Download PDF by Jacquet H., Langlands R.P.: Automofphic forms on GL(2)

By Jacquet H., Langlands R.P.

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Example text

For a given Φ in S(F 2 ) there is an ideal a such that Φ(x, y) = Φ(x, 0) ′ for y in a. If a is the complement of a Φ(x, y) |x|s+1 |y|s d× x d× y is equal to the sum of Φ(x, y) |x|s+1 |y|s d× x d× y a′ F which has no pole at s = 0 and a constant times Φ(x, 0) |x|s dx F |y|s d× y a If a = pn the second integral is equal to |̟|ns (1 − |̟|s )−1 If Φ(x, 0) dx = 0 F the first term, which defines a holomorphic function of s, vanishes at s = 0 and the product has no pole there. If ϕ0 is the characteristic function of OF set Φ(x, y) = ϕ0 (x) − |̟|−1 ϕ0 (̟ −1 x) ϕ0 (y).

1 ν(−1) σ Replacing ρ by ρ−1 ν0−1 we obtain the first part of the proposition. If ρ = ν then δ(ρν −1 ) = 1. Moreover, as is well-known and easily verified, η(ρν −1 , ̟ r ) = 1 if r ≥ −ℓ, η(ρν −1 , ̟ −ℓ−1 ) = |̟|(|̟| − 1)−1 and η(ρν −1 , ̟ r ) = 0 if r ≤ −ℓ − 2. 4) is equal to −∞ z0−p−r Cn+r (ν)Cn+r (ν −1 ν0−1 ). ν0 (−1)δn,pI+(|̟|−1)−1 z0−p+ℓ+1 Cn−ℓ−1 (ν)Cn−ℓ−1 (ν −1 ν0−1 )− r=−ℓ−2 The second part of the proposition follows. 12) (i) For every n, p, ν and ρ Cn (ν)Cp (ρ) = Cp (ρ)Cn (ν) (ii) There is no non-trivial subspace of X invariant under all the operators Cn (ν).

5) we deduce a relation of the form q λi Cn+r−i (ν) Cn+r (ν) = i=1 where r is a fixed integer and n is any integer greater than p. 3 is a consequence of the following more precise lemma. If pm is the conductor of a character ρ we refer to m as the order of ρ. 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−2ℓ (ρ) = ν2−1 ρ(−1)z0−m−ℓ η(ν1−1 ρ, ̟ −m−ℓ ) η(ν2 ρ−1 , ̟ −m−ℓ ) and Cp (ρ) = 0 if p = −2m − 2ℓ.

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

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