[Cosmo-torun] MG11 - Topology of the Universe (fwd) (and PDS - how many tiles?)

Boud Roukema boud w astro.uni.torun.pl
Pon, 6 Lut 2006, 16:58:25 CET

hi everyone,

The following is self-explanatory - you can reply to Marek directly -
here is his email in antispam format  (this list is publicly archived so
better not write people's email addresses in ways that robots can read them):

  Marek.Demianski at fuw edu pl

> ---------- Forwarded message ----------
> Date: Mon, 6 Feb 2006 15:02:38 +0100 (CET)
> From: Marek Demianski
> Subject: MG11 - Topology of the Universe
> Dear Colleague,
> I have been asked to organize a parallel session on "Topology of the
> Universe" at the XI Marcel Grossmann Meeting in Berlin, July 23
> - 29, 2006. If you are planing to attend the meeting and would like to
> present a contribution at this session please let me know. Please spread this
> information among your friends and collaborators.
> More information about the Meeting (and on line registration) you can
> find at www.icra.it/MG/mg11. Please register by May 31
> to save 50EURO on the conference fee.
> Sincerely yours,
> Marek Demianski

BTW: i have a question on PDS (Poincare Dodecahedral Space) intuition
which someone might be able to answer.

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
layer 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.

Going to a 122 tiling would save the tiling symmetry, but intuition says
that 120 is better than 122.

On the other hand, 2 * (1 + 12 + 48) = 122 seems simpler in terms of
tiling symmetry.

In the Nature paper, 120 is quoted:

(After all, we *have* made a few minor mistakes in the past in our
subject - e.g. when we said that there were uncountably many
negatively curved 3-manifolds, and eventually, at MG IX (?), Jeff
explained and showed us that the number was countably infinite - the
classification is complicated and unfinished, but the number is


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