# Quasi-isometric rigidity of Nil

Let $Nil$ be the unique simply connected non-abelian three-dimensional nilpotent Lie group, i.e. the group of upper triangular matrices with all the eigenvalues equal to 1 (this group is also known as the 3-dimensional Heisenberg group), endowed with a left-invariant Riemannian metric.

The following statement can be found in the literature:

Any finitely generated group quasi-isometric to $Nil$ is in fact virtually isomorphic to a uniform lattice in $Nil$.

Recall that two groups are virtually isomorphic if they can be obtained one from the other via a finite number of steps, where each step consists in taking a finite extension or extracting a finite-index subgroup (or viceversa). The fact that virtually isomorphic groups are quasi-isometric is elementary, and any uniform lattice is quasi-isometric to its ambient Lie group as a consequence of Milnor-Svarc Lemma.

For example, the statement above is attributed to Gromov, Asymptotic invariants of infinite groups'' (1993), in the paper

Quasi-isometries preserve the geometric decomposition of Haken manifolds'', Kapovich and Leeb, Inventiones Mathematicae (1997) and it is also stated in

 Quasi-isometric classification of graph manifolds groups'', Behrstock and Neumann, Duke Math. J. (2008).

I tried to reconstruct the proof of the quasi-isometric rigidity of $Nil$, but I am not completely sure what should be the correct attribution. By Gromov's polynomial growth Theorem, any f.g. group quasi-isometric to $Nil$ is virtually nilpotent, and it is a classical result by Malcev that every f.g. (torsion free) nilpotent group is a uniform lattice in some nilpotent Lie group. Therefore, every f.g. group quasi-isometric to $Nil$ is virtually isomorphic to a uniform lattice in a (maybe different) nilpotent Lie group. Finally, maybe using the work of Pansu in Croissance des boules et des géodésiques fermées dans les nilvariétés'' (1983) one could probably show that the $Nil$ is recognized among the class of all nilpotent Lie groups by its asymptotic cone. This should conclude the proof of the quasi-isometric rigidity of $Nil$.

My question is: Is there an established reference for the quasi-isometric rigidity of $Nil$? If not, did someone write somewhere the details (if correct!) for the tentative proof I outlined above? Or is there a shorter argument that I did not spot?

• The Heisenberg group is certainly not the unique $3$-dimensional nilpotent Lie group, e.g. $\mathbb{R}^3$ is nilpotent, as are nontrivial quotients of the Heisenberg group. It might be the unique $3$-dimensional simply connected nilpotent nonabelian Lie group. – Qiaochu Yuan Jan 18 '16 at 15:34
• Of course you are right. By "unique'' my brain probably meant "unique non-trivial (i.e. non-abelian), up to taking covnerings...''. I will edit my question accordingly. Thanks. – Roberto Frigerio Jan 19 '16 at 10:34
• In this context (QI classification) you really want to consider simply connected nilpotent Lie groups, or if you allow nilpotent Lie groups, you want to consider them not up to covering, but up to quotient by their maximal compact subgroups. But anyway it makes things unnecessarily complicated (since the discrete Heisenberg group appears a lattice in nilpotent Lie groups of arbitrary large dimension, but only in NIL if you require simply connected) – YCor Jan 19 '16 at 10:54

Here's a sketch of proof (which intersects yours):

Let $\Gamma$ be QI to NIL. As you say, by Gromov's theorem, $\Gamma$ is virtually nilpotent; let some finite index subgroup be a lattice in some simply connected nilpotent Lie group $G$.

The growth of $G$ and NIL are equivalent, hence the growth of $G$ is $\simeq r^4$. Only two simply connected nilpotent Lie groups have growth $\sim r^4$: the abelian 4-dimensional one, and NIL. (This is a baby case of results of Guivarch or Jenkins in 1973)

So either $\Gamma$ is virtually $\mathbf{Z}^4$, or it's virtually a lattice in NIL.

To exclude the 1st case, you can invoke Pansu's 1983 paper but it's indeed a baby case: it's enough to show that NIL and $\mathbf{R}^4$ are not QI. But, NIL being endowed with a Carnot-Catheodory metric, both spaces are proper spaces with nontrivial 1-parameter groups of non-isometric similarities, so are isometric to their asymptotic cones. Since they are not homeomorphic, it follows that the asymptotic cones are not isometric and hence NIL and $\mathbf{R}^4$ are not quasi-isometric. Avoiding asymptotic cones, another approach is the Dehn function, which is $\simeq n^2$ for $\mathbf{R}^4$ and $\simeq n^3$ for NIL. Also, Roe proved later that for a simply connected solvable Lie group, the dimension is a quasi-isometry invariant.

Note: While's it's not seriously needed here, Pansu's 1983 result is in general much harder because it describes the asymptotic cone for simply connected nilpotent Lie groups that are not Carnot, i.e., do not admit a Carnot-Caratheodory metric with a non-isometric 1-parameter group of dilations. It gives a description of the asymptotic cone up to bilipschitz homeomorphism, but to get strong consequences one has to combine with Pansu's 1989 Annals result, which says that the associated Carnot-graded Lie algebra (a purely algebraic object! no metric here) of a simply connected nilpotent Lie group is a QI-invariant. This allows to distinguish, for instance, some infinite families in fixed dimension.

Edit: there is a statement avoiding passing to finite index subgroups: if $\Gamma$ is a finitely generated group QI to NIL, (NIL is being endowed with a left-invariant metric invariant under a maximal compact subgroup of automorphisms), then $\Gamma$ admits a proper cocompact isometric action on NIL.

Here's another, more topological way to see that $\mathrm{Nil}$ is not quasi-isometric to $\mathbf{R}^4$: use the fact that $\mathrm{Nil}$ is homeomorphic to $\mathbf{R}^3$, so that $\pi_2^\infty(\mathrm{Nil})$ is nontrivial, whereas $\pi_2^\infty(\mathbf{R}^4)$ is trivial.

This works because $\pi_2^\infty$ is indeed a quasi-isometry invariant within a class of metric spaces which contains $\mathbf{R}^4$ and $\mathrm{Nil}$. The key property is that they are uniformly 2-connected, i.e. spheres of dimension $\le 2$ can be filled by balls with metric control.

Concretely, to prove the result by hand you can do the following: assume that there is a quasi-isometry $f:\mathrm{Nil} \to \mathbf{R}^4$. Without loss of generality, $f$ takes the origin to the origin. Let $\phi$ be an embedding of $\mathbf{S}^2$ into $\mathrm{Nil}$ with image a metric sphere of large radius centred at the origin. Put a sufficiently fine triangulation $\mathcal T$ on $\mathbf{S}^2$. By induction on the skeleta, construct a continuous map $g:\mathbf{S}^2\to \mathbf{R}^4$ which coincides with $f\circ\phi$ on the $0$-skeleton and is at bounded distance from $f\circ\phi$.

Since $\mathbf{R}^4$ is 2-connected at infinity, we can extend $g$ to a continuous map $\bar g$ from the 3-ball to $\mathbf{R}^4$ which stays far away from the origin. Composing with a coarse inverse of $f$ and using the same trick to make it continuous, we obtain a continous map from the $3$-ball to $\mathrm{Nil}\setminus\{O\}$ which extends $\phi$. This is a contradiction.

(Remark: we have actually only used that $\mathrm{Nil}$ is uniformly 2-connected and $\mathbf{R}^4$ is uniformly 1-connected.)

• We could expect that such an argument reproves the (nontrivial) fact that for a simply connected solvable Lie group, the dimension is a QI-invariant (and in particular, the Hirsch length of a polycyclic group is QI-invariant). Indeed, such a group has the property that it has a family of parallepipeds $B_n$ (=continuous images of the standard $n$-ball by a global homeomorphism) such that for every left-invariant Riemannian metric, there exists $c,C>0$ such that $B_n$ contains the $cn$-ball and is contained in the $Cn$-ball. – YCor Jan 31 '16 at 16:26
• PS: that NIL and $\mathbf{R}^4$ are not QI is only one of the steps of the proof, and not the most difficult. – YCor Jan 31 '16 at 16:39
• There is a work of Alex Eskin for Solvable groups, related to your question, math.uchicago.edu/~eskin/icm2.pdf, I think his method is useful – user21574 Feb 1 '16 at 3:13
• @HassanJolany I think your comment is addressed to Roberto and should not be here but a comment to the question. And it's not work of Eskin, but of Eskin, Fisher and Whyte. – YCor Feb 1 '16 at 3:18
• Ohhh, all right, I just checked, thanks, – user21574 Feb 1 '16 at 3:19