LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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The curl measures the antisymmetric part of the total derivative

Statement

Let $U \subseteq R^{3}$ be open, let $F : U \to R^{3}$ be $C^{1}$ and let $p \in U$ . Then for all $u, v \in R^{3}$ ,

$⟨ D F (p) u, v ⟩ - ⟨ D F (p) v, u ⟩ = ⟨ curl F (p), u \times v ⟩,$

where $D F (p)$ is the total derivative of $F$ at $p$ , whose matrix is the Jacobian matrix $J F (p) = (\partial_{j} F_{i} (p))_{i, j < 3}$ .

Facts & Assumptions

Given: The open set $U \subseteq R^{3}$ , the $C^{1}$ field $F : U \to R^{3}$ , the point $p \in U$ and vectors $u, v \in R^{3}$ , with the three coordinates named $x, y, z$ .

[F1]

The curl of a $C^{1}$ field $F$ on an open $U \subseteq R^{3}$ is $curl F = (\partial_{y} F_{z} - \partial_{z} F_{y}, \partial_{z} F_{x} - \partial_{x} F_{z}, \partial_{x} F_{y} - \partial_{y} F_{x})$ (Divergence and curl of a $C^{1}$ vector field).

[F2]

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 $R^{3}$ ).

[F3]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{k < m} x_{k} y_{k}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F4]

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).

[L1]

If $f$ is totally differentiable at $a$ then $D_{v} f (a)$ exists for every $v$ and equals $D f (a) v$ ; in particular $\partial_{j} f (a) = D f (a) e_{j}$ , and the matrix of $D f (a)$ is $J f (a)$ (A total derivative computes every directional derivative, and its matrix is the Jacobian).

[L2]

If every partial derivative of $f$ exists on a neighbourhood of $a$ and is continuous at $a$ , then $f$ is totally differentiable at $a$ and $D f (a)$ is the linear map with matrix $J f (a)$ (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[L3]

The cross product is bilinear and alternating (The cross product is bilinear, alternating, and orthogonal to both factors).

Proof

technique · direct

1.1

Since $F$ is $C^{1}$ on $U$ , its partial derivatives exist on $U$ and are continuous, so [L2] makes $F$ totally differentiable at $p$ with $D F (p)$ the linear map of matrix $J F (p)$ ; by [L1] and [F4] the entries of that matrix are $\partial_{j} F_{i} (p)$ , so $(D F (p) u)_{i} = \sum_{j < 3} \partial_{j} F_{i} (p) u_{j}$ .

givenL1L2F4

1.2

Hence, by [F3] and [F4], $⟨ D F (p) u, v ⟩ - ⟨ D F (p) v, u ⟩ = \sum_{i < 3} \sum_{j < 3} \partial_{j} F_{i} (p) u_{j} v_{i} - \sum_{i < 3} \sum_{j < 3} \partial_{j} F_{i} (p) v_{j} u_{i},$ and exchanging the names of the two summation indices in the second double sum turns it into $\sum_{i < 3} \sum_{j < 3} \partial_{i} F_{j} (p) v_{i} u_{j}$ , so the difference equals $\sum_{i < 3} \sum_{j < 3} (\partial_{j} F_{i} (p) - \partial_{i} F_{j} (p)) u_{j} v_{i}$ .

F3F4algebra

2.1

In the double sum of step 1.2 the terms with $i = j$ have coefficient $\partial_{i} F_{i} (p) - \partial_{i} F_{i} (p) = 0$ , so only the six terms with $i \neq j$ contribute, that is the three unordered index pairs ${y, z}$ , ${z, x}$ and ${x, y}$ , each occurring twice.

step 1.2algebra

3.1

Grouping the two terms of the pair ${y, z}$ gives $(\partial_{y} F_{z} (p) - \partial_{z} F_{y} (p)) u_{y} v_{z} + (\partial_{z} F_{y} (p) - \partial_{y} F_{z} (p)) u_{z} v_{y}$ , that is $(\partial_{y} F_{z} (p) - \partial_{z} F_{y} (p)) (u_{y} v_{z} - u_{z} v_{y})$ . The pair ${z, x}$ gives $(\partial_{z} F_{x} (p) - \partial_{x} F_{z} (p)) (u_{z} v_{x} - u_{x} v_{z})$ and the pair ${x, y}$ gives $(\partial_{x} F_{y} (p) - \partial_{y} F_{x} (p)) (u_{x} v_{y} - u_{y} v_{x})$ .

step 1.2step 2.1F1algebra

4.1

By [F1] the three coefficients in step 3.1 are the first, second and third coordinates of $curl F (p)$ , and by [F2] the three bracketed factors are the first, second and third coordinates of $u \times v$ . By [F3] their sum is therefore $⟨ curl F (p), u \times v ⟩$ , which with step 1.2 is the asserted identity.

step 3.1F1F2F3

5.1

As a check on the signs, take $u = e_{x}$ and $v = e_{y}$ : the left side is $⟨ D F (p) e_{x}, e_{y} ⟩ - ⟨ D F (p) e_{y}, e_{x} ⟩ = \partial_{x} F_{y} (p) - \partial_{y} F_{x} (p)$ and the right side is the third coordinate of $curl F (p)$ , since $e_{x} \times e_{y} = e_{z}$ by [F2]; the pairs $(e_{y}, e_{z})$ and $(e_{z}, e_{x})$ give the first and second coordinates in the same way. When $u = v$ both sides vanish, the left by inspection and the right because the cross product is alternating by [L3].

step 4.1L3F2∎

Remarks

The identity is what makes the curl coordinate-free enough for Stokes. Its left side is built from the total derivative and two vectors, with no reference to a coordinate system beyond the one the inner product carries; the right side reads off the coordinates. That is exactly the form in which the curl enters The curl flux integrand of a $C^{2}$ patch is a two-dimensional curl of the pulled-back field, where $u$ and $v$ are the two parameter derivatives of a patch.

Depends on

Used by

The curl flux integrand of a C² patch is a two-dimensional curl of the pulled-back field Lemma

Dependency tree · two levels

29 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 6.7 (standard reference, not scraped)
M. Corral, Vector Calculus, chapter 4 (LibreTexts) (standard reference, not scraped)