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Let $X$ be a finite dimensional $K(\pi,1)$ manifold. Recall that the space of unbased null-homotopic loops $L_0(X)$ is the space of contractible continuous maps $S^{1}\to X$. There is an obvious $S^1$-action on $L_0(X)$.

Is it true that the quotient $L_0(X)/S^1$ is homotopy equivalent to $X$? Moreover, is there is a deformation retraction from $L_0(X)/S^1$ to the space of constant loops in it? What if $X$ is a finite dimensional $CW$-complex?

I think I can prove this statement in the case when $X$ is a smooth compact negatively curved manifold and we consider smooth maps from $S^1$ it. (I believe smoothness should not be essential.)

This is the follow up of The space of contractible loops of a finite dimensional $K(\pi,1)$

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    $\begingroup$ Do you mean null-homotopic instead of contractible? The word contractible is usually kept for spaces with the same homotopy type as a point. Also, spaces are not homotopic but homotopy equivalent. $\endgroup$
    – Fernando Muro
    Commented Apr 2, 2018 at 21:22
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    $\begingroup$ @Fernando I see "contractible loop" with some frequency meaning null-homotopic loop, implicitly "contractible in M". $\endgroup$
    – mme
    Commented Apr 2, 2018 at 21:57
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    $\begingroup$ I use homotopic as a shorthand for homotopy equivalent. While it's certainly casual (and probably better to avoid in an MO question), I think it's pretty common since I've never had anyone not understand what I meant by it. I'm pretty sure I've heard other people use it, including some senior mathematicians. $\endgroup$
    – Joe Berner
    Commented Apr 2, 2018 at 22:13
  • $\begingroup$ OK, maybe I'm too purist. $\endgroup$
    – Fernando Muro
    Commented Apr 2, 2018 at 22:24

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THIS ANSWER IS WRONG (in that it claims the result in excess generality)

Suppose that $X$ is a $K(\pi,1)$ and a CW complex. Let $E$ be a universal covering space of $X$, so that $E$ is contractible and there is a free $\pi$-action on $E$ with orbit space $E/\pi\cong X$. (I really mean free and properly discontinuous -- what Hatcher's book calls a "covering space action".)

This gives $L(E)$ a $\pi$-action, again free, with orbit space $L(E)/\pi\cong L_0X$. And $L(E)$ has also $S^1$ acting, commuting with the $\pi$-action, in such a way that $$ (L(E)/S^1)/\pi=L(E)/(S^1\times\pi)=(L(E)/\pi)/S^1\cong L_0(X)/S^1. $$ On the other hand, a deformation retraction of $E$ to a point $p$ yields an $S^1$-equivariant deformation retraction of $L(E)$ to $L(p)$ (i.e. one that is compatible with the $S^1$-action), and this in turn yields a deformation retraction of $L(E)/S^1$ to $L(p)/S^1$.

So $L(E)/S^1$ is contractible and has a free action of $\pi$ with orbit space $L_0(X)/S^1$. This implies that $L_0(X)/S^1$ is a $K(\pi,1)$.

THE ERROR in the answer as written is in the assertion that the action of $\pi$ on $L(E)/S^1$ is free. But if we add the hypothesis that the group $\pi$ is torsionfree (which is true in the kinds of cases that the question is asking about) then this is all right: if $g\in\pi$ fixes the $S^1$-orbit of the loop $\gamma:S^1\to X$, then $g\circ \pi=\pi\circ R$ for some rotation $R:S^1\to S^1$; since the action of $\pi$ on $E$ is free (in the strong sense), this cannot happen unless $g$ and $R$ have (the same) finite order.

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  • $\begingroup$ Can't spot the mistake...(but unaccept it by request ) $\endgroup$
    – aglearner
    Commented Apr 3, 2018 at 0:37
  • $\begingroup$ Thanks for explaining. I wonder then if indeed $X=\mathbb RP^{\infty}$ will give a counter example with torsion. One could take in $L_0(X)/S^1$ the subset corresponding to all lines in $\mathbb RP^{\infty}$ travelled twice. This is the grassmanian $G(2,\infty)$. Does not look like it can be "contracted" to $\mathbb RP^{\infty}$ ...? $\endgroup$
    – aglearner
    Commented Apr 3, 2018 at 8:21
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    $\begingroup$ Yes. You can see explicitly that the "constant loops" map $X\to L_0(X)/S^1$ induces the trivial map in $\pi_1$ if $X$ is $\mathbb RP^\infty$ (or even $\mathbb RP^2$). $\endgroup$
    – Tom Goodwillie
    Commented Apr 3, 2018 at 14:23

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