Proof that there are at least 5 regular polyhedra?

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DavidRoss
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Proof that there are at least 5 regular polyhedra?

Postby DavidRoss » Thu Jun 02, 2011 6:52 am UTC

There is a proof that there are no more than five regular polyhedra? The upper bound proof looks at how many regular polygons of different orders can meet at a vertex point and form a convex surface. There are only five ways of doing that (3, 4, or 5 triangles, 3 squares, 3 pentagons). That's the upper bound. The lower bound is typically proven by brute force (which is admittedly not a lot of force in this case) by trotting out the five Platonic solids.

Is there a way to prove that all possible combinations MUST result in a polyhedra, besides the brute force method? In other words, suppose there is a construction rule that, starting with a point, you add q instances of a p-edged face so that their edges coincide, then move to other partially covered vertices, add more instances to get to a total of q instances meeting at that next vertex, then repeat until no more faces can be added. Can it be proven that the construction rule requires that there be no gaps? We know by looking at the five examples that if we tile a ball of clay with them according to the construction rule, all five all happen to meet up on the other side exactly, but is that a happy circumstance or a necessary consequence of the construction rule?

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silverhammermba
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Re: Proof that there are at least 5 regular polyhedra?

Postby silverhammermba » Thu Jun 02, 2011 9:01 am UTC

If such a general proof does exist, I doubt it would be particularly illuminating since we already know it only applies to five specific cases.

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Re: Proof that there are at least 5 regular polyhedra?

Postby t1mm01994 » Thu Jun 02, 2011 10:14 am UTC

Your question does not need a general proof, as you have already given the proof in the OP. You have 5 examples, so you are sure there are at least 5. Qed.
Similar to: Prove that y=(x-1)(x-2) has at least 2 solutions over the reals. No complicated stuff needed, a proof could be as follows:
To show there are at least 2, I will show there are 2, namely x=1 and x=2.
Firstly, fill in 1. this will lead to y=0(-1), thus y=0. So x=1 is a solution.
Now let's fill in 2. this will lead to y=1(0), thus y=0. So x=2 is a solution too.
So there are at least 2 solutions!
Nothing more needed.

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Re: Proof that there are at least 5 regular polyhedra?

Postby Token » Thu Jun 02, 2011 10:17 am UTC

So, the OP asks if there's a proof beside exhibiting the examples, and you tell him there's no need to have one because he can just exhibit the examples? I'm getting the feeling that you've missed the point somewhere...
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Re: Proof that there are at least 5 regular polyhedra?

Postby skeptical scientist » Thu Jun 02, 2011 11:21 am UTC

One place to look might be this book.
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Re: Proof that there are at least 5 regular polyhedra?

Postby achan1058 » Thu Jun 02, 2011 12:59 pm UTC

Token wrote:So, the OP asks if there's a proof beside exhibiting the examples, and you tell him there's no need to have one because he can just exhibit the examples? I'm getting the feeling that you've missed the point somewhere...
Indeed, one can extend this question to any dimensions d>=3.

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Re: Proof that there are at least 5 regular polyhedra?

Postby Incompetent » Thu Jun 02, 2011 2:14 pm UTC

A fancier approach to regular polyhedra is to classify the Coxeter groups of rank 3 (there are 3 of them) and show these admit faithful representations in R^3 in which the generators act as reflections. You can reconstruct the polyhedra by taking a suitably chosen point on the unit sphere and using its orbit under the group action as the set of vertices: for the tetrahedron, octahedron and icosahedron you can take the intersection of the sphere with two reflecting planes, and for the tetrahedron, cube and dodecahedron you can take the midpoint (on the sphere) between two adjacent intersection points. You can then show the polyhedron is regular by showing the group acts transitively on vertices, edges and faces (strictly speaking, you should check the action is transitive on maximal flags, where a maximal flag consists of a vertex, an edge and a face, all incident with each other).

I don't know if this is any improvement on the naïve approach, though.

The 'upper bound' you mention is a combinatorial restriction on the possible polyhedra. It happens to be strong enough in 3 dimensions to give exactly the list of regular polyhedra, but in higher dimensions, the combinatorial restriction is weaker than the geometric one (i.e. the requirement that the combinatorial symmetry is realisable by isometries): you get 'abstract polytopes' which cannot be realised as regular 'geometric polytopes' of the same dimension. So you are right, the combinatorial restriction is not by itself an existence argument.

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Re: Proof that there are at least 5 regular polyhedra?

Postby DavidRoss » Thu Jun 02, 2011 4:13 pm UTC

skeptical scientist wrote:One place to look might be this book.

Thanks. I happen to be in possession of that book (not sure how that came to be). I'll look again, but Coxeter's discussion of regular polyhedra in R^3 starts with the entirely reasonable assumption that all five do exist. His definition of a regular polyhedra is regular faces, equal faces and equal solid angles, i.e., start a vertex with a given p and q and then only use that p and q. It doesn't jump out at me why (1) you just so happen to need an integer number of faces or (2) there are no gaps.

Coxeter does show that the solid angle subtended by a face relative to the intersection of the center normals of adjacent faces is a unit fraction of 4 * pi * r^2, but a proof that the sum of the areas of the puzzle parts is equal to the area of puzzle is not a proof that the pieces can be tiled without gaps.

(Yes, I do recognize that such a proof is not useful in R^3.)

There is a symmetry argument, but I am not sure that is enough. There are no gaps at the first vertex, and by rotating one of the faces to put the first vertex at the old position of a second vertex, the rotated face is unchanged and the lack of gaps at the first vertex forces the lack of gaps at the second vertex and from there one can rotate to additional vertices with the same result. But is that enough of a proof that everything will meet on the other side?

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Re: Proof that there are at least 5 regular polyhedra?

Postby jaap » Thu Jun 02, 2011 5:29 pm UTC

DavidRoss wrote:There is a symmetry argument, but I am not sure that is enough. There are no gaps at the first vertex, and by rotating one of the faces to put the first vertex at the old position of a second vertex, the rotated face is unchanged and the lack of gaps at the first vertex forces the lack of gaps at the second vertex and from there one can rotate to additional vertices with the same result. But is that enough of a proof that everything will meet on the other side?


The symmetry argument does at least show that there is no local impediment to adding faces around each vertex. The symmetry argument applies just as well to planar infinite tilings though (e.g. 63, 44, 36), so to show that you get a polyhedron you have to show that there are only a finite number of faces resulting from that building process.

Maybe you could transfer it to a pure algebra setting, for example by showing that { a, b | a2 = bm = (ab)n = 1 } is a finite group, where n is the number sides of a face and m the number of faces at a vertex. The group elements a, b, and ab correspond to rotations at an edge midpoint, vertex, and face centre respectively.

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Re: Proof that there are at least 5 regular polyhedra?

Postby Jyrki » Thu Jun 02, 2011 5:38 pm UTC

I don't know if it helps you much, but since you have Coxeter's book at hand here is a possibility.

If you believe in the classification of finite groups generated by reflections (these are often also called Coxeter groups, and are probably covered in his book), then you get a sort of uniform construction of the regular polyhedra. Select a poin in the fundamental Weyl chamber that is equidistant from a suitable subset of the wall bounding the chamber, and is on the remaining walls. Study the orbit of that point, and start forming edges and faces.

This is a bit vague, sorry, but I think you should get all five out of this by varying the choice of the Coxeter group and subsets of walls.

As a bonus you get that the symmetry group acts transitively on edges, vertices and faces.

Edit: I think what I said is more than a tad too sweeping. IOW it's not true. :oops:

I still think that you can get all the regular polyhedra in this way, but what I said about the action of the symmetry group
being transitive on edges cannot be true. Take the icosahedral group G of 120 elements (= the group of symmetries of a regular icosahedron or a dodecahedron). A full orbit has 120 points. Each point P in the orbit then has 3 closest neighbors (reflect P with respect to one of the 3 walls of the chamber). If we connect all those neighbors we get 180 edges, so obviously the action of G is not transitive on the set of edges.

In this case the three walls of a chamber have angles Pi/2, Pi/3 and Pi/5 between them. We do get the set of vertices of an icosahedron or a dodecahedron by using the orbit of a point P that lies in the intersection of a pair of non-orthogonal walls of a chamber, because then the stabilizer under the G-action is a dihedral group of order 6 or 10 (depending on whether the pair of walls has angle Pi/3 or Pi/5 between them). Then the orbit sizes are 20 and 12 respectively, as desired. The stabilizer of P then also acts transitively on the set of closest neighbors of P, and thus we get G acting transitively on the set of edges. However, we still need a coplanarity condition for the vertices belonging to a face. That is a bit more subtle, and I don't see it right away. Coxeter is the place to look.

Sorry.

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yeyui
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Re: Proof that there are at least 5 regular polyhedra?

Postby yeyui » Tue Jun 07, 2011 5:30 pm UTC

Generally, we are not interested in non-constructive existence proofs when constructive ones are available. Counterexamples to this maxim are usually situations where a single general existence proof covers a large number of objects requiring many different constructions. In this case it really only takes three constructions to build the five platonic solids. However, this is still an interesting challenge, especially when applied to a classic problem. I think that showing the existence of at least five distinct regular polyhedra without giving a construction of any of them will be difficult.

Let me ponder the kinds of non-constructive existence proofs I've seen for polyhedra. Sometimes one can show that an object can be realized as the convex hull of points without knowing the coordinates of the points. We also need to show regularity. An algebraic approach was mentioned above. Perhaps one can show that every group with property X corresponds to a regular polyhedron. But then you have just pushed to problem to the realm of algebra. You must now show that there exist enough of these groups without identifying them explicitly (otherwise this gives a roundabout construction of the polyhedra).

An interesting approach (although perhaps not possible) would a counting argument along the lines of: There are at least X objects in class A, but at most Y are not regular polyhedra (where X-Y is at least 5).

DavidRoss
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Re: Proof that there are at least 5 regular polyhedra?

Postby DavidRoss » Wed Jun 15, 2011 11:27 pm UTC

yeyui wrote:Generally, we are not interested in non-constructive existence proofs when constructive ones are available.


Well, I could use the excuse that "hey, the non-constructive existence proof is useful for higher dimensions" but I really don't have any true justification for needing it. I just observed that "hey, the files meet around the back perfectly every time" and then looking for an existence proof that has some term or step that signals that that will be the case. For example, line up the Taylor series for sin, cos and e and it leaps out at you that e^ix = cos(x) + i*sin(x). Another example is the derivation - using a Fourier Series - of the set of all closed curves wherein the ratio of the enclosed area to the length of the curve (in 2D) is minimized and it leaps out at you that after all of the terms of the series fall away, you're left with only the terms for a circle.

I think I will have to settle for the conversion of the question into algebra and using symmetry.


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