Square Root of a Multivector of Clifford Algebras in 3D: A Game with Signs

Abstract

An algorithm is presented to extract the square root from a multivector (MV) in real Clifford algebras ${\mathit{Cl}}_{p,q}$, where $n=p+q\le 3$, in radicals. It is shown that in ${\mathit{Cl}}_{3,0}$, ${\mathit{Cl}}_{1,2}$, and ${\mathit{Cl}}_{0,3}$ algebras, there are up to four isolated square roots in a case of the most general (generic) MV. The algebra ${\mathit{Cl}}_{2,1}$ is an exception and, there, the MV can have up to 16 isolated roots. In addition, a continuum of roots has been found in all Clifford algebras except $p+q=1$. Examples which clarify computations are provided to illustrate the properties of roots in all $n=3$ algebras. The results may be useful in solving nonlinear equations, like for example, the Clifford–Riccati equation.

1. Introduction

The square root has a long history. Solution by radicals of the cubic equation was first published in 1545 by G. Cardano. Simultaneously, a concept of the square root of a negative number was developed [1]. In 1872, A. Cayley was the first to carry over the square root to matrices [2]. In the recent book by N. J. Higham [3], where an extensive literature is presented on nonlinear functions of matrices, two sections are devoted to matrix square roots. In the context of Clifford algebra (CA), the main attention up till now was concentrated on the square roots of quaternions [4,5], or their derivatives such as coquaternions (also called split quaternions) or nectarines [5,6,7]. The square root of biquaternions (complex quaternions) was considered in [8]. Quaternions and related objects are isomorphic to one of $n=2$ algebras ${\mathit{Cl}}_{0,2}$, ${\mathit{Cl}}_{1,1}$, ${\mathit{Cl}}_{2,0}$, and therefore, the quaternionic square root analysis can be easily rewritten in terms of CA (see Appendix B). In this paper, we shall mainly be interested in higher, namely, $n=3$ Clifford algebras (CAs), where the main object is the eight-component multivector (abbreviated as MV through the article).

For CAs of dimension $n\ge 3$, the investigation and understanding of square root properties is still in infancy. The most akin to the present paper are the investigation of conditions for the existence of square root of $-1$ [8,9,10]. The existence of such roots allows to extend the Fourier transform to MVs, where they are used in formulating Clifford–Fourier transform and CA-based wavelet theories [11].

Our preliminary investigation [12] on this subject was concerned with square roots of individual multivector grades such as scalar, vector, bivector, pseudoscalar, or their simple combinations. For this purpose, we have applied the Gröbner basis algorithm to analyze the system of nonlinear polynomial equations that ensue from the MV equation ${\mathsf{A}}^2=\mathsf{B}$, where $\mathsf{A}$ and $\mathsf{B}$ are the MVs. The Gröbner basis is accessible in many symbolic mathematical packages. Specifically, Mathematica commands such as Reduce[ ], Solve[ ], Eliminate[ ] and others also employ the Gröbner basis to solve nonlinear problems. With their help, we were able to find new properties of roots for the $n=3$ case, namely, that the MVs may have no roots, a single or multiple isolated roots, or even an infinite number (continuum) of roots in 4D parameter spaces or spaces of smaller dimension. A short overview of current Clifford algebra software is described in the introduction of [13].

In this paper, we continue our [12] investigations of the square root problem in real CAs for the $n=3$ case. In particular, we examine and provide explicit conditions for an MV to have discrete and continuum of roots, and how to express real root coefficients in radicals. For this purpose, a symbolic package was used [14] that appeared to be invaluable both for detecting specific solutions of the nonlinear equation ${\mathsf{A}}^2=\mathsf{B}$ and for numerical checks in general.

In Section 2, the notation is introduced. The algorithm to calculate the square root of a generic MV and special cases that follow are given in Section 3, Section 4 and Section 5 for ${\mathit{Cl}}_{3,0}\simeq {\mathit{Cl}}_{1,2}$, ${\mathit{Cl}}_{0,3}$, and ${\mathit{Cl}}_{2,1}$ algebras, respectively. The computation is illustrated by a number of examples. The conclusions are drawn in Section 6. For completeness, in Appendix A and Appendix B, the MV square roots are presented for lower-dimensional CAs. Appendix C provides summary for the two most important algebras, ${\mathit{Cl}}_{3,0}$ and ${\mathit{Cl}}_{0,3}$ algebras, which will be useful for implementation. Appendix D is devoted to MV determinants.

2. Notation

For $n=3$, a general MV can be expanded in the orthonormal basis that consists of $2^n=8$ elements listed in inverse degree lexicographic ordering (note an increasing order of digits in indices). Therefore, we write ${\mathbf{e}}_{13}$ instead of ${\mathbf{e}}_{31}=-{\mathbf{e}}_{13}$. This convention is reflected in opposite signs of some terms in formulas.

$$\{1,{\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3,{\mathbf{e}}_{12},{\mathbf{e}}_{13},{\mathbf{e}}_{23},{\mathbf{e}}_{123}\equiv I\},$$

(1)

where ${\mathbf{e}}_i$ are basis vectors and ${\mathbf{e}}_{ij}$ are the bivectors (oriented planes). The last term is the pseudoscalar. The number of subscripts indicates the grade of basis element. The scalar is a grade-0 element, the vectors ${\mathbf{e}}_i$ are the grade-1 elements, etc. In the orthonormalized basis, the geometric (Clifford) products of basis vectors satisfy the anticommutation relation,

$${\mathbf{e}}_i{\mathbf{e}}_j+{\mathbf{e}}_j{\mathbf{e}}_i=\pm 2{\delta }_{ij}.$$

(2)

For ${\mathit{Cl}}_{3,0}$ and ${\mathit{Cl}}_{0,3}$ algebras, the squares of basis vectors, correspondingly, are ${\mathbf{e}}_i^2=+1$ and ${\mathbf{e}}_i^2=-1$, where $i=1,2,3$. For mixed signature algebras such as ${\mathit{Cl}}_{2,1}$ and ${\mathit{Cl}}_{1,2}$, we have ${\mathbf{e}}_1^2={\mathbf{e}}_2^2=1$, ${\mathbf{e}}_3^2=-1$ and ${\mathbf{e}}_1^2=1$, ${\mathbf{e}}_2^2={\mathbf{e}}_3^2=-1$, respectively. The sign of squares of higher grade elements is determined by squares of vectors and property (2). For example, in ${\mathit{Cl}}_{3,0}$, we have ${\mathbf{e}}_{12}^2={\mathbf{e}}_{12}{\mathbf{e}}_{12}=-{\mathbf{e}}_1{\mathbf{e}}_2{\mathbf{e}}_2{\mathbf{e}}_1=-{\mathbf{e}}_1(+1){\mathbf{e}}_1=-{\mathbf{e}}_1{\mathbf{e}}_1=-1$. However, in ${\mathit{Cl}}_{1,2}$, a similar computation gives ${\mathbf{e}}_{12}^2=-{\mathbf{e}}_1{\mathbf{e}}_2{\mathbf{e}}_2{\mathbf{e}}_1=-{\mathbf{e}}_1(-1){\mathbf{e}}_1={\mathbf{e}}_1{\mathbf{e}}_1=+1$.

When $n=3$, an MV $\mathsf{A}$ in real CA can be expanded in the basis (1),

$$\begin{array}{cc}\hfill \mathsf{A}=& a_0+a_1{\mathbf{e}}_1+a_2{\mathbf{e}}_2+a_3{\mathbf{e}}_3+a_{12}{\mathbf{e}}_{12}+a_{23}{\mathbf{e}}_{23}+a_{13}{\mathbf{e}}_{13}+a_{123}I\hfill \\ \hfill \equiv & a_0+\mathbf{a}+\mathcal{A}+a_{123}I,\hfill \end{array}$$

(3)

where $a_i$, $a_{ij}$, and $a_{123}$ are the real coefficients, and $\mathbf{a}=a_1{\mathbf{e}}_1+a_2{\mathbf{e}}_2+a_3{\mathbf{e}}_3$ and $\mathcal{A}=a_{12}{\mathbf{e}}_{12}+a_{23}{\mathbf{e}}_{23}+a_{13}{\mathbf{e}}_{13}$ are, respectively, the vector and the bivector. We will seek a real MV $\mathsf{A}$ (with real coefficients), the square of which satisfies

$$\mathsf{A}\mathsf{A}\equiv {\mathsf{A}}^2=\mathsf{B}=b_0+\mathbf{b}+\mathcal{B}+b_{123}I.$$

(4)

The MV $\mathsf{A}$ is called a square root of $\mathsf{B}$. In Equation (4), the square ${\mathsf{A}}^2$ has been expanded in the orthonormal basis where $b_0,\mathbf{b},\mathcal{B}$ and $I\equiv I_3$ denote, respectively, a scalar, a vector ($\mathbf{b}=b_1{\mathbf{e}}_1+b_2{\mathbf{e}}_2+b_3{\mathbf{e}}_3$), a bivector ($\mathcal{B}=b_{12}{\mathbf{e}}_{12}+b_{23}{\mathbf{e}}_{23}+b_{13}{\mathbf{e}}_{13}$), and a pseudoscalar. The representation (3) is not convenient for our problem; therefore, for all 3D algebras ${\mathit{Cl}}_{3,0}$, ${\mathit{Cl}}_{0,3}$, ${\mathit{Cl}}_{1,2}$, and ${\mathit{Cl}}_{2,1}$ a more symmetric representation is introduced,

$$\mathsf{A}=s+\mathbf{v}+(S+\mathbf{V})I,$$

(5)

where now both s and S are the real scalars and both $\mathbf{v}=v_1{\mathbf{e}}_1+v_2{\mathbf{e}}_2+v_3{\mathbf{e}}_3$ and $\mathbf{V}=V_1{\mathbf{e}}_1+V_2{\mathbf{e}}_2+V_3{\mathbf{e}}_3$ are the vectors with real coefficients $v_i$ and $V_i$. The MV representation (5) allows to disentangle the coupled nonlinear equations in a regular manner for all listed algebras. To select the scalar s in (5), the grade selector $⟨\mathsf{A}⟩\equiv {⟨\mathsf{A}⟩}_0=s$ is used. The pseudoscalar part can be extracted by $⟨\mathsf{A}⟩\equiv {⟨-\mathsf{A}I⟩}_0=S$, and similarly for other grades. More about CAs and MV properties can be found, for example, in books [15,16].

When $n=1,2$, the MV square root algorithm simplifies substantially. All needed formulas are presented in Appendix A and Appendix B, respectively.

3. Square Roots in ${\mathit{Cl}}_{\mathbf{3},\mathbf{0}}$ and ${\mathit{Cl}}_{\mathbf{1},\mathbf{2}}$ Algebras

This section describes the method of substitution of variables in CA which paves a direct way to square root algorithm. The Euclidean ${\mathit{Cl}}_{3,0}$ algebra is the most simple one among $n=3$ algebras. The algebra ${\mathit{Cl}}_{1,2}$ is isomorphic to ${\mathit{Cl}}_{3,0}$; therefore, the computations for this algebra follows the same route except that, there, some notational differences appear.

The goal is to solve nonlinear MV equation ${\mathsf{A}}^2=\mathsf{B}$, where $\mathsf{B}=b_0+b_1{\mathbf{e}}_2+b_2{\mathbf{e}}_2+b_3{\mathbf{e}}_3+b_{12}{\mathbf{e}}_{23}+b_{13}{\mathbf{e}}_{13}+b_{23}{\mathbf{e}}_{23}+b_{123}I$ and $\mathsf{A}$ is unknown. The latter may have the general form (5). Expanding ${\mathsf{A}}^2$ in components and equating (real) coefficients at basis elements to respective coefficients in $\mathsf{B}$, one obtains a system of eight nonlinear equations:

$$\begin{array}{cccc}\hfill b_0& =s^2-S^2+{\mathbf{v}}^2-{\mathbf{V}}^2,\hfill & \hfill b_{123}& =2(sS+\mathbf{v}·\mathbf{V}),\hfill \end{array}$$

(6)

$$\begin{array}{cccc}\hfill b_1& =2(s{v}_1-S{V}_1),\hfill & \hfill b_{23}& =2(s{V}_1+S{v}_1),\hfill \end{array}$$

(7)

$$\begin{array}{cccc}\hfill b_2& =2(s{v}_2-S{V}_2),\hfill & \hfill b_{13}& =2(s{V}_2+S{v}_2),\hfill \end{array}$$

(8)

$$\begin{array}{cccc}\hfill b_3& =2(s{v}_3-S{V}_3),\hfill & \hfill b_{12}& =2(s{V}_3+S{v}_3).\hfill \end{array}$$

(9)

These equations were written in [12], but they were not solved properly, so the results presented there did not cover all possible cases. Here, we provide a complete solution of square root problem for $n=3$ Clifford algebras. The square root computations split into two cases: the generic case where either $s^2+S^2\ne 0$ (in ${\mathit{Cl}}_{3,0}$ and ${\mathit{Cl}}_{1,2}$) or $s^2-S^2\ne 0$ (in ${\mathit{Cl}}_{0,3}$ and ${\mathit{Cl}}_{2,1}$), and the special case where $s^2+S^2=0$ (in ${\mathit{Cl}}_{3,0}$ and ${\mathit{Cl}}_{1,2}$) or $s^2-S^2=0$ (in ${\mathit{Cl}}_{0,3}$ and ${\mathit{Cl}}_{2,1}$).

3.1. The Generic Case $s^2+S^2\ne 0$

The system of six Equations (7)–(9) is linear in new variables $v_i$ and $V_i$ in (5). It has a very simple solution which is a key to the analysis that follows. As a note, the symmetry of Equations (10) and (11) with respect to pairs ($v_2,V_2$), ($v_1,V_1$), and ($v_3,V_3$) differ as it was explained in the beginning of Section 2.

It can be restored if $b_{13}$ is replaced by $-b_{31}$.

$$\begin{array}{cccccc}\hfill v_1& ={\displaystyle \frac{b_{1}s+b_{23}S}{2(s^2+S^2)}},\hfill & \hfill v_2& ={\displaystyle \frac{b_{2}s-b_{13}S}{2(s^2+S^2)}},\hfill & \hfill v_3& ={\displaystyle \frac{b_{3}s+b_{12}S}{2(s^2+S^2)}},\hfill \end{array}$$

(10)

$$\begin{array}{cccccc}\hfill V_1& ={\displaystyle \frac{b_{23}s-b_{1}S}{2(s^2+S^2)}},\hfill & \hfill V_2& =-{\displaystyle \frac{b_{13}s+b_{2}S}{2(s^2+S^2)}},\hfill & \hfill V_3& ={\displaystyle \frac{b_{12}s-b_{3}S}{2(s^2+S^2)}}.\hfill \end{array}$$

(11)

The Equations (10) and (11) express the components of vectors $\mathbf{v}$ and $\mathbf{V}$ in terms of scalars s and S, which are to be determined from a pair of Equation (6). The solution is valid when $s^2+S^2\ne 0$, i.e., when either $s\ne 0$ or $S\ne 0$, or both s and S are non-zero scalars. If these conditions are not satisfied, we have the subcase $s=S=0$. After substitution of (10) and (11), i.e., of $(v_1,v_2,v_3)$ and $(V_1,V_2,V_3)$, into (6), we get a system of two coupled algebraic equations for two unknowns s and S,

$$\begin{array}{cc}\hfill 4(b_0-s^2+S^2){(s^2+S^2)}^2=& +{(b_{1}s+b_{23}S)}^2+{(b_{2}s-b_{13}S)}^2+{(b_{3}s+b_{12}S)}^2\hfill \\ \hfill & -{(b_{23}s-b_{1}S)}^2-{(b_{13}s+b_{2}S)}^2-{(b_{12}s-b_{3}S)}^2,\hfill \\ \hfill 2(b_{123}-2sS){(s^2+S^2)}^2=& +(b_{1}s+b_{23}S)(b_{23}s-b_{1}S)-(b_{2}s-b_{13}S)\hfill \\ \hfill & \times (b_{13}s+b_{2}S)+(b_{3}s+b_{12}S)(b_{12}s-b_{3}S).\hfill \end{array}$$

(12)

The system (12) has exactly four solutions that can be expressed in radicals. If new variables t and T are introduced and substitution

$$sS=t,\frac{1}{2}(-s^2+S^2)=T$$

(13)

is used, the system (12) reduces to

$$\begin{array}{cc}\hfill & (b_0+4T)(4t-b_{123})-b_I/2=0,\hfill \\ \hfill & b_S-(b_0-b_{123}+4T+4t)(b_0+b_{123}+4T-4t)=0.\hfill \end{array}$$

(14)

In (14), coordinate-free abbreviations $b_S$ and $b_I$ have been introduced,

$$\begin{array}{c}{b}_S={⟨\mathsf{B}\widetilde{\stackrel{⌢}{\mathsf{B}}}⟩}_0=b_0^2-b_1^2-b_2^2-b_3^2+b_{12}^2+b_{13}^2+b_{23}^2-b_{123}^2,\hfill \\ b_I={⟨\mathsf{B}\widetilde{\stackrel{⌢}{\mathsf{B}}}I⟩}_0=2b_3{b}_{12}-2b_2{b}_{13}+2b_1{b}_{23}-2b_0{b}_{123}.\hfill \end{array}$$

(15)

In (15), the MV $\widetilde{\stackrel{⌢}{\mathsf{B}}}$ denotes the Clifford conjugate of $\mathsf{B}$, where tilde is the grade reversion and cap is the grade inversion. Note that for remaining algebras ${\mathit{Cl}}_{0,3}$, ${\mathit{Cl}}_{2,1}$, and ${\mathit{Cl}}_{2,1}$, the signs of individual terms inside $b_S$ and $b_I$ all are different. As we shall see below, the square roots for all $n=3$ algebras are predetermined by four real coefficients only, namely, $b_0$, $b_{123}$, $b_S$, and $b_I$.

After substitution of (13), the resulting system of Equation (14) is of degree $\le 4$. Thus, we conclude that the initial system (12) can be explicitly solved in radicals. In particular, two real solutions of (13) have the form

$$\begin{array}{cc}\hfill & \left(s_{1,2}=\pm \sqrt{-T+\sqrt{T^2+t^2}},S_{1,2}=\pm {\displaystyle \frac{t}{\sqrt{-T+\sqrt{T^2+t^2}}}}\right),\hfill \end{array}$$

(16)

where the signs in pairs $(s_i,S_i)$ must be identical, plus or minus. The denominator of $S_{1,2}$ becomes zero if $s=S=0$. The remaining two solutions of (13), which can be obtained from (16) by the substitution $\sqrt{T^2+t^2}\to -\sqrt{T^2+t^2}$, are complex-valued due to the inequality $\sqrt{T^2+t^2}\ge T$ and therefore must be rejected.

The two real-valued solutions of Equation (14) are

$$\left\{\begin{array}{c}\left(t_{1,2}={\displaystyle \frac{1}{4}}\left(b_{123}\pm {\displaystyle \frac{1}{\sqrt{2}}}\sqrt{-b_S+\sqrt{D}}\right),T_{1,2}={\displaystyle \frac{1}{4}}\left({\displaystyle \frac{\pm b_I}{\sqrt{2}\sqrt{-b_S+\sqrt{D}}}}-b_0\right)\right),\hfill \\ \phantom{t_{1,2}={\displaystyle \frac{1}{4}}\left(b_{123}\pm {\displaystyle \frac{1}{\sqrt{2}}}\sqrt{-b_S+\sqrt{D}}\right),T_{1,2}={\displaystyle \frac{1}{4}}{b}_{123}}\mathrm{if}-b_S+\sqrt{D}>0,\hfill \\ \left(t_{1,2}=b_{123}/4,T_{1,2}={\displaystyle \frac{1}{4}}\left(\pm \sqrt{b_S}-b_0\right)\right),\mathrm{if}-b_S+\sqrt{D}=0\wedge b_S>0.\hfill \end{array}\right.$$

(17)

No additional conditions are required for the determinant $D=b_S^2+b_I^2\ge 0$ of the MV $\mathsf{B}$, since for ${\mathit{Cl}}_{3,0}$ algebra, it is always positive definite $D\ge b_S$ (see in Appendix D how to compute the MV determinant). Again, we should take the same signs for $t_i$ and $T_i$. The two complex-valued solutions of (14), which can be obtained from (17) by substitution $\sqrt{D}\to -\sqrt{D}$, must be rejected. The denominator of $T_{1,2}$ in (17) turns into zero when $b_S=\sqrt{D}$, i.e., when $b_I=0$.

To summarize, starting from (17) and then going to (16), and finally to Formulas (10) and (11), one obtains four explicit real solutions which completely determine the square root $\mathsf{A}=\sqrt{\mathsf{B}}$ of generic MV $\mathsf{B}$ in terms of radicals $\mathsf{A}=s+\mathbf{v}+(S+\mathbf{V})I$ of real Clifford algebra ${\mathit{Cl}}_{3,0}$.

3.2. The Special Case $s^2+S^2=0$

The only special case in ${\mathit{Cl}}_{3,0}$ corresponds to $s=S=0$. In the subcases $s=S\ne 0$ and $s=-S\ne 0$, one can rewrite expressions (16) in a simpler form. In particular, when $s=S\ne 0$, we have

$$\begin{array}{cc}\hfill s_{1,2}=& \left\{\begin{array}{cc}\pm \frac{1}{2}\sqrt{b_{123}+{\displaystyle \frac{b_I}{2b_0}}}\hfill & \mathrm{if}{b}_0\ne 0,\hfill \\ \pm \frac{1}{2}\sqrt{b_{123}\pm \sqrt{-b_S}}\hfill & \mathrm{if}{b}_0=0,\hfill \end{array}\right.\hfill \end{array}$$

(18)

and when $s=-S\ne 0$,

$$\begin{array}{cc}\hfill s_{1,2}=& \left\{\begin{array}{cc}\pm \frac{1}{2}\sqrt{-b_{123}-{\displaystyle \frac{b_I}{2b_0}}}\hfill & \mathrm{if}{b}_0\ne 0,\hfill \\ \pm \frac{1}{2}\sqrt{-b_{123}\pm \sqrt{-b_S}}\hfill & \mathrm{if}{b}_0=0,\hfill \end{array}\right.\hfill \end{array}$$

(19)

where all expressions inside square roots are assumed to be positive.

The case $s=S=0$ is special, because the condition implies that the number of square roots of $\mathsf{B}$ may be infinite (the case of simple MV roots is given in [12]). Indeed, in this case, Equations (7)–(9) are compatible only if the vector $(b_1,b_2,b_3)$ and bivector $(b_{12},b_{13},b_{23})$ coefficients are zeros. Then, Equation (6) reduces to

$$b_0={\mathbf{v}}^2-{\mathbf{V}}^2,b_{123}=2(\mathbf{v}·\mathbf{V}),$$

(20)

where ${\mathbf{v}}^2=v_1^2+v_2^2+v_3^2$ and ${\mathbf{V}}^2=V_1^2+V_2^2+V_3^2$ for ${\mathit{Cl}}_{3,0}$. Since, in general, we have $3+3=6$ unknowns which must satisfy Equation (20), we are left with four real arbitrary parameters as will be explicitly demonstrated in Example 1. The solution therefore makes a four-dimensional (or smaller) set of real-valued MV coefficients. It is interesting that both expressions in (20) have a very clear geometric interpretation. Indeed, if the ends of vectors $\mathbf{v}$ and $\mathbf{V}$ represent two concentric spheres, then the coefficient $b_0$ controls the lengths of radii $\left|\mathbf{v}\right|$ and $\left|\mathbf{V}\right|$, while the pseudoscalar coefficient $b_{123}$ controls the angle between the vectors $\mathbf{v}$ and $\mathbf{V}$. From this follows that, due to periodicity of the angle, one can introduce principal value for coefficient $b_{123}$. Similar properties, i.e., the multiplicity of roots and the existence of principal angle in a complex plane, are well-known in case complex numbers [17].

3.3. ${\mathit{Cl}}_{1,2}\simeq {\mathit{Cl}}_{3,0}$ Algebra

In paper [18], it is shown that “…for odd $n\ge 3$, there are three classes of isomorphic Clifford algebras what is consistent with Cartan’s classification of real Clifford algebras.” In particular, two algebras, ${\mathit{Cl}}_{3,0}$ and ${\mathit{Cl}}_{1,2}$, are represented by $2\times 2$ complex matrices $\mathbb{C}\left(2\right)$. The similarity between square root expressions obtained below also confirms that these two algebras fall into the same isomorphism class. On the other hand, the algebras ${\mathit{Cl}}_{0,3}$ and ${\mathit{Cl}}_{2,1}$ are represented by blocked $2\times 2$ and $1\times 1$ matrices, respectively, ${}^2\mathbb{R}\left(2\right)$ and ${}^2\mathbb{H}\left(1\right)$. Therefore, they belong to different classes. Indeed, as we shall show later, the analysis of roots in ${\mathit{Cl}}_{2,1}$ is only roughly similar to that in ${\mathit{Cl}}_{0,3}$. However, between ${\mathit{Cl}}_{2,1}$ and ${\mathit{Cl}}_{0,3}$, there are distinctions: they are isomorphic to different, real, and quaternionic matrices.

As far as ${\mathit{Cl}}_{1,2}$ algebra is concerned, its difference from ${\mathit{Cl}}_{3,0}$ is contained in the explicit expression for $b_S$,

$$\begin{array}{cc}\hfill & b_S={⟨\mathsf{B}\widetilde{\stackrel{⌢}{\mathsf{B}}}⟩}_0=b_0^2-b_1^2+b_2^2+b_3^2-b_{12}^2-b_{13}^2+b_{23}^2-b_{123}^2,\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill & b_I={⟨\mathsf{B}\widetilde{\stackrel{⌢}{\mathsf{B}}}I⟩}_0=2b_3{b}_{12}-2b_2{b}_{13}+2b_1{b}_{23}-2b_0{b}_{123},\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill & D=b_S^2+b_I^2,{\mathit{Cl}}_{1,2}\hfill \end{array}$$

(23)

and expressions for $v_i$ and $V_i$

$$\begin{array}{cccccc}\hfill v_1& ={\displaystyle \frac{b_{1}s+b_{23}S}{2(s^2+S^2)}},\hfill & \hfill v_2& ={\displaystyle \frac{b_{2}s+b_{13}S}{2(s^2+S^2)}},\hfill & \hfill v_3& ={\displaystyle \frac{b_{3}s-b_{12}S}{2(s^2+S^2)}},\hfill \end{array}$$

(24)

$$\begin{array}{cccccc}\hfill V_1& ={\displaystyle \frac{b_{23}s-b_{1}S}{2(s^2+S^2)}},\hfill & \hfill V_2& ={\displaystyle \frac{b_{13}S-b_{2}S}{2(s^2+S^2)}},\hfill & \hfill V_3& =-{\displaystyle \frac{b_{12}s+b_{3}S}{2(s^2+S^2)}}.\hfill \end{array}$$

(25)

The expressions for $b_I$ and D (the determinant of $\mathsf{B}$) remain the same. Note that in (20), the scalar product in ${\mathit{Cl}}_{1,2}$ has both plus/minus signs, in particular ${\mathbf{v}}^2=v_1^2-v_2^2-v_3^2$. Before considering other algebras, it is helpful to analyze a few examples.

3.4. Examples for ${\mathit{Cl}}_{3,0}$ and ${\mathit{Cl}}_{1,2}$

Example 1.

The Case $s\ne S$.

The square root of $\mathsf{B}={\mathbf{e}}_1-2{\mathbf{e}}_{23}$ in ${\mathit{Cl}}_{3,0}$. The coefficients in this case are $b_1=1$ and $b_{23}=-2$, and all remaining ones are equal to zero. Then, from (15) follows that $b_I=-4$ and $b_S=3$. The expression (17) gives $t_{1,2}=(\frac{1}{4},-\frac{1}{4})$ and $T_{1,2}=(-\frac{1}{2},\frac{1}{2})$. Finally, using (16), we find the real solutions for s and S,

$$\begin{array}{cc}\hfill & \left(s_{1,2}=\mp \frac{1}{2}{c}_1,S_{1,2}=\pm \frac{1}{2}{c}_2\right)\mathrm{and}\left(s_{3,4}=\pm \frac{1}{2}{c}_2,S_{3,4}=\pm \frac{1}{2}{c}_1\right),\hfill \end{array}$$

(26)

where $c_1=\sqrt{-2+\sqrt{5}}$ and $c_2=\sqrt{2+\sqrt{5}}$. Thus, the MV is regular. Using (10) and (11), we have the following four sets of non-zero coefficients:

$$\begin{array}{cccc}(s_1=-\frac{1}{2}{c}_1,\hfill & S_1=\frac{1}{2}{c}_2,\hfill & v_1=-\frac{1}{2}{c}_2,\hfill & V_1=-\frac{1}{2}{c}_1),\hfill \\ (s_2=\frac{1}{2}{c}_1,\hfill & S_2=-\frac{1}{2}{c}_2,\hfill & v_1=\frac{1}{2}{c}_2,\hfill & V_1=\frac{1}{2}{c}_1),\hfill \\ (s_3=\frac{1}{2}{c}_2,\hfill & S_3={\displaystyle \frac{1}{2}}{c}_1,\hfill & v_1=\frac{1}{2}{c}_1,\hfill & V_1=-\frac{1}{2}{c}_2),\hfill \\ (s_4=-\frac{1}{2}{c}_2,\hfill & S_4=-\frac{1}{2}{c}_1,\hfill & v_1=-\frac{1}{2}{c}_1,\hfill & V_1=\frac{1}{2}{c}_2).\hfill \end{array}$$

(27)

The remaining coefficients are equal to zero, $v_2=v_3=V_2=V_3=0$. Finally, inserting the coefficients (27) into (5), one can find four different roots,

$$\begin{array}{cc}\hfill & {\mathsf{A}}_{1,2}=\mp \frac{1}{2}{c}_2\left(-2+\sqrt{5}+{\mathbf{e}}_1+(-2+\sqrt{5}){\mathbf{e}}_{23}-{\mathbf{e}}_{123}\right),\hfill \\ \hfill & {\mathsf{A}}_{3,4}=\pm \frac{1}{2}{c}_1\left(2+\sqrt{5}+{\mathbf{e}}_1-(2+\sqrt{5}){\mathbf{e}}_{23}+{\mathbf{e}}_{123}\right),\hfill \end{array}$$

(28)

squares of which give the initial MV $\mathsf{B}={\mathbf{e}}_1-2{\mathbf{e}}_{23}$.

Example 2.

The Case $s=S$.

The square root of $\mathsf{B}=-1+{\mathbf{e}}_3-{\mathbf{e}}_{12}+{\displaystyle \frac{1}{2}}{\mathbf{e}}_{123}$ in ${\mathit{Cl}}_{3,0}$. Now, $b_0=-1$, $b_{123}=\frac{1}{2}$, $b_I=-1$, $b_S=\frac{3}{4}$. Then, from (16) and (17) follows real solutions for $s_i$ and $S_i$,

$$\left(s_{1,2}=\pm \frac{1}{2},S_{1,2}=\pm \frac{1}{2}\right)\mathrm{and}\left(s_{3,4}=0,S_{3,4}=\pm 1\right).$$

(29)

Then, for case $(s_{1,2},S_{1,2})$, Equations (10) and (11) yield

$$\begin{array}{cccc}(s_1=-\frac{1}{2},\hfill & v_1=v_2=v_3=0,\hfill & V_1=V_2=0,\hfill & V_3=1),\hfill \\ (s_2=\frac{1}{2},\hfill & v_1=v_2=v_3=0,\hfill & V_1=V_2=0,\hfill & V_3=-1).\hfill \end{array}$$

(30)

The case $(s_{3,4},S_{3,4})$ is treated exactly as in Example 1. The final answer consists of four roots too:

$$\begin{array}{cc}\hfill & {\mathsf{A}}_{1,2}=\pm \frac{1}{2}\left(-1+2{\mathbf{e}}_{12}-{\mathbf{e}}_{123}\right),\hfill \\ \hfill & {\mathsf{A}}_{3,4}=\pm \frac{1}{2}\left({\mathbf{e}}_3+{\mathbf{e}}_{12}-2{\mathbf{e}}_{123}\right).\hfill \end{array}$$

(31)

Example 3.

The Case $s=S=0$.

The square root of $\mathsf{B}=-1+{\mathbf{e}}_{123}$, which is the center of ${\mathit{Cl}}_{3,0}$. The coefficients $b_0=-1$, $b_{123}=1$ give $b_I=2$, $b_S=0$. Then, from expressions (17) and (16) follows

$$\begin{array}{cc}\hfill & \left(s_{1,2}=\pm c_1,S_{1,2}=\pm c_2\right)\mathrm{and}\left(s_3=0,S_3=0\right),\hfill \end{array}$$

(32)

where $c_1=\sqrt{-1/2+1/\sqrt{2}}$ and $c_2=\sqrt{1/2+1/\sqrt{2}}$.

The case $(s_{1,2},S_{1,2})$ in (32) can be computed similarly as in Example 1. The two square roots, which are obtained from case $\left(s_{1,2}=\pm c_1,S_{1,2}=\pm c_2\right)$, therefore are

$${\mathsf{A}}_{1,2}=\pm (c_1+c_2{\mathbf{e}}_{123}).$$

(33)

The set of two roots above should be extended by adding a set of roots provided by the case $(s_3=0,S_3=0)$ in (32), which is special. Indeed, some of coefficients in this case remain unspecified and therefore may be treated as free parameters that yield an uncountable number (continuum) of roots. The coefficients ($b_1,b_2,b_3$) and ($b_{12},b_{13},b_{23}$) in this case are zeroes; however, the compatibility of (7)–(9) is satisfied and the solution set is not empty. Indeed, as seen from (20), the system can be solved for an arbitrary pair of coefficients ($v_1,v_2,v_3,V_1,V_2,V_3$), for example with ($v_1,V_1$). If ($v_1,V_1$) is inserted into (5), one gets an MV with four free parameters,

$$\mathsf{A}=f_1(v_2,v_3,V_2,V_3){\mathbf{e}}_1+v_2{\mathbf{e}}_2+v_3{\mathbf{e}}_3+f_2(v_2,v_3,V_2,V_3){\mathbf{e}}_{23}-V_2{\mathbf{e}}_{13}+V_3{\mathbf{e}}_{12},$$

(34)

where $v_1=f_1(v_2,v_3,V_2,V_3)$ and $V_1=f_2(v_2,v_3,V_2,V_3)$ denote explicit solutions of (20),

$$\begin{array}{cc}\hfill & v_1=\mp {\displaystyle \frac{c_1}{\sqrt{2}}},V_1=\pm {\displaystyle \frac{1}{c_1}}{\displaystyle \frac{-b_{123}+2(v_2{V}_2+v_3{V}_3)}{\sqrt{2}}},\mathrm{with}\hfill \\ \hfill & c_1=(\pm \sqrt{{\left(b_0-v_2^2-v_3^2+V_2^2+V_3^2\right)}^2+{(b_{123}-2(v_2{V}_2+v_3{V}_3))}^2}\hfill \\ \hfill & \phantom{{\displaystyle \frac{1}{\sqrt{2}}}{b}_0-v_2^2-v_3^2(b_{123}-2(v_2{V}_2+v_3{V}_3))x}+\left(b_0-v_2^2-v_3^2+V_2^2+V_3^2\right){)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

(35)

For example, by setting all free parameters to zero, $v_2=V_2=v_3=V_3=0$, we select from a continuum two roots, which we denote

$${\mathsf{A}}_{3,4}=\pm (c_1{\mathbf{e}}_1+c_2{\mathbf{e}}_{23}).$$

(36)

It is important to realize, however, that the number of roots provided by case $(s_3=0,S_3=0)$ in general is infinite and the two roots in (36) represent the simplest choice of free parameters. All roots ${\mathsf{A}}_j$ satisfy ${\mathsf{A}}_j^2=\mathsf{B}=-1+{\mathbf{e}}_{123}$.

If, instead of $\mathsf{B}=-1+{\mathbf{e}}_{123}$, we had tried to find the square root of MV that does not belong to the center, for example, if we had worked with $\mathsf{B}={\mathbf{e}}_1+{\mathbf{e}}_{12}$, which is directly related to polarized electromagnetic wave in ${\mathit{Cl}}_{3,0}$, we would have ended up with an empty solution set. Indeed, in the latter case, $s_1=0$, $S_1=0$ and $b_0=b_{123}=b_I=b_S=0$. Then, after substitution of $s\to s_1=0$ and $S\to S_1=0$ into Equations (7)–(9), one obtains the contradiction $1=0$.

Example 4.

The case of Quaternion.

Quaternions are isomorphic to even subalgebra ${\mathit{Cl}}_{3,0}^{+}$ with elements $\{1,{\mathbf{e}}_{12},{\mathbf{e}}_{23},{\mathbf{e}}_{13}\}$; therefore, the provided formulas allow to find quaternionic square root too. Taking into account that quaternion imaginary units are $\mathbf{i}={\mathbf{e}}_{12}$, $\mathbf{j}=-{\mathbf{e}}_{13}$ and $\mathbf{k}={\mathbf{e}}_{23}$, let us compute the square root of $\mathsf{B}=1+{\mathbf{e}}_{12}-{\mathbf{e}}_{13}+{\mathbf{e}}_{23}=1+\mathbf{i}+\mathbf{j}+\mathbf{k}$. In this example, we have $b_0=1$, $b_{123}=0$, and $b_I=0$, $b_S=4$. Starting from (17) and then using Equation (16), it is easy to find that the MV represents a regular case with four different coefficients

$$\begin{array}{cc}\hfill & \left(s_{1,2}=0,S_{1,2}=\pm 1/\sqrt{2}\right)\mathrm{and}\left(s_{3,4}=\pm \sqrt{3/2},S_{3,4}=0\right).\hfill \end{array}$$

(37)

Using (10)–(11) and (5), we can write the answer:

$$\begin{array}{cc}\hfill & {\mathsf{A}}_{1,2}=\pm ({\mathbf{e}}_1+{\mathbf{e}}_2+{\mathbf{e}}_3+{\mathbf{e}}_{123})/\sqrt{2},\hfill \\ \hfill & {\mathsf{A}}_{3,4}=\pm (3+{\mathbf{e}}_{12}-{\mathbf{e}}_{13}+{\mathbf{e}}_{23})/\sqrt{6}\equiv \pm (3+\mathbf{i}+\mathbf{j}+\mathbf{k})/\sqrt{6}.\hfill \end{array}$$

(38)

The squares of all roots yield the initial MV. It should be noticed that in ${\mathsf{A}}_{3,4}$, the quaternion imaginary units have remained in even algebra only. The source of this ‘strange’ difference is related to the algebra where the square root problem is solved. In particular, here, the roots are computed in ${\mathit{Cl}}_{3,0}$ algebra rather than in ${\mathit{Cl}}_{0,2}$, i.e., algebra of quaternions. However, the formulas for ${\mathit{Cl}}_{0,2}$ (see resp. equation in Appendix B) give two roots only.

Example 5.

The regular case of ${\mathit{Cl}}_{1,2}$ algebra.

Using the same initial MV, $\mathsf{B}={\mathbf{e}}_1-2{\mathbf{e}}_{23}$, as in Example 1, one obtains the same values for $(b_S,b_I)$ and $(s,S)$. After substitution into (24), (25), and then into (5), the square roots are found to be

$$\begin{array}{cc}\hfill & {\mathsf{A}}_{1,2}=\pm \frac{1}{2}(c_2(-{\mathbf{e}}_1+{\mathbf{e}}_{123})-c_1(1+{\mathbf{e}}_{23})),\hfill \\ \hfill & {\mathsf{A}}_{3,4}=\pm \frac{1}{2}(-c_1({\mathbf{e}}_1+{\mathbf{e}}_{123})+c_2(-1+{\mathbf{e}}_{23})),\hfill \end{array}$$

(39)

where $c_1=\sqrt{-2+\sqrt{5}}$ and $c_2=\sqrt{2+\sqrt{5}}$.

4. Square Roots in ${\mathit{Cl}}_{\mathbf{0},\mathbf{3}}$ Algebra

The similar approach to the root problem allows to write down explicit square root formulas for ${\mathit{Cl}}_{0,3}$ algebra as well. Using the same notation (5) for $\mathsf{A}$ and $\mathsf{B}$ and equating coefficients at same basis elements in ${\mathsf{A}}^2=\mathsf{B}$, now we obtain the following system of equations. The formulas have the same structure and differ in signs of some constituent terms only. Below, for easier reading and application, all formulas, including those for mixed algebras, are written explicitly without introducing a large number of sign epsilons ${\epsilon }_{\pm }=\pm 1$. In fact, appearance of different signs in structurally similar expressions brings in different conditions for real root existence in distinct algebras.

$$\begin{array}{cccc}\hfill b_0& =s^2+S^2+{\mathbf{v}}^2+{\mathbf{V}}^2,\hfill & \hfill b_{123}& =2(sS+\mathbf{v}·\mathbf{V}),\hfill \end{array}$$

(40)

$$\begin{array}{cccc}\hfill b_1& =2(s{v}_1+S{V}_1),\hfill & \hfill b_{23}& =-2(s{V}_1+S{v}_1),\hfill \end{array}$$

(41)

$$\begin{array}{cccc}\hfill b_2& =2(s{v}_2+S{V}_2),\hfill & \hfill b_{13}& =2(s{V}_2+S{v}_2),\hfill \end{array}$$

(42)

$$\begin{array}{cccc}\hfill b_3& =2(s{v}_3+S{V}_3),\hfill & \hfill b_{12}& =-2(s{V}_3+S{v}_3),\hfill \end{array}$$

(43)

where, now, ${\mathbf{v}}^2=-v_1^2-v_2^2-v_3^2$ and $\mathbf{v}·\mathbf{V}=-v_1{V}_1-v_2{V}_2-v_3{V}_3$.

4.1. The Generic Case $s^2-S^2\ne 0$

The solution of Equations (41)–(43) is

$$\begin{array}{cccccc}\hfill v_1& ={\displaystyle \frac{b_{1}s+b_{23}S}{2(s^2-S^2)}},\hfill & \hfill v_2& =-{\displaystyle \frac{b_{2}s-b_{13}S}{2(s^2-S^2)}},\hfill & \hfill v_3& ={\displaystyle \frac{b_{3}s+b_{12}S}{2(s^2-S^2)}},\hfill \end{array}$$

(44)

$$\begin{array}{cccccc}\hfill V_1& =-{\displaystyle \frac{b_{23}s+b_{1}S}{2(s^2-S^2)}},\hfill & \hfill V_2& ={\displaystyle \frac{b_{13}s-b_{2}S}{2(s^2-S^2)}},\hfill & \hfill V_3& =-{\displaystyle \frac{b_{12}s+b_{3}S}{2(s^2-S^2)}},\hfill \end{array}$$

(45)

which is valid when $s^2-S^2\ne 0$, and corresponds to the generic case. After substitution of (44) and (45) into (40), one obtains two coupled nonlinear algebraic equations for two unknowns s and S,

$$\begin{array}{cc}\hfill & b_S+4s^2(-6S^2+b_0)+8sS{b}_{123}=4s^4+{(-2S^2+b_0)}^2+b_{123}^2,\hfill \\ \hfill & b_I=2(2(s^2+S^2)-b_0)(4sS-b_{123}),\hfill \end{array}$$

(46)

where, again, the coordinate-free notation is introduced,

$$\begin{array}{cc}\hfill & b_S={⟨\mathsf{B}\widetilde{\stackrel{⌢}{\mathsf{B}}}⟩}_0=b_0^2+b_1^2+b_2^2+b_3^2+b_{12}^2+b_{13}^2+b_{23}^2+b_{123}^2,\hfill \\ \hfill & b_I={⟨\mathsf{B}\widetilde{\stackrel{⌢}{\mathsf{B}}}I⟩}_0=-2b_3{b}_{12}+2b_2{b}_{13}-2b_1{b}_{23}+2b_0{b}_{123}.\hfill \end{array}$$

(47)

Note change of signs as compared to the ${\mathit{Cl}}_{3,0}$ case. The determinant D in ${\mathit{Cl}}_{0,3}$ is expressed as a difference, $D=b_S^2-b_I^2$, which is always positive, $D>0$.

To reduce the degree of the above equations, the substitution

$$sS=t;\frac{1}{2}(s^2+S^2)=T,$$

(48)

is used that transforms the system (46) into a simpler one. To eliminate s and S, Mathematica commands Eliminate[ ], GroebnerBasis[ ] have been used. They allow to rewrite the initial Equation (48) in a number of equivalent forms.

$$b_S={(4T-b_0)}^2+{(4t-b_{123})}^2,b_I=2(4T-b_0)(4t-b_{123}).$$

(49)

The solution of (49) when $(b_S\pm \sqrt{D})>0$ is

$$\begin{array}{cc}\hfill & \left\{\begin{array}{c}\left(t_{1,2}={\displaystyle \frac{1}{4}}(b_{123}\pm {\displaystyle \frac{1}{\sqrt{2}}}\sqrt{b_S-\sqrt{D}}),T_{1,2}={\displaystyle \frac{1}{4}}(b_0\pm {\displaystyle \frac{b_I}{\sqrt{2}\sqrt{b_S-\sqrt{D}}}})\right),\hfill \\ \phantom{t_{1,2}={\displaystyle \frac{1}{4}}\left(b_{123}\pm {\displaystyle \frac{1}{\sqrt{2}}}\sqrt{-b_S+\sqrt{D}}\right),T_{1,2}={\displaystyle \frac{1}{4}}{b}_{123}}\mathrm{if}{b}_S-\sqrt{D}>0,\hfill \\ (t_{1,2}={\displaystyle \frac{1}{4}}{b}_{123},T_{1,2}={\displaystyle \frac{1}{4}}(\pm \sqrt{b_S}+b_0)),\mathrm{if}{b}_S-\sqrt{D}=0\mathrm{and}{b}_S>0.\hfill \end{array}\right.\hfill \end{array}$$

(50)

The ± signs in the above formulas are mutually related; thus, there are only two possibilities that correspond to either plus or minus signs inside $t_i$ and $T_i$ formulas. The remaining two solutions of (49), which were obtained from (50) after replacement $-\sqrt{D}\to +\sqrt{D}$, yield a complex-valued expression for $T\pm \sqrt{T^2-t^2}$ (see Equation (51) below); therefore, they were dismissed in advance.

Once the equations in (50) are computed, they can be substituted back into solutions of (48),

$$\left\{\begin{array}{c}\left(s_{1,2,3,4}=\pm \sqrt{T\pm \sqrt{T^2-t^2}},S_{1,2,3,4}={\displaystyle \frac{\pm t}{\sqrt{T\pm \sqrt{T^2-t^2}}}}\right)\mathrm{if}T\ge 0,t\ne 0,\hfill \\ (s_{1,2}=S_{3,4}=\pm \sqrt{2T},S_{1,2}=s_{3,4}=0)\mathrm{if}T\ge 0,t=0.\hfill \end{array}\right.$$

(51)

In the obtained equations, the same signs must be chosen in the same index positions in $s_{1,2,3,4}$ and $S_{1,2,3,4}$ (four possibilities).

Thus, starting from pairs $(t_1,T_1)$ and $(t_2,T_2)$ in Equation (50) and then going to (51), and finally to Formulas (44), (45), and (5), one obtains explicit real solutions that completely determine the square root of equation $\mathsf{B}={\mathsf{A}}^2$ (with $\mathsf{A}=s+\mathbf{v}+(S+\mathbf{V})I$) of the generic MV $\mathsf{B}$ of real ${\mathit{Cl}}_{0,3}$ algebra in radicals. It appears that, at most, only four isolated real solutions (because the condition $T\ge 0$ in (51) selects only single sign from (50)) are possible in this algebra too, since other choices of signs in (50) and (51) yield negative expressions inside square roots.

4.2. The Special Case $s^2-S^2=0$

There are three subcases: (1) $s=S\ne 0$, (2) $s=-S\ne 0$, and (3) $s=S=0$.

4.2.1. The Subcase $s=S\ne 0$

Here, the system of Equations (41)–(43) has a special solution,

$$\begin{array}{c}\hfill v_1={\displaystyle \frac{b_1}{2s}}-V_1,v_2={\displaystyle \frac{b_2}{2s}}-V_2,v_3={\displaystyle \frac{b_3}{2s}}-V_3,\end{array}$$

(52)

if and only if the MV $\mathsf{B}$ coefficients satisfy $b_1=-b_{23}$, $b_2=b_{13}$, $b_3=-b_{12}$. In (52), $v_i$ is expressed in terms of $V_i$. Appearance of s in the denominators implies that the case $s=S=0$ must be investigated separately. After substituting the solution (52) into (40) and taking into account the mentioned conditions ($b_1=-b_{23},b_2=b_{13},b_3=-b_{12}$) one gets two equations,

$$\begin{array}{cc}\hfill & -{\displaystyle \frac{b_1^2+b_2^2+b_3^2}{4s^2}}+{\displaystyle \frac{b_1{V}_1+b_2{V}_2+b_3{V}_3}{s}}-b_0+2s^2+2(\mathbf{V}·\mathbf{V})=0,\hfill \\ \hfill & -{\displaystyle \frac{b_1{V}_1+b_2{V}_2+b_3{V}_3}{s}}-b_{123}+2s^2-2(\mathbf{V}·\mathbf{V})=0,\hfill \end{array}$$

(53)

that should be kept mutually compatible. To this end, we subtract and add the above equations to get

$$\begin{array}{cc}\hfill & -{\displaystyle \frac{4s^2\left(b_0-b_{123}-4(\mathbf{V}·\mathbf{V})\right)-8s(b_1{V}_1+b_2{V}_2+b_3{V}_3)+b_1^2+b_2^2+b_3^2}{4s^2}}=0,\hfill \\ \hfill & -{\displaystyle \frac{4s^2\left(b_0+b_{123}-4s^2\right)+b_1^2+b_2^2+b_3^2}{4s^2}}=0.\hfill \end{array}$$

(54)

Then, making use of expanded form of $b_S={⟨\mathsf{B}\widetilde{\stackrel{⌢}{\mathsf{B}}}⟩}_0=b_0^2+2(b_1^2+b_2^2+b_3^2)+b_{123}^2$, where the conditions $b_1=-b_{23},b_2=b_{13},b_3=-b_{12}$ have been taken into account, one can express the sum $b_1^2+b_2^2+b_3^2$ from the second equation in (54), $b_1^2+b_2^2+b_3^2={\displaystyle \frac{1}{2}}(b_S-b_0^2-b_{123}^2)$, and substitute the latter into the first of the equations. The result is the quadratic equations for $V_i$’s. After solving, for example, with respect to $V_1$, one can express $V_1$ in terms of, now, arbitrary free parameters $V_2$ and $V_3$,

$$\begin{array}{cc}\hfill V_1=& {\displaystyle \frac{\sqrt{2}}{8s}}(\sqrt{2}{b}_1\pm (-8s^2\left(b_0-b_{123}+4\left(V_2^2+V_3^2\right)\right)+\hfill \\ \hfill & \phantom{(-8s^2}16s(b_2{V}_2+b_3{V}_3)+b_0^2+2b_1^2+b_{123}^2-b_S{)}^{1/2}),\hfill \end{array}$$

(55)

that warrants compatibility of the system (53). Thus, further analysis may be restricted to the simplest single equation: the second equation in (54) that after introduction of shortcut $b_S$ can be cast to form

$$\begin{array}{c}\hfill b_S=-8b_0{s}^2-8b_{123}{s}^2+b_0^2+b_{123}^2+32s^4.\end{array}$$

(56)

The solution of (56) with respect to s can be expressed in radicals,

$$\begin{array}{cc}\hfill s_{1,2}=& \pm \frac{1}{2\sqrt{2}}\sqrt{\sqrt{2b_S-{(b_0-b_{123})}^2}+b_0+b_{123}},\hfill \end{array}$$

(57)

where all expressions inside square roots are assumed to be positive. The expressions (57), (55), and (52) after substitution into (5) yield the final answer for this special case under conditions for MV $\mathsf{B}$ coefficients: $b_1=-b_{23},b_2=b_{13},b_3=-b_{12}$ that in an abridged version reduce to $b_S-{(b_0-b_{123})}^2=b_I$. In conclusion, the solution set contains two free parameters, $V_2$ and $V_3$, and therefore represents continuum of roots on a two-dimensional manifold in the parameter space.

4.2.2. The Subcase $s=-S\ne 0$

Performing exactly the same analysis as in Section 4.2.1, one obtains the conditions for the existence of a solution: $b_1=b_{23}$, $b_2=-b_{13}$, and $b_3=b_{12}$, or in short, $-b_S+{(b_0+b_{123})}^2=b_I$. Similarly, expressing $v_i$ in terms of $V_i$, one gets

$$\begin{array}{c}\hfill v_1={\displaystyle \frac{b_1}{2s}}+V_1,v_2={\displaystyle \frac{b_2}{2s}}+V_2,v_3={\displaystyle \frac{b_3}{2s}}+V_3.\end{array}$$

(58)

If $V_1$ is expressed in terms of $V_2$ and $V_3$,

$$\begin{array}{cc}\hfill V_1=& {\displaystyle \frac{\sqrt{2}}{8s}}(-\sqrt{2}{b}_1\pm (-8s^2\left(b_0+b_{123}+4(V_2^2+V_3^2)\right)\hfill \\ \hfill & \phantom{(-8s^2}-16s(b_2{V}_2+b_3{V}_3)+b_0^2+2b_1^2+b_{123}^2-b_S{)}^{1/2}),\hfill \end{array}$$

(59)

we find two real solutions for s,

$$\begin{array}{cc}\hfill s_{1,2}=& \pm \frac{1}{2\sqrt{2}}\sqrt{\sqrt{2b_S-{(b_0+b_{123})}^2}+b_0-b_{123}},\hfill \end{array}$$

(60)

After substitution into (5), the above expressions again yield the final MV, provided the conditions $b_1=b_{23},b_2=-b_{13},b_3=b_{12}$ are satisfied.

4.2.3. The Subcase $s=S=0$

The analysis of this special subcase is very similar to that in ${\mathit{Cl}}_{3,0}$. Equations (41)–(43) satisfy the compatibility condition if vector $(b_1,b_2,b_3)$ and bivector $(b_{12},b_{13},b_{23})$ coefficients are equated to zero. Then, Equation (40) assumes the following form

$$b_0={\mathbf{v}}^2+{\mathbf{V}}^2,b_{123}=2(\mathbf{v}·\mathbf{V})$$

(61)

from which follows that four parameters remain unspecified. For example, if Equation (61) is solved with respect to pair $(v_1,V_1)$, one gets

$$\begin{array}{cc}\hfill & v_1=\mp {\displaystyle \frac{c_1}{\sqrt{2}}},V_1=\pm {\displaystyle \frac{1}{c_1}}{\displaystyle \frac{b_{123}+2(v_2{V}_2+v_3{V}_3)}{\sqrt{2}}},\mathrm{where}\hfill \\ \hfill & c_1=(\pm \sqrt{{\left(b_0+v_2^2+v_3^2+V_2^2+V_3^2\right)}^2-{(b_{123}+2(v_2{V}_2+v_3{V}_3))}^2}\hfill \\ \hfill & \phantom{{\displaystyle \frac{1}{\sqrt{2}}}{b}_0-v_2^2-v_3^2(b_{123}-2(v_2{V}_2+v_3{V}_3))x}-b_0-v_2^2-v_3^2-V_2^2-V_3^2{)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

(62)

The pairs $(v_2,V_2)$ and $(v_3,V_3)$ may be interpreted as free parameters that generate a continuum of roots in a four parameter space. The geometric interpretation of Equation (61) is similar to those in (20) for ${\mathit{Cl}}_{3,0}$.

4.3. Examples for ${\mathit{Cl}}_{0,3}$

Example 6.

The regular case.

As in Example 1, let the initial MV be $\mathsf{B}={\mathbf{e}}_1-2{\mathbf{e}}_{23}$, the coefficients of which are $b_1=1$, $b_{12}=-2$. The shortcuts $b_I$ and $b_S$ in (47) have the values $b_I=4$ and $b_S=5$. The formulas in (50) give $(T_1,t_1)=(\frac{1}{2},\frac{1}{4})$ and $(T_2,t_2)=(-\frac{1}{2},-\frac{1}{4})$. Since $T_1$ is positive, the pair $(T_1,t_1)$ is used in the following. Then, from the system (51), we find four values of $(s_i,S_i)$ and then from (44), (45), the coefficients $v_i$ and $V_i$,

$$\begin{array}{ccc}(s_1=V_1=\frac{1}{4}{d}_3,\hfill & S_1=v_1=-\frac{1}{2}{d}_2,\hfill & v_2=v_3=V_2=V_3=0),\hfill \\ (s_2=V_1=\frac{1}{2}{d}_1,\hfill & S_2=v_1=\frac{1}{2}{d}_2,\hfill & v_2=v_3=V_2=V_3=0),\hfill \\ (s_3=V_1=-\frac{1}{2}{d}_2,\hfill & S_3=-\frac{1}{2d_2},v_1=\frac{1}{4}{d}_3,\hfill & v_2=v_3=V_2=V_3=0),\hfill \\ (s_4=V_1=\frac{1}{2}{d}_2,\hfill & S_4=\frac{1}{2d_2},v_1=\frac{1}{2}{d}_1,\hfill & v_2=v_3=V_2=V_3=0).\hfill \end{array}$$

(63)

where $d_1=\sqrt{2-\sqrt{3}}$, $d_2=\sqrt{2+\sqrt{3}}$, and $d_3=\sqrt{2}-\sqrt{6}$. Finally, in the same way as in Example 1, we find four different square roots,

$$\begin{array}{cc}\hfill & {\mathsf{A}}_1=\frac{1}{2}\left(d_1+d_2{\mathbf{e}}_1-d_1{\mathbf{e}}_{23}+d_2{\mathbf{e}}_{123}\right),\hfill \\ \hfill & {\mathsf{A}}_2=\frac{1}{2}\left(\frac{1}{2}{d}_3-d_2{\mathbf{e}}_1-\frac{1}{2}{d}_3{\mathbf{e}}_{23}-d_2{\mathbf{e}}_{123}\right).\hfill \\ \hfill & {\mathsf{A}}_3=\frac{1}{2}(d_2+d_1{\mathbf{e}}_1-d_2{\mathbf{e}}_{23}+{\displaystyle \frac{{\mathbf{e}}_{123}}{d_2}}),\hfill \\ \hfill & {\mathsf{A}}_4=\frac{1}{2}(-d_2+\frac{1}{2}{d}_3{\mathbf{e}}_1+d_2{\mathbf{e}}_{23}-{\displaystyle \frac{{\mathbf{e}}_{123}}{d_2}}),\hfill \end{array}$$

(64)

Noting that $d_3=-2d_1$ and $d_2^{-1}=d_1$, the roots may be rewritten in the standard form

$$\begin{array}{cc}\hfill & {\mathsf{A}}_{1,2}=\pm \frac{1}{2}\left(d_1+d_2{\mathbf{e}}_1-d_1{\mathbf{e}}_{23}+d_2{\mathbf{e}}_{123}\right),\hfill \\ \hfill & {\mathsf{A}}_{3,4}=\pm \frac{1}{2}\left(d_2+d_1{\mathbf{e}}_1-d_2{\mathbf{e}}_{23}+d_1{\mathbf{e}}_{123}\right).\hfill \end{array}$$

(65)

Example 7.

The case $s=S\ne 0$.

The square root of $\mathsf{B}=-{\mathbf{e}}_3+{\mathbf{e}}_{12}+4{\mathbf{e}}_{123}$. The shortcuts in $(b_I,b_S)$ have values $b_I=2$ and $b_S=18$, and afterwards, the expression (50) gives $(T_1,t_1)=(\frac{c_1}{4},\frac{c_1}{4})$, where $c_1=(2+\sqrt{5})$. The negative T solution has been omitted. All this gives $(s_1,S_1)=(-\frac{\sqrt{c_1}}{2},-\frac{\sqrt{c_1}}{2})$ and $(s_2,S_2)=(\frac{\sqrt{c_1}}{2},\frac{\sqrt{c_1}}{2})$. The coefficients satisfy the relations $b_1=-b_{23}$, $b_2=b_{13}$, $b_3=-b_{12}$; therefore, a special solution consisting of four MVs exists:

$$\begin{array}{cc}\hfill & {\mathsf{A}}_1=-{\displaystyle \frac{1}{2}}({\mathbf{e}}_1+{\mathbf{e}}_{23})\sqrt{-4V_2^2+4(c_1-V_3)V_3+c_3}-{\mathbf{e}}_{12}{V}_3+({\mathbf{e}}_{13}-{\mathbf{e}}_2)V_2\hfill \\ \hfill & \phantom{-{\displaystyle \frac{1}{2}}({\mathbf{e}}_1+{\mathbf{e}}_{23})\sqrt{-4V_2^2+4(c_1-V_3)V_3+c_3}}+{\mathbf{e}}_3(c_1-V_3)-{\displaystyle \frac{1}{2}}{c}_2({\mathbf{e}}_{123}+1),\hfill \\ \hfill & {\mathsf{A}}_2={\displaystyle \frac{1}{2}}({\mathbf{e}}_1+{\mathbf{e}}_{23})\sqrt{-4V_2^2+4(c_1-V_3)V_3+c_3}-{\mathbf{e}}_{12}{V}_3+({\mathbf{e}}_{13}-{\mathbf{e}}_2)V_2\hfill \\ \hfill & \phantom{-{\displaystyle \frac{1}{2}}({\mathbf{e}}_1+{\mathbf{e}}_{23})\sqrt{-4V_2^2+4(c_1-V_3)V_3+c_3}}+{\mathbf{e}}_3(c_1-V_3)-{\displaystyle \frac{1}{2}}{c}_2({\mathbf{e}}_{123}+1),\hfill \\ \hfill & {\mathsf{A}}_3={\displaystyle \frac{1}{2}}(-({\mathbf{e}}_1+{\mathbf{e}}_{23})\sqrt{-4V_2^2-4V_3(V_3+c_1)+c_3}-2{\mathbf{e}}_{12}{V}_3\hfill \\ \hfill & \phantom{-{\displaystyle \frac{1}{2}}({\mathbf{e}}_1+{\mathbf{e}}_{23})\sqrt{-4V_2^2}}+2({\mathbf{e}}_{13}-{\mathbf{e}}_2)V_2-2{\mathbf{e}}_3(V_3+c_1)+c_2({\mathbf{e}}_{123}+1)),\hfill \\ \hfill & {\mathsf{A}}_4={\displaystyle \frac{1}{2}}(({\mathbf{e}}_1+{\mathbf{e}}_{23})\sqrt{-4V_2^2-4V_3(V_3+c_1)+c_3}-2{\mathbf{e}}_{12}{V}_3\hfill \\ \hfill & \phantom{-{\displaystyle \frac{1}{2}}({\mathbf{e}}_1+{\mathbf{e}}_{23})\sqrt{-4V_2^2}}+2({\mathbf{e}}_{13}-{\mathbf{e}}_2)V_2-2{\mathbf{e}}_3(V_3+c_1)+c_2({\mathbf{e}}_{123}+1)),\hfill \end{array}$$

(66)

where $c_1=\sqrt{\sqrt{5}-2}$, $c_2=\sqrt{\sqrt{5}+2}$, $c_3=-\sqrt{5}+6$. Assuming concrete values of parameters $V_2$ and $V_3$, one can check that the root formulas give real MVs.

It should be noted, however, that symbolical expressions do not guarantee that we will always be able to find real parameters $V_2$ and $V_3$, what would ensure real square roots. For example, if instead of the above MV $\mathsf{B}=-{\mathbf{e}}_3+{\mathbf{e}}_{12}+4{\mathbf{e}}_{123}$, we tried to find the square root of MV $\mathsf{B}=-{\mathbf{e}}_3+{\mathbf{e}}_{12}$ in ${\mathit{Cl}}_{0,3}$ (the MV was used earlier in the Example 2), we would find $s_1=\frac{1}{2}$ and $s_2=-\frac{1}{2}$. The first value then yields $v_1=-V_1=\frac{1}{2}\sqrt{-4V_2^2-{(1+2V_3)}^2}$, $v_2=-V_2$, $v_3=-1-V_3$, and the second one yields $V_1=-v_1=\frac{1}{2}\sqrt{-4V_2^2-{(1-2V_3)}^2}$, $v_2=-V_2$, and $v_3=1-V_3$. Taking the square of symbolical expressions, one can easily check that, formally, we indeed obtain the MV $\mathsf{B}=-{\mathbf{e}}_3+{\mathbf{e}}_{12}$. It is obvious, however, that in both cases ($s_1=\frac{1}{2}$ and $s_2=-\frac{1}{2}$), the expression under square root can be made non-negative (i.e., only zero in this case) for a single choice of parameters. In particular, in the case $s_1=\frac{1}{2}$, the requirement $-4V_2^2-{(1+2V_3)}^2\ge 0$ yields $V_2=0,V_3=-1/2$. Alternatively, in the case $s_2=-\frac{1}{2}$, from equation $-4V_2^2-(1-2V_3)\ge 0$ follows $V_2=0,V_3=1/2$. Both cases yield an isolated root $\pm {\displaystyle \frac{1}{2}}(1-{\mathbf{e}}_3+{\mathbf{e}}_{12}+{\mathbf{e}}_{123})$. Therefore, in this algebra, there exists only the isolated real square root of $\mathsf{B}=-{\mathbf{e}}_3+{\mathbf{e}}_{12}$.

5. Square Roots in ${\mathit{Cl}}_{\mathbf{2},\mathbf{1}}$ Algebra

5.1. The Generic Case $s^2-S^2\ne 0$

The system of nonlinear equations is

$$\begin{array}{cccc}\hfill b_0& =s^2+S^2+{\mathbf{v}}^2+{\mathbf{V}}^2,\hfill & \hfill b_{123}& =2(sS+\mathbf{v}·\mathbf{V}),\hfill \end{array}$$

(67)

$$\begin{array}{cccc}\hfill b_1& =2(s{v}_1+S{V}_1),\hfill & \hfill b_{23}& =2(s{V}_1+S{v}_1),\hfill \end{array}$$

(68)

$$\begin{array}{cccc}\hfill b_2& =2(s{v}_2+S{V}_2),\hfill & \hfill b_{13}& =-2(s{V}_2+S{v}_2),\hfill \end{array}$$

(69)

$$\begin{array}{cccc}\hfill b_3& =2(s{v}_3+S{V}_3),\hfill & \hfill b_{12}& =-2(s{V}_3+S{v}_3),\hfill \end{array}$$

(70)

where, now, ${\mathbf{v}}^2=v_1^2+v_2^2-v_3^2$ and $\mathbf{v}·\mathbf{V}=v_1{V}_1+v_2{V}_2-v_3{V}_3$. When $s^2-S^2\ne 0$, the solutions of systems (68)–(70) are

$$\begin{array}{cccccc}\hfill v_1& ={\displaystyle \frac{b_{1}s-b_{23}S}{2(s^2-S^2)}},\hfill & \hfill v_2& ={\displaystyle \frac{b_{2}s+b_{13}S}{2(s^2-S^2)}},\hfill & \hfill v_3& ={\displaystyle \frac{b_{3}s+b_{12}S}{2(s^2-S^2)}},\hfill \end{array}$$

(71)

$$\begin{array}{cccccc}\hfill V_1& ={\displaystyle \frac{b_{23}s-b_{1}S}{2(s^2-S^2)}},\hfill & \hfill V_2& =-{\displaystyle \frac{b_{13}s+b_{2}S}{2(s^2-S^2)}},\hfill & \hfill V_3& =-{\displaystyle \frac{b_{12}s+b_{3}S}{2(s^2-S^2)}}.\hfill \end{array}$$

(72)

Insertion of $v_i$ and $V_i$ into (67) gives two coupled equations for unknowns $s,S$

$$\begin{array}{cc}\hfill & b_S+4s^2(-6S^2+b_0)+8sS{b}_{123}=4s^4+{(-2S^2+b_0)}^2+b_{123}^2,\hfill \\ \hfill & b_I=2(2(s^2+S^2)-b_0)(4sS-b_{123}),\hfill \end{array}$$

(73)

where $b_S$ and $b_I$ are functions of coefficients in $\mathsf{B}$,

$$\begin{array}{cc}\hfill & b_S={⟨\mathsf{B}\widetilde{\stackrel{⌢}{\mathsf{B}}}⟩}_0=b_0^2-b_1^2-b_2^2+b_3^2+b_{12}^2-b_{13}^2-b_{23}^2+b_{123}^2,\hfill \\ \hfill & b_I={⟨\mathsf{B}\widetilde{\stackrel{⌢}{\mathsf{B}}}I⟩}_0=-2b_3{b}_{12}+2b_2{b}_{13}-2b_1{b}_{23}+2b_0{b}_{123}.\hfill \end{array}$$

(74)

Because Equations (73) and (47) have the same shape (the concrete equations for $b_S$ and $b_I$, of course, are different) we can make use of (48) with the purpose of lowering the order of the system. However, there arises an important difference: the determinant, $D=b_S^2-b_I^2$, in ${\mathit{Cl}}_{2,1}$ is not always positive. It may happen that for some $\mathsf{B}$, the MV determinant may become negative, $D<0$. In such a case, the solution set becomes empty. The other particularity is that in the solution (50), instead of single sign ($-\sqrt{D}$), we have to take into account both signs, i.e., $\pm \sqrt{D}$, what doubles the number of possible solutions in the case $D>0$,

$$\begin{array}{cc}\hfill & \left\{\begin{array}{c}\left(t_{1,2,3,4}={\displaystyle \frac{1}{4}}\left(b_{123}\pm {\displaystyle \frac{1}{\sqrt{2}}}\sqrt{b_S\pm \sqrt{D}}\right),T_{1,2,3,4}={\displaystyle \frac{1}{4}}\left({\displaystyle \frac{\pm b_I}{\sqrt{2}\sqrt{b_S\pm \sqrt{D}}}}+b_0\right)\right),\hfill \\ \phantom{t_{1,2}={\displaystyle \frac{1}{4}}\left(b_{123}\pm {\displaystyle \frac{1}{\sqrt{2}}}\sqrt{-b_S+\sqrt{D}}\right),T_{1,2}={\displaystyle \frac{1}{4}}{b}_{123}}\mathrm{if}{b}_S\pm \sqrt{D}>0,\hfill \\ \left(t_{1,2}=\frac{1}{4}{b}_{123},T_{1,2}=\frac{1}{4}(\pm \sqrt{b_S}+b_0)\right),\mathrm{if}{b}_S\pm \sqrt{D}=0\mathrm{and}{b}_S>0.\hfill \end{array}\right.\hfill \end{array}$$

(75)

Here, again, the sign of $t_i$ must be taken in all possible combinations, and the sign of T must follow the same upper–lower sign position as in $t_i$. The condition $b_S\pm \sqrt{D}=0$ implies that $b_I=0$. Since we already have four sign combinations in the solution for $s,S$ (as in (51)), we end up with 16 different square roots of MV in a generic case of ${\mathit{Cl}}_{2,1}$.

5.2. The Special Case $s^2-S^2=0$

The analysis again closely follows the ${\mathit{Cl}}_{0,3}$ case, except that now, different signs appear in expressions.

5.2.1. The Subcase $s=S\ne 0$

Now, the coefficients satisfy the conditions $b_1=b_{23}$, $b_2=-b_{13}$, $b_3=-b_{12}$, which allows to eliminate the singularity at $s=S$. As a result, the system of Equations (68)–(70) has a special solution,

$$\begin{array}{c}\hfill v_1={\displaystyle \frac{b_1}{2s}}-V_1,v_2={\displaystyle \frac{b_2}{2s}}-V_2,v_3={\displaystyle \frac{b_3}{2s}}-V_3,\end{array}$$

(76)

which coincides with the same solution for ${\mathit{Cl}}_{0,3}$ (see Equation (52)). Thus, after similar calculations, one finds that Equation (55) becomes

$$\begin{array}{cc}\hfill V_1=& {\displaystyle \frac{\sqrt{2}}{8s}}(\sqrt{2}{b}_1\pm (8s^2(b_0-b_{123}+4(-V_2^2+V_3^2))+16s(b_2{V}_2-b_3{V}_3)\hfill \\ \hfill & \phantom{(-8s^2\left(b_0-b_{123}+4(V_2^2+V_3^2)\right)}-b_0^2+2b_1^2-b_{123}^2+b_S{)}^{1/2}).\hfill \end{array}$$

(77)

The coefficients $s_1$ and $s_2$ are similar to (57), except that now, we have to take into account all sign combinations in inner square root,

$$\begin{array}{cc}\hfill s_{1,2}=& \pm \frac{1}{2\sqrt{2}}\sqrt{\pm \sqrt{2b_S-{(b_0-b_{123})}^2}+b_0+b_{123}}.\hfill \end{array}$$

(78)

The above listed formulas solve the square root problem in the case $s=S\ne 0$.

5.2.2. The Subcase $s=-S\ne 0$

The only formulas which differ from ${\mathit{Cl}}_{0,3}$ algebra are connected with the coefficient compatibility condition $b_1=-b_{23}$, $b_2=b_{13}$, $b_3=b_{12}$. Now, the coefficients must be replaced by

$$\begin{array}{cc}\hfill V_1=& {\displaystyle \frac{\sqrt{2}}{8s}}(-\sqrt{2}{b}_1\pm (8s^2(b_0+b_{123}+4(-V_2^2+V_3^2))\hfill \\ \hfill & \phantom{(-8s^2}-16s(b_2{V}_2-b_3{V}_3)-b_0^2+2b_1^2-b_{123}^2+b_S{)}^{1/2}),\hfill \end{array}$$

(79)

$$\begin{array}{cc}\hfill s_{1,2}=& \pm \frac{1}{2\sqrt{2}}\sqrt{\pm \sqrt{2b_S-{(b_0+b_{123})}^2}+b_0-b_{123}}.\hfill \end{array}$$

(80)

The remaining formulas which are needed for final answer exactly match the formulas in the corresponding subcase of ${\mathit{Cl}}_{0,3}$ algebra.

5.2.3. The Subcase $s=S=0$

The only distinct formulas from ${\mathit{Cl}}_{0,3}$ are listed below:

$$\begin{array}{cc}\hfill & v_1=\pm {\displaystyle \frac{c_1}{\sqrt{2}}},V_1=\pm {\displaystyle \frac{1}{c_1}}{\displaystyle \frac{b_{123}+2(-v_2{V}_2+v_3{V}_3)}{\sqrt{2}}},\mathrm{where}\hfill \\ \hfill & c_1=(\pm \sqrt{{\left(b_0-v_2^2+v_3^2-V_2^2+V_3^2\right)}^2-{(b_{123}+2(-v_2{V}_2+v_3{V}_3))}^2}\hfill \\ \hfill & \phantom{{\displaystyle \frac{1}{\sqrt{2}}}{b}_0-v_2^2-v_3^2(b_{123}-2(v_2{V}_2+v_3{V}_3))}+b_0-v_2^2+v_3^2-V_2^2+V_3^2{)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

(81)

This ends the investigation of the square root formulas for all real 3D CAs.

5.3. Examples for ${\mathit{Cl}}_{2,1}$

Example 8.

The regular case.

First, we shall show that MV $\mathsf{B}={\mathbf{e}}_1-2{\mathbf{e}}_{23}$ has no real square roots. Indeed, we have $b_S=-5$, $b_I=4$ and $D=b_S^2-b_I^2={\left(3\right)}^2$. As a result, the expression under square root in (75), namely $b_S\pm \sqrt{D}=-5\pm 3$, is always negative and therefore, there are no real-valued solutions.

Next, we shall calculate the roots of $\mathsf{B}=2+{\mathbf{e}}_1+{\mathbf{e}}_{13}$. The values of $b_I$ and $b_S$ are 0 and 2, respectively. The determinant of the MV is positive, $D=4>0$. The Equation (75) give four real values for pairs: $(T_1,t_1)=\left({\displaystyle \frac{1}{4}}(2-\sqrt{2}),0\right)$, $(T_2,t_2)=\left({\displaystyle \frac{1}{4}}(2+\sqrt{2}),0\right)$, $(T_3,t_3)=\left({\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{2\sqrt{2}}}\right)$, and $(T_4,t_4)=\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2\sqrt{2}}}\right)$. After insertion into (51), 16 pairs of scalars $(s_i,S_i)$ are found:

$$\begin{array}{ccc}(s_1=0,S_1=-{\displaystyle \frac{c_2}{\sqrt{2}}}),\hfill & (s_2=0,S_2={\displaystyle \frac{c_2}{\sqrt{2}}}),\hfill & (s_3=-{\displaystyle \frac{c_2}{\sqrt{2}}},S_3=0),\hfill \\ (s_4={\displaystyle \frac{c_2}{\sqrt{2}}},S_4=0),\hfill & (s_5=0,S_5=-{\displaystyle \frac{c_1}{\sqrt{2}}}),\hfill & (s_6=0,S_6={\displaystyle \frac{c_1}{\sqrt{2}}}),\hfill \\ (s_7=-{\displaystyle \frac{c_1}{\sqrt{2}}},S_7=0),\hfill & (s_8={\displaystyle \frac{c_1}{\sqrt{2}}},S_8=0),\hfill & (s_9=-{\displaystyle \frac{c_2}{2}},S_9={\displaystyle \frac{c_1}{2}}),\hfill \\ (s_{10}={\displaystyle \frac{c_2}{2}},S_{10}=-{\displaystyle \frac{c_1}{2}}),\hfill & (s_{11}=-{\displaystyle \frac{c_1}{2}},S_{11}={\displaystyle \frac{1}{\sqrt{2}{c}_1}}),\hfill & (s_{12}={\displaystyle \frac{c_1}{2}},S_{12}=-{\displaystyle \frac{1}{\sqrt{2}{c}_1}}),\hfill \\ (s_{13}=-{\displaystyle \frac{c_2}{2}},S_{13}=-{\displaystyle \frac{c_1}{2}}),\hfill & (s_{14}={\displaystyle \frac{c_2}{2}},S_{14}={\displaystyle \frac{c_1}{2}}),\hfill & (s_{15}=-{\displaystyle \frac{c_1}{2}},S_{15}=-{\displaystyle \frac{1}{\sqrt{2}{c}_1}}),\hfill \\ (s_{16}={\displaystyle \frac{c_1}{2}},S_{16}={\displaystyle \frac{1}{\sqrt{2}{c}_1}}),\hfill \end{array}$$

where $c_1=\sqrt{2+\sqrt{2}}$ and $c_2=\sqrt{2-\sqrt{2}}$. After substitution of $(s_i,S_i)$ into Equation (71) and then into Equation (5), we obtain 16 roots ${\mathsf{A}}_{i,j}=\pm \sqrt{2+{\mathbf{e}}_1+{\mathbf{e}}_{13}}$:

$$\begin{array}{cc}\hfill & {\mathsf{A}}_{1,2}=\pm \frac{1}{2}\left(c_1{\mathbf{e}}_2-c_1{\mathbf{e}}_{23}-\sqrt{2}{c}_2{\mathbf{e}}_{123}\right),\hfill \\ \hfill & {\mathsf{A}}_{3,4}=\pm \frac{1}{\sqrt{2}}\left(-c_1^{-1}{\mathbf{e}}_2+c_1^{-1}{\mathbf{e}}_{23}+c_1{\mathbf{e}}_{123}\right),\hfill \\ \hfill & {\mathsf{A}}_{5,6}=\pm \frac{1}{2}\left(\sqrt{2}{c}_2+c_1{\mathbf{e}}_1+c_1{\mathbf{e}}_{13}\right),\hfill \\ \hfill & {\mathsf{A}}_{7,8}=\pm \frac{1}{\sqrt{2}}\left(c_1+c_1^{-1}{\mathbf{e}}_1+c_1^{-1}{\mathbf{e}}_{13}\right),\hfill \\ \hfill & {\mathsf{A}}_{9,10}=\pm \frac{1}{2\sqrt{2}}\left(\sqrt{2}{c}_2-c_2{\mathbf{e}}_1-c_1{\mathbf{e}}_2-c_2{\mathbf{e}}_{13}+c_1{\mathbf{e}}_{23}+\sqrt{2}{c}_1{\mathbf{e}}_{123}\right),\hfill \\ \hfill & A_{11,12}=\pm \frac{1}{2\sqrt{2}}\left(\sqrt{2}{c}_1+c_1{\mathbf{e}}_1-c_2{\mathbf{e}}_2+c_1{\mathbf{e}}_{13}+c_2{\mathbf{e}}_{23}-2c_1^{-1}{\mathbf{e}}_{123}\right),\hfill \\ \hfill & {\mathsf{A}}_{13,14}=\pm \frac{1}{2\sqrt{2}}\left(\sqrt{2}{c}_1+c_1{\mathbf{e}}_1+c_2{\mathbf{e}}_2+c_1{\mathbf{e}}_{13}-c_2{\mathbf{e}}_{23}+2c_1^{-1}{\mathbf{e}}_{123}\right),\hfill \\ \hfill & {\mathsf{A}}_{15,16}=\pm \frac{1}{2\sqrt{2}}\left(-\sqrt{2}{c}_2+c_2{\mathbf{e}}_1-c_1{\mathbf{e}}_2+c_2{\mathbf{e}}_{13}+c_1{\mathbf{e}}_{23}+\sqrt{2}{c}_1{\mathbf{e}}_{123}\right).\hfill \end{array}$$

(82)

In the end, it is worth noting that the necessary (but not sufficient) condition for a square root of MV $\mathsf{B}$ to exist in real Clifford algebras ${\mathit{Cl}}_{p,q}$ requires the positivity of the multivector determinant $det\left(\mathsf{B}\right)$ [19,20]. Indeed, if the MV $\mathsf{A}$ exists and $\mathsf{A}\mathsf{A}=\mathsf{B}$, then the determinant of both sides gives $det\left(\mathsf{A}\right)det\left(\mathsf{A}\right)=det\left(\mathsf{B}\right)$, where we have used the multiplicative property of the determinant [21]. Since the determinant of $\mathsf{A}$ in real CAs is a real quantity, the condition can be satisfied if and only if $det\left(\mathsf{B}\right)\ge 0$. This is in agreement with explicit formulas for $n\le 3$.

6. Conclusions

First, we have shown analytically that the square root of general MV in $n=p+q\le 3$ Clifford algebras (CAs) can be expressed in radicals and have provided a detailed analysis and formulas to accomplish the task. For a general MV, the computation algorithm is rather complicated, where many conditions are controlled by plus/minus signs. Our first paper [12] was limited to the roots of individual grades, where it is possible to write down explicit formulas in a coordinate-free form.

Second, the paper shows that MV roots may be isolated (up to 16 roots in case of ${\mathit{Cl}}_{2,1}$) and/or continuous, or conversely, there may be no roots at all. Thus, the MV algebras may also accommodate number-free parameters that bring in a continuum of roots on respective parameter hypersurface.

Third, the described algorithm was implemented in the Mathematica system [14] and applied in checking up algorithms with purely numerical root search. For this purpose, the Mathematica universal root search algorithm was realized and used in the system function FindInstance[ ] to check whether there are cases when the isolated root algorithm fails. No such cases were found. The only complication we encountered in the algorithm programming was that Mathematica symbolic zero detection algorithm PossibleZeroQ[ ], in the more complicated cases, often switched over to numerical procedure to detect that the involved symbolic expression with nested radicals indeed represents zero. This is quite understandable, since it is well-known that a two-expression equivalence problem is, in general, undecidable.

Fourth, we found that for algebras ${\mathit{Cl}}_{3,0}$ and ${\mathit{Cl}}_{1,2}$, the square root solution in general is a union of the following sets: (1) when $s^2\ne S^2$, the set consists of (up to) four different isolated roots; (2) when $s^2=S^2\ne 0$, the set consists of two isolated roots; and (3) when $s^2=S^2=0$, there appears a continuum of roots that belong to four-or-smaller-dimensional parameter manifolds. Similar sets with minor modifications exist for remaining algebras ${\mathit{Cl}}_{2,1}$ and ${\mathit{Cl}}_{0,3}$ as well.

The proposed method is a step forward in solving general quadratic equations in CAs (examples are given in [12]), and may find new applications in the control and systems theory [22], partly because presented solutions uncover totally new properties of square roots of MVs; for example, the root multiplicity and appearance of free parameters in the roots. Due to intricacies of square root algorithms, it is recommended to perform all calculations with prepared-in-advance numerical/symbolic subroutines.

Appendix C.1. Computation Flow in ${\mathit{Cl}}_{3,0}$

${\mathit{Cl}}_{3,0}$ specific expressions:

$$\begin{array}{cc}\hfill & \sqrt{D}=\sqrt{b_S^2+b_I^2},\hfill \\ \hfill & \mathrm{bSD}=-b_S+\sqrt{D}.\hfill \end{array}$$

Computation-ready s and S expressions:

$$\begin{array}{cccc}\hfill F_1:& s=-\sqrt{-\mathrm{s}2\mathrm{S}2+\mathrm{sSsq}},\hfill & & S=-(\mathrm{stS}/\sqrt{-\mathrm{s}2\mathrm{S}2+\mathrm{sSsq}});\hfill \\ \hfill F_2:& s=\sqrt{-\mathrm{s}2\mathrm{S}2+\mathrm{sSsq}},\hfill & & S=(\mathrm{stS}/\sqrt{-\mathrm{s}2\mathrm{S}2+\mathrm{sSsq}});\hfill \\ \hfill F_3:& s=0,\hfill & & S=-\sqrt{2}\sqrt{\mathrm{s}2\mathrm{S}2};\hfill \\ \hfill F_4:& s=0,\hfill & & S=\sqrt{2}\sqrt{\mathrm{s}2\mathrm{S}2};\hfill \\ \hfill F_5:& s=-\sqrt{2}\sqrt{-\mathrm{s}2\mathrm{S}2},\hfill & & S=0;\hfill \\ \hfill F_6:& s=\sqrt{2}\sqrt{-\mathrm{s}2\mathrm{S}2},\hfill & & S=0;\hfill \\ \hfill F_7:& s=0,\hfill & & S=0,\hfill \end{array}$$

(A4)

where $\mathrm{sSsq}=\sqrt{\mathrm{s}2\mathrm{S}{2}^2+{\mathrm{stS}}^2}$. Substitutions that are used for condition checks in (A5) and for computation of final $s,S$ values in Formula (A4):

$$\begin{array}{cccc}\hfill T_1:& \mathrm{stS}=(\sqrt{\mathrm{bSD}}/\sqrt{2}+b_{123})/4,\hfill & & \mathrm{s}2\mathrm{S}2=(\left(\sqrt{2}{b}_I\right)/\sqrt{\mathrm{bSD}}-2b_0)/8,\hfill \\ \hfill T_2:& \mathrm{stS}=(-\sqrt{\mathrm{bSD}}/\sqrt{2}+b_{123})/4,\hfill & & \mathrm{s}2\mathrm{S}2=((-\sqrt{2}{b}_I)/\sqrt{\mathrm{bSD}}-2b_0)/8,\hfill \\ \hfill T_3:& \mathrm{stS}=b_{123}/4,\hfill & & \mathrm{s}2\mathrm{S}2=(-\sqrt{b_S}-b_0)/4,\hfill \\ \hfill T_4:& \mathrm{stS}=b_{123}/4,\hfill & & \mathrm{s}2\mathrm{S}2=(\sqrt{b_S}-b_0)/4.\hfill \end{array}$$

After $T_i$ quantities are determined, we start checking all conditions on the right-hand side of the piecewise function:

$$\begin{array}{cc}\hfill s,S=& \left\{\begin{array}{cc}{F}_1{|}_{T_{1,2}},\hfill & {\mathrm{bSD}>0\wedge (-\mathrm{s}2\mathrm{S}2+\mathrm{sSsq})|}_{T_{1,2}}{>0\wedge \mathrm{stS}|}_{T_{1,2}}\ne 0,\hfill \\ F_2{|}_{T_{1,2}},\hfill & {\mathrm{bSD}>0\wedge (-\mathrm{s}2\mathrm{S}2+\mathrm{sSsq})|}_{T_{1,2}}{>0\wedge \mathrm{stS}|}_{T_{1,2}}\ne 0,\hfill \\ F_3{|}_{T_{1,2}},\hfill & {\mathrm{bSD}>0\wedge \mathrm{s}2\mathrm{S}2|}_{T_{1,2}}{>0\wedge \mathrm{stS}|}_{T_{1,2}}=0,\hfill \\ F_4{|}_{T_{1,2}},\hfill & {\mathrm{bSD}>0\wedge \mathrm{s}2\mathrm{S}2|}_{T_{1,2}}{>0\wedge \mathrm{stS}|}_{T_{1,2}}=0,\hfill \\ F_5{|}_{T_{1,2}},\hfill & {\mathrm{bSD}>0\wedge -\mathrm{s}2\mathrm{S}2|}_{T_{1,2}}{>0\wedge \mathrm{stS}|}_{T_{1,2}}=0,\hfill \\ F_6{|}_{T_{1,2}},\hfill & {\mathrm{bSD}>0\wedge -\mathrm{s}2\mathrm{S}2|}_{T_{1,2}}{>0\wedge \mathrm{stS}|}_{T_{1,2}}=0,\hfill \\ F_1{|}_{T_{3,4}},\hfill & {\mathrm{bSD}=0\wedge (-\mathrm{s}2\mathrm{S}2+\mathrm{sSsq})|}_{T_{3,4}}>0\wedge b_S{\ge 0\wedge \mathrm{stS}|}_{T_{3,4}}\ne 0,\hfill \\ F_2{|}_{T_{3,4}},\hfill & {\mathrm{bSD}=0\wedge (-\mathrm{s}2\mathrm{S}2+\mathrm{sSsq})|}_{T_{3,4}}>0\wedge b_S{\ge 0\wedge \mathrm{stS}|}_{T_{3,4}}\ne 0,\hfill \\ F_3{|}_{T_{3,4}},\hfill & {\mathrm{bSD}=0\wedge \mathrm{s}2\mathrm{S}2|}_{T_{3,4}}>0\wedge b_S{\ge 0\wedge \mathrm{stS}|}_{T_{3,4}}=0,\hfill \\ F_4{|}_{T_{3,4}},\hfill & {\mathrm{bSD}=0\wedge \mathrm{s}2\mathrm{S}2|}_{T_{3,4}}>0\wedge b_S{\ge 0\wedge \mathrm{stS}|}_{{\mathrm{T}}_{3,4}}=0,\hfill \\ F_5{|}_{T_{3,4}},\hfill & {\mathrm{bSD}=0\wedge -\mathrm{s}2\mathrm{S}2|}_{T_{3,4}}>0\wedge b_S{\ge 0\wedge \mathrm{stS}|}_{T_{3,4}}=0,\hfill \\ F_6{|}_{T_{3,4}},\hfill & {\mathrm{bSD}=0\wedge -\mathrm{s}2\mathrm{S}2|}_{T_{3,4}}>0\wedge b_S{\ge 0\wedge \mathrm{stS}|}_{T_{3,4}}=0,\hfill \\ F_7,\hfill & b_1=b_2=b_3=b_{12}=b_{13}=b_{23}=b_S-b_0^2+b_{123}^2=b_I+2b_0{b}_{123}=0\hfill \end{array}\right.\hfill \end{array}$$

(A5)

The notation ${|}_{T_{1,2}}$ means that we test substitutions $T_1$ and $T_2$ independently, i.e., first, substitute $T_1$ values into (A5) conditions and if they are satisfied, select $F_i$ expression for $s,S$ and substitute $T_1$ values into the formula. Then, repeat the same procedure with $T_2$ values (a single entry represents two separate entries in the piecewise function (A5)). By checking combinations of all 25 conditions encoded in (A5), we establish all valid values of $s,S$ coefficients. If $s\ne 0$ or $S\ne 0$ (cases $F_1-F_6$), then using (10) and (11), we compute $\mathbf{v}$ and $\mathbf{V}$. When $s=S=0$, we use (20) and solve any two coefficients, for example $v_1$ and $V_1$. The remaining four coefficients $v_2,v_3,V_2,V_3$ then become free (real) parameters, provided that the solved coefficients $v_1$ and $V_1$ can acquire real values. The determination of whether these real values exist is known to be a hard problem (especially for $\mathsf{B}$ with symbolic coefficients). It can be solved by using the famous quantifier elimination algorithm, the explanation of which is out of the scope of this article (our implementation uses Mathematica command Resolve[ ] for the task).

Lastly, substitute $s,S,\mathbf{v},\mathbf{V}$ into (A2) to obtain $\mathsf{A}$, which then represents all possible square roots of $\mathsf{B}$. In ${\mathit{Cl}}_{3,0}$ algebra, we can have 0, 2, or 4 isolated roots and 4D continuum of roots (in an exceptional case, continuum can turn into an isolated real root).

Appendix C.2. Computation Flow in ${\mathit{Cl}}_{0,3}$

${\mathit{Cl}}_{0,3}$ specific notations:

$$\begin{array}{cc}\hfill & \sqrt{D}=\sqrt{b_S^2-b_I^2},\hfill \\ \hfill & \mathrm{bSD}=b_S-\sqrt{D},\hfill \end{array}$$

Formulas for s and S are as follows:

$$\begin{array}{cccc}\hfill F1:& s=-\sqrt{\mathrm{s}2\mathrm{S}2-\mathrm{sSsq}},\hfill & & S=-(\mathrm{stS}/\sqrt{\mathrm{s}2\mathrm{S}2-\mathrm{sSsq}}),\hfill \\ \hfill F2:& s=\sqrt{\mathrm{s}2\mathrm{S}2-\mathrm{sSsq}},\hfill & & S=(\mathrm{stS}/\sqrt{\mathrm{s}2\mathrm{S}2-\mathrm{sSsq}}),\hfill \\ \hfill F3:& s=-\sqrt{\mathrm{s}2\mathrm{S}2+\mathrm{sSsq}},\hfill & & S=-(\mathrm{stS}/\sqrt{\mathrm{s}2\mathrm{S}2+\mathrm{sSsq}}),\hfill \\ \hfill F4:& s=\sqrt{\mathrm{s}2\mathrm{S}2+\mathrm{sSsq}},\hfill & & S=(\mathrm{stS}/\sqrt{\mathrm{s}2\mathrm{S}2+\mathrm{sSsq}}),\hfill \\ \hfill F5:& s=0,\hfill & & S=-\sqrt{2}\sqrt{\mathrm{s}2\mathrm{S}2},\hfill \\ \hfill F6:& s=0,\hfill & & S=\sqrt{2}\sqrt{\mathrm{s}2\mathrm{S}2},\hfill \\ \hfill F7:& s=-\sqrt{2}\sqrt{\mathrm{s}2\mathrm{S}2},\hfill & & S=0,\hfill \\ \hfill F8:& s=\sqrt{2}\sqrt{\mathrm{s}2\mathrm{S}2},\hfill & & S=0,\hfill \\ \hfill F9:& s=-{\displaystyle \frac{1}{2\sqrt{2}}}\sqrt{b_0+b_{123}-\sqrt{2b_S-{(b_0-b_{123})}^2}}\hfill & & S=s,\hfill \\ \hfill F10:& s={\displaystyle \frac{1}{2\sqrt{2}}}\sqrt{b_0+b_{123}-\sqrt{2b_S-{(b_0-b_{123})}^2}}\hfill & & S=s,\hfill \\ \hfill F11:& s=-{\displaystyle \frac{1}{2\sqrt{2}}}\sqrt{b_0+b_{123}+\sqrt{2b_S-{(b_0-b_{123})}^2}}\hfill & & S=s,\hfill \\ \hfill F12:& s={\displaystyle \frac{1}{2\sqrt{2}}}\sqrt{b_0+b_{123}+\sqrt{2b_S-{(b_0-b_{123})}^2}}\hfill & & S=s,\hfill \\ \hfill F13:& s=-{\displaystyle \frac{1}{2\sqrt{2}}}\sqrt{b_0-b_{123}-\sqrt{2b_S-{(b_0+b_{123})}^2}}\hfill & & S=-s,\hfill \\ \hfill F14:& s={\displaystyle \frac{1}{2\sqrt{2}}}\sqrt{b_0-b_{123}-\sqrt{2b_S-{(b_0+b_{123})}^2}}\hfill & & S=-s,\hfill \\ \hfill F15:& s=-{\displaystyle \frac{1}{2\sqrt{2}}}\sqrt{b_0-b_{123}+\sqrt{2b_S-{(b_0+b_{123})}^2}}\hfill & & S=-s,\hfill \\ \hfill F16:& s={\displaystyle \frac{1}{2\sqrt{2}}}\sqrt{b_0-b_{123}+\sqrt{2b_S-{(b_0+b_{123})}^2}}\hfill & & S=-s,\hfill \\ \hfill F17:& s=0,\hfill & & S=0,\hfill \end{array}$$

(A6)

where $\mathrm{sSsq}=\sqrt{\mathrm{s}2\mathrm{S}{2}^2-{\mathrm{stS}}^2}$. Substitutions that are used in condition checks (A7) and $s,S$ Formula (A6) are

$$\begin{array}{cccc}\hfill T_1:& \mathrm{stS}=(\sqrt{\mathrm{bSD}}/\sqrt{2}+b_{123})/4,\hfill & & \mathrm{s}2\mathrm{S}2=(\left(\sqrt{2}{b}_I\right)/\sqrt{\mathrm{bSD}}+2b_0)/8,\hfill \\ \hfill T_2:& \mathrm{stS}=(-\sqrt{\mathrm{bSD}}/\sqrt{2}+b_{123})/4,\hfill & & \mathrm{s}2\mathrm{S}2=((-\sqrt{2}{b}_I)/\sqrt{\mathrm{bSD}}+2b_0)/8,\hfill \\ \hfill T_3:& \mathrm{stS}=b_{123}/4,\hfill & & \mathrm{s}2\mathrm{S}2=(-\sqrt{b_S}+b_0)/4,\hfill \\ \hfill T_4:& \mathrm{stS}=b_{123}/4,\hfill & & \mathrm{s}2\mathrm{S}2=(\sqrt{b_S}+b_0)/4.\hfill \end{array}$$

Once the above quantities are computed, we start checking all conditions on the right-hand side of the piecewise function:

$$\begin{array}{cc}\hfill & s,S=\hfill \\ \hfill & \left\{\begin{array}{cc}{F}_{1,2}{|}_{T_{1,2}},\hfill & {\mathrm{bSD}>0\wedge (\mathrm{s}2\mathrm{S}2-\mathrm{sSsq})|}_{T_{1,2}}>0\wedge \mathrm{s}2\mathrm{S}2{{|}_{T_{1,2}} \ge 0\wedge \mathrm{stS}|}_{T_{1,2}}\ne 0\wedge s^2-S^2\ne 0,\hfill \\ F_{3,4}{|}_{T_{1,2}},\hfill & {\mathrm{bSD}>0\wedge (\mathrm{s}2\mathrm{S}2+\mathrm{sSsq})|}_{T_{1,2}}>0\wedge \mathrm{s}2\mathrm{S}2{{|}_{T_{1,2}} \ge 0\wedge \mathrm{stS}|}_{T_{1,2}}\ne 0\wedge s^2-S^2\ne 0,\hfill \\ F_{1,2}{|}_{T_{3,4}},\hfill & {\mathrm{bSD}=0\wedge (\mathrm{s}2\mathrm{S}2-\mathrm{sSsq})|}_{T_{3,4}}>0\wedge \mathrm{s}2\mathrm{S}2{{|}_{T_{3,4}} \ge 0\wedge \mathrm{stS}|}_{T_{3,4}}\ne 0\wedge s^2-S^2\ne 0,\hfill \\ F_{3,4}{|}_{T_{3,4}},\hfill & {\mathrm{bSD}=0\wedge (\mathrm{s}2\mathrm{S}2+\mathrm{sSsq})|}_{T_{3,4}}>0\wedge \mathrm{s}2\mathrm{S}2{{|}_{T_{3,4}} \ge 0\wedge \mathrm{stS}|}_{T_{3,4}}\ne 0\wedge s^2-S^2\ne 0,\hfill \\ F_{5,6}{|}_{T_{1,2}},\hfill & {\mathrm{bSD}>0\wedge (\mathrm{s}2\mathrm{S}2-\mathrm{sSsq})|}_{T_{1,2}}>0\wedge \mathrm{s}2\mathrm{S}2{{|}_{T_{1,2}} >0\wedge \mathrm{stS}|}_{T_{1,2}}=0\wedge s^2-S^2\ne 0,\hfill \\ F_{7,8}{|}_{T_{1,2}},\hfill & {\mathrm{bSD}>0\wedge (\mathrm{s}2\mathrm{S}2+\mathrm{sSsq})|}_{T_{1,2}}>0\wedge \mathrm{s}2\mathrm{S}2{{|}_{T_{1,2}} >0\wedge \mathrm{stS}|}_{T_{1,2}}=0\wedge s^2-S^2\ne 0,\hfill \\ F_{5,6,7,8}{|}_{T_{3,4}},\hfill & \mathrm{bSD}=0\wedge b_S>0\wedge \mathrm{s}2\mathrm{S}2{{|}_{T_{3,4}}>0\wedge \mathrm{stS}|}_{T_{3,4}}=0\wedge s^2-S^2\ne 0,\hfill \\ F_{9,10,11,12},\hfill & {\mathrm{cond}}_1\wedge 2b_S-{(b_0-b_{123})}^2>0\wedge b_0+b_{123}-\sqrt{2b_S-{(b_0-b_{123})}^2}>0,\hfill \\ F_{9,10,11,12},\hfill & {\mathrm{cond}}_1\wedge 2b_S-{(b_0-b_{123})}^2>0\wedge b_0+b_{123}+\sqrt{2b_S-{(b_0-b_{123})}^2}>0,\hfill \\ F_{13,14,15,16},\hfill & {\mathrm{cond}}_2\wedge 2b_S-{(b_0+b_{123})}^2>0\wedge b_0-b_{123}-\sqrt{2b_S-{(b_0+b_{123})}^2}>0,\hfill \\ F_{13,14,15,16},\hfill & {\mathrm{cond}}_2\wedge 2b_S-{(b_0+b_{123})}^2>0\wedge b_0-b_{123}+\sqrt{2b_S-{(b_0+b_{123})}^2}>0,\hfill \\ F_{17},\hfill & b_1=b_2=b_3=b_{12}=b_{13}=b_{23}=b_S-b_0^2-b_{123}^2=b_I-2b_0{b}_{123}=0,\hfill \end{array}\right.\hfill \end{array}$$

(A7)

where ${\mathrm{cond}}_1$ denotes triple logical conjunction $\left(\right(b_1+b_{23}=0(\wedge (b_2-b_{13}=0)\wedge (b_3+b_{12}=0))$ and ${\mathrm{cond}}_2$ logical conjunction $((b_1-b_{23}=0)\wedge (b_2+b_{13}=0)\wedge (b_3-b_{12}=0))$. The $F_{i,k,\dots }{|}_{T_{1,2}}$ is the abridgment of lines $F_i{|}_{T_{1,2}}$ and $F_k{|}_{T_{1,2}}$, etc. The notation ${|}_{T_{1,2}}$ means two independent lines with substitutions ${|}_{T_1}$ and ${|}_{T_2}$. Therefore, for example, $F_{5,6,7,8}{|}_{T_{3,4}}$ is equivalent to $4\times 2=8$ piecewise function (represented as cases) entries. Note that in (A7), the condition list includes the post-selection rule $(s^2-S^2)\ne 0$, which should be checked after values s and S were (successfully) computed.

By checking combinations of all 49 entries present in (A7), we establish all valid values of $s,S$ coefficients. If $s^2-S^2\ne 0$ (first 7 entries in (A7), i.e., formulas $F_1-F_8$), then use (45) for $\mathbf{v}$ and $\mathbf{V}$.

For $s,S$ given by formulas $F_9-F_{12}$, use (52) and the solutions

$$\begin{array}{cc}\hfill V_1& =(2b_1+\sqrt{2}\sqrt{-b_S+b_0^2+2b_1^2+b_{123}^2+16s(b_2{V}_2+b_3{V}_3)-8s^2(b_0-b_{123}+4(V_2^2+V_3^2))})/\left(8s\right),\hfill \\ \hfill V_1& =(2b_1-\sqrt{2}\sqrt{-b_S+b_0^2+2b_1^2+b_{123}^2+16s(b_2{V}_2+b_3{V}_3)-8s^2(b_0-b_{123}+4(V_2^2+V_3^2))})/\left(8s\right)\hfill \end{array}$$

to find $\mathbf{v},\mathbf{V}$ (coefficients $V_2$ and $V_3$ are free by our choice). The solution represents 2D parameter manifold which consist of two pieces.

In a similar way, for $s,S$ given by formulas $F_{13}-F_{16}$, use (58) and the solutions

$$\begin{array}{cc}\hfill V_1& =(-2b_1-\sqrt{2}\sqrt{-b_S+b_0^2+2b_1^2+b_{123}^2-16s(b_2{V}_2+b_3{V}_3)-8s^2(b_0+b_{123}+4(V_2^2+V_3^2))})/\left(8s\right),\hfill \\ \hfill V_1& =(-2b_1+\sqrt{2}\sqrt{-b_S+b_0^2+2b_1^2+b_{123}^2-16s(b_2{V}_2+b_3{V}_3)-8s^2(b_0+b_{123}+4(V_2^2+V_3^2))})/\left(8s\right)\hfill \end{array}$$

The solution represents 2D parameter manifold which consist of two pieces, where parameters $V_2$ and $V_3$ are made free by our choice.

For $F_{17}$, case $s=S=0$, we again have 4D parameter solution manifold, which is obtained by solving any two parameters from (61), for example, the solution (62).

Lastly, substitute $s,S,\mathbf{v},\mathbf{V}$ into (A2) to obtain $\mathsf{A}$, which then represents all possible square roots of $\mathsf{B}$. In ${\mathit{Cl}}_{0,3}$ algebra, we can have 0, 2, or 4 isolated roots and 2D or 4D continuum of roots and/or their mixture.

Since Clifford algebra ${\mathit{Cl}}_{2,1}$ is rarely used (for this algebra, a piecewise function has 81 entries) and the algebra ${\mathit{Cl}}_{1,2}$ is isomorphic to ${\mathit{Cl}}_{3,0}$, we refer a reader to our implementation [14], from which all summary formulas were extracted.