The Lorentz Group

Here, our focus is on using Geometric Algebra (GA) to understand and implement Lorentz transformations — and to decompose those transformations into conventional pieces. In the context of spacetime and Lorentz transformations, GA allows us to work with null tetrads and the Lorentz group in a more intuitive and geometrically meaningful way. This treatment ties in surprisingly neatly with the approach via stereographic coordinates and conformal transformations [9–11], while retaining clearer connections to the underlying geometry.

The standard basis and the null tetrad

We start with a standard basis of spacetime vectors, which we denote as $(𝐭, 𝐱, 𝐲, 𝐳)$. We use signature ${-}{+}{+}{+}$, meaning that

\[𝐭² = -1, \qquad 𝐱² = 𝐲² = 𝐳² = +1,\]

and all other products are zero. We then define the null tetrad

\[\begin{aligned} \boldsymbol{\ell} &= \frac{𝐭+𝐳}{\sqrt{2}}, \\ 𝐦 &= \frac{𝐱+𝐈₃𝐲}{\sqrt{2}}, \\ 𝐦̄ &= \frac{𝐱-𝐈₃𝐲}{\sqrt{2}}, \\ 𝐧 &= \frac{𝐭-𝐳}{\sqrt{2}}, \end{aligned}\]

where $𝐈₃ = 𝐈𝐭 = 𝐱𝐲𝐳$ is the spatial pseudoscalar, and replaces the $i$ used in the Newman-Penrose formalism. See below for details about this replacement, but the important features here are that $𝐈₃²=-1$, and it commutes with spatial vectors but anticommutes with $𝐭$. This means that the null tetrad is indeed null,

\[\boldsymbol{\ell}² = 𝐦² = 𝐦̄² = 𝐧² = 0,\]

and the only nonzero inner products are

\[\boldsymbol{\ell} \cdot 𝐧 = -1, \qquad 𝐦 \cdot 𝐦̄ = 1.\]

This simple tetrad is aligned with the axes of the standard basis, but we can apply any Lorentz transformation to it to get a more general null tetrad. In particular, we can apply an ordinary rotation to "point" $\boldsymbol{\ell}$ in any direction we like. This suggests the standard factorization of the Lorentz group, beginning with rotations tied to the spherical coordinates, followed by rotations about the radial direction, followed by boosts in the radial direction, and finally followed by null rotations about the rotated $\boldsymbol{\ell}$. We will discuss these transformations in more detail below, but the important point is that we decompose a general Lorentz transformation specifically with respect to either $\boldsymbol{\ell}$ or $𝐧$.

The Lorentz group

Before we decompose the (proper orthochronous) Lorentz group, we need to understand how the Lorentz group shows up in Geometric Algebra.

Null rotations

A null rotation is a very particular type of Lorentz transformation (sometimes called a parabolic transformation) that leaves a chosen null vector invariant. For simplicity, let us choose the null vector

\[\boldsymbol{\ell} = \frac{𝐭+𝐳}{\sqrt{2}}\]

as the invariant null vector. Its complementary null vector is

\[𝐧 = \frac{𝐭-𝐳}{\sqrt{2}},\]

which will be important. Now, for any (not necessarily unit) vector $\boldsymbol{ξ}$ in the $𝐱$-$𝐲$ plane, the bivector $\boldsymbol{\ell ξ} = -\boldsymbol{ξ \ell}$ generates a null rotation. More specifically, this bivector generates a boost in the $\boldsymbol{ξ}$ direction, and simultaneously a rotation in the $\boldsymbol{ξ}$-$𝐳$ plane. Define the spinor

\[𝐑 = \exp\left[ \frac{1}{2} \boldsymbol{\ell ξ} \right].\]

Because $\boldsymbol{\ell}² = 0$, the exponential series terminates after the second term, and we have

\[𝐑 = 1 + \frac{1}{2} \boldsymbol{\ell ξ},\]

which makes calculations particularly simple. The results on the basis $(𝐭, 𝐱, 𝐲, 𝐳)$ are not enlightening, but the results on the $(\boldsymbol{\ell}, 𝐧, 𝐦, 𝐦̄)$ basis are very interesting:

\[\begin{aligned} 𝐑 \boldsymbol{\ell} \bar{𝐑} &= \boldsymbol{\ell}, \\ 𝐑 𝐦 \bar{𝐑} &= 𝐦 + 𝐱\boldsymbol{ξ\ell}, \\ 𝐑 𝐦̄ \bar{𝐑} &= 𝐦̄ - 𝐱\boldsymbol{ξ\ell}, \\ 𝐑 𝐧 \bar{𝐑} &= 𝐧 + \boldsymbol{ξ} + \frac{1}{2} ξ² \boldsymbol{\ell}. \end{aligned}\]

\[\begin{aligned} 𝐑 \boldsymbol{\ell} \bar{𝐑} &= \left(1 + \frac{1}{2} \boldsymbol{\ell ξ} \right) \boldsymbol{\ell} \left(1 - \frac{1}{2} \boldsymbol{\ell ξ} \right) \\ &= \boldsymbol{\ell} + \frac{1}{2} \boldsymbol{\ell ξ \ell} - \frac{1}{2} \boldsymbol{\ell \ell ξ} - \frac{1}{4} \boldsymbol{\ell ξ \ell \ell ξ} \\ &= \boldsymbol{\ell}, \end{aligned}\]

\[\begin{aligned} 𝐑 𝐧 \bar{𝐑} &= \left(1 + \frac{1}{2} \boldsymbol{\ell ξ} \right) 𝐧 \left(1 - \frac{1}{2} \boldsymbol{\ell ξ} \right) \\ &= 𝐧 + \frac{1}{2} \boldsymbol{\ell ξ} 𝐧 - \frac{1}{2} 𝐧 \boldsymbol{\ell ξ} - \frac{1}{4} \boldsymbol{\ell ξ 𝐧 \ell ξ} \\ &= 𝐧 - \frac{1}{2} \boldsymbol{(\ell 𝐧 + 𝐧 \ell) ξ} - \frac{1}{4} \boldsymbol{\ell 𝐧 \ell ξ²} \\ &= 𝐧 + \boldsymbol{ξ} + \frac{1}{2} \boldsymbol{ξ}² \boldsymbol{\ell}, \end{aligned}\]

\[\begin{aligned} 𝐑 𝐱 \bar{𝐑} &= \left(1 + \frac{1}{2} \boldsymbol{\ell ξ} \right) 𝐱 \left(1 - \frac{1}{2} \boldsymbol{\ell ξ} \right) \\ &= 𝐱 + \frac{1}{2} \boldsymbol{\ell ξ} 𝐱 - \frac{1}{2} 𝐱 \boldsymbol{\ell ξ} - \frac{1}{4} \boldsymbol{\ell ξ 𝐱 \ell ξ} \\ &= 𝐱 + \frac{1}{2} \boldsymbol{\ell ξ} 𝐱 + \frac{1}{2} \boldsymbol{\ell 𝐱 ξ} - \frac{1}{4} \boldsymbol{\ell² ξ 𝐱 ξ} \\ &= 𝐱 + \boldsymbol{\ell (ξ \cdot 𝐱)} \end{aligned}\]

\[\begin{aligned} 𝐑 𝐲 \bar{𝐑} &= 𝐲 + \boldsymbol{\ell (ξ \cdot 𝐲)} \end{aligned}\]

\[\begin{aligned} 𝐑 𝐈₃𝐲 \bar{𝐑} &= \left(1 + \frac{1}{2} \boldsymbol{\ell ξ} \right) 𝐈₃𝐲 \left(1 - \frac{1}{2} \boldsymbol{\ell ξ} \right) \\ &= 𝐈₃𝐲 + \frac{1}{2} \boldsymbol{\ell ξ} 𝐈₃𝐲 - \frac{1}{2} 𝐈₃𝐲 \boldsymbol{\ell ξ} - \frac{1}{4} \boldsymbol{\ell ξ 𝐈₃𝐲 \ell ξ} \\ &= 𝐈₃𝐲 + \frac{1}{2} \boldsymbol{\ell ξ} 𝐈₃𝐲 - \frac{1}{2} 𝐈₃𝐲 \boldsymbol{\ell ξ} - \frac{1}{4} \boldsymbol{\ell ξ 𝐈₃𝐲 \ell ξ} \\ \end{aligned}\]

Reinterpreting $i$

TL;DR: $i ∈ ℂ$ is replaced by $𝐈₃ = 𝐈𝐭$, the spatial pseudoscalar. It actually transforms whenever $𝐭$ transforms, but if we just write expressions in terms of $𝐈₃' = 𝐈𝐭'$ without explicitly transforming $𝐭$, that should be fine, because the interpretation of $i$ also needs to change. And at that point, it's just a bookkeeping device, so we don't need to worry about the fact that it transforms. The reason $𝐈₃$ appears is because it is central in the spatial subalgebra; it commutes with everything, which is why it can act like $i$.

The unit imaginary $i ∈ ℂ$ is a purely algebraic object that has no geometric meaning to Newman and Penrose. In Geometric Algebra, we try to identify the geometric meaning of all algebraic objects. But the replacement for $𝐦$ is not so clear. We need something that ensures $𝐦𝐦=0$, while also transforming reasonably under null rotations. The obvious guess is $i↦𝐈$, which is invariant under (proper, orthochronous) Lorentz transformations. Unfortunately, $(𝐱+𝐈𝐲)²$ simply does not have zero scalar part. The next obvious guess is $i↦𝐱𝐲$, the pseudoscalar of the "screen" space that $𝐦$ represents. Unfortunately, $(𝐱+𝐱𝐲𝐲)=2𝐱$, which also obviously does not square to zero. Finally, we come to $𝐈₃=𝐈𝐭$. This does actually work correctly, with the caveat that $𝐭$ also transforms; when we transform a quantity involving $i$, we have to remember that $i$ will have new meaning in the new frame.

This is almost the null tetrad used in, e.g., the Newman-Penrose formalism, except our definitions of $𝐦$ and $𝐦̄$ do not use the unit imaginary $i ∈ ℂ$, but rather the unit pseudoscalar $𝐈 ∈ 𝒢(ℝ^{3,1})$. In fact, with these definitions, $𝐦$ and $𝐦̄$ are not even vectors, but more general multivectors. This makes almost no difference to the calculations, but it does allow us to work entirely within the geometric algebra, without the gratuitous and geometrically meaningless use of complex numbers in just part of the tetrad.