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Is the observable universe homeomorphic to $B^3$? Where $$B^3=\{x\in \mathbb{R}^3 : |x|\leq 1 \}$$

Or is it even sensible to talk about space (rather than spacetime) as a 3 manifold?

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  • $\begingroup$ I've deleted a few inappropriate comments and responses, and moved the rest to chat if you would like to continue the discussion. $\endgroup$ – David Z Aug 18 '14 at 7:04
  • $\begingroup$ Why isn't this obviously true? $\endgroup$ – MBN Aug 19 '14 at 11:16
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I think that "observable universe" is not defined precisely enough to make such statements about it.

The spacetime events that we can see are the events on our past light cone. That light cone intersects the last-scattering surface (about 400,000 years after the big bang) in an approximate sphere. By convention the light cone is cut off there (because we can't see through the opaque plasma before last scattering—though future neutrino and gravitational-wave astronomy might change that). The matter passing through that sphere (which is also the boundary of the light cone) will, by the continuity equation, pass through the light cone at some point, while matter outside can't without exceeding the speed of light. The matter that passes through the sphere is called the observable universe.

In a perfectly uniform zero-pressure universe described exactly by an FLRW metric, and in which the last-scattering time is precisely defined, the sphere will be exactly a sphere, and the locus of observable matter will be exactly a cylinder ($\mathbb B^3 \times \mathbb R$) in FLRW coordinates. The metric breaks spacetime symmetry, giving a natural separation into space and cosmological time, and a natural correspondence between spatial points at different cosmological times. You could think of this universe as a 3D space with a geometry that's time-invariant up to an overall conformal scale factor (the inflating-balloon analogy, sort of). The observable universe is topologically and even metrically a ball in that space.

In reality, the universe went from almost opaque to almost transparent over some nonzero time, so there is an inherent ambiguity in the cutoff of the past light cone and the boundary of the observable universe. Also, the matter making up the observable universe does not stay in place relative to the FLRW "space". In the case of identical quantum particles, you can't trace the later motion of the matter that passed through the sphere even in principle. And since the geometry of spacetime is determined by the matter distribution, the FLRW metric is not exactly correct and there is no precisely defined FLRW "space".

The observable universe is still a "fuzzy ball", but I don't think "fuzzy" can be given a precise mathematical definition.

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  • $\begingroup$ +1, it should be also noted that tests of non-trivial topology in the observable universe have been done but none was found. I have described this in my answer here. $\endgroup$ – Void Aug 21 '14 at 19:46

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