Riemannian Homogeneous Spaces – Equivalence to Linear Group Orbit

algebraic-groupsgt.geometric-topologyhomogeneous-spaceslie-groupsriemannian-geometry

Let $ M $ be a smooth manifold.

Recall that a manifold $ M $ is smooth homogeneous if there exists a Lie group acting transitively on $ M $.

Recall that a manifold $ M $ is Riemannian homogeneous if it admits a metric with respect to which the isometry group is transitive, moreover this metric can always be chosen to have nonnegative curvature.

And recall that a manifold $ M $ is a linear group orbit if there exists a representation $\pi:G \to GL(V) $ and a vector $v \in V$ such that the orbit of $ v$
$$
\mathcal{O}_v:=\{ \pi(g)v:g \in G\}
$$
is diffeomorphic to $ M $.

The fundamental group of a linear group orbit is always finite by abelian (i.e. has finite commutator subgroup):

How bad can $\pi_1$ of a linear group orbit be?

And the fundamental group of a Riemannian homogeneous space is also always finite by abelian (i.e. has finite commutator subgroup):

https://www.uni-muenster.de/imperia/md/content/theoretische_mathematik/diffgeo/mr1783960.pdf

This condition on the fundamental group holds in both cases for essentially the same reason. In both cases the manifold $ M $ is the total space of a vector bundle (the vector bundle is trivial if $ M $ is Riemannian homogeneous see noncompact Riemannian homogeneous is trivial vector bundle over compact homogeneous but possibly nontrivial if $ M $ is a linear group orbit) over a compact Riemannian homogeneous base $ B $. And thus $ M $ deformation retracts onto $ B $. And a quotient of compact groups always has $ \pi_1 $ with finite commutator subgroup (https://math.stackexchange.com/questions/4321106/transitive-action-by-compact-lie-group-implies-almost-abelian-fundamental-group/4359177#4359177).

Are the following three properties equivalent?

  • $ \textbf{(1)} $ $ M $ is Riemannian homogeneous
  • $ \textbf{(2)} $ $ M $ is a linear group orbit
  • $ \textbf{(3)} $ $ M $ is smooth homogeneous and $ \pi_1(M) $ has finite commutator subgroup

How about if we assume $ M $ compact? In other words, are the following three (actually four I added one) properties equivalent:

  • $ \textbf{(cc)} $ $ M $ admits a transitive action by a compact Lie group (so a fortiori is compact)
  • $ \textbf{(1c)} $ $ M $ is compact and Riemannian homogeneous
  • $ \textbf{(2c)} $ $ M $ is compact and a linear group orbit
  • $ \textbf{(3c)} $ $ M $ is compact and smooth homogeneous and $ \pi_1(M) $ has finite commutator subgroup

Best Answer

The answer is true if $M$ is compact with finite fundamental group. A theorem of Lichnerowicz says that the isometries of a riemannian manifold $M$ is a Lie group $G$. If $G$ acts transitively on $M$, there exists a compact subgroup of $H$ of $G$ which acts transitively on $M$ (your previous question). A Lie compact group is a subgroup of $SO(n)$.

https://math.stackexchange.com/questions/3905118/what-class-of-lie-groups-embed-in-a-general-linear-group

Related Question