Square Roots of a General Element in the Biquaternion Algebra

Introduction

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

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

The method needs no Clifford algebra, no norm and no form. It is the same vector–scalar split that solves the three central cases $\tilde\Xi^2=-1,0,+1$ in Square Roots of One and the Idempotents of the Biquaternion Algebra and Square Roots of Zero and the Nilpotents of the Biquaternion Algebra, 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 $x=\Xi_t^2$ 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 results. For a radicand with non-vanishing vector part the roots come in pairs $\pm\tilde\Xi$: there are four roots generically, two in the two degenerate rows, and none for the rootless row $(\vec Q,\vec Q) = 0$, $n = 0$, $Q_t = 0$. For a complex scalar radicand $\tilde Q = Q_t e_t$ the roots are the two scalars $\pm\sqrt{Q_t}\,e_t$ together with a four-parameter family of pure $\vec\Xi$ with $(\vec\Xi,\vec\Xi) = -Q_t$. The three central values recover the three sets of Square Roots of One and the Idempotents of the Biquaternion Algebra and Square Roots of Zero and the Nilpotents of the Biquaternion Algebra. Each case below states its answer first and then proves it.

The article is algebraic. The biquaternion algebra also carries a Clifford structure, treated in Biquaternion Other Algebraic Element Representations; that structure is not used here. The three central values of Square Roots of One and the Idempotents of the Biquaternion Algebra and Square Roots of Zero and the Nilpotents of the Biquaternion Algebra are recovered as the case of a complex scalar $\tilde Q$, in §The Three Central Values, which checks this method against the direct computations of those articles. The zero-divisor cone, of which the nonzero roots of $0$ are a part, is Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors, named here only where the classification touches it. The polar and exponential decompositions that the answer must be consistent with are Modulus–Unit Polar Decomposition of Biquaternions.

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

Throughout, the quaternion basis is $e_t=1,\vec e_x,\vec e_y,\vec e_z$, the scalar imaginary is $i$, and a general biquaternion is $\tilde Q=Q_te_t+Q_x\vec e_x+Q_y\vec e_y+Q_z\vec e_z$ 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 equation $\tilde\Xi^2=\tilde Q$ names a product, so the root is relative to it. The plain square $\tilde\Xi\tilde\Xi$ and the general definition with the roots of a central value are Definitions for the Study of the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, §Powers, the Square and the Roots; throughout this article the square is the one of the general plain bilinear product $\tilde\Xi\tilde\Xi$, the multiplication of the biquaternion algebra, and the other three general products and their root problems are Square Roots of a General Element in the General Plain Sesqualgebra, Square Roots of a General Element in the General Quaternionic Algebra and Square Roots of a General Element in the General Quaternionic Sesqualgebra.

The Reduction

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

$$ \tilde\Xi = \Xi_te_t + \vec\Xi, \qquad \vec\Xi=\Xi_x\vec e_x+\Xi_y\vec e_y+\Xi_z\vec e_z, $$

$$ \tilde Q = Q_te_t + \vec Q, \qquad \vec Q=Q_x\vec e_x+Q_y\vec e_y+Q_z\vec e_z, $$

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

$$ \tilde\Xi^2 = \bigl(\Xi_t^2-(\vec\Xi,\vec\Xi)\bigr)e_t + 2\Xi_t\vec\Xi, \qquad (\vec\Xi,\vec\Xi)=\Xi_x^2+\Xi_y^2+\Xi_z^2 . $$

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

$$ 2\Xi_t\vec\Xi = \vec Q \qquad\text{(the vector part)}, $$

$$ \Xi_t^2-(\vec\Xi,\vec\Xi) = Q_t \qquad\text{(the scalar part)} . $$

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

Case A — the radicand has non-vanishing vector part ($\vec Q\neq0$). Then $\Xi_t\neq0$ and $\vec\Xi=\vec Q/(2\Xi_t)$: 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 ($\vec Q=0$). Then $2\Xi_t\vec\Xi=0$, which forces $\vec\Xi=0$ or $\Xi_t=0$: 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

$$ (\vec Q,\vec Q) = Q_x^2+Q_y^2+Q_z^2 , $$

and the conjugate scalar is

$$ n = Q_t^2+Q_x^2+Q_y^2+Q_z^2 = Q_t^2+(\vec Q,\vec Q) . $$

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

$$ \tilde Q\tilde Q^{\natural} = \bigl(Q_t^2+Q_x^2+Q_y^2+Q_z^2\bigr)e_t = n\,e_t , $$

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

For a radicand with non-vanishing vector part the roots come in opposite pairs $\pm\tilde\Xi$: four roots when $(\vec Q,\vec Q)\neq0$ and $n\neq0$, two roots in the two degenerate rows, and none when $(\vec Q,\vec Q)=0$, $n=0$ and $Q_t=0$.

Assume $\vec Q\neq0$. Then $2\Xi_t\vec\Xi=\vec Q$ forces $\Xi_t\neq0$, and the vector equation determines the vector part of every root,

$$ \vec\Xi = \frac{\vec Q}{2\Xi_t} . $$

Substituting this into the scalar equation gives

$$ \Xi_t^2-\frac{(\vec Q,\vec Q)}{4\Xi_t^2} = Q_t . $$

Put

$$ x = \Xi_t^2 . $$

Since $(\vec Q/(2\Xi_t),\vec Q/(2\Xi_t)) = (\vec Q,\vec Q)/(4x)$, the scalar equation becomes the reduced quadratic

$$ 4x^2 - 4Q_tx - (\vec Q,\vec Q) = 0 . $$

Its discriminant is

$$ 16Q_t^2 + 16(\vec Q,\vec Q) = 16n , $$

so its two solutions are

$$ \boxed{\; x_+ = \frac{Q_t+\sqrt{n}}{2}, \qquad x_- = \frac{Q_t-\sqrt{n}}{2} \;}, $$

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 $\vec Q\neq0$, 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 $\Xi_t=\pm\sqrt{x}$, the element

$$ \boxed{\; \tilde\Xi = \Xi_te_t + \frac{\vec Q}{2\Xi_t} \;} $$

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

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

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

The Roots of a Complex Scalar

For a complex scalar radicand $\tilde Q = Q_t e_t$ the roots are the two scalars $\pm\sqrt{Q_t}\,e_t$ together with the four-parameter family of pure $\vec\Xi$ with $(\vec\Xi,\vec\Xi) = -Q_t$.

Assume $\vec Q=0$, so that $\tilde Q=Q_te_t$ lies in the centre $\mathbb{C}e_t$. The vector equation is $2\Xi_t\vec\Xi=0$, and since $\mathbb{C}$ is a field this forces $\vec\Xi=0$ or $\Xi_t=0$. The two sub-cases are disjoint and exhaustive.

Sub-case 1 — the root is a complex scalar ($\vec\Xi=0$). Then $\tilde\Xi=\Xi_te_t$ and the scalar equation reduces to $\Xi_t^2=Q_t$. The roots in this case are the two elements

$$ \tilde\Xi = +\sqrt{Q_t}\,e_t, \qquad \tilde\Xi = -\sqrt{Q_t}\,e_t, $$

which coincide as the single root $\tilde\Xi=0$ when $Q_t=0$.

Sub-case 2 — the root is pure ($\Xi_t=0$). Then $\tilde\Xi=\vec\Xi=\Xi_x\vec e_x+\Xi_y\vec e_y+\Xi_z\vec e_z$ and the scalar equation reduces to

$$ (\vec\Xi,\vec\Xi) = \Xi_x^2+\Xi_y^2+\Xi_z^2 = -Q_t . $$

The roots in this case are the pure biquaternions of complex square $-Q_t$. Writing $\Xi_k=\xi_k+i\xi'_k$ with $\xi_k,\xi'_k\in\mathbb{R}$, the condition is the pair of real equations

$$ \xi_x^2+\xi_y^2+\xi_z^2-(\xi'_x)^2-(\xi'_y)^2-(\xi'_z)^2 = -\operatorname{Re}Q_t, \qquad \xi_x\xi'_x+\xi_y\xi'_y+\xi_z\xi'_z = -\tfrac12\operatorname{Im}Q_t, $$

six real coefficients carrying two constraints, hence a four-real-parameter family. For $Q_t\neq0$ the family is nonempty: taking $\Xi_y=\Xi_z=0$ leaves $\Xi_x^2=-Q_t$, which is solved by $\Xi_x=\sqrt{-Q_t}$. For $Q_t=0$ the family is the pure null cone $(\vec\Xi,\vec\Xi)=0$, whose nonzero elements are the nilpotents.

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

The Classification

Every radicand falls into exactly one of the rows below: four roots generically, two roots in the two degenerate rows, no roots for the rootless row, and a four-parameter family for a complex scalar.

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

condition on $\tilde Q=Q_te_t+\vec Q$	number of roots
$\vec Q\neq0$, $(\vec Q,\vec Q)\neq0$, $n\neq0$	four
$\vec Q\neq0$, $(\vec Q,\vec Q)\neq0$, $n=0$	two
$\vec Q\neq0$, $(\vec Q,\vec Q)=0$, $Q_t\neq0$	two
$\vec Q\neq0$, $n=0$, $Q_t=0$	none
$\vec Q=0$	a four-parameter family

The generic case is four roots. The complex square $(\vec Q,\vec Q)$ 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 $\vec Q$ itself decides which of the two cases applies.

The roots in each case. Reading the same rows:

No roots. If $\vec Q\neq0$, $n=0$ and $Q_t=0$, then $\tilde Q$ has no square root and the root set is $\varnothing$.

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

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

$$ \boxed{\; \tilde\Xi = \pm\Bigl(\sqrt{x_+}\,e_t+\frac{\vec Q}{2\sqrt{x_+}}\Bigr), \qquad \tilde\Xi = \pm\Bigl(\sqrt{x_-}\,e_t+\frac{\vec Q}{2\sqrt{x_-}}\Bigr) \;}. $$

A complex scalar. If $\tilde Q$ is a complex scalar, $\tilde Q=Q_te_t$, then the root set is the two elements $\pm\sqrt{Q_t}e_t$ together with the four-real-parameter family of pure $\vec\Xi$ with $(\vec\Xi,\vec\Xi)=-Q_t$.

The Three Central Values

The three central values reproduce the three sets of the companion articles: the roots of $-1$ and of $+1$ in the first, the roots of $0$ in the second.

The classification specialises to the three central values $\tilde Q=-1,0,+1$ of the companion articles Square Roots of One and the Idempotents of the Biquaternion Algebra and Square Roots of Zero and the Nilpotents of the Biquaternion Algebra, and reproduces them; the agreement is the check that this method and the direct computations of those articles are the same computation.

$\tilde Q=-e_t$ ($Q_t=-1$, $\vec Q=0$). A complex scalar, so the complex-scalar case. The scalar roots are $\pm\sqrt{-1}\,e_t=\pm ie_t$, the trivial roots of the companion articles. The family is $(\vec\Xi,\vec\Xi)=1$, that is $\xi_x^2+\xi_y^2+\xi_z^2-(\xi'_x)^2-(\xi'_y)^2-(\xi'_z)^2=1$ and $\xi_x\xi'_x+\xi_y\xi'_y+\xi_z\xi'_z=0$, which is exactly the pure roots of $-1$: the real roots are the members with $\vec\xi'=0$, and the non-trivial roots are the members with $\vec\xi'\neq0$.

$\tilde Q=0$ ($Q_t=0$, $\vec Q=0$). The scalar roots are the single element $\tilde\Xi=0$. The family is the pure null cone $(\vec\Xi,\vec\Xi)=0$, whose nonzero elements are the nilpotents of the algebra; its structure is Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors.

$\tilde Q=+e_t$ ($Q_t=+1$, $\vec Q=0$). The scalar roots are $\pm e_t$. The family is $(\vec\Xi,\vec\Xi)=-1$: in the notation of the companion articles a member is $\vec\Xi=\vec\xi+i\vec\xi'$, and its two real conditions, $\xi_x^2+\xi_y^2+\xi_z^2-(\xi'_x)^2-(\xi'_y)^2-(\xi'_z)^2=1$ and $\xi_x\xi'_x+\xi_y\xi'_y+\xi_z\xi'_z=0$, 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 $\tilde\Xi^2=\tilde Q$.

A pure imaginary quaternion. For $\tilde Q=-i\vec e_z$ (that is $-Ik$ in the notation of Acus and Dargys) the vector part is $\vec Q=-i\vec e_z$, with $(\vec Q,\vec Q)=(-i)^2=-1$ 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 $\Xi_t=(1+i)/2$ gives $\vec\Xi=\vec Q/(2\Xi_t)=-\tfrac{1+i}{2}\vec e_z$ and the pair $\pm\tfrac{1+i}{2}(e_t-\vec e_z)$; taking $x_-=-i/2$ and $\Xi_t=(1-i)/2$ gives $\vec\Xi=\tfrac{1-i}{2}\vec e_z$ and the pair $\pm\tfrac{1-i}{2}(e_t+\vec e_z)$. The four roots are

$$ \tilde\Xi = \pm\tfrac{1+i}{2}\bigl(e_t-\vec e_z\bigr), \qquad \tilde\Xi = \pm\tfrac{1-i}{2}\bigl(e_t+\vec e_z\bigr), $$

in agreement with the roots of $-i\vec e_z$ 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 $Q_t=-1+i$, so the row is the four-parameter one. The isolated roots are $\pm\sqrt{-1+i}\,e_t$. Since

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

the isolated roots are

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

and the family is $(\vec\Xi,\vec\Xi)=1-i$, that is $\xi_x^2+\xi_y^2+\xi_z^2-(\xi'_x)^2-(\xi'_y)^2-(\xi'_z)^2=1$ and $\xi_x\xi'_x+\xi_y\xi'_y+\xi_z\xi'_z=-\tfrac12$. Squaring the displayed root gives $-1+i$.

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

$$ (\vec e_x-i\vec e_z)^2 = \vec e_x^2+(-i)^2\vec e_z^2-i\bigl(\vec e_x \vec e_z+\vec e_z \vec e_x\bigr) = -1+1+0 = 0 . $$

This is the example of Acus and Dargys.

Four roots of a non-scalar. For $\tilde Q=-(2+i)\vec e_z$ (that is $-(2+I)k$) the vector part is $\vec Q=-(2+i)\vec e_z$, with $(\vec Q,\vec Q)=(2+i)^2=3+4i$ 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

$$ \tilde\Xi = \pm\bigl(1.0291+0.2429\,i\bigr)\bigl(e_t-\vec e_z\bigr), \qquad \tilde\Xi = \pm\bigl(0.2429-1.0291\,i\bigr)\bigl(e_t+\vec e_z\bigr), $$

whose squares are $-(2+i)\vec e_z$; the second pair is the first multiplied by $-i$, with $e_t-\vec e_z$ exchanged for $e_t+\vec e_z$.

A nilpotent root of $0$. For $\tilde Q=0$ the scalar root is $\tilde\Xi=0$ and the family is the pure null cone; $\tilde\Xi=\vec e_x+i\vec e_y$ is one of its members, with

$$ (\vec e_x+i\vec e_y)^2 = \vec e_x^2+i^2\vec e_y^2+i\bigl(\vec e_x \vec e_y+\vec e_y \vec e_x\bigr) = -1+1+0 = 0 . $$

Summary

For a general $\tilde Q\in\mathbb{B}$ the square roots are found from the vector–scalar split $\tilde Q=Q_te_t+\vec Q$ and $\tilde\Xi=\Xi_te_t+\vec\Xi$, which turns $\tilde\Xi^2=\tilde Q$ into the vector equation $2\Xi_t\vec\Xi=\vec Q$ and the scalar equation $\Xi_t^2-(\vec\Xi,\vec\Xi)=Q_t$.

When $\vec Q\neq0$, a root has $\Xi_t\neq0$ and $\vec\Xi=\vec Q/(2\Xi_t)$, and $x=\Xi_t^2$ solves the reduced quadratic $4x^2-4Q_tx-(\vec Q,\vec Q)=0$, whose discriminant is $16n$ with $n=Q_t^2+(\vec Q,\vec Q)$; each nonzero solution $x$ contributes the pair $\tilde\Xi=\pm\bigl(\sqrt{x}\,e_t+\vec Q/(2\sqrt{x})\bigr)$.

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

The number of roots is: four when $(\vec Q,\vec Q)\neq0$ and $n\neq0$; two when $(\vec Q,\vec Q)\neq0$ and $n=0$, or when $(\vec Q,\vec Q)=0$ and $Q_t\neq0$; none when $n=0$ and $Q_t=0$ with $\vec Q\neq0$; and the four-parameter family 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
$\tilde\Xi$	A square root of $\tilde Q$
$\vec Q=Q_x\vec e_x+Q_y\vec e_y+Q_z\vec e_z$	Vector part of $\tilde Q$
$(\vec Q,\vec Q)=Q_x^2+Q_y^2+Q_z^2$	Complex square of the vector part
$n=Q_t^2+(\vec Q,\vec Q)$	The scalar of $\tilde Q\tilde Q^{\natural}=ne_t$
$x=\Xi_t^2$	Square of the scalar part of a root
$x_\pm=\tfrac12\bigl(Q_t\pm\sqrt{n}\bigr)$	The two solutions of the reduced quadratic