How to select the smooth atlas to use for spacetime?

[Shirish]

I'm studying differential geometry basics for general relativity (no specific source, just googling around). I know that spacetime is modeled as a ##4##-dimensional smooth manifold. Smooth manifold means that we consider a restriction of the maximal atlas such that all charts in it are compatible.

From what I've read, it seems like the differentiability of curves depends on the atlas we're using (I could be wrong about this, please correct me in that case). So a curve may be differentiable w.r.t. one atlas and not to another.

What if there are multiple non-equivalent restrictions of the maximal atlas such that their charts are compatible? I tried searching for a result that conveniently tells us, for example, that there's a unique such restriction. On the contrary, I found that there can be uncountably many such atlases!

So then which one do we select? The differentiability of a curve ("existence of velocities") in spacetime shouldn't depend on our choice, right? Do we make any additional assumptions to make this notion well-defined w.r.t. choice of atlas?

 

[PeterDonis]

From what I've read, it seems like the differentiability of curves depends on the atlas we're using (I could be wrong about this, please correct me in that case)

You are wrong. The differentiability of a curve in spacetime is independent of any choice of coordinates.

 

[Shirish]

You are wrong. The differentiability of a curve in spacetime is independent of any choice of coordinates.

Given an atlas, yes. I'm saying the differentiability of a curve isn't independent of the choice of atlas - never said in my OP that differentiability depends on the choice of coordinates.

 

[PeterDonis]

I'm saying the differentiability of a curve isn't independent of the choice of atlas

You are wrong. It is independent of the choice of atlas. When I said "independent of any choice of coordinates" I meant any choice of coordinates, i.e., in any atlas.

 

[Shirish]

You are wrong. It is independent of the choice of atlas. When I said "independent of any choice of coordinates" I meant any choice of coordinates, i.e., in any atlas.

But there may be some curves that are differentiable under some smooth atlas but not in another smooth atlas?

Here's a video (I've linked to the timestamp where the lecturer starts talking about the issue) in which this is discussed. The relevant part is till 24:00 so it's just a 1.5 min watch.

At the 23:25 timestamp, the lecturer talks about differentiability of some curves depending on the choice of atlas. Hence the confusion. He talks about it again (very briefly) at this timestamp.

 

[PeterDonis]

At the 23:25 timestamp, the lecturer talks about differentiability of some curves depending on the choice of atlas.

He doesn't give any specific examples and I have never encountered any, or encountered any such statement in any GR text. So I'm not sure what he's referring to. It's possible that he has some "pathological" cases in mind that a mathematician might be able to show the existence of but which no physicist would ever actually use. I have never encountered any actual issue of this kind in a physics text.

You might want to reframe the question purely as a math question (without any specific reference to GR) and ask it in a new thread in the appropriate math forum.

 

[Shirish]

He doesn't give any specific examples and I have never encountered any, or encountered any such statement in any GR text. So I'm not sure what he's referring to. It's possible that he has some "pathological" cases in mind that a mathematician might be able to show the existence of but which no physicist would ever actually use. I have never encountered any actual issue of this kind in a physics text.

You might want to reframe the question purely as a math question (without any specific reference to GR) and ask it in a new thread in the appropriate math forum.

You're probably right about the pathological cases - I mean if it were a big issue, it'd be covered in mathematically rigorous GR texts, and you seem to have a good enough background in that.

Just wondering what the resolution is in case he isn't referring to pathological cases - there must be a resolution - maybe we make some additional assumptions/theorem that show that differentiability is a well-defined notion even across smooth atlases. Still not entirely certain about him talking about pathological case because of what he says here ["That's a real problem, because which one do you choose for Physics?"]

In the meanwhile, I'll follow your advice on rephrasing and asking this in a math forum. Thank you!

 

[PeterDonis]

That's a real problem, because which one do you choose for Physics?

And in practice the answer for physicists is "the ones that your simple common sense points you to, which turn out to work just fine without all this extra mathematical folderol". I simply don't see a major issue here in practice. Perhaps he elaborates on this more in some of his papers; when I have time I'll look him up on arxiv.org to see what's here.

 

[Shirish]

the ones that your simple common sense points you to, which turn out to work just fine without all this extra mathematical folderol - I'll keep this in mind. Probably not a good idea to get too lost in the mathematical minutiae.

 

[PeterDonis]

Perhaps he elaborates on this more in some of his papers; when I have time I'll look him up on arxiv.org to see what's here.

And here is the first paper of his that comes up in a Google search:

https://arxiv.org/pdf/1111.4824.pdf

The beginning of the Introduction:

The recent announcement of superluminal neutrino propagation by the OPERA collaboration is a reminder that the spacetime geometry might well not be given by a Lorentzian manifold. But by what else? Fortunately, this question has a rather comprehensive answer. For we will show in these lectures that the spectrum of tensor fields that can serve as a space-time geometry--in the sense that matter field dynamics is predictive and observers agree on the sign of particle energies--is severely restricted. Only geometries on which the dispersion relation of matter fields is encoded in a totally symmetric contravariant even-rank tensor field satisfying three simple algebraic conditions--it must be hyperbolic, time-orientable and energy-distinguishing--can present candidates for a spacetime geometry.

In other words, here he solves the problem he posed in the video ("which smooth manifolds do we use for physics?"). But note that his solution starts with the observation that Lorentzian manifolds, by themselves, might not be sufficient. Which appears to imply that if you only deal with Lorentzian manifolds (and of course all work in GR up to now has used Lorentzian manifolds), the problem he describes in the video simply doesn't exist: whatever issues there might be with non-compatible atlases and differentiability of curves only involve kinds of manifolds that aren't Lorentzian manifolds.

If the above is correct, it would explain why he discusses the problem in the video (he's talking about the general mathematical case, and it's obviously of interest to him since he published a paper on it) but the problem he describes never comes up in GR (since GR uses just one particular kind of manifold, Lorentzian manifolds, where the problem does not exist).

 

[Shirish]

And here is the first paper of his that comes up in a Google search:

https://arxiv.org/pdf/1111.4824.pdf

The beginning of the Introduction:
In other words, here he solves the problem he posed in the video ("which smooth manifolds do we use for physics?"). But note that his solution starts with the observation that Lorentzian manifolds, by themselves, might not be sufficient. Which appears to imply that if you only deal with Lorentzian manifolds (and of course all work in GR up to now has used Lorentzian manifolds), the problem he describes in the video simply doesn't exist: whatever issues there might be with non-compatible atlases and differentiability of curves only involve kinds of manifolds that aren't Lorentzian manifolds.

If the above is correct, it would explain why he discusses the problem in the video (he's talking about the general mathematical case, and it's obviously of interest to him since he published a paper on it) but the problem he describes never comes up in GR (since GR uses just one particular kind of manifold, Lorentzian manifolds, where the problem does not exist).

Oh wow, thank you so much! That perfectly resolves my doubt; I will go through the paper too just to get a general sense.

 

[Infrared]

@PeterDonis I don't think that's true. The issue doesn't have to do with metrics (whether Riemannian or Lorentzian).

An atlas is used to ##\textit{define}## which functions are smooth, and whether or not a function is smooth certainly depends on the atlas you use. For example, define ##\varphi:\mathbb{R}\to\mathbb{R}## by ##\varphi(x)=\sqrt[3]{x}##. This defines an atlas for the topological manifold ##\mathbb{R}## with a single chart, making it a smooth manifold (there are no transition functions, so they are all vacuously smooth). However, ##\mathbb{R}## with this atlas clearly has different smooth functions than ##\mathbb{R}## with the standard atlas (e.g. the cube root function is now smooth).

What is true is that once an atlas is fixed, then whether or not a function is smooth is independent of which chart is used.

In my example above, it is still true that ##\mathbb{R}## with this nonstandard atlas is diffeomorphic to the usual smooth manifold ##\mathbb{R}## (again, via with the cube root function). However, there are examples of topological manifolds with non-diffeomorphic smooth structures (see for example: https://en.wikipedia.org/wiki/Exotic_sphere)

 

[PeterDonis]

the cube root function is now smooth

Why would the cube root function not be smooth with an atlas defined by ##\varphi(x) = x## (which I assume is what you mean by the standard atlas)?

 

[Infrared]

Why would the cube root function not be smooth with an atlas defined by ##\varphi(x) = x## (which I assume is what you mean by the standard atlas)?

It's not differentiable at ##0##.

 

[PeterDonis]

It's not differentiable at ##0##.

Ah, got it. But with the alternate atlas/chart, that function just becomes ##f(x) = x##?

 

[Shirish]

@PeterDonis I don't think that's true. The issue doesn't have to do with metrics (whether Riemannian or Lorentzian).

An atlas is used to ##\textit{define}## which functions are smooth, and whether or not a function is smooth certainly depends on the atlas you use. For example, define ##\varphi:\mathbb{R}\to\mathbb{R}## by ##\varphi(x)=\sqrt[3]{x}##. This defines an atlas for the topological manifold ##\mathbb{R}## with a single chart, making it a smooth manifold (there are no transition functions, so they are all vacuously smooth). However, ##\mathbb{R}## with this atlas clearly has different smooth functions than ##\mathbb{R}## with the standard atlas (e.g. the cube root function is now smooth).

What is true is that once an atlas is fixed, then whether or not a function is smooth is independent of which chart is used.

In my example above, it is still true that ##\mathbb{R}## with this nonstandard atlas is diffeomorphic to the usual smooth manifold ##\mathbb{R}## (again, via with the cube root function). However, there are examples of topological manifolds with non-diffeomorphic smooth structures (see for example: https://en.wikipedia.org/wiki/Exotic_sphere)

Just want to confirm one thing on my end: the paper claims that there's a very narrow scope in the choice of differentials manifolds we can use to model spacetime. And those specific types of manifolds admit a well-defined notion of differentiability even across atlases (i.e. Differentiability is independent of atlas choice for such manifolds).

Is the above correct?

 

[PeterDonis]

Is the above correct?

I think so, but I haven't read the paper in detail.

 

[Infrared]

@PeterDonis Yes, that's the idea (where "becomes" refers to the induced map between charts).

The way to think about this example is that if ##f:X\to Y## is a homeomorphism of topological manifolds, and ##\mathcal{A}## is an atlas on ##Y##, then there is a naturally associated pullback atlas ##f^*\mathcal{A}## on ##X##. 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.

Just want to confirm one thing on my end: the paper claims that there's a very narrow scope in the choice of ##\bf{differential \ manifolds}## we can use to model spacetime. And those specific types of manifolds admit a well-defined notion of differentiability even across atlases (i.e. Differentiability is independent of atlas choice for such manifolds).

(emphasis mine)

If you have a differential manifold, then you've already specified your (equivalence class of) atlas, so the sheaf of smooth functions is fixed. The ambiguity in smooth functions comes from when you have a topological manifold and you are choosing how to make it a smooth manifold (i.e. choosing an atlas).

 

[Shirish]

If you have a differential manifold, then you've already specified your (equivalence class of) atlas, so the sheaf of smooth functions is fixed. The ambiguity in smooth functions comes from when you have a topological manifold and you are choosing how to make it a smooth manifold (i.e. choosing an atlas).

Yep - that latter ambiguity's what's been stumping me.

 

[vanhees71]

And here is the first paper of his that comes up in a Google search:

https://arxiv.org/pdf/1111.4824.pdf

The beginning of the Introduction:
In other words, here he solves the problem he posed in the video ("which smooth manifolds do we use for physics?"). But note that his solution starts with the observation that Lorentzian manifolds, by themselves, might not be sufficient. Which appears to imply that if you only deal with Lorentzian manifolds (and of course all work in GR up to now has used Lorentzian manifolds), the problem he describes in the video simply doesn't exist: whatever issues there might be with non-compatible atlases and differentiability of curves only involve kinds of manifolds that aren't Lorentzian manifolds.

If the above is correct, it would explain why he discusses the problem in the video (he's talking about the general mathematical case, and it's obviously of interest to him since he published a paper on it) but the problem he describes never comes up in GR (since GR uses just one particular kind of manifold, Lorentzian manifolds, where the problem does not exist).

But isn't usually a differentiable manifold defined as one and only one equivalence class of compatible atlasses? Then this problem is formally solved by definition, i.e., diff. manifolds defined with two incompatible atlasses describe different manifolds.

Since superluminal neutrinos are fortunately gone since it was just due to two technical flaws in the measurement, that's not the reason to think more about whether the right spacetime model is a Lorentzian manifold. It's rather the question, whether you need a Lorentz-Cartan manifold, i.e., a differentiable manifold with a connection and a pseudo-metric, with the connection being compatible with the pseudo-metric (due to the equivalence principle) but maybe have non-vanishing torsion. That's because if you have matter consisting of particles with spin often you necessarily get torsion (e.g., when looking at Dirac particles in curved spacetime).

If you like to check more on the issue in this direction, just look for papers by F. Hehl. One is

F. Hehl, P. Von Der Heyde, G. Kerlick and J. Nester, General
Relativity with Spin and Torsion: Foundations and Prospects,
Rev. Mod. Phys. 48, 393 (1976),
https://dx.doi.org/10.1103/RevModPhys.48.393

There's also a good summary from the point of view of GR as a gauge theory:

P. Ramond, Field Theory: A Modern Primer,
Addison-Wesley, Redwood City, Calif., 2 edn. (1989).

 

[PeterDonis]

isn't usually a differentiable manifold defined as one and only one equivalence class of compatible atlasses?

See post #18.

 

[Shirish]

So I'd asked this question on SE as well, and one of the comments was:

The standard assumption in relativity is that spacetime is locally diffeomorphic to a standard ##\mathbb{R}^4##. If you have some reason to want to weaken this assumption and allow exotic versions of ##\mathbb{R}^4##, then the possibility that the theory is flexible enough to accommodate your needs is a feature, not a bug.

Does the first part of the above comment imply that there is a unique differentiable manifold (i.e. unique equivalence class of atlases) that's locally diffeomorphic to standard ##\mathbb{R}^4##?

 

[martinbn]

##\mathbb R^4## has exotic smooth structures, but when it is said the standard one, it means the standard one. Infrared has already said everything you need for you question. But my guess is that you are still not comfortable with the notion of differentiable manifold.

 

[Shirish]

##\mathbb R^4## has exotic smooth structures, but when it is said the standard one, it means the standard one. Infrared has already said everything you need for you question. But my guess is that you are still not comfortable with the notion of differentiable manifold.

I recently started learning the subject, yes. I'm comfortable with the basics at least, but definitely not familiar with the more advanced or even intermediate aspects. Could you elaborate on the part in bold, please? It's not obvious to me where Infrared addressed it, so it'd be a great help and I can go over that part.

 

[PeterDonis]

Does the first part of the above comment imply that there is a unique differentiable manifold (i.e. unique equivalence class of atlases) that's locally diffeomorphic to standard ##\mathbb{R}^4##?

Can you give a link to the SE question and answer? Without seeing the actual SE thread where you got the response, there's no way for us to know for sure since we don't know what the poster there meant by "standard" ##\mathbb{R}^4##. My guess would be that they meant "the unique differentiable manifold given by the topological space ##\mathbb{R}^4## with the standard smooth structure on it", in which case the answer to your question would be yes, but it's just a guess without seeing the actual thread.

 

[Shirish]

Can you give a link to the SE question and answer? Without seeing the actual SE thread where you got the response, there's no way for us to know for sure since we don't know what the poster there meant by "standard" ##\mathbb{R}^4##. My guess would be that they meant "the unique differentiable manifold given by the topological space ##\mathbb{R}^4## with the standard smooth structure on it", in which case the answer to your question would be yes, but it's just a guess without seeing the actual thread.

Sure! Here's the link: https://physics.stackexchange.com/q...erentiable-manifold-to-use-to-model-spacetime

 

[Infrared]

The standard assumption in relativity is that spacetime is locally diffeomorphic to a standard ##\mathbb{R}^4##.

I think this is a very misleading answer. Every differentiable ##4##-manifold (including exotic ##\mathbb{R}^4##) is locally diffeomorphic to the standard ##\mathbb{R}^4.## You don't need to assume this as an axiom for a physical theory!

I don't know very much physics, so someone should tell me if the following is wrong. But perhaps what could be meant is that picking spatial coordinates ##x,y,z## and a time coordinate ##t## defines defines coordinates on your space-time manifold, and hence gives you a smooth structure to use.

 

[Shirish]

I think this is a very misleading answer. Every differentiable ##4##-manifold (including exotic ##\mathbb{R}^4##) is locally diffeomorphic to the standard ##\mathbb{R}^4.## You don't need to assume this as an axiom for a physical theory!

I don't know very much physics, so someone should tell me if the following is wrong. But perhaps what could be meant is that picking spatial coordinates ##x,y,z## and a time coordinate ##t## defines defines coordinates on your space-time manifold, and hence gives you a smooth structure to use.

Even in Schuller's lectures that I'm referring to for self-study, he categorically stated that for a topological manifold, we consider it to be locally homeomorphic to the standard topology ##\mathbb{R}^4## (in the context of Physics). Then the differentiable structure enters the picture. To me, that suggests that there can be multiple ways to assign a differentiable structure to a topological manifold such that it's locally diffeomorphic to the standard topology ##\mathbb{R}^4##.

And that's why I sought justification from the SE user because his comment suggested there's a unique differentiable manifold that can be locally diffeomorphic to standard ##\mathbb{R}^4##

 

[PeterDonis]

perhaps what could be meant is that picking spatial coordinates ##x, y, z## and a time coordinate ##t## defines coordinates on your space-time manifold, and hence gives you a smooth structure to use.

You don't have to pick those coordinates; any continuous assignment of 4-tuples of real numbers will do. With that caveat, yes, I think that's what the SE poster meant.

 

[Infrared]

@Shirish I agree with your summary of Schuller. I also think the discussion of exotic ##\mathbb{R}^4## is a bit of a red herring. On any topological manifold, there are many possible smooth structures (i.e. different functions are smooth). The difference for exotic ##\mathbb{R}^4## is that it has a smooth structure that is not even diffeomorphic to (standard) ##\mathbb{R}^4.##

You don't have to pick those coordinates; any continuous assignment of 4-tuples of real numbers will do. With that caveat, yes, I think that's what the SE poster meant.

You also want this assignment to define a coordinate system, but I don't think 'continuous' is enough (e.g. what if I use ##x,y,z,\sqrt[3]{t}## as my coordinates). You want them to be smooth, but that assumes you already have a smooth structure. The point I was trying to make is that space-time already has coordinates attached to it (spatial and time up to re-scaling and choosing directions) that we can use, and these coordinates define a smooth structure.

I probably shouldn't let anything I say about relativity be taken seriously, though.

 

[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.