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MODULAR FORMS-page58

MODULAR FORMS-page58 - 54 LECTURE 6 MODULAR FORMS Here we...

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54 LECTURE 6. MODULAR FORMS Here we used that φ ( TS ) = φ ( ST ) - 1 since φ ( T 2 S 2 ) = 1 and similarly φ ( TST ) = φ ( STS ) - 1 , φ ( TSTS ) = φ ( STST ) - 1 , φ ( TSTST ) = φ ( STSTS ) - 1 . Also φ ( ST ) = φ ( STST ) - 1 . Thus it is enough to verify that the elements S, ST, STS, T are not in the kernel, i.e. do not belong to Γ(2). This is verified directly. Example 6.3 . Consider the theta constants ϑ 2 2 . Applying the transformation τ = τ + 1 twice and using formulas (5.1) , we obtain ϑ 00 ( τ + 2) = ϑ 0 1 2 ( τ + 1) = ϑ 00 ( τ ) , ϑ 0 1 2 ( τ + 2) = ϑ 00 ( τ + 1) = ϑ 0 1 2 ( τ ) , ϑ 1 2 0 ( τ + 2) = e πi/ 4 ϑ 1 2 0 ( τ + 1) = e πi/ 2 ϑ 1 2 0 ( τ ) . Next, using formulas (5.11)-(5.14), we have ϑ 00 ( ST 2 ) = ϑ 00 ( ` - 1 0 2 - 1 ´ · τ ) = ϑ 00 ( - 1 ( - 1 ) + 2 ) = e 3 πi/ 4 ( - 1 τ +2) 1 / 2 ϑ 00 ( - 1 τ +2) = e 3 πi/ 4 ( - 1 τ + 2) 1 / 2 ϑ 00 ( - 1 τ ) = e 3 πi/ 2 ( - 1 τ + 2) 1 / 2 ( τ ) 1 / 2 ϑ 00 ( τ ) = - i (2 τ - 1) 1 / 2 ϑ 00 ( τ ) . Similarly we obtain ϑ 0 1 2 ( ST 2 ) = e 3 πi/ 4 ( - 1 τ + 2) 1 / 2 ϑ 1 2 0 ( - 1 τ + 2) = ie 3 πi/ 4 ( - 1 τ + 2) 1 / 2 ϑ 1 2 0 ( - 1 τ ) = (2 τ - 1) 1 / 2 ϑ 0 1 2 ( τ ) , ϑ 1 2 0 ( ST 2 ) = - i (2 τ - 1) 1 / 2 ϑ 0 1 2 ( τ ) . Applying Lemma 6.2, this shows that ϑ 00 ( τ ) 4 , ϑ 1 2 0 ( τ ) 4 , ϑ 0 1 2 ( τ ) 4 (6.11) are weak modular forms with respect to the group Γ(2). This group has three cusps
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