1. Introduction. This is the rest of the letter I promised. After making the necessary apologies for its length, the style in which it is written, and the delay in sending it let me tell you what is in it and what is not in it. There are also one or two matters about which you should be concerned.

Of course the goal is to extend the theorem of your paper to all number fields and to function fields. If I have made no mistakes such an extension is obtained in paragraph 7 (Although I am not really at home with function fields I do not think I made any blunders). Moreover as I said I do have to assume the existence of an Euler product.

If you want to see quickly what the basic idea of the proof is you should probably concentrate on function fields. For these only paragraphs 6 and 7 are necessary. Indeed in this letter the only difference between a number field and a function field is that a function field has some archimedean primes. The reason that so much space is devoted to archimedean fields is that, at the moment, I know more about the representation of GL(2, K) for such fields. As soon as I understand the representation theory of GL(2, K) for non-archimedean fields I should be able to avoid the assumption, which appears in both the letter and your paper, about the character χ. Of course ignorance of the representation theory of GL(2, K) for a non-archimedean field is not fatal. The same ignorance for an archimedean field would be.

Perhaps it will help when you read paragraph 7 if I give some idea of the relations between the notation of the letter and your paper. Associate to the function Γ of your paper the function

F_{0}(g) = \frac{(ad − bc)^{k/2}}{(ci + d)^{k}} F (\frac{ai + b}{ci + d}) g = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ GL_{+}(2, ℝ).

If K = ℚ, as we now assume, the divisor D of the letter is just the number A of your paper. Let ϵ′ be the ϵ of your paper and let δ be a character modulo A. If ( \begin{matrix}a & b \\ c & d \end{matrix} ) lies in UL_{K_{𝔭}}^{D} set

ϵ_{𝔭} ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ϵ′(a)δ(ad)

if 𝔭 | D and set

ϵ_{𝔭} ( \begin{matrix}a & b \\ c & d \end{matrix} ) = 1

if 𝔭\D. Then, for ( \begin{matrix}a & b \\ c & d \end{matrix} ) in U^{D}

ϵ ( \begin{matrix}a & b \\ c & d \end{matrix} ) = Π_{𝔭≠𝔭_{∞}} ϵ_{𝔭} ( \begin{matrix}a_{𝔭} & b_{𝔭} \\ c_{𝔭} & d_{𝔭} \end{matrix} )

is the ϵ of the letter. The relation F | γ = ϵ(γ)^{−1}Γ for γ = ( \begin{matrix}r & s \\ t & u \end{matrix} ) in Γ_{0}(A) is equivalent to

F_{0}(γg) = Π_{𝔭|D} ϵ_{𝔭}(γ_{𝔭})}F_{0}(g)

for γ in G_{K} ∩ G_{K_{𝔭^{∞}}} × U^{D}, detγ > 0. Define a function φ on G_{𝔸}^{D} by

φ(γg) = F_{0}(g_{∞}) Π_{𝔭|D} ϵ_{𝔭}(g_{𝔭})

if γ belongs to G_{K} ∩ G_{𝔸}^{D}, g belongs to G_{K_{𝔭∞}} × U^{D}, and det(g_{∞}) > 0. φ is well defined and is the φ of my letter. If we want to indicate its independence on δ we should write φ = φ_{δ}.

Now let me show that the assumption

F | ω(A) = C^{−1}i^{−k}F

implies that φ̂ = \frac{i^{k}}{C} \{ Π_{𝔭|D} ϵ_{𝔭}( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )\}φ_{ϵ′δ}. Since (ϵ′)^{2} = 1

ϵ̃_{𝔭} ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ϵ′(a)δ(ad) = ϵ′(a)(ϵ′δ)(ad)

of 𝔭 | D. If g belongs to G_{K_{𝔭}} × U^{D} and detg_{∞} > 0

φ̂(g) = φ(( \begin{matrix}0 & A^{−1} \\ 1 & 0 \end{matrix} ) g ∏_{𝔭|D} ( \begin{matrix}0 & 1 \\ A_{𝔭} & 0 \end{matrix} ))

= φ(( \begin{matrix}0 & A^{−1} \\ −1 & 0 \end{matrix} ) g ∏_{𝔭|D} ( \begin{matrix}0 & −1 \\ A_{𝔭} & 0 \end{matrix} )) ∏_{𝔭|D} ϵ̃_{𝔭}(( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )

= F_{0}(( \begin{matrix}A_{∞}^{−1} & 0 \\ 0 & −A_{∞}^{−1} \end{matrix} ) ( \begin{matrix}0 & −1 \\ A_{∞} & 0 \end{matrix} ) g_{∞})\{ ∏_{𝔭|D} ϵ̃_{𝔤}(g_{𝔭})\}\{ ∏_{𝔭|D} ϵ_{𝔭} ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) \}

= \frac{i^{k}}{C} \{ ∏_{𝔭|D} ϵ_{𝔭} ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) \}F_{0}(g_{∞})\{ ∏_{𝔭|D} ϵ̃_{𝔭}(g_{𝔭})\}

= \frac{i^{k}}{C} \{ ∏_{𝔭|D} ϵ_{𝔭} ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) \}φ_{ϵ′d}(g).

If

ω(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ))|\frac{α_{1}}{α_{2}} |^{\frac{k−1}{2}} (sgn α_{2})^{k} α_{1}, α_{2} ∈ ℝ^{k}

the representation Π_{𝔭_{∞}} is the infinite-dimensional quasi-simple irreducible representation deducible from π_{ω}, φ_{𝔭_{∞}} will have to lie in L(ξ_{𝔭_{∞}}, π_{𝔭_{∞}})_{k}. ξ will of course be the character

ξ(x) = e^{2πix_{∞}} ∏_{𝔭} e^{−2πix_{𝔭}}.

Let χ′ be one of the χ of your paper. χ′ determines a homeomorphism of Π_{𝔭|m} O_{𝔭}^{×} into ℂ^{×}. m is of course the conductor of χ. Let χ′ also denote the character of K^{×}\I which satisfies

χ′( Π_{𝔭} β_{𝔭}) = χ′( Π_{𝔭|m} β_{𝔭})

if β_{𝔭_{∞}} > 0 and β_{𝔭} ∈ O_{𝔭}^{×} for 𝔭 ≠ 𝔭_{∞}. Then χ = (ϵ′δχ′)^{−1} is one of the characters of the letter.

If g is

I × Π_{𝔭∤𝔭_{∞}} ( \begin{matrix}1 & \frac{1}{m_{𝔭}} \\ 0 & 1 \end{matrix} )

the value of the integral of Lemma 7.3 is, in your notation, ^{∗}

\frac{1}{φ(m)} ∑_{amodm} \overline{χ}′(a) ∫_{0}^{∞}∑_{n} c_{n}e^{2πin(t− \frac{a}{m})}t^{s+ \frac{k}{2}} \frac{d}{t}t.

This equals

\frac{\overline{g(χ′)}}{φ(m)}Γ(s + \frac{k}{2} ) ∑_{n=1}^{∞} \frac{χ(n)c_{n}}{(2πn)^{s+ \frac{k}{2}}} = \frac{\overline{g(χ′)}}{φ(m)m^{s+ \frac{k}{2}}} Λ_{χ′}(s + \frac{k}{2} ).

On the other hand it is equal to the product of Ξ(s, χ) and the expression (D) on page 7.20^{†}. If φ_{𝔭_{∞}} is suitably normalized then, for the g chosen, this expression equals

\frac{1}{(2π)^{s+ \frac{k}{2}}} ∏_{𝔭∈R} ∫_{O_{𝔭}^{×}} e^{−2πi \frac{α}{m}} ζ_{𝔭}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα.

This is equal to

\frac{1}{(2π)^{s+ \frac{k}{2}}} \frac{\overline{g(χ′)}}{φ(m)}.

Thus

Ξ(s, χ) = (\frac{απ}{m} )^{s+ \frac{k}{2}} Λχ′(s + \frac{k}{2} ).

Moreover

Ξ̂(s, (χη)^{−1}) = \frac{i^{k}}{c} (∏_{𝔭|D} ϵ_{𝔭} ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ))(\frac{2π}{m})^{s+ \frac{k}{2}} Λ\overline{χ}(s + \frac{k}{2} ).

The letter and the paper will be consistent of

(\frac{2π}{m})^{s+ \frac{k}{2}} A^{−s}Cϵ′(m)\frac{g(χ′)}{g(\overline{χ}′)}χ^{−1}(−A)

is equal to

\frac{i^{k}}{C} \{ ∏_{𝔭|D} ϵ_{𝔭}(( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ))\}(\frac{2π}{m})^{−s+ \frac{k}{2}} \{ ∏_{𝔭|D} ζ′_{𝔭}(( \begin{matrix}−A_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))\}ϵ(ζ_{𝔭_{∞}}ξ_{𝔭_{∞}}, π_{𝔭_{∞}})\{ ∏_{{𝔭∤0 \atop 𝔭≠𝔭∞}} ϵ(ζ_{𝔭}, ξ_{𝔭}, ω_{𝔭})\}.

∗ The second formula from the bottom on p. 150 of your paper does not look correct.

† Added - this is pagination of the original letter.

This is a consequence of the following relations.

∏_{𝔭|D} ϵ_{𝔭}(( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )) = δ(−1)
∏_{𝔭|D} ζ_{𝔭}(( \begin{matrix}A_{𝔭} & 0 \\ 0 & 1 \end{matrix} )) = ∏_{𝔭∤D} ζ_{𝔭}^{−1}(( \begin{matrix}−A_{𝔭} & 0 \\ 0 & 1 \end{matrix} )) = χ′(A)δ(−1)(−1)^{k}A^{−s}
ϵ(ζ_{𝔭_{∞}}ξ_{𝔭_{∞}}, π_{𝔭_{∞}}) = i^{k}(2π)^{2s}
ϵ(ζ_{𝔭}, ξ_{𝔭}, π_{𝔭}) = 1 of 𝔭 ∤ D and 𝔭 ∤ m
∏_{𝔭|m} ϵ(ζ_{𝔭}, ξ_{𝔭}, ω_{𝔭}) = \frac{g(χ′)}{g(\overline{χ}′)} ∏_{𝔭|m} ζ_{𝔭}(( \begin{matrix}m_{𝔭} & 0 \\ 0 & −m_{𝔭}^{−1} \end{matrix} ))
∏_{𝔭|m} ζ_{𝔭}(( \begin{matrix}m_{𝔭} & 0 \\ 0 & −m_{𝔭}^{−1} \end{matrix} )) = ∏_{𝔭|m}(ϵ′δ^{2})^{−1}(−m_{𝔭}^{−1}) ∏_{𝔭|m}(ϵ′δχ′)^{−1}(−m_{𝔭}^{2}) ∏_{𝔭|m} |m_{𝔭}|^{2s}

= {∏_{𝔭|m}(ϵ′)^{−1}(m_{𝔭})}{∏_{𝔭|m} χ′(−m_{𝔭}^{−2})}m^{−2s}

= {∏_{𝔭∤m} ϵ′(m_{𝔭})}{∏_{𝔭∤m} χ′(−m_{𝔭}^{2})}m^{−2s}

= ϵ′(m)χ′(−1)m^{−2s}.

Of course all these formulas will be meaningless to you until you have read the letter.

For lemma 2.4 and 4.3 I have referred to a paper of Harish-Chandra. These lemmas are not stated explicitly in that paper. It has been a long time since I looked at that paper and I should read it again to see that the lemmas are really implicit in it. I will do so as soon as possible. The appendix to paragraph 7 is not relevant to the rest of the paper. You should not read it. I include it only because the footnotes contain corrections to paragraph 5.

With so many formulae there are bound to be some small errors. They should show up as soon as one starts to apply the theorem.

2. Representations of GL(2, ℝ). In this paragraph the next G_{ℝ} will be GL(2, ℝ) and G_{ℝ}^{∘} will be the group of matrices in G_{ℝ} with determinant ±1. U will be O(2, ℝ) and U^{0} will be SO(2, ℝ). 𝔤 will be the Lie algebra of G_{ℝ} and 𝔤_{ℂ} its complexification. 𝔤^{o} will be the Lie algebra of G_{ℝ}^{0} and 𝔤_{ℂ}^{0} its complexification. 𝔄 and 𝔄^{0} will be the universal enveloping algebras of 𝔤_{ℂ} and 𝔤_{ℂ}^{0} respectively. Since neither G_{ℝ} nor G_{ℝ}^{0} is connected it is not sufficient for us to study representations of 𝔄 or 𝔄^{0}. Let

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

A representation π of {σ, 𝔄} on a vector space W assigns to each X in 𝔄 a linear transformation π(X) of W. It also assigns to σ a linear transformation π(σ). We demand not only that X → π(X) be a representation of 𝔄 but also that (π(σ))^{2} = I, and π(σ)π(X)π(σ^{−1}) = π(ad σ(X)) for all X in 𝔄. A representation of {σ, 𝔄^{0}} is defined in a similar manner. If π is a representation of {σ, 𝔄}, π^{0} will denote its restriction to {σ, 𝔄^{0}}

Two bases of 𝔤_{ℂ}^{0} are

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

and

U = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )  V = ( \begin{matrix}1 & i \\ i & −1 \end{matrix} )  W = ( \begin{matrix}1 & −i \\ −i & −1 \end{matrix} ) .

U is contained in the Lie algebra of the one-dimensional group U. If π is a representation of {σ, 𝔄} on W let W_{n} = {w ∈ W |π(U)w = inw}. We shall always assume that W_{n} = {0} if n is not an integer. The representation π will be called quasi-simple^{∗} if W = Σ_{n}W_{n} and π(Z) is a scalar for all Z in the centre of 𝔄. If π_{1} and π_{2} are two representations of {σ, 𝔄} on W_{1} and W_{2} respectively π_{2} will be said to be deducible from π_{1} if there are two invariant subspaces. W_{3} ⊇ W_{4} of W_{1} and π_{2} is equivalent to the representation of {σ, 𝔄} on W_{3}/W_{4}. Similar notions can be introduced for representations of {σ, 𝔄^{0}}.

If Z lies in the centre of 𝔄 then ad σ(Z) = Z. The centre of 𝔄^{0} is generated by

D = XY + Y X + \frac{1}{2}Z^{2} = 2Y X + Z + \frac{1}{2}Z^{2} = 2XY − Z + \frac{1}{2}Z^{2}.

The centre of 𝔄 is generated by D and J = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ).

If G is any Lie group and X lies in its Lie algebra ρ(X) is the left-invariant vector field defined by ρ(X)φ(g) = \frac{d}{dt}φ(g exp tx)|_{t=0} and λ(X) is the right-invariant vector field defined by λ(X)φ(g) = \frac{d}{dt}φ(exp(−tX)g)|_{t=0}. The maps X → ρ(X) and X → λ(X) extend to representations of the complex universal enveloping algebra.

Let ω be a continuous homomorphism of A_{ℝ}, the group of diagonal matrices in G_{ℝ}, into ℂ^{×}. Let ω_{1} and ω_{2} be the homomorphisms of ℝ^{×} into ℂ^{×} defined by

ω_{1}(t) = ω(( \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} ))

∗ I use the expression in a slightly different sense than Harish-Chandra. and

ω_{2}(t) = (( \begin{matrix}1 & 0 \\ 0 & t \end{matrix} )).

Let ω_{i}(t) = |t|^{s_{i}}(\frac{t}{|t|})^{m_{i}}, m_{i} = 0 or 1, and set s = s_{1} − s_{2}, m = m_{1} − m_{2}. If N_{ℝ} is the group of all matrices of the form

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

let L(ω) be the space of all infinitely differentiable U-finite functions on N_{ℝ}\G_{ℝ} satisfying φ(ag) ≡ ω(a)|\frac{α_{1}}{α_{2}} |^{1/2}φ(g) for all

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

in A_{ℝ}. If φ belongs to L(ω) and X belongs to 𝔄 then ρ(X)φ also belongs to L(ω). Of course ρ(σ)φ which is defined by (ρ(σ)φ)(g) = φ(gσ) also belongs to L(ω) and we obtain a representation π_{ω} of {σ, 𝔄} on L(ω).

Because of the Iwasawa decomposition G_{ℝ} = N_{ℝ}A_{ℝ}U^{0} the functions in L(ω) are determined by their restrictions to U^{0}. The functions φ_{n} with \frac{n−m}{2} ∈ ℤ, which are defined by

φ_{n}(g) = ω(a)|\frac{α_{1}}{α_{2}} |^{1/2}e^{inθ}

if g = ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )a( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) and a = ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ), form a basis of L(ω).

Lemma 2.1.

\begin{matrix}(i) π_{ω}(σ)φ_{n} = (−1)^{m_{2}}φ_{−n} & (ii) π_{ω}(U)φ_{n} = inφ_{n} \\ (iii) π_{ω}(V)φ_{n} = (s + 1 + n)φ_{n+2} & (iv) π_{ω}(W)φ_{n} = (s + 1 − n)φ_{n−2} \\ (v) π_{ω}(D) = \frac{s^{2} − 1}{2}I & (vi) π_{ω}(J) = (s_{1} + s_{2})I \end{matrix}

The relations (i), (ii), and (vi) are clear. To prove (iv) we observe that ρ(D) = λ(D) and that if φ ∈ L(ω), λ(D)φ = λ(Z)φ + \frac{1}{2}λ(Z^{2})φ = [−(s + 1) + \frac{1}{2}(s + 1)^{2}]φ = \frac{s^{2}−1}{2}φ. Since [U, V] = 2iV and [U, W] = 2iW, π_{ω}(V)φ_{n} is a multiple of φ_{n+2} and π_{ω}(W)φ_{n} is a multiple of φ_{n−2}. It is easily seen that (π_{ω}(V)φ_{n})(1) = (s + 1 + n) and (π_{ω}(W)φ_{n})(1) = (s + 1 − n). The relations (ii) and (iii) follow

Corollary. (i) If s − m is not an even integer the restriction of π_{ω} to 𝔄^{0} is irreducible (ii) If s − m is an odd integer and s ≥ 0 the only subspaces of L(ω) invariant under 𝔄^{0} are

M_{1}(ω) = ∑_{{n≥s+1 \atop \frac{n−m}{2} ∈ℤ}} ℂφ_{n}, M_{2}(ω) = ∑_{{n≤−(s+1) \atop \frac{n−m}{2} ∈ℤ}} ℂφ_{n},

and M(ω) = M_{1}(ω) + M_{2}(ω). The spaces M_{1}(ω), M_{2}(ω), and L(ω)/M(ω) are irreducible under 𝔄^{0}. The only subspace invariant under {σ, 𝔄^{0}} is M(ω). The representations of {σ, 𝔄^{0}} on M(ω) and L(ω)/M(ω) are irreducible.

(iii) If s − m is an odd integer and s < 0 the only subspaces of L(ω) invariant under 𝔄^{0} are

M_{1}(ω) = ∑_{{n≥s+1 \atop \frac{n−m}{2} ∈ℤ}} ℂφ_{n}, M_{2}(ω) = ∑_{{n≤−(s+1) \atop \frac{n−m}{2} ∈ℤ}} ℂφ_{n},

and M(ω) = M_{1}(ω) ∩ M_{2}(ω). The only subspace invariant under {σ, 𝔄^{0}} is M(ω) and the representations of 𝔄^{0} on M(ω) and L(ω)/M(ω) are irreducible.

This follows immediately from the lemma and and the observation that an invariant subspace of L(ω) is spanned by the φ_{n} it contains.

If π is a quasi-simple representation of {σ, 𝔄^{0}} on H then π(V)H_{n} ⊆ H_{n+2} and π(W)H_{n} ⊆ H_{n−2}. Consequently H^{0} = ∑_{n even} H_{n} and H^{1} = ∑_{n odd} H_{n} are invariant subspaces of H. We shall say that π is of type 0 if H^{1} = {0} and that π as of type 1 if H^{0} = {0}.

Lemma 2.2. Suppose π is a quasi-simple irreducible representation of {σ, 𝔄^{0}} on H which is of type m. Suppose moreover that π(D) = \frac{s^{2}−1}{2}I and s − m is not an odd integer. If n ≥ 0 let A_{n} be the restriction of

\frac{π(σ)π(W)^{n}}{∏_{k=0}^{m−1}(s + 2k − (n − 1))}

to H_{n}. If n ≤ 0 let A_{n} be the restriction of

\frac{π(σ)π(V)^{|n|}}{∏_{k=0}^{|n|−1}(s + 2k − (|n| − 1))}

to H_{n}. Then A_{n}^{2} = I for all n. Let A(π) be the operator H whose restriction to H_{n} is A_{n}. A(π) commutes with π(σ) and with π(X) if X is in 𝔄^{0}.

Using the relations Z = \frac{V +W}{2} , 2X = U − i\frac{(V −W)}{2}, 2Y = −U − i\frac{(V −W)}{2} one shows easily that

D = \frac{V W}{4} + \frac{W V}{4} − \frac{U^{2}}{2} = \frac{V W}{2} + iU − \frac{U^{2}}{2} = \frac{W V}{2} − iU − \frac{U^{2}}{2} .

Thus, if φ lies in H_{n},

\begin{aligned}π(V)π(W) & = π(2D − 2iU + U^{2})φ = (s^{2} − 1 + 2n − n^{2})φ = [s^{2} − (n − 1)^{2}]φ \\ π(W)π(V)φ & = π(2D + 2iU + U^{2})φ = (s^{2} − 1 − 2n − n^{2})φ = [s^{2} − (n + 1)^{2}]φ. \end{aligned}

In particular if 0 ≤ j < |n| and φ ∈ W_{n}

\begin{aligned}π(V)^{j+1}π(W)^{j+1}φ & = [s^{2} − (n − 2j − 1)^{2}]π(V)^{j}π(W)^{j}φ if n ≥ 0 \\ π(W)^{j+1}π(V)^{j+1}φ & = [s^{2} − (|n| − 2j − 1)^{2}]π(W)^{j}π(V)^{j}φ if n ≤ 0. \end{aligned}

Since π(σ)π(W)π(σ) = π(V) and π(σ)π(V)π(σ) = π(W) it follows that

A_{n}^{2}ρ = \frac{Π_{j=0}^{|n|−1}[s^{2} − (|n| − 2j − 1)^{2}]}{\{Π_{k=0}^{|n|−1}(s + 2k − (|n| − 1))\}^{2}} φ = φ

It is easy to see that A(π) commutes with π(σ) and π(U). Thus to prove the last assertion of the lemma we need only show that it commutes with π(V) and π(W) or that A_{n+2}π(V) = π(V)A_{n} and A_{n−2}π(W) = π(W)A_{n}. We must study various cases separately.

Suppose that n ≥ 0 and φ belongs to H_{n}.

A_{n+2}π(V)φ = \frac{1}{Π_{k=0}^{n+1}(s + 2k − (n + 1))}π(σ)π(W)^{n+2}π(V)φ

= \frac{π(V)}{Π_{k=0}^{n+1}(s + 2k − (n + 1))} · (s^{2} − (n + 1)^{2})π(σ)π(W)^{n}φ

= π(V)A_{n}φ.

If n ≥ 2

π(W)A_{n}φ = \frac{1}{Π_{k=0}^{n−1}(s + 2k − (n − 1))}π(W)π(σ)π(W)^{n}φ

= \frac{π(σ)}{Π_{k=0}^{n−1}(s + 2k − (n − 1))}(s^{2} − (−n + 1)^{2})π(W)^{n−1}φ

= A_{n−2}π(W)φ.

If n = 1

π(W)A_{n}φ = \frac{1}{s}π(W)π(σ)π(W)φ = \frac{1}{s}π(σ)π(V)π(W)φ = A_{n−2}π(W)φ.

If n = 0

A_{n−2}π(W)φ = \frac{1}{s^{2} − 1}π(σ)π(V)^{2}π(W)φ = \frac{π(W)}{s^{2} − 1}π(σ)π(V)π(W)φ = π(W)π(σ)φ = π(W)A_{n}φ.

There is no need to discuss the case n ≤ 0 because π(σ)A_{n′}π(σ) = A_{−n}, π(σ)π(W)π(σ) = π(V), and π(σ)π(V)π(σ) = π(W).

Lemma 2.3. A quasi-simple representation π of {σ, 𝔄} is irreducible if and only if π^{0} is irreducible. If π is an irreducible quasi-simple representation of {σ, 𝔄} on H there are two possibilities.

(i) The restriction \overline{π} of π to 𝔄 is irreducible and the two representations X → \overline{π}(X) and X → \overline{π}(ad σ(X)) are equivalent.

(ii) H is the direct sum of two subspaces H_{1} and H_{2} invariant under 𝔄. The representations \overline{π}_{1} and \overline{π}_{2} of 𝔄 on H_{1} and H_{2} are inequivalent but \overline{π}_{2} is equivalent to X → \overline{π}_{1}(ad σ(X)) and π(σ)H_{1} = H_{2}.

The first assertion is a matter of definition. Suppose π is irreducible. Either H is irreducible under 𝔄, when the first possibility occurs, or it is not. Suppose it is not. Let H_{1} be a proper subspace of H invariant under 𝔄 and let H_{2} = π(σ)H_{1}. Since H_{1} + H_{2} and H_{1} ∩ H_{2} are invariant under {σ, 𝔄}, H_{1} ∩ H_{2} = {0} and H = H_{1} ⊕ H_{2}. If H′_{1} were a proper subspace of H_{1} invariant under 𝔄 then H′_{1} ⊕ H′_{2}, with H′_{2} = π(σ)H′_{1}, would be a proper invariant subspace of H. \overline{π}_{2} is certainly equivalent to X → \overline{π}_{1}(ad σ(X)). To complete the proof of the lemma we have merely to show that \overline{π}_{1} and X → \overline{π}_{1}(ad σ(X)) are not equivalent. To do this we use the following lemma which is a special case of a theorem of Harish-Chandra (Representations of semi-simple Lie groups, II, T.A.M.S. v. 16, 1954).

Lemma 2.4. Let \overline{σ} be an irreducible quasi-simple representation of 𝔄 on W. There is at least one continuous homomorphism ω of A_{ℝ} into ℂ^{×} such that \overline{σ} is of type |m| and \overline{σ}(D) = \frac{s^{2}−1}{2}I and \overline{σ}(J) = (s_{1}+s_{2})I. Moreover if ω is any such homomorphism \overline{σ} is deducible from \overline{π}_{ω}, the restriction of π_{ω} to 𝔄.

As usual ω_{1}(t) = ω(( \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} )), ω_{2}(t) = ω(( \begin{matrix}1 & 0 \\ 0 & t \end{matrix} )), ω_{i}(t) = |t|^{s_{i}}(\frac{t}{|t|})^{m_{i}}, s = s_{1} − s_{2}, and m = m_{1} − m_{2}.

Although the adjectives of the lemma have only been defined for representations of {σ, 𝔄} their meaning for representations of 𝔄 is clear. The lemma implies that W_{n} is of dimension at most 1. Consequently any linear transformation leaving W_{n} invariant has an eigenvector and any linear transformation commuting with \overline{σ}(X) for all X in 𝔄 is a scalar.

If \overline{π}_{1} and X → \overline{π}_{1}(ad σ(X)) were equivalent there would be an operator A such that A^{−1}\overline{π}_{1}(X)A = \overline{π}_{1}(ad σ(X)) for all X. Thus A^{2}\overline{π}_{1}(X)A^{−2} = A(\overline{π}_{1}(ad σ(X)))A^{−1} = \overline{π}_{1}(X) and A^{2} is a scalar. We may suppose that A^{2} = I. If x lies in H_{1} and X lies in 𝔄 then π(X)(x ⊕ π(σ)Ax) = y ⊕ π(σ)π(A)y if y = π(X)x and π(σ)(x ⊕ π(σ)Ax) = y ⊕ π(σ)Ay if y = Ax so that {x ⊕ π(σ)Ax} is a proper invariant subspace.

Lemma 2.5. Suppose π is an irreducible quasi-simple representation of {σ, 𝔄} on H. There is a continuous homomorphism ω of A_{ℝ} into ℂ^{×} such that π is of type |m|, π(D) = \frac{s^{2}−1}{2}I, π(J) = (s_{1} + s_{2})I and, if s − m is not an odd integer, A(π) = (−1)^{m_{2}}I. If ω is any such homomorphism and π is infinite dimensional then π is deducible from π_{ω}.

Choose s so that π(D) = \frac{s^{2}−1}{2}I and define s_{1} and s_{2} by s_{1} − s_{2} = s and π(J) = (s_{1} + s_{2})I. Choose m_{2} to be 0 or 1 and define m_{1}, which is 0 or 1, by the condition that π is of type |m| if m = m_{2}. If s − m is not an odd integer A(π) is defined and commutes with π(σ) and all π(X). By the previous two lemmas H_{n} is finite dimensional. Consequently A(π) is a scalar. Since A^{2}(π) = I, A(π) = ±I. Choose m_{2} so that A(π) = (−1)^{m_{2}}I. If s − m is an odd integer m_{2} may be chosen to be either 0 or 1. It follows from Lemma 2.1 that if s − m is not an odd integer then A(π_{ω}) = (−1)^{m_{2}}I.

Suppose first that s − m is not an odd integer. Lemmas 2.3, 2.4 and the corollary to Lemma 2.1 imply that \overline{π}, the restriction of π to 𝔄, is irreducible and equivalent to \overline{π}_{ω}. Let B be a map from H to L(ω) such that Bπ(X) = π_{ω}(X)B for all X. I claim that Bπ(σ) = π_{ω}(σ)B. It is enough to verify that Bπ(σ)x = π_{ω}(σ)Bx for x in H_{n}. Clearly BA(π) = A(π_{ω})B. Since A^{2}(π) = I, Bπ(σ)x = Bπ(σ)A^{2}x. If n ≥ 0

Bπ(σ)x = \frac{Bπ(σ)A(π)π(σ)π(W)^{n}x}{Π_{k=0}^{n−1}(s + 2k − (n − 1))}

= \frac{bA(π)π(W)^{n}x}{Π_{k=0}^{n−1}(s + 2k − (n − 1))}

= \frac{A(π_{ω})π_{ω}(W)^{n}Bx}{Π_{k=0}^{n−1}(s + 2k − (n − 1))}

= π_{ω}(σ)Bx and if n ≤ 0

Bπ(σ)x = \frac{Bπ(σ)A(π)π(σ)π(V)^{|n|}x}{Π_{k=0}^{|n|−1}(s + 2k − (|n| − 1))}

= \frac{BA(π)π(V)^{|n|}x}{Π_{k=0}^{|n|−1}(s + 2k − (|n| − 1))}

= \frac{A(π_{ω})π_{ω}(V)^{|n|}Bx}{Π_{k=0}^{|n|−1}(s + 2k − (|n| − 1))}

= π_{ω}(σ)Bx.

If s − m is odd integer and π is infinite dimensional it follows from Lemmas 2.3, 2.4 and the corollary to Lemma 2.1 that H = H_{1} ⊕ H_{2}. Let V′ ⊇ V′′ be subspaces of L(ω) invariant under 𝔄 such that \overline{π}_{1} is equivalent to the representation of 𝔄 on V′/V′′. Let W′ be the intersection of all subspaces of L(ω) which contain V′ and are invariant under {σ, 𝔄}. Let W′′ be the union of all subspaces of L(ω) which are contained in V′′ and are invariant under {σ, 𝔄}. By the corollary to Lemma 2.1 the representation π̃_{ω} of {σ, 𝔄} on W = W′/W′′ is irreducible. By Lemma 2.3 W is the direct sum of two subspaces W_{1} and W_{2} invariant under 𝔄. We may suppose that the representation of 𝔄 on W_{1} is equivalent to \overline{π}_{1}. Let B_{1} be a map of H_{1} to W_{1} such that B_{1}π(S) = π_{ω}(X)B_{1} for X in 𝔤. Let B_{2} = π̃_{ω}(σ)B_{1}π(σ) and set B = B_{1} ⊕ B_{2}. It is immediate that Bπ(σ) = π̃_{ω}(σ)B and Bπ(X) = π̃_{ω}(X)B for all X.

It is not difficult to see that every finite dimensional representation of {σ, 𝔄} is deducible from some π_{ω}. As a consequence A(π) can be defined by the formulas of Lemma 2.6. If π is deducible from π_{ω} then A(π) = (−1)^{m_{2}}I.

Corollary. Suppose λ(D), λ(J), and m, which is to be 0 or 1, are given numbers. Let λ(D) = \frac{s^{2}−1}{2}. If s − m is not an odd integer there are two irreducible quasi-simple representations π of {σ, 𝔄} of type m for which π(D) = λ(D)I and π(J) = λ(J) = λ(J)I. For one A(π) = I and for the other A(π) = −I. If s − m is an odd integer there are three such representations. One is infinite dimensional. The other two are finite dimensional. For one of these A(π) = I and for the other A(π) = −I.

Since s is not unambiguously determined neither is A(π). However once a representation π_{ω} from which π is deducible is specified s can be taken to be s_{1} − s_{2}. Such a choice was implicit at various places in the preceding paragraph.

3. The local functional equation for GL(2, ℝ). If π is an irreducible quasi-simple representation of {σ, 𝔄} and π_{1} is a representation of {σ, 𝔄} on W we shall say that π is contained in π_{1} if there is an invariant subspace V of W such that the restriction of π_{1} to V is equivalent to π. We shall say that π is contained at most once in π_{1} if there is at most one such subspace. If V′ were another such subspace either V ∩ V′ = {0} or V = V′_{i}; thus to show that π is contained at most once in π_{1} one has merely to show that two such subspaces must have a non-zero element in common. Similar notions can be introduced for representations of {σ, 𝔄^{0}}.

If η is a continuous homomorphism of A_{ℝ} into ℂ^{×} let L(η) be the space of all infinitely differentiable U-finite functions on G_{ℝ} satisfying φ(ag) ≡ η(a)φ(g) for all a in A_{ℝ}. If φ lies in L(η) so does ρ(σ)φ and ρ(X)φ for X in 𝔄. Thus we have a representation ρ(η) of {σ, 𝔄} on L(η).

Lemma 3.1. No irreducible quasi-simple representation of {σ, 𝔄} is contained more than once in ρ(η).

Let π be an irreducible quasi-simple representation of {σ, 𝔄} and let π^{0} be its restriction to {σ, 𝔄^{0}}. Suppose π is deducible from π_{ω}. Let L^{0}(η) be the space of infinitely differentiable U-finite functions on G_{ℝ}^{0} satisfying φ(ag) ≡ η(a)φ(g) for all a in A_{ℝ} ∩ G_{ℝ}^{0} and let ρ^{0}(η) be the representation of {σ, 𝔄^{0}} on L^{0}(η). It is enough to show that π^{0} is contained at most once in ρ^{0}(η).

Suppose H ⊆ L^{0}(η) and the restriction of ρ^{0}(η) to H is equivalent to π^{0}. The integers n for which H_{n} ≠ {0} are determined by π. To prove the lemma we need only show that, for some such n, H_{n} is uniquely determined by π. Let η_{1}(t) = η(( \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} )), η_{2}(t) = η(( \begin{matrix}1 & 0 \\ 0 & t \end{matrix} )), and let η_{i}(t) = |t|^{r_{i}}(\frac{g}{|t|})^{ℓ_{i}} with ℓ_{i} = 0 or 1. If φ lies in H_{n} set ψ(x) = φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )); then

φ(g) = η(a)ψ(x)e^{inθ}

if

g = a ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} )

with a in A_{ℝ} ∩ G_{ℝ}^{0}. Consequently φ is uniquely determined by ψ. Let φ_{1} = ρ(V)φ, φ_{2} = ρ(W)φ, and let ψ_{1} and ψ_{2} be the corresponding functions on ℝ. Since

\begin{aligned}ρ(U)φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) & = inψ(x) \\ ρ(Z)φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) & = rψ(x) − 2x\frac{dψ}{dx} \\ ρ(X)φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) & = \frac{dψ}{dx} \end{aligned}

and

\begin{aligned}V & = Z + 2iX − iU \\ W & = Z − 2iX + iU \end{aligned}

one has

\begin{aligned}ψ_{1}(x) & = −2(x − i)\frac{dψ}{dx} + (r + n)ψ \\ ψ_{2}(x) & = −2(x + i)\frac{dψ}{dx} + (r − n)ψ. \end{aligned}

Moreover ρ(D)φ = ρ(\frac{W V}{2} − iU − \frac{U^{2}}{2} )φ corresponds to the function 2(x^{2} + 1)\frac{d^{2}ψ}{dx^{2}} + (4x − 2rx − 2in)\frac{dψ}{dx} + \frac{(r−1)^{2}−1}{2}ψ.

Consequently

2(x^{2} + 1)\frac{d^{2}ψ}{dx^{2}} + (4x − 2rx − 2in)\frac{dψ}{dx} + \frac{[(r − 1)^{2} − s^{2}]}{2}ψ = 0.  (A)

Finally ρ(σ)φ corresponds to (−1)^{ℓ_{2}}ψ(−x).

There are a number of separate cases to consider. If s−m is an odd integer and π is infinite dimensional take n_{0} = |s| + 1. Then H_{n_{0}} ≠ {0} and ρ(W)φ = 0 if φ ∈ H_{n_{0}}. Thus

−2(x + i)\frac{dψ}{dx} + (r − n_{0})ψ = 0.

This equation determines ψ up to a scalar factor.

If s − m is not an odd integer or π is finite-dimensional and if m = 0 then H_{0} ≠ {0}. If φ lies in H_{0} then ψ must satisfy equation (A) and the condition ψ(−x) = (−1)^{ℓ_{2}+m_{2}}ψ(x) because A(π^{0}) = (−1)^{m_{2}}I. Thus ψ is determined up to a scalar factor.

If s − m is not an odd integer or π is finite-dimensional and if |m| = 1 then H_{1} ≠ {0}. Referring to the definition of A(π) in Lemma 2.2 we see that ψ satisfies equation (A) and the equation

−2(x + i)\frac{dψ}{dx} + (r − 1)ψ(x) = (−1)^{ℓ_{2}+m_{2}}sψ(−x).

This equation implies a non-trivial linear relation between the values of ψ and its first derivative at x = 0. Thus ψ is determined up to a scalar factor.

If ξ(x) = e^{iux}, with u ≠ 0, is a non-trivial character of ℝ let L(ξ) be the space of all infinitely differentiable U-finite functions on G_{ℝ} satisfying:

(i) φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) ≡ ξ(x)φ(g) for all x in ℝ,

(ii) if g belongs to G_{ℝ} and X belongs to 𝔄 there is a constant M such that

|ρ(X)φ(( \begin{matrix}t_{1} & 0 \\ 0 & t_{2} \end{matrix} ) g) ≤ M\{|t_{1}|^{M} + |t_{2}|^{M}|\}

for |t_{1}| ≥ |t_{2}|.

Let ρ(ξ) be the representation of {σ, 𝔄} on L(ξ).

Lemma 3.2. No irreducible quasi-simple representation of {σ, 𝔄} is contained more than once in ρ(ξ).

Let π be such a representation and let π be deducible from π_{ω}. Let L^{0}(ξ) be the space of all infinitely differentiable U-finite functions on G_{ℝ}^{0} satisfying

(i) φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) ≡ ξ(x)φ(g),

(ii) If g lies in G_{ℝ}^{0} and X lies in 𝔄^{0} there is a constant M such that

|ρ(X)φ(( \begin{matrix}t^{1/2} & 0 \\ 0 & t^{−1/2} \end{matrix} ) g)| ≤ Mt^{M}

for t ≥ 1.

Let ρ^{0}(ξ) be the representation of {σ, 𝔄^{0}} on L^{0}(ξ). It is enough to show that π^{0} is contained at most once in ρ^{0}(η). The proof of this will be similar to the proof of the previous lemma.

Suppose H is an invariant subspace of L^{0}(ξ) and the restriction of ρ^{0}(ξ) to H is equivalent to π^{0}. If φ lies in H_{n} set

ψ(t) = φ(( \begin{matrix}\frac{t}{|t|^{1/2}} & 0 \\ 0 & \frac{1}{|t|^{1/2}} \end{matrix} )), t ∈ ℝ^{×}.

Since φ(g) = ξ(x)ψ(t)e^{inθ} if

g = ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}\frac{t}{|t|^{1/2}} & 0 \\ 0 & \frac{1}{|t|^{1/2}} \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} )

the function φ is determined by ψ. Let φ_{1} = ρ(V)φ, φ_{2} = ρ(W)ρ, and let ψ_{1} and ψ_{2} be the corresponding functions on ℝ^{×}. Since

\begin{aligned}ρ(U)φ(( \begin{matrix}\frac{t}{|t|^{1/2}} & 0 \\ 0 & \frac{1}{|t|^{1/2}} \end{matrix} )) & = inψ(t) \\ ρ(Z)φ(( \begin{matrix}\frac{t}{|t|^{1/2}} & 0 \\ 0 & \frac{1}{|t|^{1/2}} \end{matrix} )) & = 2t\frac{dψ}{dt} \\ ρ(X)φ(( \begin{matrix}\frac{t}{|t|^{1/2}} & 0 \\ 0 & \frac{1}{|t|^{1/2}} \end{matrix} )) & = iutψ(t) \end{aligned}

one has

\begin{aligned}ψ_{1}(t) & = 2t\frac{dψ}{dt} − (2ut − n)ψ \\ ψ_{2}(t) & = 2t\frac{dψ}{dt} + (2ut − n)ψ. \end{aligned}

Moreover ρ(D)φ corresponds to 2t\frac{d}{dt}(t\frac{dψ}{dt}) − 2t\frac{dψ}{dt} + (2nut − 2u^{2}t^{2})ψ so that

2t\frac{d}{dt}(t\frac{dψ}{dt}) − 2t\frac{dψ}{dt} + (2nut − 2u^{2}t^{2})ψ = \frac{s^{2} − 1}{2}ψ.  (B)

Finally ρ(σ)φ corresponds to (−1)^{n}ψ(−t).

Suppose that s − m is an odd integer and π is infinite dimensional. Take n_{0} = |s| + 1. Then H_{n_{0}} ≠ {0} and ρ(W)φ = 0 if φ belongs to H_{n_{0}}. Consequently

2t\frac{dψ}{dt} + (2ut − n_{0})ψ = 0.

If ψ is to satisfy this equation and the growth condition it must vanish for ut < 0 and be a multiple of |t|^{n_{0}/2}e^{−ut} for ut > 0. Thus it is determined up to a scalar factor.

Before discussing the remaining cases we should comment on equation (B). It may be written as

\frac{d^{2}ψ}{dt^{2}} + (−u^{2} + \frac{nu}{t} + \frac{(1 − s^{2})}{4t^{2}})ψ = 0.

Dropping the terms in \frac{1}{t} and \frac{1}{t^{2}} we obtain the equation \frac{d^{2}ψ}{dt^{2}} − u^{2}ψ = O. As a consequence the original equation has one solution on the positive real axis of the form t^{μ}e^{−|u|t}(1 + O(\frac{1}{t})) and one of the form t^{ν}e^{|u|t}(1 + O(\frac{1}{t})). Only the first will satisfy the growth conditions. On the negative real axis it has solutions of the forms t^{μ′}e^{|u|t}(1+O(\frac{1}{t})) and t^{ν′}e^{−|u|t}(1+O(\frac{1}{t})). Only the first satisfies the required growth conditions. Thus the space of solutions of equation (B) which satisfy the growth conditions has dimension two.

If s − m is not an odd integer or π is finite-dimensional and if m = 0 then H_{0} ≠ {0}. If φ belongs to H_{0} then ψ(−t) = (−1)^{m_{2}}ψ(t) because A(π^{0}) = (−1)^{m_{2}}I. This supplementary condition will determine ψ up to a scalar factor.

If s − m is not an odd integer or π is finite dimensional and if |m| = 1 then H_{1} ≠ {0}. If φ belongs to H_{1} then

2t\frac{dψ}{dt} + (2ut − 1)ψ(t) = (−1)^{(m_{2}+1)}sψ(−t).

This supplementary condition determines ψ up to a scalar factor.

Suppose ψ(t) satisfies equation (B) with n = 1 and

ψ′(t) = \frac{1}{s}(−1)^{m_{2}+1}\{−2t\frac{dψ}{dt} (−t) − (2ut + 1)ψ(−t)\}.

Then

\frac{(−1)^{m_{2}+1}}{s} \{−2t\frac{dψ′}{dt} (−t) − (2ut + 1)ψ′(−t)\} = \frac{(−1)^{m_{2}+1}}{s} \{2t\frac{d}{dt}(ψ′(−t)) − (2ut + 1)ψ′(−t)\}

which equals

\frac{2}{s^{2}} \{t\frac{d}{dt}[2t\frac{dψ}{dt} + (2ut − 1)ψ] − \frac{(2ut + 1)}{2}[2t\frac{dψ}{dt} + (2ut − 1)ψ(t)\}.

Simplifying we obtain

\frac{2}{s^{2}} \{2t\frac{d}{dt}(t\frac{dψ}{dt}) − 2t\frac{dψ}{dt} + (−2u^{2}t^{2} + 2ut + \frac{1}{2})ψ\}

which is just ψ itself.

Corollary. Let π be an irreducible quasi-simple representation of {σ, 𝔄}. π is contained in ρ(ξ) if and only if π is infinite dimensional.

It is enough to show that π^{0} is contained in π^{0}(ξ) if and only if π^{0} is infinite dimensional. Suppose H is a non-trivial finite dimensional subspace of L^{0}(ξ). Let τ be the representation of {σ, 𝔄} on H and let τ̃ be the contragredient representation. If

X_{x} = ( \begin{matrix}0 & x \\ 0 & 0 \end{matrix} )

the only eigenvalue of τ̃(X_{x}) is zero because τ̃ is finite dimensional. Let φ̃ be the element in the dual of H defined by φ̃(φ) = φ(1). φ̃ is not zero and

(τ̃(X_{x})φ̃)(φ) = −φ̃(τ(X_{x})φ) = −(τ(X_{x})ϖ)(1) = −iuφ(1)

so −iu is an eigenvalue of τ̃(X_{x}). This is a contradiction.

Suppose π is infinite dimensional and deducible from π_{ω}. Let L^{0}(ξ, s) be the space of functions in L^{0}(ξ) satisfying ρ(D)φ = \frac{s^{2}−1}{2}φ. The dimension of L^{0}(ξ, s)_{n} is two. Let

L^{0}(ξ, s, m) = Σ_{\frac{n−m}{2} ∈ℤ}L^{0}(ξ, s)_{n}.

and let ρ^{0}(ξ, s, m) be the representation of {σ, 𝔄} on L^{0}(ξ, s, m).

Suppose W_{1} ⊇ W_{2} are two invariant subspaces of L^{0}(ξ, s, m) and W = W_{1}/W_{2}. The representation of {σ, 𝔄} on W is quasi-simple. Choose n so that W_{n} is not empty. The dimension of W_{n} is at most two. Among all the non-zero subspaces of W_{n} obtained by intersecting W_{n} with an invariant subspace of W there is a minimal one W_{n}^{0}. Let W′ be the intersection of all invariant subspaces containing W_{n}^{0} and let W′′ be the sum of all invariant subspaces of W′ which do not contain W_{n}^{0}. W′′ does not contain W_{n}^{0} and the representation of {σ, 𝔄} on V = W′/W′′ is irreducible.

If s − m is not an odd integer Lemma 2.1 and Lemma 2.5 and its corollary imply that V_{n} ≠ \{0\} if \frac{n−m}{2} is an integer. Because the dimension of L^{0}(ξ, s)_{n} is two we conclude that there is no chain L^{0}(ξ, s, m) ⫌ W_{1} ⫌ W_{2} ⫌ {0} of invariant subspaces. The operator A(ρ^{0}(ξ, s, m)) is defined and L^{0}(ξ, s, m) is the direct sum of L^{+} = {φ | A(ρ^{0}(ξ, s, m))φ = φ} and L^{−} = {φ | A(ρ^{0}(ξ, s, m))φ = −φ}. We have seen that neither of these is empty. Consequently they are both irreducible and the corollary to Lemma 2.5 implies that the restriction of ρ^{0}(ξ, s, m) to one of them is equivalent to π^{0}.

If s − m is an odd integer the same kind of argument shows that there is no chain L^{0}(ξ, s, m) ⫌ W_{1} ⫌ W_{2} ⫌ W_{3} ⫌ W_{4} ⫌ {0} of invariant subspaces. As a consequence L^{0}(ξ, s, m) must contain an invariant irreducible subspace. The restriction of ρ^{0}(ξ, s, m) to this subspace will be equivalent to π^{0} which is the only infinite dimensional irreducible representation deducible from π_{ω}^{0}.

We return to the study of the functions ψ(t). The Mellin transforms

\begin{aligned}θ^{+}(Z) & = ∫_{R^{×}} ψ(t)|t|^{z−1}dt \\ θ^{−}(Z) & = ∫_{R^{×}} ψ(t)(sgn t)|t|^{z−1}dt \end{aligned}

are defined for Re z sufficiently large. Equations (B) are equivalent to the difference equations

\begin{matrix}[(2z + 1)^{2} − s^{2}]θ^{+}(z) & + 4nuθ^{−}(z + 1) − 4u^{2}θ^{+}(z + 2) = 0 \\ [(2z + 1)^{2} − s^{2}]θ^{−}(z) & + 4nuθ^{+}(z + 1) − 4u^{2}θ^{−}(z + 2) = 0. \end{matrix}

If, as before, ψ corresponds to φ, ψ_{1} corresponds to φ_{1} = ρ(V)ρ, and ψ_{2} corresponds to φ_{2} = ρ(W)φ let θ_{i}^{+} and θ_{i}^{−} be the Mellin transforms of ψ_{i}. Then

\begin{aligned}θ_{1}^{+}(z) & = −2zθ^{+}(Z) − 2uθ^{−}(z + 1) + nθ^{+}(z) \\ θ_{1}^{−}(z) & = −2zθ^{−}(z) − 2uθ^{+}(z + 1) + nθ^{−}(z) \\ θ_{2}^{+}(z) & = −2zθ^{+}(z) + 2uθ^{−}(z + 1) − nθ^{+}(z) \\ θ_{2}^{−}(z) & = −2zθ^{−}(z) + 2uθ^{+}(z + 1) − nθ^{−}(z). \end{aligned}  (C)

If φ is replaced by ρ(σ)φ then θ^{+}(z) is replaced by (−1)^{n}θ^{+}(z) and θ^{−}(z) is replaced by (−1)^{n+1}θ^{−}(z).

If π is an infinite dimensional irreducible quasi-simple representation of {σ, 𝔄} let L^{0}(ξ, π) be the unique subspace of L^{0}(ξ) which transforms according to π^{0}.

Lemma 3.3. Suppose π is an infinite-dimensional irreducible quasi-simple representation of {σ, 𝔄} which is deducible from π_{ω}. If L^{0}(ξ, π)_{n} ≠ 0 let θ_{n}^{+}(z) and θ_{n}^{−}(z) be the Mellin transforms corresponding to some non-zero element in L^{0}(ξ, π)_{n}.

(i) If s − m is not an odd integer, m = 0, and m_{2} = 0, then

θ_{0}^{+}(z) = α_{0}(\frac{2}{|u|})^{z}Γ(\frac{z + \frac{1}{2} + \frac{s}{2}}{2} )Γ(\frac{z + \frac{1}{2} − \frac{s}{2}}{2} )
θ_{0}^{−}(z) = 0
θ_{2}^{+}(z) = α_{1}(\frac{2}{|u|})^{z}zΓ(\frac{z + \frac{1}{2} + \frac{s}{2}}{2} )Γ(\frac{z + \frac{1}{2} − \frac{s}{2}}{2} )
θ_{2}^{−}(z) = 2α_{1}sgn u(\frac{2}{|u|})^{z}Γ(\frac{z + \frac{3}{2} + \frac{s}{2}}{2} )Γ(\frac{z + \frac{3}{2} − \frac{s}{2}}{2} ).

(ii) If s − m is not an odd integer, m = 0, and m + 2 = 1, then

θ_{0}^{+}(z) = 0
θ_{0}^{−}(z) = β_{0}(\frac{2}{|u|})^{z}Γ(\frac{z + \frac{1}{2} + \frac{s}{2}}{2} )Γ(\frac{z + \frac{1}{2} − \frac{s}{2}}{2} )
θ_{2}^{+}(z) = 2β_{2} sgn u(\frac{2}{|u|})^{z}Γ(\frac{z + \frac{3}{2} + \frac{s}{2}}{2} )Γ(\frac{z + \frac{3}{2} − \frac{s}{2}}{2} )
θ_{2}^{−}(z) = β_{1}(\frac{2}{|u|})^{z}zΓ(\frac{z + \frac{1}{2} + \frac{s}{2}}{2} )Γ(\frac{z + \frac{1}{2} − \frac{s}{2}}{2} ).

(iii) If s − m is not an odd integer, |m| = 1, and m_{2} = 0 then

\begin{aligned}θ_{1}^{+}(z) & = γ_{0}(\frac{2}{|u|})^{z}Γ(\frac{+\frac{3}{2} + \frac{s}{2}}{2} )Γ(\frac{z + \frac{1}{2} − \frac{s}{2}}{2} ) \\ θ_{1}^{−}(x) & = γ_{0} sgn u(\frac{2}{|u|})^{z}Γ(\frac{z + \frac{1}{2} + \frac{s}{2}}{2} )Γ(\frac{z + \frac{3}{2} − \frac{s}{2}}{2} ). \end{aligned}

(iv) If s − m is not an odd integer, |m| = 1, and m_{1} = 1 then

\begin{aligned}θ_{1}^{+}(z) & = γ_{1}(\frac{2}{|u|})^{z}Γ(\frac{z + \frac{1}{2} + \frac{s}{2}}{2} )Γ(\frac{z + \frac{3}{2} − \frac{s}{2}}{2} ) \\ θ_{1}^{−}(z) & = γ_{1} sgn u(\frac{2}{|u|})^{z}Γ(\frac{z + \frac{3}{2} + \frac{s}{2}}{2} )Γ(\frac{z + \frac{1}{2} − \frac{s}{2}}{2} ). \end{aligned}

(v) If s − m is an odd integer and n_{0} = |s| + 1 then

\begin{aligned}θ_{n_{0}}^{+}(z) & = \frac{δ_{0}}{|u|^{z+ \frac{n_{0}}{2}}} Γ(z + \frac{1}{2} + \frac{|s|}{2}) \\ θ_{n_{0}}^{−}(z) & = \frac{δ_{0}}{|u|^{z+ \frac{n_{0}}{2}}} sgn uΓ(z + \frac{1}{2} + \frac{|s|}{2}). \end{aligned}

The letters α_{0}, α_{1}, β_{0}, β_{1}, γ_{0}, γ_{1}, δ_{0} denote constants.

If s−m is not an odd integer and m = 0 the supplementary conditions on θ_{0}^{+}(z) and θ_{0}^{−}(z) corresponding to A(π^{0}) = (−1)^{m_{2}}I are θ_{0}^{+} = (−1)^{m_{2}}θ_{0}^{+}(z), θ_{0}^{−}(z) = (−1)^{m_{2}+1}θ_{0}^{−}(z). The first and second functions in parts (i) and (ii) of the lemma satisfy these conditions as well as the difference equations. Taking the inverse Mellin transform we obtain a function ψ(t) which satisfies the growth condition as well as the differential equation (B). The function defined by φ(g) = ψ(x)ψ(t) if

g = ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}\frac{t}{|t|^{1/2}} & 0 \\ 0 & \frac{1}{|t|^{1/2}} \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} )

will lie in L^{0}(ξ, s, m)_{0}. Moreover A(ρ^{0}(ξ, s, m))φ will equal (−1)^{m_{2}}φ so that φ will lie in L^{0}(ξ, π)_{0}. Thus the first two equations of parts (i) and (ii) are valid. The last two can be obtained from the first two by applying relations (C).

In the first four cases π is equivalent to π_{ω}. It follows from Lemma 2.5 that π is equivalent to πω̃ if

ω̃(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )) = ω(( \begin{matrix}α_{2} & 0 \\ 0 & α_{1} \end{matrix} )).

Replacing ω by ω̃ interchanges cases (iii) and (iv) so we need discuss case (iii) alone.

Substituting the function (\frac{2}{|u|})^{z}Γ(\frac{z+ \frac{3}{2} ± \frac{s}{2}}{2} )Γ(\frac{z+ \frac{1}{2} ∓ \frac{s}{2}}{2} ) (taking only all the upper signs or all the lower signs) into the expression [(2z + 1)^{2} − s^{2}]θ(z) − 4u^{2}θ(z + 2) one obtains

[\frac{(2z + 1)^{2} − s^{2}}{2} − 2(z + \frac{3}{2} ± s)(z + \frac{1}{2} ∓ s)]|u|(\frac{2}{|u|})^{z+1}Γ\frac{(z + \frac{3}{2} ± \frac{s}{2})}{2}Γ\frac{(z + \frac{1}{2} ∓ \frac{s}{2})}{2}

which equals

−4u\{ sgn u(\frac{2}{|u|})^{z+1}Γ(\frac{(z + 1) + \frac{1}{2} ± \frac{s}{2}}{2} )Γ(\frac{(z + 1) + \frac{3}{2} ∓ \frac{s}{2}}{2} )\}.

Consequently the functions of part (iii) satisfy the difference equations. The supplementary conditions on θ_{1}^{+}(z) and θ_{1}^{−}(z) correspond to the relation A(π) = I are

\begin{aligned}sθ_{1}^{+}(z) & = 2zθ_{1}^{+}(z) − 2uθ_{1}^{−}(z + 1) + θ_{1}^{+}(z) \\ −sθ_{1}^{−}(z) & = 2zθ_{1}^{−}(z) − 2uθ_{1}^{+}(z + 1) + θ_{1}^{−}(z). \end{aligned}

These will be satisfied by the functions of part (iii) because

\begin{aligned}(2z + 1 ∓ s)(\frac{2}{|u|})^{z}Γ( & \frac{z + \frac{1}{2} ∓ \frac{s}{2}}{2} )Γ(\frac{z + \frac{3}{2} ± \frac{s}{2}}{2} ) \\ & = 2u\{sgn u(\frac{2}{|u|})^{z+1}Γ(\frac{(z + 1) + \frac{1}{2} ± \frac{s}{2}}{2} )Γ(\frac{(z + 1) + \frac{3}{2} ∓ \frac{s}{2}}{2} ). \end{aligned}

The formulae of part (iii) can now be proved in the same way as those of parts (i) and (ii).

The simplest way to prove part (v) is to appeal to the explicit form for the corresponding function ψ(t) found during the proof of Lemma 3.2.

Lemma 3.4. Suppose ψ(t) corresponds to φ in L^{0}(ξ, π)_{n} and π is deducible from π_{ω}.

(i) If s − m is not an integer, m = 0, m_{2} = 0 then, m a neighbourhood of 0, ψ(t) has a convergent expansion

of the form

|t|^{\frac{s+1}{2}} ∑_{p=0}^{∞} a_{p}t^{p} + |t|^{\frac{−s+1}{2}} ∑_{p=0}^{∞} b_{p}t^{p}.

(ii) If s − m is an even integer, m = 0, and m_{2} = 0 then, in a neighbourhood of 0, ψ(t) has a convergent

expansion of the form

|t|^{\frac{−|s|+1}{2}} ∑_{p=0}^{∞} a_{p}t^{p} + (log |t|)t^{\frac{|s|+1}{2}} ∑_{p=0}^{∞} b_{p}t^{p}.

(iii) If s − m is not an integer, m = 0, and m_{2} = 1 then, in a neighbourhood of 0, ψ(t) has a convergent

expansion of the form

(sgn t)|t|^{\frac{s+1}{2}} ∑_{p=0}^{∞} a_{p}t^{p} + (sgn t)|t|^{\frac{−s+1}{2}} ∑_{p=0}^{∞} b_{p}t^{p}.

(iv) If s − m is an even integer, m = 0, and m_{2} = 1, then, in a neighbourhood of 0, ψ(t) has a convergent expansion of the form

(sgn t)|t|^{\frac{−|s|+1}{2}} ∑_{p=0}^{∞} a_{p}t^{p} + (sgn t)|t|^{\frac{|s|1}{2}} log |t| ∑_{p=0}^{∞} b_{p}t^{p}.

(v) If s − m is not an integer, |m| = 1, and m_{2} = 0 then, in a neighbourhood of 0, ψ(t) has a convergent

expansion of the form

(sgn t)|t|^{\frac{s+1}{2}} ∑_{p=0}^{∞} a_{p}t^{p} + |t|^{\frac{−s+1}{2}} ∑_{p=0}^{∞} b_{p}t^{p}.

(vi) If s − m is an even integer, |m| = 1, and m_{2} = 0 then, in a neighbourhood of 0, ψ(t) has a convergent

expansion of the form

|t|^{\frac{−s+1}{2}} ∑_{p=0}^{∞} a_{p}t^{p} + (sgn t)|t|^{\frac{s+1}{2}} log |t| ∑_{p=0}^{∞} b_{p}t^{p}

if s is positive and one of the form

(sgn t)|t|^{\frac{s+1}{2}} ∑_{p=0}^{∞} a_{p}L^{p} + |t|^{\frac{−s+1}{2}} log |t| ∑_{p=0}^{∞} b_{p}t^{p}

if s is negative.

(vii) If s − m is not an integer, |m| = 1, and m_{2} = 1, then, in a neighbourhood of 0, ψ(t) has a convergent

expansion of the form

|t|^{\frac{s+1}{2}} ∑_{p=0}^{∞} a_{p}t^{p} + (sgn t)|t|^{\frac{−s+1}{2}} ∑_{p=0}^{∞} b_{p}t^{p}.

(viii) If s − m is an even integer, |m| = 1, and m_{2} = 1 then, in a neighbourhood of 0, ψ(t) has a convergent

expansion of the form

(sgn t)|t|^{\frac{−s+1}{2}} ∑_{p=0}^{∞} a_{t}t^{p} + |t|^{\frac{s+1}{2}} log |t| ∑_{p=0}^{∞} b_{p}t^{p}

if s is positive and one of the form

|t|^{\frac{s+1}{2}} ∑_{p=0}^{∞} a_{p}t^{p} + (sgn t)|t|^{\frac{−s+1}{2}} log |t| ∑_{p=0}^{∞} b_{p}t^{p}

if s is negative.

(ix) If s − m is an odd integer then ψ(t) is zero unless nut > 0 and in this region ψ(t) has a convergent

expansion of the form

|t|^{\frac{|s|+1}{2}} ∑_{p=0}^{∞} a_{t}t^{p}.

We know that if ψ(t) corresponds to φ then 2t\frac{dψ}{dt} − (2ut − n)ψ corresponds to ρ(V)φ, 2t\frac{dψ}{dt} + (2ut − n)ψ corresponds to ρ(W)φ, and (−1)^{n}ψ(−t) corresponds to ρ(σ)φ. Because each of these operations take a function with an expansion of one of the given forms to a function with an expansion of the same form and π^{0} is irreducible it will be enough to show that there is at least one n for which the lemma is valid. If s − m is not an odd integer we shall take n = |m| and if s − m is an odd integer we shall take n = |s| + 1.

The indicial equation of the equation (B) does not depend on n. It is (2λ − 1)^{2} − s^{2} = 0 and has the roots λ_{1} = \frac{s+1}{2}, λ_{2} = \frac{−s+1}{2} with difference λ_{1} − λ_{2} = s. If n = 0 the series t^{λ_{i}} ∑_{p=0}^{∞} c_{t}t^{p}, t > 0 satisfies the equation if and only if

[(2(λ_{i} + p) − 1)^{2} − s^{2}]c_{p} = 4u^{2}c_{p−2}.

Thus if s is not an integer ψ(t) has an expansion of the form

t^{\frac{s+1}{2}} ∑_{p=0}^{∞} a_{2p}t^{2p} + t^{\frac{−s+1}{2}} ∑_{p=0}^{∞} b_{2p}t^{2p}

valid for t positive and close to 0. If s is an even integer one of the two linearly independent solutions given by the method of Frobenius must contain a logarithmic term because it will not be possible to solve these equations recursively when λ_{i} is the smaller of the roots. Since the equation is invariant under the substitution t → −t the logarithmic solution must be of the form

t^{\frac{−|s|+1}{2}} ∑_{p=0}^{∞} c_{2p}t^{2p} + t^{\frac{|s|+1}{2}} log t ∑_{p=0}^{∞} d_{2p}t^{2p}

and ψ(t) has an expansion of the form

t^{\frac{−|s|+1}{2}} ∑_{p=0}^{∞} a_{2p}t^{2p} + t^{\frac{|s|+1}{2}} log t ∑_{p=0}^{∞} b_{2p}t^{2p}

valid for t positive and close to 0. Cases (i) to (iv) of the lemma follow immediately because, since n = 0, ψ(t) is even in the first two and odd in the second two.

Just as in the previous lemma, (vii) and (viii) are redundant since they are covered already by (v) and (vi) which we now treat. If n = 1, t^{\frac{±s+1}{2}} ∑_{p=0}^{∞} c_{p}t^{p}, t > 0, satisfies equation (B), if and only if

\frac{1}{2}[(±s + 2p)^{2} − s^{2}]c_{p} + 2uc_{p−1} − 2u^{2}c_{p−2} = 0

or

(±2ps + 2p^{2})c_{p} + 2uc_{p−1} − 2u^{2}c_{p−1} = 0.

For convenience let c_{p} = 0 if p < 0. If s is not an integer choose c_{0}^{±} and define c_{p}^{±} inductively by (±s + 2p)c_{p}^{±} + 2uc_{p−1}^{±} = ±s(−1)^{p}c_{p}^{±} or, equivalently, (±s + p)c_{p}^{±} + uc_{p−1}^{±} = 0 when p is odd and pc_{p}^{±} + uc_{p−1}^{±} = 0 when p is even. This equation will be satisfied for all p if c_{p}^{±} = 0 when p is negative. If p is odd

(±ps + p^{2})c_{p}^{±} + uc_{p−1}^{±} − u^{2}c_{p−2}^{±} = −u[(p − 1)c_{p−1}^{±} + uc_{p−2}^{±}] = 0

and if p is even

(±ps + p^{2})c_{p}^{±} + uc_{p−1}^{±} − u^{2}c_{p−2}^{±} = −u[(±s + (p − 1))c_{p−1}^{±} + uc_{p−2}^{±}] = 0.

Thus, if s is not an integer, ψ(t) will have an expansion of the form

t^{\frac{s+1}{2}} ∑_{p=0}^{∞} c_{p}^{+}t^{p} + t^{\frac{−s+1}{2}} ∑_{p=0}^{∞} c_{p}^{−}t^{p}

valid for t positive and close to zero. Since m_{2} = 0

−sψ(−t) = 2t\frac{dψ}{dt} + (2ut − 1)ψ(t).

The expression

2t\frac{d}{dt}\{t^{\frac{±s+1}{2}} ∑_{p=0}^{∞} c_{t}^{±}t^{p}\} + (2ut − 1)t^{\frac{±s+1}{2}} ∑_{p=0}^{∞} c_{p}^{±}t^{p}

is equal to

t^{\frac{±s+1}{2}} ∑_{p=0}^{∞}[(±s + 2p)c_{p}^{±} + 2uc_{p−1}^{±}]t^{p} = ±st^{\frac{±s+1}{2}} ∑_{p=0}^{∞} c_{p}^{±}(−t)^{p}.

Case (v) of the lemma for n = 1 follows immediately.

Since

t\frac{d}{dt}(log tA(t)) = t\frac{dA}{dt} log t + A(t)
t\frac{d}{dt}(t\frac{d}{dt}(log tA(t))) = t\frac{d}{dt}(t\frac{dA}{dt}) + 2t\frac{dA}{dt}

the series

t^{\frac{−|s|+1}{2}} ∑_{p=0}^{∞} c_{p}t^{p} + t^{\frac{|s|+1}{2}} log t ∑_{p=0}^{∞} d_{p}t^{p}, t > 0,

will satisfy equation (B) when s is an odd integer and n = 1 if and only if

\frac{[(|s| + 2p)^{2} − s^{2}]}{2}d_{p} + 2ud_{p−1} − 2u^{2}d_{p−1} = 0

or

(|s|p + p^{2})d_{p} + ud_{p−1} − u^{2}d_{p−1} = 0

and

(−|s|p + p^{2})cp + uc_{p−1} − u^{2}c_{p−2} + (−|s| + 2p)d_{p−|s|} = 0.

Choose c_{0} and c_{|s|} and define the other coefficients by (|s| + 2p)d_{p} + 2ud_{p−1} = (−1)^{p}|s|d_{p} or pd_{p} + ud_{p−1} = 0 if p is even and (|s| + p)d_{p} + ud_{p−1} = 0 if p is odd and

(−|s| + 2p)c_{p} + 2uc_{p−1} + 2d_{p−|s|} = (−1)^{p+1}|s|c_{p}

or

pc_{p} + uc_{p−1} + d_{p−|s|} = 0

if p is even and

(−|s| + p)c_{p} + uc_{p−2} + d_{p−|s|} = 0

if p is odd. Take c_{p} and d_{p} to be 0 if p is negative. These equations are consistent and determine the remaining c_{p} and all d_{p} uniquely. We have already seen that the coefficients d_{p} will satisfy

(|s|p + p^{2})d_{p} + ud_{p−1} − u^{2}d_{p−1} = 0.

If p is even

(−|s|p + p^{2})c_{p} + uc_{p−1} − u^{2}c_{p−2} + (−|s| + 2p)d_{p−|s|}

equals

[|s| − (p − 1)]uc_{p−1} − u^{2}c_{p−2} + pd_{p−|s|} = ud_{p−|s|−1} + (|s| + (p − |s|))d_{p−|s|} = 0

and if p is odd

(−|s|p + p^{2})c_{p} + uc_{p−1} − u^{2}c_{p−2} + (−|s| + 2p)d_{p−|s|}

equals

−u[(p − 1)c_{p−1} + uc_{p−1}] + (p − |s|)d_{p−|s|} = ud_{p−|s|−1} + (p − |s|)d_{p−|s|} = 0.

Thus if c_{0} and c_{|s|} are suitably chosen

ψ(t) = t^{\frac{−|s|+1}{2}} ∑_{p=0}^{∞} c_{p}t^{p} + t^{\frac{|s|+1}{2}} log t ∑_{p=0}^{∞} d_{p}t^{p}

for t positive and close to 0.

Since m_{2} = 0

−sψ(−t) = 2t\frac{dψ}{dt} + (2ut − 1)ψ.

The right hand side is equal to

t^{\frac{−|s|+1}{2}} ∑_{p=0}^{∞} c′_{p}t^{p} + t^{\frac{|s|+1}{2}} ∑_{p=0}^{∞} d′_{p}t^{p}

with

\begin{aligned}c′_{p} & = (−|s| + 2p)c_{p} + 2uc_{p−1} + 2d_{p−|s|} = −|s|(−1)^{p}c_{p} \\ d′_{p} & = (|s| + 2p)d_{p} + 2ud_{p−1} = (−1)^{p}|s|d_{p}. \end{aligned}

Case (vi) of the lemma follows.

The assertion for case (ix) with n = |s| + 1 was established while proving Lemma 3.2.

If ψ(t) is the function of the lemma and x is a real number the functions

\begin{aligned}θ^{+}(z, x) & = ∫_{R^{×}} e^{itx}ψ(t)|t|^{z−1}dt \\ θ^{−}(z, x) & = ∫_{R^{×}} e^{itx}ψ(t)|t|^{z−1}sgn tdt \end{aligned}

are defined for Re z sufficiently large.

Lemma 3.5. θ^{±}(z, x) are meromorphic in the whole complex plane and bounded in regions of the form |Re u| ≤ constant, |Im u| ≥ constant ≫ 0

(i) If s − m is not an odd integer, m = 0, and m_{2} = 0 then \frac{θ^{+}(z,x)}{Γ( \frac{z+ \frac{1}{2} + \frac{s}{2}}{2} )Γ( \frac{z+ \frac{1}{2} − \frac{s}{2}}{2} )} and

\frac{θ^{−}(z,x)}{Γ( \frac{z+ \frac{3}{2} + \frac{s}{2}}{2} )Γ( \frac{z+ \frac{3}{2} − \frac{s}{2}}{2)}} are entire functions of z.

(ii) If s − m is not an odd integer, m = 0, and m_{2} = 1 then \frac{θ^{+}(z,x)}{Γ( \frac{z+ \frac{3}{2} + \frac{s}{2}}{2} )Γ( \frac{z+ \frac{3}{2} − \frac{s}{2}}{2} )} and

\frac{θ^{−}(z,x)}{Γ( \frac{z+ \frac{1}{2} + \frac{s}{2}}{2} )Γ( \frac{z+ \frac{1}{2} − \frac{s}{2}}{2} )} are entire functions of z.

(iii) If s − m is not an odd integer, |m| = 1 and m_{2} = 0 then \frac{θ^{+}(z,x)}{Γ( \frac{z+ \frac{3}{2} + \frac{s}{2}}{2} )Γ( \frac{z+ \frac{1}{2} − \frac{s}{2}}{2} )} and

\frac{θ^{−}(z,x)}{Γ( \frac{z+ \frac{1}{2} + \frac{s}{2}}{2} )Γ( \frac{z+ \frac{3}{2} − \frac{s}{2}}{2} )} are entire functions of z.

(iv) If s − m is not an odd integer, |m| = 1, and m_{2} = 1 then \frac{θ^{+}(z,x)}{Γ \frac{(z+ \frac{1}{2} + \frac{s}{2} )}{2} Γ \frac{(z+ \frac{3}{2} − \frac{s}{2} )}{2}} and

\frac{θ^{−}(z,x)}{Γ \frac{(z+ \frac{3}{2} + \frac{s}{2} )}{2} Γ \frac{(z+ \frac{1}{2} − \frac{s}{2} )}{2}} are entire functions of z.

(v) If s − m is an odd integer then \frac{θ^{±}(z,x)}{Γ(z+ \frac{1}{2}} + \frac{|s|}{2)} are entire functions of z.

Let m(t) be an infinity differentiable function with compact support on the line which is even and equal to 1 in a neighborhood of 0. θ^{±}(z, x) is the sum of

θ̂^{±}(z, x) = ∫_{R^{×}} e^{itx}ψ(t)|t|^{z−1}(sgn t)^{\frac{1±1}{2}} m(t)dt

and

∫_{R^{×}} e^{itx}ψ(t)|t|^{z−1}(sgn t)^{\frac{1±1}{2}} (1 − m(t))dt.

The second integral is an entire function of z which is bounded in vertical strips. Thus it is enough to prove the lemma with θ^{±}(z, x) replaced by θ̂^{±}(z, x). The function e^{itx}ψ(t) + (−1)^{\frac{1±1}{2}} e^{−itx}ψ(−t) is, for t > 0, a linear combination of convergent series of the form

t^{α}(log t)^{β} ∑_{p=0}^{α} c_{p}t^{p}

where α is \frac{s+1}{2} or \frac{−s+1}{2} and β is 0 or 1. Given a series of this form and a real number c there is a P such that

∫_{0}^{∞} t^{α}(log t)^{β}{∑_{p≥P} c_{p}t^{p}}t^{z−1}m(t)dt

is analytic for Re z > c and bounded in vertical strips of finite width contained in this region. The first assertion of the lemma is a consequence of the relations

\begin{aligned}∫_{0}^{∞} t^{α+p+z−1}m(t)dt & = \frac{−1}{α + p + z} ∫_{0}^{∞} t^{α+p+z}m′(t)dt \\ ∫_{0}^{∞} t^{α+p+z−1} log tm(t)dt & = \frac{−1}{(α + p + z)^{2}} ∫_{0}^{∞}[(α + p + z)t^{α+p+z} log t − t^{α+p+z}]m′(t)dt \end{aligned}

and the condition that m′(t) vanish near zero. To prove the remaining assertions one shows that the zeros of the denominator on the right are cancelled by the poles of the Γ-factor. This is easy but the various cases of Lemma 3.4 must be examined separately. I leave it to the reader to do so.

If η is any homomorphism of A_{ℝ} into ℂ^{×} then η̃ will be the homomorphism defined by

η̃(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )) = η(( \begin{matrix}α_{2} & 0 \\ 0 & α_{1} \end{matrix} )).

If ζ is a homomorphism of A_{ℝ}^{×} into ℂ^{×} such that

ζ(( \begin{matrix}α & 0 \\ 0 & α \end{matrix} ))ω(( \begin{matrix}α & 0 \\ 0 & α \end{matrix} )) ≡ 1

and z and ℓ are defined by

ζ(( \frac{t}{|t|^{1/2}} ; \frac{1}{|t|^{1/2}} )) = |t|^{z}(sgn t)^{ℓ},

with ℓ = 0 or 1, then ζ is determined by z and ℓ and we shall sometimes write ζ = ζ(z, ℓ).

Lemma 3.6. Suppose π is an infinite-dimensional representation of {σ, 𝔄} and π is deducible from π_{ω}. Let L(ξ, π) be the unique subspace of L(ξ) which transforms according to π. If φ belongs to L(ξ, π) and ζ = ζ(z, ℓ) the function

Φ(g, ζ, φ) = ∫_{ℝ^{×}} φ(( \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} ) g)ζ(( \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} ))d^{×}t

is defined for Re z sufficiently large

(i) If s − m is not an odd integer set

Φ′(g, ζ, φ) = \frac{Φ(g, ζ, φ)}{Γ\frac{(z+|m_{1}−ℓ|+ \frac{1}{2} + \frac{s}{2} )}{2}Γ\frac{(z+|m_{2}−ℓ|+ \frac{1}{2} − \frac{s}{2} )}{2}} .

(ii) If s − m is an odd integer set

Φ′(g, ζ, φ) = \frac{Φ(g, ζ, φ)}{Γ(z + \frac{1}{2} + \frac{|s|}{2})}.

Then Φ′(g, ζ(z, ℓ), φ) is an entire function of z and Φ(g, ζ(z, ℓ)φ) is bounded in regions of the form |Re z| ≤ constant, |Im z| ≥ constant ≫ 0. Moreover if s − m is not an odd integer

(\frac{2}{|u|})^{−z}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = (i)^{|m_{1}−ℓ|+|m_{2}−ℓ|}(sgn u)^{m}(\frac{2}{|u|})^{z}Φ′(g, ζ̃, φ)

and if s − m is an odd integer

(\frac{1}{|u|})^{−z}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = (i)^{|s|+1}(sgn u)^{m}(\frac{1}{|u|})^{z}Φ′(g, ζ̃, φ).

It is enough to prove the lemma for φ in L(ξ, π)_{n}. If φ̃ is the restriction of φ to G_{ℝ}^{∘} let ψ(t) be the function on ℝ^{×} corresponding to φ̃. Then, if

g = ( \begin{matrix}t_{1} & 0 \\ 0 & t_{2} \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) ,

φ(( \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} ) g) is equal to

e^{i \frac{tt_{1}}{t_{2}} ux}ω(( \begin{matrix}|tt_{1}t_{2}|^{1/2}sgn t_{2} & 0 \\ 0 & |tt_{1}t_{2}|^{1/2}sgn t_{2} \end{matrix} ))ψ(\frac{tt_{1}}{t_{2}} )e^{inθ}.

Thus

φ(( \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} ) g)ζ(( \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} ))

is equal to

ζ^{−1}(( \begin{matrix}t_{1} & 0 \\ 0 & t_{2} \end{matrix} ))e^{ei \frac{tt_{1}}{t_{2}} ux}ζ(( \begin{matrix}\frac{tt_{1}}{t_{2}} |\frac{t_{2}}{tt_{1}} |^{1/2} & 0 \\ 0 & |\frac{t_{2}}{tt_{1}} |^{1/2} \end{matrix} ))ψ(\frac{tt_{1}}{t_{2}} )e^{inθ}

and Φ(g, ζ(z, ℓ), φ) is equal to

\begin{matrix}ζ^{−1}(( \begin{matrix}t_{1} & 0 \\ 0 & t_{2} \end{matrix} ))θ^{+}(z, uλ)e^{inθ}, & ifℓ = 0, \\ ζ^{−1}(( \begin{matrix}t_{1} & 0 \\ 0 & t_{2} \end{matrix} ))θ^{−}(z, ux)e^{inθ}, & ifℓ = 1. \end{matrix}

All assertions of the lemma except the functional equations follow immediately from Lemma 3.5. If η = ζ̃^{−1} the maps

\begin{aligned}φ & → Φ′(g, ζ̃, φ), \\ φ & → Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, p) \end{aligned}

are {σ, 𝔄} invariant maps of L(ξ, π) into L(η). According to Lemma 3.1 one must be a scalar multiple of the other. To see what the multiple is we choose g = 1 so that Φ(g, ζ̃, φ) is equal to θ_{n}^{+}(−z) if ℓ − |m| = 0 and is equal to θ_{n}^{−}(−z) if |ℓ − |m|| = 1 and choose n in such a way that Lemma 3.3 can be applied.

Φ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ)

is equal to (i)^{n}θ_{n}^{+}(z) if ℓ = 0 and to (i)^{n}θ_{n}^{−}(z) is ℓ = 1. In the first column below we write the values of Φ′(1, ζ̃, φ) for the values of n and ℓ in the last column; in the second column we write the values of Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )ζ, φ). Comparing them we obtain the lemma. In all but the last line s − m is not an odd integer.

Φ′(1, ζ̃, φ)  Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) , ζ, φ)
α_{0}(\frac{2}{|u|})^{−z}  α_{0}(\frac{2}{|u|})^{z}  ℓ = 0, m = 0, m_{2} = 0, n = 0.
2α_{1}sgn u(\frac{2}{|u|})^{−z} −2α_{1}sgn u(\frac{2}{|u|})^{z}  ℓ = 1, m = 0, m_{2} = 0, n = 2.
2β_{1}sgn u(\frac{2}{|u|})^{−z} −2β_{1}sgn u(\frac{2}{|u|})^{z}  ℓ = 0, m = 0, m_{1} = 1, n = 2.
β_{0}(\frac{2}{|u|})^{−z}  β_{0}(\frac{2}{|u|})^{z}  ℓ = 1, m = 0, m_{2} = 1, n = 0.
γ_{0}sgn u(\frac{2}{|u|})^{−z} iγ_{0}(\frac{2}{|u|})^{z}  ℓ = 0, |m| = 1, m_{2} = 0, n = 1.
γ_{0}(\frac{2}{|u|})^{−z}  iγ_{0}sgn u(\frac{2}{|u|})^{z}  ℓ = 1, |m| = 1, m_{2} = 0, n = 1.
γ_{1}sgn u(\frac{2}{|u|})^{−z} iγ_{1}(\frac{2}{|u|})^{z}  ℓ = 0, |m| = 1, m_{2} = 1, n = 1.
γ_{1}(\frac{2}{|u|})^{−z}  iγ_{1}sgn u(\frac{2}{|u|})^{z}  ℓ = 1, |m| = 1, m_{2} = 1, n = 1.

If s − m is an odd integer the two values are

\begin{matrix}δ_{0}(\frac{1}{|u|})^{−z+n/2}(sgn u)^{ℓ} & (i)^{|s|+1}δ_{0}(\frac{1}{|u|})^{z+n/2}(sgn u)^{ℓ}  n = |s| + 1, m = 0. \\ δ_{0}(\frac{1}{|u|})^{−z+n/2}(sgn u)^{ℓ−1} & (i)^{|s|+1}δ_{0}(\frac{1}{|u|})^{z+n/2}(sgn u)^{ℓ}  n = |s| + 1, |m| = 1. \end{matrix}

4. Representations of GL(2, ℂ). In this paragraph and the next G_{ℂ} will be GL(2, ℂ) and G_{ℂ}^{0} will be SL(2, ℂ). U will be the group of unitary matrices in G_{ℂ} and U^{0} will be U ∩ G_{ℂ}^{0}. G_{ℂ} and G_{ℂ}^{0} will be considered as real Lie groups. The Lie algebra of G_{ℂ} is

𝔤 = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ⊕ ( \begin{matrix}ā & b̄ \\ c̄ & d̄ \end{matrix})\};

its complexification is

𝔤_{ℂ} = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ⊕ ( \begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix} )\} .

The Lie algebra of G_{ℂ}^{0} is

𝔤^{0} = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ⊕ ( \begin{matrix}ā & b̄ \\ c̄ & d̄ \end{matrix}) |a + d = 0\};

its complexification is

𝔤_{ℂ}^{0} = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) + ( \begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix} ) |a + d = a′ + d′ = 0\} .

The Lie algebra of U is

𝔲 = \{( \begin{matrix}a & b \\ −b̄ & d \end{matrix} ) ⊕ ( \begin{matrix}ā & b̄ \\ −b & d̄ \end{matrix}) |a = −ā, d = −d̄\};

its complexification is

𝔲_{ℂ} = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ⊕ ( \begin{matrix}−a & −c \\ −b & −d \end{matrix} )\} .

Finally 𝔲^{0} = 𝔲 ∩ y^{0} and 𝔲_{ℂ}^{0} = 𝔲_{ℂ} ∩ 𝔤_{ℂ}^{0}. When there is no risk of confusion an element of 𝔲_{ℂ} will be identified by giving its first component.

Let V_{n} be the space of binary forms of degree n and let Ṽ_{n} be its dual. We write the elements of V_{n} as

ψ(x, y) = ∑_{{|k|≤n \atop \frac{n}{2} −k∈ℤ}} ψ^{k}x^{\frac{n}{2} +k}y^{\frac{n}{2} −k}

ψ^{k} will be called the k^{th} component of ψ. If |k| > \frac{n}{2} let ψ^{k} = 0. Let σ_{n} be the representation of U^{0} on V_{n} defined by

σ_{n} ( \begin{matrix}a & b \\ c & d \end{matrix} ) ψ(x, y) = ψ(ax + cy, bx + dy).

Denote the corresponding representation of 𝔲_{ℂ}^{0} by σ_{n} also. If ψ_{1} = σ_{n}( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} )ψ then ψ_{1}^{k−1} = (\frac{n}{2} +k)ψ^{k} = c_{k}ψ^{k}

where c_{k} ≠ 0 for −\frac{n}{2} < k ≤ \frac{n}{2} and if ψ_{2} = σ_{n}( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} )ψ then ψ_{2}^{k+1} = (\frac{n}{2} − k)ψ^{k} = d_{k}ψ^{k} where d_{k} ≠ 0 for

−\frac{n}{2} ≤ k < \frac{n}{2}.

Let 𝔄 be the universal enveloping algebra of 𝔤_{ℂ} and 𝔄^{0} that of 𝔤_{ℂ}^{0}. If π is a representation of 𝔄 on a vector space W then π^{0} will be the restriction of π to 𝔄^{0}. Let W_{n} be the set of all vectors in W which transform under 𝔲_{ℂ}^{0} according to σ_{n}. π will be called quasi-simple^{∗} if

(i) W = ∑_{n} ⊕W_{n}

∗ I use the expression in a different way than Harish-Chandra.

(ii) If Z lies in the centre of 𝔄 then π(Z) is a scalar. Suppose π_{1} and π_{2} are two representations of 𝔄 on W_{1} and W_{2} respectively. π_{2} will be said to be deducible from π_{1} if there are two invariant subspaces W_{3} and W_{4} of W_{1} with W_{3} ⊇ W_{4} and π_{2} is equivalent to the representation of 𝔄 on W_{3}/W_{4}. The same notions will be used for representations of 𝔄^{0}.

Set

X = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) Y = ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) Z = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) .

The centre of the universal enveloping algebra 𝔄^{0} is generated by

D = (X ⊕ 0)(Y ⊕ 0) + (Y ⊕ 0)(X ⊕ 0) + \frac{1}{2}(Z ⊕ 0)^{2}

= 2(Y ⊕ 0)(X ⊕ 0) + Z ⊕ 0 + \frac{1}{2}(Z ⊕ 0)^{2}

= 2(X ⊕ 0)(Y ⊕ 0) − Z ⊕ 0 + \frac{1}{2}(Z ⊕ 0)^{2}

and

D′ = (0 ⊕ X)(0 ⊕ Y) + (0 ⊕ Y)(0 ⊕ X) + \frac{1}{2}(0 ⊕ Z)^{2}

= 2(0 ⊕ Y)(0 ⊕ X) + (0 ⊕ Z) + \frac{1}{2}(0 ⊕ Z)^{2}

= 2(0 ⊕ X)(0 ⊕ Y) − (0 ⊕ Z) + \frac{1}{2}(0 ⊕ Z)^{2}.

The centre of 𝔄 is generated by D, D′, J = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) ⊕ 0, and J′ = 0 ⊕ ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ).

Let ω be a continuous homomorphism of the group A_{ℂ} of diagonal matrices into ℂ^{×}. If N_{ℂ} is the group of matrices of the form ( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ) let L(ω) be the space of infinitely differentiable U-finite functions on N_{ℂ}\G_{ℂ} satisfying ψ(ag) = |\frac{α_{1}}{α_{2}} |ω(a)φ(g) if a = ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) is in A_{ℂ}. The restriction of ρ to L(ω) defines a representation π_{ω} of 𝔄 on L(ω). Define ω_{1} and ω_{2} on ℂ^{×} by ω_{1}(t) = ω(( \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} )) and ω_{2}(t) = ω(( \begin{matrix}1 & 0 \\ 0 & t \end{matrix} )). Let ω_{i}(t) = |t|^{s_{i}}(\frac{g}{|t_{i}|})^{m_{i}} and set s = \frac{s_{1}−s_{2}}{2} , m = \frac{m_{1}−m_{2}}{2} .

Lemma 4.1. L(ω)_{n} ≠ \{0\} if and only if \frac{n}{2} − |m| is a non-negative integer and then L(ω)_{n} is irreducible under 𝔲_{ℂ}^{0}. Moreover

\begin{aligned}π_{ω}(D) & = & \frac{(s + m)^{2} − 1}{2}I, & π_{ω}(J) & = & \{\frac{s_{1} + s_{2}}{2} − I(\frac{m_{1} + m_{2}}{2} )\}I, \\ π_{ω}(D′) & = & \frac{(s − m)^{2} − 1}{2}I, & π_{ω}(J′) & = & \{\frac{s_{1} + s_{2}}{2} + i(\frac{m_{1} + m_{2}}{2} )\}I. \end{aligned}

The first assertion is an immediate consequence of the Iwasawa decomposition and the Frobenius reciprocity law. Set

\begin{aligned}Z_{1} & = & ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) & ⊕ ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) , & Z_{2} & = & ( \begin{matrix}i & 0 \\ 0 & −i \end{matrix} ) & ⊕ ( \begin{matrix}−i & 0 \\ 0 & i \end{matrix} ) , \\ X_{1} & = & ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) & ⊕ ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) , & X_{2} & = & ( \begin{matrix}0 & i \\ 0 & 0 \end{matrix} ) & ⊕ ( \begin{matrix}0 & −i \\ 0 & 0 \end{matrix} ) , \\ Y_{1} & = & ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) & ⊕ ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) , & Y_{2} & = & ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) & ⊕ ( \begin{matrix}0 & 0 \\ −i & 0 \end{matrix} ) , \\ W_{1} & = & ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) & ⊕ ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) , & W_{2} & = & ( \begin{matrix}0 & i \\ i & 0 \end{matrix} ) & ⊕ ( \begin{matrix}0 & −i \\ −i & 0 \end{matrix} ) . \end{aligned}

Then

\begin{aligned}Z & ⊕ & 0 & = & \frac{Z_{1} − iZ_{2}}{2} , & 0 & ⊕ & Z & = & \frac{Z_{1} + iZ_{2}}{2} , \\ X & ⊕ & 0 & = & \frac{X_{1} − iX_{2}}{2} , & 0 & ⊕ & X & = & \frac{X_{1} + iX_{2}}{2} , \\ Y & ⊕ & 0 & = & \frac{Y_{1} − iY_{2}}{2} , & 0 & ⊕ & Y & = & \frac{Y_{2} + iY_{2}}{2} . \end{aligned}

It is clear that ρ(D) = λ(D), that ρ(D′) = λ(D′) and that λ(X_{i})φ = 0 if φ belongs to L(ω). Thus

\begin{aligned}ρ(D)φ & = λ(Z ⊕ 0)φ + \frac{1}{2}λ((Z ⊕ 0)^{2})φ \\ ρ(D′)φ & = λ(0 ⊕ Z)φ + \frac{1}{2}λ((0 ⊕ Z)^{2})φ. \end{aligned}

Combining this with the relations λ(Z_{1})φ = −2(s + 1)φ and λ(Z_{2})φ = −2imφ one obtains the assorted values for π_{ω}(D) and π_{ω}(D′). The other two relations of the lemma are very simple to verify.

Lemma 4.2. If neither −s − 1 − |m| nor s − 1 − |m| is a non-negative integer then π_{ω} is irreducible. If −s − 1 − |m| = \frac{n_{0}}{2} − |m| is a non-negative integer then

∑_{{|m|≤n≤n_{0} \atop \frac{n}{2} −|m|∈ℤ}} L(ω)_{n} = M(ω)

is invariant and the representations of 𝔄 on M(ω) and L(ω)/M(ω) are irreducible. If s − 1 − |m| = \frac{n_{0}}{2} − |m| is a non-negative integer then

∑_{{n>n_{0} \atop \frac{n}{2} −|m|∈ℤ}} L(ω)_{n} = M(ω)

is invariant and the representations of 𝔄 on M(ω) and L(ω)/M(ω) are irreducible.

Set

\begin{aligned}U^{+} & = & X & ⊕ & −Y, & U & = & Z & ⊕ & −Z, & U^{−} & = & Y & ⊕ & −X, \\ V^{+} & = & X & ⊕ & Y, & V & = & Z & ⊕ & Z, & V^{−} & = & Y & ⊕ & X. \end{aligned}

These six elements form a basis of 𝔤_{ℂ}^{−}. U^{+}, U, and U^{−} form a basis of 𝔲_{ℂ}^{0}. The space 𝔭_{ℂ} spanned by V^{+}, V, and V^{−} is invariant under the adjoint action of 𝔲_{ℂ}^{0} and the map V^{+} → x^{2}, V → −2xy, V^{−} → −y^{2} extends to a 𝔲_{ℂ}^{0}-invariant map of 𝔭_{ℂ} to V_{2}. The map W ⊗ φ → π_{ω}(W)φ, W ∈ 𝔭_{ℂ}, φ ∈ L(ω)_{n} extends to a 𝔲_{ℂ}^{0} invariant map of 𝔭_{ℂ} ⊗ L(ω)_{n} into L(ω). It follows from the existence of the Clebsch-Gordan series that the image lies in L(ω)_{n−2} + L(ω)_{n} + L(ω)_{n+2}. To prove the lemma all we need do is show that the image contains a non-zero element in L(ω)_{n+2} if and only if s ≠ −(\frac{n}{2} + 1) and that if \frac{n}{2} > |m| it contains a non-zero element in L(ω)_{n−2} if and only if s ≠ \frac{n}{2}.

Let \frac{n}{2} − k ∈ ℤ. If |k| ≤ \frac{n}{2} let δ_{k}(x, y) = x^{\frac{n}{2} +k}y^{\frac{n}{2} −k} and if |k| > \frac{n}{2} let δ_{k}(x, y) = 0. If |k| ≤ \frac{n}{2} let γ_{k} be the element of Ṽ_{n} such that γ_{k}(∑_{j} ψ^{j}x^{\frac{n}{2} +j}y^{\frac{n}{2} −j}) = ψ^{k}; if |k| > \frac{n}{2} let γ_{k} = 0. If g = ( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} )au with

a = ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) in A_{ℂ} and u in U^{0} set

φ_{n,k}(g) = |\frac{α_{1}}{α_{2}} |ω(a)γ_{m}σ_{n}(u)δ_{k} |k| ≤ \frac{n}{2}.

The function φ_{n,k} form a basis of L(ω)_{n}.

Using the method described on p. 129-130 of Weyl’s book on quantum mechanics to decompose 𝔭_{ℂ} ⊗ V_{n} or V_{2} ⊗ V_{n} into a direct sum of irreducible subspaces one finds that

(\frac{n}{2} + k)(\frac{n}{2} + k + 1)ρ(V^{+})φ_{n,k−1} − (\frac{n}{2} + k + 1)(\frac{n}{2} − k + 1)ρ(V)φ_{n,k} − (\frac{n}{2} − k)(\frac{n}{2} − k + 1)ρ(V^{−})φ_{n,k+1}

is equal to

(\frac{n}{2} + k + 1)!(\frac{n}{2} − k + 1)!a(n, ω)φ_{n+2,k}

and

ρ(V^{+})φ_{n,k−1} + ρ(V)φ_{n,k} − ρ(V^{−})φ_{n,k+1} = (\frac{n}{2} + k − 1)!(\frac{n}{2} − k − 1)!b(n, ω)φ_{n−2,k}

of |k| ≤ \frac{n}{2} − 1. The image contains a non-zero element in L(ω)_{n+2} if and only if a(n, ω) ≠ 0 and a non-zero element in L(ω)_{n−2} if and only if b(n, ω) ≠ 0. Since φ_{n+2,m}(1) = 1 and φ_{n−2,m}(1) = 1 if \frac{n}{2} > |m| all we need do to find a(n, ω) and b(n, ω) is to take k = m and evaluate the left sides of the above expressions at 1.

Now V = Z_{1}, V^{+} = (X_{1} + \frac{W_{1}}{2} ) − i(X_{2} − \frac{W_{2}}{2} ), and V^{−} = (X_{1} + \frac{W_{1}}{2} ) + i(X_{2} − \frac{W_{2}}{2} ). Since

\begin{aligned}ρ(Z_{1})φ_{n,k}(1) & = 2(s + 1)φ_{n,k}(1), \\ ρ(X_{1})φ_{n,k}(1) & = ρ(X_{2})φ_{n,k}(1) = 0, \\ ρ(W_{1})φ_{n,k}(1) & = γ_{m}σ_{n} ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) δ_{k}, \\ ρ(W_{2})φ_{n,k}(1) & = γ_{m}σ_{n} ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) δ_{k}, \end{aligned}

one has ρ(V)φ_{n,k}(1) = 2(s + 1)γ_{m}γ_{k} and

((V^{+})φ_{n,k}(1) = −γ_{m}σ_{n} ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) δ_{k}, ρ(V^{−})φ_{n,k}(1) = γ_{m}σ_{n} ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) δ_{k}.

Thus

(\frac{n}{2} + m + 1)!(\frac{n}{2} − m + 1) a(n, ω)

is equal to

−(\frac{n}{2} + m)(\frac{n}{2} + m + 1)(\frac{n}{2} − m + 1)(\frac{n}{2} + m + 1)(\frac{n}{2} − m + 1)(s + 1) − (\frac{n}{2} − m)(\frac{n}{2} − m + 1)(\frac{n}{2} + m + 1)

which equals

−2[(\frac{n}{2} + 1)^{2} − m^{2}][s + \frac{n}{2} + 1],

and

(\frac{n}{2} + m − 1)!(\frac{n}{2} − m − 1)!b(n, ω) = −(\frac{n}{2} − m + 1) + 2(s + 1) − (\frac{n}{2} + m + 1) = 2(s − \frac{n}{2}).

Lemma 4.3. Suppose π is an irreducible quasi-simple representation of 𝔄 on the vector space H. There is at least one continuous homomorphism ω of A_{ℂ}^{×} into ℂ^{×} such that

\begin{aligned}π(D) & = & \frac{(s + m)^{2} − 1}{2}I, & π(J) & = & \{\frac{s_{1} + s_{2}}{2} − i\frac{(m_{1} + m_{2})}{2}\}I, \\ π(D′) & = & \frac{(s − m)^{2} − 1}{2}I, & π(J′) & = & \{\frac{s_{1} + s_{2}}{2} + i\frac{(m_{1} + m_{2})}{2}\}I, \end{aligned}

and such that H_{n_{0}} ≠ 0 for at least one n_{0} with \frac{n_{0}}{2} − |m| a non-negative integer. If ω is any such homomorphism then π is deducible from π_{ω}.

The lemma is a special case of a theorem of Harish-Chandra (Representation of semi-simple Lie groups II, T.A.M.S., v. 76, 1954). It implies that H_{n} is irreducible under 𝔲_{ℂ}^{0}. A similar assertion isolated for 𝔄.

Lemma 4.4. Suppose λ(D), λ(D′), λ(J), and λ(J′) are four given numbers. Apart from equivalence there are at most two quasi-simple irreducible representations of 𝔄 satisfying

π(D) = λ(D)I, π(D′) = λ(D′)I, π(J) = λ(J)I, π(J′) = λ(J′)I.

If there are two, then one of them is finite dimensional.

If there is one such representation there is an ω such that λ(D) = \frac{(s+m)^{2}−1}{2}, λ(D′) = \frac{(s−m)^{2}−1}{2}, λ(J) = \frac{s_{1}+s_{2}}{2} − i\frac{(m_{1}+m_{2})}{2}, λ(J′) = \frac{s_{1}+s_{2}}{2} + i\frac{(m_{1}+m_{2})}{2}. If ω′ is such that these representations are satisfied by s′_{1}, s′_{2}, m′_{i}, m′_{2}, one must have \frac{s_{1}+s_{2}}{2} = \frac{s′_{1}+s′_{2}}{2} and \frac{m_{1}+m_{2}}{2} = \frac{m′_{1}+m′_{2}}{2} . In particular m − m′ = \frac{m_{1}−m′_{1}}{2} − \frac{m_{2}−m′_{2}}{2} = m_{1} − m′_{1} is integral. The relations (s + m)^{2} = (s′ + m′)^{2} and (s − m)^{2} = (s′ − m′)^{2} are satisfied if and only if one of the following holds.

(i)  s = s′  m = m′  (iii)  s = m′  m = s′
(ii)  s = −s′  m = −m′  (iv)  s = −m′  m = −s′.

If s−m is not integral only the first two are possible. π_{ω} and π_{ω′} are irreducible by Lemma 4.2 and equivalent by Lemma 4.3. If s − m is integral one can choose ω so that s ≥ |m|. It follows from Lemma 4.3 that every quasi-simple irreducible representation deducible from π_{ω′} is deducible from π_{ω}. There are only two such representations deducible from π_{ω} and one of them is finite-dimensional. It is clear that Lemma 4.4 could also be formulated for 𝔄^{0}.

5. The local functional equation for GL(2, ℂ). If η is a continuous homomorphism of A_{ℂ} into ℂ^{×} let L(η) be the space of all U-finite infinitely differentiable functions on G satisfying

φ(ag) = η(a)φ(g)

for all a in A_{ℂ}. If φ lies in L(η) and X lies in 𝔄 then ρ(X)φ lies in L(η) so that we have a representation ρ(η) of 𝔄 on L(η).

Lemma 5.1. No irreducible, quasi-simple representation is contained more than once in ρ(η).

Let π be an irreducible, quasi-simple representation. Suppose it is deducible from π_{ω} and suppose its restriction to 𝔲 contains σ_{n}. If π occurs in L(η) then η_{1}η_{2} = ω_{1}ω_{2} and for the proof we may as well assume that this is the case. Let L^{0}(η) be the space of infinitely differentiable U^{0}-finite functions on G_{ℂ}^{0} satisfying φ(ag) = η(a)φ(g) for a in A_{ℂ}^{0} and let ρ^{0}(η) be the representation of 𝔄^{0} on L^{0}(η). We have to show that π^{0} is contained at most once in ρ^{0}(η).

Let H ⊆ L^{0}(η) be 𝔄^{0}-invariant and suppose that the restriction of ρ^{0}(η) to H is equivalent to π^{0}. There is a map φ → Φ of H_{n} to V_{n} and a function Φ(g) on G^{0} with values in Ṽ_{n} such that φ(g) = Ψ(g)Φ and Ψ(gk) = Ψ(g)π(k). Let ω_{1}(t) = |t|^{s_{1}}(\frac{g}{|t|})^{m_{1}}, ω_{2}(t) = |t|^{s_{2}}(\frac{t}{|t|})^{m_{2}}, η_{1}(t) = |t|^{r_{1}}(\frac{t}{|t|})^{ℓ_{1}}, η_{2}(t) = |t|^{r_{2}}(\frac{t}{|t|})^{ℓ_{2}}. If z = x + iy let ψ(z) = Ψ(( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} )). Ψ is uniquely determined by ψ. Let us rewrite the equations

\begin{aligned}ρ(D)Ψ & = \frac{(\frac{(s_{1}−s_{2})+(m_{1}−m_{2})}{2})^{2} − 1}{2}Ψ = \frac{(s + m)^{2} − 1}{2}Ψ, \\ ρ(D)Ψ & = \frac{(\frac{(s_{1}−s_{2})−(m_{1}−m_{2})}{2})^{2} − 1}{2}Ψ = \frac{(s − m)^{2} − 1}{2}Ψ \end{aligned}

in terms of ψ. D may be written as

\begin{matrix}2(\frac{X_{1} − iX_{2}}{2} )(\frac{Y_{1} − iY_{2}}{2} ) − (\frac{Z_{1} − iZ_{2}}{2} ) + \frac{1}{2}(\frac{Z_{1} − iZ_{2}}{2} )^{2} \\ = \frac{(X_{1} − iX_{2})(X_{1} + iX_{2})}{2} + \frac{(X_{1} − iX_{2})(W_{1} − iW_{2})}{2} − \frac{(Z_{1} − iZ_{2})}{2} + \frac{(Z_{1} − iZ_{2})^{2}}{8} \end{matrix}

and D′ may be written as

\begin{matrix}2\frac{(X_{1} + iX_{2})}{2} \frac{(Y_{1} + iY_{2})}{2} − \frac{(Z_{1} + iZ_{2})}{2} + \frac{1}{2} \frac{(Z + iZ_{2})^{2}}{2} \\ = \frac{(X_{1} + iX_{2})(X_{1} − iX_{2})}{2} + \frac{(X_{1} + iX_{2})(W_{1} + iW_{2})}{2} − \frac{(Z_{1} + iZ_{2})}{2} + \frac{(Z_{1} + iZ_{2})^{2}}{8} . \end{matrix}

It is easily seen that

ρ(X_{1}^{2} + X_{2}^{2})Ψ(( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} )) = \frac{∂^{2}ψ}{∂x^{2}} + \frac{∂^{2}ψ}{∂y^{2}} = 4\frac{∂^{2}ψ}{∂z∂z̄}

if

\frac{∂}{∂z} = \frac{1}{2}(\frac{∂}{∂x} + \frac{1}{i} \frac{∂}{∂y}) \frac{∂}{∂z̄} = \frac{1}{2}(\frac{∂}{∂x} − \frac{1}{i} \frac{∂}{∂y})

and that r = \frac{r_{1}−r_{2}}{2} and ℓ = \frac{ℓ_{1}−ℓ_{2}}{2} .

\begin{aligned}ρ((X_{1} − iX_{2})(W_{1} − iW_{2}))Ψ(( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} )) & = 4\frac{∂ψ}{∂z} σ_{n} ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) \\ ρ((X_{1} + iX_{2})(W_{1} + iW_{2}))ψ(( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} )) & = −4\frac{∂ψ}{∂z} σ_{n} ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) \\ ρ(j_{1})Ψ(( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} )) & = (2rψ − 2x\frac{∂ψ}{∂x} − 2y \frac{∂ψ}{∂y}) \\ ρ(Z_{1})Ψ(( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} )) & = (2iℓψ − 2y \frac{∂ψ}{∂x} − 2x\frac{∂ψ}{∂y}). \end{aligned}

Putting everything together one obtains the equation

\begin{matrix}2\frac{∂^{2}ψ}{∂z∂z̄} & + 2\frac{∂ψ}{∂z} σ_{n} ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) + \frac{1}{2}[(r + ℓ − 1) + 2z \frac{∂}{∂z}]^{2}ψ = \frac{(s + m)^{2}}{2} ψ \\ 2\frac{∂^{2}ψ}{∂z∂z̄} & − 2\frac{∂ψ}{∂z} σ_{n} ( \begin{matrix}1 & 1 \\ 0 & 0 \end{matrix} ) + \frac{1}{2}[(r − ℓ − 1) + 2z̄\frac{∂}{∂z̄}]^{2}ψ = \frac{(s − m)^{2}}{2} ψ. \end{matrix}

There is an auxiliary equation corresponding to the relation

ψ(e^{2iθ}z)σ_{n} ( \begin{matrix}e^{iθ} & 0 \\ 0 & e^{−iθ} \end{matrix} ) = e^{2iℓθ}ψ(z).

It is

−2y \frac{∂ψ}{∂x} + 2x\frac{∂ψ}{∂y} + ψσ_{n} ( \begin{matrix}i & 0 \\ 0 & −i \end{matrix} ) = 2iℓψ

or

z \frac{∂ψ}{∂z} − z̄\frac{∂ψ}{∂z̄} = ψ\{ℓI − \frac{1}{2}σ_{n} ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )\}.

Since ψ(z) is an analytic function of x and y it can be expanded in a power series

∑_{p,q=0}^{∞} z^{p}z̄^{q}ψ_{p,q}.

According to the auxiliary equation,

(p − q)ψ_{p,q} = ψ_{p,q}\{ℓI − \frac{1}{2}σ_{n} ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )\}.

Thus ψ_{p,q}^{k} = 0 unless p − q = ℓ − k. Substituting in the first two equations one obtains

\begin{matrix}2(p + q + 2)ψ_{p+1,q+1}^{k} + 2c_{k}ψ_{p+1,q}^{k−1} + \frac{1}{2}[(r + ℓ − 1) + 2p]^{2}ψ_{p,q}^{k} = \frac{(s + m)^{2}}{2} ψ_{p,q}^{k} \\ 2(p + q + 2)ψ_{p+1,q+1}^{k} − 2d_{k}ψ_{p,q+1}^{k+1} + \frac{1}{2}[(r − ℓ − 1) + 2q]^{2}ψ_{p,q}^{k} = \frac{(s − m)^{2}}{2} ψ_{p,q}^{k}. \end{matrix}

Here ψ_{p,q}^{j} = 0 if |j| > \frac{n}{2} and c_{k} ≠ 0 if \frac{−n}{2} < k ≤ \frac{n}{2}. The second equation can be used to determine the numbers ψp,q^{\frac{n}{2}} inductively, then the first can be used to determine the numbers ψ_{p,q}^{k} for k < \frac{n}{2}.

Let ξ(z) = e^{iRe(wz)} with w ≠ 0 be a character of ℂ and let L(ξ) be the space of all infinitely differentiable U-finite functions on G_{ℂ} satisfying

(i) φ(( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ) g) = ξ(z)φ(g).

(ii) If k ∈ 𝔄 and g ∈ G there is a constant M such that

|φ(( \begin{matrix}t_{1} & 0 \\ 0 & t_{2} \end{matrix} ) g)| ≤ M(|t_{1}|^{M} + |t_{2}|^{M})

if |t_{1}| ≥ |t_{2}|.

Let ρ(ξ) be the representation of 𝔄 on L(ξ).

Lemma 5.2. Every quasi-simple irreducible representation of 𝔄 is contained at most once in L(ξ).

Let π be such a representation. Suppose π is deducible from π_{ω} and the restriction of π to 𝔲 contains π_{n}. Let L^{0}(ξ) be the space of all infinitely differentiable U^{0}-finite functions on G_{ℂ}^{0} such that

(i) φ(( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ) g) = ξ(z)φ(g).

(ii) If X ∈ 𝔄^{0} and g ∈ G_{ℂ}^{0} there is a constant M such that |ρ(X)φ(( \begin{matrix}t^{1/2} & 0 \\ 0 & t^{−1/2} \end{matrix} )g)| ≤ Mt^{M} for t ≥ 1.

Let ρ^{0}(ξ) be the representation of 𝔄^{0} on L^{0}(ξ). It is enough to show that π^{0} is contained at most once in ρ^{0}(ξ).

Suppose H ⊆ L^{0}(ξ) is invariant and the restriction of ρ^{0}(ξ) to H is equivalent to π^{0}. There is a function Ψ(g) on G_{ℂ}^{0} with values in Ṽ_{n} such that H_{n} is the set of functions of the form Ψ(g)Φ, Φ ∈ V_{n}. Ψ(gu) = Ψ(g)σ_{n}(u) if u ∈ U^{0}. Let ψ(t) = Ψ(( \begin{matrix}t^{1/2} & 0 \\ 0 & t^{−1/2} \end{matrix} )) for t > 0. Ψ is completely determined by ψ. It is necessary to write the equations ρ(D)Ψ = \frac{(s+m)^{2}−1}{2}Ψ and ρ(D′)Ψ = \frac{(s−m)^{2}−1}{2}Ψ in terms of ψ. It is easy to verify that

\begin{aligned}ρ(X_{1}^{2} + X_{1}^{2})Ψ(( \begin{matrix}t^{1/2} & 0 \\ 0 & t^{−1/2} \end{matrix} )) & = −t^{2}|w|^{2}ψ(t) \\ ρ((X_{1} − iX_{2})(W_{1} − iW_{2}))Ψ(( \begin{matrix}t^{1/2} & 0 \\ 0 & t^{−1/2} \end{matrix} )) & = 2tiwψ(t)σ_{n} ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) \\ ρ((X_{1} + iX_{2})(W_{1} + iW_{2}))Ψ(( \begin{matrix}t^{1/2} & 0 \\ 0 & t^{−1/2} \end{matrix} )) & = −2tiw̄ψ(t)σ_{n} ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) \\ ρ(Z_{1})Ψ(( \begin{matrix}t^{1/2} & 0 \\ 0 & t^{−1/2} \end{matrix} )) & = 2t\frac{∂ψ}{∂t} \\ ρ(Z_{2})Ψ(( \begin{matrix}t^{1/2} & 0 \\ 0 & t^{−1/2} \end{matrix} )) & = ψ(t)σ_{n} ( \begin{matrix}i & 0 \\ 0 & −i \end{matrix} ) . \end{aligned}

Thus

\frac{1}{2} t \frac{d}{dt} (t \frac{dψ}{dt})−t \frac{dψ}{dt} \{I− \frac{1}{2} σ_{n}( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )\}+ \frac{1}{2} ψ\{I− \frac{1}{2} σ_{n}( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )\}^{2}− \frac{t2|W |2}{2} ψ+tiwψσ_{n}( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} )= \frac{(s+m)2}{2} ψ

\frac{1}{2} t \frac{d}{dt} (t \frac{dψ}{dt})−t \frac{dψ}{dt} \{I+ \frac{1}{2} σ_{n}( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )\}+ \frac{1}{2} ψ\{I+ \frac{1}{2} σ_{n}( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )\}^{2}− \frac{t2|w|2}{2} ψ+tiw̄ψσ_{n}( \begin{matrix}0 & −1 \\ 0 & 0 \end{matrix} )= \frac{(s−m)2}{2} ψ.

In terms of components these equations are

\begin{matrix}\frac{1}{2}[t\frac{d}{dt} & + k − 1]^{2}ψ^{k} − \frac{t^{2}|w|^{2}}{2} ψ^{k} + c_{k}tiwψ^{k−1} = \frac{(s + m)^{2}}{2} ψ^{k}, \\ \frac{1}{2}[t\frac{d}{dt} & − k − 1]^{2}ψ^{k} − \frac{t^{2}|w|^{2}}{2} ψ^{k} − d_{k}tiw̄ψ^{k+1} = \frac{(s − m)^{2}}{2} ψ^{k}, \end{matrix}  (A)

where ψ^{j} = 0 if |j| > \frac{n}{2}. Since c_{k} ≠ 0 for −\frac{n}{2} < k ≤ \frac{n}{2} and d_{k} ≠ 0 for −\frac{n}{2} ≤ k < \frac{n}{2} these equations allow one to solve for all ψ^{k} in terms of ψ^{\frac{n}{2}} or ψ^{− \frac{n}{2}} .

For k = \frac{n}{2} the second equation is

\frac{1}{2}[t\frac{d}{dt} − \frac{n}{2} − 1]^{2}ψ^{\frac{n}{2}} − \frac{t^{2}|w|^{2}}{2} ψ^{\frac{n}{2}} = \frac{(s − m)^{2}}{2} ψ^{\frac{n}{2}}

which may be written as

\frac{1}{2} \frac{d^{2}ψ^{\frac{n}{2}}}{dt^{2}} + (−\frac{1}{2} − \frac{n}{2})\frac{1}{t} \frac{dψ^{\frac{n}{2}}}{dt} + (−\frac{|w|^{2}}{2} + \frac{(\frac{n}{2} + 1)^{2}}{2t^{2}} )ψ^{\frac{n}{2}} = \frac{(s − m)^{2}}{2t^{2}} ψ^{\frac{n}{2}} .

Dropping the terms in \frac{1}{t} and \frac{1}{t^{2}} one obtains an equation with the solutions e^{±|w|t}. Thus the given equation has one solution of the form t^{−μ}e^{−|w|t}(1 + 0(\frac{1}{t})) and one of the form t^{−ν}e^{|w|t}(1 + 0(\frac{1}{t})). Since ψ^{\frac{n}{2}} (t) = 0(t^{M}) as t → ∞ it must be a multiple of the first solution. The lemma follows.

To find μ we examine the formal solution

ψ^{\frac{n}{2}} (t) = t^{−μ}e^{−|w|t} ∑_{n=0}^{∞} a_{n}t^{−n}.

If a_{−1} = a_{−2} = 0 the first derivative is

e^{−|w|t}{∑_{n=0}^{∞} −(|w|a_{n} + (μ + n − 1)a_{n−1})t^{−n−μ}}

and the second derivative is

e^{−|w|t}{∑_{n=0}^{∞}(|w|^{2}a_{n} + |w|(2μ + 2n − 1)a_{n−1} + (μ + n − 1)(μ + n − 2)a_{n−2})t^{−n−μ}}.

Substituting into the equation, dividing be e^{−|w|t}, and equating coefficients of t^{μ−1} we obtain μ + \frac{n}{2} = 0. For k = −\frac{n}{2} the first of the equations (A) is

\frac{1}{2}[t\frac{dψ^{− \frac{n}{2}}}{dt} − \frac{n}{2} − 1]^{2}ψ^{− \frac{n}{2}} − \frac{t^{2}|w|^{2}}{2} ψ^{− \frac{n}{2}} = \frac{(s + m)^{2}}{2} ψ^{− \frac{n}{2}} .

This is the equation just discussed except that −m is replaced by m. Thus if |k| = \frac{n}{2} then ψ^{k}(t) = t^{|k|}e^{−|w|t}(1 + 0(\frac{1}{t})) as t → ∞.

During the preceding discussion we have assumed the existence of H and thus the existence of solutions of equation (A) which satisfy the required growth condition. We continue with our discussion of this assumption.

Since 0 is a singular point of the first kind for the first equations of (A) there is an N such that, for all k, ψ^{k}(t) = O(\frac{1}{t^{N}}) as t → 0. Thus

θ^{k}(u) = ∫_{0}^{∞} ψ^{k}(t)t^{u−1}dt

is defined for Re u sufficiently large. These functions satisfy the difference equations

\begin{aligned}|w|^{2}θ^{k}(u + 2) & = [(u − k + 1)^{2} − (s + m)^{2}]θ^{k}(u) + 2ic_{k}wθ^{k−1}(u + 1) \\ |w|^{2}θ^{k}(u + 2) & = [(u + k + 1)^{2} − (s − m)^{2}]θ^{k}(u) − 2id_{k}w̄θ^{k+1}(u + 1). \end{aligned}

Lemma 5.3. If |k| = \frac{n}{2} then θ^{k}(u) is a multiple of

(2/|w|)^{u}Γ\frac{(u + 1 + s + |k − m|)}{2}Γ\frac{(u + 1 − s + |k + m|)}{2}.

Since \frac{n}{2} ≥ |m| the second of the difference equations is, when k = \frac{n}{2}, just

|w|^{2}θ^{k}(u + 2) = (u + 1 + s + |\frac{n}{2} − m|)(u + 1 − s + |\frac{n}{2} + m|)θ^{k}(u)

which is an equation satisfied by the function of the lemma. Thus the inverse Mellin transform of that function, which is bounded by a power of t, satisfies the differential equation determining ψ^{\frac{n}{2}} and must be a multiple of ψ^{\frac{n}{2}} . A similar argument proves the lemma when k = −\frac{n}{2}.

Lemma 5.4. If |m| = \frac{n}{2} the functions

θ_{0}^{k}(u) = \frac{2^{u}}{|w|^{u}} (\frac{iw}{|w|})^{k− \frac{n}{2}} Γ(\frac{u + 1 + s + |k − m|}{2})Γ(\frac{u + 1 − s + |k + m|}{2})

satisfy the difference equations. They are the only solutions of the equations for which

θ0^{\frac{n}{2}} (u) = \frac{2^{u}}{|w|^{u}} Γ\frac{(u + 1 + s + |\frac{n}{2} − m|)}{2}Γ\frac{(u + 1 − s + |\frac{n}{2} + m|)}{2}.

The uniqueness is evident from the form of the equations. It is convenient to treat the cases m = \frac{n}{2} and m = −\frac{n}{2} separately when verifying that they satisfy the equations. If m = \frac{n}{2} then |w|^{2}θ_{0}^{k}(u + 2) − 2ic_{k}wθ_{0}^{k−1}(u + 1) is equal to

\frac{2^{u}}{|w|^{u}} (\frac{iw}{|w|})^{k− \frac{n}{2}} \{(u + 1 + s + \frac{n}{2} − k)(u + 1 − s + \frac{n}{2} + k) − 2(\frac{n}{2} + k)(u + 1 + s + \frac{n}{2} − k)\}

× Γ\frac{(u + 1 + s + \frac{n}{2} − k)}{2}Γ\frac{(u + 1 − s + \frac{n}{2} + k)}{2}

= (u + 1 + s + \frac{n}{2} − k)(u + 1 − s − \frac{n}{2} − k)θ_{0}^{k}(u)

and |w|^{2}θ_{0}^{k}(u + 2) + 2d_{k}iw̄θ_{0}^{k+1}(u + 1) is equal to

\frac{2u}{|w|u} ( \frac{iw}{|w|})^{k− \frac{n}{2}} \{(u+1+s+ \frac{n}{2} −k)(u+1−s+ \frac{n}{2} +k)−2( \frac{n}{2} +k)(u+1+s+ \frac{n}{2} −k)\}Γ \frac{(u+1+s+ \frac{n}{2} −k)}{2} Γ \frac{(u+1−s+ \frac{n}{2} +k)}{2}

or

(u + 1 − s + \frac{n}{2} + k)(u + 1 + s − \frac{n}{2} + k)θ_{0}^{k}(u).

It is not necessary to treat the case m = −\frac{n}{2} because the equations are not changed if θ^{k} is replaced by θ^{−k}, m by −m, and w by −w̄.

Corollary. The quasi-simple irreducible representation π is contained in ρ(ξ) if and only if π is infinite dimensional.

It is enough to show that π^{0} is contained in ρ^{0}(ξ) if and only if π^{0} is infinite dimensional. Suppose H is a finite dimensional invariant subspace of L^{0}(ξ). Let τ be the restriction of ρ^{0}(ξ) to L^{0}(ξ) and let τ̃ be the representation contragredient to τ. If X_{z} = ( \begin{matrix}0 & z \\ 0 & 0 \end{matrix} ) lies in g^{0} all the eigenvalues of τ̃(X_{z}) must be zero because τ̃ is finite dimensional. On the other hand if φ̃ is the element of H̃, the dual space of H, defined by φ̃(φ) = φ(1), φ ∈ H, then

(τ̃(X_{z})φ̃)(φ) = −φ̃(τ(X_{z})φ) = ρ(X_{z})φ(1) = −izwφ(1)

so that −izw is an eigenvalue of X_{z}. This is a contradiction.

Suppose π is deducible from π_{ω}. Let W be the set of all functions in L^{0}(ξ) satisfying ρ(D)φ =

\frac{(s+m)^{2}−1}{2}φ, ρ(D′)φ = \frac{(s−m)^{2}−1}{2}φ, and φ(g( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} )) = (−1)^{2m}φ(g). If θ = \{n|W_{n} ≠ \{0\}\} then W =

∑_{n∈θ} W_{n}. Combining the results of the previous lemma with the arguments used to prove Lemma 5.3 one sees that when \frac{n}{2} = |m| the equations (A) have a solution satisfying the desired growth conditions. Thus W_{n} is not zero for n = |\frac{m}{2}|. Although it is not important at present, I observe that if s − m is integral then W_{|2s|} is also not zero. The proof of Lemma 5.2 shows that W_{n} is irreducible under 𝔲^{0} the Lie algebra of U^{0}. Consequently every invariant subspace is of the form W (σ) = ∑_{n∈σ} W_{n} where σ is a subset of θ. Suppose σ_{0} ⫆̸ σ_{3} and W (σ_{0}) and W (σ_{3}) are invariant. Let n_{0} ∈ σ_{0}, n_{0} ∉ σ_{3}. There is a minimal element in {σ|W (σ) is invariant, σ_{3} ⊂ σ ⊂ σ_{0}, n_{0} ∈ σ_{1}}; let it be σ_{1}. There is a maximal element in {σ|W (σ) is invariant, σ_{3} ⊂ σ ⊂ σ_{1}, n_{0} ∉ σ}; let it be σ_{2}. The representation of 𝔄^{0} on W (σ_{1})/W (σ_{2}) is irreducible. Thus there is an irreducible representation deducible from the representation of 𝔄^{0} on W (σ_{0})/W (σ_{3}). Suppose W itself is not irreducible and let W (σ_{1}) be a proper invariant subspace. If W (σ_{1}) were not irreducible there would be a proper invariant subspace W (σ_{2}). No two of the irreducible representations deduced from the representations on W/W (σ_{1}), W (σ_{1})/W (σ_{2}), W (σ_{2}) could be equivalent because the restrictions to 𝔲_{ℂ} would not be equivalent. This would contradict Lemma 4.4. For the same reason the representation on W/W (σ_{1}) is irreducible. Thus either W is irreducible or W contains a proper invariant irreducible subspace W_{1} such that W/W_{1} is irreducible. Combining Lemma 4.4 with the earlier observations about finite dimensional representations one sees that if π is infinite dimensional the representation of 𝔄^{0} on W is equivalent to π^{0} if W is irreducible and that if W is not irreducible the representation of 𝔄^{0} on W_{1} is equivalent to π^{0}.

We return to the study of the functions ψ^{k}(t).

Lemma 5.5. If (s − m) is not an integer then, near 0, ψ^{k}(t) can be expanded in a series of the form

ψ^{k}(t) = t^{|k−m|+1+s} ∑_{p=0}^{∞} a_{p}^{k}t^{2p} + t^{|k+m|+1−s} ∑_{p=0}^{∞} b_{p}^{k}t^{2p}.

If s − m is an integer t and |k + m| − s ≥ |k − m| + s then

ψ^{k}(t) = t^{|k−m|+1+s} ∑_{p=0}^{∞} a_{p}^{k}t^{2p} + t^{|k+m|+1−s} log t ∑_{p=0}^{∞} b_{p}^{k}t^{2p}

but if |k + m| − s ≤ |k − m| + s then

ψ^{k}(t) = t^{|k−m|+1+s} log t ∑_{p=0}^{∞} a_{p}^{k}t^{2p} + t^{|k+m|+1−s} ∑_{p=0}^{∞} b_{p}^{k}t^{2p}.

As before when k = \frac{n}{2} the second equation of (A) is

\frac{1}{2}[t\frac{d}{dt} − \frac{n}{2} − 1]^{2}ψ^{\frac{n}{2}} − \frac{t^{2}|w|^{2}}{2} = \frac{(s − m)^{2}}{2} ψ^{\frac{n}{2}} .

The indicial equation \frac{1}{2}[λ − \frac{n}{2} − 1]^{2} = \frac{(s−m)^{2}}{2} has the roots λ_{i} = \frac{n}{2} + 1 − s + m and λ_{2} = \frac{n}{2} + 1 + s − m and λ_{1} − λ_{2} = 2(m − s). The series t^{λ_{i}} ∑_{p=0}^{∞} a_{p}t^{p} will satisfy the equation if and only if

\{[λ_{i} + p − \frac{n}{2} − 1]^{2} − (s − m)^{2}\}c_{p} = |w|^{2}c_{p−2}.

Since |\frac{n}{2} ± m| = \frac{n}{2} ± m the assertion of the lemma for k = \frac{n}{2} follows from an application of the method of Frobenius.

To prove the lemma for general k we use induction and the equation

−c_{k}iwψ^{k−1}(t) = \frac{1}{2t}\{[t\frac{d}{dt} + k − 1]^{2}ψ^{k} − (s + m)^{2}ψ^{k} − t^{2}|w|^{2}ψ^{k}\}.

The symbol A(t) will stand for a convergent series of the form ∑_{p=0}^{∞} a_{p}t^{2p} and B(t) will stand for a convergent series of the form ∑_{p=1}^{∞} b_{p}t^{2p}. The series represented by these symbols will vary but not within a given formula. One has

\begin{matrix}\frac{1}{2t}\{[t\frac{d}{dt} + k − 1]^{2}t^{|k∓m|+1±s}A(t) − (s + m)^{2}t^{|k∓m|+1±s}A(t) − t^{2}|w|^{2}A(t)\} \\ = \frac{1}{2t}\{[(|k ∓ m| + k ± s)^{2} − (s + m)^{2}]a_{0}t^{|k∓m|+1±s} + t^{|k∓m|+1±s}B(t)\}. \end{matrix}

If k > ±m so that |k − 1 ∓ m| = |k ∓ m| − 1 this is of the form t^{|k−1∓m|+1±s}A(t). If k ≤ ±m then |k ∓ m| = ±m − k and (|k ∓ m| + k ± s)^{2} − (s + m)^{2} = 0 and it is of the form t^{|k−1∓m|+1±s}B(t) because |k − 1 ∓ m| = |k ∓ m| + 1. The first statement of the lemma follows immediately.

If F (t) is any function

[t\frac{d}{dt} + k − 1] log tF (t) = F (t) + log t[t\frac{d}{dt} + k − 1]F (t)

and

[t\frac{d}{dt} + k − 1]^{2} log tF (t) = 2[t\frac{d}{dt} + k − 1]F (t) + log t[t\frac{d}{dt} + k − 1]^{2}F (t).

Thus

\frac{1}{2t}\{[t\frac{d}{dt} + k − 1]^{2}t^{|k∓m|+1±s} log t(A(t) − (s + m)^{2}t^{|k∓k|+1±s} log tA(t) − t^{2}|w|^{2}t^{|k∓m|+1±s} log tA(t)\}

is equal to the sum of a term of the form t^{|k−1∓m|+1±s} log tA(t) and

\frac{1}{t}\{(|k ∓ m| ± s + k)t^{|k∓m|+1±s}a_{0} + t^{|k∓m|+1±s}B(t)\}  (B)

Suppose s − m is an integer and the assertions of the lemma are true for a given k. Let |k ∓ m| ± s ≥ |k ± m| ∓ s (Either all the top signs or all the bottom signs are taken). If (|k ∓ m| ± s) − (|k ± m| ∓ s), which is an integer, is at least two then |k ∓ m| ± s ≥ |k − 1 ± m| + 1 ∓ s and the expression (B) is of the form t^{|k−1±m|+1∓s}. Since |k − 1 ∓ m| ± s will still be greater than or equal to |k − 1 ± m| ∓ s the induction goes through.

The remaining possibility is |k ∓ m| ± s = |k ± m| ∓ s. If k > ∓m then |k − 1 ± m| = |k ± m| − 1 so that |k − 1 ∓ m| ± s ≥ |k − 1 ± m| ∓ s and the expression (B) is of the form t^{|k−1±m|+1∓s}A(t). If k > ±m then |k − 1 ∓ m| = |k ∓ m| − 1 so that |k − 1 ± m| ∓ s ≥ |k − 1 ∓ m| ± s and the expression (B) is of the form t^{|k−1∓m|+1±s}A(t).

Thus we have only to treat the case that k ≤ ∓m, k ≤ ±m and |k ∓m|±s = |k ±m|∓s. Then |k ∓m| = ±m − k and |k ± m| = ∓m − k so ±m − k ± s = ∓m − k ∓ s or m + s = 0 and |k ∓ m| ± s + k = ±(m + s) = 0. Thus |k − 1 ∓ m| ± s = |k − 1 ± m| ∓ s and the expression (B) is of the form t^{|k−1±m|+1∓s}A(t).

Let ψ(t) be the function with components ψ^{k}(t). If 2ℓ is an integer and z is a fixed complex number set

θ(u, ℓ; z) = ∫_{0}^{∞}\{\frac{1}{4π} ∫_{0}^{4π} e^{it Re (e^{iθ}z)}ψ(t)σ_{n} ( \begin{matrix}e^{\frac{iθ}{2}} & 0 \\ 0 & e^{− \frac{iθ}{2}} \end{matrix} ) e^{−iℓθ}dθ\}t^{u−1}dt.

The integral converges for Re u sufficiently large. The k^{th} component of θ(u, ℓ; z) is

θ^{k}(u, ℓ; z) = ∫_{0}^{∞}\{\frac{1}{4π} ∫_{0}^{4π} e^{it Re (e^{iθ}z)}e^{i(k−ℓ)}dθ\}ψ^{k}(t)t^{u−1}dt.

Lemma 5.6. For each ℓ and z the function

\frac{θ^{k}(u, ℓ; z)}{Γ\frac{(u+1+s+|ℓ−m|)}{2}Γ\frac{(u+1−s+|ℓ+m|)}{2}}

is an entire function of u. Moreover θ^{k}(u, ℓ, z) is bounded in any region of the form |Re u| ≤ constant, |Im u| ≥ constant ≫ 0.

Let m(t) be an infinitely differentiable function with compact support on the real line which is 1 in a neighbourhood of 0. Then θ^{k}(u, ℓ, z) is the sum of

θ̂^{k}(u, ℓ, z) = ∫_{0}^{∞}\{\frac{1}{4π} ∫_{0}^{4π} e^{itRe(e^{iθ}z)}e^{i(k−ℓ)}dθ\}ψ^{k(t)}t^{u−1}m(t)dt

and

∫_{0}^{∞}\{\frac{1}{4π} ∫_{0}^{4π} e^{itRe(e^{iθ}z)}e^{i(k−ℓ)}dθ\}ψ^{k}(t)t^{u−1}(1 − m(t))dt.

The second integral defines an entire function of u which is bounded on vertical strips so it will be enough to prove the lemma with θ^{k}(u, ℓ, z) replaced by θ̂^{k}(u, ℓ, z).

The inner integral is equal to

∑_{r=0}^{∞} \frac{(it)^{r}}{2^{r}r!} · \frac{1}{4π} ∫_{0}^{4π}(e^{iθ}z + e^{−iθ}z̄)^{r}e^{i(k−ℓ)θ}dθ.

It is zero if k − ℓ is not integral. If k − ℓ is integral let Δ be the set of integers r satisfying (i) r ≥ |k − ℓ| and (ii) \frac{r+ℓ−k}{2} is integral. Then this expression equals

∑_{r∈Δ} \frac{(it)^{r}}{2^{r}} \frac{z^{r+ℓ−k}}{(\frac{r+ℓ−k}{2} )!} \frac{z̄^{r+k−ℓ}}{(\frac{r+k−ℓ}{2})!}.

If a real number c is given there is an R such that

∫_{0}^{∞}\{ ∑_{{r∈Δ \atop r≥R}} \frac{(it)^{r}}{2^{r}} \frac{z^{r+ℓ−k}}{(\frac{r+ℓ−k}{2} )!} \frac{z̄^{r+k−ℓ}}{(\frac{r+k−ℓ}{2})!}\}ψ^{k(t)}t^{u−1}m(t)dt

is analytic and bounded for Re u > c. We need only study the analytic properties of

∫_{0}^{∞} ψ^{k}(t)t^{r+u−1}m(t)dt r ∈ Δ.

The same observation when combined with Lemma 5.5 shows that when s − m is not an integer we need only study

∫_{0}^{∞} t^{|k±m|∓s+r+u+2p}m(t)dt r ∈ Δ, p ∈ ℤ, p ≥ 0

and that when s − m is integral and |k ∓ m| ± s ≥ |k ± m| ∓ s we need only study

∫_{0}^{∞} t^{|k±m|∓s+r+u+2p}m(t)dt r ∈ Δ, p ∈ ℤ, p ≥ 0

and

∫_{0}^{∞} t^{|k∓m|±s+r+u+2p}m(t) log tdt r ∈ Δ, p ∈ ℤ, p ≥ 0.

The second assertion of the lemma is going to be obvious and only the first will have to be dealt with explicitly. The first is going to follow from the observation that if s − m is not an integer the denominator in the lemma has poles of order 1 at −1 ∓ s − |ℓ ∓ m| − 2, q ∈ ℤ, q ≥ 0 and no zeros and that if s − m is an integer and |ℓ ∓ m| ± s ≥ |ℓ ± m| ∓ s it has poles of order at least one at −1 ± s − |ℓ ± m| − 2q and poles of order two at −1 ∓ s − |ℓ ∓ m| − 2q, q ∈ ℤ, q ≥ 0.

It has to be shown that these poles cancel the singularities of the numerator.

\begin{aligned}∫_{0}^{∞} t^{a+u}m(t)dt & = \frac{−1}{a + u + 1} ∫_{0}^{∞} t^{a+u+1}m′(t)dt \\ ∫_{0}^{∞} t^{a+u} log tm(t)dt & = \frac{−1}{(a + u + 1)^{2}} ∫_{0}^{∞}[(a + u + 1)t^{a+u+1} log t − t^{a+u+1}]m′(t)dt. \end{aligned}

Since m′(t) vanishes near 0 the first integral has at most a pole of order one at −(a + 1) and no other singularities while the second has at most a pole of order two at −(a + 1).

If s−m is not an integer the lemma will follow if it is shown that, for r ∈ Δ, |k±m|+r = |ℓ±m|+2q, q ∈ ℤ, q ≥ 0. This is so because r = |k − ℓ| + 2p, p ∈ ℤ, p ≥ 0 and |k ± m| + |k − ℓ| − |ℓ ± m| is a non-negative even integer. If s − m is an integer one has to show in addition that if |ℓ ∓ m| − |ℓ ± m| ± 2s > 0 and |k ± m| − |k ∓ m| ∓ 2s > 0 (either all upper or all lower signs are taken so there are only two possibilities) then

|k ± m| ∓ s + |k − ℓ| = |ℓ ∓ m| ± s + 2q q ∈ ℤ, q ≥ 0.

The left side is

|k ∓ m| ± s + |k − ℓ| + {|k ± m| − |k ∓ m| ∓ 2s}.

The expression in brackets, which is a non-negative integer, is by assumption positive.

If π is an infinite-dimensional irreducible quasi-simple representation of 𝔄 let L(ξ, π) be the unique subspace of L(ξ) such that the restriction of ρ(ξ) to L(ξ, π) is equivalent to π. It follows from the proof of Lemma 4.4 that there is an ω such that π is equivalent to π_{ω}. As usual let ω_{1}(α) = ω(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ), ω_{2}(α) =

ω(( \begin{matrix}1 & 0 \\ 0 & α \end{matrix} )) for α ∈ ℂ^{×} and let ω_{i}(te^{iθ}) = t^{s_{i}}e^{im_{i}θ} for t > 0.

If η is any character of A_{ℂ} then η̃ is the character defined by η̃(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )) = η(( \begin{matrix}α_{2} & 0 \\ 0 & α_{1} \end{matrix} )). If ζ is a character of A_{ℂ}^{×} such that ζ(( \begin{matrix}α & 0 \\ 0 & α \end{matrix} ))ω(( \begin{matrix}α & 0 \\ 0 & α \end{matrix} )) ≡ 1 and u and ℓ are defined by ζ(( \begin{matrix}t^{1/2}e^{iθ/2} & 0 \\ 0 & t^{−1/2}e^{−iθ/2} \end{matrix} )) = t^{u}e^{iℓθ} then ζ is uniquely determined by u and ℓ and we shall occasionally write ζ = ζ(u, ℓ).

Lemma 5.7. Suppose ζ(( \begin{matrix}α & 0 \\ 0 & α \end{matrix} ))ω(( \begin{matrix}α & 0 \\ 0 & α \end{matrix} )) ≡ 1. If φ ∈ L(ξ, π) and ζ = ζ(u, ℓ) the function

Φ(g, ζ, φ) = ∫_{0}^{∞}\{\frac{1}{2π} ∫_{0}^{2π} φ(( \begin{matrix}te^{iθ} & 0 \\ 0 & 1 \end{matrix} ) g)ζ(( \begin{matrix}te^{iθ} & 0 \\ 0 & 1 \end{matrix} ))dθ\}\frac{dt}{t}

is defined for Re u sufficiently large. Set

Φ′(g, ζ, φ) = \frac{Φ(g, ζ, φ)}{Γ(\frac{u+1+s+|ℓ+m|}{2})Γ(\frac{u+1−s+|ℓ−m|}{2})}.

Then Φ′(g, ζ(u, ℓ), φ) is an entire function of u and Φ(g, ζ(u, ℓ), φ) is bounded in regions of the form |Re u| ≤ constant, |Im u| ≥ constant ≫ 0. Moreover

(\frac{2}{|w|})^{−u}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = γ(ℓ, m)(\frac{iw}{|w|})^{−2ℓ}(\frac{2}{|w|})^{u}Φ′(g, ζ̃, φ)

if γ(ℓ, m) = (−1)^{|ℓ|+ℓ} for |ℓ| ≥ |m| and γ(ℓ, m) = (−1)^{|m|+ℓ} for |ℓ| ≤ |m|.

It is enough to prove the lemma for φ in L(ξ, π)_{n}. There is a Φ in V_{n} such that if g = a( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} )u with

a = ( \begin{matrix}t_{1}e^{iθ_{1}} & 0 \\ 0 & t_{2}e^{iθ_{2}} \end{matrix} ) and u in U then φ(( \begin{matrix}te^{iθ} & 0 \\ 0 & 1 \end{matrix} )g)ζ(( \begin{matrix}te^{iθ} & 0 \\ 0 & 1 \end{matrix} )) is equal to the product of

e^{\frac{itt_{1}}{t_{2}} Re (e^{i(θ+θ_{1}−θ_{2})_{wZ})}}ω(( \begin{matrix}\sqrt{tt_{1}t_{2}}e^{i(θ+θ_{1}+θ_{2})} & 0 \\ 0 & \sqrt{tt_{1}t_{2}}e^{i(θ+θ_{1}+θ_{2})} \end{matrix} ))ζ(( \begin{matrix}te^{iθ} & 0 \\ 0 & 1 \end{matrix} ))

and

ψ(\frac{tt_{1}}{t_{2}} )σ_{n}(( \begin{matrix}e^{\frac{i(θ+θ_{1}−θ2)}{2}} & 0 \\ 0 & e^{\frac{i(θ_{2}−θ−θ_{1})}{2}} \end{matrix} ) u)Φ

which equals the product of

ζ^{−1}(( \begin{matrix}t_{1}e^{iθ_{1}} & 0 \\ 0 & t_{2}e^{iθ_{2}} \end{matrix} ))e^{i \frac{tt_{1}}{t_{2}} Re (e^{i(θ+θ_{1}+θ_{2})_{wz)}}} ζ(( \begin{matrix}\sqrt{\frac{tt_{1}}{t_{2}}} e^{i \frac{(θ+θ_{1}−θ_{2})}{2}} & 0 \\ 0 & \sqrt{\frac{t_{2}}{tt_{1}}} e^{i \frac{(θ_{2}−θ−θ_{2})}{2}} \end{matrix} ))

and

ψ(\frac{tt_{1}}{t_{2}} )σ_{n}(( \begin{matrix}e^{i \frac{(θ+θ_{1}−θ_{2})}{2}} & 0 \\ 0 & e^{i \frac{(θ_{1}−θ−θ_{2})}{2}} \end{matrix} ) u)Φ.

Consequently Φ(g, ζ(u, ℓ), φ) is equal to

ζ^{−1}(( \begin{matrix}t_{1}e^{iθ_{1}} & 0 \\ 0 & t_{2}e^{iθ_{2}} \end{matrix} ))θ(u, −ℓ, wz)σ_{n}(u)Φ.

The first two assertions of the lemma follow immediately.

If η = ζ̃^{−1} the maps

\begin{matrix}φ & ⟶ Φ′(g, ζ̃, φ), \\ φ & ⟶ Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) \end{matrix}

are easily seen to be 𝔄-invariant maps of L(ξ, π) into L(η). It follows from Lemma 5.1 that one is a multiple of the other. To see what the multiple is choose g = 1 and φ as above with Φ = δ_{ℓ}. Then

\begin{aligned}Φ′(1, ζ̃, φ) & = \frac{θ^{ℓ}(u)}{Γ(\frac{−u+1+s+|ℓ−m|}{2})Γ(\frac{−u+1−s+|ℓ+m|}{2})} \\ & = f_{ℓ}(−u)(\frac{2}{|w|})^{−u}(\frac{iw}{|w|})^{ℓ− \frac{n}{2}} \end{aligned}

if the functions θ^{ℓ}(u) are normalized as in the appendix.

Since σ_{n}( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )δ_{ℓ} = (−1)^{\frac{n}{2} +ℓ}δ_{−ℓ},

\begin{aligned}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) , ζ, φ) & = \frac{(−1)^{\frac{n}{2} +ℓ}θ^{−ℓ}(u)}{Γ(\frac{u+1+s+|ℓ+m|}{2})Γ(\frac{u+1−s+|ℓ−m|}{2})} \\ & = (−1)^{\frac{n}{2} +ℓ}f_{−ℓ}(u)(\frac{2}{|w|})^{u}(\frac{iw}{|w|})^{−ℓ− \frac{n}{2}} . \end{aligned}

Taking \frac{n}{2} = |m| if |ℓ| ≤ |m| and \frac{n}{2} = |ℓ| if |ℓ| ≥ m we see that

(\frac{2}{|w|})^{−u}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) , ζ, φ) = γ(ℓ, m)(\frac{iw}{|w|})^{−2ℓ}(\frac{2}{|w|})^{u}Φ′(1, ζ̃, φ)

because as is shown in the Appendix, f_{± \frac{n}{2}} (u) = 1 and, as is shown in Lemma 5.4, f_{k}(u) = 1 if \frac{n}{2} = |m|.

Appendix. Unfortunately the preliminary material of this paragraph was not sufficient to give the constant occurring in the functional equation. A little more information about the functions θ^{k}(u) is necessary. Normalize them by setting

θ^{\frac{n}{2}} (u) = (\frac{2}{|w|})^{u}Γ(\frac{u + 1 + s + \frac{n}{2} − m}{2})Γ(\frac{u + 2 − s + \frac{n}{2} + m}{2}).

It is an immediate consequence of the difference equations that none of the functions θ^{k}(u), |k| ≤ \frac{n}{2}, \frac{n}{2} −k ∈ ℤ, can vanish identically.

Lemma A. Let α_{k} = min\{\frac{n}{2} − |k|, \frac{n}{2} − |m|\}. Then θ^{k}(u) is of the form

f_{k}(u)(\frac{2}{|w|})^{u}(\frac{iw}{|w|})^{k− \frac{n}{2}} Γ(\frac{u + 1 + s + |k − m|}{2})Γ(\frac{u + 1 − s + |k + m|}{2})

where f_{k}(u) is a polynomial in u of degree α_{k}. Its coefficients are polynomials in s which do not depend on w.

We shall show that if θ^{k}(u) is of this form with a polynomial of degree β_{k} the same is true of θ^{k−1}(u) with a polynomial of degree β_{k−1} where β_{k−1} − β_{k} ≤ α_{k−1} − α_{k}. This is enough to prove the lemma because β\frac{n}{2} = α\frac{n}{2} = 0 and if β_{k_{0}} were less than α_{k_{0}} for some k_{0} then β_{k} would be less than α_{k} for all succeeding k. Since α_{− \frac{n}{2}} = 0 this is impossible.

The first difference equations show that 2c_{k} \frac{|w|^{u+1}}{2^{u+1}} (\frac{iw}{|w|})^{\frac{n}{2} −(k−1)}θ^{k−1}(u + 1) is the product of

\frac{1}{2}f_{k}(u + 2)[u + 1 + s + |k − m|][u + 1 − s + |k + m|] − \frac{1}{2}f_{k}(u)[u − k + 1 + s + m][u − k + 1 − s − m]

and

Γ(\frac{u + 1 + s + |k − m|}{2})Γ(\frac{u + 1 − s + |k + m|}{2}).

If k > |m| then α_{k−1} = α_{k} + 1, |k − 1 ± m| = |k ± m| − 1, the second factor is

Γ(\frac{(u + 1) + s + |k − 1 − m|}{2})Γ(\frac{(u + 1) + 1 − s + |k − 1 + m|}{2}),

and the first factor is a polynomial in u and s of degree at most β_{k} + 1 in u.

Suppose |m| ≥ k > −|m|, such that α_{k} = α_{k−1}. Let ±m ≥ 0. The first factor is the product of \frac{(u+1±s±m−k)}{2} and

f_{k}(u + 2)[u + 1 ∓ s + |k ∓ m|] − f_{k}(u)[u − k + 1 ∓ s ∓ m]

which is a polynomial of degree at most β_{k}. Moreover |k − 1 ± m| = |k ± m| − 1, |k − 1 ∓ m| = |k ∓ m| + 1, and |k ± m| = ±m − k; so the product of \frac{u+1±s±m−k}{2} and the second factor is

Γ(\frac{(u + 1) + 1 + s + |k − 1 − m|}{2})Γ(\frac{(u + 1) + 1 − s|k − 1 + m|}{2}).

If −|m| ≥ k > \frac{n}{2} then |k − m| = m − k, |m + k| = −m − k, |k − 1 − m| = |k − m| + 1, |k − 1 + m| = |k + m| + 1, and α_{k−1} = α_{k} − 1. The first factor is the product of (\frac{u−k+1+s+m)}{2})(\frac{u−k+1−s−m}{2}) and 2(f_{k}(u + 2) − f_{k}(u)) which is either zero or a polynomial of degree at most β_{k} − 1. Moreover the product of (\frac{u−k+1+s+m}{2})(\frac{u−k+1−s−m}{2}) and the second factor is

Γ(\frac{(u + 1) + 1 + s + |k − 1 − m|}{2})Γ(\frac{(u + 1) + 1 − s + |k − 1 + m|}{2}).

It follows from the corollary to Lemma 5.4 that the equations (A) and thus the difference equations have a solution at least when s − m is not an integer. We could have used the same ideas to show that they had a solution for all s and m. This also follows from the above lemma. To indicate explicitly the dependence of f_{k}(u) on s and m we write f_{k}(u) = f_{k}(u, s, m). The function f_{− \frac{n}{2}} (u, s, m) is independent of u.

Lemma B.

f_{− \frac{n}{2}} (u, s, m) ≡ 1.

For the proof we observe that the functions

θ̂^{k}(u) = f_{−k}(u, s, −m)(\frac{2}{|w|})^{n}(−\frac{iw̄}{|w|}))^{−k− \frac{n}{2}} Γ(\frac{u + 1 + s + |kk − m|}{2})Γ(\frac{u + 1 − s + |k + m|}{2})

also satisfy the difference equations. From uniqueness and the relation

θ̂^{\frac{n}{2}} (u) = f_{− \frac{n}{2}} (u, s, −m)(−\frac{iw̄}{|w|})^{−n}θ^{\frac{n}{2}} (u)

we conclude that

f_{−k}(u, s, −m)(−\frac{iw̄}{|w|})^{−k− \frac{n}{2}} = f_{− \frac{n}{2}} (u, s, −m)f_{k}(u, s, m)(−\frac{iw̄}{|w|})^{−n}(\frac{iw}{|w|})^{k− \frac{n}{2}}

or

f_{−k}(u, s, −m) = f_{− \frac{n}{2}} (u, s, −m)f_{k}(u, s, m).

Choosing k = −\frac{n}{2} we see that

f_{− \frac{n}{2}} (u, s, −m)f_{− \frac{n}{2}} (u, s, m) ≡ 1.

Since both terms on the right are polynomials in s they must be independent of s and f_{− \frac{n}{2}} (u, s, m) = ϵ(m).

When s = 0 the difference equations do not change when m is replaced by −m. Consequently f_{k}(u, 0, −m) = f_{k}(u, 0, m) and

f_{−k}(u, 0, m) = ϵ(m)f_{k}(u, 0, m).

If m is an integer we can take k = 0 and conclude that ϵ(m) = 1. If m is a half-integer take k = \frac{1}{2}. Then c_{k} = \frac{n+1}{2}. We have just seen that if ±m ≥ 0

(n + 1)f_{− \frac{1}{2}} (u + 1, 0, m) = f_{\frac{1}{2}} (u + 2, 0, m)[u + 1 ± m + \frac{1}{2}] − f_{\frac{1}{2}} (u, 0, m)[u + 1 − \frac{1}{2} ∓ m]

= [f_{\frac{1}{2}} (u + 2, 0, m) − f_{\frac{1}{2}} (u, 0, m)][u + 1] + [f_{\frac{1}{2}} (u + 2, 0, m) + f_{\frac{1}{2}} (u, 0, m)][|m| + \frac{1}{2}].

The degree of both sides is \frac{n}{2} − |m|. Let a be the coefficient of u^{\frac{n}{2} −|m|} in f_{\frac{1}{2}} (u, 0, m). The coefficient of u^{\frac{n}{2} −|m|} in the polynomial on the left is (n + 1)ϵ(m)a. The coefficient of u^{\frac{n}{2} −|m|} in the polynomial on the right is 2(\frac{n}{2} − |m|)a + 2(|m| + \frac{1}{2})a = (n + 1)a. Thus ϵ(m) = 1.

6. The local functional equation at a non-archimedean prime. Let K be a non-archimedean local field, let O be the ring of integers in K, and let π be a generator of the prime ideal in 0. Let G_{K} = GL(2, K) and let G_{O} = GL(2, O). If A is the group of diagonal matrices and N the group of matrices of the form

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

then the Haar measure on G_{K} may be so normalized that

∫_{G_{K}} f(g)dg = ∫_{A_{K}/A_{O}} |\frac{α_{1}}{α_{2}} |^{−1}da ∫_{N_{K}} dn ∫_{G_{O}} dkf(nak)

if

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

The Hecke algebra H is just the algebra, under convolution, of functions on G_{K} which have compact support and are bi-invariant under G_{O}. Let H̃ be the algebra, under convolution, of functions with compact support on A_{K}/A_{O} which satisfy

f(a) = f(ā).

If

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

then

ā = ( \begin{matrix}α_{2} & 0 \\ 0 & α_{1} \end{matrix} ) .

Lemma 6.1. If f ∈ H and a ∈ A_{K} set

f̃(a) = |\frac{α_{1}}{α_{2}} |^{1/2} ∫_{N_{K}} f(an)dn.

The map f → f̃ is an isomorphism of H with H̃.

To show that f̃ lies in H̃ one has to show that f̃(a) = f̃(ā). This is clear if a = ā; so suppose a ≠ ā. Since a is conjugate to ā in G_{K} the integrals

∫_{A_{K}\G_{K}} f(g^{−1}ag)dg

and

∫_{A_{K}\G_{K}} f(g^{−1}āg)dg

are equal if they exist. But

\begin{aligned}∫_{A_{K}\G_{K}} f(g^{−1}ag)dg & = ∫_{N_{K}} dn ∫_{G_{O}} dk(f(k^{−1}n^{−1}ank)) \\ & = ∫_{N_{K}} f(a(a^{−1}n^{−1}an))dn. \end{aligned}

A simple change of variables shows that the last integral equals

\frac{|\frac{α_{2}}{α_{1}} |^{1/2}}{|1 − \frac{α_{2}}{α_{1}} |}f̃(a).

Combining this with the relation

\frac{|\frac{α_{2}}{α_{1}} |^{1/2}}{|1 − \frac{α_{2}}{α_{1}} |} = \frac{|\frac{α_{1}}{α_{2}} |^{1/2}}{|1 − \frac{α_{1}}{α_{2}} |}

one sees that f̃(a) = f̃(ā).

If f = f_{1} ∗ f_{2} then

f̃(b) = |\frac{β_{1}}{β_{2}} |^{1/2} ∫_{N_{K}} \{∫_{G_{K}} f_{1}(bvg)f_{2}(g^{−1})dg\}dv.

The Haar measure has been so normalized that this equals

|\frac{β_{1}}{β_{2}} |^{1/2} ∫_{A_{K}/A_{O}} da ∫_{N_{K}} du ∫_{N_{K}} dv\{f_{1}(bvua)f_{2}(a^{−1}u^{−1})|\frac{α_{1}}{α_{2}} |^{−1}\}.

Simple manipulation shows that this equals

|\frac{β_{1}}{β_{2}} |^{1/2} ∫_{A_{K}/A_{O}} da\{∫_{N_{K}} f_{1}(bav)dv\}\{∫_{N_{K}} f_{2}(a^{−1}u)du\} = f̃_{1} ∗ f̃_{2}(b).

G_{K} is the disjoint union of the double cosets

G_{O} ( \begin{matrix}π^{m} & 0 \\ 0 & π^{n} \end{matrix} ) G_{O} = G_{O}a(m, n)G_{O} m ≤ n.

The characteristic function of such a double coset will be denoted by f_{m,n}. If a(m′, n′)N_{K} meets G_{O}a(m, n)G_{O} then m + n = m′ + n′ and m ≤ m′; moreover

a(m, n)N_{K} ∩ G_{O}a(m, n)G_{O} = a(m, n)(G_{O} ∩ N_{K}).

Thus f̃_{m,n}(a(m′, n′)) = 0 unless m + n = m′ + n′ and m ≤ m! Moreover

f̃_{m,n}(a(m, n)) = 1.

It follows readily that the map f → f̃ is an isomorphism. Consequently every homomorphism of H into ℂ is of the form

χ_{ω}(f) = ∫_{A_{K}/A_{O}} f̃(a)ω(a)da

where ω is a homomorphism of A_{K}/A_{O} into ℂ^{×}.

If η is a homomorphism of A_{K} into ℂ^{×} let η_{1} and η_{2} be the functions on K^{×} defined by

η_{1}(α) = η(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} )), η_{2}(α) = η(( \begin{matrix}1 & 0 \\ 0 & α \end{matrix} )).

Lemma 6.2. Let η be a homomorphism of A_{K} into ℂ^{×} and ω a homomorphism of A_{K}/A_{O} into ℂ^{×}. There is up to a scalar factor at most one function φ on G_{K} satisfying

(i) φ(ag) = η(a)φ(g) for all a in A_{K},

(ii)

∫_{G_{K}} φ(gh)f(h)dh = χ_{ω}(f)φ(g)

for all f in H.

If there is any non-zero solution of this equation then η_{1}η_{2} = ω_{1}ω_{2} so that η_{1} and η_{2} have the same conductor. Let it be (π^{a}). φ is determined by its restriction to N_{K}. If y ∈ O^{×} then

φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) = φ(( \begin{matrix}1 & x + y \\ 0 & 1 \end{matrix} ))

and if α ∈ O^{×} then

η_{1}(α)φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) = φ(( \begin{matrix}α & αx \\ 0 & 1 \end{matrix} )) = φ(( \begin{matrix}1 & αx \\ 0 & 1 \end{matrix} )).

If x = π^{−b} and b < a there is an α in O^{×} such that α ≡ 1(modπ^{b}) and η_{1}(α) ≠ 1. Then

η_{1}(α)φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) = φ(( \begin{matrix}1 & x + (α − 1)x \\ 0 & 1 \end{matrix} )) = φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))

so that

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

To prove the lemma we need only show that if

φ(( \begin{matrix}1 & π^{−a} \\ 0 & 1 \end{matrix} ))

then

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

for b > a. If O is the disjoint union ∪_{i=1}^{q}x_{i} + (π) then G_{O}a(0, 1)G_{O} is the disjoint union

∪_{i=1}^{q} ( \begin{matrix}π & x_{i} \\ 0 & 1 \end{matrix} ) G_{O} ∪ ( \begin{matrix}1 & 0 \\ 0 & π \end{matrix} ) G_{O}.

Thus if b ≥ a

χ_{ω}(f_{0,1})φ(( \begin{matrix}1 & π^{−b} \\ 0 & 1 \end{matrix} )) = Σ_{i}φ(( \begin{matrix}π & π^{−b} + x_{i} \\ 0 & 1 \end{matrix} )) + φ(( \begin{matrix}1 & π^{−(b−1)} \\ 0 & π \end{matrix} ))

= η_{1}(π)Σ_{i}φ(( \begin{matrix}1 & π^{−(b+1)}(1 + π^{b}x_{i}) \\ 0 & 1 \end{matrix} )) + η_{2}(π)φ(( \begin{matrix}1 & π^{−(b−1)} \\ 0 & 1 \end{matrix} ))

= qη_{1}(π)φ(( \begin{matrix}1 & π^{−(b+1)} \\ 0 & 1 \end{matrix} )) + η_{2}(π)φ(( \begin{matrix}1 & π^{−(b−1)} \\ 0 & 1 \end{matrix} ))

because η_{1}(1 + π^{b}x) = 1.

Lemma 6.3. Let ξ be a non-trivial character of K and ω a homomorphism of A_{K}/A_{O} into ℂ^{×}. Apart from a scalar factor there is exactly one function φ on G_{K} which satisfies

(i)

φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )g) = ξ(x)φ(g)

(ii)

∫_{G} φ(gh) = f(h)dh = χ_{ω}(f)φ(g)

for all f in H.

Suppose φ satisfies these relations. Take an A_{K} and set φ′(g) = φ(ag). The function φ′ satisfies (ii). Moreover

φ′(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = φ(( \begin{matrix}1 & α_{1} × α_{2}^{−1} \\ 0 & 1 \end{matrix} ) ag) = ξ(α_{1} × α_{2}^{−1})φ′(g);

thus if ξ′(x) = ξ(α_{1} × α_{2}^{−1}) it satisfies (i) with ξ replaced by ξ′. Assume then for simplicity that O is the largest ideal on which ξ is trivial.

If φ is to satisfy (i) it must be of the form

φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ak) = ξ(x)Φ(a)

with Φ a function on A_{K}/A_{O}. The function φ is well-defined if and only if ξ(x)Φ(a) = Φ(a) when α_{1}^{−1} × α_{2} is in O. Thus Φ(a) = 0 unless α_{1}α_{2}^{−1} is in O. The relations (ii) will be satisfied if and only if

Φ(( \begin{matrix}α & 0 \\ 0 & α \end{matrix} ) a) = ω_{1}(α)ω_{2}(α)Φ(a)  (A)

and

∫_{G} φ(gh)f_{0,1}(h)dh = χ_{ω}(f_{0,1})φ(g).  (B)

We can satisfy (A) and the previous conditions while specifying in an arbitrary manner the value of Φ at a = ( \begin{matrix}π & 0 \\ 0 & 1 \end{matrix} ), α ≥ 0. (B) will be satisfied for all g if it is satisfied for g = ( \begin{matrix}π^{α} & 0 \\ 0 & 1 \end{matrix} ) when it becomes

Σ_{i}φ(( \begin{matrix}π^{α+1} & π^{α}x_{i} \\ 0 & 1 \end{matrix} )) + φ(( \begin{matrix}π^{α} & 0 \\ 0 & π \end{matrix} )) = q^{1/2}(ω_{1}(π) + ω_{2}(π))φ(( \begin{matrix}π^{α} & 0 \\ 0 & 1 \end{matrix} )).

If α < −1 all terms on both sides are zero. If α = −1 the right side is 0 and the left side is

\{Σ_{i}ξ(\frac{x_{i}}{π})\}φ(( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )) = 0.

If α ≥ 0 the left side is

qφ(( \begin{matrix}π^{α+1} & 0 \\ 0 & 1 \end{matrix} )) + ω_{1}(π)ω_{2}(π)φ(( \begin{matrix}π^{α−1} & 0 \\ 0 & 1 \end{matrix} )).

Some simple algebra then shows that (ii) will be satisfied if and only if

∑_{n=−∞}^{α} x^{n}q^{\frac{n}{2}} Φ(( \begin{matrix}π^{n} & 0 \\ 0 & 1 \end{matrix} )) = \frac{(1 − ω_{1}(n)x)(1 − ω_{2}(n)x)}{Φ}(( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )).  (X)

The lemma follows

If (π^{−δ}) is the largest ideal of K on which ξ is trivial let φ(g, ω, ξ) be that solution of (i) and (ii) which takes the value 1 at ( \begin{matrix}π^{δ} & 0 \\ 0 & 1 \end{matrix} ). If ξ′(x) = ξ(βx) then

φ(g; ω, ξ′) = φ(( \begin{matrix}β & 0 \\ 0 & 1 \end{matrix} ) g, ω, ξ).

Let ζ be a character of A_{K} such that ζ_{1}ζ_{2} = ω_{1}^{−1}ω_{2}^{−1}. Set ζ_{1}(α) = ζ_{0}(α)|α|^{s} where ζ_{0}(π) = 1. ζ is uniquely determined by s and ζ_{0} and we shall sometimes write ζ = ζ(s, ζ_{0}).

Lemma 6.4. Let ζ be a homomorphism of A_{K} into ℂ^{×} such that ζ_{1}ζ_{2} = ω_{1}^{−1}ω_{2}^{−1}. If ζ = ζ(s, ζ_{0}) the function

Φ(g, ζ; ω, ξ) = ∫_{K^{×}} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g; ω, ξ)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))d^{×}α

is defined for Re s sufficiently large. If ζ_{0} = 1 then

Φ′(g, ζ; ω, ξ) = (1 − \frac{ω_{1}(π)ζ_{1}(π)}{q^{1/2}})(1 − \frac{ω_{2}(π)ζ_{1}(π)}{q^{1/2}})Φ(g, ζ; ω, ξ)

is, for each g, a polynomial in q^{s} and q^{−s} and if (π^{−δ}) is the largest ideal on which ξ is trivial then

ζ_{1}(π^{δ})Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ; ω, ξ) = ζ̃_{1}(π^{+δ})Φ′(g, ζ̃, ω, ξ)

where ζ̃(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )) = ζ(( \begin{matrix}α_{2} & 0 \\ 0 & α_{1} \end{matrix} )). If δ = 0 then

Φ′(1, ζ; ω, ξ) = 1.

If the conductor of ζ_{0} is (π^{γ}), γ > 0, and

g(ξ, γ) = ∫_{O^{×}} ξ(\frac{α}{π^{γ+δ}})ζ_{1}(α)d^{×}α

then Φ(g, ζ; ω, ξ) is a polynomial in q^{s} and q^{−s} and

\frac{ζ_{1}(π^{γ+δ})}{g(ξ, ζ)}Φ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ; ω, ξ) = ζ_{1}(−1)\frac{ζ̃_{1}(π^{γ+δ})}{g(ξ, γ̃)}Φ(g, ζ̃, ω, ξ).

If ξ′(x) = ξ(π^{β}x) then
Φ(g, ζ; ω, ξ′) = ∫_{K^{×}} φ(( \begin{matrix}π^{β}α & 0 \\ 0 & 1 \end{matrix} ) g; ω, ξ)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))d^{×}α  (Y)
= ζ_{1}^{−1}(π^{β})Φ(g, ζ; ω, ξ)

and g(ξ′, ζ) = g(ξ, ζ); so it is enough to prove the lemma for δ = 0.

If g = a( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )k with a = ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) in A_{K} and k in G_{O} then

∫_{K^{×}} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g; ω, ξ)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))d^{×}α

is equal to

ζ^{−1}(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )) ∫_{K^{×}} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ); ω, ξ)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))d^{×}α.

Because φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} )( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ); ω, ξ) = ξ(αx)φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ); ω, ξ) the function

φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ); ω, ξ) − φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ); ω, ξ)

has, for a given x, compact support on K^{×}. Since the integral

ζ^{−1}(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )) ∫_{K^{×}} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ); ω, ξ)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))d^{×}α

exists for Re s sufficiently large so does that of the lemma. Moreover the difference between Φ(g, ζ; ω, ξ) and this expression is a polynomial in q^{s} and q^{−s}. If ζ_{0} = 1 the expression equals

ζ^{−1}(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )) ∑_{n=−∞}^{∞} ζ_{1}(π^{n})φ(( \begin{matrix}π^{n} & 0 \\ 0 & 1 \end{matrix} )) = \frac{ζ^{−1}(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ))}{(1 − \frac{ω_{1}(π)ζ_{1}(π)}{q^{1/2}})(1 − \frac{ω_{2}(π)ζ_{a}(π)}{q^{1/2}})}

and if the conductor is (π^{γ}) and γ > 0 it equals zero. All assertions of the lemma except the functional equations follow.

Let η = ζ̃^{−1}. If ζ_{0} = 1 then Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ; ω, ξ) and Φ′(g, ζ̃; ω, ξ) both satisfy the assumptions of Lemma 6.2. Since they both take the value 1 at g = 1 they are equal. If the conductor of ζ_{0} is (π^{γ}) and γ > 0 then

Φ(( \begin{matrix}1 & π^{−γ} \\ 0 & 1 \end{matrix} ) , ζ̃; ω, ξ) = ∫_{K^{×}} ξ(απ^{−γ})φ(( \begin{matrix}α & 1 \\ 0 & 1 \end{matrix} ) , ω, ξ)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))d^{×}α.

The last integral is easily seen to equal g(ξ, ζ̃). Since

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

the value of Φ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ), ζ; ω, ξ) is ζ_{1}^{−1}(π^{γ})ζ̃_{1}^{−1}(π^{−γ})ζ_{1}(−1)g(ξ, ζ). The functional equation again follows from Lemma 6.2.

7. The Main Theorem. Let k be either the rational number field or the field of rational functions in one variable over a finite field and let K be a finite separable extension of k. Let S_{∞} be the set of archimedean primes of K. Let 𝔸^{0} be the adèle ring of K and let I be the group of idèles.

If R is any commutative ring with unit let G_{R} be the group of 2 × 2 matrices from R which have a determinant which is a unit of R. A_{R} will be the group of diagonal matrices in G_{R}. If 𝔭 is a non-archimedean prime let U_{K_{𝔭}} be G_{O_{𝔭}}, where O_{𝔭} is the ring of integers in K_{𝔭}, and if 𝔭 is an archimedean prime let U_{K_{𝔭}} be the group of unitary matrices which lie in G_{K_{𝔭}}.

Lemma 7.1.^{∗} There is a constant c_{0} such that if g belongs to G_{𝔸} there is a γ in G_{ℚ} such that max{|c|, |d|} ≤

c_{0}|det g|^{1/2} if γ_{g} = ( \begin{matrix}a & b \\ c & d \end{matrix} ).

Fix a measure on 𝔸. This determines a measure on 𝔸 ⊕ 𝔸. K ⊕ K is a discrete subgroup of 𝔸 ⊕ 𝔸 and the quotient 𝔸 ⊕ 𝔸/K ⊕ K has finite measure c_{1}. The lattice Lg = (K ⊕ K)g, is discrete and the quotient 𝔸 ⊕ 𝔸/Lg has measure c_{1}|det g|. The non-zero elements of Lg are, for all practical purposes, the last rows of the matrices γg, γ ∈ G_{ℚ}. There is a positive constant c_{2} such that the measure of {(x, y)| max{|x|, |y|} ≤ d_{0}} is at least c_{2}d_{0}^{2}.

Let c_{0} be any number larger than 2\sqrt{\frac{c_{1}}{c_{2}}} . If Lg contained no non-zero (c, d) with max\{|c|, |d|\} ≤ c_{0}|det g|^{1/2} the measure of the projection of \{(x, y)| max\{|x|, |y|\} ≤ \frac{c_{0}}{2} |det g|^{1/2}\} on 𝔸 ⊕ 𝔸/Lg would be greater than c_{1}|det g|.

𝔏 will be the space of functions φ on G_{K}\G_{𝔸} satisfying conditions (i), (ii), and (iii) below

(i) If U = ∏_{𝔭} U_{K_{𝔭}} then φ is U-finite on the right.

(ii) If 𝔭 is an archimedean prime the function φ(hg), g ∈ G_{K_{𝔭}}, is infinitely differentiable.

If 𝔭 is any such prime let 𝔄_{𝔭} be the universal enveloping of G_{K_{𝔭}}. If, for each 𝔭, X_{𝔭} belongs to 𝔄_{𝔭} the function {∏_{𝔭} ρ(X_{𝔭})}φ is defined.

(iii) If c_{1} is any constant there are constants M_{1} and M_{2} such that^{†}

|\{π_{𝔭}ρ(X_{𝔭})\}φ(g)| ≤ M_{1}[\{|det g|\frac{1}{|det g|}\}\{\frac{|det g|^{1/2}}{max\{|c|, |d|\}}\}]^{M_{2}}

on the set max{|c|, |d|} ≤ c_{1}|det g|^{1/2}.

If 𝔭 is a non-archimedean prime the group G_{K_{𝔭}} operates on 𝔏. If 𝔭 is a complex prime 𝔄_{𝔭} acts on 𝔏. If 𝔭 is a real prime let σ_{𝔭} be the element ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) of G_{K_{𝔭}}; the pair \{σ_{𝔭}, 𝔄_{𝔭}\} acts on 𝔏.

∗ (1998) As observed in the comments this lemma is not what is needed. Indeed, neither it nor its proof make much sense. The correct lemma, which there is at this stage no need to state, would replace max {|c|, |d|} by ∏_{𝔭} max{|c|_{𝔭}, |d|_{𝔭}}. See Lemma 5.1 of the following letter.

† See previous footnote.

If 𝔭 is a non-archimedean prime a representation of G_{K_{𝔭}} on a vector space H_{𝔭} will be called quasi-simple if the isotropy group of every vector in H_{𝔭} is an open subgroup of G_{K_{𝔭}}. It follows from Lemma 6.1 that the space of vectors whose isotropy group contains U_{K_{𝔭}} has dimension at most 1 if the presentation is irreducible.

Suppose that for every prime 𝔭 we are given a quasi-simple irreducible representation of either G_{K_{𝔭}}, 𝔄_{𝔭}, or {σ_{𝔭}, 𝔄_{𝔭}}, according to the nature of the prime, on a vector space H_{𝔭}. Suppose there is a finite set S_{0} of primes which contains S_{∞} such that if 𝔭 is not in S_{0} there is a non-zero vector in H_{𝔭} which is fixed by U_{𝔭}. For each 𝔭 not in S_{0} choose such a vector x_{𝔭}^{0}. If S contains S_{0} let H_{S} = ⊗_{𝔭∈S}H_{𝔭}. If S_{2} ⊇ S_{1} ⊇ S_{0} let δ_{S_{1},S_{2}} be the injection of H_{S_{1}} into H_{S_{2}} which sends ⊗_{𝔭∈S_{1}}X_{𝔭} to

(⊗_{𝔭∈S_{1}}X_{𝔭}) · (⊗_{𝔭∈S_{2}−S_{1}}X_{𝔭}^{0})

and let H be the injective limit of the spaces H_{S}. Let 𝔄 be the system consisting of all the G_{K_{𝔭}}, 𝔭 not in S_{∞}, 𝔄_{𝔭}, 𝔭 complex, and {σ_{𝔭}, 𝔄_{𝔭}}, 𝔭 real. The system 𝔄 acts on H.

For our purposes a divisor D is just a function 𝔭 → m_{𝔭} from the non-archimedean primes to the nonnegative integers such that m_{𝔭} = 0 for almost all 𝔭. 𝔭 | D means that m_{𝔭} > 0 and 𝔭 ∤ D means that m_{𝔭} = 0 or 𝔭 ∈ S_{O}. If 𝔭 is not in S_{∞} let U_{K_{𝔭}}^{D} be the set of ( \begin{matrix}a & b \\ c & d \end{matrix} ) in U_{K_{𝔭}} for which c ≡ 0(mod𝔭^{m_{𝔭}}) and let

U^{D} = ∏_{𝔭∉S_{∞}} U_{K_{𝔭}}^{D}. Let Û_{K_{𝔭}}^{D} be the set of ( \begin{matrix}a & b \\ c & d \end{matrix} ) in U_{K_{𝔭}}^{D} for which a ≡ d ≡ 1(mod𝔭^{m_{𝔭}}) and let Û^{D} = Π_{𝔭∉S_{∞}} Û_{K_{𝔭}}^{D}.

U^{D} is in the normalizer of Û^{D}.

Lemma 7.2. There is a D such that H^{D} = {x | π(u)x = x for all u in Û^{D}} contains a non-zero vector. Moreover H^{D} is the sum of one-dimensional subspaces invariant under U^{D}.

Although we have not troubled to be explicit it is clear how Π_{𝔭∉S_{∞}} G_{K_{𝔭}} operates on H.

Given any divisor D let Ũ_{K_{𝔭}}^{D} be the set of ( \begin{matrix}a & b \\ c & d \end{matrix} ) in U_{K_{𝔭}} which are congruent to I modulo 𝔭^{m_{𝔭}} and let Ũ^{D} = Π_{𝔭∉S_{∞}} Ũ_{K_{𝔭}}^{D}. Given x ≠ 0 in H there is a D′ such that Ũ^{D′} is contained in the isotropy group of

x. Choose for each non-archimedean prime an α_{𝔭} so that (α_{𝔭}) = 𝔭^{m′_{𝔭}} and set g = Π_{𝔭∉S_{∞}}( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ). Then, if m_{𝔭} = 2m′_{𝔭} and D is the divisor {m_{𝔭}}, gÛ^{D}g^{−1} is contained in Ũ^{D′} and Û^{D} is contained in the isotropy group of π(g^{−1})x. The second assertion of the lemma is immediate because U^{D}/Û^{D} is a finite abelian group.

If ϵ is any homomorphism of U^{D} into ℂ^{×} which sends Û^{D} to 1 let H_{ϵ}^{D} = {x | π(u)x = ϵ(u)x for all u in U^{D}}. ϵ is determined by its restriction to the diagonal matrices. Let ϵ̃ be the homomorphism satisfying

ϵ̃(( \begin{matrix}a & 0 \\ 0 & d \end{matrix} )) = ϵ(( \begin{matrix}d & 0 \\ 0 & a \end{matrix} )).

If g is any matrix in G_{𝔸} such that g_{𝔭} = I, if 𝔭 ∈ S_{∞} or 𝔭 ∤ D and g_{𝔭} = ( \begin{matrix}0 & 1 \\ α_{𝔭} & 0 \end{matrix} ) with (α_{𝔭}) = 𝔭^{m_{𝔭}} if 𝔭 | D then

gU^{D}g^{−1} = U^{D} and π(g)H_{ϵ}^{D} = HD ; ϵ̃.

Let ℌ be a subspace of 𝔏 such that the representation of 𝔄 on ℌ is equivalent to that on H. We want to study some of the Dirichlet series associated to ℌ. Let ℌ_{ϵ}^{D} be the subspace of ℌ corresponding to H_{ϵ}^{D}. We suppose that H_{ϵ}^{D} is not {0}.

Choose a non-trivial character ξ of 𝔸 which is trivial on K. If φ belongs to 𝔏 set

φ_{0}(g) = \frac{1}{measure (K\𝔸)} ∫_{K\𝔸} φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g)dv,
φ_{1}(g) = \frac{1}{measure (K\𝔸)} ∫_{K\𝔸} φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g)\overline{ξ}(x)dx.

By the Fourier inversion formula

φ(g) = φ_{0}(g) + ∑_{α∈K^{×}} φ_{1}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g).

Let G_{𝔸}^{D} be the set of all g in g_{𝔸} such that g_{𝔭} ∈ U_{K_{𝔭}}^{D} if 𝔭 | D. Since G_{𝔸} = G_{K}G_{𝔸}^{D} any function in 𝔏 is determined by its restriction to G_{𝔸}^{D}.

If 𝔭 is a non-archimedean prime which does not divide D and φ belongs to ℌ_{ϵ}^{D} then φ must be an eigenfunction of the corresponding Hecke operators. Let it be an eigenfunction corresponding to the homomorphism ω_{𝔭}. Varying φ in ℌ_{ϵ}^{D} does not change ω_{𝔭}. It follows from Lemmas 3.2, 5.2, and 6.3 that ℌ_{ϵ}^{D} is spanned by functions φ for which

φ_{1}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} )g)=a_{α}\{ Π_{𝔭∈S∞} φ_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} )g_{𝔭})\}\{ Π ; {𝔭∉S_{∞} \atop 𝔭∤D} φ(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} )g_{𝔭},ω_{𝔭},ξ_{𝔭})\}ϵ(g_{D})

for g in G_{𝔸}^{D}. a_{α} is a constant which depends on α and α_{𝔭} is the image of α in K_{𝔭}. g_{D} is the projection of G on Π_{𝔭|D}G_{K_{𝔭}}. ξ_{𝔭} is the restriction of ξ to K_{𝔭} and φ_{𝔭}, 𝔭 ∈ S_{∞}, is a function in L(ξ_{𝔭}, π_{𝔭}) determined solely by

φ. Let I^{D} = \{ι ∈ I | |ι_{𝔭}| = 1 if 𝔭 | D\}. If β lies in K^{×} ∩ I^{D} then a_{αβ} = ϵ(( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ))a_{α}.

We shall only consider those φ for which the functions φ_{1}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} )g) are of the above form. φ(g) is the sum of φ_{0}(g) and

∑_{α∈K×/K×∩ID} a_{α} ∑_{β∈K×∩ID}\{ ∏_{𝔭∈S∞} φ_{𝔭}(( \begin{matrix}α_{𝔭}β_{𝔭} & 0 \\ 0 & 1 \end{matrix} )g_{𝔭})\}\{ ∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} φ(( \begin{matrix}α_{𝔭}β_{𝔭} & 0 \\ 0 & 1 \end{matrix} )g_{𝔭},ω_{𝔭},ξ_{𝔭})\}ϵ(( \begin{matrix}β_{0} & 0 \\ 0 & 1 \end{matrix} )g_{D})

if β_{D} is the projection of β on ∏_{𝔭|D} K_{𝔭}^{×}. In an appendix to this paragraph we shall discuss the form of the function ψ_{0}. Lemma E of the appendix will eventually be used to show that φ is the sum of a cusp form and a function which is represented by an Eisenstein series. For the present we consider only the case that φ_{0}(g) ≡ 0.

Then φ(g) is a cusp form. Let η be the homomorphism of K^{×}\I into ℂ^{×} defined by

φ(( \begin{matrix}α & 0 \\ 0 & α \end{matrix} ) g) = η(α)φ(g).

It is no real restriction to assume that |η(α)| = 1 and we shall do so. It then follows from the general theory of automorphic forms that φ is bounded.

If M_{1} = sup_{g∈G_{𝔸}} |φ(g)| and

M_{2} = sup_{g∈G_{𝔸}^{D}} | ∏_{𝔭∈S_{∞}} φ_{𝔭}(g_{𝔭})|| ∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} φ(g_{𝔭}, ω_{𝔭}, ξ_{𝔭})||ϵ(g_{D})|

then

|a_{α}| ≤ \frac{M_{1}}{M_{2}} .

If φ ≢ 0, as we certainly suppose, M_{2} is not zero. Of course it is not ∞ either for then all the a_{α} would be zero. In any case a_{α} is a bounded function.

If the number M_{2} is finite the function φ(g_{𝔭}, ω_{𝔭}, ξ_{𝔭}) is bounded. Appealing to the formula^{∗} at the top of p. 6.10 we see that the inequalities

|ω_{𝔭,1}(π)| ≤ |π|^{−1/2}  |ω_{𝔭,2}(π)| ≤ |π|^{−1/2}  (A)

must be satisfied.

If K is the real or complex field, π a quasi-simple irreducible representation of {σ, 𝔄} or 𝔄 respectively, and ζ a homomorphism of A_{K} into ℂ^{×} satisfying the condition of Lemma 3.6 or 5.7 let Γ(ζ, π) be the function defined by

Γ(ζ, π)Φ′(g, ζ, φ) = Φ(g, ζ, φ), φ ∈ L(ξ, π).

Φ(g, ζ, φ) and Φ′(g, ζ, φ) are the functions introduced in Lemmas 3.6 and 5.7.^{∗} Γ(ζ, π) also depends on ξ but we do not take this into account explicitly.

Let χ be a character of K^{×} ∩ I^{D}\I^{D}. If s is a complex number define ζ = ζ(s, χ) by

ζ(( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )) = η^{−1}(β)|αβ^{−1}|^{s}χ(αβ^{−1}).

Let ζ_{𝔭} be the restriction of ζ to A_{K_{𝔭}}.

Lemma 7.3. The integral

∫_{K^{×}∩I^{D}\I^{D}} φ(( \begin{matrix}α & 1 \\ 0 & 1 \end{matrix} ) g)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα

converges absolutely for Re s sufficiently large and G in G_{𝔸}^{D}. It is equal to zero if

ζ(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))ϵ(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))

∗ (1998) labeled X in this version in which the pagination differs from that of the manuscript.

∗ In the digressions to establish notation we allow ourselves to use, in a new sense, symbols whose meaning has otherwise been fixed for the course of this paragraph. is not identically 1 in 0_{𝔭}^{×}. There is a constant M > 0 such that a_{α} = 0 if |α_{𝔭}| > M for some 𝔭 | D. Consequently the series

∑_{α∈K^{×}/K^{×}∩I^{D}} a_{α} ∏_{𝔭|D} ζ_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))

converges absolutely for Re s sufficiently large. Let R be the set of non-archimedean primes which do not divide D for which ζ_{𝔭} is not trivial on A_{O_{𝔭}}. The product

∏_{{𝔭∉S_{∞}∪R \atop 𝔭∤D}} \frac{1}{(1 − ω_{𝔭,1}(π)ζ_{𝔭,1}(π)|π|^{1/2})(1 − ω_{𝔭,2}(π)ζ_{𝔭,2}(π)|π|^{1/2})}

also converges absolutely for Re s sufficiently large. The integral is the product of the above two expressions with

\{ ∏_{𝔭∈S∞} Γ(ζ_{𝔭},π_{𝔭})Φ′(g_{𝔭},ζ_{𝔭},φ_{𝔭})\}\{ ∏_{{𝔭∉S_{∞}∪R \atop 𝔭∤D}} Φ′(g_{𝔭},ζ_{𝔭},ω_{𝔭},ξ_{𝔭})\}\{ ∏_{𝔭∈R} Φ(g_{𝔭},ζ_{𝔭},ω_{𝔭},ξ_{𝔭})\}ϵ(g_{D}).

According to Lemma 6.4 only a finite number of terms in the last product are different from 1. The absolute convergence of the other infinite product follows immediately from the inequalities (A). For each 𝔭 the character ξ_{𝔭} is non-trivial. If 𝔭 | D, x ∈ O_{𝔭}, α ∈ K_{𝔭}^{×}, and g ∈ G_{𝔸}^{D}

φ_{1}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) = φ_{1}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )).

If

g_{𝔭} = ( \begin{matrix}a_{𝔭} & b_{𝔭} \\ c_{𝔭} & d_{𝔭} \end{matrix} )

this equals

φ_{1}(( \begin{matrix}1 & \frac{α_{𝔭}a_{𝔭}}{d_{𝔭}} \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) = ξ_{𝔭}(\frac{α_{𝔭}a_{𝔭}}{d_{𝔭}} x)φ_{1}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g).

Thus if a_{α} is not zero, α must lie in the largest ideal of K_{𝔭} on which ξ_{𝔭} is trivial. The existence of the constant M follows immediately.

Recalling that, for almost all 𝔭, φ(g_{𝔭}, ω_{𝔭}, ξ_{𝔭}) equals 1 if g_{𝔭} lies in U_{K_{𝔭}} we see that

∫_{K^{×}∩I^{D}\I^{D}} |φ(( \begin{matrix}γ & 0 \\ 0 & 1 \end{matrix} ) g)ζ(( \begin{matrix}γ & 0 \\ 0 & 1 \end{matrix} ))|dγ

is at most the sum over K^{×}/K^{×} ∩ I^{D} of the product of

|a_{α}| ∏_{𝔭∈S_{∞}} ∫_{K_{𝔭}^{×}} |φ_{𝔭}(( \begin{matrix}α_{𝔭}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭})ζ_{𝔭}(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))dγ_{𝔭}

and

∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} ∫_{K_{𝔭}^{×}} |φ(( \begin{matrix}α_{𝔭}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭}, ω_{𝔭}, ξ_{𝔭})|_{𝔭}(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))|dγ_{𝔭}.

Changing variables in the integral and recalling the product formula we see that the sum is the product of

∑_{K^{×}/K^{×}∩I^{D}} |a_{α}| ∏_{𝔭|D} |ζ_{𝔭}(( \begin{matrix}α_{1} & 0 \\ 0 & 1 \end{matrix} ))|

and

∏_{𝔭∈S_{∞}} ∫_{K_{𝔭}^{×}} |φ_{𝔭}(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭})ζ_{𝔭}(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))|dγ_{𝔭}

and

∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} ∫_{K_{𝔭}^{×}} |φ(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭}, ω_{𝔭}, ξ_{𝔭})ζ_{𝔭}(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))|dγ_{𝔭}.

The first term is certainly finite for Re s sufficiently large. The convergence of the integrals over K_{𝔭}^{×}, 𝔭 ∈ S_{∞}, was proved in Lemma 3.6 and 5.7. It remains to show that if Re s is sufficiently large each of the integrals in the infinite product is finite and the product converges. It was proved in Lemma 6.4 that for a given 𝔭 the integral is finite if Re s is sufficiently large. Thus we can, in our considerations, drop any finite set of terms from the product.

The first formula^{∗} on the top of p. 6.10 shows that if g_{𝔭} is a unit and O_{𝔭} is the largest ideal on which ξ_{𝔭} is trivial then

∫_{K_{𝔭}^{×}} |φ(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭}, ω_{𝔭}, ξ_{𝔭})ζ_{𝔭}(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))|dγ_{𝔭}

is at most

\frac{1}{(1 − |π|^{s})(1 − |π|^{s})}

if Re s > 0. The infinite product converges if Re s > 1.

Thus the integral is finite. A simple formal manipulation which is now justified shows that it is the product of

∑_{α∈K^{×}/K^{×}∩I^{D}} a_{α} ∏_{𝔭|D} ζ_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))

and

∏_{𝔭|D} ∫_{O_{𝔭}^{×}} ϵ(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭})ζ_{𝔭}(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))dγ_{𝔭}

and

∏_{𝔭∈S_{∞}} ∫_{K_{𝔭}^{×}} φ_{𝔭}(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭})ζ_{𝔭}(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))dγ_{𝔭}

and

∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} ∫_{K_{𝔭}^{×}} φ(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭}, ω_{𝔭}, ξ_{𝔭})ζ_{𝔭}(( \begin{matrix}γ_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))dγ_{𝔭}.

The remaining statements of the lemma are now just a matter of definition.

∗ (1998) Labeled (X) for convenience.

We shall be able to state the next lemma more succinctly if we first introduce some notation. First let K be the real or complex field and let π be a quasi simple irreducible representation of {σ, 𝔄} or 𝔄 respectively. Let ξ be a character of K and ζ a continuous homomorphism of A_{K} into ℂ^{×} which satisfies the condition of Lemma 3.6 or 5.1. Define ϵ(ζ, ξ, π) by the relation

Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ̃, φ) = ϵ(ζ, ξ, π)Φ′(g, ζ, φ).

The exact form of the factor is given in Lemma 3.6 and 5.7. If K is a non-archimedean field, ω a homomorphism of A_{K}/A_{D} into ℂ^{×}, ζ a continuous homomorphism of A_{K} into ℂ^{×} which satisfies the condition of Lemma 6.4, and ξ a character of K define ϵ(ζ, ξ, π) by the relation

Φ′(( \begin{matrix}0 & 0 \\ −1 & 0 \end{matrix} ) g, ζ̃, ω, ξ) = ϵ(ζ, ξ, ω)Φ′(g, ζ, ω, ξ)

if ζ is trivial on A_{O} and by the relation

Φ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ̂, ω, ξ) = ϵ(ζ, ξ, ω)Φ(g, ζ, ω, ξ)

if it is not. The form of this factor is given in Lemma 6.4.

Choose A in K^{×} so that (A_{𝔭}) = 𝔭^{m_{𝔭}} if 𝔭 | D and set

φ̂(g) = φ(g ∏_{𝔭|D} ( \begin{matrix}0 & 1 \\ A_{𝔭} & 0 \end{matrix} )).

φ̂ lies in ℌD ; ϵ̃. If ζ is the homomorphism of A_{𝔸} into ℂ^{×} introduced in Lemma 7.3, let Ξ(x, χ) be the product of

\{ ∑_{α∈K^{×}/K^{×}∩I^{D}} a_{α} ∏_{𝔭|D} ζ_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))\} ∏_{𝔭∈S_{∞}} Γ(ζ_{𝔭}, π_{𝔭})

and

∏_{{𝔭∉S_{∞}∪R \atop 𝔭∤D}} \frac{1}{(1 − α_{𝔭,1}(π)ζ_{𝔭,1}(π)|π|^{1/2})(1 − w_{𝔭,2}(π)|π|^{1/2}} .

Given φ the functions φ_{𝔭} are determined only up to a scalar factor. Thus there is an undetermined constant in the numbers a_{α} and hence in the function Ξ(s, χ). However we can certainly suppose that, for 𝔭 archimedean, φ̂_{𝔭}, the function associated to φ̂, is the same as φ_{𝔭}. This assumption is implicit in the statement and proof of the following lemma.

Lemma 7.4. Ξ(s, χ) is an entire function of s. It satisfies the functional equation^{∗}

Ξ(s, χ) = \{∏_{𝔭|D} ζ_{𝔭}(( \begin{matrix}−A_{𝔭} & 0 \\ 0 & 1 \end{matrix} )) ∏_{𝔭∈S_{∞}} ϵ(ζ_{𝔭}, ξ_{𝔭}, π_{𝔭}) ∏_{𝔭∤D} ϵ(ζ_{𝔭}, ξ_{𝔭}, ω_{𝔭})\}Ξ̂(−s, (χη)^{−1}).

The integral in Lemma 7.3 is the sum of

∫_{|α|≥1} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα  (B)

and

∫_{|α|≤1} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα.

The latter integral is equal to

∫_{|α|≥1} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) g)η^{−1}(α)ζ^{−1}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα

or

∫_{|α|≥1} φ̂(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) g ∏_{𝔭|D} ( \begin{matrix}0 & A_{𝔭}^{−1} \\ 1 & 0 \end{matrix} ))ζ̃(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα.  (C)

If the first integral converges for Re s sufficiently large, as it does, it must converge for all s. The resulting function of s is entire. Since the substitution of −s for s, (χη)^{−1} for χ, φ̂ for φ, and ( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} )g ∏_{m_{𝔭}>0}( \begin{matrix}0 & A_{𝔭}^{−1} \\ 1 & 0 \end{matrix} ) for g interchanges the integrals (B) and (C), the latter integral is also an entire function.

We conclude that the product of Φ(s, χ) and

\{ ∏_{𝔭∈S_{∞}} Φ′(g_{𝔭}, ζ_{𝔭}, φ_{𝔭})\}\{ ∏_{{𝔭∉S_{∞}∪R \atop 𝔭∤D}} Φ′(g_{𝔭}, ζ_{𝔭}, ω_{𝔭}, ξ_{𝔭})\}\{ ∏_{𝔭∈R} Φ(g_{𝔭}, ζ_{𝔭}, ω_{𝔭}, ξ_{𝔭})\}ϵ(g_{D})\}  (D)

is an entire function of s.

It is clear that if 𝔭 is an archimedean prime the function Φ(s, χ) is not changed if φ is replaced by a non-zero linear combination of functions obtained from φ by operations of {σ_{𝔭}, 𝔄_{𝔭}} or 𝔄_{𝔭}, according to the nature of the prime. Thus to prove the lemma we can choose the functions φ_{𝔭} in any way convenient. I claim that these functions and g in G_{𝔸}^{D} can be so chosen that almost all of the factors in (D) are 1 and the rest are of the form ae^{bs} with a ≠ 0. It will follow that Ξ(s, χ) is entire.

g_{D} may as well be taken to be I. If we take g_{𝔭} = I for 𝔭 ∉ R ∪ S_{∞}, g ∤ D then according to Lemma 6.4 and the formula^{∗} at the top of p. 6.12 each of the functions Φ′(g_{𝔭}, ζ_{𝔭}ω_{𝔭}, ξ_{𝔭}) is of this form and all but a finitely number are identically a. If 𝔭 ∈ R then, according to the formulae at the top of p. 6.12 and the bottom of p. 6.13

Φ(( \begin{matrix}1 & π^{−γ_{𝔭}} \\ 0 & 1 \end{matrix} ) , ζ_{𝔭}, ω_{𝔭}, ξ_{𝔭})

will be of this form of a suitable choice of γ_{𝔭}. For a real prime choose g_{𝔭} = I and φ_{𝔭} so that the formulae on p. 3.37^{†} can be applied. For a complex prime choose g_{𝔭} = I and φ_{𝔭} as on pp. 5.28 and 5.19.^{††}

∗ Ξ̂(−s, (χη)^{−1}) is the function obtained on replacing φ by φ̂.

∗ (1998) Labeled (Y).

† (1998) Now p. 33. †† (1998) At the very end of the chapter, just before the appendix.

Now let us see what happens to the expression (D) when the substitution mentioned above is performed. The substitution replaces ζ by ζ̃ and ϵ by ϵ̃. The factor ϵ(g_{0}) is not changed. The functions occurring in the other factors are not changed but some of the variables are g_{𝔭} is replaced by ( \begin{matrix}0 & 1 \\ A_{𝔭} & 0 \end{matrix} )g_{𝔭} and ζ_{𝔭} is replaced by ζ̃_{𝔭}. Thus the expression (D) is multiplied by

\{∏_{𝔭|D} ζ̃_{𝔭}^{−1}(( \begin{matrix}1 & 0 \\ 0 & −A_{𝔭} \end{matrix} ))\}\{ ∏_{𝔭∈S_{∞}} ϵ(ζ_{𝔭}, ξ_{𝔭}, π_{𝔭})\}\{ ∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} ϵ(ζ_{𝔭}, ξ_{𝔭}, ω_{𝔭})\}

which equals

\{∏_{𝔭|D} ζ_{𝔭}(( \begin{matrix}−A_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))\}\{ ∏_{𝔭∈S_{∞}} ϵ(ζ_{𝔭}, ξ_{𝔭}, π_{𝔭})\}\{ ∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} ϵ(ζ_{𝔭}, ξ_{𝔭}, ω_{𝔭})\}.

The lemma follows

We want to prove a converse to this lemma. Suppose we are given the divisor D and hence U^{D}, a homomorphism ϵ of U^{D}/Û^{D} into ℂ^{×}, and a non-trivial character ξ of 𝔸|K. Suppose that we are given bounded functions a_{α} and â_{α} on K^{×} such that

a_{αβ} = ϵ(( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ))a_{α}  â_{αβ} = ϵ̃(( \begin{matrix}β & 0 \\ 0 & 1 \end{matrix} ))â_{α}

if β lies in K^{×} ∩ I^{D}. Suppose moreover that a_{α} = 0 if, for some 𝔭 dividing D, α_{𝔭} does not lie in the largest ideal of K_{𝔭} on which ξ_{𝔭} is trivial. We will also have to be given, for each archimedean prime 𝔭, an irreducible quasi-simple representations of {σ_{𝔭}, 𝔄_{𝔭}} or 𝔄_{𝔭} according to the nature of the prime and, for each non-archimedean prime which does not divide D, a character ω_{𝔭} of A_{K𝔭}/A_{O𝔭} which satisfies

|ω_{𝔭,1}(π)| ≤ |π|^{−1/2} |ω_{𝔭,2}(π)| ≤ |π|^{1/2}.

If 𝔭 is archimedean let π_{𝔭} be deducible from π_{ω_{𝔭}}. We shall also suppose that the homomorphism

η(α) = ∏_{𝔭∈S_{∞}} ω_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & α_{𝔭} \end{matrix} )) ∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} ω_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & α_{𝔭} \end{matrix} )) ∏_{𝔭|D} ϵ(( \begin{matrix}α_{𝔭} & 0 \\ 0 & α_{𝔭} \end{matrix} ))

of I^{D} into ℂ^{×} is trivial on K^{×} ∩ I^{D}.

Lemma 7.5. Choose for each archimedean prime a function φ_{𝔭} in L(ξ_{𝔭}, π_{𝔭}). If g ∈ G_{𝔸}^{D} the series

∑_{α∈K^{×}} a_{α}\{ ∏_{𝔭∈S_{∞}} φ_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭})\}\{ ∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} φ ( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭}, ω_{𝔭}, ξ_{𝔭})\}ϵ(g_{D})

converges absolutely. Moreover the convergence is uniform on compact subsets of G_{𝔸}^{D}. Let φ(g) be its sum. If x belongs to K and x_{𝔭} lies in O_{𝔭} for 𝔭 | D then φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )g) = φ(g) and, if α, β lie in K^{×} ∩ I^{D}, φ(( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )g) = φ(g).

Choose a compact subset C of G_{𝔸}^{D}. According to the discussion^{∗} on p. 6.8 there is for each nonarchimedean prime 𝔭 which does not divide D a number M_{𝔭} such that if g ∈ O

φ(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭}, ω_{𝔭}, ξ_{𝔭}) = 0

if |α_{𝔭}| > M_{𝔭}. Moreover almost all of the numbers M_{𝔭} can be taken to be 1. Because of the assumption on the function {a_{α}} the sum in the lemma can be replaced by a sum over a finite set if K is a function field and by a sum over a lattice in K if K is a number field. If K is a function field the first two assertions of the lemma are immediate. Suppose K is a number field.

Combining the formula^{†} at the top of p. 6.10 with our assumptions on the magnitude of the numbers ω_{𝔭,1}(π) and ω_{𝔭,2}(π) we see that there is a positive constant b and for each non-archimedean prime 𝔭 which do not divide D a constant C_{γ} such that

|φ(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭}, ω_{𝔭}, ξ_{𝔭})| ≤ C_{𝔭}|α_{𝔭}|^{−b}

if g is in C. For all but a finite number of primes C_{𝔭} can be taken to be 1.

Because of the product formula we are reduced to considering the sum

∑ ∏_{𝔭∈S_{∞}} |α_{𝔭}|^{b}|φ′_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭})|

over the non-zero points of a lattice in K. On pages^{††} 3.9 and 5.9 we have discussed the behaviour of the functions ψ(t) and ψ^{\frac{n}{2}} (g) as t → ∞. The first of the equations (A) on p. 5.8 can be used to determine the asymptotic behaviour of all the functions ψ^{k}(t). In Lemma 3.4 and 5.4 we have discussed the behaviour of these functions as |t| → 0. Putting all the information together we see that these are positive constants c and d and a constant Q such that

|φ_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭})| ≤ Q|α_{𝔭}|^{−c}e^{−d|α_{𝔭}|}

if g is archimedean and g lies in C. The absolute and uniform convergence of the sum follows.

The last two statements of the lemma can be proved for both types of field simultaneously. If x ∈ K and x_{𝔭} ∈ O_{𝔭} for 𝔭 | D and a_{α} ≠ 0 then, by assumption, ξ_{𝔭}(α_{𝔭}x_{𝔭}) = 1, if 𝔭 | D. Thus

∏_{𝔭∤D} ξ_{𝔭}(α_{𝔭}x_{𝔭}) = 1.

∗ (1998) Now following Lemma 6.3.

† (1998) See previous footnotes. †† Between Lemma 3.2 and its corollary and just before Lemma 5.3.

The product

\{ ∏_{𝔭∈S_{∞}} φ_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & x_{𝔭} \\ 0 & 1 \end{matrix} ) g_{𝔭})\}\{ ∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} φ(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & x_{𝔭} \\ 0 & 1 \end{matrix} ) g_{𝔭}, ω_{𝔭}, ξ_{𝔭})\}ϵ(( \begin{matrix}1 & x_{D} \\ 0 & 1 \end{matrix} ) g_{D})

is equal to

∏_{𝔭∤D} ξ_{𝔭}(α_{𝔭}x_{𝔭})\{ ∏_{𝔭∈S_{∞}} φ_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭})\}\{ ∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} φ(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭}, ω_{𝔭}, ξ_{𝔭})\}ϵ(( \begin{matrix}1 & x_{D} \\ 0 & 1 \end{matrix} ) g_{D}).

Since ϵ(( \begin{matrix}1 & x_{D} \\ 0 & 1 \end{matrix} )g_{D}) = ϵ(g_{D}) the relation φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )g) = φ(g) follows.

The relation φ(( \begin{matrix}β & 0 \\ 0 & β \end{matrix} )g) = φ(g) for β ∈ K^{×} ∩ I^{D} is, essentially, one of the assumptions. To complete the proof of the lemma we need only show that φ(( \begin{matrix}β & 0 \\ 0 & 1 \end{matrix} )g) = φ(g) when β lies in K^{×} ∩ I^{D}. After replacing g by ( \begin{matrix}β & 0 \\ 0 & 1 \end{matrix} )g in the sum defining φ we can change variables in the summation, replacing α by αβ^{−1}. The sum becomes

∑_{α∈K^{×}} a_{α,β^{−1}}\{ ∏_{𝔭∈S_{∞}} φ_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭})\}\{ ∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} φ(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ) g_{𝔭}, ω_{𝔭}, ξ_{𝔭})\}ϵ(( \begin{matrix}β_{D} & 0 \\ 0 & 1 \end{matrix} ) g_{D}).

The relation φ(( \begin{matrix}β & 0 \\ 0 & 1 \end{matrix} )g) = φ(g) is thus a consequence of the assumption

a_{αβ^{−1}}ϵ(( \begin{matrix}β_{D} & 0 \\ 0 & 1 \end{matrix} )) = a_{α}.

With the same choice of functions φ_{𝔭} the function {â_{α}} determines a function φ̂. Of course ϵ must be replaced by ϵ̃.

Let χ be a character of K^{×} ∩ I^{D}\I^{D} such that

χ(α_{𝔭})ϵ(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} )) = 1

for 𝔭 | D and α_{𝔭} in O_{𝔭}^{×}. If s is a complex number define ζ = ζ(s, χ) as before by

ζ(( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )) = η(β)|αβ^{−1}|^{s}χ(αβ^{−1}).

The function Ξ(s, χ) given as the product of

∑_{x∈K^{×}/K^{×}∩I^{D}} a_{α} ∏_{𝔭|D} ζ_{𝔭}(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))\}\{ ∏_{𝔭∈S_{∞}} Γ(ζ_{𝔭}, π_{𝔭})

and

∏_{{𝔭∉S_{∞}∪R \atop 𝔭∤D}} \frac{1}{(1 − ω_{𝔭,1}(π)ζ_{𝔭,1}(π)|π|^{1/2})(1 − ω_{𝔭,2}(π)ξ_{𝔭}, ζ_{𝔭}, (π)|π|^{1/2})}

is defined for Re s sufficiently large. If {a_{α}} is replaced by {â_{α}} and χ by χ^{−1}η^{−1} we can define a similar function Ξ̂(s, χ^{−1}η^{−1}).

Lemma 7.6. If there is an A in K^{×} with (A_{𝔭}) = 𝔭^{m_{𝔭}} for 𝔭 | D and if, for all possible choices of χ, Ξ(s, χ) is an entire function of s which is bounded in vertical strips and satisfies the functional equation

Ξ(s, χ) = \{∏_{𝔭|D} ζ_{𝔭}(( \begin{matrix}−A_{𝔭} & 0 \\ 0 & 1 \end{matrix} ))\}\{ ∏_{𝔭∈S_{∞}} ϵ(ζ_{𝔭}, ξ_{𝔭}, π_{𝔭})\}\{ ∏_{{𝔭∉S_{∞} \atop 𝔭∤D}} ϵ(ζ_{𝔭}, ξ_{𝔭}, ω_{𝔭})\}Ξ̂(−s, χ^{−1}η^{−1})

then, for all g in G_{𝔸}^{D},

φ̂(( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) g) ∏_{𝔭|D} ( \begin{matrix}0 & A_{𝔭}^{−1} \\ 1 & 0 \end{matrix} )) = φ(g).

Let φ_{1}(g) be the function on the left side of this equation and let I_{0}^{D} be the idèles of norm 1 in I^{D}. We have to show that for each g in G_{𝔸}^{D}

φ_{1}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) = φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)

for all α in I_{0}^{D}. Since both sides are continuous functions on K^{×} ∩ I_{0}^{D}\I_{0}^{D} which is compact we just have to compare Fourier coefficients. Any character of I_{0}^{D} ∩ K^{×}\I_{0}^{D} is obtained by restricting a character χ of K^{×} ∩ I^{D} to I_{0}^{D}. Set

\begin{aligned}μ(χ, g) & = ∫_{K^{×}∩I^{D}\I_{0}^{D}} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)χ(α)dα, \\ μ_{1}(χ, g) & = ∫_{K^{×}∩I^{D}\I_{0}^{D}} φ_{1}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)χ(α)dα. \end{aligned}

μ(χ, g) and μ_{1}(χ, g) are both identically zero if χ(α_{𝔭})ϵ(( \begin{matrix}α_{𝔭} & 0 \\ 0 & 1 \end{matrix} )) ≠ 1 for some 𝔭 | D and some α_{𝔭} in O_{𝔭}^{×}. Thus we need only consider the χ satisfying the conditions of the lemma.

The functions

\begin{matrix}χ(α) & μ(χ, ( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) \\ χ(α) & μ_{1}(χ, ( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) \end{matrix}

are continuous functions on I_{0}^{D}\I^{D} which is isomorphic to R^{+} if K is a number field and to ℤ if K is a function field. As in the proof of Lemma 6.3 the Mellin transform

∫_{I_{0}^{D}\I^{D}} χ(α)μ(χ, ( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)|α|^{s}dα = ∫_{K^{×}∩I^{D}\I^{D}} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα

is defined for Re s sufficiently large and the Mellin transform

∫_{I_{0}^{D}\I^{D}} χ(α)μ_{1}(χ, ( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)|α|^{s}dα

which equals

∫_{K^{×}∩I^{D}\I^{D}} φ̂(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) g ∏_{𝔭|D} ( \begin{matrix}0 & A^{−1} \\ 1 & 0 \end{matrix} ))ζ̃(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα

is defined for Re s sufficiently small.

To prove the lemma in the case of a function field we need only verify that both the Mellin transforms are entire functions of s and that they are equal. In the case of a number field we must show in addition that they are bounded in each vertical strip of finite width.^{∗}

As in Lemma 7.3 the first integral is the product of Ξ(s, χ) and

\{ ∏_{𝔭∈S_{∞}} Φ′(g_{𝔭}, ζ_{𝔭}, φ_{𝔭},)\}\{ ∏_{{𝔭∉S_{∞}∪R \atop 𝔭∤D}} Φ′(g_{𝔭}, ζ_{𝔭}, ω_{𝔭}, ϵ_{𝔭})\}\{ ∏_{𝔭∈R} Φ(g_{𝔭}, ζ_{𝔭}, ω_{𝔭}, ϵ_{𝔭})\}ϵ(g_{D}).  (E)

R is the set of non-archimedean primes which do not divide D such that ζ is not trivial on A_{O_{𝔭}}. According to Lemma 3.6, 5.1, and 6.4 each of the functions occurring in the product is an entire function of s and all but finitely many are identically 1. Thus the first Mellin transform is an entire function of s. The second is the product of Ξ̂(−s, χ^{−1}η^{−1}) and the factors

∏_{𝔭∈S_{∞}} ζ̃_{𝔭}^{−1}(( \begin{matrix}1 & 0 \\ 0 & −A_{𝔭} \end{matrix} ))Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, ζ̃_{𝔭}, φ_{𝔭})

∏_{{𝔭∉S_{∞}∪R \atop 𝔭∤D}} ζ̃_{𝔭}^{−1}(( \begin{matrix}1 & 0 \\ 0 & −A_{𝔭} \end{matrix} ))Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, ζ̃_{𝔭}, ω_{𝔭}, ϵ_{𝔭})

\{ ∏_{𝔭∈R} ζ̃_{𝔭}^{−1}(( \begin{matrix}1 & 0 \\ 0 & −A_{𝔭} \end{matrix} ))Φ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, ζ̃_{𝔭}, ξ_{𝔭})\}ϵ(g_{D}).

It is also an entire function of s and, by the definitions of the factors ϵ(ζ_{𝔭}, ξ_{𝔭}, π_{𝔭}) and ε(ζ_{𝔭}, ξ_{𝔭}, ω_{𝔭}) together with the functional equation satisfied by the function Ξ(s, χ), equal to the first Mellin transform.

One of the Mellin transforms is bounded in vertical strips of a right half-plane, the other is bounded in vertical strips of a left half-plane. Thus to show they are bounded we can apply the Phragmen-Lindelöf theorem for strips. The function \frac{1}{Γ(as+b)}, a real, grows no faster than an exponential in vertical strips so it is enough to show that we can multiply the Mellin transforms by a product of functions of the form Γ(as + b), a real, and obtain a function which is bounded in regions of the form. Re s < constant, |Im s| ≫ 0. By assumption Φ(s, χ) is bounded in such regions. The factors in the product (E) corresponding to the nonarchimedean primes were shown in Lemma 6.4 to be bounded in vertical strips of finite width. If 𝔭 is an archimedean prime Γ(ζ_{𝔭}, π_{𝔭})is a function of this form and

Γ(ζ_{𝔭}, π_{𝔭})Φ′(g_{𝔭}, ζ_{𝔭}, φ_{𝔭}) = Φ(g_{𝔭}, ζ_{𝔭}, φ_{𝔭})

was shown in Lemma 3.6 and 5.7 to be bounded in regions of the form |Re s| ≤ constant, |Im s| ≫ 0.

∗ This seems to be the simplest condition which allows the application of an inversion theorem to establish the identity of the original functions.

Theorem 7.7. If the assumptions of Lemma 7.6 are satisfied the function φ is a function on G_{K} ∩ G_{𝔸}^{D}\G_{𝔸}^{D}.

The set of all ( \begin{matrix}a & b \\ c & d \end{matrix} ) in G_{D}F ∩ G_{𝔸}^{D} which satisfy

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

is a subgroup of G_{K} ∩ G_{𝔸}^{D}. By Lemma 7.5 it contains all those matrices for which c = 0. If b = 0 then

\begin{aligned}φ(( \begin{matrix}a & 0 \\ c & d \end{matrix} ) g) & = φ̂(( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) ( \begin{matrix}a & 0 \\ c & d \end{matrix} ) g ∏_{m_{𝔭}>0} ( \begin{matrix}0 & A_{𝔭}^{−1} \\ 1 & 0 \end{matrix} )) \\ & = φ̂(( \begin{matrix}d & \frac{c}{A} \\ 0 & a \end{matrix} ) ( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) g ∏_{m_{𝔭}>0} ( \begin{matrix}1 & A_{𝔭}^{−1} \\ 1 & 0 \end{matrix} )). \end{aligned}

Applying Lemma 7.5 to φ̂ we see that the last expression is equal to

φ̂(( \begin{matrix}0 & 1 \\ A & 0 \end{matrix} ) g ∏_{m_{𝔭}>0} ( \begin{matrix}0 & A_{𝔭}^{−1} \\ 1 & 0 \end{matrix} )) = φ(g).

The theorem is a consequence of the following lemma. Lemma 7.8. G_{K} ∩ G_{𝔸}^{D} is generated by the matrices in it of the form ( \begin{matrix}a & b \\ 0 & d \end{matrix} ) and ( \begin{matrix}a & 0 \\ c & d \end{matrix} ).

Indeed

( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}a & 0 \\ c & d − \frac{bc}{a} \end{matrix} ) ( \begin{matrix}1 & \frac{b}{a} \\ 0 & 1 \end{matrix} ) .

If the matrix on the right is in G_{K} ∩ G_{𝔸}^{D} so are both the matrices on the right.

Appendix. Some preliminary remarks are necessary before the nature of the function φ_{0}(g) can be determined. It is convenient to treat the various types of fields separately.

We consider the real field first and use the notation of paragraphs 2 and 3. Let L be the space of infinitely differentiable functions on N_{ℝ}\G_{R} which are U-finite on the right.

Lemma A. Let π be the infinite-dimensional irreducible quasi-simple representation of {σ, 𝔄}. Suppose π is deducible from π_{ω}. Let H be a subspace of L which transforms according to π.

(i) If s − m is not an odd integer and s ≠ 0 then ω ≠ ω̃ and H is contained in L(ω) + L(ω̃).

(ii) If s = 0 and m = 0 let ′L(ω) be the space spanned by the functions

φ′_{n}(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{1} & 0 \\ 0 & α_{1} \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ))

defined as

|\frac{α_{1}}{α_{1}} |^{1/2}ω(( \begin{matrix}α_{1} & 0 \\ 0 & α_{1} \end{matrix} ))\{log |\frac{α_{1}}{α_{1}} | + ∑_{k=1}^{\frac{n}{2} |} \frac{1}{2k − 1}\}e^{inθ},

\frac{n}{2} ∈ ℤ. L′(ω) is an invariant irreducible subspace of L and the representation of \{σ, 𝔄\} on L′(ω) is equivalent to π. H is contained in L(ω) + L′(ω).

(iii) If s − m is an odd integer suppose, as we may, that s ≥ 0. Define ω′ by

ω′(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )) = sgn (α_{2}α_{2})ω(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )).

Then H is contained in L(ω) + L(ω′).

ω′ is of course defined for any ω. In Paragraph 2 we saw that if s − m is not an odd integer and s ≠ 0 then π is equivalent to the representation of {σ, 𝔄} on L(ω) and L(ω̃) but is not contained in the representation of {σ, 𝔄} on L(ω′) or L(ω̃′). We also saw that if s − m is an odd integer and s ≠ 0 the representation π is contained once in the representation of {σ, 𝔄} on L(ω) and L(ω′) but is not contained in the representation of {σ, 𝔄} on L(ω̃) or L(ω̃′). Thus if s ≠ 0 we need only show that H is contained in L(ω) + L(ω̃) + L(ω′) + L(ω̃′).

Suppose s = 0 and m = 0. It is clear that

ρ(σ)φ′_{n} = ω(( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ))φ′_{n}, ρ(J)φ′_{n} = (s_{1} + s_{2})φ′_{n}, ρ(U)φ′_{n} = inφ′_{n}.

On the other hand taking u = 0 in the formulae ^{∗} on pg. 3.7 and 3.8 we see that

ρ(V)φ′_{n}(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ))

∗ (1998) See previous footnotes. is equal to

|\frac{α_{1}}{α_{2}} |^{1/2}ω(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ))\{(n + 1) log |\frac{α_{1}}{α_{2}} | + 1 + (n + 1) ∑_{k=1}^{| \frac{n}{2} |} \frac{1}{2k − 1}\}e^{i(n+2)Θ}

and that

ρ(W)φ′_{n}(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ))

is equal to

|\frac{α_{1}}{α_{2}} |^{1/2}ω(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ))\{(−n + 1) log |\frac{α_{1}}{α_{2}} | + 1 + (−n + 1) ∑_{k=1}^{| \frac{n}{2} |} \frac{1}{2k − 1}\}e^{i(n−2)θ}.

Thus ρ(V)φ′_{n} = (n+1)φ′_{n+2} and ρ(W)φ′_{n} = (−n+1)φ′_{n−2}. It follows from Lemma 2.1 that the representation on L′(ω) is equivalent to π_{ω} and hence to π. The representations of {σ, 𝔄} on L(ω′) and L′(ω′) are not equivalent to π. Again we need only show that H is contained in L(ω) + L′(ω) + L(ω′) + L′(ω′).

Suppose φ lies in H. There are functions φ_{n}(α_{1}, α_{2}) on ℝ^{×} × ℝ^{×}, only a finite number of which do not vanish identically, such that for α_{2} > 0.

φ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} )) = ∑_{n} φ_{n}(α_{1}, α_{2})e^{inΘ}.

Moreover there are functions ψ_{n}(L) on R^{×} such that

φ_{n}(α_{1}, α_{2}) = ω(( \begin{matrix}|α_{1}α_{2}|^{1/2} & 0 \\ 0 & |α_{1}α_{2}|^{1/2} \end{matrix} ))ψ_{n}(\frac{α_{1}}{α_{2}} ).

Since φ is in L, ρ(D)φ = λ(Z)φ + \frac{1}{2}λ(Z^{2})φ and the equation ρ(D)φ = \frac{s^{1}−1}{2}φ reduces to the equations

−2t\frac{dψ_{n}}{dt} + 2t\frac{d}{dt}(t\frac{dψ_{n}}{dt}) = \frac{s^{2} − 1}{2}ψ_{n}

or

4(t\frac{d}{dt} − \frac{1}{2})^{2}ψ_{n} = s^{2}ψ_{n}.

If s ≠ 0 four linearly independent solutions of this are (sgn t)^{a}|t|^{\frac{1±s}{2}} , a = 0 or 1 and if s = 0 four linearly independent solutions of this are (sgn t)^{a}|t|^{1/2} and (sgn t)^{a}|t|^{1/2} log |t|, a = 0 or 1. The lemma follows for all representations except the one for which s = 0 and |m| = 1.

If s = 0 and |m| = 1 the space H_{1} contains a non-zero vector. If φ lies in H_{1} the function ψ_{n} is zero if n ≠ 1. According to the first formula on p. 3.8 the equation ρ(W)φ = 0 is equivalent to

2t\frac{dψ_{1}}{dt} − ψ_{1}(t) = 0.

Thus ψ_{1}(t) is a linear combination of |t|^{1/2} and (sgn t)|t|^{1/2}. Thus H meets L(ω) + L(ω′). Since H is irreducible, H is contained in L(ω) + L(ω′).

For the complex field we use the notation of paragraphs 4 and 5. Let L be the space of infinitely differentiable functions on N_{ℂ}\G_{ℂ} which are U-finite on the right.

Lemma B. Let π be an infinite-dimensional irreducible quasi-simple representation of 𝔄. Suppose π is deducible from π_{ω}. Let H be a subspace of L which transforms according to π.

(i) If s − m is not integral then ω ≠ ω̃ and H is contained in L(ω) + L(ω̃). If s − m is integral define ω′ by

ω′(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )) = |α_{1}α_{2}|^{\frac{s_{1}+s_{2}}{2}} |\frac{α_{1}}{α_{2}} |^{m}(\frac{α_{1}}{|α_{2}|})^{\frac{m_{1}+m_{2}}{2} +s}(\frac{α_{2}}{|α_{2}|})^{\frac{m_{1}+m_{2}}{2} −s}.

π is deducible from π_{ω}, πω̃, π_{ω′}, and π′ ; ω̃.

(ii) If |s| > |m| we can assume with no loss of generality that s > |m|. Then H is contained in L(ω) + L(ω′) + L(ω̃′).

(iii) If |s| = |m| and s ≠ 0 either ω = ω′ or ω = ω̃′. In this case H is contained in L(ω) + L(ω̃).

(iv) If s = 0 and m = 0 define γ_{0} and δ_{k} as on^{∗} p. 4.8 and let c_{n} = ∑k=1^{\frac{n}{2}} \frac{1}{k} if n is a non-negative even integer. If t > 0 set ψ_{n}(t) = log t + c_{n}. Let L′(ω) be the space spanned by the functions

φ̂_{n,k}(( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) u) = ω(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ))|\frac{α_{1}}{α_{2}} |ψ_{n}(|\frac{α_{1}}{α_{2}} |)γ_{0}σ_{n}(u)δ_{k}

with \frac{n}{2} ∈ ℤ, −k ∈ ℤ, and |k| ≤ \frac{n}{2}. L′(ω) is an irreducible invariant subspace of L and the representation of 𝔄 on L′(ω) is equivalent to π. The space H must be in L(ω) + L′(ω).

The most complicated part of the lemma to verify is the assertion that L′(ω) is invariant and irreducible so we verify that first.

For convenience set φ̂_{n,k}(g) = 0 if k ∈ ℤ and |k| > \frac{n}{2}. Just as^{∗} in Paragraph 4 the existence of the Clebsch-Gordan series allows us to assert that the function

φ_{n+2,k}^{+}(( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) u)

which equals

(\frac{n}{2} + k + 1)(\frac{n}{2} + k)ρ(V^{+})φ̂_{n,k−1} − (\frac{n}{2} + k + 1)(\frac{n}{2} − k + 1)ρ(V)φ̂_{n,k} − (\frac{n}{2} − k)(\frac{n}{2} − k + 1)ρ(V^{−})φ̂_{n,k+1}

∗ (1998) Just after Lemma 4.2.

∗ The right hand sides of the formula on p. 4.9 are not correct. They should be

(\frac{n}{2} + k + 1)!(\frac{n}{2} − k + 1)!a(n, ω)φ_{n+1,k}

and

(\frac{n}{2} + k − 1)!(\frac{n}{2} − k − 1)!b(n, ω)φ_{n−1,k}

is of the form

(\frac{n}{2} + k + 1)!(\frac{n}{2} − k + 1)!ω(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ))|\frac{α_{1}}{α_{2}} |ψ_{n+2}^{+}(|\frac{α_{1}}{α_{2}} |)γ_{0}σ_{n+2}(u)δ_{k},

that the function

φ_{n,k}^{0} = (\frac{n}{2} + k)ρ(V^{+})φ̂_{n,k−1} + kρ(V)φ̂_{n,k} + (\frac{n}{2} − k)ρ(V^{−})φ̂_{n,k+1}

is of the form

φ_{n,k}^{0}(( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) u) = (\frac{n}{2} + k)!(\frac{n}{2} − k)!ω(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ))|\frac{α_{1}}{α_{2}} |ψ_{n}^{0}(|\frac{α_{1}}{α_{2}} |)γ_{0}σn(u)δ_{k},

and that the function

φ_{n−2,k}^{−} = ρ(V^{+})φ̂_{n,k−1} + ρ(V)φ̂_{n,k} − ρ(V^{−})φ̂_{n,k+1}

is of the form

φ_{n−2,k}(( \begin{matrix}1 & Z \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) u) = (\frac{n}{2}_{k} − 1)!(\frac{n}{2} − k − 1)!ω(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ))|\frac{α_{1}}{α_{2}} |ψ_{n−2}^{−}(|\frac{α_{1}}{α_{2}} |)γ_{0}σ_{n−2}(u)δ_{k}.

In these three formulae δ_{k} is respectively x^{\frac{n}{2} +1+k}y^{\frac{n}{2} +2−k}, x^{\frac{n}{2} +k}y^{\frac{n}{2} −k}, and x^{\frac{n}{2} +k−1}y^{\frac{n}{2} −k−1} and γ_{0} lies in the dual of V_{n+2}, V_{n}, and V_{n−2} respectively.

To show that L′(ω) is invariant we need only verify that ψ_{n+2}^{+} is a multiple of ψ_{n+2}, that ψ_{n}^{0} is a multiple of ψ_{n}, and that ψ_{n−2}^{−} is a multiple of ψ_{n−2}. If n = 0 only ψ_{n+2}^{+} is defined. Evaluating φ_{n+2,0}^{+} at ( \begin{matrix}t^{1/2} & 0 \\ 0 & t^{−1/2} \end{matrix} ) we see that^{∗} [(\frac{n}{2} + 1)!]^{2}tψ_{n+2}^{+}(t) is equal to the sum of three terms,

(\frac{n}{2} + 1)(\frac{n}{2})t(log t + c_{n})γ_{0}σ_{n} ( \begin{matrix}0 & −1 \\ 0 & 0 \end{matrix} ) δ_{−1},

−2(\frac{n}{2} + 1)^{2}t\frac{d}{dt}(t log t + c_{n}t),

−(\frac{n}{2})(\frac{n}{2} + 1)t(log t + c_{n})γ_{0}σ_{n} ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) δ_{1},

a sum that equals (cf. p. 4.2)

−2(\frac{n}{2} + 1)^{2}(\frac{n}{2} + 1)t(log t + c_{n} + \frac{1}{\frac{n}{2} + 1}) = −2(\frac{n}{2} + 1)^{3}tψ_{n+2}(t).

In the same way we see that (\frac{n}{2}!)^{2}tψ_{n}^{0}(t) is equal to

[−(\frac{n}{2})γ_{0}σ_{n} ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) δ_{−1} + (\frac{n}{2})γ_{0}σ_{n} ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) δ_{1}](t log t + c_{n}t),

∗ The formula at the top of p. 4.10 is not correct. It should be

V^{−} = (X_{1} + \frac{W_{1}}{2} ) + i(X_{2} − \frac{W_{2}}{2} ).

which equals

[−(\frac{n}{2})(\frac{n}{2} + 1) + (\frac{n}{2})(\frac{n}{2} + 1)](t log t + c_{n}t) = 0.

Finally [(\frac{n}{2} − 1)!]^{2}tψ_{n−2}^{−}(t) is equal to

[−γ_{0}σ_{n} ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) δ_{−1} − γ_{0}σ_{n} ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) δ_{1}](t log t + c_{n}t) + 2t\frac{d}{dt}(t log t + c_{n}t)

which equals

−n(t log t + c_{n}t − \frac{t}{\frac{n}{2}} ) = −ntψ_{n−2}(t).

If the functions φ_{n,k} are defined as on p. 4.9 then, as we have seen,^{†} when s = 0 and m = 0

(\frac{n}{2} + k)(\frac{n}{2} + k + 1)ρ(V^{+})φ_{n,k−1} − (\frac{n}{2} + k + 1)(\frac{n}{2} − k + 1)ρ(V)φ_{n−k} − (\frac{n}{2} − k)(\frac{n}{2} − k + 1)ρ(V^{−})φ_{n,k+1}

is equal to

−2\frac{(\frac{n}{2} + k + 1)!}{(\frac{n}{2} + 1)!} \frac{(\frac{n}{2} − k + 1)!}{\frac{n}{2} + 1)!}(\frac{n}{2} + 1)^{3}φ_{n+2,k}

and

ρ(V^{+})φ_{n,k−1} + ρ(V)φ_{n,k} − ρ(V^{−})φ_{n,k+1}

is equal to

\frac{(\frac{n}{2} + k − 1)!}{(\frac{n}{2} − 1)!} \frac{(\frac{n}{2} − k − 1)!}{(\frac{n}{2} − 1)!}(−n)φ_{n−2,k}.

Moreover one shows readily that

(\frac{n}{2} + k)ρ(V^{+})φ_{n,k−1} + kρ(V)φ_{n,k} + (\frac{n}{2} − k)ρ(V^{−})φ_{n,k+1}

is equal to zero. It follows immediately that the representation on L′ is equivalent to the representation on L(ω).

The remarks of the lemma can now be verified rather easily. Choose n so that H_{n} ≠ 0. There is a function Ψ(g) on G_{ℂ} with values in V̂_{n} such that H_{n} is the set of functions of the form Ψ(g)Φ, Φ ∈ V_{n}. Moreover Ψ(gu) = Φ(g)σ_{n}(u) if u ∈ U^{0} and Ψ(( \begin{matrix}α & 0 \\ 0 & α \end{matrix} )g) = ω(( \begin{matrix}α & 0 \\ 0 & α \end{matrix} ))Ψ(g). Let ψ(t) = Ψ(( \begin{matrix}t^{1/2} & 0 \\ 0 & t^{−1/2} \end{matrix} )); Ψ is determined by ψ. According to the formulae^{∗} on p. 5.8 the equations ρ(D)Ψ = \frac{(s+m)^{2}−1}{2}Ψ and ρ(D)Ψ = \frac{(s−m)^{2}−1}{2}Ψ reduce to

\begin{aligned}[t\frac{d}{dt} + k − 1]^{2}ψ^{k} & = (s + m)^{2}ψ^{k} \\ [t\frac{d}{dt} − k − 1]^{2}ψ^{k} & = (s − m)^{2}ψ^{k}. \end{aligned}

If either (s + m) ≠ 0 or (s − m) ≠ 0 these equations imply that each ψ^{k} is a power of t. Thus H_{n}, and hence H, is contained in a space of the form Σ_{i=1}^{r}L(ω_{i}) for some ω_{1}, …, ω_{r}. Parts (i), (ii) and (iii) of the lemma

† According to a remark in a previous footnote the left hand sides of the equation on p. 4.10 should be (\frac{n}{2} + m + 1)!(\frac{n}{2} − m + 1)!a(n, ω) and (\frac{n}{2} + m − 1)!(\frac{n}{2} − m − 1)!b(n, ω).

∗ (1998) Between Lemmas 5.2 and 5.3. follow from Lemma 4.2 and the proof of Lemma 4.4. If s + m = 0 and s − m = 0 then s = m = 0. Then ψ^{k} ≡ 0 if k ≠ 0 and ψ^{0}(t) is a linear combination of t and t log t. Part (iv) of the lemma above follows.

For a non-archimedean local field we use the notation of paragraph 6. If ω is a homomorphism of A_{K}/A_{O} into ℂ^{×} define the function φ_{ω} by

φ_{ω}(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) u) = |\frac{α_{1}}{α_{2}} |^{1/2}ω(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )) u ∈ G_{O}.

If ω ≠ ω̃ then φ_{ω} ≠ φω̃. If ω = ω̃ define φ′_{ω} by

φ′_{ω}(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) u) = |\frac{α_{1}}{α_{2}} |^{1/2}ω(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )) log |\frac{α_{1}}{α_{2}} |.

Lemma C. Suppose φ is a function on N_{K}\G_{K} which satisfies φ(gu) ≡ φ(g) for u in G_{O} and suppose that for all f in H

∫_{G_{K}} φ(gh)f(h)dh = χ_{ω}(f)φ(g).

If ω ≠ ω̃, φ is a linear combination of φ_{ω} and φ_{ω} and, if ω = ω̃, φ is a linear combination of φ_{ω} and φ′_{ω}.

Choosing f to be the characteristic function of a(1, 1)G_{O} we obtain the relations

\begin{aligned}φ(( \begin{matrix}πα_{1} & 0 \\ 0 & πα_{2} \end{matrix} )) & = ω(π)φ(( \begin{matrix}α_{1} & π \\ w_{1} & α_{2} \end{matrix} )) \\ qφ(( \begin{matrix}πα_{1} & 0 \\ 0 & α_{2} \end{matrix} )) + φ(( \begin{matrix}α_{1} & 0 \\ 0 & πα_{2} \end{matrix} )) & = q^{1/2}[w(( \begin{matrix}π & 0 \\ 0 & 1 \end{matrix} )) + ω(( \begin{matrix}1 & 0 \\ 0 & π \end{matrix} ))]φ(( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} )). \end{aligned}

It is easy to see that these relations are satisfied by φω, φω̃ and, if ω = ω̃, by φ_{ω}. If ω ≠ ω̃ then φ_{ω}(( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )) = φω̃(( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )) ≠ 0 but φ_{ω}(( \begin{matrix}π & 0 \\ 0 & 1 \end{matrix} )) ≠ φω̃(( \begin{matrix}π & 0 \\ 0 & 1 \end{matrix} )). Subtracting from φ a suitable linear combination of φ_{ω} and φω̃ we obtain a function ψ which satisfies these relations and vanishes at ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) and ( \begin{matrix}π & 0 \\ 0 & 1 \end{matrix} ). If ω = ω′ then φ_{ω}(( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )) ≠ ω but φ′_{ω}(( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) = 0 while φ′_{ω}(( \begin{matrix}π & 0 \\ 0 & 1 \end{matrix} )) ≠ 0. We can again subtract from φ a suitable linear combination of φ_{ω} and φ′_{ω} and obtain a function ψ which satisfies these relations and vanishes at ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) and ( \begin{matrix}π & 0 \\ 0 & 1 \end{matrix} ). To prove that, in either case, ψ vanishes identically we need only show that it vanishes at the matrices ( \begin{matrix}π^{m+n} & 0 \\ 0 & π^{n} \end{matrix} ). The first relation implies this is so if n = 0 or 1. Taking ( \begin{matrix}α_{1} & 0 \\ 0 & α_{2} \end{matrix} ) = ( \begin{matrix}π^{m+n} & 0 \\ 0 & π^{n} \end{matrix} ) and substituting in the second relation we see that if this is so if for all m and n = n_{0} and n_{0} + 1 it is true for all m and n = n_{0} − 1 and that if this is so for all m and n = n_{0} and n_{0} − 1 is for all m and n = n_{0} + 1. The lemma follows by induction.

Let S be a finite set of primes containing the archimedean primes and the primes which divide D. Let I_{s} = {ι | ι_{𝔭} is a unit if 𝔭 ∉ S}. We suppose S is so large that I^{D}(K^{×} ∩ I^{D})I_{S}^{D} if I_{S}^{D} = I_{S} ∩ I^{D}. Let G_{S} = ∏_{𝔭∈S} G_{K_{𝔭}} × ∏_{𝔭∉S} G_{O_{𝔭}} and let G_{S}^{D} = G_{𝔸}^{D} ∩ G_{S}. According to the previous three lemmas the restriction of φ_{0} to G_{S}^{D} is a linear combination of functions of the form^{∗}

\begin{matrix}ψ(\{∏_{𝔭∈S} ( \begin{matrix}1 & γ_{𝔭} \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} ) u_{𝔭}\}\{ Π_{𝔭∉S} u_{𝔭}\}) = \\ \{∏_{𝔭∈S} |\frac{α_{𝔭}}{β_{𝔭}} |^{1/2}\}η( Π_{𝔭∈S} ( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} ))ζ(∏_{𝔭∈S} u_{𝔭}) ∏_{𝔭∈S_{1}} log |\frac{α_{𝔭}}{β_{𝔭}} |. \end{matrix}

Here ( \begin{matrix}1 & x_{𝔭} \\ 0 & 1 \end{matrix} )( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} )u_{𝔭} lies in U_{K_{𝔭}}^{D} if 𝔭 | D, η is a homomorphism of the group of diagonal matrices with entries from I_{S}^{D} into ℂ^{×} such that η(( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} )) = 1 if 𝔭 ∉ S and α_{𝔭}, β_{𝔭} lie in O_{𝔭}^{×}, and S_{1} is a subset of S. If γ and δ belong to K^{×} ∩ I_{S}^{D} then φ_{0}(( \begin{matrix}γ & 0 \\ 0 & δ \end{matrix} )g) = φ_{0}(g). Moreover Σ_{𝔭∈S} log |γ_{𝔭}| = 0 is the only linear relation satisfied by all the matrices {log |γ_{𝔭}| | 𝔭 ∈ S} as γ varies over K^{×} ∩ I_{S}^{D}. A simple argument then shows that the restriction of φ_{0} to G_{S}^{D} is of the form

\begin{matrix}φ_{0}(\{∏_{𝔭∈S} ( \begin{matrix}1 & x_{𝔭} \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} ) u_{𝔭}\}) = \\ ∏_{𝔭∈S} |\frac{α_{𝔭}}{β_{𝔭}} |^{1/2} ∑_{i=1}^{r} η^{(1)}(∏_{𝔭∈S} ( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} ))\{ζ_{1}^{(i)}(∏_{𝔭∈S} u_{𝔭}) + ζ_{2}^{(i)}(∏_{𝔭∈S} u_{𝔭}) ∑_{𝔭∈S} log |\frac{α_{𝔭}}{β_{𝔭}} |\}. \end{matrix}

The homomorphisms η^{(1)}, …, η^{(r)} are to be distinct and for each i either ζ_{1}^{(i)} or ζ_{2}^{(i)} is to be different from

zero. If α and β lie in K^{×} ∩ I_{S}^{D} then η^{(i)}( ∏_{𝔭∈S}( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} )) = 1.

Each η^{(i)} determines a homomorphism of the diagonal matrices with entries from I^{D} into ℂ^{×}. This homomorphism, which will be 1 on the matrices with entries from K^{×} ∩ I^{D} we again call η^{(i)}. The value of

φ_{0} at ∏_{𝔭}( \begin{matrix}1 & x_{𝔭} \\ 0 & 1 \end{matrix} )( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} )u_{𝔭} is the same as its value at \{ ∏_{𝔭∈S}( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} )u_{𝔭}\}\{ ∏_{𝔭∉S}( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} )\} which is

\{∏_{𝔭} |\frac{α_{𝔭}}{β_{𝔭}} |^{1/2}\} ∑_{i=1}^{i} η|^{(i)}(∏_{𝔭} ( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} ))\{ζ_{1}^{(i)}(∏_{𝔭∈S} u_{𝔭}) + ζ_{2}^{(i)}(∏_{𝔭∈S} u_{𝔭}) ∑_{𝔭} log |\frac{α_{𝔭}}{β_{𝔭}} |\}.

Define η̃^{(i)} by

η̃^{(i)}(∏_{𝔭} ( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} )) = η^{(i)}(∏_{𝔭} ( \begin{matrix}β_{𝔭} & 0 \\ 0 & α_{𝔭} \end{matrix} )).

Lemma D. If i ≠ j then η^{(j)} = η̃^{(i)}

Let

\begin{aligned}η^{(i)^{−1}}η^{(j)}(( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )) & = |α|^{a}|β|^{b}χ(( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )) \\ η̃^{(i)^{−1}}η^{(j)}(( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )) & = |α|^{c}|β|^{d}χ′(( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )) \end{aligned}

∗ In this formula and the similar ones following the absolute value at the complex primes is the square of the usual absolute value. for α, β in I^{D}. Here a, b, c, d are real numbers and χ and χ′ are characters in the usual sense. Lemma C implies that if 𝔭 ∉ S the restriction of either η^{(i)^{−1}}η^{(j)} or η̃^{(i)^{−1}}η^{(j)} to \{( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} )|α_{𝔭}, β_{𝔭} ∈ K_{𝔭}^{×}\} is trivial. This can only happen if a = b = 0 or c = d = 0. Suppose that a ≠ 0 or b ≠ 0. Then c = d = 0 and η̃^{(i)^{−1}}η^{(j)} is an ordinary character. It is known that the values η̃^{(i)^{−1}}η^{(j)} takes on the matrices ( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} ), α_{𝔭}, β_{𝔭} ∈ K_{𝔭}^{×}, 𝔭 ∉ S are dense in the set of values which η̃^{(i)^{−1}}η^{(j)} takes on. It follows that η^{(j)} = η̃^{(i)}. In the same way we show that if c ≠ 0 or d ≠ 0 then η^{(i)} = η^{(j)}. This is of course excluded. It remains to treat the case a = b = c = d = 0. In this case the values taken by the vector-valued function (η^{(i)^{−1}}η^{(j)}, η̃^{(i)^{−1}}η^{(j)}) on the matrices ( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} ), α_{𝔭}, β_{𝔭} ∈ K_{𝔭}^{×}, 𝔭 ∉ S are dense in the set of all values it assumes. It follows from Lemma C that (1 − η^{(i)^{−1}}η^{(j)})(1 − η̃^{(i)^{−1}}η^{(j)}) vanishes identically. If η̃^{(i)} ≠ η^{(j)} there is an ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) such that η̃^{(i)^{−1}}η^{(j)}(( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )) ≠ 1. Thus, necessarily η^{(i)^{−1}}η^{(j)}(( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )) = 1. Since η^{(i)} ≠ η^{j)} there is a ( \begin{matrix}γ & 0 \\ 0 & δ \end{matrix} ) such that

η^{(i)^{−1}}η^{(j)}(( \begin{matrix}γ & 0 \\ 0 & δ \end{matrix} )) ≠ 1. Then η̃^{(i)^{−1}}η^{(j)}(( \begin{matrix}γ & 0 \\ 0 & δ \end{matrix} )) = 1. One sees immediately that (1−η^{(i)^{−1}}η^{(j)})(1−η̃^{(i)^{−1}}η^{(j)})

will not vanish at ( \begin{matrix}αγ & 0 \\ 0 & βδ \end{matrix} ). This is a contradiction.

Lemma E. There are two possible forms for the function φ_{0}.

(i) There is a homomorphism ω of the diagonal matrices with entries from I^{D} into ℂ^{×}, which is 1 on the matrices with entries from K^{×} ∩ I^{D}, such that ω ≠ ω̃ and two functions ζ and ζ′ on ∏_{𝔭∈S} U_{K_{𝔭}} such that

if g = ∏_{𝔭}( \begin{matrix}1 & x_{𝔭} \\ 0 & 1 \end{matrix} )( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} )u_{𝔭} lies in G_{𝔸}^{D} then φ_{0}(g) equals

\{∏_{𝔭} |\frac{α_{𝔭}}{β_{𝔭}} |^{1/2}\}\{ω(∏_{𝔭}( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} ))ζ(∏_{𝔭∈S} u_{𝔭}) + ω̃(∏_{𝔭}( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} ))ζ′(∏_{𝔭∈S} u_{𝔭})\}.

(ii) There is a homomorphism ω of the diagonal matrices with entries from I^{D} into ℂ^{×}, which is 1 on the matrices with entries from K^{×} ∩ I^{D}, such that ω = ω̃ and two functions ζ and ζ′ on ∏_{𝔭∈S} U_{K_{𝔭}} such that

if g = ∏_{𝔭}( \begin{matrix}1 & x_{𝔭} \\ 0 & 1 \end{matrix} )( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} )u_{𝔭} lies in G_{𝔸}^{D} then φ_{0}(g) equals

\{∏_{𝔭} |\frac{α_{𝔭}}{β_{𝔭}} |^{1/2}\}\{ω(∏_{𝔭}( \begin{matrix}α_{𝔭} & 0 \\ 0 & β_{𝔭} \end{matrix} ))\}\{ζ(∏_{𝔭∈S} u_{𝔭}) + ζ′(∏_{𝔭∈S} u_{𝔭}) ∑_{𝔭∈S} log |\frac{α_{𝔭}}{β_{𝔭}} |\}.