Symmetry and the Fourth Dimension (Part 8)

Surprise!

I bet you thought this series had died. But it was only snoozing.

I start projects at a rate faster than I can finish them. I wish I could do what a character in Greg Egan’s Permutation City could do, and have my ‘exoself’ adjust my personality so I would stick with just one project for an arbitrarily long time:

The workshop abutted a warehouse full of table legs—one hundred and sixty-two thousand, three hundred and twenty-nine, so far. Peer could imagine nothing more satisfying than reaching the two hundred thousand mark—although he knew that he would probably change his mind and abandon the workshop before that happened; new vocations were imposed by his exoself at random intervals, but statistically, the next one was overdue. Before taking up woodworking, he’d passionately devoured all the higher mathematics texts in the central library, run all the tutorial software, and then contributed several important new results to group theory—unconcerned by the fact that the Elysian mathematicians would never be aware of his work. Before that, he had written over three hundred comic operas, with librettos in Italian, English and French—and staged most of them, with puppet performers and audiences. Before that, he had patiently studied the structure and biochemistry of the brain for sixty-seven years; towards the end he had fully grasped, to his own satisfaction, the nature of the process of consciousness. Every one of these pursuits had been utterly engrossing, and satisfying, at the time.

But since I can’t do this, I’ve been trying to develop enough discipline to make sure I eventually come around back and finish most of what I start.

The missing solids

If you were paying attention, you should have noticed something funny when we worked our way from a Platonic solid to its dual by chopping off its corners more and more. For example, in Part 5 we started with the cube:

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—•

See what’s funny? We get 5 shapes as we go from our Platonic solid to its dual… but there are 2³ = 8 ways to mark the 3 dots in the Coxeter diagram either black (•) or white (o).

It makes sense to leave out the diagram where all dots are white, for reasons that should become clear. But that leaves two more diagrams missing from our chart!

The missing diagrams are the one with the end dots black:

•—4—o—3—•

and the one with all dots black:

•—4—•—3—•

Are there shapes corresponding to these diagrams?

YES!

And in fact, these will be some of the most complex and beautiful shapes we’ve met so far!

Coxeter diagrams and polyhedra

To get our hands on them, we have to remember the rules of the game. We’ve been dealing with Coxeter diagrams with three dots, and these dots stand for vertex, edge and face, in this order:

V—4—E—3—F

The polyhedra we’re playing with have corners that arise from the dots that are blackened. Let me remind you how, with some examples:

• a cube obviously has one corner for each vertex of the cube, so we blacken the V dot:

•—4—o—3—o

• a truncated cube has one corner for each vertex-edge flag of the cube, meaning a pair consisting of a vertex and an edge it lies on:

So, we blacken the V and E dots:

•—4—•—3—o

• a cuboctahedron has one corner for each edge of the cube:

So, we blacken just the E dot:

o—4—•—3—o

• a truncated octahedron has one corner for each edge-face flag of the cube, meaning a pair consisting of an edge and face containing it. So, we blacken the E and F dots:

o—4—•—3—•

• an octahedron has one corner for each face of the cube:

So, we blacken the F dot:

o—4—o—3—•

How to find the missing solids

Now let’s look at the two diagrams that are missing from this list! This one:

•—4—o—3—•

has the dots for ‘vertex’ and ‘face’ blackened. So, following the idea that’s worked so far, it should stand for a polyhedron that has one corner for each vertex-face flag of the cube: that is, each pair consisting of a vertex and a face that it lies on.

The cube has 6 faces and each face has 4 vertices lying on it:

So, it has 6 × 4 = 24 vertex-face flags. And if we make a shape with one corner for each of these, we get this:

This is called the rhombicuboctahedron. The corners of each red square here correspond to the 4 vertices lying on a given face of the cube. So indeed, this thing has one corner for each vertex-face flag of the cube!

Similarly, in this diagram:

•—4—•—3—•

the dots for ‘vertex’, ‘edge’ and ‘face’ are all blackened. So it should stand for a polyhedron that has one corner for each complete flag of the cube. Remember, a complete flag consists of a vertex, an edge and a face, where the vertex lies on the edge and the edge lies on the face.

Now the cube has 6 faces, each with 4 edges, each with 2 vertices. So, it has 6 × 4 × 2 = 48 complete flags. And if we make a shape with one corner for each of these, we get this:

This is called the truncated cuboctahedron, because you can also get it from truncating an cuboctahedron.

Puzzle 1. Why does that happen? Why should snipping off the corners of an cuboctahedron give a shape with one corner for each complete flag of the cube?

Next let’s go through all three families of shapes:

• the tetrahedron family,
• the cube/octahedron family, and
• the dodecahedron/icosahedron family

and list their two ‘missing members’.

Tetrahedron family

Cuboctahedron: •—3—o—3—•

 

Truncated octahedron: •—3—•—3—•

 

Cube/octahedron family

Rhombicuboctahedron: •—4—o—3—•

 

Truncated cuboctahedron: •—4—•—3—•

 

Dodecahedron/icosahedron family

Rhombicosidodecahedron: •—5—o—3—•

 

Truncated icosidodecahedron: •—5—•—3—•

 

These last two are my favorites, since they’re the fanciest. Let’s explore them a bit further.

In this diagram:

•—5—o—3—•

the dots for ‘vertex’ and ‘face’ are blackened. So, this gives a solid with one corner for each vertex-face flag of the dodecahedron.

How many flags of this sort are there? As its name suggests, the dodecahedron has 12 faces, and each face has 5 vertices lying on it:

So, it has 12 × 5 = 60 vertex-face flags. And if we make a shape with one corner for each of these, we get the rhombicosidodecahedron:

The corners of each red pentagon here correspond to the 5 vertices lying on a given face of the dodecahedron. So indeed, this thing has one corner for each vertex-face flag of the dodecahedron.

In this diagram:

•—5—•—3—•

the dots for ‘vertex’, ‘edge’ and ‘face’ are all blackened. So it should stand for a polyhedron that has one corner for each complete flag of the dodecahedron. Now the dodecahedron has 12 faces, each with 5 edges, each with 2 vertices. So, it has 12 × 5 × 2 = 120 complete flags. And if we make a shape with one corner for each of these, we get this:

This is called the truncated icosidodecahedron, because you can also get it from truncating an icosidodecahedron.

Puzzle 2. Why does that happen? Why should snipping off the corners of an icosidodecahedron give a shape with one corner for each complete flag of the dodecahedron?

Afterword

As usual, the pretty pictures of solids with brass balls at the vertices were made by Tom Ruen using Robert Webb’s Stella software.

You can see the previous episodes here:

• Part 1: Platonic solids and Coxeter complexes.

• Part 2: Coxeter groups.

• Part 3: Coxeter diagrams.

• Part 4: duals of Platonic solids.

• Part 5: Interpolating between a Platonic solid and its dual, and how to describe this using Coxeter diagrams. Example: the cube/octahedron family.

• Part 6: Interpolating between a Platonic solid and its dual. Example: the dodecahedron/icosahedron family.

• Part 7: Interpolating between a Platonic solid and its dual. Example: the tetrahedron family.