Some doubts concerning the mathematical bases of GR

[TrickyDicky]

I'm a little confused about certain assumptions usually made in GR as to how rigorous they are mathematically speaking.
For instance the assumption generally presented without proof that GR spacetime manifold is a Hausdorff space seems not to be warranted given the fact that pseudometric spaces (with no definite positive metric) are not Hausdorff. Why make that assumption then?
On the other hand the defining property of GR was explaining gravity thru curvature as an invariant, but Lorentzian manifolds, precisely due to their not being metric spaces, may be both flat and curved depending on what patch is chosen, in other words curvature is not a property of the manifold alone.
Finally the assumption that the GR manifold is smooth seems to be contradicted by the existence of singularities, the condition usually imposed that one must only look at the space and time intervals that are singularity free doesn't seem a very rigorous mathematical prescription.

 

[Dickfore]

Please justify the following claims:

...given the fact that pseudometric spaces (with no definite positive metric) are not Hausdorff.

...but Lorentzian manifolds, precisely due to their not being metric spaces, may be both flat and curved depending on what patch is chosen, in other words curvature is not a property of the manifold alone.

 

[martinbn]

I'm a little confused about certain assumptions usually made in GR as to how rigorous they are mathematically speaking.
For instance the assumption generally presented without proof that GR spacetime manifold is a Hausdorff space seems not to be warranted given the fact that pseudometric spaces (with no definite positive metric) are not Hausdorff. Why make that assumption then?

There is a brief discussion about this in Hawking-Ellis, plus some non-Hausdorff examples. I don't remember what the reason was to not consider them, but it was definitely physics not mathematics.

On the other hand the defining property of GR was explaining gravity thru curvature as an invariant, but Lorentzian manifolds, precisely due to their not being metric spaces, may be both flat and curved depending on what patch is chosen, in other words curvature is not a property of the manifold alone.

The first part is not true. It is true that curvature is not a property of the manifold. You need a connection. In this case the Levi-Chevita connection associated with the matric.

Finally the assumption that the GR manifold is smooth seems to be contradicted by the existence of singularities, the condition usually imposed that one must only look at the space and time intervals that are singularity free doesn't seem a very rigorous mathematical prescription.

That is also not true. In this context the smooth manifold is singular if it is geophysically incomplete (or some version of it).

 

[Dale]

given the fact that pseudometric spaces (with no definite positive metric) are not Hausdorff.

I agree with Dickfore, I think that this is incorrect. A neighborhood in the Hausdorff sense is a topological concept, which is more primitive than the metric. So it is not defined by the metric, so I don't know why having or not having a specific kind of metric would have anything to do with whether or not a space is Hausdorff.

For any spacetime manifold M with two distinct points x and y there exist open subsets X and Y such that:
x \in X \subseteq M
y \in Y \subseteq M
X \cap Y = \emptyset
so spacetime manifolds in GR certainly seem to be Hausdorff to me.

 

[TrickyDicky]

Please justify the following claims:

The first claim can be read in the wikipedia entry for Hausdorff spaces where it says "pseudometric spaces typically are not Hausdorff" in the paragraph about examples and counterexamples.
The second just meant that the pseudometric space induced topological structure makes curvature depend on the topology.

 

[TrickyDicky]

I agree with Dickfore, I think that this is incorrect. A neighborhood in the Hausdorff sense is a topological concept, which is more primitive than the metric. So it is not defined by the metric, so I don't know why having or not having a specific kind of metric would have anything to do with whether or not a space is Hausdorff.

For any spacetime manifold M with two distinct points x and y there exist open subsets X and Y such that:
x \in X \subseteq M
y \in Y \subseteq M
X \cap Y = \emptyset
so spacetime manifolds in GR certainly seem to be Hausdorff to me.

The metric alluded to in the OP is not exactly the metric tensor but the distance function obtained by integrating the metric tensor.

The example you give about the distinct points x, y, don't seem to apply when the points lie in a null geodesic.

 

[Dickfore]

How about this one:
Metric identification

 

[Matterwave]

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.

 

[Dale]

The metric alluded to in the OP is not exactly the metric tensor but the distance function obtained by integrating the metric tensor.

OK, but you don't need a distance function to define a neighborhood in the Hausdorff sense. You don't need any notion of distance at all since it is only required that the neighborhoods be open subsets. It is not required that the neighborhoods be open balls.

The example you give about the distinct points x, y, don't seem to apply when the points lie in a null geodesic.

I am not sure what you are trying to explain here. A path (whether null or geodesic or not) does not define an open subset, so it doesn't define a neighborhood.

 

[Dale]

The first claim can be read in the wikipedia entry for Hausdorff spaces where it says "pseudometric spaces typically are not Hausdorff" in the paragraph about examples and counterexamples.

Interesting. I don't understand that comment, but I don't consider myself expert enough to correct or remove it.

 

[TrickyDicky]

How about this one:
Metric identification

What about it?
Surely if you relax the condition that makes a metric space be a pseudometric space, you get a metric space, so?

 

[PAllen]

I think the confusion here is between pseudometric space and semi-riemannian manifold. The former uses the non-positive definite distance function part of the definition of its topology (the open sets). That latter does not, and is defined on top of differentiable manifold, which is Hausdorff as part of its definition. This is better covered (if you must use Wikipedia) in its articles on Differentiable Manifolds:

http://en.wikipedia.org/wiki/Differentiable_manifold

and also in the article on Pseudo-Riemannian manifolds:

http://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold

which refers defines it on top of differentiable manifold.

All this by way of verifying that Matterwave's post was a complete answer to the question, that unfortunately was mostly ignored.

 

[PAllen]

Here is another reference clarifying that for a pseudometric space the topology is induced by the metric. This the opposite of a pseudo-riemannian manifold, where metric is a structure added on top of a differentiable manifold which is already Hausdorff. The reference confirms the a pseudometric space need not be Hausdorff. Obviously, pseudometric spaces have nothing to do with GR.

http://igitur-archive.library.uu.nl/dissertations/1916832/c2.pdf

 

[TrickyDicky]

I think the confusion here is between pseudometric space and semi-riemannian manifold. The former uses the non-positive definite distance function part of the definition of its topology (the open sets). That latter does not, and is defined on top of differentiable manifold, which is Hausdorff as part of its definition. This is better covered (if you must use Wikipedia) in its articles on Differentiable Manifolds:

http://en.wikipedia.org/wiki/Differentiable_manifold

and also in the article on Pseudo-Riemannian manifolds:

http://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold

which refers defines it on top of differentiable manifold.

All this by way of verifying that Matterwave's post was a complete answer to the question, that unfortunately was mostly ignored.

You introduce something that could help clarify things a bit.
Could you specify how exactly a pseudo-riemannian manifold (in thi case a Lorentzian one) is not a pseudometric space?
Consider points on the light cone for instance, the distance betwen two different points can be zero, right?
Wrt the comment about all differentiable manifolds being Hausdorff, that is true in a limited sense, that is, it is true locally, forgetting about the topology. But being a Hausdorff space is usually considered a global property of a space, and here we find the problem that a manifold with singularities is not differentiable globally.

Other important limitations of pseudoriemannian manifolds are listed in the wikipedia entry

" On the other hand, there are many theorems in Riemannian geometry which do not hold in the generalized case. For example, it is not true that every smooth manifold admits a pseudo-Riemannian metric of a given signature; there are certain topological obstructions. Furthermore, a submanifold does not always inherit the structure of a pseudo-Riemannian manifold; for example, the metric tensor become zero on any light-like curve".

 

[George Jones]

Could you specify how exactly a pseudo-riemannian manifold (in thi case a Lorentzian one) is not a pseudometric space?

How, exactly, is a Riemannian manifold a metric space?

 

[TrickyDicky]

How, exactly, is a Riemannian manifold a metric space?

I believe a Riemannian manifold is a metric space because its metric is positive definite.
I think a pseudo-Riemannian manifold is a pseudometric space because its metric is not positive definite and two different points x and y in a null cone can have zero distance. Is this wrong?

 

[George Jones]

I believe a Riemannian manifold is a metric space because its metric is positive definite.

A metric space (X , d) is a set X together to together with a function d that maps pairs of elements of X to real numbers (the distances between the elements of the pairs).

For a Riemannian manifold (M , g), g is function that maps pairs of tangent vectors to real numbers, i.e., g doesn't map pairs of elements of M to real numbers. Consequently, a Riemannian manifold (M , g) is not a metric space in the sense in which you are using the term "metric space".

This is an unfortunate and confusing clash of terminology: "metric" in "Riemannian metric" and "metric" in "metric space" mean different things.

 

[TrickyDicky]

Btw, there seems to be a mistake in the wikipedia entry on differentiable manifolds, they present topological manifolds as hausdorff, but if one reads Hawking and Ellis "The large scale structure of spacetime, they devote a few pages to discuss non-Hausdorff manifolds.

 

[PAllen]

You introduce something that could help clarify things a bit.
Could you specify how exactly a pseudo-riemannian manifold (in thi case a Lorentzian one) is not a pseudometric space?
Consider points on the light cone for instance, the distance betwen two different points can be zero, right?
Wrt the comment about all differentiable manifolds being Hausdorff, that is true in a limited sense, that is, it is true locally, forgetting about the topology. But being a Hausdorff space is usually considered a global property of a space, and here we find the problem that a manifold with singularities is not differentiable globally.

Other important limitations of pseudoriemannian manifolds are listed in the wikipedia entry

" On the other hand, there are many theorems in Riemannian geometry which do not hold in the generalized case. For example, it is not true that every smooth manifold admits a pseudo-Riemannian metric of a given signature; there are certain topological obstructions. Furthermore, a submanifold does not always inherit the structure of a pseudo-Riemannian manifold; for example, the metric tensor become zero on any light-like curve".

I believe you are mixing up different fields of mathematics. That is also (I think) what George Jones is hinting at.

As I understand it:

- A metric space is just a set with a function on pairs of elements in it meeting certain properties. It is in the field of point-set topology, not differential geometry. Similarly for a pseudometric space. The key is that it's definition starts from set not topological space or manifold.

- Orthogonal to these concepts, there is a hiearchy of constructs with increase structure as follows:
topological space -> hausdorff space -> manifold -> riemannian or pseudo-reimannian manifold.

Topological space starts from imposing an arbitrary collection of subsets meeting certain properties such that we may call them 'the open sets'.

The quote you refer to above simply makes that point that any (smooth) manifold can be made riemannian, while some maniolds cannot be made pseudo-rieamannian.

 

[TrickyDicky]

A metric space (X , d) is a set X together to together with a function d that maps pairs of elements of X to real numbers (the distances between the elements of the pairs).

For a Riemannian manifold (M , g), g is function that maps pairs of tangent vectors to real numbers, i.e., g doesn't map pairs of elements of M to real numbers. Consequentl, a Riemannian manifold (M , g) is not a metric space in the sense in which you are using the term "metric space".

This is an unfortunate and confusing clash of terminology: "metric" in "Riemannian metric" and "metric" in "metric space" mean different things.

George, I am aware of the distinction between a metric tensor g and a metric as referred to in "metric space"( I even mentioned the difference in a previous post).
Every time I wrote metric I referred to the distance function, not the metric tensor.
So I am not yet clear if you agree a Riemannian manifold is a metric space, and a Lorentzian manifold is not.

 

[PAllen]

Btw, there seems to be a mistake in the wikipedia entry on differentiable manifolds, they present topological manifolds as hausdorff, but if one reads Hawking and Ellis "The large scale structure of spacetime, they devote a few pages to discuss non-Hausdorff manifolds.

This is covered here:

http://en.wikipedia.org/wiki/Non-Hausdorff_manifold

 

[George Jones]

Okay, I am finding this thread to be very confusing. Why should a pseudo-Riemannian manifold be a metric space? As Matterwave and PAllen have pointed out, pseudo-Riemannian manifolds are topological spaces via the manifold topology, even if they don't have natural metrics.

What do you want to do with a metric? Why does this introduce doubts about the mathematical basis of GR?

 

[PAllen]

George, I am aware of the distinction between a metric tensor g and a metric as referred to in "metric space"( I even mentioned the difference in a previous post).
Every time I wrote metric I referred to the distance function, not the metric tensor.
So I am not yet clear if you agree a Riemannian manifold is a metric space, and a Lorentzian manifold is not.

Seems to me you could have metric space that cannot even be a manifold (e.g. it is defined on a finite set). Similarly, a disconnected Riemannian manifold would not allow introduction of a global distance function based on integrated metric distance. Maybe every complete Riemannian manifold can be treated as a metric space.

 

[TrickyDicky]

Okay, I am finding this thread to be very confusing. Why should a pseudo-Riemannian manifold be a metric space?


?

Sorry, I thought my questions were clear, my fault.
But I'm not saying a pseudo-Riemannian manifold is a metric space, my claim is that it is not.
I can see that a metric space is not a Riemannian manifold but I had the notion a Riemannian manifold had the property of being a metric space, am I wrong?


Similarly a pseudometric space is not a pseudo-Riemannian manifold but my belief is that pseudo-Riemannian manifolds have as defining property their pseudometricity(that is the key difference wrt Riemannian manifolds) , is this also not correct?
Thanks for your help and sorry again about the misunderstanding?

 

[TrickyDicky]

Maybe I should add that I am only considering connected manifolds.

 

[PAllen]

Sorry, I thought my questions were clear, my fault.
But I'm not saying a pseudo-Riemannian manifold is a metric space, my claim is that it is not.
I can see that a metric space is not a Riemannian manifold but I had the notion a Riemannian manifold had the property of being a metric space, am I wrong?

See my earlier post for an example of metric space that is not a manifold and Riemannian manifold that cannot be treated as a metric space (at least not by building up from the metric tensor).

Similarly a pseudometric space is not a pseudo-Riemannian manifold but my belief is that pseudo-Riemannian manifolds have as defining property their pseudometricity(that is the key difference wrt Riemannian manifolds) , is this also not correct?
Thanks for your help and sorry again about the misunderstanding?

This has the same confusion of categories as the prior case.

[edit: and there may be additional problems. The pseudo-metric tensor does not, in general, lead to any unique mechanism for constructing a global pseudometric function of points. For example, for some topologies, you could have both spacelike path and a timelike path between two points. Then what pseudo-metric global function value do you assign? The minimum among spacelike paths or the minimum among timelike paths? ]

 

[TrickyDicky]

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.

 

[George Jones]

As PAallen has said, given a connected (need not be complete, though) Riemannian manifold (M,g), g can be used to define a d (in a natural way) such that (M,d) is a metric space.

When this is done, the metric topology and the manifold topology are the same.

I have never worked with pseudometric spaces (that I remember), but, off the top of my head, I don't think a corresponding statement can be made about semi-Riemannian manifolds and pseudo metric spaces. I don't think the statement

Given a connected semi-Riemannian manifold (M,g), g can be used to define a d (in a natural way) such that (M,d) is a pseudometric space.

can be made. I could very well be wrong.

 

[TrickyDicky]

See my earlier post for an example of metric space that is not a manifold

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.

and Riemannian manifold that cannot be treated as a metric space (at least not by building up from the metric tensor).

I added the connectedness assumption for Riemannian manifolds.

 

[TrickyDicky]

As PAallen has said, given a connected (need not be complete, though) Riemannian manifold (M,g), g can be used to define a d (in a natural way) such that (M,d) is a metric space.

When this is done, the metric topology and the manifold topology are the same.

I have never worked with pseudometric spaces (that I remember), but, off the top of my head, I don't think a corresponding statement can be made about semi-Riemannian manifolds and pseudo metric spaces. I don't think the statement


can be made. I could very well be wrong.

Thanks George.

 

[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.