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Suppose that M is a manifold admitting a smooth transitive action by a compact Lie group. Prove that the fundamental group of M fits into the SES $$ 1 \to \Gamma \to \pi_1(M) \to A \to 1 $$ Where $ \Gamma $ is finite and $ A $ is abelian.

This is equivalent to the commutator subgroup of $ \pi_1(M) $ (the kernel of the first Hurewicz map) being finite.

For example this criterion shows that the klein bottle does not admit a transitive action by any compact group.

Ian Gershon Teixeira
  • 10,176
  • I think you can derive a “reversed” exact sequence (ie make the $\pi_1$ an extension of a finite group by an abelian group), using Ehresman and the homotopy long exact sequence for fibrations. – Aphelli Dec 01 '21 at 14:56
  • Whoever downvoted my post could you give a suggestion for how to make it better? – Ian Gershon Teixeira Dec 01 '21 at 16:24
  • I for one think this question is fine the way it is. I'll complement the downvote :-) – Jeroen van der Meer Dec 01 '21 at 17:44
  • This is what makes stack exchange so great :-) Also @Mindlack what Ehresman theorem do you mean? – Ian Gershon Teixeira Dec 02 '21 at 03:46
  • That is a really nice question. I always assumed the Klein bottle was a model for Euclidean geometry, but I suppose the point is that $(x,y)\mapsto (x+1,-y)$ is not central in isometries of the plane. I already gave (+1) when someone bizarrely downvoted - otherwise would do so again. – tkf Dec 02 '21 at 14:54
  • A shot in the dark, but do you think if $G$ compact acts transitively on $M$, then some connected group $G'$ (not necessarily compact) acts on $\hat{M}$, some normal cover of $M$ with finite fundamental group, and $M=\hat{M}/A$ for some discrete subgroup $A\subseteq G'$ (with respect to the group action of $G'$ restricted to $A$) ? Then $A$ would necessarily be abelian, and we have: $$1 \to \pi_1(\hat{M})\to \pi_1(M)\to A\to 1 $$. – tkf Dec 02 '21 at 15:08

1 Answers1

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I think this is very similar to what tkf suggested! See theorem 2.2 of

https://www.uni-muenster.de/imperia/md/content/theoretische_mathematik/diffgeo/mr1783960.pdf

Suppose the compact group G acts transitively on M. Then the universal cover
$$ \widetilde{G}\cong K \times \mathbb{R}^n $$ also acts transitively (here $ K $ is compact and simply connected). So for some closed stabilizer group $ H $ we have
$$ M \cong \widetilde{G}/H $$ Consider the projection p of H onto the second factor $ \mathbb{R}^n $. Then we have a SES $$ 1 \to ker(p) \to H \to p(H) \to 1 $$ Observe that p(H) is abelian since it is a subgroup of $ \mathbb{R}^n $ and $ ker(p) $ is compact because it is a subgroup of $ K \times 0 $. Take $ \pi_0 $ of this SES then we have another SES of discrete groups $$ 1\to \pi_0(ker(p)) \to \pi_0(H) \to \pi_0(p(H)) \to 1 $$ The first component group is compact and discrete so it must be finite. The third component group is a discrete subgroup $ \mathbb{R}^n $ so just $ \mathbb{Z}^k $. This proves that $\pi_0(H) $ is finite by abelian. Since $ \widetilde{G} $ is simply connected then $ \pi_1(M)=\pi_0(H) $ is finite by abelian.

Related fun facts from this same reference:

If G compact acts transitively then M admits a nonnegative curvature metric with respect to which G acts by isometries.

Ian Gershon Teixeira
  • 10,176