Topology of flat spatime -metric?

[dslowik]

I am studying topology. There I learn that the open sets given by the metric can be used to define a topology, e.g. the usual metric topology on R^n given by the Euclidean metric.

Now I try to understand the topology of (flat) spacetime. Is there a metric? The Lorentz 'metric' gives negative values -thus not a metric as in topology courses, e.g. Munkres. Is spacetime a metric space (in the sense of topology)? If so, what is the metric?

 

[HallsofIvy]

There is no metric in the usual sense. You can write a "pseudo-metric" for space time as $ds^2= dx^2+ dy^2+ dz^2- dt^2$. Or you can let $\tau= it$ and write $ds^2= dx^2+ dy^2+ dz^2+ d\tau^2$.

 

[dextercioby]

A Lorentzian manifold called <flat spacetime> is essentially R^4 endowed with a pseudometric. It's still a topological space in the sense of abstract topology, because one can use the pseudometric to define distances, then open balls and thus open and closed sets, hence a topology.

 

[dslowik]

The Lorentz metric is not a pseudometric. A pseudometric is non-negative and obeys the triangle inequality. Open balls of a pseudometric generate a topology. Pseudometrics can be used to define an equivalence relation or partition of the space; the corresponding quotient space is the metric identification which is also a topological space -Hausdorff unlike the pseudometric space (which ain't even T_o).

I think the topology of spacetime is quite divorced from the Lorentz 'metric'. As far as the topology of ST goes, I think it is much the same situation as the topology of, say, GL(N,R) (where there is no inherent distance or metric) -we identify the N^2 tuples as points in Euclidean R^(N^2) space with the Euclidean metric, this makes the group continuous topological group [G X G -> G and G -> G^-1 continuous maps), in fact a topological manifold i.e., locally homeomorphic to euclidean space. So we can analogously identify the 4-tuple (x,y,z,t) as a pt in R^4, impose the Euclidean metric (thus the 'usual' topology), and ST is thus a topological manifold. but why did we choose this topology? to make it (locally)homeomorphic to euclidean space? we could have chosen the discrete metric instead. Is the justification that this is the weakest topology that makes the group of (4) translations continuous? so we again have a topological group locally euclidean. For some purposes giving ST this topology makes sense e.g. can apply calculus on the ST manifold now. But we are in a situation where the metric used to define the open sets of the ST topology is quite different than the one used to define physically invariant closeness of points.

Can/should ST be given a topology more consistent with the invariant distance (lorentz 'metric')?(for some purposes)? If it was a pseudometric, the quotient map would collapse light cones (open hyperboloid balls) to a points. But it's not and the light cones do not partition ST. I can't think of a pseudo metric that gives pts on light cones zero seperation. Future (or past) point light (half) cones could partition ST and the quotient space would be a 1D timelike curve.
Is there a topology that ties together points separated by null curves? etc

 

[lavinia]

I am studying topology. There I learn that the open sets given by the metric can be used to define a topology, e.g. the usual metric topology on R^n given by the Euclidean metric.

Now I try to understand the topology of (flat) spacetime. Is there a metric? The Lorentz 'metric' gives negative values -thus not a metric as in topology courses, e.g. Munkres. Is spacetime a metric space (in the sense of topology)? If so, what is the metric?

I believe that flat space times have restricted topology because they have zero curvature.
It seems - though I am not sure -that one can replace the Lorentz metric with a Riemannian metric by just reversing the sign of the self inner products of purely time like vectors. This metric will be positive definite and will have zero curvature - I think. If this is right it gives you a flat Riemannian manifold.

Flat Riemannian - as opposed to Lorentzian - manifolds are covered by R^n by the discontinuous action of a group of isometries. These isometry groups have specific structure. They have no elements of finite order and contain a lattice of translations which is normal and the quotient of the group by this lattice is a finite group.

Examples of flat Riemannian manifolds are Euclidean space with the flat metric, flat cylinders, flat tori, and the flat Klein bottle.

In the real world it would seem that the only possible flat space time would be Euclidean space because the others all contain circles or tori or strangely twisted spaces that are not null homologous. What would the physical cause of such topological features be?

For instance, if a flat Klein bottle were a model of space time then the time direction would form a closed loop. So one could go into the past by going into the future then return to the present. Further there would be a point in time when space would suddenly double in volume but half of it would be in the past - though it would be connected to itself in the purely space like direction.

 

[dslowik]

My question is about the 'local' topology, not global topology. It has to do with the open balls, and whether the space is locally homeomorphic to R^n. For example both (0,1) and S^1 are topological manifolds since they are both locally homeomorphic to R^1 (in there usual topologies) and to each other, even though there is no homeomorphism between the entire spaces (global homeomorphism).

The global topology will be governed by the Einstein equations relating the geometry to the matter/energy content. I am not concerned (in my question) about the flatness or curvature issues, just a topological question. I mentioned 'flat' in my original post just to (try to) simplify things...

 

[lavinia]

My question is about the 'local' topology, not global topology. It has to do with the open balls, and whether the space is locally homeomorphic to R^n. For example both (0,1) and S^1 are topological manifolds since they are both locally homeomorphic to R^1 (in there usual topologies) and to each other, even though there is no homeomorphism between the entire spaces (global homeomorphism).

The global topology will be governed by the Einstein equations relating the geometry to the matter/energy content. I am not concerned (in my question) about the flatness or curvature issues, just a topological question. I mentioned 'flat' in my original post just to (try to) simplify things...

Spacetime models start out with a differentiable manifold then add a metric to it.
But as I said in my post, you can reverse the signs of the Lorentz metric to obtain a standard Riemannian metric. And you can recover the topology from it. But... since this is a differentiable manifold you can not even define the metric without first having the topology.

I also think it should be easy to recover the topology directly from the Lorentz metric using the exponential mapping. For a Riemannian manifold, it is a theorem that the topology determined by local distance measurements using geodesics is the same as the original topology of the manifold. This theorem is nothing more than the demonstration that there exists geodesic coordinates around any point.

As far as the topology being determined by Einstein's equations, it is the geometry that is determined. the topology is given. I suppose as in the case of flat space times that there are restricitons on the topology determined by the geometry. What are they?

 

[dslowik]

"ST models start out with a diff manifold, then add a metric to it."
I think that agrees with what I was saying above(may 6 9:01pm), that the Lorentz metric is quite divorced from the topology, that the topology is given as I described there by just identifying the 4-tuple with R^n points. And yes, the motivation to make it differentiable, as in making translation group continuous where group generators are partial derivatives. I think the generation of geodesics via exp map is fundamentally related.

As far as what I said about Einsteins' eq giving the globalll topology, I don't really know what I'm saying there. Yes, the Einstein eqs give the geometry. Doesn't that mean it will say something about the global topology, e.g. whether the manifold closes back on itself eventually? maybe I'm requiring ST to be embedded in higher dimensional flat space when I think of curvature leading to a particular shape or global topology of the manifold and I'm therefore wrong. I haven't gotten that far in my studies..

So I think there is still the interesting point:
If the ST topology is given as locally euclidean, even before we impose the Lorentz metric, the physical significance of null curves does not manifest in the topology. e.g. ST is Hausdorff even though some points have zero ST interval between them. And the metric used to define the topology (Euclidean) differs from the Lorentz 'metric'.

 

[lavinia]

So I think there is still the interesting point:
If the ST topology is given as locally euclidean, even before we impose the Lorentz metric, the physical significance of null curves does not manifest in the topology. e.g. ST is Hausdorff even though some points have zero ST interval between them. And the metric used to define the topology (Euclidean) differs from the Lorentz 'metric'.

All manifolds are locally Euclidean by definition. A Lorentz or Riemannian metric determines the same local topology.





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[George Jones]

The article A new topology for curved space–time which incorporates the causal, differential, and conformal structures by Hawking, King, and McCarthy,

http://jmp.aip.org/resource/1/jmapaq/v17/i2/p174_s1?isAuthorized=no ,

might be of interest.

 

[dslowik]

Yes, I am reading in John Lee's Intro to top manifolds the definition:
1)locally homeomorphic to euclidean
2)hausdorff
3) 2nd countable

Now, Riemannian -locally euclidean -fine. But how is Lorentz locally Euclidean? It isn't a metric (not non-negative, no triangle ineq), or even a pseudo-metric! That's the thing, it seems we imposed a topology without use of a metric..

 

[yenchin]

Yes, I am reading in John Lee's Intro to top manifolds the definition:
1)locally homeomorphic to euclidean
2)hausdorff
3) 2nd countable

Now, Riemannian -locally euclidean -fine. But how is Lorentz locally Euclidean? It isn't a metric (not non-negative, no triangle ineq), or even a pseudo-metric! That's the thing, it seems we imposed a topology without use of a metric..

You might find the Appendix A1 of the following book helpful https://www.amazon.com/dp/1441931023/?tag=pfamazon01-20 It's a math book, not physics.

 

[holy_toaster]

Hi there,

that is an interesting discussion, but as a person working in that field in my PhD, I will try to clarify some points:

Firstly any spacetime has two features: It is a manifold and it supports a Lorentzian metric. In the beginning these two things are not related. Yes, as manifold the spacetime is locally euclidian. But that has nothing to do with the Lorentzian metric on the manifold or with any metric on it. The topology of the spacetime is the "manifold topology", that is independent of any metric on manifold. The "manifold topology" simply exists because any manifold is a topological space.

Now as a next step you can look at any Riemannian metric on this manifold and you will see that the open balls of the distance associated to that Riemannian metric do exactly correspond to the "manifold topology". The open balls of any Riemannian metric form a basis of the "manifold topology".

Now the Lorentzian metric that exists on the manifold is not a real metric in the usual sense. The distance function associated to it does not satisfy the triangle inequality but some kind of reverse triangle inequality. There are vectors of zero or negative length, and so on. The result is that no open balls of a Lorentzian metric can be used to form a basis of the "manifold topology" and in general the Lorentzian metric and the "manifold topology" are not related in any sense.

But, however, there are situations when specific open sets can be formed from the Lorentzian metric on the spacetime, that form a basis for the "manifold topology". Namely if the spacetime is strongly causal (see http://en.wikipedia.org/wiki/Causality_conditions" ). Now if the spacetime is strongly causal the Alexandrov topology coincides with the "manifold topolgy".

I hoped that helped a little to clarify the issue...

 

[dslowik]

Thanks holy_toaster. Given my currently limited knowledge in this area:
-I think everything you said is correct,
-You're confirming that the Lorentz 'metric' structure is distinct from the underlying topological structure, which is that of a Euclidean space (topological manifold).

The fact that it is a topological manifold means that it is a metric space; the pull-back of the euclidean metric via the defining local homeomorphisms give ST a metric whose open sets are equivalent to any Riemannian metric on it.

I hope to get to the point of appreciating the points you made in the last paragraph. And to understand the Hawking et.al paper above. It seems the Lorentz 'metric' can be used to define a topology either different from the original reimannian/euclidean of a topological manifold, or the same under certain special circumstances (strong causal).

But maybe still, fundamentally, why did we make/let ST be a topological manifold(locally euclidean) in the first place? There was no real metric to guide us.

 

[holy_toaster]

But maybe still, fundamentally, why did we make/let ST be a topological manifold(locally euclidean) in the first place? There was no real metric to guide us.

Well, from the viewpoint of the mathematician, you need to have a differentiable manifold in order to be able to do calculus with functions, vector fields, tensors and so on. We want to have a space where locally the process of taking derivatives works in the same way as in euclidian (or say Minkowski) space. But globally the space should be different, i.e. curved or something. And the underlying structure of every differentiable manifold is a topological manifold.

Einstein, when developing GR approached that from a different viewpoint: he wanted to be able to transform into all possible coordinate systems and the laws of physics to look the same after these transformations. The idea is that of general covariance. The result is the same as in the mathematical approach: You need a differentiable manifold.

 

[dextercioby]

But maybe still, fundamentally, why did we make/let ST be a topological manifold(locally euclidean) in the first place? There was no real metric to guide us.

As per definition, any manifold is locally euclidean and essentially a topological space. You don't need a metric to build a manifold, however the useful manifolds are metrizable and paracompact. Essentially, you need to reconcile the concept of <relativistic interval> which is the key point of SR with the topological notion of <metric>. And then you add the curves, the (semi)cones and all.

 

[dslowik]

Yes, differentiability and ability to do calculus is important. Is this the same as demanding the translation group be continuous?

For some reason I can read those Latex I(z) symbols in your earlier post now. That sounds similar to what I was trying to say back on my May 6 9:01pm post. Also there it goes into more detail of the differentiability of the manifold being equivalent to the continuity of the translation group.

 

[quasar987]

Well, from the viewpoint of the mathematician, you need to have a differentiable manifold in order to be able to do calculus with functions, vector fields, tensors and so on. We want to have a space where locally the process of taking derivatives works in the same way as in euclidian (or say Minkowski) space. But globally the space should be different, i.e. curved or something. And the underlying structure of every differentiable manifold is a topological manifold.

Einstein, when developing GR approached that from a different viewpoint: he wanted to be able to transform into all possible coordinate systems and the laws of physics to look the same after these transformations. The idea is that of general covariance. The result is the same as in the mathematical approach: You need a differentiable manifold.

Here's an idea. Some parts may not make sense. But I'd like your opinion on it.

Fact: there are topological 4-manifolds that admit no C^k-structure for any k>0.
Meaning that given any set of coordinates covering the whole spacetime, there is always a pair of overlapping coordiate systems for which there is a "crease" in the transition function. If the universe originated in a singularity, the existence of such a crease sounds plausible, don't you think?. If spacetime is indeed such a manifold, then we must either abandon general covariance (the belief that the laws of nature take the same form in any coordinate system), or we must forfeit differential calculus as the fundamental language in which the laws of nature are expressed. That is, we must find a new mathematical aparatus\language (discrete, or, at most, continuous in nature) in which to express the laws of nature, and their predictions must correspond in some approximation to the solutions of our actual "differential equations formulation" of the laws of nature.

Then, perhaps, in this new formalism, there is no problem between GR and QM! :D

 

[aleazk]

A differentiable manifold is a topological space that is locally homeomorphic to R^n and such that these pieces of the space that "look like" R^n can be sewn together smoothly. So, in general, the entire manifold is not homeomorphic to R^n, only pieces of it. But if your manifold has a flat metric (of whatever signature), then you can cover it entirely with riemann normal coordinates. But this implies that there exists an homeomorphism from all the manifold to R^n. So, the manifold, at least in this case, has all the topological properties of R^n. In fact, as a manifold, IS R^n. Note that the topology in the manifold is the "manifold topology", i.e., the topology that the manifold has from it's definition, induced by the maps F:OcM to UcR^n (an open set in M has the form F^-1[V], where V is an open set in R^n with the standard topology).

 

[quasar987]

A differentiable manifold is a topological space that is locally homeomorphic to R^n and such that these pieces of the space that "look like" R^n can be sewn together smoothly. So, in general, the entire manifold is not homeomorphic to R^n, only pieces of it. But if your manifold has a flat metric (of whatever signature), then you can cover it entirely with riemann normal coordinates. But this implies that there exists an homeomorphism from all the manifold to R^n. So, the manifold, at least in this case, has all the topological properties of R^n. In fact, as a manifold, IS R^n.

This is obviously false as stated. For instance the n-torus has a flat metric which it inherits from its covering by R^n. Similarly, the covering of the Klein bottle by the 2-torus allows one to put a flat metric on the bottle. Neither of these are homeomorphic to R^n.

What is true though is that if M is a connected, simply connected, complete, flat riemannian manifold, then you can cover it entirely with riemann normal coordinates so it is diffeomorphic to R^n. This is a special case of Cartan-Hadamard's theorem.

I don't know if something similar holds for pseudo-riemannian metrics though. Are you saying it does? In fact, I don't know how closely the pseudo-riemannian theory follows the riemannian one. For instance, is there a Nash embedding theorem for pseudo-riemannian manifolds (can any pseudo-riemannian manifold of signature (p,q) be embedded as a submanifold of R^n with the standard metric of signature (p,q)?)

 

[aleazk]

This is obviously false as stated. For instance the n-torus has a flat metric which it inherits from its covering by R^n. Similarly, the covering of the Klein bottle by the 2-torus allows one to put a flat metric on the bottle. Neither of these are homeomorphic to R^n.

What is true though is that if M is a connected, simply connected, complete, flat riemannian manifold, then you can cover it entirely with riemann normal coordinates so it is diffeomorphic to R^n. This is a special case of Cartan-Hadamard's theorem.

I don't know if something similar holds for pseudo-riemannian metrics though. Are you saying it does? In fact, I don't know how closely the pseudo-riemannian theory follows the riemannian one. For instance, is there a Nash embedding theorem for pseudo-riemannian manifolds (can any pseudo-riemannian manifold of signature (p,q) be embedded as a submanifold of R^n with the standard metric of signature (p,q)?)

Hi, yes, you are right. I was a little sloppy there. Suppose that the entire manifold is diffeomorphic to R^n via the exponential map (then is homeomorphic to R^n). This implies that you can cover the manifold entirely with riemann normal coordinates, and so the metric is flat. So, as you say, if the metric is flat this not necessarily implies that the manifold is homeomorphic to R^n. But if the manifold is diffeomorphic to R^n, and then homeomorphic to R^n, this implies a flat metric, which is the case of flat spacetime covered entirely with inertial coordinates.
With respect to "pseudo-riemannian following riemannian" I suppose that this result still apply because the exponential map is defined in the same way and riemann coordinates still implies that the first derivatives of the metric vanishes. I simply don't know if the Nash theorem is still valid in the form you say.

 

[holy_toaster]

What is true though is that if M is a connected, simply connected, complete, flat riemannian manifold, then you can cover it entirely with riemann normal coordinates so it is diffeomorphic to R^n. This is a special case of Cartan-Hadamard's theorem.

I don't know if something similar holds for pseudo-riemannian metrics though. Are you saying it does? In fact, I don't know how closely the pseudo-riemannian theory follows the riemannian one. For instance, is there a Nash embedding theorem for pseudo-riemannian manifolds (can any pseudo-riemannian manifold of signature (p,q) be embedded as a submanifold of R^n with the standard metric of signature (p,q)?)

There is no Cartan-Hadamard Theorem in pseudo-Riemannian geometry. I am not sure how things are for general signatures (p,q), but for the Lorentzian case of signature (1,n-1), you always need some additional conditions for theorems that connect the geometry of the manifold with its topology. Mostly what is needed is some causality condition, e.g. global hyperbolicity. A theorem similar to Nash's embedding theorem has only been proven recently in the Lorentzian case and requires stable causality (see here: http://arxiv.org/abs/0812.4439" )

 

[George Jones]

For instance, is there a Nash embedding theorem for pseudo-riemannian manifolds (can any pseudo-riemannian manifold of signature (p,q) be embedded as a submanifold of R^n with the standard metric of signature (p,q)?)

Chris Clarke* worked on this. In particular. every 4-dimensional spacetime can be embedded isometically in higher dimensional flat space, and that 90 dimensions suffices - 87 spacelike and 3 timelike. A particular spacetime may be embeddable in a flat space that has dimension less than 90, but 90 guarantees the result for all possible spacetimes.

* Clarke, C. J. S., "On the global isometric embedding of pseudo-Riemannian
manifolds," Proc. Roy. Soc. A314 (1970) 417-428

http://rspa.royalsocietypublishing.org/content/314/1518/417

 

[dextercioby]

Chris Clarke* worked on this. In particular. every 4-dimensional spacetime can be embedded isometically in higher dimensional flat space, and that 90 dimensions suffices - 87 spacelike and 3 timelike. A particular spacetime may be embeddable in a flat space that has dimension less than 90, but 90 guarantees the result for all possible spacetimes.

* Clarke, C. J. S., "On the global isometric embedding of pseudo-Riemannian
manifolds," Proc. Roy. Soc. A314 (1970) 417-428

http://rspa.royalsocietypublishing.org/content/314/1518/417

Good reference, George! I remember that the tiny GR book of Dirac (1975) used an embedding of the curved Lorentzian space-time in a flat space of dimension "n" when discussing the concept of parallel transport. I wonder if Dirac presented this in the light of the rigorous result appeared in 1970 in a journal where he published most of his masterpieces.

 

[quasar987]

Thanks George. Nice find!