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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?

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    $\begingroup$ Somewhat tangential: see the work of Guralnick on Beauville structure about obtaining smooth projective surfaces as a quotient by finite group action on product of curves. One link is arxiv.org/abs/1009.6183 $\endgroup$
    – P Vanchinathan
    Apr 10, 2014 at 14:59

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Yes.

By a result of Freitag, in general one has $q(S) = g(C_1/G)+g(C_2/G)$. Look here, Corollary 3.6.

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    $\begingroup$ Actually this goes probably back to Poincaré. If a group $G$ acts freely on a projective manifold $M$, the pull back map $H^0(M/G,\Omega ^i)\rightarrow H^0(M,\Omega_M ^i)^G$ is an isomorphism -- this is straightforward. In your case you have $H^0(C_1\times C_2,\Omega ^1)= H^0(C_1,\Omega ^1)\oplus H^0(C_2,\Omega ^1)$, so taking the invariants under $G$ gives what you want. $\endgroup$
    – abx
    Apr 10, 2014 at 15:43

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