Metric induced on linear group orbit is Riemannian homogeneous

Consider the standard representation of $ SO_3(\mathbb{R}) $ acting on $ \mathbb{R}^3 $. Then for any nonzero vector $ v \in \mathbb{R}^3 $ the orbit $$ \mathcal{O}_v=\{gv: g \in SO_3(\mathbb{R}) \} $$ is a sphere of radius $ |v| $. And the metric on the sphere, induced by the Euclidean metric on the ambient space $ \mathbb{R}^3 $, is the round metric. A round sphere is Riemannian homogeneous (in other words, the isometry group acts transitively).

This leads to my question. Let $ G $ be a compact group and $ \pi:G \to GL_n(\mathbb{R}) $ a representation. For any $ v \in \mathbb{R}^n $ define the orbit of $ v $ to be $$ \mathcal{O}_v=\{\pi(g)v: g \in G \} $$ Let $ g $ be the metric on $ \mathcal{O}_v $ induced by the Euclidean metric on $ \mathbb{R}^n $. Then is $ (\mathcal{O}_v,g) $ Riemannian homogeneous? In other words, does the isometry group $ Iso(\mathcal{O}_v,g) $ act transitively on $ \mathcal{O}_v $?

Solution 1:

This is false. As a counterexample, consider an $SO(3)$ action on $\mathbb{R}^n$ conjugate to the standard one by a non-orthogonal invertible matrix $A$. The orbits are non-spherical ellipsoids, which are not homogeneous.

The statement is true specifically of orthogonal representations, and of course every representation of a compact group is orthogonal with respect to some inner product, but not necessarily the standard one.