Introduction to tensor 3: Exterior power and Exterior algebra, determinant, cross product

AbstractExterior Powers and Alternating MapBasic properties of multiplication DeterminantInner product and Cross product

Abstract

One of the aims of this blog is to prove that

$$\begin{array}{}\text{(1)}& \det (\phi \circ \psi )=\det \left(\phi \right)\det \left(\psi \right)\end{array}$$

and point out that is the functorial property of

$$\begin{array}{}\text{(2)}& \begin{array}{c}\stackrel{n}{\bigwedge }(-)\end{array}\end{array}$$

Notice that we already view the chain rule of the differential operator as the functorial property as well. Click here to see details.

In addition, I will shows how could we using the exterior algebra and hodge star operator to deal with the inner product and cross product on ${\mathbb{R}}^3$$\mathbb R^3$, that is neat!

Exterior Powers and Alternating Map

Let $M$$M$ and $N$$N$ be two $A-$$A-$ modules. An $r-$$r-$ multilinear map $\alpha :M^r\to N$$\alpha:M^r\to N$ is called $r-$$r-$alternating if $\alpha (m_1,...,m_r)=0$$\alpha(m_1,...,m_r)=0$ whenever there exist $i\ne j$$i\ne j$ with $m_i=m_j$$m_i=m_j$.

The alternating map from $M^r$$M^r$ gives us a functor.

That is,

$$\begin{array}{}\text{(3)}& {\mathrm{Alt}}^r(M,-):A-\mathrm{Module}⟶A-\mathrm{Module}\end{array}$$

For the object

$$\begin{array}{}\text{(4)}& N⟼{\mathrm{Alt}}^r(M,N)\end{array}$$

For the morphism

$$\begin{array}{}\text{(5)}& f⟼f_{\ast }\end{array}$$

The $f_{\ast }$$f_*$ here is the pushforward.

$$\begin{array}{}\text{(6)}& f_{\ast }(\alpha ):=f\circ \alpha \end{array}$$

Let ${\mathrm{Alt}}^r(M)$$\mathrm{Alt}^r(M)$ be the image of this functor, that is, the category of alternating map from $M$$M$.

Denote the initial object of ${\mathrm{Alt}}^r(M)$$\mathrm{Alt}^r(M)$ as $\stackrel{r}{\bigwedge }(M)$$\bigwedge^r(M)$, which is the exterior power.

The element in $\stackrel{r}{\bigwedge }(M)$$\bigwedge ^r(M)$ is $x_1\wedge x_2\wedge ...\wedge x_r$$x_1\wedge x_2\wedge...\wedge x_r$

Here the exterior power gives us a functor.

$$\begin{array}{}\text{(7)}& \stackrel{r}{\bigwedge }(-):R-\mathrm{Mod}⟶R-\mathrm{Mod}\end{array}$$

$$\begin{array}{}\text{(8)}& \begin{array}{c}M⟼\stackrel{r}{\bigwedge }(M)\\ (T:x\mapsto T(x))⟼x_1\wedge x_2\wedge ...\wedge x_r\mapsto T{x}_1\wedge T{x}_2\wedge ...\wedge T{x}_r)\end{array}\end{array}$$

Let us construct it(up to isomorphism).

This form a functor because

$$\begin{array}{}\text{(9)}& \stackrel{r}{\bigwedge }(TS)x_1\wedge x_2\wedge ...\wedge x_n=(TS)x_1\wedge (TS)x_2\wedge ...\wedge (TS)x_n=T(S{x}_1)\wedge T(S{x}_2)\wedge ...\wedge T(S{x}_n)\end{array}$$

But the right handside is just ${\displaystyle \stackrel{r}{\bigwedge }(T)\circ \stackrel{r}{\bigwedge }(S)x_1\wedge x_2\wedge ...\wedge x_n}$$\displaystyle\bigwedge^r (T)\circ\bigwedge^r (S)x_1\wedge x_2\wedge...\wedge x_n$

Let $Q_r\subseteq T^r(M)$$Q_r\subseteq T^r(M)$ be the $A-$$A-$​ submodule generated by the set

$$\begin{array}{}\text{(10)}& E_r:=\{m_1\otimes ...\otimes m_r\in T^r(M);\mathrm{\exists }1\le i\ne j\le r,m_i=m_j\}\end{array}$$

Then every $r-$$r-$alternating map pass-through $T^r(M)/Q_r$$T^r(M)/Q_r$, Hence $T^r(M)/Q_r\cong \stackrel{r}{\bigwedge }(M)$$T^r(M)/Q_r\cong\bigwedge ^r(M)$.

By the universal property, we have the natural isomorphism between two functors.

$$\begin{array}{}\text{(11)}& {\mathrm{Hom}}_{R-\mathrm{Mod}}(\stackrel{r}{\bigwedge }(M),-)\cong {\mathrm{Alt}}^r(M,-)\end{array}$$

i.e. ${\mathrm{Alt}}^r(M,-)$$\mathrm{Alt}^r(M,-)$ is representable.

The direct sum

$$\begin{array}{}\text{(12)}& \underset{r\ge 0}{⨁}\stackrel{r}{\bigwedge }(M)\cong T(M)/I_{\wedge }(M)\end{array}$$

is called the exterior algebra of $M$$M$.

Here $I_{\wedge }(M):=⟨x\otimes x:x\in M⟩$$I_\wedge(M):=\langle x\otimes x:x\in M\rangle$ is the ideal in the tensor algebra $T(M)$$T(M)$.

If $M$$M$ is a free module, and $\mathrm{rank}(M)=n$$\mathrm{rank}(M)=n$, let $x_1,...,x_n$$x_1,...,x_n$ be the basis of $M$$M$.

Then

$$\begin{array}{}\text{(13)}& \mathrm{rank}(\stackrel{r}{\bigwedge }(M))=(\genfrac{}{}{0ex}{}{n}{r})\end{array}$$

When $r>n$$r> n$, the rank will be zero.

Usually, we denote the product in $\underset{r\ge 0}{⨁}\stackrel{r}{\bigwedge }(M)$$\bigoplus_{r\ge 0}\bigwedge^r (M)$ as

$$\begin{array}{}\text{(14)}& \mu :\stackrel{p}{\bigwedge }(M)\times \stackrel{q}{\bigwedge }(M)⟶\stackrel{p+q}{\bigwedge }(M)\end{array}$$

$$\begin{array}{}\text{(15)}& (\omega ,\eta )\mapsto \omega \wedge \eta \end{array}$$

Basic properties of multiplication

Form the construction

$$\begin{array}{}\text{(16)}& \underset{r\ge 0}{⨁}\stackrel{r}{\bigwedge }(M)\cong T(M)/I_{\wedge }(M)\end{array}$$

We can see that

$$\begin{array}{}\text{(17)}& \mathrm{\forall }x\in M,x\wedge x=0\end{array}$$

Hence

$$\begin{array}{}\text{(18)}& x\wedge y=x\wedge x+x\wedge y=x\wedge (x+y)=x\wedge (x+y)-(x+y)\wedge (x+y)=-y\wedge (x+y)=-y\wedge x-y\wedge y=-y\wedge x\end{array}$$

Hence we have :

$$\begin{array}{}\text{(19)}& \text{For }\omega \in \stackrel{p}{\bigwedge }(M),\eta \in \stackrel{q}{\bigwedge }(M),\omega \wedge \eta =(-1{)}^{pq}\eta \wedge \omega \end{array}$$

Proof.

Let $\omega =x_1\wedge ...\wedge x_p,\eta =y_1\wedge ...\wedge y_n$$\omega=x_1\wedge...\wedge x_p,\eta=y_1\wedge...\wedge y_n$,

$$\begin{array}{}\text{(20)}& \omega \wedge \eta =x_1\wedge ...\wedge x_n\wedge y_1\wedge ...\wedge y_n\end{array}$$

Break it step by step, it is not hard to see that

$$\begin{array}{}\text{(21)}& x_1\wedge ...\wedge x_n\wedge y_1\wedge ...\wedge y_n=(-1{)}^q{x}_1\wedge ...\wedge \eta \wedge x_n=(-1{)}^{2q}{x}_1\wedge ...\wedge \eta \wedge x_{n-1}\wedge x_n...=(-1{)}^{pq}\eta \wedge \omega \end{array}$$

Here we obtain an isomorphism

$$\begin{array}{}\text{(22)}& \stackrel{p}{\bigwedge }(M)\otimes \stackrel{q}{\bigwedge }(M)\cong \stackrel{q}{\bigwedge }(M)\otimes \stackrel{p}{\bigwedge }(M)\end{array}$$

$$\begin{array}{}\text{(23)}& \omega \wedge \eta ⟼(-1{)}^{pq}\eta \wedge \omega \end{array}$$

Remark

Let $(I,+)$$(I,+)$ be an abelian monoid and $ϵ:I\to \mathbb{Z}/2\mathbb{Z}$$\epsilon:I\to\mathbb Z/2\mathbb Z$ be a monoid homomorphism. If graded algebra $A$$A$ satisfies

$$\begin{array}{}\text{(24)}& xy=(-1{)}^{ϵ(\mathrm{deg}x)ϵ(\mathrm{deg}y)}yx\end{array}$$

Then we say that $A$$A$ is $ϵ$$\epsilon$ commutative.

If $ϵ$$\epsilon$ is zero map, then $A$$A$ is commutative.

Usually, we choose $I\subseteq \mathbb{Z}$$I\subseteq\mathbb Z$, and $ϵ:\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$$\epsilon:\mathbb Z\to\mathbb Z/2\mathbb Z$ . In this case, $A$$A$ is anti-commutative.

I will write a blog about it. It is neat. We will deal with the functorial properties of the exterior algebra in that blog.

Determinant

Let $E$$E$ be a free module over a commutative ring $R$$R$ and $\mathrm{rank}(E)=n$$\mathrm{rank}(E)=n$.

Definition. Let $\phi \in {\mathrm{End}}_R(E)$$\varphi\in\mathrm{End}_R(E)$, the functor $\stackrel{n}{\bigwedge }(-)$$\bigwedge^n(-)$ gives us a map

$$\begin{array}{}\text{(25)}& \stackrel{n}{\bigwedge }(\phi )\in \mathrm{End}(\stackrel{n}{\bigwedge }(E))=R\end{array}$$

Remark. The claim

$$\begin{array}{}\text{(26)}& \mathrm{End}(\stackrel{n}{\bigwedge }(E))=R\end{array}$$

follows from

$$\begin{array}{}\text{(27)}& \mathrm{rank}(\stackrel{n}{\bigwedge }(E))=(\genfrac{}{}{0ex}{}{n}{n})=1\end{array}$$

We define that

$$\begin{array}{}\text{(28)}& \det \phi :=\stackrel{n}{\bigwedge }(\phi )\end{array}$$

That is, choose $x_1,...,x_n$$x_1,...,x_n$ be a basis of $E$$E$.

$$\begin{array}{}\text{(29)}& \det \left(\phi \right)x_1\wedge x_2\wedge ...\wedge x_n=\phi (x_1)\wedge \phi (x_2)\wedge ...\wedge \phi (x_n)=\sum _{i=1}^n{a}_{i,1}{x}_i\wedge ...\wedge \sum _{i=1}^n{a}_{i,n}{x}_n\end{array}$$

Using the distributive law and $x_i\wedge x_i=0$$x_i\wedge x_i=0$ we get that

$$\begin{array}{}\text{(30)}& \det \left(\phi \right)x_1\wedge x_2\wedge ...\wedge x_n=\sum _{1\le j_1,...,j_n\le n}{a}_{j_{1}1}...a_{j_{n}n}{x}_{j_1}\wedge ...\wedge x_{j_n}=\sum _{\sigma \in {\mathfrak{S}}_n}\mathrm{sgn}(\sigma )a_{\sigma (1),1}...a_{\sigma (n)n}{x}_1\wedge x_2\wedge ...\wedge x_n\end{array}$$

Where $j_i\ne j_k$$j_i\ne j_k$ and $\sigma (i)=j_i$$\sigma(i)=j_i$. We get that

$$\begin{array}{}\text{(31)}& \det \left(\phi \right)=\sum _{\sigma \in {\mathfrak{S}}_n}\mathrm{sgn}(\sigma )a_{\sigma (1),1}...a_{\sigma (n)n}\end{array}$$

Corollary. $\det \left(\phi \right)=\det \left({}^t\phi \right)$$\det(\varphi)=\det(^t\varphi)$.

Proof. Since the transpose $(a_{i,j})⟼(a_{j,i})$$(a_{i,j})\longmapsto(a_{j,i})$ gives us that

$$\begin{array}{}\text{(32)}& \sum _{\sigma \in {\mathfrak{S}}_n}\mathrm{sgn}(\sigma )a_{\sigma (1),1}...a_{\sigma (n)n}⟼\sum _{\sigma \in {\mathfrak{S}}_n}\mathrm{sgn}(\sigma )a_{1,\sigma (1)}...a_{n,\sigma (n)}=\det \left({}^t\phi \right)\end{array}$$

But ${\displaystyle a_{i,j}=a_{i,\sigma (i)}=a_{{\sigma }^{-1}(j),j}}$$\displaystyle a_{i,j}=a_{i,\sigma(i)}= a_{\sigma^{-1}(j),j}$ , hence

$$\begin{array}{}\text{(33)}& \det \left({}^t\phi \right)=\sum _{\sigma \in {\mathfrak{S}}_n}\mathrm{sgn}(\sigma )a_{1,\sigma (1)}...a_{n,\sigma (n)}=\sum _{\sigma \in {\mathfrak{S}}_n}\mathrm{sgn}(\sigma )a_{{\sigma }^{-1}(1),1}...a_{{\sigma }^{-1}(n),n}\end{array}$$

Since $\mathrm{sgn}(\sigma )=\mathrm{sgn}({\sigma }^{-1})$$\mathrm{sgn}(\sigma)=\mathrm{sgn}(\sigma^{-1})$, hence we get

$$\begin{array}{}\text{(34)}& \det \left({}^t\phi \right)=\sum _{{\sigma }^{-1}\in {\mathfrak{S}}_n}\mathrm{sgn}({\sigma }^{-1})a_{{\sigma }^{-1}(1),1}...a_{{\sigma }^{-1}(n),n}=\sum _{\sigma \in {\mathfrak{S}}_n}\mathrm{sgn}(\sigma )a_{\sigma (1),1}...a_{\sigma (n)n}=\det \left(\phi \right)\end{array}$$

Proposition. $\det (\phi \circ \psi )=\det \phi \det \psi$$\det (\varphi\circ\psi)=\det\varphi\det\psi$​​​​​.

Proof.

$$\begin{array}{}\text{(35)}& \phi \circ \psi (x_1)\wedge \phi \circ \psi (x_2)\wedge ...\wedge \phi \circ \psi (x_n)=\phi (\psi (x_1))\wedge \phi (\psi (x_2))\wedge ...\wedge \phi (\psi (x_n))\end{array}$$

According to $(28)$$(28)$,

$$\begin{array}{}\text{(36)}& \det \left(\phi \right)\psi (x_1)\wedge \psi (x_2)\wedge ...\wedge \psi (x_n)=\det \left(\phi \right)\det \left(\psi \right)x_1\wedge x_2\wedge ...\wedge x_n\end{array}$$

That is the property of functor,

$$\begin{array}{}\text{(37)}& \det (\phi \circ \psi )=\stackrel{n}{\bigwedge }(\phi \circ \psi )=\stackrel{n}{\bigwedge }(\phi )\stackrel{n}{\bigwedge }(\psi )=\det \phi \det \psi \end{array}$$

Proposition. $\phi \in {\mathrm{End}}_R(E)$$\varphi\in\mathrm{End}_R(E)$ is invertible if and only if $\det \phi \in R^{\ast }$$\det\varphi\in R^*$.

Let $x_1,...,x_n$$x_1,...,x_n$ be a basis of $E$$E$.

The proof follows from

$$\begin{array}{}\text{(38)}& \phi \text{ is invertible }⟺\phi x_1\wedge ...\wedge \phi x_n\text{ is a basis of}\stackrel{n}{\bigwedge }(E)\end{array}$$

and

$$\begin{array}{}\text{(39)}& \det \phi x_1\wedge ...\wedge x_n=\phi x_1\wedge ...\wedge \phi x_n\end{array}$$

directly.

$⟹$$\Longrightarrow$

Since $\det$$\det$ is a monoid homomorphism, hence it preserve the invertible elements.

$⟸$$\Longleftarrow$

Since $\phi x_1\wedge ...\wedge \phi x_n$$\varphi x_1\wedge...\wedge\varphi x_n$ is a basis.

Hence

$$\begin{array}{}\text{(40)}& x_1\wedge x_2\wedge ....\wedge x_n=\lambda \phi x_1\wedge ...\wedge \phi x_n=\lambda \det \phi x_1\wedge x_2...\wedge x_n⟹\lambda \det \phi =1_R◻\end{array}$$

Definition. Adjugate map.

Let $E$$E$ to a free module over $R$$R$ and $\mathrm{rank}(E)=n$$\mathrm{rank}(E)=n$. Then we have ${\displaystyle \stackrel{n-1}{\bigwedge }(E)\cong E,\stackrel{n}{\bigwedge }(E)=1.}$$\displaystyle\bigwedge^{n-1}(E)\cong E,\bigwedge^n(E)=1.$

Hence we have the following isomorphism:

$$\begin{array}{}\text{(41)}& E\cong {\mathrm{Hom}}_{R-\mathrm{Mod}}(E,R)\cong {\mathrm{Hom}}_{R-\mathrm{Mod}}(\stackrel{n-1}{\bigwedge }E,\stackrel{n}{\bigwedge }E)\end{array}$$

Let $\psi :E\to {\mathrm{Hom}}_{R-\mathrm{Mod}}(\stackrel{n-1}{\bigwedge }E,\stackrel{n}{\bigwedge }E)$$\psi:E\to \mathrm{Hom}_{R-\mathrm{Mod}}(\bigwedge^{n-1}E,\bigwedge ^n E)$ be the isomorphism defined as:

$$\begin{array}{}\text{(42)}& v⟼({\varphi }_v:\stackrel{n-1}{\bigwedge }E\to \stackrel{n}{\bigwedge }E,\alpha \to v\wedge \alpha )\end{array}$$

For an $R-$$R-$module homomorphism $T$$T$, we can define the adjugate of $T$$T$, i.e. $T^{\vee }$$T^\vee$ as

$$\begin{array}{}\text{(43)}& {\varphi }_{T^{\vee }(v)}={\varphi }_v\circ \stackrel{n-1}{\bigwedge }T\end{array}$$

if $e_1,...,e_n$$e_1,...,e_n$ form a basis of $E$$E$, then

$$\begin{array}{}\text{(44)}& T^{\vee }{e}_1\wedge e_2\wedge ....\wedge e_n=e_1\wedge T(e_2)\wedge ...\wedge T(e_n)\end{array}$$

Notice that ${\varphi }_v\circ \stackrel{n-1}{\bigwedge }T\in {\mathrm{Hom}}_{R-\mathrm{Mod}}(\stackrel{n-1}{\bigwedge }E,\stackrel{n}{\bigwedge }E)$$\phi_v\circ\bigwedge^{n-1}T\in \mathrm{Hom}_{R-\mathrm{Mod}}(\bigwedge^{n-1}E,\bigwedge ^n E)$ and $\varphi$$\phi$ is an isomorphism. Hence for each $v\in E$$v\in E$, there exists an unique $T^{\vee }(v)$$T^\vee(v)$ such that ${\displaystyle {\varphi }_{T^{\vee }(v)}={\varphi }_v\circ \stackrel{n-1}{\bigwedge }}$$\displaystyle \phi _{T^\vee (v)}=\phi_v\circ\bigwedge^{n-1}$ . Hence $T^{\vee }$$T^{\vee}$ is well defined.

Proposition. $T^{\vee }T=\det T\cdot I$$T^\vee T=\det T\cdot I$.

Proof. Replace $e_1$$e_1$ by $T(e_1)$$T(e_1)$ then $T^{\vee }T(e_1)\wedge e_2\wedge ...\wedge e_n=\det T{e}_1\wedge e_2\wedge ...\wedge e_n.◻$$T^\vee T(e_1)\wedge e_2\wedge...\wedge e_n=\det T e_1\wedge e_2\wedge...\wedge e_n.\quad\square$

Inner product and Cross product

Now let us consider a special exterior algebra: ${\displaystyle \bigwedge ({\mathbb{R}}^3)}$$\displaystyle\bigwedge(\mathbb{R}^3)$, which is an excellent example.

Observe that $\begin{array}{c}\bigwedge (V)=\underset{r\ge 0}{⨁}\stackrel{r}{\bigwedge }(V)\end{array}$$\bigwedge(V) = \bigoplus_{r\geq 0}\bigwedge^r(V)\\$ and ${\displaystyle \stackrel{i}{\bigwedge }(V)\cong \stackrel{n-i}{\bigwedge }(V)}$$\displaystyle\bigwedge^i(V) \cong \bigwedge^{n-i}(V)$, where $n=\dim V$$n = \dim V$, since $\begin{array}{c}(\genfrac{}{}{0ex}{}{n}{i})=(\genfrac{}{}{0ex}{}{n}{n-i})\end{array}$$\binom{n}{i} = \binom{n}{n-i}\\$.

There exists a way to define an isomorphism between $\stackrel{i}{\bigwedge }(V)$$\bigwedge^i(V)$ and $\stackrel{n-i}{\bigwedge }(V)$$\bigwedge^{n-i}(V)$ called Hodge duality. We will deal with the general case in the future. In this essay, we will use the particular case to consider the inner and cross products in ${\mathbb{R}}^3$$\mathbb{R}^3$.

Let $e_1,e_2,e_3$$e_1, e_2, e_3$ be the standard orthogonal basis of ${\mathbb{R}}^3$$\mathbb{R}^3$.

Define the Hodge star operator ${\displaystyle \ast :\stackrel{i}{\bigwedge }(V)\to \stackrel{n-i}{\bigwedge }(V)}$$\displaystyle*: \bigwedge^i(V) \to \bigwedge^{n-i}(V)$ as follows:

$$\begin{array}{}\text{(45)}& e_i\wedge \ast e_i=e_1\wedge e_2\wedge e_3\end{array}$$

For example, $\ast e_1=e_2\wedge e_3$$*e_1 = e_2 \wedge e_3$ and $\ast (e_2\wedge e_3)=e_1$$*(e_2 \wedge e_3) = e_1$.

Remark. $\ast (\ast v)=v$$*(*v) = v$.

Let

$$\begin{array}{}\text{(46)}& \omega =a{e}_1+b{e}_2+c{e}_3,\eta =a^{\prime }{e}_1+b^{\prime }{e}_2+c^{\prime }{e}_3\in \stackrel{1}{\bigwedge }({\mathbb{R}}^3)\end{array}$$

Proposition. $\ast (\omega \wedge \ast \eta )=⟨\omega ,\eta ⟩$$*(\omega \wedge *\eta) = \langle \omega, \eta \rangle$.

Proof.

$$\begin{array}{}\text{(47)}& \ast (\omega \wedge \ast \eta )=\ast [(a{e}_1+b{e}_2+c{e}_3)\wedge (a^{\prime }\ast e_1+b^{\prime }\ast e_2+c^{\prime }\ast e_3)]\end{array}$$

i.e.,

$$\begin{array}{}\text{(48)}& \ast [a{a}^{\prime }{e}_1\wedge \ast e_1+b{b}^{\prime }{e}_2\wedge \ast e_2+c{c}^{\prime }{e}_3\wedge \ast e_3]=\ast [(a{a}^{\prime }+b{b}^{\prime }+c{c}^{\prime })e_1\wedge e_2\wedge e_3]=⟨\omega ,\eta ⟩◻\end{array}$$

Proposition. $\omega \times \eta =\ast (\omega \wedge \eta )$$\omega \times \eta = *(\omega \wedge \eta)$

Proof. One can simply prove by observation by the definition, but I will show the calculations to provide a friendly explanation for readers who are still not familiar with the Hodge star operator.

$$\begin{array}{}\text{(49)}& \ast (\omega \wedge \eta )=\ast (a{b}^{\prime }{e}_1\wedge e_2+b{a}^{\prime }{e}_2\wedge e_1+a{c}^{\prime }{e}_1\wedge e_3+c{a}^{\prime }{e}_3\wedge e_1+b{c}^{\prime }{e}_2\wedge e_3+c{b}^{\prime }{e}_3\wedge e_2)\end{array}$$

Using the fact that $e_i\wedge e_j=-e_j\wedge e_i$$e_i \wedge e_j = -e_j \wedge e_i$, we get:

$$\begin{array}{}\text{(50)}& \ast (\omega \wedge \eta )=\ast [(a{b}^{\prime }-a^{\prime }b)e_1\wedge e_2+(b{c}^{\prime }-b^{\prime }c)e_2\wedge e_3+(c{a}^{\prime }-c^{\prime }a)e_3\wedge e_1]\end{array}$$

Finally,

$$\begin{array}{}\text{(51)}& (a{b}^{\prime }-a^{\prime }b)e_3+(b{c}^{\prime }-b^{\prime }c)e_1+(c{a}^{\prime }-c^{\prime }a)e_2=\omega \times \eta ◻\end{array}$$

Proposition. $⟨\omega ,\omega \times \eta ⟩=0=⟨\eta ,\omega \times \eta ⟩$$\langle \omega, \omega \times \eta \rangle = 0 = \langle \eta, \omega \times \eta \rangle$.

Proof. We only need to prove $⟨\omega ,\omega \times \eta ⟩=0$$\langle \omega, \omega \times \eta \rangle = 0$.

Since $\ast (\omega \wedge \ast \eta )=⟨\omega ,\eta ⟩$$*(\omega \wedge *\eta) = \langle \omega, \eta \rangle$, we have $⟨\omega ,\omega \times \eta ⟩=\ast (\omega \wedge \ast (\omega \times \eta ))$$\langle \omega, \omega \times \eta \rangle = *(\omega \wedge *(\omega \times \eta))$. But $\omega \times \eta =\ast (\omega \wedge \eta )$$\omega \times \eta = *(\omega \wedge \eta)$!

Thus, we have

$$\begin{array}{}\text{(52)}& ⟨\omega ,\omega \times \eta ⟩=\ast (\omega \wedge \ast (\omega \times \eta ))=\ast (\omega \wedge \omega \wedge \eta )=\ast (0)=0◻\end{array}$$