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MODULAR FORMS-page47 - 43 where k k a b 2 2 Summarizing we...

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43 where ( a , b ) = ( δa - γb + kγδ 2 , - βa + αb - kαβ 2 ) . (5.4) Summarizing we obtain that, for any ϑ ( z, τ ) Th( k ; Λ τ ) ab , e iπkγ ( γτ + δ ) z 2 ϑ (( γτ + δ ) z ; τ ) Th( k, Λ τ ) a b . (5.5) Now let us replace α β γ δ with its inverse ` - δ - β - γ α ´ . We rewrite (5.13) and (5.14) as e - iπkγ ( - γτ + α ) z 2 ϑ (( - γτ + α ) z ; τ ) Th( k, Λ τ ) a b , (5.6) where ( a , b ) = ( αa + γb - kγα 2 , βa + δb + kδβ 2 ) . It remains to replace τ with ατ + β γτ + δ in (5.15) to obtain the assertion of the theorem. Substituting z = 0 we get Corollary 5.1. Let ϑ 1 ( z, τ ) , . . . , ϑ k ( z ; τ ) be a basis of the space Th( k ; Λ τ ) ab and ϑ 1 ( z, τ ) , . . . , ϑ k ( z ; τ ) be a basis of Th( k ; Λ τ ) a b , where ( a , b ) are defined in the Theo- rem. Then, for any M = α β γ δ SL(2 , Z ) there exists a matrix A = ( c ij ) GL( k, C ) depending on M and τ only such that ϑ i (0 , ατ + β γτ + δ ) = k X j =1 c ij ϑ j (0 , τ ) . 5.2 Let us take k = 1 and ( a, b ) = ( / 2 , η/ 2) , , η = 0 , 1. Applying the previous Theorem, we get ϑ ab ( z ; τ + 1) = a,b + a + 1 2 ( z ; τ ) for some
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