Re: triacontrahedron as projection

Hi Paul

>Can a 5D cube be projected to a triacontrahedron, or only a 6D cube can?

A close analogy is the familiar tiling that shows a sheet of cubes in a step like
arrangement using hexagon divided into three rhombohedra.

2D Penrose tilings are projections of a 2-D surface of a section through the
lattice of 5-D cubes in exactly the same way. The sheet is cut through at a
particular angle to provide the non-periodicity.

The 3-d Penrose tilings are projections of a 3D surface of a section cut
through the lattice of 6-D cubes in 6D space.

So the triacontrahedron is from this projection of a 3-D surface in 6D regular
lattice of 6-D cubes.

But how much of the surface is it? I've wondered that.

It has edges in all six directions, so must be a projection of parts of at least
two 5-d cube faces.

However it has nowhere near enough rhombohedral cells to be a projection
of even as much as a complete 5-D face.

Here are the numbers for n-dim cubes, as I've just calc. on back of envelope
as it were:

n+1 dim cube = 2 n dim cubes with lines joining all pts of one to the other.

if P k n = number of k dim faces in n dim. cube (k=0 == pts)
then it follows that

P i n+1 = 2*Pi n + P i-1 n
For i=0
P 0 n+1 = 2*P 0 n

E.g. P 2 4 = number of squares in hypercube = 2*P 2 3 + P 1 3
because you get the hypercube from two copies of the cube joined
to each other using lines to connect matching points in each,
and each line in the cube becomes a square when the two copies are
joined together, and you have all the squares from both copies as well.

So numbers of pts, lines etc are:
(2, 1) // two pts, one line

(2*2, 2*1+2, 1)
=
(4, 4, 1) // 4 pts, 4 lines, 1 square

(8, 2*4+4, 2*1+4, 1)
=
(8, 12, 6, 1) // 8 pts, 12 lines, six squares, 1 cube

// leave out calc. now:
(16, 32, 24, 8, 1) // 16 pts, 32 lines, 24 squares, 8 cubes, 1 hypercube
(32, 80, 80, 40, 10, 1) // 32 pts, 80 lines, 80 squares, 40 cubes, 10 hypercubes, 1 5-D cube
(64,192, 240, 160, 60, 12, 1) // 64 pts, 192 lines, 240 squares, 160 cubes, 60 hypercubes, 12 5-D cubes, 1 6-D cube

So the 20 golden rhombohedra in the triacontrahedron are a projection of half
the total number of 40 cubes in the 5-D face.

But we know they aren't from only one 5-D face because of the six directions.

That's as far as I've got so far,

I'll see what Coxeter has to say about triacontrahedra in projections from 6-D space.

Robert