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MODULAR FORMS-page94

# MODULAR FORMS-page94 - 90 LECTURE 8 THE MODULAR CURVE Proof...

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90 LECTURE 8. THE MODULAR CURVE Proof. We know this already when Γ = Γ(1). So we may assume that μ Γ > 1. By Theorem 8.5 we can identify the space M 6 (Γ) with L ( D ), where deg D = 6 deg K X + 6 r + 3 r 2 + 4 r 3 = 12 g - 12 + 6 r + 3 r 2 + 4 r 3 . I claim that deg D > 2 g + 1 . If g 0 this is obvious. If g = 0 we use the formula for the genus from Theorem 8.4. It easily gives that - 12 + 6 r + 3 r 2 + 4 r 3 = μ > 2 g + 1 = 1 . It follows from the proof of Theorem 8.8 that ν x ( f i ) = ν x ( f i /j 6 ) + D ( x ) . (8.28) Now we use the standard argument from the theory of algebraic curves. First of all the map is well-defined. In fact, if all functions f i vanish at the same point x , we obtain ν x ( f i ) > 0 for all i = 0 , . . . , N , and hence ν x ( f i /j 6 ) + D ( x ) - 1 0 for i = 0 , . . . , N . This implies that L ( D ) = L ( D - x ) . However, this contradicts the Riemann-Roch theorem: since deg( K X - D ) < deg( K X - ( D - x )) = 2 g - 2 - deg D + 1 < 0 , it gives dim L ( D ) = deg D + 1 - g > dim L ( D - x ) = deg D - 1 + 1 - g. Suppose f ( x ) = f ( x ) = p P N ( C ) for some x = x . Without loss of generality we may assume that p = (1 , 0 , . . . , 0) (to achieve this we make a linear transformation of coordinates).
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