Problem 1.7 (Stereographic projection)

Question

Let $N$ denote the north pole $(0, \ldots, 0,1) \in \mathbb{S}^n \subseteq \mathbb{R}^{n+1}$, and let $S$ denote the south pole $(0, \ldots, 0,-1)$. Define the stereographic projection $\sigma: \mathbb{S}^n \backslash\{N\} ightarrow \mathbb{R}^n$ by
$$
\sigma\left(x^1, \ldots, x^{n+1}ight)=\frac{\left(x^1, \ldots, x^night)}{1-x^{n+1}} .
$$
Let $\widetilde{\sigma}(x)=-\sigma(-x)$ for $x \in \mathbb{S}^n \backslash\{S\}$.
\begin{enumerate}[label=(\alph*)]
\item For any $x \in \mathbb{S}^n \backslash\{N\}$, show that $\sigma(x)=u$, where $(u, 0)$ is the point where the line through $N$ and $x$ intersects the linear subspace where $x^{n+1}=0$ (Fig. 1.13). Similarly, show that $\widetilde{\sigma}(x)$ is the point where the line through $S$ and $x$ intersects the same subspace. (For this reason, $	ilde{\sigma}$ is called stereographic projection from the south pole.)
\item  Show that $\sigma$ is bijective, and
$$
\sigma^{-1}\left(u^1, \ldots, u^night)=\frac{\left(2 u^1, \ldots, 2 u^n,|u|^2-1ight)}{|u|^2+1} .
$$
\item Compute the transition map $\widetilde{\sigma} \circ \sigma^{-1}$ and verify that the atlas consisting of the two charts $\left(\mathbb{S}^n \backslash\{N\}, \sigmaight)$ and $\left(\mathbb{S}^n \backslash\{S\}, \widetilde{\sigma}ight)$ defines a smooth structure on $\mathbb{S}^n$. (The coordinates defined by $\sigma$ or $\widetilde{\sigma}$ are called stereographic coordinates.)
\item Show that this smooth structure is the same as the one defined in Example 1.31.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(\alph*)]
    \item Fix an arbitrary $x \in \mathbb{S}^n\setminus \{N\}$. A line passing through $N$ and $x$ is given by
    \begin{equation*}
        L_{(N,x)}: \mathbb{R} \to \mathbb{R}^{n+1}, \qquad t \mapsto N + t \left(N - x \right)
    \end{equation*}
    or in vector notation,
    \begin{equation*}
        L_{(N,x)}(t) 
        %= N + t \left(N - x \right) 
        = \begin{bmatrix} 0\ \vdots\ 0\ 1\end{bmatrix} + t \left(\begin{bmatrix} 0\ \vdots\ 0\ 1\end{bmatrix} - \begin{bmatrix} x^1\ \vdots\ x^{n}\ x^{n+1}\end{bmatrix}\right) 
        = \begin{bmatrix} -tx^1\ \vdots\ -tx^n\ 1 + t(1-x^{n+1})\end{bmatrix}
    \end{equation*}   
    \begin{figure}[H]  \centering
    \begin{tikzpicture}[scale=2,mycirc/.style={circle,fill=blue!50, minimum size=0.01cm}] 
        % Axes
        \draw[->, thin] (-1.5,0) -- (2,0) node[right] {$x^{n+1}$-axis};
        \draw[->, thin] (0,-1.5) -- (0,1.5) node[above] {$x^n$-axis};
        % Circle
        \node[mycirc, label=above:{$x$}] (x) at (0.833, 0.555) {};
        \node[mycirc, label=above:{$\sigma(x)$}] (xProj) at (1.8, 0) {};
        \node[mycirc, label=45:{$N$}] (N) at (0, 1) {};
        \node[mycirc, label=135:{$S$}] (S) at (0, -1) {};
        \draw (N) -- (xProj);
        \draw[dashed, thick] (0,0) circle (1);
    \end{tikzpicture}
    \caption{Stereographic projection from the north pole in case of $\mathbb{S}^1$.}
    \label{fig:enter-label}
    \end{figure}
    Where does $L$ exactly cross the $x^{n+1}$-axis? This happens at the value $t_0$ for which the $n+1$th component of $L(t_0)$ is zero. To derive it, we set
    \begin{align*}
        1 + t_0(1-x^{n+1}) \stackrel{!}{=} 0 \implies t_0 = -\frac{1}{1-x^{n+1}}
    \end{align*}
    which results in 
    \begin{equation*}
        L(t_0) = \begin{bmatrix} \dfrac{x^1}{1-x^{n+1}}, \ldots, \dfrac{x^n}{1-x^{n+1}}, 0 \end{bmatrix} = \begin{bmatrix} \sigma(x),0 \end{bmatrix}.
    \end{equation*}
    Therefore, the intersection point is indeed $\begin{pmatrix}\sigma(x),0\end{pmatrix}$.

\item We compute the formula for $\sigma \circ \sigma^{-1}$ and $\sigma^{-1} \circ \sigma$. Before we start, we derive a formula for $\sigma^2$:

\begin{align*}
|\sigma(x)|^2 & =\dfrac{\left.\left(x^1\right)^2+\ldots+\left(x^n\right)^2\right)}{\left(1-x^{n+1}\right)^2} \[3.5pt]
& =\dfrac{\left(x^1\right)^2+\ldots+\left(x^n\right)^2+\left(x^{n+1}\right)^2-\left(x^{n+1}\right)^2}{\left(1-x^{n+1}\right)^2} \[3.5pt]
& =\dfrac{|x|-\left(x^{n+1}\right)^2}{\left(1-x^{n+1}\right)^2} \[3.5pt]
& =\dfrac{1-\left(x^{n+1}\right)^2}{\left(1-x^{n+1}\right)^2}.
\end{align*}

Using the above, we compute $\sigma^{-1} \circ \sigma(x)$:
\begin{align*}
\sigma^{-1} \circ \sigma(x) & =\sigma^{-1}\left(\frac{\left(x^1, \ldots x^n\right)}{1-x^{n+1}}\right) \[3.5pt]
& =\dfrac{\left(2 x^1, \ldots, 2 x^n,\left(|\sigma(x)|^2-1\right)\left(1-x^{n+1}\right)\right)}{\left(|\sigma(x)|^2+1\right)\left(1-x^{n+1}\right)} \[3.5pt]
& =\dfrac{\left(2 x^1, \ldots, 2 x^n,\left(1-x^{n+1}\right)-\left(1-x^{n+1}\right)\right)}{\left(\frac{1-\left(x^{n+1}\right)^2}{\left(1-x^{n+1}\right)^2}+1\right)\left(1-x^{n+1}\right)} \[3.5pt]
& =\dfrac{\left(2 x^1, \ldots 2 x^n, 2 x^{n+1}\right.}{\frac{1-\left(x^{n+1}\right)^2}{1-x^{n+1}}+1-x^{n+1}} \[3.5pt]
& =\dfrac{\left(2 x^1, \ldots 2 x^n, 2 x^{n+1}\right.}{\left(\frac{1-\left(x^{n+1}\right)^2+\left(1-x^{n+1}\right)^2}{1-x^{n+1}}\right)} \[3.5pt]
& =\dfrac{\left(2 x^1, \ldots 2 x^{n+1}\right)}{\left(\frac{2-2 x^{n+1}}{1-x^{n+1}}\right)} \[3.5pt]
& =\dfrac{\left(2 x^1, \ldots 2 x^{n+1}\right)}{2} \[3.5pt]
& =\left(x^1, \ldots, x^{n+1}\right) \[3.5pt]
& =x.
\end{align*}

Similarly, $\sigma \circ \sigma^{-1}$ is given by
\begin{align*}
\sigma \circ \sigma^{-1}(x) & =\sigma\left(\frac{\left(2 x^1, \ldots, 2 x^n,|x|^2-1\right)}{|x|^2+1}\right) \[3.5pt]
& =\dfrac{\left(2 x^1, \ldots, 2 x^n\right)}{\left(|x|^2+1\right)\left(1-\frac{|x|^2-1}{|x|^2+1}\right)} \[3.5pt]
& =\dfrac{\left(2 x^1, \ldots, 2 x^n\right)}{|x|^2+1-\left(|x|^2-1\right)} \[3.5pt]
& =\dfrac{\left(2 x^1, \ldots, 2 x^n\right)}{2)} \[3.5pt]
& =\left(x^1, \ldots x^n\right) \[3.5pt]
& =x.
\end{align*}
Thus, $\sigma$ is a bijection.

\item The transition map is given by:
    \begin{equation*}
        \widetilde{\sigma} \circ \sigma^{-1}: \sigma\left(\mathbb{S}^n \backslash\{N, S\}\right) \to \widetilde{\sigma}\left(\mathbb{S}^n \backslash\{N, S\}\right), 
        \quad 
        \left(u^1, \ldots, u^n\right) \mapsto \frac{\left(u^1, \ldots, u^n\right)}{|u|}
    \end{equation*}
    (since $\sigma(S)=0$, the point $u = \mathbf{0}$ is not in the domain of $\widetilde{\sigma} \circ \sigma^{-1}$). To see why, compute
\begin{align*}
\widetilde{\sigma} \circ \sigma^{-1}\left(u^1, \ldots u^n\right) & =\widetilde{\sigma}\left(\frac{\left(2 u^1, \ldots, 2 u^n,|u|^2-1\right)}{|u|^2+1}\right) \
& =-\sigma\left((-1)\left(\frac{\left(2 u^1, \ldots, 2 u^n,|u|^2-1\right)}{|u|^2+1}\right)\right) \
& =-\sigma\left(\frac{\left(2 u^1, \ldots, 2 u^n,|u|^2-1\right)}{-|u|^2-1}\right) \
& =-\frac{\left(2 u^1, \ldots, 2 u^n\right)}{\left(|u|^2+1\right)+\left(|u|^2-1\right)} \
& =\frac{\left(2 u^1, \ldots, 2 u^n\right)}{2|u|^2} \
& =\frac{u}{|u|^2}
\end{align*}
We can skip a similar computation for its inverse by noticing that it is its own inverse:
$$
\left(\widetilde{\sigma} \circ \sigma^{-1}\right) \circ\left(\widetilde{\sigma} \circ \sigma^{-1}\right)(u)=\widetilde{\sigma} \circ \sigma^{-1}\left(\frac{u}{|u|^2}\right)=\frac{\frac{u}{|u|^2}}{\left|\frac{u}{|u|^2}\right|^2}=\frac{\frac{u}{|u|^2}}{\frac{1}{|u|^2}}=u
$$
     
In other words, the transition map is a diffeomorphism. Same goes for $\sigma \circ \widetilde{\sigma}^{-1}$. Since $\mathbb{S}^n \backslash\{N\}$ and $\mathbb{S}^n \backslash\{S\}$ form an open cover of $\mathbb{S}^n$, we conclude that $\left(\mathbb{S}^n \backslash\{N\}, \sigma\right)$ and $\left(\mathbb{S}^n \backslash\{S\}, \widetilde{\sigma}\right)$ define a smooth structure on $\mathbb{S}^n$.
    
    \item The charts from Example 1.31 can be grouped into three categories:
    \begin{itemize}
        \item $\left(U_{n+1}^{+}, \phi_{n+1}^{+}\right)$ which contain $N$.
        \item $\left(U_{n+1}^{-}, \phi_{n+1}^{-}\right)$ which contain $S$.
        \item $\left(U_i^{ \pm}, \phi_i^{ \pm}\right)$for $i=1, \ldots, n$ which do not contain $N$ and $S$.
    \end{itemize}  
    
    We argue that each type of chart is smoothly compatible with $\sigma$; the case for $\tilde{\sigma}$ is similar. In the first and in the second case, we have
$$
\phi_{n+1}^{ \pm} \circ \sigma^{-1}\left(u^1, \ldots, u^n\right)=\frac{\left(2 u^1, \ldots, 2 u^n\right)}{|u|^2+1},
$$
Both of these functions are smooth. The inverses are given by
$$
\sigma \circ\left(\phi_{n+1}^{ \pm}\right)^{-1}\left(u^1, \ldots, u^n\right)=\frac{\left(u^1, \ldots, u^n\right)}{1 \mp \sqrt{1-|u|^2}}
$$
Thus, $\sigma \circ\left(\phi_{n+1}^{-}\right)^{-1}$ is smooth. Since the domain $\left(\phi_{n+1}^{+}\right)\left(U_{n+1}^{+} \backslash\{N\}\right)$ does not include $\mathbf{0}$, $\sigma \circ\left(\phi_{n+1}^{+}\right)^{-1}$ is smooth.

For the third case, we have
$$
\phi_i^{ \pm} \circ \sigma^{-1}\left(u^1, \ldots, u^n\right)=\frac{\left(2 u^1, \ldots, \widehat{2 u^i}, \ldots, 2 u^n,|u|^2-1\right)}{|u|^2+1},
$$
which is smooth, and
$$
\sigma \circ\left(\phi_i^{ \pm}\right)^{-1}\left(u^1, \ldots, u^n\right)=\frac{\left(u^1, \ldots, u^{i-1}, \sqrt{1-|u|^2}, u^i, \ldots, u^{n-1}\right)}{1-u^n},
$$
which is also smooth, since $u^n \neq 1$ in $U_i^{ \pm}$. In other words, the smooth atlas from Example 1.31 is smoothly compatible with the smooth atlas of stereographic projection.
\end{enumerate}

Problem 1.7 (Stereographic projection)

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Problem 1.7 (Stereographic projection)

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