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Let $ M $ be a connected manifold. Let $ \pi_1(M) $ be the fundamental group of $ M $.

Suppose there exists a compact group $ K $ that acts transitively on $ M $. Then $ \pi_1(M) $ must have a finite commutator subgroup. See Transitive action by compact Lie group implies almost abelian fundamental group.

On the other hand, suppose that a Lie group $ G $ acts transitively on a compact manifold $ M $. Furthermore suppose that $ \pi_1(M) $ has finite commutator subgroup. Then can we conclude that there exists some compact group $ K $ (probably related to the maximal compact subgroup of $ G $) which acts transitively on $ M $?

If we strengthen our assumption from $ \pi_1(M) $ has a finite commutators subgroup to just the whole group $ \pi_1(M) $ is finite, then the desired result is exactly Corollary 3 of Montgomery: Montgomery - Simply connected homogeneous spaces.

Improve this question
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  • $\begingroup$ I'm not exactly sure what finite-by-abelian means, but what about the transitive R-action (R=real numbers) on S^1 = R/Z? $\endgroup$
    – kiran
    Commented Dec 6, 2021 at 0:34
  • $\begingroup$ @kiran Here a group is said to be finite-by-abelian if its derived subgroup is finite (i.e., extension with finite kernel and abelian quotient). $\endgroup$
    – YCor
    Commented Dec 6, 2021 at 8:56

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