Integration of differential forms

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Chapter 10: Integration of differential forms

Integration can be studied on many levels. In Chap. 6, the theory was developed for reasonably well-behaved functions on subintervals of the real line. ln Chap. 11 we shall encounter a very highly developed theory of integration that can be applied to much larger classes of functions, whose domains are more or less arbitrary sets, not necessarily subsets of \mathbb{R}^n. The present chapter is devoted to those aspects of integration theory that are closely related to the geometry of Euclidean spaces, such as the change of variables formula, line integrals, and the machinery of differential forms that is used in the statement and proof of then-dimensional analogue of the fundamental theorem of calculus, namely Stokes' theorem.

Contents
Contents
 1.  Integration
 2.  Primitive mappings
 3.  Partitions of unity
 4.  Change of variables
 5.  Differential forms
 6.  Simplexes and chains
 7.  Stokes' theorem
 8.  Closed forms and exact forms
 9.  Vector analysis

1. Integration

A priori, this definition of the integral depends on the order in which the k integrations are carried out. However, this dependence is only apparent. To prove this, let us introduce the temporary notation L(f) for the integral and L^\prime (f) for the result obtained by carrying out the k integrations in some other order.

If h(\mathbf{x})=h _ 1(x _ 1)\dots h _ k(x _ k), where h _ j\in\mathcal{C}([a _ j,b _ j]), then L(h)=L^\prime (h). If \mathcal{A} is the set of all finite sums of such functions h, it follows that L(g) = L^\prime (g) for all g \in \mathcal{A}. Also, \mathcal{A} is an algebra of functions on I^k to which the Stone-Weierstrass theorem applies.

Put V=\prod(b _ i-a _ i). If f\in\mathcal{C}(I^k) and \varepsilon>0, there exists g\in\mathcal{A} such that |f-g|<\varepsilon/V, where |f| is defined as \max |f(\mathbf{x})|\, (\mathbf{x}\in I^k). Then |L(f-g)|<\varepsilon, |L^\prime (f-g)|<\varepsilon, and since L(f)-L^\prime (f)=L(f-g)+L^\prime (g-f), we conclude that |L(f)-L^\prime (f)|<2\varepsilon. Therefore L(f)=L^\prime (f).

If f is a continuous function with compact support, let I^k be any k-cell which contains the support of f, and define \int _ {\mathbb{R}^k}f=\int _ {I^k}f. The integral so defined is evidently independent of the choice of I^k, provided only that I^k contains the support of f.

It is now tempting to extend the definition of the integral over \mathbb{R}^k to functions which are limits (in some sense) of continuous functions with compact support. We do not want to discuss the conditions under which this can be done; the proper setting for this question is the Lebesgue integral. We shall merely describe one very simple example which will be used in the proof of Stokes' theorem.

{Example}
Let Q^k be the k-simplex which consists of all points \mathbf{x}=(x _ 1,\dots,x _ k) in \mathbb{R}^k for which x _ 1+\dots+x _ k\leqslant1 and x _ i\geqslant0 for i=1,\dots,k. If k=3, for example, Q^k is a tetrahedron, with vertices at \mathbf{0},\mathbf{e} _ 1,\mathbf{e} _ 2,\mathbf{e} _ 3. If f\in\mathcal{C}(Q^k), extend f to a function on I^k by setting f(\mathbf{x}) = 0 off Q^k, and define \int _ {Q^k}f=\int _ {I^k}f. Here I^k is the "unit cube" defined by 0\leqslant x _ i\leqslant1\, (1\leqslant i\leqslant k).

Since f may be discontinuous on I^k, the existence of the integral needs proof, which we omit here.

Our next goal is the change of variables formula stated in Theorem theorem1041. To facilitate its proof, we first discuss so-called primitive mappings, and partitions of unity. Primitive mappings will enable us to get a clearer picture of the local action of a \mathcal{C}^\prime-mapping with invertible derivative, and partitions of unity are a very useful device that makes it possible to use local information in a global setting.
2. Primitive mappings

If g is differentiable at some point \mathbf{a}\in E, so is \mathbf{G}. The matrix [\alpha _ {ij}] of the operator \mathbf{G^\prime (a)} has \left(D _ {1} g\right)(\mathbf{a}), \ldots,\left(D _ {m} g\right)(\mathbf{a}), \ldots,\left(D _ {n} g\right)(\mathbf{a}) as its mth row. For j\neq m, we have \alpha _ {jj}=1 and \alpha _ {ij}=0 if i\neq j. The Jacobian of \mathbf G at \mathbf{a} is thus given by J _ \mathbf{G}(\mathbf{a})=\det[\mathbf{G^\prime (a)}]=(D _ mg)(\mathbf{a}), and we see that \mathbf{G^\prime (a)} is invertible if and only if (D _ mg)(\mathbf{a})\neq0.

In the proof that follows, we shall use the projections P _ 0,\dots,P _ n in \mathbb{R}^n, defined by P _ 0\mathbf{x}=\mathbf{0} and P _ m\mathbf{x}=x _ 1\mathbf{e} _ 1+\dots+x _ m\mathbf{e} _ m for 1\leqslant m\leqslant n. Thus P _ m is the projection whose range and null space are spanned by \{\mathbf{e} _ 1,\dots,\mathbf{e} _ m\} and \{\mathbf{e} _ {m+1},\dots,\mathbf{e} _ n\}, respectively.

Briefly, (1016) represents \mathbf{F} locally as a composition of primitive mappings and flips.

In Definition 9.7, we introduce the notion of diffeomorphism. If we ignore the flips in the theorem, we can state the theorem with the notion of elementary diffeomorphism.

By Theorem theorem1021, we have:

If we ignore the flips in the theorem, we can prove Theorem theorem1022 as follows.

It is trivial when \mathbf{f} is an elementary diffeomorphism. Assume the mapping can be locally represented as the product of k-1 elementary diffeomorphisms when the mapping changes k-1 coordinates. Now suppose \mathbf{f} changes k coordinates. Since we ignore the flips, we may assume the first k coordinates are changed. Then the matrix [D _ jf _ i]\in\mathbb{R}^{k\times k} is invertible. Ignoring the flips, we may assume the submatrix [D _ jf _ i]\in\mathbb{R}^{(k-1)\times(k-1)} is invertible. Due to the continuity, there is an open set containing \mathbf{0} in which the determinant of this submatrix is non-zero. We define mapping \mathbf{G}:\, \mathbf{x}\mapsto(f _ 1(\mathbf{x}),\dots,f _ {k-1}(\mathbf{x}),x _ k,\dots,x _ n). Then the Jacobian of \mathbf{G} in a neighbourhood of \mathbf{0} is non-zero. By inverse function theorem, \mathbf{G} restricted to a neighbourhood of \mathbf{0}, say U, is a diffeomorphism.

Define mapping \mathbf{H} of an open set V=\mathbf{G}(U) by \mathbf{H}=\mathbf{F}\circ \mathbf{G}^{-1}. V contains \mathbf{u} _ 0=\mathbf{G(0)}, and \mathbf{H} is a diffeomorphism. We have
\[\mathbf{H(u)}=(u _ 1,\dots,u _ {k-1},f _ k\circ \mathbf{G}^{-1}(\mathbf{u}),u _ {k+1},\dots,u _ n).\]
\mathbf{H} is an elementary diffeomorphism. By induction hypothesis, \mathbf{G}=\mathbf{G} _ {k-1}\circ\dots\circ\mathbf{G} _ 1, where each \mathbf{G} _ j\, (j=1,\dots,k-1) is elementary diffeomorphism. Hence \mathbf{F=H\circ G=H}\circ \mathbf{G} _ {k-1}\circ\dots\circ\mathbf{G} _ 1. Our induction hypothesis holds with k in place of k-1.

{Proof.}We now turn to Theorem theorem1021. Put \mathbf{F=F} _ 1. Assume 1\leqslant m\leqslant n-1, make the following induction hypothesis (which evidently holds for m =1):

V _ m is a neighbourhood of \mathbf{0}, \mathbf{F} _ m\in\mathcal{C}^\prime (V _ m), \mathbf{F} _ m(\mathbf{0})=\mathbf{0}, \mathbf{F}^\prime _ m(\mathbf{0}) is invertible, and P _ {m-1}\mathbf{F} _ m(\mathbf{x})=P _ {m-1}(\mathbf{x})\, (\mathbf{x}\in V _ m).

(P _ 0,\dots,P _ n are projections in \mathbb{R}^n, defined by P _ 0\mathbf{x}=\mathbf{0} and P _ {m} \mathbf{x}=x _ {1} \mathbf{e} _ {1}+\cdots+x _ {m} \mathbf{e} _ {m} for 1\leqslant m\leqslant n. Thus P _ m is the projection whose range and null space are spanned by \{\mathbf{e} _ 1, \dots , \mathbf{e} _ m\} and \{\mathbf{e} _ {m+1}, \dots , \mathbf{e} _ n\}, respectively.)

We have
\[\mathbf{F} _ {m}(\mathbf{x})=P _ {m-1} \mathbf{x}+\sum _ {i=m}^{n} \alpha _ {i}(\mathbf{x}) \mathbf{e} _ {i},\]
where \alpha _ m,\dots,\alpha _ n are real \mathcal{C}^\prime-functions in V _ m. Hence
\[\mathbf{F} _ {m}^{\prime}(\mathbf{0}) \mathbf{e} _ {m}=\sum _ {i=m}^{n}\left(D _ {m} \alpha _ {i}\right)(\mathbf{0}) \mathbf{e} _ {i}.\]
Since \mathbf{F}^\prime _ m(\mathbf{0}) is invertible, the left side is not \mathbf{0}, and therefore there is a k such that m \leqslant k\leqslant n and (D _ m\alpha _ k)(\mathbf{0})\neq0. Let B _ m be the flip that interchanges m and this k (if k=m, B _ m is the identity) and define \mathbf{G} _ {m}(\mathbf{x})=\mathbf{x}+\left[\alpha _ {k}(\mathbf{x})-x _ {m}\right] \mathbf{e} _ {m} \, \left(\mathbf{x} \in V _ {m}\right). Then \mathbf{G} _ m\in\mathcal{C}^\prime (V _ m), \mathbf{G} _ m is primitive, and \mathbf{G^\prime (0)} is invertible, since (D _ m\alpha _ k)(\mathbf{0})\neq0.

The inverse function theorem shows therefore that there is an open set U _ m with 0\in U _ m \subset V _ m, such that \mathbf{G} _ m is a 1-1 mapping of U _ m onto a neighborhood V _ {m+1} of \mathbf{0}, in which \mathbf{G} _ m^{-1} is continuously differentiable. Define \mathbf{F} _ {m + 1} by
\[\mathbf{F} _ {m+1}(\mathbf{y})=B _ {m} \mathbf{F} _ {m} \circ \mathbf{G} _ {m}^{-1}(\mathbf{y}) \quad\left(\mathbf{y} \in V _ {m+1}\right).\]
Then \mathbf{F} _ {m+1}\in\mathcal{C}^\prime (V _ {m+1}), \mathbf{F} _ {m+1}\mathbf{(0)=0}, and \mathbf{F}^\prime _ {m+1}(\mathbf{0}) is invertible (by the chain rule). Also, for \mathbf{x}\in U _ m,
\[P _ {m} \mathbf{F} _ {m+1}\left(\mathbf{G} _ {m}(\mathbf{x})\right) =P _ {m} B _ {m} \mathbf{F} _ {m}(\mathbf{x}) =P _ {m}\left[P _ {m-1} \mathbf{x}+\alpha _ {k}(\mathbf{x}) \mathbf{e} _ {m}+\cdots\right] =P _ {m-1} \mathbf{x}+\alpha _ {k}(\mathbf{x}) \mathbf{e} _ {m}=P _ {m} \mathbf{G} _ {m}(\mathbf{x})\]
so that
\[P _ {m} \mathbf{F} _ {m+1}(\mathbf{y})=P _ {m} \mathbf{y} \quad\left(\mathbf{y} \in V _ {m+1}\right).\]
Our induction hypothesis holds therefore with m + 1 in place of m.

Since B _ mB _ m=I, we have \mathbf{F} _ {m}(\mathbf{x})=B _ {m} \mathbf{F} _ {m+1}\left(\mathbf{G} _ {m}(\mathbf{x})\right) \, \left(\mathbf{x} \in U _ {m}\right). If we apply this with m = 1, \dots , n - 1, we successively obtain
\[\mathbf{F}=\mathbf{F} _ {1}=B _ {1} \mathbf{F} _ {2} \circ \mathbf{G} _ {1}=B _ {1} B _ {2} \mathbf{F} _ {3} \circ \mathbf{G} _ {2} \circ \mathbf{G} _ {1}=\cdots=B _ {1} \cdots B _ {n-1} \mathbf{F} _ {n} \circ \mathbf{G} _ {n-1} \circ \cdots \circ \mathbf{G} _ {1}\]
in some neighborhood of \mathbf{0}. Note that \mathbf{F} _ n is primitive. This completes the proof.
3. Partitions of unity

Because of (3), \{\psi _ i\} is called a partition of unity, and (2) is sometimes exp1essed by saying that \{\psi _ i\} is subordinate to the cover \{V _ \alpha\}.

The point of this corollary is that it furnishes a representation of f as a sum of continuous functions \psi _ i f if with "small" supports.

Proof. Associate with each \mathbf{x}\in K an index \alpha(\mathbf{x}) so that \mathbf{x}\in V _ {\alpha(\mathbf{x})}. Then there are open balls B(\mathbf{x}) and W(\mathbf{x}), centered at \mathbf{x}, with \overline{B(\mathbf{x})} \subset W(\mathbf{x}) \subset \overline{W(\mathbf{x})} \subset V _ {\alpha(\mathbf{x})}.

Since K is compact, there are points \mathbf{x} _ 1,\dots,\mathbf{x} _ s in K such that K\subset B(\mathbf{x} _ 1)\cup\dots\cup B(\mathbf{x} _ s). There are functions \varphi _ 1\dots,\varphi _ s\in \mathcal{C}(\mathbb{R}^n), such that \varphi _ i(\mathbf{x})=1 on B(\mathbf{x} _ i), \varphi _ i(\mathbf{x})=0 outside W(\mathbf{x} _ i), and 0\leqslant\varphi _ i(\mathbf{x})\leqslant1 on \mathbb{R}^n. Define \psi _ 1=\varphi _ 1 and \psi _ {i+1}=(1-\varphi _ 1)\dots(1-\varphi _ i)\varphi _ {i+1} for i=1,\dots,s-1.

Properties (1) and (2) are clear. The relation
\begin{equation}\label{1029}
\psi _ {1}+\cdots+\psi _ {i}=1-\left(1-\varphi _ {1}\right) \cdots\left(1-\varphi _ {i}\right)
\end{equation}
is trivial for i=1. If it holds for some i<s, addition of \psi _ {i+1} and (1029) yields (1029) with i+1 in place of i. It follows that
\[\sum _ {i=1}^{s} \psi _ {i}(\mathbf{x})=1-\prod _ {i=1}^{s}\left[1-\varphi _ {i}(\mathbf{x})\right] \quad\left(\mathbf{x} \in\mathbb R^{n}\right).\]
If \mathbf{x}\in K, then \mathbf{x}\in B(\mathbf{x} _ i) for some i, hence \varphi _ i(\mathbf{x})=1, and the product in the above formula is 0. This proves (3).
4. Change of variables We can now describe the effect of a change of variables on a multiple integral. For simplicity, we confine ourselves here to continuous functions with compact support, although this is too restrictive for many applications.

We recall that J _ T is the Jacobian of T. The assumption J _ T(\mathbf{x})\neq0 implies, by the inverse function theorem, that T^{-1} is continuous on T(E), and this ensures that the integrand on the right of (1031) has compact support in E.

The appearance of the absolute value of J _ T(\mathbf{x}) in (1031) may call for a comment. Take the case k = 1, and suppose T is a 1-1 \mathcal{C}^\prime-mapping of \mathbb{R}^1 onto \mathbb{R}^1. Then J _ T(\mathbf{x})=T^\prime (x); and if T is increasing, we have \int _ {\mathbb R^{1}} f(y)\, d y=\int _ {\mathbb R^{1}} f(T(x)) T^\prime (x)\, d x for all continuous f with compact support. But if T decreases, then T^\prime (x)<0; and if f is positive in the interior of its support, the left side is positive and the right side is negative. A correct equation is obtained if T^\prime is replaced by |T^\prime |.

The point is that the integrals we are now considering are integrals of functions over subsets of \mathbb{R}^k, and we associate no direction or orientation with these subsets. We shall adopt a different point of view when we come to integration of differential forms over surfaces.

Proof. It follows from the remarks just made that (1031) is true if T is a primitive \mathcal{C}^\prime-mapping, and Theorem theorem1011 shows that (1031) is true if T is a linear mapping which merely interchanges two coordinates.

If the theorem is true for transformations P, Q, and if S(\mathbf{x}) = P(Q(\mathbf{x})), then
\[\int f(\mathbf{z})\, d \mathbf{z} =\int f(P(\mathbf{y}))\left|J _ {P}(\mathbf{y})\right|\, d \mathbf{y}=\int f(P(Q(\mathbf{x})))\left|J _ {P}(Q(\mathbf{x}))\right|\left|J _ {Q}(\mathbf{x})\right|\, d \mathbf{x}=\int f(S(\mathbf{x}))\left|J _ {S}(\mathbf{x})\right|\, d \mathbf{x},\]
since
\[J _ {P}(Q(\mathbf{x})) J _ {Q}(\mathbf{x})=\det P^\prime (Q(\mathbf{x})) \det Q^\prime (\mathbf{x}) =\det P^\prime (Q(\mathbf{x})) Q^\prime (\mathbf{x})=\det S^\prime (\mathbf{x})=J _ {S}(\mathbf{x}),\]
by the multiplication theorem for determinants and the chain rule. Thus the theorem is also true for S.

Each point \mathbf{a}\in E has a neighborhood U \subset E in which
\[T(\mathbf{x})=T(\mathbf{a})+B _ {1} \cdots B _ {k-1} \mathbf{G} _ {k} \circ \mathbf{G} _ {k-1} \circ \cdots \circ \mathbf{G} _ {1}(\mathbf{x}-\mathbf{a}),\]
where \mathbf{G} _ i and B _ i are as in Theorem theorem1021. Setting V=T(U), it follows that (1031) holds if the support of f lies in V. Thus: Each point \mathbf{y}\in T(E) lies in an open set V _ \mathbf{y}\subset T(E) such that (1031) holds for all continuous functions whose support lies in V _ \mathbf{y}.

Now let f be a continuous function with compact support K \subset T(E). Since \{V _ \mathbf{y}\} covers K, Corollary in last section shows that f=\sum\psi _ if, where each \psi _ i is continuous, and each \psi _ i has its support in some V _ \mathbf{y}. Thus (1031) holds for each \psi _ if, and hence also for their sum f.
5. Differential forms We shall now develop some of the machinery that is needed for the n-dimensional version of the fundamental theorem of calculus which is usually called Stokes' theorem. The original form of Stokes' theorem arose in applications of vector analysis to electromagnetism and was stated in terms of the curl of a vector field. Green's theorem and the divergence theorem are other special cases. These topics are briefly discussed at the end of the chapter.

It is a curious feature of Stokes' theorem that the only thing that is difficult about it is the elaborate structure of definitions that are needed for its statement. These definitions concern differential forms, their derivatives, boundaries, and
orientation. Once these concepts are understood, the statement of the theorem is very brief and succinct, and its proof presents little difficulty.

Up to now we have considered derivatives of functions of several variables only for functions defined in open sets. This was done to avoid difficulties that can occur at boundary points. It will now be convenient, however, to discuss differentiable functions on compact sets. We therefore adopt the following convention:

To say that \mathbf{f} is a \mathcal{C}^\prime-mapping (or a \mathcal{C}^{\prime\prime}-mapping) of a compact set D \subset \mathbb{R}^k into \mathbb{R}^n means that there is a \mathcal{C}^\prime-mapping (or a \mathcal{C}^{\prime\prime}-mapping) \mathbf{g} of an open set W\subset \mathbb{R}^k into \mathbb{R}^n such that D\subset W and such that \mathbf{g(x) = f(x)} for all \mathbf{x}\in D.

We shall confine ourselves to the simple situation in which D is either a k-cell or the k-simplex Q^k described in Example in Section 10.1. The reason for this is that we shall have to integrate over D, and we have not yet discussed integration over more complicated subsets of \mathbb{R}^k. It will be seen that this restriction on D (which will be tacitly made from now on) entails no significant loss of generality in the resulting theory of differential forms.

We stress that k-surfaces in E are defined to be mappings into E, not subsets of E. This agrees with our earlier definition of curves. In fact, 1-surfaces are precisely the same as continuously differentiable curves.

Note that the right side of (1035) is an integral over D, as defined in Section 10.1 and that (1035) is the definition of the symbol \int _ \Phi\omega.

A k-form \omega is said to be of class \mathcal{C}^\prime or \mathcal{C}^{\prime\prime} if the functions a _ {i _ {1}\cdots i _ {k}} in (1034) are all of \mathcal{C}^\prime or \mathcal{C}^{\prime\prime}.

A 0-form in E is defined to be a continuous function in E.

{Elementary properties} Let \omega,\omega _ 1,\omega _ 2 be k-forms in E. We write \omega _ 1=\omega _ 2 if and only if \omega _ 1(\Phi)=\omega _ 2(\Phi) for every k-surface \Phi in E. In particular, \omega = 0 means that \omega(\Phi)=0 for every k-surface \Phi in E. If c is a real number, then c\omega is the k-form defined by
\[\int _ {\Phi}c\, \omega=c\int _ {\Phi}\omega,\]
and \omega=\omega _ 1+\omega _ 2 means that
\[\int _ {\Phi}\omega=\int _ {\Phi}\omega _ 1+\int _ {\Phi}\omega _ 2\]
for every k-surface \Phi in E. As a special case, note that -\omega is defined that so that \int _ {\Phi}(-\omega)=-\int _ {\Phi}\omega.

Consider a k-form \omega=a(\mathbf{x})\, dx _ {i _ 1}\wedge\dots\wedge dx _ {i _ k} and let \bar\omega be the k-form obtained by interchanging some pair of subscripts in it. We see that \bar\omega=-\omega.

As a special case of this, note that the anticommutative relation
\begin{equation}\label{1042}
dx _ i\wedge dx _ j=-\, dx _ j\wedge dx _ i
\end{equation}
holds for all i and j. In particular, dx _ i\wedge dx _ i=0\, (i=1,\dots,n).

More generally, let us return to \omega, and assume that i _ r=i _ s for some r\neq s. If these two subscripts are interchanged, then \bar \omega = \omega, hence \omega = 0.

In other words, if \omega=a(\mathbf{x})\, dx _ {i _ 1}\wedge\dots\wedge d _ {i _ k}, then \omega=0 unless the subscripts i _ 1,\dots,i _ k are all distinct.

If \omega is as in (1034), the summands with repeated subscripts can be omitted without changing \omega.

It follows that 0 is the only k-form in any open set of \mathbb{R}^n, if k>n.

The anticommutativity expressed by (1042) is the reason for the inordinate amount of attention that has to be paid to minus signs when studying differential forms.

{Basic k-forms}If i _ 1,\dots,i _ k are integers such that 1\leqslant i _ 1<i _ 2<\dots<i _ k\leqslant n, and if I is the ordered k-tuple \{i _ 1,\dots,i _ k\}, then we call I an increasing k-index, and we use the brief notation dx _ I=dx _ {i _ 1}\wedge\dots\wedge dx _ {i _ k}. These forms dx _ I are the so-called basic k-forms in \mathbb{R}^n.

It is not hard to verify that there are precisely n!/k!(n- k)! basic k-forms in \mathbb{R}^n, we shall make no use of this, however.

Much more important is the fact that every k-form can be represented in terms of basic k-forms. To see this, note that every k-tuple \{j _ 1,\dots,j _ k\} of distinct integers can be converted to an increasing k index J by a finite number of inter changes of pairs; each of these amounts to a multiplication by -1; hence dx _ {j _ 1}\wedge\dots\wedge dx _ {j _ k}=\varepsilon(j _ 1,\dots,j _ k)dx _ J where \varepsilon(j _ 1,\dots,j _ k) is 1 or -1, depending on the number of interchanges that are needed. In fact, it is easy to see that \varepsilon(j _ 1,\dots,j _ k) is equal to the signature of the permutation j _ 1,\dots,j _ k (which is used in the definition of determinant).

If every k-tuple in (1034) is converted to an increasing k-index, then we obtain the so-called standard presentation of \omega:
\[\omega=\sum _ I b _ I(\mathbf{x})\, dx _ I.\]
The summation extends over all increasing k-indices I. [Of course, every increasing k-index arises from many (from k!, to be precise) k-tuples. Each b _ I may thus be a sum of several of the coefficients that occur in (1034).]

The following uniqueness theorem is one of the main reasons for the introduction of the standard presentation of a k-form.

Note that the analogous statement would be false for sums such as (1034), since, for example, dx _ 1\wedge dx _ 2+dx _ 2\wedge dx _ 1=0.

Proof. Assume, to reach a contradiction, that b _ J(\mathbf{v}) > 0 for some \mathbf{v} \in E and for some increasing k-index J = \{j _ 1, \dots ,j _ k\}. Since b _ J is continuous, there exists h > 0 such that b _ J(\mathbf{x}) > 0 for all \mathbf{x} \in \mathbb{R}^n whose coordinates satisfy | x _ i- v _ i | \leqslant h. Let D be the k-cell in \mathbb{R}^k such that \mathbf{u} \in D if and only if |u _ r|\leqslant h for r = 1, \dots , k. Define \Phi(\mathbf{u})=\mathbf{v}+\sum _ {r=1}^{k}u _ r\mathbf{e} _ {j _ r}\, (\mathbf{u}\in D). Then \Phi is a k-surface in E, with parameter domain D, and b _ J(\Phi(\mathbf{u})) > 0 for every \mathbf{u} \in D.

We claim that \int _ \Phi\omega=\int _ Db _ J(\Phi(\mathbf{u}))\, d\mathbf{u}. Since the right side is positive, it follows that \omega(\Phi)\neq 0. Hence this gives our contradiction. To prove it, apply (1035) to the presentation \omega=\sum _ I b _ I(\mathbf{x})\, dx _ I. More specifically, compute the Jacobians that occur in (1035). We have: \dfrac{\partial(x _ {j _ 1},\dots,x _ {j _ k})}{\partial(u _ 1,\dots,u _ k)}=1. For any other increasing k-index I \neq J, the Jacobian is 0, since it is the determinant of a matrix with at least one row of zeros.

{Products of basic k-forms}Suppose I=\{i _ 1,\dots,i _ p\}, J=\{j _ 1,\dots,j _ q\} where 1\leqslant i _ 1<\dots<i _ p\leqslant n and 1\leqslant j _ 1<\dots<j _ q\leqslant n. The product of the corresponding basic forms dx _ I and dx _ J in \mathbb{R}^n is a p+q-form in \mathbb{R}^n, denoted by the symbol dx _ I\wedge dx _ J, and defined by dx _ I\wedge dx _ J=dx _ {i _ 1}\wedge\dots\wedge dx _ {i _ p}\wedge dx _ {j _ 1}\wedge\dots\wedge dx _ {j _ q}.

If I and J have an element in common, then the previous discussion shows that dx _ I\wedge dx _ J = 0.

If I and J have no element in common, let us write [I, J] for the increasing (p + q)-index which is obtained by arranging the members of I\cup J in increasing order. Then dx _ {[I,J]} is a basic (p + q)-form. We claim that
\begin{equation}\label{1053}
dx _ I\wedge dx _ J=(-1)^\alpha dx _ {[I,J]}
\end{equation}
where \alpha is the number of differences j _ t-i _ s the are negative. (The number of positive differences is thus pq- \alpha.)

Note that the right side of (1053) is the standard presentation of dx _ I\wedge dx _ J.

Next, let K=(k _ 1,\dots,k _ r) be an increasing r-index in \{1, \dots , n\}. We shall use (1053) to prove that
\begin{equation}\label{1055}
(dx _ I\wedge dx _ J)\wedge dx _ K=dx _ I\wedge(dx _ j\wedge dx _ K).
\end{equation}

If any two of the sets I, J, K have an element in common, then each side of (1055) is 0, hence they are equal. So let us assume that I, J, K are pairwise disjoint. Let [I, J, K] denote the increasing (p + q + r )-index obtained from their union. Associate \beta with the ordered pair (J, K) and \gamma with the ordered pair (I, K) in the way that \alpha was associated with (I, J) in (1053). The left side of (1055) is then (-1)^{\alpha}\, d x _ {[I, J]} \wedge d x _ {K}=(-1)^{\alpha}(-1)^{\beta+\gamma}\, d x _ {[I, J, K]} by two applications of (1053), and the right side of (1055) is (-1)^{\beta}\, d x _ {I} \wedge d x _ {[J, K]}=(-1)^{\beta}(-1)^{\alpha+\gamma}\, d x _ {[I, J, K]}. Hence (1055) is correct.

{Multiplication}Suppose \omega and \lambda are p- and q-forms, respectively, in some open set E\subset\mathbb{R}^n, with standard presentations \omega=\sum _ I b _ I(\mathbf{x})\, dx _ I, \lambda=\sum _ J c _ J(\mathbf{x})\, dx _ J, where I and J range over all increasing p-indices and over all increasing q-indices taken from the set \{1,\dots , n\}. Their product, denoted by the symbol \omega\wedge\lambda, is defined to be
\[\omega \wedge \lambda=\sum _ {I, J} b _ {I}(\mathbf{x}) c _ {J}(\mathbf{x})\, d x _ {I} \wedge d x _ {J}.\]
In this sum, I and J range independently over their possible values, and dx _ I\wedge dx _ J is as in previous definition. Thus \omega\wedge\lambda is a (p + q)-form in E.

It Is quite easy to see (we omit the details) that the distributive laws
\begin{align*}
(\omega _ 1+\omega _ 2)\wedge\lambda & =\omega _ 1\wedge\lambda+\omega _ 2\wedge\lambda \\
\omega\wedge(\lambda _ 1+\lambda _ 2) & =(\omega\wedge\lambda _ 1)+(\omega\wedge\lambda _ 2)
\end{align*}
holds, with respect to the addition defined previously. If these distributive laws are combined with (1055), we obtain the associative law
\[(\omega\wedge\lambda)\wedge\sigma=\omega\wedge(\lambda\wedge\sigma)\]
for arbitrary forms \omega,\lambda,\sigma in E.

In this discussion it was tacitly assumed that p \geqslant 1 and q\geqslant 1. The product of a 0-form f with the p-form \omega is simply defined to be the p-form f\omega=\omega f=\sum _ I f(\mathbf{x})b _ I(\mathbf{x})\, dx _ I. It is customary to write f\omega, rather than f\wedge\omega, when f is a 0-form.

{Differentiation} We shall now define a differentiation operator d which associates a (k + 1)-form d\omega to each k-form \omega of class \mathcal{C}^\prime in some open set E\subset\mathbb{R}^n.

A 0-form of class \mathcal{C}^\prime in E is just a real function f \in\mathcal{C}^\prime (E), and we define
\begin{equation}\label{1059}
d f=\sum _ {i=1}^{n}\left(D _ {i} f\right)(\mathbf{x})\, d x _ {i}.
\end{equation}
If \omega=\sum _ I b _ I(\mathbf{x})\, dx _ I is the standard presentation of a k-form \omega, and b _ I \in \mathcal{C}^\prime (E) for each increasing k-index I, then we define
\begin{equation}\label{1060}
d\omega=\sum _ I (db _ I)\wedge dx _ I.
\end{equation}

{Change of variables}Suppose E is an open set in \mathbb{R}^n, T is a \mathcal{C}^\prime-mapping of E into an open set V\subset \mathbb{R}^m, and \omega is a k-form in V, whose standard presentation is \omega=\sum _ I b _ I(\mathbf{y})\, dy _ I. (We use \mathbf{y} for points of V, \mathbf{x} for points of E.)

Let t _ 1,\dots , t _ m be the components of T: y _ i=t _ i(\mathbf{x}). As in (1059),
\begin{equation}\label{1066}
d t _ {i}=\sum _ {j=1}^{n}\left(D _ {j} t _ {i}\right)(\mathbf{x})\, d x _ {j} \quad(1 \leqslant i \leqslant m).
\end{equation}
Thus each dt _ i is a 1-form in E.

The mapping T transforms \omega into a k-form \omega _ T in E, whose definition is
\begin{equation}\label{1067}
\omega _ {T}=\sum _ {I} b _ {I}(T(\mathbf{x}))\, d t _ {i _ {l}} \wedge \cdots \wedge d t _ {i _ {k}}.
\end{equation}
In each summand of (1067), I= \{i _ 1, \dots , i _ k\} is an increasing k-index.

Our next theorem shows that addition, multiplication, and differentiation of forms are defined in such a way that they commute with changes of variables.

Our next objective is Theorem theorem1056. This will follow directly from two other important transformation properties of differential forms, which we state first.

The final result of this section combines the two preceding theorems.

Proof. Theorem theorem1052. (1) follows if (1063) is proved for the special case \omega=f\, dx _ I, \lambda=g\, dx _ J, where f,g\in\mathcal{C}^\prime (E), dx _ I is a basic k-form, and dx _ J is a basic m-form. [If k or m or both are 0, simply omit dx _ I or dx _ J in \omega or \lambda; the proof that follows is unaffected by this.] Then \omega\wedge\lambda=fg\, dx _ I\wedge dx _ J. Let us assume that I and J have no element in common. Then, using (1053),
\[d(\omega \wedge \lambda)=d\left(f g\, d x _ {I} \wedge d x _ {J}\right)=(-1)^{\alpha} d\left(f g\, d x _ {[I, J]}\right).\]
By (1059), d(fg)=f\, dg+g\, df. Hence (1060) gives
\[d(\omega \wedge \lambda) =(-1)^{\alpha}(f\, d g+g\, d f) \wedge d x _ {[I, J]}=(g\, d f+f\, d g) \wedge d x _ {I} \wedge d x _ {J}.\]
Since dg is a 1-form and dx _ I is a k-form, we have dg\wedge dx _ I=(-1)^kdx _ I\wedge dg, by (1042). Hence
\[d(\omega \wedge \lambda)=\left(d f \wedge d x _ {I}\right) \wedge\left(g\, d x _ {J}\right)+(-1)^{k}\left(f\, d x _ {I}\right) \wedge\left(d g \wedge d x _ {J}\right)=(d \omega) \wedge \lambda+(-1)^{k} \omega \wedge d \lambda,\]
which proves (1).

Note that the associative law was used freely.

Let us prove (2) first for a 0-form f\in\mathcal{C}^{\prime\prime}:
\[d^{2} f =d\Big(\sum _ {j=1}^{n}\left(D _ {j} f\right)(\mathbf{x})\, d x _ {j}\Big)=\sum _ {j=1}^{n} d\left(D _ {j} f\right) \wedge d x _ {j}=\sum _ {i, j=1}^{n}\left(D _ {i j} f\right)(\mathbf{x})\, d x _ {i} \wedge d x _ {j}.\]
Since D _ {ij}f=D _ {ji}f and dx _ i\wedge dx _ j=-dx _ j\wedge dx _ i, we see that d^2f=0.

If \omega=fdx _ I, then d\omega=(df)\wedge dx _ I. By (1060), d(dx _ I)=0. Hence (1063) shows that
\[d^2\omega=(d^2f)\wedge dx _ I=0.\]

Theorem theorem1053. Part (1) follows immediately from the definitions. Part (2) is almost as obvious, once we realize that \left(d y _ {i _ {1}} \wedge \cdots \wedge d y _ {i _ {r}}\right) _ {T}=d t _ {i _ {1}} \wedge \cdots \wedge d t _ {i _ {r}} regardless of whether \{i _ 1, \dots , i _ r\} is increasing or not; it holds because the same number of minus signs are needed on each side of it to produce increasing rearrangements.

We turn to the proof of (3). If f is a 0-form of class \mathcal{C}^\prime in V, then f _ T(\mathbf{x})=f(T(\mathbf{x})), df=\sum _ i (D _ if)(\mathbf{y})\, dy _ i. By the chain rule, it follows that
\[d(f _ T)=\sum _ j(D _ jf _ T)(\mathbf{x})\, dx _ j=\sum _ j\sum _ i(D _ if)(T(\mathbf{x}))(D _ jt _ i)(\mathbf{x})\, dx _ j=\sum _ i(D _ if)(T(\mathbf{x}))\, dt _ i=(df) _ T.\]
If dy _ I=dy _ {i _ 1}\wedge\dots\wedge dy _ {i _ k}, then (dy _ I) _ T=dt _ {i _ 1}\wedge\dots\wedge dt _ {i _ k}, and Theorem theorem1052 shows that d((dy _ I) _ T)=0. (This is where the assumption T\in\mathcal{C}^{\prime\prime} is used.) Assume now that \omega=f\, dy _ I. Then
\[\omega _ T=f _ T(x)\, (dy _ I) _ T\]
and the preceding calculation lead to
\[d(\omega _ T)=d(f _ T)\wedge (dy _ I) _ T=(df) _ T\wedge (dy _ I) _ T=((df)\wedge dy _ I) _ T=(d\omega) _ T.\]

The general case of (3) follows from the special case just proved, if we apply (1).

Theorem theorem1054. If \omega and \lambda are forms in W, Theorem theorem1053 shows that ((\omega\wedge\lambda) _ S) _ T=(\omega _ S\wedge\lambda _ S) _ T=(\omega _ S) _ T\wedge(\lambda _ S) _ T and (\omega\wedge\lambda) _ {ST}=\omega _ {ST}\wedge\lambda _ {ST}. Thus if (1071) holds for \omega and for \lambda, it follows that (1071) also holds for \omega\wedge\lambda. Since every form can be built up from 0-forms and 1-forms by addition and multiplication, and since (1071) is trivial for 0-forms, it is enough to prove (1071) in the case \omega=dz _ q, q = 1, \dots , p. (We denote the points of E, V, W by \mathbf{x, y, z}, respectively.)

Let t _ 1,\dots,t _ m be the components of T, let s _ 1, \dots , s _ p be the components of S, and let r _ 1, \dots , r _ p be the components of ST. If \omega = dz _ q, then \omega _ S=ds _ q=\sum _ j(D _ js _ q)(\mathbf{y})\, dy _ j, so that the chain rule implies
\begin{align*}
(\omega _ S) _ T & =\sum _ j(D _ js _ q)(T(\mathbf{x}))\, dt _ j =\sum _ j(D _ js _ q)(T(\mathbf{x}))\sum _ i (D _ it _ j)(\mathbf{x})\, dx _ i \\
& =\sum _ i\Big[\sum _ j(D _ js _ q)(T(\mathbf{x})) (D _ it _ j)(\mathbf{x})\Big]\, dx _ i=\sum _ i(D _ ir _ q)(\mathbf{x})\, dx _ i=dr _ q=\omega _ {ST}.
\end{align*}

Theorem theorem1055. We need only consider the case \omega=a(\mathbf{x})\, dx _ {i _ 1}\wedge\dots\wedge dx _ {i _ k}. If \phi _ 1,\dots,\phi _ n are the components of \Phi, then \omega _ \Phi=a(\Phi(\mathbf{u}))\, d\phi _ {i _ 1}\wedge\dots\wedge d\phi _ {i _ k}. The theorem will follow if we can show that
\[d\phi _ {i _ 1}\wedge\dots\wedge d\phi _ {i _ k}=J(\mathbf{u})\, du _ 1\wedge\dots\wedge du _ k,\quad J(\mathbf{u})=\frac{\partial(x _ {i _ 1},\dots,x _ {i _ k})}{\partial(u _ 1,\dots,u _ k)},\]
since this implies
\[\int _ \Phi\omega=\int _ Da(\Phi(\mathbf{u}))J(\mathbf{u})\, d\mathbf{u}=\int _ \Delta a(\Phi(\mathbf{u}))J(\mathbf{u})\, du _ 1\wedge\dots\wedge du _ k=\int _ \Delta\omega _ \Phi.\]

Let [A] be the k by k matrix with entries \alpha(p,q)=(D _ q\phi _ {i _ p})(\mathbf{u})\, (p,q=1,\dots,k). Then d\phi _ {i _ p}=\sum _ q\alpha(p,q)\, du _ q so that
\[d \phi _ {i _ {1}} \wedge \cdots \wedge d \phi _ {i _ {k}}=\sum \alpha\left(1, q _ {1}\right) \cdots \alpha\left(k, q _ {k}\right)\, d u _ {q _ {1}} \wedge \cdots \wedge d u _ {q _ {k}}.\]
In this last sum, q _ 1, \dots , q _ k range independently over 1, \dots , k. The anticommutative relation (1042) implies that d u _ {q _ {1}} \wedge \cdots \wedge d u _ {q _ {k}}=s(q _ {1}, \ldots, q _ {k})\, d u _ {1} \wedge \cdots \wedge d u _ {k}, where s is the signature of the permutation q _ 1 ,\dots,q _ k (which is used in the definition of determinant). We see that
\[d \phi _ {i _ {1}} \wedge \cdots \wedge d \phi _ {i _ {k}}=\det[A]\, d u _ {q _ {1}} \wedge \cdots \wedge d u _ {q _ {k}};\]
and since J(\mathbf{u})=\det[A], the proof is completed.

Theorem theorem1056. Let D be the parameter domain of \Phi (hence also of T\Phi) and define \Delta as in Theorem theorem1055. Then
\[\int _ {T\Phi} \omega=\int _ {\Delta} \omega _ {T \Phi}=\int _ {\Delta}\left(\omega _ {T}\right) _ {\Phi}=\int _ {\Phi} \omega _ {T}\]
The first of these equalities is Theorem theorem1055, applied to T\Phi in place of \Phi. The second follows from Theorem theorem1054. The third is Theorem theorem1055, with \omega _ T in place of \omega.
6. Simplexes and chains

{Affine simplexes} A mapping \mathbf{f} that carries a vector space X into a vector space Y is said to be affine if \mathbf{f- f(0)} is linear. In other words, the requirement is that
\[\mathbf{f(x)=f(0)}+A\mathbf{x}\]
for some A\in L(X,Y).

An affine mapping of \mathbb{R}^k into \mathbb{R}^n is thus determined if we know \mathbf{f(0)} and \mathbf{f}(\mathbf{e} _ i) for 1 \leqslant i\leqslant k; as usual, \{\mathbf{e} _ 1, \dots , \mathbf{e} _ k\} is the standard basis of \mathbb{R}^k.

We define the standard simplex Q^k to be the set of all \mathbf{u}\in \mathbb{R}^k of the form u=\sum \alpha _ i\mathbf{e} _ i such that \alpha _ i\geqslant0 for i=1,\dots,k and \sum\alpha _ i\leqslant1.

Assume now that \mathbf{p} _ 0,\mathbf{p} _ 1,\dots,\mathbf{p} _ k are points of \mathbb{R}^n. The oriented affine k-simplex \sigma=[\mathbf{p} _ 0,\mathbf{p} _ 1,\dots,\mathbf{p} _ k] is defined to be the k-surface in \mathbb{R}^n with parameter domain Q^k which is given by the affine mapping
\begin{equation}\label{1076}
\sigma\left(\alpha _ {1} \mathbf{e} _ {1}+\cdots+\alpha _ {k} \mathbf{e} _ {k}\right)=\mathbf{p} _ {0}+\sum _ {i=1}^{k} \alpha _ {i}\left(\mathbf{p} _ {i}-\mathbf{p} _ {0}\right).
\end{equation}
Note that \sigma is characterized by
\begin{equation}\label{1077}
\sigma(\mathbf{0})=\mathbf{p} _ 0,\quad \sigma(\mathbf{e} _ i)=\mathbf{p} _ i\quad (1\leqslant i\leqslant k),
\end{equation}
and that
\begin{equation}\label{1078}
\sigma(\mathbf{u})=\mathbf{p} _ 0+A\mathbf{u}\quad(\mathbf{u}\in Q^k)
\end{equation}
where A\in L(\mathbb{R}^k,\mathbb{R}^n) and A\mathbf{e} _ i=\mathbf{p} _ i-\mathbf{p} _ 0 for 1\leqslant i\leqslant k.

We call \mathbf{u} oriented to emphasize that the ordering of the vertices \mathbf{p} _ 0,\mathbf{p} _ 1,\dots,\mathbf{p} _ k is taken into account. If \bar{\sigma}=\left[\mathbf{p} _ {i _ {0}}, \mathbf{p} _ {i _ {1}}, \dots, \mathbf{p} _ {i _ {k}}\right], where \{i _ 0,\dots,i _ k\} is a permutation of the order set \{0,1,\dots,k\}, we adopt the notation \bar\sigma=s(i _ 0,\dots,i _ k)\sigma, where s is the signature of a permutation. Thus \bar\sigma=\pm\sigma, depending on whether s=1 or s=-1. Strictly speaking, having adopted the definition of \sigma, we should not write \bar\sigma=\sigma unless i _ 0=0,\, \dots,\, i _ k=k, even if s(i _ 1,\dots,i _ k)=1; what we have here is an equivalence relation, not an equality. However, for our purposes the notation is justified by Theorem theorem1061.

If \bar\sigma=\varepsilon\sigma (using the above convention) and if \varepsilon=1, we say that \bar\sigma and \sigma have the same orientation; if \varepsilon=-1, \bar\sigma and \sigma are said to have the opposite orientations. Note that we have not defined what we mean by the "orientation of a simplex". What we have defined is a relation between pairs of simplexes having the same set of vertices, the relation being that of "having the same orientation".

There is, however, one situation where the orientation of a simplex can be defined in a natural way. This happens when n = k and when the vectors \mathbf{p} _ i- \mathbf{p} _ 0\, (1 \leqslant i \leqslant k) are independent. In that case, the linear transformation A that appears in (1078) is invertible, and its determinant (which is the same as the Jacobian of \sigma) is not 0. Then \sigma is said to be positively (or negatively) oriented if \det A is positive (or negative). In particular, the simplex [\mathbf{0}, \mathbf{e} _ 1, \dots , \mathbf{e} _ k] in \mathbb{R}^k, given by the identity mapping, has positive orientation.

So far we have assumed that k > 1. An oriented 0-simplex is defined to be a point with a sign attached. We write \sigma = +\mathbf{p} _ 0 or \sigma = -\mathbf{p} _ 0. If \sigma = \varepsilon \mathbf{p} _ 0\, (\sigma = \pm 1) and if f is a 0-form (i.e., a real function), we define \int _ \sigma f=\varepsilon f(\mathbf{p} _ 0).

Proof. For k=0, (1081) follows from the preceding definition. So we assume k \geqslant 1 and assume that \sigma is given by \sigma=[\mathbf{p} _ 0,\mathbf{p} _ 1,\dots,\mathbf{p} _ k].

Suppose 1\leqslant j\leqslant k, and suppose \bar\sigma is obtained from \sigma by interchanging \mathbf{p} _ 0 and \mathbf{p} _ j. Then \varepsilon=-1, and \bar\sigma(\mathbf{u})=\mathbf{p} _ j+B\mathbf{u}\, (\mathbf{u}\in Q^k), where B is the linear mapping of \mathbb{R}^k into \mathbb{R}^n defined by B\mathbf{e} _ i=\mathbf{p} _ i-\mathbf{p} _ j if i\neq j, B\mathbf{e} _ j=\mathbf{p} _ 0-\mathbf{p} _ j. If we write A\mathbf{e} _ i=\mathbf{x} _ i\, (1\leqslant i\leqslant k), where A is given by (1078), the column vector of B (that is, the vectors B\mathbf{e} _ i) are \mathbf{x} _ {1}-\mathbf{x} _ {j},\, \ldots,\, \mathbf{x} _ {j-1}-\mathbf{x} _ {j},\, -\mathbf{x} _ {j},\, \mathbf{x} _ {j+1}-\mathbf{x} _ {j},\, \ldots,\, \mathbf{x} _ {k}-\mathbf{x} _ {j}. If we subtract the jth column from each of the others, none of the determinants in (1035) are affected, and we obtain columns \mathbf{x} _ 1,\dots,\mathbf{x} _ {j-1},-\mathbf{x} _ j, \mathbf{x} _ {j+1},\dots,\mathbf{x} _ k. These differ from those of A only in the sign of the jth column. Hence (1081) holds for this case.

Suppose next that 0<i <j\leqslant k and that \bar\sigma is obtained from \sigma by interchanging \mathbf{p} _ i and \mathbf{p} _ j . Then \bar\sigma(\mathbf{u})=\mathbf{p} _ 0+C\mathbf{u}, where C has the same columns as A, except that the ith and jth columns have been interchanged. This again implies that (1081) holds, since \varepsilon = -1.

The general case follows, since every permutation of \{0, 1, \dots , k\} is a composition of the special cases we have just dealt with.

{Affine chain}An affine k-chain \Gamma in an open set E \subset \mathbb{R}^n is a collection of finitely many oriented affine k-simplexes \sigma _ 1,\dots,\sigma _ r in E. These need not be distinct; a simplex may thus occur in \Gamma with a certain multiplicity.

If \Gamma is as above, and if \omega is a k-form in E, we define \int _ \Gamma\omega=\sum\int _ {\sigma _ i}\omega.

We may view a k-surface \Phi in E as a function whose domain is the collection of all k-forms in E and which assigns the number \int _ \Phi\omega to \omega. Since real-valued functions can be added, this suggests the use of the
notation \Gamma=\sigma _ 1+\dots+\sigma _ r, or, more compactly, \Gamma=\sum\sigma _ i to state the fact that \int _ \Gamma\omega=\sum\int _ {\sigma _ i}\omega holds for every k-form \omega in E.

To avoid misunderstanding, we point out explicitly that the notations introduced above have to be handled with care. The point is that every oriented affine k-simplex \sigma in \mathbb{R}^n is a function in two ways, with different domains and different ranges, and that therefore two entirely different operations of addition are possible. Originally, \sigma was defined as an \mathbb{R}^n-valued function with domain Q^k; accordingly, \sigma _ 1+\sigma _ 2 could be interpreted to be the function \sigma that assigns the vector \sigma _ 1(\mathbf{u}) +\sigma _ 2(\mathbf{u}) to every \mathbf{u}\in Q^k; note that \sigma is then again an oriented affine k-simplex in \mathbb{R}^n! This is not what is meant by \Gamma=\sum\sigma _ i.

For example, if \sigma _ 2=-\sigma _ 1 (that is to say, if \sigma _ 1 and \sigma _ 2 have the same set of vertices but are oppositely oriented) and if \Gamma = \sigma _ 1 + \sigma _ 2, then \int _ \Gamma\omega = 0 for all \omega, and we may express this by writing \Gamma = 0 or \sigma _ 1 + \sigma _ 2 = 0. This does not mean that \sigma _ 1(\mathbf{u}) +\sigma _ 2(\mathbf{u}) is the null vector of \mathbb{R}^n.

{Boundaries}For k\geqslant1, the boundary of the oriented affine k-simplex \sigma=[\mathbf{p} _ 0,\mathbf{p} _ 1,\dots,\mathbf{p} _ k] is defined to be the affine (k-1)-chain
\begin{equation}\label{1085}
\partial \sigma=\sum _ {j=0}^{k}(-1)^{j}\left[\mathbf{p} _ {0}, \ldots, \mathbf{p} _ {j-1}, \mathbf{p} _ {j+1}, \ldots, \mathbf{p} _ {k}\right].
\end{equation}

For example, if \sigma=[\mathbf{p} _ 0,\mathbf{p} _ 1,\mathbf{p} _ 2], then \partial \sigma=\left[\mathbf{p} _ {1}, \mathbf{p} _ {2}\right]-\left[\mathbf{p} _ {0}, \mathbf{p} _ {2}\right]+\left[\mathbf{p} _ {0}, \mathbf{p} _ {1}\right]=\left[\mathbf{p} _ {0}, \mathbf{p} _ {1}\right]+\left[\mathbf{p} _ {1}, \mathbf{p} _ {2}\right]+\left[\mathbf{p} _ {2}, \mathbf{p} _ {0}\right], which coincides with the usual notion of the oriented boundary of a triangle.

For 1\leqslant j\leqslant k, observe that the simplex \sigma _ j=[\mathbf{p} _ {0}, \ldots, \mathbf{p} _ {j-1}, \mathbf{p} _ {j+1}, \ldots, \mathbf{p} _ {k}] which occurs in (1085) has Q^{k-1} its parameter domain and that it is defined by \sigma _ j(\mathbf{u})=\mathbf{p} _ 0+B\mathbf{u}\, (\mathbf{u}\in Q^{k-1}), where B is the linear mapping from \mathbb{R}^{k-1} to \mathbb{R}^n determined by B\mathbf{e} _ i=\mathbf{p} _ i-\mathbf{p} _ 0\, (1\leqslant i\leqslant j-1), B\mathbf{e} _ i=\mathbf{p} _ {i+1}-\mathbf{p} _ i\, (j\leqslant i\leqslant k-1).

The simplex \sigma _ 0=[\mathbf{p} _ 1,\mathbf{p} _ 2,\dots,\mathbf{p} _ k] which also occurs in (1085), is given by the mapping \sigma _ 0(\mathbf{u})=\mathbf{p} _ 1+B\mathbf{u}, where B\mathbf{e} _ i=\mathbf{p} _ {i+1}-\mathbf{p} _ 1 for 1\leqslant i\leqslant k-1.

{Differentiable simplexes and chains}Let T be a \mathcal{C}^{\prime\prime}-mapping of an open set E\subset \mathbb{R}^n into an open set V\subset \mathbb{R}^m; T need not be one-to-one. If \sigma is an oriented affine k-simplex in E, then the composite mapping \Phi= T\circ \sigma (which we shall sometimes write in the simpler form T\sigma) is a k-surface in V, with parameter domain Q^k. We call \Phi an oriented k-simplex of class \mathcal{C}^{\prime\prime}.

A finite collection \Psi of oriented k-simplexes \Phi _ 1,\dots,\Phi _ r of class \mathcal{C}^{\prime\prime} in V is called a k-chain of class \mathcal{C}^{\prime\prime} in V. If \omega is a k-form in V, we define \int _ \Psi\omega=\sum\int _ {\Phi _ i}\omega and use the corresponding notation \Psi=\sum\Phi _ i.

If \Gamma=\sum\sigma _ i is an affine chain and if \Phi _ i=T\circ\sigma _ i, we also write \Psi=T\circ\Gamma, or T\left(\sum \sigma _ i\right)=\sum T\sigma _ i.

The boundary \partial\Phi of the oriented k-simplex \Phi = T \circ\sigma is defined to be the (k-1)-chain \partial\Phi=T(\partial\sigma). In justification of it, observe that if T is affine, then \Phi=T\circ\sigma is an affine k-simplex, in which case the formula is not a matter of definition, but is seen to be a consequence of (1085). For example,
\begin{align*}
T(\partial \sigma) &= T\bigl([\mathbf{p} _ 1,\mathbf{p} _ 2,\mathbf{p} _ 3] - [\mathbf{p} _ 0, \mathbf{p} _ 2, \mathbf{p} _ 3] + [\mathbf{p} _ 0, \mathbf{p} _ 1, \mathbf{p} _ 3] - [\mathbf{p} _ 0, \mathbf{p} _ 1, \mathbf{p} _ 2])\\ &=T\circ [\mathbf{p} _ 1, \mathbf{p} _ 2, \mathbf{p} _ 3] - T\circ [\mathbf{p} _ 0, \mathbf{p} _ 2, \mathbf{p} _ 3] + T\circ [\mathbf{p} _ 0, \mathbf{p} _ 1, \mathbf{p} _ 3] - T\circ [\mathbf{p} _ 0, \mathbf{p} _ 1, \mathbf{p} _ 2]\\
&= [T(\mathbf{p} _ 1), T(\mathbf{p} _ 2), T(\mathbf{p} _ 3)] - [T(\mathbf{p} _ 0), T(\mathbf{p} _ 2), T(\mathbf{p} _ 3)] + [T(\mathbf{p} _ 0), T(\mathbf{p} _ 1), T(\mathbf{p} _ 3)] - [T(\mathbf{p} _ 0), T(\mathbf{p} _ 1), T(\mathbf{p} _ 2)]\\ &= \partial [T(\mathbf{p} _ 0), T(\mathbf{p} _ 1), T(\mathbf{p} _ 2), T(\mathbf{p} _ 3)]\\ &= \partial (T\circ \sigma)
\end{align*}
Thus the formula generalizes this special case.

It is immediate that \partial\Phi is of class \mathcal{C}^{\prime\prime} if this is true of \Phi.

Finally, we define the boundary \partial\Psi of the k-chain \Psi=\sum\Phi _ i to be the (k-1)-chain \partial\Psi=\sum\partial\Phi _ i.

{Positively oriented boundaries}So far we have associated boundaries to chains, not to subsets of \mathbb{R}^n. This notion of boundary is exactly the one that is most suitable for the statement and proof of Stokes' theorem. However, in
applications, especially in \mathbb{R}^2 or \mathbb{R}^3, it is customary and convenient to talk about "oriented boundaries" of certain sets as well. We shall now describe this briefly.

Let Q^n be the standard simplex in \mathbb{R}^n, let \sigma _ 0 be the identity mapping with domain Q^n. \sigma _ 0 may be regarded as a positively oriented n-simplex in \mathbb{R}^n. Its boundary \partial\sigma _ 0 is an affine (n - 1)-chain. This chain is called the positively oriented boundary of the set Q^n.

Now let T be a 1-1 mapping of Q^n into \mathbb{R}^n, of class \mathcal{C}^{\prime\prime}, whose Jacobian is positive (at least in the interior of Q^n). Let E = T(Q^n). By the inverse function theorem, E is the closure of an open subset of \mathbb{R}^n. We define the positively oriented boundary of the set E to be the (n - 1)-chain \partial T=T(\partial\sigma _ 0), and we may denote this (n-1)-chain by \partial E.

An obvious question occurs here: If E = T _ 1(Q^n) = T _ 2(Q^n), and if both T _ 1 and T _ 2 have positive Jacobians, is it true that \partial T _ 1 =\partial T _ 2? That is to say, does the equality \int _ {\partial T _ 1}\omega=\int _ {\partial T _ 2}\omega hold for every (n-1)-form \omega? The answer is yes, but we shall omit the proof.

One can go further. Let \Omega=E _ 1\cup\dots\cup E _ r, where E _ i=T _ i(Q^n), each T _ i has the properties that T had above, and the interiors of the sets E _ i are pairwise disjoint. Then the (n - 1)-chain \partial T _ 1+\dots+\partial T _ r=\partial \Omega is called the positively oriented boundary of \Omega.

If \Phi is a 2-surface in \mathbb{R}^m, with parameter domain I^2, then \Phi (regarded as
a function on 2-forms) is the same as the 2-chain \Phi\circ\sigma _ 1+\Phi\circ\sigma _ 2. Thus \partial \Phi =\partial\left(\Phi \circ \sigma _ {1}\right)+\partial\left(\Phi \circ \sigma _ {2}\right)
=\Phi\left(\partial \sigma _ {1}\right)+\Phi\left(\partial \sigma _ {2}\right)=\Phi\left(\partial I^{2}\right). In other words, if the parameter domain of \Phi is the square I^2, we need not refer back to the simplex Q^2, but can obtain \partial\Phi directly from \partial I^2.
7. Stokes' theorem

The case k = m = 1 is nothing but the fundamental theorem of calculus (with an additional differentiability assumption). The case k = m = 2 is Green's theorem, and k = m = 3 gives the so-called "divergence theorem" of Gauss. The case k = 2, m = 3 is the one originally discovered by Stokes. These special cases will be discussed further at the end of the present chapter.

Proof. It is enough to prove that
\begin{equation}\label{1092}
\int _ \Phi d\omega=\int _ {\partial\Phi}\omega
\end{equation}
for every oriented k-simplex \Phi of c1ass \mathcal{C}^{\prime\prime} in V.

Fix such a \Phi and put \sigma=[\mathbf{0},\mathbf{e} _ 1,\dots,\mathbf{e} _ k]. Thus \sigma is the oriented affine k-simplex with parameter domain Q^k which is defined by the identity mapping. Since \Phi is also defined on Q^k and \Phi\in\mathcal{C}^{\prime\prime}, there is an open set E\subset \mathbb{R}^k which contains Q^k, and there is a \mathcal{C}^{\prime\prime}-mapping T of E into V such that \Phi= T\circ\sigma. (By definition, \Phi=T^\prime \circ\sigma^\prime =T^\prime \circ\sigma^\prime \circ\sigma=T\circ\sigma, where T=T^\prime \circ\sigma^\prime \in\mathcal{C}^{\prime\prime}.) By Theorems theorem1056 and theorem1053(3), the left side of (1092) is equal to
\[\int _ {T \sigma} d \omega=\int _ {\sigma}(d \omega) _ {T}=\int _ {\sigma} d\left(\omega _ {T}\right).\]
Another application of Theorem theorem1056 shows that the right side of (1092) is
\[\int _ {\partial(T\sigma)} \omega=\int _ {T(\partial \sigma)} \omega=\int _ {\partial \sigma} \omega _ {T}.\]

Since \omega _ T is a (k-1)-form in E, we see that in order to prove (1092) we merely have to show that
\begin{equation}\label{1094}
\int _ \sigma d\lambda=\int _ {\partial\sigma}\lambda
\end{equation}
for the special simplex \sigma=[\mathbf{0},\mathbf{e} _ 1,\dots,\mathbf{e} _ k] and for every (k-1)-form \lambda of class \mathcal{C}^\prime in E.

If k=1, the definition of an oriented 0-simplex shows that (1094) merely asserts that \int _ {0}^{1}f^\prime (u)\, du=f(1)-f(0) for every continuously differentiable function f on [0, 1], which is true by the fundamental theorem of calculus.

From now on we assume that k > 1, fix an integer r\, (1 \leqslant r \leqslant k), and choose f\in\mathcal{C}(E). It is then enough to prove (1094) for the case
\[\lambda=f(\mathbf{x})\, d x _ {1} \wedge \cdots \wedge d x _ {r-1} \wedge d x _ {r+1} \wedge \cdots \wedge d x _ {k}.\]

By (1085), the boundary of the simplex \sigma is \partial \sigma=\left[\mathbf{e} _ {1}, \ldots, \mathbf{e} _ {k}\right]+\sum _ {i=1}^{k}(-1)^{i} \tau _ {i}, where we define
\begin{align*}
\tau _ {i} & =\left[\mathbf{0}, \mathbf{e} _ {1}, \ldots, \mathbf{e} _ {i-1}, \mathbf{e} _ {i+1}, \ldots, \mathbf{e} _ {k}\right],\quad i=1,\dots,k. \\
\tau _ {0} & =\left[\mathbf{e} _ {r}, \mathbf{e} _ {1}, \ldots, \mathbf{e} _ {r-1}, \mathbf{e} _ {r+1}, \ldots, \mathbf{e} _ {k}\right]。
\end{align*}
Note that \tau _ 0 is obtained from [\mathbf{e} _ 1, \dots , \mathbf{e} _ k] by r-1 interchanges of \mathbf{e} _ r and its left neighbors. Thus
\begin{equation}\label{1097}
\partial \sigma=(-1)^{r-1} \tau _ {0}+\sum _ {i=1}^{k}(-1)^{i} \tau _ {i}.
\end{equation}
Each \tau _ i has Q^{k-1} as parameter domain.

If 1\leqslant i\leqslant k, \mathbf{u}\in Q^{k-1}, and \mathbf{x}=\tau _ i(\mathbf{u}), then \tau _ i(u _ j\mathbf{e} _ j) =u _ j\mathbf{e} _ j for 1\leqslant j<i, and \tau _ i(u _ j\mathbf{e} _ j) =u _ j\mathbf{e} _ {j+1} for i\leqslant j\leqslant k-1. Thus
\begin{equation}\label{1099}
x _ j=
\begin{cases}
u _ j,&1\leqslant j<i,\\
0,& j=i,\\
u _ {j-1},&i<j\leqslant k.
\end{cases}
\end{equation}

If \mathbf{x}=\tau _ 0(\mathbf{u}) and \mathbf{u}\in Q^{k-1}, then \tau _ 0(\mathbf{0})=\mathbf{e} _ r, \tau _ 0(u _ j\mathbf{e} _ j)-\tau _ 0(\mathbf{0}) =u _ j(\mathbf{e} _ j-\mathbf{e} _ r) for 1\leqslant j<r, and \tau _ 0(u _ j\mathbf{e} _ j)-\tau _ 0(\mathbf{0}) =u _ j(\mathbf{e} _ {j+1}-\mathbf{e} _ r) for r\leqslant j\leqslant k-1. Hence we obtain \tau _ 0(\mathbf{u})-\tau _ 0(\mathbf{0})=\sum _ {j=1}^{r-1}u _ j(\mathbf{e} _ j-\mathbf{e} _ r)+\sum _ {j=r}^{k-1} u _ j(\mathbf{e} _ {j+1}-\mathbf{e} _ r)=(u _ 1,\dots,u _ {r-1}, -\sum _ {j=1}^{k-1}u _ j,u _ r,\dots,u _ {k-1}).
\begin{equation}\label{1098}
x _ j=
\begin{cases}
u _ j,&1\leqslant j<r,\\
1-(u _ 1+\dots+u _ {k-1}),& j=r,\\
u _ {j-1},&r<j\leqslant k.
\end{cases}
\end{equation}

For 0\leqslant i\leqslant k, let J _ i be the Jacobian of the mapping \left(u _ {1}, \ldots, u _ {k-1}\right) \mapsto\left(x _ {1}, \ldots, x _ {r-1}, x _ {r+1}, \ldots, x _ {k}\right) induced by \tau _ i. When i=0 or i=r, this is the identity mapping. Thus J _ 0=1, J _ r=1. For other i, the fact that x _ i=0 in (1099) shows that J _ i has a row of zeros, hence J _ i=0. Thus \int _ {\tau _ i}\lambda=0\, (i\neq0,i\neq r). Consequently, (1097) gives
\begin{equation}\label{10102}
\int _ {\partial \sigma} \lambda =(-1)^{r-1} \int _ {\tau _ {0}} \lambda+(-1)^{r} \int _ {\tau _ {r}} \lambda=(-1)^{r-1} \int _ {Q^{k-1}} \left[f\left(\tau _ {0}(\mathbf{u})\right)-f\left(\tau _ {r}(\mathbf{u})\right)\right] d \mathbf{u}
\end{equation}

On the other hand, d \lambda =\left(D _ {r} f\right)(\mathbf{x})\, d x _ {r} \wedge d x _ {1} \wedge \cdots \wedge d x _ {r-1} \wedge d x _ {r+1} \wedge \cdots \wedge d x _ {k} =(-1)^{r-1}\left(D _ {r} f\right)(\mathbf{x}) d x _ {1} \wedge \cdots \wedge d x _ {k} so that
\begin{equation}\label{10103}
\int _ {\sigma}\, d \lambda=(-1)^{r-1} \int _ {Q^{k}}\left(D _ {r} f\right)(\mathbf{x})\, d \mathbf{x}.
\end{equation}
We evaluate (10103) by first integrating with respect to x _ r, over the interval [0,1-(x _ 1+\dots+x _ {r-1}+x _ {r+1}+\dots+x _ k)], put (x _ 1,\dots,x _ {r-1},x _ {r+1},\dots,x _ k)=(u _ 1,\dots,u _ {k-1}), and see with the aid of (1098) that the integral over Q^k in (10103) is equal to the integral over Q^{k-1} in (10102). Thus (1094) holds, and the proof is complete.
8. Closed forms and exact forms

In certain sets E, for example in convex ones, the converse is true; this is the content of Theorem 10. (usually known as Poincaré's lemma) and Theorem 10.. However, Examples 1 and 2 will exhibit closed forms that are not exact.

{Remarks} (a) Whether a given k-form \omega is or is not closed can be verified by simply differentiating the coefficients in the standard presentation of \omega. For example, a 1-form \omega=\sum f _ i(\mathbf{x})\, dx _ i with f _ i\in\mathcal{C}^\prime (E) for some open set E\subset\mathbb{R}^n, is closed if and only if the equations (D _ jf _ i)(\mathbf{x})=(D _ if _ j)(\mathbf{x}) hold for all i,j in \{1,\dots,n\} and for all \mathbf{x}\in E. Note that this is a "pointwise" condition; it does not involve any global properties that depend on the shape of E. On the other hand, to show that \omega is exact in E, one has to prove the existence of a form \lambda, defined in E, such that d\lambda=\omega. This amounts to solving a system of partial differential equations, not just locally, but in all of E. For example, to show that \omega is exact in a set E, one has to find a function (or 0-form) \mathbf{g}\in\mathcal{C}^\prime (E) such that (D _ i\mathbf{g})(\mathbf{x})=f _ i(\mathbf{x})\, (\mathbf{x}\in E,\, 1\leqslant i\leqslant n). Of course, (D _ jf _ i)(\mathbf{x})=(D _ if _ j)(\mathbf{x}) is a necessary condition for the solvability of this.

(b) Let \omega be an exact k-form in E. Then there is a (k-1)-form \lambda in E with d\lambda=\omega, and Stokes' theorem asserts that \int _ \Psi\omega=\int _ \Psi\, d\lambda=\int _ {\partial\Psi}\lambda for every k-chain \Psi of class \mathcal{C}^{\prime\prime} in E.

If \Psi _ 1 and \Psi _ 2 are such chains, and if they have the same boundaries, it follows that
\[\int _ {\Psi _ 1}\omega= \int _ {\Psi _ 2}\omega.\]

In particular, the integral of an exact k-form in E is 0 over every k-chain in E whose boundary is 0.

As an important special case of this, note that integrals of exact 1-forms in E are 0 over closed (differentiable) curves in E.

(c) Let \omega be a closed k-form in E. Then d\omega = 0, and Stokes' theorem asserts that \int _ {\partial\Psi}\omega=\int _ \Psi\, d\omega=0 for every (k+1)-chain \Psi of class \mathcal{C}^{\prime\prime} in E.

In other words, integrals of closed k-forms in E are 0 over k-chains that are boundaries of (k + 1)-chains in E.

(d) Let \Psi be a (k + 1)-chain in E and let \lambda be a (k- 1)-form in E, both of class \mathcal{C}^{\prime\prime}. Since d^2\lambda= 0, two applications of Stokes' theorem show that \int _ {\partial \partial \Psi} \lambda=\int _ {\partial \Psi} d \lambda=\int _ {\Psi} d^{2} \lambda=0.

We conclude that \partial^2\Psi= 0. In other words, the boundary of a boundary is 0.

{Example 1} Let E=\mathbb{R}^2-\{\mathbf{0}\}, the plane with the origin removed. The 1-form \eta=\frac{x\, dy-y\, dx}{x^2+y^2} is closed in \mathbb{R}^2-\{\mathbf{0}\}. This is easily verified by differentiation. Fix r > 0, and define \gamma(t)=(r\cos t,r\sin t)\, (0\leqslant t\leqslant 2\pi). Then \gamma is a curve (an "oriented 1-simplex") in \mathbb{R}^2-\{\mathbf{0}\}. Since \gamma(0)=\gamma(2\pi), we have \partial\gamma=0.

Direct computation shows that \int _ \gamma\eta=2\pi\neq0. The discussion in Remarks shows that we can draw two conclusions from this: First, \eta is not exact in \mathbb{R}^2-\{\mathbf{0}\}, for otherwise \int _ \gamma\eta would be 0. Secondly, \gamma is not the boundary of any 2-chain in \mathbb{R}^2-\{\mathbf{0}\} (of class \mathcal{C}^{\prime\prime}), for otherwise the fact that \eta is closed would force the integral \int _ \gamma\eta to be 0.
{Example 2} Let E=\mathbb{R}^3-\{\mathbf{0}\}, 3-space with the origin removed. Define \zeta=\frac{x\, d y \wedge d z+y\, d z \wedge d x+z\, d x \wedge d y}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} where we have written (x, y, z) in place of (x _ 1, x _ 2 , x _ 3). Differentiation shows that d\zeta = 0, so that \zeta is a closed 2-form in \mathbb{R}^3-\{\mathbf{0}\}.

Let \Sigma be the 2-chain in \mathbb{R}^3-\{\mathbf{0}\} that is the usual parametrization of the unit sphere in \mathbb{R}^3. It is easy to compute \int _ \Sigma\zeta=4\pi\neq0. As in the preceding example, we can now conclude that \zeta is not exact in \mathbb{R}^3-\{\mathbf{0}\} (since \partial\Sigma = 0) and that the sphere \Sigma is not the boundary of any 3-chain in \mathbb{R}^3-\{\mathbf{0}\} (of class \mathcal{C}^{\prime\prime}), although \partial\Sigma = 0.

Suppose that \omega=\sum\omega _ i\, dx _ i is a 1-form on \mathbb{R}^n and \omega happens to equal df=\sum D _ if\, dx _ i. We can clearly assume that f(\mathbf{0})=0. We have
\[f(\mathbf{x})=\int _ {0}^{1}\frac{d}{dt}f(t\mathbf{x})\, dt=\int _ {0}^{1}\sum _ {i=1}^{n}D _ if(t\mathbf{x})\cdot x _ i\, dt= \int _ {0}^{1}\sum _ {i=1}^{n}\omega _ i(t\mathbf{x})\cdot x _ i\, dt.\]
This suggests that in order to find f, given \omega, we consider the function I\omega, defined by I\omega(\mathbf{x})=\int _ {0}^{1}\sum _ {i=1}^{n}\omega _ i(t\mathbf{x})\cdot x _ i\, dt.

Note that the definition of I\omega makes sense if \omega is defined only on an open set A\subset \mathbb{R}^n with the property that whenever \mathbf{x}\in A, the line segment from \mathbf{0} to \mathbf{x} is contained in A; such an open set is called star-shaped with respect to \mathbf{0}. A somewhat involved calculation shows that (on a star-shaped open set) we have \omega = d(I\omega) provided that \omega satisfies the necessary condition d\omega = 0. The calculation, as well as the definition of I\omega, may be generalized considerably:

Proof. We will define a function I from l-forms to (l - 1)-forms (for each l), such that I(0) = 0 and \omega = I(d\omega) + d(I\omega) for any form \omega. It follows that \omega = d(I\omega) if d\omega = 0. Let
\[\omega=\sum _ {i _ 1<\dots<i _ l}\omega _ {i _ 1,\dots,i _ l}\, dx _ {i _ 1}\wedge\dots\wedge dx _ {i _ l}=\sum _ I\omega _ I\, dx _ I.\]
Since A is star-shaped we can define
\[I\omega(\mathbf{x})=\sum _ I\sum _ {\alpha=1}^{l}(-1)^{\alpha-1}\Big(\int _ {0}^{1}t^{l-1}\omega _ I(t\mathbf{x})\, dt\Big)x _ {i _ \alpha}\, dx _ {i _ 1}\wedge\dots\wedge\widehat{dx _ {i _ \alpha}}\wedge\dots\wedge dx _ {i _ l}.\]
(The symbol \widehat{\cdot} over dx _ {i _ {\alpha}} indicates that it is omitted.) The proof that \omega=I(d\omega)+d(I\omega) is an elaborate computation: We have
\[
\begin{split}
d(I\omega) & =l\cdot\sum _ I\Big(\int _ {0}^{1}t^{l-1}\omega _ I(t\mathbf{x})\, dt\Big)\, dx _ I \\
& \quad+ \sum _ I\sum _ {\alpha=1}^{l}\sum _ {j=1}^{n}(-1)^{\alpha-1}\Big(\int _ {0}^{1}t^l(D _ j\omega _ I)(t\mathbf{x})\, dt\Big) x _ {i _ \alpha}\, dx _ j\wedge dx _ {i _ 1}\wedge\dots\wedge\widehat{dx _ {i _ \alpha}}\wedge\dots\wedge dx _ {i _ l}.
\end{split}
\]
We also have
\[d\omega=\sum _ I\sum _ {j=1}^{n}(D _ j\omega _ I)\, dx _ j\wedge dx _ {i _ 1}\wedge\dots\wedge dx _ {i _ l}.\]
Applying I to the (l + 1)-form d\omega, we obtain
\[
\begin{split}
I(d\omega) & =\sum _ I\sum _ {j=1}^{n}\Big(\int _ {0}^{1}t^l(D _ j\omega _ I)(t\mathbf{x})\, dt\Big)x _ j\, dx _ I \\
& \quad-\sum _ I\sum _ {j=1}^{n}\sum _ {\alpha=1}^{l} (-1)^{\alpha-1}\Big(\int _ {0}^{1}t^l(D _ j\omega _ I)(t\mathbf{x})\, dt\Big)x _ {i _ \alpha}\, dx _ j\wedge dx _ {i _ 1}\wedge\dots\wedge\widehat{dx _ {i _ \alpha}}\wedge\dots\wedge dx _ {i _ l}.
\end{split}
\]
Adding, the triple sums cancel, and we obtain
\begin{align*}
d(I\omega)+I(d\omega) & =\sum _ Il\cdot\Big(\int _ {0}^{1}t^{l-1}\omega _ I(t\mathbf{x})\, dt\Big)\, dx _ I+ \sum _ I\sum _ {j=1}^{n}\Big(\int _ {0}^{1}t^lx _ j(D _ j\omega _ I)(t\mathbf{x})\, dt\Big)\, dx _ I \\
& =\sum _ I\Big(\int _ {0}^{1}\frac{d}{dt}[t^l\omega _ I(t\mathbf{x})]\, dt\Big)\,dx _ I=\sum _ I\omega _ I\, dx _ I=\omega.
\end{align*}
This proves the theorem.

Proof. Let ro be a k-form in U, with d\omega=0. By Theorem theorem1053(3), \omega _ T is a k-form in E for which d(\omega _ T) = 0. Hence \omega _ T=d\lambda. For some (k-1)-form \lambda in E. We have \omega=(\omega _ T) _ S=(d\lambda) _ S=d(\lambda _ S). Since \lambda _ S is a (k-1)-form in U, \omega is exact in U.

{Remark} In applications, cells are often more convenient parameter domains than simplexes. If our whole development had been based on cells rather than simplexes, the computation that occurs in the proof of Stokes' theorem would be even simpler. The reason for preferring simplexes is that the definition of the boundary of an oriented simplex seems easier and more natural than is the case for a cell. Also, the partitioning of sets into simplexes (called "triangulation") plays an important role in topology, and there are strong connections between certain aspects of topology, on the one hand, and differential forms, on the other.

Since every cell can be triangulated, we may regard it as a chain.
9. Vector analysis We conclude this chapter with a few applications of the preceding material to theorems concerning vector analysis in \mathbb{R}^3. These are special cases of theorems about differential forms, but are usually stated in different terminology. We are thus faced with the job of translating from one language to another.

{Vector fields}Let \mathbf{F}=F _ 1\mathbf{e} _ 1+F _ 2\mathbf{e} _ 2+F _ 3\mathbf{e} _ 3 be a continuous mapping of an open set E\subset \mathbb{R}^3 into \mathbb{R}^3. Since \mathbf{F} associates a vector to each point of E, \mathbf{F} is sometimes called a vector field, especially in physics. With every such \mathbf{F} is associated a 1-form
\begin{equation}\label{10124}
\lambda _ \mathbf{F}=F _ 1\, dx+F _ 2\, dy+F _ 3\, dz
\end{equation}
and a 2-form
\begin{equation}\label{10125}
\omega _ \mathbf{F}=F _ 1\, dy\wedge dz+F _ 2\, dz\wedge dx+F _ 3\, dx\wedge dy.
\end{equation}
Here, and in the rest of this chapter, we use the customary notation (x, y, z) in place of (x _ 1, x _ 2 , x _ 3).

It is clear, conversely, that every 1-form \lambda in E is \lambda _ \mathbf{F} for some vector field \mathbf{F} in E, and that every 2-form \omega is \omega _ \mathbf{F} for some \mathbf{F}. In \mathbb{R}^3, the study of 1-forms and 2-forms is thus coextensive with the study of vector fields.

If u\in\mathcal{C}^\prime (E) is a real function, then its gradient
\[\nabla u=(D _ 1u)\mathbf{e} _ 1+(D _ 2u)\mathbf{e} _ 2+(D _ 3u)\mathbf{e} _ 3\]
is an example of a vector field in E.

Suppose now that \mathbf{F} is a vector field in E, of class \mathcal{C}^\prime. Its curl \nabla\times \mathbf{F} is the vector field defined by
\[\nabla \times \mathbf{F}=\left(D _ {2} F _ {3}-D _ {3} F _ {2}\right) \mathbf{e} _ {1}+\left(D _ {3} F _ {1}-D _ {1} F _ {3}\right) \mathbf{e} _ {2}+\left(D _ {1} F _ {2}-D _ {2} F _ {1}\right) \mathbf{e} _ {3}\]
and its divergence is the real function \nabla\cdot \mathbf{F} defined in E by
\[\nabla \cdot \mathbf{F}=D _ {1} F _ {1}+D _ {2} F _ {2}+D _ {3} F _ {3}.\]

These quantities have various physical interpretations. Here are some relations between gradients, curls, and divergences.

If we compare the definitions of \nabla u, \nabla\times \mathbf{F}, and \nabla\cdot \mathbf{F} with the differential forms \lambda _ \mathbf{F} and \omega _ \mathbf{F} given by (10124) and (10125), we obtain the following four statements:
\begin{align*}
&\quad&\mathbf{F} & =\nabla u &\Longleftrightarrow&&\lambda _ \mathbf{F}&=du.&\quad&\\
&\quad&\nabla\times \mathbf{F} & =\mathbf{0} &\Longleftrightarrow&&d\lambda _ \mathbf{F}&=0.&\quad&\\
&\quad&\mathbf{F} & =\nabla\times \mathbf{G} &\Longleftrightarrow&&\omega _ \mathbf{F}&=d\lambda _ \mathbf{G}.&\quad&\\
&\quad&\nabla\cdot \mathbf{F} & =0&\Longleftrightarrow&&d\omega _ \mathbf{F}&=0.&\quad&
\end{align*}

Now if \mathbf{F}=\nabla u, then \lambda _ \mathbf{F}=du, hence d\lambda _ \mathbf{F}=d^2u=0, which means that \nabla\times \mathbf{F}=\mathbf{0}. As regards the converse of (1), the hypothesis amounts to saying that d\lambda _ \mathbf{F}=0 in E. Then \lambda _ \mathbf{F}= du for some 0-form u. Hence \mathbf{F}=\nabla u.

The proof of (2) and the converse follow exactly the same pattern.

{Volume elements} The k-form dx _ 1\wedge\dots\wedge dx _ k is called the volume element in \mathbb{R}^k. It is often denoted by dV (or by dV _ k if it seems desirable to indicate the dimension explicitly), and the notation
\begin{equation}\label{10126}
\int _ {\Phi} f(\mathbf{x})\, d x _ {1} \wedge \cdots \wedge d x _ {k}=\int _ {\Phi} f\, d V
\end{equation}
is used when \Phi is a positively oriented k-surface in \mathbb{R}^k and f is a continuous function on the range of \Phi.

The reason for using this terminology is very simple: If D is a parameter domain in \mathbb{R}^k, and if \Phi is a 1-1 \mathcal{C}^\prime-mapping of D into \mathbb{R}^k, with positive Jacobian J _ \Phi, then the left side of (10126) is
\[\int _ {D} f(\Phi(\mathbf{u})) J _ {\Phi}(\mathbf{u}) d \mathbf{u}=\int _ {\Phi(D)} f(\mathbf{x}) d \mathbf{x}.\]

In particular, when f=1, (10126) defines the volume of \Phi. The usual notation for dV _ 2 is dA.

Put \lambda=\alpha\, dx+\beta\, dy. Then d\lambda=(D _ 2\alpha)\, dy\wedge dx+(D _ 1\beta)\, dx\wedge dy=(D _ 1\beta-D _ 2\alpha)\, dA, the formula is the same as \int _ {\partial\Omega}\lambda=\int _ \Omega d\lambda, which is true by Stokes' theorem.

With \alpha(x,y)=-y and \beta(x,y)=x, we have \frac12\int _ {\partial\Omega}(x\, dy-y\, dx)=A(\Omega), the area of \Omega. With \alpha= 0, \beta = x, a similar formula is obtained.

{Area elements in \mathbb{R}^3}Let \Phi be a 2-surface in \mathbb{R}^3, of class \mathcal{C}^\prime, with parameter domain D\subset \mathbb{R}^2. Associate with each point (u, v)\in D the vector
\begin{equation}\label{10129}
\mathbf{N}(u, v)=\frac{\partial(y, z)}{\partial(u, v)} \mathbf{e} _ {1}+\frac{\partial(z, x)}{\partial(u, v)} \mathbf{e} _ {2}+\frac{\partial(x, y)}{\partial(u, v)} \mathbf{e} _ {\mathbf{3}}.
\end{equation}
The Jacobians in it correspond to the equation (x,y,z)=\Phi(u,v).

If f is a continuous function on \Phi(D), the area integral of f over \Phi is defined to be
\begin{equation}\label{10131}
\int _ {\Phi} f\, d A=\int _ {D} f(\Phi(u, v))|\mathbf{N}(u, v)|\, d u d v.
\end{equation}
In particular, when f=1 we obtain the area of \Phi, namely,
\begin{equation}\label{10132}
A(\Phi)=\int _ D|\mathbf{N}(u,v)|\, dudv.
\end{equation}

The following discussion will show that (10131) and its special case (10132) are reasonable definitions. It will also describe the geometric features of the vector \mathbf{N}.

Write \Phi=\varphi _ 1\mathbf{e} _ 1+\varphi _ 2\mathbf{e} _ 2+\varphi _ 3\mathbf{e} _ 3, fix a point \mathbf{p} _ 0=(u _ 0,v _ 0)\in D, put \mathbf{N}=\mathbf{N}(\mathbf{p} _ 0), put \alpha _ i=(D _ 1\varphi _ i)(\mathbf{p} _ 0), beta _ i=(D _ 2\varphi _ i)(\mathbf{p} _ 0), for i=1,2,3, and let T\in L(\mathbb{R}^2,\mathbb{R}^3) be the linear transformation given by T(u,v)=\sum (\alpha _ iu+\beta _ iv)\mathbf{e} _ i. Note that T=\Phi^\prime (\mathbf{p} _ 0).

Let us now assume that the rank of T is 2 (If it is 1 or 0, then \mathbf{N=0}, and the tangent plane mentioned below degenerates to a line or to a point.) The range of the affine mapping (u,v)\mapsto \Phi(\mathbf{p} _ 0)+T(u,v) is then a plane \Pi, called the tangent plane to \Phi at \mathbf{p} _ 0. [One would like to call \Pi the tangent plane at \Phi(\mathbf{p} _ 0), rather than at \mathbf{p} _ 0; if \Phi is not one-to-one, this runs into difficulties.]

We have
\begin{gather*}
\mathbf{N}=\left(\alpha _ {2} \beta _ {3}-\alpha _ {3} \beta _ {2}\right) \mathbf{e} _ {1}+\left(\alpha _ {3} \beta _ {1}-\alpha _ {1} \beta _ {3}\right) \mathbf{e} _ {2}+\left(\alpha _ {1} \beta _ {2}-\alpha _ {2} \beta _ {1}\right) \mathbf{e} _ {3}, \\
T\mathbf{e} _ 1=\sum \alpha _ i\mathbf{e} _ i,\quad T\mathbf{e} _ 2=\sum\beta _ i\mathbf{e} _ i,\\
\mathbf{N}\cdot T\mathbf{e} _ 1= \mathbf{N}\cdot T\mathbf{e} _ 2=0.
\end{gather*}
Hence \mathbf{N} is perpendicular to \Pi. It is therefore called the normal to \Phi at \mathbf{p} _ 0.

A second property of \mathbf{N} is that the determinant of the linear transformation of \mathbb{R}^3 that takes \{\mathbf{e} _ 1, \mathbf{e} _ 2 , \mathbf{e} _ 3\} to \{T\mathbf{e} _ 1, T\mathbf{e} _ 2 , \mathbf{N}\} is |\mathbf{N}|^2>0. The 3-simplex [\mathbf{0},T\mathbf{e} _ 1, T\mathbf{e} _ 2 , \mathbf{N}] is thus positively oriented.

The third property of \mathbf{N} that we shall use is a consequence of the first two: The above-mentioned determinant, whose value is |\mathbf{N}|^2, is the volume of the parallelepiped with edges [\mathbf{0}, T\mathbf{e} _ 1], [\mathbf{0}, T\mathbf{e} _ 2], [\mathbf{0, N}]. We have seen that [\mathbf{0, N}] is perpendicular to the other two edges. The area of the parallelogram with vertices \mathbf{0},T\mathbf{e} _ 1, T\mathbf{e} _ 2, T(\mathbf{e} _ 1+\mathbf{e} _ 2) is therefore |\mathbf{N}|.

This parallelogram is the image under T of the unit square in \mathbb{R}^2. If E is any rectangle in \mathbb{R}^2, it follows (by the linearity of T) that the area of the parallelogram T(E) is A(T(E))=|\mathbf{N}|A(E)=\int _ E|\mathbf{N}(u _ 0,v _ 0)|\, dudv.

We conclude that (10132) is correct when \Phi is affine. To justify the definition (110132) in the general case, divide D into small rectangles, pick a point (u _ 0 , v _ 0) in each, and replace \Phi in each rectangle by the corresponding tangent plane. The sum of the areas of the resulting parallelograms is then an approximation to A(\Phi). Finally, one can justify (10131) from (10132) by approximating by step functions.

If we think of \mathbf{N}=\mathbf{N}(u, v) as a directed line segment, pointing from \Phi(u, v) to \Phi(u, v) + \mathbf{N}(u, v), then \mathbf{N} points outward, that is to say, away from \Psi(K). This is so because J _ \Psi > 0 when t = a.

{Integrals of 1-forms in \mathbb{R}^3}Let \gamma be a \mathcal{C}^\prime-curve in an open set E\subset \mathbb{R}^3, with parameter interval [0, 1], let F be a vector field in E, and define \lambda _ \mathbf{F} by (10124). The integral of \lambda _ \mathbf{F} over \gamma can be rewritten in a certain way which we now describe.

For any u\in[0,1], \gamma^\prime (u)=\gamma _ {1}^{\prime}(u) \mathbf{e} _ {1}+\gamma _ {2}^{\prime}(u) \mathbf{e} _ {2}+\gamma _ {3}^{\prime}(u) \mathbf{e} _ {3} is called the tangent vector to \gamma at u. We define \mathbf{t} = \mathbf{t}(u) to be the unit vector in the direction of \gamma^\prime (u). Thus \gamma^\prime (u)=|\gamma^\prime (u)|\mathbf{t}(u). [If \gamma^\prime (u) = \mathbf{0} for some u, put \mathbf{t}(u) = \mathbf{e} _ 1; any other choice would do just as well.] We have
\[\int _ {\gamma} \lambda _ {\mathbf{F}} =\sum _ {i=1}^{3} \int _ {0}^{1} F _ {i}(\gamma(u)) \gamma _ {i}^{\prime}(u)\, d u= \int _ {0}^{1} \mathbf{F}(\gamma(u)) \cdot \gamma^{\prime}(u)\, d u=\int _ {0}^{1} \mathbf{F}(\gamma(u)) \cdot \mathbf{t}(u)\left|\gamma^{\prime}(u)\right|\, d u .\]
Theorem in Chapter 6 makes it reasonable to call |\gamma^\prime (u)|\, du the element of arc length along \gamma. A customary notation for it is ds, and the formula is rewritten in the form
\begin{equation}\label{10143}
\int _ \gamma\lambda _ \mathbf{F}=\int _ \gamma(\mathbf{F\cdot t})\, ds.
\end{equation}

Since \mathbf{t} is a unit tangent vector to \gamma, \mathbf{F\cdot t} is called the tangential component of \mathbf{F} along \gamma.

The right side of (10143) should be regarded as just an abbreviation for \int _ {0}^{1} \mathbf{F}(\gamma(u)) \cdot \mathbf{t}(u)\left|\gamma^{\prime}(u)\right|\, d u. The point is that \mathbf{F} is defined on the range of \gamma, but \mathbf{t} is defined on [0, 1]; thus \mathbf{F\cdot t} has to be properly interpreted. Of course, when \gamma is one-to-one, then \mathbf{t}(u) can be replaced by \mathbf{t}(\gamma(u)), and this difficulty disappears.

{Integrals of 2-forms in \mathbb{R}^3}Let \Phi be a 2-surface in an open set E\subset \mathbb{R}^3, of class \mathcal{C}^\prime. with parameter domain D \subset \mathbb{R}^2. Let \mathbf{F} be a vector field in E, and define \omega _ \mathbf{F} by (10125). As in the preceding section, we shall obtain a different representation of the integral of \omega _ \mathbf{F} over \Phi.

By (1035) and (10129),
\begin{align*}
\int _ {\Phi} \omega _ {\mathbf{F}} &=\int _ {\Phi}\left(F _ {1}\, d y \wedge d z+F _ {2}\, d z \wedge d x+F _ {3}\, d x \wedge d y\right) \\
&=\int _ {D}\Big\{\left(F _ {1} \circ \Phi\right) \frac{\partial(y, z)}{\partial(u, v)}+\left(F _ {2} \circ \Phi\right) \frac{\partial(z, x)}{\partial(u, v)}+\left(F _ {3} \circ \Phi\right) \frac{\partial(x, y)}{\partial(u, v)}\Big\}\, dudv \\
&=\int _ {D} \mathbf{F}(\Phi(u, v)) \cdot \mathbf{N}(u, v)\, d u d v.
\end{align*}

Now let \mathbf{n} = \mathbf{n}(u, v) be the unit vector in the direction of \mathbf{N}(u, v). [If \mathbf{N}(u, v) = \mathbf{0} for some (u, v)\in D, take \mathbf{n}(u, v) = \mathbf{e} _ 1.] Then \mathbf{N} = \mathbf{|N|n}, and therefore the last integral becomes \int _ {D} \mathbf{F}(\Phi(u, v)) \cdot \mathbf{n}(u, v)|\mathbf{N}(u, v)|\, d u d v. By (10131), we can finally write this in the form
\begin{equation}\label{10144}
\int _ {\Phi} \omega _ {\mathbf{F}}=\int _ {\Phi}(\mathbf{F} \cdot \mathbf{n})\, d A.
\end{equation}
With regard to the meaning of \mathbf{F\cdot n}, the remark made at the end of last part applies here as well.

We can now state the original form of Stokes' theorem.

Put \mathbf{H}=\nabla\times\mathbf{F}. Then, we have \omega _ \mathbf{H}=d\lambda _ \mathbf{F}. Hence
\[\int _ {\Phi}(\nabla \times \mathbf{F}) \cdot \mathbf{n}\, d A =\int _ {\Phi}(\mathbf{H} \cdot \mathbf{n})\, d A=\int _ {\Phi} \omega _ {\mathbf{H}}=\int _ {\Phi}\, d \lambda _ {\mathbf{F}}=\int _ {\partial \Phi} \lambda _ {\mathbf{F}}=\int _ {\partial \Phi}(\mathbf{F} \cdot \mathbf{t})\, d s.\]

By (10125),
\begin{gather*}
d \omega _ {\mathbf{F}}=(\nabla \cdot \mathbf{F})\, d x \wedge d y \wedge d z=(\nabla \cdot \mathbf{F})\, d V. \\
\int _ {\Omega}(\nabla \cdot \mathbf{F})\, d V=\int _ {\Omega}\, d \omega _ {\mathbf{F}}=\int _ {\partial \Omega} \omega _ {\mathbf{F}}=\int _ {\partial \Omega}(\mathbf{F} \cdot \mathbf{n})\, d A.
\end{gather*}