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 $)?


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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    $\begingroup$ I think you should add the word 'compact' to the question itself because, as it stands, it's easy to overlook the one place that specifies that you are asking only about compact manifolds. In fact, I was misled this way when I gave a (now deleted) answer that didn't take this condition into account because it wasn't in the question. $\endgroup$ Jan 23, 2022 at 15:27
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    $\begingroup$ Your argument in the case of the tori is incomplete as it does not rule out exotic tori which exist in high dimensions. but it's easy to see that if a compact Riemannian homogeneous space $M$ is aspherical then $Iso(M)$ is also aspherical which does imply that $Iso_0(M)$ is a torus and so is $M$. $\endgroup$ Feb 19, 2022 at 16:42
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    $\begingroup$ @IanGershonTeixeira yes, this is easy. there are several ways to see it. The simplest is that any $M=G/H$ where $G$ is compact admits a metric of nonnegative sectional curvature (coming from a biinvariant metric on $G$). It's a standard consequence of Cheeger-Gromoll splitting theorem that any nonnegatively curved metric on a topological torus is flat. you can also give a lie group argument based on structure theory of compact Lie groups but the above suffices. $\endgroup$ Feb 19, 2022 at 22:51
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    $\begingroup$ to elaborate further Cheeger and Gromoll proved that the universal cover of any closed nonnegatively curved $M$ is isometric to $\mathbb R^k\times C$ where $C$ is simply connected and closed. If $M$ is aspherical then $C$ must be a point which means that $M$ is flat. $\endgroup$ Feb 19, 2022 at 23:00
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    $\begingroup$ @IanGershonTeixeira sorry, i was going by your main question where you assume that the metric is Riemannian homogeneous. if you only assume that it's smooth homogeneous then I don't immediately see how to show that any homogeneous torus must be standard although I am sure this is true. You might want to ask a separate question about it. $\endgroup$ Feb 19, 2022 at 23:20

1 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.

  • $\begingroup$ Note that in OP there is no requirement that $G$ is related to $Isom(M,g)$ in any way (except for both acting on the same manifold). $\endgroup$ Dec 28, 2021 at 14:44
  • $\begingroup$ Thanks MoisheKohan! In particular the spirit of my question is "If a non compact G acts transitively then when can we arrange for the maximal compact subgroup K of G to act transitively by isometries"? But I kept the wording vague because the real line acts transitive isometries on the flat circle but the maximal compact is trivial. I expect the answer is "yes when G acts transitively and the fundamental group of M has finite commutator subgroup then there is always a transitive isometric action by $ K \times T^n $ where $ K $ is maximal semisimple compact, $ n $ dimension maximal abelian $\endgroup$ Dec 28, 2021 at 14:53
  • $\begingroup$ Oh wow the mapping torus for an orientation reversing isometry of the three sphere! This is just the sort of interesting counterexample I was looking for. I'll definitely have to look in my copy of Thurston's geometry book. I guess this would be in the section on $ S^2 \times \mathbb{R} $ geometry? If you feel confident about the details I'm happy to accept your answer. If not we can wait until maybe you get a chance to look more at Thurston or someone else like @MoisheKohan comes back and say the details look good. $\endgroup$ Dec 31, 2021 at 16:25
  • $\begingroup$ I think your statement "the mapping torus for an orientation-reversing isometry on an odd-dimensional sphere $S^{2n+1}$ with the round metric... is homogeneous under the action of $\mathrm{O}(2n+2)\times\mathbb{R}$" is wrong and therefore your suggestion is not a valid counterexample. Note that the natural action of $\mathrm{O}(2n+2)\times\mathbb{R}$ on the universal cover $ S^{2n+1} \times \mathbb{R} $ does not descend to an action on the mapping torus see math.stackexchange.com/questions/4360976/… $\endgroup$ Jan 23, 2022 at 4:40
  • $\begingroup$ If you have another action of $ O(2n+2)×\mathbb{R} $ on the mapping torus $S^{2n+1} \rtimes S^1$ in mind I would love to hear about it. If not and you just made a mistake when you claimed the existence of such a homogeneous action then I would encourage you to edit your answer accordingly. For what it's worth even spheres don't work as a counterexample either because, as you anticipated by specializing to the case of a non central isometry, for even spheres the natural action factors through the action of a compact group and thus the mapping torus is Riemannian homogeneous. $\endgroup$ Jan 23, 2022 at 4:46

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