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Dec 13, 2010 at 19:16 vote accept Vivek Shende
Oct 6, 2010 at 22:19 answer added Dmitri Panov timeline score: 2
Oct 6, 2010 at 21:22 comment added damiano 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!
Oct 6, 2010 at 19:03 comment added Vivek Shende 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.
Oct 6, 2010 at 16:06 comment added damiano 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.
Oct 6, 2010 at 14:48 history edited Vivek Shende CC BY-SA 2.5
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Oct 6, 2010 at 14:20 comment added Vivek Shende I don't know any examples of anything. Does it just never happen? Can you prove it just in terms of the lattice?
Oct 6, 2010 at 13:18 comment added damiano 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"?
Oct 6, 2010 at 12:47 history asked Vivek Shende CC BY-SA 2.5