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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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The complement property is not sufficient for complex phase retrieval

Statement refuted

In C2 use a1=(1,0), a2=(0,1), and a3=(1,1), and define ⟨a,x⟩=a(1)‾x1+a(2)‾x2 for a=(a(1),a(2)). Identify signals by global phase, x≈z when z=λx for some λ∈C with ∣λ∣=1, and use intensity data (∣⟨aj,x⟩∣2)j=13. Here the complex complement property means that every subset of the three labels or its complement spans C2. Each pair spans C2, so the family has this property; nevertheless, x=(1,i) and z=(1,−i) have equal intensity data and are not related by a global phase. Thus the real complement-property criterion and its real full-spark count cannot be transferred to complex phase retrieval.

Facts & Assumptions

Given: The three vectors in C2, the Hermitian pairing and global-phase relation defined in the counterexample, and the real-field scope of the cited theorem.

[F1]

Real phase retrieval is equivalent to the complement property: the complement-property equivalence proved here has hypotheses over Rd and identifies only the pair x,−x.

[F2]

The sharp real phase-retrieval count for full-spark measurements: the 2d−1 full-spark sufficiency statement is for finite real measurement families.

[F5]

Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S, Linear subspace of a vector space: each span is a complex linear subspace containing its generators, so every displayed linear combination belongs to that span.

Counterexample

1.1givenF3F5algebra

For arbitrary (u,v)∈C2, the three identities (u,v)=ua1+va2=(u−v)a1+va3=ua3+(v−u)a2 show that each of the pairs {a1,a2}, {a1,a3}, and {a2,a3} spans C2: the span is a subspace containing the pair, so it contains the corresponding displayed combination for every (u,v). Any partition of three labels has at least two labels on one side, and a set containing a spanning pair also spans; hence the complex complement property holds.

1.2givenF3F4algebra

The measurements of x=(1,i) are (1,i,1+i) and those of z=(1,−i) are (1,−i,1−i). By [F3] and [F4], their squared moduli are (1,1,2) in both cases.

2.1step 1.1step 1.2F1F2F3algebra∎

If z=λx with ∣λ∣=1, the first coordinate forces λ=1, while the second would then force −i=i, which is false in the field [F3]. Thus x≉z, and step 1.2 proves noninjectivity despite step 1.1. By [F1] and [F2], this does not contradict either real-field result: the ambient field, phase relation, and measurements differ.

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