# Quasi-isometry groups of metric spaces

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 $$X$$?

• 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.
– YCor
Jan 27 at 7:28
• 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.
– YCor
Jan 27 at 7:30
• @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). Jan 27 at 9:22
• @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. Jan 27 at 9:27
• 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.
– YCor
Jan 27 at 10:58

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.

• 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? Jan 27 at 9:06
• 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 $\mathcal{QI}(X_0) = \{e\}$. Jan 27 at 15:17
• 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? Feb 11 at 18:30

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..