Find all finite dimensional simple quotients of a possibly infinite dimensional Lie algebra generated by three elements Let $\mathfrak{g}$ be the (possibly infinite dimensional) Lie algebra over $\mathbb{C}$ with three generators $a,b,c\in \mathfrak{g}$ and defining relations
\begin{align}
[a,[a,b]]=b, &&& [b,[b,a]]=a \tag{1}\label{1} \\
[b,[b,c]]=c, &&& [c,[c,b]]=b \tag{2}\label{2} \\
[c,[c,a]]=a, &&& [a,[a,c]]=c. \tag{3}\label{3}
\end{align}
Question: what are all possible finite-dimensional simple quotient Lie algebras of $\mathfrak{g}$?
I already know two solutions:
(a) quotient over $c=i[a,b]$, and we get $\mathfrak{sl}(2)$.
(b) $\mathfrak{sl}(3)$, in which (using fundamental representation) $a=e_{12}+e_{21}, b=e_{23}+e_{32}, c=e_{13}+e_{31}$.
I want to know if there are any other possibilities.
 A: This isn't exactly an answer to your question (as I'm not sure the answer is known), but hopefully it is a helpful extended comment. Your Lie algebra is isomorphic to the fixed point subalgebra of the affine Lie algebra $\widetilde{\mathfrak{sl}_3}$ under a Cartan involution. This is a result of Berman, see On generators and relations for certain involutory subalgebras of Kac–
Moody Lie algebras, Comm. Alg. 17 (1989) 3165-3185.
To explain: let ${\mathfrak g}$ be a simply-laced Kac–Moody Lie algebra with basis of simple positive roots $\Delta$; let $e_{\pm\alpha}$, $\alpha\in\Delta$ be generators (of the derived subalgebra of ${\mathfrak g}$) and let $\theta:{\mathfrak g}\rightarrow{\mathfrak g}$ be the involution which swaps $e_\alpha$ and $e_{-\alpha}$ and which acts as $-1$ on the Cartan subalgebra. Then Berman showed that the fixed point subalgebra ${\mathfrak k}$ is generated by $X_\alpha=e_\alpha+e_{-\alpha}$, with relations:
$$[X_\alpha,[X_\alpha,X_\beta]]=X_\beta\;\;\mbox{if $\alpha+\beta$ a root}, \; [X_\alpha,X_\beta]=0\;\;\mbox{otherwise.}$$
So your relations are exactly those for $X_\alpha,X_\beta,X_\gamma$ when $\{ \alpha,\beta,\gamma\}$ is a basis of simple positive roots in the untwisted affine $A_2$ root system.
If ${\mathfrak g}$ is a simple Lie algebra then the fixed point subalgebra for an involution (indeed, any periodic automorphism) is a reductive subalgebra of ${\mathfrak g}$. But there is no analogous property for arbitrary Kac–Moody Lie algebras.
There has been some recent interest in these isotropy subalgebras for infinite-dimensional Kac–Moody algebras (interest coming from Physics). As a starting point, see the recent papers of Ralf Köhl and his collaborators.
Note that in the case of an untwisted affine Lie algebra, it changes nothing to take the fixed points in the loop algebra ${\mathfrak g}_0\otimes {\mathbb C}[t,t^{-1}]$ where ${\mathfrak g}_0$ is the corresponding finite-dimensional simple Lie algebra. (In your case ${\mathfrak g}_0=\mathfrak{sl}_3$.) Then there is a symmetric (i.e. eigen-)space decomposition: ${\mathfrak g}_0={\mathfrak k}_0\oplus{\mathfrak p}_0$ for the action of $\theta$, and
$${\mathfrak k}={\mathfrak k}_0\otimes {\mathbb C}[t+t^{-1}]\oplus {\mathfrak p}_0\otimes (t-t^{-1}){\mathbb C}[t+t^{-1}].$$
I'm not sure, however, whether this observation provides an answer to your question. The affine case is the focus of a recent preprint of Kleinschmidt, Köhl, Lautenbacher and Nicolai, see Representations of involutory subalgebras of affine Kac–Moody algebras.
