一般性的轴对称度规可以写成(此处使用++++欧式号差)
\begin{equation} \mathrm{d}s^2 =\sum_A e^{2\mu_A}\mathrm{d}x^A +e^{2\psi}(\mathrm{d}x^1 -\sum_A q_A\mathrm{d}x^A)^2 \end{equation}
四个正交1形式是
\theta^A= e^{\mu_A}\mathrm{d}x^A,\space\theta^1 = e^{\psi}(\mathrm{d}x^1-\sum_A q_A \mathrm{d}x^A)
自然坐标用1形式表示
\mathrm{d}x^A = e^{-\mu_A}\theta^A,\space \mathrm{d}x^1=e^{-\psi}\theta^1+\sum_A e^{-\mu_A}\theta^A
由公式 0=\mathrm{d}\theta+\omega\wedge \theta ，计算联络 \omega
\begin{equation} \begin{split} \mathrm{d}\theta^A &=\sum_B \mu_{A,B}e^{-\mu_A}\mathrm{d}x^B\wedge \mathrm{d}x^A+\mu_{A,1}\mathrm{d}x^1\wedge \mathrm{d}x^A\\ &=\sum_B e^{-\mu_B}\mu_{A,B}\theta^B\wedge \theta^A +\mu_{A,1}[e^{-\psi}\theta^1+\sum_B e^{-\mu_B}q_B \theta^B]\wedge \theta^A\\ &=\sum_B \Big (\mu_{A,B}+q_B\mu_{A,1}\Big )e^{-\mu_B}\theta^B\wedge \theta^A+\mu_{A,1}e^{-\psi}\theta^1\wedge \theta^A \end{split} \end{equation}
定义 f_{:A}=f_{,A}+q_Af_{,1} 。
上式可以写作
\begin{equation} \mathrm{d}\theta^A=-\sum_B \mu_{A:B}e^{-\mu_B}\theta^A\wedge \theta^B-\mu_{A,1}e^{-\psi}\theta^A\wedge \theta^1\quad (1) \end{equation}
让我们暂时放下 \mathrm{d}\theta^A ，来计算 \mathrm{d}\theta^1 ，如下
\begin{equation} \begin{split} \mathrm{d}\theta^1 &=\sum_A e^{\psi}\psi_{,A}\mathrm{d}x^A\wedge \mathrm{d}x^1+\sum_A e^{\psi}(q_{A,1}+q_A\psi_{,1})\mathrm{d}x^A\wedge \mathrm{d}x^1 \\ &\quad-\sum_{AB}e^{\psi}(\psi_{,B}q_A+q_{A,B})\mathrm{d}x^B\wedge \mathrm{d}x^A\\ &=\sum_A (\psi_{,A}+q_A\psi_{,1}+q_{A,1})e^{\psi-\mu_A}\theta^A \wedge \Big(e^{-\psi}\theta^1+\sum_B q_B e^{-\mu_B}\theta^B\Big)\\ &\quad+\sum_{AB}e^{\psi-\mu_A-\mu_B}(\psi_{,B}q_A+q_{A,B})\theta^A\wedge \theta^B\\ &=\sum_A (\psi_{:A}+q_{A,1})e^{-\mu_A}\theta^A\wedge \theta^1+\sum_{AB}e^{\psi-\mu_A-\mu_B}\Big (q_{A,B}+q_Bq_{A,1}\\ &\qquad+\color{red}{\bcancel{\psi_{,A}q_B+\psi_{,B}q_A+q_Aq_B\psi_{,1}}}\Big )\theta^A \wedge \theta^B\\ &=\sum_A(\psi_{:A}+q_{A,1})e^{-\mu_A}\theta^A\wedge \theta^1+\sum_{AB}e^{\psi-\mu_A-\mu_B}q_{A:B}\theta^A\wedge\theta^B \end{split} \end{equation}
定义 Q_{AB}=q_{A:B}-q_{B:A} ， \Psi_A=\psi_{:A}+q_{A,1} 。
上式可以写为 \begin{equation} \begin{split} \mathrm{d}\theta^1 =-\sum_A e^{-\mu_A}\Psi_A\theta^1 \wedge \theta^A -\frac{1}{2}\sum_{AB}e^{2\psi-\mu_A-\mu_B}Q_{AB}\theta^B\wedge \theta^A \quad (2) \end{split} \end{equation}
由(2)式，可以得到
\begin{equation} \begin{split} \omega^1{}_A=e^{-\mu_A}\Psi_A\theta^1+\frac{1}{2}\sum_B e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^B+f\theta^A \end{split} \end{equation}
因为 \omega_{1A}=-\omega_{A1} ，由于我们使用的欧式号差的 \delta_{AB} ，可以忽略上下指标之间的差异。 \omega^1{}_A=-\omega^A{}_1 ，将 \omega^1{}_A 代入(1)式，得到
\begin{equation} \begin{split} &-\sum_B \omega^A{}_B \wedge \theta^B-\omega^A{}_1\wedge \theta^1 =-\sum_B \mu_{A:B}e^{-\mu_B}\theta^A\wedge \theta^B-\mu_{A,1}e^{-\psi}\theta^A\wedge \theta^1\\ \Longrightarrow &-\sum_B \omega^A{}_B\wedge \theta^B = \omega^A{}_1 \wedge \theta^1 -\mu_{A,1}e^{-\psi}\theta^A \wedge \theta^1-\sum_B \mu_{A:B}e^{-\mu_B}\theta^A\wedge \theta^B \\ \Longrightarrow &-\sum_B \omega^A{}_B \wedge \theta^B= -\Big (e^{-\mu_A}\Psi_A\theta^1+\frac{1}{2}\sum_B e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^B+f\theta^A\Big ) \wedge \theta^1\\ &\qquad\qquad\qquad\qquad -\mu_{A,1}e^{-\psi}\theta^A\wedge \theta^1 -\sum_B \mu_{A:B}e^{-\mu_B}\theta^A\wedge \theta^B \\ \Longrightarrow&-\sum_B\omega^A{}_B\wedge \theta^B = -(f+\mu_{A,1}e^{-\psi})\theta^A\wedge \theta^1\\ &\qquad\qquad\qquad\qquad-\sum_B \bigg (e^{-\mu_B}\mu_{A:B}\theta^A-\frac{1}{2}e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^1\bigg )\wedge \theta^B \end{split} \end{equation}
比较系数，可以得到 f=-\mu_{A,1}e^{-\psi} ， \omega^A{}_B = e^{-\mu_B}\mu_{A:B}\theta^A +h\theta^B -\frac{1}{2}e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^1 。
同样的，因为 \omega_{AB}=-\omega_{BA} ，所以 \omega^A{}_B =-\omega^B{}_A ，所以 h=-e^{-\mu_A}\mu_{B:A} 。
我们得到全部的联络
\begin{equation} \begin{split} \omega^1{}_A &=e^{-\mu_A}\Psi_A\theta^1-e^{-\psi}\mu_{A,1}\theta^A+\frac{1}{2}\sum_B e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^B\\ \omega^A{}_B&=-\frac{1}{2}e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^1+e^{-\mu_B}\mu_{A:B}\theta^A -e^{-\mu_A}\mu_{B:A}\theta^B \end{split} \end{equation}
其中 \Psi_A=\psi_{:A}+q_{A,1} ， Q_{AB}=q_{A:B}-q_{B:A} 。
下面是全部的联络表达式 \begin{equation} \begin{split} \omega^1{}_2 &= e^{-\mu_2}\Psi_2\theta^1-e^{-\psi}\mu_{2,1}\theta^2+\frac{1}{2}e^{\psi-\mu_2-\mu_3}Q_{23}\theta^3 +\frac{1}{2}e^{\psi-\mu_2-\mu_4}Q_{24}\theta^4 \\ \omega^1{}_3 &= e^{-\mu_3}\Psi_3\theta^1-e^{-\psi}\mu_{3,1}\theta^3+\frac{1}{2}e^{\psi-\mu_2-\mu_3}Q_{32}\theta^2 +\frac{1}{2}e^{\psi-\mu_3-\mu_4}Q_{34}\theta^4 \\ \omega^1{}_4 &= e^{-\mu_4}\Psi_4\theta^1-e^{-\psi}\mu_{4,1}\theta^4+\frac{1}{2}e^{\psi-\mu_2-\mu_4}Q_{42}\theta^2 +\frac{1}{2}e^{\psi-\mu_3-\mu_4}Q_{43}\theta^3 \\ \omega^2{}_3 &=-\frac{1}{2}e^{\psi-\mu_2-\mu_3}Q_{23}\theta^1+e^{-\mu_3}\mu_{2:3}\theta^2-e^{-\mu_2}\mu_{3:2}\theta^3\\ \omega^2{}_4 &=-\frac{1}{2}e^{\psi-\mu_2-\mu_4}Q_{24}\theta^1+e^{-\mu_4}\mu_{2:4}\theta^2-e^{-\mu_2}\mu_{4:2}\theta^4\\ \omega^3{}_4&=-\frac{1}{2}e^{\psi-\mu_3-\mu_4}Q_{34}\theta^1+e^{-\mu_4}\mu_{3:4}\theta^3-e^{-\mu_3}\mu_{4:3}\theta^4\\ \end{split} \end{equation}
接下的内容，需要计算曲率 \Omega 。
根据嘉当公式 \Omega=\mathrm{d}\omega+\omega\wedge \omega ， 就可以得到曲率 \Omega^a{}_b =\frac{1}{2}R^a{}_{bcd}\theta^c\wedge \theta^d 。
在开始之前，我们先准备两个辅助公式，如下
\begin{equation} \begin{split} \mathrm{d}\big(F\theta^1\big)&=\sum_Ae^{-\psi-\mu_A}\mathscr{D}_A\big(Fe^{\psi}\big)\theta^A\wedge \theta^1\\ &\quad+\frac{1}{2}\sum_{AB}Fe^{\psi-\mu_A-\mu_B}Q_{AB}\theta^A\wedge\theta^B\qquad(3)\\ \mathrm{d}\big (F\theta^A\big )&=\sum_Be^{-\mu_A-\mu_B}\big(Fe^{\mu_A}\big)_{:B}\theta^B\wedge \theta^A\\ &\quad+e^{-\psi-\mu_A}\big (Fe^{\mu_A}\big )_{,1}\theta^1\wedge\theta^A\qquad(4) \end{split} \end{equation}
其中， \mathscr{D}_A f:= f_{,A}+(fq_A)_{,1} 。
首先推导(3)式，如下
\begin{equation} \begin{split} \mathrm{d}\big (F\theta^1\big ) &=F_{,1}\mathrm{d}x^1\wedge \theta^1+ \sum_A F_{,A}\mathrm{d}x^A\wedge \theta^1+F\mathrm{d}\theta^1\\ &=F_{,1}\big (e^{-\psi}\theta^1+\sum_Ae^{-\mu_A}q_A\theta^A\big )\wedge \theta^1+\sum_AF_{,A}e^{-\mu_A}\theta^A\wedge \theta^1\\ &\quad-\sum_A Fe^{-\mu_A}\big (\psi_{:A}+q_{A,1}\big )\theta^1\wedge \theta^A-\frac{1}{2}\sum_{AB}Fe^{\psi-\mu_A-\mu_B}Q_{AB}\theta^B\wedge \theta^A\\ &= \sum_A e^{-\mu_A}\Big (\color{red}{q_AF_{,1}+F_{,A}+F\psi_{:A}+Fq_{A,1}}\Big )\theta^A\wedge \theta^1\\ &\quad+\frac{1}{2}\sum_{AB}Fe^{\psi-\mu_A-\mu_B}Q_{AB}\theta^A\wedge\theta^B \end{split} \end{equation}
上式中红色部分是
\begin{equation} \begin{split} \mathscr{D}_A \big (Fe^{\psi}\big )&=(Fe^{\psi})_{,A}+(q_AFe^{\psi})_{,1}\\ &=e^{\psi}\big (\color{red}{F_{,A}+F\psi_{:A}+q_{A,1}F+q_AF_{,1}}\big )\\ \Longrightarrow e^{-\psi}\mathscr{D}_A \big (Fe^{\psi}\big )&=\color{red}{F_{,A}+F\psi_{:A}+q_{A,1}F+q_AF_{,1}} \end{split} \end{equation}
将得到的结果替换红色部分，可以得到(3)式
\begin{equation} \begin{split} \mathrm{d}\big (F\theta^1\big ) &= \sum_A e^{-\psi-\mu_A}\mathscr{D}_A\big (Fe^{\psi}\big )\theta^A\wedge \theta^1\\ &\quad+\frac{1}{2}\sum_{AB}Fe^{\psi-\mu_A-\mu_B}Q_{AB}\theta^A\wedge\theta^B \end{split} \end{equation}
接着推导(4)式，
\begin{equation} \begin{split} \mathrm{d}\big (F\theta^A\big )&= F_{,1}\mathrm{d}x^1\wedge \theta^A+\sum_B F_{,B}\mathrm{d}x^B\wedge \theta^A+F\mathrm{d}\theta^A\\ &=F_{,1}\big(e^{-\psi}\theta^1+\sum_B e^{-\mu_B}q_B\theta^B\big)\wedge \theta^A+\sum_B F_{,B}e^{-\mu_B}\theta^B\wedge \theta^A\\ &\quad+\sum_BFe^{-\mu_B}\mu_{A:B}\theta^B\wedge \theta^A+Fe^{-\psi}\mu_{A,1}\theta^1\wedge \theta^A\\ &=e^{-\psi}\big (F_{,1}+F\mu_{A,1}\big )\theta^1\wedge \theta^A+\sum_B e^{-\mu_B}\big(F_{:B}+F\mu_{A:B}\big)\theta^B\wedge \theta^A\\ &=e^{-\psi-\mu_A}\big (Fe^{\mu_A}\big )_{,1}\theta^1\wedge \theta^A +\sum_B e^{-\mu_A-\mu_B}\big (Fe^{\mu_A}\big )_{:B}\theta^B\wedge \theta^A \end{split} \end{equation}
计算 \mathrm{d}\omega^1{}_A ，如下
\begin{equation} \begin{split} \mathrm{d}\omega^1{}_A&=\overbrace{\color{red}{\mathrm{d}\Big [e^{-\mu_A}\Psi_A\theta^1\Big ]}}^{\fbox{1}}\overbrace{-\color{blue}{\mathrm{d}\Big [e^{-\psi}\mu_{A,1}\theta^A\Big ]}}^{\fbox{2}}\overbrace{\color{magenta}{+\frac{1}{2}\sum_B\mathrm{d}\Big [e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^B\Big ]}}^{\fbox{3}} \end{split} \end{equation}
\begin{equation} \begin{split} \fbox{1}&=\sum_B e^{-\psi-\mu_B}\mathscr{D}_B\big (e^{\psi-\mu_A}\Psi_A\big )\theta^B\wedge \theta^1\\ &\quad+\frac{1}{2}\sum_{BC}e^{\psi-\mu_A-\mu_B-\mu_C}\Psi_AQ_{BC}\theta^B\wedge \theta^C \end{split} \end{equation}
\begin{equation} \begin{split} \fbox{2}&=-\sum_Be^{-\mu_A-\mu_B}\big (e^{\mu_A-\psi}\mu_{A,1}\big )_{:B}\theta^B\wedge \theta^A \\ &\quad-e^{-\psi-\mu_A}\big (e^{\mu_A-\psi}\mu_{A,1}\big )_{,1}\theta^1\wedge \theta^B \end{split} \end{equation}
\begin{equation} \begin{split} \fbox{3}&=\frac{1}{2}\sum_B \Big [ \sum_C e^{-\mu_B-\mu_C}\big(e^{\mu_B}e^{\psi-\mu_A-\mu_B}Q_{AB}\big)_{:C}\theta^C\wedge\theta^B\\ &\quad+e^{-\psi-\mu_B}\big(e^{\mu_B}e^{\psi-\mu_A-\mu_B}Q_{AB}\big )_{,1}\theta^1\wedge\theta^B\Big ]\\ &=\frac{1}{4}\sum_{BC}e^{-\mu_B-\mu_C}\Big [\big(e^{\psi-\mu_A}Q_{AC}\big)_{:B}-\big (e^{\psi-\mu_A}Q_{AB}\big )_{:C}\Big ]\theta^B\wedge\theta^C\\ &\quad+\frac{1}{2}\sum_Be^{-\psi-\mu_B}\big (e^{\psi-\mu_A}Q_{AB}\big )_{,1}\theta^1\wedge \theta^B \end{split} \end{equation}
\fbox{1}+\fbox{2}+\fbox{3} 就得到，
\begin{equation} \begin{split} \mathrm{d}\omega^1{}_A&=\sum_Be^{-\psi-\mu_B}\Big [\mathscr{D}_B \big (e^{\psi-\mu_A}\Psi_A\big )-\frac{1}{2}\big (e^{\psi-\mu_A}Q_{AB}\big )_{,1}\Big ]\theta^B\wedge \theta^1\\ &\quad-e^{-\psi-\mu_A}\big(e^{\mu_A-\psi}\mu_{A,1}\big)_{,1}\theta^1\wedge \theta^A\\ &\quad+\sum_Be^{-\mu_A-\mu_B}\big (e^{\mu_A-\psi}\mu_{A,1}\big )_{:B}\theta^A\wedge\theta^B \\ &\quad+\frac{1}{2}\sum_{BC}e^{-\mu_B-\mu_C}\bigg (e^{\psi-\mu_A}\Psi_AQ_{BC}\\ &\quad+\frac{1}{2}\Big [\big (e^{\psi-\mu_A}Q_{AC}\big )_{:B}-\big(e^{\psi-\mu_A}Q_{AB}\big)_{:C}\Big ]\bigg )\theta^B \wedge \theta^C \end{split} \end{equation}
计算 \sum_B\omega^1{}_B\wedge \omega^B{}_A ，如下
\begin{equation} \begin{split} \sum_B \omega^1{}_B\wedge \omega^B{}_A &= \sum_B \Big [e^{-\mu_B}\mu_{A:B}\theta^A -e^{-\mu_A}\mu_{B:A}\theta^B\\ &\quad-\frac{1}{2}e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^1\Big ]\bigwedge \Big [e^{-\mu_B}\Psi_B \theta^1\\ &\quad-e^{-\psi}\mu_{B,1}\theta^B+\frac{1}{2}\sum_C e^{\psi-\mu_B-\mu_C}Q_{BC}\theta^C\Big ]\\ \end{split} \end{equation}
我们用简写 (\omega^2)^1{}_A 表示 \sum_B \omega^1{}_B\wedge \omega^B{}_A ，
\begin{equation} \begin{split} (\omega^2)^1{}_A&=\sum_Be^{-2\mu_B}\mu_{A:B}\Psi_B\theta^A\wedge \theta^1 -\sum_Be^{-\psi-\mu_B}\mu_{A:B}\mu_{B,1}\theta^A\wedge\theta^B \\ &\quad+\frac{1}{2}\sum_{BC}e^{\psi-2\mu_B-\mu_C}\mu_{A:B}Q_{BC}\theta^A\wedge \theta^C-\sum_Be^{-\mu_A-\mu_B}\mu_{B:A}\Psi_B\theta^B\wedge\theta^1\\ &\quad-\frac{1}{4}\sum_{BC}e^{\psi-\mu_A-\mu_B-\mu_C}\big(\mu_{B:A}+\mu_{C:A}\big )Q_{BC}\theta^B\wedge\theta^C\\ &\quad+\frac{1}{2}\sum_{B}e^{-\mu_A-\mu_B}Q_{AB}\mu_{B,1}\theta^1\wedge \theta^B \\ &\quad-\frac{1}{4}\sum_{BC}e^{2\psi-\mu_A-2\mu_B-\mu_C}Q_{AB}Q_{BC}\theta^1\wedge\theta^C \end{split} \end{equation}
曲率 \Omega^1{}_A 的完整形态
\begin{equation} \begin{split} \Omega^1{}_A &=\mathrm{d}\omega^1{}_A+\sum_B \omega^1{}_B\wedge\omega^B{}_A\\ &=\theta^1\wedge \theta^A \Big[-e^{-\psi-\mu_A}\mathscr{D}_A\big(e^{\psi-\mu_A}\Psi_A\big)-e^{-\psi-\mu_A}\big(e^{\mu_A-\psi}\mu_{A,1}\big)_{,1}\\ &\qquad\qquad\quad-\sum_Be^{-2\mu_B}\mu_{A:B}\Psi_B+\frac{1}{4}\sum_Be^{2\psi-2\mu_A-2\mu_B}Q_{AB}^2\Big] \\ &\quad+\sum_B\theta^1\wedge\theta^B \Big [-e^{-\psi-\mu_B}\mathscr{D}_B\big(e^{\psi-\mu_A}\Psi_A\big)+e^{-\mu_A-\mu_B}\mu_{B:A}\Psi_B \\ &\qquad\qquad\qquad\quad+\frac{1}{2}e^{-\psi-\mu_B}\big (e^{\psi-\mu_A}Q_{AB}\big)_{,1}+\frac{1}{2}e^{-\mu_A-\mu_B}Q_{AB}\mu_{B,1}\\ &\qquad\qquad\qquad\quad+\frac{1}{4}e^{2\psi-\mu_A-\mu_B-2\mu_C}Q_{AC}Q_{BC} \Big ] \\ &\quad+\sum_B \theta^A\wedge\theta^B\Big [e^{\psi-2\mu_A-\mu_B}\big(\Psi_A-\frac{1}{2}\mu_{B:A}\big)Q_{AB}-e^{-\psi-\mu_B}\mu_{A:B}\mu_{B,1}\\ &\qquad\qquad\qquad\quad +e^{-\mu_A-\mu_B}\big (e^{\mu_A-\psi}\mu_{A,1}\big)_{:B}+\frac{1}{2}e^{-\mu_A-\mu_B}\big(e^{\psi-\mu_A}Q_{AB}\big )_{:A}\\ &\qquad\qquad\qquad\quad +\frac{1}{2}e^{\psi-\mu_B-2\mu_C}\mu_{A:C}Q_{CB}\Big] \\ &\quad+\theta^B\wedge\theta^C \bigg (\frac{1}{2}e^{\psi-\mu_A-\mu_B-\mu_C}\Big[2\Psi_A- \big(\mu_{B:A}+\mu_{C:A}\big)\Big]Q_{BC} \\ &\qquad\qquad\qquad+\frac{1}{2}e^{-\mu_B-\mu_C}\Big [\big (e^{\psi-\mu_A}Q_{AC}\big )_{:B}-\big(e^{\psi-\mu_A}Q_{AB}\big)_{:C}\Big ] \bigg)\end{split} \end{equation}
同上计算 \mathrm{d}\omega^A{}_B ，
\begin{equation} \begin{split} \mathrm{d}\omega^A{}_B&=\overbrace{\color{red}{\mathrm{d}\Big [e^{-\mu_B}\mu_{A:B}\theta^A\Big ]}}^{\fbox{1}}\overbrace{\color{blue}{-\mathrm{d}\Big [e^{-\mu_A}\mu_{B:A}\theta^B\Big ]}}^{\fbox{2}}\overbrace{\color{magenta}{-\mathrm{d}\Big [\frac{1}{2}e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^1\Big ]}}^{\fbox{3}} \end{split} \end{equation}
\begin{equation} \begin{split} \fbox{1}&=-\sum_Ce^{-\mu_A-\mu_C}\big (e^{\mu_A-\mu_B}\mu_{A:B}\big )_{:C}\theta^C\wedge \theta^B \\ &\quad-e^{-\psi-\mu_A}\big (e^{\mu_A-\mu_B}\mu_{A:B}\big )_{,1}\theta^1\wedge \theta^A \end{split} \end{equation}
\begin{equation} \begin{split} \fbox{2}&=-\sum_Ce^{-\mu_B-\mu_C}\big (e^{\mu_B-\mu_A}\mu_{B:A}\big )_{:C}\theta^C\wedge \theta^B \\ &\quad-e^{-\psi-\mu_B}\big (e^{\mu_B-\mu_A}\mu_{B:A}\big )_{,1}\theta^1\wedge \theta^B \end{split} \end{equation}
\begin{equation} \begin{split} \fbox{3}&=\sum_Ce^{-\psi-\mu_C}\mathscr{D}_C\Big [-\frac{1}{2}e^{\psi-\mu_A-\mu_B}Q_{AB}e^{\psi}\Big ]\theta^C\wedge \theta^1\\ &\quad+\frac{1}{2}\sum_{CD}\Big [ -\frac{1}{2}e^{\psi-\mu_A-\mu_B}Q_{AB}\Big ]e^{\psi-\mu_C-\mu_D}Q_{CD}\theta^C\wedge\theta^D\\ &=-\frac{1}{2}\sum_C e^{-\psi-\mu_C}\mathscr{D}_C\Big [e^{2\psi-\mu_A-\mu_B}Q_{AB}\Big ]\theta^C\wedge \theta^1 \\ &\quad-\frac{1}{4}\sum_{CD}e^{2\psi-\mu_A-\mu_B-\mu_C-\mu_D}Q_{AB}Q_{CD}\theta^C\wedge \theta^D \end{split} \end{equation}
因为这里只有四个标架，我们把指标 A 、 B 当作已知条件，这样我们可以用 C 表示剩下的不同于 1 、 A 和 B 的指标，如此操作我们可以去掉 \fbox{1}\cdots \fbox{3} 里的求和号 \sum_C 。
\begin{equation} \begin{split} \fbox{1}&=e^{-\mu_A-\mu_B}\big(e^{\mu_A-\mu_B}\mu_{A:B}\big)_{:B}\theta^B\wedge \theta^A\\ &\quad+e^{-\mu_A-\mu_C}\big(e^{\mu_A-\mu_B}\mu_{A:B}\big)_{:C}\theta^C\wedge \theta^A\\ &\quad+e^{-\psi-\mu_A}\big(e^{\mu_A-\mu_B}\mu_{A:B}\mu\big)_{,1}\theta^1\wedge \theta^A \end{split} \end{equation}
\begin{equation} \begin{split} \fbox{2}&=-e^{-\mu_A-\mu_B}\big(e^{\mu_B-\mu_A}\mu_{B:A}\big)_{:A}\theta^A\wedge \theta^B\\ &\quad-e^{-\mu_B-\mu_C}\big(e^{\mu_B-\mu_A}\mu_{B:A}\big)_{:C}\theta^C\wedge \theta^A\\ &\quad-e^{-\psi-\mu_B}\big(e^{\mu_B-\mu_A}\mu_{B:A}\mu\big)_{,1}\theta^1\wedge \theta^B \end{split} \end{equation}
\begin{equation} \begin{split} \fbox{3}&=-\frac{1}{2}e^{-\psi-\mu_A}\mathscr{D}_A\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)\theta^A\wedge\theta^1\\ &\quad-\frac{1}{2}e^{-\psi-\mu_B}\mathscr{D}_B\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)\theta^B\wedge \theta^1\\ &\quad-\frac{1}{2}e^{-\psi-\mu_C}\mathscr{D}_C\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)\theta^C\wedge \theta^1\\ &\quad-\frac{1}{2}e^{2\psi-2\mu_A-2\mu_B}Q_{AB}^2 \theta^A\wedge \theta^B\\ &\quad-\frac{1}{2}e^{2\psi-2\mu_A-\mu_B-\mu_C}Q_{AB}Q_{AC}\theta^A\wedge \theta^C\\ &\quad-\frac{1}{2}e^{2\psi-\mu_A-2\mu_B-\mu_C}Q_{AB}Q_{BC}\theta^B\wedge\theta^C \end{split} \end{equation}
\fbox{1}+\fbox{2}+\fbox{3} 得到 \mathrm{d}\omega^A{}_B 的完全表达式，如下
\begin{equation} \begin{split} \mathrm{d}\omega^A{}_B &=\theta^1\wedge\theta^A\Big [e^{-\psi-\mu_A}\big(e^{\mu_A-\mu_B}\mu_{A:B}\big)_{,1}+\frac{1}{2}e^{-\psi-\mu_A}\mathscr{D}_A\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)\Big ]\\ &\quad +\theta^1\wedge\theta^B \Big [-e^{-\psi-\mu_B}\big(e^{\mu_B-\mu_A}\mu_{B:A}\big)_{,1}+\frac{1}{2}e^{-\psi-\mu_B}\mathscr{D}_B\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)\Big ]\\ &\quad+\theta^1\wedge \theta^C \Big [\frac{1}{2}e^{-\psi-\mu_C}\mathscr{D}_C\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)\Big ]\\ &\quad+\theta^A\wedge\theta^B \bigg (-e^{-\mu_A-\mu_B}\Big[ \big(e^{\mu_A-\mu_B}\mu_{A:B}\big)_{:B}-\big(e^{\mu_B-\mu_A}\mu_{B:A}\big)_{:A}\\ &\qquad\qquad\qquad-\frac{1}{2}e^{2\psi-\mu_A-\mu_B}Q_{AB}^2\Big ]\bigg )\\ &\quad+\theta^A\wedge\theta^C \Big [-e^{-\mu_A-\mu_C}\big(e^{\mu_A-\mu_B}\mu_{A:B}\big)_{:C}-\frac{1}{2}e^{2\psi-2\mu_A-\mu_B-\mu_C}Q_{AB}Q_{AC}\Big ]\\ &\quad+\theta^B\wedge\theta^C\Big [+e^{-\mu_B-\mu_C}\big(e^{\mu_B-\mu_A}\mu_{B:A}\big)_{:C}-\frac{1}{2}e^{2\psi-\mu_A-2\mu_B-\mu_C}Q_{AB}Q_{BC}\Big ] \end{split} \end{equation}
\omega\wedge \omega 部分的计算
\begin{equation} \begin{split} \omega^A{}_1\wedge \omega^1{}_B &= -\omega^1{}_A\wedge \omega^1{}_B\\ &=-\Big [e^{-\mu_A}\Psi_A\theta^1-e^{-\psi}\mu_{A,1}\theta^A+\frac{1}{2} e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^B\\ &\qquad+\frac{1}{2}e^{\psi-\mu_A-\mu_C}Q_{AC}\theta^C\Big] \bigwedge \Big [ e^{-\mu_B}\Psi_B\theta^1-e^{-\psi}\mu_{B,1}\theta^B\\&\qquad-\frac{1}{2} e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^A+\frac{1}{2}e^{\psi-\mu_B-\mu_C}Q_{BC}\theta^C\Big] \end{split} \end{equation}
另起一行，
\begin{equation} \begin{split} \omega^A{}_1\wedge\omega^1{}_B &=\quad\theta^1\wedge \theta^A \Big [+\frac{1}{2}e^{\psi-2\mu_A-\mu_B}\Psi_AQ_{AB}-e^{-\psi-\mu_B}\mu_{A,1}\Psi_B\Big ]\\ &\quad+\theta^1\wedge\theta^B \Big [-\frac{1}{2}e^{\psi-\mu_A-2\mu_B}\Psi_BQ_{BA}+e^{-\psi-\mu_A}\mu_{B,1}\Psi_A\Big ]\\ &\quad+\theta^1\wedge \theta^C\Big [-\frac{1}{2}e^{\psi-\mu_A-\mu_B-\mu_C}\big(\Psi_AQ_{BC}-\psi_BQ_{AC}\big)\Big ]\\ &\quad+\theta^A\wedge\theta^B\Big [-e^{-2\psi}\mu_{A,1}\mu_{B,1}-\frac{1}{4}e^{2\psi-2\mu_A-2\mu_B}Q_{AB}^2\Big ]\\ &\quad+\theta^A\wedge\theta^C \Big [+\frac{1}{2}e^{-\mu_B-\mu_C}\mu_{A,1}Q_{BC}-\frac{1}{4}e^{2\psi-2\mu_A-\mu_B-\mu_C}Q_{AB}Q_{AC}\Big ]\\ &\quad+\theta^B\wedge\theta^C\Big [-\frac{1}{2}e^{-\mu_A-\mu_C}\mu_{B,1}Q_{AC}-\frac{1}{4}e^{2\psi-\mu_A-2\mu_B-\mu_C}Q_{AB}Q_{BC}\Big ] \end{split} \end{equation}
\omega^A{}_C\wedge \omega^C{}_B 的计算，如下
\begin{equation} \begin{split} \omega^A{}_C\wedge\omega^C{}_B&=-\omega^A{}_C\wedge\omega^B{}_C\\ &=-\Big [e^{-\mu_C}\mu_{A:C}\theta^A-e^{-\mu_A}\mu_{C:A}\theta^C\\ &\qquad-\frac{1}{2}e^{\psi-\mu_A-\mu_C}Q_{AC}\theta^1\Big ]\bigwedge \Big [e^{-\mu_C}\mu_{B:C}\theta^B\\ &\qquad-e^{-\mu_B}\mu_{C:B}\theta^C-\frac{1}{2}e^{\psi-\mu_B-\mu_C}\theta^1 \Big ] \end{split} \end{equation}
另起一行，
\begin{equation} \begin{split} \omega^A{}_C\wedge\omega^C{}_B &=-\frac{1}{2}e^{\psi-\mu_B-2\mu_C}\mu_{A:C}Q_{BC}\theta^1\wedge \theta^A\\ &\quad+\frac{1}{2}e^{\psi-\mu_A-2\mu_C}\mu_{B:C}Q_{AC}\theta^1\wedge\theta^B\\ &\quad-\frac{1}{2}e^{\psi-\mu_A-\mu_B-\mu_C}\big(\mu_{C:B}Q_{AC}-\mu_{C:A}Q_{BC}\big)\theta^1\wedge\theta^C\\ &\quad-e^{-2\mu_C}\mu_{A:C}\mu_{B:C}\theta^A\wedge\theta^B\\ &\quad+e^{-\mu_B-\mu_C}\mu_{A:C}\mu_{C:B}\theta^A\wedge\theta^C\\ &\quad-e^{-\mu_A-\mu_C}\mu_{B:C}\mu_{C:A}\theta^B\wedge\theta^C \end{split} \end{equation}
曲率 \Omega^A{}_B 的完整表达式，如下
\begin{equation} \begin{split} \Omega^A{}_B&=\mathrm{d}\omega^A{}_B+\Omega^A{}_1\wedge\omega^1{}_B+\Omega^A{}_C\wedge\Omega^C{}_B\\ &=\theta^1\wedge\theta^A \Big [\frac{1}{2}e^{-\psi-\mu_A}\mathscr{D}_A\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)+e^{-\psi-\mu_A}\big (e^{\mu_A-\mu_B}\mu_{A:B}\big)_{,1}\\ &\qquad\qquad\quad-e^{-\psi-\mu_B}\mu_{A,1}\Psi_B+\frac{1}{2}e^{\psi-2\mu_A-\mu_B}\Psi_AQ_{AB}\\ &\qquad\qquad\quad-\frac{1}{2}e^{\psi-\mu_B-2\mu_C}\mu_{A:C}Q_{BC}\Big]\\ &\quad+\theta^1\wedge\theta^B\Big [\frac{1}{2}e^{-\psi-\mu_B}\mathscr{D}_B\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)-e^{-\psi-\mu_B}\big(e^{\mu_B-\mu_A}\mu_{B:A}\big)_{,1}\\ &\qquad\qquad\qquad+e^{-\psi-\mu_A}\mu_{B,1}\Psi_A-\frac{1}{2}e^{\psi-\mu_A-2\mu_B}\Psi_BQ_{BA}\\ &\qquad\qquad\qquad+\frac{1}{2}e^{\psi-\mu_A-2\mu_C}\mu_{B:C}Q_{AC}\Big ]\\ &\quad+\theta^1\wedge\theta^C \Big [\frac{1}{2}e^{-\psi-\mu_C}\mathscr{D}_C\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)-\frac{1}{2}e^{\psi-\mu_A-\mu_B-\mu_C}\big(\Psi_AQ_{BC}\\ &\qquad\qquad\qquad-\Psi_BQ_{AC}\big)-\frac{1}{2}e^{\psi-\mu_A-\mu_B-\mu_C}\big(\mu_{C:B}Q_{AC}-\mu_{C:A}Q_{BC}\big)\Big ]\\ &\quad+\theta^A\wedge\theta^B \bigg \{-e^{-\mu_A-\mu_B}\Big [\big(e^{\mu_A-\mu_B}\mu_{A:B}\big)_{:B}-\big(e^{-\mu_B-\mu_A}\mu_{B:A}\big)_{:A}\Big ]\\ &\qquad\qquad\qquad-e^{-2\psi}\mu_{A,1}\mu_{B,1}-e^{-2\mu_C}\mu_{A:C}\mu_{B:C}-\frac{3}{4}e^{2\psi-2\mu_A-2\mu_B}Q_{AB}^2\bigg \}\\ &\quad+\theta^A\wedge\theta^C\Big [-e^{-\mu_A-\mu_C}\big(e^{\mu_A-\mu_B}\mu_{A:B}\big)_{:C}+e^{-\mu_B-\mu_C}\mu_{A:C}\mu_{C:B}\\ &\qquad\qquad\qquad+\frac{1}{2}e^{-\mu_B-\mu_C}\mu_{A,1}Q_{BC}-\frac{3}{4}e^{2\psi-2\mu_A-\mu_B-\mu_C}Q_{AB}Q_{AC}\Big ]\\ &\quad+\theta^B\wedge\theta^C\Big [+e^{-\mu_B-\mu_C}\big(e^{\mu_B-\mu_A}\mu_{B:A}\big)_{:C}-e^{-\mu_A-\mu_C}\mu_{B:C}\mu_{C:A}\\ &\qquad\qquad\qquad-\frac{1}{2}e^{-\mu_A-\mu_C}\mu_{B,1}Q_{AC}+\frac{3}{4}e^{2\psi-\mu_A-2\mu_B-\mu_C}Q_{AB}Q_{BC}\Big ] \end{split} \end{equation}
总结
联络
\begin{equation} \begin{split} \omega^1{}_A &=e^{-\mu_A}\Psi_A\theta^1-e^{-\psi}\mu_{A,1}\theta^A+\frac{1}{2}\sum_B e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^B\\ \omega^A{}_B&=-\frac{1}{2}e^{\psi-\mu_A-\mu_B}Q_{AB}\theta^1+e^{-\mu_B}\mu_{A:B}\theta^A -e^{-\mu_A}\mu_{B:A}\theta^B \end{split} \end{equation}
曲率
\begin{equation} \begin{split} \Omega^1{}_A &=\theta^1\wedge \theta^A \Big[-e^{-\psi-\mu_A}\mathscr{D}_A\big(e^{\psi-\mu_A}\Psi_A\big)-e^{-\psi-\mu_A}\big(e^{\mu_A-\psi}\mu_{A,1}\big)_{,1}\\ &\qquad\qquad\quad-\sum_Be^{-2\mu_B}\mu_{A:B}\Psi_B+\frac{1}{4}\sum_Be^{2\psi-2\mu_A-2\mu_B}Q_{AB}^2\Big] \\ &\quad+\sum_B\theta^1\wedge\theta^B \Big [-e^{-\psi-\mu_B}\mathscr{D}_B\big(e^{\psi-\mu_A}\Psi_A\big)+e^{-\mu_A-\mu_B}\mu_{B:A}\Psi_B \\ &\qquad\qquad\qquad\quad+\frac{1}{2}e^{-\psi-\mu_B}\big (e^{\psi-\mu_A}Q_{AB}\big)_{,1}+\frac{1}{2}e^{-\mu_A-\mu_B}Q_{AB}\mu_{B,1}\\ &\qquad\qquad\qquad\quad+\frac{1}{4}e^{2\psi-\mu_A-\mu_B-2\mu_C}Q_{AC}Q_{BC} \Big ] \\ &\quad+\sum_B \theta^A\wedge\theta^B\Big [e^{\psi-2\mu_A-\mu_B}\big(\Psi_A-\frac{1}{2}\mu_{B:A}\big)Q_{AB}-e^{-\psi-\mu_B}\mu_{A:B}\mu_{B,1}\\ &\qquad\qquad\qquad\quad +e^{-\mu_A-\mu_B}\big (e^{\mu_A-\psi}\mu_{A,1}\big)_{:B}+\frac{1}{2}e^{-\mu_A-\mu_B}\big(e^{\psi-\mu_A}Q_{AB}\big )_{:A}\\ &\qquad\qquad\qquad\quad +\frac{1}{2}e^{\psi-\mu_B-2\mu_C}\mu_{A:C}Q_{CB}\Big] \\ &\quad+\theta^B\wedge\theta^C \bigg (\frac{1}{2}e^{\psi-\mu_A-\mu_B-\mu_C}\Big[2\Psi_A- \big(\mu_{B:A}+\mu_{C:A}\big)\Big]Q_{BC} \\ &\qquad\qquad\qquad+\frac{1}{2}e^{-\mu_B-\mu_C}\Big [\big (e^{\psi-\mu_A}Q_{AC}\big )_{:B}-\big(e^{\psi-\mu_A}Q_{AB}\big)_{:C}\Big ] \bigg)\end{split} \end{equation}
\begin{equation} \begin{split} \Omega^A{}_B&=\theta^1\wedge\theta^A \Big [\frac{1}{2}e^{-\psi-\mu_A}\mathscr{D}_A\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)+e^{-\psi-\mu_A}\big (e^{\mu_A-\mu_B}\mu_{A:B}\big)_{,1}\\ &\qquad\qquad\quad-e^{-\psi-\mu_B}\mu_{A,1}\Psi_B+\frac{1}{2}e^{\psi-2\mu_A-\mu_B}\Psi_AQ_{AB}\\ &\qquad\qquad\quad-\frac{1}{2}e^{\psi-\mu_B-2\mu_C}\mu_{A:C}Q_{BC}\Big]\\ &\quad+\theta^1\wedge\theta^B\Big [\frac{1}{2}e^{-\psi-\mu_B}\mathscr{D}_B\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)-e^{-\psi-\mu_B}\big(e^{\mu_B-\mu_A}\mu_{B:A}\big)_{,1}\\ &\qquad\qquad\qquad+e^{-\psi-\mu_A}\mu_{B,1}\Psi_A-\frac{1}{2}e^{\psi-\mu_A-2\mu_B}\Psi_BQ_{BA}\\ &\qquad\qquad\qquad+\frac{1}{2}e^{\psi-\mu_A-2\mu_C}\mu_{B:C}Q_{AC}\Big ]\\ &\quad+\theta^1\wedge\theta^C \Big [\frac{1}{2}e^{-\psi-\mu_C}\mathscr{D}_C\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)-\frac{1}{2}e^{\psi-\mu_A-\mu_B-\mu_C}\big(\Psi_AQ_{BC}\\ &\qquad\qquad\qquad-\Psi_BQ_{AC}\big)-\frac{1}{2}e^{\psi-\mu_A-\mu_B-\mu_C}\big(\mu_{C:B}Q_{AC}-\mu_{C:A}Q_{BC}\big)\Big ]\\ &\quad+\theta^A\wedge\theta^B \bigg \{-e^{-\mu_A-\mu_B}\Big [\big(e^{\mu_A-\mu_B}\mu_{A:B}\big)_{:B}-\big(e^{-\mu_B-\mu_A}\mu_{B:A}\big)_{:A}\Big ]\\ &\qquad\qquad\qquad-e^{-2\psi}\mu_{A,1}\mu_{B,1}-e^{-2\mu_C}\mu_{A:C}\mu_{B:C}-\frac{3}{4}e^{2\psi-2\mu_A-2\mu_B}Q_{AB}^2\bigg \}\\ &\quad+\theta^A\wedge\theta^C\Big [-e^{-\mu_A-\mu_C}\big(e^{\mu_A-\mu_B}\mu_{A:B}\big)_{:C}+e^{-\mu_B-\mu_C}\mu_{A:C}\mu_{C:B}\\ &\qquad\qquad\qquad+\frac{1}{2}e^{-\mu_B-\mu_C}\mu_{A,1}Q_{BC}-\frac{3}{4}e^{2\psi-2\mu_A-\mu_B-\mu_C}Q_{AB}Q_{AC}\Big ]\\ &\quad+\theta^B\wedge\theta^C\Big [+e^{-\mu_B-\mu_C}\big(e^{\mu_B-\mu_A}\mu_{B:A}\big)_{:C}-e^{-\mu_A-\mu_C}\mu_{B:C}\mu_{C:A}\\ &\qquad\qquad\qquad-\frac{1}{2}e^{-\mu_A-\mu_C}\mu_{B,1}Q_{AC}+\frac{3}{4}e^{2\psi-\mu_A-2\mu_B-\mu_C}Q_{AB}Q_{BC}\Big ] \end{split} \end{equation}
联络展开式
\begin{equation} \begin{split} \omega^1{}_2 &= e^{-\mu_2}\Psi_2\theta^1-e^{-\psi}\mu_{2,1}\theta^2+\frac{1}{2}e^{\psi-\mu_2-\mu_3}Q_{23}\theta^3 +\frac{1}{2}e^{\psi-\mu_2-\mu_4}Q_{24}\theta^4 \\ \omega^1{}_3 &= e^{-\mu_3}\Psi_3\theta^1-e^{-\psi}\mu_{3,1}\theta^3+\frac{1}{2}e^{\psi-\mu_2-\mu_3}Q_{32}\theta^2 +\frac{1}{2}e^{\psi-\mu_3-\mu_4}Q_{34}\theta^4 \\ \omega^1{}_4 &= e^{-\mu_4}\Psi_4\theta^1-e^{-\psi}\mu_{4,1}\theta^4+\frac{1}{2}e^{\psi-\mu_2-\mu_4}Q_{42}\theta^2 +\frac{1}{2}e^{\psi-\mu_3-\mu_4}Q_{43}\theta^3 \\ \omega^2{}_3 &=-\frac{1}{2}e^{\psi-\mu_2-\mu_3}Q_{23}\theta^1+e^{-\mu_3}\mu_{2:3}\theta^2-e^{-\mu_2}\mu_{3:2}\theta^3\\ \omega^2{}_4 &=-\frac{1}{2}e^{\psi-\mu_2-\mu_4}Q_{24}\theta^1+e^{-\mu_4}\mu_{2:4}\theta^2-e^{-\mu_2}\mu_{4:2}\theta^4\\ \omega^3{}_4&=-\frac{1}{2}e^{\psi-\mu_3-\mu_4}Q_{34}\theta^1+e^{-\mu_4}\mu_{3:4}\theta^3-e^{-\mu_3}\mu_{4:3}\theta^4\\ \end{split} \end{equation}
全部不为0的曲率四分量，可以从上面得到 \Omega^1{}_A 和 \Omega^A{}_B 获得，如下
由 \Omega^1{}_A 的 \theta^1\wedge\theta^A 项
\begin{equation} \begin{split} R^1{}_{A1A}&=-e^{-\psi-\mu_A}\mathscr{D}_A\big(e^{\psi-\mu_A}\Psi_A\big)-e^{-\psi-\mu_A}\big(e^{\mu_A-\psi}\mu_{A,1}\big)_{,1}\\ &\quad-\sum_Be^{-2\mu_B}\mu_{A:B}\Psi_B+\frac{1}{4}\sum_Be^{2\psi-2\mu_A-2\mu_B}Q_{AB}^2\Big] \end{split} \end{equation}
可以得到 R^1{}_{212} 、 R^1{}_{313} 、 R^1{}_{414}
\begin{equation} \begin{split} R^1{}_{212}&=-e^{-\psi-\mu_2}\mathscr{D}_2\big(e^{\psi-\mu_2}\Psi_2\big)-e^{-\psi-\mu_2}\big(e^{\mu_2-\psi}\mu_{2,1}\big)_{,1}-e^{-2\mu_3}\mu_{2:3}\Psi_3\\ &\quad-e^{-2\mu_4}\mu_{2:4}\Psi_4+\frac{1}{4}e^{2\psi-2\mu_2-2\mu_3}Q_{23}^2+\frac{1}{4}e^{2\psi-2\mu_2-2\mu_4}Q_{24}^2 \end{split} \end{equation}
\begin{equation} \begin{split} R^1{}_{313}&=-e^{-\psi-\mu_3}\mathscr{D}_3\big(e^{\psi-\mu_3}\Psi_3\big)-e^{-\psi-\mu_3}\big(e^{\mu_3-\psi}\mu_{3,1}\big)_{,1}-e^{-2\mu_2}\mu_{3:2}\Psi_2\\ &\quad-e^{-2\mu_4}\mu_{3:4}\Psi_4+\frac{1}{4}e^{2\psi-2\mu_2-2\mu_3}Q_{32}^2+\frac{1}{4}e^{2\psi-2\mu_3-2\mu_4}Q_{34}^2 \end{split} \end{equation}
\begin{equation} \begin{split} R^1{}_{414}&=-e^{-\psi-\mu_4}\mathscr{D}_4\big(e^{\psi-\mu_4}\Psi_4\big)-e^{-\psi-\mu_4}\big(e^{\mu_4-\psi}\mu_{4,1}\big)_{,1}-e^{-2\mu_2}\mu_{4:2}\Psi_2\\ &\quad-e^{-2\mu_3}\mu_{4:3}\Psi_3+\frac{1}{4}e^{2\psi-2\mu_2-2\mu_4}Q_{42}^2+\frac{1}{4}e^{2\psi-2\mu_3-2\mu_4}Q_{43}^2 \end{split} \end{equation}
由 \Omega^1{}_A 的 \theta^1\wedge \theta^B 项 \begin{equation} \begin{split} R^1{}_{A1B}&=e^{-\psi-\mu_B}\mathscr{D}_B\big(e^{\psi-\mu_A}\Psi_A\big) +e^{-\mu_A-\mu_B}\mu_{B:A}\Psi_B +\frac{1}{2}e^{-\psi-\mu_B}\big (e^{\psi-\mu_A}Q_{AB}\big)_{,1}\\ &\quad+\frac{1}{2}e^{-\mu_A-\mu_B}Q_{AB}\mu_{B,1}+\frac{1}{4}e^{2\psi-\mu_A-\mu_B-2\mu_C}Q_{AC}Q_{BC} \end{split} \end{equation}
稍作化简
\begin{equation} \begin{split} R^1{}_{A1B}&=e^{-\psi-\mu_B}\mathscr{D}_B\big(e^{\psi-\mu_A}\Psi_A\big) +e^{-\mu_A-\mu_B}\mu_{B:A}\Psi_B \\ & \quad+\frac{1}{2}e^{-\mu_A-\mu_B}\Big [Q_{AB,1}+Q_{AB}(\psi+\mu_B-\mu_A\big)_{,1}\Big]\\ &\quad+\frac{1}{4}e^{2\psi-\mu_A-\mu_B-2\mu_C}Q_{AC}Q_{BC} \end{split} \end{equation}
由此可以得到 R^1{}_{213} 、 R^1{}_{214} 、 R^1{}_{314} 。
\begin{equation} \begin{split} R^1{}_{213}&=-e^{-\psi-\mu_3}\mathscr{D}_3\big (e^{\psi-\mu_2}\Psi_2\big)+e^{-\mu_2-\mu_3}\mu_{3:2}\Psi_3\\ &\quad+\frac{1}{2}e^{-\mu_2-\mu_3}\Big[Q_{23,1}+Q_{23}\big(\psi+\mu_3-\mu_2\big)_{,1}\Big]\\ &\quad+\frac{1}{4}e^{2\psi-\mu_2-\mu_3-2\mu_4}Q_{24}Q_{34} \end{split} \end{equation}
\begin{equation} \begin{split} R^1{}_{214}&=-e^{-\psi-\mu_4}\mathscr{D}_4\big (e^{\psi-\mu_2}\Psi_2\big)+e^{-\mu_2-\mu_4}\mu_{4:2}\Psi_4\\ &\quad+\frac{1}{2}e^{-\mu_2-\mu_4}\Big[Q_{24,1}+Q_{24}\big(\psi+\mu_4-\mu_2\big)_{,1}\Big]\\ &\quad+\frac{1}{4}e^{2\psi-\mu_2-2\mu_3-\mu_4}Q_{23}Q_{43} \end{split} \end{equation}
\begin{equation} \begin{split} R^1{}_{314}&=-e^{-\psi-\mu_4}\mathscr{D}_4\big (e^{\psi-\mu_3}\Psi_3\big)+e^{-\mu_3-\mu_4}\mu_{4:3}\Psi_4\\ &\quad+\frac{1}{2}e^{-\mu_3-\mu_4}\Big[Q_{34,1}+Q_{34}\big(\psi+\mu_4-\mu_3\big)_{,1}\Big]\\ &\quad+\frac{1}{4}e^{2\psi-2\mu_2-\mu_3-\mu_4}Q_{32}Q_{42} \end{split} \end{equation}
由 \Omega^1{}_A 的 \theta^A\wedge\theta^B 项
\begin{equation} \begin{split} R^1{}_{AAB}&=e^{\psi-2\mu_A-\mu_B}\big(\Psi_A-\frac{1}{2}\mu_{B:A}\big)Q_{AB}-e^{-\psi-\mu_B}\mu_{A:B}\mu_{B,1}\\ &\quad +e^{-\mu_A-\mu_B}\big (e^{\mu_A-\psi}\mu_{A,1}\big)_{:B}+\frac{1}{2}e^{-\mu_A-\mu_B}\big(e^{\psi-\mu_A}Q_{AB}\big )_{:A}\\ &\quad+\frac{1}{2}e^{\psi-\mu_B-2\mu_C}\mu_{A:C}Q_{CB}\end{split} \end{equation}
可以得到 R^1{}_{223} 、 R^1{}_{224} 、 R^1{}_{332} 、 R^1{}_{334} 、 R^1{}_{442} 、 R^1{}_{443}
\begin{equation} \begin{split} R^1{}_{223}&= e^{\psi-2\mu_2-\mu_3}\big(\Psi_2-\frac{1}{2}\mu_{3:2}\big)Q_{23}-e^{-\psi-\mu_3}\mu_{2:3}\mu_{3,1}\\ &\quad+e^{-\mu_2-\mu_3}\big(e^{\mu_2-\psi}\mu_{2,1}\big)_{:3}+\frac{1}{2}e^{-\mu_2-\mu_3}\big(e^{\psi-\mu_2}Q_{23}\big)_{:2}\\ &\quad+\frac{1}{2}e^{\psi-\mu_3-2\mu_4}\mu_{2:4}Q_{43} \end{split} \end{equation}
\begin{equation} \begin{split} R^1{}_{224}&= e^{\psi-2\mu_2-\mu_4}\big(\Psi_2-\frac{1}{2}\mu_{4:2}\big)Q_{24}-e^{-\psi-\mu_4}\mu_{2:4}\mu_{4,1}\\ &\quad+e^{-\mu_2-\mu_4}\big(e^{\mu_2-\psi}\mu_{2,1}\big)_{:4}+\frac{1}{2}e^{-\mu_2-\mu_4}\big(e^{\psi-\mu_2}Q_{24}\big)_{:2}\\ &\quad+\frac{1}{2}e^{\psi-2\mu_3-\mu_4}\mu_{2:3}Q_{34} \end{split} \end{equation}
\begin{equation} \begin{split} R^1{}_{332}&= e^{\psi-\mu_2-2\mu_3}\big(\Psi_3-\frac{1}{2}\mu_{2:3}\big)Q_{32}-e^{-\psi-\mu_2}\mu_{3:2}\mu_{2,1}\\ &\quad+e^{-\mu_2-\mu_3}\big(e^{\mu_3-\psi}\mu_{3,1}\big)_{:2}+\frac{1}{2}e^{-\mu_2-\mu_3}\big(e^{\psi-\mu_3}Q_{32}\big)_{:3}\\ &\quad+\frac{1}{2}e^{\psi-\mu_2-2\mu_4}\mu_{3:4}Q_{42} \end{split} \end{equation}
\begin{equation} \begin{split} R^1{}_{334}&= e^{\psi-2\mu_3-\mu_4}\big(\Psi_3-\frac{1}{2}\mu_{4:3}\big)Q_{34}-e^{-\psi-\mu_4}\mu_{3:4}\mu_{4,1}\\ &\quad+e^{-\mu_3-\mu_4}\big(e^{\mu_3-\psi}\mu_{3,1}\big)_{:4}+\frac{1}{2}e^{-\mu_3-\mu_4}\big(e^{\psi-\mu_3}Q_{34}\big)_{:3}\\ &\quad+\frac{1}{2}e^{\psi-2\mu_2-\mu_4}\mu_{3:2}Q_{24} \end{split} \end{equation}
\begin{equation} \begin{split} R^1{}_{442}&= e^{\psi-\mu_2-2\mu_4}\big(\Psi_4-\frac{1}{2}\mu_{2:4}\big)Q_{42}-e^{-\psi-\mu_2}\mu_{4:2}\mu_{2,1}\\ &\quad+e^{-\mu_2-\mu_4}\big(e^{\mu_4-\psi}\mu_{4,1}\big)_{:2}+\frac{1}{2}e^{-\mu_2-\mu_4}\big(e^{\psi-\mu_4}Q_{42}\big)_{:4}\\ &\quad+\frac{1}{2}e^{\psi-\mu_2-2\mu_3}\mu_{4:3}Q_{32} \end{split} \end{equation}
\begin{equation} \begin{split} R^1{}_{443}&= e^{\psi-\mu_3-2\mu_4}\big(\Psi_4-\frac{1}{2}\mu_{3:4}\big)Q_{43}-e^{-\psi-\mu_3}\mu_{4:3}\mu_{3,1}\\ &\quad+e^{-\mu_3-\mu_4}\big(e^{\mu_4-\psi}\mu_{4,1}\big)_{:3}+\frac{1}{2}e^{-\mu_3-\mu_4}\big(e^{\psi-\mu_4}Q_{43}\big)_{:4}\\ &\quad+\frac{1}{2}e^{\psi-2\mu_2-\mu_3}\mu_{4:2}Q_{23} \end{split} \end{equation}
最后 \Omega^1{}_A 只剩下一项 \theta^B\wedge\theta^C 。
\begin{equation} \begin{split} R^1{}_{ABC}&=+\frac{1}{2}e^{\psi-\mu_A-\mu_B-\mu_C}\Big[2\Psi_A- \big(\mu_{B:A}+\mu_{C:A}\big)\Big]Q_{BC} \\ &\quad+\frac{1}{2}e^{-\mu_B-\mu_C}\Big [\big (e^{\psi-\mu_A}Q_{AC}\big )_{:B}-\big(e^{\psi-\mu_A}Q_{AB}\big)_{:C}\Big ] \end{split} \end{equation}
R^1{}_{(ABC)}=0 正是比安基第一恒等式。
\begin{equation} \begin{split} R^1{}_{234}&=\frac{1}{2}e^{\psi-\mu_2-\mu_3-\mu_4}\Big[2\Psi_2-\big(\mu_{3:2}+\mu_{4:2}\big)\Big]Q_{34}\\ &\quad+\frac{1}{2}e^{-\mu_3-\mu_4}\Big [\big(e^{\psi-\mu_2}Q_{24}\big)_{:3}-\big(e^{\psi-\mu_2}Q_{23}\big)_{:4}\Big]\\ \end{split} \end{equation}
\begin{equation} \begin{split} R^1{}_{342}&=\frac{1}{2}e^{\psi-\mu_2-\mu_3-\mu_4}\Big[2\Psi_3-\big(\mu_{4:3}+\mu_{2:3}\big)\Big]Q_{42}\\ &\quad+\frac{1}{2}e^{-\mu_2-\mu_4}\Big [\big(e^{\psi-\mu_4}Q_{32}\big)_{:4}-\big(e^{\psi-\mu_4}Q_{34}\big)_{:2}\Big]\\ \end{split} \end{equation}
\begin{equation} \begin{split} R^1{}_{423}&=\frac{1}{2}e^{\psi-\mu_2-\mu_3-\mu_4}\Big[2\Psi_4-\big(\mu_{2:4}+\mu_{3:4}\big)\Big]Q_{23}\\ &\quad+\frac{1}{2}e^{-\mu_2-\mu_3}\Big [\big(e^{\psi-\mu_4}Q_{43}\big)_{:2}-\big(e^{\psi-\mu_4}Q_{42}\big)_{:3}\Big]\\ \end{split} \end{equation}
从 \Omega^A{}_B 的 \theta^1\wedge \theta^A 项、 \theta^1\wedge \theta^B 项、 \theta^1\wedge \theta^C 项，可以得到 R^A{}_{B1A} 、 R^A{}_{B1B} 和 R^A{}_{B1C} 。 R^A{}_{B1A} 和 -R^A{}_{B1B} 有相同的形式，只需要把 A 和 B 互换。它们同 \Omega^1{}_A 的 \theta^A\wedge \theta^B 项 R^1{}_{AAB} 相同。尽管表达式看上去有些不一样！
\begin{equation} \begin{split} R^A{}_{B1A}&=\frac{1}{2}e^{-\psi-\mu_A}\mathscr{D}_A\big(e^{2\psi-\mu_A-\mu_B}Q_{AB}\big)+e^{-\psi-\mu_A}\big (e^{\mu_A-\mu_B}\mu_{A:B}\big)_{,1}\\ &\quad-e^{-\psi-\mu_B}\mu_{A,1}\Psi_B+\frac{1}{2}e^{\psi-2\mu_A-\mu_B}\Psi_AQ_{AB}\\ &\quad-\frac{1}{2}e^{\psi-\mu_B-2\mu_C}\mu_{A:C}Q_{BC}\\ &\color{red}{V}\color{blue}{S}\\ R^1{}_{AAB}&=e^{\psi-2\mu_A-\mu_B}\big(\Psi_A-\frac{1}{2}\mu_{B:A}\big)Q_{AB}-e^{-\psi-\mu_B}\mu_{A:B}\mu_{B,1}\\ &\quad +e^{-\mu_A-\mu_B}\big (e^{\mu_A-\psi}\mu_{A,1}\big)_{:B}+\frac{1}{2}e^{-\mu_A-\mu_B}\big(e^{\psi-\mu_A}Q_{AB}\big )_{:A}\\ &\quad+\frac{1}{2}e^{\psi-\mu_B-2\mu_C}\mu_{A:C}Q_{CB} \end{split} \end{equation}
所以我们无需劳神列出 R^A{}_{B1A} 。
从 \Omega^A{}_B 的 \theta^A\wedge\theta^B 项
\begin{equation} \begin{split} R^A{}_{BAB}&=-e^{-\mu_A-\mu_B}\Big [\big(e^{\mu_A-\mu_B}\mu_{A:B}\big)_{:B}-\big(e^{-\mu_B-\mu_A}\mu_{B:A}\big)_{:A}\Big ]\\ &\quad-e^{-2\psi}\mu_{A,1}\mu_{B,1}-e^{-2\mu_C}\mu_{A:C}\mu_{B:C}-\frac{3}{4}e^{2\psi-2\mu_A-2\mu_B}Q_{AB}^2 \end{split} \end{equation}
可以得到 R^2{}_{323} 、 R^2{}_{424} 、 R^3{}_{434} 。
\begin{equation} \begin{split} R^2{}_{323}&=-e^{-\mu_2-\mu_3}\Big [\big(e^{\mu_2-\mu_3}\mu_{2:3}\big)_{:3}-\big(e^{\mu_3-\mu_2}\mu_{3:2}\big)_{:2}\Big ]\\ &\quad-e^{-2\psi}\mu_{2,1}\mu_{3,1}-e^{-2\mu_4}\mu_{2:4}\mu_{3:4}\\ &\quad-\frac{3}{4}e^{2\psi-2\mu_2-2\mu_3}Q_{23}^2 \end{split} \end{equation}
\begin{equation} \begin{split} R^2{}_{424}&=-e^{-\mu_2-\mu_4}\Big [\big(e^{\mu_2-\mu_4}\mu_{2:4}\big)_{:4}-\big(e^{\mu_4-\mu_2}\mu_{4:2}\big)_{:2}\Big ]\\ &\quad-e^{-2\psi}\mu_{2,1}\mu_{4,1}-e^{-2\mu_3}\mu_{2:3}\mu_{4:3}\\ &\quad-\frac{3}{4}e^{2\psi-2\mu_2-2\mu_4}Q_{24}^2 \end{split} \end{equation}
\begin{equation} \begin{split} R^3{}_{434}&=-e^{-\mu_3-\mu_4}\Big [\big(e^{\mu_3-\mu_4}\mu_{3:4}\big)_{:4}-\big(e^{\mu_4-\mu_3}\mu_{4:3}\big)_{:3}\Big ]\\ &\quad-e^{-2\psi}\mu_{3,1}\mu_{4,1}-e^{-2\mu_2}\mu_{3:2}\mu_{4:2}\\ &\quad-\frac{3}{4}e^{2\psi-2\mu_3-2\mu_4}Q_{34}^2 \end{split} \end{equation}
从 \Omega^A{}_B 的 \theta^A\wedge \theta^C 项
\begin{equation} \begin{split} R^A{}_{BAC}&=-e^{-\mu_A-\mu_C}\big(e^{\mu_A-\mu_B}\mu_{A:B}\big)_{:C}+e^{-\mu_B-\mu_C}\mu_{A:C}\mu_{C:B}\\ &\quad+\frac{1}{2}e^{-\mu_B-\mu_C}\mu_{A,1}Q_{BC}-\frac{3}{4}e^{2\psi-2\mu_A-\mu_B-\mu_C}Q_{AB}Q_{AC} \end{split} \end{equation}
稍作化简后为
\begin{equation} \begin{split} R^A{}_{BAC}&=-e^{-\mu_B-\mu_C}\Big[\mu_{A:B:C}+\mu_{A:B}\big(\mu_A-\mu_B\big)_{:C}-\mu_{A:C}\mu_{C:B}\Big]\\ &\quad+\frac{1}{2}e^{-\mu_B-\mu_C}\mu_{A,1}Q_{BC}-\frac{3}{4}e^{2\psi-2\mu_A-\mu_B-\mu_C}Q_{AB}Q_{AC} \end{split} \end{equation}
\begin{equation} \begin{split} R^2{}_{324}&=-e^{-\mu_3-\mu_4}\Big[\mu_{2:3:4}+\mu_{2:3}\big(\mu_2-\mu_3\big)_{:4}-\mu_{2:4}\mu_{4:3}\Big]\\ &\quad+\frac{1}{2}e^{-\mu_3-\mu_4}\mu_{2,1}Q_{34}-\frac{3}{4}e^{2\psi-2\mu_2-\mu_3-\mu_4}Q_{23}Q_{24} \end{split} \end{equation}
\begin{equation} \begin{split} R^3{}_{432}&=-e^{-\mu_2-\mu_4}\Big[\mu_{3:4:2}+\mu_{3:4}\big(\mu_3-\mu_4\big)_{:2}-\mu_{3:2}\mu_{2:4}\Big]\\ &\quad+\frac{1}{2}e^{-\mu_2-\mu_4}\mu_{3,1}Q_{42}-\frac{3}{4}e^{2\psi-\mu_2-2\mu_3-\mu_4}Q_{34}Q_{32} \end{split} \end{equation}
\begin{equation} \begin{split} R^4{}_{243}&=-e^{-\mu_2-\mu_3}\Big[\mu_{4:2:3}+\mu_{4:2}\big(\mu_4-\mu_2\big)_{:3}-\mu_{4:3}\mu_{3:2}\Big]\\ &\quad+\frac{1}{2}e^{-\mu_2-\mu_3}\mu_{4,1}Q_{23}-\frac{3}{4}e^{2\psi-2\mu_4-\mu_2-\mu_3}Q_{42}Q_{43} \end{split} \end{equation}
从 \Omega^A{}_B 的 \theta^B\wedge \theta^C 项 获得的 R^A{}_{BBC} 和 -R^A{}_{BAC} 形式相同。
<hr/> |
