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?



Fundamental group of a product of two curves
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](http://www.jstor.org/pss/25098976), which considers the more general situation $X=(C\_1 \times C\_2)/G$, where $G$ is a finite group acting freely on the product $C\_1 \times C\_2$. 


