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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.

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  • $\begingroup$ 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. $\endgroup$ – Victor Protsak Apr 11 '19 at 6:06
  • $\begingroup$ 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. $\endgroup$ – C.Niculescu Apr 11 '19 at 9:56

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