2

I know how to draw a Dynkin diagram and add labels on the vertices using Tikz-cd or the package for Dynkin diagram, but how to draw an automorphism intuitively as in the following picture? I don't know how to draw those curly arrows on the diagrams.

[

  • Welcome to TeX.SE! Since you are saying "I know how to draw a Dynkin diagram a ..." could you please post the code? The answer will depend on what you are doing. – user121799 Nov 26 '18 at 22:17
  • I decided that these diagrams are easier to read if you bend the Dynkin diagram around the axis of symmetry, and then just draw thicker bars between the roots t that are being interchanged. I referred to these as folded Dynkin diagrams. If you compare my list of Satake diagrams in my dynkin-diagrams package to Satake's, I hope you will agree that it is easier to read. – Benjamin McKay Dec 7 '18 at 13:49
4

There is a full-fledged package for that: dynkin-diagrams. (And the outer automorphism group of D_4 is S_3.)

\documentclass{amsart}
\usepackage[mark=o,radius=.3cm,edgeLength=1cm]{dynkin-diagrams}
\begin{document}
\begin{tikzpicture}
\begin{scope}[local bounding box=E6]
 \dynkin{E}{6} 
 \foreach \X in {1,...,6}
 {\node at (root \X) {\X};}
 \draw[latex-latex,shorten >=2mm,shorten <=2mm] (root 1) to[out=-60,in=-120] 
 node[midway,below]{$\sigma$} (root 6);
 \draw[latex-latex,shorten >=2mm,shorten <=2mm] (root 3) to[out=-60,in=-120] 
 node[midway,below]{$\sigma$} (root 5);
\end{scope}
\begin{scope}[xshift=5.5cm,yshift=-0.6cm,rotate=30,local bounding box=D4]
 \dynkin{D}{4} 
 \foreach \X in {1,...,4}
 {\node at (root \X) {\X};}
 \draw[latex-latex,shorten >=2mm,shorten <=2mm] (root 4) to[bend left] 
 node[midway,auto]{$\sigma_{14}$} (root 1);
 \draw[latex-latex,shorten >=2mm,shorten <=2mm] (root 1) to[bend left] 
 node[midway,auto]{$\sigma_{13}$} (root 3);
 \draw[latex-latex,shorten >=2mm,shorten <=2mm] (root 3) to[bend left] 
 node[midway,auto]{$\sigma_{34}$} (root 4);
\end{scope}
\node[anchor=south] at (E6.north) {$\mathrm{E}_6$};
\node[anchor=south] at (D4.north) {$\mathrm{D}_4$};
\end{tikzpicture}
\end{document}

enter image description here

  • Thank you for help! I shall learn more about this useful package. – sawdada Nov 27 '18 at 0:12
0

This is but a quick sketch for D_4.

\documentclass{article}
\usepackage{tikz}
\tikzstyle{vertex}=[circle, draw, minimum size=0pt]
\newcommand{\vertex}{\node[vertex]}


\begin{document}

\begin{tikzpicture}
\node at (0,0) {$D_4$};

\vertex (a) at (0,-1) {4};
\vertex (b) at (0,-3) {2};
\vertex (c) at (-2,-3) {1};
\vertex (d) at (2,-3) {3};
\draw[->] (b) -- (a);
\draw[->] (b) -- (c);
\draw[->] (b) --(d);
\draw[<->] (c) edge[bend right=80] node[below] {$\sigma$} (d);


\end{tikzpicture}

\end{document}

enter image description here

  • 1
    \tikzstyle is deprecated; it should be \tikzset{vertex/.style={circle, draw, minimum size=0pt}} – egreg Nov 26 '18 at 22:53
  • That's pretty good, though for E_6 it's more complicated.. – sawdada Nov 27 '18 at 0:13
0

Using the latest version (3.141) of the dynkin-diagrams package, you can compare the way that the package handles Satake diagrams by default (using a thick grey bar to show which roots are identified by an involution), with the way that you have described them (using arrows with $\sigma$ written under them). These examples are included in the package documentation.

Dynkin diagrams variously folded

\documentclass{amsart}
\usepackage{dynkin-diagrams}
\newcommand{\invol}[2]{\draw[latex-latex] (root #1) to [out=-60,in=-120] 
node[midway,below]{$\sigma$} (root #2);}
\begin{document}
\begin{tabular}{ccc}
\dynkin[ply=3]{D}{4} &
\dynkin{E}{II} & 
\dynkin{A}{IIIa} \\
{\tikzset{/Dynkin diagram/fold style/.style={stealth-stealth,thick,
shorten <=1mm,shorten >=1mm,}}
\dynkin[ply=3,edge length=.75cm]{D}{4}}
&
\begin{dynkinDiagram}[edge length=.75cm,labels*={1,...,6}]{E}{6}
\invol{1}{6}\invol{3}{5}
\end{dynkinDiagram} 
&
\begin{dynkinDiagram}[edge length=.75cm]{A}{oo.o**.**o.oo}
\invol{1}{10}\invol{2}{9}\invol{3}{8}\invol{4}{7}\invol{5}{6}
\end{dynkinDiagram} \\
\end{tabular}
\end{document}

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.