Radical of a Lie algebra

In the mathematical field of Lie theory, the radical of a Lie algebra ${\mathfrak {g}}$ is the largest solvable ideal of ${\mathfrak {g}}.$

The radical, denoted by ${\rm {rad}}({\mathfrak {g}})$, fits into the exact sequence

$0\to {\rm {rad}}({\mathfrak {g}})\to {\mathfrak {g}}\to {\mathfrak {g}}/{\rm {rad}}({\mathfrak {g}})\to 0$.

where ${\mathfrak {g}}/{\rm {rad}}({\mathfrak {g}})$ is semisimple. When the ground field has characteristic zero and ${\mathfrak {g}}$ has finite dimension, Levi's theorem states that this exact sequence splits; i.e., there exists a (necessarily semisimple) subalgebra of ${\mathfrak {g}}$ that is isomorphic to the semisimple quotient ${\mathfrak {g}}/{\rm {rad}}({\mathfrak {g}})$ via the restriction of the quotient map ${\mathfrak {g}}\to {\mathfrak {g}}/{\rm {rad}}({\mathfrak {g}}).$

A similar notion is a Borel subalgebra, which is a (not necessarily unique) maximal solvable subalgebra.

Definition

Let $k$ be a field and let ${\mathfrak {g}}$ be a finite-dimensional Lie algebra over $k$. There exists a unique maximal solvable ideal, called the radical, for the following reason.

Firstly let ${\mathfrak {a}}$ and ${\mathfrak {b}}$ be two solvable ideals of ${\mathfrak {g}}$. Then ${\mathfrak {a}}+{\mathfrak {b}}$ is again an ideal of ${\mathfrak {g}}$, and it is solvable because it is an extension of $({\mathfrak {a}}+{\mathfrak {b}})/{\mathfrak {a}}\simeq {\mathfrak {b}}/({\mathfrak {a}}\cap {\mathfrak {b}})$ by ${\mathfrak {a}}$. Now consider the sum of all the solvable ideals of ${\mathfrak {g}}$. It is nonempty since $\{0\}$ is a solvable ideal, and it is a solvable ideal by the sum property just derived. Clearly it is the unique maximal solvable ideal.

Related concepts

• A Lie algebra is semisimple if and only if its radical is $0$.
• A Lie algebra is reductive if and only if its radical equals its center.

See also

• Levi decomposition