Is the projective space a smooth manifold?

[Delong66]

Suppose you have the map $$\pi : \mathbb{R}^{n+1}-\{0\} \longrightarrow \mathbb{P}^n$$.
I need to prove that the map is differentiable.
But this map is a chart of $$\mathbb{P}^n$$ so by definition is differentiable?

MENTOR NOTE: fixed Latex mistakes double $ signs and backslashes needed for math

 

[PeroK]

You need an extra dollar for Latex to render.

 

[fresh_42]

Suppose you have the map $$\pi : \mathbb{R}^{n+1}-\{0\} \longrightarrow \mathbb{P}^n$$.
I need to prove that the map is differentiable.
But this map is a chart of $$\mathbb{P}^n$$ so by definition is differentiable?

MENTOR NOTE: fixed Latex mistakes double $ signs and backslashes needed for math

In order to prove differentiability, you need a differential structure on both. This is no problem for $\mathbb{R}^{n+1}-\{0\}$ but what is it for $\mathbb{P}^n$? In order to equip the projective space with a differential structure, we need charts, two charts to be exact.

 

[martinbn]

In order to equip the projective space with a differential structure, we need charts, two charts to be exact.

At least two. The standard way uses more (except for the projective line).

 

[WWGD]

You need an extra dollar for Latex to render.

I could use a few myself, render or not. ;).

And no compact manifold can have a single chart, as if it did, it would be homeomorphic to $\mathbb R^n$ itself, which is not compact.

 

[lavinia]

Suppose you have the map $$\pi : \mathbb{R}^{n+1}-\{0\} \longrightarrow \mathbb{P}^n$$.
I need to prove that the map is differentiable.
But this map is a chart of $$\mathbb{P}^n$$ so by definition is differentiable?

MENTOR NOTE: fixed Latex mistakes double $ signs and backslashes needed for math

The projection map from $R^{n+1}-0$ is not a chart. First of all, it is not one to one. Each ray through the origin is sent to a single point in projective space. Also a chart is technically defined as a map from an open set in the manifold into Euclidean space not from Euclidean space into the manifold.

On the other hand the restriction of this map to the unit sphere $S^{n}$ is two to one and any open subset that does not contain a pair of antipodal points (e.g. a polar ice cap) is mapped homeomorphically onto its image in projective space. One can take the inverses of these maps as coordinate charts. In this way $P^{n}$ inherits a differential structure from the n sphere.

To avoid antipodal points, more than two such charts are required to completely cover projective space.

 

[WWGD]

If we only needed one chart, that would imply a homeomorphism with Euclidean n-space. But that's not possible since the Projective Space is compact, while Euclidean n-space is not.

 

[lavinia]

If we only needed one chart, that would imply a homeomorphism with Euclidean n-space. But that's not possible since the Projective Space is compact, while Euclidean n-space is not.

What about two charts? What are the compact manifolds that can be covered by only two charts?

 

[martinbn]

What about two charts? What are the compact manifolds that can be covered by only two charts?

For example spheres.

 

[WWGD]

What about two charts? What are the compact manifolds that can be covered by only two charts?

Is there any such classification at this point?

 

[lavinia]

Is there any such classification at this point?

Not sure. If one starts with an open cover by homeomorphs of an open disk then I think the only compact manifold with two charts is the sphere.

If one requires all finite intersections of the sets in the cover to also be homeomorphs of a disk then one gets what is called a "good cover". I don't think there are any compact manifolds with a good cover with only two charts.

Question: How many arcs are needed to make a good cover of the circle? Notice that 2 arcs fail because their intersection is a pair of arcs rather than just one. How many open disks for the 2 sphere? Open n-balls for the n sphere?

Generally I suspect that finding the minimum number of open sets in a good cover of a manifold is not easy.

 

[WWGD]

Not sure. If one starts with an open cover by homeomorphs of an open disk then I think the only compact manifold with two charts is the sphere.

If one requires all finite intersections of the sets in the cover to also be homeomorphs of a disk then one gets what is called a "good cover". I don't think there are any compact manifolds with a good cover with only two charts.

Question: How many arcs are needed to to make a good cover of the circle? Notice that 2 arcs fail because their intersection would be a pair of arcs rather than just one. How many open disks for the 2 sphere. Open n-balls for the n sphere?

Generally I suspect that finding the minimum number of open sets in a good cover of a manifold is not easy.

Ah. I'm reminded of the term " finite good cover". Though unfortunately, O cant remember now where Ive heard it.

 

[lavinia]

Ah. I'm reminded of the term " finite good cover". Though unfortunately, O cant remember now where Ive heard it.

I saw an application in Bott and Tu which relates the real Cech cohomology of a good cover of a manifold to its De Rham cohomology.

 

[WWGD]

I saw an application in Bott and Tu which relates the real Cech cohomology of a good cover of a manifold to its De Rham cohomology.

Ah, yes, I think that's where I first read it. Re some type of chain complex.