Are singularities part of the manifold?

[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.[/itex]

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

 

[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.

 

[George Jones]

Years ago, there was a related discussion between vanesch and me, posts 53, 54, 58, 62, 63 in

https://www.physicsforums.com/showthread.php?p=1251928#post1251928.

I don't know how this is for the present discussion.

 

[martinbn]

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 :)

Yes, sorry about this. I was just commenting on the original question, because I think that's what he was asking, and your example may give him the wrong impression.

 

[martinbn]

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.

But it is not a chart of the same atlas. The way I understand the question is, can e chart map be homeomorphism but not a diffeomorphism, no other structures are considered.

 

[WannabeNewton]

Yes, sorry about this. I was just commenting on the original question, because I think that's what he was asking, and your example may give him the wrong impression.

Yeah but there's no need to apologize, this thread has just been really hard to sort out.

Years ago, there was a related discussion between vanesch and me, posts 53, 54, 58, 62, 63 in

https://www.physicsforums.com/showthread.php?p=1251928#post1251928.

Thanks George!

But it is not a chart of the same atlas. The way I understand the question is, can e chart map be homeomorphism but not a diffeomorphism, no other structures are considered.

Ok then I myself misunderstood the original question.

 

[micromass]

OK, let's throw in some references to end this:

From Lee's smooth manifolds:

Let $M$ be a topological $n$-manifold. A coordinate chart (or just a chart) on $M$ is a pair $(U,\varphi)$, where $U$ is an open subset of $M$ and $\varphi:U\rightarrow \tilde{U}$ is a homeomorphism from $U$ to an open subset $\tilde{U}=\varphi(U)\subseteq \mathbb{R}^n$.

So no smooth structure is required to be a chart.

If $M$ is a smooth manifold, any chart $(U,\varphi)$ contained in the given maximal smooth atlas will be called a smooth chart.

So charts don't need to be diffeomorphisms. While smooth charts are.

 

[micromass]

But it is not a chart of the same atlas. The way I understand the question is, can e chart map be homeomorphism but not a diffeomorphism, no other structures are considered.

Sure, it can be. A chart map is just defined to be a homeomorphism. The smooth structure is irrelevant.
If it turns out to be a diffeomorphism, then it's a smooth chart.

 

[PeterDonis]

The way I understand the question is, can e chart map be homeomorphism but not a diffeomorphism, no other structures are considered.

Since I asked the original question that spawned this subthread, I suppose I should clarify: I was not really trying to impose specific conditions on the question, I was just trying to understand why the distinction between homeomorphisms and diffeomorphisms is drawn at all in this connection. The various examples given show why: a chart map, by itself, does not *have* to be a diffeomorphism, only a homeomorphism. That's what I was trying to clarify; I wasn't concerned with any particular specific examples, only the general question.

 

[martinbn]

Since I asked the original question that spawned this subthread, I suppose I should clarify: I was not really trying to impose specific conditions on the question, I was just trying to understand why the distinction between homeomorphisms and diffeomorphisms is drawn at all in this connection. The various examples given show why: a chart map, by itself, does not *have* to be a diffeomorphism, only a homeomorphism. That's what I was trying to clarify; I wasn't concerned with any particular specific examples, only the general question.

Yes, that is how I understood the question, and the reason for my remark. A differentiable manifold is not just the topological space $M$, but a pair [itex](M\mathcal A)[/tex] and any question about differentiability of maps is in the context of the chosen atlas.[/itex]

 

[PeterDonis]

A differentiable manifold is not just the topological space $M$, but a pair $(M, \mathcal A)$ and any question about differentiability of maps is in the context of the chosen atlas.

This seems to me to be a question of terminology. You're not saying micromass' example is invalid, period; you're just saying it doesn't fit within the definition you're using. I wasn't concerned with any specific set of definitions; I was just looking for examples. Yours, micromass', and WN's have all helped to clarify what's going on.

 

[martinbn]

Well, yes, I agree that what micro and wbn say is correct, I was just giving my take on this. It is indeed a terminological issue. Sorry if I wasn't expressing myself clearly. English is a hard language to write in.