Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The following question might be filed under "idle curiosity", but I'm hoping some expert on Lie algebras can answer it.


One of the fundamental structural results about Lie algebras is the Levi-Malčev decomposition, which states that a finite-dimensional Lie algebra $\mathfrak{g}$ (over a field of zero characteristic) is a split extension of a semisimple subalgebra $\mathfrak{l} < \mathfrak{g}$ (called the Levi factor) by its maximal solvable ideal $\mathfrak{r}$ (called the (solvable) radical). In other words, we have a split exact sequence $$ 0 \to \mathfrak{r} \to \mathfrak{g} \to \mathfrak{l} \to 0,$$ or, what is the same, that $\mathfrak{g} \cong \mathfrak{l} \ltimes \mathfrak{r}$ is the semidirect product of its Levi factor by the radical.

The radical can be calculated explicitly using the Killing form, a fact alluded to in Steve Huntsman's answer to the question Why the Killing form?. In fact, one has that $\mathfrak{r} = [\mathfrak{g},\mathfrak{g}]^\perp$ is the perpendicular complement of the first derived ideal $[\mathfrak{g},\mathfrak{g}]$ relative to the Killing form $$\kappa(x,y) = \operatorname{Tr}(\operatorname{ad}_x \operatorname{ad}_y).$$

However, the Killing form gives rise to a solvable ideal called the Killing radical and defined by $$\mathfrak{g}^\perp = \lbrace x \in \mathfrak{g} \mid \kappa(x,y) = 0~\forall y \in \mathfrak{g} \rbrace.$$ In other words, it's the maximal subspace of $\mathfrak{g}$ on which $\kappa$ is identically zero.

We have that $\mathfrak{g}^\perp < \mathfrak{r}$, but generally they are not the same. Nevertheless, a while back I realised playing with this stuff, that one can approximate the radical by a sequence of Killing radicals.

To see this, let us define a sequence of Lie algebras $\mathfrak{g}_0 = \mathfrak{g}$ and $\mathfrak{g}_{i+1} = \mathfrak{g}_i/\mathfrak{g}_i^\perp$. In other words, $\mathfrak{g}_1$ is the quotient $\mathfrak{g}/\mathfrak{g}^\perp$ of $\mathfrak{g}$ by its Killing radical. On $\mathfrak{g}_1$ we have a Killing form $\kappa_1$ and hence we can talk about its Killing radical, so we let $\mathfrak{g}_2 = \mathfrak{g}_1/\mathfrak{g}_1^\perp$, et cetera. Eventually this will end with either $\mathfrak{g}_N$ being semisimple or else $\mathfrak{g}_N = 0$. In the latter case, $\mathfrak{g}$ is solvable and hence $\mathfrak{r} = \mathfrak{g}$, whereas in the former, $\mathfrak{r}$ is the kernel of the map $\mathfrak{g} \to \mathfrak{g}_N$ given by the composition of the successive quotients $$ \mathfrak{g} \to \mathfrak{g}_1 \to \mathfrak{g}_2 \to \cdots \to \mathfrak{g}_N .$$


  1. Is there a good conceptual explanation for what's going on?
  2. Are the $\mathfrak{g}_i$ interesting?
  3. Is there a reference for this?

Thanks in advance.

share|improve this question
References? Chapter 1 of Bourbaki's Groupes et algebres de Lie covers the foundational material: Killing form, semisimplicity, Levi decomposition, etc. They just write $\mathfrak{k}$ for what you call the Killing radical and work out some inclusions among characteristic ideals including the radical, the Killing radical, and the "nilradical" of a finite dimensional Lie algebra $\mathfrak{g}$ over a field of characteristic 0. A book I haven't seen is Lie Algebras: Theory and Algorithms by W.A. deGraaf, North-Holland, 2000. –  Jim Humphreys Aug 3 '10 at 14:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.