slug: chruscinski-2022-qdeformed-commutator-negative-q-refutation bibkey: chruscinski2022qdeformed doi: 10.48550/arXiv.2202.11520 url: https://arxiv.org/abs/2202.11520v2 triage: theorem motivation_gids:

D5/S3/QuantumBounds/QDeformedCommutatorNegativeQRefutation.result

A rank-one partner breaks the negative-q bound for a traceless matrix

Problem

D. Chruściński, G. Kimura, H. Ohno and T. Singal, Bounding the Frobenius norm of a q-deformed commutator, arXiv:2202.11520v2, state Conjecture 2: “For any $q \leq 0$ , if $A$ or $B$ is traceless, the sharp bound is $∥ [A, B]_{q} ∥_{F}^{2} \leq max [g (n) (1 - q)^{2}, 1 + q^{2}] ∥ A ∥_{F}^{2} ∥ B ∥_{F}^{2}$ ”, with $[A, B]_{q} = A B - qB A$ and $g (n) = (n^{2} - 3 n + 3) / (n (n - 1))$ . Their Proposition 3 proves the case $n = 2$ with $tr A = 0$ for every real $q$ . The verbatim statements are in the literature note.

Issue #14632 reads the inequality clause for every $n \geq 2$ , every $q \leq 0$ and all $A, B \in M_{n} (C)$ with $tr A = 0$ or $tr B = 0$ .

Motivation

The conjecture proposes the optimal Frobenius-norm constant for deformed commutators with $q \leq 0$ , which include the anticommutator ( $q = - 1$ ), under the trace condition of the companion Conjecture 1. The value $g (n) (1 - q)^{2}$ is attained by the source’s pair $A = B = diag (n - 1, - 1, \dots, - 1)$ .

Gap

Issue #14632 records the literature check before any Lean: the citing works arXiv:2208.10005, arXiv:2403.04199, arXiv:2402.01085 and arXiv:2608.03897 do not treat Conjecture 2 or a trace-constrained bound for $q \leq 0$ ; Crossref, DataCite and the formal-conjectures repository show no settlement; not-found-in-searched-scope.

Route

At $n = 3$ and $q = - 1$ , take $A = diag (2, - 1, - 1)$ (trace $0$ ) and $B = E_{11}$ . Then $A B = B A = diag (2, 0, 0)$ , $∥ [A, B]_{- 1} ∥_{F}^{2} = ∥ diag (4, 0, 0) ∥_{F}^{2} = 16$ , $∥ A ∥_{F}^{2} = 6$ , $∥ B ∥_{F}^{2} = 1$ and $max [g (3) \cdot 4, 2] = 2$ , so the bound would give $16 \leq 12$ .

Falsifier

The refutation would fail under a reading that requires both matrices to be traceless. The source writes “ $A$ or $B$ ” and proves the case $n = 2$ under the single condition $tr A = 0$ .

Evidence

The canonical source is D5/S3/QuantumBounds/QDeformedCommutatorNegativeQRefutation.lean, with public g, claim and result : ¬ claim. The squared Frobenius norm is the existing mass of D5/S3/Quantum/Entanglement/MoreauYosidaFormationSelectiveLoccRefutation, the only import. The axiom closure of result is exactly propext, Classical.choice and Quot.sound; there is no sorry, native_decide, or new axiom. The module statement is sha256:c4768e9ee56732b9fb335ec0fb91f42875bee81e6ae90d3057cc7462ae850dbd, the result statement sha256:b90da19082989d186deda8bfef7665cefa24100982255e14bc9ebf26b8d192ee, the claim statement sha256:99b3f8e7af1534e05327c07cc3146099a5c8d5fbb58e8797479f0a486dff6ba2 and the g statement sha256:277f2af201e07ae642d464db8465b0e9eb1eebab7cdfef5e945e0fcc3a03aa0a. The Freeze event is sha256:b8101b6f547af29e2c37d9770e9b6d191f33ce473a542da74d00c0bc5f6bae73; its project-level prerequisite is the Moreau–Yosida module (sha256:8acf232e43275fdfb443d70e0636ddf93b58a9a9635be8c8810b087f0de9d860).

Triage

Tier 1 conjecture of a 2022 paper, preregistered in issue #14632 before any Lean. theorem; resolution refuted.

declaration	proof_shape	escape_witness	admission_basis
result	bind-only	none	open-problem-resolution

Utility is kind=certified-instance; basis=refutes (the claim and its refutation). There is no digestion atom.

What the refutation shows

Proved by result: Conjecture 2 fails at $n = 3$ , $q = - 1$ for a traceless $A$ and a rank-one, non-traceless $B$ commuting with it.

Argued, not formalized.

Every $n \geq 3$ . For $A = diag (n - 1, - 1, \dots, - 1)$ and $B = E_{11}$ , $∥ [A, B]_{q} ∥_{F}^{2} / (∥ A ∥_{F}^{2} ∥ B ∥_{F}^{2}) = (1 - q)^{2} (n - 1) / n$ . Since $(n - 1) / n - g (n) = (n - 2) / (n (n - 1)) > 0$ , this exceeds $g (n) (1 - q)^{2}$ for every $n \geq 3$ and $q < 1$ ; at $q = - 1$ it is $4 (n - 1) / n > 2$ , so the conjecture fails at $q = - 1$ in every dimension $n \geq 3$ (exact rational check in #14632).
Mechanism. The source’s extremal family has both traces zero. A partner with nonzero trace concentrated on the large eigenvalue of $A$ raises the ratio, and the single trace condition does not prevent it.
What survives (bounded numerics, not proof). At $n = 3$ , $q = - 1$ , numerical maximization over random complex pairs found ratios approaching $2$ , the conjectured coefficient, when both matrices are traceless, and approaching $8/3$ when only $A$ is traceless; the witness above attains $8/3$ . These readings certify neither a supremum nor an upper bound.

Open. Whether the conjectured bound holds when both $A$ and $B$ are traceless, and the optimal constant under a single trace condition for $n \geq 3$ .

Effect on the paper. Conjecture 2 as printed is false for $n \geq 3$ ; the case $n = 2$ remains proved by Proposition 3.

ASSUMED-UNVERIFIED

Fang–Cheng, On some conjectures concerning the Böttcher–Wenzel inequality for weighted Frobenius norms, Linear Algebra Appl. (2025), DOI 10.1016/j.laa.2025.03.015, cites the source; its abstract and text could not be read, and by its title it concerns the weighted-norm conjectures of arXiv:2403.04199. That it does not settle Conjecture 2 is not verified firsthand. The journal version of the source was not compared with arXiv v2. The bounded literature check does not establish exhaustive worldwide novelty, priority, or the absence of an independent refutation.

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