Problem 10-7 (Transition function determined by stereographic coordinates)

Question

Compute the transition function for $T\mathbb{S}^2$ associated with the two local trivializations determined by stereographic coordinates (\href{./problem_1-7}{Problem 1.7}).

Elu Thingol · Accepted Answer

Import all of the definitions from \href{./problem_1-7}{Problem 1-7}. For the tangent bundle $\pi:T\mathbb{S}^2 \to \mathbb{S}^2$ given by $\pi(v_p) = p$, we define two smooth local trivializations constructed as in the proof of Proposition 10.4 (Eq. 10.1):
\begin{align}
    \Phi: \underbrace{\pi^{-1} (\mathbb{S}^2 \setminus \{N\})}_{\subset\, T\mathbb{S}^2} \to \mathbb{S}^2 \setminus \{N\} \times \mathbb{R}^2, 
    \quad
    &\Phi(v_p) = \Phi\left(\sum_{i=1}^{2} v^i \dfrac{\partial}{\partial x^i}  \right) := \begin{pmatrix} p,\, \displaystyle\sum_{i=1}^{2} v^i e_i \end{pmatrix}\
    \Psi: \underbrace{\pi^{-1} (\mathbb{S}^2 \setminus \{S\})}_{\subset\, T\mathbb{S}^2} \to \mathbb{S}^2 \setminus \{S\} \times \mathbb{R}^2, 
    \quad
    &\Psi(v_p) = \Psi\left(\sum_{i=1}^{2} \widetilde{v}^i \dfrac{\partial}{\partial \widetilde{x}^i}  \right) := \begin{pmatrix} p,\, \displaystyle\sum_{i=1}^{2} \widetilde{v}^i e_i \end{pmatrix}
\end{align}

Following the definition in Lemma 10.5, we compute the transition map 
\begin{equation*}
    \Psi \circ \Phi^{-1}: \mathbb{S}^2\setminus\{N,S\} \times \mathbb{R}^2 \to \mathbb{S}^2\setminus\{N,S\} \times \mathbb{R}^2
\end{equation*}
which is given by
\begin{align*}
    \Psi \circ \Phi^{-1}(p,v) &= \Psi\left(\sum_{i=1}^{2} v^i \dfrac{\partial}{\partial x^i}\right)\
    &= \Psi\left(\sum_{i=1}^{2} v^i \left(\dfrac{\partial \widetilde x^1}{\partial x^i}  \dfrac{\partial}{\partial \widetilde{x}^1}  + \dfrac{\partial \widetilde x^2}{\partial x^i}  \dfrac{\partial}{\partial \widetilde{x}^2} \right)\right)\
    &= \Psi\left(\sum_{j=1}^{2} \left(v^1 \dfrac{\partial \widetilde x^j}{\partial x^1} + v^2 \dfrac{\partial \widetilde x^j}{\partial x^2} \right) \dfrac{\partial}{\partial \widetilde x^j}\right)\
    &= \begin{pmatrix} p, \ \displaystyle\sum_{j=1}^{2} \left(v^1 \dfrac{\partial \widetilde x^j}{\partial x^1} + v^2 \dfrac{\partial \widetilde x^j}{\partial x^2} \right) e_j \end{pmatrix}\
    &= \begin{pmatrix} p, \ 
    v^1 \dfrac{\partial \widetilde x^1}{\partial x^1} + v^2 \dfrac{\partial \widetilde x^1}{\partial x^2}, \ 
    v^1 \dfrac{\partial \widetilde x^2}{\partial x^1} + v^2 \dfrac{\partial \widetilde x^2}{\partial x^2} \end{pmatrix}\
    &= \begin{pmatrix} p, \ 
    \begin{bmatrix} \dfrac{\partial \widetilde x^1}{\partial x^1} & \dfrac{\partial \widetilde x^1}{\partial x^2}\[15pt]
    \dfrac{\partial \widetilde x^2}{\partial x^1} & \dfrac{\partial \widetilde x^2}{\partial x^2} \end{bmatrix} v \end{pmatrix}
\end{align*}

We must calculate the change of coordinate coefficients $(\partial \widetilde x^j / \partial x^i)_{1\leq i,j\leq 2}$ at this point. In \href{./problem_1-7}{Problem 1-7}, we have already calculated the transition map
\begin{equation*}
    \widetilde{\sigma} \circ \sigma^{-1} = \begin{pmatrix} \widetilde{x}^1, \widetilde{x}^2 \end{pmatrix}: \sigma(\mathbb{S}^2\setminus\{N,S\}) \to \widetilde\sigma(\mathbb{S}^2\setminus\{N,S\}), \quad  \begin{pmatrix} x^1, x^2 \end{pmatrix} \mapsto \begin{pmatrix} \dfrac{u^1}{\|(u^1, u^2)\|^2}, \, \dfrac{u^2}{\|(u^1, u^2)\|^2}\end{pmatrix}.
\end{equation*}

Inserting this into the change of coordinate formula (3.11), we obtain
\begin{align*}
    \left. \dfrac{\partial}{\partial x^i} \right|_p
    %&= d(\widetilde{\sigma}^{-1})_{\widetilde{\sigma}(p)} \left( \textcolor{gray}{\left.\dfrac{\partial}{\partial x^i}\right|_{\widetilde{\sigma}(p)}} \right) \
    %&= d(\sigma^{-1})_{\sigma(p)} \circ d(\sigma \circ \widetilde{\sigma}^{-1})_{\widetilde{\sigma}(p)} \left( \textcolor{gray}{\left.\dfrac{\partial}{\partial x^i}\right|_{\widetilde{\sigma}(p)}} \right)
    &= \sum_{j=1}^{2} \dfrac{\partial \widetilde x^j}{\partial x^i}(\sigma(p)) \left. \dfrac{\partial}{\partial \widetilde x^j} \right|_p
\end{align*}
%
%\begin{align*}
%    \left. \dfrac{\partial}{\partial x} \right|_p &= \dfrac{y^2 - x^2}{(x^2 + y^2)^2} \left. \dfrac{\partial}{\partial \widetilde x} \right|_p + \dfrac{-2 x y}{(x^2 + y^2)^2} \left. \dfrac{\partial}{\partial \widetilde y} \right|_p\[5pt]
%    \left. \dfrac{\partial}{\partial y} \right|_p &= \dfrac{x^2-y^2}{(x^2 + y^2)^2} \left. \dfrac{\partial}{\partial \widetilde x} \right|_p + \dfrac{-2 x y}{(x^2 + y^2)^2} \left. \dfrac{\partial}{\partial \widetilde y} \right|_p
%\end{align*}
%
Denoting the coefficients by $(x,y)$ instead of $(x^1,x^2)$ to work with exponents, we obtain:
\begin{align*}
    % dx1/dx1
    \dfrac{\partial \widetilde x}{\partial x} &= \dfrac{y^2 - x^2}{(x^2 + y^2)^2} \quad &
    % dx1/dx2
    \dfrac{\partial \widetilde x}{\partial y} &= -\dfrac{2 x y}{(x^2 + y^2)^2}\[5pt]
    % dx2/dx1
    \dfrac{\partial \widetilde y}{\partial x} &= -\dfrac{2 x y}{(x^2 + y^2)^2} \quad &
    % dx2/dx2
    \dfrac{\partial \widetilde y}{\partial y} &= \dfrac{x^2-y^2}{(x^2 + y^2)^2}
\end{align*}

We convert these into the original coordinates $(p_1,p_2,p_3)$ in $\mathbb{R}^3$ from the north pole coordinates $(x_1,x_2)$ of $\sigma$ as to match with the representation of $p$:
\begin{align*}
    % dx1/dx1
    \dfrac{\partial \widetilde x}{\partial x} &= \left(\dfrac{1}{1+z}\right)^2 (y^2 - x^2)\quad &
    % dx1/dx2
    \dfrac{\partial \widetilde x}{\partial y} &= -\left(\dfrac{1}{1+z}\right)^2 (2 x y)\[5pt]
    % dx2/dx1
    \dfrac{\partial \widetilde y}{\partial x} &= \left(\dfrac{1}{1+z}\right)^2 (x^2-y^2) \quad &
    % dx2/dx2
    \dfrac{\partial \widetilde y}{\partial y} &= -\left(\dfrac{1}{1+z}\right)^2 (2 x y)
\end{align*}

Thus, the transition function is given by
\begin{equation*}
    \tau: \mathbb{S}^n \setminus \{N,S\} \to \mathrm{GL}(2,\mathbb{R}), \quad p \left(\dfrac{1}{1+p^3}\right)^2 \cdot \begin{bmatrix}
        (p^1)^2 - (p^2)^2 & -2 p^1p^2\
        -2 p^1p^2 & (p^1)^2 - (p^2)^2
    \end{bmatrix}.
\end{equation*}

For the sake of completeness, we also provide the transition map of $\Phi \circ \Psi^{-1}$:
\begin{equation*}
    \tau_{2}: \mathbb{S}^n \setminus \{N,S\} \to \mathrm{GL}(2,\mathbb{R}), \quad p \mapsto \left(\dfrac{1}{1-p^3}\right)^2 \begin{bmatrix}
        (p^1)^2 - (p^2)^2 & -2 p^1p^2\
        -2 p^1p^2 & (p^1)^2 - (p^2)^2
    \end{bmatrix}.
\end{equation*}

Problem 10-7 (Transition function determined by stereographic coordinates)

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Problem 10-7 (Transition function determined by stereographic coordinates)

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