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Transferred from MSE where it now received a complete answer. Maybe the following is easy, but I am not an expert in finite-dimensional Lie algebras and was stuck on the following problem.

Can you prove or disprove the following statement ?

S) Let $V$ be an indecomposable (finite-dimensional) module over a nilpotent $k$-Lie algebra ($k$ is algebraically closed), then there is a unique character $c: {\frak{g}}\to k$ ($c$ is linear and $[{\frak{g}},{\frak{g}}]\subset ker(c)$) such that, for all $g\in \frak{g}$, the operator $$ \pi(g)-c(g)Id_V\in End_k(V) $$ is nilpotent ($\pi$ is the representation morphism).

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    $\begingroup$ Math.SE link: Indecomposable modules over nilpotent Lie algebra. $\endgroup$ – Martin Sleziak Mar 17 '18 at 8:31
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    $\begingroup$ [copied as answer to MathSE post] It's true and done in Bourbaki, Groupes et algèbres de Lie (Lie groups and Lie algebras), beginning of Chap VII. See esp. Prop 9 in §1.3 (on decomposition of modules over nilpotent Lie algebras). $\endgroup$ – YCor Mar 17 '18 at 8:38
  • $\begingroup$ Oh, thank you (+1 everywhere). This seems to solve my problem, I must enter the notations though (then, I will probably accept your answer on the basis of fruitful and spotted interaction :). $\endgroup$ – Duchamp Gérard H. E. Mar 17 '18 at 8:50
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    $\begingroup$ I've added a few details to the answer there. $\endgroup$ – YCor Mar 17 '18 at 9:06
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The answer is positive. Details and link to Bourbaki, Lie groups and Lie algebras, Chap VII, are given in the answer here.

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