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
Hi Jeff, Jesper, cos-top,
On Tue, 7 Feb 2006, Gundermann, Jesper wrote:
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
Thanks for the explanation - it's the equatorial layer - "layer 5" - with 30 dodecahedrons which solves my intuitive problem of needing (2 * (layer with an odd number of dodecahedrons)).
Jeff wrote:
Note that the cells in layer 5 sit "vertically" with respect to the equatorial hyperplane (i.e. they're orthogonal to the equatorial hyperplane) which is why they appear flat in the attached image (each dark blue hexagon is the 2D shadow of a 3D cell when you project from 4D space to 3D space).
i guess another slightly confusing thing is in the picture of layer 4, where the hexagons look "flat" whereas if i understand correctly, these should be concave in order that layer 5 cells can be stuck on here.
It might be nice to have a picture of the equatorial S^2 surface, showing where layer 4 cells touch layer 6 cells (in whole faces), and the cross-sections through the layer 5 cells.
Note that all numbers are "dodecahedal numbers", respecting the dodecahedral symmetry of the whole construction.
The 120-cell is quite beautiful, isn't it?
Definitely :)
cheers boud