Realizing $\mathbb{Z}/n \mathbb{Z} \oplus \mathbb{Z}/n\mathbb{Z}$ as an isometry group

Question

Every finite group arises as the isometry group of a subspace of an euclidean space (Albertsona, Boutin, Realizing Finite Groups in Euclidean Space). What are natural examples of spaces realizing the groups

• $\mathbb{Z}/n\mathbb{Z}$
• $\mathbb{Z}/n \mathbb{Z} \oplus \mathbb{Z}/n\mathbb{Z}$ ?

For $\mathbb{Z}/n\mathbb{Z}$ we may simply take a regular $n$-gon whose edges are extended in one consistent direction so that reflections are excluded. But this is not really a space which appears naturally. Also what about $\mathbb{Z}/n \mathbb{Z} \oplus \mathbb{Z}/n\mathbb{Z}$? The more examples, the better.

Answer 1

In $\mathbb R^4$ pick regular $n$-gons around $0$ with one in the $(x,y,0,0)$ plane, and the other in the $(0,0,z,w)$ plane. So essentially the isometries of one do not affect the others.

A "natural" space with $\mathbb Z/n\mathbb Z$ symmetry is a pyramid with base the regular $n$-gon[*].

In general, if there are spaces $U\subseteq \mathbb R^n$ and $V\subseteq R^m$ which are not isometric to each other, with $0\not\in U,V$, with isometry groups $G$ and $H$ respectively, then $U\times 0\cup 0\times V\subset \mathbb R^{n+m}$ will have isometry group $G\times H$. This space will not be connected. (You can have $0\in U,V$ if the isometries of $U,V$ fix $0$ - for the example above, you can take the apex of the pyramid as $0$ since it is fixed by all isometries.)

(You might be able to use $U\times V$, but I feel like there might some additional isometries...)

In the case of $\mathbb Z/n\mathbb Z\times\mathbb Z/n\mathbb Z$, it's hard to do much better than that for the product, because the "natural" representations for $\mathbb Z/n\mathbb Z$ should be rotations, but rotations tend to not commute with each other.

If $G$ acts faithfully on a set $X$ with $m$ elements, $G$ is a subgroup of the group of isometries of the regular $m-1$-simplex. You can probably systematically remove things from that set to ensure that the resulting space has exactly $G$ as the group of isometries.

In particular, $G$ acts on itself faithfully.

[*] When $n=3$, you'd have to pick a pyramid that is not a regular tetrahedron, or you'll get more isometries, of course.

Answer 2

$\mathbb{Z}/n\mathbb{Z}$ is the fundamental group of the 3-dimensional Lens space $L(n,1)$. There is a universal covering map $S^3 \mapsto L(n,1)$ and a deck transformation action of the group $\mathbb{Z} / n \mathbb{Z}$ on the space $S^3$ which acts by isometries of the standard metric on $S^3$. This descends to a metric on $L(n,1)$ which is locally isometric to the standard metric on $S^3$. Of course, $\mathbb{Z}/n\mathbb{Z}$ is not the full isometry group of $S^3$, that would be the Lie group $O(4)$.

Now, depending on what you mean by "natural", do a little perturbation of the metric on $L(n,1)$.

Maybe leave it out in the sun for a few hours, that's a nice natural environment.

Lift the perturbed metric to the universal cover $S^3$. Now the deck transformation group $\mathbb{Z}/n\mathbb{Z}$ is the full isometry group of $S^3$ with respect to this perturbed metric.

You can do a similar construction for any group $G$ and any CW complex $X$ with fundamental group $G$. Whatever nice metric on $X$ you start with, some reasonably nice natural perturbation will yield an action of $G$ on the universal cover of $X$ which is the full isometry group. From a topologists point of view, the "most natural" example of $X$ would be a $K(G,1)$ space, and the "most natural" action of $G$ would be the deck transformation action on the universal covering space of a $K(G,1)$.

So from that topological point of view, the "most natural" action of $\mathbb{Z}/n\mathbb{Z}$ is its deck transformation action on $S^\infty$ with quotient the infinite dimensional Lens space of type $(n,1)$, which is a $K(\mathbb{Z}/n\mathbb{Z},1)$ space. One must still perturb the metric in this situation, in order to arrange that $\mathbb{Z}/n\mathbb{Z}$ the full group of isometries.