Exterior Forms

-Forms

-form

An exterior -form on a vector space is a skew-symmetric multilinear map . Skew-symmetric means that for any permutation , where is or , depending on whether the permutation is even or odd. The set of all -forms on is denoted by .

e.g.

• In particular, a -form is exactly a covector in the dual space .
• Suppose that we have chosen a linear coordinate system on a vector space , then each coordinate defines a -form.
• If a uniform force field is given on , its work along the displacement is a -form acting on .
• The oriented area of a parallelogram spanned by two vectors and in is given by the -form

Corollary

Every -form on is either zero or the oriented volume of the parallelepiped spanned by vectors in .

Proposition

Suppose . If are linearly dependent vectors in , then .

Proof Since are linearly dependent, one can assume that for scalars . Then, using the multilinearity of , we have By the skew-symmetry of , vanishes for all , which shows as desired.

Corollary

Any -form on an -dimensional vector space with is identically zero.

Proof vectors in must be linearly dependent if .

Attention

We shall from now on, assume that a -form on an -dimensional vector space satisfies without further notice.

The Exterior Product

Exterior (Wedge) Product

The exterior product of a -form and a -form is defined as the -form given by: Alternatively, we can also write

The above two expressions are equivalent, because instead of summing over all permutations of elements, we can sum over all possible partitions of and elements, and for each partition, we can permute the and elements separately. Notice that for each fixed partition , assuming , , and for each permutation , , suppose the resulting permutation of elements is , i.e. , we have as it is a group homomorphism. Thus , because the inverse of any element is itself in the group . Moreover, by skew-symmetry,So finally we get where the second equality holds as there are permutations for each fixed partition.

e.g. Given two one forms , their exterior product is a two form that satisfies

Proposition

Let be a basis for a vector space and be the dual basis. Then for any

Proof We will prove by induction on . Clearly, the base case is true. Now assume it holds for all , then where the second equality holds because is nonzero only if fixes , and the last equality holds by the induction hypothesis.

Theorem

The exterior multiplication of forms is skew-commutative, distributive, and associative:

• for any -form and -form .
• for any -forms , and -form .
• .

e.g. Let be a basis for a vector space with dual basis . Then we can calculate

Proposition

The space of -forms on an -dimensional vector space is a vector space of dimension , and the exterior product gives the space of all forms the structure of a super-commutative graded algebra over the field . In particular, if forms a basis for the 1-forms, forms a basis for the -forms.

Proof We first show that the forms are linearly independent. Suppose is a linear combination of these forms, i.e. We set to be identically . Notice that since is nonzero only if , holds for all , thus the forms are linearly independent. Now we show that every -form can be expressed as a linear combination of these forms. Suppose , then Finally, the exterior product gives the structure of a super-commutative graded algebra because of the theorem.

Pullback Invariance

Pullback Preserves Exterior Product

For a linear map , and , , there holds

Proof For all , we have $$\begin{aligned}(A^{}(\omega)\wedge A^{}(\eta))(v_{1},v_{2},\dots,v_{k+l})&=\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}\sgn(\sigma)\omega(Av_{\sigma(1)},\dots,Av_{\sigma(k)})\eta(Av_{\sigma(k+1)},\dots,Av_{\sigma(k+l)})\&=(\omega\wedge\eta)(Av_{1},\dots ,Av_{k+l})\&=A^{*}(\omega\wedge\eta)(v_{1},v_{2},\dots,v_{k+l})\end{aligned}$$$\square$

Interior Product

Interior Product

The interior product of a -form and a vector is defined as the -form given by:

Proposition

The interior product is a degree superderivation in the sense that it defines a linear map from -forms to -forms, and satisfies

Proof

Theorem

Let be the standard volume form on . Then is an isomorohism from to . Moreover, the following identities hold: