Are singularities part of the manifold?

[TrickyDicky]

http://books.google.com/books?id=Xw...skal extension covers entire manifold&f=false

http://books.google.com/books?id=k8...skal extension covers entire manifold&f=false

http://physics.stackexchange.com/qu...ition-of-a-timelike-and-spacelike-singularity

http://philsci-archive.pitt.edu/2980/1/The-singular-nature-of-space-time.pdf (page 5)

http://catdir.loc.gov/catdir/samples/cam031/72093671.pdf (page 7)

I can see all those references agree to consider the singularity is not part of the manifold because they are trying to do physics and singularities are outside the reach of physics.

 

[TrickyDicky]

Also, none of this changes the fact that the singularity is not part of the manifold. Are you now agreeing with that?

See post above.
But using logic alone, if you say the singularity is not part of the manifold, you are saing that the manifold is singularity-free. See Hawking and Penrose singularity theorems.

 

[PeterDonis]

But using logic alone, if you say the singularity is not part of the manifold, you are saing that the manifold is singularity-free.

Only if you define "singularity-free" in this way. That's not how the term is usually used; see below.

See Hawking and Penrose singularity theorems.

Which say that geodesics in such a manifold can't be extended to arbitrary values of their affine parameters; physical invariants along them increase without bound as some finite value of the affine parameter is approached. This does not require, and the singularity theorems do not state, that the point at which the affine parameter actually *achieves* the finite value that indicates the singularity (infinite values of physical invariants) is part of the manifold. It only requires that the singularity can be approached arbitrarily closely (where "closely" is defined by the value of the affine parameter along the geodesic of approach) while staying within the manifold.

The term "singularity-free", then, doesn't just mean there are no singularities in the manifold; it means something stronger, that there are no singularities in the manifold *and* that all geodesics can be extended to arbitrary values of their affine parameter with all physical invariants along them remaining finite. I believe Hawking and Ellis explicitly give this definition.

 

[TrickyDicky]

I mentioned the theorems to indicate that GR manifolds seem to have singularities, otherwise why worry about them?

 

[PeterDonis]

I mentioned the theorems to indicate that GR manifolds seem to have singularities, otherwise why worry about them?

If "singularities" is defined appropriately, I agree. But the appropriate definition does not imply that there are actual points contained in the manifold that are singularities. It only implies what I said in my last post, that a manifold with singularities has geodesics that can't be extended to arbitrary values of their affine parameters.

 

[TrickyDicky]

If "singularities" is defined appropriately, I agree. But the appropriate definition does not imply that there are actual points contained in the manifold that are singularities.

The fact is there is no definition whatsoever of singularity in GR(only of singular spacetime, which to my surprise according to many is the one that doesn't contain a singularity) unlike in mathematics, so I wonder how it can be appropriately defined at all.

 

[WannabeNewton]

Hawking and Ellis address your exact misconception in section 8.1

 

[Nugatory]

It looks to me like everyone is using different definitions of "simultaneity" and "relativity of simultaneity". Maybe we should taboo those terms in this thread, so that everyone has to explicitly define what they mean by them.

Totally completely against my better judgement I'm going to step into this thread, just long enough to say that the best definition I've been able to come up with is:

Two events are simultaneous if they have the same time coordinate. This definition makessit clear that that simultaneity is a convention based on the choice of coordinates; and it also allows for the SR definition of simultaneity in which every observer chooses to use the coordinate system in which he is at rest.

Relativity of simultaneity is just the observation that different observers using different coordinate systems will will assign different values of the time coordinate to events, and therefore may disagree about which events have the same time coordinate.

 

[stevendaryl]

The fact is there is no definition whatsoever of singularity in GR(only of singular spacetime, which to my surprise according to many is the one that doesn't contain a singularity) unlike in mathematics, so I wonder how it can be appropriately defined at all.

The sort of singularity that is associated with a black hole can be characterized as the existence of a geodesic such that the curvature increases without bound as you follow the geodesic.

 

[PeterDonis]

The sort of singularity that is associated with a black hole can be characterized as the existence of a geodesic such that the curvature increases without bound as you follow the geodesic.

You also have to include that the curvature increases without bound as a finite value of the affine parameter along the geodesic is approached.

 

[PeterDonis]

Relativity of simultaneity is just the observation that different observers using different coordinate systems will will assign different values of the time coordinate to events, and therefore may disagree about which events have the same time coordinate.

This is one way of looking at it, yes. But it's not the only one. You could define simultaneity the way Einstein originally did: two events A and B are simultaneous for a given observer O if (1) the distance from O to A is the same as the distance from O to B; and (2) a light ray from event A reaches O at the same event on O's worldline as a light ray from event B.

Simultaneity is still relative, i.e., observer-dependent, on this definition; but this definition does not require defining any coordinates.

 

[TrickyDicky]

Having understood in what sense it was claimed that several spacetimes in GR can be covered by a single global coordinate system by considering singular points not to be part of the manifold, a doubt remains for me that I would appreciate that it was addressed. GR's general covariance amounts to saying that physical laws should not be affected by arbitrary changes of coordinates. According to this it should be possible to transform the single global coordinate chart of a singular spacetime to any other coordinate system like for instance an inertial coordinate system, but it is not possible in general for a curved manifold to be covered by a single catesian coordinate system.

 

[PeterDonis]

GR's general covariance amounts to saying that physical laws should not be affected by arbitrary changes of coordinates. According to this it should be possible to transform the single global coordinate chart of a singular spacetime to any other coordinate system like for instance an inertial coordinate system

That doesn't follow, because a global "inertial" coordinate system requires a specific spacetime geometry (Minkowski spacetime). If the spacetime geometry is something else, you can make any arbitrary coordinate transformation you want, but it won't get you an inertial coordinate system.

In other words, the presence of a global inertial coordinate system is a feature of one particular solution to the Einstein Field Equation; it's not a general property that any solution must have. But the physical law, the EFE, applies to any solution; and general covariance just means that I can do any arbitrary coordinate transformation I want, without affecting the validity of the EFE or changing the geometry of any particular solution.

 

[TrickyDicky]

That doesn't follow, because a global "inertial" coordinate system requires a specific spacetime geometry (Minkowski spacetime). If the spacetime geometry is something else, you can make any arbitrary coordinate transformation you want, but it won't get you an inertial coordinate system.

In other words, the presence of a global inertial coordinate system is a feature of one particular solution to the Einstein Field Equation; it's not a general property that any solution must have. But the physical law, the EFE, applies to any solution; and general covariance just means that I can do any arbitrary coordinate transformation I want, without affecting the validity of the EFE or changing the geometry of any particular solution.

And changing to an inertial coordinate is not an arbitrary coordinate transformation? It seems to be when dealing with local charts.

 

[TrickyDicky]

Consider the standard definition of a manifold's coordinate chart:
"A coordinate map, a coordinate chart, or simply a chart, of a manifold is an invertible map between a subset of the manifold and a simple space such that both the map and its inverse preserve the desired structure.[7] For a topological manifold, the simple space is some Euclidean space Rn and interest focuses on the topological structure. This structure is preserved by homeomorphisms, invertible maps that are continuous in both directions. 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.

 

[stevendaryl]

Having understood in what sense it was claimed that several spacetimes in GR can be covered by a single global coordinate system by considering singular points not to be part of the manifold, a doubt remains for me that I would appreciate that it was addressed. GR's general covariance amounts to saying that physical laws should not be affected by arbitrary changes of coordinates. According to this it should be possible to transform the single global coordinate chart of a singular spacetime to any other coordinate system like for instance an inertial coordinate system, but it is not possible in general for a curved manifold to be covered by a single catesian coordinate system.

Saying that you can use any coordinates you like just means that

• if you have a description of spacetime in terms of coordinates x,y,z,t, and
• X(x,y,z,t), Y(x,y,z,t), Z(x,y,z,t), T(x,y,z,t) are four functions that are smooth, and
• the map \langle x,y,z,t \rangle \Rightarrow \langle X, Y, Z, T\rangle is invertible, then
• you can use X,Y,Z,T just as well.

As you say, no change of coordinates of this form will transform curved spacetime into flat spacetime.

 

[TrickyDicky]

Saying that you can use any coordinates you like just means that

• if you have a description of spacetime in terms of coordinates x,y,z,t, and
• X(x,y,z,t), Y(x,y,z,t), Z(x,y,z,t), T(x,y,z,t) are four functions that are smooth, and
• the map \langle x,y,z,t \rangle \Rightarrow \langle X, Y, Z, T\rangle is invertible, then
• you can use X,Y,Z,T just as well.

As you say, no change of coordinates of this form will transform curved spacetime into flat spacetime.

Exactly, thanks Steven. So if no diffeomorphism is possible between the global chart and R^n it is evident to me it doesn't correspond to the standard definition of a manifold's coordinate chart that I showed above and that I insisted all along that I was referring to.

 

[Dale]

Having understood in what sense it was claimed that several spacetimes in GR can be covered by a single global coordinate system by considering singular points not to be part of the manifold

Not all manifolds can be covered by a single chart, but that has little to do with singular points, it has to do with topology. The singularities only contribute insofar as they change the topology.

For example, a sphere is a 2D manifold which cannot be covered by a single chart. There are no curvature singularities on a sphere, but it cannot be covered in a single global chart. In contrast, a cone does have a curvature singularity, but it can be covered in a single global chart since it can be mapped to a plane.

 

[stevendaryl]

Exactly, thanks Steven. So if no diffeomorphism is possible between the global chart and R^n it is evident to me it doesn't correspond to the standard definition of a manifold's coordinate chart that I showed above and that I insisted all along that I was referring to.

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

 

[PeterDonis]

And changing to an inertial coordinate is not an arbitrary coordinate transformation? It seems to be when dealing with local charts.

Changing to a local inertial coordinate chart isn't a global transformation; the local inertial chart will only be valid on a small local patch 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?