Quaternions as Geometric Algebra

Quaternions arise naturally as the even subalgebra of the geometric algebra over three-dimensional Euclidean space. This page derives the conventions used throughout the package from first principles — starting from the geometric product — so that every sign choice, basis assignment, and rotation formula has a clear geometric justification.

The treatment here focuses on real quaternions. The extension to complex quaternions (needed for Lorentz boosts via the spacetime algebra) is covered on a separate page.

The geometric algebra over $ℝ³$

We start with the standard right-handed Cartesian coordinate system and its unit basis vectors $(𝐱, 𝐲, 𝐳)$. The geometric product of two vectors $𝐯$ and $𝐰$ is defined by

\[𝐯 𝐰 = 𝐯 ⋅ 𝐰 + 𝐯 ∧ 𝐰,\]

where the dot product is the usual scalar inner product and the wedge product is the antisymmetric outer product (the exterior product). The geometric product is linear, associative, and distributive, and satisfies

\[𝐯 𝐯 = 𝐯 ⋅ 𝐯.\]

Two key consequences follow immediately. Parallel vectors commute: $𝐯 𝐰 = 𝐰 𝐯$ when $𝐯 ∥ 𝐰$. Orthogonal vectors anticommute: $𝐯 𝐰 = -𝐰 𝐯$ when $𝐯 ⟂ 𝐰$. In particular, computations with an orthonormal basis become simple exercises in counting the parity of permutations.

The full algebra over $ℝ³$ has dimension $2^3 = 8$. A basis is provided by products of the Cartesian basis vectors, grouped by grade (number of vector factors):

\[\begin{array}{ll} \text{grade 0 (scalar):} & \boldsymbol{1}, \\[4pt] \text{grade 1 (vectors):} & 𝐱,\; 𝐲,\; 𝐳, \\[4pt] \text{grade 2 (bivectors):} & 𝐱𝐲,\; 𝐱𝐳,\; 𝐲𝐳, \\[4pt] \text{grade 3 (pseudoscalar):} & 𝐈 = 𝐱𝐲𝐳. \end{array}\]

Each bivector squares to $-1$. For example,

\[𝐱𝐲𝐱𝐲 = -𝐱𝐲𝐲𝐱 = -𝐱(𝐲𝐲)𝐱 = -𝐱𝐱 = -1.\]

The pseudoscalar $𝐈 = 𝐱𝐲𝐳$ also squares to $-1$, so its inverse is $𝐈^{-1} = -𝐈 = 𝐳𝐲𝐱$.

$𝐈$ has a special property in three dimensions: since moving it past any grade-1 vector costs $(-1)^{n-1} = (-1)^2 = +1$ sign changes, it commutes with every element of the algebra. Left- and right-multiplication by $𝐈$ are therefore identical, and we can unambiguously write $𝐈\,𝐯 = 𝐯\,𝐈$.

The reverse

The reverse of a multivector, denoted $\widetilde{A}$, is obtained by reversing the order of the vector factors in each basis blade:

\[\widetilde{e_{i_1} e_{i_2} \cdots e_{i_r}} = e_{i_r} \cdots e_{i_2}\, e_{i_1}.\]

Reordering the $r$ vectors back to standard order requires $r(r-1)/2$ transpositions, each costing a sign change, so a grade-$r$ blade picks up a factor of $(-1)^{r(r-1)/2}$. In $ℝ³$:

Grades 0 and 1 are unchanged: $\widetilde{𝐯} = 𝐯$.
Grades 2 and 3 negate: for example

\[\widetilde{𝐱𝐲} = 𝐲𝐱 = -𝐱𝐲, \qquad \widetilde{𝐈} = 𝐳𝐲𝐱 = -𝐱𝐲𝐳 = -𝐈.\]

Duality

The Hodge dual maps grade-$k$ elements to grade-$(n-k)$ elements. The formula^[1]

\[\star b = \widetilde{b}\, 𝐈\]

holds for blades of any grade and any metric signature. For grade-1 vectors $\widetilde{𝐯} = 𝐯$, so right-multiplication by $𝐈$ suffices:

\[\star 𝐱 = 𝐱\, 𝐈 = 𝐲𝐳, \qquad \star 𝐲 = 𝐲\, 𝐈 = 𝐳𝐱, \qquad \star 𝐳 = 𝐳\, 𝐈 = 𝐱𝐲.\]

For grade-2 bivectors the reverse introduces a sign (e.g., $\widetilde{𝐲𝐳} = 𝐳𝐲 = -𝐲𝐳$), giving

\[\star(𝐲𝐳) = (𝐳𝐲)(𝐱𝐲𝐳) = 𝐱, \qquad \star(𝐳𝐱) = 𝐲, \qquad \star(𝐱𝐲) = 𝐳.\]

Reflections and rotations

One of the most important applications of the geometric product is to represent reflections and rotations via simple products with other elements of the algebra. For a unit vector $𝐧$ ($𝐧^2 = 1$), the sandwich map $𝐯 \mapsto -𝐧\,𝐯\,𝐧$ reflects $𝐯$ through the hyperplane orthogonal to $𝐧$ — or, we might say, "along" $𝐧$. Decompose $𝐯 = 𝐯_\parallel + 𝐯_\perp$: the parallel component commutes with $𝐧$ (parallel vectors commute), giving $-𝐧\,𝐯_\parallel\,𝐧 = -𝐯_\parallel$; the perpendicular component anticommutes, giving $-𝐧\,𝐯_\perp\,𝐧 = +𝐯_\perp$. So the component along $𝐧$ is flipped and the hyperplane component is preserved — the correct reflection.

The Cartan–Dieudonné theorem states that every rotation of $\mathbb{R}^n$ is a composition of an even number of reflections. Composing two reflections along unit vectors $𝐦$ and $𝐧$:

\[-𝐦\bigl(-𝐧\,𝐯\,𝐧\bigr)𝐦 = (𝐦𝐧)\,𝐯\,(𝐦𝐧)^{-1}.\]

The product $R = 𝐦𝐧$ is an even-grade element satisfying $R \widetilde{R} = 1$. For unit vectors separated by angle $\alpha$ this product equals $\cos\alpha + \sin\alpha\,\hat{B}$, where $\hat{B}$ is the unit bivector of the plane spanned by $𝐦$ and $𝐧$. The effect of $R\,𝐯\,R^{-1}$ is to rotate $𝐯$ by an angle $2\alpha$ in the plane of $\hat{B}$. Setting $\alpha = \vartheta/2$:

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

and $R\,𝐯\,R^{-1} = R\,𝐯\,\widetilde{R}$ rotates the component of $𝐯$ in the plane of $\hat{B}$ by angle $\vartheta$.

The set of all such "rotors" $R$ forms a group under multiplication, called the spin group: $\mathrm{Spin}(3) ≅ \mathrm{SU}(2)$. The factor of $1/2$ is a hallmark of the well known double-cover structure over the rotation group $\mathrm{SO}(3)$.

Each plane is represented by either of two unit bivectors that differ by a sign. For example, the $𝐱𝐲$-plane is represented by either $𝐱𝐲$ or $𝐲𝐱 = -𝐱𝐲$. We must determine which sign gives a right-handed rotation. Consider $R = \exp(\frac{\vartheta}{2}\,𝐱𝐲) = \cos\frac{\vartheta}{2} + \sin\frac{\vartheta}{2}\,(𝐱𝐲)$ acting on $𝐱$. We can readily calculate

\[R\,𝐱\,\widetilde{R} = \left(\cos^2\!\frac{\vartheta}{2} - \sin^2\!\frac{\vartheta}{2}\right)\,𝐱 - 2\sin\frac{\vartheta}{2}\cos\frac{\vartheta}{2}\,𝐲 = \cos(\vartheta)\,𝐱 - \sin(\vartheta)\,𝐲.\]

A right-handed rotation by $\vartheta$ about $𝐳$ (curling the right hand from $𝐱$ toward $𝐲$, thumb along $+𝐳$) maps $𝐱 \to \cos\vartheta\,𝐱 + \sin\vartheta\,𝐲$. The formula above has a negative $\sin$ term: $\exp(\theta\,𝐱𝐲)$ rotates $𝐱$ toward $-𝐲$, i.e., clockwise in the $𝐱𝐲$-plane (left-handed about $𝐳$). Right-handed rotation by $\vartheta$ about $+𝐳$ requires the opposite sign: $\exp(\frac{\vartheta}{2}\,𝐲𝐱)$. The same test on the remaining coordinate planes — checking whether $\exp(\vartheta\,𝐲𝐳)$ maps $𝐲$ toward $+𝐳$ or $-𝐳$, and whether $\exp(\vartheta\,𝐳𝐱)$ maps $𝐳$ toward $+𝐱$ or $-𝐱$ — gives in each case the same conclusion: the bivector with reversed index order is the right-handed one. The results are:

\[\begin{array}{ll} \text{right-handed rotation about } 𝐱: & \exp\left(\tfrac{\vartheta}{2}\,𝐳𝐲\right), \\[4pt] \text{right-handed rotation about } 𝐲: & \exp\left(\tfrac{\vartheta}{2}\,𝐱𝐳\right), \\[4pt] \text{right-handed rotation about } 𝐳: & \exp\left(\tfrac{\vartheta}{2}\,𝐲𝐱\right). \end{array}\]

The even subalgebra: Quaternions

Products of an even number of vectors — grades 0 and 2 — form a closed subalgebra, the even subalgebra, spanned by $\{𝟏, 𝐱𝐲, 𝐲𝐳, 𝐳𝐱\}$.^[2] This four-dimensional algebra is precisely isomorphic to the algebra of quaternions.

In quaternion terminology, the bivector basis elements are canonically presented as the generators of rotations about the specific axes. Comparing with the results at the end of the previous section, the right-handed generators are

\[\begin{aligned} 𝐢 &= -\star 𝐱 = 𝐳𝐲, \\ 𝐣 &= -\star 𝐲 = 𝐱𝐳, \\ 𝐤 &= -\star 𝐳 = 𝐲𝐱. \end{aligned}\]

Each of these squares to $-1$ (as expected for a unit bivector), and one can verify the standard cyclic rules:

\[𝐢^2 = 𝐣^2 = 𝐤^2 = 𝐢𝐣𝐤 = -\boldsymbol{1}, \qquad 𝐢𝐣 = 𝐤, \quad 𝐣𝐤 = 𝐢, \quad 𝐤𝐢 = 𝐣.\]

The opposite sign choice, $𝐢' = \star 𝐱 = 𝐲𝐳$, produces left-handed rotations and satisfies $𝐢'𝐣'𝐤' = +\boldsymbol{1}$. This is the convention used in much of the engineering literature.^[3]

A general quaternion can be written

\[𝐐 = w\,\boldsymbol{1} + x\,𝐢 + y\,𝐣 + z\,𝐤,\]

and is represented in this package by the Quaternion type. Specifically, $(w, x, y, z)$ are the names of the components used throughout this package, and the symbols 𝐢, 𝐣, and 𝐤 are exported with precisely the meaning given here:

julia> using Quaternionic

julia> 𝐢^2 == 𝐣^2 == 𝐤^2 == 𝐢*𝐣*𝐤 == -1
true

julia> 𝐢*𝐣 == 𝐤 && 𝐣*𝐤 == 𝐢 && 𝐤*𝐢 == 𝐣
true

For this general quaternion, the reverse is

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

which is exactly the quaternion conjugate conj(Q) in the code. Two important consequences follow. First,

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

is a pure scalar — a non-negative real number for real quaternions. In analogy with abs2 for complex numbers, this product is abs2(Q) in the code, and abs(Q) is its square root. Second, every nonzero quaternion has an inverse

\[𝐐^{-1} = \frac{\widetilde{𝐐}}{𝐐\,\widetilde{𝐐}} = \frac{\texttt{conj(Q)}}{\texttt{abs2(Q)}} = \texttt{inv(Q)}.\]

Real vs. complex quaternions

For real quaternions, $𝐐\,\widetilde{𝐐}$ is a positive real scalar. For complex quaternions, $𝐐\,\widetilde{𝐐}$ is still a scalar element of the algebra (no $𝐢$, $𝐣$, or $𝐤$ component), but it is complex-valued. In particular, this is the spinor norm $\sum_i z_i^2$ rather than the Euclidean norm $\sum_i |z_i|^2$. The distinction matters for Lorentz boosts, as discussed on the spacetime algebra page.

The general quaternion is represented in this package by the Quaternion type. Unit quaternions (rotors, which have $𝐐\,\widetilde{𝐐} = 1$) are represented by the Rotor type. And "pure-vector" quaternions (having $w=0$) are represented by the QuatVec type.

Note that the function-call syntax R(v) implements the sandwich $𝐑\, 𝐯\, \widetilde{𝐑}$ efficiently with a dedicated formula that is roughly twice as fast as the multiplication sequence R * v * conj(R):

R * v * conj(R) ≈ R(v)   # both correct; R(v) is ~2× faster

This applies to both Quaternion and Rotor objects acting on a QuatVec.

The vector–quaternion isomorphism and rotations

We have already demonstrated an isomorphism between the vector space of $ℝ³$ and the space of "pure-vector" quaternions via the (negative) Hodge dual:

\[𝐯 \mapsto 𝐕 = -\star 𝐯.\]

Here, $𝐕$ represents the corresponding quaternion. The inverse map is given by the same formula:^[4]

\[𝐕 \mapsto 𝐯 = -\star 𝐕.\]

This isomorphism is a vector-space isomorphism. But, importantly, it is also compatible with rotation in the sense that rotating $𝐯$ is the same as mapping to $𝐕$, rotating $𝐕$, and then mapping back to the original vector space. That is,

\[R 𝐯 \widetilde{R} = -\star \left( R (-\star 𝐯) \widetilde{R} \right).\]

Thus, we can freely move back and forth between the two representations of vectors.