I Smooth coordinate chart on spacetime 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?

 

[cianfa72]

No more time...I'm definitly wrong.

 

[PeterDonis]

No more time...I'm definitly wrong.

Ok. This thread is now closed.