{Φ} where Φ is in 𝒮(ℂ^{2}) and SU(2, ℂ)-finite under right translations.

If

Φ(x, y) = e^{−2π(xx̄+yȳ)}ȳ^{a}y^{b}

then, since a + b = 0, Φ transforms under right translations by SU(2, ℂ) according to ρ_{a+b} so we need only check that f_{Φ} is not 0. Proceeding according to the definition we see that

\begin{aligned}f_{Φ}(e) & = ∫_{ℂ^{×}} Φ(0, t)(tt̄)^{s− \frac{a+b}{2}} t^{a}t̄^{b} d^{×}t \\ & = ∫_{ℂ^{×}} e^{−2πtt̄}(tt̄)^{1+s+ \frac{a+b}{2}} d^{×}t. \end{aligned}

Apart from a constant which depends on the choice of Haar measure this is

(2π)^{−s− \frac{a+b}{2}} Γ(1 + s + \frac{a + b}{2})

and is thus not 0.

Just as in the previous paragraph W (μ_{1}, μ_{2}, ψ) is spanned by functions W_{Φ} where Φ is of the form

Φ(x, y) = e^{−2π(xx̄+uūyȳ)}x^{p}x̄^{q}y^{m}ȳ^{n}

where p, q, m, and n are integers. u is determined by the relation ψ(z) = e^{4πi Re uz}. We can show that

W_{Φ} \begin{aligned}( \\ ( & \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

decreases exponentially as |t| → ∞.

To prove the uniqueness we will use a differential equation as in the previous chapter. This time the equations are a little more complicated. Suppose W_{1}(π, ψ) is a space of functions satisfying the first two conditions of the theorem. We regard ρ_{n} as acting on the space V_{n} of binary forms of degree n according to the rule

ρ_{n} \begin{aligned}( \\ ( & \begin{matrix}a & b \\ c & d \end{matrix} & ) \\ & & ) \end{aligned} φ(x, y) = φ(ax + cy, bx + dy)

If

φ(x, y) = ∑_{{|k|≤n \atop \frac{n}{2} −k∈ℤ}} φ^{k}x^{\frac{n}{2} +k}x^{\frac{n}{2} −k}

Chapter 1 122 then φ^{k} is called the k^{th} coordinate of φ. On the dual space Ṽ_{n} we introduce the dual coordinates.

If ρ_{n} is contained in π there is an injection A of V_{n} into W_{1}(π, ψ) which commutes with the action of SU(2, ℂ). Let Φ(g) be the function on G_{ℂ} with values in Ṽ_{n} defined by

⟨φ, Φ(g)⟩ = Aφ(g).

It is clear that W_{1}(π, ψ) is determined by Φ which is in turn determined by W_{1}(π, ψ) up to a scalar factor. The function Φ(g) is determined by the function

φ(t) = Φ \begin{aligned}( \\ ( & \begin{matrix}t^{\frac{1}{2}} & 0 \\ 0 & t^{− \frac{1}{2}} \end{matrix} & ) \\ & & ) \end{aligned}

on the positive real numbers. If φ^{k}(t) is the k^{th} coordinate of φ(t) and if π is a constituent of ρ(μ_{1}, μ_{2}) the differential equations

\begin{aligned}ρ(D_{1})Φ & = \frac{1}{2}\{(s + \frac{a − b}{2})^{2} − 1\}Φ \\ ρ(D_{2})Φ & = \frac{1}{2}\{(s + \frac{b − a}{2})^{2} − 1\}Φ \end{aligned}

may, if our calculations are correct, be written as

\begin{aligned}\frac{1}{2}[t\frac{d}{dt} + k − 1]^{2}φ^{k} − t^{2} \frac{|u|^{2}}{2} φ^{k} + (\frac{n}{2} + k)tiuφ^{k−1} & = \frac{1}{2}(s + \frac{a − b}{2})^{2}φ^{k} \\ \frac{1}{2}[t\frac{d}{dt} − k − 1]^{2}φ^{k} − t^{2} \frac{|u|^{2}}{2} φ^{k} − (\frac{n}{2} − k)tiūφ^{k+1} & = \frac{1}{2}(s + \frac{b − a}{2})^{2}φ^{k}. \end{aligned}

We have set φ^{k} = 0 if |k| ≥ n/2. Recall that ψ(z) = e^{4πi Re uz}. These equations allow one to solve for all φ^{k} in terms of φ^{n/2} or φ^{−n/2}.

For k = \frac{n}{2} the second equation may be written as

\frac{1}{2} \frac{d^{2}φ^{n/2}}{dt^{2}} + (−\frac{1}{2} − \frac{n}{2}) \frac{1}{t} \frac{dφ}{dt}^{n/2} + \{−\frac{|u|^{2}}{2} + \frac{(\frac{n}{2} + 1)^{2}}{2t^{2}} \} φ^{n/2} = (s − \frac{b − a}{2})^{2}φ^{n/2}. (∗)

If we have two independent solutions of this equation their Wronskian W (t) is a non-trivial solution of the equation

\frac{dW}{dt} = \frac{(n + 1)}{t}W

and therefore a non-zero multiple of t^{n+1}. Since we already have shown the existence of a solution of (∗) which decreases exponentially we see that there cannot be another solution which is bounded by a power of t as t → ∞. The uniqueness of the space W (π, ψ) follows

Every irreducible admissible representation of ℋ_{ℂ} is of the form π = π(μ_{1}, μ_{2}). Moreover π(μ_{1}, μ_{2}) = π(μ′_{1}, μ′_{2}) if and only if {μ_{1}, μ_{2}} = {μ′_{1}, μ′_{2}}. Thus we may set

L(s, π) = L(s, μ_{1}) L(s, μ_{2})

and

ϵ(s, π, ψ) = ϵ(s, μ_{1}, ψ) ϵ(s, μ_{2}, ψ).

Then

L(s, π̃) = L(s, μ_{1}^{−1}) L(s, μ_{2}^{−1}).

The local functional equation which is proved just as in the real case read as follows.

Theorem 6.4. Let π be an infinite-dimensional irreducible admissible representation of ℋ_{ℂ}. Let ω be the quasi-character of ℂ^{×} defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

for a in ℂ^{×}. If W is in W (π, ψ) the integrals

\begin{aligned}Ψ(g, s, W) & = ∫_{ℂ^{×}} W (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|_{ℂ}^{s−1/2} d^{×}a, \\ Ψ̃(g, s, W) & = ∫_{ℂ^{×}} W (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|_{ℂ}^{s−1/2}ω^{−1}(a) d^{×}a \end{aligned}

converge absolutely in some right half-plane. Set

\begin{aligned}Ψ(g, s, W) & = L(s, π) Φ(g, s, W), \\ Ψ̃(g, s, W) & = L(s, π̃) Φ̃(g, s, W). \end{aligned}

The functions Φ(g, s, W) and Φ̃(g, s, W) can be analytically continued to the whole complex plane as holomorphic functions of s. For a suitable choice of W the function Φ(e, s, W) is an exponential function of s. The functional equation

Φ̃(wg, 1 − s, W) = ϵ(s, π, ψ) Φ(g, s, W)

is satisfied. Moreover, if W is fixed |Ψ(g, s, W)| remains bounded as g varies over a compact subset of G_{ℂ} and s varies in a vertical strip of finite width from which discs about the poles of L(s, π) have been removed.

The following lemma can be verified by an explicit computation. The first assertion may also be proved by the method of Lemma 5.16.

Lemma 6.5 If σ = σ(μ_{1}, μ_{2}) and π = π(μ_{1}, μ_{2}) are defined then

\frac{L(1 − s, σ̃) ϵ(s, σ, ψ)}{L(s, σ)} = \frac{L(1 − s, π̃) ϵ(s, π, ψ)}{L(s, π)}

and the quotient

\frac{L(s, χ ⊗ σ)}{L(s, χ ⊗ π)}

is the product of a constant, a polynomial, and an exponential. Moreover the polynomial is of positive degree for some choice of the quasi-character χ.

We verify the last assertion. There is no harm in supposing that σ = π(ν_{1}, ν_{2}) and that χμ_{1}(z) = z^{a+p}z̄^{b+q}, χμ_{2}(z) = z^{a}z̄^{b}, χν_{1}(z) = z^{a+p}z̄^{b}, and χν_{2}(z) = z^{a}z^{b+q}. p ≥ 1 and q ≥ 1 are integers. Varying χ is equivalent to varying a and b through all the integers. If m_{1} is the largest of a + p and b + q and m_{2} is the largest of a and b while n_{1} is the largest of a + p and b and n_{2} is the largest of a and b + q the quotient

\frac{L(s, χ ⊗ σ)}{L(s, χ ⊗ π)}

differs from

\frac{Γ(s + n_{1}) Γ(s + n_{2})}{Γ(s + m_{1}) Γ(s + m_{2})}

by a constant times an exponential. It is clear that n_{1} and n_{2} are both greater than or equal to m_{2} and that either n_{1} or n_{2} is greater than or equal to m_{1}. Thus the quotient is a polynomial. If p ≥ q choose a and b so that b + q > a ≥ b. Then n_{1} = m + 1 and n_{2} > m_{2} so that the quotient is of positive degree. If q ≥ p choose a and b so that a + p > b ≥ a. Then n_{2} = m_{1} and n_{1} > m_{2}.

Lemma 6.6. Let π and π′ be two infinite-dimensional irreducible representations of ℋ_{ℂ}. Suppose there is a quasi-character ω of ℂ^{×} such that

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

and

π′ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

for all a in ℂ^{×}. If

ϵ(s, χ ⊗ π′, ψ)\frac{L(1 − s, χ^{−1} ⊗ π̃′)}{L(s, χ ⊗ π′)} = ϵ(s, χ ⊗ π, ψ)\frac{L(1 − s, χ^{−1} ⊗ π̃)}{L(s, χ ⊗ π)}

for all quasi-characters χ then π and π′ are equivalent.

Let π = π(μ_{1}, μ_{2}) and let π′ = (μ′_{1}, μ′_{2}). We let

μ_{i}(z) = (zz̄)^{s_{i}} \{\frac{z}{(zz̄)^{1/2}} \}^{a_{i}}

and

μ′_{i}(z) = (zz̄)^{s′_{i}} \{\frac{z}{(zz̄)^{1/2}} \}^{a_{i}}

with a_{i} and a′_{i} in ℤ. By assumptions s_{1} + s_{2} = s′_{1} + s′_{2} and a_{1} + a_{2} = a′_{1} + a′_{2}. Choose

χ(z) = \{\frac{z}{(zz̄)^{1/2}} \}^{n}

with n in ℤ. The quotient on the right has the same zeros and poles as

\frac{Γ(1 − s − s_{1} + |\frac{n+a_{1}}{2} |)}{Γ(s + s_{1} + |\frac{n+a_{1}}{2} |)} · \frac{Γ(1 − s − s_{2} + |\frac{n+a_{2}}{2} |)}{Γ(s + s_{2} + |\frac{n+a_{2}}{2} |)}.

A pole of the numerator can cancel a pole of the denominator if and only if there are two non-negative integers ℓ and m such that

s_{1} − s_{2} = 1 + ℓ + m + |\frac{n + a_{1}}{2} | + |\frac{n + a_{2}}{2} |

or

s_{2} − s_{1} = 1 + ℓ + m + |\frac{n + a_{1}}{2} | + |\frac{n + a_{2}}{2} |.

This can happen only if μ_{1}μ_{2}^{−1} is of the form μ_{1}μ_{2}^{−1}(z) = z^{p}z̄^{q} or μ_{1}μ_{2}^{−1}(z) = z^{−p}z̄^{−q} where p ≥ 1 and q ≥ 1 are integers. Since π(μ_{1}, μ_{2}) is infinite-dimensional it cannot be of either these forms and no poles cancel.

Consequently for every integer n, \{s_{1} + |\frac{n+a_{1}}{2} |, s_{2} + |\frac{n+a_{2}}{2} |\} = \{s′_{1} + |\frac{n+a′_{1}}{2} |, s′_{2} + |\frac{n+a′_{2}}{2} |\}. This can happen only if s_{1} = s′_{1}, a_{1} = a′_{1}, s_{2} = s′_{2}, and a_{2} = a′_{2} or s_{1} = s′_{2}, a_{1} = a′_{2}, s_{2} = s′_{1}, and a_{2} = a′_{1}. Thus π and π′ are equivalent.

The following proposition is an easy consequence of these two lemmas.

Proposition 6.7. Suppose π and π′ are two irreducible admissible representations of ℋ_{ℂ}. Suppose there is a quasi-character ω of ℂ^{∗} such that

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I

and

π′ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = ω(a)I.

If L(s, χ ⊗ π) = L(s, χ ⊗ π′), L(s, χ^{−1} ⊗ π̃) = L(s, χ^{−1} ⊗ π̃′) and

ϵ(s, χ ⊗ π, ψ) = ϵ(s, χ ⊗ π′, ψ)

for all quasi-characters χ the representations π and π′ are equivalent.

Chapter 1 126 §7. Characters. If F is a non-archimedean local field and π is an admissible representation of G_{F} the operator π(f) is of finite rank for every f in ℋ_{F} and therefore ha a trace Tr π(f). In this paragraph we prove that if π is irreducible there is a locally integrable function χ_{π} on G_{f} such that

Tr π(f) = ∫_{G_{F}} f(g) χ_{π}(g) dg.

Although Tr π(f) depends on the choice of the Haar measure the function χ_{π} does not. The following simple lemma shows that χ_{π} determines the class of π.

Lemma 7.1 If {π_{1}, ⋯, π_{p}} is a set of inequivalent irreducible admissible representations of ℋ_{F} the set of linear forms Tr π_{1}, Tr π_{2}, ⋯, Tr π_{p} is linearly independent.

Let π_{i} act on V_{i} and let ξ be an elementary idempotent such that none of the spaces π_{i}(ξ)V_{i}, 1 ≤ i ≤ p, are 0. Let π̄_{i} be the representation of ξℋ_{F} ξ on the finite dimensional space π_{i}(ξ)V_{i} = V_{i}(ξ). Suppose π̄_{i} and π̄_{j} are equivalent. Then there is an invertible linear map A from V_{i}(ξ) to V_{j}(ξ) which commutes with the action of ξℋ_{F} ξ. Choose a non-zero vector v_{i} in V_{i}(ξ) and let v_{j} = Av_{i}. We are going to show that π_{i} and π_{j} are equivalent. It is enough to show that, for any f in ℋ_{F}, π_{i}(f)v_{i} = 0 if and only if π_{j}(f)v_{j} = 0. But π_{i}(f)v_{i} = 0 if and only if π_{i}(ξ ∗ h)π_{i}(f)v_{i} = 0 for all h in ℋ_{F}. Since π_{i}(ξ ∗ h)π_{i}(f)v_{i} = π_{i}(ξ ∗ h ∗ f ∗ ξ)v_{i} and ξ ∗ h ∗ f ∗ ξ is in ξℋ_{F} ξ the assertion follows.

Thus the representations π̄_{1}, ⋯, π̄_{p} are inequivalent. Using this we shall show that the linear forms Tr π̄_{1}, ⋯, Tr π̄_{p} on ξℋ_{F} ξ are linearly independent. The lemma will then be proved. Take h in ξℋ_{F} ξ. Since π̄_{i} is irreducible and finite-dimensional Tr π̄_{i}(hf) = 0 for all f in ξℋ_{F} ξ if and only if π̄_{i}(h) = 0. Suppose we had h_{1}, ⋯, 𝔥_{p} in ξℋ_{F} ξ so that for at least one i the operator π̄_{i}(h_{i}) was not 0 while

∑_{i=1}^{p} Tr π̄_{i}(h_{i}f) = 0

for all f in ξℋ_{F} ξ. There must then be at least two integers j and k such that π̄_{j}(h_{j}) ≠ 0 and π̄_{k}(h_{k}) ≠ 0. Since π̄_{j} and π̄_{k} are not equivalent we can find an h in ξℋ_{F} ξ such that π̄_{j}(h) = 0 while π̄_{k}(h) is invertible. Replacing h_{i} by h_{i}h we obtain a relation of the same type in which the number of i for which π̄_{i}(h_{i}) = 0 has been increased. By induction we see that no such relation is possible. Since ξℋ_{F} ξ contains a unit the required independence follows.

For most of these notes the existence of χ_{π} is irrelevant. It is used only toward the end. The reader who is more interested in automorphic forms than in group representations will probably want to take the existence of χ_{π} for granted and, for the moment at least, skip this paragraph. To do so will cause no harm. However he will eventually have to turn back to read the first few pages in order to review the definition of the Tamagawa measure.

Choose a non-trivial additive character ψ of F. If X is an analytic manifold over F and ω is a differential form of highest degree on X we can associate to ω a measure on X which is denoted by |ω|_{F} or sometimes simply by ω. If X = F and ω = dx is the differential of the identity application the measure |ω|_{F} = dx is by definition the Haar measure on F which is self-dual with respect to ψ. In general if p belongs to X and x^{1}, ⋯, x^{n} are local coordinates near p so that

ω = a(x^{1}, ⋯, x^{n}) dx^{1} ∧ ⋯ ∧ dx^{n}

then, if f is a continuous real-valued functions with support in a small neighborhood of p,

∫_{X} f|ω|_{F} = ∫ f(x^{1}, ⋯, x^{b}) |a(x^{1}, ⋯, x^{b})| dx^{1} ⋯ dx^{n}.

Chapter 1 127 The absolute value |a(x^{1}, ⋯, x^{n})| is the normalized absolute value in the field F. To prove the existence of the measure ω one has to estabish the usual formula for a change of variable in a multiple integral. For this and other facts about these measures we refer to the notes of Weil [12].

If G is an algebraic group over F then G_{F} is an analytic space. If ω is a left-invariant form of highest degree on G_{F} the measure |ω|_{F} is a Haar measure on G_{F}. It is called the Tamagawa measure. It depends on ω and ψ.

If M is the algebra of 2 × 2 matrices over F the additive group of M is an algebraic group. If a typical element of M is

x = ( \begin{matrix}a & b \\ c & d \end{matrix} )

then

μ = da ∧ db ∧ dc ∧ dd

is an invariant form of highest degree and |μ| = dx is the additive Haar measure which is self-dual with respect to the character ψ_{M}(x) = ψ_{F} (τ(x)) if τ is the trace of x.

On the multiplicative group G of M we take the form ω(x) = (detx)^{−2}μ(x). The associated Haar measure is

|ω(x)| = |detx|_{F}^{−2} dx = |x|_{M}^{−1} dx.

An element of G is said to be regular if its eigenvalues are distinct. The centralizer in G_{F} of a regular element in G_{F} is a Cartan subgroup of G_{F}. Such a Cartan subgroup B_{F} is of course abelian. There seems to be no canonical choice for the invariant forms on B_{F}. However the centralizer of B_{F} in M_{F} is an algebra E of degree two over F. It is either isomorphic to the direct sum of F with itself or it is a separable quadratic of F. B_{F} is the multiplicative group of E. In the first paragraph we introduced a map ν from E to F. Once a form μ_{E} on E which is invariant for the additive group has been chosen we can set μ_{B}(x) = ν(x)^{−1}μ_{E}(x). μ_{b} is an invariant form on B_{F}. The associated measure is invariant under all automorphisms of E over F. We should also recall at this point that two Cartan subgroups B_{F} and B′_{F} are conjugate in G_{F} if and only if the corresponding algebras are isomorphic.

Once μ_{E} and therefore μ_{B} has been chosen we can introduce on B_{F} \ G_{F} which is also an analytic manifold the form ω_{B} which is the quotient of ω by μ_{B}. Then

∫_{G_{F}} f(g) ω(g) = ∫_{B_{F} \G_{F}} {∫_{B_{F}} f(bg) μ_{B}(b)} ω_{B}(g).

The centre of the algebra of M_{F} is isomorphic to F and the centre Z_{F} of G_{F} is isomorphic to F^{×}. On F^{×} we have the form x^{−1} dx. We take μ_{Z} to be the corresponding form on Z_{F}. μ_{B}^{0} will be the quotient of μ_{B} by μ_{Z} and ω^{0} will be the quotient of ω by μ_{Z}. The corresponding integration formulae are

∫_{B_{F}} f(b) μ_{B}(b) = ∫_{Z_{f} \B_{F}} {∫_{Z_{F}} f(zb) μ_{Z}(z)} μ_{B}^{0}(b)

and

∫_{G_{F}} f(G) ω(g) = ∫_{Z_{F} \G_{F}} {∫_{Z_{F}} f(zg) μ_{Z}(z)} ω^{0}(g).

If g belongs to G_{F} its eigenvalues α_{1} and α_{2} are the roots of the equation

X^{2} − τ(g)X + ν(g) = 0

Chapter 1 128 and

\frac{(α_{1} − α_{2})^{2}}{α_{1}α_{2}} = \frac{\{τ(g)\}^{2} − 4ν(g)}{ν(g)}

belongs to F. Set

δ(g) = |\frac{(α_{1} − α_{2})^{2}}{α_{1}α_{2}} |_{F} .

Since g is regular if and only if δ(g) ≠ 0 the set Ĝ_{F} of regular elements is open in G_{F} and its complement has measure zero.

There are two more integration formulae that we shall need. Their proof proceeds as for archimedean fields. Choose a system S of representatives of the conjugacy classes of Cartan subgroups of G_{F}. Then

∫_{G_{F}} f(g) ω(g) = ∑_{B_{F} ∈S} \frac{1}{2} ∫_{B_{F}} δ(b) \{∫_{B_{F} \G_{F}} f(g^{−1}bg) ω_{B}(g)\} μ_{B}(b)  (7.2.1)

∫_{Z_{F} \G_{F}} f(g) ω^{0}(g) = ∑_{B_{F} ∈S} \frac{1}{2} ∫_{Z_{F} \B_{F}} δ(b) \{∫_{B_{F} \G_{F}} f(g^{−1}bg) ω_{B}(g)\} μ_{B}^{0}(b)  (7.2.2)

if f is an integrable function on G_{F} or Z_{F} \G_{F}. Notice that the sum on the right is not necessarily finite. Let B̂_{F} = B_{F} ∩ Ĝ_{F} and let

B̂_{F}^{G} = {g^{−1}bg | b ∈ B̂_{F} , g ∈ G_{F} }.

Then Ĝ_{F} is the disjoint union

⋃_{B_{F} ∈S} B̂_{F}^{G}.

There is a simple lemma to be verified.

Lemma 7.2 (i) For any Cartan subgroup B_{F} the set B̂_{F}^{G} is open.

(ii) The set Ĝ_{F} is open. (iii) The set G̃_{F} of g in G_{F} whose eigenvalues do not belong to F is open.

The second statement is a consequence of the first. If B_{F} corresponds to the separable quadratic extension E then B̂_{F}^{G} is the set of matrices with distinct eigenvalues in E and if B_{F} splits and therefore corresponds to the direct sum of F with itself, B̂_{F}^{G} is the set of matrices with distinct eigenvalues in F. Thus the first assertion is a consequence of the following lemma which is a form of Hensel’s lemma or of the implicit function theorem.

Lemma 7.2.1 Let E be a separable extension of F. Assume the equation

X^{p} + a_{1}X^{p−1} + ⋯ + a_{p} = 0

with coeficients in F has a simple root λ in E. Given ε > 0 there is a δ > 0 such that whenever b_{1}, ⋯, b_{p} are in F and |b_{i} − a_{i}|_{F} < δ for 1 ≤ i ≤ p the equation

X^{p} + b_{1}X^{p−1} + ⋯ + b_{p} = 0

Chapter 1 129 has a root μ in E for which |λ − μ|_{E} < ε.

There is no need to prove this lemma. To prove the third assertion we have to show that the set of matrices with eigenvalues in F is closed. Suppose g_{n} → g and g_{n} has eigenvalues λ_{n} and μ_{n} in F. Then λ_{n} + μ_{n} → τ(g) and λ_{n}μ_{n} → ν(g). If λ_{n} and μ_{n} did not remain in a compact subset of F^{×} then, since their product does, we would have, after passing to a subsequence, |λ_{n}| → 0, |μ_{n}| → ∞ or |λ_{n}| → ∞, |μ_{n}| → 0. In either case λ_{n} + μ_{n} could not converge. Thus, again passing to a subsequence, we have λ_{n} → λ and μ_{n} → μ. λ and μ are the eigenvalues of g.

If the characteristic of F is not two the sets Ĝ_{F} and G̃_{F} are the same. We now introduce a function on G_{F} which plays an important role in the discussion of characters. If B_{F} is a split Cartan subgroup we set c(B_{F}) = 1 but if B_{F} is not split and corresponds to the quadratic extension E we set

c(B_{F}) = |ϖ|F^{\frac{t+1}{2}}

ϖ is a generator of 𝔭_{F} and 𝔭_{F}^{t+1} is the discriminant of E over F. If g in Ĝ_{F} belongs to the Cartan subgroup B_{F} we set

ξ(g) = c(B_{F})δ^{−1/2}(g).

If g is singular we set ξ(g) = ∞. The factor c(B_{F}) is important only in characteristic two when there are an infinite number of conjugacy classes of Cartan subgroups.

Lemma 7.3 The function ξ is locally constant on Ĝ_{F} and bounded away from zero on any compact subset of G_{F}. It is locally integrable on Z_{F} \ G_{F} and on G_{F}.

It is of course implicit in the satement of the lemma that ξ is constant on cosets of Z_{F}. The two previous lemmas show that ξ is locally constant on Ĝ_{F}. To prove the remaining assertions we recall some facts about orders and modules in separable quadratic extensions of non-archimedean fields.

If E is a separable quadratic extension of F an order R of E is a subring of O_{E} which contains O_{F} and a basis of E. A module I in E is a finitely generated O_{F} submodule of E which contains a basis of E. If I is a module the set

{α ∈ E | αI ⊆ I}

is an order R_{I}. It is clear that an order is a module and that R_{R} = R. Two modules I and J are said to be equivalent if there is an α in E^{×} so that J = αI. Then R_{I} = R_{J}.

Suppose the module I is contained on O_{E} and contains 1. Since I/O_{F} is a torsion-free O_{F} module the module I has a basis of the form {1, δ}. Since δ is integral δ^{2} belongs to I. Therefore I is an order and R_{I} = I. Since any module is equivalent to a module which contains 1 and lies in O_{E} the collection of modules I for which R_{I} = R forms, for a given order R, a single equivalence class.

As observed any order has a basis, over O_{F}, of the form {1, δ}. The absolute values of the numbers δ occurring in such bases are bounded below. A basis {1, δ} is said to be normal if δ has the smallest possible absolute value. It is easily seen, by considering the ramified and unramified extensions separately, that if {1, δ} is normal

R = O_{F} + δO_{E}.

Thus R determines and is determined by |δ|_{E}. It is easily seen that if E/F is unramified |δ|_{E} is any number of the form |ϖ_{E}|_{E}^{n} with n ≥ 0. ϖ_{E} is a generator of 𝔭_{E}. We set n = ω(R). If E/F is Chapter 1 130 ramified, |δ|_{E} is any number of the form |ϖ_{E}|_{E}^{2n+1} with n ≥ 0. We set ω(R) = n. In the ramified case

[E^{×} : F^{×}(U_{E} ∩ R)] = 2|ϖ_{F} |_{F}^{−ω(R)}.

In the unramified case

[E^{×} : F^{×}(U_{E} ∩ R)] = |ϖ_{F} |_{F}^{−ω(R)}(1 + |ϖ_{F} |_{F})

unless ω(R) = 0 and then

[E^{×} : F^{×}(U_{E} ∩ R)] = 1.

It is clear that R′ contains R if and only if ω(R′) ≤ ω(R). Thus ω(R) + 1 is the number of orders which contain R. If γ belongs to O_{E} but not to O_{F} let R(γ) be the order with basis {1, γ} and let ω(γ) = ω(R(γ)).

Lemma 7.3.1 Let γ̄ be the conjugate of γ in E and let

|(γ − γ̄)^{2}|_{F}^{1/2} = |ϖ_{F} |_{F}^{m(γ)}.

If 𝔭_{E}^{t+1} is the discriminant of E and γ belongs to O_{K} but not to O_{F} then

m(γ) = ω(γ) + \frac{t + 1}{2}.

Let {1, δ} be a normal basis of R(γ). Then γ = a + bδ with a and b in O_{F}. Moreover δ = c + dγ with c and d in O_{F}. Thus γ = (a + bc) + bdγ so that a + bc = 0. and bd = 1. Therefore b is a unit and |γ − γ̄| = |δ − δ̄|. We can thus replace γ by δ. Suppose first that E/F is unramified so that t + 1 = 0. We take δ = εϖ_{F}^{n} where n = ω(R(γ)) and ε is a unit of O_{E}. Since

δ − δ̄ = (ε − ε̄)ϖ_{F}^{n}

we have only to show that ε − ε̄ is a unit. ε is not congruent to an element of O_{F} modulo 𝔭_{E} and therefore {1, ε} determines a basis of O_{E}/𝔭_{E}. Since the Galois group acts faithfully on O_{E}/𝔭_{E} the number ε − ε̄ is not in 𝔭_{E}.

If E/F is ramified we may take δ = ϖ_{F}^{n} ϖ_{E} with n = ω(δ). It is well-known that

|ϖ_{E} − \overline{ϖ}_{E}| = |ϖ_{E}|_{E}^{t+1}

Thus

|(δ − δ̄)^{2}|_{F}^{1/2} = |δ − δ̄|_{E}^{1/2} = |ϖ_{F} |_{F}^{n} |ϖ_{K}|K^{\frac{t+1}{2}} = |ϖ_{F} |_{F}^{n+ \frac{t+1}{2}}

The lemma follows.

There are two more lemmas to be proved before we return to the proof of Lemma 7.3.

Chapter 1 131 Lemma 7.3.2 Let C be a compact subset of Z_{F} \ G_{F} and let χ_{C} be the characteristic function of C and of its inverse image in G_{F}. There is a constant c such that for every b in G_{F} which is contained in an anisotropic Cartan subgroup

∫_{Z_{F} \G_{F}} χ_{C}(g^{−1}bg) ω^{0}(g) ≤ cξ(b).

The assertion is trivial unless b is regular. Then the assumption is that its eigenvalues are distinct and do not lie in F. Any h in G_{F} can be written as

g_{1} ( \begin{matrix}ϖ_{F}^{p} \\ & ϖ_{F}^{q} \end{matrix} ) g_{2}

where g_{1} and g_{2} belongs to GL(2, O_{F}) and p ≤ q. The numbers ϖ_{F}^{p} and ϖ_{F}^{q} are the elementary divisors of h. Let T_{r} be the set of all those h for which q − p ≤ r. T_{r} is the inverse image of a compact subset T′_{r} of Z_{F} \ G_{F}. If r is sufficiently large C is contained in T′_{r}. Thus we may replace χ_{C} and χ_{r} the characteristic function of T′_{r}.

If h belongs to GL(2, O_{F}) then h^{−1}g^{−1}bgh belongs to T_{r} if and only if g^{−1}bg belongs to T_{r}. Thus the integral is the product of the measure of GL(2, O_{F}) ∩ Z_{F} \ GL(2, O_{F}) by the number of right cosets of Z_{F} GL(2, O_{F}) whose elements g are such that g^{−1}bg belong to T_{r}. If H is such a coset and B_{F} is the Cartan subgroup containing b then for any b′ in B_{F} the coset b′H has the same property. Thus the integral equals

measure(GL(2, O_{F}) ∩ Z_{F} \ GL(2, O_{F})) ∑[B_{F} g GL(2, O_{F}) : Z_{F} GL(2, O_{F})].

The sum is over a set of representatives of the cosets in B_{F} \ G_{F} /GL(2, O_{F}).

Let B_{F} correspond to the separable quadratic extension E. Choose a basis of O_{E} over O_{F}. It will also be a basis of E over F. By means of this basis we identify G_{F} with the group of invertible linear transformations of E over F. GL(2, O_{F}) is the stabilizer of O_{E}. Every γ in E^{×} determines a linear transformation b_{γ} : x → γx of E. The set of all such linear transformations is a Cartan subgroup conjugate to B_{F} and with no loss of generality we may assume that it is B_{F}. Choose γ so that b = b_{γ}.

Every module is of the form gO_{E} with g in G_{F}. Moreover g_{1}O_{E} and g_{2}O_{E} are equivalent if and only if g_{1} and g_{2} belong to same double coset in B_{F} \ G_{F} /GL(2, O_{F}). Thus there is a one-to-one correspondence between the collection of double cosets and the collection of orders of E. Let B_{F} g GL(2, O_{F}) correspond to the order R. The index

[B_{F} g GL(2, O_{F}) : Z_{F} GL(2, O_{F})]

is equal to

[B_{F} : B_{F} ∩ Z_{F} g GL(2, O_{F})g^{−1}]

Two elements b_{1} and b_{2} in B_{F} belong to the same coset of B_{F} ∩ Z_{F} g GL(2, O_{F})g^{−1} if and only if there is a z in Z_{F} and an h in GL(2, O_{F}) such that

b_{1}g = b_{2}zgh

This can happen if and only if

b_{1}gO_{E} = b_{2}zgO_{E}.

Chapter 1 132 Let I = gO_{E} and let b_{i} = b_{γ_{i}}. If we identify Z_{F} and F^{×} so that z may be regarded as an element of F^{×} the last relation is equivalent to

γ_{1}I = γ_{2}zI

or γ_{1}^{−1}γ_{2}z ∈ R ∩ U_{E}. Thus

[B_{F} g GL(2, O_{F}) : Z_{F} GL(2, O_{F})] = [E^{×} : F^{×}(R ∩ U_{E})].

Let |detb|_{F} = |γ|_{K} = |ϖ_{F} |_{F}^{m}. Let ϖ_{F}^{p} and ϖ_{F}^{q} with p ≤ q be the elementary divisors of g^{−1}bg. Certainly p + q = m. The matrix g^{−1}bg belongs to T_{r} if and only if q − p = m − 2p ≤ r. If s is the integral part of \frac{r−m}{2} this is so if and only if ϖ_{F}^{s} g^{−1}bg has integral coefficients, that is if and only if

ϖ_{F}^{s} g^{−1}bg O_{E} ⊆ O_{E}

or ϖ_{F}^{s} γ ∈ R.

Our integral is therefore equal to

measure(GL(2, O_{F}) ∩ Z_{F} \ GL(2, O_{F})) ∑_{ϖ_{F}^{s} γ∈R}[E^{×} : F^{×}(R ∩ U_{E})].  (∗)

The sum is over all orders of E which contains ϖ_{F}^{s} γ. ϖ_{F}^{s} γ does not lie in F. If it does not lie in O_{K} the sume is zero. If it lies in O_{K} then ϖ_{F}^{s} γ belongs to R if and only if ω(R) ≤ ω(ϖ_{F}^{s} γ). In this case the expression (∗) is bounded by

2 measure(GL(2, O_{F}) ∩ Z_{F} \ GL(2, O_{F})) ∑_{0≤k≤ω(ϖ_{F}^{s} γ)} |ϖ_{F} |_{F}^{−k}.

This in turn is bounded by a constant, which is independent of B_{F} and r, times

|ϖ_{F} |_{F}^{−ω(ϖ_{F}^{s} γ)}

We have c(B_{F}) = |ϖ_{F} |F^{\frac{t+1}{2}} , m(ϖ_{F}^{s} γ) = s + m(γ) ≤ \frac{r−m}{2} + m(γ), and

δ(b)^{1/2} = \frac{|(γ − γ̄)^{2}|_{F}^{1/2}}{|γγ̄|_{F}^{1/2}} = |ϖ_{F} |_{F}^{−m/2} |ϖ_{F} |_{F}^{m(γ)}.

To prove the lemma we have only to show that

−m(γ) + \frac{m}{2} + \frac{t + 1}{2} + ω(ϖ_{F}^{s} γ)

is bounded above by a constant which depends only on r. By the previous lemma

ω(ϖ_{F}^{s} γ) = m(ϖ_{F}^{s} γ) − \frac{t + 1}{2}

so that

−m(γ) + \frac{m}{2} + \frac{t + 1}{2} + ω(ϖ_{F}^{s} γ) ≤ \frac{r − m}{2} + \frac{m}{2} = \frac{r}{2}.

Suppose the Cartan subalgebra B_{F} corresponds to the algebra E. Once the measure μ_{E} on E has been chosen we can form the measure μ_{B} on B_{F} and the measure ω_{B} on B_{F} \ G_{F}. Once μ_{E} and therefore μ_{B} and ω_{B} are chosen we let n(B_{F}) be that constant which makes n(B_{F})μ_{E} self-dual with respect to the character x → ψ(τ(x)) on E.

Chapter 1 133 Lemma 7.3.3 If r is a non-negative integer there is a constant d_{r} such that for any Cartan subgroup B_{F} and any b in B_{F}

∫_{B_{F} \G_{F}} χ_{r}(g^{−1}bg) ω_{B}(g) ≤ d_{r}n(B_{F}) δ(b)^{−1/2}.

We may again suppose that b belongs to B̂_{F}. If B_{F} is anisotropic the left side is equal to

\frac{1}{measure(Z_{F} \ B_{F})} ∫_{Z_{F} \G_{F}} χ_{r}(g^{−1}bg) ω^{0}(g).

Suppose B_{F} corresponds to the quadratic extension E. If E/F is unramified

measure(Z_{f} \ B_{F}) = \frac{1}{n(B_{F})}(1 + |ϖ_{F} |)

because n(B_{F})μ_{E} assigns the measure 1 to O_{E}. If E/F is ramified n(B_{F})μ_{E} assigns the measure |ϖ_{F} |^{\frac{t+1}{2}} to O_{E} and

measure(Z_{F} \ B_{F}) = \frac{2}{n(B_{F})}|ϖ_{F} |^{\frac{t+1}{2}} = \frac{2}{n(B_{F})}c(B_{F})

In these cases the assertion is therefore a consequence of the previous lemma.

If the inequality of the lemma is true for one Cartan subgroup it is true for all conjugate subgroups. To complete the proof we have to verify it when B_{F} is the group A_{F} of diagonal matrices. Since we are now dealing with a fixed Cartan subgroup the choice of Haar measure on B_{F} \ G_{F} is not important. Moreover GL(2, O_{F}) T_{r} GL(2, O_{F}) = T_{r} so that, using the Iwasawa decomposition and the associated decomposition of measures, we may take the itnegral to be

∫_{F} χ_{r} (( \begin{matrix}1 & −x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) dx

if

b = ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )

The argument in the integrand is

( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & (1 − \frac{β}{α})x \\ 0 & 1 \end{matrix} )

Changing the variables in the integral we obtain

\frac{1}{|1 − \frac{β}{α} |} ∫_{F} χ_{r} (( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) dx.

Let |α| = ϖ_{F} |^{ℓ}, |β| = |ϖ_{F} |^{m}, and |x| = |ϖ_{F} |^{n}. With no loss of generality we may suppose |α| ≥ |β|. If n ≥ 0 the elementary divisors of

( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )

are ϖ_{F}^{ℓ} and ϖ_{F}^{m} so that it is in T_{r} if and only if m − ℓ ≤ r. If n < 0 its elementary divisors are ϖ_{F}^{ℓ+n} and ϖ_{F}^{m−n} so that it is in T_{r} if and only if m − ℓ − 2n ≤ r. Thus the integral is at most

measure\{x | |x| ≤ |ϖ_{F} |^{\frac{m−ℓ−r}{2}} \}

which is, apart from a factor depending on the choice of the Haar measure, |ϖ_{F} |^{\frac{m−ℓ−r}{2}} . Since

|ϖ_{F} |^{\frac{m−ℓ−r}{2}} = |\frac{β}{α}|^{\frac{1}{2}} |ϖ_{F} |^{− \frac{r}{2}}

and

\frac{|\frac{β}{α} |^{\frac{1}{2}}}{|1 − \frac{β}{α} |} = δ(b)^{− \frac{1}{2}}

the lemma follows.

We return to Lemma 7.3 and prove first that ξ is bounded away from zero on each compact subset C. In other words we show that there is a positive constant c such that ξ(h) ≥ c on C. There is a z in Z_{F} such that every matrix in zC has integral entries. Since ξ(zh) = ξ(h) we may as well assume that every matrix in C itself has integral entires. There is a constant c_{1} > 0 such that

|deth|_{F}^{1/2} ≥ c_{1}

on C and a constant c_{2} such that

|τ(h)^{2} − 4ν(h)|^{1/2} ≤ c_{2}

on C. τ and ν are the trace and determinant of h. Thus

δ^{−1/2}(h) ≥ \frac{c_{1}}{c_{2}}

on C. ξ(h) is certainly bounded away from 0 on the singular elements and the preceding inequality shows that it is bounded away from 0 on the regular elements in C which lie in a split Cartan subalgebra. Suppose h is regular and lies in the anisotropic Cartan subgroup B_{F}. Let B_{F} correspond to the field E and let h have eigenvalues γ and γ̄ in E. Then

|(γ − γ̄)^{2}|_{F}^{− \frac{1}{2}} c(B_{F}) = |ϖ_{F} |^{−m(γ)} |ϖ_{F} |^{\frac{t+1}{2}} = |ϖ_{F} |^{−ω(γ)}

Since ω(γ) ≥ 0 we have ξ(h) ≥ c_{1}.

The function ξ is certainly measurable. It is locally integrable in G_{F} if and only if it is locally integrable on Z_{F} \ G_{F}. Let C be a compact set in Z_{F} \ G_{F}. We have to show

∫_{Z_{F} \G_{F}} χ_{C}(g) ξ(g) ω^{0}(g)

is finite. As usual it will be enough to show that

∫_{Z_{F} \G_{F}} χ_{r}(g) ξ(g) ω^{0}(g)

is finite for every non-negative integer r. According to formula (7.2.2) this integral is the sum of

\frac{1}{2} ∫_{Z_{F} \A_{F}} ξ(a) δ(a) \{∫_{A_{F} \G_{F}} χ_{r}(g^{−1}ag) ω_{A}(g)\} μ_{A}^{0}(a)

and

\frac{1}{2} ∑_{B_{F} ∈S′} ∫_{Z_{F} \B_{F}} ξ(b) δ(b) \{∫_{B_{F} \G_{F}} χ_{r}(g^{−1}bg) ω_{B}(g)\} μ_{B}^{0}(b).

It is easy to see that there is a compact set C_{0} in Z_{F} \ A_{F} such that χ_{r}(g^{−1}ag) = 0 for all g unless the projection of a lies in C_{0}. Thus the first integral need only be taken over C_{0}. The inner integral is at most d_{r}n(A_{F}) δ(a)^{−1/2}. Since ξ(a) δ(a) δ(a)^{−1/2} = 1 on A_{F} the first integral causes no trouble. We can also use Lemma 7.3.3 to see that the sum over S′, which is by the way a set of representatives for the conjugacy classes of anisotropic Cartan subgroups, is less than or equal to

\frac{1}{2} ∑_{B_{F} ∈S′} d_{r}n(B_{F})c(B_{F}) ∫_{Z_{F} \B_{F}} μ_{B}^{0}(b).

If the characteristic is not two this sum is finite and there is no problem.

In general if B_{F} corresponds to the field E and 𝔭_{F}^{t_{E}+1} is the discriminant of E we have c(B_{F}) = |ϖ_{F} |^{\frac{t_{E} +1}{2}} and

n(B_{F}) ∫_{Z_{F} \B_{F}} μ_{B}^{0}(b) ≤ 2|ϖ_{F} |^{(t_{E}+1)/2}

To complete the proof we have to show that

∑_{E} |ϖ_{F} |^{t_{E}+1}

is finite if F has characteristic 2. The sum is over all separable quadratic extensions of F. Let M(t) be the number of extensions E for which t_{E} = t. Associated to any such E is a quadratic character of F^{×} with conductor 𝔭_{F}^{t+1}. Thus

M(t) ≤ [F^{×} : (F^{×})^{2}(1 + 𝔭_{F}^{t+1})] = 2[U_{F} : U_{F}^{2} (1 + 𝔭_{F}^{t+1})]

if t ≥ 0. Of course M(−1) = 1. Any element of U_{F} is congruent modulo 1 + 𝔭_{F}^{t+1} to an element of the form

a_{0} + a_{1}ϖ_{F} + ⋯ + a_{t}ϖ_{F}^{t} .

Such a number is a square if a_{i} = 0 for i odd. Thus

M(t) = 0(|ϖ_{F} |^{− \frac{t+1}{2}} )

and the series converges.

We can now begin the study of characters.

Chapter 1 136 Proposition 7.4 The character of an absolutely cuspidal representation exists as a locally integrable function whose absolute value is bounded by a multiple of ξ. It is continuous on Ĝ_{F} ∪ G̃_{F}.

If the character χ_{π} of π exists and χ is a quasi-character of F^{×} then the character of π′ = χ⊗π also exists and χ_{π′}(g) = χ(detg) χ_{π}(g). Thus the proposition has only to be proved for unitary representations π. π is then square integrable and we can make use of the following lemma for which, although it is well-known, we provide a proof.

Lemma 7.4.1 Let f belong to ℋ_{F} and let u be a vector of length one in the space on which the absolutely cuspidal unitary representation π acts. Then

Tr π(f) = d(π) ∫_{Z_{F} \G_{F}} {∫_{G_{F}} f(h)(π(g^{−1}hg)u, u) dh} dg

if d(π) is the formal degree of π.

Let Q be the operator

π(f) = ∫_{G_{F}} f(h) π(h) dh.

Let {v_{i}} be an orthonormal basis of the space on which π acts. All but a finite number of the coefficients

Q_{ij} = (Qv_{i}, v_{j})

are zero. We have

(π(g^{−1}) Qπ(g)u, u) = (Qπ(g)u, π(g)u)

The right side equals

∑_{i} (Qπ(g)u, v_{i})(v_{i}, π(g)u)

which in turn equals

∑_{i} ∑_{j} (π(g)u, v_{j}) Q_{ji}(v_{i}, π(g)u)

In both series there are only a finite number of non-zero terms. Thus

∫_{Z_{F} \G_{F}} (π(g^{−1}) Qπ(g)u, u) dg = ∑_{i,j} Q_{ji} ∫_{Z_{F} \G_{F}} (π(g)u, v_{j})(v_{i}, π(g)u) dg

The integrals on the right exist because the representation is square-integrable. Applying the Schur orthogonality relations we see that the right side is equal to

\frac{1}{d(π)} ∑_{i,j} Q_{ij}(v_{i}, v_{j}) = \frac{1}{d(π)} ∑_{i} Q_{ii} = \frac{1}{d(π)} Tr π(f).

Since

(π(g^{−1}) Qπ(g)u, u) = ∫_{G_{F}} f(h) (π(g^{−1}) π(h) π(g)u, u) dh

the lemma follows.

Observe that the integral of the lemma is an iterated and not a double integral. It is the limit as r approaches infinity of

∫_{T′_{r}} {∫_{G_{F}} f(h) (π(g^{−1}hg)u, u) dh} dg

Since T′_{r} is compact this integral is absolutely convergent and equals

∫_{G_{F}} f(h) {∫_{T′_{r}} (π(g^{−1}hg)u, u) dg} dh.

To prove the first part of the proposition we show that the sequence of functions

φ_{r}(h) = ∫_{T′_{r}}(π(g^{−1}hg)u, u) dg

is dominated locally by a multiple of ξ and converges almost everywhere on G_{F}. We shall set

χ_{π}(h) = d(π) lim_{r→∞} φ_{r}(h)

whenever the limit exists.

When proving the second part of the proposition we shall make use of the following lemma.

Lemma 7.4.2 Let C_{1} be a compact subset of G̃_{F} and let C_{2} be a compact set in G_{F}. The image in Z_{F} \ G_{F} of

{g ∈ G_{F} | g^{−1}C_{1}g ∩ Z_{F} C_{2} ≠ 0}

is compact.

The set is clearly closed so we have only to show that it is contained in some compact set. We may suppose that GL(2, O_{F}) C_{2} GL(2, O_{F}) = C_{2}. Let

g = ( \begin{matrix}α & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}β & 0 \\ 0 & β \end{matrix} ) h

with h in GL(2, O_{F}). Then

g^{−1}C_{1}g ∩ Z_{F} C_{2} ≠ 0

if and only if

( \begin{matrix}α & x \\ 0 & 1 \end{matrix} )^{−1}C_{1} ( \begin{matrix}α & x \\ 0 & 1 \end{matrix} ) ∩ Z_{F} C_{2} ≠ 0.

We have to show that this condition forces α to lie in a compact subset of F^{×} and x to lie in a compact subset of F. Since

det(g^{−1}cg) = detc

we may replace Z_{F} C_{2} by the compact set

C_{3} = {h ∈ Z_{F} C_{2} | deth ∈ detC_{1}}.

Chapter 1 138

Let

( \begin{matrix}a & b \\ c & d \end{matrix} )

be a typical element of C_{1}. The entry c is never 0 on C_{1} and therefore its absolute value is bounded below,

( \begin{matrix}α & x \\ 0 & 1 \end{matrix} )^{−1} ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}α & x \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}a − xc & y \\ cα & cx + d \end{matrix} ) .

The number y is of no interest. The matrix on the right cannot lie in C_{3} unless |cx + d| is bounded above by some number depending on C_{3}. Since |d| is bounded above and |c| is bounded below x is forced to lie in some compact set Ω of F. If C_{4} is the compact set

\{( \begin{matrix}1 & −x \\ 0 & 1 \end{matrix} ) h ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) |x ∈ Ω, h ∈ C_{1}\}

we have finally to show that if

( \begin{matrix}α^{−1} & 0 \\ 0 & 1 \end{matrix} ) C_{4} ( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ∩ C_{3} ≠ 0

then α is forced to lie in a compact subset of F^{×}. We now let

( \begin{matrix}a & b \\ c & d \end{matrix} )

be a typical element of C_{4}. On C_{4} both |b| and |c| are bounded blow. Since

( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}α^{−1} & 0 \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}a & b/α \\ cα & d \end{matrix} )

and all matrix entries are bounded above in absolute value on C_{3} the absolute value |α| must indeed be bounded above and below.

If π acts on V then for any u in V the support of the function (π(g)u, u) has been shown, in the second paragraph, to be compact modulo Z_{F}. Let C be its compact image in Z_{F} \ G_{F}. Let C_{1} be a compact subset of G_{F}. By the previous lemma the set of g in G_{F} such that

(π(g^{−1}hg)u, u) ≠ 0

for some h in C_{1} has an image in Z_{F} \ G_{F} which is contained in a compact set C_{2}. Therefore the integral

∫_{Z_{F} \G_{F}} (π(g^{−1}hg)u, u) dg = ∫_{C_{2}} (π(g^{−1}hg)u, u) dg

is convergent for h in C_{1}. Moreover if r is large enough T′_{r} contains C_{2} and

φ_{r}(h) = ∫_{Z_{f} \G_{F}} (π(g^{−1}hg)u, u) dg.

Therefore the sequence {φ_{r}} converges uniformly on any compact subset of G̃_{F} and its limit d^{−1}(π) χ_{π}(h) is continuous on G̃_{F}. We may state the following proposition.

Proposition 7.5 If h belongs to G̃_{F} then

∫_{Z_{F} \G_{F}} (π(g^{−1}hg)u, u) dg

exists and is equal to d^{−1}(π) χ_{π}(h).

Since

|(π(g)u, u)| ≤ χ_{C}(g)

it follows from Lemma 7.3.2 that, for some constant c,

|φ_{r}(h)| ≤ cξ(h)

on G̃_{F}. The set Ĝ_{F} − G̃_{F} is Â_{F}^{G} which is open. To complete the proof of Proposition 7.4 we show that on the intersection of Â_{F}^{G} with a compact subset of G_{F} the sequence {φ_{r}} is dominated by a multiple of ξ and that it converges uniformly in a compact subset of Â_{F}^{G}.

Let C_{3} be a compact subset of G_{F}. Any h in Â_{F} may be written in the form

h = h_{1}^{−1} ( \begin{matrix}1 & −x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) h_{1}

where h_{1} belongs to GL(2, O_{F}) and α ≠ β. In C_{3} ∩ Â_{F}^{G} the absolute values of α and β are bounded above and below. If C_{3} is contained in Â_{F}^{G} the absolute value of 1 − \frac{β}{α} is also bounded above and below on C_{3}. Since

( \begin{matrix}1 & −x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & (1 − \frac{β}{α})x \\ 0 & 1 \end{matrix} )

the absolute value of x will be bounded above.

Since GL(2, O_{F}) T_{r} GL(2, O_{F}) = T_{r} the integral which defines φ_{r}(h) is equal to

∫_{T′_{r}} (π(g^{−1}h′g)u, u) dg

if

h′ = ( \begin{matrix}1 & −x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )

and we may as well assume that h itself is of this form. We are going to show that there is a constant c such that

|φ_{r}(h)| ≤ c|1 − \frac{β}{α}|^{−1}

for all r and all such h and that the sequence {φ_{r}} converges uniformly if x remains in a compact subset of F and α, β and 1− \frac{β}{α} remain in a compact subset of F^{×}. Then the proof of the proposition will be complete.

Chapter 1 140

The stabilizer of u is some open subgroup U of GL(2, O_{F}). Let h_{1}, ⋯, h_{p} be a set of coset representatives for GL(2, O_{F})/U and let u_{i} = π(h_{i})u. apart from an unimportant factor coming from the Haar measure φ_{r}(h) is given by

∑_{i=1}^{p} φ_{r}^{i}(h)

with

φ_{r}^{i}(h) = ∫ (π (( \begin{matrix}γ & γx_{1} \\ 0 & 1 \end{matrix} ) h ( \begin{matrix}γ & γx_{1} \\ 0 & 1 \end{matrix} )) u_{i}, u_{i}) dx_{1} d^{×}γ.

The integral is taken over the set of all those γ and x_{1} for which

( \begin{matrix}γ & γx_{1} \\ 0 & 1 \end{matrix} )

belongs to T_{r}. Since

( \begin{matrix}γ & γx_{1} \\ 0 & 1 \end{matrix} )^{−1} h ( \begin{matrix}γ & γx_{1} \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & (1 − \frac{β}{α})(γ^{−1}x + x_{1}) \\ 0 & 1 \end{matrix} )

we can change variables in the integral to obtain

|1 − \frac{β}{α}|^{−1} ∫ (π (( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & γ^{−1}(1 − \frac{β}{α})x + x_{1} \\ 0 & 1 \end{matrix} )) u_{i}, u_{i}) dx_{1} d^{×}γ.  (7.4.3)

The integration is now taken over all those x_{1} and γ for which

( \begin{matrix}γ & γ(1 − \frac{β}{α})^{−1}x_{1} \\ 0 & 1 \end{matrix} )  (7.4.4)

is in T_{r}.

Let |1 − \frac{β}{α}| = |ϖ_{F} |^{t}, |γ| = |ϖ_{F} |^{m}, and |x| = |ϖ_{F} |^{n}. Let ϖ_{F}^{p} and ϖ_{F}^{q} be the elementary divisors of the matrix (7.4.4). We now list the possibilities for p and q together with the condition that the matrix belong to T_{r}, that is that q − p be at most r.

(a) m ≥ 0, −t + m + n ≥ 0, p = 0, q = m : 0 ≤ m ≤ r

(b) m ≥ 0, −t + m + n ≤ 0, p = −t + m + n, q = n − t : −r ≤ m + 2n − 2t

(c) m ≤ 0, −t + m + n ≤ m, p = −t + m + n, q = n − t : −r ≤ m + 2n − 2t

(d) m ≤ 0, −t + m + n ≥ m, p = m, q = 0 : −r ≤ m ≤ 0. These conditions amount to the demand −r ≤ m ≤ r and that 2n ≥ 2t − r − m. On the other hand we know that there is an integer s such that

∫_{|x|≤|ϖ_{F} |^{j}} π \begin{aligned}( \\ ( & \begin{matrix}1 & x_{1} \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} u_{i} dx = 0

for 1 ≤ i ≤ p if j ≤ s.

Thus if |γ| = |ϖ_{F} |^{m} the integral

∫ (π (( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & γ^{−1}(1 − \frac{β}{α})_{1}^{x+x_{1}} \\ 0 & 1 \end{matrix} )) u_{i}, u_{i}) dx_{1}

Chapter 1 141 taken over all x_{1} for which

( \begin{matrix}γ & γ(1 − \frac{β}{α})x_{1} \\ 0 & 1 \end{matrix} )

is in T_{r} is zero if 2t − r − m ≤ 2s. Therefore in (7.4.3) we need only take the integral over those γ and x for which |γ| = |ϖ_{F} |^{m} with 0 ≤ m_{r} ≤ 2(t − s) and |x| ≤ |ϖ_{F} |^{t− \frac{m+r}{2}} . We should also have m ≤ r but since we are about to replace the integrand by its absolute value that does not matter. For each such γ the integration with respect to x gives a result which is bounded in absolute value by a constant times |ϖ_{F} |^{t− \frac{m+r}{2}} . Integrating with respect to γ we obtain a result which is bounded in absolute value by a constant times

|ϖ_{F} |^{t} ∑_{k=0}^{2(t−s)−1} |ϖ_{F} |^{− \frac{k}{2}} ≤ |ϖ_{f}|^{s} ∑_{k=0}^{∞} |ϖ_{F} |^{\frac{k}{2}}

The right side depends on neither r nor t.

The value of |1 − \frac{β}{α}|φ_{r}^{i}(h) is

∫ (π \begin{aligned}( \\ ( & \begin{matrix}α & 0 \\ 0 & β \end{matrix} & ) \\ & & ) \end{aligned} π (( \begin{matrix}1 & ϖ_{F}^{r} γ^{−1}(1 − \frac{β}{α})x + x_{1} \\ 0 & 1 \end{matrix} )) u_{i}, u_{i}) dx_{1} d^{×}γ.

The integration is taken over those γ and x_{1} for which |γ| = |ϖ_{F} |^{m} with 0 ≤ m < 2(t − s) and |x| ≤ |ϖ_{F} |^{t− \frac{m}{2}} . Of course |1 − \frac{β}{α}| = |ϖ_{F} |^{t}. Since we are now interested in a set of α and β on which t takes only a finite number of values we may as well assume it is constant. Then the integral is taken over a fixed compact subset of F × F^{×}. The integrand converges uniformly on this set uniformly in the α, β and x under consideration as r approaches infinity.

We have still to prove the existence of the character of a representation which is not absolutely cuspidal. Most of them are taken care of by the next proposition.

Proposition 7.6 Let μ_{1} and μ_{2} be a pair of quasi-characters of F^{×}. Let χ_{μ_{1},μ_{2}} be the function which is 0 on Ĝ_{F} ∩ G̃_{F}, undefined on the singular elements, and equal to

\{μ_{1}(α) μ_{2}(β) + μ_{2}(β) μ_{1}(α)\} |\frac{αβ}{(α − β)^{2}} |^{1/2}

at an element of g of Â_{F}^{G} with eigenvalues α and β. Then χ_{μ_{1},μ_{2}} is continuous on Ĝ_{F} and is dominated in absolute value by some multiple of ξ. Moreover if π = ρ(μ_{1}, μ_{2})

Tr π(f) = ∫_{G_{F}} χ_{μ_{1},μ_{2}}(g) f(g) dg

for all f in ℋ_{F}.

Only the last assertion requires verification. Since the absolute value of χ_{μ_{1},μ_{2}} is bounded by a multiple of ξ the function χ_{μ_{1},μ_{2}} is locally integrable. Suppose f belongs to ℋ_{F}. When applied to the function χ_{μ_{1},μ_{2}}f the relation (7.2.1) shows that

∫_{G_{F}} χ_{μ_{1},μ_{2}}(g) f(g) dg  (7.6.1)

\frac{1}{2} ∫_{A_{F}} δ(a) \{∫_{A_{F} \G_{F}} χ_{μ_{1},μ_{2}}(g^{−1}ag) f(g^{−1}ag) dg\} da.

Since χ_{μ_{1},μ_{2}} is a class function this may be written as

\frac{1}{2} ∫_{A_{F}} \{μ_{1}(α) μ_{2}(β) + μ_{2}(α) μ_{1}(β)\} |\frac{(α − β)^{2}}{αβ} |^{1/2} \{∫_{A_{F} \G_{F}} f(g^{−1} ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) g) dg\} da

if

a = ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) .

Since a is conjugate to

( \begin{matrix}β & 0 \\ 0 & α \end{matrix} )

we have

∫_{A_{F} \G_{F}} f (g^{−1} ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) g) dg = ∫_{A_{F} \G_{F}} f (g^{−1} ( \begin{matrix}β & 0 \\ 0 & α \end{matrix} ) g) dg.

Thus (7.6.1) is equal to

∫_{A_{F}} μ_{1}(α) μ_{2}(β) |\frac{(α − β)^{2}}{αβ} |^{1/2} \{∫_{A_{F} \G_{F}} f (g^{−1} ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) g) dg\} da.  (7.6.2)

As long as the measure on A_{F} \ G_{F} is the quotient of the measure on G_{F} by that on A_{F} the choice of Haar measure on A_{F} and G_{F} is not relevant. Thus we may write (7.6.2) as

∫_{A_{F}} μ_{1}(α) μ_{2}(β) |\frac{(α − β)^{2}}{αβ} |^{1/2} (f (k^{−1}n^{−1} ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) nk) dk dn\} da.

The inner integral is taken over GL(2, O_{F}) × N_{F}. If

n = ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )

then

n^{−1} ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) n = ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & (1 − \frac{β}{α})x \\ 0 & 1 \end{matrix} )

Changing variables in the last integral we obtain

∫_{A_{F}} μ_{1}(α) μ_{2}(β) |\frac{α}{β}|^{1/2}) \{∫ f (k^{−1} ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) nk) dk dn\} da.  (7.6.3)

To evaluate Tr π(f) we observe that if φ belongs to ℬ(μ_{1}, μ_{2}) then, if k_{1} is in GL(2, O_{F})

π(f) φ(k_{1}) = ∫_{G_{F}} φ(k_{1}g) f(g) dg.

Chapter 1 143 Replacing g by k_{1}^{−1}g and writing the integral out in terms of the Haar measure we have chosen we obtain

∫_{GL(2,O_{F})} φ(k_{1}) \{∫ f(k_{1}^{−1} ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) nk_{2})μ_{1}(α) μ_{2}(β) |\frac{α}{β}|^{1/2} da dn\} dk_{2}.

The inner integral is taken over A_{F} × N_{F}. We have of course used the relation

φ (( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) nk_{2}) = μ_{1}(α)μ_{2}(β) |\frac{α}{β}|^{1/2}φ(k_{2}).

If

K(k_{1}, k_{2}) = ∫ f (k_{1}^{−1} ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) nk_{2}) μ_{1}(α) μ_{2}(β) |\frac{α}{β}|^{1/2} da dn

then

π(f) φ(k_{1}) = ∫_{GL(2,O_{F})} K(k_{1}, k_{2}) φ(k_{2}) dk_{2}.

ℬ(μ_{1}, μ_{2}) may be regarded as a space of functions on GL(2, O_{F}). Then π(f) is the integral operator with kernel K(k_{1}, k_{2}). It is easily seen that this operator, when allowed to act on the space of all GL(2, O_{F})-finite functions on GL(2, O_{F}), has range in ℬ(μ_{1}, μ_{2}). Thus the trace of π(f) is the same as the trace of the integral operator which is of course

∫_{GL(2,O_{F})} K(k, k) dk.

When written out in full this integral becomes (7.6.3).

Theorem 7.7 Let π be an irreducible admissible representation of ℋ_{F}. There is a function χ_{π} which is continuous on G_{F} and locally bounded in absolute value of Ĝ_{F} by a multiple of ξ such that

Tr π(f) = ∫_{G_{F}} χ_{π}(g) f(g) dg

for all f in ℋ_{F}.

The theorem has only to be verified for the one-dimensional and the special representations. If π is a one-dimensional representation associated to the quasi-character χ we may take χ_{π}(g) = χ(detg). χ_{π} is locally bounded and therefore, by Lemma 7.3, locally bounded by a multiple of ξ.

Suppose π_{1}, π_{2} and π_{3} are three admissible representations of F on space V_{1}, V_{2}, and V_{3} respectively. Suppose also that there is an exact sequence

0 → V_{1} → V_{2} → V_{3} → 0

of ℋ_{F} -modules. If f is in ℋ_{F} all the operators π_{1}(f), π_{2}(f) and π_{3}(f) are of finite rank so that

Tr π_{2}(f) = Tr π_{1}(f) + Tr π_{3}(f).

Thus if χ_{π_{1}} and χ_{π_{2}} exist so does χ_{π_{3}}. Applying this observation to π_{3} = σ(μ_{1}, μ_{2}), π_{2} = ρ(μ_{1}, μ_{2}), and π_{1} = π(μ_{1}, μ_{2}) we obtain the theorem.

If F is taken to be the real or complex field Theorem 7.7 is a special case of a general and difficult theorem of Harish-Chandra. The special case is proved rather easily however. In fact Proposition 7.6 is clearly valid for archimedean fields and Theorem 7.7 is clearly valid for archimedean fields if π is finite-dimensional. There remains only the special representations and these are taken care of as before.

Chapter 1 144 §8. Odds and ends In this paragraph various facts which will be used in the discussion of the constant term in the Fourier expansion of an automorphic form are collected together. If H is a locally compact abelian group a continuous complex-valued function f on H will be called H-finite or simply finite if the space spanned by the translates of f is finite dimensional.

Let H be a group of the form

H = H_{0} × ℤ^{m} × ℝ^{n}

where H_{0} is compact. We regard ℤ^{m} × ℝ^{n} as a subgroup of ℝ^{m+n}. The projection

ξ_{i} : h = (h_{0}, x_{1}, ⋯, x_{m+n}) → x_{i}

may be regarded as a function on H with values in ℝ. If p_{1}, ⋯, p_{m+n} is a sequence of non-negative integers and χ is a quasi-character we may introduce the function

χ ∏_{i=1}^{m+n} ξ_{i}^{p_{i}}

on H.

Lemma 8.1 For any sequence p_{1}, ⋯, p_{m+n} and any quasi-character χ the function χ ∏_{i=1}^{m+n} ξ_{i}^{p_{i}} is continuous and finite. These functions form a basis of the space of continuous finite functions on H.

If χ is a fixed quasi-character of H and p is a non-negative integer let V (χ, p) be the space spanned by the functions χ ∏_{i=1}^{m+n} ξ_{i}^{p_{i}} with 0 ≤ p_{i} ≤ p. Since it is finite-dimensional and invariant under translations the first assertion of the lemma is clear.

To show that these functions are linearly independent we shall use the following simple lemma.

Lemma 8.1.1 Suppose E_{1}, ⋯, E_{r} are r sets and ℱ_{1}, ⋯ ℱ_{r} are linearly independent sets of complexvalued functions on E_{1}, ⋯, E_{r} respectively. Let ℱ be the set of functions

(x_{1}, ⋯, x_{r}) → f_{1}(x_{1}) f_{2}(x_{2}) ⋯ f_{r}(x_{r})

on E_{1} × ⋯ × E_{r}. Here f_{i} belongs to ℱ_{i}. Then ℱ is also linearly independent.

Any relation

∑_{f_{1},···,f_{r}} a(f_{1}, ⋯, f_{r}) f_{1}(x_{1}) ⋯ f_{r}(x_{r}) ≡ 0

leads to

∑_{f_{r}} { ∑_{f_{1},···,f_{r−1}} a(f_{1}, ⋯, f_{r}) f_{1}(x_{1}) ⋯ f_{r−1}(x_{r−1})} f_{r}(x_{r}) ≡ 0

As ℱ_{r} is linearly independent this implies that

∑_{f_{1},···f_{r−1}} a(f_{1}, ⋯, f_{r}) f_{1}(x_{1}) ⋯ f_{r−1}(x_{r−1}) ≡ 0

and the lemma follows by induction.

To show that the functions χ ∏_{i=1}^{m+n} ξ_{i}^{p_{i}} span the space of continuous finite functions we use another simple lemma.

Lemma 8.1.2. Let H_{1} and H_{2} be two locally compact abelian groups and let H = H_{1} × H_{2}. Then every continuous finite function f on H is a finite linear combination of the form

f(x, y) = ∑_{i} λ_{i} φ_{i}(x) ψ_{i}(y)

where the φ_{i} and ψ_{i} are continuous functions on H_{1} and H_{2} respectively.

Let V be any finite dimensional space of continuous functions on H. We associate to any point ξ in H the linear functional f → f(ξ) on V. Since no function but zero is annihilated by all these functionals we can choose ξ_{1}, ⋯, ξ_{p} so that the corresponding functionals form a basis of the dual of V. Then we can choose a basis f_{1}, ⋯, f_{p} of V so that f_{i}(ξ_{j}) = δ_{ij}.

Now suppose V is invariant under translations. It could for example be the space spanned by the translates of a single finite continuous function. The space V_{1} of functions φ on H_{1} defined by φ(x) = f(x, 0) with f in V is finite dimensional and translation invariant. Therefore the functions in it are finite and of course continuous. We define V_{2} in a similar manner. If f is in V the function h → f(g + h) is, for any g in H, also in V. Thus

f(g + h) = ∑_{i} λ_{i}(g) f_{i}(h).

Since

λ_{i}(g) = f(g + ξ_{i})

the function λ_{i} belongs to V. If φ_{i}(x) = λ_{i}(x, 0) and ψ_{i}(y) = f_{i}(0, y) then

f(x, y) = ∑ φ_{i}(x) ψ_{i}(y)

as required.

These two lemmas show that we need prove the final assertions of Lemma 8.1 only for H compact, H = ℤ, or H = ℝ.

Suppose H is compact. If we have a non-trivial relation

∑_{i=1}^{r} a_{i} χ_{i}(h) ≡ 0

we may replace h be g + h to obtain

∑_{i=1}^{r} a_{i} χ_{i}(g) χ_{i}(h) ≡ 0.

If such a relation holds we must have r ≥ 2 and at least two coefficients say a_{1} and a_{2} must be different from zero. Choose g so that χ_{1}(g) ≠ χ_{2}(g). Multiplying the first relation by χ_{1}(g) and subtracting the second relation from the result we obtain a relation

∑_{i=2}^{r} b_{i} χ_{i}(h) ≡ 0.

Chapter 1 146 Since b_{2} = {χ_{1}(g) − χ_{2}(g)}a_{2} the new relation is non-trivial. The independence of the quasicharacters can therefore be proved by induction on r.

To prove that when H is compact the quasi-characters span the space of finite continuous functions we have just to show that any finite-dimensional space V of continuous functions which is translation invariant is spanned by the quasi-characters it contains. Choose a basis {f_{i}} of V as before and let

ρ(g) f_{i} = ∑ λ_{ij}(g) f_{j}.

We saw that the functions λ_{ij}(g) are continuous. Thus the action of H on V by right translations is continuous and V is the direct sum of one-dimensional translation invariant spaces. Each such space is easily seen to contain a character.

When applied to a locally compact abelian group the argument of the previous paragraph leads to weaker conclusions. We can then find subspaces V_{1}, ⋯, V_{r} of V and quasi-characters χ_{1}, ⋯, χ_{r} of H such that

V = ∑_{i=1}^{r} ⊕V_{i}

and, for every h in H,

{ρ(h) − χ_{i}(h)}^{dim V_{i}}

annihilates V_{i}. Now we want to take H equal to ℤ or ℝ. Then H is not the union of a finite number of proper closed subgroups. Suppose μ_{1}, ⋯, μ_{2} are quasi-characters of H and for every h in H the operator

∏_{i=1}^{s}{ρ(h) − μ_{i}(h)}  (8.1.3)

on V is singular. Then for every h in H there is an i and a j such that μ_{i}(h) = χ_{j}(h). If

H_{ij} = {h | μ_{i}(h) = χ_{j}(h)}

then H_{ij} is a closed subgroup of H. Since the union of these closed subgroups is H there must be an i and a j such that H_{ij} = H and μ_{i} = χ_{j}. If the operator (8.1.3) were zero the same argument would show that for every j there is an i such that μ_{i} = χ_{j}.

If μ is a quasi-character of H, now taken to be ℤ or ℝ, we let V (μ, p) be the space spanned by the functions μ ξ^{i}, with 0 ≤ i ≤ p. ξ is the coordinate function on H. It is clear that V (μ, p) is annihilated by {ρ(h) − μ(h)}^{p+1} for all h in H. Suppose μ, μ_{1}, ⋯, μ_{2} are distinct and

V = V (μ, p) ∑_{i=1}^{s} V (μ_{i}, p_{i})

is not zero. Decomposing V as above we see that χ_{1}, ⋯, χ_{r} must all be equal to μ on one hand and on the other that every μ_{i} is a χ_{j}. This is a contradiction. Thus if there is any non-trivial relation at all between the functions χ ξ^{i} where χ is any quasi-character and i is a non-negative integer there is one of the form

∑_{i=0}^{p} a_{i} μ ξ^{i} = 0.

Since the polynomial ∑_{i=0}^{p} a_{i} ξ^{i} would then have an infinite number of zeros this is impossible.

Chapter 1 147

To prove the functions χ ξ^{i} span the space of finite continuous functions we have only to show that if χ is a given quasi-character and V is a finite-dimensional space of continuous functions which is invariant under translations and annihilated by {ρ(h) − χ(h)}^{dim V} for all h in H then every function in V is the product of χ and a polynomial. Since we can always multiply the functions in V by χ^{−1} we may as well suppose that χ is trivial. We have only to observe that any function f annihilated by the operator {ρ(h) − 1}^{n} for all h in H is a polynomial of degree at most n. This is clear if n = 1 so by induction we can assume that ρ(h)f − f is a polynomial ∑_{i=0}^{n−1} a_{i}(h) ξ^{i}. We can certainly find a polynomial f′ of degree n such that

ρ(1)f′ − f′ = ∑_{i=0}^{n−1} a_{i}(1) ξ^{i}

and we may as well replace f by f − f′. The new f satisfies ρ(1)f = f. It is therefore bounded. Moreover ρ(h)f − f is a bounded polynomial function and therefore a constant c(h). c(h) is a bounded function of h and satisfies c(h_{1} + h_{2}) = c(h_{1}) + c(h_{2}). It is therefore zero and the new f is a constant.

Lemma 8.1 is now completely proved. Although it is trivial it is important to the notes and we thought it best to provide a proof. We might as well prove Lemma 2.16.4 at the same time. Let B be the space of all functions f on ℤ such that for some n_{0} depending on f we have f(n) = 0 for n < n_{0}. Let A_{0} be the space of functions on ℤ which vanish outside a finite set. ℤ acts on B and on A_{0} by right translations and therefore it also acts on B̄ = B/A_{0}. In particular let D = ρ(1). We have merely to show that if P is a polynomial with leading coefficient 1 then the null space of P (D) in B̄ is finite dimensional. If

P (X) = ∏_{i=1}^{r}(X − α_{i})^{p_{i}}

the null space of P (D) is the direct sum of the null spaces of the operators (D − α_{i})^{p_{i}}. The null space of (D − α)^{p} is the image in B̄ of the functions in B which are zero to the left of 0 and of the form

n → α^{n} Q(n)

to the right of 0. Q is a polynomial of degree at most p.

Lemma 8.1 is certainly applicable to the direct product of a finite number of copies of the multiplicative group of a local field F. If H = (F^{×})^{n} any finite continuous function on H is a linear combination of functions of the form

f(x_{1}, ⋯, x_{n}) = ∏_{i=1}^{n}{χ_{i}(x_{i}) (log |x_{i}|_{F})^{n_{i}}}.

Let ℬ = ℬ_{F} be the space of continuous functions f on G_{F} which satisfy the following three conditions.

(i) f is finite on the right under the standard maximal compact subgroup K of G_{F}.

(ii) f is invariant on the left under N_{F}. (iii) f is A_{F} -finite on the left. ℬ_{F} is invariant under left translations by elements of A_{F}. If f is in ℬ_{F} let V be the finitedimensional space generated by these left translates. Choose g_{1}, ⋯, g_{p} in G_{F} so that the linear functions φ → φ(g_{i}) are a basis of the dual of V and let f_{1}, ⋯, f_{p} be the dual basis. If a is A_{F} we may write

f(a, g) = ∑_{i=1}^{p} θ_{i}(a) f_{i}(g).

Then

θ_{i}(a) = f(ag_{i})

so that

θ_{i}(ab) = ∑_{j=1}^{p} θ_{j}(a) f_{j}(bg_{i}).

Thus the functions θ_{i} are continuous and finite. We may write them in the form

θ_{i}(a) = ∑ c_{m,n,μ,ν}^{i} μ(α_{1}) ν(α_{2}) (log |α_{1}|)^{m} (log |α_{2}|)^{n}

if

a = ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) .

The sum is over all quasi-characters μ and ν of F^{×} and all non-negative integers m and n. Of course only a finite number of the coefficients c_{m,n,μ,ν}^{i} are different from zero.

We may replace μ by α_{F}^{1/2}μ and ν by α_{F}^{−1/2}ν in the sum. Thus if

f_{m,n,μ,ν} = ∑_{i=1}^{p} c_{m,n,μ,ν}^{i} f_{i}

we have

f(ag) = |\frac{α_{1}}{α_{2}} |^{1/2} ∑ μ(α_{1}) ν(α_{2}) (log |α_{1}|)^{m} (log |α_{2}|)^{n} f_{m,n,μ,ν}(g).  (8.2)

Let M be a non-negative integer and S a finite set of pairs of quasi-characters of F^{×}. ℬ(S, M) will be the sset of f in for which the sum in (8.2) need only be taken over those m, n, μ, ν for which m + n ≤ M and (μ, ν) belong to S. Observe that the functions f_{m,n,μ,ν} are determined by f. ℬ is the union of the spaces ℬ(S, M); if S consists of the single pair (μ_{1}, μ_{2}) we write ℬ(μ_{1}, μ_{2}, M) instead of ℬ(S, M). If f is in (μ_{1}, μ_{2}, M)

f(ag) = |\frac{α_{1}}{α_{2}} |^{1/2} μ_{1}(α_{1}) μ_{2}(α_{2}) ∑(log |a_{1}|)^{m} (log |a_{2}|)^{n} f_{m,n}(g).

The space ℬ(μ_{1}, μ_{2}, 0) is just ℬ(μ_{1}, μ_{2}).

The functions f_{m,n,μ,ν} are uniquely determined and by their construction belong to the space spanned by left translates of f by elements of A_{F}. Thus if f belongs to ℬ(S, M) so do the functions f_{m,n,μ,ν}. We want to verify that f_{0,0,μ,ν} belongs to ℬ(μ, ν, M). If

b = ( \begin{matrix}β_{1} & 0 \\ 0 & β_{2} \end{matrix} )

and we replace a by ab in the relation (8.2) we find that

|\frac{α_{1}}{α_{2}} |^{1/2} ∑ μ(α_{1}) ν(α_{2}) (log |α_{1}|)^{m} (log |α_{2}|)^{n}) f_{m,n,μ,ν}

is equal to

|\frac{α_{1}β_{1}}{α_{2}β_{2}} |^{1/2} ∑ μ(α_{1}β_{1}) ν(α_{2}β_{2})(log |α_{1}| + log |β_{2}|)^{m} (log |α_{2}| + log |β_{2}|)^{n} f_{m,n,μ,ν}(g).

Fix b and g are regard this equality as an identity in the variable a. Because of Lemma 8.1 we can compare the coefficients of the basic finite functions. The coefficient of μ(α_{1})ν(α_{2}) on one side is f_{0,0,μ,ν}(bg). On the other it is

|\frac{β_{1}}{β_{2}} |^{1/2} ∑_{m+n≤M} μ(β_{1}) ν(β_{2}) (log |β_{1}|)^{m} (log |β_{2}|)^{n} f_{m,n,μ,ν}(g).

The resulting identity is the one we wanted to verify.

Taking a = 1 in (8.2) we see that

f(g) = ∑_{(μ,ν)∈S} f_{0,0,μ,ν}(g).

Therefore

ℬ(S, M) = ∑_{(μ,ν)∈S} ℬ(μ, ν, M).

The sum is direct.

It is fortunately possible to give a simple characterization of ℬ.

Proposition 8.3 Let φ be a continuous function on G_{F}. Assume φ is K-finite on the right and invariant under N_{F} on the left. Then φ belongs to ℬ if and only if the space

{ρ(ξf) φ | f ∈ ℋ_{F} }

is finite dimensional for every elementary idempotent in ℋ_{F}.

We have first to show that if φ belongs to ℬ

{ρ(ξf) φ | f ∈ ℋ_{F} }

is finite dimensional. φ belongs to some ℬ(S, M). Both ℬ and ℬ(S, M) are invariant under right translations by elements of ℋ_{F}. Thus we have only to show that the range of ρ(ξ) as an operator on ℬ(S, M) is finite dimensional. This is tantamount to showing that any irreducible representation of K occurs with finite multiplicity in the representation of ℬ(S, M).

Let σ be such a representation and let V be the space of continuous functions on K which transform according to σ under right translations. V is finite dimensional. If f is in ℬ(S, M) we may write

f(ag) |\frac{α_{1}}{α_{2}} |^{1/2} ∑ μ(α_{1}) ν(α_{2}) (log |α_{1}|)^{m}) (log |α_{2}|)^{n} f_{m,n,μ,ν}(g)

Chapter 1 150 The restriction of f_{m,n,μ,ν} to K lies in V. Call this restriction f̄_{m,n,μ,ν}. Moreover f is determined by its restriction to A_{F} K. Thus

f → ∑_{{(μ,ν)∈S \atop m+n≤M}} ⊕f̄_{m,n,μ,ν}

is an injection of the space of functions under consideration into the direct sum of a finite number of copies of V.

The converse is more complicated. Suppose φ is K-finite on the right, invariant under N_{F} on the left, and the space

{ρ(ξf) φ | f ∈ ℋ_{F} }

is finite dimensional for every elementary idempotent ξ. Choose ξ so that ρ(ε)φ = φ. There is actually a function f in ξℋ_{F} ξ such that ρ(f)φ = φ. If F is non-archimedean ξ is itself a function so this is clear. If F is archimedean we observe that if f_{1} is an approximation to the δ-function then ρ(f_{1})φ is close to φ. Then if f′_{1} = ξ ∗ f_{1} ∗ ξ the function f′_{1} is in ξℋ_{F} ξ and ρ(f′_{1})φ is also close to φ. The existence of f then follows from the fact that ρ(ξℋ_{F} ξ)φ is finite dimensional. This argument was used before in Paragraph 5.

Take F to be archimedean. Then φ must be an infinitely differentiable function on G_{F}. Let ℨ be the centre of the universal enveloping algebra of the Lie algebra of G_{F}. If Z is in ℨ then

ρ(Z)φ = ρ(Z) ρ(f)φ = ρ(Z ∗ f)φ

and Z ∗ f is still in ξℋ_{F} ξ. Thus φ is also ℨ-finite. For the rest of the proof in the archimedean case we refer to Chapter I of [11].

Now take F non-archimedean. We may replace ξ by any elementary idempotent ξ′ for which ξ′ξ = ξ. In particular if we choose n to be a sufficiently large positive integer and let K′ be the elements of K which are congruent to the identity modulo 𝔭^{n} we may take

ξ = ∑ ξ_{i}

where the sum is over all elementary idempotents corresponding to irreducible representations of K whose kernel contains K′. Notice that n is at least 1. Then ξℋ_{F} ξ is the space of functions on G_{F} which are constant on double cosets of K′.

Let V be the space spanned by the functions ρ(k)φ with k in K. It is finite dimensional and all the functions in V satisfy the same conditions as φ. Let φ_{i}, 1 ≤ i ≤ p, be a basis of V. If k belongs to K we may write

φ(gk) = ∑_{i=1}^{p} θ_{i}(k) φ_{k}(g)

and φ is determined by the functions θ_{i} and the restrictions of the functions φ_{i} to A_{F}. To show that φ is A_{F} -finite on the left we have merely to show that the restriction of each φ_{i} to A_{F} is finite. We may as well just show that the restriction of φ to A_{F} is finite.

Suppose f is in ξℋ_{F} ξ and ρ(f)φ = φ. If a is in Z_{F} then

λ(a)φ = ρ(a^{−1})φ = ρ(δ_{a^{−1}} ∗ f)φ

if δ_{a^{−1}} is the δ-function at a^{−1}. Since δ_{a^{−1}} ∗ f is still in ξℋ_{F} ξ the function φ is certainly Z_{F} -finite and so is its restriction φ̄ to A_{F}. If α and β are units and α ≡ β ≡ 1(mod 𝔭^{n}) then

λ \begin{aligned}( \\ ( & \begin{matrix}α & 0 \\ 0 & β \end{matrix} & ) \\ & & ) \end{aligned} φ̄ = φ̄.

Thus the translate of φ̄ by the elements of A_{f} ∩ K span a finite dimensional space and if ϖ is a generator of 𝔭 we have only to show that the translates of φ̄ by the group

H = \{( \begin{matrix}ϖp & 0 \\ 0 & 1 \end{matrix} ) | p ∈ ℤ\}

span a finite-dimensional space. Suppose the span W of

\{λ \begin{aligned}( \\ ( & \begin{matrix}ϖp & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ̄ | p ≤ 0\}

is finite dimensional. Then

λ \begin{aligned}( \\ ( & \begin{matrix}ϖ^{−1} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

maps W into itself and annihilates no vector but zero so that it has an inverse on W which must be

λ \begin{aligned}( \\ ( & \begin{matrix}ϖ & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} .

Thus W is invariant under H and φ̄ is finite.

To show that W is finite dimensional we show that if

a = ( \begin{matrix}ϖ^{−p} & 0 \\ 0 & 1 \end{matrix} )

with p > 0 there is a function f_{a} in ξℋ_{F} ξ such that

λ(a)φ̄ = φ̄′

if φ′ = ρ(f_{a})φ. There is an f in ξℋ_{F} ξ such that

φ(g) = ∫_{G_{F}} φ(gh) f(h) dh

for all g in G_{F}. Thus if b belongs to A_{F}

λ(a) φ̄(b) = φ(a^{−1}b) = ∫_{G_{F}} φ(ba^{−1}h) f(h) dh.

If f_{1}(h) = f(ah) the integral is equal to

∫_{G_{F}} φ(bh) f_{1}(h) dh.

If f_{1} were in ξℋ_{F} ξ we would be done. Unfortunately this may not be so. However f_{1}(hk) = f_{1}(h) if k belongs to K′. If

k = ( \begin{matrix}α & β \\ γ & δ \end{matrix} )

then

f_{1}(kh) = f (( \begin{matrix}α & ϖ^{−p}β \\ ϖ^{p}γ & δ \end{matrix} ) ah) .

Thus f_{1}(kh) = f_{1}(h) if α ≡ δ ≡ 1 (mod 𝔭^{n}), γ ≡ 0 (mod 𝔭^{n}), and β ≡ 0 (mod 𝔭^{n+p}). Set

f_{2}(h) = ∫_{𝔭^{n}/𝔭^{n+p}} f_{1} (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) h) dx

where the Haar measure is so chosen that the measure of the underlying space 𝔭^{n}/𝔭^{n+p} is 1. Since φ(bnh) = φ(bh) for all n in N_{F}

λ(a) φ̄(b) = ∫_{G_{F}} φ(bh) f_{2}(h) dh.

We show that f_{2} lies in ξℋ_{F} ξ.

Certainly f_{2}(hk) = f_{2}(h) if k is in K′. Moreover, because of its construction, f_{2}(kh) = f_{2}(h) if

k = ( \begin{matrix}α & β \\ 0 & δ \end{matrix} )

with α ≡ δ ≡ 1 (mod 𝔭^{n}) and β ≡ 0 (mod 𝔭^{n}). Since every element of K′ is a product

( \begin{matrix}1 & 0 \\ γ & 1 \end{matrix} ) ( \begin{matrix}α & β \\ 0 & δ \end{matrix} )

where both terms lie in K′ we have only to show that f_{2} is invariant under the first factor. If

k = ( \begin{matrix}1 & 0 \\ γ & 1 \end{matrix} )

with γ ≡ 0 (mod 𝔭^{n}) and

k_{1}(x) = ( \begin{matrix}1 & 0 \\ 0 & 1 + xγ \end{matrix} ) ( \begin{matrix}\frac{1}{1+xγ} & 0 \\ \frac{−γ}{1+xγ} & 1 \end{matrix} )

then

k_{1}(x) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) k = ( \begin{matrix}1 & \frac{x}{1+xγ} \\ 0 & 1 \end{matrix} ) .

Moreover if x is in O_{F}

f_{1}(k_{1}(x)g) = f_{1}(g).

Thus f_{2}(kg) which is given by

∫_{𝔭^{n}/𝔭^{n+p}} f_{1} (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) kh) dx

Chapter 1 153 is equal to

∫_{𝔭^{n}/𝔭^{n+p}} f_{1} (( \begin{matrix}1 & \frac{x}{1+xγ} \\ 0 & 1 \end{matrix} ) h) dx.

Since the map x → \frac{x}{1+xγ} is a one-to-one map of the finite set 𝔭^{n}/𝔭^{n+p} onto itself it is measure preserving and the above integral is equal to f_{2}(h).

Analyzing the above proof one sees that in the non-archimedean case the left translates of φ̄ are contained in the space X obtained by restricting the functions in ρ(ξℋ_{F} ξ)φ to A_{F}. Thus if Y is the space of the functions on K/K′ the left translates of φ by elements of A_{F} are contained in the space of functions on N_{F} \ G_{F} of the form

φ′(ak) = ∑ θ_{i}(k) φ_{i}(a)

with θ_{i} in Y and φ_{i} in X.

In the archimedean case Y is the space of continuous functions θ on K for which θ ∗ ξ = ξ ∗ θ = θ. It is again finite dimensional. X is defined in the same way. In this case there are a finite number of invariant differential operators D_{1}, ⋯, D_{r} on A_{F} such that the left translates of φ by elements of A_{F} are contained in the space of functions N_{F} \ G_{F} of the form

φ′(ak) = ∑ θ_{i}(k) φ_{i}(a)

with θ_{i} in Y and φ_{i} in ∑_{j=1}^{r} D_{j}X.

There is a corollary of these observations. Let F_{1}, ⋯, F_{n} be a finite collection of local fields. Let G_{i} = G_{F_{i}}, N_{i} = N_{F_{i}}, A_{i} = A_{F_{i}}, and let K_{i} be the standard maximal compact subgroup of G_{i}. We set G = ∏_{i=1}^{n} G_{i}, N = ∏_{i=1}^{n} N_{i} and so on. If ℋ_{i} = ℋ_{F_{i}} we let ℋ = ⊗_{i}ℋ_{i}. ℋ may be regarded as an algebra of measures on G.

Corollary 8.4 Let φ be a continuous function on N \ G which is K-finite on the right. If for every elementary idempotent ξ in ℋ the space

{ρ(ξf)φ | f ∈ ℋ}

is finite dimensional φ is A-finite on the left.

If φ satisfies the conditions of the lemma so does any left translate by an element of A. Thus we need only show that φ is A_{i}-finite on the left for each i. If g is in G we write g = (g_{i}, ĝ_{i}) where g_{i} is in G_{i} and ĝ_{i} is in Ĝ_{i} = ∏_{j≠i} G_{j}. We may suppose that there is a ξ′ of the form ξ′ = ⊗_{i}ξ′_{i} where ξ′_{i} is an elementary idempotent of ℋ_{i} such that ρ(ξ′)φ = φ. By means of the imbedding f → f ⊗ ∏_{j≠i} ξ′_{j} the algebra ℋ_{i} becomes a subalgebra of ℋ. The left translates of φ by A_{i} all lie in the space of functions of the form

φ(a_{i}k_{i}, ĝ_{1}) = ∑_{j} θ_{j}(k_{i}) φ(a_{i}, ĝ_{1})

where θ_{j} lie in a certain finite dimensional space determined by ξ′_{i} and the φ_{j} lie in the space obtained by restricting the functions in ρ(ξ_{i}ℋ_{i})φ to A_{i} × Ĝ_{i} or, in the archimedean case, the space obtained from this space by applying certain invariant differential operators. ξ_{i} is a certain elementary idempotent which may be different from ξ′_{i}.

With the odds taken care of we come to the ends.

Chapter 1 154 Proposition 8.5 Let ℬ(μ, ν, ∞) = ⋃_{M≥0} ℬ(μ, ν, M). If an irreducible admissible representation π of ℋ_{F} is a constituent of the representation ρ(μ, ν, ∞) on ℬ(μ, ν, ∞) it is a constituent of ρ(μ, ν).

There are two invariant subspaces V_{1} and V_{2} of ℬ(μ, ν, ∞) such that V_{1} contains V_{2} and π is equivalent to the representation on ℋ_{F} on V_{1}/V_{2}. Choose M so that V_{1} ∩ ℬ(μ, ν, M) is not contained in V_{2}. Since π is irreducible

V_{1} = V_{2} + (V_{1} ∩ ℬ(μ, ν, M))

and

V_{1}/V_{2} = {V_{2} + (V_{1} ∩ ℬ(μ, ν, M))}/V_{2}

is isomorphic as an ℋ_{F} module to

V_{1} ∩ ℬ(μ, ν, M)/V_{2} ∩ ℬ(μ, ν, M)

so that we may as well suppose that V_{1} is contained in ℬ(μ, ν, M).

Given π we choose M as small as possible. If M = 0 there is nothing to prove so assume M is positive. If φ is in ℬ(μ, ν, M) we can express

φ (( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) g)

as

|\frac{α_{1}}{α_{2}} |^{1/2} μ(α_{1}) ν(α_{2}) ∑_{m+n≤M}(log |α_{1}|)^{m} (log |α_{2}|)^{n} φ_{m,n}(g)

We can express

φ (( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) ( \begin{matrix}β_{1} & 0 \\ 0 & β_{2} \end{matrix} ) g)

in two ways because the second factor can be absorbed into the first or the third. One way we obtain

|\frac{α_{1}}{α_{2}} |^{1/2} μ(α_{1}) ν(α_{2}) ∑_{m+n≤M}(log |α_{1}|)^{m}(log |α_{2}|)^{n} φ_{m,n} (( \begin{matrix}β_{1} & 0 \\ 0 & β_{2} \end{matrix} ) g)

and the other way we obtain

|\frac{α_{1}β_{1}}{α_{2}β_{2}} |^{1/2} μ(α_{1}β_{1}) ν(α_{2}β_{2}) ∑_{m+n≤M}(log |α_{1}| + log |β_{1}|)^{m} (log |α_{2}| + log |β_{2}|)^{n}φ_{m,n}(g).

On comparing coefficients we see that if m + n = M

φ_{m,n} (( \begin{matrix}β_{1} & 0 \\ 0 & β_{2} \end{matrix} ) g) = |\frac{β_{1}}{β_{2}} |^{1/2} μ(β_{1}) ν(β_{2}) φ_{m,n}(g)

so that φ_{m,n} is in ℬ(μ, ν). Consider the map

φ → ⨁_{m+n=M} φ_{m,n}

of V_{1} into

⨁_{m+n=M} ℬ(μ, ν).

Its kernel is V_{1} ∩ ℬ(μ, ν, M − 1). Since V_{2} + (V_{1} ∩ ℬ(μ, ν, M − 1)) cannot be V_{1} the image of V_{2} is not the same as the image of V_{1}. Since the map clearly commutes with the action of ℋ_{F} the representation π is a constituent of ⨁_{m+n=M} ρ(μ, ν).

Proposition 8.5 is now a consequence of the following simple lemma.

Chapter 1 155 Lemma 8.6 Suppose π is an irreducible representation of an algebra H. Suppose ρ is a representation of H of which π is a constituent and that ρ is the direct sum of the representations ρ_{λ}, λ ∈ Ω. Then π is a constituent of at least one of the ρ_{λ}.

Let ρ_{λ} act on X_{λ} and let ρ act on X the direct sum of X_{λ}. Supose that Y_{1} and Y_{2} are invariant subspaces of X and that the representation on the quotient Y_{1}/Y_{2} is equivalent to π. There is a finite subset Λ_{0} of Λ such that

Y_{1} ∩ ( ∑_{λ∈Λ_{0}} X_{λ})

is not contained in Y_{2}. We may as well replace Y_{1} by Y_{1} ∩ (∑_{λ∈Λ_{0}} X_{λ}) and Y_{2} by Y_{2} ∩ (∑_{λ∈Λ_{0}} X_{λ}) and suppose that Λ is finite. If Λ = {λ_{1}, ⋯, λ_{p}} we have only to show that π is a constituent of ρ_{λ_{1}} or of ρ_{λ_{2}} ⊕ ⋯ ⊕ ρ_{λ_{p}} for we can then use induction. Thus we may as well take p = 2. If the projections of Y_{1} and Y_{2} on X_{λ_{1}} are not equal we can replace Y_{1} and Y_{2} by these projections to see that π is a constituent of ρ_{λ_{1}}. If they are equal Y_{1} = Y_{2} + (Y_{1} ∩ X_{λ_{2}}) and we can replace Y_{1} and Y_{2} by Y_{1} ∩ X_{λ_{2}} and Y_{2} ∩ X_{λ_{2}} to see that π is a constituent of ρ_{λ_{2}}.

References for Chapter I

The Weil representation is constructed in:

1. Weil, A., Sur certains groupes d'opérateurs unitaires, Acta Math., t. 111, 1964. One of the first to study representations of groups over non-archimedean local fields was

F. Mautner in:

2. Mautner, F., Spherical functions over 𝔭-adic fields, I. Amer. Jour. Math., vol LXXX, 1958. Absolutely cuspidal representations were first constructed by Gelfand and Graev. References

to their work and that of Kirillov will be found in

3. Gelfand, I.M., M.I. Graev, and I.I. Pyatetskii–Shapiro, Representation Theory and Automorphic Functions, W.B. Saunders Co., 1966. These representations were constructed in terms of the Weil representation by Shalika and by

Tanaka.

4. Shalika, J. Representations of the two-by-two unimodular group over local fields, Notes, Institute for Advanced Study.

5. Tanaka, S., On irreducible unitary representations of some special linear groups of the second order, Osaka Jour. Math., 1966. To classify the representations over an archimedean field we have used a theorem of Harish–

Chandra which may be found in

6. Harish–Chandra, Representations of semisimple Lie groups, II, T.A.M.S., vol 76, 1954. Our discussion of characters owes much to:

7. Sally, P.J. and J.A. Shalika, Characters of the discrete series of representations of SL(2) over a local field, P.N.A.S., 1968. Three standard references to the theory of L-functions are:

8. Lang, S., Algebraic numbers, Addison–Wesley, 1964.

9. Tate, J., Fourier analysis in number fields and Hecke’s Zeta-functions in algebraic number theory, Thompson Book Co., 1967.

10. Weil, A., Basic number theory, Springer–Verlag, 1967.

In Paragraph 8 we have used a result from:

11. Harish–Chandra, Automorphic forms on semisimple Lie groups, Springer–Verlag, 1968. Tamagawa measures are discussed in:

12. Weil, A., Adèles and algebraic groups, Institute for Advanced Study, 1961.

Chapter II: Global Theory

§9. The global Hecke algebra. Let F be a global field, that is, an algebraic number field of finite degree over the rationals or a function field in one variable over a finite field. 𝔸 will be the adèle ring of F. Before studying the representations of GL(2, 𝔸) or, more precisely, the representations of a suitable group algebra of GL(2, 𝔸) we introduce some simple algebraic notions.

Let {V_{λ} | λ ∈ Λ} be a family of complex vector spaces. Suppose that for all but a finite number of λ we are given a non-zero vector e_{λ} in V_{λ}. Let V^{0} be the set of all x = ∏_{λ} x_{λ} in ∏_{λ} V_{λ} such that x_{λ} = e_{λ} for al but a finite number of λ. Let C be the free vector space with complex coefficients over V^{0} and let D be the subspace generated by vectors of the form

{(aY_{μ} + bZ_{μ}) × ∏_{λ≠μ} x_{λ}} − a {y_{μ} × ∏_{λ≠μ} x_{λ}} − b {z_{μ} × ∏_{λ≠μ} x_{λ}}.

a and b belong to ℂ and μ is any element of Λ. The quotient of C by D is called the tensor product of the V_{λ} with respect to the family e_{λ} and is written

V = ⊗_{e_{λ}}V_{λ}

or simply ⊗V_{λ}. It has an obvious universal property which characterizes it up to isomorphism. The image of ∏ x_{λ} in V is written ⊗x_{λ}.

If Λ′ is a subset of Λ with finite complement we may form the ordinary tensor product

⊗_{λ∈Λ−Λ′}V_{λ}

and we may form

⊗_{λ∈Λ′}V_{λ}

with respect to the family e_{λ}. Then ⊗_{λ∈Λ}V_{λ} is canonically isomorphic to

{ ⊗_{λ∈Λ−Λ′} } ⊗ { ⊗_{λ∈Λ′} V_{λ}}

If S is a finite subset of Λ let

V_{S} = ⊗_{λ∈S}V_{λ}

If S is so large that e_{λ} is defined for λ not in S let φ_{S} be the map of V_{S} into V which sends ⊗_{λ∈S}x_{λ} to {⊗_{λ∈S}x_{λ}} ⊗ {⊗_{λ∉S}e_{λ}}. If S′ contains S there is a unique map φ_{S,S′} of V_{S} into V_{S′} which makes

V_{S} ⟶^{φ_{S,S′}} V′_{S}
↘^{φ_{S}}  ↙_{φ_{S′}}
V

commutative. If we use these maps to form the inductive limit of the spaces V_{S} we obtain a space which the layman is unable to distinguish from V.

Suppose that for every λ we are given a linear map B_{λ} of V_{λ} into itself. If B_{λ} e_{λ} = e_{λ} for all but a finite number of λ there is exactly one linear transformation B of ⊗V_{λ} such that

B : ⊗x_{λ} → ⊗B_{λ} x_{λ}

Chapter 2 158 B is denoted by ⊗B_{λ}.

For example if A_{λ}, λ ∈ Λ is a family of associative algebras, which may or may not have a unit, and if, for almost all λ, ξ_{λ} is a given idempotent of A_{λ} one may turn

A = ⊗_{ξ_{λ}}A_{λ}

into an algebra in such a way that

(⊗a_{λ})(⊗b_{λ}) = ⊗(a_{λ} b_{λ}).

Let V_{λ}, λ ∈ Λ, be an A_{λ} module. If for almost all λ a vector e_{λ} such that ξ_{λ}e_{λ} = e_{λ} is given we may turn V = ⊗_{e_{λ}}V_{λ} into an A = ⊗_{ξ_{λ}}A_{λ} module in such a way that

(⊗a_{λ})(⊗x_{λ}) = ⊗(a_{λ} x_{λ})

Suppose the family {e_{λ}} is replaced by a family {e′_{λ}} but that, for all but a finite number of λ, e′_{λ} = α_{λ} e_{λ} where α_{λ} is a non-zero scalar. Suppose for example that e′_{λ} = α_{λ} e_{λ} if λ is not in the finite set S. There is a unique map of ⊗_{e_{λ}}V_{λ} to ⊗_{e′_{λ}}V_{λ} which sends

{ ⊗_{λ∈S} x_{λ}} ⊗ { ⊗_{λ∉S} x_{λ}}

to

{ ⊗_{λ∈S} x_{λ}} ⊗ { ⊗_{λ∉S} α_{λ} x_{λ}}

It is invertible and commutes with the action of A. Moreover apart from a scalar factor it is independent of S.

Now suppose F is a global field. A place of F is an equivalence class of injections, with dense image, of F into a local field. If λ_{1} takes F into F_{1} and λ_{2} takes F into F_{2} they are equivalent if there is a topological isomorphism φ of F_{1} with F_{2} such that λ_{2} = φ ∘ λ_{1}. The symbol for a place will be v. If v contains the imbedding λ_{1} and a belongs to F we set |a|_{v} = |λ_{1}(a)|. To be definite we let F_{v} be the completion of F with respect to the absolute value a → |a|_{v}. v is archimedean or non-archimedean according to the nature of F_{v}. Non-archimedean places will sometimes be denoted by 𝔭.

If G_{F} = GL(2, F) we set

G_{v} = G_{F_{v}} = GL(2, F_{v}).

K_{v} will be the standard maximal compact subgroup of G_{v}. then G_{𝔸} = GL(2, 𝔸) is the restricted direct product of the groups G_{v} with respect to the subgroups K_{v}.

If v is non-archimedean we set O_{v} = O_{F_{v}} and U_{v} = U_{F_{v}}. O_{v} is the ring of integers of F_{v} and U_{v} is the group of units of O_{v}. Suppose M′ is a quaternion algebra over F. Let M′_{v} = M′_{F_{v}} = M′⊗_{F} F_{v}. For almost all v the algebra M′_{v} is split, that is, there is an isomorphism

θ_{v} : M′_{v} → M(2, F_{v})

where M(2, F_{v}) is the algebra of 2 × 2 matrices over F_{v}. For every place v at which M′_{v} is split we want to fix such an isomorphism θ_{v}. Let B be a basis of M over F and let L_{v} be the O_{v} module generated in M_{v} by B. We may and do choose θ_{v} so that for almost all v

θ_{v}(L_{v}) = M(2, O_{v}).

If B′ is another basis and {θ′_{v}} a family of isomorphism associated to B′ then for every place v at which M′_{v} splits there is a g_{v} in GL(2, F_{v}) such that

θ′_{v} θ_{v}^{−1} = g_{v} a g_{v}^{−1}

for all a in M(2, F_{v}). Moreover g_{v} belongs to K_{v} for all but a finite number of v.

Suppose the family of isomorphisms θ_{v} has been chosen. If M′_{v} is split we define a maximal compact subgroup K′_{v} of G′_{v}, the group of invertible elements of M′_{v}, by the condition

θ_{v}(K′_{v}) = K_{v}.

If M′_{v} is not split we set

K′_{v} = {x ∈ M′_{v} | |ν(x)|_{v} = 1}.

This group is compact. In any case K′_{v} is defined for all v. Since many of the constructions to be made depend on the family K′_{v} which in turn depends on the family of θ_{v} it is very unfortunate that the family of θ_{v} is not unique. We should really check at every stage of the discussion that the constructions are, apart from some kind of equivalence, independent of the initial choice of θ_{v}. We prefer to pretend that the difficulty does not exist. As a matter of fact for anyone lucky enough not to have been indoctrinated in the functorial point of view it doesn’t. We do however remark that any two choices of the family of K′_{v} lead to the same result for almost all v. G′_{𝔸} is the restricted direct product of the groups G′_{v} with respect to the subgroups K′_{v}.

We have now to introduce the Hecke algebras ℋ and ℋ′ of G_{𝔸} and G′_{𝔸}. Let ℋ_{v} be ℋ_{F_{v}}. If M′_{v} is split G′_{v} isomorphic, by means of θ_{v}, to G_{v} and we let ℋ′_{v} be the algebra of measures on G′_{v} corresponding to ℋ_{v}. Suppose M′_{v} is not split. If v is non-archimedean ℋ′_{v} is the algebra of measures defined by the locally constant compactly supported functions on G′_{v}. If v is archimedean ℋ′_{v} will be the sum of two subspaces, the space of measuures defined by infinitely differentiable compactly supported functions on G′_{v} which are K′_{v}-finite on both sides and the space of measures on K′_{v} defined by the matrix coeffients of finite-dimensional representations of K′_{v}.

Let ε_{v} and ε′_{v} be the normalized Haar measures on K_{v} and K′_{v}. ε_{v} is an elementary idempotent of ℋ_{v} and ε′_{v} is an elementary idempotent of ℋ′_{v}. We set

ℋ = ⊗_{ε_{v}}ℋ_{v}

and

ℋ′ = ⊗_{ε′_{v}}ℋ′_{v}

If S is the finite set of places at which M′_{v} does not split we may write

ℋ = { ⊗_{v∈S} ℋ_{v}} ⊗ { ⊗_{v∉S} ℋ_{v}} = ℋ_{s} ⊗ ℋ̂_{s}

and

ℋ′ = { ⊗_{v∈S} ℋ′_{v}} ⊗ { ⊗_{v∉S} ℋ′_{v}} = ℋ′_{s} ⊗ ℋ̂′_{s}

By construction, if M′_{v} is split, ℋ_{v} and ℋ′_{v} are isomorphic in such a way that ε_{v} and ε′_{v} correspond. Using these isomorphism we may construct an isomorphism of ℋ̂_{s} and ℋ̂′_{S}. We may also write

G_{𝔸} = { ∏_{v∈S} G_{v}} × { ∏_{v∉S} G_{v}} = G_{S} × Ĝ_{S}

Chapter 2 160 and

G′_{𝔸} = { ∏_{v∈S} G′_{v}} × { ∏_{v∉S} G′_{v}} = G′_{S} × Ĝ′_{S}.

The second factor is in both cases a restricted direct product. There is an isomorphism θ : Ĝ′_{S} → Ĝ_{s} defined by

θ( ∏_{v∉S} g′_{v}) = ∏_{v∉S} θ_{v}(g′_{v})

We will interpret ℋ̂_{S} and ℋ̂′_{S} as algebras of measures on Ĝ_{S} and Ĝ′_{S} and then the isomorphism between them will be that associated to θ.

We can also interpret the elements of ℋ and ℋ′ as measures on G_{𝔸} and G′_{𝔸}. For example any element of ℋ is a linear combination of elements of the form f = ⊗_{v}f_{v}. Let T be a finite set of places and suppose that f_{v} = ε_{v} for v not in T. If T′ contains T, on the group

G_{𝔸(T′)} = { ∏_{v∈T′} G_{v}} × { ∏_{v∉T′} K_{v}}

we can introduce the product of the measures f_{v}. Since G_{𝔸} is the union of these groups and the measures on them are consistent we can put the measure together to form a measure f on G_{𝔸}. If each f_{v} is the measure associated to a function then f is also. Such measures form a subalgebra ℋ_{1} and ℋ.

The notion of an elementary idempotent of ℋ or ℋ′ is defined in the obvious way. If ξ is an elementary idempotent of ℋ there is another elementary idempotent ξ_{1} of the form ξ_{1} = ⊗_{v}ξ_{v} where ξ_{v} is an elementary idempotent of ℋ_{v} and ξ_{v} = ε_{v} for almost all v so that ξ_{1}ξ = ξ.

We shall now discuss the representations of ℋ. A representation π of ℋ on the vector space V over ℂ will be called admissible if the following conditions are satisifed

(i) Every w in V is a linear combination of the form ∑ π(f_{i}) w_{i} with f_{i} in ℋ_{1}. (ii) If ξ is an elementary idempotent the range of π(ξ) is finite dimensional. (iii) Let v_{0} be an archimedean place. Suppose that for each v an elementary idempotent ξ_{v} is given

and that ξ_{v} = ε_{v} for almost all v. Let ξ = ⊗_{v}ξ_{v}. If w is in V the map

f_{v_{0}} → π(f_{v_{0}} ⊗ { ⊗_{v≠v_{0}} ξ_{v}})w

of ξ_{v_{0}}ℋ_{v_{0}}ξ_{v_{0}} into the finite dimensional space π(ξ)V is continuous.

Suppose that an admissible representation π_{v} of ℋ_{V} on V_{v} is given for each v. Assume that for almost all v the range of π_{v}(ε_{v}) is not zero. Assume also that the range of π_{v}(ε_{v}) has dimension one when it is not zero. As we saw in the first chapter this supplementary condition is satisified if the representations π_{v} are irreducible. Choosing for almost all v a vector e_{v} such that π_{v}(ε_{v}) e_{v} = e_{v} we may form V = ⊗_{e_{v}}V_{v}. Let π be the representation ⊗_{v}π_{v} on V. Because of the supplementary condition it is, apart from equivalence, independent of the choice of the e_{v}.

π will be admissible. To see this observe first of all that condition (i) has only to be verified for vectors of the form w = ⊗_{v} w_{v}. Suppose w_{v} = e_{v} when v is not in the finite set T which we suppose contains all archimedean places. If v is not in T let f_{v} = ε_{v} so that w_{v} = π(f_{v}) w_{v}. If v is in T let

w_{v} = ∑ π_{v}(f_{v}^{i}) w_{v}^{i}.

Then

w = { ⊗_{v∈T} ∑ π_{v}(f_{v}^{i}) w_{v}^{i}} ⊗ { ⊗_{v∉T} π(f_{v}) w_{v}}.

Expanding the right hand side we obtain the desired relation. The second condition has only to be verified for elementary idempotents of the form ξ = ⊗_{v}ξ_{v}. Then

π(ξ) V = ⊗π(ξ_{v}) V_{v}

Since π(ξ_{v})V_{v} is finite dimensional for all v and π(ξ_{v})V_{v} = π(ε_{v})V_{v}, which has dimension one, for almost all v the right side is finite dimensional. The last condition results from the admissibility of π_{v_{0}}.

π cannot be irreducible unless each π_{v} is. Suppose however that each π_{v} is irreducible. If ξ_{v} is an elementary idempotent of ℋ_{v} and if π_{v}(ξ_{v}) ≠ 0 we have a representation π_{ξ_{v}} of ξ_{v} ℋ_{v} ξ_{v} on π_{v}(ξ_{v}) V_{v}. Since it is irreducible π_{ξ_{v}} determines a surjective map

π_{ξ_{v}} : ξ_{v} ℋ_{v} ξ_{v} → L(ξ_{v})

if L(ξ_{v}) is the ring of linear transformations of V (ξ_{v}) = π_{v}(ξ_{v})V_{v}. To show that π is irreducible we have only to show that for every elementary idempotent of the form ξ = ⊗_{v} ξ_{v} the representation of ξℋξ on V (ξ) = π(ξ)V is irreducible. Suppose that ξ_{v} = ε_{v} if v is not in T. Then

V (ξ) = ⊗_{v}V (ξ_{v})

is isomorphic to ⊗_{v∈T} V (ξ_{v}). The full ring of linear transformations of this space is

⊗_{v∈T} L(ξ_{v})

and therefore the full ring of linear transformations of V (ξ) is

{ ⊗_{v∈T} L(ξ_{v})} ⊗ { ⊗_{v∉T} π_{v}(ε_{v})}.

This is the image under π of

{ ⊗_{v∈T} ξ_{v} ℋ_{v} ξ_{v}} ⊗ { ⊗_{v∉T} ε_{v}}

which is contained in ξℋξ.

An admissible representation equivalent to one constructed by tensor products is said to be factorizable.

Proposition 9.1 Every irreducible admissible representation of ℋ is factorizable. The factors are unique up to equivalence.

Suppose π is such a representation. Let I be the set of elementary idempotents of the form ξ = ⊗ξ_{v} for which π(ξ) is not 0. I is certainly not empty. Let V (ξ) = π(ξ)V if V is the space on which π acts. If ξ and ξ′ are elementary idempotents we write ξ ≤ ξ′ if ξ′ξ = ξ. Then ξ ξ′ will also equal ξ. If ξ = ⊗ξ_{v} and ξ′ = ⊗ξ′_{v} then ξ ≤ ξ′ if and only if ξ_{v} ξ′_{v} = ξ′_{v} ξ_{v} = ξ_{v} for all v. If ξ ≤ ξ′ and ξ belongs to I so does ξ′. ξℋξ is a subalgebra of ξ′ℋξ′. Let ι(ξ′, ξ) be the corresponding injection and let L(ξ) and L(ξ′) be the spaces of linear transformations of V (ξ) and V (ξ′). There is exactly one map

φ(ξ′, ξ) : L(ξ) → L(ξ′)

Chapter 2 162 which makes

ξℋξ ⟶^{ι(ξ′,ξ)} ξ′ ℋ ξ′
π_{ξ}↓  π_{ξ′} ↓
L(ξ) ⟶^{φ(ξ′,ξ)} L(ξ′)

commutative.

There is a map of ξ_{v}ℋ_{v}ξ_{v} into ξℋξ which sends f_{v} to f_{v} ⊗ {⊗_{w≠v}ξ_{w}}. Composing this map with π_{ξ} we obtain a map π_{ξ}^{v} of ξ_{v}ℋ_{v}ξ_{v} onto a subalgebra L_{v}(ξ) of L(ξ). L(ξ) and L_{v}(ξ) have the same unit, namely π_{ξ}(ξ). If v ≠ w the elements of L_{v}(ξ) commute with those of L_{w}(ξ). If we form the tensor product of the algebras L_{v}(ξ) with respect to the family of units there is a map from ⊗_{v}L_{v}(ξ) to L(ξ) which sends ⊗_{v}λ_{v} to ∏_{v} λ_{v}. Moreover we may identify ⊗_{v} ξ_{v}ℋ_{v}ξ_{v} and ξℋξ. Since the diagram

⊗_{v} ξ_{v} ℋ_{v}ξ_{v} ⟶ ξ ℋ ξ
⊗_{v} π_{ξ}^{v} ↓  π_{ξ}↓
⊗_{v} L_{v}(ξ) ⟶ L(ξ)

is commutative the bottom arrow is surjective.

Lemma 9.1.1 The algebras L_{v}(ξ) are simple and the map ⊗_{v}L_{v}(ξ) → L(ξ) is an isomorphism.

To show that L_{v}(ξ) is simple we need only show that the faithful L_{v}(ξ)-module V (ξ) is spanned by a family of equivalent irreducible submodules. Let M be any irreducible submodule. Then the family {T M} where T runs over the image of 1_{v} ⊗ {⊗_{w≠v}L_{w}(ξ)} spans V (ξ) and each T M is 0 or equivalent to M because T commutes with the elements of L_{v}(ξ). 1_{v} is the unit of L_{v}(ξ). We have only to show that ⊗_{v}L_{v}(ξ) ↪ L(ξ). Since ⊗_{v}L_{v}(ξ) is the inductive limit of ⊗_{v∈T} L_{v}(ξ), where T is a finite set, we have only to show that the map is injective on these subalgebras. As they are tensor products of simple algebras they are simple and the map is certainly injective on them.

If ξ ≤ ξ′ there is a commutative diagram

⊗_{v} ξ_{v} ℋ_{v} ξ_{v} ⟶^{ι(ξ′,ξ)} ⊗_{v} ξ′_{v} ℋ_{v} ξ′_{v}
↓ ↓
⊗_{v} L_{v}(ξ)  ⊗_{v} L_{v}(ξ′)
↓  ↓
L(ξ)  ⟶^{φ(ξ′,ξ)}  L(ξ′)

Moreover if ι_{v}(ξ′, ξ) is the imbedding of ξ_{v}ℋ_{v}ξ_{v} into ξ′_{v}ℋ_{v}ξ′_{v} then ι(ξ′, ξ) = ⊗_{v}ι_{v}(ξ′, ξ). We want to verify that a horizontal arrow ⊗_{v}φ_{v}(ξ′, ξ) can be inserted in the middle without destroying the commutativity. To do this we have only to show that if f_{v} is in ξ_{v}ℋ_{v}ξ_{v} and therefore in ξ′_{v}ℋ_{v}ξ′_{v} then π_{ξ}^{v}(f_{v}) = 0 if and only if π_{ξ′}^{v}(f_{v}) = 0. Let U = π_{ξ}^{v}(f_{v}) and let T = π_{ξ′}^{v}(f_{v}). If

E = π_{ξ′}(ξ′_{v} ⊗ { ⊗_{w≠v} ξ_{w}})

then

T E = π_{ξ′}(f_{v} ⊗ { ⊗_{w≠v} ξ_{w}})

Chapter 2 163 is determined by its restriction to V (ξ) and that restriction is U.

It is clear that if S is a sufficiently large finite set the map ⊗_{w∈S}L_{w}(ξ′) → L(ξ′) is an isomorphism. We suppose that S contains v. E belongs to the image M of 1_{v} ⊗ {⊗_{w≠v}L_{w}(ξ′)}. Since M is simple and E is not 0 there are A_{i}, B_{i} 1 ≤ i ≤ r in M such that

∑_{i=1}^{r} A_{i} E B_{i} = 1

Thus

T = ∑_{i} T A_{i} E B_{i} = ∑_{i} A_{i} T E B_{i}

and T = 0 if and only if U = 0.

Since the necessary compatibility conditions are satisifed we can take inductive limits, over I, to the left and right. The inductive limit of the ξℋξ is ℋ and that of the ξ_{v}ℋ_{v}ξ_{v} is ℋ_{v}. Let L_{v} be that of L_{v}(ξ) and L that of L(ξ). There is a map π^{v} : ℋ_{v} → L_{v} and, for almost all v, π^{v}(ε_{v}) = μ_{v} is not zero. We have a commutative diagram

⊗ ℋ_{v} ⟶ ℋ
⊕π^{v} ↓  ↓
⊗ L_{v} ⟶ L

in which the rows are isomophisms. Moreover L acts faithfully on V and the representation of ℋ on V can be factored through L.

If A is an algebra with a minimal left ideal J then any faithful irreducible representation of A on a vector space X is equivalent to the representation on J. In fact we can choose x_{0} in X so that Jx_{0} ≠ 0. The map j → jx_{0} of J to X gives the equivalence. Thus to prove that π is factorizable it will be enough to show that L has a minimal left ideal, that the representation of L on this minimal left ideal is a tensor product of representations σ_{v} of L_{v}, and that σ_{v} ∘ π^{v} is admissible.

Suppose A is a simple algebra and J is a left ideal in A. If a in A is not 0 and aJ = 0 then Aa AJ = AJ = 0. If J is not 0 this is impossible. Suppose e is an idempotent of A and A_{1} = eAe. Let J_{1} be a minimal left ideal of A_{1} and let J = AJ_{1}. If J were not minimal it would properly contain a non-zero ideal J′. J′ ∩ A_{1} would have to be 0. Since Je = J we must have eJ = eJe = 0. Since this is a contradiction J is minimal. Suppose for example that A is the union of a family {A_{λ}} of matrix algebras. Suppose that for each λ there is an idempotent e_{λ} in A such that A_{λ} = e_{λ}Ae_{λ} and that given λ_{1} and λ_{2} there is a λ_{3} such that A_{λ_{3}} contains A_{λ_{1}} and A_{λ_{2}}. Then A is certainly simple and, by the preceding discussion, contains a minimal left ideal.

The algebras L and L_{v} satisfy these conditions. In fact, speaking a little loosely, L is the union of the L(ξ) and L_{v} is the union of L_{v}(ξ). Choose ξ so that V (ξ) ≠ 0 and let J_{v} be a minimal left ideal in L_{v}(ξ). Since L_{v}(ξ) is one-dimensional for almost all v the ideal J_{v} = L_{v}(ξ) for almost all v. Thus J = ⊗J_{v} exists and is a minimal left ideal of L(ξ). Thus LJ = ⊗L_{v}J_{v}. LJ is a minimal left ideal of L and L_{v}J_{V} is a minimal left ideal of L_{v}. The representation of L on LJ is clearly the tensor product of the representations σ_{v} of L_{v} on L_{v}J_{V}.

Thus π is equivalent to the tensor product of the representations π_{v} = σ_{v} ∘ π^{v}. The representations π_{v} are irreducible. Since it is easily seen that a tensor product ⊗π_{v} is admissible only if each factor is admissible we may regard the first assertion of the proposition as proved.

Chapter 2 164

If π is an admissible representation of ℋ on V and v is a place we may also introduce a representation of ℋ_{v} on V which we still call π. If u is in V we choose ξ = ⊗_{w}ξ_{w} so that π(ξ)u = u. Then if f belongs to ℋ_{v} we set

π(f)u = π(f ξ_{v} ⊗ { ⊗_{w≠v} ξ_{w}})u

The second part of the proposition is a consequence of the following lemma whose proof is immediate.

Lemma 9.1.2 Suppose π = ⊗_{w}π_{w}. Then the representation π of ℋ_{v} is the direct sum of representations equivalent to π_{v}.

Let S_{a} be the set of archimedean primes. One can also associate to an admissible representation π of ℋ on V a representation of Ĝ_{S_{a}}, the group formed by the elements of G_{A} whose components at every archimedean place are 1, on V. If v is archimedean one can associate to π a representation of 𝔄_{v}, the universal enveloping algebra of the Lie algebra of G_{v}, on V. Finally π determines a representation of the group Z_{𝔸} of scalar matrices in GL(2, 𝔸). If π is irreducible there is a quasi-character η of I the group of idèles such that

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η(a) I

for all a in I. If π_{v} is associated to η_{v} and π = ⊗_{v}π_{v} then π is associated to the quasi-character η defined by

η(a) = ∏_{v} η_{v}(a_{v}).

One may define the contragredient of π and the tensor product of π with a quasi-character of I. All the expected formal relations hold. In particular π̃ is equivalent to η^{−1} ⊗ π if π is irreducible.

The above discussion applies, mutatis mutandis, to the algebra ℋ′. The next proposition, which brings us a step closer to the theory of automorphic forms, applies to ℋ alone.

Proposition 9.2 Let π = ⊗π_{v} be an irreducible admissible representation of ℋ. Suppose that π_{v} is infinite dimensional for all v. Let ψ be a non-trivial character 𝔸/F. There is exactly one space W (π, ψ) of continuous functions on G_{𝔸} with the following properties:

(i) If W is in W (π, ψ) then for all g in G_{𝔸} and all x in 𝔸

W (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = ψ(x) W (g)

(ii) W (π, ψ) is invariant under the operators ρ(f), f ∈ ℋ, and transforms according to the repre-

sentation π of ℋ. In particular it is irreducible under the action of ℋ. (iii) If F is a number field and v an archimedean place then for each W in W (π, ψ) there is a real

number N such that

W \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = O( |a|^{N})

as a → ∞ in F_{v}^{×}.

In the last assertion F_{v}^{×} is regarded as a subgroup of I. F_{v} is a subgroup of 𝔸 and the restriction ψ_{v} of ψ to F_{v} is non-trivial. Thus for each place v the space W (π_{v}, ψ_{v}) is defined and Chapter 2 165 we may suppose that π_{v} acts on it. Moreover for almost all v the largest ideal of F_{v} on which ψ_{v} if trivial is O_{v} and π_{v} contains the trivial representation of K_{v}. Thus by Proposition 3.5 there is a unique function φ_{v}^{0} in W (π_{v}, ψ_{v}) such that φ_{v}^{0}(g_{v} k_{v}) = φ_{v}^{0}(g_{v}) for all k_{v} in K_{v} and φ_{v}^{0}(I) = 1. Then φ_{v}^{0}(k_{v}) = 1 for all k_{v} in K_{v}. π acts on

⊗_{φ_{v}^{0}}W (π_{v}, ψ_{v})

If g is in G_{𝔸} and ⊗φ_{v} belongs to this space then φ_{v}(g_{v}) = 1 for almost all v so that we can define a function φ on G_{𝔸} by

φ(g) = ∏_{v} φ_{v}(g_{v}).

The map ⊗φ_{v} → φ extends to a map of ⊗W (π_{v}, ψ_{v}) into a space W (π, ψ) of functions on G_{𝔸}. W (π, ψ) certainly has the required properties. We have to show that it is characterized by these properties.

Suppose 𝔐 is another space with these properties. There is an isomorphism T of ⊗W (π_{v}, ψ_{v}) and 𝔐 which commutes with the action of ℋ. All we have to do is show that there is a constant c such that if φ = ⊗φ_{v} then

T φ(g) = c ∏_{v} φ_{v}(g_{v}).

Let S be a finite set of places and let

W_{S} = ⊗_{v∈S}W (π_{v}, ψ_{v})

and

Ŵ_{S} = ⊗_{v≠S}W (π_{v}, ψ_{v}).

Then

⊗ W (π_{v}, ψ_{v}) = W_{S} ⊗ Ŵ_{S}.

We first show that if S is given there is a function c_{S} on Ĝ_{S} × Ŵ_{S} such that if

f = T ({ ⊗_{v∈S} φ_{v}} ⊗ φ)

with φ in Ŵ_{S} then

f(gh) = c_{S}(h, φ) ∏_{v∈S} φ_{v}(g_{v})

if g is in G_{S} and h is in Ĝ_{S}.

Suppose that S consists of the single place v. If φ belongs to Ŵ_{S} and h belongs to Ĝ_{S} associate to every function φ_{v} in W (π_{v}, ψ_{v}) the function

φ′_{v}(g_{v}) = f(g_{v}h)

on G_{v}. f is T (φ_{v} ⊗ φ). By construction, if φ_{v} is replaced by ρ(f_{v})φ_{v} with f_{v} in ℋ_{v} the function φ′_{v} is replaced by ρ(f_{v})φ′_{v}. Moreover if x is in F_{v}

φ′_{v} (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g_{v}) = ψ_{v}(x) φ′_{v}(g_{v}).

Chapter 2 166 Since any conditions on rates of growth can easily be verified we see that the functions φ′_{v} are either all zero or they fill up the space W (π_{v}, ψ_{v}). In both cases the map φ_{v} → φ′_{v} is a map of W (π_{v}, ψ_{v}) into itself which commutes with the action of ℋ_{v} and therefore consists merely of multiplication by a scalar c_{S}(h, φ).

Now suppose that S′ is obtained by adjoining the place w to S and that our assertion is true for S. Take h in Ĝ_{S′} and φ in Ŵ_{S′}. If

f = T ({ ⊗_{v∈S′} φ_{v}} ⊗ φ)

then, for g in G_{S}, and g_{w} in G_{w},

f(g g_{w}h) = c_{S}(g_{w}h, φ_{w} ⊗ φ) ∏_{v∈S} φ_{v}(g_{v}).

The argument used before shows that for a given h and φ the function

g_{w} → c_{S}(g_{w}h, φ_{w} ⊗ φ)

is a multiple c_{S′}(h, φ) of φ_{w}.

To prove the existence of c we obseve first that if S is the disjoint union of S_{1} and S_{2} we may write any h_{1} in Ĝ_{S_{1}} as h_{1} = h ∏_{v∈S_{2}} h_{v} with h in Ĝ_{S}. Suppose φ_{1} = {⊗_{v∈S_{2}} φ_{v}} ⊗ φ with φ in Ŵ_{S} is in Ŵ_{S_{1}}. Then

c_{S_{1}}(h_{1}, φ_{1}) = { ∏_{v∈S_{2}} φ_{v}(h_{v})} c_{S}(h, φ)  (9.2.1)

because the right hand side has all the properties demanded of the left. If S_{1} is large enough that φ_{v}^{0} exists for v not in S_{1} then, by its definition, c_{S_{1}}(h, ⊗_{v∉S_{1}}φ_{v}^{0}) has a constant c(S_{1}) on

∏_{v∉S_{1}} K_{v}

The formula (9.2.1) shows that c(S) = c(S_{1}) if S contains S_{1}. We take c to be the common value of these constants. Given φ = ⊗ φ_{v} and g = ∏ g_{v} we choose S so that φ_{v} = φ_{v}^{0} and g_{v} ∈ K_{v} for v not in S. Then

\begin{aligned}T φ(g) & = c( ∏_{v∉S} g_{v}, ⊗_{v∉S}φ_{v}) ∏_{v∈S} φ_{v}(g_{v}) \\ & = c ∏_{v} φ_{v}(g_{v}). \end{aligned}

We observed that if π_{v} is finite dimensional the space W (π_{v}, ψ_{v}) cannot exist if v is nonarchimedean or real. Although we neglected to mention it, the argument used for real field also shows that W (π_{v}, ψ_{v}) cannot exist if v is complex. The proof of Proposition 9.2 can therefore be used, with minor changes, to verify the next proposition.

Proposition 9.3 If π = ⊗π_{v} is given and if one of the representations π_{v} is finite dimensional there can exist no space W (π, ψ) satisfying the first two conditions of the previous proposition.

An admissible representation π of ℋ on the space V is said to be unitary if there is a positive definite hermitian form (v_{1}, v_{2}) on V such that, if f^{∗}(g) = f̄(g^{−1}),

(π(f)v_{1}, v_{2}) = (v_{1}, π(f^{∗})v_{2})

for all f in ℋ.

Chapter 2 167 Lemma 9.4 If π is unitary and admissible then V is the direct sum of mutually orthogonal invariant irreducible subspaces.

The direct sum of the lemma is to be taken in the algebraic sense. We first verify that if V_{1} is an invariant subspace and V_{2} is its orthogonal complement then V = V_{1} ⊕ V_{2}. Certainly V_{1} ∩ V_{2} = 0. Let ξ be an elementary idempotent and let V (ξ), V_{1}(ξ), V_{2}(ξ) be the ranges of π(ξ) in V, V_{1}, and V_{2}. let V_{1}^{⊥}(ξ) be the range of 1 − π(ξ) acting on V_{1}. Then V (ξ) and V_{1}^{⊥}(ξ) are orthogonal and

V_{1} = V_{1}(ξ) ⊕ V_{1}^{⊥}(ξ).

Thus V_{2}(ξ) is just the orthogonal complement of V_{1}(ξ) in V (ξ). Since V (ξ) is finite dimensional

V (ξ) = V_{1}(ξ) ⊕ V_{2}(ξ).

Since every element of V is contained in some V (ξ) we have V = V_{1} + V_{2}.

To complete the proof we shall use the following lemma.

Lemma 9.4.1 If π is a unitary admissible representation of ℋ on the space V then V contains a minimal non-zero invariant subspace.

Choose an idempotent ξ so that V (ξ) = π(ξ)V ≠ 0. Since V (ξ) is finite dimensional amongst all the non-zero subspaces of it obtained by intersecting it with an invariant subspace of V there is a minimal one N. Let M be the intersection of all invariant subspaces containing N. If M is not irreducible it is the direct sum of two orthogonal invariant subspaces M_{1} and M_{2}. Then

N = M ∩ V (ξ) = π(ξ) M = π(ξ) M_{1} ⊕ π(ξ) M_{2}

The right side is

{M_{1} ∩ V (ξ)} ⊕ {M_{2} ∩ V (ξ)}

so that one of M_{1} ∩V (ξ) and M_{2} ∩V (ξ) is N. Then M_{1} or M_{2} contains M. This is a contradiction.

Let A be the set consisting of families of mutually orthogonal invariant, and irreducible subspaces of V. Each member of the family is to be non-zero. Let {V_{λ}} be a maximal family. Then V = ⊕_{λ}V_{λ}. If not let V_{1} = ⊕_{λ}V_{λ}. The orthogonal complement of V_{1} would be different from zero and therfore would contain a minimal non-zero invariant subspace which when added to the family {V_{λ}} would make it larger.

If T is a finite set of places most of the results of this paragraph are valid for representations π of ℋ̂_{T}. For example π is factorizable and W (π, ψ) exists as a space of functions on Ĝ_{T}.

Chapter 2 168 §10. Automorphic Forms. In this paragraph F is still a global field. We shall begin by recalling a simple result from reduction theory. If v is a place of 𝔸 and a is in 𝔸 then |a|_{v} is the absolute value of a_{v} the vth component of a. If a is in I

|a| = ∏_{v} |a|_{v}

Lemma 10.1 There is a constant c_{0} such that if g belongs to G_{𝔸} there is a γ in G_{F} for which

∏_{v} max\{|c|_{v}, |d|_{v}\} ≤ c_{0}|detg|^{\frac{1}{2}}

if

γ g = ( \begin{matrix}a & b \\ c & d \end{matrix} )

If F is a number field let O_{F} be the ring of integers in F and if F is a function field take any transcendental element x of F over which F is separable and let O_{F} be the integral closure in F of the ring generated by 1 and x. A place v will be called finite if |a|_{v} ≤ 1 for all a in O_{F}; otherwise it will be called infinite. If S is a finite set of places which contains all the infinite places let

\begin{aligned}𝔸(S) & = \{a ∈ 𝔸 | |a|_{v} ≤ 1 if v ≠ S\} \\ I(S) & = \{a ∈ I | |a|_{v} = 1 if v ≠ S\} \end{aligned}

Then 𝔸 = F + 𝔸(S) and if S is sufficiently large I = F^{×}I(S). We first verify that if I = F^{×}I(S) then

G_{𝔸} = G_{F} G_{𝔸(S)}

where G_{𝔸(S)} = GL(2, 𝔸(S)). If v is not in S then v is non-archimedean and we can speak of ideals of F_{v}. Any element of G_{𝔸} may be written as a product

g = ( \begin{matrix}α & β \\ 0 & γ \end{matrix} ) ( \begin{matrix}a & b \\ c & d \end{matrix} )

in which the second factor belongs to

K = ∏_{v} K_{v}

and therefore to G_{𝔸(S)}. It will be sufficient to show that the first factor is in G_{F} G_{𝔸(S)}. If α = α_{1}α_{2} and γ = γ_{1}γ_{2} with α_{1} and γ_{1} in F^{×} and α_{2} and γ_{2} in I(S)

( \begin{matrix}α & β \\ 0 & γ \end{matrix} ) = ( \begin{matrix}α_{1} & 0 \\ 0 & γ_{1} \end{matrix} ) ( \begin{matrix}1 & β/α_{1}γ_{2} \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{2} & 0 \\ 0 & γ_{2} \end{matrix} )

The first factor is in G_{F} and the third in G_{𝔸(S)}. Since \frac{β}{α_{1}γ_{2}} belongs to F + 𝔸(S) the second factor is in G_{F} G_{𝔸(S)} and the assertion follows.

There is certainly a u in O_{F} such that |u|_{v} < 1 at all finite places in S. Enlarging S if necessary we may assume that a finite place v belongs to S if and only if |u|_{v} < 1. Then

F ∩ 𝔸(S) = \{\frac{x}{u^{m}} | x ∈ O_{F} , m ∈ ℤ\}.

We identify the prime ideals of O_{F} with the places corresponding to them. By the theory of rings of quotients the proper ideals of F ∩ 𝔸(S) are the ideals of the form

(F ∩ 𝔸(S)) ∏_{𝔭∉S} 𝔭^{m_{𝔭}}

Since I = F^{×}I(S) every such ideal is principal. Thus F ∩ 𝔸(S) is a principal ideal domain.

To prove the lemma we show that there is a constant c_{0} such that if g belongs to G_{𝔸(S)} there is a γ in G_{F ∩𝔸(S)} such that

∏_{v∈S} max\{|c|_{v}, |d|_{v}\} ≤ c_{0}|detg|^{\frac{1}{2}}

if

γ g = ( \begin{matrix}a & b \\ c & d \end{matrix} ) .

Fix a Haar measure on the additive group 𝔸(S). This determines a measure on 𝔸(S) ⊕ 𝔸(S). The group L = (F ∩ 𝔸(S)) ⊕ (F ∩ 𝔸(S)) is a discrete subgroup of 𝔸(S) ⊕ 𝔸(S) and the quotient 𝔸(S) ⊕ 𝔸(S)/L is compact and has finite measure c_{1}. If g belongs to G_{𝔸(S)} the lattice Lg is also discrete and the quotient 𝔸(S) ⊕ 𝔸(S)/Lg has measure c_{1}|detg|.

Suppose (m, n) = (μ, ν)g belongs to Lg. If a ≠ 0 belongs to F ∩ 𝔸(S) then

∏_{v∈S} max{|am|_{v}, |an|_{v}} = ( ∏_{v∈S} |a|_{v})( ∏_{v∈S} max{|c|_{v}, |d|_{v}}).

Since

1 = ∏_{v} |a|_{v} = ( ∏_{v∈S} |a|_{v})( ∏_{v≠S} |a|_{v})

the product ∏_{v∈S} |a|_{v} is at least 1 and

∏_{v∈S} max{|am|_{v}, |an|_{v}} ≥ ∏_{v∈S} max{|m|_{v}, |n|_{v}}.

Let R be a positive number and consider the set

E = {(m, n) ∈ Lg | ∏_{v∈S} max{|m|_{v}, |n|_{v}} ≤ R}.

The previous inequality shows that if I contains a non-zero element of Lg it contains one (m, n) = (μ, ν)g for which μ and ν are relatively prime. Then we may choose κ and λ in F ∩ 𝔸(S) so that κν − λμ = 1. If

γ = ( \begin{matrix}κ & λ \\ μ & ν \end{matrix} )

then γ belongs to G_{F} ∩ 𝔸(S) and if

γg = ( \begin{matrix}a & b \\ c & d \end{matrix} )

Chapter 2 170 then c = m and d = n so that

∏_{v∈S} max{|c|_{v}, |d|_{v}} ≤ R.

To prove the lemma we have to show that there is a constant c_{0} such that if g is in G_{𝔸(S)} and R = c_{0}|detg|^{\frac{1}{2}} the set E is not reduced to \{0\}. We will show in fact that there is a constant c_{2} such that for all g there is a non-zero vector (m, n) in Lg with

sup_{v∈S} max\{|m|_{v}, |n|_{v}\} ≤ c_{2}|detg|^{\frac{1}{2s}}

if s is the number of elements in S. There is certainly a positive constant c_{3} such that the measure of

{(m, n) ∈ 𝔸(S) ⊕ 𝔸(S) | sup_{v∈S} max{|m|_{v}, |n|_{v}} ≤ R}

is, for any choice of R, at least c_{3}R^{2s}. Choose c_{2} so that

c_{2} > 2(\frac{c_{1}}{c_{3}} )^{\frac{1}{2s}} .

If Lg contained no non-zero vector satisfying the desired inequality the set

\{(m, n) ∈ 𝔸_{s} ⊕ 𝔸_{S} | sup_{v∈S} max\{|m|_{v}, |n|_{v}\} ≤ \frac{c_{2}}{2} |detg|^{\frac{1}{2s}} \}

would intersect none of its translates by the elements of Lg. Therefore its measure would not be changed by projection on 𝔸(S) ⊕ 𝔸(S)/Lg and we would have

c_{1} ≤ c_{3}(\frac{c_{2}}{2} )^{2s}

which is impossible.

Choose some place v of F which is to be archimedean if F is a number field. If c is any positive constant there is a compact set C in I such that

{a ∈ I | |a| ≥ c}

is contained in

{ab | a ∈ F_{v}^{×}, |a| ≥ c, b ∈ C}

If ω_{1} is a compact subset of 𝔸, ω_{2} a compact subset of I, and c a positive constant we may introduce the Siegel domain 𝔖 = 𝔖(ω_{1}, ω_{2}, c, v) consisting of all

g = ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) ( \begin{matrix}bb_{1} & 0 \\ 0 & 1 \end{matrix} ) k

with x in ω_{1}, a in I, b in ω_{2}, b_{1} in F_{v}^{×} with |b_{1}| ≥ c, and k in K. Then Z_{𝔸}𝔖 = 𝔖. If we use the Iwasawa decomposition of G_{𝔸} to calculate integrals we easily see that the projection of 𝔖 on Z_{𝔸} \G_{𝔸} has finite measure. Moreover it follows readily from the previous lemma that, for a suitable choice of ω_{1}, ω_{2}, and c,

G_{𝔸} = G_{F} 𝔖.

Chapter 2 171 Thus Z_{𝔸} G_{F} \ G_{𝔸} has finite measure.

Let φ be a continuous function on G_{F} \ G_{𝔸}. If it is Z_{𝔸}-finite the space V spanned by the functions ρ(a) φ, a ∈ Z_{𝔸}, is finite dimensional. We may choose a finite set of points g_{1}, ⋯, g_{p} and a basis φ_{1}, ⋯, φ_{p} of V so that φ_{i}(g_{j}) = δ_{ij}. Then

ρ(a) φ = ∑_{i=1}^{p} λ_{i}(a) φ_{i}.

Since λ_{i}(a) = φ(ag_{i}) the function λ_{i} are continuous and finite as functions on Z_{𝔸} or Z_{F} \Z_{𝔸}. Since Z_{F} \ Z_{𝔸} is isomorphic to F^{×} \ I it satisfies the hypothesis of Lemma 8.1 and λ_{i} is a finite linear combination of functions of the form

λ_{i} \begin{aligned}( \\ ( & \begin{matrix}α & 0 \\ 0 & α \end{matrix} & ) \\ & & ) \end{aligned} = χ(a) (log |α|)^{m}

where χ is a quasi-character of F^{×} \ I.

A continuous function φ on G_{F} \ G_{𝔸} which is Z_{𝔸}-finite will be called slowly increasing if for any compact set Ω in G_{𝔸} and any c > 0 there are constants M_{1} and M_{2} such that

|φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) | ≤ M_{2} |a|^{M_{1}}

for g in Ω, a in I, and |a| ≥ c. If such an inequality is valid, with suitable choice of M_{2}, for any M_{1} we will say, for lack of a better terminology, that φ is rapidly decreasing.

Suppose φ is a continuous function on G_{F} \ G_{𝔸}. Assume it is K-finite on the right and that for every elementary idempotent ξ in ℋ the space

{ρ(ξf)φ | f ∈ ℋ}

is finite dimensional. An argument used more than once already shows that there is a ξ and an f in ξℋ_{1}ξ such that ρ(f)φ = φ. If a belongs to Z_{𝔸}

ρ(a) φ = ρ(δ_{a} ∗ f) φ

so that φ is Z_{𝔸}-finite. Thus we can make the following definition.

Definition 10.2 A continuous function φ on G_{F} \ G_{𝔸} is said to be an automorphic form if

(i) It is K-finite on the right (ii) For every elementary idempotent ξ in ℋ the space

{ρ(ξf)φ | f ∈ ℋ}

is finite dimensional. (iii) If F is a number field ξ is slowly increasing.

We observe, with regret, in passing that there has been a tendency of late to confuse the terms automorphic form and automorphic function. If not the result it is certainly the cause of much misunderstanding and is to be deplored.

Let 𝒜 be the vector space of automorphic forms. If φ is in 𝒜 and f is in ℋ then ρ(f)φ is in 𝒜 so that ℋ operates on 𝒜. A continuous function on φ on G_{F} \ G_{𝔸} is said to be cuspidal if

∫_{F \𝔸} φ (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) dx = 0

for all g inG_{𝔸}. An automorphic form which is cuspidal is called a cusp form. The space 𝒜_{0} of cusp forms is stable under the action of ℋ.

Chapter 2 172 Proposition 10.3 Let F be a function field and let φ be a function on G_{F} \ G_{𝔸}. If φ satisfies the following three conditions it is a cusp form.

(i) φ is K-finite on the right. (ii) φ is cuspidal. (iii) There is a quasi-character η of F^{×} \ I such that

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g)

for all a in I.

If ξ is an elementary idempotent of ℋ there is an open subgroup K′ of K such that ξ is invariant under translations on either side by the elements of K′. Therefore the functions ρ(ξf)φ are invariant under right translations. To prove the proposition we show that if K′ is a given open subgroup of K and η is a given quasi-character of F^{×} \ I then the space V of all continuous functions φ on G_{F} \ G_{𝔸} which are cuspidal and satisfy φ(gk) = φ(g) for all k in K′ as well as

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g)

for all a in F^{×} \ I is finite-dimensional.

We shall show that there is a compact set C in G_{𝔸} such that the support of every φ in V is contained in G_{F} Z_{ℤ}C. Then the functions in V will be determined by their restrictions to C. Since C is contained in the union of a finite number of left translates of K′ they will actually be determined by their values on a finite set and V will be finite dimensional.

Choose a Siegel domain 𝔖 = 𝔖(ω_{1}, ω_{2}, c, v) so that G_{𝔸} = G_{F} 𝔖. If

𝔖′ = \{( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}bb_{1} & 0 \\ 0 & 1 \end{matrix} ) k | x ∈ ω_{1}, b ∈ ω_{2}, b_{1} ∈ F_{v}^{×}, |b_{1}| ≥ c, k ∈ K\}

we have just to show that the support in 𝔖′ of every φ in V is contained in a certain compact set which is independent of φ. In fact we have to show the existence of a constant c_{1} such that φ vanishes on

( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}bb_{1} & 0 \\ 0 & 1 \end{matrix} ) k

as soon as |b_{1}| ≥ c_{1}. Let k_{1}, ⋯, k_{n} be a set of representatives of the cosets of K/K′ and let φ_{i}(g) = φ(gk_{i}). If k belongs to k_{i}K′ then φ(gk) = φ_{i}(g) and it will be enough to show that there is a constant c_{2} such that, for 1 ≤ i ≤ n,

φ_{i} (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} )) = 0

if x belongs to 𝔸 and |a| > c_{2}. It is enough to show this for a single, but arbitrary, φ_{i}. Since φ_{i} satisifes the same hypothesis as φ, perhaps with a different group K′, we just prove the corresponding fact for φ.

We use the following lemma which is an immediate consequnce of the theorem of Riemann– Roch as described in reference [10] of Chapter I.

Chapter 2 173 Lemma 10.3.1 Let X be an open subgroup of 𝔸. There is a constant c_{2} such that 𝔸 = F + aX if a belongs to I and |a| > c_{2}.

Let X be the set of all y for which

( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} )

belongs to K′. Since

φ (( \begin{matrix}1 & ay \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & x \\ 0 & 1 \end{matrix} )) = φ (( \begin{matrix}a & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))

we have

φ (( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & x \\ 0 & 1 \end{matrix} )) = φ \begin{aligned}( \\ ( & \begin{matrix}a & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

if z is in aX. The equation also holds for z in F and therefore for all z in 𝔸 if |a| > c_{2}. Then

φ \begin{aligned}( \\ ( & \begin{matrix}a & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = \frac{1}{measure(F \ 𝔸)} ∫_{F \𝔸} φ (( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & x \\ 0 & 1 \end{matrix} )) dz

which by assumption is zero.

There is a corollary.

Proposition 10.4 Suppose φ is a cusp form and for some quasi-character η of F^{×} \ I

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g)

for all a in I. Then φ is compactly supported modulo G_{F} \ Z_{𝔸}. Moreover the function

a → φ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

on F^{×} \ I is compactly supported.

The first assertion has just been verified. We know moreover that there is a constant c such that

φ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

is 0 for |a| ≥ c. If

w = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )

and φ′(g) = φ(gw) then φ′ is also a cusp form. Since

φ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = φ (w^{−1} ( \begin{matrix}1 & 0 \\ 0 & a \end{matrix} ) w) = η(a) φ \begin{aligned}( \\ ( & \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

there is also a constant c_{1} such that it vanishes for |a| ≤ c_{1}.

Proposition 10.5 Let F be a function field and η a quasi-character of F^{×} \ I. Let 𝒜_{0}(η) be the space of cusp forms φ for which

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g)

for all a in I. The representation of ℋ on 𝒜_{0}(η) is the direct sum of irreducible admissible representations each occurring with finite multiplicity.

The proof of Proposition 10.3 showed that the representation π of ℋ on 𝒜_{0}(η) is admissible. Let η′(α) = |η(α)|^{−1}η(α). The map φ → φ′ is an isomorphism of 𝒜_{0}(η) with 𝒜_{0}(η′) which replaces π by η_{1} ⊗ π if η_{1}(α) = |η(α)|^{−1/2}. Thus we may as well suppose that η is a character. Then if φ_{1} and φ_{2} belong to 𝒜_{0}(η) the function φ_{1}\overline{φ_{2}} is a function on G_{F} Z_{𝔸} \ G_{𝔸}. Since it has compact support we may set

(φ_{1}, φ_{2}) = ∫_{G_{F} Z_{𝔸}\G_{𝔸}} φ_{1}(g) \overline{φ_{2}(g)} dg.

It is easily seen that

(ρ(f)φ_{1}, φ_{2}) = (φ_{1}, ρ(f^{∗})φ_{2})

so that, by Lemma 9.4, π is the direct sum of irreducible admissible representations. Since π is admissible the range of π(ξ) is finite dimensional for all ξ so that no irreducible representation occurs an infinite number of times.

The analogue of this proposition for a number field is somewhat more complicated. If φ is a continuous function on G_{𝔸}, if v is a place of F, and if f_{v} belongs to ℋ_{v} we set

ρ(f_{v})φ = ∫_{G_{v}} φ(gh_{v}) f_{v}(h_{v}) dh_{v}.

Since f_{v} may be a measure the expression on the right is not always to be taken literally. If v is archimedean and if the function φ(hg_{v}) on G_{v} is infinitely differentiable for any h in G_{𝔸} then for any X in 𝔄_{v} the universal enveloping algebra of G_{v}, we can also define ρ(X)φ. If S is a finite set of places we can in a similar fashion let the elements of

ℋ_{s} = ⊗_{v∈S}ℋ_{v}

or, if every place in S is archimedean,

𝔄_{S} = ⊗_{v∈S}𝔄_{v}

act on φ. It is clear what an elementary idempotent in ℋ_{S} is to be. If S = S_{a} is the set of archimedean places we set ℋ_{a} = ℋ_{S}.

Chapter 2 175 Proposition 10.6 Suppose F is a number field. A continuous function φ on G_{F} \ G_{𝔸} is a cusp form if it satisfies the following five conditions.

(i) φ is K-finite on the right. (ii) φ is cuspidal. (iii) There is a quasi-character η of F^{×} \ I such that

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g)

for all a in I. (iv) For any elementary idempotent ξ in ℋ_{a} the space

{ρ(ξf) φ | f ∈ ℋ_{a}}

is finite dimensional.

(v) φ is slowly increasing.

There is a ξ in ℋ_{a} such that ρ(ξ)φ = φ. Because of the fourth condition φ transforms according to a finite dimensional representation of ξℋ_{a}ξ and the usual argument shows that there is a function f in ℋ_{a} such that ρ(f) φ = φ.

Since φ is invariant under right translations by the elements of an open subgroup of ∏_{v∉S_{a}} K_{v} this implies in turn the existence of another function f in ℋ such that ρ(f)φ = φ. From Theorem 2 of [14] one infers that φ is rapidly decreasing.

As before we may assume that η is a character. Then φ is bounded and therefore its absolute value is square integrable on G_{F} Z_{𝔸} \ G_{𝔸} which has finite measure. Let L^{2}(η) be the space of measurable functions h on G_{F} \ G_{𝔸} such that

h (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) h(g)

for all g in G_{𝔸} and all a in I and

∫_{G_{F} Z_{𝔸}\G_{𝔸}} |h(g)|^{2} dg < ∞.

According to a theorem of Godement (see reference [11] to Chapter I) any closed subspace of L^{2}(η) which consists entirely of bounded functions is finite dimensional.

What we show now is that if ξ is an elementary idempotent of ℋ the space

V = {ρ(ξf) φ | f ∈ ℋ}

is contained in such a closed subspace. The functions in V itself certainly satisfy the five conditions of the proposition and therefore are bounded and in L^{2}(η). Replacing ξ by a larger idempotent if necessary we may suppose that ξ = ξ_{a} ⊗ ξ̂_{a} where ξ_{a} is an elementary idempotent in ℋ_{a}. There is a two-sided ideal 𝔞 in ξ_{a}ℋ_{a}ξ_{a} such that ρ(f)φ = 0 if f belongs to 𝔞. The elements of 𝔞 continue to annihilate V and its closure in L^{2}(η). Approximating the δ-function as usual we see that there is a function f_{1} in ℋ_{a} and a polynomial P with non-zero constant term such that P (f_{1}) belongs to 𝔞. Therefore there is a function f_{2} in ℋ_{a} such that f_{2} − 1 belongs to 𝔞. To complete the proof of the proposition we have merely to refer to Theorem 2 of [14] once again.

For a number field the analogue to Proposition 10.4 is the following.

Propositon 10.7 Suppose φ is a cusp form and for some quasi-character η of F^{×} \ I

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g)

for all a in I. Then for any real number M_{1} there is a real number M_{2} such that

|φ \begin{matrix}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{matrix}| ≤ M_{2} |a|^{M_{1}}

for all a in I. Moreover the absolute value of φ is a square integrable on G_{F} Z_{𝔸} \ G_{𝔸}.

We need another corollary of Proposition 10.6. To prove it one has just to explain the relation between automorphic forms on G_{𝔸} and G_{ℝ}, which is usually assumed to be universally known, and then refer to the first chapter of reference [11] to Chapter I. It is perhaps best to dispense with any pretence of a proof and to rely entirely on the reader’s initiative. We do not however go so far as to leave the proposition itself unstated.

Proposition 10.8 Let ℨ_{v} be the centre of 𝔄_{v} and let ℑ be an ideal of finite codimension in ℨ = ⊗_{v∈S_{a}}ℨ_{v}. Let ξ be an elementary idempotent of ℋ and η a quasi-character of F^{×} \ I. Then the space of infinitely differentiable functions φ on G_{F} \ G_{𝔸} which satisfy the following five conditions is finite dimensional.

(i) φ is cuspidal. (ii) ρ(ξ)φ = φ. (iii) If a is in I then

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g).

(iv) ρ(X)φ = 0 for all X in ℑ

(v) φ is slowly increasing.

Proposition 10.9 Let η be a quasi-charcter of F^{×} \ I and let 𝒜_{0}(η) be the space of cusp forms φ for which

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = φ(g)

for all a in I. The representation of ℋ on 𝒜_{0}(η) is the direct sum of irreducible admissible representations each occurring with finite multiplicity.

Every element of 𝒜_{0}(η) is annihilated by some ideal of finite codimension in ℨ. If ℑ is such an ideal let 𝒜_{0}(η, ℑ) be the space of functions in 𝒜_{0}(η) annihilated by ℑ. It is enough to prove the first part of the proposition for the space 𝒜_{0}(η, ℑ). Then one may use the previous proposition and argue as in the proof of Proposition 10.5. To show that every representation occurs with finite multiplicity one combines the previous proposition with the observation that two functions transforming under the same representation of ℋ are annihilated by the same ideal in ℨ.

ℋ acts on the space 𝒜. An irreducible admissible representation π of ℋ is a constituent of the representation on 𝒜 or, more briefly, a constituent of 𝒜 if there are two invariant subspaces U and V of 𝒜 such that U contains V and the action on the quotient space U/V is equivalent to π. A constituent of 𝒜_{0} is defined in a similar fashion. The constituents of 𝒜_{0} are more interesting than the constituents of 𝒜 which are not constituents of 𝒜_{0}.

Chapter 2 177 Theorem 10.10 Let π = ⊗π_{v} be an irreducible admissible representation of ℋ which is a constituent of 𝒜 but not of 𝒜_{0}. Then there are two quasi-characters μ and ν of F^{×} \ I such that for each place v the representation π_{v} is a constituent of ρ(μ_{v}, ν_{v}).

μ_{v} is the restriction of μ to F_{v}^{×}. Let ℬ be the space of all continuous functions φ on G_{𝔸} satisfying the following conditions.

(i) For all x in 𝔸

φ (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = φ(g).

(ii) For all α and β in F^{×}

φ (( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) g) = φ(g).

(iii) φ is K-finite on the right. (iv) For every elementary idempotent ξ in ℋ the space

{ρ(ξf)φ | f ∈ ℋ}

is finite dimensional.

Lemma 10.10.1 A continuous function φ on G_{𝔸} which satisfies the first three of these conditions satisfies the fourth if and only if it is A_{𝔸}-finite on the left.

A is the group of diagonal matrices. Since φ is a function on A_{F} \ G_{𝔸} it is A_{𝔸} finite if and only if it is A_{F} \ A_{𝔸} finite. If it is A_{F} \ A_{𝔸} finite there is a relation of the form

φ(ag) = ∑_{i} λ_{i}(a) φ_{i}(g)

where the λ_{i} are finite continuous functions on A_{F} \ A_{𝔸}. Since A_{F} \ A_{𝔸} is isomorphic to the direct product of F^{×} \ I with itself it is a group to which Lemma 8.1 can be applied. Thus there is a unique family φ_{m,n,μ,ν} of functions on G_{𝔸} such that

φ (( \begin{matrix}a_{1} & 0 \\ 0 & a_{2} \end{matrix} ) g) = |\frac{a_{1}}{a_{2}} | ∑ μ(a_{1}) ν(a_{2}) (log |a_{1}|)^{m} (log |a_{2}|)^{n} φ_{m,n,μ,ν}(g)

The functions φ_{m,n,μ,ν} also satisfy the first three conditions. Moreover there is a finite set S of pairs (μ, ν) and a non-negative integer M such that φ_{m,n,μ,ν} is 0 if (μ, ν) does not belong to S or m + n > M.

Given S and M let ℬ(S, M) be the space of continuous functions f on G_{𝔸} which satisfy the first three conditions and for which

f (( \begin{matrix}a_{1} & 0 \\ 0 & a_{2} \end{matrix} ) g)

can be expanded in the form

|\frac{a_{1}}{a_{2}} |^{\frac{1}{2}} ∑ μ(a_{1}) ν(a_{2}) (log |a_{1}|)^{m} (log |a_{2}|)^{n} f_{m,n,μ,ν}(g)

Chapter 2 178 where the sum is taken only over the pairs (μ, ν) in S the pairs (m, n) for which m + n ≤ M. ℬ(S, M) is invariant under ℋ. To show that if φ is A_{F} \ A_{𝔸} finite it satisfies the fourth condition we show that the range of ρ(ξ) on ℬ(S, M) is finite dimensional.

A function f in ℬ(S, M) is determined by the restriction of the finitely many functions f_{m,n,μ,ν} to K. If f is in the range of ρ(ξ) these resrictions lie in the range of ρ(ξ) acting on the continuous functions on K. That range is finite dimensional.

We have also to show that if φ satisifes the fourth condition it is A_{𝔸} finite. The space V spanned by the right translates of φ by the elements of K is finite dimensional and each element in it satisfies all four conditions. Let φ_{1}, ⋯, φ_{p} be a basis of V. We can express φ(gk) as

∑_{i=1}^{p} λ_{i}(k) φ_{i}(g).

Because of the Iwasawa decomposition G_{𝔸} = N_{𝔸}A_{𝔸}K it is enough to show that the restriction of each φ_{i} to A_{𝔸} is finite. Since φ_{i} satisfies the same conditions as φ we need only consider the restriction of φ.

Since φ is K finite there is a finite set S of places such that φ is invariant under right translations by the elements of ∏_{v∉S} K_{v}. Let

I_{S} = ∏_{v∈S} F_{v}^{×}.

We regard I_{S} as a subgroup of I. If we choose S so large that I = F^{×}I(S) then every element α of I is a product of α = α_{1}α_{2}α_{3} with α_{1} in F^{×}, α_{2} in I_{S}, and α_{3} in I(S) such that its component at any place in S is 1. If β in I is factored in a similar fashion

φ \begin{aligned}( \\ ( & \begin{matrix}α & 0 \\ 0 & β \end{matrix} & ) \\ & & ) \end{aligned} = φ \begin{aligned}( \\ ( & \begin{matrix}α_{2} & 0 \\ 0 & β_{2} \end{matrix} & ) \\ & & ) \end{aligned} .

Thus we need only show that the restriction of φ to

A_{S} = \{ ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) | α, β ∈ I_{S}\}

is finite. This is a consequence of Corollary 8.4 since the restriction of φ to G_{S} clearly satisfies the conditions of the corollary.

The next lemma explains the introduction of ℬ.

Lemma 10.10.2 If π is a constituent of 𝒜 but not of 𝒜_{0} then it is a constituent of ℬ.

If φ belongs to 𝒜 the functions

φ_{0}(g) = \frac{1}{measure(F \ 𝔸)} ∫_{F \𝔸} φ (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) dx

belongs to ℬ. The map φ → φ_{0} commutes with the action of ℋ and its kernel is 𝒜_{0}. Suppose U and V are two invariant subspaces of 𝒜 and π occurs on the quotient of U by V. Let U_{0} be the image of U and V_{0} be the image of V in ℬ. Since π is irreducible there are two possibilities. Either U_{0} ≠ V_{0} in which case π is equivalent to the representation on U_{0}/V_{0} and is a constituent of ℬ or U_{0} = V_{0}. In the latter case

U = V + U ∩ 𝒜_{0}

and π is equivalent to the representation on

U ∩ 𝒜_{0}/V ∩ 𝒜_{0}

which is precisely the possibility we have excluded.

Chapter 2 179 Lemma 10.10.3 If π is a constituent of ℬ then there is a pair of quasi-characters μ, ν and a non-negative integer M such that π is a constituent of ℬ(μ, ν, M).

If S consists of the single pair (μ, ν) then, by definition, ℬ(μ, ν, M) = ℬ(S, M). Suppose π occurs on the quotient of U by V. Choose the finite set S of pairs of quasi-characters and the non-negative integer M so that U ∩ ℬ(S, M) is different from V ∩ ℬ(S, M). Then π occurs on the quotient of U ∩ ℬ(S, M) by V ∩ ℬ(S, M) and we may as well assume that U is contained in ℬ(S, M). The argument used in the eighth paragraph in an almost identical context shows that

ℬ(S, M) = ⊕_{(μ,ν)∈S}ℬ(μ, ν, M)

so that the lemma is a consequence of Lemma 8.6.

The next lemma is proved in exactly the same way as Proposition 8.5

Lemma 10.10.4 If π is a constituent of ℬ(μ, ν, M) for some M then it is a constituent of ℬ(μ, ν) = ℬ(μ, ν, 0).

Let μ_{v} and ν_{v} be the restrictions of μ and ν to F_{v}^{×}. For almost all v the quasi-characters μ_{v} and ν_{v} are unramifed and there is a unique function φ_{v}^{0} in ℬ(μ_{v}, ν_{v}) such that φ_{v}^{0}(g_{v}k_{v}) = φ_{v}^{0}(g_{v}) for all k_{v} in K_{v} while φ_{v}^{0}(e) = 1. We can form

⊗_{φ_{v}^{0}}ℬ(μ_{v}, ν_{v})

There is clearly a linear map of this space into ℬ(μ, ν) which sends ⊗φ_{v} to the function

φ(g) = ∏_{v} φ_{v}(g_{v})

It is easily seen to be surjective and is in fact, although this is irrelevant to our purposes, an isomorphism. In any case an irreducible constituent of ℬ(μ, ν) is a constituent of ⊗_{v}ρ(μ_{v}, ν_{v}).

With the following lemma the proof of Theorem 10.10 is complete.

Lemma 10.10.5 If the irreducible admissible representation π = ⊗_{v}π_{v} is a constituent of ρ = ⊗ρ_{v}, the tensor product of admissible representations which are not necessarily irreducible, then, for each v, π_{v} is a constituent of ρ_{v}.

As in the ninth paragraph π and ρ determine representations π and ρ of ℋ_{v}. The new π will be a constituent of the new ρ. By Lemma 9.12 the representation π of ℋ_{v} is the direct sum of representations equivalent to π_{v}. Thus π_{v} is a constituent of π and therefore of ρ. Since ρ is the direct sum of representations equivalent to ρ_{v}, Lemma 8.6 shows that π_{v} is a constituent of ρ_{v}.

The considerations which led to Proposition 8.5 and its proof will also prove the following proposition.

Proposition 10.11 If π is an irreducible constituent of the space 𝒜_{0} then for some quasi-character η it is a constituent of 𝒜_{0}(η).

Observe that if π is a constituent of 𝒜_{0}(η) then

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η(a) I

for all a in I. There are two more lemmas to be proved to complete the preparations for the Hecke theory.

Chapter 2 180 Lemma 10.12 Suppose there is a continuous function φ on G_{𝔸} with the following properties.

(i) φ is K finite on the right. (ii) For all α and β in F^{×} and all x in 𝔸

φ (( \begin{matrix}α & x \\ 0 & β \end{matrix} ) g) = φ(g).

(iii) There is a quasi-character η of F^{×} \ I such that

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g)

for all a in I. (iv) There is a finite set S of non-archimedean places such that the space

V = ρ(ℋ̂_{S})φ

transforms under ℋ̂_{S} according to the irreducible admissible representation π = ⊗_{v∉S}π_{v}. Then V is a subspace of ℬ and there are two quasi-characters μ and ν of F^{×} \ I such that π_{v}

is a constituent of ρ(μ_{v}, ν_{v}) for all v not in S.

If one observes that there is a finite set T of places which is disjoint from S such that I = F^{×}I_{T} one can proceed as in Lemma 10.10.1 to show that φ is 𝔸-finite on the right. Thus there is a finite set R of pairs of quasi-characters and a non-negative integer M such that V is contained in ℬ(R, M). The same reduction as before shows that π is a constituent of the representation of ℋ̂_{S} on some ℬ(μ, ν) and that π_{v} is a constituent of ρ(μ_{v}, ν_{v}) if v is not in S.

Lemma 10.13 Let φ be a continuous function on G_{F} \ G_{𝔸}. If φ satisfies the four following conditions it is an automorphic form.

(i) φ is K finite on the right. (ii) There is a quasi-character η of F^{×} \ I such that

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g)

for all a in I. (iii) There is a finite set S of non-archimedean places such that ρ(ℋ̂_{S})φ transforms according to

an irreducible admissible representation of ℋ̂_{S}. (iv) If F is a number field φ is slowly increasing.

We have to show that for every elementary idempotent ξ in ℋ the space ρ(ξℋ)φ is finite dimensional. If f is a continuous function on G_{F} \ G_{𝔸} let

f_{0}(g) = \frac{1}{measure(F \ 𝔸)} ∫_{F \𝔸} f (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) dx.

The map f → f_{0} commutes with the action of ℋ or of ℋ̂_{S}. Consquently φ_{0} satisfies the conditions of the previous lemma and belongs to a space ℬ(R, M) invariant under ℋ on which ρ(ξ) has a finite dimensional range.

We need only show that

V = {f ∈ ρ(ξℋ) φ | f_{0} = 0}

is finite dimensional. If F is a function field then, by Proposition 10.3, V is contained in 𝒜_{0}(η). More precisely it is contained in the range of ρ(ξ), as an operator on 𝒜_{0}(η), which we know is finite dimensional. Suppose F is a number field. Since every place of S is non-archimedean the third condition guarantees that φ is an eigenfunction of every element of ℨ. In particular there is an ideal ℑ of finite codimension in ℨ which annihilates φ and therefore every element of ρ(ξℋ)φ. By Proposition 10.6 the space V is contained in 𝒜_{0}(η) and therefore in 𝒜_{0}(η, ℑ). By Proposition 10.8 the range of ρ(ξ) in 𝒜_{0}(η, ℑ) is finite dimensional.

Chapter 2 182 §11. Hecke theory. The preliminaries are now complete and we can broach the central topic of these notes. Let ψ be a non-trivial character of F \ 𝔸. For each place v the restriction ψ_{v} of ψ to F_{v} is non-trivial. Let π = ⊗_{v}π_{v} be an irreducible admissible representation of ℋ. The local L-functions L(s, π_{v}) and the factors ε(s, π_{v}, ψ_{v}) have all been defined. Since for almost all v the representation π_{v} contains the trivial representation of K_{v} and O_{v} is the largest ideal on which ψ_{v} is trivial almost all of the factor ε(s, π_{v}, ψ_{v}) are identically 1 and we can form the product

ε(s, π) = ∏_{v} ε(s, π_{v}, ψ_{v}).

In general it depends on ψ. Suppose however that

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η(a)I

and that η is trivial on F^{×}. If ψ is replaced by the character x → ψ(αx) with α in F^{×} then ε(s, π_{v}, ψ_{v}) is multiplied b η_{v}(α)|α|_{v}^{2s−1} so that ε(s, π) is multiplied by

∏_{v} η_{v}(α) |α|_{v}^{2s−1} = η(α)|α|^{2s−1} = 1

The product

∏_{v} L(s, π_{v})

does not converge and define a function L(s, π) unless π satisfies some further conditions.

Theorem 11.1 Suppose the irreducible admissible representation π = ⊗π_{v} is a constituent of 𝒜. Then the infinite products defining L(s, π) and L(s, π̃) converge absolutely in a right half-plane and the functions L(s, π) and L(s, π̃) themselves can be analytically continued to the whole complex plane as meromorphic functions of s. If π is a constituent of 𝒜_{0} they are entire. If F is a number field they have only a finite number of poles and are bounded at infinity in any vertical strip of finite width. If F is a function field with field of constants 𝔽_{q} they are rational functions of q^{−s}. Finally they satisfy the functional equation

L(s, π) = ε(s, π) L(1 − s, π̃).

Observe that if π = ⊗_{v}π_{v} then π̃ = ⊗_{v}π̃_{v}. Consider first a representation π which is a constituent of 𝒜 but not of 𝒜_{0}. There are quasi-characters μ and ν of F^{×} \ I such that π_{v} is a constituent of ρ(μ_{v}, μ_{v}) for all v. Since π_{v} has to contain the trivial representation of K_{v} for all but a finite number of v it is equal to π(μ_{v}, ν_{v}) for almost all v.

Consider first the representation π′ = ⊗_{v}π(μ_{v}, ν_{v}). Recall that

\begin{aligned}L(s, π(μ_{v}, ν_{v})) & = L(s, μ_{v}) L(s, ν_{v}) \\ L(s, π̃(μ_{v}, ν_{v})) & = L(s, μ_{v}^{−1}) L(s, ν_{v}^{−1}) \end{aligned}

and

ε(s, π(μ_{v}, ν_{v}), ψ_{v}) = ε(s, μ_{v}, ψ_{v}) ε(s, ν_{v}, ψ_{v})

If χ is any quasi-character of F^{×} \ I the product

∏_{v} L(s, χ_{v})

is known to converge in a right half plane and the funtion L(s, χ) it defines is known to be analytically continuable to the whole plane as a meromorphic function. Moreover if

ε(sχ) = ∏_{v} ε(s, χ_{v}, ψ_{v})

the functional equation

L(s, χ) = ε(s, χ) L(1 − s, χ^{−1})

is satisifed. Since

L(s, π′) = L(s, μ) L(s, ν)

and

L(s, π̃′) = L(s, μ^{−1}) L(s, ν^{−1})

they too are defined and meromorphic in the whole plane and satisfy the functional equation

L(s, π′) = ε(s, π′) L(1 − s, π̃′).

The other properties of L(s, π′) demanded by the lemma, at least when π′ is a constituent of 𝒜, can be inferred from the corresponding properties of L(s, μ) and L(s, ν) which are well known.

When π_{v} is not π(μ_{v}, μ_{v}) it is σ(μ_{v}, ν_{v}). We saw in the first chapter that

\frac{L(s, σ(μ_{v}, ν_{v}))}{L(s, π(μ_{v}, ν_{v}))}

is the product of a polynomial and an exponential. In particular it is entire. If we replace π(μ_{v}, ν_{v}) by π_{v} we change only a finite number of the local factors and do not disturb the converge of the infinite product. If S is the finite set of places v at which π_{v} = σ(μ_{v}, ν_{v}) then

L(s, π) = L(s, π′) ∏_{v∈S} \frac{L(s, σ(μ_{v}, ν_{v}))}{L(s, π(μ_{v}, ν_{v}))}

and therefore is meromorphic with no more poles that L(s, π′). For L(s, π̃) the corresponding equation is

L(s, π̃) = L(s, π̃′) ∏_{v∈S} \frac{L(s, σ(μ_{v}^{−1}, ν_{v}^{−1}))}{L(s, π(μ_{v}^{−1}, ν_{v}^{−1}))}.

The functional equation of L(s, π) is a consequence of the relations

\frac{L(s, σ(μ_{v}, ν_{v}))}{L(s, π(μ_{v}, ν_{v}))} = \frac{ε(s, σ(μ_{v}, ν_{v}), ψ_{v}) L(1 − s, σ(μ_{v}^{−1}, ν_{v}^{−1}))}{ε(s, π(μ_{v}, ν_{v}), ψ_{v}) L(1 − s, π(μ_{v}^{−1}, ν_{v}^{−1}))}

Chapter 2 184 which were verified in the first chapter. It also follows from the form of the local factors that L(s, π) and L(s, π̃) are rational functions of q^{−s} when F is a function field. If F is a number field L(s, π) is bounded in vertical strips of finite width in a right half-plane and, because of the functional equation, in vertical strips in a left half-plane. Its expression in terms of L(s, π′) prevents it from growing very fast at infinity in any vertical strip of finite width. The Phrágmen–Lindelöf principle implies that it is bounded at infinity in any such strip.

Now suppose π is a constituent of 𝒜_{0}. It is then a constituent of 𝒜_{0}(η) if

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η(a) I

for a in I. Since the representation of ℋ in 𝒜_{0}(η) is the direct sum of invariant irreducible subspaces there is an invariant subspace U of 𝒜_{0}(η) which transforms according ot π. Let φ belong to U. If g is in G_{𝔸}

φ_{g}(x) = φ (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g)

is a function on F \ 𝔸. Since φ_{g} is continuous it is determined by its Fourier series. The constant term is

\frac{1}{measure F \ 𝔸} ∫_{F \𝔸} φ (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) dx

which is 0 because φ is a cusp form. If ψ is a given non-trivial character of F \ 𝔸 the other non-trivial characters are the functions x → ψ(αx) with α in F^{×}. Set

φ_{1}(g) = \frac{1}{measure F \ 𝔸} ∫_{F \𝔸} φ (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) ψ(−x) dx.

Since φ is a function on G_{F} \ G_{𝔸}.

φ_{1} (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) = \frac{1}{measure F \ 𝔸} ∫_{F \𝔸} φ (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) ψ(−αx) dx

if α belongs to F^{×}. Thus, formally at least,

φ(g) = φ_{g}(e) = ∑_{α∈F^{×}} φ_{1} (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) .

In any case it is clear that φ_{1} is not 0 unless φ is.

Let

U_{1} = {φ_{1} | φ ∈ U}.

Since the map φ → φ_{1} commutes with the action of ℋ the space U_{1} is invariant and transforms according to π under right translation by ℋ. Moreover

φ_{1} (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = ψ(x) φ_{1}(g)

if x is in 𝔸. If F is a number field φ is slowly increasing. Therefore if Ω is a compact subset of G_{𝔸} there is a real number M such that

φ_{1} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) = O(|a|^{M})

as |a| → ∞ for all g in Ω. Propositions 9.2 and 9.3 imply that all π_{v} are infinite dimensional and that U_{1} is W (π, ψ). Therefore U_{1} is completely determined by π and ψ and U is completely determined by π. We have therefore proved the following curious proposition.

Chapter 2 185 Proposition 11.1.1 If an irreducible represntation of ℋ is contained in 𝒜_{0}(η) it is contained with multiplicity one.

For almost all v there is in W (π_{v}, ψ_{v}) a function φ_{v}^{0} such that φ_{v}^{0}(g_{v}k_{v}) = φ_{v}^{0}(g_{v}) for all k_{v} in K_{v} while φ_{v}^{0}(e) = 1. W (π, ψ) is spanned by functions of the form

φ_{1}(g) = ∏_{v} φ_{v}(g_{v})  (11.1.2)

where φ_{v} is in W (π_{v}, ψ_{v}) for all v and equal to φ_{v}^{0} for almost all v.

Suppose φ corresponds to a function φ_{1} of the form (11.1.2). Suppose φ = φ_{v}^{0} so that π_{v} contains the trivial representation of K_{v}. If ε_{v} is the normalized Haar measure on K_{v} let λ_{v} be the homomorphism of ε_{v}ℋ_{v}ε_{v} into ℂ associated to π_{v}. If f_{v} is in ε_{v}ℋ_{v}ε_{v} then

λ_{v}(f_{v}) φ(g) = ∫_{G_{v}} φ(gh) f_{v}(h) dh

and if λ′_{v} is the homomorphism associated to |η_{v}|^{−1/2} ⊗ π_{v}

λ′_{v}(f_{v}) |η(detg)|^{− \frac{1}{2}} φ(g)

is equal to

∫_{G_{v}} |η(detgh)|^{− \frac{1}{2}} φ(gh) f_{v}(h) dh.

Since φ is a cusp form the function |η(detg)|^{−1/2} φ(g) is bounded and λ′_{v} satisifies the conditions of Lemma 3.10. Thus if π_{v} = π(μ_{v}, ν_{v}) both μ_{v} and ν_{v} are unramified and

\begin{aligned}|η(ϖ_{v})|^{\frac{1}{2}} |ϖ_{v}|^{\frac{1}{2}} & ≤ |μ_{v}(ϖ_{v})| ≤ |η(ϖ_{v})|^{\frac{1}{2}} |ϖ_{v}|^{− \frac{1}{2}} \\ |η(ϖ_{v})|^{\frac{1}{2}} |ϖ_{v}|^{\frac{1}{2}} & ≤ |ν_{v}(ϖ_{v})| ≤ |η(ϖ_{v})|^{\frac{1}{2}} |ϖ_{v}|^{− \frac{1}{2}} \end{aligned}

if ϖ_{v} is the generator of the maximal ideal of O_{v}. Consequently the infinite products defining L(s, π) and L(s, π̃) converge absolutely for Re s sufficiently large.

We know that for any v and any φ_{v} in W (π_{v}, ψ_{v}) the integral

∫_{F_{v}^{×}} φ_{v} (( \begin{matrix}a_{v} & 0 \\ 0 & 1 \end{matrix} ) g_{v}) |a_{v}|^{s− \frac{1}{2}} d^{×}a_{v}

converges absolutely for Re s large enough. Suppose that, for all a in I, |η(a)| = |a|^{r} with r real. Applying Lemma 3.11 we see that if s + r > \frac{1}{2} and φ_{v}^{0} is defined

∫_{F_{v}^{×}} |φ_{v}^{0} (( \begin{matrix}a_{v} & 0 \\ 0 & 1 \end{matrix} ) g_{v})| |a_{v}|^{s− \frac{1}{2}} d^{×}a_{v}

is, for g_{v} in K_{v}, at most

\frac{1}{(1 − |ϖ_{v}|^{s+r− \frac{1}{2}} )^{2}} .

Thus if φ_{1} is of the form (11.1.2) the integral

Ψ(g, s, φ_{1}) = ∫_{I} φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s− \frac{1}{2}} d^{×}a

is absolutely convergent and equal to

∏_{v} Ψ(g_{v}, s, φ_{v})

for Re s sufficiently large. Since Φ(g_{v}, s, φ_{v}) is, by Proposition 3.5, equal to 1 for almost all v we can set

Φ(g, s, φ_{1}) = ∏_{v} Φ(g_{v}, s, φ_{v})

so that

Ψ(g, s, φ_{1}) = L(s, π) Φ(g, s, φ_{1}).

We can also introduce

Ψ̃(g, s, φ_{1}) = ∫_{I} φ_{1} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) η^{−1}(a) |a|^{s− \frac{1}{2}} d^{×}a

and show that

Ψ̃(g, s, φ_{1}) = L(s, π̃) Φ̃(g, s, φ_{1})

if

Φ̃(g, s, φ_{1}) = ∏_{v} Φ̃(g_{v}, s, φ_{v}).

Lemma 11.1.3 There is a real number s_{0} such that for all φ_{1} in W (π, ψ) the integrals

\begin{aligned}Ψ(g, s, φ_{1}) & = ∫_{I} φ_{1} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s− \frac{1}{2}} d^{×}a \\ Ψ̃(g, s, φ_{1}) & = ∫_{I} φ_{1} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) η^{−1}(a) |a|^{s− \frac{1}{2}} d^{×}a \end{aligned}

are absolutely convergent for Re s > s_{0}. The functions Ψ(g, s, φ_{1}) and Ψ̃(g, s, φ_{1}) can both be extended to entire functions of s. If F is a number field they are bounded in vertical strips and if F is function field they are rational functions of q^{−s}. Moreover

Ψ̃(wg, 1 − s, φ_{1}) = Ψ(g, s, φ_{1}).

We have seen that the first assertion is true for functions of the form (11.1.2). Since they form a basis of W (π, ψ) it is true in general. To show that

φ(g) = ∑_{α∈F^{×}} φ_{1} (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)

Chapter 2 187 we need only show that the series on the right is absolutely convergent. We will do this later on in this paragraph. At the moment we take the equality for granted. Then, for all φ_{1}, Ψ(g, s, φ_{1}) which equals

∫_{F^{×}\I} \{ ∑_{α∈F^{×}} φ_{1} (( \begin{matrix}αa & 0 \\ 0 & 1 \end{matrix} ) g) \} |a|^{s− \frac{1}{2}} d^{×}a

is equal to

∫_{F^{×}\I} φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s− \frac{1}{2}} d^{×}a

for Re s sufficiently large. Also Ψ̃(g, s, φ_{1}) is equal to

∫_{F^{×}\I} φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) η^{−1}(a) |a|^{s− \frac{1}{2}} d^{×}a.

We saw in the previous paragraph that, for a given g and any real number M,

|φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g)| = O(|a|^{M})

as |a| approaches 0 or ∞. Thus the two intregals define entire functions of s which are bounded in vertical strips. If F is a function field the function

a → φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g)

has compact support on F^{×} \ I so that the integral can be expressed as a finite Laurent series in q^{−s}.

The function Ψ̃(wg, 1 − s, φ_{1}) is equal to

∫ φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) wg) η^{−1}(a) |a|^{\frac{1}{2} −s} d^{×}a.

Since w is in G_{F} the equality φ(wh) = φ(h) holds for all h in G_{𝔸} and this integral is equal to

∫ φ (( \begin{matrix}1 & 0 \\ 0 & a \end{matrix} ) g) η^{−1}(a) |a|^{\frac{1}{2} −s} d^{×}a.

Since

( \begin{matrix}1 & 0 \\ 0 & a \end{matrix} ) = ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) ( \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} )

we can change variables in the integral to obtain

∫ φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s− \frac{1}{2}} d^{×}a

which is Ψ(g, s, φ_{1}).

If we choose φ_{1} of the form (11.1.2) we see that L(s, π) Φ(g, s, φ_{1}) is entire and bounded in vertical strips of finite width. For almost all v the value of Φ(g_{v}, s, φ_{v}^{0}) at the identity e is 1 and for Chapter 2 188 such v we choose φ_{v} = φ_{v}^{0}. At the other places we choose φ_{v} so that Φ(e, s, φ_{v}) is an exponential e^{a_{v}s} with real a_{v}. Then Φ(e, s, φ_{1}) is an exponential. Consequently L(s, π) is also entire and bounded in vertical strips of finite width. If F is a number field Φ(e, s, φ_{1}) will be a power of q^{−s} so that L(s, π) will be a finite Laurent series in q^{−s}. Similar considerations apply to L(s, π̃).

To prove the functional equation we start with the relation

L(s, π) ∏_{v} Φ(e, s, φ_{v}) = L(1 − s, π̃) ∏_{v} Φ̃(w, 1 − s, φ_{v}).

By the local functional equation the right hand side is

L(1 − s, π̃) ∏_{v} {ε(s, π_{v}, ψ_{v}) Φ(e, s, φ_{v})}.

Cancelling the term ∏_{v} Φ(e, s, φ_{v}) we obtain

L(s, π) = ε(s, π) L(1 − s, π̃).

Corollary 11.2 Suppose π = ⊗_{v}π_{v} is a constituent of 𝒜. For any quasi-character ω of F^{×} \ I the products

∏_{v} L(s, ω_{v} ⊗ π_{v})

and

∏_{v} L(s, ω_{v}^{−1} ⊗ π̃_{v})

are absolutely convergent for Re s sufficiently large. The functions L(s, ω ⊗ π) and L(s, ω^{−1} ⊗ π̃) they define can be analytically continued to the whole complex plane as meromorphic functions which are bounded at infinity in vertical strips of finite width and have only a finite number of poles. If F is a function field they are rational functions of q^{−s}. If π is a constituent of 𝒜_{0} they are entire. In all cases they satisfy the functional equation

L(s, ω ⊗ π) = ε(s, ω ⊗ π) L(1 − s, ω^{−1} ⊗ π̃)

if

ε(s, ω ⊗ π) = ∏_{v} ε(s, ω_{v} ⊗ π_{v}, ψ_{v}).

If π = ⊗_{v}π_{v} is a constituent of 𝒜 or 𝒜_{0} and ω is a quasi-character of F^{×} \ I so is ω ⊗ π. Moreover ω ⊗ π = ⊗_{v}(ω_{v} ⊗ π_{v}).

The converses to the corollary can take various forms. We consider only the simplest of these. In particular, as far as possible, we restrict ourselves to cusp forms.

Theorem 11.3 Let π = ⊗π_{v} be a given irreducible representation of ℋ. Suppose that the quasicharacter η of I defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η(a) I

is trivial on F^{×}. Suppose there is a real number r such that whenever π_{v} = π(μ_{v}, ν_{v}) the inequalities

|ϖ_{v}|^{−r} ≤ |μ_{v}(ϖ_{v})| ≤ |ϖ_{v}|^{r}

and

|ϖ_{v}|^{−r} ≤ |ν_{v}(ϖ_{v})| ≤ ϖ_{v}|^{r}

are satisfied. Then for any quasi-character ω of F^{×} \ I the infinite products

L(s, ω ⊗ π) = ∏_{v} L(s, ω_{v} ⊗ π_{v})

and

L(s, ω^{−1} ⊗ π̃) = ∏_{v} L(s, ω_{v}^{−1} ⊗ π̃_{v})

are absolutely convergent for Re s large enough. Suppose L(s, ω ⊗ π) and L(s, ω^{−1} ⊗ π̃) are, for all ω, entire functions of s which are bounded in vertical strips and satisfy the functional equation

L(s, ω ⊗ π) = ε(s, ω ⊗ π) L(1 − s, ω^{−1} ⊗ π̃)

If the π_{v} are all infinite dimensional π is a constituent of 𝒜_{0}.

The absolute convergence of the infinite products is clear. We have to construct a subspace U of 𝒜_{0} which is invariant under ℋ and transforms according to the representation π. W (π, ψ) transforms according to π. If φ_{1} belongs to W (π, ψ) set

φ(g) = ∑_{α∈F^{×}} φ_{1} (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)

We shall see later that this series converges absolutely and uniformly on compact subsets of G_{𝔸}. Thus φ is a continuous function on G_{𝔸}. Since the map φ_{1} → φ commutes with right translations by elements of ℋ we have to show that, for all φ_{1}, φ is in 𝒜_{0} and that φ is not zero unless φ_{1} is.

Since ψ is a character of F \ 𝔸

φ (( \begin{matrix}1 & ξ \\ 0 & 1 \end{matrix} ) g) = φ(g)

for all ξ in F. Thus, for each g,

φ (( \begin{matrix}1 & x \\ 0 & a \end{matrix} ) g)

is a function on F \ 𝔸. The constant term of its Fourier expansion is

\frac{1}{measure F \ 𝔸} ∫_{F \𝔸} φ (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) dx.

Chapter 2 190 The integral is equal to

∑_{α} ∫_{F \𝔸} φ_{1} (( \begin{matrix}1 & αx \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) dx.

A typical term of this sum is

φ_{1} (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) ∫_{F \𝔸} ψ(αx) dx = 0.

In particular φ is cuspidal. Another simple calculation shows that if β belongs to F^{×}

\frac{1}{measure F \ 𝔸} ∫_{F \𝔸} φ (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) ψ(−βx) dx

is equal to

φ_{1} (( \begin{matrix}β & 0 \\ 0 & 1 \end{matrix} ) g) .

Thus φ_{1} is zero if φ is.

By construction

φ (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) = φ(g)

if α is in F^{×}. Moreover, for all a in I,

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g).

If a is in F^{×} the right side is just φ(g). Thus φ is invariant under left translations by elements of P_{F}, the group of super-triangular matrices in G_{F}. Since G_{F} is generated by P_{F} and w = ({0 \atop −1} {1 \atop 0}) all we need do to show that φ is a function on G_{F} \ G_{𝔸} is to show that

φ(wg) = φ(g).

By linearity we need only establish this when φ_{1} has the form (11.1.2). The hypothesis implies as in the direct theorem that the integrals

Ψ(g, s, φ_{1}) = ∫_{I} φ_{1} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s− \frac{1}{2}} d^{×}a

and

Ψ̃(g, s, φ_{1}) = ∫_{I} φ_{1} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) η^{−1}(a) |a|^{s− \frac{1}{2}} d^{×}a

converge absolutely for Re s sufficiently large. Moreover

Ψ(g, s, φ_{1}) = ∏_{v} Ψ(g_{v}, s, φ_{v}) = L(s, π) ∏_{v} Φ(g_{v}, s, φ_{v}).

Almost all factors in the product on the right are identically one so that the product, and therefore Ψ(g, s, φ_{1}), is an entire function of s. In the same way

Ψ̃(g, s, φ_{1}) = L(s, π̃) ∏_{v} Φ̃(g_{v}, s, φ_{v})

and is entire. Since

Φ̃(wg_{v}, 1 − s, φ_{v}) = ε(s, π_{v}, ψ_{v}) Φ(g_{v}, s, φ_{v})

the function Ψ̃(wg, 1 − s, φ_{1}) is equal to

L(1 − s, π̃) ε(s, π) ∏_{v} Φ(g_{v}, s, φ_{v})

which, because of the functional equation assumed for L(s, π), is equal to Ψ(g, s, φ_{1}).

¿From its integral representation the function Ψ(g, s, φ_{1}) is bounded in any vertical strip of finite width contained in a certain right half-plane. The equation just established shows that it is also bounded in vertical strips of a left half-plane. To verify that it is bounded in any vertical strip we just have to check that it grows sufficiently slowly that the Phrágmen–Lindelöf principle can be applied,

Ψ(g, s, φ_{1}) = L(s, π) ∏_{v} Φ(g_{v}, s, φ_{v}).

The first term is bounded in any vertical strip by hypothesis. Almost all factors in the infinite product are identically 1. If v is non-archimedean Φ(g_{v}, s, φ_{v}) is a function of |ϖ_{v}|^{s} and is therefore bounded in any vertical strip. If v is archimedean

Φ(g_{v}, s, φ_{v}) = \frac{Ψ(g_{v}, s, φ_{v})}{L(s, π_{v})}

We have shown that the numerator is bounded at infinity in vertical strips. The denominator is, apart from an exponential factor, a Γ-function. Stirling’s formula shows that it goes to 0 sufficiently slowly at infinity.

If Re s is sufficiently large

Ψ(g, s, φ_{1}) = ∫_{F^{×}\I} ∑_{α∈F^{×}} φ_{1} (( \begin{matrix}αa & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s− \frac{1}{2}} d^{×}a

which is

∫_{F^{×}\I} φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s− \frac{1}{2}} d^{×}a.

This integral converges absolutely when Re s is sufficiently large. If Re s is large and negative

Ψ̃(wg, 1 − s, φ_{1}) = ∫_{F^{×}\I} φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) wg) η^{−1}(a) |a|^{\frac{1}{2} −s} d^{×}a

which equals

∫_{F^{×}\I} φ (w ( \begin{matrix}1 & 0 \\ 0 & a \end{matrix} ) g) η^{−1}(a) |a|^{\frac{1}{2} −s} d^{×}a.

Chapter 2 192 Using the relation

( \begin{matrix}1 & 0 \\ 0 & a \end{matrix} ) = ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) ( \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} )

and changing variables we see that this integral is equal to

∫_{F^{×}\I} φ (w ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) |a|^{s− \frac{1}{2}} d^{×}a.

Set

f_{1}(a) = φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g)

and

f_{2}(s) = φ (w ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) .

We are trying to show that for any g the functions f_{1} and f_{2} are equal. The previous discussion applies to ω ⊗ π as well as to π. If φ_{1} is in W (π, ψ) the function

φ′_{1}(g) = ω(detg) φ_{1}(g)

is in W (ω ⊗ π, ψ). When φ_{1} is replaced by φ′_{1} the function φ is replaced by

φ′(g) = ω(detg) φ(g)

and f_{i} is replaced by

f′_{i}(a) = ω(detg) ω(a) f_{i}(a).

Thus for any quasi-character ω of F^{×} \ I the integral

∫_{F^{×}\I} f_{1}(a) ω(a) |a|^{s− \frac{1}{2}} d^{×}a

is absolutely convergent for Re s sufficiently large and the integral

∫_{F^{×}\I} f_{2}(s) ω(a) |a|^{s− \frac{1}{2}} d^{×}a

is absolutely convergent for Re s large and negative. Both integrals represent functions which can be analytically continued to the same entire function. This entire function is bounded in vertical strips of finite width.

The equality of f_{1} and f_{2} is a result of the following lemma.

Lemma 11.3.1 Let f_{1} and f_{2} be two continuous functions on F^{×} \ I. Assume that there is a constant c such that for all characters of ω of F^{×} \ I the integral

∫_{F^{×}\I} f_{1}(a) ω(a) |a|^{s} d^{×}a

is absolutely convergent for Re s > c and the integral

∫_{F^{×}\I} f_{2}(a) ω(a) |a|^{s} d^{×}a

is absolutely convergent for Re s < −c. Assume that the functions represented by these integrals can be analytically continued to the same entire function and that this entire function is bounded in vertical strips of finite width. Then f_{1} and f_{2} are equal.

Let I_{0} be the group of idèles of norm 1. Then F^{×} \ I_{0} is compact. It will be enough to show that for each b in I the functions f_{1}(ab) and f_{2}(ab) on F^{×} \ I_{0} are equal. They are equal if they have the same Fourier expansions. Since any character of F^{×} \ I_{0} can be extended to a character of F^{×} \ I we have just to show that for every character ω of F^{×} \ I

f̂_{1}(ω, b) = ω(b) ∫_{F^{×}\I_{0}} f_{1}(ab) ω(a) d^{×}a

is equal to

f̂_{2}(ω, b) = ω(b) ∫_{F^{×}\I_{0}} f_{2}(ab) ω(a) d^{×}a.

These two functions are functions on I_{0} \ I which is isomorphic to ℤ if F is a number field and to ℝ if F is a function field.

If F is a function field we have only to verify the following lemma.

Lemma 11.3.2 Suppose {a_{1}(n) | n ∈ ℤ} and {a_{2}(n) | n ∈ ℤ} are two sequences and q > 1 is a real number. Suppose

∑_{n} a_{1}(n) q^{−ns}

converges for Re s sufficiently large and

∑_{n} a_{2}(n) q^{−ns}

converges absolutely for Re s large and negative. If the functions they represent can be analytically continued to the same entire function of s the two sequences are equal.

Once stated the lemma is seen to amount to the uniqueness of the Laurent expansion. If F is a number field the lemma to be proved is a little more complicated.

Chapter 2 194 Lemma 11.3.3 Suppose g_{1} and g_{2} are two continuous functions on ℝ. Suppose there is a constant c such that

ĝ_{1}(s) = ∫_{ℝ} g_{1}(x) e^{sx} dx

converges absolutely for Re s > c and

ĝ_{2}(s) = ∫_{ℝ} g_{2}(x) e^{sx} dx

converges absolutely for Re s < −c. If ĝ_{1} and ĝ_{2} represent the same entire function and this function is bounded in vertical strips then g_{1} = g_{2}.

All we need do is show that for every compactly supported infinitely differentiable function g the functions g ∗ g_{1} and g ∗ g_{2} are equal. If

ĝ(s) = ∫_{ℝ} g(x) e^{sx} dx

is the Laplace transform of g the Laplace transform of g ∗ g_{i} is ĝ(s) ĝ_{i}(s). By the inversion formula

g ∗ g_{i}(x) = \frac{1}{2πi} ∫_{b−i∞}^{b+i∞} ĝ(s) ĝ_{i}(s) e^{−xs} ds

where b > c if i = 1 and b < −c if i = 2. The integral converges because ĝ goes to 0 faster than the inverse of any polynomial in a vertical strip. Cauchy’s integral theorem implies that the integral is independent of b. The lemma follows.

To complete the proof of Theorem 11.3, and Theorem 11.1, we have to show that for any φ_{1} in W (π, ψ) the series

∑_{α∈F^{×}} φ_{1} (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)

is uniformly absolutely convergent for g in a compact subset of G_{𝔸} and that if φ(g) is its sum then, if F is a number field, for any compact subset Ω of G_{𝔸} and any c > 0 there are constants M_{1} and M_{2} such that

|φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g)| ≤ M_{1} |a|^{M_{2}}

for g in Ω and |a| ≥ c. We prefer to prove these facts in a more general context which will now be described.

For us a divisor is just a formal product of the form

D = ∏ 𝔭^{m_{𝔭}}.

It is taken over all non-archimedean places. The integers m_{𝔭} are non-negative and all but a finite number of them are 0. Let S be a finite set of non-archimedean places containing all the divisors of D, that is, all places 𝔭 for which m_{𝔭} > 0.

If a belongs to I we can write a in a unique manner as a product a_{S}â_{S} where the components of a_{S} outisde S are 1 and those of â_{S} inside S are 1. a_{S} belongs to I_{s} = ∏_{v∈S} F_{v}^{×}. Let I_{D}^{S} be Chapter 2 195 the set of idèles a such that, for any 𝔭 in S, a_{𝔭} is a unit which satisfies a_{𝔭} ≡ 1(mod 𝔭^{m_{𝔭}}). Then I = F^{×} I_{D}^{S} and F^{×} \ I is isomorphic to F^{×} ∩ I_{D}^{S} \ I_{D}^{S}.

If 𝔭 is in S let K_{𝔭}^{D} be the subgroup of all

( \begin{matrix}a & b \\ c & d \end{matrix} )

in K_{𝔭} for which c ≡ 0 (mod 𝔭^{m_{𝔭}}). Let K̂_{𝔭}^{D} be the subgroup of such matrices for which a ≡ d ≡ 1(mod 𝔭^{m_{𝔭}}). Set

K_{S}^{D} = ∏_{𝔭∈S} K_{𝔭}^{D}

and set

K̂_{S}^{D} = ∏_{𝔭∈S} K̂_{𝔭}^{D}

K̂_{S}^{D} is a normal subgroup of K_{S}^{D} and the quotient K_{S}^{D}/K̂_{S}^{D} is abelian.

Let G_{D}^{S} be the set for all g in G_{𝔸} such that g_{𝔭} is in the group K_{𝔭}^{D} for all 𝔭 in S. Any g in G_{𝔸} may be written as a product g_{S}ĝ_{S} where g_{S} has component 1 outside of S and ĝ_{S} has component 1 inside S. G_{S} is the set of g_{S} and Ĝ_{S} is the set of ĝ_{S}. In particular

G_{D}^{S} = K_{S}^{D} · Ĝ_{S}.

It is easily seen that

G_{𝔸} = G_{F} G_{D}^{S}.

In addition to D and S we suppose we are given a non-trivial character ψ of F \ 𝔸, two characters ε and ε̂ of K_{S}^{D}/K̂_{S}^{D}, two complex valued functions α → a_{α} and α → â_{α} on F^{×}, an irreducible representation π of ℋ̂_{S} = ⊗_{v∉S}ℋ_{v}, and a quasi-character η of F^{×} \ I.

There are a number of conditions to be satisifed. If

( \begin{matrix}a & 0 \\ 0 & b \end{matrix} )

belongs to K_{S}^{D} then

ε̂\begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & b \end{matrix} & ) \\ & & ) \end{aligned} = ε \begin{aligned}( \\ ( & \begin{matrix}b & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} .

If α belongs to F^{×} and β belongs to F^{×} ∩ I_{D}^{S}

a_{αβ} = ε \begin{aligned}( \\ ( & \begin{matrix}β_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} a_{α}

and

â_{αβ} = ε̂\begin{aligned}( \\ ( & \begin{matrix}β_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} â_{α}.

Chapter 2 196 The functions α → a_{α} and α → â_{α} are bounded. Moreover a_{α} = â_{α} = 0 if for some v in S the number α regarded as an element of F_{v} does not lie in the largest ideal on which ψ_{v} is trivial. If v belongs to S and a is a unit in O_{v}

ε \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η_{v}(a).

Let π = ⊗_{v∉S}π_{v}. Then for a in F_{v}^{×}

π_{v} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η_{v}(a) I.

Because of these two conditions η is determined by π and ε. There is a real number r such that if π_{v} = π(μ_{v}, ν_{v})

|ϖ_{v}|^{r} ≤ |μ_{v}(ϖ_{v})| ≤ |ϖ_{v}|^{−r}

and

|ϖ_{v}|^{r} ≤ |ν_{v}(ϖ_{v})| ≤ |ϖ_{v}|^{−r}.

Finally we suppose that π_{v} is infinite dimensional for all v not in S.

These conditions are rather complicated. Nonetheless in the next paragraph we shall find ourselves in a situation in which they are satisifed. When S is empty, D = 0, a_{α} = â_{α} = 1 for all α they reduce to those of Theorem 11.3. In particular with the next lemma the proof of that theorem will be complete. We shall use the conditions to construct a space U of automorphic forms on G_{𝔸} such that U transforms under ℋ̂_{S} according to π while each φ in U satisifes

φ(gh) = ε(h) φ(g)

for h in K_{S}^{D}. If U is such a space then for any φ in U and any a in I

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g).

This is clear if a belongs ot I_{D}^{S} and follows in general from the relation I = F^{×} I_{D}^{S}.

Recall that W (π, ψ) is the space of functions on Ĝ_{S} spanned by functions of the form

φ_{1}(g) = ∏_{v∉S} φ_{v}(g_{v})

where φ_{v} belongs to W (π_{v}, ψ_{v}) for all v and is equal to φ_{v}^{0} for almost all v.

Lemma 11.4 Suppose φ_{1} belongs to W (π, ψ).

(i) For any g in G_{D}^{S} the series

φ(g) = ∑_{α∈F^{×}} a_{α} ε(g_{S}) φ_{1} (( \begin{matrix}α̂_{S} & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S})

converges absolutely. The convergence is uniform on compact subsets of G_{D}^{S}. (ii) The function φ defined by this series is invariant under left translation by the matrices in

G_{F} ∩ G_{D}^{S} of the form

( \begin{matrix}α & β \\ 0 & δ \end{matrix} ) .

(iii) Suppose F is a number field. Let Ω be a compact subset of G_{D}^{S}. Then there are positive constants M_{1} and M_{2} such that

|φ(g)| ≤ M_{1}{|a| + |a|^{−1}}^{M_{2}}

if

g = ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) h

with h in Ω, a in I_{D}^{S}, and ({1 \atop 0} {x \atop 1}) in G_{D}^{S}.

It is enough to prove these assertions when φ_{1} has the form

φ_{1}(g) = ∏_{v∉S} φ_{v}(g_{v}).

To establish the first and third assertions we need only consider the series

∑_{α∈F^{×}} δ(α) ∏_{v∉S} |φ_{v} (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g_{v})|  (11.4.1)

where δ(α) = 0 if for some v in S the number α regarded as an element of F_{v} is not the largest ideal on which ψ_{v} is trivial and δ(α) = 1 otherwise.

We need only consider compact sets Ω of the form

Ω = K_{S}^{D} ∏_{v∉S} Ω_{v}  (11.4.2)

where Ω_{v} is a compact subset of G_{v} and Ω_{v} = K_{v} for almost all v.

Lemma 11.4.3 Suppose Ω is of the form (11.4.2). There is a positive number ρ such that for each non-archimedean place v which is not in S there is a constant M_{v} such that

|φ_{v} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) h)| ≤ M_{v}|a|^{−ρ}

for a in F_{v}^{×} and h in Ω_{v} and a constant c_{v} such that

|φ_{v} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) h)| = 0

if |a| > c_{v} and h is in Ω_{v}. Moroever one may take M_{v} = c_{v} = 1 for almost all v.

Since φ_{v} is invariant under an open subgroup of K_{v} for all v and is invariant under K_{v} for almost all v while Ω_{v} = K_{v} for almost all v it is enough to prove the existence of M_{v}, c_{v}, and ρ such that these relations are satisifed when h = 1. Since the function

a → φ_{v} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

belongs to the space of the Kirillov model the existence of c_{v} is clear. c_{v} can be taken to be 1 when O_{v} is the largest ideal of F_{v} on which ψ_{v} is trivial and φ_{v} = φ_{v}^{0}.

The existence of M_{v}, for a given v and sufficiently large ρ, is a result of the absolute convergence of the integral defining Ψ(e, s, φ_{v}). Thus all we need do is show the existence of a fixed ρ such that the inequality

|φ_{v} \begin{matrix}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix}| ≤ |a|^{−ρ}

is valid for almost all v. For almost all v the representation π_{v} is of the form π(μ_{v}, ν_{v}) with μ_{v} and ν_{v} unramified, O_{v} is the largest ideal of F_{v} on which ψ_{v} is trivial, and φ_{v} = φ_{v}^{0}. Thus, for such v,

φ_{v} \begin{aligned}( \\ ( & \begin{matrix}εa & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = φ_{v} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

if ε is a unit in O_{v} and

∑_{n} φ_{v} \begin{matrix}( \\ ( & \begin{matrix}ϖ_{v}^{n} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix} |ϖ_{v}|^{n(s− \frac{1}{2})} = L(s, π_{v}).

If ρ_{v} = μ_{v}(ϖ_{v}) and σ_{v} = ν_{v}(ϖ_{v})

L(s, π_{v}) = \frac{1}{(1 − ρ_{v}|ϖ_{v}|^{s})} \frac{1}{(1 − σ_{v}|ϖ_{v}|^{s})}.

Since |ρ_{v}| ≤ |ϖ_{v}|^{−r} and |σ_{v}| ≤ |ϖ_{v}|^{−r}

|φ_{v} \begin{matrix}( \\ ( & \begin{matrix}ϖ_{v}^{n} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix}| = |\frac{ρ_{v}^{n+1} − σ_{v}^{m+1}}{ρ_{v} − σ_{v}} | ≤ (n + 1) |ϖ_{v}|^{−rn}.

Since |ϖ_{v}| ≤ \frac{1}{2} there is a constant ε > 0 such that

(n + 1) ≤ |ϖ_{v}|^{−εn}

for all v and all n ≥ 0.

If v is archimedean the integral representations of the functions in W (π_{v}, ψ_{v}) show that there are positive constants c_{v}, d_{v}, and M_{v} such that

|φ_{v} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) h)| ≤ M_{v} |a|^{−c_{v}} exp ( − d_{v} |a|_{v}^{ε_{v}})

for a in F_{v}^{×} and h in Ω_{v}. ε_{v} is 1 if v is real and \frac{1}{2} if v is complex.

Since we want to prove not only the first assertion but also the third we consider the sum

f (( \begin{matrix}b & 0 \\ 0 & 1 \end{matrix} ) g) = ∑_{α∈F^{×}} δ(α) ∏_{v∉S} |φ_{v} (( \begin{matrix}b_{v}α & 0 \\ 0 & 1 \end{matrix} ) g_{v})|

where g lies in the set (11.4.2) and b is an idèle such that b_{v} = 1 for all non-archimedean v. We also suppose that there is a positive number t such that b_{v} = t for all archimedean v. If Λ is a set of α in F for which |α|_{v} ≤ c_{v} for all non-archimedean v not in S and δ(α) ≠ 0 then

f (( \begin{matrix}b & 0 \\ 0 & 1 \end{matrix} ) g)

is bounded by

∑_{{α∈Λ \atop α≠0}} \{ ∏_{v∈S_{a}} M_{v}|αt|_{v}^{−c_{v}} exp(−d_{v}t |α|_{v}^{ε_{v}})\}\{ ∏_{v∉S∪S_{a}} M_{v}|α_{v}|^{−ρ}\}.

If F is a function field Λ is a finite set and there is nothing more to prove. If it is a number field choose for each v in S a constant c_{v} such that δ(α) = 0 unless |α|_{v} ≤ c_{v}. Since

\begin{aligned}∏_{v} |α|_{v} & = 1, \\ ∏_{v∉S∪S_{a}} |α|_{v}^{−ρ} & ≤ \{ ∏_{v∈S} c_{v}^{ρ}\}\{ ∏_{v∈S_{a}} |α|_{v}^{ρ}\}. \end{aligned}

Thus our sum is bounded by a constant times the product of ∏_{v∈S_{a}} t^{−c_{v}/ε_{v}} and

∑_{{α∈Λ \atop α≠0}} ∏_{v∈S_{a}} \{|α|_{v}^{ρ−c_{v}} exp ( − d_{v}t|α|_{v}^{ε_{v}})\}.

The product ∏_{v∈S_{a}} |α|_{v} is bounded below on Λ − {0}. Multiplying each term by the same sufficiently high power of ∏_{v∈S_{a}} |α|_{v} we dominate the series by another series

∑_{α∈Λ} ∏_{v∈S_{a}} {|α|_{v}^{ρ_{v}} exp(−d_{v}t |α|_{v}^{ε_{v}})}

in which the exponents ρ_{v} are non-negative. This in turn is dominated by ∏_{v∈S_{a}} t^{−ρ_{v}/ε_{v}} times

∑_{α∈Λ} ∏_{v∈S_{a}} exp (\frac{−d_{v}}{2} t |α|_{v}^{ε_{v}}).

Λ may be regarded as a lattice in ∏_{v∈S_{a}} F_{v}. if λ_{1}, ⋯, λ_{n} is a basis of Λ there is a constant d such that if α = ∑ a_{i} λ_{i}

∑_{v∈S_{a}} \frac{d_{v}}{2} |α|_{v}^{ε_{v}} ≥ d ∑ |a_{i}|.

Thus

f (( \begin{matrix}b & 0 \\ 0 & 1 \end{matrix} ) g)

is dominated by some power of t times a multiple of

{ ∑_{a=−∞}^{∞} e^{−dt|a|}}^{n}

which is bounded by a multiple of (1 + \frac{1}{t})^{n}.

The first assertion is now proved and the third will now follow from the second and the observation that every element of I_{D}^{S} is the product of an element of F^{×}, an idèle whose components are 1 at all non-archimedean places and equal to the same positive number at all archimedean places, and an idèle which lies in a certain compact set.

Suppose ξ is in F and

( \begin{matrix}1 & ξ \\ 0 & 1 \end{matrix} )

belong to G_{D}^{S}. Then ξ is integral at each prime of S and ψ_{v}(αξ) = 1 if a_{α} ≠ 0. If g belongs to G_{D}^{S} and

h = ( \begin{matrix}1 & ξ \\ 0 & 1 \end{matrix} ) g

then ε(h_{S}) = ε(g_{S}) and if v is not in S

φ_{v} (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) h_{v}) = ψ_{v}(αξ) φ_{v} (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g_{v}) .

If a_{α} ≠ 0

∏_{v∉S} ψ_{v}(αξ) = ∏_{v} ψ_{v}(αξ) = 1.

Consequently

φ (( \begin{matrix}1 & ξ \\ 0 & 1 \end{matrix} ) g) = φ(g).

If b belongs to I_{D}^{S} then

ε (( \begin{matrix}b_{S} & 0 \\ 0 & b_{S} \end{matrix} ) g_{S}) = η(b_{S}) ε(g_{S})

Chapter 2 201 and

φ_{1} (( \begin{matrix}b̂_{S}α̂_{S} & 0 \\ 0 & b̂_{S} \end{matrix} ) ĝ_{S}) = η(b̂_{S}) φ_{1} (( \begin{matrix}α̂_{S} & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S})

so that

φ (( \begin{matrix}b & 0 \\ 0 & b \end{matrix} ) g) = η(b) φ(g).

In particular if β belongs to F^{×} ∩ I_{D}^{S}

φ (( \begin{matrix}β & 0 \\ 0 & β \end{matrix} ) g) = φ(g).

If β belongs to F^{×} ∩ I_{D}^{S} and

h = ( \begin{matrix}β & 0 \\ 0 & 1 \end{matrix} ) g

then

ε(h_{S}) = ε \begin{aligned}( \\ ( & \begin{matrix}β_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} ε(g_{S})

and φ(h) is equal to

∑_{α} a_{α}ε \begin{aligned}( \\ ( & \begin{matrix}β_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} ε(g_{S}) φ_{1} (( \begin{matrix}α̂_{S}β̂_{S} & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S}) .

Since

a_{αβ} = ε \begin{aligned}( \\ ( & \begin{matrix}β_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} a_{α}

we can change variables in the summation to see that φ(h) = φ(g).

The lemma is now proved. The function

φ̂(g) = ∑_{α∈F^{×}} â_{α} ε̂(g_{S}) φ_{1} (( \begin{matrix}α̂_{S} & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S})

can be treated in the same fashion.

Theorem 11.5 If ω is a quasi-characters of F^{×} \ I such that

ω_{v}(a_{v}) ε \begin{aligned}( \\ ( & \begin{matrix}a_{v} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = 1

for all units a_{v} of O_{v} set

Λ(s, ω) = \{ ∑_{F^{×}∩I_{D}^{S}\F^{×}} a_{α} ω(α_{S}) |α_{S}|^{s− \frac{1}{2}} \} ∏_{v∉S} L(s, ω_{v} ⊗ π_{v}).

If

ω_{v}(a_{v}) ε̂\begin{aligned}( \\ ( & \begin{matrix}a_{v} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = 1

for all units a_{v} in O_{v} set

Λ̂(s, ω) = \{ ∑_{F^{×}∩I_{D}^{S}\F^{×}} â_{α} ω(α_{S}) |α_{S}|^{s− \frac{1}{2}} \} ∏_{v∉S} L(s, ω_{v} ⊗ π_{v}).

Then Λ(s, ω) and Λ̂(s, ω) are defined for Re s sufficiently large. Suppose that whenever ω is such that Λ(s, ω) or Λ̂(s, ω) is defined they can be analytically continued to entire functions which are bounded in vertical strips. Assume also that there is an A in F^{×} such that |A|_{𝔭} = |ϖ_{𝔭}|^{m_{𝔭}} for any 𝔭 in S and

Λ(s, ω) = { ∏_{v∈S} ω_{v}(−A) |A|_{v}^{s−1/2}}{ ∏_{v∉S} ε(s, ω_{v} ⊗ π_{v}, ψ_{v})} Λ̂(1 − s, η^{−1}ω^{−1})

whenever Λ(s, ω) is defined. Then for any φ_{1} in W (π, ψ) there is an automorphic form φ on G_{𝔸} such that

φ(g) = ∑ a_{α} ε(g_{S}) φ_{1} (( \begin{matrix}α̂_{S} & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S})

on G_{D}^{S}.

The infinite products occurring in the definition of Λ(s, ω) and Λ̂(s, ω) certainly converge for Re s sufficiently large. To check that the other factors converge one has to check that

∑ |α_{S}|^{s− \frac{1}{2}}

converges for Re s sufficiently large if the sum is taken over those elements α of a system of coset representatives of F^{×} ∩ I_{D}^{S} \ F^{×} for which |α|_{v} ≤ c_{v} for v in S. This is easily done.

A_{S} is the idèle whose components are 1 outside of S and A in S. Since

( \begin{matrix}0 & 1 \\ A_{S} & 0 \end{matrix} ) ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}0 & A_{S}^{−1} \\ 1 & 0 \end{matrix} ) = ( \begin{matrix}d & cA_{S}^{−1} \\ A_{S}b & a \end{matrix} )

the matrix

( \begin{matrix}0 & 1 \\ A_{S} & 0 \end{matrix} )

normalizes K_{S}^{D}. In particular if g belongs to G_{D}^{S} so does

( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) g ( \begin{matrix}0 & A_{S}^{−1} \\ 1 & 0 \end{matrix} ) .

Lemma 11.5.1 If φ_{1} is in W (π, ψ) and g is in G_{D}^{S} then, under the hypotheses of the theorem,

φ̂ (( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) g ( \begin{matrix}0 & A_{S}^{−1} \\ 1 & 0 \end{matrix} )) = φ(g).

Let φ′(g) be the function on the left. As before all we need do is show that for every character ω of F^{×} ∩ I_{D}^{S} \ I_{D}^{S} and every g in G_{D}^{S} the integral

∫_{F^{×}∩I_{D}^{S}\I_{D}^{S}} φ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) ω(a) |a|^{s− \frac{1}{2}} d^{×}a  (11.5.2)

is absolutely convergent for Re s large and positive. The integral

∫_{F^{×}∩I_{D}^{S}\I_{D}^{S}} φ′ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g) ω(a) |a|^{s− \frac{1}{2}} d^{×}a  (11.5.3)

is absolutely convergent for Re s large and negative, and they can be analytically continued to the same entire function which is bounded in vertical strips.

If for any v in S the character

a → ω_{v}(a) ε \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned}

on the group of units of O_{v} is not trivial the integrals are 0 when they are convergent. We may thus assume that

ω_{v}(a) ε \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = 1

for all units in O_{v} if v is in S.

We discuss the first integral in a formal manner. The manipulations will be justified by the final result. The integrand may be written as a double sum

∑ ∑ a_{αγ} ε (( \begin{matrix}a_{S} & 0 \\ 0 & 1 \end{matrix} ) g_{S}) φ_{1} (( \begin{matrix}â_{S}α̂_{S}γ̂_{S} & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S}) ω(a) |a|^{s− \frac{1}{2}} .

The inner sum is over γ in F^{×}∩I_{D}^{S} and the outer over a set of coset representatives α of F^{×}∩I_{D}^{S}\F^{×}. Since

a_{αγ} ε \begin{aligned}( \\ ( & \begin{matrix}a_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = a_{α} ε \begin{aligned}( \\ ( & \begin{matrix}γ_{S}a_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

and

ω(a) |a|^{s− \frac{1}{2}} = ω(γa) |γa|^{s− \frac{1}{2}}

the integral is equal to ε(g_{S}) times the sum over α of

a_{α} ∫_{I_{D}^{S}} φ_{1} (( \begin{matrix}α̂_{S}â_{S} & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S}) ε \begin{aligned}( \\ ( & \begin{matrix}a_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} ω(a) |a|^{s− \frac{1}{2}} d^{×}a.

Since I_{D}^{S} is the direct product of

Î_{S} = {a ∈ I | a_{S} = 1}

and a compact group under which the integrand is invariant the previous expression is equal to

a_{α} ∫Î_{S} φ_{1} (( \begin{matrix}α̂_{S}a & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S}) ω(a) |a|^{s− \frac{1}{2}} d^{×}a.

Changing variables to rid ourselves of the α̂_{S} in the integrand and taking into account the relation

1 = ω(α) |α|^{s− \frac{1}{2}} = ω(α_{S}) ω(α̂_{S}) |α_{S}|^{s− \frac{1}{2}} |α̂_{S}|^{s− \frac{1}{2}}

we can see the original integral is equal to

ε(g_{S}) ∑ a_{α} ω(α_{S}) |α_{S}|^{s− \frac{1}{2}} ∫Î_{S} φ_{1} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S}) ω(a) |a|^{s− \frac{1}{2}} d^{×}a.

There is no harm in supposing that φ_{1} is of the form

φ_{1}(ĝ_{S}) = ∏_{v∉S} φ_{v}(g_{v}).

We have already seen that, in this case,

∫Î_{S} φ_{1} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S}) ω(a) |a|^{s− \frac{1}{2}} d^{×}a

is convergent for Re s large and positive and is equal to

∏_{v∉S} ∫_{F_{v}^{×}} φ_{v} (( \begin{matrix}a_{v} & 0 \\ 0 & 1 \end{matrix} ) g_{v}) ω_{v}(a_{v}) |a_{v}|^{s− \frac{1}{2}} d^{×}a_{v}.

If φ′_{v} is the function

φ′_{v}(h) = ω_{v}(h) φ_{v}(h)

in W (ω_{v} ⊗ π_{v}, ψ_{v}) this product is

∏_{v∉S} {L(s, ω_{v} ⊗ π_{v}) Φ(g_{v}, s, φ′_{v}) ω_{v}^{−1}(detg_{v})}.

Thus the integral (11.5.2) is absolutely convergent for Re s large and positive and is equal to

ε(g_{S}) ω(detĝ_{S}) Λ(s, ω) ∏_{v∉S} Φ(g_{v}, s, φ′_{v}).

The argument used in the proof of Theorem 11.3 shows that this function is entire. If

h = ( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g ( \begin{matrix}0 & A_{S}^{−1} \\ 1 & 0 \end{matrix} )

then

ε̂(h_{S}) = ε (( \begin{matrix}a_{S} & 0 \\ 0 & 1 \end{matrix} ) g_{S}) .

Thus the integrand in (11.5.3) is equal to

∑ â_{α} ε (( \begin{matrix}a_{S} & 0 \\ 0 & 1 \end{matrix} ) g_{S}) φ_{1} (( \begin{matrix}α̂_{S} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ Â_{S} & 0 \end{matrix} ) ( \begin{matrix}â_{S} & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S}) ω(a) |a|^{s− \frac{1}{2}} .

The sum can again be written as a double sum over γ and α. Since

â_{αγ} = ε \begin{aligned}( \\ ( & \begin{matrix}a_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = â_{α} ε \begin{aligned}( \\ ( & \begin{matrix}1 & 0 \\ 0 & γ_{S} \end{matrix} & ) \\ & & ) \end{aligned} ε \begin{aligned}( \\ ( & \begin{matrix}a_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

Chapter 2 205 which equals

â_{α} η(γ_{S}) ε \begin{aligned}( \\ ( & \begin{matrix}γ_{S}^{−1}a_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned}

and

φ_{1} (( \begin{matrix}α̂_{S}γ̂_{S} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ Â_{S} & 0 \end{matrix} ) ( \begin{matrix}â_{S} & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S})

is equal to

η(γ̂_{S}) φ_{1} (( \begin{matrix}α̂_{S} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ Â_{S} & 0 \end{matrix} ) ( \begin{matrix}γ̂_{S}^{−1}â_{S} & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S})

we can put the sum over F^{×} ∩ I_{D}^{S} and the integration over F^{×} ∩ I_{D}^{S} \ I_{D}^{S} together to obtain ε(g_{S}) times the sum over F^{×} ∩ I_{D}^{S} \ F^{×} of

â_{α} ∫_{I_{D}^{S}} ε \begin{aligned}( \\ ( & \begin{matrix}a_{S} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ_{1} (( \begin{matrix}α̂_{S} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ Â_{S} & 0 \end{matrix} ) ( \begin{matrix}â_{S} & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S}) ω(a) |a|^{s− \frac{1}{2}} d^{×}a.

We write

( \begin{matrix}α̂_{S} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ Â_{S} & 0 \end{matrix} ) ( \begin{matrix}â_{S} & 0 \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}α̂_{S} & 0 \\ 0 & α̂_{S} \end{matrix} ) ( \begin{matrix}−α̂_{S}^{−1}Â_{S}â_{S} & 0 \\ 0 & 1 \end{matrix} )

and then change variables in the integration to obtain the product of ω(−A_{S}) |A_{S}|^{s− \frac{1}{2}} and

â_{α} η^{−1}(α_{S}) ω^{−1}(α_{S}) |α_{S}|^{\frac{1}{2} −s} ∫Î_{S} φ_{1} (( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) ĝ_{S}) ω(a) |a|^{s− \frac{1}{2}} d^{×}a.

Replacing a by a^{−1} and making some simple changes we see that the integral is equal to

∫Î_{S} φ_{1} (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ĝ_{S}) η^{−1}(s) ω^{−1}(a) |a|^{\frac{1}{2} −s} d^{×}a

which converges for Re s large and negative and is equal to

∏_{v∉S} {L(1 − s, η_{v}^{−1} ω_{v}^{−1} ⊗ π_{v}) Φ(wg_{v}, 1 − s, φ′_{v}) ω_{v}(detg_{v})}.

Thus the integral 11.5.3 is equal to

ε(g_{S}) ω(detĝ_{S}) ω(−A_{S}) |A_{S}|^{s− \frac{1}{2}} Λ̂(1 − s, η^{−1}ω^{−1}) ∏_{v∉S} Φ(wg, 1 − s, φ′_{v})

which is entire.

Since

Φ(wg_{v}, 1 − s, φ′_{v}) = ε(s, ω_{v} ⊗ π_{v}, ψ) Φ(g_{v}, s, φ′_{v})

the analytic continuations of (11.5.2) and (11.5.3) are equal. We show as in the proof of Theorem 11.3 that the resultant entire function is bounded in vertical strips of finite width.

There is now a simple lemma to be proved.

Lemma 11.5.3 The group G_{F} ∩ G_{D}^{S} is generated by the matrices in it of the form

( \begin{matrix}a & β \\ 0 & δ \end{matrix} )

and

( \begin{matrix}α & 0 \\ γ & δ \end{matrix} ) .

This is clear if S is empty. Suppose that S is not empty. If

g = ( \begin{matrix}α & β \\ γ & δ \end{matrix} )

belongs to G_{F} ∩ G_{D}^{S} and |α|_{v} = 1 for all v in S then

g = ( \begin{matrix}α & 0 \\ γ & δ − \frac{βγ}{α} \end{matrix} ) ( \begin{matrix}1 & \frac{β}{γ} \\ 0 & 1 \end{matrix} )

and both matrices belong to G_{F} ∩G_{D}^{S}. In general if g is in G_{F} ∩G_{D}^{S} then, for each v in S, |α|_{v} ≤ 1, |γ|_{v} ≤ 1 and either |α|_{v} or |γ|_{v} is 1. Choose ξ in F so that, for every v in S, |ξ_{v}| = 1 if |α|_{v} < 1 and |ξ_{v}| < 1 if |α|_{v} = 1. Then

( \begin{matrix}1 & ξ \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) = ( \begin{matrix}α + ξγ & β + ξδ \\ γ & δ \end{matrix} )

and |α + ξγ|_{v} = 1 for all v in S. The lemma follows.

We know that if

h = ( \begin{matrix}α & β \\ 0 & δ \end{matrix} )

belongs to G_{F} ∩ G_{D}^{S} then φ(hg) = φ(g) and φ̂(hg) = φ̂(g). Suppose

( \begin{matrix}α & 0 \\ γ & δ \end{matrix} )

is in G_{F} ∩ G_{D}^{S}. Then

φ (( \begin{matrix}α & 0 \\ γ & δ \end{matrix} ) g) = φ̂ (( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) ( \begin{matrix}α & 0 \\ γ & δ \end{matrix} ) g ( \begin{matrix}0 & A_{S}^{−1} \\ 1 & 0 \end{matrix} )) .

Since the argument on the right can be written

( \begin{matrix}δ & γA^{−1} \\ 0 & α \end{matrix} ) ( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) g ( \begin{matrix}0 & A_{S}^{−1} \\ 1 & 0 \end{matrix} )

and the first term of this product lies in G_{F} ∩ G_{D}^{S} the right side is equal to

φ̂ (( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) g ( \begin{matrix}0 & A_{S}^{−1} \\ 1 & 0 \end{matrix} )) = φ(g).

Chapter 2 207

Thus φ is invariant under G_{F} ∩ G_{D}^{S}. Since G_{𝔸} = G_{F} G_{D}^{S} the function φ extends in a unique manner to a function, still denoted φ, on G_{F} \ G_{𝔸}. It is clear that φ is K-finite and continuous and that

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g)

for all a in I. It is not quite so clear that φ is slowly increasing. If Ω is a compact subset of G_{𝔸} there is a finite set γ_{1}, ⋯, γ_{ℓ} in G_{F} such that

Ω = ∪_{i=1}^{ℓ}Ω ∩ γ_{i}^{−1} G_{D}^{S}.

What we have to show then is that if γ belongs to G_{F} and c > 0 is given there are constants M_{1} and M_{2} such that for all g in Ω ∩ γ^{−1} G_{D}^{S} and all a in I for which |a| ≥ c

|γ (( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g)| ≤ M_{1}|a|^{M_{2}}.

If v is a place of F, which is not in S and is archimedean if F is a number field, there is a compact set C in I such that

{a ∈ I | |a| ≥ c} ⊆ F^{×}{a ∈ F_{v}^{×} | |a| ≥ c}C

Thus the inequality has only to be verified for a in F_{v}^{×} — of course at the cost of enlarging Ω. If

γ = ( \begin{matrix}α & β \\ 0 & δ \end{matrix} )

then

γ ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g = ( \begin{matrix}a & x \\ 0 & 1 \end{matrix} ) γg

with x = (1 − α)\frac{β}{δ} and the conclusion results from Lemma 11.4 and the relation

B_{𝔸} = (B_{𝔸} ∩ G_{F}) (B_{𝔸} ∩ G_{D}^{S})

if

B_{𝔸} = \{ ( \begin{matrix}b & y \\ 0 & 1 \end{matrix} ) ∈ G_{𝔸}\}.

Otherwise we write

γ = ( \begin{matrix}α_{1} & β_{1} \\ 0 & δ_{1} \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}1 & β_{2} \\ 0 & 1 \end{matrix} ) .

Then

γ ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ) g = ( \begin{matrix}1 & \frac{β_{1}}{δ_{1}} \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{1} & 0 \\ 0 & δ_{1}a \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}1 & \frac{β_{2}}{a} \\ 0 & 1 \end{matrix} ) g.

The matrix

( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}1 & \frac{β_{2}}{a} \\ 0 & 1 \end{matrix} ) g

lies in a certain compact set which depends on Ω, c, and γ. The required inequality again follows from Lemma 11.4

The space U of functions φ corresponding to φ_{1} in W (π, ψ) transforms under ℋ̂_{S} according to π. Lemma 10.13 implies that every element of U is an automorphic form. If it is not contained in 𝒜_{0}, Lemma 10.12 applied to the functions

φ_{0}(g) = \frac{1}{measure F \ 𝔸} ∫_{F \𝔸} φ (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) dx

with φ in U shows that there are two quasi-characters μ and ν on F^{×} \ I such that π_{v} = π(μ_{v}, ν_{v}) for almost all v.

Chapter 2 208 Corollary 11.6 Suppose there does not exist a pair μ, ν of quasi-characters of F^{×} \ I such that π_{v} = π(μ_{v}, ν_{v}) for almost all v. Then there is a constituent π′ = ⊗π′_{v} of 𝒜_{0} such that π_{v} = π′_{v} for all v not in S.

Since U transforms under ℋ̂_{S} according to π it is, if v is not in S, the direct sum of subspaces transforming under ℋ_{v} according to π_{v}. By assumption U is contained in 𝒜_{0} and therefore in 𝒜_{0}(η). 𝒜_{0}(η) is the direct sum of subspaces invariant and irreducible under ℋ. Choose one of these summands V so that the projection of U on V is not 0. If π′ = ⊗π′_{v} is the representation of ℋ on V it is clear that π′_{v} = π_{v} if v is not in S.

Another way to guarantee that U lies in the space of cusp forms and therefore that the conclusion of the corollary holds is to assume that for at least one v not in S the representation π_{v} is absolutely cuspidal.

Chapter 2 209 §12. Some extraordinary representations. In [18] Weil has introduced a generalization of the Artin L-functions. To define these it is necessary to introduce the Weil groups. These groups are discussed very clearly in the notes of Artin–Tate but we remind the reader of their most important properties. If F is a local field let C_{F} be the multiplicative group of F and if F is a global field let C_{F} be the idèle class group F^{×} \ I. If K is a finite Galois extension of F the Weil group W_{K/F} is an extension of 𝔊(K/F), the Galois group of K/F, by C_{K}. Thus there is an exact sequence

1 → C_{K} → W_{K/F} → 𝔊(K/F) → 1.

If L/F is also Galois and L contains K there is a continous homomorphism τ_{L/F,K/F} of W_{L/F} onto W_{K/F}. It is determined up to an inner automorphism of W_{K/F} by an element of C_{K}. In particular W_{F/F} = C_{F} and the kernel of τ_{K/F,F/F} is the commutator subgroup of W_{K/F}. Also if F ⊆ E ⊆ K we may regard W_{K/F} as a subgroup of W_{K/F}. If F is global and v a place of F we also denote by v any extension of v to K. There is a homomorphism α_{v} of W_{K_{v}/F_{v}} into W_{K/F} which is determined up to an inner automorphism by an element of C_{K}.

A representation σ of W_{K/F} is a continuous homomorphism of W_{K/F} into the group of invertible linear transformations of a finite-dimensional complex vector space such that σ(w) is diagonalizable for all w in W_{K/F}. If K is contained in L then σ ∘ τ_{L/F,K/F} is a representation of W_{L/F} whose equivalence class is determined by that of σ. In particular if ω is a generalized character of C_{F} then ω ∘ τ_{K/F,F/F} is a one-dimensional representation of W_{K/F} which we also call ω. If σ is any other representation ω ⊗ σ has the same dimension as σ. If F ⊆ E ⊆ K and ρ is a representation of W_{K/E} on X let Y be the space of functions φ on W_{K/F} with values in X which satisfy

φ(uw) = ρ(u) φ(w)

for all u in W_{K/E}. If v ∈ W_{K/F} and φ ∈ Y let σ(v)φ be the function

W → φ(wv)

σ(v)φ also belongs to Y and v → σ(v) is a representation of W_{K/F}. We write

σ = Ind(W_{K/F} , W_{K/E}, ρ).

If F is global and σ is a representation of W_{K/F} then, for any place v, σ_{v} = σ ∘ α_{v} is a representation of W_{K_{v}/F_{v}} whose class is determined by that of σ.

Now we remind ourselves of the definition of the generalized Artin L-functions. Since we are going to need a substantial amound of detailed information about these functions the best reference is probably [19]. In fact to some extent the purpose of [19] is to provide the background for this chapter and the reader who wants to understand all details will need to be quite familiar with it. If F is a local field then to every representation σ of W_{K/F} we can associate a local L-function L(s, σ). Moreover if ψ_{F} is a non-trivial additive character of F we can define a local factor ε(s, σ, ψ_{F}). The L-function and the factor ε(s, σ, ψ_{F}) depend only on the equivalance class of σ.

If F is a global field we set

L(s, σ) = ∏_{v} L(s, σ_{v})

Chapter 2 210 The product converges in a right half-plane and L(s, σ) can be analytically continued to a function meromorphic in the whole complex plane. If ψ_{F} is a non-trivial character of F \ 𝔸 the functions ε(s, σ_{v}, ψ_{v}) are identically 1 for all but a finite number of v. If

ε(s, σ) = ∏_{v} ε(s, σ_{v}, ψ_{v})

and σ̃ is the representation contragredient to σ the functional equation

L(s, σ) = ε(s, σ) L(1 − s, σ̃)

is satisfied. For all but finitely many places v the representation σ_{v} is the direct sum of d, the dimension of σ, one-dimensional representations. Thus there are generalized characters μ_{v}^{1}, ⋯ μ_{v}^{d} of C_{F_{v}} such that σ_{v} is equivalent to the direct sum of the one dimensional representations

w ↦ μ_{v}^{i}(τ_{K_{v}/F_{v},F_{v}/F_{v}}(w)).

Moreover, for all but finitely many of these v, μ_{v}^{1}, ⋯, μ_{v}^{d} are unramified and there is a constant r, which does not depend on v, such that

|μ_{v}^{i}(ϖ_{v})| ≤ |ϖ_{v}|^{r} 1 ≤ i ≤ d.

If F is a global or a local field and σ is a representation of W_{K/F} then w → detσ(w) is a one-dimensional representation and therefore corresponds to a generalized character of C_{F}. We denote this character by detσ.

If F is a local field, σ is a two-dimensional representation of W_{K/F}, and ψ_{F} is a non-trivial additive character of F then, as we saw in the first chapter, there is at most one irreducible admissible representation π of ℋ_{F} such that

π \begin{aligned}( \\ ( & \begin{matrix}α & 0 \\ 0 & α \end{matrix} & ) \\ & & ) \end{aligned} = detσ(α) I

and, for all generalized characters ω of C_{F},

\begin{aligned}L(s, ω ⊗ π) & = L(s, ω ⊗ σ) \\ L(s, ω^{−1} ⊗ π̃) & = L(s, ω^{−1} ⊗ σ) \\ ε(s, ω ⊗ π, ψ_{F}) & = ε(s, ω ⊗ σ, ψ_{F}). \end{aligned}

If ψ′_{F} (x) = ψ_{F} (βx) then

ε(s, ω ⊗ σ, ψ′_{F}) = detω ⊗ σ(β) ε(s, ω ⊗ σ, ψ_{F})

and, since

π \begin{aligned}( \\ ( & \begin{matrix}α & 0 \\ 0 & α \end{matrix} & ) \\ & & ) \end{aligned} = detσ(α) I

one also has

ε(s, ω ⊗ π, ψ′_{F}) = detω ⊗ σ(β) ε(s, ω ⊗ π, ψ_{F}).

Chapter 2 211 Thus π, if it exists at all, is independent of ψ_{F}. We write π = π(σ).

There are a number of cases in which the existence of π(σ) can be verified simply by comparing the definitions of the previous chapter with those of [19]. If μ and ν are two quasi-characters of C_{F} and σ is equivalent to the representation

w → ( \begin{matrix}μ(τ_{K/F,F/F} (w)) & 0 \\ 0 & ν(τ_{K/F,F/F} (w)) \end{matrix} )

then π(σ) = π(μ, ν). If K/F is a separable quadratic extension, χ is a quasi-character of C_{K} = W_{K/K}, and

σ = Ind(W_{K/F} , W_{K/K}, χ)

then π(σ) = π(χ). Observe that π(χ) is alway infinite-dimensional.

Suppose F is a global field and K is a separable quadratic extension of F. Let χ be a quasicharacter of C_{K} and let

σ = Ind(W_{K/F} , W_{K/K}, χ).

If v does not split in K

σ_{v} = Ind(W_{K_{v}/F_{v}}, W_{K_{v}/K_{v}}, χ_{v}),

but if v splits in K the representation σ_{v} is the direct sum of two one-dimensional representations corresponding to quasi-characters μ_{v} and ν_{v} such that μ_{v}ν_{v}^{−1} is a character. Thus π(σ_{v}) is defined and infinite-dimensional for all v.

Proposition 12.1 If there is no quasi-character μ of C_{F} such that χ(α) = μ(N_{K/F} α) for all α in C_{K} the representation ⊗_{v}π(σ_{v}) is a constituent of 𝒜_{0}.

If ω is a generalized character of F then

(ω ⊗ σ)_{v} = ω_{v} ⊗ σ_{v}.

Define a generalized charcter ω_{K/F} of C_{K} by

ω_{K/F} (α) = ω(N_{K/F} (α)).

Then

ω ⊗ σ = Ind(W_{K/F} , W_{K/K}, ω_{K/F} χ)

and

L(s, ω ⊗ σ) = L(s, ω_{K/F} χ).

The L-function on the right is the Hecke L-function associated to the generalized character ω_{K/F} χ of C_{K}. It is entire and bounded in vertical strips unless there is a complex number r such that

ω_{K/F} (α) χ(α) = |α|^{r} = |N_{K/F} α|^{r}.

But then

χ(α) = ω^{−1}(N_{K/F} α) |N_{K/F} α|^{r}

which is contrary to assumption. The function

L(s, ω^{−1} ⊗ σ̃) = L(s, ω_{K/F}^{−1} χ^{−1})

is also entire and bounded in vertical strips. It follows immediately that the collection {π(σ_{v})} satisfies the conditions of Theorem 11.3.

This proposition has a generalization which is one of the principal results of these notes.

Chapter 2 212 Theorem 12.2 Suppose F is a global field and σ is a two-dimensional representation of W_{K/F}. Suppose also that for every generalized character ω of C_{F} both L(s, ω ⊗ σ) and L(s, ω^{−1} ⊗ σ̃) are entire functions which are bounded in vertical strips. Then π(σ_{v}) exists for every place v and ⊗_{v}π(σ_{v}) is a constituent of 𝒜_{0}.

We observe that the converse to this theorem is an immediate consequence of Theorem 11.1. We are going to apply Corollary 11.6. There are a large number of conditions which must be

verified. We know that π(σ_{v}) is defined for all but a finite number of v. In particular it is defined for v archimedean for then σ_{v} is either induced from a quasi-character of a quadratic extension of F_{v} or is the direct sum of two one-dimensional representations. If σ_{v} is equivalent to the direct sum of two one-dimensional representations corresponding to quasi-charcters μ_{v} and ν_{v} then μ_{v}ν_{v}^{−1} is a character so that π(σ_{v}) is infinite-dimensional. Let S be the set of places for which π(σ_{v}) is infinite-dimensional. Let S be the set of places for which π(σ_{v}) is not defined or, since this is still conceivable, finite dimensional. We are going to show that S is empty but at the moment we are at least sure that it is finite. If v is not in S set π_{v} = π(σ_{v}).

If v is in S the representation σ_{v} must be irreducible so that

L(s, ω_{v} ⊗ σ_{v}) = L(s, ω_{v}^{−1} ⊗ σ̃_{v}) = 1

for every generalized character ω_{v} of F_{v}^{×}. The Artin conductor 𝔭_{v}^{m_{v}} of σ_{v} is defined in the Appendix to [19]. There is a constant c_{v}, depending on σ_{v}, such that if ω_{v} is unramifed

ε(s, ω_{v} ⊗ σ_{v}, ψ_{v}) = c_{v} ω_{v}(ϖ_{v})^{m_{v}+2n} |ϖ_{v}|^{(m_{v}+2n_{v})(s− \frac{1}{2} )}

if 𝔭_{v}^{−n_{v}} is the largest ideal on which ψ_{v} is trivial. ψ_{v} is the restriction to F_{v} of a given non-trivial character of F \ 𝔸.

We take

D = ∏_{𝔭∈S} 𝔭^{m_{𝔭}}

and η = detσ. We define ε and ε̂ by

ε \begin{aligned}( \\ ( & \begin{matrix}a_{v} & 0 \\ 0 & b_{v} \end{matrix} & ) \\ & & ) \end{aligned} = detσ_{v}(b_{v})

and

ε̂\begin{aligned}( \\ ( & \begin{matrix}a_{v} & 0 \\ 0 & b_{v} \end{matrix} & ) \\ & & ) \end{aligned} = detσ_{v}(a_{v})

if v belongs to S and a_{v} and b_{v} are units of O_{v}. If α belongs to F^{×} and |α|_{v} = |ϖ_{v}|^{−n_{v}} for every v in S we set a_{α} = 1 and â_{α} = ∏_{v∈S} c_{v} detσ_{v}(α); otherwise we set a_{α} = â_{α} = 0.

The function Λ(s, ω) of Theorem 11.5 is defined only if ω_{v} is unramified at each place of S and then it equals

\{ ∏_{v∈S} ω_{v}(ϖ_{v}^{−n_{v}}) |ϖ_{v}|^{−n_{v}(s− \frac{1}{2} )}\} \{ ∏_{v∉S} L(s, ω_{v} ⊗ π_{v})\}

which is

\{ ∏_{v∈S} ω_{v}(ϖ_{v}^{−n_{v}}) |ϖ_{v}|^{−n_{v}(s− \frac{1}{2} )}\} L(s, ω ⊗ σ).

Λ̂(s, ω^{−1}η^{−1}) is also defined if ω_{v} is unramifed at each place of S and is equal to

\{ ∏_{v∈S} c_{v} ω_{v}(ϖ_{v}^{n_{v}}) |ϖ_{v}|^{−n_{v}(s− \frac{1}{2} )}\} L(s, ω^{−1} ⊗ σ̃).

Choose A in F^{×} so that |A_{v}| = |ϖ_{v}|^{m_{v}} for every v in S. Then

∏_{v∈S} ω_{v}(−A) |A|_{v}^{s− \frac{1}{2}} = ∏_{v∈S} ω_{v}(ϖ_{v})^{m_{v}} |ϖ_{v}|^{m_{v}(s− \frac{1}{2} )}.

The functional equation asserts that L(s, ω ⊗ σ) is equal to

{ ∏_{v∈S} ε(s, ω_{v} ⊗ σ_{v}, ψ_{v})}{ ∏_{v∉S} ε(s, ω_{v} ⊗ σ_{v}, ψ_{v})} L(1 − s, ω^{−1} ⊗ σ̃).

The first factor is equal to

\{ ∏_{v∈S} c_{v} ω_{v}(ϖ_{v})^{2n_{v}} |ϖ_{v}|^{2n_{v}(s− \frac{1}{2} )}\}\{ ∏_{v∈S} ω_{v}(−A) |A|_{v}^{s− \frac{1}{2}} \}.

Therefore Λ(s, ω) is equal to

\{ ∏_{v∈S} ω_{v}(−A) |A|_{v}^{s− \frac{1}{2}} \}\{ ∏_{v∉S} ε(s, ω_{v} ⊗ σ_{v}, ψ_{v})\}Λ̂(1 − s, ω^{−1}η^{−1}).

The assumptions of Theorem 11.5 are now verified. It remains to verify that of Corollary 11.6. It will be a consequence of the following lemma.

Lemma 12.3 Suppose F is a global field, K is a Galois extension of F, and ρ and σ are two representations of the Weil group W_{K/F}. If for all but a finite number of places v of F the local representations ρ_{v} and σ_{v} are equivalent then ρ and σ are equivalent.

We set

L_{0}(s, σ) = ∏_{𝔭} L(s, σ_{𝔭}).

The product is taken over all non-archimedean places. We first prove the following lemma.

Lemma 12.4 If σ is unitary the order of the pole of L_{0}(s, σ) at s = 1 is equal to the multiplicity with which the trivial representation is contained in σ.

There are fields E_{1}, ⋯, E_{r} lying between F and K, characters χ_{E_{1}}, ⋯, χ_{E_{r}}, and integers m_{1}, ⋯, m_{r} such that σ is equivalent to

⊕_{i=1}^{r}m_{i} Ind(W_{K/F} , W_{K/E}, χ_{E_{i}})

Let δ_{i} = 1 if χ_{E_{i}} is trivial and 0 otherwise. Since

L_{0}(s, σ) = ∏_{i=1}^{r} L_{0}(s, χ_{E_{i}})^{m_{i}}

the order of its pole at s = 1 is ∑_{i=1}^{r} m_{i} δ_{i}. However

Ind(W_{K/F} , W_{K/E}, χ_{E_{i}})

contains the trivial representation if and only if χ_{E_{i}} is trivial and then it contains it exactly once. Thus ∑_{i=1}^{r} m_{i} δ_{i} is also the number of times the trivial representation occurs in σ.

Observe that if T is any finite set of non-archimedean primes the order of the pole of

∏_{𝔭∉T} L(s, σ_{𝔭})

at s = 1 is the same as that of L_{0}(s, σ).

The first step of the proof of Lemma 12.3 is to reduce it to the case that both ρ and σ are unitary. Then ρ and σ certainly have the same degree d. Let ρ act on X and let σ act on Y. Under the restriction ρ to C_{K} the space X decomposes into the direct sum of invariant one-dimensional subspaces X_{1}, ⋯, X_{d} which transform according to quasi-characters μ^{1}, ⋯, μ^{d} of C_{K}. If a is a real number let

M(a) = {i | |μ^{i}(α)| = |α|^{a} for all α in C_{K}}

and let

X(a) = ∑_{i∈M(a)} X_{i}

X(a) is invariant under W_{K/F} and X = ∑_{a} ⊕X(a). Let ρ(a) be the restriction of ρ to X(a). Replacing ρ by σ and X by Y we can define ν^{1}, ⋯, ν^{d} and Y (a) in a similar fashion.

We now claim that if ρ_{v} is equivalent to σ_{v} then ρ_{v}(a) is equivalent σ_{v}(a) for each a. To see this we need only verify that any linear transformation from X to Y which commutes with the action of W_{K_{v}/F_{v}} or even of C_{K_{v}} takes X(a) to Y (a). Observe that under the restriction of ρ_{v} to C_{K_{v}} the space X_{i} transforms according to the character μ_{v}^{i} and that |μ_{v}^{i}(α)| = |α|^{a} for all α in C_{K_{v}} if and only if |μ^{i}(α)| = |α|^{a} for all α in C_{K}. Thus X(a) and Y (a) can be defined in terms of ρ_{v} and σ_{v} alone. The assertion follows.

Thus we may as well assume that for some real number a

|μ^{i}(α)| = |ν^{i}(α)| = |α|^{a}

for all i and all α in C_{K}. Replacing σ by α → |α|^{−a} σ(α) and ρ by α → |α|^{−a} ρ(α) if necessary we may even assume that a = 0. Then ρ and σ will be equivalent to unitary representations and we now suppose them to be unitary.

If τ is irreducible and ρ ≃ τ ⊕ ρ′ and σ ≃ τ ⊕ σ′ then ρ′_{v} is equivalent to σ′_{v} whenever ρ_{v} is equivalent to σ_{v}. Since we can use induction on d it is enough to show that if τ is irreducible and unitary and contained in ρ then it is contained in σ. Let ρ̃ and σ̃ be the representations contragredient to ρ and σ. Certainly (ρ̃⊗ τ)_{v} = ρ̃_{v} ⊗ τ_{v} is equivalent to (σ̃ ⊗ τ)_{v} for all but a finite number of v. Moreover ρ̃ ⊗ τ contains τ̃ ⊗ τ which contains the identity. If σ̃ ⊗ τ contains the identity then, as is well-known and easily verified, σ contains τ. On the other hand the orders of the poles of L_{0}(s, ρ̃⊗ τ) and L_{0}(s, σ̃ ⊗ τ) at s = 1 are clearly equal so that, by Lemma 12.4, σ̃ ⊗ τ contains the trivial representations if ρ̃⊗ τ does.

Chapter 2 215

We return to the proof of Theorem 12.2. It follows from Lemma 12.3 that if the assumptions of Corollary 11.6 are not satisfied σ is equivalent to the direct sum of two one-dimensional representations associated to quasi-characters μ and ν of C_{F}. Then

L(s, ω ⊗ σ) = L(s, ωμ) L(s, ων).

The two functions on the right are Hecke L-functions. The function on the left is entire for every choice of ω. Taking ω = μ^{−1} and ω = ν^{−1} we see that L(s, μ^{−1}ν) and L(s, ν^{−1}μ) have a zero at s = 1. Let μ^{−1} ν(α) = |α|^{r} χ(α) where χ is a character. Then

\begin{aligned}L(s, μ^{−1}ν) & = L(s + r, χ) \\ L(s, ν^{−1}μ) & = L(s − r, χ^{−1}). \end{aligned}

Now neither L(s, χ) nor L(s, χ^{−1}) has a zero in the set Re s ≥ 1. Therefore 1+r < 1 and 1−r < 1. This is impossible.

We can now apply Corollary 11.6 to assert that there is a constituent π′ = ∏_{v} ⊗π′_{v} of 𝒜_{0} such that π′_{v} = π(σ_{v}) for v not in S. To prove the theorem we need only show that π′_{v} = π(σ_{v}) for v in S. Taking the quotient of the two functional equations

L(s, ω ⊗ σ) = { ∏_{v} ε(s, ω_{v} ⊗ σ_{v}, ψ_{v}} L(1 − s, ω^{−1} ⊗ σ̃)

and

L(s, ω ⊗ π′) = { ∏_{v} ε(s, ω_{v} ⊗ π′_{v}, ψ_{v})} L(1 − s, ω^{−1} ⊗ π̃′),

we find that

∏_{v∈S} \frac{L(s, ω_{v} ⊗ σ_{v})}{L(s, ω_{v} ⊗ π′_{v})}

is equal to

\{∏_{v∈S} \frac{ε(s, ω_{v} ⊗ σ_{v}ψ_{v})}{ε(s, ω_{v} ⊗ π′_{v}, ψ_{v})}\} \{∏_{v∈S} \frac{L(1 − s, ω_{v}^{−1} ⊗ σ̃_{v})}{L(1 − s, ω_{v}^{−1} ⊗ π̃′_{v})}\} .

We need one more lemma. If v is a non-archimedean place and ω_{v} is a quasi-character of F_{v}^{×} let m(ω_{v}) be the smallest non-negative integer such that ω_{v} is trivial on the units of O_{v} congruent to 1 modulo 𝔭_{v}^{m(ω_{v})}.

Lemma 12.5 Suppose S is a finite set of non-archimedean places and v_{0} ∈ S. Suppose that we are given a quasi-character χ_{v_{0}} of F_{v_{0}}^{×} and for each v ≠ v_{0} in S a non-negative integer m_{v}. Then there is a quasi-character ω of C_{F} such that ω_{v_{0}} = χ_{v_{0}} and m(ω_{v}) ≥ m_{v} if v ≠ v_{0} belongs to S.

Suppose χ_{v_{0}}(α) = |α|_{v_{0}}^{a} χ′_{v_{0}}(α) where χ′_{v_{0}} is a character. If ω′ is a character of C_{F} and ω′_{v_{0}} = χ′_{v_{0}} while m(ω′_{v}) ≥ m_{v} for v ≠ v_{0} in S we may take ω to be the generalized character α → |α|^{r} ω′(α) of C_{F}. In other words we may assume initially that χ_{v_{0}} is a character. Let A be the group of idèles whose component at places not in S is 1, whose component of a place v ∉ v_{0} in S is congruent to 1 modulo 𝔭_{v}^{m_{v}}, and whose component at v_{0} is arbitrary. Certainly F^{×} ∩ A = {1}. We claim that F^{×}A is closed in I. Indeed if α ∈ I there is a compact neighborhood X of α on which the norm is bounded above by 1/ε and below by ε where ε is a positive constant. If β ∈ F^{×} and γ ∈ A then |βγ| = |γ|. Moreover

A_{ε} = \{γ ∈ A | ε ≤ |γ| ≤ \frac{1}{ε}\}

is compact. Since F^{×} is discrete F^{×}A_{ε} is closed. Since any point has a compact neighborhood whose intersection with F^{×}A is closed the set F^{×}A is itself closed.

We can certainly find a character of A which equals χv_{0} on F_{v_{0}}^{×} and, for any v ≠ v_{0} in S, is non-trivial on the set of units in O_{v} congruent to 1 modulo 𝔭_{v}^{m_{v}}. Extend this character to F^{×}A by setting it equal to 1 on F^{×}. The result can be extended to a character of I which is necessarily 1 on F^{×}. We take ω to be this character.

Let π′_{v}(({α_{v} \atop 0} {0 \atop α_{v}} )) = η_{v}(α_{v}). If η(α) = ∏_{v} η_{v}(α_{v}) then η is a quasi-character of F^{×} \ I. Since, by construction, η = detσ on I_{D}^{s} the quasi-characters η and detσ are equal. Therefore η_{v} = detσ_{v} for all v. We know that if m(ω_{v}) is sufficiently large

L(s, ω_{v} ⊗ σ_{v}) = L(s, ω_{v} ⊗ π′_{v}) = 1

and

L(1 − s, ω_{v}^{−1} ⊗ σ̃_{v}) = L(1 − s, ω_{v}^{−1} ⊗ π̃′_{v}) = 1.

Moreover, by Proposition 3.8

ε(s, ω_{v} ⊗ π′_{v}, ψ_{v}) = ε(s, ω_{v}η_{v}, ψ_{v}) ε(s, ω_{v}, ψ_{v}).

It is shown in the Appendix of [19] that if m(ω_{v}) is sufficiently large

ε(s, ω_{v} ⊗ σ_{v}, ψ_{v}) = ε(s, ω_{v}detσ_{v}, ψ_{v}) ε(s, ω_{v}, ψ_{v}).

Applying Lemma 12.5 and the equality preceding it we see that if v is in S and ω_{v} is any quasi-character of F_{v}^{×}

\frac{L(s, ω_{v} ⊗ σ_{v})}{L(s, ω_{v} ⊗ π′_{v})} = \{\frac{ε(s, ω_{v} ⊗ σ_{v}, ψ_{v})}{ε(s, ω_{v} ⊗ π′_{v}, ψ_{v})}\} \{\frac{L(1 − s, ω_{v}^{−1} ⊗ σ̃)}{L(1 − s, ω_{v}^{−1} ⊗ π̃′_{v})}\} .

Recalling that

L(s, ω_{v} ⊗ σ_{v}) = L(1 − s, ω_{v}^{−1} ⊗ σ̃_{v}) = 1

for v in S we see that

\frac{L(1 − s, ω_{v}^{−1} ⊗ π̃′_{v})}{L(s, ω_{v} ⊗ π′_{v})} = \frac{ε(s, ω_{v} ⊗ σ_{v}, ψ_{v})}{ε(, ω_{v} ⊗ π′_{v}, ψ_{v})}.  (12.5.1)

The theorem will follow if we show that

L(s, ω_{v} ⊗ π′_{v}) = L(1 − s, ω_{v}^{−1} ⊗ π′_{v}) = 1

for all choices of ω_{v}.

If not, either π′_{v} is a special representation or there are two quasi-characters μ_{v} and ν_{v} of F_{v}^{×} such that π′_{v} = π(μ_{v}, ν_{v}). According to (12.5.1) the quotient

\frac{L(1 − s, ω_{v}^{−1} ⊗ π′_{v})}{L(s, ω_{v} ⊗ π′_{v})}

is an entire function of s for every choice of ω_{v}. If π′_{v} = π(μ_{v}, ν_{v}) and m(μ_{v}^{−1}ν_{v}) is positive

\frac{L(1 − s, μ_{v} ⊗ π̃′_{v})}{L(s, μ_{v}^{−1} ⊗ π′_{v})} = \frac{1 − |ϖ_{v}|^{s}}{1 − |ϖ_{v}|^{1−s}}

which has a pole at s = 1. If m(μ_{v}^{−1}ν_{v}) = 0

\frac{L(1 − s, μ_{v} ⊗ π̃′_{v})}{L(s, μ_{v}^{−1} ⊗ π′_{v})} = \{\frac{1 − |ϖ_{v}|^{s}}{1 − |ϖ_{v}|^{1−s}} \} \{\frac{1 − μ_{v}^{−1} ν_{v}(ϖ_{v}) |ϖ_{v}|^{s}}{1 − μ_{v} ν_{v}^{−1}(ϖ_{v}) |ϖ_{v}|^{1−s}} \}

which has a pole at s = 1 unless μ_{v} ν_{v}^{−1}(ϖ_{v}) = |ϖ_{v}|. But then it has a pole at s = 2. If π′_{v} is the special representation associated to the pair of quasi-characters

α → μ_{v}(α) |α|^{\frac{1}{2}} α → μ_{v}(α) |α|^{− \frac{1}{2}}

of F_{v}^{×} then

\frac{L(1 − s, μ_{v} ⊗ π̃′_{v})}{L(s, μ_{v}^{−1} ⊗ π′_{v})} = \frac{1 − |ϖ_{v}|^{s+ \frac{1}{2}}}{1 − |ϖ_{v}|^{\frac{1}{2} −s}}

which has a pole at s = \frac{1}{2}.

There is a consequence of the theorem which we want to observe.

Proposition 12.6 Suppose E is a global field and that for every separable extension F of E, every Galois extension K of F, and every irreducible two-dimensional representation σ of W_{K/F} the function L(s, σ) is entire and bounded in vertical strips. Then if F_{1} is the completion of E at some place, K_{1} is a Galois extension of F_{1}, and σ_{1} is a two-dimensional representation of W_{K_{1}/F_{1}} the representation π(σ_{1}) exists.

We begin with a simple remark. The restriction of σ_{1} to C_{K_{1}} is the direct sum of two onedimensional representations corresponding to generalized characters χ_{1} and χ_{2} of C_{K_{1}}. If τ belongs to G = 𝔊(K_{1}/F_{1}) either χ_{1}(τ(α)) = χ_{1}(α) for all α in C_{K} or χ_{1}(τ(α)) = χ_{2}(α) for all α in C_{K}. If the representation σ_{1} is irreducible there is at least one τ for which χ_{1}(τ(α)) = χ_{2}(α). If χ_{1} ≠ χ_{2}, the fixed field L_{1} of

H = {τ ∈ G | χ_{1}(τ(α)) ≡ χ_{1}(α)}

is a quadratic extension of F. The restriction of σ_{1} to W_{K_{1}/L_{1}} is the direct sum of two onedimensional representations and therefore is trivial on the commutator subgroup W_{K_{1}/L_{1}}^{c} which is the kernel of τ_{K_{1}/F_{1},L_{1}/F_{1}}. With no loss of generality we may suppose that K_{1} equals L_{1} and is therefore a quadratic extension of F_{1}. Then σ_{1} is equivalent to the representation

Ind(W_{K_{1}/F_{,}}, W_{K_{1}/K_{1}}, χ_{1}).

If σ_{1} is reducible π(σ_{1}) is defined. The preceding remarks show that it is defined if σ_{1} is irreducible and σ_{1}(α) is not a scalar matrix for some α in C_{K_{1}}. The proposition will therefore follow from Theorem 12.2 and the next lemma.

Chapter 2 218 Lemma 12.7 Suppose F_{1} is the completion of the field E at some place, K_{1} is a Galois extension of F_{1}, and σ_{1} is an irreducible two-dimensional representation such that σ_{1}(α) is a scalar matrix for all α in C_{K_{1}}. Then there is a separable extension F of E, a Galois extension K of F, a place v of K, and isomorphism φ of K_{v} with K_{1} which takes F_{v} to F_{1}, and an irreducible two-dimensional representation σ of W_{K/F} such that σ_{v} is equivalent to σ_{1} ∘ φ.

Observe that the existence of σ_{1} forces F_{1} to be non-archimedean. We establish a further sequence of lemmas.

Lemma 12.8 Suppose V is a finite dimensional real vector space, G is a finite group of linear transformations of V, and L is a lattice in V invariant under G. If χ is a quasi-character of L invariant under G there is a quasi-character χ′ of V invariant under G and a positive integer m such that the restrictions of χ′ and χ to mL are equal.

Let V̂ be the dual of V and V̂_{ℂ} its complexification. There is a y in V̂_{ℂ} such that χ(x) = e^{2πi⟨x,y⟩} for all x in L. If z belongs to V̂_{C} the generalized character x → e^{2πi⟨x,z⟩} is trivial on L if and only if z belongs to \frac{_{L}}{m}. L̂ is the lattice

{v ∈ V̂ | ⟨x, v⟩ ∈ Z for all x in L}.

Let Ĝ be the group contragredient to G. We have to establish the existence of an m and a z in \frac{_{L}}{m} such that y − z is fixed by Ĝ. If σ belongs to Ĝ then σy − y = w_{σ} belongs to L̂. Clearly σw_{τ} + w_{σ} = w_{στ}. Set

z = \frac{1}{[G : 1]} ∑_{τ} w_{τ} .

If m is taken to be [G : 1] this is the required element.

Lemma 12.9 Suppose F is a global field, K is a Galois extension of it, and v is a place of K. Suppose also that [K_{v} : F_{v}] = [K : F] and let χ_{v} be a quasi-character of C_{K_{v}} invariant under G = 𝔊(K_{v}/F_{v}) = 𝔊(K/F). There is a closed subgroup A of finite index in C_{K} which is invariant under G and contains C_{K_{v}} and a quasi-character χ of A invariant under G whose restriction to C_{K_{v}} is χ_{v}.

Suppose first that the fields have positive characteristic. We can choose a set of non-negative integers n_{w}, w ≠ v, all but a finite number of which are zero, so that the group

B = C_{K_{v}} × ∏_{w≠v} U_{K_{w}}^{n_{w}}

is invariant under G and contains no element of K^{×} except 1. U_{K_{w}}^{n_{w}} is the group of units of O_{K_{w}} which are congruent to 1 modulo 𝔭_{K_{w}}^{n_{w}}. We extend χ_{v} to B by setting it equal to 1 on

∏_{w≠v} U_{K_{w}}^{n_{w}}

and then to A = K^{×}B/K^{×} by setting it equal to 1 on K^{×}.

Chapter 2 219

Now let the fields have characteristic 0. Divide places of K different from v into two sets, S, consisting of the archimedean places, and T, consisting of the non-archimedean ones. Choose a collection of non-negative integers π′_{w}, w ∈ T, all but a finite number of which are zero, so that

B′ = C_{K_{v}} × ∏_{w∈S} C_{K_{w}} × ∏_{w∈T} U_{K_{w}}^{n′_{w}}.

is invariant under G and contains no roots of unity in K except 1. If w is archimedean let U_{K_{w}} be the elements of norm 1 in K_{w} and set

B′_{a} = ∏_{w∈S} U_{K_{w}} × ∏_{w∈T} U_{K_{w}}^{n′_{w}}.

B′/B′_{a} is isomorphic to the product of C_{K_{v}} and

V = ∏_{w∈S} C_{K_{w}}/U_{K_{w}}

which is a vector group. The projection L of

M = B′_{1}(B′ ∩ K^{×})/B′_{1}

on V is a lattice in V and the projection is an isomorphism. Define the quasi-character μ of L so that if m in M projects to m_{1} in C_{K_{v}} and to m_{2} in V then

χ_{v}(m_{1}) μ(m_{2}) = 1.

μ is invariant under G. Choose a quasi-character μ′ of V and an integer n so that μ′ and μ are equal on nL. Let ν′ be the quasi-character obtained by lifting χ_{v} ×μ′ from C_{K_{v}} ×V to B′. It follows from a theorem of Chevalley ([20] Theorem 1) that we can choose a collection of non-negative ingegers {n_{w} | w ∈ T } all but a finite number of which are zero so that n_{w} ≥ n′_{w} for all w in T, so that

B = C_{K_{v}} × ∏_{w∈S} C_{K_{w}} × ∏_{w∈T} U_{K_{w}}^{n_{w}}

is invariant under G, and so that every element of B ∩ K^{×} is an nth power of some element of B′ ∩ K^{×}. The restriction ν of ν′ to B is trivial on B ∩ K^{×}. We take A = K^{×}B/K^{×} and let χ be the quasi-character which is 1 on K^{×} and ν on B.

Lemma 12.10 Suppose F_{1} is a completion of the global field E, K_{1} is a finite Galois extension of F_{1} with Galois group G_{1}, and χ_{K_{1}} is a quasi-character of C_{K_{1}} invariant under G_{1}. There is a separable extension F of E, a Galois extension K of F, a place v of K such that [K_{v} : F_{v}] = [K : F], an isomorphism φ of K_{v} with K_{1} which takes F_{v} to F_{1}, and a quasi-character χ of C_{K} invariant under 𝔤(K/F) such that χ_{v} = χ_{K_{1}} ∘ φ.

We may as well suppose that F_{1} = E_{w}, where w is some place of E. It is known ([8], p.31) that there is a polynomial with coefficients in E such that if θ is a root of this polynomial E_{w}(θ)/E_{w} is isomorphic to K_{1}/F_{1}. Let L be the splitting field of this polynomial and extend w to a place of L. The extended place we also call w. Replacing E by the fixed field of the decomposition group of w Chapter 2 220 if necessary we may suppose that F_{1} = E_{w}, K_{1} = L_{w} and [L_{w} : E_{w}] = [L : E]. Now set χ_{w} = χ_{K_{1}} and extend χ_{w} to a quasi-character χ′ of A as in the previous lemma.

Let K be the abelian extension of L associated to the subgroup A. Since A is invariant under 𝔊(L/E) the extension K/E is Galois. Let v be a place of K dividing the place w of L. Since A contains C_{L_{w}} the fields K_{v} and L_{w} are equal. Let F be the fixed field of the image of 𝔊(K_{v}/E_{w}) in 𝔊(K/E). Let v also denote the restriction of v to F. The fields F_{v} and E_{w} are the same. The mapping N_{K/L} : C_{K} → C_{L} maps C_{K} into A. Let χ = χ′ ∘ N_{K/L}. Then χ is clearly invariant under 𝔊(K/F). Since N_{K/L} restricted to K_{v} is an isomorphism of K_{v} with L_{w} which takes F_{v} onto E_{w} the lemma is proved.

To prove Lemma 12.7 we need only show that if F is a global field, K is a Galois extension of F, χ is a quasi-character of C_{K} invariant under 𝔊(K/F), v is a place of K such that [K : F] = [K_{v} : F_{v}], and σ_{1} is an irreducible two-dimensional representation of W_{K_{v}/F_{v}} such that σ_{v}(α) = χ_{v}(α)I for all α in C_{K_{v}} then there is a two-dimensional representation σ of W_{K/F} such that σ_{v} is equivalent to σ_{1}. σ will be irreducible because σ_{1} is.

Let σ_{1} act on X. Let ρ_{v} be the right regular representation of W_{K_{v}/F_{v}} on the space V_{v} of functions f on W_{K_{v}/F_{v}} satisfying

f(αw) = χ_{v}(α) f(w)

for all α in C_{K_{v}} and all w in W_{K_{v}/F_{v}}. If λ is a non-zero linear functional on X the map from x to the function λ(σ_{1}(w)x) is a W_{K_{v}/F_{v}}-invariant isomorphism of X with a subspace Y of V_{v}.

Let V be the space of all functions f on W_{K/F} satisfying

f(αw) = χ(α) f(w)

for all α in C_{K} and all w in W_{K/F}. Since [K : F] = [K_{v} : F_{v}] the groups 𝔊(K/F) and 𝔊(K_{v}/F_{v}) are equal. Therefore

W_{K/F} = C_{K} W_{K_{v}/F_{v}}

Moreover C_{K_{v}} = C_{K} ∩W_{K_{v}/F_{v}}. Thus the restriction of functions in V to W_{K_{v}/F_{v}} is an isomorphism of V with V_{v}. For simplicity we identify the two spaces. Let ρ be the right regular representation of W_{K/F} on V. If α belongs to C_{K} then

f(wα) = χ(wαw^{−1}) f(w) = χ(α) f(w)

because χ is 𝔊(K/F) invariant. Therefore ρ(α) = χ(α) I and a subspace V is invariant under W_{K/F} if and only if it is invariant under W_{K_{v}/F_{v}}. If we take for σ the restriction of ρ to Y then σ_{v} will be equivalent to σ_{1}.

Chapter 2 221 References for Chapter II

Automorphic forms are discussed in terms of group representations in [3] and [11] as well as:

13. Godement, R., Analyse spectrale des fonctions modulaires, Seminaire Bourbaki, No. 278.

14. Godement, R., Introduction à la theorie de Langlands, Seminaire Bourbaki, No. 321. As its name implies the Hecke theory is a creation of Hecke.

15. Hecke E., Mathematische Werke.

Maass seems to have been the first to consider it outside the classical context.

16. Maass, H., Über eine neue Art von nichtanalytischen automorphen Funktionen und die Bestimmung Dirichletscher Reihen durch Funktionalgleichungen, Math. Ann. 121 (1944). It seems to have been Weil who first used several L-functions to prove a converse theorem.

17. Weil, A., Über die Bestimmung Dirichletscher Reihen durch Funktionalgleichungen, Math. Ann. 168 (1967). His generalizations of the Artin L-functions are introduced in:

18. Weil, A., Sur la théorie du corps de classes, Jour. Math. Soc. Japan, vol. 3 (1951). For various technical facts used in the twelfth paragraph we refer to:

19. Langlands R., On the functional equation of the Artin L-functions, Notes, Yale University (in preparation). We have also had occasion to refer to:

20. Chevalley, C., Deux théorèmes d'arithmetique, Jour. Math. Soc. Japan, vol. 3 (1951). A result more or less the same as Proposition 12.1 is proved in:

21. Shalika, J.A. and S. Tanaka, On an explicit construction of a certain class of automorphic forms, Preprint.

Chapter III: Quaternion Algebras

§13. Zeta-functions for M(2, F). In this paragraph F is again a local field and A = M(2, F) is the algebra of 2 × 2 matrices with entries from F. The multiplicative group A^{×} of A is just G_{F} = GL(2, F). If g is in G_{F} we set

|g|_{A} = α_{A}(g) = |detg|_{F}^{2} .

Let π be an admissible representation of ℋ_{F} on the space V. Let the contragredient representation π̃ act on on Ṽ. If v belongs to V and ṽ to Ṽ the function

⟨π(g)v, ṽ⟩ = ⟨v, π̃(g^{−1})ṽ⟩

is characterized by the relation

∫ ⟨π(gh), v, ṽ⟩ f(h) dh = ⟨π(g) π(f)v, ṽ⟩

for all f in ℋ_{F}.

If Φ belongs to the Schwartz space 𝒮(A) and v belongs to V and ṽ to Ṽ we set

Z(π, Φ, v, ṽ) = ∫_{G_{F}} Φ(g) ⟨π(g)v, ṽ⟩ d^{×}g

and

Z(π̃, Φ, v, ṽ) = ∫_{G_{F}} Φ(g) ⟨v, π̃(g)ṽ⟩ d^{×}g

The choice of Haar measure is not important provided that it is the same for both integrals. If ω is a quasi-character of F^{×}

Z(ω ⊗ π, Φ, v, ṽ) = ∫_{G_{F}} Φ(g) ω(detg)⟨π(g)v, ṽ⟩ d^{×}g

The purpose of this paragraph is to prove the following theorem.

Theorem 13.1 Let π be an irreducible admissible representation of ℋ_{F} and π̃ its contragredient. Let π act on V and π̃ on Ṽ.

(i) For every v in V, ṽ in Ṽ, and Φ in 𝒮(A) the integrals defining Z(α_{F}^{s} ⊗ π, Φ, v, ṽ) and Z(α_{F}^{s} ⊗ π̃, Φ, v, ṽ) converge absolutely for Re s sufficiently large.

(ii) Both functions can be analytically continued to functions which are meromorphic in the whole

plane and bounded at infinity in vertical strips of finite width. (iii) If

Z(α_{F}^{s+ \frac{1}{2}} ⊗ π, Φ, v, ṽ) = L(s, π) Ξ(s, Φ, v, ṽ)

and

Z(α_{F}^{s+ \frac{1}{2}} ⊗ π̃, Φ, v, ṽ) = L(s, π̃) Ξ̃(s, Φ, v, ṽ)

then Ξ(s, Φ, v, ṽ) and Ξ̃(s, Φ, v, ṽ) are entire. (iv) There exist ϕ, v_{1}, ⋯, v_{n} and ṽ_{1}, ⋯, ṽ_{n} such that ∑_{i=1}^{n} Ξ(s, Φ, v_{i}, ṽ_{i}) is of the form a e^{bs} with

a ≠ 0.

(v) If Φ′ is the Fourier transform of Φ with respect to the character ψ_{Z}(x) = ψ_{F} (tr x) then

Ξ̃(1 − s, Φ′, v, ṽ) = ε(s, π, ψ_{F}), Ξ(s, Φ, v, ṽ).

We suppose first that F is non-archimedean and π is absolutely cuspidal. Then we may take π in the Kirillov form so that V is just 𝒮(F^{×}). Since an additive character ψ_{F} = ψ is given we will of course want to take the Kirillov model with respect to it. The next lemma is, in the case under consideration, the key to the theorem.

Lemma 13.1.1 If φ belongs to 𝒮(F^{×}), v belongs to V, and ṽ belongs to Ṽ set

Φ(g) = φ(detg)⟨v, π̃(g)ṽ⟩ |detg|_{F}^{−1}

if g belongs to G_{F} and set Φ(g) = 0 if g in A is singular. Then Φ belongs to 𝒮(A) and its Fourier transform is given by

Φ′(g) = φ′(detg)⟨π(g)v, ṽ⟩ |detg|_{F}^{−1} η^{−1}(detg)

if g belongs to G_{F} and

Φ′(g) = 0

if g is singular. Here η is the quasi-character of F^{×} defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η(a) I

and

φ′ = π \begin{aligned}( \\ ( & \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} & ) \\ & & ) \end{aligned} φ.

This lemma is more easily appreciated if it is compared with the next one which is simpler but which we do not really need.

Lemma 13.1.2 Let 𝒮_{0}(A) be the space of all Φ in 𝒮(A) that vanish on the singular elements and satisfy

∫ ϕ (g_{1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g_{2}) dx = 0

for g_{1} and g_{2} in G_{F}. If Φ is in 𝒮_{0}(A) so is its Fourier transform.

Since 𝒮_{0}(A) is stable under left and right translations by the elements of G_{F} it is enough to show that

Φ′ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 0 \end{matrix} & ) \\ & & ) \end{aligned} = 0

for a in F and that

∫_{F} Φ′ \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} dx = 0

To verify these relations we just calculate the left sides!

Φ′ \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 0 \end{matrix} & ) \\ & & ) \end{aligned} = ∫_{A} Φ(g) ψ_{A} (g ( \begin{matrix}a & 0 \\ 0 & 0 \end{matrix} )) dg

The right side is a positive multiple of

∫_{G_{F}} Φ(g) ψ_{A} (g ( \begin{matrix}a & 0 \\ 0 & 0 \end{matrix} )) |detg|^{2} d^{×}g

which equals

∫_{G_{F} /N_{F}} ψ_{A} (g ( \begin{matrix}a & 0 \\ 0 & 0 \end{matrix} )) |detg|^{2}\{ ∫_{F} Φ (g ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) dx\} d^{×}g

This is 0 because the inner integral vanishes identically.

∫_{F} Φ′ \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} dx

is equal to

∫ \{ ∫ Φ \begin{aligned}( \\ ( & \begin{matrix}α & β \\ γ & δ \end{matrix} & ) \\ & & ) \end{aligned} ψ_{F} (α + δ + γx) dα dβ dγ dδ\} dx

which, by the Fourier inversion fomula, is equal to

∫ Φ \begin{aligned}( \\ ( & \begin{matrix}α & β \\ 0 & δ \end{matrix} & ) \\ & & ) \end{aligned} ψ_{F} (α + δ) dα dδ dβ

which equals

∫ |α| ψ_{F} (α + δ)\{Φ (( \begin{matrix}α & 0 \\ 0 & δ \end{matrix} ) ( \begin{matrix}1 & β \\ 0 & 1 \end{matrix} )) dβ\} dα dδ

and this is 0.

We return to the proof of Lemma 12.1.1 for absolutely cuspidal π. Since ⟨v, π̃(g)ṽ⟩ has compact support on G_{F} modulo Z_{F} the function Φ(g) belongs to 𝒮(A). Moreover

∫_{F} Φ (g ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) h) dx

is equal to

φ(detgh) |detgh|_{F}^{−1} ∫ ⟨π(g^{−1}v, π̃ \begin{aligned}( \\ ( & \begin{matrix}1 & x \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π̃(h) v⟩ dx.

Since π is absolutely cuspidal this integral is 0. Thus Φ belongs to 𝒮_{0}(A) and, in particular, Φ′ vanishes at the singular elements.

Suppose we can show that for all choice of φ, v, and ṽ

Φ′(e) = φ′(e) ⟨v, ṽ⟩.  (13.1.3)

If h belongs to G_{F} set Φ_{1}(g) = Φ(h^{−1}g). If a = deth, φ_{1}(x) = |a| φ(a^{−1}x), and v_{1} = π(h) v,

Φ_{1}(g) = φ_{1}(detg)⟨v_{1}, π̃(g)ṽ⟩ |detg|_{F}^{−1}.

Then Φ′_{1}(e) is equal to

φ′_{1}(e) ⟨v_{1}, v⟩.

On the other hand

φ′_{1} = π(w) φ = |a| π(w) π \begin{aligned}( \\ ( & \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ

which equals

|a| π \begin{aligned}( \\ ( & \begin{matrix}a^{−1} & 0 \\ 0 & a^{−1} \end{matrix} & ) \\ & & ) \end{aligned} π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} π(w) φ.

Thus Φ′(h), which equals Φ′_{1}(e) |deth|^{−2}, is

φ′(deth)⟨π(h) v, ṽ⟩ η^{−1}(deth) |deth|^{−1}.

The formula (13.1.3) will be a consequence of the next lemma.

Chapter 3 225 Lemma 13.1.4 Let dε be the normalized Haar measure on the group U = U_{F}. If ν is a character of U set

η(ν, x) = ∫_{U} ν(ε) ψ(εx) dε

if x is in F. Let dx be the Haar measure on F which is self-dual with respect to ψ. Then

∫_{F} η(ν, xϖ^{n}) ψ(ax) dx = 0

unless |a| = |ϖ|^{n} but if a = ζϖ^{n} with ζ in U

∫_{F} η(ν, xϖ^{n}) ψ(ax) dx = ν(−ζ) |ϖ|^{−n} c^{−1}

if c is the measure of U with respect to dx.

The general case results from the case n = 0 by a change of variable; so we suppose n = 0. In this case the formulae amount to a statement of the Fourier inversion formula for the function which is 0 outside of U and equal to c^{−1}ν(ε) on U.

Suppose we could show that there is a positive constant d which does not depend on π such that for all φ, v, and ṽ

Φ′(e) = dφ′(e) ⟨v, ṽ⟩.

Then we would have

Φ′(g) = dφ′(detg) ⟨π(g)v, ṽ⟩ η^{−1}(detg).

Exchanging π and π̃ and recalling that π̃ = η^{−1} ⊗ π we see that Φ′′, the Fourier transform of Φ′, is given by

Φ′′(g) = d^{2}φ′′(detg)⟨v, π̃(g)ṽ⟩ |detg|_{F}^{−1} η(detg),

where φ′′ = π̃(w)φ_{1} if φ_{1}(a) = φ′(a) η^{−1}(a). According to the remarks preceding the statement of Theorem 2.18, φ′′ is the product of π(w) φ′ = η(−1) φ and η^{−1}(detg). Thus

Φ′′(g) = η(−1) d^{2}φ(detg) ⟨v, π̃(g)ṽ⟩ |detg|_{F}^{−1}.

Since Φ′′ = Φ(−g) = η(−1) Φ(g) the numbers d^{2} and d are both equal to 1. The upshot is that in the proof of the formula (13.1.3) we may ignore all positive constants and in particular do not need to worry about the normalizaton of Haar measures.

Moreover it is enough to prove the formula for φ, v, ṽ in a basis of the spaces in which they are constrained to lie. Oddly enough the spaces are all the same and equal to 𝒮(F^{×}). Assume φ_{1} = v, φ_{2} = ṽ, and φ are supported respectively by ϖ^{n_{1}}U, ϖ^{n_{2}}U, and ϖ^{n}U and that, for all ε in U, φ_{1}(ϖ^{n_{1}}ε) = ν_{1}^{−1}(ε), φ_{2}(ϖ^{n_{2}}ε) = ν_{2}^{−1}(ε) and φ(ϖ^{n}ε) = ν^{−1}(ε). ν, ν_{1} and ν_{2} are three characters of U.

The formal Mellin transforms of these three functions are φ̂_{1}(μ, t) = δ(μν_{1}^{−1})t^{n_{1}}, φ̂_{2}(μ, t) = δ(μν_{2}^{−1})t^{n_{2}}, and φ̂(μ, t) = δ(μν^{−1})t^{n}. Recall that, for example,

φ̂(μ, t) = ∑_{n} t^{n} ∫_{U} φ(ϖ^{n}ε) μ(ε) dε.

The scalar product ⟨φ_{1}, φ_{2}⟩ is equal to

∫ φ_{1}(a) φ_{2}(−a) d^{×}a = δ(ν_{1}ν_{2}) δ(n_{1} − n_{2}) ν_{2}(−1).

If η(εϖ^{n}) = ν_{0}(ε) z_{0}^{n} then

φ′(μ, t) = C(μ, t) φ̂(μ^{−1}ν_{0}^{−1}, t^{−1}z_{0}^{−1})

which equals

δ(νμν_{0}) ∑_{m} C_{m}(ν^{−1}ν_{0}^{−1})t^{m−n}z_{0}^{−n}.

Consequently

φ′(1) = C_{n}(ν^{−1}ν_{0}^{−1}) z_{0}^{−n}.

Thus the formula to be proved reads

Φ′(e) = C_{n}(ν^{−1}ν_{0}^{−1}) z_{0}^{−n} ν_{2}(−1) δ(ν_{1}ν_{2}) δ(n_{1} − n_{2}).

Almost all g in A can be written in the form

g = ( \begin{matrix}b & 0 \\ 0 & b \end{matrix} ) ( \begin{matrix}1 & −x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} )

with a and b in F^{×} and x and y in F. The additive Haar measure dg on A may be written as

dg = |detg|_{F}^{2} d^{×}g = |b^{4}| d^{×}b dx|a| d^{×}a dy

and for any g of this form

ψ_{A}(g) = ψ_{F} (b(x − y))

while Φ(g) is equal to

η^{−1}(b) |b^{2}a|^{−1} φ(b^{2}a)⟨π (( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ( \begin{matrix}a^{−1} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) φ_{1}, π̃ \begin{aligned}( \\ ( & \begin{matrix}1 & y \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} φ_{2}⟩.

Let f_{1} and f_{2} be the two functions which appear in the scalar product. Their formal Mellin transforms can be calculated by the methods of the second paragraph,

f̂_{1}(μ, t) = ν_{0}(−1) C(μ, t) η(μ^{−1}ν_{0}^{−1}ν_{1}^{−1}, ϖ^{n_{1}}x) μ^{−1}ν_{0}^{−1}(ζ) z_{0}^{−r−n_{1}}t^{−r−n_{2}}

if a = ζφ^{r} and

f̂_{2}(μ, t) = η(μν_{2}^{−1}, ϖ^{n_{2}}y) t^{n_{2}}.

The scalar product of f_{1} and f_{2} is equal to

∫ f_{1}(a) f_{2}(−a) d^{×}a

which, by the Plancherel theorem for F^{×}, is equal to

∑_{μ} μ(−1)\frac{1}{2π} ∫_{0}^{2π} f̂_{1}(μ, e^{iθ}) f̂_{2}(μ^{−1}, e^{−iθ}) dθ.

A typical integral is equal to the product of ν_{0}(−1) μ^{−1}ν_{0}^{−1}(ζ)z_{0}^{−r−n_{1}} and

∫_{0}^{2π} C(μ, e^{iθ}) e^{−i(r+n_{1}+n_{2})θ} η(μ^{−1}ν_{0}^{−1}ν_{1}^{−1}, ϖ^{n_{1}}x) η(μ^{−1}ν_{2}^{−1}, ϖ^{n_{2}}y) dθ

which equals

2πC_{r+n_{1}+n_{2}}(μ) η(μ^{−1}ν_{0}^{−1}ν_{1}^{−1}, ϖ^{n_{1}}x) η(μ^{−1}ν_{2}, ϖ^{n_{2}}y).

Also if a = ζϖ^{r}

η^{−1}(b) |b^{2}a|^{−1} φ(b^{2}a) = φ(b^{2}ϖ^{r}) ν^{−1}(ζ) η^{−1}(b) |b^{2}|^{−1}|a|^{−1}.

If we put all this information together we get a rather complicated formula for Φ(g) which we have to use to compute Φ′(e). Φ′(e) is expressed as an integral with respect to a, b, x, and y. We will not try to write down the integrand. The integral with respect to a is an integration over ζ followed by a sum over r. The integrand is a sum over μ. The integration over ζ annihilates all but one term, that for which μνν_{0} = 1. We can now attempt to write down the resulting integrand, which has to be integrated over b, x, and y, and summed over r. It is the product of

η^{−1}(b) |b|^{2} ν(−1) z_{0}^{−r−n_{1}} φ(b^{2}ϖ^{r}) C_{r+n_{1}+n_{2}}(ν^{−1}ν_{0}^{−1})

and

η(νν_{1}^{−1}, ϖ^{n_{1}}x) η(νν_{0}ν_{2}^{−1}, ϖ^{n_{2}}y) ψ_{F} (b(x − y)).

The second expression can be integrated with respect to x and y. Lemma 13.1.4 shows that the result is 0 unless |b| = |ϖ|^{n_{1}} = |ϖ|^{n_{2}}. In particular Φ′(e) = 0 if n_{1} ≠ n_{2}. If n_{1} = n_{2} the integration over b need only be taken over ϖ^{n_{1}}U. Then the summation over r disappears and only the term for which r + 2n_{1} = n remains. Apart from positive constants which depend only on the choices of Haar measure Φ′(e) is equal to

z_{0}^{−n}ν_{1}(−1) C_{n}(ν^{−1}ν_{0}^{−1}) ∫_{U} ν_{1}^{−1}ν_{2}^{−1}(ε) dε.

Since

∫_{U} ν_{1}^{−1}ν_{2}^{−1}(ε) dε = δ(ν_{1}ν_{2})

the proof of Lemma 13.1.1 is complete.

Since L(s, π) = L(sπ̃) = 1 if π is absolutely cuspidal the first three assertions of the theorem are, for such π, consequences of the next lemma.

Lemma 13.1.5 Suppose Φ belongs to 𝒮(A), v belongs to V, and ṽ belongs to Ṽ. If π is absolutely cuspidal the integral

∫ Φ(g)⟨π(g)v, ṽ⟩ |detg|^{s+ \frac{1}{2}} d^{×}g

is absolutely convergent for Re s sufficiently large and the functions it defines can be analytically continued to an entire function.

Suppose the integral is convergent for some s. If ξ is an elementary idempotent such that π(ξ)v = v the integral is not changed if Φ is replaced by

Φ_{1}(g) = ∫_{GL(2,O_{F})} Φ(gh^{−1}) ξ(h) dh.

Chapter 3 228 Since π is absolutely cuspidal it does not contain the trivial representation of GL(2, O_{F}) and we can choose ξ to be orthogonal to the constant functions on GL(2, O_{F}). Then Φ_{1}(0) = 0. Thus, when proving the second assertion of the lemma we can suppose that Φ(0) = 0.

The support of ⟨π(g)v, ṽ⟩ is contained in a set Z_{F} C with C compact. Moreover there is an open subgroup K′ of GL(2, O_{F}) such that the functions Φ(g) and ⟨π(g)v, ṽ⟩ are invariant under right translations by the elements of K′. If

C ⊆ U_{i=1}^{p}g_{i}K′

the integral is equal to

∑_{i=1}^{p}⟨π(g_{i})v, ṽ⟩ |detg_{i}|^{s+ \frac{1}{2}} ∫_{F^{×}} Φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g_{i}) η(a) |a|^{2s+1} d^{×}a,

if each of the integrals in this sum converges. They are easily seen to converge if Re s is sufficiently large and if Φ(0) = 0 they converge for all s. The lemma is proved.

Now we verify a special case of the fifth assertion.

Lemma 13.1.6 Suppose φ is in 𝒮(F^{×}) and

Φ(g) = φ(detg)⟨v, π̃(g)ṽ⟩ |detg|^{−1}.

Then for all u in V and all ũ in Ṽ

Ξ̃(1 − s, Φ′, u, ũ) = ε(s, π, ψ) Ξ(s, Φ, u, ũ).

The expression Ξ(s, Φ, u, ũ) is the integral over G_{F} of

|detg|^{s− \frac{1}{2}} φ(detg) ⟨π(g)u, ũ⟩ ⟨v, π̃(g)ṽ⟩.

The integral

∫_{SL(2,F)}⟨π(gh)u, ũ⟩ ⟨v, π̃(gh)ṽ⟩ dh

depends only on detg. Set it equal to F (detg). Then Ξ(s, Φ, u, ũ) is equal to

∫_{F^{×}} φ(a) F (a) |a|^{s− \frac{1}{2}} d^{×}a.

By Lemma 13.1.1

Φ′(g) = φ′(detg) |detg|^{−1} η^{−1}(detg) ⟨π(g)v, ṽ⟩

so that Ξ(s, Φ′, u, ũ) is equal to

∫_{F^{×}} φ′(a) F̃(a) |a|^{s− \frac{1}{2}} η^{−1}(a) d^{×}a

if

F̃(a) ∫_{SL(2,F)}⟨u, π̃(gh)ũ⟩ ⟨π(gh)v, ṽ⟩ dh

whenever a = detg. Since the integrand is not changed when g is replaced by

( \begin{matrix}b & 0 \\ 0 & b \end{matrix} ) g

we have F̃(b^{2}a) = F̃(a) and F̃(a) = F̃(a^{−1}). The same relations are valid for F. Also F̃(a) = F (a^{−1}) so that F = F̃.

We remind ourselves that we are now trying to show that

∫_{F^{×}} φ′(a) F̃(a) η^{−1}(a) |a|^{\frac{1}{2} −s} d^{×}a

is equal to

ε(s, π, ψ) ∫_{F^{×}} φ(a) F (a) |a|^{s− \frac{1}{2}} d^{×}a.

If U′ is an open subgroup of U_{F} such that

π̃ \begin{aligned}( \\ ( & \begin{matrix}ε & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} ũ = ũ

and

π \begin{aligned}( \\ ( & \begin{matrix}ε & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} v = v

for ε in U′ then F and F̃ are constant on cosets of (F^{×})^{2}U′ which is of finite index in F^{×}. Write

F (a) = ∑_{i=1}^{p} c_{i}χ_{i}(a)

where χ_{i} are characters of F^{×}/(F^{×})^{2}U′. We may assume that all c_{i} are different from 0. Then

F (a^{−1}) = ∑_{i=1}^{p} c_{i} χ_{i}(a^{−1}).

The factor ε(s, π ⊗ χ_{i}, ψ) was defined so that

∫_{F^{×}} φ′(a) χ_{i}^{−1} η^{−1}(a) |a|^{\frac{1}{2} −s} d^{×}a

would be equal to

ε(s, χ_{i} ⊗ π, ψ) ∫_{F^{×}} φ(a) χ_{i}(a) |a|^{s− \frac{1}{2}} d^{×}a.

All we need do is show that π and χ_{i} ⊗ π are equivalent, so that

ε(s, χ_{i} ⊗ π, ψ) = ε(s, π, ψ).

A character χ is one of the χ_{i} if and only if χ is trivial on (F^{×})^{2} and

∫_{F^{×}/(F^{×})^{2}} F (a) χ(a) d^{×}a ≠ 0.

This integral is equal to

∫_{G_{F} /Z_{F}} χ(g)⟨π(g)u, ũ⟩ ⟨v, π̃(g)ṽ⟩ dg

which equals

∫_{G_{F} /Z_{F}} ⟨χ ⊗ π(g)u, ũ⟩ ⟨v, π̃(g)ṽ⟩ dg.

The integral does not change if π is replaced by ω ⊗ π. Thus the Schur orthogonality relations imply that it is non-zero only if π and χ ⊗ π are equivalent.

If Φ belongs to 𝒮_{0}(A) the functions Φ(g) |detg|^{s+ \frac{1}{2}} belongs to ℋ_{F} and we can form the operator

T (s, Φ) = ∫_{G_{F}} Φ(g) |detg|^{s+ \frac{1}{2}} π(g) d^{×}g.

If Φ has the form of the previous lemma the functional equation may be written as

T (1 − s, Φ′) = ε(s, π, ψ) T (s, Φ).

Lemma 13.1.7 Given a non-zero w in V, the set of all u in V such that for some Φ of the form

Φ(g) = φ(detg) ⟨v, π̃(g)ṽ⟩ |detg|_{F}^{−1}

the vector T (s, Φ)w is of the form e^{bs}u is a set that spans V.

If the function Φ is of this form so is the function Φ′(g) = Φ(hg) and

T (s, Φ′)w = |deth|^{−(s+ \frac{1}{2} )}π(h^{−1}) T (s, Φ) w

Since π is irreducible we need only show that there is at least one non-zero vector in the set under consideration. Moreover there is an r such that α_{F}^{r} ⊗ π is unitary and we may as well suppose that π itself is unitary. Let (u, v) be a positive invariant form on V.

Choose v = w and ṽ so that ⟨u, ṽ⟩ = (u, w) for all u. Let φ be the characteristic function of U_{F}. Then

Φ(g) = (w, π(g)w)

if |detg| = 1 and is 0 otherwise. If

H = {g ∈ G_{F} | |detg| = 1}

then

T (s, Φ)w = ∫_{H} (w, π(g)w) π(g)w d^{×}g

is independent of s and is non-zero because

(T (s, Φ)w, w) = ∫_{H} |(π(g)w, w)|^{2} d^{×}g.

The fourth assertion follows immediately and the fifth will now be a consequence of the following lemma.

Lemma 13.1.8 Suppose Φ belongs to 𝒮(A) and Ψ belongs to 𝒮_{0}(A). There is a vertical strip in which the integrals

∫∫ Φ(g) Ψ′(h) ⟨π(g)v, π̃(h)ṽ⟩ |detg|^{s+ \frac{1}{2}} |deth|^{\frac{3}{2} −s} d^{×}g d^{×}h

and

∫∫ Φ′(g) Ψ(h) ⟨π^{−1}(g)v, π̃(h^{−1})ṽ⟩ |detg|^{\frac{3}{2} −s} |deth|^{s+ \frac{1}{2}} d^{×}g d^{×}h

exist and are equal.

A little juggling shows that there is no harm in supposing that the quasi-character η defined by

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η(a) I

is a character. Fix v and ṽ. Let C be a compact subset of G_{F} which contains the support of Ψ and Ψ′. The set

{π̃(h)ṽ | h ∈ C}

is finite. Thus there is a compact set in G_{F} such that for any h in C the function

g → ⟨π(g)v, π̃(h)ṽ⟩

has its support in Z_{F} C′. Moreover these functions are uniformly bounded. The first integral is therefore absolutely convergent for Re s > −\frac{1}{2}. The second is convergent for Re s < \frac{3}{2}.

If −\frac{1}{2} < Re s < \frac{3}{2} the first integral is equal to

∫ Ψ′(h) |deth|^{\frac{3}{2} −s}\{ ∫ Φ(g)⟨π(g)v, π̃(h)ṽ⟩ |detg|^{s+ \frac{1}{2}} d^{×}g\} d^{×}h.

Replacing g by hg we obtain

∫ Ψ′(h) |deth|^{2}\{ ∫ Φ(gh)⟨π(g)v, ṽ⟩ |detg|^{s+ \frac{1}{2}} d^{×}g\} d^{×}h.

If we take the additive Haar measure to be dh = |deth|^{2} d^{×}h this may be written as

∫ ⟨π(g)v, ṽ⟩ |detg|^{s+ \frac{1}{2}} \{ ∫ Φ(hg) Ψ′(h) dh\} d^{×}g.

The second integral is

∫ Ψ(h) |deth|^{s+ \frac{1}{2}} \{ ∫ Φ′(g)⟨π^{−1}(g)v, π̃^{−1}(h)ṽ⟩ |detg|^{\frac{3}{2} −s} d^{×}g\} d^{×}h.

After a change of variables this becomes

∫ ⟨π^{−1}(g)v, ṽ⟩ |detg|^{\frac{3}{2} −s}\{ ∫ Φ′(gh) Ψ(h) dh\}d^{×}g.

Replacing g by g^{−1} we obtain

∫ ⟨π(g)v, ṽ⟩ |detg|^{s+ \frac{1}{2}} \{|detg|^{−2} ∫ Φ′(g^{−1}h) Ψ(h) dh\} d^{×}g.

Since

∫ Φ(hg) Ψ′(h) dh

is equal to

|detg|^{−2} ∫ Φ′(g^{−1}h) Ψ(h) dh

the lemma follows.

The theorem is now proved when π is absolutely cuspidal. Suppose that it is a constituent of τ = ρ(μ_{1}, μ_{2}). In this case the field may be archimedean. Although τ is not necessarily irreducible it is admissible and its matrix coefficients are defined. The contragredient representation τ̃ is ρ(μ_{1}^{−1}, μ_{2}^{−1}) and the space of τ is ℬ(μ_{1}, μ_{2}) while that of τ̃ is ℬ(μ_{1}^{−1}, μ_{2}^{−1}). If f belongs to ℬ(μ_{1}, μ_{2}) and f̃ belongs to ℬ(μ_{1}^{−1}, μ_{2}^{−1}) then

⟨τ(g)f, f̃⟩ = ∫_{K} f(kg) f̃(k) dk

and

⟨f, τ̃(g)f̃⟩ = ∫_{K} f(k) f̃(kg) dk

if K is the standard maximal compact subgroup of G_{F}.

If we set

\begin{aligned}L(s, τ) & = L(s, μ_{1}) L(s, μ_{2}) \\ L(s, τ̃) & = L(s, μ_{1}^{−1}) L(s, μ_{2}^{−1}) \end{aligned}

and

ε(s, τ, ψ) = ε(s, μ_{1}, ψ) ε(s, μ_{2}, ψ)

the theorem may be formulated for the representation τ. We prove it first for τ and then for the irreducible constituents of τ.

We use a method of R. Godement. If Φ belongs to 𝒮(A) then for brevity the function x → Φ(gxh) which also belongs to 𝒮(A) will be denoted by hΦg. Also let

φ_{Φ}(a_{1}, a_{2}) = ∫_{F} Φ \begin{aligned}( \\ ( & \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} & ) \\ & & ) \end{aligned} dx

where dx is the measure which is self dual with respect to ψ. φ_{Φ} belongs to 𝒮(F^{2}). The map Φ → φ_{Φ} of 𝒮(A) into 𝒮(F^{2}) is certainly continuous.

We are now going to define a kernel K_{Φ}(h, g, s) on K × K. We set

K_{Φ}(e, e, s) = Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ_{Φ}).

Recall that the right-hand side is

∫∫ φ_{Φ}(a_{1}, a_{2}) μ_{1}(a_{1}) |a_{1}|^{s} μ_{2}(a_{2}) |a_{2}|^{s} d^{×}a_{1} d^{×}a_{2}.

In general

K_{Φ}(h, g, s) = K_{gΦh^{−1}}(e, e, s).

We also set

K̃_{Φ}(e, e, s) = Z(μ_{1}^{−1}α_{F}^{s} , μ_{2}^{−1}α_{F}^{s} , φ_{Φ})

and

K̃_{Φ}(h, g, s) = K̃_{gΦh^{−1}}(e, e, s).

The kernels are defined for Re s sufficiently large and are continuous in h, g, and s and, for fixed h and g, holomorphic in s.

We now make some formal computations which will be justified by the result. The expression Z(α_{F}^{s+ \frac{1}{2}} ⊗ τ, Φ, f, f̃) is equal to

∫_{G_{F}} Φ(g)\{ ∫_{K} f(kg) f̃(k) dk\} |detg|^{s+ \frac{1}{2}} d^{×}g

which is

∫_{K} f̃(k)\{ ∫_{G_{F}} Φ(g) f(kg) |detg|^{s+ \frac{1}{2}} d^{×}g\} dk.

Changing variables in the inner integral we obtain

∫_{K} f̃(k)\{ ∫_{G_{F}} Φ(k^{−1}g) f(g) |deth|^{s+ \frac{1}{2}} d^{×}g\} dk.

Using the Iwasawa decomposition to evaluate the integral over G_{F} we see that this is equal to

∫_{K×K} K_{Φ}(k_{1}, k_{2}, s) f(k_{2}) f̃(k_{1}) dk_{1} dk_{2}.

Since we could have put in absolute values and obtained a similar result all the integals are convergent and equal for Re s sufficiently large. A similar computation shows that

Z(α_{F}^{s+ \frac{1}{2}} ⊗ τ̃, Φ, f, f̃)

is equal to

∫_{K×K} K̃_{Φ}(k_{1}, k_{2}, s) f(k_{1}) f̃(k_{2}) dk_{1} dk_{2}

if Re s is large enough.

If ξ is an elementary idempotent such that τ(ξ)f = f and τ̃(ξ)f̃ = f̃ then Z(α_{F}^{s+ \frac{1}{2}} ⊗ τ, Φ, f, f̃) is not changed if Φ is replaced by

Φ_{1}(g) = ∫∫ Φ(k_{1}gk_{2}^{−1}) ξ(k_{1}) ξ(k_{2}) dk_{1} dk_{2}.

Thus, at least when proving the second and third assertions, we may suppose that Φ is K-finite on both sides and, in fact, transforms according to a fixed finite set of irreducible representations of K. Then, as s varies, the functions

K_{Φ}(k_{1}, k_{2}, s)

stay in some fixed finite dimensional space U of continuous functions on K × K. The map

F → ∫∫ F (k_{1}, k_{2}) f(k_{2}) f̃(k_{1}) dk_{1} dk_{2}

is a linear form on this space and we can find g_{1}, ⋯, g_{n} and h_{1}, ⋯, h_{n} in K such that it can be represented in the form

F → ∑_{i=1}^{n} λ_{i} F (g_{i}, h_{i}).

Thus

Z(α_{F}^{s+ \frac{1}{2}} ⊗ τ, Φ, f, f̃) = ∑ λ_{i} K_{Φ}(g_{i}, h_{i}, s).

Thus to prove the second and third assertions we need only show that for each g and h in K the function

\frac{K_{Φ}(g, h, s)}{L(s, τ)}

is entire and K_{Φ}(g, h, s) itself is bounded at infinity in vertical strips. There is certainly no harm in supposing that g = h = e so that

K_{Φ}(e, e, s) = Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ_{Φ})

Thus the desired facts are consequences of the results obtained in paragraphs 3, 5, and 6 when proving the local functional equation for constituents of τ. Replacing τ by its contragredient representation we obtain the same results for Z(α_{F}^{s+ \frac{1}{2}} ⊗ τ̃, Φ, f, f̃).

To prove the functional equation we have to see what happens to the Fourier transform when we pass from the function Φ to to Φ_{1}. The answer is simple:

Φ′_{1}(g) = ∫∫ Φ′(k_{1}gk_{2}^{−1}) ξ(k_{1}) ξ(k_{2}) dk_{1} dk_{2}.

Thus in proving the functional equation we may suppose that Φ is K-finite on both sides. We may also suppose that if F (k_{1}, k_{2}) is in U so is F′(k_{1}, k_{2}) = F (k_{2}, k_{1}). Then Z(α_{F}^{s+ \frac{1}{2}} ⊗ τ, Φ′, f, f̃) = ∑ λ_{i} K̃_{Φ′}(h_{i}, g_{i}, s). To prove the functional equation we have to show that

\frac{K̃_{Φ′}(h, g, 1 − s)}{L(1 − s, τ)} = ε(s, τ, ψ) \frac{K_{Φ}(g, h, s)}{L(s, τ)}

for any h and g in K. Since the Fourier transform of gΦh^{−1} is hΦ′g it will be enough to do this for h = g = e. Then the equality reduces to

\frac{Z(μ_{1}^{−1}α_{F}^{1−s}, μ_{2}^{−1}α_{F}^{1−s}, φ_{Φ′})}{L(1 − s, τ̃)} = ε(s, τ, ψ) \frac{Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ_{Φ})}{L(s, τ)}

and is a result of the facts proved in the first chapter and the next lemma.

Chapter 3 235 Lemma 13.2.1 The Fourier transform of the function φ_{Φ} is the function φ_{Φ′}.

The value of Φ′ at

( \begin{matrix}α & β \\ γ & δ \end{matrix} )

is

∫ Φ \begin{aligned}( \\ ( & \begin{matrix}x & y \\ z & t \end{matrix} & ) \\ & & ) \end{aligned} ψ(αx + βz + γy + δt) dx dy dz dt

if dx, dy, dz, and dt are self-dual with respect to ψ. Thus φ_{Φ′}(α, δ) is equal to

∫ \{ ∫ Φ \begin{aligned}( \\ ( & \begin{matrix}x & y \\ z & t \end{matrix} & ) \\ & & ) \end{aligned} ψ(αx + δt) ψ(βz) dx dy dz dt\} dβ

Applying the Fourier inversion formula to the pair of variables β and z we see that this is equal to

∫ Φ \begin{aligned}( \\ ( & \begin{matrix}x & y \\ 0 & t \end{matrix} & ) \\ & & ) \end{aligned} ψ(αx + δt) dx dy dt

which is the value of the Fourier transform of φ_{Φ} at (α, δ).

The theorem, with the exception of the fourth assertion, is now proved for the representation τ. We will now deduce it, with the exception of the fourth assertion, for the constituents of τ. We will return to the fourth assertion later.

If π is a constituent of τ either π = π(μ_{1}, μ_{2}) or π = σ(μ_{1}, μ_{2}). In the first case there is nothing left to prove. In the second only the third assertion remains in doubt. If F is the complex field even it is alright because we can always find another pair of quasi-characters μ′_{1} and μ′_{2} such that π = π(μ′_{1}, μ′_{2}). We ignore this case and suppose that F is real or non-archimedean.

First take F to be non-archimedean. We may suppose that μ_{1} and μ_{2} are the form μ_{1} = χαF^{\frac{1}{2}}

and μ_{2} = χα_{F}^{− \frac{1}{2}} . The one-dimensional representation g → χ(detg) is contained in τ̃ = ρ(μ_{1}^{−1}, μ_{2}^{−1}) and acts on the function g → χ(detg). The matrix elements for π are the functions

g → ⟨τ(g)f, f̃⟩ = ⟨π(g)f, f̃⟩

where f̃ belongs to ℬ(μ_{1}^{−1}, μ_{2}^{−1}) and

∫_{K} f(k) χ(detk) dk = 0.

For such an f there is an elementary idempotent ξ such that τ(ξ)f = f while

∫_{K} ξ(k) dk = 0

The value of Z(α_{F}^{s+ \frac{1}{2}} ⊗ π, Φ, f, f̃) is not changed if we replace Φ by

Φ_{1}(g) = ∫_{K} Φ(gh^{−1}) ξ(h) dh.

Lemma 13.2.2 If g_{1} and g_{2} belong to G_{F} then

∫∫ Φ_{1} (g_{1} ( \begin{matrix}x & y \\ 0 & 0 \end{matrix} ) g_{2}) dx dy = 0.

It will be enough to prove this when g_{1} is the identity. Let

φ(x, y) = Φ_{1} \begin{aligned}( \\ ( & \begin{matrix}x & y \\ 0 & 0 \end{matrix} & ) \\ & & ) \end{aligned} .

If g_{1} is the identity then, after a change of variables, the integral becomes

|detg_{2}|^{−1}∫∫ φ(x, y) dx dy

so that we can also assume g_{2} is the identity. Then the integral equals

∫_{K} \{∫∫ Φ (( \begin{matrix}x & y \\ 0 & 0 \end{matrix} ) k) dx dy\} ξ(k^{−1}) dk.

Changing variables as before we see that the inner integral does not depend on K. Since

∫_{K} ξ(k^{−1}) dk = 0

the lemma follows.

To establish the third assertion for the representation π all we need do is show that for any g and h in K the function

\frac{K_{Φ}(g, h, s)}{L(s, π)}

is entire provided

∫∫ Φ (g_{1} ( \begin{matrix}x & y \\ 0 & 0 \end{matrix} ) g_{2}) dx dy = 0

for all g_{1} and g_{2} in G_{F}. As usual we need only consider the case that g = h = e. Since

∫ φ_{Φ}(x, 0) dx = 0

and

K_{Φ}(e, e, s) = Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ_{Φ})

we need only refer to Corollary 3.7.

If F is the field of real numbers the proof is going to be basically the same but a little more complicated. We may assume that μ_{1}μ_{2}^{−1}(x) = |x|^{2p+1−m}(sgn x)^{m}, where p is a non-negative integer and m is 0 or 1, and that π acts on ℬ_{S}(μ_{1}, μ_{2}). The restriction of π to SO(2, ℝ) contains only those representations κ_{n} for which n ≡ 1 − m (mod 2) and |n| ≥ 2p + 1 − m. Let ξ_{n} be the elementary idempotent corresponding to the representation κ_{n}. As before we may suppose that

∫_{SO(2,ℝ)} Φ(xk^{−1}) ξ_{n}(k) dk = 0  (13.2.3)

if κ_{n} does not occur in the restriction of π to SO(2, ℝ).

Lemma 13.2.4 If Φ satisfies (13.2.3), if g_{1} and g_{2} belong to G_{F}, and φ = φ_{g_{1}Φg_{2}} then

∫_{ℝ} x^{i} \frac{∂^{j}}{∂y^{j}} φ(x, 0) dx = 0

if i ≥ 0, j ≥ 0 and i + j = 2p − m.

We may assume that g_{2} = e. If φ = φ_{Φ} let

L(Φ) = ∫_{ℝ} x^{i} \frac{∂^{j}}{∂y^{j}} φ(x, 0) dx

and let

F (g) = L(gΦ).

We have to show that, under the hypothesis of the lemma, F (g) = 0 for all g. However F is defined for all Φ in 𝒮(A) and if Φ is replaced by hΦ the function F is replaced by F (gh). Thus to establish the identity

F (( \begin{matrix}a_{i} & z \\ 0 & a_{2} \end{matrix} ) g) = η_{1}(a_{2}) η_{2}(a_{2}) F (g)

where η_{1}(a_{1}) = a_{1}^{−i} |a_{1}|^{−1} and η_{2}(a_{2}) = a_{2}^{j} |a_{2}|^{−1}, we need only establish it for g = e.

Let

h = ( \begin{matrix}a_{1} & z \\ 0 & a_{2} \end{matrix} ) .

Then

hΦ \begin{aligned}( \\ ( & \begin{matrix}x & u \\ 0 & y \end{matrix} & ) \\ & & ) \end{aligned} = Φ \begin{aligned}( \\ ( & \begin{matrix}a_{1}y & xz + a_{2}u \\ 0 & a_{2}y \end{matrix} & ) \\ & & ) \end{aligned} .

If φ = φ_{Φ} and φ_{1} = φ_{hΦ} then φ_{1}(x, y), which is given by

∫ Φ \begin{aligned}( \\ ( & \begin{matrix}a_{1}x & xz + a_{2}u \\ 0 & a_{2}y \end{matrix} & ) \\ & & ) \end{aligned} du,

is equal to

|a_{2}|^{−1} ∫ Φ \begin{aligned}( \\ ( & \begin{matrix}a_{1}x & u \\ 0 & a_{2}y \end{matrix} & ) \\ & & ) \end{aligned} du = |a_{2}|^{−1} φ(a_{1}x, a_{2}y).

Moreover F (h) is equal to

∫ x^{i} \frac{∂^{j}φ_{1}}{∂y^{j}} (x, 0) dx

which equals

a_{1}^{−i} |a_{1}|^{−1} a_{2}^{j} |a_{2}|^{−1} ∫ x^{i} \frac{∂^{j}φ}{∂y^{j}}(x, 0) dx

as required.

Finally if

g = ( \begin{matrix}α & β \\ γ & δ \end{matrix} )

and φ = φ_{gΦ} then F (g) is equal to

∫ x^{i} \frac{∂^{j}φ}{∂y^{j}}(x, 0) dv

and

φ(x, y) = ∫ Φ \begin{aligned}( \\ ( & \begin{matrix}αx + γu & βx + δu \\ yγ & yδ \end{matrix} & ) \\ & & ) \end{aligned} du.

Since we can interchange the orders of differentiation and integration,

\frac{∂^{j}φ}{∂y^{j}}(x, 0) = ∑_{n=0}^{j} λ_{n}γ^{n}δ^{j−n} ∫ φ_{n}(αx + γu, βx + δu) du

where

φ_{n}(x, y) = \frac{∂^{j}Φ}{∂γ^{n} ∂δ^{j−n}} \begin{aligned}( \\ ( & \begin{matrix}x & y \\ 0 & 0 \end{matrix} & ) \\ & & ) \end{aligned}

and the number λ_{n} are constants. Thus F (g) is a linear combination of the functions

γ^{n}δ^{j−n}∫∫ x^{i} φ_{n}(αx + γu, βx + δu) dx du.

If α ≠ 0 we may substitute x − \frac{γu}{2} for x to obtain

γ^{n}δ^{j−n}∫∫ \frac{(x − γu)^{i}}{2} φ_{n}(αx, βx + \frac{Δu}{2}) dx du

where Δ = detg. Substituting u − \frac{αβ}{Δ} x for u we obtain

γ^{n}δ^{j−n}∫∫ (x + \frac{βγ}{Δ} x − \frac{γu}{2})^{i} φ_{n}(αx, \frac{Δu}{2}) dx du.

After one more change of variables this becomes

Δ^{−i} |Δ|^{−1} γ^{n}δ^{j−n}∫∫ (δx − γu)^{i} φ_{n}(x, u) dx du.

In conclusion F (g) is a function of the form

F \begin{aligned}( \\ ( & \begin{matrix}α & β \\ γ & δ \end{matrix} & ) \\ & & ) \end{aligned} = Δ^{−i} |Δ|^{−1} P (α, β, γ, δ)

where P is a polynomial.

Thus the right translates of F by the elements of G_{F} span a finite-dimensional space. In particular it is O(2, ℝ) finite and if η_{1} = μ′_{1}αF^{\frac{1}{2}} while η_{2} = μ′_{2}α_{F}^{− \frac{1}{2}} it lies in a finite-dimensional invariant subspace of ℬ(μ′_{1}, μ′_{2}). Thus it lies in ℬ_{F} (μ′_{1}, μ′_{2}). Since μ′_{1}μ′_{2}^{−1} = μ_{1}^{−1}μ_{2} no representation of SO(2, ℝ) occurring in π(μ′_{1}, μ′_{2}) can occur in π = σ(μ_{1}, μ_{2}). If F is not zero then for at least one such representation κ_{n}

F_{1}(g) = ∫_{SO(2,ℝ)} f(gk^{−1}) ξ_{n}(k) dk

is not identically 0. But F_{1} is the result of replacing Φ by

Φ_{1}(x) = ∫_{SO(2,ℝ)} Φ(xk^{−1}) ξ_{n}(k) dk

in the definition of F. In particular if Φ satisfies the conditions of the lemma both Φ_{1} and F_{1} are zero. Therefore F is also zero and the lemma is proved.

The third assertion can now be verified as in the non-archimedean case by appealing to Lemma 5.17. The fourth has still to be proved in general.

If F is the real field let 𝒮_{1}(A) be the space of functions of the form

Φ \begin{aligned}( \\ ( & \begin{matrix}a & b \\ c & d \end{matrix} & ) \\ & & ) \end{aligned} = exp ( − π(a^{2} + b^{2} + c^{2} + d^{2})) P (a, b, c, d)

where P is a polynomial. If F is the complex field 𝒮_{1}(A) will be the space of functions of the form

Φ \begin{aligned}( \\ ( & \begin{matrix}a & b \\ c & d \end{matrix} & ) \\ & & ) \end{aligned} = exp ( − π(aā + bb̄ + cc̄ + dd̄)) P (a, ā, b, b̄, c, c̄, d, d̄)

where P is again a polynomial. If F is non-archimedean 𝒮_{1}(A) will just be 𝒮(A). The space 𝒮_{1}(F^{2}) is defined in a similar manner.

Lemma 13.2.5 Suppose φ belongs to 𝒮_{1}(F^{2}). Then there is a Φ in 𝒮_{1}(A) such that

K_{Φ}(e, e, s) = Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ)

and f_{1}, ⋯, f_{n} in ℬ(μ_{1}, μ_{2}) together with f̃_{1}, ⋯, f̃_{n} in ℬ(μ_{1}^{−1}, μ_{2}^{−1}) such that

∑_{i=1}^{n} ∫_{K×K} K_{Φ}(h, g, s) f_{i}(g) f̃_{i}(h) dg dh = K_{Φ}(e, e, s).

Since there is a φ in 𝒮_{1}(F^{2}) such that

Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ) = ae^{bs} L(s, τ)

this lemma will imply the fourth assertion for the representation τ.

Given φ the existence of Φ such that φ = φ_{Φ} and therefore

K_{Φ}(e, e, s) = Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ_{Φ})

is a triviality and we worry only about the existence of f_{1}, ⋯, f_{n} and f̃_{1}, ⋯, f̃_{n}.

It is easily seen that if

( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} )

and

( \begin{matrix}b_{1} & y \\ 0 & b_{2} \end{matrix} )

Chapter 3 240 belong to K then

K_{Φ} (( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} ) h, ( \begin{matrix}b_{1} & y \\ 0 & b_{2} \end{matrix} ) g, s)

is equal to

μ_{1}(a_{1}) μ_{2}(a_{2}) μ_{1}^{−1}(b_{1}) μ_{2}^{−1}(b_{2}) K_{Φ}(h, g).

Also

K_{Φ}(hh_{1}, gg_{1}, s) = K_{g_{1}Φh_{1}^{−1}}(h, g, s).

Since Φ belongs to 𝒮_{1}(A) it is K-finite on the left and right. Thus there is a finite set S of irreducible representations of K such that if U_{1} is the space of functions F on K which satisfy

F (( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} ) h) = μ_{1}(a_{1}) μ_{2}(a_{2}) F (h)

for all

( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} )

in K and can be written as a linear combination of matrix elements of representations in S and U_{2} is the space of functions F′ on K which satisfy

F′ (( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} ) h) = μ_{1}^{−1}(a_{1}) μ_{2}^{−1}(a_{2}) F′(h)

and can be written as a linear combination of matrix elements of representations in S then, for every s, the function

(h, g) → K_{Φ}(h, g, s)

belongs to the finite dimensional space U spanned by functions of the form (h, g) → F (h) F′(g) with F in U_{1} and F′ in U_{2}.

Choose F_{1}, ⋯, F_{n} and F′_{1}, ⋯, F′_{n} so that for every function F in U

F (e, e) = ∑_{i=1}^{n} λ_{i} ∫_{K×K} F (h, g) F̄_{i}(h) F̄′_{i}(g) dh dg.

Since F̄_{i} is the restriction to K of an element of ℬ(μ_{1}^{−1}, μ_{2}^{−1}) while F̄′_{i} is the restriction to K of an element of ℬ(μ_{1}, μ_{2}) the lemma follows.

Unfortunately this lemma does not prove the fourth assertion in all cases. Moreover there is a supplementary condition to be verified.

Chapter 3 241 Lemma 13.2.6 Suppose F is non-archimedean and π is of the form π = π(μ_{1}, μ_{2}) with μ_{1} and μ_{2} unramified. Suppose Φ is the characteristic function of M(2, O_{F}) in M(2, F). If v and ṽ are invariant under K = GL(2, O_{F}) and if

∫_{K} d^{×}g = 1

then

Z(α_{F}^{s+ \frac{1}{2}} ⊗ π, Φ, v, ṽ) = L(s, π) ⟨v, ṽ⟩.

Suppose f belongs to ℬ(μ_{1}, μ_{2}) and is identically 1 on K while f̃ belongs to ℬ(μ_{1}^{−1}, μ_{2}^{−1}) and is identically 1 on K. Then

⟨f, f̃⟩ = ∫_{K} f(k) f̃(k) dk = 1

and if τ = ρ(μ_{1}, μ_{2}) we are trying to show that

Z(α_{F}^{s+ \frac{1}{2}} ⊗ τ, Φ, f, f̃) = L(s, τ).

The left side is equal to

∫_{K×K} K_{Φ}(h, g, s) f(h) f̃(g) dh dg.

Since Φ is invariant on both side under K this is equal to

K_{Φ}(e, e, s) = Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ)

if

φ(x, y) = ∫ Φ \begin{aligned}( \\ ( & \begin{matrix}x & z \\ 0 & y \end{matrix} & ) \\ & & ) \end{aligned} dz.

Since we have so normalized the Haar measure on G_{F} that

∫_{G_{F}} F (g) dg = ∫_{K} \{ ∫ F (( \begin{matrix}a_{1} & 0 \\ 0 & a_{2} \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) k) d^{×}a_{1} d^{×}a_{2} dx\} dk

where dk is the normalized measure on K, dx is the measure on F which assigns the measure 1 to O_{F}, and d^{×}a is the measure on F^{×} which assigns the measure 1 to U_{F} the function φ is the characteristic function of O_{F} × O_{F} and

Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ) = L(s, μ_{1}) L(s, μ_{2})

as required.

This lemma incidentally proves the fourth assertion for the one-dimensional representation g → χ(detg) if χ is unramified. If χ is ramified and π corresponds to χ then π = π(μ_{1}, μ_{2}) if μ_{1}(a) = χ(a) |a|^{\frac{1}{2}} and μ_{2}(a) = χ(a) |a|^{− \frac{1}{2}} . Thus L(s, π) = 1. If Φ is the restriction of the function χ^{−1} to K then

Z(π, Φ, v, ṽ) = ⟨v, ṽ⟩ ∫_{K} d^{×}g

and the fourth assertion is verified in this case.

Take μ_{1} and μ_{2} of this form with χ possibly unramified and suppose that π = σ(μ_{1}, μ_{2}). Suppose first that χ is unramified. Let φ_{0} be the characteristic function of O_{F} in F and let

φ_{1}(x) = φ_{0}(x) − |ϖ^{−1}| φ_{0}(ϖ^{−1}x).

It has O_{F} for support. Set

Φ \begin{aligned}( \\ ( & \begin{matrix}a & b \\ c & d \end{matrix} & ) \\ & & ) \end{aligned} = φ_{1}(a) φ_{0}(b) φ_{0}(c) φ_{0}(d).

It has M(2, O_{F}) for support and depends only on the residues of a, b, c, and d modulo 𝔭_{F}. If

K^{1} = {k ∈ K | k ≡ e(mod 𝔭)}

then K_{Φ}(h, g, s) depends only on the cosets of h and g modulo K^{1}. Also

K_{Φ} (e, w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) , s) = 0

if x is in O_{F}. To see this we observe first that if

Φ_{1}(g) = Φ (gw ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))

then φ_{Φ_{1}}(a_{1}, a_{2}) is equal to

∫_{F} Φ \begin{aligned}( \\ ( & \begin{matrix}−y & a_{1} − xy \\ −a_{2} & −a_{2}x \end{matrix} & ) \\ & & ) \end{aligned} dy

which equals

φ_{0}(a_{2}) φ_{0}(a_{2}x) ∫_{O_{F}} φ_{1}(y) φ_{0}(a_{1} − xy) dy.

Since x is in O_{F} the function φ_{0}(a_{1} −xy) equals φ_{0}(a_{1}) for y in O_{F} and this expression is 0 because

∫_{O_{F}} φ_{1}(y) dy = 0.

We choose f in ℬ_{S}(μ_{1}, μ_{2}) so that f(gk) = f(g) if k belongs to K_{1}, f(e) = 1, and

f(e) + ∑_{x∈O_{F} /𝔭} f (w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) = 0.

We choose f̃ in ℬ(μ_{1}, μ_{2}) so that f̃(gk) = f̃(g) if k belongs to K^{1}, f̃(e) = 1, and

f̃(w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) = 0

if x belongs to O_{F}. Then

∫_{K×K} K_{Φ}(h, g, s) f̃(h) f(g) dh dg

is equal to

∫_{K} K_{Φ}(e, g, s) f(g) dg = K_{Φ}(e, e, s)

which equals

Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ_{Φ}).

Moreover

φ_{Φ}(a_{1}, a_{2}) = φ_{1}(a) φ_{0}(a_{2})

so that, as we saw when proving Corollary 3.7, L(s, π) is a constant times Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ_{Φ}). If χ is ramified L(s, π) = 1. If Φ has support in K then Z(α_{F}^{s+1/2} ⊗ π, Φ, v, ṽ) is equal to

∫_{K} Φ(k)⟨π(k)v, ṽ⟩ dk

and we can certainly choose v, ṽ and Φ so that this is not 0.

We are not yet finished. We have yet to take care of the representations not covered by Lemma 13.2.5 when the field is archimedean. If F is the complex field we have only the finite dimensional representations to consider. There is a pair of characters μ_{1} and μ_{2} such that π is realized on the subspace ℬ_{f}(μ_{1}, μ_{2}) of ℬ(μ_{1}, μ_{2}). There will be positive integers p and q such that μ_{1}μ_{2}^{−1}(z) = z^{−p}z̄^{−q}. The representations σ = ρ_{|q−p|} of SU(2, ℂ) which is of degree |q − p| + 1 is contained in the restriction of π to SU(2, ℂ). In particular ℬ_{f}(μ_{1}, μ_{2}) contains all functions f in ℬ(μ_{1}, μ_{2}) whose restrictions to SU(2, ℂ) satisfy

f (( \begin{matrix}a_{1} & 0 \\ 0 & a_{2} \end{matrix} ) k) = μ_{1}(a_{1}) μ_{2}(a_{2}) f(k)

and transform on the right according to σ.

We are going to use an argument like that used to prove Lemma 13.2.5. Suppose we can find a function Φ in 𝒮_{1}(A) such that

Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ_{Φ})

differs from L(s, π) by an exponential factor and such that Φ transforms on the right under SU(2, ℂ) according to the representation σ. Then K_{Φ}(h, g, s) will satisfy the same conditions as in Lemma 13.2.5. Moreover the functions F′ in the space we called U_{2} can be supposed to transform on the right under SU(2, ℂ) according to σ so that the functions F̄′_{i} will correspond to functions f_{i} in ℬ_{f}(μ_{1}, μ_{2}). Then

∫_{K×K} K_{Φ}(h, g, s) f̃_{i}(h) f_{i}(g) dh dg = Z(α_{F}^{s+ \frac{1}{2}} ⊗ τ, Φ, f_{i}, f̃_{i})

is equal to

Z(α_{F}^{s+ \frac{1}{2}} ⊗ π, Φ, v_{i}, ṽ_{i})

if v_{i} = f_{i} and ṽ_{i} is the restriction of f̃_{i}, regarded as a linear functional, to ℬ_{f}(μ_{1}, μ_{2}).

There are four possible ways of writing μ_{1} and μ_{2}.

(i) μ_{1}(z) = z^{m_{1}}(zz̄)^{s_{1}}, μ_{2}(z) = z^{m_{2}}(zz̄)^{s_{2}}, m_{1} − m_{2} = q − p. (ii) μ_{1}(z) = z^{m_{1}}(zz̄)^{s_{1}}, μ_{2}(z) = z̄^{m_{2}}(zz̄)^{s_{2}}, m_{1} + m_{2} = q − p. (iii) μ_{1}(z) = z̄^{m_{1}}(zz̄)^{s_{1}}, μ_{2}(z) = z^{m_{2}}(zz̄)^{s_{2}}, −m_{1} − m_{2} = q − p.

Chapter 3 244 (iv) μ_{1}(z) = z̄^{m_{1}}(zz̄)^{s_{1}}, μ_{2}(z) = z̄^{m_{2}}(zz̄)^{s_{2}}, m_{2} − m_{1} = q − p.

In all four cases m_{1} and m_{2} are to be non-negative integers. Φ is the product of exp(−π(aā + bb̄ + cc̄ + dd̄)) and a polynomial. We write down the polynomial in all four cases and leave the verifications to the reader. (i.a) m_{1} ≥ m_{2} : ā^{m_{1}−m_{2}}(ād̄− b̄c̄)^{m_{2}}. (i.b) m_{1} ≤ m_{2} : (ād̄− b̄c̄)^{m_{1}} d̄^{n_{2}−m_{1}}. (ii) ā^{m_{1}} d^{m_{2}}. (iii) a^{m_{1}} d̄^{m_{2}}. (iv.a) m_{1} ≥ m_{2} : a^{m_{1}−m_{2}}(ad − bc)^{m_{2}}. (iv.b) m_{2} ≥ m_{1} : (ad − bc)^{m_{1}} d^{m_{2}−m_{1}}.

For the real field the situation is similar. Suppose first that π = π(μ_{1}, μ_{2}) is finite dimensional. If μ_{1}μ_{2}(−1) = 1 then π contains the trivial representation of SO(2, ℝ) and if μ_{1}μ_{2}(−1) = 1 it contains the representation

κ : ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) → e^{iθ}.

We list the four possibilities for μ_{1} and μ_{2} and the polynomial P by which exp (−π(a^{2}+b^{2}+c^{2}+d^{2})) is to be multiplied to obtain Φ.

(i) μ_{1}(−1) = μ_{2}(−1) = 1: P (a, b, c, d) = 1. (ii) μ_{1}(−1) = μ_{2}(−1) = 1: P (a, b, c, d) = ad − bc. (iii) μ_{1}(−1) = 1, μ_{2}(−1) = −1: P (a, b, c, d) = c − id. (iv) μ_{1}(−1), μ_{2}(−1) = 1: P (a, b, c, d) = a − ib.

Only the special representations remain to be considered. We may suppose that π = σ(μ_{1}, μ_{2}) where μ_{1} and μ_{2} are of the form μ_{1}(x) = |x|^{r+ \frac{q}{2}} and μ_{2}(x) = |x|^{r− \frac{q}{2}} (sgn t)^{m} with q = 2p + 1 − m and with p a non-negative integer. Moreover m is 0 or 1. L(s, π) differs from

Γ(\frac{s + r + \frac{q}{2}}{2} ) Γ(\frac{s + r + \frac{q}{2} + 1}{2})

by an exponential as does

Z(μ_{1}α_{F}^{s} , μ_{2}α_{F}^{s} , φ)

if

φ(a_{1}, a_{2}) = e^{−π(a_{1}^{2}+a_{2}^{2})} a_{2}^{q+1}.

Since the representation of κ_{q+1} occurs in the restriction of π to SO(2, ℝ) we may take

Φ \begin{aligned}( \\ ( & \begin{matrix}a & b \\ c & d \end{matrix} & ) \\ & & ) \end{aligned} = exp ( − π(a^{2} + b^{2} + c^{2} + d^{2})) (c + id)^{q+1}.

Chapter 3 245 §14. Automorphic forms and quaternion algebras. Let F be a global field and let M′ be a quaternion algebra over F. The multiplicative group G′ of M′ may be regarded as an algebraic group over F. In the ninth paragraph we have introduced the group G′_{𝔸} and the Hecke algebra ℋ′. A continuous function φ on G′_{F} \ G′_{𝔸} is said to be an automorphic form if for every elementary idempotent ξ in ℋ′ the space

{ρ(ξf) φ | f ∈ ℋ′}

is finite dimensional.

If φ is an automorphic form it is Z′_{𝔸} finite on the left if Z′ is the centre of G′. Let 𝒜′ be the space of automorphic forms on G′_{𝔸} and if η is a quasi-character of F^{×} \ I let 𝒜′(η) be the space of φ in 𝒜′ for which φ′(ag) = η(a) φ′(g) for all a in Z′_{𝔸} which, for convenience, we identify with I. The first assertion of the following lemma is easily proved by the methods of the eighth paragraph. The second is proved by the methods of the tenth. The proof is however a little simpler because G′_{F} Z′_{𝔸} \ G′_{𝔸} is compact. Since, at least in the case of number fields, the proof ultimately rests on general facts from the theory of automorphic forms nothing is gained by going into details.

Lemma 14.1 (i) If an irreducible admissible representation π of ℋ′ is a constituent of 𝒜′ then for some η it is a constituent of 𝒜′(η). (ii) The space 𝒜′(η) is the direct sum of subspaces irreducible and invariant under ℋ′. The rep-

resentation of ℋ′ on each of these subspaces is admissible and no representation occurs more than a finite number of times in 𝒜′(η).

Now we have to remind ourselves of some facts whose proofs are scattered throughout the previous paragraphs. Suppose π = ⊗_{v}π_{v} is an irreducible admissible representation of ℋ′. For each v the representation π_{v} of ℋ′_{v} is irreducible and admissible. Suppose ψ is a non-trivial additive character of F \A and ψ_{v} is its restriction to F_{v}. We have defined L(s, π_{v}), L(s, π̃_{v}), and ε(s, π_{v}, ψ_{v}). If u_{v} is in the space of π_{v} and ũ_{v} in the space of π̃_{v} we have set

Z(α_{F}^{s+ \frac{1}{2}} ⊗ π_{v}, Φ, u_{v}, ũ_{v})

equal to

∫_{G′_{Fv}} Φ(g) ⟨π_{v}(g)u_{v}, ũ_{v}⟩ |ν(g)|^{s+ \frac{1}{2}} d^{×}g.

We know that

h_{v} ε(s, π_{v}, ψ_{v}) \{\frac{Z(α_{F}^{s+ \frac{1}{2}} ⊗ π_{v}, Φ, u_{v}, ũ_{v})}{L(s, π_{v})}\}

is entire and equals

\frac{Z(α_{F}^{\frac{1}{2} −s} ⊗ π̃_{v}, Φ′, u_{v}, ũ_{v})}{L(1 − s, π̃_{v})}.

The factor h_{v} is 1 of G′_{F_{v}} is isomorphic to GL(2, F_{v}) and is −1 otherwise. The case that G′_{F_{v}} is isomorphic to GL(2, F_{v}) was treated in the previous paragraph. The other cases were treated in the fourth and fifth paragraphs.

Chapter 3 246 Theorem 14.2 Suppose π is a constituent of the space of automorphic forms on G′_{𝔸}. The infinite products

∏_{v} L(s, π_{v})

and

∏_{v} L(s, π̃_{v})

are absolutely convergent for Re s sufficiently large. The functions L(s, π) and L(s, π̃) defined by them can be analytically continued to the whole complex plane as meromorphic functions. If F is a number field they will have only a finite number of poles and will be bounded at infinity in vertical strips of finite width. If

ε(s, π) = ∏_{v} ε(s, π_{v}, ψ_{v})

the functional equation

L(s, π) = ε(s, π) L(1 − s, π̃)

will be satisfied.

We may suppose that π acts on the subspace V of 𝒜′(η). Let φ be a non-zero function in V. For almost all v the algebra M′_{v} = M′ ⊗_{F} F_{v} is split and G′_{F_{v}} = G′_{v} is isomorphic to GL(2, F_{v}). Moreover for almost all such v, say for all v not in S, φ is an eigenfunction of the elements of ℋ′_{v} = ℋ′_{F_{v}} which are invariant on both sides under translations by the elements of K′_{v}. Thus if f is such an element and φ(g) ≠ 0 the corresponding eigenvalue λ_{v}(f) is

λ_{v}(f) = φ(g)^{−1} ∫_{G′_{v}} φ(gh) f(h) dh.

To prove the absolute convergence of the infinite products we have only to refer to Lemma 3.11 as in the proof of Theorem 11.1.

The representation π̃ contragredient to π can be defined. If π = ⊗π_{v} acts on V = ⊗_{u_{v}^{0}}V_{v} then π̃ = ⊗π̃_{v} acts on Ṽ = ⊗_{u_{v}^{0}} Ṽ_{v} where ũ_{v}^{0} is, for almost all v, fixed by K′_{v} and satisfies ⟨u_{v}^{0}, ũ_{v}^{0}⟩ = 1. The pairing between V and Ṽ is defined by

⟨⊗u_{v}, ⊗ũ_{v}⟩ = ∏_{v}⟨u_{v}, ũ_{v}⟩.

Almost all terms in the product are equal to 1. If u is in V and ũ is in Ṽ the matrix element ⟨π(g)u, ũ⟩ can also be introduced. If f is in ℋ′

⟨π(f)u, ũ⟩ = ∫_{G′_{𝔸}} f(g) ⟨π(g)u, ũ⟩ d^{×}g.

If F (g) is a linear combination of such matrix elements and Φ belongs to the Schwartz space on A′_{𝔸} we set*

Z(α_{F}^{s+ \frac{1}{2}} , Φ, F ) = ∫_{G′_{𝔸}} Φ(g) F (g) |ν(g)|^{s+ \frac{1}{2}} d^{×}g.

* Unfortunately the symbol F plays two quite different roles on this page!

Chapter 3 247 The function F̃(g) = F (g^{−1}) is a linear combination of matrix coefficients for the representation π̃. We set

Z(α_{F}^{s+ \frac{1}{2}} , Φ, F̃) = ∫_{G′_{𝔸}} Φ(g) F̃(g) |ν(g)|^{s+ \frac{1}{2}} d^{×}g.

Before stating the next lemma we observe that if χ is a quasi-character of F^{×} \ I the one dimensional representation g → χ(ν(g)) is certainly a constituent of 𝒜′.

Lemma 14.2.1 If π is a constituent of 𝒜′ the integrals defining Z(α_{F}^{s− \frac{1}{2}} , Φ, F) and Z(α_{F}^{s− \frac{1}{2}} , Φ, F̃) are absolutely convergent for Re s large enough. The two functions can be analytically continued to the whole complex plane as meromorphic functions with only a finite number of poles. If π is not of the form g → χ(ν(g)) they are entire. If F is a number field they are bounded at infinity in vertical strips of finite width. In all cases they satisfy the functional equation

Z(α_{F}^{s+ \frac{1}{2}} , Φ, F ) = Z(α_{F}^{\frac{3}{2} −s}, Φ′, F̃)

if Φ′ is the Fourier transform of Φ.

There is no harm in assuming that F is of the form

F (g) = ∏_{v}⟨π(g_{v})u_{v}, ũ_{v}⟩ = ∏_{v} F_{v}(g_{v})

and that Φ is of the form

Φ(x) = ∏_{v} Φ_{v}(x_{v})

where, for almost all v, Φ_{v} is the characteristic function of M(2, O_{v}). Recall that for almost all v we have fixed an isomorphism θ_{v} of M′_{v} with M(2, F_{v}).

We know that each of the integrals

∫_{G′_{v}} Φ_{v}(g_{v}) F_{v}(g_{v}) |ν(g_{v})|^{s+ \frac{1}{2}} d^{×}g_{v}

converges absolutely for Re s sufficiently large. Let S be a finite set of primes which contains all archimedean primes such that outside of S the vector u_{v} is u_{v}^{0}, the vector ũ_{v} is ũ_{v}^{0}, Φ_{v} is the characteristic function of M(2, O_{v}), and π_{v} = π_{v}(μ_{v}, ν_{v}) where μ_{v} and ν_{v} are unramified. Let π′_{v} = π_{v}(|μ_{v}|, |ν_{v}|). If v is not in S the integral

∫_{K′_{v}} Φ(g_{v}) F_{v}(g_{v}) |ν(g_{v})|^{s+ \frac{1}{2}} d^{×}g_{v} = 1

and if σ = Re s

∫_{G′_{v}} |Φ_{v}(g_{v})| |F_{v}(g_{v})| |ν(g_{v})|^{σ+ \frac{1}{2}} d^{×}g_{v}

is, as we see if we regard π_{v} as acting on ℬ(μ_{b}, ν_{v}), at most

∫_{G′_{v}} Φ_{v}(g_{v}) ⟨π′_{v}(g_{v}) f_{v}, f̃_{v}⟩ |ν(g_{v})|^{σ+ \frac{1}{2}} d^{×}g_{v}

Chapter 3 248 if f_{v} and f̃_{v} are the unique K′_{v}-invariant elements in ℬ(|μ_{v}|, |ν_{v}|) and ℬ(|μ_{v}|^{−1}, |ν_{v}|^{−1}) which take the value 1 at the identity. We suppose that the total measure of K′_{v} is 1 so that ⟨f_{v}, f̃_{v}⟩ = 1. According to Lemma 13.2.6 the integral is equal to L(σ, π′_{v}). Since

∏_{v∈S} L(σ, π′_{v})

is absolutely convergent for σ sufficiently large the integral defining Z(α_{F}^{s+ \frac{1}{2}} , Φ, F) is also and is equal to

∏_{v} Z(α_{F_{v}}^{s+ \frac{1}{2}} ⊗ π_{v}, Φ_{v}, u_{v}, ũ_{v})

and to

L(s, π) ∏_{v} Ξ(s, Φ_{v}, u_{v}, ũ_{v}).

Notice that Ξ(s, Φ_{v}, u_{v}, ũ_{v}) is identically 1 for almost all v. Z(α_{F}^{s+ \frac{1}{2}} , Φ, F̃) may be treated in a similar fashion. If we take π to be the trivial representation we see that

∫_{G′_{A}} Φ(g) |ν(g)|^{s+ \frac{1}{2}} d^{×}g

is absolutely convergent for Re s sufficiently large.

It will be enough to prove the remaining assertions of the lemma when η is a character. We may also assume that if η is of the form η(a) = |a|^{r} then r = 0. We have identified V with a subspace of 𝒜′(η). We may take Ṽ to be {φ̃ | φ ∈ V }. To see this observe that this space is invariant under ℋ′ and that

⟨φ_{1}, φ_{2}⟩ = ∫_{G′_{F} Z′_{𝔸}\G′_{𝔸}} φ_{1}(g) φ_{2}(g) dg

is a non-degenerate bilinear form. Here φ_{1} belongs to V and φ̃_{2} belongs to V. The remaining assertions need only be verified for functions of the form

F (g) = ∫_{G′_{F} Z′_{𝔸}\G′_{𝔸}} φ(hg) φ̃(h) dh

with φ in V and φ̃ in Ṽ.

For such an F the function Z(α_{F}^{s+ \frac{1}{2}} , Φ, F) is equal to

∫ Φ(g)\{ ∫ φ(hg) φ̃(h) dh\} |ν(g)|^{s+ \frac{1}{2}} d^{×}g.

Since φ and φ̃ are bounded this double integral converges absolutely for Re s sufficiently large. We first change variables by substituting h^{−1}g for g. The integration with respect to g can then be carried out in three steps. We first sum over G_{F′}, then we integrate over Z′_{F} \ Z′_{𝔸} which we identify with F^{×} \ I, and finally we integrate over G′_{F} Z′_{𝔸} \ G′_{𝔸}. Thus if K_{Φ}(h_{1}, h_{2}, s) is

|ν(h_{1}^{−1}|^{s+ \frac{1}{2}} |ν(h_{2})|^{s+ \frac{1}{2}} ∫_{F^{×}\I} ∑_{G′_{F}} Φ(h_{1}^{−1}ξah_{2}) η(a) |a|^{2s+1} d^{×}a

the function Z(α_{F}^{s+ \frac{1}{2}} , Φ, F) is equal to

∫∫ φ(h_{2}) φ̃(h_{1}) K_{Φ}(h_{1}, h_{2}, s) dh_{1} dh_{2}.

The integrations with respect to h_{1} and h_{2} are taken over G′_{F} Z′_{𝔸} \ G′_{𝔸}. A similar result is of course valid for Z(α_{F}^{s+ \frac{1}{2}} , Φ, F̃). If K̃_{Φ}(h_{1}, h_{2}, s) is

|ν(h_{1}^{−1})|^{s+ \frac{1}{2}} |ν(h_{2})|^{s+ \frac{1}{2}} ∫_{F^{×}\I} ∑_{G′_{F}} Φ(h_{1}^{−1}ξah_{2}) η^{−1}(a) |a|^{2s+1} d^{×}a

then Z(α_{F}^{s+ \frac{1}{2}} , Φ, F̃) is equal to

∫∫ φ(h_{2}) φ̃(h_{2}) K̃_{Φ}(h_{1}, h_{2}, s) dh_{1} dh_{2}.

We first study

θ(s, Φ) = ∫_{F^{×}\I} ∑_{ξ≠0} Φ(ξa) η(a) |a|_{F}^{2s+1} d^{×}a

and

θ̃(s, Φ) = ∫_{F^{×}\I} ∑_{ξ≠0} Φ(ξa) η^{−1}(a) |a|_{F}^{2s+1} d^{×}a.

The sums are taken over G′_{F} the set of non-zero elements of M′. Choose two non-negative continuous functions F_{0} and F_{1} on the positive real numbers so that F_{0}(t) + F_{1}(t) = 1, F_{1}(t) = F_{0}(t^{−1}), and so that F_{0} vanishes near zero while F_{1} vanishes near infinity. If

θ_{i}(s, Φ) = ∫_{F^{×}\I} ∑_{ξ≠0} Φ(ξ_{a}) η(a) |a|^{2s+1} F_{i}(|a|) d^{×}a

we have

θ(s, Φ) = θ_{0}(s, Φ) + θ_{1}(s, Φ).

In the same way we may write

θ̃(s, Φ) = θ̃_{0}(s, Φ) + θ̃_{1}(s, Φ)

θ_{0}(s, Φ) and θ̃_{0}(s, Φ) are entire functions of s which are bounded in vertical strips.

Applying the Poisson formula we obtain

Φ(0) + ∑_{ξ≠0} Φ(ξa) = |a|_{F}^{−4} {Φ′(0) + ∑_{ξ≠0} Φ′(ξa^{−1})}.

Thus, for Re s sufficiently large, θ_{1}(s, Φ) is equal to the sum of

∫_{F^{×}\I} ∑_{ξ≠0} Φ′(ξa^{−1}) η(a) |a|^{2s−3} F_{1}(|a|) d^{×}a,

which, after the substitution of a^{−1} for a, is seen to equal θ̃_{0}(1 − s, Φ′), and

∫_{F^{×}\I} {Φ′(0) |a|^{−4} − Φ(0)} η(a) |a|^{2s+1} F_{1}(|a|) d^{×}a.

Thus if

λ(s) = ∫_{F^{×}\I} |a|^{s} η(a) F_{1}(|a|) d^{×}a

the function θ(s, Φ) is equal to

θ_{0}(s, Φ) + θ̃_{0}(1 − s, Φ′) + Φ′(0) λ(2s − 3) − Φ(0) λ(2s + 1).

A similar result is valid for θ̃(s, Φ). The function

θ_{0}(s, Φ) + θ̃_{0}(1 − s, Φ′)

is entire and bounded in vertical strips and does not change when s and Φ are replaced by 1 − s and Φ′.

If η is not of the form η(a) = |a|^{r} the function λ(s) vanishes identically. If η is trivial and I_{0} is the group of idèles of norm 1

λ(s) = ∫_{F^{×}\I} |a|^{2s+1} F_{j}(|a|) d^{×}a.

It is shown in [10] that this function is meromorphic in the whole plane and satisfies λ(s)+λ(−s) = 0. If F is a number field, its only pole is at s = 0 and is simple. Moreover it is bounded at infinity in vertical strips of finite width. If F is a function field its poles are simple and lie at the zeros of 1 − q^{−s}. Here q is the number of elements in the field of constants.

Thus θ(s, Φ) is meromorphic in the whole plane and is equal to θ̃(1 − s, Φ′). If hΦg is the function x → Φ(gxh) then

K_{Φ}(h_{1}, h_{2}, s) = |ν(h_{1}^{−1})|^{s+ \frac{1}{2}} |ν(h_{2})|^{s+ \frac{1}{2}} θ(s, h_{2}Φh_{1}^{−1})

while

K̃_{Φ}(h_{1}, h_{2}, s) = |ν(h_{1}^{−1})|^{s+ \frac{1}{2}} |ν(h_{2})|^{s+ \frac{1}{2}} θ̃(s, h_{2}Φh_{1}^{−1}).

Since the Fourier transform of h_{2}Φh_{1}^{−1} is

|ν(h_{2})|^{−2} |ν(h_{1})|^{2} h_{1}Φ′h_{2}^{−1}

we have

K_{Φ}(h_{1}, h_{2}, s) = K̃_{Φ′}(h_{2}, h_{1}, s).

The functional equation of the lemma follows. So do the other assertions except the fact that the functions Z(α_{F}^{s+ \frac{1}{2}} , Φ, F) and Z(α_{F}^{s+ \frac{1}{2}} , Φ, F̃) are entire when η is trivial and π is not of the form g → χ(ν(g)). In this case the functions φ and φ̃ are orthogonal to the constant functions and the kernels K_{Φ}(h_{1}, h_{2}, s) and K̃_{Φ}(h_{1}, h_{2}, s) may be replaced by

K′_{Φ′}(h_{1}, h_{2}, s) = K̃_{Φ′}(h_{1}, h_{2}, s) + Φ(0) λ(2s + 1) − Φ′(0) λ(2s − 3)

Chapter 3 251 and

K̃′_{Φ′}(h_{1}, h_{2}, s) = K̃_{Φ′}(h_{1}, h_{2}, s) + Φ′(0) λ(2s + 1) − Φ(0) λ(2s − 3).

The functional equation of the kernels is not destroyed but the poles disappear.

The theorem follows easily from the lemma. In fact suppose that the finite set of places S is so chosen that for v not in S

Ξ(s, Φ_{v}^{0}, u_{v}^{0}, ũ_{v}^{0}) = 1

if Φ_{v}^{0} is the characteristic function of M(2, O_{v}). If v is in S choose Φ′_{v}, u_{v}^{i}, ũ_{v}^{i}, 1 ≤ i ≤ n_{v}, so that

∑_{i=1}^{n_{v}} Ξ(s, Φ_{v}^{i}, u_{v}^{i}, ũ_{v}^{i}) = e^{b_{vs}}

where b_{v} is real. If α is a function from S to the integers and, for each v in S, 1 ≤ α(v) ≤ n_{v}, set

Φ_{α}(g) = { ∏_{v∈S} Φ_{v}^{α(v)}(g_{v})} { ∏_{v∈S} Φ_{v}^{0}(g_{v})}

and set

F_{α}(g) = { ∏_{v∈S}⟨π_{v}(g_{v})u_{v}^{α(v)}, ũ_{v}^{α(v)}⟩} { ∏_{v∉S}⟨π_{v}(g_{v})u_{v}^{0}, ũ_{v}^{0}⟩}.

Then

∑_{α} Z(α_{F}^{s+ \frac{1}{2}} , Φ_{α}, F_{α}) = c^{bs} L(s, π)

where b is real. The required analytic properties of L(s, π) follow immediately.

To prove the functional equation choose for each v the function Φ_{v} and the vectors u_{v} and ũ_{v} so that

Ξ(s, Φ_{v}, u_{v}, ũ_{v})

is not identically 0. We may suppose that, for almost all v, Φ_{v} = Φ_{v}^{0}, u_{v} = u_{v}^{0}, and ũ_{v} = ũ_{v}^{0}. Let

Φ(g) = ∏_{v} Φ_{v}(g_{v})

and let

F (g) = ∏_{v}⟨π_{v}(g_{v})uv, ũ_{v}⟩.

Then

Z(α_{F}^{s+ \frac{1}{2}} , Φ, F ) = L(s, π) ∏_{v} Ξ(s, Φ_{v}, u_{v}, ũ_{v})

and

Z(α_{F}^{\frac{3}{2} −s}, Φ′, F̃) = L(1 − s, π̃) ∏_{v} Ξ̃(1 − s, Φ′_{v}, u_{v}, ũ_{v}).

Since

Ξ̃(1 − s, Φ′_{v}, u_{v}, ũ_{v}) = h_{v}ε(s, π_{v}, ψ_{v}) Ξ(s, Φ_{v}, u_{v}, ũ_{v})

the functional equation of the lemma implies that

L(s, π) = { ∏_{v} h_{v}} ε(s, π) L(1 − s, π̃).

Since, by a well-known theorem, the algebra M′ is split at an even number of places the product ∏_{v} h_{v} equals 1.

Chapter 3 252 Corollary 14.3 If π is a constituent of 𝒜′ which is not of the form g → χ(ν(g)) then for any quasi-character ω of F^{×} the functions L(s, ω ⊗ π) and L(s, ω^{−1} ⊗ π̃) are entire and bounded in vertical strips of finite width. Moreover they satisfy the functional equation

L(s, ω ⊗ π) = ε(s, ω ⊗ π) L(1 − s, ω^{−1} ⊗ π̃).

We have only to observe that if π is a constituent of 𝒜′ then ω ⊗ π is also.

Now we change the notation slightly and let π′ = ⊗π′_{v} be an irreducible admissible representation of ℋ′. We want to associate to it a representation π = ⊗π_{v} of ℋ the Hecke algebra of GL(2, 𝔸). If M′_{v} is split then π_{v} is just the representation corresponding to π′_{v} by means of the isomorphism θ_{v} of G_{F_{v}} and G′_{F_{v}}. If M′_{v} is not split π_{v} is the representation π_{v}(π′_{v}) introduced in the fourth and fifth paragraphs. In both cases π_{v} is defined unambiguously by the following relations

\begin{aligned}L(s, ω_{v} ⊗ π_{v}) & = L(s, ω_{v} ⊗ π′_{v}) \\ L(s, ω_{v} ⊗ π̃_{v}) & = L(s, ω_{v} ⊗ π̃′_{v}) \\ ε(s, ω_{v} ⊗ π_{v}, ψ_{v}) & = ε(s, ω_{v} ⊗ π′_{v}, ψ_{v}) \end{aligned}

which holds for all quasi-characters ω_{v} of F_{v}^{×}.

Applying the previous corollary and Theorem 11.3 we obtain the following theorem.

Theorem 14.4 If π′ is a constituent of 𝒜′ and π′_{v} is infinite dimensional at any place where M′ splits then π is a constituent of 𝒜_{0}.

Some comments on the assumptions are necessary. If π′ is a constituent of 𝒜′ we can always find a quasi-character of ω of F^{×} \ I such that ω ⊗ π′ is unitary. If π′ = ⊗π′_{v} the same is true of the representations π′_{v}. In particular if M′ splits at v the representation π′_{v} will not be finite dimensional unless it is one dimensional. Various density theorems probably prevent this from happening unless π′ is of the form g → χ(ν(g)). If π′ is of this form then all but a finite number of the representations π_{v} are one dimensional. But if M′ does not split at v the representation π_{v} is infinite dimensional. Thus π cannot act on a subspace of 𝒜. However it can still be a constituent of 𝒜. This is in fact extremely likely. Since the proof we have in mind involves the theory of Eisenstein series we prefer to leave the question unsettled for now.

Chapter 3 253 §15. Some orthogonality relations. It is of some importance to characterize the range of the map π′ → π from the constituents of 𝒜′ to those of 𝒜 discussed in the last chapter. In this paragraph we take up the corresponding local question. Suppose F is a local field and M′ is the quaternion algebra over F. G′_{F} is the group of invertible elements of M′. We know how to associate to every irreducible admissible representation π′ of ℋ′_{F} an irreducible admissible representation π = π(π′) of ℋ_{F} the Hecke algebra of GL(2, F).

Theorem 15.1 Suppose F is non-archimedean. Then the map π′ → π is injective and its image is the collection of special representations together with the absolutely cuspidal representations.

The proof requires some preparation. We need not distinguish between representations of G′_{F} and ℋ′_{F} or between representations of G_{F} and ℋ_{F}. An irreducible admissible representation π of G_{F} is said to be square-integrable if for any two vectors u_{1} and u_{2} in the space of π and any two vectors ũ_{1} and ũ_{2} in the space of π̃ the integral

∫_{Z_{F} \G_{F}} ⟨π(g)u_{1}, ũ_{1}⟩ ⟨u_{2}, π̃(g)ũ_{2}⟩ dg

is absolutely convergent. Since π̃ is equivalent to η^{−1} ⊗ π if

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η(a) I

this is equivalent to demanding that

∫_{Z_{F} \G_{F}} |⟨π(g)u_{1}, ũ_{1}⟩|^{2} |η^{−1}(detg)| dg

be finite for every u_{1} and u_{2}.

If π is square-integrable and ω is a quasi-character of F^{×} then ω ⊗ π is square integrable. We can always choose ω so that ω^{2}η is a character. If η is a character choose u_{0} different from 0 in the space V of π. Then

(u_{1}, u_{2}) = ∫_{Z_{F} \G_{F}} ⟨π(g)u_{1}, u_{0}⟩ \overline{⟨π(g)u_{2}, u_{0}⟩} dg

is a positive-definite form on the space V of π so that π is unitary and square-integrable in the usual sense.

The Schur orthogonality relations when written in the form

∫_{Z_{F} \G_{F}} ⟨π(g)u_{1}, ũ_{1}⟩ ⟨u_{2}, π̃(g)ũ_{2}⟩ dg = \frac{1}{d(π)} ⟨u_{2}, ũ_{1}⟩ ⟨u_{1}, ũ_{2}⟩

are valid not only for representations which are square-integrable in the usual sense but also for represntations which are square-integrable in our sense. The formal degree d(π) depends on the choice of Haar measure. Notice that d(ω ⊗ π) = d(π).

The absolutely cuspidal representations are certainly square-integrable because their matrix elements are compactly support modulo Z_{F}.

Lemma 15.2 The special representations are square-integrable.

Suppose σ = σ(αF^{\frac{1}{2}} , α_{F}^{− \frac{1}{2}} ). Since

χ ⊗ σ = σ(χαF^{\frac{1}{2}} , χα_{F}^{− \frac{1}{2}} )

it is enough to show that σ is square-integrable. If φ belongs to ℬ_{s}(αF^{\frac{1}{2}} , α_{F}^{− \frac{1}{2}} ) and φ̃ belongs to

ℬ(α_{F}^{− \frac{1}{2}} , αF^{\frac{1}{2}} ) then

f(g) = ⟨φ, ρ(g^{−1})φ̃⟩

is the most general matrix coefficient of σ. ℬ(α_{F}^{− \frac{1}{2}} , αF^{\frac{1}{2}} ) is the space of locally constant functions on N_{F} A_{F} \ G_{F} and ℬ_{s}(αF^{\frac{1}{2}} , α_{F}^{− \frac{1}{2}} ) is the space of locally constant functions φ on G_{F} that satisfy

φ (( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} ) g) = |\frac{a_{1}}{a_{2}} | φ(g)

and

∫ φ (w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) dx = 0.

Since

G_{F} = ⋃_{n≥0} Z_{F} K ( \begin{matrix}ϖ^{n} & 0 \\ 0 & 1 \end{matrix} ) K

we can choose the Haar measure on Z_{F} \ G_{F} so that

∫_{Z_{F} \G_{F}} |f(g)|^{2} dg

is equal to

∑_{n≥0} c(n) ∫ |f (k_{1} ( \begin{matrix}ϖ^{−n} & 0 \\ 0 & 1 \end{matrix} ) k_{2})|^{2} dk_{1} dk_{2}

where c(0) = 1 and

c(n) = q^{n}(1 + \frac{1}{q})

if n > 0. Here q = |ϖ|^{−1}. Since f is K-finite on both sides and its translates are also matrix coefficients we need only show that

∑_{n=0}^{∞} |f \begin{matrix}( \\ ( & \begin{matrix}ϖ^{−n} & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{matrix}|^{2} q^{n}

is finite. It will be more than enough to show that

Φ(a) = f \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} & ) \\ & & ) \end{aligned} = O(|a|^{−1})

as a → ∞.

We recall that

Φ(a) = ∫_{F} φ (w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) φ̃ (w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} )) dx.

The function

φ_{1}(x) = φ (w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))

is integrable and the function

φ_{2}(x) = φ̃ (w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))

is bounded and locally constant. Moreover

Φ(a) = ∫_{F} φ_{1}(x) φ_{2}(a^{−1}x) dx.

Suppose φ_{2}(x) = φ_{2}(0) for |x| ≤ M. If |a| ≥ 1

Φ(a) = φ_{2}(0) ∫_{{x | |x|≤|a|M}} φ_{1}(x) dx + ∫_{{x | |x|>|a|M}} φ_{1}(x) φ_{2}(a^{−1}x) dx.

Since

∫_{F} φ_{1}(x) dx = 0

Φ(a) is equal to

∫_{{x | |x|>|a|M}} (φ_{2}(a^{−1}x) − φ_{2}(0)) φ_{1}(x) dx.

The function φ_{2} is bounded so we need only check that

∫_{{x | |x|>|a>}} |φ_{1}(x)| dx = O(|a|^{−1})

as |a| → ∞. The absolute value of the function φ is certainly bounded by some multiple of the function φ′ in ℬ(αF^{\frac{1}{2}} , α_{F}^{− \frac{1}{2}} ) defined by

φ′ (( \begin{matrix}a_{1} & x \\ 0 & a_{2} \end{matrix} ) k) = |\frac{a_{1}}{a_{2}} |

if k is in GL(2, O_{F}). Since

w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}0 & 1 \\ −1 & −x \end{matrix} ) = ( \begin{matrix}x^{−1} & 0 \\ 0 & x \end{matrix} ) ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ) k

with y in F and k in GL(2, O_{F}) if |x| > 1

∫_{{x | |x|>|ϖ|^{−n}}} |φ_{1}(x)| dx = O( ∑_{k=n+1}^{∞} |ϖ|^{k}) = O(|ϖ|^{−n}).

Chapter 3 256

Since we need to compare orthogonality relations on the two groups G_{F} = GL(2, F) and G′_{F} we have to normalize their Haar measure simultaneously. There are two ways of doing this. We first describe the simplest. Choose a non-trivial additive character ψ of F. Then ψ_{M}(x) = ψ(tr x) and ψ_{M′}(x) = ψ(τ(x)) are non-trivial additive characters of M = M(2, F) and M′. Let dx and dx′ be the Haar measures on M and M′ self-dual with respect to ψ_{M} and ψ_{M′}. Then

d^{×}x = |x|_{M}^{−1} dx = |detx|_{F}^{−2} dx

and

d^{×}x′ = |x′|_{M′}^{−1} dx′ = |ν(x′)|_{F}^{−2} dx′

are Haar measure on G_{F} and G′_{F}.

The second method takes longer to describe but is more generally applicable and for this reason well worth mentioning. Suppose G and G′ are two linear groups defined over F and suppose there is an isomorphism φ of G′ with G defined over the finite Galois extension K. Suppose the differential form ω on G is defined over F. In general the form ω′ = φ_{∗}ω on G′ is not defined over F. Suppose however that ω is left and right invariant and under an arbitrary isomorphism it is either fixed or changes sign. Suppose moreover that for every σ in 𝔊(K/F) the automorphism σ(φ)φ^{−1} of G is inner. Then

σ(ω′) = σ(φ)_{∗} σω = σ(φ)_{∗} ω = φ_{∗}(σ(φ) φ^{−1})_{∗}ω = φ_{∗}ω = ω′

and ω′ is also defined over F. If ξ is another such isomorphism of G′ with G then

ξ_{∗}(ω) = φ_{∗}(ξφ^{−1})_{∗}ω = ±φ_{∗}ω = ±ω′

and the measures associated to φ_{∗}ω and ξ_{∗}ω are the same. Thus a Tamagawa measure on G_{F} determines one on G′_{F}.

We apply this method to the simple case under consideration. If

x = ( \begin{matrix}a & b \\ c & d \end{matrix} )

is a typical element of M then

μ = da ∧ db ∧ dc ∧ dd

is a differential form invariant under translations and the associated measure is self-dual with respect to ψ_{M}. If ω = (detx)^{−2}μ then ω is an invariant form on G and the associated measure is d^{×}x.

If K is any separable quadratic extension of F we may imbed K in both M and M′. Let σ be the non-trivial element of 𝔊(K/F). There is a u in M and a u′ in M′ such that M = K + Ku and M′ = K + Ku′ while uxu^{−1} = x^{σ} and u′xu′^{−1} = x^{σ} for all x in K. u^{2} is a square in F^{×} and u′^{2} = γ is an element of F^{×} which is not the norm of any element of K. We may suppose that u^{2} = 1. If we let K act to the right the algebra L = K ⊗_{F} K is an algebra over K. σ acts on L through its action on the first factor. There is an isomorphism L → K ⊕ K which transforms σ into the involution (x, y) → (y, x). In particular every element of K ⊗ 1 is of the form δδ^{σ} with δ in L. Choose δ so that γ = δδ^{σ}. If

M′_{K} = M′ ⊗_{F} K = L ⊗ Lu′

and

M_{K} = M ⊗_{F} K = L ⊗ Lu,

let φ be the linear map from M′_{K} to M_{K} which sends x + yu′ to x + yδu. φ is easily seen to be an isomorphism of M′_{K} and M_{K} as algebras over K. σ(φ)φ^{−1} takes x + yu to

x + yδ^{σ}δ^{−1}u = δ^{−1}(x + yu)δ

and is therefore inner. Thus φ determines an isomorphism of G′ the multiplicative group of M′ with G the multiplicative group of M. φ is defined over K and σ(φ)φ^{−1} is inner. Let |ω′| be the Haar measure on G′_{F} associated to the Haar measure |ω| = d^{×}x of G_{F}. We want to show that |ω′| is just d^{×}x′.

Let θ be an invariant form on K. The obvious projections of M = K ⊕ Ku on K define differential forms θ_{1} and θ_{2} on M. Let θ_{1} ∧ θ_{2} = cμ. In the same way the projections of M′ = K ⊕ Ku′ on K define differential forms θ′_{1} and θ′_{2} on M′. If we extend the scalars from F to K we can consider the map x → xδ of L into itself. θ is a form on L and its inverse image is N(δ)θ = γθ. Thus

φ_{∗}(θ_{1} ∧ θ_{2}) = γθ′_{1} ∧ θ′_{2}.

Thus if μ′ = φ_{∗}(μ)

cμ′ = γθ′_{1} ∧ θ′_{2}.

Suppose c_{1}|θ| is self-dual with respect to the character ψ_{K}(x) = ψ(τ(x)) on K. Then

∫ { ∫ Φ(a, b) ψ_{K}(ax + by^{σ}) |θ(a)| |θ(b)|} |θ(x)| |θ(y)| = c_{1}^{−4}Φ(0, 0)

and

|γ|_{F}^{2} ∫ { ∫ Φ(a, b) ψ_{K}(ax + by^{σ}γ) |θ(a)| |θ(b)|} |θ(x)| |θ(y)| = c_{1}^{−4}Φ(0, 0).

If x + yu belongs to M with x and y in K then, since τ(u) = 0,

τ(x + yu) = τ(x) = Tr_{K/F} (x).

In the same way

τ(x + yu′) = Tr_{K/F} (x).

Thus

\begin{aligned}ψ_{M}((x + yu)(a + bu)) & = ψ_{K}(xa + yb^{σ}) \\ ψ_{M′}((x + yu′)(a + bu′)) & = ψ_{K}(xa + yb^{σ}γ). \end{aligned}

Thus c_{1}^{2}|θ_{1} ∧ θ_{2}| is self-dual with respect to ψ_{M} and c_{1}^{2}|γ|_{F} |θ′_{1} ∧ θ′_{2}| is self-dual with respect to ψ_{M′}. Since c_{1}^{2} = |c|_{F} the measure |μ′| is self-dual with respect to ψ_{M′}. Finally ω′ = ν(x′)^{−2} dx′ so that |ω′| is just d^{×}x′. Thus the two normalizations lead to the same result.

If b is in M or M′ the eigenvalues of b are the roots α_{1} and α_{2} of the equation

X^{2} − τ(b) X + ν(b) = 0.

Chapter 3 258 If b is in G_{F} or G′_{F} it is said to be regular of α_{1} and α_{2} are distinct; otherwise it is singular. We set

δ(b) = |\frac{(α_{1} − α_{2})^{2}}{α_{1}α_{2}} |_{F} .

The set of singular elements is of measure 0. If b is regular the subalgebra of M or M′ generated by b is a separable quadratic extension E of F and the multiplicative group of E is a Cartan subgroup of G_{F} or G′_{F}. To obtain a set of representatives for the conjugacy classes of Cartan subgroups of G_{F} or G′_{F} we choose once and for all a set S′ of representatives for the classes of separable quadratic extensions of F. We also choose for each E in S′ an imbedding of E in M and in M′. The multiplicative group of E may be regarded as a Cartan subgroup B_{F} of either G_{F} or G′_{F}. S′ will also stand for the collection of Cartan subgroups obtained in this way. It is a complete set of representatives for the conjugacy classes of Cartan subgroups of G′_{F}. If S is the result of adjoining to S′ the group A_{F} of diagonal matrices then S is a complete set of representatives for the conjugacy classes of Cartan subgroups of G_{F}. If B_{F} is in S′ we choose the Tamagawa measure μ_{B} on B_{F} as in the seventh paragraph. The analogue for G′_{F} of the formula (1.2.2) is

∫_{Z′_{F} \G′_{F}} f(g) ω′_{0}(g) = ∑_{S′} \frac{1}{2} ∫_{Z_{F} \B_{F}} δ(b)\{ ∫_{B_{F} \G′_{F}} f(g^{−1}bg) ω′_{B}(g)\} μ_{B}^{0}(b).

Let B̂_{F} be the set of regular elements in B_{F} and let

C = ⋃_{S′} Z_{F} \ B̂_{F} .

We may regard C as the discrete union of the spaces Z_{F} \ B̂_{F}. We introduce on C the measure μ(c) defined by

∫_{C} f(c) μ(c) = \frac{1}{2} ∑_{S′} \frac{1}{measure(Z_{F} \ B_{F})} ∫_{Z_{F} \_{B_{F}}} f(b) δ(b) μ_{B}^{0}(b).

Lemma 15.3 Let η be a quasi-character of F^{×} and let Ω′(η) be the set of equivalence classes of irreducible representations π of G′_{F} such that π(a) = η(a) for a in Z′_{F}, which we identify with F^{×}. If π_{1} and π_{2} belong to Ω′(η) and

f(g) = χ_{π_{1}}(g) χπ̃_{2}(g)

where χ_{π}(g) = Tr π(g) then

∫_{C} f(c) μ(c) = 0

if π_{1} and π_{2} are not equivalent and

∫_{C} f(c) μ(c) = 1

if they are.

Since Z′_{F} \ G′_{F} is compact we may apply the Schur orthogonality relations for characters to see that

\frac{1}{measure Z′_{F} \ G′_{F}} ∫_{Z′_{F} \G′_{F}} f(g) ω′_{0}(g)

is 0 if π_{1} and π_{2} are not equivalent and is 1 if they are. According to the integration formula remarked above this expression is equal to

\frac{1}{measure Z′_{F} \ G′_{F}} ∑_{S′} \frac{1}{2} ∫_{Z_{F} \B_{F}} f(b) δ(b) (measure B_{F} \ G′_{F}) μ_{b}^{0}(b).

Since

measure Z′_{F} \ G′_{F} = (measure Z_{F} \ B_{F})(measure B_{F} \ G′_{F})

the lemma follows. Observe that Z_{F} and Z′_{F} tend to be confounded.

There is form of this lemma which is valid for G_{F}.

Lemma 15.4 Let η be a quasi-character of F^{×}. Let Ω_{0}(η) be the set of equivalence classes of irreducible admissible representations π of G_{F} which are either special or absolutely cuspidal and satisfy π(a) = η(a) for all a in Z_{F}. Suppose π_{1} and π_{2} belong to Ω_{0}(η). Let f = f_{π_{1},π_{2}} be the function

f(b) = χ_{π_{1}}(b) χπ̃_{2}(b)

on C. Then f is integrable and

∫_{C} f(c) μ(c)

is 1 if π_{1} and π_{2} are equivalent and 0 otherwise.

It is enough to prove the lemma when η is a character. Then χπ̃ is the complex conjugate of χ_{π} and f_{π,π} is positive. If the functions f_{π,π} are integrable then by the Schwarz inequality all the functions f_{π_{1},π_{2}} are integrable.

Let Ω(η) be the set of irreducible admissible representations π of G_{F} such that π(a) = η(a) for a in Z_{F}. If φ is a locally constant function on G_{F} such that

φ(ag) = η^{−1}(a) φ(g)

for a in Z_{F} and such that the projection of the support of φ on Z_{F} \ G_{F} is compact then we define π(φ), if π is in Ω(η), by

π(φ) = ∫_{Z_{F} \G_{F}} φ(g) π(g) ω^{0}(g).

It is easily seen that π(φ) is an operator of finite rank and that the trace of π(φ) is given by the convergent integral

∫_{Z_{F} \G_{F}} φ(g) χ_{π}(g) ω^{0}(g).

In fact this follows from the observation that there is a φ_{1} in ℋ_{F} such that

φ_{1}(g) = ∫_{Z_{F}} φ_{1}(ag) η(a) μ_{Z}(a)

and the results of the seventh paragraph.

Suppose π_{1} is absolutely cuspidal and unitary and acts on the space V_{1}. Suppose also that π_{1}(a) = η(a) for a in Z_{F}. Choose a unit vector u_{1} and V_{1} and set

φ(g) = d(π_{1}) (u_{1}, π_{1}(g)u_{1}).

Chapter 3 260 Since π_{1} is integrable it follows from the Schur orthogonality relations that π_{2}(φ) = 0 if π_{2} in Ω(η) is not equivalent to π_{1} but that π_{2}(φ) is the orthogonal projection on ℂu_{1} if π_{2} = π_{1}. In the first case Tr π_{1}(φ) = 0 and in the second Tr π_{2}(φ) = 1.

On the other hand

Tr π_{2}(φ) = ∫_{Z_{F} \G_{F}} χ_{π_{2}}(g) φ(g) ω^{0}(g).

We apply formula (7.2.2) to the right side to obtain

∑_{S} \frac{1}{2} ∫_{Z_{F} \B_{F}} χ_{π_{2}}(b) δ(b)\{ ∫_{B_{F} \G_{F}} φ(g^{−1}bg) ω_{B}(g)\} μ_{B}^{0}(g).

If B_{F} belongs to S′ the inner integral is equal to

\frac{1}{measure Z_{F} \ B_{F}} d(π_{1}) ∫_{Z_{F} \G_{F}} (u_{1}, π_{1}(g^{−1}bg)u_{1}) ω_{B}(g)

which by Proposition 7.5 is equal to

\frac{1}{measure Z_{F} \ B_{F}} χπ̃_{1}(b).

If B_{F} is A_{F} the group of diagonal matrices the inner integral is, apart from a constant relating Haar measures, the product of d(π_{1}) and the integral over GL(2, O_{F}) of

∫_{F} (π_{1} (( \begin{matrix}1 & −x \\ 0 & 1 \end{matrix} ) b ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) π_{1}(k)u_{1}, π_{1}(k)u_{1}) dx.

If

b = ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )

this is

|1 − \frac{α_{2}}{α_{1}} |^{−1} ∫_{F} (π_{1}(b) π_{1} (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) π_{1}(k)u_{1}, π_{1}(k)u_{1}) dx

which we know is 0. Collecting these facts together we see that f = f_{π_{2},π_{1}} is integrable on C if π_{1} is absolutely cuspidal and that its integral has the required value.

To complete the proof all we need do is show that if π = σ(χα_{F}^{\frac{1}{2}} , χ_{F}^{− \frac{1}{2}} ) is a special representation then f = f_{π,π} is integrable on C and

∫_{C} f(c) μ(c) = 1.

If π′ is the one-dimensional representation g → χ(ν(g)) of G′_{F} then π = π(π′). To prove the existence of χ_{π} we had to show in effect that if B_{F} was in S′ and b was in B̂_{F} then

χ_{π}(b) = −χ_{π′}(b).

Thus f_{π,π} = f_{π′,π′} and the assertion in this case follows from the previous lemma.

The relation just used does not seem to be accidental.

Proposition 15.5 Suppose π′ is an irreducible admissible representation of G′_{F} and π = π(π′) the corresponding representation of G_{F}. If B_{F} is in S′ and b is in B̂_{F}

χ_{π′}(b) = −χ_{π}(b).

We may suppose that π′ is not one-dimensional and that π is absolutely cuspidal. We may also suppose that they are both unitary. We take π in Kirillov form with respect to some additive character ψ. If φ is in 𝒮(F^{×}) the function

φ′ = π \begin{aligned}( \\ ( & \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} & ) \\ & & ) \end{aligned} φ

is also.

Since the measure μ and μ′ are self-dual with respect to the characters ψ_{M} and ψ_{M′} Lemma 13.1.1 and Proposition 4.5 show us that for any λ in F^{×}

∫_{G_{F}} φ(detg)(π(g^{−1})u, u) |detg| ψ_{M}(λg) ω(g)  (15.5.1)

is equal to

φ′(λ^{2}) η^{−1}(λ) |λ|^{−2}

and that

∫_{G′_{F}} φ(detg)(π′(g^{−1})u′, u′) |detg| ψ_{M′}(λg) ω′(g)  (15.5.2)

is equal to

−φ′(λ^{2}) η^{−1}(λ) |λ|^{−2}.

Here u is a unit vector in the space of π and u′ a unit vector in the space π′. In any case (15.5.1) is just the negative of (15.5.2).

If we use formula (7.2.1) to express the integral (15.5.1) as a sum over S we obtain

\frac{1}{2} ∑_{S′} \frac{1}{measure Z_{F} \ B_{F}} ∫_{B_{F}} φ(detb) |detb| \frac{χ_{π}(b^{−1})}{d(π)} δ(b) ψ_{M}(λb) μ_{B}(b).

The contribution from A_{F} vanishes as in the previous lemma. The other integrals have been simplified by means of Proposition 7.5. There is of course an obvious analogue for the group G′_{F} of the formula (7.2.1). If we apply it we see that (15.5.2) is equal to

\frac{1}{2} ∑_{S′} \frac{1}{measure Z_{F} \ B_{F}} ∫_{B_{F}} φ(detb) |ν(b)| \frac{χ_{π′}(b^{−1})}{d(π′)} δ(b) ψ_{M′}(λb) μ_{B}(b)

if ν(b) is the reduced norm. Of course on B_{F} the functions ν(b) and detb are the same. Choose B_{F}^{0} in S′ and b_{0} in B̂_{F}^{0} . We shall show that

\frac{χ_{π′}(b_{0}^{−1})}{d(π′)} = \frac{−χ_{π}(b_{0}^{−1})}{d(π)}.

Chapter 3 262 The orthogonality relations of the previous two lemmas will show that d(π) = d(π′) and we will conclude that

χ_{π′}(b_{0}^{−1}) = −χ_{π}(b_{0}^{−1}).

The norm and the trace of b_{0} are the same whether it is regarded as an element of M or of M′. In fact if B̂_{F}^{0} is the multiplicative group of E in S′ the norm and the trace are in both cases the norm and the trace of b_{0} as an element of E. Since b_{0} and its conjugate in E are conjugate in G_{F} and G′_{F} we can choose an open set U in E^{×} containing both b_{0} and its conjugate so that

|ν(b)| χ_{π′}(b^{−1}) δ(b) = |ν(b_{0})| χ_{π′}(b_{0}^{−1}) δ(b_{0})

if b is in U. Lemma 7.4.2 shows that χ_{π} is locally constant in B̂_{F}^{0} . Thus we can also suppose that

|detb| χ_{π}(b^{−1}) δ(b) = |detb_{0}| χ_{π}(b_{0}^{−1}) δ(b_{0})

if b is in U. Suppose α_{0} and β_{0} are the trace and norm of b_{0}. We can choose a positive integer m so that if α − α_{0} and β − β_{0} belong to 𝔭_{F}^{m} the roots of

X^{2} − αX + b

belong to E and in fact lie in U.

Let ξ(λ) be the expression (15.5.1) regarded as a function of λ. Keeping in mind the fact that

ψ_{M}(λb) = ψ_{M′}(λb) = ψ(λ tr b),

we compute

\frac{1}{measure 𝔭_{F}^{−m−n}} ∫_{𝔭_{F}^{−m−n}} ξ(λ) ψ(−λα_{0}) dλ  (15.5.3)

where 𝔭_{F}^{−n} is the largest ideal on which ψ is trivial. Since

\frac{1}{measure 𝔭_{F}^{−m−n}} ∫_{𝔭_{F}^{−m−n}} ψ(λ(tr b − α_{0})) dλ

is 0, unless tr b − α_{0} belongs to 𝔭_{F}^{m} when it is 1, the integral (15.5.3) is equal to

\frac{1}{2} ∑_{S′} \frac{1}{measure Z_{F} \ B_{F}} ∫_{V (B_{F})} φ(detb) |detb| \frac{χ_{π}(b^{−1})}{d(π)} δ(b) μ_{B}(b)

if

V (B_{F}) = {b ∈ B_{F} | tr b − α_{0} ∈ 𝔭_{F}^{m}}.

If we take φ to be the characteristic function of

{β ∈ F | β − β_{0} ∈ 𝔭_{F}^{m}}

the summation disappears and we are left with

\frac{1}{2} · \frac{1}{measure Z_{F} \ B_{F}} |detb_{0}| \frac{χ_{π}(b_{0}^{−1})}{d(π)} δ(b_{0}) ∫_{V (B_{F}^{0} )} φ(detb) μ_{B}(b).

If we replace ξ(λ) by the expression (15.5.2) the final result will be

\frac{1}{2} · \frac{1}{measure Z_{F} \ B_{F}} |ν(b_{0})| \frac{χ_{π′}(b_{0}^{−1})}{d(π)} δ(b_{0}) ∫_{V (B_{F}^{0} )} φ(detb) μ_{B}(b).

Since these differ only in sign the proposition follows.

We are now in a position to prove Theorem 15.1. The orthogonality relations and the previous lemma show that the map π′ → π is injective because the map takes Ω′(η) into Ω_{0}(η). It is enough to verify that V is surjective when η is unitary. Let ℒ^{2}(η) be the space of all measurable functions f on

∪_{S′} B̂_{F}

such that f(ab) = η(a) f(b) if a is in Z_{F} and

∫_{C} |f(c)|^{2} μ(c)

is finite. By the Peter–Weyl theorem the set of functions χ_{π′}, π′ ∈ Ω′(η), form an orthonormal basis of ℒ^{2}(η). The family χ_{π}, π ∈ Ω_{0}(η), is an orthonormal family in ℒ^{2}(η). By the previous proposition the image of Ω′(η) in Ω_{0}(η) is actually an orthonormal basis and must therefore be the whole family.

We observe that it would be surprising if the relation d(π) = d(π′) were not also true when π′ is one-dimensional. The facts just discussed are also valid when F is the field of real numbers. They follow immediately from the classification and the remarks at the end of the seventh paragraph.

We conclude this paragraph with some miscellaneous facts which will be used elsewhere. F is again a non-archimedean field. Let K = GL(2, O_{F}) and let K_{0} be the set of all matrices

( \begin{matrix}a & b \\ c & d \end{matrix} )

in K for which c ≡ 0(mod 𝔭_{F}). Suppose π is an irreducible admissible representation of G_{F} in the space V. We are interested in the existence of a non-zero vector v in V such that

π \begin{aligned}( \\ ( & \begin{matrix}a & b \\ c & d \end{matrix} & ) \\ & & ) \end{aligned} v = ω_{1}(a) ω_{2}(d) v

for all matrices in K_{0} while

π \begin{aligned}( \\ ( & \begin{matrix}0 & 1 \\ ϖ & 0 \end{matrix} & ) \\ & & ) \end{aligned} v = ω_{0} v

ω_{0} is a constant and ω_{1} and ω_{2} two characters of U_{F}. ϖ is a generator of 𝔭_{F}. Since

( \begin{matrix}0 & ϖ^{−1} \\ 1 & 0 \end{matrix} ) ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}1 & 1 \\ ϖ & 0 \end{matrix} ) = ( \begin{matrix}d & ϖ^{−1}c \\ ϖb & a \end{matrix} )

such a vector can exist only if ω_{1} = ω_{2} = ω.

Chapter 3 264 Lemma 15.6 Suppose ω and ω_{0} are given. Let π be ρ(μ_{1}, μ_{2}) which may not be irreducible. There is a non-zero vector φ in ℬ(μ_{1}, μ_{2}) satisfying the above conditions if and only if the restrictions of μ_{1} and μ_{2} to U_{F}, the group of units of O_{F}, are equal to ω and

ω_{0}^{2} = μ(−ϖ) μ_{2}(−ϖ)

Moreover φ if it exists is unique apart from a scalar factor.

It is easily seen that K is the disjoint union of K_{0} and

K_{0} ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) K_{0} = K_{0}wK_{0}

Let φ_{1} be the function which is 0 on K_{0}wK_{0} and on K_{0} is given by

φ_{1} \begin{aligned}( \\ ( & \begin{matrix}a & b \\ c & d \end{matrix} & ) \\ & & ) \end{aligned} = ω(ad).

Let φ_{2} be the function which is 0 on K_{0} and takes the value ω(a′d′ad) at

( \begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}a & b \\ c & d \end{matrix} ) .

If φ in ℬ(μ_{1}, μ_{2}) satisfies

π \begin{aligned}( \\ ( & \begin{matrix}a & b \\ c & d \end{matrix} & ) \\ & & ) \end{aligned} φ = ω(ad)φ

for all matrices in K_{0} then the restrictions of φ to K must be a linear combination of φ_{1} and φ_{2}. This already implies that ω is the restriction of μ_{1} and μ_{2} to U_{F}. Suppose φ = aφ_{1} + bφ_{2}. Since

π \begin{aligned}( \\ ( & \begin{matrix}0 & 1 \\ ϖ & 0 \end{matrix} & ) \\ & & ) \end{aligned} φ_{1} = |ϖ|^{\frac{1}{2}} μ_{1}(ϖ) μ_{2}(−1) φ_{2}

and

π (( \begin{matrix}0 & 1 \\ ϖ & 0 \end{matrix} )) φ_{2} = |ϖ|^{− \frac{1}{2}} μ_{2}(−ϖ) φ_{1}

while μ_{1}(−1) = μ_{2}(−1) = ω(−1), we have

ω_{0}b = |ϖ|^{\frac{1}{2}} μ_{1}(−ϖ)a

and

ω_{0}a = |ϖ|^{− \frac{1}{2}} μ_{2}(−ϖ)b

Apart from scalar factors there is at most one solution of this equation. There is one non-trivial solution if and only if ω_{0}^{2} = μ_{1}(−ϖ) μ_{2}(−ϖ).

Lemma 15.7 Suppose π = σ(μ_{1}, μ_{2}) is the special representation corresponding to the quasichacters μ_{1} = χα_{F}^{− \frac{1}{2}} and μ_{2} = χαF^{\frac{1}{2}} . There is a non-zero vector v in the space of π such that

π \begin{aligned}( \\ ( & \begin{matrix}a & b \\ c & d \end{matrix} & ) \\ & & ) \end{aligned} v = ω(ad)v

for all matrices in K_{0} while

π \begin{aligned}( \\ ( & \begin{matrix}0 & 1 \\ ϖ & 0 \end{matrix} & ) \\ & & ) \end{aligned} v = ω_{0}v

if and only if ω is the restriction of χ to U_{F} and ω_{0} = −χ(−ϖ). If v exists it is unique apart from a scalar.

We first let π act on ℬ_{s}(μ_{2}, μ_{2}) a subspace of ℬ(μ_{2}, μ_{1}). The condition on ω follows from the previous lemma which also shows that ω_{0} must be ±χ(−ϖ). If we take the plus sign we see that v must correspond to the function whose restriction to K is constant. Since this function does not lie in ℬ_{s}(μ_{2}, μ_{1}) only the minus sign is possible. To see the existence we let π act on

ℬ_{s}(μ_{1}, μ_{2}) = ℬ(μ_{1}, μ_{2})/ℬ_{f}(μ_{1}, μ_{2})

In ℬ(μ_{1}, μ_{2}) there are two functions satisfying the conditions of the lemma. One with ω_{0} = −χ(−ϖ) and one with ω_{0} = χ(−ϖ). One of the two, and we know which, must have a non-zero projection on ℬ_{s}(μ_{1}, μ_{2}).

The above lemmas together with the next one sometimes allow us to decide whether or not a given representation is special.

Lemma 15.8 If the absolutely cuspidal representation π acts on V there is no non-zero vector v in V such that

π \begin{aligned}( \\ ( & \begin{matrix}0 & 1 \\ ϖ & 0 \end{matrix} & ) \\ & & ) \end{aligned} v = ω_{0}v

and

π \begin{aligned}( \\ ( & \begin{matrix}a & b \\ c & d \end{matrix} & ) \\ & & ) \end{aligned} v = ω(ad)v

for all matrices in K_{0}.

We may suppose that π is the Kirillov form with respect to an additive character ψ such that O_{F} is the largest ideal on which ψ is trivial. Then v is a function φ in 𝒮(F^{×}). If a is in U_{F} and b is in F^{×} we must have φ(ab) = ω(a) φ(b). Moreover if b is in F^{×} and x is in O_{F} then φ(b) = ψ(xb) φ(b). Thus φ(b) = 0 if b is not in O_{F}. Consequently φ̂(ν, t) is 0 if ν ≠ ω^{−1} but φ̂(ω^{−1}, t) is a polynomial of the form

a_{m}t^{m} + ⋯ + a_{n}t^{n}

with a_{m}a_{n} ≠ 0. If φ_{1}(b) = φ(−ϖb) then

φ̂_{1}(ω^{−1}, t) = \frac{ω(−1)}{t} φ̂(ω^{−1}, t).

Let

π \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η(a) I

Chapter 3 266 and let ν_{0} be the restriction of η to U_{F} while z_{0} = η(ϖ). The character ν_{0} will have to be equal to ω^{2}. The relation

ω_{0}φ = π \begin{aligned}( \\ ( & \begin{matrix}0 & 1 \\ ϖ & 0 \end{matrix} & ) \\ & & ) \end{aligned} φ = π \begin{aligned}( \\ ( & \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} & ) \\ & & ) \end{aligned} φ_{2}

implies that

ω_{0}φ̂(ω^{−1}, t) = C(ω^{−1}, t) ω(−1) z_{0}t φ̂(ω^{−1}, z_{0}^{−1}t^{−1}).

By Proposition 2.23, C(ω^{−1}, t) is of the form ct^{−ℓ} with ℓ ≥ 2. Thus the right side has a pole at 0 not shared by the left. This is a contradiction.

Chapter 3 267 §16. An application of the Selberg trace formula. In the fourteenth paragraph we saw that if π′ = ⊗_{v}π′_{v} is a constituent of 𝒜′ and π′ is not of the form g → χ(ν(g)) where χ is a quasi-character of F^{×} \ I then π = ⊗_{v}π_{v}, with π_{v} = π(π′_{v}), is a constituent of 𝒜_{0}. Let S be the set of places at which the quaternion algebra M′ does not split. Given the results of the previous paragraph it is tempting to conjecture that the following theorem is valid.

Theorem 16.1 Suppose π = ⊗π_{v} is a constituent of 𝒜_{0}. If for every v in S the representation π_{v} is special or absolutely cuspidal then for every v there is a representation π′_{v} such that π_{v} = π(π′_{v}) and π′ = ⊗π′_{v} is a constituent of 𝒜′.

The existence of π′_{v} has been shown. What is not clear is that π′ is a constituent of 𝒜′. It seems to be possible to prove this by means of the Selberg trace formula. Unfortunately a large number of analytical facts need to be verified. We have not yet verified them. However the theorem and its proof seem very beautiful to us; so we decided to include a sketch of the proof with a promise to work out the analytical details and publish them later. We must stress that the sketch is merely a formal argument so that the theorem must remain, for the moment, conjectural.

We first review some general facts about traces and group representations. Suppose G is a locally compact unimodular group and Z is a closed subgroup of the centre of G. Let η be a character of Z. We introduce the space L^{1}(η) of all measurable functions f on G which satisfy f(ag) = η^{−1}(a) f(g) for all a in Z and whose absolute values are integrable on Z \ G. If f_{1} and f_{2} belong to L^{1}(η) so does their product f_{1} ∗ f_{2} which is defined by

f_{1} ∗ f_{2}(g) = ∫_{Z\G} f_{1}(gh^{−1}) f_{2}(h) dh

If f belongs to L^{1}(η) let f^{∗} be the function f^{∗}(g) = f̄(g^{−1}). It also belongs to L^{1}(η). A subalgebra B of L^{1}(η) will be called ample if it is dense and closed under the operation f → f^{∗}.

Let π be a unitary representation of G on the Hilbert space H such that π(a) = η(a) I for all a in Z. We do not suppose that π is irreducible. If f belongs to L^{1}(η) we set

π(f) ∫_{Z\G} f(g) π(g) dg

If π(f) is compact for all f in some ample subalgebra B then π decomposes into the direct sum of irreducible representations no one of which occurs more than a finite number of times.

Lemma 16.1.1 Suppose π_{1} and π_{2} are two unitary representations of G such that π_{1}(a) = η(a) I and π_{2}(a) = η(a) I for all a in Z. Suppose there is an ample subalgebra B of L^{1}(η) such that π_{1}(f) and π_{2}(f) are of Hilbert–Schmidt class for all f in B.

(i) If for every f in B

trace π_{1}(f) π_{1}(f^{∗}) ≥ trace π_{2}(f) π_{2}(f^{∗})

then π_{2} is equivalent to a subrepresentation of π_{1}. (ii) If for every f in B

trace π_{1}(f) π_{1}(f^{∗}) = trace π_{2}(f) π_{2}(f^{∗})

then π_{2} is equivalent to π_{1}.

Let π_{1} act on H_{1} and let π_{2} act on H_{2}. A simple application of Zorn’s lemma shows that we can choose a pair of closed invariant subspaces M_{1} and M_{2}, of H_{1} and H_{2} respectively, such that Chapter 3 268 the restrictions of π_{1} to M_{1} and π_{2} to M_{2} are equivalent and such that the pair M_{1}, M_{2} is maximal with respect to this property. Replacing H_{1} and H_{2} by the orthogonal complements of M_{1} and M_{2} we may suppose that M_{1} = 0 and that M_{2} = 0. To prove the first assertion of the lemma we have to show that with this assumption H_{2} = 0. If the second condition is satisfied we can reverse the roles of π_{1} and π_{2} to see that H_{1} is also 0.

Before beginning the proof we make a simple remark. Suppose σ is an irreducible unitary representation of G on L and σ_{α}, α ∈ A, is an irreducible unitary representation of G on L_{α}. Suppose that σ(a) = η(a) I for all a in Z and σ_{α}(a) = η(a) I for all a in Z and all α in A. Suppose that σ is equivalent to none of the σ_{α} and that a non-zero vector x in L and vectors x_{α} in L_{α} are given. Finally suppose that

∑_{α} ∥σ_{α}(f) x_{α}∥^{2}

is finite for every f in B. Then if ε is any positive number there is an f in B such that

∑_{α} ∥σ_{α}(f)x_{α}∥^{2} < ε∥σ(f)x∥^{2}.

Suppose the contrary and let L′ be the closure in ⊕_{α}L_{α} of

{⊗σ_{α}(f) x_{α} | f ∈ B}

L′ is invariant under G and the map

⊕σ_{α}(f)x_{α} → σ(f)x

may be extended to a continuous G-invariant map A′ of L′ into L. If A′ were 0 then σ(f)x = 0 for all f in B which is impossible. Let A be the linear transformation from ⊕L_{α} to L which is A′ on L′ and 0 on its orthogonal complement. A commutes with G and is not 0. Let A_{α} be the restriction of A to L_{α}. A_{α} is a G-invariant map of L_{α} into L and is therefore 0. Thus A is 0. This is a contradiction.

Suppose H_{2} is not 0. There is an h in B such that π_{1}(h) ≠ 0. If f = h∗h^{∗} then π_{2}(f) is positive semi-definite and of trace class. It has a positive eigenvalue and with no loss of generality we may suppose that its largest eigenvalue is 1. Let π_{2} = ⊗π_{2}^{β}, where π_{2}^{β} acts on H_{2}^{β}, be a decomposition of π_{1} into irreducible representations. There is a β_{0} and a unit vector x in H_{2}^{β_{0}} such that π_{2}(f)x = x. Let π_{1} = ⊕π_{1}^{α}, where π_{1}^{α} acts on H_{1}^{α}, be a decomposition of π_{1} into irreducible representations. Choose an orthogonal basis {x^{α,γ} | γ ∈ Γ_{α}} of H_{1}^{α} consisting of eigenvectors of π_{1}(f). Since

trace π_{1}(f) ≥ trace π_{2}(f)

the largest eigenvalue of π_{1}(f) is positive. Let it be λ.

If f_{1} belongs to B,

∑_{α} ∑_{γ} ∥π_{1}^{α}(f_{1}) x^{α,γ}∥^{2}

is the Hilbert–Schmidt norm of π_{1}(f_{1}) and is therefore finite. By assumption π_{2}^{β_{0}} is not equivalent to any of the representations π_{1}^{α} so that we can apply our earlier remark to the vector x and the Chapter 3 269 family of representations π_{1}^{α,γ} = π_{1}^{α} together with the family of vectors x^{α,γ} to infer the existence of an f_{1} in B such that

∑_{α} ∑_{γ} ∥π_{1}(f_{1}) x^{α,γ}∥^{2} < \frac{1}{2λ} ∥π_{2}(f_{1}) x∥^{2}.

Then

trace π_{1}(f_{1}f) π_{1}^{∗}(f_{1}f) = trace π_{1}^{∗}(f_{1}f) π_{1}(f_{1}f)

is equal to

∑_{α,γ} ∥π_{1}(f_{1}) π_{1}(f) x^{α,γ}∥^{2} ≤ λ ∑_{α,γ} ∥π_{1}(f_{1}) x^{α,γ}∥^{2}.

The right side is less than

\frac{1}{2}∥π_{2}(f_{1}) x∥^{2} = \frac{1}{2}∥π_{2}(f_{1}f) x∥^{2}

which is at most

\frac{1}{2} trace π_{2}(f_{1}f) π_{2}^{∗}(f_{1}f).

This is a contradiction.

The next lemma is a consequence of the results of [35].

Lemma 16.1.2 Suppose η is trivial so that L^{1}(η) = L^{1}(Z \ G). Suppose that B is an ample subalgebra of L^{1}(η) which is contained in L^{2}(Z \ G). If there is a positive constant γ and a unitary representation π of Z \ G such that π(f) is of Hilbert–Schmidt class for all f in B and

trace π(f) π(f^{∗}) = γ ∫_{Z\G} |f(g)|^{2} dg

then Z \ G is compact.

In proving the theorem it is better to deal with representations in the adèle groups than to deal with representations of the global Hecke algebras. We have to assume that the reader is sufficiently well acquainted with the theory of group representations to pass back and forth unaided between the two viewpoints.

If F is a global field, 𝔸 is the adèle ring of F, G = GL(2), and η is a character of the idèle class group F^{×} \ I the space A(η) of all measurable functions φ on G_{F} \ G_{𝔸} that satisfy

φ (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η(a) φ(g)

for all a in I and whose absolute values are square-integrable on G_{F} Z_{𝔸} \ G_{𝔸} is a Hilbert space. If φ belongs to this space

∫_{N_{F} \N_{𝔸}} φ(ng) dn

is defined for almost all g. If it is 0 for almost all g the function φ is said to be a cusp form. The space A_{0}(η) of all such cusp forms is closed and invariant under G_{𝔸}. It is in fact the closure of 𝒜_{0}(η). It decomposes in the same way but now into a direct sum of closed orthogonal subspaces V on which G_{𝔸} acts according to an irreducible representation π = ⊗π_{v}. Thus V is now isomorphic Chapter 3 270 to a tensor product of Hilbert spaces. Of course the same representations occur now as occurred before. Similar remarks apply to the multiplicative group G′ of quaternion algebra M′ over F.

It will be enough to prove the theorem when π is a constituent of some 𝒜_{0}(η) or A_{0}(η) and η is a character because we can always take the tensor product of π with a suitable quasi-character. Suppose η is given. Let S be the set of places at which M′ does not split. Suppose that for each v in S we are given an irreducible unitary representation σ′_{v} of G′_{F_{v}} = G′_{v} such that

σ′_{v}(a) = η_{v}(a) I

for all a in F_{v}^{∗} which we identify with Z′_{v} = Z′_{F_{v}}. Let σ_{v} = π(σ′_{v}) be the representation of G_{v} corresponding to σ′_{v}. We may take σ_{v} unitary. Let σ_{v} act on U_{v} and let σ′_{v} act on U′_{v}. Fix a unit vector u′_{v} in U_{v} and a unit vector u_{v} in U_{v} which is K_{v}-finite. u′_{v} is automatically K′_{v}-finite.

Write A_{0}(η) as the direct sum, in the Hilbert space sense, of mutually orthogonal invariant irreducible subspaces V^{1}, V^{2}, · · ·. Let the factorization of the representation π^{i} on V^{i} be ⊗π_{v}^{i}. Let π_{v}^{i} act on V_{v}^{i}. For simplicity of notation we identify V^{i} with ⊗V_{v}^{i}. We also suppose that if v is in S and π_{v}^{i} is equivalent to σ_{v} then U_{v} = V_{v}^{i} and π_{v}^{i} = σ_{v}. Let X be the set of all i such that π_{v}^{i} = σ_{v} for all v in S and if i belongs to X let

M^{i} = {⊗_{v∈S}u_{v}} ⊗ {⊗_{v∈S}V_{v}^{i}}.

M^{i} is invariant and irreducible under the action of

Ĝ_{S} = {g = (g_{v}) | g_{v} = 1 for all v in S}.

Let

M = ⊕_{i∈X}M^{i}.

M is a Hilbert space and Ĝ_{S} acts on M. If at least one of the representations σ′_{v}, v ∈ S, is not onedimensional set N = M. If they are all one-dimensional let N be the subspace of A_{0}(η) spanned, in the Hilbert space sense, by M and the functions g → χ(detg) where χ is a character of F^{×} \ I such that χ^{2} = η and σ′_{v}(g) = χ_{v}(ν(g)) for all g in G′_{v} if v is in S. If v is non-archimedean this last condition determines χ_{v} uniquely. If v is real it only determines it on the positive numbers.

Let A′(η) be the space of all measurable functions φ on G′_{F} \G′_{𝔸} that satisfy φ(ag) = η(a) φ(g) for all a in I and whose absolute values are square integrable on G′_{F} Z′_{𝔸} \ G′_{𝔸}. Replacing σ_{v} by σ′_{v} and u_{v} by u′_{v} we define N′ in the same way as we defined M. If at least one of the representations σ′_{v}, v ∈ S, is not one-dimensional we set M′ = N′. However if they are all one-dimensional and χ is a character of F^{×} \ I such that χ^{2} = η and σ′_{v}(g) = χ_{v}(ν(g)) for all G in G′_{V} if v is in S then the function g → χ(ν(g)) belongs to N′. We let M′ be the orthogonal complement in N′of the set of such functions. The group Ĝ′_{S} acts on M′ and N′. However by means of the local isomorphisms θ_{v} we can define an isomorphism of Ĝ_{S} and Ĝ′_{S}. Thus Ĝ_{S} acts on M and M′. To prove the theorem we need only show that the representations on these two spaces are equivalent. To do this we combine Lemma 16.1.1 with the Selberg trace formula.

To apply Lemma 16.1.1 we have to introduce an algebra B. B will be the linear span of B_{0}, the set of functions f on Ĝ_{S} of the form

f(g) = ∏_{v∉S} f_{v}(g_{v})

Chapter 3 271 where the functions f_{v} satisfy the following conditions.

(i) If a_{v} belongs to F_{v}^{×} then

f_{v}(a_{v}g_{v}) = η_{v}^{−1}(a_{v}) f_{v}(g_{v}).

(ii) f_{v} is K_{v}-finite on both sides and the projection of the support of f_{v} on Z_{v} \ G_{v} is compact. (iii) If v is archimedean f_{v} is infinitely differentiable. (iv) If v is non-archimedean f_{v} is locally constant.

(v) For almost all non-archimedean v the functions f_{v} is 0 outside of Z_{v}K_{v} but on Z_{v}K_{v} is given

by

f_{v}(g) = ω_{v}^{−1}(detg)

where ω_{v} is unramified and satisfies ω_{v}^{2} = η_{v}.

We introduce B′ in the same way. We may identify B and B′ and to verify the conditions of the lemma we need only show that if f = f_{1} ∗ f_{2} with f_{1} and f_{2} in B_{0} then

trace σ(f) = trace σ′(f)

if σ is the representation on M and σ′ that on M′. Let τ be the representation on N and τ′ that on N′. Since

trace τ(f) = trace σ(f) + ∑ ∫_{_{Z_{S}}\_{G_{S}}} χ(g) f(g) dg

and

trace τ′(f) = trace σ′(f) + ∑ ∫_{_{Z_{S}}\_{G_{S}}} χ(g) f(g) dg

we need only show that

trace τ(f) = trace τ′(f).

Before beginning the proof we had better describe the relation between the Haar measures on the groups Z_{𝔸} \ G_{𝔸} and Z′_{𝔸} \ G′_{𝔸}. Choose a non-trivial character ψ of F \ 𝔸. If ω_{0} is any invariant form of maximal degree on Z \ G defined over F and therefore over each F_{v} we can associate to ω_{0} and ψ_{v} a Haar measure ω_{0}(v) on Z_{v} \ G_{v}. Then ∏_{v∉S} ω_{0}(v) determines a Haar measure ω_{0} on Ẑ_{S} \ Ĝ_{S} and ∏_{v} ω_{0}(v) determines a Haar measure ω_{0} on Z_{𝔸} \ G_{𝔸}. The measure on Z_{𝔸} \ G_{𝔸} is independent of ψ and is called the Tamagawa measure. As in the previous paragraph we can associate to ω_{0}(v) a measure ω′_{0}(v) on Z′_{v} \ G′_{v} and therefore to ω_{0} a measure ω′_{0} on Ẑ′_{S} \ Ĝ′_{S} or Z′_{𝔸} \ G′_{𝔸}.

We first take f = f_{1} ∗ f_{2} in B′ and find a formula for trace τ′(f). Let d(σ′_{v}) be the formal degree of σ′_{v} with respect to the measure ω′_{0}(v) and let ξ′_{v} be the function

ξ′_{v}(g) = d(σ′_{v}) \overline{(σ′_{v}(g)u′_{v}, u′_{v})}

on G′_{v}. Let Φ′ = Φ′_{f} be the function

Φ′(g) = { ∏_{v∈S} ξ′_{v}(g_{v})} f(ĝ_{S})

on G′_{𝔸}. Here ĝ_{S} is the projection of g on Ĝ′_{S}. If ρ′ is the representation of G′_{𝔸} on A′(η) the restriction of ρ′(Φ′) to N′ is τ′(f) and ρ′(Φ) annihilates the orthogonal complement of N′. Thus

trace ρ′(Φ′) = trace τ′(f).

If φ is in A′(η) then ρ′(Φ′) φ(g) is equal to

∫_{Z′_{𝔸}\G′_{𝔸}} φ(gh) Φ′(h) ω′_{0}(h) = ∫_{Z′_{A}\G′_{A}} φ(h) Φ′(g^{−1}h) ω′_{0}(h).

The integration on the right can be performed by first summing over Z′_{F} \ G′_{F} and then integrating over Z_{𝔸}G′_{F} \ G′_{𝔸}. If

Φ′(g, h) = ∑_{Z′_{F} \G′_{F}} Φ′(g^{−1}γh)

the result is

∫_{Z′_{𝔸}G′_{F} \G′_{𝔸}} φ(h) Φ′(g, h) ω′_{0}(h).

Thus the trace of ρ′(Φ) is equal to

∫_{Z′_{A}G′_{F} \G′_{A}} Φ′(g, g) dg.

If we write out the integrand and perform the usual manipulations (cf [29]) we see that this integral is

∑_{{γ}} measure (Z′_{𝔸} G′_{F} (γ) \ G′_{𝔸}(γ)) ∫_{G′_{A}(γ)\G′_{A}} Φ′(g^{−1}γg).  (16.1.3)

The sum is over a set of representatives of the conjugacy classes in G′_{F}. G′_{𝔸}(γ) is the centralizer of γ in G′_{𝔸} and G′_{F} (γ) is its centralizer in G′_{F}.

Let Q′ be a set of representatives for the equivalence classes of quadratic extensions E of F such that E ⊗_{F} F_{v} is a field for all v in S. For each E in Q′ fix an imbedding of E in the quaternion algebra M′. Let B_{F} = B_{F} (E) be the multiplicative group of E, considered as a subalgebra of M′, or what is the same the centralizer of E in G′_{F}. Let B_{𝔸} = B_{𝔸}(E) be the centralizer of E in G′_{𝔸}. Let Q′_{1} be the separable extensions in Q′ and Q′_{2} the inseparable ones if they exist. Then (16.1.3) is the sum of

measure(Z′_{𝔸} G′_{F} \ G′_{𝔸}) Φ′(e),  (16.1.4)

if e is the identity,

\frac{1}{2} ∑_{Q′_{1}} ∑_{{γ∈Z{′ \atop F} \B_{F} \atop γ∉Z{′ \atop F}}} measure(Z′_{𝔸} B_{F} \ B_{𝔸}) ∫_{B_{𝔸}\G_{𝔸}} Φ′(g^{−1}γg) ω_{B}(g)  (16.1.5)

and

∑_{Q′_{2}} ∑_{{γ∈Z{′ \atop F} \B_{F} \atop γ∉Z_{F}}} measure(Z′_{𝔸} B_{F} \ B_{𝔸}) ∫_{B_{𝔸}\G_{𝔸}} Φ′(g^{−1}γg) ω_{B}(g).  (16.1.6)

The last sum is deceptive because Q′_{2} has at most one element. ω_{B} is the quotient of the measure on Z′_{𝔸} \ G′_{𝔸} by that on Z′_{𝔸} \ B_{𝔸}. The choice of the measure on Z′_{𝔸} \ B_{𝔸} is not too important. We do suppose that it is a product measure.*

* In (16.1.5) the factor \frac{1}{2} is not quite correct. If we want to leave it in, both γ and its conjugate must be counted, even if they differ only by an element of F.

The expression (16.1.4) is equal to

measure(Z′_{𝔸} G′_{F} \ G′_{𝔸}){ ∏_{v∈S} d(σ′_{v})} f(e).

The integrals of (16.1.5) and (16.1.6) are equal to the product

∏_{v∈S} \frac{χ_{σ′_{v}}(γ^{−1})}{measure(Z′_{v} \ B_{v})}

and

∫_{_{B_{S}}\_{G′_{S}}} f(g^{−1}γg) ω_{B}.

Now regard f = f_{1} ∗ f_{2} as an element of B. We can still introduce for each v in S the function

ξ_{v}(g) = d(σ_{v})\overline{(σ_{v}(g)u_{v}u_{v})}

on G_{v}. d(σ_{v}) is the formal degree of σ_{v} with respect to the measure ω_{0}(v). If σ′_{v} is not onedimensional ξ_{v} is integrable and we can use it to define a function Φ to which we can hope to apply the trace formula. When σ′_{v} is one-dimensional the function ξ_{v} is not even integrable so it is of no use to us. However in this case we can find an integrable function ζ_{v} with the following properties.

(i) For all a in F_{v}

ζ_{v} (( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) g) = η_{v}^{−1}(a) ζ_{v}(g).

(ii) For a suitable choice of u_{v} the operator σ_{v}(ζ_{v}) is the orthogonal projection on the space ℂu_{v}. (iii) If χ_{v} is a character of F_{v}^{×} such that χ_{v}^{2} = η_{v} then

∫_{Z_{v}\G_{v}} χ_{v}(detg) ζ_{v}(g) ω_{0}(v)

is −1 if σ′_{v}(h) = χ_{v}(ν(h)) for all h in G′_{v} and is 0 otherwise. (iv) If π_{v} is a unitary infinite-dimensional irreducible admissible representation of G_{v} which is not

equivalent to σ_{v} but satisfies

π_{v} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η_{v}(a) I

for all a in F_{v}^{×} then

trace π_{v}(ζ_{v}) = 0.

If v is real we cannot describe ζ_{v} without a great deal more explanation than is desirable at present. However after a few preliminary remarks we will be able to describe it when v is non-archimedean.

Suppose σ′_{v}(g) = χ_{v}(ν(g)) for g in G′_{v} and π_{v} is a representation of G_{v} such that

π_{v} \begin{aligned}( \\ ( & \begin{matrix}a & 0 \\ 0 & a \end{matrix} & ) \\ & & ) \end{aligned} = η_{v}(a) I

Chapter 3 274 for all a in F_{v}^{×}. Applying Lemma 3.9 to χ_{v}^{−1} ⊗ π_{v} we see that the restriction of π_{v} to K_{v} contains the representation k → χ_{v}(detk) if and only if π_{v} = π(μ_{v}, ν_{v}), μ_{v}ν_{v} = η_{v}, and the restrictions of μ_{v} and ν_{v} to U_{v}, the group of units of F_{v}, are both equal to the restriction of χ_{v}. Let ζ′_{v} be the function on G_{v} which is 0 outside of Z_{v}K_{v} but on K_{v} is equal to

\frac{1}{measure(Z_{v} \ Z_{v} K_{v})} χ_{v}^{−1}(detg).

Let H_{v} be the group generated by Z_{v}, the matrices

( \begin{matrix}a & b \\ c & d \end{matrix} )

in K_{v} for which c ≡ 0(mod 𝔭_{v}), and

( \begin{matrix}0 & 1 \\ ϖ_{v} & 0 \end{matrix} ) .

Let ω_{v} be the character ω_{v}(a) = (−1)^{n} χ_{v}(a) if |a| = |ϖ_{v}|^{n}. According to the concluding lemmas of the previous paragraph there is a non-zero vector u in the space of π_{v} such that

π_{v}(g) u = ω_{v}(detg) u

for all g in H_{v} if and only if π_{v} is equivalent to σ_{v}, π_{v} = π(μ_{v}, ν_{v}) is infinite-dimensional, μ_{v}ν_{v} = η_{v}, and the restrictions of μ_{v} and ν_{v} to U_{v} are equal to the restriction of χ_{v}, or π_{v} is the one-dimensional representation

g → ω_{v}(detg).

Let ζ′′_{v} be the function which is 0 outside of H_{v} and equal to

\frac{1}{measure Z_{v} \ H_{v}} ω_{v}^{−1}(detg)

on H_{v}. We may take

ζ_{v} = ζ′′_{v} − ζ′_{v}.

There are some consequences of the four conditions on ζ_{v} which we shall need. If μ_{v} and ν_{v} are two characters of F_{v}^{×} such that μ_{v}ν_{v} = η_{v}, the trace of ρ(ζ_{v}, μ_{v}, ν_{v}) is a multiple of

∫_{Z_{v}\A_{v}} μ_{v}(α) ν_{v}(β) |\frac{α}{β}|^{\frac{1}{2}} \{ ∫_{N_{v}} ∫_{K_{v}} ζ_{v}(k^{−1}ank) dn dk\} da

if

a = ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) .

Since this is 0 for all possible choice of μ_{v} and ν_{v}

∫_{N_{v}} ∫_{K_{v}} ζ_{v}(k^{−1}ank) dk dn = 0

for all a. We also observe that if σ′_{v} is not one-dimensional then

∫_{N_{v}} ∫_{K_{v}} ξ_{v}(k^{−1}ank) dk dn = 0

for all a.

If π_{v} is special or absolutely cuspidal trace π_{v}(ζ_{v}) is therefore equal to

\frac{1}{2} ∑_{S′} ∫_{Z_{v}\B_{v}} \{ ∫_{B_{v}\G_{v}} ζ_{v}(g^{−1}bg) ω_{b}(v)\} χ_{π_{v}}(b) δ(b)μ_{B}^{0}(b).

Since trace π_{v}(ζ_{v}) is 1 if π_{v} is equivalent to σ_{v} and 0 otherwise the orthogonality relations imply that

∫_{B_{v}\G_{v}} ζ_{v}(g^{−1}bg) ω_{B}(v) = \frac{−1}{measure Z_{v} \ B_{v}} χ_{σ_{v}}(b^{−1})

for all regular b and therefore, by continuity, for all b whose eigenvalues do not lie in F_{v}. It probably also follows from the Plancherel theorem that ζ_{v}(e) = d(σ_{v}). We do not need this but we shall eventually need to know that ζ_{v}(e) = d(σ′_{v}). For the moment we content ourselves with observing that if ω_{v} is a character of F_{v}^{×} and σ′_{v} is replaced by ω_{v} ⊗ σ′_{v} the formal degree does not change and ζ_{v} is replaced by the function g → ω_{v}^{−1}(detg) ζ_{v}(g) so that ζ_{v}(e) does not change. Thus the relation ζ_{v}(e) = d(σ′_{v}) need only be proved when σ′_{v} is trivial.

Let S_{1} be the subset of v in S for which σ′_{v} is one-dimensional and let S_{2} be the complement of S_{1} in S. Given f = f_{1} ∗ f_{2} in B we set

Φ(g) = { ∏_{v∈S_{1}} ζ_{v}(g_{v})}{ ∏_{v∈S_{2}} ξ_{v}(g_{v})} f(ĝ_{S}).

Let ρ_{0}^{+} be the representation of G_{𝔸} on A_{0}^{+}(η) the sum, in the Hilbert space sense, of A_{0}(η) and the functions χ : g → χ(detg) where χ is a character of F^{×} \ I such that χ^{2} = η and let ρ be the representation on A(η). If at least one of the representations ρ′_{v} is not one-dimensional ρ_{0}^{+}(Φ) annihilates the orthogonal complement of A_{0}(η). If they are all one-dimensional we apply the third condition on the functions ζ_{v} together with the fact that the number of places in S is even to see that ρ_{0}^{+}(Φ) χ = 0 unless σ′_{v}(h) = χ_{v}(ν(h)) for all h in G′_{v} and all v in S but that if this is so

ρ_{0}^{+}(Φ) χ = τ(f) χ.

Recall that A_{0}(η) is the direct sum of spaces V^{i} on which G_{𝔸} acts according to representations π^{i} = ⊗π_{v}^{i}. If at least one of the representations σ′_{v} is not one-dimensional ρ_{0}^{+}(Φ) is equal to σ(f) on M and annihilates the orthogonal complement of M in A_{0}(η). Suppose they are all one-dimensional. If i belongs to X the restrictions of ρ_{0}^{+}(Φ) and σ(f) or τ(f) to M^{i} are equal and ρ_{0}^{+}(Φ) annihilates the orthogonal complement of M^{i} in V^{i}. If i is not in X the trace of the restriction of ρ_{0}^{+}(Φ) to V^{i} is

{ ∏_{v∈S} trace π_{v}^{i}(ζ_{v})}{trace π̂_{S}^{i}(f)}

in π̂_{S}^{i} = ⊗_{v∉S}π_{v}^{i}. Since π_{v}^{i}, v ∈ S, are all infinite-dimensional and for at least one such v the representation π_{v}^{i} is not equivalent to σ_{v},

∏_{v∈S} trace π_{v}^{i}(ζ_{v}) = 0.

We conclude that

trace ρ_{0}^{+}(Φ) = trace τ(f).

To show that trace τ(f) = trace τ′(f)

we have to apply the trace formula to find a suitable expression for trace ρ_{0}^{+}(Φ). In order to describe the formula we need to state some results in the theory of Eisenstein series.

Consider the collection of pairs of character μ, ν of F^{×} \ I such that μν = η. To such pairs, μ, ν and μ′, ν′ are said to be equivalent if there is a complex number r such that μ′ = μα_{F}^{r} and ν′ = να_{F}^{−r}. If a belongs to I then α_{F}^{r} (a) = |a|^{r}. Let P be a set of representatives for these equivalence classes.

Suppose (μ, ν) belongs to P. If s is a complex number the space ℬ(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) of functions on N_{𝔸} \ G_{𝔸} is defined as in the tenth paragraph. Since the functions in this space are determined by their restrictions to K we may think of it as a space of functions on K in which case it is independent of s. Thus we have isomorphisms

T_{s} : ℬ(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) → ℬ(μ, ν).

The theory of Eisenstein series provides us with a function (φ, s) → E(φ, s) from ℬ(μ, ν) × ℂ to 𝒜(η). E(g, φ, s) is the value of E(φ, s) at g. For a given φ the function E(g, φ, s) is continuous in g and meromorphic in s. Moreover there is a discrete set of points in ℂ such that outside of this set it is holomorphic in s for all g and φ. If s is not in this set the map φ → E(T_{s}φ, s) of ℬ(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) into 𝒜(η) commutes with the action of ℋ.

If the total measure of N_{F} \ N_{𝔸} is taken to be 1 the integral

∫_{N_{F} \N_{𝔸}} E(ng, T_{s}φ, s) dn

is equal to

φ(g) + (M(s)φ)(g).

M(s) is a linear transformation from ℬ(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) to ℬ(να_{F}^{− \frac{s}{2}} , μαF^{\frac{s}{2}} ) which commutes with the action of ℋ. It is mermorphic in the sense that

⟨M(s)T_{s}^{−1}φ_{1}, T_{s}^{−1}φ_{2}⟩

is meromorphic if φ_{1} belongs to ℬ(μ, ν) and φ_{2} belongs to ℬ(ν^{−1}, μ^{−1}). The quotient of M(s) by

\frac{L(1 − s, νμ^{−1})}{L(1 + s, μν^{−1})} = ε(1 − s, νμ^{−1}) \frac{L(s, μν^{−1})}{L(1 + s, μν^{−1})}

Chapter 3 277 is holomorphic for Re s ≥ 0. Since the analytic behavior of E(g, φ, s) is controlled by that of M(s) it should be possible, as we observed before, to use the Eisenstein series to show that a constituent of ℬ(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) is also a constituent of 𝒜(η).

To indicate the dependence of M(s) on μ and ν we write M(μ, ν, s). Then

M(μ, ν, s)M(ν, μ, −s) = I.

If s is purely imaginary we can introduce the inner product

(φ_{1}, φ_{2}) = ∫_{K} φ_{1}(k) φ̄_{s}(k) dk

on ℬ(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ). Let B(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) be its completion with respect to this inner product. B(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) may be thought of as a function space on G_{𝔸} on which G_{𝔸} acts by right translations. The representation of G_{𝔸} on B(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) is unitary. Let g correspond to the operator ρ(g, μ, ν, s) and if f is in L^{1}(η) let

ρ(f, μ, ν, s) = ∫_{Z_{𝔸}\G_{𝔸}} f(g) ρ(g, μ, ν, s) ω_{0}(g)

T_{s} extends to an isometry, from B(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) to B(μ, ν) and M(μ, ν, s) extends to an isometry

from B(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) to B(να_{F}^{− \frac{s}{2}} , μαF^{\frac{s}{2}} ). In particular

M^{∗}(μ, ν, s) = M(ν, μ, −s).

Suppose (μ, ν) is in P and, for some r, ν = μα_{F}^{r} and μ = να_{F}^{−r}. Replacing μ by μαF^{\frac{r}{2}} and ν by να_{F}^{− \frac{r}{2}} if necessary we may suppose that μ = ν. We may also suppose that if (μ, ν) is in P and is not equivalent to (ν, μ) then (ν, μ) is also in P. Let L be the Hilbert space sum

⊕_{P} B(μ, ν)

and let ℒ be the algebraic sum

⊕_{P} ℬ(μ, ν).

If we define L(s) to be

⊕_{p}B(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} )

and ℒ(s) to be

⊕_{P} ℬ(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} )

we can introduce the map

T_{s} : L(s) → L.

The representation g → ρ(g, s) is the representation

g → ⊕ρ(g, μ, ν, s)

on L(s). M(s) will be the operator on L(s) which takes ⊕φ(μ, ν) to ⊕φ_{1}(μ, ν) with

φ_{1}(ν, μ) = M(μ, ν, s) φ(μ, ν).

It is unitary.

If F has characteristic O let H be the space of all square integrable functions φ from the imaginary axis to L such that

T_{−s}^{−1}φ(−s) = M(s) T_{s}^{−1}φ(s)

with the norm

\frac{c}{π} ∫_{−i∞}^{i∞} ∥φ(s)∥^{2} d|s|

c is a positive constant relating various Haar measures. It will be defined more precisely later. If F is a function field with field of constants 𝔽_{q} the functions in H are to be periodic of period \frac{log q}{2π} i and the norm is to be

\frac{c log u}{π} ∫0^{\frac{2π}{log q}} ∥φ(s)∥^{2} d|s|.

On the whole we shall proceed as though F had characteristic O merely remarking from time to time the changes to be made when the characteristic is positive.

If φ = ⊕φ(μ, ν) is in ℒ we set

E(g, φ, s) = ∑ E(g, φ(μ, ν), s).

If φ in H takes values in ℒ

lim_{T →∞} \frac{1}{2π} ∫_{−iT}^{iT} E(g, φ(s), s) d|s| = φ̃(g)

exists in A(η). The map φ → φ̃ extends to an isometry of H with a subspace A_{1}(η) of A(η). If g is in G_{𝔸} and φ′ is defined by

φ′(s) = T_{s} ρ(g, s) T_{s}^{−1}φ(s)

then φ̃′ is ρ(g)φ̃.

The orthogonal complement of A_{1}(η) is A_{0}^{+}(η). Thus if E is the orthogonal projection of A(η) on A_{1}(η) the trace of ρ_{0}^{+}(Φ) is the trace of ρ(Φ) − Eρ(Φ) which, according the the Selberg trace formula, is the sum of the following expressions which we first write out and then explain. (i)

measure(Z_{𝔸} G_{F} \ G_{𝔸}) Φ(e).

(ii)

\frac{1}{2} ∑_{Q_{1}} ∑_{{γ∈Z_{F} \B_{F} \atop γ∉Z_{F}}} measure(Z_{𝔸} B_{F} \ B_{𝔸}) ∫_{B_{𝔸}\G_{𝔸}} Φ(g^{−1}γg) ω_{B}(g).

(iii)

∑_{Q_{2}} ∑_{{γ∈Z_{F} \B_{F} \atop γ∉Z_{F}}} measure(Z_{𝔸} B_{F} \ B_{𝔸}) ∫_{B_{𝔸}\G_{𝔸}} Φ(g^{−1}γg) ω_{b}(g).

Chapter 3 279 (iv)

−c ∑_{{γ∈Z_{F} \A_{F} \atop γ∉Z_{F}}} ∑_{v} \{ ∏_{w≠v} ω(γ, f_{w})\} ω(γ, f_{v}).

(v)

c[λ_{0} ∏_{v} θ(0, f_{v}) + λ_{−1}{ ∑_{v} θ′(0, f_{v}) ∏_{w≠v} θ(0, f_{w})}].

(vi) If F is a number field

−\frac{1}{4} trace M(0) ρ(Φ, 0),

but

−\frac{log q}{4} \{trace M(0) ρ(Φ, 0) + trace M(\frac{π}{log q} ) ρ(Φ, \frac{π}{log q} )\}

if F is a function field. (vii) If F is a number field

\frac{1}{4π} ∫_{−i∞}^{i∞} trace m^{−1}(s) m′(s) ρ(Φ, s) d|s|,

but

\frac{log q}{4π} ∫0^{\frac{2π}{log q}} trace m^{−1}(s) m′(s) ρ(Φ, s) d|s|

if F is a function field. (viii) The sum over (μ, ν) and v of

\frac{1}{4π} ∫_{−i∞}^{i∞} \{ tr R^{−1}(μ_{v}, ν_{v}, s) R′(μ_{v}, ν_{v}, s) ρ(f, μ_{v}, ν_{v}, s)\}\{ ∏_{w≠v} tr ρ(f_{w}, μ_{w}, ν_{w}, s)\} d|s|

if F is a number field and of

\frac{log q}{4π} ∫0^{\frac{2π}{log q}} \{ tr R^{−1}(μ_{v}, ν_{v}, s) R′(μ_{v}, ν_{v}, s) ρ(f, μ_{v}, ν_{v}, s)\}\{ ∏_{w≠v} tr ρ(f_{w}, μ_{w}, ν_{w}, s)\} d|s|

if F is a function field. The function Φ is of the form

Φ(g) = ∏_{v} f_{v}(g_{v}).

Let Q be a set of representatives for the equivalence clases of quadratic extensions of F. For each E in Q fix an imbedding of E in the matrix algebra M = M(2, F). Let B_{F} = B_{F} (E) be the multiplicative group of E, considered as a subalgebra of M. B_{F} is the centralizer of E in G_{F}. Let B_{𝔸} = B_{𝔸}(E) be the centralizer of E in G_{𝔸}. Q_{1} is the collection of separable extensions in Q and Q_{2} is the collection of inseparable extensions. A_{F} is the group of diagonal matrices in G_{F}.

Choose on N_{𝔸} that Haar measure which makes the measure of N_{F} \ N_{𝔸} equal to 1. Choose on K the normalized Haar measure. On the compact group H obtained by taking the quotient of

\{( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ∈ A_{𝔸} | |α| = |β|\}

by Z_{𝔸}A_{F} choose the normalized Haar measure. H is the kernel of the map

( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) → log |\frac{α}{β}|^{\frac{1}{2}}

of A_{F} Z_{𝔸} \ A_{𝔸} onto ℝ or log q ℤ. On ℝ one has the standard measure dx and on log q ℤ one has the standard measure which assigns the measure 1 to each point. The measures on H and on H \ (A_{F} Z_{𝔸} \ A_{𝔸}) together with the measure on Z_{𝔸} \ A_{F} Z_{A} which assigns the measure 1 to each point serve to define a measure da on Z_{𝔸} \ A_{𝔸}. The constant c is defined by demanding that

∫_{Z_{𝔸}\G_{𝔸}} f(g) ω_{0}(g)

be equal to

c ∫_{Z_{𝔸}\A_{𝔸}} ∫_{N_{𝔸}} ∫_{K} f(ank) da dn dk

if f is an integrable function on Z_{𝔸} \ G_{𝔸}. We may suppose that the measures on Z_{𝔸} \ A_{𝔸}, N_{𝔸}, and K are given as product measures and in particular that

∫_{K_{v}} dk_{v} = 1

and

∫_{N_{v}} χ(n_{v}) dn_{v} = 1

for almost all v if χ is the charcteristic function of

\{( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) | x ∈ O_{F_{v}}\} .

The factors ω(γ, f_{v}) and ω_{1}(γ, f_{v}) appearing in the fourth expression are defined by

ω(γ, f_{v}) = ∫_{N_{v}} ∫_{K_{v}} f_{v}(k_{v}^{−1}n_{v}^{−1}γn_{v}k_{v}) dn_{v} dk_{v}

and

ω_{1}(γ, f_{v}) = ∫_{N_{v}} ∫_{K_{v}} f_{v}(k_{v}^{−1}n_{v}^{−1}γn_{v}k_{v}) log λ(n_{v}) dn_{v} dk_{v}.

If

( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) n = ( \begin{matrix}α′ & 0 \\ 0 & β′ \end{matrix} ) n′k′

then

λ(n) = |\frac{α′}{β′} |.

Set θ(s, f_{v}) equal to

\frac{1}{L(1 + s, 1_{v})} ∫_{Z_{v}\A_{v}} ∫_{K_{v}} f_{v}(k_{v}^{−1}a_{v}^{−1}n_{0}a_{v}k_{v})|\frac{α_{v}}{β_{v}} |^{−1−s} da_{v} dk_{v}

Chapter 3 281 where

a_{v} ( \begin{matrix}α_{v} & 0 \\ 0 & β_{v} \end{matrix} )

and

n_{0} = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) .

1_{v} is the trivial character of F_{v}^{×}. Then θ(s, f_{v}) is analytic at least for Re s > −1. It derivative at 0 is θ′(0, f_{v}). If

L(1 + s, 1_{F}) = ∏_{v} L(1 + s, 1_{v})

the Laurent expansion of L(1 + s, 1_{F}) about s = 0 is

\frac{λ_{−1}}{s} + λ_{0} + · · · .

The operator m(s) is the operator on L(s) which for each (μ, ν) multiplies every element of

B(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) by

\frac{L(1 − s, νμ^{−1})}{L(1 + s, μν^{−1})}.

We may represent B(μαF_{v}^{\frac{s}{2}}, να_{F}^{− \frac{s}{2}} ) as

⊗_{v}B(μ_{v}αF_{v}^{\frac{s}{2}}, ν_{v}α_{F_{v}}^{− \frac{s}{2}} )

when s is purely imaginary. If Re s > 0 let R(μ_{v}, ν_{v}, s) be the operator from ℬ(μ_{v}αF_{v}^{\frac{s}{2}}, ν_{v}α_{F_{v}}^{− \frac{s}{2}} ) to

ℬ(ν_{v}α_{F_{v}}^{− \frac{s}{2}} μ_{v}αF_{v}^{\frac{s}{2}}) defined by setting

R(μ_{v}, ν_{v}, s) φ(g)

equal to

ε(1 − s, μ_{v}^{−1}ν_{v}, ψ_{v}) \frac{L(1 + s, μ_{v}ν_{v}^{−1})}{L(s, μ_{v}ν_{v}^{−1})} ∫_{N_{v}} φ (( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ng) dn.

These operators can be defined for s purely imaginary by analytic continuation. They are then scalar multiples of unitary operators and for a given μ, ν are in fact unitary for almost all v. Thus R(μ_{v}, ν_{v}, s) can be defined as an operator B(μ_{v}αF_{v}^{\frac{s}{2}}, ν_{v}α_{F_{v}}^{− \frac{s}{2}} ) when s is purely imaginary and

M(s) = ∑_{(μ,ν)} \{ ⊗_{v} R(μ_{v}, ν_{v}, s)\} \frac{L(1 − s, νμ^{−1})}{L(1 + s, μν^{−1})}.

Set

N(s) = T_{S}M(s) T_{S}^{−1}

and if N′(s) is the deriviative of N(s) set

M′(s) = T_{s}^{−1} N′(s) T_{s}

Define R′(μ_{v}, ν_{v}, s) in a similar fashion. Then

trace M^{−1}(s) M′(s) ρ(Φ, s)

is the sum of

trace m^{−1}(s) m′(s) ρ(Φ, s)

and

∑_{(μ,ν)} ∑_{v} { tr R^{−1}(μ_{v}, ν_{v}, s) R′(μ_{v}, ν_{v}, s) ρ(f_{v}, μ_{v}, ν_{v}, s)}{ ∏_{w≠v} tr ρ(f_{w}, μ_{w}, ν_{w}, s)},

where ρ(f_{v}, μ_{v}, ν_{v}, s) is the restriction of ρ(f_{v}) to B(μ_{v}αF_{v}^{\frac{s}{2}}, ν_{v}α_{F_{v}}^{− \frac{s}{2}} ).

If E(μ, ν, s) is the projection of L(s) on B(μαF^{\frac{s}{2}} , να_{F}^{− \frac{s}{2}} ) we can write
m(s) = ∑ a(μ, ν, s) E(μ, ν, s)

where the a(μ, ν, s) are scalars. Thus

trace m^{−1}(s) m′(s) ρ(Φ, s)

is equal to

∑ \frac{a′(μ, ν, s)}{a(μ, ν, s)}\{ ∏_{v} trace ρ(f_{v}, μ_{v}, ν_{v}, s)\}.

We can also write

M(0) = ∑ a(μ, ν) E(μ, ν, 0)

so that

trace M(0) ρ(Φ, 0)

is equal to

∑ a(μ, ν){ ∏_{v} trace ρ(f_{v}, μ_{v}, ν_{v}, 0)}.

If F is a function field

M(\frac{π}{log q} ) = ∑ B(μ, ν) E(μ, ν, \frac{π}{log q} ).

If

∫_{K_{v}} ∫_{N_{v}} f_{v}(k^{−1}ank) dn dk = 0  (16.1.7)

for all a in A_{v} = A_{F_{v}} then ω(γ, f_{v}) = 0 for all γ, θ(0, f_{v}) = 0, and

trace ρ(f_{v}, μ_{v}, ν_{v}, s) = 0

for all μ_{v}, ν_{v}, and s. In particular if (16.1.7) is satisifed for at least two v the expressions (iv) to (viii) vanish and the trace formula simplifies considerably.

We now apply this formula to the function

Φ(g) = { ∏_{v∈S_{1}} ζ_{v}(g_{v})}{ ∏_{v∈S_{2}} ξ_{v}(g_{v})} f(ĝ_{S})

where f = f_{1} ∗ f_{2} with f_{1} and f_{2} in B is of the form

f(ĝ_{S}) = ∏_{v∉S} f_{v}(g_{v}).

Since S has at least two elements and the functions ζ_{v} and ξ_{v} satisfy (16.1.7) only the expressions (i) to (iii) do not vanish identically. The expression (i) is now equal to

{ ∏_{v∈S_{1}} ζ_{v}(e)}{ ∏_{v∈S_{2}} d(σ_{v})} f(e)

We recall that d(σ_{v}) = d(σ′_{v}) if v is in S_{2}.

We may suppose that Q_{2} is equal to Q′_{2} and that Q′_{1} is a subset of Q_{1}. If E is in Q_{1} or Q_{2} and γ is in B_{F} = B_{F} (E) but not in Z_{F}

∫_{B_{𝔸}\G_{𝔸}} Φ(g^{−1}γg) ω_{B}(g)

is equal to the product of

{ ∏_{v∈S_{1}} ∫_{B_{v}\G_{v}} ζ_{v}(g_{v}^{−1}γg_{v}) ω_{B}(v)}{ ∏_{v∈S_{2}} ∫_{B_{v}\G_{v}} ξ_{v}(g_{v}^{−1}γg_{v}) ω_{B}(v)}

and

∫_{_{B_{S}}\_{G_{S}}} f(g^{−1}γg) ω_{B}.

If v is in S and E ⊗_{F} F_{v} is not a field so that B_{v} is conjugate to A_{v} the corresponding factor in the first of these two expressions vanishes. Thus the sum in (ii) need only be taken over Q′_{1}. If E is in Q′_{1} or Q_{2} the first of these two expressions is equal to

∏_{v∈S} \frac{χ_{σ_{v}}(γ^{−1})}{measure Z_{v} \ B_{v}} .

Thus, in the special case under consideration, (ii) is equal to (16.1.5) and (iii) is equal to (16.1.6) so that

trace τ(f) − { ∏_{v∈S_{1}} ζ_{v}(e)}{ ∏_{v∈S_{2}} d(σ)} measure(Z_{𝔸} G_{F} \ G_{𝔸}) f(e)

is equal to

trace τ′(f) − { ∏_{v∈S} d(σ′_{v})} measure(Z′_{𝔸} G′_{F} \ G′_{𝔸}) f(e).

We may take η to be trivial and apply Lemmas 16.1.1 and 16.1.2 to see that, in this case,

trace τ(f) = trace τ′(f)

Chapter 3 284 and

{ ∏_{v∈S} d(σ′_{v})} measure(Z′_{𝔸} G′_{F} \ G′_{𝔸})

is equal to

{ ∏_{v∈S_{1}} ζ_{v}(e)}{ ∏_{v∈S_{2}} d(σ_{v})} measure(Z_{𝔸} G_{F} \ G_{𝔸}).

Still taking η trivial we choose the σ′_{v} so that none of them are one-dimensional and conclude that

measure(Z′_{𝔸} G′_{F} \ G′_{𝔸}) = measure(Z_{𝔸} G_{F} \ G_{𝔸}).  (16.1.8)

Then we take exactly one of them to one-dimensional and conclude that ζ_{v}(e) = d(σ′_{v}). Thus ζ_{v}(e) = d(σ′_{v}) and

trace τ(f) = trace τ′(f)

in general.

The relation (16.1.8) is well-known. One can hope however that the proof of it just given can eventually be used to show that the Tamagawa numbers of two groups which differ only by an inner twisting are the same or at least differ only by an explicitly given factor. Since the method of [33] can probably be used to evaluate the Tamagawa numbers of quasi-split groups the problem of evaluating the Tamagawa numbers of reductive groups would then be solved. However a great deal of work on the representation theory of groups over local fields remains to be done before this suggestion can be carried out.

To complete our formal argument we need to sketch a proof of the trace formula itself. One must use a bootstrap method. The first step, which is all we shall discuss, is to prove it for some simple class of functions Φ. We take Φ of the form Φ = f′ ∗ f′′ with

f′(g) = ∏_{v} f′_{v}(g_{v})

and

f′′(g) = ∏_{v} f′′_{v}(g_{v})

where f′_{v} and f′′_{v} satisfy the five conditions on page 325. The function f_{v} is f′_{v} ∗ f′′_{v}.

Suppose φ is a K-finite compactly supported function in A(η). For each purely imaginary s define φ̃(s) in ℒ by demanding that

\frac{1}{2c} ∫_{G_{F} Z_{𝔸}\G_{𝔸}} φ(g) Ē(g, φ′, s) ω_{0}(g) = (φ̃(s), φ′)

be valid for all φ′ in ℒ. The map φ → φ̃(s) extends to a continuous map of A(η) onto H. φ̃(s) is the function in H corresponding to Eφ in A_{1}(η).

For each (μ, ν) in P choose an orthonormal basis {φ_{i}(μ, ν)} of ℬ(μ, ν). We may suppose that any elementary idemotent in ℋ annihilates all but finitely many elements of this basis. If

φ̃(s) = ∑_{(μ,ν)} ∑_{i} a_{i}(μ, ν, s) φ_{i}(μ, ν)

Chapter 3 285 then

a_{i}(μ, ν, s) = \frac{1}{2c} ∫_{G_{F} Z_{𝔸}\G_{𝔸}} φ(g) Ē(g, φ_{i}(μ, ν), s) ω_{0}(g).

Let

ρ(Φ, s) T_{S}^{−1} φ_{i}(μ, ν) = ∑_{j} ρ_{ji}(Φ, μ, ν, s) T_{S}^{−1} φ_{j}(μ, ν).

For all but finitely many μ, ν, i and j the functions ρ_{ji}(Φ, μ, ν, s) vanish identically. Eρ(Φ)φ is equal to

lim_{T →∞} ∑_{μ,ν} ∑_{i,j} ∫_{−iT}^{iT} ρ_{ij}(Φ, μ, ν, s) a_{j}(μ, ν, s) E(g, φ_{i}(μ, ν), s) d|s|.

A typical one of these integrals is equal to the integral over G_{F} Z_{𝔸} \ G_{𝔸} of the product of φ(g) and

∫_{−iT}^{iT} ρ_{ij}(Φ, μ, ν, s) E(g, φ_{i}(μ, ν), s) Ē(h, φ_{j}(μ, ν), s) d|s|.

Thus the kernel of E ρ(Φ) is the sum over (μ, ν) and i, j of

\frac{1}{4πc} ∫_{−i∞}^{i∞} ρ_{ij}(Φ, μ, ν, s) E(g, φ_{i}(μ, ν), s) Ē(h, φ_{j}(μ, ν), s) d|s|.

The kernel of ρ(Φ) is

Φ(g, h) = ∑_{Z_{F} \G_{F}} Φ(g^{−1}γh).

To compute the trace of ρ(Φ) − E ρ(Φ) we integrate the difference of these two kernels over the diagonal.

The function Φ(g, g) may be written as the sum of

∑_{δ∈P_{F} \G_{F}} ∑_{{γ∈N_{F} \atop γ≠e}} Φ(g^{−1}δ^{−1}γ δg),  (16.2.1)

where P_{F} is the group of super-triangular matrices in G_{F},

\frac{1}{2} ∑_{{γ∈Z_{F} \A_{F} \atop γ∉Z_{F}}} ∑_{δ∈A_{F} \G_{F}} Φ(g^{−1}δ^{−1}γδg),  (16.2.2)

where A_{F} is the group of diagonal matrices in G_{F},

\frac{1}{2} ∑_{Q_{1}} ∑_{{γ⊂Z_{F} \B_{F} \atop γ∉Z_{F}}} ∑_{δ∈B_{F} \G_{F}} Φ(g^{−1}δ^{−1}γδg)  (16.2.3)

and

∑_{Q_{2}} ∑_{{γ∈Z_{F} \B_{F} \atop γ∉Z_{F}}} ∑_{δ∈B_{F} \G_{F}} Φ(g^{−1}δ^{−1}γδg)  (16.2.4)

together with

Φ(e).  (16.2.5)

The constant Φ(e) can be integrated over G_{F} Z_{𝔸} \ G_{A} immediately to give the first term of the trace formula. The standard manipulations convert (16.2.3) and (16.2.4) into the second and third terms of the trace formula.

The expressions (16.2.1) and (16.2.2) have to be treated in a more subtle fashion. We can choose a constant e_{1} > 0 so that if

g = ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) k

with x in 𝔸, α and β in I such that |\frac{α}{β}| ≥ c_{1}, and k in K and if

γg = ( \begin{matrix}1 & x′ \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α′ & 0 \\ 0 & β′ \end{matrix} ) k′

with γ in G_{F}, x′ in 𝔸, α′ and β′ in I such that |\frac{α′}{β′} | ≥ c_{1}, and k′ in K then γ belongs to P_{F}. Let χ be the characteristic function of

\{( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) k | |\frac{α}{β}| ≥ c_{1}\} .

The expression (16.2.2) is the sum of

\frac{1}{2} ∑_{δ∈P_{F} \G_{F}} ∑_{{γ∈Z_{F} \P_{F} \atop γ∉Z_{F} N_{F}}} Φ(g^{−1}δ^{−1}γδg)(χ(δg) + χ(ε(γ)δg))

and

\frac{1}{2} ∑_{δ∈P_{F} \G_{F}} ∑_{{γ∈Z_{F} \P_{F} \atop γ∉Z_{F} N_{F}}} Φ(g^{−1}δ^{−1}γδg)(1 − χ(δg) − χ(ε(γ)δg)).

Here ε(γ) is any element of G_{F} not in P_{F} such that

ε(γ) γ ε^{−1}(γ) ∈ P_{F} .

There is always at least one such ε(γ). The integral of the second sum over G_{F} Z_{𝔸} \ G_{𝔸} converges. It is equal to

\frac{1}{2} ∫_{Z_{𝔸}P_{F} \G_{𝔸}} ∑_{{γ⊂Z_{F} \P_{F} \atop γ∉Z_{F} N_{F}}} Φ(g^{−1}γg)(1 − χ(g) − χ(ε(γ)g)) ω_{0}(g).

Every γ occurring in the sum can be written as δ^{−1}γ_{0}δ with γ_{0} in A_{F} and δ in P_{F}. Then

(δ^{−1}ε(γ_{0})δ)(δ^{−1}γ_{0}δ)(δ^{−1}ε(γ_{0})δ)^{−1} = δ^{−1}(ε(γ_{0})γ_{0}ε^{−1}(γ_{0}))δ

so that we can take ε(γ) = δ^{−1}ε(γ_{0})δ. We take

ε(γ_{0}) = w = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) .

Since χ(δg) = χ(g) and

χ(δ^{−1}wδg) = χ(wδg)

the integrand is

∑_{{γ∈Z_{F} \A_{F} \atop γ∉Z_{F}}} ∑_{δ∈A_{F} \P_{F}} Φ(g^{−1}δ^{−1}γδg)(1 − χ(δg) − χ(wδg)).

The integral itself is equal to

\frac{1}{2} ∑_{{γ∈Z_{F} \A_{F} \atop γ∉Z_{F}}} ∫_{Z_{𝔸}A_{F} \G_{𝔸}} Φ(g^{−1}γg)(1 − χ(g) − χ(wg)) ω_{0}(g).

All but a finite number of the integrals in this sum are 0.

It is convenient to write each of them in another form. If

g = ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) nk

then χ(g) is 1 if |\frac{α}{β}| ≥ c_{1} and is 0 if |\frac{α}{β}| < c_{1}. If

wn = ( \begin{matrix}α′ & 0 \\ 0 & β′ \end{matrix} ) n′k′

and λ(n) is |\frac{α′}{β′} | then χ(wg) is 1 if |\frac{α}{β}| ≤ \frac{λ(n)}{c_{1}} and is 0 if |\frac{α}{β}| > \frac{λ(n)}{c_{1}}. It is easily seen that λ(n) ≤ 1. Thus if c_{1} > 1, as we may suppose, one of χ(g) and χ(wg) is always 0. The integral

∫_{Z_{𝔸}A_{F} \G_{𝔸}} Φ(g^{−1}γg)(1 − χ(g) − χ(wg)) ω_{0}(g)

is equal to

c ∫_{N_{𝔸}} ∫_{K} Φ(k^{−1}n^{−1}γnk)(2 log c_{1} − log λ(n)) dn dk

which we write as the sum of

2c log c_{1} ∫_{N_{𝔸}} ∫_{K} Φ(k^{−1}n^{−1}γnk) dn dk  (16.2.6)

and

− ∑_{v} c ∫_{N_{𝔸}} ∫_{K} Φ(k^{−1}n^{−1}γnk) log λ(n_{v}) dn dk.

if we express each of the integrals in the second expression as a product of local integrals we obtain the fourth term of the trace formula. All but a finite number of the integrals are 0 so that the sum is really finite. We will return to (16.2.6) later. If F is a function field over 𝔽_{q} it is best to take c_{1} to be a power of q^{n} of q. Then 2 log c_{1} is replaced by 2n − 1.

The expression (16.2.1) is the sum of

∑_{δ∈P_{F} \G_{F}} ∑_{{γ∈N_{F} \atop γ≠e}} Φ(g^{−1}δ^{−1}γδg) χ(δg)

Chapter 3 288 and

∑_{δ∈P_{F} \G_{F}} ∑_{{γ∈N_{F} \atop γ≠e}} Φ(g^{−1}δ^{−1}γδg)(1 − χ(δg)).

The integral of the second expression over G_{F} Z_{𝔸} \ G_{𝔸} converges. It is equal to

∫_{P_{F} Z_{𝔸}\G_{𝔸}} ∑_{{γ∈N_{F} \atop γ≠e}} Φ(g^{−1}γg)(1 − χ(g)) ω_{0}(g).

If

n_{0} = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} )

the integrand is equal to

∑_{N_{F} Z_{F} \P_{F}} Φ(g^{−1}δ^{−1}n_{0}δg)(1 − χ(δg))

so that the integral itself is equal to

∫_{N_{F} Z_{𝔸}\G_{𝔸}} Φ(g^{−1}n_{0}g)(1 − χ(g)) ω_{0}(g)

which is

c ∫_{Z_{𝔸}\A_{𝔸}} ∫_{K} Φ(k^{−1}a^{−1}n_{0}ak)(1 − χ(a)) |\frac{α}{β}|^{−1} da dk

if

a = ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) .

The integrand vanishes outside of a compact set. Thus the integral is the limit as s approaches 0 from above of

c ∫_{Z_{𝔸}\A_{𝔸}} ∫_{K} Φ(k^{−1}a^{−1}n_{0}ak)(1 − χ(a)) |\frac{α}{β}|^{−1−s}da dk,

which is the difference of

c ∫_{Z_{𝔸}\A_{𝔸}} ∫_{K} Φ(k^{−1}a^{−1}n_{0}ak)|\frac{α}{β}|^{−1−s} da dk

and

c ∫_{Z_{𝔸}\A_{𝔸}} ∫_{K} Φ(k^{−1}a^{−1}n_{0}ak)|\frac{α}{β}|^{−1−s}χ(a) da dk.

The first of these two expressions is equal to

c\{ ∏_{v} ∫_{Z_{v}\A_{v}} ∫_{K_{v}} f_{v}(k_{v}^{−1}a_{v}^{−1}n_{0}a_{v}k_{v})|\frac{α_{v}}{β_{v}} |^{−1−s} da_{v} dk_{v}\}

which is

c L(1 + s, 1_{F}){ ∏_{v} θ(s, f_{v})}.  (16.2.7)

Observe that if v is non-archimedean and f_{v} is 0 outside of Z_{v} K_{v} and is 1 on the elements of Z_{v} K_{v} of determinant 1 then

∫_{Z_{v}\A_{v}} ∫_{K_{v}} f_{v}(k_{v}^{−1}a_{v}^{−1}n_{0}a_{v}k_{v})|\frac{α_{v}}{β_{v}} |^{−1−s} da_{v} dk_{v}

is the product of the measure of

\{( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ∈ Z_{v} \ A_{v} | |α| = |β|\}

and

∑_{n=0}^{∞} |ϖ_{v}^{n}|^{1+s} = L(1 + s, 1_{v}),

so that

∏_{v} θ(s, f_{v}) = θ(s, Φ)

is analytic for Re s > −1 and its derivative at 0 is

∑_{v} θ′(s, f_{v}){ ∏_{w≠v} θ(s, f_{w})}.

The function (16.2.7) has a simple pole at s = 0. The constant term in its Laurent expansion is

c[λ_{0} ∏_{v} θ(0, f_{v}) + λ_{−1}{ ∑_{v} θ′(0, f_{v}) ∏_{w≠v} θ(0, f_{w})}]

which is the fifth term of the trace formula.

The expression

c ∫_{Z_{𝔸}\A_{𝔸}} ∫_{K} Φ(k^{−1}a^{−1}n_{0}ak)|\frac{α}{β}|^{−1−s} χ(a) da dk

is equal to

c ∫_{Z_{𝔸}A_{F} \A_{𝔸}} ∫_{K} ∑_{{γ∈N_{F} \atop γ≠e}} Φ(k^{−1}a^{−1}γak)|\frac{α}{β}|^{−1−s} χ(a) da dk.

Choose a non-trivial character ψ of F \ 𝔸 and let

Ψ(y, g) = ∫_{𝔸} Φ (g^{−1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) ψ(xy) dx.

Then

Ψ(y, ag) = |\frac{β}{α}|^{−1} Ψ(\frac{α}{β}y, g).

Moreover by the Poisson summation formula

∑_{{γ∈N_{F} \atop γ≠e}} Φ(k^{−1}a^{−1}γak)

Chapter 3 290 is equal to

∑_{y≠0} |\frac{β}{α}|^{−1} Ψ(\frac{α}{β}y, k) + |\frac{β}{α}|^{−1} Ψ(0, k) − Φ(e).

The integral

c ∫_{Z_{𝔸}A_{F} \A_{𝔸}} ∫_{K} |\frac{α}{β}|^{−s} χ(a) \{ ∑_{y≠0} Ψ(\frac{α}{β}y, k)\} da dk

is a holomorphic function of s and its value at s = 0 approaches 0 as c_{1} approaches ∞. Since we shall eventually let c_{1} approach ∞ it contributes nothing to the trace formula. If F is a number field

c ∫_{Z_{𝔸}A_{F} \A_{𝔸}} ∫_{K} Φ(e) |\frac{α}{β}|^{−1−s} χ(a) da dk

is a multiple of

\frac{1}{1 + s} · \frac{1}{c_{1}^{1+s}}

which is defined at s = 0. Its value there approaches 0 as c_{1} approaches ∞. Finally

c ∫_{Z_{𝔸}A_{F} \A_{𝔸}} ∫_{K} Ψ(0, k) |\frac{α}{β}|^{−s} χ(a) da dk

is equal to

\frac{c}{s c_{1}^{s}} ∫_{K} Ψ(0, k) dk.

The pole of this function at s = 0 must cancel that of (16.2.7). Consequently

∫_{K} Ψ(0, k) dk = λ_{−1} θ(0, Φ).

The constant term in its Laurent expansion about 0 is

−c log c_{1} ∫_{K} Ψ(0, k) dk.

Not this expression but its negative

c log c_{1} ∫_{K} Ψ(0, k) dk  (16.2.8)

enters into the integral of the kernel of ρ(Φ) − E ρ(Φ) over the diagonal. If F is a function field \frac{c}{s c_{1}^{s}} is to be replaced by

\frac{cq^{−ns}}{1 − q^{−s}}

and log c_{1} by n − \frac{1}{2}.

The Poisson summation formula can be used to simplify the remaining part of (16.2.1). We recall that it is

∑_{δ∈P_{F} \G_{F}} ∑_{{γ∈N_{F} \atop γ≠e}} Φ(g^{−1}δ^{−1}γδg) χ(δg).

Chapter 3 291 We subtract from this

∑_{δ∈P_{F} \G_{F}} Ψ(0, δg) χ(δg)

to obtain the difference betwen

∑_{δ∈P_{F} \G_{F}} ∑_{y≠0} Ψ(y, δg) χ(δg)

and

∑_{δ∈P_{F} \G_{F}} Φ(e) χ(δg).

The integrals of both these functions over Z_{𝔸} G_{F} \ G_{𝔸} converge and approach 0 as c_{1} approaches ∞. They may be ignored.

The remaining part of (16.2.2) is the sum of

\frac{1}{2} ∑_{δ∈P_{F} \G_{F}} ∑_{{γ∈Z_{F} \P_{F} \atop γ∉Z_{F} N_{F}}} Φ(g^{−1}δ^{−1}γδg) χ(δg)

and

\frac{1}{2} ∑_{δ∈P_{F} \G_{F}} ∑_{{γ∈Z_{F} \P_{F} \atop γ∉Z_{F} N_{F}}} Φ(g^{−1}δ^{−1}γδg) χ(ε(γ)δg).

These two sums may be written as

\frac{1}{2} ∑_{{γ∈Z_{F} \A_{F} \atop γ∉Z_{F}}} ∑_{A_{F} \G_{F}} Φ(g^{−1}δ^{−1}γδg) χ(δg)

and

\frac{1}{2} ∑_{{γ∈Z_{F} \A_{F} \atop γ∉Z_{F}}} ∑_{A_{F} \G_{F}} Φ(g^{−1}δ^{−1}γδg) χ(wδg).

Replacing δ by w^{−1}δ in the second sum we see that the two expressions are equal. Their sum is equal to twice the first which we write as

∑_{{γ_{1}∈Z_{F} \A_{F} \atop γ_{1}∉Z_{F}}} ∑_{δ∈P_{F} \G_{F}} ∑_{γ_{2}∈N_{F}} Φ(g^{−1}δ^{−1}γ_{1}γ_{2}δg) χ(δg).

For a given Φ all but finitely many of the sums

∑_{δ∈P_{F} \G_{F}} ∑_{γ_{2}∈N_{F}} Φ(g^{−1}δ^{−1}γ_{1}γ_{2}δg) χ(δg)  (16.2.9)

are zero. Set

Ψ(y, γ_{1}, g) = ∫_{𝔸} Φ (g^{−1}γ_{1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) ψ(xy) dx.

The expression (16.2.9) is the sum of

∑_{δ∈P_{F} \G_{F}} ∑_{y≠0} Ψ(y, γ_{1}, δg) χ(δg)

and

∑_{δ∈P_{F} \G_{F}} Ψ(0, γ_{1}, δg) χ(δg).

The first of these two expressions is integrable on G_{F} Z_{𝔸} \ G_{𝔸} and its integral approaches 0 as c_{1} approaches ∞.

Since Ψ(0, g) = Ψ(0, e, g) we have expressed Φ(g, g) as the sum of

∑_{δ∈P_{F} \G_{F}} ∑_{γ∈Z_{F} \A_{F}} Ψ(0, γ, δg) χ(δg)  (16.2.10)

and a function which can be integrated over G_{F} Z_{𝔸} \ G_{𝔸} to give the first five terms of the trace formula, the sum of (16.2.8) and one-half of the sum over γ in Z_{F} \ A_{F} but not in Z_{F} of (16.2.6) which is

c log c_{1} ∑_{γ∈Z_{F} \A_{F}} ∫_{N_{𝔸}} ∫_{K} Φ(k^{−1}γnk) dn dk,  (16.2.11)

and an expression which goes to 0 as c_{1} approaches ∞.

Now we discuss the kernel of E ρ(Φ) in the same way. Set H(g; μ, ν, i, j, s) equal to

ρ_{ij}(Φ, μ, ν, s) E(g, φ_{i}(μ, ν), s) Ē(g, φ_{j}(μ, ν), s).

On the diagonal the kernel of E ρ(Φ) is equal to

∑_{μ,ν} ∑_{i,j} \frac{1}{4πc} ∫_{−i∞}^{i∞} H(g; μ, ν, i, j, s) d|s|

if F is a number field and to

∑_{μ,ν} ∑_{i,j} \frac{log q}{4πc} ∫0^{\frac{2π}{log q}} H(g; μ, ν, i, j, s) d|s|

if F is a function field. We set E_{1}(g, φ, s) equal to

∑_{P_{F} \G_{F}} {T_{s}^{−1} φ(δg) + M(s) T_{S}^{−1} φ(δg)} χ(δg)

and let

E_{2}(g, φ, s) = E(g, φ, s) − E_{1}(g, φ, s).

If, for m = 1, 2, n = 1, 2, H_{mn}(g; μ, ν, i, j, s) is

ρ_{ij}(Φ, μ, ν, s) E_{m}(g, φ_{i}(μ, ν), s) Ē_{n}(g, φ_{j}(μ, ν), s)

and Φ_{mn}(g) is, at least when F is a number field,

∑_{μ,ν} ∑_{i,j} \frac{1}{4πc} ∫_{−i∞}^{i∞} H_{m,n}(g; μ, ν, i, j, s) d|s|,

the kernel of E ρ(Φ) is

∑_{m=1}^{n} ∑_{n=1}^{2} Φ_{mn}(g)

on the diagonal.

If m or n is 2

∫_{G_{F} Z_{𝔸}\G_{𝔸}} Φ_{mn}(g) ω_{0}(g)

is equal to

\frac{1}{4πc} ∫_{−i∞}^{i∞} ∑_{μ,ν} ∑_{i,j} \{ ∫_{G_{F} Z_{𝔸}\G_{𝔸}} H_{m,n}(g, μ, ν, i, j, s) ω_{0}(g)\} d|s|.  (16.2.12)

Take first m = n = 2. If F is a number field a formula for the inner product

∫_{G_{F} Z_{𝔸}\G_{𝔸}} E_{2}(g, φ_{1}, s) Ē_{2}(g, φ_{2}, s) ω_{0}(g)

can be inferred from the formulae of [26] and [27]. The result is the sum of

c lim_{t→0} \frac{1}{2t}\{c_{1}^{2t}(φ_{1}, φ_{2}) − c_{1}^{−2t}(N(t + s)φ_{1}, N(t + s)φ_{2})\},

where N(t + s) = T_{t+s} M(t + s) T_{t+s}^{−1}, and

c lim_{t→0} \frac{1}{2s}\{c_{1}^{2s}(φ_{1}, N(t + s)φ_{2}) − c_{1}^{−2s}(N(t + s)φ_{1}, φ_{2})\}.

The second expression is equal to

\frac{c}{2s}\{c_{1}^{2s}(φ_{2}, N(s)φ_{2}) − c_{1}^{−2s}(N(s)φ_{1}, φ_{2})\}.

The first is the sum of

2c log c_{1}(φ_{1}, φ_{2})

and

−\frac{c}{2}\{(N^{−1}(s) N′(s)φ_{1}, φ_{2}) + (φ_{1}, N^{−1}(s) N′(s) φ_{2})\}.

If F is a function field over 𝔽_{q} and c_{1} = q^{n}the inner product is the sum of

c log q \{\frac{1 − q^{s} + q^{−s}}{1 − q^{−2s}} (φ_{1}, N(s)φ_{2}) q^{2(n−1)s} + \frac{1 − q^{−s} + q^{s}}{1 − q^{2s}} (N(s)φ_{1}, φ_{2}) q^{−2(n−1)s}\}

and

(2n − 1)c(φ_{1}, φ_{2})

Chapter 3 294 and

−\frac{c}{2}\{(N^{−1}(s) N′(s) φ_{1}, φ_{2}) + (φ_{1}, N^{−1}(s) N′(s) φ_{2})\}.

Certainly

∑_{μ,ν} ∑_{i,j} ρ_{ij}(Φ, μ, ν, s) (φ_{i}(μ, ν), φ_{j}(μ, ν)) = trace ρ(Φ, s)

which equals

∑_{μ,ν} c ∫_{N_{𝔸}} ∫_{Z_{𝔸}\A_{𝔸}} ∫_{K} Φ(k^{−1}ank) μ(α) ν(β)|\frac{α}{β}|^{\frac{s+1}{2}} dn da dk

or

∑_{μ,ν} c ∫_{N_{𝔸}} ∫_{Z_{𝔸}A_{F} \A_{𝔸}} ∫_{K} ∑_{γ∈Z_{F} \A_{F}} Φ(k^{−1}aγnk) μ(α) ν(β)|\frac{α}{β}|^{\frac{s+1}{2}} .

Thus if H is the set of all

( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )

in Z_{𝔸}A_{F} \ A_{𝔸} for which |α| = |β|

\frac{1}{4πc} ∫_{−i∞}^{i∞} trace ρ(Φ, s) d|s|  (16.2.13)

is equal to

\frac{1}{2} ∑_{μ,ν} ∫_{H} ∫_{N_{𝔸}} ∫_{K} ∑_{γ} Φ(k^{−1}aγnk) μ(α) ν(β) dn dk da

which is

\frac{1}{2} ∫_{N_{𝔸}} ∫_{K} Φ(k^{−1}γnk) dn dk.

When multiplied by 2c log c_{1} the effect of this is to cancel the term (16.2.11). If F is a function field (16.2.13) is said to be replaced by

\frac{log q}{4πc} ∫0^{\frac{2π}{log q}} trace ρ(Φ, s) d|s|

but the conclusion is the same.

The expression

∑_{μ,ν} ∑_{i,j} ρ_{ij}(Φ, μ, ν, s) (φ_{i}(μ, ν), N(s)φ_{j}(μ, ν))

is equal to

trace M^{−1}(s) ρ(Φ, s)

when s is purely imaginary and

∑_{μ,ν} ∑_{i,j} ρ_{ij}(Φ, μ, ν, s) (N(s)φ_{i}(μ, ν), φ_{j}(μ, ν))

is equal to

trace M(s) ρ(Φ, s).

Since M(0) = M^{−1}(0)

lim_{c_{1}→∞} \frac{1}{8π} ∫_{−i∞}^{i∞} \frac{1}{2}\{c_{1}^{2s} trace M^{−1}(s) ρ(Φ, s) − c_{1}^{−2s} trace M(s) ρ(Φ, s)\} d|s|

is equal to

\frac{1}{4} trace M(0) ρ(Φ, 0).

When multiplied by −1 this is the sixth term of the trace formula. For a function field it is to be replaced by

\frac{log q}{4} \{ trace M(0) ρ(Φ, 0) + trace M(\frac{π}{log q} ) ρ(Φ, \frac{π}{log q} )\}.

When s is purely imaginary

(N^{−1}(s) N′(s) φ_{1}, φ_{2}) = (φ_{1}, N^{−1}(s) N′(s) φ_{2}).

Moreover

∑_{μ,ν} ∑_{i,j} ρ_{ij}(Φ, μ, ν, s)(N^{−1}(s) N′(s) φ_{i}(μ, ν), φ_{j}(μ, ν))

is equal to

trace M^{−1}(s) M′(s) ρ(Φ, s).

Thus

\frac{1}{4π} ∫_{−i∞}^{i∞} trace M^{−1}(s) M′(s) ρ(Φ, s) d|s|

is to be added to the trace formula. It gives the seventh and eighth terms.

Next we consider (16.2.12) when m = 2 and n = 1. If φ′_{2} = T_{S}^{−1}φ_{2} and φ′′_{2} = M(s) T_{s}^{−1} φ_{2} the integral

∫_{G_{F} Z_{𝔸}\G_{𝔸}} E_{2}(g, φ_{1}, s) Ē_{1}(g, φ_{2}, s) ω_{0}(g)  (16.2.14)

is the sum of

∫_{P_{F} Z_{𝔸}\G_{𝔸}} E_{2}(g, φ_{1}, s) φ̄′_{2}(g) χ(g) ω_{0}(g)

and

∫_{P_{F} Z_{𝔸}\G_{𝔸}} E_{2}(g, φ_{1}, s) φ̄′′_{2}(g) χ(g) ω_{0}(g).

Since φ′_{2}, φ′′_{2} and χ are all functions on Z_{𝔸}N_{𝔸}P_{F} \ G_{𝔸} while, as is known,

χ(g) ∫_{N_{𝔸}} E_{2}(ng, φ_{1}, s) dn = 0

when c_{1} is sufficiently large, the integral (16.2.14) is 0. Thus (16.2.12) is 0 when m = 2 and n = 1 and also when m = 1 and n = 2.

Set

F (g, φ, s) = T_{s}^{−1} φ(g) + M(s) T_{s}^{−1} φ(g)

and set H_{0}(g, μ, ν, i, j, s) equal to

ρ_{ij}(Φ, μ, ν, s) F (g, φ_{i}(μ, ν), s) F̄(g, φ_{j}(μ, ν), s) χ(g).

If c_{1} is so large that χ(δ_{1}g) χ(δ_{2}g) = 0 when δ_{1} and δ_{2} do not belong to the same coset of P_{F} the function Φ_{1,1}(g) is equal to

∑_{μ,ν} ∑_{i,j} ∑_{P_{F} \G_{F}} \frac{1}{4πc} ∫_{−i∞}^{i∞} H_{0}(δg, μ, ν, i, j, s) d|s|.

If φ′_{i}(g, μ, ν) is the value of T_{S}^{−1} φ_{i}(μ, ν) at g then

∑_{i,j} ρ_{ij}(Φ, μ, ν, s) φ′_{i}(h, μ, ν) φ̄′_{j}(g, μ, ν)

is the kernel of ρ(Φ, μ, ν, s) which is

c ∫_{N_{𝔸}} ∫_{Z_{𝔸}\A_{𝔸}} Φ(g^{−1}anh)|\frac{α}{β}|^{\frac{s+1}{2}} μ(α) ν(β) dn da.

If we set h = g, divide by 4πc, integrate from −i∞ to i∞, and then sum over μ and ν we obtain

\frac{1}{2} ∑_{γ∈Z_{F} \A_{F}} Ψ(0, γ, g).

If φ′′_{i}(g, μ, ν) is the value of M(s) T_{S}^{−1} φ_{i}(μ, ν) at g

∑_{i,j} ρ_{ij}(Φ, μ, ν, s) φ′′_{i}(h, μ, ν) φ̄′′_{j}(g, μ, ν)

is the kernel of

M(μ, ν, s) ρ(Φ, μ, ν, s) M(ν, μ, −s) = ρ(Φ, ν, μ, −s).

Thus Φ_{1,1}(g) is the sum of

∑_{δ∈P_{F} \G_{F}} ∑_{γ∈Z_{F} \A_{F}} Ψ(0, γ, δg) χ(δg)  (16.2.15)

and

∑_{μ,ν} ∑_{i,j} ∑_{P_{F} \G_{F}} \frac{χ(δg)}{4πc} ∫_{−i∞}^{i∞} \{H_{1}(δg, μ, ν, i, j, s) + H_{2}(δg, μ, ν, i, j, s)\} d|s|

where H_{1}(g, μ, ν, i, j, s) is

ρ_{ij}(Φ, μ, ν, s) φ′_{i}(g, μ, ν) φ̄′′_{j}(g, μ, ν)

and H_{2}(g, μ, ν, i, j, s) is

ρ_{ij}(Φ, μ, ν, s) φ′′_{i}(g, μ, ν) φ̄′_{j}(g, μ, ν).

The expression (16.2.15) cancels (16.2.10). If g = nak with

a = ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) ,

H_{1}(g, μ, ν, i, j, s) is equal to

ρ_{ij}(Φ, μ, ν, s) μ(\frac{α}{β}) |\frac{α}{β}|^{s+1} φ′_{i}(k) φ̄′′_{j}(k).

The functions ρ_{ij}(Φ, μ, ν, s) are infinitely differentiable on the imaginary axis. Thus

\frac{1}{4πc} ∫_{−i∞}^{i∞} H_{1}(g, μ, ν, i, j, s) d|s|

is, O(|\frac{α}{β}|^{M}) as |\frac{α}{β}| → ∞ for any real M. Thus if this expression is multiplied by χ(g) and averaged over P_{F} \ G_{F} the result is integrable on Z_{𝔸} G_{F} \ G_{𝔸} and its integral approaches 0 as c_{1} approaches ∞. Thus it contributes nothing to the trace. Nor do the analogous integrals for H_{2}(g, μ, ν, i, j, s).

References for Chapter III

The Dirichlet series associated to automorphic forms on a quaternion algebra are discussed in:

22. Godement, R., Les functions ζ des algèbres simples, I, II Seminaire Bourbaki, 1958/1959 23. Shimura, G., On Dirichlet series and abelian varieties attached to automorphic forms, Ann of

Math., vol 76(1962)

24. Tamagawa, T., On ζ-functions of a division algebra, Ann of Math., vol 77(1963) They are also referred to, but rather obliquely, in:

25. Selberg, A., Discontinuous groups and harmonic analysis, Proc. Int. Cong. Math. (1962) The theory of Eisenstein series is discussed in [11], [13], [14], and [25] as well as in:

26. Langlands, R.P., On the functional equations satisifed by Eisenstein series, Mimeographed notes

27. Langlands, R.P., Eisenstein series, in Algebraic Groups and Discontinuous Subgroups, Amer. Math. Soc. (1966)

28. Selberg, A., Harmonic analysis and discontinuous groups in weakly symmetric Riemannian spaces with applications to Dirichlet series, Jour. Ind. Math. Soc., vol XX (1956)

[25] and [28] are of course the basic references for the Selberg trace formula. Some of its formal aspects are also described in:

29. Langlands, R.P., Dimension of spaces of automorphic forms, in Algebraic Groups and Discontinuous Subgroups, Amer. Math. Soc. (1966) The theorem of §16 can still be stated and proved if M is replaced by a quaternion algebra

which splits everywhere that M′ does. The proof is in fact rather easier. However these apparently more general theorems are immediate consequences of the proof of the original theorem. Theorems very similar to that of §16 and its extensions have been proved by Shimizu. Our methods differ little from his.

30. Shimizu, H., On discontinuous groups operating on the product of the upper half planes, Ann. of Math., vol 77(1963)

31. Shimizu, H., On traces of Hecke operators, Jour. Fac. Sci. Univ. Tokyo, vol. 10 (1963)

32. Shimizu, H., On zeta functions of quaternion algebras, Ann. of Math., vol 81(1965) We have also had occasion to refer to:

33. Langlands, R.P., The volume of the fundamental domain for some arithmetical subgroups of Chevalley groups, in Algebraic Groups and Discontinuous Subgroups, Amer. Math. Soc. (1966) Orthogonality relations for the characters of non-compact groups first appeared in:

34. Harish–Chandra, Discrete series for semisimple Lie groups, II, Acta Math., vol. 116 (1966) We thank R. Lipsman for the reference to:

35. Rickert, Neil. W., Convolution of L^{2}-functions, Coll. Math, v. XIX (1968)