by **wendy** » Sat Aug 06, 2016 10:50 am

You would get these directly out of the 'station-diagrams', ie 'standing-point tables' of the lamina in A_n. It's pretty clear.

A_n is the singly marked ring x(n+1 3's)z. eg x3o3o3z, x3o3o3o3o3z, u.s.w. Each of the nodes in the ring corresponds to a kind of hole in the tiling, the cell is given by removing the inntial x, and placing a new x there. So, relative to a given x3o3o3o3...z the o3x3o3o3..z is 'upwards simplex, o3o3x3o3..z is 'rectified upwards simplex, etc. The points in x3o3o3...z are the point itself, and then the runcinate, viz x3o3o..o3x.

So we get now a layer of spheres (or mootly half-edges), all in A_n, and we simply stack these up, layer after layer. If we one station (ie to the tetrahedron), we get then A_{n+1}, with two stations, we get B_{1+n}, with three E_{1+n} and four is possible too. These are primitives of k_01, k_11, k_21, and k_31 resp.

Since the produced lattices are the complete eutactic stars of the groups, (ie from a point, it is all possible points where one might move in unit steps perpendicular to the mirrors: ie a mirror-edge-star), we find all of the polytope vertices in here, given a sphere large enough.

So let's look at the example by Klitzing, which is based on A_5, or a six-node ring.

The nodal distances are 2n, 1.(n-1), 2.(n-2), 3.(n-3) &c, eg 12, 5, 8, 9, 8, 5.

If we now place the second layer on top of the first, the height is then 12-v, where v is the value in the row. But for the first case, the hole is not empty, so it sits at 12. So the thickness of the strata are then 12, 7, 4, 3, 4, 7... for the various lattices. The subsequent layers are then at the squares of these, eg h, 4h, 9h, 16h, &c, because we are using squares of distances.

Gosset-Elke figures, on these axies, are characterised by each layer being moved three to the right of the previous.

So when one plots the vertices of some simplex-figure on the base, the figure falls in the same column as where the central kernel is. For example, the group o3o3o has points (1)3(2)3(3), modulo 4. A figure x3x3o adds 1+2=3, and the central is o3o3x (3). When you put a double-edge length etc, then you simply multiply the node-value by the length, eg a tetrahedron of edge 2, has a central figure of 2,0,0 = 2 gives octahedron. A tetrahedron of edge 3 (which gives the tT too), gives (3,0,0) = 3 inverted figures.

So when you are placing the figures of .3.3.3.3.3. over a symmetry, the central cell is found entirely by the modulo of the nodes, ie o3o3x3o3o is in column 3, and x3o3o3o3x is 1+5, is in column 0. The progression of the three groups, for any shape, is that each subsequent layer must proceed by a modulo-sum of 1, 2, 3.

So we get eg oxo3xoo3[ooo3]oox3oxo&#xt gives the runcinated simplex in the next dimension.

Likewise oxo3ooo3xoo3[ooo3]oox3oxo&#xt gives the truncated orthotope of the next dimension,

The gosset figure can proceed to four layers, but the middle layer is here x3o..3x runcinated simplex, and the upper and lower are runcinates shifted by three.

So it's a simple matter, given the circumradius, to simply add height with radius. In the example here, we have

height = 0 ; x3o3o3o3o3x 12 together 12

height = 3 ; o3o3x3o3o3o 9 together 12

height = 12; o3o3xo3o3o3o 0 together 12

This makes the 1_22.

Gosset's polytope of diam 8, is satisfied in the same layers as

height 0 = o3x3o3o3o3o 8 together 8

height 3 = o3o3o3o3o3x 5 together 8

height 12 = (already too big).

Note that the figures have a column-modulus here of 2 and 5, This is one of the two orientations of the 2_21 in the off-positions (rather like the up and down triangles in the A2.)

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the dream we dream together is reality.