Kelvin–Stokes theorem

The Kelvin–Stokes theorem^{[1][2][3][4][5]} (named Lord Kelvin and George Stokes), also known as the curl theorem,^[6] is a theorem in vector calculus on R³. Given a vector field, the theorem relates the integral of the curl of the vector field over some surface, to the line integral of the vector field around the boundary of the surface. The Kelvin–Stokes theorem is a special case of the “generalized Stokes' theorem.”^[7][8] In particular, a vector field on R³ can be considered as a 1-form in which case curl is the exterior derivative.

Theorem

Let γ: [a, b] → R² be a Piecewise smooth Jordan plane curve. The Jordan curve theorem implies that γ divides R² into two components, a compact one and another that is non-compact. Let D denote the compact part that is bounded by γ and suppose ψ: D → R³ is smooth, with S := ψ(D). If Γ is the space curve defined by Γ(t) = ψ(γ(t))^{[note 1]} and F is a smooth vector field on R³, then:^[1][2][3]

$\oint_\Gamma \mathbf{F}\, \cdot\, d{\mathbf{\Gamma}} = \iint_S \nabla\times\mathbf{F}\, \cdot\, d\mathbf{S}$

Proof

The proof of the theorem consists of 4 steps.^{[2][3][note 2]} We assume Green's theorem, so what is of concern is how to boil down the three-dimensional complicated problem (Kelvin–Stokes theorem) to a two-dimensional rudimentary problem (Green's theorem). When proving this theorem, mathematicians normally use the differential form. The "pull-back^{[note 2]} of a differential form" is a very powerful tool for this situation, but learning differential forms requires substantial background knowledge. So, the proof below does not require knowledge of differential forms, and may be helpful for understanding the notion of differential forms.

First step of the proof (defining the pullback)

Define

$\mathbf{P}(u,v) = (P_1(u,v), P_2(u,v))$

so that P is the pull-back^{[note 2]} of F, and that P(u, v) is R²-valued function, dependent on two parameters u, v. In order to do so we define P₁ and P₂ as follows.

$P_1(u,v)=\left\langle \mathbf{F}(\psi(u,v)) \bigg| \frac{\partial \psi}{\partial u} \right\rangle, \qquad P_2(u,v)=\left\langle \mathbf{F}(\psi(u,v)) \bigg| \frac{\partial \psi}{\partial v} \right\rangle$

Where $\langle \ |\ \rangle$ is the normal inner product (for Euclidean vectors, the dot product; see Bra-ket notation) of R³ and hereinafter, $\langle \ |A|\ \rangle$ stands for the bilinear form according to matrix A.^{[note 3]}

Second step of the proof (first equation)

According to the definition of a line integral,

$\begin{align} \oint_{\Gamma} \mathbf{F} \cdot d\mathbf{\Gamma} &=\int_a^b \left\langle (\mathbf{F}\circ \Gamma(t))\bigg|\frac{d\Gamma}{dt}(t) \right\rangle \, dt \\ &= \int_a^b \left\langle (\mathbf{F}\circ \Gamma(t))\bigg|\frac{d(\psi\circ\gamma)}{dt}(t) \right\rangle \, dt \\ &= \int_a^b \left\langle (\mathbf{F}\circ \Gamma(t))\bigg|(J\psi)_{\gamma(t)}\cdot \frac{d\gamma}{dt}(t) \right\rangle \, dt \end{align}$

where, Jψ stands for the Jacobian matrix of ψ, and the clear circle denotes function composition. Hence,^{[note 3]}

$\begin{align} \left\langle (\mathbf{F}\circ \Gamma(t))\bigg|(J\psi)_{\gamma(t)}\frac{d\gamma}{dt}(t) \right\rangle &= \left\langle (\mathbf{F}\circ \Gamma (t))\bigg|(J\psi)_{\gamma(t)} \bigg|\frac{d\gamma}{dt}(t) \right\rangle \\ &= \left\langle ({}^{t}\mathbf{F}\circ \Gamma (t))\cdot(J\psi)_{\gamma(t)}\ \bigg|\ \frac{d\gamma}{dt}(t) \right\rangle \\ &= \left\langle \left( \left\langle (\mathbf{F}(\psi(\gamma(t))))\bigg|\frac{\partial\psi}{\partial u}(\gamma(t)) \right\rangle , \left\langle (\mathbf{F}(\psi(\gamma(t))))\bigg |\frac{\partial\psi}{\partial v}(\gamma(t)) \right\rangle \right) \bigg|\frac{d\gamma}{dt}(t)\right\rangle \\ &= \left\langle (P_1(u,v) , P_2(u,v))\bigg|\frac{d\gamma}{dt}(t)\right\rangle\\ &= \left\langle \mathbf{P}(u,v) \bigg|\frac{d\gamma}{dt}(t)\right\rangle \end{align}$

So, we obtain the following equation

$\oint_{\Gamma} \mathbf{F} \cdot d\mathbf{\Gamma} = \oint_{\gamma} \mathbf{P} d\gamma$

Third step of the proof (second equation)

First, calculate the partial derivatives, using the Leibniz rule (product rule):

$\begin{align} \frac{\partial P_1}{\partial v} &= \left\langle \frac{\partial (\mathbf{F}\circ \psi)}{\partial v} \bigg | \frac{\partial \psi}{\partial u} \right\rangle + \left\langle \mathbf{F}\circ \psi \bigg | \frac{\partial^2 \psi}{ \partial v \partial u} \right\rangle \\ \frac{\partial P_2}{\partial u} &= \left\langle \frac{\partial (\mathbf{F}\circ \psi)}{\partial u} \bigg | \frac{\partial \psi}{\partial v} \right\rangle + \left\langle \mathbf{F}\circ \psi \bigg | \frac{\partial^2 \psi}{\partial u \partial v} \right\rangle \end{align}$

So,^{[note 3][note 4]}

$\begin{align} \frac{\partial P_1}{\partial v} - \frac{\partial P_2}{\partial u} &= \left\langle \frac{\partial (\mathbf{F}\circ \psi)}{\partial v} \bigg| \frac{\partial \psi}{\partial u} \right\rangle - \left\langle \frac{\partial (\mathbf{F}\circ \psi)}{\partial u} \bigg| \frac{\partial \psi}{\partial v} \right\rangle \\ &= \left\langle (J\mathbf{F})_{\psi(u,v)}\cdot \frac{\partial \psi}{\partial v} \bigg |\frac{\partial \psi}{\partial u} \right\rangle - \left\langle (J\mathbf{F})_{\psi(u,v)}\cdot \frac{\partial \psi}{\partial u} \bigg|\frac{\partial \psi}{\partial v} \right\rangle && \text{ chain rule}\\ &= \left\langle \frac{\partial \psi}{\partial u} \bigg|(J\mathbf{F})_{\psi(u,v)} \bigg| \frac{\partial \psi}{\partial v} \right\rangle - \left\langle \frac{\partial \psi}{\partial u} \bigg |{}^{t}(J\mathbf{F})_{\psi(u,v)} \bigg| \frac{\partial \psi}{\partial v} \right\rangle \\ &= \left\langle \frac{\partial \psi}{\partial u}\bigg |(J\mathbf{F})_{\psi(u,v)} - {}^{t}{(J\mathbf{F})}_{\psi(u,v)} \bigg| \frac{\partial \psi}{\partial v} \right\rangle \\ &= \left\langle \frac{\partial \psi}{\partial u}\bigg |\left ((J\mathbf{F})_{\psi(u,v)} - {}^t (J\mathbf{F})_{\psi(u,v)} \right )\cdot \frac{\partial \psi}{\partial v} \right\rangle \\ &= \left\langle \frac{\partial \psi}{\partial u}\bigg |(\nabla\times\mathbf{F})\times\frac{\partial \psi}{\partial v} \right\rangle && \left ( (J\mathbf{F})_{\psi(u,v)} - {}^{t} (J\mathbf{F})_{\psi(u,v)} \right ) \cdot \mathbf{x} = (\nabla\times\mathbf{F})\times \mathbf{x} \\ &=\det \left [ (\nabla\times\mathbf{F})(\psi(u,v)) \quad \frac{\partial\psi}{\partial u}(u,v) \quad \frac{\partial\psi}{\partial v}(u,v) \right ] && \text{ Scalar Triple Product} \end{align}$

On the other hand, according to the definition of a surface integral,

$\begin{align} \iint_S (\nabla\times\mathbf{F}) \, dS &=\iint_D \left\langle (\nabla\times\mathbf{F})(\psi(u,v)) \bigg |\frac{\partial\psi}{\partial u}(u,v)\times \frac{\partial\psi}{\partial v}(u,v)\right\rangle \,du\,dv\\ &= \iint_D \det \left [ (\nabla\times\mathbf{F})(\psi(u,v)) \quad \frac{\partial\psi}{\partial u}(u,v) \quad \frac{\partial\psi}{\partial v}(u,v) \right ] \,du \,dv && \text{ scalar triple product} \end{align}$

So, we obtain

$\iint_S (\nabla\times\mathbf{F}) \,dS =\iint_D \left( \frac{\partial P_2}{\partial u} - \frac{\partial P_1}{\partial v} \right) \,du\,dv$

Fourth step of the proof (reduction to Green's theorem)

Combining the second and third steps, and then applying Green's theorem completes the proof.

Application for conservative vector fields and scalar potential

In this section, we will discuss the lamellar vector field based on Kelvin–Stokes theorem.

First, we define the notarization map as follows.

$\begin{cases} \theta_{[a,b]}:[0,1]\to[a,b] \\ \theta_{[a,b]}=s(b-a)+a \end{cases}$

$\theta_{[a,b]}$ is a strictly increasing function. For all piece-wise smooth paths c: [a, b] → R³ and all smooth vector fields F, the domain of which includes c([a, b]), one has:

$\int_c \mathbf{F} dc =\int_{c\circ\theta_{[a,b]}}\ \mathbf{F}\ d(c\circ\theta_{[a,b]})$

So, we can assume the domain of the curve to be [0, 1].

The Lamellar vector field

Definition 2-1 (Lamellar vector field). A smooth vector field, F on an open U ⊆ R³ is called a Lamellar vector field if ∇ × F = 0.

In fluid dynamics, it is often referred to as a vortex-free or irrotational vector field. Furthermore, if the domain of F is simply connected, then in mechanics, it can be identified as a conservative force.

Helmholtz's theorems

In this section, we will introduce a theorem that is derived from the Kelvin–Stokes theorem and characterizes vortex-free vector fields. In fluid dynamics it is called Helmholtz's theorems.^{[note 5]}

That theorem is also important in the area of Homotopy theorem.^[7]

Theorem 2-1 (Helmholtz's Theorem in Fluid Dynamics).^[7] and see p142 of Fujimoto^[5]

Let U ⊆ R³ be an open subset with a Lamellar vector field F, and piecewise smooth loops c₀, c₁: [0, 1] → U. If there is a function H: [0, 1] × [0, 1] → U such that

• [TLH0] H is piecewise smooth,
• [TLH1] H(t, 0) = c₀(t) for all t ∈ [0, 1],
• [TLH2] H(t, 1) = c₁(t) for all t ∈ [0, 1],
• [TLH3] H(0, s) = H(1, s) for all s ∈ [0, 1].

Then,

$\int_{c_0} \mathbf{F} \, dc_0=\int_{c_1} \mathbf{F} \, dc_1$

Some textbooks such as Lawrence^[7] call the relationship between c₀ and c₁ stated in Theorem 2-1 as “homotope” and the function H: [0, 1] × [0, 1] → U as “homotopy between c₀ and c₁”. However, “homotope” or “homotopy” in above-mentioned sense are different (stronger than) typical definitions of “homotope” or “homotopy”.^{[note 6]} So from now on we refer to homotopy (homotope) in the sense of Theorem 2-1 as tube-like-homotopy (homotope).^{[note 7]}

Proof of the theorem

The definitions of γ₁, ..., γ₄

Hereinafter, the ⊕ stands for joining paths^{[note 8]} the $\ominus$ stands for backwards of curve^{[note 9]}

Let D = [0, 1] × [0, 1]. By our assumption, c₁ and c₂ are piecewise smooth homotopic, there are the piecewise smooth homogony H: D → M

$\begin{align} &\begin{cases}\gamma_1:[0,1] \to D \\ \gamma_1(t) := (t, 0) \end{cases} \\ &\begin{cases}\gamma_2:[0,1] \to D \\ \gamma_2(s) := (1, s) \end{cases} \\ &\begin{cases}\gamma_3:[0,1] \to D \\ \gamma_3(t) := (-t+0+1, 1)\end{cases} \\ &\begin{cases}\gamma_4:[0,1] \to D \\ \gamma_4(s) := (0, 1-s) \end{cases} \\ [6pt] \gamma(t) &:= (\gamma_1 \oplus \gamma_2 \oplus \gamma_3 \oplus \gamma_4)(t) \\ \Gamma_i(t) &:= H(\gamma_{i}(t)) && i=1, 2, 3, 4 \\ \Gamma(t) &:= H(\gamma(t)) =(\Gamma_1 \oplus \Gamma_2 \oplus \Gamma_3 \oplus \Gamma_4)(t) \\ \end{align}$

And, let S be the image of D under H. Then,

$\oint_{\Gamma} \mathbf{F}\, d\Gamma = \iint_S \nabla\times\mathbf{F}\, dS$

will be obvious according to the Theorem 1 and, F is Lamellar vector field that, right side of that equation is zero, so,

$\oint_{\Gamma} \mathbf{F}\, d\Gamma =0$

Here,

$\oint_{\Gamma} \mathbf{F}\, d\Gamma =\sum_{i=1}^{4} \oint_{\Gamma_i} \mathbf{F} d\Gamma$ ^{[note 8]}

and, H is Tubeler-Homotopy that,

$\Gamma_{2}(s)= {\Gamma}_{4}(1-s)=\ominus{\Gamma}_{4}(s)$

that, line integral along Γ₂(s) and line integral along Γ₄(s) are compensated each other^{[note 9]} so,

$\oint_{{\Gamma}_{1}} \mathbf{F} d\Gamma +\oint_{\Gamma_3} \mathbf{F} d\Gamma =0$

On the other hand,

$c_{1}(t)=H(t,0)=H({\gamma}_{1}(t))={\Gamma}_{1}(t)$

$c_{2}(t)=H(t,1)=H(\ominus{\gamma}_{3}(t))=\ominus{\Gamma}_{3}(t)$

that, subjected equation is proved.

Application for conservative force

Helmholtz's theorem, gives an explanation as to why the work done by a conservative force in changing an object's position is path independent. First, we introduce the Lemma 2-2, which is a corollary of and a special case of Helmholtz's theorem.

Lemma 2-2.^[7][8] Let U ⊆ R³ be an open subset, with a Lamellar vector field F and a piecewise smooth loop c₀: [0, 1] → U. Fix a point p ∈ U, if there is a homotopy (tube-like-homotopy) H: [0, 1] × [0, 1] → U such that

• [SC0] H is piecewise smooth,
• [SC1] H(t, 0) = c₀(t) for all t ∈ [0, 1],
• [SC2] H(t, 1) = p for all t ∈ [0, 1],
• [SC3] H(0, s) = H(1, s) = p for all s ∈ [0, 1].

Then,

$\int_{c_0} \mathbf{F} \, dc_0=0$

Lemma 2-2, obviously follows from Theorem 2-1. In Lemma 2-2, the existence of H satisfying [SC0] to [SC3] is crucial. It is a well-known fact that, if U is simply connected, such H exists. The definition of Simply connected space follows:

Definition 2-2 (Simply Connected Space).^[7][8] Let M ⊆ Rⁿ be non-empty, connected and path-connected. M is called simply connected if and only if for any continuous loop, c: [0, 1] → M there exists H: [0, 1] × [0, 1] → M such that

• [SC0'] H is continuous,
• [SC1] H(t, 0) = c(t) for all t ∈ [0, 1],
• [SC2] H(t, 1) = p for all t ∈ [0, 1],
• [SC3] H(0, s) = H(1, s) = p for all s ∈ [0, 1].

You will find that, the [SC1] to [SC3] of both Lemma 2-2 and Definition 2-2 is same.

So, someone may think that, "for a conservative force, the work done in changing an object's position is path independent" is elucidated. However, there are very large gaps between following two:

• There are continuous H such that it satisfies [SC1] to [SC3]
• There are piecewise smooth H such that it satisfies [SC1] to [SC3]

To fill that gap, the deep knowledge of Homotopy Theorem is required. For example, the following resources may be helpful for you.

• Lee teaches Whitney approximation theorem (^[8] page 136) and "How to use that theorem to this isuue" (^[8] page 421).
• More general statements appear in^[9] (see Theorems 7 and 8).

Considering above-mentioned fact and Lemma 2-2, we will obtain following theorem.

Theorem 2-2.^[7][8] Let U ⊆ R³ be a simply connected and open with a Lamellar vector field F. For all piecewise smooth loops, c: [0, 1] → U we have:

$\int_{c_0} \mathbf{F} \, dc_0=0$

Kelvin–Stokes theorem on singular 2-cube and cube subdivisionable sphere

Singular 2-cube and boundary

Definition 3-1 (Singular 2-cube)^[10] Set D = [a₁, b₁] × [a₂, b₂] ⊆ R² and let U be a non-empty open subset of R³. The image of D under a piecewise smooth map ψ: D → U is called a singular 2-cube. Moreover, we define the notarization map of D

$\begin{cases} \theta_D: I \times I \to D \\ \theta_D({u}_{1},{u}_{2})= \begin{pmatrix} {u}_{1}(b_1 - a_1) + {a}_{1}\\ {u}_{2}(b_2 - a_2) +{a}_{2} \end{pmatrix} \end{cases}$

where I = [0, 1]. Then θ_D has the following property:

$\forall (u_1, u_2)\in\mathbf{R}^3 \qquad \det(J(\theta_{D})_{(u_1, u_2)})>0.$

Lemma 3-1 (Notarization map of singular two cube). Let D be a singular 2-cube with map ψ and U ⊆ R³ open and non-empty. Suppose the image of I × I under a piecewise smooth map $\varphi\circ \theta_D, \widetilde{S}:=\varphi\circ \theta_D (I \times I)$ be a singular 2-cube. If F is a smooth vector field on U we have:

$\int_S \mathbf{F} \, dS =\int_{\widetilde{S}} \mathbf{F} \, d\widetilde{S}$

We omit the proof of the lemma. Using the lemma from now we consider all singular 2-cubes to be notarized. In other words, we assume that the domain of all singular 2-cubes is I × I.

In order to facilitate the discussion of boundary, we define

$\begin{cases} \delta_{[k,j,c]}:\mathbf{R}^k \to \mathbf{R}^{k+1} \\ \delta_{[k,j,c]}(t_1,\cdots,t_k):= \left (t_1,\cdots,t_{j-1},c,t_{j+1},\cdots,t_k \right )\end{cases}$

γ₁, ..., γ₄ are the one-dimensional edges of the image of I × I.Hereinafter, the ⊕ stands for joining paths^{[note 8]} and, the $\ominus$ stands for backwards of curve.^{[note 9]}

$\begin{align} &\begin{cases} \gamma_{1}:[0, 1] \to I^2 \\ \gamma_{1}(t) := {\delta}_{[1,2,0]}(t)=(t,0) \end{cases} \\ &\begin{cases} \gamma_{2}:[0, 1] \to I^2 \\ \gamma_{2}(t) :={\delta}_{[1,1,1]}(t)=(1, t) \end{cases} \\ &\begin{cases} \gamma_{3}:[0, 1] \to I^2 \\ \gamma_{3}(t) := \ominus{\delta}_{[1,2,1]}(t)= (-t+0+1, 1)\end{cases} \\ &\begin{cases} \gamma_{4}:[0, 1] \to I^2 \\ \gamma_{4}(t) := \ominus{\delta}_{[1,1,0]}(t) =(0, 1-t)\end{cases} \\ [6pt] \gamma(t) &:= (\gamma_{1} \oplus \gamma_{2} \oplus \gamma_{3} \oplus \gamma_{4})(t) \\ \Gamma_{i}(t) &:= \varphi(\gamma_{i}(t)) && i=1, 2, 3, 4 \\ \Gamma(t) &:= \varphi(\gamma(t)) =(\Gamma_{1} \oplus \Gamma_{2} \oplus \Gamma_{3} \oplus \Gamma_{4})(t) \end{align}$

Cube subdivision

Definition 3-2 (Cube subdivisionable sphere). (see Iwahori^[4] p. 399) A non-empty subset S ⊆ R³ is said to be a "Cube subdivisionable sphere" when there are at least one Indexed family of singular 2-cube

$\left \{(I^2, \varphi_\lambda, S_\lambda) \right \}_{\lambda\in\Lambda}$

such that

• Λ is a finite set.
• [CSS0] For all λ ∈ Λ,
  • φ_λ is an Injective function on I × I and,
  • for almost all (u₁, u₂) ∈ Int(I × I),

$\det((J\varphi)_{(u_1, u_2)})\neq 0$

• [CSS1] S is equal to the union of S_λ as λ varies in Λ.
• [CSS2] λ₁ ≠ λ₂ implies $\varphi_{\lambda_1}(Int(I^2))\cap \varphi_{\lambda_2}(Int(I^2)) =\varnothing$
• [CSS3] If c₁, c₂ = 0 or 1, j₁, j₂ = 1 or 2 and,

$\varphi_{\lambda_1}\circ\delta_{[1, j_1,c_1]}(I) \cap \varphi_{\lambda_2} \circ\delta_{[1, j_2, c_2 ]}(I)\neq \varnothing$

then

$\varphi_{\lambda_1}\circ\delta_{[1, j_1, c_1]}(I) = \varphi_{\lambda_2}\circ\delta_{[1, j_2, c_2]}(I)$

Then

$\left \{(I^2,\varphi_\lambda, S_{\lambda}) \right\}_{\lambda\in\Lambda}$

is called a Cube subdivision of the S.

Definitions 3-3 (Boundary of a Cube Subdivision Sphere). (see Iwahori^[4] p. 399) Let S ⊆ R³ be a cube subdivisionable sphere with cube subdivision:

$\left \{(I^2, \varphi_\lambda, S_\lambda) \right \}_{\lambda\in\Lambda}.$

(1) The $\varphi_{\lambda_1}\circ\delta_{[1, j_1, c_1]}$ are said to be an edge of the cube subdivision if

$\varphi_{\lambda_1}\circ\delta_{[1, j_1, c_1]}(I) = \varphi_{\lambda_2}\circ\delta_{[1, j_2, c_2]}(I)$

implies $(\lambda_1, j_1, c_1)=(\lambda_2, j_2, c_2)$. That means, l is said to be an edge of $\{(I^2, \varphi_\lambda, S_\lambda)\}_{\lambda\in\Lambda}$ if and only if there are unique λ, c and j such that, $l=\varphi_\lambda\circ\delta_{[1, j, c]}$.

(2) The boundary of the cube subdivision is a collection of edges in the above sense. $\partial \{(I^2, \varphi_\lambda, S_\lambda)\}_{\lambda\in\Lambda}$ means the boundary of $\{(I^2, \varphi_\lambda, S_\lambda)\}_{\lambda\in\Lambda}$

(3) The notation for an edge l is as follows:

$l \prec \partial \left \{( I^2, \varphi_\lambda, S_\lambda) \right \}_{\lambda\in\Lambda}$

The definition of the boundary of the Definitions 3-3 is apparently depends on the cube subdivision. However, considering the following fact, the boundary is not depends on the cube subdivision.

Lemma 3-4. (see Iwahori^[4] p. 399) Let S ⊆ R³ be a cube subdivisionable sphere with cube subdivisions:

$\left \{(I^2, \varphi_\lambda, S_\lambda) \right \}_{\lambda\in\Lambda}, \quad \left \{(I^2, \psi_\mu, L_\mu) \right \}_{\mu\in M}.$

Then

$\partial \left \{(I^2, \varphi_\lambda, S_\lambda) \right \}_{\lambda\in\Lambda}=\partial \left \{(I^2, \psi_\mu, L_\mu) \right \}_{\mu\in M}$

therefore the definition of boundary is not dependent on the choice of cube subdivision.

Therefore, the following Definition is well-defined:

Definitions 3-5 (Boundary of Surface). (see Iwahori^[4] p399) Let S ⊆ R³ be a cube subdivisionable sphere and, $\{(I^2, \varphi_\lambda, S_\lambda)\}_{\lambda\in\Lambda}$, then

(1)

$\partial S:= \partial \left \{(I^2, \varphi_\lambda, S_\lambda) \right \}_{\lambda\in\Lambda}$

(2) If

$l \prec \partial \left \{(I^2, \varphi_\lambda, S_\lambda) \right \}_{\lambda\in\Lambda}$

then

$l \prec \partial S$

and then such l are said to be an edge of S.

Notes

1. γ and Γ are both loops, however, Γ is not necessarily a Jordan curve
2. Knowledge of differential forms and identification of the vector field A = (a₁, a₂, a₃),

   $\mathbf{A} = \omega_{\mathbf{A}} = a_1 dx_1+a_2 dx_2+a_3 dx_3$

   $\mathbf{A} = {}^* \omega_{\mathbf{A}} = a_1 dx_2 \wedge dx_3+a_2dx_3 \wedge dx_1+a_3dx_1 \wedge dx_2$

   admits a proof similar to the proof using the pullback of ω_F. Under the identification ω_F = F the following equations are satisfied.

   $\begin{align} \nabla\times\mathbf{F} &= d\omega_{\mathbf{F}} \\ \psi^{*}\omega_{\mathbf{F}} &= P_1du +P_2dv \\ \psi^{*}(d \omega_{\mathbf{F}}) &= \left (\frac{\partial P_2}{\partial u} - \frac{\partial P_1}{\partial v} \right ) du\wedge dv \end{align}$

   Here, d stands for exterior derivative of the differential form, ψ^∗ stands for pull back by ψ and, P₁ and P₂ are same as P₁ and P₂ of the body text of this article respectively.
3. Given a n × m matrix A we define a bilinear form:

   $\mathbf{x} \in \mathbf{R}^m, \mathbf{y} \in \mathbf{R}^n \ : \qquad \left\langle \mathbf{x} |A| \mathbf{y} \right\rangle ={}^{t}\mathbf{x}A\mathbf{y}$

   which also satisfies (^tA is the transpose of the matrix A):

   $\left \langle \mathbf{x} |A| \mathbf{y} \right \rangle= \left \langle \mathbf{y} \left |{}^t A \right| \mathbf{x} \right \rangle.$

   $\left \langle \mathbf{x} |A| \mathbf{y} \right \rangle + \left \langle \mathbf{x} |B| \mathbf{y}\right \rangle = \left \langle \mathbf{x} |A+B|\mathbf{y}\right \rangle$

4. We prove following (★0).

   $({(J\mathbf{F})}_{\psi(u,v)} - {}^{t}{(J\mathbf{F})}_{\psi(u,v)}) \mathbf{x} =(\nabla\times\mathbf{F})\times \mathbf{x}, \quad \text{for all}\, \mathbf{x}\in\mathbf{R}^{3}$　(★0)

   First, we note the linearity of the algebraic operator a × and we obtain its matrix representation (see linear map). Let both of a and x be 3-dimensional column vectors, represented as follows,

   $\mathbf{a}= \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix}, \quad \mathbf{x}= \begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}$

   Then, according to the definition of the cross product, a × x are represented as follows.

   $\mathbf{a}\times\mathbf{x}= \begin{pmatrix} a_2 x_3-a_3 x_2\\ a_3 x_1-a_1 x_3\\ a_1 x_2-a_2 x_1\end{pmatrix}$

   Therefore, when we apply the operator a ×　to each of the standard basis, we obtain following, and thus, the matrix representation of a × as shown in (★1).

   $\mathbf{a}\times\mathbf{e}_1 = \begin{pmatrix} 0 \\ a_3\\ -a_2 \end{pmatrix}, \quad \mathbf{a}\times \mathbf{e}_{2}= \begin{pmatrix} -a_3\\ 0\\ a_1 \end{pmatrix}, \quad \mathbf{a}\times\mathbf{e}_{3}= \begin{pmatrix} a_2\\ -a_1\\ 0 \end{pmatrix}$　

   $\mathbf{a}\times\mathbf{x}= \begin{pmatrix} 0&-a_3&a_2\\ a_3&0&-a_1\\ -a_2&a_1&0 \end{pmatrix} \begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}$　(★1)

   Next, let $A=(a_{ij})$ is a 3 × 3 matrix and using the following substitution, (a₁, ... , a₃ are components of a),

   ${a}_{1}={a}_{32}-{a}_{23}$ (★2-1)

   ${a}_{2}={a}_{13}-{a}_{31}$ (★2-2)

   ${a}_{3}={a}_{21}-{a}_{12}$ (★2-3)

   Then we obtain following (★3) from the (★1).

   $(A-{}^t A)\mathbf{x}=\mathbf{a}\times\mathbf{x}$　(★3)

   We substitute the JF to above mentioned A, under the substitution of (★2-1), (★2-2), and (★2-3), we obtain the following (★4)

   $\mathbf{a}=(\nabla\times\mathbf{F})$　(★4)

   The (★0) is obvious from (★3) and (★4).
5. There are a number of theorems with the same name, however they are not necessarily the same.
6. Typical definition of homotopy and homotope are as follows.
   Definition (Homotopy and Homotope). Suppose Z and W are topological spaces, with continuous maps f₀, f₁: Z → W. (1) The continuous map H: Z × [0, 1] → W is said to be a "Homotopy between f₀ and f₁" if

   • [H1] H(t, 0) = f₀(t) for all t ∈ Z,
   • [H2] H(t, 1) = f₁(t) for all t ∈ Z.

   (2) If there is a homotopy between f₀ and f₁", f₀ and f₁" are said to be homotope. (3) Suppose f₀ and f₁ are homotope and H is a homotopy between them. f₀ and f₁ are said to be piecewise homotope, if f₀, f₁, and H are piecewise smooth. H is then said to be the piecewise homotopy between f₀ and f₁.
7. In some textbooks such as Conlon, Lawrence (2008). Differentiable Manifolds. Modern Birkhauser Classics. Boston: Birkhaeuser. use the term of homotopy and homotope in Theorem 2-1 sense. homotopy and homotope in Theorem 2-1 sense Indeed, it is convenience to adopt such sense to discuss conservative force. However, homotopy in Theorem 2-1 sense and homotope in Theorem 2-1 sense are different from and stronger than homotopy in typical sense and homotope in typical sense. So there are no appropriate terminology which can discriminate between homotopy in typical sense and sense of Theorem 2-1. In this article, to avoid ambiguity and to discriminate between them, we will define two “just-in-time term”, tube-like homotopy and tube-like homotope as follows.
   Definition (tube-like homotopy and tube-homotope). Suppose c₀, c₁ satisfy the following:

   • [A] M is differentiable manifold,
   • [B] The domain of c₀: [0, 1] → M and c₁ : [0, 1] → M are the same,
   • [C] Both c₀, and c₁ are continuous curves.

   Then, (1) Tube-Like-Homotopy: A homotopy H: [0, 1] × [0, 1] → M is "Tube-Like", if

   • [TLH0] H is continues
   • [TLH1] H(t, 0) = c₀(t)
   • [TLH2] H(t, 1) = c₁(t)
   • [TLH3] H(0, s) = H(1, s) for all s ∈ [0, 1]

   (2) Tube homotope: c₀, and c₁ are "Tube Homotope" if and only if "there are H such that there is a Tube-like-Homotopy between c₀ and c₁. (3) Tube like and piecewise smooth homotopy: The homotopy H of (1) is Tube like and piecewise smooth homotopy when that H is piecewise smooth. And the relation of (1) is “Piecewise smooth Tube Homotope” when that H is piecewise smooth (so, it is “Piecewise smooth Tube Homotope”).
8. If the two curves α: [a₁, b₁] → M, β: [a₂, b₂] → M, satisfy α(b₁) = β(a₂) then, we can define new curve α ⊕ β so that, for all smooth vector field F (if domain of which includes image of α ⊕ β)

   $\int_{\alpha\oplus \beta} \mathbf{F} \, d(\alpha\oplus \beta)= \int_\alpha \mathbf{F} \, d\alpha + \int_\beta \mathbf{F}\, d\beta$

   which is also used when we define the fundamental group. To do so, accurate definition of the “joint of paths” is as follows.
   Definition (Joint of paths). Let M be a topological space and α: [a₁, b₁] → M, β: [a₂, b₂] → M, be two paths on M. If α and β satisfy α(b₁) = β(a₂) then we can join them at this common point to produce new curve α ⊕ β : [a₁, b₁+(b₂-a₂)] → M defined by:

   $(\alpha\oplus \beta)(t) =\begin{cases} \alpha(t) & a_1 \le t \le \ b_1, \\ \beta(t+(a_2-b_1)) & b_1 < t \le b_1+(b_2-a_2)\end{cases}$

9. Given curve on M, α: [a₁, b₁] → M, we can define new curve $\ominus$α so that, for all smooth vector field F (if domain of which includes image of α)

   $\int_{\ominus\alpha} \mathbf{F} \, d(\ominus\alpha)= -\int_{\alpha}\mathbf{F} \, d\alpha,$

   which is also used when we define fundamental group. To do so, accurate definition of the “backwards of curve” is as follows.
   Definition (Backward of curve). Let M be a topological space and α : [a₁, b₁] → M, be path on M. We can define backward thereof, $\ominus$α : [a₁, b₁] → M by:

   $\ominus\alpha(t)=\alpha(b_1+a_1-t)$

   And, given two curves on M, α: [a₁, b₁] → M, β: [a₂, b₂] → M, which satisfy α(b₁ = β(b₂) (that means α(b₁) = $\ominus$ β(a₂), we can define $\alpha\ominus\beta$ as following manner.

   $\alpha\ominus\beta:=\alpha\oplus(\ominus\beta)$

References

1. Stewart, James (2010). Essential Calculus: Early Transcendentals. Cole.
2. This proof is based on the Lecture Notes given by Prof. Robert Scheichl (University of Bath, U.K) , please refer the
3. This proof is also same to the proof shown in
4. Nagayoshi Iwahori, et.al:"Bi-Bun-Seki-Bun-Gaku" Sho-Ka-Bou(jp) 1983/12 ISBN 978-4-7853-1039-4 (Written in Japanese)
5. Atsuo Fujimoto;"Vector-Kai-Seki Gendai su-gaku rekucha zu. C(1)" Bai-Fu-Kan(jp)(1979/01) ISBN 978-4563004415 (Written in Japanese)
6. http://mathworld.wolfram.com/CurlTheorem.html
7. Lawrence Conlon; "Differentiable Manifolds (Modern Birkhauser Classics)" Birkhaeuser Boston (2008/1/11)
8. John M. Lee; "Introduction to Smooth Manifolds (Graduate Texts in Mathematics, 218) "Springer (2002/9/23)
9. L. S. Pontryagin, Smooth manifolds and their applications in homotopy theory, American Mathematical Society Translations, Ser. 2, Vol. 11, American Mathematical Society, Providence, R.I., 1959, pp. 1–114. MR 0115178 (22 #5980 )
10. Spivak, Michael (1971). Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus. Westview Press.