MODULAR FORMS-page117

# MODULAR FORMS-page117 - 113 ( + )2k f (Z + Z) = ( + )2k f (...

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113 ( γτ + δ ) - 2 k ˜ f ( Z + τ Z ) = ( γτ + δ ) - 2 k f ( τ ) . By property (iii) of Lemma 1, we obtain that T ( n ) leave the set of functions ˜ f on H satisfying (11.6) invariant. Let F k be the space of functions on L of the form ˜ f where f ∈ M (Γ(1)) k . Theorem 11.1. For any positive integer n and any non-negatve integer k , T n ( F k ) ⊂ F k . Proof. Let f ∈ M (Γ(1)) k and ˜ f ∈ F k . We know that T ( n ) ˜ f ( c Λ) = X [Λ:Λ 0 ]= n ˜ f ( c Λ 0 ) = c - 2 k X [Λ:Λ 0 ]= n ˜ f 0 ) . This shows that T ( n ) ˜ f = ˜ g , where g is a function on H satisfying g ( α β γ δ τ ) = ( γτ + δ ) - 2 k g ( τ ) for any α β γ δ « Γ(1). We have to check that g is a holomorphic function on H and at inﬁnity. Applying Lemma 2, we have g ( τ ) = T ( n ) ˜ f ( Z + τ Z ) = X A ∈A 0 n ˜ f (( + b ) Z + d Z ) = X A ∈A 0 n d - 2 k f ( + b d ) . Thus g ( τ ) = X A ∈A 0 n d - 2 k f ( + b d ) = X A ∈A 0 n f | k A. (11.11) Clearly, g is holomorphic on H as soon as f is holomorphic. It remains to ﬁnd its behavior at inﬁnity. Let f = X m =0 c m e 2 πimτ be the Fourier expansion of f at . Then
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