where is a -surface in , is a -form and is a surface (or a collection of surfaces) in that make up the boundary of .
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
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 .
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 3 Recall 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 4 A compact subset is a -surface if there is a covering of by finitely many relatively open subsets such that for each we have
where is open, is one-to-one and for each .
Remark 5 Recall 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 .
Remark 7 The functions are called the transition functions of the coordinate chart . One can define an abstract 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 8 Prove 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 small there is a unique such that
Likewise 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 9 Let be a -surface and let , , be a coordinate chart for . We say the chart is oriented if each transition function satisfies for every . The -surface is called orientable if an oriented coordinate chart exists. An oriented -surface is a -surface together with a choice of an oriented chart.
Remark 10 There are non-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
Exercise 12 Prove 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 be oriented, as otherwise we might get a minus sign in the change of variables formula.
More generally, we have the following
Theorem 13 Let 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 14 1) What this says is that there are exactly two possible orientations for a connected surface. (It is essential that we assumed the surface to be connected, as otherwise each component could have its own orientation.) 2) If case (1) holds we say that and have the same orientation and if case (2) holds we say that they have the opposite orientation.
Exercise 15 Prove Theorem 13. This is a bit long and involved. Here is a sketch: The key is to show the uniform positivity or negativity of the determinants. The rest follows from change of variables as in the proof of Lemma 11. To see that the determinants have a definite sign proceed as follows. Fix . Suppose for . So and
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
for any continuous -form .
Theorem 16 Let 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 17 An -surface with boundary in is a compact subset together with a covering of by relatively open subsets , such that for each we have
where is open and is a one-to-one map with for each . The surface is oriented if 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:
and 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 19 If is an oriented -surface with boundary then the boundary of is the oriented -surface
with 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