Let $V$ be a finite dimensional vector space over a field $k$ of characteristic zero. Suppose that $\mathfrak{g}$ is a Lie subalgebra of $\mathfrak{gl}(V)$, and suppose that the representation ad$:\mathfrak{g}\to\mathfrak{gl}(\mathfrak{g})$ is faithful, i.e. $\mathfrak{g}$ has trivial centre.
Recall that for all $x\in\mathfrak{g}$, $x=x_s+x_n$ for some commuting $x_s,x_n\in\mathfrak{gl}(V)$ such that ad$(x_s)$ is diagonalisable, ad$(x_n)$ is nilpotent, and both act as derivations on $\mathfrak{g}$. This is the Jordan decomposition for $x$. We say that $\mathfrak{g}$ is decomposable if $x_s$ and $x_n$ both lie in $\mathfrak{g}$ for all $x\in\mathfrak{g}$.
It is true that if $\mathfrak{g}$ is semisimple, then $\mathfrak{g}$ is always decomposable, using the fact that all derivations of $\mathfrak{g}$ are of the form ad$(y)$ for some $y\in\mathfrak{g}$, but I don't think this holds in general for solvable Lie algebras.
My question is, does anyone know of any examples of solvable Lie algebras with trivial centre that are not decomposable? Or if not, then does anyone know a proof that they always are?
A Lie algebra $L$ over a field of characteristic zero has an abstract Jordan-Chevalley decomposition if and only if $L$ is perfect. Hence solvable Lie algebras do not have such a decomposition.
Reference: Jordan-Chevalley decomposition.
Example: Let $L$ be the non-abelian Lie algebra of dimension $2$, with basis $x,y$ and bracket $[x,y]=x$.
If we are not assuming $[x_s,x_n]=0$ for the Jodan decomposition, then we have the following result:
Theorem: Let $L$ be a solvable Lie algebra of matrices, let $A ∈ L$ and assume that $A = S + N$ with $S, N ∈ L$, where $S$ is semisimple, $N$ is nilpotent (we are not assuming $[S, N ] = 0$). Then the semisimple and nilpotent summands of the JCD of $A$ belong to $L$.
Reference: This article.