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.
[Math] Application of Riemann-Roch
complex-analysisriemann-surfaces
Best Answer
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.