# Solvable Lie algebra

(Redirected from Derived algebra)

In mathematics, a Lie algebra g is solvable if its derived series terminates in the zero subalgebra. The derived Lie algebra is the subalgebra of g, denoted

${\displaystyle [{\mathfrak {g}},{\mathfrak {g}}]}$

that consists of all Lie brackets of pairs of elements of g. The derived series is the sequence of subalgebras

${\displaystyle {\mathfrak {g}}\geq [{\mathfrak {g}},{\mathfrak {g}}]\geq [[{\mathfrak {g}},{\mathfrak {g}}],[{\mathfrak {g}},{\mathfrak {g}}]]\geq [[[{\mathfrak {g}},{\mathfrak {g}}],[{\mathfrak {g}},{\mathfrak {g}}]],[[{\mathfrak {g}},{\mathfrak {g}}],[{\mathfrak {g}},{\mathfrak {g}}]]]\geq ...}$

If the derived series eventually arrives at the zero subalgebra, then the Lie algebra is solvable.[1] The derived series for Lie algebras is analogous to the derived series for commutator subgroups in group theory.

Any nilpotent Lie algebra is solvable, a fortiori, but the converse is not true. The solvable Lie algebras and the semisimple Lie algebras form two large and generally complementary classes, as is shown by the Levi decomposition.

A maximal solvable subalgebra is called a Borel subalgebra. The largest solvable ideal of a Lie algebra is called the radical.

## Characterizations

Let g be a finite-dimensional Lie algebra over a field of characteristic 0. The following are equivalent.

• (i) g is solvable.
• (iii) There is a finite sequence of ideals ai of g:
${\displaystyle {\mathfrak {g}}={\mathfrak {a}}_{0}\supset {\mathfrak {a}}_{1}\supset ...{\mathfrak {a}}_{r}=0,\quad \forall i[{\mathfrak {a}}_{i},{\mathfrak {a}}_{i}]\subset {\mathfrak {a}}_{i+1}.}$
• (iv) [g, g] is nilpotent.[2]
• (v) For g n-dimensional, there is a finite sequence of subalgebras ai of g:
${\displaystyle {\mathfrak {g}}={\mathfrak {a}}_{0}\supset {\mathfrak {a}}_{1}\supset ...{\mathfrak {a}}_{n}=0,\quad \forall i\operatorname {dim} {\mathfrak {a}}_{i}/{\mathfrak {a}}_{i+1}=1,}$
with each ai + 1 an ideal in ai.[3] A sequence of this type is called an elementary sequence.
• (vi) There is a finite sequence of subalgebras gi of g,
${\displaystyle {\mathfrak {g}}={\mathfrak {g}}_{0}\supset {\mathfrak {g}}_{1}\supset ...{\mathfrak {g}}_{r}=0,}$
such that gi + 1 is an ideal in gi and gi/gi + 1 is abelian.[4]

## Properties

Lie's Theorem states that if V is a finite-dimensional vector space over an algebraically closed field K of characteristic zero, and g is a solvable linear Lie algebra over a subfield k of K, and if π is a representation of g over V, then there exists a simultaneous eigenvector vV of the matrices π(X) for all elements Xg. More generally, the result holds if all eigenvalues of π(X) lie in K for all Xg.[6]

• Every Lie subalgebra, quotient and extension[clarification needed] of a solvable Lie algebra is solvable.
• A solvable nonzero Lie algebra has a nonzero abelian ideal, the last nonzero term in the derived series.[7]
• A homomorphic image of a solvable Lie algebra is solvable.[7]
• If a is a solvable ideal in g and g/a is solvable, then g is solvable.[7]
• If g is finite-dimensional, then there is a unique solvable ideal rg containing all solvable ideals in g. This ideal is the radical of g, denoted rad g.[7]
• If a, bg are solvable ideals, then so is a + b.[1]
• A solvable Lie algebra g has a unique largest nilpotent ideal n, the set of all Xg such that adX is nilpotent. If D is any derivation of g, then D(g) ⊂ n.[8]

## Completely solvable Lie algebras

A Lie algebra g is called completely solvable or split solvable if it has an elementary sequence[when defined as?] of ideals in g from 0 to g. A finite-dimensional nilpotent Lie algebra is completely solvable, and a completely solvable Lie algebra is solvable. Over an algebraically closed field a solvable Lie algebra is completely solvable, but the 3-dimensional real Lie algebra of the group of Euclidean isometries of the plane is solvable but not completely solvable.

A solvable Lie algebra g is split solvable if and only if the eigenvalues of adX are in k for all X in g.[7]

## Examples

${\displaystyle X=\left({\begin{matrix}0&\theta &x\\-\theta &0&y\\0&0&0\end{matrix}}\right),\quad \theta ,x,y\in \mathbb {R} .}$
Then g is solvable, but not split solvable.[7] It is isomorphic with the Lie algebra of the group of translations and rotations in the plane.

## Solvable Lie groups

Because the term "solvable" is also used for solvable groups in group theory, there are several possible definitions of solvable Lie group. For a Lie group G, there is

• termination of the usual derived series of the group G (as an abstract group);
• termination of the closures of the derived series;
• having a solvable Lie algebra.

To have equivalence one needs to assume G connected. For connected Lie groups, these definitions are the same, and the derived series of the Lie algebra is the Lie algebra of the derived series of (closed) subgroups.