My question is very basic: where can I find a complete (and hopefully self-contained) proof of the classification of Platonic solids? In all the references that I found, they use Euler's formula $v-e+f=2$ to show that there are exactly five possible triples $(v,e,f)$. But of course this is not a complete proof because it does not rule out the possibility of different configurations or deformations. Has anyone ever written up a complete proof of this statement?!
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$\begingroup$ If you're ruling out Euler's formula you might be starting off with a not-entirely-standard definition of a Platonic solid. Are you not taking Platonic solids to mean a regular convex polyhedron? $\endgroup$– Ryan BudneyCommented Oct 23, 2012 at 20:49
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6$\begingroup$ @Ryan: I'm reading her questions not as objecting to the use of Euler's formula, but rather objecting to stopping with determining the combinatorial type of the polytopes. (Basically, determining the f-vector.) E.g.: why is the cube the only way to fit 6 4-faces together? $\endgroup$– Russ WoodroofeCommented Oct 23, 2012 at 21:07
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3$\begingroup$ Here is a sketch argument for uniqueness: From the combinatorics you know the number of faces and the number of sides on each face (say n). Once you know the number of faces, that determines the spherical area of each face. Up to rotation of the sphere, there is then a unique regular spherical n-gon of the given area. The configuration of faces is unique because there is at most one way to tile the sphere so that the edges match up: start with one face and then reflect repeatedly in its sides. $\endgroup$– Colin ReidCommented Oct 23, 2012 at 21:07
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$\begingroup$ Are you not happy with Euclid's original proof (well, an adaptation of that)? en.wikipedia.org/wiki/Platonic_solid#Geometric_proof $\endgroup$– Marco GollaCommented Oct 23, 2012 at 21:10
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3 Answers
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This is a classical question. Here is my reading of it: Why is there a unique polytope with given combinatorics of faces, which are all regular polygons? Of course, for simple polytopes (tetrahedron, cube, dodecahedron) this is clear, but for the octahedron and icosahedron this is less clear.
The answer lies in the Cauchy's theorem. It was Legendre, while writing his Elements of Geometry and Trigonometry, noticed that Euclid's proof is incomplete in the Elements. Curiously, Euclid finds both radii of inscribed and circumscribed spheres (correctly) without ever explaining why they exist. Cauchy worked out a proof while still a student in 1813, more or less specifically for this purpose. The proof also had a technical gap which was found and patched up by Steinitz in 1920s.
The complete (corrected) proof can be found in the celebrated Proofs from the Book, or in Marcel Berger's Geometry. My book gives a bit more of historical context and some soft arguments (ch. 19). It's worth comparing this proof with (an erroneous) pre-Steinitz exposition, say in Hadamard's Leçons de Géométrie Elémentaire II, or with an early post-Steinitz correct but tedious proof given in (otherwise, excellent) Alexandrov's monograph (see also ch.26 in my book which compares all the approaches).
P.S. Note that Coxeter in Regular Polytopes can completely avoid this issue but taking a different (modern) definition of the regular polytopes (which are symmetric under group actions). For a modern exposition and the state of art of this approach, see McMullen and Schulte's Abstract Regular Polytopes.
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I believe Hermann Weyl's classic book 'Symmetry' discusses this question but I can't recall what his approach is (ie, whether he makes use of the Euler formula).
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$\begingroup$ I don't think he uses Euler's formula (and the Greeks did not either...) $\endgroup$ Commented Oct 24, 2012 at 2:47
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$\begingroup$ Thanks for the clarification, I will need to have a look back to see what he does say! $\endgroup$ Commented Oct 24, 2012 at 3:22
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One place I have seen such results is in Conway et al., Symmetry of things. I am not enough of an expert to be able to comment on the details of the proof.
answered Oct 23, 2012 at 21:15