I'm familiar with the idea that there are very strong reasons to believe that possible spacetimes [itex](M,g)[/itex] for the universe can have restricted topologies. For example, I believe Hawking proved during the 70s that, given a four-dimensional manifold [itex]M[/itex] and a Lorentzian metric [itex]g[/itex], then [itex](M,g)[/itex] can be regarded as a spacetime if and only if [itex]M[/itex] is non-compact.(adsbygoogle = window.adsbygoogle || []).push({});

However, we also know that we don't really need to deal with spacetime concepts when looking at general relativity. We can, for example, propose that the spacetime is topologically identified as [itex]M\simeq\Sigma\times I[/itex], where [itex]I[/itex] is some interval in [itex]\mathbb{R}[/itex] and [itex]\Sigma[/itex] is some three-dimensional manifold.

The question I have is this. If we take [itex]M[/itex] as being non-compact, surely that doesn't imply that [itex]\Sigma[/itex] also has to be non-compact? For example, we could presumably take [itex]M\simeq S^3\times I[/itex] as being a spacetime since [itex]S^3[/itex], which is compact, can be usually be used to foliate [itex]M[/itex].

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# Topology of spacetime

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