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Re: [math-fun] Rational Group
by asimovd＠aol.com 10 Oct '03

10 Oct '03

Jon Perry writes: << ... if we consider n-manifolds as spaces constructed from n-1-manifolds, then the n-manifold is formed by a continous expansion of varying n-1-mainfolds. ... >> There are any number of things that this could mean. Could you spell out what you mean by each of these two clauses? Thanks, Dan

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RE: [math-fun] The spatial universe is finite !!!???
by Thomas Colthurst 10 Oct '03

10 Oct '03

Perhaps the first thing to say is that this is experimental physics, so everything is model dependent and everything has error bars. For example, one of the central pieces of Weeks et al.'s evidence for the dodecahedral model are the low quadrupole and octopole WMAP amplitudes. But a recent paper by G. Efstathiou (http://xxx.lanl.gov/abs/astro-ph/0310207) shows that if you do a better statistical estimation of the poles that is more robust to foreground galaxy (aka the Milky Way) contamination, then the pole amplitudes are higher than originally estimated: ~200 uK^2 for the quadropole (vs. the original 123 uK^2) and ~1000 uK^s for the octopole (vs. the original 611 uK^2). [Naive explanation: if you do the simple thing and just "cut out" the Milky Way portion of the WMAP from the sky, you are going to heavily bias the low spherical harmonics toward low energies. The cleverer you are, the more you are going to remove that bias.] These numbers make both the dodecahedral model and the standard flat universe model fit the quadrople better (though the dodacehdral is still much closer), and makes the flat universe model fit the octopole better than the dodecahedral model. The probability of the data given the flat universe model used to be around 0.2%; now it's 3-4%. Depending on your priors for the two models, that difference can be a big deal. Another example: the WMAP data will never prove (in the Popperian sense) that the universe is R^3; in fact, that's the one and only geometry that it can't conclusively rule out! An R^3 universe will never show circles on the sky or other topological effects, and it's flatness will never be measured to be "1 exactly", merely as "1 + 1.x * 10^(-n) +/- y * 10^(-m)" for large n and m. Now, if your prior on the universe said that there was a 33% chance of flatness, then n need only be 2 or 3 before you are pretty sure that the universe is flat. On the other hand, if your prior said that curvature = 1 exactly was just as probable as curvature = 1.0000000007 exactly, then you still wouldn't know. The second thing to say is that a rebuttal to the docahedral model by Tegmark et al. is up on the arXiv: http://xxx.lanl.gov/abs/astro-ph/0307282 It's mostly about testing (and ruling out) a toroidal universe model with one dimension (in the direction of Virgo) smaller than the horizon scale. It's also the first circles in the sky analysis of the WMAP data that I've seen. The fact that it's been over half a year and the first analysis appeared just yesterday should give some indication of the computational and modelling dificulties involved: there is foreground noise again (that darn Milky Way!), the late integrated Sachs-Wolfe effect, the Doppler effect, detector noise, pixelization effects, six dimensional search space, ... Anyway, Tegmark et al. only look for "back-to-back" matched circles, and they don't find any. The Weeks et al. dodecahedral model predicts matched circles, so that should be enough to rule it out. Still, the difficulties of actually finding even matched circles is big enough that I wouldn't take this as the last word just yet. David Spergel of the WMAP team is also going to post a rebuttal, including the "definitive" circles in the sky analysis, on the arXiv soon, but it's not there yet. Finally, I should point out that the presence of a distinguished spatial section (aka, the big bang) makes it hard to physically motivate a factorization of large scale space-time into anything other than space times (some interval of R). I'm not saying it can't be done; I'm just saying it's hard to do in a way that a cosmologist would take seriously. My own question for math fun about manifold factorizations is this: let's say we take the string theorists seriously for a second and pretend that space is locally ten dimensional. Are there any "nice" ten dimensional manifolds that factorize "naturally" as a very very large dodecahedral (say) four manifold times a very very small Calabi-Yau six manifold? -Thomas C

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Re: [math-fun] The spatial universe is finite !!!???
by asimovd＠aol.com 10 Oct '03

10 Oct '03

Megachiropteran cinematographer: marchantiaceous centauromachias. Bedroom boredom? Brighten berthing! << Oh! -- but which direction you get rotated in invariant! That is, the Poincare Dodecahedron is not its own mirror image. So such a universe would have a large-scale handedness. Wouldn't that be neat... >> I have the impression that the Poincaré dodecahedral manifold D created by a 1/10 rotation identifying opposite faces, and the one using a -1/10 rotation, are isometric to each other -- much as a right- and a left-handed H opf fibration of the 3-sphere are isometric. What I'm wondering now is whether D double-covers a non-orientable manifold. * * * By the way, a cool thing about D is that it was the first counterexample to something Poincaré thought he had proved -- that any 3-manifold M with trivial (i.e., zero) first homology group H_1(M) must be homeomorphic to the 3-sphere. He discovered that the fundamental group pi_1(D), when abelianized (to obtain H_1(D)) becomes the zero group. So Poincaré, chastened, then suggested that *maybe* if the fundamental group of a 3manifold is trivial, *then* it must be homeomorphic to the 3-sphere. (He didn't actually couch this as a conjecture, but only as a possibility.) --Dan

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Antwort: Re: [math-fun] The spatial universe is finite !!!???
by dirk.lattermann＠bgs-ag.de 10 Oct '03

10 Oct '03

Hello, forgive my naive thoughts, but wouldn't the argument used in the article of wavelengths needing to fit in the space (don't even know if this is a commonly agreed one, even if it sounds plausible) prevent non-orientable finite spaces at all? At least, it would only allow an uneven number of zero-crossings of a wave, whereas an orientable finite space allows only an even number? Could this be astronomically observed somehow? (Just tried to draw some waves on a moebius band) Greetings, Dirk Lattermann [...] Boy, I'd be delighted if the universe were a Poicare Dodecahedron! (Though I'd been hoping for a nonorientable space: if Weeks is right, then you could go in a straight line for 74 billion light-years and get back to where you started, just rotated by 2pi/5. I'd been hoping you'd get back to where you started mirror imaged.) --Michael Kleber kleber(a)brandeis.edu _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] The spatial universe is finite !!!???
by Shel Kaphan 09 Oct '03

09 Oct '03

At 08:48 PM 10/9/2003 -0400, Michael Kleber wrote: >if Weeks is right, then >you could go in a straight line for 74 billion light-years and get back to >where you started, just rotated by 2pi/5. ... but after 74 billion years (minimum) you'd hardly recognize the place! In fact, doesn't it become a bit difficult even to define what is meant by "where you started"?

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[math-fun] finite Q-series
by Richard Schroeppel 09 Oct '03

09 Oct '03

rwg> The integerhood of a 3F2 summand I mentioned follows from the conjecture that n /===\ k - j + 1 k + 2 j + 2 k + 2 j + 3 | | (1 - q ) (1 - q ) (1 - q ) | | ------------------------------------------------------ | | j j + 1 j + 2 j = 1 (1 - q ) (1 - q ) (1 - q ) is a polynomial in q for all n,k >0. This seems to be hard!? I'll bet there's a theorem that for an expression like this, you only need to check a dozen or so consecutive n and k --rwg Chorasmian maraschino harmonicas inspissate antisepsis. Should the exponent on the first q in the numerator be k+j+1? As it stands, it looks like k=1 n=3 (and hence the j=3 term) will create a q^-1 that won't disappear. --- Wrt the pyramid puzzle, I recall seeing an article about it Science News, maybe 5-10 years ago. It may have been the Math SAT? Rich rcs(a)cs.arizona.edu

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[math-fun] Rational Group
by Jon Perry 09 Oct '03

09 Oct '03

{Q,+} (the rationals with the addition operator) is a group. How come the infinite sum of rationals is not then necessarily a member of this group, e.g. zeta(2)? Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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[math-fun] nonobviously integer sequence
by R. William Gosper 09 Oct '03

09 Oct '03

Let a and c be integers. "Shew that" F := 1, F := a c - 1, F := 0 1 n 2 (2 n + a - 3) ((2 c - 1) (2 n + a - 4) (2 n + a - 2) + (a - 2) ) F n - 1 (----------------------------------------------------------------------- 2 - (n - 1) (n + a - 3) (2 n + a - 2) F )/(n (n + a - 2) (2 n + a - 4)) n - 2 (note cubic denominator) is an integer sequence. Distantly related: it appears that [ n + 4 n + 5 ] [ - n, -----, ----- | ] F [ 2 2 | -4] = 1,6,57,701,10147,164317,2888282,54047434, 3 2 [ | ] [ 3, 2 ] is an integer sequence. The smallest prime factors go {},2,3,701,73,37,2,2,7,2,3,3,5,2,23,37,65306671610636210891,2, 31,3,3,19,2,2,3,2,3,7,3,3,2,2,7,2,67,5,... . Likewise, [ n + 5 n + 6 ] [ - n, -----, ----- | ] 4 F [ 2 2 | -4] = 4,25,228,2620,35164,527663,8613004, 3 2 [ | ] [ 4, 2 ] [ n + 6 n + 7 ] [ - n, -----, ----- | ] 15 F [ 2 2 | -4] = 15,99,891,9825,125085,1772775,27303603, 3 2 [ | ] [ 5, 2 ] [ n + 7 n + 8 ] [ - n, -----, ----- | ] 5 F [ 2 2 | -4] = 5,29,230,2260,25921,334105,4717653, 3 2 [ | ] [ 5, 3 ] More generally, [ n + k + 1 n + k + 2 ] [ n + k + 2 n + k + 3 ] [-n,---------,---------| ] [-n,---------,---------| ] F [ 2 2 |-4] & F [ 2 2 |-4] 3 2[ | ] 3 2[ | ] [ k, 2 ] [ k, 3 ] appear to have bounded denominators. --rwg Does OEIS list sequences not proven integral?

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[math-fun] manners
by Richard Schroeppel 09 Oct '03

09 Oct '03

Let's keep the discussion civil, please. Rich rcs(a)cs.arizona.edu

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RE: [math-fun] Rational Group
by Jon Perry 09 Oct '03

09 Oct '03

'But it doesn't give you a way to add *infinitely many* elements together. The usual way to define infinite sums is as a limit of finite sums. But a group structure is not enough to be able to define limits; you need an additional, topological, structure, to do that. And even once you have such a topological structure, that doesn't guarantee that *every* series or *every* sequence converges. Some do, some don't. If your topological structure comes from a metric structure (The standard topological structure on Q does), and the metric space is complete (Q isn't), then any Cauchy sequence will converge. Since the Reals are complete, this guarantees that the series for zeta(2) converges to some Real number (but not necessarily to a number in Q, since Q is not complete).' So how do we know that the incomplete nature of Q produces Reals, or indeed how can we determine the 'completetion' of an incomplete group (R is the completion of Q). Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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