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The BGG complex cannot be used unchanged at a singular weight

Statement refuted

The BGG construction works unchanged for every weight $λ$ : with $C_{k} = ⨁_{ℓ (w) = k} M (w \circ λ)$ and the signed cover maps it is a complex and resolves $L (λ)$ ; in particular the hypothesis $λ \in Λ^{+}$ is not needed.

Facts & Assumptions

Given: $g = {s l}_{2}$ with positive root $α$ , Weyl vector $ρ = α / 2$ , $W = {e, s}$ , and the singular (non-dominant) weight $λ = - ρ = - ω$ .

[F1]

The dot action is $w \circ λ = w (λ + ρ) - ρ$ . Here $λ + ρ = 0$ is fixed by $W$ , so $s \circ λ = λ$ ; consequently $C_{0} (λ) = M (λ)$ , $C_{1} (λ) = M (s \circ λ) = M (λ)$ and $C_{k} (λ) = 0$ for $k \geq 2$ , and the only arrow of the Bruhat graph is the cover $s ⊳ e$ (The Bruhat graph and the BGG Verma sum in degree k, Finite Weyl root system, lattice and chamber conventions, The Weyl vector rho for a chosen positive system).

[F2]

The mimic of the BGG differential takes for the arrow $s \to e$ the unique-up-to-scalar nonzero homomorphism $M (s \circ λ) \to M (e \circ λ)$ ; here this is an endomorphism of $M (λ)$ , ${Hom}_{g} (M (λ), M (λ))$ is one-dimensional and every nonzero element of it is injective, and $d_{0} = π ⁣ : M (λ) ↠ L (λ)$ is the canonical surjection (The BGG differential from signed Verma maps, Homomorphism spaces between Verma modules have dimension at most one, A nonzero homomorphism between Verma modules is injective, A Verma module has a unique simple quotient).

[F3]

$M (λ)$ is simple: the irreducibility criterion $⟨ λ + ρ, α^{\lor} ⟩ \notin Z_{> 0}$ holds because $λ + ρ = 0$ ; note that the strict antidominant hypothesis $⟨ λ + ρ, α^{\lor} ⟩ < 0$ of Antidominant regular Verma modules are simple is not met here, so that supplier alone would not cover this weight (The Verma irreducibility criterion from Shapovalov determinants).

Counterexample

1.1F1F2

Because $s \circ λ = λ$ , source and target of the only differential coincide: $d_{1} = ε (s, e) ι$ for a sign $ε (s, e) = \pm 1$ and a nonzero homomorphism $ι ⁣ : M (λ) \to M (λ)$ , and $d_{0} = π$ ; the unaugmented chain condition $d_{1} \circ d_{2} = 0$ holds vacuously since $C_{2} (λ) = 0$ .

2.1F2step 1.1

By [F2] the one-dimensional space ${Hom}_{g} (M (λ), M (λ))$ is spanned by the identity and every nonzero element is injective; hence $ι = c \cdot id$ with $c \neq 0$ , so $d_{1} = c^{'} \cdot id$ with $c^{'} \neq 0$ and $im d_{1} = M (λ) \neq 0$ .

3.1F2F3step 2.1

By [F3] $M (λ)$ is simple, so the canonical surjection is an isomorphism and $\ker d_{0} = \ker π = 0$ . Therefore $\ker d_{0} = 0 \neq M (λ) = im d_{1}$ : the sequence fails to be exact at $C_{0}$ , $L (λ)$ is not $coker (d_{1})$ (that cokernel is $0$ ), and even the augmented square condition fails because $d_{0} \circ d_{1} = c^{'} \cdot id \neq 0$ .

4.1step 3.1∎

Hence the mimic of the BGG construction at the singular weight $λ = - ρ$ does not resolve $L (λ)$ , so the theorem cannot be extended unchanged to arbitrary weights. Dominant integrality implies regularity of $λ + ρ$ ; regularity alone does not imply dominant integrality.

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Sources

A. Rocha-Caridi, Splitting criteria for modules induced from a subalgebra of a semisimple Lie algebra, Trans. AMS 262 (1980), Sec. 10, p. 353 (the hypothesis $\lambda\in P^+$ in the construction) (standard reference, not scraped)
Fan Zhou, The classical and the functorial BGG resolutions (Columbia thesis 2021), Part I Sec. 3.2, p. 11 (the construction requires $\lambda\in\Lambda^+$) (standard reference, not scraped)