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Summary: What are the possible regular polygons that can be created by the orbit of a 3D point under $O_h$ or $I_h$?

I've been reading about uniform polyhedra for a while now. Something that has stricken me as odd is that, excluding prisms and antiprisms, there are only certain regular polygons that can appear on a uniform polyhedron. When the symmetry group of the figure is a subgroup of $O_h$, the possible number of sides are 3, 4, 6 and 8. Likewise, when the symmetry group is a subgroup of $I_h$, the corresponding numbers of sides are 3, 4, 5, 6 and 10. And moreover, polyhedra with more than 4 sides are always invariant under some rotation of the group. One could prove this by looking at the classification of uniform polyhedra, but that feels like cheating. Instead, I'd like to prove the more general fact that the orbit of a point under one of these groups can only create the corresponding regular polygons.

I've tried looking at the individual symmetries of a regular $n$-gon creating a counterexample, trying to prove they must be contained in the greater symmetry group, to no avail. I've also tried more combinatorial approaches, as well as stuff with vectors, but those just seem to lead to humongous casework. I'm completely stuck at the moment. Any help?

ViHdzP
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$O_h = [3,4]$ has 2-gonal, 3-gonal and 4-gonal rotational symmetry axes. So clearly you could generate a regular triangle and a square ($n$-gon). But when placing the edges somewhat more apart you also can generate a hexagon and an octagon ($2n$-gon).

Same for $I_h = [3,5]$, it has 2-gonal, 3-gonal and 5-gonal rotational symmetry axes. So clearly you could generate a regular triangle and a regular pentagon ($n$-gon). But when placing the edges somewhat more apart you also can generate a square, a hexagon and a decagon ($2n$-gon).

--- rk

Dr. Richard Klitzing
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  • That's all true, but that doesn't guarantee I can't get a regular heptagon somehow, for example. I'm asking if that's the complete list. – ViHdzP Oct 31 '19 at 17:54
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    In any Wythoffian polyhedron (constructable via kaleidoscopical device out of the corresponding Coxeter-Dynkin graph) each $m$-gonal face corresponds exactly to a axis of $n$-fold rotational symmetry where either $m=n$ or $m=2n$. Neither $O_h$ nor $I_h$ has eg. a $C_7$ subgroup. – Dr. Richard Klitzing Feb 20 '20 at 15:39
  • As for snubs the argument is similar. These are obtained secundarily via hemiating (alternated faceting) of (topological variations of) the omnitruncated Wythoffians. Thus their faces are either according hemiated faces or are vertex figures of the former. As the latter are triangles only (for spherical geometry) you're done here too. – Dr. Richard Klitzing Feb 20 '20 at 15:40
  • In fact there is just a single uniform polyhedron, which is neither Wythoffian nor a snub. That one is known as Miller's monster, cf. https://en.wikipedia.org/wiki/Great_dirhombicosidodecahedron . - And that specific one too has no heptagonal face. – Dr. Richard Klitzing Feb 20 '20 at 15:47
  • I actually was trying to find an alternate argument for the enumeration of uniform polyhedra. If we could prove that the possible faces were limited, as well as the number of possible polygons per vertex figure, a more systematic argument than Skilling's original one might be possible. But alas, this doesn't seem at all easier. – ViHdzP Mar 09 '20 at 19:31
  • The answer to this quest https://math.stackexchange.com/questions/3498904/what-are-the-main-ideas-needed-to-prove-that-only-92-johnson-solids-exist?rq=1 links the original paper of Johnson which proves that only a quite limited set of n-gons can participate. – Dr. Richard Klitzing Mar 10 '20 at 06:35