10
$\begingroup$

Suppose $G$ has a finite index subgroup $N$ such that $N$ acts properly and cocompactly on a CAT(0)-cube complex. Does $G$ also act properly and cocompactly on a CAT(0)-cube complex?

Edit: After searching the web a little I found that the answer to this question is no, in general. There exists for example a 3-dimensional torsion-free crystallographic group that does not act freely and cocompactly on a CAT(0)-cube complex. (see Example 16.11 in 'THE STRUCTURE OF GROUPS WITH A QUASICONVEX HIERARCHY', by Dani Wise).

So my new question becomes: what conditions can one impose on $G$, such that the answer to the original question is yes? For example, what if $G$ is hyperbolic (as suggested by HW, in comments below)?

$\endgroup$
5
  • $\begingroup$ You probably already know that one gets an induced action of $G$ on the direct product of $|G:N|$ copies of the cube complex. The difficulty is to find a convex subcomplex on which $G$ acts cocompactly. In general, of course, this doesn't exist (consider, for instance, $2\mathbb{Z}\subseteq\mathbb{Z}$. Probably it's OK when $G$ is word-hyperbolic. $\endgroup$
    – HJRW
    Apr 16, 2013 at 12:29
  • 3
    $\begingroup$ Specifically, I think you can deduce the word-hyperbolic case from Sageev's theorem (see, for instance, Theorem 7.1 of arXiv:1209.1074v2). $\endgroup$
    – HJRW
    Apr 16, 2013 at 12:56
  • $\begingroup$ HW: I think, you are right about the hyperbolic case, it simply a corollary of Sageev's theorem. In the general case, I think, affine Coxeter group $\tilde{A}_2$ should give a counter-example. $\endgroup$
    – Misha
    Apr 16, 2013 at 17:46
  • $\begingroup$ HW: why not post the hyperbolic case as an answer? $\endgroup$
    – Dieter
    Apr 17, 2013 at 7:26
  • $\begingroup$ Dieter: I will when I have time. $\endgroup$
    – HJRW
    Apr 17, 2013 at 11:00

2 Answers 2

9
$\begingroup$

The answer is 'yes' when your group $G$ is word-hyperbolic. This can be deduced from Sageev's theorem. I'll explain this here, but a good reference is Hruska--Wise's paper 'Finiteness properties of cubulated groups' (arXiv:1209.1074v2).

The first step is to notice that $G$ admits a proper, but not cocompact, action on a cube complex.

Lemma: Suppose $|G:N|<\infty$ and $N$ acts properly on a CAT(0) cube complex $X$. Then $G$ acts properly on $X^{|G:N|}$.

The proof of this is the usual messing around with the induced representation, so I'll leave it as an exercise. The one extra fact about this action that we will need is that the new hyperplane stabilizers are precisely the $G$-conjugates of the hyperplane stabilizers in $N$.

When $G$ is word-hyperbolic, the Schwarz--Milnor Lemma implies that $X$ is Gromov-hyperbolic. We also have the following facts about hyperplane stabilizers in $G$.

  1. Hyperplane stabilizers in $N$ are codimension-one subgroups. Since this is a coarse property, the same is true of hyperplane stabilizers in $G$.

  2. Hyperplanes in $X$ are convex. Since quasigeodesics in Gromov-hyperbolic spaces are uniformly close to geodesics, it follows that hyperplane stabilizers are quasiconvex in $N$, and hence in $G$.

Associated to any finite collection $\lbrace H_i\rbrace$ of codimension-one subgroups in a group $G$, Sageev constructed a CAT(0) cube complex on which $G$ acts by isometries. In this case, we will take $\{H_i\}$ to be a set of $G$-conjugacy representatives for the hyperplane stabilizers in $G$.

Sageev's theorem: If the $\{H_i\}$ are all quasiconvex then $G$ acts cocompactly on the associated cube complex.

By 1 and 2 above, Sageev's theorem applies. It remains to prove that $G$ also acts properly. This is a matter of making sure that we have chosen a large enough collection of hyperplane stabilizers.

The Hruska--Wise paper contains some useful criteria for checking that the action is proper. Roughly speaking, their Theorem 5.4 says that, as long as the axis of every infinite-order element crosses some hyperplane, the action is proper. But, indeed, if $g\in G$ has infinite order then it shares an axis with $g^{|G:N|}\in N$, and this certainly crosses a hyperplane, since the action of $N$ on $X$ was proper.

This completes the proof in the word-hyperbolic case (modulo filling in some details).

Finally, I'll just mention that much of the purpose of the Hruska--Wise paper is to generalize these ideas to the relatively hyperbolic setting. In particular, something similar should be true there, subject to suitable restrictions on the action of the parabolic subgroups.

$\endgroup$
4
  • $\begingroup$ I think you can just use the hyperplane stabilizers of the cubulation of $N$ to cubulate $G$ a la Bergeron-Wise. $\endgroup$
    – Ian Agol
    Apr 19, 2013 at 16:56
  • $\begingroup$ I guess I mean that the Hruska-Wise result is not new (2012) for hyperbolic groups, but is probably new in the relatively hyperbolic setting. $\endgroup$
    – Ian Agol
    Apr 19, 2013 at 16:59
  • $\begingroup$ Ian - I agree with both your comments. As I say, the $H_i$ that we use are precisely the hyperplane stabilizers in $N$. And I didn't mean to suggest that you need to use the Hruska--Wise result, only that it's a convenient statement to apply. I suppose, strictly speaking, you could do it without starting with the induced action on $X^{|G:N|}$, but I find that a helpful way of thinking about it. $\endgroup$
    – HJRW
    Apr 19, 2013 at 19:28
  • $\begingroup$ n the final paragraph, I point out that the main purpose of the Hruska--Wise paper is to develop these ideas in the relatively hyperbolic setting. That said, I think the paper was around in one form or another a lot earlier than 2012. Indeed, it's cited (as in preparation) in Bergeron--Wise. $\endgroup$
    – HJRW
    Apr 19, 2013 at 19:30
4
$\begingroup$

A paper of Caprace-Muhlherr (pdf) characterizes the Coxeter groups that act cocompactly on the CAT(0) cube complex associated to the group by Niblo-Reeves. It seems possible that this gives a characterization of the Coxeter groups which may act cocompactly on any CAT(0) cube complex, although I think this may still be open.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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