2
$\begingroup$

Let $S$ be a complex projective non-singular surface. Couldd you explain the following implication:

If the canonical divisor $K_S$ is nef then there exists a number $m>>0$ such that $mK_S$ is equivalent to an effective divisor.

I remember that a divisor $D$ is called nef if $D.C\ge0$ for every irrducible curve $C\subset S$.

$\endgroup$
2
$\begingroup$

It is a general fact, let me try to sketch a proof of it.

If a divisor $D$ is nef then $D^2\geq0$.

If $D^2>0$ just apply the Riemann-Roch inequality to $mD$ and get $h^0(mD)>0$ for $m$ big enough.

If $D^2=0$ then $D$, being nef, is numerically trivial. Then it might be that $D$ is itself equivalent to zero (in which case you are done) or it is not. So you only are left with the latter case to deal with. It is a non-trivial fact that any numerically trivial $D$ is such that $mD\sim0$ for $m$ big enough. (See Remark 1.1.20 in Lazarsfeld' book on Positivity). Surely for the special case $D=K$ it has to be simpler but I can't think of a straightforward argument right now.

Edit: as Schemer suggests in the comments the cited Remark of Lazarsfeld is false as stated. The precise content is: a line bundle $L$ is numerically trivial if and only if there is some $m\neq0$ such that $L^{\otimes m}\in\mathrm{Pic}^0(X)$. Therefore the end of the above proof does not apply to general $D$. However when $D=K$ it follows e.g. by the Enriques-Castelnuovo classification of smooth projective surfaces. Indeed, if $K$ is nef then $S$ is not rational. Hence, by a rationality criterion of Enriques you get $h^0(12K)\neq0$. Therefore $12K$ is equivalent to an effective divisor.

$\endgroup$
  • 2
    $\begingroup$ It isn't true that a multiple of a numerically trivial bundle is trivial. (Pic^0 is a torus, with lots of non-torsion points!) I cannot find the remark you refer to in Lazarsfeld; the closest thing I see is Remark 1.1.20. But if you meant that remark, you left out the key word deformation. Nevertheless it is true that if $K$ is nef and $K^2=0$, then $K$ is torsion; the only argument I see is classification, which is unsatsifying. $\endgroup$ – Schemer Oct 27 '15 at 21:32
  • $\begingroup$ @Schemer: oops! You are right. Let me edit. $\endgroup$ – Heitor Fontana Oct 27 '15 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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