Solvable Lie algebra

Let ${\displaystyle {\mathfrak {g}}}$ be a finite-dimensional Lie algebra over a field of characteristic 0. The following are equivalent.

• (i) ${\displaystyle {\mathfrak {g}}}$ is solvable.
• (ii) ${\displaystyle {\rm {ad}}({\mathfrak {g}})}$, the adjoint representation of ${\displaystyle {\mathfrak {g}}}$, is solvable.
• (iii) There is a finite sequence of ideals ${\displaystyle {\mathfrak {a}}_{i}}$ of ${\displaystyle {\mathfrak {g}}}$:

  ${\displaystyle {\mathfrak {g}}={\mathfrak {a}}_{0}\supset {\mathfrak {a}}_{1}\supset ...{\mathfrak {a}}_{r}=0,\quad [{\mathfrak {a}}_{i},{\mathfrak {a}}_{i}]\subset {\mathfrak {a}}_{i+1}\,\,\forall i.}$

• (iv) ${\displaystyle [{\mathfrak {g}},{\mathfrak {g}}]}$ is nilpotent.[3]
• (v) For ${\displaystyle {\mathfrak {g}}}$ ${\displaystyle n}$-dimensional, there is a finite sequence of subalgebras ${\displaystyle {\mathfrak {a}}_{i}}$ of ${\displaystyle {\mathfrak {g}}}$:

  ${\displaystyle {\mathfrak {g}}={\mathfrak {a}}_{0}\supset {\mathfrak {a}}_{1}\supset ...{\mathfrak {a}}_{n}=0,\quad \operatorname {dim} {\mathfrak {a}}_{i}/{\mathfrak {a}}_{i+1}=1\,\,\forall i,}$

with each ${\displaystyle {\mathfrak {a}}_{i+1}}$ an ideal in ${\displaystyle {\mathfrak {a}}_{i}}$.[4] A sequence of this type is called an elementary sequence.

• (vi) There is a finite sequence of subalgebras ${\displaystyle {\mathfrak {g}}_{i}}$ of ${\displaystyle {\mathfrak {g}}}$,

  ${\displaystyle {\mathfrak {g}}={\mathfrak {g}}_{0}\supset {\mathfrak {g}}_{1}\supset ...{\mathfrak {g}}_{r}=0,}$

such that ${\displaystyle {\mathfrak {g}}_{i+1}}$ is an ideal in ${\displaystyle {\mathfrak {g}}_{i}}$ and ${\displaystyle {\mathfrak {g}}_{i}/{\mathfrak {g}}_{i+1}}$ is abelian.[5]

• (vii) The Killing form ${\displaystyle B}$ of ${\displaystyle {\mathfrak {g}}}$ satisfies ${\displaystyle B(X,Y)=0}$ for all X in ${\displaystyle {\mathfrak {g}}}$ and Y in ${\displaystyle [{\mathfrak {g}},{\mathfrak {g}}]}$.[6] This is Cartan's criterion for solvability.

Lie's Theorem states that if ${\displaystyle V}$ is a finite-dimensional vector space over an algebraically closed field of characteristic zero, and ${\displaystyle {\mathfrak {g}}}$ is a solvable Lie algebra, and if ${\displaystyle \pi }$ is a representation of ${\displaystyle {\mathfrak {g}}}$ over ${\displaystyle V}$, then there exists a simultaneous eigenvector ${\displaystyle v\in V}$ of the endomorphisms ${\displaystyle \pi (X)}$ for all elements ${\displaystyle X\in {\mathfrak {g}}}$.[7]

• Every Lie subalgebra and quotient of a solvable Lie algebra are solvable.[8]
• Given a Lie algebra ${\displaystyle {\mathfrak {g}}}$ and an ideal ${\displaystyle {\mathfrak {h}}}$ in it,

  ${\displaystyle {\mathfrak {g}}}$ is solvable if and only if both ${\displaystyle {\mathfrak {h}}}$ and ${\displaystyle {\mathfrak {g}}/{\mathfrak {h}}}$ are solvable.[8][2]

The analogous statement is true for nilpotent Lie algebras provided ${\displaystyle {\mathfrak {h}}}$ is contained in the center. Thus, an extension of a solvable algebra by a solvable algebra is solvable, while a central extension of a nilpotent algebra by a nilpotent algebra is nilpotent.

• A solvable nonzero Lie algebra has a nonzero abelian ideal, the last nonzero term in the derived series.[2]
• If ${\displaystyle {\mathfrak {a}},{\mathfrak {b}}\subset {\mathfrak {g}}}$ are solvable ideals, then so is ${\displaystyle {\mathfrak {a}}+{\mathfrak {b}}}$.[1] Consequently, if ${\displaystyle {\mathfrak {g}}}$ is finite-dimensional, then there is a unique solvable ideal ${\displaystyle {\mathfrak {r}}\subset {\mathfrak {g}}}$ containing all solvable ideals in ${\displaystyle {\mathfrak {g}}}$. This ideal is the radical of ${\displaystyle {\mathfrak {g}}}$.[2]
• A solvable Lie algebra ${\displaystyle {\mathfrak {g}}}$ has a unique largest nilpotent ideal ${\displaystyle {\mathfrak {n}}}$, called the nilradical, the set of all ${\displaystyle X\in {\mathfrak {g}}}$ such that ${\displaystyle {\rm {ad}}_{X}}$ is nilpotent. If D is any derivation of ${\displaystyle {\mathfrak {g}}}$, then ${\displaystyle D({\mathfrak {g}})\subset {\mathfrak {n}}}$.[9]

A Lie algebra ${\displaystyle {\mathfrak {g}}}$ is called completely solvable or split solvable if it has an elementary sequence{(V) As above definition} of ideals in ${\displaystyle {\mathfrak {g}}}$ from ${\displaystyle 0}$ to ${\displaystyle {\mathfrak {g}}}$. 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 ${\displaystyle 3}$-dimensional real Lie algebra of the group of Euclidean isometries of the plane is solvable but not completely solvable.

A solvable Lie algebra ${\displaystyle {\mathfrak {g}}}$ is split solvable if and only if the eigenvalues of ${\displaystyle {\rm {ad}}_{X}}$ are in ${\displaystyle k}$ for all ${\displaystyle X}$ in ${\displaystyle {\mathfrak {g}}}$.[2]

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 ${\displaystyle G}$, there is

• termination of the usual derived series of the group ${\displaystyle G}$ (as an abstract group);
• termination of the closures of the derived series;
• having a solvable Lie algebra

• Cartan's criterion
• Killing form
• Lie-Kolchin theorem
• Solvmanifold
• Dixmier mapping

• EoM article Lie algebra, solvable
• EoM article Lie group, solvable

• Fulton, W.; Harris, J. (1991). Representation theory. A first course. Graduate Texts in Mathematics. Vol. 129. New York: Springer-Verlag. ISBN 978-0-387-97527-6. MR 1153249.
• Humphreys, James E. (1972). Introduction to Lie Algebras and Representation Theory. Graduate Texts in Mathematics. Vol. 9. New York: Springer-Verlag. ISBN 0-387-90053-5.
• Knapp, A. W. (2002). Lie groups beyond an introduction. Progress in Mathematics. Vol. 120 (2nd ed.). Boston·Basel·Berlin: Birkhäuser. ISBN 0-8176-4259-5..
• Jean-Pierre Serre: Complex Semisimple Lie Algebras, Springer, Berlin, 2001. ISBN 3-5406-7827-1