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We found previously that a complex pair representation of a GA(2,0) multivector had a compact geometric product realization. Now that we know the answer, let’s work backwards from that representation to verify that everything matches our expectations.

We are representing a multivector of the form

\begin{equation}\label{eqn:bicomplexCl20:20}

M = a + b \Be_1 \Be_2 + x \Be_1 + y \Be_2,

\end{equation}

as the pair of complex numbers

\begin{equation}\label{eqn:bicomplexCl20:40}

M \sim \lr{ a + i b, x + i y }.

\end{equation}

Given a pair of multivectors with this complex representation

\begin{equation}\label{eqn:bicomplexCl20:60}

\begin{aligned}

M &= \lr{ z_1, z_2 } \\

N &= \lr{ q_1, q_2 },

\end{aligned}

\end{equation}

we found that our geometric product representation was

\begin{equation}\label{eqn:bicomplexCl20:80}

M N \sim

\lr{ z_1 q_1 + z_2^\conj q_2, z_2 q_1 + z_1^\conj q_2 }.

\end{equation}

Our task is now to verify that this is correct. Let’s set

\begin{equation}\label{eqn:bicomplexCl20:100}

\begin{aligned}

z_1 &= a + i b \\

q_1 &= a’ + i b’ \\

z_2 &= x + i y \\

q_2 &= x’ + i y’,

\end{aligned}

\end{equation}

and proceed with an expansion of the even grade components

\begin{equation}\label{eqn:bicomplexCl20:120}

\begin{aligned}

z_1 q_1 + z_2^\conj q_2

&=

\lr{ a + i b } \lr{ a’ + i b’ }

+

\lr{ x – i y } \lr{ x’ + i y’ } \\

&=

a a’ – b b’ + x x’ + y y’

+ i \lr{ b a’ + a b’ + x y’ – y x’ } \\

&=

x x’ + y y’ + i \lr{ x y’ – y x’ } + \quad a a’ – b b’ + i \lr{ b a’ + a b’ }.

\end{aligned}

\end{equation}

The first terms is clearly the geometric product of two vectors

\begin{equation}\label{eqn:bicomplexCl20:140}

\lr{ x \Be_1 + y \Be_2 } \lr{ x’ \Be_1 + y’ \Be_2 }

=

x x’ + y y’ + i \lr{ x y’ – y x’ },

\end{equation}

and we are able to verify that the second parts can be factored too

\begin{equation}\label{eqn:bicomplexCl20:160}

\lr{ a + b i } \lr{ a’ + b’ i }

=

a a’ – b b’ + i \lr{ b a’ + a b’ }.

\end{equation}

This leaves us with

\begin{equation}\label{eqn:bicomplexCl20:180}

\gpgrade{ M N }{0,2} = \gpgradeone{ M } \gpgradeone{ N } + \gpgrade{ M }{0,2} \gpgrade{ N }{0,2},

\end{equation}

as expected. This part of our representation checks out.

Now, let’s look at the vector component of our representation. First note that to convert from our complex representation of our vector \( z = x + i y \) to the standard basis representation of our vector, we need only multiply by \( \Be_1 \) on the left, for example:

\begin{equation}\label{eqn:bicomplexCl20:220}

\Be_1 \lr{ x + i y } = \Be_1 x + \Be_1 \Be_1 \Be_2 y = \Be_1 x + \Be_2 y.

\end{equation}

So, for the vector component of our assumed product representation, we have

\begin{equation}\label{eqn:bicomplexCl20:200}

\begin{aligned}

\Be_1 \lr{ z_2 q_1 + z_1^\conj q_2 }

&=

\Be_1 \lr{ x + i y } \lr{ a’ + i b’ }

+

\Be_1 \lr{ a – i b } \lr{ x’ + i y’ } \\

&=

\Be_1 \lr{ x + i y } \lr{ a’ + i b’ }

+

\lr{ a + i b } \Be_1 \lr{ x’ + i y’ } \\

&=

\gpgradeone{ M } \gpgrade{ N}{0,2}

+ \gpgrade{ M }{0,2} \gpgradeone{ N},

\end{aligned}

\end{equation}

as expected.

Our complex-pair realization of the geometric product checks out.