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Throughout, let $X$ be a connected finite CW-complex. If the universal covering of $X$ is contractible, then $\pi_n(X)=0$ for all $n \geq 2$. In this case $X$ is a model for $B\pi_1(X)$.

I am wondering whether this is the only reason why higher homotopy groups vanish above a certain degree. More precisely:

Question: Let $k \geq 3$ be an integer. Can it happen that $\pi_n(X) = 0$ for all $n \geq k$ and $\pi_{k-1}(X)\neq 0$?

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No, it cannot happen. In a paper by McGibbon and Neisendorfer, it is proven that if X is a 1-connected space and its mod-p-homology is non-zero in some degree, but zero in all higher degrees, then the $\pi_n X$ contain a subgroup of order p for infinitely many n. This can be applied to the universal cover of your X.

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    $\begingroup$ This shows that for a finite dimensional simply connected space with finitely many homotopy groups, those groups are vector spaces over $\mathbb Q$. Since there are finitely presented groups containing $\mathbb Q$, my guess is that this can actually happen. $\endgroup$
    – Ben Wieland
    Commented Sep 20, 2022 at 17:32
  • $\begingroup$ @BenWieland If $X$ is simply-connected and all its homotopy groups are $\mathbb{Q}$-vector spaces, then also all of its reduced homology groups are $\mathbb{Q}$-vector spaces (see e.g. Theorem 9.3 in Rational Homotopy Theory by Felix, Halperin and Thomas). For a finite complex this cannot happen. If you only require finite-dimensional, I don't know, however. $\endgroup$
    – Lennart Meier
    Commented Jun 9, 2023 at 15:02

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