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Given a metric space $(X, d)$, we can consider the set of all quasi-isometries $f: X \to X$, and quotient out by the equivalence relation identifying $f$ and $g$ if $\sup_{x \in X}d(f(x), g(x))$ is finite. Doing so, we obtain a set of equivalence classes $\mathcal{QI}(X)$ that is a group under composition.

In the same spirit as the questions Every group is a fundamental group and Is every group the automorphism group of a group?, we can ask: for which groups $G$ does there exist a metric space $X$ such that $\mathcal{QI}(X) \cong G?$

Surprisingly, someone told me today that basically nothing is known about this question. According to them, we do not even know how to construct a metric space $X$ such that $\mathcal{QI}(X)$ is a finite cyclic group.

Given this, my question is: what do we know about the quasi-isometry groups of metric spaces? For example, what are some metric spaces $X$ for which $\mathcal{QI}(X)$ has been computed? Do we know of any groups $G$ which are not isomorphic to $\mathcal{QI}(X)$ for any $X$?

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    $\begingroup$ I'm pretty sure every group is isomorphic to $QI(X)$ for some $X$. This should be lengthy to prove, like the cousin questions, but the class of metric space is flexible enough. $\endgroup$
    – YCor
    Jan 27 at 7:28
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    $\begingroup$ The other open-ended question "what are some $X$ for which $QI(X)$ is known then goes in another direction. There are many known examples (with natural $X$), and many examples for which a lot is known. I'm afraid these are too distinct questions to fit in a single one. $\endgroup$
    – YCor
    Jan 27 at 7:30
  • $\begingroup$ @YCor Do you know where I could find a proof that every group $G$ is isomorphic to $\mathcal{QI}(X)$ for some $X$? Alex Nolte has provided a nice construction to produce symmetric groups as quasi-isometry groups, but I can't imagine the same idea extending to any group $G$ (unlike the proof showing every group is a fundamental group). $\endgroup$
    – ckefa
    Jan 27 at 9:22
  • $\begingroup$ @YCor Also, do you have any books/papers you would recommend on quasi-isometry groups, especially ones where they compute a bunch of $\mathcal{QI}(X)$'s for various $X$? So far, I've only been able to find a few papers talking about various properties of $\mathcal{QI}(\mathbb{R}),$ but maybe I'm not searching in the right place or using the right key words; any help would be greatly appreciated. $\endgroup$
    – ckefa
    Jan 27 at 9:27
  • $\begingroup$ I think it has never been proved because nobody really tried so far. Of course one can imagine some specific constructions working in some very special cases, and this is a reasonable starting point. $\endgroup$
    – YCor
    Jan 27 at 10:58

2 Answers 2

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A first observation is that $\mathcal{QI}$ is quite complicated for most natural spaces. For instance, any two linear maps $x \mapsto \lambda x, x \mapsto \lambda' x$ are equal in $\mathcal{QI}(\mathbb{R})$ if and only if $\lambda = \lambda'$. A corollary of this is that $\mathcal{QI}(\mathbb{N})$ is uncountable.

In order to get small $\mathcal{QI}(X)$ to be small, one must spread apart points in $X$ to sabotage the flexibility of the quasi-isometry condition. A helpful building block and motivating example is $X_0 = \{ n! \mid n \in \mathbb{N} \} \subset \mathbb{N}$. Any $(K,C)$-quasi isometry $f$ of $X_0$ must have $f(n!) = n!$ for all $n$ large enough, for instance $n > \text{max}(K, C) + 1$, by considering $d(f(n!), f((n+1)!))$. So $\mathcal{QI}(X_0)$ is trivial. One obtains a space with $\mathcal{QI}(X_n) = S_n$ (with $S_n$ the symmetric group on $n$ letters) by taking a pinwheel of $n$ copies of $X_0$.

An addendum: this construction also gives direct products of symmetric groups by mixing growth rates in building blocks. For instance, let $Y_0 = \{ (n!)! \, | n \in \mathbb{N}\}$. Then gluing the pinwheel of $n$ copies of $X_0$ and the pinwheel of $m$ copies of $Y_0$ together at $1$ gives a space with $\mathcal{QI}(X) = S_n \times S_m$. This is because for a $(K, C)$ quasi-isometry, one can not map sufficiently large elements of $X_0$ into $Y_0$ or vice-versa. One sees this by comparing the distances between $3$ consecutive elements in $X_0$ or $Y_0$.

It seems quite unclear how to build many other groups with explicit examples.

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  • $\begingroup$ Can you elaborate on what you mean by "a pinwheel of $n$ copies of $X_{0}$"? I am interpreting this to be the space with $n$ copies of $X_{0}$, glued together at the point $1 \in X_{0}$ for each copy. We then get $\mathcal{QI}(X_{n}) = S_{n}$ because any permutation of the copies of $X_{0}$ works? $\endgroup$
    – ckefa
    Jan 27 at 9:06
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    $\begingroup$ Yes, this is what I meant: glue $n$ copies of $X_0$ together at $1$, and specify the metric by the distance between $(n!)_{X_0^i}, (m!)_{X_0^j}$ being $n! + m! -2$ for $i \neq j$. (Shortest paths between spokes are through the center). One can permute copies of $X_0$, but can do nothing else due to the argument for $\mathcal{QI}(X_0) = \{e\}$. $\endgroup$
    – Alex Nolte
    Jan 27 at 15:17
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    $\begingroup$ It seems to me that you can actually permute each set of $k$ copies of $n!$ separately under a (2,0)-quasi-isometry, so you get a QI group of the form $\prod_{n=1}^\infty S_k/\oplus_{n=1}^\infty S_k$. Nice explicitly computable group, but decidedly not finite. Am I missing something? $\endgroup$
    – Fedya
    Feb 11 at 18:30
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Regarding the question of spaces $X$ for which $QI(X)$ is known, a good keyword is quasi-isometric rigidity. One reference would be the survey in Chapter 25 of the Druţu–Kapovich book "Geometric group theory".

A space $X$ is called strongly QI rigid if the map $\operatorname{Isom}(X) \to QI(X)$ is surjective. Often $X$ has no non-trivial isometries that are bounded distance from the identity, so we get an isomorphism. For example, Pansu showed that quaternionic hyperbolic spaces are strongly QI rigid.

Pansu, Pierre, Carnot-Carathéodory metrics and quasiisometries of symmetric spaces of rank 1, Ann. Math. (2) 129, No. 1, 1-60 (1989). ZBL0678.53042..

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