Exercise 16, § 1

Let $\mathfrak{g}$ be a Lie algebra and $\mathfrak{D}$ the Lie algebra of derivations of $\mathfrak{g}$. The identity mapping of $\mathfrak{D}$ defines a semi-direct product $\mathfrak{h}$ of $\mathfrak{D}$ by $\mathfrak{g}$ called the holomorph of $\mathfrak{g}$. $\mathfrak{g}$ is identified with an ideal and $\mathfrak{D}$ with a subalgebra of $\mathfrak{h}$.

(a) Show that the centralizer $\mathfrak{g}^*$ of $\mathfrak{g}$ in $\mathfrak{h}$ is the set of $\operatorname{ad}_g x - x(x \in \mathfrak{g})$.

(b) For every element $x + d$ of $\mathfrak{h}$ ($x \in \mathfrak{g}$, $d \in \mathfrak{D}$), we write

Show that $\theta$ is an automorphism of $\mathfrak{h}$ of order 2 which maps $\mathfrak{g}$ onto $\mathfrak{g}^*$, so that $\mathfrak{h}$ can be considered as the holomorph of $\mathfrak{g}^*$.

(c) For every ideal $\mathfrak{k} \neq \{0\}$ of $\mathfrak{h}$, $\mathfrak{k} \cap (\mathfrak{g} + \mathfrak{g}^*) \neq \{0\}$. (If $\mathfrak{k} \cap \mathfrak{g} = \{0\}$, $\mathfrak{k} \subset \mathfrak{g}^*$.)

(d) If $\mathfrak{g}$ is commutative, every ideal $\mathfrak{k} \neq \{0\}$ of $\mathfrak{h}$ contains $\mathfrak{g}$. If $K$ is a field and $\dim_K \mathfrak{g} < +\infty$, deduce that, if $\mathfrak{k} \neq \mathfrak{g}$, $\mathfrak{h}$ cannot be the holomorph of $\mathfrak{k}$ (use (b)).