Added 28 January 2019 - further notes from Wendy Krieger
The model i have been working with is D4D4. D4 equates to a form of
body-centred tesseract tiling.
D4 has four 'stations', which equate to where the D4 is sitting, along
with three seperate D4 cells in the centres of the three different
orientations of the 16ch that make this lattice.
When we take this D4D4 as an eight-dimensional lattice, the three
remaining sets of deep holes in square product form the three-by-three
matrix, from which the various E8's might be drawn.
E6 has a construction in D4H2 (H = hexagonal lattice)
If one takes a hexagonal lattice, it is possible to colour cells that
are on opposite sides of a neighbour in the same colour. This results
in a four-colouring of the hexagonal lattice, which we label X, O, E,
N. In D4H2, there is a distribution of 2_21, where the X-cells contain
just one vertex, and the O,E,N contain 8 vertices. Given a vertex in
any of the four colours of H2, we get the other three cells representing
a distribution of 16ch in the same arrangement.
Since the H2 maps onto D4 as a van Oss polygon, it comes that diameters
of the 3,4,3 give rise to a hexagon as described above, and then there
are 16 such girthing hexagons on the {3,4,3}, that are notionally in a
group AA4 (bi-alternating group 4), represented by the even alternation
of rows+columns of a 4*4 matrix. Any row or column represents a set of
four hexagons that cover the full hexagon
At any point in E8, it is thus possible to have four crossing E6's that
do not contain any other point. The arrangement is chiral, and there are
all-together, eight sets of four that do this, given this particular
arrangement.
The complex-polygon equivalent of this is that 3{3}3{3}3{3}3 has
diametric 3-spaces, of the form 3{3}3{3}3{4}2, which four of these cross
only at the centre.
