8.3 Exterior product

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The vector space $\Lambda (V)$ is defined to be the direct sum

$$\Lambda (V)={\Lambda }^0(V)\oplus {\Lambda }^1(V)\oplus {\Lambda }^2(V)\oplus \cdots \oplus {\Lambda }^n(V).$$

Elements of $\Lambda (V)$ are called multivectors, written

$$A=A_0+A_1+\cdots +A_n\text{ where }{A}_r\in {\Lambda }^r(V).$$

As shown in Section 6.4,

$$\dim (\Lambda (V))=\sum _{r=0}^n\left(\genfrac{}{}{0ex}{}{n}{r}\right)=2^n.$$

For any $r$-vector $A$ and $s$-vector $B$ we define their exterior product or wedge product $A\wedge B$ to be the ($r$ $s$)-vector

$$A\wedge B=\mathcal{A}(A\otimes B),\text{\hspace{1em}(8.9)}$$

and extend to all of $\Lambda (V)$ by linearity,

$$(aA+bB)\wedge C=aA\wedge C+bB\wedge C,A\wedge (aB+bC)=aA\wedge B+bA\wedge C.$$

The wedge product of a 0-vector, or scalar, $a$ with an $r$-vector $A$ is simply scalar multipli cation, since

$$a\wedge A=\mathcal{A}(a\otimes A)=\mathcal{A}(aA)=a\mathcal{A}A=aA.$$

For general multivectors $({\sum }_a{A}_a)$ and $({\sum }_b{B}_b)$ ) we have

$$\left(\sum _a{A}_a\right)\wedge \left(\sum _b{B}_b\right)=\sum _a\sum _b{A}_a\wedge B_b.$$

The associative law holds by Theorem 8.1 and the associative law for tensor products,

$$\begin{array}{rl}A\wedge (B\wedge C)& =\mathcal{A}(A\otimes \mathcal{A}(B\otimes C))\\ & =\mathcal{A}(A\otimes (B\otimes C))\\ & =\mathcal{A}((A\otimes B)\otimes C)\\ & =\mathcal{A}(\mathcal{A}(A\otimes B)\otimes C)\\ & =(A\wedge B)\wedge C.\end{array}$$

The space $\Lambda (V)$ with wedge product is therefore an associative algebra, called the exterior algebra over $V$. There is no ambiguity in writing expressions such as $A\wedge B\wedge C$ . Since the exterior product has the property

$$\wedge :{\Lambda }^r(V)\times {\Lambda }^s(V)\to {\Lambda }^{r+s}(V),$$

it is called a graded product and the exterior algebra $\Lambda (V)$ is called a graded algebra.

Example 8.1

If $u$ and $v$ are vectors then their exterior product has the property

$$\begin{array}{rl}(u\wedge v)(\omega ,\rho )& =\mathcal{A}(u\otimes v)(\omega ,\rho )\\ & =\mathcal{A}(u\otimes v)(\omega ,\rho )\\ & =\frac{1}{2}(u(\omega )v(\rho )-u(\rho )v(\omega )),\end{array}$$

whence

$$u\wedge v=\frac{1}{2}(u\otimes v-v\otimes u)=-v\wedge u.\text{\hspace{1em}(8.10)}$$

Obviously $u\wedge u=0$ . Setting $\omega ={\epsilon }^i$ and $\rho ={\epsilon }^j$ in the derivation of Eq. (8.10) gives

$$(u\wedge v{)}^{ij}=\frac{1}{2}(u^i{v}^j-u^j{v}^i).$$

In many textbooks exterior product $A\wedge B$ is defined as $\frac{(r+s)!}{r!s!}\mathcal{A}(A\otimes B)$ , in which case the factor $\frac{1}{2}$ does not appear in these formulae.

The anticommutation property Eq. (8.10) is easily generalized to show that for any permu tation $\pi$ of $1,2,\dots ,r,$

$$u_{\pi (1)}\wedge u_{\pi (2)}\wedge \cdots \wedge u_{\pi (r)}=(-1{)}^{\pi }{u}_1\wedge u_2\wedge \cdots \wedge u_r.\text{\hspace{1em}(8.11)}$$

The basis $r$-vectors $e_{i_1\dots i_r}$ defined in Eq. (8.4) can clearly be written

$$e_{i_1{i}_2\dots i_r}=e_{i_1}\wedge e_{i_2}\wedge \cdots \wedge e_{i_r},\text{\hspace{1em}(8.12)}$$

and the permutation property Eq. (8.5) is equivalent to Eq. (8.11).

For any pair ofbasis elements $e_{i_1\dots i_r}\in {\Lambda }^r(V)$ and $e_{j_1\dots j_s}\in {\Lambda }^s(V)$ it follows immediately from Eq. (8.12) that

$$e_{i_1\dots i_r}\wedge e_{j_1\dots j_s}=e_{i_1\dots i_r{j}_1\dots j_s}.\text{\hspace{1em}(8.13)}$$

These expressions permit us to give a unique expression for the exterior products ofarbitrary multivectors, for if $A$ is an $r$-vector and $B$ an $s$-vector,

$$A=r!\underbrace{\sum _{i_1<i_2<\cdots <i_r}\cdots \sum _{i_r}}{A}^{i_1\dots i_r}{e}_{i_1\dots i_r},B=s!\underbrace{\sum _{j_1<j_2<\cdots <j_s}\cdots \sum _{j_s}}{B}^{j_1\dots j_s}{e}_{j_1\dots j_s},$$

then

$$A\wedge B=r!s!\underset{i_1<\cdots <i_r}{\underbrace{\sum \cdots \sum }}\underset{j_1<\cdots <j_s}{\underbrace{\sum \cdots \sum }}{A}^{i_1\dots i_r}{B}^{j_1\dots j_s}{e}_{i_1\dots i_r{j}_1\dots j_s}.\text{\hspace{1em}(8.14)}$$

Alternatively, the formula for wedge product can be written

$$\begin{array}{rl}A\wedge B& =A^{i_1{i}_2\dots i_r}{e}_{i_1{i}_2\dots i_r}\wedge B^{j_1{j}_2\dots j_s}{e}_{j_1{j}_2\dots j_s}\\ & =A^{i_1\dots i_r}{B}^{j_1\dots j_s}{e}_{i_1\dots i_r{j}_1\dots j_s}\\ & =A^{[i_1\dots i_r}{B}^{j_1\dots j_s]}{e}_{i_1}\otimes \cdots \otimes e_{i_r}\otimes e_{j_1}\otimes \cdots \otimes e_{j_s}\end{array}$$

on using Eq. (8.4). The tensor components of $A\wedge B$ are thus

$$(A\wedge B{)}^{i_1\dots i_r{j}_1\dots j_s}=A^{[i_1\dots i_r}{B}^{j_1\dots j_s]}.\text{\hspace{1em}(8.15)}$$

Example 8.2

If $u$ and $v$ are 1-vectors, then

$$(u\wedge v{)}^{ij}=u^{[i}{v}^{j]}=\frac{1}{2}(u^i{v}^j-u^j{v}^i)$$

as in Example 8.1. For exterior product of a 1-vector $u$ and a 2-vector $A$ we find, using the skew symmetry $A^{jk}=-A^{kj}$ ,

$$\begin{array}{l}(u\wedge A{)}^{ijk}=u^{[i}{A}^{jk]}\\ =\frac{1}{6}(u^i{A}^{jk}-u^i{A}^{kj}+u^j{A}^{ki}-u^j{A}^{ik}+u^k{A}^{ij}-u^k{A}^{ji})\\ =\frac{1}{3}(u^i{A}^{jk}+u^j{A}^{ki}+u^k{A}^{ij}).\end{array}$$

The wedge product of three vectors is

$$\begin{array}{c}u\wedge v\wedge w=\frac{1}{6}(u\otimes v\otimes w+v\otimes w\otimes u+w\otimes u\otimes v\\ -u\otimes w\otimes v-w\otimes v\otimes u-v\otimes u\otimes w).\end{array}$$

In components,

$$\begin{array}{l}(u\wedge v\wedge w{)}^{ijk}=u^{[i}{v}^j{w}^{k]}\\ =\frac{1}{6}(u^i{v}^j{w}^k-u^i{v}^k{w}^j+u^j{v}^k{w}^i-u^j{v}^i{w}^k+u^k{v}^i{w}^j-u^k{v}^j{w}^i).\end{array}$$

Continuing in this way, the wedge product of any $r$ vectors $u_1,u_2,\dots ,u_r$ results in the $r$-vector

$$u_1\wedge u_2\wedge \cdots \wedge u_r=\frac{1}{r!}\sum _{\pi }(-1{)}^{\pi }{u}_{\pi (1)}\otimes u_{\pi (2)}\otimes \cdots \otimes u_{\pi (r)}$$

which has components

$${\left(u_1\wedge u_2\wedge \cdots \wedge u_r\right)}^{i_1{i}_2\dots i_r}=u_1^{[i_1}{u}_2^{i_2}\dots u_r^{i_r]}.$$

The anticommutation rule for vectors, $u$ $\wedge v=-v\wedge u.$ , generalizes for an $r$-vector $A$ and $s$-vector $B$ to

$$A\wedge B=(-1{)}^{rs}B\wedge A.\text{\hspace{1em}(8.16)}$$

This result has been proved in Section 6.4, Eq. (6.20). It follows from

$$\begin{array}{c}A\wedge B=A^{i_1\dots i_r}{B}^{j_1\dots j_s}{e}_{i_1\dots i_r{j}_1\dots j_s}\\ =(-1{)}^{rs}{B}^{j_1\dots j_s}{A}^{i_1\dots i_r}{e}_{j_1\dots j_s{i}_1\dots i_r}\end{array}$$

since $rs$ interchanges are needed to bring the indices $j_1,\dots ,j_r$ in front of the indices $i_1,\dots ,i_s$

Ifr is even then A commutes with all multivectors, while a pair ofodd degree multivectors always anticommute. For example if $u$ is a 1-vector and $A$ a 2-vector, then $u$ $\wedge A=A\wedge u$ , since

$$(u\wedge A{)}^{ijk}=u^{[i}{A}^{jk]}=A^{[jk}{u}^{i]}=A^{[ij}{u}^{k]}=(A\wedge u{)}^{ijk}.$$

The space of multiforms is defined in a totally analagous manner,

$${\Lambda }^{\ast }(V)=\Lambda (V^{\ast })={\Lambda }_0^{\ast }(V)\oplus {\Lambda }_1^{\ast }(V)\oplus {\Lambda }_2^{\ast }(V)\oplus \cdots \oplus {\Lambda }_n^{\ast }(V),$$

with an exterior product

$$\alpha \wedge \beta =\mathcal{A}(\alpha \otimes \beta )\text{\hspace{1em}(8.17)}$$

having identical properties to the wedge product on multivectors

$$\alpha \wedge (\beta \wedge \gamma )=(\alpha \wedge \beta )\wedge \gamma \text{ and }\alpha \wedge \beta =(-1{)}^{rs}\beta \wedge \alpha .$$

The basis $r$-forms defined in Eq. (8.7) can be written as

$${\epsilon }^{i_1\dots i_r}={\epsilon }^{i_1}\wedge \cdots \wedge {\epsilon }^{i_r},$$

and the component expression for exterior product of a pair of forms is

$$(\alpha \wedge \beta {)}_{i_1\dots i_r{j}_1\dots j_s}={\alpha }_{[i_1\dots i_r}{\beta }_{j_1\dots j_s]}.$$

Simple $p$-vectors and subspaces

A simple $p$-vector is one that can be written as a wedge product of 1-vectors,

$$A=v_1\wedge v_2\wedge \cdots \wedge v_p,v_i\in {\Lambda }^1(V)=V.$$

Similarly a simple $p$-form $\alpha$ is one that is decomposable into a wedge product of 1-forms,

$$\alpha ={\alpha }^1\wedge {\alpha }^2\wedge \cdots \wedge {\alpha }^p,{\alpha }^i\in {\Lambda }^{\ast 1}(V)=V^{\ast }.$$

Let $W$ be a $p$-dimensional subspace of $V$. For any basis $e_1,e_2,\dots ,e_p$ of $W_{;}$ , define the $p$-vector $E_W=e_1\wedge e_2\wedge \cdot \cdot \cdot \wedge e_p$ . If $e_1^{\prime },e_2^{\prime },\dots ,e_p^{\prime }$ is a second basis then for some coefficients $B_{i^{\prime }}^i$

$$e_{i^{\prime }}^{\prime }=\sum _{i=1}^p{B}_{i^{\prime }}^i{e}_i,$$

and the $p$-vector corresponding to this basis is

$$\begin{array}{l}{E}_W^{\prime }=e_1^{\prime }\wedge \cdots \wedge e_p^{\prime }\\ ={\sum }_{i_1}\cdots {\sum }_{i_p}{B}_1^{i_1}{B}_2^{i_2}\dots B_p^{i_p}{e}_{i_1}\wedge e_{i_2}\wedge e_{i_p}\\ ={\sum }_{\pi }(-1{)}^{\pi }{B}_1^{i_{\pi }(1)}{B}_2^{i_{\pi }(2)}\dots B_p^{i_{\pi }(p)}{e}_1\wedge e_2\wedge \cdots \wedge e_p\\ =\det [B_{j^{\prime }}^i]E_W.\end{array}$$

Hence the subspace $W$ corresponds uniquely, up to a multiplying factor, to a simple $p$-vector $E_W$

Theorem 8.2 · The subspace determined by a simple multivector

A vector $u$ belongs to $W$ ifand only if $u$ $\wedge E_W=0$

Proof

This statement is an immediate corollary of Theorem 6.2.

Problems

Problem 8.1

Express components of the exterior product of two 2-vectors $A=A^{ij}{e}_{ij}$ and $B=B^{kl}{e}_{kl}$ as a sum of six terms,

$$(A\wedge B{)}^{ijkl}=\frac{1}{6}(A^{ij}{B}^{kl}+A^{ik}{B}^{lj}+\dots ).$$

How many terms would be needed for a product of a 2-vector and a 4-vector? Show that in genera the components of the exterior product of an $r$-vector and an $s$-vector can be expressed as a sum of $\frac{(r+s)!}{r!s!}\mathrm{terms}.$

Problem 8.2

Let $V$ be a four-dimensional vector space with basis $\{e_1,e_2,e_3,e_4\}$ , and $A$ $a$ 2-vecto on $V$.

(a) Show that a vector $u$ satisfies the equation

$$A\wedge u=0$$

if and only if there exists a vector $v$ such that

$$A=u\wedge v.$$

[Hint: Pick a basis such that $e_1=u.]$

(b) If

$$A=e_2\wedge e_1+a{e}_1\wedge e_3+e_2\wedge e_3+c{e}_1\wedge e_4+b{e}_2\wedge e_4$$

write out explicitly the equations $A\wedge u=0$ where $u=u^1{e}_1+u^2{e}_2+u^3{e}_3+u^4{e}_4$ and show that they have a solution if and only $i$ $c=ab$ . In this case find two vectors $u$ and $v$ such tha $A=u\wedge v.$

(c) In general show that the 4-vector $A$ $\wedge A=8\alpha e_1\wedge e_2\wedge e_3\wedge e_4$ where

$$\alpha =A^{12}{A}^{34}+A^{23}{A}^{14}+A^{31}{A}^{24},$$

and

$$\det [A^{ij}]={\alpha }^2.$$

(d) Show that $A$ is the wedge product of two vectors $A=u\wedge$ $v$ if and only if $A\wedge A=0$

Problem 8.3

Prove Cartan’s lemma, that if $u_1,u_2,\dots ,u_p$ are linearly independent vectors and $v_1,\dots ,v_p$ are vectors such that

$$u_1\wedge v_1+u_2\wedge v_2+\cdots +u_p\wedge v_p=0,$$

then there exists a symmetric set of coefficients $A_{ij}=A_{ji}$ such that

$$v_i=\sum _{j=1}^p{A}_{i}j{u}_j.$$

[Hint: Extend the $u_i$ to a basis for the whole vector space $V$.]

Problem 8.4

If $V$ is an $n$-dimensional vector space and $A$ a 2-vector, show that there exists a basi $e_1,e_2,\dots ,e_n$ of $V$ such that

$$A=e_1\wedge e_2+e_3\wedge e_4+\dots e_{2r-1}\wedge e_{2r},$$

for some number 2$r$, called the rank of $A$.

(a) Show that the rank only depends on the 2-vector $A$, not on the choice of basis, by showing tha ${\mathbf{A}}^r\ne 0$ and $A^{r+1}=0$ where

$$A^p=\underset{p}{\underbrace{A\wedge A\wedge \cdots \wedge A}}.$$

(b) If $f_1,f_2,\dots ,f_n$ is any basis of $V$ and $\mathcal{A}={\mathcal{A}}^{ij}{f}_i\otimes f_j$ where $A^{ij}=-A^{ji}$ , show that the rank of the matrix of components $\mathsf{A}=[A^{ij}]$ coincides with the rank as defined above.

Problem 8.5

Let $V$ be an $n$-dimensional space and $A$ an arbitrar $(n-1)-\mathrm{vector}.$

(a) Show that the subspace $V_A$ of vectors $u$ such that $u$ $\wedge A=0$ has dimension $n-1$

(b) Show that every ($n$ 1)-vector $A$ is decomposable, $A=v_1\wedge v_2\wedge \cdot \cdot \cdot \wedge v_{n-1}$ for some vector $v_1,\dots ,v_{n-1}\in V$ . [Hint: Take a basis for $e_1,\dots ,e_n$ of $V$ such that the first $n-1$ vectors span the subspace $V_A$ , which is always possible by Theorem 3.7, and expand $A$ in terms of this basis.]