3
$\begingroup$

Let $\mathcal{O}^{\mathfrak{p}}$ be a parabolic subcategory (see, Humphrey's BGG's Ch.9) of the BGG category $\mathcal{O}$ (w.r.t the Cartan $\mathfrak{h}$ and Borel $\mathfrak{b}$) over a finite-dimensional,complex, semisimple Lie algebra $\mathfrak{g}$.

First, assume that $\mathfrak{p} = \mathfrak{b}$. It follows from Bott's theorem that for all $n\in \mathbb{N}$ \begin{align} Ext^n_{\mathcal{O}}(M(\mu),N) \cong H^n(\mathfrak{n},N)_{\mu}, \end{align} where $N\in \mathcal{O}$, and $M(\mu)$ is the Verma module, and $\mathfrak{n}$ is the nilradical, also see Humphreys's BGG's Section 6.15.

Namely, this is a special case in the following equality by letting $\mathfrak{l} = \mathfrak{h}$

\begin{align} Ext^n_{\mathcal{O}}(M(\mu),N) \cong Hom_{\mathfrak{l}}(L(\mathfrak{l}, \mu),H^n(\mathfrak{n},N)), \end{align} for all $n\in \mathbb{N}$.

$\textbf{Question: Is it true for general parabolic subalgebra?}$

Thanks!

Improve this question
$\endgroup$
1
  • $\begingroup$ I never got far enough into this literature to give you an expert answer, but a basic source is Kostant's fundamental 1961 Annals paper: ams.org/mathscinet-getitem?mr=0142696 Even though some of his notation is now obsolete, he influenced a great many people who wrote related papers afterward. The key point, I think, is that he studied cohomology for nilradicals of parabolic subalgebras in a way that explained Bott's observations when the parabolic is a Borel subalgebra. Later work of Knapp and Wallach might be most helpful in rewriting Kostant's ideas. $\endgroup$
    – Jim Humphreys
    Commented May 18, 2016 at 22:29

1 Answer 1

Reset to default
1
$\begingroup$

The answer appears to be yes if the block is regular -- I just read this statement the other day in the paper by Boe-Hunziker on BGG resolutions of Kostant modules in parabolic category O: http://arxiv.org/pdf/math/0604336.pdf. You have to replace Verma modules with generalized Vermas on the one side, and take the Hom with respect to the Levi and take the cohomology of the nilradical of the parabolic on the other side, then you're good to go. They say it's well-known and the proof follows from an extension of the proof in usual category O and they give a reference for that. See Section 3.4 of their paper.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .