Let $\mathfrak{g}$ be a finite dimensional complex Lie algebra. A derivation $D: \mathfrak{g} \rightarrow \mathfrak{g}$ is $\mathbb{C}$-linear map $\mathfrak{g} \rightarrow \mathfrak{g}$ such that for any $a, b \in \mathfrak{g}$, we have $D[a, b] = [Da, b] + [a, Db]$.
Let $\hbox{Rad}(\mathfrak{g})$ be the radical of $\mathfrak{g}$. My question is: does $D$ map $\hbox{Rad}(\mathfrak{g})$ into itself, i.e. is it true that $D(\text{Rad}(\mathfrak{g})) \subset \hbox{Rad}(\mathfrak{g})$?
On
In the case of finite-dimensional Lie algebras over $\mathbb{R}$ or $\mathbb{C}$ we can argue very straightforwardly as follows:
The radical is clearly characteristic in the sense that it is preserved by all automorphisms of $\mathfrak{g}$. Now observe that $D : \mathfrak{g} \to \mathfrak{g}$ is a derivation iff the exponential $\exp(tD) : \mathfrak{g} \to \mathfrak{g}$ is a one-parameter group of automorphisms of $\mathfrak{g}$ (in fact $\text{Der}(\mathfrak{g})$ is the Lie algebra of $\text{Aut}(\mathfrak{g})$), so this exponential preserves the radical and hence so does its derivative.
This is true. Any derivation maps the solvable radical into the nilradical, hence we have $$ D(rad(\mathfrak{g}))\subseteq nil(\mathfrak{g})\subseteq rad(\mathfrak{g}). $$ The proof can be found in Jacobson's book on Lie algebras, see Corollary $2$ to Theorem $13$ in chapter $II$, section $7$. Ideals which are invariant under all derivations are called characteristic ideals.