Link btw manifolds and space-time

[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.

 

[vanesch]

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

I think that the "dangers" of CTC are overstated. They don't lead to any paradoxes IMO, because there is no way to keep any "memory" of a previous passage. You cannot have monotonous entropy increase along a CTC, obviously, which would be necessary in order for such a memory to exist, which could be used then to do "paradoxial" things. In other words, by the time you come back to your own past in any way which could influence your actual self, you're long (thermodynamically) dead !

 

[robphy]

I think that the "dangers" of CTC are overstated. They don't lead to any paradoxes IMO, because there is no way to keep any "memory" of a previous passage. You cannot have monotonous entropy increase along a CTC, obviously, which would be necessary in order for such a memory to exist, which could be used then to do "paradoxial" things. In other words, by the time you come back to your own past in any way which could influence your actual self, you're long (thermodynamically) dead !

Playfully paraphrasing a common saying: "It's not all about you [humans]"..
One has to consider if solutions to field equations [e.g. Maxwell equations] are consistent [possibly without violating some physically imposed condition] in that spacetime. (This is akin to (say) studying the solutions to Schrodinger's equation in a potential well with periodic boundary conditions.)

 

[masudr]

Advanced and retarded potentials, anyone?

 

[vanesch]

One has to consider if solutions to field equations [e.g. Maxwell equations] are consistent [possibly without violating some physically imposed condition] in that spacetime. (This is akin to (say) studying the solutions to Schrodinger's equation in a potential well with periodic boundary conditions.)

Yes, but if that's the case (that one cannot obtain any consistent field solution on a spacetime with CTCs) then it should be a *theorem* and not a postulate that CTCs are forbidden, no ?

 

[Haelfix]

Take a FRW universe with negative cosmological constant. One topology that is consistent with the metric (though not the only one) is S^4. You do not necessarily get CtCs.

Now consider a FRW with zero cosmological constant. Usually we take the topology to be simply R^4 (the infinite flat plane). We claim the big bang then happens simultaneously everywhere. However, S^4 is also consistent with this metric. Write S^4 as two copies of R^4 with north and south pole identified. Homotopically, the space contracts to a point. We then say the big bang happens at a point! A true singularity, not just of the metric, but of the topology.

 

[George Jones]

Take a FRW universe with negative cosmological constant. One topology that is consistent with the metric (though not the only one) is S^4. You do not necessarily get CtCs.

But this violates the well-known theorem that every compact spacetime contains closed timelike curves. The simple proof of this theorem is independent of energy conditions and of the cosmological constant; see Theorem 3.3.11 of Naber's Spacetime and Singularities: an Introduction or Proposition 6.4.2 of Hawking and Ellis.

 

[coalquay404]

Take a FRW universe with negative cosmological constant. One topology that is consistent with the metric (though not the only one) is S^4. You do not necessarily get CtCs.

You do get closed timelike curves whenever the spacetime is compact. This is not debatable. See, for example, chapter six of Hawking & Ellis

 

[Haelfix]

Can you list the assumptions in the proof? I do not have the book handy here. AFAIK ctcs do not occur in closed FRW universes b/c the recurrence time is much larger than the big crunch.

So let me change the topology to S^3 * R^1. Satisfied?

 

[vanesch]

So let me change the topology to S^3 * R^1. Satisfied?

You guys are beyond me... but isn't this a non-compact manifold ?

 

[coalquay404]

I'll reply to Haelfix tomorrow (it's ridiculously late here, and there's rugby on tomorrow), but yes, S^3\times\mathbb{R} is non-compact.

 

[George Jones]

AFAIK ctcs do not occur in closed FRW universes b/c the recurrence time is much larger than the big crunch.

So let me change the topology to S^3 * R^1. Satisfied?

The topology of a (simply connected) closed FRW universe is S^3 x R, which is non-compact, and which doesn't contains CTCs. (Note: the theorem does not say that non-compact spacetimes don't contain closed timelike curves). Even FRW universes that have a cosmological constant/dark energy don't have closed CTCs. For example, observations are consistent with (but don't prove) our universe being a closed S^3 x R universe that expands forever. Galaxies (negelecting peculiar velocities) are given fixed comoving coordinates (in space) on S^3, while the R coordinate, cosmological time, increases. (Proper distances are found by using the time-dependent scale factor.) This is the foliation of spacetime that coalquay404 mentioned.

It is possible to go on journey through space, and, without "turning around", return to the same position in space, i.e., to return the starting S^3 coordinates. But time will have advanced, so the R coordinate will be different.

 

[George Jones]

Can you list the assumptions in the proof? I do not have the book handy here.

After its proof, Hawking and Ellis makes an interesting point. No compact spacetime is simply connected, so all compact spacetimes can be formed by identifying points of non-compact spacetimes.

Now I don't have my books handy, but I do have a copy of Naber's proof that I made when my books were in storage. Only the concepts of past and future are used, so I suppose orientability is necessary.

Definition: p << q iff there exists a smooth, future-directed timelike curve from p to q.

Lemma: << is transitive. (If p << q << r, connect p to q by a smooth curve, connect q to r by a smooth curve, and "smooth the corner" at q to obtain a smooth curve from p to r.)

Definition: The chronological future of any event p is I+(p) = {q in M | p << q}.

Lemma: For any p, I+(p) is open. (It's the interior of a future lightcone.)

Theorem: Any compact spacetime M contains closed timelike curves.

Proof: {I+(p) | p in M} is an open cover for M since any q in M is in the future of some p (technical details omitted). Since M is compact, this cover must admit a finite subcover {I+(p_1), I+(p_2), ... , I+(p_n)}. We may assume that I+(p_1) is not contained in any I+(p_j) for j >= 2, otherwise I+(p_1) could be removed from the subcover. But then p_1 is not in I+(p_j) for any j>=2 by the transitivity of <<. Consequently, p_1 is in I+(p_1), i.e., there exists a smooth, future-directed timelike curve from p_1 to p_1.

I think the proof is quite beautiful.

 

[MeJennifer]

Sorry for the late response, I am currently in Thailand and have less access to the internet than normal.

(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?

I don't disagree that space-time is not a Riemann manifold, I was merely paraphrasing the, in my view, incorrect idea that space-time is Riemann and that the (pseudo) metric in some way operates onto this as some sort of algebra.

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

Explain to me how a manifold with non positive definite metric can possibly be Hausdorff?

(3) is also completely incorrect.

So what are you saying here, that the shape of the manifold is not determined by the EFE? What then determines the shape according to you?

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.

It is a plain and simple fact of SR and GR that the distance in space-time between an emmited and absorbed photon is 0. You can foliate the two events in 3 dimensions, but in GR there is no preferred foliation. Note that ether theorists have the opinion that space-time is definitively 3D+1.
Sorry Pervect but to say that two events are "far away" from each other when in fact the ds is zero is complete nonsense and counter to the first principles of the theory of relativity.

Einstein demonstrated that both distance and duration are not absolute concepts, they cannot be taken in isolation! What for one observer is "far away" might be "nearby" for another one and what for one observer is "a long time" might be "a short time" for another one.


Again I am not saying that GR is in some way wrong, I am saying that when we want to use a geometric mathematical model we must admit that the mathematics are incomplete. That does not mean we cannot make any calculations but nevertheless it is incomplete.

And we have not even started to consider how it is logical how EFE equations can lead to singularities on a manifold shaped by only smooth deformations.

 

[Hurkyl]

Explain to me how a manifold with non positive definite metric can possibly be Hausdorff?

How about an example?

The smooth manifold R^2 is Hausdorff. Agreed?

(In the canonical basis) g := dx^2 - dy^2 is a nondegenerate symmetric bilinear form that is not positive definite. Agreed?

Therefore, the pseudo-Riemannian manifold (R^2, g) is Hausdorff pseudo-Riemannian manifold with a metric tensor that is not positive definite.

 

[coalquay404]

I have a reply in progress but since I can't get the LaTeX preview to work, I'll do it offline and upload it in a bit

 

[coalquay404]

Actually, screw the long explanation. MeJennifer, before I continue, you are presumably aware of the differences between topological metricity and tensorial metricity?

(And before you start, yes, I'm perfectly well aware that Hausdorff isn't compatible with pseudometricity in the topological sense.)

 

[MeJennifer]

Actually, screw the long explanation. MeJennifer, before I continue, you are presumably aware of the differences between topological metricity and tensorial metricity?

Well that was my earlier point. It is just a way to avoid the issue.

If we intepret GR as some algebraic system using tensor algebra then we obviously avoid the whole problem, but then there is no manifold and arguably not even a notion of curvature!

You can’t have your cake and eat it!

 

[MeJennifer]

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.

Sorry but forgive me think this is simply a trick to get around the issue of our current mathematical limitations. Our mathematical knowledge is afteral limited right?

Furthermore, this is actually an imposed limitation on the scope of a non definite positive metric. Feel free to demonstrate, by using physical, not mathematical arguments why this imposed limitation is justified.

 

[vanesch]

If we intepret GR as some algebraic system using tensor algebra then we obviously avoid the whole problem, but then there is no manifold and arguably not even a notion of curvature!

In order for there to even be a tensor algebra, you need a manifold to start with. If you have your manifold, and if you have such a tensor algebra, which allows for the definition of a symmetric 2-tensor, you have introduced a (pseudo) metric.
In what way is this incompatible ?

The manifold doesn't come from the metric ; the metric is defined over the manifold. It's a 2-tensor.

But even before the 2-tensor was there, there are topological properties to a manifold (by definition of a manifold). These topological properties are not defined by the metric (they only need to be compatible, in that the topology of the manifold needs to allow for the 2-tensor which defines the metric with necessary properties such as smoothness and signature etc...).

 

[MeJennifer]

Not sure why there is so much resistance to the idea that space-time is non-Hausdorff. I think it is a great feature, but a feature that is obviously beyond our current mathematical understanding.
Many simply don't want to deal with it.

For instance the idea of bifurcating curves opens perspectives for embedding MWI of quantum theory in space-time as mentioned both by Visser and Penrose.