MODULAR FORMS-page93

# MODULAR FORMS-page93 - 89 Let us compute We know that 8 >2...

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

89 Let us compute ν τ (Φ) . We know that ν τ ( j - j ( τ )) = 8 > < > : 2 if i Γ(1) τ , 3 if e 2 πi/ 3 Γ(1) τ , 1 otherwise. This immediately implies that ν τ ( j 0 ) = 8 > < > : 1 if i Γ(1) · τ , 2 if e 2 πi/ 3 Γ(1) · τ , 0 otherwise. Thus ν x (Φ) = k ( e x ( j ) - 1) /e ( x ) . Now, let x = c i be a cusp represented by c P 1 ( Q ). We used the local parameter e 2 πiτ/h to deﬁne ν c ( F ). Since j admits the Fourier expansion e - 2 πiτ +744+ c 1 e 2 πiτ + ... at , we see that j 0 has the expansion - 2 πie - 2 πiτ + c 2 2 πie 2 πiτ + ... at the cusp c . This shows that ν c (Φ) = - kh . So we get div( F/ Φ) = div( F ) - k X x ( e x ( j ) - 1 e ( x ) ) x + r X i =1 h i c i . Comparing this with the computation of div( dj ) in the proof of Theorem 8.5, we get div( F ) = div( F/ Φ) + k div( dj ) + k X elliptic x (1 - e ( x ) - 1 ) x + k r X i =1 c i . Since div( F ) 0 we obtain that F/ Φ L ( D 0 ), where D 0 is linearly equivalent to kK X + b kD c as in the assertion of the theorem. Conversely, if Ψ L ( D 0 ) we easily get that F = ΨΦ ∈ M k (Γ). Finally, if F is a cuspidal modular form, we have ν x ( F ) > 0 at cusps. This easily implies that
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online