# 共形平坦的黎曼曲面的共形函数所满足的方程

$\newcommand{\rd}{\mathrm{d}}\newcommand{\set}[1]{\left\{#1\right\}}\newcommand{\Lp}{\Delta\,}$事实上, 假设$\rd s^2=g_{ij}\rd x^i\rd x^j$是$M^2=(\Omega,g)$上的Riemann度量. 要使$M^2$ 是共形平坦的, 那么
$\rd s^2=g^{ij}\rd x_i\rd x_j=e^{2\lambda u}\left((\rd x_1)^2+(\rd x_2)^2\right).$

$\rd\omega_i=e^{\lambda u}\rd x_i,\quad i=1,2,$

\begin{align*}
\rd\omega_1&=-\lambda e^{\lambda u}u_2\rd x_1\wedge\rd x_2=-\lambda u_2\rd x_1\wedge\omega_2\\
\rd\omega_2&=\lambda e^{\lambda u}u_1\rd x_1\wedge\rd x_2=-\lambda u_1\rd x_2\wedge\omega_1,
\end{align*}

$\begin{cases} \rd\omega_i=-\omega_{ij}\wedge\omega_j\\ \rd\omega_{ij}=-\omega_{ik}\wedge\omega_{kj}+\Omega_{ij}, \end{cases}$

$\omega_{12}=\lambda\left(u_2\rd x_1-u_1\rd x_2\right)=-\omega_{21}.$

$\rd\omega_{12}=\frac{1}{2}R_{12kl}\omega_k\wedge\omega_l=-\lambda\Lp u\rd x_1\wedge\rd x_2=-\lambda\frac{\Lp u}{e^{2\lambda u}}\omega_1\wedge\omega_2,$

$K=R_{1212}=-\lambda\frac{\Lp u}{e^{2\lambda u}},$

$\Lp u+e^{2\lambda u}K/\lambda=0.$

$\Lp u+2K e^{u}=0.$