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Let $M = G/K$ be a compact homogeneous space, i.e. $G$ a connected compact Lie group, and $K$ a closed subgroup inside $G$ that contains no nontrivial normal subgroup of $G$.

  1. If $r(G) > r(K)$ (here, $r$ means the rank of maximal torus), is it true that $G/K$ admits free $S^1$-action? If so, why? If not, what is a counterexample?
  2. What if we further assume $K$ is connected? (Note: In “Homogeneous Spaces with Non-Vanishing Euler Characteristics” by HC Wang (1949), it was proven that $\chi(G/K) \ne 0 \Leftrightarrow r(G) = r(K)$.)
  3. If 1. is unknown, then, restricting to case when $G$ semisimple, is 1. known?

Any input would be really helpful. Thanks a lot!

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    $\begingroup$ If this were true then the manifold would be a principle circle bundle over the quotient. If $G/K$ is a 3-manifold with $b_1 = 0$, this implies $G/K$ is diffeomorphic to a lens space $L(n,1)$. Thus a great many spherical 3-manifolds give counterexamples, such as the Poincare homology sphere where $G = SO_3$ and $K = A_5$ is the discrete group of symmetries of the icosahedron. Maybe you have better luck with K connected? $\endgroup$
    – mme
    Commented Nov 8, 2022 at 14:22
  • $\begingroup$ Thanks for your answer! Yes, I was considering $K$ connected, but nevertheless what you gave is great. I'll update the question $\endgroup$
    – abracadabra12345
    Commented Nov 8, 2022 at 16:57

1 Answer 1

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Here is a counterexample where $K$ is connected.

Let $W = SU_3/SO_3$ be the Wu manifold. This 5-dimensional manifold is simply-connected and has $H_2(W;\Bbb Z) = \Bbb Z/2$, and in particular, $\pi_2(W) = \Bbb Z/2$ by the Hurewicz theorem. Note that $r(SU_3) = 2$ and $r(SO_3) = 1$.

Now I claim that there is no free circle action on $W$. For if $M$ is any manifold with free circle action and quotient $X$, the projection defines a fiber-bundle $$S^1 \to M \to X.$$

  • If $M$ is simply-connected, the long exact sequence of homotopy groups implies $X$ is, too.
  • If $M$ is further a simply-connected 5-manifold, so that $X$ is now a simply-connected 4-manifold, observe that by Poincare duality $H_2(X;\Bbb Z)$ is free abelian. The long exact sequence of homotopy groups gives an injection of $H_2(W;\Bbb Z) \cong \pi_2(W)$ into $\pi_2(X) \cong H_2(X;\Bbb Z)$. Thus if $M$ is a simply-connected 5-manifold which supports a free circle action, $\pi_2(M)$ must be free abelian.

Because the Wu manifold has $\pi_2(W) = \Bbb Z/2$, it does not support a free circle action. This answers your question in the negative, and unfortunately I do not see a way to salvage it by adding more conditions.

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    $\begingroup$ www2.math.ou.edu/~shankar/papers/freeactl_eks.pdf contains a complete classification of which compact symmetric spaces admit free isometric circle actions, just give to give the OP some more information. $\endgroup$
    – Jason DeVito - on hiatus
    Commented Nov 8, 2022 at 19:14
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    $\begingroup$ In addition, archive.ymsc.tsinghua.edu.cn/pacm_download/613/… Corollary 5.3 indicates that the Berger space $Sp(2)/Sp(1)_{max}$ also does not admit a free $S^1$ action. (The notation $Sp(1)_{max}$ denotes the subgroup coming from the unique $4$-dim irrep of $SU(2) = Sp(1)$.) $\endgroup$
    – Jason DeVito - on hiatus
    Commented Nov 8, 2022 at 19:39
  • $\begingroup$ @JasonDeVito So one may even take $K$ is simply connected. That's nice. $\endgroup$
    – mme
    Commented Nov 9, 2022 at 0:58
  • $\begingroup$ Awesome. Thanks! $\endgroup$
    – abracadabra12345
    Commented Nov 10, 2022 at 15:33

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