MODULAR FORMS-page58

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ( 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 0 = 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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online