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

• D5/S3/QuantumBounds/QDeformedCommutatorTracelessRefutation.result

A traceless counterexample to the q-deformed Böttcher–Wenzel bound

Problem

D. Chruściński, G. Kimura, H. Ohno and T. Singal, Bounding the Frobenius norm of a q-deformed commutator, arXiv:2202.11520v2, prove $\Vert AB-qBA{\Vert }_F^2\le (1+q^2)\Vert A{\Vert }_F^2\Vert B{\Vert }_F^2$ for $q>0$ when $A$ or $B$ is normal, note that it fails in general, and state Conjecture 1: “For any $q>0$, if $A$ or $B$ is traceless, the inequality … holds and is sharp.” They prove it for $n=2$. The verbatim statements are in the literature note.

Issue #13930 reads the inequality clause for every $n$, every $q>0$ and all $A,B\in M_n(\mathbb{C})$ with $\mathrm{tr}A=0$ or $\mathrm{tr}B=0$.

Motivation

The conjecture would extend the Böttcher–Wenzel inequality, which is the case $q=1$, to deformed commutators by the same trace condition under which the ordinary commutator is insensitive to traces.

Gap

Issue #13930 records the literature check before any Lean: the follow-up arXiv:2608.03897 does not treat the traceless case; the Zenodo records on the same authors’ one-parameter Böttcher–Wenzel functional concern a different conjecture; the repository had no result. not-found-in-searched-scope.

Route

At $q=2$ and $n=5$, the pair $A=\left[\begin{array}{cc}6& 42\\ 0& -3\end{array}\right]\oplus (-I_3)$, $B=\left[\begin{array}{cc}6& 0\\ -42& -3\end{array}\right]\oplus (-I_3)$ is traceless, $\Vert A{\Vert }_F^2=\Vert B{\Vert }_F^2=1812$, and $\Vert AB-2BA{\Vert }_F^2=16417164>16416720=5\cdot 1812^2$.

Falsifier

The refutation would fail if the conjecture were read for $n=2$ only; the paper states it for general $n$ and proves only $n=2$.

Evidence

The canonical source is D5/S3/QuantumBounds/QDeformedCommutatorTracelessRefutation.lean, with public 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:f4d28dfa9e64a255bf7676cc918bda5b227d569a443c0f50bd8f8b4981052a2a, the result statement sha256:bcddb28149b4dceac91841750aabb71d24dd118a582bbf4db8c2f5936a356ff9 and the claim statement sha256:61f533309ab5da634cf0b059cb95cf42e68bd5e2e618f40afd51a8e2dff37331. The Freeze event is sha256:7570e47bab63b7eede934aad91f3ad4652bde1f20e9d34ffbac64080f1fdaacc; its project-level prerequisite is the Moreau–Yosida module (sha256:5ba5226c1d185e37836d461616c2928638177220bb70a5360c8d549ad8283566).

Triage

Tier 1 conjecture of a 2022 paper, preregistered in issue #13930 before any Lean. theorem; resolution refuted.

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 1 fails at $q=2$ for a pair of traceless $5\times 5$ integer matrices; both traces vanish, so even the stronger hypothesis “both traceless” does not rescue it.

Argued, not formalized.

• Mechanism (trace dilution). For any $A,B\in M_r(\mathbb{C})$ with traces $a,b$, put $A_m=A\oplus (-\frac{a}{m}{I}_m)$ and $B_m=B\oplus (-\frac{b}{m}{I}_m)$. Then $\mathrm{tr}{A}_m=\mathrm{tr}{B}_m=0$, $\Vert A_m{\Vert }_F^2=\Vert A{\Vert }_F^2+|a{|}^2/m$, $\Vert B_m{\Vert }_F^2=\Vert B{\Vert }_F^2+|b{|}^2/m$ and $\Vert [A_m,B_m{]}_q{\Vert }_F^2=\Vert [A,B{]}_q{\Vert }_F^2+(1-q{)}^2|ab{|}^2/m^3$. As $m\to \infty$ the ratio $\Vert [A_m,B_m{]}_q{\Vert }_F^2/(\Vert A_m{\Vert }_F^2\Vert B_m{\Vert }_F^2)$ tends to that of $(A,B)$, so for every $q>0$ the traceless condition does not lower the optimal constant of the unrestricted problem uniformly in $n$. With the pair $A=\left[\begin{array}{cc}2& 14\\ 0& -1\end{array}\right]$, $B=\left[\begin{array}{cc}2& 0\\ -14& -1\end{array}\right]$ at $q=2$ (the paper’s non-normal family $A=\left[\begin{array}{cc}q& \sqrt{t}\\ 0& -1\end{array}\right]$, $B=\left[\begin{array}{cc}q& 0\\ -\sqrt{t}& -1\end{array}\right]$ at $t=196$; the paper’s displayed example is $t=64$), the excess is $676-2010/m-5/m^2+1/m^3>0$ for every $m\ge 3$ (exact rational evaluation in #13930), so the conjecture fails in every dimension $n\ge 5$.
• What survives. The bound holds when one matrix is normal (the paper’s Proposition 1) and for $n=2$; the trace condition is the wrong replacement for normality, because a trace can be moved into a block where it costs only $O(1/m)$ in norm.

Open. Whether Conjecture 1 holds for $n=3$ and $n=4$, and the optimal constant of the traceless problem in fixed dimension, are not determined here.

Effect on the paper. Conjecture 1 is false for $n\ge 5$; the traceless case does not give a q-deformed Böttcher–Wenzel inequality with constant $1+q^2$.

ASSUMED-UNVERIFIED

The bounded literature check does not establish exhaustive worldwide novelty, priority, or the absence of an independent proof.