By H. Jacquet, R. P. Langlands

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Example text

If Φ is a function on F 2 define Φι by Φι (x, y) = Φ(y, x). To prove the proposition we show that, if Φ is in S(F 2 ), µ1 (detg) |detg|1/2 θ µ1 , µ2 ; r(g)Φι = µ2 (detg) |detg|1/2 θ µ2 , µ1 ; r(g)Φ . If g is the identity this relation follows upon inspection of the definition of θ(µ1 , µ2 ; Φι ). It is also easily seen that r(g)Φι = [r(g)Φ]ι if g is in SL(2, F ) so that it is enough to prove the identity for g= a 0 0 1 . It reduces to µ1 (a) × Φι (at, t−1 )µ1 (t)µ−1 2 (t) d t = µ2 (a) × Φ(at, t−1 )µ2 (t)µ−1 2 (t) d t.

Chapter 1 29 Since 1 x 0 1 ϕ−π ϕ is in V0 the right side is equal to λϕ(1) − λψ(x)ϕ(1) + ψ(x)L(ϕ) so that 1 − ψ(x) L(ϕ) = λ 1 − ψ(x) ϕ(1) which implies that L(ϕ) = λϕ(1). To prove the second lemma we have only to show that ϕ(1) = 0 implies L(ϕ) = 0. If we set ϕ(0) = 0 then ϕ becomes a locally constant function with compact support in F . Let ϕ be its Fourier transform so that ϕ(a) = ψ(ba) ϕ (−b) db. F Let Ω be an open compact subset of F × containing 1 and the support of ϕ. There is an ideal a in F so that for all a in Ω the function ϕ (−b)ψ(ba) is constant on the cosets of a in F .

Suppose V is another such space of functions and π a representation of GF on V which is equivalent to π . We suppose of course that π (b)ϕ = ξψ (b)ϕ if b is in BF and ϕ is in V . Let A be an isomorphism of V with V such that Aπ(g) = π (g)A for all g . Let L be the linear functional L(ϕ) = Aϕ(1) on V . Then a 0 0 1 L π ϕ = Aϕ(a) so that A is determined by L. If we could prove the existence of a scalar λ such that L(ϕ) = λϕ(1) it would follow that Aϕ(a) = λϕ(a) for all a such that Aϕ = λϕ. This equality of course implies the theorem.

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