Fundamental group of a product of two curves 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?
 A: All two-dimensional 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 two-dimensional complex tori are not biholomorphically isomorphic to a product of elliptic curves.
Shafarevich's ``Basic algebraic geometry"  contains examples of two-dimensional 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), 11--29).
As  for complex abelian surfaces $A$ (i.e., algebraizable  two-dimensional 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 $C:y^2=x^5-x-1$. 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.
A: The answer in no, because of the following result:
Theorem 1. Let $X$ be a non-ruled  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 $C_1 \times C_2$. 
