**2D**

Given m,n >= 3, we can try to build a 2-dimensional tiling where m regular n-gons meet at each vertex. The angle of a regular n-gon is alpha = (1 - 2/n) pi, and we have the following three cases:

- m alpha < 2 pi. This yields a regular polyhedron (a Platonic solid). There are 5 cases like that:

a. n = 3, m = 3: tetrahedron, a self-dual polyhedron, the 3-dimensional simplex

b. n = 3, m = 4: octahedron, the 3-dimensional cross-polytope

c. n = 3, m = 5: icosahedron

d. n = 4, m = 4: cube, dual to octahedron, the 3-dimensional hypercube

e. n = 5, m = 3: dodecahderon, dual to icosahedron

Each of those defines a finite subgroup of SO(3), the 3-dimensional rotation group and of O(3) the 3-dimensional rotation-and-reflection group. These subgroups are, of course, the symmetry groups of the polyhedra. - m alpha = 2 pi. This yields a regular tiling of the Euclidean plane. There are 3 cases like that:

a. n = 3, m = 6

b. n = 4, m = 4 A self-dual tiling

c. n = 6, m = 3 dual to a

Each of those defines a discrete subgroup of the group of isometries (rotations, translations and reflections) of the Euclidean plane. Alternatively we can use orientation-preserving isometries (rotations and translations only). - m alpha > 2 pi. This yields a regular tiling of the hyperbolic plane. There's an infinite number of cases. Each of them defines a discrete subgroup of SO(2, 1). The later group has various geometric realizations:

a. Orientation-preserving isometries of the hyperbolic plane.

b. Lorentz transformations of special relativity in 3-dimensional spacetime (2 space dimensions and 1 time dimension).

**3D**

Given A, B regular polyhedrons, we can try to build a 3-dimensional tiling where #{faces of B} A-polyhedra meet at a B-type vertex. What do I mean by a B-type vertex? Imagine the vertex being in the ceter O of a B-polyhedron Y. Fix a face F of

Y. F corresponds to an A-polyhedron X of the tiling. The lines passing through the O and the vertices of F correspond to edges of X.

This wouldn't work for any A, B. For purely combinatorial reasons, we need

Y. F corresponds to an A-polyhedron X of the tiling. The lines passing through the O and the vertices of F correspond to edges of X.

This wouldn't work for any A, B. For purely combinatorial reasons, we need

#{faces meeting at a vertex of A} = #{sides of a face of B}

Geometrically, we again have three cases, depending on

- alpha, the dihedral angle of A, that is, it is the angle between its two adjacent faces.
- m, the number of faces meeting at a vertex of B.

- m alpha < href="http://en.wikipedia.org/wiki/Convex_regular_4-polytope">polychoron). There are 6 cases like that:

a. A = tetrahedron, B = tetrahedron: pentachoron, a self-dual polychoron. It is the 4-dimensional simplex.

b. A = tetrahedron, B = octahedron: hexadecachoron. It is the 4-dimensional cross-polytope.

c. A = tetrahedron, B = icosahedron: hexacosichoron.

d. A = cube, B = tetrahedron: tesseract, dual to the hexadecahoron. It is the 4-dimensional hypercube.

e. A = octahedron, B = cube: icositetrachoron, a self-dual polychoron.

f. A = dodecahedron, B = tetrahedron: hecatonicosachoron, dual to the hexaicosohoron.

Each of those defines a finite subgroup of SO(4), the group of 4-dimensional rotations. It also defines a finite subgroup of O(4), the group of 4-dimensional rotations-and-reflections. - m alpha = 2 pi. This yields a regular tiling of the Euclidean space. There is only 1 case like that: A = cube, B = octahedron. It defines a discrete subgroup of the group of isometries (rotations, translations and reflections) of the Euclidean space, or of the group of orientation-preserving isometries (no reflections).
- m alpha > 2 pi. This yields a regular tiling of the 3-dimensional hyperbolic space. There are 4 cases like that:

a. A = cube, B = icosahedron

b. A = dodecahedron, B = octahedron: dual to a

c. A = dodecahedron, B = icosahedron: self-dual

d. A = icosahedron, B = dodecahedron: self-dual

Each of those defines a discrete subgroup of SO(3, 1). The later group has various geometric realizations:

a. The group of orientation-preserving isometries of the 3-dimensional hyperbolic space.

b. The group of Lorentz transformations in special relativity.

c. The group of orientation-preserving conformal transformations of the

2-sphere.

Realization b is intriguing since it makes me wonder whether these discrete subgroups appear in any physically-interesting situation.

Realization c is intriguing for the following reason. Each such transformation has one or two fixed points. Consider the set of fixed points of all transformations belonging to a given discrete subgroup. This is a countable subset of the sphere, invariant under the discrete subgroup (due to conjugation). Clearly it must be either dense everywhere or a sort of fractal, but I don't know which.

**4D**

This time we take A, B to be regular polychorons. We want to construct a 4-dimensional tiling of A-polychorons which meet at a B-type vertex. The combinatorial compatibility condition is

vertex polyhedron of A = hyperface polyhedron of B

We have 3 geometric cases:

- A regular tiling of the 4-sphere, that is, a 5-dimensional regular polytope. There are 3 cases like that:

a. A = pentachoron, B = pentachoron: the self-dual 5-dimensional simplex.

b. A = pentachoron, B = hexadecahoron: the 5-dimensional cross-polytope.

c. A = tesseract, B = pentachoron: the 5-dimensional hypercube, dual to a. - A regular tiling of the 4-dimensional Euclidean space. One example is

A = tessaract, B = hexadecahoron, which is self-dual - A regular tiling of the 4-dimensional hyperbolic space.

- A = pentachoron, B = hexacosichoron
- A = hexadecachoron, B = icositetrachoron
- A = tesseract, B = hexacosichoron
- A = icositetrachoron, B = tesseract: dual to 2
- A = hecatonicosachoron, B = pentachoron: dual to 1
- A = hecatonicosachoron, B = hexadecachoron: dual to 3
- A = hecatonicosachoron, B = hexacosichoron: self-dual

Higher dimension

We take A, B to be regular n-dimensional polytopes. We want to construct an n-dimensional tiling of A-polytopes which meet at a B-type vertex. The combinatorial compatibility condition is

n-1-dimensional vertex polytope of A = n-1-dimensional hyperface polytope of B

Once again, we have 3 geometric cases:

- A regular tiling of the n-sphere, that is an n+1-dimensional regular polytope. There are 3 cases like that:

a. A = n-dimensional simplex, B = n-dimensional simplex. This is the self-dual

n+1-dimensional simplex.

b. A = n-dimensional hypercube, B = n-dimensional simplex. This is the n+1-dimensional hypercube.

c. A = n-dimensional simplex, B = n-dimensional cross-polytope. This is the

n+1-dimensional cross-polytope, dual to the n+1-dimensional hypercube. - A regular tiling of the n-dimensional Euclidean space. There is only 1 case:

A = n-dimensional hypercube, B = n-dimensional cross-polytope. - A regular tiling of the n-dimensional hyperbolic space. There are none!

As we said, we construct n-dimensional tilings out of a pair A, B of n-dimensional polytopes. Now, a polytope is a tiling of the n-1-sphere. What if we take A, B to be tilings of the n-1-dimensional hyperbolic space instead? Logically, we should get a tiling of a space of Lorentzian signature, since the hyperbolic space plays the same role in Minkowski space that the sphere plays in Euclidean space. I'm not sure how would such a tiling look like, it appears it would be

self-interecting. As before, such tilings would come with different curvatures. That is, we should get

self-interecting. As before, such tilings would come with different curvatures. That is, we should get

- Positive curvature: a tiling of de Sitter space.
- Zero curvature: a tiling of Minkowski space.
- Negative curvature: a tiling of anti-de Sitter space.

## 2 comments:

Haven't read it all yet, though it looks bloody interesting! Just a small comment so far: in the 2D section in 2.a I think you meant n = 3.

You are right, thx!

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