Is every bi-invariant Finsler metric on $SU(N)$ necessarily Riemannian?
If possible I'd also like to know if the right translation of the Shatten $p-norm$ on the Lie algebra gives rise to a bi-invariant metric?
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1$\begingroup$ 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. $\endgroup$– alvarezpaivaCommented May 29, 2014 at 14:13
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$\begingroup$ It doesn't, I'll fix the wording $\endgroup$– BenjaminCommented May 29, 2014 at 16:40
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1$\begingroup$ By alvarezpaiva's comment, the answer to the first question is yes for $N=2$ and no for $N\ge 3$. $\endgroup$– YCorCommented May 29, 2014 at 18:01
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1$\begingroup$ 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 $p$. $\endgroup$– YCorCommented May 29, 2014 at 21:07
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1$\begingroup$ 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. $\endgroup$– YCorCommented May 29, 2014 at 21:10
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