# Question about Jantzen-Zuckerman's translation princinple

Section 2.3, I quote " Assume from now on that $\mu$ is an integral weight and $\mu+\rho$ is antidominant; i.e., $(\mu+\rho,\alpha^\lor) \in \mathbb{Z}_{\le 0}$ for all $\alpha\in\Delta$; (if it is not antidominant, we can replace it by a W -translate, so we are justified in making this assumption). Let $\Phi_\mu = \{ \alpha\in\Phi: (\mu+\rho,\alpha^\lor)=0\}$.

If $\Phi_\mu = \emptyset$, then $\mu + \rho$ is a regular weight.

If $\mu+\rho$ and $\nu+\rho$ are both regular weights, then $\mathcal{O}^\mu_S$ is equivalent to $\mathcal{O}^\nu_S$ by the Jantzen–Zuckerman translation principle."

I would like to know why the last assertion is true? What is Jantzen–Zuckerman translation principle?

• I don't have time to look at the details of the paper right now, but I would assume that this is a reference to the fact that under the given conditions, composing the relevant inclusion, translation and Zuckerman functors gives an equivalence. – Tobias Kildetoft Sep 11 '18 at 13:01
• Two quick comments: 1) Your way of referencing a paper by Kenyon Platt is too cumbersome. It's better to write <a href="sciencedirect.com/science/article/pii/S0021869309002671">*here*</a>. 2) The tag 'translation' usually lists requests for English translations, while the tag 'ct.category-theory' is usually inappropriate for the BGG category $\mathcal{O}$ and the like. – Jim Humphreys Sep 11 '18 at 13:31
• Maybe this question, is too elementary to be asked here, but the more analytic (Lie group) version by Gregg Zuckerman in English and the more algebraic (Lie algebra) version by Jens Carsten Jantzen in German need to be sorted out. For the latter, see my 2008 AMS graduate textbook, Chapter 7 (modulo numerous corrections posted on my webpage). The category equivalence for regular weights is treated in 7.8. – Jim Humphreys Sep 11 '18 at 15:00
• In chapter 7.8 of your 2008 AMS graduate textbook, the theorem answers my question in the setting of BGG category $\mathcal{O}$. But how to see it is also true for the parabolic BGG category $\mathcal{O}^\mathfrak{p}$? – James Cheung Sep 14 '18 at 10:16
• Sorry for the delay in responding to your comment. My information a decade ago when the book was written (see section 9.15 there) was that the blocks of parabolic categories weren't yet known in general. If one knew the blocks one might be able to settle the equivalence question as in 7.8. (What is true in Platt's special cases?) – Jim Humphreys Sep 16 '18 at 17:23