Missing steps+intuition for geometric algebra manipulations

Question

I'm working my way through Doran and Lasenby's Geometric Algebra for Physicists, but I've run into some trouble in chapter three with a couple of derivations where I can't follow the steps.

The first one is equation (3.126): $$A\cdot(x\wedge(x\cdot B)) = \langle Ax(x\cdot B)\rangle = \langle (A\cdot x)x B\rangle = B\cdot(x\wedge(x\cdot A)),$$ where $A$ and $B$ are bivectors (in 3D Euclidean space) and $x$ is a vector, and $\langle \rangle$ picks out the scalar component. I don't quite understand how the equalities follow. For example, I understand that the result of $A\cdot(x\wedge(x\cdot B))$ should be a scalar, and that the dot and wedge products can be expanded in terms of geometric products, but why do we make the choice to write it as $\langle Ax(x\cdot B)\rangle$ instead of, say, $\langle AxxB\rangle$? And then in the second equality how do we shift the dot product from $x$ and $B$ to $A$ and $x$? I feel like there's some shortcut I'm missing. Do I just need to expand out everything in terms of geometric products?

The second, related, question comes in equation (3.128), where the text says: $$a_{i}a_{j}\mathcal{I}_{ij} = -\int d^{3}x \rho(Ia)\cdot(x\wedge(x\cdot(Ia)))\\ =\int d^{3}x \rho (x\cdot(Ia))^{2} >0,$$ where $I$ is the unit pseudoscalar in three dimensions, and $a$ is a vector. $\rho$ is the density, a scalar function, and $\mathcal{I}_{ij}$ is the inertia tensor matrix. I don't understand how we go so quickly from the first line to the second; I feel like there is some result or identity that can be used that I am missing.

Any tips/resources would be appreciated!

Answer 1

$ \newcommand\grade[1]{\langle#1\rangle} \newcommand\lcontr{{\rfloor}} \newcommand\rcontr{{\lfloor}} $

I will answer the first question in two ways: the way they did it, and the better way. By then we will have the answer to the second question.

How they did it

$A$ and $B$ are bivectors, and $x$ is a vector. We get $$ A\cdot(x\wedge(x\cdot B)) = \grade{A(x\wedge(x\cdot B))} = \grade{A\,x{\wedge}(x\cdot B)} $$ since we know the result is a scalar, and then we drop the first set of parentheses since we give the geometric product less precedence. Now, the geometric product $x(x\cdot B)$ has a grade 0 part $x\cdot(x\cdot B)$ and a grade 2 part $x\wedge(x\cdot B)$. But $\grade{A(\text{not grade 2})} = 0$ because $A$ is grade 2 (the product of a $k$-vector and an $l$-vector gives a least possible grade of $|k-l|$). So in this case it's perfectly safe to replace $x\wedge(x\cdot B)$ with $x(x\cdot B)$: $$ \grade{A\,x{\wedge}(x\cdot B)} = \grade{Ax(x\cdot B)} = \grade{(Ax)(x\cdot B)}. $$ The last equality is by associativity of the geometric product (which is key in a lot of manipulations like this). The term $x\cdot B$ is grade 1, so $\grade{(\text{not grade 1})(x\cdot B)} = 0$, meaning we only get the grade 1 part of $Ax$ which is $A\cdot x$. Hence $$ \grade{(Ax)(x\cdot B)} = \grade{(A\cdot x)(x\cdot B)}. $$ I don't know why they decided to write $\grade{(A\cdot x)xB}$ instead, since what I would next is $$ \grade{(A\cdot x)(x\cdot B)} = (A\cdot x)\cdot(x\cdot B) = -(x\cdot A)\cdot(x\cdot B). $$ The first equality comes from the fact that $A\cdot x$ and $x\cdot B$ are vectors. Now we see that this expression is symmetric in $A$ and $B$ (since the dot of two vectors is), so we can simply swap them in the original expression: $$ A\cdot(x\wedge(x\cdot B)) = B\cdot (x\wedge(x\cdot A)). $$

The better way

Read The Inner Products of Geometric Algebra (2002) by Leo Dorst. The left and right contractions $\lcontr$ and $\rcontr$ are variants of Doran an Lasenby's dot product which are better theoretically and practically. They could be defined as $$ A_r\lcontr B_s = \begin{cases} \grade{A_rB_s}_{s-r} &\text{if } s \geq r, \\ 0 &\text{otherwise}, \end{cases} $$$$ A_r\rcontr B_s = \begin{cases} \grade{A_rB_s}_{r-s} &\text{if } r \geq s, \\ 0 &\text{otherwise}, \end{cases} $$ where here subscripts indicate grade. The important identities for our purposes are $$ (A\wedge B)\lcontr C = A\lcontr(B\lcontr C),\quad A\rcontr(B\wedge C) = (A\rcontr B)\rcontr C, $$ which hold for all multivectors $A, B, C$. We translate the dots into contractions and then everything is simple. The dot with $B$ could be $\lcontr$ or $\rcontr$, but we judiciously choose $\rcontr$. $$ A\cdot(x\wedge(x\cdot B)) = A\rcontr(x\wedge(x\lcontr B)) = (A\rcontr x)\rcontr(x\lcontr B) = (A\rcontr x)\cdot(x\lcontr B). $$ In the last step I switch $\rcontr$ to $\cdot$ since the arguments have the same grade and I like to emphasize the symmetry. Same as before we get $$ (A\rcontr x)\cdot(x\lcontr B) = -(x\lcontr A)\cdot(x\lcontr B), $$ which shows that $A$ and $B$ are symmetric in this expression, hence $$ A\rcontr(x\wedge(x\lcontr B)) = B\rcontr(x\wedge(x\lcontr A)). $$

Question 2

The expression $$ (Ia)\cdot(x\wedge(x\cdot(Ia))) $$ is the same one as before with $A = B = Ia$, and I've shown in two ways that $$ (Ia)\cdot(x\wedge(x\cdot(Ia))) = -(x\cdot(Ia))\cdot(x\cdot(Ia)) $$ But then we see $$ -(x\cdot(Ia))\cdot(x\cdot(Ia)) = -(x\cdot(Ia))^2, $$ and there's a minus sign in front the integral which cancels with this one.

Answer 2

We can replace the wedge product from the right bivector term with a geometric product $$\begin{aligned}x \wedge \left( { x \cdot B } \right) &= x \left( { x \cdot B } \right) - x \cdot \left( { x \cdot B } \right) \\ &= x \left( { x \cdot B } \right),\end{aligned}$$ since $ x $ and $ x \cdot B $ are perpendicular ( $ x \cdot B $ is the projection of $ x $ onto $ B $, but rotated 90 degrees in that plane.) As pointed out by @mr_e_man in the comments, this also follows from the identity $ x \cdot (y \cdot B) = (x \wedge y) \cdot B $.

The dot product in the original expression, can now be written as a grade selection $$\begin{aligned} A \cdot \left( { x \wedge \left( { x \cdot B } \right) } \right) &= \left\langle{{ A \left( { x \wedge \left( { x \cdot B } \right) } \right) }}\right\rangle \\ &= \left\langle{{ A x \left( { x \cdot B } \right) }}\right\rangle \\ &= \left\langle{{ \left( { A \cdot x + A \wedge x } \right) \left( { x \cdot B } \right) }}\right\rangle \\ &= \left\langle{{ \left( { A \cdot x } \right) \left( { x \cdot B } \right) }}\right\rangle + \left\langle{{ \left( { A \wedge x } \right) \left( { x \cdot B } \right) }}\right\rangle.\end{aligned}$$ Within the second grade selection operator, is a product of a trivector and a vector, which is a bivector, and has no grade-0 components, so that grade selection is zero. We are left with $$\begin{aligned} A \cdot \left( { x \wedge \left( { x \cdot B } \right) } \right) &= \left\langle{{ \left( { A \cdot x } \right) \left( { x \cdot B } \right) }}\right\rangle \\ &= \left( { A \cdot x } \right) \cdot \left( { x \cdot B } \right) \\ &= -\left( { x \cdot A } \right) \cdot \left( { x \cdot B } \right).\end{aligned}$$

Now we apply a reverse operation, noting that $ \left( {M N} \right)^{\dagger} = N^\dagger M^\dagger $, for any multivectors $ M, N $, and that scalars and vectors are their own reverses. Both $ A \cdot x , x \cdot B $ are vectors, so $$\begin{aligned} \left\langle{{ \left( { A \cdot x } \right) \left( { x \cdot B } \right) }}\right\rangle^\dagger &= \left\langle{{ \left( { x \cdot B } \right) \left( { A \cdot x } \right) }}\right\rangle \\ &= -\left( { x \cdot B } \right) \cdot \left( { x \cdot A } \right).\end{aligned}$$ This shows that $ A \cdot \left( { x \wedge \left( { x \cdot B } \right) } \right) $ is symmetric in $ A, B $, which provides the end result of (3.126). We didn't really need the second-last intermediate step $ \left\langle{{ \left( { A \cdot x } \right) x B }}\right\rangle $, but it's fairly simple to show that from the above, should you want to.

For the second question, from the above, note that $$\begin{aligned}A \cdot \left( { x \wedge \left( { x \cdot A } \right) } \right) &= -\left( { x \cdot A } \right) \cdot \left( { x \cdot A } \right) \\ &= -\left( { x \cdot A } \right)^2,\end{aligned}$$ since $ x \cdot A $ is a vector that, when wedged with itself, is zero. Substitution of $ A = I a $ provides the transition from the first to second line in the second question.