Cos-top February 2006

cos-top@cosmo.torun.pl

4 participants
6 discussions

binary polyhedral groups
by Jeff Weeks 10 Feb '06

10 Feb '06

Dear Cos-Top, Jesper pointed out that the final sentence of my earlier message, namely This same relation holds, of course, for the other single-action spherical manifolds. implies something that is not correct. To make that statement more precise (and correct!) please note that in the construction if you travel in the direction of a symmetry axis [of Polyhedron A] of order n, you'll find 2n translates of the fundamental domain [Polyhedron B] spaced pi/n radians apart the polyhedra should be interpreted as follows: Polyhedron A is the polyhedron for which the group is named (that is, a tetrahedron for the binary tetrahedral group, an octahedron for the binary octahedral group, or an icosahedrdon for the binary icosahedral group). Polyhedron B is the fundamental domain itself (respectively an octahedron, a truncated cube or a dodecahedron in the preceding three examples). In the case of the binary icosahedral group, the symmetries of the icosahedron are exactly the symmetries of the dodecahedron, so no confusion is possible. In the case of the binary octahedral group, the symmetries of the octahedron are exactly the symmetries of the truncated cube, so again no confusion is possible. But... in the case of the binary tetrahedral group, the symmetries of the tetrahedron are a proper subset of the symmetries of the fundamental domain (the octahedron). In other words, the fundamental domain (the octahedron) acquires some "accidental" symmetries that are not a priori forced upon it by the symmetries of the basic tetrahedron. In this case, to enumerate the elements of the binary tetrahedral group, it's crucial that we start with the basic tetrahedron, and not with the octahedral fundamental domain. (The reason the binary tetrahedral group's fundamental domain acquires extra symmetry is that the tetrahedron is self-dual. In effect the octahedral fundamental domain is the intersection of the basic tetrahedron and its dual. Thus the doubling of the symmetry.) For an elementary explanation of binary polyhedral groups, see Section 3 (pages 5161-5163) of "Topological lensing in spherical spaces", Classical and Quantum Gravity 18 (2001) 5155-5186 online at www.arxiv.org/gr-qc/0106033 (Someday I plan to write up a different explanation of these groups, but for now I hope the cited reference will serve well enough.) In any case, many thanks to Jesper for pointing out the error in the earlier e-mail. Best wishes to all, Jeff www.geometrygames.org/contact.html

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Jesper tells why
by Jeff Weeks 07 Feb '06

07 Feb '06

P.P.S. I like Jesper's response because it hints at *why* the various levels contain the number of dodecahedra that they do. In effect, Jesper is showing the relation between the binary icosahedral group (which is the holonomy group of the Poincaré dodecahedral space) and the ordinary dodecahedral group (the symmetries of the original dodecahedron at layer 1). The relationship, which Jesper's table clearly shows, is that if you travel in the direction of a symmetry axis (of the original layer 1 dodecahedron) of order n, you'll find 2n translates of the dodecahedron spaced pi/n radians apart. This same relation holds, of course, for the other single-action spherical manifolds.

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120 cell
by Jeff Weeks 07 Feb '06

07 Feb '06

Dear Boud et al., You intuition is correct -- there are some additional in-between layers. The attached images show layers 1 through 5. Layers 6 through 9 are of course symmetrical as you work your way towards the south pole. 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). Anyhow, here's the accounting: layer 1 1 layer 2 12 layer 3 20 layer 4 12 layer 5 30 layer 6 12 layer 7 20 layer 8 12 layer 9 1 ----------- total 120 Note that all numbers are "dodecahedal numbers", respecting the dodecahedral symmetry of the whole construction. The 120-cell is quite beautiful, isn't it? Best wishes to all of you, Jeff

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"vertical" faces
by Jeff Weeks 07 Feb '06

07 Feb '06

Boud, >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. No, the hexagons on the surface of layer 4 really are flat (not concave) ! Indeed they must be flat, because the 30 dodecahedra in layer 5 are completely flat (zero volume!) in this projection. The situation is analogous to placing a cube on a tabletop and then projecting it orthogonally onto that tabletop. The cube's four side faces project to line segments. Those projections have zero area, even though the side faces themselves have positive area. The analogous thing happens with the orthogonal projection of the dodecahedron from 4D to 3D. The 30 dodecahedra near the equator have positive volume, but because they are perpendicular to the 3D space we're projecting onto, their shadows (projections) have zero volume. I hope that's clear! Best wishes, Jeff P.S. I'm not sure whether my message which bounced yesterday has made it to the list yet or not, so for sake of completeness I'll reattach those images to the present message for the sake of the cc readers. Please forgive the duplication.

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SV: [Cos-top] MG11 - Topology of the Universe (fwd) (and PDS - how many tiles?)
by Gundermann, Jesper 07 Feb '06

07 Feb '06

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

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MG11 - Topology of the Universe (fwd) (and PDS - how many tiles?)
by Boud Roukema 06 Feb '06

06 Feb '06

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

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Cos-top February 2006