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