MODULAR FORMS-page65

# MODULAR FORMS-page65 - 61 Example 6.4 We apply the previous...

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61 Example 6.4 . We apply the previous theorem to Φ( z ; τ ) = ( z ) and z = 1 2 . In this case Γ = Γ 0 (2) = { α β γ δ « Γ(1) : 2 | γ } . Now, replacing z with z/ ( γτ + δ ) in (6.22), we get ( z γτ + δ ; ατ + β γτ + δ ) = ( γτ + δ ) 2 [ z - 2 + X ( m,n ) 6 =(0 , 0) 1 z - m ( γτ + δ ) + n ( ατ + β ) ] . Since Z + Z τ = Z ( γτ + δ ) + Z ( ατ + β ) , we get ﬁnally that ( z γτ + δ ; ατ + β γτ + δ ) = ( γτ + δ ) 2 ( z ; τ ) . Thus ( z ; τ ) satisﬁes the assumption of the Lemma with m = 2. Let M Γ(1). Since (0 , 1 2 ) · M - ( 1 2 , 0) Z 2 if and only if M Γ 0 (2) we obtain that the 0-th coeﬃcient g 0 ( τ ) = ( 1 2 ) of the Taylor expansion of ( z ) at 1 2 satisﬁes ( 1 2 ; ατ + β γτ + δ ) = ( γτ + δ ) 2 ( 1 2 ; τ ) . Similarly, if we replace 1 2 with τ 2 and τ 2 + 1 2 we get that ( τ 2 ; ατ + β γτ + δ ) = ( γτ + δ ) 2 ( τ 2 ; τ ) , α β γ δ « Γ 0 (2) , where Γ 0 (2) = { α β γ δ « Γ(1) : 2 | β } = 1 1 - 1 0 « Γ 0 (2) 0 - 1 1 0 « . ( τ 2 + 1 2 ; ατ + β γτ + δ ) = ( γτ + δ ) 2 ( τ 2 + 1 2 ; τ ) . We skip the veriﬁcation that ( τ 2 ) and ( 1 2 ) satisfy the regularity condition at the cusps. Since both Γ 0 (2) and Γ 0 (2) contain Γ(2) as its subgroup, we see that
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