如何在 LaTeX 中的 align* 多个长方程式中使用大括号？

Question

\left(这是我的代码，但是当我将和更改\right)为\Big)和时它不起作用\Big)

\begin{align*}
\bar{h}_{m}^{jk}&=\omega_{k}h_{m}^{jk}(a)\\
&=\left(  \omega_{k}\left\{
\zeta_{h}\mid\zeta_{h}\in p_{1}\circ h_{m}^{jk}(a)\right\}  ,\omega
_{k}\left\{  \zeta_{h}\mid\zeta_{h}\in p_{2}\circ h_{m}^{jk}(a)\right\}
,\dots, \right.\\
 &\qquad \left.\omega_{k}\left\{  \zeta_{h}\mid\zeta_{h}\in p_{m}\circ h_{m}%
^{jk}(a)\right\}  \right)  \\
&=\left(  \left\{  \bar{\zeta}_{h}\mid\bar{\ze
}_{h}\in p_{1}\circ h_{m}^{jk}(a)\right\}  ,\left\{  \bar{\zeta}_{h}\mid
\bar{\zeta}_{h}\in p_{2}\circ h_{m}^{jk}(a)\right\}  ,\dots,\left\{  \bar{\zeta
}_{h}\mid\bar{\zeta}_{h}\in p_{m}\circ h_{m}^{jk}(a)\right\}  \right)  .
\end{align*}

Answer 1

每个\left总是与 配对\right，并且这种配对不能跨越&或\\。无论谁编写了你的​​代码，实际上（正确地）将\left(和\right.配对\left.；\right)这.意味着 TeX 不会排版栅栏，但现在知道要测量多少尺寸。

更改\left(为\Big(和\right)意味着\Big)您将保留\left.和\right.，而 TeX 无法理解您要配对的内容。解决方案是，当您更改\left(为\Big(和\right)时\Big)，您还需要删除现在多余的\left.和\right.。

顺便说一句，出于间距目的，您可能需要使用\Bigl(and \Bigr)。这告诉 TeX 它正在处理左或右栅栏。

Answer 2

像这样？

\documentclass{article}
\usepackage{mathtools}

\begin{document}
\begin{align*}
\bar{h}_{m}^{jk}
    & = \omega_{k}h_{m}^{jk}(a)\\
    & = \begin{multlined}[t]
    \biggl(\omega_{k}\Bigl\{
        \zeta_{h}\mid\zeta_{h}\in p_{1}\circ h_{m}^{jk}(a)\Bigr\},\omega_{k}
            \Bigl\{\zeta_{h}\mid\zeta_{h}\in p_{2}\circ h_{m}^{jk}(a)
                    \Bigr\}, \dots \\
    \dots, 
    \omega_{k}\Bigl\{\zeta_{h}\mid\zeta_{h}\in p_{m}\circ h_{m}^{jk}(a)
              \Bigr\}
    \biggr)
    \end{multlined} \\
    & = \begin{multlined}[t]
    \biggl(\Bigl\{\bar{\zeta}_{h}\mid\bar{\zeta}_{h}\in p_{1}\circ h_{m}^{jk}(a)\Bigr\},
           \Bigl\{\bar{\zeta}_{h}\mid\bar{\zeta}_{h}\in p_{2}\circ h_{m}^{jk}(a)
           \Bigr\},\dots         \\
    \dots,
           \Bigl\{\bar{\zeta}_{h}\mid\bar{\zeta}_{h}\in p_{m}\circ h_{m}^{jk}(a)\Bigr\}
    \biggr).
    \end{multlined}
\end{align*}
\end{document}

（红线表示文字边框）

附录： 使用aligned代替multilined可以定义锚点，其中 是多线方程的对齐线：

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\begin{align*}
\bar{h}_{m}^{jk}
    & = \omega_{k}h_{m}^{jk}(a)\\
    & = \begin{aligned}[t]
        \biggl(\omega_{k}\Bigl\{
        \zeta_{h}\mid\zeta_{h}\in p_{1}\circ h_{m}^{jk}(a)\Bigr\},
        &   \;\omega_{k}\Bigl\{\zeta_{h}\mid\zeta_{h}\in p_{2}\circ h_{m}^{jk}(a)\Bigr\}, \dots \\
    \dots,
        &   \;\omega_{k}\Bigl\{\zeta_{h}\mid\zeta_{h}\in p_{m}\circ h_{m}^{jk}(a)\Bigr\}\biggr)
        \end{aligned} \\
    & = \begin{aligned}[t]
        \biggl(\Bigl\{\bar{\zeta}_{h}\mid\bar{\zeta}_{h}\in p_{1}\circ h_{m}^{jk}(a)\Bigr\},
        &\; \Bigl\{\bar{\zeta}_{h}\mid\bar{\zeta}_{h}\in p_{2}\circ h_{m}^{jk}(a)\Bigr\},\dots \\
    \dots,
        &\; \Bigl\{\bar{\zeta}_{h}\mid\bar{\zeta}_{h}\in p_{m}\circ h_{m}^{jk}(a)\Bigr\}\biggr).
    \end{aligned}
\end{align*}
\end{document}

Answer 3

我提出了这个变体，其代码稍微简单一些，基于 fleqnfrom nccmath、\DeclarePairedDelimiter命令 frommathtools和multlined环境（同一个包）。

\documentclass[svgnames]{article}
\usepackage{showframe}
\renewcommand{\ShowFrameLinethickness}{0.4pt}
\renewcommand{\ShowFrameColor}{\color{Coral}}
 \usepackage{nccmath, mathtools}
 \DeclarePairedDelimiter\set\{\}

\begin{document}
\leavevmode\vskip 1cm

\begin{fleqn}
\begin{align*}
\bar{h}_{m}^{jk} &=\omega_{k}h_{m}^{jk}(a) \\
&=\begin{multlined}[t][0.9\linewidth]\Bigl( \omega_{k}\set*{\zeta_{h}\mid\zeta_{h}\in p_{1}\circ h_{m}^{jk}(a)} ,\omega
_{k}\set{ \zeta_{h}\mid\zeta_{h}\in p_{2}\circ h_{m}^{jk}(a)},\dots \\[-1.5ex]
\dots, \omega_{k}\set*{ \zeta_{h}\mid\zeta_{h}\in p_{m}\circ h_{m}^{jk}(a)}\Bigr) \end{multlined} \\
&= \begin{multlined}[t][0.9\linewidth]\Bigl( \set*{ \bar{\zeta}_{h}\mid\bar{\zeta}_{h}\in p_{1}\circ h_{m}^{jk}(a)} ,%
\set*{ \bar{\zeta}_{h}\mid \bar{\zeta}_{h}\in p_{2}\circ h_{m}^{jk}(a)} ,\dots \phantom{.} \\[-1.5ex]
\dots, \set*{ \bar{\zeta}_{h}\mid\bar{\zeta}_{h}\in p_{m}\circ h_{m}^{jk}(a)} \Bigr).
\end{multlined}
\end{align*}
\end{fleqn}

\end{document}