Let $C\_1,C\_2$ be smooth, projective curves of genera $g\_1,g\_2 \geq 2$. Assume that a group $G$ of order $(g\_1 - 1)(g\_2 - 1)$ acts on $C\_1$ and $C\_2$ such that $C\_1/G \cong \mathbb{P}^1$ and $C\_2/G \cong \mathbb{P}^1$. Assume that $G$ acts freely on $C\_1 \times C\_2$, then the quotient $S = (C\_1 \times C\_2)/G$ is smooth.


I know that $q(S) = p\_g(S)$: we have $\chi(C\_1 \times C\_2) = (g\_1 - 1)(g\_2 - 1)$, and therefore $\chi(S) = 1$, which implies that $q(S) = p\_g(S)$. Is it also true that both these numbers are equal to $0$? If so, why?



Quotient of product of curves
Yes.


By a result of Freitag, in general one has $q(S) = g(C\_1/G)+g(C\_2/G)$. Look [here](http://arxiv.org/abs/0805.1424), Corollary 3.6. 


