Some doubts concerning the mathematical bases of GR

1. Aug 15, 2012

TrickyDicky

I'm a little confused about certain assumptions usually made in GR as to how rigorous they are mathematically speaking.
For instance the assumption generally presented without proof that GR spacetime manifold is a Hausdorff space seems not to be warranted given the fact that pseudometric spaces (with no definite positive metric) are not Hausdorff. Why make that assumption then?
On the other hand the defining property of GR was explaining gravity thru curvature as an invariant, but Lorentzian manifolds, precisely due to their not being metric spaces, may be both flat and curved depending on what patch is chosen, in other words curvature is not a property of the manifold alone.
Finally the assumption that the GR manifold is smooth seems to be contradicted by the existence of singularities, the condition usually imposed that one must only look at the space and time intervals that are singularity free doesn't seem a very rigorous mathematical prescription.

Last edited: Aug 15, 2012
2. Aug 15, 2012

Dickfore

3. Aug 15, 2012

martinbn

There is a brief discussion about this in Hawking-Ellis, plus some non-Hausdorff examples. I don't remember what the reason was to not consider them, but it was definitely physics not mathematics.
The first part is not true. It is true that curvature is not a property of the manifold. You need a connection. In this case the Levi-Chevita connection associated with the matric.
That is also not true. In this context the smooth manifold is singular if it is geophysically incomplete (or some version of it).

4. Aug 15, 2012

Staff: Mentor

I agree with Dickfore, I think that this is incorrect. A neighborhood in the Hausdorff sense is a topological concept, which is more primitive than the metric. So it is not defined by the metric, so I don't know why having or not having a specific kind of metric would have anything to do with whether or not a space is Hausdorff.

For any spacetime manifold M with two distinct points x and y there exist open subsets X and Y such that:
$x \in X \subseteq M$
$y \in Y \subseteq M$
$X \cap Y = \emptyset$
so spacetime manifolds in GR certainly seem to be Hausdorff to me.

Last edited: Aug 15, 2012
5. Aug 15, 2012

TrickyDicky

The first claim can be read in the wikipedia entry for Hausdorff spaces where it says "pseudometric spaces typically are not Hausdorff" in the paragraph about examples and counterexamples.
The second just meant that the pseudometric space induced topological structure makes curvature depend on the topology.

6. Aug 15, 2012

TrickyDicky

The metric alluded to in the OP is not exactly the metric tensor but the distance function obtained by integrating the metric tensor.

The example you give about the distinct points x, y, don't seem to apply when the points lie in a null geodesic.

7. Aug 15, 2012

Dickfore

8. Aug 15, 2012

Matterwave

The topology of a manifold is determined by the local mappings to a Euclidean space and the standard topology in the Euclidean space (which is Hausdorff).

Therefore, all (topological) manifolds (of which differentiable manifolds are a subsection) are Hausdorff.

This has nothing to do with the metric.

9. Aug 15, 2012

Staff: Mentor

OK, but you don't need a distance function to define a neighborhood in the Hausdorff sense. You don't need any notion of distance at all since it is only required that the neighborhoods be open subsets. It is not required that the neighborhoods be open balls.

I am not sure what you are trying to explain here. A path (whether null or geodesic or not) does not define an open subset, so it doesn't define a neighborhood.

Last edited: Aug 15, 2012
10. Aug 15, 2012

Staff: Mentor

Interesting. I don't understand that comment, but I don't consider myself expert enough to correct or remove it.

11. Aug 15, 2012

TrickyDicky

Surely if you relax the condition that makes a metric space be a pseudometric space, you get a metric space, so?

12. Aug 15, 2012

PAllen

I think the confusion here is between pseudometric space and semi-riemannian manifold. The former uses the non-positive definite distance function part of the definition of its topology (the open sets). That latter does not, and is defined on top of differentiable manifold, which is Hausdorff as part of its definition. This is better covered (if you must use Wikipedia) in its articles on Differentiable Manifolds:

http://en.wikipedia.org/wiki/Differentiable_manifold

and also in the article on Pseudo-Riemannian manifolds:

http://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold

which refers defines it on top of differentiable manifold.

All this by way of verifying that Matterwave's post was a complete answer to the question, that unfortunately was mostly ignored.

13. Aug 15, 2012

PAllen

Here is another reference clarifying that for a pseudometric space the topology is induced by the metric. This the opposite of a pseudo-riemannian manifold, where metric is a structure added on top of a differentiable manifold which is already Hausdorff. The reference confirms the a pseudometric space need not be Hausdorff. Obviously, pseudometric spaces have nothing to do with GR.

http://igitur-archive.library.uu.nl/dissertations/1916832/c2.pdf [Broken]

Last edited by a moderator: May 6, 2017
14. Aug 15, 2012

TrickyDicky

You introduce something that could help clarify things a bit.
Could you specify how exactly a pseudo-riemannian manifold (in thi case a Lorentzian one) is not a pseudometric space?
Consider points on the light cone for instance, the distance betwen two different points can be zero, right?
Wrt the comment about all differentiable manifolds being Hausdorff, that is true in a limited sense, that is, it is true locally, forgetting about the topology. But being a Hausdorff space is usually considered a global property of a space, and here we find the problem that a manifold with singularities is not differentiable globally.

Other important limitations of pseudoriemannian manifolds are listed in the wikipedia entry

" On the other hand, there are many theorems in Riemannian geometry which do not hold in the generalized case. For example, it is not true that every smooth manifold admits a pseudo-Riemannian metric of a given signature; there are certain topological obstructions. Furthermore, a submanifold does not always inherit the structure of a pseudo-Riemannian manifold; for example, the metric tensor become zero on any light-like curve".

15. Aug 15, 2012

George Jones

Staff Emeritus
How, exactly, is a Riemannian manifold a metric space?

16. Aug 15, 2012

TrickyDicky

I believe a Riemannian manifold is a metric space because its metric is positive definite.
I think a pseudo-Riemannian manifold is a pseudometric space because its metric is not positive definite and two different points x and y in a null cone can have zero distance. Is this wrong?

17. Aug 15, 2012

George Jones

Staff Emeritus
A metric space (X , d) is a set X together to together with a function d that maps pairs of elements of X to real numbers (the distances between the elements of the pairs).

For a Riemannian manifold (M , g), g is function that maps pairs of tangent vectors to real numbers, i.e., g doesn't map pairs of elements of M to real numbers. Consequently, a Riemannian manifold (M , g) is not a metric space in the sense in which you are using the term "metric space".

This is an unfortunate and confusing clash of terminology: "metric" in "Riemannian metric" and "metric" in "metric space" mean different things.

Last edited: Aug 15, 2012
18. Aug 15, 2012

TrickyDicky

Btw, there seems to be a mistake in the wikipedia entry on differentiable manifolds, they present topological manifolds as hausdorff, but if one reads Hawking and Ellis "The large scale structure of spacetime, they devote a few pages to discuss non-Hausdorff manifolds.

19. Aug 15, 2012

PAllen

I believe you are mixing up different fields of mathematics. That is also (I think) what George Jones is hinting at.

As I understand it:

- A metric space is just a set with a function on pairs of elements in it meeting certain properties. It is in the field of point-set topology, not differential geometry. Similarly for a pseudometric space. The key is that it's definition starts from set not topological space or manifold.

- Orthogonal to these concepts, there is a hiearchy of constructs with increase structure as follows:
topological space -> hausdorff space -> manifold -> riemannian or pseudo-reimannian manifold.

Topological space starts from imposing an arbitrary collection of subsets meeting certain properties such that we may call them 'the open sets'.

The quote you refer to above simply makes that point that any (smooth) manifold can be made riemannian, while some maniolds cannot be made pseudo-rieamannian.

20. Aug 15, 2012

TrickyDicky

George, I am aware of the distinction between a metric tensor g and a metric as referred to in "metric space"( I even mentioned the difference in a previous post).
Every time I wrote metric I referred to the distance function, not the metric tensor.
So I am not yet clear if you agree a Riemannian manifold is a metric space, and a Lorentzian manifold is not.

21. Aug 15, 2012

PAllen

This is covered here:

http://en.wikipedia.org/wiki/Non-Hausdorff_manifold

22. Aug 15, 2012

George Jones

Staff Emeritus
Okay, I am finding this thread to be very confusing. Why should a pseudo-Riemannian manifold be a metric space? As Matterwave and PAllen have pointed out, pseudo-Riemannian manifolds are topological spaces via the manifold topology, even if they don't have natural metrics.

What do you want to do with a metric? Why does this introduce doubts about the mathematical basis of GR?

23. Aug 15, 2012

PAllen

Seems to me you could have metric space that cannot even be a manifold (e.g. it is defined on a finite set). Similarly, a disconnected Riemannian manifold would not allow introduction of a global distance function based on integrated metric distance. Maybe every complete Riemannian manifold can be treated as a metric space.

24. Aug 15, 2012

TrickyDicky

Sorry, I thought my questions were clear, my fault.
But I'm not saying a pseudo-Riemannian manifold is a metric space, my claim is that it is not.
I can see that a metric space is not a Riemannian manifold but I had the notion a Riemannian manifold had the property of being a metric space, am I wrong?

Similarly a pseudometric space is not a pseudo-Riemannian manifold but my belief is that pseudo-Riemannian manifolds have as defining property their pseudometricity(that is the key difference wrt Riemannian manifolds) , is this also not correct?