Link btw manifolds and space-time

[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:

-----'-----

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

-----.-----


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.

 

[quasar987]

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

I of course am not that "better versed" person, but I was under the impression that there were two types of singularities: real ones and those that can be made to disappear under a proper change of coordinates. For example, when one changes from spherical coordinates to the Kruskal-Szekeres coordinates in the black hole solution, the singularity at the horizon disappears but not the one at the center of the BH.

 

[vanesch]

I of course am not that "better versed" person, but I was under the impression that there were two types of singularities: real ones and those that can be made to disappear under a proper change of coordinates. For example, when one changes from spherical coordinates to the Kruskal-Szekeres coordinates in the black hole solution, the singularity at the horizon disappears but not the one at the center of the BH.

Yes, but that is a singularity in the metric. That is, some part of the metric tensor blows up (and if it blows up in one coordinate system then it blows up in all). But the metric is just a 2-tensor defined over the manifold. This is like having, in Euclidean space, say, a vector field which takes on the form:

v(p) = (x+y) 1_x + (z+x) 1_y + (z^3 + y^3)/(x^2+y^2) 1_z

Now, this vector field blows up on the z-axis, and this is not a coordinate system problem: in any coordinate system, this vector field blows up. However, does this mean that the Euclidean space itself, over which the vector field is defined, blows up ?

In the same way, it is not because a 2-tensor (which happens to be called a pseudo-riemanian metric) blows up at a certain point of a manifold, that the manifold itself has a problem there.

Look for instance at the Kerr BH, in the Kerr form. The metric goes singular for x^2 + y^2 = a^2 ; z = 0. But the coordinate mapping is still smooth there.

Again, caveat: I'm no expert at all in this!

 

[George Jones]

Maximally extended Schwarzschild (and Kerr) are both timelike geodeically incomplete. There exist worldlines of freely falling observers that end after a finite amount of proper time - these freely falling observers "fall off of spacetime" after a finite amounts of time have elapsed on their watches.

Some spacetimes exhibit this type of behaviour without having curvature invariants blow up. Other spacetimes that are geodesically complete are classified as singular because they have inextenable worldlines that have bounded 4-acceleration - observers in rockets that have finite thrusts "fall off the edge of spacetime".

Of course all of this - geodesics, 4-acceleration, etc. - is determined by the metric.

 

[vanesch]

Maximally extended Schwarzschild (and Kerr) are both timelike geodeically incomplete. There exist worldlines of freely falling observers that end after a finite amount of proper time - these freely falling observers "fall off of spacetime" after a finite amounts of time have elapsed on their watches.

Some spacetimes exhibit this type of behaviour without having curvature invariants blow up. Other spacetimes that are geodesically complete are classified as singular because they have inextenable worldlines that have bounded 4-acceleration - observers in rockets that have finite thrusts "fall off the edge of spacetime".

Of course all of this - geodesics, 4-acceleration, etc. - is determined by the metric.

Could we picture this "falling off spacetime" as just a quirk of the metric, as in the following sense:
picture the Euclidean plane as a manifold (on which we can define a traditional (x,y) chart). Now define the metric tensor ds^2 = 1/(1+x^2 + y^2)^2(dx^2 + dy^2). Clearly, any radial line going "all the way" on the manifold will just have a total accumulated length of pi/2 (from the top of my head, if I'm not mistaking). So the whole Euclidean plane just "looks like a disk" from the metric PoV (of course, in this case, we could probably extend it).

 

[MeJennifer]

Firstly, what is the problem you think needs to be resolved?

Well it must be my lack of intelligence to see how the following is consistent:

1. Gravity is curvature of space-time
2. Space-time is a Riemannian manifold with a positive definite metric
3. The shape of this manifold is determined by the EFE in terms of a metric tensor that is not a valid metric.

I suppose the consistency of the emperor's fabric is beyond my level.

 

[cesiumfrog]

1. Gravity is curvature of space-time
2. Space-time is a Riemannian manifold with a positive definite metric
3. The shape of this manifold is determined by the EFE in terms of a metric tensor that is not a valid metric.

I suppose the consistency of the emperor's fabric is beyond my level.

(2) is mistaken. Space-time is a not a Riemannian manifold (instead it is a different particular type of manifold, specifically the type named psuedo-Riemannian), with a metric that has a Lorentzian signature (hence is not positive definite).

I thought you'd seen this earlier along the thread, but nonetheless.. Now that you know the 2nd fibre that the emperor really uses, do you still have any issue with the fabric's consistency?

 

[coalquay404]

Well it must be my lack of intelligence to see how the following is consistent:

1. Gravity is curvature of space-time
2. Space-time is a Riemannian manifold with a positive definite metric
3. The shape of this manifold is determined by the EFE in terms of a metric tensor that is not a valid metric.

I suppose the consistency of the emperor's fabric is beyond my level.

I wouldn't say you lack intelligence, but you're definitely confused. (2) is incorrect. It should read "Spacetime $(M,g)$ is a four-dimensional paracompact, connected smooth Hausdorff manifold $M$ without boundary, and with an indefinite (or, if you like, pseudo-Riemannian) metric structure $g$."

(3) is also completely incorrect.

 

[George Jones]

Could we picture this "falling off spacetime" as just a quirk of the metric, as in the following sense:
picture the Euclidean plane as a manifold (on which we can define a traditional (x,y) chart). Now define the metric tensor ds^2 = 1/(1+x^2 + y^2)^2(dx^2 + dy^2). Clearly, any radial line going "all the way" on the manifold will just have a total accumulated length of pi/2 (from the top of my head, if I'm not mistaking). So the whole Euclidean plane just "looks like a disk" from the metric PoV (of course, in this case, we could probably extend it).

What is the metric when expressed in terms of coordinates rho and theta, where r = tan(rho), and r and theta are related to x and y in the standard way?

 

[pervect]

I don't know if this is the source of the confusion, but I hope we can all agree that points connected by null curves, are not necessarily neighbors in the sense used by topological spaces.

Thus if I am looking out at the night sky at Andromeda, I see some particular event happening that is not "close" to me, even though it is connected to me by a light ray which ideally has a Lorentz interval of zero.

What makes the Lorentz interval interesting is not that it defines points which are "neighbors", "close" or "nearby", (it doesn't do that function) - rather, it is interesting because it is an invariant of space-time, something that different observers can agree on.

 

[vanesch]

What makes the Lorentz interval interesting is not that it defines points which are "neighbors", "close" or "nearby", (it doesn't do that function) - rather, it is interesting because it is an invariant of space-time, something that different observers can agree on.

yes, that was my point. The topology over the spacetime manifold is NOT given by the lorentz metric, but rather by the standard topology over R^4 (which is, in itself, given by the Euclidean metric over R^4). This is because we consider "coordinate transformations" (which are nothing else but transition maps between charts) to be continuous, when they are continuous in the standard calculus sense.

 

[MeJennifer]

Thus if I am looking out at the night sky at Andromeda, I see some particular event happening that is not "close" to me, even though it is connected to me by a light ray which ideally has a Lorentz interval of zero.

Sorry but that is complete nonsense, physically you see a photon hitting your retina. It is impossible to observe a photon from a distance.

 

[vanesch]

What is the metric when expressed in terms of coordinates rho and theta, where r = tan(rho), and r and theta are related to x and y in the standard way?

Hmmm, let me think:

in r and theta, ds^2 = 1/(1+r^2)^2 (dr^2 + r^2 d theta^2)

Now, if we transform r = tan(rho), then we have dr = sec(rho)^2 drho = (1+r^2) d rho, so:

ds^2 = 1/sec^4(rho) (sec^4(rho) drho^2 + tan^2(rho) d theta^2)

ds^2 = drho^2 + 1/4 sin^2(rho) d theta^2

(if I didn't make a mistake).

Now, the point is that the original chart, which was supposed to cover the entire manifold, was the entire (x,y) plane. The (r,theta) chart (actually, we'd need two charts here) was 0 < r and 0<theta < 2pi.

In the (rho,theta) chart, this is 0 < rho < pi/2 and 0 < theta < 2pi
(we'd also need at least a second chart to be strictly correct, which covers the positive x-axis and the origin).

Now, I hear you coming: we could extend the manifold by considering now the chart to be "bigger" (rho > pi/2). But then we are, strictly speaking, talking about ANOTHER manifold, right ?

 

[George Jones]

Now, I hear you coming: we could extend the manifold by considering now the chart to be "bigger" (rho > pi/2). But then we are, strictly speaking, talking about ANOTHER manifold, right ?

A Riemannian manifold (M' , g') is an extension of (M , g) if M can be regarded (by a suitably differentiable embedding) as a proper subset of M', and g' restricted to M "looks like" g (the embedding is isometric). In relativity, the spacetime manifold is usually taken to inextendable.

So, to me, this example looks similar the coordinate singularity of Schwarzschild in Schwarzschild coordinates, i.e., t -> infinity corresponds to a geodesic approaching the event horizon after a finite amount of proper time.

For a singular inextendable spacetime, the luxury of removing the singularity by extending the spacetime is removed.

 

[pervect]

Sorry but that is complete nonsense, physically you see a photon hitting your retina. It is impossible to observe a photon from a distance.

The photon hitting your retina is indeed close to you, but the event where the photon originated at Andromeda is the one I was referring to. I hope you and everyone will agree that that event (the one in Andromeda) is "far away", even though it is connected by a curve of zero Lorentz interval (the null curve, in this case a null geodesic, of the path of the photon) to you.

 

[Haelfix]

People are getting very confused about two different concepts.

Distinguish the topology of R^4
and
the topology of spacetime in the sense of the global visible universe.

The latter is essentially glued together from the former, and I use the word 'glue' rather nonstandardly here b/c there's several maps going on before this even takes place. The former is purely formal in the sense that we construct a *local* homeomorphism from R^4 --> R^4 to identify our local neighborhood with the usual Lorentzian one. Note, this identification is already topologically restricting, eg we now have a 4 dimensional manifold, and not just any manifold, but one with a special relativity isometry group acting on points. So amongst all possible topologies, we have restricted the set theoretic structure to be both the usual one, as well as endowed a diffeomorphic manifold structure on it.

Now, we again *choose* to add more structure (further restricting the topology) b/c this isn't physically interesting yet. Namely we choose a connection. Again no god given choice, but there is a preferred one for a Riemmanian metric, or a pseudo riemanian metric. So we go ahead and choose this levi Cevita connection (or the Christofel connection) and we now have a metric space. Note we could have already guessed the connection a priori simply by requiring the isometry group action to be faithful.

This construction then automatically satisfies the usual nice properties of being Hausdorf, paracompact w/o boundary (tho again you are free not to make this particular choice, but it has afaik no immediate physical relevance)

So what are we left with topologically... Still a dazzling array of possibilities. And indeed, it is still an open question as to exactly what the topology of the universe is. It could be R^4, but then again it need not be. It could be S^4.

This does make a difference.. Physically even! Astronomers actively look for galaxies that are basically copies of each other.

 

[vanesch]

People are getting very confused about two different concepts.

Distinguish the topology of R^4
and
the topology of spacetime in the sense of the global visible universe.

The latter is essentially glued together from the former, and I use the word 'glue' rather nonstandardly here b/c there's several maps going on before this even takes place. The former is purely formal in the sense that we construct a *local* homeomorphism from R^4 --> R^4 to identify our local neighborhood with the usual Lorentzian one.

Yes, that was what I meant to say. Of course, the global topology of a spacetime manifold doesn't need to be exactly that of R^4 (otherwise, there wouldn't even be a point in talking about a manifold, as it would be trivial), however, locally, "local neighbourhoods" are in 1-1 relationship with "local neighbourhoods" in (a chunk of) R^4, in which these "local neighbourhoods" in R^4 are to be understood in the usual standard topology (which is generated by the Euclidean metric over R^4), which has nothing to do a priori with the metric tensor over the manifold we're going to define.

Note, this identification is already topologically restricting, eg we now have a 4 dimensional manifold, and not just any manifold, but one with a special relativity isometry group acting on points. So amongst all possible topologies, we have restricted the set theoretic structure to be both the usual one, as well as endowed a diffeomorphic manifold structure on it.

Actually, I don't know much about this stuff. I would be of the naive opinion that any smooth (!) 4-dim manifold could be arbitrarily endowed with a symmetric 2-tensor which we could call "metric", no ? Is the requirement of fixed signature a limitation on the possibilities of choice of the 4-dim manifold ?

 

[George Jones]

It could be R^4, but then again it need not be. It could be S^4.

This does make a difference.. Physically even!

S^4 is compact, so if the topology of the universe is S^4, ithe universe contains closed timelike curves.

 

[George Jones]

Actually, I don't know much about this stuff. I would be of the naive opinion that any smooth (!) 4-dim manifold could be arbitrarily endowed with a symmetric 2-tensor which we could call "metric", no ? Is the requirement of fixed signature a limitation on the possibilities of choice of the 4-dim manifold ?

In fact, most manifolds admit inequivalent Lorentzian metrics. It is even possible to have non-zero Riemann curvature tensor tensor for (M,g), while (M,h')
is completely (intrinsically) flat (see https://www.physicsforums.com/showpost.php?p=1234845&postcount=20" for examples)! Intrinsic curvature is not a property of the manifold alone.

 

[vanesch]

In fact, most manifolds admit inequivalent Lorentzian metrics. It is even possible to have non-zero Riemann curvature tensor tensor for (M,g), while (M,h')
is completely (intrinsically) flat

That doesn't surprise me in fact. Take an arbitrary manifold, with arbitrary atlas. Define a zero 4-tensor on it (always possible to have a smooth constant function!), and call that the Riemann tensor.

 

[coalquay404]

S^4 is compact, so if the topology of the universe is S^4, ithe universe contains closed timelike curves.

You beat me to it! Perhaps he meant that compactness is allowed in the following sense. If we have a spacetime $(M,g)$ such that one can topologically identify the spacetime manifold as $M=I\times\Sigma$ where $I\subseteq\mathbb{R}$ and $\Sigma$ is some smoothly embedded three-manifold, then $\Sigma$ (which is a spatial section) is allowed to be compact (a common example is of course $\Sigma=S^3$).

The claim that $M$ itself can be compact is of course wrong if one wants to avoid closed (or nearly closed) timelike curves.