align* 中的空白来自哪里？

Question

我正在尝试使用align*环境来对齐两个简单方程：

\begin{align*}
\phi(\gedge{\tilde{v}_{i_1}}{\tilde{u}_i}) & = \phi(\gedge{\tilde{v}_{i_2}}{\tilde{u}_i}) & = \gedge{v}{u} \\
\phi(\gedge{\tilde{u}_i}{\tilde{v}_{i_1}}) & = \phi(\gedge{\tilde{u}_i}{\tilde{v}_{i_2}}) & = \gedge{u}{v} & \qquad\mathrm{(z\,wl.\,2)}\\
\end{align*}

不幸的是我得到的结果很糟糕：

为什么第二个标志前有这么多的水平空间=?!?

不要担心\gedge，它只是一个箭头：

% graph edge
\newcommand{\gedge}[2]{%
#1 \to #2%
}%

Answer 1

所以你有两个选择，

1. 使用对齐

   \begin{align*}
   \phi(\gedge{\tilde{v}_{i_1}}{\tilde{u}_i}) & = \phi(\gedge{\tilde{v}_{i_2}}{\tilde{u}_i}) = \gedge{v}{u} \\
   \phi(\gedge{\tilde{u}_i}{\tilde{v}_{i_1}}) & = \phi(\gedge{\tilde{u}_i}{\tilde{v}_{i_2}}) = \gedge{u}{v}  \qquad\mathrm{(z\,wl.\,2)}\\
   \end{align*}
   

2. 使用 alignat

   \begin{alignat*}{4}
   \phi(\gedge{\tilde{v}_{i_1}}{\tilde{u}_i}) & = \phi(\gedge{\tilde{v}_{i_2}}{\tilde{u}_i}) && = \gedge{v}{u} \\
   \phi(\gedge{\tilde{u}_i}{\tilde{v}_{i_1}}) & = \phi(\gedge{\tilde{u}_i}{\tilde{v}_{i_2}}) && = \gedge{u}{v} & \qquad\mathrm{(z\,wl.\,2)}\\
   \end{alignat*}