MODULAR FORMS-page46

MODULAR FORMS-page46 - 42 LECTURE 5. TRANSFORMATIONS OF...

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Unformatted text preview: 42 LECTURE 5. TRANSFORMATIONS OF THETA FUNCTIONS Thus for any ϑ ∈ Th(k; Λτ )ab , we have ϑ(z (γτ + δ )) ∈ Th(eλ ; Z + τ Z), where τ= ατ + β , γτ + δ em+nτ (z ) = e(m+nτ )(γτ +δ ) (z (γτ + δ )). We have, using (5.1), e1 (z ) = eγτ +δ (z (γτ + δ )) = e2πi(aδ−bγ ) e−πik(2γz(γτ +δ)+γ e−πikγ ((γτ +δ)(z+1) 2 2 −(γτ +δ )z ) πikγδ 2πi(aδ −bγ ) e e 2 τ) = . This shows that 2 eπikγ (γτ +δ)z Th({eλ (z )}; Z + τ Z) = Th({eλ (z )}; Z + τ Z), where e1 (z ) = eπi[kγδ+2(aδ−bγ )] . (5.2) Now comes a miracle! Let us compute eτ (z ). We have eτ (z ) = eπikγ (γτ +δ)((z+τ )2 −z 2 ) eτ πikγ (γτ +δ )(2zτ +τ 2 ) e eπik[γ (γτ +δ)(2zτ +τ 2 2 )−(2αz (γτ +δ )+α τ ) 2πi(−bα+βa)] e (γτ +δ ) (z (γτ + δ )) = eβ +ατ (z (γτ + δ )) = = eiπikG e2πi(−bα+βa) , where G = γ (γτ + δ )(2zτ + τ 2 ) − 2αz (γτ + δ ) − α2 τ = ατ + β ατ + β 2 γ (γτ + δ )(2z ( )+( ) ) − 2αz (γτ + δ ) − α2 τ = γτ + δ γτ + δ 2zγ (ατ + β ) + γ (ατ + β )2 − 2αz (γτ + δ ) − α2 τ = γτ + δ γ (ατ + β )2 − α2 τ (γτ + δ ) . γτ + δ Here we used that αδ − βγ = 1. Now −2z + γ (ατ + β )2 − α2 τ (γτ + δ ) = 2γαβτ + γβ 2 − δα2 τ = −α(αδ − βγ )τ + αβ (γτ + δ ) − β (αδ − βγ ) = −(ατ + β ) + αβ (γτ + δ ). This allows us to rewrite G in the form G = −2z − ατ + β + αβ = −2z − τ + αβ. γτ + δ Putting G back in the expression (5.3) we get eτ (z ) = e−πik(2z+τ ) eπi[kαβ −2(βa−αb)] . Together with (5.2) this shows that Th({eτ (z )}, Λτ ) = Th(k, Λτ )a b , (5.3) ...
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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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