Solvable Lie algebra

See @olver86 pag. 151.
Throughout, let $g \subseteq gl (n, K)$ be a finite‑dimensional matrix Lie algebra over a field $K$ of characteristic zero (typically $K = R$ or $C$ ).

Define the derived series,by using the commutator subalgebra, by

g^{(0)} = g, g^{(k + 1)} = [g^{(k)}, g^{(k)}], k \geq 0.

We say $g$ is solvable if

\exists r s.t. g^{(r)} = {0} .

In the matrix setting one has complete splitting of the adjoint action. The following are equivalent:

$g$ is solvable.
There exists a chain of ideals

{0} = I_{0} ⊲ I_{1} ⊲ I_{2} ⊲ \dots ⊲ I_{n} = g, \dim I_{k} = k .

There exists a sequence of subalgebras

{0} = a^{0} \subset a^{1} \subset \dots \subset a^{k} = g

such that $a^{i}$ is an ideal of $a^{i + 1}$ and the quotient $a^{i + 1} / a^{i}$ is abelian.
4. Equivalently, there exists a basis ${v_{1}, \dots, v_{n}}$ of $g$ such that

[v_{i}, v_{j}] \in span {v_{1}, \dots, v_{j - 1}}, \forall 1 \leq i < j \leq n .

Equivalently, in that same basis each adjoint ${ad}_{v}$ is represented by an upper‑triangular $n \times n$ matrix.

It can be shown that a connected Lie group $G \subset GL (n, K)$ is solvable if and only if its Lie algebra is solvable.

On the other hand, if $g$ is a finite dimensional Lie algebra, there exists a unique maximal solvable ideal called the radical of a Lie algebra.