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Nov 18, 2010 at 18:24 comment added diverietti 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
Nov 18, 2010 at 11:05 comment added Francesco Polizzi If $A$ is a complex torus and $X$ an irreducible subvariety of $A$ which is non-degenerated, locally complete intersection and such that $2 \dim X > \dim A$, then $\pi_1(X)=\pi_1(A)$. So the fundamental group of $X$ is Abelian, hence amenable, in particular $X$ is not Kahler hyperbolic. If you know that $X$ is Kobayashi hyperbolic, this actually gives an example.
Nov 18, 2010 at 9:20 comment added diverietti Recently it has been proved by Damian Brotbek that generic projective complete intersections of high multi-degree are hyperbolic. These should give other examples. Otherwise, what about subvarieties of complex tori which do not contain any translate of a sub-tours ?
Nov 17, 2010 at 17:19 history edited Francesco Polizzi CC BY-SA 2.5
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Nov 17, 2010 at 16:05 comment added sfilip Thanks so much! I guess I need to learn how to do computations, not only the statements of the theorems :)
Nov 17, 2010 at 16:01 vote accept sfilip
Nov 17, 2010 at 10:57 history edited Francesco Polizzi CC BY-SA 2.5
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Nov 17, 2010 at 10:54 comment added Francesco Polizzi Fixed. By the way Diverietti, do you know any other examples?
Nov 17, 2010 at 10:47 history edited Francesco Polizzi CC BY-SA 2.5
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Nov 17, 2010 at 10:26 comment added diverietti 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.
Nov 17, 2010 at 10:11 history edited Francesco Polizzi CC BY-SA 2.5
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Nov 17, 2010 at 10:04 comment added Francesco Polizzi 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.
Nov 17, 2010 at 9:49 comment added diverietti 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 $n=4$.
Nov 17, 2010 at 9:35 history edited Francesco Polizzi CC BY-SA 2.5
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Nov 17, 2010 at 9:29 history edited Francesco Polizzi CC BY-SA 2.5
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Nov 17, 2010 at 9:11 history answered Francesco Polizzi CC BY-SA 2.5