Why does this vector field approach zero near the north pole? In this question, Raziel's answer builds a vector field over $S^2$. The vector field is built from the push forward of the stereographic projection on $N$. Let $p \in S^2 \setminus \{N\}$, and let's say that $X_p = U_p \partial_u + V_p \partial_v$. I understant that as $p$ approaches the north pole, by the change of coordinates given in the other post, it can be seen that $X_p$ approaches zero (and thus can be extended at $N$ with zero). However, I don't understand it at an intuitive level. (I want to understand why this happens.)
If we fix $f \in C^{\infty}(S^2)$, should I understand that $X_p f$ approaches zero as $p \to N$ because you are deriving $f$ respect to a larger vector, or is that nonsense? I think that expressing $X_p$ in function of $\partial_x$ and $\partial_y$ would help me, but I'm not sure how to compute that.
 A: Maybe you can visualize the vector field $X_p$ using the usual identifications $T_pS^2 \subseteq T_p\mathbb{R}^3 = \mathbb{R}^3$. We have that
$$
\phi^{-1}_N(u,v)=\bigg{(}\frac{2u}{u^2+v^2+1},\frac{2v}{u^2+v^2+1},\frac{u^2+v^2-1}{u^2+v^2+1}\bigg{)}
$$
So, under the identifications above, we compute
$$
d_{(u,v)}\phi^{-1}_N\bigg{(}\frac{\partial}{\partial u}\bigg{)} = \frac{\partial \phi^{-1}_N(u,v)}{\partial u} = \bigg{(}\frac{2+2v^2-2u^2}{(u^2+v^2+1)^2},\frac{-4uv}{(u^2+v^2+1)^2},\frac{4u}{(u^2+v^2+1)^2}\bigg{)}
$$
Which is your vector field $X$ at the point $\phi^{-1}_N(u,v)$. You can verify that it is tangent taking the dot product. This formula works for every point in the sphere except the north pole $N$. But you can verify taking the limit $(u,v)\to \infty$ that it can be extended continuously at $N$ by $(0,0,0)$. And it is an interesting exercise to prove that all its derivatives (of all orders) tends to $(0,0,0)$ as $(u,v)\to\infty$. So the extension is smooth.
Regarding your comment about why $X_pf$ tends to $0$ as $p\to N$ this is simply because it is a continuos function and $X_N=0$ is the zero derivation so $X_Nf=0$.
