Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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.

An affine conditional path has the stated velocity field

Statement

Let d≥1, fix x1∈Rd, and let t↦μt(x1)∈Rd and t↦σt(x1)∈(0,∞) be continuously differentiable on [0,1], with one-sided derivatives at the endpoints. For z∈Rd, define the affine map ψt(z∣x1):=σt(x1)z+μt(x1). For each t, this map is a diffeomorphism with inverse ψt−1(x∣x1)=(x−μt(x1))/σt(x1). The unique vector field satisfying ddtψt(z∣x1)=ut(ψt(z∣x1)∣x1) for every z∈Rd is ut(x∣x1)=σ˙t(x1)σt(x1)(x−μt(x1))+μ˙t(x1).

Facts & Assumptions

Given: d, x1, and the continuously differentiable functions μt(x1) and σt(x1)>0 in the Statement.

[F1]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: vectors in Rd have coordinatewise addition and scalar multiplication.

Proof

Given: The hypotheses in the Statement.

1.1givenF1algebra

Fix t∈[0,1]. Since σt(x1)>0, solving x=σt(x1)z+μt(x1) gives the unique inverse z=(x−μt(x1))/σt(x1). Both the map and inverse are continuously differentiable in z, so ψt(⋅∣x1) is a diffeomorphism.

2.1givenstep 1.1algebra

For fixed z, differentiation gives ddtψt(z∣x1)=σ˙t(x1)z+μ˙t(x1). Substitute the inverse expression from step 1.1 to obtain ddtψt(z∣x1)=σ˙t(x1)σt(x1)(ψt(z∣x1)−μt(x1))+μ˙t(x1). Thus the displayed ut satisfies the flow equation.

3.1step 1.1step 2.1algebra∎

If another field u~t satisfies the same equation for every z, then for any x choose the explicit inverse value z=ψt−1(x∣x1). The flow equation forces u~t(x∣x1)=σ˙t(x1)σt(x1)(x−μt(x1))+μ˙t(x1), proving uniqueness.

Source notes

The proof is the explicit affine inversion and differentiation argument. The cited paper and two independent treatments were inspected for the same formula and schedule convention.

Depends on

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Sources