By rational homotopy theory, $H(\Lambda M; \mathbb{Q})$ is infinite-dimensional over $\mathbb{Q}$ if $M$ is simply-connected. Are there (non-simply-connected) examples when $H(\Lambda M; \mathbb{Q})$ is finite-dimensional? I am most interested when $M$ is a manifold. Also, note that $H_0(\Lambda M; \mathbb{Q})$ is the set of conjugacy classes of $\pi_1(M)$, ie. loops up to homotopy.
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1Your title says "based loop space", but usually $\Lambda M$ is the space of free loops (i.e. unbased). Which is it...? – Najib Idrissi Feb 22 '17 at 09:43
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@NajibIdrissi Thanks for pointing this out. It should be free loop space; I fixed it. – user39598 Feb 23 '17 at 00:45
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Take a nontrivial (say, infinite) group $G$ which has exactly two conjugacy classes (e.g. one given by the HNN construction). Now, take $X=K(G,1)$. By working harder one can construct examples of such groups which have finite cohomological dimension and, hence, are fundamental groups of (noncompact) aspherical manifolds.
Edit. In fact, if you apply the HNN construction to the infinite cyclic group $G_0$, the result will an (infinitely generated) countable group $G$ of cohomological dimension 2 with exactly two conjugacy classes (the presentation complex $X$ of $G$ will be 2-dimensional and aspherical). Hence, by Stallings embedding theorem, there exists a 4-dimensional aspherical manifold $M=K(G,1)$ (obtained by embedding $X'$ homotopy-equivalent to $X$ into $R^4$ and then taking a regular neighborhood there).
Lastly, there are no known examples of infinite finitely presented groups with finitely many conjugacy classes.
Moishe Kohan
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Do you have examples of closed (compact) manifolds with the property tha $H(\Lambda M)$ is finite-dimensional? In this case, $\pi_1(M)$ is finitely-generated. – user39598 Apr 28 '17 at 07:36
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@user39598: As far as I know, there are no known examples of infinite finitely presented groups with finitely many conjugacy classes. Finitely generated examples were only relatively recently constructed by Osin (Annals of Math. 2010). – Moishe Kohan Apr 28 '17 at 14:13
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I looked at the Osin article but I'm not sure that it applies to manifolds. I am interested in a closed manifold such that $H(\Lambda M)$ is finite-dimensional. One potential example is when $M$ is hyperbolic and $\pi_1(M)$ has finitely many conjugacy classes (but $\pi_1(M)$ is itself infinite). The examples from Osin are finitely-generated groups with finitely many conjugacy classes but I'm not sure that these groups are the fundamental groups of closed hyperbolic manifolds - for example, $\mathbb{Z}/2$ is not since it is not infinite. – user39598 Apr 30 '17 at 04:59
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@user39598 Infinite hyperbolic groups have infinitely many conjugacy classes. To get a compact manifold one needs a finitely presented group, but such infinite groups are unknown. – Moishe Kohan Apr 30 '17 at 05:02
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@MoiseCohen Thank you. So just to clarify, is my question still unknown? Whether there are closed manifold such that $H(\Lambda M)$ is finite-dimensional? – user39598 Apr 30 '17 at 05:07
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@user39598: If you assume that the manifold is connected with infinite fundamental group then yes, unknown. – Moishe Kohan Apr 30 '17 at 11:54
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I don't understand these two sentences in your reply: "If $\pi_1(M)$ is finite then $H_∗(\Lambda M;\mathbb{Q})$ will be finite dimensional since it embeds in the rational cohomology ring of the free loop space of the universal cover of M." This is false since if M is simply-connected, then $H_∗(\Lambda M;\mathbb{Q})$ is always infinite-dimensional. Also you say " Lastly, there are no known examples of finitely presented groups with finitely many conjugacy classes." What about any abelian group? I think you mean, infinite groups that are finitely presented. – user39598 May 04 '17 at 01:37