# {7,3,3} Honeycomb Meets Plane at Infinity

{7,3,3} Honeycomb Meets Plane at Infinity – Roice Nelson

This picture by Roice Nelson shows the boundary of the {7,3,3} honeycomb, the shape featured in our last article:

The black circles are holes, not contained in the boundary of the {7,3,3} honeycomb. There are infinitely many holes, and the actual boundary, shown in white, is a fractal with area zero.

The {7,3,3} honeycomb lives in 3-dimensional hyperbolic space, a space that can be fit inside a ball using the Poincaré ball model. By the ‘boundary’ of the {7,3,3} honeycomb, we mean the set of points on the surface of the Poincaré ball that are limits of points in the {7,3,3} honeycomb.

Roice Nelson used stereographic projection to draw part of the surface of the Poincaré ball as a plane: the plane at infinity. So, the black region on the outside of the picture is also a hole in the boundary of the {7,3,3} honeycomb.

The boundary of the {7,3,3} honeycomb is topologically interesting, because it is homeomorphic to the Sierpinski carpet. See also:

We can see this using a result proved by Gordon Whyburn in 1958. A continuum is a nonempty connected compact metric space. Suppose $X$ is a continuum embedded in the plane. Suppose the complement $\mathbb{R}^2 – X$ has countably many connected components $C_1, C_2, C_3, \dots$ and suppose:

• the diameter of $C_i$ goes to zero as $i \to \infty$;

• the boundary of $C_i$ and the boundary of $C_j$ are disjoint if $i \ne j$;

• the boundary of $C_i$ is a simple closed curve for each $i$;

• the union of the boundaries of $C_i$ is dense in $X$.

Then $X$ is homeomorphic to the Sierpinski carpet!

To apply this result, note that the all the black circles in this picture, and also the black region on the outside, are the connected components $C_i$.

In general, for any $n \ge 7$, the boundary of the {$n$,3,3} honeycomb is homeomorphic to the Sierpinski carpet. For more details, see Section 7 of this paper:

• Danny Calegari and Henry Wilton, 3-manifolds everywhere.

They prove that spaces of this sort occur very often inside the Gromov boundaries of hyperbolic groups.

Roice Nelson, the creator of this image, has a blog with lots of articles about geometry, and he makes plastic models of interesting geometrical objects using a 3d printer:

Roice.

Visual Insight is a place to share striking images that help explain advanced topics in mathematics. I’m always looking for truly beautiful images, so if you know about one, please drop a comment here and let me know!

## 3 comments on “{7,3,3} Honeycomb Meets Plane at Infinity”

1. roice on said:

Hi John, I like your explanation of Whyburn’s result. I was pondering it as I fell asleep last night, and think it can be used to show a wider class of honeycombs have boundaries homeomorphic to the Sierpinski carpet. Reading the bulleted criteria, I’m confident that for a {p,q,r} honeycomb, the same result holds as long as the {p,q} cell is a hyperbolic tiling and the {q,r} vertex figure is a spherical tiling. As another example besides the {7,3,3}, here is an image of the {7,5,3}.

I wonder if this could also be true for Euclidean {q,r}, for example the {7,3,6}. See this picture of a single cell of this honeycomb in the ball model. It’s the ideal points of the cell meeting the boundary which I’m not sure about. If some of the C_i components are points instead of circular areas, does that fail to meet one or more of your bulleted crieteria? Probably. For example, it seems like you can’t really talk about the boundary of an isolated point (3rd bullet).

When {q,r} is a hyperbolic tiling, I think the honeycomb boundary must no longer be a Sierpinski carpet because the limit set has positive area.

• johnbaez on said:

It would be nice to straighten this out. I’d start by reading Danny Calegari and Henry Wilton’s paper, if you haven’t yet; I forget if they only handle {7,3,3}, {8,3,3} and so on, or also other cases.

For example, it seems like you can’t really talk about the boundary of an isolated point.

The boundary of an isolated point in the plane is that point itself. In general the boundary of a subset $X \subseteq \mathbb{R}^n$ is the subset of $\mathbb{R}^n$ consisting of points $p$ such that for any $\epsilon > 0$ there’s a point $x \in X$ whose distance to $p$ is less than $\epsilon$, and also a point $y$ in the complement $\mathbb{R}^n – X$ whose distance to $y$ is less than $\epsilon$.

A set with isolated points can’t be homeomorphic to the Sierpinski carpet, both because of bullet point 3 and because the Sierpinski carpet is connected.

• roice on said:

Thanks John, you answered the question about Euclidean {q,r}. So the boundaries of such honeycombs are NOT Sierpinski carpets. For fun, here is a boundary image of the {4,6,3} in a similar “white lace” style. It does look different.

{4,6,3} White Lace

I checked the 3-Manifolds Everywhere paper, and they only discuss {p,3,3}. They are showing that subgroups of random groups are “commensurable” with {p,3,3} groups. Since the tetrahedral group is a subgroup of the octahedral and icosahedral groups, {p,3,3} groups are subgroups of {p,q,r} groups with spherical {q,r}, and so it seems natural they would work with just the simplest {p,3,3} subgroups. Even though the paper does not discuss honeycombs like the {7,5,3}, I think their conclusions don’t conflict with the following claim:

Boundaries of noncompact {p,q,r} honeycombs with spherical {q,r} are all Sierpinski carpets.

And I do think this follows from Whyburn’s result as you describe it, but please correct any errors you see in the reasoning!

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