Groups of isometries

Question

I recently read the popular scientific book "Symmetry and the Monster" and it emphasizes groups as the sets of symmetries of geometrical objects. So I was wondering, do all groups appear as symmetry groups of geometrical objects?

I think the question can be formalized like this: Let $M$ be a metric space and let $S\subseteq M$. The set of all isometries $f$ in $M$ with $f(S)=S$ form a group $G_S$ with composition. Let $G(M) = \{G_S : S \subseteq M \}$. This raises many questions, all up to isomorphism, here are some of them:

1. Do all (finite) groups appear as some $G_S$?
2. If $M=\mathbb{R}^d$ with Euclidean metric, what groups appear in $G(M)$?
3. For $d \in \mathbb{N}$, is there always a group $G$ such that $G \in G(\mathbb{R}^d)$ but $G \notin G(\mathbb{R}^{d-1})$?

I think I could continue this list for hours, but maybe you could help me finding some literature for these problems? I would also be happy about partial answers.

Answer 1

Yes, every finite group is (isomorphic to) the symmetry group of a convex polytope in Euclidean space $\mathbb{E}^d$ for some $d$. I just read about a nice construction of such a polytope for an arbitrary group $G$ within the past couple of days, but frustratingly, I can't remember where I read it! Here's how I recall it going:

Any finite group $G$ is isomorphic to a subgroup of a symmetric group $S_m$ (Cayley's theorem.) You can always do it with $m = |G|$, but generally a smaller $m$ will suffice; the minimal such $m$ is called the permutation degree of $G$.

$S_m$ is (isomorphic to) the symmetry group of a regular simplex in $(m-1)$-space. Say $d = m-1$ and let $S \cong S_m$ be such a symmetry group (with Coxeter diagram $A_d$); it operates by permuting the $m$ vertices of a $d$-simplex, and is generated by $d$ reflections in hyperplanes. (We'll assume the $d$-simplex is centered at the origin.) Choose a point $p$ not on any of the reflection hyperplanes, and let $K$ be the convex hull of all its images under $S$. This is known as an omnitruncated $d$-simplex, and has the property that $S$ acts simply transitively on its vertices: i.e., for any two vertices $v$ and $w$, there is a unique isometry in $S$ taking $v$ to $w$. (This is a consequence of the fact that $p$ is in the interior of a fundamental domain for $S$, which is bounded by reflection hyperplanes.)

The dual of $K$, $K^*$, instead has the property that $S$ acts simply transitively on its facets (each facet of the dual corresponds to a vertex $v$ of $K$; it's bounded by the hyperplane normal to $v$.) Let $q$ be a point above the center of one of the facets $F$ of $K^*$, but beneath all the other facets. (The term "beneath" means that, for a facet $H$, $q$ lies on the same side of the hyperplane $\operatorname{aff}(H)$ that the interior of $K^*$ does. In other words, $q$ is sufficiently close to $F$ that adding it to the convex hull doesn't affect any facets besides $F$.) Let $P$ be the convex hull of $K^*$ together with all the images of $q$ under $G$ (viewing $G$ as a subgroup of $S$): $P = \operatorname{conv}(K^* \cup G \cdot q)$.

Any isometry of $S$ which is not in $G$ will carry $q$ to a point over some facet which is not a point of $G \cdot q$, hence is not a symmetry of $P$. There is some concern that $K^*$ might have additional symmetries outside $S$—for instance, with the omnitruncated 3-simplex, the initial point $p$ can be chosen such that you get a truncated octahedron with the 48 octahedral symmetries instead of just the 24 tetrahedral symmetries. I don't remember how my source dealt with this, but I can suggest two ways that might work:

• For $d > 3$, the symmetry group of a simplex is not a subgroup of any finite reflection group in $\mathbb{E}^d$.
• Almost any choice for $p$ won't result in the extra symmetries, so just don't pick one that does.

So, $G$ is exactly the symmetry group of $P$. Thus, to answer your questions:

1. Yes, all finite groups appear as some $G_S$.
2. Every finite group appears in $G(\mathbb{R}^d)$ for some $d$. The construction shows that every group with permutation degree $d$ or less appears in $G(\mathbb{R}^d)$. In general, I would expect that most finite groups can be realized as the symmetry group of some polytope in $\mathbb{R}^k$ for some $k$ less than the permutation degree. The characterization of the minimal dimension $k$ such that $G$ is the symmetry group of some $k$-polytope is, as far as I know, unexamined; it probably will turn out to be related to representation theory.
3. I would expect so. I think the symmetry group of the $d$-simplex (the symmetric group $S_{d+1}$) might appear in $G(\mathbb{R}^d)$, but no lower dimension. But I haven't thought about it too much.

The same question was asked on Math Overflow and several of the answers there give different constructions. The accepted answer and a comment appear to answer the question I raised in #2: the minimal $k$ such that $G \in G(\mathbb{R}^k)$ is the least positive integer such that $G$ has a faithful representation in $O(k,\mathbb{R})$.