Timeline for Right invariant Killing fields of Right invariant Riemanian metrics
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 21, 2014 at 9:00 | vote | accept | Benjamin | ||
| May 14, 2014 at 19:00 | answer | added | José Figueroa-O'Farrill | timeline score: 2 | |
| May 14, 2014 at 18:57 | history | edited | Benjamin | CC BY-SA 3.0 |
added 25 characters in body
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| May 14, 2014 at 18:56 | comment | added | José Figueroa-O'Farrill | Indeed. But I’m just pointing out, perhaps unhelpfully, that your first question as written has a trivial answer. Hence, perhaps you’d like to ask a more precise question. | |
| May 14, 2014 at 18:52 | comment | added | Benjamin | I should have clarified, right invariant but not bi-invariant. Sorry for the confusion. $B$ is bi-invariant but $w$ is only right invariant. As far as I know such a metric cannot be bi-invariant as $w$ would need to be bi-invariant and there are no such vector fields on $SU(n)$. | |
| May 14, 2014 at 18:51 | comment | added | José Figueroa-O'Farrill | The answer to your question is trivially yes, because a bi-invariant metric is right-invariant. | |
| May 14, 2014 at 17:41 | answer | added | Peter Michor | timeline score: 3 | |
| May 14, 2014 at 17:05 | history | asked | Benjamin | CC BY-SA 3.0 |