Trying to compute the formula for the triple cross product $$ a \times (b \times c) = \langle c, a\rangle b - \langle b, a \rangle c $$ using wedge, hodge star and contraction operations I get \begin{align}a \times (b \times c) &= *(a \wedge *(b \wedge c)) = \\ &= i_a(b \wedge c) =\\ &= \langle a, b \rangle c - \langle a, c\rangle b \end{align} which is the opposite of the currect result. Where is the error?
The formulas I used are
- $ a \times b = *(a \wedge b)$;
- $ i_x(v) = \langle v, x \rangle$ for $x, y$ 1-vectors and $i_x(v \wedge a) = i_x(v) a - v \wedge i_x(a)$ when $a$ is a $k\geq 1$-vector;
- $i_x(a) = *(x \wedge *(a))$.
Edit.
Apparently the currect identity is
$$*(v \wedge w) = i_w(* v)$$
and using the fact that $*^2 = 1$ in $\mathbb{R}^3$ the derivation would be
\begin{align}a \times (b \times c) &= *(a \wedge *(b \wedge c)) = \\ &= (-1)^{1 \cdot 1} *(*(b \wedge c) \wedge a) = \\ &= - i_a(*^2(b \wedge c)) = \\ &= - i_a(b \wedge c) = \\ &= - \langle a, b \rangle c + \langle a, c\rangle b \end{align}
I think the mistake is that the dual of the dual in dimension 3, is -1 ($*\circ * = (-1)\text{id}$).E.g. I believe you should have$i_x(a) = -(*(x\wedge *(a)))$.
Note: the original point is wrong since I believed that $*(a) \leftrightarrow Ia$ in geometric algebra. The final result is correct, though, since it does map directly onto GA.