Usually a CAT(0) group is defined to be a group acting properly isometrically and cocompactly on a CAT(0) space, but I would like to consider only those groups that act properly, isometrically and cocompactly on a finite-dimensional CAT(0) space.

So is there a group I have to leave out?

Not every CAT(0) space with a proper isometric cocompact group action is finite-dimensional. For example the trivial group acts on the compact CAT(0)-space $[0;1]^\mathbb{N}$.

  • $\begingroup$ What CAT(0) metric are you taking on $[0,1]^{\mathbb{N}}$? $\endgroup$
    – Ian Agol
    Commented Jan 5, 2012 at 18:22
  • $\begingroup$ For example $d(x,y):=\sum_{n\in \mathbb{N}} 2^{-n} |x_n-y_n|$, so maybe I should have written $\prod_{n\in\mathbb{N}} [0;2^{-n}]$. $\endgroup$ Commented Jan 5, 2012 at 21:02
  • 1
    $\begingroup$ I think you want the $l^2$ instead of $l^1$ metric on $\prod [0,2^{-n}]$, since the $l^1$ metric is not CAT(0) (e.g. there are not unique geodesics). $\endgroup$
    – Ian Agol
    Commented Jan 5, 2012 at 23:20
  • $\begingroup$ @Agol: oh of course. $\endgroup$ Commented Jan 6, 2012 at 8:26

2 Answers 2


First, I suppose that by proper action you mean the one in the sense of Bridson and Haefliger, otherwise you would have to regard ${\mathbb R}$ as a $CAT(0)$ groups. Now, it follows from Eric Swenson's paper "A cut point theorem for CAT(0) groups" (Journal of Diff. Geometry, 1999) that the ideal boundary of the $CAT(0)$ space $X$ (on which a group $G$ acts geometrically) is finite-dimensional. This suffices for many practical purposes. For instance, it follows (from Bestvina's work) that $G$ has finite cohomological dimension over ${\mathbb Q}$ and, if you consider torsion-free groups, over ${\mathbb Z}$ as well. (This immediately excludes Thompson's group, etc.) In particular, geometric dimension of $G$ is finite, $G$ has finite type, etc. From this you can make pretty much the same algebraic conclusions about $G$ as in the case when $G$ acts geometrically on a finite-dimensional $CAT(0)$ space. Thus, in the torsion-free case, I do not think you are missing (or gaining) much by restricting to finite-dimensional $CAT(0)$ spaces. (For instance, I do not see how assuming finite dimension of $X$ would help with proving that $G$ has finite asymptotic dimension.)

I am not sure what happens in the case of groups with torsion: It is conjectured by Swenson that a $CAT(0)$ group $G$ cannot contain infinite torsion subgroups. Maybe it would be easier to exclude some infinite torsion subgroups (say, the infinite permutation group) using the assumption that $G$ acts geometrically on a finite-dimensional $CAT(0)$ space, but I do not see how.

Swenson's work had a follow-up paper by Geoghegan and Ontaneda http://arxiv.org/abs/math/0407506 where they weaken some of his assumptions and strengthen some of his conclusions.

Note: In view of Swenson's result it is tempting to say: Take the closed convex hull (in $X$) of the ideal boundary of the $CAT(0)$ space $X$ and show that it is finite-dimensional. It might work, but, in general, convex hulls in $CAT(0)$ spaces tend to be much bigger than expected.

  • $\begingroup$ concerning the last paragraph: There might be many geodesics from a boundary point to another, if one takes all the space might be too big (example $\mathbb{R}\times I^\infty$) so one has to make a suitable choice. Maybe it works for CAT(-1) spaces. Furthermore if it works we would have found a finite dim. convex set containing the orbit of a point. That was another approach discussed here, but it was not clear how to find such a point. $\endgroup$ Commented Mar 27, 2012 at 8:02

I think you will miss the notorious Thompson's group $F$: it acts properly isometrically on a $CAT(0)$ cube complex (see Dan Farley, http://www.users.muohio.edu/farleyds/Far1.pdf); but it cannot act properly isometrically on a finite-dimensional complex, as this would make it of finite cohomological dimension.

EDIT: Indeed as Mark pointed out, the action of $F$ on this $CAT(0)$-cube complex is NOT co-compact, hence my answer should be discarded. Don't trust MO too much.

  • $\begingroup$ I think the action is not cocompact, right ? $\endgroup$ Commented Jan 5, 2012 at 15:24
  • $\begingroup$ That action is not co-compact. $\endgroup$
    – user6976
    Commented Jan 5, 2012 at 15:33
  • $\begingroup$ An action on a cube complex that is not finite-dimensional can never be cocompact. The cube complex that Farley uses is not finite-dimensional. $\endgroup$
    – IJL
    Commented May 16, 2018 at 10:31

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.