Let $ (M,g) $ be a homogeneous Riemannian manifold. That is, the isometry group $ Iso(M,g) $ acts transitively on $ M $. Let $ \pi_1(M) $ be the fundamental group of $ M $. Then $ \pi_1(M) $ has finite commutator subgroup.

I am looking for a converse of sorts. Let $ M $ be a **compact** manifold. Suppose that $ M $ is smooth homogeneous. In other words, there exists some finite dimensional Lie group $ G $ **not necessarily compact** acting smoothly **but not necessarily by isometries** and transitively on $ M $. Furthermore suppose that $ \pi_1(M) $ has finite commutator subgroup. Then can we conclude that there exists some homogeneous metric $ g $ on $ M $ (i.e. the isometry group $ Iso(M,g) $ acts transitively on $ M $)?

EDIT:

A counter example cannot be simply connected (or even have finite fundamental group) because then it would be the case for the smooth transitive action of the noncompact group $ G $ that the action of the maximal compact subgroup is still transitive [Corollary 3, Montgomery, Simply Connected Homogeneous Spaces]. So a counterexample must have infinite fundamental group. In particular that means for any counter example the universal cover cannot be compact

Also no aspherical manifold can be a counterexample. To see why, suppose that $ M $ is a smooth homogeneous aspherical closed manifold whose fundamental group has finite commutator subgroup. An aspherical manifold has torsion free fundamental group. So if the commutator subgroup is finite then the fact that the fundamental group is torsion free implies that the commutator subgroup vanishes and the fundamental group is abelian. Every finitely generated torsion free abelian group is isomorphic to $ \mathbb{Z}^n $, the fundamental group of the torus. A smooth homogeneous aspherical closed manifold with abelian (even nilpotent, even virtually nilpotent) fundamental group is determined up to diffeomorphism by its fundamental group see

So $ M $ must be diffeomorphic to the standard torus $ T^n $, which admits a flat metric. In conclusion, any smooth homogeneous aspherical closed manifold whose fundamental group has finite commutator subgroup is diffeomorphic to the standard torus and thus admits a Riemannian homogeneous metric (the flat metric).

Also, all counterexamples have to be in at least dimension $ 4 $.

In $ n \leq 2 $ every compact smooth homogeneous space is Riemannian homogeneous except the Klein bottle which is aspherical so cannot be a counterexample.

For $ n=3 $ most compact smooth homogeneous spaces are aspherical. The ones that aren't are all Riemannian homogeneous with the exception of $ \mathbb{R}P^3 \# \mathbb{R}P^3$ which has infinite fundamental group (the free product $ C_2 * C_2 $) so it is ruled out by the fundamental group condition since the abelianization is finite $ C_2 \oplus C_2 $ so the commutator subgroup of $ \pi_1 $ is not finite.

EDIT 2: Since abelian groups have trivial commutator subgroup then certainly the commutator subgroup is finite and thus the simplest possible fundamental group for a counterexample is $$ \pi_1(M) \cong\mathbb{Z} $$ So the simplest counterexample would be a compact (non-aspherical) 4 manifold with fundamental group $ \pi_1(M) \cong\mathbb{Z} $ that admits a transitive action by a noncompact group $ G $ but does not admit any Riemannian homogeneous metric. Note that this is exactly what Robin Goodfellow attempted in his answer. The issue is that his manifold is not even smooth homogeneous to begin with, therefore not a counterexample.

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