I Maximal atlas of topological manifold

[cianfa72]

TL;DR

Is the maximal atlas of a topological manifold unique?

From my understanding, a topological manifold $M$ comes with, by definition, a locally euclidean topology and a (topological) atlas $\mathcal A_1$.

From this atlas one can construct the maximal atlas $\mathcal A$ throwing in all the chart maps $(U,\varphi)$ each from one of the open sets in $M$ (any of these chart maps are actually homemorphism).

Now the question: is such maximal atlas $\mathcal A$ unique ? My answer is yes, since it contains by definition all the possible chart maps from any of the possible open sets in $M$ and they are all $C^0$-compatible each other.

If the above is correct, then one can speak of the maximal atlas of $M$, right ?

 

[fresh_42]

Theoretically, yes, but this concept leads nowhere. There are simply far too many open sets, e.g. take $M=\mathbb{R}.$ The minimal atlas is more interesting.

 

[cianfa72]

Theoretically, yes, but this concept leads nowhere. There are simply far too many open sets, e.g. take $M=\mathbb{R}.$ The minimal atlas is more interesting.

Ok, so for instance on $M = \mathbb R$ with standard topology a minimal atlas for it as topological manifold has a countable number of charts $(U, \varphi)$ in it.

 

[fresh_42]

Ok, so for instance on $M = \mathbb R$ with standard topology a minimal atlas for it as topological manifold has a countable number of charts $(U, \varphi)$ in it.

Yes, since $\mathcal{A}_{min}=\{\mathbb{R},\operatorname{id}\}$ and $\left|\mathcal{A}_{min}\right|=1$ which is countable. A sphere requires at least two charts, one around a pole and one for the rest.

 

[cianfa72]

Theoretically, yes, but this concept leads nowhere.

For instance, starting from the maximal atlas of a topological manifold, one can throw away non $C^1$-compatible charts to get a $C^1$ differentiable manifold.

 

[pasmith]

TL;DR Summary: Is the maximal atlas of a topological manifold unique?

From my understanding, a topological manifold $M$ comes with, by definition, a locally euclidean topology and a (topological) atlas $\mathcal A_1$.

From this atlas one can construct the maximal atlas $\mathcal A$ throwing in all the chart maps $(U,\varphi)$ each from one of the open sets in $M$ (any of these chart maps are actually homemorphism).

Now the question: is such maximal atlas $\mathcal A$ unique ? My answer is yes, since it contains by definition all the possible chart maps from any of the possible open sets in $M$ and they are all $C^0$-compatible each other.

If the above is correct, then one can speak of the maximal atlas of $M$, right ?

A maximal atlas is [the union of every atlas in] an equivalance class of atlases, and is therefore uniquely determined by the choice of atlas.

Compatability of charts is not an equivalance relation; compatability of atlases is. (Two C^k atlases are C^k compatible iff every chart of one atlas is C^k compatible with every chart of the other atlas.)

 

[fresh_42]

For instance, starting from the maximal atlas of a topological manifold, one can throw away non $C^1$-compatible charts to get a $C^1$ differentiable manifold.

Not necessarily. It requires that there are diffeomorphisms at all, i.e. that $M$ is differentiable.

 

[cianfa72]

A maximal atlas is [the union of every atlas in] an equivalance class of atlases, and is therefore uniquely determined by the choice of atlas.

Ok, but in case of $C^0$-compatibility there is only one equivalence class (all $C^0$ atlases are in same equivalence class). Therefore any atlas determinates the same (unique) maximal $C^0$-topological atlas.

 

[cianfa72]

Not necessarily. It requires that there are diffeomorphisms at all, i.e. that $M$ is differentiable.

Not sure I get it. Do you mean that starting from the maximal topological atlas it is not guaranteed one can "extract" a differentiable atlas from it ?

 

[fresh_42]

Not sure I get it. Do you mean that starting from the maximal topological atlas is not guaranteed one can "extract" a differentiable atlas from it ?

Yes. What makes you think that you won't be left with nothing? Or with charts that do not cover the entire manifold? Consider $\{(x,|x|)\}.$ It is continuous but not differentiable.

 

[cianfa72]

Yes. What makes you think that you won't be left with nothing? Or with charts that do not cover the entire manifold?

Ah ok. Nevertheless a theorem from Whitney claims that any $C^k, k \ge 1$ atlas of a topological manifold contains a $C^{\infty}$ atlas (i.e. one can "extract" a $C^{\infty}$ atlas from it).

Btw, which is the use of "at all" in a positive sentence (like yours in post#7) ?

 

[fresh_42]

Ah ok. Nevertheless a theorem from Whitney claims that any $C^k, k \ge 1$ atlas of a topological manifold contains a $C^{\infty}$ atlas (i.e. one can "extract" a $C^{\infty}$ atlas from it).

Do you have a reference? Whitney's strong embedding theorem requires a smooth manifold, his weak embedding theorem only claims an approximation.

Btw, which is the use of "at all" in a positive sentence (like yours in post#7) ?

Maybe a translation error of mine. I meant, what if there aren't any differentiable charts, i.e. no differentiable charts at all?

 

[cianfa72]

Do you have a reference?

See for instance MSE.

Maybe a translation error of mine. I meant, what if there aren't any differentiable charts, i.e. no differentiable charts at all?

ok, got it.

 

[fresh_42]

See for instance MSE.

MSE and the video do not contain proofs so I cannot check the exact preliminaries, e.g. whether they mean a real or complex manifold. Both refer to the overlap functions and not necessarily on each chart.

I haven't found the claim in Hirsch's book that they quoted on MSE. Hirsch only states the two Whitney theorems as in Wikipedia. https://en.wikipedia.org/wiki/Whitney_embedding_theorem.

I would still be interested in a reference with proof. I do not say it's wrong. For example, once complex differentiable implies infinitely often complex differentiable, and I can understand Whitney's weak embedding theorem. But what to do with a real function that is once continuously differentiable but not twice and its graph as the manifold?

 

[cianfa72]

Consider $\{(x,|x|)\}.$ It is continuous but not differentiable.

Sorry, is your $\{(x,|x|)\}$ an atlas for $M=\mathbb R$ with standard topology made of one chart alone ? In this case $|x|$ isn't a valid chart map since it isn't injective on $\mathbb R$.

 

[fresh_42]

Sorry, is your $\{(x,|x|)\}$ an atlas for $M=\mathbb R$ with standard topology made of one chart alone ? In this case $|x|$ isn't a valid chart map since it isn't injective on $\mathbb R$.

It is a one-dimensional closed, continuous submanifold of $\mathbb{R}^2.$ The map $\varphi\, : \,M\longrightarrow \mathbb{R}$ defined by $\varphi(p)=\varphi(x_p,\left|x_p\right|)=x_p$ is injective.

 

[cianfa72]

It is a one-dimensional closed submanifold of $\mathbb{R}^2.$

Ah ok, $\{(x,|x|)\}$ is the set in $\mathbb R^2$. To me it is a smooth manifold when one doesn't look at how it is embedded in $\mathbb R^2$ (a manifold with an atlas of one chart alone is by definition $C^{\infty}$ differentiable - the compatibility condition is vacuosly fulfilled).

 

[fresh_42]

Ah ok, $\{(x,|x|)\}$ is the set in $\mathbb R^2$. To me it is a smooth manifold when one doesn't look at how it is embedded in $\mathbb R^2$ (a manifold with an atlas of one chart alone is by definition $C^{\infty}$ differentiable - the compatibility condition is vacuosly fulfilled).

The chart $(-\varepsilon,\varepsilon)\subseteq \mathbb{R}$ around $(0,0)\in M$ is not differentiable. Maybe that's the difference between a $\mathcal{C}^\infty$ atlas and a $\mathcal{C}^\infty$ manifold which I don't know. But if so, then it is a rather confusing name.

 

[lavinia]

There are topological manifolds that can not be given a smooth structure.

 

[cianfa72]

There are topological manifolds that can not be given a smooth structure.

Ah ok, so from their (unique) maximal topological atlas one can't extract a subset of smooth/$C^{\infty}$ compatible charts covering the manifold itself.

Can you kindly provide an example?

 

[cianfa72]

The chart $(-\varepsilon,\varepsilon)\subseteq \mathbb{R}$ around $(0,0)\in M$ is not differentiable. Maybe that's the difference between a $\mathcal{C}^\infty$ atlas and a $\mathcal{C}^\infty$ manifold which I don't know. But if so, then it is a rather confusing name.

Actually what isn't differentiable is the map $$\begin{align} (-\varepsilon,\varepsilon) \to \mathbb R^2 \nonumber \\ x \mapsto (x,|x|) \nonumber \end{align}$$ w.r.t. the standard differential structure in the domain and the target. However in this context such a notion is not involved at all.

What is true is that the set $M=\{(x,|x|)\}$ isn't a smooth submanifold of $\mathbb R^2$ w.r.t. its standard differential structure. Nevertheless $M$ as set can be turned into a smooth/$C^{\infty}$ one-dimensional differentiable manifold by means of an atlas containing only one chart, e.g. the chart defined above.

 

[lavinia]

Ah ok, so from their (unique) maximal topological atlas one can't extract a subset of smooth/$C^{\infty}$ compatible charts covering the manifold itself.

Can you kindly provide an example?

This is a difficult subject which I know nothing about except a few factoids. I think the lowest dimension of a non-smoothable manifold is 8. Here are a couple of links.

https://math.stackexchange.com/questions/408221/the-easiest-non-smoothable-manifold

https://encyclopediaofmath.org/wiki/Non-smoothable_manifold

If you would like to learn this stuff, I would be happy to learn it with you.

 

[lavinia]

Ah ok, so from their (unique) maximal topological atlas one can't extract a subset of smooth/$C^{\infty}$ compatible charts covering the manifold itself.

Can you kindly provide an example?

Also some manifolds have more than one differentiable structure. This means that the underlying topological manifolds are homeomorphic but there is no diffeomorphism between them. I believe that the first examples discovered were exotic 7 spheres. Milnor was stunned when he realized what he had discovered. The 7 sphere has 27 distinct non-diffeomorphic differentiable structures.

Later on other examples were found. Euclidean 4 space has uncountably many differentiable structures. All other Euclidean spaces have only one. Go figure.

https://mathoverflow.net/questions/24970/exotic-differentiable-structures-on-r4

Milnor's Characteristic Classes proves that there is an exotic 7 sphere. As I remember it involves 3 sphere bundles over the 4 sphere.

Here is a paper

https://www.math.utoronto.ca/tiozzo/docs/exotic.pdf

I just found this paper

https://www.sciencedirect.com/science/article/pii/S0926224503000123

In the introduction it says that 15 of the 27 exotic spheres are 3 sphere bundles over the 4 sphere. All have Euler class +/- 1. Milnor's Characteristic Classes and Bott and Tu explain Euler classes of sphere bundles..

I would also be willing to learn this stuff with you.

One cannot understand these ideas without Algebraic Topology. I could guide you through what you need. Bott and Tu uses De Rham Cohomology whereas Milnor uses Singular Cohomology.

 

[lavinia]

Actually what isn't differentiable is the map $$\begin{align} (-\varepsilon,\varepsilon) \to \mathbb R^2 \nonumber \\ x \mapsto (x,|x|) \nonumber \end{align}$$ w.r.t. the standard differential structure in the domain and the target. However in this context such a notion is not involved at all.

What is true is that the set $M=\{(x,|x|)\}$ isn't a smooth submanifold of $\mathbb R^2$ w.r.t. its standard differential structure. Nevertheless $M$ as set can be turned into a smooth/$C^{\infty}$ one-dimensional differentiable manifold by means of an atlas containing only one chart, e.g. the chart defined above.

The embedding is not smooth but what do you mean that it is not a smooth submanifold? If you define this precisely then the answer will pop out.

 

[cianfa72]

The embedding is not smooth but what do you mean that it is not a smooth submanifold?

I mean that since the embedding in $\mathbb R^2$ is not smooth w.r.t. the $\mathbb R^2$ standard differential structure (it is just a topological embedding), then it is not an embedded/regular submanifold according to the definition here.

 

[cianfa72]

Also some manifolds have more than one differentiable structure. This means that the underlying topological manifolds are homeomorphic but there is no diffeomorphism between them

I would say that they actually share the same underlying topological manifold structure (i.e. the differentiable atlases defining the different (even though diffeomorphic) differentiable structures are all $C^0$-compatible each other hence any of them determines the same (unique) topological/$C^0$ maximal atlas).

 

[jbergman]

From Lee's example 1.23, consider the homeomorphism $\psi(x) = x^3$

The atlas consisting of that single chart is not smoothy compatible with the standard smooth structure defined by the identity function $id(x) = x$.

Therefore we can construct different smooth structures on any given positive dimensional topological manifold if it supports a smooth structure.

See also problem 1-6.

 

[cianfa72]

Therefore we can construct different smooth structures on any given positive dimensional topological manifold if it supports a smooth structure.

It follows that if a topological manifold admits an atlas consisting of one (global) chart alone then for sure it supports a smooth structure.

 

[fresh_42]

It follows that if a topological manifold admits an atlas consisting of one (global) chart alone then for sure it supports a smooth structure.

Reference? This time with proof, please.

I think you misunderstood Whtney's weak embedding theorem.

 

[cianfa72]

Reference? This time with proof, please.

I believe it is straightforward. A smooth atlas is a collection of $C^{\infty}$-compatible charts covering the topological manifold. Now if the atlas consists of a single chart alone then the $C^{\infty}$-compatibility condition is vacuosly satisfied. Such an atlas defines a smooth structure on the topological manifold (actually it identifies the unique maximal smooth atlas it belongs to).

Edit: see also the corollary of Proposition 1.17 in Lee's book.