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I have a strong feeling that, for a compact connected Riemann surface $X$ of genus $g>0$, the Euler characteristic of the Weierstrass divisor $W$ equals $$\chi(X,\mathcal{O}_X(W)) = (g-1)^2.$$ Is this true?

Answer:

By Riemann-Roch, the Euler characteristic is given by $$ \chi(X,W) = g^3 -g + 1- g = g^3-2g+1.$$

This is not equal to $(g-1)^2$ for $g>1$.

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closed as too localized by Angelo, Qiaochu Yuan, Dan Petersen, Felipe Voloch, Mark Sapir Jan 7 '12 at 19:12

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Just apply Riemann-Roch: the divisor $W$ has degree $g^3 - g$. I voted to close as "Too localized". –  Angelo Jan 7 '12 at 7:19
    
You're right. I asked this question too quickly. –  Harized Jan 7 '12 at 7:20

1 Answer 1

up vote 0 down vote accepted

I'm not sure I've heard the term Weierstrass divisor before, but I take it you mean the sum of the Weierstrass points, with multiplicities given by the weights. In this case, the sum of the weights, and hence the degree of the divisor, is given by

$$\sum_{p\in X} w(p) = (g-1)g(g+1).$$

By Riemann-Roch, the Euler characteristic then equals $g^3-2g+1$, which is different from your formula.

See Arbarello-Cornalba-Griffiths-Harris, Geometry of Algebraic Curves, exercise batch I.E.

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