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.