![跨越多行的方程中的多重对齐](https://linux22.com/image/429798/%E8%B7%A8%E8%B6%8A%E5%A4%9A%E8%A1%8C%E7%9A%84%E6%96%B9%E7%A8%8B%E4%B8%AD%E7%9A%84%E5%A4%9A%E9%87%8D%E5%AF%B9%E9%BD%90.png)
答案1
两种可能性aligned
:
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{align*}
R(X,Y) e_j&=\nabla_X \nabla_Y e_j - \nabla_Y \nabla_X e_j - \nabla_{[X,Y]} e_j \\
& =\begin{aligned}[t] \bigl( X \omega^i_j(Y) & - Y \omega^i_j(X) - \omega^i_j([X,Y]) \bigr) e_i \\
& + \bigl(\omega^i_k(X) \omega^k_j(Y) - \omega^i_k(Y) \omega^k_j(X) \bigr) e_i
\end{aligned}
\end{align*}
\begin{align*}
R(X,Y) e_j&= \nabla_X \nabla_Y e_j - \nabla_Y \nabla_X e_j - \nabla_{[X,Y]} e_j \\
& =\begin{aligned}[t] \bigl( X \omega^i_j(Y) - Y \omega^i_j(X) & - \omega^i_j([X,Y]) \bigr) e_i \\
{} + \bigl(\omega^i_k(X) \omega^k_j(Y) & - \omega^i_k(Y) \omega^k_j(X) \bigr) e_i
\end{aligned}
\end{align*}
\end{document}