August 16, 2000

This Week's Finds in Mathematical Physics (Week 155)

John Baez

It's a hot summer day here in Riverside, so I just want to have fun. Break out the Klein bottles and Platonic solids!

I still remember the day as a kid when I first made a Möbius strip, and saw how it didn't fall apart when cut in half. I could see it, but I couldn't quite grok it. I was fascinated - and more than a little annoyed when it turned out my dad already knew about it.

I don't remember exactly when I first saw a Klein bottle, but I loved it at first sight:

            A mathematician named Klein
            Thought the Möbius strip was divine.
            Said he: "If you glue 
            The edges of two
            You'll get a weird bottle like mine!"

1) Acme Klein bottles sliced in half, http://www.kleinbottle.com/sliced_klein_bottles.htm

Meanwhile, I've been thinking about the Platonic solids lately, and also their generalizations to higher dimensions - the so-called "regular polytopes". To really learn about regular polytopes, you have to go to the source: the king of geometry, Harold Scott Macdonald Coxeter. But for some reason I didn't get around to reading his books until just recently:

2) H. S. M. Coxeter, Regular Polytopes, 3rd edition, Dover, New York, 1973.

Regular Complex Polytopes, 2nd edition, Cambridge U. Press, Cambridge, 1991.

Now my head is full of neat facts about regular polytopes, so I want to rattle some off before I forget!

Let's start in 3 dimensions. I assume you're friends with the tetrahedron, cube, octahedron, dodecahedron and icosahedron. But you might not know all the nice relationships between them!

For example, there's a nice way to fit a tetrahedron snugly into a cube: if you take every other vertex of the cube, you get the vertices of a tetrahedron. And of course I mean a regular tetrahedron - I'm not interested in any other kind, here. There are two ways to do this, and if you put both these tetrahedra inside the cube, they combine to form a star-shaped solid called the "stella octangula". This was discovered and given its name by Kepler, who was really fond of this sort of thing.

Here's a picture:

3) Eric Weisstein, stella octangula, http://mathworld.wolfram.com/StellaOctangula.html

You can rotate it by grabbing it with your mouse!

Similarly, there is a nice way to fit a cube in a dodecahedron. The dodecahedron has 20 vertices, and we can use 8 of these as the vertices of a cube. This becomes obvious once we realize that these points are the vertices of a dodecahedron:

                (+-1/G, +-G, 0),
                (+-G, 0, +-1/G),
                (0, +-1/G, +-G),
                (+-1, +-1, +-1),

By rotating the whole picture, we get some other ways of putting a cube in a dodecahedron: 5 in all. Any rotation of the dodecahedron permutes these 5 cubes, and we get all even permutations of the cubes this way: this is one nice way to prove that the rotational symmetry group of the dodecahedron is A₅ (the group of even permutations of 5 things).

If we put all 5 cubes inside the dodecahedron, we get a fancy shape that would make a marvelous Christmas tree decoration - I don't know what it's called, but you can see a picture of it in Coxeter's "Regular Polytopes", and also here:

4) Eric Weisstein, cube 5-compound, http://mathworld.wolfram.com/Cube5-Compound.html

Now let's combine these two tricks. If we put a tetrahedron in a cube, and then put the cube in a dodecahedron, we get a way of fitting the tetrahedron snugly into the dodecahedron! If we choose one way of doing this and then rotate the picture to get other ways, we get 5 tetrahedra in the dodecahedron. Putting these all together gives a scary-looking shape:

5) Eric Weisstein, tetrahedron 5-compound, http://mathworld.wolfram.com/Tetrahedron5-Compound.html

but the coolest thing about this shape is that it has an inherent handedness - like a sugar molecule, it comes in "levo" and "dextro" forms! If we reflect it, we get 5 other ways to put a tetrahedron into a dodecahedron, for a total of 10. All of these tetrahedra taken together form a mirror-symmetric shape:

6) Eric Weisstein, tetrahedron 10-compound, http://mathworld.wolfram.com/Tetrahedron10-Compound.html

Okay. So far we've related the tetrahedron, the cube and the dodecahedron. What about the other two Platonic solids: the octahedron and icosahedron? Well, from the point of view of symmetry groups these guys are redundant. The octahedron is dual to the cube, so it has the same rotational symmmetry group. Similarly, the icosahedron is dual to the dodecahedron and has the same symmetry group.

From the group-theoretic viewpoint, here's what's really going on. Our trick for fitting the tetrahedron in the cube lets us turn any symmetry of the tetrahedron into a symmetry of the cube. The rotational symmetry group of the tetrahedron is A₄ - that is, all even permutations of the 4 vertices. The symmetry group of the cube is S₄ - that is, all permutations of the 4 lines connecting opposite vertices. So what we've got is a trick for making A₄ into a subgroup of S₄.

(This immediately leads to a little puzzle. There's an obvious way to find A₄ as a subgroup of S₄, since even permutations are a special case of permutations. So: does the above trick give this obvious way, or some other way?)

Anyway, it's also true that any way of fitting the tetrahedron in the dodecahedron lets us turn any symmetry of the tetrahedron into a symmetry of the dodecahedron. So we've also got a trick for making A₄ into a subgroup of A₅.

(You might also think that our trick for fitting the cube in the dodecahedron gives a way to turn any symmetry of the cube into a symmetry of the dodecahedron. I thought this for a while, but it's not true! For starters, if it were true, we'd get a trick for making S₄ into a subgroup of A₅ - which is impossible, since the order of the group S₄ doesn't divide that of A₅. And the problem turns out to be this: a 90 degree rotation of the cube does not correspond to a symmetry of the dodecahedron.) Playing with this stuff would be a nice way to start learning group theory - but even if you already know group theory, it's sort of fun. For more along these lines, try:

7) John Baez, Some thoughts on the number 6, http://math.ucr.edu/home/baez/six.html

Check out the new link to Greg Egan's website illustrating some of the concepts!

Now, despite their redundancy from the group-theoretic viewpoint, it's unfair to leave the octahedron and icosahedron out in the cold. These guys should be related somehow. After all, the octahedron has 12 edges, while the icosahedron has 12 vertices. Is there any way we can exploit this fact?

Yes! The octahedron is the only Platonic solid whose faces can be colored black and white so that no two faces of the same color share an edge. So go ahead: grab a regular octahedron and color it like that. Next, pick an edge and start marching along it with the white face to your left and the black face to your right. Go 1/Gth of the way, where G is the golden ratio again, and mark this point with a dot. Now do this for all the edges. You'll get 12 dots - and these dots form the vertices of a regular icosahedron!

Next, let's take a quick tour of the 4th dimension. This is the most exciting dimension for regular polytopes. In all higher dimensions there are only three - analogues of the tetrahedron, cube and octahedron. But in 4 dimensions, there are six.

I won't describe these systematically here. For that you should read Coxeter's books, or if you're in a rush, my webpage:

8) John Baez, Platonic solids in all dimensions, http://math.ucr.edu/home/baez/platonic.html

Instead, I'll just talk about a cool relationship between my two favorite 4d regular polytopes: the 24-cell and the 600-cell.

First let me set the stage, by reminding you what these look like. A 24-cell looks like this:

9) Eric Weisstein, 24-cell, http://mathworld.wolfram.com/24-Cell.html

To visualize it on your own, first imagine a hypercube with vertices

                   (+-1, +-1, +-1, +-1).

                     (+-2, 0, 0, 0),
                     (0, +-2, 0, 0),
                     (0, 0, +-2, 0),
                     (0, 0, 0, +-2).

But the 24-cell does not get its name from having 24 vertices. It gets its name from having 24 faces! It has 24 octahedral faces, 96 triangles, 96 edges and 24 vertices. The symmetry here comes from the fact that the 24-cell is self-dual - which comes from the fact that we've built it from two polytopes that are dual to each other: the hypercube and the cross-polytope.

What would happen if we had tried this trick in 3 dimensions? Let's see! Take a cube and take an octahedron. Center them both at the origin, line them up nicely, and rescale them so all their vertices are the same distance from the origin: say

                      (+-1, +-1, +-1)

                     (+-sqrt(3), 0, 0),
                     (0, +-sqrt(3), 0),
                     (0, 0, +-sqrt(3)).

10) Kevin Brown, Kepler's rhombic dodecahedron, http://www.seanet.com/~ksbrown/coinc2.htm

Here's another way to think about this stuff. Take two cubes, equal in size. Chop one up into 6 pyramids, each having one face of the cube as its base, and each having the cube's center as its apex. Now take these 6 pyramids and glue their bases onto the faces of the other cube. What do you get? A rhombic dodecahedron! If can't visualize this, go here:

11) Mark Newbold's rhombic dodecahedron page, http://dogfeathers.com/mark/rhdodec.html

Now do the same thing in 4 dimensions. Take two hypercubes, equal in size. Chop one up into 8 "hyperpyramids", each having one face of the hypercube as its base, and each having the hypercube's center as its apex. Now take these hyperpyramids and glue their bases onto the faces of the other hypercube. What do you get? The 24-cell!

(Of course, one can play this game in any dimension, but it works best in dimension 4. I could explain why, but it's probably better to figure it out yourself.)

Okay. Now for the 600-cell. This one is harder: it has 600 tetrahedral faces, 1200 triangles, 720 edges, and 120 vertices. When Buckminster Fuller died and went to heaven, he probably took up residence in one of these. It looks like this:

12) Eric Weisstein, 600-cell, http://mathworld.wolfram.com/600-Cell.html

Here's how you build one. Start with 600 regular tetrahedra. Take 20 of them and glue them together so they all meet at one vertex and the outside looks just like an icosahedron. Of course you can't do this in flat 3-dimensional space: there's "wiggle room" left over when you try! So you have to bend the whole setup a little bit into the 4th dimension, like a piece of a 4d geodesic dome. Then keep adding more tetrahedra, always making sure that 20 meet at each vertex in an icosahedral pattern. By the time you've used up all of them, your 600-cell will be complete - a nice rigid structure.

Of course, if you're a mathematician, there are other more elegant ways to build your 600-cell. For example: start with an icosahedron. Its rotational symmetry group is a 60-element subgroup of SO(3). Using the double cover SU(2) → SO(3) lift this to a 120-element subgroup of SU(2). But SU(2) is isomorphic to the unit quaternions, so we get 120 points on the unit sphere in 4 dimensions. These are the vertices of the 600-cell!

In fact we can construct the 24-cell in the same way, as I explained in week91. Here we start with the tetrahedron, whose rotational symmetry group is a 12-element subgroup of SO(3), and we get 24 points on the unit sphere in 4 dimensions, which are the vertices of the 24-cell.

But if you really want to show off, you can build a 600-cell starting from a 24-cell! Here's how. It's a bit like the trick where we started with an octahedron, systematically marked a point 1/Gth of the way along each edge, and got the vertices of an icosahedron. But it's fancier.

Start with the 24-cell. Take any edge, start walking down it, and when you've gone 1/Gth of the way, mark that point with a dot. Of course these instructions are ambiguous, since I didn't tell you which end of the edge to start at! I could tell you, but I won't - I'll just say that if you do it the right way, you'll get 96 dots which are the vertices of a marvelous polytope in 4 dimensions. It's not a regular polytope, but it's "semiregular": it has 24 regular icosahedra and 120 regular tetrahedra as faces. Coxeter calls it s{3,4,3}, but it really deserves a more glamorous name.

Now as we've seen, in 4 dimensions there is a way to glue 20 tetrahedra together in an icosahedral pattern. You can picture this as a squat pyramid-shaped gadget with a regular icosahedron as base and 20 tetrahedral faces all meeting at the apex.

So: glue one of these pyramid-shaped gadgets onto each of the 24 icosahedral faces of our s{3,4,3}. We get a polytope which has 20 x 24 new tetrahedral faces in addition to the 120 original tetrahedral faces of our s{3,4,3}, for a total of 600. Voila - it's the 600-cell!

For the proof that all this works as advertised, read Coxeter's "Regular Polytopes". Note that it's really easier to work backwards: start with the 600-cell, then truncate it to get s{3,4,3}.

Okay, now for one last trick. I actually thought of this myself - though I can't believe it's new. It gives a way to see the vertices of the 24-cell as a subset of the vertices of the 600-cell.

As I already said, the rotational symmetry group of the tetrahedron has a "double cover" consisting of 24 unit quaternions, which happen to be precisely the vertices of the 24-cell.

Similarly, the rotational symmetry group of the dodecahedron has a "double cover" consisting of 120 unit quaternions, which happen to be precisely the vertices of the 600-cell.

Any way of fitting the tetrahedron snugly into the dodecahedron therefore gives a way of making the vertices of the 24-cell into a subset of the vertices of the 600-cell!

Now, we've already seen 10 ways of snugly fitting the tetrahedron into the dodecahedron: 5 which make the "levo" form of that scary-looking shape, and 5 which make the "dextro" form. The first 5 give 5 different ways of stuffing the 24-cell into the 600-cell. But the second 5 give nothing new.

So this trick actually gives us 5 ways of making the vertices of the 24-cell into a subset of the vertices of the 600-cell. And all these ways have one vertex in common, corresponding to the element 1 of the unit quaternions.

Okay, that's it for this week. No serious stuff this time. I just want to mention that in addition to the above websites, there are a lot that show polyhedra in a way that requires red-blue 3d glasses or a VRML plugin. Since I don't have either of these, and you might not either, I've avoided links to those pages. By the way, VRML stands for "virtual reality modelling language", but it's really just a language for delivering interactive 3d objects over the web. If you can handle VRML, you can probably have a lot of fun here:

13) George W. Hart's Pavilion of Polyhedrality, http://www.georgehart.com/pavilion.html

If you don't, you can still enjoy the annotated bibliography and links to other websites. You can also get a lot out of Vladimir Bulatov's collection of polyhedra without VRML, but again, it's better if you have it:

14) Victor Bulatov's Polyhedra Collection, http://www.physics.orst.edu/~bulatov/polyhedra/index.html

Finally, if you're good at crossing your eyes, you can see some 4-dimensional polytopes at this website, which also has a lot of cool information on how the 4d regular polytopes are related to other branches of math:

15) Tony Smith, 24-cell animation, 120-cell, 600-cell, http://www.innerx.net/personal/tsmith/24anime.html

I thank Jim Heckman and Noam Elkies for helping me fix some errors in the original version of this article.