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An Enriques surface (in characteristic zero) is an algebraic surface which is the quotient of a K3 surface by a fixed-point-free involution. Such a surface has a rank 10 lattice of divisors.

(1) What are the ample and effective cones of an Enriques surface?

In particular,

(2) Is it the case that there are ample divisors with arbitrarily large numbers of global sections which do not split nontrivially as the sum of two effective divisors?

Edit: As per Damiano's comment,

(3) Is there any surface for which the answer to (2) is yes?

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    $\begingroup$ Do you know an example of a surface and ample divisors on it for which the answer to the second question you ask is "yes"? $\endgroup$
    – damiano
    Commented Oct 6, 2010 at 13:18
  • $\begingroup$ I don't know any examples of anything. Does it just never happen? Can you prove it just in terms of the lattice? $\endgroup$
    – Vivek Shende
    Commented Oct 6, 2010 at 14:20
  • $\begingroup$ Vivek, I think that the answer to (3) is no. First, pass to a subseq to reduce to the case in which the direction of the ample divisors $\\{A_n\\}$ converges (in the real cone of curves); call $A$ any divisor with this limiting direction. Second, use the fact that the ample cone is open to choose an effective curve $C$ such that $A-tC$ is ample for small enough $t>0$. For large enough $n$ (roughly $n \approx 1/t$), the divisor $A_n-C$ is ample. For even larger $n$, it will be also effective. I will not fill in the details, but I think that they are not difficult. $\endgroup$
    – damiano
    Commented Oct 6, 2010 at 16:06
  • $\begingroup$ Damiano, I don't understand why the sequence you write should converge to something ample. Indeed, the picture I had in my head of such a sequence was one of ample divisors getting ever closer to the boundary of the ample cone. $\endgroup$
    – Vivek Shende
    Commented Oct 6, 2010 at 19:03
  • $\begingroup$ Vivek, I agree with you: the divisor $A$ is nef, but not nec ample. It is nevertheless true that the ample cone is open (and non-empty for projective vars!) and that you can always "perturb" in some direction an element in the closure of the ample cone so that the resulting perturbation will still be ample. This is why I required $t$ to be small and positive, rather than simply small in absolute value. Hope this clarifies your doubts! $\endgroup$
    – damiano
    Commented Oct 6, 2010 at 21:22

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The answer to question 2 is negative thanks to Kamawata-Morrison conjecture about nef cones of Calabi-Yau varieties that is proven for surfaces (a recent reference is here http://arxiv.org/abs/1008.3825), and to the comment of Damiano below.

Indeed, it is known that on a minimal two-dimesnional surface of Kodaira dimesnion 0 the fundamental domain of the action of automorphism of the surface on the nef cone is rational polyhedral (see the above reference). So it is sufficient to consider only ample classes that belong to one rational polyhedral fundamental domain. Notice that every nef integral class (apart from $K$) on each Enriques surface is effective (see the proof of Damiano). Now notice that since the fundamental domain is rational polyhedral, the semi-group of integer vectors in it is finitely generated, so only finitely many integer vectors can not be presented as a sum of two others. In other words, only finite number of ample divisors in a given fundamental domain are not linearly equivalent to a sum of several divisors. Since by definition all domains are the same, the statement is proven.

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    $\begingroup$ On an Enriques surface, a nef divisor class $N$ is either effective or it is the canonical divisor $K$. Indeed, by Riemann-Roch, $\chi(N)=N^2/2+1>0$, and hence either $N$ or $K-N$ is effective. On the other hand, $N$ is nef and hence pseudo-effective, while $K$ is torsion and hence numerically trivial, so that $K-N$ is the opposite of a pseudo-effective class and it can be effective only if it is numerically trivial. Thus $K-N$ is effective only if it is zero, i.e. only if $N=K$. $\endgroup$
    – damiano
    Commented Oct 7, 2010 at 7:38
  • $\begingroup$ Damiano, huge thanks!! This settles the question $\endgroup$
    – Dmitri Panov
    Commented Oct 7, 2010 at 11:30

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