Kelvin–Stokes theorem

The Kelvin–Stokes theorem, also known as the curl theorem, 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.” In particular, the vector field on R³ can be considered as a 1-form in which case curl is the exterior derivative.

The Theorem

Kelvin–Stokes Theorem. Let γ : [a, b] → R² be a Piecewise smooth Jordan plane curve, that bounds the domain D ⊂ R². Suppose ψ : D → R³ is smooth, with S := ψ[D] , and Γ is the space curve defined by Γ(t) = ψ(γ(t)). If F a smooth vector field on R³, then

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

Proof

The proof of the Theorem consists of 4 steps The proof below does not require background information on differential form, and may be helpful for understanding the notion of differential form, especially pull-back of differential form.

First Step of Proof (Defining the Pullback)

Define

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

so that P is the pull-back of F, and that P(u, v) is R²-valued function, depends on two parameter 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 of R³ and hereinafter, $\langle \ |A|\ \rangle$ stands for bilinear form according to matrix A .

Second Step of Proof (First Equation)

According to the definition of line integral,

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

where, Jψ stands for the Jacobian matrix of ψ. Hence,

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

So, we obtain following equation

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

Third Step of Proof (Second Equation)

First, calculate the partial derivatives, using Leibniz rule of inner product

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

So,

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

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

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

So, we obtain

$\iint_S (\nabla\times\mathbf{F}) dS =\iint_{D} \left( \frac{\partial P_2}{\partial u} - \frac{\partial P_1}{\partial v} \right) dudv$

Fourth Step of Proof (Reduction to Green's Theorem)

According to the result of Second step, and according to the result of Third step, and further considering the Green's theorem, subjected equation is proved.

Application for Conservative force and Scalar potential

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

First, we define the notarization map, ${\theta}_{[a,b]}:[0,1]\to[a,b]$ as follows.

${\theta}_{[a,b]}=s(b-a)+a$

Above mentioned ${\theta}_{[a,b]}$is strongly increase function that, for all piece wise sooth path c:[a,b]→R³, for all smooth vector field F, domain of which includes $c[[a,b]]$ (image of [a,b] under c.), following equation is satisfied.

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

So, we can unify the domain of the curve from the beginning to [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 mechanics a lamellar vector field is called a Conservative force; in Fluid dynamics, it is called a Vortex-free vector field. So, lamellar vector field, conservative force, and vortex-free vector field are the same notion.

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,.

That theorem is also important in the area of Homoropy thorem.

Theorem 2-1 (Helmholtz's Theorem in Fluid Dynamics). and see p142 of Fujimoto

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,

$\int_{c_0} \mathbf{F} d{c}_{0}=\int_{c_1} \mathbf{F} dc_{1}$

Some textbooks such as Lawrence 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 toward (stronger than) typical definitions of “Homotope” or “Homotopy”.

So there are no appropriate terminology which can discriminate between homotopy in typical sense and sense of Theorem 2-1. So, in this article, to discriminate between them, we say “Theorem 2-1 sense homotopy as Tube-like-Homotopy and, we say “Theorem 2-1 sense Homotope” as Tube-like-Homotope.

Proof of the Theorem

The definitions of γ₁, ..., γ₄

Hereinafter, the ⊕ stands for joining paths the $\ominus$ stands for backwards of curve

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

$\begin{align} \begin{cases} \gamma_{1}:[0, 1]\to D \\ \gamma_{1}(t) := (t,0) \end{cases}, \qquad &\begin{cases}\gamma_{2}:[0,1] \to D \\ \gamma_{2}(s) := (1, s) \end{cases} \\ \begin{cases} \gamma_{3}:[0, 1] \to D \\ \gamma_{3}(t) := (-t+0+1, 1)\end{cases}, \qquad &\begin{cases}\gamma_{4}:[0,1] \to D \\ \gamma_{4}(s) := (0, 1-s)\end{cases} \end{align}$

$\gamma(t):= (\gamma_{1} \oplus \gamma_{2} \oplus \gamma_{3} \oplus \gamma_{4})(t)$

$\Gamma_{i}(t):= H(\gamma_{i}(t)), \qquad i=1, 2, 3, 4$

$\Gamma(t):= H(\gamma(t)) =(\Gamma_{1} \oplus \Gamma_{2} \oplus \Gamma_{3} \oplus \Gamma_{4})(t)$

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,

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

and, H is Tubeler-Homotopy that ,

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

that, line integral along $\Gamma_{2}(s)$ and line integral along $\Gamma_{4}(s)$ are compensated each other 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. 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). 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 contenious,
• [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, the issue, "when the Conservative Force, the work done in changing an object's position is path independent" is elucidated. However there are very large gap 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, to fill the gap, following resources may be helpful for you.

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

Considering above mentioned fact and Lemma 2-2, we will obtain following theorem. That theorem is anser for subjecting issue.

Theorem 2-2. 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) 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.

Given $D:=[{a}_{1},{b}_{1}]\times[{a}_{2},{b}_{2}]$, we define the notarization map of sngler two cube ${\theta}_{D}:{I}^{2}\to D$

${\theta}_{D}({u}_{1},{u}_{2})= \left( \begin{array}{c} {u}_{1}({b}_{1}-{a}_{1}) +{a}_{1}\\ {u}_{2}({b}_{2}-{a}_{2}) +{a}_{2} \end{array} \right)$

here, the I:=[0,1] and I² stands for ${I}^{2}=I\times I$.

Above mentioned is strongly increase function (that means $det(J(\theta_{D})_{({u}_{1},{u}_{2})})>0$ (for all $({u}_{1},{u}_{2})\in\mathbb{R}^{3}$) that, following lemma is satisfied.

Lemma 3-1(Notarization map of sngler two cube). Set D = [a₁, b₁]× [a₂, b₂] ⊆ R² and let U be a non-empty open subset of R³. Let the image of D under a piecewise smooth map ψ : D → U ,S:= ψ[D] be a singular 2-cube. Let the image of I² under a piecewise smooth map $\varphi\circ{\theta}_{D}$ , $\tilde{S}:=\varphi\circ{\theta}_{D}[{I}^{2}]$ be a singular 2-cube.then, For all $\mathbf{F}$,smooth vector field on U,

${\int}_{S} \mathbf{F} dS ={\int}_{\tilde{S}} \mathbf{F} d\tilde{S}$

Above mentioned lemma is obverse that, we neglects the proof. Acceding to the above mentioned lemma, hereinafter, we consider that, domain of all singular 2-cube are notarized (that means, hereinafter, we consider that domain of all singular 2-cube are from the beginning, I².

In order to facilitate the discussion of boundary, we define ${\delta}_{[k,j,c]}:\mathbb{R}^{k}\to \mathbb{R}^{k+1}$ by

${\delta}_{[k,j,c]}({t}_{1},\cdots,{t}_{k}):= ({t}_{1},\cdots,{t}_{j-1},c,{t}_{j+1},\cdots ,{t}_{k})$

γ₁, ..., γ₄ are the one-dimensional edges of the image of I².Hereinafter, the ⊕ stands for joining paths and,　 the $\ominus$ stands for backwards of curve .

$\begin{align} \begin{cases} \gamma_{1}:[0, 1]\to {I}^{2} \\ \gamma_{1}(t) := {\delta}_{[1,2,0]}(t)=(t,0) \end{cases}, \qquad &\begin{cases}\gamma_{2}:[0,1] \to { I }^{2} \\ \gamma_{2}(t) :={\delta}_{[1,1,1]}(t)=(1, t) \end{cases} \\ \begin{cases} \gamma_{3}:[0, 1] \to {I}^{2} \\ \gamma_{3}(t) := \ominus{\delta}_{[1,2,1]}(t)= (-t+0+1, 1)\end{cases}, \qquad &\begin{cases}\gamma_{4}:[0,1] \to {I}^{2} \\ \gamma_{4}(t) := \ominus{\delta}_{[1,1,0]}(t) =(0, 1-t)\end{cases} \end{align}$

$\gamma(t):= (\gamma_{1} \oplus \gamma_{2} \oplus \gamma_{3} \oplus \gamma_{4})(t)$

$\Gamma_{i}(t):= \varphi(\gamma_{i}(t)), \qquad i=1, 2, 3, 4$

$\Gamma(t):= \varphi(\gamma(t)) =(\Gamma_{1} \oplus \Gamma_{2} \oplus \Gamma_{3} \oplus \Gamma_{4})(t)$

Cube subdivision

Definition 3-2(Cube subdivisionable sphere).(see Iwahori p399) Let S ⊆ R³ be a non empty subset then, that S is said to be a "Cube subdivisionable sphere" when there are at least one Indexed family of singular 2-cube $\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$ such that

• [CSS0] For all $\lambda\in\Lambda$, $({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})$ are Legular that means,
  • $\Lambda$ is finite set.
  • ${\varphi}_{\lambda}$ are Injective function on ${I}^{2}$ and,
  • for almost all $({u}_{1},{u}_{2})\in Int({I}^{2})$, $det((J\varphi)_{({u}_{1},{u}_{2})})\neq 0$
• [CSS1]$S = \bigcup_{\lambda \in \Lambda} S_{\lambda}$
• [CSS2]${\lambda}_{1}\neq{\lambda}_{2}$ ⇒ ${\varphi}_{{\lambda}_{1}}[Int({I}^{2})]\cap{\varphi}_{{\lambda}_{2}}[Int({I}^{2})]=\varnothing$
• [CSS3]If ${c}_{1},{c}_{2}\ =0\ or\ 1$, ${j}_{1},{j}_{2} =1\ or\ 2$ 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,${\varphi}_{{\lambda}_{1}}\circ\delta_{[1,{j}_{1},{c}_{1}]}[I] = {\varphi}_{{\lambda}_{2}}\circ\delta_{[1,{j}_{2},{c}_{2}]}[I]$

and then abovementioned $\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$ are said to be a Cube subdivision of the S.

Definitions 3-3(Boundary of $\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$ ).(see Iwahori p399)

Let S ⊆ R³ be a "Cube subdivisionable sphere" and, Let $\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$ be a Cube subdivision of the S.,then

(1)The ${\varphi}_{{\lambda}_{1}}\circ\delta_{[1,{j}_{1},{c}_{1}]}$ are said to be an edge of $\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$ if ${\varphi}_{{\lambda}_{1}}\circ\delta_{[1,{j}_{1},{c}_{1}]}[I]$ satisfies

"${\varphi}_{{\lambda}_{1}}\circ\delta_{[1,{j}_{1},{c}_{1}]}[I]= {\varphi}_{{\lambda}_{2}}\circ\delta_{[1,{j}_{2},{c}_{2}]}[I]$then,$({\lambda}_{1},{j}_{1},{c}_{1})=({\lambda}_{2},{j}_{2},{c}_{2})$"

that means "although not line contact even if the point contact with other ridge line" and above mentioned "=" stands for equal as a set.
That means, l is said to be an edge of $\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$ iff "There is only one $\lambda$ only one c and only one j such that, $l={\varphi}_{{\lambda}}\circ\delta_{[1,{j},{c}]}$"

(2)Boundary of $\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$ is a collection of edges in the sense of "(1)". $\partial\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$ means the boundary of $\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$

(3)If l is an edge in the sense of "(1)", then, we described as follows.

$l \prec \partial\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$

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

Fact(Boundary of $\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$ ).(see Iwahori p399)

Let S ⊆ R³ be a "Cube subdivisionable sphere" and, Let both $\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$ and $\{({I}^{2},{\psi}_{\mu},{L}_{\mu})\}_{\mu\in M}$ be a Cube subdivision of the S, then

$\partial\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}=\partial\{({I}^{2},{\psi}_{\mu},{L}_{\mu})\}_{\mu\in M}$

that means, the definition of boundary is not depends on the cube subdivision.

So, considerting above mentiond fact, following "Definition3-4" is well-defined.

Definitions 3-4(Boundary of Surface(see Iwahori p399) Let S ⊆ R³ be a "Cube subdivisionable sphere" and, Let $\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$, then

(1)

$\partial S:= \partial\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$

(2)If

$l \prec \partial\{({I}^{2},{\varphi}_{\lambda},{S}_{\lambda})\}_{\lambda\in\Lambda}$

then

$l \prec \partial S$

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