B How to select the smooth atlas to use for spacetime?

[PeterDonis]

I don't think 'continuous' is enough

You're right, it isn't. "Define a smooth structure" would be a better way to say it, as you do in your post.

 

[PeterDonis]

space-time already has coordinates attached to it (spatial and time up to re-scaling and choosing directions)

This is a bit misstated.

Spacetime, considered as a geometric object, is independent of any choice of coordinates, so it's not really correct to say that it "already has coordinates attached to it".

However, it is true that spacetime, considered as a manifold with (pseudo) metric, must have a Lorentzian signature, which means it is always possible to find an open neighborhood of any event in which you can assign three "space" coordinates and one "time" coordinate in the way that is familiar from special relativity. Or, to put it another way, every event has an open neighborhood in which there is a causal structure (spacelike, timelike, and null vectors, light cone) that is the same as the causal structure of Minkowski spacetime, and that causal structure can be used to make the choice of coordinates just described on the open neighborhood.

Once we have the above, all we need in addition is the condition that where any open neighborhoods defined as above overlap, the coordinates on each open neighborhood must be compatible; there must be a valid coordinate transformation from one to the other, meeting appropriate conditions, which would include having the same smooth structure. So once we've chosen a smooth structure on any open neighborhood, the same smooth structure will end up having to be chosen on the entire spacetime. And of course the choice of smooth structure actually made in GR is the standard smooth structure on $\mathbb{R}^4$.

 

[Dragon27]

Smooth manifold means that we consider a restriction of the maximal atlas such that all charts in it are compatible.

What if there are multiple non-equivalent restrictions of the maximal atlas such that their charts are compatible?

It has already been said, but I just wanted to repeat it explicitly. In smooth manifold theory one usually considers maximal smooth atlas and it's already unique for a given smooth atlas (any smooth atlas is contained in a unique smooth maximal atlas). If a chart is compatible with (the smooth structure of) the smooth maximal atlas, then it should already be contained in it (otherwise we could just add it to the atlas and make it bigger). So a chart from one smooth maximal atlas can't be compatible with a smooth maximal atlas that is distinct from it.

 

[Shirish]

And of course the choice of smooth structure actually made in GR is the standard smooth structure on $\mathbb{R}^4$.

The rest of your post is clear to me, but could you elaborate on this just a bit? I'm assuming by "standard" in the above you don't mean $\mathbb{R}^4$ with standard topology, but the standard Minkowski space that we consider in special relativity. So does the above quote imply that there's a unique differentiable structure on the general spacetime manifold that's compatible with the "standard" (in the sense of being compatible with the SR Minkowski spacetime) locally flat $\mathbb{R}^4$?

 

[PeterDonis]

I'm assuming by "standard" in the above you don't mean $\mathbb{R}^4$ with standard topology, but the standard Minkowski space that we consider in special relativity.

I mean the standard smooth structure on $\mathbb{R}^4$, which does not require any particular metric. It just has to be compatible with having a locally Minkowski metric with standard inertial coordinates in every sufficiently small open neighborhood.

does the above quote imply that there's a unique differentiable structure on the general spacetime manifold that's compatible with the "standard"

I believe there is a unique smooth structure that is compatible with having a locally Minkowski metric with standard inertial coordinates in every sufficiently small open neighborhood.

 

[vanhees71]

Interesting! Do you have a reference for this? I thought to define the standard topology on $\mathbb{R}^n$ you need a proper metric. A pseudo-metric won't do I think.

 

[PeterDonis]

I thought to define the standard topology on $\mathbb{R}^4$ you need a proper metric.

Yes, the standard topology is defined using the standard Euclidean metric. I don't think I have said anything that contradicts that.

What I said is that the smooth structure has to be compatible with having a locally Minkowski pseudo-metric. That does not require that the pseudo-metric be used to define the topology (which of course it can't be).

 

[vanhees71]

I see, that's of course right.

 

[George Jones]

What I said is that the smooth structure has to be compatible with having a locally Minkowski pseudo-metric.

Actually, this isn't very restrictive. Since all differentiable manifolds are topological spaces, a 4-dimenional manifold (4-manifold) is either compact, or it isn't compact. Not all compact 4-manifolds admit Lorentzian metrics (even in math books, "pseudo" is often dropped, as it is understood by context). If a compact 4-manifold does admit a Lorentzian metric, then it necessarily has closed timelike curves, and thus (most likely) is not suitable as a model for spacetime. Every non-compact 4-manifold admits a Lorentzian metric.

 

[cianfa72]

Take $X=Y$ to be your favorite smooth manifold, and $f:X\to X$ a homeomorphism that is not smooth. Then the pullback of the smooth structure on $X$ by $f$ gives you a new smooth structure.

Or for example, if you fix a homeomorphism of a square with a circle, this gives you a smooth structure on the square, etc.

I take that like this:

1. take the circle and the square as subsets of $\mathbb R^2$ equipped with the subset topology -- inherited from $\mathbb R^2$ standard topology
2. as such the circle and the square are topological manifolds
3. the circle is also a submanifold of $\mathbb R^2$ inheriting its smooth structure whereas the square is not

Endow then the square with a smooth structure -- basically define an atlas for it -- and define an homeomorphism *not* smooth with the circle (endowed as said before with the smooth structure as submanifold of $\mathbb R^2$).

Now the pullback of the 'circle smooth structure' through the given homeomorphism gives the square a new smooth structure turning that homeomorphism in a diffeomorphism

Make sense ?

 

[Infrared]

@cianfa72 All I mean in the second paragraph is that is that if $f$ is a homeomorphism from a square to a circle, then $f$ induces a smooth structure on the square by pullback. I didn't mean to do any of this:

Endow then the square with a smooth structure -- basically define an atlas for it -- and define an homeomorphism *not* smooth with the circle (endowed as said before with the smooth structure as submanifold of $\mathbb R^2$).

The business of finding a homeomorphism that isn't smooth in the first paragraph was meant to find two distinct smooth structures on $X$. If this is what you want to do for $X=\text{Square}$, then the above is right, but you should explain how you are picking a smooth structure on the square in the first place. There do exist topological manifolds with no smooth structure. What you could do in this case is let $g:S^1\to S^1$ be a homeomorphism that isn't smooth, and let $f$ be a homeomorphism from a square to a circle. Then $f$ and $g\circ f$ give two different pullback smooth structures on the square.

 

[cianfa72]

The business of finding a homeomorphism that isn't smooth in the first paragraph was meant to find two distinct smooth structures on X. If this is what you want to do for X=Square, then the above is right, but you should explain how you are picking a smooth structure on the square in the first place.

The following two charts should define a smooth atlas for the square (note the two regions of overlap are open)

Anyway I believe it is not the 'restriction' of the $\mathbb R^2$ smooth structure just because the square is not a submanifold of $\mathbb R^2$

 

[BiGyElLoWhAt]

He doesn't give any specific examples and I have never encountered any, or encountered any such statement in any GR text.

Aren't coordinate singularities a thing? Wouldn't that be an example of what the OP is talking about? Functions that are "sharp" at a point in one coordinate system but not another?

 

[vanhees71]

No, if you want to investigate the differentiability of a curve at a certain point you have of course to choose a proper chart containing this point in its domain.

 

[cianfa72]

The following two charts should define a smooth atlas for the square (note the two regions of overlap are open)

View attachment 267036

Anyway I believe it is not the 'restriction' of the $\mathbb R^2$ smooth structure just because the square is not a submanifold of $\mathbb R^2$

Any comment, is that correct ?

 

[cianfa72]

Actually I've another doubt about what he said at this minute . It seems that the sphere in $\mathbb R^3$ and the sphere where I make a 'little edge' (basically folded a little) are no longer diffeomorphic

My concern is that he said in dimension 2 there exist basically just one smooth structure - up to diffeomorphism -- so why then a such diffeomorphism does not exist between them ?

 

[Dragon27]

What is this sphere with a little edge? Isn't it a different object altogether?

 

[cianfa72]

What is this sphere with a little edge? Isn't it a different object altogether?

I believe he just mean a 'not smooth sphere' with let me say a sort of 'corner' in some part

 

[martinbn]

Actually I've another doubt about what he said at this minute . It seems that the sphere in $\mathbb R^3$ and the sphere where I make a 'little edge' (basically folded a little) are no longer diffeomorphic

My concern is that he said in dimension 2 there exist basically just one smooth structure - up to diffeomorphism -- so why then a such diffeomorphism does not exist between them ?

He wasn't being careful. He probably meant that the two are the same topological space with two different smooth structures. But in dimension 2 the to different structures are diffeomorphic.

 

[Dragon27]

I believe he just mean a 'not smooth sphere' with let me say a sort of 'corner' in some part

Well, I'm not sure what he meant exactly. I know (as is explained and the end of that lecture) that a topological manifold in dimension less than 4 has a unique differential structure. So I would suppose that two smooth manifolds that are homeomorphic should be automatically diffeomorphic (for $n<4$)? And what he meant is that a sphere with an edge isn't diffeomorphic to the regular sphere because it's not smooth in the first place?

Is wiki correct?
https://en.wikipedia.org/wiki/Diffeomorphism#Homeomorphism_and_diffeomorphism

In dimensions 1, 2, 3, any pair of homeomorphic smooth manifolds are diffeomorphic.

 

[cianfa72]

He wasn't being careful. He probably meant that the two are the same topological space with two different smooth structures. But in dimension 2 the to different structures are diffeomorphic.

Watching it again, I believe as follows:

• for both the regular and the sphere with 'edge' he takes the subspace topology inherited from $\mathbb R^3$
• for the regular sphere all the smooth charts in $\mathbb R^3$ 'restricted' to the sphere are compatible and thus define a maximal atlas for it
• however for the sphere with 'edge' we cannot take all those charts because when 'restricted' to it are not longer mutually compatible

Thus the sphere with 'edge' and the regular one have not the same smooth structure (actually the regular sphere is a submanifold of $\mathbb R^3$ whereas the other is not) nevertheless are diffeomorphic as manifolds of dimension 2