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The BGG Resolution — Examples

1 · Prerequisites

2 · Summary

These five leaves make the resolution concrete. The rank-one example writes the two-term resolution of L(mω) for sl2 with the singular-vector embedding, and the A2 example displays the six Verma summands with their eight signed cover maps and verifies the diamond square condition term by term; the sign-cancellation example computes one diamond entry in full.

The two counterexamples mark the boundaries of the construction: with all signs equal to +1 the edge sums fail to square to zero already in the smallest non-abelian diamond, and the construction fails at a singular (non-dominant) weight, where the dot translates collide and the augmentation no longer has the simple module as its cokernel.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The BGG resolution for sl2

Example

Assume the Axiom of Choice (The Axiom of Choice). Let g=sl2 with W={1,s}, positive root α, ρ=α/2 and λ=mω with m≥0 dominant integral. Then s∘λ=−λ−2ρ=−(m+2)ω and the BGG resolution of The BGG resolution of a finite-dimensional simple module reads

0→M(−(m+2)ω)→M(mω)→L(mω)→0,

with the injection the canonical embedding of the Verma submodule generated by the singular vector fm+1vλ and the surjection the canonical projection. Both end degrees are computed: ∣Φ+∣=1, w0=s, C0=M(λ), C1=M(s∘λ).

Facts & Assumptions

Given: The Axiom of Choice, g=sl2 with its standard positive root α and ρ=α/2=ω, the Weyl group W={1,s}, a dominant integral weight λ=mω with m≥0, and the BGG complex C∙(λ).

[F1]

∣Φ+∣=1, α=2ω, s(ω)=−ω and w0=s; for λ=mω one has λ+ρ=(m+1)ω and s∘λ=s(λ+ρ)−ρ=−(m+1)ω−ω=−(m+2)ω=−λ−2ρ (Finite Weyl root system, lattice and chamber conventions, The Weyl vector rho for a chosen positive system, The Bruhat graph and the BGG Verma sum in degree k).

[F2]

C0(λ)=M(λ), C1(λ)=M(s∘λ), and Ck(λ)=0 for k≥2; the only arrow is the cover s⊳e, the first differential is d1=ε(s,e)ιs→e with ε(s,e)=±1, and d0=π ⁣:M(λ)↠L(λ) (The Bruhat graph and the BGG Verma sum in degree k, The BGG differential from signed Verma maps).

[F3]

The canonical embedding ιs→e ⁣:M(s∘λ)↪M(λ) is a nonzero g-homomorphism, injective with image the submodule generated by the singular vector f⟨λ+ρ,α∨⟩vλ=fm+1vλ of weight s∘λ; it is the canonical submodule M(s∘λ)⊆M(λ) and every element of the one-dimensional space Hom⁡g(M(s∘λ),M(λ)) is a scalar multiple of it (The simple-root singular vector in a Verma module, Dominant integral dot translates embed canonically in the Verma module, Bruhat covers give canonical Verma embeddings, and composites are inclusions, Simple-reflection embeddings of Verma modules).

[F4]

For λ∈Λ+ the full augmented sequence is exact: im⁡d1=ker⁡d0=ker⁡π, coker⁡d1=L(λ), and here ker⁡d1=0 because C2(λ)=0 and d1 is injective (The BGG resolution of a finite-dimensional simple module, The augmentation kernel is the sum of the simple-reflection Verma submodules).

[F5]

The irreducible sl2-module with highest weight mω has dimension m+1 (Finite-dimensional representations of sl_2); the weight −(m+2)ω satisfies ⟨−(m+2)ω+ρ,α∨⟩=−(m+1)<0, so M(−(m+2)ω) is simple (Antidominant regular Verma modules are simple).

Verification

1.1F1F2

The terms: by [F1] and [F2], C0=M(mω) and C1=M(s∘mω)=M(−(m+2)ω), and Ck=0 for k≥2; hence the complex is concentrated in degrees 0 and 1, and the sequence displayed in the Example is the BGG complex with d0=π.

1.2F3F5

The injection: by [F3] the map d1 is, up to the sign ±1, the canonical embedding M(−(m+2)ω)↪M(mω) with image the Verma submodule generated by fm+1vλ, and it is injective; its source is even a simple Verma module by [F5], but simplicity is not needed for the injection.

1.3F4F5

Exactness at C0: by [F4], im⁡d1=ker⁡π=ker⁡d0 and coker⁡d1=L(mω); equivalently the middle quotient M(mω)/M(−(m+2)ω)≅L(mω) is the standard finite-dimensional quotient of dimension m+1 by [F5].

2.1step 1.2

Exactness at C1: ker⁡d1=0=im⁡d2, since C2=0 and d1 is injective by step 1.2; this is the injectivity of a nonzero Verma map.

3.1step 1.1step 1.3step 2.1∎

Combining the computed end degrees of step 1.1, the identification of the injection in step 1.2, and the exactness in steps 1.3 and 2.1, the BGG resolution of L(mω) is exactly 0→M(−(m+2)ω)→M(mω)→L(mω)→0 with the stated maps, and ∣Φ+∣=1 is the length of the resolution.

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The A2 BGG resolution with six Verma summands

Example

Assume the Axiom of Choice (The Axiom of Choice). Let g=sl3 with simple roots α1,α2, W=S3={e,s1,s2,s1s2,s2s1,w0}, and let λ∈Λ+ be dominant integral. The BGG complex has

C0=M(λ),C1=M(s1∘λ)⊕M(s2∘λ),C2=M(s1s2∘λ)⊕M(s2s1∘λ),C3=M(w0∘λ),

with d1 given by the two embeddings M(si∘λ)↪M(λ), d2 given by the four cover embeddings (s1s2⊳s1, s1s2⊳s2, s2s1⊳s2, s2s1⊳s1) with compatible signs, and d3 given by the two cover embeddings w0⊳s1s2 and w0⊳s2s1 with compatible signs. All Bruhat intervals of rank two are diamonds, so d2=0 holds by the square condition, and the complex is exact by The BGG resolution of a finite-dimensional simple module. The weights w∘λ=w(λ+ρ)−ρ are pairwise distinct and the listed embeddings are the canonical submodules inside M(λ).

Facts & Assumptions

Given: The Axiom of Choice, the A2 root system with simple reflections s1,s2 and longest element w0=s1s2s1=s2s1s2, a dominant integral weight λ, and the BGG complex C∙(λ).

[F1]

For type A2 the Weyl group is W=S3={e,s1,s2,s1s2,s2s1,w0} with w0=s1s2s1=s2s1s2 and lengths 0,1,1,2,2,3. The length-adjacent pairs are s1⊳e, s2⊳e, s1s2⊳s1, s1s2⊳s2, s2s1⊳s2, s2s1⊳s1, w0⊳s1s2, w0⊳s2s1; each longer element has a reduced expression containing the shorter one as a subword (w0=s1s2s1 contains s1s2 in positions 1,2 and s2s1 in positions 2,3, and s1s2 contains s1 and s2), so each pair is a Bruhat cover, and since these are all the length-adjacent pairs they are exactly the covers of the A2 Bruhat graph (The Bruhat graph and the BGG Verma sum in degree k, Bruhat covers are right multiplication by positive-root reflections, Finite Weyl root system, lattice and chamber conventions, Finite Weyl positive roots and simple reflections).

[F2]

Ck(λ)=⨁ℓ(w)=kM(w∘λ), so the terms are the four displayed sums, with 1+2+2+1=6 Verma summands in total; the differential dk has (w,w′)-component ε(w,w′)ιw→w′ for a cover w⊳w′ and 0 otherwise (The Bruhat graph and the BGG Verma sum in degree k, The BGG differential from signed Verma maps).

[F3]

For every cover x⊳y the map ιx→y ⁣:M(x∘λ)↪M(y∘λ) is the canonical inclusion of the Verma submodule of M(y∘λ) generated by the singular vector of weight x∘λ, and for a saturated path x⊳m⊳y the composite is the canonical inclusion M(x∘λ)↪M(y∘λ), independent of m (Dominant integral dot translates embed canonically in the Verma module, Bruhat covers give canonical Verma embeddings, and composites are inclusions).

[F4]

A rank-two Bruhat interval {y<m<x} with ℓ(x)=ℓ(y)+2 has exactly two middle elements, both covered by x and covering y; a compatible sign function exists, with opposite total signs on the two saturated paths of every such interval (Bruhat intervals of rank two are diamonds, Compatible signs exist on the Bruhat graph).

[F5]

For a complex built from compatible signs as in [F2] one has dk−1∘dk=0 for all k≥2, because each component is a sum over the (zero or two) middle elements of a rank-two interval and the two path contributions cancel by [F4] (The BGG differential squares to zero).

[F6]

For λ∈Λ+ the augmented BGG complex is exact, and the weights w∘λ are pairwise distinct (The BGG resolution of a finite-dimensional simple module, Positive coroot pairings of a dominant integral weight).

Verification

1.1F1F2

The six summands: by [F1] the lengths are 0,1,1,2,2,3, so C0=M(λ), C1=M(s1∘λ)⊕M(s2∘λ), C2=M(s1s2∘λ)⊕M(s2s1∘λ) and C3=M(w0∘λ): six Verma summands in total, as displayed.

1.2F1F2F3

The differentials: the covers of [F1] fall into 2 (from length 1 to 0), 4 (from length 2 to 1) and 2 (from length 3 to 2); by [F2] the components of d1,d2,d3 are exactly the signed cover embeddings listed in the Example, and by [F3] these listed embeddings are the canonical submodules of the ambient M(λ) along the respective paths.

1.3F6

Exactness: by [F6] the full sequence 0→C3→C2→C1→C0→L(λ)→0 is exact; this is the assertion that the displayed six-summand complex resolves L(λ). The weights w∘λ for the six elements w are pairwise distinct by [F6], so the six summands are pairwise non-isomorphic labelled Verma modules.

2.1F4F5step 1.2

The rank-two intervals of the Bruhat poset of type A2 are [e,s1s2] with middles s1,s2, [e,s2s1] with middles s1,s2, [s1,w0] with middles s1s2,s2s1, and [s2,w0] with middles s1s2,s2s1; each has exactly two middle elements by [F4]. Hence every component of d2 either is zero (no middle) or cancels by the square condition, so dk−1∘dk=0 for all k≥2 by [F5].

3.1step 1.1step 1.2step 2.1step 1.3∎

Each listed structural claim is verified: the terms in step 1.1, the signed cover differentials in step 1.2, vanishing of d2 in step 2.1, and exactness and weight distinctness in step 1.3.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Sign cancellation in an A2 Bruhat diamond

Example

Assume the Axiom of Choice (The Axiom of Choice). In the A2 setting of The A2 BGG resolution with six Verma summands, the interval [e,s1s2] has exactly two saturated paths s1s2⊳s1⊳e and s1s2⊳s2⊳e, and the interval [s1,w0] has exactly two saturated paths w0⊳s1s2⊳s1 and w0⊳s2s1⊳s1. In each case the two composites of canonical inclusions M(x∘λ)→M(y∘λ) coincide, while the two products of signs are opposite, so the corresponding component of d2 vanishes: for the (w0,s1)-component of the composite of the last two differentials one computes

ε(w0,s1s2)ε(s1s2,s1)+ε(w0,s2s1)ε(s2s1,s1)=0,

and analogously for the (s1s2,e)-component. This makes the cancellation mechanism of The BGG differential squares to zero explicit on the smallest non-abelian diamond.

Facts & Assumptions

Given: The Axiom of Choice, the A2 data of The A2 BGG resolution with six Verma summands with dominant integral weight λ, and the differentials dk built from a compatible sign function ε.

[F1]

In type A2 the interval {e,s1,s2,s1s2} has ℓ(s1s2)=2 and ℓ(e)=0, and both s1 and s2 lie strictly between because each is a reduced subword of s1s2 and covers e; hence its two intermediate elements are s1,s2 and the two saturated paths are s1s2⊳s1⊳e and s1s2⊳s2⊳e. Likewise {s1,s1s2,s2s1,w0} has ℓ(w0)=3, ℓ(s1)=1, and both s1s2 and s2s1 lie strictly between because w0=s2s1s2 contains s2s1 as a subword and s1 is a subword of both s1s2 and s2s1; hence its two saturated paths are w0⊳s1s2⊳s1 and w0⊳s2s1⊳s1. The diamond lemma identifies these as the only saturated paths of the two intervals (Bruhat intervals of rank two are diamonds, Bruhat covers are right multiplication by positive-root reflections, The Bruhat graph and the BGG Verma sum in degree k).

[F2]

For a cover x⊳y the (x,y)-component of the relevant differential is ε(x,y)ιx→y; consequently a two-step component is the sum over the intermediate elements, and for a saturated path x⊳m⊳y the composite ιm→y∘ιx→m is the canonical inclusion M(x∘λ)↪M(y∘λ), the same for all m (The BGG differential from signed Verma maps, Bruhat covers give canonical Verma embeddings, and composites are inclusions).

[F3]

For every square the product of the four signs is −1, so the two saturated paths of a diamond carry opposite total signs: ε(x,m1)ε(m1,y)=−ε(x,m2)ε(m2,y) (Compatible signs exist on the Bruhat graph).

Verification

1.1F1F2F3

The two diamonds and their paths are as displayed by [F1]; the composites along the two paths in each diamond are equal by [F2], and the two sign products are opposite by [F3].

2.1F2F3step 1.1

For the (s1s2,e)-component of d1∘d2 the two contributions come from the middles s1 and s2: the component equals ε(s1s2,s1)ε(s1,e) ι+ε(s1s2,s2)ε(s2,e) ι, where ι is the common composite M(s1s2∘λ)↪M(λ); since the two coefficients are opposite by [F3], the whole component is (ε(s1s2,s1)ε(s1,e)+ε(s1s2,s2)ε(s2,e))ι=0.

2.2F2F3step 1.1

For the (w0,s1)-component of d2∘d3 the two contributions come from the middles s1s2 and s2s1: the component equals (ε(w0,s1s2)ε(s1s2,s1)+ε(w0,s2s1)ε(s2s1,s1))ι′, and this vanishes because the two path products are opposite by [F3].

3.1step 2.1step 2.2∎

The two computations exhibit the cancellation explicitly in the two entries that involve both intermediate elements of a diamond: the coincidence of the composites lets the two terms be added, and the opposite signs make the sum zero. This is exactly the mechanism by which the signed differential squares to zero in these components.

CounterexampleConstruction: AI-generatedVerification: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Unsigned Bruhat edge sums need not square to zero

Statement refuted

Assume the Axiom of Choice (The Axiom of Choice). In the A2 BGG complex the signs can all be taken equal to +1: with dkuns defined by using the canonical cover embeddings with coefficient +1 on every arrow, the unsigned edge sums satisfy dk−1uns∘dkuns=0 and form a complex.

Facts & Assumptions

Given: The Axiom of Choice, the A2 setting g=sl3, simple roots α1,α2, W=S3={e,s1,s2,s1s2,s2s1,w0}, a dominant integral weight λ∈Λ+, and the unsigned edge sums dkuns ⁣:Ck(λ)→Ck−1(λ) whose (w,w′)-component is ιw→w′ for every cover w⊳w′ and 0 otherwise (every sign +1).

[F1]

C0(λ)=M(λ), C1(λ)=M(s1∘λ)⊕M(s2∘λ), C2(λ)=M(s1s2∘λ)⊕M(s2s1∘λ), and the covers of the A2 Bruhat graph are w0⊳s1s2, w0⊳s2s1, s1s2⊳s1, s1s2⊳s2, s2s1⊳s2, s2s1⊳s1, s1⊳e, s2⊳e (The Bruhat graph and the BGG Verma sum in degree k, Bruhat intervals of rank two are diamonds).

[F2]

For a cover x⊳y the canonical embedding ιx→y ⁣:M(x∘λ)↪M(y∘λ) is nonzero and injective; for a saturated path x⊳m⊳y the composite ιm→y∘ιx→m is the canonical inclusion of M(x∘λ) into M(y∘λ) and is independent of the middle element m (Bruhat covers give canonical Verma embeddings, and composites are inclusions, Dominant integral dot translates embed canonically in the Verma module).

[F3]

The four nonzero components of d2uns ⁣:C2→C1 are the cover embeddings s1s2→s1, s1s2→s2, s2s1→s1, s2s1→s2, all with coefficient +1. The two nonzero components of d1uns ⁣:C1→C0 are s1→e and s2→e, again with coefficient +1. Composition sums the component composites over intermediate summands (The BGG differential from signed Verma maps).

Counterexample

1.1F1F3

The (s1s2,e)-component of d1uns∘d2uns is the sum ιs1→e∘ιs1s2→s1+ιs2→e∘ιs1s2→s2. These are exactly the two saturated paths s1s2⊳s1⊳e and s1s2⊳s2⊳e of the rank-two interval [e,s1s2].

2.1F2step 1.1

Both composites are the same canonical inclusion ι ⁣:M(s1s2∘λ)↪M(λ) by [F2]. Their coefficients are both +1, so the component equals 2ι.

3.1F2step 2.1∎

The inclusion ι is nonzero and the base field C has characteristic zero, so 2ι≠0. Hence d1uns∘d2uns≠0, and the unsigned sums do not form a complex. Compatible signs are needed to make the two equal path maps cancel.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The BGG complex cannot be used unchanged at a singular weight

Statement refuted

The BGG construction works unchanged for every weight λ: with Ck=⨁ℓ(w)=kM(w∘λ) and the signed cover maps it is a complex and resolves L(λ); in particular the hypothesis λ∈Λ+ is not needed.

Facts & Assumptions

Given: g=sl2 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∘λ=w(λ+ρ)−ρ. Here λ+ρ=0 is fixed by W, so s∘λ=λ; consequently C0(λ)=M(λ), C1(λ)=M(s∘λ)=M(λ) and Ck(λ)=0 for k≥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→e the unique-up-to-scalar nonzero homomorphism M(s∘λ)→M(e∘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 d0=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∨⟩∉Z>0 holds because λ+ρ=0; note that the strict antidominant hypothesis ⟨λ+ρ,α∨⟩<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∘λ=λ, source and target of the only differential coincide: d1=ε(s,e) ι for a sign ε(s,e)=±1 and a nonzero homomorphism ι ⁣:M(λ)→M(λ), and d0=π; the unaugmented chain condition d1∘d2=0 holds vacuously since C2(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⋅id⁡ with c≠0, so d1=c′⋅id⁡ with c′≠0 and im⁡d1=M(λ)≠0.

3.1F2F3step 2.1

By [F3] M(λ) is simple, so the canonical surjection is an isomorphism and ker⁡d0=ker⁡π=0. Therefore ker⁡d0=0≠M(λ)=im⁡d1: the sequence fails to be exact at C0, L(λ) is not coker⁡(d1) (that cokernel is 0), and even the augmented square condition fails because d0∘d1=c′⋅id⁡≠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.

Sources