Topological constrains for the solutions of EFE

[cianfa72]

TL;DR

About the constrains for the topology of spacetime given by Robertson - Walker metric

I'm keep watching the lectures on GR from F.P. Shuller and D. Giulini -- International Winter School on Gravity and Light.

As far as I can tell, the spacetime metric $g_{a b}$ in the Robertson-Walker (RW) form relies on the hypothesis of spatial homogeneity & isotropy (which provide constraints for the spacelike hypersurfaces of the relevant spacetime foliations). This implies a constrain on the topology of the spacetime itself. Namely its topology must be the product topology of a maximally symmetric 3D Riemman manifold $\Sigma$ times $\mathbb R$ -- see Lecture 18.

In the definition of product topology enter the topology of $\Sigma$ as topological 3D manifold plus the standard topology of $\mathbb R$. So far so good.

I was thinking that assumption might be too restrictive, putting a too strong constrain on the possible solutions of Einstein Field Equations (EFEs).

What do you think about ?

 

[PeterDonis]

I was thinking that assumption might be too restrictive, putting a too strong constrain on the possible solutions of Einstein Field Equations (EFEs).

The topology constraint you refer to is global, not local. But the EFE is local. A global topological constraint cannot restrict the solutions of a local equation.

To put this another way: the EFE is a tensor equation, and tensor equations are equations in the tangent space at a point in spacetime. A "solution" of the EFE is a solution of the tensor equation that is valid in the tangent space at every point in some open region of spacetime. Every possible global topology has such open regions. So the global topology can't restrict what EFE solutions you can get.

 

[cianfa72]

The topology constraint you refer to is global, not local. But the EFE is local. A global topological constraint cannot restrict the solutions of a local equation.

So why does the lecturer (D. Giulini) claim that condition/restriction on the (global) topology of spacetime in that lecture ?

 

[martinbn]

I was thinking that assumption might be too restrictive, putting a too strong constrain on the possible solutions of Einstein Field Equations (EFEs).

What do you think about ?

Which assumption?

 

[cianfa72]

Which assumption?

The lecturer claims that, to write down EFE's solutions in a simplified form (like RW metric), the spacetime must be globally hyperbolic. This is equivalent to say that its topology must be the product $\mathbb R \times \Sigma$ where $\Sigma$ is a 3D maximally symmetric manifold.

 

[martinbn]

The lecturer claims that, to write down EFE's solutions in a simplified form (like RW metric), the spacetime must be globally hyperbolic. This is equivalent to say that its topology must be the product $\mathbb R \times \Sigma$ where $\Sigma$ is a 3D maximally symmetric manifold.

Ok, but i dont understand your question, or werher you have a question.

 

[PeterDonis]

So why does the lecturer (D. Giulini) claim that condition/restriction on the (global) topology of spacetime in that lecture ?

In technical language, he's talking about the possible global topology of the maximal analytic extension of a spacetime whose local solution of the EFE meets the conditions he stated.

The lecturer claims that, to write down EFE's solutions in a simplified form (like RW metric), the spacetime must be globally hyperbolic.

That's a much more general condition, which is believed to apply to any physically realistic spacetime.

This is equivalent to say that its topology must be the product $\mathbb R \times \Sigma$ where $\Sigma$ is a 3D maximally symmetric manifold.

No, it's not. A globally hyperbolic spacetime can have other topologies. For example, maximally extended Schwarzschild spacetime is globally hyperbolic, and does not have a topology of this form.

 

[PeterDonis]

its topology must be the product $\mathbb R \times \Sigma$ where $\Sigma$ is a 3D maximally symmetric manifold.

It's worth noting that this is not purely a topological restriction; the concept of a "maximally symmetric manifold" requires a metric and is not a topological property.

 

[cianfa72]

Ok, but i dont understand your question, or werher you have a question.

My goal was to better understand the point he made there. At minue 37:30 he claims that in 4D, the global hyperbolicity condition (i.e. spacetime admits a Cauchy hypersurface) is equivalent to be homeomorphic to the product $\mathbb R \times \Sigma$ where $\Sigma$ is a 3D topological manifold.

 

[cianfa72]

It's worth noting that this is not purely a topological restriction; the concept of a "maximally symmetric manifold" requires a metric and is not a topological property.

Ah yes, definitely. Indeed in those lectures maximally symmetric condition enter later to write down the metric in RW form.

 

[PeterDonis]

At minue 37:30 he claims that in 4D, the global hyperbolicity condition (i.e. spacetime admits a Cauchy hypersurface) is equivalent to be homeomorphic to the product $\mathbb R \times \Sigma$ where $\Sigma$ is a 3D topological manifold.

I don't see this claim being made in the lecture. I strongly suspect you are misinterpreting something.

As I have already said in post #7, the claim you are stating in the quote above is false; global hyperbolicity is a much more general condition than the topological criterion you give. I even gave you an example of a globally hyperbolic spacetime with a different topology.

It is true that the 3D surfaces with topology $\Sigma$ in a homogeneous and isotropic spacetime are Cauchy surfaces for the spacetime. But that is a much weaker statement than the claim you are making in the quote above.

 

[martinbn]

No, it's not. A globally hyperbolic spacetime can have other topologies. For example, maximally extended Schwarzschild spacetime is globally hyperbolic, and does not have a topology of this form.

Hm, why not?

My goal was to better understand the point he made there. At minue 37:30 he claims that in 4D, the global hyperbolicity condition (i.e. spacetime admits a Cauchy hypersurface) is equivalent to be homeomorphic to the product $\mathbb R \times \Sigma$ where $\Sigma$ is a 3D topological manifold.

I think this is true in one direction. In the other you need more, $\Sigma$ has to be Cauchy surface.

 

[PeterDonis]

why not?

The global topology of maximally extended Schwarzschild spacetime is $R^2 \times S^2$. The simplest way to see this is to look at a diagram of it in Kruskal coordinates and note that the locus $r = 0$ is not part of the spacetime manifold at all; every point in the spacetime in Kruskal coordinates has $r > 0$. So every Cauchy surface has topology $R \times S^2$, and the global topology is $R$ times that. And $R \times S^2$ is not the global topology of any maximally symmetric manifold (the only two possibilities there are $S^3$ and $R^3$).

I think this is true in one direction. In the other you need more, $\Sigma$ has to be Cauchy surface.

The claim you are quoting is not true in either direction. The spacetime admitting a Cauchy surface does not logically imply that its topology is homeomorphic to $R \times \Sigma$. And, as you appear to realize, the global topology being $R \times \Sigma$ does not logically imply that the spacetime has a Cauchy surface. Of course you can assume that $\Sigma$ is a Cauchy surface, but that's not proving anything at all, it's just assuming the condition you want.

 

[PAllen]

It seems to me that global homogeneity and isotropy, formalized as requiring the product of R and a maximally symmetric, homeneous 3-manifold, is both a global topology constraint and a local geometric constraint. Of course it limits the EFE solutions since only those of the FLRW metric (in a coordinate independent sense) could satisfy the geometric constraint. In addition, it limits the topology. For example, in the case of flat spatial slices, a 3-cylinder topology for them would satisfy the geometry, but would not be isotropic.

 

[PeterDonis]

It seems to me that global homogeneity and isotropy, formalized as requiring the product of R and a maximally symmetric, homeneous 3-manifold, is both a global topology constraint and a local geometric constraint.

I'm not sure I agree that there is just one constraint here.

First, we have a global topological constraint: the global topology must be $R \times \Sigma$, where $\Sigma$ is some 3-manifold topology. But topology alone can't restrict $\Sigma$ any further.

Second, we have the constraint that $\Sigma$ must be a maximally symmetric manifold. But as I've already commented, that's a geometric constraint, since "maximally symmetric" is a geometric concept. Geometrically, I would say that constraint is local--it constrains what EFE solutions will work.

Third, we have the consequence of the second constraint above on the global topology of $\Sigma$--it has to be $S^3$ or $R^3$, because those are the only two possible global topologies for a maximally symmetric 3-manifold.

Conflating all of the above does not, IMO, mean we have just one constraint. It means the lecturer was glossing over details.

 

[PAllen]

I'm not sure I agree that there is just one constraint here.

First, we have a global topological constraint: the global topology must be $R \times \Sigma$, where $\Sigma$ is some 3-manifold topology. But topology alone can't restrict $\Sigma$ any further.

To me, the idea is that formalizing the notion of global isotropy and homogeneity leads to both topological constraints and geometric constraints. The single statement that the spacetime must be a product of R and a maximally symmetric 3-manifold, contains both. A single statement can have multiple consequences (in this case, 2 topological constraints and a geometric constraint).

Second, we have the constraint that $\Sigma$ must be a maximally symmetric manifold. But as I've already commented, that's a geometric constraint, since "maximally symmetric" is a geometric concept. Geometrically, I would say that constraint is local--it constrains what EFE solutions will work.

Maximally symmetric 3-manifold also contains a topological constraint on the 3-manifolds, viz the example I gave of a flat 3-cylinder. The FLRW metric would apply everywhere, but the spatial isotropy would be violated.

 

[PeterDonis]

Maximally symmetric 3-manifold also contains a topological constraint on the 3-manifolds

Yes, I've already agreed with that; the only possible topologies are $S^3$ and $R^3$. The cylinder you describe would be $S^2 \times R$--you can put a flat metric on that manifold, but it won't be maximally symmetric (it won't have enough Killing vector fields).

 

[PeterDonis]

A single statement can have multiple consequences

Yes, agreed. I think we're saying basically the same thing.

 

[cianfa72]

Another point related to this. The Lie commutator of any pair of coordinate vector fields (say X,Y,Z,T) vanishes hence the frame field they define is integrable.

Take for instance the FLRW metric in comoving coordinates:
$$ds^2 = - dt^2 + a(t) d\Sigma^2$$
The vector field $T = {\partial} / {\partial_t}$ is pointwise orthogonal to the span defined pointwise by the X,Y,Z spacelike coordinate vector fields. Hence the timelike congruence "at rest" in those coordinates is hypersurface orthogonal.

Now, given the expression of the metric tensor in a coordinate system, from this piece of information alone, nothing can be said about the extension of such metric expression to spacetime region.

 

[PeterDonis]

given the expression of the metric tensor in a coordinate system, from this piece of information alone, nothing can be said about the extension of such metric expression to spacetime region.

What do you mean by this?

 

[cianfa72]

What do you mean by this?

I mean....if I give you an expression for the metric tensor field in some chart and nothing else, you can't say in which region of spacetime such expression applies.

Now if the metric components in that (implied) chart have no singularities at all then, in principle, does make sense to extend it to the whole of spacetime ?

 

[PeterDonis]

if I give you an expression for the metric tensor field in some chart and nothing else, you can't say in which region of spacetime such expression applies.

Yes you can. It applies in whatever region of spacetime is covered by the chart. That's part of the specification of the chart.

if the metric components in that (implied) chart have no singularities at all then, in principle, does make sense to extend it to the whole of spacetime ?

You're thinking of it backwards. As above, when you define the chart, you define the region of spacetime that it applies to. Whether or not that region is "the whole of spacetime" depends on what physical scenario you are trying to model and how you define the chart in relation to that scenario.

Given a chart and a metric on it, you can ask what the maximum analytic extension of that metric is; in that sense, you can ask whether the chart covers "the whole of spacetime" or just a portion of it. (For example, you can discover in this way that the Schwarzschild chart on the region of Schwarzschild spacetime outside the horizon only covers a portion of "the whole of spacetime" given the Schwarzschild metric and assuming it applies everywhere.) But that maximal analytic extension might not be physically reasonable (as the maximal analytic extension of Schwarzschild spacetime is not).

 

[cianfa72]

Yes you can. It applies in whatever region of spacetime is covered by the chart. That's part of the specification of the chart.

Ah ok. So for example it is part of Schwarzschild chart's specification that it covers the open region of spacetime outside the horizon.

Given a chart and a metric on it, you can ask what the maximum analytic extension of that metric is; in that sense, you can ask whether the chart covers "the whole of spacetime" or just a portion of it. (For example, you can discover in this way that the Schwarzschild chart on the region of Schwarzschild spacetime outside the horizon only covers a portion of "the whole of spacetime" given the Schwarzschild metric and assuming it applies everywhere.) But that maximal analytic extension might not be physically reasonable (as the maximal analytic extension of Schwarzschild spacetime is not).

You mean: the attempt to compute the maximum analytic extension of the SC chart that is defined on the open region of SC spacetime outside the horizon (given the SC metric expression in that chart and assuming it applies everywhere) is doomed to fail to cover the entire spacetime.

 

[PeterDonis]

So for example it is part of Schwarzschild chart's specification that it covers the open region of spacetime outside the horizon.

More precisely, it is a part of the exterior Schwarzschild chart's specification that it covers the coordinate range $2m < r < \infty$ and $- \infty < t < \infty$ (and the proper range of angular coordinates as well, of course). The fact that this is the open region outside the horizon is something you discover when you discover that the locus $r = 2m$ is an event horizon.

You mean: the attempt to compute the maximum analytic extension of the SC chart that is defined on the open region of SC spacetime outside the horizon (given the SC metric expression in that chart and assuming it applies everywhere) is doomed to fail to cover the entire spacetime.

No. The maximum analytic extension is an extension of the spacetime geometry, not the chart. Obviously you can't extend the Schwarzschild chart any further, because of the coordinate singularity at $r = 2m$. But you can extend the spacetime geometry further. The maximal analytic extension tells you how much further. Of course, in order to do this, you have to find another chart.

 

[Ibix]

You mean: the attempt to compute the maximum analytic extension of the SC chart that is defined on the open region of SC spacetime outside the horizon (given the SC metric expression in that chart and assuming it applies everywhere) is doomed to fail to cover the entire spacetime.

Also, if you define the exterior Schwarzschild chart only for $r>R$, where $R$ is some constant greater than the Schwarzschild radius, you can stitch lots of different things into the spherical hole. A non-rotating planet, a star, a Dyson sphere, an Oppenheimer-Snyder black hole... Anything spherically symmetric can be dropped in. Knowing that some region is accurately described by the Schwarzschild metric tells you only that the stress-energy distribution is spherically symmetric everywhere, and vacuum in the radius range you know is covered by your chart.

 

[cianfa72]

No. The maximum analytic extension is an extension of the spacetime geometry, not the chart. Obviously you can't extend the Schwarzschild chart any further, because of the coordinate singularity at $r = 2m$. But you can extend the spacetime geometry further. The maximal analytic extension tells you how much further. Of course, in order to do this, you have to find another chart.

Ok, so in the case of SC spacetime geometry, does exist a maximal analytic extension that can be obtained starting from the SC chart and metric in exterior region outside the horizon ?

 

[PeterDonis]

Ok, so in the case of SC spacetime geometry, does exist a maximal analytic extension that can be obtained starting from the SC chart

Not if you try to use that chart. Go read my post again, carefully.

and metric in exterior region outside the horizon ?

For the Schwarzschild spacetime geometry, yes, of course there is a maximal analytic extension. Hint: Kruskal.

 

[cianfa72]

For the Schwarzschild spacetime geometry, yes, of course there is a maximal analytic extension. Hint: Kruskal.

Let's say we transform from SC exterior chart (that's defined for $r > 2m, - \infty < t < \infty$) coordinates to Kruskal-Szekeres (KS) coordinates $(T,X)$ (suppressing the inessential others).

Now, as far as I can understand your point, instead of restricting the target to the image of the domain under the transformation, we can consider it in the open region $$- \infty < X < \infty, - \infty < T^2 - X^2 < 1$$ In this open region the expression of the metric tensor field has not singularities. What we get this way is the maximum analytic extension for SC geometry.

 

[PeterDonis]

What we get this way is the maximum analytic extension for SC geometry.

Yes, that's correct.

 

[Pavel Dubov]

I think the point Giulini is making in his lecture is more about the requirements for a well-posed initial value problem rather than a restriction of the Einstein Field Equations (EFE) themselves. While the EFE is indeed a local tensor equation, the global hyperbolicity is a physical 'must' if we want the universe to be predictable (admitting a Cauchy surface).

It's an interesting tension -how local physics handles these global topological requirements. I've been looking at a similar 'bridge' where internal topological constraints of a matter field (like in certain soliton solutions) seem to influence the local geometry and stiffness of compact objects. It's fascinating how much 'structure' we have to assume to make the math work.