1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ This is neat, thanks! $\endgroup$ – fedlemming Dec 26 '16 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.