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I have started working on group growth earlier this year, mainly using Drutu and Kapovich's notes. This morning I found myself wondering if I could find an example of groups that are not quasi-isometric but have the same growth rate. Spontaneously, I thought about finite groups, groups of linear growth and free groups. All those cannot provide such an example.

I firmly believe this to be possible but I have been unable to find one. Google searches have not helped me either so maybe some of you can.

AKadoc12
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  • @DietrichBurde Different growth rates imply non-quasi-isometry, but I'm not sure the converse is obvious. I feel like I once saw a counter example, but I can't remember where or what it was unfortunately – David Sheard Jun 19 '22 at 14:26
  • You could try asking this on mathoverflow. – Derek Holt Jun 19 '22 at 20:03
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    My answer here gives the simplest example in the realm of groups of polynomial growth: both $\mathbb Z^4$ and the integer Heisenberg group have polynomial growth of degree 4. – Lee Mosher Jun 19 '22 at 20:16
  • Thank you @LeeMosher, it looks exactly like what I need – AKadoc12 Jun 19 '22 at 20:49
  • The class of groups with exponential growth is quite huge and contains plenty of non-quasi-isometric groups. For instance, $F_2$ and $F_2\times F_2$ are not quasi-isometric. There are many subclasses that are known to be QI-closed, e.g., (non-elementary hyperbolic) $\subsetneq$ (finitely presented of exponential growth) $\subsetneq$ (exponential growth). Polynomial growth examples are more restricted but anyway do exist, as indicated by Lee Mosher, although the full classification of QI among polynomial growth groups is still conjectural. – YCor Jul 01 '22 at 17:17

2 Answers2

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To augment the answer of @QiaochuYuan, I'll turn my comment into an answer: In the realm of groups of polynomial growth, the simplest example is that both $\mathbb Z^4$ and the integer Heisenberg group have polynomial growth of degree $4$.

One still has to prove that those groups are not quasi-isometric to each other. The simplest proof that I know is that $\mathbb Z^4$ acts freely, properly and cocompactly on the topological space $\mathbb R^4$ hence its cohomological dimension is $4$; whereas the integer Heisenberg group acts freely, properly and cocompactly on $\mathbb R^3$ hence its cohomological dimension is $3$. And in the realm of groups that act freely, properly and cocompactly on Euclidean spaces (or even on contractible CW complexes), the cohomological dimension is a quasi-isometry invariant.

Lee Mosher
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This is not at all an area I know about, but some quick googling gives the following:

  • The fundamental group of any closed hyperbolic $n$-manifold is quasi-isometric to $\mathbb{H}^n$, and any two such fundamental groups (for $n \ge 2$) have exponential growth rate.
  • $\mathbb{H}^n$ and $\mathbb{H}^m$ are not quasi-isometric if $n \neq m$; they can be distinguished by their Gromov boundaries, which are $S^{n-1}$ and $S^{m-1}$ respectively.

So we can take the fundamental groups of a closed hyperbolic $n$-manifold and $m$-manifold for $n \neq m \ge 2$.

Qiaochu Yuan
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