Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Got the theorem, having trouble with the proof

  1. Jun 12, 2005 #1
    Got the theorem, having trouble with the proof... [SOLVED]

    Hi all. OK, so I am trying to prove a theorem that I have for some time been just using as-is. Long story short, it occured to me that I needed to prove it. So, I have almost done it, but am stuck near the end. The theorem is:

    Suppose [itex] \mathcal{X} [/itex] is a smooth vector field on a manifold [itex]\mathcal{M}[/itex]. Assuming that [itex]\mathcal{X}_p\neq0[/itex] at a point [itex]p\in\mathcal{M}[/itex], then there exists a coordinate neighborhood [itex]\left(\mathcal{W};w^i\right)[/itex] about [itex]p[/itex] such that
    \left.\mathcal{X}\right\vert_\mathcal{W}=\dfrac{\partial}{\partial w^1}

    Proof: ????

    Now, it's not that I have nothing for the proof, it's just that I'm stuck. As well, since there is more than one way to skin a cat, I figured it would be better to leave the proof empty, rather than potentially confuse anyone with the technique I have employed thus far.

    That said, a big thanks in advance for all the help!
    Last edited: Jun 13, 2005
  2. jcsd
  3. Jun 12, 2005 #2
    lets see if it can be contradicted.

    what the proof asks, is whether there we can setup a coordinate system around the point [itex]\displaystyle p[/itex] s.t. the vector [itex]\displaystyle X_p[/itex] points in the same direction as one of our tangent space basis vectors [tex]\frac{\partial}{\partial w}[/tex]

    Well, let's assume that we cannot orient our coordinate system s.t. [tex]X_p = \frac{\partial}{\partial w}[/tex]. Then this says that either [itex]X_p = 0[/itex] (which you have already told us is not allowed) or [itex]X_p[/itex] does not lie in the tangent space [tex]M_p[/tex].

    If it does not lie in the tangent space, then [tex]X[/tex] is not a smooth vector field at the point [tex]p[/tex], a contradiction.
  4. Jun 12, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    he want's it at every point nearby. not just one, so it is probably the existence theorem for differential equations i guess.
  5. Jun 12, 2005 #4

    Yeah, I have had to invoke the existance and uniqueness of solutions of ODE's for this one so far...
  6. Jun 13, 2005 #5
    I have the proof in a book of mine, it has as a necessary and sufficient condition for 2 linearally independant vector fields to be a coordinate basis, on a 2 dimensional manifold, that the Lie bracket vanishes (i.e. they commute). Can't be bothered to write out the proof but you can find it in the section titled "when is a basis a coordinate basis?" in chapter 2 of "Geometrical Methods for Mathematical Physics"-B. Schutz. The proof obviously can be extended to higher dimensions.
    Last edited by a moderator: Jun 13, 2005
  7. Jun 13, 2005 #6
    OK, I think I got it. The indexing terms were giving me the businsess, but it seems to be all good now. Submitted for your approval:


    Attached Files:

    Last edited: Jun 13, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook