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Agol's recent VHC paper gave a characterization of virtually special groups in terms of being $\mathcal{QVH}$. He remarks that this may be taken as the defining property of virtually special groups which can be used in this paper (and so presumably, in the proof of the the main theorem).

My concern: Recall that the goal is to show that given a cube complex $X$ and hyperbolic group $G$ (satisfying the hypotheses of the theorem) there exists a finite index subgroup $G'$ such that $X/G'$ is special. Now if one was to just show that $G$ is in $\mathcal{QVH}$ (and hence virtually special), then we would know that it acts virtually specially on some cube complex $Y$. But a priori this cube complex may not be the same as $X$.

Even though it may not be necessary for the VHC itself, it seems like we still need the full strength of the original result (i.e that we can take $X$ to be $Y$) since the proof is by induction.

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  • $\begingroup$ Your question is quite focused, you should include some context and explain or (or give a reference if too technical) for VHC, QVH, special, etc. $\endgroup$
    – YCor
    Jan 31, 2013 at 12:18

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Being virtually special is actually a group-theoretic property, independent of the cube complex (at least in the word-hyperbolic case of interest to Agol). More precisely, Haglund and Wise, in their seminal GAFA paper 'Special cube complexes', proved the following.

Theorem: Let $X$ be a finite cube complex with $\pi_1X$ word-hyperbolic. Then $X$ is virtually special if and only of every quasiconvex subgroup of $\pi_1X$ is separable.

(Recall that separable means 'closed in the profinite topology' or, more simply, 'is an intersection of finite-index subgroups'.)

The 'only if' direction is Theorem 7.3, and the 'if' direction Theorem 8.13, in Haglund--Wise's paper.

The upshot of this is that you can change cube complexes without worrying.

Proof outline

The idea of the proof of is quite straightforward (though there are some technical details). Let me quickly outline it here.

For the 'only if' direction, because $X$ is virtually special, any quasiconvex subgroup $H$ is also a quasiconvex subgroup of a right-angled Coxeter group. This is then separable by an argument that goes back to Scott's 1979 paper 'Subgroups of surface groups are almost geometric', worked out in the general right-angled Coxeter case by Haglund in his paper 'Finite-index subgroups of graph products'.

For the 'if' direction, note that hyperplane stabilizers are quasiconvex, and therefore separable by hypothesis. We need to eliminate certain pathologies: self-intersection, self-osculation etc. It follows from separability that, for each hyperplane $Y$, there is a finite-sheeted cover of $X$ in which the unit-cube neighbourhood of $Y$ lifts to an embedded, trivial interval bundle over $Y$. After doing this for all hyperplanes, there is no self-intersection and no self-osculation.

Inter-osculation is similar, but as two hyperplanes are involved, we need to use separability of double cosets of hyperplanes. The easiest thing to do here is to quote a result of Minasyan (in 'Separable subsets of GFERF negatively curved groups'), who proved that if a word-hyperbolic group has the property that every quasiconvex subgroup is separable, then it follows that all double cosets of quasiconvex subgroups are also separable. So we have the separability of double cosets that we require.

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  • $\begingroup$ Thanks! I was initially reluctant to work in terms of fundamental groups since the actions on a cube complex we're interested in may not be free. However virtually special groups have finite index subgroups which embed in RAAGs (which are torsion free) so of course this goes through. $\endgroup$ Feb 2, 2013 at 6:15
  • $\begingroup$ No, you certainly need the action to be free. Others have made this mistake too - evidently it's not made clear enough in the literature. See this MO question: mathoverflow.net/questions/116178/… . $\endgroup$
    – HJRW
    Feb 2, 2013 at 8:06
  • $\begingroup$ Let me put it another way. I call a group $G$ 'special' if it is the fundamental group of a special cube complex. In particular, $G$ acts freely etc on the universal cover. $\endgroup$
    – HJRW
    Feb 2, 2013 at 13:18
  • $\begingroup$ Of course, we want free actions when we want to work in terms of fundamental groups. My point was that the hypothesis in Agol's main theorem of the paper does NOT require the action of $G$ on $X$ to be free (the finite index subgroup $G'$ such that $X/G'$ is a special cube complex will be torsion free and hence act freely on $X$). $\endgroup$ Feb 3, 2013 at 3:43
  • $\begingroup$ Ah, I see what your worry was now. As you say, the point is that both conditions (virtually special and separable quasiconvex subgroups) imply virtually torsion-free. $\endgroup$
    – HJRW
    Feb 3, 2013 at 8:20

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