Let $X$ be a connected CW-complex with $\pi_k(X)$ trivial for $k >2$. Is it known under which circumstances $X$ is an $H$-group?

I have been able only to deduce the necessary condition that $\pi_1(X)$ has to be abelian and act trivially on $\pi_2(X)$. Furthermore, if the necessary condition holds, vanishing of the Postnikov invariant $\beta \in H^3( \pi_1(X), \pi_2(X))$ is a sufficient condition. Thus the interesting case is that of $\beta$ nonzero.

  • 5
    $\begingroup$ A slightly stronger and more natural condition (I don't know if it's equivalent to being an $H$-group in this case) is that $X$ has a delooping $BX$; this means not only that $\pi_1$ is abelian and acts trivially on $\pi_2$ but that the Postnikov invariant must arise from a Postnikov invariant in $H^4(B^2 \pi_1, \pi_2)$. This is exactly the space of quadratic maps $\pi_1 \to \pi_2$, although I don't know how to describe the map to $H^3(B \pi_1, \pi_2)$ in these terms. $\endgroup$ – Qiaochu Yuan Oct 7 '20 at 19:37

Your Answer

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

Browse other questions tagged or ask your own question.