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

Finally, if

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:

**Proposition 2:**

- If then
- If then
- If then

**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:

**Proposition 4:**

- If
and
, then
provided that

- If
, and
then

**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
.

- Each point
, has a
**tangent space**whose elements are of the form where . So the map can be viewed as assigning to each point an vector in its tangent space. When we view in this way, we call it a**vector field**. - To each point
, we can associate a copy
denoted
whose elements are alternating k-tensors
in the tangent space at
. When we view the function
as mapping
points to alternating k-tensors of the tangent space at
, we call
a
**k-form**(or differential form) on .

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

- If
, 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:

**Proposition 9:**

- If
is a
-form and
is an
-form, then
- If is a -form on and is differentiable, then

**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:

**Definition 8:**

- If
is a
-form on
, then
for some unique function
. Define
which can be written

In the case of a 0-form, define the integral to be

- If
is a
-form on
and
is a singular
-cube in
,
define
In the case of a 0-form, define the integral to be

- If
is a
-form on
and
is a
-chain in
, define

**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

One has

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:

And so,

Finally, in the general case where is a -chain, one has