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).

  • 1
    $\begingroup$ Math.SE link: Indecomposable modules over nilpotent Lie algebra. $\endgroup$ – Martin Sleziak Mar 17 '18 at 8:31
  • 2
    $\begingroup$ [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
  • 1
    $\begingroup$ I've added a few details to the answer there. $\endgroup$ – YCor Mar 17 '18 at 9:06

The answer is positive. Details and link to Bourbaki, Lie groups and Lie algebras, Chap VII, are given in the answer here.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.