Radical of a Lie algebra

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

The radical, denoted by $\mathrm{r}\mathrm{a}\mathrm{d}(𝔤)$, fits into the exact sequence

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

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

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

Definition

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

Firstly let $𝔞$ and $𝔟$ be two solvable ideals of $𝔤$. Then $𝔞+𝔟$ is again an ideal of $𝔤$, and it is solvable because it is an extension of $(𝔞+𝔟)/𝔞\simeq 𝔟/(𝔞\cap 𝔟)$ by $𝔞$. Now consider the sum of all the solvable ideals of $𝔤$. 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