TITLE: homotopy type of loop space components
QUESTION [5 upvotes]: 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?

REPLY [8 votes]: 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.