In most texts on GR we are first introduced to a formal and rigorous definition of a manifold.

We then learn the point that in GR "any coordinate system" might be used for the 4D spacetime metric. The Einstein equation relating curvature and energy is invariant regardless because of tensor transformation law. Put it another way, we are not limited to only Lorentz linear transformations as in SR, hence GR is indeed the general theory of relativity.

But I have not seen examples in GR texts in which the rigorous definition of the manifold, i. e., that there must be smooth charts in an atlas that covers the entire manifold is used explicitly every time we introduce a new coordinate system.

The question is when and where one needs to check this out and does not count on an intuitive sense?

Or can you suggest an example in which we build a new arbitrary coordinate system and the entire manifold cannot be covered through a maximal atlas?

Should we be alarmed about the violation of the manifold's definition only where we see metric singularities in all possible candidate coordinate systems, e.g. at Big Bang or that inside a black hole?

  • 4
    $\begingroup$ Your question does not make sense to me. When GR is done in a single coordinate system, then the assertion is simply that the "manifold" is the manifold defined by that chart. When the assertion is that the total manifold cannot be covered in one chart (e.g. spheres), then you must give an atlas for the entire manifold (in the case of the spheres, you can choose e.g. the two stereographic projections). Can you perhaps give an example of a case where you think this "check" you're talking about has not been performed, but should be? $\endgroup$
    – ACuriousMind
    Apr 21, 2016 at 14:37
  • $\begingroup$ My point: A rigorous definition is introduced for the manifold. This provides, then, a complete freedom to mix the spacetime in Cartesian or polar coordinates. But we never check that definition again throughout GR texts. Am interested in examples in GR metrics that show the manifold definition breaks down. For instance, can we show that big bang is not in the manifold or cannot find an atlas to cover that point in spacetime? $\endgroup$
    – user56963
    Apr 21, 2016 at 17:18
  • $\begingroup$ Related: physics.stackexchange.com/q/251867/2451 $\endgroup$
    – Qmechanic
    Apr 30, 2016 at 5:24
  • $\begingroup$ I'm having trouble figuring out what the OP would want as an answer to this question, because it's not clear what they're ruling out. In the question, they seem to rule out the big bang and black holes, but in a comment they give the big bang as an example of something they would like addressed. $\endgroup$
    – user4552
    Oct 22, 2019 at 19:25

1 Answer 1


Existence of smooth structures

You're asking when does a (topological) manifold $M$ fail to be covered by a smooth atlas. Another way to phrase this is "when does a manifold admit a smooth structure".

This is a well-studied problem. It turns out examples of manifolds which do not admit smooth structures only occur in dimensions $\ge 4$ (see e.g. differential structures, wikipedia). In particular you cannot find a manifold in dimension $< 4$ which does not admit a smooth structure.

Why spacetime is always smooth

As far as I know examples of manifolds which do not admit smooth structures do not show up in general relativity. Here's why:

In general relativity you solve Einstein's equations on some neighborhood homeomorphic to a subset of $\mathbb R^n$ (possibly with simple periodic identifications), then call a submanifold of that neighborhood where the solution is well-behaved our spacetime. Hence our spacetime inherits a smooth structure from the embedding in $\mathbb R^n$.

  • $\begingroup$ For every manifold with a $C^1$ structure, there exists a compatible smooth structure on that manifold. Therefore only strictly $C^0$ manifolds do not admit one, and those do not have a well-defined tangent bundle, which is necessary for a spacetime. $\endgroup$
    – Slereah
    Nov 28, 2020 at 11:25

Your Answer

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