问题及解答

计算黎曼流形 $(\mathbb{H}^2,h)$ 的截面曲率, 其中

\[
\mathbb{H}^2=\{(x,y)\in\mathbb{R}^2\mid y > 0\},\quad h=\frac{1}{y^2}(dx\otimes dx+dy\otimes dy).
\]

余切标架为 $\omega^1=\frac{1}{y}dx$, $\omega^2=\frac{1}{y}dy$. 因此

\[
d\omega^1=d(\frac{1}{y}dx)=-\frac{1}{y^2}dy\wedge dx,\qquad d\omega^2=d(\frac{1}{y}dy)=0.
\]

由 Cartan 方程,

\[
\begin{aligned}
d\omega^1&=\omega^2\wedge\omega_{2}^{1}=\frac{1}{y}dy\wedge\omega_{2}^{1},\\
d\omega^2&=\omega^1\wedge\omega_{1}^{2}=-\frac{1}{y}dx\wedge\omega_{2}^{1}.
\end{aligned}
\]

由此解出 $\omega_{2}^{1}=-\frac{1}{y}dx$. 于是

\[
\Omega_{2}^{1}=d\omega_{2}^{1}+\omega_{k}^{1}\wedge\omega_{2}^{k}=d\omega_{2}^{1}=d(-\frac{1}{y}dx)=\frac{1}{y^2}dy\wedge dx=-(\frac{1}{y}dx)\wedge(\frac{1}{y}dy)=-\omega^1\wedge\omega^2.
\]

又 $\Omega_{2}^{1}=\frac{1}{2}R_{1212}\omega^1\wedge\omega^2$, 故 $R_{1212}=-1$. 因此截面曲率 $K=-1$.