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?

  • $\begingroup$ Manifolds are locally homeomorphic to some $\mathbb{R}^d$ but in general not globally. Black hole solutions provide such examples. $\endgroup$
    – Gold
    Commented Oct 26, 2022 at 12:09
  • $\begingroup$ This must have been asked before on Phys.SE. $\endgroup$
    – Qmechanic
    Commented Oct 26, 2022 at 16:33

2 Answers 2


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.

  • $\begingroup$ Oof didn't expect that. Should have been a comment. Hopefully the idea still makes sense $\endgroup$
    – Babu
    Commented Oct 26, 2022 at 18:00
  • 3
    $\begingroup$ @TrystwithFreedom The point of my answer is that the premise (namely that global topology is unobservable) from which you want to argue that we don't actually need to consider manifolds in GR is incorrect. I do not know what "idea" you still want to make sense of here. $\endgroup$
    – ACuriousMind
    Commented Oct 26, 2022 at 18:24

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.

  • $\begingroup$ Exactlly. You understood my question perfectly. $\endgroup$
    – Babu
    Commented Oct 26, 2022 at 14:35
  • $\begingroup$ But then the question is still there, instead of space time manifold as a general placeholder, just write R^4 and then say that the covariant derivative and metric is different than standard (to acc. for curvature) $\endgroup$
    – Babu
    Commented Oct 26, 2022 at 14:37
  • 1
    $\begingroup$ @TrystwithFreedom You can use $R^4$ only locally (about some point for some particular choice of coordinates). Consider an ant living on 2-sphere. Say, the ant is small enough that the region around the ant is approx. a flat land. If ant moves from P to Q (arbitrarily far) on the sphere, the region around the ant will still look flat. But it doesn't mean that the flat land around P and Q are the same flat land. It shows that you can't just replace the 2-sphere with one common $R^2$ (flat land). Intrinsic curvature plays an important role here. Same is true for a general space-time manifold $\endgroup$
    – paul230_x
    Commented Oct 27, 2022 at 10:22
  • $\begingroup$ Curvature is derived from the connection. My thought is that one can modify the connection such that R^4 works @KP99 $\endgroup$
    – Babu
    Commented Oct 27, 2022 at 12:42
  • 1
    $\begingroup$ @TrystwithFreedom there is no connection $\nabla$ on the sphere $S^2$ such that the Riemann curvature $R^{\nabla}$ of the connection vanishes identically. In other words, the sphere does not admit a flat connection. The plane $\Bbb{R}^2$ obviously has a flat connection. So, there are genuine differences in these manifolds; this is just one instance where the topology has implications for the curvature. $\endgroup$
    – peek-a-boo
    Commented Oct 28, 2022 at 3:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.