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The Bloch conjecture states the following:

Bloch's conjecture. 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.

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 $f(\mathbb C)\subset Y$.

This conjecture has been proven, thanks to the works of Ochiai, by Kawamata and, independently, by Wong.

A standard Albanese map argument permits to reduce the conjecture to the following statement:

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.

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.

Question 1. Is there any geometric characterization or sufficient condition in order to insure that the Albanese variety of a projective algebraic manifold is simple?

Question 2. Is there any geometric characterization or sufficient condition, other than having big irregularity, in order to insure that the image of a projective algebraic manifold via the Albanese map is a proper subvariety?

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.

N.B. 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.

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  • $\begingroup$ What is an analytically degenerate curve? $\endgroup$
    – aglearner
    Commented Oct 24, 2012 at 16:42
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    $\begingroup$ I define this at line 7 from above. $\endgroup$
    – diverietti
    Commented Oct 24, 2012 at 16:42
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    $\begingroup$ A subvariety $X$ of abelian variety $A$ is of general type unless there is an abelian subvariety $B$ of $A$, translation by which preserves $X$; this is proved in Ueno's Springer Lecture Notes on classification theory, but I do not have the precise reference right now. If the abelian variety is not simple, then using products you can construct subvarieties with arbitrarily large irregularity but not of general type. $\endgroup$
    – naf
    Commented Oct 24, 2012 at 18:43
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    $\begingroup$ By "Abelian Variety", do you mean a complete, irreducible and reduced group scheme of finite type over $\mathbb{C}$? $\endgroup$
    – Filippo Alberto Edoardo
    Commented Oct 25, 2012 at 8:37
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    $\begingroup$ Ahahahahahahahahahahahah. Exactly! $\endgroup$
    – diverietti
    Commented Oct 25, 2012 at 8:53

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