In General Relativity, spacetime is a $4$-dimensional manifold with one Lorentzian metric tensor defined on it. In the Special Relativity case what manifold is spacetime is quite clear: it is essentially $\mathbb{R}^4$ endowed with the metric tensor $\eta_{\mu\nu}=\operatorname{diag}(-1,1,1,1)$.

On the other hand, on General Relativity I can't understand exactly what manifold spacetime is. I'll try to make my point clearer. Some people say: "you can't know this beforehand, the Einstein Field Equations are the source of this information". Now, the Einstein equations are equations for the metric tensor, not for the manifold (this wouldn't even make sense).

But the metric tensor is a tensor field. It is a function defined on spacetime. It only makes sense talking about it, if we know beforehand its domain!

The equation itself is one differential equation for functions defined on $M$, how can we work with those function, if the domain was never defined?

I understand that the field equations give the metric, but I also understand that it doesn't make sense talking about the metric without any knowledge about the manifold where it is being defined.

In that my question is: what manifold $M$ is spacetime in General Relativity?

  • 2
    $\begingroup$ I'm not an expert on GR, but I believe the answer is very typical for physicists: It's not fixed! You start with some equations given in a certain coordinate patch, and then you can maximally extend this to a spacetime that's complete except for singularities. If you prefer complete spacetimes, you use this one, but it's just as valid to use a "smaller" one if you just omit causally disconnected regions. $\endgroup$
    – ACuriousMind
    Commented Apr 24, 2016 at 20:55
  • 3
    $\begingroup$ Related: physics.stackexchange.com/q/1787/2451 , physics.stackexchange.com/q/111670/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Apr 24, 2016 at 21:05
  • $\begingroup$ For the same metric, you might be able to have it defined on several different manifolds. In 4 dimensions, you can have 10 different manifolds, if I recall correctly, for flat space (I think it reduces to 6 if you require it to be orientable). Manifolds in general can have the same metric as their cover manifolds over some region. If your only information is the EFE, the manifold will be ambiguous. $\endgroup$
    – Slereah
    Commented Apr 24, 2016 at 21:10
  • $\begingroup$ I have had a similar question and I think one of the more clear and better stated answers to the question is covered in section 2.2 of Sean Carroll's book "Introduction to General Relativity Spacetime and Geometry". There are other good descriptions but this one is short and easy to understand for the context of EFE. Note that by itself it is not as complete as some other texts but I think more clear and sufficient. $\endgroup$
    – K7PEH
    Commented Apr 24, 2016 at 21:44
  • $\begingroup$ Here is a version of that description of Manifolds without having to buy the text if you do not have it (and, this treatment is more in depth mathematically than the book's section 2.2): preposterousuniverse.com/wp-content/uploads/grnotes-two.pdf $\endgroup$
    – K7PEH
    Commented Apr 24, 2016 at 21:47

2 Answers 2


Your intuition that

the Einstein equations are equations for the metric tensor, not for the manifold

is mostly on the right track, but the details are wrong. That core bit of intuition is best phrased, I think, as saying that the Einstein equations are local equations for the geometry of the manifold. That is, they tell you that, whatever manifold your spacetime is, its geometry must obey that specific constraint at each specific event in the manifold.

Of course, these local constraints are very powerful, and they severely restrict what the manifold can do; this is particularly true once you specify how one patch of spacetime looks and you start to continue the geometry from there onwards.

However, as local constraints they stop short of having a determining effect on the topology; the Einstein field equations are compatible both with manifolds as bland as $\mathbb R^4$ or more interesting ones that may be multiply connected and so on. The simplest example is probably the open/flat/closed trichotomy the topology of spacetime: if you assume that the universe is homogeneous and isotropic, then you get rather constrained local dynamics for the geometry, with three possible types of curvature (negative, zero, or positive) which have direct effects on the topology of spacetime.

The goal of GR, then, is to find spacetimes whose geometry locally obeys the field equations and whose other properties - including the global topology but also the type and distribution of matter and so on - are consistent with our expectations for physical systems. (We already do this, for example, when we rule out spacetimes with exotic matter.)

In general, though, the Einstein field equations are local and that makes it perfectly possible to talk about a metric even when we have yet to determine the shape of the metric's domain: we simply look at the metric one coordinate patch at a time, and there the domain is perfectly well defined. Once we have the patches, though, we do need to stitch them together to make up the whole manifold and check that they do come together to make one coherent entity, and there's nothing in the books that says this will not be every bit as challenging as the original local solutions.


The general story

Here's an attempt to formalize how physicists build spacetime manifolds.

Let $N$ be one of

  • $\mathbb R^n$
  • Some dimension $n$ product manifold of $S^1$ and $\mathbb R$ (corresponding to periodic solutions).

Pick one such $N$. Now

  1. Take stress tensor $T$ defined on some open subset $U \subset N$ and some boundary conditions (e.g. falloff at neighborhood of boundary of $U$). Consider Einstein's equations on $U$.

  2. Let $V$ denote the maximal domain on which a smooth solution $g$ exists. Call $V$ the "spacetime" manifold.

In summary, you should think about the Einstein equations as being a relation between tensor fields on an open subset of a simple manifold (our $U$ above):

  • $U$ is well-defined as a smooth manifold on which you can consider smooth tensor fields.
  • "Spacetime" $V \subset U$ is a submanifold which you can equip with a Riemannian structure given by the Einstein equations.

In particular topology of $V$ is irrelevant to defining the Einstein equations.

Just to beat this to death, here's another way to put it

  • Einstein's equation is a equation between smooth tensor fields. You do not need a Riemannian structure on your manifold to have the Einstein equations be well-defined. All you need is a smooth structure.
  • Observe once you solve Einstein's equations, you are free to do Riemannian geometry on $V$, on which you have a well-defined Riemannian structure.


Special Relativity

You mentioned the case of special relativity. It might be instructive to see how this construction works in that. Take

  • $N \equiv \mathbb R^n$
  • $U = \mathbb R^n$
  • Let $T \equiv 0 \in \otimes^2T^*M$ be the zero tensor field.

Einstein's equations yield $\eta$, which is well-defined on all of $\mathbb R$. So in this case we have $V = U =N$. We endow $V$ with a Riemann structure, and do geometry on $(V, \eta)$.


In the case of Schwarzschild, it is well known that the solution diverges in some places, so the Einstein equation solution $g_{Schwarzschild}$ is only well-defined on a strict subset $V \subset U$. $(V, g_{Schwarzschild})$ is a well-defined Riemannian manifold, while $g_{Schwarzschild}$ does not give a well-defined Riemannian structure to $U$. However the Einstein equations are still well-defined on $U$, which is what enabled you to solve them in the first place.


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.