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 $\mathfrak{g}.$^[1]

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.

References

1. Hazewinkel, Michiel; Gubareni, Nadiya; Kirichenko, V. V. (2010), Algebras, Rings and Modules: Lie Algebras and Hopf Algebras, Mathematical Surveys and Monographs 168, Providence, RI: American Mathematical Society, p. 15, doi:10.1090/surv/168, ISBN 978-0-8218-5262-0, MR 2724822.