cross product not associative, outer product associative The cross product is not associative. If $i=(1,0,0)$, $j=(0,1,0)$ and
$k=(0,0,1)$, then 
\begin{eqnarray}
 i \times (i \times j) = i \times k = -j \\
 (i \times i) \times j = 0
\end{eqnarray}
However in Geometric Algebra, if $e_1=(1,0,0)$, $e_2=(0,1,0)$ and
$e_3=(0,0,1)$ then 
\begin{eqnarray}
  e_1 \wedge (e_1 \wedge e_2) = (e_1 \wedge e_1 ) \wedge e_2 = 0
\end{eqnarray}
According to D. Hestenes, New Foundations for Classical Mechanics
(equation 3.13)
\begin{eqnarray}
   a \times b = -i a \wedge b  \quad, i=\sqrt{-1} 
\end{eqnarray}
So, where is the flaw here? Except for a complex scalar both definitions in $\mathbb{R}^3$ are the same. But.....one is associative and the other is not?
Thanks.
 A: The repeated wedge product $a \wedge (b \wedge c)$ has nothing to do with the repeated cross product $a \times (b \times c)$. Instead it corresponds (via the duality between trivectors and scalars in three-dimensional geometric algebra) to the scalar triple product $a \cdot (b \times c)$.
A: The basic point is that a difference of associative bilinear products will still be bilinear but need not be associative at all. Still, it's instructive to get an explicit expression for the associator of the cross product.
Let $I := e_1 e_2 e_3$, so that
$$
 \forall a,b \in \mathbb{R}^3, \quad a \times b = -I(ab - a \cdot b) = -\frac{1}{2}I(ab-ba).
$$
Then for any $a, b, c \in \mathbb{R}^3$,
\begin{align}
 (a \times b) \times c &= -\frac{1}{2}I((a \times b)c - c(a \times b))\\ &= -\frac{1}{2}\left(\left(-\frac{1}{2}I(ab-ba)\right)c - c \left(-\frac{1}{2}I(ab-ba)\right) \right)\\ &= -\frac{1}{4}(abc-bac-cab+cba),
\end{align}
whilst
\begin{align}
 a \times (b \times c) &= -\frac{1}{2}I(a(b \times c) - (b \times c)a)\\
&= -\frac{1}{2}I\left(a\left(-\frac{1}{2}I(bc-cb)\right) - \left(-\frac{1}{2}I(bc-cb)\right)a\right)\\
&= -\frac{1}{4}(abc-acb-bca+cba),
\end{align}
so that
\begin{align*}
 (a \times b) \times c - a \times (b \times c) &= -\frac{1}{4}(abc-bac-cab+cba) + \frac{1}{4}(abc-acb-bca+cba)\\ &=\frac{1}{4}(bac+cab-acb-bca)\\
&= \frac{1}{2}(b(a \wedge c)-(a \wedge c)b)\\
&= b \wedge (a \wedge c)\\
&= -I(b \wedge (-I(a \wedge c)))\\
&= -b \times (a \times c),
\end{align*}
so that $(a \times b) \times c = a \times (b \times c)$ if and only if $b \times (a \times c) = 0$. 
Anyhow, for example, $e_1 \times (e_1 \times e_2) = e_1 \times e_3 = -e_2 \neq 0$, so that $(e_1 \times e_1) \times e_2 \neq e_1 \times (e_1 \times e_2)$; more explicitly,
\begin{align*}
 (e_1 \times e_1) \times e_2 = -\frac{1}{2}I(e_1e_1-e_1e_1) \times e_2 = 0 \times e_2 = 0,
\end{align*}
where, of course, $e_1 \times e_1 = -\frac{1}{2}I(e_1 \wedge e_1) = 0$, whilst
\begin{align*}
 e_1 \times (e_1 \times e_2) &= -\tfrac{1}{2}I(e_1(-\tfrac{1}{2}I(e_1e_2-e_2e_1)) - (-\tfrac{1}{2}I(e_1e_2-e_2e_1))e_1)\\
&= -\frac{1}{4}(e_1e_1e_2-e_1e_2e_1-e_1e_2e_1+e_2e_1e_1)\\
&= -e_1^2e_2\\
&= -e_2 \neq 0.
\end{align*}
