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Given a differentiable manifold $M$, a Riemannian metric $g$ on $M$ is called homogeneous if its isometry group acts transitively on $M$. In that case, given any group $G$ acting transtitively on $(M,g)$ by isometries, we have the diffeomorphism $M\simeq G/H$ where $H$ is the stabilizer of any point $o\in M$.

We now fix $M=G$, a compact connected Lie group (I am mostly interested in the semisimple case). It turns out that $G$ acts on itself by multiplication on the left, and this action is obviously transtive. Thus, one has a natural space of homogeneous metrics on $G$: the space of left-invariant metrics, which are the Riemannian metrics on $G$ such that this action is an isometry.

Is any homogeneous Riemannian metric on $G$ necessarily isometric to a left-invariant metric on $G$?

Although a right-invariant metric on $G$ is in general not left-invariant, it is true that $(G,g_r)$ with $g$ a right-invariant metric on $G$ is isometric to $(G,g_l)$ for some left-invariant metric $g_l$ on $G$.

Edit on April 18, 2022:

The general problem of classifying all homogeneous metrics on a fixed differentiable manifold $M$ seems to be extremely difficult. As far as I understand, one has to know all transitive Lie group actions on $M$ and (i.e., roughly, all possible realization $M=G/H$), for each of them, obtain the invariant metrics.

This problem is solved for tori, the (underlying manifolds) compact rank one symmetric spaces, and for certain Grassmannian spaces.

A more general question than previous one is

Are the homogeneous metrics on a compact connected Lie group $M=G$ classified?

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    $\begingroup$ I guess that using that (see mathoverflow.net/q/67030/14094) there are non-isomorphic diffeomorphic compact Lie groups, it shouldn't be too hard to cook up counterexamples. $\endgroup$
    – YCor
    Commented Apr 13, 2022 at 9:51
  • $\begingroup$ As far as I understand, two non-isomorphic and diffeomorphic compact Lie groups share the same universal cover, which is again a compact Lie group. Since left-invariant metrics depends on the Lie algebra level, perhaps such counterexample does not exist. $\endgroup$
    – emiliocba
    Commented Apr 13, 2022 at 9:59
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    $\begingroup$ the map $G\to G$ given by $g\mapsto g^{-1}$ turns any left invariant metric into a right invariant one. $\endgroup$ Commented Apr 13, 2022 at 12:52

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