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 $\mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{𝔤})$, fits into the exact sequence

$0\to \mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{𝔤})\to \mathfrak{𝔤}\to \mathfrak{𝔤}/\mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{𝔤})\to 0$.

where $\mathfrak{𝔤}/\mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{𝔤})$ 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 $\mathfrak{𝔤}/\mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{𝔤})$ via the restriction of the quotient map $\mathfrak{𝔤}\to \mathfrak{𝔤}/\mathrm{r}\mathrm{a}\mathrm{d}(\mathfrak{𝔤}).$

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