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The question is actually from Löh's Geometric Group Theory: an introduction on page 40:

Exercise 2.E.10: Is there for every group $G$, an $n\in \mathbb{N}$ and a subset $X\subset \mathbb{R}^n$ such that the isometry group of $X$ is isomorphic to $G$?

I am aware of the paper Realizing Finite Groups in Euclidean Space by Albertson and Boutin, which says the answer is yes if $|G|<\infty$. But for $|G|$ infinite, say countably infinite, I do not have a quick answer yet.

Most likely, the answer is no. If so, can anyone help give an example of such $G$ with $G$ countable?

hinamizawa
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2 Answers2

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No. There are a great many countably infinite groups which are not even subgroups of $\mathrm{GL}_n(\mathbf R)$ for any $n \in \mathbf N$ (they are "not linear").

For example let $G$ be the group of all finitely supported permutations of $\mathbf Z$. This group $G$ contains an isomorphic copy of every finite group, so it cannot embed in $\mathrm{GL}_n(\mathbf R)$ for any $n$. This follows for example from Jordan's theorem on finite linear groups. To get a finitely generated example you can take $H$ to be the subgroup of $\mathrm{Sym}(\mathbf Z)$ generated by the transposition $(0,1)$ and the infinite cycle $n \mapsto n+1$. Note that $G \le H$.

For a finitely presented example one can take Higman's group $$G = \langle a, b, c, d \mid a^b = a^2, b^c = b^2, c^d = c^2, d^a = d^2\rangle,$$ which has no nontrivial finite quotient. By contrast, finitely generated linear groups are residually finite (a theorem of Mal'cev).

Edit: In the comments it was pointed out that the OP could have in mind the looser definition of isometry group as the group of all permutations $\pi : X \to X$ preserving Euclidean distance. In this case some more explanation is required.

Let $X \subset \mathbf R^n$ and let $G$ be the isometry group of $X$ in the above sense. By translating $X$ we may assume that $0 \in X$, and by reducing the dimension if necessary we may assume that $X$ spans $\mathbf R^n$.

We claim that every isometry of $X$ extends uniquely to an (affine linear) isometry of $\mathbf R^n$. Let $B \subset X$ be a basis. Let $\pi$ be an isometry of $X$. It follows from the law of cosines that $$\langle \pi(x) - \pi(0), \pi(y) - \pi(0)\rangle = \langle x, y \rangle \qquad (*)$$ for all $x, y \in X$. Applied to $x, y \in B$ and considering a Gram determinant, it follows that $B' = \pi(B) - \pi(0)$ is also a basis. Let $f$ be the unique (affine linear) isometry of $\mathbf R^n$ such that $f(0) = \pi(0)$ and $f(b) = \pi(b)$ for all $b \in B$. Now applying $(*)$ with $y \in B$ and the same identity with $f$ in place of $\pi$ shows that $$\langle \pi(x) - \pi(0), b'\rangle = \langle f(x) - f(0), b' \rangle$$ for all $x \in X$ and $b' \in B'$, which implies that $f(x) = \pi(x)$ for all $x \in X$.

Thus $G$ is isomorphic to a subgroup of $\mathrm{Isom}(\mathbf R^n) \le \mathrm{AGL}_n(\mathbf R) \le \mathrm{GL}_{n+1}(\mathbf R)$.

Sean Eberhard
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  • Does every group of isometries of any $X\subseteq\mathbb{R}^m$ embed into some $GL_n(\mathbb{R})$? – freakish May 10 '23 at 10:08
  • @freakish Yes of course! By definition, the group of isometries of $X \subset \mathbf R^m$ is the group of all $g \in \mathrm{SO}(m)$ such that $g(X)=X$. Note that $\mathrm{SO}(m) \le \mathrm{GL}_m(\mathbf R)$. – Sean Eberhard May 10 '23 at 10:09
  • It is possible you have a looser definition of isometry group in mind, and if so you should specify. My answer will still apply but there may be some more explanation required. – Sean Eberhard May 10 '23 at 10:10
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    Well, for me the group of isometries is defined as the group of all $X\to X$ which are isometries, i.e. they preserve the distance. I find your definition non-standard. Unless every isometry $X\to X$ can be extended to isometry $\mathbb{R}^n\to\mathbb{R}^n$? Which I doubt. – freakish May 10 '23 at 10:11
  • And for the avoidance of doubt, you mean the usual Euclidean distance, right? Again, if you choose a nonstandard distance, I still think my answer will apply, but even more explanation may be required. – Sean Eberhard May 10 '23 at 10:14
  • Yeah, the OP talks about $\mathbb{R}^n$, I assume about standard $\mathbb{R}^n$ with the standard Euclidean distance. – freakish May 10 '23 at 10:16
  • Isn't the OP asking about the isometry group of $X$ w.r.t. the Riemannian metric induced from $\mathbb R^n$ ? If that's the case, then I believe the answer is yes. For, 1) every (say, countable) group $G$ is the fundamental group of a 2-manifold $\overline X$, which admits a Riemannian metric, 2) take $X$ to be the (Riemannian) universal covering of $\overline X$; then $G$ acts isometrically on $X$, 3) embeb $X$ in $\mathbb R^n$ isometrically via Nash (with $n\le51$), 4) then $G$ is the isometry group of $X$ with respect to the Riemannian metric induced from $\mathbb R^n$. Is there any flaw? – Dry Bones May 10 '23 at 11:11
  • @DryBones It seems far from clear that was the OP's intention, but still maybe you should post your comment as another answer. – Sean Eberhard May 10 '23 at 11:27
  • Hi SeanEberhard and @DryBones. Thanks for you both. Both your answers are what I desire. As you pointed out, the answer depends on the question's/author's intention. I did not realize such a group is a subgroup of $SO(n,\mathbb{R})$ in the first case and failed to recognize the fact that every group is the fundamental group of a $4$-manifold (I think not 2, is it?) in the second case. – hinamizawa May 10 '23 at 12:27
  • @SeanEberhard In your second answer, don't you need to also ensure that $\pi(B)$ spans the whole $\mathbf{R}^n$ so that you can conclude that $\langle f(x), f(b) \rangle = \langle \pi(x), f(b) \rangle$ for all $b$ implies $f(x)=\pi(x)$? – Salini Mendisi May 10 '23 at 14:43
  • Hm, perhaps 2 works for CW complexes, but for manifolds one needs 4 because of smoothness, is that right? Here a construction of such a 4-manifold is presented; then one has to adjust the Nash upper bound to $n\le 230$. Also, the link above reminded me that there are countable groups which are not finitely presentable (i.e., do not admit a finite set of generators), so this should also be taken into account. Anything else? – Dry Bones May 10 '23 at 15:23
  • @SaliniMendisi That was an oversight, along with another oversight to do with affine linearity (isometries need not have fixed points if $X$ is not finite). Should be fixed now. – Sean Eberhard May 10 '23 at 15:50
  • @DryBones Finitely generated = finite set of generators. Finitely presented = finite set of generators with a finite basis of relations. – Sean Eberhard May 10 '23 at 15:52
  • By the way, the fact that every isometry is an affine linear orthogonal map is a theorem of Mazur-Ulam. See https://en.wikipedia.org/wiki/Mazur%E2%80%93Ulam_theorem – Brauer Suzuki May 10 '23 at 16:08
  • @BrauerSuzuki Mazur--Ulam is about isometries of the full vector space with respect to an arbitrary norm. The special case of the Euclidean norm is just the law of cosines. – Sean Eberhard May 10 '23 at 16:13
  • @DryBones How do you ensure that the universal cover $X$ of $\overline X$ does not have any isometries apart from deck transformations? Is the idea that this will be true if $\overline X$ has no nontrivial isometries, and maybe that is true if the Riemannian metric on $\overline X$ is somehow generic? – Sean Eberhard May 10 '23 at 18:25
  • Exactly, nontrivial isometries of Riemannian manifolds are very rare, even if rather restricting conditions are imposed, for example constant curvature. In other words, we can always deform $\overline X$ in order to get an isometry free manifold, and then construct $X$, which shall only have the prescribed deck transformations. – Dry Bones May 10 '23 at 19:52
  • @Freakish: In fact it is true that every isometry of a subset of $\mathbb R^n$ extends to an isometry of $\mathbb R^n$, which is unique if the set is not "coplanar" in the sense that it does not lie in an affine $n-1$ dimensional plane. For 3 noncollinear points in $\mathbb R^2$ this is the SSS-theorem of classical geometry, and that special case in $\mathbb R^2$ quickly implies the general case in $\mathbb R^2$. An induction argument kicks in for higher dimensions. (And of course Sean's edit give a more high powered proof). – Lee Mosher May 11 '23 at 12:34
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As you are reading Löh's book on GGT, it is sensible to give a standard, important, and famous example of a non-linear group, and which is one of a family of groups which appears throughout her book, and throughout GGT.

In particular, consider the Baumslag-Solitar group $$\operatorname{BS}(2, 3)=\langle a, b\mid ba^2b^{-1}=a^3\rangle.$$

A group is Hopfian if every surjective endomorphism is an automorphism, and $\operatorname{BS}(2, 3)$ is the standard example of a finitely presented non-Hopfian group (this is Löh's Exercises 2.E.21).

Finitely generated residually finite groups are Hopfian (this is Löh's Exercise 4.E.28), and so $\operatorname{BS}(2, 3)$ is non-residually finite. Then apply @SeanEberhard's answer/Mal'cev's theorem to get that the group is non-linear (see here for a sketch proof).

user1729
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