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Let $X$ be a smooth projective surface over $\mathbb{C}$. We know that nef line bundles form a closed convex cone whose interior is the ample cone. My doubt is the other direction, is it possible to have an ample class $C$ such that $C=C_1+C_2$ where neither $C_i$ is nef? In fact I know that $C\in |L|$ where $L$ is very ample.

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I want $C_i$ to be effective. I am interested in the case when $X$ is the Jacobian of a genus 2 curve.

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  • $\begingroup$ You're asking $C_i$ to be effective, I guess? $\endgroup$
    – Mattia Talpo
    Nov 14, 2015 at 5:24
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    $\begingroup$ It is never possible if $X$ has Picard rank $1$ (e.g. on $\mathbb P^2$). If the Picard rank is at least $2$, it is always possible to write $C$ as the sum of two classes that are not even pseudoeffective. If you want the $C_i$ to be effective, this is a more subtle question about the shape of the relevant cones. $\endgroup$
    – user47305
    Nov 14, 2015 at 5:33
  • $\begingroup$ @Mark , yes! I want $C_i$ to be effective. $\endgroup$
    – gradstudent
    Nov 14, 2015 at 6:25
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    $\begingroup$ If your $X$ is the Jacobian of a genus 2 curve, then any effective divisor will be nef. This happens on any abelian variety $X$ (in fact, on any homogeneous variety). See Example 1.4.7 in Lazarsfeld's Positivity book. $\endgroup$
    – pgraf
    Nov 14, 2015 at 18:56

3 Answers 3

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Take an arbitrary surface which has lots of non-nef effective curves. In particular, take one that has two non-nef effective curves on "opposite" sides of the cone of effective curves. Then some linear combination of these two curves will lie in the interior of the cone and will be ample.

In particular, take a K3 surface with at least 2 smooth rational curves on it. This will satisfy the above condition.

For an explicit example take a general quartic surface in $\mathbb P^3$ that contains a (smooth) plane conic. It is relatively easy to see that the intersection of this quartic with the plane containing the given conic is the union of two smooth conics (the original and another one), so you have two $(-2)$-curves whose sum is the hyperplane class of this embedding to $\mathbb P^3$. Voilà.

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On $\mathbb{P}^2$, take a cubic $C $ with a double point $p$. Blow up $p$ and 6 other general points of $C$. The linear system $|L|$ given by cubics through the 7 points is ample, but contains $E+C'$, where $E$ is the exceptional divisor above $p$ and $C'$ the strict transform of $C$. Since $E^2=C'^2=-1$, neither is nef.

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This question already has two answers, but let me add one more example where the Kodaira dimension of $X$ can be arbitrary.

Take $S = C \times D$, where $C$ and $D$ are smooth projective curves. Let $X$ be the blowup of $S$ in an arbitrary point $(p, q) \in S$. Let $E \subset X$ be the exceptional divisor and $F \subset X$ the strict transform of the fibre $\{ p \} \times D \subset S$. Then $E^2 = F^2 = -1$, so $E$ and $F$ are not nef. But $E + F$ is, since $(E + F) \cdot E = (E + F) \cdot F = 0$. Now pick an ample divisor $H$ on $X$. For sufficiently small $0 < \varepsilon \ll 1$, the divisor $E + (F + \varepsilon H)$ will be ample although $F + \varepsilon H$ is not nef.

By varying the genus of $C$ and $D$, you can get examples of arbitrary Kodaira dimension.

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