Biquaternion Square Roots of a General Element

Introduction

This article solves the square-root problem in the biquaternion algebra $\mathbb{B}$: given $\tilde Q\in\mathbb{B}$, find every $\tilde P\in\mathbb{B}$ with

$$ \tilde P^2 = \tilde Q . $$

The method needs no Clifford algebra, no norm and no form. It is the same vector–scalar split that solves the three central cases $\tilde P^2=-1,0,+1$ in Biquaternion Square Roots of Minus One, Zero and Plus One, carried one step further: the split turns the equation into one vector equation and one scalar equation, and the scalar equation is a quadratic in the square $x=P_0^2$ of the scalar part of a root. The roots are then read off the solutions of that quadratic, and the answer is a finite set — generically four elements, falling to two — or a four-parameter family, or empty.

The article is algebraic. The biquaternion algebra also carries a Clifford structure, treated in the topology part of the corpus in The Clifford Structure of the Biquaternion Algebra; that structure is not used here. The three central values of Biquaternion Square Roots of Minus One, Zero and Plus One are recovered as the case of a complex scalar $\tilde Q$, in §The Three Central Values, which checks this method against the direct computation of that article. The zero-divisor cone, of which the nonzero roots of $0$ are a part, is Biquaternion Zero Divisors, named here only where the classification touches it. The polar and exponential decompositions that the answer must be consistent with are Biquaternion Polar Element Representation.

The treatment is mathematically honest: every claim is either proved or stated as a definition. No physics is invoked.

Throughout, the quaternion basis is $e_0=1,e_1,e_2,e_3$, the scalar imaginary is $i$, and a general biquaternion is $\tilde Q=Q_0e_0+Q_1e_1+Q_2e_2+Q_3e_3$ with $Q_k\in\mathbb{C}$. The quaternion conjugate is $\tilde Q^{\natural}$, the complex conjugate is denoted $\bar{\tilde Q}$, and the Hermitian conjugate is $\tilde Q^{*}=\overline{\tilde Q^{\natural}}$.

The Reduction

Write the root and the radicand in scalar and vector parts,

$$ \tilde P = P_0e_0 + \boldsymbol{P}, \qquad \boldsymbol{P}=P_1e_1+P_2e_2+P_3e_3, $$

$$ \tilde Q = Q_0e_0 + \boldsymbol{Q}, \qquad \boldsymbol{Q}=Q_1e_1+Q_2e_2+Q_3e_3, $$

with all coefficients in $\mathbb{C}$. The product formula for biquaternions gives the square of $\tilde P$ as

$$ \tilde P^2 = \bigl(P_0^2-(\boldsymbol{P},\boldsymbol{P})\bigr)e_0 + 2P_0\boldsymbol{P}, \qquad (\boldsymbol{P},\boldsymbol{P})=P_1^2+P_2^2+P_3^2 . $$

Equating $\tilde P^2$ to $\tilde Q$, component by component, splits the single equation into two:

$$ 2P_0\boldsymbol{P} = \boldsymbol{Q} \qquad\text{(the vector part)}, $$

$$ P_0^2-(\boldsymbol{P},\boldsymbol{P}) = Q_0 \qquad\text{(the scalar part)} . $$

The vector equation involves only the vector part $\boldsymbol{Q}$ of the radicand, and it decides the shape of the answer. It has two outcomes.

Case A — the radicand has non-vanishing vector part ($\boldsymbol{Q}\neq0$). Then $P_0\neq0$ and $\boldsymbol{P}=\boldsymbol{Q}/(2P_0)$: a root is fixed by its scalar part, and the problem collapses onto the scalar equation. This case is §The Roots with Non-Vanishing Vector Part.

Case B — the radicand is a complex scalar ($\boldsymbol{Q}=0$). Then $2P_0\boldsymbol{P}=0$, which forces $\boldsymbol{P}=0$ or $P_0=0$: a root is either a complex scalar or pure. This case is §The Roots of a Complex Scalar.

Two data are used throughout. The complex square of the vector part is

$$ (\boldsymbol{Q},\boldsymbol{Q}) = Q_1^2+Q_2^2+Q_3^2 , $$

and the conjugate scalar is

$$ n = Q_0^2+Q_1^2+Q_2^2+Q_3^2 = Q_0^2+(\boldsymbol{Q},\boldsymbol{Q}) . $$

The conjugate scalar is not a norm here: it is the algebraic identity

$$ \tilde Q\tilde Q^{\natural} = \bigl(Q_0^2+Q_1^2+Q_2^2+Q_3^2\bigr)e_0 = n\,e_0 , $$

in which $\tilde Q^{\natural}$ is the quaternion conjugate. This uses only the product and the conjugate, both of which the algebra carries before any form is introduced.

The Roots with Non-Vanishing Vector Part

Assume $\boldsymbol{Q}\neq0$. Then $2P_0\boldsymbol{P}=\boldsymbol{Q}$ forces $P_0\neq0$, and the vector equation determines the vector part of every root,

$$ \boldsymbol{P} = \frac{\boldsymbol{Q}}{2P_0} . $$

Substituting this into the scalar equation gives

$$ P_0^2-\frac{(\boldsymbol{Q},\boldsymbol{Q})}{4P_0^2} = Q_0 . $$

Put

$$ x = P_0^2 . $$

Since $(\boldsymbol{Q}/(2P_0),\boldsymbol{Q}/(2P_0)) = (\boldsymbol{Q},\boldsymbol{Q})/(4x)$, the scalar equation becomes the reduced quadratic

$$ 4x^2 - 4Q_0x - (\boldsymbol{Q},\boldsymbol{Q}) = 0 . $$

Its discriminant is

$$ 16Q_0^2 + 16(\boldsymbol{Q},\boldsymbol{Q}) = 16n , $$

so its two solutions are

$$ x_+ = \frac{Q_0+\sqrt{n}}{2}, \qquad x_- = \frac{Q_0-\sqrt{n}}{2}, $$

where $\sqrt{n}$ is either square root of the complex number $n$; the choice of square root only exchanges $x_+$ and $x_-$.

Proposition (the roots with non-vanishing vector part). Let $\boldsymbol{Q}\neq0$, and let $x_+$ and $x_-$ be the two solutions of the reduced quadratic. For each solution $x\in\{x_+,x_-\}$ with $x\neq0$, and for each of the two square roots $P_0=\pm\sqrt{x}$, the element

$$ \tilde P = P_0e_0 + \frac{\boldsymbol{Q}}{2P_0} $$

is a square root of $\tilde Q$, and every square root of $\tilde Q$ with $\boldsymbol{Q}\neq0$ arises this way.

Proof. The derivation above shows that a root with $\boldsymbol{Q}\neq0$ has $P_0\neq0$, has $\boldsymbol{P}=\boldsymbol{Q}/(2P_0)$, and has $x=P_0^2$ solving the reduced quadratic. Conversely, given such an $x$ and $P_0=\pm\sqrt{x}\neq0$, the two equations hold by construction, so $\tilde P^2=\tilde Q$. The two values $P_0=+\sqrt{x}$ and $P_0=-\sqrt{x}$ give opposite roots, so each nonzero solution contributes the pair $\pm\tilde P$. $\square$

Corollary (the number of roots). The solution $x=0$ occurs if and only if $(\boldsymbol{Q},\boldsymbol{Q})=0$, since the quadratic reads $-(\boldsymbol{Q},\boldsymbol{Q})=0$ at $x=0$. Hence there are four roots when $(\boldsymbol{Q},\boldsymbol{Q})\neq0$ and $n\neq0$, the two solutions being then distinct and nonzero; two roots when $(\boldsymbol{Q},\boldsymbol{Q})\neq0$ and $n=0$, the quadratic having the double nonzero solution $x=Q_0/2$; two roots when $(\boldsymbol{Q},\boldsymbol{Q})=0$ and $Q_0\neq0$, the solutions being $0$ and $Q_0$ and only $x=Q_0$ contributing; and no roots when $(\boldsymbol{Q},\boldsymbol{Q})=0$ and $Q_0=0$, the only solution being $x=0$.

The Roots of a Complex Scalar

Assume $\boldsymbol{Q}=0$, so that $\tilde Q=Q_0e_0$ lies in the centre $\mathbb{C}e_0$. The vector equation is $2P_0\boldsymbol{P}=0$, and since $\mathbb{C}$ is a field this forces $\boldsymbol{P}=0$ or $P_0=0$. The two sub-cases are disjoint and exhaustive.

Sub-case 1 — the root is a complex scalar ($\boldsymbol{P}=0$). Then $\tilde P=P_0e_0$ and the scalar equation reduces to $P_0^2=Q_0$. The roots in this case are the two elements

$$ \tilde P = +\sqrt{Q_0}\,e_0, \qquad \tilde P = -\sqrt{Q_0}\,e_0, $$

which coincide as the single root $\tilde P=0$ when $Q_0=0$.

Sub-case 2 — the root is pure ($P_0=0$). Then $\tilde P=\boldsymbol{P}=P_1e_1+P_2e_2+P_3e_3$ and the scalar equation reduces to

$$ (\boldsymbol{P},\boldsymbol{P}) = P_1^2+P_2^2+P_3^2 = -Q_0 . $$

The roots in this case are the pure biquaternions of complex square $-Q_0$. Writing $P_k=p_k+ip'_k$ with $p_k,p'_k\in\mathbb{R}$, the condition is the pair of real equations

$$ p_1^2+p_2^2+p_3^2-p_1'^2-p_2'^2-p_3'^2 = -\operatorname{Re}Q_0, \qquad p_1p'_1+p_2p'_2+p_3p'_3 = -\tfrac12\operatorname{Im}Q_0, $$

six real coefficients carrying two constraints, hence a four-real-parameter family. For $Q_0\neq0$ the family is nonempty: taking $P_2=P_3=0$ leaves $P_1^2=-Q_0$, which is solved by $P_1=\sqrt{-Q_0}$. For $Q_0=0$ the family is the pure null cone $(\boldsymbol{P},\boldsymbol{P})=0$, whose nonzero elements are the nilpotents.

So when $\boldsymbol{Q}=0$ the root set is the two complex scalars $\pm\sqrt{Q_0}e_0$ — one element when $Q_0=0$ — together with this four-real-parameter family.

The Classification

Collecting the two cases, every $\tilde Q\in\mathbb{B}$ falls into exactly one of the following rows.

condition on $\tilde Q=Q_0e_0+\boldsymbol{Q}$	number of roots
$\boldsymbol{Q}\neq0$, $(\boldsymbol{Q},\boldsymbol{Q})\neq0$, $n\neq0$	four
$\boldsymbol{Q}\neq0$, $(\boldsymbol{Q},\boldsymbol{Q})\neq0$, $n=0$	two
$\boldsymbol{Q}\neq0$, $(\boldsymbol{Q},\boldsymbol{Q})=0$, $Q_0\neq0$	two
$\boldsymbol{Q}\neq0$, $n=0$, $Q_0=0$	none
$\boldsymbol{Q}=0$	a continuum

The generic case is four roots. The complex square $(\boldsymbol{Q},\boldsymbol{Q})$ of the vector part decides whether a solution $x$ is lost at $x=0$; the conjugate scalar $n$ decides whether the two solutions of the reduced quadratic are distinct; and the vector part $\boldsymbol{Q}$ itself decides which of the two cases applies.

The roots in each case. Reading the same rows:

No roots. If $\boldsymbol{Q}\neq0$, $n=0$ and $Q_0=0$, then $\tilde Q$ has no square root and the root set is $\varnothing$.

Two roots. If $\boldsymbol{Q}\neq0$ and either $(\boldsymbol{Q},\boldsymbol{Q})\neq0$ and $n=0$, or $(\boldsymbol{Q},\boldsymbol{Q})=0$ and $Q_0\neq0$, then the root set is a single pair $\tilde P=\pm\bigl(\sqrt{x}\,e_0+\boldsymbol{Q}/(2\sqrt{x})\bigr)$, with $x=Q_0/2$ in the first instance and $x=Q_0$ in the second.

Four roots. If $\boldsymbol{Q}\neq0$, $(\boldsymbol{Q},\boldsymbol{Q})\neq0$ and $n\neq0$, then the root set is two pairs, one for each solution of the reduced quadratic:

$$ \tilde P = \pm\Bigl(\sqrt{x_+}\,e_0+\frac{\boldsymbol{Q}}{2\sqrt{x_+}}\Bigr), \qquad \tilde P = \pm\Bigl(\sqrt{x_-}\,e_0+\frac{\boldsymbol{Q}}{2\sqrt{x_-}}\Bigr). $$

A continuum. If $\tilde Q$ is a complex scalar, $\tilde Q=Q_0e_0$, then the root set is the two elements $\pm\sqrt{Q_0}e_0$ together with the four-real-parameter family of pure $\boldsymbol{P}$ with $(\boldsymbol{P},\boldsymbol{P})=-Q_0$.

The Three Central Values

The classification specialises to the three central values $\tilde Q=-1,0,+1$ of Biquaternion Square Roots of Minus One, Zero and Plus One and reproduces them; the agreement is the check that this method and the direct computation of that article are the same computation.

$\tilde Q=-e_0$ ($Q_0=-1$, $\boldsymbol{Q}=0$). A complex scalar, so the continuum row. The scalar roots are $\pm\sqrt{-1}\,e_0=\pm ie_0$, the trivial roots of the companion article. The family is $(\boldsymbol{P},\boldsymbol{P})=1$, that is $p_1^2+p_2^2+p_3^2-p_1'^2-p_2'^2-p_3'^2=1$ and $p_1p'_1+p_2p'_2+p_3p'_3=0$, which is exactly the pure roots of $-1$: the real roots are the members with $\boldsymbol{p}'=0$, and the non-trivial roots are the members with $\boldsymbol{p}'\neq0$.

$\tilde Q=0$ ($Q_0=0$, $\boldsymbol{Q}=0$). The scalar roots are the single element $\tilde P=0$. The family is the pure null cone $(\boldsymbol{P},\boldsymbol{P})=0$, whose nonzero elements are the nilpotents of the algebra; its structure is Biquaternion Zero Divisors.

$\tilde Q=+e_0$ ($Q_0=+1$, $\boldsymbol{Q}=0$). The scalar roots are $\pm e_0$. The family is $(\boldsymbol{P},\boldsymbol{P})=-1$: in the notation of the companion article a member is $\boldsymbol{P}=\boldsymbol{p}i-\boldsymbol{p}'$, and its two real conditions, $p_1^2+p_2^2+p_3^2-p_1'^2-p_2'^2-p_3'^2=1$ and $p_1p'_1+p_2p'_2+p_3p'_3=0$, are exactly the conditions on the corresponding root of $-1$. These are the non-trivial roots of $+1$.

The three sets are therefore the degenerate data of the general classification, and this article is where the four-root and two-root cases live.

Worked Examples

The examples are computed with the classification and verified by squaring; the verification is the multiplication check $\tilde P^2=\tilde Q$.

A pure imaginary quaternion. For $\tilde Q=-ie_3$ (that is $-Ik$ in the notation of Acus and Dargys) the vector part is $\boldsymbol{Q}=-ie_3$, with $(\boldsymbol{Q},\boldsymbol{Q})=(-i)^2=-1$ and $n=-1$. Both are nonzero, so there are four roots, from $x_\pm=\pm\sqrt{-1}/2=\pm i/2$. Taking $x_+=i/2$ and $P_0=(1+i)/2$ gives $\boldsymbol{P}=\boldsymbol{Q}/(2P_0)=-\tfrac{1+i}{2}e_3$ and the pair $\pm\tfrac{1+i}{2}(e_0-e_3)$; taking $x_-=-i/2$ and $P_0=(1-i)/2$ gives $\boldsymbol{P}=\tfrac{1-i}{2}e_3$ and the pair $\pm\tfrac{1-i}{2}(e_0+e_3)$. The four roots are

$$ \tilde P = \pm\tfrac{1+i}{2}\bigl(e_0-e_3\bigr), \qquad \tilde P = \pm\tfrac{1-i}{2}\bigl(e_0+e_3\bigr), $$

in agreement with the roots of $-ie_3$ computed directly, and with the source's $\sqrt{-Ik}$.

A complex scalar. For $\tilde Q=-1+i$ (that is $-1+I$) the radicand is a complex scalar with $Q_0=-1+i$, so the row is the continuum. The isolated roots are $\pm\sqrt{-1+i}\,e_0$. Since

$$ \sqrt{-1+i} = \sqrt{\tfrac{\sqrt2-1}{2}} + i\sqrt{\tfrac{\sqrt2+1}{2}}, $$

the isolated roots are

$$ \tilde P = \pm\Bigl(\sqrt{\tfrac{\sqrt2-1}{2}} + i\sqrt{\tfrac{\sqrt2+1}{2}}\Bigr)e_0, $$

and the family is $(\boldsymbol{P},\boldsymbol{P})=1-i$, that is $p_1^2+p_2^2+p_3^2-p_1'^2-p_2'^2-p_3'^2=1$ and $p_1p'_1+p_2p'_2+p_3p'_3=-\tfrac12$. Squaring the displayed root gives $-1+i$.

A rootless element. For $\tilde Q=e_1-ie_3$ (that is $i-Ik$) the vector part is $\boldsymbol{Q}=e_1-ie_3$, with $(\boldsymbol{Q},\boldsymbol{Q})=1+(-i)^2=0$ and $n=0$. The row is the rootless one, and $\tilde Q$ has no square root. The element is itself a nilpotent,

$$ (e_1-ie_3)^2 = e_1^2+(-i)^2e_3^2-i\bigl(e_1e_3+e_3e_1\bigr) = -1+1+0 = 0 . $$

This is the example of Acus and Dargys.

Four roots of a non-scalar. For $\tilde Q=-(2+i)e_3$ (that is $-(2+I)k$) the vector part is $\boldsymbol{Q}=-(2+i)e_3$, with $(\boldsymbol{Q},\boldsymbol{Q})=(2+i)^2=3+4i$ and $n=3+4i$, both nonzero, so there are four roots. Since $\sqrt{3+4i}=2+i$, the two solutions are $x_\pm=\pm\tfrac{2+i}{2}$, and the four roots are

$$ \tilde P = \pm\bigl(1.0291+0.2429\,i\bigr)\bigl(e_0-e_3\bigr), \qquad \tilde P = \pm\bigl(0.2429-1.0291\,i\bigr)\bigl(e_0+e_3\bigr), $$

whose squares are $-(2+i)e_3$; the second pair is the first multiplied by $-i$, with $e_0-e_3$ exchanged for $e_0+e_3$.

A nilpotent root of $0$. For $\tilde Q=0$ the scalar root is $\tilde P=0$ and the family is the pure null cone; $\tilde P=e_1+ie_2$ is one of its members, with

$$ (e_1+ie_2)^2 = e_1^2+i^2e_2^2+i\bigl(e_1e_2+e_2e_1\bigr) = -1+1+0 = 0 . $$

Summary

For a general $\tilde Q\in\mathbb{B}$ the square roots are found from the vector–scalar split $\tilde Q=Q_0e_0+\boldsymbol{Q}$ and $\tilde P=P_0e_0+\boldsymbol{P}$, which turns $\tilde P^2=\tilde Q$ into the vector equation $2P_0\boldsymbol{P}=\boldsymbol{Q}$ and the scalar equation $P_0^2-(\boldsymbol{P},\boldsymbol{P})=Q_0$.

When $\boldsymbol{Q}\neq0$, a root has $P_0\neq0$ and $\boldsymbol{P}=\boldsymbol{Q}/(2P_0)$, and $x=P_0^2$ solves the reduced quadratic $4x^2-4Q_0x-(\boldsymbol{Q},\boldsymbol{Q})=0$, whose discriminant is $16n$ with $n=Q_0^2+(\boldsymbol{Q},\boldsymbol{Q})$; each nonzero solution $x$ contributes the pair $\tilde P=\pm\bigl(\sqrt{x}\,e_0+\boldsymbol{Q}/(2\sqrt{x})\bigr)$.

When $\boldsymbol{Q}=0$, the radicand is a complex scalar and the roots are the two elements $\pm\sqrt{Q_0}e_0$ (one when $Q_0=0$) together with the four-real-parameter family of pure $\boldsymbol{P}$ with $(\boldsymbol{P},\boldsymbol{P})=-Q_0$.

The number of roots is: four when $(\boldsymbol{Q},\boldsymbol{Q})\neq0$ and $n\neq0$; two when $(\boldsymbol{Q},\boldsymbol{Q})\neq0$ and $n=0$, or when $(\boldsymbol{Q},\boldsymbol{Q})=0$ and $Q_0\neq0$; none when $n=0$ and $Q_0=0$ with $\boldsymbol{Q}\neq0$; and the four-parameter continuum when and only when $\tilde Q$ is a complex scalar.

Summary of Notation

Symbol	Meaning
$\mathbb{B}$	Biquaternion algebra
$\tilde Q$	The element whose square roots are sought
$\tilde P$	A square root of $\tilde Q$
$\boldsymbol{Q}=Q_1e_1+Q_2e_2+Q_3e_3$	Vector part of $\tilde Q$
$(\boldsymbol{Q},\boldsymbol{Q})=Q_1^2+Q_2^2+Q_3^2$	Complex square of the vector part
$n=Q_0^2+(\boldsymbol{Q},\boldsymbol{Q})$	The scalar of $\tilde Q\tilde Q^{\natural}=ne_0$
$x=P_0^2$	Square of the scalar part of a root
$x_\pm=\tfrac12\bigl(Q_0\pm\sqrt{n}\bigr)$	The two solutions of the reduced quadratic