6.1 Stokes' theorem

One of the main tools in the study of de Rham cohomology groups is Stokes’ theorem with which most physicists are familiar from electromagnetism. Gauss’ theorem and Stokes’ theorem are treated in a unified manner here.

6.1.1 Preliminary consideration

Let us define an integration of an $r$ -form over an $r$ -simplex in a Euclidean space. To do this, we need first to define the standard $n$ -simplex ${\bar{\sigma }}_r=(p_0{p}_1\dots p_r)$ in ${\mathbb{R}}^r$ where

$$p_0=(0,0,\dots ,0)$$

$$p_1=(1,0,\dots ,0)$$

$$\cdot \cdot \cdot$$

$$p_r=(0,0,\dots ,1)$$

see figure 6.1. If $\{x^{\mu }\}$ is a coordinate of ${\mathbb{R}}^r$ , ${\bar{\sigma }}_r$ is given by

Figure 6.1. The standard 2-simplex ${\bar{\sigma }}_2=(p_0{p}_1{p}_2)$ and the standard 3-simplex ${\bar{\sigma }}_3=(p_0{p}_1{p}_2{p}_3)$ .

$${\bar{\sigma }}_r=\left\{(x^1,\dots ,x^r)\in {\mathbb{R}}^r\mid x^{\mu }\ge 0,\sum _{\mu =1}^r{x}^{\mu }\le 1\right\}.\text{\hspace{1em}(6.1)}$$

An $r$ -form $\omega$ (the volume element) in ${\mathbb{R}}^r$ is written as

$$\omega =a(x)\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \cdots \wedge \mathrm{d}{x}^r.$$

We define the integration of $\omega$ over ${\bar{\sigma }}_r$ by

$${\int }_{{\bar{\sigma }}_r}\omega \equiv {\int }_{{\bar{\sigma }}_r}a(x)d{x}^{1}d{x}^2\dots d{x}^r\text{\hspace{1em}(6.2)}$$

where the RHS is the usual $r$ -fold integration. For example, if $r=2$ and $\omega =\mathrm{d}x\wedge \mathrm{d}y$ , we have

$${\int }_{{\bar{\sigma }}_2}\omega ={\int }_{{\bar{\sigma }}_2}\mathrm{d}x\mathrm{d}y={\int }_0^1\mathrm{d}x{\int }_0^{1-x}\mathrm{d}y=\frac{1}{2}.$$

Next we define an $r$ -chain, an $r$ -cycle and an $r$ -boundary in an $m$ -dimensional manifold $M$ . Let ${\sigma }_r$ be an $r$ -simplex in ${\mathbb{R}}^r$ and let $f:{\sigma }_r\to M$ be a smooth map. [To avoid the subtlety associated with the differentiability of $f$ at the boundary of ${\sigma }_r$ , $f$ may be defined over an open subset $U$ of ${\mathbb{R}}^r$ , which contains ${\sigma }_r$ .] Here we assume $f$ is not required to have an inverse. For example, $\mathrm{im}f$ may be a point in $M$ . We denote the image of ${\sigma }_r$ in $M$ by $s_r$ and call it a (singular) $r$ -simplex in $M$ . These simplexes are called singular since they do not provide a triangulation of $M$ and, moreover, geometrical independence of points makes no sense in a manifold (see section 3.2). If $\{s_{r,i}\}$ is the set of $r$ -simplexes in $M$ , we define an $r$ -chain in $M$ by a formal sum of $\{s_{r,i}\}$ with $\mathbb{R}$ -coefficients

$$c=\sum _i{a}_i{s}_{r,i}{a}_i\in \mathbb{R}.\text{\hspace{1em}(6.3)}$$

In the following, we are concerned with $\mathbb{R}$ -coefficients only and we omit the explicit quotation of $\mathbb{R}$ . The $r$ -chains in $M$ form the chain group $C_r(M)$ . Under $f:{\sigma }_r\to M$ , the boundary $\partial {\sigma }_r$ is also mapped to a subset of $M$ . Clearly, $\partial s_r\equiv f(\partial {\sigma }_r)$ is a set of $(r-1)$ -simplexes in $M$ and is called the boundary of

$s_r$ . $\partial s_r$ corresponds to the geometrical boundary of $s_r$ with an induced orientation defined in section 3.3. We have a map

$$\partial :C_r(M)\to C_{r-1}(M).\text{\hspace{1em}(6.4)}$$

The result of section 3.3 tells us that $\partial$ is nilpotent; ${\partial }^2=0$ .

Cycles and boundaries are defined in exactly the same way as in section 3.3 (note, however, that $\mathbb{Z}$ is replaced by $\mathbb{R}$ ). If $c_r$ is an $r$ -cycle, $\partial c_r=0$ while if $c_r$ is an $r$ -boundary, there exists an $(r+1)$ -chain $c_{r+1}$ such that $c_r=\partial c_{r+1}$ . The boundary group $B_r(M)$ is the set of $r$ -boundaries and the cycle group $Z_r(M)$ is the set of $r$ -cycles. There are infinitely many singular simplexes which make up $C_r(M),B_r(M)$ and $Z_r(M)$ . It follows from ${\partial }^2=0$ that $Z_r(M)\supset B_r(M)$ ; cf theorem 3.3. The singular homology group is defined by

$$H_r(M)\equiv Z_r(M)/B_r(M).\text{\hspace{1em}(6.5)}$$

With mild topological assumptions, the singular homology group is isomorphic to the corresponding simplicial homology group with $\mathbb{R}$ -coefficients and we employ the same symbol to denote both of them.

Now we are ready to define an integration of an $r$ -form $\omega$ over an $r$ -chain in $M$ . We first define an integration of $\omega$ on an $r$ -simplex $s_r$ of $M$ by

$${\int }_{s_r}\omega ={\int }_{{\bar{\sigma }}_r}{f}^{\ast }\omega \text{\hspace{1em}(6.6)}$$

where $f:{\bar{\sigma }}_r\to M$ is a smooth map such that $s_r=f({\bar{\sigma }}_r)$ . Since $f^{\ast }\omega$ is an $r$ -form in ${\mathbb{R}}^r$ , the RHS is the usual $r$ -fold integral. For a general $r$ -chain $c={\sum }_i{a}_i{s}_{r,i}\in C_r(M)$ , we define

$${\int }_c\omega =\sum _i{a}_i{\int }_{s_{r,i}}\omega .\text{\hspace{1em}(6.7)}$$

6.1.2 Stokes’ theorem

Theorem 6.1 · Stokes' theorem

Let $\omega \in {\Omega }^{r-1}(M)$ and $c\in C_r(M)$ . Then

$${\int }_c\mathrm{d}\omega ={\int }_{\partial c}\omega .\text{\hspace{1em}(6.8)}$$

Proof

Since $c$ is a linear combination of $r$ -simplexes, it suffices to prove (6.8) for an $r$ -simplex $s_r$ in $M$ . Let $f:{\bar{\sigma }}_r\to M$ be a map such that $f({\bar{\sigma }}_r)=s_r$ . Then

$${\int }_{s_r}\mathrm{d}\omega ={\int }_{{\bar{\sigma }}_r}{f}^{\ast }(\mathrm{d}\omega )={\int }_{{\bar{\sigma }}_r}\mathrm{d}(f^{\ast }\omega )$$

where (5.75) has been used. We also have

$${\int }_{\partial s_r}\omega ={\int }_{\partial {\bar{\sigma }}_r}{f}^{\ast }\omega .$$

Note that $f^{\ast }\omega$ is an $(r-1)$ -form in ${\mathbb{R}}^r$ . Thus, to prove Stokes’ theorem

$${\int }_{s_r}\mathrm{d}\omega ={\int }_{\partial s_r}\omega \text{\hspace{1em}(6.9a)}$$

it suffices to prove an alternative formula

$${\int }_{{\bar{\sigma }}_r}\mathrm{d}\psi ={\int }_{\partial {\bar{\sigma }}_r}\psi \text{\hspace{1em}(6.9b)}$$

for an $(r-1)$ -form $\psi$ in ${\mathbb{R}}^r$ . The most general form of $\psi$ is

$$\psi =\sum a_{\mu }(x)d{x}^1\wedge \cdots \wedge d{x}^{\mu -1}\wedge d{x}^{\mu +1}\wedge \cdots \wedge d{x}^r.$$

Since an integration is distributive, it suffices to prove (6.9b) for $\psi =a(x)\mathrm{d}{x}^1\wedge$ $\dots \wedge \mathrm{d}{x}^{r-1}$ . We note that

$$\mathrm{d}\psi =\frac{\partial a}{\partial x^r}\mathrm{d}{x}^r\wedge \mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^{r-1}=(-1{)}^{r-1}\frac{\partial a}{\partial x^r}\mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^{r-1}\wedge \mathrm{d}{x}^r.$$

By direct computation, we find, from (6.2), that

$$\begin{array}{rl}& {\int }_{{\bar{\sigma }}_r}\mathrm{d}\psi =(-1{)}^{r-1}{\int }_{{\bar{\sigma }}_r}\frac{\partial a}{\partial x^r}\mathrm{d}{x}^1\dots \mathrm{d}{x}^{r-1}\mathrm{d}{x}^r\\ & =(-1{)}^{r-1}{\int }_{x^{\mu }\ge 0,{\sum }_{\mu =1}^{r-1}{x}^{\mu }\le 1}d{x}^1\dots d{x}^{r-1}{\int }_0^{1-{\sum }_{\mu =1}^{r-1}{x}^{\mu }}\frac{\partial a}{\partial x^r}d{x}^r\\ & =(-1{)}^{r-1}\int \mathrm{d}{x}^1\dots \mathrm{d}{x}^{r-1}\\ & \times \left[a\left(x^1,\dots ,x^{r-1},1-\sum _{\mu =1}^{r-1}{x}^{\mu }\right)-a\left(x^1,\dots ,x^{r-1},0\right)\right].\end{array}$$

For the boundary of ${\bar{\sigma }}_r$ , we have

$$\begin{array}{rl}& \partial {\bar{\sigma }}_r=(p_1,p_2,\dots ,p_r)-(p_0,p_2,\dots ,p_r)\\ & +\cdots +(-1{)}^r\left(p_0,p_1,\dots ,p_{r-1}\right).\end{array}$$

Note that $\psi =a(x)\mathrm{d}{x}^1\wedge \dots \wedge \mathrm{d}{x}^{r-1}$ vanishes when one of $x^1,\dots ,x^{r-1}$ is constant. Then it follows that

$${\int }_{(p_0,p_2,\dots ,p_r)}\psi =0$$

since $x^1\equiv 0$ on $(p_0,p_2,\dots ,p_r)$ . In fact, most of the faces of $\partial {\bar{\sigma }}_r$ do not contribute to the RHS of (6.9b) and we are left with

$${\int }_{\partial {\bar{\sigma }}_r}\psi ={\int }_{(p_1,p_2,\dots ,p_r)}\psi +(-1{)}^r{\int }_{(p_0,p_1,\dots ,p_{r-1})}\psi .$$

Since $(p_0,p_1,\dots ,p_{r-1})$ is the standard $(r-1)$ -simplex ( $x^{\mu }\ge 0$ , ${\sum }_{\mu =1}^{r-1}{x}^{\mu }\le 1$ ), on which $x^r=0$ , the second term is

$$(-1{)}^r{\int }_{(p_0,p_1,\dots ,p_{r-1})}\psi =(-1{)}^r{\int }_{{\bar{\sigma }}_{r-1}}a(x^1,\dots ,x^{r-1},0)\mathrm{d}{x}^1\dots \mathrm{d}{x}^{r-1}.$$

The first term is

$$\begin{array}{rl}{\int }_{(p_1,p_2,\dots ,p_r)}\psi & ={\int }_{(p_1,\dots ,p_{r-1},p_0)}a\left(x^1,\dots ,x^{r-1},1-\sum _{\mu =1}^{r-1}{x}^{\mu }\right)d{x}^1\dots d{x}^{r-1}\\ & =(-1{)}^{r-1}{\int }_{{\bar{\sigma }}_{r-1}}a\left(x^1,\dots ,x^{r-1},1-\sum _{\mu =1}^{r-1}{x}^{\mu }\right)d{x}^1\dots d{x}^{r-1}\end{array}$$

where the integral domain $(p_1,\dots ,p_r)$ has been projected along $x^r$ to the $(p_1,\dots ,p_{r-1},p_0)$ -plane, preserving the orientation. Collecting these results, we have proved (6.9b). [The reader is advised to verify this proof for $m=3$ using figure 6.1.]

Exercise 6.1

Let $M={\mathbb{R}}^3$ and $\omega =a\mathrm{d}x+b\mathrm{d}y+c\mathrm{d}z$ . Show that Stokes’ theorem is written as

$${\int }_S\mathrm{curl}\mathbf{\omega }\cdot \mathrm{d}\mathbf{S}={\oint }_C\mathbf{\omega }\cdot \mathrm{d}\mathbf{S}(\text{S t o k e s ＂ t h e o r e m})\text{\hspace{1em}(6.10)}$$

where $\omega =(a,b,c)$ and $C$ is the boundary of a surface $S$ . Similarly, for $\psi =\frac{1}{2}{\psi }_{\mu \nu }\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }$ , show that

$${\int }_V\mathrm{div}\mathbf{\psi }dV={\oint }_S\mathbf{\psi }\cdot dS(\text{G a u s s ’ t h e o r e m})$$

where ${\psi }^{\lambda }={\epsilon }^{\lambda \mu \nu }{\psi }_{\mu \nu }$ and $S$ is the boundary of a volume $V$ .

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