Link btw manifolds and space-time

[quasar987]

Fact: Spacetime is a curved pseudo-Riemannian manifold with a metric of signature (-+++).

Fact: A manifold is a set together with a topology that is locally homeomorphic to R^n.

Question: In the case of space-time, what is the set, what is the topology and what is n?

 

[cesiumfrog]

Set: every event (even the boring ones)
n: 4
topology: isn't answering this what all the work is about?
Also, the word "curved" seems redundant in the context of a these many-folded things..

 

[quasar987]

What are events if not 4-tuples?

 

[cesiumfrog]

well.. you additionally must specify which coordinate-chart the 4-tuple of coordinates is associated with.

 

[pervect]

Well, the set in space-time is the set of all possible events. You can think of an event as a 4-tuple because for space-time, n=4.

This leaves the question of "what is the topology".

Let's talk about the topology of R, first.

R is the set of all real numbers.

A topological space (R,T) consists of the set R together with a collection of all subsets of R that satisfy the following three properties

1) The union of an arbitrary collection of subsets (often called open balls) each of which is in T, is also in T

2) The intersection of a finite number of subsets in T is also in T

3) R is in T, and so is the empty set

The usual topology of R is just (a,b), i.e. an "open ball" is defined as the set of points between a and b but not including the endpoints. Basically, the idea is that an open ball is the set of points "in the neighborhood" of some other point.

If we consider R^2, I think we can consider open balls to be either circular (like an actual ball) or square, but you might want to get a mathemetican to advise you more fullly if you care about the details.

You might also want to look up "homeomorphism", first one defines continuous maps as those maps f: X->Y such that every open set O in Y has an inverse image that is an open set in X. If f is continuous, one-one, onto, and its inverse is also continuous, f is a "homeomorphism".

 

[quasar987]

Hey pervect,

I am quite familiar with the maths of GR up to covariant derivatives of a general tensor field*. It is precisely how the link "math --> physics" is made that I'm wondering about.

So now you two have made it clear that space-time is indeed R^4, but with an a priori unknown topology, but be make the hypothesis that is it locally homeomorphic to R^4 with the...standard open-ball topology. Is that it?

*which I studied from "Relativity on Curved Manifold" by de Felice and Clarke. Do you know about it, and if so, what is your opinion of it?

 

[mjsd]

Fact: Spacetime is a curved pseudo-Riemannian manifold with a metric of signature (-+++).

Fact: A manifold is a set together with a topology that is locally homeomorphic to R^n.

So, I believe the connection is (which could be wrong of course)
Spacetime is a curved pseudo-Riemannian manifold ...etc
means spacetime is locally flat (from your definition that it is locally homomorphic to R^n, ie. Euclidean space)... but since the signature is -+++ not ++++ so it is "pseudo".

 

[quasar987]

Hi mjsd,

You bring up another concern of mine. I don't quite get why people translate "is locally homeomorphic" into "is locally flat". An homeomorphism is only a topological isomorphism, i.e. opens are preserved under application of the map and its inverse, and hence also some properties that rely on the notion of open sets. But it does not say anything about how the metric on the two manifolds is linked (supposing our R^4 with the standard topology is equipped with a metric). So even if we equip our R^4 with a flat metric (minkowski-flat or euclidian-flat), we cannot say anything about the "local flatness" of space-time.

I think this connection is made by the principle of equivalence, but I'm not sure, I don't really understand what it says.

 

[vanesch]

You bring up another concern of mine. I don't quite get why people translate "is locally homeomorphic" into "is locally flat". An homeomorphism is only a topological isomorphism, i.e. opens are preserved under application of the map and its inverse, and hence also some properties that rely on the notion of open sets. But it does not say anything about how the metric on the two manifolds is linked (supposing our R^4 with the standard topology is equipped with a metric). So even if we equip our R^4 with a flat metric (minkowski-flat or euclidian-flat), we cannot say anything about the "local flatness" of space-time.

You are right: it would be silly to say that spacetime is locally flat ! Flatness of a manifold is essentially measured by the Riemann tensor (which is in 1-1 relationship with the metric on torsionless connections if I'm not mistaking), and that Riemann tensor can be non-vanishing. However, there's a way in which one can understand the statement, and that is that Riemanian curvature is "second-order" in displacements. So everything that is only concerned with first order in displacements will not see the difference between a curved and a flat piece of spacetime. Think of a "sphere which is locally flat", as a justification for the fact that Earth looks flat to us. The curvature is not 0 nowhere on a sphere, but a small enough piece can manifest some kind of "flatness" for certain quantities, as long as they are first order in displacement.

 

[mjsd]

Hi mjsd,

You bring up another concern of mine. I don't quite get why people translate "is locally homeomorphic" into "is locally flat". An homeomorphism is only a topological isomorphism, i.e. opens are preserved under application of the map and its inverse, and hence also some properties that rely on the notion of open sets. But it does not say anything about how the metric on the two manifolds is linked (supposing our R^4 with the standard topology is equipped with a metric). So even if we equip our R^4 with a flat metric (minkowski-flat or euclidian-flat), we cannot say anything about the "local flatness" of space-time.

I think this connection is made by the principle of equivalence, but I'm not sure, I don't really understand what it says.

i think you mean "is locally homeomorphic to R^4" being translated to "being locally flat"... don't think the translation is that direct, as u suggested you need more... but perhaps the metric is assumed (ie.usual physicist's abuse of terminologies). btw, remember when Einstein did all these, he did so using only tensor algebras and very little modern differential geometry... In my opinion, it is more of the subject of differential geometry swallowing GR (making it more formal) rather than GR being "grow out of formal mathematics" in the first place. (ie. Top to bottom approach, rather than bottom to top)

principle of equivalence (strong): laws of physics in a freely falling inertial frame are identical to their laws in Special Relativity

Back to the topology/local flatness issue:
start with understanding the logic behind the construction of a Reimanian manifold (semi-layman perspective)

you begin with a set of things, then introduce the notion of an open set (ie. topology is introduced), then make sure that it is topological space (ie. satisfying all relevant axioms) . Next, force each open sets to look like a piece of R^n and that your coordinate charts are smoothly put together (ie. no weird singularities etc.) and now you have a Manifold.
Give this manifold a metric (which automatically means that you have the affine connections), then you have now arrived at the Riemannian manifold ...and spacetime is pseudo-Riemannian because of the -+++...phew!

In a sense, local flatness does seems to remind us of the equivalent principle because Special Relativity operates in flat spacetime. Whether the link between the two is direct or accidental, I can't remember off top of my head.

Anyway, going back to your original question of what is the topology... as u said, it will depend on what is our notion of an open set... mmm...i don't think I know enough to answer it

 

[robphy]

Here's a good reference:

"Global Structure of Spacetimes", by R. Geroch and G. Horowitz,
in "General Relativity: An Einstein Centenary Survey", edited by
Hawking and Israel, 1979. [http://www.worldcatlibraries.org/wcpa/top3mset/27ed62c8845ce767.html]

With regard to issue of "topology of spacetime", let me comment that there is usual "manifold topology", which can be argued isn't very physical from a spacetime viewpoint. Some alternative topologies (which are equivalent to the manifold topology under suitable causality conditions) have been studied by Alexandrov, Zeeman, and Hawking-King-McCarthy [see also Malament]

for Hawking-King http://dx.doi.org/10.1063/1.522874 ;
for Malament http://dx.doi.org/10.1063/1.523436
for worldcat http://www.worldcat.org/oclc/749335181

 

[quasar987]

Could it be that no topology is ever explicitely mentionned? We just assume out of physical consideration that space-time is 4-manifold with some topology that makes it paracompact, connected, Hausdorff and without boundaries. Furthermore, since quantum field theories should be described on them, we demand that it admits a spinor structure (Geroch, 1986). [all these conditions I copy/pasted from p.129 of De Felice & Clarke's 'Relativity on Curved Manifolds']

After all, what does it matter to know these things explicitely. All we are concerned about are the measurable effects, i.e. the curvature of this manifold.

It's just a guess. Comments?

 

[pervect]

Could it be that no topology is ever explicitely mentionned? We just assume out of physical consideration that space-time is 4-manifold with some topology that makes it paracompact, connected, Hausdorff and without boundaries. Furthermore, since quantum field theories should be described on them, we demand that it admits a spinor structure (Geroch, 1986). [all these conditions I copy/pasted from p.129 of De Felice & Clarke's 'Relativity on Curved Manifolds']

After all, what does it matter to know these things explicitely. All we are concerned about are the measurable effects, i.e. the curvature of this manifold.

It's just a guess. Comments?

I meant to get back to this thread, but I got busy and forgot :-(.

AFAIK we don't have any evidence on what the global topology of the universe is. If we see "circles in the sky" http://www.citebase.org/abstract?id=oai%3AarXiv.org%3Aastro-ph%2F9801212
for instance, we'll have some experimental evidence about the topology of the universe, but it appears that there is no clear evidence at this point for the existence of such circles. (There were some papers suggesting that there might be such circles, but IIRC after re-analysis the authors feel more data is needed).

Observing a wormhole would also be an example of a measurement that would show us that the universe has a non-trivial global topology.

I think that the "global toplogy", i.e. R^4 vs S x R^3, etc, is of what's interest to the original poster.

As has been remarked, we can map a section of a curved sphere (S^2) to a section of a flat plane via a homeomorphism. Another way of saying this - curvature doesn't enter the picture with just a topology, one needs a metric to define the curvature.

I'm not sure if it's been remarked yet that the formal defintion of a toplogy isn't a single mapping, but a collection of (local) homeomorphic mappings, i.e. a topology is defined so that it can contain many 'charts'. There are some conditions on the "seams" of how the charts are "glued together" in regions where they overlap as well - basically, the seams have to be continuous.

 

[quasar987]

AFAIK we don't have any evidence on what the global topology of the universe is. If we see "circles in the sky" http://www.citebase.org/abstract?id=oai%3AarXiv.org%3Aastro-ph%2F9801212
for instance, we'll have some experimental evidence about the topology of the universe, but it appears that there is no clear evidence at this point for the existence of such circles. (There were some papers suggesting that there might be such circles, but IIRC after re-analysis the authors feel more data is needed).

Observing a wormhole would also be an example of a measurement that would show us that the universe has a non-trivial global topology.

Hi pervect,

What do you (and the authors of "circles in the sky") mean by "topology of the universe"? I feel there is a meaning that elludes me because to me, the question of the topology of space-time is purely abstract; it is just a specification of what subsets we choose to call "open sets". This is purely mathematical and I don't see how the topology could ever be "observed" experimentally !

I think that the "global toplogy", i.e. R^4 vs S x R^3, etc, is of what's interest to the original poster.

The question of wheter space-time is actually the whole of R^4 or just S^1 x R^3 sounds to me like the question of "what is the set" and not "what is the topology"! But what would be the physical implication of space-time being just this "cylinder" ?

As has been remarked, we can map a section of a curved sphere (S^2) to a section of a flat plane via a homeomorphism. Another way of saying this - curvature doesn't enter the picture with just a topology, one needs a metric to define the curvature.

I hear what you're saying: this is just a restatement of the fact that a manifold can be homeomorphic to flat space in a neighborhood of a point and yet have a non-vanishing curvature there. I.e. the statement "a manifold is a topological space that is locally flat" is not right.

I'm not sure if it's been remarked yet that the formal defintion of a toplogy isn't a single mapping, but a collection of (local) homeomorphic mappings, i.e. a topology is defined so that it can contain many 'charts'. There are some conditions on the "seams" of how the charts are "glued together" in regions where they overlap as well - basically, the seams have to be continuous.

This sounds like the definition of a (C^0) differentiable manifold, rather than that of a topology does it not?!? You yourself gave the definition of a topology in post #5.

 

[robphy]

What do you (and the authors of "circles in the sky") mean by "topology of the universe"? I feel there is a meaning that elludes me because to me, the question of the topology of space-time is purely abstract; it is just a specification of what subsets we choose to call "open sets". This is purely mathematical and I don't see how the topology could ever be "observed" experimentally !

Apparently, you didn't follow up on the references I posted earlier.

At the small-scale, topology is about what points are "close enough" to a given point. In principle, one can ask how one might go about determining experimentally what spacetime events are close to a given event. It would seem that your tools for measuring separations between spacetime events are of a different nature than those used to measure separation of points in (say) a Euclidean plane.

The question of wheter space-time is actually the whole of R^4 or just S^1 x R^3 sounds to me like the question of "what is the set" and not "what is the topology"! But what would be the physical implication of space-time being just this "cylinder" ?

Consider an (interval I) x R³ vs. S¹ x R³.
They describe the same underlying point set. However, they have different topologies... they are connected differently.

If your spacetime has the topology of S¹ x R³, you might be describing a universe with a periodic time, which allows closed timelike curves.

 

[quasar987]

Apparently, you didn't follow up on the references I posted earlier.

I checked them out and saved them on my computer for later because they are over my head right now. Thank you for those.

At the small-scale, topology is about what points are "close enough" to a given point. In principle, one can ask how one might go about determining experimentally what spacetime events are close to a given event. It would seem that your tools for measuring separations between spacetime events are of a different nature than those used to measure separation of points in (say) a Euclidean plane.

Which events are close to which other events... isn't that the job of the metric?? (which we agreed had nothing to do with the topology)

 

[robphy]

Which events are close to which other events... isn't that the job of the metric?? (which we agreed had nothing to do with the topology)

The metric tells you "how close".

 

[pervect]

Hi pervect,

What do you (and the authors of "circles in the sky") mean by "topology of the universe"? I feel there is a meaning that elludes me because to me, the question of the topology of space-time is purely abstract; it is just a specification of what subsets we choose to call "open sets". This is purely mathematical and I don't see how the topology could ever be "observed" experimentally !

Hi - actually, I was trying to guess what aspect of topology you were interested in - and it looks like I might have guessed wrong.

The results I was talking about intially were "point set topoology" results. See for instance http://mathworld.wolfram.com/Point-SetTopology.html

This is the sort to define the fundamental structure of manifolds, where are defined by a set of charts (or coordinate systems). It is, as robphy points out, fundamentally driven by the concept of the "neighborhood" of a point, which is formalized by the notion of "open balls" or "open sets".

These very basic defintions are needed as the starting point.

But point set topology isn't the end of the story. While locally one can (by defintion) pick a particular chart of a topology and have the geometry of that part of the topology be equal to R^n, it's not in general possible to represent an arbitrary topology by a single chart. There's a fairly simple proof, for example, that it takes at least 2 charts to cover a sphere with a 1:1 mapping. Note that using lattitude and longitude as coordinates fails to be a 1:1 mapping at the poles, the pole corresponds to 0 degrees lattitude and many longitudes, not just one longitude.

So I thought you might have been interested in the more global aspects. Some famous results in this area (I'm not quite sure what the correct name for this subtype of topology) would be Euler characteristic numbers and problems like the "Seven bridgnes of Konigsberg".

 

[pervect]

Since I'm still not sure what you're interested in (perhaps you are just interested in learning more about a lot of unrelated things), I'll take the opportunity to present some more motivational material about the original topic of "point set" topology and why it's set up the way it is.

First some history. It has been proven by Cantor that there are the same number of points on a line as points on a plane - i.e. there is a 1:1 reverisble mapping between points on a line and points on a plane.

See for instance http://www.math.okstate.edu/mathdept/dynamics/lecnotes/node21.html

This idea (with a little elaboration) proves that the points in R and R^2 may be placed in one-to-one correspondence. Cantor had already accepted the idea of ``one-to-one correspondence'' as the means for deciding when two infinite sets had the same number of elements.

This mapping between R and R^2 highly artificial in the sense that points which are near one another in R may be unthreaded into two points in R^2 ose to one another. That is to say, Cantor's correspondence is not continuous. There remained the question of whether or not there is a continuous mapping.

Avoding these sorts of mappings of points on a line to points on a plane is the motivation behind the "open balls" formalism and the defintion of homeomorphisms. Explaining exactly how it avoids it would take too much time to do well and probably be confusing, but if you want something to think about, try thinking about why Cantor's mapping from R to R^2 (or if you are really ambitious, look up "space filling curves", another different mapping from R to R^2) do not satisfy the defintion of a homeomorphism.

All of point set topology as I have outlined it above is ultimately based on ZFC, the branch of mathematics that can handle infinite sets. http://en.wikipedia.org/wiki/ZFC

 

[George Jones]

The question of wheter space-time is actually the whole of R^4 or just S^1 x R^3 sounds to me like the question of "what is the set" and not "what is the topology"!

I woundn't say this. R^4 and S^1 x R^3 are isomorphic in the category of sets, but they (with their usual topologies) are not isomorphic in the category of topological spaces.

In other words there is a bijection between R^4 and S^1 x R^3, but there is no homeomorphism between them. As robphy says, R^4 is simply connected, while S^1 x R^3 is not, and simply-connectedness is a topological property, i.e., simply-connectedness is preserved by homeomorphisms.

I hear what you're saying: this is just a restatement of the fact that a manifold can be homeomorphic to flat space in a neighborhood of a point and yet have a non-vanishing curvature there.

Not just in the neighbourhood of a point - this can be true for the entire manifold. Examples: Minkowski space with a point removed is the topological space S^3 x R, the underlying space for the manifold of closed Friedmann-Robertson-Walker universes, and Minkowski space with a straight line removed is S^2 x R^2, the underlying space for the manifold of extended Schwarzschild.

 

[MeJennifer]

Could it be that no topology is ever explicitely mentionned? We just assume out of physical consideration that space-time is 4-manifold with some topology that makes it paracompact, connected, Hausdorff and without boundaries. Furthermore, since quantum field theories should be described on them, we demand that it admits a spinor structure (Geroch, 1986). [all these conditions I copy/pasted from p.129 of De Felice & Clarke's 'Relativity on Curved Manifolds']

After all, what does it matter to know these things explicitely. All we are concerned about are the measurable effects, i.e. the curvature of this manifold.

It's just a guess. Comments?

The mathematical model of special and general relativity is most certainly incomplete.
Space-time, as modeled by a Riemann manifold, is not Hausdorff. In fact the concept of a manifold being Hausdorff did not even exist when SR and GR were developed and no-one, not even Cartan, has resolved it since. Space-time, as modeled by a Riemann manifold, does not even have a valid metric.

Sure some like to shove it under the rug by simply placing the term pseudo in front of everything and then claiming that all is well. But that obviously won't do anything for those who like to think exact!
For them it is like someone saying "Well, admittedly it is not true but for sure it is pseudo-true".

 

[cesiumfrog]

The mathematical model of special and general relativity is most certainly incomplete.
Space-time, as modeled by a Riemann manifold, is not Hausdorff. In fact the concept of a manifold being Hausdorff did not even exist when SR and GR were developed and no-one, not even Cartan, has resolved it since. Furthermore space-time, as modeled by a Riemann manifold, does not have a valid metric.

Sure some like to shove it under the rug by simply placing the term pseudo in front of everything and then claiming that all is well. But that obviously won't do anything for those who like to think exact!
For them it is like someone saying "Well, admittedly it is not true but for sure it is pseudo-true".

Huh? When we describe space-time, as a psuedo-Riemannian (Hausdorff) manifold, the "psuedo" just refers to the Lorentzian signature (whereas otherwise, Riemannian manifolds have positive definite metrics, which cannot have time-like and null distances). We don't then assume theorems only proven for completely Riemannian manifolds; "psuedo-Riemannian" has it's own precise mathematical definition.

As for the Hausdorff part, don't modern treatments explicitly always choose that the manifold is defined in terms of Hausdorff's concept of "topological space"? So what if the very mathematical terminology was slightly different before the first important application was found? How does this make GR incomplete?

 

[MeJennifer]

When we describe space-time, as a psuedo-Riemannian (Hausdorff) manifold...

Then those who do that are making a mistake.
Feel free to provide or give a reference to the proof that a pseudo-Riemann manifold is Hausdorff. It is not.

How does this make GR incomplete?

I did not write that GR is incomplete, I wrote that the mathematical modeling of GR is incomplete. Something entirely different!

 

[cesiumfrog]

I'm not saying that psuedo-Riemannian implies Hausdorff. I'm saying that the accepted mathematical model of space-time in GR is a "psuedo-Riemannian Hausdorff manifold" (at least, this is the starting point from which additional constraints may be added, like EFE, energy conditions, asymptotic metric, no timelike loops, etc).

 

[MeJennifer]

I'm not saying that psuedo-Riemannian implies Hausdorff. I'm saying that the accepted mathematical model of space-time in GR is a "psuedo-Riemannian Hausdorff manifold" (at least, this is the starting point from which additional constraints may be added, like EFE, energy conditions, asymptotic metric, no timelike loops, etc).

So let's cut to the chase, is a pseudo-Riemann manifold Hausdorff or not?
I say no, what do you say?

"psuedo-Riemannian" has it's own precise mathematical definition.

Any references you can give me that defines it without a bunch of pseudo's?

 

[pervect]

So let's cut to the chase, is a pseudo-Riemann manifold Hausdorff or not?

Is a red car fast?

 

[masudr]

So let's cut to the chase, is a pseudo-Riemann manifold Hausdorff or not?
I say no, what do you say?

Some pseudo-Riemannian manifolds are Hausdorff, others aren't. But the ones we choose to use as models of spacetime are Hausdorff.

If you can show that any pseudo-Riemannian manifold has to be, by a mathematical proof, non-Hausdorff, I (and I'm sure others) would be very interested in seeing it.

 

[robphy]

The mathematical model of special and general relativity is most certainly incomplete.
Space-time, as modeled by a Riemann manifold, is not Hausdorff. In fact the concept of a manifold being Hausdorff did not even exist when SR and GR were developed and no-one, not even Cartan, has resolved it since. Space-time, as modeled by a Riemann manifold, does not even have a valid metric.

Certainly, these geometrical structures are convenient in the mathematical modeling of physical systems. They provide a framework in which a mathematical analysis (e.g. proofs of mathematical theorems) arising from the modeling can be carried out, which hopefully still reasonably describe the physics.

While much of the progress [theoretical/mathematical results and experiemental results] to date in relativity has been built on this framework, it is possible that the model will have to be revised to allow for (say) quantum gravity. However, these revisions probably wouldn't invalidate what has already been done [within the limits of the model it used].

Since you question the current model of spacetime [which is okay to do],
here are two questions for you:

- Can you share your physical reasoning behind your claim:
"Space-time, as modeled by a Riemann manifold, is not Hausdorff." ?
(That is, what specific physical effects are being neglected by imposing the Hausdorff condition?)

- Can you share your physical or mathematical reasoning behind your claim:
"Space-time, as modeled by a Riemann manifold, does not even have a valid metric." ?
Are you referring to a semi- or pseudo-Riemannian metric? What is physically or mathematically in"valid" about it?

 

[AlphaNumeric]

Then those who do that are making a mistake.
Feel free to provide or give a reference to the proof that a pseudo-Riemann manifold is Hausdorff. It is not.

Maybe not in general, but where is the point or set of points in Minkowski space-time which fail to be Hausdorff?

I don't think it's been swept under the rug and as pointed out, 'pseudo' isn't a prefix to mean 'it's half true', it refers to a specific class of metrics, those which are not positive definite. The distinction is required because not all Riemannian manifold results hold for Pseudo-Riemannian, namely those resting on the requirement that the metric is positive definite.

While not all Riemannian manifolds are Hausdorff, the collection of ones used in physics almost always are Hausdorff.

 

[MeJennifer]

- Can you share your physical reasoning behind your claim:
"Space-time, as modeled by a Riemann manifold, is not Hausdorff." ?
(That is, what specific physical effects are being neglected by imposing the Hausdorff condition?)

Any topological model that fully describes SR and GR must actually remove the Hausdorff constraint. The problem is that Riemannian geometry requires a manifold to be Hausdorff, a fact, by the way, that was unknown at the time SR and GR was developed.

- Can you share your physical or mathematical reasoning behind your claim:
"Space-time, as modeled by a Riemann manifold, does not even have a valid metric." ?
Are you referring to a semi- or pseudo-Riemannian metric? What is physically or mathematically in"valid" about it?

One of the requirements for a metric on a Riemannian manifold is that the triangle inequality must hold. The Minkowski "metric" which has a negative definite signature is obviously not a metric by that definition.

Now we could simply put the term "pseudo" in front of anything and be done with it but that does not make the real mathematical issue go away. It is a pseudo solution.

There is another way around it, by defining an "Einstein algebra", but then we cannot any longer think in terms of a topological model of space-time.

So in short, this is not an SR or GR problem, it is a mathematics problem.
Nevertheless and obviously, mathematics can still supply a workable framework for SR and GR, but clearly not an mathematically complete one.

While not all Riemannian manifolds are Hausdorff, the collection of ones used in physics almost always are Hausdorff.

Could you give me one single reference to a mathematical textbook or paper that does not require a Riemannian manifold to be Hausdorff?

 

[robphy]

Any topological model that fully describes SR and GR must actually remove the Hausdorff constraint. The problem is that Riemannian geometry requires a manifold to be Hausdorff, a fact, by the way, that was unknown at the time SR and GR was developed.

Why? Are you claiming that Riemannian requires [and not merely chooses for convenience] the Hausdorff condition? Can you provide a reference?

Historically, lots of things were unknown at the time SR and GR were being developed. For a long time, the emphasis was on systems-of-PDEs in coordinate patches and not the global structure underlying the modern formulations of spacetime [including causal structures].

One of the requirements for a metric on a Riemannian manifold is that it the triangle inequality must hold. The Minkowski "metric" which has a negative definite signature is obviously not a metric by that definition.

The triangle-inequality has some nice properties... which is needed for certain properties of Riemannian geometry. But, the clock effect tells us that that triangle-inequality is not satisfied by triangle with timelike-vectors... we have instead the reverse-triangle inequality.

FYI, Minkowski is NOT http://mathworld.wolfram.com/NegativeDefiniteMatrix.html" . It has signature (-+++) or (+---), depending on your convention.

In any case, as Riemannian generalizes Euclidean, Lorentzian (or more generally semi-riemannian) generalizes Riemannian. One can further generalize to Finslerian manifolds, complex-manifolds, non-metric manifolds, non-manifold topological spaces, etc... In the mathematical hierarchy, is there is anything so sacred about Riemannian geometry with its positive-definite metric?

The point for modeling the physical world is: which best models spacetime... [with hopefully each mathematical structure in model having some physical interpretation]?

(By the way, from a projective-geometric viewpoint [in the spirit of Felix Klein], one can see that Minkowskian geometry is a completely consistent geometric theory [as one does with Elliptic and Hyperbolic geometries].)

Now we could simply put the term "pseudo" in front of anything and be done with it but that does not make the real mathematical issue go away. It is a pseudo solution.

This sounds like the comments that regard the complex numbers as "strange" number systems.

There is another way around it, by defining an "Einstein algebra", but then we cannot any longer think in terms of a topological model of space-time.

Geroch's Einstein Algebra? Using that does NOT say that you "cannot any longer think in terms of a topological model of space-time"... rather, you don't have to think in that way. You can [and do] use the Einstein Algebra for ordinary spacetime. (By the way, some noncommutative-geometric approaches use a similar approach.)

So in short, this is not an SR or GR problem, it is a mathematics problem.
Nevertheless and obviously, mathematics can still supply a workable framework for SR and GR, but clearly not an mathematically complete one.

It may be a mathematics problem... but don't lose sight of the physicist's goal: find the best model for spacetime [sufficient for the problem under study].

 

[MeJennifer]

FYI, Minkowski is NOT http://mathworld.wolfram.com/NegativeDefiniteMatrix.html" . It has signature (-+++) or (+---), depending on your convention.

Ops, of course you are correct, it can be positive, negative and 0. I should have written it is not a positive-definite.

It may be a mathematics problem... but don't lose sight of the physicist's goal: find the best model for spacetime [sufficient for the problem under study].

I fully agree!
And clearly a pseudo-Riemannian manifold is currently the best we have. To me, and perhaps of my lack of knowledge in this area, it is a workable but not a completely satisfactory solution to model SR and GR on.

 

[AlphaNumeric]

Sorry, my error. Not all manifolds are Hausdorff but all Riemannian ones are because they are differentiable and you need a notion of 'neighbourhoods' around each point in the manifold to create the differentiable structure (at least that seems to be my applied maths understanding of it).

A pseudo-Riemannian manifold still satisfies the Hausdorff requirement, since you can pick open charts which don't intersect around any two points you like, which gives you the Hausdorff property.

Besides, weren't you just saying that (pseudo)Riemannian manifolds aren't Hausdorff, particularly those used in GR?! You didn't answer where the non-Hausdorff points in Minkowski space-time are.

Nakahara has the following requirements for a Riemannian manifold :

1. It's differentiable
2. It's metric g is defined at every point and has the property g_{p}(V,U) = g_{p}(U,V)
3a. g_{p}(U,U) \geq 0 for all U and equality iff U=0.

He then extends this to a pseudo-Riemannian manifold by changing 3a. to the following :

3b. if g_{p}(U,V)=0 for any U in the tangent space, then V=0.

It's not quite the same, but does share some of the properties of 3a. It's not that Pseudo-Riemannian are totally invalid, it's that their metric isn't held to the same restriction as the Riemannian ones.

The Minkowski "metric" which has a negative definite signature is obviously not a metric by that definition.?

No, it's obviously not a Riemannian metric. It's still a metric if you slacken your constraints somewhat.

There's nothing wrong with doing that provided you're consistent. Take SUSY for instance. By slackening the restrictions on Lie algebras to allow graded Lie algebras you develop a new, rich algebraic system.

Yes, it's important you're consistent with your usage of Riemannian and Pseudo-R. results, even Hawking once got them confused and tried to publish a paper using Riemannian results but applied to GR manifolds (at least so said my old GR lecturer).

 

[cesiumfrog]

Any topological model that fully describes SR and GR must actually remove the Hausdorff constraint.

Why? (It seems like you didn't answer the question: What physics are you neglecting if you constrain the model to be Hausdorff? Personally, seems like a non-Hausdorff space would be the one behaving unphysically.)

 

[pervect]

BTW, does anyone have an example of a manifold (which must by definition be covered by charts diffeomorphic to the familiar R^n) that is not Hausdorff?

I gather it's supposed to be possible, but I've never seen an example.

Wikipedia says that any metric space is Hausdorff, for whatever it's worth, so presumably our non-Hausdorff manifold won't have a metric (unless the wiki is wrong).

I don't understand all the excitement over the word "psuedo" in "Psuedo-Riemannian manifold" - are we being trolled?

 

[Doodle Bob]

Wikipedia says that any metric space is Hausdorff, for whatever it's worth, so presumably our non-Hausdorff manifold won't have a metric (unless the wiki is wrong).

i believe that the proof for this goes something like: Let x and y be distinct points in the metric space. then d=dist(x,y)>0. Let U={z: dist(x,z)<d/3} and V={z: dist(z,y)<d/3}, both open sets by the metric topology. By the triangle inequality, U and V are disjoint.(oops, i probably just gave away a homework problem there...)

So, if a manifold has a Riemannian metric that defines a distance on it, then it is necessarily Hausdorff. However, i do believe there are examples of topological manifolds (maybe even differentiable manifolds) that do not admit a partition of unity and hence do not have even a well-defined global Riemannian metric.

 

[quasar987]

BTW, does anyone have an example of a manifold (which must by definition be covered by charts diffeomorphic to the familiar R^n) that is not Hausdorff?

I gather it's supposed to be possible, but I've never seen an example.

Wikipedia says that any metric space is Hausdorff, for whatever it's worth, so presumably our non-Hausdorff manifold won't have a metric (unless the wiki is wrong).

The key word is 'non-metrizability'. See

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

for a plenitude of metrizability theorems and also an example of a non metrizable (and hence, not Hausdorff) manifold:

An example of a space that is not metrizable is the real line with the lower limit topology.

 

[Hurkyl]

Manifolds are explicitly stated to be Hausdorff. The reason is that it is useful. A space such as the line with a double point at the origin

------:------

satisfies all the other conditions of being a manifold, and can even have a differential structure. But the extra point at the origin essentially contributes nothing beyond its identity: any continuous function, vector field, tensor field, or whatever has exactly the same value at both of the points at the origin.


Metrizability implies Hausdorff. Nonmetrizability does not imply non-Hausdorff. The famous example of a nonmetrizable manifold, the long line, is clearly a Hausdorff space.

Incidentally, some authors require that manifolds must also be metrizable, so the long line wouldn't be a manifold.


In pseudoRiemannian geometry, the word "metric" refers to the metric tensor, rather than the metric one would study in the context of metric spaces.

 

[cesiumfrog]

does anyone have an example of a manifold (which must by definition be covered by charts diffeomorphic to the familiar R^n) that is not Hausdorff? I gather it's supposed to be possible, but I've never seen an example.

It seems trivial that Euclidean space satisfies the Hausdorff axioms, right? And by the definition of a manifold, around every point in a manifold is a region "topologically equivalent" to Euclidean space. Sound's like a Diff. Geom. homework question now, "prove homeomorphisms preserve the Hausdorff condition".

if a manifold has a Riemannian metric[..]

From above, I think "manifold" implies Hausdorff, which is stronger (ie. doesn't depend on there actually being a metric, let alone on whether it is Riemannian).

an example of a non metrizable (and hence, not Hausdorff) manifold

I think that example of a topological space is not a manifold.

 

[Hurkyl]

It seems trivial that Euclidean space satisfies the Hausdorff axioms, right? And by the definition of a manifold, around every point in a manifold is a region "topologically equivalent" to Euclidean space. Sound's like a Diff. Geom. homework question now, "prove homeomorphisms preserve the Hausdorff condition".

It would, if being Hausdorff was a local property. (It's not)

Again, I cite the line with two origins:

-----:-----

This space is locally homeomorphic to Euclidean space. Here's a neighborhood of the top point at the origin that is homeomorphic to an interval:

--[/color]---'---[/color]--[/color]

Here is a neighborhood of the bottom point at the origin that is homeomorphic to an interval:

--[/color]---.---[/color]--[/color]


And, of course, any other point can be covered with an interval that misses the origin entirely.

 

[pervect]

Manifolds are explicitly stated to be Hausdorff. The reason is that it is useful. A space such as the line with a double point at the origin

------:------

satisfies all the other conditions of being a manifold, and can even have a differential structure. But the extra point at the origin essentially contributes nothing beyond its identity: any continuous function, vector field, tensor field, or whatever has exactly the same value at both of the points at the origin.


Metrizability implies Hausdorff. Nonmetrizability does not imply non-Hausdorff. The famous example of a nonmetrizable manifold, the long line, is clearly a Hausdorff space.

Incidentally, some authors require that manifolds must also be metrizable, so the long line wouldn't be a manifold.


In pseudoRiemannian geometry, the word "metric" refers to the metric tensor, rather than the metric one would study in the context of metric spaces.

OK, thanks, this is just the sort of info I was looking for. I don't believe that Wald, for instance, defines a manifold as necessarily being "Hausdorff" in "General Relativity".

What Wald does say is "Viewed as topological spaces, we shall consider in this book only manifolds which are Hausdorff and paracompact; these terms are defined in Appendix A". So he basically avoids the issue without discussing the details.

As far as a "metric space" goes - for the most part in GR I'm interested in spaces with a metric tensor. Is it correct, mathematically, to view a space with a metric tensor as a specific example of a more general "metric space"?

 

[Hurkyl]

The definition I'm going by for manifold came from Spivak, and Wikipedia agrees too; maybe some authors permit an even more definition of manifold that permit them to be non-Hausdorff, but I haven't seen it.

A Riemannian metric tensor (on a sufficiently "small" manifold) can be used to create a metric space structure, and I imagine the metric tensor can be recovered from the metric space structure.

But I don't know of a similar thing that can be said for a non-Riemannian metric tensor. (Of course, that doesn't mean such a thing doesn't exist; I just don't know of it)

 

[vanesch]

Maybe this is a moot point, but I had the impression that MeJennifer's claim comes from the following: the metric on a pseudo-Riemannian manifold is not a positive-definite metric, and as such, cannot be used to define a generating set of open balls which generates a topology. The topology on space time, however, is not THIS one. It is not the one that "follows from the pseudo-riemanian metric", but another one, that follows from the local equivalence to an Euclidean space R^4 (a chart).
The topology on R^4 comes from the Euclidean metric on R^4, and not from any pseudo-riemanian metric induced on R^4.

At least, that's how I understand it.

 

[quasar987]

The topology on space time, however, is not THIS one. It is not the one that "follows from the pseudo-riemanian metric", but another one, that follows from the local equivalence to an Euclidean space R^4 (a chart).

Can you say that again please? The topology on space-time is not the open ball aka metric topology induced by the metric tensor, instead it is...what? What is this topology "that follows from the local equivalence to an Euclidean space R^4 (a chart)"?

 

[vanesch]

Can you say that again please? The topology on space-time is not the open ball aka metric topology induced by the metric tensor, instead it is...what? What is this topology "that follows from the local equivalence to an Euclidean space R^4 (a chart)"?

Well, a smooth manifold is defined by an atlas, which contains diffeomorphisms between open sets in R^n (with the usual Euclidean metric + topology) and the abstract manifold M, which is, in order to even be able to define "continuous", also equipped with a topology. The charts are hence local "isomorphisms of topology" (homeomorphisms is the name I think). Now, once we have the manifold, we can define a symmetric tensor over it which will give us the semi-Riemanian metric. However, the "balls" of the semi-riemanian metric are not a generator for the topology on the manifold which was used to define the charts!

Indeed, look at simple Minkowski space M. There is one single chart in an atlas, which is a mapping from M to R^4: c: p->(x0,x1,x2,x3).

If we take as "fundamental balls" with radius epsilon in the Minkowski metric, we have, for instance,

(x0-y0)^2 - (x1-y1)^2 - (x2-y2)^2 - (x3-y3)^2 < epsilon

as its image through the chart in R^4. But that's not even a compact set in R^4 under its usual topology! It means that points which are (nearly) lightlike connected, are in "close neighbourhood". But, using an intermediate point, ANY TWO POINTS can be lightlike connected! So this means that any two points are in the neighbourhood of each other. This means that continuous functions must be constant over M, and even the coordinate functions are not continuous functions.

 

[MeJennifer]

Maybe this is a moot point, but I had the impression that MeJennifer's claim comes from the following: the metric on a pseudo-Riemannian manifold is not a positive-definite metric, and as such, cannot be used to define a generating set of open balls which generates a topology.

Correct, furthermore sequences do not neccesarily converge to one single point in non-Hausdorff spaces.

The topology on space time, however, is not THIS one. It is not the one that "follows from the pseudo-riemanian metric", but another one, that follows from the local equivalence to an Euclidean space R^4 (a chart).
The topology on R^4 comes from the Euclidean metric on R^4, and not from any pseudo-riemanian metric induced on R^4.

At least, that's how I understand it.

I know what you mean, but there is absolutely nothing in SR and GR that implies that! Forgive me for taking the liberty to think that that "solution" is simply "mathematical spielerei" to avoid, rather than adress, the issue.

 

[vanesch]

I know what you mean, but there is absolutely nothing in SR and GR that implies that!

I think there is, but it is implicit. The point is that in GR, one requires "smooth" transition functions between "coordinates", but in order to say what "smooth" is, one uses the standard 4 real variable into 4 real variable diffeomorphisms, as in calculus. That uses implicitly the standard topology on R^4 which is generated by the open balls of the Euclidean metric, but one doesn't say so explicitly. It is essentially the topology used to decide which transition functions are diffeomorphisms which defines the topology on the manifold, and the "accepted" diffeomorphisms in GR (the "changes of coordinates") are smooth functions between R^4 -> R^4 with standard (Euclidean) topology.
One doesn't use any bizarre "Minkowski topology" (whatever that may mean!) to find out whether coordinate transformations are "smooth".

EDIT: in fact, when limiting oneself to topology, one should just use "continuous", but I had the impression that in relativity, one wants to use smooth manifolds, and not just topological manifolds.

 

[MeJennifer]

but I had the impression that in relativity, one wants to use smooth manifolds, and not just topological manifolds.

Sure, how otherwise are you going to use differential calculus.

But, however, how can a smooth manifold ever have singularities if only smooth deformations are allowed?

 

[coalquay404]

The mathematical model of special and general relativity is most certainly incomplete.
Space-time, as modeled by a Riemann manifold, is not Hausdorff.

This is technically correct, but only technically. If you want to talk about spacetime solutions to the EFEs it is customary to utter the following incantation before one begins:

Spacetime is a four-dimensional paracompact, connected smooth Hausdorff manifold without boundary.

Non-Hausdorff spacetimes don't arise directly in the context of the EFEs. Where they do appear is when one looks at certain quotient spaces in general relativity. The example with which I am most familiar is that of the quotient spaces involved in the maximally extended covering spaces for Taub-NUT space. It's a reasonably interesting area (if you like that sort of thing) but not a particularly fruitful one. Petr Hajicek wrote a nice paper on causality in non-Hausdorff spaces in the seventies, when such things were popular, but I'm not aware of any more recent work.

In fact the concept of a manifold being Hausdorff did not even exist when SR and GR were developed and no-one, not even Cartan, has resolved it since. Space-time, as modeled by a Riemann manifold, does not even have a valid metric.

There are several confusing things here. Firstly, what is the problem you think needs to be resolved? Secondly, there may perhaps be some difficulty with your terminology. Don't refer to things such as Riemannian or pseudo-Riemannian manifolds; it will only confuse people when you try to talk to somebody who's not familiar with the subject area. It's better in the context of general relativity to talk simply of a differentiable manifold (with the explicit degree of differentiability being largely unimportant unless you're interested in studying reductions of the EFEs in the context of, say, weighted Sobolev spaces), and then to specify the type of metric structure that you're working with on that manifold. For example, classical general relativity (I'm talking about Einstein's way of viewing it) deals with differentiable manifolds which are endowed with a pseudo-Riemannian metric structure, i.e., a space of indefinite metric forms. Modern GR deals with topological identifications between a spacetime and a foliation by a three-dimensional manifold endowed with positive definite metric structure.

Sure some like to shove it under the rug by simply placing the term pseudo in front of everything and then claiming that all is well. But that obviously won't do anything for those who like to think exact!
For them it is like someone saying "Well, admittedly it is not true but for sure it is pseudo-true".

I can assure you that nobody who knows what they're talking about "shoves [anything] under the rug". In the field, when people use pseudo-Riemannian or Riemannian, their usage is quite clear.

 

[vanesch]

Sure, how otherwise are you going to use differential calculus.

Right, but differential calculus is based upon standard R^4 topology.

But, however, how can a smooth manifold ever have singularities if only smooth deformations are allowed?

I'm not 100% sure that the manifold itself becomes non-smooth at GR singularities. Rather, the METRIC becomes singular, no ? But I'm no expert, maybe someone better versed in this can correct me.