Split-Quaternion Norm and Invertibility

Introduction

This article studies the split-quaternion norm, its isotropy, and the invertibility theory it determines. It proves the criterion that an element is invertible exactly when its split-quaternion norm does not vanish, describes the group of units, classifies the elements, and describes how the invertible elements are distributed among the distinguished subspaces.

The split-quaternion algebra, its basis, its conjugation ${}^{\natural}$, its split-quaternion norm $N$, its idempotents $\tilde\pi_\pm$ and its subspaces $S$, $V$, $\mathbb{D}_2$, $\mathbb{D}_3$ are assumed from Split-Quaternion Algebra and are not redefined. The zero divisor set is treated separately in Split-Quaternion Zero Divisors, and the roots of $-1$ in Split-Quaternion Roots of Minus One. Nothing physical is invoked.

The Split-Quaternion Norm

The Split-Quaternion Norm

Definition. The split-quaternion norm of $\tilde q = q_0 e_0 + q_1 e_1 + q_2 e_2 + q_3 e_3$ is

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

where $\tilde{q}^{\natural} = q_0 e_0 - q_1 e_1 - q_2 e_2 - q_3 e_3$ is the conjugation of (Split-Quaternion Algebra, §The Conjugation).

The form is a real quadratic form of signature $(2,2)$, read directly from the diagonal expression as two positive and two negative squares. It is multiplicative:

$$ N(\tilde q \tilde p) = N(\tilde q)N(\tilde p) \qquad (\tilde q, \tilde p \in \mathbb{H}_{\mathrm{s}}). $$

Proof. The product $\tilde q\tilde{q}^{\natural}$ is fixed by the conjugation, hence central; therefore, for all $\tilde q, \tilde p$,

$$ N(\tilde q \tilde p) = (\tilde q \tilde p)((\tilde q \tilde p))^{\natural} = \tilde q\, \tilde p \tilde p^{\natural}\, \tilde{q}^{\natural} = \tilde q \tilde{q}^{\natural}\, \tilde p \tilde p^{\natural} = N(\tilde q)N(\tilde p), $$

using $(\tilde q \tilde p)^{\natural} = \tilde p^{\natural} \tilde{q}^{\natural}$ and the centrality of $\tilde p \tilde p^{\natural}$.

Its polarisation is the bilinear form

$$ B(\tilde q, \tilde p) = \tfrac{1}{2}\big(N(\tilde q+\tilde p) - N(\tilde q) - N(\tilde p)\big) = q_0 p_0 + q_1 p_1 - q_2 p_2 - q_3 p_3 $$

for $\tilde q = q_0 e_0 + q_1 e_1 + q_2 e_2 + q_3 e_3$ and $\tilde p = p_0 e_0 + p_1 e_1 + p_2 e_2 + p_3 e_3$. The matrix of $B$ in the basis $1, e_1, e_2, e_3$ is $\operatorname{diag}(+1, +1, -1, -1)$.

Multiplicativity and the Sign

The multiplicativity of $N$ has an immediate consequence for the sign.

Proposition. The set $\{N > 0\}$ and the set $\{N < 0\}$ are each closed under multiplication, and the product of an element of $\{N>0\}$ with an element of $\{N<0\}$ has $N < 0$. The scalar line and the $e_1$-direction have positive norm, while the $e_2$- and $e_3$-directions have negative norm.

Proof. If $N(\tilde q)$ and $N(\tilde p)$ are both positive, then $N(\tilde q \tilde p) = N(\tilde q)N(\tilde p) > 0$, and similarly in the other cases. The signs of the basis elements are $N(1) = N(e_1) = +1$ and $N(e_2) = N(e_3) = -1$.

Isotropy

Definition. The form $N$ is isotropic: there exist nonzero $\tilde q$ with $N(\tilde q) = 0$. A nonzero element with $N(\tilde q) = 0$ is an isotropic vector, and a one-dimensional subspace $\mathbb{R}\tilde q$ spanned by an isotropic vector is an isotropic line.

Theorem (The Isotropic Vectors). The isotropic vectors of $\mathbb{H}_{\mathrm{s}}$ are the nonzero quadrivectors ((q_0, q_1, q_2, q_3)) with

$$ q_0^2 + q_1^2 = q_2^2 + q_3^2 . $$

Writing $A = q_0 + iq_1$ and $B = q_2 + iq_3$ with $i^2 = -1$, the condition is $|A| = |B|$. The isotropic vectors are therefore parametrised by a pair $(A, B)$ of complex numbers of equal modulus.

Proof. The equation $N(\tilde q) = 0$ is $q_0^2 + q_1^2 = q_2^2 + q_3^2$, which in the notation of the statement is $|A|^2 = |B|^2$.

Proposition (Explicit Isotropic Lines). Write $\tilde q = q_0 e_0 + q_1 e_1 + q_2 e_2 + q_3 e_3$. If $\tilde q$ is isotropic then $(q_0,q_1) \neq (0,0)$ and $(q_2,q_3) \neq (0,0)$, and $\tilde q$ is a positive multiple of

$$ (\cos\alpha, \sin\alpha, \cos\beta, \sin\beta) $$

for a unique pair of angles $\alpha, \beta$ modulo the simultaneous replacement $(\alpha,\beta) \mapsto (\alpha+\pi, \beta+\pi)$. The isotropic lines are therefore parametrised by $S^1 \times S^1$ modulo the identification $(\alpha,\beta)\sim(\alpha+\pi,\beta+\pi)$ of pairs of angles, and the isotropic lines lying in the vector subspace $V$ are exactly

$$ \mathbb{R}\big(e_1 + \cos\theta\, e_2 + \sin\theta\, e_3\big), \qquad \theta \in [0, 2\pi), $$

a circle's worth of lines; in particular $\mathbb{R}(e_1 + e_2)$, $\mathbb{R}(e_1 - e_2)$, $\mathbb{R}(e_1 + e_3)$ and $\mathbb{R}(e_1 - e_3)$ are isotropic.

Proof. The equation $N(\tilde q) = 0$ is $q_0^2 + q_1^2 = q_2^2 + q_3^2$. If $(q_0,q_1) = (0,0)$ then $q_2 = q_3 = 0$ and $\tilde q = 0$, contrary to the definition of an isotropic vector; the same argument applies to $(q_2,q_3)$. Hence both pairs are nonzero, and there are $r > 0$ and angles $\alpha, \beta$ with $(q_0,q_1) = r(\cos\alpha, \sin\alpha)$ and $(q_2,q_3) = r(\cos\beta, \sin\beta)$; the common radius is forced by the equation. Multiplying $\tilde q$ by a positive scalar does not change either angle, and multiplying by $-1$ adds $\pi$ to both, so the line determines $(\alpha,\beta)$ modulo the simultaneous replacement. For the vector subspace, $q_0 = 0$ and the equation is $q_1^2 = q_2^2 + q_3^2$ with $q_1 \neq 0$; normalising $q_1 = 1$ and writing $(q_2,q_3) = (\cos\theta,\sin\theta)$ gives the displayed family.

Theorem (The Isotropic Lines Are Doubly Ruled). The isotropic lines of $\mathbb{H}_{\mathrm{s}}$ are the points of the projective null quadric $\{N = 0\} \subset \mathbb{P}(\mathbb{H}_{\mathrm{s}})$, and that quadric is doubly ruled: its lines fall into two families, such that two lines of one family are skew while a line of one family meets a line of the other in exactly one point unless the two are parallel. The two families are defined and computed in Split-Quaternion Zero Divisors, §The Two Families.

Proof. A nonzero element $\tilde q$ is isotropic exactly when $N(\tilde q) = 0$, so the isotropic lines are the points of the projective null quadric of the non-degenerate form $N$ of signature $(2,2)$. A non-degenerate quadric of signature $(2,2)$ in $\mathbb{P}^3$ is doubly ruled. The explicit ruling is proved in Split-Quaternion Zero Divisors, §The Two Families.

The isotropic lines are also visible in the vector subspace: the isotropic lines lying in $V$ are the lines of the three-dimensional light cone $q_1^2 = q_2^2 + q_3^2$. Isotropic lines not lying in $V$ have a nonzero scalar part; an example is $\mathbb{R}(1 + e_2)$, since $N(1+e_2) = 1 - 1 = 0$, and another is $\mathbb{R}(1 + e_3)$.

The Invertibility Criterion

Theorem (The Invertibility Criterion). Let $\tilde q \in \mathbb{H}_{\mathrm{s}}$ be nonzero. The following are equivalent.

1. $\tilde q$ is invertible: there exists $\tilde p$ with $\tilde q \tilde p = \tilde p\tilde q = 1$.
2. $N(\tilde q) \neq 0$.

When these hold, the inverse is

$$ \tilde q^{-1} = \frac{\tilde{q}^{\natural}}{N(\tilde q)} . $$

Proof. Suppose first that $N(\tilde q) \neq 0$. Then $\tilde{q}^{\natural}/N(\tilde q)$ is a real multiple of $\tilde{q}^{\natural}$, and

$$ \tilde q \cdot \frac{\tilde{q}^{\natural}}{N(\tilde q)} = \frac{\tilde q\tilde{q}^{\natural}}{N(\tilde q)} = \frac{N(\tilde q)}{N(\tilde q)} = 1, \qquad \frac{\tilde{q}^{\natural}}{N(\tilde q)} \cdot \tilde q = \frac{\tilde{q}^{\natural}\tilde q}{N(\tilde q)} = 1, $$

so $\tilde q$ is invertible with the displayed inverse. Conversely, suppose $\tilde q$ is invertible, say $\tilde q \tilde p = 1$. Applying $N$ and using multiplicativity, $N(\tilde q)N(\tilde p) = N(1) = 1$, so $N(\tilde q) \neq 0$. This proves the equivalence of (1) and (2).

Corollary (Zero Divisors). A nonzero element is a zero divisor if and only if $N(\tilde q) = 0$. If $N(\tilde q) = 0$ and $\tilde q \neq 0$, then $\tilde q\tilde{q}^{\natural} = 0$ with $\tilde{q}^{\natural} \neq 0$, so $\tilde q$ is a zero divisor; conversely a zero divisor is not a unit, so $N(\tilde q) = 0$ by the criterion.

The criterion has the form the menu names: invertibility is $N \neq 0$, and the boundary is the null cone of the split-quaternion norm. Since the split-quaternion norm is multiplicative, the multiplicative structure of the algebra and the invertibility theory are governed by one quadratic form.

The Group of Units

Definition. The group of units of the split-quaternion algebra is

$$ \mathbb{H}_{\mathrm{s}}^{\times} = \{\tilde q \in \mathbb{H}_{\mathrm{s}} : N(\tilde q) \neq 0\}, $$

with multiplication inherited from the algebra.

Theorem. The split-quaternion norm restricts to a surjective group homomorphism

$$ N : \mathbb{H}_{\mathrm{s}}^{\times} \longrightarrow \mathbb{R}^{\times}. $$

Proof. Multiplicativity of $N$ makes it a group homomorphism from the units to $\mathbb{R}^{\times}$, and it is surjective because $N(\lambda) = \lambda^2$ and $N(\lambda e_2) = -\lambda^2$ for real $\lambda \neq 0$.

Corollary (The Norm-One Groups). The kernel of $N$ is the group of unit split-quaternions

$$ U = \{\tilde q \in \mathbb{H}_{\mathrm{s}} : N(\tilde q) = 1\} \cong \mathrm{SL}_2(\mathbb{R}), $$

and the union of the two split-quaternion norm levels $\pm 1$ is the disjoint union $\{\tilde q : N(\tilde q) = \pm 1\} = U \sqcup (-U)$ of two connected components, namely $U$ and its negative. The group of units has two connected components as well, the open sets $\{N > 0\}$ and $\{N < 0\}$; the norm-one subgroup $U$ is connected.

Proof. The kernel of $N$ is $\{N = 1\}$, which is the group of unit split-quaternions and is connected, as recalled in Matrix Groups and Classical Groups; the second norm level is its negative coset $-U$. The group of units is the disjoint union of the two open sign sets $\{N > 0\}$ and $\{N < 0\}$.

This is the exact point at which the indefinite norm changes the group theory. The quaternion unit sphere is the compact group $Sp(1) \cong SU(2)$, the kernel of a positive-definite norm on a division algebra. The split-quaternion norm-one set is the non-compact $\mathrm{SL}_2(\mathbb{R})$, and the passage from $Sp(1)$ to $\mathrm{SL}_2(\mathbb{R})$ is the passage from the double cover of the rotation group of three-space to the double cover of the Lorentz group of signature $(2,1)$. The rotations themselves are treated in Split-Quaternion Rotations and the Lorentz Group.

The Three-Way Classification

The menu's three-way classification of the elements is the following.

Theorem (Classification). Every element of $\mathbb{H}_{\mathrm{s}}$ falls into exactly one of the three classes:

There is no fourth class, and in particular the third class is not empty: the isotropic vectors of Isotropy are zero divisors by the corollary of The Invertibility Criterion. The two classes partition $\mathbb{H}_{\mathrm{s}} \setminus \{0\}$; the invertible class is open, and the zero divisor class is closed there.

Proof. Let $\tilde q$ be nonzero. The real number $N(\tilde q)$ is either zero or a unit of $\mathbb{R}$; there is no third possibility, because $\mathbb{R}$ is a field. If $N(\tilde q) = 0$ then $\tilde q$ is a zero divisor by the corollary of the criterion; if $N(\tilde q) \neq 0$ then $\tilde q$ is invertible. The two cases are exclusive and exhaust the nonzero elements.

Remark. In the split-biquaternion case the corresponding classification genuinely has three nonzero classes, because there the norm takes values in a ring with zero divisors rather than in a field, so that $N(\tilde q)$ can be a nonzero non-unit. The three-way classification of the present article is therefore a dichotomy plus the zero element, and its third row is the single element $0$. The contrast is developed in Comparison with the Quaternion and Split-Biquaternion Cases.

Corollary (The Refinement by Sign). The invertible class splits into the two open sets

$$ P = \{\tilde q : N(\tilde q) > 0\}, \qquad Q = \{\tilde q : N(\tilde q) < 0\}, $$

each of which is closed under multiplication, while $P \cdot Q \subseteq Q$ and $Q \cdot Q \subseteq P$. The identity lies in $P$, and $P$ is the identity component of the group of units.

Distribution of the Invertible Elements

The invertible elements are distributed over the distinguished subspaces as follows. The subspaces are those of (Split-Quaternion Algebra, §Conjugations and Fixed-Point Subspaces and §The Idempotents and the Split-Complex Subspaces).

The Scalar Subspace

On $S = \mathbb{R} \cdot 1$, an element is $\tilde q = q_0$ with $N(\tilde q) = q_0^2$. Every nonzero scalar is a unit, and every such unit lies in $P$. The only non-unit of $S$ is $0$.

The Vector Subspace

On $V$, an element is $u = q_1 e_1 + q_2 e_2 + q_3 e_3$ with

$$ N(u) = q_1^2 - q_2^2 - q_3^2 . $$

The invertible elements of $V$ are the vectors with $q_1^2 \neq q_2^2 + q_3^2$: the spacelike vectors with $q_1^2 < q_2^2 + q_3^2$, on which $N < 0$, and the timelike vectors with $q_1^2 > q_2^2 + q_3^2$, on which $N > 0$. The non-invertible nonzero elements of $V$ are the lightlike vectors, the cone $q_1^2 = q_2^2 + q_3^2$. This is the trichotomy of the Lorentzian geometry of the vector subspace, and it is the reason the geometry of the system is the hyperbolic plane; see Split-Quaternion Rotations and the Lorentz Group.

The Split-Complex Subspaces

On $\mathbb{D}_2 = \operatorname{span}\{1, e_2\}$, an element is $\tilde q = q_0 + q_2 e_2$ with

$$ N(\tilde q) = q_0^2 - q_2^2 . $$

The invertible elements are those with $q_0^2 \neq q_2^2$; the non-invertible nonzero elements are the real multiples of $1 + e_2$ and of $1 - e_2$, which are the zero divisors of the split-complex algebra $\mathbb{D}$ studied in Split-Complex Algebra. On $\mathbb{D}_3 = \operatorname{span}\{1, e_3\}$ the identical statement holds with $e_3$ in place of $e_2$.

The Minimal Left and Right Ideals

On the two minimal left ideals $\mathbb{H}_{\mathrm{s}} \tilde\pi_\pm$, every element is a zero divisor or zero.

Proposition. For every $\tilde q \in \mathbb{H}_{\mathrm{s}}$, $N(\tilde q\tilde\pi_\pm) = N(\tilde q)N(\tilde\pi_\pm) = 0$. Hence $\mathbb{H}_{\mathrm{s}} \tilde\pi_+$ and $\mathbb{H}_{\mathrm{s}} \tilde\pi_-$ are totally isotropic: they contain no invertible element other than the origin. The same statement holds for the two minimal right ideals $\tilde\pi_+ \mathbb{H}_{\mathrm{s}}$ and $\tilde\pi_- \mathbb{H}_{\mathrm{s}}$.

Proof. $N(\tilde\pi_\pm) = \tfrac14 N(1 \pm e_2) = \tfrac14(1 - 1) = 0$, and multiplicativity gives $N(\tilde q\tilde\pi_\pm) = N(\tilde q) \cdot 0 = 0$.

So each of the four two-dimensional subspaces $\mathbb{H}_{\mathrm{s}} \tilde\pi_\pm$, $\tilde\pi_\pm \mathbb{H}_{\mathrm{s}}$ consists entirely of zero divisors together with the origin.

Summary of the Distribution

The invertible elements are the complement of the null cone $\{N = 0\}$, an open dense set of full measure. They form two connected components, $\{N > 0\}$ and $\{N < 0\}$.

Comparison with the Quaternion and Split-Biquaternion Cases

The Quaternion Case

For $\mathbb{H}$ the quaternion norm is $N(\tilde q) = q_0^2 + q_1^2 + q_2^2 + q_3^2$, positive definite by (Quaternion Algebra, §Basic Properties). It vanishes only at the origin, so every nonzero quaternion is invertible, the algebra is a division algebra, the invertible class is the whole of $\mathbb{H} \setminus \{0\}$, and the classification has the single nonzero class. The norm-one group is the compact $Sp(1) \cong SU(2)$, and the group of units is $\mathbb{R}_{>0} \times Sp(1)$, which is connected. The change from $\mathbb{H}$ to $\mathbb{H}_{\mathrm{s}}$ is the change of the norm from signature $(4,0)$ to signature $(2,2)$; it empties no class away, but it inserts the null cone and the two-component structure.

The Split-Biquaternion Case

The eight-dimensional algebra $\mathbb{H}_{\mathbb{D}} = \mathbb{D} \otimes_{\mathbb{R}} \mathbb{H}$ of the notation table is a real algebra of dimension eight, and it is treated later in Part V, under Split-Biquaternions; nothing of it is used here. The structural difference that decides the comparison is visible from the conventions alone: its coefficients lie in the split-complex ring $\mathbb{D}$, so its norm takes values in $\mathbb{D}$, and $\mathbb{D}$ has zero divisors of its own. An invertibility criterion in that system is a criterion in a ring with zero divisors, and its classification therefore has a third nonzero class — the elements whose norm is a nonzero zero divisor — which does not exist in the present article. The present classification, by contrast, is the dichotomy of The Three-Way Classification, and the reason is exactly that the coefficient field here is $\mathbb{R}$.

Summary

The split-quaternion norm is $N(\tilde q) = q_0^2 + q_1^2 - q_2^2 - q_3^2$, of signature $(2,2)$ and multiplicative. The form is isotropic; its isotropic vectors satisfy $q_0^2 + q_1^2 = q_2^2 + q_3^2$, its isotropic lines are the points of a doubly ruled projective null quadric, and on the vector subspace the isotropic lines are the lines of the light cone $q_1^2 = q_2^2 + q_3^2$.

A nonzero element is invertible exactly when $N(\tilde q) \neq 0$, and then $\tilde q^{-1} = \tilde{q}^{\natural}/N(\tilde q)$; it is a zero divisor exactly when $N(\tilde q) = 0$. The group of units is $\{N \neq 0\}$, the norm-one subgroup is $U = \{N = 1\} \cong \mathrm{SL}_2(\mathbb{R})$, and $\{N = \pm 1\} = U \sqcup (-U)$ has two components. The units form the two connected components $\{N > 0\}$ and $\{N < 0\}$.

The classification of the elements is a dichotomy plus the zero element: invertible, or zero divisor, or zero; there is no further class, because the split-quaternion norm takes values in the field $\mathbb{R}$. The invertible elements are distributed as follows: all nonzero scalars are units; in $V$ the units are the spacelike and timelike vectors and the zero divisors are the light cone; in each split-complex subalgebra the units avoid the two isotropic lines; and the four minimal ideals $\mathbb{H}_{\mathrm{s}} \tilde\pi_\pm$, $\tilde\pi_\pm \mathbb{H}_{\mathrm{s}}$ are totally isotropic. In the eight-dimensional $\mathbb{H}_{\mathbb{D}}$ the norm takes values in a ring with zero divisors, and the corresponding classification has a genuinely third nonzero class; that system is treated later under Split-Biquaternions.

Summary of Notation

Further Reading

• Pertti Lounesto, Clifford Algebras and Spinors, 2nd ed. (Cambridge University Press, 2001), for the split-quaternion norm and determinant forms of the low-dimensional Clifford algebras.
• Ian R. Porteous, Clifford Algebras and the Classical Groups (Cambridge University Press, 1995), for $SL_2(\mathbb{R})$ and the indefinite orthogonal groups in their matrix models.
• T. Y. Lam, Introduction to Quadratic Forms over Fields (American Mathematical Society, 2005), for isotropic forms, their null cones and their maximal totally isotropic subspaces.
• John H. Conway and Derek A. Smith, On Quaternions and Octonions (A K Peters, 2003), for the split forms and the comparison with the division algebra case.