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Let $f:Y\to X$ be a proper birational map of normal varieties over an algebraically closed field which is an isomorphism over the regular locus.

Q1: Is it the case that the pullback $f^*\operatorname{Pic}(X)\to\operatorname{Pic}(Y)$ is always injective?
(Update: this was answered in the affirmative by Jason Starr.)

Q2: Suppose that there exists a finite set $\{C_i \mid i\in I\}$ of projective curves in $Y$ such that a line bundle is ample (respectively nef) on $Y$ if and only if its restriction to each $C_i$ has positive (respectively non-negative) degree. Is it the case that a line bundle on $X$ (interpreted via pullback as a line bundle on $Y$) is ample (respectively nef) if and only if its restriction to each $C_i$ that isn't contracted by $f$ has positive (respectively non-negative) degree?

For example, suppose that $Y$ is a crepant resolution of a Kleinian singularity, with exceptional divisor $E\subset Y$ consisting of a finite union of projective lines. Suppose that $X$ is a partial resolution, obtained from $Y$ by contracting some (but not all) of the components of $E$. Then the Picard group of $X$ should be isomorphic to the subgroup of the Picard group of $Y$ consisting of line bundles whose restrictions to the contracted curves have degree zero, and its ample cone should be those line bundles whose restrictions to the un-contracted curves have positive degree.

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    $\begingroup$ If $X$ is normal, then in particular it is S2. So if you have a trivialization of the pullback to $Y$ of an invertible sheaf, and thus you have a trivialization on an open subset of $X$ whose complement has codimension $2$, then that trivialization extends to all of $X$. $\endgroup$
    – Jason Starr
    Commented Oct 26, 2015 at 19:04
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    $\begingroup$ As you observe, if $X$ is quasi-projective yet $Y$ is not, then it is impossible to say much about the ample cones. There is an example of such a proper, birational morphism in the appendices to Hartshorne's "Algebraic Geometry". Of course you do know that the image under pullback of the nef cone of $X$ is contained in the nef cone of $Y$. But without Kleiman's criterion (which requires a projectivity hypothesis), this does not imply anything about ample cones. $\endgroup$
    – Jason Starr
    Commented Oct 26, 2015 at 19:12
  • $\begingroup$ Jason: you're right, I shouldn't expect to be able to say anything at all about ample cones when $f$ is not projective. I've edited Q2 to assume that $f$ is projective, but I definitely do not want to assume that $X$ and $Y$ are themselves quasiprojective. Is there a relative version of Kleiman's criterion? $\endgroup$
    – Nicholas Proudfoot
    Commented Oct 26, 2015 at 19:49

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There is a relative version of the ample cone, discussed a bit in Section 1.6.2 of the draft http://www.math.utah.edu/~defernex/book.pdf . There is indeed a relative version of Kleiman's criterion telling you that the interior of the relative ample cone is the relative nef cone.

But just knowing the relative ample cone of $f$ and the ample cone of $X$ doesn't really tell you very much about the ample cone of $Y$, unless you are only looking for extremely coarse information. For example, take $X$ the blow-up of $\mathbb P^2$ at eight general points. The nef cone is a rational polyhedral cone, since this is a del Pezzo surface. Now let $Y$ be the blow-up of $X$ at a single general point. The relative nef cone is a cone in a 1-dimensional vector space, and just a single ray. But the nef cone of $Y$ is a wild thing with infinitely many extremal rays.

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  • $\begingroup$ How can the interior of the relative ample cone equal the relative nef cone in a situation where the relative ample cone is empty, yet the relative nef cone is not empty? I think there must be a projectivity hypothesis, just as in Kleiman's original statement of his theorem. $\endgroup$
    – Jason Starr
    Commented Oct 26, 2015 at 20:44
  • $\begingroup$ Sorry, yes, I mean all this only in the case when everything in sight is projective. Without that there are bigger issues, as you observe. $\endgroup$
    – user47305
    Commented Oct 26, 2015 at 21:34
  • $\begingroup$ In my examples, I understand the ample cone of $Y$ already, and I'm trying to compute the ample cone of $X$. I would like to say something along the lines of the statement that ample line bundles on $X$ are those that pull back to line bundles on $Y$ that have degree zero on all projective curves that get contracted and positive degree on all projective curves that don't get contracted. But this doesn't quite sound right, since Nagata showed that having positive degree on every projective curve is not sufficient for being ample. So I'm looking for the correct modification of this statement. $\endgroup$
    – Nicholas Proudfoot
    Commented Oct 26, 2015 at 22:37
  • $\begingroup$ Here's the formulation that I'm looking for: suppose that I have a finite set of projective curves on $Y$ with the property that a line bundle on $Y$ is ample if and only if it is positive on each of these curves. I would like to say that a line bundle on $X$ (interpreted via pullback as a line bundle on $Y$) is ample if and only if it is positive on every one of these curves that does not get contracted. $\endgroup$
    – Nicholas Proudfoot
    Commented Oct 26, 2015 at 23:28

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