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

### Background

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 $$\mathfrak{g} \to \mathfrak{g}_1 \to \mathfrak{g}_2 \to \cdots \to \mathfrak{g}_N .$$

### Questions

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?

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