MODULAR FORMS-page53

MODULAR FORMS-page53 - g ( g · τ ) j g ( τ ) . (6.4)...

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Lecture 6 Modular Forms 6.1 We have seen already in Lecture 5 (5.2) and Corollary 5.3 that the functions θ ( τ ) 4 k = ϑ 00 (0; τ ) 4 k (resp. η ( τ ) 24 ) satisfy the functional equation f ( τ + 2) = f ( τ ) , f ( - 1 ) = τ 2 f ( τ ) , (resp. f ( τ + 1) = f ( τ ) , f ( - 1 ) = τ 12 f ( τ )) . In fact, they satisfy a more general equation f ( ατ + β γτ + δ ) = ( γτ + δ ) 2 k f ( τ ) , α β γ δ « Γ , (6.1) where Γ is the subgroup of SL(2 , Z ) generated by the matrices ± ( 0 - 1 1 0 ) , ± ( 1 2 0 1 ) (resp. ± ( 0 - 1 1 0 ) , ± ( 1 1 0 1 )) . To see this we first rewrite (6.1) in the form f ( g · τ ) j g ( τ ) k = f ( τ ) , (6.2) where j g ( τ ) = d ατ + β γτ + δ = ( γτ + δ ) - 2 . (6.3) By the chain rule j gg 0 ( τ ) = j
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Unformatted text preview: g ( g · τ ) j g ( τ ) . (6.4) Thus replacing τ with g · τ in (6.2), we get ( f ( g · ( g · τ )) j k g ( g · τ )) j k g ( τ ) = f ( gg · τ ) j k gg ( τ ) = f ( τ ) . This shows that f ( τ ) | k g := f ( g · τ ) j g ( τ ) k (6.5) satisfies f ( τ ) | k ( gg ) = ( f ( τ ) | k g ) | k g , ∀ g,g ∈ Γ . In other words (6.5) defines a linear representation ρ : Γ → GL ( O ( H ) hol ) 49...
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This note was uploaded on 01/08/2012 for the course MATH 300 taught by Professor Ontonkong during the Fall '09 term at SUNY Stony Brook.

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