SV: [Cos-top] MG11 - Topology of the Universe (fwd) (and PDS - how many tiles?)

[Gundermann, Jesper]

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