The BGG differential from signed Verma maps

Assume the Axiom of Choice (The Axiom of Choice). Fix $\lambda \in {\mathrm{\Lambda }}^{+}$ and a compatible sign function $\epsilon$, i.e. a function from the arrows $x\to y$ of the Bruhat graph to $\{\pm 1\}$ whose product over the four edges of every square is $-1$ (Compatible signs exist on the Bruhat graph). Write $\epsilon (w,w^{\mathrm{\prime }})$ for the value of $\epsilon$ on the arrow $w\to w^{\mathrm{\prime }}$ whenever $w⊳w^{\mathrm{\prime }}$ are elements of $W$, and let ${\iota }_{w\to w^{\mathrm{\prime }}}\text{\hspace{0.17em}⁣}:M(w\circ \lambda )\hookrightarrow M(w^{\mathrm{\prime }}\circ \lambda )$ be the canonical cover embedding of Bruhat covers give canonical Verma embeddings, and composites are inclusions.

For $k\ge 1$ define a $\mathfrak{g}$-homomorphism

$d_k\text{\hspace{0.17em}⁣}:C_k(\lambda )\to C_{k-1}(\lambda )$

out of the degree-$k$ Verma sum $C_k(\lambda )={⨁}_{\mathrm{\ell }(w)=k}M(w\circ \lambda )$ of The Bruhat graph and the BGG Verma sum in degree k by requiring that its component

$(d_k{)}_{w,w^{\mathrm{\prime }}}\text{\hspace{0.17em}⁣}:M(w\circ \lambda )\to M(w^{\mathrm{\prime }}\circ \lambda )$

from the summand indexed by $w$ (with $\mathrm{\ell }(w)=k$) into the summand indexed by $w^{\mathrm{\prime }}$ (with $\mathrm{\ell }(w^{\mathrm{\prime }})=k-1$) is

$(d_k{)}_{w,w^{\mathrm{\prime }}}=\{\begin{array}{ll}\epsilon (w,w^{\mathrm{\prime }}){\iota }_{w\to w^{\mathrm{\prime }}},& w⊳w^{\mathrm{\prime }},\\ 0,& w⋫w^{\mathrm{\prime }},\end{array}$

and that all components between pairs of summands with $\mathrm{\ell }(w)\mathrm{\ne }k$ or $\mathrm{\ell }(w^{\mathrm{\prime }})\mathrm{\ne }k-1$ are zero. Since a finite direct sum in an abelian category is a biproduct, a family of morphisms between finitely many summands and vanishing outside the $(w,w^{\mathrm{\prime }})$ pairs just described determines a unique $\mathfrak{g}$-homomorphism $C_k(\lambda )\to C_{k-1}(\lambda )$; each $C_k(\lambda )$ lies in $\mathcal{O}$ and $C_k(\lambda )=0$ for $k>\mathrm{\mid }{\mathrm{\Phi }}^{+}\mathrm{\mid }$, so $d_k=0$ for $k>\mathrm{\mid }{\mathrm{\Phi }}^{+}\mathrm{\mid }+1$ as a map out of the zero object. The map $d_k$ is a morphism of degree $-1$ for the grading by $k$ of the graded object $C_{\bullet }(\lambda )$ (Chain complex in an abelian category).

Set $d_0=\pi \text{\hspace{0.17em}⁣}:C_0(\lambda )=M(\lambda )\twoheadrightarrow L(\lambda )$, the canonical surjection of A Verma module has a unique simple quotient, and set $d_k=0$ for $k<0$. The maps $d_k$ are the differentials of the BGG complex. For another compatible signing ${\epsilon }^{\mathrm{\prime }}$, Compatible signs exist on the Bruhat graph supplies vertex signs $c(w)$ with $c(e)=1$ and ${\epsilon }^{\mathrm{\prime }}(w,w^{\mathrm{\prime }})\mathrm{/}\epsilon (w,w^{\mathrm{\prime }})=c(w)\mathrm{/}c(w^{\mathrm{\prime }})$. The automorphism $T_k$ that multiplies the summand indexed by $w$ by $c(w)$ satisfies $T_{k-1}{d}_k=d_k^{\mathrm{\prime }}{T}_k$: the two component coefficients agree because $c(w{)}^2=c(w^{\mathrm{\prime }}{)}^2=1$. Also $T_0$ is the identity, so this intertwines the augmentations. Hence the resulting complexes are isomorphic. Whether $d_{k-1}\circ d_k=0$ depends only on the square condition on $\epsilon$ and is proved in The BGG differential squares to zero.