Regular compounds in 4D

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Regular compounds in 4D

Postby wendy » Tue Apr 30, 2019 1:40 pm

The current list of regular compounds in 4d is 186 items, that is 140 more than that of Coxeter's regular polytopes.

The list is like this.

1: 4h

This is in the simplex group, where each number from 1 to x represents a polytope, and the compounds are identified with the vertices, edges, tringles, etc of a simplex.

Coxeter's notation, which we severly modify is v{vertex-shell}[p {figure}]f{face-shell},

A compound of 'p' figures may have all their vertices in a vertex-shell, each vertex of the shell taken v times, and likewise, may use the faces of an inner core, the face of the inner core yielding a compound of f elements or faces of the outer core. In 3d, the compound of ten tetrahedra (40 vertices), is 2{5,3}[10{3,3}]2{3,5}, uses the vertices of the outer dodecahedron twice each, and the faces lie by pairs in the face-planes of the core icosahedron, though this is not a regular hexagon (it is x3f ),

[2] Order-2 of this is the {4,p}[2 {p,p}] {p,4} eg stella octangula and {2p/2} eg hexagram.
[3] Order-3 is represented by subgroups of {3,4,3} and {6,3}. We have 1={3,3,4}, 2 = {4,3,3} and 3 = {3,4,3} or 1 = h3o6o, 2 = x6o3o, and 3 = x3o6o. {2}[2{1,2}], [2{2,1}]2 and the two regular stars 2{3}[3{2,1}]{3} and {3}[3[1,2]2{3},
[4] Order-4 occurs in the {3,4,3,3}, contains the various combinations of vertex, edge, triangle, tetrahedron. Note that you can have just two edges or four edges of the tetrahedron as a valid compound. The pair of opposite edges corresponds to the self-dual ((4,3,3,4)), ie a cubic and its body-centre. The four edges form a compound representing the four quarter-cubics.

The h/f or LR group

The totem for the LR groups 3r/f and 4h/f is a array of 5×5 squares. The whole array is one figure, each row and column is a copy of a second figure, and the squares form a third figure. In 3d, the central inversion transforms rows into columns, and so there is a definite diagonal not present in 2D or 4D.

In 3D, the cells are vertices, the rows or columns are tetrahedra, the row and column that passes through the same diagonal cell form a cube. The total is the dodecahedron vertices or icosahedron cells. The five diagonals represent the 5 copies of 3r = [2,2] that make 3f = [3,5].

In 4D, each square represents the point-group of 24 (ie vertices or faces of the {3,4,3}, the actual {3,4,3} is (24,24), meaning that the vertex-rays and face-rays point in different directions.) The vertex-ray and face-rays of the stella-prismata and stella-tegmata point in the same direction, so these are represented by (24). A row or column is 120, representing the 120 verticies of the {3,3,5} and most of the stars, or the faces of {5,3,3} and most of the stars. In other words, what we write as (120) represents 13 different polyhedra.

So as before, we have (five cells in a row), or (five rows in a grid) or (ten rows in a grid), or (25 cells in a grid).

This gives (120)[5(24,24)](120) = (120,120), (600)[5(120,600)], 2(600)[10(120,600)], and (600)[25(24,24)]5(120) = 600[5(120,120)]5(120). of 3, 12, 12, 6 compounds respectively.

The (120,600) represent the vertex-rays of the {3,3,5} and {5,3,3}, as well as the face-rays of {3,3,5/2} and {5/2,3,3}. The remaining stars {5/2,5,3}, {3,5,5/2}, {5,5/2,5}, {5/2,3,5}. {5,3,5/2}, {5/2,5,5/2}, {5,5/2,3}, and {3,5/2,5} have vertex and face-rays pointing to the same (120).

The point-group of (24) represents the vertex-ray of a x3o4o3o, the face-ray of o3o4o3x, and the vertex- and face-rays of x3+3o4o and o3+3o4x (+ = klitzing-sytle branch *a4/2*c). So if the (24,24) is in a compound with numbers at both ends, then 3 are given if the two ends are the same, and 6 otherwise. The face-rays of one direction become the vertex-rays of the other.

The compound of five 24ch, represents the five squares in a row. But note that there is a compound of (600)[5(120,120)]5(120), This means, that in the first half, there is a unique 25-colouring of the (600), each representing one of the inscribed (600)[25(24,..., and at the same time, each of the vertices is represented five-times over in the 5(120)[25(24.. . In other words, for a given diametric axis, it appears once in each row and column. The common element between two cells, not in the same row or column is a hexagon. One can see by fixing one cell, that there is a 4×4 array representing the 96 edges of the {3,4,3} as two sets of separate hexagons.

Since each row of the array represents both a subset of 600 and a group of 120, we can take n rows to get

n(120)[5n(24,24)]. four each for n=2,3,4. 1 and 5 have already bean dealt with.

The figures with 24n vertices on the right-hand side, are respectively, the figure bounded by 72 weimholt-hexahedra, of 48 teddys, and the snub-24 cell, none of which are regular. This is 12 new compounds. None of these are in coxeter.

Given further that each row and column represents (120), we can 'flip' the array so it stands above the plane with five vertical towers (intersecting at a row), and downwards on the columns. This creates nine rows, the combination of 5400 verticies makes nothing unless only the middle row is taken. But the compounds are all set in the 600 rays of the central layer.

The symmetry is that each set of above and below are chirally alike, (ie 100 = 001), and there is no ordered series here. In other words, the combinations that represent different compounds here are [1-4][0-1][0-a], where 'a' is the value in the first set. So 100 represents a layer above the grid, while 001 is a layer below the grid. These two are identical, and are counted once. 010 is the layer of the grid, and is entirely different.

So we get 100, 110, 101, 200, 111, 210, 201, 300, 211, 310, 202, 301, 400, 212, 311, 410, 302, 401, 302, 411, 312, 303, 402, 313, 412, 403, 413, 404, 414, or 28 different sets.

Coxeter list only 400 and 404. The transcription of his list gives 410 and 414, the first by the process of (600)[(5(120,... and 5(120)[25(24,.. gives 5(600)[125(24,...).

None of these vertices of the duals point to any regular, so we get 4 compounds for each, of the form

n(600)[25n(24,24)] = n{5,3,3}[25n{3,4,3}], n{5,3,3}[75{3,3,4}]2n{3,3,5}, 2n{5,3,3}[75{4,3,3}]n{3,3,5} and [25{3,4,3}]n{3,3,5}.

where n is the sum of digits in the string above, eg n=5 for 212, 311, 410, amd 302. Coxeter's notation represents these five distinct compounds.

The group 4s/f "mete-star" (3,3,3).

The root compound here is n(600)[120n(5) = n{5,3,3}[120n{3,3,3}]{3,3,5}, for n = 1 to 7 (some repeats). The totem here for the non-special group is a six-by-six array.

Each row and column of the array represents 100 vertices, being a bi-decagonal prism. It is the same as the centres of the 100 tetrahedra of the grand antiprism, that do not touch the pentagonal antiprisms. The six array is represented by the six diagonals of the icosahedron, we can take these left- and right- parallel.

Since each row of the totem is a partition of the same (600), we can take 1, 2, 3, 4, 5, 6 of these, and add in the special to each, to give 1s, 2s, 3s, 4s, 5s, 6s.

We now investigate the totem, and see that all 720 possible simplexes are indicated in right-turning of the 36 inscribed bidecagon-prisms. That means that for a row, the 120 corresponding left-turning figures are scattered through the remaining five rows. We can, for example, take it that if '1' represents the partition of (600) into 6 sets of (100), and that set is further partitioned into 5(20), the common element of intersection of a row and column is (0), and the cells off the intersection is then (20) = five pentachora, apparing left in one direction and right in the other. Then we can take the left- and right- of a set of 6 clifford-parallel prisms (ie a row and column), and these form a compound of n=2, different to 2 rows.

We now have 1x, and its complement 5x, different to the rows of 2 and 4 above. Each of these can also couple with the special group s, giving 1xs, 5xs,

All together, we add here 16 new compounds, over and above coxeter's listing of 's'.

All together, there are 4 + 33 + 12 + 116 + 1 + 16, giving 186 compounds, and a lesson in colour-groups or ray-sets, to boot.
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Re: Regular compounds in 4D

Postby ndl » Wed May 08, 2019 2:31 am

Would there be a good way to reduce these compounds into categories, lead by the compound that represents all the iterations of a sub-symmetric polychoron in the larger symmetry?
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Re: Regular compounds in 4D

Postby ndl » Sun May 19, 2019 2:32 am

Would all the 4D compounds be able to be created by faceting the regular and rectified polychora and stellating the duals like can be done in 3D? Klitzing already has the facetings of the regulars on his website and that does provide many of the compounds already.
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Re: Regular compounds in 4D

Postby wendy » Sun May 19, 2019 12:08 pm

Some are faceting, some are stellations (ie extending the faces), and some are both.

The greater part come from that a 24-point group contains the vertices and faces of 3 tesseracts, and the vertices and faces of 3 16chora, or the group of 5 contains 5 vertices and 5 faces of a pentachoron. But the greatest of these has 5400 vertices, which really correspond to nothing in particular, nor do anything 600n, where n>1.

The real supprise here is that the bulk of the s/f groups (pentachora), pass through a bidecagon duoprism, the vertices of {5,3,3} split neatly into six of these in twelve different ways.

Only some of the simpler ones would appear as simple facetings. For example, if you take a {5,3,3}, and replace its faces with 5 cubes or 10 tetrahedra, you will end up with the compounds of 75 {3,3,4} and {4,3,3}. The five octahedra will lead to 25 {3,4,3}. But those in the non-special groups don't lead to much. Some of the vertex-lists are given: the compounds of 5n {3,4,3} are given as known CRF's. But the situation is worse than in 3d, as many of the figures in the extraordinary groups have many more vertices.

You can derive these by faceting, which is what Coxeter did. But unless you wrangle the colour-groups, you won't find anything like 1/2 of the total.
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Re: Regular compounds in 4D

Postby ndl » Tue Jun 04, 2019 12:29 am

I found two very interesting uniform compounds while playing around with Stella4D: A 25 and 50 hex compounds. Both have the convex hull of gap and the 50hex is also a compound of 25 haddets. The vertex figures are compounds of 2 and 4 octs respectively. They really are subsets of the 25 stico compound removing 1 or 2 hexes from each stico. It's just cool they have a uniform hull. Anyone have any good names for them?
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