Can we just take the underlying set of the spacetime manifold as $\mathbb{R^4}$ for all practical purposes?

Question

In mathematical GR and also in some informal GR presentations (eg: MTW), manifolds are always mentioned before talking about GR... but now I am starting to wonder.. if it even actually neccesary?

In this answer, it is said that it doesn't really matter what topological manifold we use to model a situation in space time because all of them are homeomorphic to some subset of $R^4$ by definition of manifold and it's apparently impossible to actually check the topology at a global level due to the censorship theorem.

All of this tells me that other than getting Physicists and Mathematicians to use similar terminology, the manifolds in it's full generality self is probably not relevant to GR except at the highest levels of study at very specialized research (beyond grad school for instance). Is this conclusion correct or am I missing something?

Answer 1

Topological censorship is a theorem from the 1993 paper "Topological censorship" by Friedman, Schleich and Witt. It is a technical statement about certain manifolds (!), and it does not say that "it's apparently impossible to actually check the topology at a global level" as the question claims.

The paper explicitly says on the implications of its theorem:

Thus general relativity prevents one from actively probing the topology of spacetime. However, note that one can passively observe that topology by detecting light that originates at a past singularity.

What follows in the paper is further discussion of what restrictions, if any, there are on such passive observation.

Answer 2

You do need the concept of manifolds for GR. Locally homeomorphic to $\mathbb{R}^4$ just means that for every point $p \in \mathcal{M}$ there exists some coordinate chart $\Phi : \mathcal{U} \subseteq \mathcal{M} \to \mathbb{R}^4$ where $\Phi(p) = (x^0,x^1,x^2,x^3)$ which describes the manifold on some open set (nearby $p$).

If I understand your question right, it sounds as if you're asking why we need manifolds at all since we can describe them locally with $\mathbb{R}^4$. Even if the overall topology of the spacetime can be hard to nail down, we still need to describe GR with manifolds because they are the language for studying curvature.

Curvature is encoded in the metric $g_{\mu\nu}$ which tells you how to take inner products of vectors at every point of the spacetime. This changes from point to point with the curvature.

As a simple example; the 2-sphere $\mathbb{S}^2 = \{ \mathbf{x} \in \mathbb{R}^3 : |\mathbf{x}| = R \}$ is a manifold - just because you describe points $p \in \mathbb{S}^2$ using coordinates $\Phi(p) = (\theta,\varphi)$ which are in $\mathbb{R}^2$ doesn't mean the space isn't curved.