# Is the observable universe homeomorphic to $B^3$?

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?

• I've deleted a few inappropriate comments and responses, and moved the rest to chat if you would like to continue the discussion. – David Z Aug 18 '14 at 7:04
• Why isn't this obviously true? – MBN Aug 19 '14 at 11:16

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.