## Hexagonal Hyperbolic Honeycombs

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

●—6—o—3—o—3—o

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:

o—6—o—3—o—3—o

This diagram says there are four generators $s_1, \dots, s_4$ obeying relations encoded in the edges of the diagram:

$(s_1 s_2)^6 = 1$
$(s_2 s_3)^3 = 1$
$(s_3 s_4)^3 = 1$

together with relations

$s_i^2 = 1$

and

$s_i s_j = s_j s_i \; \textrm{ if } \; |i - j| > 1$

Marking the Coxeter diagram in different ways lets us describe many honeycombs with the same symmetry group as the hexagonal tiling honeycomb—in fact, 24 – 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:

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—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 $\mathrm{PSL}(2,\mathbb{C})$. 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 $\mathrm{PSL}_2(\mathbb{Z}[e^{2 \pi i / 3}])$, 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 $\mathrm{P\overline{S}L}_2(\mathbb{E})$, where $\mathbb{E}$ is their notation for the Eisenstein integers, also called $\mathbb{Z}[e^{2 \pi i / 3}]$. The weird little line over the $\mathrm{S}$ is a notation of theirs: $\mathrm{SL}_2$ stands for 2 × 2 matrices with determinant 1, but $\mathrm{\overline{S}L}_2$ is their notation for 2 × 2 matrices whose determinant has absolute value 1.

### 13 Responses to Hexagonal Hyperbolic Honeycombs

1. Phil Gossett says:

I was waiting for someone more knowledgeable to chime in, but here’s a somewhat informal attempt at the first few puzzles:

Puzzle 1: On a hyperbolic plane, the sum of the angles in a triangle is less than 180 degrees. So it’s not surprising that the internal angles of a hexagon can be less than 120 degrees.

Puzzle 2: According to Wikipedia:

http://en.wikipedia.org/wiki/Random_walk

a random walk on a Euclidean 3D “Manhattan” (or {4,3,4}) grid has less than probability 1 of returning to the origin. Since this {6,3,5} grid has both a higher branching factor (11 vs 5) and a longer minimum simple cycle (6 vs 4) than a {4,3,4} grid, seems obvious it must be less likely to return to the origin. So the probability must be less than 1.

Puzzle 3: For {6,3,7}, the vertex figure must itself be hyperbolic. So it’s “doubly hyperbolic” in that sense. Cheating (by looking it up):

https://en.wikipedia.org/wiki/Infinite-order_hexagonal_tiling_honeycomb

that apparently makes it “noncompact”. Though in some sense, it doesn’t seem all that different. Is there some simple way to state the difference between paracompact and noncompact?

Puzzle 4: Way over my head…

I’m glad you’re back to this series!

• John Baez says:

Thanks for tackling these puzzles! I agree with your answer to puzzle 2 though I don’t have a proof. I have a lot to say about puzzle 3 but not this moment… I don’t entirely understand it. As for this one:

Puzzle 1: On a hyperbolic plane, the sum of the angles in a triangle is less than 180 degrees. So it’s not surprising that the internal angles of a hexagon can be less than 120 degrees.

The problem is this. The corners of the hexagons in one of the ‘sheets’ here

lie not on a projective plane but on an ordinary flat plane that happens to be embedded in hyperbolic 3-space! This kind of flat plane is called a horosphere, and it’s tough to find people admitting that it’s flat, but there are papers that prove this. Here’s a picture of a hexagonal tiling on a horosphere, taken from Wikipedia:

So the puzzle is: how is all this consistent with a situation where all the internal angles of the hexagons are 90°, as in the first picture here?

(Beware: the second picture is more misleading than helpful, when it comes to this puzzle! In the second picture the internal angles are 120°.)

• Marek14 says:

Puzzle 1: your confusion is caused by not specifying what are the EDGES of the hexagons. If you draw the hexagon on the horosphere, its edges will be horocyclic arcs and the angles will be 120 degrees.
But to make the tiling, the edges of the hexagons are line segments which do NOT lie on the horosphere. That’s why the true angle will be smaller.

For puzzle 3, the lines in {6,3,7} will diverge. In {6,3,6}, you can cut a set of lines with a horosphere so that it shows {3,6} tiling; for {6,3,7}, you can cut a set of lines with plane or equidistant surface so it shows a {3,7} pattern.

2. Phil Gossett says:

Puzzle 1: Hmmm. So I guess the edges of the hexagons meet at 90 degrees in (hyperbolic) 3-space, but at 120 degrees in (flat) 2-space. Or something like that. Hard to see in what sense the 2-space can be flat while the 3-space is hyperbolic, but in these matters one must check ones intuition at the door.

Puzzle 2: While I’m too lazy to actually do it, it seems like the argument could be formalized by comparing cycles of the {4,3,4} grid with cycles of {6,3,5} order-by-order (which happily in both cases must always be even, I think). If the probability of the latter is always less than the former (which I’m pretty sure it is), you’re done.

Puzzle 3: Looking forward to that!

Thanks!

• John Baez says:

Phil wrote:

Hard to see in what sense the 2-space can be flat while the 3-space is hyperbolic, but in these matters one must check ones intuition at the door.

It’s really no more odd than a round 2-sphere sitting inside a flat 3-dimensional space.

Hmmm. So I guess the edges of the hexagons meet at 90 degrees in (hyperbolic) 3-space, but at 120 degrees in (flat) 2-space. Or something like that.

You don’t get to have it both ways quite so easily! The question is about the hexagons in the {6,3,4} lattice, and the edges of these hexagons really meet at 90° angles, as you can see:

To solve the puzzle it’s probably good to pay attention to what I’m not saying, and don’t assume that’s true.

3. Phil Gossett says:

Let me make sure I’m interpreting this picture correctly.

Take the nearest central hexagon. That hexagon appears to be shared by two hexagonally tessellated sheets, one going into the plane of the picture at 90 degrees, and one coming out. That there can be two such sheets seems plausible enough, since the space is hyperbolic. And indeed, that gives you your 90 degree internal angles, with three hexagons of any one sheet meeting at a vertex. And these sheets meet in interesting ways to form the overall lattice. Seems consistent enough.

But here’s where I seem to be misunderstanding something basic:

http://en.wikipedia.org/wiki/Angular_defect

And you’re saying this sheet (a horosphere) is in fact a (flat) plane. So that’s (obviously) the (apparent) contradiction. I tried to evade that by saying the angles in the flat 2-sheet are different from the angles in hyperbolic 3-space, but that’s not allowed. Another way out (which seems rather far-fetched, and I don’t think works) is along the lines of 3D gravity:

http://www.staff.science.uu.nl/~hooft101/gthpub/Deser_Jack_tH_ThreeD_Grav_1984.pdf (in particular, part V)

Which is locally flat with point defects (representing particles). So at the points, angles sum to 270 degrees instead of the usual 360. So each vertex would be such a defect.

But I suspect this is not what you’re getting at, since these sheets with point defects are only locally flat, but globally not flat…

• John Baez says:

Phil wrote:

Take the nearest central hexagon. That hexagon appears to be shared by two hexagonally tessellated sheets, one going into the plane of the picture at 90 degrees, and one coming out.

First of all, I have a lot of trouble seeing the sheet or sheets which the biggest hexagon in this picture lies in—it’s so big and close that it’s confusing:

It’s much easier to see the sheet of smaller hexagons inside this biggest hexagon. This sheet recedes to infinity near the middle of the picture.

Second of all, I don’t see how the big hexagon, which is roughly parallel to the plane of the picture, could be lying in a sheet ‘going into the plane of the picture at 90 degrees’, i.e. at right angles to the picture.

But while writing this your comment has started to make some sense to me. I think the biggest hexagon lies in the sheet of smaller hexagons that we see inside it! This sheet is a horosphere, roughly like this:

‘Away’ in the first picture corresponds to ‘up’ in this second one.
The big hexagon at the ‘front’ of the first picture would be at the bottom of the second one. The clearly visible sheet of hexagons in the first one is the horosphere in the second. The hard part to comprehend is that this horosphere is actually a flat plane.

I believe you’re also trying to tell me that the big hexagon also lies on a second horosphere, which would go ‘down’ in the second picture.

I have a lot more to say but maybe we should see if we agree so far.

• John Baez says:

Phil wrote:

Another way out (which seems rather far-fetched, and I don’t think works) is along the lines of 3D gravity:

http://www.staff.science.uu.nl/~hooft101/gthpub/Deser_Jack_tH_ThreeD_Grav_1984.pdf (in particular, part V)

Which is locally flat with point defects (representing particles). So at the points, angles sum to 270 degrees instead of the usual 360. So each vertex would be such a defect.

But I suspect this is not what you’re getting at, since these sheets with point defects are only locally flat, but globally not flat…

This is a very clever attempt to get around the problem… I don’t know if you’re aware of my work on point defects in 3d gravity, which is part of a big line of developments stemming from the paper you cite.

However, as you note, this attempt doesn’t really make sense, since hyperbolic space doesn’t have any ‘defects’: if you zoom in close to any point it looks flatter and flatter, more like Euclidean space. A horosphere in hyperbolic space is perfectly flat, even at large distance scales. So if we have a horosphere tessellated by polygons in any way, the sum of the internal angles at a point must be 360 degrees. And as we zoom in closer, a horosphere looks more and more like a flat plane in Euclidean space.

• Marek14 says:

You say that a horosphere in hyperbolic space is perfectly flat, but that is not true. Horosphere is a curved surface.

Horosphere can be approximated in two different ways:

Imagine yourself on a surface of a sphere. Now move the center of the sphere away from you, increasing its radius.

The bigger the radius of the sphere, the closer its surface geometry will be to Euclidean. With infinite radius, the surface geometry will be exactly Euclidean.

In Euclidean space, a sphere with infinite radius will be a flat plane; not so in hyperbolic space.

Imagine yourself on an equidistant surface hovering a bit above some plane, each point of the surface fixed distance above the plane. Now move the plane away while keeping the surface equidistant.
This is harder to imagine because we don’t really have equidistant surfaces of this kind in Euclidean space, but what happens is that the further the plane is, the less “hyberbolic” the geometry of the surface will be. In limit, when the plane is infinitely far away, the geometry will be Euclidean.

Horosphere is, therefore, less curved than any sphere, but simultaneously more curved than any equidistant surface.

4. Phil Gossett says:

Yes, that’s the interpretation I’m trying to describe. And yes, I knew of your work on quantum gravity. (Though I knew about that paper for unrelated reasons. I don’t recall you mentioning it in TWF when it was still iMP.)

On the two horosphere interpretation of that picture, it seems inevitable if you think of this as a limit of a polyhedral honeycomb. There, you have two polyhedra sharing each polygonal facet. So this is essentially the same case, with two horospheres sharing each hexagon.

What’s more than a little weird about this particular case is that the neighboring hexagons of the two sheets head off at 90 degrees from the shared one, in exactly opposite directions. Which you can see (on the sheet going into the picture) as being somewhat distorted hexagons, even though they’re mostly occluded by the edges of the shared central hexagon. And yet those neighboring hexagons are still in some sense “parallel” (as opposed to “antiparallel”). But I guess that can happen in a sufficiently extreme hyperbolic space.

(And more than a little misleadingly, the plane of the picture with that closest central hexagon is *not* a sheet of this lattice. If it were, there would be 4 hexagons around each vertex, which would break the regularity of the lattice.)

By the symmetries of the mutually perpendicular rods at each vertex, the only way (I think) you can get three hexagons on a sheet sharing a vertex is for those hexagons to be on the planes dividing the octants formed by those rods. The two sheets from that central hexagon form a “tunnel” going off to infinity into the picture in this projection, and another one heading off behind the eyepoint (and hence not visible in this projection). For some reason, I find what’s happening near that central hexagon easier to see than what’s happening on the walls of that “tunnel”. There’s more symmetry to lean on.

Does any of that make sense? Or am I even more confused than I think I am? ;-)

5. Phil Gossett says:

So this {6,3,4} case now seems less weird to me. The other {6,3,n} cases seem similar, with the angle being 360/n degrees. In all cases, the sheets around any particular edge just fold by that angle, with their “backs” being “glued together” for the hexagons adjacent to that edge. At n>4, the angle becomes even more acute.

And that pattern seems to hold for arbitrary n>2, even for n>6. At n=6, the vertex figure has an infinite number of points (being a triangular tiling of the plane). That can, of course, be mapped to a sphere. Which gives the {6,3,6} picture. But at n=7, the vertex figure is a tiling of a hyperbolic plane. But that can be mapped to a Poincare disk. Which can in turn be mapped a sphere. So it seems (naively) you could continue the pattern indefinitely.

Coming back to something like problem 2, at n=6 the probability of a random walk returning to the origin will be zero, since the branching factor will be infinite. But that’s still considered (according to Wikipedia) paracompact. So I’m still not getting the distinction implied by problem 3.

Anyway, while I think I understand things somewhat better than I did, it seems unlikely I’ll get the answer to problem 1. Perhaps it’s time to just give the answer.