1. Gaussian sums In the second paragraph I shall discuss the representations of the group of 2 × 2 non-singular matrices over a non-archimedean field. In the discussion a number of identities for Gaussian sums will be required. In this paragraph the necessary identities, trivial or not, are stated and proved.

Let k be a non-archimedean local field, let o be the ring of integers in k, let 𝔭 be the maximal ideal of o, and let π be a generator of 𝔭. Let k^{×} be the multiplicative group of k and let o^{×} be the group of units. If n ≥ 0 then o_{n}^{×} = {α ∈ o^{×}|α − 1 ∈ 𝔭^{n}}. Fix a character ξ_{0} of k with the property that o is the largest ideal of k on which ξ_{0} is trivial.

If μ is a character of o^{×} and x belongs to k set

Δ(μ, x) = ∫_{o^{×}} ξ_{0}(αx)μ(α)dα.

It is clear that if β belongs to o^{×}

Δ(μ, βx) = μ^{−1}(β)Δ(μ, x).

Lemma 1.1 Let 𝔭^{n} be the conductor of μ.

(i) If n = 0 then Δ(μ, π^{m}) = 1 if m ≥ 0, Δ(μ, π^{−1}) = \frac{|π|}{|π|−1}, and Δ(μ, π^{m}) = 0 if m < −1.

(ii) If n > 0 then Δ(μ, π^{m}) = 0 if m ≠ −n but

|Δ(μ, π^{−n}| = \frac{|π|^{\frac{n}{2}}}{1 − |π|}.

If n = 0 then μ is trivial and it is clear that Δ(μ, π^{m}) = 1 for m ≥ 0. It is also clear that if m < 0

1 + ∑_{k=m}^{−1} \frac{1 − |π|}{|π|^{k}}Δ(μ, π^{k}) = 0.

The first part of the lemma follows immediately.

Certainly

∫_{o^{×}} ξ_{0}(απ^{m})μ(α)dα = ∫_{o^{×}/o_{n}^{×}} μ(α){∫_{o_{n}^{×}} ξ_{0}(αβπ^{m})dβ}dα.

If n > 0 the inner integral is equal to

\frac{ξ_{0}(απ^{m})}{1 − |π|} ∫_{𝔭^{n}} ξ_{0}(απ^{m}y)dy.

This is zero if the character y → ξ_{0}(απ^{m}y) is not trivial on 𝔭^{n}, that is, if m < −n. On the other hand if m > −n so that for some ℓ, with 0 ≤ ℓ < −n, m + ℓ ≥ 0 then

∫_{o^{×}} ξ_{0}(απ^{m})μ(α)dα = ∫_{o^{×}/o_{ℓ}^{×}} ξ_{0}(απ^{m})μ(α){∫_{o_{ℓ}^{×}} μ(β)dβ}dα

The inner integral on the right is zero.

Finally

|Δ(μ, π^{−n})|^{2} = ∫_{o^{×}} dα ∫_{o^{×}} dβ ξ_{0}((α − β)π^{−n})μ (\frac{α}{β})
= ∫_{o^{×}} dα ∫_{o^{×}} dβ ξ_{0}(β(α − 1)π^{−n})μ(α).

By part (i) of the lemma the integral with respect to β is 1 if α ∈ o_{n}^{×}, \frac{|π|}{|π|−1} if α ∈ o_{n−1}^{×} − o_{n}^{×}, and zero otherwise. Since

\frac{|π|}{|π| − 1} ∫_{o_{n−1}^{×}−o_{n}^{×}} μ(α)dα = \frac{|π|}{1 − |π|} ∫_{o_{n}^{×}} μ(α)dα,

we have

|Δ(μ, π^{−n})|^{2} = \frac{1}{1 − |π|}(measure o_{n}^{×}) = \frac{|π|^{n}}{(1 − |π|)^{2}} .

If the conductor of μ is 𝔭^{n} we shall refer to n as the order of μ.

Lemma 1.2 Suppose μ and ν are characters of o^{×}. Let the order of μν be r. If r ≥ 1 then

\frac{Δ(μ, π^{m})Δ(ν, π^{n})}{Δ(μν, π^{−r})} = ∫_{\{α∈o^{×}|π^{r+m}α+π^{r+m}∈o^{×}\}} μ(α)(μν)^{−1}(π^{r+m}α + π^{r+m})dα.

If r = 0 then Δ(μ, π^{m})Δ(ν, π^{n}) is equal to

∫_{\{α∈o^{×}|π^{m}α+π^{n}∈o\}} μ(α)dα + \frac{|π|}{|π| − 1} ∫_{\{α∈o^{×}|π^{m+1}α+π^{n+1}∈o^{×}\}} μ(α)dα.

The product Δ(μ, π^{m})Δ(ν, π^{n}) is equal to

∫_{o^{×}} ∫_{o^{×}} ξ_{0}(π^{m}α + π^{n}β)μ(α)ν(β)dαdβ = ∫_{o^{×}} ∫_{o^{×}} ξ_{0}(β(π^{m}α + 1))dβ dα.

If r ≥ 1 the right side is equal to

Δ(μν, π^{−r}) ∫_{{α∈o^{×}|π^{r+m}α+π^{r+m}∈o^{×}}} μ(α)(μν)^{−1}(π^{r+m}α + π^{r+m})dα.

If r = 0 the right side equals

∫_{\{α|π^{m}α+π^{n}∈o\}} μ(α)dα + \frac{|π|}{|π| − 1} ∫_{\{α|π^{m+1}α+π^{n+1}∈o^{×}\}} μ(α)dα.

Now let K be a two dimensional commutative algebra over k with a non-degenerate trace. There are two possibilities for K. Either it is the direct sum of k with itself or it is a separable quadratic extension of k. In both cases k has exactly one non-trivial automorphism over k. We will denote this automorphism by s. If x ∈ K then Sx = x + x^{s} and Nx = xx^{s}. Let O be the elements of k integral over o and let O^{∗} be the group of units of O. If K = k ⊕ k set Π = π ⊗ π and if n_{1} and n_{2} are any two integers set π^{n_{1},n_{2}} = π^{n_{1}} ⊕ π^{n_{2}}. If K is an unramified extension of k set Π = π and if n_{1} = n_{2} set π^{n_{1},n_{2}} = π^{n_{1}}. If K is a ramified extension choose π and Π so that NΠ = π, if n_{2} = 0 set π^{n_{1},n_{2}} = Π^{n_{1}}. Thus the symbol π^{n_{1},n_{2}} has a meaning only for certain values of n_{1} and n_{2}. We shall adhere to the convention that any expression in which the symbol π^{n_{1},n_{2}} occurs with values of n_{1} and n_{2} for which it has no meaning is equal to zero. If n_{1} ≥ 0, n_{2} ≥ 0 and π^{n_{1},n_{2}} is defined set O_{n_{1},n_{2}}^{×} = {α ∈ O^{×}|α − 1 ∈ π^{n_{1},n_{2}}O}. If M is a character of O^{×} then amongst all groups of this type on which M is trivial there is a maximal one O_{m_{1},m_{2}}^{×}. (m_{1}, m_{2}) will be called the order of M.

If K = k ⊗ k or K is an unramified extension we set f = 0. Otherwise (Π^{−f}) is the inverse different. The index of NK^{×} in k^{×} is either 1 or 2. If it is 1 let χ be the trivial character of k^{×}; if it is 2 let χ be the unique non-trivial character of k^{×} whose restriction to NK^{×} is trivial. Let χ_{0} be the restriction of χ to o^{×}. The order of χ_{0} is f.

Before going on I recall some facts whose proofs are either completely trivial or are to be found in the book “Corps Locaux” of Serre.

Lemma 1.3 (i) Let n_{1} and n_{2} be non-negative integers. If K = k ⊕ k the map x → Sx takes π^{n_{1},n_{2}}O onto 𝔭^{r} with r = min{n_{1}, n_{2}}. The map x → Nx maps O_{n_{1},n_{2}}^{×} onto o_{r}^{×}.

(ii) If K is an unramified extension of k the map x → Sx maps π^{n,n}O onto 𝔭^{n}. The map x → Nx takes O_{n,n}^{×} onto o_{n}^{×}.

(iii) If K is a ramified extension of k the map x → Sx maps π^{n,0}O onto 𝔭^{r} with r = [\frac{n+f}{2} ]. If n ≥ f the smallest number m such that N(O_{m,0}^{×}) = o_{n}^{×} is 2n − f; the largest such number is 2n − f + 1. If n < f then N(O_{n,0}^{×}) is contained in o_{n}^{×} and if 0 ≤ m < n the map N: O_{m,0}^{×}/O_{n,0}^{×} → o_{m}^{×}/o_{n}^{×} is an isomorphism. If m < f the kernel and the cokernel of the map

N: O_{n,0}^{×}/O_{f,0}^{×} → o_{n}^{×}/o_{f}^{×} both have order two.

If μ is a character of o^{×} let μ^{1+s} be the character of O^{×} defined by μ^{1+s}(α) = μ(αα^{s}). Let n be the order of μ. If K = k ⊗ k or K is unramified the order of μ^{1+s} is (n, n). If K is ramified the order of μ^{1+s} is (2n − f, 0)if n > f; it is (n, 0) if n < f, but if n = f all one can say is that it is (r, 0) with r ≤ f.

If M_{0} is a character of O^{×} set

Δ(M_{0}, π^{n_{1},n_{2}}) = ∫_{O^{×}} ξ_{0}(S(απ^{n_{1},n_{2}}))M_{0}(α)dα.

The following lemma is an immediate consequence of Lemma 1.2 but it is convenient to state it explicitly.

Lemma 1.4 Suppose M_{0} and H_{0} are two characters of O^{∗}. Let the order of M_{0}H_{0} be (r_{1}, r_{2}). If r_{1} > 0 and r_{2} + f > 0 then

\frac{Δ(M_{0}, π^{m_{1},m_{2}})Δ(H_{0}, π^{n_{1},n_{2}})}{Δ(M_{0}H_{0}, π^{−r_{1}−f,−r_{2}})} = ∫_{\{α∈O^{×}|π^{r_{1}+m_{1}+f,r_{2}+m_{2}} α+π^{r_{1}+n_{1}+f,r_{2}+n_{2}} ∈O^{∗}\}} M_{0}(α)(M_{0}H_{0})^{−1}(π^{r_{1}+m_{1}+f,r_{2}+m_{2}}α + π^{r_{1}+m_{1}+f,r_{2}+n_{2}})dα.

If K = k ⊕ k and r_{1} = 0 and r_{2} = 0 the left hand side is equal to the sum of

∫_{{α∈O^{×}|π^{m_{1},m_{2}} α+π^{n_{1},n_{2}} ∈O}} M_{0}(α)dα

and \frac{|π|}{|π|−1} times

∫_{{α∈O^{×}|π^{m_{1}+1,m_{2}} α+π^{n_{1}+1,n_{2}} ∈o^{×}⊕o}} M_{0}(α)dα + ∫_{{α∈O^{×}|π^{m_{1},m_{2}+1}α+π^{n_{1},n_{2}+1}∈o⊕o}} M_{0}(α)dα

and (\frac{|π|}{|π|−1})^{2} times

∫_{{α∈O^{×}|π^{m_{1}+1,m_{2}+1}α+π^{n_{1}+1,n_{2}+1}∈O^{×}}} M_{0}(α)dα.

If K is an unramified extension and r_{1} = r_{2} = 0 it is the sum of

∫_{{α∈O^{∗}|π^{m_{1},m_{2}} α+π^{n_{1},n_{2}} ∈O}} M_{0}(α)dα

and

\frac{|π|^{2}}{|π|^{2} − 1} ∫_{\{α∈O^{×}|π^{m_{1}+1,m_{2}+1}α+π^{n_{1}+1,n_{2}+1}∈O^{×}\}} M_{0}(α)dα.

If K is a ramified extension and r_{1} = 0 it is the sum of

∫_{{α∈O^{×}|π^{m_{1}+f,m_{2}} α+π^{n_{1}+f,n_{2}} ∈O}} M_{0}(α)dα

and

\frac{|π|}{|π| − 1} ∫_{\{α∈O^{×}|π^{m_{1}+1+f,m_{2}} α+π^{n_{2}+1+f,n_{2}} ∈O^{×}\}} M_{0}(α)dα.

Lemma 1.5 Let M_{0} be a character of O^{×} of order (m_{1}, m_{2}) and let μ and ν be characters of o^{×} of orders n_{1} and n_{2} respectively. Suppose that M_{0} = χ_{0}μν on o^{×} and that the order of M_{0}^{−1}ν^{1+s} is (ℓ_{1}, ℓ_{2}) with n_{1} ≥ ℓ_{1} + ℓ_{2} + f. If ℓ_{1} ≥ ℓ_{2}, n_{1} ≥ n_{2}, and n_{1} + n_{2} = m_{1} + m_{2} + f then

\frac{Δ(M_{0}, π^{−m_{1}−f,−m_{2}})Δ(χ_{0}, π^{−f})}{|Δ(M_{0}, π^{−n_{1}−f,−m_{2}})Δ(χ_{0}, π^{−f})|} = \frac{M_{0}(Π^{(f_{+}m_{1}−n_{2})+s(m_{2}−n_{2})})}{χ(π^{n_{1}})} \frac{Δ(μ, π^{−n_{1}})Δ(ν, π^{−n_{2}})}{|Δ(μ, π^{−n_{1}})Δ(ν, π^{−n_{2}})|}.

Since both sides of this identity have the same absolute value all we need do is show that

Δ(M_{0}, π^{−m_{1}−f,−m_{2}})\overline{Δ(μ, π^{−n_{1}})} \overline{Δ(ν, π^{−n_{2}})}

is equal to the product of

\overline{Δ(χ_{0}, π^{−f})}M_{0}(Π^{(f+m_{1}−n_{1})+s(m_{2}−n_{2})})χ^{−1}(π^{n_{1}})

and a positive constant. As a start observe that it is equal to

∫_{O^{×}} dα ∫_{o^{×}} dβ ∫_{o^{×}} dγξ_{0}(S(π^{−m_{1}−f,−m_{2}}α) − π^{−n_{2}}β − π^{−n_{1}}γ)M_{0}(α)ν^{−1}(β)μ^{−1}(γ)

which equals

∫_{O^{×}} dα ∫_{o^{×}} dβ\{∫_{o^{×}} ξ_{0} [\frac{γ}{π^{n_{1}}} (π^{n_{1}}S(π^{−m_{1}−f,−m_{2}}α) − π^{n_{1}−n_{2}}β − 1)] χ_{0}(γ)dγ\}M(α)ν^{−1}(β). (A)

If f > 0 the integral with respect to γ is zero unless

π^{n_{1}}S(π^{−m_{1}−f,−m_{2}}α) − π^{n_{1}−n_{2}}β − 1 ∈ 𝔭^{n_{1}−f} − 𝔭^{n_{1}−f+1}.

However if this last condition is satisfied it is equal to

Δ(χ_{0}, π^{−f})χ_{0} (\frac{S(Π^{(n_{1}−m_{1}−f)+(n_{1}−m_{2})s}α) − π^{n_{1}−n_{2}}β − 1}{π^{n_{1}−f}})

Changing variables we see that the integral is equal to the product of

Δ(χ_{0}, π^{−f})M_{0}^{−1}(Π^{(n_{1}−m,−f)+(n_{2}−m_{2})s})

and

∫_{\{(α,β)|S(Π^{s(n_{1}−n_{2})}α)−π^{n_{1}−n_{2}} β−1∈𝔭^{n_{1}−f} −𝔭^{n_{1}−f+1}\}} M_{0}(α)ν^{−1}(β)χ_{0} (\frac{S(Π^{s(n_{1}−n_{2})}α) − π^{n_{1}−n_{2}}β − 1}{π^{n_{1}−f}}) dα dβ

If n_{1} > f and n_{1} > n − 2 then the restriction of M to o^{×} has order n_{1}. Thus m_{1} ≥ 2n_{1} and m_{1} + m_{2} + f > 2n_{1} > n_{1} + n_{2} contrary to assumption. Consequently we need only consider the case that n_{1} = f or n_{1} = n_{2}. If n_{1} > f or n_{1} > n_{2} then S(Π^{σ(n_{1}−n_{2})}α) − π^{n_{1}−n_{2}}β − 1 can belong to 𝔭^{n_{1}−f} − 𝔭^{n_{1}−f+1} only if S(Π^{σ(n_{1}−n_{2})}α) − 1 belongs to o^{×}.

Suppose that n_{1} = n_{2} = f and S(α) − 1 ∈ 𝔭. Replacing β by \frac{1}{β} in

∫_{{β|Sα−β−1∈o^{×}}} M_{0}(α)ν^{−1}(β)χ_{0}(S(α) − β − 1)dβ

we obtain

M_{0}(α)χ_{0}(−1) ∫_{o^{×}} ν(β)χ_{0}(β)χ_{0}(1 − β(S(α) − 1))dβ.

Since n_{1} = n_{2} = f, ℓ_{1} = 0 and M_{0} = ν^{1+σ}. Since M = χ_{0}μν on o^{×}, νχ_{0} = μ and the order of νχ_{0} is f. If β ∈ o^{×} and ν ∈ o_{f−1}^{×} then

1 − βγ(Sα − 1) ≡ 1 − β(Sα − 1) (mod 𝔭^{f})

Thus the above expression is equal to

M_{0}(α)χ_{0}(−1) ∫_{o^{×}/o_{f−1}^{×}} ν(β)χ_{0}(β)χ_{0}(1 + β(Sα − 1)){∫_{o_{f−1}^{×}} ν(γ)χ_{0}(γ)dγ}dβ = 0.

In all cases we can take the integral over

{(α, β)|S(Π^{σ(n_{1}−n_{2})}α) − 1 ∉ 𝔭, S(Π^{σ(n_{1}−n_{2})}α) − π^{n_{1}−n_{2}}β − 1 ∈ 𝔭^{n_{1}−f} − 𝔭^{n_{1}−f+1}} .

Replacing β by [S(Π^{σ(n_{1}−n_{2})}α) − 1]β we obtain

\{∫_{\{α|S(Π^{σ(n_{1}−n_{2})}α)−1∉𝔭\}} M_{0}(α)ν^{−1}χ_{0}(S(Π^{σ(n_{1}−n_{2})}α) − 1)dα\}\{∫_{\{β|π^{n_{1}−n_{2}} β−1∈𝔭^{n_{1}−f} −𝔭^{n_{1}−f+1}\}} ν^{−1}(β)χ_{0}(\frac{1 − π^{n_{1}−n_{2}}β}{π^{n_{1}−f}} )d\},

an expression we label (B).

Suppose n_{1} > f and consider the first integral. Replacing α by α(1 + v) with v ∈ Π^{n_{1}}O does not change the value of the integral. The integrand becomes

M_{0}(α)ν^{−1}(Sα − 1)χ_{0}(Sα − 1)M_{0}(1 + v)ν^{−1} (1 + \frac{S(αv)}{Sα − 1}) χ_{0} (1 + \frac{S(αv)}{Sα − 1}) .

Since n_{1} ≥ ℓ_{1}, M_{0}(1 + v) = v(1 + Sv + Nv) = ν(1 + Sv). Moreover [\frac{n_{1}+f}{2} ] ≥ [\frac{2f}{2} ] = f so that χ_{0} (1 + \frac{S(αν)}{Sα−1}) = 1. Also [\frac{n_{1}+f}{2} ] ≥ [\frac{n_{1}+1}{2}] ≥ \frac{n_{1}}{2} so that

v^{−1} (1 + \frac{S(αv)}{Sα − 1}) = ν (1 − \frac{S(αv)}{Sα − 1})

and

v(1 + Sv)ν^{−1} (1 + \frac{S(αv)}{Sα − 1}) = ν(1 + S(δv))

if δ = 1 − \frac{α}{Sα−1}. Integrating over Π^{n_{1}}O we obtain 0 unless |δ| = |π|^{s} and [\frac{s+n_{1}+f}{2} ] ≥ n_{1}, that is, s + n_{1} + f ≥ 2n_{1} or s ≥ n_{1} − f when we obtain |π|^{n_{1}}. Since |δ| = |α − 1| we can in all cases write the first integral of (B) as

∫_{{α∈O_{n_{1}−f,0}^{×}|S(Π^{σ(n_{1}−n_{2})}α)−1∉𝔭}} M_{0}(α)ν^{−1}χ_{0}(S(Π^{σ(n_{1}−n_{2})}α) − 1)dα.
Since n_{1} − f ≥ ℓ_{1} and χ_{0}(Nα) = 1 this may be written as

∫_{\{α∈O_{n_{1}−f,0}^{×}|S(Π^{σ(n_{1}−n_{2})}α)−1∉𝔭\}} ν^{−1}χ_{0} (π^{n_{1}−n_{2}} − N (\frac{Π^{σ(n_{1}−n_{2})}α − 1}{α})) dα.

Set Π^{n_{1}−f}γ = \frac{Π^{σ(n_{1}−n_{2})}α−1}{α} so that α = \frac{1}{Π^{σ(n_{1}−n_{2})}−Π^{n_{1}−f} γ}. The integral is the product of a positive constant and

∑_{{γ∈O/Π^{f} O|π^{n_{1}−n_{2}} −π^{n_{1}−f} Nγ∉𝔭}} ν^{−1}χ_{0}(π^{n_{1}−n_{2}} − π^{n−1−f}Nγ).  (C)

If n_{1} > n_{2} every γ appearing in this sum is a unit and the sum is equal to

∑_{\{β∈O^{∗}/O_{n_{1}}^{∗} |π^{n_{1}−n_{2}} −β∈𝔭^{n_{1}−f} −𝔭^{n_{1}−f+1}\}} ν^{−1}χ_{0}(β) [1 + χ_{0} (\frac{π^{n_{1}−n_{2}} − β}{π^{n_{1}−f}} )] .

Since this sum is taken over all of o^{×} it is equal to

∑_{\{β∈o^{×}/o_{n_{1}}^{×} |π^{n_{1}−n_{2}} −β∈𝔭^{n_{1}−f} −𝔭^{n_{1}−f+1}\}} ν^{−1}χ_{0}(β)χ_{0} (\frac{π^{n_{1}−n_{2}} − β}{π^{n_{1}−f}} ) .

If n_{1} = n_{2} then (C) is the sum of

∑_{{γ∈O^{×}/O_{f,0}^{×}|1−π^{n_{1}−f} Nγ∉𝔭}} ν^{−1}χ_{0}(1 − π^{n_{1}−f}Nγ)

and

∑_{r=1}^{f} ∑_{γ∈O^{×}/O_{f−r,0}^{×}} ν^{−1}χ_{0}(1 − π^{n_{1}+r−f}Nγ).

Since the map γ → Nγ defines an isomorphism of O^{×}/O_{f−r,0}^{×} and o^{×}/o_{f−r}^{×} the latter sum is equal to

∑_{𝔭/𝔭^{f}} ν^{−1}χ_{0}(1 − w).

Since

∑_{o^{×}/o_{f}^{×}} ν^{−1}χ_{0}(β) = 0

we can subtract it from (C) without changing (C). The result is

∑_{\{β∈o^{×}/o_{n_{1}}^{×} |π^{n_{1}−n_{2}} −β∈𝔭^{n_{1}−f} −𝔭^{n_{1}−f+1}\}} ν^{−1}χ_{0}(β) (\frac{π^{n_{1}−n_{2}} − β}{π^{n_{1}−f}} ) .  (D)

Thus (C) and (D) are equal in all cases.

Replace β by \frac{1}{β} in the second integral of (B) to see that it is equal to the product of a positive constant and

χ_{0}(−1) \{∑_{\{β∈o^{×}/o_{n_{1}}^{×} |π^{n_{1}−n_{2}} −β∈𝔭^{n_{1}−f} −𝔭^{n_{1}−f+1}\}} νχ_{0}^{−1}(β)χ_{0} (\frac{π^{n_{1}−n_{2}} − β}{π^{n_{1}−f}} )\} .

This is the product of χ_{0}(−1) and the complex conjugates of (D). Since \overline{Δ(χ_{0}, π^{−f})} = χ_{0}(−1)Δ(χ_{0}, π^{−f}) the lemma is proved for f > 0.

If f = 0 then in the integral (A) we may replace χ_{0}(γ) by 1. If n_{1} = 0 then n_{2} = 0 and m_{1} = m_{2} = 0 so that μ, ν and M_{0} are all trivial. The lemma is also; so we suppose n_{1} > 0. If n_{1} > n_{2} then K = k ⊕ k. Let M_{0}(α ⊕ β) = μ_{1}(α)ν_{1}(β). Then m_{1} is the order of μ_{1} and m_{2} is the order of ν_{1}. Since μ_{1}ν_{1} = μν either m_{1} ≥ n_{1} or m_{2} ≥ n_{1}. if m_{1} ≥ n_{1} then ℓ_{1} = m_{1} so that ℓ_{2} = 0. Then ν_{1} = ν and μ_{1} = μ. If m_{2} ≥ n_{1} then ℓ_{2} = m_{2} so that ℓ_{1} = 0 which is contrary to the assumption that ℓ_{1} ≥ ℓ_{2}. Thus the lemma is trivial if n_{1} > n_{2}; so we suppose that n_{1} = n_{2}. Then m_{1} = m_{2} = n_{1}.

The integral (A) is equal to

∫_{O^{×}} dα ∫_{o^{×}} dβ\{∫_{o^{×}} ξ_{0} [\frac{γ}{π^{n_{1}}} (Sα − β − 1)] dγ\}M_{0}(α)ν^{−1}(β).

The inner integral is different from zero if and only if Sα − β − 1 ∈ 𝔭^{n−1}. If n_{1} > 1 this mplies that Sα − 1 ∉ 𝔭. Set

M_{0}(α)ν^{−1}(β) ∫_{o^{×}} ξ_{0} [\frac{γ}{π^{n_{1}}} (Sα − β − 1)] dγ = ψ(α, β).

If Sα − 1 ∈ 𝔭 and n_{1} = 1 then ψ(α, β) = \frac{|π|}{|π|−1}M_{0}(α)ν^{−1}(β). Since n_{1} = n_{2}

∫_{o^{×}} ψ(α, β)dβ = 0

if Sα − 1 ∈ 𝔭. Thus if

φ(x) = ∫_{o^{×}} ξ_{0} (\frac{γx}{π^{n_{1}}} ) dγ

the integral (A) is equal to

∫_{{α|Sα−1∉𝔭}} ∫_{o^{×}} M_{0}(α)ν^{−1}(Sα − 1)ν^{−1}(β)φ((1 − β)(Sα − 1))dβ dα

If Sα − 1 ∉ 𝔭 then φ((1 − β)(Sα − 1)) = φ(1 − β). Moreover

∫_{o^{×}} M_{0}(α)ν^{−1}(Sα − 1)ν^{−1}(β)φ(1 − β)dβ

is equal to the product of M_{0}(α)ν^{−1}(Sα − 1) and

∫_{o_{n_{1}}^{×}} ν^{−1}(β)dβ + \frac{|π|}{|π| − 1} ∫_{o_{n_{1}−1}^{t}imeso_{n_{1}}^{×}} ν^{−1}(β)dβ.

The first integral is equal to the measure of o_{n_{1}}^{×}. The second is equal to

−\frac{|π|}{|π| − 1} ∫_{o_{n_{1}}^{×}} ν^{−1}(β)dβ = \frac{|π|}{1 − |π|} measure o_{n_{1}}^{×}.

Thus the integral (A) is the product of a positive constant and

∫_{\{α|Sα−1∉𝔭\}} M_{0}(α)ν^{−1}(Sα − 1)dα = ∫_{\{α|Sα−1∉𝔭\}} M_{0}(α)ν^{−1}(Nα)ν^{−1} (1 − N (\frac{α − 1}{α})) dα.

If K = k ⊕k and ℓ_{2} = 0 the lemma is trivial. Suppose K = k ⊕k and ℓ_{2} > 0. Let α = α_{1} ⊕α_{2}. If y is in 𝔭^{ℓ_{2}} then replacing α by α_{1} ⊕ α_{2}(1 + y) in the integrand does not change the value of the integral. The integrand becomes

M_{0}(α)ν^{−1}(Nα)ν^{−1} (1 − N (\frac{α − 1}{α}) − \frac{α_{1} − 1}{α_{1}} \frac{y}{α_{2}(1 + y)}) .

The integral of this over 𝔭^{ℓ_{2}} is the measure of 𝔭^{ℓ_{2}} or zero according as α_{1} − 1 ∈ 𝔭^{n_{1}−ℓ_{2}} or not. The same observation applies to the first variable. Thus the integral is equal to

∫_{O_{n_{1}−ℓ_{2},n_{2}−ℓ_{1}}^{×}} M_{0}(α)ν^{−1}(α)ν^{−1} (1 − N (\frac{α − 1}{α})) dα.

Since n_{1} − ℓ_{2} ≥ ℓ_{1}, n_{2} − ℓ_{1} ≥ ℓ_{2} and n_{1} − ℓ_{2} + n_{2} − ℓ_{1} ≥ n_{2} the integrand is identically one. Thus the lemma is proved if K = k ⊕ k.

If K is an unramified extension let k_{1} = k_{2} be the smallest integer greater than or equal to \frac{n_{1}}{2} . Let y ∈ π^{k_{1},k_{2}}O. Replacing α by α(1 + y) in the integrand does not change the value of the integral. Since k_{1} ≥ ℓ_{1} and 2k_{1} ≥ n_{1} the integrand becomes

M_{0}(α)ν^{−1}(Nα)ν^{−1} (1 − N (\frac{α − 1}{α})) ν^{−1} (1 + γS (\frac{(α^{s} − 1}{1 + y}))

if γ = −_{\frac{1}{Nα(1−N(\frac{α−1}{α}))}}. The integral of this expression over π^{k_{1},k_{2}}O is the measure of π^{k_{1},k_{2}}O or zero according as α ∈ O_{n_{1}−k_{1},n_{2}−k_{2}}^{×} or not. Thus our integral is equal to

∫_{\{α∈O_{n_{1}−k_{1},n_{2}−k_{2}}^{×} |Sα−1∉𝔭\}} M_{0}(α)ν^{−1}(Nα)ν^{−1} (1 − N (\frac{α − 1}{α})) dα.

Since n_{1} − k_{1} ≥ ℓ_{1} this is equal to

∫_{\{α∈O_{n_{1}−k_{1},n_{2}−k_{2}}^{×} |Sα−1∈𝔭\}} ν^{−1} (1 − N (\frac{α − 1}{α})) dα

If n_{1} is even, k_{1} = \frac{n_{2}}{2} and the integrand is identically one. Thus the lemma is proved in this case. If n_{1}is odd set \frac{α−1}{α} = π^{n_{1}−k_{2}}β so that α = \frac{1}{1−π^{n_{1}−k_{1}}} β. Since 2(n_{1} − k_{1}) = n_{1} − 1 when n_{1} is odd this integral is the product of a positive constant and

∑_{{β∈O/Π_{O}|π^{n_{1}−1}Nβ≠1}} ν^{−1}(1 − π^{n_{1}−1}Nβ)

If x ≢ (mod 𝔭) the equation Nβ = x(mod 𝔭) has \frac{|π|+1}{|π|} solutions modulo ΠO, otherwise it has just one. Thus if n_{1} > 1 the sum equals

\frac{|π| + 1}{|π|} ∑_{x∈o/𝔭} ν^{−1}(1 − π^{n_{1}−1}x) − \frac{1}{|π|}ν^{−1}(1) = −\frac{1}{|π|}

and if n_{1} = 1 it equals

\frac{|π| + 1}{|π|} ∑_{o^{×}/o_{1}^{×}} ν^{−1}(x) − \frac{1}{|π|}ν^{−1}(1) = −\frac{1}{|π|}

The lemma is completely proved.

If K = k ⊕ k we set ε = 1; if K is an unramified extension of k we set ε = −1, and if K is a ramified extension of k we set ε = 0. If M is generalized a character of K^{×}, if M_{0} is its restriction to O^{×}, and ν is a character of o^{×} set

T (M, ν, n) = (1 − |π|)(1 − ε|π|)\frac{Δ(χ_{0}, π^{−f})}{|Δ(χ_{0}, π^{−f}|}|π|^{\frac{n+f}{2}} ∑_{n_{1}+n_{2}=n} M((π^{n_{1},n_{2}})^{s})Δ(M_{0}^{−1}ν^{1+s}, π^{n_{1},n_{2}})

where the sum is taken over all n_{1}, n_{2} for which π^{n_{1},n_{2}} is defined.

Lemma 1.6 Let ω and M be homomorphisms of k^{×} and K^{×} respectively into ℂ^{×}. Suppose that the restriction of M to k^{×} is ωχ. Let ν and η be characters of o^{×} and let ω_{0} be the restriction of ω to o^{×}. Suppose that the order n of νηω_{0}^{−1} is positive. Then, for all integers k and ℓ,

ω(π^{n})T (M, η; k − n)T (M, ν, ℓ − n) = \frac{Δ(νηω_{0}^{−1}, π^{−n})}{|Δ(νηω_{0}^{−1}, π^{−n})|^{2}} ∑_{ρ} Δ(ηρ^{−1}, π^{k})Δ(νρ^{−1}, π^{ℓ})T (M, ρ, k + ℓ)

where the sum is over all characters of o^{∗}.

The formula of the lemma will be referred to as formula (E). Notice that all but a finite number of terms in the sum on the right are zero. The sum on the right is the product of (1 − |π|)(1 − ε|π|)|π|^{\frac{k+ℓ+f}{2}} \frac{Δ(χ_{0},π^{−f})}{|Δ(χ_{0},π^{−f})|} and

∑_{n_{1}+n_{2}=k+ℓ} M((π^{n_{1},n_{2}} )^{s}) ∑_{ρ} ∫_{O^{×}} dα ∫_{o^{×}} dβ ∫_{o^{∗}} dαξ_{0}(S(π^{n_{1},n_{2}} α)+π^{ℓ}β+π^{k}γ)M_{0}^{−1}(α)ρ(\frac{Nα}{βγ})ν(β)η(γ).

Given ν, η, M, k, and ℓ there is a number m such that this integral is zero if the order of ρ is greater than m. Thus we may restrict the sum to a sum over the characters of o^{×}/o_{m}^{×}. Replace α by βα, γ by βγ, and take one of the summations under the integral sign to obtain, if μ is the restriction of M_{0} to o^{×},

∑_{n_{1}+n_{2}=k+ℓ} M((π^{n_{1},n_{2}} )^{s})∫_{O^{×}} dα∫_{o^{×}} dβ∫_{o^{×}} dγξ_{0}[(S(π^{n_{1},n_{2}} α)+π^{ℓ}+π^{k}γ)]M^{−1}(α)η(Nα)νημ^{−1}(β)\{∑_{ρ} ρη^{−1}(\frac{Nα}{γ})\}.

The summation over ρ is different from zero if and only if γ ≡ Nα(mod 𝔭^{m}). If K = k ⊕ k set λ(α) = N(π^{n_{1}−ℓ,n_{2}}α + π^{0,ℓ}); if K is an unramified extension of k set λ(α) = π^{k}N (α + π^{\frac{ℓ−k}{2} , \frac{ℓ−k}{2}} ); if K is a ramified extension set λ(α) = π^{k}N(α + Π^{−k+sℓ}). The above expression is equal to

∑_{n_{1}+n_{2}=k+ℓ} M((π^{n_{1},n_{2}})^{s}) ∫_{O^{×}} dα ∫_{o^{×}} dβξ_{0}(βλ(α))M_{0}^{−1}(α)η(Nα)νημ^{−1}(β).

If the order r of νημ^{−1} is not zero this is equal to

Δ(νημ^{−1}, π^{−r}) ∑_{n_{1}+n_{2}=k+ℓ} M((π^{n_{1},n_{2}})^{s}) ∫_{{α∈O^{×}|π^{r}λ(α)∈o^{×}}} M_{0}^{−1}(α)η(Nα)μν^{−1}η^{−1}(π^{r}λ(α))dα. (F)

If the order of νημ^{−1} is zero it is equal to

∑_{n_{1}+n_{2}=k+ℓ} M((π^{n_{1},n_{2}})^{s}) \{∫_{\{α∈O^{×}|λ(α)∈o\}} M_{0}^{−1}(α)η(Nα)dα + \frac{|π|}{|π| − 1} ∫_{\{α∈O^{×}|πλ(α)∈o^{×}\}} M_{0}^{−1}(α)η(Nα)dα ,

an expression that will be labelled (G).

If K is an unramified extension of k, r = n > 0 and the expression (F) is zero unless k − ℓ and k − n are even. There is only one term in (F) and the corresponding integral is

∫_{\{α∈O^{×}|π^{r+ \frac{k−r}{2} ,r+ \frac{k−r}{2}} α+π^{r+ \frac{ℓ−r}{2} , \frac{ℓ−r}{2}} ∈O^{×}\}} M_{0}^{−1}(α)η(Nα)μν^{−1}η^{−1} (N (π^{r+ \frac{k−r}{2} ,r+ \frac{k−r}{2}} α + π^{r+ \frac{ℓ−r}{2} ,r+ \frac{ℓ−r}{2}} )) dα

Set M_{0}^{s}(α) = M_{0}(α^{s}). Since M_{0}^{−1}η^{1+s} · M_{0}^{−s}ν^{1+s} = (μ^{−1}νη)^{1+s} this integral is equal to

\frac{Δ (M_{0}^{−1}η^{1+s}, π^{\frac{k−n}{2} , \frac{k−n}{2}} ) Δ (M_{0}^{−1}ν^{1+s}, π^{\frac{ℓ−n}{2} , \frac{ℓ−n}{2}} )}{Δ((μ^{−1}νη)^{1+s}, π^{−r,−r}).}

Putting everything together and appealing to Lemmas 1.1 and 1.5, we see that the right side of (E) is equal to

χ(π^{n})M ((π^{\frac{k+ℓ}{2} , \frac{k+ℓ}{2}} )^{s}) (1 − |π|^{2})^{2}|π|^{\frac{k+ℓ}{2} −n}Δ (M_{0}^{−1}η^{1+s}, π^{\frac{k−n}{2} , \frac{k−n}{2}} ) Δ (M_{0}^{−1}ν^{1+s}, π^{\frac{ℓ−n}{2} , \frac{ℓ−n}{2}} )

Since Δ(M^{−σ}, π^{m,m}) = Δ(M^{−1}, π^{m,m}) it is equal to the left side.

If K is a ramified extension of k and r > 0 there is only one term in the sum (F) and the integral appearing in that term is

∫_{{α∈O^{×}|Π^{r+k}α+Π^{r+sℓ}∈O^{×}}} M_{0}^{−1}(α)η(Nα)μν^{−1}η^{−1} (N (Π^{r+k}α + Π^{r+sℓ})) dα.

Replace α by Π^{−ℓ+sℓ}α to obtain

M (Π^{ℓ−sℓ}) ∫_{{α∈O^{×}|Π^{r+k}α+Π^{r+ℓ}∈O^{×}}} M_{0}^{−1}(α)η(Nα)μν^{−1}η^{−1} (N (Π^{r+k}α + Π^{r+ℓ})) dα.

If r > f then r = n, the order of (μν^{−1}η^{−1})^{1+s} is 2n − f, 0, r + k = 2n − f + (k − n + f), r + ℓ = 2n − f + (k − n + f). If r < f then n = f, the order of (μν^{−1}η^{−1})^{1+s} is (r, 0), and r + k = r + (k − n + f), r + ℓ = r + (ℓ − n + f). If r = f then n ≤ f, the order of (μν^{−1}η^{−1})^{1+s} is (n, 0), r + k = n + (k − n + f), r + ℓ = n + (k − ℓ + f). According to Lemma 1.4 the above expression is equal to

M (Π^{ℓ−sℓ}) \frac{Δ(M_{0}^{−1}η^{1+s}, π^{k−n,0})Δ(M_{0}^{−s}ν^{1+s}, π^{ℓ−n,0})}{Δ((μ^{−1}νη)^{1+s}, π^{−r_{1},−r_{2}})}

if (r_{1}, r_{2}) is the order of (μ^{−1}νη)^{1+s}. Observe that

Δ(M_{0}^{−1}ν^{1+s}, π^{ℓ−n,0}) = M (Π^{(n−ℓ)(1−s)}) Δ(M_{0}^{−1}ν^{1+s}, π^{(ℓ−n,0)}).

Appealing to Lemmas 1.1 and 1.5, we see that, if r > 0, the right side of (E) is equal to

M(Π^{n(1+s)})M(Π^{(k+ℓ−2n)s})(1−|π|)^{2}|π|^{a} \{\frac{Δ(χ_{0}, π^{−f})}{|Δ(χ_{0}π^{−f})|}\}^{2} Δ(M_{0}^{−1}η^{1+s}, π^{k−n,0})Δ(M_{0}^{−1}ν^{1+s}, π^{ℓ−n,0})

with a = \frac{k+ℓ+f}{2} − \frac{n}{2} + \frac{r}{2} − \frac{r_{1}}{2} = \frac{k−n+f}{2} + \frac{ℓ−n+f}{2} . This is obviously equal to the left side. If r = 0 the expression (G) is equal to the product of M(Π^{(k+ℓ)s}) and

∫_{\{α∈O^{×}|Π^{k}α+Π^{sℓ}∈O\}} M_{0}^{−1}(α)η(Nα)dα + \frac{|π|}{|π| − 1} ∫_{\{α∈O^{×}|Π^{k+1}α+Π^{1+sℓ}∈O^{×}\}} M_{0}^{−1}(α)η(Nα)dα

After a change of variables this becomes

M(Π^{ℓ−sℓ}) \{∫_{\{α∈O^{×}|Π^{k}α+Π^{ℓ}α∈O\}} M_{0}^{−1}(α)(η(Nα)dα +∫_{\{α∈O^{×}|Π^{k+1}α+Π^{ℓ+1}∈O^{×}\}} M_{0}^{−1}(α)η(Nα)dα\frac{|π|}{|π| − 1}\} .

Since (μ^{−1}νη)^{1+s} will also be trivial this is equal to

M(Π^{ℓ−2s})Δ(M_{0}^{−1}η^{1+s}, π^{k−n,0})Δ(M_{0}^{−s}η^{1+s}, π^{ℓ−n,0})

because n = f in this case. Thus the right side of (E) is equal to

M(Π^{n(1+s)})M(Π^{(k+ℓ−2n)s})(1−|π|)^{2}|π|^{\frac{k+ℓ}{2}}\{\frac{Δ(χ_{0}, π^{−f})}{|Δ(χ_{0}, π^{−f})|}\}^{2}Δ(M_{0}^{−1}η^{1+s}, π^{k−n,0})Δ(M_{0}^{−1}ν^{1+s}, π^{ℓ−n,0}).

Since χ(Π^{n(1+s)}) = 1 and Π^{n(1+s)} = π^{n}, it is equal to the left side.

It remains to consider the case that K = k ⊕ k. Then r − n is not zero and (F) is equal to the product of Δ(νημ^{−1}, π^{n}) and ∑_{n_{1}+n_{2}=k+ℓ} ∑_{m_{1}+m_{2}=r} M(π^{n_{1},n_{2}}) ∫_{{α∈O^{×}|π^{n_{1}+m_{1}−ℓ,n_{2}+m_{2}} α+π^{m_{1},m_{2}+ℓ}∈O^{×}}} M_{0}^{−1}(α)η(Nα)μν^{−1}η^{−1}(N(π^{n_{1}+m_{1}−ℓ,n_{2}+m_{2}}α+π^{m_{1},m_{2}+ℓ}))dα. This is equal to ∑_{n_{1}+n_{2}=k+ℓ} ∑_{m_{1}+m_{2}=r} M(π^{n_{2},n_{1}})Δ(M_{0}^{−1}η^{1+s}, π^{n_{1}+m_{1}−n−ℓ,n_{2}+m_{2}−n})Δ(M_{0}^{−1}ν^{1+s}, π^{m_{2}+ℓ−n,m_{1}−n)} divided by Δ((μ^{−1}νη)^{1+s}, π^{−n,−n}). Replace m_{1} by m_{1} + n, m_{2} by m_{2} − ℓ + n, interchange the order of summation and replace n_{1} by n_{1} − m_{1} + ℓ, n_{2} by n_{2} − m_{2} + ℓ to see that the sum is equal to

∑_{m_{1}+m_{2}=ℓ−n} ∑_{n_{2}+n_{1}=k−n} M(π^{n_{1},n_{2}})M(π^{m_{1},m_{2}})M(π^{n,n})Δ(M_{0}^{−1}η^{1+s}, π^{n_{1},n_{2}})Δ(M_{0}^{−1}η^{1+s}, π^{m_{2},m_{1}})

Appealing to Lemmas 1.1 and 1.5, we see that the right side of (E) is equal to

ω(π^{n})(1−|π|)^{4}|π|^{\frac{k+ℓ}{2} +n} ∑_{n_{1}+n_{2}=k−n} ∑_{m_{1}+m_{2}=ℓ−n} M(π^{n_{2},n_{1}} )M(π^{m_{2},m_{1}} )Δ(M_{0}^{−1}η^{1+s},π^{n_{1},n_{2}} )Δ(M_{0}^{−1}ν^{1+s},π^{m_{1},m_{2}} ).

This is of course just the left side.

Lemma 1.7 Let ω and M be homomorphisms of k^{×} and K^{×} respectively into ℂ^{×}. Suppose that the restriction of M to k^{×} is ωχ. Let ν and η be characters of o^{×} and let ω_{0} be the restriction of ω to o^{×}. If νηω_{0}^{−1} is trivial then for all integers k and ℓ

\begin{matrix}∑_{−∞}^{−2} −ω(π^{−m}) & T (M, η, k+m)T (M, ν, ℓ+m)+\frac{1}{|π|−1}ω(π)T (M, η, k−1)T (M, ν, ℓ−1)+ω_{0}(−1)δ_{ℓ,k}ω(π^{ℓ}) \\ = ∑_{ρ} Δ(η, ρ^{−1}, π^{k})Δ(νρ^{−1}, π^{ℓ})T (M, ρ, k + ℓ) \end{matrix}

δ_{ℓ,k} is of course Kroneckev’s delta. For brevity denote the left side by L_{k,ℓ} and the right side by R_{k,ℓ}. Suppose at first that k ≪ 0 and ℓ ≪ 0. Then L_{k,ℓ} = ω_{0}(−1)ω(π^{ℓ})δ_{ℓ,k}. The only terms which contribute anything to the right hand side are those for which order(ρ) = −k and order(ρ) = −ℓ. Thus the right side is zero if k ≠ ℓ. Suppose order(ρ) = −ℓ and k = ℓ. The order, (r_{1}, r_{2}), of M_{0}^{−1}ρ^{1+s} is (−ℓ, −ℓ) if K = k ⊕ k or K is an unramified extension of k. It is (−2ℓ − f, 0) if K is a ramified extension of k. Moreover if n_{1} + n_{2} = k + ℓ

Δ(M_{0}^{−1}ρ^{1+s}, π^{n_{1},n_{2}}) = 0

if −n_{1} ≠ r_{1}+f. The orders of η^{−1}ρ and ν^{−1}ρ are both −ℓ. The orders of (M_{0}^{−1}ρ^{1+s})^{−1}(η^{−1}ρ)^{1+s} = M_{0}η^{−1−s} and (M_{0}^{−1}ρ^{1+s})^{−1}(ν^{−1}ρ)^{1+s} = M_{0}ν^{−1−s} are independent of ℓ. Moreover the restriction of M_{0}^{−1}ρ^{1+s} to o^{×} is equal to χ_{0}(η^{−1}ρ)(ν^{−1}ρ). According to Lemma 1.5

Δ(M_{0}^{−1}ρ^{1+s}, π^{−r_{1}−f_{1}−r_{2}}) = \frac{M(Π^{ℓ−sℓ})}{χ(π^{ℓ})}Δ(η^{−1}ρ, π^{ℓ})Δ(ν^{−1}ρ, π^{ℓ})\frac{\overline{Δ(χ_{0}, π^{−f})}}{|Δ(χ_{0}, π^{−f})|} \frac{1 − |π|}{1 − ε|π|}|π|^{\frac{r_{1}+r_{2}+2ℓ}{2}} .

Since Δ(η^{−1}ρ, π^{ℓ}) = ηρ^{−1}(−1)\overline{Δ(ηρ^{−1}, π^{ℓ})} the term corresponding to a ρ with order ρ = −ℓ is

\frac{M(Π^{ℓ−sℓ})M((π^{−r_{1}−f,−r_{2}})^{s})}{χ(π^{ℓ})}νη(−1)\frac{1}{|π|^{ℓ}(1 − |π|)^{2}} .

This is clearly equal to

\frac{M(π^{ℓ})}{χ(π^{ℓ})}ω_{0}(−1)\frac{1}{|π|^{ℓ}(1 − |π|)^{2}} = \frac{ω(π^{ℓ})ω_{0}(−1)}{|π|^{ℓ}(1 − |π|)^{2}}

Since the number of such characters is |π|^{ℓ}(1 − |π|)^{2} the lemma is valid if k ≪ 0 and ℓ ≪ 0. Thus to prove the lemma it is enough to show that

L_{k+1,ℓ+1} − ω(π)L_{k,ℓ} = R_{k+1,ℓ+1} = ω(π)R_{k,ℓ}

for all k and ℓ. The left-hand side is equal to

\frac{ω(π)}{|π| − 1}T (M, η, k)T (M, ν, ℓ) − \frac{|π|}{|π| − 1}ω^{2}(π)T (M, η, k − 1)T (M, ν, k − 1).

Suppose K is an unramified extension. If k − ℓ is odd both of these terms are zero and so is the right side. We suppose then that k − ℓ is even. If k is even only the first of these two terms can be different from zero. If k is odd only the second can be. Remembering that χ(π) = −1 so that ω(π) = −M(π^{1,1}) we apply formula (G) to see that the right side is the product of (1 − |π|^{2})|π|^{\frac{k+ℓ}{2}} M(π^{\frac{k+ℓ}{2} , \frac{k+ℓ}{2} +1}) and

\begin{matrix}|π|∫_{\{α∈O^{×}|π^{k+1}N(α+π^{\frac{ℓ−k}{2} , \frac{ℓ−k}{2}} )∈o\}} M_{0}^{−1}(α)η(Nα)dα + \frac{|π|^{2}}{|π|−1}∫_{\{α∈O^{×}|π^{\frac{k+2}{2} , \frac{k+2}{2}} α+π^{\frac{ℓ+2}{2} , \frac{ℓ+2}{2}} ∈O^{×}\}} M_{0}^{−1}(α)η(Nα)dα \\ +∫_{\{α∈O^{×}|π^{k}N(α+π^{\frac{ℓ−k}{2} , \frac{ℓ−k}{2}} )∈o\}} M_{0}^{−1}(α)η(Nα)dα + \frac{|π|}{|π|−1}∫_{\{α∈O^{×}|π^{\frac{k+1}{2} , \frac{k+1}{2}} α+π^{\frac{ℓ+1}{2} , \frac{ℓ+1}{2}} ∈O^{×}\}} M_{0}^{−1}(α)η(Nα)dα. \end{matrix}

If k is even π^{k+1}N (α + π^{\frac{ℓ−k}{2} , \frac{ℓ−k}{2}} ) ∈ o if and only if π^{k}N (α + π^{\frac{ℓ−k}{2} , \frac{ℓ−k}{2}} ) ∈ o so that, if k is even, this expression equals

(|π|+1) \{∫_{\{α∈O^{×}|π^{\frac{k}{2} , \frac{k}{2}} α+π^{\frac{ℓ}{2} , \frac{ℓ}{2}} ∈O^{×}\}} M_{0}^{−1}(α)η(Nα)dα + \frac{|π|^{2}}{|π|^{2} − 1} ∫_{\{α∈O^{×}|π^{\frac{k+2}{2} , \frac{k+2}{2}} α+π^{\frac{ℓ+2}{2} , \frac{ℓ+2}{2}} ∈O^{×}\}} M_{0}^{−1}(α)η(Nα)dα

which equals

(|π| + 1)Δ (M_{0}^{−1}η^{1+s}, π^{\frac{k}{2} , \frac{k}{2}} ) Δ (M_{0}^{−1}ν^{1+s}, π^{\frac{ℓ}{2} , \frac{ℓ}{2}} ) .

The identity, for even k, follows immediately. If k is odd the expression above simplifies to

(|π|+1) \{∫_{\{α∈O^{×}|π^{\frac{k−1}{2} , \frac{k−1}{2}} α+π^{\frac{ℓ−1}{2} , \frac{ℓ−1}{2}} ∈O\}} M_{0}^{−1}(α)η(Nα)dα + \frac{|π|^{2}}{|π|^{2} − 1} ∫_{\{α∈O^{×}|π^{\frac{k+1}{2} , \frac{k+1}{2}} α+π^{\frac{ℓ+1}{2} , \frac{ℓ+1}{2}} ∈O^{×}\}} M_{0}^{−1}(α)η(Nα)dα\}

which equals

(|π| + 1)Δ (M_{0}^{−1}η^{1+s}, π^{\frac{k−1}{2} , \frac{k−1}{2}} ) Δ (M_{0}^{−1}ν^{1+s}, π^{\frac{ℓ−2}{2} , \frac{ℓ−2}{2}} ) .

The identity, for odd k, follows immediately.

Suppose f > 0. If μ is the restriction of M_{0} to o^{×} then νημ^{−1} = χ_{0}. According to (F), R_{k,ℓ} is equal to the product of

(1 − |π|)|π|^{\frac{k+ℓ+f}{2}} \frac{Δ(χ_{0}, π^{−f})}{|Δ(χ_{0}, π^{−f})|}Δ(χ_{0}, π^{−f})M(Π^{s(k+ℓ)}) = χ_{0}(−1)|π|^{\frac{k+ℓ}{2} +f}M(Π^{s(k+ℓ)})

and

M(Π^{ℓ−sℓ}) ∫_{{α∈O^{×}|Π^{f+k}α+Π^{f+ℓ}∈O^{×}}} M_{0}^{−1}(α)η(Nα)dα.

On the other hand

\frac{ω(π)}{|π| − 1}T (M, η, k)T (M, ν, ℓ) − \frac{|π|}{|π| − 1}ω^{2}(π)T (M, η, k − 1)T (M, ν, ℓ − 1)

is equal to the product of (1 − |π|)^{2}|π|^{\frac{k+ℓ}{2} +f}χ_{0}(−1)M(Π^{ℓ+1+s(k+1))} and

\begin{matrix}\frac{1}{|π| − 1} & \{∫_{\{α∈O^{×}|Π^{f+k}α+π^{f+ℓ}∈O\}} M_{0}^{−1}(α)η(Nα)dα + \frac{|π|}{|π| − 1} ∫_{\{α∈O^{×}|Π^{f+k+1}α+π^{f+ℓ+1}∈O^{×}\}} M_{0}^{−1}(α)η(Nα)dα\} \\ − \frac{1}{|π| − 1} \{∫_{\{α∈O^{×}|Π^{f+k−1}α+Π^{f+ℓ−1}∈O\}} M_{0}^{−1}(α)η(Nα)dα + \frac{|π|}{|π| − 1} ∫_{\{α∈O^{×}|Π^{f+k}α+Π^{f+ℓ}∈O^{×}\}} M_{0}^{−1}(α)η(Nα)dα\} \end{matrix}

Some simple rearrangements show that this is equal to

\frac{−1}{(|π| − 1)^{2}} ∫_{\{α∈O^{×}|Π^{f+k}α+Π^{f+ℓ}∈O^{×}\}} M_{0}^{−1}(α)η(Nα)dα + \frac{|π|}{(|π| − 1)^{2}} ∫_{\{α∈O^{×}|Π^{f+k+1}α+Π^{f+ℓ+1}∈O^{×}\}} M_{0}^{−1}(α)η(Nα)dα.

The identity follows immediately.

Finally we have to treat the case that K = k ⊕ k. It is enough to verify that

∑_{p=0}^{∞} |π|^{p}ω(π^{p}) {L_{k+1−p,ℓ+1−p} − ω(π)L_{k−p,ℓ−p}} = ∑_{p=0}^{∞} |π|^{p}ω(π^{p}) {R_{k+1−p,ℓ+1−p} − ω(π)R_{k−p,ℓ−p}} .

The left side is equal to

\frac{ω(π)}{|π| − 1}T (M, η, k)T (M, ν, ℓ).

For brevity set

ψ(n_{1}, n_{2}; m_{1}, m_{2}) = ∫_{{α∈O^{×}|π^{n_{1},n_{2}} α+π^{m_{1},m_{2}} ∈O^{×}}} M_{0}^{−1}(α)η(Nα)dα.

Apply formula (G) to see that R_{k,ℓ} is the product of (1 − |π|)^{2}|π|^{\frac{k+ℓ}{2}} and

\begin{aligned}& ∑_{n_{1}+n_{2}=k+ℓ} M(π^{n_{2},n_{1}}) ∑_{m_{1}+m_{2}=0} ∑_{q≤0} ψ(n_{1} + m_{1} + q − ℓ, n_{2} + m_{2}; m_{1} + q, m_{2} + ℓ) \\ + & \frac{|π|}{|π| − 1} ∑_{n_{1}+n_{2}=k+ℓ} M(π^{n_{2},n_{1}}) ∑_{m_{1}+m_{2}=0} ψ(n_{1} + m_{1} + 1 − ℓ, n_{2} + m_{2}; m_{1} + 1, m_{2} + ℓ) \end{aligned}

Thus, ∑_{p=0}^{∞} |π|^{p}ω(π^{p})R_{k−p,ℓ−p} is the product of (1 − |π|)^{2}|π|^{\frac{k+ℓ}{2}} and

\begin{aligned}& ∑_{n_{1}+n_{2}=k+ℓ} M(π^{n_{2},n_{1}} ∑_{m_{1}+m_{2}=0} ∑_{p≤0} ∑_{q≤0} ψ(n_{1} + m_{1} + q − ℓ, n_{2} + m_{2} + p; m_{1} + q, m_{2} + ℓ + p) \\ + & \frac{|π|}{|π| − 1} ∑_{n_{1}+n_{2}=k+ℓ} M(π^{n_{2},n_{1}}) ∑_{m_{1}+m_{2}=0} ∑_{p≤0} ψ(n_{1} + m_{1} + 1 − ℓ, n_{2} + m_{2} + p; m_{1} + 1, m_{2} + ℓ + p) \end{aligned}

Now

\frac{ω(π)}{|π| − 1}T (M, η, k)T (M, ν, ℓ)

is equal to the product of (|π| − 1)^{3}ω(π)|π|^{\frac{k+ℓ}{2}} and

∑_{m_{1}+m_{2}=ℓ} ∑_{n_{1}+n_{2}=k} M(π^{n_{2},n_{1}})M(π^{m_{2},m_{1}})Δ(M_{0}^{−1}η^{1+s}, π^{n_{1},n_{2}})Δ(M_{0}^{−1}ν^{1+s}, π^{m_{1},m_{2}}). Replace n_{1} by n_{1} − m_{2} = n_{1} + m_{1} − ℓ, n_{2} by n_{2} − m_{1} = n_{1} + m_{2} − ℓ, and then m_{2} by m_{2} + ℓ to obtain

∑_{n_{1}+n_{2}=k+ℓ} M(π^{n_{2},n_{1}}) ∑_{m_{1}+m_{2}=0} Δ(M_{0}^{−1}η^{1+s}, π^{n_{1}+m_{1}−ℓ,n_{2}+m_{2}})Δ(M_{0}^{−1}ν^{1+s}, π^{m_{1},m_{2}+ℓ}). According to Lemma 1.4 this is the sum of ∑_{n_{1}+n_{2}=k+ℓ} ∑_{m_{1}+m_{2}=0} M(π^{n_{2},n_{1}}) ∑_{p≤0} ∑_{q≤0} ψ(n_{1} + m_{1} + q − ℓ, n_{2} + m_{2} + p; m_{1} + q, m_{2} + ℓ + p) and

\frac{|π|}{|π|−1}∑_{n_{1}+n_{2}=k+ℓ} ∑_{m_{1}+m_{2}=0} M(π^{n_{1},n_{2}}) ∑_{p≤0} ψ(n_{1} + m_{1} + 1 − ℓ, n_{2} + m_{2} + p, m_{1} + 1, m_{2} + ℓ + p)

and

\frac{|π|}{|π|−1}∑_{n_{1}+n_{2}=k+ℓ} ∑_{m_{1}+m_{2}=0} M(π^{n_{2},n_{1}}) ∑_{q≤0} ψ(n_{1} + m_{1} + q − ℓ, n_{2} + m_{2} + 1, m_{1} + q, m_{2} + ℓ + 1)

and

(\frac{|π|}{|π|−1})^{2}∑_{n_{1}+n_{2}=k+ℓ} ∑_{m_{1}+m_{2}=0} M(π^{n_{2},n_{1}})ψ(n_{1} + m_{1} + 1 − ℓ, n_{2} + m_{2} + 1; m_{1} + 1, m_{2} + ℓ + 1).

On the other hand

∑_{p=0}^{∞} |π|^{p}ω(π^{p})R_{k+1−p,ℓ+1−p} − ω(π) ∑_{p=0}^{∞} |π|^{p}ω(π^{p})R_{k−p,ℓ−p}

is equal to (1 − |π|)^{2}|π|^{\frac{k+ℓ}{2}} ω(π) times the sum of

\frac{|π|^{2}}{|π|−1}∑_{n_{1}+n_{2}=k+ℓ} M(π^{n_{2},n_{1}}) ∑_{m_{1}+m_{2}=0} ψ(n_{1} + m_{1} + 1 − ℓ, n_{2} + m_{2} + 1, m_{1} + 1, m_{2} + ℓ + 1)

and

\frac{|π|^{2}}{|π|−1}∑_{n_{1}+n_{2}=k+ℓ} M(π^{n_{2},n_{1}}) ∑_{m_{1}+n_{2}=0} ∑_{p≤0} ψ(n_{1} + m_{1} + 1 − ℓ, n_{2} + m_{2} + p; m_{1} + 1, m_{2} + ℓ + p)

and

|π|∑_{n_{1}+n_{2}=k+ℓ} M(π^{n_{2},n_{1}}) ∑_{m_{1}+m_{2}=0} ∑_{q≤0} ψ(n_{2} + m_{1} + q − ℓ, n_{2} + m_{2} + 1, m_{1} + q, m_{2} + ℓ + 1)

and

|π|∑_{n_{1}+n_{2}=k+ℓ} M(π^{n_{2},n_{1}}) ∑_{m_{1}+m_{2}=0} ∑_{p≤0} ∑_{q≤0} ψ(n_{1} + m_{1} + q − ℓ, n_{2} + m_{2} + p; m_{1} + q, m_{2} + ℓ + p),

the contributions of the first infinite series, and

−\frac{|π|}{|π|−1} ∑_{n_{1}+n_{2}=k+ℓ} M(π^{n_{2},n_{1}}) ∑_{m_{1}+m_{2}=0} ∑_{p≤0} ψ(n_{1} + m_{1} + 1 − ℓ, n_{2} + m_{2} + p; m_{1} + 1, m_{2} + ℓ + p)

and

− ∑_{n_{1}+n_{2}=k+ℓ} M(π^{n_{2},n_{1}}) ∑_{m_{1}+m_{2}=0} ∑_{p≤0} ∑_{q≤0} ψ(n_{1} + m_{1} + q − ℓ, n_{2} + m_{2} + p; m_{1} + q, m_{2} + ℓ + p),

the contributions of the second. The identity can now be verified by inspection.

2. Representations of the general linear group in two variables over a non-archimedean field. This paragraph is, in its essentials, a recapitulation of work of Gelfand, Graev, and Kirillov. We adhere to the notation of the previous paragraph. Let G_{k} = GL(2, k) and let G_{O} = GL(2, O). A is the group of diagonal matrices and N is the group of matrices of the term \binom{1 x}{0 1}.

A representation σ of G_{k} on a vector space V, over ℂ, will be called quasi-simple if

(i) The stabilizer of every vector in V is an open subgroup of G_{k} (ii) If α ∈ k^{×} then σ(\binom{α 0}{0 α}) is a scalar multiple of the identity.

Lemma 2.1 Suppose σ is a quasi-simple irreducible representation of G_{k} on the vector space V. V contains a non-zero vector invariant under N_{k} if and only if V is finite dimensional.

First of all suppose that V contains a non-zero vector v whose stabilizer contains N_{k}. Let H = {g ∈ G_{k}|σ(g)v = λv with λ ∈ ℂ}. Since V is spanned by the set {σ(g)v|g ∈ G_{k}} it is sufficient to show that H is of finite index in G_{k}. Since H contains the diagonal matrices together with an open subgroup of G_{k} the image of H under the determinant function is of finite index in k^{×}. Thus it is sufficient to show that H_{0} = {g ∈ G_{k}|σ(g)v = v} contains all matrices of determinant 1.

Let W be the space of column vectors of length 2 with entries from k. Let us show first that if w ∈ W and w ≠ 0 there is an h in H and an x in k^{×} such that

w = h\binom{x}{0}

If the second coordinate of w is zero this is clear. Since the stabilizer of v is open in G_{k} there is g in H_{0} such that

g\binom{1}{0} = \binom{α}{β}

with β ≠ 0. Then

\binom{1 x}{0 1}\binom{α}{β} = \binom{α + βx}{β}.

If the second coordinate of w is not 0 we can choose x so that w is a scalar multiple of the vector on the right.

In particular H_{0} contains a matrix of the form \binom{0 b}{c d}. Since

\binom{0 b}{c d} ( \begin{matrix}1 & −d/c \\ 0 & 1 \end{matrix} ) = \binom{0 b}{c 0}

and

\binom{0 b}{c 0}\binom{1 x}{0 1} ( \begin{matrix}0 & 1/c \\ 1/b & 0 \end{matrix} ) = \binom{0 b}{c 0} ( \begin{matrix}x/b & 1/c \\ 1/b & 0 \end{matrix} ) = ( \begin{matrix}1 & 0 \\ cx/b & 1 \end{matrix} ) ,

H_{0} contains all matrices of the form \binom{1 0}{y 1}. Since

\binom{1 z}{0 1}\binom{1 0}{x 1} = ({1 + xz \atop x} {z \atop 1}) = ({1 \atop x/(1 + xz)} {0 \atop 1}) ({1 + xz \atop 0} {0 \atop 1/(1 + xz)}) ({1 \atop 0} {z/(1 + xz) \atop 1}) ,

if 1 + xz ≠ 0, H_{0} contains all diagonal matrices of determinant 1. Since

\binom{1 0}{x 1}\binom{α 0}{0 β}\binom{1 y}{0 1} = ({α \atop αx} {0 \atop β} ) \binom{1 y}{0 1} = ({α \atop αx} {αy \atop β + αxy} )

H_{0} contains all matrices \binom{a b}{c d}, a ≠ 0, which have determinant 1. Since

\binom{1 x}{0 1}\binom{0 b}{c d} = ({cx \atop c} {b + dx \atop d}) ,

H_{0} contains all matrices of determinant 1.

Conversely if V is finite dimensional the kernel of σ is an open subgroup of G_{k} and there is an ε > 0 such that \binom{1 x}{0 1} belongs to this kernel if |x| < ε. Since

\binom{α 0}{0 1}\binom{1 x}{0 1}\binom{\frac{1}{α} 0}{0 1} = ({1 \atop 0} {αx \atop 1})

and for any x there is an α in k^{×} such that |αx| < ε, the kernel of σ contains N_{k}.

Corollary. If σ is a finite-dimensional quasi-simple irreducible representation of G_{k} then σ is onedimensional and there is a continuous homomorphism ρ of k^{×} into ℂ^{×} such that σ(g) = ρ(det g)

Since the kernel of σ contains N_{k} together with an open subgroup of G_{k} the above discussion shows that it contains every matrix of determinant 1. Also the inverse image of the group of non-zero matrices is of finite index in G_{k}. Thus if g ∈ G_{k} there is a λ in ℂ^{∗} and a positive integer n such that σ(g)^{n} − λ = 0. Thus σ(g) is semi-simple. The corollary follows immediately.

Again we fix a character ξ_{0} of the additive group of k such that the largest ideal on which ξ_{0} is trivial is 0.

Lemma 2.2 Suppose σ is an infinite dimensional quasi-simple irreducible representation of G_{k} on V. Let W be the set of all vectors v in V such that for some ideal 𝔞 of k

∫_{𝔞} \overline{ξ_{0}(x)}σ (\binom{1 x}{0 1}) v dx = 0.

Then W is a subspace of V. Let U = V/W. If v ∈ V let φ_{v} be the function k^{×} with values in U defined by

φ_{v}(α) = ψ (π (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} )) v)

where φ is the natural mapping from V to U. The map v → φ_{v} is an injection of V into the space of functions on k^{×} with values in U.

Since the stabilizer of v in N_{k} is an open subgroup of N_{k} the function σ(\binom{1 x}{0 1})v takes only a finite number of values on 𝔞. Thus the integral involves no limiting processes and is well defined. If 𝔞 ⊂ 𝔟 then

∫_{𝔟} \overline{ξ_{0}(x)}σ(\binom{1 x}{0 z})v dx = ∑_{y} \overline{ξ_{0}(y)}σ(\binom{1 y}{0 1})[∫_{𝔞} \overline{ξ_{0}(x)}σ(\binom{1 x}{0 1})v dx]

where the sum is taken over a system of representatives of 𝔟/𝔞. It follows immediately that if the integral vanishes for a given ideal then it vanishes for all larger ideals. A simple argument now shows that W is a subspace of V.

If φ_{v} vanishes identically then for every α in K^{×} there is an ideal 𝔞(α) such that

∫_{𝔞(α)} \overline{ξ_{0}(αx)}σ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))v dx = 0.

If β ∈ o^{×} and σ(\binom{β 0}{0 1})v = v then

∫_{𝔞} \overline{ξ_{0}(αβx)}σ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))v dx = σ(( \begin{matrix}β^{−1} & 0 \\ 0 & 1 \end{matrix} )) ∫_{𝔞} \overline{ξ_{0}(αx)}σ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))v dx

Since the set of all β in k^{×} such that σ(\binom{β 0}{0 1})v = v is an open subgroup of k^{×}, there is for each integer n an ideal 𝔞_{n} such that if 𝔞_{n} ⊆ 𝔞 and α ∈ o^{×}

∫_{𝔞} \overline{ξ_{0}(απ^{n}x)}σ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))v dx = 0.

There certainly is an integer n_{0} such that the function σ(\binom{1 x}{0 1})v is constant on cosets of 𝔭^{−n_{0}}. Let us show that if φ_{v} vanishes identically and this function is constant on left cosets of 𝔭^{−n} then it is constant on left cosets of 𝔭^{−n−1}. This will show that σ(\binom{1 x}{0 1})v = v for all x. It will then follow from Lemma 2.1 that v = 0.

Take any ℓ such that ℓ ≥ n + 1 and 𝔭^{−ℓ} ⊇ 𝔞_{n}. if x ∈ 𝔭^{−ℓ} then

σ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))v = |π|^{ℓ} ∑_{m=n}^{ℓ} ∑_{α∈o^{×}/o_{ℓ−m}} ξ_{0}(απ^{m}x) ∫_{𝔭^{−ℓ}} \overline{ξ_{0}(απ^{n}y)}σ(( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))v dy

By assumption the terms of this sum corresponding to m = n are zero. Since ξ_{0}(απ^{m}x) is constant on left cosets of 𝔭^{−n−1} if m > n the assertion follows.

Lemma 2.3 (i) If w = σ(\binom{α 0}{0 1})v then φ_{w}(β) = φ_{v}(βα) (ii) If w = σ(\binom{1 x}{0 1})v then φ_{w}(β) = ξ_{0}(βα)φ_{v}(β) (iii) If v is in V there is an integer k and a non-negative integer n such that φ_{v}(α) = 0 if |α| > |π|^{k}

and φ_{v}(βα) = φ_{v}(α) if β ∈ o_{n}^{×}.

The first assertion is a matter of definition. To prove the second we have to show that

σ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))σ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))v − ξ_{0}(αx)σ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))v = z

is in W. Let u = σ(\binom{α 0}{0 1})v. Then

∫_{𝔞} \overline{ξ_{0}(y)}σ(( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))z dy

is equal to

∫_{𝔞} \overline{ξ_{0}(y)}σ(( \begin{matrix}1 & y + αx \\ 0 & 1 \end{matrix} ))u dy − ξ_{0}(αx) ∫_{𝔞} \overline{ξ_{0}(y)}σ(( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))u dy

If αx ∈ 𝔞 we can change variables in the first integral to see that it equals the second term. Finally it is clear that if σ(\binom{1 x}{0 1})v = v for |x| ≤ |π|^{−k} and σ(\binom{β 0}{0 1})v = v for β ∈ o_{n}^{×} then φ_{v}(α) = 0 if |α| > |π|^{k} and φ_{v}(βα) = φ_{v}(α) if β ∈ o_{n}^{×}.

Let ν be a character of o^{×} and let V_{ν} = \{v ∈ V |σ(\binom{α 0}{0 1})v = ν(α)v for all α ∈ o^{×}\}. It is clear that V is the direct sum of the spaces V_{ν}. Let V̂ be the set of all v in V such that, for some k ≥ 0, φ_{v}(α) = 0 if |α| > |π|^{−k} or |α| < |π|^{k}. Let V̂_{ν} = V_{ν} ∩ V̂. It is also clear that V̂ is the direct sum of the spaces V̂_{ν}. Finally let V^{0} be the set of all v in V such that φ_{v}(α) ≠ 0 if |α| ≠ 1 and let V_{ν}^{0} = V_{ν} ∩ V^{0}. V^{0} is the direct sum of the spaces V_{ν}^{0}.

Lemma 2.4 (i) For each ν the restriction of ψ to V_{ν}^{0} defines an isomorphism of V_{ν}^{0} and U (ii) V̂_{ν} is the direct sum of the spaces σ(\binom{π^{k} 0}{0 1})V_{ν}^{0}, k ∈ ℤ. (iii) If v is in V_{ν} there is a unique v_{k} in V_{ν}^{0} such that if u = v − σ(\binom{π^{−k} 0}{0 1})v_{k} then φ_{u}(α) = 0 if

|α| = |π|^{k}. (iv) V is spanned by V̂ and the vectors of the form

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

with v in V̂.

We start with (iii) of which (ii) is an obvious consequence. The uniqueness of v_{k} is clear. If k is negative and |k| is sufficiently large we can take v_{k} = 0. Thus the proof can proceed by induction on k. Set

w = v − ∑_{ℓ<k} σ(( \begin{matrix}π^{−ℓ} & 0 \\ 0 & 1 \end{matrix} ))v_{ℓ}.

φ_{w}(α) = 0 if |α| > |π^{k}| and φ_{w}(α) = φ_{v}(α) if |α| ≤ |π|^{k}. Set

v_{k} = |π|^{−k−1}σ(\binom{π^{k} 0}{0 1}) ∫_{𝔭^{−k−1}} \{w − σ(\binom{1 x}{0 1})w\} dx

Then,

φ_{v_{k}}(π^{−k}α) = |π|^{−k−1} ∫_{𝔭^{−k−1}} {1 − ξ_{0}(αx} dx φ_{w}(α).

The right side is zero if |α| ≤ |π|^{k+1} or |α| > |π^{k}|. It is φ_{w}(α) if |α| = |π^{k}|. Part (iii) of the lemma follows.

It is clear that the restriction of ψ to V_{ν}^{0} is an injection. It follows from (iii) that the restriction of ψ to V^{0} is a surjection. Thus U = ∑_{μ} ψ(V_{μ}^{0}). To prove part (i) it is sufficient to show that if u ∈ U and u = ψ(v) for a v in some V_{μ}^{0} then there is a w in V_{ν}^{0} such that u = ψ(w). Given v set

z = ∫_{o^{×}} ν̄(β)σ(( \begin{matrix}β & 0 \\ 0 & 1 \end{matrix} ))σ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))v dβ

where x is yet to be determined. Then z is in V_{ν} and since

φ_{z}(α) = φ_{v}(α) ∫_{o^{×}} μν̄(β)ξ_{0}(αβx)dβ

it is in V_{ν}^{0}. In particular

ψ(z) = φ_{z}(1) = {∫_{o^{×}} μν̄(β)ξ_{0}(βx)dβ}u.

Choose x so that this integral is not zero and set

w = {∫_{o^{×}} μν̄(β)ξ_{0}(βx)dβ}^{−1}z

It follows from (i) that V̂ ≠ {0}. Choose w different from zero in V̂. Since σ is irreducible V is spanned by the vectors σ(g)w, g ∈ G_{k}. Either g = \binom{a b}{0 d} or g = \binom{1 x}{0 1}\binom{0 1}{−1 0}\binom{a b}{0 d}. In the first case σ(g)w is in V̂. In the second case σ(g)w is of the form σ(\binom{1 x}{0 1})σ(\binom{0 1}{−1 0})u with u in V̂. It is easily seen that if v belongs to V and x belongs to k then σ (\binom{1 x}{0 1}) v − v belongs to V̂. The last assertion of the lemma follows.

If v is in V let v = ∑_{ν} v_{ν} with v_{ν} in V_{ν}. Choose v_{k,ν} so that if u = v_{ν} − σ(\binom{π^{−k} 0}{0 1})v_{k,ν} then φ_{u}(α) = 0 if |α| = |π^{k}|. Set u_{k,ν} = ψ(v_{k,ν}) and write, purely formally,

v ∼ ∑_{ν} ∑_{ℓ} u_{ℓ,ν}z^{ℓ}.

Let σ (\binom{α 0}{0 α}) = ω(α)I for α ∈ k^{×} and let ω_{0} be the restriction of ω to o^{×}. Let ṽ(α) = ω_{0}(α)ν^{−1}(α) if α ∈ o^{×}. If v is in V_{ν}^{0}, then σ(\binom{0 1}{−1 0})v is in Vν̂. Let

σ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))v ∼ ∑_{k} u_{k}z^{k}.

If ψ(v) = u the map u → u_{k} is a linear transformation from U to U. Denote it by T_{k,ν}. If v is in V̂ and

v ∼ ∑_{ν} ∑_{ℓ} u_{ℓ,ν}z^{ℓ}

then

σ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )) ∑_{ν} ∑_{ℓ}\{∑_{m−k=ℓ} ω(π^{−k})T_{m,ν̂}u_{k,ν̃}\}z^{ℓ}.

It follows from the third part of Lemma 2.4 that if v ∈ V there is a unique v_{k} in V^{0} such that if u = v − σ(\binom{π^{−k} 0}{0 1})v_{k} then φ_{u}(α) = 0 if |α| = |π|^{k}. If w = σ(\binom{1 x}{0 1})v then v_{k} is replaced by w_{k} = σ(\binom{1 π^{k}x}{0 1})v_{k}. If v_{k} = ∑_{ν} v_{k,ν} with v_{k,ν} in V_{ν}^{0} and w_{k} = ∑_{ν} w_{k,ν} with w_{k,ν} in V_{ν}^{0} then

w_{k,μ} = ∑_{ν} ∫_{o^{×}} νμ̄(β)σ(( \begin{matrix}1 & βπ^{k}x \\ 0 & 1 \end{matrix} ))v_{k,ν}dβ.

Consequently

ψ(w_{k,μ}) = ∑_{ν} Δ(νμ^{−1}, π^{k}x)ψ(v_{k,ν}).

Thus if

\begin{aligned}v & ∼ ∑_{ν} ∑_{ℓ} u_{ℓ,ν}z^{ℓ}, \\ w & ∼ ∑_{ν} ∑_{ℓ}\{∑_{μ} Δ(μν^{−1}, π^{ℓ}x)u_{ℓ,μ}\}. \end{aligned}

It is also easily seen that

σ(( \begin{matrix}π^{n} & 0 \\ 0 & 1 \end{matrix} ))v ∼ ∑_{ν} ∑_{ℓ} u_{ℓ+n,ν}z^{ℓ}.

The identity

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

for x ≠ 0 is easily verified. If v is in V̂ and

v ∼ ∑_{ν} ∑_{ℓ} u_{ℓ,v}z^{ℓ}

then

\begin{matrix}π(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))v ∼ ∑_{ν} ∑_{k}\{∑_{ℓ} ω(π^{−ℓ})T_{k+ℓ,ν̃}u_{ℓ,ν̃}\}z^{k}, \\ σ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) σ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))v ∼ ∑_{μ} ∑_{k}\{∑_{ℓ} ∑_{ν} ω(π^{−ℓ})Δ(νμ^{−1}, π^{k}x)T_{k+ℓ,ν̃}u_{ℓ,ν̃}\}z^{ℓ}. \end{matrix}

If δ_{ν} = 1 if ν is trivial and δ_{ν} = 0 otherwise, then σ(\binom{1 x}{0 1})(\binom{0 1}{−1 0})v − σ(\binom{0 1}{−1 0})v which belongs to V̂ corresponds to

∑_{μ} ∑_{k}{∑_{ℓ} ∑_{ν}[Δ(νμ^{−1}, π^{k}x) − δ_{νμ^{−1}}]ω(π^{−ℓ})T_{k+ℓ,ν̃}u_{ℓ,ν̃}}z^{ℓ}.

Finally σ(\binom{0 1}{−1 0})(\binom{1 x}{0 1})(\binom{0 1}{−1 0})v corresponds to

\begin{matrix}∑_{μ} ∑_{k}\{∑_{ℓ,m,ν} [Δ(νμ̃^{−1}, π^{m}x) − δ_{νμ̃^{−1}}] ω(π^{−ℓ−m}) & T_{k+m,μ̃}T_{m+ℓ,ν̃}u_{ℓ,ν̃}\}z^{k} \\ + ∑_{μ} ∑_{k} ω(−1)u_{k,μ}z^{k}. \end{matrix}

On the other hand σ(\binom{1 −1/x}{0 1})v corresponds to

∑_{μ} ∑_{ℓ}{∑_{ν} Δ(νμ^{−1}, −π^{ℓ}/x)u_{ℓ,ν}}z^{ℓ}

and σ(\binom{0 1}{−1 0}\binom{1 −1/x}{0 1})v corresponds to ∑_{μ} ∑_{k}{∑_{ℓ} ∑_{k} ω(π^{−ℓ})Δ(νμ̃^{−1}, −π^{ℓ}/x)T_{k+ℓ,μ̃}u_{ℓ,ν}}z^{k}.

Letting σ(\binom{1 −x}{0 1}) operate we obtain a vector corresponding to

∑_{μ} ∑_{k}{∑_{ℓ,ν,η} ω(π^{−ℓ})Δ(ημ^{−1}, −π^{k}x)Δ(νη̃^{−1}, −π^{ℓ}/x)T_{k+ℓ,η̃}u_{ℓ,ν}}z^{k}.

Finally if \frac{1}{x} = π^{r}β with β ∈ o^{×} we apply σ(\binom{−1/x 0}{0 −x}) to obtain

∑_{μ} ∑_{k}{∑_{ℓ,ν,η} μ(β^{2})ω(−x)ω(π^{−ℓ})Δ(ημ^{−1}, −π^{k+2r}x)Δ(νη̃^{−1}, −π^{ℓ}x)T_{k+2r+ℓ,η̃}u_{ℓ,ν}}z^{k}.

Thus we obtain the identities

\begin{matrix}∑_{m}\{Δ(ν̃μ̃^{−1}, π^{m}x) & ω(π^{−ℓ−m})T_{k+m,μ̃}T_{m+ℓ,ν} − δν̃_{}_{μ̃^{−1}}ω(π^{−ℓ−m})T_{k+m,μ̃}T_{m+ℓ,ν}\} + ω(−1)δ_{νμ^{−1}}δ_{ℓ,k} \\ = ∑_{η} μ(β^{2})ω(−x)ω(π^{−ℓ})Δ(ημ^{−1}, −π^{k+2r}x)Δ(νη̃^{−1}, −π^{ℓ}/x)T_{k+2r+ℓ,η̃}. \end{matrix}

For all we know at present both these sums are infinite. However all but a finite number of the operators on each side send a given vector in U to zero. Thus as an operational equation the identity has a sense.

We can rewrite the identities as

\begin{aligned}∑_{m}\{[Δ(ν^{−1}μ^{−1} & ω_{0}, π^{m}β^{−1}) − δ_{νμω_{0}^{−1}}]ω(π^{−ℓ−m})T_{k+m,μ}T_{m+ℓ,ν}\} + ω_{0}(−1)δ_{νμω_{0}^{−1}}Δ_{ℓ,k} \\ & = ∑_{η} μ̃(β^{2})ω_{0}^{−1}(−β)ω(π^{−ℓ})Δ(μη^{−1}, −π^{k}β^{−1})Δ(νη^{−1}, −π^{ℓ}β)T_{k+ℓ,η}. \end{aligned}

Recalling that Δ(ν, βy) = ν^{−1}(β)Δ(ν, y) we simplify the identities to

\begin{aligned}∑_{m}\{[Δ(ν^{−1}μ^{−1} & ω_{0}, π^{m}) − δ_{νμω_{0}^{−1}}]ω(π^{−m})T_{k+m,μ}T_{m+ℓ,ν}\} + ω_{0}(−1)δ_{νμω_{0}^{−1}}ω(π^{ℓ}) \\ & = μνω_{0}^{−1}(−1) ∑_{η} Δ(μη^{−1}, π^{k})Δ(vη^{−1}, π^{ℓ})T_{k+ℓ,η}. \end{aligned}

Making use of Lemma 1.1 we can simplify these identities further. If the order n of νμω_{0}^{−1} is positive the identity becomes

Δ(ν^{−1}μ^{−1}ω_{0}, π^{−n})ω(π^{n})T_{k−n,μ}T_{ℓ−n,ν} = μνω_{0}^{−1}(−1) ∑_{η} Δ(μη^{−1}, π^{k})Δ(νη^{−1}, π^{ℓ})T_{k+ℓ,η}. (A)

If νμω_{0}^{−1} = 1 the identity becomes

\begin{matrix}∑_{m=−∞}^{−2} −ω(π^{−n}) & T_{k+m,μ}T_{m+ℓ,ν} + \frac{1}{|π| − 1}ω(π)T_{k−1,μ}T_{ℓ−1,ν} + ω_{0}(−1)δ_{ℓ,k}ω(π^{ℓ}) \\ = ∑_{η} Δ(μη^{−1}, π^{k})Δ(νη^{−1}, π^{ℓ})T_{k+ℓ,η} \end{matrix}  (B)

Lemma 2.5 (i) For all k, ℓ, μ, and ν, T_{k,μ}T_{ℓ,ν} = T_{ℓ,ν}T_{k,μ}.

(ii) There is no non-trivial subspace of u left invariant by all the operators T_{k,μ}.

If νμω_{0}^{−1} is not trivial the identity

T_{k,μ}T_{ℓ,ν} = T_{ℓ,ν}T_{k,μ}

follows immediately from (A). If νμω_{0}^{−1} is trivial let u be in U. For a given k and ℓ and for m ≪ 0 both T_{k+m,μ}u and T_{ℓ+m,ν}u are zero. For such m

T_{k+m,μ}T_{ℓ+m,ν}u = T_{ℓ+m,ν}T_{k+m,μ}u.

Using the identity (B) and induction on m one shows readily that this relation iw valid for all m.

Suppose that U′ is a nontrivial subspace of U left invariant by all the operators T_{k,μ}. Let V′ be the set of all v in V such that φ_{v}(α) ∈ U′ for all α. If v ∈ V_{ν}^{0} then v ∈ V′ if and only if ψ(v) ∈ U′. Thus V′ is neither {0} nor V and V′ ∩ V_{ν}^{0} ≠ {0}. It is clear that V′ and V′ ∩ V̂ are left fixed by the operators σ(\binom{a b}{0 d}). Since V is irreducible it is spanned by V′ ∩ V̂ together with the set σ(\binom{1 x}{0 1})σ(\binom{0 1}{−1 0})v, v ∈ V′ ∩ V̂. Thus to obtain a contradiction we need only show that if v is in V′ ∩ V̂ then σ(\binom{0 1}{−1 0})v is in V′. This is however an obvious consequence of the assumption.

It follows from this lemma that each T_{k,μ} is either zero or an invertible linear transformation. Thus for each μ there is an integer k(μ) such that T_{k,μ} = 0 if k < k(μ). Moreover one of these operators can have a non-trivial eigenvector if and only if it is a scalar.

Now I would like to make some remarks which are not relevant to the main purpose of the letter. First of all let me observe that if k, ℓ, μ, ν are arbitrary there is a scalar a and scalars a_{m,ρ} all but a finite number of which are zero such that

T_{k,μ}T_{ℓ,ν} = a + ∑_{ρ} ∑_{m} a_{m,ρ}T_{m,ρ}

If μνω_{0}^{−1} is not trivial this follows immediately from identity (A). If μνω_{0}^{−1} is trivial consider the set of integers p for which T_{k+p,μ}T_{ℓ+p,ν} is a linear combination of the identity and the operators T_{m,≤}. If p ≪ 0, T_{k+p,μ}T_{ℓ+p,ν} = 0 and p belongs to this set. Using identity (B) and a simple induction argument one shows that the set contains all integers. It follows from this observation and the previous lemma that if u ≠ 0 belongs to U then U is spanned by u and the set {T_{m,ρ}u}.

Choose a fixed ν and let the order n of μ be positive and so large that the orders of ν^{−1}μ^{−1}ω_{0} and νμ^{−1} are also n and T_{r_{0},ν} ≠ 0 for some r_{0} ≥ −n. Take ℓ = r_{0} + n ≥ 0 in identity (A) and cancel T_{r_{0},ν} to obtain

Δ(ν^{−1}μ^{−1}ω_{0}, π^{−n})ω(π^{n})T_{k−n,μ} = μνω_{0}^{−1}(−1)Δ(μν^{−1}, π^{k})

As a consequence for all but a finite number of characters of o^{×} the operator T_{k,μ} is a scalar for all k. If, for all ρ, T_{m,ρ} = 0 if m ≥ −1 the there are only a finite number of operators in the set {T_{m,ρ}} which are not scalars. Consequently U is finite dimensional and each of the operators T_{m,ρ} has a non-trivial eigenvector and is thus a scalar. It follows that U has dimension 1.

It is very unlikely that our assumptions (i) and (ii) together with irreducibility imply that U is one-dimensional. Consequently we make the further assumption which can certainly be useful in the case of interest to us at present.

(iii) No representation of G_{o} occurs more than a finite number of times in the restriction of σ to G_{o}.

If ρ is a representation of G_{o} let V_{ρ} be the set of all vectors in V which transform according to ρ. Any operator on V which commutes with all the operators σ(g) must leave each of the finite dimensional spaces V_{ρ} invariant. Thus it must have a non-trivial eigenvector and, because of the irreducibility, must be a scalar. It follows immediately from the first part of Lemma 2.4 that the map v → φ_{v} maps V̂ onto the set of all locally constant functions on k^{×} with values in U which vanish outside of some compact set. Suppose T is an operator on U which commutes with all the operators T_{m,ρ}. If φ is a function on k^{×} with values in U define T φ by (T φ)(α) = T (φ(α)). If v ∈ V̂ and

v ∼ ∑_{μ} ∑_{k} u_{k,μ}z^{k}

then T φ_{v} = φ_{w} where

w ∼ ∑_{μ} ∑_{k} T u_{k,μ}z^{k}.

Then

\begin{aligned}σ (( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )) v & ∼ ∑_{μ} ∑_{k}\{∑_{m−ℓ=k} ω(π^{−ℓ})T_{m,μ̃}u_{ℓ,μ̃}\}z^{k}, \\ σ (( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )) w & ∼ ∑_{μ} ∑_{k} T \{∑_{m−ℓ=k} ω(π^{−ℓ})T_{m,μ̃}u_{ℓ,μ̃}\}z^{k}. \end{aligned}

It follows immediately that T takes the image of V to itself. Thus it determines a linear transformation of V which is easily shown to commute with all the operators. As a consequence of assumption (iii) this linear transformation is a scalar. Thus T is a scalar. In particular all the operators T_{m,ρ} are scalars and U is one-dimensional.

In the next two lemmas it is assumed that U is one dimensional. Thus the operators T_{m,ρ} are taken to be complex numbers.

Lemma 2.7 Suppose that there is a character μ_{1} of o^{×} and a k_{1} ≥ −1 such that T_{k_{1},μ_{1}} ≠ 0. Let K = k ⊕ k. There is a continuous homomorphism M of K^{×} into ℂ^{×} such that for all ℓ and ν

T_{ℓ,ν} = T (M, ν; ℓ)

Let me observe immediately that it was shown in the previous paragraph that if the restriction of M to k^{×} is ω the identities (A) and (B) are satisfied if T_{ℓ,ν} is replaced by T (M, ν; ℓ). Set μ_{2} = μ_{1}^{−1}ω_{0}. It will perhaps require less mental effort if the cases μ_{1} = μ_{2} and μ_{1} ≠ μ_{2} are treated separately.

Suppose first that μ_{1} = μ_{2}. In identity (A) take ν = μ_{1}, μ ≠ μ_{1}, and take ℓ = k_{1} + n ≥ 0 to obtain

Δ(μ_{1}μ^{−1}, π^{−n})ω(π^{n})T_{k−n,μ}T_{k_{1},μ_{1}} = μμ_{1}^{−1}(−1)Δ(μμ_{1}^{−1}, π^{k})T_{k+ℓ,μ_{1}}.

The right side is zero unless k = −n but if k = −n we can cancel T_{k_{1},μ_{1}} from both sides to obtain

T_{−2n,μ} = \frac{μμ_{1}^{−1}(−1)}{ω(π^{n})} \frac{Δ(μμ_{1}^{−1}, π^{−n})}{Δ(μ_{1}μ^{−1}, π^{−n})} = (1 − |π|)^{2}|π|^{−n}ω(π^{−n}) \{Δ(μ_{1}^{−1}μ, π^{−n}\}^{2} .

Thus if ω_{1} and ω_{2} are two complex numbers such that ω_{1}ω_{2} = ω(π) and M is defined by M(π^{p}α ⊕ π^{q}β) = ω_{1}^{p}ω_{2}^{q}μ_{1}(αβ) for α ∈ o^{×}, β ∈ o^{×} then T_{k,μ} = T (M, k, μ) for μ ≠ μ_{1}.

Take μ = ν = μ_{1} and k = ℓ in identity (B). If k < −1 the right side is a sum over those η such that the conductor of μ_{1}η^{−1} is 𝔭^{−ℓ}. For such η

Δ(μ_{1}η^{−1}, π^{k})Δ(μ_{1}η^{−1}, π^{k})T_{2k,η} = ω(π^{k})μμ_{1}^{−1}(−1)Δ(ημ_{1}^{−1}, π^{k})Δ(μ_{1}η^{−1}, π^{k}) = \frac{ω(π^{k})|π|^{k}}{(1 − |π|)^{2}}

Since the number of such characters is |π|^{−k}(1 − |π|)^{2} the right side of (B) is equal to ω(π^{k}). Since ω_{0}(−1) = μ_{1}^{2}(−1) = 1 we have, for k < −1,

∑_{m=−∞}^{−2} −ω(π^{−m})T_{k+m,μ_{1}}T_{k+m,μ_{1}} + \frac{ω(π)}{|π| − 1}T_{k−1,μ_{1}}T_{k−1,μ_{1}} = 0

It follows by induction that T_{m,μ_{1}} = 0 if m < −2.

Now take μ = ν = μ_{1}, ℓ = −1, and k ≥ 0 in (B) to obtain

\frac{ω(π)}{|π| − 1}T_{k−1,μ_{1}}T_{−2,μ_{2}} = \frac{|π|}{|π| − 1}T_{k−1,μ_{1}}

Since T_{k−1,μ_{1}} ≠ 0 for some k ≥ 0 we conclude that ω(π)T_{−2,μ_{1}} = |π|.

Choose ω_{1} and ω_{2} to be the two solutions of the equation

(|π| − 1)\{\frac{x}{ω(π)}|π|^{1/2} + \frac{|π|^{1/2}}{x} \} = T_{−1,μ_{1}}.

It is easy to see that

∑_{m=−∞}^{∞} T (M, μ_{1}, m)x^{m} = \frac{(1 − ω_{1}^{−1}x^{−1}|π|^{1/2})(1 − ω_{2}^{−1}x^{−1}|π|^{1/2})}{(1 − ω_{1}x|π|^{1/2})(1 − ω_{2}x|π|^{1/2})}

if |x| > 0 and |x| is sufficiently small. Thus T_{m,μ_{1}} = T (M, μ_{1}, m) if m < 0. Taking μ = ν = μ_{1}, k = 0, and ℓ ≥ 0 in (B) we obtain

∑_{m=−∞}^{−2} −ω(π^{−m})T_{m,μ_{1}}T_{ℓ+m,μ_{1}} + \frac{ω(π)}{|π| − 1}T_{−1,μ_{1}}T_{ℓ−1,μ_{1}} + ω_{0}(−1)δ_{ℓ,k}ω(π^{ℓ}) = T_{ℓ,μ_{1}}

Since the same formula is valid if T_{p,μ_{1}} is replaced by T (M, μ_{1}, p) we can show inductively that T_{m,μ_{1}} = T (M, μ_{1}, m) for all m.

Now suppose μ_{1} ≠ μ_{2}. Let n be the order of μ_{1}μ_{2}^{−1}. Take μ = ν = μ_{1}, ℓ = k_{1} + 1, and k = −1 in identity (A) to obtain

Δ(μ_{2}μ_{1}^{−1}, π^{−n})ω(π^{n})T_{−1−n,μ_{1}}T_{ℓ−n,μ_{1}} = μ_{1}μ_{2}^{−1}(−1)\frac{|π|}{|π| − 1}T_{k_{1},μ_{1}}.

Thus T_{−1−n,μ_{1}} ≠ 0. Now take ℓ = k_{1} + n, k < −1 to obtain

Δ(μ_{2}μ_{1}^{−1}, π^{−n})ω(π^{n})T_{k−n,μ_{1}}T_{k,μ_{1}} = 0.

Thus T_{k−n,μ_{1}} = 0 if k < −1.

Now let us look at the identity (B) with μ = μ_{1}, ν = μ_{2}. If k > −n and ℓ > −n the right side is zero because the order of either μ_{1}η^{−1} or μ_{2}η^{−1} is at least n. Thus in this case

∑_{m=−∞}^{−2} −ω(π^{−m})T_{k+m,μ_{1}}T_{ℓ+m,μ_{2}} + \frac{ω(π)}{|π| − 1}T_{k−1,μ_{1}}T_{ℓ−1,μ_{2}} + ω_{0}(−1)δ_{ℓ,k}ω(π^{ℓ}) = 0. (C)

In particular take ℓ = n + 1 to see that if k > −n + 1 and T_{k−1,μ_{2}} = 0 so does T_{k−2,μ_{2}}.

If k < −n and ℓ ≠ k the right side of (B) is zero for μ = μ_{1}, ν = μ_{2} because k ≤ −2 and if the order of μ_{1}η^{−1} is −k so is the order of μ_{2}η^{−1}. Thus in this case

∑_{m=−∞}^{−2} −ω(π^{−m})T_{k+m,μ_{1}}T_{ℓ+m,μ_{2}} + \frac{ω(π)}{|π| − 1}T_{k−1,μ_{1}}T_{ℓ−1,μ_{2}} = 0.

The same result is valid if ℓ < −n and k ≠ ℓ. Take k = −n to see that T_{m,μ_{2}} = 0 if m < −n − 1.

Thus if T_{m,μ_{2}} = 0 for all m ≥ −1 then the only m for which T_{m,μ_{2}} ≠ 0 is m = −n − 1. Taking ℓ = −n − 1 in (C) we would find that T_{m,μ_{1}} = 0 for m ≥ −n which is contrary to assumption. At this point μ_{1} and μ_{2} play identical roles.

Taking k = −n + 1 in (C) we see that if ℓ ≥ −n

T_{ℓ+1,μ_{2}}T_{−n,μ_{1}} = (|π| − 1)ω(π)T_{ℓ,μ_{2}}T_{−1−n,μ_{1}}.

Thus T_{−n,μ_{1}} ≠ 0 and T_{ℓ,μ_{2}} ≠ 0 if ℓ ≥ −n. Set, if ℓ ≥ −n,

ω_{1}|π|^{1/2} = T_{\frac{ℓ+1,μ_{2}}{T_{ℓ,μ_{2}}}} = (|π| − 1)ω(π)T_{\frac{−1−n,μ_{1}}{T_{−n,μ_{1}}}} .

Similarly T_{−n,μ_{2}} ≠ 0 and T_{ℓ,μ_{1}} ≠ 0 if ℓ ≥ −n. If ℓ ≥ −n, set

ω_{2}|π|^{1/2} = T_{\frac{ℓ+1,μ_{1}}{T_{ℓ,μ_{1}}}} = (|π| − 1)ω(π)T_{\frac{−1−n,μ_{2}}{T_{−n,μ_{2}}}} .

Now take ν = μ = μ_{1}, ℓ ≥ 0, k = −1 in (A) to obtain

Δ(μ_{1}^{−1}μ_{2}, π^{−n})ω(π^{n})T_{−1−n,μ_{1}}T_{ℓ−n,μ_{1}} = μ_{1}μ_{2}^{−1}(−1)\frac{|π|}{|π| − 1}T_{ℓ−1,μ_{1}}.

Thus

T_{−1−n,μ_{1}} = \frac{|π| − 1}{|π|^{\frac{n−1}{2}}} \frac{ω_{2}^{n−1}}{ω(π^{n})}Δ(μ_{1}μ_{2}^{−1}, π^{−n}).

In the same way

T_{1−n,μ_{2}} = \frac{|π| − 1}{|π|^{\frac{n−1}{2}}} \frac{ω_{1}^{n−1}}{ω(π^{n})}Δ(μ_{2}μ_{1}^{−1}, π^{−n})

Thus if γ = \frac{ω_{1}ω_{2}}{ω(π)} ,

T_{−1−n,μ_{1}} = γ^{n} \frac{(|π| − 1)}{|π|^{\frac{n−1}{2}}} \frac{1}{ω_{2}ω_{1}^{n}} Δ(μ_{2}^{−1}μ_{1}, π^{−n}),
T_{ℓ,μ_{1}} = γ^{n−1}(1 − |π|)^{2}|π|^{\frac{ℓ}{2}} ω_{1}^{ℓ+n}ω_{1}^{−n}Δ(μ_{2}^{−1}μ_{1}, π^{−n}),  ℓ ≥ −n,
T_{−1−n,μ_{2}} = γ^{n} \frac{(|π| − 1)}{|π|^{\frac{n−1}{2}}} \frac{1}{ω_{2}^{n}ω_{1}} Δ(μ_{1}^{−1}μ_{2}, π^{−n}),
T_{ℓ,μ_{2}} = γ^{n−1}(1 − |π|)^{2}|π|^{\frac{ℓ}{2}} ω_{2}^{−n}ω_{1}^{ℓ+n}Δ(μ_{1}^{−1}μ_{2}, π^{−n}),  ℓ ≥ −n.

If we take μ different from μ_{1} and μ_{2}, ν = μ_{1} and ℓ ≥ 0 in identity (A) we obtain

Δ(μ_{2}μ_{1}^{−1}, π^{−n_{2}})ω(π^{n_{2}})T_{k−n_{2},μ}T_{ℓ−n_{2},μ_{1}} = μμ_{2}^{−1}(−1)Δ(μμ_{1}^{−1}, π^{k})T_{k+ℓ,μ_{1}}

if n_{1} is the order of μ_{1}μ^{−1} and n_{2} is the order of μ_{2}μ^{−1}. Thus T_{m,μ} = 0 if m ≠ −n_{1} − n_{2} but

T_{−n_{1}−n_{2},μ} = (1 − |π|)^{2}|π|^{\frac{−n_{1}−n_{2}}{2}} γ^{n_{2}}ω_{2}^{−n_{1}}ω_{1}^{−n_{2}}Δ(μ_{1}^{−1}μ, π^{−n_{1}})Δ(μ_{2}^{−1}μ, π^{−n_{2}}).

If we can show that γ = 1 we will have proved that if M(π^{p}α ⊕ π^{q}β) = ω_{1}^{p}ω_{2}^{p}μ_{1}(α)μ_{2}(β) then T (M, μ, m) = T_{m,μ} for all μ and all m.

Take μ = μ_{1}, ν = μ_{2} and k = ℓ = −n in (B). If the order of both ημ_{1}^{−1} and ημ_{2}^{−1} is n, the value of the corresponding term on the right side is

μ_{1}μ_{2}(−1)\frac{|π|^{n}}{(1 − |π|)^{2}} \frac{1}{ω(π^{n})}.

If n > 1 there are \frac{1−|π|}{|π|^{n}}(1 − 2|π|) such characters η. The terms corresponding to the other characters are all zero so the right-hand side is \frac{μ_{1}μ_{2}(−1)}{ω(π^{n})} \frac{1−2|π|}{1−|π|}. If n = 1 there are \frac{1−3|π|}{|π|} such characters. However the terms corresponding to η = μ_{1} and η = μ_{2} give a total contribution of

\frac{μ_{1}μ_{2}(−1)}{ω(π)} \frac{|π|^{2}}{(1 − |π|)^{2}} + \frac{μ_{1}μ_{2}(−1)}{ω(π)} \frac{|π|^{2}}{(1 − |π|)^{2}} .

Thus the right side is again \frac{μ_{1}μ_{2}(−1)}{ω(π^{n})} \frac{1−2|π|}{1−|π|}. The left side is

\frac{γ^{n−1}}{ω(π^{n})}μ_{1}μ_{2}(−1)\frac{|π|}{|π| − 1} + \frac{μ_{1}μ_{2}(−1)}{ω(π^{n})}.

Consequently γ^{n−1} = 1. Now take k = ℓ = −n + 1 in (C) to obtain

\frac{−μ_{1}μ_{2}(−1)}{ω(π^{n−1})} |π| + \frac{μ_{1}μ_{2}(−1)}{ω(π^{n−1})}γ^{n−2}(|π| − 1) + \frac{μ_{1}μ_{2}(−1)}{ω(π^{n−1})} = 0

Thus γ^{n−2} = 1 and γ = 1.

It will be convenient to record here the closed expressions for

∑_{n=−∞}^{∞} T (M, μ, n)z^{n} = T (M, μ, z)

The series of course converges for |z| > 0 and sufficiently small.

Lemma 2.8 (i) Let K = k ⊕ k and let M(π^{p}α ⊕ π^{q}β) = ω_{1}^{p}ω_{2}^{p}μ_{1}(α)μ_{2}(β) if α ∈ o^{×} and β ∈ o^{×}.

(a) If μ_{1} = μ_{2} then

T (M, μ_{1}, z) = \frac{(1 − ω_{1}^{−1}z^{−1}|π|^{1/2})}{(1 − ω_{1}z|π|^{1/2})} \frac{(1 − ω_{2}^{−1}z^{−1}|π|^{1/2})}{(1 − ω_{2}z|π|^{1/2})}

and if μ ≠ μ_{1} and the order of μ^{−1}μ_{1} is n

T (M, μ, z) = (1 − |π|)^{2}|π|^{−n}ω_{1}^{−n}ω_{2}^{−n}Δ(μ_{1}^{−1}μ, π^{−n})Δ(μ_{1}^{−1}, μ, π^{−n})z^{−2n}.

(b) If μ_{1} ≠ μ_{2} then

\begin{aligned}T (M, μ_{1}, z) & = (1 − |π|)|π|^{− \frac{n}{2}} ω_{1}^{−n}Δ(μ_{2}^{−1}μ_{1}, π^{−n})\frac{(1 − ω_{2}^{−1}z^{−1}|π|^{1/2})}{(1 − ω_{2}z|π|^{1/2})}z^{−n}; \\ T (M, μ_{2}, z) & = (1 − |π|)|π|^{− \frac{n}{2}} ω_{2}^{−n}Δ(μ_{1}^{−1}μ_{2}, π^{−n})\frac{(1 − ω_{1}^{−1}z^{−1}|π|^{1/2})}{(1 − ω_{1}z|π|^{1/2})}z^{−n} \end{aligned}

if n is the order of μ_{1}^{−1}μ_{2}. If μ is different from μ_{1} and μ_{2} and the order of μ^{−1}μ_{1} is n_{1} and the order of μ^{−1}μ_{2} is n_{2} then

T (M, μ, z) = (1 − |π|)^{2}|π|^{\frac{−n_{1}−n_{2}}{2}} ω_{2}^{−n_{1}}ω_{1}^{−n_{2}}Δ(μ_{1}^{−1}μ, π^{−n_{1}})Δ(μ_{2}^{−1}μ, π^{−n_{2}})z^{−n_{1}−n_{2}}.

(ii) Let K be an unramified extenson of k.

(a) Suppose there is a generalized character M_{1} of k^{×} such that M(α) = M_{1}(α^{1+s}). Let M_{1}(π^{p}β) = ω_{1}^{p}μ_{1}(β) for β ∈ o^{×}. Then

T (M, μ_{1}; z) = \frac{1 − ω_{1}^{−2}z^{−2}|π|}{1 − ω_{1}^{2}z^{2}|π|} = \frac{1 − ω_{1}^{−1}z^{−1}|π|^{1/2}}{1 − ω_{1}z|π|^{1/2}} \frac{1 + ω_{1}^{−1}z^{−1}|π|^{1/2}}{1 + ω_{1}z|π|^{1/2}} .

If μ ≠ μ_{1} and the order of μμ_{1}^{−1} is n so that the order of μ^{1+s}M_{0}^{−1} is also n then

\begin{aligned}T (M, μ; z) & = (1 − |π|^{2})|π|^{−n}ω_{1}^{−2n}Δ(M_{0}^{−1}μ^{1+s}, Π^{−n})z^{−2n} \\ & = (1 − |π|)^{2}|π|^{−n}ω_{1}^{−n}(−ω_{1})^{−n} \{Δ(μ_{1}^{−1}μ, π^{−n})\}^{2} z^{−2n}. \end{aligned}

(b) If there is no such character then for all μ

T (M, μ, z) = (1 − |π|^{2})|π|^{−n}M(Π^{−n})Δ(M_{0}^{−1}μ^{1+s}, Π^{−n})z^{−2n}

if n is the order of M_{0}^{−1}μ^{1+s}. (iii) Let K be a ramified extension of k.

(a) Suppose there is a character M_{1} of k^{×} such that M(α) = M_{1}(α^{1+s}). Let M_{1}(π^{p}β) = ω_{1}^{p}μ_{1}(β) if β ∈ o^{×}. Then

\begin{aligned}T (M, μ_{1}; z) & = (1 − |π|)|π|^{− \frac{f}{2}} ω_{1}^{−f}Δ(χ_{0}, π^{−f})\frac{1 − ω_{1}^{−1}z^{−1}|π|^{1/2})}{(1 − ω_{1}z|π|^{1/2})}z^{−f} \\ T (M, χ_{0}μ_{1}, z) & = (1 − |π|)|π|^{− \frac{f}{2}} ω_{1}^{−f}Δ(χ_{0}, π^{−f})\frac{(1 − ω_{1}^{−1}z^{−1}|π|^{1/2})}{(1 − ω_{1}z|π|^{1/2})}z^{−f} \end{aligned}

and if μ is different from μ_{1} and χ_{0}μ_{1} then

\begin{aligned}T (M, μ, z) & = (1 − |π|)^{2}Δ(χ_{0}, π^{−f})|π|^{− \frac{n−f}{2}} ω_{1}^{−n−f}Δ(M_{0}^{−1}μ^{1+s}, Π^{−n−f})z^{−n−f} \\ & = (1 − |π|)^{2}|π|^{\frac{−n_{1}−n_{2}}{2}} ω_{1}^{−n_{1}−n_{2}}Δ(μ_{1}^{−1}μ, π^{−n_{1}})Δ(μ_{2}^{−1}μ, π^{−n_{2}})z^{−n_{1}−n_{2}} \end{aligned}

if μ_{2} = χ_{0}μ_{1}, n is the order of M_{0}^{−1}μ^{1+s}, n_{1} is the order of μ_{1}^{−1}μ, and n_{2} is the order of μ_{2}^{−1}μ.

(b) If there is no such character M_{1} then, for all μ,

T (M, μ, z) = (1 − |π|)^{2}|π|^{\frac{−n−f}{2}} Δ(χ_{0}, π^{−f})M(Π^{−s(n+f)})Δ(M_{0}^{−1}μ^{1+s}, Π^{−n−f})z^{−n−f}

if n is the order of M_{0}^{−1}μ^{1+s}.

The formulas of this lemma follow from the definitions together with Lemmas 1.1 and 1.5. I would like to observe in cases (ii, a) and (iii, a) that if M′ is the character of (k ⊕ k)^{∗} defined by M′(α ⊕ β) = M(α)M_{1}(β)χ(β), the, for all μ

T (M, μ, z) = T (M′, μ, z).

It follows from Lemmas 2.7 and 2.8 that if the collection {T_{m,μ}} satisfies identities (A) and (B) the series

∑_{m} T_{m,μ}z^{m}

converges for |z| > 0 and sufficient small and its sum T_{μ}(z) is a rational function. If we return to the discussion of the representation σ we can choose some isomorphism of U with ℂ and regard the functions φ_{v} as scalars. Let L′ be the set of all locally constant complex-valued functions, i.e. invariant under some open subgroup, on k^{×}. If ν is a character of o^{×} let L′_{ν} be the set of all functions φ in L′ such that φ(βα) = ν(β)φ(α) if β ∈ o^{×}. It is clear that L′ is the direct sum of the spaces L′_{ν}. If φ ∈ L′ we write φ = ∑_{ν} φ_{ν} with φ_{ν} ∈ L′_{ν} and set u_{k,ν} = φ_{ν}(π^{k}). Let L be the set of all functions φ in L′ such that, for each ν, u_{k,ν} = 0 for k ≪ 0 and

φ_{ν}(z) = ∑_{k} u_{k,ν}z^{k}

converges for |z| > 0 and sufficiently small and represents a rational function. If Ĥ is the set of all functions in L′ with compact support then Ĥ ⊆ L. Ĥ is clearly the image of V̂. By the way, it will not conflict with our previous notation if when φ = ∑ φ_{ν} lies in L′ and u_{k,ν} = φ_{ν}(π^{k}) we set

φ ∼ ∑_{ν} ∑_{k} u_{k,ν}z^{k}.

Now suppose {T_{m,μ}} is the collection corresponding to the representation σ. If v ∈ V̂ and

φ = φ_{ν} ∑_{ν} ∑_{k} u_{k,ν}z^{k}

then

w = σ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))v ∼ ∑_{ν} ∑_{ℓ} \{∑_{m+k=ℓ} ω(π^{k})T_{m,ν̂}u_{−k,ν̂}\} z^{ℓ}.

Thus ψ = φ_{w} is also in L and

ψ_{ν}(z) = Tν̂(z)φν̂(ω^{−1}(π)z^{−1})
If T_{m,μ} = 0 whenever m ≥ −1 then V = V̂ so that
φ_{ν}(z) = ω_{0}(−1)Tν̂(z)ψν̂(ω^{−1}(π)z^{−1}).

Thus, in this case,

ω_{0}(−1)T_{ν}(z)Tν̂(ω^{−1}(π)z^{−1}) = 1  (D)

On the other hand if one notices that MM^{s} = ω^{1+s} so that M^{−1}ω^{1+s}μ^{−1−s} = (M^{−1}μ^{1+s})^{−s} one can verify by inspection that

ω_{0}(−1)T (M, ν, z)T (M, ν̂, ω^{−1}(π)z^{−1}) = 1.

Thus the identity (D) is valid whenever σ is an irreducible representation satisfying (i), (ii), and (iii).

Now let us suppose that ω is a continuous homomorphism of k^{×} into ℂ^{×} and that the family \{T_{m,μ}\} satisfies the relations (A), (B), and (D). If φ belongs to L′ and \binom{a b}{0 d} belongs to G_{k} let τ(\binom{a b}{0 d})φ be the function whose value at α is ω(d)ξ_{0} (\frac{αb}{d}) φ (\frac{αa}{d}). τ is a representation of the group of upper triangular matrices in G_{k} on L′. Ĥ is an invariant subspace of L′ for τ. It is clear that the operators τ(\binom{a 0}{0 b}) leave L invariant. If φ ∈ L then, for all x ∈ k, the function ψ defined by ψ(α) = ξ_{0}(αx)φ(α) − φ(α) lies in Ĥ. Thus the operators τ(\binom{1 x}{0 1}) leave any subspace of L containing Ĥ invariant. Define τ(\binom{0 1}{−1 0}) by the condition that if φ ∈ L and ψ = τ(\binom{0 1}{−1 0})φ then

ψ_{ν}(z) = Tν̂(z)φν̂(ω^{−1}(π)z^{−1}).

It is easy to verify that

τ(( \begin{matrix}a & 0 \\ 0 & d \end{matrix} ))τ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )) = τ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))τ(( \begin{matrix}d & 0 \\ 0 & a \end{matrix} )).

Thus the operators τ(\binom{a b}{0 d}) and τ(\binom{0 1}{−1 0}) leave the space spanned by Ĥ and the functions τ(\binom{0 1}{−1 0})φ, φ ∈ Ĥ invariant. Call this space H. Every matrix in G_{k} which is not supertriangular can be written in exactly one way as

g = ( \begin{matrix}β & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} )

Set τ(g) = τ(\binom{β 0}{0 β})τ(\binom{1 x}{0 1})τ(\binom{0 1}{−1 0})τ(\binom{α 0}{0 1})τ(\binom{1 y}{0 1}). Thus τ(g) is defined for all g in G_{k}.

Let us verify that τ(g_{1}g_{2}) = τ(g_{1})τ(g_{2}). This is clear if g_{1} ∼ g_{2} is a supertriangular matrix. Thus it is enough to verify this when

\begin{aligned}g_{1} & = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ) \\ g_{2} & = ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) \end{aligned}

The case x + y = 0 is taken care of by identity (D) so suppose x + y = u ≠ 0. Then

\begin{aligned}g_{1}g_{2} & = ( \begin{matrix}−\frac{1}{u} & 0 \\ 0 & −u \end{matrix} ) ( \begin{matrix}1 & −u \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}1 & −1/u \\ 0 & 1 \end{matrix} ) , \\ τ(g_{1})τ(g_{2}) & = τ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))τ(( \begin{matrix}1 & u \\ 0 & 1 \end{matrix} ))τ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )), \\ τ(g_{1}g_{2}) & = τ(( \begin{matrix}−\frac{1}{u} & 0 \\ 0 & −u \end{matrix} ))τ(( \begin{matrix}1 & −u \\ 0 & 1 \end{matrix} ))τ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))τ(( \begin{matrix}1 & −1/u \\ 0 & 1 \end{matrix} )). \end{aligned}

However if one examines the derivation of the identities (A) and (B) one sees that they are equivalent to the assertion that these two operators have the same effect on an element of Ĥ. To verify that the two operators are equal we need to show that if φ ∈ Ĥ then

τ(g_{1}g_{2})τ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))φ = τ(g_{1})τ(g_{2})τ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))φ

The left side is equal to

τ(g_{1}g_{2} ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))φ = τ(g_{1})τ(g_{2} ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))φ = τ(g_{1})τ(g_{2})τ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))φ

The representation τ on H certainly satisfies condition (ii). If \binom{a b}{c d} ∈ G_{0} and c ∈ 𝔭

( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}1 & 0 \\ −\frac{c}{a} & 1 \end{matrix} ) ( \begin{matrix}a & b \\ c & d − \frac{bc}{a} \end{matrix} ) .

It is clear that for any φ in H the sets

{g|τ(g)φ = φ},

\{g|τ(g)τ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))φ = τ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ))φ\} .

both contain an open subgroup of th group of upper triangular matrices. Thus the first set contains an open subgroup of the group of lower triangular matrices. It follows from the simple identity above that it contains an open subgroup of G_{k}.

To prove that the third condition is satisfied we need only show that if U is an open subgroup of the group of upper triangular matrices then the set X_{0} of all φ in H such that U is contained in the isotropy group of both φ and τ(\binom{0 1}{−1 0})φ is finite dimensional. If φ belongs to H then φ_{ν} has poles only at 0 and ∞. In general the poles of φ_{ν} at any point besides 0 and ∞ are of no higher order than those of Tν̂(z). It is clear that, if φ ∈ X, φ_{ν} = 0 for all but a finite number of ν. Thus to prove the assertion all we need to do is obtain, for each ν, a bound on the order of the pole of φ_{ν} at 0 and ∞ which is valid for all φ in X. A glance at the form of the operator τ(\binom{1 x}{0 1}) convinces one that there is a number N such that if U is in the isotropy group of φ then φ(α) = 0 if |α| > |π|^{N}. Thus the order of the pole of φ_{ν}(z) at 0 is at most −N. If φ is in X the order of the pole of

T_{ν}(z)φ_{ν}(ω^{−1}(π)z^{−1})

at 0 is also at most −N. The assertion follows.

Arguments similar to those used to prove Lemma 2.4 show that any invariant subspace of H different from {0} contains a non-zero vector in Ĥ and that Ĥ is irreducible under the action of the upper-triangular matrices. It follows immediately that τ is an irreducible representation of G_{k} on H.

Thus to completely classify all irreducible representations of G_{k} satisfying (i), (ii), and (iii) all we need to is study the families {T_{m,μ}} of complex numbers which satisfy (A), (B), and (D) and have the property that, for all μ, T_{m,μ} = 0 if m ≥ −1. In this case, which is the case we shall discuss in the rest of this chapter, H = Ĥ.

Before going on let me observe that if ζ is another homomorphism of k^{×} into ℂ^{×} and ω is replaced by ωζ^{2} and T_{m,μ} is replaced by ζ(π^{m})T_{m,ζ_{0}^{−1}μ} the relations (A), (B), and (D) continue to be satisfied. Thus, for our purposes, there is no harm in assuming that ω is a character.

Define an inner product on Ĥ by

(φ, ψ) = ∫_{k^{×}} φ(α)ψ(α)dα.

It is clear that, if g is an upper triangular matrix, (τ(g)φ, τ(g)ψ) = (φ, ψ). It is also clear that if ⟨φ, ψ⟩ is another inner product with this property it is of the form

⟨φ, ψ⟩ = ∑_{ν} a_{ν}(φ_{ν}, ψ_{ν}).

Thus if T is the operator on Ĥ defined by

T (∑ φ_{ν}) = ∑ a_{ν}φ_{ν}
(τ(g)T ψ, τ(g)ψ) = (T τ(g)φ, τ(g)ψ),

so that τ(g)T = T τ(g) for all upper triangular matrices g. Thus each eigenspace of T is invariant under τ(g); so T is a scalar.

Let φ_{ℓ,ν} be the function in Ĥ_{ν} = Ĥ ∩ L′_{ν} satisfying φ_{ℓ,ν}(π^{ℓ}) = 1 and φ_{ℓ,ν}(α) = 0 if |α| ≠ |π|^{ℓ}. The collection {φ_{ℓ,ν}} is an orthonormal basis of Ĥ. If φ is in Ĥ and

φ ∼ ∑_{ν} ∑_{ℓ} u_{ℓ,ν}z^{ℓ}

then

φ = ∑_{ν} ∑_{ℓ} u_{ℓ,ν}φ_{ℓ,ν}.

If

g = ( \begin{matrix}β & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}π^{n} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} )

with α ∈ o^{×} let us find the effect of τ(g) on φ. We iterate the effect of the various factors entering into the expression of τ(g) as a product.

τ(( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))φ = ∑_{ℓ,ν}\{∑_{μ} Δ (μν^{−1}, π^{ℓ}y) u_{ℓ,μ}\}φ_{ℓ,ν}.

Applying τ(\binom{α 0}{0 1}) to this one obtains

∑_{ℓ,ν} {ν(α) ∑_{μ} Δ(μν^{−1}, π^{ℓ}y)u_{ℓ,μ}} φ_{ℓ,ν}.

τ(\binom{π^{n} 0}{0 1}) sends this to

∑_{ℓ,ν} {ν(α) ∑_{μ} Δ(μν^{−1}, π^{ℓ+n}y)u_{ℓ+n,μ}} φ_{ℓ,ν}.

Now apply τ(\binom{1 x}{0 1}) to obtain

∑_{ℓ,ν} {∑_{m−k=ℓ} T_{m,ν̂}ω(π^{−k})ν̂(α) ∑_{μ} Δ(μν̂^{−1}, π^{k+n}y)u_{k+n,μ}} φ_{ℓ,ν}

Finally τ(\binom{β 0}{0 β})τ(\binom{1 x}{0 1}) transforms this to

∑_{ℓ,ν} {ω(β) ∑_{m−k=ℓ} ∑_{ρ,μ} ρ̃(α)ω(π^{−k})T_{m,ρ̃}Δ(ρν^{−1}, π^{ℓ}x)Δ(μρ̃^{−1}, π^{k+n}y)u_{k+n,μ}} φ_{ℓ,ν}.

Thus if g has the above form the matrix element (τ(g)φ_{k,μ}φ_{ℓ,ν}) is equal to

ω(β)ω(π^{n−k}) ∑_{ρ} ρ̃(α)T_{k+ℓ−n,ρ̃}Δ(ρν^{−1}, π^{ℓ}x)Δ(μρ̃^{−1}, π^{k}y)

If g = \binom{β 0}{0 β}\binom{π^{n} 0}{0 1}\binom{α 0}{0 1}\binom{1 y}{0 1} then (τ(g)φ_{k,μ}, φ_{ℓ,ν}) is equal to 0 if k ≠ ℓ + n but if k = ℓ + n it equals

ω(β)ν(α)Δ(μν^{−1}, π^{k}y).

A subset X of G_{k} will be called pseudo-compact if there is a compact subset Y of G_{k} such that X ⊆ ∪_{α∈k^{×}} \binom{α 0}{0 α}Y.

Lemma 2.9 If T_{m,μ} = 0 for m ≥ 1 the functions (τ(g)φ_{k,μ}, φ_{ℓ,ν}) have their support in a pseudocompact set.

It is clear that the intersection of the support of (τ(g)φ_{k,μ}, φ_{ℓ,ν}) with the group of uppertriangular matrices is a pseudo-compact set. If

g = ( \begin{matrix}β & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}π^{n} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} )

then

g = ( \begin{matrix}β & 0 \\ 0 & β \end{matrix} ) ( \begin{matrix}−π^{n}xα & 1 − π^{n}αxy \\ −π^{n}α & −π^{n}αy \end{matrix} ) .

Thus, if N > 0 and n varies over \{n|n ≥ −N\} while x and y vary over \{z ∈ k | |π|^{\frac{n}{2}} |z| < N\} and β varies over k^{×} the matrix g varies over a pseudo-compact set.

For a g of this form set

f_{ρ}(g) = ω(β)ω(π^{n−k})ρ̃(α)T_{k+ℓ−n,ρ̃}Δ(ρν^{−1}, π^{ℓ}x)Δ(μρ̃^{−1}, π^{k}y).

The support of f_{ρ} is certainly contained in a pseudo-compact set. As we saw some time ago, if the order of ρ is sufficiently large,

T_{k−m,ρ} = ω(π^{−m}) \frac{Δ(ρν_{0}^{−1}, π^{k})Δ(ω_{0}^{−1}ν_{0}ρ, π^{−m})}{|Δ(ω_{0}^{−1}ν_{0}ρ, π^{−m})|^{2}}

where ν_{0} is a fixed character and m is the order of ρ. Thus, if the order of ρ is sufficiently large, f_{ρ}(g) = 0 unless n = k + ℓ + 2m, |π^{m+ℓ}x| = 1, and |π^{k+ℓ}y| = 1. The lemma follows.

If ℓ and ν are fixed and C_{k} = \{(\binom{α 0}{0 α}) | α ∈ k^{×}\}

⟨φ, ψ⟩ = ∫_{G_{k}/C_{k}}(τ(g)φ, φ_{ℓ,ν})(\overline{τ(g)ψ, φ_{ℓ,ν}})dg

is a non-degenerate inner product on Ĥ. Clearly ⟨τ(g)φ, τ(g)ψ⟩ = ⟨φ, ψ⟩ for all g in G_{k} and in particular for the upper-triangular matrices. Thus there is a positive constant C_{ℓ,ν} such that ⟨φ, ψ⟩ ≡ C_{ℓ,ν}(φ, ψ). Consequently the representation τ is unitary.

Lemma 2.10 If the family {T_{m,μ}} of complex numbers satisfies the relations (A), (B), and (D) there is a two-dimensional semi-simple algebra K over k and a homomorphism M of K^{×} into ℂ^{×} such that

T_{m,μ} = T (M, μ, m)

for all m and μ.

Because of Lemma 2.7 we need only prove this when the associated representation τ acts on Ĥ, is unitary and the matrix element (τ(g)φ_{k,μ}, φ_{ℓ,ν}) have compact support. To do this we need the Plancherel formula of Gelfand and Graev which will require a paragraph by itself. For now let us assume Lemma 2.10 and go on to its applications to the theory of automorphic forms.

3. The local functional equation for non-archimedean fields. For the sake of brevity we shall call an irreducible representation σ of G_{k} which satisfies (i), (ii), and (iii) of the previous chapter a simple representation.

If η is a continuous homomorphism of A_{k}, the group of diagonal matrices in G_{k}, into ℂ^{×}, let L(η) be the space of all locally constant functions on G_{k} satisfying φ(ag) ≡ η(a)φ(g) for all a in A. Since L(η) is invariant under right translations we obtain a representation g → ρ(g) of G_{k} on L(η).

Lemma 3.1 No infinite-dimensional simple representation of G_{k} is contained more than once in the restriction of ρ to L(η).

We may take the simple representation to be the representation τ on H considered in the previous paragraph. Suppose V is a subspace of L(η) and T is an isomorphism of H with V such that

T (τ(g)φ) = ρ(g)T φ

for all φ. Set λ(φ) = T φ(1). Then T φ(g) = (ρ(g)T φ)(1) = λ(τ(g)φ). Thus T is completely determined by λ. If a ∈ A_{k} then

λ(τ(a)φ) = η(a)λ(φ).

Let us verify that up to a scalar factor there is at most one linear function on H with this property. Let η(\binom{a 0}{0 b}) = η_{1}(a)η_{2}(b) and, assumed, let τ(\binom{α 0}{0 α}) = ω(α)I. There is no such function unless η_{1}η_{2} = ω. If φ ∈ H_{ν} = H ∩ L′_{ν} and α ∈ o^{×} then η_{1}(α)λ(φ) = ν(α)λ(φ). Thus λ vanishes on H_{ν} unless ν = ν_{0}, the restriction of η_{1} to o^{×}. If φ ∈ H_{ν_{0}} and ψ = τ(\binom{π^{−1} 0}{0 1})φ − η_{1}(π)^{−1}φ or, what is the same, if

ψ(z) = (z − η_{1}^{−1}(π))φ(z)  (A)

then λ(ψ) = 0.

If H_{ν_{0}} = Ĥ_{ν_{0}} then {ψ(z) | ψ ∈ H_{ν_{0}}} consists of all rational functions with poles nowhere but at 0 and ∞. Then ψ(z) can be put in the above form if and only if η_{1}^{−1}(π) is a zero of ψ(z). The assertion follows in this case. If H_{ν_{0}} ≠ Ĥ_{ν_{0}} either

Tν̃_{0}(z) = cz^{−k} \frac{(z − γ_{1})}{(z − δ_{1})} \frac{(z − γ_{2})}{(z − δ_{2})}

or

Tν̃_{0}(z) = cz^{−k} \frac{z − γ_{1}}{z − δ_{1}}

Here c is a complex constant, k is an integer, and γ_{1}, γ_{2}, δ_{1}, δ_{2} are complex constants. In the first case we may suppose that γ_{i} ≠ δ_{j} for i, j = 1 or 2 and in the second case we may suppose that γ_{1} ≠ δ_{1}. In the first case {ψ(z) | ψ ∈ H_{ν_{0}}} consists of all rational functions with poles of arbitrary order at 0 and ∞, poles of order at most 1 at δ_{1} and δ_{2} and no other poles. In the second case it consists of all rational functions with poles of arbitrary order at 0 and ∞, a pole of order at most 1 at δ_{1}, and no other poles. In any case ψ(z) is of the form (A) if and only if the order of its pole at η_{1}^{−1}(π) is 1 less than the maximum allowable. This completes the proof of the lemma.

If ξ(x) is a non-trivial character of k let L(ξ) be the set of all locally constant functions on G_{k} satisfying φ(\binom{1 x}{0 1}g) ≡ ξ(x)φ(g) for all x ∈ k. Let ρ(ξ) be the restriction of the right regular representation to L(ξ).

Lemma 3.2 Every infinite-dimensional simple representation of G_{k} occurs exactly once in ρ(ξ).

Choose γ in k^{×} so that ξ(x) = ξ_{0}(γx). Let the simple representation τ act onH, as before. Suppose there is a homomorphism T of H into L(ξ) such that T (τ(g)φ) = ρ(g)(T φ). Set λ(φ) = T φ(1). Then λ(τ(\binom{1 x}{0 1})φ) = ξ_{0}(γx)λ(φ). Since T φ(g) = λ(τ(g)φ), T is detrmined by λ. Conversely if λ is such a linear function and T φ is defined by T φ(g) = λ(τ(g)φ) then T commutes with the action of G_{k}.

Such a linear function must annihilate all functions in H of the form ψ(α) = {ξ_{0}(γx) − ξ_{0}(αx)}φ(α) with φ in H. Since any function in H which vanishes at γ is a linear combination of such functions the assertion follows.

Suppose τ is a simple representation of G_{k}. Let K be a two-dimensional semi-simple algebra over k and let M be a homomorphism of K^{×} into ℂ^{×}. Suppose τ is associated to the family {T (M, μ, n)}. Let the restriction of M to k^{×} be χω. Suppose ζ is a continuous homomorphism of A_{k} into ℂ^{×} such that ζ(\binom{α 0}{0 α})ω(α) ≡ 1. Let ζ(\binom{α 0}{0 β}) = ζ_{1}(α)ζ_{2}(β). Let ζ_{0} be the restriction of ζ_{1} to o^{×} and let ζ_{1}(απ^{n}) = ζ_{0}(α)|π|^{s} for α ∈ o^{×}. ζ is uniquely determined by ζ_{0} and s and we shall occasionally write ζ = ζ(s, ζ_{0}). let L(ξ, τ) be the unique subspace of L(ξ) which transforms according to the representation τ.

If η is any continuous homomorphism of A_{k} into C^{×} let η̃ be the homomorphism defined by η̃(\binom{a 0}{0 b}) = η(\binom{b 0}{0 a}).

Lemma 3.3 If τ is given there is a number N such that if φ belongs to L(ξ, τ) and ζ = ζ(s, ζ_{0}) the integral

Φ(g, ζ, φ) = ∫_{k^{×}} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα

is defined for Re(s) > N.

(i) Suppose K = k ⊕ k and M(π^{p}α ⊕ π^{q}β) = ω_{1}^{p}ω_{2}^{q}μ_{1}(αβ) if α, β ∈ o^{×}. Suppose also that neither \frac{ω_{1}}{ω_{2}} nor \frac{ω_{2}}{ω_{1}} is equal to |π|. If μ_{1} = ζ_{0}^{−1} set

Φ′(g, ζ, φ) = (1 − ω_{1}|π|^{s+1/2})(1 − ω_{2}|π|^{s+1/2})Φ(g, ζ, φ).

Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover if 𝔭^{−d} is the largest ideal on which ξ is trivial

ζ_{1}(π^{d})Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, η, φ) = ζ̃_{1}(π^{d})Φ′(g, ζ̃, φ)

If μ_{1} ≠ ζ_{0}^{−1} set Φ′(g, ζ, φ) = Φ(g, ζ, φ). Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{−s} and |π|^{s} and, for a suitable choice of g and φ it is a constant. Moreover, if ξ(x) = ξ_{0}(γx),

ζ_{1}(γ)Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = (1 − |π|)^{2}|π|^{−n−2ns}ω_{1}^{−n}ω_{2}^{−n}ζ̃_{1}(γ) \{Δ(μ_{1}ζ_{0}, π^{−n})\}^{2} Φ′(g, ζ̃, φ)

if n is the order of μ_{1}ζ_{0}.

(ii) Suppose K = k ⊕k and M(π^{p}α⊕π^{q}β) = ω_{1}^{p}ω_{2}^{q}μ_{1}(αβ) if α, β ∈ o^{×}. Suppose also that \frac{ω_{1}}{ω_{2}} = |π|. If μ_{1} = ζ_{0}^{−1} set

Φ′(g, ζ, φ) = (1 − ω_{1}|π|^{s+1/2})Φ(g, ζ, φ)

Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

ζ_{1}(π^{d})Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = −\frac{|π|^{− \frac{1}{2} −s}}{ω_{2}} ζ̃_{1}(π^{d})Φ′(g, ζ̃, φ)

If μ_{1} ≠ ζ_{0}^{−1} set Φ′(g, ζ, φ) = Φ(g, ζ, φ). Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

ζ_{1}(γ)Φ′(( \begin{matrix}0 & 1 \\ −1 & 1 \end{matrix} ) g, ζ, φ) = (1 − |π|)^{2}|π|^{−n−2ns}ω_{1}^{−n}ω_{2}^{−n}ζ̃_{1}(γ) \{Δ(μ_{1}ζ_{0}, π^{−n})\}^{2} Φ′(g, ζ̃, φ)

(iii) Suppose K = k⊕k and M(π^{p}α⊕π^{q}β) = ω_{1}^{p}ω_{2}^{q}μ_{1}(αβ) if α, β ∈ o^{×}. Suppose also that \frac{ω_{2}}{ω_{1}} = |π|. If μ_{1} = ζ_{0}^{−1} set

Φ′(g, ζ, φ) = (1 − ω_{2}|π|^{s+1/2})Φ(g, ζ, φ).

Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

ζ_{1}(π^{d})Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = −\frac{|π|^{−1/2−s}}{ω_{1}} ζ̃_{1}(π^{d})Φ′(g, ζ̃, φ).

If μ_{1} ≠ η_{0}^{−1} set Φ′(g, ζ, φ) = Φ(g, ζ, φ). Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

ζ_{1}(γ)Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = (1 − |π|)^{2}|π|^{−n−2ns}ω_{1}^{−n}ω_{2}^{−n}ζ̃_{1}(γ)\{Δ(μ_{1}ζ_{0}, π^{−n})\}^{2}Φ′(g, ζ̃, ρ).

(iv) Suppose K = k ⊕ k and M(π^{p}α ⊕ π^{q}β) = ω_{1}^{p}ω_{2}^{q}μ_{1}(α)μ_{2}(β) if α, β ∈ o^{×} where μ_{1} ≠ μ_{1}. If μ_{1} = ζ_{0}^{−1} set

Φ′(g, ζ, φ) = (1 − ω_{1}|π|^{s+1/2})Φ(g, ζ, φ)

Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

ζ_{1}(γ)Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = (1 − |π|)|π|^{− \frac{n}{2} −ns}ω_{2}^{−n}Δ(ζ_{0}μ_{2}, π^{−n})ζ̃_{1}(γ)Φ′(g, ζ̃, φ)

if n is the order of μ_{1}^{−1}μ_{2}. If μ_{2} = ζ_{0}^{−1} set

Φ′(g, ζ, φ) = (1 − ω_{2}|π|^{s+1/2})Φ(g, ζ, φ).

Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover

ζ_{1}(γ)Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = (1 − |π|)|π|^{− \frac{n}{2} −ns}ζ̃_{1}(γ)ω_{1}^{−n}Δ(ζ_{0}μ_{1}, π^{−n})Φ′(g, ζ̃, φ)

If ζ_{0}^{−1} is different from both μ_{1} and μ_{2} set Φ′(g, ζ, φ) = Φ(g, ζ, φ). Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

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

is equal to

(1 − |π|)^{2}|π|^{(−n_{1}−n_{2})( \frac{1}{2} +s)}ζ̃_{1}(γ)ω_{1}^{−n_{2}}ω_{2}^{−n_{1}}Δ(μ_{2}ζ_{0}, π^{−n_{1}})Δ(μ_{1}ζ_{0}, π^{−n_{2}})Φ′(g, ζ̃, φ)

if n_{1} is the order of μ_{2}ζ_{0} and n_{2} is the order of μ_{1}ζ_{0}.

(v) Suppose K is an unramified extension of k and there is a homomorphism M_{1} of k^{×} into ℂ^{×} such that M(α) = M_{1}(Nα). Let M_{1}(π^{p}β) = ω_{1}^{p}μ_{1}(β) for p in o^{×}. If μ_{1} = ζ_{0}^{−1} set

Φ′(g, ζ, φ) = (1 − ω_{1}^{2}|π|^{2s+1})Φ(g, ζ, φ).

Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

ζ_{1}(π^{d})Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = ζ̃_{1}(π^{d})Φ′(g, ζ̃, φ)

If μ_{1} ≠ ζ_{0}^{−1} set Φ′(g, ζ, φ) = Φ(g, ζ, φ). Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

ζ_{1}(γ)Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = ζ̃_{1}(γ)(1 − |π|^{2})|π|^{−n−2ns}ω_{1}^{−2n}Δ((μ_{1}ζ_{0})^{1_{s}}, Π^{−n})Φ′(g, ζ̃, φ)

if n is the order of μ_{1}ζ_{0}.

(vi) Suppose K is an unramified extension of k and there is no homomorphism M_{1} of k^{×} into ℂ^{×} such that M(α) ≡ M_{1}(Nα). Set Φ′(g, ζ, φ) = Φ(g, ζ, φ). Then, for all g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

ζ_{1}(γ)Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = ζ̃_{1}(γ)(1 − |π|^{2})|π|^{−n−2ns}M(Π^{−n})Δ(M_{0}^{−1}(ω_{0}ζ_{0})^{1+s}, Π^{−n})Φ′(g, ζ̃, φ).

(vii) Suppose K is a ramified extension of k and there is a generalized character M_{1} of k^{×} such that M(α) ≡ M_{1}(Nα). Let M_{1}(π^{p}β) = ω_{1}^{p}μ_{1}(β) if β ∈ o^{×}. If μ_{1} = ζ_{0}^{−1} set

Φ′(g, ζ, φ) = (1 − ω_{1}|π|^{s+1/2})Φ(g, ζ, φ)

Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

ζ_{1}(γ)Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = (1 − |π|)|π|^{− \frac{f}{2} −fs}ω_{1}^{−f} ζ̃_{1}(γ)Δ(χ_{0}, π^{−f})Φ′(g, ζ̃, φ)

If μ_{1}χ_{0} = ζ_{0}^{−1} set

Φ′(g, ζ, φ) = (1 − ω_{1}|π|^{s+1/2})Φ(g, ζ, φ)

Then, for for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

ζ_{1}(γ)Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) = (1 − |π|)|π|^{− \frac{f}{2} −fs}ω_{1}^{−f} ζ̃_{1}(γ)Δ(χ_{0}, π^{−f})Φ′(g, ζ̃, φ)

If ζ_{0}^{−1} is equal to neither μ_{1} nor μ_{1}χ_{0} set Φ′(g, ζ, φ) = Φ(g, ζ, φ) Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s} and for a suitable choice of g and φ it is a constant. Moreover,

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

is equal to

(1 − |π|)^{2}|π|^{− \frac{n−f}{2} −(n+f)s}ω_{1}^{−n−f} ζ̃_{2}(γ)Δ(M_{0}^{−1}(ω_{0}ζ_{0})^{1+s}, Π^{−n−f})Δ(χ_{0}, π^{−f})Φ′(g, ζ̃, φ).

(viii) Suppose K is a ramified extension of k and there is no homomorphism M_{1} of k^{×} into ℂ^{×} such that M(α) ≡ M_{1}(Nα). Set Φ′(g, ζ, φ) = Φ(g, ζ, φ). Then, for each g, Φ′(g, ζ, φ) is a polynomial in |π|^{s} and |π|^{−s}. Moreoever,

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

is equal to (1 − |π|)^{2}|π|^{− \frac{n−f}{2} −(n−f)s}ζ̃_{1}(γ)M(Π^{−s(n+f)})Δ(χ_{0}, π^{−f})Δ(M_{0}^{−1}(ω_{0}ζ_{0})^{1+s}, Π^{−n−f})Φ′(g, ζ, φ).

Of course, ζ̃(\binom{a 0}{0 b}) = ζ̃_{1}(a)ζ̃_{2}(b). Thus ζ̃_{1} = ζ_{2}. Since ζ_{1}ζ_{2} = ω^{−1}, ζ̃_{1} = ω^{−1}ζ_{1}^{−1}. In particular, ζ̃_{0} = ω_{0}^{−1}ζ_{0}^{−1} so that ζ̃_{0} = ζ_{0} if ζ_{0}^{−2} = ω_{0}. If ξ(x) = ξ_{0}(γx) then the map φ → ψ with ψ(g) = φ(\binom{γ 0}{0 1}g) is an isomorphism of L(ξ_{0}, τ) with L(ξ, τ).

∫_{k^{×}} ψ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα = ζ_{1}^{−1}(γ) ∫_{k^{×}} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα.

This, together with the previous observation that ζ̃_{0} = ζ_{0} if ζ_{0}^{−1} = ω_{0}, makes it clear that it is enough to prove the lemma for ξ = ξ_{0}.

Since L(ξ_{0}, τ) is invariant under right translations it is enough to prove the assertions of the lemma for g = 1. The map ψ → T ψ where

T ψ(g) = (τ(g)ψ)(1)

is an isomorphism of H and L(ξ_{0}, τ). If φ = T ψ then

∫_{k^{×}} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))ζ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ))dα = ∫_{k^{×}} ψ(α)ζ_{1}(α)dα.

Since H ⊆ L the integral on the right converges if Re(s) is sufficiently large and

Φ(1, ζ, φ) = ψ_{ζ_{0}^{−1}}(|π|^{s})

The proof of Lemma 3.1, together with Lemma 2.8, shows that there are at most two points, which are independent of ν and ψ, besides 0 and ∞ where ψ_{ν}(z) can have a pole. This shows that for Re(s) sufficiently large the integral on the right converges for all ψ. Let ψ′ = τ(\binom{0 1}{−1 0})ψ. Then

\begin{aligned}Φ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, ζ, φ) & = ψ′_{ζ_{0}^{−1}}(|π|^{s}) \\ & = T_{ω_{0}ζ_{0}}(|π|^{s})ψ_{ω_{0}ζ_{0}}(ω^{−1}(π)|π|^{−s}) \end{aligned}

and

Φ(1, ζ̃, φ) = ψ_{ω_{0}ζ_{0}}(ω^{−1}(π)|π|^{−s}).

The lemma follows from these two relations, the formulae of Lemma 2.8, and the observations about {ψ(z) | ψ ∈ H_{ν}} made while proving Lemma 3.1. It is a matter of inspection which must be left to the reader.

Lemma 3.4 There is a vector in H whose isotropy group contains G_{o} only if ω_{0} is the trivial character. If ω_{0} is trivial the only cases of the previous lemma for which H contains such a vector are (i) and (v). In cases (i) and (v) H contains such a vector if and only if μ_{1} is trivial.

It is clear that such a function (or vector) can exist only if ω_{0} is trivial and that if ν_{0} is the trivial character of o^{×} it must lie in Hν_{0}. Suppose there is a function φ in Hν_{0} invariant under G_{o}. Then φ(z) has no pole at zero and

φ(z) = T ν_{0}(z)φ(ω^{−1}(π)z^{−1})

In all cases, T_{ν_{0}}(Z) has a pole of order at least two at 0. Thus φ(ω^{−1}(π)z^{−1}) has a zero of order at least two at 0 and φ(z) has a zero of order at least two at ∞. Consequently it has at least two poles in the finite plane. The discussion during the proof of Lemma 3.1 shows that this is possible only in the cases mentioned. Besides these two poles there can be no others. Thus the only zeros are at infinity and φ(z) is a constant multiple of

\frac{1}{(1 − ω_{1}z|π|^{1/2})} \frac{1}{(1 − ω_{2}z|π|^{1/2})}

in the first case and of

\frac{1}{1 − ω_{1}^{2}z^{2}|π|}

in the fifth.

Conversely if ω_{0} is trivial, φ lies in Hν_{0} and φ(z) has this form, the isotropy group of φ contains \binom{0 1}{−1 0} and the upper triangular matrices in G_{o}. However G_{o} is generated by \binom{0 1}{−1 0} ad the upper triangular matrices in it.

Lemma 3.5 No one-dimensional simple representation of G_{k} is continual in ρ(ξ).

According to the corollary to Lemam 2.1 any function on G_{k} which transformed according to a one-dimensional simple representation of G_{k} would be invariant on the right, and therefore on the left, under the group of matrices in G_{k} of determinant 1. In particular it would satisfy φ(\binom{1 x}{0 1}g) ≡ φ(g) for all x in k. Such a function could not possibly lie in L(ξ).

Let L_{0} be the space of all functions on N_{k} ∖ G_{k} which are G_{o} finite on the right.

Lemma 3.6 (i) Let K = k ⊕ k, let M(α_{1} ⊕ α_{2}) = χ_{1}(α_{1})χ_{2}(α_{2}) be a continuous homomorphism of K^{×} into ℂ^{×}, and let τ be the representation associated with the family {T (M, μ, m)}.

(a) Suppose χ_{1}χ_{2}^{−1} is not one of the characters α → 1, α → |α|, α → |α|^{−1}. Then there are two subspaces H_{1} and H_{2} of L_{0} which transform according to the representation τ and have the property that φ(\binom{α 0}{0 β}g) ≡ |\frac{α}{β} |^{1/2}χ_{1}(α)χ_{2}(β)φ(g) if φ ∈ H_{1} and φ(\binom{α 0}{0 β}g) ≡ |\frac{α}{β} |^{1/2}χ_{1}(β)χ_{2}(α)φ(g) if φ ∈ H_{2}. Moreover, any subspace H of L_{0} which transforms according to τ is contained in H_{1}+H_{2}.

(b) Suppose χ_{1} = χ_{2}. Then there are two subspaces H_{1} and H_{2} of L_{0} which transform according to the representation τ and an isomorphism T of H_{2} into H_{1} which commutes with the action of G_{k} and is such that φ(\binom{α 0}{0 β}g) ≡ |\frac{α}{β} |^{1/2}χ_{1}(α)χ_{2}(β)φ(g) if φ ∈ H_{1} and φ(\binom{α 0}{0 β}g) ≡ |\frac{α}{β} |^{1/2}χ_{1}(α)χ_{2}(β)\{φ(g) +T φ(g) log|\frac{α}{β} |\} if φ ∈ H_{2}. Moreover, any subspace of L_{0} which transforms according to τ is contained in H_{1} + H_{2}.

(c) Suppose χ_{1}χ_{2}^{−1}(α) ≡ |α|. Then there is a subspace H_{1} of L_{0} which transforms according to the representation τ and has the property that φ(\binom{α 0}{0 β}g) ≡ |\frac{α}{β} |^{1/2}χ_{1}(α)χ_{2}(β)φ(g) if φ ∈ H. Moreover H_{1} is the only subspace of L_{0} which transforms according to τ.

(ii) Let K be a separable extension of k and let M be a continuous homomorphism of K^{×} into ℂ^{×}. Let τ be the representation associated to the family {T (M, m, μ)}. If there is no continuous homomorphism M_{1} of k^{×} into ℂ^{×} such that M(α) ≡ M_{1}(Nα) then there is no subspace of L_{0} which transforms according to τ.

As in the proofs of Lemmas 3.1 and 3.2, there is a one:one correspondence between G-invariant homomorphisms T of H, the space on which τ acts as in paragraph 2, into L_{0} and linear functions λ on H satisfying λ (τ(\binom{1 x}{0 1})φ) = λ(φ) for all φ in H and all x ∈ k. Given such a linear function (T φ)(g) = λ(τ(g)φ). A linear function λ is of the required type if and only if it annihilates all functions of the form

ψ(α) = (ξ(αx) − 1)φ(α) φ ∈ H, x ∈ k.

The space spanned by such functions is just Ĥ. Now Ĥ_{ν} = H_{ν} for all but one or two characters ν. Moreover if H_{ν} ≠ Ĥ_{ν} then λ(φ_{ν}), φ_{ν} ∈ H_{ν} can depend only on the coefficients of the principal parts of φ_{ν}(z) at the finite poles different from 0.

Part (ii) of the lemma follows immediately. For part (i) let χ_{i}(π^{p}a) = ω_{i}^{p}μ_{i}(α) if α ∈ o^{×}. If we are in case (i,a) set

\begin{aligned}λ_{1}(φ) & = λ_{1}(∑ φ_{ν}) = Re(s)φ_{μ_{1}}(z) _{\frac{1}{ω_{1}|π|^{1/2}}} , \\ λ_{2}(φ) & = λ_{2}(∑ φ_{ν}) = Re(s)φ_{μ_{2}}(z) _{ω_{2}|π|^{1/2}}. \end{aligned}

Then λ is a linear combiantion of λ_{1} and λ_{2}. If we are in case (i,b) let

\frac{a_{1}}{(z − \frac{1}{ω_{2}|π|^{1/2}} )^{2}} + \frac{a_{2}}{(z − \frac{1}{ω_{1}|π|^{1/2}} )}

be the principal part of φ_{μ_{1}}(z) at \frac{1}{ω_{1}|π|^{1/2}} and set λ_{1}(φ) = λ_{1}(∑ φ_{ν}) = a_{1} and λ_{2}(φ) = λ_{2}(∑ φ_{ν}) = a_{2}. Then λ is a linear combination of λ_{1} and λ_{2}. If we are in case (i,c) let

λ_{1}(φ) = λ_{1}(∑ φ_{ν}) = Re(s)φ_{μ_{1}}(z) _{\frac{1}{ω_{1}|π|^{1/2}}} .

In all cases H_{i} is the image of H under the map T_{i} associated to λ_{i}. In case (i,b) take T_{0} = −\frac{1}{log |π|}T_{1}T_{2}^{−1}. The other assertions of the lemma follow from the form of the mapping associated to a given linear function, the fact that τ(\binom{α 0}{0 α})φ = χ_{1}(α)χ_{2}(α)φ, and the fact that if ψ = τ(\binom{π^{p}α 0}{0 1})φ with α ∈ o^{×} then ψ_{ν}(z) = ν(α)z^{−p}φ_{ν}(z).

4. The local functional equations reconsidered. In mathematics also “our beginnings never know our ends.” In order to give the main theorem a more striking form than was previously possible I want to reformulate the local functional equations. First of all let me recall the functional equations of the Hecke L-series.

Suppose K is a local field. We shall associate to each generalized character χ of K^{×} a function ξ(s, χ) of the complex variable s. We shall introduce a local factor ε(s, χ). ε(s, χ) will depend upon the choice of a character ξ of K. (Notice that the symbol ξ, like the symbol s, is used to denote two different objects.)

If K is a global field, χ a generalized character of K^{×}\I, and ξ a character of K\𝔸 let χ_{𝔭} and ξ_{𝔭} be the restrictions of χ and ξ to K_{𝔭}^{×} and K_{𝔭}, respectively. Define ξ(s, χ_{𝔭}) and ε(s, χ_{𝔭}) to be the local factors corresponding to ξ_{𝔭}. The (modified) zeta function associated to χ will be

Π_{𝔭}ξ(s, χ_{𝔭}) = ξ(s, χ)

It will satisfy the functional equation

\begin{aligned}ξ(s, χ) & = ε(s, χ)ξ(1 − s, χ^{−1}), \\ ε(s, χ) & = Π_{𝔭}ε(s, χ_{𝔭}), \end{aligned}

both products are taken over all primes, both finite and infinite.

Let us describe the functions ξ(s, χ) and ε(s, χ) for local fields.

(i) K = ℝ. Let χ(α) = (sgn α)^{M}|α|^{r}, with m = 0 or 1, and let ξ(x) = e^{2πiux}. Then

\begin{aligned}ξ(s, χ) & = π^{− \frac{1}{2} (s+r+m)}Γ(\frac{s + r + m}{2}), \\ ε(s, χ) & = \frac{(i sgn n)^{m}}{|u|^{1/2−s−r}} . \end{aligned}

(ii) K = ℂ. Let |α| be the square of the ordinary absolute value. Let χ(α) = |α|^{r}(\frac{α^{m}ᾱ^{n}}{|α|^{\frac{m+n}{2}}} ) with mn = 0 and m + n ≥ 0. Let ξ(z) = e^{4πi Re wz}. Then

\begin{aligned}ξ(s, χ) & = 2(2π)^{−(s+r+ \frac{m+n}{2} )}Γ(s + r + \frac{m + n}{2}) \\ ε(s, χ) & = i^{m+n}χ(w)|w|^{s− \frac{1}{2}} \end{aligned}

(iii) K is non-archimedean. Let 𝔓^{−d} be the largest ideal on which ξ is trivial. If Π is a generator of 𝔓 and the conductor of χ is 0

\begin{aligned}ξ(s, χ) & = \frac{1}{1 − χ(Π)|Π|^{s}} \\ ε(s, χ) & = χ(Π^{d})|Π^{d}|^{s− \frac{1}{2}} \end{aligned}

If the conductor of χ is 𝔓^{n} with n > 0

ξ(s, χ) = 1,

ε(s, χ) = χ(Π^{d+n})|Π^{d}|^{s− \frac{1}{2}} \frac{1 − |Π|}{|Π|^{n/2}} ∫_{O^{×}} ξ(\frac{α}{Π^{d+n}} )χ^{−1}(α)dα.

Before restating the local functional equations let me introduce some conventions. Let k be a local field. Let us introduce some language which, though rather bizarre, will be useful. If k = ℝ a simple representation of G_{k} is an irreducible quasi-simple representation of {σ, 𝔄} (the notation is that of paragraph 2 of my letter to Weil). If k = ℂ a simple representation of G_{k} is an irreducible quasi-simple representation of 𝔄 (the notation is that of paragraph 4 of my letter). If k is nonarchimedean the simple representations of G_{k} have been introduced in the previous paragraph. If τ is an infinite-dimensional simple representation of G_{k} and ξ is a character of k the space L(ξ, τ) has been defined.

If χ is a homomorphism of k^{×} into ℂ^{×} and s a complex number and φ belongs to L(ξ, τ) set

Φ(g, s, φ, χ) = ∫_{k^{×}} φ (( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) χ(α)|α|^{s}dα.

The integral converges for Re(s) sufficiently large. We shall introduce factors^{†} ξ(s, τ, χ) and ε(s, τ, χ) and set

Φ′(g, s, φ, χ) = \frac{Φ(g, s, φ, χ)}{ξ(s, τ, χ)}

Then the local functional equation will be

Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g, −s, φ, (ηχ)^{−1}) = ε(s, τ, χ)Φ′(g, s, φ, χ).

if^{‡} τ(\binom{α 0}{0 α}) ≡ η(α)I. I shall write down the factors ξ(s, τ, χ) and ε(s, τ, χ) but I will leave to the reader the task of verifying that the local functional equation takes the above form. He will probably require paper and pencil. Since the analytical properties of the functions Φ′(g, s, τ, χ) follow immediately from previous results I shall not formulate them explicitly either.

(i) k = ℝ

(a) Let M be a continuous homomorphism of ℝ^{×} × ℝ^{×} into ℂ^{×}. Let

M((t_{1}, t_{2})) = |t_{1}|^{s_{1}}|t_{2}|^{s_{2}}(\frac{t_{1}}{|t_{1}|})^{m_{1}}(\frac{t_{2}}{|t_{2}|})^{m_{2}}

with m_{1} and m_{2} equal to 0 or 1. Suppose (s_{1} − s_{2}) − (m_{1} − m_{2}) is not an odd integer. Set χ_{1}(t) = M((t, 1)), χ_{2}(t) = M((1, t)). Let τ = τ_{M} be the simple representation π_{M} introduced in paragraph 2 of my letter to Weil. Set

\begin{aligned}ξ(s, τ, χ) & = ξ(\frac{1}{2} + s, χ_{1}χ)ξ(\frac{1}{2}, s, χ_{2}χ), \\ ε(s, τ, χ) & = ε(\frac{1}{2} + s, χ_{1}χ)ε(\frac{1}{2} + s, χ_{2}χ). \end{aligned}

(Notice when verifying this that there is an error in part (i) on page 3.34 of the letter to Weil. ^{∗} The second factor in the denominator on the right should be Γ(z + |m_{2} − ℓ| + \frac{1}{2} − \frac{s}{2}).)

† They, too, will depend on the choice of a character of k .

‡ I leave it to the reader to give a meaning to τ(\binom{α 0}{0 α}) in the case of the real or complex field.

∗ in Lemma 3.6 (1998)

(b) Let M be a continuous homomorphism of ℂ^{×} into ℂ^{×}. Suppose

M(α) = (Nα)^{r} \frac{α^{m}α^{−n}}{|α|^{\frac{m+n}{2}}}

with mn = 0, m + n ≥ 0. Let ω be the homomorphism (t_{1}, t_{2}) → |t_{1}t_{2}|^{r}|\frac{t_{1}}{t_{2}} |^{\frac{m+n}{2}} sgn t_{1} of ℝ^{×} × ℝ^{×} into ℂ^{×} and let τ = τ_{M} be the unique infinite-dimensional irreducible representation deducible from π_{ω}. If ξ is a character of ℝ then ξ′(z) = ξ(z + z̄) is a character of ℂ. If χ is a homomorphism of ℝ^{×} into ℂ^{×} then χ′(α) = χ(Nα) = χ(αᾱ) is a homomorphism of ℂ into ℂ^{×}. Set

\begin{aligned}ξ (s, τ, χ) & = ξ (s + \frac{1}{2}, Mχ′) , \\ ε(s, τ, χ) & = (i sgn u)ε (s + \frac{1}{2}, Mχ′) . \end{aligned}

Of course the expressions on the left are for the character ξ and those on the right are for the character ξ′.

(c) Suppose M is a continuous homomorphism of ℝ^{×} × ℝ^{×} into ℂ^{×} of the form (t_{1}, t_{2}) → |t_{1}t_{2}|^{r}sgn t_{1} or (t_{1}, t_{2}) → |t_{1}t_{2}|^{r}sgn t_{2}. In the first case let χ_{1}(t) = |t|^{r}sgn t, χ_{2}(t) = |t|^{r}; in the second case set χ_{1}(t) = |t|^{r}, χ_{2}(t) = |t|^{r}sgn t. the representation π_{M} introduced in paragraph 2 of my letter to Weil is irreducible. Let τ = τ_{M} be π_{M}. Set

\begin{aligned}ξ(s, τ, χ) & = ξ(\frac{1}{2} + s, χ_{1}χ)ξ(\frac{1}{2} + s, χ_{2}χ), \\ ε(s, τ, χ) & = ε(\frac{1}{2} + s, χ_{1}χ)ε(\frac{1}{2} + s, χ_{2}χ). \end{aligned}

(ii) k = ℂ. Let M be continuous homomorphism of ℂ^{×} × ℂ^{×} into ℂ^{×}. Let M((t_{1}, t_{2})) = |t_{1}|^{s_{1}}|t_{2}|^{s_{2}}(\frac{t_{1}}{|t_{1}|^{1/2}} )^{m_{1}}(\frac{t_{2}}{|t_{2}|^{1/2}} )^{m_{2}} and suppose that neither \frac{s_{1}−s_{2}}{2} − (1 + \frac{|m_{1}−m_{2}|}{2}) nor \frac{s_{2}−s_{1}}{2} − (1 + \frac{|m_{1}−m_{2}|}{2}) is a non-negative integer. The representation π_{M} introduced in paragraph 4 of my letter to Weil is irreducible. Let τ = τ_{M} be π_{M}. Set

\begin{aligned}ξ(s, τ, χ) & = ξ(s + \frac{1}{2}, χ_{1}χ)ξ(s + \frac{1}{2}, χ_{2}χ), \\ ε(s, τχ) & = ε(s + \frac{1}{2}, χ_{1}, χ)ε(s + \frac{1}{2}, χ_{2}χ), \end{aligned}

if χ_{1}(t) = M((t, 1)) and χ_{2}(t) = M((1, t)).

(iii) k is a non-archimedean field.

(a) Let M be a continuous homomorphism of k^{×} × k^{×} into ℂ^{×}. Let M((α, β)) = χ_{1}(α)χ_{2}(β). Suppose that neither χ_{1}χ_{2}^{−1} nor χ_{1}χ_{1}^{−1} is the character α → |α|. Let τ = τ_{M} be the simple representation associated to the family {T (M, μ, m)}. Set

\begin{aligned}ξ(s, τ, χ) & = ξ(s + \frac{1}{2}, χ_{1}χ)ξ(s_{\frac{1}{2}} , χ_{2}χ) \\ ε(s, τ, χ) & = ε(s + \frac{1}{2}, χ_{1}χ)ε(s + \frac{1}{2}, χ_{2}χ) \end{aligned}

(b) Suppose K is an unramified extension of k and M is a continuous homomorphism of K^{×} into ℂ^{×}. Let τ = τ_{M} be the representation associated to the family {T (M, μ, m)}. If ξ is a character of k then ξ′(x) = ξ(Sx) is a character of K. If χ is a continuous homomorphism of k^{×} into ℂ^{×} let χ′ be the homomorphism α → χ(Nα) of K^{×} into ℂ^{×}. Set^{†}

\begin{aligned}ξ(s, τ, χ) & = ξ(s + \frac{1}{2}, Mχ′) \\ ε(s, τ, χ) & = ρ(K/k)ε(s + \frac{1}{2}, Mχ′) \end{aligned}

The factors on the left are taken with respect to ξ and those on the right with respect to ξ′.

(c) Suppose K is a ramified extension of k and M is a continuous homomorphism of K^{×} into ℂ^{×}. Let τ = τ_{M} be the representation associated to the family {T (M, μ, m)}. If ξ is a character of k then ξ′(x) = ξ(Sx) is a character of K. If χ is a continuous homomorphism of k^{×} into ℂ^{×} let χ′ be the homomorphism α → χ(Nα) of K^{×} into ℂ^{×}. Set

\begin{aligned}ξ(s, τ, χ) & = ξ(s + \frac{1}{2}, Mχ′), \\ ε(s, τ, χ) & = ρ(K/k)ε(s + \frac{1}{2}, Mχ′), \\ ρ(K/k) & = (1 − |π|)|π|^{− \frac{f}{2}} χ_{0} (π^{f+d}) ∫_{O^{×}} ξ(\frac{α}{π^{f+d}} )χ_{0}^{−1}(α)dα, \end{aligned}

if 𝔭^{−d} is the largest ideal on which ξ is trivial. Notice that this expression is independent of the choice of π but not of ξ. χ_{0} is of course the unique non-trivial character of k^{×}/NK^{×}.

(d) Suppose M((t_{1}, t_{2})) = χ_{1}(t_{1})χ_{2}(t_{2}) is a continuous homomorphism of k^{×} × k^{×} into ℂ^{×} and suppose χ_{1}χ_{2}^{−1}(α) ≡ |α|. Let τ be the representation associated to the family {T (M, μ, m)}

\begin{aligned}ξ(s, τ, χ) & = ξ(s + \frac{1}{2}, χ_{1}χ) \\ ε(s, τ, χ) & = −χχ_{1}(π^{2d+1})|π|^{(2d+1)(s−1/2)} \end{aligned}

† If 𝔭^{−d} is the largest ideal of k on which ξ is trivial and if χ_{0} is the unique non-trivial character of k^{×}/NK^{×} then ρ(K/k) = χ_{0}(π^{d}). It is independent of the choice of π. if the conductor of χχ_{1} is o and

ε(s, τ, χ) = (1 − |π|)^{2}|π|^{−d−2n}χχ_{1}(π^{2d+2n})|π|^{(2d+2n)s}\{∫_{o^{×}} ξ(\frac{α}{π^{d+n}} )χ^{−1}χ_{1}^{−1}(α)dα\}^{2}

if the order of χχ_{1} is n.

(e) Suppose M((t_{1}, t_{2})) = χ_{1}(t_{1})χ_{2}(t_{2}) is a continuous homomorphism of k^{×} × k^{×} into ℂ^{×} and suppose χ_{1}^{−1}χ_{2}(α) ≡ |α|. Let τ be the representation associated to the family {T (M, μ, m)}. Set

\begin{aligned}ξ(s, τ, χ) & = ξ(s + \frac{1}{2}, χ_{2}χ), \\ ε(s, τ, χ) & = −χχ_{2}(π^{2d+1})|π|^{(2d+1)(s−1/2)}, \end{aligned}

if the conductor of χχ_{1} is o and

ε(s, τ, χ) = (1 − |π|)^{2}|π|^{−d−2n}χχ_{2}(π^{2d+2n})|π|^{(2d+2n)s}\{∫_{o^{×}} ξ(\frac{α}{π^{d+n}} )χ^{−1}χ_{2}^{−1}(α)dα\}^{2}

if the order of χχ_{2} is n.

The representations of (d) and (e) are anomalous. I do not know if they have any role to play in the theory of automorphic forms. Before coming to the main theorem there is an observation we should make. Suppose k is a local field, K a two-dimensional semi-simple algebra over k, and ξ a character of k. If k is non-archimedean and K is a field we have introduced the symbol ρ(K/k) = ρ(K/k, ξ). If k = ℝ and K = ℂ and ξ(x) = e^{2πiux} set ρ(K/k, ξ) = i sgn u. If K is not a field set ρ(K/k, ξ) = 1. Now let k be a global field, K a two-dimensional semi-simple algebra over k, and ξ a character of 𝔸/k. If 𝔭 is a prime of k let K_{𝔭} = K ⊗_{k} k_{𝔭} and let ξ_{𝔭} be the restriction of ξ to k_{𝔭}. I claim that

Π_{𝔭}ρ(K_{𝔭}/k_{𝔭}, ξ_{𝔭}) = 1

This is clear if Kis not a field. If K is a field the (modified) zeta function of K is

Π_{𝔓}ξ(s, 1_{𝔓}) = ξ_{K}(s, 1)

On the other hand if χ is the unique non-trivial character of I_{k}/k^{×}NI_{K} it is

Π_{𝔭}ξ(s, 1_{𝔭})ξ(s, χ_{𝔭})

Taking as our character of 𝔸_{K}/K the character x → ξ(Sx) we find that

\frac{ξ_{K}(s, 1)}{ξ_{K}(1 − s, 1)} = Π_{𝔓}ε(s, 1_{𝔓}) = Π_{𝔭}ε(s, 1_{𝔭})ε(s, χ_{𝔭})

Checking things case by case we find that, for all 𝔭,

{Π_{𝔓|𝔭}ε(s, 1_{𝔓})}ρ(K_{𝔭}/k_{𝔭}, ξ_{𝔭}) = ε(s, 1_{𝔭})ε(s, χ_{𝔭})

The result follows. it is of course well known. I remark it because it shows immediately that the main theorem is applicable to the Hecke L-series over a quadratic extension of the ground field.

Appendix. There are a few facts which it will be useful to have at our disposal when proving the main theorem. For lack of a better place I record them here. Suppose τ(\binom{α 0}{0 α}) = η(α)I. Let ζ = ζ(χ, s) be defined by

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

Then the map

φ → Φ′( · , s, φ, χ)

is a homomorphism of L(ξ, τ) into the unique subspace of L(ζ) transforming according to the representation τ (cf. Lemma 3.1 and Lemmas 3.1 and 5.1 of the previous letter). Since we know that, for a suitable choice of g and φ, Φ′(g, s, φ, χ) is a non-zero exponential in s, this homomorphism can never be zero.

On the other hand we know (cf. Lemma 3.5 and the appendix to paragraph 7 of the previous letter) that for some τ and some continuous homomorphisms ω of A_{k} = \{\binom{α 0}{0 β} | α ∈ k^{×}, β ∈ k^{×}\} into ℂ^{×} there is a “G_{k}-invariant” map of L(ξ, τ) into the space of function on G_{k} satisfying φ(\binom{1 x}{0 1}\binom{α 0}{0 β}g) = ω(\binom{α 0}{0 β})φ(g). The image of L(ξ, τ) will, in particular lie in L(ω′) if ω′(\binom{α 0}{0 β}) = |\frac{α}{β} |^{1/2}ω(\binom{α 0}{0 β}). Thus if ω′ = ζ(s, χ) it must be a constant multiple of the map φ → Φ′( · , s, φ, χ).

Suppose L(ζ τ) is an invariant subspace of L(ω′) which transforms according to the representation τ. Suppose N_{1} and N_{2} are two spaces of functions on G_{k} invariant under the right regular representation (of {σ, 𝔄}, 𝔄, or G_{k} according as k is real, complex, or non-archimedean). Suppose N_{1} and N_{2} are irreducible and transform according to τ. Suppose also that there are isomorphisms T_{1} and T_{2} of N_{1} and N_{2}, respectively, with L(ζ, τ) such that if φ ∈ N_{i}

φ (( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) g) = ζ (( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )) \{φ(g) + c_{1} log |\frac{α}{β}| T_{1}φ(g)\}

where c_{i}, i = 1, 2 is a non-zero constant. Set T = T_{2}^{−1}T_{1}. Then, if φ ∈ H_{1}, c_{2}φ − c_{1}T φ ∈ L(ζ, τ). Thus N_{1} + L(ζ, τ) = N_{2} + L(ζ, τ). if ζ(\binom{α 0}{0 β}) = η(β)χ(βα^{−1})|βα^{−1}|^{2} then the set of functions

\frac{d}{ds}Φ′( · s, φ, χ), φ ∈ L(ζ, τ),

would be a possible choice for N_{1}. On the other hand if τ = τ_{M} where M is a homomorphism of k^{×} × k^{×} into ℂ^{×} of the form M((α, β)) = χ(αβ) and ω′(\binom{α 0}{0 β}) = |\frac{α}{β} |^{1/2}χ(αβ) then both L(ω′, τ) and N_{2} can be taken^{†} to lie in the space of functions on G_{k} satisfying φ(\binom{1 x}{0 1}g) ≡ φ(g).

5. The Main Theorem. Now let k be a global field and let 𝔸 be the adèle ring of k. The corrected form of Lemma 7.1 of the previous letter is

† Notice that in part (ii) of Lemma A in the appendix to paragraph 7 of the previous letter one should have s = 0 and m = 0.

Lemma 5.1 There is a constant c_{0} such that if g belongs to G_{𝔸} there is a γ in G_{k} such that Π_{𝔭} max(|c|_{𝔭}, |d_{𝔭}|) ≤ c_{2}|det g|^{1/2} if γ_{g} = \binom{a b}{c d}.

There seems little point in including a proof of this.

Let us take the space ℒ as in the previous letter except for making the modification in condition (iii) required by the change in Lemma 7.1.

Suppose that V is a complex vector space and for each real prime 𝔭 we have a representation of {σ_{𝔭}, 𝔄_{𝔭}} in V, for each complex prime a representation of 𝔄_{𝔭} on V. If any two operators associated to distinct primes commute we shall, for the purposes of this paragraph, say that we have a “representation” of G_{𝔸} on V.

Suppose in particular that for each prime 𝔭 we are given a simple representation τ_{𝔭} of G_{k_{𝔭}} (in the sense of the previous paragraph) on a vector space V_{𝔭}. Suppose moreover that for almost all non-archimedean primes V_{𝔭} contains a non-zero vector invariant under G_{o_{𝔭}}. since this vector is determined up to a scalar factor we have in all but finitely many of the V_{𝔭} a distinguished onedimensional subspace and we can form the tensor product ⊗_{𝔭}V_{𝔭}. The natural “representation” of G_{𝔸} on V will be denoted ⊗_{𝔭}τ_{𝔭}. A “representation” of G_{𝔸} equivalent to such a representation will be called a simple representation of G_{𝔸}.

Certainly we have a “representation” of G_{𝔸} on ℒ. An invariant subspace of ℒ which transforms according to a simple representation of G_{𝔸} will be called a characteristic space of automorphic forms. Suppose L is a characteristic space of automorphic forms and let ξ be a character of k/𝔸. If φ ∈ L set

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

As before

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

Suppose the “representation” of G_{𝔸} on Lis equivalent to ⊗_{𝔭}τ_{𝔭}. If one of the τ_{𝔭} is finite dimensional it follows rather easily from Lemma 3.5 of this letter and the corollaries to Lemma 3.2 and 5.4 of the previous letter that, for all φ in L, φ_{1}(g) ≡ 0. Then φ(hg) ≡ φ(g) if h ∈ G_{k} or h = (\binom{1 x}{0 1}) with x ∈ 𝔸. The argument used in the proof of Lemma 2.1 shows rather easily that, if G^{0} is the group of matrices of determinant 1 in G, φ is a function on G_{𝔸}\G_{𝔸}^{0}. Consequently L is one dimensional. We exclude this case from the following discussion.

With this case excluded the function φ_{1} can never vanish identically. For a suitable choice of φ it is of the form

φ_{1}(g) = φ_{1}(Π_{𝔭}g_{𝔭}) = Π_{𝔭}φ_{𝔭}(g_{𝔭})

with φ_{𝔭} in L(ξ_{𝔭}, τ_{𝔭}). Moreover we can suppose that for almost all non-archimedean primes φ_{𝔭}(1) = 1

Lemma 5.2 Suppose φ_{0} is different from zero for some φ in L. Then there is a continuous homomorphism M of k^{×}\I × k^{×}\I such that τ_{𝔭} = τ_{M_{𝔭}} for any prime for which τ_{M_{𝔭}} is defined. If τ_{M_{𝔭}} is not defined and 𝔭 is archimedean then τ_{𝔭} is the unique infinite-dimensional simple representation deducible from π_{M_{𝔭}}. if τ_{M_{𝔭}} is not defined and 𝔭 is non-archimedean τ_{𝔭} is the simple representation associated to the family {T (M_{𝔭}, μ, m)}. Let M((α, α)) = η(α).

(i) Suppose M((α, β)) = η(β)χ(βα^{−1})|βα^{−1}|^{s_{0}}. If M((α, 1)) ≢ M((1, α)) there are constants c_{1} and c_{2} such that when φ_{1} is of the above form

φ_{0}(g) = c_{1}Π_{𝔭}Φ′(g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) + c_{2}Π_{𝔭}Φ′(g_{𝔭}, −\frac{1}{2} − s_{0}, φ_{𝔭}, (η^{−1}χ^{−1})_{𝔭})

If M((α, 1)) ≡ M((1, α)) there are constants c_{1} and c_{2} such that when φ_{1} is of the above form

φ_{0}(g) = c_{1}Π_{𝔭}Φ′(g_{𝔭}, s_{0} + \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) + c_{2} \frac{d}{ds}Π_{𝔭}Φ′(g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}).

(ii) Suppose M((β, α)) = η(β)χ(βα^{−1})|βα^{−1}|^{s_{0}+1/2}. If M((α, 1)) ≢ M((1, α)) there are constants^{†} c_{1} and c_{2} such that when φ_{1} is of the above form

φ_{0}(g) = c_{2}Π_{𝔭}Φ′(g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) + c_{1}Π_{𝔭}Φ′(g_{𝔭}, −\frac{1}{2} − s_{0}, φ_{𝔭}, (η^{−1}χ^{−1})_{𝔭})

If M((α, 1)) ≡ M((1, α) there are constants c_{1} and c_{2} such that when φ_{1} is of the above form

φ_{0}(g) = c_{2}Π_{𝔭}Φ′(g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) + c_{1} \frac{d}{ds}Π_{𝔭}Φ′(g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭})

The proof of this lemma will be based on the appendix to paragraph 4 and Lemma E of the appendix to paragraph 7 of the previous letter. However the proof of that lemma was written up rather hastily so I do not have complete confidence in it. I will examine it more carefully later. If it turns out to be unsatisfactory I shall let you know. In order to get on to the main point I will take Lemma 5.2 for granted.

† The constants of parts (i) and (iii) are the same.

In proving the main theorem I shall not enter into questions of convergence. Anything which is not discussed in the previous letter is taken care of by Lemma 5.2 Thus if χ is a continuous homomorphism of k^{×}\I into ℂ^{×} and φ_{1} is of the above form

∫_{I} φ_{1}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)χ(α)|α|^{s}dα

converges absolutely for Re(s) sufficiently large. It is equal to

{Π_{𝔭}ξ(s, τ_{𝔭}, χ_{𝔭})} {Π_{𝔭}Φ′(g_{𝔭}, s, φ_{𝔭}, χ_{𝔭})}

On the other hand it is equal to

∫_{k^{×}\I} \{φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) − φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)\} χ(α)|α|^{s}dα

This is equal to the sum of^{†}

∫_{\{α||α|≥1\}} \{φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) − φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)\} χ(α)|α|^{s}dα

and

∫_{\{α||α|≤1\}} \{φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) − φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)\} χ(α)|α|^{s}dα

The first of these integrals is an entire function of s On the other hand if φ(\binom{α 0}{0 α}g) ≡ η(α)φ(g) for α ∈ I

∫_{\{α||α|≥1\}}\{φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} )( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g)−φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} )( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g)\} (ηχ)^{−1}(α)|α|^{s}dα

= ∫_{\{α||α|≤1\}}\{φ(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )( \begin{matrix}α^{−1} & 0 \\ 0 & 1 \end{matrix} )( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g) − φ_{0}(( \begin{matrix}α^{−1} & 0 \\ 0 & 1 \end{matrix} )( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g)\} ηχ(α)|α|^{s}dα

= ∫_{\{α||α|≤1\}}\{φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)−φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)\} χ(α)|α|^{s}dα

+ ∫_{\{α|α|≤1\}}\{φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)−η(α)φ_{0}(( \begin{matrix}α^{−1} & 0 \\ 0 & 1 \end{matrix} ) g)\} χ(α)|α|^{s}dα

Let us suppose that φ_{0} is not zero for all φ and consider the last integral. Let M be the homomorphism of Lemma 5.2 and let M((α, β)) = χ_{1}(α)χ_{2}(β). If neither χ_{1}χ nor χ_{2}χ is trivial

† At first we shall discuss the case of a number field. Afterwards the necessary modifications for a function field will be indicated. The argument of the previous letter was not correct for a function field. on the idèles of norm one this integral is zero. Suppose that χ_{1}χ is trivial on the idèles of norm one but χ_{2}χ is not. Let χ_{1}χ(α) = |α|^{−s_{0}}. Then the integral is equal to

c_{1}Π_{𝔭}Φ′ (g_{𝔭}, s_{0}− \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) ∫_{0}^{1}t^{s−s_{0}+ \frac{1}{2}} \frac{dt}{t} −c_{2}Π_{𝔭}Φ′(\binom{0 1}{−1 0}g_{𝔭}, −s_{0}− \frac{1}{2}, φ_{𝔭}, (η^{−1}χ^{−1})_{𝔭})∫_{0}^{1}t^{s−s_{0}− \frac{1}{2}} \frac{dt}{t}

= \frac{c_{1}}{\frac{1}{2} +s−s_{0}} Π_{𝔭}Φ′(g_{𝔭}, s_{0}− \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) + \frac{c_{2}}{\frac{1}{2} −s+s_{0}} Π_{𝔭}Φ′(\binom{0 1}{−1 0}g_{𝔭}, −s_{0} − \frac{1}{2}φ_{𝔭}, (η^{−1}χ^{−1})_{𝔭}). On the other hand if χ_{2}χ is trivial on the idèles of norm one and χ_{1}χ is not, let χ_{2}χ(α) = |α|^{−s_{0}}. Then the integral is equal to

\frac{c_{2}}{\frac{1}{2} + s − s_{0}} Π_{𝔭}Φ′(g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) + \frac{c_{1}}{\frac{1}{2} − s + s_{0}} Π_{𝔭}Φ′(( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) g_{𝔭}, −s_{0} − \frac{1}{2}, φ_{𝔭}, (η^{−1}χ^{−1})_{𝔭}).

Since

Π_{𝔭}Φ′(( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) = Π_{𝔭}Φ′(g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭})

it is clear, in this case at least, that these expressions do not change if g is replaced by \binom{0 1}{−1 0}g, χ by η^{−1}χ^{−1} and s by −s.

Now suppose that χ_{1} ≠ χ_{2} but both χ_{1}χ and χ_{2}χ are trivial on the idèles of norm 1. Let χ_{1}χ(α) = |α|^{−s_{1}}, χ_{2}χ(α) = |α|^{−s_{2}}. Then the integral is equal to

\frac{c_{1}}{\frac{1}{2} +s−s_{1}} Π_{𝔭}Φ′ (g_{𝔭}, s_{1} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) + \frac{c_{2}}{\frac{1}{2} +s−s_{2}} Π_{𝔭}Φ′ (g_{𝔭}, s_{2} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) + \frac{c_{1}}{\frac{1}{2} −s+s_{2}} Π_{𝔭}Φ′(\binom{0 1}{−1 0}g_{𝔭}, −s_{2}− \frac{1}{2}, φ_{𝔭}, (η^{−1}χ^{−1})_{𝔭})+\frac{c_{2}}{\frac{1}{2} −s+s_{1}} Π_{𝔭}Φ′(\binom{0 1}{−1 0}g_{𝔭}, −s_{1}− \frac{1}{2}, φ_{𝔭}, (η^{−1}χ^{−1})_{𝔭}). When χ is replaced by η^{−1}χ^{−1}, s_{1} is replaced by −s_{2} and s_{2} is replaced by −s_{1}. Thus this expression is not changed if s is replaced by −s, χ by η^{−1}χ^{−1}, and g by \binom{0 1}{−1 0}g.

Finally suppose that χ_{1} = χ_{2} and χ_{1}χ(α) = |α|^{−s_{0}}. Then the integral is equal to

c_{2} \frac{d}{ds}Π_{𝔭}Φ′(g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) ∫_{0}^{1} t^{\frac{1}{2} +s−s_{0}} \frac{dt}{t}

+ Π_{𝔭}Φ′(g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) ∫_{0}^{1}(c_{1} − c_{2} log t)t^{\frac{1}{2} +s−s_{0}} \frac{dt}{t}

− c_{2} \frac{d}{ds}Π_{𝔭}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, −s_{0} − \frac{1}{2}, φ_{𝔭}, (ηχ)_{𝔭}^{−1}) ∫_{0}^{1} t^{s−s_{0}− \frac{1}{2}} \frac{dt}{t}

− Π_{𝔭}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, −s_{0} − \frac{1}{2}φ_{𝔭}, (ηχ)_{𝔭}^{−1}) ∫_{0}^{1}(c_{1} + c_{2} log t)t^{s−s_{0}− \frac{1}{2}} \frac{dt}{t}.

This is of course equal to

\frac{c_{2}}{\frac{1}{2} +s−s_{0}} \frac{d}{ds}Π_{𝔭}Φ′(g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) + \frac{c_{2}}{\frac{1}{2} −s+s_{0}} \frac{d}{ds}Π_{𝔭}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, −s_{0} − \frac{1}{2}, φ_{𝔭}, (ηχ)_{𝔭}^{−1})

+ \{ \frac{c_{1}}{\frac{1}{2} +s−s_{0}} + _{\frac{c_{2}}{(\frac{1}{2} +s−s_{0})^{2}}} \} Π_{𝔭}Φ′ (g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭})

+ \{ \frac{c_{1}}{\frac{1}{2} −s+s_{0}} + _{\frac{c_{2}}{(\frac{1}{2} −s+s_{0})^{2}}} \} Π_{𝔭}, Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, −s_{0} − \frac{1}{2}, φ_{𝔭}, (ηχ)_{𝔭}^{−1})

It is clear that this does not change if s is replaced by −s, χ by η^{−1}χ^{−1}, and g by \binom{0 1}{−1 0}g. Putting everything together we see that

{Π_{𝔭}ξ(s, τ_{𝔭}, χ_{𝔭})} {Π_{𝔭}Φ′(g_{𝔭}, −s, φ_{𝔭}, χ_{𝔭})}

is meromorphic in the whole complex plane and equals

\{Π_{𝔭}ξ(−s, τ_{𝔭}, (ηχ)_{𝔭}^{−1})\} \{Π_{𝔭}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, −s, φ_{𝔭}, (ηχ)_{𝔭}^{−1})\}

The second factor is equal to

{Π_{𝔭}ε(s, τ_{𝔭}, χ_{𝔭})} {Π_{𝔭}Φ′(g_{𝔭}, s, φ_{𝔭}, χ_{𝔭})}

Thus if

\begin{aligned}ξ(s, L, χ) & = Π_{𝔭}ξ(s, τ_{𝔭}, χ_{𝔭}), \\ ε(s, L, χ) & = Π_{𝔭}ε(s, τ_{𝔭}, χ_{𝔭}), \end{aligned}

ξ(s, L, χ) is meromorphic in the entire complex plane and satisfies the functional equation

ξ(−s, L, (ηχ)^{−1})ε(s, L, χ) = ξ(s, L, χ).

To investigate its poles we use the fact that for a suitable choice of φ and g

Π_{𝔭}Φ′(g_{𝔭}, s, φ_{𝔭}, χ_{𝔭})

is an exponential in s. Thus if neither χ_{1}χ nor χ_{2}χ is trivial on the idèles of norm 1 it has no poles. If φ_{0} = 0 for all φ in L then it has no poles for any choice of χ. To find the principal parts at the poles in the other cases we observe that

\frac{1}{\frac{1}{2} + s − s_{0}} \{Π_{𝔭}Φ′(g_{𝔭}, s, φ_{𝔭}, χ_{𝔭}) − Π_{𝔭}Φ′ (g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭})\}

and

\frac{1}{(\frac{1}{2} + s − s_{0})^{2}} \{Π_{𝔭}Φ′(g_{𝔭}, s, φ_{𝔭}, χ_{𝔭}) − Π_{𝔭} (g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭})

− (\frac{1}{2} + s − s_{0}) \frac{d}{ds}Π_{𝔭}Φ′ (g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭})\}

are entire functions of s.

Thus if χ_{1}χ is trivial on the idèles of norm 1 there are simple poles at s_{0} − \frac{1}{2} and s_{0} + \frac{1}{2} with residues −c_{1} and c_{2}ε (s_{0} + \frac{1}{2}, L, χ) respectively. If χ_{2}χ is trivial on the idèles of norm 1 but χ_{1}χ is not there are simple poles at s_{0} − \frac{1}{2} and s_{0} + \frac{1}{2} with residues −c_{2} and c_{1}ε(s_{0} + \frac{1}{2}, L, χ) respectively. If χ_{1} ≠ χ_{2} but both χ_{1}χ and χ_{2}χ are trivial on the idèles of norm 1 there are simple poles at s_{1} − \frac{1}{2}, s_{2} − \frac{1}{2}, s_{1} + \frac{1}{2}, and s_{2} + \frac{1}{2} with residues −c_{1}, −c_{2}, c_{2}ε (s_{1} + \frac{1}{2}, L, χ), c_{1}ε (s_{2} + \frac{1}{2}, L, χ) respectively. If χ_{1} = χ_{2} there are poles of order two at s_{0} − \frac{1}{2} and s_{0} + \frac{1}{2}. The principal part at s_{0} − \frac{1}{2} is

−\frac{c_{2}}{(s − s_{0} + 1/2)^{2}} − \frac{c_{1}}{s − s_{0} + 1/2}

The principal part at s_{0} + \frac{1}{2} is determined by the functional equation.

For a function field we write our integral as the sum of

\begin{matrix}∫_{\{α||α|>1\}}\{φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g) − φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)\}χ(α)|α|^{s}dα \\ +∫_{\{α||α|>1\}}\{φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g) − φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g)\}(ηχ)^{−1}(α)|α|^{−s}dα \end{matrix}

and

∫_{|α|≤1} φ(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)χ(α)|α|^{s}dα

and

− ∫_{\{α||α|=1\}} φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)χ(α)|α|^{s}dα + ∫_{\{α||α|<1\}} φ_{0}(( \begin{matrix}α^{−1} & 0 \\ 0 & 1 \end{matrix} ) g)η(α)χ(α)|α|^{s}dα.

The first two of these expressions are clearly entire functions of s which do not change when g is replaced by \binom{0 1}{−1 0}g, s by −s, and χ by η^{−1}χ^{−1}.

Again let us consider the last expression when φ_{0} is not zero for all φ and at least one of χ_{1}χ or χ_{2}χ is trivial on the idèles of norm 1. If χ_{1}χ is but χ_{2}χ is not, let χ_{1}χ(α) = |α|^{−s_{0}}. The expression equals

−\frac{c_{1}}{1 − q^{− \frac{1}{2} −s+s_{0}}} Π_{𝔭}Φ′(g_{𝔭}, s_{0}− \frac{1}{2}, φ_{𝔭}, χ_{𝔭})− \frac{c_{2}}{1 − q^{− \frac{1}{2} +s−s_{0}}} Π_{𝔭}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, −s_{0}− \frac{1}{2}, φ_{𝔭}, (η^{−1}χ^{−1})_{𝔭})

If χ_{2}χ is trivial but χ_{1}χ is not, and χ_{2}χ(α) = |α|^{−s_{0}} it equals

\frac{c_{2}}{1 − q^{− \frac{1}{2} −s+s_{0}}} Π_{𝔭}Φ′(g_{𝔭}, s_{0} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) − \frac{c_{1}}{1 − q^{− \frac{1}{2} +s−s_{0}}} Π_{𝔭}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, −s_{0} − \frac{1}{2}, φ_{𝔭}, (η^{−1}χ^{−1})_{𝔭})

If χ_{1} ≠ χ_{2} but both χ_{1}χ and χ_{2}χ are trivial on the idèles of norm 1 let χ_{1}χ(α) = |α|^{−s_{1}} and χ_{2}χ(α) = |α|^{−s_{2}}. The expression equals

− \frac{c_{1}}{1 − q^{− \frac{1}{2} −s+s_{1}}} Π_{𝔭}Φ′(g_{𝔭}, s_{1} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭}) − \frac{c_{2}}{1 − q^{− \frac{1}{2} −s+s_{2}}} Π_{𝔭}Φ′(g_{𝔭}, s_{2} − \frac{1}{2}, φ_{𝔭}, χ_{𝔭})

− \frac{c_{1}}{1 − q^{− \frac{1}{2} +s−s_{2}}} Π_{𝔭}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, −s_{2} − \frac{1}{2}, φ_{𝔭}(η^{−1}χ^{−1})_{𝔭})

− \frac{c_{2}}{1 − q^{− \frac{1}{2} +s−s_{1}}} Π_{𝔭}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, −s_{1} − \frac{1}{2}, φ_{𝔭}, (η^{−1}χ^{−1})_{𝔭})

Finally suppose that χ_{1} = χ_{2} and χ_{1}χ(α) = |α|^{−s_{0}}. The expression yields

\begin{matrix}−_{\frac{c2}{1−q− \frac{1}{2} −s+s0}} \frac{d}{ds} Π_{𝔭}Φ′(g_{𝔭},s_{0}− \frac{1}{2} ,φ_{𝔭},χ_{𝔭})−_{\frac{c2}{1−q− \frac{1}{2} +s−s0}} \frac{d}{ds} Π_{𝔭}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )g_{𝔭},−s_{0} \frac{1}{2} ,φ_{𝔭},(ηχ)_{𝔭}^{−1}) \\ −\{_{\frac{c1}{1−q− \frac{1}{2} −s+s0}} +c_{2} ^{\begin{matrix}\frac{d}{ds} & (1−q− \frac{1}{2} −s+s0) \\ (1−q− \frac{1}{2} −s+s0) & 2 \end{matrix}} \}Π_{𝔭}Φ′(g_{𝔭},s_{0}− \frac{1}{2} ,φ_{𝔭},χ_{𝔭})−\{_{\frac{c1}{1−q− \frac{1}{2} +s−s0}} −c_{2} ^{\begin{matrix}\frac{d}{ds} & (1−q− \frac{1}{2} +s−s0) \\ (1−q−1/2+s−s0) & 2 \end{matrix}} \} \end{matrix}

The functional equation follows as before. The principal parts at the poles can also be detrmined. Since I am principally interested in the case of a number field I shall not bother to discuss them explicitly. Moreover for the converse theorem I shall limit myself to the case of a number field. The statement and the proof for a function field will differ only in minor points.

For the converse theorem we suppose that, for each prime 𝔭, we are given an infinite-dimen-sional simple representation τ_{𝔭} of G_{k_{𝔭}} on V_{𝔭}. We suppose that for almost all non-archimedean primes there is a non-zero vector in V_{𝔭} whose isotropy group contains G_{o_{𝔭}}. For such a prime there will be a continuous homomorphism M_{𝔭}((α, β)) = χ′_{𝔭}(α)χ′_{𝔭}(β) of k_{𝔭}^{×} × k_{𝔭}^{×} into ℂ^{×} such that τ_{𝔭} = τ_{M_{𝔭}}. We suppose that there is a constant N > 0 such that for all such 𝔭

|χ′_{𝔭}(π)| ≤ |π|^{−N} |χ′′_{𝔭}(π)| = |π|^{−N}

if π is a generator of the maximal ideal of o_{𝔭}. Let τ_{𝔭}(\binom{α 0}{0 α}) = η_{𝔭}(α)I if α ∈ k_{𝔭}^{×}. We suppose that

η(α) = η(Π_{𝔭}α_{𝔭}) = Π_{𝔭}η_{𝔭}(α_{𝔭})

which is a continuous homomorphism of I into ℂ^{×} is trivial on k^{×}.

If χ is a continuous homomorphism of k^{×}\I into ℂ^{×} the product

Π_{𝔭}ξ(s, τ_{𝔭}, χ_{𝔭}) = ξ_{0}(s, χ)

converges for Re(s) sufficiently large. We suppose that for each χ it is meromorphic in the whole plane, that it has only a finite number of poles, that it is bounded in the regions obtained by removing circles about its poles from any vertical strip of finite width, and that the functional equations

ξ_{0}(−s, (ηχ)^{−1})ξ_{0}(s, χ) = ξ_{0}(s, χ),

with

ε_{0}(s, χ) = Π_{𝔭}ε(s, τ_{𝔭}, χ_{𝔭}),

are satisfied.

We suppose that there are two continuous homomorphisms χ_{1} and χ_{2} of k^{×}\I into ℂ^{×} with χ_{1}χ_{2} = η and two complex numbers c_{1} and c_{2} such that ξ_{0}(s, χ) has no poles unless either χ_{1}χ or χ_{2}χ is trivial on the idéles of norm 1.

(i) If χ_{1}χ is trivial on the idéles of norm 1 but χ_{2}χ is not and if χ_{1}χ(α) = |α|^{−s_{0}} there are simple poles at s_{0} − \frac{1}{2} and s_{0} + \frac{1}{2} with residues −c_{1} and c_{2}ε_{0} (s_{0} + \frac{1}{2}, χ) respectively. (ii) If χ_{2}χ is trivial on the idéles of norm 1 but χ_{1}χ is not and χ_{2}χ(α) = |α|^{−s_{0}} there are simple poles at s_{0} − \frac{1}{2} and s_{0} + \frac{1}{2} with residues −c_{2} and c_{1}ε_{0} (s_{0} + \frac{1}{2}, χ) respectively. (iii) If χ_{1}χ(α) = |α|^{−s_{1}} and χ_{2}χ(α) = |α|^{−s_{2}} withs_{1} ≠ s_{2} there are simple poles at s_{1} − \frac{1}{2}, s_{2} − \frac{1}{2}, s_{1} + \frac{1}{2}, s_{2} + \frac{1}{2} with residues −c_{1}, −c_{2}, c_{2}ε_{0} (s_{1} + \frac{1}{2}, χ), c_{1}ε_{0} (s_{2} + \frac{1}{2}, χ) respectively. (iv) If χ_{1}χ(α) = χ_{2}χ(α) = |α|^{−s_{0}} there are poles of order two at s_{0} − \frac{1}{2} and s_{0} + \frac{1}{2}. The principal part at s_{0} − \frac{1}{2} is

−\frac{c_{2}}{(s − s_{0} + \frac{1}{2})^{2}} − \frac{c_{1}}{s − s_{0} + 1/2}.

The principal part at s_{0} + \frac{1}{2} is determined by the functional equation.

We allow the possibility that c_{1} or c_{2} or both are zero. In particular if

ψ_{1}(g) = ψ_{1}(Π_{𝔭}g_{𝔭}) = Π_{𝔭}Φ′ (g_{𝔭}, −\frac{1}{2}, φ_{𝔭}, χ_{1,𝔭}^{−1})

is not, for any choice of the collection {φ_{𝔭}} with φ_{𝔭} in L(ξ_{𝔭}, τ_{𝔭}) such that G_{o_{𝔭}} lies in the isotropy group of φ_{𝔭} for almost all non-archimedean primes and φ_{𝔭}(1) = 1 for almost all non-archimedean primes, a function satisfying ψ_{1}(\binom{1 x}{0 1}) = ψ_{1}(g) for all x in 𝔸 we demand thatc_{1} = 0. Also if χ_{1} ≠ χ_{2} we demand that c_{2} = 0 if for the same choices of the collection {φ_{𝔭}} the functions

ψ_{2}(g) = Π_{𝔭}Φ′ (g_{𝔭}, −\frac{1}{2}, φ_{𝔭}, χ_{2,𝔭}^{−1})

do not all satisfy ψ_{2}(\binom{1 x}{0 1}g)≡ ψ_{2}(x) for all x ∈ 𝔸. If χ_{1} = χ_{2} we demand that c_{2} = 0 if

ψ_{2}(g) = \frac{d}{ds}Π_{𝔭}Φ′ (g_{𝔭}, −\frac{1}{2}, φ_{𝔭}, χ_{1,𝔭}^{−1})

does not satisfy this condition. Notice that given χ_{1} and χ_{2} and the collection {τ_{𝔭}} we can, according ot the appendix to the previous paragraph, decide whether or not ψ_{1} and ψ_{2} satisfy these conditions. Notice also that our theorem will be most interesting when both c_{1} and c_{2} are zero.

In any case the converse theorem states that when all these conditions are satisfied there is a characteristic space of automorphic forms which transforms according to the “representation” ⊗_{𝔭}τ_{𝔭}. To prove it we show that if the collection {ψ_{𝔭}} is chosen as above and

φ_{1}(g) = Π_{𝔭}φ_{𝔭}(g_{𝔭})

while

φ_{0}(g) = c_{1}Π_{𝔭}Φ′ (g_{𝔭}, −\frac{1}{2}, φ_{𝔭}, χ_{1,𝔭}^{−1}) + c_{1}Π_{𝔭}Φ′ (g_{𝔭}, −\frac{1}{2}, φ_{𝔭}, χ_{2,𝔭}^{−1})

if χ_{1} ≠ χ_{2} and

φ_{0}(g) = c_{1}Π_{𝔭}Φ′ (g_{𝔭}, −\frac{1}{2}, φ_{𝔭}, χ_{1,𝔭}^{−1}) + c_{2} \frac{d}{ds}Π_{𝔭}Φ′ (g_{𝔭}, −\frac{1}{2}, φ_{𝔭}, χ_{1,𝔭}^{−1})

if χ_{1} = χ_{2} then

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

is a function on G_{k}\G_{𝔸}.

By its very construction it is invariant under left translations by upper triangular matrices in G_{k} so the only problem is to show that φ(\binom{0 1}{−1 0}g) ≡ φ(g). Let us show that for each g the functions φ(\binom{0 1}{−1 0}\binom{α 0}{0 1}g) and φ(\binom{α 0}{0 1}g) on I are equal. Let φ_{1}(α) be the function obtained from the second of these functions by subtracting φ_{0}(\binom{α 0}{0 1}g) if |α| ≥ 1 and φ_{0}(\binom{0 1}{−1 0}\binom{α 0}{0 1}g) if |α| ≤ 1. Let ψ_{2}(α) be the function obtained from the other function by the same process. It is enough to show that ψ_{1}(α) ≡ ψ_{2}(α). Now if χ is any character of k^{×}\I

∫_{k^{×}\I} ψ_{1}(α)χ(α)|α|^{s}dα = μ_{1}(s, χ)

is defined for Re(s) sufficiently large and, as we shall see,

∫_{k^{×}\I} ψ_{2}(α)χ(α)|α|^{s}dα = μ_{2}(s, χ)

is defined for Re(s) sufficiently small. It is enough to show that, for each χ, μ_{1}(s, χ) and μ_{2}(s, χ) are entire functions of s which equal each other. We must also show that they are bounded in vertical strips.

The first integral is equal to the sum of

ξ_{0}(s, χ)Π_{𝔭}Φ′(g_{𝔭}, s, φ_{𝔭}, χ_{𝔭})

and

∫_{|α|≤1} φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)χ(α)|α|^{s}dα − ∫_{|α|≤1} φ_{0} (( \begin{matrix}α^{−1} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g) η(α)χ(α)|α|^{s}dα.

The second integral is the sum of

∫_{k^{×}\I} φ_{1}(( \begin{matrix}α^{−1} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g)ηχ(α)|α|^{s}dα,

which equals

ξ_{0}(−s, η^{−1}χ^{−1})Π_{𝔭}Φ′(( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g_{𝔭}, −s, φ_{𝔭}, (ηχ)_{𝔭}^{−1}),

and of

∫_{|α|≥1} φ_{0}(( \begin{matrix}α^{−1} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g)ηχ(α)|α|^{s} − φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) g)χ(α)|α|^{s}dα,

which equals the sum of

− ∫_{|α|≤1} φ_{0}(( \begin{matrix}α^{−1} & 0 \\ 0 & 1 \end{matrix} ) g)χ^{−1}(α)|α|^{−s}dα

and

∫_{|α|≤1} φ_{0}(( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) g)η^{−1}χ^{−1}(α)|α|^{−s}dα.

The functional equation assumed for ξ_{0}(s, χ) together with the local functional equations show that the first term in the expression for μ_{2}(s, χ) is the same as the first term in the expression for μ_{2}(s, χ). The second term in the expression for μ_{1}(s, χ) is an integral we have already investigated. We know that its poles cancel the assumed poles of the first term and that it is given by an analytical expression which does not change when g is replaced by (\binom{0 1}{−1 0})g, s is replaced by −s, and χ is replaced by η^{−1}χ^{−1}. But the second term in the expression for μ_{2}(s, χ) is given by the same analytical expression except that s is replaced by −s, g by (\binom{0 1}{−1 0})g, and χ by η^{−1}χ^{−1}. One shows as in the previous letter that μ_{1}(s, χ) and μ_{2}(s, χ) are bounded in vertical strips. The converse theorem is thus proved.