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Untitled

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I created this stub as I started some articles on higher dimensional uniform polytopes. It's crude, and I appreciate any help to improve it. Thanks!

Tom Ruen 04:41, 6 September 2006 (UTC)[reply]

I fleshed out some of the expressions of convex uniform polytopes and moved content under Schläfli symbol to here, since it doesn't well belong there. Still in progress, and I'll remove/reduce the text under Schläfli symbol when the restructuring is completed. Tom Ruen 07:32, 15 September 2006 (UTC)[reply]

Confusing

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I added a {{confusing}} tag and merged some single-sentence paragraphs in the WP:LEAD. This article needs help from WP:MSM, specifically, it is seriously in need of {{prose}}. For example, there is a section, with eight sub-sections and dozens of diagrams, with only a single sentence of prose. That's unencyclopedic. Orange Knight of Passion (talk) 00:05, 14 September 2008 (UTC)[reply]

circumradii and v.f.

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Uniform polytopes whose circumradius is equal to the edge length can be used as vertex figures for uniform tessellations. For example, the regular hexagon divides into 6 equilateral triangles and is the vertex figure for the regular triangular tiling. Also the cuboctahedron divides into 8 regular tetrahedra and 6 square pyramids (half octahedron), and it is the vertex figure for the alternated cubic honeycomb.

How useful is this paragraph? Any polytope that can be inscribed in a sphere is a candidate for a vertex figure, provided that its edge lengths are in appropriate ratios; but is it proven that all polytopes described above are v.f. of uniform tilings?

Five nonconvex polyhedra fit the description: octahemioctahedron, dodecadodecahedron, cubohemioctahedron, small dodecahemicosahedron, great dodecahemicosahedron. Are all of these known to be v.f. of (nonconvex) honeycombs? Also, 31 uniform polyhedra have circumradii smaller than their edge-length; are they all known to correspond to triangle-faced nonconvex polychora? —Tamfang (talk) 03:40, 3 June 2010 (UTC)[reply]

Quite obviously, the polytopes with H3 and H4 symmetry cannot be used as the vertex figures of a uniform tiling of Euclidean space. Even the regular pentagon cannot be the vertex figure of Euclidean space.
Additionally, the vertex figure of a tiling must be the dual polytope of the cell of the dual tiling. So, the snub dodecahedron can be the vertex figure of a uniform hyperbolic tiling if and only if the pentagonal hexecontahedron can tile hyperbolic space. Is that possible? It certainly seems implausible. Calcyman (talk) 15:50, 3 June 2010 (UTC)[reply]
Quite obviously, a hyperbolic tiling has no finite circumsphere, so it's not at issue here.
The regular pentagon's circumradius is less than its edge, so it is the v.f. of a tiling of S2, not E2. —Tamfang (talk) 18:25, 3 June 2010 (UTC)[reply]
I have been looking at uniform tessellations of Euclidean spaces, and found many had vertex figures as uniform polytopes of circumradius 1 (edge length). I scanned all the uniform polytopes of Klitzing's lists, took the subset with circumradius 1, and mapped them all onto single-ringed uniform tessellations. These subsets of uniform tessellations also correspond to sphere packings, with one sphere centered at each vertex, and the kissing number of each sphere packing is equal to the number of vertices in the vertex figure polytope. My test-list is at User:Tomruen/Uniform_lattices, and that was what I was getting at. Tom Ruen (talk) 19:38, 3 June 2010 (UTC)[reply]

The cubohemioctahedron and octahemioctahedra are indeed vertex figures of nonconvex Euclidean honeycombs. In the tetrahedral-octahedral honeycomb you can (as the name suggests) find tetrahedra (corresponding to the triangles of the cuboctahedron), octahedra (corresponding to the squares), but also triangular tilings as flat surfaces (not actually cells, corresponding to the equatorial hexagons). The honeycomb with octahemioctahedra as vertex figures has the tetrahedra and the triangular tilings; the one with cubohemioctahedra as vertex figures has the octrahedra and the triangular tilings. They're #4 and #5 on Eric Binnendyk's list of uniform honeycombs. The octahemioctahedron-verf one is on Richard Klitzing's site, but I can't find the cubohemioctahedron-verf one. Double sharp (talk) 08:10, 8 September 2022 (UTC)[reply]

Special case prismatic reductions

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Special cases of products become hypercubes: ...

True, but the notation shown is for the symmetry groups, not for the polytopes themselves; this is wrong — the symmetry group [4,3] is bigger than the product []×[]×[]. —Tamfang (talk) 19:09, 12 June 2010 (UTC)[reply]

Rewrite if you like. By symmetry they are all listable, but by geometry (of uniform solutions) and topological result, they're identical, so the question is how far you want to expand the lists. Tom Ruen (talk) 03:08, 13 June 2010 (UTC)[reply]

single-ring v.f.

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Vertex figures for single-ringed Coxeter-Dynkin diagrams can be constructed from the diagram by removing the ringed node, and ringing neighboring nodes. Such vertex figures are also uniform polytopes, being prismatic if the ringed node was in contact with more than one node.

I'm going to remove the latter sentence. Such figures are vertex-transitive, but not uniform (faces are not regular) if the arcs from the ringed node are not alike; and they're not prismatic if the parent CD is multiply connected. —Tamfang (talk) 16:27, 6 May 2011 (UTC)[reply]

Summary table

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I'm still working on organization for the higher dimensional uniform polytopes, but here's a summary of finite and convex uniform polytope articles. Tom Ruen (talk) 22:43, 3 June 2011 (UTC)[reply]

n Coxeter
group
Regular
t0
Quasiregular
t1
Truncated
t0,1
Cantellated
t0,2
Runcinated
t0,3
Stericated
t0,4
Pentellated
t0,5
Hexicated
t0,6
Heptellated
t0,7
3 A3=[3,3]
BC3=[3,4]
H3=[3,5]
4 A4=[3,3,3]
F4=[3,4,3]
BC4=[3,3,4]
H4=[3,3,5]
5 A5=[3,3,3,3]
BC5=[3,3,3,4]
D5=[32,1,1]
6 A6=[35]
BC6=[34,4]
D6=[33,31,1]
E6=[32,2,1]
7 A7=[36]
BC7=[35,4]
D7=[34,31,1]
E7=[33,2,1]
8 A8=[37]
BC8=[36,4]
D8=[35,31,1]
E8=[34,2,1]
9 A9=[38]
BC9=[37,4]
D9=[36,31,1]
10 A10=[39]
BC10=[38,4]
D10=[37,31,1]

Definition/Lede

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"A uniform polytope is a vertex-transitive polytope made from uniform facets of a lower dimension. The uniform polytopes in two dimensions are the regular polygons."

If the definition is the first sentence only, a requirement that all the edges must congruent should be added for consistency within the article. I'm not aware if there's a competing definition of uniformity that permits faces with an even number of edges to have two different alternating lengths, but if there is it's not the most commonly accepted one. If the second sentence is part of the definition, this should be rephrased to make that more clear. 173.227.48.5 (talk) 23:40, 11 April 2013 (UTC)[reply]

Yes, wording could be improved. Its a recursive definition with defining uniform polygons as regular polygons as the smallest nontrivial dimension, so congruent edges are implied from 2 and up. Tom Ruen (talk) 02:57, 12 April 2013 (UTC)[reply]
This might be excessively pedantic, but the article appears to consider the line segment to be a uniform polytope. A single vertex and the null polytope might be considered (uniform) polytopes as well. The revised definition only refers to dimensions 2 and up. 173.227.48.5 (talk) 17:07, 12 April 2013 (UTC)[reply]
Actually a apeirogon can be considered uniform (rather than regular) if it has alternating 2 colors (types) of edges, which is also a simple truncation t{∞}=t0,1{∞}=. Tom Ruen (talk) 03:55, 26 September 2013 (UTC)[reply]

"Scaliform"

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The "scaliform" term is really only used by Bowers – unless it is adapted by someone more notable, I don't believe it belongs on this article. I deleted the corresponding section with this justification. Hope it isn't an issue. – OfficialURL (talk) 21:53, 6 April 2020 (UTC)[reply]