# Got the theorem, having trouble with the proof

1. Jun 12, 2005

### sambo

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 $\mathcal{X}$ is a smooth vector field on a manifold $\mathcal{M}$. Assuming that $\mathcal{X}_p\neq0$ at a point $p\in\mathcal{M}$, then there exists a coordinate neighborhood $\left(\mathcal{W};w^i\right)$ about $p$ 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. Jun 12, 2005

### quetzalcoatl9

lets see if it can be contradicted.

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

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

If it does not lie in the tangent space, then $$X$$ is not a smooth vector field at the point $$p$$, a contradiction.

3. Jun 12, 2005

### mathwonk

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

4. Jun 12, 2005

### sambo

Yeah, I have had to invoke the existance and uniqueness of solutions of ODE's for this one so far...

5. Jun 13, 2005

### Dave1989

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
6. Jun 13, 2005

### sambo

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:

<proof.pdf>

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Last edited: Jun 13, 2005