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## The exterior algebra of a vector space.

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.

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Michael Murray
1998-09-16