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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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You should change $\pi_k(X)$ to $\pi_{k-1}(X)$. – Ćukasz Grabowski Oct 29 2010 at 12:19
Thanks, I corrected it. – Andreas Thom Oct 29 2010 at 12:30

1 Answer

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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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Thanks. Very nice. – Andreas Thom Oct 29 2010 at 12:32

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