Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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.

Serre presentation of a kac moody algebra

Statement

For a finite symmetrizable GCM over C, r± is the ideal of the free half n~± generated by (adei)1aijej or (adfi)1aijfj, respectively, for ij. Hence g(A) has exactly the Cartan relations and both Serre families as a presentation. Its halves have the corresponding separate Serre presentations, and multiplication gives the vector-space isomorphism U(n)U(h)U(n+)U(g).

Facts & Assumptions

Given: The symmetrizable Serre quotient and its vanishing residual kernel.

[F1]

The quotient by both Serre families is g(A). (The serre quotient has weyl symmetry and no residual kac moody kernel).

[F2]

PBW applies to the homogeneous bases. (PBW for countably presented Kac Moody Lie algebras).

Proof

1.1

Let S+ be the ideal in the free positive half generated by the positive Serre vectors. The adjoint rank-one calculation gives [fk,sij]=0 for each such generator already in g~: its coefficient for k=i is (1aij)(aij(1aij)+1)=0, and for k=j the only exponent-one boundary uses aji=0. For other k it is zero by the mixed simple relations. Cartan brackets scale each homogeneous generator. Jacobi then proves by induction on positive adjoints that S+ is stable under all negative simple generators; hence it is an ideal of the whole universal algebra contained in its positive half. The same holds for S. Therefore the ideal generated in the whole algebra by both families is exactly S+S.

F1given
2.1

F1 identifies that whole ideal with r. Intersect its direct sum in step 1.1 with each free half to obtain S±=r±. Quotienting the original triangular direct sum therefore gives the stated separate half presentations and the full Cartan–Serre presentation. Order the resulting homogeneous bases negative, Cartan, positive. F2 identifies the tensor product of their three ordered monomial bases bijectively with the ordered basis of U(g), proving the multiplication isomorphism.

F1F2step 1.1

Sources

Source comparison: Kleshchev, Theorem 9.3.5, pp.125–126; half-ideal and PBW consequences.

Depends on

Used by

Dependency tree · two levels

7 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