Alternating multilinear forms

1 Permutations

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

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

Definition 1.

A map $f\in {ℒ}_p(E;ℝ)$ is called alternating if $(x_1,\mathrm{\dots },x_p)\in E^p$ with $x_i=x_{i+1}$ for some implies $f(x_1,\mathrm{\dots },x_p)=0$. Let ${𝒜}_p(E;ℝ)$ be the set of alternating elements of ${ℒ}_p(E;ℝ)$.

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

Definition 2.

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

$$(\sigma f)(x_1,\mathrm{\dots },x_p)=f(x_{\sigma (1)},\mathrm{\dots },x_{\sigma (p)}),(x_1,\mathrm{\dots },x_p)\in E^p.$$

Theorem 3.

For a function $f:E^p\to ℝ$ and for $\sigma ,\tau \in S_p$,

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

Proof.

Define $g=\sigma f$. For $(x_1,\mathrm{\dots },x_p)\in E^p$ and for $y_i=x_{\tau (i)}$, we have

$\tau (\sigma f)(x_1,\mathrm{\dots },x_p)$ $=\tau (g)(x_1,\mathrm{\dots },x_p)$

$=g(x_{\tau (1)},\mathrm{\dots },x_{\tau (p)})$

$=g(y_1,\mathrm{\dots },y_p)$

$=(\sigma f)(y_1,\mathrm{\dots },y_p)$

$=f(y_{\sigma (1)},\mathrm{\dots },y_{\sigma (p)})$

$=f(x_{\tau (\sigma (1))},\mathrm{\dots },x_{\tau (\sigma (p))})$

$=(\tau \sigma )(f)(x_1,\mathrm{\dots },x_p).$

Thus

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

∎

For $1\le i,j\le p$, define $(i,j)\in S_p$ by

called a transposition. Define

$${\tau }_i=(i,i+1),$$

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

$(i,j)$ $=(j-1,j)(j-2,j-1)\mathrm{\cdots }(i+1,i+2)(i,i+1)(i+1,i+2)\mathrm{\cdots }(j-1,j)$

$={\tau }_{j-1}\mathrm{\cdots }{\tau }_{i+1}{\tau }_i{\tau }_{i+1}\mathrm{\cdots }{\tau }_{j-1}.$

Theorem 4.

For $\sigma ,\tau \in S_p$,

$$\mathrm{sgn}(\sigma \tau )=\mathrm{sgn}(\sigma )\mathrm{sgn}(\tau ).$$

Theorem 5.

Let $f\in {ℒ}_p(E;ℝ)$. $f\in {𝒜}_p(E;ℝ)$ if and only if $\sigma f=(\mathrm{sgn}\sigma )f$ for all $\sigma \in S_p$.

Proof.

(i) Suppose that $f\in {𝒜}_p(E;ℝ)$ and let $\sigma \in S_p$; we have to show that $\sigma f=(\mathrm{sgn}\sigma )f$. Let $(x_1,\mathrm{\dots },x_p)\in E^p$ and for define $g_i:E^2\to ℝ$ by

$$g_i(y_1,y_2)=f(x_1,\mathrm{\dots },\underset{i}{\underbrace{y_1}},\underset{i+1}{\underbrace{y_2}},\mathrm{\dots },x_p),(y_1,y_2)\in E^2.$$

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

$g_i(x_i+x_{i+1},x_i+x_{i+1})$ $=g_i(x_i,x_i)+g_i(x_i,x_{i+1})+g_i(x_{i+1},x_i)+g_i(x_{i+1},x_{i+1})$

$=g_i(x_i,x_{i+1})+g_i(x_{i+1},x_i).$

Therefore

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

that is,

$$f(x_1,\mathrm{\dots },x_p)=-f(x_1,\mathrm{\dots },x_p).$$

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 $(x_1,\mathrm{\dots },x_p)\in E^p$ with $x_i=x_{i+1}$ for some ; we have to show that $f(x_1,\mathrm{\dots },x_p)=0$. On the one hand,

$${\tau }_{i}f(x_1,\mathrm{\dots },x_p)=(\mathrm{sgn}{\tau }_i)f(x_1,\mathrm{\dots },x_p)=-f(x_1,\mathrm{\dots },x_p).$$

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

${\tau }_{i}f(x_1,\mathrm{\dots },x_p)$ $=f(x_{{\tau }_i(1)},\mathrm{\dots },x_{{\tau }_i(i)},x_{{\tau }_i(i+1)},\mathrm{\dots },x_{{\tau }_i(p)})$

$=f(x_1,\mathrm{\dots },x_{i+1},x_i,\mathrm{\dots },x_p)$

$=f(x_1,\mathrm{\dots },x_i,x_{i+1},\mathrm{\dots },x_p).$

Hence

$$-f(x_1,\mathrm{\dots },x_p)=f(x_1,\mathrm{\dots },x_p),$$

which implies that $f(x_1,\mathrm{\dots },x_p)=0$. This shows that $f\in {𝒜}_p(E;ℝ)$. ∎

Theorem 6.

Let $f\in {𝒜}_p(E;ℝ)$. If $(x_1,\mathrm{\dots },x_p)\in E^p$ with $x_i=x_j$ for some $i\ne j$, then $f(x_1,\mathrm{\dots },x_p)=0$.

Proof.

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

$(\sigma f)(x_1,\mathrm{\dots },x_p)$ $=f(x_i,x_j,x_{\sigma (3)},\mathrm{\dots },x_{\sigma (p)})$

$=f(x_i,x_i,x_{\sigma (3)},\mathrm{\dots },x_{\sigma (p)})$

$=0.$

But $(\sigma f)=(\mathrm{sgn}\sigma )f$, so $(\mathrm{sgn}\sigma )f(x_1,\mathrm{\dots },x_p)=0$. Therefore $f(x_1,\mathrm{\dots },x_p)=0$. ∎

Definition 7.

For $f\in {ℒ}_p(E;ℝ)$, 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 ${ℒ}_p(E;ℝ)\to {𝒜}_p(E;ℝ)$.

Proof.

Let $f\in {ℒ}_p(E;ℝ)$. For $\sigma \in S_p$, $\sigma f\in {ℒ}_p(E;ℝ)$, hence $A_{p}f\in {ℒ}_p(E;ℝ)$. 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 $S_p\to S_p$,

$\sigma (A_{p}f)$ $={\displaystyle \frac{1}{p!}}{\displaystyle \sum _{\tau \in S_p}}(\mathrm{sgn}\tau )\sigma \tau f$

$=(\mathrm{sgn}\sigma ){\displaystyle \frac{1}{p!}}{\displaystyle \sum _{\tau \in S_p}}(\mathrm{sgn}\tau )\tau f$

$=(\mathrm{sgn}\sigma )Af,$

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

Theorem 9.

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

Proof.

Suppose $f\in {𝒜}_p(E;ℝ)$. 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 $A_{p}f\in {𝒜}_p(E;ℝ)$, hence $f\in {𝒜}_p(E;ℝ)$. ∎

2 Wedge products

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

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

Lemma 10.

$|S_{p,q}|=\left(\genfrac{}{}{0pt}{}{p+q}{p}\right)=\frac{(p+q)!}{p!q!}$.

Let ${𝒜}_{p,q}(E;ℝ)$ be the set of those $h\in {ℒ}_{p+q}(E;ℝ)$ such that (i) for each $(y_1,\mathrm{\dots },y_q)\in E^q$, the map

$$(x_1,\mathrm{\dots },x_p)\mapsto h(x_1,\mathrm{\dots },x_p,y_1,\mathrm{\dots },y_q),E^p\to ℝ,$$

belongs to ${𝒜}_p(E;ℝ)$, and (ii) for $(x_1,\mathrm{\dots },x_p)\in E^p$, the map

$$(y_1,\mathrm{\dots },y_q)\mapsto h(x_1,\mathrm{\dots },x_p,y_1,\mathrm{\dots },y_q),E^q\to ℝ,$$

belongs to ${𝒜}_q(E;ℝ)$.

Definition 11.

For $h\in {𝒜}_{p,q}(E;ℝ)$ define

$${\varphi }_{p,q}(h)=\sum _{\sigma \in S_{p,q}}(\mathrm{sgn}\sigma )(\sigma h).$$

Theorem 12.

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

Proof.

Let $h\in {𝒜}_{p,q}(E;ℝ)$, and say $(x_1,\mathrm{\dots },x_{p+q})\in E^{p+q}$ with $x_k=x_{k+1}$ for some .

Let $A_1$ be those $\sigma \in S_{p,q}$ such that $i={\sigma }^{-1}(k),j={\sigma }^{-1}(k+1)\le p$. For $\sigma \in A_1$, by Theorem 6,

$$(\sigma h)(x_1,\mathrm{\dots },x_{p+q})=h(x_{\sigma (1)},\mathrm{\dots },x_{\sigma (p)},\mathrm{\dots },x_{\sigma (p+q)})=0.$$

Let $A_2$ be those $\sigma \in S_{p,q}$ such that ${\sigma }^{-1}(k),{\sigma }^{-1}(k+1)\ge p+1$. For $\sigma \in A_2$, by Theorem 6,

$$(\sigma h)(x_1,\mathrm{\dots },x_{p+q})=h(x_{\sigma (1)},\mathrm{\dots },x_{\sigma (p)},\mathrm{\dots },x_{\sigma (p+q)})=0.$$

Thus

$$\sum _{\sigma \in A_1}(\mathrm{sgn}\sigma )(\sigma h)(x_1,\mathrm{\dots },x_{p+q})=0$$

and

$$\sum _{\sigma \in A_2}(\mathrm{sgn}\sigma )(\sigma h)(x_1,\mathrm{\dots },x_{p+q})=0.$$

Let $A_3$ be those $\sigma \in S_{p,q}$ for which and ${\sigma }^{-1}(k+1)\ge p+1$ and let $A_4$ be those $\sigma \in S_{p,q}$ for which ${\sigma }^{-1}(k)\ge p+1$ and ${\sigma }^{-1}(k+1)\le p$. If $\sigma \in A_3$ then

$${({\tau }_k\sigma )}^{-1}(k)={\sigma }^{-1}{\tau }_k^{-1}(k)={\sigma }^{-1}(k+1)\ge p+1$$

and

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 and $j\ge p+1$. Then, as $x_k=x_{k+1}$,

Therefore

$$\sum _{\sigma \in A_3\cup A_4}(\mathrm{sgn}\sigma )(\sigma h)(x_1,\mathrm{\dots },x_{p+q})=0.$$

But $S_{p,q}=A_1\cup A_2\cup A_3\cup A_4$, so

$${\varphi }_{p,q}(h)(x_1,\mathrm{\dots },x_{p+q})=0.$$

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

Definition 13.

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

$$(f{\otimes }_{p,q}g)(x_1,\mathrm{\dots },x_{p+q})=f(x_1,\mathrm{\dots },x_p)g(x_{p+1},\mathrm{\dots },x_{p+q}).$$

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

$${\wedge }_{p,q}:{𝒜}_p(E;ℝ)\times {𝒜}_q(E;ℝ)\to {𝒜}_{p+q}(E;ℝ)$$

by, for $f\in {𝒜}_p(E;ℝ),g\in {𝒜}_q(E;ℝ)$,

$$f{\wedge }_{p,q}g={\varphi }_{p,q}(f\otimes g),$$

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

$(f{\wedge }_{p,q}g)(x_1,\mathrm{\dots },x_{p+q})$ $={\displaystyle \sum _{\sigma \in S_{p,q}}}(\mathrm{sgn}\sigma )(\sigma h)$

$={\displaystyle \sum _{\sigma \in S_{p,q}}}(\mathrm{sgn}\sigma )h(x_{\sigma (1)},\mathrm{\dots },x_{\sigma (p+q)})$

$={\displaystyle \sum _{\sigma \in S_{p,q}}}(\mathrm{sgn}\sigma )f(x_{\sigma (1)},\mathrm{\dots },x_{\sigma (p)})g(x_{\sigma (p+1)},\mathrm{\dots },x_{\sigma (p+q)}).$

Theorem 15.

For $f\in {𝒜}_p(E;ℝ)$ and $g\in {𝒜}_q(E;ℝ)$,

$$f{\wedge }_{p,q}g=\frac{(p+q)!}{p!q!}{A}_{p+q}(f\otimes g).$$

Proof.

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

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

$f{\wedge }_{p,q}g$ $=$

∎

Theorem 16.

For $f\in {𝒜}_p(E;ℝ)$ and $g\in {𝒜}_q(E;ℝ)$,

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

Proof.

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

$$\alpha (i)=q+i,1\le i\le p,\alpha (p+i)=i,1\le i\le q.$$

Then

$$\alpha =\prod _{1\le i\le p}\prod _{1\le j\le q}(i+q-j,i+q-j+1).$$

Thus

$$\mathrm{sgn}\alpha =\prod _{1\le i\le p}\prod _{1\le j\le q}(-1)={(-1)}^{pq}.$$

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

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

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

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

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

thus

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

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

and

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

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

Hence for $(x_1,\mathrm{\dots },x_{p+q})\in E^{p+q}$,

Thus

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

∎

Let ${𝒜}_{p,q,r}(E;ℝ)$ be the set of those $u\in {ℒ}_{p+q+r}(E;ℝ)$ such that (i) for each $(y_1,\mathrm{\dots },y_q,z_1,\mathrm{\dots },z_r)\in E^{q+r}$, the map

$$(x_1,\mathrm{\dots },x_p)\mapsto u(x_1,\mathrm{\dots },x_p,y_1,\mathrm{\dots },y_q,z_1,\mathrm{\dots },z_r),E^p\to ℝ,$$

belongs to ${𝒜}_p(E;ℝ)$, (ii) for $(x_1,\mathrm{\dots },x_p,z_1,\mathrm{\dots },z_r)\in E^{p+r}$, the map

$$(y_1,\mathrm{\dots },y_q)\mapsto u(x_1,\mathrm{\dots },x_p,y_1,\mathrm{\dots },y_q,z_1,\mathrm{\dots },z_r),E^q\to ℝ,$$

belongs to ${𝒜}_q(E;ℝ)$, and (iii) for $(x_1,\mathrm{\dots },x_p,y_1,\mathrm{\dots },y_q)\in E^{p+q}$, the map

$$(z_1,\mathrm{\dots },z_r)\mapsto u(x_1,\mathrm{\dots },x_p,y_1,\mathrm{\dots },y_q,z_1,\mathrm{\dots },z_r),E^r\to ℝ,$$

belongs to ${𝒜}_r(E;ℝ)$.

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

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

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

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

and

It follows that

and

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

Thus $\sigma \tau \in S_{p,q,r}$. ∎

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

$${\varphi }_{p,q,\overline{r}}(u)=\sum _{\sigma \in S_{p,q,\overline{r}}}(\mathrm{sgn}\sigma )(\sigma u),u\in {𝒜}_{p,q,r}(E;ℝ)$$

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

$${\varphi }_{\overline{p},q,r}(u)=\sum _{\sigma \in S_{\overline{p},q,r}}(\mathrm{sgn}\sigma )(\sigma u),u\in {𝒜}_{p,q,r}(E;ℝ)$$

Lemma 18.

For $u\in {𝒜}_{p,q,r}(E;ℝ)$,

$$({\varphi }_{p+q,r}\circ {\varphi }_{p,q,\overline{r}})u=\sum _{\rho \in S_{p,q,r}}(\mathrm{sgn}\rho )\rho u$$

and

$$({\varphi }_{p,q+r}\circ {\varphi }_{\overline{p},q,r})u=\sum _{\rho \in S_{p,q,r}}(\mathrm{sgn}\rho )\rho u,$$

and so

$${\varphi }_{p+q,r}\circ {\varphi }_{p,q,\overline{r}}={\varphi }_{p,q+r}\circ {\varphi }_{\overline{p},q,r}.$$

Proof.

Applying Lemma 17 we get

$({\varphi }_{p+q,r}\circ {\varphi }_{p,q,\overline{r}})u$ $={\displaystyle \sum _{\sigma \in S_{p+q,r}}}(\mathrm{sgn}\sigma )\sigma {\varphi }_{p,q,\overline{r}}(u)$

$={\displaystyle \sum _{\sigma \in S_{p+q,r}}}(\mathrm{sgn}\sigma )\sigma {\displaystyle \sum _{\tau \in S_{p,q,\overline{r}}}}(\mathrm{sgn}\tau )(\tau u)$

$={\displaystyle \sum _{\sigma \in S_{p+q,r}}}(\mathrm{sgn}\sigma \tau ){\displaystyle \sum _{\tau \in S_{p,q,\overline{r}}}}\sigma \tau u$

$={\displaystyle \sum _{\rho \in S_{p,q,r}}}(\mathrm{sgn}\rho )\rho u$

and similarly

$({\varphi }_{p,q+r}\circ {\varphi }_{\overline{p},q,r})u$ $={\displaystyle \sum _{\sigma \in S_{p,q+r}}}(\mathrm{sgn}\sigma )\sigma {\varphi }_{\overline{p},q,r}(u)$

$={\displaystyle \sum _{\sigma \in S_{p,q+r}}}(\mathrm{sgn}\sigma )\sigma {\displaystyle \sum _{\tau \in S_{\overline{p},q,r}}}(\mathrm{sgn}\tau )(\tau u)$

$={\displaystyle \sum _{\sigma \in S_{p,q+r}}}(\mathrm{sgn}\sigma \tau ){\displaystyle \sum _{\tau \in S_{\overline{p},q,r}}}\sigma \tau u$

$={\displaystyle \sum _{\rho \in S_{p,q,r}}}(\mathrm{sgn}\rho )\rho u.$

Thus

$$({\varphi }_{p+q,r}\circ {\varphi }_{p,q,\overline{r}})u=({\varphi }_{p,q+r}\circ {\varphi }_{\overline{p},q,r})u,$$

from which the claim follows. ∎

Theorem 19.

If $f\in {𝒜}_p(E;ℝ)$, $g\in {𝒜}_q(E;ℝ)$, and $h\in {𝒜}_r(E;ℝ)$, 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,

$({\varphi }_{p+q,r}\circ {\varphi }_{p,q,\overline{r}})(f\otimes g\otimes h)$ $={\varphi }_{p+q,r}({\varphi }_{p,q,\overline{r}}((f\otimes g)\otimes h)$

$={\varphi }_{p+q,r}((f{\wedge }_{p,q}g)\otimes h)$

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

On the other hand,

$({\varphi }_{p,q+r}\circ {\varphi }_{\overline{p},q,r})(f\otimes g\otimes h)$ $={\varphi }_{p,q+r}({\varphi }_{\overline{p},q,r})(f\otimes (g\otimes h))$

$={\varphi }_{p,q+r}(f\otimes (g{\wedge }_{q,r}h))$

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

But by Lemma 18,

$${\varphi }_{p+q,r}\circ {\varphi }_{p,q,\overline{r}}={\varphi }_{p,q+r}\circ {\varphi }_{\overline{p},q,r},$$

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 $E^{*}={ℒ}_1(E;ℝ)$, the dual space of $E$, whose elements we call linear forms. It is immediate that ${𝒜}_1(E;ℝ)={ℒ}_1(E;ℝ)=E^{*}$.

Theorem 20.

If $f_1,\mathrm{\dots },f_n\in E^{*}$ then for $(x_1,\mathrm{\dots },x_n)\in E^n$,

$$(f_1\wedge \mathrm{\cdots }\wedge f_n)(x_1,\mathrm{\dots },x_n)=\sum _{\sigma \in S_n}(\mathrm{sgn}\sigma )f_1(x_{\sigma (1)})\mathrm{\cdots }{f}_n(x_{\sigma (n)}).$$

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 $n\ge 2$ and let $(f_1,\mathrm{\dots },f_n,f_{n+1})\in E^{*}$ and $(x_1,\mathrm{\dots },x_n,x_{n+1})\in E^{n+1}$. Then, setting $u=f_1\wedge \mathrm{\cdots }\wedge f_n\in {𝒜}_n(E;ℝ)$, we have

thus the claim is true for $n+1$. ∎

Let $f_1,\mathrm{\dots },f_n\in E^{*}$ and $x_1,\mathrm{\dots },x_n\in E$ and put

$$a_{i,j}=f_i(x_j),1\le i,j\le n;$$

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

$$deta=\sum _{\sigma \in S_n}(\mathrm{sgn}\sigma )\prod _{i=1}^n{a}_{i,\sigma (i)}=\sum _{\sigma \in S_n}(\mathrm{sgn}\sigma )\prod _{i=1}^n{f}_i(x_{\sigma (j)}).$$

Then Theorem 20 gives

$$det{(f_i(x_j))}_{1\le i,j\le n}=(f_1\wedge \mathrm{\cdots }\wedge f_n)(x_1,\mathrm{\dots },x_n).$$

Lemma 21.

If $f_1,\mathrm{\dots },f_n\in E^{*}$ are linearly independent then there are $x_1,\mathrm{\dots },x_n\in E$ such that

$$f_i(x_j)={\delta }_{i,j},1\le i,j\le n.$$

Theorem 22.

$f_1,\mathrm{\dots },f_n\in E^{*}$ are linearly dependent if and only if

$$f_1\wedge \mathrm{\cdots }\wedge f_n=0.$$

Proof.

Suppose $f_1,\mathrm{\dots },f_n$ are linearly dependent, say, for some ${\lambda }_i\in ℝ$, $i\ne k$,

$$f_k=\sum _{i\ne k}{\lambda }_i{f}_i.$$

Then, as $f_i\wedge f_i=0$,

$$f_1\wedge \mathrm{\cdots }\wedge f_n=0.$$

Suppose that $f_1,\mathrm{\dots },f_n\in E^{*}$ are linearly independent. By Lemma 21, there are $x_1,\mathrm{\dots },x_n\in E$ such that

$$f_i(x_j)={\delta }_{i,j},1\le i,j\le n.$$

Then $det(f_i(x_j))=1$, and hence

$$(f_1\wedge \mathrm{\cdots }\wedge f_n)(x_1,\mathrm{\dots },x_n)=1,$$

so $f_1\wedge \mathrm{\cdots }\wedge f_n$ is not identically $0$. ∎

4 R^k

We now take $E={ℝ}^k$. For $1\le i\le k$ define ${\xi }_i\in {({ℝ}^k)}^{*}$ by

$${\xi }_i(x_1,\mathrm{\dots },x_k)=x_i,(x_1,\mathrm{\dots },x_k)\in {ℝ}^k.$$

Let $e_i=(0,\mathrm{\dots },\underset{i}{\underbrace{1}},\mathrm{\dots },0)\in {ℝ}^k$ for $1\le i\le k$, in other words,

$${\xi }_i(e_j)={\delta }_{i,j},1\le i,j\le k.$$

For $x\in {ℝ}^k$,

$$x=\sum _{1\le i\le k}{\xi }_i(x)e_i.$$

Theorem 23.

(i) If $f\in {ℒ}_p({ℝ}^k;ℝ)$ then for $(x_1,\mathrm{\dots },x_k)\in {({ℝ}^k)}^p$,

$$f(x_1,\mathrm{\dots },x_p)=\sum _{1\le i_1,\mathrm{\dots },i_p\le k}f(e_{i_1},\mathrm{\dots },e_{i_p}){\xi }_{i_1}(x_1)\mathrm{\cdots }{\xi }_{i_p}(x_p).$$

(ii) If $f\in {𝒜}_p({ℝ}^k;ℝ)$ then

(iii)

$$dim{𝒜}_p({ℝ}^k;ℝ)=\left(\genfrac{}{}{0pt}{}{k}{p}\right).$$

(iv) If $f\in {𝒜}_k({ℝ}^k;ℝ)$ then

$$f=f(e_1,\mathrm{\dots },e_n){\xi }_1\wedge \mathrm{\cdots }\wedge {\xi }_k.$$

Proof.

(i) Let $f\in {ℒ}_p({ℝ}^k;ℝ)$. For $(x_1,\mathrm{\dots },x_p)\in {({ℝ}^k)}^p$, because $f:{({ℝ}^k)}^p\to ℝ$ is multilinear,

$f(x_1,\mathrm{\dots },x_p)$ $=f({\displaystyle \sum _{1\le i_1\le k}}{\xi }_{i_1}(x_1)e_{i_1},\mathrm{\dots },{\displaystyle \sum _{1\le i_p\le k}}{\xi }_{i_p}(x_p)e_{i_p})$

$={\displaystyle \sum _{1\le i_1,\mathrm{\dots },i_p\le k}}{\xi }_{i_1}(x_1)\mathrm{\cdots }{\xi }_{i_p}(x_p)f(e_{i_1},\mathrm{\dots },e_{i_p}).$

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

$$f=\frac{1}{p!}\sum _{\sigma \in S_p}(\mathrm{sgn}\sigma )\sigma f,$$

so for $(x_1,\mathrm{\dots },x_p)\in {({ℝ}^k)}^p$, applying Theorem 20,

$f(x_1,\mathrm{\dots },x_p)$ $={\displaystyle \frac{1}{p!}}{\displaystyle \sum _{\sigma \in S_p}}(\mathrm{sgn}\sigma )f(x_{\sigma (1)},\mathrm{\dots },x_{\sigma (p)})$

$={\displaystyle \frac{1}{p!}}{\displaystyle \sum _{\sigma \in S_p}}(\mathrm{sgn}\sigma ){\displaystyle \sum _{1\le i_1,\mathrm{\dots },i_p\le k}}{\xi }_{i_1}(x_{\sigma (1)})\mathrm{\cdots }{\xi }_{i_p}(x_{\sigma (p)})f(e_{i_1},\mathrm{\dots },e_{i_p})$

$={\displaystyle \frac{1}{p!}}{\displaystyle \sum _{1\le i_1,\mathrm{\dots },i_p\le k}}f(e_{i_1},\mathrm{\dots },e_{i_p}){\displaystyle \sum _{\sigma \in S_p}}(\mathrm{sgn}\sigma ){\xi }_{i_1}(x_{\sigma (1)})\mathrm{\cdots }{\xi }_{i_p}(x_{\sigma (p)})$

$={\displaystyle \frac{1}{p!}}{\displaystyle \sum _{1\le i_1,\mathrm{\dots },i_p\le k}}f(e_{i_1},\mathrm{\dots },e_{i_p})({\xi }_{i_1}\wedge \mathrm{\cdots }\wedge {\xi }_{i_p})(x_1,\mathrm{\dots },x_p).$

Since $f$ is alternating, $i_r=i_s$ for $r\ne s$ implies $f(e_{i_1},\mathrm{\dots },e_{i_p})=0$. Let

$${ℐ}_{p,k}=\{I\subset \{1,\mathrm{\dots },k\}:|I|=p\};$$

For $I\in {ℐ}_{p,k}$, define $I_1,\mathrm{\dots },I_p$ by $I=\{I_1,\mathrm{\dots },I_p\}$ and . Then, applying Theorem 16, as $|S_I|=p!$,

$f$ $={\displaystyle \frac{1}{p!}}{\displaystyle \sum _{I\in {ℐ}_{p,k}}}{\displaystyle \sum _{\tau \in S_I}}f(e_{\tau (I_1)},\mathrm{\dots },e_{\tau (I_p)}){\xi }_{\tau (I_1)}\wedge \mathrm{\cdots }\wedge {\xi }_{\tau (I_p)}$

$={\displaystyle \frac{1}{p!}}{\displaystyle \sum _{I\in {ℐ}_{p,k}}}{\displaystyle \sum _{\tau \in S_I}}(\mathrm{sgn}\tau )f(e_{I_1},\mathrm{\dots },e_{I_p})(\mathrm{sgn}\tau ){\xi }_{I_1}\wedge \mathrm{\cdots }\wedge {\xi }_{I_p}$

$={\displaystyle \sum _{I\in {ℐ}_{p,k}}}f(e_{I_1},\mathrm{\dots },e_{I_p}){\xi }_{I_1}\wedge \mathrm{\cdots }\wedge {\xi }_{I_p}.$

proving the claim.

(iii) $|{ℐ}_{p,k}|=\left(\genfrac{}{}{0pt}{}{k}{p}\right)$.

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