By H. Jacquet, R. P. Langlands

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**Example text**

It is clear that Ψ(g, s, W ) = ω(detg)Ψ(g, s, W ) so that if the third part of the theorem is valid when π is replaced by π the function Φ(g, s, W ) is a holomorphic function of s. Combining the functional equation for π and for π one sees that (s, π, ψ) (1 − s, π, ψ) = ω(−1). Let V be the space on which the Kirillov model of π acts. For every W in W (π, ψ) there is a unique ϕ in V such that W a 0 0 1 = ϕ(a). Chapter 1 39 If π is itself the canonical model π(w)ϕ(a) = W where a 0 0 1 0 1 −1 0 w= w .

5. In particular they converge to the right of some vertical line and if W = WΦ Ψ(e, s, W ) = Z(µ1 αsF , µ2 αsF , Φ) −1 s s Ψ(e, s, W ) = Z(µ−1 2 αF , µ1 αF , Φ). Moreover Ψ(g, s, W ) L(s, π) is a holomorphic function of s and Ψ(g, s, W ) Ψ(g, 1 − s, W ) = ε(s, π, ψ) . 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.

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. The left side equals µ1 (a) × Φ(t−1 , at)µ1 (t)µ−1 2 (t) d t which, after changing the variable of integration, one sees is equal to the right side. Chapter 1 52 −1 If µ1 µ−1 2 is not αF or αF so that ρ(µ1 , µ2 ) is irreducible we let π(µ1 , µ2 ) be any representation in the class of ρ(µ1 , µ2 ).