1
$\begingroup$

It follows from Exercise 1.13 in Humphreys' Category $\mathcal{O}$ book that $M\in\mathcal{O}^\mathfrak{p}_{\chi_\lambda}$ has a direct sum decomposition $M=\oplus M_i$ such that all weights of each $M_i$ are contained in a single coset of root lattice $\Lambda_r$ in $\mathfrak{h}^*$. Then, the category $\mathcal{O}^\mathfrak{p}_{\chi_\lambda}$ decomposes as a direct sum of full subcategories, which can be indexed by the nonempty intersection of the orbit $W\cdot\lambda$ with the cosets $\mathfrak{h}^*/\Lambda_r$. From now on, we use the anti-dominant weight $\mu$ in the intersection to parameterize the corresponding subcategory of $\mathcal{O}^\mathfrak{p}_{\chi_\lambda}$. We denote this subcategories by $\mathcal{O}^\mathfrak{p}_{\mu}$. In particular, if $I=\emptyset$, then the parabolic subalgebra $\mathfrak{p}$ collapses to a Borel subalgebra $\mathfrak{b}$ of $\mathfrak{g}$ and $\mathcal{O}^\mathfrak{p}_{\mu}$ turns into the ordinary block $\mathcal{O}_{\mu}$ of the BGG category $\mathcal{O}$.

I would like to know whether $\mathcal{O}^\mathfrak{p}_{\mu}=\mathcal{O}^\mathfrak{p}_{\chi_\mu}$ for $\mu$ being regular integral?

$\endgroup$
0
$\begingroup$

$\mathcal{O}_\chi^{\mathfrak{p}}$ is a full subcategory of $\mathcal{O}_\chi$, so the answer is yes. That is:

  • $\mathcal{O}_\mu = \mathcal{O}_{\chi_\mu}$, for all $\mu$
  • $\mathcal{O}_\mu^{\mathfrak{p}}$ is the full subcategory of $\mathcal{O}_\mu$ of locally $\mathfrak{l}$-finite modules.
  • $\mathcal{O}_{\chi_\mu}^{\mathfrak{p}}$ is the full subcategory of $\mathcal{O}_{\chi_\mu}$ of locally $\mathfrak{l}$-finite modules.

Hence $\mathcal{O}_\mu^{\mathfrak{p}} = \mathcal{O}_{\chi_\mu}^{\mathfrak{p}}$ for all $\mu$.

$\endgroup$
  • $\begingroup$ I think the right argument requires the fact that $\mu$ is integral. $\endgroup$ – James Cheung Sep 27 '18 at 13:31
  • $\begingroup$ Actually, reading the question again I don't think I understand it. Using your notation, the $\mu$ in $\mathcal{O}_\mu^{\mathfrak{p}}$ is required to be anti-dominant, you define it as such. How is $\mathcal{O}_\mu^{\mathfrak{p}}$ different from $\mathcal{O}_{\chi_\mu}^{\mathfrak{p}}$? $\endgroup$ – Johan Kåhrström Sep 28 '18 at 7:48
  • $\begingroup$ $\mathcal{O}_{\chi_\mu}^\mathfrak{p}$ can be decomposed into direct sum of full.subcategories, which can be indexed by the nonempty intersection of the orbit $W\cdot\mu $ with the cosets $\mathfrak{h}^*/\Lambda_r$. While $\mathcal{O}_{\mu}^\mathfrak{p}$ is just one direct summand of direct sum of full.subcategories. I would like to know whether the direct sum of full.subcategories is just trivial direct sum when $\mu$ is integral. $\endgroup$ – James Cheung Sep 28 '18 at 19:21
  • $\begingroup$ Ah, I see what you mean. Humphreys discusses this in Section 9.15. $\endgroup$ – Johan Kåhrström Sep 29 '18 at 9:04

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.