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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, 2022 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, 2022 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, 2022 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, 2022 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, 2022 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. If one allows metrics that obtain $\infty$ as a distance, one obtains a space with $\mathcal{QI}(X_n) = S_n$ (with $S_n$ the symmetric group on $n$ letters) by taking $n$ copies of $X_0$, with infinite distance between any two copies.

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 taking the disjoint union of $n$ copies of $X_0$ and $m$ copies of $Y_0$ as above 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 $Y_0$.

A correction: as pointed out by Fedya below, a previous version of this answer incorrectly asserted that spaces $X_k$ obtained as pinwheels of $k$ copies of $X_0$ have $\mathcal{QI}(X_k) \cong S_k$. One can independently permute each of the $k$ copies of $n!$ while remaining a quasi-isometry, and finite-distance maps allow one to freely move around finitely many points freely. This yields the quasi-isometry group of $X_k$ to be isomorphic to $(\prod_{k=1}^{\infty}S_k) / (\bigoplus_{k=1}^\infty S_k).$

It seems quite unclear how to build many other groups with explicit examples. In particular, one question I think is interesting but do not know how to answer is if there exist metric spaces (with only finite distances allowed) with finite and nontrivial $\mathcal{QI}$ group.

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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, 2022 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, 2022 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, 2022 at 18:30
  • $\begingroup$ @Fedya You're entirely correct, my bad. This computes the QI group of these spaces correctly. The original construction can be modified to work as intended if one allows infinite distances, and specifies that the prongs are infinite distance apart, I'll edit the answer. $\endgroup$
    – Alex Nolte
    May 22, 2022 at 17:58
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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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