Throughout, let $X$ be a connected finite CW-complex.

Question: If $X$ is of dimension $n$. Is there some integer $n'$ (maybe depending only on $n$), such that all homotopy groups $\pi_k(X)$ for $k \geq n'$ are finite?

For the spheres $S^n$, $n'=2n+1$ works by Freudenthal's Suspension Theorem and Serre's result that the stable homotopy groups in that range are finite. More generally, if $\pi_1(X)=0$, then the Milnor-Moore theorem relates the rational homotopy groups to the rational homology of the loop space of $X$ and I believe that this can be used to get a similar conclusion. But what if $\pi_1(X) \neq 0$?

EDIT: Igor Belegradek (besides answering the question) pointed out that what I stated in the last three lines is not correct.

  • $\begingroup$ Take any CW complex whose 1-connected cover is a sphere... $\endgroup$ – David Roberts Oct 29 '10 at 10:26
  • $\begingroup$ @David: I could take a sphere. But probably, you had something interesting in mind. What is the relation to the question? $\endgroup$ – Andreas Thom Oct 29 '10 at 10:44
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    $\begingroup$ A direct example: rationally, the homotopy groups of $S^3 \vee S^3$ form the free (shifted) Lie algebra on two generators under the Whitehead product, and this is nontrivial in every odd degree greater than or equal to 3. $\endgroup$ – Tyler Lawson Oct 29 '10 at 11:11
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    $\begingroup$ I do not understand the edit. I think my answer is not about the last 3 lines; it is about the highlighted question. Specifically any rationally hyperbolic gives a "no" answer to the highlighed question. $\endgroup$ – Igor Belegradek Oct 29 '10 at 13:23
  • $\begingroup$ You are right. But in addition, I claimed something for the simply-connected case which is wrong. $\endgroup$ – Andreas Thom Oct 29 '10 at 13:47

The answer is no in a very strong way even for simply-connected complexes. In rational homotopy theory there is a famous dichotomy between elliptic and hyperbolic spaces: a simply-connected finite complex is either elliptic or hyperbolic. Elliptic means that all but finitely many homotopy groups are finite. Hyperbolic means that the sum of ranks of first $k$ homotopy groups grows exponentially with $k$. In some sense most spaces are hyperbolic. If I remember correctly, $m$-fold connected sum of $S^2\times S^2$ with itself is hyperbolic if $m>1$. You can read more of this in the book "Rational homotopy theory" by Felix-Halperin-Thomas.

  • $\begingroup$ Thanks for the answer. Do you prove this dichotomy using the Milnor-Moore theorem and say the Serre spectral sequence? $\endgroup$ – Andreas Thom Oct 29 '10 at 10:45
  • $\begingroup$ I have not been through the proof of the dichotomy, but my impression was that it uses more delicate and modern ideas. It first appeared in early 80s in [Félix, Yves; Halperin, Stephen; Thomas, Jean-Claude The homotopy Lie algebra for finite complexes. Inst. Hautes Études Sci. Publ. Math. No. 56 (1982), 179–202 (1983).] $\endgroup$ – Igor Belegradek Oct 29 '10 at 13:19
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    $\begingroup$ You can discover the existence of rationally hyperbolic spaces by yourself. Try to do a rational homotopy calculation of a manifold that it not a homogeneous space, say hypersurfaces in $CP^3$, where you know everything about the real cohomology ring. Moreover, these space are simply-connected and formal; so it looks like an easy exercise. After you filled several pages with generators and differentials, you begin to realize that the computation will not close off as elegantly as in the case of $CP^n$ or $S^n$. As a grad student, I spent a long night doing that -:)) $\endgroup$ – Johannes Ebert Oct 29 '10 at 14:31

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