Uniform tilings in H3 and their possible cuts.

Higher-dimensional geometry (previously "Polyshapes").

Uniform tilings in H3 and their possible cuts.

Postby Marek14 » Wed Jan 11, 2012 2:26 am

And just for fun, how it works in hyperbolic space (convex and concave borders might be the other way around sometimes):

There are 3 kinds of C-D diagrams, 3 linear ones, 1 branching and 5 cyclical.

Linears are 353, 435 and 535, corresponding to 4 regular tilings of H3, icosahedral, cubical and two kinds of dodecahedral tilings.

353:
Regular: Made from icosahedra, 12 to vertex. You can cut pentagonal pyramids off 12 icosahedra and fit a dodecahedral cell inside. You can also border it with pseudohedra, in three different ways:
1: {5,4} pseudohedron. Both convex and concave border is formed by alternating pentagonal pyramids and gyroelongated pentagonal pyramids, but concave border has more icosahedra fit between them, while on convex border they fit together snugly.
2: {5,5} pseudohedron. The convex border is made purely of gyroelongated pentagonal pyramids, while the concave border is made of pentagonal pyramids.
3: (3,5,3,5,3,5) pseudohedron. The convex border is made of icosahedra and pentagonal pyramids, the concave border is made of icosahedra and gyroelongated pentagonal pyramids.
Rectified: Made from dodecahedra and icosidodecahedra. Can be bordered by a (5,10,10) pseudohedron: convex border is made of dodecahedra and pentagonal rotundas, concave border is made of icosidodecahedra and pentagonal rotundas.
Truncated: Made from dodecahedra and truncated icosahedra. Can't be diminished.
Small rhombated: Made from rhombicosidodecahedra, triangular prisms and icosidodecahedra. An icosidodecahedral cell can be enlarged to truncated dodecahedron by swallowing 20 nearby triangular prisms and cutting pentagonal cupolas from 12 nearby rhombicosidodecahedra. In addition, the tiling can be bordered by a (4,10,10) pseudohedron, convex border is made of triangular prisms, diminished rhombicosidodecahedra and pentagonal rotundas, while concave border is bordered by rhombicosidodecahedra, pentagonal cupolas and pentagonal rotundas.
Small prismated: Made from icosahedra and triangular prisms. An icosahedral cell can be enlarged to rhombicosidodecahedron by cutting pentagonal pyramids from 12 nearby icosahedra and swallowing 20+30 nearby triangular prisms. It can be bordered by a (4,5,4,5) plane, the border is made of triangular prisms, gyroelongated pentagonal pyramids and pentagonal pyramids.
Bitruncated: Made from truncated dodecahedra. Can't be diminished.
Great rhombated: Made from truncated icosidodecahedra, triangular prisms and truncated dodecahedra. Can't be diminished.
Prismatotruncated: Made from truncated icosahedra, hexagonal prisms, triangular prisms and rhombicosidodecahedra. A truncated icosahedral cell can be enlarged to truncated icosidodecahedron by cutting pentagonal cupolas from 12 nearby rhombicosidodecahedra and swallowing 20 nearby hexagonal prisms and 30 nearby triangula prisms.
Great prismated: Made from truncated icosidodecahedra and hexagonal prisms. Can't be diminished.

435 group:
Regular 435: Made from cubes. Can be bordered by a {4,5} pseudohedron, both convex and concave border are formed by cubes, but concave border has extra cubes fit to the edges.
Regular 534: Made from dodecahedra. Can be bordered by a {5,4} plane, border formed by dodecahedra.
Rectified 435: Made from cuboctahedra and icosahedra. Icosahedral cell can be enlarged to truncated icosahedron by cutting pentagonal pyramid from 12 nearby icosahedra and cutting 20 nearby cuboctahedra in half. In addition, the tiling can be bordered by a (3,6,5,6) pseudohedron; convex border consists of cuboctahedra, gyroelongated pentagonal pyramids and triangular cupolas, concave border consists of icosahedra, pentagonal pyramids and triangular cupolas. It can be also bordered by a (4,5,4,5) pseudohedron; convex border consists of cuboctahedra and pentagonal pyramids, concave border consists of cuboctahedra and gyroelongated pentagonal pyramids.
Rectified 534: Made from icosidodecahedra and octahedra. Can be bordered by a (4,10,10) pseudohedron, convex border consists of square pyramids and pentagonal rotundas, concave border consists of those as well, but with additional icosidodecahedra fit to edges. Can be also bordered by a (3,10,3,10) plane, border consisting of octahedra, icosidodecahedra and pentagonal rotundas, or by a (4,5,4,5) plane, border consisting of icosidodecahedra and square pyramids.
Truncated 435: Made from truncated cubes and icosahedra. Can be bordered by a (5,8,8) pseudohedron, convex border consisting of pentagonal pyramids and truncated cubes, concave border consisting of gyroelongated pentagonal pyramids and truncated cubes.
Truncated 534: Made from truncated dodecahedra and octahedra. Can be bordered by a (4,10,10) plane, border consisting of square pyramids and truncated dodecahedra.
Small rhombated 435: Made from rhombicuboctaherda, pentagonal prisms and icosidodecahedra. Can be bordered by a (5,8,8) pseudohedron, convex border consisting of pentagonal prisms and gyroelongated square cupolas, concave border consisting of icosidodecahedra and square cupolas. Can be also bordered by a (4,8,10) pseudohedron, convex border consisting of pentagonal prisms, square cupolas and pentagonal rotundas, convave border consisting of rhombicuboctahedra, elongated square cupolas and pentagonal rotundas.
Small rhombated 534: Made from rhombicosidodecahedra, cubes and cuboctahedra. A rhombicosidodecahedral cell can be enlarged to truncated icosidodecahedron by cutting pentagonal cupolas from 12 nearby rhombicosidodecahedra, cutting 20 nearby cuboctahedra in half and swallowing 30 nearby cubes. The tiling can be also bordered by a (4,10,10) pseudohedron, convex border consisting of cubes and diminished rhombicosidodecahedra, concave border consisting of cuboctahedra and pentagonal cupolas.
Small prismated 435: Made from cubes of two kinds, pentagonal prisms and dodecahedra. Can be bordered by a (4,4,4,5) pseudohedron, convex border consisting of cubes and pentagonal prisms, concave border consisting of cubes, pentagonal prisms and dodecahedra.
Bitruncated 435: Made from truncated octahedra and truncated icosahedra. Can't be diminished.
Great rhombated 435: Made from truncated cuboctahedra, pentagonal prisms and truncated icosahedra. Can't be diminished.
Great rhombated 534: Made from truncated icosidodecahedra, cubes and truncated octahedra. Can't be diminished.
Prismatotruncated 435: Made from truncated cubes, octagonal prisms, pentagonal prisms and rhombicosidodecahedra. Can be bordered by a (4,8,10) pseudohedron, convex border consisting of octagonal prisms, truncated cubes and pentagonal cupolas, concave border consisting of pentagonal prisms, octagonal prisms and diminished rhombicosidodecahedra.
Prismatotruncated 534: Made from truncated dodecahedra, decagonal prisms, cubes and rhombicuboctahedra. Can be bordered by a (4,8,10) pseudohedron, convex border consisting of decagonal prisms, truncated dodecahedra and square cupolas, concave border consisting of cubes, decagonal prisms and elongated square cupolas.
Great prismated 435: Made from truncated cuboctahedra, octagonal prisms, decagonal prisms and truncated icosidodecahedra. Can't be diminished.

535 group:
Regular: Made from dodecahedra. Can be bordered by a {5,5} pseudohedron, convex border consisting of dodecahedra, concave border as well, but with extra dodecahedra fit to edges.
Rectified: Made from icosidodecahedra and icosahedra. Can be bordered by a (5,10,10) pseudohedron, convex border consisting of pentagonal pyramids and pentagonal rotundas, concave border consisting of gyroelongated pentagonal pyramids and pentagonal rotundas. Can be also bordered by a {5,4} pseudohedron, convex border consisting of pentagonal pyramids and icosidodecahedra, concave border consisting of gyroelongated pentagonal pyramids and icosidodecahedra. Can be also bordered by a (3,10,5,10) pseudohedron, convex border consisting of icosidodecahedra, gyroelongated pentagonal pyramids and pentagonal rotundas, concave border consisting of icosahedra, pentagonal pyramids and pentagonal rotundas.
Truncated: Made from truncated dodecahedra and icosahedra. Can be bordered by a (5,10,10) pseudohedron, convex border consisting of pentagonal pyramids and truncated dodecahedra, concave border consisting of gyroelongated pentagonal pyramids and truncated dodecahedra.
Small rhombated: Made from rhombicosidodecahedra, pentagonal prisms and icosidodecahedra. Can be bordered by a (4,10,10) pseudohedron, convex border consisting of cubes, pentagonal cupolas and pentagonal rotundas, concave border consisting of rhombicosidodecahedra, diminished rhombicosidodecahedra and pentagonal rotundas. Can be also bordered by a (5,10,10) pseudohedron, convex border consisting of pentagonal prisms and diminished rhombicosidodecahedra, concave border consisting of icosidodecahedra and pentagonal cupolas.
Small prismated: Made from dodecahedra and pentagonal prisms. Can be bordered by a (4,5,4,5) plane, border consisting of pentagonal prisms and dodecahedra.
Bitruncated: Made from truncated icosahedra. Can't be diminished.
Great rhombated: Made from truncated icosidodecahedra, pentagonal prisms and truncated icosahedra. Can't be diminished.
Prismatotruncated: Made from truncated dodecahedra, decagonal prisms, pentagonal prisms and rhombicosidodecahedra. Can be bordered by a (4,10,10) pseudoherdon, convex border consisting of decagonal prisms, truncated dodecahedra and pentagonal cupolas, concave border consisting of pentagonal prisms, decagonal prisms and diminished rhombicosidodecahedra.
Great prismated: Made from truncated icosidodecahedra and decagonal prisms. Can't be diminished.

Branching {3,3,5} group:
Semiregular (alternating): Made from tetrahedra and icosahedra. A vertex can be enlarged to icosidodecahedral cell, cutting pentagonal pyramid from 12 nearby icosahedra and swallowing 20 nearby tetrahedra. The tiling can be also bordered by a (3,5,3,5,3,5) pseudohedron, convex border consisting of tetrahedra and gyroelongated pentagonal pyramids, concave border consisting of icosahedra and pentagonal pyramids. It can be also bordered by a {5,5} pseudohedron, convex border consisting of pentagonal pyramids, concave border consisting of gyroelongated pentagonal pyramids. It can be also bordered by a {5,6} plane, border consisting of pentagonal pyramids and gyroelongated pentagonal pyramids. It can be also bordered by a {3,10} plane, border consisting of tetrahedra and icosahedra.
Truncated: Made from truncated tetrahedra, truncated icosahedra and icosidodecahedra. It can be bordered by a (6,6,10) plane, border consisting of truncated tetrahedra, truncated icosahedra and pentagonal rotundas.
Small rhombated: Made from rhombicosidodecahedra, tetrahedra and dodecahedra. A dodecahedral cell can be enlarged to truncated dodecahedron by cutting a pentagonal cupola from 12 nearby rhombicosidodecahedra and swallowing 20 nearby tetrahedra. The tiling can be also bordered by a (5,10,10) pseudohedron, convex border consisting of dodecahedra and diminished rhombicosidodecahedra, concave border consisting of rhombicosidodecahedra and pentagonal cupolas.
Great rhombated: Made from truncated icosidodecahedra, truncated tetrahedra and truncated dodecahedra. Can't be diminished.

Cyclical (3,3,3,4):
(3,3)-Semiregular: Made of tetrahedra and two kinds of octahedra. A vertex can be enlarged to a rhombicuboctahedral cell by cutting 6+12 nearby octahedra in half and swallowing 8 nearby tetrahedra. The tiling can be also bordered by a (3,4,3,4,3,4) pseudohedron, convex border consisting of octahedra and square pyramids, concave border consisting of tetrahedra and square pyramids. It can be also bordered by a {3,8} pseudohedron, convex border consisting of tetrahedra and octahedra, concave border consisting of octahedra only.
(3,4)-Semiregular: Made of tetrahedra, cubes and cuboctahedra. Can be bordered by a (3,6,4,6) pseudohedron, convex border consisting of tetrahedra, cubes and triangular cupolas, concave border consisting of cuboctahedra and triangular cupolas. Can be also bordered by a (3,4,3,4,3,4) pseudohedron, convex border consisting of tetrahedra and cuboctahedra, concave border consisting of cuboctahedra and cubes.
Truncated (343): Made of tetrahedra and truncated cubes. Can be bordered by a (3,8,3,8) plane, border consisting of tetrahedra and truncated cubes.
Truncated (334): Made of tetrahedra, cubes, truncated tetrahedra and truncated octahedra. Can be bordered by a (3,6,4,6) pseudohedron, convex border consisting of tetrahedra, truncated octahedra and truncated tetrahedra, concave border consisting of truncated tetrahedra, cubes and truncated octahedra.
Truncated (333): Made of octahedra and truncated tetrahedra. Can be bordered by a (3,6,4,6) pseudohedron, convex border consisting of truncated tetrahedra and square pyramids, concave border consisting of octahedra, square pyramids and truncated tetrahedra.
Small rhombated: Made of octahedra, two kinds of cuboctahedra and rhombicuboctahedra. A cuboctahedral cell can be enlarged to a truncated cuboctahedron by cutting square cupolas from 6 nearby rhombicuboctahedra, cutting 8 nearby cuboctahedra in half and cutting 12 nearby octahedra in half. The tiling can be also bordered by a (6,6,8) pseudohedron, convex border consisting of triangular cupolas and elongated square cupolas, concave border consisting of triangular cupolas and square cupolas. It can also be bordered by a (3,8,3,8) pseudohedron, convex border consisting of cuboctahedra, octahedra and square cupolas, concave border consisting of cuboctahedra and elongated square cupolas. It can also be bordered by a (3,6,4,6) pseudohedron, convex border consisting of octahedra, rhombicuboctahedra and triangular cupolas, concave border consisting of rhombicuboctahedra, cuboctahedra and triangular cupolas. It can also be bordered by a (4,4,4,6) plane, border consisting of square pyramids, cuboctahedra, rhombicuboctahedra and triangular cupolas.
Great rhombated (33): Made from two kinds of truncated octahedra, truncated tetrahedra and rhombicuboctahedra. Can be bordered by a (6,6,8) pseudohedron, convex border consisting of truncated tetrahedra, truncated octahedra and square cupolas, concave border consisting of truncated octahedra and elongated square cupolas.
Great rhombated (34): Made from truncated tetrahedra, truncated cubes, truncated cuboctahedra and cuboctahedra. Can be bordered by a (6,6,8) pseudohedron, convex border consisting of truncated tetrahedra, triangular cupolas and truncated octahedra, concave border consisting of truncated cuboctahedra, triangular cupolas and truncated cubes.
Great prismated: Made from truncated octahedra and truncated cuboctahedra. Can't be diminished.

Cyclical (3,3,3,5):
(3,3)-Semiregular: Made of tetrahedra, octahedra and icosahedra. A vertex can be enlarged to a rhombicosidodecahedral cell, cutting pentagonal pyramids from 12 nearby icosahedra, cutting 30 nearby octahedra in half and swallowing 20 nearby tetrahedra. Can be bordered by a (3,5,3,5,3,5) pseudohedron, convex border consisting of octahedra and pentagonal pyramids, concave border consisting of tetrahedra and gyroelongated pentagonal pyramids. Can be also bordered by a {3,10} pseudohedron, convex border consisting of octahedra and tetrahedra, concave border consisting of icosahedra and octahedra. Can be also bordered by a (3,5,3,5,3,5,3,5,3,5) pseudohedron, convex border consisting of octahedra and gyroelongated pentagonal pyramids, concave border consisting of tetrahedra and pentagonal pyramids. Can be also bordered by a (3,4,4,3,4,4,3,4,4) pseudohedron, convex border consisting of icosahedra and square pyramids, concave border consisting of octahedra and square pyramids.
(3,5)-Semiregular: Made of tetrahedra, dodecahedra and icosidodecahedra. Can be bordered by a (3,10,5,10) pseudohedron, convex border consisting of tetrahedra, dodecahedra and pentagonal rotundas, concave border consisting of icosidodecahedra and pentagonal rotundas. Can be also bordered by a (3,5,3,5,3,5) pseudohedron, convex border consisting of tetrahedra and icosidodecahedra, concave border consisting of icosidodecahedra and dodecahedra.
Truncated (353): Made of tetrahedra and truncated dodecahedra. Can be bordered by a (3,10,3,10) plane, border consisting of tetrahedra and truncated dodecahedra.
Truncated (335): Made of tetrahedra, dodecahedra, truncated tetrahedra and truncated icosahedra. Can be bordered by a (3,6,10,6) pseudohedron, convex border consisting of tetrahedra, truncated icosahedra and truncated tetrahedra, concave border consisting of truncated tetrahedra, dodecahedra and truncated icosahedra.
Truncated (333): Made of icosahedra and truncated tetrahedra. Can be bordered by a (3,6,5,6) pseudohedron, convex border consisting of truncated tetrahedra and pentagonal pyramids, concave border consisting of icosahedra, gyroelongated pentagonal pyramids and truncated tetrahedra. Can be also bordered by a (5,6,5,6) plane, border consisting of pentagonal pyramids, gyroelongated pentagonal pyramids and truncated tetrahedra.
Small rhombated: Made of octahedra, icosidodecahedra, cuboctahedra and rhombicosidodecahedra. An icosidodecahedral cell can be enlarged to a truncated icosidodecahedron by cutting pentagonal cupolas from 12 nearby rhombicosidodecahedra, cutting 20 nearby cuboctahedra in half and cutting 30 nearby octahedra in half. The tiling can be also bordered by a (6,10,10) pseudohedron, convex border consisting of triangular cupolas, diminished rhombicosidodecahedra and pentagonal rotundas, concave border consisting of triangular cupolas, pentagonal cupolas and pentagonal rotundas. It can also be bordered by a (3,10,3,10) pseudohedron, convex border consisting ofcuboctahedra, octahedra and pentagonal cupolas, concave border consisting of icosidodecahedra, cuboctahedra and diminished icosidodecahedra. It can also be bordered by a (3,6,5,6) pseudohedron, convex border consisting of octahedra, rhombicosidodecahedra and triangular cupolas, concave border consisting of rhombicosidodecahedra,icosidodecahedra and triangular cupolas. It can also be bordered by a (4,4,4,10) plane, border consisting of square pyramids, cuboctahedra, rhombicosidodecahedra and pentagonal rotundas.
Great rhombated (3,3): Made of truncated tetrahedra, truncated octahedra, truncated icosahedra and rhombicosidodecahedra. Can be bordered by a (6,6,10) pseudohedron, convex border consisting of truncated tetrahedra, truncated octahedra and pentagonal cupolas, concave border consisting of truncated octahedra, truncated icosahedra and diminished rhombicosidodecahedra.
Great rhombated (3,5): Made of truncated tetrahedra, truncated dodecahedra, truncated icosidodecahedra and cuboctahedra. Can be bordered by a (6,6,10) pseudohedron, convex border consisting of truncated tetrahedra, triangular cupolas and truncated icosidodecahedra, concave border consisting of truncated octahedra, triangular cupolas and truncated dodecahedra.
Great prismated: Made from truncated octahedra and truncated icosidodecahedra. Can't be diminished.

Cyclical (3,4,3,4):
Semiregular: Made of cubes, octahedra and cuboctahedra. Can be bordered by a (4,6,4,6) pseudohedron, convex border consisting of cubes, square pyramids and triangular cupolas, concave border consisting of cuboctahedra, square pyramids and triangular cupolas. Can be also bordered by a {4,6} pseudohedron, convex border consisting of cuboctahedra and square pyramids, concave borded consisting of cubes and square pyramids. Can be also bordered by a (3,4,3,4,3,4,3,4) pseudohedron, convex border consisting of cuboctahedra and cubes, concave border consisting of octahedra and cuboctahedra.
Truncated (434): Made of cubes and truncated octahedra. Can be bordered by a (4,6,4,6) plane, border consisting of cubes and truncated octahedra.
Truncated (343): Made of octahedra and truncated cubes. Can be bordered by a (3,8,4,8) pseudohedron, convex border consisting of truncated cubes and square pyramids, concave border consisting of octahedra, square pyramids and truncatedcubes.
Small rhombated: Made of cuboctahedra and rhombicuboctahedra. Can be bordered by a (6,8,8) pseudohedron, convex border consisting of triangular cupolas, square cupolas and elongated square cupolas, concave border consisting of triangular cupolas, elongated square cupolas and square cupolas, but with more polyhedra fit to edges. Can also be bordered by a (3,8,4,8) pseudohedron, convex border consisting of cuboctahedra, rhombicuboctahedra and square cupolas, concave border consisting of cuboctahedra, rhombicuboctahedra and elongated square cupolas. Can also be bordered by a (4,6,4,6) plane, border consisting of rhombicuboctahedra and triangular cupolas.
Great rhombated: Made of truncated cubes, truncated octahedra, truncated cuboctahedra and rhombicuboctahedra. Can be bordered by a (6,8,8) pseudohedron, convex border consisting of truncated octahedra, truncated cuboctahedra and elongated square cupolas, concave border consisting of truncated cuboctahedra, truncated cubes and square cupolas.
Great prismated: Made from truncated cuboctahedra. Can't be diminished.

Cyclical (3,4,3,5):
(3,4)-Semiregular: Made of cubes, cuboctahedra and icosahedra. Can be bordered by a (4,6,5,6) pseudohedron, convex border consisting of cubes, pentagonal pyramids and triangular cupolas, concave border consisting of cuboctahedra, gyroelongated pentagonal pyramids and triangular cupolas. Can be also bordered by a (4,5,4,5,4,5) pseudohedron, convex border consisting of cuboctahedra and pentagonal pyramids, concave border consisting of cubes and gyroelongated pentagonal pyramids. Can be also bordered by a (3,4,3,4,3,4,3,4,3,4) pseudohedron, convex border consisting of cuboctahedra and cubes, concave border consisting of icosahedra and cuboctahedra. Can be also bordered by a (4,5,4,5,4,5,4,5,4,5) pseudohedron, convex border consisting of cuboctahedra and gyroelongated pentagonal pyramids, concave border consisting of cubes and pentagonal pyramids.
(3,5)-Semiregular: Made of dodecahedra, octahedra and icosidodecahedra. Can be bordered by a (4,10,5,10) pseudohedron, convex border consisting of square pyramids, pentagonal rotundas and dodecahedra, concave border consisting of square pyramids, pentagonal rotundas and icosidodecahedra. Can be also bordered by a (4,5,4,5,4,5) pseudohedron, convex border consisting of square pyramids and icosidodecahedra, concave border consisting of square pyramids and dodecahedra. Can be also bordered by a (3,5,3,5,3,5,3,5) pseudohedron, convex border consisting of icosidodecahedra and dodecahedra, concave border consisting of octahedra and icosidodecahedra.
Truncated (435): Made of cubes, dodecahedra, truncated octahedra and truncated icosahedra. Can be bordered by a (4,6,5,6) pseudohedron, convex border consisting of cubes, truncated icosahedra and truncated octahedra, concave border consisting of truncated octahedra, dodecahedra and truncated icosahedra.
Truncated (353): Made of octahedra and truncated dodecahedra. Can be bordered by a (3,10,4,10) pseudohedron, convex border consisting of truncated dodecahedrahedra and square pyramids, concave border consisting of octahedra, square pyramids and truncated dodecahedra.
Truncated (343): Made of icosahedra and truncated cubes. Can be bordered by a (3,8,5,8) pseudohedron, convex border consisting of truncated cubes and pentagonal pyramids, concave border consisting of icosahedra, gyroelongated pentagonal pyramids and truncated cubes. Can be also bordered by a (5,8,5,8) plane, border consisting of pentagonal pyramids, gyroelongated pentagonal pyramids and truncated cubes.
Small rhombated: Made of cuboctahedra, icosidodecahedra, rhombicuboctahedra and rhombicosidodecahedra. Can be bordered by a (6,8,10) pseudohedron, convex border consisting of triangular cupolas, elongated square cupolas and pentagonal cupolas, concave border consisting of triangular cupolas, square cupolas and diminished rhombicosidodecahedra. Can be also bordered by a (8,10,10) pseudohedron, convex border consisting of square cupolas, diminished rhombicosidodecahedra and pentagonal rotundas, concave border consisting of elongated square cupolas, pentagonal cupolas and pentagonal rotundas. Can also be bordered by a (3,10,4,10) pseudohedron, convex border consisting of rhombicuboctahedra, cuboctahedra and pentagonal cupolas, concave border consisting of icosidodecahedra, rhombicuboctahedra and diminished rhombicosidodecahedra. Can also be bordered by a (3,8,5,8) pseudohedron, convex border consisting of cuboctahedra, rhombicosidodecahedra and square cupolas, concave border consisting of rhombicosidodecahedra,icosidodecahedra and elongated square cupolas. Can also be bordered by a (4,6,4,10) plane, border consisting of rhombicuboctahedra, rhombicosidodecahedra, triangular cupolas and pentagonal rotundas.
Great rhombated (3,4): Made of truncated cubes, truncated icosahedra, truncated cuboctahedra and rhombicosidodecahedra. Can be bordered by a (6,8,10) pseudohedron, convex border consisting of truncated icosahedra, truncated cuboctahedra and diminished rhombicosidodecahedra, concave border consisting of truncated cuboctahedra, truncated cubes and pentagonal cupolas.
Great rhombated (3,5): Made of truncated octahedra, truncated dodecahedra, truncated icosidodecahedra and rhombicuboctahedra. Can be bordered by a (6,8,10) pseudohedron, convex border consisting of truncated octahedra, elongated square cupolas and truncated icosidodecahedra, concave border consisting of truncated icosidodecahedra, square cupolas and truncated dodecahedra.
Great prismated: Made from truncated cuboctahedra and truncated icosidodecahedra. Can't be diminished.

Cyclical (3,5,3,5):
Semiregular: Made of dodecahedra, icosidodecahedra and icosahedra. Can be bordered by a (5,10,5,10) pseudohedron, convex border consisting of dodecahedra, pentagonal pyramids and pentagonal rotundas, concave border consisting of icosidodecahedra, gyroelongated pentagonal pyramids and pentagonal rotundas. Can be also bordered by a {5,6} pseudohedron, convex border consisting of icosidodecahedra and pentagonal pyramids, concave border consisting of dodecahedra and gyroelongated pentagonal pyramids. Can be also bordered by a (3,5,3,5,3,5,3,5,3,5) pseudohedron, convex border consisting of icosidodecahedra and dodecahedra, concave border consisting of icosahedra and icosidodecahedra. Can be also bordered by a {5,10} pseudohedron, convex border consisting of icosidodecahedra and gyroelongated pentagonal pyramids, concave border consisting of dodecahedra and pentagonal pyramids.
Truncated (535): Made of dodecahedra and truncated icosahedra. Can be bordered by a (5,6,5,6) plane, border consisting of dodecahedra and truncated icosahedra.
Truncated (353): Made of icosahedra and truncated dodecahedra. Can be bordered by a (3,10,5,10) pseudohedron, convex border consisting of truncated dodecahedra and pentagonal pyramids, concave border consisting of icosahedra, gyroelongated pentagonal pyramids and truncated dodecahedra. Can be also bordered by a (5,10,5,10) plane, border consisting of pentagonal pyramids, gyroelongated pentagonal pyramids and truncated dodecahedra.
Small rhombated: Made of icosidodecahedra and rhombicosidodecahedra. Can be bordered by a {10,3} pseudohedron, convex border consisting of pentagonal rotundas, pentagonal cupolas and diminished rhombicosidodecahedra, concave border consisting of pentagonal rotundas, diminished rhombicosidodecahedra and pentagonal cupolas, but with more polyhedra fit to edges. Can also be bordered by a (3,10,5,10) pseudohedron, convex border consisting of rhombicosidodecahedra, icosidodecahedra and pentagonal cupolas, concave border consisting of icosidodecahedra, rhombicosidodecahedra and diminished rhombicosidodecahedra. Can also be bordered by a (4,10,4,10) plane, border consisting of rhombicosidodecahedra and pentagonal rotundas.
Great rhombated: Made of truncated dodecahedra, truncated icosahedra, truncated icosidodecahedra and rhombicosidodecahedra. Can be bordered by a (6,10,10) pseudohedron, convex border consisting of truncated icosahedra, truncated icosidodecahedra and diminished rhombicosidodecahedra, concave border consisting of truncated icosidodecahedra, truncated dodecahedra and pentagonal cupolas.
Great prismated: Made from truncated icosidodecahedra. Can't be diminished.
Marek14
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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Fri Jan 25, 2013 9:45 am

Great work Marek!
Do you have a similar analysis for H4 too?

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Re: Uniform tilings in H3 and their possible cuts.

Postby Marek14 » Fri Jan 25, 2013 11:20 pm

Klitzing wrote:Great work Marek!
Do you have a similar analysis for H4 too?

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Well, not really, considering that you're the first reply, and more than a year after my post :)
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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Sat Jan 26, 2013 9:22 am

So you've got time enough, isn't it? :lol:

Well, to be more ernest, at the moment I'm esp. interested in the possible (resp. impossible) subtilings of x3o3o3o5o. As this is the tiling Wendy uses for basis in her thread bid.3.3.3.5 . There she tries to apply the different known diminishings of ex onto that hyperbolic beast. But somehow we encounter problems when elaborating those. (Might be that we aren't clever enough, but could be that there is an intrinsic reasen which does not allow swirlprismatic subsymmetry, for instance.)

Even so, in the full elaboration on all those tilings I would be interested too, if accessible one time...

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Re: Uniform tilings in H3 and their possible cuts.

Postby Marek14 » Sat Jan 26, 2013 6:09 pm

Not sure how much time I'll have, but we'll see :)
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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Wed Jan 30, 2013 5:02 pm

... that bid.3.3.3.5 thread needs esp. the support on whether there is a 6-coloring of the vertices of x3o3o3o5o or not.
The idea there was to inscribe the (tau-scaled) chiral 5x 24-cell compound into the vertex figure (600-cell) of that tiling. Locally this clearly is possible. Question remains whether this extends consitently across whole x3o3o3o5o. And, in positive case, hopefully getting additional clues on that to be determined structure.

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Re: Uniform tilings in H3 and their possible cuts.

Postby Marek14 » Wed Jan 30, 2013 9:12 pm

Let's see...

3,3,3,5 has a 600-cell as a verf. The various cuts are based on possible hyperplanes through the verf, of course. Too bad Stella doesn't support 4D facetings :)

And here is where my imagination falters for now, as I don't have a good idea of the hyperplanes I'm looking for (they should be cuts that never split an edge). How many different distances does even exist between 600-cell vertices?

But let's talk the basics. You can, obviously, remove an icosahedral pyramid from a 600-cell. If that 600-cell is a vertex figure of a 3,3,3,5, you are removing a single vertex from the figure, resulting in 600 pentachora being removed and replaced by a single 600-cell. Not sure what other configurations are possible...
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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Wed Jan 30, 2013 11:38 pm

If you look at the icosahedron (ike) you would have kind an analogy to the hexacosachoron (ex). Next look at an inscribed, tau-scaled small stellated dodecahedron (sissid), this would resemble the ex-inscribed tau-scaled sishi. Finally you clearly can further inscribe an great icosahedron (gike) into sissid. There you see 5 triangles meeting each vertex. This resembles dox, the compound of 25 icosatetrachora (ico). Dox likewise could be told to be a compound of 5 chi, where chi itself is the chiral compound of 5 icoes. This latter compound has the advantage to be incident to ex's vertices just once. (Kind of like using just one triangle of gike at each vertex of ike...)

Well this is a mere dimensional analogy, but it might help, perhaps. You might also refer to http://bendwavy.org/klitzing/explain/regular.htm#4d, where also the different vertex distances within ex are listed.

In fact, there is an interesting connection between "ex" and sadi, that snub of the icositetrachoron (s3s4o3o): just vertex inscribe a tau-scaled ico into ex. Then diminish from ex exactly those 24 vertices of incidence, i.e. chop off an ike-pyramid each.

You further could use 2 icoes (of that compound chi) instead and diminish ex at the corresponding inscribed incident vertices. The result is the scaliform, swirl-symmetric, and noble polychoron with the Bowers acronym bidex (bi-icositetra-diminished ex). Its cells are nothing but 48 tridiminished icosahedra. That figure once was discovered by Weimholt. The vertex figure of that bidex is topologically a triangular prism with one of its squares diagonally dissected. Even so, it rather is a hexa-diminished ike! (You could find a picture of that vertex figure together with a short description of bidex at this page of hedrondude.

Well this stuff can be run further. - The idea of Wendy's bid.3.3.3.5 thread is to find somehow an analogy between these iterated diminishings of ike (x3o5o), those of ex (x3o3o5o), and extrapolating that then onto x3o3o3o5o.

The idea was fascinating. Even so Wendy and me still have problems to derive the correct incidence matrices of those supposed hyperbolic diminishings.

The polytopal 3D and 4D stuff essentially is known so far. Now we are exploring 4D hyperbolic tiling space in this context. I thought you might come in and help us out a bit - if you'd like. :)

The point here is: x3o3o3o5o has for vertex figure . x3o3o5o, i.e. ex. We are looking for exactly that iterated diminishing process here. Thus we consider the ex-inscribed tau-scaled chi, or rather one ico after the other. Accordingly we could start with, say, a red vertex. That one will have 120 neighbouring vertices (of ex), which clearly can be divided into 5 * 24, right in the way of that chiral compound, e.g. coloring those vertices in orange, yellow, green, blue, and purple. This gives us a local start configuration of a supposed 6-coloring of the x3o3o3o5o vertices. Thus the question is whether this very start configuration could be continued uniformely.

If the answer would come out to be "yes, it exists", we clearly would like to learn more about that structure. (Esp. investigating why our aims so far did not succeed.) So then we should continue Wendy's initial research. - If the answer would be "no, it doesn't", then we could understand our (so far) fail. (Sure a constructive proof/disproof, instead of an abstract one, would be better for that purpose.)

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Re: Uniform tilings in H3 and their possible cuts.

Postby Marek14 » Thu Jan 31, 2013 7:30 am

So, basically: 3,3,3,5 has a certain symmetry group and you are trying to find its subgroups.

My first idea would be: do we know the structure of the basic group? Is there an abstract mathematical structure with same structural properties as this tesselation?
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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Thu Jan 31, 2013 11:40 am

Marek14 wrote:So, basically: 3,3,3,5 has a certain symmetry group and you are trying to find its subgroups.

Right, this is what it should amount to, in the first run.
Later we might look for geometry, then.

My first idea would be: do we know the structure of the basic group? Is there an abstract mathematical structure with same structural properties as this tesselation?

Well, I'm no crack in group theory. So I don't know what is known there. I just could provide some public references, to start with:

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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Thu Jan 31, 2013 11:46 am

In fact, this hyperbolic research, and now esp. that mere group theoretical aspect, is why I switched this discussion of Wendy's thread from in here, i.e. from "polytopes" into "other geometry"...

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Re: Uniform tilings in H3 and their possible cuts.

Postby wendy » Fri Feb 01, 2013 8:55 am

The subject of groups gets quite heavy and ugly quite fast.

The first three are by Tom Ruen. He sends me many emails full of questions, i can't see how 'geometric folding' works outside what is known by mirror insertion, but he seems to derive things like 3,5 from 3,3,3 (their rotation group calls down to A_5). Still, there are things one is not meant to know. He does seem to want to reduce things to 'coxeter groups', when such things are not mayly.

I have not figured out what the are differences between 'lorentzian lattices' and 'bollis polytopes' . Lorentz appears to be the 'time-like' axis that hyperbolic systems are supposed to have: John Conway used this time-like thing too. I'm hardly sure what is meant by 'lattice', certianly there are things i would not count as a lattice, however ill defined.

I had a quick glance at davis, but it is replete in mathematics.

Groups are not polytopes. It's a kind of description, like 'three' in 'three sheep' and 'three tins'. One can run into all sorts of problems when dealing with numbers like '27' in 2_22, and 120 in [3,3,5], which exists as 5*24 also.

The underlying group in the bid.3335, etc, is that at {3,3,5}, the group is [3,4,3+], of order 576 (496). This correpsonds to a 24-choron, its faces divided into eight octants, giving 192 (172). These octants are like cube-corners, there is an order 3 rotation around the triangle to give a complete order of 576 (496). Introducing a mirror here gives an order of 1152 (972). Both the octahedron and icosahedron share the pyritohedral group (order 24), but the mirrors that cross the octant in the icosahedron, do not correspond to six mirrors that the octahedral group provides.

Wendy could have moved the thread. It is in her powers.

With 3,3,3,5, there seem to be shared symmetries with 4,3,3,5 and 5,3,3,5, but the groups are not simple. The second has a cell of 8 1/2 of the first, the third as 26. But there is no simple multiple of 17 or 26 of 3,3,3,5 that gives E,3,3,5 or 5,3,3,5. Even a group like [3,9] and [3,18] have a common subgroup [3,3,9:], of orders 2 and 4 respectively, but the mirrors that divide the common subgroup don't match up.

The number of distances that exist in a polytope is pretty much defined by the vertex arrangement. For polytopes of the {3,..,5} series, the icosahedral series is used. For {5,...,3}, one uses the dodecahedral series.

The series consist of numbers, in base phi, a kind of binary, where the digit pair -11- is forbidden everywhere. This gives, then

0, 1, 10, 100, 101, 1000, 1001, 1010, 10000, etc.

The icosahedral series forbids the last two digits as -10. This represents a kind of unallowed shift in the isomorph. Actually, the forbidden point is the last '1' in an 'odd power', eg -10 00 00 is also forbidden. This is because the isomorph is negative, while the geometric isomorph has positive radii throughout.

The dodecahedral series forbids also, the last four digits -0101, as well as -10. This is due to properties of {5/2,3,...} Note that -01 01 00 is allowed, because the size of 5/2,3,... is less than 1. 01 00 ...

There are 31 series from 0 to '1 00 01 00 01', the diameter of the {5,3,3}, in the dodecahedral series, and 9 in the icosahedral series 0 to 1 01 01 for the {3,3,5}. The rings of the {3,3,3,5} belong to the icosahedral series. The actual length, is found by evaluating these base phi (eg 100 as phi-squared), and then taking the square root. The ring of 5,3,3 at 10 10 00 01, contains some 28 vertices, 4 belonging to an inscribed {3,3,3} with full symmetry, and the remaining 24 belong to 6 (12) tetrahedra, which belong to the compound of lesser regular pentachora.
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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Fri Feb 01, 2013 11:43 am

Thanks for commenting on my provided reference list.
Folding so far is kind of dubious to me too. At least I've not completely understood what's going on there in detail...

I had a quick glance at davis, but it is replete in mathematics.
True.

Wendy could have moved the thread. It is in her powers.
Hehe. I rather had in mind to kind of open a secondary issue on the same context. Here to deal with hyperbolics (this thread) and "other Geometry" (this section), while things concerning the actual polytopes and the polytopes being contained within those k-id.3335 structures, to my opinion, well remain there. (I even posted an addition there too, yesterday, you might have seen...)

The series consist of numbers, in base phi, a kind of binary, where the digit pair -11- is forbidden everywhere. This gives, then

0, 1, 10, 100, 101, 1000, 1001, 1010, 10000, etc.
Not clear what's exactly the base you are using there. Is it a thing like {..., tau^3, tau^2, tau^1, tau^0}, or something in the sense of {..., tau*100, 100, tau*10, 10, tau*1, 1} (with integers being base 2 here, i.e. 1=2^0, 10=2^1, 100=2^2, ...), or what else do you mean?

In fact, because of tau^2=tau+1, you should be able to reduce everything into m*tau+n*1.

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Re: Uniform tilings in H3 and their possible cuts.

Postby wendy » Sat Feb 02, 2013 9:14 am

The importance of base phi in the process is that's where the boundaries lie.

We will start with "Reskalierungssymmetrien quasiperiodischer Struckturen" by R. Klitzing. The title translates as 'rescaling-symmetry of quasiperiodic structures'. What we're interested here is not rescaling, but isomorphism.

Consider for example the Fibonacci Substitution L -> LS and S -> L. This produces a general series LSLLSLSL... It's not easy to type, the point here is that the first 'L' is at point '0'. Where L = 1, S = 0.618033&c, then this series gives the important lines in order, of a pentagonal tiling, like 5,3,3,3 &c. When S is replaced by -1.618033&c. then the whole series is restricted to a range between -1 < x < 1.61803398875, it zigzags between them. It is this isomorphism that interests us, since this equates to replacing, eg 3,3,3,5 by 3,3,3,5/2, &c.

The series is better represented by numbers in tau, or better, tau², where in the latter, one has digits 0, 1, f=1.61803398875, and the combination 1f = f0 is used everywhere. If one uses this counting rule, one produces the fibonacci series directly in the last digit, 0 and f produce a L, and 1 produces an S.

0=L, 1=S, f=L, 10=L, 11=S, f0=L, f1=S, ff=L, 100=L, 101 = S, 10f = L, 110 = L, 111 = S, f00 = L, etc.

It's quite possible to break into this series at any point, because these numbers can be expressed as fibonacci numbers. This is just a count of allowed positions, not the values of the things.

1 = 1, f=2, 10=3, f0=5, 100=8, f00 = 13, 1000=21, f000 = 34, 10000 = 55, f0000 = 89, 100000 = 144 &c.

We then write 120 = 89 + 21 + 8 + 2 = f110f, gives L, 121 f1110 gives L, 122 = f1111 = S, 123 = ff000 = L, 124 = ff001 = S, 125 = ff00f = L, etc.

More interestingly, is the isomorphic structure. If steps are allowed in one, it is allowed in the other. The range from -1 to +f is covered in three ranges, representing the final non-zero digit of f (negative), or 1 (positive). Since we can't take the square root of a negative number, we don't count bands ending in an 'f'. Since also we are restricted by the physical size of 5/2,3,3...,3, then we don't count bands that end in -11 at the end.

All of the rest do evaluate, there is no restriction on (3,3,3..,5/2), so we get, eg this table.

Here are the first 64 rings. The second column is the base-phi2 item. The third column is the square-root, of the number, base phi2. The final columns give the rings of 3335, 53, and 5333.

Rule 'a' is numbers ending in 'f', which gives an negative isomorph. Rule 'd' is the dodecahedral rule, which implies that the isomorph is less than, 1.1, ie 1.381967. g=sqrt(f.r5), rN = sqrt(N). fN = f^N, (p) is a divisor of a number p. () is an unevaluated divisor: it's not hard, but neither Coxeter nor i have bothered.

Code: Select all
                          335     53    533
  0     0  L       0         1      1     1
  1     1  S       1        12      3     4
  2     f  L  a
  3    10  L       f        20      6    12
  4    11  S  d    g        12
  5    f0  L  a
  6    f1  S       f.r2     30      6    24
  7    ff  L  a
  8   100  L       f^2      12      3    12
  9   101  S       f.r3     20      1     4
10   10f  L  a
11   110  L       f.g      12           24
12   111  S  d    2f        1
13   f00  L  a
14   f01  S       (11)                  24
15   f0f  L  a
16   f10  L       2.f^2                 32
17   f11  S  d
18   ff0  L  a
19   ff1  S       (11)                  24
20   fff  L  a
21  1000  L       f^3                   12
22  1001  S       fg.r2                 24
23  100f  L  a
24  1010  L       f2.r3                 28
25  1011  S  d
26  10f0  L  a
27  10f1  L       (19)                  12
28  10ff  S
29  1100  L  a    f2.g                  24
30  1101  S
31  110f  L  a
32  1110  L       2f2                   54
33  1111  S  d
34  f000  L
35  f001  S       ()                    24
36  f00f  L  a
37  f010  L       (11)                  12
38  f011  S  d
39  f0f0  L  a
40  f0f1  S       f2.r5                 28
41  f0ff  L  a
42  f100  L       f3.r2                 24
43  f101  S       ()                    12
44  f10f  L  a
45  f110  L       (11)                  24
46  f111  L  d
47  ff00  S  a
48  ff01  S       f2.r6                 32
49  ff0f  L  a
50  ff10  L       (11)                  24
51  ff11  S  d
52  fff0  L  a
53  fff1  S       ()                    24
54  ffff  L  a
55 10000  L       f4                     4
56 10001  S       f2.r7                 12
57 1000f  L  a
58 10010  L       f2.g.r2               24
59 10011  S  d
60 100f0  L  a
61 100f1  S       ()                    12
62 100ff  L  a
63 10100  L       f3.r2                  4
64 10101  S  a    2f2r2                  1
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Re: Uniform tilings in H3 and their possible cuts.

Postby wendy » Sat Feb 02, 2013 11:36 am

On 'folding groups'

The only 'folding' known to me, is the geometric inserting and removal of mirrors.

The simplex group

For a set of nodes, one can test for an equality, by lettering the nodes to be tested, and numbering the balance. Then

For lettered nodes, A,B,... then x = AB = AC = BC = AD = BD = CD = ..., for each numbered node, A1 = B1 = C1 = D1 = ... One can then replace said nodes by a chain A-(2x)--b----c----... The process is reversable. b, c, d... are new mirrors replacing B, C, D, ... such that the image of A in b is B, the image of B in c is C, etc. This rule suffices to find most of the known subgroups.

The group x5o3o3o, can be represented in the form A5b3c3d, (using the letters above), and thence into a group of four mirrors, where every pair is joined by a 5/2 branch. By grouping the mirrors into 1+3 or 2+2, one can derive subgroups of 4!/1! 3! = 4 (ie 5/2, 5, 3), and 4!/2! 2! = 6, (ie 5, 5/2, 5)

The arms of the groups like 1_22 are not equal by this rule, since eg if the ends were A and B, and the numbers 3 and 4 were applied to the branches with A, B, then A3 = 3, while B3 = 2. The symmetry that swaps the arms of 1_22 is not produced by mirrors.

The antiprism group

A group of the form o--(2n)--o--(n)--o--(2n)--o=====z, being a loop of '3' branches, except for a chain 2n,n,2n, and ==== being an unmarked chain of 3's, of length 0 or more, can be replaced by a group o--2n--o=====o--4--o, of sub-order equal to the number of faces of o====o-4--o,

The group o6o3o6o3z is a subgroup of order eight of [6,3,4]

The group o---a---o---b---o---b---o---a---o======z, being a loop abba and any number of three's, gives a subgroup o--b--o--a--o=====o--4--o, of sub-order equal to the number of faces of o=====o---4--o

Equity of figures

If two Wythoff groups (ie mirror-walled simplex), give the same figure, in every geometry other then euclidean, then there is a common subgroup, the order being by way of the common group. 334A and 434A share a common figure, being o3o3o4xAo = x4o3o4xAo, and so their groups stand in the ratios against the vertices, viz o3o3o3o = 100, (120), vs o3o3o = (24), the first is a fifth of the second.

Coxeter's Sums

Coxeter does a set of fancy sums of tetrahedra in 'regular polytopes', but the nature of this evaluation is beyond my means. However, it is understood how to reverse a lune (a pie slice: this is the english word that RK was searching for in the slices of 3,3,5). This is used to evaluate the density of the octagrammy (ie 73 for o3x4/3x3o, and because a face of octagrammy is the vertex-figure of the o3x4x3o octagonny, the vertex angle of that figure = 30.50 = 73/288. One can verify from 5/2,3,3 (which produces a simplex of order 24*1.71 = 38.24), that 5/2,5,5/2, and 5,5/2,3, have the correct densities at 66 and 76 respectively.

These same densities arise from the weighted form of Euclid's equation, when applied to the discrete lattices 533V, V53V, 53V5, 35V3, and V5V5. (V=5/2).
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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Sat Feb 02, 2013 3:24 pm

Have to gaze at your mails for a while on...
But to begin with, I'll start to comment on an evident point. :D
wendy wrote:[...]
Coxeter's Sums

Coxeter does a set of fancy sums of tetrahedra in 'regular polytopes', but the nature of this evaluation is beyond my means.
[...]


Quite easy. Any Dynkin diagram is nothing but the dual of some topological simplex (Schläfli's orthoscheme): every node represents a mirror, every link (including the ones being dropped in display, i.e. the ones which would have to be marked by 2) represents an dihedral angle of size pi/k (if k would be that mark).

That arrangement of mirrors finally divides space into pieces, each being a copy of the fundamental cell of symmetry.

Now let us consider linear Dynkin diagrams first. Here the relevant fundamental domains would be called elementary domains too. In fact, those pieces would be nothing but the Voronoi domains of the structure being represented by that Dynkin diagram, if all nodes would be ringed.

For any other Dynkin diagram it will turn out, that the relevant fundamental domain is nothing but a larger topological simplex, assembeled from some number of some former elemntary ones. That number then will provide the order (= multiplicity) of the considered Coxeter group in terms of the easier one with the linear diagram. Group theoretical this then is just the index of the quotient of these groups.

For 3D groups those duals of the Dynkin diagram of the symmetry groups are known as Schwarz triangles. In 4D the relevant name is Goursat tetrahedra.

Now consider 2 adjoined triangles:
Code: Select all
   a_
  / \ - _
 /   \    - _
b-----c-------d
This shows already 2 conditions:
If you want to attach 2 fundamental domains (abc and acd) to become a potential larger one (i.e. abd), then 1) the former angles at a will have to be added, while 2) the former angles at c have to sum to pi.
In terms of link-numbers of the dual Dynkin diagrams this amounts to: 1) the inverse of the relevant new number will be the sum of the inverses of the respective former ones (because of: alpha = pi/k). And, 2), if you would have a link number k=n/d for the first triangle, then the relvant link-number of the other triangle should be k'=n/(n-d) (because of: pi = alpha + alpha' = pi/k + pi/k' = pi [d/n + (n-d)/n]).

You could find an outline of that here (for 3D, i.e. Schwarz triangles). The similar thing for 4D (or Goursat tetrahedra), and esp. the full table of additions, to my knowledge was the subject of N. Johnson's PHD. I recompiled that, mainly independent thereof for my own (even so having some few mails with him - so far I never managed to get a glimpse into his PHD). The results together with the detailed describtions of the advice are shown here.

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Re: Uniform tilings in H3 and their possible cuts.

Postby wendy » Wed Jul 24, 2013 10:35 am

I actually spent a good deal of time trying to build W4, the 4-dimensional version of the 'wythoff symbol'.

The letters R=2, V=5/2. S=3, Q=4, F=5 are used. An "i" produces the supplement, ie Vi= 5/3, Qi=4/3, Fi = 5/4. These then represent the angles, in twelfty, of R=30, V=24, S=20, Q=15, F=12, and the supplements are 60-angle.

The actual symbol represents pairs of letters, as opposite angles. So FS/SR/RR is the group {5,3,3}. You can reverse pairs of letters, but the top represents a base, and the bottom represents an apex. So if Fs/Sr/Rr represents a face of a simplex, one can make a vertex by looking at Fs/Sr/rR = SF/RS/RR. This is the lune-slice that represents the subgroup [5,3].

For example, if ab/cd/ef is a tetrahedron, then ab/dc/fe, ba/cd/fe, and ba/dc/ef represents different representations of the same group, but single reversals of pairs do not work. Likewise, ii = 1, ie 60-(60-a), = 60-60+a = a. Leaving the i's in on a row shows the move made from the previous row.

Anti-lunes are formed by supplementing a face, so the anti-lune (ie the slice from pole to pole, less the base), is found by supplementing a face, so eg FiS/ SiR / RiR is a supplement, as is FiS/SRi/RRi. But since R = Ri, one simply removes these.

One can use it for calculations, eg the density of o4x4/3x4o, the dual of which has faces whose dual is that of o3x4x3o.

Code: Select all
   SS/RQ/RR     ;0.2    base 48  (most appropriate for this group).
   SSi/RQi/RR   lune-complement of SRR = 0;4   gives ;03.46   SRR is order 12
   SSii/RiQi/RRi   lune-complement of RSQi  = 0;7   gives 0;03,02 = 3/48 1/24 = 73/1152. 


But that's as far as it got. One needs to be rather tidier than is my want to make this work.
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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Wed Jul 24, 2013 11:37 am

You're right, that 4D Goursat tetrahedra addition theme is quite a mass. In fact, it once was, as I was told, the PHD of Norman Johnson, so it's sure that it won't be an afternoon walk.

On the other hand, it surely asks for possibilities of improved representation and thus directer access.

The outcome, on the other hand, is known and now also readily accessible at my webpage (links I'd posted in my last mail of this thread).

The main issue however is the handy representation of the respective fundamental domains (Goursat tetrahedra) by means of the required link marks of the Dynkin diagrams. And when all links are displayed (not reduced) we clearly have a tetrahedral diagram whenever we consider 4 nodes. Moreover, the Dynkin diagram itself is somehow dual to the respective fundamental domain. But this duality is somewhat different to the 3D case of Schwarz triangles:

In 3D we represent corner angles by link marks. In 4D we represent dihedral angles (i.e. around an edge) by those link marks. Thus the (unreduced) Dynkin diagram notation is surely essential (when refering to those domains by means of link marks), but have to be translated freely into the corresponding real world domain, and back again.

So if you now add some further level, by representing those tetrahedral Dynkin diagrams by inline sequences of characters("AB/CD/EF") this most probably would require even more translation work to deal with.

So far I have not completely fiddled out your notation, but I suppose that those pairs of characters would represent the opposite edges of the domain resp. opposite links of the diagram.

Such an notation adds furtheron the disadvantage to be supplemented by some devices how a mere rotation of your domain under consideration would effect your notation.

Inline notation clearly would be preferable, for sure. At least with respect to the space needed to represent graphical displays (even when being done by ASCII-art, as I did). But I don't know whether this truely is worth that other issue. In fact, any grapfical display has the same problem of equating rotated copies, but there this is much more obvious by a mere look onto the graphic.

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Re: Uniform tilings in H3 and their possible cuts.

Postby wendy » Thu Jul 25, 2013 6:55 am

I gave W4 a bit of a bash today. It's actually quite a good representation of the tetrahedron.

The three pairs of numbers represent the three opposite edges of the tetrahedron, which by themselves make an octahedron. So in essence, the three pairs can be thought of as (+x, -x),( +y, -y), (+z, -z). The full set of permutations is retained, so like W3 (wythoff notation for polyhedra), you can put the symbols in any order. So x,y,z or y,z,x or whatever all work.

The reversal of one axis would place the vertex of a tetrahedron against a face, so this is not allowed. Instead, you have to reverse these in pairs. So (+x, -x) (-y, +y), (-z, +z) is a vertex too. It's pretty easy to flip pairs of these at a time, since you still keep an odd number of + signs in the top row.

Now, suppose we want a W3 (schwarz-triangle), on the top. This means that +++ is a vertex, and --- is a face. In the original form, these were written as top / bottom, so we speak of top for first, and bottom for second. So if there is an odd number of X in the first position, it's a top-like figure.

Because the four axies like 111, 1-1-1, etc represent both the four vertices and four axies, we can associate these axies with a special off-set one (1,1,1), and the others are associated with the positive axis (1,-1,-1) is the x-vertex. Likewise, the special face is (-1,-1,-1), and the x-face is (-1,1,1).

When we come to encode dynkin symbols, such as used by Coxeter, the node or dot, comes as a face, so it has to go to the bottom. This means that the direct connection between node t and the nodes x,y,z appear as the second of the two axial angles. So the X axis has the angles yz, tx, and then the mark for the X node. Likewise, the y node run xz, ty Y.

So, given a dynkin-graph like f3v3x5o, we can now associate the nodes, in no particular order, to TXYZ.

This gives by TXYZ T=f , X = FSv, Y = RRx, Z = SRo or "x FSo RRx SRo".

If we suppose instead, Zf3Xv3Tx5Y, we get T=x, X=RSv, Y=SFo, Z=RRf ie "x RSv SFo RRf".

We can see that these are equal, by lining first the axies to match opposites, ie x SFo RRf RSv

The next bit is axis-reversals of the first and third points. This involves swapping the marks for both the reversed pair, and for the non-reversed and chief. Here we're reversing 1 and 3, and swapping the names on 0 and 2, so

f FSv RRx SRo, which is the same.

Note also, that it is easy to recover the four schwarz-triangles, as the top, and one top, two bottom, gives

FRS, FRR, SRR, and SRS, where the top node is the all, first, second, third angles.

The lunes-complements are easy to find, if the angles are written as fractions of all space, since the full lune from pole to pole of FRS is 0;01 (twelfty), and the angle on this tetrahedron is 0;00,01, so

the lune-complement is a reversal of the bottom, with 'i',s so

FS RR SR (0.00.01) reversed against FRS (0.01), gives FSi RRi SRi (0;00.E9).

One uses a similar calculation to demonstrate that o3o4/3o3o has a density of 73. When we calculate the reversal, the lune of reversal is the one we don't take complements of. I have included the i against R, to show in each row, which R is being reversed. Of course, R = Ri.

The final bit is a calculation to show that the first angle does go into the outcome, and that it's 73.

Code: Select all
    o 3 o 4 o 3 o                 reversal gives A - angle
     SS  RR  QR      0;00.12.60                          A
     SiS RRi QiR     0;09.V7.60   reversal against SRR (;10)
     SS  RiR QiRi    0;07,72.60   reversal against SQiR (;17.60)

                           73
            1.05.00 -----------------
              12.60 ) 7.72.60
                      7.35.00     7*1.05.00 = 70 * 12.60
                      -------
                        37.60
                        37.60     3*12.60
                        -----
                            -
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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Wed Sep 18, 2013 10:33 am

Did you already realize that x5o3o3o5o (admitted, in H4) and x5o10o (itself in H2, but sure be usable as bollohedron too) both have the same curvature? (There purely imaginary circumradius calculate as sqrt[(5-3 sqrt(5))/40] = 0.206652 i each.)

Thus it should occur as true bollohedral pseudo facet, right? And, taken the other way round, there might be some corresponding faceting...

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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Wed Sep 18, 2013 10:57 am

Further I just calculated that the circumradii of x3o3o3o5o, x3o3o5o5/2o, x3o5o5/2o5o and o3o5o5/2o5x are all sqrt[(1-sqrt(5))/8] = 0.393076 i, and that those of o4o3o3o5x and o3o3o5o5/2x are both sqrt[2-sqrt(5)]/2 = 0.242934 i.

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Re: Uniform tilings in H3 and their possible cuts.

Postby wendy » Thu Sep 19, 2013 7:35 am

This is something i did as filler text for LaTeX last century.

\alpha = 1+sqrt(2) = j(6), a,b are the heptagonal chords 1.801937, 2.246979, \omega is (\sqrt(6)+\sqrt(2))/2 = 1.93185165,

In the table of pentagonal chords on pp3-4, the numbers like A12, A20, etc, are chords of a circle spanning ;12 = 36 degrees. It represents the chords of a {3,3,5} or o3x5o. A measure like 2A12, means that it's a straignt line, twice the length of A12.

pp6 and 7 give pentagonal chords, eg relative to the {5,3,3} and inscribed figures. Then various entries for the octagonal, dodecagonal, heptagonal, and enneagonal systems, some of which produce interesting numbers.
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Re: Uniform tilings in H3 and their possible cuts.

Postby wendy » Thu Sep 19, 2013 10:57 am

Klitzing wrote:Further I just calculated that the circumradii of x3o3o3o5o, x3o3o5o5/2o, x3o5o5/2o5o and o3o5o5/2o5x are all sqrt[(1-sqrt(5))/8] = 0.393076 i, and that those of o4o3o3o5x and o3o3o5o5/2x are both sqrt[2-sqrt(5)]/2 = 0.242934 i.

--- rk


Seeing x3o3o5o5/2o, x3o5o5/2o5o and x5o5/2o5o3x actually have the same edge, etc x3o3o3o5o, in the same way as o3x5o has the edges of x5o5/2x etc. These are the magic 6.472135955 in my table, along with xEo3o5o.

It's not obvious that x5/2o5o3o3o ought have the same edge by laws of symmetry, but they do. This was rather amazing considering the smallness of the edge of x5o3o3o3o (which it stellates).
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Re: Uniform tilings in H3 and their possible cuts.

Postby Klitzing » Thu Sep 19, 2013 11:37 am

wendy wrote:
Klitzing wrote:Further I just calculated that the circumradii of x3o3o3o5o, x3o3o5o5/2o, x3o5o5/2o5o and o3o5o5/2o5x are all sqrt[(1-sqrt(5))/8] = 0.393076 i, and that those of o4o3o3o5x and o3o3o5o5/2x are both sqrt[2-sqrt(5)]/2 = 0.242934 i.

--- rk


Seeing x3o3o5o5/2o, x3o5o5/2o5o and x5o5/2o5o3x actually have the same edge, etc x3o3o3o5o, in the same way as o3x5o has the edges of x5o5/2x etc. These are the magic 6.472135955 in my table, along with xEo3o5o.

It's not obvious that x5/2o5o3o3o ought have the same edge by laws of symmetry, but they do. This was rather amazing considering the smallness of the edge of x5o3o3o3o (which it stellates).


No, Wendy, x5/2o5o3o3o and x5o3o3o3o do not have the same circumradius (that of the latter actually is sqrt[-10-5 sqrt(5)]/2 = 2.301105 i).

But me, I've brought in a typo too, sorry :oops: . The second part of my above sentence should corrected as
and that those of x4o3o3o5o and o3o3o5o5/2x are both sqrt[2-sqrt(5)]/2 = 0.242934 i.

and this should be even more surprising, as on the first glance squares and pentagrams would have nothing in common...

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Re: Uniform tilings in H3 and their possible cuts.

Postby wendy » Fri Sep 20, 2013 7:25 am

I made a typo here.

It's not obvious that x5/2o5o3o3o ought have the same edge as x4o3o3o5o by laws of symmetry, but they do. This was rather amazing considering the smallness of the edge of x5o3o3o3o (which it stellates).


In fact, the 5/2 would reduce the 5,3,3 by a factor of phi, the 4 increases 3,3,5 by a factor of sqrt(2), so the vertex-figures are the same size. It's just then to show that 5,3,3 contains the vertices of 3,3,5.
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