# 黎曼球面

${\displaystyle 1/0=\infty .}$

• 複射影直线，记为 ${\displaystyle \mathbb {CP} ^{1}}$，和
• 扩充复平面，记为 ${\displaystyle \mathbb {\hat {C}} }$或者${\displaystyle \mathbb {C} \cup \{\infty \}}$.

## 作为复流形

${\displaystyle \zeta =1/\xi ,}$
${\displaystyle \xi =1/\zeta .}$

## 作为複射影线

${\displaystyle (\alpha ,\beta )=(\lambda \alpha ,\lambda \beta )}$

${\displaystyle (\alpha ,\beta )=(\zeta ,1).}$

${\displaystyle (\alpha ,\beta )=(1,\xi ).}$

${\displaystyle (1,\xi )=(1/\xi ,1)=(\zeta ,1)}$

## 作为球面

${\displaystyle \zeta ={\frac {x+iy}{1-z}}=\cot(\varphi /2)\;e^{i\theta }.}$

${\displaystyle \xi ={\frac {x-iy}{1+z}}=\tan(\varphi /2)\;e^{-i\theta }.}$

(两份复平面和平面${\displaystyle z=0}$的对应方式不同。必须使用定向翻转来保证球面上定向的一致性，实际上复共轭使得变换映射成为全纯函数。）${\displaystyle \zeta }$-座标和${\displaystyle \xi }$-座标之间的变换函数可以通过将其中一个映射和另一个的逆的复合得到。它们就是如上所述的${\displaystyle \zeta =1/\xi }$${\displaystyle \xi =1/\zeta }$。因此单位球面和黎曼球面微分同胚

## 度量

${\displaystyle ds^{2}=\left({\frac {2}{1+|\zeta |^{2}}}\right)^{2}\,|d\zeta |^{2}={\frac {4}{\left(1+\zeta {\bar {\zeta }}\right)^{2}}}\,d\zeta d{\bar {\zeta }}.}$

${\displaystyle ds^{2}={\frac {4}{\left(1+u^{2}+v^{2}\right)^{2}}}\left(du^{2}+dv^{2}\right).}$

## 自同构

${\displaystyle f(\zeta )={\frac {a\zeta +b}{c\zeta +d}},}$

${\displaystyle f(\alpha ,\beta )=(a\alpha +b\beta ,c\alpha +d\beta )={\begin{pmatrix}\alpha &\beta \end{pmatrix}}{\begin{pmatrix}a&c\\b&d\end{pmatrix}}.}$