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CAT(0) groups are groups that act on a CAT(0) space properly and cocompactly. If a group acts on a CAT(0) cubical complex properly and cocompactly, then of course it is a CAT(0) Group. I am wondering the other direction,

Quesion: Given a CAT(0) group G , will one always be able to find a CAT(0) cubical complex such that G acts properly and cocompactly.

I assume the answer is no, but I could not find any counter example by google. If the answer is indeed no, one can further ask does there exists a CAT(0) group that can not act on a CAT(0) cubical complex properly.

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    $\begingroup$ Groups with property (T) do not act on CAT(0) cube complexes, so any lattice in a simple Lie group of higher rank (or $Sp(n,1)$) is an example. $\endgroup$ Sep 26, 2014 at 13:06
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    $\begingroup$ Small correction: any uniform lattice (since the action is required to be cocompact). $\endgroup$
    – HJRW
    Sep 26, 2014 at 16:07
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    $\begingroup$ A more difficult class of examples comes from Kahler groups. (Say, uniform lattices in PU(n,1), $n\ge 2$, cannot act discretely on CAT(0) cube complexes.) This is due to Pierre Py. $\endgroup$
    – Misha
    Sep 26, 2014 at 18:00
  • $\begingroup$ @Misha don't you mean "don't act geometrically"? I don't think Property PW (or its failure) is known for lattices in $PU(n\ge 2,1)$. $\endgroup$
    – YCor
    Aug 18, 2019 at 13:01
  • $\begingroup$ @YCor: Yves, in their paper "Cubulable Kahler groups" Delzant and Py prove a theorem that does not require cocompactness. It suffices to assume (in addition to proper discontinuity) that there is no fixed point at infinity, is essential and does not preserve a flat factor. $\endgroup$
    – Misha
    Aug 18, 2019 at 22:34

1 Answer 1

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Many CAT(0) groups cannot act geometrically on CAT(0) cube complexes. For instance:

  • CAT(0) groups satisfying Kazhdan's property (T), eg. uniform lattices in simple Lie groups of higher rank or in quaternionic hyperbolic spaces (as mentioned by Jean Raimbault in the comments).
  • Some Kähler groups (as mentioned by Misha in the comments), including uniform lattices in complex hyperbolic spaces.
  • Some crystallographic groups, as proved by Mark Hagen.
  • Some Coxeter groups (see Niblo and Reeves' article for more details).
  • Some braid groups, eg. $B_n$ for $n=4,5,6$ (see Haettel's article Virtually cocompactly cubulated Artin-Tits groups and references therein).
  • The group $\mathrm{Aut}(B_4) \simeq \mathrm{Aut}(\mathbb{F}_2)$ (see Piggott, Ruane and Walsh's article The automorphism group of the free group of rank two is a CAT(0) group).

Probably, the easiest example of such a group is the triangle group $$T=\langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(bc)^3=(ac)^3=1 \rangle.$$ It coincides with the symmetry group of the regular tilling of the Euclidean plane by equilateral triangles. So it acts geometrically on the Euclidean plane: it is a CAT(0) group.

As observed by Wise, if a virtually $\mathbb{Z}^n$ group $G$ acts geometrically on a CAT(0) cube complex, then it acts geometrically on $\mathbb{E}^n$ endowed with its usual cubulation. (See Lemma 16.12 in his long paper The structure of groups with a quasiconvex hierarchy. Essentially, the idea is the following: apply the flat torus theorem to find a $G$-invariant flat $\mathbb{R}^n$ inside your cube complex $C$, and look at the CAT(0) cube complex obtained by cubulating the wallspace defined by the intersections of the hyperplanes of $C$ with your flat.) Consequently, if $T$ were cubulable then it would act geometrically on the square complex $\mathbb{E}^2$.

Notice that $\mathrm{Isom}(\mathbb{E}^2)= (D_{\infty} \times D_{\infty}) \rtimes \mathbb{Z}_2$ does not contain elements of order three, so, for any action $T \curvearrowright \mathbb{E}^2$ by isometries, the elements $ab$, $bc$ and $ac$ must be trivial. Consequently, such an action must factorise through the quotient $T \twoheadrightarrow \mathbb{Z}_2$ sending all the generators to $1$. A fortiori, the action cannot be geometric (and even proper).

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