Let $L$ be a Lie algebra. Let $Ad$ be the adjoin representation of $L$. If $Ad$ is completely reducible, then $$Ad(L)\cong L/Z(L) \cong L_1 \oplus \dots \oplus L_t$$ Now from this I think it's clear than $L$ it's completely reducible. So if $I$ is an ideal of $L$, then exists $J$ ideal of $L$ such that $$L=I \oplus J$$ For me this is clear from a "logical point of view" but I don't find a way to prove this in a formal way. some ideas?
Asked
Active
Viewed 55 times
0
-
1See here. Ideals are submodules of $L$ - see here. Your question seems related to this one. – Dietrich Burde Jun 18 '23 at 15:18