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?

  • 1
    $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

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.