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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.

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    $\begingroup$ 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. $\endgroup$ – Rafael Mrđen Nov 23 '17 at 22:55
  • $\begingroup$ Yeah i agree (as i said in my question) but do you have a reference? Or an idea for the argument? :-) $\endgroup$ – bob Nov 23 '17 at 23:11
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    $\begingroup$ 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). $\endgroup$ – Rafael Mrđen Nov 24 '17 at 9:57
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Modulo the messy bookkeeping sometimes encountered when passing to dual modules, what Kempf does in his paper is to exploit the geometric setting of Cousin complexes in order to find an independent approach to the BGG resolution. Though this is mostly limited to working over a sufficiently large field of characteristic 0, the result is (essentially) the dual of the usual BGG resolution but translated into the language of algebraic groups. In particular, the guess by bob about the next term of the sequence is correct, though with the usual Verma modules here replaced by their duals in the BGG category.

Note that the type of algebraic group which is of primary interest here is semisimple, though the reductive case (with nontrivial semisimple derived group) is then easily treated. But characteristic 0 is above all essential in order to start with a simple module, one affording an irreducible representation. Here the induction process, or equivalently the functor of global sections of line bundles on a suitable flag variety, yields such a simple module for the group or its Lie algebra: this is essentially the Borel-Weil Theorem. Of course, these ideas originated in differential geometry relative to Lie groups, but the Chevalley classification of semisimple algebraic groups shows that this is the same result as for semisimple algebraic groups (and in fact almost the same in any characteristic). So one can shift without extra difficulty into algebraic geometry, at least in characteristic 0.

Which approach you prefer depends mostly on the language with which you are most comfortable and the further steps you have in mind. Kempf himself thought mostly in geometric terms, even when proving his important 1976 vanishing theorem for higher cohomology groups of dominant line bundles in prime characteristic (analogue of Kodaira vanishing); but soon afterward both Andersen and Haboush found a simpler algebraic proof (incorporated in Jantzen's book Representations of Algebraic Groups, Chapter II.4, Academic Press, 1987; Amer. Math. Soc., 2nd ed., 2003). To understand what is going on, it's probably easiest in any case to start with rank 1, where the ideas in Kempf's version of the BGG resolution (or else the original Lie algebra version) are transparent.

One other remark is that there is of course a reasonable generalization of Kempf's set-up to the parabolic framework (as rafaelm indicates in a comment): here a Borel subgroup is replaced by an arbitrary parabolic subgroup. This dualizes the way Alvany Rocha (in her Rutgers thesis work with Wallach) generalized the classical BGG resolution to the parabolic case for semisimple Lie algebras.

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  • $\begingroup$ 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"). $\endgroup$ – bob Nov 25 '17 at 14:39
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    $\begingroup$ 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). $\endgroup$ – Jim Humphreys Nov 25 '17 at 21:15
  • $\begingroup$ 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. $\endgroup$ – Jim Humphreys Nov 26 '17 at 21:19

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