Let $S$ be a complex surface whose fundamental group is isomorphic to the fundamental group of a product of two curves of genera $>1$. Does $S$ have to be a product of two curves?

The answer in no, because of the following result: Theorem 1. Let $X$ be a nonruled minimal surface. Then there exists a finite ramified covering $S \to X$ of degree $>1$, such that $S$ is minimal of general type with $K_S$ very ample, $\pi_1(S) \cong \pi_1(X)$ and $S$ is not birationally equivalent to $X$. We can moreover assume that $S$ has negative index, i.e. $K_S^2  8 \chi(\mathcal{O}_S) <0$. So the fundamental group $\pi_1(X)$ alone does not determine the birational type of $X$, and in general not even its diffeomorphism type. When $X$ is the product of two curves, however, something more can be said, provided that one also knows the topological Euler number. More precisely one proves the following Theorem 2. Let $C_1$, $C_2$ be smooth curves of genus $g_1$, $g_2$, with $g_i \geq 2$, and let $X=C_1 \times C_2$. Then any surface $S$ such that $\pi_1(S) \cong \pi_1(X)$ and $e(S)=e(X)$ is isomorphic to a product of two curves of the same genera. Theorems 1 and 2 were proven by F. Catanese in his paper Fibred surfaces, varieties isogenous to a product and related moduli spaces, which considers the more general situation $X=(C_1 \times C_2)/G$, where $G$ is a finite group acting freely on the product $C_1 \times C_2$. 


All twodimensional complex tori $T$ have the same fundamental group, because such a torus is homeomorphic to a product of four copies of the unit circle $S^1$. Among them there are all the products of $E_1 \times E_2$ of elliptic curves. Since each elliptic curve is homeomorphic to a product of two copies of $S^1$, the fundamental groups of $T$ and $E_1 \times E_2$ are isomorphic (for all $T, E_1,E_2$). However, almost all twodimensional complex tori are not biholomorphically isomorphic to a product of elliptic curves. Shafarevich's ``Basic algebraic geometry" contains examples of twodimensional complex tori that do not contain complete complex curves at all and therefore are not the products of two curves. One may also get an explicit example of such a torus (without curves), starting with a totally complex quartic number field $F$ that does not contain an imaginary quadratic subfield, choosing a rank 4 discrete lattice $\Gamma$ in the realification $F_R$ of $F$ and putting $T=F_R/\Gamma$ (Math. Ann. 303 (1995), 1129). As for complex abelian surfaces $A$ (i.e., algebraizable twodimensional complex tori), almost all of them are also not isomorphic to a product of elliptic curves. An explicit example is provided by the jacobian $J(C)$ of the genus 2 curve $C:y^2=x^5x1$. Actually, it is known (arXiv:math/9909052 [math.AG]) that $J(C)$ has no nontrivial endomorphisms and therefore is not isomorphic to a product of elliptic curves. 

