The Quaternion Pairing of the Polar Decomposition of Biquaternions

Introduction

Every biquaternion of non-vanishing norm is the ordered product of a positive real scale, a central phase and two unit factors,

$$ \tilde{Q} = r\,\exp(i\alpha)\,\tilde U_{\mathrm b}\,\tilde u_{\mathrm r} = r\,\exp(i\alpha)\,\exp\Bigl(i\tfrac{\psi}{2}\vec u_{\mathrm b}\Bigr)\,\exp\Bigl(\tfrac{\theta}{2}\vec u_{\mathrm r}\Bigr) . $$

The four factors are those of Modulus–Unit Polar Decomposition of Biquaternions, §The Four Factors, and their exponential forms those of Exponential Polar Decomposition of Biquaternions, §The Decomposition: the scale $r>0$, the central phase $\exp(i\alpha)$ with $\alpha\in(-\pi/2,\pi/2]$, the Hermitian positive boost $\tilde U_{\mathrm b}$ of rapidity $\psi\ge0$ and the unit real rotor $\tilde u_{\mathrm r}$ of angle $\theta\in[0,2\pi]$. They are unique, so they can be grouped in pairs in exactly three ways. This article reads the grouping that puts the scale with the rotor.

The quaternion pairing is the grouping that joins the scale to the rotor and the phase to the boost, and writes the element with a real quaternion modulus $\tilde q$ and a real quaternion angle $\tilde\Psi$:

$$ \tilde{Q} = \tilde q\,\exp(i\tilde\Psi), \qquad \tilde q = r\,\tilde u_{\mathrm r}, \qquad \exp(i\tilde\Psi) = \exp(i\alpha)\,\tilde u_{\mathrm r}^{-1}\tilde U_{\mathrm b}\tilde u_{\mathrm r} . $$

The fourth factor of the word, the rotor $\tilde u_{\mathrm r}$, is moved to the left, past the central phase, next to the scale, and the boost is conjugated by it so that the two non-central factors can stand together on the right as one exponential. The modulus $\tilde q = r\tilde u_{\mathrm r}$ is a real quaternion, and the angle is the real quaternion $\tilde\Psi = \alpha+\theta_{\mathrm b}\vec u_{\mathrm b}'$, with $\theta_{\mathrm b} = \psi/2$ half the rapidity and $\vec u_{\mathrm b}' = \tilde u_{\mathrm r}^{-1}\vec u_{\mathrm b}\tilde u_{\mathrm r}$ the boost axis rotated by the rotor. The element is thus determined by the two real quaternions $\tilde q$ and $\tilde\Psi$. The split itself is made by the real quaternion part $\tilde q_r$ of the element, which the construction has to invert. The pairing is not named for the complex numbers. Its only complex object is the central phase, and $\mathbb{C}$ has a polar decomposition of its own in Modulus–Unit Polar Decomposition of Complex Numbers; the older name complex representation invited a confusion that the name here avoids. One reading of the first factor is worth stating at the outset: written $\tilde q = r\tilde u_{\mathrm r}$, the modulus $\tilde q$ is the polar decomposition of an ordinary quaternion in the sense of Modulus–Unit Polar Decomposition of Quaternions, with the same two factors and the same symbols. The pairing therefore carries the polar word of the quaternion algebra as its first factor, and only the second factor leaves that algebra.

The pairing is partial: its modulus is a composite object, the product of the scale and the rotor, and the pairing as constructed is available only where that object can be read from the parts of the element. The other partial pairing, the Hamilton one, groups the scale with the phase and the boost with the rotor; the third grouping, in which the scale joins the boost and the phase joins the rotor, is not partial and is The Cartan Hermitian Pairing of the Polar Decomposition of Biquaternions.

The approach. The article follows the skeleton shared by the three pairing articles: the results imported, the pairing and its two factors, existence and domain, the comparison of the three bracketings, the three conjugations, the degenerate limits, and the worked examples. Each result is collected in a table, and each imported proof is cited rather than repeated. The four factors are taken from Modulus–Unit Polar Decomposition of Biquaternions and Exponential Polar Decomposition of Biquaternions; the only proof of the article is the identity that ties the construction to the polar modulus. No physics is invoked.

The result. Writing $\tilde Q=\tilde q_r+i\tilde q_i$ with real quaternion parts, the element is the product $\tilde q\exp(i\tilde\Psi)$ of the quaternion modulus and the quaternion angle. The factorisation is a regrouping of the polar word on $\langle\tilde Q\rangle_{\natural}\neq0$; when $\tilde q_r$ is invertible, the factors are $\tilde q=\tilde q_r(\cos\tilde\Psi)^{-1}$ and $\tan\tilde\Psi=\tilde q_r^{-1}\tilde q_i$. The pairing is collected in the table.

The Results Used

The article takes the following results as established and does not prove them again.

The key identity proved here is that of the construction with the polar modulus, $\tilde q=\tilde q_r(\cos\tilde\Psi)^{-1}$; it is proved in §The Construction From the Parts.

The Pairing

The Definition

The quaternion pairing is the pairing of the scale with the rotor against the phase with the boost:

$$ \tilde{Q} = \tilde q\,\exp(i\tilde\Psi), \qquad \tilde q = r\,\tilde u_{\mathrm r}, \qquad \exp(i\tilde\Psi) = \exp(i\alpha)\,\tilde u_{\mathrm r}^{-1}\tilde U_{\mathrm b}\tilde u_{\mathrm r} . $$

The modulus is a real quaternion, of norm $\langle\tilde q\rangle_{\natural} = r^2$, and the exponential is central in its phase but not in its boost: the boost has been conjugated by the rotor, so that it can stand to the right of the modulus in the canonical order. Writing the conjugated boost as

$$ \tilde u_{\mathrm r}^{-1}\tilde U_{\mathrm b}\tilde u_{\mathrm r} = \cosh\theta_{\mathrm b}\,e_t + i\sinh\theta_{\mathrm b}\,\vec u_{\mathrm b}', \qquad \theta_{\mathrm b} = \frac{\psi}{2}, $$

with $\psi$ the rapidity of $\tilde U_{\mathrm b}$ and $\vec u_{\mathrm b}'$ its axis rotated by the rotor, the exponent is

$$ \boxed{\; \tilde\Psi = \alpha + \theta_{\mathrm b}\,\vec u_{\mathrm b}' \;}, $$

a real quaternion with scalar part $\alpha$ and pure quaternion part $\theta_{\mathrm b}\vec u_{\mathrm b}'$. This is the arrangement in which the pairing is stated: its $\tilde\Psi$ has a scalar part and a pure part, and the two travel together.

The Statement

The pairing is stated as follows. Write $\tilde{Q} = \tilde q_r + i\tilde q_i$ with $\tilde q_r,\tilde q_i\in\mathbb{H}$ the real and imaginary quaternion parts, suppose $\tilde q_r \neq 0$, put $\tan\tilde\Psi = \tilde q_r^{-1}\tilde q_i$ and $\tilde q = \tilde q_r(\cos\tilde\Psi)^{-1}$. Then $\tilde{Q} = \tilde q\exp(i\tilde\Psi)$ with $\tilde q$ and $\tilde\Psi$ real quaternions. The verification is the computation $\tilde q\exp(i\tilde\Psi) = \tilde q_r(\cos\tilde\Psi)^{-1}(\cos\tilde\Psi+i\sin\tilde\Psi) = \tilde q_r + i\tilde q_r\tan\tilde\Psi = \tilde q_r+i\tilde q_i$, in which the two trigonometric factors commute because they are functions of the single quaternion $\tilde\Psi$.

The Construction From the Parts

The modulus $\tilde q = r\tilde u_{\mathrm r}$ of the pairing is exactly the modulus $\tilde q_r(\cos\tilde\Psi)^{-1}$ of the construction from the element's parts. The identity is not a coincidence of two constructions; it follows from the polar decomposition. Since $\tilde q = r\tilde u_{\mathrm r}$ is a real quaternion and $\exp(i\tilde\Psi)$ is the exponential of $i$ times a real quaternion, the real quaternion part of $\tilde{Q} = \tilde q\exp(i\tilde\Psi)$ is

$$ \tilde q_r = \tilde q\,\frac{\exp(i\tilde\Psi)+\exp(-i\tilde\Psi)}{2} = \tilde q\cos\tilde\Psi , $$

because the complex conjugation of $\exp(i\tilde\Psi)$ is $\exp(-i\tilde\Psi)$ and the modulus $\tilde q = r\tilde u_{\mathrm r}$ is fixed by it. Hence $\tilde q = \tilde q_r(\cos\tilde\Psi)^{-1}$, which is the construction formula. The identity is therefore the equality of the modulus of the pairing with the modulus of the construction, not an agreement of two independent constructions.

The First Factor: The Quaternion Modulus

The factor is the modulus of the ordinary quaternion, and its data are these.

The Two-Factor Word of the Ordinary Quaternions

The modulus $\tilde q=r\tilde u_{\mathrm r}$ is not a new object. Read inside the ordinary quaternions $\mathbb{H}$, it is exactly the polar decomposition $\tilde q=r\tilde u_{\mathrm r}$ of Modulus–Unit Polar Decomposition of Quaternions, §The Two Factors: the same positive real $r$ and the same unit factor $\tilde u_{\mathrm r}\in\mathrm{Sp}(1)$. The quaternion pairing therefore reads

$$ \tilde Q=\bigl(r\,\tilde u_{\mathrm r}\bigr)\exp(i\tilde\Psi), $$

the polar form of an ordinary quaternion followed by the exponential of a real quaternion; only the second factor leaves $\mathbb{H}$.

The companion articles give the unit factor two forms, and the pairing inherits both. The formulas are those of Exponential Polar Decomposition of Quaternions, §The Two Factors and §The Rotor Part, and are reproduced only to name the objects.

The nested word has the properties that the second factor lacks. Its two factors are a positive real and a unit element, so the decomposition is unique and needs no construction and no choice of branch. It is fixed by complex conjugation, $\bar{\tilde q}=\tilde q$. And it is closed under the other two conjugations in the same shape, $\tilde q^{\natural}=r\tilde u_{\mathrm r}^{\natural}$ and $\tilde q^{*}=r\tilde u_{\mathrm r}^{\natural}$, again the polar decomposition of an ordinary quaternion with the same modulus $r$ and the conjugated unit factor.

The correspondence is a correspondence of objects, not of names: the modulus of the pairing is the ordinary quaternion written in the two-factor form, while the Hamilton modulus is the complex scalar $\Rho=r\exp(i\alpha)$, an object of the centre. This is one of the two senses, independent of domain, in which the two named pairings are complementary; the other is recorded in §Behaviour Under the Three Conjugations.

The Second Factor: The Exponential

The second factor is the exponential of a real quaternion,

$$ \exp(i\tilde\Psi) = \exp(i\alpha)\,\tilde u_{\mathrm r}^{-1}\tilde U_{\mathrm b}\tilde u_{\mathrm r}, \qquad \tilde\Psi = \alpha+\theta_{\mathrm b}\vec u_{\mathrm b}' , $$

with $\theta_{\mathrm b}=\psi/2$ and $\vec u_{\mathrm b}'=\tilde u_{\mathrm r}^{-1}\vec u_{\mathrm b}\tilde u_{\mathrm r}$. It is not central and not real in general: its scalar part is $\exp(i\alpha)\cosh\theta_{\mathrm b}$ and its vector part $\exp(i\alpha)\,i\sinh\theta_{\mathrm b}\,\vec u_{\mathrm b}'$. It is nevertheless the element divided on the left by the modulus,

$$ \boxed{\; \exp(i\tilde\Psi)=\tilde q^{-1}\tilde Q \;}, $$

because $\tilde q=r\tilde u_{\mathrm r}$ is invertible. The modulus carries the scale, $\langle\tilde q\rangle_{\natural}=r^{2}$, and the second factor the phase and the rotor-conjugated boost, $\langle\exp(i\tilde\Psi)\rangle_{\natural}=\exp(2i\alpha)$; the product is the norm of the element.

The exponential and the logarithm of a biquaternion are those of Biquaternion Elementary Functions, §The Exponential and §The Logarithm. The logarithm of the second factor is $i\tilde\Psi$ up to the kernel of the exponential; the article uses that result and does not compute the kernel.

Existence, Uniqueness and the Domain

The Theorem

Theorem (the quaternion pairing). Let $\tilde Q\in\mathbb{B}$ with $\langle\tilde Q\rangle_{\natural}\neq0$. Then $\tilde Q=\tilde q\exp(i\tilde\Psi)$, with $\tilde q=r\tilde u_{\mathrm r}$ a real quaternion and $\tilde\Psi$ a real quaternion, as a regrouping of the polar word. If in addition $\tilde q_r\neq0$, the factors are $\tan\tilde\Psi=\tilde q_r^{-1}\tilde q_i$ and $\tilde q=\tilde q_r(\cos\tilde\Psi)^{-1}$.

The factorisation $\tilde Q=(r\tilde u_{\mathrm r})\exp(i\tilde\Psi)$ exists whenever the four-factor word exists, that is, whenever $\langle\tilde Q\rangle_{\natural}\neq0$: it is a regrouping of unique factors, so it inherits their existence and their uniqueness. The construction from the parts is a separate statement, with a smaller domain. The two are separated below.

Uniqueness and the Branch

The factorisation is unique because the four factors are: the two factors $\tilde q=r\tilde u_{\mathrm r}$ and $\exp(i\tilde\Psi)$ are the scale-times-rotor and the phase-times-boost of a unique word, so their product is determined by the element. The construction is determined only up to $(\tilde q,\tilde\Psi)\mapsto(-\tilde q,\tilde\Psi+\pi)$, the branch ambiguity of the tangent, since $\exp(i(\tilde\Psi+\pi))=-\exp(i\tilde\Psi)$ and the two signs cancel. The sign of $\tilde q$ is fixed by the polar word: $\tilde q=r\tilde u_{\mathrm r}$ with $r>0$, so the representative is the one whose norm is the positive scale $r$.

The Domain

The construction from the element's parts needs $\tilde q_r^{-1}$, so it is available where the real quaternion part is invertible,

$$ \{\tilde{Q} : \tilde q_r\neq0\} , $$

the condition $\tilde q_r\neq0$ being equivalent to invertibility because $\mathbb{H}$ is a division algebra.

This condition is necessary but not sufficient. The construction needs both $\langle\tilde Q\rangle_{\natural}\neq0$ and $\tilde q_r\neq0$, and its domain is their intersection. The two conditions are independent. The element $i\vec e_x+i\vec e_y$ has $\langle\tilde Q\rangle_{\natural} = -2$ and $\tilde q_r=0$: the factorisation exists and the construction fails. The element $e_t+i\vec e_x$ has $\tilde q_r=e_t$ and $\langle\tilde Q\rangle_{\natural}=1+i^{2}=0$: the construction has $\tilde q_r$ to divide by and the factorisation does not exist.

On the domain of the word, however, $\tilde q_r\neq0$ is also sufficient. The tangent of a real quaternion $a+b\vec u$ is $(\tan a+\tanh b\,\vec u)/(1-\tan a\tanh b\,\vec u)$, so its image omits exactly the pure unit quaternions: the equation $\tan\tilde\Psi=T$ has a real solution unless $T^{2}=-e_t$. When it fails, $\tilde q_i=\tilde q_rT$ and $$ \langle\tilde Q\rangle_{\natural}=\tilde Q\tilde Q^{\natural}=\tilde q_r\,(e_t+iT)(e_t-iT)\,\tilde q_r^{\natural}=\tilde q_r\,(e_t+T^{2})\,\tilde q_r^{\natural}=0 . $$ That failure is therefore on the null cone and nowhere else. Off the cone the tangent has a real solution, and the construction produces $\tilde q$ and $\tilde\Psi$ with $\tilde Q=\tilde q\exp(i\tilde\Psi)$. Taking norms and writing $\alpha=\operatorname{Sc}\tilde\Psi$ gives $\langle\tilde Q\rangle_{\natural}=\langle\tilde q\rangle_{\natural}\exp(2i\alpha)=r^{2}\exp(2i\alpha)\neq0$, since $\tilde q$ is a nonzero real quaternion. The condition $\tilde q_r\neq0$ is therefore exactly the extra condition on the word's domain, and the set $\{\tilde q_r\neq0\}$ is not contained in the word's domain.

The Hamilton domain and the quaternion construction domain are not nested either: $1+\vec e_x+i\vec e_y$ has $\tilde q_r = 1+\vec e_x \neq 0$ but no Hamilton axis, and $i\vec e_x+i\vec e_y$ has a Hamilton axis but no quaternion construction.

The Factorisation and the Construction

The factorisation and the construction are therefore different statements. The factorisation is the regrouping of the four factors and exists on the whole complement of the null cone; the construction is the formula $\tan\tilde\Psi=\tilde q_r^{-1}\tilde q_i$ and needs the word to exist and $\tilde q_r$ invertible. On $i\vec e_x+i\vec e_y$ the factorisation is explicit,

$$ \boxed{\; i\vec e_x+i\vec e_y = \left(\vec e_x+\vec e_y\right)\,\exp\!\left(i\frac{\pi}{2}\right) \;}, $$

with modulus $\vec e_x+\vec e_y$, a real quaternion, and exponent $\pi/2$, real, while the construction has no $\tilde q_r$ to invert.

The Three Pairings

The Three Bracketings

The four factors can be bracketed in exactly three ways, and each bracketing is a pairing. The table records the two factors and the norm each carries; the domains are the subject of the next subsection.

The three rows split the same norm $\Rho^{2}=r^{2}\exp(2i\alpha)$ in three ways. The Hamilton row puts the whole norm in a central first factor and leaves a unit second factor. The quaternion and Cartan rows put the scale in the first factor and the phase in the second; they differ in the kind of the first factor, a real quaternion in one and a Hermitian positive element in the other. The three moduli are therefore a complex scalar, a real quaternion and a Hermitian positive element, and the four-factor word is the reference in which the scale stands alone.

The Domains and the Witnesses

Five elements separate the cases. The Hamilton axis needs $B^{2}=\langle\vec Q,\vec Q\rangle_{\natural}\neq0$, the quaternion construction needs $\langle\tilde Q\rangle_{\natural}\neq0$ and $\tilde q_r\neq0$, and the Cartan pairing needs only $\langle\tilde Q\rangle_{\natural}\neq0$.

The first line is the generic case, in which all three pairings exist and regroup the same four factors in three ways. The third line separates the construction of this article: the norm is $-2$, so the factorisation exists, while $\tilde q_r=0$, so the construction fails. The second line separates the Hamilton pairing, and the fourth line is missed by both partial pairings. The fifth line separates the two conditions on the quaternion construction: $\tilde q_r=e_t$ is invertible and the element lies on the cone, so neither the word nor the construction exists.

The two senses of partial are separated in the table.

The quaternion row is partial in its construction only: the factorisation inherits the word, while the construction must invert $\tilde q_r$. The Hamilton row is partial in both senses, and the Cartan row in neither.

The Hamilton domain and the quaternion construction domain are not nested, and the Cartan domain contains both; The Cartan Hermitian Pairing of the Polar Decomposition of Biquaternions, §The Domains and the Witnesses records the exact cover of the complement of the null cone.

What This Pairing Carries

The quaternion row is the only one whose first factor is a real quaternion: it is the polar decomposition of the ordinary quaternions, and the only modulus that lies in $\mathbb{H}_{\mathbb{B}}$. Its second factor is the only one whose exponent is a real quaternion. These two facts, and not the domain, are what the pairing contributes to the series: the factorisation is a regrouping of the word, while the construction is the formula from the parts, and §The Domains and the Witnesses separates their conditions.

Behaviour Under the Three Conjugations

The algebra has three nontrivial conjugations: quaternion conjugation ${}^{\natural}$, the anti-automorphism fixing the centre $\mathbb{C}_{\mathbb{B}}$; complex conjugation $\bar{\cdot}$, the automorphism conjugating every coefficient; and Hermitian conjugation ${}^{*} = {}^{\natural}\circ\bar{\cdot}$, the two operations commuting because one acts on the coefficients and the other on the units. The three act on the quaternion pairing below as a computation rather than as a contrast of styles; their action on the Hamilton data is computed in The Hamilton Root Pairing of the Polar Decomposition of Biquaternions, and the contrast between the two is what the closing subsection records.

The Quaternion Data

In $\tilde{Q} = \tilde q\exp(i\tilde\Psi)$ the modulus $\tilde q$ and the angle $\tilde\Psi$ are real quaternions, so $\bar{\tilde q} = \tilde q$ and $\bar{\tilde\Psi} = \tilde\Psi$, while $\tilde q^{\natural}\neq \tilde q$ and $\tilde\Psi^{\natural}\neq\tilde\Psi$ in general; the two factors need not commute, and it is this that makes the quaternion pairing the less symmetric of the two under conjugation.

Complex conjugation. Since $\bar{\cdot}$ is an automorphism fixing $\tilde q$ and $\tilde\Psi$ and conjugating the scalar imaginary,

$$ \bar{\tilde{Q}} = \tilde q\exp(-i\tilde\Psi) : $$

the modulus and the angle are both unchanged and only the sign of the exponent changes. Equivalently, the canonical angle of $\bar{\tilde{Q}}$ is $-\tilde\Psi$, because the real and imaginary quaternion parts of $\bar{\tilde{Q}}$ are $(\tilde q_r,-\tilde q_i)$, whence $\tan\tilde\Psi(\bar{\tilde{Q}}) = \tilde q_r^{-1}(-\tilde q_i) = -\tan\tilde\Psi$.

Quaternion conjugation. Since ${}^{\natural}$ is an anti-automorphism, the two factors exchange order:

$$ (\tilde{Q})^{\natural} = \exp(i\tilde\Psi^{\natural})\,\tilde q^{\natural} , $$

which is not the canonical shape, the modulus of the quaternion pairing being the left factor. Recovering the canonical shape requires the parts of the conjugated element, and the result is not the conjugate angle in general. The parts of $(\tilde{Q})^{\natural}$ are $(\tilde q_r^{\natural},\tilde q_i^{\natural})$, so its angle $\tilde\Psi'$ satisfies

$$ \tan\tilde\Psi' = (\tilde q_r^{\natural})^{-1}\tilde q_i^{\natural} , \qquad\text{whereas}\qquad (\tan\tilde\Psi)^{\natural} = \tilde q_i^{\natural}(\tilde q_r^{\natural})^{-1} , $$

and the two agree exactly when $\tilde q_r^{\natural}$ and $\tilde q_i^{\natural}$ commute, equivalently when $\tilde q_r$ and $\tilde q_i$ commute. The commuting case is exact and the non-commuting case is genuinely different: the element

$$ \tilde{Q} = (1+\vec e_x) + i(\vec e_x+\vec e_y), \qquad \langle\tilde Q\rangle_{\natural} = 2i, \qquad \tilde q_r = 1+\vec e_x\ \text{invertible} , $$

has $\tan\tilde\Psi = \tfrac12(1+\vec e_x+\vec e_y-\vec e_z)$, conjugate $\tfrac12(1-\vec e_x-\vec e_y+\vec e_z)$, and angle of the conjugated element $\tan\tilde\Psi' = \tfrac12(1-\vec e_x-\vec e_y-\vec e_z)$; the two differ in the $\vec e_z$ coefficient alone, the commutator being $\tilde q_r\tilde q_i-\tilde q_i\tilde q_r = 2\vec e_z$.

Hermitian conjugation. Combining the two, $\tilde{Q}^{*} = \exp(-i\tilde\Psi^{\natural})\,\tilde q^{\natural}$, again with the factors in the reversed order.

The three actions are collected in the table, in the order quaternion, complex, Hermitian.

The two anti-automorphisms reverse the order and the automorphism preserves it; the canonical angle of the conjugate is a quaternion conjugate only in the commuting case.

What the Comparison Shows

The two named representations are distinguished by conjugation as sharply as by domain. For the Hamilton representation all three conjugations act on the parameters, ${}^{\natural}$ reversing the axis, $\bar{\cdot}$ conjugating the three parameters and ${}^{*}$ doing both, and the representation is closed under every one of them, the computation being in The Hamilton Root Pairing of the Polar Decomposition of Biquaternions. For the quaternion pairing, complex conjugation touches only the scalar imaginary of the exponent, while quaternion conjugation and Hermitian conjugation exchange the order of the two factors, and quaternion conjugation carries the canonical angle to the quaternion conjugate of the angle only when the two quaternion parts of the element commute. A statement about a biquaternion and its conjugates is therefore cheapest in the representation whose conjugation action it needs, and this is a second sense, independent of domain, in which the Hamilton and quaternion pairings are complementary.

The Two Degenerate Limits

Two exact statements identify which classical polar form each named pairing generalises. Both are cases in which the element lies in a proper subalgebra of $\mathbb{B}$: a real quaternion, or a complex scalar $z e_t$ with $z\in\mathbb{C}$ and $\operatorname{Re}z\neq0$.

A real quaternion. If all four coefficients of $\tilde{Q}$ are real and the vector part is nonzero then $\tilde{Q}\in\mathbb{H}_{\mathbb{B}}$, the biquaternion norm $\langle\tilde Q\rangle_{\natural} = \sum_\mu q_\mu^2$ is a positive real, and with the principal branch of the square root

$$ \Rho = \sqrt{\langle\tilde Q\rangle_{\natural}} \in \mathbb{R}_{>0}, \qquad \cos\Theta = \frac{q_t}{\Rho}\in[-1,1], \qquad \sin\Theta = \frac{|\vec q|}{\Rho}\in[0,1] , $$

so both the modulus and the angle are real and $\vec U = \vec q/|\vec q|$ is a unit pure real quaternion. The Hamilton representation is then the quaternion polar decomposition of the companion articles Modulus–Unit Polar Decomposition of Quaternions and Exponential Polar Decomposition of Quaternions, in the two forms they give it: the two-factor form $\tilde q = r\tilde u_{\mathrm r}$ with $r = \Rho$ and $\tilde u_{\mathrm r} = \exp(\vec U\Theta)$, and the exponential form $\exp(\ln r)\exp(\mu\tfrac{\theta}{2})$ with $\mu = \vec U$ and $\theta = 2\Theta$, the exponential article's $\theta$ being twice the angle of the unit factor, which is $\Theta$ here. The identification is exact rather than asymptotic. The quaternion pairing is available on the same elements and is degenerate there: $\tilde q_i = 0$, hence $\tan\tilde\Psi = 0$, $\tilde\Psi = 0$ and $\tilde q = \tilde q_r = \tilde{Q}$, the element itself in the two-factor form $\tilde q = r\tilde u_{\mathrm r}$ of the modulus–unit article, with $r = \sqrt{\langle\tilde Q\rangle_{\natural}}$ and $\tilde u_{\mathrm r} = \tilde{Q}/r$; the second factor is trivial. In this limit the two named pairings are two readings of the same two-factor word of the modulus–unit article: the quaternion pairing gives the word itself, with its second factor trivial, and the Hamilton pairing gives the exponential form of its unit factor.

A complex scalar. If $\tilde{Q} = ze_t$ with $z\in\mathbb{C}$ and $\operatorname{Re}z\neq0$ then the vector part vanishes, so $B = 0$ and the Hamilton representation is unavailable. This is the scalar vertex of Spectral Regularity and Degeneracy in the Biquaternion Algebra, §The Behaviour on the Cone: the element is semisimple with a repeated eigenvalue, and its eigenprojections are not unique. The quaternion pairing gives

$$ \tilde q_r = (\operatorname{Re}z)e_t, \qquad \tilde q_i = (\operatorname{Im}z)e_t, \qquad \tan\tilde\Psi = \frac{\operatorname{Im}z}{\operatorname{Re}z}\,e_t, \qquad \tilde q = \frac{\operatorname{Re}z}{\cos\tilde\Psi}\,e_t = |z|\,e_t , $$

so that $\tilde\Psi = (\arg z)e_t$ is a scalar and $\tilde{Q} = |z|\exp\!\big(i(\arg z)e_t\big)$: the quaternion pairing reduces to the ordinary complex polar form of $\mathbb{C}$, with the central scalar imaginary of the algebra in the role of the scalar imaginary of $\mathbb{C}$ and the quaternion modulus collapsed to a scalar.

The two limits are disjoint, and together they are the reason the two named pairings are complementary rather than competing: the Hamilton representation is the one whose whole word reduces to the quaternion polar form, and the quaternion pairing is the one that carries the complex polar form.

Worked Examples

The four elements are collected in the table, with the status of the factorisation and of the construction and with the two factors. The subsections give the computations and the readings.

The Generic Element

Take $\tilde{Q} = (1+i)e_t+\vec e_x+2\vec e_y$, the worked example of the companion article. Its polar decomposition is

$$ r = 2.320595787, \quad \alpha = 0.190253189, \quad \tilde U_{\mathrm b} = 1.072349609\,e_t - 0.173166789\,i\,\vec e_x - 0.346333577\,i\,\vec e_y, \quad \tilde u_{\mathrm r} = 0.470592172\,e_t+0.394599292\,\vec e_x+0.789198585\,\vec e_y . $$

The regrouping. The modulus is the real quaternion

$$ \tilde q = r\tilde u_{\mathrm r} = 1.092054\,e_t + 0.915705\,\vec e_x + 1.831411\,\vec e_y , $$

matching the modulus $\tilde q_r(\cos\tilde\Psi)^{-1}$ of the construction with $\tilde q_r = 1+\vec e_x+2\vec e_y$, and the exponent is

$$ \tilde\Psi = \alpha + \frac{\psi}{2}\vec u_{\mathrm b}' = 0.190253 - 0.169108\,\vec e_x - 0.338216\,\vec e_y , $$

where $\psi = 0.756273220$ is the rapidity of $\tilde U_{\mathrm b}$ and $\vec u_{\mathrm b}' = -(1,2,0)/\sqrt5$ is its axis, unchanged by the rotor because the rotor's axis is parallel to it.

The two partial regroupings of the same four factors are therefore displayed on one element, the Hamilton regrouping of it being computed in The Hamilton Root Pairing of the Polar Decomposition of Biquaternions, and their moduli are visibly different objects: a complex scalar and a real quaternion. The third pairing, the Cartan pairing, groups the four factors differently again and is treated in The Cartan Hermitian Pairing of the Polar Decomposition of Biquaternions.

The Element with a Null Vector Part

Take $\tilde{Q} = 1+\vec e_x+i\vec e_y$. Then $\langle\tilde Q\rangle_{\natural} = 1+1+i^2 = 1$ and $\langle\vec Q,\vec Q\rangle_{\natural} = 1+i^2 = 0$, so the Hamilton representation does not exist: $B = 0$ and the axis is undefined. The quaternion pairing does exist, with

$$ r = 1, \qquad \alpha = 0, \qquad \tilde U_{\mathrm b} = 1.414213562\,e_t + 0.707106781\,i\,\vec e_y + 0.707106781\,i\,\vec e_z, \qquad \tilde u_{\mathrm r} = 0.707106781\left(e_t+\vec e_x\right) , $$

so that the quaternion modulus is $\tilde q = \tilde u_{\mathrm r} = 0.707106781(e_t+\vec e_x)$. The modulus computed from the parts, $\tilde q_r(\cos\tilde\Psi)^{-1}$, is the same quaternion, and this equality is the identity proved in the section on the pairing above; the element is also the non-scalar example of that section.

The Element with a Vanishing Real Part

Take $\tilde{Q} = i\vec e_x+i\vec e_y$. Then $\langle\tilde Q\rangle_{\natural} = -2$ and $\tilde q_r = 0$, so the construction of the quaternion pairing from the element's parts fails. The polar decomposition gives

$$ r = \sqrt{2}, \qquad \alpha = \frac{\pi}{2}, \qquad \tilde U_{\mathrm b} = e_t, \qquad \tilde u_{\mathrm r} = \frac{\vec e_x+\vec e_y}{\sqrt{2}} , $$

so the Hamilton representation exists, with $B = \sqrt{-2} = i\sqrt{2}$, axis $\vec U = (\vec e_x+\vec e_y)/\sqrt2$, $\Rho = i\sqrt2$, $\cos\Theta = 0$ and $\sin\Theta = 1$:

$$ \tilde{Q} = i\sqrt{2}\,\exp\!\left(\frac{\vec e_x+\vec e_y}{\sqrt2}\frac{\pi}{2}\right) . $$

The quaternion pairing, on the other hand, exists with modulus $r\tilde u_{\mathrm r} = \vec e_x+\vec e_y$ and exponent $\tilde\Psi = \alpha = \pi/2$, as recorded above. The example therefore exhibits both the failure of the construction and the existence of the factorisation.

The Scalar Imaginary

Take $\tilde{Q} = i$. Then $\langle\tilde Q\rangle_{\natural} = -1$, so the element is invertible and the polar decomposition is $i = 1\cdot \exp(i\pi/2)\cdot e_t\cdot e_t$: the scale is one and the boost and the rotor are trivial. Both partial constructions fail, since $\vec Q = 0$ and $\tilde q_r = 0$; the element is nevertheless in the domain of the polar decomposition, and the Cartan pairing, which needs only $\langle\tilde Q\rangle_{\natural}\neq0$, is available as $i = e_t\cdot i$. The example is the clearest single illustration of the difference between having a representation and being able to construct one from the element's parts.

Summary

Main Results

The quaternion pairing groups the scale with the rotor into the real quaternion modulus $\tilde q=r\tilde u_{\mathrm r}$ and the phase with the rotor-conjugated boost into the exponential $\exp(i\tilde\Psi)$. Its first factor is exactly the polar decomposition of the ordinary quaternions. The construction reads the pairing off the parts of the element, $\tan\tilde\Psi=\tilde q_r^{-1}\tilde q_i$ and $\tilde q=\tilde q_r(\cos\tilde\Psi)^{-1}$, and the two readings of the modulus agree.

The construction needs $\tilde q_r$ invertible; the pairing itself needs only $\langle\tilde Q\rangle_{\natural}\neq0$, because it is a regrouping of the four factors, which exist on the whole complement of the null cone. This is the difference between the factorisation and the construction. The Hamilton domain and the quaternion construction domain are not nested; the Cartan pairing is the third, total in both senses.

Under the three conjugations complex conjugation fixes the modulus and flips the sign of the exponent, while quaternion and Hermitian conjugation reverse the order of the two factors. In the degenerate limits a non-scalar real quaternion gives a trivial second factor, and a complex scalar with nonzero real part gives the ordinary complex polar form.

Summary of Notation

Further Reading

• The Biquaternion Eigenvalue Decomposition, §The Eigenvalues, for the eigenvalues $\lambda_\pm=\Rho\exp(\pm i\Theta)$ and the pure unit element $\vec U=\vec Q/B$ of the Hamilton row.
• Biquaternion Elementary Functions, for the exponential and the logarithm of which the second factor of the pairing is built, and which carry the branch ambiguity of the construction.
• Exponential Polar Decomposition of Biquaternions, for the four exponentials and the existence and uniqueness of the word that the pairing regroups.
• Modulus–Unit Polar Decomposition of Biquaternions, for the four factors, the modulus $\Rho = \sqrt{\langle\tilde Q\rangle_{\natural}}$, the domain, and §The Algorithm of that article for the procedure itself, the factors in the coordinates of $\tilde{Q}$ and the exponents with their positions.
• The Hamilton Root Pairing of the Polar Decomposition of Biquaternions, for the Hamilton grouping, its complex modulus and its non-central root of $-1$, the exponential of the element, the non-uniqueness of its parameters, and the reading of the same word at the roots of one.
• The Cartan Hermitian Pairing of the Polar Decomposition of Biquaternions, for the third pairing of the four factors, its Hermitian and anti-Hermitian exponent split, its full domain and its identification with the matrix polar decomposition.
• The Generalized Polar Decomposition of a Biquaternion: Partial Isometries and the Null Cone, for the two-factor Cartan form on the null cone, where the construction of this article stops, and for the positive semidefinite factor and partial isometry that replace the Cartan factors there.
• Modulus–Unit Polar Decomposition of Quaternions and Exponential Polar Decomposition of Quaternions, for the polar decomposition $\tilde q = r\tilde u_{\mathrm r}$ of the ordinary quaternions, which is the first factor of the pairing, for its exponential form $\exp(\ln r)\exp(\mu\tfrac{\theta}{2})$, which the quaternion pairing reproduces on the real quaternions, and for the degenerate limit in which the two-factor form is read on the element itself.
• The Biquaternion Algebra and Its Vector Spaces, for the real, complex and quaternionic coordinate systems and the convention $Q_\mu = q_\mu+iq'_\mu$ used in the derivation above.
• Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, for the null cone, where the four-factor word and the unitary pairings stop and the generalized pairing of the companion article begins.
• Exponential Polar Decomposition of Complex Numbers and Exponential Polar Decomposition of Split-Complex Numbers, for the two-dimensional members of the series and the rule of the exponential $\exp(\nu\theta)$ that the factors of these pairings are built from.