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If
is a vector space we define a -linear
map to be a map
where there are
copies of , which is linear
in each factor. That is

We define
a -linear map
to be totally antisymmetric if
for all vectors
and all .
Note that it follows that
and if
is a permutation of
letters then
sgn

where
sgn
is the sign of the permutation . We denote the
vector space of all -linear, totally antisymmetric maps by
. and call them
forms. If
the
is just
the space of all linear functions on
and if
we make the convention that
. We need to
collect some results on the linear algebra of these spaces.
Assume that
has dimension
and that
is a basis
of . Let
be a
form. Then if
are arbitrary vectors and we
expand them in the basis as

then we have
so that it follows that
is completely determined by
its values on basis vectors. In particular if
then
.
If
and
are two linear maps in
then we
define an element
,
called the wedge product of
and , in
by

More generally if
and
we define
by

Assume that
. Then we leave as an exercise the following proposition.

**Proposition 5.1**
The direct sum

with the wedge product is an associative algebra.

We call
the exterior
algebra of . We call an element
an element of degree .
Because of associativity we can
repeatedly wedge and disregard brackets.
In particular we can define the wedge product of
elements in
and we leave it as an exercise to show that

sgn

Notice that
and that
Still assuming that
is
dimensional
choose a basis
of .
Define the dual basis of ,
,
by

for all
and . We want to define a basis of
.
Define elements of
by choosing
numbers
between
and
and
considering
As we are trying to form a basis
we may as well keep
the
distinct and ordered
. We show first
that these elements span
. Let
be an element of
. Notice that
equals zero unless there is a permutation
such that
for all
and equals
sgn
if
there is such a permutation. Consider vectors
and expand them in the basis as
Then we have
so that it follows that
is completely determined by
its values on basis vectors. For any ordered -tuple
define
and consider
We show that
. It suffices to apply both sides to
vectors
for any
and show that they are equal but that
is clear from previous discussions. So
is
spanned by the basis vectors
.
We have

**Proposition 5.2**
The vectors

where

are a basis for

.

*Proof*.
We have already seen that these vectors span. It suffices to
show that they are linearly independent. We do this by induction
on

. If

then the result is clear as the only non-trivial
case is

when the result is straightforward. More generally
assume we have a linear relation amongst some of the
basis vectors. There has to be an index

such that the corresponding

does not occur in all the vectors in that linear relation.
Otherwise there is only one vector in the linear relation and that
is not possible. Then wedge the whole relation with

. The
terms containing

disappear and we obtain a relation between
the vectors constructed for the case of a dimension less so by
induction that is not possible.

It is sometimes useful to sum over all -tuples
not just ordered ones. We can do this -- an
keep the uniqueness of the coefficients
--
if we demand that they be antisymmetric. That is

Then we have

We will need one last piece of linear algebra called *contraction*.
Let
and . Then we define
a
form
, the contraction of
and
by

where
are any
elements of
.

**Example 5.1**
Consider the vector space

. Then we know that
zero forms and one forms are just real numbers and
linear maps respectively. Notice that in the case
of

we can identify any linear map

with
the vector

where

Let
be the basis of linear functions defined
by
. We have seen that every two
form
on
has the form

Every three-form

takes the form

It follows that in
we can identify three-forms
with real numbers by identifying
with
and we can identify two-forms with vectors
by identifying
with
.

It is easy to check that with these identifications
the wedge product of two vectors
and
is identified
with the vector
. In other words wedge
product corresponds to cross product.

** Next:** Differential forms and the
** Up:** Differential forms.
** Previous:** Differential forms.
** Contents**
*Michael Murray*

*1998-09-16*