The connected components of the free loop space 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?
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.
 A: You might like to look at my preprint
http://arxiv.org/abs/1003.5617
on the homotopy 2-type of a free loop space $LX$. It assumes that $X$ is a 2-type, i.e. the classifying space of a crossed module, and then gives precise formulae for crossed modules representing the 2-types of the components. 
I am aware that the main interest in free loop spaces seems to be their homology, and I can't see how these results help on that. 
The paper is planned to be revised with Chris Wensley and to include specific computer calculations, hence the delay. 
I might as well quote the theorem. 
Let  $\mathcal M$ be the crossed module of groups  $\delta: M \to P$
and let  $X=B\mathcal M$ be the classifying space of $\mathcal M$. Then the
components of $LX$, the free loop space on $X$,  are determined by
equivalence classes of elements $a \in P$ where $a,b$ are equivalent
if and only if there are elements $m \in M, p \in P $ such that
$$b= p + a + \delta m -p. $$ Further the homotopy $2$-type of a
component of $LX$ given by $a \in P$ is determined by the crossed
module of groups  $L\mathcal M [a]=(\delta_a: M \to P(a))$ where
(i)  $P(a)$ is the group of   elements $(m,p)\in M \times P$
such that $\delta m= [a,p]$, with composition $(n,q)+(m,p)=
(m+n^p,q+p)$;
(ii)  $\delta_a(m)= ( -m^a + m,\delta m)$, for $m \in M$;
(iii) the action of $P(a)$ on $M$ is given by $n^{(m,p)}= n^p$ for $n \in M, (m,p) \in P(a)$.
In particular $\pi_1(LX,a)$ is isomorphic to Cok $ \delta_a$, and
$\pi_2(LX,a) \cong \pi_2(X,*)^{\bar{a}}$, the elements of
$\pi_2(X,*)$ fixed under the action of $\bar{a}$, the class of $a$
in $G=\pi_1(X,*)$.
A: 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.
