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I am trying to understand why semi-simple Lie algebra cannot have a abelian ideal. More specifically I search for an argument as direct as possible saying if an algebra is a direct some of simple Lie algebras then it cannot have abelian ideals, and possibly an argument for the converse. I would prefer an argument which does not involve solvable algebras (i did not understand them well enough yet..) Thanks you.

Chevallier
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2 Answers2

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A semi-simple Lie algebra is a sum $S_1\oplus...\oplus S_n$ of simple ideals $S_i$, by definition, $S_i$ is simple if and only if it is non abelian and has no non zero proper ideal. Let $I$ be an abelian ideal of $S$. Let $p:I\rightarrow S_i$ such that $p(u)=u_i$, where $u=u_1+...+u_n, u_i\in S_i$,$p_i(I)$ is an ideal of $S_i$: for every $s\in S_i, [s,u]\in I, [s,u]=[s,u_1+..+u_n]=[s,u_i]$ since $[S_i,S_j]=0$ for $i\neq j$. Since $S_i$ is simple, $p_i(I)=0$ and $I=0$.

Tsemo Aristide
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  • Ah thanks, it is actually simple. Is there such an argument for the converse? – Chevallier Mar 09 '20 at 11:32
  • A possible argument for the converse, which however uses solvable ideals: by definition, $L$ is semisimple iff $Rad(L) = \left{ 0 \right}$, where $Rad(L)$ is a maximal solvable ideal; by contradiction, if $I \neq \left{0\right}$ is an abelian ideal of $L$, then $I$ is also solvable, so $I \subseteq Rad(L)$. So $Rad(L) \neq \left{0 \right}$, which is a contradiction. – cip Dec 30 '20 at 15:58
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All abelian ideals of any Lie Algebra are solvable and since, you are taking your Lie Algebra say, $L$ to be semi-simple. So, Radical of $L$ or maximal solvable ideal has to be zero. Hence, all abelian ideals are zero.

Vats Y
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