1
$\begingroup$

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 $G$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – HJRW Dec 14 '15 at 13:53
  • $\begingroup$ 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? $\endgroup$ – Samarkand Dec 15 '15 at 11:30
  • $\begingroup$ 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. $\endgroup$ – Samarkand Dec 15 '15 at 11:34
  • $\begingroup$ 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... $\endgroup$ – HJRW Dec 15 '15 at 12:41
  • $\begingroup$ ...You could look at arxiv.org/abs/1003.5117 for one version of a proof of undecidability and related things. $\endgroup$ – HJRW Dec 15 '15 at 12:42

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.