$\mathcal{T}=\frac{1}{2}\int_{V}\,dV\,\rho\left(\boldsymbol{r}_{0}^{\prime}\right)\Big{(}\boldsymbol{v}_{G}^{2}+2\,\boldsymbol{v}_{G}\cdot\boldsymbol{v}^{\prime}+\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)^{2}\Big{)}$
(234)

Using the mass (216), the kinetic energy (234) becomes,

$\mathcal{T}=\frac{1}{2}\,m\,\boldsymbol{v}_{G}^{2}+\boldsymbol{v}_{G}\cdot\int_{V}\,dV\,\rho\left(\boldsymbol{r}_{0}^{\prime}\right)\boldsymbol{v}^{\prime}+\frac{1}{2}\int_{V}\,dV\,\rho\left(\boldsymbol{r}_{0}^{\prime}\right)\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)^{2}$
(235)

By definition of the centre of mass, the relative momentum of the rigid body
vanishes,

$\boldsymbol{p}^{\prime}=\int_{V}\,dV\,\rho\left(\boldsymbol{r}_{0}^{\prime}\right)\boldsymbol{v}^{\prime}=0$
(236)

which implies that the kinetic energy (235) reduces to,

$\mathcal{T}=\frac{1}{2}\,m\,\boldsymbol{v}_{G}^{2}+\frac{1}{2}\int_{V}\,dV\,\rho\left(\boldsymbol{r}_{0}^{\prime}\right)\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)^{2}$
(237)

In view of the dualities (378), (386) and (390), we obtain the identity,

$\begin{split}&\left(\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)^{2}\right)^{\ast}=\Big{(}\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)\cdot\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)\Big{)}^{\ast}=\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)\wedge\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)^{\ast}\\\
&\phantom{\left(\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)^{2}\right)^{\ast}\,}=\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)\wedge\boldsymbol{r}^{\prime}\wedge\boldsymbol{\Omega}^{\ast}=-\,\boldsymbol{r}^{\prime}\wedge\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)\wedge\boldsymbol{\Omega}^{\ast}\\\
&\phantom{\left(\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)^{2}\right)^{\ast}\,}=-\,\Big{(}\\!\left(\boldsymbol{r}^{\prime}\wedge\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)\right)\cdot\boldsymbol{\Omega}\,\Big{)}^{\ast}\end{split}$
(238)

The dual of the identity (238) reads,

$\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)^{2}=-\,\Big{(}\boldsymbol{r}^{\prime}\wedge\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)\\!\Big{)}\cdot\boldsymbol{\Omega}$
(239)

Using the identity (239), the kinetic energy (237) becomes,

$\mathcal{T}=\frac{1}{2}\,m\,\boldsymbol{v}_{G}^{2}-\,\frac{1}{2}\int_{V}\,dV\,\rho\left(\boldsymbol{r}_{0}^{\prime}\right)\Big{(}\boldsymbol{r}^{\prime}\wedge\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)\\!\Big{)}\cdot\boldsymbol{\Omega}$
(240)

In the continuum limit, the angular momentum bivector (169) evaluated at the
centre of mass $G$ is recast as,

$\boldsymbol{L}_{G}=\int_{V}\,dV\,\rho\left(\boldsymbol{r}_{0}^{\prime}\right)\Big{(}\boldsymbol{r}^{\prime}\wedge\left(\boldsymbol{r}^{\prime}\cdot\boldsymbol{\Omega}\right)\\!\Big{)}$
(241)

In view of the total momentum of the rigid body,

$\boldsymbol{p}=m\,\boldsymbol{v}_{G}$ (242)

and the angular momentum (241), the kinetic energy (240) reduces to,

$\mathcal{T}=\frac{1}{2}\ \boldsymbol{p}\cdot\boldsymbol{v}_{G}-\,\frac{1}{2}\
\boldsymbol{L}_{G}\cdot\boldsymbol{\Omega}$ (243)

In view of the rotation (170) of the initial angular velocity (182) and the
basis vectors (227) of the principal body frame, the angular velocity is
written in this frame as,

$\boldsymbol{\Omega}=R\
\boldsymbol{\Omega}_{0}\,R^{{\dagger}}=\frac{1}{2}\sum_{i,j=1}^{3}\,\Omega_{ij}\,R\left(\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\right)R^{{\dagger}}=\frac{1}{2}\sum_{i,j=1}^{3}\,\Omega_{ij}\,\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}$
(244)

The scalar product of the angular momentum and the angular velocity (244) is
written in the principal body frame as,

$\boldsymbol{L}_{G}\cdot\boldsymbol{\Omega}=\frac{1}{4}\sum_{i,j,k,\ell=1}^{3}\,I_{G,ij}\,\Omega_{ij}\,\Omega_{k\ell}\left(\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}\right)\cdot\left(\boldsymbol{\hat{f}}_{k}\wedge\boldsymbol{\hat{f}}_{\ell}\right)$
(245)

where the unit vectors satisfy the scalar identity,

$\begin{split}&\left(\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}\right)\cdot\left(\boldsymbol{\hat{f}}_{k}\wedge\boldsymbol{\hat{f}}_{\ell}\right)=\frac{1}{4}\left(\boldsymbol{\hat{f}}_{i}\,\boldsymbol{\hat{f}}_{j}-\,\boldsymbol{\hat{f}}_{j}\,\boldsymbol{\hat{f}}_{i}\right)\cdot\left(\boldsymbol{\hat{f}}_{k}\,\boldsymbol{\hat{f}}_{\ell}-\,\boldsymbol{\hat{f}}_{\ell}\,\boldsymbol{\hat{f}}_{k}\right)\\\
&=\delta_{jk}\,\delta_{i\ell}-\,\delta_{ik}\,\delta_{j\ell}\end{split}$ (246)

Using the scalar identity (246), the scalar product (247) reduces to,

$\boldsymbol{L}_{G}\cdot\boldsymbol{\Omega}=-\,\frac{1}{2}\sum_{i,j=1}^{3}\,I_{G,ij}\,\Omega_{ij}^{2}$
(247)

In view of the identities (242) and (247), the kinetic energy is recast as,

$\mathcal{T}=\frac{1}{2}\
m\,\boldsymbol{v}_{G}^{2}+\frac{1}{4}\sum_{i,j=1}^{3}\,I_{G,ij}\,\Omega_{ij}^{2}$
(248)

In view of the dualities (210) and (223) and the property
$\varepsilon_{ijk}^{2}=1$ of the Levi-Civita symbol (73), the kinetic energy
(248) is recast as,

$\mathcal{T}=\frac{1}{2}\
m\,\boldsymbol{v}_{G}^{2}+\frac{1}{4}\sum_{i,j,k=1}^{3}\varepsilon_{ijk}\,i_{G,k}\,\omega_{k}^{2}$
(249)

The antisymmetry of the Levi-Civita symbol (73) implies that the sum over the
different indices $i$ and $j$ yields a factor $2$. Thus, the kinetic energy
(249) reduces to,

$\mathcal{T}=\frac{1}{2}\
m\,\boldsymbol{v}_{G}^{2}+\frac{1}{2}\sum_{k=1}^{3}\,i_{G,k}\,\omega_{k}^{2}$
(250)

as expected in vector algebra.

##  14 Euler equations

The rotational dynamics of a rigid body is described by the Euler equations.
In order to establish this equation in geometric algebra, we begin by
considering the angular momentum theorem (160) evaluated at the centre of mass
$G$,

$\sum\,\boldsymbol{T}_{G}^{\,\text{ext}}=\boldsymbol{\dot{L}}_{G}$ (251)

Since the moments of inertia $I_{G,ij}$ are constant for a rigid body, the
time derivative of the angular momentum (219) is given by,

$\begin{split}&\boldsymbol{\dot{L}}_{G}=\frac{1}{2}\sum_{i,j=1}^{3}\,I_{G,ij}\,\dot{\Omega}_{ij}\,R\left(\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\right)R^{{\dagger}}\\\
&\phantom{\boldsymbol{\dot{L}}_{G}=}+\frac{1}{2}\sum_{i,j=1}^{3}\,I_{G,ij}\,\Omega_{ij}\left(\dot{R}\left(\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\right)R^{{\dagger}}+R\left(\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\right)\dot{R}^{{\dagger}}\right)\end{split}$
(252)

In view of the dynamical equations for the rotor (89) and its reverse (90),
the time derivative of the angular momentum (252) becomes,

$\begin{split}&\boldsymbol{\dot{L}}_{G}=\frac{1}{2}\sum_{i,j=1}^{3}\,I_{G,ij}\,\dot{\Omega}_{ij}\,R\left(\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\right)R^{{\dagger}}\\\
&\phantom{\boldsymbol{\dot{L}}_{G}=}+\frac{1}{4}\sum_{i,j=1}^{3}\,I_{G,ij}\,\Omega_{ij}\Big{(}R\left(\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\right)R^{{\dagger}}\,\boldsymbol{\Omega}-\,\boldsymbol{\Omega}\,R\left(\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\right)R^{{\dagger}}\Big{)}\end{split}$
(253)

In view of the rotation (228) and the angular momentum (219), the time
derivative of the angular momentum (253) is recast as,

$\boldsymbol{\dot{L}}_{G}=\frac{1}{2}\sum_{i,j=1}^{3}\,I_{G,ij}\,\dot{\Omega}_{ij}\,\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}+\frac{1}{2}\left(\boldsymbol{L}_{G}\,\boldsymbol{\Omega}-\,\boldsymbol{\Omega}\,\boldsymbol{L}_{G}\right)$
(254)

In view of the commutator (A) of the bivectors $\boldsymbol{L}_{G}$ and
$\boldsymbol{\Omega}$, the time derivative of the angular momentum (254) is
recast as,

$\boldsymbol{\dot{L}}_{G}=\frac{1}{2}\sum_{i,j=1}^{3}\,I_{G,ij}\,\dot{\Omega}_{ij}\,\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}+\boldsymbol{L}_{G}\times\boldsymbol{\Omega}$
(255)

In view of the initial angular velocity (182) and the rotation (228), the
angular velocity pseudovector is written as,

$\boldsymbol{\Omega}=R\
\boldsymbol{\Omega}_{0}\,R^{{\dagger}}=\frac{1}{2}\,\sum_{k,\ell=1}^{3}\,\Omega_{k,\ell}\,R\left(\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\right)R^{{\dagger}}=\frac{1}{2}\,\sum_{k,\ell=1}^{3}\,\Omega_{k,\ell}\,\boldsymbol{\hat{f}}_{k}\wedge\boldsymbol{\hat{f}}_{\ell}$
(256)

The commutator of the angular momentum bivector (229) and the angular velocity
bivector (256) is written in components as,

$\begin{split}&\boldsymbol{L}_{G}\times\boldsymbol{\Omega}=\frac{1}{2}\left(\boldsymbol{L}_{G}\,\boldsymbol{\Omega}-\,\boldsymbol{\Omega}\,\boldsymbol{L}_{G}\right)\\\
&\phantom{\boldsymbol{\Omega}\times\boldsymbol{L}_{G}}=\frac{1}{8}\sum_{i,j,k,\ell=1}^{3}\left(I_{G,ij}-\,I_{G,k\ell}\right)\Omega_{ij}\,\Omega_{k\ell}\left(\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}\right)\left(\boldsymbol{\hat{f}}_{k}\wedge\boldsymbol{\hat{f}}_{\ell}\right)\end{split}$
(257)

where the unit vectors satisfy the bivectorial identity,

$\begin{split}&\left(\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}\right)\left(\boldsymbol{\hat{f}}_{k}\wedge\boldsymbol{\hat{f}}_{\ell}\right)=\frac{1}{4}\left(\boldsymbol{\hat{f}}_{i}\,\boldsymbol{\hat{f}}_{j}-\,\boldsymbol{\hat{f}}_{j}\,\boldsymbol{\hat{f}}_{i}\right)\left(\boldsymbol{\hat{f}}_{k}\,\boldsymbol{\hat{f}}_{\ell}-\,\boldsymbol{\hat{f}}_{\ell}\,\boldsymbol{\hat{f}}_{k}\right)\\\
&=\delta_{jk}\,\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{\ell}-\,\delta_{ik}\,\boldsymbol{\hat{f}}_{j}\wedge\boldsymbol{\hat{f}}_{\ell}+\delta_{i\ell}\,\boldsymbol{\hat{f}}_{j}\wedge\boldsymbol{\hat{f}}_{k}-\,\delta_{j\ell}\,\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{k}\\\
&=\delta_{jk}\,\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{\ell}+\delta_{ik}\,\boldsymbol{\hat{f}}_{\ell}\wedge\boldsymbol{\hat{f}}_{j}-\,\delta_{j\ell}\,\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{k}-\,\delta_{i\ell}\,\boldsymbol{\hat{f}}_{k}\wedge\boldsymbol{\hat{f}}_{j}\end{split}$
(258)

Thus, the commutator (257) becomes,

$\begin{split}&\boldsymbol{L}_{G}\times\boldsymbol{\Omega}=\frac{1}{8}\sum_{i,j,k=1}^{3}\left(I_{G,ij}-\,I_{G,jk}\right)\Omega_{ij}\,\Omega_{jk}\
\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{k}\\\
&\phantom{\boldsymbol{L}_{G}\times\boldsymbol{\Omega}=}+\frac{1}{8}\sum_{i,j,k=1}^{3}\left(I_{G,ij}-\,I_{G,ik}\right)\Omega_{ij}\,\Omega_{ik}\
\boldsymbol{\hat{f}}_{k}\wedge\boldsymbol{\hat{f}}_{j}\\\
&\phantom{\boldsymbol{L}_{G}\times\boldsymbol{\Omega}=}-\,\frac{1}{8}\sum_{i,j,k=1}^{3}\left(I_{G,ij}-\,I_{G,kj}\right)\Omega_{ij}\,\Omega_{kj}\
\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{k}\\\
&\phantom{\boldsymbol{L}_{G}\times\boldsymbol{\Omega}=}-\,\frac{1}{8}\sum_{i,j,k=1}^{3}\left(I_{G,ij}-\,I_{G,ki}\right)\Omega_{ij}\,\Omega_{ki}\
\boldsymbol{\hat{f}}_{k}\wedge\boldsymbol{\hat{f}}_{j}\end{split}$ (259)

where the indices $\ell$ and $k$ were exchanged in the first two terms on the
right hand side. In view of the antisymmetric components of the angular
velocity bivector, i.e. $\Omega_{jk}=-\,\Omega_{kj}$ and
$\Omega_{ik}=-\,\Omega_{ki}$, the commutator (259) reduces to,

$\begin{split}&\boldsymbol{L}_{G}\times\boldsymbol{\Omega}=\frac{1}{4}\sum_{i,j,k=1}^{3}\left(I_{G,ij}-\,I_{G,jk}\right)\Omega_{ij}\,\Omega_{jk}\
\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{k}\\\
&\phantom{\boldsymbol{L}_{G}\times\boldsymbol{\Omega}=}+\frac{1}{4}\sum_{i,j,k=1}^{3}\left(I_{G,ij}-\,I_{G,ik}\right)\Omega_{ij}\,\Omega_{ik}\
\boldsymbol{\hat{f}}_{k}\wedge\boldsymbol{\hat{f}}_{j}\end{split}$ (260)

and is recast as,

$\begin{split}&\boldsymbol{L}_{G}\times\boldsymbol{\Omega}=\frac{1}{4}\sum_{i,j,k=1}^{3}\left(I_{G,ik}-\,I_{G,kj}\right)\Omega_{ik}\,\Omega_{kj}\
\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}\\\
&\phantom{\boldsymbol{L}_{G}\times\boldsymbol{\Omega}=}+\frac{1}{4}\sum_{i,j,k=1}^{3}\left(I_{G,kj}-\,I_{G,ki}\right)\Omega_{kj}\,\Omega_{ki}\
\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}\end{split}$ (261)

where the indices $j$ and $k$ were exchanged in the first term and the indices
$i$ and $k$ were exchanged in the first term on the right hand side. In view
of the antisymmetric components of the angular velocity bivector (223), and
the antisymmetric moments of inertia (210), the commutator (259) reduces to,

$\boldsymbol{L}_{G}\times\boldsymbol{\Omega}=\frac{1}{2}\sum_{i,j=1}^{3}\left(\,\sum_{k=1}^{3}\left(I_{G,ik}-\,I_{G,kj}\right)\Omega_{ik}\,\Omega_{kj}\right)\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}$
(262)

In view of the commutator (262), the time derivative of the angular momentum
(255) is recast in components as,

$\boldsymbol{\dot{L}}_{G}=\frac{1}{2}\sum_{i,j=1}^{3}\left(I_{G,ij}\,\dot{\Omega}_{ij}+\sum_{k=1}^{3}\left(I_{G,ik}-\,I_{G,kj}\right)\Omega_{ik}\,\Omega_{kj}\right)\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}$
(263)

The net external torque bivector $\boldsymbol{T}_{G}^{\,\text{ext}}$ is
written in components in the principal body frame as,

$\sum\,\boldsymbol{T}_{G}^{\,\text{ext}}=\sum\,\frac{1}{2}\sum_{i,j=1}^{3}T_{G,ij}^{\,\text{ext}}\
\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}$ (264)

According to the angular momentum theorem (251), the components of the
bivectors (263) and (264) are equal,

$\sum\,T_{G,ij}^{\,\text{ext}}=I_{G,ij}\,\dot{\Omega}_{ij}+\sum_{k=1}^{3}\left(I_{G,ik}-\,I_{G,kj}\right)\Omega_{ik}\,\Omega_{kj}$
(265)

For a given unit bivector
$\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}$, where $i\neq j$, the
index $k$ is the remaining index and no summation over $k$ is needed in
equation (265). This equation for the bivector
$\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}$ describes the
rotation of the rigid body in the oriented plane spanned by this bivector,

$\sum\,T_{G,ij}^{\,\text{ext}}=I_{G,ij}\,\dot{\Omega}_{ij}+\left(I_{G,ik}-\,I_{G,kj}\right)\Omega_{ik}\,\Omega_{kj}$
(266)

Using the antisymmetric components of the angular velocity bivector (223), and
the antisymmetric moments of inertia (210), equation (266) evaluated for the
couples of indices $ij\in\\{12,23,31\\}$ yields the Euler equations in
geometric algebra,

$\begin{split}&\sum\,T_{G,12}^{\,\text{ext}}=I_{G,12}\,\dot{\Omega}_{12}+\left(I_{G,23}-\,I_{G,31}\right)\Omega_{23}\,\Omega_{31}\\\
&\sum\,T_{G,23}^{\,\text{ext}}=I_{G,23}\,\dot{\Omega}_{23}+\left(I_{G,31}-\,I_{G,12}\right)\Omega_{31}\,\Omega_{12}\\\
&\sum\,T_{G,31}^{\,\text{ext}}=I_{G,31}\,\dot{\Omega}_{31}+\left(I_{G,12}-\,I_{G,23}\right)\Omega_{12}\,\Omega_{23}\end{split}$
(267)

In order to deduce from the Euler equations in geometric algebra the Euler
equations in vector algebra, we use the duality of the components of the
external torques,

$T_{G,ij}=-\,T_{G,ji}=\varepsilon_{ijk}\,\tau_{G,k}$ (268)

the duality of the components of the angular acceleration,

$\dot{\Omega}_{ij}=-\,\dot{\Omega}_{ji}=\varepsilon_{ijk}\,\dot{\omega}_{k}$
(269)

and the duality of the components of the moments of inertia (210). Thus, the
Euler equations (267) are recast as,

$\begin{split}&\sum\,\varepsilon_{123}\,\tau_{G,3}^{\,\text{ext}}=\varepsilon^{2}_{123}\,i_{G,3}\,\dot{\omega}_{3}+\varepsilon^{2}_{231}\,\varepsilon^{2}_{312}\left(i_{G,1}-\,i_{G,2}\right)\omega_{1}\,\omega_{2}\\\
&\sum\,\varepsilon_{231}\,\tau_{G,1}^{\,\text{ext}}=\varepsilon^{2}_{231}\,i_{G,1}\,\dot{\omega}_{1}+\varepsilon^{2}_{312}\,\varepsilon^{2}_{123}\left(i_{G,2}-\,i_{G,3}\right)\omega_{2}\,\omega_{3}\\\
&\sum\,\varepsilon_{312}\,\tau_{G,2}^{\,\text{ext}}=\varepsilon^{2}_{312}\,i_{G,2}\,\dot{\omega}_{2}+\varepsilon^{2}_{123}\,\varepsilon^{2}_{231}\left(i_{G,3}-\,i_{G,1}\right)\omega_{3}\,\omega_{1}\end{split}$
(270)

Taking into account the invariance of the Levi-Civita symbols (73) under
cyclic permutation,

$\varepsilon_{123}=\varepsilon_{231}=\varepsilon_{312}$ (271)

the Euler equations in vector algebra (270) reduce to,

$\begin{split}&\sum\,\tau_{G,1}^{\,\text{ext}}=i_{G,1}\,\dot{\omega}_{1}+\left(i_{G,2}-\,i_{G,3}\right)\omega_{2}\,\omega_{3}\\\
&\sum\,\tau_{G,2}^{\,\text{ext}}=i_{G,2}\,\dot{\omega}_{2}+\left(i_{G,3}-\,i_{G,1}\right)\omega_{3}\,\omega_{1}\\\
&\sum\,\tau_{G,3}^{\,\text{ext}}=i_{G,3}\,\dot{\omega}_{3}+\left(i_{G,1}-\,i_{G,2}\right)\omega_{1}\,\omega_{2}\end{split}$
(272)

##  15 Symmetric spinning disk

We now consider the torque free motion of a symmetric spinning disk (Fig. 16).
The disk is a gyroscope that can freely rotate around its center of mass $G$.
The principal body frame
$\\{\boldsymbol{\hat{f}}_{1},\boldsymbol{\hat{f}}_{2},\boldsymbol{\hat{f}}_{3}\\}$
is oriented such that the unit vector $\boldsymbol{\hat{f}}_{3}$ is along the
symmetry axis of the disk. It coincides initially with the fixed orthonormal
frame
$\\{\boldsymbol{\hat{e}}_{1},\boldsymbol{\hat{e}}_{2},\boldsymbol{\hat{e}}_{3}\\}$.
The plane spanned by the unit bivector
$\boldsymbol{\hat{f}}_{1}\wedge\boldsymbol{\hat{f}}_{2}$ that is orthogonal to
the unit vector $\boldsymbol{\hat{f}}_{3}$ is the plane of symmetry of the
disk. By symmetry, the moments of inertia in the orthogonal planes spanned by
the unit bivectors $\boldsymbol{\hat{f}}_{2}\wedge\boldsymbol{\hat{f}}_{3}$
and $\boldsymbol{\hat{f}}_{3}\wedge\boldsymbol{\hat{f}}_{1}$ containing the
unit vector $\boldsymbol{\hat{f}}_{3}$ are equal,

Figure 16: Symmetric disk spinning around its centre of mass $G$. The
principal body frame
$\\{\boldsymbol{\hat{f}}_{1},\boldsymbol{\hat{f}}_{2},\boldsymbol{\hat{f}}_{3}\\}$
is oriented such that the unit vector $\boldsymbol{\hat{f}}_{3}$ is along the
symmetry axis of the disk. The orthonormal frame
$\\{\boldsymbol{\hat{e}}_{1},\boldsymbol{\hat{e}}_{2},\boldsymbol{\hat{e}}_{3}\\}$
is fixed.

$I_{G,\parallel}\equiv I_{G,12}\qquad\text{and}\qquad I_{G,\perp}\equiv
I_{G,23}=I_{G,31}$ (273)

where the indices $\parallel$ and $\perp$ indicate the moment of inertia is
defined with respect to a principal axis plane that is either parallel or
perpendicular to the plane of symmetry. There is no net external torque
bivector evaluated at the center of mass of the freely rotating disk,

$\sum\,\boldsymbol{T}_{G}^{\,\text{ext}}=\boldsymbol{0}$ (274)

We now identify the constant bivectors in order to find the rotor that
describes entirely the rotator of the symmetric disk. Firstly, in view of the
angular momentum theorem (251), the angular momentum bivector is constant,

$\boldsymbol{L}_{G}=\textbf{const}$ (275)

Secondly, in the absence of an external torque (274), using the symmetry
(273), the projection of the Euler equations (267) in the plane of symmetry
reads,

$I_{G,\parallel}\,\dot{\Omega}_{12}=0$ (276)

which means that the velocity bivector $\boldsymbol{\Omega}_{0,\parallel}$
that is initially in the plane of symmetry is constant,

$\boldsymbol{\Omega}_{0,\parallel}\equiv\Omega_{12}\,\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}=\textbf{const}$
(277)

In of view of relations (229) and (273), the angular momentum bivetcor is
given by,

$\boldsymbol{L}_{G}=I_{G,\parallel}\,\Omega_{12}\,\boldsymbol{\hat{f}}_{1}\wedge\boldsymbol{\hat{f}}_{2}+I_{G,\perp}\,\Omega_{23}\,\boldsymbol{\hat{f}}_{2}\wedge\boldsymbol{\hat{f}}_{3}+I_{G,\perp}\,\Omega_{31}\,\boldsymbol{\hat{f}}_{3}\wedge\boldsymbol{\hat{f}}_{1}$
(278)

and recast as,

$\begin{split}&\boldsymbol{L}_{G}=I_{G,\perp}\left(\Omega_{12}\,\boldsymbol{\hat{f}}_{1}\wedge\boldsymbol{\hat{f}}_{2}+\Omega_{23}\,\boldsymbol{\hat{f}}_{2}\wedge\boldsymbol{\hat{f}}_{3}+\Omega_{31}\,\boldsymbol{\hat{f}}_{3}\wedge\boldsymbol{\hat{f}}_{1}\right)\\\
&\phantom{\boldsymbol{L}_{G}=}+\left(I_{G,\parallel}-\,I_{G,\perp}\right)\Omega_{12}\,\boldsymbol{\hat{f}}_{1}\wedge\boldsymbol{\hat{f}}_{2}\end{split}$
(279)

In view of the angular velocity (244), the angular momentum bivector (279)
reduces to,

$\boldsymbol{L}_{G}=I_{G,\perp}\,\boldsymbol{\Omega}+R\,\Big{(}\\!\left(I_{G,\parallel}-\,I_{G,\perp}\right)\Omega_{12}\,\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}\Big{)}R^{{\dagger}}$
(280)

Taking into account the bivector (277), the angular momentum bivector (280)
becomes,

$\boldsymbol{L}_{G}=I_{G,\perp}\,\boldsymbol{\Omega}+R\left(I_{G,\parallel}-\,I_{G,\perp}\right)\boldsymbol{\Omega}_{0,\parallel}\,R^{{\dagger}}$
(281)

Thus, the angular velocity bivector $\boldsymbol{\Omega}$ is expressed in
terms of the constant bivectors $\boldsymbol{L}_{G}$ and
$\boldsymbol{\Omega}_{0,\parallel}$ as,

$\boldsymbol{\Omega}=\frac{1}{I_{G,\perp}}\Big{(}\boldsymbol{L}_{G}+R\left(I_{G,\parallel}-\,I_{G,\perp}\right)\boldsymbol{\Omega}_{0,\parallel}\,R^{{\dagger}}\Big{)}$
(282)

Using the angular momentum bivector (282) and the normalisation condition
(83), the time evolution equation for the rotor (89) is recast as,

$\dot{R}=-\,\frac{1}{2}\,\boldsymbol{\Omega}\,R=-\,\frac{1}{2\,I_{G,\perp}}\Big{(}\boldsymbol{L}_{G}\,R+R\left(I_{G,\parallel}-\,I_{G,\perp}\right)\boldsymbol{\Omega}_{0,\parallel}\Big{)}$
(283)

Using the two constant bivectors defined as,

$\boldsymbol{\Omega}_{L}\equiv-\,\frac{1}{2\,I_{G,\perp}}\
\boldsymbol{L}_{G}\qquad\text{and}\qquad\boldsymbol{\Omega}_{R}\equiv-\,\frac{I_{G,\parallel}-\,I_{G,\perp}}{2\,I_{G,\perp}}\
\boldsymbol{\Omega}_{0,\parallel}$ (284)

the time evolution equation of the rotor (283) is recast as,

$\dot{R}=-\,\frac{1}{2}\,\boldsymbol{\Omega}_{L}\,R-\,\frac{1}{2}\,R\
\boldsymbol{\Omega}_{R}$ (285)

The rotor satisfies the trivial initial condition,

$R\left(0\right)=1$ (286)

The solution of the time evolution equation (285) that satisfies the initial
condition (286) is the rotor,

$R\left(t\right)=\exp\left(-\,\frac{1}{2}\,\boldsymbol{\Omega}_{L}\,t\right)\exp\left(-\,\frac{1}{2}\,\boldsymbol{\Omega}_{R}\,t\right)$
(287)

as we now show. The time derivative of the rotor (287) is,

$\begin{split}&\dot{R}\left(t\right)=\left(-\,\frac{1}{2}\,\boldsymbol{\Omega}_{L}\right)\exp\left(-\,\frac{1}{2}\,\boldsymbol{\Omega}_{L}\,t\right)\exp\left(-\,\frac{1}{2}\,\boldsymbol{\Omega}_{R}\,t\right)\\\
&\phantom{\dot{R}\left(t\right)=}+\exp\left(-\,\frac{1}{2}\,\boldsymbol{\Omega}_{L}\,t\right)\exp\left(-\,\frac{1}{2}\,\boldsymbol{\Omega}_{R}\,t\right)\left(-\,\frac{1}{2}\,\boldsymbol{\Omega}_{R}\right)\end{split}$
(288)

which reduces to the dynamical equation (285) in view of the rotor (287), as
it should.

##  16 Conclusion

Rotations can always be expressed as the composition of two reflections of
intersecting planes. These planes containing the origin $O$ are determined by
their respective orthogonal unit vectors $\boldsymbol{\hat{n}}_{1}$ and
$\boldsymbol{\hat{n}}_{2}$. The geometric product of these vectors is the
rotor
$R_{\theta}=\boldsymbol{\hat{n}}_{2}\,\boldsymbol{\hat{n}}_{1}=e^{-\,\boldsymbol{\hat{B}}\,\theta/2}=e^{-\,\boldsymbol{\hat{B}}\,\phi}$
where $\boldsymbol{\hat{B}}$ is the unit bivector and $\phi=\theta/2$ is the
angle between the unit vectors $\boldsymbol{\hat{n}}_{1}$ and
$\boldsymbol{\hat{n}}_{2}$ in the rotation plane. The rotation of a vector
$\boldsymbol{v}$ by an angle $\theta=2\,\phi$ in the rotation plane spanned by
the unit bivector $\boldsymbol{\hat{B}}$ is given by the transformation law,

$\mathsf{R}_{\boldsymbol{\hat{B}}\,\theta}\left(\boldsymbol{v}\right)=R\,\boldsymbol{v}\,R^{{\dagger}}=e^{-\,\boldsymbol{\hat{B}}\theta/2}\,\boldsymbol{v}\,e^{\boldsymbol{\hat{B}}\theta/2}$
(289)

where $R$ is the rotor of angle $\theta$ in the rotation plane spanned by the
unit bivector $\boldsymbol{\hat{B}}$. Similarly, the rotation of a multivector
$M=s+\boldsymbol{v}+\boldsymbol{A}+s^{\prime}\,I$ is given by the
transformation law,

$\mathsf{R}_{\boldsymbol{\hat{B}}\,\theta}\left(M\right)=R\,M\,R^{{\dagger}}=e^{-\,\boldsymbol{\hat{B}}\theta/2}\,M\,e^{\boldsymbol{\hat{B}}\theta/2}$
(290)

The time evolution of the rotor $R$ and its reverse $R^{{\dagger}}$ are
expressed in terms of the angular velocity bivector
$\boldsymbol{\Omega}=\dot{\theta}\,\boldsymbol{\hat{B}}$ in the rotation plane
as,

$\dot{R}=-\,\frac{1}{2}\,\boldsymbol{\Omega}\,R\qquad\text{and}\qquad\dot{R}^{{\dagger}}=\frac{1}{2}\,R^{{\dagger}}\,\boldsymbol{\Omega}$
(291)

where the angular bivector $\boldsymbol{\Omega}$ is the dual of the angular
pseudovector $\boldsymbol{\omega}$, i.e.
$\boldsymbol{\omega}=\boldsymbol{\Omega}^{\ast}$. In geometric algebra, the
time evolution of the rotor yields the Poisson formula describing the time
evolution of the rotating basis vectors $\boldsymbol{\hat{f}}_{i}$,

$\boldsymbol{\dot{\hat{f}}}_{i}=\boldsymbol{\hat{f}}_{i}\cdot\boldsymbol{\Omega}$
(292)

The geometric meaning of the inner product
$\boldsymbol{\hat{f}}_{i}\cdot\boldsymbol{\Omega}$ is that the angular
velocity bivector $\boldsymbol{\Omega}$ rotates the basis vector
$\boldsymbol{\hat{f}}_{i}$ by a $90^{\circ}$ angle in the rotation direction
defined by the unit bivector $\boldsymbol{\hat{B}}$.

Rotational dynamics is described by the angular momentum theorem. For a rigid
body, this theorem is expressed in the rotation plane in geometric algebra in
terms of bivectors evaluated with respect to the centre of mass $G$ as,

$\sum\,\boldsymbol{T}_{G}^{\,\text{ext}}=\boldsymbol{\dot{L}}_{G}$ (293)

where the angular momentum bivector $\boldsymbol{L}_{G}$ is the dual of the
angular momentum pseudovector $\boldsymbol{\ell}_{G}$, i.e.
$\boldsymbol{\ell}_{G}=\boldsymbol{L}^{\ast}_{G}$, and the external torque
bivector $\boldsymbol{T}_{G}^{\,\text{ext}}$ which is the dual of the external
torque pseudovector $\boldsymbol{\tau}_{G}^{\,\text{ext}}$. According to the
transformation law (290) for a multivector under rotation, the time evolution
of an angular momentum bivector $\boldsymbol{L}_{G}$ describing the intrinsic
rotation of a rigid body around its centre of mass is written as,

$\boldsymbol{L}_{G}=R\ \boldsymbol{L}_{G,0}\,R^{{\dagger}}$ (294)

where the initial angular momentum bivector $\boldsymbol{L}_{G,0}$ is a linear
map $\boldsymbol{I}_{G}\left(\boldsymbol{\Omega}_{0}\right)$ of the initial
angular velocity bivector,

$\boldsymbol{\Omega}_{0}=\frac{1}{2}\sum_{i,j=1}^{3}\,\Omega_{ij}\
\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\qquad\text{with}\qquad\Omega_{ji}=-\,\Omega_{ij}$
(295)

that can be expressed as,

$\boldsymbol{L}_{G,0}=\boldsymbol{I}_{G}\left(\boldsymbol{\Omega}_{0}\right)=\frac{1}{2}\sum_{i,j=1}^{3}\,I_{G,ij}\,\Omega_{ij}\
\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\qquad\text{with}\qquad
I_{G,ji}=I_{G,ij}$ (296)

where $\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}$ are the unit
bivectors in the initial principal axis frame of the rigid body,
$I_{G,ij}=\left(\varepsilon_{ijk}\right)^{2}i_{G,k}$ is the dual of the moment
of inertia $i_{G,k}$ and $\Omega_{ij}$ is the antisymmetric component of the
initial angular moment bivector $\boldsymbol{\Omega}_{0}$ in the rotation
plane spanned by the bivectors
$\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}$. The vectors
$\boldsymbol{\hat{e}}_{i}$ and bivectors
$\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}$ of the principal axis
frame of the rigid body are related to the vectors $\boldsymbol{\hat{f}}_{i}$
and bivectors $\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}$ of the
initial principal axis frame of the rigid body through the rotational
transformation laws (290),

$\boldsymbol{\hat{f}}_{i}=R\,\boldsymbol{\hat{e}}_{i}\,R^{{\dagger}}\qquad\text{and}\qquad\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}=R\,\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\,R^{{\dagger}}$
(297)

In view the linear inertia map (296) and the rotational transformation laws
(297), the angular momentum bivector is recast in the principal axis frame of
the rigid body in terms of the unit bivectors
$\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}$ as,

$\boldsymbol{L}_{G}=\frac{1}{2}\sum_{i,j=1}^{3}\,I_{G,ij}\,\Omega_{ij}\
\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}$ (298)

In geometric algebra, the Huygens-Steiner theorem for a rigid body of mass $m$
in the plane spanned by the unit bivector frame
$\boldsymbol{\hat{f}}_{i}\wedge\boldsymbol{\hat{f}}_{j}$ reads,

$I_{A,ij}=I_{G,ij}+m\,\boldsymbol{r}_{AG}^{2}$ (299)

and the kinetic energy is given by,

$\mathcal{T}=\frac{1}{2}\
m\,\boldsymbol{v}_{G}^{2}+\frac{1}{4}\sum_{i,j=1}^{3}\,I_{G,ij}\,\Omega_{ij}^{2}$
(300)

The Euler equations for the rigid body are expressed in terms of the
components of the dynamical bivectors in three orthogonal planes spanned by
the unit bivectors $\boldsymbol{\hat{f}}_{1}\wedge\boldsymbol{\hat{f}}_{2}$,
$\boldsymbol{\hat{f}}_{2}\wedge\boldsymbol{\hat{f}}_{3}$ and
$\boldsymbol{\hat{f}}_{3}\wedge\boldsymbol{\hat{f}}_{1}$ as,

$\begin{split}&\sum\,T_{G,12}^{\,\text{ext}}=I_{G,12}\,\dot{\Omega}_{12}+\left(I_{G,23}-\,I_{G,31}\right)\Omega_{23}\,\Omega_{31}\\\
&\sum\,T_{G,23}^{\,\text{ext}}=I_{G,23}\,\dot{\Omega}_{23}+\left(I_{G,31}-\,I_{G,12}\right)\Omega_{31}\,\Omega_{12}\\\
&\sum\,T_{G,31}^{\,\text{ext}}=I_{G,31}\,\dot{\Omega}_{31}+\left(I_{G,12}-\,I_{G,23}\right)\Omega_{12}\,\Omega_{23}\end{split}$
(301)

The dynamics of rigid bodies in geometric algebra can be applied to the torque
free motion of a gyroscope consisting of a symmetric spinning disk around its
centre of mass $G$. It is described by the rotor,

$R\left(t\right)=\exp\left(-\,\frac{1}{2}\,\boldsymbol{\Omega}_{L}\,t\right)\exp\left(-\,\frac{1}{2}\,\boldsymbol{\Omega}_{R}\,t\right)$
(302)

where the angular velocity bivectors $\boldsymbol{\Omega}_{L}$ and
$\boldsymbol{\Omega}_{R}$ are given by,

$\boldsymbol{\Omega}_{L}\equiv-\,\frac{1}{2\,I_{G,\perp}}\
\boldsymbol{L}_{G}\qquad\text{and}\qquad\boldsymbol{\Omega}_{R}\equiv-\,\frac{I_{G,\parallel}-\,I_{G,\perp}}{2\,I_{G,\perp}}\
\boldsymbol{\Omega}_{0,\parallel}$ (303)

$I_{G,\parallel}$ is the moment of inertia and
$\boldsymbol{\Omega}_{0,\parallel}$ is the angular velocity in the plane of
the disk, and $I_{G,\perp}$ are the moments of inertia in the planes
orthogonal to the plane of the disk.

## Acknowledgements

The author would like to thank his PhD thesis adviser Anthony Lasenby for
introducing him to geometric algebra.

##  Appendix A Geometric algebra (GA)

We consider an orthonormal vector spatial frame
$\\{\boldsymbol{\hat{e}}_{1},\boldsymbol{\hat{e}}_{2},\boldsymbol{\hat{e}}_{3}\\}$.
The geometric product of two basis vectors reads, [5]

$\boldsymbol{\hat{e}}_{i}\,\boldsymbol{\hat{e}}_{j}=\boldsymbol{\hat{e}}_{i}\cdot\boldsymbol{\hat{e}}_{j}+\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}\qquad\text{with}\qquad
i,j=1,2,3$ (304)

where,

$\boldsymbol{\hat{e}}_{1}^{2}=\boldsymbol{\hat{e}}_{2}^{2}=\boldsymbol{\hat{e}}_{3}^{2}=1\qquad\text{and}\qquad\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{i}=\boldsymbol{0}$
(305)

The inner product of two basis vectors is symmetric and defined as,

$\boldsymbol{\hat{e}}_{i}\cdot\boldsymbol{\hat{e}}_{j}=\boldsymbol{\hat{e}}_{j}\cdot\boldsymbol{\hat{e}}_{i}=\frac{1}{2}\left(\boldsymbol{\hat{e}}_{i}\,\boldsymbol{\hat{e}}_{j}+\boldsymbol{\hat{e}}_{j}\,\boldsymbol{\hat{e}}_{i}\right)$
(306)

and the outer product is antisymmetric,

$\boldsymbol{\hat{e}}_{i}\wedge\boldsymbol{\hat{e}}_{j}=-\,\boldsymbol{\hat{e}}_{j}\wedge\boldsymbol{\hat{e}}_{i}=\frac{1}{2}\left(\boldsymbol{\hat{e}}_{i}\,\boldsymbol{\hat{e}}_{j}-\,\boldsymbol{\hat{e}}_{j}\,\boldsymbol{\hat{e}}_{i}\right)$
(307)

The $8$ basis elements of the geometric algebra (GA) $\mathbb{G}^{3}$ called
the geometric algebra are :

  * 1.

one scalar : $\ 1$

  * 2.

three vectors : $\ \boldsymbol{\hat{e}}_{1}$, $\ \boldsymbol{\hat{e}}_{2}$, $\
\boldsymbol{\hat{e}}_{3}$

  * 3.

three bivectors : $\ \boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}$,
$\ \boldsymbol{\hat{e}}_{2}\wedge\boldsymbol{\hat{e}}_{3}$, $\
\boldsymbol{\hat{e}}_{3}\wedge\boldsymbol{\hat{e}}_{1}$

  * 4.

one trivector or pseudoscalar : $\
\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}\wedge\boldsymbol{\hat{e}}_{3}$

Figure 17: A scalar, a vector, a bivector and a pseudoscalar are illustrated
graphically. Their orientation and modulus are independent of their shape.

where the number of elements of each type $1$, $3$, $3$, $1$ are the binomial
coefficients in three dimensions. In $\mathbb{G}^{3}$, there are $2^{3}$ basis
elements. The geometric interpretation of the basis elements is clear (Fig.
17). A scalar $s$ is an oriented point, a vector is an oriented line element,
a bivector is an oriented plane element and a pseudoscalar is an oriented
volume element. The notation used is the following : scalars $s$ are written
in lower case letters, vectors $\boldsymbol{v}$ are written in lower case bold
letters, bivectors $\boldsymbol{B}$ are written in upper case bold letters and
pseudoscalars are written as $s^{\prime}I$. A multivector $M$, which is linear
combination of elements of $\mathbb{G}^{3}$, is written in upper case letters,

$M=s+\boldsymbol{v}+\boldsymbol{B}+s^{\prime}I$ (308)

Taking into account the definition (304), the bivectors and the pseudoscalar
are also written as geometric products of the basis unit vectors
$\boldsymbol{\hat{e}}_{1}$, $\boldsymbol{\hat{e}}_{2}$,
$\boldsymbol{\hat{e}}_{3}$ :

  * 1.

bivectors : $\ \boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}$, $\
\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}$, $\
\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}$

  * 2.

pseudoscalar : $\
\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}$

The pseudoscalar
$I=\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}$
behaves as an imaginary number since,

$I^{2}=\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)=\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\left(\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{3}\right)=-\,\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{1}\right)\left(\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{2}\right)=-\,1$
(309)

The geometric product of two basis vectors $\boldsymbol{u}$ and
$\boldsymbol{v}$ reads,

$\boldsymbol{u}\,\boldsymbol{v}=\boldsymbol{u}\cdot\boldsymbol{v}+\boldsymbol{u}\wedge\boldsymbol{v}$
(310)

The inner product between the two basis vectors is symmetric,

$\boldsymbol{u}\cdot\boldsymbol{v}=\boldsymbol{v}\cdot\boldsymbol{u}=\frac{1}{2}\left(\boldsymbol{u}\,\boldsymbol{v}+\boldsymbol{v}\,\boldsymbol{u}\right)$
(311)

and the outer product is antisymmetric,

$\boldsymbol{u}\wedge\boldsymbol{v}=-\,\boldsymbol{v}\wedge\boldsymbol{u}=\frac{1}{2}\left(\boldsymbol{u}\,\boldsymbol{v}-\,\boldsymbol{v}\,\boldsymbol{u}\right)$
(312)

The geometric product of a vector $\boldsymbol{v}$ and a bivector
$\boldsymbol{B}$ is the sum of the inner product of the outer product,

$\boldsymbol{v}\,\boldsymbol{B}=\boldsymbol{v}\cdot\boldsymbol{B}+\boldsymbol{v}\wedge\boldsymbol{B}$
(313)

and the geometric product of a bivector $\boldsymbol{B}$ and a vector
$\boldsymbol{v}$ is the sum of the inner product of the outer product,

$\boldsymbol{B}\,\boldsymbol{v}=\boldsymbol{B}\cdot\boldsymbol{v}+\boldsymbol{B}\wedge\boldsymbol{v}$
(314)

To determine the symmetry of the these products, the spatial frame is oriented
such that the bivector reads
$\boldsymbol{B}=B_{12}\,\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}=B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}$
and the orientation of the vector
$\boldsymbol{v}=v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}+v_{3}\,\boldsymbol{\hat{e}}_{3}$
is arbitrary. The geometric product of the vector $\boldsymbol{v}$ and the
bivector $\boldsymbol{B}$ reads,

$\begin{split}&\boldsymbol{v}\,\boldsymbol{B}=\left(v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}+v_{3}\,\boldsymbol{\hat{e}}_{3}\right)\,\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\\\
&\phantom{\boldsymbol{v}\,\boldsymbol{B}}=v_{1}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+v_{2}\,B_{12}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+v_{3}\,B_{12}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\\\
&\phantom{\boldsymbol{v}\,\boldsymbol{B}}=B_{12}\,v_{1}\,\boldsymbol{\hat{e}}_{2}-\,B_{12}\,v_{2}\,\boldsymbol{\hat{e}}_{1}+B_{12}\,v_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\end{split}$
(315)

The geometric product of the bivector $\boldsymbol{B}$ and the vector
$\boldsymbol{v}$ reads,

$\begin{split}&\boldsymbol{B}\,\boldsymbol{v}=\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\left(v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}+v_{3}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\boldsymbol{B}\,\boldsymbol{v}}=B_{12}\,v_{1}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{1}+B_{12}\,v_{2}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{2}+B_{12}\,v_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\\\
&\phantom{\boldsymbol{B}\,\boldsymbol{v}}=-\,B_{12}\,v_{1}\,\boldsymbol{\hat{e}}_{2}+B_{12}\,v_{2}\,\boldsymbol{\hat{e}}_{1}+B_{12}\,v_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\end{split}$
(316)

The outer product of the vector $\boldsymbol{v}$ and the bivector
$\boldsymbol{B}$ reads,

$\displaystyle\boldsymbol{v}\wedge\boldsymbol{B}=\left(v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}+v_{3}\,\boldsymbol{\hat{e}}_{3}\right)\wedge\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}\right)$
$\displaystyle\phantom{\boldsymbol{v}\wedge\boldsymbol{B}}=v_{1}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}+v_{2}\,B_{12}\,\boldsymbol{\hat{e}}_{2}\wedge\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}+v_{3}\,B_{12}\,\boldsymbol{\hat{e}}_{3}\wedge\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}$
$\displaystyle\phantom{\boldsymbol{v}\wedge\boldsymbol{B}}=v_{3}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}\wedge\boldsymbol{\hat{e}}_{3}$
(317)

The outer product of the bivector $\boldsymbol{B}$ and the vector
$\boldsymbol{v}$ reads,

$\displaystyle\boldsymbol{B}\wedge\boldsymbol{v}=\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}\right)\wedge\left(v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}+v_{3}\,\boldsymbol{\hat{e}}_{3}\right)$
$\displaystyle\phantom{\boldsymbol{B}\wedge\boldsymbol{v}}=B_{12}\,v_{1}\,\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}\wedge\boldsymbol{\hat{e}}_{1}+B_{12}\,v_{2}\,\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}\wedge\boldsymbol{\hat{e}}_{2}+B_{12}\,v_{3}\,\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}\wedge\boldsymbol{\hat{e}}_{3}$
$\displaystyle\phantom{\boldsymbol{B}\wedge\boldsymbol{v}}=B_{12}\,v_{3}\,\boldsymbol{\hat{e}}_{1}\wedge\boldsymbol{\hat{e}}_{2}\wedge\boldsymbol{\hat{e}}_{3}$
(318)

From relations (A) and (A) we conclude that the outer product of a vector
$\boldsymbol{v}$ and a bivector $\boldsymbol{B}$ is symmetric,

$\boldsymbol{v}\wedge\boldsymbol{B}=\boldsymbol{B}\wedge\boldsymbol{v}$ (319)

In view of relations(313), (315) and (A),

$\boldsymbol{v}\cdot\boldsymbol{B}=\boldsymbol{v}\,\boldsymbol{B}-\,\boldsymbol{v}\wedge\boldsymbol{B}=-\,B_{12}\,v_{2}\,\boldsymbol{\hat{e}}_{1}+B_{12}\,v_{1}\,\boldsymbol{\hat{e}}_{2}$
(320)

In view of relations(314), (316) and (A),

$\boldsymbol{B}\cdot\boldsymbol{v}=\boldsymbol{B}\,\boldsymbol{v}-\,\boldsymbol{B}\wedge\boldsymbol{v}=B_{12}\,v_{2}\,\boldsymbol{\hat{e}}_{1}-\,B_{12}\,v_{1}\,\boldsymbol{\hat{e}}_{2}$
(321)

From relations (320) and (321) we conclude that the inner product of a vector
$\boldsymbol{v}$ and a bivector $\boldsymbol{B}$ is antisymmetric,

$\boldsymbol{v}\cdot\boldsymbol{B}=-\,\boldsymbol{B}\cdot\boldsymbol{v}$ (322)

In view of relations (315) and (316), (319) and (322),

$\displaystyle\boldsymbol{v}\cdot\boldsymbol{B}=\frac{1}{2}\left(\boldsymbol{v}\,\boldsymbol{B}-\,\boldsymbol{B}\,\boldsymbol{v}\right)$
(323)
$\displaystyle\boldsymbol{v}\wedge\boldsymbol{B}=\frac{1}{2}\left(\boldsymbol{v}\,\boldsymbol{B}+\boldsymbol{B}\,\boldsymbol{v}\right)$
(324)

To determine the geometric product of two bivectors $\boldsymbol{A}$ and
$\boldsymbol{B}$, we write them as
$\boldsymbol{A}=A_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+A_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+A_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}$
and
$\boldsymbol{B}=B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+B_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+B_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}$.
The geometric product of the bivectors $\boldsymbol{A}$ and $\boldsymbol{B}$
reads,

$\displaystyle\boldsymbol{A}\,\boldsymbol{B}=\left(A_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+A_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+A_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\right)\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+B_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+B_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\right)$
$\displaystyle\phantom{\boldsymbol{A}\,\boldsymbol{B}}=A_{12}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+A_{12}\,B_{23}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+A_{12}\,B_{31}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}$
(325)
$\displaystyle\phantom{\boldsymbol{A}\,\boldsymbol{B}=}+A_{23}\,B_{12}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+A_{23}\,B_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+A_{23}\,B_{31}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}$
$\displaystyle\phantom{\boldsymbol{A}\,\boldsymbol{B}=}+A_{31}\,B_{12}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+A_{31}\,B_{23}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+A_{31}\,B_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}$

which reduces to,

$\begin{split}&\boldsymbol{A}\,\boldsymbol{B}=-\,\left(A_{12}\,B_{12}+A_{23}\,B_{23}+A_{31}\,B_{31}\right)+\left(A_{31}\,B_{23}-\,A_{23}\,B_{31}\right)\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\\\
&\phantom{\boldsymbol{A}\,\boldsymbol{B}=}+\left(A_{12}\,B_{31}-\,A_{31}\,B_{12}\right)\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+\left(A_{23}\,B_{12}-\,A_{12}\,B_{23}\right)\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\end{split}$
(326)

The geometric product of the bivectors $\boldsymbol{B}$ and $\boldsymbol{A}$
reads,

$\displaystyle\boldsymbol{B}\,\boldsymbol{A}=\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+B_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+B_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\right)\left(A_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+A_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+A_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\right)$
$\displaystyle\phantom{\boldsymbol{B}\,\boldsymbol{A}}=A_{12}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+A_{23}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+A_{12}\,B_{31}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}$
(327)
$\displaystyle\phantom{\boldsymbol{B}\,\boldsymbol{A}=}+A_{12}\,B_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+A_{23}\,B_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+A_{31}\,B_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}$
$\displaystyle\phantom{\boldsymbol{B}\,\boldsymbol{A}=}+A_{12}\,B_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+A_{23}\,B_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+A_{31}\,B_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}$

which reduces to,

$\begin{split}&\boldsymbol{B}\,\boldsymbol{A}=-\,\left(A_{12}\,B_{12}+A_{23}\,B_{23}+A_{31}\,B_{31}\right)+\left(A_{23}\,B_{31}-\,A_{31}\,B_{23}\right)\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\\\
&\phantom{\boldsymbol{B}\,\boldsymbol{A}=}+\left(A_{31}\,B_{12}-\,A_{12}\,B_{31}\right)\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+\left(A_{12}\,B_{23}-\,A_{23}\,B_{12}\right)\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\end{split}$
(328)

The geometric product (326) of two bivectors $\boldsymbol{A}$ and
$\boldsymbol{B}$ is the sum symmetric product and an antisymmetric product,

$\boldsymbol{A}\,\boldsymbol{B}=\boldsymbol{A}\cdot\boldsymbol{B}+\boldsymbol{A}\times\boldsymbol{B}$
(329)

According to relations (326) and (328), the inner product, or anticommutator,
of two bivectors $\boldsymbol{A}$ and $\boldsymbol{B}$ yields a scalar,

$\boldsymbol{A}\cdot\boldsymbol{B}=\boldsymbol{B}\cdot\boldsymbol{A}=\frac{1}{2}\left(\boldsymbol{A}\,\boldsymbol{B}+\boldsymbol{B}\,\boldsymbol{A}\right)=-\,\left(A_{12}\,B_{12}+A_{23}\,B_{23}+A_{31}\,B_{31}\right)$
(330)

According to relations (326) and (328), the cross product, or commutator, of
two bivectors $\boldsymbol{A}$ and $\boldsymbol{B}$ yields a bivector,

$\displaystyle\boldsymbol{A}\times\boldsymbol{B}=-\,\boldsymbol{B}\times\boldsymbol{A}=\frac{1}{2}\left(\boldsymbol{A}\,\boldsymbol{B}-\,\boldsymbol{B}\,\boldsymbol{A}\right)$
$\displaystyle=\left(A_{31}\,B_{23}-\,A_{23}\,B_{31}\right)\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+\left(A_{12}\,B_{31}-\,A_{31}\,B_{12}\right)\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}$
(331)
$\displaystyle\phantom{=}+\left(A_{23}\,B_{12}-\,A_{12}\,B_{23}\right)\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}$

The fact that the anticommutator of two bivectors yields a scalar and the
commutator of two bivectors yields a bivector is specific to the geometric
algebra $\mathbb{G}^{3}$. To determine the geometric product of a vector
$\boldsymbol{v}$ and the pseudoscalar $I$, we write them as
$\boldsymbol{B}=B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+B_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+B_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}$
and
$I=\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}$.
The geometric product of the vector $\boldsymbol{v}$ and the pseudoscalar $I$
reads,

$\displaystyle\boldsymbol{v}\,I=\left(v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}+v_{3}\,\boldsymbol{\hat{e}}_{3}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)=v_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+v_{2}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}+v_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}$
(332) $\displaystyle
I\,\boldsymbol{v}=\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\left(v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}+v_{3}\,\boldsymbol{\hat{e}}_{3}\right)=v_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+v_{2}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}+v_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}$
(333)

According to relation (332) and (333), the vector $\boldsymbol{v}$ commutes
with the pseudoscalar $I$,

$\boldsymbol{v}\,I=I\,\boldsymbol{v}$ (334)

To determine the geometric product of a vector $\boldsymbol{v}$ and the
pseudoscalar $I$, we write them as
$\boldsymbol{v}=v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}+v_{3}\,\boldsymbol{\hat{e}}_{3}$
and
$I=\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}$.
The geometric product of the bivector $\boldsymbol{B}$ and the pseudoscalar
$I$ reads,

$\displaystyle\boldsymbol{B}\,I=\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+B_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+B_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)$
$\displaystyle\phantom{\boldsymbol{B}\,I}=-\left(B_{23}\,\boldsymbol{\hat{e}}_{1}+B_{31}\,\boldsymbol{\hat{e}}_{2}+B_{12}\,\boldsymbol{\hat{e}}_{3}\right)$
(335) $\displaystyle
I\,\boldsymbol{B}=\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+B_{23}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}+B_{31}\,\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{1}\right)=-\,B_{12}\,\boldsymbol{\hat{e}}_{3}$
$\displaystyle\phantom{I\,\boldsymbol{B}}=-\left(B_{23}\,\boldsymbol{\hat{e}}_{1}+B_{31}\,\boldsymbol{\hat{e}}_{2}+B_{12}\,\boldsymbol{\hat{e}}_{3}\right)$
(336)

According to relation (A) and (A), the bivector $\boldsymbol{B}$ commutes with
the pseudoscalar $I$,

$\boldsymbol{B}\,I=I\,\boldsymbol{B}$ (337)

##  Appendix B Duality in geometric algebra

The reverse of the scalar $s$, the vector
$\boldsymbol{v}=v_{3}\,\boldsymbol{\hat{e}}_{3}$, of the bivector
$\boldsymbol{B}=B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}$
and of the pseudoscalar $I$ is obtained by reversing the order of the basis
vectors, [6]

$\displaystyle\boldsymbol{s}^{{\dagger}}=s$ (338)
$\displaystyle\boldsymbol{v}^{{\dagger}}=v_{3}\,\boldsymbol{\hat{e}}_{3}=\boldsymbol{v}$
(339)
$\displaystyle\boldsymbol{B}^{{\dagger}}=B_{12}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{1}=-\,B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}=-\,\boldsymbol{B}$
(340) $\displaystyle
I^{{\dagger}}=\boldsymbol{\hat{e}}_{3}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{1}=-\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}=-\,I$
(341)

Thus, the reverse of the multivector (308) is given by,

$M^{{\dagger}}=s^{{\dagger}}+\boldsymbol{v}^{{\dagger}}+\boldsymbol{B}^{{\dagger}}+s^{\prime}I^{{\dagger}}=s+\boldsymbol{v}-\,\boldsymbol{B}-\,s^{\prime}I$
(342)

The square of the modulus of a vector $\boldsymbol{v}$, a bivector
$\boldsymbol{B}$ and a pseudovector $I$ are defined as,

$\displaystyle\left|\boldsymbol{v}\right|^{2}=\boldsymbol{v}^{{\dagger}}\cdot\boldsymbol{v}=\boldsymbol{v}\cdot\boldsymbol{v}=\boldsymbol{v}^{2}$
(343)
$\displaystyle\left|\boldsymbol{B}\right|^{2}=\boldsymbol{B}^{{\dagger}}\cdot\boldsymbol{B}=-\,\boldsymbol{B}\cdot\boldsymbol{B}=-\,\boldsymbol{B}^{2}$
(344) $\displaystyle\left|I\right|^{2}=I^{{\dagger}}\cdot I=-\,I\cdot
I=-\,I^{2}=1$ (345)

The geometric interpretation of the modulus is clear : the modulus of a vector
$\left|\boldsymbol{v}\right|$ is the length of a line element, the modulus of
a bivector $\left|\boldsymbol{B}\right|$ is the surface of a plane element and
the modulus of a pseudoscalar $\left|s^{\prime}I\right|$ is the volume of a
space element. Thus, the modulus of the bivector obtained by taking the outer
product of a vector $\boldsymbol{u}$ and a vector $\boldsymbol{v}$ is the
surface of the parallelogram spanned by these vectors,

$\left|\boldsymbol{u}\wedge\boldsymbol{v}\right|=\left|\boldsymbol{u}\right|\,\left|\boldsymbol{v}\right|\,\sin\theta$
(346)

where $\theta$ is the acute angle between these vectors. This modulus is the
same as the modulus of the cross product of these vectors,

$\left|\boldsymbol{u}\times\boldsymbol{v}\right|=\left|\boldsymbol{u}\right|\,\left|\boldsymbol{v}\right|\,\sin\theta$
(347)

Thus,

$\left|\boldsymbol{u}\wedge\boldsymbol{v}\right|=\left|\boldsymbol{u}\times\boldsymbol{v}\right|$
(348)

The inverse of a vector $\boldsymbol{v}$, a bivector $\boldsymbol{B}$ and a
pseudovector $I$ are defined as,

$\displaystyle\boldsymbol{v}^{-1}=\frac{\boldsymbol{v}}{\boldsymbol{v}^{2}}=\frac{\boldsymbol{v}^{{\dagger}}}{\left|\boldsymbol{v}\right|^{2}}$
(349)
$\displaystyle\boldsymbol{B}^{-1}=\frac{\boldsymbol{B}}{\boldsymbol{B}^{2}}=\frac{\boldsymbol{B}^{{\dagger}}}{\left|\boldsymbol{B}\right|^{2}}$
(350) $\displaystyle
I^{-1}=\frac{I}{I^{2}}=\frac{I^{{\dagger}}}{\left|I\right|^{2}}=-\,I$ (351)

The dual of a vector $\boldsymbol{v}$, a bivector $\boldsymbol{B}$ and a
pseudovector $I$ are defined as, [6]

$\displaystyle\boldsymbol{v}^{\ast}=\frac{\boldsymbol{v}}{I}=\boldsymbol{v}\,I^{-1}=-\,\boldsymbol{v}\,I$
(352)
$\displaystyle\boldsymbol{B}^{\ast}=\frac{\boldsymbol{B}}{I}=\boldsymbol{B}\,I^{-1}=-\,\boldsymbol{B}\,I$
(353) $\displaystyle I^{\ast}=\frac{I}{I}=I\,I^{-1}=1$ (354)

This duality is the transformation as the Hodge duality in differential forms.
The dual of the dual of a vector $\boldsymbol{v}$, a bivector $\boldsymbol{B}$
and a pseudovector $I$ are their opposite,

$\displaystyle\left(\boldsymbol{v}^{\ast}\right)^{\ast}=-\,\boldsymbol{v}^{\ast}\,I=\boldsymbol{v}\,I^{2}=-\,\boldsymbol{v}$
(355)
$\displaystyle\left(\boldsymbol{B}^{\ast}\right)^{\ast}=-\,\boldsymbol{B}^{\ast}\,I=\boldsymbol{B}\,I^{2}=-\,\boldsymbol{B}$
(356) $\displaystyle\left(I^{\ast}\right)^{\ast}=-\,I^{\ast}\,I=I\,I^{2}=-\,I$
(357)

The dual of the vector $\boldsymbol{v}=v_{3}\,\boldsymbol{\hat{e}}_{3}$ is,

$\boldsymbol{v}^{\ast}=-\,\boldsymbol{v}\,I=-\,\left(v_{3}\,\boldsymbol{\hat{e}}_{3}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)=-\,v_{3}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}$
(358)

Defining the bivector $\boldsymbol{V}$ as,

$\boldsymbol{V}=v_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\qquad\text{where}\qquad\left|\boldsymbol{V}\right|=\left|\boldsymbol{v}\right|\qquad\text{and
thus}\qquad v_{12}=v_{3}$ (359)

it satisfies the duality,

$\boldsymbol{v}^{\ast}=-\,\boldsymbol{V}=\left(\boldsymbol{V}^{\ast}\right)^{\ast}\qquad\text{and
thus}\qquad\boldsymbol{V}^{\ast}=\boldsymbol{v}$ (360)

The dual of the bivector
$\boldsymbol{B}=B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}$
is,

$\boldsymbol{B}^{\ast}=-\,\boldsymbol{B}\,I=-\,\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)=B_{12}\,\boldsymbol{\hat{e}}_{3}$
(361)

Defining the vector $\boldsymbol{b}$ as,

$\boldsymbol{b}=b_{3}\,\boldsymbol{\hat{e}}_{3}\qquad\text{where}\qquad\left|\boldsymbol{b}\right|=\left|\boldsymbol{B}\right|\qquad\text{and
thus}\qquad b_{3}=B_{12}$ (362)

it satisfies the duality,

$\boldsymbol{B}^{\ast}=\boldsymbol{b}=-\,\left(\boldsymbol{b}^{\ast}\right)^{\ast}\qquad\text{and
thus}\qquad\boldsymbol{b}^{\ast}=-\,\boldsymbol{B}$ (363)

There is a duality between a vector and a bivector of same modulus and there
is a duality between a scalar and a pseudoscalar of same modulus. In view of
relations (352) and (363),

$\boldsymbol{B}=-\,\boldsymbol{b}^{\ast}=\boldsymbol{b}\,I$ (364)

which implies that by duality the multivector (308) is recast as, [7]

$M=\left(s+s^{\prime}I\right)+\left(\boldsymbol{v}+\boldsymbol{b}\,I\right)$
(365)

and the reverse of the multivector (342) is recast as,

$M^{{\dagger}}=\left(s-\,s^{\prime}I\right)+\left(\boldsymbol{v}-\,\boldsymbol{b}\,I\right)$
(366)

which means that the reverse of a multivector is like a complex conjugate
where the pseudoscalar $I$ is like the imaginary number $i$. The bivector
$\boldsymbol{W}$ and the vector $\boldsymbol{w}$ are defined as the wedge
product and the cross product of two vectors $\boldsymbol{u}$ and
$\boldsymbol{v}$ respectively,

$\boldsymbol{W}=\boldsymbol{u}\wedge\boldsymbol{v}\qquad\text{and}\qquad\boldsymbol{w}=\boldsymbol{u}\times\boldsymbol{v}$
(367)

According to relation (348) the modulus of the bivector $\boldsymbol{W}$ and
vector $\boldsymbol{w}$ are equal, which means that the vector
$\boldsymbol{w}$ is the dual of the bivector $\boldsymbol{W}$,

$\left|\boldsymbol{W}\right|=\left|\boldsymbol{w}\right|\qquad\text{and
thus}\qquad\boldsymbol{W}^{\ast}=\boldsymbol{w}\qquad\text{and}\qquad\boldsymbol{w}^{\ast}=-\,\boldsymbol{W}$
(368)

Thus, the cross product of the vectors $\boldsymbol{u}$ and $\boldsymbol{v}$
is the dual of the wedge product of these vectors,

$\left(\boldsymbol{u}\wedge\boldsymbol{v}\right)^{\ast}=\boldsymbol{u}\times\boldsymbol{v}\qquad\text{and}\qquad\left(\boldsymbol{u}\times\boldsymbol{v}\right)^{\ast}=-\,\boldsymbol{u}\wedge\boldsymbol{v}$
(369)

To establish the duality between the inner and outer product of two vectors,
we choose a vector
$\boldsymbol{u}=u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}$
and a vector
$\boldsymbol{v}=v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}$
in the same plane. The dual of the outer product of the vectors
$\boldsymbol{u}$ and $\boldsymbol{v}$ yields,

$\begin{split}&\left(\boldsymbol{u}\wedge\boldsymbol{v}\right)^{\ast}=-\,\Big{(}\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}\right)\wedge\left(v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}\right)\Big{)}\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\left(\boldsymbol{u}\wedge\boldsymbol{v}\right)^{\ast}}=-\,\left(u_{1}\,v_{2}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}+u_{2}\,v_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{1}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\left(\boldsymbol{u}\wedge\boldsymbol{v}\right)^{\ast}}=\left(u_{1}\,v_{2}-\,u_{2}\,v_{1}\right)\boldsymbol{\hat{e}}_{3}\end{split}$
(370)

The inner product of the vector $\boldsymbol{u}$ and $\boldsymbol{v}^{\ast}$
yields,

$\begin{split}&\boldsymbol{u}\cdot\boldsymbol{v}^{\ast}=-\,\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}\right)\cdot\Big{(}\left(v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\Big{)}\\\
&\phantom{\boldsymbol{u}\cdot\boldsymbol{v}^{\ast}}=-\,\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}\right)\cdot\left(v_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}-\,v_{2}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\boldsymbol{u}\cdot\boldsymbol{v}^{\ast}}=\left(u_{1}\,v_{2}-\,u_{2}\,v_{1}\right)\boldsymbol{\hat{e}}_{3}\end{split}$
(371)

The identification of relations (370) and (371) yields the vectorial duality,

$\left(\boldsymbol{u}\wedge\boldsymbol{v}\right)^{\ast}=\boldsymbol{u}\cdot\boldsymbol{v}^{\ast}$
(372)

which is recast in terms of the dual bivector
$\boldsymbol{V}=-\,\boldsymbol{v}^{\ast}$ as,

$\left(\boldsymbol{u}\wedge\boldsymbol{v}\right)^{\ast}=-\,\boldsymbol{u}\cdot\boldsymbol{V}=\boldsymbol{V}\cdot\boldsymbol{u}$
(373)

In view of relations (369) and (372), we obtain,

$\boldsymbol{u}\times\boldsymbol{v}=\boldsymbol{u}\cdot\boldsymbol{v}^{\ast}$
(374)

which is recast in terms of the dual bivector
$\boldsymbol{V}=-\,\boldsymbol{v}^{\ast}$ as,

$\boldsymbol{u}\times\boldsymbol{v}=-\,\boldsymbol{u}\cdot\boldsymbol{V}=\boldsymbol{V}\cdot\boldsymbol{u}$
(375)

The dual of the inner product of the vectors $\boldsymbol{u}$ and
$\boldsymbol{v}$ yields,

$\begin{split}&\left(\boldsymbol{u}\cdot\boldsymbol{v}\right)^{\ast}=-\,\Big{(}\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}\right)\cdot\left(v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}\right)\Big{)}\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\left(\boldsymbol{u}\cdot\boldsymbol{v}\right)^{\ast}}=-\,\left(u_{1}\,v_{1}+u_{2}\,v_{2}\right)\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\end{split}$
(376)

The outer product of the vectors $\boldsymbol{u}$ and $\boldsymbol{v}^{\ast}$
yields,

$\begin{split}&\boldsymbol{u}\wedge\boldsymbol{v}^{\ast}=-\,\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}\right)\wedge\Big{(}\left(v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\Big{)}\\\
&\phantom{\boldsymbol{u}\wedge\boldsymbol{v}^{\ast}}=-\,\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}\right)\wedge\left(v_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}-\,v_{2}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\boldsymbol{u}\wedge\boldsymbol{v}^{\ast}}=\left(u_{1}\,v_{1}+u_{2}\,v_{2}\right)\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\end{split}$
(377)

The identification of relations (376) and (377) yields the pseudoscalar
duality,

$\left(\boldsymbol{u}\cdot\boldsymbol{v}\right)^{\ast}=\boldsymbol{u}\wedge\boldsymbol{v}^{\ast}$
(378)

which is recast in terms of the dual bivector
$\boldsymbol{V}=-\,\boldsymbol{v}^{\ast}$ as,

$\left(\boldsymbol{u}\cdot\boldsymbol{v}\right)^{\ast}=-\,\boldsymbol{u}\wedge\boldsymbol{V}$
(379)

To establish the duality between the inner and outer product of a bivector and
a vector, the spatial frame is oriented such that the bivector is given by
$\boldsymbol{B}=B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}$
and the vector is written as
$\boldsymbol{v}=v_{1}\,\boldsymbol{\hat{e}}_{1}+v_{2}\,\boldsymbol{\hat{e}}_{2}+v_{3}\,\boldsymbol{\hat{e}}_{3}$.
The dual of the outer product of the vector $\boldsymbol{u}$ and the bivector
$\boldsymbol{B}$ yields,

$\begin{split}&\left(\boldsymbol{u}\wedge\boldsymbol{B}\right)^{\ast}=-\,\Big{(}\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}+u_{3}\,\boldsymbol{\hat{e}}_{3}\right)\wedge\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\Big{)}\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\left(\boldsymbol{u}\wedge\boldsymbol{B}\right)^{\ast}}=-\,\left(u_{3}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\left(\boldsymbol{u}\wedge\boldsymbol{B}\right)^{\ast}}=u_{3}\,B_{12}\end{split}$
(380)

The inner product of the vector $\boldsymbol{u}$ and the bivector
$\boldsymbol{B}^{\ast}$ yields,

$\begin{split}&\boldsymbol{u}\cdot\boldsymbol{B}^{\ast}=-\,\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}+u_{3}\,\boldsymbol{\hat{e}}_{3}\right)\cdot\Big{(}\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\Big{)}\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\Big{)}\\\
&\phantom{\boldsymbol{u}\cdot\boldsymbol{B}^{\ast}}=\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}+u_{3}\,\boldsymbol{\hat{e}}_{3}\right)\cdot\left(B_{12}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\boldsymbol{u}\cdot\boldsymbol{B}^{\ast}}=u_{3}\,B_{12}\end{split}$
(381)

The identification of relations (380) and (381) yields the scalar duality,

$\left(\boldsymbol{u}\wedge\boldsymbol{B}\right)^{\ast}=\boldsymbol{u}\cdot\boldsymbol{B}^{\ast}$
(382)

which is recast in terms of the dual vector
$\boldsymbol{b}=\boldsymbol{B}^{\ast}$ as,

$\left(\boldsymbol{u}\wedge\boldsymbol{B}\right)^{\ast}=\boldsymbol{u}\cdot\boldsymbol{b}$
(383)

and is the dual of the pseudoscalar duality (379) for
$\boldsymbol{b}=\boldsymbol{v}$ and $\boldsymbol{B}=\boldsymbol{V}$. The dual
of the inner product of the vector $\boldsymbol{u}$ and the bivector
$\boldsymbol{B}$ yields,

$\begin{split}&\left(\boldsymbol{u}\cdot\boldsymbol{B}\right)^{\ast}=-\,\Big{(}\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}+u_{3}\,\boldsymbol{\hat{e}}_{3}\right)\cdot\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\Big{)}\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\left(\boldsymbol{u}\cdot\boldsymbol{B}\right)^{\ast}}=-\,\left(u_{1}\,B_{12}\,\boldsymbol{\hat{e}}_{2}-\,u_{2}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\left(\boldsymbol{u}\cdot\boldsymbol{B}\right)^{\ast}}=u_{1}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{3}+u_{2}\,B_{12}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\end{split}$
(384)

The outer product of the vector $\boldsymbol{u}$ and the dual of the bivector
$\boldsymbol{B}^{\ast}$ yields,

$\begin{split}&\boldsymbol{u}\wedge\boldsymbol{B}^{\ast}=-\,\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}+u_{3}\,\boldsymbol{\hat{e}}_{3}\right)\wedge\Big{(}\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\Big{)}\\\
&\phantom{\boldsymbol{u}\wedge\boldsymbol{B}^{\ast}}=\left(u_{1}\,\boldsymbol{\hat{e}}_{1}+u_{2}\,\boldsymbol{\hat{e}}_{2}+u_{3}\,\boldsymbol{\hat{e}}_{3}\right)\wedge\left(B_{12}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\boldsymbol{u}\wedge\boldsymbol{B}^{\ast}}=u_{1}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{3}+u_{2}\,B_{12}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\end{split}$
(385)

The identification of relations (384) and (385) yields the bivectorial
duality,

$\left(\boldsymbol{u}\cdot\boldsymbol{B}\right)^{\ast}=\boldsymbol{u}\wedge\boldsymbol{B}^{\ast}$
(386)

which is recast in terms of the dual vector
$\boldsymbol{b}=\boldsymbol{B}^{\ast}$ as,

$\left(\boldsymbol{u}\cdot\boldsymbol{B}\right)^{\ast}=\boldsymbol{u}\wedge\boldsymbol{b}$
(387)

and is the dual of the vectorial duality (373) for
$\boldsymbol{b}=\boldsymbol{v}$ and $\boldsymbol{B}=\boldsymbol{V}$. To
establish the duality between the inner and outer product of bivectors in the
same plane, the spatial frame is oriented such that the bivectors
$\boldsymbol{A}$ and $\boldsymbol{B}$ are written as
$\boldsymbol{A}=A_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}$
and
$\boldsymbol{B}=B_{12}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}$.
The dual of the inner product of the bivectors $\boldsymbol{A}$ and
$\boldsymbol{B}$ yields,

$\begin{split}&\left(\boldsymbol{A}\cdot\boldsymbol{B}\right)^{\ast}=-\,\Big{(}\left(A_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\cdot\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\Big{)}\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\left(\boldsymbol{A}\cdot\boldsymbol{B}\right)^{\ast}}=A_{12}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\end{split}$
(388)

The outer product of the bivectors $\boldsymbol{A}$ and
$\boldsymbol{B}^{\ast}$ yields,

$\begin{split}&\boldsymbol{A}\wedge\boldsymbol{B}^{\ast}=-\,\left(A_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\wedge\Big{(}\left(B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\left(\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\right)\Big{)}\\\
&\phantom{\boldsymbol{A}\wedge\boldsymbol{B}^{\ast}}=\left(A_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\right)\wedge\left(B_{12}\,\boldsymbol{\hat{e}}_{3}\right)\\\
&\phantom{\boldsymbol{A}\wedge\boldsymbol{B}^{\ast}}=A_{12}\,B_{12}\,\boldsymbol{\hat{e}}_{1}\,\boldsymbol{\hat{e}}_{2}\,\boldsymbol{\hat{e}}_{3}\end{split}$
(389)

The identification of relations (388) and (389) yields the pseudoscalar
duality,

$\left(\boldsymbol{A}\cdot\boldsymbol{B}\right)^{\ast}=\boldsymbol{A}\wedge\boldsymbol{B}^{\ast}$
(390)

The pseudoscalar duality (390) is expressed in terms of the dual vector
$\boldsymbol{b}=\boldsymbol{B}^{\ast}$,

$\left(\boldsymbol{A}\cdot\boldsymbol{B}\right)^{\ast}=\boldsymbol{A}\wedge\boldsymbol{b}$
(391)

In view of the identities (319), (386), the inner product of two bivectors
$\boldsymbol{A}$ and $\boldsymbol{B}$ is expressed in terms of the dual
vectors $\boldsymbol{a}=\boldsymbol{A}^{\ast}$ and
$\boldsymbol{b}=\boldsymbol{B}^{\ast}$ as,

$\begin{split}&\left(\boldsymbol{A}\cdot\boldsymbol{B}\right)^{\ast}=\boldsymbol{A}\wedge\boldsymbol{b}=\boldsymbol{b}\wedge\boldsymbol{A}=-\,\boldsymbol{b}\wedge\boldsymbol{a}^{\ast}\\\
&\phantom{\left(\boldsymbol{A}\cdot\boldsymbol{B}\right)^{\ast}}=-\,\left(\boldsymbol{b}\cdot\boldsymbol{a}\right)^{\ast}=-\,\left(\boldsymbol{a}\cdot\boldsymbol{b}\right)^{\ast}\end{split}$
(392)

The dual of identity (392) yields,

$\boldsymbol{A}\cdot\boldsymbol{B}=-\,\boldsymbol{a}\cdot\boldsymbol{b}$ (393)

To establish the duality of the inner product of a vector $\boldsymbol{v}$ and
a trivector $T=t\,I$, we write that trivector as $T=t\,I$, where $t$ is a
scalar. Thus, in view of identity (354),

$T^{\ast}=\left(t\,I\right)^{\ast}=-\,\left(t\,I\right)I=t\qquad\text{and}\qquad
t^{\ast}=-\,T$ (394)

Use the dualities (353) and (394), and the commutation (334) between the
vector $\boldsymbol{v}$ and the pseudoscalar $I$ or the trivector $T=t\,I$,
the dual of bivector $\boldsymbol{v}\cdot T$ is written as,

$\left(\boldsymbol{v}\cdot
T\right)^{\ast}=\left(\boldsymbol{v}\,T\right)^{\ast}=-\,\left(\boldsymbol{v}\,t\,I\right)I=\boldsymbol{v}\,t=\boldsymbol{v}\,T^{\ast}$
(395)

##  Appendix C Algebraic identities in geometric algebra

The double cross product of the three vectors $\boldsymbol{u}$,
$\boldsymbol{v}$ and $\boldsymbol{w}$ is written as,

$\boldsymbol{u}\times\left(\boldsymbol{v}\times\boldsymbol{w}\right)=\left(\boldsymbol{u}\cdot\boldsymbol{w}\right)\boldsymbol{v}-\,\left(\boldsymbol{u}\cdot\boldsymbol{v}\right)\boldsymbol{w}$
(396)

In view of the duality (369) between the cross and wedge products, the
vectorial duality (374), the double duality (356) and the antisymmetry of the
outer product (312), the left-hand side of relation (396) is recast as,

$\boldsymbol{u}\times\left(\boldsymbol{v}\times\boldsymbol{w}\right)=\boldsymbol{u}\times\left(\boldsymbol{v}\wedge\boldsymbol{w}\right)^{\ast}=\boldsymbol{u}\cdot\Big{(}\left(\boldsymbol{v}\wedge\boldsymbol{w}\right)^{\ast}\Big{)}^{\ast}=-\,\boldsymbol{u}\cdot\left(\boldsymbol{v}\wedge\boldsymbol{w}\right)=\boldsymbol{u}\cdot\left(\boldsymbol{w}\wedge\boldsymbol{v}\right)$
(397)

According to the identity (397), the double cross product (396) yields the
triple product,

$\boldsymbol{u}\cdot\left(\boldsymbol{v}\wedge\boldsymbol{w}\right)=\left(\boldsymbol{u}\cdot\boldsymbol{v}\right)\boldsymbol{w}-\,\left(\boldsymbol{u}\cdot\boldsymbol{w}\right)\boldsymbol{v}$
(398)

This triple product (398) represents the projection of the vector
$\boldsymbol{u}$ on the bivector $\boldsymbol{v}\wedge\boldsymbol{w}$. The
vectors $\boldsymbol{u}$, $\boldsymbol{v}$ and $\boldsymbol{w}$ satisfy the
identity,

$\boldsymbol{u}\cdot\left(\boldsymbol{v}\times\boldsymbol{w}\right)=\boldsymbol{v}\cdot\left(\boldsymbol{w}\times\boldsymbol{u}\right)=\boldsymbol{w}\cdot\left(\boldsymbol{u}\times\boldsymbol{v}\right)$
(399)

In view of the duality (369) between cross product and wedge product and the
scalar duality (382), the three terms of identity (399) are recast as,

$\begin{split}&\boldsymbol{u}\cdot\left(\boldsymbol{v}\times\boldsymbol{w}\right)=\boldsymbol{u}\cdot\left(\boldsymbol{v}\wedge\boldsymbol{w}\right)^{\ast}=\left(\boldsymbol{u}\wedge\boldsymbol{v}\wedge\boldsymbol{w}\right)^{\ast}\\\
&\boldsymbol{v}\cdot\left(\boldsymbol{w}\times\boldsymbol{u}\right)=\boldsymbol{v}\cdot\left(\boldsymbol{w}\wedge\boldsymbol{u}\right)^{\ast}=\left(\boldsymbol{v}\wedge\boldsymbol{w}\wedge\boldsymbol{u}\right)^{\ast}\\\
&\boldsymbol{w}\cdot\left(\boldsymbol{u}\times\boldsymbol{v}\right)=\boldsymbol{w}\cdot\left(\boldsymbol{u}\wedge\boldsymbol{v}\right)^{\ast}=\left(\boldsymbol{w}\wedge\boldsymbol{u}\wedge\boldsymbol{v}\right)^{\ast}\end{split}$
(400)

According to the relations (400), the scalar identity (399) becomes,

$\left(\boldsymbol{u}\wedge\boldsymbol{v}\wedge\boldsymbol{w}\right)^{\ast}=\left(\boldsymbol{v}\wedge\boldsymbol{w}\wedge\boldsymbol{u}\right)^{\ast}=\left(\boldsymbol{w}\wedge\boldsymbol{u}\wedge\boldsymbol{v}\right)^{\ast}$
(401)

The dual of the scalar identity (401) yields the pseudoscalar identity,

$\boldsymbol{u}\wedge\boldsymbol{v}\wedge\boldsymbol{w}=\boldsymbol{v}\wedge\boldsymbol{w}\wedge\boldsymbol{u}=\boldsymbol{w}\wedge\boldsymbol{u}\wedge\boldsymbol{v}$
(402)

This identity represents the fact the oriented volume of the parallelepiped
spanned by the vectors $\boldsymbol{u}$, $\boldsymbol{v}$, $\boldsymbol{w}$ is
the same as that of the parallelepiped spanned by the vectors
$\boldsymbol{v}$, $\boldsymbol{w}$, $\boldsymbol{u}$ and also the same as that
of the parallelepiped spanned by the vectors $\boldsymbol{w}$,
$\boldsymbol{u}$, $\boldsymbol{v}$ because these three parallelepipeds have
the same orientation. According to relations (323) and (324) for
$\boldsymbol{B}=\boldsymbol{v}\wedge\boldsymbol{w}$, the inner and outer
products of the vector $\boldsymbol{u}$ and the bivector
$\boldsymbol{v}\wedge\boldsymbol{w}$ are given by,

$\displaystyle\boldsymbol{u}\cdot\left(\boldsymbol{v}\wedge\boldsymbol{w}\right)=\frac{1}{2}\Big{(}\boldsymbol{u}\left(\boldsymbol{v}\wedge\boldsymbol{w}\right)-\,\left(\boldsymbol{v}\wedge\boldsymbol{w}\right)\boldsymbol{u}\Big{)}$
(403)
$\displaystyle\boldsymbol{u}\wedge\left(\boldsymbol{v}\wedge\boldsymbol{w}\right)=\frac{1}{2}\Big{(}\boldsymbol{u}\left(\boldsymbol{v}\wedge\boldsymbol{w}\right)+\left(\boldsymbol{v}\wedge\boldsymbol{w}\right)\boldsymbol{u}\Big{)}$
(404)

To determine the mixed product of two vectors $\boldsymbol{u}$,
$\boldsymbol{v}$ and a bivector $\boldsymbol{B}$, we use the dual vector
defined as $\boldsymbol{b}=-\,\boldsymbol{B}\,I$. Using the identity (322) and
the triple product of vectors (399), we obtain,

$\left(\boldsymbol{u}\wedge\boldsymbol{v}\right)\cdot\boldsymbol{B}=\left(\boldsymbol{u}\wedge\boldsymbol{v}\right)\cdot\boldsymbol{b}\,I=-\,\boldsymbol{b}\cdot\left(\boldsymbol{u}\wedge\boldsymbol{v}\right)\,I=-\,I\left(\boldsymbol{b}\cdot\boldsymbol{u}\right)\boldsymbol{v}+I\left(\boldsymbol{b}\cdot\boldsymbol{v}\right)\boldsymbol{u}$
(405)

which reduces to,

$\left(\boldsymbol{u}\wedge\boldsymbol{v}\right)\cdot\boldsymbol{B}=\left(\boldsymbol{B}\cdot\boldsymbol{u}\right)\cdot\boldsymbol{v}-\,\left(\boldsymbol{B}\cdot\boldsymbol{v}\right)\cdot\boldsymbol{u}=\boldsymbol{u}\cdot\left(\boldsymbol{v}\cdot\boldsymbol{B}\right)-\,\boldsymbol{v}\cdot\left(\boldsymbol{u}\cdot\boldsymbol{B}\right)$
(406)

To determine the first kind of mixed product of two bivectors
$\boldsymbol{A}$, $\boldsymbol{B}$ and a vector $\boldsymbol{v}$, we use the
dual vectors defined as $\boldsymbol{a}=-\,\boldsymbol{A}\,I$ and
$\boldsymbol{b}=-\,\boldsymbol{B}\,I$. Using the identities (322) and
$I^{2}=-1$, we obtain,

$\left(\boldsymbol{A}\wedge\boldsymbol{v}\right)\cdot\boldsymbol{B}=\left(\boldsymbol{a}\,I\wedge\boldsymbol{v}\right)\cdot\boldsymbol{b}\,I=-\,\left(\boldsymbol{a}\wedge\boldsymbol{v}\right)\cdot\boldsymbol{b}=\boldsymbol{b}\cdot\left(\boldsymbol{a}\wedge\boldsymbol{v}\right)$
(407)

Using the triple product of vectors (399), the mixed product is recast as,

$\begin{split}&\left(\boldsymbol{A}\wedge\boldsymbol{v}\right)\cdot\boldsymbol{B}=\boldsymbol{b}\cdot\left(\boldsymbol{a}\wedge\boldsymbol{v}\right)=\left(\boldsymbol{a}\cdot\boldsymbol{b}\right)\boldsymbol{v}-\,\left(\boldsymbol{b}\cdot\boldsymbol{v}\right)\boldsymbol{a}\\\
&\phantom{\left(\boldsymbol{A}\wedge\boldsymbol{v}\right)\cdot\boldsymbol{B}}=-\,\left(\boldsymbol{A}\cdot\boldsymbol{B}\right)\boldsymbol{v}+\left(\boldsymbol{B}\cdot\boldsymbol{v}\right)\cdot\boldsymbol{A}\end{split}$
(408)

To determine the first kind of mixed product of two bivectors
$\boldsymbol{A}$, $\boldsymbol{B}$ and a vector $\boldsymbol{v}$, we use the
dual vector defined as
$\boldsymbol{a}=\boldsymbol{A}^{\ast}=-\,\boldsymbol{A}\,I$ and the duality
$\left(\boldsymbol{v}\wedge\boldsymbol{a}\right)^{\ast}=-\left(\boldsymbol{v}\wedge\boldsymbol{a}\right)I$.
Using the identities (322), (372) and $I^{2}=-1$, we obtain,

$\begin{split}&\left(\boldsymbol{A}\cdot\boldsymbol{v}\right)\wedge\boldsymbol{B}=-\,\left(\boldsymbol{v}\cdot\boldsymbol{A}\right)\wedge\boldsymbol{B}=\left(\boldsymbol{v}\cdot\boldsymbol{a}^{\ast}\right)\wedge\boldsymbol{B}=\left(\boldsymbol{v}\wedge\boldsymbol{a}\right)^{\ast}\wedge\boldsymbol{B}\\\
&\phantom{\left(\boldsymbol{A}\cdot\boldsymbol{v}\right)\wedge\boldsymbol{B}}=-\,\left(\boldsymbol{v}\wedge\boldsymbol{a}\right)I\wedge\boldsymbol{B}=\boldsymbol{v}\wedge\left(\boldsymbol{A}\wedge\boldsymbol{B}\right)\end{split}$
(409)

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We know that if $\tau_b$ is a hyperbolic translation then $\mathcal{U}[z] \circ \tau_b,$ also solves (<ref>) and hence the kernel of the linearized operator contains non-trivial elements. It was shown in [62] that the degeneracy happens only along an $n$-dimensional subspace characterized by the vector fields

\begin{align*}
V_j(x) := (1 + |x|^2) \frac{\partial}{\partial x_j} \, - \, 2x_j \sum_{l =1}^{n} x_l \, \frac{\partial}{\partial x_l}.
\end{align*}

for $j = 1, \ldots, n.$ More precisely, we define

\begin{align*}
\Phi_j(x) := \frac{d}{dt} \Big|_{t=0} \mathcal{U}[z] \circ \tau_{t e_j} , \ \ 1 \leq i \leq n,
\end{align*}

then $\Phi_{j}(x) = V_j(\calu[z]),$ $\Phi_j$ solves the eigenvalue problem

\begin{align} \label{eigenvalue}
-\Delta_{\bn} \Phi_j - \lambda \Phi_j = p \,  \mathcal{U}[z]^{p-1} \, \Phi_j.
\end{align}

and the degeneracy in the solution space to (<ref>) can occur only along the directions $\Phi_j, 1 \leq j \leq n.$
[[62] ]
Let $V_j$ be the vector fields in $\bn$ defined above and $\Phi_j = V_j(\calu[z]).$ Then $\{\Phi_j\}_{j=1}^n$ forms a basis for the kernel of $(-\Delta_{\bn} - \lambda - p\calu[z]^{p-1}).$

As a result, we obtain complete information on the first and the second eigenvalues and corresponding eigenspaces of the operator $\mathcal{L}.$ We recall the following result from our earlier work [14]

The first and the second eigenvalues of the operator
$$\mathcal{L} := (-\Delta_{\bn} - \lambda)/\mathcal{U}[z]^{p - 1}$$

are respectively $1$ and $p.$ Moreover, the first eigenspace is one dimensional and spanned by $\mathcal{U}[z]$ and the second eigenspace is $n$-dimensional and spanned by $\{\Phi_i\}_{1 \leq i \leq n}.$

In this sequel, we also recall another relevant result used in this article whose proof can be found in [62].

The first Dirichlet eigenvalue of the operator
$$\mathcal{L} := (-\Delta_{\bn} - \lambda)/\mathcal{U}[z]^{p - 1}$$

in the negative half $\bn_{-}:= \{ x = (x_1, \ldots, x_n) \in \bn \ : \ x_j <0 \}$ is  $p$ for every $j = 1,\ldots,n$ and the corresponding eigenspace is one dimensional spanned by $V_j(\calu[z]).$

The research of M. Bhakta is partially supported by the SERB WEA grant (WEA/2020/000005) and DST Swarnajaynti fellowship (SB/SJF/2021-22/09). D. Ganguly is partially supported by the INSPIRE faculty fellowship (IFA17-MA98). D. Karmakar acknowledges the support of the Department of Atomic Energy, Government of India, under project no. 12-R&D-TFR-5.01-0520. S. Mazumdar is partially supported by IIT Bombay SEED Grant RD/0519-IRCCSH0-025.

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