Some Results in Geometric Algebra

$\newcommand{\abs}[1] {\left\vert #1 \right\vert}$

Cramer's Rule

The roots of Cramer's rule have always seemed a little mysterious. If we have a system of equations

$$a_1 x_1  + b_1 x_2 = c_1$$
$$a_2 x_1 + b_2 x_2 = c_2$$

Then Cramer says
$$x_1 = \frac{\det  \begin{bmatrix} c_1 & b_1 \\  c_2 & b_2  \end{bmatrix}}{\det  \begin{bmatrix} a_1 & b_1 \\  a_2 & b_2  \end{bmatrix}}$$

and

$$x_2 = \frac{\det  \begin{bmatrix} a_1 & c_1 \\  a_2 & c_2  \end{bmatrix}}{\det  \begin{bmatrix} a_1 & b_1 \\  a_2 & b_2  \end{bmatrix}}$$

That determinants play any role in isolating linear parameters is not obvious. So it's delightful how geometric algebra shines light on this.

A note on notation: in this section, latin letters are vectors and greek letters are scalars, so that

$$x = \alpha a + \beta b$$

is a vector equation relating vector $x$ to vectors $a$ and $b$. The gift we are given by geometric algebra is that vectors become invertible, so we can write, since $a \wedge a = 0$:

$$a \wedge x = \beta a \wedge b$$
$$\beta = \frac {a \wedge x}{a \wedge b}$$

And similarly,

$$\alpha = \frac {x \wedge b}{a \wedge b}$$

so that

$$x = \left( \frac {x \wedge b}{a \wedge b} \right) a +  \left( \frac {a \wedge x}{a \wedge b} \right) b$$

Now, in $\mathbb{R_2}$, 2-vectors are pseudoscalars, and pseudoscalars are scalar multiples of $I$, the unique (up to a scalar) highest-grade multivector. The determinant is just that scalar (say, $\lambda$): $a \wedge b = \lambda I$. So we can write:

$$x = \left(\frac {\det (x \wedge b) I}{\det (a \wedge b)  I} \right) a +   \left( \frac {\det (a \wedge x) I}{\det (a \wedge b) I} \right) b$$


$$x = \frac {\det (x \wedge b) }{\det (a \wedge b)  } a +   \frac {\det (a \wedge x) }{\det (a \wedge b) }  b$$

That's the fastest path to Cramer's Rule anywhere.

BAC-CAB

Consider the product of a vector and bivector: $a (b \wedge c)$.

$$\frac{1}{2} a (bc - cb)$$

$$\frac{1}{2} (abc - acb)$$

Since $ab =2 a \cdot b -ba$, we can write

$$\frac{1}{2} (2 a \cdot b \; c - bac -2a \cdot c \; b -cab )$$

$$\begin{equation} \label{eq1} a \cdot b c - a \cdot c b -\frac{1}{2} (bac-cab) \end{equation}$$

The last two terms can be expanded as before:

$$- \frac12(bac-cab) = -b a \cdot c + \frac{bca}{2}  + c a \cdot b - \frac{cba}{2}$$

So in the expanded $\eqref{eq1}$, the dot products combine to give us

$$2  a \cdot b  c - 2  a \cdot c  b  + \frac{(bc-cb) a}{2}$$

So finally

$$a (b \wedge c)  = 2  ( a \cdot b  c -  a \cdot c b ) + (b \wedge c) a$$

$$\frac{ a (b \wedge c) - (b \wedge c) a}{2} =  a \cdot b c - a \cdot c b$$

$$a \cdot (b \wedge c) = a \cdot b c - a \cdot c b$$

This is a handy result.  And, not restricted to $\mathbb{R^3}$.

Using the Dual to Exchange Dot and Wedge Products

We will develop a nice result with the dual operation that lets us exchange dot and wedge products.

For any multivector $A$, we can write $A = A^\perp I$, where $A^\bot$ is whatever multivector you need to multiply with $I$ to get $A$, called the dual of $A$. In other words, taking the dual is multiplying on right with $I^{-1}$. Now consider the dual of $AB$

$$(AB)^\bot = AB^\bot$$

This can be confirmed by examining $(AB)I^{-1} = ABI^{-1}$.

Now compare grades of this expression, and conclude that the largest grades give us

$$(A \cdot B)^{\bot} = A \wedge B^\bot$$

and the smallest grades give us

$$(A \wedge B)^{\bot} = A \cdot B^\bot$$

Gibbs' Vector Product

The dual lets us move between Gibbs-style cross products $a \times b$ and geometric products:
$$a \times b \equiv (a \wedge b)^{\bot}$$

Of course Gibbs' cross product is only defined in $\mathbb{R^3}$.

Here's an example, which for some readers may complete the BAC-CAB story:
$$\begin{align} a \times (b \times c) & = \{ a \wedge  (b \times c) \} ^\bot \\ & = a \cdot (b \times c)^\bot \\ & = -a \cdot b \wedge c \end{align}$$
 
Another:
$$\begin{align} a \cdot b \times c & = a \cdot (b\wedge c)^\bot \\ & = (a \wedge b \wedge c)^\bot \end{align}$$

Rotations

On our way to describing angular momentum under a geometric algebra framework, we need to establish a preliminary object known as a rotor. And on the way to the rotor, we need to work through reflections. Consider a reflection axis $n$ ($n$ is a unit vector) and a vector $v$.

$$v = v n n^{-1}$$
$$= [ v \cdot n + v \wedge n ] n^{-1}$$

The first term is the component of $v$ along $n$.  So, the reflected $v$ is just

$$v' =  -v \cdot n n^{-1} + v \wedge n  n^{-1}$$

A small manipulation lets us write

$$v' =  -n \cdot v n^{-1} - n \wedge v  n^{-1}$$
$$=  - [nv] n^{-1} = - nvn^{-1}$$

Kinda cute. And, due to Hamilton, this gets us halfway to rotations, because he noticed that a rotation is simply two reflections. Say we reflect along $b$, followed by $a$:

$$v_{rotated} = (ab)v(ab)^{-1} = RvR^{-1}$$

$R$ is known as a rotor: a product of two unit vectors.

$$R = ab = a \cdot b + a \wedge b$$

We can find the magnitude of $a \wedge b$ by writing

$$ba =  b \cdot a + b \wedge a = a \cdot b - a \wedge b$$

and then considering the following product:

$$abba = a^2 b^2 = 1 = (a \cdot b)^2 - (a \wedge b)^2$$

Since $a \cdot b$ is $\cos \theta$ we can write

$$1 - \cos^2 \theta = \sin^2 \theta = - (a \wedge b)^2$$

So $a \wedge b$ is a bivector whose magnitude is $\sin \theta$ and square is negative. Let $I$ be the unit bivector

$$I = \frac{a \wedge b}{\sin \theta}$$

and write

$$ab = R = \cos \theta + I \sin \theta = \exp (I \theta)$$

Looks like the common exponential phase factor $e^{i \theta}$. Aside: what are the complex numbers?  In two dimensions, vector $(a,b)$ is related to complex $a + ib$ like so: $a \hat{x} + b \hat{y} = \hat{x} (a + \hat{x} \hat{y} b) = \hat{x} (a + I b)$. Just pre-multiply by $\hat{x}$, selecting the real axis. You always knew there was something like this going on behind the scenes.

Moving along: let's use the dagger operator to reverse the order of vector products: $(ab)^\dagger = ba$. Since unit vectors are their own inverse, we can write $y$ as a rotated version of $x$ like so:

$$y = R x R^{-1} = R x R^\dagger$$

If $R$ is in fact a function of time, we can consider $\dot{y}$:

$$\begin{align} \dot{y} &= \dot{R} x R^\dagger + R x \dot{R^\dagger} \\ &= \dot{R} R^\dagger y + y R  \dot{R^\dagger} \\ &= (\dot{R} R^\dagger) y - y (\dot{R}R^\dagger) \end{align}$$

For the last step, take the time derivative of $R R^\dagger = 1$ and show $\dot{R} R^\dagger = - R \dot{R}^\dagger$.

Let $\Omega= - {1 \over 2} \dot{R} R^\dagger$. $\Omega$ is apparently of even grade, and has no scalar part. $\Omega$ must be a bivector, and we can write:

$$\dot{y} = - \Omega\cdot y = y \cdot \Omega$$

Orbital Dynamics

With rotational motion in our toolkit, let's develop the first few results in central-force dynamics. A nice trick for describing this is to shift the dynamics to a rotor. We accomplish this by considering a vector $x$ as a rotation of some basis vector, plus a stretch.

$$x= U e_1 U^{\dagger}$$

We are now allowing a stretch as well as rotation, so U is a little more than a rotor: we are now considering a general even element. In three dimensions this is a scalar plus bivector [side note: the spinor and quaternion concepts fall out of this]. This presents us with a slight issue right away, in that U has four degrees of freedom, but we only need three for $x$. This allows us to apply a constraint on $U.$ But before deciding on that constraint, consider the rate of change of $x$:

$$\dot{x} = \dot{U} e_1 U^\dagger + U e_1 \dot{U}^\dagger$$

Under the special case where $\dot{U} e_1 U^\dagger = U e_1 \dot{U}^\dagger$ (this is our constraint) we can write

$$\dot{x} = 2 \dot{U} e_1 U^\dagger$$

Introduce a new variable s, such that ${dt \over ds} =r= U U^\dagger$

$$2 {dU \over ds} = \dot{x} U e_1$$

$$2 {d^2 U \over {ds}^2} = r \ddot{x} U e_1+ \dot{x} {dU \over ds} e_1 =  (\ddot{x} x+{1 \over 2} \dot{x}^2) U$$

The last step follows because $r  U e_1 = x U$. If we have inverse square central force motion

$$m \ddot{x} = -{k x\over r^3}$$

then we have

$$2 {d^2 U \over {ds}^2} =  (\ddot{x} x+{1 \over 2} \dot{x}^2) U$$

$$= ( -{k \over m} {x^2 \over r^3} + {1 \over 2} \dot{x}^2) U$$

$$=  {1 \over m} ( -{k \over r} + {1 \over 2}m \dot{x}^2) U$$

$${d^2 U \over {ds}^2} = \left( {E \over 2m}  \right) U$$

Harmonic motion. Nice.

Orbit Shape

A mass $m$, at $r$, moving with respect to some origin, defines a plane described by the bivector $L$, the angular momentum:

$$\begin{align*} L &\equiv r \wedge p \\ &= m r \wedge \dot{r}\\ &= m r \wedge (r \dot{\hat{r}} + \dot{r}\hat{r})\\ &= m r^2 (\hat{r} \wedge \dot{\hat{r}}) = m r^2 \hat{r} \dot{\hat{r}} \end{align*}$$

For motion in a potential $V= -{k \over r}$ we have

$$\dot{v} = -{k \over m r^2} \hat{r}$$

Form the product of this with $L$:

$$L\dot{v} = -mr^2\dot{\hat{r}}\hat{r}  \left( {-k \over mr^2} \right) \hat{r} = k\dot{\hat{r}}$$

$${d \over {dt}} (Lv - k \hat{r}) = 0$$

And we see that $Lv- k \hat{r}$ is a constant of the motion. This vector, which sits in the plane of motion $(L \wedge v = 0)$, is known as the Runge-Lenz vector. Let's be explicit about the conserved vector and call it $e$:
$$Lv = k \hat{r} + ke$$

With this definition, $e$ is dimensionless. If we multiply through with $r$ we have

$$\begin{equation} \label{eq2} Lvr = k \hat{r} r + ker  \end{equation}$$

The left hand side is

$$\begin{align} Lvr &= L (v \cdot r + v \wedge r)\\ &= L \left(v \cdot r + \frac{L^\dagger}{m} \right) \\ &= L (v \cdot r) + \frac{l^2}{m} \label{eq3} \end{align}$$

The right hand side is

$$\begin{align*} k \hat{r} r + ker &= k \abs{r} + k(e \cdot r + e \wedge r)   \\ &= k \abs{r} + \abs{e} \abs{r} \cos \theta + e \wedge r \end{align*}$$

So we have, finally

$$l^2/m + (r \cdot v) L = k \abs{r} + k \abs{e} \abs{r} \cos \theta + e \wedge r$$

The scalar part is

$$l^2/m = k \abs{r} + k \abs{e} \abs{r} \cos \theta$$

or

$$\abs{r} = \frac{l^2/mk}{1+\abs{e} \cos \theta}$$

and the bivector part is

$$(r \cdot v) L = e \wedge r$$

or

$$L = \frac{e \wedge r }{r \cdot v}$$

Returning to $\eqref{eq2}$ we can develop a relation between $E$, and $l$ and $e$

$$\begin{align*} (Lv - k \hat{r})^2 &= k^2 e ^2 \\ (Lv)^2 - 2k(Lv) \cdot \hat{r} + k^2 &= k^2 e^2 \\ l^2 v^2 -2k \frac{\left< Lvr \right>}{r} &= k^2 (e^2 -1) \end{align*}$$

Using $\eqref{eq3}$

$$\begin{align*} l^2 v^2 - \frac{2kl^2}{mr} &= k^2 (e^2 -1)\\ \frac{2 l^2}{m} \left( \frac12  m v^2 - \frac kr \right) &= k^2 (e^2-1) \\ E &= \frac{m k^2}{2 l^2} (e^2 -1) \end{align*}$$