# Tensoring Harish-Chandra bimodules with Verma modules

The question is about the functor $$T_\lambda$$ defined by Bernstein and Gelfand in the paper Tensor Products of Finite and Infinite Dimensional Representations of Semisimple Lie Algebras.

Setup: Let $$\mathfrak{g}$$ be a complex semisimple Lie algebra, $$U=U(\mathfrak{g})$$ its enveloping algebra, let $$\lambda$$ be a dominant weight and $$\theta$$ the central character corresponding to $$\lambda$$.

Let $$U_{\theta}=U/\ker({\theta}) U$$. In section 5, Bernstein and Gelfand consider the category $$\mathcal{H}(\theta)$$ of finitely generated $$U-U_\theta$$ Harish-Chandra bimodules. They define a functor $$T_\lambda$$ from $$\mathcal{H}(\theta)$$ to the BGG category $$\mathcal{O}$$ by $$T_\lambda(H)=H \otimes_{U_{\theta}} M(\lambda).$$

In case $$\lambda$$ is also regular, the functor $$T_\lambda$$ is also exact. Is this an if and only if or is it possible for $$T_\lambda$$ to be exact for $$\lambda$$ singular? Can one characterise when $$T_\lambda$$ is exact?

In general, $$T_\lambda$$ defines an equivalence between $$\mathcal{H}(\theta)$$ and the category of $$P(\lambda)$$-presentable modules. If $$\lambda$$ is also regular this is equal to $$\mathcal{O}_{\lambda+ \Lambda}$$ (here $$\Lambda$$ is the lattice of integer weights in $$\mathfrak{h}^*$$).

One can observe that of $$T_\lambda$$ is exact iff the category of $$P(\lambda)$$-presentable modules is Abelian. Does this happen only for $$\lambda$$ regular or can it happen for $$\lambda$$ being a singular dominant weight?

Any hints, partial answers and references will be appreciated.

• A chapter in Jantzen's monograph on universal enveloping algebras of semisimple Lie algebras (in German) is devoted to Harish-Chandra bimodules and relations with category $\mathcal{O}$, with full proofs. I am not sure whether the answer to this question is contained there. – Victor Protsak Apr 11 '19 at 6:06
• Thank you for the input. I assume you are refering the book Einhüllende Algebren halbeinfacher Lie-Algebren by Jantzen. As far as I understand German, the answer is not contained in there, but it might be worth checking with a native German. – C.Niculescu Apr 11 '19 at 9:56