# Groups quasi-isometric to $\mathbb{Z}^n$

I am interested by the following result: A groups quasi-isometric to $\mathbb{Z}^n$ is virtually $\mathbb{Z}^n$.

I know the article Harmonic analysis, cohomology, and the large-scale geometry of amenable groups by Yehuda Shalom, but currently I don't have any access to it.

Do you know another document on this subject?

-

This can be proved using the fact that the Hirsch length of a polycyclic group is a quasi-isometric invariant (this is a result of Bridson). Let's denote the Hirsch length of a group by $h(G)$.

Now Bass proved that a virtually nilpotent group $G$ has polynomial growth, and the degree of that growth is given by $$d(G) = \displaystyle\sum rk(G_i/G_{i+1})i.$$

Here $G_i$ is the $i$-th term of the lower central series.

The converse is Gromov's polynomial growth theorem, that groups with polynomial growth are virtually nilpotent.

So now let's put all this together. Let $G$ be a finitely generated group quasi-isometric to $\mathbb{Z}^n$. Then by Gromov's theorem there is a subgroup of finite index $H\le G$, which is nilpotent (and so polycyclic).

Now it is easy to check that $h(\mathbb{Z}^n)=d(\mathbb{Z}^n)=n$. So $h(H)=d(H)=n$. But this implies (using Bass's formula) that $H'$ is finite. It is easy to then see that $[H:Z(H)]$ is finite (see below).

Now $Z(H)$ is an abelian group, also quasi-isometric to $\mathbb{Z}^n$. Thus, again by Bass's formula, $Z(H)\cong \mathbb{Z}^n\times T$, for some finite abelian group $T$.

But then $[G:\mathbb{Z}^n] = |T|\cdot[H:Z(H)]\cdot[G:H]$, which is finite.

Proof that $|H'|$ finite implies $[H:Z(H)]$ finite.

Let $H$ be generated by $h_1,h_2,\ldots,h_m$. Now the conjugates of $h_i$ are contained in the set $h_iH'$, which by assumption is finite. Thus $h_i$ has finitely many conjugates, and so $C_H(h_i)$ has finite index in $H$. But then $Z(H)=\cap C_H(h_i)$ is also finite index.

-
The article I mentioned is "The optimal isoperimetric inequality for torus bundles over the circle" by Bridson and Gersten. – user641 Aug 9 '12 at 19:19
I should also mention a more geometric proof can be gotten by letting $G$ act by isometries on the asymptotic cone of $\mathbb{Z}^n$, namely $\mathbb{R}^n$. This embeds $G$ in the isometry group of $\mathbb{R}^n$, and then you can finish with Gromov's theorem and the Flat Torus theorem. – user641 Aug 9 '12 at 19:21
#SteveD: I don't understand your additional proof (in the comments): if $G$ is QI to $\mathbf{R}^n$, it's not clear why the action of $G$ on the asymptotic cone should be by isometries, and also it's not clear why it should be faithful. – YCor Aug 24 '12 at 12:04
@Yves: because G and Zn are QI, their asymptotic cones are isometric. – user641 Aug 24 '12 at 17:57
Steve, you don't answer my question, please read it again. – YCor Feb 23 '13 at 1:36