Derived subalgebra of the radical of a Lie algebra is contained in the radical of the Killing form

Let $\mathfrak{g}$ be a finite dimensional Lie algebra over $\mathbb{C}$, $R(\mathfrak{g})$ the radical of $\mathfrak{g}$ and $\mathfrak{g}^{\bot}$ the radical of the Killing form. Show that $[R(\mathfrak{g}),R(\mathfrak{g})] \subset \mathfrak{g}^{\bot}$.

I'm kind of stuck on this one. I know basic theorems about solvable Lie algebras, like Lie's theorem and Cartan's criterion. The first part of the exercise was to show that $\mathfrak{g}^{\bot} \subset R(\mathfrak{g})$ which wasn't hard, it follows straight from Lie's theorem. Thanks for any help.

Let $L$ be a solvable ideal.

Let $f:\mathfrak{g}\to\mathfrak{gl}(W)$ be a nonzero finite-dimensional representation of $\mathfrak{g}$. By Lie's theorem, $L$ has an eigenspace. This is a nonzero subspace $V$ of $W$ such that for every $h\in L$ and $v\in V$, we have $g\dot v=t(g)v$. It is straightforward that $t$ is a Lie algebra homomorphism; in particular $M=[L,L]$ kills $V$. The set of elements killed by the ideal $M$ is a subrepresentation (because $M$ is an ideal). If $W$ is irreducible, then we deduce that $M$ kills $W$. In particular, for all $g\in \mathfrak{g}$ and $m\in M$ we have $f(g)f(m)=0$, and hence $\mathrm{Tr}(f(g)f(m))=0$. This latter consequence on traces only depends on diagonal blocks in any upper triangular block decomposition, and hence holds for an arbitrary finite-dimensional representation $W$ as a consequence of the irreducible case. In particular, it applies to the adjoint representation, and hence the Killing form vanishes.

Edit: I can give some more precise statements.

1) the above proof can be adapted to show the following: let $\mathfrak{r}$ be the solvable radical. Then for every finite-dimensional representation $\rho$, the ideal $[\mathfrak{g},\mathfrak{g}]\cap\mathfrak{r}$ belongs to the kernel of bilinear form $(x,y)\mapsto\mathrm{Tr}(\rho(x)\rho(y))$.

The argument is the same: the point is that in the irreducible case, $\mathfrak{r}$ acts by scalar matrices, while the trace on this representation is a homomorphism defined on all $\mathfrak{g}$, so the scalar vanishes on $[\mathfrak{g},\mathfrak{g}]\cap\mathfrak{r}$, which thus acts trivially. Note that $[\mathfrak{g},\mathfrak{g}]\cap\mathfrak{r}$ is often larger than $[\mathfrak{r},\mathfrak{r}]$ (typical example: $\mathfrak{sl}_n(\mathbf{C})\ltimes\mathbf{C}^n$, $n\ge 2$).

Conversely, it is not hard to produce a finite-dimensional representation whose associated bilinear form has kernel precisely $[\mathfrak{g},\mathfrak{g}]\cap\mathfrak{r}$.

2) It actually holds that if $\mathfrak{n}$ is the nilpotent radical of $\mathfrak{g}$ (so $[\mathfrak{r},\mathfrak{r}]\subset[\mathfrak{g},\mathfrak{g}]\cap\mathfrak{r}\subset \mathfrak{n}$; the latter inclusion can also be strict, e.g., when $\mathfrak{g}$ is nilpotent), then $\mathfrak{n}$ itself belongs to the kernel of the Killing form. Indeed, $\mathfrak{n}$ acts nilpotently on $\mathfrak{g}$ (because it acts trivially on $\mathfrak{g}/\mathfrak{n}$ and nilpotently on $\mathfrak{n}$), so the scalars through which it acts on irreducible subquotients of the adjoint representation are zero, and the previous arguments go through.

• This is neat, thanks! – fedlemming Dec 26 '16 at 8:45