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## The topology of spacetimes

 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).

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 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)?

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 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.

 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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 Quote by kevinferreira 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.

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 Quote by kevinferreira 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.

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 Quote by kevinferreira 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.
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.

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 Quote by atyy 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.

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 Quote by micromass 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 ?

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 Quote by atyy 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.

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 Quote by Fredrik 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.

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 Quote by George Jones 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 lets 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.

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 Quote by zonde No. In relativity it is not reasonable to believe that spacetime events connected with null geodesics are separable. Or lets 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...

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 Quote by zonde No. In relativity it is not reasonable to believe that spacetime events connected with null geodesics are separable. Or lets 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.

 Quote by kevinferreira 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.
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.

 Quote by atyy 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.

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 Quote by WannabeNewton 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.

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