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Let $L\ne \{0\}$ be a non-abelian simple Lie algebra, then the only ideals are $L$ and $\{0\}$. We know that $L'$ is the smallest ideal of $L$ such that $L/L'$ is abelian. If $L'=\{0\}$, then $L/L'=L$, but $L$ is non-abelian. So $L'=L$ and hence $L^{(k)}=L\ne \{0\}$ for all $k$, i.e. $L$ is not solvable. So simplicity implies non-solvability.

My question is, can we extend this idea to semi-simple Lie algebras? In other words, is it true that semi-simplicity implies non-solvability?

My definition for semi-simple: $L$ has no non-zero solvable ideals.

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  • $\begingroup$ It should probably be Lie algebras, I edited accordingly. $\endgroup$ Commented May 14, 2013 at 17:44
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    $\begingroup$ The definition of semisimplicity I know asks for a Lie algebra with no solvable ideals (besides $0$ of course), that includes the Lie algebra itself... You better tell us your definition of semisimplicity. $\endgroup$ Commented May 14, 2013 at 17:49
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    $\begingroup$ This is a weird question! :-) $\endgroup$ Commented May 14, 2013 at 17:53
  • $\begingroup$ My definition for semi-simple: $L$ has non non-zero solvable ideals. $\endgroup$
    – Phil-ZXX
    Commented May 14, 2013 at 18:04
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    $\begingroup$ @Thomas Aha, but with this definition the statement has a much simpler proof (as Olivier Bégassat already said), namely $L$ is an ideal of $L$ and it is non-zero by assumption. Thus it is non-solvable as $L$ is semi-simple. $\endgroup$ Commented May 14, 2013 at 18:21

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As noted in the comments, the answer I give may not coincide with the definitions/assumptions of the book/lecture course you are following.

Suppose $L$ is a non-abelian semi-simple Lie algebra. Then by definition of semi-simplicity (or it is a theorem depending on the approach you take) $L=\bigoplus_{i=1}^n L_i$ for some simple Lie algebras. And because $L$ is non-abelian at least one of the $L_i$ has to be non-abelian.

Because the argument is not really different for bigger $n$, let me for simplicity of notation assume that $L=L_1\oplus L_2$ with $L_1$ non-abelian (and we don't know whether $L_2$ is abelian or not). This means that $L=L_1\oplus L_2$ as vector spaces and additionally $[l_1,l_2]=[l_2,l_1]=0$ for all $l_1\in L_1$ and $l_2\in L_2$.

Now, by definition $L'=[L,L]=[L_1\oplus L_2,L_1\oplus L_2]=[L_1,L_1]+[L_1,L_2]+[L_2,L_1]+[L_2,L_2]$. The last equality follows as the Lie bracket is bilinear. But since we have a direct sum decomposition of Lie algebras, we have $[L_1,L_2]=[L_2,L_1]=0$. Hence $L'=L_1'\oplus L_2'$. Now since $L_1'$ is non-abelian simple, we have that $L_1'=L_1$. Thus $L^{(k)}=L_1\oplus L_2^{(k)}$. In particular, $L$ is not solvable.

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  • $\begingroup$ This looks good. I just don't quite understand the part about $[L_1,L_2]=[L_2,L_1]=0$ and $L'=L_1'\oplus L_2'$. $\endgroup$
    – Phil-ZXX
    Commented May 14, 2013 at 18:11
  • $\begingroup$ I mean, does $L'=L_1'\oplus L_2'$ not follow from the definition itself? If $(l_1,l_2),(l_1',l_2')\in L_1\oplus L_2$ then $[(l_1,l_2),(l_1',l_2')]=([l_1,l_1'],[l_2,l_2'])$. $\endgroup$
    – Phil-ZXX
    Commented May 14, 2013 at 18:16
  • $\begingroup$ Yes, I was overly complicated. I included another approach. Yours also works (and is more easy). $\endgroup$ Commented May 14, 2013 at 18:18

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