The Deal With These Ideals

The study of complex semisimple Lie algebras is a very well understand and marvelous thing. From the study of the structure of a semisimple Lie algbera, to their classification, and to their representation theory via highest weights; I recommend Humphreys for the interested reader. To those with familiarity whether through a formal class or self-studying, there are certain structures along the way that should have made an impression on you. In particular, certain ideals of a Lie algebra are very prolific and determine much of the structure of the overall Lie algebra. These include

1. The maximal solvable radical, commonly denoted \(\text{Rad}(\mathfrak{g})\).
2. The Killing form radical, commonly denoted \(\mathfrak{g}^{\perp}\).
3. The nilradical, or maximal nilpotent ideal, commonly denoted \(\text{Nil}(\mathfrak{g})\).
4. The nilpotent radical defined as the intersection of all kernels of irreducible finite-dimensional representations of \(\mathfrak{g}\), commonly denoted \(\mathfrak{s}\).

Admittedly I was not aware of the fourth ideal during my initial reading of Humphreys. However, I came across it on a mathoverflow post, and its relation with the other ideals were nice.

This post is meant to explore certain relations satisfied between these ideals and their proofs, ultimately leading to the interesting chain of ideals

\begin{equation} \mathfrak{s}\subseteq \text{Nil}(\mathfrak{g})\subseteq \mathfrak{g}^{\perp}\subseteq \text{Rad}(\mathfrak{g}) \end{equation}

Throughout this post, I am assuming my Lie algebra \(\mathfrak{g}\) is finite-dimensional and over a characteristic zero field \(\mathbb{K}\), which may have to be algebraically closed at times.

Proposition The maximal solvable ideal is the orthogonal complement to the derived subalgebra with respect to the Killing form i.e. \(\text{Rad}(\mathfrak{g}) = [\mathfrak{g},\mathfrak{g}]^{\perp}\).

Proof This is the result of two exercises in Humphreys. However, proposition 5 in Further Properties of the Radical in Bourbaki is far more enlightening.

Proposition \([\mathfrak{g},\text{Rad}(\mathfrak{g})] = [\mathfrak{g},\mathfrak{g}]\cap\text{Rad}(\mathfrak{g})\).

Proof

Proposition \([\mathfrak{g},\mathfrak{g}]\cap\text{Rad}(\mathfrak{g}) = \mathfrak{s}\).

Proof Every linear form on a Lie algebra which is zero on its derived subalgebra is a simple representation, so s is contained in the derived subalgebra. It is also “contained in the largest nilpotent ideal of the Lie algebra.” That gives one inclusion. For the reverse inclusion, it suffices to show that every simple representation of the Lie algebra is zero on this intersection.

Proposition \(\mathfrak{s}\subseteq \text{Nil}(\mathfrak{g})\).

Proof It clearly suffices to prove that \(\mathfrak{s}\) is nilpotent, which by Engel’s theorem is equivalent to proving each element is ad-nilpotent.

Proposition \(\text{Nil}(\mathfrak{g})\subseteq \mathfrak{g}^{\perp}\).

Proof To proceed we need only prove that every element of the nilradical is orthogonal to all of \(\mathfrak{g}\).

Proposition \(\mathfrak{g}^{\perp}\subseteq \text{Rad}(\mathfrak{g})\).

Proof It suffices to prove that the Killing form radical is a solvable ideal.

Note that when the Lie algebra is semisimple all of the ideals coincide: they are in fact all zero! However, there are cases for which some ideals are zero while others are nonzero. For instance,