Alas that I lack the sophistimacation to make these figures (and a couple hundred others) myself, to my own taste.

**Puzzle 1.** Note that if n is even the faces of {6,3,n} tile hyperbolic planes, because if n faces meet at an edge then for each face and edge there’s a corresponding face n/2 steps around. {6,3,4} has {6,4} planes, and {6,3,6} has {6,∞} planes.

**Puzzle 3.** Vladimir Bulatov, for one, has depicted similar things. I can’t say I understand the images.

If you pack N>>12 discs in S2, you typically get an approximate hexagonal packing with 12 defects where the local symmetry is fivefold.

If you pack N>>120 balls in S3, I guess you get the usual FCC packing, but what are the defects like? Are they found at discrete points, or along edges, or …?

]]>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.

]]>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.

]]>I would still like to understand your answer for puzzle 1.

Thanks for getting me to think about this stuff!

]]>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.

(Sheepishly slinking away…)

]]>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? ;-)

]]>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.

]]>