Timeline for Irreducibility of quotient stacks.
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 5, 2011 at 16:25 | vote | accept | ginevra86 | ||
| Jun 21, 2011 at 10:41 | comment | added | Daniel Bergh | Mike, I assume that you still mean that $G$ acts trivially on $X$, since $[Y/G]$ isn't defined unless $Y$ is $G$-invariant. | |
| Jun 21, 2011 at 10:30 | answer | added | Daniel Bergh | timeline score: 3 | |
| Jun 20, 2011 at 23:24 | comment | added | Mike Skirvin | Supplementing shenghao's comment, you can show that for any quotient stack $X/G$, the open and closed substacks of $X/G$ are in bijection with the open and closed subschemes of $X$ (i.e., they're all of the form $Y/G$ for Y open or closed in $X$). | |
| Jun 20, 2011 at 21:37 | comment | added | shenghao | I guess $[X/G]$ is always irreducible, as long as $X$ is (namely one doesn't even need to assume the $G$ action is trivial). If $[X/G]$ is the union of two proper closed substacks, or equivalently, there exist two nonempty open substacks that do not meet, then their inverse images in $X$ do not meet, too, contradicting the assumption. | |
| Jun 20, 2011 at 17:18 | history | edited | David Carchedi |
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| Jun 20, 2011 at 16:28 | history | asked | ginevra86 | CC BY-SA 3.0 |