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My question is related to several notions of hyperbolicity, applied to Kahler manifolds (projective, in general). Kahler hyperbolicity was introduced in this paper of Gromov's. He calls a Kahler manifold Kahler hyperbolic if the lift to the universal cover of the symplectic form(=imaginary part of hermitian metric) is the differential of a bounded $1$-form. A $1$-form is bounded if its norm is pointwise absolutely bounded (the norm is induced by the pull-back of the metric to the tangent space).

As he notes in the introduction, this notion implies Kobayashi hyperbolicity. My question is regarding the converse, namely to find an example of a Kobayashi hyperbolic manifold that isn't Kahler hyperbolic.


My guess is that an example of an algebraic variety with the properties described below exists (so it would be nice if an algebraic geometer could say a few things here).

For a Kahler hyperbolic manifold, Gromov proves that its Euler characteristic is $(-1)^n$, where $n$ is its complex dimension. He also shows a Kahler hyperbolic manifold has quasi-ample canonical bundle (namely, it has Kodaira dimension $n$). Another observation is that a Kahler hyperbolic manifold cannot have amenable fundamental group. So any algebraic variety which fails one of the above tests but still is Kobayashi hyperbolic would fit the bill.

For somewhat different reasons, I would be happier if this would be a projective variety, which also doesn't have any $(2,0)$ cohomology. As a side question, just to check my understanding, is it true that for a variety with no $(2,0)$ classes in cohomology, any homology class that comes from a map of a surface (pushing forward the fundamental class) can in fact be realized by some algebraic curve in the variety?

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I think you mean "the sign of its Euler characteristic is $(-1)^n$" –  Francesco Polizzi Nov 17 '10 at 9:40
    
You might have a look on a survey of Voisin math.jussieu.fr/~voisin/Articlesweb/harvard.pdf, On some problems of Kobayashi and Lang –  Dmitri Nov 17 '10 at 12:10
    
It is a conjecture by Kobayashi (open starting from dimension three) that projective hyperbolic manifolds have ample canonical bundle. In particular, conjecturally you won't be able to find such examples by finding hyperbolic manifolds with non maximal Kodaira dimension. –  diverietti Nov 19 '10 at 15:58

1 Answer 1

up vote 6 down vote accepted

Take a very general hypersurface $j \colon X \hookrightarrow \mathbb{P}^n$ of sufficiently high degree, $3 \leq n \leq 4$. Then $X$ is Kobayashi hyperbolic (Kobayashi actually conjectured that this is true for all $n$ and Siu recently outlined a strategy for proving this, see Diverietti's comment below).

On the other hand, Lefschetz hyperplane theorem says that the natural map

$\pi_1(X) \stackrel{j_*}{\longrightarrow} \pi_1(\mathbb{P}^n)$

is an isomorphism. Then $X$ is simply connected, in particular its fundamental group is amenable and so $X$ cannot be Kahler hyperbolic.

If $n=4$, the projective variety $X$ also satisfies your second request. If fact, again by Lefschetz theorem, see for instance [Dimca, Singularities and topology of hypersurfaces, Theorem 2.6 p. 151], for all $n \geq 4$ we also have an isomorphism

$H^2(\mathbb{P}^n) \stackrel{j^*}{\longrightarrow} H^2(X)$.

Since $H^2(\mathbb{P}^n)=\mathbb{C}$, Hodge decomposition yields

$H^{2,0}(X)=H^{0,2}(X)=0, \quad H^{1,1}(X)=\mathbb{C}$.

If Kobayashi's conjecture were true, then the hypersurfaces $X$ would provide examples in all dimensions.

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Let's say that Siu indicated a quite precise strategy to prove that. Up to now, we have a full proof only for surfaces in $\mathbb P^3$ and threefolds in $\mathbb P^4$. So your exemple works for the moment just for $n=4$. –  diverietti Nov 17 '10 at 9:49
    
Ok, thank you for pointing this out. By the way, Lefschetz is true also for $n=3$, so surfaces in $\mathbb{P}^3$ are examples too. However, the computations on $H^{2,0}$ works only for $n \geq 4$. I have edited the answer. –  Francesco Polizzi Nov 17 '10 at 10:04
    
Just a last comment, to be precise. This is not Siu's conjecture, this is the Kobayashi conjecture. The strategy to prove it is by Siu. –  diverietti Nov 17 '10 at 10:26
    
Fixed. By the way Diverietti, do you know any other examples? –  Francesco Polizzi Nov 17 '10 at 10:54
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That's exactly what I meant. A subvariety of a complex torus is Kobayashi hyperbolic if and only if it does not contain any translate of a subtorus –  diverietti Nov 18 '10 at 18:24

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