I would like to draw the following image using the package dynkin-diagrams:

enter image description here


This is not easy, because the package is only designed to deal with Dynkin diagrams, particularly those which are most frequently encountered. I was not able to change the label locations to match your image without a lot more work. I also did not match the angle you wanted; the package is designed to use angles that display most clearly the symmetries of Dynkin diagrams.

diagram requested

\advance\dynkin@nodes by 1
\begin{dynkinDiagram}[labels={v_p,z,x_r,y_q},mark=o,edge length=1cm]{D}{4}
  • If it is too big, or too small, change the line edge length=1cm to edge length=.5cm or edge length=1.5cm, etc. – Benjamin McKay Dec 1 '19 at 11:09

Welcome! If you can live with different angles, you can just use the extended Dynkin diagram of E_6.

\dynkin[edge length=.5cm,%
indefinite edge ratio=3,%
make indefinite edge={2-0},%
make indefinite edge={5-6},%
make indefinite edge={1-3},

enter image description here

D_4/so(8) would have the angle of your screen shot but you somehow would need to add the inner nodes. If you insist on that angle, maybe plain TikZ is easier.

\begin{tikzpicture}[circ/.style={circle,draw,inner sep=2pt,label=above:$#1$,
 node contents={}},scale=1.5]
 \path (-3,0) node[circ=v_1] (-2,0) coordinate (p-1) (-1,0) node[circ=v_p] 
   (45:3) node[circ=x_1] (45:2) coordinate (p-3) (45:1) node[circ=x_r] 
   (-45:3) node[circ=y_1] (-45:2) coordinate (p-5) (-45:1) node[circ=y_q] 
   (0,0) node[circ=z];
 \draw foreach \X in {1,3,5} {(c-\X) -- (p-\X) (c-\the\numexpr\X+1) -- (c-7)};
 \draw[densely dotted] foreach \X in {1,3,5} {(c-\the\numexpr\X+1) -- (p-\X)};

enter image description here

  • Thank you very much Schrodinger's cat, but i would like to change the size of the second tikz picture, how can i do it? – Adam Hamam Nov 30 '19 at 11:58
  • @AdamHamam Replace 1.5 in scale=1.5 by another factor. – Schrödinger's cat Nov 30 '19 at 14:25
  • I appreciate your wise response Schrödinger's cat. Well thanks. – Adam Hamam Nov 30 '19 at 21:35

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