# Automofphic forms on GL(2) by Jacquet H., Langlands R.P.

By Jacquet H., Langlands R.P.

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Additional resources for Automofphic forms on GL(2)

Example text

We have −1 defined a non-degenerate pairing between B(µ1 , µ2 ) and B(µ−1 1 , µ2 ). All elements of the orthogonal −1 complement of X are invariant under NF . Thus if µ1 µ2 is not αF the orthogonal complement is 0 and −1 X is B(µ1 , µ2 ) so that the representation is irreducible. The contragredient representation ρ(µ−1 1 , µ2 ) is also irreducible. 1/2 −1/2 If µ1 µ2−1 = αF we write µ1 = χαF , µ2 = χαF . In this case X is the space of the functions −1 −1 −1 orthogonal to the function χ−1 in B(µ−1 1 , µ2 ).

We have to verify that π(w)T ϕ = T π(w)ϕ at least for ϕ on V0 and for ϕ = π(w)ϕ0 with ϕ0 in V0 . We have already seen that the identity holds for ϕ in V0 . Thus if ϕ = π(w)ϕ0 the left side is π(w)T π(w)ϕ0 = π 2 (w)T ϕ0 = ν0 (−1)T ϕ0 and the right side is T π 2 (w)ϕ0 = ν0 (−1)T ϕ0. Because of this proposition we can identify X with C and regard the operators Cn (ν) as complex numbers. For each r the formal Laurent series C(ν, t) has only finitely many negative terms. We now want to show that the realization of π on a space of functions on F × is, when certain simple conditions are imposed, unique so that the series C(ν, t) are determined by the class of π and that conversely the series C(ν, t) determine the class of π .

Then, for all characters ν, Cn (ν) = 0 if n ≥ −1. Take a character ν and choose n1 such that Cn1 (ν) = 0. Then Cn (ν) = 0 for n = n1 . If ν = ν −1 ν0−1 then, as we have seen, C(ν, t)C(ν, t−1 z0−1 ) = ν0 (−1) so that Cn (ν) = 0 for n = n1 and Cn1 (ν)Cn1 (ν) = ν0 (−1)z0n1 . 11 take n = p = n1 + 1 to obtain η(σ −1 ν, ̟ n1 +1 )η(σ −1 ν, ̟ n1 +1 )C2n1 +2 (σ) = z0n1 +1 ν0 (−1) + (|̟| − 1)−1 z0 Cn1 (ν)Cn1 (ν). σ The right side is equal to z0n1 +1 ν0 (−1) · |̟| . |̟| − 1 Assume n1 ≥ −1. Then η(σ −1 ν, ̟ n1 +1 ) is 0 unless σ = ν and η(σ −1 ν, ̟ n1 +1 ) is 0 unless σ = ν .