Are singularities part of the manifold?

[PeterDonis]

For a differentiable manifold the structure is preserved by diffeomorphisms between the manifold's subset and Euclidean space."

I just can't see how a diffeomorphism is possible between a single chart in GR covering the whole manifold and Euclidean space.

A diffeomorphism preserves the differentiable structure, but that is not the same as the metric. Preserving the metric is a much stronger condition than just preserving the differentiable structure.

 

[TrickyDicky]

I think that there might be some confusion about what it means to map something to $R^n$. If you have coordinates for a region of spacetime, then you ALREADY have a mapping from that region to $R^n$. That's what coordinates ARE. So every chart can be mapped to $R^n$.

That's what I'm saying is the standard definion.

But these mappings don't mean that you get to use the usual metric on $R^n$.

I didn't mention metrics that I recall. But since we are in the category of pseudoriemannian manifolds the metric is included in the isomorphism.

 

[PeterDonis]

I didn't mention metrics that I recall.

No, but you implicitly included them when you talked about a global inertial coordinate system, since, as I noted in a prior post, you can only have one of those for a particular spacetime geometry (Minkowski spacetime), i.e., a particular metric.

But since we are in the category of pseudoriemannian manifolds the metric is included in the isomorphism.

Which isomorphism? If you're talking about an arbitrary coordinate transformation as that term is used when defining general covariance, then yes, the transformation has to preserve the metric (i.e., the geometry), but that doesn't mean you can take a chart on any metric you like and do a coordinate transformation to a global inertial chart on the same metric; obviously you can only do that if the geometry was Minkowski to begin with. But you can still do coordinate transformations on other metrics; you just can't do one that gets you a global inertial chart if the geometry doesn't admit one.

 

[PeterDonis]

I just can't see how a diffeomorphism is possible between a single chart in GR covering the whole manifold and Euclidean space.

You didn't give a reference for where you were quoting this definition from, but many sources are sloppy about using the term "Euclidean space"; they use it both to refer to the topological manifold, R4, without any metric structure added, *and* to refer to R4 with a Euclidean metric on it. Those are different mathematical objects. Since the quote says "interest focuses on the topological structure", that indicates the first usage, where "Euclidean space" only means the topological space R4, with no metric structure assumed or used.

As stevendaryl said, and you agreed, if you have a coordinate chart on some manifold, you already have a mapping between that manifold and R4, the topological space. The metric doesn't come into it. But obviously such a chart can only be a global inertial chart if the metric of the manifold you're charting is Minkowski.

 

[TrickyDicky]

You didn't give a reference for where you were quoting this definition from, but many sources are sloppy about using the term "Euclidean space"; they use it both to refer to the topological manifold, R4, without any metric structure added, *and* to refer to R4 with a Euclidean metric on it. Those are different mathematical objects. Since the quote says "interest focuses on the topological structure", that indicates the first usage, where "Euclidean space" only means the topological space R4, with no metric structure assumed or used.

As stevendaryl said, and you agreed, if you have a coordinate chart on some manifold, you already have a mapping between that manifold and R4, the topological space. The metric doesn't come into it. But obviously such a chart can only be a global inertial chart if the metric of the manifold you're charting is Minkowski.

Yes, I agree with this, so here is the thing, I think general covariance means diffeomorphism invariance, so it includes all diffeomorphisms, those that preserve the metric and those that don't.
Then it is clear as you said that a change of coordinates from say Kruskal to Minkowskian coordinates wouldn't preserve the metric, but it would still be a diffeomorphism, so it would be included under the general diffeomorphism invariance of general covariance, but you probably agree with me such a transformation would change the physics (after all it takes us from curved to flat metric).

 

[PeterDonis]

I think general covariance means diffeomorphism invariance, so it includes all diffeomorphisms, those that preserve the metric and those that don't.

The term "general covariance" is another one that sources tend to get sloppy about. There are two possible ways to interpret it:

(1) If I take a particular spacetime geometry, I can pick any coordinate chart I like on that geometry, and write down the metric for that geometry in my chosen chart, and the metric I write down will be a solution of the Einstein Field Equation. The transformation between any two such charts will be a diffeomorphism. Furthermore, when I compute physical invariants, they will be the same regardless of which chart I pick (since the underlying geometry is the same).

(2) If I take a particular solution of the Einstein Field Equation, expressed in a particular chart, I can do a coordinate transformation that changes the metric, but still ends up with a (different) metric that is a solution of the Einstein Field Equation. Any such transformation will also be a diffeomorphism.

The purported transformation from Kruskal to Minkowski coordinates that you refer to would be an example of #2, if it actually existed; however, it doesn't as you state it, at least not as a diffeomorphism, because the underlying topological spaces are different (Kruskal coordinates are on R2 X S2, the underlying manifold of maximally extended Schwarzschild spacetime, while Minkowski coordinates are on R4). (You could restate it as a transformation between a patch of Kruskal/Schwarzschild spacetime and a patch of Minkowski spacetime, and have it be a diffeomorphism.)

I personally don't find transformations of type #2 very interesting, because you're basically changing everything that's of physical interest, so what's the point? But mathematically, they do exist, and "general covariance" can be interpreted as including them, yes.

 

[TrickyDicky]

The term "general covariance" is another one that sources tend to get sloppy about. There are two possible ways to interpret it:

(1) If I take a particular spacetime geometry, I can pick any coordinate chart I like on that geometry, and write down the metric for that geometry in my chosen chart, and the metric I write down will be a solution of the Einstein Field Equation. The transformation between any two such charts will be a diffeomorphism. Furthermore, when I compute physical invariants, they will be the same regardless of which chart I pick (since the underlying geometry is the same).

(2) If I take a particular solution of the Einstein Field Equation, expressed in a particular chart, I can do a coordinate transformation that changes the metric, but still ends up with a (different) metric that is a solution of the Einstein Field Equation. Any such transformation will also be a diffeomorphism.

The purported transformation from Kruskal to Minkowski coordinates that you refer to would be an example of #2, if it actually existed; however, it doesn't as you state it, at least not as a diffeomorphism, because the underlying topological spaces are different (Kruskal coordinates are on R2 X S2, the underlying manifold of maximally extended Schwarzschild spacetime, while Minkowski coordinates are on R4). (You could restate it as a transformation between a patch of Kruskal/Schwarzschild spacetime and a patch of Minkowski spacetime, and have it be a diffeomorphism.)

I personally don't find transformations of type #2 very interesting, because you're basically changing everything that's of physical interest, so what's the point? But mathematically, they do exist, and "general covariance" can be interpreted as including them, yes.

I see what you say about Kruskal to Minkowski not being a diffeomorphism.But that kind of complicates things further since it seems it is not even a homeomorphism to R^4, so I'm not sure how does the single Kruskal chart qualify as a manifold chart that demands it to be at least a homeomorphism to R^n.
Let's take another example spatially flat FRW coordinates to minkowskian coordinates transformation, this one seems to qualify as a diffeomorphism, which would apparently put general covariance in trouble.

 

[George Jones]

I see what you say about Kruskal to Minkowski not being a diffeomorphism.But that kind of complicates things further since it seems it is not even a homeomorphism to R^4, so I'm not sure how does the single Kruskal chart qualify as a manifold chart that demands it to be at least a homeomorphism to R^n.

According to "Introduction to Smooth Manifolds" by Lee,

equivalent definitions of locally Euclidean spaces are obtained if instead of requiring $U$ to be homeomorphic to an open subset of $\mathbb{R}^n$, we rquire it to be homeomorpic to an open ball in $\mathbb{R}^4$, or to $\mathbb{R}^n$ itself.

Let's take another example spatially flat FRW coordinates to minkowskian coordinates transformation, this one seems to qualify as a diffeomorphism, which would apparently put general covariance in trouble.

Open FLRW universes and Minkowski spacetime are all homeomorphic to $\mathbb{R}^4$. How does this "put general covariance in trouble"?

 

[TrickyDicky]

According to "Introduction to Smooth Manifolds" by Lee,

So in your opinion a coordinate transformation from Kruskal to Minkowski coordinates is a diffeomorphism?

Open FLRW universes and Minkowski spacetime are all homeomorphic to $\mathbb{R}^4$. How does this "put general covariance in trouble"?

Again if it qualifies as a diffeomorphism to change from FRW to Minkowski coordinates and we define general covarince as invariance of the general physics laws under diffeomorphisms(diffeomorphism invariance) it would seem that such a transformation that doesn't preserve the metric from curved geomtry to flat one doesn't leave the physics invariant.

 

[George Jones]

So in your opinion a coordinate transformation from Kruskal to Minkowski coordinates is a diffeomorphism.

There is a diffeomrphism from the domain of a Kruskal chart onto $\mathbb{R}^4$.

 

[TrickyDicky]

There is a diffeomrphism from the domain of a Kruskal chart onto $\mathbb{R}^4$.

Ok, thanks , Minkowski coordinates are of course also diffeomorphic to $\mathbb{R}^4$.

 

[WannabeNewton]

A diffeomorphism only preserves the smooth structure. This is trivial stuff. An isometry preserves the Riemannian structure in the context of differential geometry, and it preserves the metric structure in the context of real analysis. Diffeomorphism invariance only refers to the smooth atlas. Seriously all of these unnecessarily long discussions can be solved by just reading a proper text because this is very basic stuff.

 

[TrickyDicky]

A diffeomorphism only preserves the smooth structure. This is trivial stuff. An isometry preserves the Riemannian structure in the context of differential geometry, and it preserves the metric structure in the context of real analysis. Diffeomorphism invariance only refers to the smooth atlas. Seriously all of these unnecessarily long discussions can be solved by just reading a proper text because this is very basic stuff.

Great, go ahead and clarify my question regarding general covariance.
Maybe you could also clarify if a coordinate chart in the pseudoriemannian manifolds category includes a mapping from the manifold subset to Euclidean space that preserves the Euclidean metric.

 

[PeterDonis]

if ... we define general covarince as invariance of the general physics laws under diffeomorphisms(diffeomorphism invariance) it would seem that such a transformation that doesn't preserve the metric from curved geomtry to flat one doesn't leave the physics invariant.

Preserving the physical law is not the same as preserving the metric. The physical law is the EFE, so a transfomation that changes the metric to another metric that still satisfies the EFE meets the requirements of general covariance.

 

[TrickyDicky]

Preserving the physical law is not the same as preserving the metric. The physical law is the EFE, so a transfomation that changes the metric to another metric that still satisfies the EFE meets the requirements of general covariance.

Ok, if the physical law is the EFE( that is compatible with all kind of unphysical situations by the way)this point is solved.

 

[WannabeNewton]

Maybe you could also clarify if a coordinate chart in the pseudoriemannian manifolds category includes a mapping from the manifold subset to Euclidean space that preserves the Euclidean metric.

Coordinate maps are just homeomorphisms.

Remember that general covariance is a statement about the gauge invariance of GR under diffeomorphisms, not isometries (which are stronger).

 

[Dale]

What is the difference between homeomorphisms, isomorphisms, and diffeomorphisms? I think I am getting them mixed up in my head now.

 

[WannabeNewton]

Isomorphism is a very general term for a structure preserving map. A homeomorphism is a map between two topological spaces that preserves the topological structure i.e. it is a continuous map between two topological spaces with a continuous inverse. A diffeomorphism is a map between two smooth manifolds that preserves the smooth structure i.e. it is a smooth map with a smooth inverse. An isometry between two Riemannian manifolds is a map that preserves the Riemannian structure i.e. it is a diffeomorphism that also preserves the metric tensor.

 

[Dale]

OK, so for a manifold to have a global coordinate chart there must exist a diffeomorphism between the entire manifold and an open subset of Rn?

 

[martinbn]

Isomorphism is a very general term for a structure preserving map.

Be careful! That's a morphism. Isomorphism has to be bijective and the inverse has to be a morphism as well.

 

[WannabeNewton]

OK, so for a manifold to have a global coordinate chart there must exist a diffeomorphism between the entire manifold and an open subset of Rn?

Just a homeomorphism. A chart is a pair $(U,\varphi)$ where $U\subseteq M$ is open and $\varphi: U \rightarrow \mathbb{R}^{n}$ is a homeomorphism.

Be careful! That's a morphism. Isomorphism has to be bijective and the inverse has to be a morphism as well.

Certainly! My mistake.

 

[TrickyDicky]

Coordinate maps are just homeomorphisms.

Sure, but in the category of differentiable manifolds they are diffeomorphisms(transition maps) aren't they?
Since we are dealing here with Riemannian manifolds I wondered if they had to also be isometries in order to maintain in the same category.

 

[PeterDonis]

Sure, but in the category of differentiable manifolds they are diffeomorphisms(transition maps) aren't they?

Every discussion I've read in the literature on general covariance has talked about diffeomorphisms, not just homeomorphisms. Since you need a differentiable structure in order to do calculus, and since tensor calculus plays a central role in the math of GR, I'm not sure how useful it would be to consider homeomorphisms that were not diffeomorphisms.

 

[WannabeNewton]

No Tricky you're sort of mixing up two different things. Say $M$ is a smooth manifold and let $p\in M$. There is a neighborhood $U$ of $p$ and a map $\varphi :U \rightarrow \mathbb{R}^{n}$ such that $\varphi$ is a homeomorphism. Transition maps are something extra. If $(U,\varphi)$ and $(V,\psi)$ are two charts with $U$ and $V$ having non-empty intersection then we require that the transition map $\psi \circ \varphi^{-1}:\varphi(U \cap V)\rightarrow \psi(U\cap V)$ be smooth which will imply that the transition map is also a diffeomorphism.

 

[PeterDonis]

There is a neighborhood $U$ of $p$ and a map $\varphi :U \rightarrow \mathbb{R}^{n}$ such that $\varphi$ is a homeomorphism.

Are there cases where $\varphi$ would be a homeomorphism but not a diffeomorphism?

 

[TrickyDicky]

No Tricky you're sort of mixing up two different things. Say $M$ is a smooth manifold and let $p\in M$. There is a neighborhood $U$ of $p$ and a map $\varphi :U \rightarrow \mathbb{R}^{n}$ such that $\varphi$ is a homeomorphism. Transition maps are something extra. If $(U,\varphi)$ and $(V,\psi)$ are two charts with $U$ and $V$ having non-empty intersection then we require that the transition map $\psi \circ \varphi^{-1}:\varphi(U \cap V)\rightarrow \psi(U\cap V)$ be smooth which will imply that the transition map is also a diffeomorphism.

Thanks, mate.

 

[WannabeNewton]

Sure. Let $M = \mathbb{R}$ and let $\varphi: \mathbb{R}\rightarrow \mathbb{R}$ be given by $\varphi(x) = x^3$. The pair $(\mathbb{R},\varphi)$ is a smooth chart for $\mathbb{R}$ (in fact a global chart) because $\varphi(x)$ is a homeomorphism and $\varphi\circ \varphi^{-1} = id_{\mathbb{R}}$ is of course smooth (and a diffeomorphism as well). However $\varphi$ itself is not a diffeomorphism because $\varphi^{-1}(x) = x^{1/3}$ is not smooth.

 

[Dale]

Sure. Let $M = \mathbb{R}$ and let $\varphi: \mathbb{R}\rightarrow \mathbb{R}$ be given by $\varphi(x) = x^3$. The pair $(\mathbb{R},\varphi)$ is a smooth chart for $\mathbb{R}$ (in fact a global chart) because $\varphi(x)$ is a homeomorphism and $\varphi\circ \varphi^{-1} = id_{\mathbb{R}}$ is of course smooth (and a diffeomorphism as well). However $\varphi$ itself is not a diffeomorphism because $\varphi^{-1}(x) = x^{1/3}$ is not smooth.

Thanks for the example.

 

[WannabeNewton]

No problem!

 

[TrickyDicky]

After checking the mainstream textbook "Analysis, Manifolds and Physics, Part 1: Basics" -page 410 it turns out that R^2XS^2 (Kruskal topology) is not homeomorphic to R^4 after all but to R^4-{line}, just like closed FRW is not homeomorphic to R^4 but to R^4-{point} .
Does this make any difference when it comes to consider whether the single charts that cover those manifolds comply with the formal definition of coordinate chart that includes a homeomorphism from any open subset U ⊂ M (in this case U is the whole manifold) to R^n?

 

[WannabeNewton]

I assume by line you mean a standard line in euclidean space. In this case if $L$ represents the subset containing the line, then $\mathbb{R}^{4}\setminus L$ is still an open subset of $\mathbb{R}^{4}$. For a topological manifold $M$, locally euclidean defined as every point of $M$ having a neighborhood homeomorphic to an open subset of $\mathbb{R}^{n}$ is equivalent to every point of $M$ having a neighborhood homeomorphic to $\mathbb{R}^{n}$ itself which is equivalent to every point of $M$ having a neighborhood homeomorphic to an open ball in $\mathbb{R}^{n}$.

 

[PeterDonis]

After checking the mainstream textbook "Analysis, Manifolds and Physics, Part 1: Basics" -page 410 it turns out that R^2XS^2 (Kruskal topology) is not homeomorphic to R^4 after all but to R^4-{line}, just like closed FRW is not homeomorphic to R^4 but to R^4-{point} .

That matches what I remember from the last time I looked such things up (which was a while ago ).

Does this make any difference when it comes to consider whether the single charts that cover those manifolds comply with the formal definition of coordinate chart that includes a homeomorphism from any open subset U ⊂ M (in this case U is the whole manifold) to R^n?

I believe I pointed out several posts ago that, because of the issue with spheres not being coverable by a single chart, I was wrong to say that there are single charts covering all of closed FRW spacetime and maximally extended Schwarzschild spacetime. You do need multiple charts because of the S2 or S3 factor in the underlying topological space. A single chart can only cover an open proper subset of these spacetimes.

 

[George Jones]

After checking the mainstream textbook "Analysis, Manifolds and Physics, Part 1: Basics" -page 410 it turns out that R^2XS^2 (Kruskal topology) is not homeomorphic to R^4 after all but to R^4-{line}, just like closed FRW is not homeomorphic to R^4 but to R^4-{point} .
Does this make any difference when it comes to consider whether the single charts that cover those manifolds comply with the formal definition of coordinate chart that includes a homeomorphism from any open subset U ⊂ M (in this case U is the whole manifold) to R^n?

I am lost. What single charts that cover what manifolds?

 

[Dale]

R4 minus a line or a point or any closed subset is still an open subset of R4, so a valid chart. The problem would be if you had to remove the point or the line from the manifold to obtain the homeomorphism (which is what I thought was the case for R2xS2). S2 is not homeomorphic to R2, but S2 minus a point is.

 

[TrickyDicky]

I am lost. What single charts that cover what manifolds?

Kruskal coordinates for extended Schwarzschild and FRW coordinates for closed FRW universe.