{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MODULAR FORMS-page48

MODULAR FORMS-page48 - 44 LECTURE 5 TRANSFORMATIONS OF...

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

View Full Document Right Arrow Icon
44 LECTURE 5. TRANSFORMATIONS OF THETA FUNCTIONS Now take M = ( 0 - 1 1 0 ) . We have e - iπz 2 ϑ 00 ( z/τ ; - 1 ) = 00 ( z ; τ ) for some B depending only on τ . Plugging in z = 0 and applying (4.2), we get B = - iτ, (5.10) where the square root takes positive values on τ i R . In particular, ϑ 00 (0; - 1 ) = - iτϑ 00 (0; τ ) . (5.11) Applying the formula (3.14) we have e - iπz 2 ϑ 0 1 2 ( z τ ; - 1 τ ) = e - iπz 2 ϑ 00 ( z τ + 1 2 ; - 1 τ ) = e - iπz 2 ϑ 00 ( z + τ 2 τ ; - 1 τ ) = Be - iπz 2 e ( z + τ 2 ) 2 ϑ 00 ( z + τ 2 ; τ ) = Be ( τ/ 4+ z ) ϑ 00 ( z + τ 2 ; τ ) = 1 2 0 ( z ; τ ) . (5.12) In particular, ϑ (0; - 1 ) 0 1 2 = - iτϑ (0; τ ) 1 2 0 . (5.13) Replacing here τ with - 1 , we obtain ϑ (0; - 1 ) 1 2 0 = iτϑ (0; τ ) 0 1 2 . (5.14) This shows that e - iπz 2 ϑ ( z/τ ; - 1 ) 1 2 0 = iτϑ ( z ; τ ) 0 1 2 . (5.15) Finally, using (5.13), (5.14) and (5.15) and applying the Jacobi theorem, we obtain ϑ (0; - 1 ) 0 1 2 1 2 = - τ - iτϑ 1 2 1 2 (0; τ ) 0 . (5.16) We know from Theorem 5.1 that e - iπz 2 ϑ ( z/τ ; - 1 ) 1 2 1 2 = B 0 ϑ ( z ; τ ) 1 2 - 1 2 = - B 0 ϑ ( z ; τ ) 1 2 1 2 . for some constant B 0 depending only on τ . Differentiating in
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online