Ample vector bundles on complex tori - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:32:20Z http://mathoverflow.net/feeds/question/102681 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102681/ample-vector-bundles-on-complex-tori Ample vector bundles on complex tori diverietti 2012-07-19T16:13:28Z 2012-07-20T07:03:08Z <p>Let $X$ be a $n$-dimensional complex torus and $\omega$ a Kähler form on $X$. Then, it is well known that a real $(1,1)$-class $[\alpha]\in H^{1,1}(X,\mathbb R)$ is a Kähler class if and only if for all $1\le j\le n$ one has $$\int_X\alpha^j\wedge\omega^{n-j}>0.$$ In particular, if $L\to X$ is a holomorphic line bundle, then $L$ is ample (and hence $X$ is an abelian variety) if and only if $$c_{1}(L)^j\cdot[\omega]^{n-j}>0,\quad 1\le j\le n.$$</p> <p><strong>Question.</strong> Is it known any similar numerical criterion for higher rank holomorphic vector bundle on complex tori?</p> <p>I would be interested also in weaker forms of this question. For instance, adding the hypothesis of semi-stability for the vector bundle, or possibly just requiring the vector bundle to be big instead of ample.</p> <p>Any hint or comment is very welcome! </p> http://mathoverflow.net/questions/102681/ample-vector-bundles-on-complex-tori/102691#102691 Answer by Sándor Kovács for Ample vector bundles on complex tori Sándor Kovács 2012-07-19T17:49:57Z 2012-07-19T23:59:13Z <p>Let $\mathscr E$ be a vector bundle on $X$ (not even necessarily a torus) and consider $P=\mathbb P(\mathscr E)$ the associated projective bundle with the corresponding natural relatively ample line bundle $\mathscr L:= \mathscr O_{\mathbb P(\mathscr E)}(1)$.</p> <p>Then $\mathscr E$ is ample on $X$ if and only if $\mathscr L$ is ample on $P$ (it is always ample over $X$, but this means ample over the base field (or scheme)).</p> <p>By the Nakai-Moishezon criterion $\mathscr L$ is ample on $P$ if and only if for any irreducible and reduced subvariety $Y\subseteq P$ of dimension $j$ for any $1\le j\le n=\dim X$, $$c_{1}(\mathscr L)^j\cdot Y>0.$$</p> <p>If $X$ is a torus, then you might be able to simplify this by limiting the list of $Y$ that you have to look at.</p> http://mathoverflow.net/questions/102681/ample-vector-bundles-on-complex-tori/102717#102717 Answer by Gunnar Magnusson for Ample vector bundles on complex tori Gunnar Magnusson 2012-07-20T06:04:27Z 2012-07-20T06:04:27Z <p>Dear Simone,</p> <p>This is just a comment.</p> <p>The answer to your question, if one exists, is surely that there exist universal polynomials $p_j^r$ ($1 \leq j \leq r = {\rm rank} E$) in the Chern classes of the vector bundle $E \to X$ such that $$\int_X p_j{}^r(c_1(E), \dots, c_j(E)) \wedge \omega^{n-j} > 0$$ for all $j$ implies that the bundle $E$ is ample. Here "universal" means that the polynomials in question only depend on the dimension $n$, but otherwise not on the variety $X$. For line bundles, these polynomials are known and are, as you wrote, $p_j{}^1(x) = x^j$ for all $j$.</p> <p>The problem is that the condition you wrote for line bundles is a corollary of a more general theorem of father, one that characterizes Kahler classes amongst real $(1,1)$-classes. Our hope of approaching the problem should thus be to find suitable positivity criterion for higher degree classes. The ideal outcome would be a higher rank version of the Kodaira condition; so we'd know that if a "degree vector" $(u_1, \dots,u_r)$ of integral cohomology classes satisfies some conditions, then there exists an ample vector bundle $E$ of rank $r$ such that $c_j(E) = u_j$ (compare with $L$ ample iff $c_1(L)$ Kahler and integral). The trouble is that finding these conditions amounts to finding the universal polynomials $p_j{}^r$, and thus answering a much more general question.</p> <p>In short, we have no idea what a "positivity condition" for a collection of cohomology classes $(u_1, \ldots, u_r)$ looks like. I agree with your approach of simplifying the problem and starting the search for these on complex tori. However, I think that if we knew the answer on complex tori, we'd know it on general complex manifolds too.</p>