Timeline for Kodaira dimension of co-adjoint orbit
Current License: CC BY-SA 3.0
25 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2014 at 13:30 | vote | accept | CommunityBot | ||
| Nov 3, 2014 at 14:21 | comment | added | Allen Knutson | $G/P$ is birationally equivalent to its open Bruhat cell, a vector space. So birational invariants of $G/P$s are boring. | |
| Nov 2, 2014 at 14:13 | comment | added | user21574 | Allen Knutson@ Can you explain more, please, thanks for your nice comment. | |
| Nov 2, 2014 at 10:48 | comment | added | Allen Knutson | By the Bruhat decomposition they are rational varieties, so your motivation is not particularly motivating. | |
| Oct 31, 2014 at 17:01 | history | edited | user21574 | CC BY-SA 3.0 |
deleted 913 characters in body
|
| Oct 30, 2014 at 17:12 | comment | added | user21574 | Peter Dalakov@ Yes, You right, thanks | |
| Oct 30, 2014 at 17:09 | comment | added | Peter Dalakov | @Hassan: Your question is about ${\it compact}$ Lie groups, in which case coadjoint orbits are ${\it not}$ hyperkaehler - consider for example $S^2$, a coadjoint orbit for $su(2)$. There is no need for Robert Bryant to revise his statement. The coadjoint orbits of a complex simple Lie algebra are another thing. | |
| Oct 30, 2014 at 16:12 | comment | added | user21574 | Robert Bryant@, all coadjoint orbits of complex reductive groups admit hyperkähler structures. So, in non-compact reductive Lie Groups, your statement need to revision. This is know fact of Olivier Biquard | |
| Oct 30, 2014 at 15:46 | comment | added | user21574 | See Peter Crooks's answer in mathoverflow.net/questions/156394/is-g-t-a-projective-variety | |
| Oct 30, 2014 at 15:38 | comment | added | user21574 | $\mathcal O_a\cong G/G_a\cong G^{\mathbb C}/P$ and see mathoverflow.net/questions/156394/is-g-t-a-projective-variety | |
| Oct 30, 2014 at 15:36 | answer | added | YangMills | timeline score: 4 | |
| Oct 30, 2014 at 15:35 | comment | added | Ben McKay | Do you mean the projectivized coadjoint orbits? The coadjoint orbits are subvarieties of $\mathfrak{g}^*$, so not compact and so not projective. The orbits in $\mathbb{P}(\mathfrak{g}^*)$ are usually not called coadjoint orbits, I think. | |
| Oct 30, 2014 at 15:18 | history | edited | user21574 |
edited tags
|
|
| Oct 30, 2014 at 14:45 | history | edited | user21574 | CC BY-SA 3.0 |
added 812 characters in body
|
| Oct 30, 2014 at 14:38 | history | edited | user21574 | CC BY-SA 3.0 |
added 812 characters in body
|
| Oct 30, 2014 at 13:26 | history | edited | user21574 | CC BY-SA 3.0 |
added 1 character in body
|
| Oct 29, 2014 at 17:14 | history | edited | user21574 | CC BY-SA 3.0 |
added 45 characters in body
|
| Oct 29, 2014 at 14:19 | history | edited | user21574 |
edited tags
|
|
| Oct 29, 2014 at 14:11 | history | edited | user21574 | CC BY-SA 3.0 |
added 126 characters in body
|
| Oct 29, 2014 at 14:05 | comment | added | user21574 | Thanks , I will edit my question. So, I think we must use of Kirillov's dimension formula and Borel–Weil theorem | |
| Oct 29, 2014 at 13:59 | history | edited | user21574 | CC BY-SA 3.0 |
deleted 9 characters in body
|
| Oct 29, 2014 at 13:56 | comment | added | Robert Bryant | The coadjoint orbits are not hyperKähler varieties. In fact, they are homogeneous and have positive Ricci curvature. | |
| Oct 29, 2014 at 13:19 | comment | added | user21574 | In fact, coadjoint orbit is hyper-kahler variety, so normally the canonical bundle is trivial and Kodaira dimension must be 0. Am I wrong? | |
| Oct 29, 2014 at 13:15 | history | edited | user21574 | CC BY-SA 3.0 |
edited body
|
| Oct 29, 2014 at 13:08 | history | asked | user21574 | CC BY-SA 3.0 |