Let $\mathfrak{g}$ be a finite-dimensional semisimple Lie algebra over $\mathbb{C}$ with a fixed Cartan subalgebra $\mathfrak{h}$ and a fixed system of simple roots. It is stated in Exercise 3.11 of Humphreys' book "Representations of semisimple Lie algebras in the BGG category $\mathcal{O}$" that

The Verma $U(\mathfrak{g})$-module $M(\lambda)$ of highest weight $\lambda$ is projective only if $\lambda$ is dominant.

Can anyone give me some hint to prove this proposition? It seems related to the BGG reciprocity.

  • $\begingroup$ Hint: If $\lambda$ is not dominant, then there is some $\mu\neq \lambda$ such that $L(\lambda)$ is a composition factor of $M(\mu)$. Now apply BGG reciprocity. $\endgroup$ – Tobias Kildetoft Oct 29 '15 at 9:28
  • $\begingroup$ @TobiasKildetoft Thanks for the comments. I know that if $L(\mu)$ is a composition factor of $M(\lambda)$, then $\mu\leq \lambda$. But why $M(\lambda)$ has a composition factor $L(\mu)$ with $\mu\neq\lambda$? $\endgroup$ – Zhihua Chang Oct 29 '15 at 9:44
  • $\begingroup$ Not $M(\lambda)$ but $M(\mu)$. Think of what submodules are in Verma modules. $\endgroup$ – Tobias Kildetoft Oct 29 '15 at 9:46
  • $\begingroup$ @TobiasKildetoft Indeed, this is my question. I do not know which $L(\mu)$ does appear as a composition factor of $M(\lambda)$. Is it possible that $M(\lambda)$ is irreducible when $\lambda$ is dominant? $\endgroup$ – Zhihua Chang Oct 29 '15 at 10:09
  • $\begingroup$ Hmm, I was thinking of using Proposition 1.4, but you need some integrality for that. $\endgroup$ – Tobias Kildetoft Oct 29 '15 at 10:14

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