How Does the Topology of Spacetimes Influence the Structure of Curved Manifolds?

[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)?

 

[WannabeNewton]

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.

 

[kevinferreira]

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
<br /> \int_a^bds=\int_a^b\sqrt{\pm g_{\mu\nu}dx^{\mu}dx^{\nu}}\geq 0<br />
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.

 

[WannabeNewton]

Why not?

I don't know Kevin; someone else would have to answer that.

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.

The terminology makes things ambiguous. Hawking and Elis clears this stuff up pretty nicely I would say. The null cone is a subset of the tangent space. The image of the null cone under the exponential map is the set of all null geodesics in M going through p which is of course a subset of M.

 

[micromass]

So we can endow spacetime itself with a metric distance, but we don't usually do it. Why not?

My guess is that the distance is just not a very useful one as it won't agree with the pseudo-Riemannian metric.

For example, take the sphere in \mathbb{R}^3. We can endow this with a metric as follows. Take two points on the sphere, draw a straight line through those points and measure the length of the line. So we take the distance on \mathbb{R}^3 and restrict it to the sphere. This defines a good distance on the sphere that agrees with the topology. However, this distance is not a very useful one as it relies on the embedding in \mathbb{R}^3.
What we want is a distance on the sphere that measures the length of the paths on the sphere. So we don't want a distance that comes from straight lines (which are not on the sphere), but rather a distance that comes from path (=great circles) on the sphere. This distance is a distance coming from a metric tensor and this is much more useful.

In the same way, we can endow a distance on a spacetime. But nothing tells us that this distance actually has a physical significance or that it agrees with some metric tensor.

 

[Fredrik]

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
<br /> \int_a^bds=\int_a^b\sqrt{\pm g_{\mu\nu}dx^{\mu}dx^{\nu}}\geq 0<br />
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.

You don't integrate "from a to b", you integrate along a curve. This way you can define the length of a spacelike curve. You can also (by changing the sign under the square root in the definition) use this method to define the "length" of a timelike curve, but we call it "proper time", not "length". Since there are always infinitely many spacelike curves connecting two given spacelike separated events, and infinitely many timelike curves connecting two given timelike separated events, this doesn't immediately lead to a well-defined notion of "distance" between the two events. You could try to define the distance between two events as the length or proper time along a geodesic connecting the two events. But I think that in some spacetimes, there can be many such geodesics. And even in spacetimes where the geodesics are unique, you have to deal with events that are null separated from each other. I doubt that there's a way to define the distance between those that would give you a distance function that satisfies the requirements in the definition, like the triangle equality.

It seems to me that there is a meaningful notion of "lightcone" on the manifold as well. (Not sure what the standard terminology is though). This would be the union of all the timelike and null geodesics through the given point.

 

[micromass]

Is a Hausdorff space necessarily a metric space?

No, wbn gave the counterexample of the long line. This is a non-metric space that is Hausdorff.

Wikipedia just says thart pseudometric spaces are typically not Hausdorff, but that seems to allow that Hausdorff spaces can be neither metric nor pseudometric.

A pseudometric space is actually Hausdorff if and only if it is a metric space. So a pseudometric space that is not a metric space can never be Hausdorff.

If that is possible, then wouldn't it be possible that Hausdorff manifolds with pseudo-Riemannian metric tensors need not be metric spaces?

Well, a manifold is usually defined as a topological space that is

• Locally Euclidean
• Hausdorff
• Second countable

It can be proven (but the proof is not easy by far), that if we have these three topological conditions, then our space is metrizable. So we can always find a metric. Moreover, we can always embed our manifold in \mathbb{R}^n.

So even without a smooth structure or a metric tensor, we already have that our manifold is metrizable. Again: the metric of the metric space might not be physical or might not have anything to do with a metric tensor!

If we drop one of the conditions from our list, then the space is not metrizable anymore. For example, if we would define a manifold as just locally euclidean and Hausdorff, then it might not be metrizable (as the long line shows). If we define a manifold as just locally euclidean and second countable, then it might also not be metrizable (as the line with two origins shows). In fact: it might not even be pseudo-metrizable.

 

[atyy]

So even without a smooth structure or a metric tensor, we already have that our manifold is metrizable. Again: the metric of the metric space might not be physical or might not have anything to do with a metric tensor!

Would it be right to paraphrase this way: you could put a metric on a Hausdorff pseudo-Riemannian manifold (eg. via a Riemannian metric tensor or some other means not involving a metric tensor at all), but it is physically irrelevant ?

 

[micromass]

Would it be right to paraphrase this way: you could put a metric on a Hausdorff pseudo-Riemannian manifold (eg. via a Riemannian metric tensor or some other means not involving a metric tensor at all), but it is physically irrelevant ?

I think that is correct.

 

[WannabeNewton]

This would be the union of all the timelike and null geodesics through the given point.

Hi Fredrik! Correct me if I'm wrong but I'm pretty sure the "light cone" itself is just the set of all null geodesics through p and the interior consists of the time - like geodesics.

 

[zonde]

For most situations, spacetime Hausdorffness seems to be a reasonable, physical separation axiom.

No. In relativity it is not reasonable to believe that spacetime events connected with null geodesics are separable. Or let's rather say that their separability does not depend on spacetime properties but rather on distribution of content within spacetime. There is a lot of matter around one particular state of motion and that determines separability of events not spacetime properties.

 

[micromass]

No. In relativity it is not reasonable to believe that spacetime events connected with null geodesics are separable. Or let's rather say that their separability does not depend on spacetime properties but rather on distribution of content within spacetime. There is a lot of matter around one particular state of motion and that determines separability of events not spacetime properties.

Well, then I guess that Wald's textbook must be completely wrong. Do you think so? Since Wald seems to let spacetimes be Hausdorff...

 

[WannabeNewton]

No. In relativity it is not reasonable to believe that spacetime events connected with null geodesics are separable. Or let's rather say that their separability does not depend on spacetime properties but rather on distribution of content within spacetime. There is a lot of matter around one particular state of motion and that determines separability of events not spacetime properties.

I'm not sure if you are understanding what it means for a topological space to be Hausdorff. Sure two events connected by a null geodesic represent a light pulse being able to get from one to the other but what does that have to do with Hausdorff? Hausdorff simply states there exist a pair of neighborhoods, for the two (distinct) events, that are disjoint but you seem to be thinking that this implies we could not anymore connect the two events with the aforementioned null geodesic. If the null geodesic connects the two events then that is that; the Hausdorff property won't break anything.

 

[TrickyDicky]

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
<br /> \int_a^bds=\int_a^b\sqrt{\pm g_{\mu\nu}dx^{\mu}dx^{\nu}}\geq 0<br />
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.

So let's make a distinction between the different tangent vectors (timelike,null, spacelike) in the tangent space at a point of a manifold with a pseudoRiemannian metric tensor field, that define the structure of a light cone in the tangent space, versus the different paths in a manifold that are also called timelike, spacelike or null according to what the tangent vector is at every point in the curve.
As you notice, in physics the length of the curve is always computed as if the tangent vector at every point where inside the light cone(in the limit at infinity) , regardless of what it is called, i.e. photon's null paths are never considered to have null length. There are no physical examples of spacelike paths so we can leave those out for now.
This is the logic thing to do since after all we are working with a smooth manifold that doesn't alter its topology nor its distance function(in the Riemannian manifold case) by the introduction of a pseudoRiemannian metric tensor.
The only problem I see is that this seems to be forgotten when applying GR to specific solutions of the EFE.

 

[TrickyDicky]

Would it be right to paraphrase this way: you could put a metric on a Hausdorff pseudo-Riemannian manifold (eg. via a Riemannian metric tensor or some other means not involving a metric tensor at all), but it is physically irrelevant ?

I'm not sure what you mean here, but I'd say the metric(distance function) is never physically irrelevant in GR. One of the pillars of the theory is the invariance of length across arbitrarily long distances, think of cosmological redshifts. If we didn't care about metrics (distances) in GR there would be no need for a curvature concept or unique connections.

 

[Fredrik]

Hi Fredrik! Correct me if I'm wrong but I'm pretty sure the "light cone" itself is just the set of all null geodesics through p and the interior consists of the time - like geodesics.

That seems to make more sense than what I said, and Wikipedia agrees with you. This sort of thing happens a lot when I post just before going to bed.

 

[atyy]

I'm not sure what you mean here, but I'd say the metric(distance function) is never physically irrelevant in GR. One of the pillars of the theory is the invariance of length across arbitrarily long distances, think of cosmological redshifts. If we didn't care about metrics (distances) in GR there would be no need for a curvature concept or unique connections.

In cosmology, 4D spacetime is cut into 3D spatial slices that change with time. On the 4D spacetime, there is a pseudo-Riemannian metric tensor, and no physically relevant metric space metric. On each 3D spatial slice there is a Riemannian metric tensor, which can be used to define a metric space metric.

It seems to me that there is a meaningful notion of "lightcone" on the manifold as well. (Not sure what the standard terminology is though). This would be the union of all the timelike and null geodesics through the given point.

Hi Fredrik! Correct me if I'm wrong but I'm pretty sure the "light cone" itself is just the set of all null geodesics through p and the interior consists of the time - like geodesics.

That seems to make more sense than what I said, and Wikipedia agrees with you. This sort of thing happens a lot when I post just before going to bed.

"achronal boundary"? "chronological future" or "timelike future"?

http://www.math.miami.edu/~galloway/beijing.pdf
http://en.wikipedia.org/wiki/Causal_structure

 

[stevendaryl]

... since after all we are working with a smooth manifold that doesn't alter its topology nor its distance function(in the Riemannian manifold case) by the introduction of a pseudoRiemannian metric tensor.
The only problem I see is that this seems to be forgotten when applying GR to specific solutions of the EFE.

What do you mean? In what sense does anything done in GR contradict the claim that the topology isn't altered by introducing a metric tensor?

 

[TrickyDicky]

In cosmology, 4D spacetime is cut into 3D spatial slices that change with time. On the 4D spacetime, there is a pseudo-Riemannian metric tensor,

Ok.

and no physically relevant metric space metric.

Do you really think the spacetime invariant interval between events is physically irrelevant? That's odd.

 

[TrickyDicky]

What do you mean? In what sense does anything done in GR contradict the claim that the topology isn't altered by introducing a metric tensor?

For instance, the topology of a smooth manifold (that by definition has no discontinuities) is altered by introducing singularities based precisely on the peculuarities of the pseudoriemannian metric tensor.

 

[micromass]

For instance, the topology of a smooth manifold (that by definition has no discontinuities) is altered by introducing singularities based precisely on the peculuarities of the pseudoriemannian metric tensor.

I don't get this. A manifold has a topology. Only then do we introduce a metric tensor. So the metric tensor is an extra structure.
How can an extra structure possibly change the topology of a manifold??