We know $sl(4,\mathbb{C})$ has dimension $15$ and $sl(3,\mathbb{C})$ has dimension $8$. Is it possible to write $sl(4,\mathbb{C})$ as the vector space sum of two Lie subalgebras that are isomorphic to $sl(3,\mathbb{C})$?
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asked Oct 10, 2019 at 15:31
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8$\begingroup$ No. Up to conjugation, a copy of $sl_3$ in $sl_4$ is (the derived subalgebra of) the stabilizer of a $3+1$ decomposition $H\oplus L$. The intersection of any two such stabilizers contains a copy of $sl_2$. Hence the sum has dimension $\le 8+8-3=13$. $\endgroup$– YCorCommented Oct 10, 2019 at 15:40
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A. L. Onishchik has determined such semisimple decompositions (not necessarily direct sums) in his article Decompositions of reductive Lie groups. For example, it is true that for simple Lie algebras $$ A_{2n-1}=A_{2n-2}+C_n. $$ For $n=2$ this gives $A_3=A_2+C_2$, i.e., $\mathfrak{sl}(4)=\mathfrak{sl}(3)+\mathfrak{sp}(4)$. However, $A_3=A_2+A_2$ is not possible.
answered Oct 18, 2019 at 9:01