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Let $J=\begin{pmatrix}0&0&0&0&0&1 \\ 0&0&0&0&1&0 \\ 0&0&0&1&0&0 \\ 0&0&1&0&0&0 \\ 0&1&0&0&0&0 \\ 1&0&0&0&0&0\end{pmatrix}$

Let $L=\mathfrak{so}(6,\mathbb{C})=\{x \in \mathfrak{gl}(6,\mathbb{C}):x^tJ+Jx=0\}$ be the semisimple Lie algebra associated to $J$ with basis $\mathcal{B}=\{e_{ij}-e_{kl} :i+l=j+k=7\}$. Consider the maximal total subalgebra of $L$ : $$H=\{\alpha(e_{11}-e_{66})+\beta(e_{22}-e_{55})+\gamma(e_{33}-e_{44}):\alpha, \beta, \gamma \in \mathbb{C}\}.$$ Respect to the basis $\{h_1=e_{11}-e_{66},h_2=e_{22}-e_{55},h_3=e_{33}-e_{44}\}$ of $H$ I obtain the root system $\Phi=\{\pm h_i^* \pm h_j^*: i \neq j\}$ and the basis of the root system $\Delta=\{\alpha_1:=h_1^*-h_2^*,\alpha_2:=h_2^*-h_3^*,\alpha_3:=h_2^*+h_3^*\}$. Now I can compute the Cartan matrix and I obtain: $\begin{pmatrix}2 & -1 & -1 \\ -1 &2 &0\\ -1 &0 &2\end{pmatrix}$. Now I have to describe the Dynkin diagram, but how can I do it?

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    $\begingroup$ The Dynkin diagarm assosiated to $\mathfrak{so}(3\cdot 2,\mathbf{C})$ can be obtained by the Cartan matrix $A$, which you provided. The nodes of the Dynkin diagram are labeled by the roots $h_1,h_2,h_3$, one draws $a_{i,j}a_{j,i}$ arrows from $h_i$ to $h_j$. In this case, the Dynkin diagram is equal to the Dynkin diagram of type $A_3$. $\endgroup$ Commented Jul 14, 2023 at 13:48

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Once you have computed the elements $\alpha_i$ of the base $\Delta$ and the Cartan matrix, to describe the Dynkin diagram recall that:

  • the nodes are the $\alpha_i \in \Delta$;
  • the edges are given by $\langle \alpha_i, \alpha_j \rangle \langle \alpha_j, \alpha_i \rangle$ for $i \neq j$;
  • if we have more than one root length, the arrow points to the shortest root.

In this case, since $\langle \alpha_i, \alpha_j \rangle = \langle \alpha_j, \alpha_i \rangle$ always holds, we do not have to worry about arrows (all the roots have the same length). Notice that $\langle \alpha_2, \alpha_3 \rangle = 0$, hence $\alpha_2$ and $\alpha_3$ are not connected in the diagram. Now, check what happens for $\alpha_1 , \alpha_2$: $\langle \alpha_1, \alpha_2 \rangle \langle \alpha_2, \alpha_1 \rangle = (-1)(-1)=1$, thus the two nodes are connected by a single line. The same happens for $\alpha_1, \alpha_3$.

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