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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