Answer to Question #226040 in Complex Analysis for smi

Question

Prove or disapprove that the complex sphere is compact.

Answer 1

The complex sphere $\bar\mathbb{C}$ is a complex manifold with two charts $(U_1,z_1)$ and $(U_2,z_2)$, both with domain equal to the complex number plane $\mathbb{C}$. The intersection of domains $U_1$ and $U_2$ is

$U_1\cap U_2=U_1\setminus\{z_1=0\}=U_2\setminus\{z_2=0\}$, the transition map $z_2=z_1^{-1}$.

Let $K_1=\{z_1\in U_1:|z_1|\leq 1\}$, $K_2=\{z_2\in U_2:|z_2|\leq 1\}$ be two closed unit disks. They are compact as closed bounded subsets of $\mathbb{C}$.

We will prove that the complex sphere $\bar\mathbb{C}$ is compact by showing that $\bar\mathbb{C}=K_1\cup K_2$. It is evident that $\bar\mathbb{C}\supset K_1\cup K_2$.

Consider any point $P\in U_1\cap U_2$.

If $z_1(P)\leq 1$, then $P\in K_1$.

If $|z_1(P)|> 1$, then $|z_2(P)|=|z_1(P)|^{-1}\leq 1$ and $P\in K_2$. Therefore, $U_1\cap U_2\subset K_1\cup K_2$.

Since the point $\{z_1=0\}\in K_1$ and the point $\{z_2=0\}\in K_2$, it follows that$\bar\mathbb{C}=U_1\cup U_2=(U_1\cap U_2)\cup\{z_1=0\}\cup\{z_2=0\}\subset K_1\cup K_2$.

Therefore $\bar\mathbb{C}=K_1\cup K_2$ and $\bar\mathbb{C}$ is compact.