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MODULAR FORMS-page87

# MODULAR FORMS-page87 - x 6 = ∞-e x f-1 if f x = ∞(8.20...

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83 Proof. Replacing D with D + div( f ), we do not change the dimensions of the spaces L ( D ) and L ( K X - D ) but change deg( D ) by deg( D + div( f )) = deg D + deg(div( f )). It follows from Riemann-Roch that deg(div( f )) = 0. Corollary 8.2. deg K X = 2 g - 2 . Proof. Take D = 0 and use that L (0) = O ( X ) = C . Here we use that a holomorphic function on compact Riemannian surface is constant. This gives g = dim L ( K X ) . (8.16) Now take D = K X and get deg( K X ) = 2 g - 2. Theorem 8.3. Let b i ( X ) = dim H i ( X, R ) be the Betti numbers of X . Then b 1 = 2 g, b 0 = b 2 = 1 . Proof. Since X is a connected compact manifold of dimension 2, this is equivalent to e ( X ) = 2 X i =0 ( - 1) i b i ( X ) = 2 - 2 g = - deg( K X ) . (8.17) Let f be a non-constant meromorphic function on X (its existence follows from the Riemann-Roch theorem). It defines a holomorphic map f : X P 1 ( C ). For any point x X set e x ( f ) = ( ν x ( f - z ) if f ( x ) = z = - ν x ( f ) if f ( x ) = . (8.18) It is a positive integer. Since deg(div( f - z )) = 0 we obtain X x f - 1 ( z ) e x ( f ) = X x f - 1 ( ) e x ( f ) . (8.19) Notice that, for any x X , ν x ( df
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Unformatted text preview: ( x ) 6 = ∞-e x ( f )-1 if f ( x ) = ∞ . (8.20) Here df is the meromorphic diﬀerential deﬁned locally by df dt dt , where t is a local parameter at x . Since the degree of df is ﬁnite we obtain that there are only ﬁnitely many points x ∈ X such that e x ( f ) > 1. In particular, there is a ﬁnite subset of points S = { y 1 ,...,y s } in P 1 ( C ) such that, for any y 6∈ S X x : f ( x )= y e x ( f ) = n = # f-1 ( y ) . (8.21) Taking into account the formulas ( ?? )-(8.21), we obtain 2 g-2 = X x ∈ X ν x (div( df )) = X y : f ( y ) 6 = ∞ ( e x ( f )-1) + X y : f ( y )= ∞ (-e x ( f )-1) = X x ∈ X ( e x ( f )-1)-2 X f ( x )= ∞ e x = X x ∈ X ( e x ( f )-1)-2 n = X y ∈ Y ( n-# f-1 ( y ))-2 n. (8.22)...
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