Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Three times three for sl3

Example

Assume the Axiom of Choice. Take g=sl3 with simple roots α1,α2, positive roots Φ+={α1,α2,α1+α2}, Weyl vector ρ=α1+α2, fundamental weights ω1,ω2, and let V=L(ω1) be the standard three-dimensional simple module (Root systems of the classical complex Lie algebras, Classical complex matrix Lie algebras, Fundamental weights, Highest-weight classification). Then the tensor-product multiplicities of Tensor-product multiplicities for finite-dimensional simple modules are cω1ω12ω1=cω1ω1ω2=1 and all others are zero, that is L(ω1)⊗L(ω1)≅L(2ω1)⊕L(ω2)=Sym⁡2V⊕Λ2V,3⊗3=6⊕3ˉ, with summands of dimensions 6 and 3; in particular the trivial module L(0) is not a summand. The same answer is obtained from Tensor product with a minuscule representation: ω1 is minuscule (Minuscule weights), its weight orbit is Wω1={ω1, ω2−ω1, −ω2}, the three weights of V, each of multiplicity one (Minuscule weights have exactly the Weyl orbit as their weights), and among the three translates ω1+γ exactly 2ω1 and ω2 are dominant integral while ω1−ω2 is not. Two consistency checks fix the omissions: L(0) is excluded because −ω1∉Wω1, and the dimension count is dim⁡Sym⁡2V+dim⁡Λ2V=6+3=9=dim⁡(V⊗V).

Facts & Assumptions

Given: AC, g=sl3 with its standard positive system and fundamental weights, and V=L(ω1) of dimension 3 with weights ω1,ω2−ω1,−ω2, each of multiplicity one.

[F1]

On the diagonal Cartan, the standard basis vectors of V=C3 have weights ε1=ω1, ε2=ω2−ω1 and ε3=−ω2, since ω1=ε1, ω2=ε1+ε2 and ε1+ε2+ε3=0. The positive roots are ε1−ε2, ε2−ε3, ε1−ε3; their coroot pairings with ω1 are 1,0,1, so ω1 is minuscule. Root reflections exchange the corresponding coordinates, giving the displayed three-element orbit. The matrix units Eij show that V is simple: applying them to a nonzero vector produces every basis vector; e1 is killed by upper-triangular root vectors and has highest weight ω1. The minuscule tensor rule applies (Root systems of the classical complex Lie algebras, Fundamental weights, Minuscule weights, Highest-weight classification, Minuscule weights have exactly the Weyl orbit as their weights, Tensor product with a minuscule representation).

[F2]

The trivial module L(0) is the one-dimensional module of highest weight 0, and a dominant integral weight has all simple-coroot pairings ≥0, whence 0≠ω1, 0≠2ω1, 0≠ω2. Since ⟨ωi,αj∨⟩=δij, the weight ω1−ω2 has pairings ⟨ω1−ω2,α1∨⟩=1 and ⟨ω1−ω2,α2∨⟩=−1, so it is not dominant (Integral, dominant, and strictly dominant weights, Fundamental weights).

[F3]

The flip τ(v⊗w)=w⊗v commutes with the diagonal Lie action. The projections (I+τ)/2 and (I−τ)/2 split V⊗V into symmetric and alternating subspaces, canonically isomorphic to the quotient powers of Symmetric and exterior powers over an arbitrary field via these projections. Their bases are ei⊗ei and ei⊗ej+ej⊗ei for i<j, respectively ei⊗ej−ej⊗ei for i<j, giving dimensions six and three. The nonzero vectors e1⊗e1 and e1⊗e2−e2⊗e1 are killed by every upper-triangular root vector and have weights 2ω1 and ω2. Complete reducibility therefore supplies a copy of each corresponding simple module in its respective subspace (Direct-sum, dual, Hom, and tensor representations, Weyl's complete reducibility theorem, Highest-weight classification).

Verification

1.1F1F2givenalgebra

By [F1] the tensor product L(ω1)⊗L(ω1) is the direct sum of the L(ω1+γ) over the three elements γ∈Wω1={ω1,ω2−ω1,−ω2}, with terms labeled by non-dominant weights dropped. The three translates are ω1+ω1=2ω1, ω1+(ω2−ω1)=ω2 and ω1−ω2; by [F2] the first two are dominant integral and the third is not. Hence cω1ω12ω1=cω1ω1ω2=1 and all other tensor-product multiplicities vanish.

2.1F1F3step 1.1algebra

Identification with symmetric and exterior squares: by [F3] the submodule Sym⁡2V⊆V⊗V is nonzero of dimension 6 and has highest weight 2ω1, so it contains L(2ω1); similarly Λ2V is nonzero of dimension 3 with highest weight ω2 and contains L(ω2). The decomposition of step 1.1 has exactly the two summands L(2ω1) and L(ω2), so 9=dim⁡(V⊗V)=dim⁡L(2ω1)+dim⁡L(ω2) with dim⁡L(2ω1)≤6 and dim⁡L(ω2)≤3; hence dim⁡L(2ω1)=6=dim⁡Sym⁡2V and dim⁡L(ω2)=3=dim⁡Λ2V, and the inclusions are equalities: Sym⁡2V=L(2ω1) and Λ2V=L(ω2).

3.1F1F2step 1.1step 2.1algebra∎

The trivial module is not a summand: it would have to be one of the L(ω1+γ) with ω1+γ=0, i.e. γ=−ω1∈Wω1; but the three elements of Wω1 listed in [F1] are distinct from −ω1 (equality would force ω2=0, ω1=ω2 or ω1=0). Hence L(0) does not occur, consistent with the dimension count 6+3=9.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

76 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources