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Can the Lie algebra $\mathfrak {sl}_{3}\mathbb(R)$ be represented as the sum of two of its nontrivial subalgebras, i.e $\mathfrak {sl}_{3}\mathbb(R) = A{\displaystyle \oplus }B$ such that ${\displaystyle \forall }a \in A$, ${\displaystyle \forall }b \in B$ $[a,b] = 0$ ? I chose the standard basis in $\mathfrak {sl}_{3}\mathbb(R)$ (the dimension of this algebra is 8) and calculated all of commutators (28 non-trivial), but I do not know how to use it here.

user444075
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    Onishchik (1966) has classified decompositions of simple Lie algebras $L=A+B$, as a (non-direct) sum of subalgebras, e.g., we have $A_{2n-1}=C_n+A_{2n-2}$ for $n\ge 2$. Of course, for ideals this cannot happen. – Dietrich Burde May 08 '17 at 08:00

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There is no need to use explicit commutator relations. The conditions you impose imply that both $A$ and $B$ are ideals in $\mathfrak{sl}_3(R)$, so the result follows from the fact that $\mathfrak{sl}_3(R)$ is simple.

Edit: If you want to do this directly, it is a slightly tedious computation (imitating the proof that $\mathfrak{sl}_3(R)$ is simple): Observe that your assumbptions imply that for $a\in A$ and $x\in\mathfrak{sl}_3(R)$ you get $[a,x]\in A$ and similarly for $B$. Now without loss of generality, one assumes that $A$ contains a matrix $M$ with non-zero entry in the top right corner and proves that $A=\mathfrak{sl}_3(R)$. Forming the commutators of $M$ with the two diagonal matrices in the standard basis, you get two more matrices in $A$, for which the components above the main diagonal are the corresponding entries of $M$ mutliplied with different factors. Forming appropriate linear combinations of these three matrices, you find that $A$ for each of the three positions above the main diagonal, $A$ contains a matrix whose only non-zero entry above the main diagonal is in that position. Forming appropriate commutators with strictly lower triangular matrixes, you next get lower triangular matrices contained in $A$ with prescribed elements on the main diagonal, and forming linear combinations you can remove the diagonal parts in the three matrices from above and get two new matrices, which are lower triangular, with prescribed nice diagonal parts. (So you see that $A$ has dimension at least $5$ already.) This can be continued in a simliar way untily you see that $A$ has dimension at least $8$ and hence equals $\mathfrak{sl}_3(R)$.

Andreas Cap
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