where is a -surface in , is a -form and is a surface (or a collection of surfaces) in that make up the *boundary of .*

A PDF version of these notes is available here.

We will obtain this result in a two stages. First we will consider two very simple cases, namely integration of the derivative of a compactly supported form over and over the *upper half space*

Second on a general *oriented **-surfaces with boundary *we will use a partition of unity to reduce the computation of to the simple case of integration over . Of course, before we do that we should *define* what exactly an oriented -surface with boundary is!

The topological boundary of is the set

which we may think of as a parameterized -surface via the map

This last identity is a special case of (1) — where we think of as if is even and the surface *with opposite orientation *if is odd.

** Two simple cases of Stokes theorem **

Recall that the *support* of a -form , denoted , is the closure of the set of such that . If is compact we say that is compactly supported. If is a compactly supported -form in we have

for any large enough that .

*Proof:* By linearity of the exterior derivative and the integral it suffices to prove this for where has size , i.e., for some . But then

Integrating first with respect to and applying the fundamental theorem of calculus we see that the integral vanishes.

Theorem 1 is already a special case of Stokes theorem — the “-surface” has no boundary. A special case of a “-surface” with boundary is the upper half space Given a compactly supported -form in we define

where denotes the standard basis of .

Remark 3Recall the definition (2) of . Since is compactly supported we have

*Proof:* As above it suffice to consider for which we have (4). If then and by the fundamental theorem of calculus On the other hand if then

and integrating first with respect to we see that

** Oriented surfaces and Stokes’ Theorem **

To state Stokes Theorem we need a more general definition of a *-surface*. Have in mind the unit circle in or the unit sphere in as you read the following.

Definition 4A compact subset is a-surfaceif there is a covering of by finitely many relatively open subsets such that for each we havewhere is open, is one-to-one and for each .

Remark 5Recall that the rank of a linear transformation is the dimension of the range .

The definition says that every point of has a neighborhood *in *that is *diffeomorphic *to an open subset of . Another name for a -surface defined in this way is an *embedded manifold*. The collection is called a *coordinate chart *for .

Proposition 6For each pair , , let and define on . If , then is , one-to-one, onto and satisfies for each point .

Remark 7The functions are called thetransition functionsof the coordinate chart . One can define anabstract manifold(not embedded in ) as a compact space with an open cover and one-to-one coordinate maps such that the transition functions are , one-to-one and have non-singular derivative.

Exercise 8Prove proposition 6. Here is a sketch. Fix with and let and . So on and on . Thus are one-to-one, onto and, if they are , have non-singular derivative. To show that they are , fix and let . Then for each sufficiently smallthere is a unique such thatLikewise for each small there is a unique such that (5) holds. Since we have

Because and have full rank you can show that for and sufficiently small we have

where . Thus and dividing by length of and taking we see from (6) that the range of is equal to the range of . Since the two linear transformations are full rank we can now solve for and in terms of each other, up to terms that are .

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

Definition 9Let be a -surface and let , , be a coordinate chart for . We say the chart isorientedif each transition function satisfies for every . The -surface is calledorientableif an oriented coordinate chart exists. Anoriented -surfaceis a -surface together with a choice of an oriented chart.

Remark 10There arenon-orientable-surfaces; the Möbius strip is an example.

Given an oriented -surface, the sets of the open cover are intersections of with open sets in :

Thus we can find a partition of unity on subordinate to . As mentioned in the proof of the partition of unity theorem we can choose the functions to be as smooth as we like. In particular they can be . Using the partition of unity we now define the *integral of* *a *-form over to be

Here we think of as a parameterized -surface. The pullback is compactly supported in so

We state without proof the following basic fact

Lemma 11Let be an oriented -surface and let be a form in . If for some then

Exercise 12Prove Lemma 11. Sketch: You need to show that for each . This follows from the change of variables formula using the transition function . It is to facilitate this argument that we need the coordinate chart to beoriented, as otherwise we might get a minus sign in the change of variables formula.

More generally, we have the following

Theorem 13Let be a connected orientable -surface and let , be two oriented coordinate charts on . Choose two partitions of unity on , and subordinate to and respectively, where and for each , . For each pair with and , define a transition function where and . Then either

- for all and for each , in which case
for any continuous -form on ; or

- for all and for each , in which case
for any continuous -form on .

Remark 141) What this says is that there are exactly two possible orientations for a connected surface. (It isessentialthat we assumed the surface to be connected, as otherwise each component could have its own orientation.) 2) If case (1) holds we say that and have the same orientation and if case (2) holds we say that they have the opposite orientation.

Exercise 15Prove Theorem 13. This is a bit long and involved. Here is a sketch: The key is to show the uniform positivity or negativity of the determinants. The rest follows from change of variables as in the proof of Lemma 11. To see that the determinants have a definite sign proceed as follows. Fix . Suppose for . So andThus

where is a transition function for the chart . Since is oriented, the corresponding determinant is positive and we see that for have the same sign. Similarly if for then and

and we see that for have the same sign. Finally let depending on whether this sign is positive or negative.

Now consider the sets

It is easy to show that these sets are both relatively open — if then it is positive on a ball centered at . On the other hand , because never vanishes since the coordinate functions have derivatives of full rank. Thus are both closed. Since is connected it follows that one of the sets is empty and one is .

Given an oriented -surface , let denote the -surface with opposite orientation (using any coordinate chart that gives it the opposite orientation). Explicitly, if is an oriented chart for , we may obtain an oriented chart for using

where with

Note that

for any continuous -form .

** Stokes Theorem **

Theorem 16Let be an oriented -surface in and let be a form in a neighborhood of . Then

*Proof:* Choose a partition of unity on subordinate to as above. Note that and so and that for each , is compactly supported in . Thus by Lemma11

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

To obtain a non-zero value for we need to consider a surface with boundary. A surface with boundary is like a surface but the the functions making up the coordinate chart are restricted to the upper half space :

Definition 17An-surface with boundary inis a compact subset together with a covering of by relatively open subsets , such that for each we havewhere is open and is a one-to-one map with for each . The surface is

orientedif for each pair the transition function , defined on , satisfies for all .

The definition of integrals over an oriented-surface with boundary is just like that over a -surface: pick a partition of unity on subordinate to where and set

As before there are two classes of oriented charts and the value of the integral does not depend on the choice of chart within an orientation class but changes sign for the opposite orientation. The *boundary * of a surface with boundary is the set of points that lie on the boundary of one of the coordinate charts — that is points of the form . To make a precise definition recall the map given by

We state without proof:

Proposition 18Let be a -surface with boundary and for each letand the functions make a coordinate chart that turns into a surface. Furthermore is oriented if is.

The proof is pretty straightforward — one simply has to verify that the transition function are the restriction to of the corresponding transition function for .

Definition 19If is an oriented -surface with boundary then theboundary ofis the oriented -surfacewith as in Prop. 18.

Thm 20 (Stokes’ Theorem )Let be an oriented -surface with boundary in and let be a -form defined in a neighborhood of . Then

*Proof:* As above it suffices, by linearity of the integral and the exterior derivative, to prove this for such that For such we have by Theorem 2