[Click here for a PDF version of this post]

I’ve been enjoying XylyXylyX’s QED Prerequisites Geometric Algebra: Spacetime YouTube series, which is doing a thorough walk through of [1], filling in missing details. The last episode QED Prerequisites Geometric Algebra 15: Complex Structure, left things with a bit of a cliff hanger, mentioning a “canonical” form for STA bivectors that was intriguing.

The idea is that STA bivectors, like spacetime vectors can be spacelike, timelike, or lightlike (i.e.: positive, negative, or zero square), but can also have a complex signature (squaring to a 0,4-multivector.)

The only context that I knew of that one wanted to square an STA bivector is for the electrodynamic field Lagrangian, which has an $F^2$ term. In no other context, was the signature of $F$, the electrodynamic field, of interest that I knew of, so I’d never considered this “Canonical form” representation.

Here are some examples:
$$\begin{equation}\label{eqn:canonicalbivectors:20} \begin{aligned} F &= \gamma_{10}, \quad F^2 = 1 \\ F &= \gamma_{23}, \quad F^2 = -1 \\ F &= 4 \gamma_{10} + \gamma_{13}, \quad F^2 = 15 \\ F &= \gamma_{10} + \gamma_{13}, \quad F^2 = 0 \\ F &= \gamma_{10} + 4 \gamma_{13}, \quad F^2 = -15 \\ F &= \gamma_{10} + \gamma_{23}, \quad F^2 = 2 I \\ F &= \gamma_{10} – 2 \gamma_{23}, \quad F^2 = -3 + 4 I. \end{aligned} \end{equation}$$
You can see in this table that all the $F$’s that are purely electric, have a positive signature, and all the purely magnetic fields have a negative signature, but when there is a mix, anything goes. The idea behind the canonical representation in the paper is to write
$$\begin{equation}\label{eqn:canonicalbivectors:40} F = f e^{I \phi}, \end{equation}$$
where $f^2$ is real and positive, assuming that $F$ is not lightlike.

The paper gives a formula for computing $f$ and $\phi$, but let’s do this by example, putting all the $F^2$’s above into their complex polar form representation, like so
$$\begin{equation}\label{eqn:canonicalbivectors:60} \begin{aligned} F &= \gamma_{10}, \quad F^2 = 1 \\ F &= \gamma_{23}, \quad F^2 = 1 e^{\pi I} \\ F &= 4 \gamma_{10} + \gamma_{13}, \quad F^2 = 15 \\ F &= \gamma_{10} + \gamma_{13}, \quad F^2 = 0 \\ F &= \gamma_{10} + 4 \gamma_{13}, \quad F^2 = 15 e^{\pi I} \\ F &= \gamma_{10} + \gamma_{23}, \quad F^2 = 2 e^{(\pi/2) I} \\ F &= \gamma_{10} – 2 \gamma_{23}, \quad F^2 = 5 e^{ (\pi – \arctan(4/3)) I} \end{aligned} \end{equation}$$

Since we can put $F^2$ in polar form, we can factor out half of that phase angle, so that we are left with a bivector that has a positive square. If we write
$$\begin{equation}\label{eqn:canonicalbivectors:80} F^2 = \Abs{F^2} e^{2 \phi I}, \end{equation}$$
we can then form
$$\begin{equation}\label{eqn:canonicalbivectors:100} f = F e^{-\phi I}. \end{equation}$$

If we want an equation for $\phi$, we can just write
$$\begin{equation}\label{eqn:canonicalbivectors:120} 2 \phi = \mathrm{Arg}( F^2 ). \end{equation}$$
This is a bit better (I think) than the form given in the paper, since it will uniformly rotate $F^2$ toward the positive region of the real axis, whereas the paper’s formula sometimes rotates towards the negative reals, which is a strange seeming polar form to use.

Let’s compute $f$ for $F = \gamma_{10} – 2 \gamma_{23}$, using
$$\begin{equation}\label{eqn:canonicalbivectors:140} 2 \phi = \pi – \arctan(4/3). \end{equation}$$
The exponential expands to
$$\begin{equation}\label{eqn:canonicalbivectors:160} e^{-\phi I} = \inv{\sqrt{5}} \lr{ 1 – 2 I }. \end{equation}$$

Multiplying each of the bivector components by $1 – 2 I$, we find
$$\begin{equation}\label{eqn:canonicalbivectors:180} \begin{aligned} \gamma_{10} \lr{ 1 – 2 I} &= \gamma_{10} – 2 \gamma_{100123} \\ &= \gamma_{10} – 2 \gamma_{1123} \\ &= \gamma_{10} + 2 \gamma_{23}, \end{aligned} \end{equation}$$
and
$$\begin{equation}\label{eqn:canonicalbivectors:200} \begin{aligned} – 2 \gamma_{23} \lr{ 1 – 2 I} &= – 2 \gamma_{23} + 4 \gamma_{230123} \\ &= – 2 \gamma_{23} + 4 \gamma_{23}^2 \gamma_{01} \\ &= – 2 \gamma_{23} + 4 \gamma_{10}, \end{aligned} \end{equation}$$
leaving
$$\begin{equation}\label{eqn:canonicalbivectors:220} f = \sqrt{5} \gamma_{10}, \end{equation}$$
so the canonical form is
$$\begin{equation}\label{eqn:canonicalbivectors:240} F = \gamma_{10} – 2 \gamma_{23} = \sqrt{5} \gamma_{10} \frac{1 + 2 I}{\sqrt{5}}. \end{equation}$$

It’s interesting here that $f$, in this case, is a spatial bivector (i.e.: pure electric field), but that clearly isn’t always going to be the case, since we can have a case like,
$$\begin{equation}\label{eqn:canonicalbivectors:260} F = 4 \gamma_{10} + \gamma_{13} = 4 \gamma_{10} + \gamma_{20} I, \end{equation}$$
from the table above, that has both electric and magnetic field components, yet is already in the canonical form, with $F^2 = 15$. The canonical $f$, despite having a positive square, is not necessarily a spatial bivector (as it may have both grades 1,2 in the spatial representation, not just the electric field, spatial grade-1 component.)

References

[1] Justin Dressel, Konstantin Y Bliokh, and Franco Nori. Spacetime algebra as a powerful tool for electromagnetism. Physics Reports, 589:1–71, 2015.