Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08
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.

Unsymmetrized parameters break the coproduct of the Serre ideal

Statement refuted

The following assertion is false: for every generalized Cartan matrix and every assignment of node parameters qi, the standard Drinfeld–Jimbo coproduct formulas descend to the full node-toral presentation, including its mixed and Serre relations, with KiEjKi−1=qiaijEj.

Take A=(2−1−22), whose standard symmetrizer is diag⁡(2,1), but assign q1=q2=q, with q indeterminate. Precisely, let B be the Q(q)-algebra on E1,E2,F1,F2,K1±1,K2±1, with commuting invertible toral generators, relations KiEjKi−1=qaijEj,KiFjKi−1=q−aijFj,[Ei,Fj]=δijKi−Ki−1q−q−1, and both symmetric quantum Serre families with the common parameter q. This explicitly changed parameter assignment is the test presentation, not the symmetrized algebra of The Drinfeld-Jimbo quantized enveloping algebra by generators and relations. Put S=E12E2−[2]qE1E2E1+E2E12=0 in B.

The assignments Δ(Ei)=Ei⊗Ki−1+1⊗Ei, Δ(Fi)=Fi⊗1+Ki⊗Fi, Δ(Ki)=Ki⊗Ki fail to define an algebra map B→B⊗B. In an explicit representation of B below, the proposed image of S acts on v0⊗v1 by q−2(1−q)(1+q2)v2⊗v2≠0. Thus the previously detected positive-Borel defect survives in the full quotient. The same proposed formulas already fail the off-diagonal mixed relation [E1,F2]=0.

Facts & Assumptions

Given: The explicitly stated unsymmetrized full presentation B, with q indeterminate.

[F1]

The symmetric Gaussian coefficients are [2]q=q+q−1 and [3]q=q2+1+q−2 (Quantum integers, factorials, Gaussian binomials and divided powers at qi).

[F2]

The displayed inverse-K positive coproduct and its negative counterpart are the normalized Drinfeld–Jimbo convention; with the correct node parameters their Serre mixed terms cancel (The coproduct preserves the positive and negative quantum Serre ideals, The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F3]

For this matrix the correct symmetry is q1=q2, q2=q; the assignment tested above violates q1a12=q2a21 (Symmetrizable Cartan data for quantum groups).

Counterexample

1.1givenconstruct

On a four-dimensional Q(q)-space with basis v0,v1,v2,v3, let K1 have eigenvalues (1,q−1,q,1) and K2 have eigenvalues (q−1,q,q−1,q). Define E1v1=v2, F1v2=v1, E2v0=v1, E2v2=v3, F2v1=v0, F2v3=v2, and make all other E/F actions zero. Both toral operators are commuting and invertible. An E1 arrow changes the pair of toral exponents by (2,−2) and each E2 arrow changes it by (−1,2); the F arrows reverse these changes. Hence all toral-action relations hold.

2.1step 1.1algebra

The diagonal entries of [E1,F1] are (0,−1,1,0), and those of [E2,F2] are (−1,1,−1,1). Each toral exponent is 0 or ±1, so these are precisely the entries of (Ki−Ki−1)/(q−q−1). For each off-diagonal pair, both operator products E1F2,F2E1 and E2F1,F1E2 are zero: none of their two-arrow sequences is composable on a basis vector. Thus all four mixed relations hold.

2.2F2step 1.1algebra

On its tensor square put Di=Ei⊗Ki−1+1⊗Ei. Direct computation gives D1(v0⊗v1)=v0⊗v2 and D2(v0⊗v1)=q−1v1⊗v1. Also D1(v1⊗v1)=qv2⊗v1+v1⊗v2, so D12D2(v0⊗v1)=(1+q−2)v2⊗v2. Applying D2 to v0⊗v2 gives qv1⊗v2+v0⊗v3, whose D1 image is v2⊗v2; thus D1D2D1(v0⊗v1)=v2⊗v2. Finally D12(v0⊗v1)=0, so the last Serre term contributes zero.

3.1F1step 1.1step 2.1algebra

All four squares E12,E22,F12,F22 are zero. The compositions E1E2E1 and F1F2F1 are zero as well: after the first color-1 arrow, the color-2 arrow either vanishes or leads to a vector on which the final color-1 arrow vanishes. These identities give both length-three Serre relations S12±=0. Every term of S21± contains a square or cube of the color-2 operator, since its four words are X23X1, X22X1X2, X2X1X22, and X1X23. Hence both length-four Serre relations vanish too. Steps 1.1 and 2.1 and these checks verify every defining relation of B, so the free-generator assignment factors through a representation of the full quotient.

4.1F1F2F3step 1.1step 3.1step 2.2algebra∎

By [F1] and step 2.2, the proposed coproduct image of S acts as (1+q−2−q−q−1)v2⊗v2=q−2(1−q)(1+q2)v2⊗v2, which is nonzero over Q(q). Since S=0 in the full represented algebra B, this contradicts the relation preservation required of a coproduct algebra map B→B⊗B. Independently, expanding the off-diagonal mixed commutator gives [Δ(E1),Δ(F2)]=(q−1)K2E1⊗F2K1−1: the cross scalar is q1a12q2−a21−1=q−1. Its action on v1⊗v1 is (q−1)v2⊗v0≠0. Thus both the full Serre and mixed-relation failures are detected without an assumption of triangular decomposition.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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