The topology of spacetimes

TrickyDicky

That is the point, it is hard to find (at least for me) mathematical justification for deriving a causal structure for the whole manifold only from the local action of the pseudoriemannian metric at the tangent space, when the distance function that prevails in smooth manifolds is not even the same as the one that integrates from the pseudoriemannian metric tensor.

George Jones

As has been indicated, some care is needed with respect to the meaning of "lightcone". Depending on the context and reference, "lighcone at [itex]p[/itex]" can either mean a subset of [itex]T_p \left(M\right)[/itex], or it can mean a subset of [itex]M[/itex]. I think that you mean the latter. In this case, [itex]M[/itex] is a neighbourhood of [itex]p[/itex] that contains its light cone.

George Jones

We want to model physics. For most situations, spacetime Hausdorffness seems to be a reasonable, physical separation axiom. Two distinct physical events always admit distinct neighbouhoods.

Having said this, we have strayed far off-topic with respect to the original post. Physics Forums rules advises that, instead of posts that are off-topic, new threads should be started.

dextercioby

IMHO, off-topic or not, this is by far the best thread in the Relativity section for quite some time. :)

micromass

I have moved the off-topic posts to a new thread so we can keep discussing this.

George Jones

Almost. Consider [itex]\mathbb{R}^2[/itex] with its standard positive-definite norm. Now obtain a new Riemannian manifold M by removing the origin. In this new manifold M, what is the distance between the points (-1 , 0) and (1 , 0)? There is no geodesic in M that joins these points. There isn't even a shortest path in M that joins these points points, i.e., if someone gives me a path in M between (-1 , 0) and (1 , 0), I can always find a shorter path in M.

This leads to a slightly subtle definition of distance in a Riemannian manifold. The distance between points p and q is the greatest bound on the lengths of all "nice" paths between p and q.

In my example, 2 is greatest lower bound of the lengths of paths between, even though there is no path of length 2, and 2 is the distance between (-1 , 0) and (1 , 0).

DaleSpam

OK, from my understanding a metric space must have a unique distance between any two points in the space. A metrizable space seems to be one that can be given a metric, not necessarily one that has a metric. So a differentiable manifold is metrizable, but by itself that doesn't make it a metric space. You know that you can equip it with a metric, and once you do so then it is a metric space, not before. Does that agree with your understanding?

Thanks, that helps my understanding. So what happens for spacelike paths? Also, you should be able to connect any pair of events with a null path, how are those avoided?

micromass

I wonder what exactly the problem is in this example. Intuitively, the problem is of course the hole at the origin. But is there a condition that we can place on our manifold such that this situation doesn't arise? I guess I'm asking for a condition where there always exists a shortest path.

WannabeNewton

Yessir.

micromass

I guess the Hopf-Rinow theorem partially answers this. Any connected and complete Riemannian manifold has length-minimizing geodesics: http://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem
But this is not an iff-condition. For example, (0,1) also has length-minimizing geodesics but is not complete.

George Jones

Completeness, i.e., convergence of Cauchy sequences and/or geodesic completenss.

"Riemannian Manifolds: An Introduction to Curvature" by Lee has interesting stuff (again!) about this on pages 108-111. For example:
[edit]Didn't see the previous post.[/edit]

George Jones

Aha! I coudn't see anything wrong with this, so I looked up the errata for Lee's Book. The second Lee statement that I quoted is wrong! Not iff. See reference to page 111 in

http://www.math.washington.edu/~lee/...ian/errata.pdf

George Jones

I make a distinction between "Riemannian" and "semi-Riemannian". What I wrote only applies to Riemannian manifolds.

kevinferreira

Let me just clear up these definitions, from wikipedia/metric:

So, a metric and our metric tensor are not the same thing, as already said before in this thread, a metric or distance is a map [itex] M\times M\longrightarrow \mathbb{R} [/itex] while the metric tensor is a map [itex] T_pM\times T_pM\longrightarrow \mathbb{R} [/itex].

A topological manifold is it metrizable, i.e. can its topology be described by a distance (may we use an atlas and [itex]\mathbb{R}^n[/itex] euclidean distance)? If yes, which distance? If not, what is then the topology of space time? Secondly, how can we use the fact that spacetime is not only a topological manifold, but a (pseudo-)Riemannian one, to help us on this task?

TrickyDicky

Well, its topology must look locally Euclidean if it is to be called a manifold, the global geometry(topology) doesn't have to.
But the important thing here is to separate the distance function from the topology, it is true that a metric distance function can induce a topology on a metrizable space, but this is not the case with manifolds, wich carry their own topology.
See above.

No need, it so happens that differentiable manifolds always admit a Riemannian metric.

TrickyDicky

This is confusing. I thought we agreed the distance function on the manifold doesn't distinguish Riemannian metric tensor from semi-riemannian metric tensor since they act locally on the tangent space rather than on the global manifold, and differentiable manifolds topological requirements only allow them to be metric spaces (can't be semimetric nor pseudometric spaces by definition, first of all because they are required to be Haussdorf). So I think Dalespam's question are relevant here.

This is related to what I commented in posts #41, #44 and #52. So far have been ignored, care to give it a try and address them? Thanks George.

atyy

Is a Hausdorff space necessarily a metric space? Wikipedia just says thart pseudometric spaces are typically not Hausdorff, but that seems to allow that Hausdorff spaces can be neither metric nor pseudometric. If that is possible, then wouldn't it be possible that Hausdorff manifolds with pseudo-Riemannian metric tensors need not be metric spaces?