1
$\begingroup$

Suppose you want to construct a perfect Lie algebra with a nonabelian solvable radical $\mathfrak{r}$, say with a commutator series of length 2. What are the conditions that guarantee the Lie algebra will be perfect? Is it sufficient to have an irreducible representation of a semisimple Lie algebra $\mathfrak{s}$ on $\mathfrak{r}/[\mathfrak{r},\mathfrak{r}]$ and $[\mathfrak{r},\mathfrak{r}]$?

$\endgroup$
2
  • $\begingroup$ It is necessary and sufficient that the representation of $\mathfrak{s}$ on $\mathfrak{r}/[\mathfrak{r},\mathfrak{r}]$ does not contain the trivial representation. (I assume the Lie algebras are in char 0 and finite-dimensional, which I guess is implicit is your question but not for everybody.) $\endgroup$
    – YCor
    Oct 28, 2014 at 10:58
  • $\begingroup$ See for a related post here. $\endgroup$ Oct 28, 2014 at 16:27

1 Answer 1

1
$\begingroup$

Let $K$ be a field of char. zero, $\mathfrak{g}$ a f.dim. Lie algebra over $K$. Let $\mathfrak{r}$ be its solvable radical, $\mathfrak{s}=\mathfrak{g}/\mathfrak{r}$ (which is semisimple) and $V=\mathfrak{r}/[\mathfrak{r},\mathfrak{r}]$ (which is naturally an $\mathfrak{s}$-module). Then $\mathfrak{g}$ is perfect iff $V^\mathfrak{s}=\{0\}$, by a simple verification using the existence of a Levi factor. (In particular, no assumption is needed on the action of $\mathfrak{s}$ on $[\mathfrak{r},\mathfrak{r}]$, and irreducibility is not the relevant condition.)

$\endgroup$
1
  • 1
    $\begingroup$ To illustrate that the nilradical can contain trivial $\mathfrak{s}$ modules, consider the semidirect product of $\mathfrak{s}=\mathfrak{sl}_2(\mathbb{R})$ and the $3D$ Heisenberg algebra $\mathfrak{h}^3$. Then as an $\mathfrak{s}$ module, $\mathfrak{h}^3=V^2+\mathbb{R}$ and $\mathbb{R}$ is trivial, but $\mathfrak{g}=\mathfrak{s}\ltimes \mathfrak{h}^3$ is perfect. $\endgroup$ Oct 30, 2014 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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