I have been reading through Wise's lecture notes on cubical complexes, which summarises the proof of the virtual Haken conjecture and the proof that all one-relator groups with torsion are residually finite.

My understanding of this stuff seems to have hit a wall.

Specifically, my understanding of (special) cubical complexes is contradicted by a result of Anatolin-Minasyan (Tit's alternative for graph products), which says that every non-abelian subgroup of a right-angled artin group (RAAG) surjects onto $F_2$ (and so is "very large"). According to my understanding of special cube complexes, every two-generated group which acts on a tree without fixed point should be a subgroup of a RAAG. Clearly, these are incompatible (and my understanding is by far and away the most likely thing to be wrong here!).

I will explain what I understand what it means for a group to be the fundamental group of a (special) cubical complex, and where I think my problem lies. However, if my problem is where I think it is, I do not know how to fill the gap! Any amendments to my understanding would be much appreciated (my question is basically "where have I gone wrong?").

So, cubical complexes were studied by Sageev in his 1994 paper "Ends of group pairs and non-positively curved cube complexes", where he motivated their study as being a generalisation of Bass-Serre theory. A non-positively curved cube complex $X$ correspond to graphs of groups, and the associated $\operatorname{CAT}(0)$ cube complex $\widetilde{X}$ corresponds to the Bass-Serre tree.

I think my problem is simply: what are the $n$-cell stabilisers?

Now, graphs are non-positively curved cube complexes, and their corresponding $\operatorname{CAT}(0)$ cube complex is a tree. Moreover, graphs are special, because their hyperplanes are simply the midpoints of edges. So, the fundamental group of a graph of groups is always special, so should embed into a RAAG. A contradiction.

I hope that all makes sense.

I wonder if my problem is simply that we are not dealing with $\pi_1$ of a graph of groups, but simply of a graph. Then this fits in with the Anatolin-Minasyan result, but not with the theory being a generalisation of Bass-Serre theory (anyway, the group acts on the $\operatorname{CAT}(0)$ cube complex $\widetilde{X}$ and the non-positively curved complex is the quotient by the action $X=\widetilde{X}/G$, so just being the normal fundamental group doesn't make sense).


1 Answer 1


I think your problem is solved if I tell you that the group is supposed to act freely (and specially, if you like) on a CAT(0) cube complex, or equivalently to be the fundamental group of a non-positively curved cube complex. In that sense, the theory only directly generalizes the fundamental groups of graphs, not graphs of groups.

  • $\begingroup$ Hmm - I really didn't pick up on this! (But it certainly makes sense - thanks.) Sageev talks about "essential" actions (which corresponds to not messing up the hyperplanes, and in the finite dimensional case this corresponds to an action with an unbounded orbit) - was this dropped somewhere along the way? $\endgroup$
    – ADL
    Dec 12, 2012 at 17:03
  • $\begingroup$ Not lost, but strengthened! 'Free' is much stronger than 'essential'. If you only assume that the action is 'essential' then the theory does indeed generalize Bass--Serre theory. $\endgroup$
    – HJRW
    Dec 12, 2012 at 17:37
  • $\begingroup$ (cont'd)... The example to bear in mind is an immersed, non-embedded, curve $gamma$ on a surface $\Sigma$. It generates a codimension-one subgroup $\langle\gamma\rangle$ (in the sense that $\langle\gamma\rangle$ coarsely separates $\pi_1\Sigma$), but you can't cut along it, so you can't realize it as the stabilizer of an edge in an action on a tree. But you can realize is as the stabilizer of a hyperplane in an action on a square complex. $\endgroup$
    – HJRW
    Dec 12, 2012 at 17:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .