Is the [imath]\mathbb{S}^2[/imath] sphere diffeomorphic to [imath]P^2R[/imath] ?

[MathNugget]

Here, [imath]\mathbb{S}^2[/imath] is the 2-dimensional sphere, and [imath]P^2R[/imath] is the real projective plane.

My idea: finding diffeomorphisms between these 2 objects and open sets of [imath]\mathbb{R}^n[/imath] and [imath]\mathbb{R}^m[/imath], apparently if n=m then they're diffeomorphic, (the diffeomorphism should be a linear izomorphism, I seem to find online). If [imath]n\neq m[/imath], then they're not diffeomorphic.

This idea is related to a result (which I cannot prove), that a composition of 2 diffeomorphisms is a diffeomorphism.

I think [imath]\mathbb{S}^2[/imath] is diffeomorphic to [imath]\mathbb{R}^2[/imath] ; it takes 2 maps, using the stereographic projection. (It has a pretty ugly form in coordinates, here's a link to it https://en.wikipedia.org/wiki/Stereographic_projection ). Now, I am fairly certain I am mixing the concept of diffeomorphism and smooth manifold (I believe a smooth manifold needs to have at least 1 diffeomorphism), I know these 2 things are smooth manifolds (can't prove it), fairly certain the sphere is diffeo- to [imath]\mathbb{R}^2[/imath]. Could I get some advice for pursuing this question, please?

 

[blamocur]

[imath]\mathbb S^2[/imath] is diffeomorphic to neither [imath]\mathbb P^2[/imath] nor to [imath]\mathbb R^2[/imath]. It is not even homeomorphic to either of them. For example [imath]\pi_2 (\mathbb R^2)[/imath] is trivial, but [imath]\pi_2(\mathbb S^2)[/imath] is not.

a result (which I cannot prove), that a composition of 2 diffeomorphisms is a diffeomorphism.

It shouldn't be difficult to find a proof in any textbook on differential geometry. Probably online too.

I think S2\mathbb{S}^2S2 is diffeomorphic to R2\mathbb{R}^2R2 ; it takes 2 maps, using the stereographic projection.

You have to have 1-to-1 global map between the manifolds, then show that local maps exist for every point. Just having two local maps is not enough, even if they cover the whole thing.

 

[MathNugget]

[imath]\mathbb S^2[/imath] is diffeomorphic to neither [imath]\mathbb P^2[/imath] nor to [imath]\mathbb R^2[/imath]. It is not even homeomorphic to either of them. For example [imath]\pi_2 (\mathbb R^2)[/imath] is trivial, but [imath]\pi_2(\mathbb S^2)[/imath] is not.

It shouldn't be difficult to find a proof in any textbook on differential geometry. Probably online too.

You have to have 1-to-1 global map between the manifolds, then show that local maps exist for every point. Just having two local maps is not enough, even if they cover the whole thing.

True, I need an atlas, a set of maps that cover the whole thing and 'go well together', sort of. (my terminology isn't very accurate, I know) . Meaning, the intersection of 2 different open sets through a map doesn't go to 2 different places...

Who's [imath]\pi_2[/imath] here? is it projection on 2nd component?

 

[blamocur]

True, I need an atlas, a set of maps that cover the whole thing and 'go well together', sort of. (my terminology isn't very accurate, I know) . Meaning, the intersection of 2 different open sets through a map doesn't go to 2 different places...

Not sure I understand this statement, but it does sound insufficient because it does not mention 1-to-1 global mapping.

Who's π2\pi_2π2 here? is it projection on 2nd component?

No, it is the second homotopy group. When two manifolds are homeomorphic the corresponding groups must be the same. For [imath]\mathbb S^2[/imath] and [imath]\mathbb R^2[/imath] they are not.

 

[MathNugget]

Not sure I understand this statement, but it does sound insufficient because it does not mention 1-to-1 global mapping.


No, it is the second homotopy group. When two manifolds are homeomorphic the corresponding groups must be the same. For [imath]\mathbb S^2[/imath] and [imath]\mathbb R^2[/imath] they are not.

It's not clear to me how I'd calculate (any) homotopy group.
It says:

[imath]\pi_n(X)[/imath] is the set of homotopy classes of maps [imath]f:S^n \rightarrow X \vert f(a)=b[/imath],

where [imath]a[/imath] is a fixed point on [imath]S^n[/imath], and [imath]b[/imath] is a fixed point on X.

Why wouldn't they have the same homotopy group, then? is it because of the (0, 0) point of the projective plane?

I'm attempting to read (parts of) lee's smooth manifolds book, hopefully it will make some sense of this exercise...

 

[blamocur]

I brought up the homotopy group as a proof -- or illustration -- that 2d sphere is not diffeomorphic to 2d plane. Details about homotopy wouldn't fit the format of this forum (they can be found in any introductory course on algebraic topology).

The important part is not proving that diffeomorphism does not exist, but that you do not have a valid one in your post.

 

[blamocur]

I'm attempting to read (parts of) lee's smooth manifolds book, hopefully it will make some sense of this exercise...

It is a good idea to develop clearer understanding of what diffeomorphism is before proceeding with smooth manifolds.

 

[MathNugget]

It is a good idea to develop clearer understanding of what diffeomorphism is before proceeding with smooth manifolds.

I can't manage to get past memorizing definitions, when it comes to a totally abstract geometry. It just...makes no sense, so I am trying to get at least a practical pattern of solving such questions.

A diffeomorphism is a differentiable and bijective map between 2 differential (=smooth?) manifolds, with differentiable inverse. While:

A differentiable manifold is a Hausdorff and second countable topological space M, together with a maximal differentiable atlas on M.

The maps of said atlas send open sets of M to open sets of [imath]\mathbb{R}^n[/imath].

Aren't both [imath]S^2[/imath] and [imath]\mathbb{P}^2[/imath] locally homeomorphic to (open sets of) [imath]\mathbb{R}^2[/imath]? I realize that maps in general aren't diffeomorphic, just continuous with continuous inverse, but here, I think, they are...

 

[MathNugget]

Also, the map here (and its inverse) seem differentiable to me... https://en.wikipedia.org/wiki/Stereographic_projection if this is what you were mentioning here:

The important part is not proving that diffeomorphism does not exist, but that you do not have a valid one in your post.

 

[blamocur]

Aren't both S2S^2S2 and P2\mathbb{P}^2P2 locally homeomorphic to (open sets of) R2\mathbb{R}^2R2?

Locally yes.

Aren't both S2S^2S2 and P2\mathbb{P}^2P2 locally homeomorphic to (open sets of) R2\mathbb{R}^2R2?

They are, just as is every 2-dimensional manifold. But locally is the important part here. But do you understand that what is missing is a 1-to-1 map between all points of two manifolds?

 

[MathNugget]

Locally yes.

They are, just as is every 2-dimensional manifold. But locally is the important part here. But do you understand that what is missing is a 1-to-1 map between all points of two manifolds?

Well, I don't really understand why it would be missing? Say, why wouldn't we fix 1 point on the sphere, N. Why wouldn't sending all the big circles that go through N into non parallel lines in the projective plane work?

The whole thing is very complicated...I m supposed to propose a function [imath]f: S^2\rightarrow \mathbb{P}^2[/imath] between the manifolds, and then check the [imath]\psi(f(\phi^{-1}))[/imath], where [imath]\phi[/imath] is the homeomorphism from [imath]S^2[/imath] to [imath]R^n[/imath], [imath]\psi[/imath] is the same thing for the projective plane, then check if this composition of 3 functions is differentiable...
And to be closer to correct, [imath]\phi \vert_x[/imath] and [imath]\psi \vert_x[/imath] should be used, because the atlases likely have more than 1 map, so we have to work in a neighbourhood of an arbitrary point...

 

[blamocur]

Well, I don't really understand why it would be missing?

A 1-to-1 map between the points. I cannot see it defined anywhere.

Say, why wouldn't we fix 1 point on the sphere, N. Why wouldn't sending all the big circles that go through N into non parallel lines in the projective plane work?

I don't understand how this would work -- can you be more specific? Where is point N mapped? Which definition of projective plane are you using? Where is an arbitrary point with spheric coordinates [imath]\theta, \zeta[/imath] mapped ?

Also, since you believe that [imath]\mathbb S^2[/imath] is diffeomorphic to [imath]\mathbb R^2[/imath] why don't you define the global map for that pair -- might be easier to discuss than projective plane.

 

[MathNugget]

Ok, let's forget that
How would proving they're not diffeomorphic go? I suppose I use the fact that a diffeomorphism is also a homeomorphism, and saying the sphere (for some reason) cannot be continually deformed into the projective plane?

 

[blamocur]

How would proving they're not diffeomorphic go?

I believe that algebraic topology has the best tools -- like homotopy groups -- for that.

I suppose I use the fact that a diffeomorphism is also a homeomorphism,

Yes.

saying the sphere (for some reason) cannot be continually deformed into the projective plane?

A typical reason is the difference of algebraic constructs -- usually groups -- from algebraic topology. When two spaces are homeomorphic their groups must be identical. Conversely, when the groups are different the spaces cannot be homeomorphic.