#coxeter groups

I encourage anyone reading this to read as much or as little as they want. I hope that anyone can understand the detailed explanation, and that most math undergrads can understand most of the observations I make. The generalisations might not be so accessible all the time.

To put it tersely, its the projection of the barycentric subdivision of a tetrahedron onto the 2-sphere, visualised on a rectangle via the mercator projection.

This is closely related to Coxeter groups, the classification of polytopes, the classification of straight line Coxeter groups, the classification of regular tilings of surfaces of constant curvature, and Schl\"afli symbols.

A more detailed explanation:

Imagine taking a tetrahedron, putting a dot on the middle of each vertex, edge, and face, and connecting them all up with straight lines along its surface. This is the so called "barycentric subdivision". Then consider the origin to be in the middle of the tetrahedron, and then project the tetrahedron (and the lines we drew on it) onto a sphere. We use the mercator projection to view it like a map, but we still think of it as lying on the sphere (see below mp4 of said sphere with the barycentric subdivision drawn on it). Each face of the tetrahedron could be imagined to be coloured a certain colour, so v_2 in the diagram on my header is the vertex at the centre of the yellow face, v_1 is the vertex in the centre of one of the yellow faces edges, and similarly for v_0. These appear to be connected by curved lines, but these are straight lines on the surface of the sphere.

Assume the tetrahedron and sphere are embedded in R^3 and share a common centre at the origin.

Some interesting observations:

each of these lines we drew now gets turned into a great circle, which corresponds to a plane going through the origin

reflections in these great circles preserves the lines we drew, and correspond to automorphisms of the tetrahedron

each triangle in the subdivision has angles pi/3 radians, pi/3 radians, and pi/2 radians.

the symmetry group of the tetrahedron is S_4, where the adjacent transpositions correspond to permutations of the faces (or if you like, vertices)

the symmetric group S_4 has a presentation , where the s_i are adjacent transpositions of 4-tuples

The exponents of the (s_is_j) terms above exactly match the denominators of the angles of the triangle mentioned above

we can pick one triangle on the sphere and consider the reflections in (the faces corresponding to) its edges, denoted s_0, s_1, and s_2. These reflections permute the coloured faces of the tetrahedron, or if you like, its vertices.

we can repeatedly apply these reflections to flip across an edge or vertex of the yellow face, rotate about the centre of the yellow face (e.g. s_0 s_1), and transpose the yellow face with any other face

In this manner we can represent every symmetry of the barycentric subdivision, and by extension, the tetrahedron, in terms of these three reflections. If you don't see this, consider the effect of conjugation.

The sphere is a surface of constant curvature

This tiling generated by the tetrahedron is a regular tiling of the sphere

In short, the tetrahedron has a symmetry group S_4 (often called A_4 in analogy with the Dynkin Diagram) that has a presentation in terms of three reflections, which act transitively on this barycentric subdivision. The angles of the barycentric subdivision correspond to the relations of the presentation. One can generalise this observation and use it to classify polyhedra.

Some theoretical results.

A Coxeter group is a group W accompanied by a set of generators S = {s_1, s_2, ..., s_n} \subseteq W, such that W = <s_i | (s_is_j)^m(i,j) = 1>, where m(i,j) is an integer at least 1, m(i,i) = 1, and m(i,j) > 1 if i != j. These relations turn out to exactly correspond to the relations necessary to define a finite system of reflections in (n+1)-dimensional space.

By polytope, I mean a bounded convex polytope.

The regular tilings of the sphere correspond to regular polytopes, which correspond to the finite irreducible Coxeter groups whose Dynkin diagrams have straight lines

We can define a polytope to be regular if the automorphism group of the polytope acts transitively on the regions of the barycentric subdivision (or equivalently, its "flags"), which corresponds to chains of i-faces of the polytope ordered by inclusion

To go from a polytope to its Coxeter group, you take its automorphism group to get the group structure, and do a similar thing to above to find the generators, you arrange some hyperplanes so that their reflections satisfy the relations of the Coxeter group, generate a system of hyperplanes closed under reflection, and intersect this with an (n-1)-sphere to get the barycentric subdivision, from which you can recover a polytope and its dual polytope, which have isomorphic Coxeter groups

The regular tilings of the plane correspond to the affine irreducible Coxeter groups with straight line Dynkin diagrams.

One can study the regular tilings of hyperbolic space and classify those Coxeter groups too.

The E_8 lattice, which gives solution to 8-dimensional sphere packing has a a load of other interesting properties, corresponds to the Coxeter group E_8, via a certain semiregular polytope which is the convex hull of some lattice points.

The classification of regular (n-dimensional) polytopes and regular tilings of R^n is via the classification of Coxeter groups (and by extension Dynkin diagrams with certain properties)

There is an elegant classification according to Bourbaki that resembles the typical intuitive classification of regular polyhedra and regular tilings of R^2

Fun extensions

There are a lot of ways to represent a symmetry of a polytope/element of a Coxeter group in terms of the reflections/generators s_i. Is there an easy way to determine whether your representation of the symmetry/group element is the shortest? Yes! In fact, you can construct a DFA on the generators in the finite case.

The Cayley graph of a Coxeter group is Hamiltonian

My pfp shows the duality between an octahedron and a cube. If you draw a vertex at the centre of each face of the cube, and take the convex hull of the vertices, you get an octahedron. Note that vertices of the octahedron correspond to faces of the cube, edges of the octahedron correspond to edges of the cube, and faces of the octahedron correspond to vertices of the cube. Two vertices of the octahedron are incident with each other when the corresponding faces of the cube share an edge, and so on. To put it formally, the poset of i-faces of the octahedron and the poset of i-faces of the cube, both under inclusion, have an anti-isomorphism between them. This causes them to have isomorphic symmetry groups.

The cube has its Coxeter group with relations (s_0s_1)^4 = (s_1s_2)^3 = (s_0s_2)^2 = s_i^2 = 1. Note that here the 4 and 3 are different numbers, and the cube has a dual of an octahedron. In the case of the tetrahedron, the exponents are the same, and the tetrahedron is self dual. In general, finite irreducible Coxeter groups with straight line Dynkin diagrams correspond to self dual polytopes exactly when their Dynkin diagrams are "reversible".