Consider the vector field $\frac{\partial}{\partial x_1}$ on $\mathbb{R^2}$. Let $\psi_N : S^2 \setminus\{N\} \to \mathbb{R^2} $ and $\psi_S : S^2 \setminus\{S\} \to \mathbb{R^2} $ be the stereographic projection charts (where $N = (0, 0, 1)$ and $S = (0, 0,-1)$ are the North and South poles). Show that the vector field $(\psi_N)^*(\frac{\partial}{\partial x_1})$ extends to a smooth vector field $X$ on the entire sphere $S^2$, with the property that $X_p = 0$ if and only if $p = N$.
I am new to the subject and I have some elementary questions, First by definition $$X : = (\psi_N)^*(\frac{\partial}{\partial x_1})=(T\psi_N)^{-1} \circ \frac{\partial}{\partial x_1} \circ \psi_N $$
is a vector field on $S^2 \setminus\{N\} $. Given $p\in S^2 \setminus\{N\} $, $$\frac{\partial}{\partial x_1} \circ \psi_N(p)$$ is a unit vector in the $x_1$ direction at $\psi_N(p)$; call it $v$.
Now my (additional) question is : is there a way to write down an expression for the tangent vector $(T\psi_N)^{-1}(v)$? ( Noting that $\psi_N(x,y,z)=( \frac{x}{1-z},\frac{y}{1-z})$ and $\psi_N^{-1}(X,Y)=(\frac{2X}{X^2+Y^2+1},\frac{2Y}{X^2+Y^2+1},\frac{X^2+Y^2-1}{X^2+Y^2+1})$)
Furthermore, if $p=N$, then $\psi_N(p)=\infty$, I can not use the definition of $X$ here, should I extend this vector field by just "defining" $X_N=0$ ?
I already appreciate your help.
First note that you use two letters for the same thing: $X=x_1$. This can be confusing.
As for your ¨additional question¨: yes, there is a way. Write the Jacobian matrix (matrix of partial derivatives) of the formula you wrote for $\psi_N^{-1}$. You should get a 3 by 2 matrix. The first column is your $(T\psi_N)^{-1}(v)$.
As for the main question: use $\psi_S$ (say) to examine $(\psi_N)^*(\frac{\partial}{\partial x_1})$ in a neighborhood of $N$. To do this, write a formula for $\psi_N\circ\psi_S^{-1}$ an take the first column of the inverse of its Jacobian matrix.
You should get a vector field in the complement of the origin, ie an expression of the form $a(X,Y)\frac{\partial}{\partial X}+b(X,Y)\frac{\partial}{\partial Y}$. You then need to show that the functions $a,b$ extend smoothly by $0$ to $X=Y=0$.
By the way, there is a slick way to do this excercise using complex variables.