Timeline for Do fixed point sets in equivariant crepant resolutions have the same cohomology? How about for the specific case of Nakajima quiver varieties?
Current License: CC BY-SA 4.0
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| Apr 21, 2023 at 16:02 | history | edited | Glorfindel | CC BY-SA 4.0 |
broken link fixed, cf. https://meta.mathoverflow.net/q/5301/70594
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| Jun 26, 2012 at 11:21 | comment | added | Hiraku Nakajima | I understand the point of the argument. If $r$ is divisible by $a+b$, the action of $a+b$-th roots of $t$ is well-defined. So the symplectic form is of weight $1$, and the above argument works. | |
| Jun 26, 2012 at 9:21 | comment | added | Hiraku Nakajima | It seems that arxiv.org/abs/1206.5640 gives a positive answer when $a+b=r$. | |
| Jun 15, 2012 at 21:17 | history | answered | Hiraku Nakajima | CC BY-SA 3.0 |