Let $D$ be a divisor on a (complex) K3 surface.
Suppose $D^2\geq0$. In general, $D$ is nef if $D\cdot C\geq0$ for all irreducible curves on the surface.
Is it sufficient in our case to check this for smooth rational curves (i.e. the (-2) curves) ?
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4$\begingroup$ What if there are no $(-2)$-curves? A very general polarized K3 surface over $\mathbb{C}$ contains no smooth rational curve. $\endgroup$– Jason StarrNov 12, 2013 at 12:38
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$\begingroup$ Jason, you are of course correct in general, but on a K3 if there are no $(-2)$ curves, then the cone of curves is the same as the (closure of the) ample cone and the same as the positive part of the cone defined by $D^2\geq 0$ so for any $D$ with $D^2\geq 0$ $D$ or $-D$ is nef and so with the additional assumption that $-D$ is not effective or not nef, it follows that it is enough to check on $(-2)$-curves (because there is nothing to check). :) $\endgroup$– Sándor KovácsNov 12, 2013 at 23:55
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$\begingroup$ @user875280: It seems that you have gone on an editing spree recently. Kindly please stop editing posts only to perform minor corrections. Thank you. $\endgroup$– Alex M.Apr 7, 2021 at 16:14
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2 Answers
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First off, you obviously have to assume something about $-D$ not being effective, because otherwise you could take a negative ample class.
The cone of curves of a K3 surface is pretty well described in this paper. And there is a newer version of it that works in positive characteristic as well here.
Here is what you get out of this:
- It is possible that there are no $(-2)$-curves on a K3 surface, but in this case for every divisor with $D^2\geq 0$ either $D$ or $-D$ is both nef and effective.
- If the Picard number is $2$, it is possible that there is only one $(-2)$-curve. In this case there are actually (effective) curves with non-negative and even positive self-intersection which are not nef. However, they are negative on the sole $(-2)$-curve. (I think I will leave this for the reader for now).
- In all other cases the $(-2)$-curves generate a cone which is dense in the cone of curves, so any divisor that is non-negative on the $(-2)$-curves is non-negative on every effective curve.
So, it actually looks like that what you want is true.
answered Nov 12, 2013 at 21:10
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Yes, as long as $D$ is effective, it's enough to check this for $C$ running over all smooth rational curves. See Corollary 1.7 of Chapter 8 of Huybrechts's notes http://www.math.uni-bonn.de/people/huybrech/K3Global.pdf and the discussion around it.
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$\begingroup$ Wait a moment. But if $D$ is effective and $D^2\geq 0$ then $D$ is nef on any surface, isn't it? $\endgroup$– HeitorNov 12, 2013 at 17:15
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1$\begingroup$ Not if the divisor $D$ is reducible - else take $D = C + f$ on a Hirzebruch surface with $C^2 = -2, f^2 = 0, C \cdot f = 1$, and notice $D \cdot C = -1$. If you don't assume $D$ is effective, then Jason's comment shows you that there's no way to distinguish $D$ from its negative. $\endgroup$ Nov 12, 2013 at 17:54