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The Riemann-Roch theorem states $l(D)-l(K-D)=1 + deg(D) - g$ where $D$ is a divisor on a compact Riemann surface $X$ and $K$ is the divisor of meromorphic 1-form. In the case $g=2$ and $D=K$ this is $l(K)=2$. So the space of holomorphic 1-forms has dimension 2. Call $\omega_1, \omega_2$ a basis for this space. I don't understand why $\frac{\omega_1}{\omega_2}$ should be a ramified covering space $X \longrightarrow \mathbb{P}^1$ (this is what I wrote in my notes) and why every surface of genus 2 is double covering space of $\mathbb{P}^1$ ramified in 6 points. Thanks for any help.

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Once you figure out how to get the map, the fact that it is ramified in 6 points follows immediately from the Riemann-Hurwitz formula: $2g(X)-2=deg(f)(2g(\mathbb{P}^1)-2)+deg R$, so plugging in gives $deg R=6$. –  Matt Jun 1 '12 at 22:06

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The quotient $\frac {s}{t}$ of any two sections $ t\neq0,s\in \Gamma(X,L)$ of any holomorphic line bundle on $X$ is a meromorphic function $f$, non constant if $s,t$ are linearly independent in $\Gamma(X,L)$.

If this is the case $f$ may be seen as a nonconstant holomorphic map $X\to \mathbb P^1$.
Such holomorphic maps are ramified coverings of degree the degree $d=deg(L)=c_1(L)$ of $L$, in the sense that they are genuine covering spaces of degree $d$ over the complement of a finite number of points in $\mathbb P^1$.
(These points are the images of the ramification points of $f$ and are called the critical values of $f$)

All this applies to your case, where $L=K_X,s=\omega_1, t=\omega_2, d=2$, and the ramification number is given (as Matt commented ) by Hurwitz's formula $$2g(X)-2=deg(f)\cdot (2g(\mathbb P^1)-2) +r $$
yielding $r=6$ here.

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Ok, but how can I divide two such sections, (in this case two holomorphic 1-forms)? We didn't define such an operation. Should I consider to divide deleting the $dz$ part? –  balestrav Jun 2 '12 at 10:08
    
Dear balestrav, at $a\in X$ choose a coordinate $z$ . Write $s=g\text {d}z, t=h\text {d}z$ in a neighbourhood of $a$. The meromorphic function we're talking about is defined by $f=g/h$ near $a$. It is well defined, independently of the choice of the chosen coordinate $z$. –  Georges Elencwajg Jun 4 '12 at 18:57
    
Ok, thanks, i should've figured it out myself! –  balestrav Jun 4 '12 at 19:09

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