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I'm writing a Bachelor thesis on the symmetries of the Platonic Solids. However, I'm not very strong in geometry and I start the thesis by proving that there are only 5 Platonic Solids. When I define a regular polytope (the majority of the thesis deals with polyhedrons of course, but I'm also showing a few examples of higher dimensional regular polytopes such as the $n$-simplex) I'm uncertain about how to properly define it without including too much geometry, since it's not my main focus. If I define the regular polygon, is it then enough to say that a regular $n$-polytope is an $n$-polytope whose facets are regular $(n-1)$-polytopes, which implies recursion down to $n=2$?

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No. The angles between the facets also have to be equal. Otherwise you could, for example, glue two regular tetrahedra together and get a polytope with six triangular faces.

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  • $\begingroup$ Right, of course. I'll add that as well, but would it then be a sufficient "explanation"? Maybe definition is too strong a word for what I intend to write. $\endgroup$
    – Auclair
    Mar 31 '16 at 21:20
  • $\begingroup$ I'm not sure if that is sufficient for all dimensions or not. The Wikipedia defines them in terms of symmetries: en.m.wikipedia.org/wiki/Regular_polytope $\endgroup$ Mar 31 '16 at 21:42
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It is not enough to say that all its $n-1$ facets are regular. This includes for example the uniform and Johnson solids. It is sufficient to add that they must be also be congruent and the same number meet around any vertex.

A more succinct definition is based on the idea of flags. For this purpose a flag is a connected or incident set of $j$-faces, for one face of each $0 \leq j \leq (n-1)$. We can then say that:

A polytope is regular if its symmetries are transitive on its flags.

That is to say, under any symmetry transformation of the polytope, the flags are permuted into each other.

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