11
$\begingroup$

I heard people mentioned this in one sentence, but don't see the reason.

Why a (smooth) variety of general type, i.e. an algebraic variety X with K_X big, is hyperbolic, i.e. has no non-constant map from the complex number into it?

I don't know what are the necessary assumption on the variety, do we need properness or smoothness?

Edit: according to David Lehavi's reply, we should certainly put some more condition on it. What's the correct statement of the fact?

Improve this question
$\endgroup$
0

3 Answers 3

Reset to default
10
$\begingroup$

You must be thinking of Lang's conjecture which predicts that a smooth projective variety is (Brody) hyperbolic if and only if all of its irreducible subvarieties are of general type.

This is still not known in general but there are many special cases that are known. A good example is McQuillan's theorem - a smooth surface of general type which satisfies $c_{1}^{2} > c_{2}$ and does not contain any rational or elliptic curves is Brody hyperbolic.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yes, the speaker mentioned one version of Lang's conjecture which says Brody hyperbolic implies rational points are not dense. The version I saw says general type implies rational points are not dense. Now the question is how to relate these two conjecture. $\endgroup$
    – Yuhao Huang
    Commented Dec 5, 2009 at 22:20
  • 1
    $\begingroup$ You have to combine the above conjecture that says that hyperbolic varieties only have subvarieties of general type with the conjecture that says that general type varieties can not have dense rational points. The two together tell you that a hyperbolic variety defined over the rationals must have only finitely many points over a number field $K$. Indeed, if you have infinitely many $K$-rational points, their Zariski closure will be a subvariety which can not be of general type which contradicts the fact that there are no such subvarieties. $\endgroup$
    – Tony Pantev
    Commented Dec 6, 2009 at 3:25
8
$\begingroup$

you can blow up a point on a general type surface, and get a general type surface containing a copy of C.

Improve this answer
$\endgroup$
7
$\begingroup$

Surely you need some assumptions on your variety, for example for every $n$ there is a smooth surface of degree $n$ in $CP^3$ that contains a line, for $n>3$ such surfaces are minimal and of general type. So for large classes of varieties of general type you need an additional assumption that the variety should be generic.

For hypersufaces in $CP^n$ the most optimistically you can hope that a generic hypersurface of degree $2n+1$ is hyperbolic (hypersurfaces of degree $2n-1$ and less always contain lines). This is related to a conjecture of Kobayshi. There is a very nice review of Claire Vosin on different aspects of hyperbolicity of complex projective manifold that you can find here http://people.math.jussieu.fr/~voisin/Articlesweb/harvard.pdf (this also contains the result mentioned by Tony Pantev)

Recently there was a genuing progress in proving of Kobayashi conjecture http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.2346v4.pdf though the obtained bound is very far fron optimal, worse than a triple exponent of n. At the conference due to 80 birthday of Atiyah Kirwan annonced that she can get a much more realistic bound.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .