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mme
mme
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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.

Here is a counterexample where $K$ is connected.

Let $W = SU_3/SO_3$ be the Wu manifold. This 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.

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.

Source Link
mme
mme
  • 9.6k
  • 5
  • 48
  • 73

Here is a counterexample where $K$ is connected.

Let $W = SU_3/SO_3$ be the Wu manifold. This 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.