Timeline for Does $G/H$ (quotient of a real semisimple Lie group by a Cartan subgroup) have a natural symplectic structure?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 16, 2012 at 3:30 | history | edited | Allen Knutson | CC BY-SA 3.0 |
added 6 characters in body
|
| Sep 1, 2012 at 6:41 | comment | added | Zhaoting Wei | Yes it is very natural from the viewpoint of coadjoint orbits. Maybe I should think more carefully before asking this question. Nevertheless, thank you all! | |
| Aug 31, 2012 at 21:51 | comment | added | Eugene Lerman | It may be worth pointing out that for semisimple Lie groups the adjoint and the coadjoint orbits are the "same": the Killing form gives an equivariant identification of the Lie algebra with its dual. | |
| Aug 31, 2012 at 19:39 | comment | added | Victor Protsak | A word of caution: a non-compact semisimple Lie group always has more than one conjugacy class of Cartan subgroups (e.g. maximally split and maximally compact), so "generically" here does not have the usual meaning. | |
| Aug 31, 2012 at 15:14 | vote | accept | Zhaoting Wei | ||
| Aug 31, 2012 at 14:50 | comment | added | Robert Bryant | Yes, Jonathan is absolutely correct. The essential point is that, when the rank of $G$ is $r$ there will be an $r$-parameter family of $\xi\in\frak{g}^\ast$ that have the same stabilizer subgroup (which is a Cartan subgroup) under the coadjoint action. Thus, you will have an $r$-parameter family of 'natural' symplectic structures on $G/T$. | |
| Aug 31, 2012 at 13:22 | history | answered | Jonathan | CC BY-SA 3.0 |