Quaternions and the Spacetime Algebra

The geometric algebra page showed that real quaternions $ℍ$ arise as the even subalgebra of the geometric algebra over $ℝ³$. This page extends that picture to four-dimensional Minkowski spacetime $ℝ^{3,1}$, where we simply include a fourth basis vector $𝐭$ which squares to $-1$. The geometric algebra over this space is a simple extension of the three-dimensional case, producing the "Spacetime Algebra" (STA).

By sheer coincidence the even subalgebra happens to be isomorphic to the complexified quaternions $ℍ(ℂ)$ — sometimes also called "biquaternions". The isomorphism also happens to be very simple: the unit imaginary scalar in the complexified quaternions corresponds to the pseudoscalar $𝐈 = 𝐭𝐱𝐲𝐳$ in the spacetime algebra, and the complex components of the bivectors represent timelike bivectors, which correspond to $𝐈$ times spatial bivectors. That is, we just think of the unit imaginary $i$ as being equivalent to the pseudoscalar $𝐈$.

The basic rotation machinery (reflections, the Cartan–Dieudonné theorem, the sandwich product) carries over unchanged from the three-dimensional case. And just as rotations are represented by unit quaternions in the three-dimensional case, general Lorentz transformations are represented by unit complexified quaternions in this more general algebra.

One crucial new feature is the spinor norm: for complex quaternions, the GA reverse gives $𝐐𝐐̃ = w² + x² + y² + z²$, which is a complex scalar, distinct from the squared Euclidean norm $|w|² + |x|² + |y|² + |z|²$, which is always real. The surprising point is that the normalization condition $𝐐𝐐̃ = 1$ is now two real conditions, because this says that the "imaginary" part is zero. This reduces the eight real degrees of freedom in a complex quaternion down to six, which is the correct number for a Lorentz transformation in four dimensions.

The spacetime algebra

The spacetime algebra (STA) is the geometric algebra over four-dimensional Minkowski space, and we use the metric signature ${-}{+}{+}{+}$. The basis vectors $𝐭, 𝐱, 𝐲, 𝐳$ satisfy

\[𝐭^2 = -1, \qquad 𝐱^2 = 𝐲^2 = 𝐳^2 = +1,\]

and all other products between distinct basis vectors anticommute. The full algebra has dimension $2^4 = 16$, with basis elements grouped by grade:

\[\begin{array}{lll} \text{grade 0:} & 1 \text{ element} & \boldsymbol{1} \\[4pt] \text{grade 1:} & 4 \text{ elements} & 𝐭,\; 𝐱,\; 𝐲,\; 𝐳 \\[4pt] \text{grade 2:} & 6 \text{ elements} & 𝐭𝐱,\; 𝐭𝐲,\; 𝐭𝐳,\; 𝐱𝐲,\; 𝐱𝐳,\; 𝐲𝐳 \\[4pt] \text{grade 3:} & 4 \text{ elements} & 𝐭𝐱𝐲,\; 𝐭𝐱𝐳,\; 𝐭𝐲𝐳,\; 𝐱𝐲𝐳 \\[4pt] \text{grade 4:} & 1 \text{ element} & 𝐈 = 𝐭𝐱𝐲𝐳 \end{array}\]

The grade-2 elements split into two qualitatively different groups. The three spatial bivectors $𝐱𝐲$, $𝐱𝐳$, $𝐲𝐳$ involve only spatial directions and square to $-1$, exactly as in $ℝ³$. The three boost bivectors $𝐭𝐱$, $𝐭𝐲$, $𝐭𝐳$ each involve the timelike direction, and they square to $+1$ because of the sign in $𝐭^2 = -1$:

\[(𝐭𝐱)^2 = 𝐭𝐱𝐭𝐱 = -𝐭𝐭𝐱𝐱 = -(𝐭^2)(𝐱^2) = -(-1)(+1) = +1.\]

This sign difference is what distinguishes boosts from rotations.

The pseudoscalar in four dimensions

The pseudoscalar $𝐈 = 𝐭𝐱𝐲𝐳$ has several properties that differ markedly from its three-dimensional counterpart:

It still squares to $-1$. Though the inclusion of an extra basis factor gives the permutation a different sign, the fact that $𝐭² = -1$ means that the overall result is unchanged.

Its reverse is itself. The reverse of a grade-$r$ blade picks up a sign $(-1)^{r(r-1)/2}$. For $r = 4$, this is $(-1)^6 = +1$, so

\[\widetilde{𝐈} = +𝐈.\]

Contrast the (grade 3) three-dimensional pseudoscalar $𝐈_{3}$, for which $\widetilde{𝐈}_{3} = -𝐈_{3}$.

It does not commute with everything. Whereas $𝐈_{3}$ commutes with everything in its three-dimensional algebra, $𝐈$ does not commute with everything in the four-dimensional algebra. In particular, $𝐈$ anti-commutes with vectors and trivectors. However, it does commute with every element of the even subalgebra (grades 0, 2, and 4), which is an important property.

The even subalgebra: Complexified quaternions

Products of an even number of vectors — grades 0, 2, and 4 — form a closed subalgebra, the even subalgebra, spanned by $\{1, 𝐱𝐲, 𝐲𝐳, 𝐳𝐱, 𝐭𝐱, 𝐭𝐲, 𝐭𝐳, 𝐈\}$. This algebra happens to be precisely isomorphic to the one spanned by $\{1, 𝐱𝐲, 𝐲𝐳, 𝐳𝐱, i𝐱𝐲, i𝐲𝐳, i𝐳𝐱, i\}$, where $i$ is the unit imaginary — which is of course precisely the algebra of complexified quaternions $ℍ(ℂ)$.

Pseudoscalar → imaginary unit. Since $𝐈^2 = -1$ and $𝐈$ commutes with all scalars, bivectors, and the pseudoscalar, it plays the role of the imaginary unit $i = \sqrt{-1}$ on the even subalgebra. Thus, we have this basic identification:

\[i \;\leftrightarrow\; 𝐈 = 𝐭𝐱𝐲𝐳.\]

We will abuse notation and terminology to write complex numbers interchangeably as both $a + i b$ and $a + 𝐈 b$, and speak of linear combinations of $\{1,i\}$ and $\{1,𝐈\}$ as complex numbers.

Spatial bivectors → quaternion units. The three spatial bivectors that generated right-handed rotations in $ℝ³$ are unchanged:

\[𝐢 = 𝐳𝐲, \qquad 𝐣 = 𝐱𝐳, \qquad 𝐤 = 𝐲𝐱.\]

They still square to $-1$ and satisfy $𝐢𝐣𝐤 = -\boldsymbol{1}$.

Boost bivectors → $i \times$ quaternion units. The three boost bivectors are products of $𝐈$ with the spatial bivectors. For example:

\[𝐈\,𝐢 = (𝐭𝐱𝐲𝐳)(𝐳𝐲) = 𝐭𝐱𝐲(𝐳𝐳)𝐲 = 𝐭𝐱𝐲𝐲 = 𝐭𝐱(𝐲^2) = 𝐭𝐱,\]

and similarly $𝐈\,𝐣 = 𝐭𝐲$ and $𝐈\,𝐤 = 𝐭𝐳$. In the complexified quaternion picture these are $i𝐢$, $i𝐣$, $i𝐤$. Consistent with $(𝐭𝐱)^2 = +1$, we have $(i𝐢)^2 = i^2 𝐢^2 = (-1)(-1) = +1$, and so on.

A general element of the even subalgebra is therefore a quaternion

\[\begin{aligned} 𝐐 &= w\,\boldsymbol{1} + x\,𝐢 + y\,𝐣 + z\,𝐤 \\ &= \Re\{w\}\,\boldsymbol{1} + \Re\{x\}\,𝐳𝐲 + \Re\{y\}\,𝐱𝐳 + \Re\{z\}\,𝐲𝐱 + \Im\{x\}\,𝐭𝐱 + \Im\{y\}\,𝐭𝐲 + \Im\{z\}\,𝐭𝐳 + \Im\{w\}\,𝐈 \end{aligned}\]

with complex coefficients $(w, x, y, z) \in \mathbb{C}$, represented by the Quaternion{Complex{T}} type in this package.

The reverse and the spinor norm

As usual in Geometric Algebra, the reverse of a multivector is computed by reversing the order of all products. Obviously, the reverse of a scalar is itself, and — as mentioned above — the reverse of the pseudoscalar is itself. The reverse applied to a bivector is just negation. Therefore, the reverse applied to a general complex quaternion is exactly the same as for a real quaternion: the scalar part is unchanged, and the vector part is negated. The complex components themselves are not conjugated:

\[\widetilde{𝐐} = w - x\,𝐢 - y\,𝐣 - z\,𝐤 = \texttt{conj(Q)},\]

exactly as for real quaternions.

Now, we have to be careful to distinguish between two different notions of "norm" for complex quaternions: the spinor norm and the Euclidean norm. The (squared) spinor norm,

\[𝐐\,\widetilde{𝐐} = w^2 + x^2 + y^2 + z^2,\]

is the one that arises from the GA reverse, and it is the physically meaningful notion of "unit" for Lorentz transformations. Notably, the result is a complex number. On the other hand, the squared Euclidean norm, which treats the four complex components as a vector in $ℂ⁴$,

\[|w|^2 + |x|^2 + |y|^2 + |z|^2,\]

is always a nonnegative real number, and does not have a direct geometric meaning in this context. (It equals the scalar part of the product of $𝐐$ with its reverse after complex conjugation of the components, but the other parts of that product are generally nonzero.) The relevant condition that defines a spinor (in this case, a Lorentz rotor) is that the spinor norm equals 1:

\[𝐐\,\widetilde{𝐐} = 1 + 0i.\]

The code functions abs2(Q) and abs(Q) compute the spinor norm and its (complex) square root.

One possibly surprising consequence of this is that a pure phase $\exp(𝐈\alpha)$ is not generally a Lorentz rotor. Its spinor norm is

\[\exp(𝐈\alpha)\,\widetilde{\exp(𝐈\alpha)} = \exp(𝐈\alpha)^2 = \exp(2𝐈\alpha),\]

which is not equal to $1$ in general. Phase factors therefore do not belong to the Lorentz rotor group, even though they are unit elements under the Euclidean norm ($|\exp(i\alpha)| = 1$).

Rotors in the STA

A rotation rotor is an element of the even subalgebra with purely real components and spinor norm 1. It has the same form as in the three-dimensional case:

\[R = \exp\!\left(\frac{\vartheta}{2}\,\hat{B}\right) = \cos\frac{\vartheta}{2} + \sin\frac{\vartheta}{2}\,\hat{B},\]

where $\hat{B}$ is any unit spatial bivector, such as $𝐳𝐲$, $𝐱𝐳$, $𝐲𝐱$, or unit-norm combinations of them. The components are real, and the spinor norm happens to equal the Euclidean norm: $R\widetilde{R} = \cos^2(\vartheta/2) + \sin^2(\vartheta/2) = 1$.

A boost in the $𝐯$-direction by rapidity $\varphi$ uses the boost bivector $𝐭𝐯$:

\[B = \exp\!\left(\frac{\varphi}{2}\,𝐭𝐯\right) = \cosh\frac{\varphi}{2} + \sinh\frac{\varphi}{2}\,𝐭𝐯.\]

The sign convention $+\varphi/2$ makes a positive rapidity correspond to a boost in the positive $𝐯$-direction.^[boost] The spinor norm is

\[B\,\widetilde{B} = \cosh^2\!\frac{\varphi}{2} + \left(𝐭𝐯\sinh\frac{\varphi}{2}\right)^{\!2} = \cosh^2\!\frac{\varphi}{2} - \sinh^2\!\frac{\varphi}{2} = 1,\]

while the squared Euclidean norm is $\cosh^2(\varphi/2) + \sinh^2(\varphi/2) = \cosh\varphi \neq 1$ for $\varphi \neq 0$. This concretely shows why the spinor norm is the physically correct notion of "unit" for Lorentz transformations.

The transformations are active: $B\,𝐰\,\widetilde{B}$ is the boosted vector $𝐰$, expressed in the original frame. For example, the boost with velocity $\vec{v}$ maps the 4-velocity $𝐭$ of a particle at rest to the 4-velocity $\gamma\,(𝐭 + \vec{v})$ of a particle moving with velocity $\vec{v}$. The components of a fixed vector as seen by an observer moving with velocity $\vec{v}$ are instead given by the inverse boost. Products act from right to left: $(Λ_1 Λ_2)\, 𝐰\, \widetilde{(Λ_1 Λ_2)} = Λ_1 \bigl(Λ_2\, 𝐰\, \widetilde{Λ_2}\bigr) \widetilde{Λ_1}$.

Iwasawa's $KA\,N$ decomposition

The Iwasawa decomposition is a factorization of the Lorentz group or its double cover $\mathrm{Spin}^+(3,1)$ (or any semisimple Lie group) into three subgroups:

\[G = {KA\,N}.\]

Our implementation of this decomposition relies on a choice of basis unit vectors: we specifically use the $𝐭$ and $𝐳$ basis vectors, and the boost plane $𝐭𝐳$ that they span. We use those to pick out boosts along that spatial direction, and null rotations that fix the corresponding null vector. Specifically, we have the following subgroups:

• $K$ is the maximal compact subgroup, which in the case of the Lorentz group is isomorphic to the rotation group $\mathrm{SO}(3)$, and in the case of $\mathrm{Spin}^+(3,1)$ is isomorphic to $\mathrm{Spin}(3)$.
• $A$ is the abelian subgroup of boosts in a fixed direction. We take this direction to be the z-axis, so that we have $A = \left\{ \exp\left[\tfrac{φₐ}{2} \, 𝐭𝐳 \right] \mid φₐ ∈ ℝ \right\}$.
• $N$ is the nilpotent subgroup of null rotations, which consists of all Lorentz transformations that can be represented as null rotations about a fixed null vector. We take this vector to be the null vector $\boldsymbol{ℓ}$, so that we have $N = \left\{ \exp\left[\tfrac{1}{2} \boldsymbol{ℓ ξ}\right] \mid \boldsymbol{ξ} = ξˣ𝐱 + ξʸ𝐲 \right\}$, where $\boldsymbol{ℓ} = (𝐭+𝐳)/√2$.

Note that $φₐ$ here is not the rapidity of the overall boost if we factor a transformation just as a boost and a rotation. Rather, it is the rapidity of the particular boost picked out as $A$ by this construction. $N$ also contributes to the overall boost, so the overall rapidity is not just $φₐ$.

One nice feature of this decomposition is that we can compute the factorization of a given Lorentz transformation $Λ$ fairly simply. We first introduce the idempotent

\[u₊ = \frac{1}{2} (1 + 𝐭𝐳).\]

It is easy to verify $u₊𝐭𝐳 = 𝐭𝐳u₊ = u₊$ and hence the idempotent identity: $u₊² = u₊$. Next, we write

\[Λ = Rₖ Rₐ Rₙ,\]

where $Rₖ ∈ K$, $Rₐ ∈ A$, and $Rₙ ∈ N$. Starting with the rightmost factor, we have

\[Rₙ = \exp\left[\tfrac{1}{2} \boldsymbol{ℓ ξ}\right] = 1 + \frac{1}{2} \boldsymbol{ℓ ξ},\]

because $\boldsymbol{ℓ}$ and $\boldsymbol{ξ}$ anticommute, and because $\boldsymbol{ℓ}² = 0$, their product $\boldsymbol{ℓ ξ}$ itself is also nilpotent. (Remember, that's what the "N" stands for.) We can simply compute the product $Rₙ u₊$ by using the fact that $\boldsymbol{ξ} u₊ = u₊ \boldsymbol{ξ}$ and expanding terms in our basis, then find

\[Rₙ u₊ %&= \left(1 + \frac{1}{2} \boldsymbol{ℓ ξ}\right) u₊, \\ % &= u₊ + \frac{1}{2} \boldsymbol{ℓ ξ} u₊, \\ % &= u₊ + \frac{1}{2} \boldsymbol{ℓ} u₊ \boldsymbol{ξ}, \\ % &= u₊ + \frac{1}{4\sqrt{2}} (𝐭+𝐳) (1+𝐭𝐳) \boldsymbol{ξ}, \\ % &= u₊ + \frac{1}{4\sqrt{2}} (𝐭+𝐳+𝐭𝐭𝐳+𝐳𝐭𝐳) \boldsymbol{ξ}, \\ % &= u₊ + \frac{1}{4\sqrt{2}} (𝐭+𝐳-𝐳-𝐭) \boldsymbol{ξ}, \\ = u₊.\]

Now, with

\[Rₐ = \exp\left[\tfrac{φₐ}{2} \, 𝐭𝐳 \right] = \cosh\left(\frac{φₐ}{2}\right) + \sinh\left(\frac{φₐ}{2}\right) 𝐭𝐳,\]

and the fact that $𝐭𝐳u₊ = u₊$, we have

\[\begin{aligned} Rₐ u₊ &= \left[\cosh\left(\frac{φₐ}{2}\right) + \sinh\left(\frac{φₐ}{2}\right)\right] u₊ \\ &= e^{φₐ/2} u₊. \end{aligned}\]

Putting these together, we have

\[Λ u₊ = e^{φₐ/2} Rₖ u₊.\]

Recall that $e^{φₐ/2}$ is a strictly positive real number, and $Rₖ$ is a pure rotation so it is "ℂ-real" — meaning that it is a linear combination of only basis elements that do not have a factor of $𝐭$ — whereas $u₊$ is just a combination of 1 and a "ℂ-imaginary" part. Therefore, we can take the real part of this expression to find

\[ℂ\Re\{Λ u₊\} = \frac{e^{φₐ/2}}{2} Rₖ.\]

This is just a strictly positive real number times $Rₖ$, which we know must have unit magnitude, so we can find $Rₖ$ by normalizing this real part:

\[Rₖ = \mathrm{normalize}\left(ℂ\Re\{Λ u₊\}\right).\]

We can continue solving for the remaining factors $Rₐ$ and $Rₙ$. One option would be to take the norm of the expression above and solve for $φₐ$, but it requires no transcendental functions and is more numerically stable to extract the components directly from the remaining product. We can evaluate $RₐRₙ = Rₖ^{-1} Λ$. But now, let us examine the left-hand side analytically:

\[Rₐ Rₙ = \left[\cosh\left(\frac{φₐ}{2}\right) + \sinh\left(\frac{φₐ}{2}\right) 𝐭𝐳\right] \left[1 + \frac{1}{2} \boldsymbol{ℓ ξ}\right]\]

The $\boldsymbol{ℓ ξ}$ term is entirely within the span of $\{𝐭𝐱, 𝐭𝐲, 𝐳𝐱, 𝐳𝐲\}$; therefore, it is not hard to show that the scalar and $𝐭𝐳$ terms in this product equal $Rₐ$. As usual, we write $⟨·⟩₀$ to extract the scalar (grade‑0) part, and we have

\[Rₐ = \left⟨ Rₖ^{-1} Λ \right⟩₀ + \left⟨ Rₖ^{-1} Λ\, 𝐭𝐳 \right⟩₀\, 𝐭𝐳.\]

Finally, we can find the remaining factor by rearranging the original factorization:

\[Rₙ = Rₐ^{-1} Rₖ^{-1} Λ.\]

This decomposition is implemented in the Quaternionic.KAN function.

The functions that implement these constructions are described on the Lorentz transformations page.

Further reading

The spacetime algebra and its application to relativistic physics are developed in detail in Doran and Lasenby [7].