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Let $X$ be a Calabi-yau 3-fold, that is, $X$ is a smooth projective 3-fold such that $K_X$ is trivial and $h^1(X, \mathcal{O}_X)=0$.

Question Is it easy to find $X$ whose nef cone is not "rational", that is, the nef cone does not coincide with the convex hull of rational points on the nef boundary?

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    $\begingroup$ It might help to clarify a bit what you mean by "a ray which is not spanned by a Q-divisor class". Even if the nef cone is rational polyhedral there will be many such rays in general, since any open cone of dimension at least 2 contains such a ray. I think what you are really asking for is a ray which lies outside the convex hull of the rational rays. $\endgroup$ – user5117 May 7 '12 at 21:34
  • $\begingroup$ Thanks for your remark. I edited my question. I hope it makes sense. $\endgroup$ – tarosano May 7 '12 at 22:25
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I think the following construction could work: Let $X$ be a Calabi-Yau threefold without any rational curves (such threefolds do exist). Then $Nef(X)=\overline{Eff}(X)$, i.e., the nef cone equals the pseudoeffecitve cone (by the log cone theorem) and the nef boundary is given by the null cone, that is, the set of divisor classes $D$ such that $D^3=0$. Note that this null cone is given by the zero locus of a degree 3 polynomial in the Neron Severi group, and the non-rational points of this will give you the example.

I should mention that the nef cone of a Calabi-Yau threefold is a very interesting object even though it is often non-rational polyhedral. Indeed, the Kawamata-Morrison cone conjecture states that the nef cone and movable cone should instead have a rational polyhedral fundamental domain for the action of $im(Aut(X)\to GL(N^1(X)))$. In some sense, this is the next best thing compared to the Fano case ($K<0$) where everything is rational polyhedral. So in particular if the automorphism group is infinite, then one would expect a non-rational polyhedral nef cone.

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  • $\begingroup$ Hi John! I'm not sure I fully follow your answer, e.g. the reference to Grassi--Morrison's paper. If I skimmed correctly, they show that there is a rational polyhedral fundamental domain for the action of automorphisms on the (full) nef cone. So the whole cone is contained in the convex hull of the rational rays. If i understand the OP's question correctly, he's looking for an example where that isn't the case (so, stronger than just not rational polyhedral). $\endgroup$ – user5117 May 7 '12 at 21:25
  • $\begingroup$ Oh yes, I misread the question, as if he wanted a 'non-(rational polyhedral)' example. I think the other construction works though? $\endgroup$ – J.C. Ottem May 7 '12 at 21:37
  • $\begingroup$ I think so, as long as the intersection form is not too reducible (for instance a product of linear forms with rational coefficients). $\endgroup$ – user5117 May 7 '12 at 21:38
  • $\begingroup$ That is not so clear I must admit. All examples I know of such threefolds are quotients of abelian threefolds by finite groups, and in that case I would expect the nef cone to be rational. I'll try to dig out some references though. $\endgroup$ – J.C. Ottem May 7 '12 at 22:30
  • $\begingroup$ Thanks for the comments. Actually, as far as I know, I only know examples of $X$ whose nef cone is generated by rational extremal rays. I think that typical examples of CY3 without rational curve are etale quotient of abelian 3-folds. Oguiso-Sakurai classified such 3-folds and they see that their nef cones are rational polyhedral. By the way, John, why the null cone give the nef boundary? $\endgroup$ – tarosano May 7 '12 at 22:37

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