Timeline for (Co)tangent complexes of quotient stacks
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 28, 2016 at 22:00 | comment | added | DamienC | @BrianFitzpatrick: if $X$ is smooth affine and $G$ is finite then yes, $[X/G]$ has no non-trivial deformations (we are working in characteristic zero). | |
| Mar 16, 2016 at 7:15 | comment | added | Brian Fitzpatrick | @QiaochuYuan So this would imply that $X$ smooth affine implies that $[X/G]$ has no non-trivial deformations? | |
| Feb 11, 2016 at 4:54 | comment | added | Qiaochu Yuan | @Brian: it's the tangent sheaf of $X$ regarded as a $G$-equivariant $\mathcal{O}_X$-module, as DamienC says. | |
| Feb 11, 2016 at 4:05 | comment | added | Brian Fitzpatrick | @QiaochuYuan So $G$ finite implies that the tangent complex of $[X/G]$ is the same as the tangent sheaf of $X$? | |
| Feb 11, 2016 at 4:03 | comment | added | Qiaochu Yuan | @Brian: no, if $G$ is finite then $\mathfrak{g}$ vanishes. In any case there isn't a natural map from $\mathcal{O}_X$ to $\mathcal{T}_X$. | |
| Jan 26, 2016 at 21:29 | comment | added | Brian Fitzpatrick | Is $\frak g$ the Lie algebra of $G$? So $G$ finite would mean that the tangent complex of $[X/G]$ is $\cal O_X\to T_X$, correct? | |
| Feb 6, 2015 at 19:31 | vote | accept | Sinan Yalin | ||
| Feb 6, 2015 at 13:19 | history | answered | DamienC | CC BY-SA 3.0 |