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This is one of these questions where it's tempting to just leave it at the title, but let me try to define the objects in question.

A conical symplectic resolution is a projective resolution of singularities $X \to Y$ such that

  • $X$ is algebraically symplectic,
  • $Y$ is affine, and
  • there are compatible $\mathbb{G}_m$-actions on the two varieties which make $Y$ into a cone and act on the symplectic form with positive weight $n$.

Examples include the Springer resolution, a minimal resolution of a rational double point, the Hilbert scheme of points in that space (via the Hilbert-Chow resolution), a hypertoric variety or a Nakajima quiver variety.

All of these spaces have something in common: they are (relative) Mori dream spaces. (For a definition of "relative Mori dream space," see this paper).

Thus, I am inclined to wonder:

Are all conical symplectic resolutions relative Mori dream spaces? Or am I just not original enough to come up with counter examples?

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    $\begingroup$ Hi Ben, maybe a silly question: in my understanding (and the link you give) a Mori dream space is projective. But your examples are not all projective. What am I missing? $\endgroup$
    – user5117
    Commented Nov 7, 2011 at 20:48
  • $\begingroup$ Artie- I probably should have said "relative Mori dream space." Rather than a projective variety, I'm thinking about the projective map $X\to Y$. $\endgroup$
    – Ben Webster ♦
    Commented Nov 7, 2011 at 21:01
  • $\begingroup$ I suspected the answer was something along those lines. Thanks for the clarification. $\endgroup$
    – user5117
    Commented Nov 7, 2011 at 21:49
  • $\begingroup$ In case anyone still cares: I came across this question again because I was reading the paper of Andreatta--Wisniewski referred to by Ben. There they have a theorem (3.2) that all 4-dimensional symplectic contractions X -> Y are relative MDS, but I must admit I find their proof hard to follow. It uses the fact that X is symplectic, but not that much, it seems to me. So it seems as though one might be able to get somewhere with the original question using recent progress in MMP. More specifically, if one can find an effective Q-Cartier divisor on X which is anti-ample over Y, then... $\endgroup$
    – user5117
    Commented Feb 7, 2013 at 23:22
  • $\begingroup$ X/Y is a relative MDS, by BCHM. This is possible if Y is Q-factorial, as explained by Sandor here: mathoverflow.net/questions/86123/…. I don't know how much that helps with the original question, but I thought I'd mention it. $\endgroup$
    – user5117
    Commented Feb 7, 2013 at 23:27

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