Timeline for Does the semi-stable set determine the linearization of a GIT quotient?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 24, 2011 at 1:35 | comment | added | Allen Knutson | Let $X = ({\mathbb P}^1)^2$, $Y = pt$, $G = T^2$. Let ${\mathcal O}(a) \boxtimes {\mathcal O}(b)$ carry the natural action. Then for all $a,b>0$, the stable locus is the open $T^2$-orbit. | |
| Sep 21, 2011 at 12:55 | vote | accept | IMeasy | ||
| Sep 19, 2011 at 0:22 | answer | added | Arend Bayer | timeline score: 4 | |
| Sep 18, 2011 at 17:49 | comment | added | IMeasy | I think I have found an example that shows that the answer is YES. Two differente linearizations may give the same (semi-)stable locuses. I would be glad if someone could prove me wrong, though! | |
| Sep 18, 2011 at 16:46 | history | asked | IMeasy | CC BY-SA 3.0 |