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?](https://mathoverflow.net/questions/206618/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](https://mathoverflow.net/questions/410334/riemannian-homogeneous-always-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



Riemannian homogeneous equivalent to linear group orbit
$ \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](https://mathoverflow.net/questions/410334/riemannian-homogeneous-always-trivial-vector-bundle-over-compact-homogeneous)). Compact Riemannian homogeneous spaces are linear group orbits ([Compact linear group orbit equivalent to linear compact group orbit](https://mathoverflow.net/questions/409511/compact-linear-group-orbit-equivalent-to-linear-compact-group-orbit/409532#409532)) 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?](https://mathoverflow.net/questions/206618/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](https://mathoverflow.net/questions/409511/compact-linear-group-orbit-equivalent-to-linear-compact-group-orbit/409532#409532) . 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!


