Let $S$ be a projective surface with pseudoeffective anticanonical divisor $K_S$. Is it true that if $C$ is an integral curve with $C^2<0$ and $C \cdot K_S >0$, then $\max_C (C \cdot K_S)$ is finte?

$\begingroup$ Yes: any curve with positive intersection with the canonical divisor must be a component of any effective representative of a (multiple of) the anticanonical divisor. In particular, there are only finitely many integral curves $C$ with $C \cdot K > 0$. $\endgroup$– M PCommented Sep 16, 2011 at 15:34

$\begingroup$ Dear MP: but here the assumption is only that K is pseudoeffective. It's not obvious to me that that implies effective... $\endgroup$– user5117Commented Sep 16, 2011 at 15:36

$\begingroup$ Ah, I see. The issue seems to be with ruled nonrational surfaces (use Zariski decomposition and RiemannRoch on the nef part to deduce effectivity of a multiple). I need to think more about this case... $\endgroup$– M PCommented Sep 16, 2011 at 15:38

2$\begingroup$ It seems that the Zariski decomposition of K will work: the negative part is the only place where the intersection with K is negative and this part is effective (up to a multiple). $\endgroup$– M PCommented Sep 16, 2011 at 15:45
1 Answer
Let $K=N+E$ be the Zariski decomposition of $K$, so that $N,E$ are $\mathbb{Q}$divisors with $N$ nef, $E$ effective, such that $N \cdot E = 0$ and the restriction of the intersection pairing to the irreducible components of $E$ is negative definite. It follows that if $C$ is an integral curve such that $C \cdot (K)<0$, then we must have $C \cdot E < 0$, so that $C$ is a component of $E$. Since $E$ has only finitely many components, we deduce that there are only a finite number of integral curves with negative intersection with $K$, and the required boundedness follows.

1$\begingroup$ Note that the assumption $C^2 < 0$ is superfluous: the integral curves $C$ satisfying $K \cdot C > 0$ must have negative square by the argument above. $\endgroup$– M PCommented Sep 16, 2011 at 16:07