Timeline for Riemannian metric on a flag variety
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 28, 2012 at 18:25 | answer | added | 314159. | timeline score: 0 | |
| Jan 24, 2012 at 10:52 | answer | added | Deane Yang | timeline score: 4 | |
| Jan 24, 2012 at 10:32 | answer | added | diverietti | timeline score: 9 | |
| Jan 23, 2012 at 22:47 | vote | accept | Ryan Reich | ||
| Jan 23, 2012 at 22:47 | history | edited | Ryan Reich | CC BY-SA 3.0 |
update
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| Jan 23, 2012 at 9:53 | comment | added | Deane Yang | It's obvious after you work out the details. In particular, $SU(n)$ acts transitively on each of these spaces, and the natural Riemannian metric is the unique one (up to a constant scale factor) that is invariant under this action. So you just have to check that the Fubini-Study metric is indeed invariant under the group action. | |
| Jan 23, 2012 at 3:11 | comment | added | Will Sawin | Is it obvious that this restricts to the correct metrick on $G(1,n)$? | |
| Jan 22, 2012 at 21:56 | answer | added | Will Sawin | timeline score: 4 | |
| Jan 22, 2012 at 21:51 | comment | added | Deane Yang | The answer is "yes". Just view the Grassmannian or flag manifold as a quotient of $SU(n)$ by the appropriate subgroup. The bi-invariant metric on $SU(n)$ induces a natural Riemannian metric on the flag manifold. For me it's all easiest to work out using moving frames, as presented, say, in Griffiths's paper (Duke Math. J. 41 (1974), 775–814). | |
| Jan 22, 2012 at 21:07 | history | asked | Ryan Reich | CC BY-SA 3.0 |