Riemann Sphere and Stereographic Projection

Sometimes while studying complex valued functions, it is sometimes useful to consider the extended complex plane $\mathbb{C} \cup \{ \infty \}$, where we append an additional point $\infty$ to the set of usual complex numbers. This is especially helpful in study of Mobius maps $z \mapsto \frac{az+b}{cz+d}$ where $ad-bc \not = 0$.

Riemann sphere provides us with a model of extended complex plane. This makes it in useful in complex analysis because it allows us to make sense of division by zero in some circumstance ($\frac{0}{0}$ is still undefined).

We start with the unit sphere $S^2$ in $\mathbb{R}^3$ ie. the spherical surface with center at origin and unit radius (set of points $(x,y,z)$ given by $x^2 +y^2 +z^2=1$). First, we shall show there is a bijection between the $R^2$ and $S^2$ minus a single point. (We identity $xy$-plane with $\mathbb{R}^2$)

Start with a point $P$ on $S^2$ other than the north pole $N = (0,0,1)$. Draw a ray starting from $N$ passing through $P$. This ray will intersect the $xy$-plane at a unique point $Q$. The map that takes $P$ to $Q$ is called as stereographic projection and is denoted as $\Pi$.

We shall use some properties of vectors to derive stereographic projection explicitly. $Q - N$ is parallel to $P - N$, so $P - N = k(Q - N)$ for some scalar $k$. Denote $P=(x,y,z)$ and $Q=(u,v,0)$. Then, we get $(x,y,z) = (ku,kv,1-k)$ so that $k=1-z$. Note $z \not = 1$ since $P \not = N$.

$$\therefore \Pi (x,y,z) = (u,v,0) = \left( \frac{x}{1-z}, \frac{y}{1-z}, 0 \right)$$

Since $x^2 + y^2 + z^2 = 1$, we have $k^2u^2 + k^2v^2 + (1-k)^2 = 1$. Simplifying (while noting $k \not = 0$ since $z \not = 1$),

$$k = \frac{2}{u^2+v^2 +1}$$

$$\therefore P = (x,y,z) = \left(  \frac{2u}{u^2+v^2 +1}, \frac{2v}{u^2+v^2 +1}, \frac{u^2 +v^2 -1}{u^2+v^2 +1} \right)$$

It takes some simple calculations to verify that $\Pi : S^2 \setminus \{ N \} \to \mathbb{R}^2$ is a bijection and the map $(u,v,0) \mapsto \left(  \frac{2u}{u^2+v^2 +1}, \frac{2v}{u^2+v^2 +1}, \frac{u^2 +v^2 -1}{u^2+v^2 +1} \right)$ is its inverse.

Interact with GeoGebra applet to see how stereographic projection actually works.

In fact our map $\Pi$ is conformal (ie. it preserves angles) but proof of this assertion requires some tools from differential geometry.

We can extend $\Pi$ to the whole sphere $S^2$ by simply defining $\Pi (N) = \infty$. Here, $\infty$ is just a symbol. And similarly extend $\Pi^{-1} :  \mathbb{R}^2 \cup \{ \infty \} \mapsto S^2$. Since, we can interpret $\mathbb{C}$ as  $\mathbb{R}^2$ (from a topological point of view), we can also define $\Pi^{-1} :  \mathbb{C} \cup \{ \infty \} \to S^2$ as

$$z \mapsto \left(  \frac{2 \Re(z)}{|z|^2 +1}, \frac{2 \Im(z)}{|z|^2 +1}, \frac{|z|^2 -1}{|z|^2 +1} \right)$$

where $\Re(z)$ and $\Im(z)$ are real and imaginary part of $z$ respectively, and

$$\infty \mapsto (0,0,1)$$

PS: From a topological point of view, we can define a topology $\tau_1$ on $\mathbb{C} \cup \{\infty\}$ such that our usual topology $\tau_2$ on $\mathbb{C}$ is a subspace of  $\tau_1$ and $\tau_1$ is one point compactification of $\tau_2$.

References:

• A Pathway to Complex Analysis by S. Kumaresan
• Topology by J. Munkres
• A Comprehensive introduction to Differential Geometry Vol. 2 by M. Spivak