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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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The real sign quotient and its induced measurement maps

Statement

Let d≥1, m≥0, and Φ=(ai)i∈[m] be a family in Rd. Define x∼z by z=x or z=−x, let Qd=Rd/∼, and put d±([x],[z])=min⁡{∥x−z∥2,∥x+z∥2}. Let Ym={b:[m]→[0,∞)} carry the metric induced from zero-based Euclidean Rm by the fixed relabeling i↦i−1 when m≥1, and the zero metric on the singleton Y0 when m=0. Define MΦ([x])=(∣⟨ai,x⟩∣)i∈[m],IΦ([x])=(⟨ai,x⟩2)i∈[m]. Then ∼ is an equivalence relation, d± is a metric on Qd, Ym is a metric space, and both displayed measurement maps are well defined. Moreover, IΦ is coordinatewise the square of MΦ, so the two maps are injective simultaneously.

Facts & Assumptions

Given: The definitions and hypotheses in the statement.

[F1]

The sign relation is x∼z exactly when z=x or z=−x, the quotient map sends x to [x], and the amplitude and intensity formulas use the real Euclidean pairing (Real phase retrieval via finite real measurements).

[F3]

A metric satisfies separation, symmetry, and the triangle inequality; these axioms can be restricted to a subset (Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric).

[F4]

For real t, ∣t∣≥0, ∣−t∣=∣t∣, and ∣t∣=0 exactly when t=0 (Absolute value in an ordered field, Basic properties of the absolute value).

[F6]

Every real square is nonnegative (Squares of nonzero elements are positive).

[F7]

The real order is total, so every two-element set of real numbers has a least element, which is its minimum (Ordered field, Maximum and minimum of a set).

Proof

Given: A finite real family Φ, d≥1, and m≥0.

1.1givenF1algebra

The relation ∼ is reflexive because x=x, symmetric because z=x implies x=z and z=−x implies x=−z, and transitive because multiplying the two signs in z=ϵx and w=δz gives w=(δϵ)x with δϵ∈{1,−1}. Thus Qd is the quotient set and q(x)=[x] is onto by the definition of equivalence classes.

1.2givenF2F3algebra

If m≥1, the relabeling J:Ym→[0,∞)m given by J(b)k=bk+1 is a bijection and ρm(b,c)=d2(Jb,Jc), so the metric axioms of [F2] restrict to Ym. If m=0, Y0 has one element and the zero function separates its only pair and is symmetric and triangular. Hence Ym is a metric space in both cases.

1.3givenF1F7algebra

Replacing representatives by x′=ϵx and z′=δz, with ϵ,δ∈{1,−1}, changes the pair (∥x−z∥2,∥x+z∥2) only by possibly interchanging its two entries, since ∥x′−z′∥2=∥x−(ϵδ)z∥2 and ∥x′+z′∥2=∥x+(ϵδ)z∥2. By [F7] the minimum is attained, so its value is independent of representatives.

2.1step 1.1F1F4F6algebra

Replacing x by −x changes each amplitude coordinate from ∣⟨ai,x⟩∣ to ∣−⟨ai,x⟩∣=∣⟨ai,x⟩∣, and changes each intensity coordinate from ⟨ai,x⟩2 to (−⟨ai,x⟩)2=⟨ai,x⟩2. Every intensity coordinate is nonnegative by [F6], so both maps have the stated codomain and are well defined on Qd, including zero rows and the zero signal. For m=0 both are the unique map to the empty tuple.

2.2step 1.3F2F7choosealgebra

Choose signs ϵ,δ∈{1,−1} attaining the two minima for d±([x],[y]) and d±([y],[z]); [F7] ensures these minima exist, and the choices are from finite two-element sets. The Euclidean triangle inequality and sign-invariance of the norm give d±([x],[z])≤∥x−ϵδz∥2≤∥x−ϵy∥2+∥y−δz∥2=d±([x],[y])+d±([y],[z]). Thus the triangle inequality holds.

3.1step 1.3step 2.2F1F2F5

Both candidate norms are nonnegative by the nonnegative-root convention [F5], and their minimum is symmetric under interchanging x and z. The minimum is zero exactly when x=z or x=−z, which is exactly [x]=[z] by [F1] and separation of the Euclidean metric in [F2]. Together with the triangle inequality in step 2.2, this proves that d± is a metric.

4.1step 2.1F5algebra∎

Each amplitude coordinate is nonnegative and its square is the corresponding intensity coordinate. If two nonnegative amplitudes have the same square, uniqueness of nonnegative square roots [F5] makes the amplitudes equal; conversely equal amplitudes have equal squares. Coordinatewise, the amplitude and intensity maps therefore have the same fibers and are injective simultaneously.

Depends on

Used by

Cited to discharge well-definedness by Real phase retrieval via finite real measurements.

Dependency tree · two levels

36 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources