by **wendy** » Thu Feb 10, 2011 9:41 am

Let's see.

Here is the three dimensional ones, that one might understand the notation.

The group 3h is a simplex group of order 2, and weight 4. This means that there are elements of order × weight, where order lies between 1 and the maximum. The group corresponds to how many distinct ways you can put m into n, where m < n < order+1. With say, 2 into 4, you can have three distinct patterns: two opposite edges, the complement of that, and all six edges.

We just have one compound, since there is only one way of putting one into two.

stella octangula (8)[2(4,= The equal says that the same applies opposite, by central inversion

{3+3} This group is transitive on its flags, and so is regular in every sense.

The group 3r/f comprises of four elements. The common symmetry is 3r (rectangular), the total symmetry is 3f (icosahedral)

The underlying point group is a five-by-five square. The thing is an example of an LD group, consisting of a chiral division into eg 5. It's rather like the left-turning knight vs right-turning knight at chess, or the multiples of 2+i vs 2-i in complex numbers (gaussian integers).

Because the central inversion turns left into right, there is a linkage between the five left and five right elements.

The 5×5 square, less the diagonal, represents the 20 vertices of the dodecahedron. Each row, or each column, individually is a tetrahedron. A column and row together, makes a cube. The five on the diagonal, represent the vertices of the 5 octahedra in the icosadodecahedron.

We then have

c / sq (20)[5(4,= five tetrahedra ,

2c / sq 2(20)[10(4, = ten tetrahedra, 2 at each vertex

c+r / sq 2(20)[5(8,6)](30) ten cubes, the dual of five octahedra have vertices at a 30 (ID).

The 4D groups.

In four dimensions, there are four groups, being

1. 4h 4

2. 4h/f 33 ordinary

3 4h/f 124 extraordinary

4 4s/f 15 or something.

The stella simplexa (10)[2(5,=, being inverted simplexes in 4d, do not have the vertices or faces of a regular solid, although it is symmetry transitive on its flags.

The group 4h corresponds to the third-order simplex, also exampled, eg by [1] {3,6} and [2] {6,3}.

1/2 (16)[2(8,16)] two tesseracts in a cross (f), or two cross in a tesseract) (v)

1/3 v 2/3 (24)[3(8,16)]2(24) two completely regular stars (symmetry transitive on the flags).

{3+3,4} stella tegmata faces are stella octangula

{4,3+3} stella prismata vertex figure is stella octangula.

We see that these are like the vertex-compound of two {3,6} in {6,3}, where the hexagrons have inscribed hexagrams, but other faces point elsewhere. It's only in the compounds of three {3,6} in {3,6} (by drawing the long diagonal of every rhombus formed by two triangles), that we see a completely regular figure, with every triangle turned into a hexagram {6/2}.

The ordinary group of 4h/f.

The point group here is a 5X5 square, where each cell is a 24-cell, each row or column is 120, and the grid is 600.

c/r (120)[5(24,24)](120) 3 examples

r/g (600)[5(120,120)]5(120) 12 examples

2r/g 2(600)[10(120,120)]10(120) 12 examples

c/r (600)[25(24,24)]5(120) 6 examples.

The first group is (120,120), that is, the vertices of the two 120-groups do not coincide. In effect, it is self recriprocal in the manner of the 24ch itself, so we can replace (24) or (24,24) by the 24 ch, or the stella prismata or stella tegmata.

The last group is that the division of the 600 into 24 recriprocates into a five-fold matching of 5 24 into 120. That is, the dodecahedral faces of the 120ch give 5 octahedra, while the tetrahedra of the 600ch give only one face each. Each (24) by itself can be replaced by a stella prismata or stella tegmata, ie 3 tesseracts or 3 cross-terids.

The groups of 12 examples come from (120) by itself represent the 120 vertices and/or faces of the 12 pentagonal figures. Since there is only one way to put 120 into 120, the second half of the compound is not used here.

The extraordinary group 4h/f.

This is by far the biggest of the groups, but is simple building.

Consider, for example the compound 5(120)[5(120,120)](600). The point group here is, say 2 five×five squares.

When we substitute (120,120) with 5(24,24), we see that on the left, we fill up the whole set of vertices, but on the right, we add just one row. There exists then a set of compounds that fill, eg 1r, 2r, 3r, 4r, 5r, of which 1r and 5r is already given above. The compounds 2r, 3r, 4r are entirely new.

We see then that 5x(120)[5x(24,24)](120x) will give 2/5, 3/5 4/5 of the group of 600, (ie two to four rows of the 25 groups of 24), but will completely fill the 120 ch, where each face of the dodecahedron is stellated into 2, 3, 4 octahedra (eg).

each 5x(120)[5x(24) then represents the vertices, the faces of the 24ch, and a stella prismata or stella tegmata: ie 12 new compounds.

The second, and older of the two groups is the partition of the 600.

Consider first, that we can replace (600)[5(120,600), we can put the (120) into 5(24) which gives 5(600)[125(24). This means that each of the 600 faces of {3,3,5}, instead of becoming a dodecahedron, now becomes 5 octahedra. The vertices of a set of 25 {3,4,3} now form a single set of 120 vertices in 600. Since the 3000 vertices of the dual point nowhere, we only look at the left here.

Each row and column of the 24>600 figure is represented by a full 5.24>120, of which exactly one row represents a common 24ch. Since each cell of the table belongs to a row, and a column, with the common cell representing the special group in the ordinary cluster, we have 4+1+4 octahedra at each face of the 600ch, of which 1 is the special one, and the remaining 4 are the four columns, and four rows that do not map onto the special group.

The notation of xyz represents how many of the 4+1+4 appear in the compound. Each cluster represents n(600). The relevant section of the 600 is a pair of twinned dodecahedra, rotated by 90 degrees. For the relevant ID, the action rotates by 90 degrees around a vertex-vertex axis, which means that four of the octahedra not on the axis are duplicated.

Coxeter lists 400 and 404 in his list. The nature of the reflection group means that we can select any number of elements in both groups, with and without the special group, and get a valid compound. However, something like 201 is the same as 102, (by reflection), but 012 is different (because it contains the right group and special one.

So we get

1 010, 100

2 110, 200, 101

3 210, 111, 300, 201

4 310, 211, 400, 301, 202

5 410, 311, 212, 401, 302

6 411, 312, 402, 303

7 412, 313, 403

8 413, 404

9 414

Of these 010 is the special group listed in the ordinary group, the rest are all new, all to gether, 28 sets of four (vertex, face, tegmata, prismata) compounds.

We have here 12 + 112 = 124 members of this group.

The groups 4s/f.

The special form is (600)[120(5,=. This is the mete star, or stella metrica. The name refers to the orders of symmetry being used as measures: a glome = 120 s, an s = 120 f. The star is regular in the sense that it is symmetry is transitive on its flag, but the overall effect is like (Q)[9(Q,Q)]Q, that the stellation of the square-lattice includes one that has cells of size 9. There is no division of the vertex into lesser symmetries.

The extraordinary group is 2f2f / 4f. The point group here is a grid of 6×6, representing the six pairs of orthogonal great circles in (120). The actual construct is a decagonal-decagonal prism, of which there there are 18, ie 3(600)[18(100, One can take a 10*10 grid, and divide into jumps of 4+2i, to get the vertex of an inscribed pentachoron.

Each bi-decagon prism represents 20 such pentachoron, one can divide 600 into 6 entirely separate bidecagon prisms, which gives a 120 (entirely separate from the above).

It's then a matter of finding exactly how many different bidecagons clusters are possible.