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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?

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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.

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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

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