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如何用图形微积分水平对齐方程式？

tag-icon • tag-icon tikz-pgf horizontal-alignment align
如何用图形微积分水平对齐方程式？

我有一个用 TikZ 和 align 环境编写的类别理论图形证明（感谢 Henri 的回答这个问题）。每行上的图表都有一个重复的特征，即标有“A”的垂直线。我希望这些线水平对齐。

图形演算

\documentclass{article}

\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{knots}
\usetikzlibrary{positioning}
\usepackage{mathtools}

\begin{document}

    \tikzset{dot/.style={circle,fill,inner sep=1pt},diagram/.style={knot=black,thick}}
    \newcommand{\tikzb}[1]{\tikz[baseline=(B.base)]{#1}}
    \newcommand{\qdim}[1]{\left<#1\right>}
    \newcommand{\Aid}{\draw[diagram] (0,-1.5) -- node[right] (B) {A} (0,1.5);}
    \newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right|\left. \vphantom{#1} #2 \right>}
    \pgfkeys{/encircle/.cd,label/.initial=$\Omega_{\mathcal{C}}$,angle/.initial=20}
    \newcommand{\encircle}[2][]{%
        \pgfkeys{/encircle/.cd,#1}
        \draw[diagram] (1,0) arc (0:\pgfkeysvalueof{/encircle/angle}:1 and 0.4) node[above] {\pgfkeysvalueof{/encircle/label}} arc (\pgfkeysvalueof{/encircle/angle}:180:1 and 0.4);
        #2
        \draw[diagram] (-1,0) arc (-180:0:1 and 0.4);
    }

    \begin{align}
        & & \tikzb{\draw[diagram] (0,-1.5) -- node[right] (B) {X} (0,1.5);}
        \tikzb{\encircle{\Aid}}\\
        & = \sum_{Y \in \Lambda_\mathcal{C}} \qdim{Y}
        & \tikzb{\draw[diagram] (0,-1.5) -- node[right] (B) {X} (0,1.5);}
        \tikzb{\encircle[label=Y]{\Aid}}
        & \qquad \text{By the definition of $\Omega_{\mathcal{C}}$} \\
        & = \sum_{\mathclap{\substack{Y \in \Lambda_\mathcal{C} \\ Z \in \Lambda_\mathcal{C} \\ \iota_i \in \mathcal{C}(Z, X \otimes Y) \\ \braket{\iota_i}{\iota_j} = \delta_{i,j}}}} \qdim{Y}\qdim{Z}
        & \tikzb{
            \draw[diagram] (-1,-1.5) to[out=90, in=245] node[left] {X} (-0.7,-0.3) coordinate[dot,label={right:$\iota_i^*$}] (A) to node[left] (B) {Z} (-0.7,0.3) coordinate[dot,label={right:$\iota_i$}] (C) to[out=115,in=270] (-1,1.5);
            \draw[diagram] (1,0) to[out=90,in=65] (C);
            \Aid
            \draw[diagram] (A) to[out=295,in=270] (1,0) node[right] {Y};
        }
        & \qquad \text{Insertion of a further $\Omega_{\mathcal{C}}$}
    \end{align}
\end{document}

您不一定需要knotstikz 库来编译此代码。我尝试在 tikz 代码中添加 &，但无济于事。

答案1

也许这更像是一种解决方法，但我会将 Tikz 部分视为图像。然后我会将所有内容放入表格中，以使所有内容都整齐地排列在列中。

然而，按图像特征进行对齐很不方便，但可靠的手动方法是将空节点添加为绘图中最左侧的项目，相对于要对齐的特征进行定位。然后可以在表格列中将其左对齐，甚至可以在对齐中工作。

通过垂直定位对齐节点，您也可以根据需要很好地匹配基线。

编辑：示例已准备好——但请注意，我使用手动定位作为左停止点的空节点，只是因为我发现调整原始节点更容易

\documentclass{article}

\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{knots}
\usetikzlibrary{positioning}
\usepackage{mathtools}

\begin{document}

    \tikzset{dot/.style={circle,fill,inner sep=1pt},diagram/.style={knot=black,thick}}
    \newcommand{\tikzb}[1]{\tikz[baseline=(B.base)]{#1}}
    \newcommand{\qdim}[1]{\left<#1\right>}
    \newcommand{\Aid}{\draw[diagram] (0,-1.5) -- node[right] (B) {A} (0,1.5);}
    \newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right|\left. \vphantom{#1} #2 \right>}
    \pgfkeys{/encircle/.cd,label/.initial=$\Omega_{\mathcal{C}}$,angle/.initial=20}
    \newcommand{\encircle}[2][]{%
        \pgfkeys{/encircle/.cd,#1}
        \draw[diagram] (1,0) arc (0:\pgfkeysvalueof{/encircle/angle}:1 and 0.4) node[above] {\pgfkeysvalueof{/encircle/label}} arc (\pgfkeysvalueof{/encircle/angle}:180:1 and 0.4);
        #2
        \draw[diagram] (-1,0) arc (-180:0:1 and 0.4);
    }

    \begin{tabular}{lll}
         & \tikzb{\draw[diagram] (0,-1.5) -- node[right] (B) {X} (0,1.5);}
        \tikzb{\encircle{\Aid}}\\
          $\displaystyle=\sum_{Y \in \Lambda_\mathcal{C}} \qdim{Y}$
        & \tikzb{\draw[diagram] (0,-1.5) -- node[right] (B) {X} (0,1.5);}
        \tikzb{\encircle[label=Y]{\Aid}}
        & By the definition of $\Omega_{\mathcal{C}}$ \\
         $\displaystyle=\sum_{\mathclap{\substack{Y \in \Lambda_\mathcal{C} \\ Z \in \Lambda_\mathcal{C} \\ \iota_i \in \mathcal{C}(Z, X \otimes Y) \\ \braket{\iota_i}{\iota_j} = \delta_{i,j}}}} \qdim{Y}\qdim{Z}$
        & \tikzb{
            \draw[diagram] (-1,-1.5) to[out=90, in=245] node[left] {X} (-0.7,-0.3) coordinate[dot,label={right:$\iota_i^*$}] (A) to node[left] (B) {Z} (-0.7,0.3) coordinate[dot,label={right:$\iota_i$}] (C) to[out=115,in=270] (-1,1.5);
            \draw[diagram] (1,0) to[out=90,in=65] (C);
            \Aid
            \draw[diagram] (A) to[out=295,in=270] (1,0) node[right] {Y};
            \node[draw=none,fill=none]   at (-1.6,0) {};%this node is a left stop.
        }
        & Insertion of a further $\Omega_{\mathcal{C}}$
    \end{tabular}
\end{document}

上述代码的输出

答案2

我只是在两张图片后面加了一些字距调整。当然，这需要手动干预。

\usepackage{mathtools}

\begin{document}

    \tikzset{dot/.style={circle,fill,inner sep=1pt},diagram/.style={knot=black,thick}}
    \newcommand{\tikzb}[1]{\tikz[baseline=(B.base)]{#1}}
    \newcommand{\qdim}[1]{\left<#1\right>}
    \newcommand{\Aid}{\draw[diagram] (0,-1.5) -- node[right] (B) {A} (0,1.5);}
    \newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right|\left. \vphantom{#1} #2 \right>}
    \pgfkeys{/encircle/.cd,label/.initial=$\Omega_{\mathcal{C}}$,angle/.initial=20}
    \newcommand{\encircle}[2][]{%
        \pgfkeys{/encircle/.cd,#1}
        \draw[diagram] (1,0) arc (0:\pgfkeysvalueof{/encircle/angle}:1 and 0.4) node[above] {\pgfkeysvalueof{/encircle/label}} arc (\pgfkeysvalueof{/encircle/angle}:180:1 and 0.4);
        #2
        \draw[diagram] (-1,0) arc (-180:0:1 and 0.4);
    }

    \begin{align}
        & & \tikzb{\draw[diagram] (0,-1.5) -- node[right] (B) {X} (0,1.5);}
        \tikzb{\encircle{\Aid}}\mkern12mu\\
        & = \sum_{Y \in \Lambda_\mathcal{C}} \qdim{Y}
        & \tikzb{\draw[diagram] (0,-1.5) -- node[right] (B) {X} (0,1.5);}
        \tikzb{\encircle[label=Y]{\Aid}}\mkern16.5mu
        & \qquad \text{By the definition of $\Omega_{\mathcal{C}}$} \\
        & = \sum_{\mathclap{\substack{Y \in \Lambda_\mathcal{C} \\ Z \in \Lambda_\mathcal{C} \\ \iota_i \in \mathcal{C}(Z, X \otimes Y) \\ \braket{\iota_i}{\iota_j} = \delta_{i,j}}}} \qdim{Y}\qdim{Z}
        & \tikzb{
            \draw[diagram] (-1,-1.5) to[out=90, in=245] node[left] {X} (-0.7,-0.3) coordinate[dot,label={right:$\iota_i^*$}] (A) to node[left] (B) {Z} (-0.7,0.3) coordinate[dot,label={right:$\iota_i$}] (C) to[out=115,in=270] (-1,1.5);
            \draw[diagram] (1,0) to[out=90,in=65] (C);
            \Aid
            \draw[diagram] (A) to[out=295,in=270] (1,0) node[right] {Y};
        }
        & \qquad \text{Insertion of a further $\Omega_{\mathcal{C}}$}
    \end{align}
\end{document}

在此处输入图片描述

答案3

为了完整起见，我将根据 Chris H 的建议添加自己的解决方案。它在每个图像的相同坐标处添加一个空节点。坐标系对齐以将 A 置于原点，因此如果此节点是图像最右边的元素，则它就是图像要对齐的边框。这样，只需手动调整一个坐标即可将所有图像对齐到同一点。这对我来说尤其重要，因为实际校样覆盖了整个页面，所以我不想摆弄每一张图片。

\documentclass{article}

\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{knots}
\usetikzlibrary{positioning}
\usepackage{mathtools}

\begin{document}

    \tikzset{dot/.style={circle,fill,inner sep=1pt},diagram/.style={knot=black,thick}}
    \newcommand{\tikzb}[1]{\tikz[baseline=(B.base)]{#1}}
    \newcommand{\qdim}[1]{\left<#1\right>}
    \newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right|\left. \vphantom{#1} #2 \right>}
    \pgfkeys{/encircle/.cd,label/.initial=$\Omega_{\mathcal{C}}$,angle/.initial=20}
    \newcommand{\encircle}[2][]{%
        \pgfkeys{/encircle/.cd,#1}
        \draw[diagram] (1,0) arc (0:\pgfkeysvalueof{/encircle/angle}:1 and 0.4) node[above] {\pgfkeysvalueof{/encircle/label}} arc (\pgfkeysvalueof{/encircle/angle}:180:1 and 0.4);
        #2
        \draw[diagram] (-1,0) arc (-180:0:1 and 0.4);
    }

    \newcommand{\tikzoffset}{\node at (1.63,0) {};}
    \newcommand{\Aid}{\draw[diagram] (0,-1.5) -- node[right] (B) {A} (0,1.5); \tikzoffset}

    \begin{align}
        & & \tikzb{
            \draw[diagram] (-1.2,-1.5) -- node[left] (B) {X} (-1.2,1.5);
            \encircle{\Aid}
        }\\
        & = \sum_{\mathclap{Y \in \Lambda_\mathcal{C}}} \qdim{Y}
        & \tikzb{
            \draw[diagram] (-1.2,-1.5) -- node[left] (B) {X} (-1.2,1.5);
            \encircle[label=Y]{\Aid}
        }
        & \qquad \text{By the definition of $\Omega_{\mathcal{C}}$} \\
        & = \sum_{\mathclap{\substack{Y, Z \in \Lambda_\mathcal{C} \\ \iota_i \in \mathcal{C}(Z, X \otimes Y^*) \\ \braket{\iota_i}{\iota_j} = \delta_{i,j}}}} \qdim{Y}\qdim{Z}
        & \tikzb{
            \draw[diagram] (-1.2,-1.5) to[out=90, in=245] node[left] {X} (-0.7,-0.3) coordinate[dot,label={right:$\iota_i^*$}] (A) to node[left] (B) {Z} (-0.7,0.3) coordinate[dot,label={right:$\iota_i$}] (C) to[out=115,in=270] node[left] {X} (-1.2,1.5);
            \draw[diagram] (1,0) to[out=90,in=65] (C);
            \Aid
            \draw[diagram] (A) to[out=295,in=270] (1,0) node[right] {Y};
        }
        & \qquad \text{Insertion of a further $\Omega_{\mathcal{C}}$}
    \end{align}
\end{document}

同时我也改进/纠正了一些其他细节。

在此处输入图片描述

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