Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
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
3  
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. –  Yves Cornulier Jul 1 '13 at 18:06
1  
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
1  
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. –  Yves Cornulier 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
show 6 more comments

1 Answer

up vote 0 down vote accepted

From the comments: 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 Z) has no such proper action.

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 Z and cannot be the fundamental group of a locally CAT(0) cubing, nor even act properly on a CAT(0) cubing.

Another obstruction is the Dehn function, which is cubic. Any CAT(0) group has quadratic Dehn function.

Finally, Nil group is not hyperbolic; it is not even semihyperbolic. Thus, integer Heisenberg group cannot be cubulated in any meaningful way.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.