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: http://arxiv.org/abs/astro-ph/0310253 (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 countable.) cheers boud