A friend and I are writing a paper that uses the BGG resolution of $L(\lambda)$ (where $\mathfrak g$ is a semisimple complex Lie algebra, $\lambda \in P^+$ is a dominant integral weight, and $L(\lambda)$ the simple finite module with highest weight $\lambda$). 


A point is unclear for us : one needs to choose appropriate signs for each edge $w \to w'$ in the Bruhat graph. It's probably easy but we can't see why the resulting complex is independent of the choice of the signs. Of course the maps will be different, but we expect that two different choices of signs give isomorphic complexes. 


There is a discussion of unicity of signs in Humphrey's book "Representations of Semisimple Lie Algebras in the BGG Category $\mathcal O$" chapter 6. It seems to me the conclusion is that the signs $\sigma(w',w)$ can be actually chosen in $\{-1,1\}$. I don't understand why it clarifies the uniqueness question, i.e why two different choices of such signs give the same complex up to isomorphism. 


I've probably misunderstood the argument or I don't see something obvious, but in any case I would appreciate clarification about it. Thanks in advance !



Unicity of the BGG complex
Theorem 33 in the preprint [1] gives the uniqueness of BGG resolutions (= direct sums of Verma modules resolving a simple module) in category $\mathcal{O}$, both in regular and singular blocks. (However, this does not apply directly to parabolic versions of category $\mathcal{O}$).




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[1]: Mazorchuk-Mrđen: BGG complexes in singular blocks of category $\mathcal{O}$, <https://arxiv.org/abs/1907.04121>


