Are singularities part of the manifold?

[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.

 

[TrickyDicky]

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 $\mathbb{R}^{n}$.

Ok, so I suppose that means that either $\mathbb{R}^{4}\setminus L$ is homeomorphic to $\mathbb{R}^{4}$ or there is really no need for U to be homeomorphic to R^n in order to qualify as a coordinate chart.

 

[WannabeNewton]

No it doesn't mean that. It means that being homeomorphic to an open subset of euclidean space, being homeomorphic to euclidean space, and being homeomorphic to an open ball in euclidean space are equivalent. No one said that the neighborhoods in the manifold for each of the equivalent statements have to be the same.

 

[TrickyDicky]

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.

You mean here?

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.)

It wasn't very explicit any admission of error on your part here and I missed it, but since you say it...
I'm not sure if others agree with this(someone promised something to me if this was the case).
Are these manifolds covered by a single chart or not?

 

[PeterDonis]

You mean here?

I also mentioned the closed FRW case in post #47. I thought I'd mentioned the Kruskal case in another post besides the one you quoted, but I can't find one.

 

[TrickyDicky]

there are no global coordinates (defined as it is standard as those that cover the whole manifold) in curved manifolds.

For the record, I want to correct this misconception of mine from post #33.
As stated it is not correct, as Dalespam said the main thing to look out for here(among other circumstances that might also make it impossible to cover them with a single chart) is the topology, since coordinate charts are homeomorphisms(thanks WN for reminding this basic fact). A compact topology like a sphere can not be covered by a single chart for that reason. So even though curved manifolds cannot be said to be uncoverable by a single chart in general those that have a compact topology can.
Now going back to GR I think it would be fair to say that the 2 most important solutions due to its cosmological and solar system consequences and empirical confirmation, are the Schwarzschild spacetime and the FRW one.
The extended Schwarzschild case has a compact component, and so does the closed FRW.
Spatially flat and negatively curved FRW don't have it, I'm not sure about the negative case, but certainly spatially flat FRW spacetime seems a good example of a curved manifold that if defined without including its singularity as it is done in mainstream GR can be covered by a single chart.

As an example, Kruskal coordinates for the Schwarzschild space-time cover the entire manifold.

WN, remember the quote from MTW we found a bit cryptic about covering "nearly all" the manifold, I think I got the meaning, being strict it can't be covered for the reasons already explained, but it is true that the graphic representation is very often given in R^2 which obviates the compact component and certainly in R^2 one just have too lok and see that (if one uses the convenient definition of GR singular manifolds that define them as not containing the singular points you referenced from Hawking and Ellis and others) it is completely covered.

Are these manifolds covered by a single chart or not?

So to answer my own question directly, can we agree(well Peter already did) that the answer is negative in the Kruskal, and closed FRW cases?

 

[WannabeNewton]

I can agree with that. Wald says a similar thing (he only mentions that the usual $U,V$ coordinates take on all values except the singularity, but doesn't mention anything about the $S^2$ part-again suppressing those dimensions). So yeah I was wrong in only thinking about the space leftover when the dimensions of spherical symmetry are suppressed.

 

[Dale]

So to answer my own question directly, can we agree(well Peter already did) that the answer is negative in the Kruskal, and closed FRW cases?

I can definitely agree that the answer is negative for closed FRW and positive for open FRW. I am not certain about Kruskal, but I strongly suspect that the answer is negative and WBN seems to think so also.

 

[TrickyDicky]

I can agree with that. Wald says a similar thing (he only mentions that the usual $U,V$ coordinates take on all values except the singularity, but doesn't mention anything about the $S^2$ part-again suppressing those dimensions). So yeah I was wrong in only thinking about the space leftover when the dimensions of spherical symmetry are suppressed.

Well, I wouldn't say you were wrong, you just weren't exhaustively right at that point

I was indeed misleading and incorrect in #33 and apologize for it.

I'm now wondering about the negatively curved case(without singularities), for instance I know hyperbolic space is topologically like Euclidean space being non-compact and simply connected(at least locally, but unlike it it has a boundary at infinity, so I'm not sure if that would be an obstacle in order to being able to be covered by a single chart, any hint?

 

[TrickyDicky]

I can definitely agree that the answer is negative for closed FRW and positive for open FRW. I am not certain about Kruskal, but I strongly suspect that the answer is negative and WBN seems to think so also.

Thanks for answering, Dale.

 

[stevendaryl]

So to answer my own question directly, can we agree(well Peter already did) that the answer is negative in the Kruskal, and closed FRW cases?

I haven't been following this discussion closely, but what I thought was true was that K-S coordinates (I'm not sure how they differ from Kruskal coordinates) describe the entire manifold except for the singularity, which isn't on the manifold. K-S coordinates don't manage to be one chart for the same reason that spherical coordinates aren't one chart--the angular coordinates aren't single-valued.

Is my understanding wrong?

 

[George Jones]

Some spacetime manifolds are not coverable by a single chart. So what. What is the big deal?

 

[TrickyDicky]

I haven't been following this discussion closely, but what I thought was true was that K-S coordinates describe the entire manifold except for the singularity, which isn't on the manifold.

Well, it certainly does when we make the usual graphical representation that supresses 2 dimensions. But in full 4 dimension which we can't visualize there is
a compact component S^2 that cannot be covered for topological reasons. Ultimately I am not sure this fact have any practical or physical implication other than to be (maybe pedantically) completely mathematically rigorous one should mention it.

(I'm not sure how they differ from Kruskal coordinates)

They are the same thing, yes.

K-S coordinates don't manage to be one chart for the same reason that spherical coordinates aren't one chart--the angular coordinates aren't single-valued.

I'm not sure I follow the part about "single-valued".

 

[TrickyDicky]

Some spacetime manifolds are not coverable by a single chart. So what. What is the big deal?

Hi, George, it is no big deal I guess, the issue was triggered by a disagreement about simultaneity of relativity in GR, see the first post of the thread.

 

[stevendaryl]

Well, it certainly does when we make the usual graphical representation that supresses 2 dimensions. But in full 4 dimension which we can't visualize there is
a compact component S^2 that cannot be covered for topological reasons. Ultimately I am not sure this fact have any practical or physical implication other than to be (maybe pedantically) completely mathematically rigorous one should mention it.

They are the same thing, yes.


I'm not sure I follow the part about "single-valued".

On the surface of a unit sphere, using coordinates $\theta, \phi$, the coordinates $\theta, \phi$ and the coordinates $\theta, \phi + 2 \pi$ refer to the same point. The point $\theta=0, \phi = 0$ is the same point ast $\theta = 0, \phi > 0$. The point $\theta=\pi, \phi = 0$ is the same point ast $\theta = \pi, \phi > 0$. But within a 2D coordinate chart, the mapping between the manifold and $R^2$ must be one-to-one. So you have to break the sphere into at least two charts: for example:

1. Chart 1: the region $0 < \theta < \pi$, $0 < \phi< 2 \pi$
2. Chart 2: the region formed by a long thin rectangle enclosing all the points with $\phi = 0$

 

[TrickyDicky]

On the surface of a unit sphere, using coordinates $\theta, \phi$, the coordinates $\theta, \phi$ and the coordinates $\theta, \phi + 2 \pi$ refer to the same point. The point $\theta=0, \phi = 0$ is the same point ast $\theta = 0, \phi > 0$. The point $\theta=\pi, \phi = 0$ is the same point ast $\theta = \pi, \phi > 0$. But within a 2D coordinate chart, the mapping between the manifold and $R^2$ must be one-to-one. So you have to break the sphere into at least two charts: for example:

1. Chart 1: the region $0 < \theta < \pi$, $0 < \phi< 2 \pi$
2. Chart 2: the region formed by a long thin rectangle enclosing all the points with $\phi = 0$

Yes, certainly, this is what we have been talking about, the K-S chart can't cover completely S2XR2 due to this.