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Let $\overline{NE}(X)$ be the closure of the cone generated by the numerical classes of effective curves and $\overline{\mathrm{Mov}}(X)$ the closure of the cone of moving curves.

(Q) Is there an example of a smooth projective variety $X$ such that

  • $\overline{NE}(X)$ is (finite) polyhedral, but
  • $\overline{\mathrm{Mov}}(X)$ is not?

Here are some trivial observation:


If $X$ is a surface then the two cones are dual to each other and hence they are either both polyhedral or not.


If $X$ is a Fano variety then

  • $\overline{NE}(X)$ is polyhedral by the Cone Theorem, and
  • $\overline{\mathrm{Mov}}(X)$ is polyhedral by a result of Barkowski if $\dim \leq4$ (see here) and by [BCHM] in general.
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By the way, Artie Prendergast-Smith has an example here of a rational threefold whose nef cone is rational polyhedral but the effective movable cone is not. Would this be an answer to your question? – J.C. Ottem Oct 25 '11 at 8:34
@JC: yes, this would be an answer. Why don'y you put it as an answer? (In the mean time I realized that this is unfortunately not enough for my initial idea for that other question I started the question with) – Sándor Kovács Oct 25 '11 at 11:34
p.s: I didn't know about this example, but I was expecting Artie to answer the question. :) – Sándor Kovács Oct 25 '11 at 11:35
I've added his example in the answer below. I'd also be interested in hearing Artie's thoughts on this, so I hope he joins the discussion. – J.C. Ottem Oct 25 '11 at 12:12
Another comment: isn't Barkowski's result true in all dimensions because of BDPP (Mov is dual to the pseudoeffective cone) and BCHM (Fanos are Mori dream spaces)? – user5117 Oct 25 '11 at 14:27
up vote 3 down vote accepted

As J.C. indicates in the comments, an example for Q1 can be gotten from the variety considered in this paper. This isn't spelled out in the paper, so let me explain it here.

First let's change the question into its dual form. The cone of curves is dual to the nef cone, and Boucksom--Demailly--Peternell--Paun showed that the cone of moving curves is dual to the cone of pseudoeffective divisors. So we want to find an example of $X$ such that $Nef(X)$ is rational polyhedral but the cone of pseudoeffective divisors $PsEff(X)$ isn't.

I claim the variety $X$ in the linked paper is such an example. Here, $X$ is constructed by blowing up $\mathbf{P}^3$ at the base locus of a general net of quadrics.

The variety $X$ is then elliptically fibred over $\mathbf{P^2}$, with the generic fibre having an infinite abelian group (more precisely, rank 8) of sections. Call this group $MW(X)$. Translating by differences of sections gives an action of $MW(X)$ on $X$ by so-called pseudo-automorphisms (meaning birational automorphisms that are isomorphisms in codimension 1). This group action preserves effective divisors, and hence the cone $PsEff(X)$. One can calculate the action fairly explicitly, and in particular one sees the orbit of a divisor $E_i$ (the exceptional divisor of one of the blowups) in N^1(X) is infinite. Now it is easy to see that each $E_i$ spans an extremal ray of $PsEff(X)$, and hence so does any $MW$-translate of $E_i$; since there are infinitely many of these, $PsEff(X)$ has infinitely many extremal rays.

On the other hand, it's not hard to show that $Nef(X)$ is rational polyhedral. This is done more or less by brute force: enumerate some curve classes, find the dual cone to the convex hull of those classes (which is then an upper bound for $Nef(X)$), and check that it's spanned by nef classes. Details are in the paper.

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Thanks Artie, I'm glad your example worked after all! – J.C. Ottem Oct 25 '11 at 22:29
Me too; thanks for the publicity! – user5117 Oct 26 '11 at 0:11
Artie, nice example! – Sándor Kovács Oct 26 '11 at 21:29
JC, you didn't have to delete your answer! – Sándor Kovács Oct 26 '11 at 21:29

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