Let $D$ be a nef divisor (moreover suppose it is effective if you prefer) on a normal projective variety of dimension $n$. Let $k\in[1,n-1]$. If $D^k\cdot V=0$ for generic subvarieties $V\subseteq X$ of dimension $k$, can I conclude that $D^k\cdot V=0$ for all subvarieties $V$ of dimension $k$? In other words can I conclude that the numerical dimension of $D$ is less than $k$?
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3$\begingroup$ Perhaps you can write the class of any subvariety $V$ as a rational linear combination of the generic subvarieties? $\endgroup$– J.C. OttemCommented Sep 26, 2012 at 10:31
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1$\begingroup$ A couple of comments: First, the first parenthetical remark seems a little strange to me: If D is effective, and if A is a very ample divisor say, then we must have D^k.A^{n-k} >0 by Nakai--Moishezon. So if "generic" includes "selfintersection of very ample divisors", the hypothesis can't hold for D effective. Second, I think it might not be possible to answer the question without having a precise definition of "generic" for subvarieties of a given dimension. $\endgroup$– user5117Commented Sep 26, 2012 at 13:52
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2$\begingroup$ @ Artie Prendergast-Smith: About your first remark: The point is that D^k might be numerically 0 (that is in fact what I want to prove). About your second remark: you are right, I wasn't clear. I mean that D^k \cdot V=0 for all irreducible subvarieties V of dimension k not contained in a proper closed subset of X. Maybe to prove this one can just using an ample. In fact If D^k is not numerically 0, then, as you say, D^k.A^{n-k} >0, and we find a contradiction. Do you agree? $\endgroup$– Gianni BelloCommented Sep 26, 2012 at 14:38
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$\begingroup$ Gianni: oops, you're right about the first point; I typed too hastily. About your last assertion: I don't think that works, unfortunately, because D^k might not be effective even if D is. (Think about a (-1) curve on a surface.) $\endgroup$– user5117Commented Sep 26, 2012 at 15:18
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$\begingroup$ Right. But maybe one can use that D is nef. For example this should work for k=n-1, as one has that D^{n-1} is in the closure of NE(X) because it is in the dual of the nef cone (by Lazarsfeld 1.4.16), so that D^{n-1}.A>0 (Lazarsfeld 1.4.29). $\endgroup$– Gianni BelloCommented Sep 26, 2012 at 15:59
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