Let $\mathfrak{g}$ be a finite dimensional complex semisimple Lie algebra with a choice of Cartan subalgebra $\mathfrak{h}$, Borel $\mathfrak{b}$, and nilpotent radical $\mathfrak{n}$. Let $\mathcal{O}$ denote the BGG category.

Let $\mathcal{C}(\mathfrak{b})$ be the category of finite dimensional weight modules for $\mathfrak{b}$. Then, by the PBW theorem, the induction functor $$U(\mathfrak{g})\otimes_{U(\mathfrak{b})}-\colon \mathcal{C}(\mathfrak{b})\to \mathcal{O}$$ is exact. (cf. [Humphreys: Representations of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$, Remark 1.3])

A special object in the image of this functor is $\Delta(\lambda)=U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\mathbb{C}_\lambda$, the Verma module corresponding to $\lambda\in \mathfrak{h}^*$. By Lie's theorem and exactness of the functor, the image of the functor is contained in the category $\mathcal{F}(\Delta)$ of modules of $\mathcal{O}$ having a Verma flag. The following questions arise from certain cases of exact Borel subalgebras of a quasi-hereditary algebra, cf. [Koenig-K-Ovsienko: Quasi-hereditary algebras, exact Borel subalgebras, $A_\infty$-categories and boxes].

  1. Is the functor $U(\mathfrak{g})\otimes_{U(\mathfrak{b})}-\colon \mathcal{C}(\mathfrak{b})\to\mathcal{F}(\Delta)$ dense?
  2. Does this functor preserve $\operatorname{Ext}$-groups for $i\geq 2$ (or even $i\geq 1$), i.e. is $\operatorname{Ext}_{\mathcal{C}(\mathfrak{b})}^i(M,N)\cong \operatorname{Ext}_{\mathcal{O}}^i(U(\mathfrak{g})\otimes_{U(\mathfrak{b})} M, U(\mathfrak{g})\otimes_{U(\mathfrak{b})} N)$?

I somewhat expect the answer to be negative, but don't know enough Lie theory to produce counterexamples. The second question seems to be related to Delorme's result that $\operatorname{Ext}^i_{\mathcal{O}}(\Delta(\lambda),N)\cong H^i(\mathfrak{n},N)_\lambda$, see e.g. [Humphreys, Section 6.15].


Your Answer

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

Browse other questions tagged or ask your own question.