# Maps to $K(\pi,1)$ spaces from manifolds with $S^1$-action

Suppose $M$ is a connected smooth manifold with a smooth $S^1$-action that fixes a point in $M$. Let $X$ be a $K(\pi,1)$-space and let $\varphi: M\to X$ be a continuous map.

Question. How to prove that $\varphi$ can be homotoped to a map $\varphi': M\to X$ that sends each orbit of the $S^1$ action to a point in $X$?

I would be grateful for a short proof or a reference.

Remark. I can prove the statement when $X$ is a negatively curved manifold, but I am confident that the statement holds as well when $X$ is $K(\pi,1)$.

• What if one orbit is sent by $\varphi$ to a nontrivial conjugacy class in $\pi$? – Denis Nardin Mar 28 '17 at 15:04
• All the orbits will be sent to contractible loops since there is one orbit (a fixed point) that is sent to a point. – aglearner Mar 28 '17 at 15:26
• First, please add connectivity assumption for $M$. Then, can't you just argue that if the $S^1$-action is reasonably nice (say, smooth, on a smooth manifold) then the projection map $M\to Y=M/S^1$ induces an isomorphism of $\pi_1$'s (since each $S^1$-orbit is null-homotopic in $M$)? This part might be in Bredon's book "Compact transformation groups". Then use Whitehead's theorem for the maps $Y\to X$ (again, assuming that $X$ is, say, a CW complex). – Misha Mar 28 '17 at 21:40
• Misha, thanks for the remark! Unfortunately $\pi_1(M)$ does not need to be equal to $\pi_1(Y)$. Indeed, even in the case when $M$ is a $3$-dimensional manifold, the quotient $M/S^1=Y$ can be a topological $2$-disc with non-trivial orbi-structure. Then $\pi_1(Y)=0$, but $\pi_1(M)$ is huge – aglearner Mar 28 '17 at 22:05
• As soon as you have an example in which $\pi_1(M)\to \pi_1(M/S^1)$ is not injective, you have a counterexample. Let $\pi$ be $\pi_1(M)$ and let $M\to K(\pi,1)$ induce isomorphism in $\pi_1$. – Tom Goodwillie Mar 29 '17 at 14:23

I would like to explain why the answer to the question is negative in general, following the suggestion of Tom Godwillie (however see PS for a positive answer if $dimX<\infty$)

Let us note first that there is an $S^1$-action on $\mathbb RP^2=M$ such that the quotient is a the segment.

Now, apply the idea of Tom Goodwille. Namely take $\Gamma=\mathbb Z_2$, and consider the inclusion $\mathbb RP^2\to \mathbb RP^{\infty}=K(\mathbb Z_2,1)$ that induces an isomorphism on $\pi_1$. Clearly, this map is not contractible, so it gives a counterexample to the claim in the question.

PS. However, using the remark of Misha it is also possible to prove that the answer to the question is positive in the case when the group $\pi=\Gamma=\pi_1(X)$ has no torsion elements. In particular, it holds when $X$ is finite-dimensional.

Sketch proof. Denote $M/S^1$ by $Y$ and let $p$ be the projection map. I claim that the kernel of the homomorphism $\phi^M_Y:\pi_1(M)\to \pi_1(Y)$ induced by $p$ is generated by torsion elements. Since $\Gamma$ has no torsion elements, it follows that any homomorphism $\phi^M_X:\pi_1(M)\to \Gamma$ factors through a homomorphism $\phi^M_Y$, i.e. there exists $\phi^Y_X:\pi_1(Y)\to \pi_1(X)$ such that $$\phi^M_X=\phi^Y_X\circ\phi^M_Y.$$

Now, since $X$ is aspherical, the homomorphism $\phi^Y_X$ is induced by a map $\psi: Y\to X$. Finally note that the original map $\varphi: M\to X$ and the composed map $\psi\circ p$ are homotopic, since they induce the same homomorphism on $\pi_1(M)$.

See this paper, theorem 3.5

Gottlieb, Daniel H.; Lee, Kyung B.; Ozaydin, Murad, Compact group actions and maps into $K(\pi,1)$-spaces, Trans. Am. Math. Soc. 287, 419-429 (1985). ZBL0524.57026.

• @aglearner Oops! Reading comprehension issues... I will look more. – Igor Rivin Mar 28 '17 at 16:04
• Yes, unfortunately this answer doesn't address the question... – aglearner Mar 28 '17 at 19:44