Timeline for Is every bi-invariant Finsler metric on $SU(N)$ necessarily Riemannian?
Current License: CC BY-SA 3.0
11 events
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| Aug 23, 2014 at 13:02 | comment | added | Benjamin | @alvarezpaiva Unless I'm making a very foolish error, and this is highly possible, then I can't understand why you are saying that in the $\mathfrak{su}(2)$ case the Shatten $p$-norm doesn't depend on $p$. If $|A| = (|a_1|^p + |a_2|^2)^{1/p}$ and $|a_1| = |a_2| = a$ (as it does in this case as you pointed out) then I calculate: $|A| = (|a_1|^p + |a_2|^2)^{1/p} = 2^{1/p}a$ which does depend on $p$. Did I just make a silly error? | |
| May 29, 2014 at 21:10 | comment | added | YCor | Actually the only invariant quadratic form on $\mathfrak{su}(n)$ is the standard one, up to scalar multiplication. It follows that for $n\ge 3$ and $p\neq 2$, the Shatten $p$-norm is not Euclidean, and hence the corresponding bi-invariant Finsler metric is not Riemannian. | |
| May 29, 2014 at 21:07 | comment | added | YCor | The special property of $\mathfrak{su}(2)$ is that for every element, the two eigenvalues are opposite, hence of the same norm, and that's why the Shatten $p$-norm does not depend on $p$. | |
| May 29, 2014 at 19:07 | comment | added | alvarezpaiva | Try it by hand, you'll see you only get the Euclidean norm on the Lie algebra of $SU(2)$. | |
| May 29, 2014 at 18:17 | comment | added | Benjamin | What about the norm on $\mathfrak{su}(2)$ $Tr(|A|^p)^{1/p}$? Is the right (or left) translation of this not a bi invairant Finsler metric that is not Riemanian? Perhaps I'm being stupid here. What property of $SU(2)$ is it that distinguishes it? | |
| May 29, 2014 at 18:01 | comment | added | YCor | By alvarezpaiva's comment, the answer to the first question is yes for $N=2$ and no for $N\ge 3$. | |
| May 29, 2014 at 17:05 | history | edited | alvarezpaiva | CC BY-SA 3.0 |
added 30 characters in body; edited title
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| May 29, 2014 at 16:40 | history | edited | Benjamin | CC BY-SA 3.0 |
added 15 characters in body
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| May 29, 2014 at 16:40 | comment | added | Benjamin | It doesn't, I'll fix the wording | |
| May 29, 2014 at 14:13 | comment | added | alvarezpaiva | A bi-invariant metric must be Riemmannian if and only if the adjoint representation acts transitively on the projectivized of the Lie algebra. If not you can always find a non-euclidean Ad-invariant norm on the Lie algebra that gives rise to a bi-invariant Finsler metric. | |
| May 29, 2014 at 13:42 | history | asked | Benjamin | CC BY-SA 3.0 |