I am trying to understand the topology (in terms of homology groups) of the free loop space $\Lambda M$ of nice spaces (Complete Riemannian connected finite dimensional manifolds $M$). I see the free loop space (of H^1 loops) as a Hilbert manifold, cf. Klingenbergs book. If the manifold $M$ has a non-trivial fundamental group, the free loop space has as many connected components as there are conjugacy classes in $\pi\_1(M)$. How much do these components of $\Lambda M$ differ? Are these components all homotopy equivalent? For the circle the answer is yes, because all components of the free loop space are homotopy equivalent to the circle itself.


The following question is related to my question


[Are the path components of a loop space homotopy equivalent?](https://mathoverflow.net/questions/34927/are-the-path-components-of-a-loop-space-homotopy-equivalent)


However, I cannot seem to use the answer to this question directly, because I cannot concatenate two free loops, but maybe I am missing something obvious.



The connected components of the free loop space
The different components are, indeed, not all homotopy equivalent, and you are quite right in noting that the argument that works for $\Omega M$ (via concatenation of loops) does not hold here.


This is best illustrated when $M = BG = K(G,1)$ is an Eilenberg-MacLane space with precisely one (non-abelian) homotopy group $G$ in dimension 1. Geometrically, we are simply assuming $M$ is aspherical. Admittedly, not all manifolds fit this description (nor are all of these spaces manifolds), but this case captures the important part of the failure of these components to be homotopy equivalent.


As you say, the set of components of $\Lambda M$ are indexed by conjugacy classes in $\pi\_1(M) = G$. This statement may be promoted to the general claim that $\Lambda M$ is homotopy equivalent to the Borel construction 


$$\Lambda M \simeq G^{ad} \times\_G EG,$$


where $G^{ad}$ is $G$, regarded as a $G$-space via the conjugation action, $EG$ is a contractible space with a free $G$ action (e.g., the universal cover of $M$), and the notation indicates the quotient by the diagonal action of $G$ on the cross product. 


Since $G$ is discrete, one may write it as a disjoint union of orbits. These are, of course, just conjugacy classes of elements in $G$. I'll write $(g)$ for the conjugacy class of $g \in G$, so


$$\Lambda M \simeq \coprod\_{(g)} (g) \times\_G EG.$$


Then an individual component of $\Lambda M$ is of the form $(g) \times\_G EG$. What is the topology of this space? Well, for one, it has fundamental group given by the centralizer of $g$ in $G$, $C(g)$. This is because it is the quotient of $(g)$ copies of the universal cover of $M$ by an action which permutes the copies via conjugation (transitively, by assumption), and the stabilizer of a given copy (say the one indexed by $g$) is simply the set of elements that commute with $g$. 


In fact, since $M$ was aspherical (i.e., $EG$ was contractible), this is the *only* homotopy group of this space. We conclude:


$$\Lambda M \simeq \coprod\_{(g)} K(C(g), 1) = \coprod\_{(g)} BC(g).$$


Now, as long as $G$ is not abelian, the centralizers of elements of $G$ will not all be isomorphic. Consequently, these components are not homotopy equivalent, as they have different fundamental groups.


