The Topology of Spacetimes: Exploring the Global Structure of Curved Manifolds

[micromass]

Let me try such question:
How do you define neighborhood if you have distance function that can give zero for two distinct points?

Let (X,d) be a pseudometric space. A set V is a neighborhood of $x\in X$ if there exists an $\varepsilon>0$ such that $B(x,\varepsilon)\subseteq V$.

 

[micromass]

Another good reference is "Fundamental of differential geometry" by Serge Lang. He covers pseudo-Riemannian metrics on page 175. It's a fun book to read, so I recommend it.

 

[zonde]

Let (X,d) be a pseudometric space. A set V is a neighborhood of $x\in X$ if there exists an $\varepsilon>0$ such that $B(x,\varepsilon)\subseteq V$.

What is $\varepsilon$ - a point or a set or an open set? And B()?

According to wikipedia http://en.wikipedia.org/wiki/Neighbourhood_(mathematics) neighborhood should contain an open set containing the point. Given spacetime properties neighborhood of any event in spcetime should include it's lightcones. But for any two distinct points there will be some place where their lightcones (future or past or future with past) will intersect. So they can't have disjoint neighbourhoods which is required to say they belong to Hausdorff space.

 

[WannabeNewton]

What is $\varepsilon$ - a point or a set or an open set? And B()?

According to wikipedia http://en.wikipedia.org/wiki/Neighbourhood_(mathematics) neighborhood should contain an open set containing the point. Given spacetime properties neighborhood of any event in spcetime should include it's lightcones. But for any two distinct points there will be some place where their lightcones (future or past or future with past) will intersect. So they can't have disjoint neighbourhoods which is required to say they belong to Hausdorff space.

A good book on topology would probably clear up much of the confusion here. Hausdorff property states there exists a pair of neighborhoods for two distinct points that are themselves disjoint; a point on a manifold will have multiple neighborhoods. As for your comment on the causal structure of space - time, please take a look at chapter 8 of Wald which should clear up confusion or something else if anyone else has another reference. In particular note that a light cone emanating from a point p on a space - time M is a subset of Tp(M) not M itself.

 

[atyy]

All the usual spacetimes are of course Hausdorff. But just for interest, Hawking and Ellis mention one example of a non-Hausdorff spacetime, and mention a paper by Hajicek.

 

[TrickyDicky]

A pseudo-Riemannian metric is a function

$$g:T_pM\times T_pM\rightarrow \mathbb{R}$$

for each p.

A pseudometric is a function

$$d:M\times M\rightarrow \mathbb{R}$$

So how can they be the same thing??

This is a key distinction IMO. The tangent space at a point and the manifold itself are two very different objects, and this is manifested even more clearly when the manifold is curved.
One shouldn't be able to draw conclusions about the global spacetime features from the purely local effect of the pseudoriemannian metric at a point, more so when the distance metric function that acts on the manifold doesn't coincide with the one that would be derived from the pseudoriemannian metric tensor, due to the smooth structure of the manifold.
When I mention the global structure of the manifold I refer to things like its maximal extended form, its singularities or its Killing vector fields nature(timelike, spacelike,lightlike).

 

[zonde]

If we say that spacetime is Hausdorff then we can't include complete lightcones in the neighborhood of an event. But then we should relay on some concept of nearness that is positive-definite and rather unrelated to spacetime distances.

It seems like a kind of double standard.

 

[micromass]

If we say that spacetime is Hausdorff then we can't include complete lightcones in the neighborhood of an event. But then we should relay on some concept of nearness that is positive-definite and rather unrelated to spacetime distances.

It seems like a kind of double standard.

Please read this again:

In particular note that a light cone emanating from a point p on a space - time M is a subset of Tp(M) not M itself.

 

[TrickyDicky]

Please read this again:

In particular note that a light cone emanating from a point p on a space - time M is a subset of Tp(M) not M itself.

Exactly, but a complete light cone structure is usually attributed in GR not only to the point p and its neighbourhood, but to the whole manifold. This is the double standard IMO.

 

[WannabeNewton]

Exactly, but a complete light cone structure is usually attributed in GR not only to the point p and its neighbourhood (TpM), but to the manifold. This is the double standard IMO.

Tp(M) is not a neighborhood it is the tangent space to M at p.

 

[zonde]

Please read this again:

In particular note that a light cone emanating from a point p on a space - time M is a subset of Tp(M) not M itself.

Done.

So what was the point? There is no analog of light cone on spacetime itself? And all spacetime distances are positive-definite? Or what?

 

[WannabeNewton]

Done.

So what was the point? There is no analog of light cone on spacetime itself? And all spacetime distances are positive-definite? Or what?

That is the subset of M generated by null geodesics emanating from p but you are talking about light cones as they relate to causal structure. Also, I'm not sure how you are concluding that the metric tensor must suddenly be positive - definite.

 

[micromass]

Done.

So what was the point? There is no analog of light cone on spacetime itself? And all spacetime distances are positive-definite? Or what?

I'm saying that the topology of $T_pM$ determined by the pseudo-Riemannian metric is of course non-Hausdorff. But this is a topology on $T_pM$ and not on M. The topology on M is Hausdorff and has nothing to do with the metric tensor.

 

[TrickyDicky]

Tp(M) is not a neighborhood it is the tangent space to M at p.

Yes, strictly you are right, but note that the whole justification of the concept of manifold depends upon the possibility of making the neighbourhood of a point and its tangent space "equivalent" in the sense of homeomorphic to R^n.

 

[zonde]

The topology on M is Hausdorff and has nothing to do with the metric tensor.

So we do relay on some positive-definite concept of nearness when we speak about topology of M, right?

 

[micromass]

So we do relay on some positive-definite concept of nearness when we speak about topology of M, right?

A topology has nothing to do with "positive-definiteness". Positive-definite is a property about inner products.

 

[TrickyDicky]

That is the subset of M generated by null geodesics emanating from p but you are talking about light cones as they relate to causal structure.

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]

If we say that spacetime is Hausdorff then we can't include complete lightcones in the neighborhood of an event.

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

 

[George Jones]

Fine there is someone else who thinks like you.
Now can you provide arguments? In that link there is only definition (belief) and no arguments.

Why do you believe that all spacetimes are Hausdorff?

All the usual spacetimes are of course Hausdorff. But just for interest, Hawking and Ellis mention one example of a non-Hausdorff spacetime, and mention a paper by Hajicek.

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]

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).

 

[Dale]

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?

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?

 

[micromass]

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.

 

[WannabeNewton]

Does that agree with your understanding?

Yessir.

 

[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–Rinow_theorem
But this is not an iff-condition. For example, (0,1) also has length-minimizing geodesics but is not complete.

 

[George Jones]

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.

[edit]Didn't see the previous post.[/edit]

 

[George Jones]

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/Books/Riemannian/errata.pdf

 

[George Jones]

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.

 

[kevinferreira]

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?

 

[TrickyDicky]

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, which 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.

 

[TrickyDicky]

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.

 

[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?

 

[WannabeNewton]

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).

 

[atyy]

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)?