TITLE: Quasi-isometric rigidity of Nil
QUESTION [10 upvotes]: 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?

REPLY [6 votes]: 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.)