Maximal atlas of topological manifold

[fresh_42]

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

It is not and I doubt it is true. If you have an atlas of homoeomorphisms, how will you make one with diffeomorphisms? How do you guarantee that it even has a differentiable structure? If it were true, what meaning would a topological manifold even have if it would be smooth anyway? I see that $(x,x^{2n})$ is homeomorphic to $(x,|x|)$ but that would not be the same manifold. It would be an approximation as in Whitney's weak embedding theorem, or a homeomorphic image.

You cannot simply claim things. And if asked about a reference, you need to deliver and not just say "straightforward". This is an adjective I would avoid in topology by all means.

 

[jbergman]

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

That statement is true from the definition of a smooth atlas, but it doesn't follow from the quoted text you responded to. The point of my post was that we have to be careful when talking about atlases. You could have a $C^k$ atlas where some of the charts are only $C^k$ on the overlap. And such that could split into two different smooth atlases depending on which charts are removed.

 

[cianfa72]

The point of my post was that we have to be careful when talking about atlases. You could have a $C^k$ atlas where some of the charts are only $C^k$ on the overlap. And such that could split into two different smooth atlases depending on which charts are removed.

Ah ok, basically you mean that, for instance, starting from a $C^k$ atlas one could get a $C^{k+1}$ atlas throwing away some charts from it.

 

[jbergman]

It is not and I doubt it is true. If you have an atlas of homoeomorphisms, how will you make one with diffeomorphisms? How do you guarantee that it even has a differentiable structure? If it were true, what meaning would a topological manifold even have if it would be smooth anyway? I see that $(x,x^{2n})$ is homeomorphic to $(x,|x|)$ but that would not be the same manifold. It would be an approximation as in Whitney's weak embedding theorem, or a homeomorphic image.

You cannot simply claim things. And if asked about a reference, you need to deliver and not just say "straightforward". This is an adjective I would avoid in topology by all means.

See page 14, of Introduction to Smooth Manifolds by John Lee. "If a topological manifold M can be covered by a single chart, the smooth compatibility condition is trivially satisfied..."

 

[fresh_42]

See page 14, of Introduction to Smooth Manifolds by John Lee. "If a topological manifold M can be covered by a single chart, the smooth compatibility condition is trivially satisfied..."

That's what I already said: the compatibility condition, yes, since there are no overlappings. But that doesn't make a homeomorphism a diffeomorphism.

 

[jbergman]

That's what I already said: the compatibility condition, yes, since there are no overlappings. But that doesn't make a homeomorphism a diffeomorphism.

A chart map doesn't have to be a diffeomorphism, only a homeomorphism. If you reread Lee's quote he says a topological manifold covered by a single chart, i.e., a homeomorphism to an open subset of $\mathbb R^n$, trivially has a smooth structure.

The definition of an atlas is a set of chart maps that are homeomorphisms that also satisfy the compatibility condition. You can see the bottom p.49 of Tu's introduction to manifolds.

Basically we start with a topological manifold and it's atlas and throw out charts that aren't compatible on overlaps. I can't find a reference that states it super clearly.

Different smooth structures just change which functions are considered as smooth on the manifold.

 

[fresh_42]

A chart map doesn't have to be a diffeomorphism, only a homeomorphism. If you reread Lee's quote he says a topological manifold covered by a single chart, i.e., a homeomorphism to an open subset of $\mathbb R^n$, trivially has a smooth structure.

The definition of an atlas is a set of chart maps that are homeomorphisms that also satisfy the compatibility condition. You can see the bottom p.49 of Tu's introduction to manifolds.

Basically we start with a topological manifold and it's atlas and throw out charts that aren't compatible on overlaps. I can't find a reference that states it super clearly.

Different smooth structures just change which functions are considered as smooth on the manifold.

I think I got it. Took a while, sorry.

I simply ignored that differentiability on a manifold itself isn't there without charts, silly me.

 

[cianfa72]

Basically we start with a topological manifold and it's atlas and throw out charts that aren't compatible on overlaps.

Just to be super clear: the compatibility on overlaps we are talking about is at least $C^1$ since $C^0$ compatibility between charts is always satisfied by definition of atlas.