Biquaternion Square Roots of Minus One, Zero and Plus One

Introduction

This article determines the square roots of the three central values $-1$, $0$ and $+1$ in the biquaternion algebra $\mathbb{B}$ — the elements $\xi\in\mathbb{B}$ satisfying $\xi^2 = -1$, $\xi^2 = 0$ and $\xi^2 = +1$ — and the relations among the three sets. It relates to Biquaternion Idempotents and Projections through the bijection between the roots of $-1$ and the idempotents. The square roots of an arbitrary element $Q$, and the Clifford-algebra algorithm that computes them, are Biquaternion Square Roots of a General Element; the three sets found here are the degenerate data of that classification, and the algorithm is deliberately not reproduced.

Physically, a root of $-1$ is an imaginary unit of the algebra, and the imaginary units are what the framework's transformations are generated from. Each root generates a one-parameter subgroup, $\exp(\theta\xi) = \cos\theta\,e_0+\sin\theta\,\xi$, and the three families of roots are three physically distinct kinds of generator:

• the trivial roots $\pm i$ generate the central phase, the abelian gauge group the framework supplies canonically;
• the real roots $\pm\mu$, with $\mu$ a pure real quaternion with $\mu^2 = -1$, generate the spatial rotations: $\mu$ is a rotation axis, and the associated idempotents are the pure states;
• the non-trivial roots $b\mu+d\nu i$ generate the mixed transformations, an elliptic combination of a rotation about $\mu$ and a boost along $\nu$, with $b$ and $d$ the hyperbolic functions of a rapidity.

The roots of $+1$ are the involutions: the reflections and the parity-type operators, of which the fermion parity $(-1)^F = ie_3$ is an example, and each of which splits the algebra into a complementary pair of projectors. Generators whose square is neither $-1$ nor $+1$ but $0$ — the parabolic generators — are exactly the null elements.

Conventions. The quaternion basis is $e_0 = 1,e_1,e_2,e_3$ and the scalar imaginary is $i$, so that it does not collide with the quaternion units; a general element is $\tilde{Q} = \sum_{\mu=0}^{3}Q_\mu e_\mu$ with $Q_\mu\in\mathbb{C}$. The conjugations are $\tilde{Q}^{\natural}$, $\bar{\tilde{Q}}$ and $\tilde{Q}^{*} = \overline{\tilde{Q}^{\natural}}$. The material coordinate is $ict\,e_0+\mathbf{x}$ and the informational coordinate is $ct'\,e_0+i\mathbf{x}'$. Throughout, $\mu$ and $\nu$ denote pure real quaternions with $\mu^2 = \nu^2 = -1$, and such a pair anticommutes when $\mu\nu + \nu\mu = 0$.

The Problem and Its Reduction

The three problems of this article — the equations $\xi^2 = -1$, $\xi^2 = 0$ and $\xi^2 = +1$ — share one reduction, carried out once here and applied three times. Write $\xi$ in scalar-vector form,

$$ \xi = Q_0+\mathbf{Q}, \qquad Q_0\in\mathbb{C}, \qquad \mathbf{Q} = Q_1e_1+Q_2e_2+Q_3e_3 , $$

with $Q_0$ the complex scalar part and $\mathbf{Q}$ the complex vector part. The square is given by the product formula,

$$ \xi^2 = \left(Q_0^2-(\mathbf{Q},\mathbf{Q})\right)e_0+2Q_0\mathbf{Q}, \qquad (\mathbf{Q},\mathbf{Q}) = Q_1^2+Q_2^2+Q_3^2 , $$

so the scalar part of $\xi^2$ is $(Q_0^2-(\mathbf{Q},\mathbf{Q}))e_0$ and the vector part is $2Q_0\mathbf{Q}$, so that equating $\xi^2$ to a central value splits the equation into a vector part and a scalar part.

The Roots of Minus One

Statement

A root of $-1$ in $\mathbb{B}$ is an element $\xi$ with $\xi^2 = -1$.

Reduction to Two Cases

Equating $\xi^2$ to $-e_0$ requires

$$ 2Q_0\mathbf{Q} = 0, \qquad Q_0^2-(\mathbf{Q},\mathbf{Q}) = -1 . $$

Since $\mathbb{C}$ is a field, $Q_0\mathbf{Q} = 0$ holds if and only if $Q_0 = 0$ or $\mathbf{Q} = 0$, so the roots split into two cases.

Case 1: $\mathbf{Q} = 0$. Then $\xi = Q_0e_0$ is a complex scalar and $Q_0^2 = -1$, so $Q_0 = \pm i$:

$$ \xi = \pm i , $$

the trivial roots.

Case 2: $Q_0 = 0$. Then $\xi = \mathbf{Q}$ is pure and $(\mathbf{Q},\mathbf{Q}) = 1$. The pure roots are the pure biquaternions whose square is $-1$.

The cases are disjoint: the trivial roots have vanishing vector part, the pure roots vanishing scalar part.

The Pure Roots

Write $Q_k = q_k+iq'_k$ with $q_k,q'_k\in\mathbb{R}$. Then

$$ (\mathbf{Q},\mathbf{Q}) = \sum_{k=1}^{3}(q_k^2-(q'_k)^2)+2i\sum_{k=1}^{3}q_kq'_k , $$

so equating real and imaginary parts to $1$ and $0$ gives

$$ \sum_{k=1}^{3} q_k^2-\sum_{k=1}^{3} (q'_k)^2 = 1, \qquad \sum_{k=1}^{3} q_k q'_k = 0 . $$

The first condition gives $\sum_k q_k^2 = 1+\sum_k (q'_k)^2\geq1>0$, so the triple $(q_1,q_2,q_3)$ is nonzero.

Sub-case 2a: $\mathbf{q}' = 0$. Then $\sum_k q_k^2 = 1$. Let $\mu = q_1e_1+q_2e_2+q_3e_3$, a pure real quaternion with $\mu^2 = -\bigl(\sum_k q_k^2\bigr)e_0 = -1$, so $\xi = \mu$. These are the real roots; the family is invariant under $\mu\mapsto-\mu$, and it may also be written in the redundant form $\xi = \pm\mu$.

Sub-case 2b: $\mathbf{q}'\neq0$. Put $b = \bigl(\sum_k q_k^2\bigr)^{1/2}>0$ and $d = \bigl(\sum_k (q'_k)^2\bigr)^{1/2}>0$, and define the pure real quaternions $\mu = \sum_k (q_k/b)e_k$ and $\nu = \sum_k (q'_k/d)e_k$. Then $\mu^2 = \nu^2 = -1$, and $\sum_k\mu_k\nu_k = \frac{1}{bd}\sum_k q_kq'_k = 0$, so $\mu\nu+\nu\mu = -2\bigl(\sum_k\mu_k\nu_k\bigr)e_0 = 0$: the two anticommute. With $\mathbf{q} = b\mu$, $\mathbf{q}' = d\nu$ and $b^2-d^2 = 1$,

$$ \mathbf{Q} = \mathbf{q}+i\mathbf{q}' = b\mu+d\nu i . $$

These are the non-trivial roots.

Statement of the Theorem

Theorem. The roots of $-1$ in $\mathbb{B}$ are exactly the elements

$$ \xi = \pm i, \qquad \xi = \pm\mu, \qquad \xi = b\mu+d\nu i , $$

with $\mu,\nu$ pure real quaternions with $\mu^2 = \nu^2 = -1$ and $\mu\nu+\nu\mu = 0$, and $b,d>0$ satisfying $b^2-d^2 = 1$.

Verification

Trivial root. $(\pm i)^2 = -1$ since $i^2 = -1$. ✓

Real roots. For a pure real quaternion $\mu$ with $\mu^2 = -1$, $(\pm\mu)^2 = \mu^2 = -1$. ✓

Non-trivial roots. Compute

$$ \xi^2 = (b\mu+d\nu i)^2 = b^2\mu^2+d^2\nu^2i^2+bd(\mu\nu+\nu\mu)i . $$

Here $\mu^2 = \nu^2 = -1$ and $i^2 = -1$, so the first two terms sum to $-(b^2-d^2)e_0$. The cross term $bd(\mu\nu+\nu\mu)i$ vanishes because $\mu$ and $\nu$ anticommute. Therefore

$$ \xi^2 = -(b^2-d^2)e_0 = -1 , $$

using the constraint. ✓

Degenerate Cases

The real roots as the case $d = 0$. In the non-trivial family the constraint $b^2-d^2 = 1$ is carried with $d>0$. If the condition $d>0$ is relaxed to $d\geq0$, then $d = 0$ forces $b = 1$ and $\xi = \mu$, a real root; the element $\nu$ no longer enters. So the real roots are the case $d = 0$ of the non-trivial formula, not a separate construction.

The trivial roots are separate. The trivial roots have vanishing vector part, and the real and non-trivial roots all have $\mathbf{Q}\neq0$. So the roots with vanishing vector part are exactly $\pm i$, which form a separate family.

The Roots as Generators

Because $\xi^2 = -1$, the exponential of $\xi$ closes:

$$ \exp(\theta\xi) = \cos\theta\,e_0+\sin\theta\,\xi, \qquad \theta\in\mathbb{R}, $$

so each root generates a one-parameter subgroup of the group of units. For a pure generator $\xi$ the square is a scalar, $\xi^2 = -(\xi,\xi)e_0$, and the three cases physics distinguishes are:

• $\xi^2 = -1$: a root of $-1$, an elliptic generator, whose exponential $\exp(\theta\xi) = \cos\theta\,e_0+\sin\theta\,\xi$ is a one-parameter subgroup;
• $\xi^2 = +1$: a root of $+1$, a hyperbolic generator, whose exponential $\exp(\theta\xi) = \cosh\theta\,e_0+\sinh\theta\,\xi$ is a one-parameter subgroup;
• $\xi^2 = 0$: a parabolic (null) generator: it is nilpotent, its exponential is the polynomial subgroup $\exp(\theta\xi) = e_0+\theta\xi$, and $\xi$ itself is a zero divisor rather than a unit.

Every pure element with $(\xi,\xi)\neq0$ is a complex scale multiple of a root of $-1$, and a pure element is nilpotent exactly when $(\xi,\xi) = 0$. Up to scale, therefore, the pure generators are the two root families together with the nilpotent cone.

Physical reading. The elliptic generators are the rotations: a real root $\mu$ is a rotation axis, and the rotation through an angle is conjugation by $\exp(\tfrac{\theta}{2}\mu)$, which sends a vector $\mathbf{v}$ to its rotation about $\mu$ by $\theta$. The hyperbolic generators are the boosts: a pure boost is generated by $i\nu$, whose square is $+\nu^2 = +1$, a root of $+1$ and not of $-1$. The parabolic generators are the null directions, and they are the zero divisors. So the trichotomy elliptic/hyperbolic/parabolic of Lorentz-type generators is, in this algebra, exactly the trichotomy of the square of a pure element: $-1$, $+1$, $0$.

Parametrization by Hyperbolic Functions

The constraint $b^2-d^2 = 1$ with $b>0$ is parametrised by

$$ b = \cosh t, \qquad d = \sinh t, \qquad t\in\mathbb{R}, $$

giving

$$ \xi = \cosh t\,\mu+\sinh t\,\nu i, \qquad \mu^2 = \nu^2 = -1, \quad \mu\nu+\nu\mu = 0 . $$

The parameter $t$ is the rapidity of the root, by analogy with the rapidity of a hyperbolic rotation.

Physical reading. The rapidity is not an analogy but the same parameter: the pair $(b,d) = (\cosh t,\sinh t)$ is the standard hyperbolic parametrisation of a boost, and the mixed root is the generator of a transformation that is a rotation about $\mu$ composed with a boost along $\nu$. The real roots are the case $t = 0$; at $d = 0$ the element $\nu$ drops out and the root is $\mu$. A mixed root is therefore the algebra's way of writing "a rotation and a boost taken together, with the generator squaring to $-1$".

Status of the Roots

The trivial roots are the two elements $\pm i$. The real roots are the elements $\pm\mu$ with $\mu$ a pure real quaternion of square $-1$; the single constraint $\sum_k\mu_k^2 = 1$ on the three real coefficients leaves two free real parameters. The non-trivial roots are parametrised by the pair $(\mu,\nu)$ with $\mu^2 = \nu^2 = -1$ and $\mu\nu+\nu\mu = 0$ — two free real parameters for $\mu$, and one for $\nu$, its two constraints $\sum_k\nu_k^2 = 1$ and $\sum_k\mu_k\nu_k = 0$ acting on three real coefficients — together with the pair $(b,d)$ with $b^2-d^2 = 1$ and $b,d>0$, one further parameter: four free real parameters in all.

All roots except the trivial ones are pure, hence lie in the six-real-dimensional vector subspace of pure biquaternions. The roots are neither idempotents ($\xi^2 = -1\neq\xi$) nor zero divisors (§The Roots as Invertible Elements): they lie in the group of units $\mathbb{B}^\times$. Physically: an imaginary unit is a group element, never a state and never a null direction.

The Roots of Zero

Reduction to Two Cases

A root of $0$ is an element with $\xi^2 = 0$. With $\xi = Q_0+\mathbf{Q}$ the square is $(Q_0^2-(\mathbf{Q},\mathbf{Q}))e_0+2Q_0\mathbf{Q}$, and equating it to $0$ forces the same choice $Q_0 = 0$ or $\mathbf{Q} = 0$ as for the roots of $-1$.

• Case $\mathbf{Q} = 0$. Then $\xi = Q_0e_0$ and $Q_0^2 = 0$; since $\mathbb{C}$ is a field, $Q_0 = 0$. This gives the single element $\xi = 0$.
• Case $Q_0 = 0$. Then $\xi = \mathbf{Q}$ is pure with $(\mathbf{Q},\mathbf{Q}) = 0$, that is $Q_1^2+Q_2^2+Q_3^2 = 0$.

Statement of the Theorem

Theorem. The roots of $0$ in $\mathbb{B}$ are $\xi = 0$ together with the pure elements $\xi = Q_1e_1+Q_2e_2+Q_3e_3$ with $Q_1^2+Q_2^2+Q_3^2 = 0$.

Verification

For such a pure element both the scalar part $-(\mathbf{Q},\mathbf{Q})$ and the vector part $2Q_0\mathbf{Q}$ of the square vanish. ✓

The nonzero roots of $0$ are the nilpotents of the algebra, and they form the nilpotent cone of the pure subspace; the instance $\xi = e_1+ie_2$ gives $\xi^2 = e_1^2+i^2e_2^2+i(e_1e_2+e_2e_1) = -1+1+0 = 0$. The set is the zero-divisor set of the algebra, whose structure, its two families and the criterion in terms of the scalar part, is Biquaternion Zero Divisors; it is named here only to complete the list of the three central values and not developed.

Physical reading: the parabolic generators and the null momenta. A root of $0$ is a parabolic generator: its exponential is the polynomial subgroup $\exp(\theta\xi) = e_0+\theta\xi$, and the element itself is a zero divisor rather than a unit. The three central values are therefore the three kinds of one-parameter generator the framework uses: $-1$ the elliptic generators (the central phase and the rotations), $+1$ the hyperbolic ones (the boosts and the parities), and $0$ the parabolic ones (the null translations). The physical instance is the photon: a null momentum $p$ with $p^2 = 0$ is a parabolic generator, and the pure null elements such as $e_1+ie_2$ are the two sectors' null directions. The parabolic generators sit at the boundary of the group of units, which is the light cone of the material and informational sectors; the null readings are developed in Biquaternion Zero Divisors.

The Roots of Plus One

Statement of the Theorem

The analogous problem for $+1$, an element $\eta$ with $\eta^2 = 1$, is also recorded. The roots of $+1$ are related to those of $-1$ by

$$ \eta = \xi i , $$

where $\xi$ is a root of $-1$. Applying the classification:

• Non-trivial roots of $+1$: $\eta = b\mu i+d\nu i^2 = b\mu i-d\nu$ with $b^2-d^2 = 1$ and $\mu\nu+\nu\mu = 0$;
• Trivial roots of $+1$: $\eta = \pm1$ (from $\xi = \pm i$), the only roots of $+1$ lying in the centre $\mathbb{C}$;
• Real roots of $+1$: $\eta = \pm\mu i$ with $\mu$ a pure real quaternion with $\mu^2 = -1$.

Verification

That $\eta = \xi i$ satisfies $\eta^2 = 1$ follows since $\xi^2i^2 = (-1)(-1) = 1$. The map $\xi\mapsto\xi i$ is a bijection: it is injective since $i$ is invertible, and if $\eta^2 = 1$ then $\xi = -\eta i$ satisfies $\xi^2 = -1$ and $\xi i = \eta$. ✓

A root of $+1$ is an involution, and every involution splits the algebra: if $\eta^2 = 1$ then $\eta\neq1$ gives the idempotents $\tfrac12(e_0\pm\eta)$ with $\tfrac12(e_0+\eta)+\tfrac12(e_0-\eta) = e_0$ and $\tfrac12(e_0+\eta)\cdot\tfrac12(e_0-\eta) = 0$, which is the bijection above read in the other direction.

Physical reading: the reflections and the parities. The roots of $+1$ are the algebra's reflections. The real roots $\pm\mu i$ are the Hermitian ones — $(\mu i)^{*} = \mu i$ — so they are exactly the involutions that split the algebra into orthogonal projectors, and they are the parity-type operators of the framework. The fermion parity of the field-theory articles is an instance: $(-1)^F = ie_3$ is a real root of $+1$ (with $\mu = e_3$), Hermitian, and its associated projectors are $\tfrac12(e_0\pm ie_3)$, the vacuum projector $\tilde\Pi_1$ and its complement. The trivial roots $\pm1$ are the two central involutions (the identity and the total sign), and the non-trivial roots are the non-Hermitian involutions, which split the algebra into complementary projectors that are not orthogonal. Beyond the framework, the roots of $+1$ generate the hyperbolic one-parameter subgroups that carry the boosts.

The Three Sets Compared

The three root sets are related by their invertibility. The roots of $-1$ and of $+1$ are units: a root $\xi$ of $\pm1$ satisfies $\xi^{-1} = \pm\xi$, in $\mathbb{B}^\times$. The nonzero roots of $0$ are the exception: they are nilpotents, hence zero divisors and not units. So of the three central values only $0$ has non-unit roots.

The three sets also differ in kind. The roots of $-1$ are two isolated elements together with a two-parameter family and a four-parameter family; the roots of $+1$ are the bijective image of that set under $\xi\mapsto\xi i$; and the roots of $0$ are the single element $0$ together with one four-parameter family, the nilpotent cone. In particular the roots of $0$ contain no unit, and the only non-pure roots of the three values are the trivial ones, $\pm i$ for $-1$ and $\pm1$ for $+1$.

Physical reading. The three sets are the three kinds of one-parameter generator: the elliptic ones from $-1$, the hyperbolic ones from $+1$, and the parabolic ones from $0$. Only the first two lie in the group of units, and only the third lies on the light cone of the two sectors; a generator is therefore classified by the sign of its square.

The Relation to the Idempotents

The classification of the roots of $-1$ gives the classification of the idempotents: the map

$$ \xi\longmapsto \tilde\Pi(\xi) = \tfrac{1}{2}(e_0+\xi i) $$

is a bijection from the roots of $-1$ onto the idempotents, under which the complementary pairs $\{\tilde\Pi,e_0-\tilde\Pi\}$ correspond to the classes $\{\xi,-\xi\}$, and under which the three families of roots give the trivial idempotents, the Hermitian idempotents in $\mathbb{M}_+$, and the idempotents lying in none of the four four-dimensional subspaces. The construction of the idempotent, the proof of the bijection and the projection interpretation are the subject of Biquaternion Idempotents and Projections.

Physical reading. The correspondence is why the framework can speak of a state direction as an algebraic datum of the same kind as a generator: a root $\xi$ is an imaginary unit, its associated projector $\tilde\Pi(\xi)$ is a pure state, and the family of real roots is the family of pure states. The rotation axis and the state direction are the same direction.

The Relation to the Zero Divisors

The Idempotents as Zero Divisors

Every non-trivial idempotent is a zero divisor, $\tilde\Pi(e_0-\tilde\Pi) = 0$ with both factors nonzero (the construction of $\tilde\Pi$ is in Biquaternion Idempotents and Projections). The trivial idempotents $0$ and $e_0$ are not zero divisors: $0$ is excluded by the definition, and $e_0$ is a unit.

The Roots as Invertible Elements

A root of $-1$ is a unit and not a zero divisor. Indeed, $\xi^2 = -1$ gives

$$ \xi\,(-\xi) = (-\xi)\,\xi = e_0, $$

so $\xi$ is invertible, with inverse $\xi^{-1} = -\xi$; in particular a pure imaginary unit is its own negative inverse. A unit is not a zero divisor: if $\xi R = 0$ then $R = (-\xi)(\xi R) = 0$, and likewise $R\xi = 0$ forces $R = 0$. The roots therefore lie in the group of units $\mathbb{B}^\times$. The central scalar $\xi\bar{\xi}$ is $-1$ for the trivial roots and $+1$ for a pure root.

The Remaining Pure Elements

The non-trivial roots lie in the six-real-dimensional vector subspace of pure biquaternions; the trivial roots $\pm i$ do not. The remaining pure elements, those with $(\xi,\xi) = 0$, are the parabolic (null) elements.

Summary

This article determines the square roots of the three central values $-1$, $0$ and $+1$ in the biquaternion algebra, by one reduction applied three times, and compares the three sets in §The Three Sets Compared.

The roots of $-1$ in the biquaternion algebra are exactly:

1. Non-trivial roots: $\xi = b\mu+d\nu i$, with $\mu,\nu$ pure real quaternions with $\mu^2 = \nu^2 = -1$ and $\mu\nu+\nu\mu = 0$, and $b,d>0$ satisfying $b^2-d^2 = 1$;
2. Trivial roots: $\xi = \pm i$;
3. Real roots: $\xi = \pm\mu$ with $\mu$ a pure real quaternion with $\mu^2 = -1$.

The proof writes $\xi = Q_0+\mathbf{Q}$, squares using the biquaternion product formula and equates to $-1$: the vector part $2Q_0\mathbf{Q}$ forces $Q_0 = 0$ or $\mathbf{Q} = 0$; the scalar part $Q_0^2-(\mathbf{Q},\mathbf{Q})$ then gives the trivial root in the scalar case and, in the pure case, reduces to $(\mathbf{Q},\mathbf{Q}) = 1$, which splits into the real and non-trivial families according to whether the imaginary part of the pure element vanishes.

The trivial roots are the two elements $\pm i$; the real roots are the family $\pm\mu$ cut out by $\sum_k\mu_k^2 = 1$; the non-trivial roots carry the pairs $(\mu,\nu)$ and $(b,d)$ as above, four free real parameters in all. All roots except the trivial ones are pure, hence lie in the six-real-dimensional vector subspace of pure biquaternions. Physically, the roots are the imaginary units of the algebra and generate its one-parameter subgroups: $\pm i$ the central phase, the real roots the rotations about the spatial axes, the non-trivial roots the mixed rotation–boost generators with rapidity $t$ in $b = \cosh t$, $d = \sinh t$. The three families of roots classify the idempotents through $\xi\mapsto\tfrac12(e_0+\xi i)$; the associated projectors are the pure states for the real roots and non-orthogonal projectors for the non-trivial ones. The parabolic generators, whose square is $0$ rather than $\pm1$, are the null elements and belong to the zero-divisor theory.

The roots of $0$ are $\xi = 0$ together with the pure elements $Q_1e_1+Q_2e_2+Q_3e_3$ with $Q_1^2+Q_2^2+Q_3^2 = 0$; the nonzero ones are the nilpotents, instance $e_1+ie_2$. Physically they are the parabolic generators, whose exponential is the polynomial subgroup $e_0+\theta\xi$ and whose physical instance is the photon's null momentum; they lie at the boundary of the group of units, unlike the roots of $\pm1$, which are units. The classification of $\xi^2 = Q$ for a general $Q$ is Biquaternion Square Roots of a General Element, and the structure of the zero-divisor set they belong to is Biquaternion Zero Divisors.

The roots of $+1$ are the involutions, obtained from the roots of $-1$ by multiplication by $i$; the real ones are the Hermitian reflections, of which the fermion parity $ie_3$ is an instance, and each splits the algebra into a complementary pair of projectors.

Summary of Notation

Further Reading

• William Rowan Hamilton, Lectures on Quaternions (Hodges and Smith, Dublin, 1853), for the original discovery of the biquaternions.
• S. J. Sangwine, "Biquaternion (complexified quaternion) roots of $-1$", Advances in Applied Clifford Algebras 16 (2006) 63–68, for the original classification.
• S. J. Sangwine and D. Alfsmann, "Determination of the biquaternion divisors of zero, including the idempotents and nilpotents", Advances in Applied Clifford Algebras 20 (2010) 401–416, for the relation to the idempotents and the zero divisors.
• S. J. Sangwine, T. A. Ell and N. Le Bihan, "Fundamental representations and algebraic properties of biquaternions or complexified quaternions", Advances in Applied Clifford Algebras 21 (2011) 607–636, for the representation theory and the constraint verification.
• J. P. Ward, Quaternions and Cayley Numbers: Algebra and Applications (Kluwer, Dordrecht, 1997), for the semi-norm and the algebraic properties of the biquaternions.
• Pertti Lounesto, Clifford Algebras and Spinors (Cambridge, 2001), for the connection to Clifford algebras.
• A. Acus and A. Dargys, Square roots of complexified quaternions, arXiv:2601.08391 (2026), for the square roots of an arbitrary complexified quaternion and the Clifford-algebra algorithm treated in Biquaternion Square Roots of a General Element.