This post is just for fun.

Roice Nelson likes geometry, and he makes plastic models of interesting objects using a 3d printer. He recently created some great pictures of ‘hexagonal hyperbolic honeycombs’. With his permission, I wrote about them on my blog *Visual Insight*. Here I’ve combined those posts into a single more polished article.

But the pictures are the star of the show. They deserve to be bigger than the 450-pixel width of this blog, so please click on them and see the full-sized versions!

### The {6,3,3} honeycomb

This is the **{6,3,3} honeycomb**.

How do you build this structure? Take 4 rods and glue them together so their free ends lie at the corners of a regular tetrahedron. Make lots of copies of this thing. Then stick them together so that as you go around from one intersection to the next, following the rods, the shortest possible loop is always a *hexagon!*

This is impossible in ordinary flat 3-dimensional space. But you can succeed if you work in hyperbolic space, a non-Euclidean space where the angles of a triangle add up to less than 180°. The result is the {6,3,3} honeycomb, shown here.

Of course, this picture has been projected onto your flat computer screen. This distorts the rods, so they look curved. But they’re actually straight… *inside curved space*.

The {6,3,3} honeycomb is an example of a ‘hyperbolic honeycomb’. In general, a 3-dimensional honeycomb is a way of filling 3d space with polyhedra. It’s the 3-dimensional analogue of a tiling of the plane. Besides honeycombs in 3d Euclidean space, we can also have honeycombs in 3d hyperbolic space. The {6,3,3} honeycomb is one of these.

But actually, when I said a honeycomb is a way of filling 3d space with polyhedra, I was lying slightly. It’s often true—but not in this example!

For comparison, in the {5,3,4} honeycomb, space really *is* filled with polyhedra:

You can see a lot of pentagons, and if you look carefully you’ll see these pentagons are faces of dodecahedra:

In the honeycomb, these dodecahedra fill hyperbolic space.

But in the {6,3,3} honeycomb, all the hexagons lie on *infinite sheets*. You can see one near the middle of this picture:

These sheets of hexagons are not polyhedra in the usual sense, because they have *infinitely many* polygonal faces! So, the {6,3,3} honeycomb is called a paracompact honeycomb.

But what does the symbol {6,3,3} mean?

It’s an example of a Schläfli symbol. It’s defined in a recursive way. The symbol for the hexagon is {6}. The symbol for the hexagonal tiling of the plane is {6,3} because 3 hexagons meet at each vertex. Finally, the hexagonal tiling honeycomb has symbol {6,3,3}, because 3 hexagonal tilings meet at each edge of this honeycomb.

So, we can build a honeycomb if we know its Schläfli symbol. And there’s a lot of information in this symbol.

For example, just as the {6,3} inside {6,3,3} describes the hexagonal tilings inside the {6,3,3} honeycomb, the {3,3} describes the vertex figure of this honeycomb: that is, the way the edges meet at each vertex. {3,3} is the Schläfli symbol for the regular tetrahedron, and in the {6,3,3} honeycomb each vertex has 4 edges coming out, just like the edges going from the center of a tetrahedron to its corners!

### The {6,3,4} honeycomb

This is the **{6,3,4} honeycomb**.

How do you build this structure? Make 3 intersecting rods at right angles to each other. Make lots of copies of this thing. Then stick them together so that as you go around from one intersection to the next, following the rods, the shortest possible loop is always a hexagon!

This is impossible in ordinary flat 3-dimensional space. You can only succeed if the shortest possible loop is a square. Then you get the familiar cubic honeycomb, also called the the {4,3,4} honeycomb:

To get hexagons instead of squares, space needs to be curved! You can succeed if you work in hyperbolic space, where it’s possible to create a hexagon whose internal angles are all 90°. In ordinary flat space, only a square can have all its internal angles be 90°.

Here’s the tricky part: the hexagons in the {6,3,4} honeycomb form infinite sheets where 3 hexagons meet at each corner. You can see one of these sheets near the center of the picture. The corners of the hexagons in one sheet lie on a flat plane in hyperbolic space, called a horosphere.

That seems to make sense, because in flat space hexagons can have all their internal angles be 120°… so three can meet snugly at a corner. But I just said these hexagons have 90° internal angles!

**Puzzle 1.** What’s going on? Can you resolve the apparent contradiction?

The Schläfli symbol of this honeycomb is {6,3,4}, and we can see why using ideas I’ve already explained. It’s made of hexagonal tilings of the plane, which have Schläfli symbol {6,3} because 3 hexagons meet at each vertex. On the other hand, the vertex figure of this honeycomb is an octahedron: if you look at the picture you can can see that each vertex has 6 edges coming out, just like the edges going from the center of an octahedron to its corners. The octahedron has Schläfli symbol {3,4}, since it has 4 triangles meeting at each corner. Take {6,3} and {3,4} and glue them together and you get {6,3,4}!

We can learn something from this. Since this honeycomb has Schläfli symbol {6,3,4}, it has 4 hexagonal tilings meeting at each edge! That’s a bit hard to see from the picture.

All the honeycombs I’ve been showing you are ‘regular’. This is the most symmetrical kind of honeycomb. A **flag** in a honeycomb is a vertex lying on an edge lying on a face lying on a **cell** (which could be a polyhedron or an infinite sheet of polygons). A honeycomb is **regular** if there’s a symmetry sending any flag to any other flag.

The {6,3,3} and {6,3,4} honeycombs are also ‘paracompact’. Remember, this means they have infinite cells, which in this case are the hexagonal tilings {6,3}. There are 15 regular honeycombs in 3d hyperbolic space, of which 11 are paracompact. For a complete list of regular paracompact honeycombs, see:

• Regular paracompact honeycombs, Wikipedia.

### The {6,3,5} honeycomb

This is the **{6,3,5} honeycomb**. It’s built from sheets of regular hexagons, and 5 of these sheets meet along each edge of the honeycomb. That explains the Schläfli symbol {6,3,5}.

If you look very carefully, you’ll see 12 edges coming out of each vertex here, grouped in 6 opposite pairs. These edges go out from the vertex to its 12 neighbors, which are arranged like the corners of a regular icosahedron!

In other words, the vertex figure of this honeycomb is an icosahedron. And even if you can’t see this in the picture, you can deduce that it’s true, because {3,5} is the Schläfli symbol for the regular icosahedron, and it’s sitting inside {6,3,5}, at the end.

But now for a puzzle. This is for people who like probability theory:

**Puzzle 2.** Say you start at one vertex in this picture, a place where edges meet. Say you randomly choose an edge and walk down it to the next vertex… each edge being equally likely. Say you keep doing this. This is the most obvious random walk you can do on the {6,3,5} honeycomb. Is the probability that eventually you get back where you started equal to 1? Or is it less than 1?

If that’s too hard, try the same sort of question with the usual cubical honeycomb in ordinary flat 3d space. Or the square lattice on the plane!

In one dimension, where you just take steps back and forth on the integers, with equal chances of going left or right each time, you have a 100% chance of eventually getting back where you started. But the story works differently in different dimensions—and it also depends on whether space is flat, spherical or hyperbolic.

### The {6,3,6} honeycomb

This is the **{6,3,6} honeycomb**. It has a lot of sheets of regular hexagons, and 6 sheets meet along each edge of the honeycomb.

The {6,3,6} honeycomb has a special property: it’s ‘self-dual’. The tetrahedron is a simpler example of a self-dual shape. If we draw a vertex in the middle of each face of the tetrahedron, and draw an edge crossing each edge, we get a new shape with a face for each vertex of the tetrahedron… but this new shape is again a tetrahedron!

If we do a similar thing one dimension up for the {6,3,6} honeycomb, this amounts to creating a new honeycomb with:

• one vertex for each infinite sheet of hexagons in the original honeycomb;

• one edge for each hexagon in the original honeycomb;

• one hexagon for each edge in the original honeycomb;

• one infinite sheet of hexagons for each vertex in the original honeycomb.

But this new honeycomb turns out to be another {6,3,6} honeycomb!

This is hard to visualize, at least for me, but it implies something cool. Just as each sheet of hexagons has infinitely many hexagons on it, each vertex has infinitely many edges going through it.

This self-duality comes from the symmetry of the Schläfli symbol {6,3,6}: if you reverse it, you get the same thing!

Okay. I’ve showed you regular hyperbolic honeycombs where 3, 4, 5, or 6 sheets of hexagons meet along each edge. Sometimes in math patterns go on forever, but sometimes they end—just like life itself. And indeed, we’ve reached the end of something here! You can’t build a regular honeycomb in hyperbolic space with 7 sheets of hexagons meeting at each edge.

**Puzzle 3.** What do you get if you try?

I’m not sure, but it’s related to a pattern we’ve been seeing. The hexagonal hyperbolic honeycombs I’ve shown you are the ‘big brothers’ of the tetrahedron, the octahedron, the icosahedron and the triangular tiling of the plane! Here’s how it goes:

• You can build a tetrahedron where 3 triangles meet at each corner:

For this reason, the Schläfli symbol of the tetrahedron is {3,3}. You can build a hyperbolic honeycomb where the edges coming out of any vertex go out to the corners of a tetrahedron… and these edges form hexagons. This is the {6,3,3} honeycomb.

• You can build an octahedron where 4 triangles meet at each corner:

The Schläfli symbol of the octahedron is {3,4}. You can build a hyperbolic honeycomb where the edges coming out of any vertex go out to the corners of an octahedron… and these edges form hexagons. This is the {6,3,4} honeycomb.

• You can build an icosahedron where 5 triangles meet at each corner:

The Schläfli symbol of the icosahedron is called {3,5}. You can build a hyperbolic honeycomb where the edges coming out of any vertex go out to the corners of an icosahedron… and these edges form hexagons. This is the {6,3,5} honeycomb.

• You can build a tiling of a flat plane where 6 triangles meet at each corner:

This triangular tiling is also called {3,6}. You can build a hyperbolic honeycomb where the edges coming out of any vertex go out to the corners of a triangular tiling… and these edges form hexagons. This is the {6,3,6} honeycomb.

The last one is a bit weird! The triangular tiling has *infinitely many* corners, so in the picture here, there are *infinitely many* edges coming out of each vertex.

But what happens when we get to {6,3,7}? That’s the puzzle.

### Coxeter groups

I’ve been telling you about Schläfli symbols, but these are closely related to another kind of code, which is deeper and in many ways better. It’s called a Coxeter diagram. The Coxeter diagram of the {6,3,3} honeycomb is

What does this mean? It looks a lot like the Schläfli symbol, and that’s no coincidence, but there’s more to it.

The symmetry group of the {6,3,3} honeycomb is a discrete subgroup of the symmetry group of hyperbolic space. This discrete group has generators and relations summarized by the *unmarked* Coxeter diagram:

This diagram says there are four generators obeying relations encoded in the edges of the diagram:

together with relations

and

Marking the Coxeter diagram in different ways lets us describe many honeycombs with the same symmetry group as the hexagonal tiling honeycomb—in fact, 2^{4} – 1 = 15 of them, since there are 4 dots in the Coxeter diagram! For the theory of how this works, illustrated by some simpler examples, try this old post of mine:

• Symmetry and the Fourth Dimension (Part 9).

or indeed the whole series. The series is far from done; I have a pile of half-written episodes that I need to finish up and publish. This post should, logically, come after all those… but life is not fully governed by logic.

Similar remarks apply to all the hexagonal hyperbolic honeycombs I’ve shown you today:

{6,3,3} honeycomb

{6,3,3} honeycomb

●—6—o—3—o—3—o

3 hexagonal tilings meeting at each edge

vertex figure: tetrahedron

**{6,3,4} honeycomb**

●—6—o—3—o—4—o

4 hexagonal tilings meeting at each edge

vertex figure: octahedron

**{6,3,5} honeycomb**

●—6—o—3—o—5—o

5 hexagonal tilings meeting at each edge

vertex figure: icosahedron

**{6,3,6} honeycomb**

●—6—o—3—o—6—o

6 hexagonal tilings meeting at each edge

vertex figure: hexagonal tiling

Finally, one more puzzle, for people who like algebra and number theory:

**Puzzle 4.** The symmetry group of 3d hyperbolic space, not counting reflections, is . Can you explicitly describe the subgroups that preserve the four hexagonal hyperbolic honeycombs?

For the case of {6,3,3}, Martin Weissman gave an answer on G+:

Well, it’s , of course!

Since he’s an expert on arithmetic Coxeter groups, this must be about right! Theorem 10.2 in this paper he showed me:

• Norman W. Johnson and Asia Ivic Weiss, Quadratic integers and Coxeter Groups, *Canad. J. Math. Vol.* **51** (1999), 1307–1336.

is a bit more precise. It gives a nice description of the **even part** of the Coxeter group discussed in this article, that is, the part generated by *products of pairs* of reflections. To get this group, we start with 2 × 2 matrices with entries in the **Eisenstein integers**: the integers with a cube root of -1 adjoined. We look at the matrices where the *absolute value* of the determinant is 1, and then we ‘projectivize’ it, modding out by its center. That does the job!

They call the even part of the Coxeter group [3,3,6]^{+}, and they call the group it’s isomorphic to , where is their notation for the Eisenstein integers, also called . The weird little line over the is a notation of theirs: stands for 2 × 2 matrices with determinant 1, but is their notation for 2 × 2 matrices whose determinant has absolute value 1.

Can you say more about this case? What about the other cases?