RE: [math-fun] The spatial universe is finite !!!???
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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Thomas Colthurst