Split-Biquaternion Idempotents and Projections

Introduction

The algebra article defined the split biquaternion algebra $\mathbb{H}_{\mathbb{D}} = \mathbb{D} \otimes_{\mathbb{R}} \mathbb{H}$, its four conjugations and its four distinguished real subspaces. This article treats the idempotents of $\mathbb{H}_{\mathbb{D}}$ — the elements satisfying $\tilde P^2 = \tilde P$ — together with the projections and the direct sum decompositions they carry.

The treatment is purely mathematical. Every claim is either proved or stated as a definition. No physics is invoked. The split biquaternion algebra $\mathbb{H}_{\mathbb{D}}$ is assumed from the basic algebra article, together with its four conjugations, its four fixed-point subspaces $\mathbb{D}_{\mathbb{H}_{\mathbb{D}}}$, $\mathbb{H}_{\mathbb{H}_{\mathbb{D}}}$, $\mathbb{M}_+$, $\mathbb{M}_-$ and its idempotent decomposition. The split complex algebra $\mathbb{D}$ is assumed from the article on split complex algebra, together with its idempotents $\tilde\Pi_+ = \tfrac{1}{2}(1 + j)$ and $\tilde\Pi_- = \tfrac{1}{2}(1 - j)$ and the isomorphism $\mathbb{D} \cong \mathbb{R} \oplus \mathbb{R}$. The quaternion algebra $\mathbb{H}$ is assumed to be a division algebra.

Throughout, a split biquaternion is written $\tilde{Q} = \sum_{\mu=0}^{3} Q_\mu e_\mu$ with $Q_\mu = q_\mu + j q'_\mu$ and $q_\mu, q'_\mu \in \mathbb{R}$. The quaternion conjugate is $\tilde{Q}^{\natural} = Q_0 e_0 - \mathbf{Q}$, the split complex conjugate is $\bar{\tilde{Q}} = \bar{Q_0} e_0 + \mathbf{Q}^*$ with $Q_{\bar{\mu}} = q_\mu - j q'_\mu$, the Hermitian conjugate is $\tilde{Q}^{*} = \overline{\tilde{Q}^{\natural}}$, and the anti-Hermitian conjugate is $\tilde{Q}^\flat = -\tilde{Q}^{*}$. The split-biquaternion norm $N(\tilde{Q}) = \tilde{Q} \tilde{Q}^{\natural}$ is the form of Split-Biquaternion Norm and Invertibility, which this article names and does not use.

The result that shapes the whole article is that $\mathbb{H}_{\mathbb{D}}$ has exactly four idempotents, and that all of them are central. This is a strict contrast with the biquaternion algebra $\mathbb{B} = \mathbb{C} \otimes_{\mathbb{R}} \mathbb{H}$, whose idempotents form a four-dimensional family in bijection with the roots of $-1$. The reason is structural: $\mathbb{H}_{\mathbb{D}} \cong \mathbb{H} \oplus \mathbb{H}$ is a product of two division algebras, and a product of division algebras admits only the idempotents that are constant on each factor. This article records that rigidity and its consequences; the parallel study of the ideals of the algebra, and of the resulting Peirce decomposition and $\mathbb{H} \oplus \mathbb{H}$ splitting, is the subject of Split-Biquaternion Ideals and Peirce Decomposition.

Idempotents in an Algebra

Let $A$ be an associative unital algebra. An element $e \in A$ is an idempotent if $e^2 = e$. Idempotents encode direct summands: for an idempotent $e$,

$$ A = Ae \oplus A(1 - e) \quad (\text{left}), \qquad A = eA \oplus (1 - e)A \quad (\text{right}), $$

and conversely every decomposition of $A$ into complementary left ideals (or complementary right ideals) arises from an idempotent in this way. Two idempotents $e, f$ are orthogonal if $ef = fe = 0$; then $e + f$ is again idempotent. A family $\{e_1, \dots, e_n\}$ is pairwise orthogonal if $e_i e_j = 0$ for $i \neq j$, and complete if in addition $\sum_i e_i = 1$. A nonzero idempotent $e$ is primitive if it is not a sum of two nonzero orthogonal idempotents. The criterion used throughout, valid for a semisimple algebra $A$, is

$$ e \text{ primitive} \iff Ae \text{ is a minimal left ideal} \iff eAe \text{ is a division ring}. $$

The algebra $\mathbb{H}_{\mathbb{D}} \cong \mathbb{H} \oplus \mathbb{H}$ is a product of two division algebras and is therefore semisimple, so the criterion applies to it. The idempotents of a product of division algebras are easy to describe: if $e = (e_1, e_2) \in D_1 \oplus D_2$ with $D_1, D_2$ division rings, then $e^2 = e$ forces $e_1^2 = e_1$ and $e_2^2 = e_2$, and a division ring has only the idempotents $0$ and $1$. Hence a product of two division algebras has exactly four idempotents, all central. This is the phenomenon that the rest of the article realises in $\mathbb{H}_{\mathbb{D}}$.

The Idempotents of the Split Complex Centre

The idempotents of the split complex algebra $\mathbb{D}$ are

$$ \tilde\Pi_+ = \tfrac{1}{2}(1 + j), \qquad \tilde\Pi_- = \tfrac{1}{2}(1 - j). $$

They satisfy

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

So $\{\tilde\Pi_+, \tilde\Pi_-\}$ is a complete family of pairwise orthogonal idempotents of $\mathbb{D}$. Since $j$ is central in $\mathbb{H}_{\mathbb{D}}$, so are $1$ and $j$, and therefore so are $\tilde\Pi_+$ and $\tilde\Pi_-$: they lie in the centre of $\mathbb{H}_{\mathbb{D}}$, which is the split complex subspace $\mathbb{D}_{\mathbb{H}_{\mathbb{D}}}$. The pair $\{\tilde\Pi_+, \tilde\Pi_-\}$ is a basis of the centre over $\mathbb{R}$, and the map

$$ \mathbb{D}_{\mathbb{H}_{\mathbb{D}}} \to \mathbb{R} \oplus \mathbb{R}, \qquad a \tilde\Pi_+ + b \tilde\Pi_- \mapsto (a, b) $$

is an isomorphism of real algebras. The two idempotents are the units of the two copies of $\mathbb{R}$ in this decomposition, and they are the only nontrivial idempotents of the centre.

The idempotents $\tilde\Pi_\pm$ are exchanged by the split complex conjugation and fixed by the quaternion conjugation. Indeed $\bar{\tilde\Pi_\pm} = \tilde\Pi_\pm$ because the coefficients of $\tilde\Pi_\pm$ are real, while $j^* = -j$ gives

$$ \tilde\Pi_+^* = \tilde\Pi_-, \qquad \tilde\Pi_-^* = \tilde\Pi_+. $$

Consequently

$$ \tilde\Pi_+^\dagger = \tilde\Pi_-, \qquad \tilde\Pi_-^\dagger = \tilde\Pi_+, \qquad \tilde\Pi_+^\flat = -\tilde\Pi_-, \qquad \tilde\Pi_-^\flat = -\tilde\Pi_+, $$

so the Hermitian and anti-Hermitian conjugations swap the two idempotents (up to sign). The idempotents $\tilde\Pi_\pm$ are therefore not Hermitian: $\tilde\Pi_\pm^\dagger = \tilde\Pi_\mp \neq \tilde\Pi_\pm$.

The Classification of the Idempotents

Theorem. The idempotents of $\mathbb{H}_{\mathbb{D}}$ are exactly $0$, $\tilde\Pi_+$, $\tilde\Pi_-$ and $1$. There are no others.

Proof. Write an idempotent in the idempotent basis as $\tilde P = \tilde P_+ \tilde\Pi_+ + \tilde P_- \tilde\Pi_-$ with $\tilde P_\pm = \tilde P \tilde\Pi_\pm \in \mathbb{H}$. Since $\tilde\Pi_+^2 = \tilde\Pi_+$, $\tilde\Pi_-^2 = \tilde\Pi_-$ and $\tilde\Pi_+ \tilde\Pi_- = 0$, the square is

$$ \tilde P^2 = \tilde P_+^2 \tilde\Pi_+ + \tilde P_-^2 \tilde\Pi_-. $$

Hence $\tilde P^2 = \tilde P$ is equivalent to the pair of equations $\tilde P_+^2 = \tilde P_+$ and $\tilde P_-^2 = \tilde P_-$ in the quaternion algebra $\mathbb{H}$. Each $\tilde P_\pm$ is therefore an idempotent of $\mathbb{H}$.

The quaternion algebra is a division algebra, so its only idempotents are $0$ and $1$: if $\tilde q = a + \mathbf{v}$ with $a \in \mathbb{R}$ and $\mathbf{v}$ pure, then $\tilde q^2 = \tilde q$ gives $2a\mathbf{v} = \mathbf{v}$ and $a^2 - |\mathbf{v}|^2 = a$; if $\mathbf{v} = 0$ then $a^2 = a$, so $a \in \{0, 1\}$; if $\mathbf{v} \neq 0$ then $a = 1/2$ and then $|\mathbf{v}|^2 = -1/4$, which is impossible. Hence $\tilde P_\pm \in \{0, 1\}$, and the four combinations give

$$ \tilde P = 0, \qquad \tilde P = \tilde\Pi_+, \qquad \tilde P = \tilde\Pi_-, \qquad \tilde P = \tilde\Pi_+ + \tilde\Pi_- = 1. $$

These four are idempotent, and there are no others.

Corollary. Every idempotent of $\mathbb{H}_{\mathbb{D}}$ is central, and every idempotent lies in the split complex subspace $\mathbb{D}_{\mathbb{H}_{\mathbb{D}}}$. The set of idempotents is $\{0, \tilde\Pi_+, \tilde\Pi_-, 1\}$, a discrete set of four points.

Proof. The four idempotents are real combinations of $1$ and $j$, hence lie in the centre $\mathbb{D}_{\mathbb{H}_{\mathbb{D}}}$.

A nonzero idempotent is primitive exactly when it cannot be written as a sum of two nonzero orthogonal idempotents. From the four idempotents, the only orthogonal pairs are $\{\tilde\Pi_+, \tilde\Pi_-\}$ and $\{0, \tilde\Pi_\pm\}$. Hence $\tilde\Pi_+$ and $\tilde\Pi_-$ are primitive, while $0$ is excluded (it is not nonzero) and $1 = \tilde\Pi_+ + \tilde\Pi_-$ is not primitive. Thus $\{\tilde\Pi_+, \tilde\Pi_-\}$ is a complete family of primitive orthogonal idempotents, and it is the only one.

The Absence of a Bijection With the Roots of Minus One

In the biquaternion algebra $\mathbb{B}$, the idempotents are in bijection with the roots of $-1$: every idempotent has the form $\tfrac{1}{2}(e_0 + \xi i)$ for a unique root $\xi$ of $-1$, and the root set is large. That correspondence fails completely in $\mathbb{H}_{\mathbb{D}}$, and the failure is quantitative.

Theorem. The roots of $-1$ in $\mathbb{H}_{\mathbb{D}}$ form the set

$$ \{\xi = \mu_+ \tilde\Pi_+ + \mu_- \tilde\Pi_- : \mu_\pm \text{ unit pure real quaternions}\}, $$

a set of real dimension $4$, parametrised by a pair of unit pure real quaternions. That the set is the topological product $S^2 \times S^2$, and its manifold structure, are established in Split-Biquaternion Analysis. The idempotents of $\mathbb{H}_{\mathbb{D}}$ number $4$. Hence there is no bijection between the two sets.

Proof. The classification of the roots of $-1$ is the content of Split-Biquaternion Roots of Minus One: the equation $\xi^2 = -1$, written in the idempotent basis as $\xi_+^2 \tilde\Pi_+ + \xi_-^2 \tilde\Pi_- = -\tilde\Pi_+ - \tilde\Pi_-$, is equivalent to $\xi_\pm^2 = -1$ in $\mathbb{H}$, whose solutions are the unit pure real quaternions, a two-sphere. The product is two-dimensional over each of the two components, so the root set has real dimension $2 + 2 = 4$. Since the idempotent set is the four-point set of the previous section, no bijection exists.

The reason for the failure is the same rigidity that produced the classification theorem. In $\mathbb{B}$ the construction $\tilde P = \tfrac{1}{2} e_0 \pm \tfrac{1}{2} \xi i$ turns a root $\xi$ of $-1$ into an idempotent because the central unit $i$ is available to convert the vector part into a scalar direction. In $\mathbb{H}_{\mathbb{D}}$ there is no central scalar imaginary — the only central units are $1$ and $j$, with $j^2 = +1$ — and the obstruction $|\mathbf{v}|^2 = -1/4$ in the proof above is unavailable. The idempotents of $\mathbb{H}_{\mathbb{D}}$ are therefore confined to the centre, and they cannot be parametrised by the roots of $-1$.

Idempotents as Projections

An idempotent $\tilde P$ satisfies $\tilde P^2 = \tilde P$, and its complement $1 - \tilde P$ is again an idempotent with

$$ \tilde P(1 - \tilde P) = \tilde P - \tilde P^2 = 0. $$

Because every idempotent of $\mathbb{H}_{\mathbb{D}}$ is central, the pair $\{\tilde P, 1 - \tilde P\}$ gives a direct sum decomposition of the algebra as a two-sided decomposition:

$$ \mathbb{H}_{\mathbb{D}} = \mathbb{H}_{\mathbb{D}} \tilde P \oplus \mathbb{H}_{\mathbb{D}}(1 - \tilde P) = \tilde P \mathbb{H}_{\mathbb{D}} \oplus (1 - \tilde P) \mathbb{H}_{\mathbb{D}}, $$

and the two summands are ideals. This is the algebraic content of the statement that an idempotent is a projection: the idempotent is the projection, its complement is the complementary projection, and the algebra splits into the image of one and the image of the other. The decomposition attached to $\tilde P = \tilde\Pi_+$ is the idempotent decomposition

$$ \mathbb{H}_{\mathbb{D}} = \mathbb{H} \tilde\Pi_+ \oplus \mathbb{H} \tilde\Pi_-, $$

with $\mathbb{H} \tilde\Pi_+ = \mathbb{H}_{\mathbb{D}} \tilde\Pi_+$ and $\mathbb{H} \tilde\Pi_- = \mathbb{H}_{\mathbb{D}} \tilde\Pi_-$.

A Hermitian idempotent is one satisfying $\tilde{P}^{*} = \tilde P$; in $\mathbb{B}$ these are the orthogonal projections for the Hermitian form and the ones that occur in the spectral decomposition of a Hermitian element. In $\mathbb{H}_{\mathbb{D}}$ there is no nontrivial Hermitian idempotent: the calculation $\tilde\Pi_\pm^\dagger = \tilde\Pi_\mp$ of the second section shows that the two nontrivial idempotents are interchanged by ${}^{*}$, and neither is fixed. The idempotent decomposition of $\mathbb{H}_{\mathbb{D}}$ is therefore a decomposition into two ideals that are exchanged by the Hermitian conjugation, not a decomposition into orthogonal projections.

Idempotents and the Zero Divisors

A nontrivial idempotent is a zero divisor, because

$$ \tilde\Pi_+ (1 - \tilde\Pi_+) = \tilde\Pi_+ \tilde\Pi_- = 0, $$

and both factors are nonzero. Explicitly, the annihilator of $\tilde\Pi_+$ contains $\tilde\Pi_-$, and the annihilator of $\tilde\Pi_-$ contains $\tilde\Pi_+$; since $\mathbb{H}_{\mathbb{D}} = \mathbb{H} \tilde\Pi_+ \oplus \mathbb{H} \tilde\Pi_-$, the two annihilators are exactly

$$ \mathrm{Ann}(\tilde\Pi_+) = \mathbb{H} \tilde\Pi_-, \qquad \mathrm{Ann}(\tilde\Pi_-) = \mathbb{H} \tilde\Pi_+, $$

each of real dimension $4$. The idempotents are therefore the two most symmetric members of the zero divisor set: the zero divisors of $\mathbb{H}_{\mathbb{D}}$ are precisely the elements with a vanishing idempotent component, $\tilde{Q}_+ = 0$ or $\tilde{Q}_- = 0$, that is, the union $\mathbb{H} \tilde\Pi_- \cup \mathbb{H} \tilde\Pi_+$ with the origin deleted.

This is a different situation from the biquaternion case. In $\mathbb{B}$ every non-pure zero divisor is a complex multiple of an idempotent, so the idempotents in a sense parametrise the zero divisors. In $\mathbb{H}_{\mathbb{D}}$ the zero divisor set is the union of two four-dimensional real subspaces, while the idempotents are only four points: almost no zero divisor is a scalar multiple of an idempotent. The classification and the structure of the zero divisor set are the subject of Split-Biquaternion Zero Divisors.

Idempotents and the Minimal Left Ideals

Because $\tilde\Pi_+$ and $\tilde\Pi_-$ are central, the left ideal generated by $\tilde\Pi_+$ coincides with the two-sided ideal it generates:

$$ \mathbb{H}_{\mathbb{D}} \tilde\Pi_+ = \tilde\Pi_+ \mathbb{H}_{\mathbb{D}} = \tilde\Pi_+ \mathbb{H}_{\mathbb{D}} \tilde\Pi_+ = \mathbb{H} \tilde\Pi_+. $$

Proposition. The ideals $\mathbb{H} \tilde\Pi_+$ and $\mathbb{H} \tilde\Pi_-$ satisfy $\mathbb{H}_{\mathbb{D}} = \mathbb{H} \tilde\Pi_+ \oplus \mathbb{H} \tilde\Pi_-$, each has real dimension $4$, and each is a minimal left ideal; moreover $\tilde\Pi_\pm \mathbb{H}_{\mathbb{D}} \tilde\Pi_\pm = \mathbb{H} \tilde\Pi_\pm \cong \mathbb{H}$ as a ring.

Proof. Every $\tilde{Q}$ satisfies $\tilde{Q} = \tilde{Q}(\tilde\Pi_+ + \tilde\Pi_-) = \tilde{Q} \tilde\Pi_+ + \tilde{Q} \tilde\Pi_-$, and the intersection is zero: if $\tilde{Q} \tilde\Pi_+ = \tilde P \tilde\Pi_-$, then multiplying on the right by $\tilde\Pi_+$ gives $\tilde{Q} \tilde\Pi_+ = 0$. In the idempotent basis $\tilde{Q} \tilde\Pi_+ = \tilde{Q}_+ \tilde\Pi_+$ with $\tilde{Q}_+ \in \mathbb{H}$, so $\mathbb{H} \tilde\Pi_+$ is the image of the projection $\tilde{Q} \mapsto \tilde{Q}_+ \tilde\Pi_+$, a real vector space of dimension $4$; the same holds for $\tilde\Pi_-$, and $4 + 4 = 8$ accounts for the whole algebra. For minimality, the map $\mathbb{H}_{\mathbb{D}} \to \mathbb{H} \tilde\Pi_+$, $\tilde{Q} \mapsto \tilde{Q} \tilde\Pi_+$, is onto with kernel $\mathbb{H} \tilde\Pi_-$; any nonzero left ideal contained in $\mathbb{H} \tilde\Pi_+$ therefore has a preimage that is a left ideal strictly containing $\mathbb{H} \tilde\Pi_-$, and since the quotient $\mathbb{H}_{\mathbb{D}} / \mathbb{H} \tilde\Pi_- \cong \mathbb{H}$ is a division algebra, that ideal must be all of $\mathbb{H} \tilde\Pi_+$. Equivalently, $\tilde\Pi_+ \mathbb{H}_{\mathbb{D}} \tilde\Pi_+ = \mathbb{H} \tilde\Pi_+ \cong \mathbb{H}$ is a division ring, so $\tilde\Pi_+$ is primitive.

The counterpart statement in the biquaternion algebra is different in kind. There the standard idempotents $p = \tfrac{1}{2}(e_0 + ie_3)$ and $q = \tfrac{1}{2}(e_0 - ie_3)$ are not central, the ideals $\mathbb{B}p$ and $\mathbb{B}q$ are minimal left ideals only, and each is isomorphic to $\mathbb{C}^2$. Here centrality upgrades the left ideals to two-sided ideals and replaces $\mathbb{C}^2$ by the division algebra $\mathbb{H}$. The ideals themselves, their lattice, the distinction between the two summands, and the Peirce decomposition associated with them are developed in Split-Biquaternion Ideals and Peirce Decomposition.

The Dimension of the Set of Idempotents

The set of idempotents of $\mathbb{H}_{\mathbb{D}}$ is the four-point set $\{0, \tilde\Pi_+, \tilde\Pi_-, 1\}$, of real dimension $0$. It is a discrete subset of the centre $\mathbb{D}_{\mathbb{H}_{\mathbb{D}}}$, whose real dimension is $2$; the four points are its extremal points, and they are exactly the idempotents that the two-dimensional algebra $\mathbb{D} \cong \mathbb{R} \oplus \mathbb{R}$ possesses.

This is the sharpest form of the rigidity noted in the introduction. In $\mathbb{B}$ the non-trivial idempotents form a real four-dimensional family with a two-dimensional boundary stratum inherited from the roots of $-1$; in $\mathbb{H}_{\mathbb{D}}$ the corresponding set has collapsed to two isolated points. The collapse is a consequence of the classification theorem: because the algebra is a product of two division algebras, its idempotent set is finite.

Summary

An idempotent of $\mathbb{H}_{\mathbb{D}}$ is an element $\tilde P$ with $\tilde P^2 = \tilde P$. The algebra has the two nontrivial idempotents

$$ \tilde\Pi_+ = \tfrac{1}{2}(1 + j), \qquad \tilde\Pi_- = \tfrac{1}{2}(1 - j), \qquad \tilde\Pi_+ \tilde\Pi_- = 0, \qquad \tilde\Pi_+ + \tilde\Pi_- = 1, $$

which are central, lie in the split complex centre $\mathbb{D}_{\mathbb{H}_{\mathbb{D}}}$, and give the idempotent decomposition $\mathbb{H}_{\mathbb{D}} = \mathbb{H} \tilde\Pi_+ \oplus \mathbb{H} \tilde\Pi_-$ into two minimal left ideals, each isomorphic to the quaternion division algebra $\mathbb{H}$.

There are exactly four idempotents, namely $0, \tilde\Pi_+, \tilde\Pi_-, 1$, and all of them are central; equivalently, $\mathbb{H}_{\mathbb{D}} \cong \mathbb{H} \oplus \mathbb{H}$ is a product of two division algebras, and a product of two division algebras has a four-point idempotent set. The nontrivial idempotents $\tilde\Pi_+$ and $\tilde\Pi_-$ are primitive and orthogonal, and together with $0$ and $1$ they exhaust the primitive idempotents. Because there is no bijection with the roots of $-1$ — that set is $S^2 \times S^2$, of dimension $4$ — the biquaternion correspondence between idempotents and roots of $-1$ does not transfer, and the idempotents are confined to the centre. The Hermitian conjugation interchanges $\tilde\Pi_+$ and $\tilde\Pi_-$, so no nontrivial idempotent is Hermitian and the idempotent decomposition is not a decomposition into orthogonal projections. Each nontrivial idempotent is a zero divisor, with annihilator the complementary ideal $\mathbb{H} \tilde\Pi_\mp$ of real dimension $4$; the zero divisors themselves form the union of the two ideals, and are treated in Split-Biquaternion Zero Divisors.

Summary of Notation

Symbol	Meaning
$\tilde P, e, f$	Idempotents, $\tilde P^2 = \tilde P$
$\tilde\Pi_+ = \tfrac{1}{2}(1 + j)$, $\tilde\Pi_- = \tfrac{1}{2}(1 - j)$	The two nontrivial idempotents, $\tilde\Pi_+ + \tilde\Pi_- = 1$
$\mathbb{H}_{\mathbb{D}} = \mathbb{D} \otimes_{\mathbb{R}} \mathbb{H}$	Split biquaternion algebra, $\cong \mathbb{H} \oplus \mathbb{H}$
$\mathbb{D}_{\mathbb{H}_{\mathbb{D}}}$	Centre, the split complex subspace; contains every idempotent
$\mathbb{H} \tilde\Pi_+, \mathbb{H} \tilde\Pi_-$	The two minimal left (and two-sided) ideals, each $\cong \mathbb{H}$
$\tilde\Pi_\pm^\dagger = \tilde\Pi_\mp$	The Hermitian conjugation exchanges the idempotents
$\mathrm{Ann}(\tilde\Pi_\pm) = \mathbb{H} \tilde\Pi_\mp$	Annihilator of an idempotent
$\xi$, $\mu_\pm$	Root of $-1$; unit pure real quaternion components
$S^2 \times S^2$	The topological product form of the root set, established in Split-Biquaternion Analysis
$N(\tilde{Q}) = \tilde{Q}\tilde{Q}^{\natural}$	Split-Biquaternion norm, named only; the subject of Split-Biquaternion Norm and Invertibility