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Section 0.2 Stereographic projection

Subsection 0.2.1 Stereographic projection \(S^1\to \R^+\)

Let \(S^1\) denote the unit circle in the \(x,y\)-plane.

\begin{equation} S^1 = \{(x,y)\colon x^2+y^2=1\}\label{s1defn}\tag{0.2.1} \end{equation}

Let \(N=(0,1)\) denote the "north pole" (that is, the point at the "top" of the unit circle). Given a point \(P=(x,y)\neq N\) on the unit circle, let \(s(P)\) denote the intersection of the line \(\overline{NP}\) with the \(x\)-axis. See Figure 0.2.1. The map \(s\colon S^1\setminus \{N\}\to \R\) given by this rule is called stereographic projection. Using similar triangles, it is easy to see that \(s(x,y)=\frac{x}{1-y}\text{.}\)

Figure 0.2.1. Stereographic projection

Draw the relevant similar triangles and verify the formula \(s(x,y) = \frac{x}{1-y}\text{.}\)

We extend stereographic projection to the entire unit circle as follows. We call the set

\begin{equation} \R^+=\R\cup \{\infty\}\label{extendedrealsdefn}\tag{0.2.2} \end{equation}

the extended real numbers, where "\(\infty\)" is an element that is not a real number. Now we define stereographic projection \(s\colon S^1 \to \R^+\) by

\begin{equation} s(x,y) = \left\{ \begin{array}{cc} \frac{x}{1-y} \amp y\neq 1\\ \infty \amp y=1 \end{array} \right..\label{stereoproj1defn}\tag{0.2.3} \end{equation}

Subsection 0.2.2 Stereographic projection \(S^2\to \C^+\)

The definitions in the previous subsection extend naturally to higher dimensions. Here are the details for the main case of interest.

Let \(S^2\) denote the unit sphere in \(\R^3\text{.}\)

\begin{equation} S^2 = \{(a,b,c)\in \R^3\colon a^2+b^2+c^2=1\}\label{s2defn}\tag{0.2.4} \end{equation}

Let \(N=(0,0,1)\) denote the "north pole" (that is, the point at the "top" of the sphere, where the postive \(z\)-axis is "up"). Given a point \(P=(a,b,c)\neq N\) on the unit sphere, let \(s(P)\) denote the intersection of the line \(\overline{NP}\) with the \(x,y\)-plane, which we identify with the complex plane \(\C\text{.}\) See See Figure 0.2.3. The map \(s\colon S^2\setminus \{N\}\to \C\) given by this rule is called stereographic projection. Using similar triangles, it is easy to see that \(s(a,b,c)=\frac{a+ib}{1-c}\text{.}\)

Figure 0.2.3. Stereographic projection

We extend stereographic projection to the entire unit sphere as follows. We call the set

\begin{equation} \C^+=\C\cup \{\infty\}\label{extendedcomplexsdefn}\tag{0.2.5} \end{equation}

the extended complex numbers, where "\(\infty\)" is an element that is not a complex number. Now we define stereographic projection \(s\colon S^2 \to \C^+\) by

\begin{equation} s(a,b,c) = \left\{ \begin{array}{cc} \frac{a+ib}{1-c} \amp c\neq 1\\ \infty \amp c=1 \end{array} \right..\label{stereoprojdefn}\tag{0.2.6} \end{equation}

Exercises 0.2.3 Exercises

Formulas for inverse stereographic projection.

It is intuitively clear that stereographic projection is a bijection. Make this rigorous by finding a formula for the inverse.

1.

For \(s\colon S^1\to \R^+\text{,}\) find a formula for \(s^{-1}\colon \R^+\to S^1\text{.}\) Find \(s^{-1}(3)\text{.}\)

2.

For \(s\colon S^2\to \C^+\text{,}\) find a formula for \(s^{-1}\colon \C^+\to S^2\text{.}\) Find \(s^{-1}(3+i)\text{.}\)

Liftings.
Let \(\pi \colon X\to Y\) and \(f\colon Y\to Y\) be functions. We say that a function \(\tilde{f}\colon X\to X\)\(\;\)lifts (or is a lifting of) \(f\) via \(\pi\) if \(\pi \circ \tilde{f} = f\circ \pi\text{.}\)
3.

Show that the map \(S^1\to S^1\) given by \((x,y)\to (x,-y)\) is a lifting of the map \(\R^+\to \R^+\) given by \(x\to 1/x\text{.}\)

4.

Show that the map \(S^2\to S^2\) given by \((a,b,c)\to (a,-b,-c)\) is a lifting of the map \(\C^+\to \C^+\) given by \(z\to 1/z\text{.}\)