我试图让 3 个很长的等式（我已将其拆分）以某种方式更优雅地与它们各自的等号对齐。缩短这 3 个等式的 LHS（也许将其拆分 3 次？）也有助于将完整的推导更对齐到页面的中心。我尊重其他人对“更优雅”的品味。
我尝试了这里讨论的方法以及附加的问题，但无济于事对齐环境中的一个多线方程和多个单线方程
\begin{align}
0&= 0+0+0 \\ \notag
0 &= R_{\mu\nu}{}^a e_{\rho a} + R_{\rho\mu}{}^a e_{\nu a} -R_{\nu\rho}{}^a e_{\mu a} \\ \notag
R_{\nu\rho}{}^a e_{\mu a} &= R_{\mu\nu}{}^a e_{\rho a} + R_{\rho\mu}{}^a e_{\nu a} \\ \notag
(\partial_{[\nu} e_{\rho]}{}^a - \omega_{[\nu}{}^{ab} e_{\rho]}{}_b) e_{\mu a} &= (\partial_{[\mu} e_{\nu]}{}^a - \omega_{[\mu}{}^{ab} e_{\nu]}{}_b) e_{\rho a} + (\partial_{[\rho} e_{\mu]}{}^a - \omega_{[\rho}{}^{ab} e_{\mu]}{}_b) e_{\nu a} \\ \notag
\partial_{[\nu} e_{\rho]}{}^a e_{\mu a} - \omega_{[\nu}{}^{ab} e_{\rho]}{}_b e_{\mu a}&= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} - \omega_{[\mu}{}^{ab} e_{\nu]}{}_b e_{\rho a} + \partial_{[\rho} e_{\mu]}{}^a e_{\nu a} - \omega_{[\rho}{}^{ab} e_{\mu]}{}_b e_{\nu a} \\ \notag
\omega_{[\mu}{}^{ab} e_{\nu]}{}{}_b e_{\rho a} + \omega_{[\rho}{}^{ab} e_{\mu]}{}_b e_{\nu a} - \omega_{[\nu}{}^{ab} e_{\rho]}{}_b e_{\mu a} &= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} + \partial_{[\rho} e_{\mu]}{}^a e_{\nu a} - \partial_{[\nu} e_{\rho]}{}^a e_{\mu a} \\ \notag
\frac{1}{2}\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} - \frac{1}{2}\omega_{\nu}{}^{ab} e_{\mu}{}{}_b e_{\rho a} +\frac{1}{2} \omega_{\rho}{}^{ab} e_{\mu}{}_b e_{\nu a} \\ \notag -\frac{1}{2}\omega_{\mu}{}^{ab} e_{\rho}{}_b e_{\nu a}- \frac{1}{2}\omega_{\nu}{}^{ab} e_{\rho}{}_b e_{\mu a} +\frac{1}{2}\omega_{\rho}{}^{ab} e_{\nu}{}_b e_{\mu a}&= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} + \partial_{[\rho} e_{\mu]}{}^a e_{\nu a} - \partial_{[\nu} e_{\rho]}{}^a e_{\mu a}
\\ \notag
\bigg(\frac{1}{2}\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} -\frac{1}{2}\omega_{\mu}{}^{ba} e_{\rho}{}_a e_{\nu b}\bigg) + \bigg( \frac{1}{2}\omega_{\rho}{}^{ab} e_{\nu}{}_b e_{\mu a} \\ \notag + \frac{1}{2} \omega_{\rho}{}^{ba} e_{\mu}{}_a e_{\nu b}\bigg)+ \bigg(-\frac{1}{2}\omega_{\nu}{}^{ab} e_{\mu}{}{}_b e_{\rho a} - \frac{1}{2}\omega_{\nu}{}^{ba} e_{\rho}{}_a e_{\mu b}\bigg) &= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} + \partial_{[\rho} e_{\mu]}{}^a e_{\nu a} - \partial_{[\nu} e_{\rho]}{}^a e_{\mu a}
\\ \notag
\bigg(\frac{1}{2}\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} +\frac{1}{2}\omega_{\mu}{}^{ab} e_{\rho}{}_a e_{\nu b}\bigg) + \bigg( \frac{1}{2}\omega_{\rho}{}^{ab} e_{\nu}{}_b e_{\mu a} \\ \notag - \frac{1}{2} \omega_{\rho}{}^{ab} e_{\mu}{}_a e_{\nu b}\bigg)+ \bigg(-\frac{1}{2}\omega_{\nu}{}^{ab} e_{\mu}{}{}_b e_{\rho a} + \frac{1}{2}\omega_{\nu}{}^{ab} e_{\rho}{}_a e_{\mu b}\bigg) &= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} + \partial_{[\rho} e_{\mu]}{}^a e_{\nu a} - \partial_{[\nu} e_{\rho]}{}^a e_{\mu a}
\\ \notag
\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} + 0 + 0 &= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} + \partial_{[\rho} e_{\mu]}{}^c e_{\nu c} - \partial_{[\nu} e_{\rho]}{}^c e_{\mu c}
\\ \notag
\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} e^{\rho a} e^{\nu b}&= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a}e^{\rho a} e^{\nu b} + \partial_{[\rho} e_{\mu]}{}^c e_{\nu c}e^{\rho a} e^{\nu b} - \partial_{[\nu} e_{\rho]}{}^c e_{\mu c} e^{\rho a} e^{\nu b}
\\ \notag
\omega_{\mu}{}^{ab} &= \partial_{[\mu} e_{\nu]}{}^a e^{\nu b} + \partial_{[\rho} e_{\mu]}{}^c \delta_c^b e^{\rho a} - \partial_{[\nu} e_{\rho]}{}^c e_{\mu c} e^{\rho a} e^{\nu b}
\\ \notag
\omega_{\mu}{}^{ab} &= \partial_{[\mu} e_{\nu]}{}^a e^{\nu b} + \partial_{[\rho} e_{\mu]}{}^b e^{\rho a} - \partial_{[\nu} e_{\rho]}{}^c e_{\mu c} e^{\rho a} e^{\nu b}
\\ \notag
\omega_{\mu}{}^{ab} &= e^{\nu b}\partial_{[\mu} e_{\nu]}{}^a + e^{\nu a}\partial_{[\nu} e_{\mu]}{}^b -e^{\rho a} e^{\lambda b} e_{\mu c} \partial_{[\lambda} e_{\rho]}{}^c
\\ \notag
\omega_{\mu}{}^{ab} &= 2 e^{\nu [b}\partial_{[\mu} e_{\nu]}{}^{a]} +e^{\rho a} e^{\lambda b} e_{\mu c} \partial_{[\rho} e_{\lambda]}{}^c
\\ \notag
\omega_{\mu}{}^{ab} &= -2 e^{\nu [a}\partial_{[\mu} e_{\nu]}{}^{b]} + e^{\rho a} e^{\lambda b} e_{\mu c} \partial_{[\rho} e_{\lambda]}{}^c
\end{align}
我的第一反应是使用对齐，但我不知道如何实现这一点
有人能帮忙使我上面代码中的 3 个多行（如果不是很长）等式更漂亮吗？
答案1
一些建议：
省略所有
\bigg大小指令并省略相关的左括号和右括号。将三个长方程的左边部分分成三部分而不是两部分，然后使用
\qquad和\quad指令将第一行和第二行“推”到左边，形成略微交错的外观。始终使用
\tfrac{1}{2}而不是\frac{1}{2}。在三个三行方程式的组前后添加一些空格。
不要在任何给定行上放置大量材料（例如较长的数学表达式）。这样，错误消息中给出的行信息将有助于加快调试过程，因为任何给定行上可能出错的内容较少。
\documentclass{article}
\usepackage[margin=2.5cm]{geometry} % set page size parameters suitably
\usepackage{amsmath}
\allowdisplaybreaks
\begin{document}
\begin{align*}
0 &= 0+0+0 \refstepcounter{equation} \tag{\theequation}
\\
0 &= R_{\mu\nu}{}^a e_{\rho a} +
R_{\rho\mu}{}^a e_{\nu a} -
R_{\nu\rho}{}^a e_{\mu a}
\\
R_{\nu\rho}{}^a e_{\mu a}
&= R_{\mu\nu}{}^a e_{\rho a} + R_{\rho\mu}{}^a e_{\nu a}
\\
(\partial_{[\nu} e_{\rho]}{}^a -
\omega_{[\nu}{}^{ab} e_{\rho]}{}_b) e_{\mu a}
&= (\partial_{[\mu} e_{\nu]}{}^a -
\omega_{[\mu}{}^{ab} e_{\nu]}{}_b) e_{\rho a} +
(\partial_{[\rho} e_{\mu]}{}^a -
\omega_{[\rho}{}^{ab} e_{\mu]}{}_b) e_{\nu a}
\\
\partial_{[\nu} e_{\rho]}{}^a e_{\mu a} -
\omega_{[\nu}{}^{ab} e_{\rho]}{}_b e_{\mu a}
&= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} -
\omega_{[\mu}{}^{ab} e_{\nu]}{}_b e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^a e_{\nu a} -
\omega_{[\rho}{}^{ab} e_{\mu]}{}_b e_{\nu a}
\\
\omega_{[\mu}{}^{ab} e_{\nu]}{}{}_b e_{\rho a} +
\omega_{[\rho}{}^{ab} e_{\mu]}{}_b e_{\nu a} -
\omega_{[\nu}{}^{ab} e_{\rho]}{}_b e_{\mu a}
&= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^a e_{\nu a} -
\partial_{[\nu} e_{\rho]}{}^a e_{\mu a}
\\[1ex]
\tfrac{1}{2}\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} -
\tfrac{1}{2}\omega_{\nu}{}^{ab} e_{\mu}{}{}_b e_{\rho a} \qquad& \\
{}+\tfrac{1}{2} \omega_{\rho}{}^{ab} e_{\mu}{}_b e_{\nu a}
-\tfrac{1}{2}\omega_{\mu}{}^{ab} e_{\rho}{}_b e_{\nu a} \quad& \\
{}-\tfrac{1}{2}\omega_{\nu}{}^{ab} e_{\rho}{}_b e_{\mu a}
+\tfrac{1}{2}\omega_{\rho}{}^{ab} e_{\nu}{}_b e_{\mu a}
&= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^a e_{\nu a} -
\partial_{[\nu} e_{\rho]}{}^a e_{\mu a}
\\
\tfrac{1}{2}\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} -
\tfrac{1}{2}\omega_{\mu}{}^{ba} e_{\rho}{}_a e_{\nu b} \qquad& \\
{}+\tfrac{1}{2}\omega_{\rho}{}^{ab} e_{\nu}{}_b e_{\mu a}
+\tfrac{1}{2} \omega_{\rho}{}^{ba} e_{\mu}{}_a e_{\nu b} \quad& \\
{}-\tfrac{1}{2}\omega_{\nu}{}^{ab} e_{\mu}{}{}_b e_{\rho a}
-\tfrac{1}{2}\omega_{\nu}{}^{ba} e_{\rho}{}_a e_{\mu b}
&= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^a e_{\nu a} -
\partial_{[\nu} e_{\rho]}{}^a e_{\mu a}
\\
\tfrac{1}{2}\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} +
\tfrac{1}{2}\omega_{\mu}{}^{ab} e_{\rho}{}_a e_{\nu b} \qquad& \\
{}+\tfrac{1}{2}\omega_{\rho}{}^{ab} e_{\nu}{}_b e_{\mu a}
-\tfrac{1}{2} \omega_{\rho}{}^{ab} e_{\mu}{}_a e_{\nu b} \quad& \\
{}-\tfrac{1}{2}\omega_{\nu}{}^{ab} e_{\mu}{}{}_b e_{\rho a}
+\tfrac{1}{2}\omega_{\nu}{}^{ab} e_{\rho}{}_a e_{\mu b}
&= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^a e_{\nu a} -
\partial_{[\nu} e_{\rho]}{}^a e_{\mu a}
\\[1ex]
\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} + 0 + 0
&= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^c e_{\nu c} -
\partial_{[\nu} e_{\rho]}{}^c e_{\mu c}
\\
\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} e^{\rho a} e^{\nu b}
&= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a}e^{\rho a} e^{\nu b} +
\partial_{[\rho} e_{\mu]}{}^c e_{\nu c}e^{\rho a} e^{\nu b} -
\partial_{[\nu} e_{\rho]}{}^c e_{\mu c} e^{\rho a} e^{\nu b}
\\
\omega_{\mu}{}^{ab}
&= \partial_{[\mu} e_{\nu]}{}^a e^{\nu b} +
\partial_{[\rho} e_{\mu]}{}^c \delta_c^b e^{\rho a} -
\partial_{[\nu} e_{\rho]}{}^c e_{\mu c} e^{\rho a} e^{\nu b}
\\
\omega_{\mu}{}^{ab}
&= \partial_{[\mu} e_{\nu]}{}^a e^{\nu b} +
\partial_{[\rho} e_{\mu]}{}^b e^{\rho a} -
\partial_{[\nu} e_{\rho]}{}^c e_{\mu c} e^{\rho a} e^{\nu b}
\\
\omega_{\mu}{}^{ab}
&= e^{\nu b}\partial_{[\mu} e_{\nu]}{}^a +
e^{\nu a}\partial_{[\nu} e_{\mu]}{}^b -
e^{\rho a} e^{\lambda b} e_{\mu c} \partial_{[\lambda} e_{\rho]}{}^c
\\
\omega_{\mu}{}^{ab}
&= 2 e^{\nu [b}\partial_{[\mu} e_{\nu]}{}^{a]} +
e^{\rho a} e^{\lambda b} e_{\mu c} \partial_{[\rho} e_{\lambda]}{}^c
\\
\omega_{\mu}{}^{ab}
&= -2 e^{\nu [a}\partial_{[\mu} e_{\nu]}{}^{b]}
+e^{\rho a} e^{\lambda b} e_{\mu c} \partial_{[\rho} e_{\lambda]}{}^c
\end{align*}
\end{document}
答案2
如果没有任何解释，推导过程就很难理解；在我看来，对齐等号并没有真正的帮助。
我建议始终采用左对齐，将较长的方程式在等号处分开，并稍微向右移动。
\documentclass{article}
\usepackage[a4paper,margin=2.5cm]{geometry}
\usepackage{amsmath,mathtools}
\begin{document}
\begin{equation}
\begin{aligned}[t]
& 0 = 0+0+0 \\[1ex]
& 0 = R_{\mu\nu}{}^a e_{\rho a} + R_{\rho\mu}{}^a e_{\nu a} -R_{\nu\rho}{}^a e_{\mu a}
\\[1ex]
& R_{\nu\rho}{}^a e_{\mu a} = R_{\mu\nu}{}^a e_{\rho a} + R_{\rho\mu}{}^a e_{\nu a}
\\[1ex]
& (
\partial_{[\nu} e_{\rho]}{}^a -
\omega_{[\nu}{}^{ab} e_{\rho]}{}_b
) e_{\mu a}
= (
\partial_{[\mu} e_{\nu]}{}^a - \omega_{[\mu}{}^{ab} e_{\nu]}{}_b
) e_{\rho a} +
(
\partial_{[\rho} e_{\mu]}{}^a - \omega_{[\rho}{}^{ab} e_{\mu]}{}_b
) e_{\nu a}
\\[1ex]
& \partial_{[\nu} e_{\rho]}{}^a e_{\mu a} -
\omega_{[\nu}{}^{ab} e_{\rho]}{}_b e_{\mu a}
= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} -
\omega_{[\mu}{}^{ab} e_{\nu]}{}_b e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^a e_{\nu a} -
\omega_{[\rho}{}^{ab} e_{\mu]}{}_b e_{\nu a}
\\[1ex]
& \omega_{[\mu}{}^{ab} e_{\nu]}{}{}_b e_{\rho a} +
\omega_{[\rho}{}^{ab} e_{\mu]}{}_b e_{\nu a} -
\omega_{[\nu}{}^{ab} e_{\rho]}{}_b e_{\mu a}
= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^a e_{\nu a} -
\partial_{[\nu} e_{\rho]}{}^a e_{\mu a} \\[1ex]
& \tfrac{1}{2}\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} -
\tfrac{1}{2}\omega_{\nu}{}^{ab} e_{\mu}{}{}_b e_{\rho a} +
\tfrac{1}{2} \omega_{\rho}{}^{ab} e_{\mu}{}_b e_{\nu a} -
\tfrac{1}{2}\omega_{\mu}{}^{ab} e_{\rho}{}_b e_{\nu a} -
\tfrac{1}{2}\omega_{\nu}{}^{ab} e_{\rho}{}_b e_{\mu a} +
\tfrac{1}{2}\omega_{\rho}{}^{ab} e_{\nu}{}_b e_{\mu a} \\
& \qquad = \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^a e_{\nu a} -
\partial_{[\nu} e_{\rho]}{}^a e_{\mu a}
\\[1ex]
&\bigl(
\tfrac{1}{2}\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} -
\tfrac{1}{2}\omega_{\mu}{}^{ba} e_{\rho}{}_a e_{\nu b}
\bigr) +
\bigl(
\tfrac{1}{2}\omega_{\rho}{}^{ab} e_{\nu}{}_b e_{\mu a} +
\tfrac{1}{2} \omega_{\rho}{}^{ba} e_{\mu}{}_a e_{\nu b}
\bigr) +
\bigl(
-\tfrac{1}{2}\omega_{\nu}{}^{ab} e_{\mu}{}{}_b e_{\rho a} -
\tfrac{1}{2}\omega_{\nu}{}^{ba} e_{\rho}{}_a e_{\mu b}
\bigr)\\
& \qquad= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^a e_{\nu a} -
\partial_{[\nu} e_{\rho]}{}^a e_{\mu a}
\\[1ex]
&\bigl(
\tfrac{1}{2}\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} +
\tfrac{1}{2}\omega_{\mu}{}^{ab} e_{\rho}{}_a e_{\nu b}
\bigr) +
\bigl(
\tfrac{1}{2}\omega_{\rho}{}^{ab} e_{\nu}{}_b e_{\mu a} -
\tfrac{1}{2} \omega_{\rho}{}^{ab} e_{\mu}{}_a e_{\nu b}
\bigr) +
\bigl(
-\tfrac{1}{2}\omega_{\nu}{}^{ab} e_{\mu}{}{}_b e_{\rho a} +
\tfrac{1}{2}\omega_{\nu}{}^{ab} e_{\rho}{}_a e_{\mu b}
\bigr)\\
& \qquad= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^a e_{\nu a} -
\partial_{[\nu} e_{\rho]}{}^a e_{\mu a}
\\[1ex]
&\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} + 0 + 0
= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} +
\partial_{[\rho} e_{\mu]}{}^c e_{\nu c} -
\partial_{[\nu} e_{\rho]}{}^c e_{\mu c}
\\[1ex]
&\omega_{\mu}{}^{ab} e_{\nu}{}{}_b e_{\rho a} e^{\rho a} e^{\nu b}
= \partial_{[\mu} e_{\nu]}{}^a e_{\rho a}e^{\rho a} e^{\nu b} +
\partial_{[\rho} e_{\mu]}{}^c e_{\nu c}e^{\rho a} e^{\nu b} -
\partial_{[\nu} e_{\rho]}{}^c e_{\mu c} e^{\rho a} e^{\nu b}
\\[1ex]
&\omega_{\mu}{}^{ab}
= \partial_{[\mu} e_{\nu]}{}^a e^{\nu b} +
\partial_{[\rho} e_{\mu]}{}^c \delta_c^b e^{\rho a} -
\partial_{[\nu} e_{\rho]}{}^c e_{\mu c} e^{\rho a} e^{\nu b}
\\[1ex]
&\omega_{\mu}{}^{ab}
= \partial_{[\mu} e_{\nu]}{}^a e^{\nu b} +
\partial_{[\rho} e_{\mu]}{}^b e^{\rho a} -
\partial_{[\nu} e_{\rho]}{}^c e_{\mu c} e^{\rho a} e^{\nu b}
\\[1ex]
&\omega_{\mu}{}^{ab}
= e^{\nu b}\partial_{[\mu} e_{\nu]}{}^a +
e^{\nu a}\partial_{[\nu} e_{\mu]}{}^b -
e^{\rho a} e^{\lambda b} e_{\mu c} \partial_{[\lambda} e_{\rho]}{}^c
\\[1ex]
&\omega_{\mu}{}^{ab}
= 2 e^{\nu [b}\partial_{[\mu} e_{\nu]}{}^{a]} +
e^{\rho a} e^{\lambda b} e_{\mu c} \partial_{[\rho} e_{\lambda]}{}^c
\\[1ex]
&\omega_{\mu}{}^{ab}
= -2 e^{\nu [a}\partial_{[\mu} e_{\nu]}{}^{b]} +
e^{\rho a} e^{\lambda b} e_{\mu c} \partial_{[\rho} e_{\lambda]}{}^c
\end{aligned}
\end{equation}
\end{document}
方程式之间的一些额外间距有助于区分它们。
不过，我真的希望每个主要步骤都有评论。
