# BGG resolution for characters of reductive groups

Take $G$ a split reductive group (over a field of char 0) with Borel $B$, opposite Borel $\overline{B}$ and maximal split torus $T\subset B$. We write $X = G/\overline{B}$. Let $O \subset X$ be the big cell of $X$ for the Bruhat decomposition, it is a a dense open subset of $X$ which is isomorphic to $U$ the unipotent radical of $B$.

Let $\chi$ be a character of $T$ which is dominant with respect to $B$ and let $\mathcal{L}_\chi$ be the locally free sheaf on $X$ associated to $\chi$. We have that $\Gamma(X,\mathcal{L}_\chi) = Ind_{\overline{B}}^G\chi$ and $\Gamma(O,\mathcal{L}_\chi) = Ind_{T}^B \chi$. Restriction of sections gives us the inclusion : $$0 \to Ind_{\overline{B}}^G\chi \to Ind_{T}^B \chi$$

This is the beginning of the BGG resolution. I would like to understand what the third term is. The only reference I know for the BGG resolution is Humphreys book on the BGG category but he works with lie algebras and the category O and I have not been able to translate his result in the language above. Looking at his book i'm guessing we must find something along the lines of $\oplus_{\alpha \in \Delta} Ind_{T}^B (s_{\alpha} \bullet \chi)$ where $\Delta$ is the basis of the root system associated to $\overline{B}$ and $\bullet$ means the dot action. But i'm not sure so if someone could help that would be great.

In fact I'm only interested in the weights of the third term of the sequence of the BGG resolution.

I think the paper "The Grothendieck-Cousin complex of an induced representation" of Kempf, which gives a geometric approach to the BGG resolution, might be what I'm looking for but again I wasn't able to extract a simple formula. You can find the paper here the important results for me I think are Theorem 12.5, lemma 12.6 and lemma 12.8.

• The k-th term in the resolution should be $\oplus_{w \in W, l(w)=k} Ind_{T}^B (w\bullet \chi)$, where $W$ is the Weyl group and $l$ the usual length function. – Rafael Mrđen Nov 23 '17 at 22:55
• Yeah i agree (as i said in my question) but do you have a reference? Or an idea for the argument? :-) – bob Nov 23 '17 at 23:11
• There is a statement in books.google.hr/books/about/… chapter 7. There is a construction for "curved" versions of G/P, arxiv.org/abs/math/0001164 and arxiv.org/abs/math/0001158 (I am not sure if it is also difficult to extract from). – Rafael Mrđen Nov 24 '17 at 9:57

• Thank you very much for your answer Prof. Humphreys ! Just to be sure are you saying that the sequence $0 \to Ind_{\overline{B}}^G \chi \to Ind_T^B \chi \to \oplus_{\alpha \in \Delta} s_\alpha \bullet \chi$ is indeed exact or do I have to modify it somehow ? (I am a bit confused because you wrote : "though with the usual Verma modules here replaced by their duals in the BGG category"). – bob Nov 25 '17 at 14:39
• Note that you are omitting the notation Ind$_T^B$ in the last term. This gives dual Verma modules, so the part of the sequence you write is obtained from BGG by dualizing and is indeed exact. For most simple types it continues further to the right, of course. Here the relevant simple modules occur in the dual Verma modules as submodules rather than quotients, but otherwise this behaves much like the BGG resolution in characteristic 0. The full resolution formally recovers Kostant's multiplicity formula as an alternating sum (hence Weyl's character formula). – Jim Humphreys Nov 25 '17 at 21:15
• P.S. I've tried to avoid too much notation, but this geometric (dual) approach still isn't my ideal way to write down the BGG resolution. In effect, working inside the group puts the emphasis on the big cell in the Bruhat decomposition of $G$, which is similar to working with the Lie algebra of $G$ but seems more cumbersome. I guess I'd need to have more motivation in order to think in terms of algebraic groups here. – Jim Humphreys Nov 26 '17 at 21:19