Is it true that every group quasi-isometric to the Heisenberg group admits a proper cocompact action by isometries on Nil?



Discrete cocompact group of isometries of Nil
This is true. The first step is that the group $\Gamma$ then has a subgroup of finite index that embeds as a lattice in the Heisenberg group $H$. This step is sketched [in this answer](https://mathoverflow.net/a/228717/14094) (I repeated the argument here below "original answer"). To deduce the general case, let me relax assumptions.$\DeclareMathOperator\Aut{Aut}$



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> **Proposition.** Let $\Gamma$ be a discrete group with a finite index subgroup $\Lambda$ that embeds as a lattice into a simply connected nilpotent Lie group $G$. Let $K$ be a maximal compact subgroup of $\Aut(G)$. Let $W$ be the finite radical of $\Gamma$ (= the maximal finite normal subgroup, which exists here). Then $\Gamma/W$ embeds as a lattice into $K\ltimes G$.
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Proof: first step: there exists a Lie group $H$ of the form $H^0\rtimes L$ with $L$ finite, $H^0$ simply connected nilpotent, in which $\Gamma$ maps as a lattice with finite kernel.


Fix an embedding of $G$ into the upper unipotent subgroup of $\mathrm{GL}\_n(\mathbf{R}$ for some $n$. This defines a representation of $\Lambda$. Induce it to $\Gamma$. We obtain a representation $f$ of $\Gamma$, for which $\Lambda$ acts unipotently. Then the Zariski closure of $f(\Lambda)$ is a simply connected nilpotent Lie group in which $f(\Lambda)$ is a lattice. Let $H$ be the Zariski closure of $f(\Gamma)$. Then $H^0$ is the Zariski closure of $f(\Lambda)$, and has finite index in $H$. Let $L$ be a maximal compact subgroup of $H$. Then $LH^0=H$ (this is true for every virtually connected Lie group, result of Mostow). Since $L\cap H^0=1$, we deduce $H=H^0\rtimes L$ and $L$ is finite. So the first step is proved.


Now conclude. Consider $H=H^0\rtimes L$ as in the first step. Without loss of generality, we can suppose that $L$ acts faithfully on $H$ (otherwise, mod out by the kernel). So $L\subset\Aut(H)$. Hence $L$ is contained in a maximal compact subgroup of $\Aut(H)$. We are now done (the assumption then now force the kernel to be exactly the finite radical of $\Gamma$).



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> **Corollary.** There exists a left-invariant Riemannian metric on $G$ and a geometric (=proper isometric cocompact) action of $\Gamma$ on $G$.
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Proof: choose a left-$G$-invariant right-$K$-invariant Riemannian metric on $G$. Then the action of $G\ltimes K$ on $G$ is geometric (=continuous proper isometric cocompact), and hence so is that of $\Gamma$.




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Original answer (which passes to a finite index subgroup)
It (essentially) directly follows from Gromov's theorem that a f.g. group quasi-isometric to $\mathsf{Nil}$ is virtually isomorphic to the integral Heisenberg group. (No need to assume the existence of an action.)


Indeed, such a group has polynomial growth, so is virtually nilpotent and hence virtually isomorphic to a lattice in some simply connected nilpotent Lie group $G$.


Once we know that $G$ is the 3-dimensional Heisenberg group, one is done since every lattice therein contains with finite index a copy of the integral Heisenberg group.


One can directly invoke Pansu's 1988 theorem to deduce that the associated Carnot Lie algebra of $G$ is isomorphic to the Heisenberg group, and hence $G$ itself as well (since $G$ is 2-step-nilpotent it is Carnot).


Alternatively, one can avoid Pansu's theorem, saying that $G$ has growth $n^4$ which leaves only 2 candidates: $\mathbf{R}^4$ and the Heisenberg group. So one just has to know that they are not quasi-isometric and this follows form various arguments, e.g., using asymptotic cones (which is homeomorphic to $\mathbf{R}^4$ for $\mathbf{R}^4$ and to $\mathbf{R}^3$ for the other).


