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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 $[0;1]^\mathbb{N}$.

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  • $\begingroup$ 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 $\prod_{n\in\mathbb{N}} [0;2^{-n}]$. $\endgroup$
    – HenrikRüping
    Commented Jan 5, 2012 at 21:02
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    $\begingroup$ 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. $\endgroup$
    – HenrikRüping
    Commented Jan 6, 2012 at 8:26

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – HenrikRüping
    Commented Mar 27, 2012 at 8:02
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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.

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  • $\begingroup$ I think the action is not cocompact, right ? $\endgroup$
    – HenrikRüping
    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

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