# 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.

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 $\lambda$.

• Perfect ! This is exactly what I needed. – Duchamp Gérard H. E. Mar 17 '18 at 9:53

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.

• I added what I would like to know within the question (+1) – Duchamp Gérard H. E. Mar 12 '18 at 7:51
• I accepted your answer rather quickly, but it doesn't help me much. Indeed a lot is known for Lie algebras linked (Borel and nilpotent radical) to semi-simple ones. I do not see how all these techniques could prove (or disprove) my statement (S) in the general case. Can you help ? – Duchamp Gérard H. E. Mar 15 '18 at 20:03