Grade operations in Geometric Algebra

Question

I'm going through the book Geometric Algebra for Physicists by Doran and Lasenby, and I have found myself lost when the authors use the grade operator to change between products and switch the order. For example, in Equation (3.126) on page 75,

$$A \cdot(\boldsymbol{x} \wedge(\boldsymbol{x} \cdot B))=\langle A \boldsymbol{x}(\boldsymbol{x} \cdot B)\rangle=\langle(A \cdot \boldsymbol{x}) \boldsymbol{x} B\rangle=B \cdot(\boldsymbol{x} \wedge(\boldsymbol{x} \cdot A))$$

where $A$ and $B$ are multivectors and $x$ a vector.

In Equation (4.154) on page 110, a similar operation is performed, $$\begin{aligned} \left\langle A_{r} \mathrm{~F}\left(B_{r}\right)\right\rangle &=\left\langle A_{r} B_{r}\right\rangle+\alpha\left\langle A_{r}\left(B_{r} \cdot f_{1}\right) \wedge f_{2}\right\rangle \\ &=\left\langle A_{r} B_{r}\right\rangle+\alpha\left\langle f_{2} \cdot A_{r} B_{r} f_{1}\right\rangle \end{aligned}$$ where $A_r$ and $B_r$ are multivectors of grade $r$ and $f_1, f_2$ are arbitrary vectors.

Can someone explain the rationale behind these examples and/or provide a source where these operations are explained?

Answer 1

The first result stems from a couple definitions:

1. The lack of a subscript implies grade 0 term: $$\langle A\rangle=\langle A\rangle_0$$which is found in Sec 2.5 of Doran & Lasenby's text (hereafter D&L). This also implies that $$\langle ab\rangle=\langle a\cdot b\rangle$$
2. The grade 0 projection is symmetric and cyclic: $$\langle AB\rangle=\langle BA\rangle,\quad\langle AB\cdots C\rangle=\langle B\cdots CA\rangle$$ which is also in D&L Sec 2.5.
3. Equations (2.52) and (2.55) in D&L, $$a\cdot B\underset{(2.52)}{=}\frac12\left(aB-Ba\right)\underset{(2.55)}{=}-B\cdot a$$ for vector $a$ and bivector $B$. Note that this operation results in a vector (i.e., $a\cdot B=c_i\mathrm{e}_i$).
4. We also find that, $$a\left(a\cdot B\right)=a\cdot\left(a\cdot B\right)+a\wedge\left(a\cdot B\right)=a\wedge\left(a\cdot B\right)$$ as $a\cdot\left(a\cdot B\right)=0$ (see Sec 2.4.1 of D&L where $a=a_\perp+a_\parallel$ and $b\perp a_\parallel$; consider also explicit computation if further proof needed).

So by using #1 and #4, we can immediately find that for vector $x$ and bivectors $A,\,B$, $$ \langle A\cdot \left(x\wedge\left(x\cdot B\right)\right)\rangle=\langle Ax\left(x\cdot B\right)\rangle $$ And then by #2 and #3, we can swap the order to give us, $$ \langle Ax\left(x\cdot B\right)\rangle=\langle\left(B\cdot x\right)xA\rangle $$ In order to swap the inner and geometric product, we use #3 again, $$\langle\left(B\cdot x\right)xA\rangle=\langle\frac{1}{2}\left(Bx-xB\right)xA\rangle=\langle\frac{1}{2}\left(BxxA-xBxA\right)\rangle $$ Then by swapping the orders of the products on the right term, we find, $$\langle\frac{1}{2}\left(BxxA-xBxA\right)\rangle=\langle\frac{1}{2}\left(BxxA-BxAx\right)\rangle=\langle Bx\frac{1}{2}\left(xA-Ax\right)\rangle=\langle Bx\left(x\cdot A\right)\rangle. $$ The final result is then obtained by removing the projection operator by inserting the inner and outer products that were removed in the first line.

I have not worked through the second equation, but I suspect it's basically the same work.

Answer 2

Apparently, capital letters denote bivectors, lower case letters denote vectors. Then, for example,

The term inner product is reserved for the lowest grade part of the geometric product of two homogeneous multivectors. For two homogeneous multivectors of the same grade the inner product and scalar product reduce to the same thing. (pp.39-40)

Therefore, $\mathbf{A}\cdot(\mathbf{x}\wedge(\mathbf{x}\cdot\mathbf{B}))=\langle\mathbf{A}\mathbf{x}\wedge(\mathbf{x}\cdot\mathbf{B}))\rangle$.

On the other hand, $\mathbf{x}(\mathbf{x}\cdot\mathbf{B})=\mathbf{x}\cdot(\mathbf{x}\cdot\mathbf{B})+\mathbf{x}\wedge(\mathbf{x}\cdot\mathbf{B})$, and $\mathbf{x}\cdot(\mathbf{x}\cdot\mathbf{B})=0$ (see Figure 2.4 and Eq. (2.55)), therefore, $\langle\mathbf{A}\mathbf{x}\wedge(\mathbf{x}\cdot\mathbf{B}))\rangle=\langle\mathbf{A}\mathbf{x}(\mathbf{x}\cdot\mathbf{B}))\rangle$.