Is any finite subset of $S^2$ satisfying the following properties, the vertex set of a regular convex polyhedron?

Question

Let $S^2 \subset \mathbb{R}^3$ denote the standard unit sphere.

Let $A \subset S^2$ be a finite set satisfying each of the following properties:

For each $a,a' \in A$ there exists a rotation $R$ (the action of a 3x3 orthogonal matrix) such that $R(A) = A$ and $R(a) = a'$.

The set $A$ is not contained in any one plane. (Such a plane need not pass through the origin.)

My question is, in the above situation, is $A$ necessarily the set of vertices of some regular convex polyhedron (embedded in $\mathbb{R}^3$)?

Note: thanks to a comment on my earlier question for pointing out the issues that arise if condition 2 above is not applied — I.A.S. Tambe, Mar 29 '23 at 20:39
The convex hull of such a set is an isogonal polyhedron. This needn't be regular or even uniform; for example, it could be a square prism ($a\times a\times b$ where $a\neq b$). — mr_e_man, Mar 30 '23 at 17:47

score 1 · Accepted Answer · answered Mar 29 '23 at 20:42

1

Look at the Archimedean solids for which

A global isometry of the entire solid takes one vertex to the other while laying the solid directly on its initial position.

There may be other examples.

answered Mar 29 '23 at 20:42

Ethan Bolker

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Thanks very much for your help! – I.A.S. Tambe Mar 29 '23 at 20:51

Is any finite subset of $S^2$ satisfying the following properties, the vertex set of a regular convex polyhedron?

1 Answers1