The theory only deals with the local curvatures, not the global topology. Hence any manifold with an allowed metric is allowed. These can be infinitely many, especially for negative curvature space-times.


So I suppose for the same physical situation, there would be many manifolds from which it can be studied, then, how do we choose on which manifold that we actually do the calculations?/ find the manifold that fits for that scenario?

  • $\begingroup$ The question as it stands now is somewhat vague. Maybe clarify a little more of it. Are you asking what kinds of manifolds are "allowed" in GR? $\endgroup$
    – VaibhavK
    Commented Sep 24, 2023 at 10:53
  • $\begingroup$ How is this question different from this earlier question of yours? $\endgroup$
    – ACuriousMind
    Commented Sep 24, 2023 at 12:03
  • $\begingroup$ That's a physical modelling question, here it's like more of a calculation question. I mean, would it be more ideal to choose one manifold than other for doing an actual calculation? If so, how would we choose the ideal manifold for a physical scenario @ACuriousMind $\endgroup$ Commented Sep 24, 2023 at 15:43
  • $\begingroup$ I don't really understand why you think we can "choose" the manifold. You probably misunderstand the linked answer: The point is that the equations of general relativity do not constraint the global topology of the manifold (beyond "there has to be a Lorentzian metric") because the EFE are local equations. This doesn't mean we're somehow free to choose any manifold we want, spacetime is still supposed to be one specific manifold - it's just that the EFE don't tell us which one. Also, what kind of "calculations" are you thinking of here? Most calculations just happen in a single chart. $\endgroup$
    – ACuriousMind
    Commented Sep 24, 2023 at 15:51

1 Answer 1


As I understand your question, you are asking - "What local effects if any, are present in GR using which we can distinguish between manifolds (representing spacetime) that are same locally but not globally."

Let me state the definition of a manifold quickly - A topological manifold $M$ is a topological space where every point $p$ of $M$ lies in some open set $U$ such that that open set can be mapped to some region of $\mathbb{R}^{d}$. This map called $x$ should be invertible and both $x$ and $x^{-1}$ should be continuous. In other words if $M$ is locally homeomorphic to $\mathbb{R}^{d}$ then M is a topological manifold.

If $M$ represents spacetime, then we can only probe $M$ within this finite region $U$. We can then probe the topology of $U$ and, due to the homeomorphic nature discussed above, understand the local topology of $M$ itself. But we can never know the global topology of a generic manifold $M$.

In GR, further to locally homeomorphic property, we demand an even stronger condition - smoothness and locally diffeomorphic.

Now, if a collection of manifolds are locally diffeomorphic to $\mathbb{R}^d$ (and therefore to each other), then we (local observers) would not be able to distinguish one from the other. So, to answer your question - in my opinion, there are no effects that we can observe locally that will inform us about the global nature of $M$ (with a possible exception of globally hyperbolic spacetimes). Hence, all local effects would be identical.


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