MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a closed manifold, and let $X^{S^1}$ denote the free loop space of $X$, that is, the set of continuous maps $S^1 \rightarrow X$. Let $Y$ denote a component of $X^{S^1}$.

What conditions ensure that $Y$ is homotopic to a finite CW complex? (by finite, I mean, the total number of cells is finite).

Edit - thanks to Somnath, sufficient conditions are certainly when $X$ is an Eilenberg-Maclane space, as then the based loop space is contractible. Are there any other different sufficient conditions?

share|cite|improve this question
If the total number of cells in a CW complex is finite, then the space it locally compact, but in general $Y$ is not locally compact. – Mariano Suárez-Alvarez Apr 11 '11 at 16:44
(you will have more luck if you look for finite CW complexes with the homotopy type of $Y$) – Mariano Suárez-Alvarez Apr 11 '11 at 17:01
@ Ricardo - It's true for $S^1$ since it's an example of a $K(\mathbb{Z},1)$-space. More generally, take $X$ to be a hyperbolic $3$-manifold or a surface of non-zero genus. These are all $K(\pi_1(X),1)$'s. Why is this relevant? Because $X^{S^1}$ fibres over $X$ with fibre $\Omega X$, the based loop space. The connected components of $X^{S^1}$ are labelled by $\pi_1(X)$. For such an $X$, $\Omega X$ has the homotopy type of $\pi_1(X)$, which is discrete. Hence, $X^{S^1}$ looks like a collection of covering spaces of $X$. If all the covers are finite and $C_\ast(X)$ is finite, then you're done! – Somnath Basu Apr 11 '11 at 18:45
For a path connected space the loopspace has one component for each conjugacy of the fundamental group of $X$, and the fundamental group of each component is the centralizer of a representative for that conjugacy class. In the cases that Somnath mentioned, these subgroups have infinite index but the corresponding noncompact covering space often has the homotopy type of a finite complex (such as a circle) anyway. – Tom Goodwillie Apr 11 '11 at 20:46
@ Mark, Mariano: I edited the question to be more meaningful, now we are only asking for the homotopy type of a finite CW complex. – Ricardo Apr 11 '11 at 21:43

There are some negative results about. For instance, a theorem of Sullivan and Vigué-Poirrier states that if $M$ is a closed manifold with $\pi_1(M)$ finite, and if the cohomology algebra $H^*(M;\mathbb{R})$ requires at least two generators, then the Betti numbers of $M^{S^1}$ are unbounded. The finiteness assumption implies that $M^{S^1}$ has only finitely many path components, and so at least one of them is not a finite CW-complex.

share|cite|improve this answer
Even in the case when $M$ is simply connected and $H^\ast(M;\mathbb{R})$ is monogenic, the Betti numbers of $M^{S^1}$ are bounded but the sum is unbounded. Of course, in this case no such $Y$ as the OP asked exists since $M^{S^1}$ is connected. – Somnath Basu Apr 11 '11 at 21:19
This is true. Whenever $M$ is a simply-connected, non-contractible manifold, the loop space $M^{S^1}$ is not homotopy equivalent to a finite complex (since it has infinite Lusternik-Schnirelmann category). – Mark Grant Apr 12 '11 at 7:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.