Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space

Introduction

Let $\star$ be a product on the biquaternion algebra $\mathbb{B}$. A nonzero $\tilde Q_2\in\mathbb{B}$ is a zero divisor of $\star$ when it kills a nonzero partner: when $\tilde Q_2\star\tilde Q_1=0$ or $\tilde Q_1\star\tilde Q_2=0$ for some nonzero $\tilde Q_1$. The first case makes it a left zero divisor, the second a right zero divisor, and an element of both kinds a two-sided zero divisor.

The twelve names. The algebra carries twelve products, so the generic definition poses twelve questions; this article answers them once for all twelve. Each of the twelve products has a three-letter code: the first letter is the operation, $\mathrm G$ the general product and $\mathrm S$, $\mathrm A$ its symmetric and antisymmetric parts; the second is the first slot, $\mathrm P$ plain and $\mathrm Q$ quaternionic; the third is the family, $\mathrm A$ the algebra over $\mathbb{C}$ and $\mathrm S$ the sesqualgebra over $\mathbb{C}$. The twelve are read in the four families and the three operations of The Algebraic Structures over the Biquaternion $\mathbb{C}$ Space,

and every table of the article follows the new order: the four general structures first, then their four symmetric parts, then their four antisymmetric parts; and inside each group the plain algebra, then the plain sesqualgebra, then the quaternionic algebra, then the quaternionic sesqualgebra.

The Zero-Divisor Cone

Definition. The zero-divisor cone is the vanishing set of the norm,

$$ \mathcal{N}=\{\tilde Q_2\in\mathbb{B}:\mathrm{Norm}(\tilde Q_2)=0\} , $$

the generic norm of The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, after Conventions in Mathematics §The form convention and the norm convention, the norm of the statement at hand. This article instantiates it by the quaternionic norm, $\mathrm{Norm}(\tilde Q_2)=Q_{2t}^{2}+Q_{2x}^{2}+Q_{2y}^{2}+Q_{2z}^{2}$. It is the common zero-divisor set of the four general products.

Definition. A subset of $\mathbb{B}$ is a complex cone when it contains $\lambda\tilde Q_2$ for every $\tilde Q_2$ it contains and every $\lambda\in\mathbb{C}$. Its apex is the origin.

The Cone Is the Complement of the Group of Units

The elements of $\mathbb{B}$ fall into three classes. The invertible elements of the plain product, the units, are exactly those of non-vanishing norm, $\mathbb{B}^{\times}=\{\tilde Q_2:\mathrm{Norm}(\tilde Q_2)\neq0\}$ (Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors); the origin is never a zero divisor; and the remaining elements are the zero divisors of the four general products. Hence

$$ \mathbb{B}=\{0\}\cup\mathbb{B}^{\times}\cup\bigl(\mathcal{N}\setminus\{0\}\bigr) . $$

So an element is a unit exactly when it lies off the cone, and a zero divisor exactly when it lies on the cone and not at the apex. The cone is the complement of the group of units, and the zero-divisor set together with the origin is that complement read as the cone.

The Twelve Structures and the Cone

The four general products split into the twelve structures through the class-preserving exchange (Symmetric and Antisymmetric Decomposition of the Four General Products). A zero divisor names its product, so each of the twelve carries a zero-divisor set of its own; the table of §Results records them, one family to a block and, inside a block, the general structure first, the product written by its defining formula.

Theorem (the cone is the mark of the four general products). Among the twelve structures the zero-divisor set is the punctured cone $\mathcal{N}\setminus\{0\}$ exactly for the four general structures $\mathrm{GPA}$, $\mathrm{GPS}$, $\mathrm{GQA}$ and $\mathrm{GQS}$, and for no other. The eight derived sets are strictly larger: the six antisymmetric and scalar-valued symmetric structures $\mathrm{SPS}$, $\mathrm{SQA}$, $\mathrm{APA}$, $\mathrm{APS}$, $\mathrm{AQA}$ and $\mathrm{AQS}$ have the whole punctured space, and the two remaining symmetric structures $\mathrm{SPA}$ and $\mathrm{SQS}$ have the enlarged set $(\{Q_{2t}=0\}\cup\mathcal{N})\setminus\{0\}$.

The twelve structures are read one by one below; each names the product, its zero-divisor set and the reason. The four general rows are read together in §Results.

Corollary (the cone is the common core). The punctured cone $\mathcal{N}\setminus\{0\}$ is contained in the zero-divisor set of each of the twelve structures, and it is equal to the set of exactly four of them, the four general products.

Proof. Every row of the table of §Results contains the cone: the four general rows equal it, the rows of $\mathrm{SPA}$ and $\mathrm{SQS}$ contain it by their second clause, and the remaining six rows are the whole punctured space.

Two consequences follow. First, because the cone lies in every set, an element of vanishing norm admits an annihilating factor in all twelve products, and the cone is the largest subset on which the twelve agree; the converse fails, since the derived sets are strictly larger. Second, the ring-theoretic picture breaks on the derived structures: the unit $\vec e_x$, whose norm is $1$, is a zero divisor of $\mathrm{SPA}$ and of $\mathrm{SQS}$, because $\vec e_x\bullet \vec e_y=\tfrac12(\vec e_x \vec e_y+\vec e_y \vec e_x)=0$ for the Jordan product and $\vec e_x\star \vec e_y=0$ for the symmetric quaternionic sesquilinear product, and it is invertible in the algebra. A zero divisor need not be a non-unit unless the product has a two-sided unit.

$\mathrm{GPA}$, the general plain algebra

The product is $\tilde Q_1\tilde Q_2$, its conjugate is $\tilde Q_1^{\natural}$, and the form of the product is

$$ \mathrm{Form}(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_2)=Q_{1t}Q_{2t}+(\vec Q_1,\vec Q_2) . $$

The norm is the form at the two equal arguments, $\mathrm{Norm}(\tilde Q_1)=\mathrm{Form}(\tilde Q_1,\tilde Q_1)$, the scalar of the central element $\tilde Q_1^{\natural}\tilde Q_1$. Write $\tilde Q_1=Q_{1t}e_t+\vec Q_1$ with $\vec Q_1=Q_{1x}\vec e_x+Q_{1y}\vec e_y+Q_{1z}\vec e_z$ and $\tilde Q_1^{\natural}=Q_{1t}e_t-\vec Q_1$. Since $\vec e_\lambda \vec e_\rho+\vec e_\rho \vec e_\lambda=0$ for $\lambda \neq \rho$ and $\vec e_\lambda^{2}=-e_t$, the vector part squares to $\vec Q_1^{2}=-(Q_{1x}^{2}+Q_{1y}^{2}+Q_{1z}^{2})e_t$, so

$$ \tilde Q_1^{\natural}\tilde Q_1=(Q_{1t}e_t-\vec Q_1)(Q_{1t}e_t+\vec Q_1)=Q_{1t}^{2}e_t-\vec Q_1^{2}=(Q_{1t}^{2}+Q_{1x}^{2}+Q_{1y}^{2}+Q_{1z}^{2})e_t , $$

and its scalar is the norm $\mathrm{Norm}(\tilde Q_1)=Q_{1t}^{2}+Q_{1x}^{2}+Q_{1y}^{2}+Q_{1z}^{2}$.

The zero-divisor set is the punctured cone $\mathcal{N}\setminus\{0\}$. The cone is the vanishing set of the norm, $\mathcal{N}=\{\tilde Q_1:Q_{1t}^{2}+Q_{1x}^{2}+Q_{1y}^{2}+Q_{1z}^{2}=0\}$, one complex quadric in the four coordinates, of complex dimension three. On $\mathcal{N}$ the product $\tilde Q_1\tilde Q_1^{\natural}$ vanishes with $\tilde Q_1^{\natural}\neq0$, so every nonzero element of the cone is a zero divisor; off $\mathcal{N}$ the norm is nonzero and $\tilde Q_1^{-1}=\tilde Q_1^{\natural}/\mathrm{Norm}(\tilde Q_1)$ is a two-sided inverse.

The units are the elements off the cone, $\mathbb{B}^{\times}=\{\tilde Q_1:\mathrm{Norm}(\tilde Q_1)\neq0\}$, and this is the one structure of the twelve whose invertible elements form a genuine group, with the two-sided unit $e_t$ and the two-sided inverse $\tilde Q_1^{-1}=\tilde Q_1^{\natural}/\mathrm{Norm}(\tilde Q_1)$; the complement of the zero-divisor set is that group together with the origin, $\mathbb{B}\setminus(\mathcal{N}\setminus\{0\})=\mathbb{B}^{\times}\cup\{0\}$.

$\mathrm{GPS}$, the general plain sesqualgebra

The product is $\tilde Q_1\tilde Q_2^{*}$, its conjugate is $\tilde Q_1^{*}$, and the form of the product is

$$ \mathrm{Form}(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1\tilde Q_2^{*})=Q_{1t}\overline{Q_{2t}}+(\vec Q_1,\overline{\vec Q_2}) . $$

The norm is the form at the two equal arguments, $\langle\tilde Q_1\rangle_{*}=\mathrm{Form}(\tilde Q_1,\tilde Q_1)=\mathrm{Sc}(\tilde Q_1\tilde Q_1^{*})=|Q_{1t}|^{2}+|\vec Q_1|^{2}$, definite.

The zero-divisor set is the punctured cone $\mathcal{N}\setminus\{0\}$. The definite norm $\langle\cdot\rangle_{*}$ vanishes at the origin alone and cuts no cone. Then $\tilde Q_1\tilde Q_2^{*}$ is the plain product, whose form is $\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_2^{*})$ and whose scalar at the two equal arguments is the quaternionic norm $\mathrm{Norm}(\tilde Q_1)=Q_{1t}^{2}+Q_{1x}^{2}+Q_{1y}^{2}+Q_{1z}^{2}$ of §The General Plain Algebra (GPA). The criterion is therefore $\mathrm{Norm}(\tilde Q_1)=0$, the witness is $\tilde Q_2=(\tilde Q_1^{\natural})^{*}$, since $\tilde Q_1\tilde Q_2^{*}=\tilde Q_1\tilde Q_1^{\natural}=\mathrm{Norm}(\tilde Q_1)e_t=0$ on the cone.

The element $e_t$ is a right unit alone, $e_t\star\tilde Q_2=\tilde Q_2^{*}$, and an element off the cone has the two-sided solution $(\tilde Q_1^{-1})^{*}$ of the inverse equations; the invertible elements are the complement of the cone and, the product being non-associative, they form no group.

$\mathrm{GQA}$, the general quaternionic algebra

The product is $\tilde Q_1^{\natural}\tilde Q_2$, its conjugate is $\tilde Q_1^{\natural}$, and the form of the product is

$$ \mathrm{Form}(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_2)=Q_{1t}Q_{2t}+(\vec Q_1,\vec Q_2) . $$

The norm is the form at the two equal arguments, $\mathrm{Norm}(\tilde Q_1)=\mathrm{Form}(\tilde Q_1,\tilde Q_1)$, the scalar of the central element $\tilde Q_1^{\natural}\tilde Q_1$ of §The General Plain Algebra (GPA), $\mathrm{Norm}(\tilde Q_1)=Q_{1t}^{2}+Q_{1x}^{2}+Q_{1y}^{2}+Q_{1z}^{2}$; the natural conjugation sends $\vec Q_1$ to $-\vec Q_1$, so $\mathrm{Norm}(\tilde Q_1^{\natural})=\mathrm{Norm}(\tilde Q_1)$. Here the norm is the criterion.

The zero-divisor set is the punctured cone $\mathcal{N}\setminus\{0\}$. The cone is the quadric $\mathcal{N}$ of §The General Plain Algebra (GPA). If $\mathrm{Norm}(\tilde Q_1)=0$ the partner $\tilde Q_2=\tilde Q_1^{\natural}$ annihilates, since $\tilde Q_1^{\natural}\tilde Q_1^{\natural}=\mathrm{Norm}(\tilde Q_1^{\natural})e_t=0$; conversely, multiplying $\tilde Q_1^{\natural}\tilde Q_2=0$ on the left by $\tilde Q_1$ gives $\mathrm{Norm}(\tilde Q_1)\tilde Q_2=0$.

The element $e_t$ is a left unit alone, and an element off the cone has the two-sided solution $(\tilde Q_1^{-1})^{\natural}$ of the inverse equations; the invertible elements are the complement of the cone, and since the product is not associative they form no group.

$\mathrm{GQS}$, the general quaternionic sesqualgebra

The product is $\tilde Q_1^{\natural}\tilde Q_2^{*}$, its conjugates are $\natural$ and $*$, and the form of the product is

$$ \mathrm{Form}(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_2^{*})=Q_{1t}\overline{Q_{2t}}-(\vec Q_1,\overline{\vec Q_2}) . $$

The norm is the form at the two equal arguments, $\langle\tilde Q_1\rangle_{\natural*}=\mathrm{Form}(\tilde Q_1,\tilde Q_1)=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_1^{*})=|Q_{1t}|^{2}-|\vec Q_1|^{2}$, indefinite.

The zero-divisor set is the punctured cone $\mathcal{N}\setminus\{0\}$. The indefinite norm $\langle\cdot\rangle_{\natural*}$ vanishes on a real cone of its own, distinct from $\mathcal{N}$. Then $\tilde Q_1^{\natural}\tilde Q_2^{*}$ is the plain product of $\tilde Q_1^{\natural}$, whose form is $\mathrm{Sc}(\tilde Q_1\tilde Q_2^{*})$ and whose scalar at the two equal arguments is $\mathrm{Norm}(\tilde Q_1^{\natural})=\mathrm{Norm}(\tilde Q_1)$ of §The General Plain Algebra (GPA). The criterion is therefore $\mathrm{Norm}(\tilde Q_1)=0$, the witness is $\tilde Q_2=(\tilde Q_1^{\natural})^{*}$.

There is no unit on either side, and an element off the cone has the two-sided solution $(\tilde Q_1^{-1})^{\natural*}$ of the inverse equations; the elements off the cone satisfy both inverse equations, but they form no group, and the structure carries no unit.

$\mathrm{SPA}$, the symmetric plain algebra

The product is the Jordan product $\tfrac12(\tilde Q_1\tilde Q_2+\tilde Q_2\tilde Q_1)$, the symmetrisation of the plain product, so its form is the same,

$$ \mathrm{Form}(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_2)=Q_{1t}Q_{2t}+(\vec Q_1,\vec Q_2) , $$

The norm is the form at the two equal arguments, $\mathrm{Norm}(\tilde Q_1)=\mathrm{Form}(\tilde Q_1,\tilde Q_1)$, the quaternionic norm $\mathrm{Norm}(\tilde Q_1)=Q_{1t}^{2}+Q_{1x}^{2}+Q_{1y}^{2}+Q_{1z}^{2}$ of §The General Plain Algebra (GPA).

The zero-divisor set is the enlarged set

$$ \boxed{\; \bigl(\{Q_{1t}=0\}\cup\mathcal{N}\bigr)\setminus\{0\} \;}. $$

Read the Jordan product from the quaternion table, $\tilde Q_1\circ\tilde Q_2=Q_{1t}Q_{2t}e_t+Q_{1t}\vec Q_2+Q_{2t}\vec Q_1-(\vec Q_1,\vec Q_2)e_t$. If $Q_{1t}\neq0$ the vector part forces $\vec Q_2=-(Q_{2t}/Q_{1t})\vec Q_1$, and the scalar part becomes $Q_{2t}\,\mathrm{Norm}(\tilde Q_1)/Q_{1t}$; a nonzero annihilator therefore exists exactly when $\mathrm{Norm}(\tilde Q_1)=0$, and it is then $\tilde Q_1^{\natural}$. If $Q_{1t}=0$ and $\tilde Q_1\neq0$, any nonzero $\vec Q_2$ orthogonal to $\vec Q_1$ annihilates.

The plane $Q_{1t}=0$ of the pure elements is a second family beside the cone.

The unit $e_t$ is two-sided, as for $\mathrm{GPA}$, and the invertible elements are again the elements of non-vanishing norm; but the product is not associative, so they form no group, and the complement of the zero-divisor set, $\{Q_{1t}\neq0,\ \mathrm{Norm}(\tilde Q_1)\neq0\}\cup\{0\}$, is not the invertible set: the pure element $\vec e_x$ is a zero divisor and invertible at once, $\vec e_x\bullet(-\vec e_x)=e_t$, the failure of zero divisor if and only if non-unit.

$\mathrm{SPS}$, the symmetric plain sesqualgebra

The product is $\tfrac12(\tilde Q_1\tilde Q_2^{*}+\overline{\tilde Q_2\tilde Q_1^{*}})$, central, and the form of the product is

$$ \mathrm{Form}(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1\tilde Q_2^{*})=Q_{1t}\overline{Q_{2t}}+(\vec Q_1,\overline{\vec Q_2}) , $$

The norm is the form at the two equal arguments, $\langle\tilde Q_1\rangle_{*}=\mathrm{Form}(\tilde Q_1,\tilde Q_1)=|Q_{1t}|^{2}+|\vec Q_1|^{2}$, definite.

The zero-divisor set is the whole punctured space $\mathbb{B}\setminus\{0\}$. The value is the central read of the form, $\mathrm{Sc}(\tilde Q_1\tilde Q_2^{*})e_t$. The norm $\langle\cdot\rangle_{*}$ is definite and does not decide; the equation $\mathrm{Sc}(\tilde Q_1\tilde Q_2^{*})=0$ is one complex linear equation in the conjugated components of $\tilde Q_2$, and its solutions fill a complex hyperplane.

There is no unit, the values lying in the centre, so there is no group of units and the complement of the zero-divisor set, the whole punctured space, is the origin alone.

$\mathrm{SQA}$, the symmetric quaternionic algebra

The product is $\tfrac12(\tilde Q_1^{\natural}\tilde Q_2+\tilde Q_2^{\natural}\tilde Q_1)$, central, and the form of the product is

$$ \mathrm{Form}(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_2)=Q_{1t}Q_{2t}+(\vec Q_1,\vec Q_2) , $$

The norm is the form at the two equal arguments, $\mathrm{Norm}(\tilde Q_1)=\mathrm{Form}(\tilde Q_1,\tilde Q_1)$, the quaternionic norm $\mathrm{Norm}(\tilde Q_1)=Q_{1t}^{2}+Q_{1x}^{2}+Q_{1y}^{2}+Q_{1z}^{2}$.

The zero-divisor set is the whole punctured space $\mathbb{B}\setminus\{0\}$. The value is the central read of the form, $\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_2)e_t$; check on $\tilde Q_1=\tilde Q_2=\vec e_x$: $\tilde Q_1^{\natural}\tilde Q_1=e_t$, of scalar $1=0+1$. The form is nondegenerate, its matrix being the identity in the four coordinates, so for $\tilde Q_1\neq0$ the equation $\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_2)=0$ is one complex linear equation in the four components of $\tilde Q_2$ and its solutions fill a complex hyperplane of nonzero vectors.

There is no unit — the values lie in the centre and $e_t$ acts as $\langle e_t,\tilde Q_2\rangle_{\natural}e_t=Q_{2t}e_t$, not as an identity — so there is no group of units and the complement of the zero-divisor set, the whole punctured space, is the origin alone.

$\mathrm{SQS}$, the symmetric quaternionic sesqualgebra

The product is $\tfrac12(\tilde Q_1^{\natural}\tilde Q_2^{*}+\overline{\tilde Q_2^{\natural}\tilde Q_1^{*}})$, central, and the form of the product is

$$ \mathrm{Form}(\tilde Q_1,\tilde Q_2)=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_2^{*})=Q_{1t}\overline{Q_{2t}}-(\vec Q_1,\overline{\vec Q_2}) , $$

The norm is the form at the two equal arguments, $\langle\tilde Q_1\rangle_{\natural*}=\mathrm{Form}(\tilde Q_1,\tilde Q_1)=|Q_{1t}|^{2}-|\vec Q_1|^{2}$, indefinite.

The zero-divisor set is the enlarged set $\left(\{Q_{1t}=0\}\cup\mathcal{N}\right)\setminus\{0\}$. The two conjugations commute and the product becomes the Jordan product of $\tilde Q_1^{\natural}$ with $\tilde Q_2^{*}$,

$$ \tfrac12(\tilde Q_1^{\natural}\tilde Q_2^{*}+\overline{\tilde Q_2^{\natural}\tilde Q_1^{*}})=\tfrac12(\tilde Q_1^{\natural}\tilde Q_2^{*}+\tilde Q_2^{*}\tilde Q_1^{\natural}). $$

By §The Symmetric Plain Algebra (SPA) the criterion is read on $\tilde Q_1^{\natural}$: the natural conjugation leaves $Q_{1t}$ unchanged and $\mathrm{Norm}(\tilde Q_1^{\natural})=\mathrm{Norm}(\tilde Q_1)$, so the condition is $Q_{1t}=0$ or $Q_{1t}^{2}+Q_{1x}^{2}+Q_{1y}^{2}+Q_{1z}^{2}=0$, (Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space).

There is no unit, both symmetrised terms conjugating a factor, so $e_t$ acts as $\tilde Q_2^{*}$ on one side and $\tilde Q_2^{\natural}$ on the other; the complement of the enlarged zero-divisor set is $\{Q_{1t}\neq0,\ \mathrm{Norm}(\tilde Q_1)\neq0\}\cup\{0\}$, which is no group of units.

$\mathrm{APA}$, the antisymmetric plain algebra

The product is the commutator $\tfrac12(\tilde Q_1\tilde Q_2-\tilde Q_2\tilde Q_1)$. It is alternating, so at the two equal arguments it vanishes, $\tilde Q_1\star\tilde Q_1=\tfrac12(\tilde Q_1\tilde Q_1-\tilde Q_1\tilde Q_1)=0$; there is no form there and no norm.

The zero-divisor set is the whole punctured space $\mathbb{B}\setminus\{0\}$. Every nonzero element is annihilated by itself.

There is no unit, the values lying in the vector subspace, so there is no group of units and the complement of the zero-divisor set — the whole punctured space — is the origin alone.

$\mathrm{APS}$, the antisymmetric plain sesqualgebra

The product is $\tfrac12(\tilde Q_1\tilde Q_2^{*}-\overline{\tilde Q_2\tilde Q_1^{*}})=\mathrm{Vect}(\tilde Q_1\tilde Q_2^{*})$. At the two equal arguments it gives the vector part of $\tilde Q_1\tilde Q_1^{*}$, not a scalar; there is no norm.

The zero-divisor set is the whole punctured space $\mathbb{B}\setminus\{0\}$. Off the cone, $\mathrm{Norm}(\tilde Q_1)\neq0$ and $\tilde Q_2=(\tilde Q_1^{-1})^{*}$ gives $\tilde Q_1\tilde Q_2^{*}=e_t$, whose vector part is zero; on the cone, $\tilde Q_2=\tilde Q_1^{\natural}$ gives $\tilde Q_1\tilde Q_2^{*}=0$. Every nonzero element is a zero divisor.

There is no unit, the values lying in the vector subspace, so there is no group of units and the complement of the zero-divisor set is the origin alone.

$\mathrm{AQA}$, the antisymmetric quaternionic algebra

The product is the commutator $\tfrac12(\tilde Q_1^{\natural}\tilde Q_2-\tilde Q_2^{\natural}\tilde Q_1)$. It is alternating, $\tilde Q_1\star\tilde Q_1=\tfrac12(\tilde Q_1^{\natural}\tilde Q_1-\tilde Q_1^{\natural}\tilde Q_1)=0$, so at the two equal arguments there is no form and no norm.

The zero-divisor set is the whole punctured space $\mathbb{B}\setminus\{0\}$. Every nonzero element is annihilated by itself.

There is no unit, the values lying in the vector subspace, so there is no group of units and the complement of the zero-divisor set is the origin alone.

$\mathrm{AQS}$, the antisymmetric quaternionic sesqualgebra

The product is $\tfrac12(\tilde Q_1^{\natural}\tilde Q_2^{*}-\overline{\tilde Q_2^{\natural}\tilde Q_1^{*}})=\vec Q_1\times\overline{\vec Q_2}$. At the two equal arguments it gives $\vec Q_1\times\overline{\vec Q_1}$, a pure vector; there is no norm.

The zero-divisor set is the whole punctured space $\mathbb{B}\setminus\{0\}$. If $\vec Q_1\neq0$, take $\tilde Q_2=\overline{\tilde Q_1}$: its vector part $\overline{\vec Q_1}$ is parallel to $\vec Q_1$, and a cross product of parallel vectors vanishes. If $\vec Q_1=0$ the element is central and the product vanishes for every $\tilde Q_2$. Every nonzero element is a zero divisor.

There is no unit, the values lying in the vector subspace, so there is no group of units and the complement of the zero-divisor set is the origin alone.

Results

The four general products share one zero-divisor set, the punctured cone $\mathcal{N}\setminus\{0\}$ of §The Zero-Divisor Cone. The twelve structures are tabulated against it, the four that coincide are separated out, and the cone itself is read in coordinates and in its two families.

The Twelve Zero-Divisor Sets

Read against the cone $\mathcal{N}$ and against the whole space, the twelve sets are

The four general products have one and the same set, the punctured cone. The six structures $\mathrm{SPS}$, $\mathrm{SQA}$, $\mathrm{APA}$, $\mathrm{APS}$, $\mathrm{AQA}$ and $\mathrm{AQS}$ have the whole punctured space. The two remaining symmetric structures $\mathrm{SPA}$ and $\mathrm{SQS}$ have the enlarged set $(\{Q_{2t}=0\}\cup\mathcal{N})\setminus\{0\}$. The cone is therefore the zero-divisor set of exactly four of the twelve structures, and it is contained in all twelve, so that it is the common core of the twelve. The theorem below and §The Twelve Structures and the Cone prove the table, and the sections that follow read its consequences.

The Four General Products Coincide

Theorem (the four zero-divisor sets coincide, and they are the cone). For the four general products the zero-divisor sets coincide. A nonzero element $\tilde Q_1$ is a left, a right or a two-sided zero divisor of one of the four general products exactly when $\mathrm{Norm}(\tilde Q_1)=0$, that is exactly when $\tilde Q_1$ lies on the cone $\mathcal{N}$, and then it is a zero divisor of all four, on both sides. In symbols,

$$ \boxed{\; \{\tilde Q_1\neq0:\tilde Q_1\text{ admits an annihilating factor in }\star\}=\mathcal{N}\setminus\{0\} \qquad\text{for each of the four general products }\star \;}. $$

Proof. Read each product in the model $\mathsf{M}_2$. Each general product is the plain product with a conjugation in one or both slots, a bijection of $\mathbb{B}$; a bijection in a slot identifies the kernel of the left multiplication by $\tilde Q_1$ with the kernel of the plain left multiplication by $\tilde Q_1$ or by $\tilde Q_1^{\natural}$. Now $\mathsf{M}_2(\tilde Q_1^{\natural})$ is $\mathsf{M}_2(\tilde Q_1)$ up to conjugation by a fixed invertible matrix, and $\mathrm{Norm}(\tilde Q_1^{\natural})=\mathrm{Norm}(\tilde Q_1)$, so the plain left multiplication is singular exactly when $\mathrm{Norm}(\tilde Q_1)=0$. In each case the deciding condition is the vanishing of the norm, read from the singularity of $\mathsf{M}_2(\tilde Q_1)$ or of its conjugate $\mathsf{M}_2(\tilde Q_1^{\natural})$. For the cone itself and for a nonzero $\tilde Q_1$ with $\mathrm{Norm}(\tilde Q_1)=0$, the witness is the natural conjugate, $\tilde Q_1\tilde Q_1^{\natural}=\tilde Q_1^{\natural}\tilde Q_1=\mathrm{Norm}(\tilde Q_1)e_t=0$, on both sides.

The four conditions are thus one condition, and the reason is structural. Each general product is the plain product with a conjugation in one or both slots; each conjugation is a bijection that either fixes or conjugates the norm; and the norm is the single invariant that decides the invertibility of the element. Composing a one-sided multiplication with a bijection changes its kernel by the bijection, so the singularity is transported from product to product.

The four rows are identical, and that is the theorem. The word is coincide, not independent. The zero divisor is defined by an equation that names a product, so the notion belongs to the product and the four products are four definitions, not one. What the theorem says is that in the biquaternion algebra the four definitions select one and the same subset of $\mathbb{B}$; it does not say that they must, and it does not extend to the eight derived structures of §The Twelve Structures and the Cone.

The Cone in Coordinates and Its Apex

Written on the coordinates, the cone common to the four general products is one complex equation in the four complex coordinates,

$$ \boxed{\; Q_{2t}^{2}+Q_{2x}^{2}+Q_{2y}^{2}+Q_{2z}^{2}=0 \;}. $$

Proposition (the cone and its singular point). The set $\mathcal{N}$ is a complex cone with apex at the origin, and the origin is its only singular point. Away from the apex it is a complex manifold of complex dimension three, hence a real manifold of real dimension six.

Proof. The norm is homogeneous of degree two, $\mathrm{Norm}(\lambda\tilde Q_2)=\lambda^{2}\mathrm{Norm}(\tilde Q_2)$, so its vanishing set is a cone. One complex equation in four complex unknowns cuts a complex hypersurface of $\mathbb{C}^{4}$, of complex dimension three wherever its gradient does not vanish; the gradient of $\mathrm{Norm}$ is proportional to $(Q_{2t},Q_{2x},Q_{2y},Q_{2z})$, which vanishes only at the origin, so the complement of the apex is smooth of complex dimension three. The origin is singular because the only term of $\mathrm{Norm}$ of degree two is the leading term, so the tangent cone at the origin is $\mathcal{N}$ itself and not a hyperplane.

The Two Families

The scalar part cuts the cone into two families, and the criterion and the square read differently on each.

Definition. An element is pure when its scalar part vanishes, $Q_{2t}=0$, and non-pure when $Q_{2t}\neq0$.

On a pure element the scalar square $Q_{2t}^{2}$ drops out of the criterion, which becomes $(\vec Q_2,\vec Q_2)=0$; on a non-pure element the criterion is the full vanishing of the norm.

The square of a pure zero divisor vanishes because $\tilde Q_2^{2}=-(Q_{2x}^{2}+Q_{2y}^{2}+Q_{2z}^{2})e_t=0$; the square of a non-pure one satisfies $\tilde Q_2^{2}=2Q_{2t}\tilde Q_2-\mathrm{Norm}(\tilde Q_2)e_t=2Q_{2t}\tilde Q_2$, and the element $\tilde\Pi=\tilde Q_2/(2Q_{2t})$ is then a nontrivial idempotent of the plain product. The two families are read in the remarkable subspaces in Introduction to the Remarkable Subspaces: the pure ones fill the nilpotent cone of the vector subspace, the non-pure ones the double cones of the two Hermitian subspaces, and a generic point of the cone lies in none of the remarkable subspaces.

The Cone in the Wider Mathematics

The results above are the cone read inside the grid. The corpus proves five further statements about the cone that name no product, and they are gathered here because they are the cone's own geometry and topology rather than a property of any of the twelve.

One cone among four null sets. The space carries four forms — the two bilinear $\langle\cdot,\cdot\rangle$ and $\langle\cdot,\cdot\rangle_{\natural}$ and the two sesquilinear $\langle\cdot,\cdot\rangle_{*}$ and $\langle\cdot,\cdot\rangle_{\natural*}$ — and their null sets have four different shapes. $\langle\cdot,\cdot\rangle$ has the complex cone $\sum_\mu\varepsilon_\mu Q_{2,\mu}^{2}=0$ of real dimension six, the Hermitian form is positive definite and has the origin alone, $\langle\cdot\rangle_{*}$ has the zero-divisor cone $\mathcal{N}$ of real dimension six, and the Krein form is indefinite and has the Krein cone $\mathcal{K}$ of real dimension seven (The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, The Krein Level Sets and the Hyperbolic Structure). The two dimension-six null sets do not agree, so the cone of this article is the null set of $\langle\cdot,\cdot\rangle_{\natural}$ and of no other form.

On the remarkable subspaces the four null sets are cut to two shapes (Remarkable Subspaces and the Four General Products, Forms and Norms). On the centre all four vanish at the origin alone. On the vector subspace the plain and the quaternionic forms vanish on the complex cone of real dimension four, the pure zero divisors, while the Hermitian and the Krein forms vanish at the origin. On the quaternion subspace the plain and the Krein forms vanish on the real null cone of real dimension three. On each sector the two indefinite forms $\langle\cdot,\cdot\rangle_{\natural}$ and $\langle\cdot,\cdot\rangle_{\natural*}$ vanish on the null cone of real dimension three, while the two definite forms $\langle\cdot,\cdot\rangle$ and $\langle\cdot,\cdot\rangle_{*}$ vanish at the origin. The cone's real dimension therefore drops from six to four to three as the subspace shrinks, and at every stage the cone is carried by the indefinite forms alone.

The cone as a topological space. The full cone is contractible, its apex is its only singular point, and its link, the intersection with the unit sphere $S^{7}_{E}$, is a closed connected real five-manifold. The link is $S^{2}\times S^{3}$; equivalently it is an $S^{1}$-bundle over $\mathbb{P}^{1}\times\mathbb{P}^{1}\cong S^{2}\times S^{2}$, with $\pi_1=0$, $\pi_2\cong\mathbb{Z}$ and $\pi_3\cong\mathbb{Z}\oplus\mathbb{Z}$; the projectivised cone is the Segre quadric, whose two rulings are the two families of chiral null planes (The Topology of the Zero-Divisor Cone). On a Lorentzian slice the cone cuts the three-dimensional null cone, whose link is the disjoint union $S^{2}\sqcup S^{2}$, one sphere per nappe (The Topology of the Zero-Divisor Cone). The one cone carries two links, the complex one $S^{2}\times S^{3}$ and the real one $S^{2}\sqcup S^{2}$.

The cone is the complement of the unit group. The units are exactly the elements off the cone, and the group they form is $\mathbb{B}^{\times}\simeq S^{1}\times S^{3}$, with $\pi_1(\mathbb{B}^{\times})\cong\mathbb{Z}$, $\pi_2(\mathbb{B}^{\times})=0$ and $\pi_3(\mathbb{B}^{\times})\cong\mathbb{Z}$, covered two-to-one by the centre and the determinant-one group (The Biquaternion Unit Group as a Topological Group). The cone and the unit group are complementarily read in homotopy: the link has $\pi_2\cong\mathbb{Z}$, and the unit group has $\pi_2=0$.

The cone's isotropic lines are its Peirce lines. A doubly null line is an isotropic line that lies in the norm cone as well, and the doubly null lines are exactly the Peirce lines $\mathbb{C}(e_t\pm i\vec u)$ spanned by the rank-one Hermitian idempotents; each generates a minimal left and a minimal right ideal of complex dimension two all of whose elements are zero divisors, and the intersection of the cone with the Krein null set is the union of those lines, a cone of real dimension four (The Krein Norm-Degenerate Sector, Ideals and Peirce Decompositions of the Biquaternion Algebra). The maximal totally isotropic subspaces are the objects of the Witt index, and the maximal one is not confined to the cone: the complex line $\mathbb{C}(e_t+\vec e_x)$ is totally isotropic but has nonzero norm and is not a zero divisor.

The cone is the boundary of the Hermitian level sets. The level sets of the Hermitian form are the real hyperboloids, the hyperboloid model of hyperbolic space, against the complex hypersurfaces of $\langle\cdot,\cdot\rangle_{\natural}$; on a Lorentzian slice the intersection degenerates to the two nappes of the null cone, and the future nappe is the Hermitian cone (The Krein Level Sets and the Hyperbolic Structure). The cone is thus the boundary of the real hyperbolic picture attached to the definite form, and it is the largest nontrivial vanishing set of a square: the natural square $\tilde Q_2\tilde Q_2^{\natural}$ vanishes exactly on the cone, while the sesquilinear square $\tilde Q_2\star\tilde Q_2=\tilde Q_2\tilde Q_2^{*}$ vanishes at the origin alone (The Squares and the Positive Cone of the Biquaternion Sesqualgebra).

These five statements are not about the twelve products. They are about the one quadratic form the twelve inherit, and they are the reason the cone is the meeting point of the grid rather than one of its cells.

The One-Sided Inverse Equations

Only the plain product has a two-sided unit, so the classical equivalence zero divisor if and only if not a unit is available for it alone, and there it is exact: for a nonzero $\tilde Q_1$ the statements $\tilde Q_1$ is a zero divisor of the plain product, $\mathrm{Norm}(\tilde Q_1)=0$ and $\tilde Q_1$ is not invertible are equivalent (Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors). The substitute for the other three products is the one-sided inverse criterion.

Theorem (the inverse equations). Let $\star$ be one of the four general products and let $\tilde Q_1\neq0$. Then $\tilde Q_1$ is a zero divisor of $\star$ exactly when neither of the two equations

$$ \tilde Q_1\star\tilde Q_3=e_t , \qquad \tilde Q_3\star\tilde Q_1=e_t $$

has a solution; equivalently, both have a solution exactly when $\mathrm{Norm}(\tilde Q_1)\neq0$.

Proof. If $\tilde Q_1\star\tilde Q_3=e_t$ holds, then in the model the equation is $\mathsf{M}_2(\tilde Q_1)M=\mathrm{I}_2$ or $\mathsf{M}_2(\tilde Q_1^{\natural})M=\mathrm{I}_2$ for an invertible $M$, so $\mathsf{M}_2(\tilde Q_1)$ is invertible and $\mathrm{Norm}(\tilde Q_1)\neq0$. Conversely, if $\mathrm{Norm}(\tilde Q_1)\neq0$ then $\tilde Q_1$ is invertible and the four solutions are those of the table below.

The solutions are written with the inverse $\tilde Q_1^{-1}=\tilde Q_1^{\natural}/\mathrm{Norm}(\tilde Q_1)$ of the algebra, which exists for $\mathrm{Norm}(\tilde Q_1)\neq0$. The two conjugations commute, ${}^{\natural}\circ{}^{*}={}^{*}\circ{}^{\natural}=\overline{\cdot}$, so the order of the two superscripts in the last row is immaterial.

This criterion uses the fixed element $e_t$, invariant under both conjugations, and makes no appeal to a unit of the product; it reads the same in the four products, and it says that the zero divisors are exactly the elements that cannot be inverted on either side.

Summary

A zero divisor is defined by an equation that names a product, so the twelve products of the biquaternion algebra define twelve zero-divisor sets. For the four general products the sets coincide: for each of them, and for a nonzero element, being a left zero divisor, being a right zero divisor and lying on the cone are the same condition, the cone being the vanishing set of the norm $\mathrm{Norm}(\tilde Q_2)=\sum_\mu Q_{2,\mu}^{2}=\mathrm{Sc}(\tilde Q_2^{\natural}\tilde Q_2)$ of The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra.

The common set is the zero-divisor set $\mathcal{N}\setminus\{0\}$, a complex cone with apex at the origin, smooth away from the apex, of complex dimension three and real dimension six, whose nonzero members are the rank-one elements of the matrix model and whose left ideals are the minimal left ideals. Its complement is the group of units together with the origin, and it splits by the scalar part into the pure family of the nilpotents, of real dimension four, and the non-pure family of the complex multiples of the nontrivial idempotents, which fills the cone in real dimension six.

The annihilators of a fixed element of the cone, two complex planes per product, are Annihilators in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space; the square-zero sets are Nilpotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space. The cone is the zero-divisor set of the four general products alone: the four antisymmetric structures and the two scalar-valued symmetric structures $\mathrm{SPS}$ and $\mathrm{SQA}$ have the whole punctured space, and $\mathrm{SPA}$ and $\mathrm{SQS}$ have $(\{Q_{2t}=0\}\cup\mathcal{N})\setminus\{0\}$. The cone is contained in all twelve sets and is the largest subset on which they agree. What a product does not change is the detection of the units: an element off the cone is invertible, and the one-sided inverse equations replace the classical equivalence for the three products that have no two-sided unit.

Beyond the grid the same set is one of four null sets. The bilinear form $\langle\cdot,\cdot\rangle$ has a complex cone of real dimension six different from $\mathcal{N}$, the Hermitian form $\langle\cdot,\cdot\rangle_{*}$ has the origin alone, and the Krein form $\langle\cdot,\cdot\rangle_{\natural*}$ has the Krein cone of real dimension seven; on the remarkable subspaces the cone drops from real dimension six to four on the vector subspace and to three on the quaternion subspace and on the sectors, carried at every stage by the indefinite forms. As a topological space the cone is contractible with the apex as its only singular point, its link over $S^{7}_{E}$ is an $S^{1}$-bundle over $S^{2}\times S^{2}$ with $\pi_y\cong\mathbb{Z}$ against the link $S^{2}\sqcup S^{2}$ of its Lorentzian slice, and its projectivisation is the Segre quadric. Its complement is the unit group $\mathbb{B}^{\times}\simeq S^{1}\times S^{3}$; its isotropic lines are the Peirce lines, the doubly null lines, and the cone meets the Krein null set in their union; its level-set companion is the real hyperboloid of the Hermitian form. These statements name no product, and they are the content of §The Cone in the Wider Mathematics.

Summary of Notation

Further Reading

• Definitions for the Study of the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the general vocabulary: the zero divisor, the annihilator, the idempotent, the nilpotent and the rest, each stated once for a general product.
• The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, for the form $\langle\cdot,\cdot\rangle_{\natural}$ and the four algebraic norms read at the two equal arguments.
• Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, for the plain-product criterion, the classification of the cone's elements and the parametrisation of the pure family.
• The Four General Products and Their Relations over the Biquaternion Algebra, for the property table, the four idempotent sets, the four square-root problems and the four unit entries.
• Symmetric and Antisymmetric Decomposition of the Four General Products, for the four general products and their symmetric and antisymmetric parts.
• The Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the twelve names, the three letters of each and the class of each.
• Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the four idempotent sets of the four general products and their reading among the twelve.
• Square Roots of a General Element in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the four root problems.
• Annihilators in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the two one-sided annihilators of one element of the cone and their relation by the conjugations.
• Square Roots of Zero and the Nilpotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the four square-zero sets and the twelve square-zero readings.
• Introduction to the 2×2 Matrix Element Representation $M_2(\mathbb{C})$ of Biquaternions, for the isomorphism and the singular matrices.
• The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, for the general plain bilinear form and its isotropic cone.
• Introduction to the Remarkable Subspaces, for the subspaces, the pure cone and the non-pure families.
• Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the criterion $Q_{2t}=0$ or $\mathrm{Norm}(\tilde Q_2)=0$ of the enlarged set of $\mathrm{SPA}$.
• Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the same criterion of the enlarged set of $\mathrm{SQS}$ and its two families.
• Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the two scalar-valued symmetric structures whose zero-divisor set is the whole punctured space.
• Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the four alternating structures whose zero-divisor set is the whole punctured space.
• Zero Divisors of the General Quaternionic Algebra, for the natural product's criterion and the coincidence of its two annihilators.
• The Square of the General Quaternionic Sesquilinear Product and the Two Halves, for the equation $\overline{\tilde Q_1}\tilde Q_1=0$ and its two halves.
• The Squares and the Positive Cone of the Biquaternion Sesqualgebra, for the definiteness of the general plain sesquilinear square.
• Ideals and Peirce Decompositions of the Biquaternion Algebra, for the minimal left ideals of the rank-one elements.
• The Isotropic Structure of the General Quaternionic Algebra, for the isotropic cones of the two bilinear forms, which are not the cone $\mathcal{N}$.
• The Topology of the Zero-Divisor Cone, for the cone as a topological space, the Segre quadric, its two rulings, the projectivised cone and the Lorentzian slice.
• Remarkable Subspaces and the Four General Products, Forms and Norms and The Krein Level Sets and the Hyperbolic Structure, for the four null sets cut by the remarkable subspaces, the Krein cone and the real hyperboloids whose boundary the cone is.
• The Biquaternion Unit Group as a Topological Group, for the units as the complement of the cone, the covering and the homotopy groups.
• The Krein Norm-Degenerate Sector, for the Peirce lines, the minimal ideals and the doubly null lines in which the cone meets the Krein null set.