About the Bloch conjecture on entire curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:32:31Z http://mathoverflow.net/feeds/question/110557 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110557/about-the-bloch-conjecture-on-entire-curves About the Bloch conjecture on entire curves diverietti 2012-10-24T16:21:54Z 2012-10-26T17:48:21Z <p>The Bloch conjecture states the following:</p> <p><strong>Bloch's conjecture.</strong> Let $X$ be a compact complex Kähler variety such that the irregularity $q = h^0(X,\Omega^1_X)$ is larger than the dimension $n = \dim X$. Then, every entire curve drawn in $X$ is analytically degenerate.</p> <p>Here $X$ may be singular and $\Omega^1_X$ can be defined in any reasonable way (direct image of the $\Omega^1_{\widetilde X}$ of a desingularization $\widetilde X$ or direct image of $\Omega^1_U$ where $U$ is the set of regular points in the normalization of $X$). By an entire curve I mean a non constant holomorphic map $f\colon\mathbb C\to X$, and analytically degenerate means that there exists a closed analytic subset $Y\subsetneq X$ such that $f(\mathbb C)\subset Y$.</p> <p>This conjecture has been proven, thanks to the works of Ochiai, by Kawamata and, independently, by Wong. </p> <p>A standard Albanese map argument permits to reduce the conjecture to the following statement:</p> <p><em>Let $A$ be an abelian variety and $f\colon\mathbb C\to A$ an entire curve. Then, the Zariski closure $\overline{f(\mathbb C)}$ is a translate of a subtorus.</em> </p> <p>In particular a subvariety of an abelian variety does not have any entire curve (Brody hyperbolicity) if and only if it does not contain any translate of a subtorus. Thus, in a simple abelian variety every subvariety is hyperbolic. More generally, if a subvariety of an abelian variety is not a translate of a subtorus, then every entire curve in $X$ is analytically degenerate.</p> <p><strong>Question 1.</strong> Is there any geometric characterization or sufficient condition in order to insure that the Albanese variety of a projective algebraic manifold is simple?</p> <p><strong>Question 2.</strong> Is there any geometric characterization or sufficient condition, <em>other than having big irregularity</em>, in order to insure that the image of a projective algebraic manifold via the Albanese map is a proper subvariety?</p> <p>Notice that by the universal property of the Albanese map, if the image of a projective algebraic manifold via the Albanese map is a proper subvariety then this image is necessarily not a translate of a subtorus.</p> <p><strong>N.B.</strong> I changed the last part of my post. Now, Question 2 is no more as in the previously, not reedited post. In particular the comment of ulrich refers to my previous Question 2. </p>