Krdytkyu

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.

Maybe you can visualize the vector field $X_p$ using the usual identifications $T_pS^2 subseteq T_pmathbb{R}^3 = mathbb{R}^3$. 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

$$
d_{(u,v)}phi^{-1}_Nbigg{(}frac{partial}{partial u}bigg{)} = frac{partial phi^{-1}_N(u,v)}{partial u} = bigg{(}frac{2+2v^2-2u^2}{(u^2+v^2+1)^2},frac{-4uv}{(u^2+v^2+1)^2},frac{4u}{(u^2+v^2+1)^2}bigg{)}
$$

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 a nice exercise to prove that all its derivatives (of all orders) tends to $(0,0,0)$ as $(u,v)toinfty$. So the extension is smooth.

Regarding your comment about why $X_pf$ tends to $0$ as $pto N$ this is simply because it is a continuos function and $X_N=0$ is the zero derivation so $X_Nf=0$.