## Symmetry and the Fourth Dimension (Part 7)

Lately I’ve been showing you what happens if you start with a Platonic solid and start chopping off its corners, more and more, until you get the dual Platonic solid. There’s just one I haven’t done yet: the tetrahedron.

This is the simplest Platonic solid, so why did I wait and do this it last?

Because the tetrahedron is self-dual. Remember, the Coxeter diagram of this shape looks like this:

V—3—E—3—F

And remember what this diagram means. It means the tetrahedron has

• 3 vertices and 3 edges touching each face,
• 3 edges and 3 faces touching each vertex.

When we take the dual of a solid, we

• replace vertices by faces;
• replace edges by edges;
• replace faces by vertices.

So when we do this to the tetrahedron, we get back a tetrahedron! This ‘self-duality’ is reflected in the symmetry of its Coxeter diagram. If we switch the letters V and F, we get the same thing back—drawn backwards, but that doesn’t matter.

This self-duality also means that when we take a tetrahedron and keep cutting off the corners more and more deeply, we wind up where we started. And in fact, when we reach the halfway point of this process, we start retracing our steps… going backwards!

Let’s see how it goes. Remember, we have a system of diagrams for drawing the most important solids we meet along the way.

### Truncated tetrahedron: •—3—•—3—o

Then we get the truncated tetrahedron:

### Octahedron: o—3—•—3—o

Then, halfway through, we get a shape we’ve seen before! It’s our friend the octahedron:

Like the other ‘halfway through’ shapes we’ve seen—the cuboctahedron and icosidodecahedron—every edge of the octahedron lies on a great circle’s worth of edges:

Puzzle 1. Why does it always work this way?

I don’t actually know!

### Truncated tetrahedron: o—3—•—3—•

Then we get back to the truncated tetrahedron:

### Tetrahedron: o—3—o—3—•

At the end, we get back where we started… the tetrahedron:

### Where are we?

We’ve begun to explore the three great families of semiregular polyhedra:

• the tetrahedron family shown here,

• the cube/octahedron family shown in Part 5,

• and the dodecahedron/icosahedron family shown in Part 6.

We’ve seen that for each family, we have a Coxeter complex, which is summarized by a Coxeter diagram. By coloring the dots in this diagram either white or black, we get different polyhedra in our family.

Our goal is to explore how this works in 4 dimensions. It’s very similar, but much more rich! We can still use Coxeter diagrams, but they’ll have four dots, so there will be more ways to label them, and we’ll get more shapes. And, of course, we’ll have the fun of learning to visualize 4-dimensional shapes!

But before we can explore the 4d story, there’s a hole in the story so far, that I need to fill.

Puzzle 2. Can you guess what it is?

Maybe you’ll see it if you look over our results so far.

### The tetrahedron family

Here are the shapes related to the tetrahedron. It has some repeats, because the tetrahedron is its own dual! It also repeats some shapes we’ll see in other families.

 tetrahedron •—3—o—3—o truncated tetrahedron •—3—•—3—o octahedron o—3—•—3—o truncated tetrahedron o—3—•—3—• tetrahedron o—3—o—3—•

And here’s the Coxeter complex that runs the show:

This has one right triangle for each element in the group that acts as symmetries of all these shapes. This group has 24 elements, and it’s called the tetrahedral finite reflection group, or A3. So, we can also call this collection of polyhedra the A3 family.

### The cube/octahedron family

Here are the shapes related to the cube and the octahedron:

 cube •—4—o—3—o truncated cube •—4—•—3—o cuboctahedron o—4—•—3—o truncated octahedron o—4—•—3—• octahedron o—4—o—3—•

And here’s the Coxeter complex:

Again, this has one right triangle for each element in the group that acts as symmetries of all these shapes. This group has 48 elements, and it’s called the octahedral finite reflection group, or B3. So, we can call this collection of polyhedra the B3 family.

### The dodecahedron/icosahedron family

And here are the shapes related to dodecahedron and icosahedron:

 dodecahedron •—5—o—3—o truncated dodecahedron •—5—•—3—o icosidodecahedron o—5—•—3—o truncated icosahedron o—5—•—3—• icosahedron o—5—o—3—•

And here’s the Coxeter complex:

Yet again, this has one right triangle for each element in the group that acts as symmetries of all these shapes. This group has 120 elements, and it’s called the icosahedral finite reflection group, or H3. So, we can call this collection of polyhedra the H3 family.

### 12 Responses to Symmetry and the Fourth Dimension (Part 7)

1. I get a glimpse of understanding here and there which is really exciting for me, encouraging me to plow through your posts. :) I had no idea how fascinating mathematical cutting up space is.

• John Baez says:

If the words and occasional formulas ever get you down, just look at the pictures for a while. Especially in this post, they’re organized in complex and interesting patterns—like the 3 families here. The words and formulas are our way of getting the verbal side of our brain fully engaged with these patterns. But the visual side is equally important (at least for me).

2. John Baez says:

So, can anyone guess the gaping hole that’s visible from the charts near the end of this post?

• Phil Gossett says:

Not sure this is what you’re looking for, but if you take the Archimedean solids, and exclude those with more than two types of polygon and those that are chiral, you get the solids in your charts, plus one: rhombicuboctahedron.

• Phil Gossett says:

More likely (and more generally), I suspect you’re looking for the cases where the two ends, or all three nodes of the Coxeter diagram are filled in. Those (I think) complete the list of all but the chiral Archimedian solids. (Which seemed to be snubbed in this scheme… ;-)

• John Baez says:

You got it! The truncation process I’ve been talking about doesn’t get us the polyhedra with Coxeter diagrams like

•—4—o—3—•

or

•—4—•—3—•

(and the similar ones with the numbers 3 or 5 replacing the 4 here). So, that’s our next order of business: we want to meet those members of our three families!

The snub polyhedra will remain snubbed.

By the way, I guess you’re not related to Thorold Gosset, discoverer of the famous ‘Gosset polytopes’ in 6, 7 and 8 dimensions. But it would be cool if you were…

3. Tobias Fritz says:

A typo: “When we take the dual of a solid, we replace faces by edges”.

Let me give an attempt at saying something about Puzzle 1. One thing I noticed about those ‘halfway through’ polytopes is their high degree of symmetry: fixing a vertex and rotating the whole thing around that vertex halfway around maps the polytope to itself. Therefore, for every edge hitting that vertex there is an opposite edge also hitting that vertex. Now we can consider that new edge and apply the same argument at its other vertex. In this way, we end up with a sequence of edges lying on a great circle.

So in order to solve the puzzle, I would try to find a good explanation for that symmetry.

• John Baez says:

Thanks! I’ll fix the typo… but more importantly, I think you’ve made some real progress here on Puzzle 1. Somehow this symmetry should arise from interpolating between a regular polyhedron and its dual. I don’t have time to think about this right now… maybe someone else can give it a try?

Another specially nice feature of these ‘halfway through’ polytopes is that ‘every edge looks alike':

In other words, their symmetry group acts transitively on their set of edges. This is not true for any of other semiregular polyhedra in the tables above (except, of course, the regular ones):

• Phil Gossett says:

A bit sketchy, but…

From the Wikipedia article for “Quasiregular polyhedron”, these polyhedra (which must be edge-transitive) have a Schlafli symbol of {p over q}, with p=3, q=3,4,5 (octahedron (really “tetratetrahedron”), cuboctahedron and icosidodecahedron, respectively).

From Coxeter, “Regular Polytopes”, page 65 (Dover), for a quasi-regular polyhedron with a Schlafli symbol of {p over q} with odd p, each vertex must be opposite to the midpoint of an edge. Hence, you get a great circle. Which by edge-transitivity implies every edge is on a great circle. (Basically, the odd p thing excludes cases like the “zig-zag equator” of a cube with the “poles” at two opposite vertices.)

So the only thing remaining is to show that these “halfway through” polyhedra must be quasiregular. Or equivalently, edge-transitive.

And I think you can get there by noting that these polyhedra are each (by construction) the common core of a dual pair of regular polyhedra.

• Phil Gossett says:

I think I got that somewhat wrong.

It’s a “plane of symmetry” which connects each vertex with the midpoint of the opposite side (in the odd p case). Vertices are actually opposite other vertices…

• Phil Gossett says:

Actually, I think you can read this straight off your Coxeter complex pictures. In particular, that the pictures for duals are always the same…

4. […] Part 7: Interpolating between a Platonic solid and its dual. Example: the tetrahedron […]