Are singularities part of the manifold?

[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.

 

[Dale]

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?

That is my understanding also. (doesn't imply that your understanding is right, just that you aren't alone)

 

[PeterDonis]

That is my understanding also. (doesn't imply that your understanding is right, just that you aren't alone)

It's my understanding as well.

 

[PeterDonis]

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?

I believe the open (k = -1) FRW case (hyperbolic spatial slices) can be covered by a single chart, just like the flat (k = 0) FRW case. It's basically like covering one "wedge" of Minkowski spacetime (say the interior of the future light cone of the origin) by hyperbolic coordinates (similar to Rindler coordinates).

I'm not sure what you mean by "a boundary at infinity".

 

[TrickyDicky]

I believe the open (k = -1) FRW case (hyperbolic spatial slices) can be covered by a single chart, just like the flat (k = 0) FRW case. It's basically like covering one "wedge" of Minkowski spacetime (say the interior of the future light cone of the origin) by hyperbolic coordinates (similar to Rindler coordinates).

You're probably right.

I'm not sure what you mean by "a boundary at infinity".

I think this is more related with negatively curved spaces outside the scope of GR, so it might be too off-topic here, I might take it over to the differential geometry subforum.

 

[martinbn]

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

My take on this is no, if $(U,\varphi)$ is a chart, then the map is always a diffeomorphism. This is tautological since this map is part of the differentiable structure of $M$.

WBN gives an example of homeomorphism which is not a diffeomorphism, but $\mathbb R$ and the map are not part of the atlas that gives the standard differentiable structure of $\mathbb R$. It is not a chart.

 

[WannabeNewton]

The example I gave is certainly an atlas on $\mathbb{R}$; it is trivial to check that it is. There is nothing that says an atlas has to be the "standard" atlas on a given smooth manifold. Every smooth manifold has uncountably many different distinct atlases.

 

[micromass]

My take on this is no, if $(U,\varphi)$ is a chart, then the map is always a diffeomorphism. This is tautological since this map is part of the differentiable structure of $M$.

WBN gives an example of homeomorphism which is not a diffeomorphism, but $\mathbb R$ and the map are not part of the atlas that gives the standard differentiable structure of $\mathbb R$. It is not a chart.

I know this is a terminology issue. But to me a chart is a pair $(U,\varphi)$ where $U$ is open and $\varphi:U\rightarrow \mathbb{R}^n$ is an open embedding. So the smooth structure doesn't enter here.

However, a smooth structure is a collection of charts that are pairswise compatible. Not all charts need to be a part of the smooth structure, so there might as well a chart that is a homeomorphism and a diffeomorphism.

This probably depends on the author though.

 

[micromass]

Every smooth manifold has uncountably many different distinct atlases.

Smooth manifold of dimension >1

/end pedantry

 

[WannabeNewton]

Smooth manifold of dimension >1

/end pedantry

 

[martinbn]

The example I gave is certainly an atlas on $\mathbb{R}$; it is trivial to check that it is. There is nothing that says an atlas has to be the "standard" atlas on a given smooth manifold. Every smooth manifold has uncountably many different distinct atlases.

Yes, but for $\mathbb R$ with this atlas, the map above is a diffeomorphism.

 

[micromass]

Yes, but for $\mathbb R$ with this atlas, the map above is a diffeomorphism.

No, because the codomain of a chart is always $\mathbb{R}^n$ with the Euclidean smooth structure.

 

[martinbn]

No, because the codomain of a chart is always $\mathbb{R}^n$ with the Euclidean smooth structure.

But in the example the domain has the different structure.

 

[micromass]

But in the example the domain has the different structure.

Yes, and it's a valid manifold. If we equip $\mathbb{R}$ with the $x^3$-structure, then this forms a smooth manifold. A chart is a map from this manifold to $\mathbb{R}$ with the Euclidean structure.

 

[martinbn]

Yes, and it's a valid manifold. If we equip $\mathbb{R}$ with the $x^3$-structure, then this forms a smooth manifold. A chart is a map from this manifold to $\mathbb{R}$ with the Euclidean structure.

Yes, and with this structure the map is a diffeomorphism.

 

[micromass]

Yes, and with this structure the map is a diffeomorphism.

No, take $M = \mathbb{R}$ with the $x^3$-structure. Take $\mathbb{R}$ with the Euclidean structure. Take $\phi:M\rightarrow \mathbb{R}:x\rightarrow x$. This is not a diffeomorphism. But it is (to me) a valid chart.

The manifolds are diffeomorphic however.

 

[WannabeNewton]

But in the example the domain has the different structure.

I see what you're saying, and I don't disagree but micromass is speaking of something different from what I said. That should sort out the pandemonium of this already chaotic thread :)

 

[martinbn]

No, take $M = \mathbb{R}$ with the $x^3$-structure. Take $\mathbb{R}$ with the Euclidean structure. Take $\phi:M\rightarrow \mathbb{R}:x\rightarrow x$. This is not a diffeomorphism. But it is (to me) a valid chart.

The manifolds are diffeomorphic however.

This is not the example, the map is $\varphi:M\rightarrow \mathbb R$ with $x\mapsto x^3$. That is the point, the chart map is diffeomorphism for the stracture it gives.

 

[micromass]

This is not the example, the map is $\varphi:M\rightarrow \mathbb R$ with [itex]x\mapsto x^3[/tex]. That is the point, the chart map is diffeomorphism for the stracture it gives.

Sure, no problem there. But $\varphi:M\rightarrow \mathbb{R}:x\rightarrow x$ is also a valid chart map for $M$. But it doesn't agree with the smooth structure we put on $M$.

 

[martinbn]

Sure, no problem there. But $\varphi:M\rightarrow \mathbb{R}:x\rightarrow x$ is also a valid chart map for $M$. But it doesn't agree with the smooth structure we put on $M$.

Yes, of course, but this is not in question here.

 

[micromass]

Yes, of course, but this is not in question here.

It is in question here. We put on $M$ the $x^3$-structure. I say that $\varphi:M\rightarrow \mathbb{R}:x\rightarrow x$ is a valid chart (and thus a homeomorphism), but not a diffeomorphism.