Timeline for G-invariant vector bundle over schemes?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 29, 2014 at 23:47 | comment | added | S. Carnahan♦ | Any choice of $\mathbb{G}_m$-torsor gives you the sort of object you want. See the last section in Vistoli's notes on descent (on the arXiv). | |
| Sep 29, 2014 at 7:45 | history | edited | KylinChen | CC BY-SA 3.0 |
added 81 characters in body
|
| Sep 29, 2014 at 7:44 | comment | added | KylinChen | yes i mean G-equivariant...sorry | |
| Sep 29, 2014 at 7:37 | comment | added | abx | You probably mean $G$-equivariant vector bundles. Then your $Y$ will just be the quotient stack $[X/G]$. | |
| Sep 29, 2014 at 7:34 | comment | added | Piotr Achinger | "$G$-invariant" per se doesn't make much sense: it would mean $g^* E = E$ for every $g\in G$, but there is a choice involved! This leads to the notion of linearization of a vector bundle $E$: it's an isomorphism $f:\pi^* E \to \mu^* E$, where $\pi, \mu:G\times X\to X$ are the projection resp. the action, satisfying a certain "cocycle condition". The pair $(E, f)$ is called a $G$-equivariant bundle. This $f$ may be non-unique, as the example of $E=\mathcal{O}_X$ on $X=\mathbb{A}^1$ with the usual $\mathbb{G}_m$-action shows (here one has $\mathbb{Z}$ worth of possible choices). | |
| Sep 29, 2014 at 7:30 | comment | added | Piotr Achinger | I think the claim about $A^n$ is false: vector bundles on $P^{n-1}$ correspond to $G_m$-equivariant vector bundles on $A^n \setminus \{0 \}$, and these don't have to extend to vector bundles on $A^n$. The tangent bundle of $P^{n-1}$ is a counterexample. | |
| Sep 29, 2014 at 6:48 | history | asked | KylinChen | CC BY-SA 3.0 |