Split-Quaternion Algebra

Introduction

This article introduces the split-quaternion algebra as an algebraic structure. It defines the algebra, describes the conjugations and their fixed-point subspaces, the idempotents and the distinguished subspaces, and the Lie algebra structure. It closes with a comparison with the quaternions $\mathbb{H}$ and with the split-biquaternions $\mathbb{H}_{\mathbb{D}}$.

The treatment is mathematically honest: every claim is either proved or stated as a definition. No physics is invoked. The inner product of the algebra is formed and evaluated here as an algebraic pairing — its scalar, its vanishing and its non-degeneracy; nothing is measured with it, and the theory of norms, forms and signatures belongs to the Topology group (Split-Quaternion Norm and Invertibility). The quaternion algebra is assumed from Quaternion Algebra; the split-complex algebra and its idempotents from Split-Complex Algebra; the Clifford algebra $\mathrm{Cl}_{p,q}$ and its low-dimensional classification from Clifford Algebras in Finite Dimensions and The Number Systems as Clifford Algebras. The rotations, the Lorentz group and the hyperbolic geometry that the algebra carries are not treated here; they belong to Split-Quaternion Rotations and the Lorentz Group, Split-Quaternion Geometry and Split-Quaternions and Hyperbolic Geometry.

The Split-Quaternion Algebra

Definition

The split-quaternion algebra $\mathbb{H}_{\mathrm{s}}$ is the four-dimensional real algebra with basis

$$ 1, \qquad e_1, \qquad e_2, \qquad e_3, $$

and multiplication rules

$$ e_1^2 = -1, \qquad e_2^2 = +1, \qquad e_3 = e_1 e_2, \qquad e_1 e_2 = -e_2 e_1. $$

A general split-quaternion is written in developed form as

$$ \tilde q = q_0 e_0 + q_1 e_1 + q_2 e_2 + q_3 e_3, \qquad q_\mu \in \mathbb{R}, $$

or, more compactly, as

$$ \tilde q = \sum_{\mu=0}^{3} q_\mu e_\mu, \qquad e_0 = 1. $$

The real number $q_0$ is the scalar part, and the triple $(q_1, q_2, q_3)$ is the vector part. We also write

$$ \tilde q = q_0 + \mathbf{v}, \qquad \mathbf{v} = q_1 e_1 + q_2 e_2 + q_3 e_3, $$

and we write $\operatorname{Sc}(\tilde q) = q_0$ and $\operatorname{Vec}(\tilde q) = \mathbf{v}$ for the two components.

The four elements $1, e_1, e_2, e_3$ are linearly independent over $\mathbb{R}$ by definition, so $\dim_{\mathbb{R}} \mathbb{H}_{\mathrm{s}} = 4$. The unit is $1$, and every element is a real linear combination of the basis elements.

The Multiplication Table

The rules of the definition determine the whole multiplication table. From $e_3 = e_1 e_2$ and $e_1 e_2 = -e_2 e_1$ one computes

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

$$ e_2 e_3 = e_2 e_1 e_2 = -e_1 e_2^2 = -e_1, \qquad e_3 e_2 = e_1 e_2 e_2 = +e_1, $$

$$ e_3 e_1 = e_1 e_2 e_1 = -e_1^2 e_2 = +e_2, \qquad e_1 e_3 = e_1 e_1 e_2 = -e_2. $$

The products are collected in the following table, whose entry in row $i$ and column $j$ is $e_i e_j$.

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

The table is read off from the definition; the entry $e_1 e_3 = -e_2$, for instance, is the identity $e_1 e_3 = e_1 e_1 e_2 = -e_2$.

Basic Properties

Associative. Split-quaternion multiplication is associative. It suffices to check the associativity of the generating products, since the product is defined by bilinear extension of the table; the four elements $1, e_1, e_2, e_3$ satisfy the relations of the Clifford algebra $\mathrm{Cl}_{1,1}$, and every Clifford algebra is associative.

Non-commutative. Split-quaternion multiplication is not commutative: $e_1 e_2 = e_3$ while $e_2 e_1 = -e_3$. The centre is computed in The Centre and Simplicity below, and it is one-dimensional.

Not a division algebra. The element $1 + e_2$ is nonzero and

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

so $1 + e_2$ and $1 - e_2$ are nonzero zero divisors. The algebra therefore has zero divisors and is not a division algebra. Frobenius' theorem, recalled in Normed Division Algebras and the Hurwitz Theorem, states that the only finite-dimensional associative real division algebras are $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$; the split-quaternion algebra is a fourth four-dimensional associative real algebra, and the failure of the division property is exactly the appearance of the zero divisors above. The zero divisor set is the subject of Split-Quaternion Zero Divisors.

The matrix model. The algebra is isomorphic to the algebra of $2\times2$ real matrices: the low-dimensional Clifford classification gives

$$ \mathbb{H}_{\mathrm{s}} \cong \mathrm{Cl}_{1,1} \cong M_2(\mathbb{R}) $$

(The Number Systems as Clifford Algebras), and this abstract identification is used throughout the category, in the ideal theory of Split-Quaternion Ideals and Peirce Decomposition, in the idempotent theory of Split-Quaternion Idempotents and Projections, and in the spectral and exponential theory of the later articles.

Conjugations and Fixed-Point Subspaces

The Conjugation

Definition. The split-quaternion conjugation is the map

$$ \tilde{q}^{\natural} = q_0 - q_1 e_1 - q_2 e_2 - q_3 e_3 . $$

It negates the three vector basis elements and fixes the scalars.

Proposition. The conjugation is an involutive algebra anti-automorphism: it is $\mathbb{R}$-linear, it satisfies $(\tilde q \tilde p)^{\natural} = \tilde{p}^{\natural}\, \tilde{q}^{\natural}$ and $(\tilde{q}^{\natural})^{\natural} = \tilde q$, and its fixed-point set is the scalar line $\mathbb{R}$.

Proof. Linearity and the second identity are immediate from the definition. For the anti-automorphism property it suffices to check the generators: $(e_1 e_2)^{\natural} = e_3^{\natural} = -e_3$, while $e_2^{\natural} e_1^{\natural} = (-e_2)(-e_1) = e_2 e_1 = -e_3$, and the other products are similar. The fixed points satisfy $q_1 e_1 + q_2 e_2 + q_3 e_3 = -(q_1 e_1 + q_2 e_2 + q_3 e_3)$, hence $q_1 = q_2 = q_3 = 0$.

The Two Eigenspaces

Since ${}^{\natural}$ is an involutive linear map, the algebra decomposes into its $+1$ and $-1$ eigenspaces:

$$ \mathbb{H}_{\mathrm{s}} = S \oplus V, \qquad S = \{\tilde q : \tilde{q}^{\natural} = \tilde q\} = \mathbb{R} \cdot 1, \qquad V = \{\tilde q : \tilde{q}^{\natural} = -\tilde q\} = \operatorname{span}\{e_1, e_2, e_3\}. $$

The scalar subspace $S$ is one-dimensional and is a subalgebra isomorphic to $\mathbb{R}$; it is the centre, as The Centre and Simplicity below shows. The vector subspace $V$ is three-dimensional, it is not a subalgebra, and it is the natural home of the geometry of the system. Every element decomposes uniquely as

$$ \tilde q = \tfrac{1}{2}(\tilde q + \tilde{q}^{\natural}) + \tfrac{1}{2}(\tilde q - \tilde{q}^{\natural}), $$

the first summand lying in $S$ and the second in $V$.

The Other Two Involutions

Two further involutions act on the algebra, and they are recorded here because the subspaces of Split-Quaternion Subspaces and the Involutions are cut out by them. The principal involution $\alpha$ is the algebra automorphism defined on the generators by

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

and the reversal $\rho$ is the algebra anti-automorphism defined by

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

Both are involutions, they commute, and their composite is the conjugation: $\tilde{q}^{\natural} = \alpha(\rho(\tilde q)) = \rho(\alpha(\tilde q))$. Because they commute, each preserves the eigenspaces of the other, and the algebra decomposes into their common eigenspaces. The two involutions do not separate $e_1$ from $e_2$, however: $\alpha$ acts as $-1$ on the whole plane $\operatorname{span}\{e_1,e_2\}$, and $\rho$ acts as $+1$ on that plane, so the common eigenspaces are

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

of dimensions $1, 2, 1$, the sign pair $(\alpha,\rho)$ being $(+,+)$ on $\mathbb{R}\cdot 1$, $(-,+)$ on $\operatorname{span}\{e_1,e_2\}$ and $(+,-)$ on $\mathbb{R}e_3$. This is the finest decomposition produced by the three involutions: on $\operatorname{span}\{e_1,e_2\}$ all three act as $-1$, so no eigenspace of any of them splits that plane. The eigenspaces of the involutions taken one at a time are the ones used in Split-Quaternion Analysis: the principal involution has $+1$ on $\operatorname{span}\{1,e_3\}$ and $-1$ on $\operatorname{span}\{e_1,e_2\}$; the reversal has $+1$ on $\operatorname{span}\{1,e_1,e_2\}$ and $-1$ on $\mathbb{R}e_3$; and the conjugation has $+1$ on $\mathbb{R}\cdot 1$ and $-1$ on $V$.

The Inner Product

The inner product of two split-quaternions is the real scalar

$$ B(\tilde q, \tilde p) = \operatorname{Sc}(\tilde q\, \tilde{p}^{\natural}) = q_0 p_0 + q_1 p_1 - q_2 p_2 - q_3 p_3, $$

for $\tilde q = q_0 e_0 + q_1 e_1 + q_2 e_2 + q_3 e_3$ and $\tilde p = p_0 e_0 + p_1 e_1 + p_2 e_2 + p_3 e_3$. It is symmetric and bilinear, and it is non-degenerate: if $B(\tilde q, \tilde p) = 0$ for every $\tilde p$, then $\tilde q = 0$, since testing against the units gives $q_\mu = 0$.

Forming the inner product, evaluating it to a scalar and asking when it vanishes is all that is done with it here. Its diagonal value is the product

$$ B(\tilde q, \tilde q) = \tilde q \tilde{q}^{\natural} = q_0^2 + q_1^2 - q_2^2 - q_3^2, $$

which lies in the centre $S$ and is multiplicative, $(\tilde q \tilde p)((\tilde q \tilde p))^{\natural} = (\tilde q\tilde{q}^{\natural})(\tilde p \tilde{p}^{\natural})$, because $\tilde q\tilde{q}^{\natural}$ is central. The equation $\tilde q\tilde{q}^{\natural} = 0$, equivalently $B(\tilde q,\tilde q) = 0$, is the algebraic condition that decides the zero divisors, and the elements on which it holds are studied in Split-Quaternion Zero Divisors. No length, no sign and no orthogonal or orthonormal decomposition is read from the inner product in this article; the metrical reading — the signature of the pairings, their isotropy, the resulting classification of the elements and the group of units — is in Split-Quaternion Norm and Invertibility.

The Idempotents and the Split-Complex Subspaces

The Idempotents

Definition. The two elements

$$ \tilde\pi_+ = \tfrac{1}{2}(1 + e_2), \qquad \tilde\pi_- = \tfrac{1}{2}(1 - e_2) $$

are the split-quaternion idempotents.

Theorem. The elements $\tilde\pi_+$ and $\tilde\pi_-$ satisfy

$$ \tilde\pi_+^2 = \tilde\pi_+, \qquad \tilde\pi_-^2 = \tilde\pi_-, \qquad \tilde\pi_+ \tilde\pi_- = \tilde\pi_- \tilde\pi_+ = 0, \qquad \tilde\pi_+ + \tilde\pi_- = 1, $$

and they are not central. Moreover

$$ \mathbb{H}_{\mathrm{s}} = \mathbb{H}_{\mathrm{s}} \tilde\pi_+ \oplus \mathbb{H}_{\mathrm{s}} \tilde\pi_- $$

is a direct sum of two minimal left ideals of real dimension $2$, and the corresponding statement holds on the right, $\mathbb{H}_{\mathrm{s}} = \tilde\pi_+ \mathbb{H}_{\mathrm{s}} \oplus \tilde\pi_- \mathbb{H}_{\mathrm{s}}$. The decomposition is a decomposition of left modules and not of algebras: the subalgebra generated by $\tilde\pi_+$ and $\tilde\pi_-$ is two-dimensional and is not the whole algebra.

Proof. The four identities are the computation

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

and its analogue for $\tilde\pi_-$, together with $\tilde\pi_+\tilde\pi_- = \tfrac14(1 - e_2^2) = 0$ and $\tilde\pi_+ + \tilde\pi_- = 1$. For non-centrality, $e_1 \tilde\pi_+ = \tfrac12(e_1 + e_3)$ while $\tilde\pi_+ e_1 = \tfrac12(e_1 - e_3)$, so $e_1 \tilde\pi_+ \neq \tilde\pi_+ e_1$. An element $\tilde q \tilde\pi_+$ of the first summand is fixed by right multiplication by $\tilde\pi_+$, since $\tilde q \tilde\pi_+ \tilde\pi_+ = \tilde q \tilde\pi_+$, and the map $\tilde q \mapsto \tilde q \tilde\pi_+$ has image of dimension $2$ because its kernel is $\mathbb{H}_{\mathrm{s}} \tilde\pi_-$. The sum $\mathbb{H}_{\mathrm{s}} \tilde\pi_+ + \mathbb{H}_{\mathrm{s}} \tilde\pi_-$ is all of $\mathbb{H}_{\mathrm{s}}$, since $\tilde q = \tilde q(\tilde\pi_+ + \tilde\pi_-)$, and the intersection is zero: if $\tilde q \tilde\pi_+ = \tilde p \tilde\pi_-$ then multiplying on the right by $\tilde\pi_+$ gives $\tilde q \tilde\pi_+ = 0$. Hence the sum is direct, and each summand is a minimal left ideal, as the idempotent theory of Split-Quaternion Idempotents and Projections shows. The final claim holds because $\tilde\pi_+ \tilde\pi_- = 0$ and the span of $\tilde\pi_+, \tilde\pi_-$ has dimension $2$.

The Split-Complex Subspaces

The line $\mathbb{R} \cdot 1$ together with either of the two square-$+1$ generators spans a copy of the split-complex algebra inside $\mathbb{H}_{\mathrm{s}}$:

$$ \mathbb{D}_2 = \operatorname{span}\{1, e_2\} \cong \mathbb{D}, \qquad \mathbb{D}_3 = \operatorname{span}\{1, e_3\} \cong \mathbb{D}, $$

where $\mathbb{D} = \mathbb{R}[j]/(j^2-1)$ is the split-complex algebra of Split-Complex Algebra. Both are commutative subalgebras, both have the two idempotents $\tfrac12(1 \pm e_2)$ and $\tfrac12(1 \pm e_3)$ respectively, and both carry the zero divisors $1 \pm e_2$ and $1 \pm e_3$. The whole algebra is generated by $\mathbb{D}_2$ together with the single element $e_1$:

$$ \mathbb{H}_{\mathrm{s}} = \mathbb{D}_2 \oplus \mathbb{D}_2 e_1 , \qquad e_1 d = \bar{d} e_1 \ \ (d \in \mathbb{D}_2), $$

which exhibits $\mathbb{H}_{\mathrm{s}}$ as a two-dimensional module over the split-complex algebra with a conjugation-twisted multiplication. The idempotents $\tilde\pi_\pm$ of the algebra are the idempotents of the subalgebra $\mathbb{D}_2$; they are non-central in $\mathbb{H}_{\mathrm{s}}$ precisely because $e_1$ does not commute with $e_2$.

The Centre and Simplicity

Theorem (The Centre). The centre of $\mathbb{H}_{\mathrm{s}}$ is the scalar line:

$$ Z(\mathbb{H}_{\mathrm{s}}) = \{\tilde q : \tilde q \tilde p = \tilde p\tilde q \ \text{for all} \ \tilde p\} = \mathbb{R} \cdot 1 . $$

Proof. The scalars are central. Conversely, suppose $\tilde q = q_0 + \mathbf{v}$ is central. Commuting $\tilde q$ with $e_1$ gives

$$ 0 = \tilde q e_1 - e_1 \tilde q . $$

The scalar part of $\tilde q$ commutes with everything, so the condition is $\mathbf{v} e_1 - e_1 \mathbf{v} = 0$. Writing $\mathbf{v} = q_1 e_1 + q_2 e_2 + q_3 e_3$ and using the table,

$$ \mathbf{v} e_1 = -q_1 - q_2 e_3 + q_3 e_2, \qquad e_1 \mathbf{v} = -q_1 + q_2 e_3 - q_3 e_2, $$

so $\mathbf{v} e_1 - e_1 \mathbf{v} = -2q_2 e_3 + 2q_3 e_2 = 0$, giving $q_2 = q_3 = 0$. Commuting with $e_2$ similarly gives $q_1 = 0$. Hence $\mathbf{v} = 0$ and $\tilde q = q_0$ is scalar.

Theorem (Simplicity). The split-quaternion algebra is simple: it has no two-sided ideal other than $0$ and the algebra itself.

Proof. Let $I \neq 0$ be a two-sided ideal and let $0 \neq \tilde q \in I$. If $\tilde q \tilde{q}^{\natural} \neq 0$ then $\tilde q$ is a unit, by the invertibility criterion of Split-Quaternion Norm and Invertibility, so $I = \mathbb{H}_{\mathrm{s}}$. If $\tilde q \tilde{q}^{\natural} = 0$, then for every real $\lambda$ the element $\tilde q + \lambda$ lies in $I$ and

$$ (\tilde q + \lambda)\overline{(\tilde q + \lambda)} = \tilde q \tilde{q}^{\natural} + 2\lambda \operatorname{Sc}(\tilde q) + \lambda^2 = \lambda\big(2\operatorname{Sc}(\tilde q) + \lambda\big), $$

which is nonzero for every $\lambda$ outside the two-element set $\{0, -2\operatorname{Sc}(\tilde q)\}$. Choosing such a $\lambda$ exhibits a unit in $I$, so again $I = \mathbb{H}_{\mathrm{s}}$. Hence the only two-sided ideals are $0$ and the algebra.

The algebra is thus associative, non-commutative, simple, with centre $\mathbb{R}$, and it is not a division algebra. The combination — associative, non-commutative, simple, centre $\mathbb{R}$, and not a division algebra — is the one the comparisons below set against the quaternions and the split-biquaternions.

The Lie Algebra Structure

The algebra carries the commutator bracket $[\tilde q, \tilde p] = \tilde q \tilde p - \tilde p\tilde q$, which makes it a real Lie algebra. The bracket of two elements of $V$ lies in $V$, since

$$ [e_1, e_2] = 2 e_3, \qquad [e_2, e_3] = -2 e_1, \qquad [e_3, e_1] = 2 e_2 . $$

So $V$ is a three-dimensional Lie subalgebra of $\mathbb{H}_{\mathrm{s}}$, and the displayed relations are those of the three-dimensional simple real Lie algebra $\mathrm{SL}_2(\mathbb{R})$; the bracket algebra on $V$ is therefore $\mathrm{SL}_2(\mathbb{R})$. This is the algebraic origin of the relation between the split-quaternions and the Lorentz group: the vector subspace is a Lie algebra of infinitesimal Lorentz transformations, and the exponential of the bracket gives the rotations of Split-Quaternion Rotations and the Lorentz Group.

The scalar line $S$ is the centre of $\mathbb{H}_{\mathrm{s}}$ and therefore contributes nothing to the bracket. The full Lie algebra $\mathbb{H}_{\mathrm{s}}$ is the direct sum of the bracket algebra on $V$ and a central scalar line.

Comparison with $\mathbb{H}$ and with $\mathbb{H}_{\mathbb{D}}$

Comparison with the Quaternions

The split-quaternions and the quaternions are the two real forms of the same complexified algebra, and they differ in the sign of one generator. The following table collects the contrast; the quaternion column is supported by Quaternion Algebra.

	$\mathbb{H}$	$\mathbb{H}_{\mathrm{s}}$
basis and squares	$e_1^2=e_2^2=e_3^2=-1$	$e_1^2=-1$, $e_2^2=e_3^2=+1$
diagonal product $\tilde q\tilde{q}^{\natural}$	$q_0^2+q_1^2+q_2^2+q_3^2$	$q_0^2+q_1^2-q_2^2-q_3^2$
zero divisors	none	yes
nonzero nilpotents	none	yes
division algebra	yes	no
centre	$\mathbb{R}$	$\mathbb{R}$
vector subspace	$\mathbb{R}^3$, $\mathrm{SO}(3)$	$V$, $\mathrm{SL}_2(\mathbb{R})$
scalar group attached	$Sp(1) \cong SU(2)$	$\mathrm{SL}_2(\mathbb{R})$

The decisive difference is the sign in the diagonal product. For a nonzero quaternion the diagonal product is positive, so the elements of diagonal product $1$ form the compact three-sphere $Sp(1) \cong SU(2)$, double-covering $\mathrm{SO}(3)$. For a nonzero split-quaternion the diagonal product takes both signs, so the elements of diagonal product $1$ form the non-compact group $\mathrm{SL}_2(\mathbb{R})$. The change of a single sign turns a compact three-sphere into a non-compact three-dimensional group. The metrical statements that follow from this — the signatures, the definiteness in the quaternion case and the indefiniteness in the split case, and the groups they define — are treated in Split-Quaternion Norm and Invertibility and Split-Quaternion Rotations and the Lorentz Group.

Comparison with the Split-Biquaternions

The name split quaternions is used in the classical literature for the four-dimensional algebra $\mathrm{Cl}_{1,1}$ of this article, and it is not the algebra that the corpus writes $\mathbb{H}_{\mathbb{D}}$. The corpus reserves $\mathbb{H}_{\mathbb{D}}$ for the eight-dimensional tensor product $\mathbb{D} \otimes_{\mathbb{R}} \mathbb{H}$, the split-biquaternions; the distinction is stated once, in The Number Systems as Clifford Algebras, and is recorded again in Examples of Algebras and in List of Algebras by Dimension. The two algebras must not be identified, and the following table records why.

	$\mathbb{H}_{\mathrm{s}}$	$\mathbb{H}_{\mathbb{D}}$
dimension over $\mathbb{R}$	$4$	$8$
isomorphic to	$\mathrm{Cl}_{1,1}$	$\mathrm{Cl}_{0,3}$
simple	yes	no
centre	$\mathbb{R}$	$\mathbb{D}$
idempotents	non-central only	central, $e_\pm$
nonzero nilpotents	yes	no

The two rows that separate the algebras most sharply are the last two. A nilpotent is an element $\tilde q \neq 0$ with $\tilde q^2 = 0$, and the split-quaternion algebra has them: $(e_1 + e_3)^2 = e_1^2 + e_1e_3 + e_3e_1 + e_3^2 = -1 + 0 + 1 = 0$, since $e_1 e_3 = -e_3 e_1$. An algebra that is a product of two division algebras, as $\mathbb{H}_{\mathbb{D}}$ is by the dictionary of The Number Systems as Clifford Algebras, has no nilpotents at all, because a nilpotent would have a nilpotent component in one of the factors and a division algebra has none. Simultaneously, the split-quaternion algebra is simple, while a product of two algebras is not. A simple algebra with nilpotents and non-central idempotents is therefore entirely different from a semisimple, non-simple product of two division algebras with central idempotents and no nilpotents.

The eight-dimensional relative of the corpus is treated later in Part V, under Split-Biquaternions; nothing in it is used here. What is stated here is stated from the algebra of this article and from the conventions: a four-dimensional simple real algebra with zero divisors and non-central idempotents, on the one hand, and the eight-dimensional $\mathbb{D} \otimes_{\mathbb{R}} \mathbb{H}$ of the notation table, on the other.

Summary

The split-quaternion algebra $\mathbb{H}_{\mathrm{s}}$ is the four-dimensional real algebra with basis $1, e_1, e_2, e_3$, the products $e_1^2 = -1$, $e_2^2 = e_3^2 = +1$, the anticommutation $e_1 e_2 = -e_2 e_1$ and $e_3 = e_1 e_2$. It is associative, non-commutative and simple, with centre $\mathbb{R}$, and it is not a division algebra: $1 + e_2$ and $1 - e_2$ are nonzero and multiply to zero.

The algebra is the Clifford algebra $\mathrm{Cl}_{1,1}$, and its conjugation $\tilde{q}^{\natural}$ is the Clifford conjugation. The diagonal product $\tilde q\tilde{q}^{\natural} = q_0^2 + q_1^2 - q_2^2 - q_3^2$ takes both signs and vanishes exactly on the zero divisors; its metrical reading, the associated group of units and the action of the group on $V$ are treated in Split-Quaternion Norm and Invertibility and Split-Quaternion Rotations and the Lorentz Group.

The conjugation $\tilde{q}^{\natural} = q_0 e_0 - q_1 e_1 - q_2 e_2 - q_3 e_3$ is an involutive anti-automorphism whose fixed-point subspaces are the scalar line $S$ and the vector space $V$. The principal involution and the reversal commute and cut the algebra into the scalar line, the plane $\operatorname{span}\{e_1,e_2\}$ and the line $\mathbb{R}e_3$. The idempotents $\tilde\pi_\pm = \tfrac12(1 \pm e_2)$ are non-central, sum to $1$, multiply to zero, and split the algebra as a direct sum of two minimal left ideals, and not as an algebra. The split-complex subalgebras $\operatorname{span}\{1,e_2\}$ and $\operatorname{span}\{1,e_3\}$ are copies of $\mathbb{D}$ and carry the zero divisors of the algebra.

The system is not the eight-dimensional $\mathbb{H}_{\mathbb{D}} = \mathbb{D} \otimes_{\mathbb{R}} \mathbb{H}$ of the notation table, which is neither simple nor free of nilpotents; the two are distinguished once, in The Number Systems as Clifford Algebras.

Summary of Notation

Symbol	Meaning	Article
$\mathbb{H}_{\mathrm{s}}$	the split-quaternion algebra, $\mathrm{Cl}_{1,1}$	this article
$1, e_1, e_2, e_3$	the basis, with $e_1^2=-1$, $e_2^2=e_3^2=+1$, $e_3=e_1e_2$	this article
$\tilde q = q_0 e_0 + q_1 e_1 + q_2 e_2 + q_3 e_3$	a general split-quaternion	this article
$q_0 = \operatorname{Sc}(\tilde q)$	the scalar part	this article
$\mathbf{v} = \operatorname{Vec}(\tilde q)$	the vector part	this article
$S = \mathbb{R}\cdot1$	the scalar subspace	this article
$V = \operatorname{span}\{e_1,e_2,e_3\}$	the vector subspace	this article
$\tilde{q}^{\natural} = q_0 e_0 - q_1 e_1 - q_2 e_2 - q_3 e_3$	the split-quaternion conjugation	this article
$\alpha$	the principal involution, $e_1,e_2 \mapsto -e_1,-e_2$, $e_3 \mapsto e_3$	this article
$\rho$	the reversal, $e_1,e_2 \mapsto e_1,e_2$, $e_3 \mapsto -e_3$	this article
$B(\tilde q,\tilde p) = \operatorname{Sc}(\tilde q \tilde{p}^{\natural}) = q_0p_0+q_1p_1-q_2p_2-q_3p_3$	the inner product, formed and evaluated algebraically	this article
$N(\tilde q) = \tilde q\tilde{q}^{\natural} = q_0^2+q_1^2-q_2^2-q_3^2$	the split-quaternion norm, named here only by forward reference	Split-Quaternion Norm and Invertibility
$\tilde\pi_\pm = \tfrac12(1 \pm e_2)$	the non-central idempotents	this article
$\mathbb{D}_2 = \operatorname{span}\{1,e_2\}$, $\mathbb{D}_3 = \operatorname{span}\{1,e_3\}$	the split-complex subalgebras	this article
$[\tilde q,\tilde p] = \tilde q \tilde p - \tilde p\tilde q$	the commutator bracket	this article
$\mathrm{SL}_2(\mathbb{R})$	the three-dimensional simple Lie algebra of $V$	this article
$\mathbb{D}$	the split-complex numbers, $j^2=+1$	Split-Complex Algebra
$\mathbb{H}$	the real quaternions	Quaternion Algebra
$\mathbb{H}_{\mathbb{D}}$	the split-biquaternions, $\mathbb{D}\otimes_{\mathbb{R}}\mathbb{H}$, a later Part V system	The Number Systems as Clifford Algebras