5
$\begingroup$

Let $X$ be a finite dimensional $K(\pi,1)$ manifold. Is it true that the space contractible loops of this manifold can be contracted to the space of constant loops on $X$? What if $X$ is a finite dimensional $CW$-complex?

I understand that the answer is positive if $X$ is a negatively curved manifold, since in this case there is a geometric argument.

Improve this question
$\endgroup$
3
  • 4
    $\begingroup$ This is correct and you will find it in a lot of literature on string topology. For example a stronger claim is made in Section 10.1 of arxiv.org/pdf/math/9911159.pdf. I'm sure you can find more detailed discussion in for example arxiv.org/pdf/math/0511181.pdf $\endgroup$
    – algebrachallenged
    Commented Mar 31, 2018 at 11:48
  • $\begingroup$ Thanks for the reference! I wonder how one proves this $\endgroup$
    – aglearner
    Commented Mar 31, 2018 at 12:00
  • $\begingroup$ There is a paper at arxiv.org/abs/1003.5617 on "The homotopy 2-type of a free loop space". $\endgroup$
    – Ronnie Brown
    Commented Mar 31, 2018 at 17:02

1 Answer 1

Reset to default
10
$\begingroup$

The statement is true for a $K(\pi,1)$ but not true for other $X$. Finite dimensionality is not relevant.

Here's a sketch: let $X = K(\pi,1)$ and I will assume $X$ has the homotopy type of a CW complex. Let $\Omega_0 X$ be the space of contractible based loops in $X$. It's easy to show that the homotopy groups of this space are trivial in every degree (in degree 0 by definition and in higher degrees since $X$ is a $K(\pi,1)$ space, using the fact that $\pi_j(\Omega_0X) = \pi_{j+1}(X)$ for $j >0$).

Let $L_0 X$ be the space of contractible unbased loops. Then the sequence $$ \Omega_0 X \to L_0 X \to X $$ is a fibration, where the second map is evaluation at the basepoint of the circle. Since the fiber has trivial homotopy groups, it follows that the map $L_0X \to X$ is a weak homotopy equivalence. It follows that the section $X\to L_0X$ given by the constant loops is also a weak homotopy equivalence. It's therefore a homotopy equivalence since the spaces in question have the homotopy type of a CW complex.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Dear John, thank you for the answer. I would like to ask one more thing. If you quotient $L_0X$ by $S^1$ and $X$ is $K(\pi,1)$, is it correct, that the resulting space is still homotopic to $X$? $\endgroup$
    – aglearner
    Commented Apr 2, 2018 at 9:23
  • $\begingroup$ I'm not sure. The action on the circle isn't free on $L_0X$. Taking the quotient by a non-free action isn't something I am able to identify. On the other hand, if we take the homotopy quotient, i.e., the Borel construction $ES^1 \times_{S^1} L_0X$, then one gets $\Bbb CP^\infty \times X$ up to homotopy. $\endgroup$
    – John Klein
    Commented Apr 2, 2018 at 10:31
  • $\begingroup$ Thanks John! Maybe then I'll ask a followup question. For a negatively curved manifold the quotient $L_0X/S^1$ is still homotopic to $X$ I believe (by a geometric argument). $\endgroup$
    – aglearner
    Commented Apr 2, 2018 at 10:56
  • $\begingroup$ And what is the geometric argument? Where can I find it? $\endgroup$
    – John Klein
    Commented Apr 2, 2018 at 16:45
  • 1
    $\begingroup$ I discovered it myself, so don't know a reference. It works if the loops are smooth, but I guess this is a technicality. The idea is the following: if we have a smooth contractible map from a circle to a manifold of negative curvature, we can uniquely contract it to a point. For example go to the universal cover, take a preimage of the circle and then find the "centre of mass" of the circle in the cover. The unique centre of mass is the point $m$ such that $\int_{S^1}d(m,\phi(x)) $ attains its minimum. Now, contract the circle to $m$ along geodesics. Finally, project it back to the manifold $\endgroup$
    – aglearner
    Commented Apr 2, 2018 at 17:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .