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?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
|
3
1
|
|||||||||||||||
|
|
7
|
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. |
||
