Inverse of a Versor

Question

Let $A=a_1a_2\ldots a_N$ be a versor, where $a_i$ is a vector for all $i$. Let $A^\dagger$ denote the reversion of $A$. Let $a_ia_j$ denote the geometric product of vectors $a_i$ and $a_j$.

According to the book on Geometric Algebra for Computer Games by John Vince, the following is true:

\begin{align} A^\dagger A &= (a_Na_{N-1}\ldots a_2a_1)(a_1a_2\ldots a_{N-1}a_N) \\ &= ( a_N ( a_{N-1} \ldots ( a_2 ( a_1 a_1 ) a_2 ) \ldots a_{N-1} ) a_N ) \\ &= |a_1|^2 ( a_N ( a_{N-1} \ldots ( a_2 a_2 ) \ldots a_{N-1} ) a_N ) \\ &= |a_1|^2 + |a_2|^2 + \cdots |a_N|^2. \end{align}

Here, I get confused. I think that $a_2 a_2=|a_2|^2$, which is a scalar. Therefore, I would have guessed that $A^\dagger A = |a_1|^2 |a_2|^2 \cdots |a_N|^2$. Why do the scalars get added versus multiplied?

I've included an image of the relevant portion fo the page here:

Answer 1

Looks like a typo. Consider the simplest example $\mathbb{R}^2$ with $A = \mathbf{e}_1 \mathbf{e}_2$

$$A^\dagger A = \mathbf{e}_2 \mathbf{e}_1 \mathbf{e}_1 \mathbf{e}_2 = 1 \ne 2.$$

Some additional confirmation can be had by looking at the product of a wedge with its reverse. For a wedge of two vectors $A = \mathbf{x} \wedge \mathbf{y}$, we have the following fairly well known identity: $$\begin{aligned}\left\langle{{A^\dagger A}}\right\rangle&= \left\langle{{(\mathbf{y} \wedge \mathbf{x})(\mathbf{x} \wedge \mathbf{y})}}\right\rangle \\ &= \mathbf{y} \cdot \left( { \mathbf{x} \cdot (\mathbf{x} \wedge \mathbf{y})} \right) \\ &= \mathbf{y}^2 \mathbf{x}^2 - \left( {\mathbf{x} \cdot \mathbf{y}} \right)^2.\end{aligned}$$

Since a versor (ref: Geometric Algebra for Computer Science) is a product of vectors, if the wedged vectors above are perpendicular, then $\mathbf{x} \wedge \mathbf{y} = \mathbf{x} \mathbf{y}$ and we are left with

$$\begin{aligned}\left\langle{{A^\dagger A}}\right\rangle&= \left\langle{{(\mathbf{y} \mathbf{x})(\mathbf{x} \mathbf{y})}}\right\rangle \\ &= \mathbf{y}^2 \mathbf{x}^2,\end{aligned}$$

which is also a product of squared magnitudes, and not their sum.