Vertical align with TikZ and booktabs

Question

I want to write one line of text in a booktabs table containing a TikZ drawing that is higher than the line-height. Per default the text is vertically aligned with the baseline of the TikZ drawing. I want it to be vertically aligned with the middle of the TikZ drawing.

Here is my code:

\documentclass[a4paper,11pt]{scrartcl}

\usepackage{xcolor}
\definecolor{HighlightColor}{RGB}{255,173,97}

\usepackage{booktabs}
\usepackage[radius=.09cm,edgeLength=1.5cm]{dynkin-diagrams}
\usepackage[left=3cm, right=3cm, top=4cm]{geometry}
\usepackage{mathtools}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{tikz-cd}


\newcommand{\R}{\mathbb{R}}
\newcommand{\SL}{\operatorname{SL}}
\newcommand{\SO}{\operatorname{SO}}

\usepackage{graphbox}

\begin{document}


\begin{center}
\begin{tabular}[c]{p{1cm}cp{3cm}}
\toprule
$\Phi$  & 
Dynkin diagram & 
Symmetric space 
\\
\midrule
$\Phi_3$ & \vspace{1cm}
\begin{tikzpicture}
\draw[HighlightColor,densely dotted,thick] (0,0.5) arc (90:270:0.5);
\draw[HighlightColor,densely dotted,thick] (1.5,-0.5) arc (270:450:0.5);
\draw[HighlightColor,densely dotted,thick] (0,0.5) -- (1.5,0.5);
\draw[HighlightColor,densely dotted,thick] (0,-0.5) -- (1.5,-0.5);
\dynkin[mark=o,text/.style={scale=1.2}]{B}{3};
\end{tikzpicture}
$\rightarrow$
\begin{tikzpicture}
\dynkin[mark=o,text/.style={scale=1.2}]{A}{2};
\end{tikzpicture}
&
$\SL(3,\R)/\SO(3)$
\\
\bottomrule
\end{tabular}
\end{center}

\end{document}

What I have:

What I want:

dynkin-diagrams.sty found here.

Answer 1

You can vertically center any picture without guessing values by a simple macro: \newcommand{\ctikz}[1]{$\vcenter{\hbox{#1}}$}. Just use \ctikz{<your picture>}. Also, no need for manual \vspace{1cm}.

\newcommand{\ctikz}[1]{$\vcenter{\hbox{#1}}$}

\begin{center}
\begin{tabular}{p{1cm} c p{3cm}}
\toprule
$\Phi$   & Dynkin diagram & Symmetric space \\
\midrule
$\Phi_3$ & %\vspace{1cm}
\ctikz{%
\begin{tikzpicture}
\draw[HighlightColor,densely dotted,thick] (0,0.5) arc (90:270:0.5);
\draw[HighlightColor,densely dotted,thick] (1.5,-0.5) arc (270:450:0.5);
\draw[HighlightColor,densely dotted,thick] (0,0.5) -- (1.5,0.5);
\draw[HighlightColor,densely dotted,thick] (0,-0.5) -- (1.5,-0.5);
\dynkin[mark=o,text/.style={scale=1.2}]{B}{3};
\end{tikzpicture}
}
$\rightarrow$
\ctikz{%
\begin{tikzpicture}
\dynkin[mark=o,text/.style={scale=1.2}]{A}{2};
\end{tikzpicture}
}
&
$\SL(3,\R)/\SO(3)$
\\
\bottomrule
\end{tabular}
\end{center}

Answer 2

Use the baseline option for the first picture, as described here:

\begin{tikzpicture}[baseline=-0.7ex] %<--- adjust as desired
\draw[HighlightColor,densely dotted,thick] (0,0.5) arc (90:270:0.5);
\draw[HighlightColor,densely dotted,thick] (1.5,-0.5) arc (270:450:0.5);
\draw[HighlightColor,densely dotted,thick] (0,0.5) -- (1.5,0.5);
\draw[HighlightColor,densely dotted,thick] (0,-0.5) -- (1.5,-0.5);
\dynkin[mark=o,text/.style={scale=1.2}]{B}{3};
\end{tikzpicture}

Remove (or adjust) the \vspace{1cm} that you added if you want to get rid of the extra space below the picture.

Answer 3

an alternative, as exercise of use dynkin package:

\documentclass[a4paper,11pt]{scrartcl}
\usepackage[hmargin=3cm, top=4cm]{geometry}

\usepackage{xcolor}
\definecolor{HighlightColor}{RGB}{255,173,97}
\usepackage{booktabs}
\usepackage[radius=.09cm,edgeLength=1.5cm]{dynkin-diagrams}
\usepackage{mathtools}  % it also loads amsmath
\usepackage[utf8]{inputenc}
\usepackage{amsfonts}


\newcommand{\R}{\mathbb{R}}
\newcommand{\SL}{\operatorname{SL}}
\newcommand{\SO}{\operatorname{SO}}

\usepackage{graphbox}

\begin{document}
    \begin{center}
\begin{tabular}{p{1cm}cp{3cm}}
    \toprule
$\Phi$      &   Dynkin diagram  &   Symmetric space     \\
\midrule
$\Phi_3$    &   \begin{tikzpicture}[baseline=-0.75ex]
                \draw[HighlightColor,densely dotted,thick]  (0.0,-0.5) arc (270:90:0.5) --
                                                            (1.5, 0.5) arc (90:-90:0.5) -- cycle;
                \dynkin[mark=o,text/.style={scale=1.2}] {B}{3};
                \draw[->] (3.3,0) -- + (0.5,0);
                \scoped[xshift=41mm]\dynkin[mark=o,text/.style={scale=1.2}]{A}{2};
                \end{tikzpicture}
                                &   $\SL(3,\R)/\SO(3)$  \\
    \bottomrule
\end{tabular}
    \end{center}
\end{document}

Answer 4

I made a few changes to the dynkin-diagrams package recently, which sadly mess up Zarko's code above. Here are the little changes you need to make to get it working again.

\documentclass[a4paper,11pt]{scrartcl}
\usepackage{xcolor}
\definecolor{HighlightColor}{RGB}{255,173,97}
\usepackage{booktabs}
\usepackage[radius=.09cm,edge-length=1.5cm,mark=o]{dynkin-diagrams}
\newcommand{\R}{\mathbb{R}}
\newcommand{\SL}[1]{\operatorname{SL}(#1,\R{})}
\newcommand{\SO}[1]{\operatorname{SO}(#1)}
\begin{document}
\begin{center}
\begin{tabular}{p{1cm}cp{3cm}}
\toprule
$\Phi$      &   Dynkin diagram  &   Symmetric space     \\
\midrule
$\Phi_3$    &   
                \begin{dynkinDiagram}{B}{3}
                \draw[HighlightColor,densely dotted,thick]  
                (0.0,-0.4) arc (270:90:0.5) --
                (1.5, 0.6) arc (90:-90:0.5) -- cycle;
                \end{dynkinDiagram}
                \tikz[baseline=-0.5ex] \draw[->] (3.3,0) -- + (0.5,0);
                \dynkin{A}{2}
                                &   $\SL{3}/\SO{3}$  \\
\bottomrule
\end{tabular}
\end{center}
\end{document}

Answer 5

considering Benjamin McKay's answer above package, where he inform us (at least me), that new version of this package have changes which hasn't back compatibility ... and some possibilities of use of this package, the new solution can be:

\documentclass[a4paper,11pt]{scrartcl}
\usepackage[hmargin=3cm, top=4cm]{geometry}

\usepackage{xcolor}
\definecolor{HighlightColor}{RGB}{255,173,97}
\usepackage{booktabs}
\usepackage[radius=.09cm,
            edge-length=1.5cm,    % <--- new syntax for edge lengths
            mark=o]{dynkin-diagrams}
\usetikzlibrary{fit, shapes.misc}

\newcommand{\R}{\mathbb{R}}
\newcommand{\SL}{\operatorname{SL}}
\newcommand{\SO}{\operatorname{SO}}

\begin{document}
    \begin{center}
\begin{tabular}{p{1cm}cp{3cm}}
    \toprule
$\Phi$      &   Dynkin diagram  &   Symmetric space     \\
\midrule
$\Phi_3$    &   \begin{tikzpicture}[baseline]
                \dynkin{B}{3}
                \node[rounded rectangle, draw=HighlightColor,
                     densely dotted, thick, fit=(root 1) (root 2)] {};
                \draw[->] (root 3) ++ (4mm,0) -- ++ (0.5,0) node[-,right]{\dynkin{A}{2}};
                \end{tikzpicture}
                                &   $\SL(3,\R)/\SO(3)$  \\
    \bottomrule
\end{tabular}
    \end{center}
\end{document}

which gives almost the same result as is obtained in other answers: