Indecomposable modules over nilpotent Lie algebra Is something known about indecomposable (finite-dimensional) modules over nilpotent Lie algebras (the ground field is algebraically closed) ? In particular, I wonder:

Is the following statement true ? 
S) Let $V$ be an indecomposable (finite-dimensional) module over nilpotent $k$-Lie algebras ($k$ is algebraically closed), then there is a unique character $c: \frak{g}\to \mathbb{k}$ such that, for all $g\in \frak{g}$, the operator 
  $$
g_V-c(g)Id_V\in End_k(V)
$$
  is nilpotent.

 A: It's true and done in Bourbaki, Lie groups and Lie algebras, beginning of Chap VII. See esp. Prop 9 in §1.3 (on decomposition of modules over nilpotent Lie algebras).
A few details. Fix $K$ a field of characteristic zero. Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over $K$ and $V$ a finite-dimensional $\mathfrak{g}$-module, given by a Lie algebra homomorphism $\pi:\mathfrak{g}\to\mathrm{End}_K(V)$. For a Lie algebra homomorphism $\lambda:\mathfrak{g}\to K$, define $$M^\lambda(V)=\{x\in V\mid \exists n:\forall g\in\mathfrak{g}:\;(\pi(g)-\lambda(g)\mathrm{Id}_V)^nx=0\}.$$
Then Proposition 9 of the above-mentioned reference states that when $\lambda$ varies, these are $\mathfrak{g}$-submodules, they generate their direct sum, and that if $\pi(g)$ is $K$-trigonalizable for every $g$ (which holds when $K$ is algebraically closed), then the direct sum is $V$. 
In particular when $K$ is algebraically closed and $M$ is indecomposable, this forces $V=M^\lambda(V)$ for some $\lambda$.
A: Yes, there is a lot known. What do you want to know? In low dimensions we can obtain indecomposable modules like here,
Indecomposable representations of Lie algebra
by passing to the nilpotent radical $\mathfrak{n}$ of the Borel subalgebra.
