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?

$\begingroup$ 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. $\endgroup$– J.C. OttemCommented Jan 18, 2011 at 18:52

16$\begingroup$ @J.C.: if you really can, you will be famous... $\endgroup$– Sándor KovácsCommented Jan 18, 2011 at 20:49
4 Answers
I think that the best result in this direction is the following theorem due to Bogomolov:
Theorem. Let $S$ be a surface of general type with $c_1^2(S) > c_2(S)$. Then 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 $2$.

$\begingroup$ 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? $\endgroup$– TongCommented Jan 18, 2011 at 19:07

$\begingroup$ I think it is. It is the Picard number. $\endgroup$– TongCommented Jan 18, 2011 at 19:25

$\begingroup$ 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... $\endgroup$ Commented Jan 18, 2011 at 19:39

2$\begingroup$ ...So Bogomolov's theorem is really a nontrivial 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). $\endgroup$ Commented Jan 18, 2011 at 19:41

$\begingroup$ 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? $\endgroup$ Commented Jan 18, 2011 at 19:42
I am surprised nobody mentioned the result of Lu and Miyaoka (Math. Res. Letter 2, 663676 (1995)) which implies indeed that there are only finitely many smooth rational curves on a surface of general type, thus answering the question of the OP.
The conjecture Francesco is referring to as the "hyperbolicity conjecture" is actually the GreenGriffithsLang 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 nonconstant 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 $c_1^2c_2>0$.
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 $13c_1^29c_2>0$.

1$\begingroup$ The resolution of singularities of holomorphic foliations by curves on 3folds is due to McQuillanPanazzolo (preprint IHES). Cano proved the resolution for codimension one foliations on 3folds $\endgroup$ Commented Jan 18, 2011 at 22:38

1$\begingroup$ 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! $\endgroup$ Commented Jan 18, 2011 at 23:25
Some related additional comments:
SiuYau showed that when the bisectional curvature of a compact K\"ahler variety is positive then there exists a rational curve.
Birkar, et al, showed that when $X$ is a Moishezon manifold which is not projective. Then $X$ contains a rational curve. Moreover we have the same result when $X$ is a normal Moishezon variety with analytic $\mathbb Q$factorial singularities.
Existence of rational curve is related to GromovWitten Invariants de to RuanTian.
Let $X$ be a smooth projective variety over $\mathbb C$. Fix a point $x\in X$, a homology class $A\in H_2(X(\mathbb C),\mathbb Z)$ and very ample divisors in general position $H_i\subset X$, $i=1,...,k$.
Let $y_0,...,y_k\in \mathbb CP^1$ be general points. Take maps $f:\mathbb CP^1\to X$ such that $f_*[\mathbb CP^1]=A$, $f(y_0)=x$ and $f(y_i)\in H_i$
RuanTian, introduced an invariant $$\tilde F_{A,X}(x,H_1,...,H_k;y_0,...,y_k):={\text{number of such maps}}$$ GromovWitten theory of pseudoholomorphic curves shows that one can make a similar definition where $X$ is replaced by a symplectic manifold $(M,\omega)$. Then the corresponding invariant is denoted by $$\tilde \Phi_{A,X}(x,H_1,...,H_k;y_0,...,y_k):={\text{number of such maps}}$$
RuanTian, showed that if $\tilde \Phi_{A,X}(x,H_1,...,H_k;y_0,...,y_k):={\text{number of such maps}}$ then there is a rational map $f:\mathbb CP^1\to X$ such that $f_*[\mathbb CP^1]=A$, $f(y_0)=x$ and $f(y_i)\in H_i$
Moreover, $\tilde \Phi_{A,X}(x,H_1,...,H_k;y_0,...,y_k)=\tilde F_{A,X}(x,H_1,...,H_k;y_0,...,y_k)$ if the following conditions are satisfied.
If $C_1,...,C_m\subset X$ are rational curves such that $\sum [C_i]=A$ and $x\in C_1$, then $m=1$
If $g:\mathbb CP^1\to X$ is any map such that $g_*[\mathbb CP^1]=A$, $g(y_0)=x$ then $H^1(\mathbb CP^1,g^*TX)=0$
Existence of rational curves are very important in the classification of the solution of K\"ahler Ricci flows .
i.e an effective way to consider GreenGriffithsLang conjecture is using classification of the solutions of relative Kahler Ricci flow and study of existence of rational curve appears,

$\begingroup$ Let $X$ be a Moishezon 3manifold, then $X$ is actually projective if it does not contain any rational curve, this result is due to Peternell, Thomas 98, and Kollar 91 $\endgroup$– user21574Commented Jul 23, 2017 at 3:18