Is a good automatic presentation known for the 'right dodecahedral' honeycomb?

Question

To make it clear, I'm talking about the group corresponding to the tiling with Schlafli symbol {5,3,4}, where each vertex has eight dodecahedra clustered around it with cubical / octahedral symmetry; see https://en.wikipedia.org/wiki/Order-4_dodecahedral_honeycomb for some more information and beautiful diagrams.

The corresponding planar tiling {5,4}, with right-angled pentagons meeting four at a vertex, has a really nice presentation in terms of reflections around the five edges of the 'root' pentagon; writing the edges as $a, b, c, d, e$ then the identities of the presentation are e.g. $a^2, b^2, \ldots, (ab)^2, (bc)^2, \ldots$; this plus a lexicographic ordering allows for a pretty straightforward canonical 'automation' of the group, which in particular gives a very nice way of addressing each cell in the tiling by a word in $a, b, c, d, e$; this is possibly the closest noncommutative analog to the way of addressing e.g. cells in the square tiling of the plane by $\mathbb{Z}^2$).

Is there any analagous way of presenting the tiling by dodecahedra of $\mathbb{H}^3$ that makes it easy to 'canonicalize' each cell of the tiling? My understanding is that when a visualization of this tiling was originally generated, duplicated cells were eliminated in a more ad hoc fashion based on just comparing coordinates against already generated cells. I'd love to have a means of generating cells of the tiling that avoids such an approach and also allows for canonical indexing akin to the two-dimensional case.

Answer 1

I am not certain, but it might be this group:

$$\langle a,b,c,d,e,f \mid a^4,b^4,c^4,d^4,e^4,f^4,aba^{-1}e, bcb^{-1}f, cdc^{-1}a, ded^{-1}b, efe^{-1}c, faf^{-1}d \rangle,$$

where the four generators are rotations about the six edges adjacent to a vertex. Notice that the final $6$ relators all say that each generator conjugates one other generator to another.

This is the symmetry group of the dodecahedral tesselation of hyperbolic space that features in the movie Not Knot.

I am not sure exactly what you mean by ``automatic presentation''. Is that a technical term? But this group is hyperbolic and so certainly automatic, and it is not very hard to compute its automatic structure.

Answer 2

This tiling is the same as the tiling by fundamental domains of the reflection group generated by reflections in the four faces of a regular, right angled dodecahedron. The standard "Coxeter group presentation" has 12 generators and 32 relators, as follows:

• There are 12 generators, one per face, which represents reflection across that face.
• There are also 12 relators, one per face, saying that the square of the reflection is the identity.
• Finally there are 20 relators, one per edge, saying that the two reflections across the two adjacent faces commute with each other (equivalently, the product of those two reflections is the identity).

You can then represent each tile uniquely using the ShortLex automatic structure, once you pick an order of the generators: use standard Coxeter group ideas to determine which words represent geodesics; and then represent each tile uniquely by the first element ordered lexicographically.

By the way, I'm pretty sure those images were not generated in an ad hoc fashion. Instead, if those images are derived from Geometry Center software then they were indeed generated using ShortLex automatic structures.