# Alternating multilinear forms

Jordan Bell
August 21, 2018

## 1 Permutations

We follow Cartan [2] and Abraham and Marsden [1].

Let $E$ be a real vector space. Let $\mathscr{L}_{p}(E;\mathbb{R})$ be the set of multilinear maps $E^{p}\to\mathbb{R}$.

###### Definition 1.

A map $f\in\mathscr{L}_{p}(E;\mathbb{R})$ is called alternating if $(x_{1},\ldots,x_{p})\in E^{p}$ with $x_{i}=x_{i+1}$ for some $1\leq i implies $f(x_{1},\ldots,x_{p})=0$. Let $\mathscr{A}_{p}(E;\mathbb{R})$ be the set of alternating elements of $\mathscr{L}_{p}(E;\mathbb{R})$.

For a set $X$, let $S_{X}$ be the group of bijections $X\to X$, and let $S_{p}=S_{\{1,\ldots,p\}}$. For $\sigma,\tau\in S_{X}$, write $\sigma\tau=\sigma\circ\tau$.

###### Definition 2.

For a function $f:E^{p}\to\mathbb{R}$ and a permutation $\sigma\in S_{p}$, define the function $\sigma f:E^{p}\to\mathbb{R}$ by

 $(\sigma f)(x_{1},\ldots,x_{p})=f(x_{\sigma(1)},\ldots,x_{\sigma(p)}),\qquad(x_% {1},\ldots,x_{p})\in E^{p}.$
###### Theorem 3.

For a function $f:E^{p}\to\mathbb{R}$ and for $\sigma,\tau\in S_{p}$,

 $\tau(\sigma f)=(\tau\sigma)f.$
###### Proof.

Define $g=\sigma f$. For $(x_{1},\ldots,x_{p})\in E^{p}$ and for $y_{i}=x_{\tau(i)}$, we have

 $\displaystyle\tau(\sigma f)(x_{1},\ldots,x_{p})$ $\displaystyle=\tau(g)(x_{1},\ldots,x_{p})$ $\displaystyle=g(x_{\tau(1)},\ldots,x_{\tau(p)})$ $\displaystyle=g(y_{1},\ldots,y_{p})$ $\displaystyle=(\sigma f)(y_{1},\ldots,y_{p})$ $\displaystyle=f(y_{\sigma(1)},\ldots,y_{\sigma(p)})$ $\displaystyle=f(x_{\tau(\sigma(1))},\ldots,x_{\tau(\sigma(p))})$ $\displaystyle=(\tau\sigma)(f)(x_{1},\ldots,x_{p}).$

Thus

 $\tau(\sigma f)=(\tau\sigma)f.$

For $1\leq i,j\leq p$, define $(i,j)\in S_{p}$ by

 $(i,j)(k)=\begin{cases}j&k=i,\\ i&k=j,\\ k&k\neq i,j,\end{cases}$

called a transposition. Define

 $\tau_{i}=(i,i+1),$

called an adjacent transposition. We can write a transposition $(i,j)$, $i, as a product of $2j-2i-1$ adjacent transpositions:

 $\displaystyle(i,j)$ $\displaystyle=(j-1,j)(j-2,j-1)\cdots(i+1,i+2)(i,i+1)(i+1,i+2)\cdots(j-1,j)$ $\displaystyle=\tau_{j-1}\cdots\tau_{i+1}\tau_{i}\tau_{i+1}\cdots\tau_{j-1}.$
###### Theorem 4.

For $\sigma,\tau\in S_{p}$,

 $\mathrm{sgn}\,(\sigma\tau)=\mathrm{sgn}\,(\sigma)\mathrm{sgn}\,(\tau).$
###### Theorem 5.

Let $f\in\mathscr{L}_{p}(E;\mathbb{R})$. $f\in\mathscr{A}_{p}(E;\mathbb{R})$ if and only if $\sigma f=(\mathrm{sgn}\,\sigma)f$ for all $\sigma\in S_{p}$.

###### Proof.

(i) Suppose that $f\in\mathscr{A}_{p}(E;\mathbb{R})$ and let $\sigma\in S_{p}$; we have to show that $\sigma f=(\mathrm{sgn}\,\sigma)f$. Let $(x_{1},\ldots,x_{p})\in E^{p}$ and for $1\leq i define $g_{i}:E^{2}\to\mathbb{R}$ by

 $g_{i}(y_{1},y_{2})=f(x_{1},\ldots,\underbrace{y_{1}}_{i},\underbrace{y_{2}}_{i% +1},\ldots,x_{p}),\qquad(y_{1},y_{2})\in E^{2}.$

Because $f$ is multilinear and alternating, on the one hand

 $g_{i}(x_{i}+x_{i+1},x_{i}+x_{i+1})=0,$

and on the other hand

 $\displaystyle g_{i}(x_{i}+x_{i+1},x_{i}+x_{i+1})$ $\displaystyle=g_{i}(x_{i},x_{i})+g_{i}(x_{i},x_{i+1})+g_{i}(x_{i+1},x_{i})+g_{% i}(x_{i+1},x_{i+1})$ $\displaystyle=g_{i}(x_{i},x_{i+1})+g_{i}(x_{i+1},x_{i}).$

Therefore

 $g_{i}(x_{i+1},x_{i})=-g_{i}(x_{i},x_{i+1}),$

that is,

 $f(x_{1},\ldots,x_{p})=-f(x_{1},\ldots,x_{p}).$

Thus, as $\mathrm{sgn}\,\tau_{i}=-1$,

 $\tau_{i}f=(\mathrm{sgn}\,\tau_{i})f.$

Because $\sigma$ is equal to a product of adjacent transpositions, it then follows from Theorem 3 and Theorem 4 that $\sigma f=(\mathrm{sgn}\,\sigma)f$.

(ii) Suppose that $\sigma f=(\mathrm{sgn}\,\sigma)f$ for all $\sigma\in S_{p}$. Let $(x_{1},\ldots,x_{p})\in E^{p}$ with $x_{i}=x_{i+1}$ for some $1\leq i; we have to show that $f(x_{1},\ldots,x_{p})=0$. On the one hand,

 $\tau_{i}f(x_{1},\ldots,x_{p})=(\mathrm{sgn}\,\tau_{i})f(x_{1},\ldots,x_{p})=-f% (x_{1},\ldots,x_{p}).$

On the other hand, using that $x_{i}=x_{i+1}$,

 $\displaystyle\tau_{i}f(x_{1},\ldots,x_{p})$ $\displaystyle=f(x_{\tau_{i}(1)},\ldots,x_{\tau_{i}(i)},x_{\tau_{i}(i+1)},% \ldots,x_{\tau_{i}(p)})$ $\displaystyle=f(x_{1},\ldots,x_{i+1},x_{i},\ldots,x_{p})$ $\displaystyle=f(x_{1},\ldots,x_{i},x_{i+1},\ldots,x_{p}).$

Hence

 $-f(x_{1},\ldots,x_{p})=f(x_{1},\ldots,x_{p}),$

which implies that $f(x_{1},\ldots,x_{p})=0$. This shows that $f\in\mathscr{A}_{p}(E;\mathbb{R})$. ∎

###### Theorem 6.

Let $f\in\mathscr{A}_{p}(E;\mathbb{R})$. If $(x_{1},\ldots,x_{p})\in E^{p}$ with $x_{i}=x_{j}$ for some $i\neq j$, then $f(x_{1},\ldots,x_{p})=0$.

###### Proof.

Check that there is some $\sigma\in S_{p}$ satisfying $\sigma(1)=i$ and $\sigma(2)=j$. For this $\sigma$,

 $\displaystyle(\sigma f)(x_{1},\ldots,x_{p})$ $\displaystyle=f(x_{i},x_{j},x_{\sigma(3)},\ldots,x_{\sigma(p)})$ $\displaystyle=f(x_{i},x_{i},x_{\sigma(3)},\ldots,x_{\sigma(p)})$ $\displaystyle=0.$

But $(\sigma f)=(\mathrm{sgn}\,\sigma)f$, so $(\mathrm{sgn}\,\sigma)f(x_{1},\ldots,x_{p})=0$. Therefore $f(x_{1},\ldots,x_{p})=0$. ∎

###### Definition 7.

For $f\in\mathscr{L}_{p}(E;\mathbb{R})$, define

 $A_{p}f=\frac{1}{p!}\sum_{\sigma\in S_{p}}(\mathrm{sgn}\,\sigma)\sigma f.$
###### Lemma 8.

$A_{p}$ is a linear map $\mathscr{L}_{p}(E;\mathbb{R})\to\mathscr{A}_{p}(E;\mathbb{R})$.

###### Proof.

Let $f\in\mathscr{L}_{p}(E;\mathbb{R})$. For $\sigma\in S_{p}$, $\sigma f\in\mathscr{L}_{p}(E;\mathbb{R})$, hence $A_{p}f\in\mathscr{L}_{p}(E;\mathbb{R})$. Namely, $A_{p}f$ is multilinear. It remains to show that it is alternating.

For $\sigma\in S_{p}$, as $\tau\mapsto\sigma\tau$ is a bijection $S_{p}\to S_{p}$,

 $\displaystyle\sigma(A_{p}f)$ $\displaystyle=\frac{1}{p!}\sum_{\tau\in S_{p}}(\mathrm{sgn}\,\tau)\sigma\tau f$ $\displaystyle=(\mathrm{sgn}\,\sigma)\frac{1}{p!}\sum_{\tau\in S_{p}}(\mathrm{% sgn}\,\tau)\tau f$ $\displaystyle=(\mathrm{sgn}\,\sigma)Af,$

showing that $A_{p}f$ is alternating by Theorem 5, so $A_{p}f\in\mathscr{A}_{p}(E;\mathbb{R})$. ∎

###### Theorem 9.

Let $f\in\mathscr{L}_{p}(E;\mathbb{R})$. $f\in\mathscr{A}_{p}(E;\mathbb{R})$ if and only if $A_{p}f=f$.

###### Proof.

Suppose $f\in\mathscr{A}_{p}(E;\mathbb{R})$. Then $\sigma f=(\mathrm{sgn}\,\sigma)f$ for each $\sigma\in S_{p}$, by Theorem 5. Then

 $A_{p}f=\frac{1}{p!}\sum_{\sigma\in S_{p}}\sigma f=\frac{1}{p!}\sum_{\sigma\in S% _{p}}f=f.$

Suppose $A_{p}f=f$. Lemma 8 tells us $A_{p}f\in\mathscr{A}_{p}(E;\mathbb{R})$, hence $f\in\mathscr{A}_{p}(E;\mathbb{R})$. ∎

## 2 Wedge products

A permutation $\sigma\in S_{p+q}$ is called a $(p,q)$-riffle shuffle if

 $\sigma(1)<\cdots<\sigma(p),\qquad\sigma(p+1)<\cdots<\sigma(p+q).$

Denote by $S_{p,q}$ those elements of $S_{p+q}$ that are $(p,q)$-riffle shuffles.

###### Lemma 10.

$|S_{p,q}|=\binom{p+q}{p}=\frac{(p+q)!}{p!q!}$.

Let $\mathscr{A}_{p,q}(E;\mathbb{R})$ be the set of those $h\in\mathscr{L}_{p+q}(E;\mathbb{R})$ such that (i) for each $(y_{1},\ldots,y_{q})\in E^{q}$, the map

 $(x_{1},\ldots,x_{p})\mapsto h(x_{1},\ldots,x_{p},y_{1},\ldots,y_{q}),\qquad E^% {p}\to\mathbb{R},$

belongs to $\mathscr{A}_{p}(E;\mathbb{R})$, and (ii) for $(x_{1},\ldots,x_{p})\in E^{p}$, the map

 $(y_{1},\ldots,y_{q})\mapsto h(x_{1},\ldots,x_{p},y_{1},\ldots,y_{q}),\qquad E^% {q}\to\mathbb{R},$

belongs to $\mathscr{A}_{q}(E;\mathbb{R})$.

###### Definition 11.

For $h\in\mathscr{A}_{p,q}(E;\mathbb{R})$ define

 $\phi_{p,q}(h)=\sum_{\sigma\in S_{p,q}}(\mathrm{sgn}\,\sigma)(\sigma h).$
###### Theorem 12.

$\phi_{p,q}$ is a linear map $\mathscr{A}_{p,q}(E;\mathbb{R})\to\mathscr{A}_{p+q}(E;\mathbb{R})$.

###### Proof.

Let $h\in\mathscr{A}_{p,q}(E;\mathbb{R})$, and say $(x_{1},\ldots,x_{p+q})\in E^{p+q}$ with $x_{k}=x_{k+1}$ for some $1\leq k.

Let $A_{1}$ be those $\sigma\in S_{p,q}$ such that $i=\sigma^{-1}(k),j=\sigma^{-1}(k+1)\leq p$. For $\sigma\in A_{1}$, by Theorem 6,11 1 $i,j$ are distinct and $1\leq i,j\leq p$; they need not be adjacent.

 $(\sigma h)(x_{1},\ldots,x_{p+q})=h(x_{\sigma(1)},\ldots,x_{\sigma(p)},\ldots,x% _{\sigma(p+q)})=0.$

Let $A_{2}$ be those $\sigma\in S_{p,q}$ such that $\sigma^{-1}(k),\sigma^{-1}(k+1)\geq p+1$. For $\sigma\in A_{2}$, by Theorem 6,

 $(\sigma h)(x_{1},\ldots,x_{p+q})=h(x_{\sigma(1)},\ldots,x_{\sigma(p)},\ldots,x% _{\sigma(p+q)})=0.$

Thus

 $\sum_{\sigma\in A_{1}}(\mathrm{sgn}\,\sigma)(\sigma h)(x_{1},\ldots,x_{p+q})=0$

and

 $\sum_{\sigma\in A_{2}}(\mathrm{sgn}\,\sigma)(\sigma h)(x_{1},\ldots,x_{p+q})=0.$

Let $A_{3}$ be those $\sigma\in S_{p,q}$ for which $\sigma^{-1}(k) and $\sigma^{-1}(k+1)\geq p+1$ and let $A_{4}$ be those $\sigma\in S_{p,q}$ for which $\sigma^{-1}(k)\geq p+1$ and $\sigma^{-1}(k+1)\leq p$. If $\sigma\in A_{3}$ then

 $(\tau_{k}\sigma)^{-1}(k)=\sigma^{-1}\tau_{k}^{-1}(k)=\sigma^{-1}(k+1)\geq p+1$

and

 $(\tau_{k}\sigma)^{-1}(k+1)=\sigma^{-1}\tau_{k}^{-1}(k+1)=\sigma^{-1}(k)

so $\tau_{k}\sigma\in A_{4}$. Likewise, if $\sigma\in A_{4}$ then $\tau_{k}\sigma\in A_{3}$. Thus $A_{4}=\tau_{k}A_{3}$. For $\sigma\in A_{3}$, let $i=\sigma^{-1}(k)$ and $j=\sigma^{-1}(k+1)$, for which $i and $j\geq p+1$. Then, as $x_{k}=x_{k+1}$,

 $\begin{split}&\displaystyle(\mathrm{sgn}\,\sigma)(\sigma h)(x_{1},\ldots,x_{p+% q})+(\mathrm{sgn}\,\tau_{k}\sigma)(\tau_{k}\sigma h)(x_{1},\ldots,x_{p+q})\\ \displaystyle=&\displaystyle(\mathrm{sgn}\,\sigma)h(x_{\sigma(1)},\ldots,x_{% \sigma(p+q)})-(\mathrm{sgn}\,\sigma)h(x_{\tau_{k}\sigma(1)},\ldots,x_{\tau_{k}% \sigma(p+q)})\\ \displaystyle=&\displaystyle(\mathrm{sgn}\,\sigma)\big{(}h(x_{\sigma(1)},% \ldots,x_{\sigma(p+q)})-h(x_{\tau_{k}\sigma(1)},\ldots,x_{\tau_{k}\sigma(i)},% \ldots,x_{\tau_{k}\sigma(j)},\ldots,x_{\tau_{k}\sigma(p+q)})\big{)}\\ \displaystyle=&\displaystyle(\mathrm{sgn}\,\sigma)\big{(}h(x_{\sigma(1)},% \ldots,x_{\sigma(p+q)})-h(x_{\tau_{k}\sigma(1)},\ldots,x_{\tau_{k}(k)},\ldots,% x_{\tau_{k}(k+1)},\ldots,x_{\tau_{k}\sigma(p+q)})\big{)}\\ \displaystyle=&\displaystyle(\mathrm{sgn}\,\sigma)\big{(}h(x_{\sigma(1)},% \ldots,x_{\sigma(p+q)})-h(x_{\sigma(1)},\ldots,x_{k+1},\ldots,x_{k},\ldots,x_{% \sigma(p+q)})\big{)}\\ \displaystyle=&\displaystyle(\mathrm{sgn}\,\sigma)\big{(}h(x_{\sigma(1)},% \ldots,x_{\sigma(p+q)})-h(x_{\sigma(1)},\ldots,x_{k},\ldots,x_{k+1},\ldots,x_{% \sigma(p+q)})\big{)}\\ \displaystyle=&\displaystyle(\mathrm{sgn}\,\sigma)\big{(}h(x_{\sigma(1)},% \ldots,x_{\sigma(p+q)})-h(x_{\sigma(1)},\ldots,x_{\sigma(p+q)})\big{)}\\ \displaystyle=&\displaystyle 0.\end{split}$

Therefore

 $\sum_{\sigma\in A_{3}\cup A_{4}}(\mathrm{sgn}\,\sigma)(\sigma h)(x_{1},\ldots,% x_{p+q})=0.$

But $S_{p,q}=A_{1}\cup A_{2}\cup A_{3}\cup A_{4}$, so

 $\phi_{p,q}(h)(x_{1},\ldots,x_{p+q})=0.$

Thus $\phi_{p,q}(h)\in\mathscr{A}_{p+q}(E;\mathbb{R})$. ∎

###### Definition 13.

For $f\in\mathscr{L}_{p}(E;\mathbb{R})$ and $g\in\mathscr{L}_{q}(E;\mathbb{R})$, define the tensor product $f\otimes_{p,q}g\in\mathscr{L}_{p+q}(E;\mathbb{R})$ by

 $(f\otimes_{p,q}g)(x_{1},\ldots,x_{p+q})=f(x_{1},\ldots,x_{p})g(x_{p+1},\ldots,% x_{p+q}).$

It is apparent that

 $(f\otimes_{p,q}g)\otimes_{p+q,r}h=f\otimes_{p,q+r}(g\otimes_{q,r}h),$

and thus it makes sense to write the tensor product without indices.

###### Definition 14.

Define the wedge product

 $\wedge_{p,q}:\mathscr{A}_{p}(E;\mathbb{R})\times\mathscr{A}_{q}(E;\mathbb{R})% \to\mathscr{A}_{p+q}(E;\mathbb{R})$

by, for $f\in\mathscr{A}_{p}(E;\mathbb{R}),g\in\mathscr{A}_{q}(E;\mathbb{R})$,

 $f\wedge_{p,q}g=\phi_{p,q}(f\otimes g),$

i.e., for $h=f\otimes g$,

 $\displaystyle(f\wedge_{p,q}g)(x_{1},\ldots,x_{p+q})$ $\displaystyle=\sum_{\sigma\in S_{p,q}}(\mathrm{sgn}\,\sigma)(\sigma h)$ $\displaystyle=\sum_{\sigma\in S_{p,q}}(\mathrm{sgn}\,\sigma)h(x_{\sigma(1)},% \ldots,x_{\sigma(p+q)})$ $\displaystyle=\sum_{\sigma\in S_{p,q}}(\mathrm{sgn}\,\sigma)f(x_{\sigma(1)},% \ldots,x_{\sigma(p)})g(x_{\sigma(p+1)},\ldots,x_{\sigma(p+q)}).$
###### Theorem 15.

For $f\in\mathscr{A}_{p}(E;\mathbb{R})$ and $g\in\mathscr{A}_{q}(E;\mathbb{R})$,

 $f\wedge_{p,q}g=\frac{(p+q)!}{p!q!}A_{p+q}(f\otimes g).$
###### Proof.

For $\sigma\in S_{p,q}$,

 $\sigma(1)<\cdots<\sigma(p),\qquad\sigma(p+1)<\cdots<\sigma(p+q).$

Let $I_{\sigma}=\{\sigma(i):1\leq i\leq p\}$ and $J_{\sigma}=\{\sigma(i):p+1\leq i\leq p+q\}$.

 $\displaystyle f\wedge_{p,q}g$ $\displaystyle=$

###### Theorem 16.

For $f\in\mathscr{A}_{p}(E;\mathbb{R})$ and $g\in\mathscr{A}_{q}(E;\mathbb{R})$,

 $g\wedge_{q,p}f=(-1)^{pq}f\wedge_{p,q}g.$
###### Proof.

Define $\alpha\in S_{p,q}$ by

 $\alpha(i)=q+i,\quad 1\leq i\leq p,\qquad\alpha(p+i)=i,\quad 1\leq i\leq q.$

Then22 2 For example, take $p=3$ and $q=2$. Then $\alpha(1)=3,\alpha(2)=4,\alpha(3)=5,\alpha(4)=1,\alpha(5)=2.$ Here $\displaystyle\prod_{1\leq i\leq p}\prod_{1\leq j\leq q}(i+q-j,i+q-j+1)$ $\displaystyle=\prod_{1\leq i\leq 3}\prod_{1\leq j\leq 2}(i-j+2,i-j+3)$ $\displaystyle=\prod_{1\leq i\leq 3}(i+1,i+2)(i,i+1)$ $\displaystyle=(2,3)(1,2)(3,4)(2,3)(4,5)(3,4)$ $\displaystyle=\alpha.$

 $\alpha=\prod_{1\leq i\leq p}\prod_{1\leq j\leq q}(i+q-j,i+q-j+1).$

Thus

 $\mathrm{sgn}\,\alpha=\prod_{1\leq i\leq p}\prod_{1\leq j\leq q}(-1)=(-1)^{pq}.$

Let $\tau\in S_{q,p}$, then for $1\leq i\leq p$,

 $(\tau\alpha)(i)=\tau(q+i)$

and for $1\leq i\leq q$,

 $(\tau\alpha)(p+i)=\tau(i),$

But $\tau\in S_{q,p}$ so

 $\tau(1)<\cdots<\tau(q),\qquad\tau(q+1)<\cdots<\tau(q+p),$

thus

 $(\tau\alpha)(1)<\cdots<(\tau\alpha)(p),\qquad(\tau\alpha)(p+1)<\cdots<(\tau% \alpha)(p+q),$

which means that $\tau\alpha\in S_{p,q}$. Likewise, if $\sigma\in S_{p,q}$ then

 $(\sigma\alpha^{-1})(1)=\sigma(q+1),\ldots,(\sigma\alpha^{-1})(q)=\sigma(p+q)$

and

 $(\sigma\alpha^{-1})(q+1)=\sigma(1),\ldots,(\sigma\alpha^{-1})(q+p)=\sigma(p),$

and because $\sigma\in S_{p,q}$ it follows that $\sigma\alpha^{-1}\in S_{q,p}$.

Hence for $(x_{1},\ldots,x_{p+q})\in E^{p+q}$,

 $\begin{split}&\displaystyle(g\wedge_{q,p}f)(x_{1},\ldots,x_{p+q})\\ \displaystyle=&\displaystyle\sum_{\tau\in S_{q,p}}(\mathrm{sgn}\,\tau)g(x_{% \tau(1)},\ldots,x_{\tau(q)})f(x_{\tau(q+1)},\ldots,x_{\tau(q+p)})\\ \displaystyle=&\displaystyle\sum_{\sigma\in S_{p,q}}(\mathrm{sgn}\,\sigma% \alpha^{-1})g(x_{(\sigma\alpha^{-1})(1)},\ldots,x_{(\sigma\alpha^{-1})(q)})f(x% _{(\sigma\alpha^{-1})(q+1)},\ldots,x_{(\sigma\alpha^{-1})(q+p)})\\ \displaystyle=&\displaystyle(\mathrm{sgn}\,\alpha^{-1})\sum_{\sigma\in S_{p,q}% }(\mathrm{sgn}\,\sigma)g(x_{\sigma(p+1)},\ldots,x_{\sigma(p+q)})f(x_{\sigma(1)% },\ldots,x_{\sigma(p)})\\ \displaystyle=&\displaystyle(-1)^{pq}\sum_{\sigma\in S_{p,q}}(\mathrm{sgn}\,% \sigma)f(x_{\sigma(1)},\ldots,x_{\sigma(p)})g(x_{\sigma(p+1)},\ldots,x_{\sigma% (p+q)})\\ \displaystyle=&\displaystyle(-1)^{pq}(f\wedge_{p,q}g)(x_{1},\ldots,x_{p+q}).% \end{split}$

Thus

 $g\wedge_{q,p}f=(-1)^{pq}f\wedge_{p,q}g.$

Let $\mathscr{A}_{p,q,r}(E;\mathbb{R})$ be the set of those $u\in\mathscr{L}_{p+q+r}(E;\mathbb{R})$ such that (i) for each $(y_{1},\ldots,y_{q},z_{1},\ldots,z_{r})\in E^{q+r}$, the map

 $(x_{1},\ldots,x_{p})\mapsto u(x_{1},\ldots,x_{p},y_{1},\ldots,y_{q},z_{1},% \ldots,z_{r}),\qquad E^{p}\to\mathbb{R},$

belongs to $\mathscr{A}_{p}(E;\mathbb{R})$, (ii) for $(x_{1},\ldots,x_{p},z_{1},\ldots,z_{r})\in E^{p+r}$, the map

 $(y_{1},\ldots,y_{q})\mapsto u(x_{1},\ldots,x_{p},y_{1},\ldots,y_{q},z_{1},% \ldots,z_{r}),\qquad E^{q}\to\mathbb{R},$

belongs to $\mathscr{A}_{q}(E;\mathbb{R})$, and (iii) for $(x_{1},\ldots,x_{p},y_{1},\ldots,y_{q})\in E^{p+q}$, the map

 $(z_{1},\ldots,z_{r})\mapsto u(x_{1},\ldots,x_{p},y_{1},\ldots,y_{q},z_{1},% \ldots,z_{r}),\qquad E^{r}\to\mathbb{R},$

belongs to $\mathscr{A}_{r}(E;\mathbb{R})$.

Let $S_{p,q,\overline{r}}$ be those $\sigma\in S_{p+q+r}$ such that

 $\sigma(1)<\cdots<\sigma(p),\quad\sigma(p+1)<\cdots<\sigma(p+q),\quad\sigma(p+q% +i)=p+q+i,1\leq i\leq r.$

Let $S_{\overline{p},q,r}$ be those $\sigma\in S_{p+q+r}$ such that

 $\sigma(i)=i,1\leq i\leq p,\quad\quad\sigma(p+1)<\cdots<\sigma(p+q),\quad\sigma% (p+q+1)<\cdots<\sigma(p+q+r).$

Let $S_{p,q,r}$ be those $\sigma\in S_{p+q+r}$ such that

 $\sigma(1)<\cdots<\sigma(p),\quad\sigma(p+1)<\cdots<\sigma(p+q),\quad\sigma(p+q% +1)<\cdots<\sigma(p+q+r).$
###### Lemma 17.
 $S_{p+q,r}S_{p,q,\overline{r}}=S_{p,q,r}$

and

 $S_{p,q+r}S_{\overline{p},q,r}=S_{p,q,r}.$
###### Proof.

Let $\sigma\in S_{p+q,r}$ and $\tau\in S_{p,q,\overline{r}}$. Then

 $\sigma(1)<\cdots<\sigma(p+q),\quad\sigma(p+q+1)<\cdots<\sigma(p+q+r)$

and

 $\tau(1)<\cdots<\tau(p),\quad\tau(p+1)<\cdots<\tau(p+q),\quad\tau(p+q+i)=p+q+i,% 1\leq i\leq r.$

It follows that

 $(\sigma\tau)(1)<\cdots<(\sigma\tau)(p)$

and

 $(\sigma\tau)(p+1)<\cdots<(\sigma\tau)(p+q)$

and for $1\leq i\leq r$, $(\sigma\tau)(p+q+i)=\sigma(p+q+i)$, so

 $(\sigma\tau)(p+q+1)<\cdots<\sigma(p+q+r).$

Thus $\sigma\tau\in S_{p,q,r}$. ∎

Define $\phi_{p,q,\overline{r}}:\mathscr{A}_{p,q,r}(E;\mathbb{R})\to\mathscr{A}_{p+q,r% }(E;\mathbb{R})$ by

 $\phi_{p,q,\overline{r}}(u)=\sum_{\sigma\in S_{p,q,\overline{r}}}(\mathrm{sgn}% \,\sigma)(\sigma u),\qquad u\in\mathscr{A}_{p,q,r}(E;\mathbb{R})$

and define $\phi_{\overline{p},q,r}:\mathscr{A}_{p,q,r}(E;\mathbb{R})\to\mathscr{A}_{p,q+r% }(E;\mathbb{R})$ by

 $\phi_{\overline{p},q,r}(u)=\sum_{\sigma\in S_{\overline{p},q,r}}(\mathrm{sgn}% \,\sigma)(\sigma u),\qquad u\in\mathscr{A}_{p,q,r}(E;\mathbb{R})$
###### Lemma 18.

For $u\in\mathscr{A}_{p,q,r}(E;\mathbb{R})$,

 $(\phi_{p+q,r}\circ\phi_{p,q,\overline{r}})u=\sum_{\rho\in S_{p,q,r}}(\mathrm{% sgn}\,\rho)\rho u$

and

 $(\phi_{p,q+r}\circ\phi_{\overline{p},q,r})u=\sum_{\rho\in S_{p,q,r}}(\mathrm{% sgn}\,\rho)\rho u,$

and so

 $\phi_{p+q,r}\circ\phi_{p,q,\overline{r}}=\phi_{p,q+r}\circ\phi_{\overline{p},q% ,r}.$
###### Proof.

Applying Lemma 17 we get

 $\displaystyle(\phi_{p+q,r}\circ\phi_{p,q,\overline{r}})u$ $\displaystyle=\sum_{\sigma\in S_{p+q,r}}(\mathrm{sgn}\,\sigma)\sigma\phi_{p,q,% \overline{r}}(u)$ $\displaystyle=\sum_{\sigma\in S_{p+q,r}}(\mathrm{sgn}\,\sigma)\sigma\sum_{\tau% \in S_{p,q,\overline{r}}}(\mathrm{sgn}\,\tau)(\tau u)$ $\displaystyle=\sum_{\sigma\in S_{p+q,r}}(\mathrm{sgn}\,\sigma\tau)\sum_{\tau% \in S_{p,q,\overline{r}}}\sigma\tau u$ $\displaystyle=\sum_{\rho\in S_{p,q,r}}(\mathrm{sgn}\,\rho)\rho u$

and similarly

 $\displaystyle(\phi_{p,q+r}\circ\phi_{\overline{p},q,r})u$ $\displaystyle=\sum_{\sigma\in S_{p,q+r}}(\mathrm{sgn}\,\sigma)\sigma\phi_{% \overline{p},q,r}(u)$ $\displaystyle=\sum_{\sigma\in S_{p,q+r}}(\mathrm{sgn}\,\sigma)\sigma\sum_{\tau% \in S_{\overline{p},q,r}}(\mathrm{sgn}\,\tau)(\tau u)$ $\displaystyle=\sum_{\sigma\in S_{p,q+r}}(\mathrm{sgn}\,\sigma\tau)\sum_{\tau% \in S_{\overline{p},q,r}}\sigma\tau u$ $\displaystyle=\sum_{\rho\in S_{p,q,r}}(\mathrm{sgn}\,\rho)\rho u.$

Thus

 $(\phi_{p+q,r}\circ\phi_{p,q,\overline{r}})u=(\phi_{p,q+r}\circ\phi_{\overline{% p},q,r})u,$

from which the claim follows. ∎

###### Theorem 19.

If $f\in\mathscr{A}_{p}(E;\mathbb{R})$, $g\in\mathscr{A}_{q}(E;\mathbb{R})$, and $h\in\mathscr{A}_{r}(E;\mathbb{R})$, then

 $(f\wedge_{p,q}g)\wedge_{p+q,r}h=f\wedge_{p,q+r}(g\wedge_{q,r}h).$
###### Proof.

On the one hand,

 $\displaystyle(\phi_{p+q,r}\circ\phi_{p,q,\overline{r}})(f\otimes g\otimes h)$ $\displaystyle=\phi_{p+q,r}(\phi_{p,q,\overline{r}}((f\otimes g)\otimes h)$ $\displaystyle=\phi_{p+q,r}((f\wedge_{p,q}g)\otimes h)$ $\displaystyle=(f\wedge_{p,q}g)\wedge_{p+q,r}h.$

On the other hand,

 $\displaystyle(\phi_{p,q+r}\circ\phi_{\overline{p},q,r})(f\otimes g\otimes h)$ $\displaystyle=\phi_{p,q+r}(\phi_{\overline{p},q,r})(f\otimes(g\otimes h))$ $\displaystyle=\phi_{p,q+r}(f\otimes(g\wedge_{q,r}h))$ $\displaystyle=f\wedge_{p,q+r}(g\wedge_{q,r}h).$

But by Lemma 18,

 $\phi_{p+q,r}\circ\phi_{p,q,\overline{r}}=\phi_{p,q+r}\circ\phi_{\overline{p},q% ,r},$

hence

 $(f\wedge_{p,q}g)\wedge_{p+q,r}h=f\wedge_{p,q+r}(g\wedge_{q,r}h).$

## 3 Linear forms

Let $E^{*}=\mathscr{L}_{1}(E;\mathbb{R})$, the dual space of $E$, whose elements we call linear forms. It is immediate that $\mathscr{A}_{1}(E;\mathbb{R})=\mathscr{L}_{1}(E;\mathbb{R})=E^{*}$.

###### Theorem 20.

If $f_{1},\ldots,f_{n}\in E^{*}$ then for $(x_{1},\ldots,x_{n})\in E^{n}$,

 $(f_{1}\wedge\cdots\wedge f_{n})(x_{1},\ldots,x_{n})=\sum_{\sigma\in S_{n}}(% \mathrm{sgn}\,\sigma)f_{1}(x_{\sigma(1)})\cdots f_{n}(x_{\sigma(n)}).$
###### Proof.

For $n=1$ the claim is immediate. For $n=2$, on the one hand, using the definition of the wedge product,

 $(f_{1}\wedge f_{2})(x_{1},x_{2})=\sum_{\sigma\in S_{1,1}}(\mathrm{sgn}\,\sigma% )f_{1}(x_{\sigma(1)})f_{2}(x_{\sigma(2)}),$

and as $S_{1,1}=S_{2}$ the claim is true for $n=2$. Suppose the claim is true for some $n\geq 2$ and let $(f_{1},\ldots,f_{n},f_{n+1})\in E^{*}$ and $(x_{1},\ldots,x_{n},x_{n+1})\in E^{n+1}$. Then, setting $u=f_{1}\wedge\cdots\wedge f_{n}\in\mathscr{A}_{n}(E;\mathbb{R})$, we have

 $\begin{split}&\displaystyle(f_{1}\wedge\cdots\wedge f_{n}\wedge f_{n+1})(x_{1}% ,\ldots,x_{n},x_{n+1})\\ \displaystyle=&\displaystyle(u\wedge_{n,1}f_{n+1})(x_{1},\ldots,x_{n},x_{n+1})% \\ \displaystyle=&\displaystyle\sum_{\sigma\in S_{n,1}}(\mathrm{sgn}\,\sigma)u(x_% {\sigma(1)},\ldots,x_{\sigma(n)})f_{n+1}(x_{\sigma(n+1)})\\ \displaystyle=&\displaystyle\sum_{\sigma\in S_{n,1}}(\mathrm{sgn}\,\sigma)% \left(\sum_{\tau\in S_{n}}(\mathrm{sgn}\,\tau)f_{1}(x_{(\sigma\tau)(1)})\cdots f% _{n}(x_{(\sigma\tau)(n)})\right)f_{n+1}(x_{\sigma(n+1)})\\ \displaystyle=&\displaystyle\sum_{\rho\in S_{n+1}}(\mathrm{sgn}\,\rho)f_{1}(x_% {\rho(1)})\cdots f_{n}(x_{\rho(n)})f_{n+1}(x_{\rho(n+1)}),\end{split}$

thus the claim is true for $n+1$. ∎

Let $f_{1},\ldots,f_{n}\in E^{*}$ and $x_{1},\ldots,x_{n}\in E$ and put

 $a_{i,j}=f_{i}(x_{j}),\qquad 1\leq i,j\leq n;$

$a\in\mathrm{Mat}_{n}(\mathbb{R})$. The Leibniz formula for the determinant of an $n\times n$ matrix tells us

 $\det a=\sum_{\sigma\in S_{n}}(\mathrm{sgn}\,\sigma)\prod_{i=1}^{n}a_{i,\sigma(% i)}=\sum_{\sigma\in S_{n}}(\mathrm{sgn}\,\sigma)\prod_{i=1}^{n}f_{i}(x_{\sigma% (j)}).$

Then Theorem 20 gives

 $\det(f_{i}(x_{j}))_{1\leq i,j\leq n}=(f_{1}\wedge\cdots\wedge f_{n})(x_{1},% \ldots,x_{n}).$
###### Lemma 21.

If $f_{1},\ldots,f_{n}\in E^{*}$ are linearly independent then there are $x_{1},\ldots,x_{n}\in E$ such that

 $f_{i}(x_{j})=\delta_{i,j},\qquad 1\leq i,j\leq n.$
###### Theorem 22.

$f_{1},\ldots,f_{n}\in E^{*}$ are linearly dependent if and only if

 $f_{1}\wedge\cdots\wedge f_{n}=0.$
###### Proof.

Suppose $f_{1},\ldots,f_{n}$ are linearly dependent, say, for some $\lambda_{i}\in\mathbb{R}$, $i\neq k$,

 $f_{k}=\sum_{i\neq k}\lambda_{i}f_{i}.$

Then, as $f_{i}\wedge f_{i}=0$,

 $f_{1}\wedge\cdots\wedge f_{n}=0.$

Suppose that $f_{1},\ldots,f_{n}\in E^{*}$ are linearly independent. By Lemma 21, there are $x_{1},\ldots,x_{n}\in E$ such that

 $f_{i}(x_{j})=\delta_{i,j},\qquad 1\leq i,j\leq n.$

Then $\det(f_{i}(x_{j}))=1$, and hence

 $(f_{1}\wedge\cdots\wedge f_{n})(x_{1},\ldots,x_{n})=1,$

so $f_{1}\wedge\cdots\wedge f_{n}$ is not identically $0$. ∎

## 4 Rk

We now take $E=\mathbb{R}^{k}$. For $1\leq i\leq k$ define $\xi_{i}\in(\mathbb{R}^{k})^{*}$ by

 $\xi_{i}(x_{1},\ldots,x_{k})=x_{i},\qquad(x_{1},\ldots,x_{k})\in\mathbb{R}^{k}.$

Let $e_{i}=(0,\ldots,\underbrace{1}_{i},\ldots,0)\in\mathbb{R}^{k}$ for $1\leq i\leq k$, in other words,

 $\xi_{i}(e_{j})=\delta_{i,j},\qquad 1\leq i,j\leq k.$

For $x\in\mathbb{R}^{k}$,

 $x=\sum_{1\leq i\leq k}\xi_{i}(x)e_{i}.$
###### Theorem 23.

(i) If $f\in\mathscr{L}_{p}(\mathbb{R}^{k};\mathbb{R})$ then for $(x_{1},\ldots,x_{k})\in(\mathbb{R}^{k})^{p}$,

 $f(x_{1},\ldots,x_{p})=\sum_{1\leq i_{1},\ldots,i_{p}\leq k}f(e_{i_{1}},\ldots,% e_{i_{p}})\xi_{i_{1}}(x_{1})\cdots\xi_{i_{p}}(x_{p}).$

(ii) If $f\in\mathscr{A}_{p}(\mathbb{R}^{k};\mathbb{R})$ then

 $f=\sum_{1\leq i_{1}<\cdots

(iii)

 $\dim\mathscr{A}_{p}(\mathbb{R}^{k};\mathbb{R})=\binom{k}{p}.$

(iv) If $f\in\mathscr{A}_{k}(\mathbb{R}^{k};\mathbb{R})$ then

 $f=f(e_{1},\ldots,e_{n})\xi_{1}\wedge\cdots\wedge\xi_{k}.$
###### Proof.

(i) Let $f\in\mathscr{L}_{p}(\mathbb{R}^{k};\mathbb{R})$. For $(x_{1},\ldots,x_{p})\in(\mathbb{R}^{k})^{p}$, because $f:(\mathbb{R}^{k})^{p}\to\mathbb{R}$ is multilinear,

 $\displaystyle f(x_{1},\ldots,x_{p})$ $\displaystyle=f\left(\sum_{1\leq i_{1}\leq k}\xi_{i_{1}}(x_{1})e_{i_{1}},% \ldots,\sum_{1\leq i_{p}\leq k}\xi_{i_{p}}(x_{p})e_{i_{p}}\right)$ $\displaystyle=\sum_{1\leq i_{1},\ldots,i_{p}\leq k}\xi_{i_{1}}(x_{1})\cdots\xi% _{i_{p}}(x_{p})f(e_{i_{1}},\ldots,e_{i_{p}}).$

(ii) Let $f\in\mathscr{A}_{p}(\mathbb{R}^{k};\mathbb{R})$. Then $f=A_{p}f$ (Theorem 9),

 $f=\frac{1}{p!}\sum_{\sigma\in S_{p}}(\mathrm{sgn}\,\sigma)\sigma f,$

so for $(x_{1},\ldots,x_{p})\in(\mathbb{R}^{k})^{p}$, applying Theorem 20,

 $\displaystyle f(x_{1},\ldots,x_{p})$ $\displaystyle=\frac{1}{p!}\sum_{\sigma\in S_{p}}(\mathrm{sgn}\,\sigma)f(x_{% \sigma(1)},\ldots,x_{\sigma(p)})$ $\displaystyle=\frac{1}{p!}\sum_{\sigma\in S_{p}}(\mathrm{sgn}\,\sigma)\sum_{1% \leq i_{1},\ldots,i_{p}\leq k}\xi_{i_{1}}(x_{\sigma(1)})\cdots\xi_{i_{p}}(x_{% \sigma(p)})f(e_{i_{1}},\ldots,e_{i_{p}})$ $\displaystyle=\frac{1}{p!}\sum_{1\leq i_{1},\ldots,i_{p}\leq k}f(e_{i_{1}},% \ldots,e_{i_{p}})\sum_{\sigma\in S_{p}}(\mathrm{sgn}\,\sigma)\xi_{i_{1}}(x_{% \sigma(1)})\cdots\xi_{i_{p}}(x_{\sigma(p)})$ $\displaystyle=\frac{1}{p!}\sum_{1\leq i_{1},\ldots,i_{p}\leq k}f(e_{i_{1}},% \ldots,e_{i_{p}})(\xi_{i_{1}}\wedge\cdots\wedge\xi_{i_{p}})(x_{1},\ldots,x_{p}).$

Since $f$ is alternating, $i_{r}=i_{s}$ for $r\neq s$ implies $f(e_{i_{1}},\ldots,e_{i_{p}})=0$. Let

 $\mathscr{I}_{p,k}=\{I\subset\{1,\ldots,k\}:|I|=p\};$

For $I\in\mathscr{I}_{p,k}$, define $I_{1},\ldots,I_{p}$ by $I=\{I_{1},\ldots,I_{p}\}$ and $I_{1}<\cdots. Then, applying Theorem 16, as $|S_{I}|=p!$,

 $\displaystyle f$ $\displaystyle=\frac{1}{p!}\sum_{I\in\mathscr{I}_{p,k}}\sum_{\tau\in S_{I}}f(e_% {\tau(I_{1})},\ldots,e_{\tau(I_{p})})\xi_{\tau(I_{1})}\wedge\cdots\wedge\xi_{% \tau(I_{p})}$ $\displaystyle=\frac{1}{p!}\sum_{I\in\mathscr{I}_{p,k}}\sum_{\tau\in S_{I}}(% \mathrm{sgn}\,\tau)f(e_{I_{1}},\ldots,e_{I_{p}})(\mathrm{sgn}\,\tau)\xi_{I_{1}% }\wedge\cdots\wedge\xi_{I_{p}}$ $\displaystyle=\sum_{I\in\mathscr{I}_{p,k}}f(e_{I_{1}},\ldots,e_{I_{p}})\xi_{I_% {1}}\wedge\cdots\wedge\xi_{I_{p}}.$

proving the claim.

(iii) $|\mathscr{I}_{p,k}|=\binom{k}{p}$.

(iv) This follows from (ii) and the fact that $|\mathscr{I}_{p,k}|=1$ with $\mathscr{I}_{p,k}=\{\{1,\ldots,k\}\}$. ∎

## References

• [1] R. Abraham and J. E. Marsden (2008) Foundations of mechanics. second edition, AMS Chelsea Publishing, Providence, RI. Cited by: §1.
• [2] H. Cartan (1970) Differential forms. Hermann, Paris. Cited by: §1.