Can the integer Heisenberg group be cubulated?

I've been interested in abstract polyhedral decompositions of 3-manifolds for a long time. One thing I've tried to do a lot is to get nice polyhedral decompositions of manifolds with Nil geometry. It has been difficult, however, to find ones without some kind of irregularity (e.g. the boundary graph contains vertices of valence 2).

Cubulation is a strong from of regularity for a group. Can the integer Heisenberg group be cubulated? If so, is there a simple, exact description of nonpositively curved cube complex with fundamental group quasi-isometric to Nil?

-
Nil group is not hyperbolic; it is not even semihyperbolic. Thus, integer Heisenberg group cannot be cubulated in any meaningful way. – Misha Jul 1 '13 at 17:54
Haglund proved something stronger: a discrete group with a proper action on an arbitrary CAT(0) cube complex has no distorted cyclic subgroups. Thus Heisenberg (which has a quadratically distorted $\mathbf{Z}$) has no such proper action. – YCor Jul 1 '13 at 18:06
Thank you for your answers; in a day or two I will post them as a CW answer to take this off the unanswered questions list, unkess someone else posts an answer. – Brian Rushton Jul 1 '13 at 18:07
PS: about "quasi-isometric to Nil": it follows from Pansu's results that if $G$ is a discrete f.g. group QI to Heisenberg, then it has a finite index subgroup isomorphic to Nil. Thus it also has a distorted $\mathbf{Z}$ and cannot be the fundamental group of a locally CAT(0) cubing, nor even act properly on a CAT(0) cubing. – YCor Jul 1 '13 at 18:08
It's almost completely sorted out which closed aspherical 3-manifolds are cubulated: this should be true if and only if it admits a non-positively curved metric. There remains only certain graph manifolds which are not NPC (Yi Liu proved they are not virtually special cubulated, but the may still act properly and freely on a CAT(0) cube complex). front.math.ucdavis.edu/1110.1940 – Ian Agol Jul 1 '13 at 19:14