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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}.{{citation

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.

References

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