Getting the Lorentz algebra through the Clifford algebra

This post shows how to derive the Lorentz algebra from the Clifford algebra $\mathrm{Cl}_{1,3}$. I assume Lorentz transformations take the form $x \mapsto L x L^{-1}$ within the Clifford algebra, and require $L$ to map vectors to vectors. By looking at $L$ close to identity, and imposing this vectors-to-vectors requirement, you can show that the generators must be bivectors. The Lorentz algebra then just follows from the Clifford algebra's multiplication rules.

Intro

The Lorentz group is a group of matrix transformations on four-vectors that preserve their norm under the Minkowski spacetime metric. A four-vector can be represented as a list of four components: $$ V = \begin{bmatrix} V^0 \\ V^1 \\ V^2 \\ V^3 \end{bmatrix} \,. $$ We typically use upper indices for four-vector components. The 0th component is considered a temporal component since it is a time duration in a spacetime displacement, and the other components are considered spatial. The Minkowski spacetime metric is a $4\times4$ matrix: $$ g = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 &-1 & 0 \\ 0 & 0 & 0 &-1 \end{bmatrix} \,. $$ The norm of the four-vector is given by $$ || V || = V^T g V = (V^0)^2 - (V^1)^2 - (V^2)^2 - (V^3)^2 \,. $$ This has important physical interpretations (depending on the four-vector in question), and matrix transformations on four-vectors that preserve this norm are extremely important. A matrix transformation $$ V \mapsto V' = \Lambda V $$ preserves this norm if $$ \Lambda^T g \Lambda = g \,. $$ So the Lorentz group is the group of matrix transformations that preserve the metric—or an isometry group. (iso=same, metry=metric.) If $g$ were the identity matrix (or any constant times identity), we'd have the four-dimensional orthogonal group. By analogy, the Lorentz group is sometimes called the pseudo-orthogonal group on three space and one time dimensions, and is often written $\mathrm{O}(3,1,\mathbb{R})$.

As a manifold, the Lorentz group consists of four connected components. These respectively contain the identity, parity reversal $$ P : (ct,x,y,z) \mapsto (ct,-x,-y,-z) \,, $$ time reversal $$ T : (ct,x,y,z) \mapsto (-ct,x,y,z) $$ and the composition of $PT$. As usual, a nice way of characterizing the group and deriving its representations is to construct the algebra of generators for the part connected to identity. This note will be about a rather indirect way of deriving the generator algebra: by going through the Clifford algebra.

Clifford algebra in 3D

As a warm-up, I'll first show how the 3D rotation group can be built using the 3D Clifford algebra. The algebra can be called $\mathrm{Cl}_3(\mathbb{R})$. Some folks refer to the 3D Clifford algebra as geometric algebra (or vanilla geometric algebra). It's the subject of some lovely YouTube videos; see sudgylacmoe: An Overview of the Operations in Geometric Algebra and Freya Holmér: Why can't you multiply vectors?. (I especially recommend the latter!)

The basic idea is that you treat vectors as algebraic objects that you can simply multiply together. The product of two vectors $V$ and $U$ is defined so that it reduces to the usual dot product when $V=U$: $$ V^2 \equiv VV = V\cdot V \,. $$ If we define unit vectors $e_1, e_2, e_3$, then: $$ \displaylines{ V = V_1 e_1 + V_2 e_2 + V_3 e_3 \\ V^2 = V_1^2 e_1^2 + V_2^2 e_2^2 + V_3^2 e_3^2 \\ + (e_1 e_2 + e_2 e_1) V_1 V_2 + (e_2 e_3 + e_3 e_2) V_2 V_3 + (e_3 e_1 + e_1 e_3) V_3 V_1 \,. } $$ By necessity, then, we require: $$ \displaylines{ e_1^2 = e_2^2 = e_3^2 = 1 \\ e_1 e_2 = -e_2 e_1 \qquad e_2 e_3 = -e_3 e_2 \qquad e_3 e_1 = -e_1 e_3 \,. } $$ These requirements can be summed up in an anticommutation rule: $$ \{e_a, e_b\} \equiv e_a e_b + e_b e_a = 2 \delta_{ab} \,. $$ Under vector multiplication, basis vectors anticommute—you get an negative sign when you swap the order of multiplication—unless they're one and the same basis vector.

Now, when you multiply two different vectors, you no longer just get a dot product: $$ UV = U\cdot V + U \wedge V \,. $$ The second, extra term is called a wedge product. Since it's missing when $U=V$, the wedge product of a vector with itself must be zero. This implies it's antisymmetric: $$ (U+V) \wedge (U+V) = U \wedge V + V \wedge U = 0 \,. $$ The wedge product is a new kind of object: it's not really a vector (since it isn't a linear sum of basis vectors), and it's not a scalar. The set of all wedge products is spanned by products of dissimilar basis vectors, namely $$ \displaylines{ f_1 = e_2 e_3 \qquad f_2 = e_3 e_1 \qquad f_3 = e_1 e_2 \\ f_{a} = \frac{1}{2} \epsilon_{abc} e_b e_c \,. } $$ Since they are specified by dissimilar vectors, they are called bivectors. Although bivectors are three-dimensional, they are different from vectors. They do not transform the same way under parity reversal. Under parity, the basis vectors all transform as $e_a \mapsto -e_a$. The basis bivectors, being a product of two basis vectors, remain invariant under parity. While vectors specify a direction in space, bivectors specify a plane.

Our algebra is not complete yet. Bivectors can be multiplied together, and multiplied by vectors. But there's a limit to the number of objects we can build. In fact, there's just one more basis object we can add—the unit trivector: $$ e_5 = e_1 e_2 e_3 \,. $$ I'll explain why I call this $e_5$ later. Since it's described by all three unit vectors, it can be thought of as describing a volume element. (See Freya Holmér's video for a really nice explanation.) A neat thing about the trivector, by the way, is that it lets us rewrite the basis bivectors as: $$ f_a = e_a e_5 \,. $$ Another really neat thing is that $$ e_5^2 = -1 \,. $$ The trivector behaves something like an imaginary unit! It also commutes with all the basis vectors $e_a$, so the bivectors also square to $-1$: $$ (e_a e_5)^2 = - 1 \,. $$ The bivectors also mutually anticommute, and in fact are isomorphic to the quaternions. I won't dwell on this; I again recommend Freya Holmér's video for more info.

Every element of our algebra is then a linear sum—with scalar coefficients—of the eight basis objects:

Basis elements	Object type	Interpretation	Dimension
$1$	Scalar	Point	1
$e_a$	Vector	Line	3
$e_a e_5$	Bivector	Plane	3
$e_5$	Trivector	Volume	1

Why is this list complete? There are only so many unique strings of basis vectors you can form when you're allowed to swap dissimilar basis vectors (by the anticommutation rule) and annihilate similar basis vectors (by $e_a^2=1$). You can't get a string of more than three basis vectors. For a string of $n$ basis vectors, you have $3$-choose-$n$ combinations. So that's one scalar ($n=0$), three vectors ($n=1$), three bivectors ($n=2$), and one trivector ($n=3$).

The 3D rotation group

So, let's get to rotations. Intuitively, rotations are transformations of spatial objects that preserve their geometry. This includes the norms of vectors and the angles between them. It also includes which planes are generated by which combinations of vectors (although the planes and vectors are all rotated). This preservation of geometry means, in the Clifford algebra formalism, the preservation of multiplication rules: $$ T(xy) = T(x) T(y) \,. $$ One way to achieve this (though not the only way) is for the transformation to take the form $$ T(x) = R x R^{-1} $$ for some invertible Clifford algebra object $R$, which represents the particular rotation. This is called the rotor formula, and $R$ is called a rotor. So scalars, vectors, bivectors and trivectors are all transformed according to the same rule. In fact, scalars remain invariant, since $R$ or $R^{-1}$ can be moved past the scalar: $$ T(s) = R s R^{-1} = R R^{-1} s = s \,. $$ And so the square of any vector—and thus its norm—remains invariant.

There's actually one more requirement to make on rotations. Rotations must map vectors to vectors, and not to other objects such as bivectors. While this might seem too obvious to state, this actually limits the allowed objects $R$.

So our aim is to build a group of invertible Clifford algebra objects $R$ for which $R x R^{-1}$ is a vector whenever $x$ is a vector. How do we do that? Well, we start by considering transformations close to identity: $$ R = e^{-i \theta_a J_a} \approx 1 - i \theta_a J_a \,, $$ where $\theta_a$ are a collection of small parameters and $J_a$ are the rotation generators. Under such a transformation, a vector $V$ transforms as $$ V \mapsto V' = V - i\theta_a [J_a, V] \,. $$ This is where the requirement that $R$ maps vectors to vectors comes in. $V$ and $V'$ must both be vectors, and therefore the commutator $[J_a,V]$ must also be a vector. The elements we can build $J_a$ from are listed in the table above. We can work out the relevant commutation rules to be: $$ \displaylines{ [1, e_b] = 0 \\ [e_a, e_b] = 2 \epsilon_{abc} e_c e_5 \\ [e_a e_5, e_b] = - 2 \epsilon_{abc} e_c \\ [e_5, e_b] = 0 \,. } $$ Now, anything that commutes with $e_b$ gives $V'=V$ in the rotor formula, so cannot be a generator. $e_a$ cannot be a generator, because it gives a bivector. Thus, the bivectors are generators of rotations: $$ J_a \propto e_a e_5 \,. $$ This makes some amount of sense: bivectors represent planes, and rotations occur in planes. Again, see Freya Holmér's video for an especially nice explanation. We can also figure out the proportionality constant since we already know the rotation algebra: $$ [J_a, J_b] = i \epsilon_{abc} J_c \,. $$ Being very careful about minus signs from $e_5^2$ and $i^2$, you can show that: $$ \big[ (-\tfrac{1}{2} i e_a e_5), (-\tfrac{1}{2} i e_b e_5) \big] = i \epsilon_{abc} \big(-\tfrac{1}{2} i e_c e_5\big) $$ so $$ J_a = - \tfrac{1}{2} i e_a e_5 \,. $$ So, we've got generators of the rotation group from the 3D Clifford algebra.

Explicit matrix representation

Here's an explicit matrix representation of the 3D Clifford algebra: $$ \displaylines{ 1 = 1 \\ e_a = \sigma_a \\ e_a e_5 = i \sigma_a \\ e_5 = i } $$ where $\sigma_a$ are the Pauli matrices: $$ \displaylines{ \sigma_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \\ \sigma_2 = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \\ \sigma_3 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \,, } $$ In this representation, we have $$ J_a = \frac{\sigma_a}{2} \,, $$ which is the usual formula in the fundamental rep of $\mathfrak{su}(2,\mathbb{C})$.

In this representation, $e_5=i$ really is the imaginary unit. Regarding the isomorphism between bivectors and quaternions, however, the mapping is actually $$ \mathbf{i} = -e_1 e_5 = - i\sigma_1 \qquad \mathbf{j} = -e_2 e_5 = - i\sigma_2 \qquad \mathbf{k} = -e_3 e_5 = - i\sigma_3 \,. $$ That minus sign is needed to ensure the last part of the famous $$ \mathbf{i}^2 = \mathbf{j}^2 = \mathbf{k}^2 = \mathbf{i} \mathbf{j} \mathbf{k} = -1 $$ is obeyed.

Parity reversal

I mentioned that $x \mapsto R x R^{-1}$ is not the only way to preserve the geometry. Complex conjugation also preserves multiplication rules: $$ (xy)^* = x^* y^* \,. $$ It's a peculiar transformation to be sure. Technically it's only well-defined for a particular matrix representation of the Clifford algebra, rather than for $\mathrm{Cl}_3(\mathbb{R})$ in the abstract.

In the matrix representation above, $1$, $e_1$, $e_3$ and $e_2 e_5$ are real, while $e_5$, $e_1e_5$, $e_3e_5$ and $e_2$ are imaginary. Complex conjugation then flips the sign of the $y$-direction basis vector, as well as the bivectors normal to the $x$- and $z$-axes and the volume basis element. The physical interpretation of this transformation is mirror reflection across the $xz$ plane.

If the mirror reflection is accompanied by a 180 degree rotation in the $xz$ plane, then the signs of all basis vectors will be flipped. This is the parity reversal transformation. The rotor for this transformation is $\pm \sigma_2$, so the parity transformation (in this specific matrix representation) is: $$ P(x) = \sigma_2 x^* \sigma_2 \,. $$ You can check that this transformation flips the signs of all vectors and the trivector, but leaves scalar and bivectors invariant.

To be sure, parity reversal and mirror reflections are not rotations. The rotor formula $x \mapsto R x R^{-1}$ does exhaustively describe the rotation group. The parity transformations that the rotor formula leaves out are isometries, however.

Clifford algebra in 4D

The Clifford algebra $\mathrm{Cl}_{1,3}$ is defined in 4D spacetime. The $1,3$ subscript here signifies 1 time and 3 space dimensions. A four-vector $V$ is a sum of components $V^\mu$ multiplied by basis vectors $\gamma_\mu$: $$ V = V^\mu \gamma_\mu \,. $$ Just as in 3D, we can simply multiply vectors together, with the rules ultimately supervening on the multiplication rules for the basis vectors $\gamma_\mu$.

A quick review about the upper and lower indices. The basis vectors have a lower index, and the vector components have an upper index. A vector can be transformed (e.g. by a Lorentz transformation) either by transforming its components: $$ V^\mu \mapsto V'^\mu = \Lambda^\mu_{\phantom{\mu}\nu} V^\nu $$ or by transforming the basis vectors: $$ \gamma_\mu \mapsto \gamma'_\mu = \Lambda^\nu_{\phantom{\mu}\mu} \gamma_\nu \,. $$ Note that this rule differs, since the first index of the transformation matrix is contracted with the basis vector. In either case we get $$ V' = \Lambda^{\mu}_{\phantom{\mu}\nu} V^\nu \gamma_\mu \,, $$ so these are equivalent. We say that the basis vectors (and anything that transforms the same way) transform covariantly, and the components transform contravariantly. More broadly: lower index = covariant, and upper index = contravariant.

Just as in the 3D case, we want vector multiplication to just give the dot product when we multiply a vector by itself. In this case, the dot product is defined though the Minkowski metric: $$ V\cdot U = g_{\mu\nu} V^\mu U^\nu \,. $$ The product of a vector with itself is: $$ V^2 = V^\mu V^\nu \gamma_\mu \gamma_\nu \,. $$ I could write out all sixteen terms like in the 3D case, but a faster and more compact option is to note that since $V^\mu V^\nu$ is symmetric when $\mu$ and $\nu$ are swapped, $$ V^2 = \frac{1}{2} V^\mu V^\nu \big( \gamma_\mu \gamma_\nu + \gamma_\nu \gamma_\mu \big) = \frac{1}{2} V^\mu V^\nu \{ \gamma_\mu, \gamma_\nu \} \,. $$ Requiring $V^2 = V\cdot V$ then means: $$ \{ \gamma_\mu, \gamma_\nu \} = 2 g_{\mu\nu} \,. $$ This is the fundamental anticommutation rule for the 4D Clifford algebra.

Just as in the 3D case, we can build more basis elements by stringing the basis vectors $\gamma_\mu$ together. There is of course 1 basis scalar and the 4 basis vectors. There are 6 independent bivectors—4 choose 2. Then there are 4 independent trivectors, and 1 quadvector. So there are 16 basis elements altogether. The conventional names for the basis elements (as in the textbooks by Bjorken and Drell, or Itzykson and Zuber) are below:

Basis elements	Object type	Interpretation	Dimension
$1$	Scalar	Point	1
$\gamma_\mu$	Vector	Line	4
$\sigma_{\mu\nu} = \frac{i}{2} [\gamma_\mu, \gamma_\nu]$	Bivector	Plane	6
$\gamma_\mu \gamma_5$	Trivector	3D hypersurface	4
$\gamma_5 = -i\gamma_0 \gamma_1 \gamma_2 \gamma_3$	Quadvector	4D hypervolume	1

The unit hypervolume element is called $\gamma_5$ basically as a historical artifact. Way back in the day, many physicists used a Euclidean metric when dealing with spacetime, but made the temporal component—which back then was called the 4th instead of the 0th component—imaginary. Hence one would have a $\gamma_4$ as a basis vector instead of $\gamma_0$. $5$ was just the next available label. It's because $\gamma_5$ is the hypervolume element in the 4D Clifford algebra that I called the 3D volume element $e_5$. I'm kinda just habituated to thinking of a subscript 5 that way.

Lorentz transformations

So let's get to Lorentz transformations. We'll consider transformations analogous to the 3D rotor formula: $$ \gamma_\mu \mapsto L \gamma_\mu L^{-1} $$ for some matrix $L$, representing the particular Lorentz transformation. We should be able to build $L$ from the Clifford algebra. An $L$ close to identity can be expressed in terms of generators as: $$ L = e^{-i \omega_a \lambda_a} \approx 1 - i \omega_a \lambda_a $$ where $\omega_a$ are a collection of small parameters and $\lambda_a$ are the Lorentz group generators. The basis vectors thus transform, under an $L$ close to identity, as: $$ \gamma_\rho \mapsto \gamma_\rho' = \gamma_\rho - i \omega_a [\lambda_a, \gamma_\rho] \,. $$ Since $\gamma_\rho$ and $\gamma'_\rho$ are both vectors, so is $[\lambda_a, \gamma_\rho]$. This is what will allow us to determine the generators.

The following commutation rules can be eked out, with a little effort, from the basic anticommutation rule and multiplication rules: $$ \displaylines{ [1, \gamma_\rho] = 0 \\ [\gamma_\mu , \gamma_\rho] = -2i \sigma_{\mu\rho} \\ [\sigma_{\mu\nu} , \gamma_\rho] = 2 i \big( g_{\nu\rho} \gamma_\mu - g_{\mu\rho} \gamma_\nu \big) \\ [\gamma_\mu \gamma_5 , \gamma_\rho] = - 2 g_{\mu\rho} \gamma_5 \\ [\gamma_5 , \gamma_\rho] = -2 \gamma_\rho \gamma_5 \,. } $$ The rule for $[1,\gamma_\rho]$ is trivial, so $1$ does not generate any non-trivial transformations. Otherwise, the only of these rules that produces a vector on the right-hand side is the rule for $ [\sigma_{\mu\nu} , \gamma_\rho] $. Thus, $\sigma_{\mu\nu}$ are up to a factor the generators of the Lorentz group. The factor is chosen to be $\frac{1}{2}$, because rotations form a subgroup of the Lorentz group, and this factor will reproduce the standard algebra for this subgroup.

The Lorentz group generators are thus: $$ \Sigma_{\mu\nu} = \frac{1}{2} \sigma_{\mu\nu} = \frac{i}{4} [\gamma_\mu, \gamma_\nu] \,. $$ From the multiplication rules, one can work out the following commutation rules, which define the Lorentz algebra: $$ [ \Sigma_{\mu\nu}, \Sigma_{\alpha\beta} ] = i \Big( g_{\mu\beta} \Sigma_{\nu\alpha} + g_{\nu\alpha} \Sigma_{\mu\beta} - g_{\mu\alpha} \Sigma_{\nu\beta} - g_{\nu\beta} \Sigma_{\mu\alpha} \Big) \,. $$ There are six Lorentz group generators, since there are six bivectors, making the Lorentz group six-dimensional.

The Lorentz group generators can be broken up into rotation generators and boost generators, depending on whether they transform $\gamma_0$. The rotation generators leave $\gamma_0$ alone, and are thus effectively just transformations on space. The rotation generators are given by $$ J_a = \frac{1}{2} \epsilon_{abc} \Sigma_{bc} \,, $$ where the Latin indices range only over spatial components (1,2,3). You can work out from the general Lorentz algebra commutation rule that $$ [J_a, J_b] = i \epsilon_{abc} J_c $$ as expected. The boost generators are then defined as $$ K_a = \Sigma_{a0} \,, $$ and you can work out from the general Lorentz algebra rule that $$ \displaylines{ [K_a, K_b] = - i \epsilon_{abc} J_c \\ [J_a, K_b] = i \epsilon_{abc} K_c \,. } $$ This completes the derivation of the Lorentz algebra through the 4D Clifford algebra.

Parity and time reversal

Though I've already completed the main point of the post, this little epilogue will address the discrete transformations—parity and time reversal—that can't be achieved by compounding transformations arbitrarily close to identity. Parity in the 4D Clifford algebra is nice. The fact that $\gamma_0^2 = 1$ and that $\gamma_0$ anticommutes with the other basis vectors actually makes two-sided multiplication by $\gamma_0$ a perfect parity transformation: $$ P(x) = \gamma_0 x \gamma_0 \,. $$ This flips the sign of any Clifford basis element with an odd number of spatial indices. Since $\gamma_0^2 = 1$, $\gamma_0$ is its own inverse and the parity transformation falls into the same $x \mapsto LxL^{-1}$ pattern as the continuous Lorentz transformations.

Time reversal, like parity in the 3D Clifford algebra, requires complex conjugation, and thus picking out a matrix representation for the Clifford algebra.

There are multiple matrix representations of the 4D Clifford algebra in common use. The appendix of Itzykson and Zuber lists three: The Dirac representation, the chiral representation, and the Majorana representation. My preferred representation is the chiral representation, for reasons I'll probably explore in a future blog post. (In short: it comes out naturally from taking the direct sum of fundamental representations of two inequivalent defining representations of $\mathrm{SL}(2,\mathbb{C})$, which is the universal cover of the part of the Lorentz group connected to identity.)

In the chiral representation, the basis vectors take the form $$ \gamma_0 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \,, \qquad \gamma_a = \begin{bmatrix} 0 & \sigma_a \\ -\sigma_a & 0 \end{bmatrix} \,, $$ where I'm using block matrix notation. The Pauli matrices $\sigma_a$ are $2\times 2$ matrices, each $0$ above is a $2\times2$ matrix with all 0s for components, and $1$ signifies a $2\times2$ identity matrix.

In the chiral representation, complex conjugation flips the sign of $\gamma_2$, but leaves the other basis vectors unchanged. Time reversal, however, should be an operation that flips only the sign of $\gamma_0$. We can get creative here by combining basis vectors to build an operator. $\gamma_0 \gamma_2$ is its own inverse, and it's straightforward to check (using the basic anticommutation rule) that: $$ \displaylines{ (\gamma_0 \gamma_2) \gamma_0 (\gamma_0 \gamma_2) = - \gamma_0 \\ (\gamma_0 \gamma_2) \gamma_1 (\gamma_0 \gamma_2) = \gamma_1 \\ (\gamma_0 \gamma_2) \gamma_2 (\gamma_0 \gamma_2) = - \gamma_2 \\ (\gamma_0 \gamma_2) \gamma_3 (\gamma_0 \gamma_2) = \gamma_3 \,, } $$ so composing two-sided multiplication by $\gamma_0\gamma_2$ with complex conjugation achieves the desired result: $$ T(x) = \gamma_0 \gamma_2 x^* \gamma_0 \gamma_2 \,. $$ And the composition $PT$ will then be: $$ PT(x) = - \gamma_2 x^* \gamma_2 \,, $$ which will flip the signs of all the basis vectors. The action of time reversal and $PT$ on the bivectors, trivectors and quadvector will be a bit peculiar because of the factors $i$ introduced into their conventional definitions. Time reversal is a antilinear transformation, meaning it modifies the usual rule for linearity by taking the complex conjugates of any scalar coefficients: $$ T(ax + by) = a^* T(x) + b^* T(y) \,. $$ This is just something we need to be mindful of when building physical quantities out of the Clifford algebra.