I have the following problem:
Can the vector field $X(x,y,z)=(-y,x,0)$ on $S^2$ be the gradient (on the sphere) of a function $f:S^2\rightarrow \mathbb{R}$ with respect to the standard euclidean metric?
I believe the answer is no, and my argument which I am unsure of is as follows:
One can see that if such an $f$ does exist then $df=-ydx+xdy$. By spherical coordinates we have:
$z=\cos \theta, \quad y=\sin \theta \sin \phi, \quad x=\sin \theta \cos \phi$
Then, I conclude that:
$dx= \cos \theta \cos \phi \cdot d\theta -\sin \theta \sin \phi \cdot d\phi$, $dy= \cos \theta \sin \phi \cdot d\theta +\sin \theta \cos \phi \cdot d\phi$ and $dz= -\sin \theta \cdot d\theta$.
Using these identities I obtain that:
$ df =\sin^2\theta \cdot d\phi $
And therefore $d^2f=\sin 2\theta \cdot d\theta \wedge d\phi \neq 0$. This is a contradiction to the fact that $d^2\equiv0$.
My problem is whether the spherical coordinates I wrote are indeed a smooth chart, and whether I am working in only one chart? I know that the complement is of measure zero in $S^2$, but I am still uncertain.
Best Answer
Spherical coordinates are indeed a smooth chart, valid on a region that excludes at least the poles and a branch cut between them.
And then - if $g=df$, then $d^2f=0$ identically and it's zero in every chart. You have found that $dg$ is not zero in this chart, so it's not identically zero. We don't need the bit about the chart covering almost everything; even a chart covering a tiny patch on the sphere would be enough.
Your argument is valid.