The Riemann Sphere

Def: The set $\mathbb{C}\cup \{\mathrm{\infty }\}$ will be denoted as $\hat{\mathbb{C}}$ , $\bar{\mathbb{C}}$, ${\mathbb{C}}^{\ast }$ or ${\mathbb{C}}_{\mathrm{\infty }}$. We can also define if $z\in \mathbb{C}$, then $z+\mathrm{\infty }=\mathrm{\infty }$ and if $z\ne 0$, then $z\cdot \mathrm{\infty }=\mathrm{\infty }$ and $z/\mathrm{\infty }=0$. Analogously, we cannot define consistently the expression such that $0\cdot \mathrm{\infty }$, $0/\mathrm{\infty }$, $\mathrm{\infty }/0$, $\mathrm{\infty }/\mathrm{\infty }$ and even the identity $\mathrm{\infty }+\mathrm{\infty }=\mathrm{\infty }$, but $\mathrm{\infty }\cdot \mathrm{\infty }=\mathrm{\infty }$ is well defined.

This is the one-point compactification or Alexandroff compactification

We get $U\subseteq \hat{\mathbb{C}}$ is open iff, $U\cap \mathbb{C}$ is open or if $\mathrm{\infty }\in U$, $\mathbb{C}\setminus U$ is compact.

Instead, we can think of it a stereographic projection. It is done considering the unit sphere centered at the origin, ${\mathbb{S}}^2\subseteq {\mathbb{R}}^3$. The way the projection works is by considering the line that passes through $N=(0,0,1)$, and any point of the sphere, only intersects the $xy$ plane at a single point.

This correspondence is known as the stereographic projection of the points of the sphere ${\mathbb{S}}^2\setminus \{N\}$ into the $xy$ plane. There’s no point corresponding to $N$ in the $xy$ plane.

The way we can define the steoregraphic projection we can define $E:{\mathbb{S}}^2\to \hat{\mathbb{C}}$, defined as

$E$ is a bijection from ${\mathbb{S}}^2$ into $\hat{\mathbb{C}}$. Then we can get $E^{-1}:\hat{\mathbb{C}}\to {\mathbb{S}}^2$ , defined as

${\mathbb{S}}^2$ is called the Riemann Sphere.

If we have a circle on the Riemann Sphere then we know that if the circle contains $N$, then the image of the circle is a straight line. If is doesn’t contain $N$, then the image is also a circle.