# Lectures on Stoke’s Theorem — MTH 429H

Stokes Theorem states that

$\displaystyle \int_{S}d\omega=\int_{\partial S}\omega \ \ \ \ \ (1)$

where ${S}$ is a ${j}$-surface in ${\mathbb{R}^{k}}$, ${\omega}$ is a ${C^{1}}$ ${j-1}$-form and ${\partial S}$ is a ${j-1}$ surface (or a collection of ${j-1}$ surfaces) in ${\mathbb{R}^{d}}$ that make up the boundary of ${S}$.

We will obtain this result in a two stages. First we will consider two very simple cases, namely integration of the derivative of a compactly supported ${j-1}$ form over ${\mathbb{R}^{j}}$ and over the upper half space

$\displaystyle \mathbb{H}^{j}=\left\{ \mathbf{x}\in\mathbb{R}^{j}\ :\ x_{j}\ge0\right\} .$

Second on a general oriented ${j}$-surfaces with boundary ${S}$ we will use a partition of unity to reduce the computation of ${\int_{S}d\omega}$ to the simple case of integration over ${\mathbb{H}^{j}}$. Of course, before we do that we should define what exactly an oriented ${j}$-surface with boundary is!

The topological boundary of ${\mathbb{H}^{j}}$ is the set

$\displaystyle \left\{ \mathbf{x}\in\mathbb{R}^{j}\ :\ x_{j}=0\right\}$

which we may think of as a parameterized ${j-1}$-surface via the map

$\displaystyle J_{j-1}\left(y_{1},\ldots,y_{j-1}\right)=\left(y_{1},\ldots,y_{j-1},0\right). \ \ \ \ \ (2)$

We will show ${\int_{\mathbb{R}^{j}}d\omega=0}$ and

$\displaystyle \int_{\mathbb{H}^{j}}d\omega=(-1)^{j}\int_{J_{j-1}}\omega. \ \ \ \ \ (3)$

This last identity is a special case of (1) — where we think of ${\partial\mathbb{H}^{j}}$ as ${J_{j-1}}$ if ${j}$ is even and the surface with opposite orientation if ${j}$ is odd.

Two simple cases of Stokes theorem

Recall that the support of a ${j}$-form ${\omega}$, denoted ${\mathrm{supp}\omega}$, is the closure of the set of ${\mathbf{x}}$ such that ${\omega\left(\mathbf{x}\right)\neq0}$. If ${\mathrm{supp}\omega}$ is compact we say that ${\omega}$ is compactly supported. If ${\omega}$ is a compactly supported ${j}$-form in ${\mathbb{R}^{j}}$ we have

$\displaystyle \int_{\mathbb{R}^{j}}\omega=\int_{\left[-R,R\right]^{j}}\omega\left(\mathbf{x}\right)\left[\mathbf{e}_{1},\ldots,\mathbf{e}_{j}\right]d\mathbf{x}$

for any ${R}$ large enough that ${[-R,R]^{j}\supset\mathrm{supp}\omega}$ .

Theorem 1 Let ${\omega}$ be a ${C^{1}}$ compactly supported ${j-1}$ form in ${\mathbb{R}^{j}}$. Then

$\displaystyle \int_{\mathbb{R}^{j}}d\omega=0.$

Proof: By linearity of the exterior derivative and the integral it suffices to prove this for ${\omega=f\left(\mathbf{x}\right)d\mathbf{x}_{S}}$ where ${S\subset\left\{ 1,\ldots,j\right\} }$ has size ${j-1}$, i.e., ${S=\left\{ \alpha\right\} ^{c}}$ for some ${\alpha}$. But then

$\displaystyle d\omega=\left(-1\right)^{\alpha-1}\frac{\partial f}{\partial x_{\alpha}}\left(\mathbf{x}\right)d\mathbf{x}_{\left\{ 1,\ldots,j\right\} }. \ \ \ \ \ (4)$

Integrating first with respect to ${x_{\alpha}}$ and applying the fundamental theorem of calculus we see that the integral vanishes. $\Box$

Theorem 1 is already a special case of Stokes theorem — the “${j}$-surface” ${\mathbb{R}^{j}}$ has no boundary. A special case of a “${j}$-surface” with boundary is the upper half space ${\mathbb{H}^{j}.}$ Given a compactly supported ${j}$-form ${\omega}$ in ${\mathbb{R}^{j}}$ we define

$\displaystyle \int_{\mathbb{H}^{j}}\omega:=\int_{\mathbb{R}^{j-1}}\int_{0}^{\infty}\omega((\mathbf{y},t))\left[\mathbf{e}_{1},\ldots,\mathbf{e}_{j}\right]dtd\mathbf{y}$

where ${\mathbf{e}_{1},\ldots,\mathbf{e}_{j}}$ denotes the standard basis of ${\mathbb{R}^{j}}$.

Theorem 2 Let ${\omega}$ be a ${C^{1}}$ compactly supported ${j-1}$ form in ${\mathbb{R}^{j}}$. Then

$\displaystyle \int_{\mathbb{H}^{j}}d\omega=\left(-1\right)^{j}\int_{J_{j-1}}\omega.$

Remark 3 Recall the definition (2) of ${T_{j-1}}$. Since ${\omega}$ is compactly supported we have

$\displaystyle \begin{array}{rcl} \int_{J_{j-1}}\omega: & = & \int_{\mathbb{R}^{j-1}}\omega_{J_{j-1}}=\int_{\mathbb{R}^{j-1}}\omega\left(J_{j-1}\mathbf{y}\right)\left[\mathbf{e}_{1},\ldots,\mathbf{e}_{j-1}\right]d\mathbf{y}\\ & = & \int_{\mathbb{R}^{j-1}}\omega\left(y_{1},\ldots,y_{j-1},0\right)\left[\mathbf{e}_{1},\ldots,\mathbf{e}_{j-1}\right]d\mathbf{y}. \end{array}$

Proof: As above it suffice to consider ${\omega=f\left(\mathbf{x}\right)d\mathbf{x}_{\left\{ \alpha\right\} ^{c}}}$ for which we have (4). If ${\alpha\neq j}$ then ${\omega_{J_{j-1}}=0}$ and by the fundamental theorem of calculus ${\int_{\mathbb{R}^{j}}d\omega=0.}$ On the other hand if ${\alpha=j}$ then

$\displaystyle \omega_{J_{j-1}}=f\left(J_{j-1}\mathbf{y}\right)dy_{1}\wedge\cdots\wedge dy_{j-1}$

and integrating first with respect to ${x_{j}}$ we see that

$\displaystyle \begin{array}{rcl} \int_{\mathbb{H}^{j}}d\omega & = & (-1)^{j-1}\int_{\mathbb{H}^{j}}\frac{\partial f}{\partial x_{j}}\left(\mathbf{x}\right)dx_{1}\wedge\cdots\wedge dx_{j}\\ & = & \left(-1\right)^{j-1}\int_{\mathbb{R}^{j-1}}\int_{0}^{\infty}\frac{\partial f}{\partial x_{j}}\left(\left(\mathbf{y},t\right)\right)dtd\mathbf{y}\\ & = & \left(-1\right)^{j}\int_{\mathbb{R}^{j-1}}f\left(y_{1},\dots,y_{j-1},0\right)dy_{1}\cdots dy_{j-1}\\ & = & \left(-1\right)^{j}\int_{\mathbb{R}^{j-1}}f\left(J_{j-1}\mathbf{y}\right)d\mathbf{y}=\left(-1\right)^{j}\int_{J_{j-1}}\omega. \end{array}$

$\Box$

Oriented surfaces and Stokes’ Theorem

To state Stokes Theorem we need a more general definition of a ${j}$-surface. Have in mind the unit circle ${S^{1}}$ in ${\mathbb{R}^{2}}$ or the unit sphere ${S^{2}}$ in ${\mathbb{R}^{3}}$ as you read the following.

Definition 4 A compact subset ${S\subset\mathbb{R}^{d}}$ is a ${j}$-surface if there is a covering of ${S}$ by finitely many relatively open subsets ${S_{1},\ldots,S_{p}}$ such that for each ${\alpha=1,\ldots,p}$ we have

$\displaystyle S_{\alpha}=T_{\alpha}\left(V_{\alpha}\right)$

where ${V_{\alpha}\subset\mathbb{R}^{j}}$ is open, ${T_{\alpha}\in C^{1}\left(V_{\alpha},\mathbb{R}^{d}\right)}$ is one-to-one and ${\mathrm{rank}dT_{\alpha}\left(\mathbf{y}\right)=j}$ for each ${\mathbf{y}\in V_{\alpha}}$.

Remark 5 Recall that the rank ${\mathrm{rank}T}$ of a linear transformation ${T\in L\left(\mathbb{R}^{j},\mathbb{R}^{d}\right)}$ is the dimension of the range ${T\left(\mathbb{R}^{j},\right)}$.

The definition says that every point of ${S}$ has a neighborhood in ${S}$ that is diffeomorphic to an open subset of ${\mathbb{R}^{j}}$. Another name for a ${j}$-surface defined in this way is an embedded manifold. The collection ${\left\{ T_{\alpha}\right\} _{\alpha=1}^{p}}$ is called a coordinate chart for ${S}$.

Proposition 6 For each pair ${\alpha\neq\beta}$, , let ${V_{\alpha;\beta}=T_{\alpha}^{-1}\left(S_{\beta}\right)}$ and define ${T_{\alpha;\beta}:=T_{\beta}^{-1}\circ T_{\alpha}}$ on ${V_{\alpha;\beta}}$. If ${V_{\alpha;\beta}\neq\emptyset}$, then ${T_{\alpha;\beta}:V_{\alpha;\beta}\rightarrow V_{\beta;\alpha}}$ is ${C^{1}}$, one-to-one, onto and satisfies ${\det dT_{\alpha;\beta}\left(\mathbf{y}\right)\neq0}$ for each point ${\mathbf{y}\in V_{\alpha;\beta}}$.

Remark 7 The functions ${T_{\alpha;\beta}}$ are called the transition functions of the coordinate chart ${\left\{ T_{\alpha}\right\} }$. One can define an abstract ${C^{1}}$ manifold (not embedded in ${\mathbb{R}^{d}}$) as a compact space ${S}$ with an open cover ${S_{1},\ldots,S_{p}}$ and one-to-one coordinate maps ${T_{\alpha}:V_{\alpha}\rightarrow S_{\alpha}}$ such that the transition functions are ${C^{1}}$, one-to-one and have non-singular derivative.

Exercise 8 Prove proposition 6. Here is a sketch. Fix ${\alpha,\beta}$ with ${V_{\alpha;\beta}\neq\emptyset}$ and let ${\Phi=T_{\alpha;\beta}}$ and ${\Psi=T_{\beta;\alpha}}$. So ${\Psi\circ\Phi=\mathrm{Id}}$ on ${V_{\alpha;\beta}}$ and ${\Phi\circ\Psi=\mathrm{Id}}$ on ${V_{\beta;\alpha}}$. Thus ${\Phi,\Psi}$ are one-to-one, onto and, if they are ${C^{1}}$, have non-singular derivative. To show that they are ${C^{1}}$, fix ${\mathbf{y}_{0}\in V_{\alpha;\beta}}$ and let ${\mathbf{\zeta}_{0}=\Phi\left(\mathbf{y}_{0}\right)}$. Then for each sufficiently small ${\mathbf{h}\in\mathbb{R}^{j}}$ there is a unique ${\mathbf{k}\in\mathbb{R}^{j}}$ such that

$\displaystyle T_{\alpha}\left(\mathbf{y}_{0}+\mathbf{h}\right)=T_{\beta}\left(\mathbf{\zeta_{0}}+\mathbf{k}\right). \ \ \ \ \ (5)$

Likewise for each small ${\mathbf{k}}$ there is a unique ${\mathbf{h}}$ such that (5) holds. Since ${T_{\alpha}\left(\mathbf{y}_{0}\right)=T_{\beta}\left(\mathbf{\zeta}_{0}\right)}$ we have

$\displaystyle dT{}_{\alpha}\left(\mathbf{y_{0}}\right)\mathbf{h}+o(\left|\mathbf{h}\right|)=dT_{\beta}\left(\mathbf{\zeta}_{0}\right)\mathbf{k}+o(\left|\mathbf{k}\right|). \ \ \ \ \ (6)$

Because ${dT_{\alpha}\left(\mathbf{y}_{0}\right)}$ and ${dT_{\beta}\left(\mathbf{\zeta_{0}}\right)}$ have full rank you can show that for ${\mathbf{h}}$ and ${\mathbf{k}}$ sufficiently small we have

$\displaystyle \frac{1}{C}\left|\mathbf{h}\right|\le\left|\mathbf{k}\right|\le C\left|\mathbf{h}\right|$

where ${C>1}$. Thus ${o\left(\left|\mathbf{h}\right|\right)=o\left(\left|\mathbf{k}\right|\right)}$ and dividing by length of ${\left|\mathbf{h}\right|}$ and taking ${\left|\mathbf{h}\right|\rightarrow0}$ we see from (6) that the range of ${dT_{\alpha}\left(\mathbf{y}_{0}\right)}$ is equal to the range of ${dT_{\beta}\left(\mathbf{\zeta}_{0}\right)}$. Since the two linear transformations are full rank we can now solve for ${\mathbf{h}}$ and ${\mathbf{k}}$ in terms of each other, up to terms that are ${o\left(\left|\mathbf{h}\right|\right)}$.

It is not possible to integrate a ${j}$-form over an arbitrary ${j}$-surface. The trouble is that the integral of a form could change sign under a change of variables. To make the integral of forms well defined we need to make sure that all of the functions in the coordinate chart are consistently oriented. To this end we will consider oriented surfaces:

Definition 9 Let ${S}$ be a ${j}$-surface and let ${T_{\alpha}:V_{\alpha}\rightarrow S_{\alpha}}$, ${\alpha=1,\ldots,p}$, be a coordinate chart for ${S}$. We say the chart is oriented if each transition function satisfies ${\det dT{}_{\alpha;\beta}\left(\mathbf{y}\right)>0}$ for every ${\mathbf{y}\in V_{\alpha;\beta}}$. The ${j}$-surface ${S}$ is called orientable if an oriented coordinate chart exists. An oriented ${j}$-surface is a ${j}$-surface ${S}$ together with a choice of an oriented chart.

Remark 10 There are non-orientable ${j}$-surfaces; the Möbius strip is an example.

Given an oriented ${j}$-surface, the sets of the open cover are intersections of ${S}$ with open sets in ${\mathbb{R}^{d}}$:

$\displaystyle S_{\alpha}=S\cap U_{\alpha}.$

Thus we can find a partition ${\left\{ \phi_{\alpha}\right\} }$ of unity on ${S}$ subordinate to ${U_{\alpha}}$. As mentioned in the proof of the partition of unity theorem we can choose the functions ${\phi_{\alpha}}$ to be as smooth as we like. In particular they can be ${C^{1}}$. Using the partition of unity ${\left\{ \phi_{\alpha}\right\} }$ we now define the integral of a ${j}$-form ${\omega}$ over ${S}$ to be

$\displaystyle \int_{S}\omega=\sum_{\alpha}\int_{T_{\alpha}}\phi_{\alpha}\omega,$

Here we think of ${T_{\alpha}}$ as a parameterized ${j}$-surface. The pullback ${\left[\phi_{\alpha}\omega\right]_{T_{\alpha}}}$ is compactly supported in ${V_{\alpha}}$ so

$\displaystyle \int_{S}\omega=\sum_{\alpha}\int_{\mathbb{R}^{j}}\phi_{\alpha}\left(T\left(\mathbf{y}\right)\right)\omega_{T_{\alpha}}\left(\mathbf{y}\right).$

We state without proof the following basic fact

Lemma 11 Let ${S}$ be an oriented ${j}$-surface and let ${\omega}$ be a ${j}$ form in ${\mathbb{R}^{d}}$. If ${\mathrm{supp}\omega\cap S\subset S_{\alpha}}$ for some ${\alpha=1,\ldots,p}$ then

$\displaystyle \int_{S}\omega=\int_{T_{\alpha}}\omega.$

Exercise 12 Prove Lemma 11. Sketch: You need to show that ${\int_{T_{\beta}}\phi_{\beta}\omega=\int_{T_{\alpha}}\phi_{\beta}\omega}$ for each ${\beta\neq0}$. This follows from the change of variables formula using the transition function ${T_{\alpha;\beta}}$. It is to facilitate this argument that we need the coordinate chart to be oriented, as otherwise we might get a minus sign in the change of variables formula.

More generally, we have the following

Theorem 13 Let ${S}$ be a connected orientable ${j}$-surface and let ${\left\{ T_{\alpha}:V_{\alpha}\rightarrow S_{\alpha}\right\} _{\alpha=1}^{p}}$, ${\left\{ \widetilde{T}_{\beta}:\widetilde{V}_{\beta}\rightarrow\widetilde{S}_{\beta}\right\} _{\beta=1}^{q}}$ be two oriented coordinate charts on ${S}$. Choose two partitions of unity on ${S}$, ${\left\{ \phi_{\alpha}\right\} _{\alpha=1}^{p}}$ and ${\left\{ \widetilde{\phi}_{\beta}\right\} _{\beta=1}^{q}}$ subordinate to ${\left\{ U_{\alpha}\right\} _{\alpha=1}^{p}}$ and ${\left\{ \widetilde{U}_{\beta}\right\} _{\beta=1}^{q}}$ respectively, where ${S_{\alpha}=U_{\alpha}\cap S}$ and ${\widetilde{S}_{\beta}=U_{\beta}\cap S}$ for each ${\alpha}$, ${\beta}$. For each pair ${\alpha,\beta}$ with ${\alpha\in\left\{ 1,\ldots,p\right\} }$ and ${\beta\in\left\{ 1,\ldots,q\right\} }$, define a transition function ${\Phi_{\alpha;\beta}=\widetilde{T}_{\beta}^{-1}\circ T_{\alpha}:W_{\alpha;\beta}\rightarrow\widetilde{W}_{\beta;\alpha}}$ where ${W_{\alpha;\beta}=T_{\alpha}^{-1}\left(S_{\alpha}\right)}$ and ${\widetilde{W}_{\beta;\alpha}=\widetilde{T}_{\beta}^{-1}\left(\widetilde{S}_{\alpha}\right)}$. Then either

1. ${\det d\Phi_{\alpha;\beta}\left(\mathbf{y}\right)>0}$ for all ${\mathbf{y}\in W_{\alpha;\beta}}$ and for each ${\alpha,\beta}$, in which case

$\displaystyle \sum_{\alpha}\int_{T_{\alpha}}\phi_{\alpha}\omega=\sum_{\beta}\int_{\widetilde{T}_{\beta}}\widetilde{\phi}_{\beta}\omega$

for any continuous ${j}$-form on ${S}$; or

2. ${\det d\Phi_{\alpha;\beta}\left(\mathbf{y}\right)<0}$ for all ${\mathbf{y}\in W_{\alpha;\beta}}$ and for each ${\alpha,\beta}$, in which case

$\displaystyle \sum_{\alpha}\int_{T_{\alpha}}\phi_{\alpha}\omega=-\sum_{\beta}\int_{\widetilde{T}_{\beta}}\widetilde{\phi}_{\beta}\omega$

for any continuous ${j}$-form on ${S}$.

Remark 14 1) What this says is that there are exactly two possible orientations for a connected surface. (It is essential that we assumed the surface to be connected, as otherwise each component could have its own orientation.) 2) If case (1) holds we say that ${\left\{ T_{\alpha}\right\} }$ and ${\left\{ \widetilde{T}_{\beta}\right\} }$ have the same orientation and if case (2) holds we say that they have the opposite orientation.

Exercise 15 Prove Theorem 13. This is a bit long and involved. Here is a sketch: The key is to show the uniform positivity or negativity of the determinants. The rest follows from change of variables as in the proof of Lemma 11. To see that the determinants have a definite sign proceed as follows. Fix ${\mathbf{x}\in S}$. Suppose ${\mathbf{x}\in S_{\alpha}\cap S_{\beta_{j}}}$ for ${j=1,2}$. So ${\mathbf{x}=T_{\alpha}\left(\mathbf{y}\right)}$ and

$\displaystyle \Phi_{\alpha;\beta_{2}}\left(\mathbf{y}\right)=\widetilde{T}_{\beta_{2}}^{-1}\left(T_{\alpha}\left(\mathbf{y}\right)\right)=\widetilde{T}_{\beta_{2}}^{-1}\circ\widetilde{T}_{\beta_{1}}\left(\Phi_{\alpha;\beta_{1}}\left(\mathbf{y}\right)\right).$

Thus

$\displaystyle \det d\Phi_{\alpha;\beta_{2}}\left(\mathbf{y}\right)=\det d\widetilde{T}_{\beta_{1};\beta_{2}}\left(\mathbf{x}\right)\det d\Phi_{\alpha;\beta_{1}}\left(\mathbf{y}\right)$

where ${\widetilde{T}_{\beta_{1};\beta_{2}}=\widetilde{T}_{\beta_{2}}^{-1}\circ\widetilde{T}_{\beta_{1}}}$ is a transition function for the chart ${\left\{ \widetilde{T}_{\beta}\right\} }$. Since ${\left\{ \widetilde{T}_{\beta}\right\} }$ is oriented, the corresponding determinant is positive and we see that ${\det d\Phi{}_{\alpha;\beta_{j}}\left(\mathbf{y}\right)}$ for ${j=1,2}$ have the same sign. Similarly if ${\mathbf{x}\in S_{\alpha_{j}}\cap S_{\beta}}$ for ${j=1,2}$ then ${\mathbf{x}=\widetilde{T}_{\beta}\left(\mathbf{y}\right)}$ and

$\displaystyle \Phi_{\alpha_{2};\beta}^{-1}\left(\mathbf{y}\right)=T_{\alpha_{1};\alpha_{2}}\left(\Phi_{\alpha_{1};\beta}^{-1}\left(\mathbf{y}\right)\right)$

and we see that ${\det d\Phi_{\alpha_{j};\beta}^{-1}\left(\mathbf{y}\right)}$ for ${j=1,2}$ ${ }$have the same sign. Finally let ${\sigma\left(\mathbf{x}\right)=\pm1}$ depending on whether this sign is positive or negative.

Now consider the sets

$\displaystyle S_{\pm}=\left\{ \mathbf{x}\in S\ :\ \sigma\left(\mathbf{x}\right)=\pm1\right\} .$

It is easy to show that these sets are both relatively open — if ${\det d\Phi_{\alpha,\beta}\left(\mathbf{y}_{0}\right)>0}$ then it is positive on a ball centered at ${\mathbf{y}_{0}}$. On the other hand ${S_{+}\cup S_{-}=S}$, because ${\det d\Phi_{\alpha;\beta}}$ never vanishes since the coordinate functions have derivatives of full rank. Thus ${S_{\pm}=S\setminus S_{\mp}}$ are both closed. Since ${S}$ is connected it follows that one of the sets is empty and one is ${S}$.

Given an oriented ${j}$-surface ${S}$, let ${\neg S}$ denote the ${j}$-surface with opposite orientation (using any coordinate chart that gives it the opposite orientation). Explicitly, if ${\left\{ T_{\alpha}:V_{\alpha}\rightarrow S_{\alpha}\right\} }$ is an oriented chart for ${S}$, we may obtain an oriented chart for ${\neg S}$ using

$\displaystyle \widetilde{T}_{\alpha}\left(y_{1},\ldots,y_{j}\right)=T_{\alpha}\left(-y_{1},y_{2},\ldots,y_{j}\right)$

where ${\widetilde{T}_{\alpha}:\widetilde{V}_{\alpha}\rightarrow S_{\alpha}}$ with

$\displaystyle \widetilde{V}_{\alpha}=\left\{ \left(-y_{1},y_{2},\ldots,y_{j}\right)\ :\ \mathbf{y}\in V_{\alpha}\right\} .$

Note that

$\displaystyle \int_{\neg S}\omega=-\int_{S}\omega$

for any continuous ${j}$-form ${\omega}$.

Stokes Theorem

Theorem 16 Let ${S}$ be an oriented ${j}$-surface in ${\mathbb{R}^{d}}$ and let ${\omega}$ be a ${C^{1}}$ ${j-1}$ form in a neighborhood of ${S}$. Then

$\displaystyle \int_{S}d\omega=0.$

Proof: Choose a ${C^{1}}$ partition of unity on ${S}$ subordinate to ${\left\{ U_{\alpha}\right\} }$ as above. Note that ${\omega=\sum_{\alpha}\phi_{\alpha}\omega}$ and so ${d\omega=\sum_{\alpha}d\left(\phi_{\alpha}\omega\right)}$ and that for each ${\alpha}$, ${d\left(\phi_{\alpha}\omega\right)}$ is compactly supported in ${U_{\alpha}}$. Thus by Lemma11

$\displaystyle \int_{S}d\omega=\sum_{\alpha}\int_{S}d\left(\phi_{\alpha}\omega\right)=\sum_{\alpha}\int_{T_{\alpha}}d\left(\phi_{\alpha}\omega\right)=\sum_{\alpha}\int_{\mathbb{R}^{j-1}}d\left(\phi_{\alpha}\omega\right)_{T_{\alpha}},$

since the pullback of a derivative is the derivative of the pullback. By Theorem 1 the integral is zero. $\Box$

To obtain a non-zero value for ${\int_{S}d\omega}$ we need to consider a surface ${S}$ with boundary. A surface ${S}$ with boundary is like a surface but the the functions ${T_{\alpha}}$ making up the coordinate chart are restricted to the upper half space ${\mathbb{H}^{j}}$:

Definition 17 An ${j}$-surface with boundary in ${\mathbb{R}^{d}}$ is a compact subset ${S\subset\mathbb{R}^{d}}$ together with a covering of ${S}$ by relatively open subsets ${S_{1},\ldots,S_{p}}$, such that for each ${\alpha=1,\ldots,p}$ we have

$\displaystyle S_{\alpha}=T_{\alpha}\left(V_{\alpha}\cap\mathbb{H}^{j}\right)$

where ${V_{\alpha}\subset\mathbb{R}^{j}}$ is open and ${T_{\alpha}\in C^{1}\left(V_{\alpha},\mathbb{R}^{d}\right)}$ is a one-to-one map with ${\mathrm{rank}dT_{\alpha}\left(\mathbf{y}\right)=j}$ for each ${\mathbf{y}\in V_{\alpha}}$. The surface is oriented if for each pair ${\alpha,\beta}$ the transition function ${T_{\alpha;\beta}=T_{\beta}^{-1}\circ T_{\alpha}}$, defined on ${V_{\alpha;\beta}=T_{\alpha}^{-1}\left(T_{\beta}\left(V_{\beta}\right)\right)}$, satisfies ${\det dT_{\alpha;\beta}\left(\mathbf{y}\right)>0}$ for all ${\mathbf{y}\in V_{\alpha;\beta}}$.

The definition of integrals over an oriented${j}$-surface with boundary is just like that over a ${j}$-surface: pick a partition of unity ${\left\{ \phi_{\alpha}\right\} }$ on ${S}$ subordinate to ${U_{\alpha}}$ where ${U_{\alpha}\cap S=S_{\alpha}}$ and set

$\displaystyle \int_{S}\omega=\sum_{j=1}^{p}\int_{V_{\alpha}\cap\mathbb{H}^{j}}\phi_{\alpha}\circ T_{\alpha}\omega_{T_{\alpha}}.$

As before there are two classes of oriented charts and the value of the integral does not depend on the choice of chart within an orientation class but changes sign for the opposite orientation. The boundary ${\partial S}$ of a surface ${S}$ with boundary is the set of points that lie on the boundary of one of the coordinate charts — that is points of the form ${T_{\alpha}\left(y_{1},\ldots,y_{j-1},0\right)}$. To make a precise definition recall the map ${J_{j-1}:\mathbb{R}^{j-1}\rightarrow\mathbb{H}^{j}}$ given by

$\displaystyle J_{j-1}\left(y_{1},\ldots,y_{j-1}\right)=\left(y_{1},\ldots,y_{j-1},0\right).$

We state without proof:

Proposition 18 Let ${S}$ be a ${j}$-surface with boundary and for each ${\alpha=1,\ldots,p}$ let

$\displaystyle W_{\alpha}=J_{j-1}^{-1}\left(V_{\alpha}\right)=\left\{ \left(y_{1},\ldots,y_{j-1}\right)\ :\ \left(y_{1},\ldots,y_{j-1},0\right)\in V_{\alpha}\right\} ,$

$\displaystyle L_{\alpha}=T_{\alpha}\circ J_{j-1}\quad\text{on }W_{\alpha},$

$\displaystyle \Gamma_{\alpha}=L_{\alpha}\left(W_{\alpha}\right).$

Then ${\left\{ \Gamma_{\alpha}\right\} }$ is an open cover for

$\displaystyle \Gamma:=\left\{ \mathbf{x}\ :\ \mathbf{x}=L_{\alpha}\left(W_{\alpha}\right)\text{ for some }\alpha\right\} \ \ \ \ \ (7)$

and the functions ${\left\{ L_{\alpha}\right\} }$ make a coordinate chart that turns ${\Gamma}$ into a ${C^{1}}$ ${j-1}$ surface. Furthermore ${\Gamma}$ is oriented if ${S}$ is.

The proof is pretty straightforward — one simply has to verify that the transition function ${L_{\beta}^{-1}\circ L_{\alpha}}$ are the restriction to ${\mathbb{R}^{j-1}}$ of the corresponding transition function for ${S}$.

Definition 19 If ${S}$ is an oriented ${j}$-surface with boundary then the boundary of ${S}$ is the oriented ${j-1}$-surface

$\displaystyle \partial S=\begin{cases} \Gamma & \mbox{if }j\mbox{ is even,}\\ \neg\Gamma & \mbox{if }j\mbox{ is odd,} \end{cases}$

with ${\Gamma}$ as in Prop. 18.

Thm 20 (Stokes’ Theorem ) Let ${S}$ be an oriented ${j}$-surface with boundary in ${\mathbb{R}^{d}}$ and let ${\omega}$ be a ${C^{1}}$ ${j}$-form defined in a neighborhood of ${S}$. Then

$\displaystyle \int_{S}d\omega=\int_{\partial S}\omega.$

Proof: As above it suffices, by linearity of the integral and the exterior derivative, to prove this for ${\omega}$ such that ${\mathrm{supp}\omega\cap S\subset S_{\alpha}.}$ For such ${\omega}$ we have by Theorem 2

$\displaystyle \int_{S}d\omega=\int_{T_{\alpha}}d\omega=\int_{\mathbb{H}^{j-1}}d\omega_{T_{\alpha}}=\left(-1\right)^{j}\int_{\mathbb{R}^{j-1}}\omega_{T_{\alpha}\circ J}=\left(-1\right)^{j}\int_{T_{\alpha}\circ J_{j-1}}\omega=\int_{\partial S}\omega.$

$\Box$