## Complex-pair representation of GA(2,0) multivectors

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
\label{eqn:bicomplexCl20:20}
M = a + b \Be_1 \Be_2 + x \Be_1 + y \Be_2,

as the pair of complex numbers
\label{eqn:bicomplexCl20:40}
M \sim \lr{ a + i b, x + i y }.

Given a pair of multivectors with this complex representation
\label{eqn:bicomplexCl20:60}
\begin{aligned}
M &= \lr{ z_1, z_2 } \\
N &= \lr{ q_1, q_2 },
\end{aligned}

we found that our geometric product representation was
\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 }.

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

and proceed with an expansion of the even grade components
\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}

The first terms is clearly the geometric product of two vectors
\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’ },

and we are able to verify that the second parts can be factored too
\label{eqn:bicomplexCl20:160}
\lr{ a + b i } \lr{ a’ + b’ i }
=
a a’ – b b’ + i \lr{ b a’ + a b’ }.

This leaves us with
\label{eqn:bicomplexCl20:180}

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:
\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.

So, for the vector component of our assumed product representation, we have
\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’ } \\
&=
\end{aligned}

as expected.

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

## Interior angles of a regular n-sided polygon: a strange way to find them.

Fig 1. Regular polygon, interior angles.

For reasons that I can’t explain, I woke up this morning dreaming about the interior angles of regular polygons. i.e. the angles $$\pi – \theta$$, as illustrated in fig. 1.

The logical way to calculate that angle would be to slice the polygon up into triangles from the center, since each slice would have an interior angle would be $$2 \pi / N$$, and then the problem is just trigonometric. However, in my dream, I was going around the outside, each time rotating by a constant angle, until reaching the original starting point. This was a vector algebra problem, instead of a trigonometric problem, as illustrated in
fig. 2.

fig. 2. Polygon vertex iteration.

I didn’t have the computational power in my dream to solve the problem, and had to write it down when I woke up, to do so. The problem has the structure of a recurrence relation:
\label{eqn:regularNgon:20}
\Bp_k = \Bp_{k-1} + a \Be_1 \lr{ e^{i\theta} }^{k-1},

where
\label{eqn:regularNgon:40}
\Bp_N = \Bp_0.

We can write these out explicitly for the first few $$k$$ to see the pattern
\label{eqn:regularNgon:60}
\begin{aligned}
\Bp_2
&= \Bp_{1} + a \Be_1 \lr{ e^{i\theta} }^{2-1} \\
&= \Bp_{0} + a \Be_1 \lr{ e^{i\theta} }^{1-1} + a \Be_1 \lr{ e^{i\theta} }^{2-1} \\
&= \Bp_{0} + a \Be_1 \lr{ 1 + \lr{ e^{i\theta} }^{2-1} },
\end{aligned}

or
\label{eqn:regularNgon:80}
\Bp_k = \Bp_{0} + a \Be_1 \lr{ 1 + e^{i\theta} + \lr{ e^{i\theta} }^{2-1} + \cdots + \lr{ e^{i\theta} }^{k-1} },

so the equation to solve (for $$\theta$$) is
\label{eqn:regularNgon:100}
\Bp_N = \Bp_0 + a \Be_1 \lr{ 1 + \cdots + \lr{ e^{i\theta} }^{N-1} } = \Bp_0,

or
\label{eqn:regularNgon:120}
1 + \cdots + \lr{ e^{i\theta} }^{N-1} = 0.

The LHS is a geometric series of the form
\label{eqn:regularNgon:140}
S_N = 1 + \alpha + \cdots \alpha^{N-1}.

Recall that the trick to solve this is noting that
\label{eqn:regularNgon:160}
\alpha S_N = \alpha + \cdots \alpha^{N-1} + \alpha^N,

so
\label{eqn:regularNgon:180}
\alpha S_N – S_N = \alpha^N – 1,

or
\label{eqn:regularNgon:200}
S_N = \frac{\alpha^N – 1}{\alpha – 1}.

For our polygon, we seek a zero numerator, that is
\label{eqn:regularNgon:220}
e^{N i \theta} = 1,

and the smallest $$\theta$$ solution to this equation is
\label{eqn:regularNgon:240}
N \theta = 2 \pi,

or
\label{eqn:regularNgon:260}
\theta = \frac{2 \pi}{N}.

The interior angle is the complement of this, since we are going around the outside edge. That is
\label{eqn:regularNgon:280}
\begin{aligned}
\pi – \theta &= \pi – \frac{2 \pi}{N} \\ &= \frac{ N – 2 }{N} \pi,
\end{aligned}

and the sum of all the interior angles is
\label{eqn:regularNgon:300}
N \lr{ \pi – \theta } = \lr{N – 2 } \pi.

Plugging in some specific values, for $$N = 3, 4, 5, 6$$, we find that the interior angles are $$\pi/3, \pi/2, 3 \pi/5, 4 \pi/6$$, and the respective sums of these interior angles for the entire polygons are $$\pi, 2 \pi, 3 \pi, 4 \pi$$.

Like, I said, this isn’t the simplest way to solve this problem. Instead, we could solve for $$2 \mu$$ with respect to interior triangle illustrated in
fig. 3, where

fig. 3. Polygon interior geometry.

\label{eqn:regularNgon:320}
2 \mu + \frac{ 2\pi}{N} = \pi,

or
\label{eqn:regularNgon:340}
2 \mu = \frac{N – 2}{N} \pi,

as found the hard way. The hard way was kind of fun though.

The toughest problem to solve would be “why on earth was my brain pondering this in the early morning?” I didn’t even go to bed thinking about anything math or geometry related (we finished the night with the brain-dead activity of watching an episode of “Stranger things”.)