Chapter 7: Integration

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Chapter 7: Integration

Contents
Contents
 1.  Manifolds with boundary
   1.1.  Smooth structure
   1.2.  Boundary as submanifold
   1.3.  Outward-pointing vector fields
 2.  Orientations
   2.1.  Orientations of vector spaces
   2.2.  Orientations of manifolds
   2.3.  Orientations and boundary
 3.  Integration of differential forms
 4.  Stokes' theorem

In this chapter we build the integration theory on manifolds. In calculus we integrate a function defined on \mathbb R^n, but on manifolds what we integrate are differential forms, which have been studied a lot in Chapter 6. In this chapter the concept of manifolds has been generalized first, so we can talk about manifolds with boundary. Next, we discuss the orientations of a manifold with which the integral over a manifold is well defined. After developing the theory of integration of differential forms, we shall encounter one of the most important theorems in differential geometry, Stokes's theorem, the n-dimensional analogue of the fundamental theorem of calculus.

1. Manifolds with boundary To define a manifold with boundary, we consider the closed upper half-space \mathbb H^n\subseteq\mathbb R^n, that is,
\[
\mathbb H^n=\{(x^1,\dots,x^n)\in\mathbb R^n\mid x^n\geqslant0\}.
\]Then a topological manifold with boundary is a topological manifold that locally looks like \mathbb H^n.

As in Chapter 1, the conditions that require M to be second countable and Hausdorff have been omitted. They are not explored too much here and thought to be satisfied by default.

If a point is contained in an interior chart, then it is an interior point of M. If it is in a boundary chart such that \varphi(p)\in\partial\mathbb H^n, then it is a boundary point. In this way we can see that we classify all the points in M into two categories. Such a classification is well defined in the sense that a point cannot be an interior point in one chart and a boundary point in another chart. For simplicity, we will consider this invariant property later on manifolds endowed with a smooth structure. See Theorem 4.

In general topology, the notions are a little bit different. A point of a set is an interior point if it has an open neighborhood contained in the set, and is an exterior point if it has an open neighborhood contained in the complement of the set. If a point in A is such that every neighborhood of it has both a point in A and a point not in A, then it is called a boundary point. The topological boundary may be different from manifold boundary. For example, the open unit disk in \mathbb R^2 has the unit circle as its topological boundary but has the empty set as its manifold boundary.

Note that a (topological/smooth) manifold generally refers to a manifold without boundary, as we defined in Chapter 1. A manifold is a manifold with boundary with empty boundary but a manifold with boundary is not necessarily a manifold.
1.1. Smooth structure In case of Euclidean spaces, a function is said to be smooth if in a neighborhood of each point in the domain it admits an extension to a smooth function defined on an open subset of \mathbb R^n. If the function F is defined on an open subset U of \mathbb H^n, then it is smooth if for each x\in U, there is an open subset \widetilde{U}\subseteq\mathbb R^n containing x and a smooth map \widetilde{F} defined on \widetilde{U} that agrees with F on \widetilde{U}\cap\mathbb H^n. By continuity, for points in U\cap\partial \mathbb H^n, all partial derivatives of F are determined by their values in \operatorname{Int}(\mathbb H^n). Therefore they are independent of the choice of extension.

Now we can consider the smooth invariance of the boundary. Before the theorem is given, a lemma is needed, which states that a diffeomorphism between an open subset U of \mathbb R^n and an arbitrary subset S of \mathbb R^n implies S is open in \mathbb R^n. This lemma requires the crucial condition that both U,S have the same dimension n. If this is not the case the conclusion is no longer true: For example the interval (0,1)\in\mathbb R is mapped by a diffeomorphism to the segment \{(x,0)\mid 0<x<1\} in \mathbb R^2, which however is not open in \mathbb R^2.

Proof. Let x _ 0\in U and f(x _ 0)=y _ 0\in S. There is some neighborhood W of y _ 0 in \mathbb R^n and a smooth function g:W\to\mathbb R^n that agrees with f^{-1} on W\cap S. There is also an open ball B centered at x _ 0 and contained in U. Then f is smooth on B. We may assume the ball is so small that B\subseteq f^{-1}(W\cap S), and then we have g\circ f| _ B=\mathbf 1 _ B.

By the chain rule, Jg(f(x)) Jf(x) is identity on B, so the square matrix Jf(x) is nonsingular. By the consequence of inverse function theorem, f(B) is an open subset of \mathbb R^n, which contains y _ 0=f(x _ 0) and is contained in S=f(U). Thus, S is open in \mathbb R^n.

It suffices to show it is impossible that p in one chart is an interior point but in another chart is a boundary point. Suppose p is in an interior chart (U,\varphi) and also in a boundary chart (V,\psi) such that \psi(p)\in\partial\mathbb H^n. We consider the transition map \psi\circ\varphi^{-1} which is a diffeomorphism. By the lemma, for a small ball B containing x=\varphi(p), (\psi\circ\varphi^{-1})(B) is an open subset of \mathbb R^n that contains y=\psi(p) and is contained in \psi(V). This is impossible for \psi(p)\in\partial\mathbb H^n.

Smooth maps

If M,N are manifolds with boundary and F:M\to N is an arbitrary map. The smoothness of F is defined in exactly the same way in terms of transition maps, with the usual understanding that a map whose domain is a subset of \mathbb H^n is smooth if it admits an extension to a smooth map in a neighborhood of each point, and a map whose codomain is a subset of \mathbb H^n is smooth if it is smooth as a map into \mathbb R^n.

By considering the coordinate representation of a map, now it is not difficult to prove the following invariance theorem of the boundary.

Most of the discussion in previous chapters applies verbatim. A few results need more or less adjustments. For example: the structure of T _ pM (as an n-dimensional vector space); local diffeomorphisms; the rank theorem; the submanifold theory; partition of unity; flows.
1.2. Boundary as submanifold Let M be a manifold of dimension n with boundary \partial M. If (U,\varphi) is a chart for M, then we have the following result:

First, \partial M has a topological (n-1)-manifold structure. For any point p\in\partial M, there is a boundary chart (U,\varphi) such that \varphi(p)\in\partial\mathbb R^{n-1}\times\{0\}. And the restriction \varphi| _ {U\cap \partial M} maps U \cap \partial M homeomorphically to an open subset of \mathbb{R}^{n-1}, giving \partial M a local Euclidean structure. Therefore, since p is any point in \partial M, \partial M is an (n-1)-dimensional manifold.

Second, \partial M has an atlas induced from the one of M. The restricted charts (U \cap \partial M, \phi| _ {U \cap \partial M}) form a smooth atlas. Transition maps are smooth because they are restrictions of the smooth transition maps of M to \mathbb{R}^{n-1} \times \{0\}.

Third, \partial M is an embedded submanifold. It can be seen that the inclusion map \iota:\partial M\hookrightarrow M, which in local coordinates maps (x _ 1,\dots,x _ {n-1}) to (x _ 1,\dots,x _ {n-1},0) , is smooth. The differential is injective so \iota is an immersion. Since \partial M has the subspace topology, \iota is a homeomorphism onto its image, hence an embedding.
1.3. Outward-pointing vector fields As a preparation of next section on orientations, we introduce the notions of inward-pointing and outward-pointing vectors. For a point in \partial M, it is intuitively evident that the tangent vectors in T _ pM can be classified into three classes: those tangent to the boundary, those pointing inward, and those pointing outward.

The following proposition gives another characterization which is usually much easier to check.

From this proposition, a tangent vector v for M is inward-pointing iff -v is outward-pointing.

The proof is not difficult. Let (U,\varphi) be a chart centered at p. We write v=v^i\frac{\partial}{\partial x^i}| _ p in terms of coordinate basis and define locally a curve \gamma with coordinate functions (tv^i). Then it can be verified that \gamma(0)=p and \gamma'(0)=v. If the x _ n-component v^n is positive, this curve is defined in M only if tv^n\geqslant0. For t\in[0,\varepsilon), v is inward-pointing. If v^n is negative, then tv^n\leqslant0. For t\in(-\varepsilon,0], v is outward-pointing. For v^n=0, this is a curve in \partial M, so \gamma\in T _ p\partial M.

Conversely, for an inward-pointing vector, since v^n is given by the limit of \gamma^n(t)/t and \gamma^n(t)\geqslant0 by the fact that the curve is in M, we have v^n\geqslant0, and the equality cannot hold because v is required not to be in T _ {p}\partial M. Similarly, an outward-pointing vector implies v^n<0. For any tangent vector v, it obviously satisfies v^n=0.
2. Orientations In calculus, it is known that line integrals and surface integrals depend on the orientation of the curve or surface. The orientation determines the sign of the integral. Similarly, we have to consider the orientation of the manifold if we want to integrate on it. We will explore first the orientation of a finite-dimensional vector space, which turns out to be a choice of two equivalence classes. Then possibly we can give an orientation to a manifold by considering the orientations of all tangent spaces.
2.1. Orientations of vector spaces Usually the orientation of a vector space is first considered. For \mathbb R^1, there is a positive direction and a negative direction. For \mathbb R^2, there is a clockwise rotation and a counterclockwise rotation. For \mathbb R^3 an orientation is either right-handed or left-handed. From these examples, we can see an orientation is specified by an ordered basis. For \mathbb R^1, the positive direction is (E _ 1). For \mathbb R^2, the counterclockwise rotation is (E _ 1,E _ 2) and the clockwise rotation is (E _ 2,E _ 1). For \mathbb R^3, the right-handed orientation is (E _ 1,E _ 2,E _ 3) and the left-handed orientation is (E _ 2,E _ 1,E _ 3).

For an arbitrary n-dimensional vector space, it makes no sense to say an ordered basis represents a "correct" orientation. What we can do is to figure out if two ordered bases have the same orientation. For two ordered bases (E _ 1,\dots,E _ n) and (\widetilde{E} _ 1,\dots,\widetilde{E} _ n), they are consistently oriented if the transition matrix between them has positive determinant. To be specific, the transition matrix is the unique matrix T such that (E _ 1,\dots,E _ n)=(\widetilde{E} _ 1,\dots,\widetilde{E} _ n)T, and the two bases are consistently oriented if \det (T)>0.

It is easy to check that this is actually an equivalence relation on the set of all ordered bases for V. It therefore partitions ordered bases into two equivalence classes. Each equivalence class is called an orientation of V. The orientation that (E _ 1,\dots,E _ n) determines is denoted by [E _ 1,\dots,E _ n]. If V is oriented, then any ordered basis in the given orientation is said to be oriented or positively oriented. If \dim V=0, V has no basis, and we define an orientation to be simply a choice of one of the numbers \pm1.

Orientations and n-covectors

For any n-dimensional vector space V, the space \Lambda^n(V^\ast) is 1-dimensional. This facts suggests another way to specify an orientation to V.

Let \omega\in \Lambda^n(V^\ast). Suppose (E _ i) and (\widetilde{E} _ i) are two ordered bases for V, and T:V\to V is a linear map determined by \widetilde{E} _ i=TE _ i for all i. By Proposition 6 in Chapter 6,
\[
\omega(\widetilde{E} _ 1,\dots,\widetilde{E} _ n)=(\det B)\omega(E _ 1,\dots,E _ n).
\]It follows that (\widetilde{E} _ i) is consistently oriented with (E _ i) iff \omega(\widetilde{E} _ 1,\dots,\widetilde{E} _ n) and \omega(E _ 1,\dots,E _ n) have the same sign. The set of all bases such that \omega takes positive values on them is then exactly an equivalence class. Hence, \omega determines an orientation on V.

Moreover, two n-covectors determines the same orientation iff each is a positive multiple of the other. Therefore choosing an orientation for V is equivalent to choose one of the two components of \Lambda^n(V^\ast)\setminus\{0\}.
2.2. Orientations of manifolds Having defined the orientations of vector spaces, we can orient a manifold M by orient the tangent space at each point in M. Of course this kind of pointwise orientation makes no sense because the orientations of nearby points have no relations to each other, so there is something more to be done.

On an open set U\subseteq M, we call an n-tuple of rough vector fields (X _ 1,\dots,X _ n) a frame if at every point p these vector fields provide an ordered basis for the tangent space T _ pM. A local frame about p\in M is a frame defined on some neighborhood of p and a global frame is a frame defined on M.

If p\in M has a neighborhood U on which the pointwise orientation is represented by a frame that is continuous, then the orientation is said to be continuous at p. Formally, the pointwise orientation is continuous at p if there exist continuous vector fields X _ 1,\dots, X _ n on U such that the pointwise orientation is exactly [X _ 1| _ q,\dots, X _ n| _ q] for all q\in U. A pointwise orientation is continuous on M if it is continuous at every point in M.

The Möbius strip is a famous example that is not orientable.

Let M be oriented. A frame (E _ i) for TM is said to be oriented or positively oriented if (E _ 1| _ p,\dots, E _ n| _ p) is a positively oriented basis for T _ pM at each point p. A negatively oriented frame is defined analogously.

Note that a continuous pointwise orientation need not be represented by a continuous global frame; it just need to be locally representable by a continuous local frame. With some knowledge of algebraic topology, any global frame that represents the pointwise orientation of the sphere \mathbb S^2 is necessarily discontinuous, although it can be shown the sphere is orientable.

We have the following proposition, in which the connectedness is required. If M is not connected, we can consider each connected component separately. Obviously the proposition cannot hold without connectedness.

To see the correctness of this proposition, we shall prove two orientations on a connected component are the same. Suppose such two orientations are given. Define a function f, which takes value 1 if they are the same and takes -1 if they are opposite. We shall prove f is locally constant, and then by some common analysis of topology (that is, to prove f^{-1}(c) for any c is both open and closed set), f is constant on the connected component, which means on it the two orientations are always the same or always opposite.

To prove f is locally constant. Let p\in M. Then there exists a connected neighborhood U of p on which the two orientations are represented by two continuous vector fields X _ i,Y _ i: [X _ 1,\dots, X _ n] and [Y _ 1,\dots,Y _ n]. Let T be a matrix function, such that (Y _ 1,\dots, Y _ n)=(X _ 1,\dots,X _ n)T. It can be seen that \det (A) is continuous and always nonzero, so by the intermediate value theorem, \det (A) on the connected U is everywhere positive or everywhere negative. Thus f\equiv1 or f\equiv-1 on U, which means f is locally constant on every connected components of M.

Orientations and differential forms

In practice, it is more useful to specify an orientation with an differential form, called an orientation form, than using definition. Previously we orient a vector field with an n-covector, so an n-form provides a pointwise orientation. To make it continuous, we need the n-form to be smooth and nowhere-vanishing.

First assume \omega is a smooth and nowhere-vanishing n-form on M. At each point p\in M, we can choose an ordered basis (X _ 1| _ p,\dots,X _ n| _ p) for T _ pM such that \omega(X _ 1| _ p,\dots,X _ n| _ p)>0. Then let (U,(x^i)) be a connected coordinate neighborhood of p. On U, \omega=f\, \mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n} for a nowhere-vanishing function f. Since f is continuous and nowhere-vanishing on a connected set, f>0 on U or f<0 on U. This implies that there is a chart (U,(\tilde x^i)) such that on U we have (\mathrm d\tilde x^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d\tilde x^{n})(X _ 1,\dots,X _ n)>0: If f>0 on U we just take the original coordinate maps; If f<0 on U, we take the coordinate maps (-x^1,x^2,\dots,x^n). For the chart (U,(\tilde x^i)), we will still use (U,(x^i)) to denote it.

On the chart (U,(x^i)), if we write X _ j=T^i _ j\frac{\partial}{\partial x^i}, then
\[
(\mathrm d x^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d x^{n})(X _ 1,\dots,X _ n =\det(T)>0.
\]Thus [X _ 1,\dots,X _ n] is always [\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}] on U, indicating the continuity of the pointwise orientation at p. Since p was arbitrary, the pointwise orientation is continuous on M.

Conversely, assume [X _ 1,\dots,X _ n] is a continuous pointwise orientation on M, which means for every point p\in M there is a neighborhood W containing p such that the pointwise orientation is represented by a continuous frame (Y _ 1,\dots,Y _ n). Let (U,(x^i)) be a connected coordinate neighborhood of p contained in W. Suppose Y _ j=T^i _ j\frac{\partial}{\partial x^i}, then
\[
(\mathrm d x^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d x^{n})(Y _ 1,\dots,Y _ n)=\det(T),
\]which is never zero. As a continuous nowhere-vanishing function on a connected set, it is everywhere positive or everywhere negative on U. By reversing x^1 if necessary, we may assume the function (\mathrm d x^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d x^{n})(Y _ 1,\dots,Y _ n) is always positive on U.

On U since [X _ 1,\dots,X _ n]=[Y _ 1,\dots,Y _ n], they are consistently oriented, i.e., the transition matrix between X basis and Y basis on U always has positive determinant, so for this p\in M, the chart (U,(x^i)) satisfies
\[
(\mathrm d x^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d x^{n})(X _ 1,\dots,X _ n)>0.
\]
Let \{(U _ \alpha,(x _ \alpha^i)\} be a collection of these charts that covers M, and let \{\phi _ \alpha\} be a partition of unity subordinate to the open cover \{U _ \alpha\}. We define a smooth n-form on M by \omega=\sum _ \alpha\phi _ \alpha\, \mathrm d x^{1} _ \alpha{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d x _ \alpha^{n}. At every point it is locally finite so it is well defined. Finally, we have
\[
\omega _ p(X _ 1| _ p,\dots,X _ n| _ p)=\sum _ \alpha\phi _ \alpha(p)\, (\mathrm d x _ \alpha^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d x _ \alpha^{n}) _ p(X _ 1| _ p,\dots,X _ n| _ p)>0.
\]which means \omega is a smooth and nowhere-vanishing n-form on M.

Orientations and atlases

It is also practical to use an atlas to specify an orientation to a manifold. An atlas is said to be consistently oriented if for any pair of charts (U _ \alpha,\varphi _ \alpha),(U _ \beta,\varphi _ \beta) in the atlas the transition map \varphi _ \beta\circ\varphi _ \alpha^{-1} has positive determinant everywhere on \varphi _ \alpha(U _ \alpha\cap U _ \beta).

In addition, an coordinate chart on an oriented manifold is said to be oriented or positively oriented if the coordinate frame (\frac{\partial}{\partial x^i}) is positively oriented, and negatively oriented if the coordinate frame (\frac{\partial}{\partial x^i}) is negatively oriented.

First, suppose M has a consistently oriented atlas and let p\in M. For any chart (U,(x^i)) containing p an orientation for T _ pM is given by [\frac{\partial}{\partial x^1}| _ p,\dots,\frac{\partial}{\partial x^n}| _ p]. If there are two charts containing p, the transition matrix between their respective coordinate frames is the Jacobian of the transition map, which has positive determinant by hypothesis, so they give the same orientation at p. Thus the atlas provides a pointwise orientation at each point, which is continuous because each point is in the coordinate neighborhood (U,(x^i)) on which the orientation is represented by a continuous frame [\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}].

Conversely, let [X _ 1,\dots,X _ n] be an orientation on M and p\in M. We have known p has a coordinate neighborhood (U,(x^i)) such that on U we have (\mathrm d x^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d x^{n})(X _ 1,\dots,X _ n)>0. It can be proved that such these charts for all p\in M form a consistently oriented atlas: For two charts (U,(x^i)),(V,(y^i)) from this atlas, it holds that
\begin{gather*}
(\mathrm d x^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d x^{n})(X _ 1,\dots,X _ n)>0, \\
(\mathrm d y^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d y^{n})(X _ 1,\dots,X _ n)>0.
\end{gather*}Therefore on U\cap V the Jacobian \det(\frac{\partial y^i}{\partial x^j}) is always positive, implying the atlas is consistently oriented.

The proof is done. In addition, if we define an equivalence relation on the set of oriented atlases, by the following rule: two atlases \{(U _ \alpha,\varphi _ \alpha)\} and \{(V _ \beta,\psi _ \beta)\} are equivalent if the transition maps \psi _ \beta\circ \varphi^{-1} _ \alpha have positive Jacobian determinant for all \alpha,\beta; then it can be proved that the equivalent classes have a bijection to the orientations to the manifold. This means we can also specify an orientation of M by an equivalence class of oriented atlases.

Orientation-preserving and reversing

Let manifolds M,N be oriented and suppose F:M\to N is a local diffeomorphism.

Proof. Assume F is orientation preserving. By definition, for every p \in M, the differential \mathrm{d}F _ p maps positively oriented bases of T _ pM to positively oriented bases of T _ {F(p)}N. Let (\frac{\partial }{\partial x^i}| _ p) be a positively oriented coordinate chart for T _ pM and (\frac{\partial }{\partial y^i}| _ p) be a positively oriented coordinate chart for T _ {F(p)}N. In Chapter 2 we have derived
\begin{align*}
& \begin{bmatrix}
\mathrm dF _ p(\frac{\partial}{\partial x^1}| _ p)&\cdots&\mathrm dF _ p(\frac{\partial}{\partial x^n}| _ p)
\end{bmatrix} \\
={} & \begin{bmatrix}
\frac{\partial }{\partial y^1}\Big| _ {F(p)}&\cdots&\frac{\partial }{\partial y^m}\Big| _ {F(p)}
\end{bmatrix}
\begin{bmatrix}
\frac{\partial F^1}{\partial x^1}(p) & \cdots &\frac{\partial F^1}{\partial x^n}(p) \\
\vdots & \ddots & \vdots \\
\frac{\partial F^m}{\partial x^1}(p) & \cdots &\frac{\partial F^m}{\partial x^n}(p)
\end{bmatrix}
\end{align*}The two bases above are consistently oriented, implying the determinant of Jacobian is positive. Conversely, the Jacobian of F with positive determinant implies the two bases are consistently oriented, so F is oriented preserving.

The equivalence between (3) and the other two can be seen through:
\[
F^\ast(u\mathrm d y^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d y^{n})=(u\circ F)\det\Big(\frac{\partial F^j}{\partial x^i}\Big)\mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n}.
\]Here the pullback formula from Proposition 16 in Chapter 6 is used. In local oriented coordinates, \omega can be represented in the form of u\; \mathrm d y^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm d y^{n} where u is a positive function.

Thus F^\ast\omega is a positive multiple of \mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n} if and only if the Jacobian has positive determinant.

From this proposition we can construct a pullback orientation induced by a local diffeomorphism F. It is not difficult to see the following proposition holds.

For a manifold with boundary, the previous discussion on orientations go through word for word, except for the case in which \dim M=1. We used the trick of reversing x _ 1, which does not work for boundary charts when \dim M=1, since the last coordinate in \mathbb H^n has to be nonnegative.

For \dim M>1, we can still use a smooth and nowhere-vanishing n-form or use a consistently oriented atlas to orient a manifold with boundary.

By Theorem 6, for an oriented manifold M with boundary, the boundary \partial M is an embedded submanifold. It turns out that in this case \partial M is also orientable. With an outward-pointing vector field along \partial M, we can use an orientation on M to induce an orientation on \partial M.

First, we show the existence of an outward-pointing vector field along the boundary. A vector field along \partial M is a map from \partial M to TM such that the image of each point p is a tangent vector in T _ pM.

Proof. Let \{U _ \alpha\} be an open cover of \partial M by boundary charts (U _ \alpha,(x^i _ \alpha)). On each chart define the local outward vector fields by X _ \alpha=-\frac{\partial}{\partial x^n _ \alpha}, which is smooth and along U _ \alpha\cap \partial M. Then we use a partition of unity \{\rho _ \alpha\} _ \alpha on \partial M subordinate to the open cover \{U _ \alpha\cap \partial M\}. Define the global vector field on \partial M by X=\sum _ \alpha \rho _ \alpha X _ \alpha. It can be seen that this is a smooth vector field along \partial M which is outward-pointing.

Interior multiplication

We will use interior multiplication to define an orientation on \partial M. As well as wedge product and pullback, the interior multiplication is also an important operation on differential forms.

In other words, i _ v\omega is obtained from \omega by inserting v into the first slot.

We calculate the left hand side directly. Evaluated on (v _ 2,\dots,v _ k), by Proposition 11 in Chapter 6 it is
\[
(\omega^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\omega^{k})(v,v _ 2,\dots,v _ k)=\det
\begin{bmatrix}
\omega^1(v) & \omega^2(v _ 2) &\cdots & \omega^1(v _ k) \\
\vdots & \vdots & & \vdots \\
\omega^k(v) & \omega^k(v _ 2) &\cdots & \omega^k(v _ k)
\end{bmatrix}.
\]Expand along the first column and use Proposition 11 in Chapter 6 again. The right hand side in the lemma is obtained.

The first part is immediate by the fact that an alternating tensor takes value zero if there is a repeated argument. For the second part, since both sides are linear in \omega and in \eta, we may assume
\[\omega=\omega^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\omega^{k},\quad \omega=\omega^{k+1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\omega^{k+l},\]where \omega^i are all covectors. We write i _ v(\omega\wedge\eta)=i _ v(\omega^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\omega^{k+l}), and use the previous lemma. It can be seen that the k+l terms obtained have the first k terms being (i _ v\omega)\wedge\eta and last l terms being (-1)^k\omega\wedge(i _ v\eta).

It can be proved if \omega is a smooth form and X is a smooth vector field, then i _ X\omega is smooth.

Induced orientation on boundary

With an orientation form on M and an outward-pointing vector field along \partial M, we can induce an orientation on \partial M, called induced orientation, or Stokes orientation on \partial M.

Let \iota _ {\partial M}:\partial M\hookrightarrow M be the inclusion map. Recall that in Chapter 3, we showed when S is an embedded submanifold of M, we can identify T _ pS with \mathrm d\iota _ p(T _ pS) under the linear map \mathrm d\iota _ p. Therefore we can think of T _ p\partial M as a certain linear subspace of T _ pM and pretend \mathrm d\iota _ p be an identity map.

Recall that in Chapter 6 we defined the pullback F^\ast\omega by F:M\to N of a differential k-form in N to be a differential n-form in M:
\[
(F^\ast\omega) _ p(v _ 1,\dots,v _ k)=\omega _ {F(p)}(\mathrm dF _ p(v _ 1),\dots,\mathrm dF _ p(v _ k)).
\]Now let F be the inclusion map \iota _ {\partial M}. We can see \iota _ {\partial M}^\ast:\Omega^k(M)\to\Omega^k(\partial M) is really just restriction to T(\partial M). Now we can state the following proposition:

Proof. Let \omega be an orientation form for M. Then \iota^\ast _ {\partial M}(N\mathbin{\lrcorner}\omega) is an (n-1)-form on \partial M.

Given any basis (E _ 1,\dots,E _ {n-1}) for T _ p\partial M, the fact that N is outward-pointing implies that (N _ p,E _ 1,\dots,E _ {n-1}) is a basis for T _ pM. Since \omega is nowhere-vanishing,
\[
[\iota^\ast _ {\partial M}(N\mathbin{\lrcorner}\omega)] _ p(E _ 1,\dots,E _ {n-1})=\omega _ p(N _ p,E _ 1,\dots,E _ {n-1})\neq0.
\]Hence \iota^\ast _ {\partial M}(N\mathbin{\lrcorner}\omega) does not vanish at p, implying it is nowhere-vanishing.

It follows that \iota^\ast _ {\partial M}(N\mathbin{\lrcorner}\omega) is an orientation form for \partial M.

For each p\in \partial M, (E _ 1,\dots, E _ {n-1}) is an oriented basis for T _ {p}\partial M iff (N _ p,E _ 1,\dots,E _ {n-1}) is an oriented basis for T _ pM.

Finally, we shall prove the induced orientation is independent of the choice of N. Let p\in\partial M and (x^i) be boundary coordinates for M on a neighborhood of p. If N,\widetilde N are two different outward-pointing vector fields along \partial M, then by Proposition 8 their last components N^n(p) and \widetilde N^n(p) are both negative.

Both (N _ p,\frac{\partial}{\partial x^1}| _ p,\dots,\frac{\partial}{\partial x^{n-1}}| _ p) and (\widetilde N _ p,\frac{\partial}{\partial x^1}| _ p,\dots,\frac{\partial}{\partial x^{n-1}}| _ p) are bases for T _ pM, and the determinant of the transition matrix between them is just N^n(p)\mathbin/\widetilde N^n(p)>0. Thus, both bases determine the same orientation for T _ pM, so N and \widetilde N determine the same orientation for T _ p\partial M.

Example: We can find the induced orientation on \partial \mathbb H^n. An orientation form for the standard orientation on \mathbb H^n is \omega=\mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n}, and a smooth outward-pointing vector field on \mathbb H^n is -\frac{\partial}{\partial x^n}. Then an orientation form for the boundary orientation on \partial\mathbb H^n is given by
\begin{align*}
\iota _ {\partial\mathbb H^n}^\ast\Big(-\frac{\partial}{\partial x^n}\mathbin\lrcorner\omega\Big) & =(-1)^{n-1}\mathrm dx^n\Big({-}\frac{\partial}{\partial x^n}\Big)\mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n-1} \\
& =(-1)^n\mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n-1}.
\end{align*}
Thus, the boundary orientation on \mathbb H^1=\{0\} is -1. On \partial\mathbb H^2 it is \mathrm dx^1, the usual orientation on \mathbb R, and on \partial \mathbb H^3 it is -\, \mathrm dx^1\wedge\mathrm dx^2, the clockwise orientation in the (x _ 1,x _ 2)-plane.
3. Integration of differential forms In calculus, if we want to integrate a function over a bounded subset of \mathbb R^n, usually the volume of the rectangle \prod[a _ i,b _ i] is first defined, that is \prod(b _ i-a _ i), and then we calculate the product of some value of f and the volume of some small rectangles. The value of the integral is obtained by taking their sum and the limit. However in manifolds we have no such "rectangles" and their volume naturally. Instead we shall consider the "signed volume" of the parallelepiped spanned by some tangent vectors. Differential forms turn out to have just the right properties of "signed volume" of the these parallelepiped.

As usual we begin here by defining integrals of n-forms in Euclidean spaces. Since every n-form can be written as \omega=f(x) \mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n} for a unique function f(x), n-forms on \mathbb R^n can be identified with functions on \mathbb R^n. Therefore the ordinary integration theory in calculus can transfer easily to integration of differential forms.

Let D be a subset of \mathbb R^n and f:D\to\mathbb R be a function. If we want to take multiple integral of f over D, there are some integrability conditions to be satisfied. For D, we require it should be a bounded set and the boundary \partial D should have measure zero. For f, we simply consider the case in which f is continuous and bounded on D. If these conditions are satisfied, f is integrable over D. Such D, i.e., a bounded set with boundary of measure zero, is called a domain of integration.

Let D be a domain of integration, and \omega=f\mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n} be an n-form on D such that f is continuous and bounded on D. The integral of \omega over D is defined to be
\[
\int _ D\omega=\int _ Df\mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n}=\int _ D f\, \mathrm dx^1\cdots\mathrm dx^n.
\]
If \omega is supported on an open subset U of \mathbb R^n or \mathbb H^n. The integral of \omega over U is defined to be the integral over D that is any domain of integration containing \operatorname{supp}(\omega) and \omega is extended by zero. Clearly this definition does not depend on what domain D is chosen.

Let us see how the integral of an n-form on an open set transforms under a change of variables.

The proof can be found at the section end.

Now we can consider the manifold case. Let M be a manifold with or without boundary. If an n-form \omega is compactly supported in the domain of a single oriented chart (U,\varphi), the integral can be defined naturally with the above results.

It can be seen (\varphi^{-1})^\ast\omega is a compactly supported n-form on the open subset U\subseteq\mathbb R^n or \mathbb H^n, so its integral is defined as discussed above.

With the help of a partition of unity, we can define the integral over an entire manifold. Now suppose \omega is compactly supported on M. Let \{U _ i\} be a finite open cover of \operatorname{supp}(\omega) by domains of positively or negatively oriented charts, and let \{\psi _ i\} be a subordinate partition of unity.

Each term in the definition is well defined since \psi _ i\omega is compactly supported in U _ i. Hence this integral is well defined if it is independent of the open cover and the partition of unity.

Integral over 0-manifolds

Actually we implicitly assume \dim M>0 previously. The integral of a compactly supported 0-form f over an oriented 0-manifold M is defined to be the sum
\[
\int _ Mf=\sum _ {p\in M}\pm f(p).
\]The positive sign is taken if the orientation at p is positive, and the negative sign is taken if it is negative. The assumption that f is compactly supported implies that there are only finitely many nonzero terms in this sum.

Integral over a submanifold

Let S\subseteq M be an oriented submanifold with or without boundary. Suppose \dim S=k, \omega is a k-form on M which is compactly supported in S, and \iota:S\hookrightarrow M is the inclusion. The integral of \omega over S is just
\[
\int _ S\omega=\int _ S\iota _ S^\ast\omega.
\]
Properties of integrals

Suppose \omega,\eta are compactly supported on M. The following propositions are some basic properties of the integrals over M.

Local parameterizations

Our definition is useless for computation due to the partition of unity in definition. The following proposition is usually useful.

We now prove the previous statements (except for Proposition 31).

First consider Proposition 22. It can be proved easily with the following lemmas.

Denote the standard coordinates on E by (y^1,\dots,y^n) and those on D by (x^1,\dots,x^n). Then \omega=f\mathrm dy^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dy^{n}. If G is orientation-preserving, then by the change of variables formula and the formula for pullbacks of n-forms,
\begin{align*}
\int _ E\omega & =\int _ E f\mathrm dy^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dy^{n} \\
& =\int _ D(f\circ G)|\det(JG)|\mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n} \\
& =\int _ D(f\circ G)(\det(JG))\mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n} \\
& =\int _ D G^\ast\omega.\end{align*}
If G is orientation-reversing, only a negative sign is introduced when the absolute value signs are removed.

To extend this result to compactly supported n-forms defined on open subsets, we need the next lemma to construct a domain of integration.

Since K is a compact set, there are finite open balls or half balls B _ 1,\dots, B _ m covering K and contained in U. Then D=B _ 1\cup\dots\cup B _ m can be verified to be the required domain of integration. (Note that the boundary of a ball has zero measure.)

Return back to Proposition 22. It is a fact (which is not difficult to prove) that a Lipschitz continuous function maps a set of measure zero to a set of measure zero. Since a smooth function on a compact set is Lipschitz continuous, it maps a set of measure zero to a set of measure zero. Let E be an open domain of integration constructed in the previous lemma such that \operatorname{supp}(\omega)\subseteq E\subseteq\overline{E}\subseteq V. We have known diffeomorphisms take interiors to interiors and boundaries to boundaries. Then it is not difficult to see that G^{-1} maps the topological boundary of E=G(D) to the topological boundary of E, indicating D=G^{-1}(E)\subseteq U is an open domain of integration containing \operatorname{supp}(G^\ast\omega). The result follows from the first lemma.

Next we prove the integral of \omega over M is well-defined. Suppose (U,\varphi) and (\widetilde{U},\widetilde{\varphi}) are two charts and \omega is compactly supported in U\cap \widetilde U. If both charts are positively oriented or both are negatively oriented, then the transition map \widetilde \varphi\circ\varphi:\varphi(U\cap\widetilde U)\to\widetilde \varphi(U\cap\widetilde U) is an orientation-preserving diffeomorphism. By Proposition 22 and Lemma 21 in Chapter 6,
\begin{align*}
\int _ {\widetilde{\varphi}(\widetilde{U})} (\widetilde{\varphi}^{-1} )^{\ast}\omega&=\int _ {\widetilde{\varphi} (U\cap\widetilde{U} )} (\widetilde{\varphi}^{-1} )^\ast\omega=\int _ {\varphi (U\cap\widetilde{U} )} (\widetilde{\varphi}\circ\varphi^{-1} )^\ast (\widetilde{\varphi}^{-1} )^\ast\omega\\&=\int _ {\varphi (U\cap\widetilde{U} )} (\varphi^{-1} )^{\ast} (\widetilde{\varphi} )^{\ast} (\widetilde{\varphi}^{-1} )^{\ast}\omega\\
&=\int _ {\varphi (U )} (\varphi^{-1} )^{\ast}\omega.
\end{align*}If the charts are oppositely oriented, the transition map is orientation-reversing, so the above equality differs by an extra negative sign, which is taken into account in definition. Thus in either case they give the same definition.

Now we prove the integral of \omega over M is well defined when \omega is compactly supported on M. Let \{U _ i\},\{\widetilde{U} _ i\} be finite open covers of \operatorname{supp}(\omega) by domains of positively or negatively oriented charts, and let \{\psi _ i\},\{\widetilde{\psi} _ j\} be subordinate partitions of unity respectively. Then
\begin{gather*}
\int _ M\psi _ i\omega =\int _ M\Big(\sum\nolimits _ j\widetilde{\psi} _ j\Big)\psi _ i\omega=\sum\nolimits _ j\int _ M\widetilde{\psi} _ j\psi _ i\omega, \\
\sum\nolimits _ i\int _ M\psi _ i\omega=\sum _ {i,j}\int _ M\widetilde{\psi} _ j\psi _ i\omega=\sum _ {i,j}\int _ M\psi _ i\widetilde{\psi} _ j\omega =\sum\nolimits _ j\int _ M\widetilde{\psi} _ j\omega.
\end{gather*}Thus the integral of \omega over M is well defined.

We then prove the properties of integrals. The linearity is due to the linearity of multiple integrals and pullbacks. For the integral over -M, by definition it is enough to consider the case in which \omega is compactly supported in a chart (U,\varphi) in M, because any compactly supported n-form
on M can be written as a finite sum of such forms by means of a partition of unity. The orientation of this chart in -M is opposite with the one in M. If this is a positively oriented chart in M, then it is a negatively oriented chart in -M, so by definition
\[
\int _ {-M} \omega = -\int _ {\varphi(U)} (\varphi^{-1})^\ast \omega = -\int _ M \omega.
\]If this is a negatively oriented chart in M, then it is a positively oriented chart in -M, so
\[
\int _ {-M} \omega = \int _ {\varphi(U)} (\varphi^{-1})^\ast \omega = -\int _ M \omega.
\]In either case, the integral over -M is equal to the negative of the integral over M.

If \omega is a positively oriented orientation form on M, it can be proved the integral is positive. In this case, if (U,\varphi) is a positively oriented chart, then (\varphi^{-1})^\ast\omega is f\mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n} where f>0 on U; and if it is a negative oriented chart f is negative: f<0 on U. Hence each term defining the integral of \omega over M is nonnegative, with at least one strictly positive term, thus proving the proposition.

To prove the diffeomorphism invariance, we can still assume \omega is compactly supported in a single positively or negatively oriented chart. Assume (U,\varphi) is such a chart that is positively oriented. When F is oriented-preserving, it can be seen that (F^{-1}(U),\varphi\circ F) is an oriented chart on N whose domain contains \operatorname{supp}F^\ast\omega, and the result follows immediately from Proposition 22. The cases in which the chart is negatively oriented or F is orientation-reversing are discussed similarly.

4. Stokes' theorem Our goal of this section is to state and prove Stokes' theorem, which is the central result in the integration theory on manifolds.

As usual, the integral of \omega over \partial M means to integrate \iota _ {\partial M}^\ast\omega.

Proof

We prove this theorem in three steps. First we consider the special case in which M=\mathbb H^n or \mathbb R^n. Next we consider the manifold case but \omega is compactly supported in the domain of a single chart. Finally the general case is proved.

Step 1. Suppose M=\mathbb H^n. The (n-1)-form \omega is compactly supported, so there is a positive number R such that
\[
\operatorname{supp}\omega\subseteq\prod _ {i=1}^{n-1}[-R,R]\times[0,R]=:A.
\]We can write \omega in standard coordinates and calculate the exterior derivative:
\begin{align*}
\omega & =\sum _ {i=1}^n\omega _ i\mathrm dx^1{\wedge}{\cdots}{\operatorname{\wedge}{}}\widehat{\mathrm dx^i}{\wedge}{\cdots}{\operatorname{\wedge}{}} \mathrm dx^n \\
\mathrm d\omega & =\sum _ {i=1}^n\mathrm d\omega _ i\wedge \mathrm dx^1{\wedge}{\cdots}{\operatorname{\wedge}{}}\widehat{\mathrm dx^i}{\wedge}{\cdots}{\operatorname{\wedge}{}} \mathrm dx^n
\\&=\sum _ {i,j=1}^n\frac{\partial\omega _ i}{\partial x^j}\mathrm dx^j\wedge \mathrm dx^1{\wedge}{\cdots}{\operatorname{\wedge}{}}\widehat{\mathrm dx^i}{\wedge}{\cdots}{\operatorname{\wedge}{}} \mathrm dx^n\\
&=\sum _ {i=1}^n(-1)^{i-1}\frac{\partial\omega _ i}{\partial x^i}\mathrm dx^1{\wedge}{\cdots}{\operatorname{\wedge}{}} \mathrm dx^n.
\end{align*}Thus
\begin{align*}
\int _ {\mathbb{H}^n}\mathrm d\omega&=\sum _ {i=1}^n(-1)^{i-1}\int _ A\frac{\partial\omega _ i}{\partial x^i}\mathrm dx^1{\wedge}{\cdots}{\operatorname{\wedge}{}} \mathrm dx^n\\
&=\sum _ {i=1}^n(-1)^{i-1}\int _ 0^R\mathrm dx^n\cdots\int _ {-R}^R\frac{\partial\omega _ i}{\partial x^i}(x)\, \mathrm dx^1\\
&=\sum _ {i=1}^n(-1)^{i-1}\int\, \mathrm dx^1\cdots\widehat{\mathrm dx^i}\cdots\mathrm dx^n\int _ {-R}^R\frac{\partial\omega _ i}{\partial x^i}(x)\, \mathrm dx^i \\
&=\sum _ {i=1}^n(-1)^{i-1}\int\Big(\omega _ i(x)\Big|^{x^i=R} _ {x^i=-R\text{ or }0}\Big)\, \mathrm dx^1\cdots\widehat{\mathrm dx^i}\cdots\mathrm dx^n \\
&=(-1)^n\int\omega _ n(x^1,\dots,x^{n-1},0)\, \mathrm dx^1\cdots\mathrm dx^{n-1}.
\end{align*}In the above calculation, we changed the order of integration in each term so as to do the x^i integration first and then applied the fundamental theorem of calculus. Since R is chosen such that \omega=0 when x^i=\pm R (for i\neq n), the only term that might not be zero is the one for which i=n.

We turn to calculate the integral of \omega:
\[
\int _ {\partial \mathbb H^n}\omega=\sum _ {i=1}^{n}\int _ {A\cap\partial\mathbb H^n}\omega _ i\mathrm dx^1{\wedge}{\cdots}{\operatorname{\wedge}{}}\widehat{\mathrm dx^i}{\wedge}{\cdots}{\operatorname{\wedge}{}} \mathrm dx^n.
\]On \partial \mathbb H^n, it can be seen that \mathrm dx^1{\wedge}{\cdots}{\operatorname{\wedge}{}}\widehat{\mathrm dx^i}{\wedge}{\cdots}{\operatorname{\wedge}{}} \mathrm dx^n(x^1,\dots,x^{n-1},0)=0. Hence the only nonzero term is the one for which i=n. Recall that the induced orientation form for \partial\mathbb H^n is (-1)^n\mathrm dx^{1}{\wedge}{\cdots}{\operatorname{\wedge}{}}\mathrm dx^{n-1}. Therefore
\begin{align*}
\int _ {\partial \mathbb H^n}\omega & =\int _ {A\cap\partial\mathbb H^n}\omega _ n(x^1,\dots,x^{n-1},0)\mathrm dx^1{\wedge}{\cdots}{\operatorname{\wedge}{}} \mathrm dx^{n-1} \\
& =(-1)^n\int\omega _ n(x^1,\dots,x^{n-1},0)\, \mathrm dx^1\cdots\mathrm dx^{n-1}\\
& = \int _ {\mathbb H^n}\mathrm d\omega.
\end{align*}
Next, suppose M=\mathbb R^n. In this case \operatorname{supp}(\omega)\subseteq[-R,R]^n. By some similar calculation, the integral of \mathrm d\omega is zero. The integral of \omega over the boundary \varnothing is also zero. Thus they are equal.

Step 2. Now let M be an arbitrary manifold with boundary, but consider an (n-1)-form \omega that is compactly supported in the domain of a single chart (U,\varphi). Suppose it is a positively oriented boundary chart. By the above results,
\[
\int _ M\mathrm d\omega=\int _ {\mathbb H^n}(\varphi^{-1})^\ast\, \mathrm d\omega=\int _ {\mathbb H^n}\mathrm d((\varphi^{-1})^\ast\omega)=\int _ {\partial \mathbb H^n}(\varphi^{-1})^\ast\omega,
\]where \partial\mathbb H^n is given the induced orientation.

Suppose p\in \partial M and (\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^{n}}) is the coordinate frame. The orientation of (\frac{\partial}{\partial x^1}| _ p,\dots,\frac{\partial}{\partial x^{n-1}}| _ p) for T _ p\partial M is determined by the orientation of (-\frac{\partial}{\partial x^n}| _ p,\frac{\partial}{\partial x^1}| _ p,\dots,\frac{\partial}{\partial x^{n-1}}| _ p) for T _ pM, which is just the orientation of (-\frac{\partial}{\partial x^n}| _ {\varphi(p)},\frac{\partial}{\partial x^1}| _ {\varphi(p)},\dots,\frac{\partial}{\partial x^{n-1}}| _ {\varphi(p)}) for T _ p\mathbb H^n. Thus the orientation of (\frac{\partial}{\partial x^1}| _ p,\dots,\frac{\partial}{\partial x^{n-1}}| _ p) for T _ p\partial M is determined by the orientation of (\frac{\partial}{\partial x^1}| _ {\varphi(p)},\dots,\frac{\partial}{\partial x^{n-1}}| _ {\varphi(p)}) for T _ p\partial\mathbb H^n. We conclude that \varphi| _ {U\cap \partial M} is an orientation-preserving diffeomorphism onto \varphi(U)\cap\partial \mathbb H^n, so
\[
\int _ M\mathrm d\omega=\int _ {\partial \mathbb H^n}(\varphi^{-1})^\ast\omega=\int _ {\partial M}\omega.
\]
If the chart (U,\varphi) is negatively oriented, (-\frac{\partial}{\partial x^n}| _ p,\frac{\partial}{\partial x^1}| _ p,\dots,\frac{\partial}{\partial x^{n-1}}| _ p) for T _ pM has the opposite orientation to (-\frac{\partial}{\partial x^n}| _ {\varphi(p)},\frac{\partial}{\partial x^1}| _ {\varphi(p)},\dots,\frac{\partial}{\partial x^{n-1}}| _ {\varphi(p)}) for T _ p\mathbb H^n. Therefore \varphi| _ {U\cap \partial M} is an orientation-reversing diffeomorphism, implying
\[
\int _ M\mathrm d\omega=-\int _ {\partial \mathbb H^n}(\varphi^{-1})^\ast\omega=\int _ {\partial M}\omega.
\]
If (U,\varphi) is an interior chart, just replace \mathbb H^n above with \mathbb R^n and apply the same argument.

Step 3. Let \omega be an arbitrary compactly supported smooth (n-1)-form. Choosing a cover of \operatorname{supp}(\omega) by finitely many domains \{U _ i\} _ {i=1}^k of charts, and choosing a subordinate partition of unity \{\psi _ i\} _ {i=1}^k, we obtain
\begin{align*}
\int _ {\partial\boldsymbol{M}}\omega&=\sum _ {i=1}^k\int _ {\partial M}\psi _ i\omega=\sum _ {i=1}^k\int _ M\mathrm d(\psi _ i\omega)=\sum _ {i=1}^k\int _ M\mathrm d\psi _ i\wedge\omega+\psi _ i\, \mathrm d\omega\\
&=\int _ {M}\mathrm d\Big(\sum _ {i=1}^k\psi _ {i}\Big)\wedge\omega+\int _ {M}\Big(\sum _ {i=1}^k\psi _ {i}\Big)d\omega=0+\int _ {M}\mathrm d\omega.
\end{align*}
The theorem is proved.

Stokes' theorem can be used to deduce the fundamental theorem of calculus and line integral and the major theorems in vector analysis, namely Green's theorem, the divergence theorem and the classical Stokes' theorem.

More discussion on vector analysis can be found in other materials. [https://gaomj.cn/pdfjs/web/viewer.html?file=pma.pdf#54]

(Manifolds with corners)

Stokes' theorem can be generalized further to manifolds with "corners" such as triangles, squares and cubes. The model for the type of corners we are concerned with is the space
\[
\overline{\mathbb R} _ +^n=\{(x^1,\dots,x^n)\in\mathbb R^n\mid x^1\geqslant0,\dots,x^n\geqslant0\}.
\]It can be shown that this is homeomorphic to \mathbb H^n, so from the topological point of view if we define smooth manifolds with corners with it there is no difference between manifolds with boundary and manifolds with corners. The difference is in the smooth structure, because in dimensions greater than 1, the compatibility condition for charts with corners is different from that for boundary charts.

The boundary of \overline{\mathbb R} _ +^n is the set of points at which at least one coordinate vanishes. The points in \overline{\mathbb R} _ +^n at which more than one coordinate vanishes are called its corner points. It is said that we can prove the invariance of corner points.

Similar to the definition of manifolds with boundary, the definition of manifolds with corners, as well as its tangent vectors, differential forms, integrals, can be given easily. The details are omitted.

The boundary of a smooth manifold with corners is in general not a smooth manifold with corners (e.g., think of the boundary of a cube). However note that
\begin{gather*}
\partial\overline{\mathbb R} _ +^n=H _ 1\cup\dots\cup H _ n, \\
H _ i=\{(x^1,\dots,x^n)\in\overline{\mathbb R} _ +^n\mid x^i=0\}.
\end{gather*}The boundary of \overline{\mathbb R} _ +^n is a union of some (n-1)-dimensional manifold with corners contained in the subspace defined by x^i=0. When defining the integral of an (n-1)-form \omega over \partial M that is compactly supported in a single chart (U,\varphi) with corners, we can define separately over these H _ i:
\[
\int _ {\partial M}\omega=\sum _ {i=1}^{n}\int _ {H _ i}(\varphi^{-1})^\ast\omega.
\]Here H _ i is given the induced orientation as part of the boundary of the set where x^i\geqslant0. We then can define the integral of a general compactly supported \omega with the help of a partition of unity, as usual.

The corresponding Stokes' theorem is stated without proof, which is almost identical to the original one.