Riemannian homogeneous equivalent to linear group orbit

Question

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

Answer 1

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

Answer 2

$ \textbf{(1)} \implies \textbf{(2)} $

Up to diffeomorphism a Riemannian homogeneous space is just a trivial vector bundles over a compact Riemannian homogeneous space (noncompact Riemannian homogeneous is trivial vector bundle over compact homogeneous). Compact Riemannian homogeneous spaces are linear group orbits (Compact linear group orbit equivalent to linear compact group orbit) and vector spaces are linear group orbits and cartesian products of linear group orbits are linear group orbits so $ M $ Riemannian homogeneous implies $ M $ a linear group orbit

$ \textbf{(2)} $ is strictly weaker. The tangent bundle of $ S^2 $ is nontrivial so cannot be Riemannian homogeneous. However it arises as a linear group orbit by consider the orbit of the vector $ (1,0,0) \in \mathbb{C}^3 $ with respect to the standard representation of $ SO_3(\mathbb{C}) $.

$ \textbf{(2)} \implies \textbf{(3)} $

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

$ \textbf{(3)} $ is strictly weaker. The Moebius band has abelian $ \pi_1 $ and a transitive action by $ SE_2 $ but is not a linear group orbit (I think, actually that claim is open here https://math.stackexchange.com/questions/4359525/is-it-possible-to-realize-the-moebius-strip-as-a-linear-group-orbit but at least we can be sure that the Moebius strip is not Riemannian homogeneous since it is a nontrivial vector bundle)

$ \textbf{(cc)} \implies \textbf{(1c)} $

Since $ M $ admits a transitive action by a compact group $ K $ then $ M $ is compact and $ K $ can always be taken to act by isometries ( by pushing forward the biinvariant metric on $ K $ to a left invariant metric on $ M $).

$ \textbf{(1c)} \implies \textbf{(2c)} $

Since $ M $ is compact then $ K=Iso(M) $ is compact so $ M $ is a quotient of the compact group $ K $ and thus by Mostow-Palais $ M $ is a linear group orbit.

$ \textbf{(2c)} \implies \textbf{(cc)} $

If $ M $ is assumed compact and $ M $ is a linear group orbit of some group $ G $ then the maximal compact subgroup of $ G $ acts transitively see Compact linear group orbit equivalent to linear compact group orbit . Thus we have

$$ \textbf{(cc)}=\textbf{(1c)}=\textbf{(2c)} $$

Now for

$ \textbf{(cc)}=\textbf{(1c)}=\textbf{(2c)} \implies \textbf{(3c)} $

The three equivalent conditions above imply that $ M $ is compact, admits a transitive action by a Lie group, and has finite by abelian fundamental group, see either of the first two links in the question or see https://math.stackexchange.com/questions/4321106/transitive-action-by-compact-lie-group-implies-almost-abelian-fundamental-group

$ \textbf{(3c)} \implies \textbf{(cc)}=\textbf{(1c)}=\textbf{(2c)} $

True for dimension 2. True for dimension 3. Not sure in general!