This section is concerned with some algebraic preliminaries needed to give a formal definition of what we mean by a differenial form.
Definition 1: Let be a vector space over . A function is said to be k-multilinear (or to be a k-tensor ) if and only if it is linear in each of its k variables. The set of all k-tensors over is denoted .
Note: with the obvious operations is a vector space over . One can define the tensor product function by
This is clearly a bilinear map, and the tensor product is associative.
Proposition 1: The vector space is of dimension where is the dimension of . In particular, if is a basis of , and is the corresponding dual basis, then the form a basis of .
Proof: Clearly the tensor products are elements of . If is a k-tensor, then it is easy to verify that
then evaluating the left side at shows that
Definition 2: A k-tensor over is said to alternating if interchanging any two of its variables changes the sign of the functional value, i.e.
The set of alternating k-tensors over is denoted .
Note: Clearly, is a vector subspace of For any k-tensor , one could make a symmetric k-tensor
where is the set of all permutations of If we let denote the sign of the permutation (i.e. it is 1 or -1 depending on whether the permutation is a product of an even or odd number of transpositions), then it is not surprising that
should be alternating. In fact, one has:
Proof: This is a straightforward consequence of the definitions.
Definition 3: The wedge product is the map defined by
For any linear map , there is a natural map defined by Restricting this to alternating k-tensors gives another map (also denoted by the same symbol) The following is an easy exercise:
Proposition 3: The wedge product is a bilinear map satisfying
The wedge product is also associative:
Proof: Let be the group of all permutations on and be the subgroup of permutations which fix . Then the cosets of in are mutually disjoint and so one can write as a disjoint union for a certain set of elements of One has
For the second assertion, apply the first result with and The fact that follows from the fact that is idempotent. Similarly, one can apply the first result with and
The final assertion follows from the second and the definition of the wedge product.
Proposition 5: The vector space is of dimension where is the dimension of the vector space . In particular, if is a dual basis, then the where form a basis for
Proof: To show that the set spans , apply to an expression for the alternating k-tensor in terms of the basis of Proposition 1. The linear independense is shown analogously to the argument of Proposition 1.
Proposition 6: Let be a basis of the vector space and . If then
Proof: Proposition 5 says that is of dimension 1, and the determinant function is the a non-zero alternating n-tensor. But can be thought of as an alternating n-tensor on . So it is a constant multiple of . Substituting , shows that the constant must be .
In both cases we can express the vectors in terms of a standard basis, and so we have a certain number of coodinate functions. When these coordinate functions are continuous, differential, etc., we call the vector field or k-form continuous, differentiable, etc. To simplify the hypotheses, we will henceforth use the word differentiable to mean .
Definition 4: If is differentiable, the 1-form is defined by
for . In the special case of the projection , one denotes as .
Note: and so is the dual basis to In particular, we will usually write k-forms as
An immediate consequence of this notation is:
Proposition 7: If is differentiable, then
The next definition tells us how to do substitutions in k-forms:
Definition 5: Let is a differentiable function and be the linear transformation defined by its derivative at . This map can be viewed as mapping the tangent spaces, i.e. we have a function defined by
Now, defines . If is a k-form on , then is the k-form on defined by
Note: To be more explicit, for one has
The next proposition tells us how to compute substitutions:
Proposition 8: If is differentiable, then
Proof: Should be here.
Definition 4 can be extended to higher order forms as follows:
Definition 6: Given a differentiable k-form
the differential of is the (k+1)-form defined by
To calculate differentials of k_forms, one has:
Proof: The first assertion is obvious from the definition, and the second asesertion follows from the formula for differentiating a product and assertion (1) of Proposition 3. The third assertion follows from assertion (1) of Proposition 3 and the equality of mixed partials:
This leaves the last assertion. The case of 0-forms is just the chain rule. Assume that the assertion is true for all -forms ; then it suffices to show it for . One has
On the other hand,
This proves the result since
where we have used equality of mixed partials.
Definition 7: A singular n-cube in is a continuous function . The convention in case , is that and are both the set . A singular 1-cube is called a curve. The standard n-cube is the inclusion map of An n-chain a linear combination with integer coefficients of singular n-cubes (i.e. an element of the free Abelian group generated by the singular n-cubes).
Given the standard n-cube , for and the singular (n-1) cube by where is defined by In the case where , the singular n-cube is called the -face of . The boundary of is defined by
The -face of a singular n-cube is the composition The boundary of is the (n-1)-chain
Finally, the boundary of an n-chain is defined to be the (n-1)-chain
The signs have been set up to guarantee
Proposition 10: If is an -chain in , then
Proof: The proof is straightforward. The idea is that taking an -face of a -face is the same as taking the -face of an -face. So, it is just a matter of checking that the signs come out right.
We are now ready to define integrals of differential forms over chains:
which can be written
In the case of a 0-form, define the integral to be
In the case of a 0-form, define the integral to be
Note: In the special case of a standard -cube , the integral of over is equal to the integral of over the An integral of a 1-form (respectively 2-form) over a 1-chain (respectively 2-chain) is often referred to as a line integral (respectively surface integral).
The main theorem of the course is:
Theorem 1: (Stokes' Theorem) If is a -form on an open set and is a -chain in , then
Proof: First consider the case where and is a -form on Using additivity, it is clearly enough to consider the case where
It follows that
Considering the other side of the equation,
Using Fubini's Theorem and the one dimensional Fundamental Theorem of Calculus, to evaluate this:
which is the same result as we had before. So the result holds in this case.
Now consider the case of a singular -cube . One has by definition:
Finally, in the general case where is a -chain, one has