Take the tour ×

Let $S$ be a complex surface of general type. Are there infinitely many smooth rational curves on $S$? And more general, what if $V$ is a variety of general type?

share|improve this question
No. Clemens showed there are no rational curves on a general quintic surface (which is of general type). –  J.C. Ottem Jan 18 '11 at 18:25
Actually, it seems that a general surface of general type over $\mathbb{C}$ should have at most finitely many rational curves, but I guess you could also make some example with infinitely many. –  J.C. Ottem Jan 18 '11 at 18:52
@J.C.: if you really can, you will be famous... –  Sándor Kovács Jan 18 '11 at 20:49
add comment

2 Answers

up vote 6 down vote accepted

I think that the best result in this direction is the following result of Bogomolov:

Theorem Let $S$ be a surface of general type with $c_1^2(S) > c_2(S)$. Thern for any $g$ the curves of geometric genus $g$ on $S$ form a bounded family.

In particular, since a surface of general type cannot be covered by rational or elliptic curves, these curves cannot deform. So Bogomolov's result implies that if $c_1^2(S) > c_2(S)$ then $S$ contains only finitely many rational or elliptic curves.

In general, it is conjectured than rational curves are never Zariski dense on a variety $V$ of general type, and more precisely it is expected that they are contained in a proper subvariety (hyperbolicity conjecture).

If $\dim V \geq 3$, you can obviously have infinitely many of them: for instance, take $V= S\times C$, where $S$ is a surface of general type containing a smooth rational curve and $C$ is a curve of genus at least $2$.

share|improve this answer
Great answer! Thanks! Well, if only assuming that any of these smooth rational curve has negetive intersection number without the condition on $c_1$ and $c_2$, can I say that the number is finite using the finiteness of the rank of Neron Severi group? –  Michael Zhang Jan 18 '11 at 19:07
I think it is. It is the Picard number. –  Michael Zhang Jan 18 '11 at 19:25
If $C$ is a smooth rational curve on a minimal surface of general type, then $C^2 \leq -2$ by the genus formula. The rank of NS(S) is always finite, but in order to prove finiteness you should also prove that the rational curves are independent in the Neron Severi. This is easy if $C^2$b is bounded from below; but if you allows $C^2$ to vary, than a priori you could have countably many smooth rational curves... –  Francesco Polizzi Jan 18 '11 at 19:39
...So Bogomolov's theorem is really a non-trivial result. At any rate, if hyperbolicity conjecture is true for surfaces of general type, then one always has only finitely many rational curves (i.e. the numerical condition on $c_1^2$ and $c_2$ would be superflous). –  Francesco Polizzi Jan 18 '11 at 19:41
I don't get this question. If $C^2<0$ then by the genus formula $C.K<-2$ and so C is a fixed component in $|K|$? Surely, there must be finitely many such curves (if any). What does this have to do with the Picard number? –  J.C. Ottem Jan 18 '11 at 19:42
show 6 more comments

The conjecture Francesco is referring to as the "hyperbolicity conjecture" is actually the Green-Griffiths-Lang conjecture. It states that on any given smooth projective manifold of general type $X$ there should exist a proper subvariety $Y\subsetneq X$ such that for all non-constant holomorphic map $f\colon\mathbb C\to X$ one has $f(\mathbb C)\subset Y$.

This would imply in particular that the same proper subvariety should contain every rational or elliptic curve (or, more generally, every image of a complex torus).

Beside the result of Bogomolov (which treats "only" the algebraic part of this conjecture) one should cite also Mc Quillan's theorem, which prove this conjecture for surfaces under the same assumption on the second Segre number $c_1^2-c_2>0$.

This condition is a technical hypothesis which guarantees the existence of an algebraic (multi)foliation on the surface. The core of Mc Quillan's proof is then in showing that an algebraic (multi)foliation on a surface of general type does not admit any dense parabolic leaf.

In higher dimensions, very little is known even for the algebraic part of the conjecture, apart from the case of generic complete intersections of high (multi)degree (Clemens, Ein, Voisin, Pacienza...).

To finish with, thanks to very recent results of Cano about resolution of singularities of holomorphic foliations by curves on threefolds, probably Mc Quillan will be soon able to improve his previous result with the weaker assumption $13c_1^2-9c_2>0$.

share|improve this answer
The resolution of singularities of holomorphic foliations by curves on 3-folds is due to McQuillan-Panazzolo (preprint IHES). Cano proved the resolution for codimension one foliations on 3-folds –  jvp Jan 18 '11 at 22:38
Definitely right Jorge! I meant adapting the strategy of Cano... And I forgot Panazzolo, because Michael Mc Quillan told me that personally quite long ago... Thanks for having given the precision! –  diverietti Jan 18 '11 at 23:25
Thanks. It is quite helpful. –  Michael Zhang Jan 19 '11 at 0:04
add comment

Your Answer


