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 $[\mathfrak{r},\mathfrak{r}]$?
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$\begingroup$ 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.) $\endgroup$– YCorOct 28, 2014 at 10:58
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$\begingroup$ See for a related post here. $\endgroup$– Dietrich BurdeOct 28, 2014 at 16:27
1 Answer
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.)
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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