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.



  • $\begingroup$ Is $k$ fixed here? $\endgroup$
    – Mark Grant
    Sep 16, 2015 at 10:42
  • 3
    $\begingroup$ 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. $\endgroup$ Sep 16, 2015 at 10:58
  • $\begingroup$ No, it isn't fixed. $\endgroup$
    – The-Mick
    Sep 16, 2015 at 11:49

1 Answer 1


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.

  • 1
    $\begingroup$ 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). $\endgroup$ 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
    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
    Feb 28, 2022 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.