TITLE: A smooth vector field on $S^2$ vanishing at a single point
QUESTION [25 upvotes]: I need to find a vector field as described in the title.
I was given a couple hints, and this is what I have so far. Let $\varphi:S^2\setminus\{N\}\to\mathbb{R}^2$ be stereographic projection ($N$ is the north pole). Let $Y$ be a smooth, non-zero vector field on $\mathbb{R}^2$ that dies off to $0$ "at infinity." Since $\varphi$ is a diffeomorphism, we can push $Y$ forward along $\varphi^{-1}$ to get a smooth vector field $(\varphi^{-1})_*Y_{\varphi(p)}$ on $S^2$. But this vector field is not zero at the north pole... it's undefined. So can I just define my vector field on $S^2$, call it Z, to be the pushforward of $Y$ (as above) away from $N$ but $0$ at $N$? And if so, do we still have smoothness?

REPLY [29 votes]: Short Version:
There is no need to use "complicated" functions to kill the field at "infinity". Standard coordinate fields will do the trick, thanks to the stereographic coordinates structure.
Let $(u,v)$ be the stereographic projection coordinates on $\mathbb{S}^2-\{N\}$. Then consider the vector field $\partial_u$ (or also $\partial_v$). A priori, it is defined only on $\mathbb{S}^2-\{N\}$
Now try to compute the local expression of this vector field in stereographic coordinatates but from the south pole. You will notice that the field vanishes at zero (i.e. at the south pole $S$).
Therefore, you can extend the field $\partial_u$ to a well defined vector field on $\mathbb{S}^2$ which vanishes exactly at one point (i.e. the south pole $S$).
Long Version:
More precisely, let us denote $(u,v)$ the stereographic coordinates relative to the projection from the north pole, that is the map $\phi_N : \mathbb{S}^2-\{N\} \to \mathbb{R}^2$. Let us denote $(\overline{u},\overline{v})$ the stereographic coordinates relative to the projection from the south pole, which is the map $\phi_S : \mathbb{S}^2-\{S\} \to \mathbb{R}^2$.
Now, consider the vector field $\partial_u$ (in coordinates), defined on $\mathbb{S}^2-\{N\}$. On the intersection of the two charts, $\mathbb{S}^2-\{N,S\}$, we can compute the expression of $\partial_u$ in $(\overline{u},\overline{v})$ coordinates. The result is:
$\partial_u = (\overline{v}^2-\overline{u}^2)\partial_\overline{u} - 2\overline{u}\overline{v}\partial_\overline{v} \qquad (1)$
Where, of course, $(\overline{u},\overline{v}) = \phi_S \circ \phi_N^{-1} (u,v)$. You can easily see that this vector field can be extended at the north pole, by formula (1).
Therefore, a field with the property you required is:
$X_p = \begin{cases}
\left(\phi_N^{-1}\right)_* \left(\frac{\partial}{\partial u}\right) & p \in \mathbb{S}^2-\{N\} \\ \left(\phi_S^{-1}\right)_* \left((\overline{v}^2-\overline{u}^2)\frac{\partial}{\partial \overline{u}} - 2\overline{u}\overline{v}\frac{\partial}{\partial \overline{v}}\right) & p \in \mathbb{S}^2-\{S\}
\end{cases}$
$X_p$ is a well defined vector field on the whole $\mathbb{S}^2$. It is also obviousvly smooth, since it is smooth in local coordinates. Moreover $X_N =0$, and $X_p = \partial_u \neq 0$ on $\mathbb{S}-\{N\}$, as required.
Explicit calculation for the change of coordinates
The change of coordinates map (and its inverse) can be computed explicitly:
$(u,v) = \phi_N \circ \phi_S^{-1}(\overline{u},\overline{v}) = \frac{(\overline{u},\overline{v})}{\overline{u}^2+\overline{v}^2}$
$(\overline{u},\overline{v}) = \phi_S \circ \phi_N^{-1}(u,v) = \frac{(u,v)}{u^2+v^2}$
By using these explicit expression it's easy to express $\partial_u$ in terms of barred coordinates:
$\frac{\partial}{\partial u} = \frac{\partial \overline{u}}{\partial u}\frac{\partial}{\partial \overline{u}} + \frac{\partial \overline{v}}{\partial u}\frac{\partial}{\partial \overline{v}}$
Now, I leave to you the last step (i.e. to explicitly compute the derivatives). REMEMBER, when computing the derivatives of the barred coordinates with respect to $u$, to express them in terms of BARRED coordinates.