By H. Jacquet, R. P. Langlands
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Additional info for Automorphic Forms on GL(2): Part 1
Any special representation σ is of the form σ(µ1 , µ2 ) with µ1 = χαF and µ2 = χαF . The −1 −1 contragredient representation of σ is σ(µ2 , µ1 ). This choice of µ1 and µ2 is implicit in the following proposition. 6 W (σ, ψ) is the space of functions W = WΦ in W (µ1 , µ2 ; ψ) for which Φ(x, 0) dx = 0. 18 will be valid if we set L(s, σ) = L(s, σ) = 1 and ε(s, σ, ψ) = ε(s, µ1 , ψ) ε(s, µ2 , ψ) when χ is ramiﬁed and we set L(s, σ) = L(s, µ1 ), L(s, σ) = L(s, µ−1 2 ), and ε(s, σ, ψ) = ε(s, µ1 , ψ) ε(s, µ2 , ψ) L(1 − s, µ−1 1 ) L(s, µ2 ) when χ is unramiﬁed.
It follows immediately that ω is identically 1 and that µ1 and µ2 have the desired form. The next lemma is the key to the theorem. 2. If |µ 1 µ2 ( )| = | |s with s > −1 there is a minimal non-zero invariant subspace X of B(µ1 , µ2 ). For all f in B(µ1 , µ2 ) and all n in NF the diﬀerence f − ρ(n)f belongs to X. 2 it is enough to prove the lemma when B(µ1 , µ2 ) is replaced by W (µ1 , µ2 ; ψ). Associate to each function W in W (µ1 , µ2 ; ψ) a function ϕ(a) = W a 0 0 1 on F × . If ϕ is 0 so is W .
If π were contained in the representation on W (ψ) there would be a nonzero function W on GF such that 1 x 0 1 W g = ψ(x)χ(detg)W (e) In particular taking g = e we find that W 1 x 0 1 = ψ(x)W (e) However it is also clear that 1 x 0 1 W = χ det 1 x 0 1 W (e) = W (e) so that ψ(x) = 1 for all x. This is a contradiction. We shall see however that π is a constituent of the representation on W (ψ). That is, there are two invariant subspaces W1 and W2 of W (ψ) such that W1 contains W2 and the representation of the quotient space W1 /W2 is equivalent to π .