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seydunas
Oct3-11, 02:18 PM
Hi,

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

lavinia
Oct3-11, 03:09 PM
Hi,

I have two questions: how can we prove a closed ball in R^n is manifold with boundry only using the definition being manifold with boundry. 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
Oct3-11, 03:25 PM
Much like you need two charts to cover the sphere, you will need two charts for the "unit ball with boundary".

quasar987
Oct3-11, 03:50 PM
Yes, C^{\infty}(M) is an infinite-dimensional vector space.

Bacle
Oct3-11, 06:48 PM
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
Oct4-11, 03:31 PM
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
Oct4-11, 03:33 PM
For manifold with boundary, how can we write the charts precisely?

Bacle
Oct4-11, 09:51 PM
You write the charts just like you do for manifolds without boundary, only that you have interior charts and boundary charts.

Jamma
Oct5-11, 04:00 AM
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.