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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.

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(After reading Craig's answer, I realised I might have misunderstood your comment about not being able to concatenate two loops so I've deleted my former comment.) –  Loop Space Feb 10 '12 at 12:03
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2 Answers 2

up vote 15 down vote accepted

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.

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Very nice! Another (perhaps more elementary) calculation of the fundamental groups of the components of the loop space is given in: Hansen, V L, On the fundamental group of a mapping space. An example. Compositio Math. 28 (1974), 33–36. –  Mark Grant Feb 10 '12 at 12:48
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Thanks! By the way, this argument points out a route to the original problem posed (for computing the homology of these components) -- they are the group homology of these centralizers. –  Craig Westerland Feb 10 '12 at 14:02
    
Thanks Mark for the reference to Hansen's paper; see my answer below. –  Ronnie Brown Feb 10 '12 at 17:47
    
@craig Westerland: Thank you very much for your answer. Do you know any reference where these computations are actually in detail? –  Thomas Rot Feb 12 '12 at 15:04
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Thomas, this thread: mathoverflow.net/questions/20671/… gives some references that could be helpful. –  Craig Westerland Feb 13 '12 at 5:42
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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 $G=\pi_1(X,*)$.

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I've remembered the paper Ellis, G.J. Homology of 2-types, J. London Math. Soc. (2) 46 (1992), no. 1, 1–-27. which was followed by a correction in the same journal (2) 52 (1995), no. 3, 447–448. –  Ronnie Brown Feb 10 '12 at 22:21
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