2
$\begingroup$

After reading Timaeus answer here: https://math.stackexchange.com/q/1302672/, I got an idea that spacetime we usually talk about in GR can be described as a manifold.

Firstly, let's address coordinates, how to switch and why in General Relativity you sometimes are forced to switch coordinates. We will start with a single coordinate system. For an event $m$ (a point in your manifold) in a region of spacetime $M_i⊂M$, where $M$ is the total spacetime, then there can be a coordinate map $ϕ_i$ which is a one-to-one mapping from all of $M_i$ to $\mathbb{R}^4$.

The fact that they mentioned in the last sentence

one-to-one mapping from all of $M_i$ to $\mathbb{R}^4$

means that a subsection of spacetime $M$ can be seen to be have Euclidean topology locally, even though we know that spacetime in GR is Lorentzian.

Even if we disregarded Timaeus answer there, we all have read GR books that started with talking about Manifolds and how manifolds can be defined locally as Euclidean. Afterwards, authors start to define Vectors, (covariant and contravariant), tensors and so on... thus preparing the notions one use in GR. My question is: Why do we need to introduce students to Manifolds before teaching them about GR, if the latter has Lorentzian signature meanwhile the former is locally described as Euclidean?

$\endgroup$
  • 3
    $\begingroup$ I believe th confusion comes from two uses of the word Euclidean. The first one is when talking about inner products: an inner product is Euclidean if its positive definite and Lorentzian if its not. The second use of Euclidean (the one that is important here) is that an Euclidean space is just $\mathbb R^n$, i.e., tuples of real numbers. The second use has nothing to do with any inner product nor any metric, so it makes no sense to state that a chart $M\to\mathbb R^4$ is Lorentzian. Local coordinates are always points in $\mathbb R^4$, ie, points in an Euclidean space (tuples of numbers) $\endgroup$ – AccidentalFourierTransform Jan 12 '16 at 12:23
  • $\begingroup$ So, when we say: Spacetime in GR is a manifold, we are indirectly saying Spacetime in GR can locally be viewed as Euclidean, i.e., events taking place in open sets in that spacetime can be mapped to $\mathbb{R}^4$. Is this a correct sentences? @AccidentalFourierTransform $\endgroup$ – 4567439 Jan 12 '16 at 12:26
  • $\begingroup$ What is the actual question? The universe is a non-euclidean manifold, and that exhausts why GR needs manifolds. Or is it about how manifolds are taught? $\endgroup$ – gented Jan 12 '16 at 12:31
  • $\begingroup$ yes: and the map $M\to \mathbb R^4$ just means that any event can be described, locally, by four real numbers. The map $M\to\mathbb R^4$ need not preserve any inner product: remember that you can define manifolds without a metric. $\endgroup$ – AccidentalFourierTransform Jan 12 '16 at 12:31
  • $\begingroup$ @GennaroTedesco I am confused about an idea which is the following: We know that spacetime is 4 dimensional differentiable manifold. We know also that spacetime is Lorentzian. Meanwhile, manifold locally is described as being Euclidean and not Lorentzian (or Minkowski). Thus, how can we say a spacetime is a manifold if this is the case? $\endgroup$ – 4567439 Jan 12 '16 at 12:35
1
$\begingroup$

Space-time is not simply a manifold, but a differentiable and pseudo-Riemannian manifold.

The topology of the manifold is already fixed by the (topological) manifold structure (that is: local homeomorphism to $\mathbb R^n$ without assuming any inner product). In this sense sufficiently small open sets from the manifold carry the topology of a subset of the Euclidean space. But this construction does not define a metric on the manifold (even the notion of differentiation requires more structure, namely a differentiable structure which is an atlas, where you require that the transition maps between charts are differentiable).

While a Riemannian structure allows one to define a topology (as the Riemannian structure induces a metric, which induces a topology) this topology is always consistent with the topology of the manifold generated by the local homeomorphisms to $\mathbb R^n$.

A pseudo-Riemannian structure, which the manifold carries in general relativity, does not allow to define a topology (as the distance function is no longer a metric, as it is not positive definite). This metric does, however, allow the construction of a geometry, and a connection (or "covariant derivative") $\Gamma^\mu_{\phantom{\mu}\nu\kappa}$ which in turn leads to geometric invariants (as the Riemann curvature $R^\mu_{\phantom{\mu}\nu\kappa\lambda}$) and geometric objects (such as geodesics).

Note, that the dynamical field in GR is not the topology of the manifold, but the pseudo-metric $g_{\nu\nu}$, which lives on the manifold. So in a sense the topological manifold is a given background, on which the metric evolves. (Note, that there is some interplay, as the curvature can constrain the global structure of the manifold. There are even manifolds that cannot be given a pseudo-Riemannian metric).

In conclusion: You have to keep the notions apart. The property of $\mathbb R^n$ that is modelled by a topological manifold is only its topology not its metric or geometry. To define a metric and geometry the manifold has to be equipped with extra structure. This structure can lead to a different local geometry from Euclidean space.

$\endgroup$
  • $\begingroup$ Ok. So, let me see if I understand this. You're saying that the usual definition of a manifold which says that points belonging to $M_i \in M$ can be mapped to $\mathbb{R}^D$ is not the complete way to define a pseudo-Riemannian manifold. This usual defintion might help us construct scalar functions, co(ntra)variant vector fields and the transformations of these from one coordinate system to another (gct). These could be carried on to GR's manifold. However, this is not the whole story. One should then add a metric (inner product) to constrain the general manifold to become the GR one. Right? $\endgroup$ – 4567439 Jan 14 '16 at 7:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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