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The Lie algebra $\mathfrak{so}_3$ has a basis $x_1,x_2,x_3$ with the multiplication table $[x_1,x_2]=x_3$, $[x_2,x_3]=x_1$, $[x_3,x_1]=x_2$. Moreover there is an isomorphism $\mathfrak{so}_3(\mathbb C)\cong\mathfrak{sl}_2(\mathbb C)$.

There is also an isomorphism $\mathfrak{so}_6(\mathbb C)\cong\mathfrak{sl}_4(\mathbb C)$.

It turns out that the algebra $\mathfrak{so}_6(\mathbb C)\cong\mathfrak{sl}_4(\mathbb C)$ also has a basis $x_1$, ..., $x_{15}$ with the following property: there are $20$ ordered triples of numbers $1$, ..., $15$, namely, \begin{array}{ccc} 1 & 2 & 3 \\ 1 & 4 & 5 \\ 1 & 6 & 7 \\ 1 & 8 & 9 \\ 2 & 6 & 10 \\ 2 & 8 & 11 \\ 2 & 12 & 5 \\ 3 & 4 & 12 \\ 3 & 7 & 10 \\ 3 & 9 & 11 \\ 4 & 7 & 14 \\ 4 & 15 & 9 \\ 5 & 8 & 15 \\ 5 & 14 & 6 \\ 6 & 8 & 13 \\ 7 & 9 & 13 \\ 10 & 11 & 13 \\ 10 & 14 & 12 \\ 11 & 12 & 15 \\ 13 & 15 & 14 \\ \end{array} such that $[x_i,x_j]=x_k$, $[x_j,x_k]=x_i$, $[x_k,x_i]=x_j$ if $(i,j,k)$ is one of these $20$ triples, while all other brackets of the basis elements are zero.

Clearly every root vector $e_\alpha$ of a semisimple Lie algebra is included in the $\mathfrak{sl}_2$-subalgebra spanned by $e_\alpha$, $e_{-\alpha}$ and $[e_\alpha,e_{-\alpha}]$, which can be then transformed to obtain an $\mathfrak{so}_3$-basis of this subalgebra. So for each semisimple Lie algebra it is also possible to find a basis which can be subdivided into triples $(x_i,x_j,x_k)$ with $[x_i,x_j]=x_k$, $[x_j,x_k]=x_i$, $[x_k,x_i]=x_j$ for each triple. However it is not clear whether the remaining brackets between elements not in the same triple can be also made similarly neat. For example, in the above case of $\mathfrak{so}_6(\mathbb C)\cong\mathfrak{sl}_4(\mathbb C)$, there are only $12$ roots, grouped into $6$ pairs of opposite roots, so this approach would only give $6$ of the $20$ needed triples.

Questions.

Do any other (semi)simple Lie algebras have bases $x_1, \dotsc, x_n$ with the property that each $[x_i,x_j]$ is either some $\pm x_k$ or zero?

Does the above system of $20$ ordered triples of a $15$-element set have some combinatorial name? Can it be constructed geometrically? I tried to use edges and faces of an icosahedron — there are 20 faces and 15 lines through the origin parallel to edges and I tried to use these but could not obtain explicit correspondence. Is it possible?

Later

As explained by Peter Wu, such basis exists for every $\mathfrak{so}_n$: the basis corresponds to all 2-element subsets of $\{1,...,n\}$ and the nonzero brackets of basis elements correspond to 3-element subsets. So only the first question for algebras different from these remains unanswered.

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    $\begingroup$ Do you know if this can be done for $\mathfrak{sl}_3$? $\endgroup$
    – LSpice
    Feb 20, 2023 at 0:16
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    $\begingroup$ For $so(n)$ this can be done. Say $M_{ij}$ is the generator with $1$ at $i,j$, $-1$ at $j,i$ otherwise $0$. Only $[M_{ij},M_{jk}]=M_{ik}$ and otherwise zero. $\endgroup$
    – Peter Wu
    Feb 20, 2023 at 1:49
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    $\begingroup$ @PeterWu Indeed! In my case, $15=\binom62$ and $20=\binom63$, it was stupid not to notice this! Thus combinatorics in general is simply that of 2- and 3-element subsets of $\{1,...,n\}$. Do you know if there is something similar for other simple algebras? I mean, even if not or you don't know, I still think your comment contains a legitimate answer. Maybe somebody else proposes a more complete answer, but so far this one is the only thing we have, right? $\endgroup$
    – მამუკა ჯიბლაძე
    Feb 20, 2023 at 6:02

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