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Let $X$ be a vector field on $\mathbb{S}^2\subset \mathbb{R}^3$ and $\pi : (x,y,z) \mapsto (u,v)$ the stereographic projection from the north pole, and let $(u,v)$ the corresponding coordinates. Find $X : \mathbb{S}^2 \rightarrow \mathbb{R}^3$ knowing that $X=v\frac{\partial}{\partial u} - u\frac{\partial}{\partial v}$ in those coordinates.

I don't know if what I did makes much sense. The projection $\pi$ is given by $$\pi(x,y,z)=\left(\frac{x}{1-z}, \frac{y}{1-z}\right) = (u,v)$$ so $$\frac{\partial}{\partial u} = \frac{1}{1-z}\frac{\partial}{\partial x}+ \frac{x}{(1-z)^2}\frac{\partial}{\partial z}$$ and $$\frac{\partial}{\partial v} = \frac{1}{1-z}\frac{\partial}{\partial y} + \frac{y}{(1-z)^2}\frac{\partial}{\partial z}$$ And plugging those values on the expression for $X$ I get: $$X = \frac{1}{(1-z)^2} \left( y\frac{\partial}{\partial x} - x \frac{\partial}{\partial y} \right)$$ but it feels weird that the terms with $\frac{\partial}{\partial z}$ cancel out, so I guess this is not the right way to proceed. Any help or tip that points me into the right direction would be appreciated. Thank you!

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    It's not weird at all :) if you integrate this vector field at plane, you'll obtain concentric circles around the origin as integral curves. If you project them back to the sphere, you'll obtain circles on sphere which lie on planes parallel to $z = 0$. Obviously all of them have tangent vectors that are parallel to $z = 0$ and hence the $z$-component is missing :) – 2017-02-09

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