I am not sure if I'm taking the derivative of a transition function correctly. This is coming from exercise 1.28 of Riemannian Geometry by Gallot-Hulin-Lafontaine.
Given the sphere $S^2$ and two stereographic charts $(V_1=S^2\setminus \{N\}, \psi_1=\frac{(x,y)}{1-z}), (V_2=S^2\setminus\{S\}, \psi_2=\frac{(x,y)}{1+z})$. I want to compute $\psi_2 \circ \psi_1^{-1})'_{\psi_1(m)} \cdot u=v$, where $u=(u_1, u_2)$ is a vector in $\mathbb{R}^2$, $\xi\in T_mS^2$, and $\xi$ is represented by $u$ in the first chart $(V_1, \psi_1).$
I have $(\psi_2\circ \psi_1^{-1})(u)=\frac{u}{||u||^2}$. Now I'm wondering if $(\psi_2\circ \psi_1^{-1})'_{\psi_{1}(m)}\cdot u= \left(\frac{||u||-u_1}{||u||^4},\frac{||u||-u_2}{||u||^4} \right)$?
I don't think this is right because the next question says to compute this when $m=(\frac{\sqrt{3}}{2}, \frac{1}{2})$ and to compute the coordinates of $\xi\in \mathbb{R}$, but I thought in the situation above, $\xi\in T_{m}S^2\cong \mathbb{R}^2$.
Please forgive me for my misunderstandings + thanks in advance!
I have also seen: Differential of transition map.
For starters, on page 12 of the book, it confirms that $(\psi_2 \circ \psi_1^{-1})(u)=\frac{u}{||u||^2}.$ Then we should first compute the vector $(\psi_2 \circ \psi_1^{-1})'_{\psi_{1}(m)}$. If we write $\psi_1(m)$ as a vector $(p_1, p_2)$, then we get $(\psi_2 \circ \psi_1^{-1})'_{\psi_{1}(m)}=(1/||\psi_1(m)||^2, 1/||\psi_1(m)||^2)$. So dotting this with $u$, we get $(\psi_2 \circ \psi_1^{-1})'_{\psi_1(m)}\cdot u=(\frac{p_1}{||\psi_1(m)||^2},\frac{p_2}{||\psi_1(m)||^2}) .$
Later the point $m$ can be viewed as a point in the equator of $S^2$, i.e. when $z=0$. I suspect that $\xi$ should be in $\mathbb{R}^2$.