Re: More vector fields question

From: Stuart M Newberger (smnewberger_at_comcast.net)
Date: 02/20/05 

Date: 20 Feb 2005 02:13:55 -0800

Tony wrote:
> Let M be a nonempty manifold of dimension n >= 1. Show that T(M) is
> infinite dimensional.
>
> 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....
>
> Thanks for any input,
>
> Tony

T(M) is not a vector space ,it is a vector bundle over M and thus a
manifold ,this manifold has dimension 2n.Regards,Stuart M Newberger

Quantcast