Exterior Forms

$k$-Forms

$k$-form

An exterior $k$-form on a vector space $V$ is a skew-symmetric multilinear map \newcommand{\d}{\mathrm{d}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\R}{\mathbb{R}}\newcommand{\D}{\mathrm{D}}\omega:V^{k}\to \R. Skew-symmetric means that $\omega (v_1,\dots ,v_k)=\mathrm{sgn}(\sigma )\omega (v_{\sigma (1)},\dots ,v_{\sigma (k)})$ for any permutation $\sigma \in S_k$, where $\text{sgn}(\sigma )$ is $1$ or $-1$, depending on whether the permutation is even or odd. The set of all $k$-forms on $V$ is denoted by ${\Lambda }^k(V)$.

e.g.

• In particular, a $1$-form is exactly a covector in the dual space $V^{\ast }$.
• Suppose that we have chosen a linear coordinate system $(x^1,x^2,\dots ,x^n)$ on a vector space $V$, then each coordinate $x^i$ defines a $1$-form.
• If a uniform force field $\mathbf{F}$ is given on ${\mathbb{R}}^3$, its work $A$ along the displacement $\mathbf{r}$ is a $1$-form acting on $\mathbf{r}$.
• The oriented area of a parallelogram spanned by two vectors $\mathbf{u}$ and $\mathbf{v}$ in ${\mathbb{R}}^2$ is given by the $2$-form $S(\mathbf{u},\mathbf{v})=\det (\left[\begin{array}{cc}\mathbf{u}& \mathbf{v}\end{array}\right])$

Remark

The idea of k-form comes from the idea the measuring some “oriented volume” in certain dimension.

Corollary

Every $k$-form on ${\mathbb{R}}^k$ is either zero or the oriented volume of the parallelepiped spanned by $k$ vectors in ${\mathbb{R}}^k$.

Proposition

Suppose $\omega \in {\Lambda }^k(V)$. If $\{u_1,u_2,\dots ,u_k\}$ are linearly dependent vectors in $V$, then $\omega (u_1,u_2,\dots ,u_k)=0$.

Proof Since $\{u_1,u_2,\dots ,u_k\}$ are linearly dependent, one can assume that $u_k=a_1{u}_1+\cdots +a_{k-1}{u}_{k-1}$ for scalars $a_1,\dots ,a_{k-1}$. Then, using the multilinearity of $\omega$, we have $\begin{array}{rl}\omega (u_1,u_2,\dots ,u_k)& =\omega (u_1,u_2,\dots ,a_1{u}_1+\cdots +a_{k-1}{u}_{k-1})\\ & =a_1\omega (u_1,u_2,\dots ,u_1)+\cdots +a_{k-1}\omega (u_1,u_2,\dots ,u_{k-1}).\end{array}$By the skew-symmetry of $\omega$, $\omega (u_1,u_2,\dots ,u_i)$ vanishes for all $i\le k-1$, which shows $\omega (u_1,u_2,\dots ,u_k)=0$ as desired. $\square$

Corollary

Any $k$-form on an $n$-dimensional vector space $V$ with $k>n$ is identically zero.

Proof $k$ vectors in $V$ must be linearly dependent if $k>n$. $\square$

Attention

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

The Exterior Product

Exterior (Wedge) Product

The exterior product of a $k$-form $\omega \in {\Lambda }^k(V)$ and a $l$-form $\eta \in {\Lambda }^l(V)$ is defined as the $(k+l)$-form $\omega \wedge \eta$ given by: $(\omega \wedge \eta )(v_1,\dots ,v_{k+l}):=\frac{1}{k!l!}{\sum }_{\sigma \in S_{k+l}}\text{sgn}(\sigma )\omega (v_{\sigma (1)},\dots ,v_{\sigma (k)})\eta (v_{\sigma (k+1)},\dots ,v_{\sigma (k+l)}).$Alternatively, we can also write $(\omega \wedge \eta )(v_1,\dots ,v_{k+l})={\sum }_{\begin{array}{c}{i}_1<\cdots <i_k,\\ j_1<\cdots <j_l\end{array}}\text{sgn}(i_1,\dots ,i_k,j_1,\dots ,j_l)\omega (v_{i_1},\dots ,v_{1_k})\eta (v_{j_1},\dots ,v_{j_l}).$

The above two expressions are equivalent, because instead of summing over all permutations of $k+l$ elements, we can sum over all possible partitions of $k$ and $l$ elements, and for each partition, we can permute the $k$ and $l$ elements separately. Notice that for each fixed partition $\{i_1,\dots ,i_k\}\cup \{j_1,\dots ,j_l\}$, assuming $i_1<\cdots <i_k$, $j_1<\cdots <j_l$, and for each permutation $\tau \in S_k$, $\theta \in S_l$, suppose the resulting permutation of $k+l$ elements is $\sigma \in S_{k+l}$, i.e. $\sigma =i_{\tau (1)},\dots ,i_{\tau (k)},j_{\theta (1)},\dots j_{\theta (l)}$, we have $\text{sgn}(\sigma )=\text{sgn}(i_1,\dots ,i_k,j_1,\dots ,j_l)\text{sgn}(\tau )\text{sgn}(\theta ),$as it is a group homomorphism. Thus $\text{sgn}(i_1,\dots ,i_k,j_1,\dots ,j_l)=\text{sgn}(\sigma )\text{sgn}(\tau )\text{sgn}(\theta )$, because the inverse of any element is itself in the group $\{\pm 1\}\cong {\mathbb{Z}}_2$. Moreover, by skew-symmetry,$\omega (v_{\sigma (1)},\dots ,v_{\sigma (k)})=\text{sgn}(\tau )\omega (v_{i_1},\dots ,v_{i_k}),\eta (v_{\sigma (k+1)},\dots ,v_{\sigma (k+l)})=\text{sgn}(\theta )\eta (v_{j_1},\dots ,v_{j_l}).$So finally we get $\begin{array}{rl}(\omega \wedge \eta )(v_1,\dots ,v_{k+l})& =\frac{1}{k!l!}\sum _{\sigma \in S_{k+l}}\text{sgn}(\sigma )\omega (v_{\sigma (1)},\dots ,v_{\sigma (k)})\eta (v_{\sigma (k+1)},\dots ,v_{\sigma (k+l)})\\ & =\frac{1}{k!l!}\sum _{\begin{array}{c}{i}_1<\cdots <i_k,\\ j_1<\cdots <j_l\end{array}}k!l!\cdot \text{sgn}(i_1,\dots ,i_k,j_1,\dots ,j_l)\omega (v_{i_1},\dots ,v_{1_k})\eta (v_{j_1},\dots ,v_{j_l})\\ & =\sum _{\begin{array}{c}{i}_1<\cdots <i_k,\\ j_1<\cdots <j_l\end{array}}\text{sgn}(i_1,\dots ,i_k,j_1,\dots ,j_l)\omega (v_{i_1},\dots ,v_{1_k})\eta (v_{j_1},\dots ,v_{j_l}).\end{array},$where the second equality holds as there are $k!l!$ permutations for each fixed partition. $\square$

e.g. Given two one forms $\omega ,\eta \in {\Lambda }^1(V)$, their exterior product $\omega \wedge \eta$ is a two form that satisfies $(\omega \wedge \eta )(v,u)=\omega (v)\eta (u)-\omega (u)\eta (v)=\det \left(\left[\begin{array}{cc}\omega (v)& \eta (v)\\ \omega (u)& \eta (u)\end{array}\right]\right).$

Proposition

Let $\{e_i{\}}_{i=1}^n$ be a basis for a vector space $V$ and $\{e_i^{\ast }{\}}_{i=1}^n$ be the dual basis. Then for any $l\le m\le n,$ $(e_l^{\ast }\wedge e_{l+1}^{\ast }\wedge \cdots \wedge e_m^{\ast })(e_l,e_{l+1},\dots ,e_m)=1.$

Proof We will prove by induction on $m$. Clearly, the base case $l=m$ is true. Now assume it holds for all $q\le k$, then $\begin{array}{rl}& (e_l^{\ast }\wedge e_{l+1}^{\ast }\wedge \cdots \wedge e_k^{\ast }\wedge e_{k+1}^{\ast })(e_l,e_{l+1},\dots ,e_{k+1})\\ =& \sum _{\sigma (l)<\cdots <\sigma (k)}\text{sgn}(\sigma )(e_l^{\ast }\wedge e_{l+1}^{\ast }\wedge \cdots \wedge e_k^{\ast })(e_{\sigma (l)},e_{\sigma (l+1)},\dots e_{\sigma (k)})\cdot e_{k+1}^{\ast }(e_{\sigma (k+1)})\\ =& (e_l^{\ast }\wedge e_{l+1}^{\ast }\wedge \cdots \wedge e_k^{\ast })(e_l,e_{l+1},\dots e_k)\cdot 1\\ =& 1,\end{array}$where the second equality holds because $e_{k+1}^{\ast }(e_{\sigma (k+1)})$ is nonzero only if $\sigma$ fixes $k+1$, and the last equality holds by the induction hypothesis. $\square$

Theorem

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

• $\omega \wedge \eta =(-1{)}^{kl}\eta \wedge \omega$ for any $k$-form $\omega$ and $l$-form $\eta$.
• $({\lambda }_1{\omega }_1+{\lambda }_2{\omega }_2)\wedge \eta ={\lambda }_1{\omega }_1\wedge \eta +{\lambda }_2{\omega }_2\wedge \eta$ for any $k$-forms ${\omega }_1$, ${\omega }_2$ and $l$-form $\eta$.
• $(\omega \wedge \eta )\wedge \beta =\omega \wedge (\eta \wedge \beta )$.

e.g. Let $e_i$ be a basis for a vector space $V$ with dual basis $e_i^{\ast }$. Then we can calculate $\begin{array}{rl}& (e_1^{\ast }\wedge e_2^{\ast }+e_2^{\ast }\wedge e_3^{\ast })(e_1+e_2,e_3+e_1)\\ =& (e_1^{\ast }\wedge e_2^{\ast }+e_2^{\ast }\wedge e_3^{\ast })(e_1,e_3+e_1)+(e_1^{\ast }\wedge e_2^{\ast }+e_2^{\ast }\wedge e_3^{\ast })(e_2,e_3+e_1)\\ =& (e_1^{\ast }\wedge e_2^{\ast }+e_2^{\ast }\wedge e_3^{\ast })(e_1,e_3)+(e_1^{\ast }\wedge e_2^{\ast }+e_2^{\ast }\wedge e_3^{\ast })(e_2,e_3)+(e_1^{\ast }\wedge e_2^{\ast }+e_2^{\ast }\wedge e_3^{\ast })(e_2,e_1)\\ =& (e_1^{\ast }\wedge e_2^{\ast })(e_1,e_3)+(e_2^{\ast }\wedge e_3^{\ast })(e_1,e_3)+(e_1^{\ast }\wedge e_2^{\ast })(e_2,e_3)\\ & +(e_2^{\ast }+e_3^{\ast })(e_2,e_3)+(e_1^{\ast }\wedge e_2^{\ast })(e_2,e_1)+(e_2^{\ast }\wedge e_3^{\ast })(e_2,e_1)\\ =& 0+0+0+1-1+0=0\end{array}$

Proposition

The space of $k$-forms on an $n$-dimensional vector space $V$ is a vector space of dimension $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, and the exterior product $\wedge$ gives the space of all forms $\Lambda (V):={\oplus }_{k=1}^n{\Lambda }^k(V)$ the structure of a super-commutative graded algebra over the field $\mathbb{R}$. In particular, if $\{e_j^{\ast }{\}}_{j=1}^n$ forms a basis for the 1-forms, $\{e_{i_1}^{\ast }\wedge e_{i_2}^{\ast }\wedge \cdots \wedge e_{i_k}^{\ast }:i_1<\cdots <i_k\}$ forms a basis for the $k$-forms.

Proof We first show that the forms are linearly independent. Suppose $\eta \in {\Lambda }^k(V)$ is a linear combination of these forms, i.e. $\eta ={\sum }_{i_1<\cdots <i_k}{a}_{i_1,\cdots ,i_k}{e}_{i_1}^{\ast }\wedge e_{i_2}^{\ast }\wedge \cdots \wedge e_{i_k}^{\ast }.$We set $\eta$ to be identically $0$. Notice that since $(e_{i_1}^{\ast }\wedge \cdots \wedge e_{i_k}^{\ast })(e_{j_1},e_{j_2},\dots ,e_{j_k})$ is nonzero only if $i=j$, $\eta (e_{j_1},e_{j_2},\dots ,e_{j_k})=a_{j_1,\dots ,j_k}=0$ holds for all $j_1<\cdots <j_k$, thus the forms are linearly independent. Now we show that every $k$-form can be expressed as a linear combination of these forms. Suppose $\omega \in {\Lambda }^k(V)$, then $\omega ={\sum }_{i_1<\cdots <i_k}\omega (e_{i_1},e_{i_2},\dots ,e_{i_k})e_{i_1}^{\ast }\wedge e_{i_2}^{\ast }\wedge \cdots \wedge e_{i_k}^{\ast }.$Finally, the exterior product gives $\Lambda (V)$ the structure of a super-commutative graded algebra because of the theorem. $\square$

Pullback Invariance

Pullback Preserves Exterior Product

For a linear map $A:V\to W$, and $\omega \in {\Lambda }^k(W)$, $\eta \in {\Lambda }^l(W)$, there holds $A^{\ast }(\omega \wedge \eta )=A^{\ast }(\omega )\wedge A^{\ast }(\eta ).$

Proof For all $v_1,\dots ,v_{k+l}\in V$, 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 $k$-form $\omega \in {\Lambda }^k(V)$ and a vector $v\in V$ is defined as the $(k-1)$-form $i_v\omega$ given by: $i_v\omega (v_1,\dots ,v_{k-1}):=\omega (v,v_1,\dots ,v_{k-1}).$

Proposition

The interior product is a degree $-1$ superderivation in the sense that it defines a linear map from $k$-forms to $(k-1)$-forms, and satisfies $i_v(\omega \wedge \eta )=i_v\omega \wedge \eta +(-1{)}^{\mathrm{deg}\omega }\omega \wedge i_v\eta .$

Proof

Theorem

Let $\eta$ be the standard volume form on ${\mathbb{R}}^3$. Then $v\mapsto i_v\eta$ is an isomorohism from ${\mathbb{R}}^3$ to ${\Lambda }^2({\mathbb{R}}^3)$. Moreover, the following identities hold: $⟨u\times v,w⟩=\eta (u,v,w),v^{\ast }\wedge w^{\ast }=i_{v\times w}\eta ,v^{\ast }\wedge i_w\eta =⟨v,w⟩\eta .$