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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Physical constraints and manifold-valued inversion

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Admissible constraint classes and restricted identifiability

Definition

Let X and Y be real vector spaces, let U⊆X be nonempty, and let A:X→Y be linear. Supply metrics dX on X and dY on Y; for a purely algebraic instance with no specified metrics, use the discrete metrics dX(x,x′)=0 if x=x′ and 1 otherwise, and likewise on Y. These metrics require no compatibility with the vector operations, and no continuity of A is assumed. The discrete triangle inequality holds because, if two endpoints differ, at least one of the two successive pairs differs. Give U and its subsets their restricted metrics. Thus every pair of real vector spaces still admits this convention. The corresponding metric-space inverse-problem model has admissible truths U and forward map A∣U. An admissible constraint class is a nonempty subset C⊆U; its constrained forward map is the restriction A∣C:C→Y. The class C is identifiable for A if A∣C is injective, meaning Ax=Ax′ for x,x′∈C implies x=x′. Define the difference set and kernel by

C−C:={x−x′:x,x′∈C},ker⁡A:={x∈X:Ax=0Y}.

PropositionStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A constraint removes kernel ambiguity exactly when its differences avoid the kernel

Statement

Let X and Y be real vector spaces, let A:X→Y be linear, and let C⊆X be nonempty. The restriction A∣C:C→Y is injective if and only if

(C−C)∩ker⁡A={0}.

This condition characterizes exact uniqueness on C; it gives no stability bound for perturbed data.

Facts & Assumptions

Given: Real vector spaces X,Y, a linear map A:X→Y, and a nonempty constraint class C⊆X.

[F1]

Admissible constraint classes and restricted identifiability with U=X and the discrete metrics on X,Y just specified there (used only to instantiate the model; the argument is algebraic): C is nonempty, C−C={x−x′:x,x′∈C}, and restricted identifiability means Ax=Ax′ for x,x′∈C implies x=x′.

[F2]

Linear map between vector spaces over the same field: linearity gives A(x−x′)=Ax−Ax′.

[F3]

Kernel and image of a linear map: ker⁡A={x∈X:Ax=0Y}.

Proof

1.1givenF1F2F3choosealgebra

Suppose A∣C is injective and take v∈(C−C)∩ker⁡A. By [F1], write v=x−x′ with x,x′∈C. By [F2] and [F3], Ax−Ax′=Av=0Y, so Ax=Ax′. Injectivity gives x=x′, hence v=0. Since C is nonempty, choose x∈C; then 0=x−x∈C−C. Linearity also gives A(0X)=A(0X+0X)=A(0X)+A(0X), hence A(0X)=0Y and 0X∈ker⁡A. Therefore (C−C)∩ker⁡A={0}.

2.1givenF1F2F3algebra∎

Conversely, suppose (C−C)∩ker⁡A={0} and take x,x′∈C with Ax=Ax′. Then x−x′∈C−C by [F1], and A(x−x′)=Ax−Ax′=0Y by [F2], so x−x′∈ker⁡A by [F3]. Thus x−x′=0, hence x=x′. Therefore A∣C is injective.

AlgorithmAlgorithm: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Finite-budget regularized gradient on a Stiefel constraint

Definition

For integers 1≤p≤n, define the real Stiefel constraint set

St⁡(n,p):={X∈Rn×p:XTX=Ip}.

Equip Rn×p with the Frobenius inner product ⟨U,V⟩F=tr⁡(UTV) and norm ∥U∥F=⟨U,U⟩F. Let H=Rn×p, let Y be a real Hilbert space, let U⊆H be open and contain St⁡(n,p), and let F:U→Y be continuously Fréchet differentiable. Given data y∈Y, a reference matrix Xref∈H, and λ≥0, define the residual and regularized objective by

r(X):=∥F(X)−y∥Y,Φ(X):=12∥F(X)−y∥Y2+λ2∥X−Xref∥F2.

Supply an ambient gradient oracle G:U→H satisfying DΦ(X)[Z]=⟨G(X),Z⟩F for every X∈U and Z∈H, an initial frame X0∈St⁡(n,p), a residual tolerance τ≥0, a finite integer budget K≥0, and a step schedule α0,…,αK−1 with every αk>0. For a square matrix M, put sym⁡(M):=(M+MT)/2. The tangent-projection and metric-dual calculation for this C1 objective, well-definedness of every tangent polar call, and finite-budget feasibility are justified by the finite-budget return and feasibility lemma. No smoothness beyond C1 is required. The procedure is:

  1. Set the current frame to X0 and the schedule index to k=0. If r(X)≤τ, return (X,r(X),target-met).

  2. While k<K, compute G(X) and the Stiefel gradient g:=G(X)−Xsym⁡(XTG(X)). If g=0, return (X,r(X),stationary).

  3. Set η:=−αkg and form the polar-retracted candidate

    Xtrial:=(X+η)((X+η)T(X+η))−1/2.

  4. Evaluate Φ(Xtrial) and r(Xtrial). If Φ(Xtrial)>Φ(X), return (X,r(X),objective-rejected). Otherwise accept Xtrial as the current frame. If its residual is at most τ, return (X,r(X),target-met); otherwise increment k by one and repeat the loop.

  5. When k=K, return (X,r(X),budget-exhausted).

This procedure is restricted to the stated finite-dimensional real Stiefel constraint and the displayed polar retraction. It gives no convergence, global-optimality, or noisy-data stability guarantee, and it makes no finite-precision claim. It does not apply to positivity constraints or other nonlinear manifolds without a separately specified compatible retraction and proof.

LemmaStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

The Stiefel gradient procedure returns a feasible iterate within its finite budget

Statement

For the supplied C1 objective and ambient gradient oracle, the displayed projected vector is the Frobenius metric-dual gradient of the restricted objective, and every polar trial is defined. For every input satisfying the finite-budget Stiefel gradient procedure, it returns after at most K trial evaluations with an output frame Xout∈St⁡(n,p). Every accepted update has nonincreasing objective value, and the returned status is target-met exactly when r(Xout)≤τ; every other returned status has r(Xout)>τ.

Facts & Assumptions

Given: The data and inputs of Finite-budget regularized gradient on a Stiefel constraint, including X0∈St⁡(n,p), a finite integer K≥0, and positive step sizes αk for 0≤k<K.

[F1]

Finite-budget regularized gradient on a Stiefel constraint supplies a C1 objective on the ambient open set and an oracle with DΦ(X)[Z]=⟨G(X),Z⟩F. At each current frame X, the procedure forms g=G−Xsym⁡(XTG) and trials η=−αkg.

[F2]

Tangent spaces and Frobenius projections on Stiefel and Grassmann: TXSt⁡(n,p)={Z:XTZ+ZTX=0}.

[F3]

The QR and polar maps are first-order Stiefel and Grassmann retractions: the polar map on a Stiefel tangent input is defined and takes values in St⁡(n,p).

[F4]

Finite-budget regularized gradient on a Stiefel constraint: an accepted candidate becomes the current frame, a rejected candidate returns the previous frame, the finite schedule index increases after each accepted nonterminal trial, and the residual test controls the returned status.

Proof

1.1givenF1F2algebra

For a current X∈St⁡(n,p), put M=XTG. From [F1] and XTX=Ip, the projected gradient satisfies XTg+gTX=M+MT−2sym⁡(M)=0. Thus g∈TXSt⁡(n,p) by [F2], and multiplication by −αk keeps η tangent.

2.1givenF1F2step 1.1algebra

Put S=sym⁡(XTG), so ST=S. If Z∈TXSt⁡(n,p), then A=XTZ is skew-symmetric by [F2]. Hence ⟨XS,Z⟩F=tr⁡(SA)=0: transposing and cyclically reordering gives tr⁡(SA)=tr⁡(ATST)=−tr⁡(AS)=−tr⁡(SA). The derivative of the restriction on a tangent Z is therefore DΦ(X)[Z]=⟨G,Z⟩F=⟨g,Z⟩F. Together with step1.1 this makes g its tangent metric-dual vector. It is unique, since two such tangent vectors differ by a tangent vector whose pairing with itself is zero. This uses only the supplied first derivative, so C1 suffices.

2.2givenF3F4step 1.1algebra

The initial frame is in St⁡(n,p) by the input condition. Whenever a current frame is in St⁡(n,p), step1.1 makes the trial tangent. Then (X+η)T(X+η)=Ip+ηTη, so for v≠0 its quadratic form is ∥v∥22+∥ηv∥22>0. Thus X+η has full column rank, the polar call is in the declared domain of [F3], and [F3] puts its candidate in St⁡(n,p)⊂U. In particular every subsequent objective, residual and oracle evaluation has an input in its supplied domain. An accepted candidate is the next current frame; a rejected candidate returns the previous frame. Hence every current frame and every possible returned frame remains in St⁡(n,p) by finite induction.

3.1givenF4step 2.2algebra∎

The index begins at zero, increases by one after each accepted nonterminal trial, and the loop ends when k=K; every other branch returns immediately. Therefore at most K trial evaluations occur. The procedure accepts only candidates with objective no greater than the current value, and its residual test labels an output target-met exactly when its residual is at most τ; every other return branch follows a failed target test and has residual greater than τ.

4 · Examples, counterexamples and false statements

None yet.

Sources