Some doubts concerning the mathematical bases of GR

[PAllen]

Pallen, do you agree that a Lorentzian manifold has a pseudometric space structure?
Furthermore it also has a semimetric space structure due to their triangle inequality axiom being the reverse of the usual.

No, because of the problem of no unique way to go from semi-riemannian metric tensor to global pseudometric function.

The concepts seem completely orthogonal in this case. A pseudometric space need not even be a topological space (let alone a manifold, or Hausdorff). A pseudo-Riemannian manifold is necessarily Hausdorff (by normal definitions of manifold), but there is no natural way to define a global pseudometric function (at minimum, several definitions would need to be added, as well - I am pretty sure - additional topological restrictions (beyond connectedness)).

To clarify this, I believe I can construct connected, pseudo-riemannian manifods such that there exist points connected both by a spacelike geodesic and a timelike geodesic. How then, do you define the global pseudometric function of points?

 

[PAllen]

Yes, that is why I wrote that a metric space is not the same as a manifold, I'm not sure if you are reading my answers.

I added the connectedness assumption for Riemannian manifolds.

It's the classic question of timing - composing posts while you are composing, but mine post later.

 

[PAllen]

I think there is an even more basic issue going pseudo-Riemannian to pseudometric.

Pseudometric simply allows a global distance function that is zero. It has no concept of spacelike versus timelike (e.g. positive or negative interval squared). Thus, I think (despite the name similarity), there is no meaningful connection between pseudometric spaces and pseudoriemannian manifolds.

 

[TrickyDicky]

The concepts seem completely orthogonal in this case. A pseudometric space need not even be a topological space (let alone a manifold, or Hausdorff).

I thought we had clarified this misunderstanding, I'm not saying anything about a pseudometric space being a manifold. It is the other way around, why do you keep bringing it up?

A pseudo-Riemannian manifold is necessarily Hausdorff (by normal definitions of manifold),

It is explained in the wikipedia link you provided that the "normal" definition ignores the general topology.

To clarify this, I believe I can construct connected, pseudo-riemannian manifods such that there exist points connected both by a spacelike geodesic and a timelike geodesic. How then, do you define the global pseudometric function of points?

This doesn't seem to be connected to what I have been talking about.
First you would have to address in what sense two different points in a null light cone in a Lorentzian manifold can have zero distance between them (the definition of pseudometric space according to wikipedia) and not have pseudometric space structure.

 

[PAllen]

It is explained in the wikipedia link you provided that the "normal" definition ignores the general topology.

The normal definition involving second countable hausdorff assumption is the one used 99% of the time.

This doesn't seem to be connected to what I have been talking about.
First you would have to address in what sense two different points in a null light cone in a Lorentzian manifold can have zero distance between them (the definition of pseudometric space according to wikipedia) and not have pseudometric space structure.

A pseudometric space requires a global function of two points satsifying some axioms. I have explained that (unlike the Riemannian -> metric space case) there is no unique natural definition of this function at all.

 

[TrickyDicky]

The normal definition involving second countable hausdorff assumption is the one used 99% of the time.

Maybe even 99.9%, but that doesn't prove anything mathematically.

A pseudometric space requires a global function of two points satsifying some axioms. I have explained that (unlike the Riemannian -> metric space case) there is no unique natural definition of this function at all.

Maybe, but that would mean even less chance to be Hausdorff, as we get even farther from the metric space structure.

 

[TrickyDicky]

Let's stick to the definition of Hausdorff space.
"Points x and y in a topological space Xcan be separated by neighbourhoods if there exists a neighbourhood Uof x and a neighbourhood V of y such that Uand V are disjoint (U∩ V = ∅). X is a Hausdorff space if any two distinct points of X can be separated by
neighborhoods. "
How are distinct points on the light cone of a Lorentzian manifold separated by neighbourhoods if their distance is zero?

 

[PAllen]

Maybe even 99.9%, but that doesn't prove anything mathematically.Maybe, but that would mean even less chance to be Hausdorff, as we get even farther from the metric space structure.

A definition is not a proof. The definition of pseudo-riemannian manifold used by essentiallly all authors is based on the normal mathematical definition of manifold, which includes the requirement of 'second countable hausdorff'.

Your second point is total nonsense. If something is part of the definition, there is no 'chance not to be hausdorff'. If I define natural numbers as integers greater than zero, what is the chance of natural number < 0?

 

[PAllen]

Let's stick to the definition of Hausdorff space.
"Points x and y in a topological space Xcan be separated by neighbourhoods if there exists a neighbourhood Uof x and a neighbourhood V of y such that Uand V are disjoint (U∩ V = ∅). X is a Hausdorff space if any two distinct points of X can be separated by
neighborhoods. "
How are distinct points on the light cone of a Lorentzian manifold separated by neighbourhoods if their distance is zero?

A pseudo-riemannian manifold is first of all a manifold. The normal definition of manifold requires the hausdorff property before even defining a metric. This is getting ridiculous. The sequence of definitions is:

topological space -> Hausdorff -> manifold-> semi-riemannian manifold.

The open sets and Hausdorff property are defined without any reference to the metric, which is not even defined yet.

This is getting just ridiculous as you ignore everything matterwave, myself, and George Jones are saying.

 

[Dale]

Let's stick to the definition of Hausdorff space.
"Points x and y in a topological space Xcan be separated by neighbourhoods if there exists a neighbourhood Uof x and a neighbourhood V of y such that Uand V are disjoint (U∩ V = ∅). X is a Hausdorff space if any two distinct points of X can be separated by
neighborhoods. "
How are distinct points on the light cone of a Lorentzian manifold separated by neighbourhoods if their distance is zero?

This is the definition I was referring to above in post 4. Note that nowhere in that definition does it refer to distance, and that distances are not required to define a neighborhood in topological spaces. The facts that some points in the neighborhood have zero distance and some points outside of the neighborhood also have zero distance is not relevant.

 

[Haelfix]

How are distinct points on the light cone of a Lorentzian manifold separated by neighbourhoods if their distance is zero?

As others have explained, you are taking 3 separate mathematical structures and trying to mix them outside of their domains of definition in a single sentence.

'neighborhoods' refer to a concept in point set topology.

'Distance' typically involves concepts at the level of a metric (distance function) but in fact, can mean several different things even when that is understood.

A 'light cone' is yet even more structure, involving a specific choice of observer and a causal structure.

 

[DrGreg]

My understanding is that the underlying topology of a manifold as used in GR is nothing to do with the metric tensor, but is simply the topology inherited via the coordinate maps from Euclidean \mathbb{R}^4 space. I.e. an open set in the manifold is just the coordinate image of an open set in Euclidean \mathbb{R}^4 coordinate space.

 

[TrickyDicky]

My understanding is that the underlying topology of a manifold as used in GR is nothing to do with the metric tensor, but is simply the topology inherited via the coordinate maps from Euclidean \mathbb{R}^4 space. I.e. an open set in the manifold is just the coordinate image of an open set in Euclidean \mathbb{R}^4 coordinate space.

Yes, I know that is the usual assumption, the OP was asking if this assumption is rigorously backed mathematically, not only based on authority amd physical convenience.
We know the toplogy of manifold in GR is not necessarily R^4.
And on the other hand metrics induce topologies, a fact that is being ignored by most posters.

 

[TrickyDicky]

As others have explained, you are taking 3 separate mathematical structures and trying to mix them outside of their domains of definition in a single sentence.

'neighborhoods' refer to a concept in point set topology.

'Distance' typically involves concepts at the level of a metric (distance function) but in fact, can mean several different things even when that is understood.

A 'light cone' is yet even more structure, involving a specific choice of observer and a causal structure.

You don't think metrics can induce topological structures in pseudo-Riemannian manifolds?

 

[TrickyDicky]

This is the definition I was referring to above in post 4. Note that nowhere in that definition does it refer to distance, and that distances are not required to define a neighborhood in topological spaces. The facts that some points in the neighborhood have zero distance and some points outside of the neighborhood also have zero distance is not relevant.

We are not talking about topological spaces in general but about pseudo-Riemannian manifolds that are by definition endowed with a metric, that induces a certain topology on the manifold that imply distance.

 

[TrickyDicky]

A pseudo-riemannian manifold is first of all a manifold. The normal definition of manifold requires the hausdorff property before even defining a metric. This is getting ridiculous. The sequence of definitions is:

topological space -> Hausdorff -> manifold-> semi-riemannian manifold.

The open sets and Hausdorff property are defined without any reference to the metric, which is not even defined yet.

This is getting just ridiculous as you ignore everything matterwave, myself, and George Jones are saying.

You have the sequence wrong, at least according to Hawking and Ellis, whom I trust more than the wikipedia.
All your arguments are based on authority like percentages of authors and definitions without mathematical proof, that is ok in itself but ignores that in the OP I was asking for mathematical rigor rather than authority or physical convenience arguments.

 

[TrickyDicky]

This excerpt might help clarify some confusion that keeps being posted about metrics and metric tensors:

From the wikipedia entry on Metric:
" Important cases of generalized metrics In differential geometry, one considers metric tensors, which can be thought of as "infinitesimal" metric functions. They are defined as inner products on the tangent space with an appropriate differentiability requirement. While these are not metric functions as defined in this article, they induce metric functions by integration. A manifold with a metric tensor is called a Riemannian manifold. If one drops the positive definiteness requirement of inner product spaces, then one obtains a pseudo-Riemannian metric tensor, which integrates to a pseudo-semimetric . These are used in the geometric study of the theory of relativity, where the tensor is also called the "invariant distance"."

 

[TrickyDicky]

The topology of a manifold is determined by the local mappings to a Euclidean space and the standard topology in the Euclidean space (which is Hausdorff).

Therefore, all (topological) manifolds (of which differentiable manifolds are a subsection) are Hausdorff.

This has nothing to do with the metric.

Sorry to answer this late.
What you are saying applies to general differentiable manifolds rather than general manifolds, (see Hawking and Ellis "The large scale structure of spacetime").
And precisely what is being asked for in this thread is a rigorous mathematical proof (or a reference to it), that a pseudo-Riemannian manifold keeps that property, or if it looses it due to its pseudometricity.

 

[TrickyDicky]

A possible route to understand this is that as I commented usually (see singularity theory I am wikipedia)manifolds are defined as spaces without singularities(discontinuities), in the case of GR they occur thru degeneration of the manifold structure, precisely due to its pseudometricity.

 

[Dale]

We are not talking about topological spaces in general but about pseudo-Riemannian manifolds that are by definition endowed with a metric, that induces a certain topology on the manifold that imply distance.

That is not how I learned it. The metric does not induce the topology, the manifold does. At least, that was the way I learned GR, and it appears to be better than your approach for exactly the reason that you are running into here.

 

[TrickyDicky]

Yes, a topological manifold has a topological structure just by being a manifold, this in no way contradicts the fact that when a manifold has a metric, this metric induces the topology.

You are right that the issues I'm bringing up are not mentioned in the usual GR texts, that is why I'm asking for some mathematical proofs, not just convenient assumptions.

 

[Dale]

Yes, a topological manifold has a topological structure just by being a manifold, this in no way contradicts the fact that when a manifold has a metric, this metric induces the topology.

It sounds like a contradiction to me. If the manifold already has a topology then if you introduce a metric which has a topology then it seems that you could get situations where the manifold topology and the metric topology contradict each other. In fact, I think that is exactly this contradiction that is causing your concerns. The topology you are trying to induce via the metric includes all points on the light cone as indistinguishable from a given point, while the topolgy of the manifold distinguishes them. The latter defines the topology of a pseudo-Riemannian manifold, not the former.

On the other hand, if you believe (which I don't) that both the manifold and the metric induce separate topologies and that there is no contradiction in that fact, then simply use the manifold topology rather than the metric topology for defining the manifold as Hausdorff. Presumably, if you have both then you can use either wherever convenient.

 

[PAllen]

You have the sequence wrong, at least according to Hawking and Ellis, whom I trust more than the wikipedia.
All your arguments are based on authority like percentages of authors and definitions without mathematical proof, that is ok in itself but ignores that in the OP I was asking for mathematical rigor rather than authority or physical convenience arguments.

Proof is not relevant at this point. These are definitions. If one defines natural numbers as integers greater than zero, there is no such thing as proving natural numbers are positive.

It is true that there are two definitions of manifold, the common one and the more general one. Most books on GR base it on the common one. It is a definitional choice.

I gave wikipedia links because they are easy to find. However, my GR books that use manifolds all start from the common definition.

This link clarifies some things, and mentions Hawking and Ellis less common usage:

http://mathworld.wolfram.com/TopologicalManifold.html

This link clarifies the usual usage (e.g. that manifold assumes T2-space = hausdorff space property). See the description 'all manifolds'. This clearly means 'under the common definition', otherwise it would be wrong (as opposed to just being shorthand).

http://mathworld.wolfram.com/ParacompactSpace.html

Obviously, for their investigations, Hawking and Ellis have chosen to start from less common definitions. They deliberately start from a manifold that is not necessarily a topological manifold. As I don't have their book, I can't say much more on this.

So, trying to rephrase what the OP is possibly getting at:

If one uses the definitional scheme of Hawking and Ellis, it is then meaningful to ask about proving the Hausdorff property under some particular conditions. Other questions which are true by definition in the common definitional framework also become interesting.

At this point, having clarified that the OP specifically refers the Hawking and Ellis sheme, it would useful for a re-statement of the specific questions the OP wants to discuss.

Unfortunately, I can't contribute further, as I have only studied the more common framework and don't have a copy of Hawking and Ellis.

 

[PAllen]

This excerpt might help clarify some confusion that keeps being posted about metrics and metric tensors:

From the wikipedia entry on Metric:
" Important cases of generalized metrics In differential geometry, one considers metric tensors, which can be thought of as "infinitesimal" metric functions. They are defined as inner products on the tangent space with an appropriate differentiability requirement. While these are not metric functions as defined in this article, they induce metric functions by integration. A manifold with a metric tensor is called a Riemannian manifold. If one drops the positive definiteness requirement of inner product spaces, then one obtains a pseudo-Riemannian metric tensor, which integrates to a pseudo-semimetric . These are used in the geometric study of the theory of relativity, where the tensor is also called the "invariant distance"."

Note that this talks about integrating the pseudo-riemannian metric into a pseudo-semimetric not a pseudometric. Semi-metric is defined in the same article, and pseudo-semimetric is obvious by context. This validates my argument that a pseudometric structure cannot (generally) be imposed (by integration) on pseudo-riemannian manifold, because the pseudo-metric tensor does not integrate into pseudometric. Even for integrating to a pseudo-semimetric, one would have to add some definitions to specify the integration (e.g. minimum interval (any type - timelike, spacelike, or null) over all geodesics [parallel transport definition] connecting two points). It is clear this definition (which is not unique) would, indeed, produce a pseuo-semimetric but not either a semimetric or a pseudometric.

Thus, it is true, as I claimed, that there simply no connection between pseudo-Riemannian manifolds and psuedometric spaces. This is contrast to the fact that any connected Riemannian manifold can be treated as a metric space via natural, unique, integration.

 

[Matterwave]

The topology on a manifold should be induced from the mappings to the Euclidean space. I.e. open sets in the Euclidean space are mapped to open sets in the manifold.

I can always introduce a metric (or more precisely a distance function) to this manifold (since it's a set of points after all). I can also always introduce a distance function that would induce a topology that contradicts the topology induced by the mappings. I can introduce the trivial distance function d(m,n)=0 for all m and n, and this will certainly induce a non-hausdorff topology; however, this is not how we define a topology on a manifold. At least, I have always seen the manifold topology defined the first way and not this second, more convoluted way.

 

[micromass]

I don't see how a pseudometric could ever come into play here.

Every Riemannian manifold or pseudo-Riemannian manifold is by definition a manifold. A manifold is almost always taken to be a locally Euclidean, second countable Hausdorff space.

By Whitney's embedding theorem, a smooth manifold can always be embedded in \mathbb{R}^n for a suitable n. As such, a smooth manifold is metrizable. Thus every smooth manifold can be given the structure of a metric space.

If we define a pseudometric on a smooth manifold, then this pseudometric is always a metric. Indeed, a pseudometric space is a metric space if and only if it is Hausdorff.

It is true that there are non-Hausdorff manifolds, these are topological spaces that are locally Euclidean and second countable. But these are usually not regarded as topological manifolds.

Note that a non-Hausdorff manifold is not even necessarily pseudometrizable. For example, the line with two origins is perhaps the most famous example of a non-Hausdorff manifold. But this line with two origins is T_1 and not Hausdorff, therefore it cannot be pseudometrizable.

We can even go further, a non-Hausdorff manifold is always a T_1. Indeed: given two point x and y in our manifold M. Take a Euclidean neighborhood U of x. If y is not in our neighborhood, then U is a neighborhood of x that does not contain y and as such the T_1 axiom is satisfied. If y is in our neighborhood, then (since our neighborhood is locally Euclidean), we can find a smaller neighborhood around x which does not contain y. Again, then T_1-axiom is satisfied.

So a non-Hausdorff manifold is always T_1 and non-Hausdorff. As such, a non-Hausdorff manifold is never pseudo-metrizable. If it were pseudometrizable, then it would be either Hausdorff or not T_1.

So talking about pseudometrizable non-Hausdorff manifolds is useless, since there are no such things.

 

[TrickyDicky]

Thanks Dalespam, Pallen an matterwave for your interesting contributions, they're surely helpful.
Micromass, that is a great , really informative post, thanks , I value it even more coming from a mathematician rather than a physicist or relativist
that might be contaminated by old habits in their thinking about GR.

 

[TrickyDicky]

Note that this talks about integrating the pseudo-riemannian metric into a pseudo-semimetric not a pseudometric. Semi-metric is defined in the same article, and pseudo-semimetric is obvious by context. This validates my argument that a pseudometric structure cannot (generally) be imposed (by integration) on pseudo-riemannian manifold, because the pseudo-metric tensor does not integrate into pseudometric. Even for integrating to a pseudo-semimetric, one would have to add some definitions to specify the integration (e.g. minimum interval (any type - timelike, spacelike, or null) over all geodesics [parallel transport definition] connecting two points). It is clear this definition (which is not unique) would, indeed, produce a pseuo-semimetric but not either a semimetric or a pseudometric.

This is right, the non-positive definite metric tensor integrates to give a both pseudo- and semi-metric, this means it doesn't only relax the point separation axiom of metric spaces, but the triangle inequality axiom. Pseudosemimetric spaces are also tipically non-Hausdorff, but see the post by micromass.
We must bear in mind that the change of signature of the metric tensor (and therefore its distance function upon integration) is the only difference between a Riemannian and a Pseudo-Riemannian manifold. So anything that eliminates that difference makes them indistinguishable.

This is contrast to the fact that any connected Riemannian manifold can be treated as a metric space via natural, unique, integration.

See micromass answer.

 

[TrickyDicky]

...every smooth manifold can be given the structure of a metric space.

If we define a pseudometric on a smooth manifold, then this pseudometric is always a metric. Indeed, a pseudometric space is a metric space if and only if it is Hausdorff.
So talking about pseudometrizable non-Hausdorff manifolds is useless, since there are no such things.

These are both great points.
In the interest of rigor and as PAllen points out let's call the distance function in question a pseudosemimetric for the Lorentzian manifold used in GR case.
Now if defining a pseudosemimetric in a smooth manifold we actually define a metric, I guess because the smooth manifold topology is the one that rules, mathematically (at least topologically) how do we make a distinction between Riemannian and Pseudo-Riemannian manifolds if their only difference is in the metric tensor that in one case integrates to a metric and in the other to a pseudosemimetric?

 

[Ben Niehoff]

how do we make a distinction between Riemannian and Pseudo-Riemannian manifolds if their only difference is in the metric tensor that in one case integrates to a metric and in the other to a pseudosemimetric?

That is the only distinction. They are both manifolds. One has been given a positive-definite metric tensor; the other has been given an indefinite metric tensor.

Given any (connected) manifold, it is always possible to put a positive-definite metric tensor on it, which can always be integrated to a metric (taking the infimum of multiple geodesics, if need be).

Given any manifold, it is not always possible to put an indefinite-signature metric tensor on it. There are topological obstructions. For example, a torus can have a Lorentzian metric tensor, but a sphere cannot.