Finally, you can make "lenses": take a portion of a horosphere or pseudosphere, tile it in any way you want, and then just glue two such portions with matching vertices together.
mr_e_man wrote::nod: The tower of cubes can also be thought of as a tiling (or truncation) of a cylinder.
...So the number of finite hyperbolic CRFs is at least countably infinite. I believe it must be countable: For any number n, we can make a finite list of all possible polyhedra with at most n faces, each face having at most n edges; then augment the list with that for n+1, etc. This list will eventually reach any such polyhedron, so they are countable.
And I've already shown, by "Euclidean" tilings of a horosphere, that the number of infinite hyperbolic CRFs is uncountably infinite.
The 3D Euclidean regular-celled honeycombs, which have only 3 types of cells (cube, tet, oct), are also uncountable. The argument is similar to the 2D case. Instead of rows of triangles and squares, we use the planes between layers of the octet honeycomb; a plane where octs connect to tets is a 0, and a plane where octs connect to octs is a 1.Finally, you can make "lenses": take a portion of a horosphere or pseudosphere, tile it in any way you want, and then just glue two such portions with matching vertices together.
Could you show a picture of this, or further explanation? I think it would be rare to be able to match the vertices.
Marek14 wrote: And I thought of this: If you take a piece of convex triangular pseudohedron, doesn't it have enough degrees of freedom to press the edges along a plane and then reflect it and join with a second copy?
wendy wrote: To get an idea of the infiniteness of hyperbolic CRF...
mr_e_man wrote: I'm also interested in more general psEuc spaces like (2,2). This is a 4-dimensional space with a null cone made of rays from the origin through a Clifford torus: x2 + y2 - z2 - w2 = 0. I wonder what types of symmetries a polytope could have here. The orthogonal group O(2,2) contains the general linear group GL(2) which is represented by all invertible 2x2 matrices.
mr_e_man wrote:It appears that we're missing some generalization or modification of the pentagonal rotunda.
3.5.n, n > 30; not seen yet
mr_e_man wrote:In hyperbolic space, a horosphere is both convex and intrinsically flat. So Euclidean plane tilings with regular tiles can be made convex. I suppose these should be excluded from CRFs, simply because they're infinite. Not only does each tiling have infinitely many tiles, but there are (uncountably!*) infinitely many different types of tilings.
mr_e_man wrote:Are the Johnson solids the same in spherical or hyperbolic space? I would expect that the curvature, changing the angles, adds or removes some possibilities. Of course, the more symmetric solids (like the cube) exist in all three spaces, though the shape depends on size.
Marek14 wrote:In spherical space, Johnson solids are the same, but hyperbolic space has infinitely many.
mr_e_man wrote:And x=0 is for an ideal polyhedron (vertices at infinity); those aren't rigid, so I exclude them.
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