# Acyclicity of covering space

Suppose we have some 2-dimensional non-aspherical finite CW-complex $K$ with $\pi_1(K)=G$. Is there any sufficient condition on $H\leq G$ (and maybe on the group $G$ itself) which allows to conclude that the covering complex $K_H$ corresponding to the subgroup $H$ has $H_2(K_H) = 0$?

For example, take any $G=\langle x_1,...,x_n|r_1,...,r_m\rangle$ with $\text{cd}(G)>2$ and such that the words $r_1,...,r_m$ become linearly independent in abelianized free group on generators $x_1,...,x_n$. Now, let $K$ be a standard complex associated to the presentation of $G$, then $H_2(K) = 0$. So this gives a lot of examples of such $K$ for the easiest case $H=G$.

Can we get more examples with $H$ a proper subgroup of $G$?

• Well, there are lots of sufficient conditions. For instance, 'generically', $K$ is aspherical and $H$ is free, so $H_2(K_H)$ is certainly zero (but much more is true). On the other hand, the general problem of computing $H_2(K_H)$ (even for aspherical $K$) is algorithmically undecidable. So I don't think you're going to get a useful answer without being more specific about the situation you have in mind. – HJRW Dec 14 '15 at 13:53
• Dear HJRW, thanks for the comment. First of all, I do not understand why 'generically' $K$ is aspherical and $H$ is free. Second, even if $H$ is free I do not understand why in this case $K_H$ should have trivial second homology group. For example, union of several circles and a 2-sphere has free fundamental group but obviously $H_2\neq 0$ in this case. Third, is there any reference explaining in detail that the problem of computing $H_2(K_H)$ is algoritmically undecidable? – Samarkand Dec 15 '15 at 11:30
• However, I am not interested in 'generical' situations. I am seeking examples of non-asperical (and this is very important condition) complexes $K$ such that $H_2(K_H) = 0$ for some proper subgroup $H \subset \pi_1(K)$. This is exactly the situation which I have in mind, no more specificity can be added. – Samarkand Dec 15 '15 at 11:34
• Apologies, I didn't notice the `aspherical' hypothesis in the question. A randomly chosen presentation complex $K$ for a group $G$ (in the few-relators model, say) is small cancellation and hence $K$ is aspherical and $G$ is hyperbolic, and a randomly chosen subgroup of a hyperbolic group is free (the first assertion is well known; the second was I think proved by Arzhantseva). If $K$ is aspherical then $K_H$ is aspherical too, and so $H_2(K_H)\cong H_2(H)=0$. The fact that $H_2(K_H)$ is not algorithmically computable is originally due to Cameron Gordon... – HJRW Dec 15 '15 at 12:41
• ...You could look at arxiv.org/abs/1003.5117 for one version of a proof of undecidability and related things. – HJRW Dec 15 '15 at 12:42