Fundamental group of a product of two curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T15:29:57Z http://mathoverflow.net/feeds/question/53816 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53816/fundamental-group-of-a-product-of-two-curves Fundamental group of a product of two curves G Gonzalez 2011-01-30T20:14:19Z 2011-01-31T02:02:31Z <p>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?</p> http://mathoverflow.net/questions/53816/fundamental-group-of-a-product-of-two-curves/53823#53823 Answer by Francesco Polizzi for Fundamental group of a product of two curves Francesco Polizzi 2011-01-30T21:26:51Z 2011-01-30T23:03:16Z <p>The answer in <strong>no</strong>, because of the following result:</p> <p><strong>Theorem 1.</strong> 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 <em>not</em> birationally equivalent to $X$. We can moreover assume that $S$ has negative index, i.e. $K_S^2 - 8 \chi(\mathcal{O}_S) &lt;0$. </p> <p>So the fundamental group $\pi_1(X)$ alone does not determine the birational type of $X$, and in general not even its diffeomorphism type. </p> <p>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</p> <p><strong>Theorem 2.</strong> 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. </p> <p>Theorems 1 and 2 were proven by F. Catanese in his paper <a href="http://www.jstor.org/pss/25098976" rel="nofollow">Fibred surfaces, varieties isogenous to a product and related moduli spaces</a>, which considers the more general situation $X=(C_1 \times C_2)/G$, where $G$ is a finite group acting freely on the product $C_1 \times C_2$. </p> http://mathoverflow.net/questions/53816/fundamental-group-of-a-product-of-two-curves/53849#53849 Answer by Yuri Zarhin for Fundamental group of a product of two curves Yuri Zarhin 2011-01-31T02:02:31Z 2011-01-31T02:02:31Z <p>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.</p> <p>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).</p> <p>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.</p>