Is C^∞(M) infinite dimensional for a smooth manifold?

[seydunas]

Hi,

I have two questions: how can we prove a closed ball in R^n is manifold with boundary only using the definition being manifold with boundary. Also i want to ask C^/inf(M) is infinite dimensional where M is smooth manifold of dimension n>0.

 

[lavinia]

Hi,

I have two questions: how can we prove a closed ball in R^n is manifold with boundary only using the definition being manifold with boundary. Also i want to ask C^/inf(M) is infinite dimensional where M is smooth manifold of dimension n>0.

Would this work?

Stand the ball on a tangent n-1 plane an subtract the height of the lower half of the boundary from the n-1 plane from each point in the lower half ball.

 

[Jamma]

Much like you need two charts to cover the sphere, you will need two charts for the "unit ball with boundary".

 

[quasar987]

Yes, $C^{\infty}(M)$ is an infinite-dimensional vector space.

 

[Bacle]

I'm not sure of the definition, but why not just produce charts for both the interior points and for the boundary points, i.e., show that the points in the (topological) boundary are also (in this case) part of the manifold boundary?

 

[seydunas]

C/inf(M) is infinite dimensional but how? I thought that for all point on M (one point is closed set) there exist open nhd, and by using partitions of unity we can extend the function on M , now i wonder that the set of theese functions is linearly independent or not? IF so, we are done.

 

[seydunas]

For manifold with boundary, how can we write the charts precisely?

 

[Bacle]

You write the charts just like you do for manifolds without boundary, only that you have interior charts and boundary charts.

 

[Jamma]

To see that C^/infty(M) is infinite dimensional, just find an infinite set of linearly independent functions. For example, any "bump function" around a point with different variations of "steepness" could generate infinitely many such functions.