LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge 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.

Pullback gradients and induced surface metrics

Statement

Assume Countable Choice. Let $F : \hat{K} \to K$ be an admissible $C^{1}$ orientation-preserving cell diffeomorphism and let $u$ be $C^{1}$ on $K$ . With $\hat{u} = u \circ F$ , one has $\nabla_{x} u \circ F = D F^{- T} \nabla_{ξ} \hat{u}$ and $\int_{K} ∣ \nabla u ∣^{2} d x = \int_{\hat{K}} J_{F} \nabla {\hat{u}}^{T} (D F^{- 1} D F^{- T}) \nabla \hat{u} d ξ$ . For a regular $k$ -dimensional parametrized face $f : U \subset R^{k} \to R^{d}$ with metric $G_{f} = D f^{T} D f$ , the induced area factor is $\sqrt{\det G_{f}}$ and the tangential gradient of $u \circ f$ is $D f G_{f}^{- 1} \nabla (u \circ f)$ ; its squared norm is $\nabla (u \circ f)^{T} G_{f}^{- 1} \nabla (u \circ f)$ .

Facts & Assumptions

Given: Countable Choice, an admissible orientation-preserving $C^{1}$ cell map, and a $C^{1}$ scalar field.

[F1]

Admissible isoparametric curved cells: $J_{F} = \det D F > 0$ , $G_{F} = D F^{T} D F$ , and the induced facet metric is the Gram matrix of the facet differential.

[F2]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : $D (u \circ F) = D u (F) D F$ .

[F3]

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: under Countable Choice a $C^{1}$ diffeomorphism changes variables with factor $∣ \det D F ∣$ .

[A1]

Countable Choice is assumed exactly for [F3] (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

1.1F1F2givenalgebra

The chain rule [F2] gives $\nabla_{ξ} \hat{u} = D F^{T} (\nabla_{x} u \circ F)$ . Since $D F$ is invertible by admissibility, multiplication by $D F^{- T}$ yields $\nabla_{x} u \circ F = D F^{- T} \nabla_{ξ} \hat{u}$ .

1.2F1F2algebra

For a regular face parametrization, the differential maps a coordinate vector $z$ to $D f z$ and its squared Euclidean length is $z^{T} G_{f} z$ . The vector $G_{f}^{- 1} \nabla (u \circ f)$ is therefore the coordinate vector whose image under $D f$ has the defining inner products with every tangent vector; its image is the tangential gradient. Taking its squared norm gives $\nabla (u \circ f)^{T} G_{f}^{- 1} \nabla (u \circ f)$ . The Gram determinant gives the $k$ -volume scaling factor.

2.1F1F3A1givenstep 1.1∎

The energy identity follows by substituting step 1.1 into $∣ \nabla u ∣^{2}$ and applying [F3] to the nonnegative measurable energy density. Since $J_{F} > 0$ , the absolute determinant in [F3] equals $J_{F}$ .

Depends on

Used by

Dependency tree · two levels

22 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

Jean-Luc Guermond, MATH 661 Fall 2023, Part III, Chapter 8 (standard reference, not scraped)
Gerald Dziuk and Charles M. Elliott, Finite element methods for surface PDEs, Acta Numerica 22 (2013) (standard reference, not scraped)