The kernel criterion for witness optimality does not imply the trace criterion

F. vom Ende and S. Cichy, Simple Sufficient Criteria for Optimality of Entanglement Witnesses, arXiv:2505.15615v2, Section IV (p. 12):

Next: is our trace-based criterion (Coro. 2) actually equivalent to our kernel criterion (Thm. 2)? This is the question mark in Fig. 2, and while the kernel criterion seems stronger we were not able to find a witness for which only the kernel criterion holds.

The kernel criterion asks for a vector of Schmidt rank $\min{m,n}$ in the kernel of $W+{\rm tr}_2(W)\otimes{\bf1}$ (if $m\le n$) or of $W+{\bf1}\otimes{\rm tr}_1(W)$ (if $m\ge n$); the trace criterion asks for a maximally entangled $\Omega$ with $\langle\Omega|W|\Omega\rangle=-{\rm tr}(W)/\min{m,n}$. The verbatim texts are in the literature note. Issue #14763 preregisters the reading and the refutation.

Optimal entanglement witnesses are the extreme detectors of entanglement; the source derives two cheap sufficient conditions for optimality from the spanning property, one a kernel condition and one an expectation-value condition that numerics can test by a minimization over one local unitary group. The source proves that the trace criterion implies the kernel criterion and asks whether the two have the same reach.

The source reports no witness that satisfies only the kernel criterion. Its citing works arXiv:2510.06863 and arXiv:2604.02420 do not address the question; the literature check in #14763 found no answer; not-found-in-searched-scope.

In the basis $(00,01,10,11)$ of $\mathbb C^2\otimes\mathbb C^2$ take

$$W=\begin{pmatrix}0&0&0&-2\0&1&0&0\0&0&4&0\-2&0&0&0\end{pmatrix} =(X\otimes{\bf1})W_{\rm red}(X^\dagger\otimes{\bf1}),\qquad X=\operatorname{diag}(1,2),$$

the reduction witness $W_{\rm red}={\bf1}-|\Gamma\rangle\langle\Gamma|$ filtered on the first factor.

1. $\langle x\otimes y|W|x\otimes y\rangle=|\bar x_0y_1-2\bar x_1y_0|^2\ge0$, and ${\rm tr}(W|\Omega_0\rangle\langle\Omega_0|)=-2<0$ for $\Omega_0=(|00\rangle+|11\rangle)/\sqrt2$: $W$ is a witness.
2. ${\rm tr}2(W)=\operatorname{diag}(1,4)$ and $z^*(W+{\rm tr}2(W)\otimes{\bf1})z=|z{00}-2z{11}|^2+2|z_{01}|^2+8|z_{10}|^2$, so $(2,0,0,1)$, of Schmidt rank $2$, lies in the kernel: the kernel criterion holds.
3. $z^*(W+2\cdot{\bf1})z=2|z_{00}-z_{11}|^2+3|z_{01}|^2+6|z_{10}|^2\ge0$, so every unit vector has $\langle\Omega|W|\Omega\rangle\ge-2>-5/2=-{\rm tr}(W)/2$: the trace criterion fails.

An error in the two sums of squares or in the kernel, or a reading of block positivity, Schmidt rank or maximal entanglement different from the source. Positive controls in #14763: the reduction witness and the two-qubit flip satisfy both criteria.

D5/S3/Quantum/Entanglement/WitnessKernelTraceCriteriaRefutation.lean defines blockPositive (the product-vector expectation is nonnegative in the complex order), IsWitness, the partial traces trTwo, trOne (the frozen partialTraceRight, partialTraceLeft), schmidtRank (the rank of the coefficient matrix), kernelCriterion, maximallyEntangled (an equal-coefficient Schmidt sum over orthonormal families of size $\min{m,n}$), traceCriterion and claim, and proves result : ¬ claim. IsWitness states ${\rm tr}(W\sigma)<0$ in the complex order, literally as in the source. The axiom closure of result is exactly propext, Classical.choice and Quot.sound; there is no sorry or native_decide. The module statement is sha256:21e7e69d5fb3b0823b89be6d5b2e6e9581a42a77d933e7864a0833cbe22fe2e1; result is sha256:01e9450248d36bfbcb9028325842962e0e2af3d48208a203adf28173fee94911 and claim is sha256:7240a2d88de7184ca142279ac4a6119a08de03475d76f0286cb91c55acb53644. The Freeze event is sha256:119b1c06df26ac2ebd73265a208ed6218be64e15bcfe75d96728a13c743b33fd.

Tier 1: an explicit question of a 2025 paper, preregistered in #14763 before any Lean. theorem; resolution refuted. result and its private lemmas are bind-only (the explicit matrix, its sums of squares and a $2\times2$ determinant, by normalization); admission basis open-problem-resolution. Utility kind=certified-instance; basis=refutes. There is no digestion atom.

What the refutation shows

• Proved by result: the kernel criterion of Theorem 2 does not imply the trace criterion of Corollary 2, already for two qubits; the implication in the source's Fig. 2 is strict.
• Mechanism (derived, not formalized): the kernel criterion is invariant under local invertible filtering. For branch (i) ($m\le n$), $W\mapsto(Y\otimes{\bf1})W(Y^\dagger\otimes{\bf1})$ transforms $W+{\rm tr}_2(W)\otimes{\bf1}$ by the same congruence and its kernel by $(Y^\dagger)^{-1}\otimes{\bf1}$, which preserves Schmidt rank; for branch (ii) ($m\ge n$) the same holds with filters ${\bf1}\otimes Y$ on the second factor and $W+{\bf1}\otimes{\rm tr}1(W)$. The trace criterion is not: it needs a kernel vector with equal Schmidt coefficients. The family $W_t=(\operatorname{diag}(1,t)\otimes{\bf1})W{\rm red}(\operatorname{diag}(1,t)\otimes{\bf1})$, $t>0$, has trace $1+t^2$ and least eigenvalue $-t$, so the trace criterion holds exactly at $t=1$; the gap is $(t-1)^2/2$ (computed, not formalized).
• Derived, not formalized: up to filtering the two criteria agree. If $m\le n$ and $v\in\ker(W+{\rm tr}_2(W)\otimes{\bf1})$ has coefficient matrix $M$ of rank $m$, then $Y^\dagger=(m,MM^\dagger)^{1/2}$ makes $(Y^{-\dagger}\otimes{\bf1})v$ maximally entangled, so $W'=(Y\otimes{\bf1})W(Y^\dagger\otimes{\bf1})$ satisfies the trace criterion; conversely the trace criterion implies the kernel criterion (source). So a witness satisfies the kernel criterion if and only if some local filtering of it satisfies the trace criterion (the case $m\ge n$ with filters on the second factor is symmetric).
• Open: the source's other two Section IV questions (an explicit entanglement-breaking channel of full Choi rank detecting the optimality of the flip in odd dimensions; multipartite versions).

The literature check in #14763 is bounded and does not establish worldwide novelty or priority; the journal status of the source was checked only through Crossref.