# The topology of spacetimes

by kevinferreira
Tags: spacetimes, topology
 Sci Advisor HW Helper P: 11,927 IMHO, off-topic or not, this is by far the best thread in the Relativity section for quite some time. :)
 Mentor P: 18,323 I have moved the off-topic posts to a new thread so we can keep discussing this.
Mentor
P: 6,246
 Quote by DaleSpam OK, I guess they must just use the length of the shortest path, regardless of whether or not there are multiple geodesics.
Almost. Consider $\mathbb{R}^2$ 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).
Mentor
P: 17,322
 Quote by WannabeNewton Indeed even though the two topological spaces mentioned are homeomorphic, they need not have same distance functions. Metrizable implies there exists some metric for the set but it doesn't state there is a single, unique metric. By the way, I think there is some confusion arising here in the terminology.
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?

 Quote by George Jones Almost. Consider $\mathbb{R}^2$ 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).
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?
Mentor
P: 18,323
 Quote by George Jones Almost. Consider $\mathbb{R}^2$ 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).
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.
C. Spirit
Thanks
P: 5,633
 Quote by DaleSpam Does that agree with your understanding?
Yessir.
Mentor
P: 18,323
 Quote by 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.
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.
Mentor
P: 6,246
 Quote by 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.
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:
 A connected Riemannian manifold is geodesically complete if and only if it is complete as as a metric space. M is complete if and only if any two points of M can be joined by a minimizing geodesic segment.
Didn't see the previous post.[/edit]
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P: 6,246
 Quote by micromass For example, (0,1) also has length-minimizing geodesics but is not complete.
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
Mentor
P: 6,246
 Quote by DaleSpam 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?
I make a distinction between "Riemannian" and "semi-Riemannian". What I wrote only applies to Riemannian manifolds.
P: 123
 Quote by WannabeNewton Any topological manifold is metrizable. As the requirement is a topological manifold, this is done before a riemannian or pseudo riemannian metric is even equipped to the manifold.
Let me just clear up these definitions, from wikipedia/metric:

 In mathematics, a metric or distance function is a function which defines a distance between elements of a set. A set with a metric is called a metric space. A metric induces a topology on a set but not all topologies can be generated by a metric. A topological space whose topology can be described by a metric is called metrizable. In differential geometry, the word "metric" may refer to a bilinear form that may be defined from the tangent vectors of a differentiable manifold onto a scalar, allowing distances along curves to be determined through integration. It is more properly termed a metric tensor.
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 $M\times M\longrightarrow \mathbb{R}$ while the metric tensor is a map $T_pM\times T_pM\longrightarrow \mathbb{R}$.

A topological manifold is it metrizable, i.e. can its topology be described by a distance (may we use an atlas and $\mathbb{R}^n$ 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?
P: 3,043
 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 $M\times M\longrightarrow \mathbb{R}$ while the metric tensor is a map $T_pM\times T_pM\longrightarrow \mathbb{R}$. A topological manifold is it metrizable, i.e. can its topology be described by a distance (may we use an atlas and $\mathbb{R}^n$ euclidean distance)?
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.
 If yes, which distance? If not, what is then the topology of space time?
See above.

 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?
No need, it so happens that differentiable manifolds always admit a Riemannian metric.
P: 3,043
 Quote by George Jones I make a distinction between "Riemannian" and "semi-Riemannian". What I wrote only applies to Riemannian manifolds.

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.
 Sci Advisor P: 8,655 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?
C. Spirit
Thanks
P: 5,633
 Quote by atyy Is a Hausdorff space necessarily a metric space?
The long line is Hausdorff but not a metric space because if it was a metric space then the fact that it is sequentially compact would imply it would be compact as well but the long line is not compact (it isn't even Lindelof).
P: 8,655
 Quote by WannabeNewton The long line is Hausdorff but not a metric space because if it was a metric space then the fact that it is sequentially compact would imply it would be compact as well but the long line is not compact (it isn't even Lindelof).
So there is no need for a Hausdorff pseudo-Riemannian manifold to be a metric space (ie. is it a red herring to be concerned about metric spaces in GR)?
C. Spirit
Thanks
P: 5,633
 Quote by atyy So there is no need for a Hausdorff pseudo-Riemannian manifold to be a metric space (ie. is it a red herring to be concerned about metric spaces in GR)?
Well manifolds are metrizable so in principle you can endow the manifold with a metric. The pseudo - Riemannian structure won't change that because the proof that topological manifolds are metrizable is, as stated, for topological manifolds which don't have any prescribed pseudo - Riemannian structure or Riemannian structure if that is what you are asking. I don't think it particularly matters in the context of GR because I've never seen a metric (as opposed to the metric tensor) ever being used in any textbook I've seen. Someone else could probably comment on that.
P: 123
 Quote by WannabeNewton Well manifolds are metrizable so in principle you can endow the manifold with a metric. The pseudo - Riemannian structure won't change that because the proof that topological manifolds are metrizable is, as stated, for topological manifolds which don't have any prescribed pseudo - Riemannian structure or Riemannian structure if that is what you are asking. I don't think it particularly matters in the context of GR because I've never seen a metric (as opposed to the metric tensor) ever being used in any textbook I've seen. Someone else could probably comment on that.
So we can endow spacetime itself with a metric distance, but we don't usually do it. Why not?

I mean, we do have an invariant $ds^2$, but this may be positive, negative or null. Then, what is done in physics is that we define our distances simply by
$$\int_a^bds=\int_a^b\sqrt{\pm g_{\mu\nu}dx^{\mu}dx^{\nu}}\geq 0$$
and this only makes sense if b is inside the lightcone defined by a, so that (by using the proper convention sign $\pm$) we always get a distance properly defined. This may be a simple trick by using the lightcone, but it can be supported by causality arguments that you want to include in our mode, I think. Does this seems good to you? I have some problems, as I'm thinking of the lightcone as being on the manifold, and not in the tangent space as has been argued.

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