I Smooth coordinate chart on spacetime manifold

[cianfa72]

TL;DR Summary

About the definition of smooth coordinate chart/system in an invariant way; is it actually possible ?

Hi,
I'm puzzled about the definition of smooth coordinate chart for a manifold (e.g. spacetime).

From my point of view there is no invariant way to define a smooth coordinate chart since a coordinate system is smooth only w.r.t. another coordinate system already defined on the given manifold.

Does it make sense ? Thank you.

 

[Orodruin]

All coordinate transformations should be smooth.

 

[martinbn]

TL;DR Summary: About the definition of smooth coordinate chart/system in an invariant way; is it actually possible ?

Hi,
I'm puzzled about the definition of smooth coordinate chart for a manifold (e.g. spacetime).

From my point of view there is no invariant way to define a smooth coordinate chart since a coordinate system is smooth only w.r.t. another coordinate system already defined on the given manifold.

Does it make sense ? Thank you.

What is the definition? Can you give a reference? Otherwise we have to guess what exactly bothers you.

 

[cianfa72]

What is the definition? Can you give a reference? Otherwise we have to guess what exactly bothers you.

That's the point. I've no precise definition of smooth coordinate chart for a manifold.

In other words I give you a chart for a manifold: how can you claim it is a smooth one ?

 

[martinbn]

That's the point. I've no precise definition of smooth coordinate chart for a manifold.

In other words I give you a chart for a manifold: how can you claim it is a smooth one ?

Then may be your question should be if there is such a notion as "smooth chart".

 

[cianfa72]

Then may be your question should be if there is such a notion as "smooth chart".

Yes, indeed. Perhaps the point is that we need to start from the definition of smooth manifold which explicitly include the maximal Atlas. By definition the transformations between all charts in the maximal Atlas are smooth (at the common intersection where both are defined).

 

[Hill]

For each point in the manifold there is a neighborhood which is homeomorphic with Euclidean space. Take for example Cartesian coordinates in that Euclidean space. The homeomorphism determines smooth coordinates in that neighborhood of the manifold. Continue from there.

 

[cianfa72]

The homeomorphism determines smooth coordinates in that neighborhood of the manifold. Continue from there.

Therefore we can claim a coordinate chart is smooth only w.r.t. a given definition of smooth manifold structure.

 

[Hill]

Therefore we can claim a coordinate chart is smooth only w.r.t. a given definition of smooth manifold structure.

Existence of such a homeomorphism is THE definition of manifold. Otherwise, it is not a manifold.

 

[cianfa72]

So just to fix ideas: take the surface of a cube in 3D euclidean space. Can we turn it in a smooth manifold ? In other words can we define a smooth atlas for it ?

 

[Orodruin]

It is homeomorphic to a sphere. However, the embedding into $\mathbb R^3$ is not smooth so the induced metric won’t be either.

 

[cianfa72]

It is homeomorphic to a sphere.

Hence the homeomorphism allows to transport to the cube's surface the sphere's smooth manifold structure.

However, the embedding into $\mathbb R^3$ is not smooth so the induced metric won’t be either.

ok, so it will not be a smooth (embedded) submanifold of $\mathbb R^3$ endowed with its standard smooth structure, right ?

 

[PeterDonis]

From my point of view there is no invariant way to define a smooth coordinate chart

Since "invariant" means "independent of any choice of coordinates", it seems odd to view what you state here as a problem.

 

[PeterDonis]

a coordinate system is smooth only w.r.t. another coordinate system already defined on the given manifold.

No, it isn't. "Smooth" refers to the existence of continuous derivatives to any degree. But derivatives with respect to what? Since a coordinate chart is a mapping from points in a manifold to tuples of real numbers, the derivatives would be with respect to points in the manifold. Those can be defined for any coordinate chart without reference to other coordinate charts. All you need is the topological structure of the manifold.

 

[PeterDonis]

In other words I give you a chart for a manifold: how can you claim it is a smooth one ?

It is smooth if each of the coordinates is a smooth function on the manifold. What's the problem?

 

[cianfa72]

It is smooth if each of the coordinates is a smooth function on the manifold. What's the problem?

Yes, but to say a coordinate function is a smooth function on the manifold, we must have an already given smooth structure for the manifold itself.

 

[PeterDonis]

to say a coordinate function is a smooth function on the manifold, we must have an already given smooth structure for the manifold itself

Which, as @Hill pointed out, will always exist for any manifold, by the construction he gave.

Are you concerned that there might be multiple incompatible smooth structures on a particular manifold? I don't think that's possible.

 

[vanhees71]

The standard construction of differentiable manifolds starts with a Hausdorff topological space $M$. Its elements are named points.

Then you define a chart as a continuous function bijective function, whose inverse function is also continuous, (in short a homeomorphism) between an open set $U \subseteq M$ and an open set $U' \subseteq \mathbb{R}^d$, $h:U \rightarrow U'$.

Two charts $h_1$ and $h_2$ are compatible, if the homeomorphism $h_1 \circ h_2^{-1}:U_2' \rightarrow U_1'$ restricted to $U_1' \cap U_2'$ is a diffeomorphism.

An atlas $A$ is a set of compatible charts, $h_j$ for which $\cup_{j} U_j =M$.

Two atlasses $A_1$ and $A_2$ are equivalent, if also $A_1 \cup A_2$ is an atlas. A differentiable manifold is then defined by the equivalence class $[A]$ of some atlas $A$.

Then let $f:U \rightarrow M$. It's called differentiable if the function $f \circ h^{-1}:U' \rightarrow \mathbb{R}$ is differentiable for an arbitrary chart in the set of compatible charts defined on $U$ that are compatible with the all corresponding charts in $[A]$.

 

[cianfa72]

Which, as @Hill pointed out, will always exist for any manifold, by the construction he gave.

ok, through neighborhoods of the manifold homeomorphic with the Euclidean space.

Are you concerned that there might be multiple incompatible smooth structures on a particular manifold? I don't think that's possible.

I don't know...for example can we endow the 3d cube's surface with different incompatible smooth structures ?

 

[cianfa72]

Then let $f:U \rightarrow M$. It's called differentiable if the function $f \circ h^{-1}:U' \rightarrow \mathbb{R}$ is differentiable for an arbitrary chart in the set of compatible charts defined on $U$ that are compatible with the all corresponding charts in $[A]$.

That was basically my point: we need a smooth structure for the manifold in the first place; only then you can check if a given function defined on the manifold (e.g. a coordinate function) is indeed a differentiable smooth function or is not.

 

[cianfa72]

Those can be defined for any coordinate chart without reference to other coordinate charts. All you need is the topological structure of the manifold.

If you only have a topological structure on the manifold, can you actually use it to define derivatives of functions on the manifold (e.g. derivatives of coordinate functions) ?

 

[PeterDonis]

If you only have a topological structure on the manifold, can you actually use it to define derivatives of functions on the manifold (e.g. derivatives of coordinate functions) ?

A topological structure means you can use the construction @Hill describes, which is sufficient to define derivatives.

 

[cianfa72]

What about the 3d cube's surface as manifold as in my post #19 ? Thanks.

 

[PeterDonis]

What about the 3d cube's surface as manifold as in my post #19 ? Thanks.

When you say "3d cube", you are already assuming more than just a topological structure (i.e., a topological 2-sphere). You are assuming a specific embedding in Euclidean 3-space with a specific metric. And if your question is, is there a smooth chart that is compatible with that embedding and that metric, I believe the answer is no: there will be no smooth chart that is both compatible with that embedding and that metric, and includes the corners of the cube.

Note that issue here is not being able to find a "smooth structure" of the manifold itself; we know how to do that for a topological 2-sphere. The issue is wanting something more than just a smooth structure or a smooth coordinate chart (or atlas of such charts)--in this case, compatibility with a particular embedding into Euclidean 3-space with a particular metric. In general there is no guarantee that such a thing will be possible.

 

[cianfa72]

You are assuming a specific embedding in Euclidean 3-space with a specific metric. And if your question is, is there a smooth chart that is compatible with that embedding and that metric, I believe the answer is no: there will be no smooth chart that is both compatible with that embedding and that metric, and includes the corners of the cube.

ok, so the point is that the 3d cube surface is not an embedded submanifold of $\mathbb R^3$ ($\mathbb R^3$ endowed with its standard smooth structure and metric).

Note that issue here is not being able to find a "smooth structure" of the manifold itself; we know how to do that for a topological 2-sphere.

ok, therefore we can define a smooth structure on the 3d cube surface turning it in a smooth manifold (leveraging the homeomorphism with the 2-sphere).

 

[PeterDonis]

the point is that the 3d cube surface is not an embedded submanifold of $\mathbb R^3$ ($\mathbb R^3$ endowed with its standard smooth structure).

I believe that is what @Orodruin's statement in post #11 implies, yes. You can define the embedding, but it won't define an embedded submanifold (although each face of the cube, considered as an open set excluding the corners and edges, will be, taken on its own).

ok, therefore we can define a smooth structure on the 3d cube surface (leveraging the homeomorphism with the 2-sphere).

Yes. It just won't be compatible with the embedding in $\mathbb{R}^3$.

 

[fresh_42]

ok, so the point is that the 3d cube surface is not an embedded submanifold of $\mathbb R^3$ ($\mathbb R^3$ endowed with its standard smooth structure and metric).

It is an embedded submanifold, but its chart mappings are only continuous, not smooth, since you cannot differentiate at the corners since left and right derivatives do not coincide.

ok, therefore we can define a smooth structure on the 3d cube surface turning it in a smooth manifold (leveraging the homeomorphism with the 2-sphere).

No. Only if you blow it up to a 2-sphere and "smoothen" the corners.

Smooth means: All smooth paths on the manifold can smoothly be transported to the chart, differentiated there, and smoothly returned to the tangent space of the manifold. The last step is basically an identity since chart and tangent space coincide. If you have a corner in the path on the manifold, why should it magically go away? "Smooth charts" could mean that the mappings to the chart and back are smooth. It doesn't actually make much sense to speak of smooth charts. Smooth belongs to a mapping, chart to a piece of the tangent space. The mapping to the chart can be smooth, a path on the manifold can be smooth (or not), but what should a smooth Euclidean (or Minkowski) space be?

 

[PeterDonis]

and "smoothen" the corners

Are the corners defined in the topology of the "cube"? Or do you need the embedding into Euclidean 3-space to define them?

 

[fresh_42]

Are the corners defined in the topology of the "cube"? Or do you need the embedding into Euclidean 3-space to define them?

The two are topologically identical as topology only knows continuity. We already need a differentiable manifold if we want to speak of a differentiable structure. And, this differentiable structure is defined by the charts. This whole subject is circular reasoning. It attributes a property of mappings (infinite differentiability) to a piece of a flat space. The cube with its natural embedding doesn't allow a differentiable structure. If we transform it homeomorphic into a sphere, then we can use the new embedding for a differentiable structure. However, the homeomorphism isn't a diffeomorphism. It breaks down at the corners and vertices.

The chart mappings of a (Riemannian) manifold are (local) homeomorphisms.
The chart mappings of a smooth manifold are (local) diffeomorphisms.
(My paperback needs a page to describe it technically correctly. The main idea is that two overlapping charts allow smooth transformations: smoothly onto one chart and smoothly back from the other.)

Since most physical laws are smooth and nature usually hasn't any corners, most manifolds are assumed to be smooth.

 

[PeterDonis]

The cube with its natural embedding doesn't allow a differentiable structure.

Ah, got it. So that's the point where things break down--once we pick "cube with its natural embedding" we are no longer compatible with any smooth structure on the 2-sphere topological manifold.

 

[cianfa72]

Which is the natural embedding of the cube in $\mathbb R^3$, how is it defined?

 

[PeterDonis]

Which is the natural embedding of the cube in $\mathbb R^3$, how is it defined?

What embedding did you have in mind in post #10? You're the one that introduced the idea of a cube in 3D Euclidean space.

 

[cianfa72]

What embedding did you have in mind in post #10?

I was thinking about the "visual" image of the 3d cube surface inside the 3-d Euclidean space. However to formally define an embedding we need a topology already defined on the cube and an homeomorphism onto its image in $\mathbb R^3$.

 

[PeterDonis]

to formally define an embedding we need a topology already defined on the cube and an homeomorphism onto its image in $\mathbb R^3$.

If that's all you have, you don't have a "cube". The word "cube" specifically implies more than that. It implies a specific geometric shape, with 6 identical faces, 12 identical edges, and 8 identical corners.

 

[fresh_42]

Which is the natural embedding of the cube in $\mathbb R^3$, how is it defined?

You get the same situation with the absolute value function $M=\{(x,|x|)\,|\,-1<x<1\} \subseteq \mathbb{R}^2.$ The cube has only a lot more cases to define it.

$M\subseteq \mathbb{R}^2$ is a (continuous) embedded manifold, $D=\{(x, 1-\sqrt{1-x^2})\,|\,-1<x<1\}\subseteq \mathbb{R}^2$ is a smooth embedded manifold. The point is that going from one chart to the manifold and back from the manifold on another chart where they overlap has to be a smooth real function. You can do this with the circle $D$ at $(0,0)$ but not with the lines $M$ at $(0,0)$ for two different charts containing the origin. In this case, you cannot even define differentiability on one chart containing the origin. Smooth in this context actually means a $C^\infty$ compatible atlas. $D$ has tangents everywhere, $M$ has not. The definition of tangents needs either an embedding that provides natural charts, or it requires a $C^1$ compatible atlas. In this case, $D$ has even a $C^\infty$ compatible atlas and is thus a smooth manifold.

Instead of speaking of a smooth chart (which only makes sense for smooth chart mappings on their overlapping area), it would be better to speak of smooth manifolds, or even better better, to speak of a $C^\infty$ compatible atlas.

The basic idea behind manifolds is simple: We open the atlas, choose a chart, do whatever we want to do in that real or complex Euclidean space the chart represents, take our result, and go back to the manifold. What we do on the chart is a calculation in local coordinates. If what we wanted to do is infinitely often differentiating and that is compatible with the overlappings of our atlas, then we call the manifold smooth. Minkowski spaces make no difference. All they require is to deal with an additional sign.

It is not by chance that the wording is the same as having a street map on your lap navigating in an unfamiliar region.

 

[vanhees71]

That was basically my point: we need a smooth structure for the manifold in the first place; only then you can check if a given function defined on the manifold (e.g. a coordinate function) is indeed a differentiable smooth function or is not.

The manifold has a priori no "smooth structure", i.e., there's no definition of derivatives there. The aim in the theory of differentiable manifolds is to define such an idea, and the construction principle is to map the set of points, making up a Hausdorff topological space, to the $\mathbb{R}^d$ and then via these maps you are able to define differentiation for functions defined on open subsets of the manifold in terms of differentiation of functions defined on open subsets of the $\mathbb{R}^d$ (with the standard topology). To make this consistent you need these constructions of charts and atlasses and all that.

 

[cianfa72]

The set $M=\{(x,|x|)\,|\,-1<x<1\} \subseteq \mathbb{R}^2$ is the image of a continuous injective function $f$ from $\mathbb R$ to $\mathbb R^2$ (both with their Euclidean topology). $f$ is also homeomorphism onto its image i.e. it is a topological embedding. Using such homeomorphism $f$, we can endow the set $M$ with a smooth structure (only one chart) turning it in a smooth manifold. However this smooth structure is not compatibile with the given smooth structure from $\mathbb R ^2$.

Coming back to the 3d cube surface I think the same applies to it: starting from a topological 2-sphere we can define the standard/canonical topological embedding of the cube within $\mathbb R^3$. Even in this case we cannot define a smooth structure on the topological embedding's image in $\mathbb R^3$ compatibile with the standard smooth structure of $\mathbb R^3$.

 

[cianfa72]

The cube with its natural embedding doesn't allow a differentiable structure. If we transform it homeomorphic into a sphere, then we can use the new embedding for a differentiable structure.

Sorry, the homeomorphism between the 2-sphere and the natural embedding of the cube in $\mathbb R^3$ does not allow to define a smooth structure on the cube embedding's image ?

 

[fresh_42]

Sorry, the homeomorphism between the 2-sphere and the natural embedding of the cube in $\mathbb R^3$ does not allow to define a smooth structure on the cube embedding's image ?

The homeomorphism ($C^0$) between the cube and the sphere is not differentiable ($C^1$).
They are topologically the same object but they are not differential geometrically identical.
The cube itself has points where no tangent can be defined and is therefore not a smooth manifold.

You can put it the other way around: folding a sheet of paper is not smooth.

Why are you concerned about cubes? AFAIK they do not occur in physics. They have points where the derivatives flip their sign.
(Maybe in symmetry breaks of gauge field theories which I do not know enough about.)

 

[PeterDonis]

The set $M=\{(x,|x|)\,|\,-1<x<1\} \subseteq \mathbb{R}^2$ is the image of a continuous injective function $f$ from $\mathbb R$ to $\mathbb R^2$ (both with their Euclidean topology).

Yes, the function is continuous but it is not everywhere differentiable (the derivative is undefined at $x = 0$).

we can endow the set $M$ with a smooth structure

No, we can't, because of the lack of differentiability described above. "Smooth" implies differentiability everywhere on the domain.

You can endow the set $\{ x, -1 < x < 1 \}$ with a smooth structure (to the extent that even makes sense for a one-dimensional set), but that set is not $M$; $M$, by your own definition, includes a specific embedding into $\mathbb{R}^2$ which, as above, is not differentiable at $x = 0$.

 

[cianfa72]

No, we can't, because of the lack of differentiability described above. "Smooth" implies differentiability everywhere on the domain.

I'm not sure about this. We've just one chart (the homeomorphism with $\mathbb R$) and so by definition the above set $M$ endowed with that chart is a smooth manifold. As said before that homeomorphism is not however a smooth immersion in $\mathbb R^3$, so it is not a smooth embedding.

 

[fresh_42]

The set $M=\{(x,|x|)\,|\,-1<x<1\} \subseteq \mathbb{R}^2$ is the image of a continuous injective function $f$ from $\mathbb R$ to $\mathbb R^2$ (both with their Euclidean topology). $f$ is also homeomorphism onto its image i.e. it is a topological embedding.

Yes, it is the graph of the continuous function $x\mapsto |x|.$ This graph is a one-dimensional manifold.

Using such homeomorphism $f$, we can endow the set $M$ with a smooth structure (only one chart) turning it in a smooth manifold.

Do it. I'm curious how your tangent at $(0,0)$ will be defined. Homeomorphisms are not diffeomorphisms. If this was true, we could use the chain rule and construct a tangent of $x\rightarrow |x|$ at $x=0.$

However this smooth structure is not compatibile with the given smooth structure from $\mathbb R^2$.

This doesn't make any sense. How do you even define "smooth" without relating to $\mathbb{R}^n$?

Coming back to the 3d cube surface I think the same applies to it:

If the same applies, why do you want to complicate the issue?

starting from a topological 2-sphere we can define the standard/canonical topological embedding of the cube within $\mathbb R^3$. Even in this case we cannot define a smooth structure on the topological embedding's image in $\mathbb R^3$ compatibile with the standard smooth structure of $\mathbb R^3$.

You are repeatedly confusing several different concepts. You mix surfaces with functions, properties of functions with properties of spaces, you use embeddings without need, and I think (my impression) you do not know what a differential manifold is. Forget about the atlas, forget the embedding. Take a path on the manifold and define its tangent bundle.

And I am still curious how you would do this at $(0,0)\in M$ for, say for the path $\gamma :[0,1]\rightarrow M$ defined by $\gamma (t)=(t-1/2\ ,\ |t-1/2|).$ What is $\dot\gamma (0)$?

 

[PeterDonis]

We've just one chart (the homeomorphism with $\mathbb R$)

If that's all you have, you don't have a smooth structure. As has already been pointed out, a smooth structure requires differentiability. A homeomorphism by itself does not give you differentiability.

 

[cianfa72]

See for example https://math.stackexchange.com/questions/4703741/is-y-x-a-differential-manifold

 

[PeterDonis]

See for example https://math.stackexchange.com/questions/4703741/is-y-x-a-differential-manifold

That discussion is mixing up two different things. The "projection map" that is defined there does not give you a "one-dimensional manifold" that matches your definition of $M$. The projection is $(x, y) \to x$, which throws away information that is necessary for your definition of $M$. All you are left with is the open set $(-1, 1)$ on the real line, with no other structure at all. You can of course define a "one-dimensional manifold" on this topological set, but this manifold will not meet your definition of $M$.

I already said all this in post #40. @fresh_42 gave a more rigorous version of the same thing in post #42.

 

[fresh_42]

You can say that it is not a differentiable submanifold of $\mathbb{R}^2$ but, as a topological space, admits a smooth structure, i.e. is homeomorphic to a smooth manifold.

This is a smoke grenade. What does admit mean? Obviously a homeomorphism here. But a homeomorphism isn't differentiable. The "as a topological space" means

Only if you blow it up to a 2-sphere and "smoothen" the corners.

Let $h\, : \,C\longrightarrow S$ be the homeomorphism from the cube $C$ onto the sphere $S$ and $p\in S$ such that $h^{-1}(p)=c\in C$ where $c$ is a corner of the cube. Let $\gamma$ be a path through $p$ on the sphere. Then $\dot\gamma (0)$ is the tangent vector at $p.$ We then get the tangent of the corresponding point $c\in C$ of $h^{-1}\circ \gamma$ by
\begin{align*}
D_0(h^{-1}\circ \gamma)\stackrel{!}{=}\left. \dfrac{d}{dt}\right|_{t=0}\left(h^{-1}\circ \gamma \right)&=\left. \dfrac{d}{dx}\right|_{x=p}h^{-1}(x) \cdot\left. \dfrac{d}{dt}\right|_{t=0} \gamma (t)\\
&=\underbrace{\left(\left. \dfrac{d}{dx}\right|_{x=p}h^{-1}(x)\right)}_{=D_{p}\,h^{-1}}\cdot \dot\gamma (0)
\end{align*}
And here is the problem: $D_p\,h^{-1}$ does not exist. A contradiction.

However, if we had smoothened it, then we are only talking about the sphere $S$ and $\gamma$ anymore. The cube is history! Lost in "as a topological space".

 

[fresh_42]

I remember that I had a discussion with some students here at PF about differentiability and its many perspectives. They asked me whether I knew an article or a walkthrough of this labyrinth. I did not. So they convinced me to write one. Here is what came out:
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/

 

[cianfa72]

You can use a single chart, for that matter. So, yes, you have a smooth structure on M, but with that structure it is not an embedded (smooth) submanifold of $\mathbb R^2$.

Surely I'm wrong, but to me this claim is clear.

 

[PeterDonis]

Surely I'm wrong, but to me this claim is clear.

Yes, it's clear, and wrong. You are claiming that you can have a smooth structure on $M$. But as you have defined $M$, that's not possible. You can have a smooth structure on the open set $(-1, 1)$ of real numbers, but that set is not $M$ as you defined it.

 

[PeterDonis]

to me this claim is clear

You have made the same claim several times now, and each time you have received the same explanation of why it is wrong in response. How many times are we going to go around this merry-go-round?