Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The regular integral sl2 block

Example

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=hn+. Write Q+=iZ0αi, μλ when λμQ+, and wλ=w(λ+ρ)ρ.

For g=sl2 identify a highest weight with its value on h, so ρ=1 and sλ=λ2. For every integer n0, the regular integral block has simple labels n and n2. Its standards are M(n) and M(n2)=L(n2), and

0L(n2)M(n)L(n)0

is nonsplit. Its costandards are (n) and (n2)=L(n2), with the nonsplit sequence 0L(n)(n)L(n2)0. The linkage order is n2<n.

Facts & Assumptions

Given: The setting above and the hypotheses in the example.

[F1]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=hn+. Write Q+=iZ0αi, μλ when λμQ+, and wλ=w(λ+ρ)ρ. For a linkage class C=Wλλ, let OC be the full subcategory of objects all of whose simple composition factors have labels in C. Then O=COC, and each nonzero OC is indecomposable as a categorical direct summand. These are precisely the blocks. Each OC lies in Oχλ; a central-character summand can contain several blocks. Independently, grouping weights by cosets of the root lattice Q gives a canonical coarser decomposition by weight cosets. (Central-character summands refine into linkage blocks)

[F2]

Fix a finite-dimensional complex semisimple Lie algebra g, a Cartan subalgebra h, and a positive Borel b=hn+. Write Q+=iZ0αi, μλ when λμQ+, and wλ=w(λ+ρ)ρ. Restricted Chevalley duality is an exact contravariant equivalence D:OOop, with a natural isomorphism D2id. 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=hn+. Write Q+=iZ0αi, μλ when λμQ+, and wλ=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,αiZ>0, there is an embedding M(siλ)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]

M(λ) is simple if and only if λ+ρ,αZ>0 for every αΦ+. (The Verma irreducibility criterion from Shapovalov determinants)

[F7]

The weights of M(λ) are exactly λβ for βQ+; 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

1.1

The integral pairing is n+1>0, and the two distinct dot-orbit labels are n and n2. The integral Weyl group is the full order-two Weyl group, so these labels form one block. At n2 the shifted pairing is n1<0, and the irreducibility criterion makes its Verma simple.

F1F6
2.1

Write vk=fkvn in M(n). The relations [h,f]=2f and [e,f]=h give hvk=(n2k)vk and evk=k(nk+1)vk1 for k1, by commuting e past the k copies of f; ev0=0. The negative nilpotent algebra is one dimensional, so its PBW monomials give one-dimensional Verma weight spaces. The singular vector vn+1 generates the embedded M(n2), which is also the embedding supplied by F4.

F4F7algebrastep 1.1
3.1

The quotient has basis v0,,vn. A nonzero submodule contains a weight vector, and applying e repeatedly reaches v0, since k(nk+1)0 for 1kn. 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.

F5algebrastep 2.1
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(n2)=L(n2), F8 and F9 also give (n2)=L(n2).

F2F3F8F9step 3.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources