Does the set of bi-invariant Finsler metrics on $SU(N)$ exactly coincide the set of Finsler metrics with the one-parameter subgroups as their geodesics through the identity?


I know that being bi-invariant implies that the geodesics through the identity are exactly the one parameter subgroups. That is to say, every element of the Lie algebra is a geodesic vector. However, I can't seem to settle whether or not the converse holds.



A property of bi-invarient Finsler metrics on SU(N)
I am re-editing this response because I got it wrong the first time around.


The following result, due to myself and José Barbosa Gomes, is (a small) part of the paper [Periodic solutions of Hilbert's fourth problem.](https://arxiv.org/abs/1809.02783)


**Theorem.** If a C2 reversible Finsler metric $F$ on a compact, connected Lie group which is neither $SU(2)$ nor $SO(3)$ has the same *unparametrized* geodesics as a bi-invariant Riemannian metric, then it is a bi-invariant metric.


So the answer to your question is **yes** for $N > 2$. Let me recall that a reversible Finsler metric is one for which the length of each tangent vector $v\_x$ equals the length of its opposite $-v\_x$.


If you admit the metric to be smooth (= as smooth as needed with the smoothness depending on the dimension) I can give you a sketchy shortcut to the proof in the paper.


First one proves that *a smooth reversible Finsler metric whose geodesics are straight lines on a torus of dimension $k$ $(k > 1)$ is flat (= locally isometric to a normed space).* This is done by looking a bit under the hood at the Busemann-Pogorelov solution of Hilbert's fourth problem.


Now if $G$ is neither $SU(2)$ nor $SO(3)$, then it has rank greater than one as a symmetric space when provided with a bi-invariant Riemannian metric. This means every geodesic belongs to a flat torus of dimension at least two. If you have a Finsler metric $F$ with the same unparametrized geodesics, then those tori will be totally geodesic for $F$ and, by the result above, the restriction of F to those tori will also be a flat metric. From this we deduce that $F$ is not just projectively equivalent to the bi-invariant Riemannian metric, but also affinely equivalent to it (i.e., the midpoints of geodesic segments are the same). By work of Z.I. Szabo (<https://arxiv.org/abs/math/0601522>), this is known to imply that $(G,F)$ is a symmetric space and so that the metric $F$ is bi-invariant.


