6
$\begingroup$

I am reading Humphrey's book "Representations of semisimple Lie algebras in the BGG Category O" on Page 189, Proposition 9.6, where he remarked that "Note that if we had developed the full BGG resolution of $L_I(\lambda)$, this would correspond to an exact sequence completing the one in Theorem 9.4(b)." I am interested in this idea because this would prove directly that $$ ch M_I(\lambda)=\sum_{w\in W_I}(-1)^{\ell(w)}ch M(w\cdot\lambda) $$ I wonder if it is clear that we do have a full BGG resolution of $L_I(\lambda)$ or it is still an open question for some reason. Why can't we just get the full BGG resolution of $L_I(\lambda)$ by applying Corollary 6.5 (Page 114) in that book to the Levi subalgebra? Did I miss some important points here?

$\endgroup$
  • $\begingroup$ A short comment, which I may try to fill in later: BGG resolution works classically when $L(\lambda)$ has finite dimension but otherwise seems very difficult to sort out. See my section 9.16 for a parabolic adaptation in the finite dimensional case. (This chapter is a survey, partly because there are many loose ends still under study. Notation also gets very complicated.) Meanwhile, I added tags. $\endgroup$ – Jim Humphreys Jan 6 '14 at 16:56
  • $\begingroup$ Thanks a lot for Jim's comments. Let me put in this way. In Page 189, Proposition 9.6 of your book, $\lambda\in\Lambda_I^{+}$, so $L_I(\lambda)$ is a finite dimensional module over the Levi subalgebra (which is reductive) corresponding to $I$. Applying Corollary 6.5, we get a BGG resolution of the simple module $L_I(\lambda)$ over this Levi subalgebra. Now applying the induction-inflation functor (which is exact), and using the fact this functor sends $V_I(w\cdot\lambda)$ to $M(w\cdot\lambda)$ and $L_I(\lambda)$ to $M_I(\lambda)$ we can get the desired resolution. Is my argument correct? $\endgroup$ – Hu Jun Jan 6 '14 at 23:52
  • $\begingroup$ I can't follow your last sentence. This needs much more detail, including specification of which simple module $L(\lambda)$ for $\mathfrak{g}$ (typically infinite dimensional) is to be resolved. Certainly you can resolve your f.d. $L_I(\lambda)$ for $\mathfrak{g}_I$, but what about $\mathfrak{g}$-modules when $\dim L(\lambda) = \infty$? $\endgroup$ – Jim Humphreys Jan 7 '14 at 11:45
  • 1
    $\begingroup$ I did not resolve any simple $\mathfrak{g}$-module. Instead, I resolve the parabolic Verma module $M_I(\lambda)$ using the usual Verma modules. By definition, $M_I(\lambda)=U(g)\otimes_{U(p)}L_I(\lambda)$. That says, I apply the functor $U(g)\otimes_{U(p)}?$, and I think this functor sends $V_I(w\cdot\lambda)$ to the Verma module $M(w\cdot\lambda)$ in the usual category $O$. In other words, from the BGG resolution of $L_{I}(\lambda)$ over the Levi subalgebra, we get a resolution of $M_I(\lambda)$ as indicated in the remark below Proposition 9.6 in your book. Did I make any mistake here? $\endgroup$ – Hu Jun Jan 7 '14 at 12:32
  • $\begingroup$ I got sidetracked by your reference to resolutions, since what you are actually doing is OK. My comment on page 189 just indicated this alternative line of reasoning, though it doesn't add any real information. Sorry if that wasn't clear. $\endgroup$ – Jim Humphreys Jan 7 '14 at 14:22

Your Answer

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

Browse other questions tagged or ask your own question.