by **wendy** » Mon Jul 30, 2018 9:36 am

Most of the videos are correct. The two videos i have are an earlier and later draft of the same presentation, the former to a single person, the second to a group of people.

The terminology is very confusing, but she helps the show along with a power-point presentation.

The coordinates she gives for the E8 in the first slide, belong to a semi-cubic where the spheres are 1/sqrt(2) in diameter. Given that she talks about 4D4D4, we should had imagined that she would have used the hammard-coordinates, which gives a sphere of diameter 1, and 16 different points inside the sphere.

The partition of the 240 into shells of 112 and 128, is a relation between E8 and B8, it means that the E8 corresponds to the union of a half-8cube and the rectified 8-cross. This particular partition can be done in 135 different ways. The hammard coordinates divides the half-eight=cube into a pair of bi-16choron (334) prisms, and the 8cross into a third bi-16choron prism + two orthogonal 24chora (343).

The 3d+4d fibulation of 8d, is produced by the QE2, or quarterion-euclidean-2, where multiplication again provides the shape of fibers, as CE2 does in 4d. In essence, CE2 produces circles that make a sphere, and QE2 produces a 4-sphere, where every point is a 3d multiplication-sphere.

The quasi-crystals involving 335, is an oblique projection of E8 onto 4 dimensions. I really have not looked too deep at this as yet, but something tells me i ought.

The E8 divides into a body-centred packing of the prismatic lattices (9-n) E(n) A(8-n), ie 9-n means how many copies are repeated. En is the gosset lattice in n dimensions. An is the simplex-lattice in n dimensions. If you multiply En by A(8-n), there are 9-n deep holes to fill per cell, and this produces the lattice.

4D4D4 is a version of n=5 in the equations above, but realised as 4d * 4d, rather than 3d * 5d. They are both the same. D4 is the 3,3,4,3. This produces deep holes in the centres of the cross-polytopes, where the prism is big enough to hold a sphere of diameter = edge. In the D4D4, there are 16 deep holes, but you can only occupy four of these at distance 1 from each other. It's a 4*4 square, you can only put one dot in one row/column.

The Hammard code is like this. You take the four hexadecimal numbers 0, 3, 5, 6, and subtract them from 15, to get 15, 12, 10, 9. These team into pairs, as (0,15), (3,12), (5,10), and (6,9). If you write out a coordinate using the binary expansions in one given pair (the same, or any order), then they will be four bits changed, eg

00 = 0000 0000 ; 0E = 0000 1111 ; E0 = 1111, 0000 EE = 1111 1111 all vary by four bits between them.

Likewise, between two different sets, like 0E and 5A, we have 0000 1111 to 0101 1010 is likewise four bits.

Since the tesseract packing 0000, gives on body-centering 0000, 1111, the individual sets like 0E etc, correspond to a D4 D4 product, and the four taken together gives 4D4D4.

The dream you dream alone is only a dream

the dream we dream together is reality.