Timeline for The group of isometries of Shahshahani metric
Current License: CC BY-SA 4.0
10 events
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| Jan 28, 2023 at 13:01 | history | edited | YCor | CC BY-SA 4.0 |
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| Jan 28, 2023 at 11:17 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Jun 15, 2018 at 8:39 | vote | accept | Ali Taghavi | ||
| Jun 14, 2018 at 10:24 | comment | added | Robert Bryant | Actually, what I said above in the case $n=2$ is not correct because, in the case $n=2$, there are more Killing fields. (See the note in my answer below.) However, when $n>2$ there are no more Killing fields than those generated by the obvious $\mathrm{SO}(n)$-action. | |
| Jun 13, 2018 at 21:56 | answer | added | Robert Bryant | timeline score: 6 | |
| Jun 13, 2018 at 20:50 | comment | added | Ali Taghavi | @RobertBryant I can not understand why Killing vector fields is isometric to the Lie algebra of $SO(n)$. For $n=2$, what is a precise Killing vector field for that metric? This may help me to find all Killing vector fields. | |
| Jun 13, 2018 at 19:14 | comment | added | Ali Taghavi | @RobertBryant Yes I was asking for global isometries however your answer about Lie algebra of Killing vector field is very helpful. I try to understand the details of both part. Thanks for this very helpfull comment. | |
| Jun 13, 2018 at 13:53 | comment | added | Robert Bryant | Are you asking about global isometries rather than infinitesimal isometries? In the former case, it's just the symmetric group on $n$-letters acting as permutations of the $x_i$ (when $n>1$). The Lie algebra of Killing vector fields, though, has dimension $\tfrac12n(n{-}1)$ (when $n>1$) and is isomorphic to the Lie algebra of $\mathrm{SO}(n)$. | |
| Jun 13, 2018 at 13:40 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Jun 13, 2018 at 13:24 | history | asked | Ali Taghavi | CC BY-SA 4.0 |