= \delta_{kn}e_{j,m} - \delta_{jm}e_{n,k}$$we get the relations $$\label{eq:::}
[E_{j,k}, E_{k,n}] = E_{j,n},\qquad [F_{j,k}, F_{k,n}] = -E_{j,n},\qquad [E_{j,k}, F_{k,n}] = F_{j,n},$$ and $$\label{eq:1231}
   [E_{j,k}, F_{j,k}] = 2D_{j,k}, \quad \mbox{ and  } \quad  [F_{j,k}, D_{j,k}] = 2E_{j,k}.$$ The relations above can be interpreted following a “triangle rule”: the bracket between an operator coupling the states $j$ and $k$ and an operator coupling the states $k$ and $n$ couples the states $j$ and $n$. On the other hand, the bracket is zero if two operators couple no common states, that is, $$\label{E000}
[Y_{j,k}, Z_{j',k'}]=0 \quad \mbox{ if } \{j,k\} \cap \{j',k'\} = \emptyset,$$ with $Y,Z \in \{E,F,D\}$.

Controllability in $\mathfrak{su}(4{\ell}+4)$
---------------------------------------------

\[prop:liemolecule\] The Lie algebra $L$ generated by $B_{1}^{(4{\ell}+4)}, B_{2}^{(4{\ell}+4)}, B_{3}^{(4{\ell}+4)}$, $J_{i}(B_{1}^{(4{\ell}+4)})$, $ J_{i}(B_{2}^{(4{\ell}+4)}),$ $J_{i}(B_{3}^{(4{\ell}+4)})$ is equal to $\mathfrak{su}(4{\ell}+4)$.

The first step of the proof consists in showing that the Lie algebra $L$ contains the elementary matrices $$\label{Evicini}
E_{({\ell},k),({\ell}+1,k+j)}\quad \mbox{ for }k=-{\ell},\dots,{\ell},\ j=-1,0,1.$$

With a slight abuse of notation and for the sake of readability, let us write $B_{j} = B_{j}^{(4{\ell}+4)}$, $j=1,2,3$. Let us also write ${\operatorname{ad}}_\alpha \beta$ for $[\alpha,\beta]$ and ${\operatorname{ad}}_\alpha^{j+1} \beta$ for $[\alpha,{\operatorname{ad}}_\alpha^{j} \beta]$.

Notice that $$\begin{aligned}
J_{i}(B_{3})  = \sum_{{\ell}=-m}^{m} p_{{\ell},m} J_{i}(F_{({\ell},m),({\ell}+1,m)}) =  \sum_{{\ell}=-m}^{m} p_{{\ell},m} E_{({\ell},m),({\ell}+1,m)}.\end{aligned}$$ By induction on $j\geq 0$ 