Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$S$: 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?

share|improve this question

1 Answer 1

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 $K_SL >0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.