We now turn to the study of hyper-surfaces in . A hyper-surface is something like a curve or a surface in or their higher dimensional analogues. A differential form is the something we can integrate over a hyper-surface.
For example, a parameterized curve in is a map , we will call this a curve if is . Given such a curve with and a function we can form the integral of “ along :”
We will call the expression
a differential one form. We will generalize this notion to make differential -forms and integrate these over hypersurfaces of dimension .
Definition 1 A parameterized -surface in is a map where is open. If with an open set, we say that is a parameterized -surface in .
Remark 2 We will call a -surface for short. In principle we should define a -surface as an equivalence class of parameterized -surface where we make if with a one-to-one map with at all points of it’s domain. Think of a curve, say where . This is the closed unit circle in . As a parametrized curve this is distinct from defined on the domain , but we see that the difference is simply one of reparameterizing the domain. We should keep this in mind, but for the sake of simplicity we will refer to parameterized -surfaces simply as -surfaces.
Definition 3 Let be an open set and an integer. A differential -form is a function from to the space of alternating -forms on . If with a non-negative integer we say that is a –form or that is of class (for we also say that is a continuous -form).
Remark 4 Since by definition, a differential -form on is just a real valued function on .
We will develop some special notation for -forms in a bit. For the moment we will denote the value of at a point by . Note that for each is an alternating -form, which in particular is a function taking as an argument -vectors in . We will denote the action of on the vectors by
Remark 5 is if and only if for each the map is .
Lemma 6 Suppose is a -form on and is a parameterized -surface. Then
is a -form on , which is continuous if is.
Proof: That is a differential -form follows easily from the fact that is. The continuity follows easily from the continuity of and .
whenever the integral on the right hand side is defined.
Remark 8 1) For the most part, it will suffice to consider only in case has compact support in . (Here the support of a -form is the closure of the set on which it is not identically zero.) In that case the integral on the right hand side of (1) is well defined, since we may continuously extend to all of by taking it to be zero off of and set
2) Sometimes it is useful to consider a -surface defined on a -cell in . If extends continuously to the boundary points of then the integral on the right hand side of (1) is well defined as an iterated integral. 3) For a -form in we will simply write
with the standard basis and a -cell. This amounts to identifying the -cell with the -surface given by the identity map.
Why is (1) a good definition? The answer comes from considering change of variables. Indeed, suppose and are parameterized surfaces in with where is a one-to-one map with everywhere. Then
If the domain of is connected then we either have or for each . Thus, by the change of variables formula,
with the sign corresponding to the sign of .
What does this sign mean? Remember that for the fundamental theorem of calculus in it was useful to define for , thus making the integral into an oriented integral that depends on the orientation of the interval. In the same way, a -surface in has two possible orientations and switching orientations results in a minus sign. You may have encountered this in vector calculus when integrating over surfaces in — there are two choices of unit normal vectors and the choice determines the sign of integrals like .
Wedge product and elementary forms
where the sum runs over all permutations of .
Remark 11 The “wedge” product of a zero form and a -form is the -form
We will typically denote this product as just .
Exercise 13 Let and be -forms and let be an -form. Show that
To proceed it is useful to introduce a notation for forms. First we define the elementary forms on by
so if and if . The for each we define the elementary forms to be the following
where — this makes sense without parentheses by exercise 12. For instance
Note that if for any then the resulting form is zero and more generally that
for any permutation of . Hence, up to sign, there are only elementary forms, given by
Let us introduce a compact notation for these elementary -forms. Given we can write its elements in increasing order where . Let
Note that given of size ,
with is the unique permutation of that puts in increasing order.
Furthermore is if and only if the functions for each .
Proof: Given of size with let
Clearly if is then is also. Furthermore
Now for each we have
where Note also that
if . Thus
Since is constant as a function of , it is so that if is for each then is too.
Definition 15 Let be a parameterized -surface in . Let points in be denoted and points in be denoted , so . Given and define the Jacobian
Given and let denote the correspnding Jacobian with ‘s and ‘s in increasing order.
Proposition 16 Let be a parameterized -surface in and a -form defined on a neighborhood of . Then
Proof: Let denote the standard basis of and the standard basis of . Then
Thus given ,
where in the second to last line Together with the previous theorem this proves the result.
A particular case of this result is when , so for we get the explicit expression
Let be an open set in . If we define
So is a continuous -form on . Similarly if is a -form we define
So is a continuous -form on .
Example 17 Let be a -form. Then
Example 18 Let be a -surface in , that is a curve in . Suppose then
Lemma 19 Let be a -form on an open set . Then .
Proof: Let . So
Exercise 21 Prove Lemma 20. (Hint: it is essentially the product rule.)
Note that the notation for the elementary one forms is consistent with the exterior derivative: is indeed the exterior derivative of the function .
Theorem 22 Let be a -form on an open set and let be a parameterized -surface in . Then
Proof: First consider a zero form, namely a function . Then and we see that
Specializing to a coordinate function we see that
where denotes the -th coordinate function of . Since (see remark 10) we see that
for any elementary -form . In particular, it follows from the product rule (Lemma 20) that
for any elementary -form .