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 $L$?


1 Answer 1


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 $L_2$.

  • $\begingroup$ o_O. I didn't notice this. $\endgroup$
    – roi_saumon
    Commented Jun 14, 2019 at 19:07
  • $\begingroup$ Could $[L_1,L_1]$ be ${0}$ instead of $L_1$? $\endgroup$
    – roi_saumon
    Commented Jun 14, 2019 at 19:08
  • $\begingroup$ No, $L_1$ is simple, hence perfect, i.e., $[L_1,L_1]=L_1$. $\endgroup$ Commented Jun 14, 2019 at 19:09
  • $\begingroup$ 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? $\endgroup$
    – roi_saumon
    Commented Jun 14, 2019 at 19:16
  • 1
    $\begingroup$ Yes, exactly... $\endgroup$ Commented Jun 14, 2019 at 21:02

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