Worked Examples in the Split-Quaternion Algebra

Introduction

This article collects explicit computations in the split-quaternion algebra $\mathbb{H}_{\mathrm{s}}$, in the order in which the theory is built: the basis products; the three involutions and their eigenspaces; the idempotents and the two minimal left ideals; explicit pairs of zero divisors; and the unit criterion. Every number below is computed from the multiplication table of Split-Quaternion Algebra.

The article is a companion to the structural articles of the category. Its purpose is to put the abstract statements on concrete elements, so that the reader can carry each of them back to a computation. It introduces no new result; the statements it illustrates are those of Split-Quaternion Algebra, Split-Quaternion Idempotents and Projections, Split-Quaternion Zero Divisors, Split-Quaternion Norm and Invertibility and Split-Quaternion Rotations and the Lorentz Group.

Conventions. A split-quaternion is $\tilde q = q_0 e_0 + q_1 e_1 + q_2 e_2 + q_3 e_3$, with $q_0, q_1, q_2, q_3 \in \mathbb{R}$ and the products $e_1^2 = -1$, $e_2^2 = +1$, $e_3 = e_1 e_2$, $e_1 e_2 = -e_2 e_1$. Its conjugation is $\tilde{q}^{\natural} = q_0 e_0 - q_1 e_1 - q_2 e_2 - q_3 e_3$ and its central product is $N(\tilde q) = \tilde q\tilde{q}^{\natural} = q_0^2 + q_1^2 - q_2^2 - q_3^2$, formed and evaluated algebraically and read metrically in Split-Quaternion Norm and Invertibility. The elements $\tilde\pi_\pm = \tfrac{1}{2}(1 \pm e_2)$ are the standard idempotents.

The Basis Products

The multiplication table, whose entry in row $i$ and column $j$ is $e_i e_j$, is

	$1$	$e_1$	$e_2$	$e_3$
$1$	$1$	$e_1$	$e_2$	$e_3$
$e_1$	$e_1$	$-1$	$e_3$	$-e_2$
$e_2$	$e_2$	$-e_3$	$+1$	$-e_1$
$e_3$	$e_3$	$e_2$	$e_1$	$+1$

Example (a product worked in full). The entry $e_3^2 = +1$ follows from $e_3 = e_1 e_2$ and anticommutativity:

$$ e_3^2 = e_1 e_2 e_1 e_2 = -e_1^2 e_2^2 = -(-1)(+1) = +1 . $$

The product of two general elements, worked from the table, is

$$ (q_0 e_0 + q_1 e_1 + q_2 e_2 + q_3 e_3)(q_0' e_0 + q_1' e_1 + q_2' e_2 + q_3' e_3) = (q_0 q_0' - q_1 q_1' + q_2 q_2' + q_3 q_3') + (q_0 q_1' + q_0' q_1 - q_2 q_3' + q_2' q_3)e_1 + (q_0 q_2' + q_0' q_2 + q_1 q_3' - q_1' q_3)e_2 + (q_0 q_3' + q_0' q_3 + q_1 q_2' - q_1' q_2)e_3 . $$

The product is associative and distributive, so the multiplication table of the four basis elements determines it completely; the table is collected in the summary. The products of the vector basis elements are

$$ e_1^2 = -1, \quad e_2^2 = 1, \quad e_3^2 = 1, \qquad e_1e_2 = e_3, \quad e_2e_1 = -e_3, \qquad e_1e_3 = -e_2, \quad e_3e_1 = e_2, \qquad e_2e_3 = -e_1, \quad e_3e_2 = e_1 . $$

Example. For $\tilde q = 2 + e_1$ and $\tilde p = 1 + e_2 + e_3$, the table gives

$$ (2 + e_1)(1 + e_2 + e_3) = 2 + 2e_2 + 2e_3 + e_1 + e_1e_2 + e_1e_3 = 2 + e_1 + 2e_2 + 2e_3 - e_2 + e_3 = 2 + e_1 + e_2 + 3e_3 , $$

so the product is $\tilde q \tilde p = 2 + e_1 + e_2 + 3e_3$; the coefficient of $e_1$ also follows from the general product formula displayed above.

The Three Involutions and Their Eigenspaces

Three involutions act on the algebra. The conjugation ${}^{\natural}$ negates the vector part, the principal involution $\alpha$ is the automorphism

$$ \alpha(e_1) = -e_1, \qquad \alpha(e_2) = -e_2, \qquad \alpha(e_3) = e_3, $$

and the reversal $\rho$ is the anti-automorphism

$$ \rho(e_1) = e_1, \qquad \rho(e_2) = e_2, \qquad \rho(e_3) = -e_3 . $$

On a general element they read

$$ \tilde{q}^{\natural} = q_0 e_0 - q_1 e_1 - q_2 e_2 - q_3 e_3, \qquad \alpha(\tilde q) = q_0 - q_1 e_1 - q_2 e_2 + q_3 e_3, \qquad \rho(\tilde q) = q_0 + q_1 e_1 + q_2 e_2 - q_3 e_3, $$

and the three are related by $\tilde{q}^{\natural} = \alpha(\rho(\tilde q)) = \rho(\alpha(\tilde q))$.

Example (eigenspaces). Each involution is diagonalisable with eigenvalues $\pm 1$.

involution	$+1$ eigenspace	$-1$ eigenspace	dimensions
${}^{\natural}$	$S = \mathbb{R}\cdot 1$	$V = \operatorname{span}\{e_1,e_2,e_3\}$	$1, 3$
$\alpha$	$\operatorname{span}\{1, e_3\}$	$\operatorname{span}\{e_1, e_2\}$	$2, 2$
$\rho$	$\operatorname{span}\{1, e_1, e_2\}$	$\mathbb{R} e_3$	$3, 1$

Example (the common eigenspaces). Since $\alpha$ and $\rho$ commute, they are simultaneously diagonalisable, and the common eigenspaces of the pair $(\alpha, \rho)$ are

$$ \mathbb{H}_{\mathrm{s}} = \mathbb{R}\cdot 1 \oplus \operatorname{span}\{e_1, e_2\} \oplus \mathbb{R} e_3, $$

on which the sign pair $(\alpha, \rho)$ is $(+,+)$, $(-,+)$ and $(+,-)$ respectively. The plane $\operatorname{span}\{e_1,e_2\}$ is not split further: on it the conjugation also acts as $-1$, so no eigenspace of any of the three involutions cuts it. This is the finest decomposition produced by the involutions, and it is recorded in Split-Quaternion Subspaces and the Involutions.

The Idempotents and the Two Minimal Left Ideals

Example (the standard idempotents). With $\tilde\pi_+ = \tfrac{1}{2}(1 + e_2)$ and $\tilde\pi_- = \tfrac{1}{2}(1 - e_2)$,

$$ \tilde\pi_+^2 = \tfrac{1}{4}(1 + 2e_2 + e_2^2) = \tfrac{1}{4}(2 + 2e_2) = \tilde\pi_+, $$

and $\tilde\pi_-^2 = \tilde\pi_-$ likewise. Moreover

$$ \tilde\pi_+ \tilde\pi_- = \tfrac{1}{4}(1 - e_2^2) = 0, \qquad \tilde\pi_+ + \tilde\pi_- = 1, \qquad N(\tilde\pi_+) = N(\tilde\pi_-) = 0 . $$

They are not central, since $e_1 \tilde\pi_+ = \tfrac{1}{2}(e_1 + e_3)$ while $\tilde\pi_+ e_1 = \tfrac{1}{2}(e_1 - e_3)$.

Example (the two minimal left ideals). The reductions $e_2 \tilde\pi_+ = \tilde\pi_+$ and $e_3 \tilde\pi_+ = \tfrac{1}{2}(e_3 + e_1) = e_1 \tilde\pi_+$ give

$$ \mathbb{H}_{\mathrm{s}} \tilde\pi_+ = \operatorname{span}\{\tilde\pi_+, e_1 \tilde\pi_+\} = \operatorname{span}\big\{\tfrac{1}{2}(1 + e_2),\, \tfrac{1}{2}(e_1 + e_3)\big\}, $$

$$ \mathbb{H}_{\mathrm{s}} \tilde\pi_- = \operatorname{span}\{\tilde\pi_-, e_1 \tilde\pi_-\} = \operatorname{span}\big\{\tfrac{1}{2}(1 - e_2),\, \tfrac{1}{2}(e_1 - e_3)\big\}, $$

each of real dimension $2$, with $\mathbb{H}_{\mathrm{s}} = \mathbb{H}_{\mathrm{s}} \tilde\pi_+ \oplus \mathbb{H}_{\mathrm{s}} \tilde\pi_-$. The four spanning elements are the vectors $\tilde\pi_+$, $\tilde p = \tfrac{1}{2}(e_1+e_3)$, $\tilde\pi_-$, $-\tilde q = \tfrac{1}{2}(e_1-e_3)$ of the matrix-unit basis $\{\tilde\pi_+, \tilde q, \tilde p, \tilde\pi_-\}$ of Split-Quaternion Ideals and Peirce Decomposition.

Example (the split-complex subalgebra). The idempotents lie in the subalgebra $\operatorname{span}\{1, e_2\} \cong \mathbb{D}$, and in that commutative algebra they are central; in $\mathbb{H}_{\mathrm{s}}$ they are not, and the decomposition they give is a decomposition of modules and not of algebras, because $\tilde\pi_+ e_3 \tilde\pi_- = \tfrac{1}{2}(e_3 - e_1) \neq 0$.

Explicit Zero-Divisor Pairs

Example (two split-complex pairs). Let $\lambda, \mu$ be real. Then

$$ (1 + e_2)(1 - e_2) = 1 - e_2^2 = 0, \qquad (1 + e_3)(1 - e_3) = 1 - e_3^2 = 0, $$

so $(1+e_2)$ with $(1-e_2)$, and $(1+e_3)$ with $(1-e_3)$, are explicit pairs of nonzero zero divisors.

Example (a nilpotent pair). The element $e_1 + e_3$ is nonzero and

$$ (e_1 + e_3)^2 = e_1^2 + e_1 e_3 + e_3 e_1 + e_3^2 = -1 + (-e_2) + e_2 + 1 = 0, $$

so $e_1 + e_3$ is a nonzero nilpotent and the pair $(e_1+e_3, e_1+e_3)$ is a zero-divisor pair.

Every nilpotent lies in the vector subspace $V$ and on the level set $N = 0$, $q_1^2 = q_2^2 + q_3^2$; the element $1 + e_2$ shows that the zero divisor set is strictly larger than the nilpotent set, since $(1+e_2)^2 = 2(1+e_2) \neq 0$.

Example (a mixed pair). The idempotent $\tilde\pi_-$ annihilates $1 + e_2$:

$$ (1 + e_2) \tilde\pi_- = 2 \tilde\pi_+ \tilde\pi_- = 0, $$

a pair in which neither factor is a nilpotent.

The Unit Criterion Worked

Criterion (from Split-Quaternion Norm and Invertibility). A nonzero $\tilde q$ is invertible if and only if $N(\tilde q) \neq 0$, and then $\tilde q^{-1} = \tilde{q}^{\natural}/N(\tilde q)$. It is recalled here only to be applied.

Example (a unit of positive norm). For $\tilde q = 2 + e_1$, the split-quaternion norm is $N(\tilde q) = 4 + 1 = 5$, and

$$ \tilde q^{-1} = \frac{\tilde{q}^{\natural}}{N(\tilde q)} = \frac{2 - e_1}{5}, \qquad (2 + e_1)\frac{2 - e_1}{5} = \frac{4 - e_1^2}{5} = \frac{5}{5} = 1 . $$

Example (a unit of norm one). For $\tilde p = 1 + e_1 + e_2$, the split-quaternion norm is $N(\tilde p) = 1 + 1 - 1 = 1$, so $\tilde p^{-1} = \tilde p^{\natural} = 1 - e_1 - e_2$; indeed $\tilde p \tilde p^{\natural} = (1+e_1+e_2)(1-e_1-e_2) = 1$. Since $N(\tilde p) = 1$, the element is a unit split-quaternion.

Example (a unit of negative norm). For $\tilde r = e_2$, the split-quaternion norm is $N(\tilde r) = -1$, and $\tilde r^{-1} = \bar{\tilde r}/N(\tilde r) = (-e_2)/(-1) = e_2$, consistent with $e_2^2 = 1$.

Example (a non-unit). For $w = 1 + e_3$, the split-quaternion norm is $N(w) = 1 - 1 = 0$, so $w$ is not invertible; and indeed $(1+e_3)(1-e_3) = 0$.

The four examples realise the three-way classification of Split-Quaternion Norm and Invertibility by the value of $N$: a positive value ($N = 5$ and $N = 1$), a negative value ($N = -1$), and the value zero ($N = 0$, the zero divisors).

The Action of the Unit Group

The action of a unit on the vector subspace by conjugation, the elliptic and hyperbolic one-parameter subgroups it produces, and the classification of the vectors of $V$ by their orbits, use the multiplication table above and nothing else; they are computations of the Geometry group and are carried out in Split-Quaternion Rotations and the Lorentz Group. They are not repeated here, and the multiplication table of §The Basis Products is what they consume.

Summary

The basis products are collected in one table, and the multiplication table determines the algebra. The three involutions — conjugation, the principal involution $\alpha$ and the reversal $\rho$ — are diagonalisable; their individual eigenspaces and their common refinement $\mathbb{R}\cdot 1 \oplus \operatorname{span}\{e_1,e_2\} \oplus \mathbb{R} e_3$ are listed, and the plane $\operatorname{span}\{e_1,e_2\}$ does not split further.

The idempotents $\tilde\pi_\pm$ are verified to be orthogonal, complete and of zero norm, and they give the two minimal left ideals $\mathbb{H}_{\mathrm{s}} \tilde\pi_\pm = \operatorname{span}\{\tilde\pi_\pm, e_1 \tilde\pi_\pm\}$. Explicit zero-divisor pairs are $(1+e_2)(1-e_2) = 0$, $(1+e_3)(1-e_3) = 0$, the nilpotent $(e_1+e_3)^2 = 0$, and the mixed pair $(1+e_2)\tilde\pi_- = 0$. The unit criterion is carried through on four elements: $2+e_1$ (positive norm), $1+e_1+e_2$ (norm one), $e_2$ (negative norm), and the non-unit $1+e_3$ (zero norm). The action of the unit group on the vector subspace is not worked here; it is the subject of Split-Quaternion Rotations and the Lorentz Group.

Summary of Notation

Symbol	Meaning	Article
$\mathbb{H}_{\mathrm{s}}$	the split-quaternion algebra	Split-Quaternion Algebra
$\tilde q = q_0 e_0 + q_1 e_1 + q_2 e_2 + q_3 e_3$	a general split-quaternion	Split-Quaternion Algebra
${}^{\natural}$, $\alpha$, $\rho$	conjugation, principal involution, reversal	Split-Quaternion Algebra
$\tilde\pi_\pm = \tfrac{1}{2}(1\pm e_2)$	the standard idempotents	Split-Quaternion Idempotents and Projections
$\mathbb{H}_{\mathrm{s}} \tilde\pi_\pm$	the two minimal left ideals	Split-Quaternion Idempotents and Projections
$N(\tilde q) = \tilde q\tilde{q}^{\natural} = q_0^2+q_1^2-q_2^2-q_3^2$	the central product, formed and evaluated algebraically	Split-Quaternion Algebra
$\tilde q^{-1} = \tilde{q}^{\natural}/N(\tilde q)$	the inverse of a unit	Split-Quaternion Norm and Invertibility