Problem 8-13 (A smooth vector field on the sphere vanishing at a single point)

Proof. Define our vector field by

1.

$X$ does not vanish on ${𝕊}^2\setminus \{N\}$
Since $d_{\hat{p}}({\sigma }^{-1})$ is a vector space homomorphism for all $\hat{p}\in {ℝ}^2$, it will not map the standard basis derivations on $T_{\hat{p}}{ℝ}^2$ to zero derivations on $T_p{𝕊}^2$. In other words, $X$ cannot vanish anywhere on ${𝕊}^2\setminus \{N\}$.

2.

$X$ is smooth
Recall Proposition 8.1 (Smoothness Criterion for Vector Fields), which states that given a smooth chart $(U,\phi )=(U,x^1,\dots ,x^n)$, the restriction of $X$ to $U$ is smooth if and only if its component functions with respect to this chart are smooth¹ . This, by construction, makes $X$ automatically smooth on ${𝕊}^2\setminus \{N\}$. Thus, it is only left to demonstrate that $X$ is smooth at $N$.

3.

$X$ is smooth at the north pole
Consider the second chart $({ℝ}^2\setminus \{S\},\tilde{\sigma })=({ℝ}^2\setminus \{S\},{\tilde{x}}^1,{\tilde{x}}^2)$ in our atlas which covers the north pole, i.e., the south pole projection $\tilde{\sigma }=-\sigma (-u)$, given by

$$\begin{array}{cc}\tilde{\sigma }:{𝕊}^2\setminus \{S\}\to {ℝ}^2,\hfill & \tilde{\sigma }(u^1,u^2,u^3)=\frac{1}{1+u^3}\cdot (u^1,u^2)\hfill \end{array}$$

The change of coordinates $\tilde{\sigma }\circ {\sigma }^1:\sigma \left({ℝ}^2\setminus \{N,S\}\right)\to \tilde{\sigma }\left({ℝ}^2\setminus \{N,S\}\right)$ is given by

$$\begin{array}{llll}\hfill (\tilde{\sigma }\circ {\sigma }^{-1}):{ℝ}^2\setminus \{0\}\to {ℝ}^2\setminus \{0\},x& \mapsto \frac{1}{1+u^3}\left(\begin{array}{cc}\hfill u^1\hfill & \hfill u^2\hfill \end{array}\right)\circ \left(\begin{array}{c}\hfill \frac{2x^1}{|x{\text{|}}^2+1},\frac{2x^2}{|x{\text{|}}^2+1},\frac{|x{\text{|}}^2-1}{|x{\text{|}}^2+1}\hfill \end{array}\right)& \hfill & \\ \hfill & =\frac{1}{1+\frac{|x{\text{|}}^2-1}{|x{\text{|}}^2-1}}\left(\begin{array}{cc}\hfill \frac{2x^1}{|x{\text{|}}^2+1},\hfill & \hfill \frac{2x^2}{|x{\text{|}}^2+1}\hfill \end{array}\right)& \hfill & \\ \hfill & =\left(\begin{array}{cc}\hfill \frac{x^1}{|x{\text{|}}^2},\hfill & \hfill \frac{x^2}{|x{\text{|}}^2}\hfill \end{array}\right)& \hfill & \end{array}$$

We compute the coordinate representation of $X_p$ in terms of the smooth chart $({𝕊}^2\setminus \{S\},\tilde{\sigma })$:

$$\begin{array}{llll}\hfill X& =1\cdot \frac{\partial }{\partial x^i}={\tilde{X}}^1\frac{\partial }{\partial {\tilde{x}}^1}+{\tilde{X}}^2\frac{\partial }{\partial {\tilde{x}}^2},& \hfill & \end{array}$$

where ${\tilde{x}}^i$ denotes the coordinates with respect to the south pole projections.

Since $({\tilde{x}}^1,{\tilde{x}}^2)=1∕|x{\text{|}}^2(x^1,x^2)$, we have

$$\begin{array}{llll}\hfill \frac{\partial }{\partial x^i}& =\frac{\partial {\tilde{x}}^1}{\partial x^i}\frac{\partial }{\partial {\tilde{x}}^1}+\frac{\partial {\tilde{x}}^2}{\partial x^i}\frac{\partial }{\partial {\tilde{x}}^2}& \hfill & \\ \hfill & =\left(\frac{{\delta }_{1,i}}{\parallel x{\text{∥}}^2}+x^1\cdot (-1)\cdot \frac{1}{\parallel x{\text{∥}}^4}\cdot 2x^i\right)\frac{\partial }{\partial {\tilde{x}}^1}+\left(\frac{{\delta }_{2,i}}{\parallel x{\text{∥}}^2}+x^2\cdot (-1)\cdot \frac{1}{\parallel x{\text{∥}}^4}\cdot 2x^i\right)\frac{\partial }{\partial {\tilde{x}}^2}& \hfill & \\ \hfill & =\left(\frac{{\delta }_{1,i}}{\parallel x{\text{∥}}^2}-\frac{2x^1{x}^i}{\parallel x{\text{∥}}^4}\right)\frac{\partial }{\partial {\tilde{x}}^1}+\left(\frac{{\delta }_{2,i}}{\parallel x{\text{∥}}^2}-\frac{2x^2{x}^i}{\parallel x{\text{∥}}^4}\right)\frac{\partial }{\partial {\tilde{x}}^2}& \hfill & \end{array}$$

We see that the component functions are smooth.

In other words, the representation agrees on ${𝕊}^2\setminus \{N,S\}$ and is smooth on ${𝕊}^2\setminus \{S\}$ since the component functions $X^i(p)$ are zero at $p=N$. Thus, $X$ is smooth at $N$.