3. The Exterior Algebra of Multicovectors

Dual Space

Let $V$ be a vector space. We define the dual space of $V$ by

$$V^{\vee }=\text{Hom}(V,\mathbb{R}).$$

We call elements of $V^{\vee }$ 1-covectors on $V$.

We assume that $V$ is finite-dimensional with basis $e_1,\dots ,e_n$, i.e., for all $v\in V$ there is a unique linear combination $v=\sum v^i{e}_i$ with $v^i\in \mathbb{R}$. Define ${\alpha }^i:V\to \mathbb{R}$ by

$${\alpha }^i(e_j)={\delta }_j^i=\{\begin{array}{ll}1& \text{if }i=j\\ 0& \text{if }i\ne j.\end{array}$$

Prp

The functions ${\alpha }^1,\dots ,{\alpha }^n$ form a basis for $V^{\vee }$.

Proof: We need to show they span and are linearly independent. Let $f\in V^{\vee }$ and $v\in V$. Notice that

$$f(v)=\sum v^{i}f(e_i)=\sum f(e_i){\alpha }^i(v).$$

So

$$f=\sum f(e_i){\alpha }^i.$$

Thus, ${\alpha }^1,\dots ,{\alpha }^n$ spans $V^{\vee }$. Next let $v^1,\dots ,v^n\in \mathbb{R}$ such that

$$\sum v^i{\alpha }^i=0.$$

Then it follows that

$$\sum v^i{\alpha }^i(e_j)=v^j=0$$

for all $j=1,\dots ,n$. Therefore, $v=0$.

Q.E.D.

Cor

The dual space $V^{\vee }$ has the same dimension as $V$.

Permutations

A permutation of a set $A=\{1,\dots ,k\}$ is a bijection $\sigma :A\to A$. It is common to visualize permutations as matrices. For example, the permutation

$$\left[\begin{array}{ccccc}1& 2& 3& 4& 5\\ 2& 4& 5& 1& 3\end{array}\right]$$

maps $1\mapsto 2$, $2\mapsto 4$, $3\mapsto 5$, $4\mapsto 1$ and $5\mapsto 3$.

The product of two permutations $\tau$ and $\sigma$ of $A$ is the permutation $\tau \circ \sigma :A\to A$ given by composition. The cyclic permutation, $(a_1{a}_2\cdots a_r)$, where each $a_i$ is distinct, is the permutation $\sigma$ such that $\sigma (a_1)=a_2$, $\sigma (a_2)=a_3$, …, $\sigma (a_{r-1})=a_r$ and $\sigma (a_r)=a_1$. Any permutation can be written as a product of disjoint transpositions. For example,

$$(12345)=(15)(14)(13)(12).$$

We define $S_k$ to be the group of all permutations on the set $\{1,\dots ,k\}$ with the group operation defined by composition. A permutation $\sigma \in S_k$ is either even or odd depending on the number of transpositions in $\sigma$‘s decomposition. We define the sign of $\sigma$ as +1 if it is even, or -1 if it is odd. One may show that

$$\text{sgn}(\sigma \tau )=\text{sgn}(\sigma )\text{sgn}(\tau )$$

for all $\sigma ,\tau \in S_k$.

An inversion in a permutation $\sigma$ is an ordered pair $(\sigma (i),\sigma (j))$ such that $i<j$ but $\sigma (i)>\sigma (j)$. For example, the permutation

$$\left[\begin{array}{ccccc}1& 2& 3& 4& 5\\ 2& 4& 5& 1& 3\end{array}\right].$$

has inversions $(2,1)$, $(4,1)$, $(5,1)$, $(4,3)$ and $(5,3)$.

Prp

A permutation is even if and only if it has an even number of inversions.

Multilinear Functions

A function $f:V^k=V\times \cdots \times V\to \mathbb{R}$ is $k$-linear if it linear in each of its arguments:

$$f(\dots ,av+bw,\dots )=af(\dots ,v,\dots )+bf(\dots ,w,\dots )$$

A $k$-linear function is also called a **$k$-tensor.

Example: With respect to the standard basis $e_1,\dots ,e_n$ for ${\mathbb{R}}^n$, the dot product

$$f(v,w)=v\cdot w=\sum v^i{w}^i$$

is bilinear.

Def

A $k$-linear function $f:V^k\to \mathbb{R}$ is symmetric if

$$f(v_{\sigma (1)},\dots ,v_{\sigma (k)})=f(v_1,\dots ,v_k)$$

for all permutations $\sigma \in S_k$; it is alternating if

$$f(v_{\sigma (1)},\dots ,v_{\sigma (k)})=(\text{sgn}(\sigma ))f(v_1,\dots ,v_k).$$

We define the space $A_k(V)$ to be all alternating $k$-linear functions on a vector space $V$. We call such functions alternating k-tensors or $k$-covectors.

The Permutation Action on Multilinear Functions

If $G$ is a group and $X$ is a set, the map $G\times X\mapsto X$ defined by

$$(\sigma ,x)\mapsto \sigma x$$

is called a left action of $G$ on $X$ if

1. $e\cdot x=x$ for all $x\in X$
2. $\tau \cdot (\sigma \cdot x)=(\tau \sigma )\cdot x$ for all $\tau ,\sigma \in G$ and $x\in X$. A right action of $G$ on $X$ is defined similarily.

The orbit of an element $x\in X$ is the set

$$Gx:=\{\sigma \cdot x\mid \sigma \in G\}$$

As we saw in the definition above, we can define a group action on $k$-linear functions by $S_k$ by defining

$$(\sigma f)(v_1,\dots ,v_k)=f(v_{\sigma (1)},\dots ,v_{\sigma (k)}).$$

Thus, $f$ is symmetric if and only if $\sigma f=f$ and alternating if and only if $\sigma f=(\text{sgn}(\sigma ))f$.

In the case $k=1$, $S_1$ is the trivial group and all 1-linear functions are both symmetric and alternating.

Lemma

If $\sigma ,\tau \in S_k$ and $f$ is a $k$-linear function on $V$, then $\tau (\sigma f)=(\tau \sigma )f$.

Proof: Notice that

$$\begin{array}{rl}\tau (\sigma f)(v_1,\dots ,v_k)& =\tau f(v_{\sigma (1)},\dots ,v_{\sigma (k)})\\ & =f(v_{\tau (\sigma (1))},\dots ,v_{\tau (\sigma (k))})\\ & =f(v_{\tau \sigma (1)},\dots ,v_{\tau \sigma (k)})\\ & =(\tau \sigma )f(v_1,\dots ,v_k).\end{array}$$

Q.E.D.

The Symmetrizing and Alternating Operators

From any $k$-linear function $f$ on a vector space $V$, we can construct a symmetric $k$-linear function $Sf$ by adding all possible permutations:

$$Sf=\sum _{\sigma \in S_k}\sigma f.$$

Indeed, notice that for any $\tau \in S_k$,

$$\begin{array}{rl}\tau (Sf)& =\sum _{\sigma \in S_k}\tau (\sigma f)\\ & =\sum _{\sigma \in S_k}(\tau \sigma )f\\ & =\sum _{{\sigma }^{\prime }\in S_k}{\sigma }^{\prime }f\\ & =Sf.\end{array}$$

Similarly, we can construct an alternating $k$-linear function

$$Af=\sum _{\sigma \in S_k}(\text{sgn}\sigma )\sigma f.$$

Notice that for any $\tau \in S_k$,

$$\begin{array}{rl}\tau (Af)& =\sum _{\sigma \in S_k}(\text{sgn}\sigma )\tau (\sigma f)\\ & =\sum _{\sigma \in S_k}(\text{sgn}\sigma )(\tau \sigma )f\\ & =(\text{sgn}\tau )\sum _{\sigma \in S_k}(\text{sgn}\tau \sigma )(\tau \sigma )f\\ & =(\text{sgn}\tau )\sum _{{\sigma }^{\prime }\in S_k}(\text{sgn}{\sigma }^{\prime }){\sigma }^{\prime }f\\ & =(\text{sgn}\tau )Af.\end{array}$$

Lemma

If $f$ is an alternating $k$-linear function on a vector space $V$, then $Af=(k!)f$.

Proof: Notice that $\sigma f=(\text{sgn}\sigma )f$ and $\text{sgn}\sigma \in \{-1,1\}$. Thus,

$$Af=\sum _{\sigma \in S_k}(\text{sgn}\sigma )\sigma f=\sum _{\sigma \in S_k}(\text{sgn}\sigma {)}^{2}f=\sum _{\sigma \in S_k}f=(k!)f.$$

Q.E.D.

The Tensor Product

Let $f$ be a $k$-linear function and $g$ an $\ell$-linear function. Then their tensor product is the $(k+\ell )$-linear function defined by

$$(f\otimes g)(v_1,\dots ,v_{k+\ell })=f(v_1,\dots ,v_k)g(v_{k+1},{\dots }_{k+\ell }).$$

Example: Let $e_1,\dots ,e_n$ be the basis for $V$ and ${\alpha }^1,\dots ,{\alpha }^n$ the basis for $V^{\vee }$ and $⟨\cdot ,\cdot ⟩:V\times V\to \mathbb{R}$ a bilinear map. So for $v=\sum v^i{e}_i$ and $w=\sum w^i{e}_i$ we get that

$$⟨v,w⟩=\sum v^i{w}^j⟨e_i,e_j⟩=\sum {\alpha }^i(v){\alpha }^j(w)⟨e_i,e_j⟩=\sum ⟨e_i,e_j⟩({\alpha }^i\otimes {\alpha }^j)(v,w).$$

Hence $⟨\cdot ,\cdot ⟩=\sum ⟨e_i,e_j⟩{\alpha }^i\otimes {\alpha }^j$.

Prp

The tensor product is associative:

$$(f\otimes g)\otimes h=f\otimes (g\otimes h).$$

Proof: Let $f$ be $k$-linear, $g$ be $\ell$-linear and $h$ be $m$-linear. Notice that

$$\begin{array}{rl}((f\otimes g)\otimes h)(v_1,\dots ,v_{k+\ell +m})& =(f\otimes g)(v_1,\dots ,v_{k+\ell })h(v_{k+\ell +1},\dots ,v_{k+\ell +m})\\ & =f(v_1,\dots ,v_k)g(v_{k+1},\dots ,v_{k+\ell })h(v_{k+\ell +1},\dots ,v_{k+\ell +m})\\ & =f(v_1,\dots ,v_k)(g\otimes h)(v_{k+1},\dots ,v_{k+\ell +m})\\ & =(f\otimes (g\otimes h))(v_1,\dots ,v_{k+\ell +m}).\end{array}$$

Q.E.D.

The Wedge Product

Let $f\in A_k(V)$ be an alternating $k$-linear function and $g\in A_{\ell }(V)$ be an alternating $\ell$-linear function. The wedge product or exterior product $f\wedge g$ is the alternating $(k+\ell )$-linear function

$$f\wedge g=\frac{1}{k!\ell !}A(f\otimes g),$$

i.e.,

$$(f\wedge g)(v_1,\dots ,v_{k+\ell })=\frac{1}{k!\ell !}\sum _{\sigma \in S_{k+\ell }}(\text{sgn}\sigma )f(v_{\sigma (1)},\dots ,v_{\sigma (k)})g(v_{\sigma (k+1)},\dots ,v_{\sigma (k+\ell )}).$$

Example: For $f\in A_2(V)$ and $g\in A_1(V)$,

$$\begin{array}{rl}A(f\otimes g)(v_1,v_2,v_3)& =f(v_1,v_2)g(v_3)-f(v_1,v_3)g(v_2)+f(v_2,v_3)g(v_1)\\ & -f(v_2,v_1)g(v_3)+f(v_3,v_1)g(v_2)-f(v_3,v_2)g(v_1).\end{array}$$

Since $f$ is alternating $f(v_1,v_2)=-f(v_2,v_1)$ and so on. So

$$(f\wedge g)(v_1,v_2,v_3)=\frac{A(f\otimes g)(v_1,v_2,v_3)}{2}=f(v_1,v_2)g(v_3)-f(v_1,v_3)g(v_2)+f(v_2,v_3)g(v_1).$$

To avoid duplicate terms in the sum, one may only consider $(k,\ell )$-shuffles where

$$\sigma (1)<\cdots <\sigma (k)\text{ and }\sigma (k+1)<\cdots <\sigma (k+\ell ).$$

This reduces the wedge product to a sum of $\left(\genfrac{}{}{0ex}{}{k+\ell }{k}\right)$ terms instead of $(k+\ell )!$, i.e.,

$$(f\wedge g)(v_1,\dots ,v_{k+\ell })=\sum _{(k,\ell )\text{-shuffles}}(\text{sgn}\sigma )f(v_{\sigma (1),\dots ,v_{\sigma (k)}})g(v_{{\sigma }_{k+1}},\dots ,v_{\sigma (k+\ell )}).$$

Example: If $f$ and $g$ are 1-covectors on $V$, then

$$(f\wedge g)(v_1,v_2)=f(v_1)g(v_2)-f(v_2)g(v_1).$$

Example: For $f,g\in A_2(V)$,

$$\begin{array}{rl}(f\wedge g)(v_1,v_2,v_3,v_4)& =f(v_1,v_2)g(v_3,v_4)-f(v_1,v_3)g(v_2,v_4)+f(v_1,v_4)g(v_2,v_3)\\ & +f(v_2,v_3)g(v_1,v_4)-f(v_2,v_4)g(v_1,v_3)+f(v_3,v_4)g(v_1,v_2).\end{array}$$

Anticommutativity of the Wedge Product

Prp

The wedge prouct is anticommuatative: if $f\in A_k(V)$ and $g\in A_{\ell }(V)$, then

$$f\wedge g=(-1{)}^{k\ell }g\wedge f.$$

Proof: Define $\tau \in S_{k+\ell }$ by

$$\left[\begin{array}{cccccc}1& \cdots & \ell & \ell +1& \cdots & \ell +k\\ k+1& \cdots & k+\ell & 1& \cdots & k\end{array}\right].$$

Notice that

$$\sigma (1)=\sigma \tau (\ell +1),\dots ,\sigma (k)=\sigma \tau (\ell +k)$$

and

$$\sigma (k+1)=\sigma \tau (1),\dots ,\sigma (k+\ell )=\sigma \tau (\ell ).$$

Thus,

$$\begin{array}{rl}A(f\otimes g)(v_1,\dots ,v_{k+\ell })& =\sum _{\sigma \in S_{k+\ell }}(\text{sgn}\sigma )f(v_{\sigma (1)},\dots ,v_{\sigma (k)})g(v_{\sigma (k+1)},\dots ,v_{\sigma (k+\ell )})\\ & =\sum _{\sigma \in S_{k+\ell }}(\text{sgn}\sigma )f(v_{\sigma \tau (\ell +1)},\dots ,v_{\sigma \tau (\ell +k)})g(v_{\sigma \tau (1)},\dots ,\sigma \tau (\ell ))\\ & =(\text{sgn}\tau )\sum _{\sigma \in S_{k+\ell }}(\text{sgn}\sigma \tau )f(v_{\sigma \tau (\ell +1)},\dots ,v_{\sigma \tau (\ell +k)})g(v_{\sigma \tau (1)},\dots ,\sigma \tau (\ell ))\\ & =(\text{sgn}\tau )(g\otimes f)(v_1,\dots ,v_{k+\ell }).\end{array}$$

Dividing by $k!\ell !$ gives

$$f\wedge g=(\text{sgn}\tau )g\wedge f$$

and $\text{sgn}\tau =(-1{)}^{k\ell }$.

Q.E.D.

Cor

If $f$ is a multicovector of odd degree on $V$, then $f\wedge f=0$.

Proof: Let $f$ be $k$-linear where $k$ is odd. Then

$$f\wedge f=(-1{)}^{k^2}f\wedge f=-f\wedge f.$$

Therefore, $f\wedge f=0$.

Q.E.D.

Associativity of the Wedge Product

Prp

Let $V$ be a real vector space and $f,g,h$ alternating multilinear functions on $V$ of degrees $k,\ell ,m$ respectively. Then

$$(f\wedge g)\wedge h=f\wedge (g\wedge h).$$

As a result, one may show that

$$f\wedge g\wedge h=\frac{1}{k!\ell !m!}A(f\otimes g\otimes h)$$

which naturally extends to arbitrary many alternating multilinear functions $f_i\in A_{d_i}(V)$,

$$f_1\wedge \cdots \wedge f_r=\frac{1}{(d_1)!\cdots (d_r)!}A(f_1\otimes \cdots \otimes f_r).$$

We denote by $[b_j^i]$ the matrix whose $(i,j)$-entry is $b_j^i$.

Prp

If ${\alpha }_1,\dots ,{\alpha }^k$ are linear functions on a vector space $V$, then

$$({\alpha }^1\wedge \cdots \wedge {\alpha }^k)(v_1,\dots ,v_k)=\det [{\alpha }^i(v_j)].$$

Proof: By Leibniz formula:

$$\begin{array}{rl}({\alpha }^1\wedge \cdots \wedge {\alpha }^k)(v_1,\dots ,v_k)& =\frac{1}{1!\cdots 1!}A({\alpha }^1\otimes \cdots \otimes {\alpha }^k)(v_1,\dots ,v_k)\\ & =\sum _{\sigma \in S_k}(\text{sgn}\sigma ){\alpha }^1(v_{\sigma (1)})\cdots {\alpha }^k(v_{\sigma (k)})\\ & =\det [{\alpha }^i(v_j)].\end{array}$$

Q.E.D.

A Basis for $k$-Covectors

We use the multi-index notation

$$I=(i_1,\dots ,i_k)$$

and write $e_I$ for $(e_{i_1},\dots ,e_{i_k})$ and ${\alpha }^I$ for ${\alpha }^{i_1}\wedge \cdots \wedge {\alpha }^{i_k}$ where $e_1,\dots ,e_n$ is the basis for $V$ and ${\alpha }^1,\dots ,{\alpha }^n$ is the basis for $V^{\vee }$.

Lemma

Let $e_1,\dots ,e_n$ be a basis for a vector space $V$ and ${\alpha }^1,\dots ,{\alpha }^n$ its dual basis for $V^{\vee }$. If $I=(1\le i_1<\cdots <i_k\le n)$ and $J=(1\le j_1<\cdots <j_k\le n)$ are strictly ascending multi-indices, then

$${\alpha }^I(e_J)=({\alpha }^{i_1}\wedge \cdots \wedge {\alpha }^{i_k})(e_{j_1},\dots ,e_{j_k})={\delta }_j^I=\{\begin{array}{ll}1& \text{if }I=J\\ 0& \text{if }I\ne J.\end{array}$$

Proof: Notice that

$${\alpha }^I(e_J)=\det [{\alpha }^i(e_j){]}_{i\in I,j\in J}.$$

If $I=J$, then $[{\alpha }^i(e_j)]$ is the identity matrix. Assume that $I\ne J$ and let $\ell$ indicate the first terms that differ, i.e.,

$$i_1=j_1,\dots ,i_{\ell -1}=j_{\ell -1},i_{\ell }\ne j_{\ell },\dots$$

If $i_{\ell }<j_{\ell }$, then $i_{\ell }\ne j_1,\dots ,i_{\ell }\ne j_{\ell -1}$ and $i_{\ell }\ne j_{l+1},\dots ,i_{\ell }\ne j_k$. As a result, ${\alpha }^{i_{\ell }}(e_{j_1})=\cdots ={\alpha }^{i_{\ell }}(e_{j_k})=0$. Hence, the matrix has a zero row. A similar argument holds if $i_{\ell }>j_{\ell }$.

Q.E.D.

Prp

The alternating $k$-linear functions ${\alpha }^I$ for $I=(i_1<\cdots <i_k)$ form a basis for the space $A_k(V)$ of alternating $k$-linear functions on $V$.

Proof: We need to show spanning and linear independence. Assume that $\sum c_I{\alpha }^I=0$ where the sum runs across all possible strictly ascending multi-indices $I$. Notice then that

$$0=\sum c_I{\alpha }^I(e_J)=c_I.$$

Thus, they are linearly independent.

Now let $f\in A_k(V)$. We claim that

$$f=\sum f(e_I){\alpha }^I=g$$

Notice that

$$g(e_J)=\sum f(e_I){\alpha }^I(e_J)=f(e_J),$$

thus $f$ and $g$ agree over the basis elements. Therefore, $f=g$.

Q.E.D.

Cor

If the vector space $V$ has dimension $n$, then $A_k(V)$ has dimension $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$.

Cor

If $k>\dim V$, then $A_k(V)=0$.

Problems

3.1. Tensor product of covectors

Let $e_1,\dots ,e_n$ be a basis for a vector space $V$ and ${\alpha }^1,\dots ,{\alpha }^n$ its dual basis in $V^{\vee }$. Let $[g_{ij}]\in {\mathbb{R}}^{n\times n}$ and define the bilinear function $f:V\times V\to \mathbb{R}$

$$f(v,w)=\sum _{1\le i,j\le n}{g}_{ij}{v}^i{w}^j$$

for $v=\sum v^i{e}_i$ and $w=\sum w^i{e}_i$ in $V$. Describe $f$ in terms of the tensor products of ${\alpha }^i$ and ${\alpha }^j$, $1\le i,j\le n$.

Answer:

$$\begin{array}{rl}f(v,w)& =\sum _{1\le i,j\le n}{g}_{ij}{\alpha }^i(v){\alpha }^j(w)=\sum _{1\le i,j\le n}{g}_{ij}({\alpha }^i\otimes {\alpha }^j)(v,w)\end{array}$$

3.2. Hyperplanes

(a) Let $V$ be a vector space of dimension $n$ and $f:V\to \mathbb{R}$ a non-zero linear functional. Show that $\dim \text{ker}f=n-1$, i.e., $\text{ker}f$ is a hyperplane in $V$.

Answer: By rank-nullity,

$$\text{rank}f+\text{nullity}f=\text{dim}V=n.$$

Since $\text{rank}f\le 1$ and $f$ is non-zero, it follows that $\text{rank}f=1$.

(b) Show that a nonzero linear functional on a vector space $V$ is determined up to a multiplicative constant by its kernel, i.e., if $f,g:V\to \mathbb{R}$ are nonzero linear functionals and $\text{ker}f=\text{ker}g$ then $g=cf$ for some $c\in \mathbb{R}$.

Answer: Since $f$ is nonzero, there is some $v\in V$ such that $f(v)=r$ where $r\ne 0$. We may assume WLOG that $r=1$, otherwise take $f(v/r)$. Notice that

$$f(x-f(x)v)=f(x)-f(x)f(v)=0$$

implies that $g(x-f(x)v)=0$. Thus,

$$g(x)=g(x-f(x)v)+g(f(x)v)=f(x)g(v).$$

3.4. A characterization of alternating $k$-tensors

Let $f$ be a $k$-tensor on a vector space $V$. Prove that $f$ is alternating if and only if $f$ changes sign whenever two successive arguments are interchanged:

$$f(\dots ,v_{i+1},v_i,\dots )=-f(\dots ,v_i,v_{i+1},\dots )$$

for $i=1,\dots ,k-1$.

Answer: Let $f$ be alternating. Notice that

$$f(\dots ,v_{i+1},v_i,\dots )=\text{sgn}(ii+1)f(\dots ,v_i,v_{i+1},\dots )=-f(\dots ,v_i,v_{i+1},\dots ).$$

For the converse, let $\sigma \in S_k$ and consider the decomposition of $\sigma$

$$\sigma ={\tau }_1\cdots {\tau }_m$$

where each ${\tau }_i$ is a transposition. Notice that

$$\begin{array}{rl}f(v_1,\dots ,v_k)& =-f(v_{{\tau }_1(1)},\dots ,v_{{\tau }_1(k)})\\ & =f(v_{{\tau }_2{\tau }_1(1)},\dots ,v_{{\tau }_2{\tau }_1(k)})\\ & ⋮\\ & =(-1{)}^{m}f(v_{\sigma (1)},\dots ,v_{\sigma (k)})\\ & =(\text{sgn}\sigma )f(v_{\sigma (1)},\dots ,v_{\sigma (k)}).\end{array}$$

Therefore, $f$ is alternating.