The idea of a Solvable Lie algebra hinges on the definition of the sequence:
$$ g \ge [g,g] \ge [[g,g],[g,g]] \ge [ [[g,g],[g,g]] , [[g,g],[g,g]] ] \ge \ldots $$
and its limiting group. If an element on the sequence is provably the trivial algebra, then subsequent subgroups on the sequence will also be trivial. This form of idempotence means that the Lie algebra is solvable.
why is this series special for the idea of solvability? What happens with intermediate sequences like:
$$ [g, [g,g]] \ge [g, [g, [g,g]]] \ge [[g, [g,g]], [g, [g,g]] ] \ge \ldots $$
It seems a priori possible that the above subsequence becomes trivial even when the original sequence does not. Is that not the case?
The idea of a solvable Lie algebra does not hinge on the definition of the commutator series alone. A common definition of solvability is also that there is a finite sequence of subalgebras $\mathfrak{g}_i$ of $\mathfrak{g}$, $$ {\displaystyle {\mathfrak {g}}={\mathfrak {g}}_{0}\supset {\mathfrak {g}}_{1}\supset \ldots \supset {\mathfrak {g}}_{r}=0,} $$ such that $\mathfrak{g}_{i + 1}$ is an ideal in $\mathfrak{g}_i$ and $\mathfrak{g}_i/\mathfrak{g}_{i + 1}$ is abelian. One can show that these two definitions are equivalent. Furthermore, any subalgebra and any quotient algebra of a solvable Lie algebra is solvable, and any extension of solvable Lie algebras is solvable. In particular, a Lie algebra is solvable if and only if it can be obtained by successive extensions of abelian Lie algebras.