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.