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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?

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Do you know anything about the higher homotopy groups? If they are non-zero then it isn't the product of two curves. If they are all zero then you have a fighting chance. –  David Roberts Jan 30 '11 at 20:30
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What do you mean by 'complex surface'? Is it compact? projective? Is $\mathbb{C}^2$ minus a suitable submanifold a candidate? –  J.C. Ottem Jan 30 '11 at 20:40
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Since the fundamental group is a birational invariant, I assume that you want the word "be" to mean "is birational to", right? Otherwise you could eg take the product of two curves and blow it up at some point. I suspect that the answer to the corrected question is "no", but I don't know an example off the top of my head. –  Andy Putman Jan 30 '11 at 20:43
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2 Answers

up vote 15 down vote accepted

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 $C_1 \times C_2$.

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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.

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OOPS! Sorry, somehow I did not notice that the curves must have genus greater than 1. –  Yuri Zarhin Jan 31 '11 at 2:06
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