The definition I mean can be found in the tag Wiki of Platonic solid tag and also in Wikipedia:

Definition 1:

A Platonic solid is a regular, convex polyhedron with congruent faces of regular polygons and the same number of faces meeting at each vertex.

Here the role of the first 'regular' adjective is a bit obscure for me. Take the following definition:

Definition 2:

A Platonic solid is a convex polyhedron with congruent faces of regular polygons and the same number of faces meeting at each vertex.

Is there any polyhedron that Platonic by definition 2 and isn't Platonic by definition 1?

  • $\begingroup$ You are right. Following the proofs one sees it is not used to show there are no more than five. $\endgroup$
    – Alamos
    Apr 24, 2015 at 2:33

1 Answer 1


You're absolutely right, the two definitions are equivalent. Here's my perspective on the differences in definitions of regularity, and why flag-transitivity is superior, in general.

In two dimensions:

Regular polygons are those that have equal edges and equal angles.

Then, we use this definition to define "regular" in three dimensions:

Regular polyhedra are those with congruent regular faces, with a constant number meeting at each vertex.

How should we define "regular" $4$-dimensional polytopes?

We could try and mimic the definitions above; its $3$-dimensional faces (facets) should be congruent regular polytopes of one smaller dimension, but what should we make of the vertices? Is it enough to say that "an equal number of facets meet at each vertex", or could the configuration of facets start to play a role, in dimension $4$ or higher? Worse still, do we need to mention anything besides the vertices, in order to ensure the symmetry we wish to describe?

So, rather than search for a theorem that says, "The configuration of facets meeting at a vertex doesn't affect regularity, only the number. Further, only vertices matter, to build a regular polytope from smaller-dimensional regular polytopes", assuming such a theorem even exists, we adopt a more succinct (albeit more abstract) definition of regularity.

We define a flag of a $d$-dimensional polytope $\mathcal{P}$ to be a sequence of faces

$$F_0 \subseteq F_1 \subseteq \ldots f_{d - 1} \subseteq F_d$$

of $\mathcal{P}$, the face $F_k$ having dimension $k$. Then we say that

A polytope is regular if its symmetry group acts transitively on its set of flags.

If we take the time to really understand this definition, it's describing exactly the property the more verbose definitions captures. Not only does every $k$-dimensional face of our polytope "look the same", but every "chain" of faces does: Each $k$-dimensional face has exactly the same relationship with its constituent $(k-1)$-dimensional faces, and our polytope is as symmetric as possible.

So, while it may take a little more time to parse the definition initially, it can describe regular polytopes of any dimension with ease, without resorting to a recursive definition of regularity of facets and some kind of "gluing" instructions on how the facets must be assembled.

  • 1
    $\begingroup$ I think you say, that not the 'regular' is that superfluous is in definition 1, but all the rest :) $\endgroup$
    – mma
    Apr 25, 2015 at 3:17
  • 1
    $\begingroup$ @mma Exactly! For polyhedra, the convex regular ones are exactly those "with congruent faces of regular polygons and the same number of faces meeting at each vertex", regularity (as defined by transitivity) is equivalent to those other criteria. So you could drop "regular", or you could drop the other conditions; they're just superfluous together. $\endgroup$
    – pjs36
    Apr 25, 2015 at 3:21
  • $\begingroup$ So, your favourite definition would be the following. >Platonic solids are the regular, convex 3-polytopes. Or expanded this into human readable language >A convex polyhedron is called Platonic iff for every pair of flags there is a symmetry transformation that brings them into each other. Flags are $vertex \cup edge \cup face$ sets where $vertex$ is in the boundary of $edge$ and $edge$ is in the boundary of $face$. Am I right? $\endgroup$
    – mma
    Apr 25, 2015 at 3:44
  • $\begingroup$ Yeah, that's a good way to phrase it! I guess the point of my answer was that flag-transitivity is really good to describe $n$-dimensional polytopes. It's certainly good to use the "old" definitions (about vertices and congruent faces) when they're reasonable (in small dimensions, $2$ and $3$), and even better to use this understanding to make flag-transitivity feel natural. $\endgroup$
    – pjs36
    Apr 25, 2015 at 4:04
  • $\begingroup$ That's a neat definition. But what is the "symmetry group"? Does it mean the subgroup of O(n,R) that permutes the vertices? Or is this group (and the vertices themselves) something more abstract? Also, it seems that this definition permits infinite tilings as well as platonic solids. How does one fix that? $\endgroup$ May 7, 2022 at 16:24

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