1
$\begingroup$

$S$: a smooth projective surface over $\mathbb{C}$ which has non-negative Kodaira dimension.

$L$: an ample divisor on $S$

Why $K_S.L\ge0?$

I know that :

for some $m\ge1$, there is an effective divisor $E \in |mK_S|$ s.t.$0< E.L=(mK_S).L=m(K_S.L).$

Why is "$=$" possible?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

Since some positive multiple of $K_S$ is effective, by Nakai-Moishezon criterion the case $K_S L=0$ is possible only if $K_S$ is a torsion line bundle, i.e. $E=\mathcal{O}_S$ in your notation.

This implies that $S$ is minimal and $\textrm{kod}(S)=0$. In fact, in this situation one has $12 K_S = \mathcal{O}_S$, hence $K_SL = \frac{1}{12} \mathcal{O}_S L =0$.

In the case $\textrm{kod}(S) \geq 1$, instead, some positive multiple of $K_S$ is an effective, non-trivial line bundle hence if $L$ is ample one has $K_SL >0$.

Improve this answer
$\endgroup$

You must log in to answer this question.

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