More vector fields question

From: Tony (Ttiger222_at_hotmail.com)
Date: 02/19/05 

Date: Fri, 18 Feb 2005 21:38:23 -0500

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

Relevant Pages

  • Re: One more vector fields question
    ... >> Show that there is a smooth vector field on S^2 that vanishes at exactly ... >> is the north pole....is this vector field smooth? ... Such a vector field will not be smooth at infinity. ...
    (sci.math)
  • One more vector fields question
    ... Show that there is a smooth vector field on S^2 that vanishes at exactly ... On S^2 minus the north pole and south pole, ... Clearly Y= 0 when p is the south pole. ...
    (sci.math)
  • Re: One more vector fields question
    ... > Show that there is a smooth vector field on S^2 that vanishes at exactly ... > where are the coordinates of the stereographic projection of p onto ... > is the north pole....is this vector field smooth? ... the hint is suggesting that you consider any constant vector field on ...
    (sci.math)
  • Re: More vector fields question
    ... On 20.02.2005 11:13, Stuart M Newberger wrote: ... >>vector field on U, then I can extend it to a smooth vector field on M ... The vector space of globally defined vector fields is not finite ...
    (sci.math)