Solvable Lie algebra

Let $g$ be a Lie algebra (finite or infinite dimensional). By using the commutator subalgebra we can construct the _derived series_ of $g$

g \geq [g, g] \geq [[g, g], [g, g]] \geq [[[g, g], [g, g]], [[g, g], [g, g]]] \geq \dots

If this series arrives to 0 in a finite number of steps it is said that $g$ is solvable.

Equivalently:

Definition (@olver86). A Lie algebra $g$ is a solvable Lie algebra if there exists a chain of subalgebras

{0} = g_{0} \subseteq g_{1} \subseteq g_{2} \subseteq \dots \subseteq g_{r - 1} \subseteq g_{r} = g

such that for each $k$ , $\dim g_{k - 1} = k$ and $g_{k - 1}$ is a normal subalgebra of $g_{k}$ :

[g_{k - 1}, g_{k}] \subseteq g_{k - 1} .

The requirement for solvability is equivalent to the existence of a basis ${v_{1}, \dots, v^{r}}$ of $g$ such that

[v_{i}, v_{j}] = \sum_{k}^{j - 1} c_{i j}^{k} v_{k}

whenever $i < j$ .

$◼$

Other characterizations:

Theorem

A Lie algebra $g$ is solvable if and only if 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. $◼$

See Ruiz_2014, page 7.

Theorem

Let $g$ be a solvable Lie algebra. There exists a sequence of ideals of $g$

{0} \subset L^{1} \subset \dots \subset L^{n} = g

such that $dim (L^{i}) = i$ for $i = 1, \dots, n$ . $◼$

See Ruiz_2014, page 8.

It can be shown that a connected Lie group 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.

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Author of the notes: Antonio J. Pan-Collantes

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