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

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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 (I repeated the argument here below "original answer"). To deduce the general case, let me relax assumptions.$\DeclareMathOperator\Aut{Aut}$

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 $K\ltimes G$.

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 $\Gamma$).

Corollary. There exists a left-invariant Riemannian metric on $G$ and a geometric (=proper isometric cocompact) action of $\Gamma$ on $G$.

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 $\Gamma$.


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).

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  • $\begingroup$ Thank you, I read this in your other answer as well. But is it sufficient that a group is virtually isomorphic to the integral Heisenberg group to be a discrete cocompact group of isometries of Nil? For example, if I'm not mistaken, in $\mathbb{E}^3$ the crystallographic groups not only have $\mathbb{Z}^3$ as a normal subgroup of finite index, but also $\mathbb{Z}^3$ is its own centraliser. So, is there no similar additional requirement in Nil apart from being virtually isometric to the integral Heisenberg group? Or has one been implicitly mentioned in your answer and I've missed it? $\endgroup$
    – George K
    Mar 15, 2022 at 18:40
  • $\begingroup$ Oops, indeed this makes this a duplicate of the other one... I didn't remember. What you're asking further in the comment, however, is not covered. $\endgroup$
    – YCor
    Mar 15, 2022 at 18:44
  • $\begingroup$ It's indeed true that every group QI to Nil has a geometric action on Nil (with a suitable metric). This follows from a refined version, embedding (well, with finite kernel) the group without passing to a finite index subgroup. I would need to search a bit before gathering ingredients, which definitely exist. $\endgroup$
    – YCor
    Mar 15, 2022 at 18:46
  • $\begingroup$ Thank you for your help. I would love an answer on this. $\endgroup$
    – George K
    Mar 15, 2022 at 18:48
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    $\begingroup$ @GeorgeK I'd suggest to edit the question then, reformulating to take into account the information from the previous one and focussing on this version that doesn't pass to a finite index subgroup. $\endgroup$
    – YCor
    Mar 15, 2022 at 19:25

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