Dear Boud

A tiling of S^3 by the PDS has 120 copies of the PDS.

If we think of this tiling starting from "here" at the "North Pole" (sorry Marcelo, this is North bias ;) (image 1) and we "stick" a copy (image 2) of the PDS next to itself along a North-Pole-South-Pole axis (e.g. X), and then moving along the X axis, we stick another copy (image 3) next to image 2, and then another and so on, then it seems to me that image 6 covers the "South Pole" and is the "antipode" copy of image 1.

In other words, the "distance" between the poles is 5 copies of the FD.

This allows an easy order of magnitude check on the total number of copies, since the "radius" of one hemi-hypersphere is 2.5 "copies", so that the total volume is roughly (using flat space arithmetic):

2* (4\pi/3) (2.5)^3 \sim 2* 2^2 * (5/2)^3 = 5^3 = 125

which is a good approximation to 120.

(AFAIR, Jeff Weeks showed me how to do this - this is not original.)

Now for my question:

We have 6 "layers": North pole - 1 FD layer 2 - 12 FD's layer 3 - n FD's L>ayer 4 - n FD's layer 5 - 12 FD's South pole - 1

Clearly layer 3 or 4 has more than 12 copies, there are some spaces that need to be filled.

But the value required for the total to be 120, i.e. n= 47, is an odd number. Intuition says that everything should be even and symmetric.

Can someone give a nice intuitive explanation for why n is odd and why this 12-symmetry is broken? Maybe a nice diagram?

In fact, i have a partial answer: it seems to me there are some in-between layers, which we might call 2.5 and 4.5 and i guess also 3.5. But it's difficult for me to imagine them, and it's still difficult for me to imagine how the 12-symmetry is broken.

It would be nice if someone had a nice diagram or intuitive explanation of how the 12-symmetry is broken.

You were right about the intermediate layers, in fact, there are 9 layers Note that the central dodecahedron has 12 faces, 30 edges and 20 vertices. The distance of the centers of The 120 dodecahedrons are Distance from origin: direction number 0 1 pi/5 midpoint faces 12 Pi/3 vertices 20 2*pi/5 midpoint faces 12 Pi/2 midpoint edges 30 3*pi/5 midpoint faces 12 2*pi/3 vertices 20 4*pi/5 midpoint faces 12 Pi 1 Sum 120 Best regards Jesper Gundermann