Blattner's conjecture

In mathematics, a Lie algebra $\mathfrak{g}$ is solvable if its derived series terminates in the zero subalgebra. That is, writing

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

for the derived Lie algebra of $\mathfrak{g}$, generated by the set of values

[x,y]

for x and y in $\mathfrak{g}$, the derived series

$\mathfrak{g}\ge [\mathfrak{g},\mathfrak{g}]\ge [[\mathfrak{g},\mathfrak{g}],[\mathfrak{g},\mathfrak{g}]]\ge [[[\mathfrak{g},\mathfrak{g}],[\mathfrak{g},\mathfrak{g}]],[[\mathfrak{g},\mathfrak{g}],[\mathfrak{g},\mathfrak{g}]]]\ge ...$

becomes constant eventually at 0.

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 is called the radical.

Properties

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

• (i) $\mathfrak{g}$ is solvable.
• (ii) $\mathrm{ad}(\mathfrak{g})$, the adjoint representation of $\mathfrak{g}$, is solvable.
• (iii) There is a finite sequence of ideals ${\mathfrak{a}}_i$ of $\mathfrak{g}$ such that:

  $\mathfrak{g}={\mathfrak{a}}_0\supset {\mathfrak{a}}_1\supset ...{\mathfrak{a}}_r=0$ where $[{\mathfrak{a}}_i,{\mathfrak{a}}_i]\subset {\mathfrak{a}}_{i+1}$ for all $i$.

• (iv) $[\mathfrak{g},\mathfrak{g}]$ is nilpotent.

Lie's Theorem states that if $V$ is a finite-dimensional vector space over an algebraically closed field of characteristic zero, and $\mathfrak{g}$ is a solvable linear Lie algebra over $V$, then there exists a basis of $V$ relative to which the matrices of all elements of $\mathfrak{g}$ are upper triangular.

Completely solvable Lie algebras

A Lie algebra $\mathfrak{g}$ is called completely solvable if it has a finite chain of ideals from 0 to $\mathfrak{g}$ such that each has codimension 1 in the next. A finite-dimensional nilpotent Lie algebra is completely solvable, and a completely solvable Lie algebra is solvable. Over an algebraically closed field and 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.

Example

• Every abelian Lie algebra is solvable.
• Every nilpotent Lie algebra is solvable.
• Every Lie subalgebra, quotient and extension of a solvable Lie algebra is solvable.
• Let ${\mathfrak{b}}_k$ be a subalgebra of ${\mathfrak{g}\mathfrak{l}}_k$ consisting of upper triangular matrices. Then ${\mathfrak{b}}_k$ is solvable.

Solvable Lie groups

The terminology arises from the solvable groups of abstract group theory. There are several possible definitions of solvable Lie group. For a Lie group G, there is

• termination of the usual derived series, in other words taking 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 Lie algebras are the Lie algebra of the derived series of (closed) subgroups.

See also

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

External links

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

References

• Humphreys, James E. Introduction to Lie Algebras and Representation Theory. Graduate Texts in Mathematics, 9. Springer-Verlag, New York, 1972. ISBN 0-387-90053-5