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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-24 (gpt-6-sol)
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Simple extensions cannot cross linkage classes

Statement

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.

A short exact sequence 0→L(μ)→E→L(λ)→0 in O splits whenever λ and μ belong to distinct integral-reflection linkage classes.

Facts & Assumptions

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

[F1]

Under 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. 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)

[F2]

For a g-module V, sending a homomorphism T:M(λ)→V to T(vλ) is a bijection onto the vectors v∈V of weight λ annihilated by n+. Here M(λ) is def-verma-module. The nonzero vectors in this target are precisely the highest-weight vectors of weight λ from def-highest-weight-vector-and-cyclic-highest-weight-module; the zero vector corresponds to the zero homomorphism. (The universal property of Verma modules)

[F3]

Under the Axiom of Choice, if [M(λ):L(μ)]≠0, then μ↑λ. (The strong linkage principle for Verma modules)

[F4]

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 set Φλ is preserved by its root reflections. If μ∈Wλ⋅λ, then Φμ=Φλ and Wμ=Wλ. The equivalence classes generated by moves η↦sα⋅η with α∈Φη are exactly Wλ⋅λ. Every strong-linkage chain stays in one such class. (Integral reflection linkage is an equivalence relation)

[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)

Proof

1.1F1

If λ<μ, dualize the sequence; exact self-duality of simples reverses the two labels, and splitting of the dual implies splitting of the original by biduality. Thus it is enough to treat the orientation λ≮μ, which includes incomparable labels.

2.1F2F5algebrastep 1.1

Lift the highest vector of L(λ) to a vector v∈Eλ. Such a weight lift exists because the sequence consists of weight modules. If a positive-root operator acted nontrivially on v, its image would lie in the submodule L(μ) at weight λ+α, forcing λ+α≤μ and hence λ<μ. Therefore v is singular, and the Verma universal property gives a map M(λ)→E whose image H surjects onto L(λ). Here the support bound for each simple follows from its being the highest-weight Verma quotient.

3.1F3F4algebrastep 2.1∎

The intersection H∩L(μ) is either zero or all of the simple submodule. In the latter case H=E, so E is a length-two quotient of M(λ) and L(μ) is a composition factor of that Verma. Strong linkage would imply μ↑λ, and hence equality of integral-reflection classes, contrary to the hypothesis. Thus the intersection is zero, and H→L(λ) is an isomorphism whose inverse is a section.

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