CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Equal representation structure need not fix energy

Statement refuted

Fix $W = V (1) \otimes V (1)$ with its tensor-product inner product and its diagonal representation $ρ : {s l}_{2} (C) \to {End}_{C} (W)$ from the spin convention, together with the rank-one Hall data at $q_{H} = 2$ and the type- $A_{1}$ KLR negative-half Grothendieck isomorphism over $Q$ . Let $H_{ρ} = {A \in {End}_{C} (W) : A = A^{*}, [A, ρ (X)] = 0 for every X \in {s l}_{2} (C)},$ where $A^{*}$ is the adjoint for the stated inner product. For an endomorphism $D$ of $W$ , set $\exp (D) = \sum_{r = 0}^{\infty} D^{r} / r!$ , convergent in operator norm, and let ${Tr}_{W}$ denote the basis-independent trace on endomorphisms of $W$ . For $C \in H_{ρ}$ and $β \geq 0$ , define its partition trace by $Z_{C} (β) : = {Tr}_{W} (\exp (- β C)) .$ The claim that the fixed algebraic data uniquely determine the spectrum and partition trace of every member of $H_{ρ}$ means that any $A, B \in H_{ρ}$ have the same eigenvalues counted with algebraic multiplicity and satisfy $Z_{A} (β) = Z_{B} (β) for every β \geq 0.$ This uniqueness claim is false. If the endomorphisms are interpreted as energies, $β$ has inverse-energy units.

Statement references

Spin weight spaces and rank-one Hall realization

Exchange preserves total Lie spin

KLR grading gives integral multiplicity bookkeeping

Finite-dimensional matrix exponential

The basis-independent trace of an endomorphism of a finite-dimensional vector space

Facts & Assumptions

Given: Use the Hall definition at $q_{H} = 2$ , so $ν = \sqrt{2}$ , and the one-vertex type- $A_{1}$ KLR datum $D$ over $k = Q$ (its off-diagonal $Q_{i j}$ family is empty). Let $W = V (1) \otimes V (1)$ with the diagonal action from the spin definition, and let $P (v \otimes w) = w \otimes v$ . Choose an energy $J > 0$ and $β = (\ln 2) / J > 0$ ; then $β J$ is dimensionless, and $β$ has inverse-energy units if $J$ is interpreted as an energy.

[A1]

Adopt the Axiom of Choice (The Axiom of Choice) inherited through the Hall/KLR supplier chain; the finite-dimensional witness uses no additional choice.

[F1]

Spin weight spaces and rank-one Hall realization: $V (1)$ has basis $∣ ↑ ⟩, ∣ ↓ ⟩$ and the global spin generators are the sums of $E, F, H / 2$ over the two tensor factors. At $q_{H} = 2$ its one-vertex Hall composition algebra is identified with $U_{\sqrt{2}}^{+} ({s l}_{2})$ by $E \mapsto [S]$ . This Hall identification assigns no endomorphism of $W$ .

[F2]

Exchange preserves total Lie spin: for $N = 2$ , the swap $P$ commutes with all three generators of the diagonal ${s l}_{2}$ action.

[F3]

KLR grading gives integral multiplicity bookkeeping: for the stated type- $A_{1}$ datum $D$ over $Q$ , the canonical KLR Grothendieck isomorphism identifies the divided-power negative half with the graded projective Grothendieck algebra; it assigns no Hamiltonian to $W$ .

[F4]

Finite-dimensional matrix exponential: for every finite complex matrix $A$ , $\exp (A) = \sum_{r \geq 0} A^{r} / r!$ converges in operator norm.

[F5]

The basis-independent trace of an endomorphism of a finite-dimensional vector space: for a finite-dimensional endomorphism, the trace is the matrix trace in any chosen basis and is independent of the basis.

Counterexample

1.1givenA1F1F2algebra

In the ordered tensor basis, define $v_{+ +} = ∣ ↑ ↑ ⟩$ , $v_{0} = (∣ ↑ ↓ ⟩ + ∣ ↓ ↑ ⟩) / \sqrt{2}$ , $v_{- -} = ∣ ↓ ↓ ⟩$ , and $v_{s} = (∣ ↑ ↓ ⟩ - ∣ ↓ ↑ ⟩) / \sqrt{2}$ . These four vectors are pairwise orthonormal by the tensor-product inner product, and they span the four-dimensional space. The swap $P$ fixes the first three and sends $v_{s}$ to $- v_{s}$ , so the same space decomposes into the symmetric eigenspace $W_{+} = \ker (P - I)$ of dimension three and the antisymmetric eigenspace $W_{-} = \ker (P + I)$ of dimension one. The swap is an involutive permutation of an orthonormal basis, so $P = P^{*}$ . By [F2], it commutes with $ρ (X)$ for every $X \in {s l}_{2} (C)$ ; hence both eigenspaces are invariant under the action.

1.2givenF1F2F4F5algebra

Let $A_{0} = 0$ and $A_{1} = J P$ . Both are self-adjoint because $J$ is real and $P = P^{*}$ , and both belong to $H_{ρ}$ : the zero map commutes with every endomorphism and [F2] gives the required commutation for $P$ . The spectrum of $A_{0}$ is $0$ with multiplicity four; that of $A_{1}$ is $J$ with multiplicity three and $- J$ with multiplicity one. Therefore, by [F4] and [F5], $Z_{A_{0}} (β) = {Tr}_{W} (\exp (- β A_{0})) = 4, Z_{A_{1}} (β) = {Tr}_{W} (\exp (- β A_{1})) = 3 e^{- β J} + e^{β J} .$

2.1givenstep 1.1step 1.2algebra

Since $β J = \ln 2$ , the second trace is $3 / 2 + 2 = 7 / 2$ , which differs from $4$ . Hence the same spin representation decomposition admits distinct symmetry-preserving spectra and Gibbs traces.

3.1givenF1F3step 1.1step 1.2step 2.1∎

The Hall data in [F1] and the KLR Grothendieck isomorphism in [F3] are fixed for both $A_{0}$ and $A_{1}$ ; neither specifies a selected endomorphism of $W$ . Since both operators lie in the quantified class but have different spectra and different partition traces, they refute the stated uniqueness claim.

Remarks

The example is an exact finite-dimensional counterexample. It does not use experimental evidence or identify the Hall or KLR categories with a tensor-product categorification of the spin space.