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!



Free $S^1$-action on compact homogeneous spaces
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.


