By H. Jacquet, R. P. Langlands

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**Example text**

It equals |Φ(t, a)| |t|s+1 d× t da F× F which is finite because s is greater than −1. To show that A is surjective we show that every function f in B(µ1 , µ2 ) is of the form fΦ for some Φ in S(F 2 ). Given f let Φ(x, y) be 0 if (x, y) is not of the form (0, 1)g for some g in GL(2, O F ) but if (x, y) is of this form let Φ(x, y) = µ−1 1 (detg)f (g). It is easy to see that Φ is well-defined and 2 belongs to S(F ). To show that f = fΦ we need only show that f (g) = fΦ (g) for all g in GL(2, OF ).

L(1 − s, π ˜) L(s, π) Therefore Φ(g, s, W ) = Ψ(g, s, W ) L(s, σ) Φ(g, s, W ) = Ψ(g, s, W ) L(s, σ) and are meromorphic functions of s and satisfy the local functional equation Φ(wg, 1 − s, W ) = ε(s, σ, ψ) Φ(g, s, W ). To compete the proof of the theorem we have to show that ε(s, σ, ψ) is an exponential function of s and we have to verify the third part of the corollary. The first point is taken care of by the observation −1 that µ−1 1 ( ) | | = µ2 ( ) so that 1 − µ2 ( ) | | s L(1 − s, µ−1 1 ) = = −µ1 ( ) | |s−1 .

2. 11 η(σ −1 ν, n )η(σ −1 ν, p )Cp+n (σ) σ is equal to −∞ z0p ν0 (−1)δn,p + (| | − 1) −1 z0+1 Cn−1− z0−r Cn+r (ν)Cp+r (ν). (ν)Cp−1− (ν) − −2− Remember that p− is the largest ideal on which ψ is trivial. Suppose first that ν = ν . Chapter 1 35 Take p = − and n > − . Then δ(n − p) = 0 and η(σ −1 ν, n )η(σ −1 ν, p )=0 unless σ = ν . Hence −∞ Cn− (ν) = (| | − 1) −1 z0+1 Cn−1− (ν)C−2 −1 (ν) z0−r Cn+r (ν)C− − +r (ν) −2− which, since almost all of the coefficients C− +r (ν) in the sum are zero, is the relation required.