Re: More vector fields question

From: Jannick Asmus (jannick.news_at_web.de)
Date: 02/19/05 

Date: Sat, 19 Feb 2005 09:34:21 +0100

On 19.02.2005 03:38, Tony wrote:
> Let M be a nonempty manifold of dimension n >= 1. Show that T(M) is
> infinite dimensional.

I assume you mean as vector space over R.

>
> Here is my proof, and I was wondering if someone could let me know if it is
> valid/correct :
>
> Let M be a manifold of dimension n. Let Y : M ---> TM be a smooth vector
> field on M. Then, if p is in M, by the construction of TM, there is an open
> set U around p, and an induced map
>
> F : U ----> U x R^n.
>
> This gives a map F : U ---> R^n, where F(p) = (F_1 (p), ..., F_n (p))
>
> Now each F_i is an element of C^(infinity) (U)
>
> U is a (sub)manifold of M, and I know that C^(infinity)(M) is infinite
> dimensional for any manifold M.
>
> So, I have concluded that T(U) is infinite dimensional. Is this enough to
> say that T(M) is infinite dimensional? I know that if I have a smooth
> vector field on U, then I can extend it to a smooth vector field on M using
> bump functions....

No, not in general. You first change the vector field by multiplication
with a bump function! The resulting vector field is extendable. Does
this induce a map between T(U) and T(M)?

What about assuming that T(M) is finite dimensional and taking a basis
X1,...,Xm (with m = dim_R T(M))? Then use the fact (you quoted) that
C^00(M) is an infinite dimensional vector space over R.

I perfer thinking in maps and check their properties. This gives me a
clear insight into the problem most of the time.

J.

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