Exact condition for smooth homogeneous to imply Riemannian homogeneous for compact manifolds

Question

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

https://math.stackexchange.com/questions/1848792/fundamental-group-of-a-compact-space-with-compact-universal-covering-space

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

https://math.stackexchange.com/questions/4386449/transitive-lie-group-actions-and-exotic-smooth-tori/4386739#4386739

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.

Answer 1

The isometry group of a compact Riemannian manifold is always compact, so the answer to this question doesn’t really depend on the fundamental group at all: if $G$ is noncompact, then it doesn’t act isometrically on a compact $M$ for any choice of metric, but if $G$ is compact, then we can get an invariant Riemannian metric using the usual averaging trick.

Assuming you want a transitive isometric action of some Lie group other than $G$ itself, this is a bit trickier, though it does still come down to finding a compact Lie group acting transitively on $M$ (for the same reasons given above). I imagined that there would be many solvmanifold counterexamples, but all the ones I can think of do not satisfy your fundamental group condition.

One possible counterexample would be the mapping torus for an orientation-reversing isometry on an odd-dimensional sphere $S^{2n+1}$ with the round metric. This is homogeneous under the action of $\mathrm{O}(2n+2)\times\mathbb{R}$ and has fundamental group $\mathbb{Z}$, but any transitive isometric Lie group action on it must descend from a transitive isometric action on the universal cover $S^{2n+1}\times\mathbb{R}$, and—assuming things work similar to the case in dimension 3, though I don’t have my copy of Thurston’s geometry book on hand to recall the argument—the maximal Lie group acting effectively and transitively on $S^{2n+1}\times\mathbb{R}$ with compact stabilizers is $\mathrm{O}(2n+2)\times\mathbb{R}\rtimes\mathbb{Z}/2\mathbb{Z}$; for odd-dimensional spheres, we don’t have central orientation-reversing isometries, so the largest normalizer (in $\mathrm{O}(2n+2)$) of a subgroup generated by such an isometry is probably small enough that the action of $\mathrm{O}(2n+2)\times\mathbb{R}\rtimes\mathbb{Z}/2\mathbb{Z}$ can’t descend to a transitive action on the mapping torus.