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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 DemianskiBTW: 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

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