Why do all the Platonic Solids exist?

Question

In three dimensions it is quite easy to prove that there exist at most five Platonic Solids. Each has to have at least three polygons meeting at each vertex, and the angles of these polygons have to add up to less than $2\pi$. This narrows down the possibilities to three, four or five triangles, three squares or three pentagons.

But the proof is not quite complete. One also has to show that each of these possibilities is actually realised. Of course it turns out that they all are. I've been wondering about how to prove this without having to construct each of them individually. After reading this answer, I've managed to reconstruct the following proof. I'm wondering if it's valid.

Suppose we want a polyhedron where $m$ $n$-gons meet at each vertex. Take any sphere. By Gauss-Bonnet we can draw a regular $n$-gon on the sphere with angles $2\pi/m$. Draw congruent $n$-gons along each edge of this one, and continue to extend the tiling in this way. Because of our choice of angle these polygons must join up locally. We want to verify that they join up locally.

Consider the topological space with one $n$-gon for each $n$-gon drawn on the sphere, joined along the edges whenever the corresponding $n$-gons share that edge. Then this topological space is a covering space of the sphere. But the sphere is already simply connected, so our covering space must be the sphere itself. So we do have a regular tiling of the sphere. Now create an actual regular polyhedron by taking the convex hull of the vertices.

If this argument does work, can it be simplified so that it could be understood by someone with no knowledge of algebraic topology?

---

Below this line is an attempt by David Speyer to restate the question. I like simplicial complexes better than CW complexes, so I am going to subdivide the polygons in the original question. Instead of a spherical $m$-gon with angles $2 \pi/n$, I'm going to place a vertex in the center of the polygon and connect it to all the vertices and the midpoints of all the edges. So I have $2m$ spherical triangles with angles $\pi/m$, $\pi/n$ and $\pi/2$.

So, here is my rephrasing. Let $(a,b,c)$ be positive integers with $1/a+1/b+1/c > 1$ (in our case, $(2,m,n)$). We form a two dimensional simplicial complex $\Delta$ whose vertices are colored amber, blue and crimson, with two triangles on each edge and $2a$, $2b$, $2c$ triangles around the amber, blue and crimson vertices respectively. One way to make this more precise is to define $W$ to be the group generated by $s_1$, $s_2$, $s_3$ subject to $s_1^2=s_2^2=s_3^2=(s_1 s_2)^a = (s_1 s_3)^b = (s_2 s_3)^c = 1$. Our vertices correspond to cosets of the subgroups $H_a:=\langle s_1, s_2 \rangle$, $H_b:=\langle s_1, s_3 \rangle$ and $H_c:=\langle s_2, s_3 \rangle$, with vertices in the same triangle if they are of the form $(w H_a, w H_b, w H_c)$.

Then $\Delta$ maps to the $2$-sphere, sending our base simplex to the spherical triangle $T$ with angles $(\pi/a, \pi/b, \pi/c)$, and choosing the images of all the other vertices by making $s_1$, $s_2$, $s_3$ act by reflections over the sides of $T$.

Anyone who has taught a course on Coxeter groups knows it is true, but a pain to prove, that the abstractly defined $\Delta$ maps isomorphically to the sphere $S^2$ and, in particular, $W$ is finite.

How much can we reduce the pain by knowing that $S^2$ is simply connected?

Answer 1

I would address this question from the point of view of group theory. Namely, if one starts to classify finite subgroups $G$ of $\mathrm{SO}(3)$ (say, first looking at thier poles – points where an rotation axis intersects the sphere $S^2$), then it's not difficult to prove by orbit counting that there are following possibilities for finite subgroups of $\mathrm{SO}(3)$:

• cyclic
• dihedral
• some group $\mathbf T$ of order 12 with three types of pole orbits of sizes 4, 6 and 4 (hence with stabilisers of sizes 3, 2 and 3);
• some group $\mathbf O$ of order 24 with three types of pole orbits of sizes 8, 12 and 6 (hence with stabilisers of sizes 3, 2 and 4);
• some group $\mathbf I$ of order 60 with three types of pole orbits of sizes 12, 30 and 20 (hence with stabilisers of sizes 5, 2 and 3).

Of course, the tricky point is to check that $\mathbf T$, $\mathbf O$ and $\mathbf I$ exist without appealing to the existence of Platonic solids :) This can be done in at least two ways:

• using explicit presentations of them as described, for instance, in § 57 of the book G. A. Miller, H. F. Blichfeldt, L. E. Dickson, Theory and applications of finite groups, Dover, New York, 1916; for instance, for the tetrahedron it reads $s_1^3=s_2^3 = (s_1s_2)^2 = 1$ which is easily realisable by rotations;
• or (amazingly!) using Riemann surfaces.

Now, if we take any orbit whose stabiliser has size more than 2, (meaning an orbit of size 4 for $\mathbf T$, an orbit of size 8 or 6 for $\mathbf O$, an orbit of size 12 or 20 for $\mathbf I$), then its points will define vertices of a regular polytope (whose edges can be defined as connecting a vertex with nearest vertices).

Each vertex can be rotated by $G$ to another one by construction, and the stabiliser of each vertex consists of rotations around it, which clearly have to permute edges going out of this vertex. Now, the sizes of stabilisers guarantee that there are exactly as many edges going out of each vertex as the order of the stabiliser, and therefore every edge can be rotated to every other edge by an element of $G$. As $G$ clearly preserves faces of our polyhedron, they are forced to be regular polygons. An inspection of orders of stabilisers thus gives the list of Platonic solids.

Notice that the “exceptional” orbits have stabilisers of order 2, so the above construction doesn't work there because there's not enough rotations around these vertices (however, we were only concerned with existence of Platonic solids anyway).