# How to show that if two Platonic solids have the same number of edges, vertices, and faces, then they are similar in $\mathbb{R}^{3}$?

Note: It appears that some of the terms here do not have standardized definitions, so some sources may give conflicting info.

I was looking into the proof that there are only five Platonic solids in Basic Concepts of Algebraic Topology by F.H. Croom at page 29, Theorem 2.7. To clarify,

• We define a Platonic solid as a simple, regular polyhedron homeomorphic to $$S^{2}$$.
• We define a simple polyhedron to be a polyhedron that does not self-intersect itself.
• We define a regular polyhedron to be a polyhedron whose faces are regular polygons all congruent to each other and whose local regions near the vertices are all congruent to each other.

Using homology theory, one can prove that the Euler formula $$V-E+F=2$$ must hold for Platonic solids. Then by using Euler's formula and invoking a counting argument, we find that there are five possible tuples $$(V, E, F)$$. This is a beautiful proof, but I am unsatisfied with a question: How do we know there can't be two non-similar Platonic solids that have the same $$(V, E, F)$$-tuple?

Almost all sources I've looked at seem to assume it is obvious that two Platonic solids with the same $$(V, E, F)$$-tuple are similar, and it is not obvious to me.

Does anyone have any suggestions for how to prove this? Alternatively, does anyone know of a reference where this is proved rigorously?

Edit 1: It's not completely clear, but it seems like the definition I used for "regular polyhedra" is different than the one commonly used. Note that I am not assuming any global symmetry, so if any global symmetry is to be invoked, it needs to be proven.

Edit 2: I've been made aware of Cauchy's rigidity theorem, which is proven in, e.g., Proofs From the BOOK by Aigner & Zeigler. One can show that any two Platonic solids that have the same $$(V, E, F)$$-tuple must be combinatorically equivalent. However, in order for the theorem to apply, we need to show that our Platonic solids are convex. I can't seem to think of any rigorous argument for why the Platonic solids have to be convex.

And actually, you don't need to show that the entire polyhedron is convex. If I'm not mistaken, the proof for Cauchy's rigidity theorem only relies on the vertices of the polyhedron being locally convex. So really it suffices to show the vertices are convex.

• What are you really uncertain about? That it's combinatorics (what faces are incident in what way etc.) might not be unique, or thats it's geometry (exact relative location of the vertices) might not be unique? If you accept that the combinatorics is unique, then the unique geometry follows from Cauchy's rigidity theorem. – M. Winter Nov 13 '20 at 21:26
• @M.Winter I accept that the combinatorics is unique; it's the geometry that is unclear to me. I've been made aware of Cauchy's rigidity theorem recently, but it only applies to convex polyhedra. My trouble now is that I can't find any rigorous argument for why the properties I outlined imply the polyhedron has to be convex. – Maximal Ideal Nov 13 '20 at 21:48
• I would then be interested in your definition of polyhedron. Apparently it does not include convexity. The uniqueness of the Platonic solids however requires convexity, as otherwise there are also the Kepler-Poinsot polyhedra (the "non-convex regular polyhedra"). For example, the great dodecahedron satisfies your definition. – M. Winter Nov 13 '20 at 22:15
• It seems reasonable to only require the weaker condition that the polyhedron is not self-intersecting; can one infer convexity from this somehow? – RavenclawPrefect Nov 14 '20 at 0:48
• Yeah, I'll have to think about this more. I very much appreciate the clarifications. I was actually willing to let this go and assume convexity. However, I'm now really curious if we can prove convexity (or vertex-convexity) by assuming no self-intersections. (I'm starting to wonder whether this is even an answered question. I may rephrase my entire question some time soon.) – Maximal Ideal Nov 14 '20 at 1:33

## 4 Answers

Edit: This answer invokes rotational symmetry about the vertices, which isn't required in the OP, so this is not a complete answer without a proof that such symmetry ought to hold.

Given $$V,E,F$$, you can compute the number of sides $$n$$ of each face ($$2E/F$$) and the degree $$k$$ of each vertex ($$2E/V$$).

By regularity, a Platonic solid with degree $$k$$ should be invariant under a rotation by $$360/k$$ degrees about a vertex. This forces the arrangement of regular $$n$$-gons at a vertex to be "rigid", because this rotational symmetry along with a specific internal angle of each polygonal face forces a unique relative position of the edges from a vertex, up to rotation of the whole configuration in space. See Eric Wofsey's comment for more elaboration. (Also note that the uniqueness of such a configuration is reliant on there being no self-intersections between the polygons around the vertex; if you allow this, the Kepler-Poinsot polyhedra result.)

Furthermore, this rigid configuration about a vertex is fully determined by the position of three of the edges in the configuration; two edges are enough to determine the configuration up to a reflection, and the third tells us whether the vertex is convex or not (whether the corner points "out" or "in").

So, if we start with a vertex and an orientation of its local neighborhood, we get a rigid structure of the polygons meeting at a single vertex. This gives us a ring of "boundary" vertices, where the neighboring polygons have not yet been fully specified. We repeatedly choose a boundary vertex where two boundary polygons meet, which has at least 3 fixed edges attached to it and therefore forces a unique attachment of polygons from that vertex by the above paragraph. By adding in these polygons to our rigid structure, this ring of boundary vertices grows outward, and we can repeat the process. So long as we perform this extension on the boundary vertices closest (in the graph-theoretic sense) to our starting vertex, we will eventually specify the local neighborhood of every vertex in the graph (possibly after infinitely many operations; we aren't assuming that things will naturally close up here.)

The result of this procedure is a surface without boundary, so if the polyhedron is homeomorphic to the $$2$$-sphere, this process must have terminated after finitely many points and closed up into a unique finite solid.

(Of course, this does not show that any such polyhedron does exist - see this question for an algebraic topology proof of existence. Essentially, the vertex-extending process described above forms a covering space of $$S^2$$, and so must actually yield the sphere.)

• Note that it is not quite obvious that the rotational symmetry gives you local rigidity at each vertex. If you view the axis of symmetry around a vertex as being vertical, then to determine the edges coming out of the vertex (up to rotation), you need to know the angle $\theta$ they form with the axis of symmetry. Now the key point is that there is a unique value of $\theta$ that will make $k$ equally spaced edges around the axis form the angles of a regular $n$-gon between each other (since the angles formed between the edges strictly decrease as you move $\theta$ away from a right angle). – Eric Wofsey Nov 10 '20 at 22:06
• Note also that OP is not using the symmetry-definition of Platonic solid, but uses that all faces and vertices are locally congruent. It needs quite some hand waving (or even much more rigor) to conclude symmetry (see e.g. the Local Theorem for tilings). – M. Winter Nov 13 '20 at 21:29
• Ah, so they are - thanks for pointing this out! I've fixed the answer's reliance on convexity, but I don't see how to remove the assumption of rotational symmetry. – RavenclawPrefect Nov 14 '20 at 1:16

Another way to interpret the $$5$$ Platonic solids is that they are the only configurations of at least $$3$$ regular polygons around each vertex satisfying that the total sum of angles at that vertex is less than $$180^{\circ}$$ Also note that each Platonic solid is uniquely determined by the number of faces around each vertex and the number of sides of each face because all Platonic solids are vertex-transitive and rotationally symmetric about each vertex. Then, note that (a) the number of sides to each face is determined by $$n_{f} = \frac{2e}{f}$$, and (b) the number of faces around each vertex is determined by $$n_{v} = \frac{2e}{v}$$. Thus, each Platonic solid is uniquely determined by the choice of $$(v, e, f)$$, or any two of them given Euler's relation. $$\blacksquare$$

• It seems that this is glossing over the essential point: why do vertex-transitivity and rotational symmetry about each vertex mean the Platonic solid is determined up to similarity by the numerical data? – Eric Wofsey Nov 10 '20 at 22:14

If I understand you correctly, then you want your polyhedron to be an embedding of the sphere with a well-defined interior. So we can consider it as the boundary of this interior (this will be important later).

We can now speak about the (generalized) Gaussian curvature of this boundary. By the Gauss-Bonnett theorem, the total curvature of this surface is $$2\pi$$ times the Euler characteristic of the sphere, hence positive. But in a polyhedron, the curvature is concentrated in the vertices (i.e. is zero everywhere else). The "curvature" at a vertex is better know as its angular defect, and since all vertices are locally the same, they all must have the same angular defect. But since these values add up to a positive value, each such angular defect must be positive.

In sum, the curvature is non-negative everywhere. Now there seems to be the following theorem: the boundary of a shape is convex if and only if it has non-negative curvature everywhere (see the answer to this question). Hence your polyhedron is convex.

Once we know this (and since you already know that the combinatorics is unique) you can apply Cauchy's rigidity theorem to conclude the uniqueness of the Platonic solids.

• This is very close to what I'm looking for. I hope you have the patience to bare with me, but it seems like that convexity theorem doesn't apply to non-smooth surfaces. You can have a non-convex polyhedron whose vertices all have the same positive angular defect. For example, take an icosahedron, then take the "top dome" (i.e. five faces touching a common vertex), and then "punch the dome inward" (i.e. reflect it so that the shape is now concave). Each vertex still has angular defect = $2\pi-5\cdot \pi/3 = \pi/3$ which is positive but the shape is not convex. – Maximal Ideal Nov 16 '20 at 18:38
• @MaximalIdeal That is an unexpected observation you make there. I do not yet have a response to that and it might invalidate my approach. I will have to come back to you when I have an idea how to resolve this. – M. Winter Nov 16 '20 at 18:44

OK so we can imagine two solids, each with the same triple of $$(V,E,F)$$ each has the same number of faces. And you calculate the number of edges that each face has as $$2E/F$$ (each edge can be in two faces). So they have the same number of faces and each face has the same number of edges. How many edges meet at each vertex? This is $$2E/V$$ (each edge hits two vertices). So these two polyhedra have the same number of faces, each with the same number of sides and the same number of edges meeting at each vertex so they must be the same polyhedron. We could build each to the precise formula of faces, number of edges per face and number of edges per vertex and we have no room to make choices. The recipe precisely defines the polyhedron.