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 $\pi_{k-1}(X)\neq 0$?

  • 1
    $\begingroup$ You should change $\pi_k(X)$ to $\pi_{k-1}(X)$. $\endgroup$ Oct 29 '10 at 12:19

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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