# On radicals of a lie algebra

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!

• $\mathfrak{s}$ vanishes iff $\mathfrak{g}$ is reductive, but it's easy to find reductive Lie algebras for which $\mathfrak{n}$ fails to vanish (e.g. abelian ones). Nov 20, 2013 at 5:28
• @Si-Qi Yu: Those parts of Bourbaki have a running hypothesis that char$(k)=0$ (look at the start of various sections); e.g., for $\mathfrak{g}=\mathfrak{sl}_p$ in characteristic $p>0$ the Killing form vanishes and the subalgebra ${\rm{Lie}}(\mu_p)$ of diagonal scalars form a nonzero central ideal yet if $p>2$ then $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, etc. Nov 20, 2013 at 14:52
• @Qiaochu Yuan: What is your definition of "reductive Lie algebra" in positive characteristic? Nov 20, 2013 at 14:53
• @Marguax: I don't have one (I assume you're hinting that the equivalence between various possible definitions fails). Is positive characteristic an important issue here? Nov 20, 2013 at 18:50
• @Qiaochu Yuan: The assertions in Bourbaki totally break down in positive characteristic, so it is more serious than a failure of equivalences. I am not aware of a good definition of "reductive Lie algebra" outside characteristic 0, so it was unclear to me if you had a definition. In the setting of Lie algebras (unlike for algebraic groups) it is safest to explicitly assume char. 0 for such things unless one provides a clear definition or reference for the terminology being used. (Bourbaki demands char. 0 in their definition of "reductive Lie algebra"; look at the start of section 6 of Ch. I.) Nov 20, 2013 at 19:56

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.
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.
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.