Square Roots of a General Element in the Quaternion Algebra

Introduction

This article solves the equation $\tilde q_2^2=\tilde q_1$ in $\mathbb{H}$ for an arbitrary $\tilde q_1$. It is the general case of which Square Roots of One and the Idempotents of the Quaternion Algebra and Square Roots of Zero and the Nilpotents of the Quaternion Algebra, above this article in this category, treat the three central values; the method is the split of a quaternion into its scalar and its vector part and the resulting pair of a quadratic and a linear equation, and it uses neither Clifford algebras nor forms. The result is a classification by the vanishing of the vector part of $\tilde q_1$: away from the scalar line there are exactly two roots, and on the scalar line the count is two roots, one root or a sphere of roots according as the scalar is positive, zero or negative. The quaternion norm and the invertibility criterion are Norm and Invertibility in the Quaternion Algebra, above this article in this category, and they are neither used nor needed.

The Reduction

Throughout, $\tilde q_1=q_{1t}e_t+\vec q_1$ and $\tilde q_2=q_{2t}e_t+\vec q_2$ with $q_{1t},q_{2t}\in\mathbb{R}$ and $\vec q_1,\vec q_2$ in the vector subspace $V=\mathbb{R}\vec e_x+\mathbb{R}\vec e_y+\mathbb{R}\vec e_z$, and $|\vec q_2|^2=q_{2x}^2+q_{2y}^2+q_{2z}^2$ is the sum of the squares of the three coordinates, a non-negative real number.

Proposition (the square). For $\tilde q_2=q_{2t}e_t+\vec q_2$,

$$ \tilde q_2^2=(q_{2t}^2-|\vec q_2|^2)e_t+2q_{2t}\vec q_2. $$

Proof. Expand: $q_{2t}^2+2q_{2t}\vec q_2+\vec q_2^2$. The square of a vector is the scalar $\vec q_2^2=-|\vec q_2|^2$, because the vector coordinates anticommute in pairs and $\vec e_\mu^2=-e_t$; the cross terms cancel in pairs.

Corollary. The equation $\tilde q_2^2=\tilde q_1$ is the pair

$$ 2q_{2t}\vec q_2=\vec q_1,\qquad q_{2t}^2-|\vec q_2|^2=q_{1t}. $$

The second is the reduced quadratic in the unknown $t=q_{2t}^2$; when $\vec q_1\neq0$ the first determines $\vec q_2=\vec q_1/(2q_{2t})$ and the second becomes the quadratic

$$ \boxed{\; t^2-q_{1t}t-\tfrac14|\vec q_1|^2=0,\qquad \Delta=q_{1t}^2+|\vec q_1|^2 \;} ,$$

with discriminant $\Delta\geq0$ and $\sqrt\Delta=|\tilde q_1|$.

The Two Cases

Theorem (non-scalar case). Let $\vec q_1\neq0$. Then $\tilde q_1$ has exactly two square roots, and they are

$$ \boxed{\; \tilde q_{2\pm}=\pm\Bigl(\sqrt{\tfrac12\bigl(q_{1t}+|\tilde q_1|\bigr)}\;e_t+\frac{\vec q_1}{2\sqrt{\tfrac12(q_{1t}+|\tilde q_1|)}}\Bigr) \;} .$$

Proof. The vector equation $2q_{2t}\vec q_2=\vec q_1$ forces $q_{2t}\neq0$ and $\vec q_2=\vec q_1/(2q_{2t})$. Substituting in the reduced quadratic gives $4q_{2t}^4-4q_{1t}q_{2t}^2-|\vec q_1|^2=0$, that is $t^2-q_{1t}t-\tfrac14|\vec q_1|^2=0$ with roots $\tfrac12(q_{1t}\pm\sqrt\Delta)$. The minus root is $\tfrac12(q_{1t}-|\tilde q_1|)$, negative because $\vec q_1\neq0$ gives $|\tilde q_1|>|q_{1t}|$; it yields no real $q_{2t}$. The plus root is $\tfrac12(q_{1t}+|\tilde q_1|)>0$, and it yields the two values $q_{2t}=\pm\sqrt{\tfrac12(q_{1t}+|\tilde q_1|)}$ and the two displayed roots. Since $q_{2t}\neq0$ and $\vec q_2$ is determined by $q_{2t}$, there are no others.

Theorem (scalar case). Let $\tilde q_1=q_{1t}e_t$, so $\vec q_1=0$. Then

$$ \boxed{\; \tilde q_2^2=q_{1t}e_t\iff \begin{cases} \tilde q_2=\pm\sqrt{q_{1t}}\,e_t, & q_{1t}>0,\\ \tilde q_2=0, & q_{1t}=0,\\ \tilde q_2=\vec q_2\in V,\quad |\vec q_2|^2=-q_{1t}, & q_{1t}<0. \end{cases} \;} $$

Proof. With $\vec q_1=0$ the vector equation reads $q_{2t}\vec q_2=0$, so $q_{2t}=0$ or $\vec q_2=0$. If $\vec q_2=0$ the scalar equation is $q_{2t}^2=q_{1t}$, solvable only for $q_{1t}\geq0$, giving $\pm\sqrt{q_{1t}}e_t$ for $q_{1t}>0$ and $0$ for $q_{1t}=0$. If $q_{2t}=0$ the scalar equation is $-|\vec q_2|^2=q_{1t}$, solvable only for $q_{1t}\leq0$, giving $|\vec q_2|^2=-q_{1t}$; for $q_{1t}=0$ this is again $\tilde q_2=0$, and for $q_{1t}<0$ it is the level set of a positive-definite quadratic form on the three-dimensional space $V$, a sphere of roots.

Theorem (classification). Every $\tilde q_1\in\mathbb{H}$ has at least one square root, and the root set is

Case	Root set	Cardinality
$\vec q_1\neq0$	$\{\tilde q_{2+},\tilde q_{2-}\}$	two
$\vec q_1=0$, $q_{1t}>0$	$\{\pm\sqrt{q_{1t}}e_t\}$	two
$\vec q_1=0$, $q_{1t}=0$	$\{0\}$	one
$\vec q_1=0$, $q_{1t}<0$	$\{\vec q_2\in V : \|\vec q_2\|^2=-q_{1t}\}$	a sphere

Proof. The two preceding theorems.

The minus root of the reduced quadratic is what the positive-definiteness of the sum of squares discards.

The Specialisation to the Central Values

The three central equations of Square Roots of One and the Idempotents of the Quaternion Algebra and Square Roots of Zero and the Nilpotents of the Quaternion Algebra, above this article in this category, are the cases $\vec q_1=0$, $q_{1t}=-1,0,1$ of the classification:

$q_{1t}=-1$: the sphere $\{\vec q_2\in V : |\vec q_2|^2=1\}$, the unit sphere of the vector subspace, matching the roots of $-1$;
$q_{1t}=0$: the single root $0$, matching the root set of $0$;
$q_{1t}=1$: the pair $\{\pm e_t\}$, matching the roots of $+1$.

The count of the roots of $-1$ is a two-dimensional sphere, that of $+1$ is two points, and that of $0$ is one point; these are the three values of the classification table, and the agreement is a consistency check on the two articles rather than a new computation.

Worked Examples

Each root below is checked by squaring.

$\tilde q_1=4e_t$: the roots are $\pm2e_t$, and $(-2e_t)^2=4e_t$.
$\tilde q_1=0$: the only root is $0$.
$\tilde q_1=-4e_t$: the roots are the pure quaternions $\vec q_2$ with $q_{2x}^2+q_{2y}^2+q_{2z}^2=4$, for instance $\tilde q_2=2\vec e_y$, and $(2\vec e_y)^2=-4e_t$.
$\tilde q_1=\vec e_x$, a pure unit quaternion: $|\tilde q_1|=1$ and $\tfrac12(q_{1t}+|\tilde q_1|)=\tfrac12$, so $\tilde q_{2\pm}=\pm\bigl(\tfrac{1}{\sqrt2}e_t+\tfrac{1}{\sqrt2}\vec e_x\bigr)$; squaring gives $\tfrac12-\tfrac12+2\cdot\tfrac12 \vec e_x=\vec e_x$.
$\tilde q_1=3e_t+2\vec e_x+2\vec e_y$: $|\tilde q_1|=\sqrt{9+4+4}=\sqrt{17}$, so $q_{2t}^2=\tfrac12(3+\sqrt{17})\approx3.56$, so $\tilde q_2\approx1.89\,e_t+0.53\,\vec e_x+0.53\,\vec e_y$; squaring returns $\tilde q_1$ to the precision of the arithmetic.
$\tilde q_1=-3e_t+4\vec e_z$: $|\tilde q_1|=5$, $q_{2t}^2=\tfrac12(-3+5)=1$, and the roots are $\pm(e_t+2\vec e_z)$, since $(e_t+2\vec e_z)^2=e_t+4\vec e_z-4e_t=-3e_t+4\vec e_z$.

Summary

Writing $\tilde q_1=q_{1t}e_t+\vec q_1$ and $\tilde q_2=q_{2t}e_t+\vec q_2$, the square is $\tilde q_2^2=(q_{2t}^2-|\vec q_2|^2)e_t+2q_{2t}\vec q_2$, so the equation $\tilde q_2^2=\tilde q_1$ is the vector equation $2q_{2t}\vec q_2=\vec q_1$ and the reduced quadratic $t^2-q_{1t}t-\tfrac14|\vec q_1|^2=0$ in $t=q_{2t}^2$, whose discriminant is $\Delta=q_{1t}^2+|\vec q_1|^2\geq0$. If the vector part of $\tilde q_1$ does not vanish there are exactly two roots, $\pm\tilde q_2$ with $q_{2t}=\sqrt{\tfrac12(q_{1t}+|\tilde q_1|)}$ and $\vec q_2=\vec q_1/(2q_{2t})$; if it vanishes the roots are $\pm\sqrt{q_{1t}}e_t$ for $q_{1t}>0$, the single root $0$ for $q_{1t}=0$, and the sphere $\{|\vec q_2|^2=-q_{1t}\}$ for $q_{1t}<0$. Every quaternion therefore has at least one square root over $\mathbb{R}$, and the three central equations $q_{1t}=-1,0,1$ reproduce the roots of $-1$, of $0$ and of $+1$.

Summary of Notation

Symbol	Meaning
$\tilde q_1=q_{1t}e_t+\vec q_1$	Quaternion split into scalar part $q_{1t}$ and vector part $\vec q_1$
$\tilde q_2=q_{2t}e_t+\vec q_2$	The unknown root, similarly split
$V=\mathbb{R}\vec e_x+\mathbb{R}\vec e_y+\mathbb{R}\vec e_z$	The vector subspace
$\|\vec q_2\|^2=q_{2x}^2+q_{2y}^2+q_{2z}^2$	Sum of the squares of the vector coordinates
$t=q_{2t}^2$	The unknown of the reduced quadratic
$\Delta=q_{1t}^2+\|\vec q_1\|^2$	Discriminant of the reduced quadratic, $=\|\tilde q_1\|^2$