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I know that every finite dimensional lie algebra over a field $\mathcal{F}$ has a unique maximal solvable ideal, all subalgebras of a solvable lie algebra are also solvable, and a sum of solvable lie algebras is solvable.

For a solvable lie algebra L, its maximal ideal is unique. By induction, the set of ideals of L is a chain relative to the order $\subset$. The maximal ideal of L is a sum of all proper ideals. I want to know if the maximal ideal is [LL]. Let $L^{(1)}=[LL]$, $L^{(i)}=[L^{(1-1)}L^{(1-1)}]$. Suppose that $L^{(n)}=0$, $n \in \mathbb{Z}_{+}$.What is the connection between
\begin{align*} 0 \subset [L^{n-1}L^{n-1}] \subset \ldots \subset [L^{1}L^{1}] \subset [LL] \subset L \end{align*} and the chain of ideals of $L$.

bing
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  • See: https://www.researchgate.net/post/The_set_of_ideals_of_a_solvable_lie_algebra_L_is_a_chain – bing Dec 22 '15 at 13:37

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This is not true, suppose $L$ is commutative of dimension 2, every 1-dimensional subspace is a maximal ideal, so there is not a unique maximal ideal and $[L,L]=0$.

Tsemo Aristide
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  • You are right. But it is the fact that every finite dimensional lie algebra over a field $\mathcal{F}$ has a unique maximal solvable ideal, see https://en.wikipedia.org/wiki/Radical_of_a_Lie_algebra. If we accept the fact, then the maximal ideal of a solvable lie algebra $L$ is $L$. Thank you very much. – bing Dec 20 '15 at 09:02
  • @bing: you can accept the fact; but it's better for you if you prove it to be convinced. – YCor Dec 21 '15 at 22:15
  • Every finite dimensional lie algebra $L$ over a field $\mathcal{F}$ has a unique maximal solvable ideal. It follows since a sum of solvable ideas is solvable, we have $I+J$ is a solvable ieal, where $I$ is a maximal solvable ideal of $L$ and $J$ is any other solvable ideal of $L$. By maximality of $I$, $I+J=I$, i.e. $J \subset I$. See Humphreys's book 'Introduction to Lie algebras and Representation Theory'. – bing Dec 22 '15 at 02:49