Topology of Spacetime: Non-Compactness Not Required?

[shoehorn]

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

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 M\simeq\Sigma\times I, where I is some interval in \mathbb{R} and \Sigma is some three-dimensional manifold.

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

 

[George Jones]

For example, I believe Hawking proved during the 70s that, given a four-dimensional manifold M and a Lorentzian metric g, then (M,g) can be regarded as a spacetime if and only if M is non-compact.

Geroch showed that if M is compact, then (M,g) has closed timelike curves.

We can, for example, propose that the spacetime is topologically identified as M\simeq\Sigma\times I, where I is some interval in \mathbb{R} and \Sigma is some three-dimensional manifold.

Geroch showed that this can be done if spacetime is globally hyperbolic.

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

Sure, this is the usual topology taken for closed universe Friedmann-Robertson-Walker spacetimes

 

[shoehorn]

Thanks!

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