This continues chapter 2 of NFCM.Next.

This time I’ve chosen beauty over clarity; the equations are aligned nicely, but there are no hyperlinks as it seems to be impossible to escape from the align-environment-in-mathjax-in-hugo Inception style nightmare I have created. I think the awful alignment was better.

$\let\oldtimes\times \renewcommand{\times}{\kern-.2em\oldtimes\kern-.2em} \let\oldcdot\cdot \renewcommand{\cdot}{\kern-.2em\oldcdot\kern-.2em} \let\oldwedge\wedge \renewcommand{\wedge}{\kern-.2em\oldwedge\kern-.2em} \renewcommand{\vec}[1]{\mathbf{#1}} \newcommand{\e}[1]{(-1)^{#1}} \newcommand{\eh}{(-1)^{r(r-1)/2}} \newcommand{\gt}{\mathcal{G}_3} \newcommand{\hf}{\frac{1}{2}} \newcommand{\rn}{\rangle^\hf_0}$

This chapter is set in familiar 3-space, $\gt$, and sees the introduction of the unit pseudoscalar $i$, a unit trivector with complex unit-like properties. Multiplying a vector/multivector by the pseudoscalar $i$, results in an inner product because the outer product is “full”, and in odd dimensions this product commutes (in even dimensions it anticommutes). The dual of a multivector is defined to be its product with the pseudoscalar, and in a twist, the classical vector cross product is revealed to be the dual of the bivector outer product of vectors and many difficult identities follow easily.



Show $\vec{x}$ is in the $\vec{B}$-plane $\Leftrightarrow \vec{x}\vec{B} = -\vec{B}\vec{x}$

$\Rightarrow$: if $\vec{x}$ is in the $\vec{B}$-plane then $\vec{x}\wedge\vec{B} = \vec{B}\wedge\vec{x} = 0$,

so $\vec{x}\vec{B} = \vec{x}\cdot\vec{B} = -\vec{B}\cdot\vec{x} = -\vec{B}\vec{x}$

$\Leftarrow$: if $\vec{x}\vec{B} = -\vec{B}\vec{x}$ then $\vec{x}\cdot\vec{B}+\vec{x}\wedge\vec{B} = -\vec{B}\cdot\vec{x} - \vec{B}\wedge\vec{x} = \vec{x}\cdot\vec{B} - \vec{B}\wedge\vec{x}$,

so $\vec{x}\wedge\vec{B} = -\vec{B}\wedge\vec{x} = -\vec{x}\wedge\vec{B}$ or $\vec{x}\wedge\vec{B}=0$ and $\vec{x}$ is in the $\vec{B}$-plane.


Let $\vec{x}’ = \vec{x}\vec{B}$. Show that $\vec{x’}$ is a vector, $\lvert\vec{x}’\rvert = \lvert\vec{B}\rvert\lvert\vec{x}\rvert$ and $\vec{x}\cdot\vec{x}’ = 0$.

In 3.1a we saw that $\vec{x}\wedge\vec{B} = 0$, so $\vec{x}’ = \vec{x}\vec{B} = \vec{x}\cdot\vec{B}$, and $\vec{x’}$ is a vector.

$\lvert\vec{x’}\rvert = \lvert\vec{x}\vec{B}\rvert = \langle (\vec{x}\vec{B})^\sim\vec{x}\vec{B} \rn$ by definition of the multivector norm (1.29).

$ = \langle\vec{B}^\sim\vec{x}^\sim\vec{x}\vec{B}\rn$ reversion definition for products (1.24a)

$ = \langle\vec{B}^\sim\vec{x}\vec{x}\vec{B}\rn$ as vectors are defined to be their own reverse (1.24d)

$ = \langle\vec{B}^\sim\vec{B}\rn(\vec{xx})^\hf$ as $\vec{xx} = \vec{x}^2$ is a scalar

$ = \langle\vec{B}^\sim\vec{B}\rn\langle\vec{xx}\rn$ scalars are already grade zero

$ = \langle\vec{B}^\sim\vec{B}\rn\langle\vec{x}^\sim\vec{x}\rn$ vectors are their own reverse

$ = \lvert\vec{B}\rvert\lvert\vec{x}\rvert$ multivector norm definition (1.29)

3.2 TODO

Show $\vec{a}\wedge\vec{b}/\vec{i} = TODO$ for $\vec{a}$, $\vec{b}$ in the $i$-plane, where $\vec{i} = \vec{\sigma}_1\vec{\sigma}_2$.



This question asks you to prove four identities involving inner and outer products, the vector cross products and the unit pseudoscalar, $i$.

The grade of expressions involving the pseudoscalar are not yet obvious to me, and converting the products to inner products helps, as inner products reduce grade.

This is shown in the general inner product definition, (1.17a): $\vec{A}_r\cdot\vec{B}_s \equiv \langle\vec{A}_r\vec{B}_s\rangle_{\lvert r-s\rvert}$, so dotting the pseudoscalar for example would reduce the pseudoscalar’s grade by one, giving you a bivector.

We use the definition of the pseudoscalar $i$, $$\vec{x}\wedge i = 0$$ for all $\vec{x}$ in $\gt$ (3.1) and the definition of the vector cross product

$$i\vec{a}\times\vec{b} \equiv \vec{a}\wedge\vec{b}$$ (3.13)

The pseudoscalar property $i^2 = -1$ (3.11b) and $\vec{A}i = i\vec{A}$ for all $\vec{A}$ in $\gt$ (3.11c) are used.

Two useful identities

All of these identities come easily by expanding the inner and outer products into geometric products and letting the pseudoscalar commute out, but it’s simpler and briefer to use the following identities, derived like so

$$ \begin{alignat}{1} &&\vec{a}(i\vec{A}_r) &= \vec{a}\cdot(i\vec{A}_r)+\vec{a}\wedge(i\vec{A}_r)\
= &&i(\vec{a}\vec{A}_r) &= i\vec{a}\cdot\vec{A}_r + i\vec{a}\wedge\vec{A}_r \end{alignat} $$

Equating $n-r+1$ and $n-r-1$ grade components gives

$$ \begin{alignat}{1} &&\vec{a}\cdot(i\vec{A}_r) &= i\vec{a}\wedge\vec{A}_r\
&&\vec{a}\wedge(i\vec{A}_r) &= i\vec{a}\cdot\vec{A}_r \end{alignat} $$

The above identities let you easily exchange inner and outer products involving $i$, and work in odd dimensions (this chapter chooses $n=3$), for even dimensions $i$ anticommutes and you get the negative.


Prove $\vec{a}\cdot\vec{b} = -i (\vec{a}\wedge(i\vec{b}))$

$$\begin{align} \vec{a}\cdot\vec{b} &= -ii\vec{a}\cdot\vec{b}\
&= -i\vec{a}\wedge(i\vec{b}) \end{align}$$


Prove $\vec{b}\times\vec{a} = i(\vec{a}\wedge\vec{b}) = \vec{a}\cdot(i\vec{b}) = -(i\vec{b})\cdot\vec{a}$

$$\begin{align} \vec{b}\times \vec{a} &= -i\vec{b}\wedge\vec{a}\
&= i\vec{a}\wedge\vec{b}\
&= \vec{a}\cdot(i\vec{b})\
&= -(i\vec{b})\cdot\vec{a} \text{ since } i\vec{b} \text{ is a bivector} \end{align}$$


Prove $\vec{a}\cdot(\vec{b}\wedge\vec{c}) = -\vec{a}\times (\vec{b}\times \vec{c}) = \vec{a}\cdot\vec{b}\vec{c}-\vec{a}\cdot\vec{c}\vec{b}$

$$ \begin{align} \vec{a}\cdot(\vec{b}\wedge\vec{c}) &= \vec{a}\cdot(i\vec{b}\times \vec{c})\
&= i\vec{a}\wedge(\vec{b}\times \vec{c})\
&= i^2\vec{a}\times (\vec{b}\times \vec{c})\
&= -\vec{a}\times (\vec{b}\times \vec{c}) \end{align} $$

and $\vec{a}\cdot\vec{b}\wedge\vec{c} = \vec{a}\cdot\vec{b}\vec{c}-\vec{a}\cdot\vec{c}\vec{b}$ by 1.14.


Prove $\vec{a}\wedge(\vec{b}\wedge\vec{c}) = i\vec{a}\cdot(\vec{b}\times \vec{c}) = \frac{1}{2}(\vec{a}\vec{b}\vec{c}-\vec{c}\vec{b}\vec{a})$

$$\begin{align} \vec{a}\wedge(\vec{b}\wedge\vec{c}) &= \vec{a}\wedge(i\vec{b}\times \vec{c})\
&= i\vec{a}\cdot(\vec{b}\times \vec{c}) \end{align}$$

The second part seems a little more high powered than it should be, using 1.13 to eliminate the reverse. I probably should have used the 3 dimensional reverse from this chapter. If anyone knows a simpler way, please let me know:

$$\begin{align} \vec{a}\vec{b}\vec{c}-\vec{c}\vec{b}\vec{a} &= \vec{a}\vec{b}\vec{c}-(\vec{a}\vec{b}\vec{c})^\sim\
&= \langle\vec{a}\vec{b}\vec{c}\rangle_1+\langle\vec{a}\vec{b}\vec{c}\rangle_3-\langle(\vec{a}\vec{b}\vec{c})^\sim\rangle_1-\langle(\vec{a}\vec{b}\vec{c})^\sim\rangle_3\
&= \langle\vec{a}\vec{b}\vec{c}\rangle_1+\langle\vec{a}\vec{b}\vec{c}\rangle_3-\langle\vec{a}\vec{b}\vec{c}\rangle_1+\langle\vec{a}\vec{b}\vec{c}\rangle_3\
&= 2\langle\vec{a}\vec{b}\vec{c}\rangle_3 = 2\vec{a}\wedge\vec{b}\wedge\vec{c} \end{align}$$

Dividing through by $2$ gives the result.

3.4 TODO

3.5 TODO


Show $\vec{A}^\sim = \alpha - i\beta + \vec{a} -i\vec{b}$.

$$\begin{align} \vec{A}^\sim &= (\alpha + i\beta + \vec{a} + i\vec{b})^\sim\
&= \alpha^\sim + (i\beta)^\sim + \vec{a}^\sim + (i\vec{b})^\sim\
&= \alpha + \beta^\sim i^\sim + \vec{a} + \vec{b}^\sim i^\sim\
&= \alpha + \beta(-i) + \vec{a} + \vec{b}(-i)\
&= \alpha - i\beta + \vec{a} -i\vec{b} \end{align}$$

Show $\lvert\vec{A}\rvert^2 = \alpha^2 + \beta^2 + \vec{a}^2 + \vec{b}^2$.

$$\begin{align} \vec{A}\vec{A}^\sim &= (\alpha + i\beta + \vec{a} + i\vec{b})(\alpha - i\beta + \vec{a} - i\vec{b})\
&= ((\alpha + \vec{a}) + i(\beta + \vec{b}))((\alpha + \vec{a}) - i(\beta + \vec{b}))\
&= (\alpha + \vec{a})^2 -i^2(\beta + \vec{b})^2\
&= (\alpha + \vec{a})^2 + (\beta + \vec{b})^2\
&= \alpha^2 + \beta^2 + \vec{a}^2 + \vec{b}^2 + 2(\alpha\vec{a} + \beta\vec{b}) \end{align}$$

By definition of the multivector norm, taking the scalar component of the above gives the result: $\lvert\vec{A}\rvert^2 \equiv \langle\vec{A}\vec{A}^\sim\rangle_0 = \alpha^2 + \beta^2 + \vec{a}^2 + \vec{b}^2$

3.8 TODO

The following two questions use the Einstein summation notation, where repeated indices imply summation over that index.



Show that $\vec{\sigma}_k\vec{a}\vec{\sigma}_k = -\vec{a}$

This answer uses the properties $\vec{\sigma}_k\cdot\vec{\sigma}_k = 3$ and $\vec{a} = a_i \vec{\sigma}_i = \vec{a}\cdot\vec{\sigma}_i\vec{\sigma}_i$

$$ \begin{align} \vec{\sigma}_k\vec{a}\vec{\sigma}_k &= (\vec{\sigma}_k\cdot\vec{a} + \vec{\sigma}_k\wedge\vec{a})\vec{\sigma}_k\
&= \vec{a}\cdot\vec{\sigma}_k\vec{\sigma}_k + \vec{\sigma}_k\wedge\vec{a}\cdot\vec{\sigma}_k + \vec{\sigma}_k\wedge\vec{a}\wedge\vec{\sigma}_k\
&= \vec{a} + \vec{\sigma}_k\cdot\vec{a}\vec{\sigma}_k - \vec{\sigma}_k\cdot\vec{\sigma}_k\vec{a}\
&= 2\vec{a} - 3\vec{a}\
&= -\vec{a} \end{align} $$


Show that $\vec{\sigma}_k\vec{a}\wedge\vec{b}\vec{\sigma}_k = \vec{b}\wedge\vec{a}$

Writing the bivector as the product of the pseudoscalar and a vector $\vec{a}\wedge\vec{b} = i\vec{a}\times \vec{b}$ and using the property $i\vec{A} = \vec{A}i$ in $\gt$ makes this fairly painless:

$$ \begin{align} \vec{\sigma}_k\vec{a}\wedge\vec{b}\vec{\sigma}_k &= \vec{\sigma}_ki\vec{a}\times \vec{b}\vec{\sigma}_k\
&= i\vec{\sigma}_k\vec{a}\times \vec{b}\vec{\sigma}_k\
&= -i\vec{a}\times \vec{b} \text{ by question 3.9a}\
&= -\vec{a}\wedge\vec{b}\
&= \vec{b}\wedge\vec{a} \end{align} $$


Show that $\vec{\sigma}_k\vec{a}\wedge\vec{b}\wedge\vec{c}\vec{\sigma}_k = 3\vec{a}\wedge\vec{b}\wedge\vec{c}$

Since a $3$-blade in $\gt$ must be a scalar multiple of the pseudoscalar, and the pseudoscalar commutes with multivectors

$$ \begin{align} \vec{\sigma}_k\vec{a}\wedge\vec{b}\wedge\vec{c}\vec{\sigma}_k &= \vec{\sigma}_k\lambda i\vec{\sigma}_k\
&= \vec{\sigma}_k\vec{\sigma}_k\lambda i\
&= 3\vec{a}\wedge\vec{b}\wedge\vec{c} \end{align}$$