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let $M$ be a compact, simply connected Riemannian manifold with dimension $< \infty$. I'm looking for a reference that $$ \dim H_k(\Lambda M, \mathbb{Z}) < \infty, $$

is true in that case. Here $$\Lambda M= \{c :S^1 \rightarrow M| c\text{ is square summable and absolutely continous}\}$$ is the free loop space.

Thanks,

Mick

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  • $\begingroup$ Is $k$ fixed here? $\endgroup$
    – Mark Grant
    Commented Sep 16, 2015 at 10:42
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    $\begingroup$ This can be proved using the Serre spectral sequences for the fibrations $\Omega M\to\Lambda M\to M$ and $\Omega M\to PM\to M$, but I do not know where this is spelled out. $\endgroup$
    – Neil Strickland
    Commented Sep 16, 2015 at 10:58
  • $\begingroup$ No, it isn't fixed. $\endgroup$
    – The-Mick
    Commented Sep 16, 2015 at 11:49

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As Neil Strickland points out, for a fixed $k$ the Betti number of the free loop space is indeed finite. However it is worth pointing out that the sequence of Betti numbers $b_k(\Lambda M)$ is unbounded whenever the rational cohomology algebra of $M$ requires at least two generators. This is the main result of

D. Sullivan, M. Vigue-Poirrier, The homology theory of the closed geodesic problem. J. Differential Geometry 11 (1976), no. 4, 633–644,

and was an early application of Sullivan's theory of minimal models.

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    $\begingroup$ Concerning the growth of Betti numbers of free loop spaces, it is worth mentionning also the papers of Pascal Lambrechts On the Betti numbers of the free loop space of a coformal space (JPAA 2001) and The Betti numbers of the free loop space of a connected sum ( J. London Math. Soc. 2001). $\endgroup$
    – Sinan Yalin
    Commented Sep 16, 2015 at 12:53
  • $\begingroup$ Hi Mark, can I ask a question - does this finiteness of Betti number hold only for simply connected manifolds? How about non-simply connected ones? $\endgroup$
    – user188722
    Commented Feb 27, 2022 at 23:15
  • $\begingroup$ In the non-simply connected case you can easily have infinite $0$-th and $1$-st Betti numbers, e.g. if $\pi_1(M)=\mathbb{Z}$ then $\Lambda M$ has infinitely many path components. Not sure about the higher Betti numbers. $\endgroup$
    – Mark Grant
    Commented Feb 28, 2022 at 7:47

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