我正在尝试排版这样的内容:
\begin{align} f_{**p}(X_p,Y_p) = X_p(Y(f)) = \sum_{i,j = 1} ^n \alpha_i
\beta_j E_{pi}(E_j(f)) f
\end{align}
short remark
\begin{align}
E_{pi}(E_j(f)) &= \frac{\partial}{\partial x_i}(E_j(f) \circ \phi^{-1} )|_{\phi(p)} \\
&= \frac{\partial}{\partial x_i} ...
\end{align}
但我想在\shortintertext
同一个对齐环境中提出简短的评论,即:
\begin{align} f_{**p}(X_p,Y_p) = X_p(Y(f)) = \sum_{i,j = 1} ^n \alpha_i
\beta_j E_{pi}(E_j(f)) f
\shortintertext{short remark}
E_{pi}(E_j(f)) &= \frac{\partial}{\partial x_i}(E_j(f) \circ \phi^{-1} )|_{\phi(p)} \\
&= \frac{\partial}{\partial x_i} ...
\end{align}
但是由于某种原因,最后一个的间距太糟糕了。我该如何解决这个问题?
谢谢你的帮助。
答案1
如果您不想让第一个方程与其他方程对齐,请使用:
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{gather}
f_{**p}(X_p,Y_p) = X_p(Y(f)) = \sum_{i,j = 1} ^n \alpha_i
\beta_j E_{pi}(E_j(f)) f
\\
\shortintertext{short remark}
\begin{align}
E_{pi}(E_j(f)) &= \frac{\partial}{\partial x_i}(E_j(f) \circ \phi^{-1} )|_{\phi(p)} \\
&= \frac{\partial}{\partial x_i}
\end{align}
\end{gather}
\end{document}
答案2
你漏掉了&
第一个等式中的 a 和\\
之前的a\shortintertext{short remark}
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{align}
f_{**p}(X_p,Y_p) &= X_p(Y(f)) = \sum_{i,j = 1} ^n \alpha_i
\beta_j E_{pi}(E_j(f)) f\\
\shortintertext{short remark}
E_{pi}(E_j(f)) &= \frac{\partial}{\partial x_i}(E_j(f) \circ \phi^{-1} )|_{\phi(p)} \\
&= \frac{\partial}{\partial x_i}
\end{align}
\end{document}
根据 Op 的要求:
您可以\span\omit
如下使用:
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{equation}
f_{**p}(X_p,Y_p) = X_p(Y(f)) = \sum_{i,j = 1} ^n \alpha_i
\beta_j E_{pi}(E_j(f)) f
\end{equation}
\begin{align}
f_{**p}(X_p,Y_p) = X_p(Y(f)) = \sum_{i,j = 1} ^n \alpha_i
\beta_j E_{pi}(E_j(f)) f \span\omit\\
\shortintertext{short remark}
E_{pi}(E_j(f)) &= \frac{\partial}{\partial x_i}(E_j(f) \circ \phi^{-1} )|_{\phi(p)} \\
&= \frac{\partial}{\partial x_i}
\end{align}
\end{document}