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Suppose $M$ is a compact Riemann surface of genus 2. Then $M$ is hyperelliptic.

This is an old question and there have been some relevant posts on MSE. But I wonder if there is a more direct method. For example, I want to construct the meromorphic function giving the degree-two cover of $\Bbb P^1$. It seems if $\alpha_1, \alpha_2$ form a basis of the space of holomorphic differentials $H^0(M, \Omega^1)$, then $f=\frac{\alpha_1}{\alpha_2}$ is of degree 2. $f$ is obviously a meromorphic function since it is independent of chart choice. And it is not a constant and with degree $>1$, otherwise $M$ is isomorphic to the Riemann sphere. But I don't know hw to carry on to the degree is exactly 2, maybe counting $f^{-1}(\infty)$?

I am just learning Riemann-Roch theorem and I unfamiliar with the language of algebraic geometry. Could you explain in the language of Riemann surfaces? I would appreciate any help or hint! A proper reference is also OK.

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  • $\begingroup$ I think you ignored the genus condition. Only compact Riemann surfaces of genus $2$ are guaranteed to be hyperelliptic. $\endgroup$
    – Zerox
    Commented Jun 11, 2022 at 12:12
  • $\begingroup$ I exactly want to prove this proposition. And the genus condition is used at the dimension of holomorphic differential space. $\endgroup$
    – user823011
    Commented Jun 11, 2022 at 13:07

1 Answer 1

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Let $M$ be a genus $g$ compact Riemann surface. Then every abelian differential of first kind (i.e., an element of $H^0(M,\Omega_M^1)$) has $2g -2$ zeros (Gauss-Bonnet: $\int_M c_1(\Omega^{1}_M)= 2g - 2$). Therefore, when $g=2$, your function $f$ has two poles.

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  • $\begingroup$ What do your $c_1$ mean and could you suggest a reference of the theorem you use? $\endgroup$
    – user823011
    Commented Jun 11, 2022 at 13:45
  • $\begingroup$ I am using the fact that the degree of the canonical bundle is 2g-2. It should be in every book on Riemann surfaces. For example, Forster's book, Theorem 17.12 $\endgroup$
    – Function
    Commented Jun 11, 2022 at 14:07
  • $\begingroup$ That is, for a canonical divisor $K$ on a compact Riemann surface of genus $g$, $\deg K=2g-2$, isn't it? Then combining the degree of $f$ cannot be $1$, we reach the conclusion. $\endgroup$
    – user823011
    Commented Jun 11, 2022 at 14:19
  • $\begingroup$ Yes, precisely! $\endgroup$
    – Function
    Commented Jun 11, 2022 at 14:20

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