# Show that a compact Riemann surface of genus two is hyperelliptic

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.

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

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.
• What do your $c_1$ mean and could you suggest a reference of the theorem you use? Commented Jun 11, 2022 at 13:45
• 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. Commented Jun 11, 2022 at 14:19