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Category O Finiteness Duality and Blocks — Examples

1 · Prerequisites

2 · Summary

Assuming AC for the block classifications, the rank-one examples separate a regular integral block from a nonintegral central-character summand with two blocks; the type A2 calculation tests a singular weight. The counterexamples isolate finite generation, bounded upward support, Cartan semisimplicity, restricted duality and the finite-dimensional tensor-factor hypothesis. Throughout, b=h⊕n+ and the Chevalley dual preserves weights.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-09-23 (gpt-6-sol)Open item page →

The regular integral sl2 block

Example

Assume the Axiom of Choice.

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ.

For g=sl2 identify a highest weight with its value on h, so ρ=1 and s⋅λ=−λ−2. For every integer n≥0, the regular integral block has simple labels n and −n−2. Its standards are M(n) and M(−n−2)=L(−n−2), and

0⟶L(−n−2)⟶M(n)⟶L(n)⟶0

is nonsplit. Its costandards are ∇(n) and ∇(−n−2)=L(−n−2), with the nonsplit sequence 0→L(n)→∇(n)→L(−n−2)→0. The linkage order is −n−2<n.

Facts & Assumptions

Given: The Axiom of Choice (The Axiom of Choice), the setting above and the hypotheses in the example.

[F1]

The category O is the categorical direct sum of its generalized central-character subcategories, and each object has only finitely many nonzero character components (Generalized central-character decomposition of O).

[F10]

Under the Axiom of Choice, χλ=χμ if and only if μ∈W⋅λ (Central characters are dot-Weyl orbits).

[F11]

Under the Axiom of Choice, every object of O has finite length (Every object of O has finite length), and its simple objects are exactly the distinct L(η) (The simple objects of O).

[F2]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. Restricted Chevalley duality is an exact contravariant equivalence D:O→Oop, with a natural isomorphism D2≅id⁡. It preserves each weight-space dimension, the formal character, and every simple composition multiplicity. (Restricted duality is exact and involutive on O)

[F3]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. The costandard object ∇(λ) has a unique simple submodule, isomorphic to L(λ). Its socle, the sum of all simple submodules, is that submodule. (The simple socle of a costandard object)

[F4]

If ⟨λ+ρ,αi∨⟩∈Z>0, there is an embedding M(si⋅λ)↪M(λ). (Simple-reflection embeddings of Verma modules)

[F5]

The proper submodule J(λ) which is the sum of all proper submodules is the unique maximal submodule of M(λ). The quotient L(λ):=M(λ)/J(λ) is simple and is its unique simple quotient. (A Verma module has a unique simple quotient)

[F6]

The standard sl2 generators satisfy [e,f]=h and [h,f]=−2f (The special linear Lie algebra sl_2). The induced Verma highest vector satisfies evη=0 and hvη=ηvη (Verma modules), and PBW makes vk=fkvη (k≥0) a basis (The PBW model of a Verma module).

[F7]

The weights of M(λ) are exactly λ−β for β∈Q+; every weight space is finite dimensional, and M(λ)λ=Cvλ. (Weights of a Verma module lie below lambda)

[F8]

For every highest weight η, D(L(η))≅L(η) as g-modules. (Restricted self-duality of simple highest-weight modules)

[F9]

The standard and costandard objects are Δ(η)=M(η) and ∇(η)=D(M(η)). (Standard and costandard objects)

Verification

1.1F6inductionalgebra

For any highest weight η, the PBW basis in [F6] and [h,f]=−2f give hvk=(η−2k)vk. Also ev0=0, and [e,f]=h gives the induction step evk+1=f(evk)+hvk. If evk=k(η−k+1)vk−1, this step yields evk+1=[k(η−k+1)+η−2k]vk=(k+1)(η−k)vk. Therefore evk=k(η−k+1)vk−1 for all k≥1.

2.1F6F10F11step 1.1

The integral pairing is n+1>0, and the two distinct dot-orbit labels are n and −n−2. Under Choice, [F10] says that L(n) and L(−n−2) are exactly the simple labels in the central-character summand Oχn: by [F11] every simple is a highest-weight L(η) and hence has the scalar character χη. For η=−n−2, every coefficient k(η−k+1)=−k(n+k+1) from step 1.1 is nonzero when k≥1. A nonzero submodule of M(η) contains a nonzero weight vector, hence some basis vector vk; repeated application of e reaches v0, which generates the entire Verma module. Thus M(−n−2)=L(−n−2) directly.

3.1F4F6F7step 1.1step 2.1

Write vk=fkvn in M(n). The action derived in step 1.1 gives hvk=(n−2k)vk and evk=k(n−k+1)vk−1 for k≥1, with ev0=0. The weight spaces are one dimensional. The singular vector vn+1 generates the embedded M(−n−2), which is also the embedding supplied by F4.

4.1F5algebrastep 1.1step 3.1

The quotient has basis v0,…,vn. A nonzero submodule contains a weight vector, and applying e repeatedly reaches v0, since k(n−k+1)≠0 for 1≤k≤n by step 1.1. Applying f then generates the whole quotient. It is simple, hence is L(n). A split sequence would make M(n) a sum of two proper submodules, contrary to its unique maximal submodule. For n=0 the quotient consists just of v0 and the same argument holds.

5.1F2F3F8F9step 2.1step 4.1

By F8, exact duality fixes both simples; it reverses the sequence, and F9 identifies the middle term as ∇(n), giving the displayed costandard sequence. If it split, its dual would split the original. The costandard socle is L(n) by F3. Since M(−n−2)=L(−n−2), F8 and F9 also give ∇(−n−2)=L(−n−2).

6.1F1F5F11step 2.1step 4.1∎

By [F1], Oχn is a categorical direct summand. It has exactly the two simple labels from step 2.1. If this summand decomposed into two nonzero categorical summands, finite length [F11] would place a simple in each; the two labels would therefore lie in different summands. Then M(n), which has both as factors by steps 3.1 and 4.1, would split as a direct sum of two nonzero submodules, contradicting the unique-maximal-submodule property [F5]. Hence Oχn is indecomposable and is the asserted regular integral block. The integral Weyl group is the full order-two Weyl group, giving the stated linkage order −n−2<n.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-09-23 (gpt-6-sol)Open item page →

Nonintegral sl2 central characters split into two simple blocks

Example

Assume the Axiom of Choice (The Axiom of Choice).

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ.

For g=sl2 and λ∈C∖Z, Oχλ has exactly two distinct singleton linkage blocks, with labels λ and −λ−2. Each block is equivalent to finite-dimensional complex vector spaces. In particular the central-character summand is semisimple but is not one indecomposable block.

Facts & Assumptions

Given: The Axiom of Choice, the setting above, and the hypotheses in the example.

[F1]

Every object of O is a direct sum of weight spaces, is finitely generated, and is stable under root operators, which shift weights by elements of the root lattice Q (The classical BGG category O).

[F2]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. Every short exact sequence 0→M(λ)→E→pM(λ)→0 in O splits. (Verma self-extensions in O split)

[F3]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. For M,N∈O, Hom⁡O(M,N) is finite dimensional. For every weight λ, End⁡O(L(λ))=Cid⁡. (Finite-dimensional Hom spaces in O)

[F4]

The Verma PBW model gives the basis frvη (r≥0) of M(η) (The PBW model of a Verma module).

[F5]

Let Z=Z(U(g)), let χ:Z→C be a unital complex-algebra character, and put mχ=ker⁡χ. For M∈O, the generalized central-character submodule is Mχ={v∈M:mχNv=0 for some N≥1}, and Oχ consists of the objects with M=Mχ. (Generalized central-character subcategories)

[F6]

The simple objects of O are exactly the modules L(μ), μ∈h∗, and L(μ)≅L(ν) if and only if μ=ν. (The simple objects of O)

[F7]

If a cyclic highest-weight module has highest weight μ, then every z∈Z(U(g)) acts on it by the scalar pr⁡(z)(μ)=χμ(z). (The Harish-Chandra projection computes the highest-weight scalar)

[F8]

Under the given Choice premise, the highest-weight central characters satisfy χμ=χλ if and only if μ∈W⋅λ. (Central characters are dot-Weyl orbits)

[F9]

Under the given Choice premise, every object of O has a finite composition series and is both Noetherian and Artinian. The length of zero is zero. (Every object of O has finite length)

[F10]

The central-character subcategories Oχ are categorical direct summands of O, with exact projections (Generalized central-character decomposition of O).

Verification

1.1F4F5F6F7F8

For sl2, the dot orbit of λ is exactly {λ,−λ−2}. Its two labels are nonintegral and distinct: equality would imply λ=−1. By F8 they have the same central character. Conversely, if a simple object belongs to Oχλ, F6 writes it as L(μ). Its highest vector is killed by a power of every z−χλ(z) by F5, while F7 says that z acts on it as χμ(z). Hence χμ=χλ, and F8 forces μ∈{λ,−λ−2}. To see that both Vermas are simple, fix either nonintegral label η. The relations [h,f]=−2f and [e,f]=h give hfrvη=(η−2r)frvη and, by induction using efr+1vη=fefrvη+hfrvη, the coefficient formula efrvη=r(η−r+1)fr−1vη for r≥1. Every nonzero submodule of M(η) contains a nonzero weight vector: project a finite sum of distinct weight components using a polynomial in h. Choose its least occurring PBW index r. If r>0, the derived formula makes efrvη=r(η−r+1)fr−1vη≠0, contradicting minimality; hence r=0 and the submodule contains vη, so it equals M(η). The shifted simple-root pairings of the two labels are λ+1 and −(λ+1), both nonintegral. Hence their integral-reflection linkage classes are singletons by The integral Weyl group of a weight.

2.1F1F9step 1.1algebra

The two labels lie in distinct root-lattice cosets: their difference is 2(λ+1) in h-weight coordinates, whereas the sl2 root lattice has even integral coordinates; λ+1∉Z. For each X∈Oχλ, group its weight spaces by every coset of Q. Root operators preserve each group, so F1 makes these groups submodules whose direct sum is X. Any nonzero group has a simple composition factor by F9. Step 1.1 allows only the two labels, whose simple modules have weights in their respective cosets, so all other groups vanish. Hence X=Xλ+Q⊕X−λ−2+Q functorially, and every nonzero summand has factors only of its corresponding label. Both summands are nonzero because they contain their simple Verma modules.

3.1F2F9algebrastep 1.1step 2.1

Fix one label η and put S=M(η)=L(η). Every object in its coset summand has a finite composition series by F9, and all its factors are S by step 2.1. Every extension of S by itself splits by F2. More generally an extension 0→Sr→E→S→0 splits: push out along each coordinate projection Sr→S, obtaining Ei=(E⊕S)/{(a,−ai):a∈Sr}. Each has a retraction to its kernel S. Composing these retractions with E→Ei and collecting coordinates gives a retraction E→Sr, hence a splitting. The case r=0 is immediate.

4.1F3F10algebrastep 2.1step 3.1∎

Induction on a composition series now expresses every object in either coset summand as a finite direct sum of its sole simple S. Since End⁡(S)=C, maps between Sr and Ss are exactly complex matrices. Thus V↦S⊗V (with trivial action on the finite-dimensional multiplicity space) and X↦Hom⁡(S,X) are inverse equivalences for each summand. Zero corresponds to the zero-dimensional vector space. Each summand is indecomposable as a category because every nonzero object contains its sole simple S and all objects are sums of S. By F10, Oχλ itself is a direct summand of O, so these are precisely its two blocks in O.

ExampleConstruction: AI-adaptedVerification: AI-adaptedverified 2026-09-23 (gpt-6-sol)Open item page →

A singular integral A2 central-character summand

Example

Assume the Axiom of Choice (The Axiom of Choice).

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ.

In type A2, take λ=−ω1, with fundamental-weight coordinates ρ=(1,1). Then λ+ρ=ω2 has Weyl stabilizer ⟨s1⟩. The dot orbit consists of

−ω1,−2ω2,−2ω1−ω2.

These three simple labels form a single singular integral central-character block. Only labels and stabilizer are computed here.

Facts & Assumptions

Given: The Axiom of Choice, the setting above, and the hypotheses in the example.

[F1]

The category O is the categorical direct sum of its generalized central-character subcategories Oχ, with exact component projections (Generalized central-character decomposition of O).

[F2]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. For a weight λ, define Φλ={α∈Φ:⟨λ+ρ,α∨⟩∈Z},Wλ=⟨sα:α∈Φλ⟩⊂W. Reflections and coroots are those of def-root-reflections-and-the-weyl-group-action, with the shift from def-weyl-vector-rho-for-a-chosen-positive-system. The integral-reflection linkage class through λ is Wλ⋅λ. The word integral includes zero and negative integral pairings. If Φλ is empty the generated group is {1}. This definition uses generating reflections; no identification with a root-lattice-coset stabilizer is assumed. (The integral Weyl group of a weight)

[F3]

Under the given Choice premise, let χλ and χμ be the central characters obtained from highest weights λ and μ. Then χλ=χμif and only ifμ∈W⋅λ, where W⋅λ:={w(λ+ρ)−ρ:w∈W}. (Central characters are dot-Weyl orbits)

[F4]

Let Φ be the root system from thm-the-root-set-is-a-reduced-crystallographic-root-system. For a root α, define the corresponding coroot by α∨:=2Hαα(Hα)∈h, where Hα is the vector from def-killing-dual-vector-attached-to-a-root. The associated root reflection is the linear map sα(λ):=λ−λ(α∨)α(λ∈h∗). The subgroup of GL⁡(h∗) generated by these reflections is the Weyl group W, acting on h∗ in the usual way. (Root reflections and the Weyl group action)

[F5]

Every simple object of O is L(η) for a unique highest weight η (The simple objects of O). Its generalized-character membership means each z−χ(z) acts locally nilpotently (Generalized central-character subcategories), while z acts on any cyclic highest-weight module by the scalar χη(z) (The Harish-Chandra projection computes the highest-weight scalar).

[F6]

Under the given Choice premise, every object of O has a finite composition series (Every object of O has finite length). Every Verma module M(η) has a unique simple quotient L(η) (A Verma module has a unique simple quotient).

[F7]

A positive integral simple-root pairing gives a nonzero Verma embedding in the reflected direction (Simple-reflection embeddings of Verma modules).

Verification

1.1F4algebra

With α1=2ω1−ω2 and α2=−ω1+2ω2, the reflection formula gives s1(a,b)=(−a,a+b) and s2(a,b)=(a+b,−b). Acting on (0,1) produces exactly (0,1),(1,−1),(−1,0): both reflection formulas permute this set and the three displayed points are reachable. Subtracting (1,1) gives (−1,0),(0,−2),(−2,−1), the claimed labels.

2.1algebrastep 1.1

Type A2 has six Weyl elements, as is also seen by the six distinct matrices 1,s1,s2,s1s2,s2s1,s1s2s1: multiplication by either generator permutes these six matrices. Exactly 1 and s1 fix (0,1), by evaluating them. Thus the stabilizer is ⟨s1⟩, and the orbit is singular.

3.1F2F3F5step 1.1step 2.1

The pairings of (0,1) with the positive coroots are 0,1,1. All are integers, so Φλ=Φ and Wλ=W by F2; the zero pairing does not remove its reflection from the integral group. By F3 the three computed labels have one central character χ. Conversely, a simple object of Oχ is L(η) by F5. On it every z acts both by χη(z) and locally with generalized eigenvalue χ(z), so χη=χ; F3 restricts η to the three labels in step 1.1. Thus these are exactly the simple objects of Oχ.

4.1F1F6F7step 3.1algebra∎

Write the labels as λ=−ω1, μ=−2ω2=s2⋅λ, and ν=−2ω1−ω2=s1⋅μ. Their shifted simple-root pairings are ⟨λ+ρ,α2∨⟩=1 and ⟨μ+ρ,α1∨⟩=1, so F7 gives M(ν)↪M(μ)↪M(λ). By F3 and the highest-weight scalar action in F5, all three Vermas lie in Oχ. The unique-simple-quotient property in F6 makes M(λ) indecomposable: in a decomposition into two nonzero submodules both would be proper, and their sum would lie in its unique maximal proper submodule. In any categorical direct-sum refinement of Oχ, M(λ) must lie in one component; its subquotients L(λ),L(μ),L(ν) therefore lie there too. These are all the simple objects by step 3.1, and F6 makes every nonzero object have a simple factor. Hence no other nonzero component exists: Oχ is indecomposable. F1 makes it a direct summand of O, so it is one singular integral block with exactly the three stated simple labels.

Notes

The source gives a typical three-element singular A2 class with an antidominant representative. This item chooses the explicit representative -omega1 of such a class and computes its coordinates; statement provenance is ai-altered to record that specialization.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The full algebraic Verma dual is too large

Statement refuted

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ.

False claim: replacing the restricted sum in Chevalley duality by the full algebraic dual always gives an object of O.

For sl2 and any λ∈C, the full Chevalley-twisted dual Hom⁡C(M(λ),C) is not h-semisimple and hence is not in O.

Facts & Assumptions

Given: The setting above and the hypotheses in the statement refuted.

[F1]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. For an h-semisimple module M with finite-dimensional weight spaces, its restricted Chevalley dual is D(M)=⨁μ∈h∗Mμ∗,(xφ)(m)=φ(τ(x)m)(x∈U(g)). Here each functional is extended by zero on the other weight spaces and τ is the fixed anti-involution of def-chevalley-contravariant-form. In particular τ(h)=h and D(M)μ=Mμ∗. A map f:M→N induces D(f):D(N)→D(M) by precomposition. The action law follows from τ(xy)=τ(y)τ(x); a root vector of weight α sends Mμ∗ to Mμ+α∗, so the restricted sum is stable. This is a complex-linear algebraic dual, with no conjugation. Ordinary Lie-module duality has a minus sign and reverses weights; twisting that dual by the Lie automorphism x↦−τ(x) gives the convention used here. (Restricted Chevalley dual)

[F2]

For λ∈h∗, the Verma module is the induced g-module M(λ):=U(g)⊗U(b)Cλ, where U(g) is the quotient algebra of def-universal-enveloping-algebra-as-a-tensor-quotient and Cλ is def-one-dimensional-borel-module-of-weight-lambda. Write vλ:=1⊗cλ. Thus M(λ) is the quotient of U(g) by the left ideal generated by x for x∈n+ and h−λ(h) for h∈h; in particular, n+vλ=0 and hvλ=λ(h)vλ. (Verma modules)

[F3]

The weights of M(λ) are exactly λ−β for β∈Q+; every weight space is finite dimensional, and M(λ)λ=Cvλ. (Weights of a Verma module lie below lambda)

Counterexample

1.1F1F2F3construct

The induced Verma model has basis vk=fkvλ for k≥0 and hvk=(λ−2k)vk. Define φ(vk)=1 for all k and extend linearly. This is an algebraic functional because vectors of M(λ) are finite sums, although its nonzero weight coordinates are infinite. On the full twisted dual use the same formula (hφ)(v)=φ(hv) as for restricted duality.

2.1algebrastep 1.1∎

For every polynomial p, (p(h)φ)(vk)=p(λ−2k). The values λ−2k are pairwise distinct over C, so a polynomial annihilating φ has infinitely many distinct zeros and must be the zero polynomial. Thus the C[h]-orbit of φ is infinite dimensional. A vector in a direct sum of eigenspaces has only finitely many eigencomponents and is annihilated by the product of their linear eigenvalue polynomials. Therefore this full dual cannot be a weight module.

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Finite weight spaces and bounded support do not replace finite generation

Statement refuted

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ.

False claim: local n+-finiteness, finite-dimensional weight spaces and support in finitely many downward cones suffice for membership in O without finite generation.

For sl2, X=⨁n≥0M(−2n) has these three local properties, with supp⁡X=−2Z≥0 and dim⁡X−2k=k+1 for k≥0, but is not finitely generated.

Facts & Assumptions

Given: The setting above and the hypotheses in the statement refuted.

[F1]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. The BGG category O is the full subcategory of left U(g)-modules M satisfying all three conditions: M is finitely generated; M=⨁μ∈h∗Mμ where Mμ={v:hv=μ(h)v for all h∈h}; and U(n+)v is finite dimensional for each v∈M. Its morphisms are all g-linear maps. The zero module is included, generated by the empty set. The enveloping and triangular conventions are those of thm-pbw-ordered-monomial-basis-for-the-enveloping-algebra and thm-triangular-decomposition-from-a-chosen-positive-root-system. (The classical BGG category O)

[F2]

The weights of M(λ) are exactly λ−β for β∈Q+; every weight space is finite dimensional, and M(λ)λ=Cvλ. (Weights of a Verma module lie below lambda)

Counterexample

1.1F2algebra

Each Verma has one basis vector at each weight −2n−2j, j≥0: the negative root algebra has the single generator f, so the induced Verma basis is fjv−2n. At weight −2k the contributing summands are exactly n=0,…,k, proving the count k+1. The direct sum is a weight module supported in the single cone 0−Q+.

2.1algebrastep 1.1

The operator e raises the Verma weight, so it kills every vector after sufficiently many applications inside each summand. A vector of the direct sum has nonzero coordinates in finitely many summands; taking the maximum of their bounds proves local C[e]-finiteness.

3.1F1algebrastep 2.1∎

Any finite list of vectors is supported in a finite set of summands. Those summands form a g-submodule, so the submodule generated by the list remains there. There is always a further nonzero Verma summand outside that finite set. Hence the list cannot generate X, which fails the finite-generation axiom of O. The empty list generates zero, not X.

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Finite weight spaces alone do not give category O

Statement refuted

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ.

False claim: a finitely generated weight module with finite-dimensional weight spaces must belong to O.

For sl2, induce the weight-zero one-dimensional module from the negative Borel Ch⊕Cf. The resulting lowest-weight Verma module X is cyclic, has one-dimensional weight spaces at 2k for k≥0, and is not in O.

Facts & Assumptions

Given: The setting above and the hypotheses in the statement refuted.

[F1]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. The BGG category O is the full subcategory of left U(g)-modules M satisfying all three conditions: M is finitely generated; M=⨁μ∈h∗Mμ where Mμ={v:hv=μ(h)v for all h∈h}; and U(n+)v is finite dimensional for each v∈M. Its morphisms are all g-linear maps. The zero module is included, generated by the empty set. The enveloping and triangular conventions are those of thm-pbw-ordered-monomial-basis-for-the-enveloping-algebra and thm-triangular-decomposition-from-a-chosen-positive-root-system. (The classical BGG category O)

[F2]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. Let M be a finitely generated h-semisimple g-module. Then M∈O if and only if supp⁡M⊂⋃i=1r(λi−Q+) for some finite list of weights. In either case every Mμ is finite dimensional. The list may be empty for M=0; finite generation is an independent hypothesis. (The support description of category O with finite generation)

[F3]

Let x1,…,xr be an ordered basis of a finite-dimensional complex Lie algebra g. Then the monomials x1a1x2a2⋯xrar(ai∈N0) form a basis of U(g). In particular, multiplication identifies gr⁡U(g) with the symmetric algebra S(g) on the symbols of the xi. (PBW gives an ordered monomial basis for the enveloping algebra)

Counterexample

1.1F3construct

Let w=1⊗1 in X=U(sl2)⊗U(Ch⊕Cf)C0, where h and f kill 1. Ordering e before the negative Borel in PBW gives the basis wk=ekw, k≥0. The relation [h,e]=2e gives hwk=2kwk. This proves cyclicity and the one-dimensional weight-space assertion.

2.1F1F2algebrastep 1.1∎

The vectors ekw=wk are linearly independent, so U(n+)w=C[e]w is infinite dimensional and violates the local-finiteness axiom. Moreover no finite union of sets λi−2Z≥0 can contain all 2k: any cone meeting this real arithmetic progression has a fixed finite upper bound there, and finitely many bounds cannot contain its unbounded sequence. Thus the support criterion fails as well.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

An ambient extension can leave category O

Statement refuted

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ.

False claim: O is extension closed in the category of all g-modules.

For sl2 and any λ∈C, let V=Cv⊕Cw be a positive-Borel module with ev=ew=0, hv=λv and hw=λw+v. Then E=U(g)⊗U(b)V is an extension of M(λ) by itself in ambient modules, but is not in O.

Facts & Assumptions

Given: The setting above and the hypotheses in the statement refuted.

[F1]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. The BGG category O is the full subcategory of left U(g)-modules M satisfying all three conditions: M is finitely generated; M=⨁μ∈h∗Mμ where Mμ={v:hv=μ(h)v for all h∈h}; and U(n+)v is finite dimensional for each v∈M. Its morphisms are all g-linear maps. The zero module is included, generated by the empty set. The enveloping and triangular conventions are those of thm-pbw-ordered-monomial-basis-for-the-enveloping-algebra and thm-triangular-decomposition-from-a-chosen-positive-root-system. (The classical BGG category O)

[F2]

Let x1,…,xr be an ordered basis of a finite-dimensional complex Lie algebra g. Then the monomials x1a1x2a2⋯xrar(ai∈N0) form a basis of U(g). In particular, multiplication identifies gr⁡U(g) with the symmetric algebra S(g) on the symbols of the xi. (PBW gives an ordered monomial basis for the enveloping algebra)

[F3]

For λ∈h∗, the Verma module is the induced g-module M(λ):=U(g)⊗U(b)Cλ, where U(g) is the quotient algebra of def-universal-enveloping-algebra-as-a-tensor-quotient and Cλ is def-one-dimensional-borel-module-of-weight-lambda. Write vλ:=1⊗cλ. Thus M(λ) is the quotient of U(g) by the left ideal generated by x for x∈n+ and h−λ(h) for h∈h; in particular, n+vλ=0 and hvλ=λ(h)vλ. (Verma modules)

Counterexample

1.1F2F3construct

The Borel relation [h,e]=2e holds on V because e acts by zero. There is a short exact sequence 0→Cλ→V→Cλ→0, with submodule Cv and quotient generated by the image of w. PBW makes U(g) right free over U(b), so induction preserves this exact sequence. Its two ends are the Verma modules by definition.

2.1F1F2algebrastep 1.1∎

PBW identifies E with C[f]⊗V as a vector space. Thus 1⊗v≠0, while (h−λ)(1⊗w)=1⊗v and (h−λ)2(1⊗w)=0. In a weight module, ker⁡(h−λ)2=ker⁡(h−λ), by evaluating on its eigenspace decomposition. This Jordan pair proves that E is not a weight module. Its end terms do satisfy the category-O axioms: their PBW basis has weights λ−2k and locally nilpotent e action, and they are cyclic. Consequently the ambient extension leaves O.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

O is not closed under arbitrary tensor products

Statement refuted

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ.

False claim: the tensor product of two objects of O always belongs to O.

For sl2, X=M(0)⊗M(0) with diagonal action has support −2Z≥0, weight multiplicities dim⁡X−2k=k+1, and locally nilpotent e action, but is not finitely generated over U(g).

Facts & Assumptions

Given: The setting above and the hypotheses in the statement refuted.

[F1]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. The BGG category O is the full subcategory of left U(g)-modules M satisfying all three conditions: M is finitely generated; M=⨁μ∈h∗Mμ where Mμ={v:hv=μ(h)v for all h∈h}; and U(n+)v is finite dimensional for each v∈M. Its morphisms are all g-linear maps. The zero module is included, generated by the empty set. The enveloping and triangular conventions are those of thm-pbw-ordered-monomial-basis-for-the-enveloping-algebra and thm-triangular-decomposition-from-a-chosen-positive-root-system. (The classical BGG category O)

[F2]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=h⊕n+. Write Q+=∑iZ≥0αi, μ≤λ when λ−μ∈Q+, and w⋅λ=w(λ+ρ)−ρ. Every M∈O is a quotient of a module with a finite Verma flag. Consequently it has a finite filtration whose nonzero factors are quotients of Verma modules, and is finitely generated over U(n−). The empty filtration is allowed for zero. No truncation hypothesis is needed, and a Verma flag of M itself is not asserted. (Finite filtrations by highest-weight quotients)

[F3]

Let x1,…,xr be an ordered basis of a finite-dimensional complex Lie algebra g. Then the monomials x1a1x2a2⋯xrar(ai∈N0) form a basis of U(g). In particular, multiplication identifies gr⁡U(g) with the symmetric algebra S(g) on the symbols of the xi. (PBW gives an ordered monomial basis for the enveloping algebra)

Counterexample

1.1F3algebra

The Verma PBW bases identify X with C[x,y], where xiyj corresponds to fiv0⊗fjv0. Its h-weight is −2(i+j), so its weight multiplicities are k+1. The diagonal f acts by multiplication by x+y. Each tensor of two basis vectors is killed by a high enough power of the diagonal e: expanding (e⊗1+1⊗e)N, every term vanishes once N>i+j. Hence e acts locally nilpotently on X.

2.1F1F2step 1.1

If X were finitely generated over U(g), its weight decomposition and local e-finiteness would put it in O. F2 then implies it is finitely generated over U(n−)=C[f], hence over C[x+y] in the displayed model.

3.1algebrastep 2.1∎

But C[x,y]/(x+y)C[x,y]≅C[x] is infinite dimensional over C. A module generated by r<∞ elements over C[x+y] has quotient by x+y generated by their r images over C, so that quotient has dimension at most r. This proves that X is not finitely generated and therefore is not in O, although each factor is cyclic with locally nilpotent e and is a weight module.

Sources