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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


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    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:


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    Last edited: Jun 13, 2005
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