Timeline for If a subgroup H of a finite group G acts freely on a variety, can the G-Hilbert scheme be computed by iteration?
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| Oct 18, 2016 at 7:11 | comment | added | Bernie | @Jason Starr: Thanks a lot. This answer really enlightens the situtaion. | |
| Oct 17, 2016 at 14:15 | comment | added | Jason Starr | Yes, you do have that isomorphism. You can check this directly, but you can also use the fact that $G-\text{Hilb}(X)$ equals the Hilbert scheme of the Deligne-Mumford stack $[X/G]$ with the "Hilbert polynomial" $p(t)$ equal to a constant polynomial corresponding to a copy of the group ring $\mathbb{Z}[G]$ as an element in $K^0(G)[t]$, cf. Olsson-Starr. For the quotient algebraic space $Y=X/H$ and for $\Gamma=G/H$ with its induced action on $Y$, the Deligne-Mumford stack $[X/G]$ is naturally isomorphic to $[Y/\Gamma]$. Thus, the Hilbert schemes of these stacks are naturally isomorphic. | |
| Oct 17, 2016 at 14:02 | history | asked | Bernie | CC BY-SA 3.0 |