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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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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.

Depends on

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