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Google searches for "local ample cone" and "local Kähler cone" yield no results, but maybe there is a different term.

Let $\pi : \hat X \to X$ be a resolution of an isolated singularity on the (complex) projective variety $X$, with exceptional set $E$, which may have components of various dimensions. Assume that we can find a set of generators for the divisor class group Cl$(\hat X)$ consisting of two types of divisors:

  1. linear combinations of components of $E$, or
  2. divisors which do not intersect any component of $E$.

It seems to me that an ample divisor might be built as follows. Find a divisor $D'$ of type (1) which has positive intersection numbers with all sub-varieties of $E$ (this may not exist). Choose an ample divisor $D$ on the singular space $X$. This pulls back to a divisor $\pi^*D$ which of type (2). Then $D' + n\pi^*D$ should be ample for large enough $n$.

The space of possible $D'$ is what I want to call the "local ample cone", as it consists of divisors in the exceptional set which are 'locally ample'. In the language of Kähler geometry, we are staying close to the wall of the Kähler cone of $\hat X$ at which $E$ shrinks to zero volume, so from this point of view we are considering the local structure of the Kähler cone near this wall. I suppose I have two questions for the experts out there:

Is the above construction sensible?

Does this notion already exist in the literature in some form?

In case it matters, I am particulary interested in the case of Calabi-Yau threefolds (including the condition $h^{0,1} = h^{0,2} = 0$).

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    $\begingroup$ A couple of comments: 1. As Karl says, the standard version of the notion you want is called the relatively ample cone. It is well-defined for any projective morphism, I guess, not just a resolution. On the other hand, it's different from your notion, because it does not consist only of divisors supported on E. 2. You probably know this, but a divisor D' of type (1) which has positive intersection with all subvarieties of E cannot be effective; indeed, it must be the negative of an effective divisor. 3. Another reference for this kind of thing is in Chapter 2 of Koll'ar--Mori. $\endgroup$
    – user5117
    Jul 11, 2012 at 14:19
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    $\begingroup$ Oh, and finally, you said "I think this is equivalent to saying that there are no divisors on X which intersect the singularity." This can't be what you meant to write. $\endgroup$
    – user5117
    Jul 11, 2012 at 14:20
  • $\begingroup$ No, I guess not. I think the two conditions I wrote down are what I need, and this must imply some condition on $X$, but I don't know what. $\endgroup$ Jul 11, 2012 at 14:55
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    $\begingroup$ I think the two conditions you write down are pretty close to the notion of factoriality of the singularity. I'll explain more in my answer. $\endgroup$ Jul 11, 2012 at 16:10

1 Answer 1

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I think this should be called the relatively ample cone and its closure the relatively nef cone. Yes, this already exists and is sensible.

This dissertation might be somewhat useful (although things are done for smooth $X$ instead of singular $X$).

UMich dissertation

EDIT: With regards to your two conditions (and the comments above), I think this condition is basically factoriality for the singularity. Suppose $x \in X$ is an isolated singularity as you describe and suppose that we are working with only one singular point.

Claim: Suppose that $X$ is factorial, this means that every divisor through $x$ on $X$ is going to be linearly equivalent to a divisor not passing through $x$.

Proof of claim: Suppose that $D > 0$ is an effective divisor passing through $x \in X$ (the effectivity assumption is harmless since we can write $D = D'-D''$ for effective $D'$ and $D''$). Choose a very ample and sufficiently ample Cartier divisor $A$ such that $\Gamma(X, O_X(A - D))$ is globally generated. Choose a global section $s$ generating the stalk $O_{X,x}(A-D)$ (this is possible since $D$ is Cartier at $x$, since $O_{X,x}$ is a UFD). Consider the inclusion $$\Gamma(X, O_X(A-D)) \subseteq \Gamma(X, O_X(A))$$ and so view $s$ as a global section of $O_X(A)$. It follows that $s$ determines a divisor $S \sim A$. Furthermore, $S$ and $D$ coincide in a neighborhood of $x \in X$ by construction. Further choose $T \sim A$ to be any divisor not passing through $x$ (this is possible since $A$ is very ample). Now then $D - S + T \sim D$, but $D - S + T$ is zero near $x \in X$. This proves the claim.

It follows from the claim that there exists a set of generating divisors of the type you want.

On the other hand, if you have divisors on $\hat{X}$ that satisfies those two conditions, then the images (pushforwards) of those divisors also generate the divisor class group of $X$. In particular, the divisor class group of $X$ is generated by divisors not passing through $x$. In particular, it is generated by Cartier divisors (since $x \in X$ was the only singular point). This implies that $X$ is factorial.

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  • $\begingroup$ Thank-you for your answer, Karl. It has at least helped me to clarify my problem, but I'm not quite there yet! "...every divisor through $x$ on $X$ is going to be linearly equivalent to a divisor not passing through $x$" is exactly what I meant. I'm afraid I can't see why that is equivalent to factoriality though (sorry if I'm being dense; I'm a little out of my depth here). If $x$ is the only singular point, doesn't factoriality depend only on $\mathcal{O}_{X,x}$? I think I have examples of the same singularity in different threefolds, once with the conditions satisfied, and once not. $\endgroup$ Jul 12, 2012 at 9:41
  • $\begingroup$ Dear Rhys Davies, I added some explanation. Let me know what you think. $\endgroup$ Jul 12, 2012 at 16:59
  • $\begingroup$ Thank-you again, that's very clear now. I have some thinking to do, because of the conflicting examples I thought I had... $\endgroup$ Jul 13, 2012 at 9:49

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