The connected components of the free loop space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:50:06Z http://mathoverflow.net/feeds/question/88090 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88090/the-connected-components-of-the-free-loop-space The connected components of the free loop space Thomas Rot 2012-02-10T10:32:30Z 2012-02-10T17:58:33Z <p>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.</p> <p>The following question is related to my question</p> <p><a href="http://mathoverflow.net/questions/34927/are-the-path-components-of-a-loop-space-homotopy-equivalent" rel="nofollow">http://mathoverflow.net/questions/34927/are-the-path-components-of-a-loop-space-homotopy-equivalent</a></p> <p>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.</p> http://mathoverflow.net/questions/88090/the-connected-components-of-the-free-loop-space/88092#88092 Answer by Craig Westerland for The connected components of the free loop space Craig Westerland 2012-02-10T11:50:27Z 2012-02-10T11:50:27Z <p>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.</p> <p>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.</p> <p>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 </p> <p>$$\Lambda M \simeq G^{ad} \times_G EG,$$</p> <p>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. </p> <p>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</p> <p>$$\Lambda M \simeq \coprod_{(g)} (g) \times_G EG.$$</p> <p>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 $g$. </p> <p>In fact, since $M$ was aspherical (i.e., $EG$ was contractible), this is the <em>only</em> homotopy group of this space. We conclude:</p> <p>$$\Lambda M \simeq \coprod_{(g)} K(C(g), 1) = \coprod_{(g)} BC(g).$$</p> <p>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.</p> http://mathoverflow.net/questions/88090/the-connected-components-of-the-free-loop-space/88120#88120 Answer by Ronnie Brown for The connected components of the free loop space Ronnie Brown 2012-02-10T17:41:34Z 2012-02-10T17:58:33Z <p>You might like to look at my preprint <a href="http://arxiv.org/abs/1003.5617" rel="nofollow">http://arxiv.org/abs/1003.5617</a> 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. </p> <p>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. </p> <p>The paper is planned to be revised with Chris Wensley and to include specific computer calculations, hence the delay. </p> <p>I might as well quote the theorem. </p> <p>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</p> <p>(i) $P(a)$ is the group of elements $(m,p)\in M \times P$ such that $\delta m= [a,p]$, with composition $(n,q)+(m,p)= (m+n^p,q+p)$;</p> <p>(ii) $\delta_a(m)= ( -m^a + m,\delta m)$, for $m \in M$;</p> <p>(iii) the action of $P(a)$ on $M$ is given by $n^{(m,p)}= n^p$ for $n \in M, (m,p) \in P(a)$.</p> <p>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 $G=\pi_1(X,*)$.</p>