LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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.

Each coordinate of the oriented area vector is the Jacobian determinant of the matching cyclic projection

Statement

Let $O \subseteq R^{2}$ be open and let $φ = (φ_{x}, φ_{y}, φ_{z}) : O \to R^{3}$ be $C^{1}$ , with parameters named $u, v$ and $φ_{u} : = \partial_{u} φ$ , $φ_{v} : = \partial_{v} φ$ . Then at every point of $O$ ,

$(φ_{u} \times φ_{v})_{k} = \det D (π_{k} \circ φ)$

for each of the three coordinate directions $k \in {x, y, z}$ , where $π_{k}$ is the cyclic coordinate projection of Simple solid regions in a coordinate direction and their cyclic coordinate projection.

Facts & Assumptions

Given: The open set $O \subseteq R^{2}$ and the $C^{1}$ map $φ : O \to R^{3}$ of the Statement.

[F1]

For $u = (u_{x}, u_{y}, u_{z})$ and $v = (v_{x}, v_{y}, v_{z})$ in $R^{3}$ , $u \times v = (u_{y} v_{z} - u_{z} v_{y}, u_{z} v_{x} - u_{x} v_{z}, u_{x} v_{y} - u_{y} v_{x})$ (The cross product in $R^{3}$ ).

[F2]

For a $C^{1}$ map $g$ of an open subset of $R^{n}$ into $R^{n}$ , its Jacobian determinant is $\det D g (x)$ , the determinant of its Jacobian matrix (The Jacobian determinant of a square-dimensional $C^{1}$ map is the determinant of its Jacobian matrix).

[F3]

If every partial derivative $\partial_{j} f_{i} (a)$ of $f$ exists, the Jacobian matrix is $J f (a) = (\partial_{j} f_{i} (a))_{i < n, j < m}$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F4]

For a commutative ring $R$ , $n \geq 1$ and $A = (a_{i j}) \in M_{n} (R)$ , $\det (A) = \sum_{σ \in S_{n}} sgn (σ) \prod_{i < n} a_{σ (i), i}$ , with columns indexed by $i < n$ and rows by $σ (i)$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F5]

An inversion of $σ \in S_{n}$ is a pair $(i, j)$ with $i < j < n$ and $σ (i) > σ (j)$ , and $sgn (σ) = (- 1)^{inv (σ)}$ (Inversions, inversion number, the sign $sgn (σ) = (- 1)^{inv (σ)}$ , and even and odd permutations).

[F6]

The cyclic coordinate projections are $π_{x} (p) = (p_{y}, p_{z})$ , $π_{y} (p) = (p_{z}, p_{x})$ and $π_{z} (p) = (p_{x}, p_{y})$ (Simple solid regions in a coordinate direction and their cyclic coordinate projection).

[F7]

A map $f : U \to R^{q}$ is of class $C^{k}$ when each component is of class $C^{k}$ ( $C^{k}$ Euclidean maps and diffeomorphisms).

Proof

technique · direct

1.1givenF2F3F4F5F7

Each $π_{k} \circ φ$ is a map of the two parameters into $R^{2}$ whose two components are components of $φ$ , hence $C^{1}$ by [F7], so by [F2] and [F3] it has a Jacobian matrix $(\begin{array}{cc} a_{00} & a_{01} \\ a_{10} & a_{11} \end{array})$ with $a_{i 0} = \partial_{u}$ and $a_{i 1} = \partial_{v}$ of its $i$ th component. There are exactly two elements of $S_{2}$ : the identity, with no inversion and sign $+ 1$ , contributing $a_{00} a_{11}$ , and the transposition sending $0$ to $1$ and $1$ to $0$ , with the single inversion $(0, 1)$ and sign $- 1$ , contributing $- a_{10} a_{01}$ . So [F4] and [F5] give $\det D (π_{k} \circ φ) = a_{00} a_{11} - a_{10} a_{01}$ .

1.2givenF1F3

By [F1] with $u = φ_{u}$ and $v = φ_{v}$ , whose coordinates are the partial derivatives named in [F3], the oriented area vector has coordinates $(φ_{u} \times φ_{v})_{x} = \partial_{u} φ_{y} \partial_{v} φ_{z} - \partial_{u} φ_{z} \partial_{v} φ_{y},$ $(φ_{u} \times φ_{v})_{y} = \partial_{u} φ_{z} \partial_{v} φ_{x} - \partial_{u} φ_{x} \partial_{v} φ_{z},$ $(φ_{u} \times φ_{v})_{z} = \partial_{u} φ_{x} \partial_{v} φ_{y} - \partial_{u} φ_{y} \partial_{v} φ_{x} .$

2.1step 1.1step 1.2F6

By [F6] the projection $π_{x}$ retains the coordinates $y$ then $z$ , so $π_{x} \circ φ = (φ_{y}, φ_{z})$ and step 1.1 gives $\det D (π_{x} \circ φ) = \partial_{u} φ_{y} \partial_{v} φ_{z} - \partial_{u} φ_{z} \partial_{v} φ_{y}$ , which is the first coordinate computed in step 1.2.

2.2step 1.1step 1.2F6

By [F6] the projection $π_{y}$ retains the coordinates $z$ then $x$ , in that cyclic order, so $π_{y} \circ φ = (φ_{z}, φ_{x})$ and step 1.1 gives $\det D (π_{y} \circ φ) = \partial_{u} φ_{z} \partial_{v} φ_{x} - \partial_{u} φ_{x} \partial_{v} φ_{z}$ , the second coordinate computed in step 1.2. Retaining $x$ then $z$ in increasing order instead would exchange the two rows and give the opposite sign, which is why the cyclic order is part of the projection.

2.3step 1.1step 1.2F6

By [F6] the projection $π_{z}$ retains the coordinates $x$ then $y$ , so $π_{z} \circ φ = (φ_{x}, φ_{y})$ and step 1.1 gives $\det D (π_{z} \circ φ) = \partial_{u} φ_{x} \partial_{v} φ_{y} - \partial_{u} φ_{y} \partial_{v} φ_{x}$ , the third coordinate computed in step 1.2.

3.1step 2.1step 2.2step 2.3∎

Steps 2.1, 2.2 and 2.3 are the three asserted identities, valid at every point of $O$ ; in particular all three determinants vanish exactly where the oriented area vector does.

Remarks

The identity holds where the patch is not regular. Nothing above uses $φ_{u} \times φ_{v} \neq 0$ . That matters because the lateral faces of a boundary presentation are exactly the patches whose $k$ th projected Jacobian determinant vanishes, and the identity is what turns that analytic condition into a geometric one.

Depends on

Used by

Dependency tree · two levels

33 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

G. Strang and E. Herman, Calculus Volume 3 (OpenStax), section 6.8 (standard reference, not scraped)