On radicals of a lie algebra

Question

Let $\mathbb{k}$ be a field, $\mathfrak{g}$ be a finite-dimensional Lie algebra over $\mathbb{k}$. In Bourbaki's "Lie Groups and Lie Algebras", Ch I, he defines four radical-like ideals of $\mathfrak{g}$:

1. the radical $\mathfrak{r}$, i.e. the maximal solvable ideal;
2. the radical of Killing form $\mathfrak{k}$, i.e. $\mathfrak{k}=\mathfrak{g}^\perp$;
3. the maximal nilpotent ideal $\mathfrak{n}$;
4. the nilpotent radical $\mathfrak{s}$, i.e. intersection of all kernels of irreducible finite-dimensional representations of $\mathfrak{g}$.

He also shows that, $\mathfrak{r}=[\mathfrak{g},\mathfrak{g}]^\perp$, $\mathfrak{s}=[\mathfrak{g}, \mathfrak{g}]\cap\mathfrak{r}=[\mathfrak{g},\mathfrak{r}]$, and the following inclusion relations: $$\mathfrak{r} \supset \mathfrak{k} \supset \mathfrak{n} \supset \mathfrak{s}.$$

On the other hand, in Jacobson's book, the nilpotent radical is defined as $\mathfrak{n}$.

My question is, does $\mathfrak{s}$ coincide with $\mathfrak{n}$? or, are Bourbaki's nilradical and Jacobson's nilradical equivalent?

I guess the answer is NO (otherwise Bourbaki should have proved it), but I cannot find any example. Could any one give me an example? Thank you!

Answer 1

If we let $\mathbb{k}$ be the field itself thought of as a Lie algebra with trivial bracket, then $\mathfrak{n}=\mathbb{k}$ since this is a nilpotent Lie algebra but $\mathbb{s}=\{0\}$, since the obvious representation of $\mathbb{k}$ be scalar multiplication is faithful. In general, these will never coincide for a nilpotent Lie algebra.

Answer 2

The two definitions do never agree for a non-trivial nilpotent Lie algebra, as Ben has remarked. Indeed, if $\mathfrak{g}$ is nilpotent then $\mathfrak{s}=[\mathfrak{g},\mathfrak{r}]=[\mathfrak{g},\mathfrak{g}]$, whereas $\mathfrak{n}=\mathfrak{g}$. Since $\mathfrak{g}$ is nilpotent, $\mathfrak{g}\neq [\mathfrak{g},\mathfrak{g}]$. For non-trivial abelian Lie algebras we have in particular $\mathfrak{s}=0$ and $\mathfrak{n}=\mathfrak{g}$.
If $\mathfrak{g}$ is solvable, then $\mathfrak{s}=[\mathfrak{g},\mathfrak{g}]$ and $\mathfrak{s}\subseteq \mathfrak{n}$, and both equality and strict inclusion can happen. For equality consider the $2$-dimensional non-abelian Lie algebra.

Answer 3

An extended comment might be useful, though Ben and Dietrich have addressed the mathematical issues concisely. The question is implicitly about the history of Lie theory, which was still somewhat unsettled around 1960 when Bourbaki's Chap. I was published. (Jacobson's book appeared in 1962 but developed earlier in his career from writing up Weyl's lectures at IAS where the term "Lie algebra" began to replace "infinitesimal group". Jacobson's reference list doesn't include Bourbaki, by the way.)

Jacobson does mention the notion of "reductive" Lie algebra in later exercises, but early in his book he focuses mostly on the ideal structure including his versions of solvable and nil radicals. Indeed, his lifelong interest was in rings and nonassociative algebras, with emphasis on their structure theory. He had studied with Wedderburn at Princeton and never got deeply involved in representation theory or group theory. (When I took his 1963-64 course based on the recently published Curtis-Reiner book, the emphasis was definitely ring-theoretic, with characters of finite groups skipped over.)

On the other hand, Serre and others in the Bourbaki group (at the time their treatise on Lie groups and Lie algebras began) were more steeped in Lie groups and linear algebraic groups. Even though Chevalley's classification seminar in the late 1950s mostly bypassed Lie algebras, it was becoming clear that the intrinsic Jordan decomposition in algebraic groups and their Lie algebras was an important new tool. Similarly, contemporary work on "reductive" Lie groups was making reductive Lie algebras natural to study alongside semisimple ones. And representation theory of many kinds was becoming more prominent. So it's not surprising that there is some divergence in Bourbaki and Jacobson when the various ideals mentioned here come into play. For Jacobson, only $\mathfrak{r}$ and $\mathfrak{n}$ played a major role.

A final remark is that abelian Lie algebras are somewhat problematic, since their role in the absence of an intrinsic Jordan decomposition is ambiguous. This makes the contrast between $\mathfrak{n}$ and $\mathfrak{s}$ inevitable if you only discuss Lie algebras and their representations abstractly.