# Decomposition of semi-simple Lie algebra into simple lie algebra (or ideal?)

A semi-simple Lie algebra $$L$$ can by definition be decomposed into simple Lie algebras :

$$L=L_1\oplus \ldots \oplus L_n$$. Are these $$L_i$$ necessarily ideals of $$L$$?

Yes, the direct summands are ideals by definition, since a Lie algebra direct sum $$L=L_1\oplus L_2$$ is defined with the Lie bracket $$[L_1,L_2]=0$$. Hence we have $$[L_1,L]=[L_1,L_1\oplus L_2]=[L_1,L_1]\oplus[L_1,L_2]=L_1,$$ hence $$L_1$$ is an ideal. The same is true for $$L_2$$.

• o_O. I didn't notice this. Commented Jun 14, 2019 at 19:07
• Could $[L_1,L_1]$ be ${0}$ instead of $L_1$? Commented Jun 14, 2019 at 19:08
• No, $L_1$ is simple, hence perfect, i.e., $[L_1,L_1]=L_1$. Commented Jun 14, 2019 at 19:09
• Oh, right, by definition $L_1$ is not Abelian, I always forget. So $L_1$ and $L_2$ inside $L$ are in direct sum as vector spaces means $L_1\cap L_2=0$ and $L_1+L_2=L$ and to be direct sum of Lie algebra means we also have $[L_1,L_2]=0$ right? Commented Jun 14, 2019 at 19:16
• Yes, exactly... Commented Jun 14, 2019 at 21:02