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).

uniformlattice (since the action is required to be cocompact). $\endgroup$ – HJRW Sep 26 '14 at 16:07