Show that $\tfrac{1}{2}(b B\cdot (c \wedge a) - a B\cdot (c \wedge b)) = (c\cdot b) a \cdot B $ in geometric algebra

Question

I would like to show that

$$\frac{1}{2}\left( b B\cdot (c \wedge a) - a B\cdot (c \wedge b)\right) = (c\cdot b) a \cdot B $$

where B is a bivector, and the others are vectors. I've tried to coerce every term into having a factor of $a\cdot B$, but this just introduces other terms.

Edit: some background: Hestenes claims

\begin{aligned} G^\beta &=\tfrac{1}{2} (g^\beta \wedge g^\mu \wedge g^\nu) \cdot \omega_{\mu\nu} \\ &= g^\mu \cdot \omega_{\mu\nu} g^{\nu \beta} -\tfrac{1}{2}g^\beta (g^\nu \wedge g^\mu)\cdot \omega_{\mu\nu} \end{aligned}

Apply Equ (1.40) (Hestenes & Sobczyk 1984) \begin{aligned} B_r \cdot (a_1 \wedge \cdots \wedge a_n) = \sum_{j_1 < \cdots < j_r} \epsilon(j_1, \dots, j_n) B_r \cdot (a_{j_1} \wedge \cdots \wedge a_{j_r}) a_{j_{r+1}} \cdots a_{j_n} \end{aligned}

Switch the order of the inner product (gives factor of +1) and then apply the formula,

\begin{aligned} G^\beta = \tfrac{1}{2} \left( g^\nu \omega_{\mu\nu} \cdot (g^\beta \wedge g^\mu) - g^\mu \omega_{\mu\nu} \cdot (g^\beta \wedge g^\nu) + g^\beta \omega_{\mu\nu} \cdot (g^\mu \wedge g^\nu) \right) \end{aligned}

The last term is equal to $-\tfrac{1}{2}g^\beta (g^\nu \wedge g^\mu) \cdot \omega_{\mu\nu}$; the remainer is the problem in question, with $a=g^\mu$, $b=g^\nu$, $c=g^\beta$, $B=\omega_{\mu\nu}$.

Answer 1

The quantity

\begin{aligned} \tfrac{1}{2} (c \wedge a \wedge b) \cdot B \end{aligned}

must contain a projection between $B$ and $c$, but the following does not:

\begin{aligned} a \cdot B (b\cdot c) - \tfrac{1}{2}c (b \wedge a) \cdot B \end{aligned}

Therefore they cannot be equal, contrary to what Hestenes claims. Therefore the subproblem statement given in the question cannot be correct either.

Answer 2

The generalized identity you present is almost certainly false, and instead the derivation you seek relies on the specific problem at hand.

Begin with $G^\beta = \frac12(g^\beta\wedge\omega_{\mu\nu})\cdot(g^\mu\wedge g^\nu)$ as Hestenes does. Then apply the two identities he suggests successively to get $$ G^\beta = \frac12(g^\beta\wedge\omega_{\mu\nu})\cdot(g^\mu\wedge g^\nu) = \frac12[g^\beta\wedge(\omega_{\mu\nu}\cdot g^\mu) + g^{\beta\mu}\omega_{\mu\nu}]\cdot g^\nu $$$$ = \frac12[g^\beta(\omega_{\mu\nu}\cdot g^\mu)\cdot g^\nu - g^{\beta\nu}\omega_{\mu\nu}\cdot g^\mu + g^{\beta\mu}\omega_{\mu\nu}\cdot g^\nu]. \tag{$*$} $$

The first term of ($*$) becomes $$ g^\beta(\omega_{\mu\nu}\cdot g^\mu)\cdot g^\nu = g^\beta\omega_{\mu\nu}\cdot(g^\mu\wedge g^\nu) = -g^\beta(g^\nu\wedge g^\mu)\cdot\omega_{\mu\nu} $$ using the commutivity of the dot product and anticommutivity of the wedge product in this case.

The second term becomes $$ -g^{\beta\nu}\omega_{\mu\nu}\cdot g^\mu = g^\mu\cdot\omega_{\mu\nu}g^{\beta\nu} $$ using the anticommutivity of the dot product in this case and the fact that $g^{\beta\nu}$ is a scalar.

For the third term of ($*$) we do the same thing but then relabel indices and use the antisymmetry of $\omega_{\mu\nu}$ to get $$ g^{\beta\mu}\omega_{\mu\nu}\cdot g^\nu = -g^\nu\cdot\omega_{\mu\nu}g^{\beta\mu} = -g^\mu\cdot\omega_{\nu\mu}g^{\beta\nu} = g^\mu\cdot\omega_{\mu\nu}g^{\beta\nu}. $$

Combining all the terms now yields $$ G^\beta = -\frac12g^\beta(g^\nu\wedge g^\mu)\cdot\omega_{\mu\nu} + g^\mu\cdot\omega_{\mu\nu}g^{\beta\nu} $$ as desired.

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I do not know how Hestenes concluded that $$ G^\beta = \frac12(g^\beta\wedge g^\mu\wedge g^\nu)\cdot\omega_{\mu\nu} = \frac12(g^\beta\wedge\omega_{\mu\nu})\cdot(g^\mu\wedge g^\nu) $$ nor how his Eq. (147) could at all be relevant. What we can derive easily is $$ \frac12(g^\beta\wedge g^\mu\wedge g^\nu)\cdot\omega_{\mu\nu} = -\frac12(g^\beta\cdot\omega_{\mu\nu})\cdot(g^\mu\wedge g^\nu) $$ which makes me feel like the previous equation is incorrect, but I have no proof of this.