By H. Jacquet, R. P. Langlands

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6 Let m0 be of the order ν0 and let m1 be an integer greater than m0 . Write ν0 in any manner in the form ν0 = ν1−1 ν2−1 where the orders of ν1 and ν2 are strictly less than m1 . If the order m of ρ is large enough C−2m−2 (ρ) = ν2−1 ρ(−1)z0−m− η(ν1−1 ρ, η(ν2 ρ−1 , −m− −m− ) ) and Cp (ρ) = 0 if p = −2m − 2 . Suppose the order of ρ is at least m1 . Then ρν1 ν0 = ρν2−1 is still of order m. 11 we see that η(σ −1 ν1 , n+m+ )η(σ −1 ρ, p+m+ )Cp+n+2m+2 (σ) σ is equal to η(ν1−1 ρ−1 ν0−1 , −m− )z0m+ ν1 ρν0 (−1)Cn−m− (ν)Cp−m− (ρ) for all integers n and p.

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 . Thus the left side is 0 unless ν = ν . However if ν = ν the left side equals C2n1 +2 (ν).

1/2 −1/2 (ii) If µ1 µ−1 . Then B(µ1 , µ2 ) contains a unique proper 2 = αF write µ1 = χαF , µ2 = χαF invariant subspace Bs (µ1 , µ2 ) which is irreducible. B(µ 2 , µ1 ) also contains a unique proper invariant subspace Bf (µ2 , µ1 ). It is one-dimensional and contains the function χ(detg). Moreover the GF -modules Bs (µ1 , µ2 ) and B(µ2 , µ1 )/Bf (µ2 , µ1 ) are equivalent as are the modules B(µ1 , µ2 )/Bs (µ1 , µ2 ) and Bf (µ2 , µ1 ). We start with a simple lemma. 1 Suppose there is a non-zero function f in B(µ 1 , µ2 ) invariant under right translations −1/2 by elements of NF .