Suppose I have a space $X$ which is connected, simply connected, CAT(0) of dimension 2 and a group $G$ which acts on $X$ freely, isometrically, properly discontinuously and cocompactly. What can be said about the group $G$? I would be very interested in group-theoretic properties which follow in this situation. Since there are so few groups acting properly and cocompactly on $\mathbb{R}^2$, I would expect that the groups in my situation above are also fairly special. For instance, and to be slightly more specific, is there anything that can be said about nilpotent quotients or the lower central series of such a group? Probably too much to hope for, but could there be a classification result of some sort for such groups?

  • $\begingroup$ Hi Matthias, I think that it is also worth mentioning that that 2D CAT(0) groups form a class of groups of the type considered in this paper: arxiv.org/pdf/1307.2640.pdf $\endgroup$
    – NWMT
    Commented Jun 27, 2016 at 20:24

1 Answer 1


A very wide array of properties are compatible with these hypotheses. Burger--Mozes famously gave examples of infinite simple groups of this form. Earlier, Wise and Bhattacharjee had independently given examples of such groups with no proper finite quotients. In particular, these groups have no non-trivial nilpotent quotients.

Let's call such an example $S$. Bridson and I used $S$ to prove the following result (see Proposition 7.5 of arXiv:1401.2273).

Proposition 1: Given any finitely presented group $G$ there exists a compact, non-positively curved square complex $X$ with a surjection

$\eta:\pi_1X\to G$

such that $\eta$ induces an isomorphism on profinite completions.

Sketch proof: Take a finite presentation complex $Y$ for $G$. Replace each 2-cell of $Y$ with an annulus, and glue the other end of the annuli to npc square complexes with fundamental groups $S$. If this is done carefully, the resulting complex $X$ is a non-positively curved square complex. The map $\pi_1X\to G$ is obtained by killing the copies of $S$ inside $\pi_1X$. QED

Notice that, since $S$ is also has no nilpotent quotients, one obtains in the same way:

Proposition 2: Given any finitely presented group $G$ there exists a compact, non-positively curved square complex $X$ with a surjection

$\eta:\pi_1X\to G$

such that $\eta$ induces an isomorphism on pro-nilpotent completions.

[That is, every nilpotent quotient of $\pi_1X$ factors through $\eta$.]

The group $\pi_1X$ acting on the universal cover $\widetilde{X}$ satisfies the hypotheses of the question, and the conclusion is that the nilpotent quotients of such groups are exactly as general as the nilpotent quotients of an arbitrary finitely presented group.

So nothing can be said about the nilpotent quotients of such a group without some additional information.

  • $\begingroup$ The groups you mention are a bit pathological but the spaces on which they act are not. So I wouldn't conclude that the question is hopeless in its generality, at least just from these examples. $\endgroup$
    – YCor
    Commented Jun 12, 2016 at 22:20
  • 1
    $\begingroup$ @YCor, I didn't really understand your objection, but anyway I've now edited to give a much more precise result. $\endgroup$
    – HJRW
    Commented Jun 13, 2016 at 8:39

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