Timeline for Does every reductive group scheme admit a maximal torus?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 6, 2014 at 13:28 | comment | added | Kestutis Cesnavicius | One can deduce that $T$ is split in this setting for, e.g., $S = \mathbb{A}^3_{\mathbb{C}} - \{(0, 0, 0)\}$ and $E$ a nontrivial vector bundle on $S$ (for the existence of which, see Sasha's answer in mathoverflow.net/questions/35788/…). I am not sure though how Zariski local decomposition into line bundles is of help for the desired counterexample. | |
| May 6, 2014 at 7:21 | comment | added | Daniel Loughran | @Laurent: Yes I see your point. Are you able to fix it? | |
| May 5, 2014 at 11:20 | vote | accept | Daniel Loughran | ||
| May 6, 2014 at 7:24 | |||||
| May 5, 2014 at 9:50 | comment | added | Laurent Moret-Bailly | @Marty: I don't get it . This only works if $T$ is split. | |
| May 5, 2014 at 8:20 | vote | accept | Daniel Loughran | ||
| May 5, 2014 at 11:20 | |||||
| May 2, 2014 at 23:44 | comment | added | Marty | A maximal torus T in GL(E) over S (with E a vector bundle) gives a decomposition of the vector bundle into line bundles. You can see this, even working with S a smooth variety over the complex numbers, since the eigenspaces for T give a local decomposition of E into 1-dim spaces. | |
| May 2, 2014 at 16:11 | comment | added | Daniel Loughran | Are you able to prove that this is the case? | |
| May 2, 2014 at 15:58 | history | answered | Victor Petrov | CC BY-SA 3.0 |