Norm and Invertibility in the Symmetric Plain Sesqualgebra
Introduction
The symmetric plain sesqualgebra is the operation $\tilde Q_1\star\tilde Q_2 = \langle\tilde Q_1,\tilde Q_2\rangle_{*}e_t$ on the biquaternion space, the Hermitian form $\langle\cdot,\cdot\rangle_{*}$ read as a multiplication (Introduction to the Symmetric Plain Sesqualgebra). This article studies the algebraic norm that the block carries and the invertibility that it does not.
A norm and an inverse name a product, and the algebraic norm of a product is the diagonal of its scalar-part form, the definition of The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra. For the block the form is $\langle\tilde Q_1,\tilde Q_2\rangle_{*}=\mathrm{Sc}(\tilde Q_1\star\tilde Q_2)$ and its diagonal is
$$ \langle\tilde Q\rangle_{*}=\langle\tilde Q,\tilde Q\rangle_{*}=\sum_{\mu\in\{t,x,y,z\}}\lvert Q_\mu\rvert^{2}, $$
the symmetric plain sesquilinear algebraic norm, real and positive definite, one of the four algebraic norms of the space and not one of the other three. The two arities of the letter are one object: with two arguments the form, with one argument its diagonal.
The invertibility is where the article records an absence. The block is not unital: every value of the product is central, so it has no two-sided unit, and consequently no inverse, no group of units and no division structure. Its algebraic norm does not repair the absence: $\langle\cdot\rangle_{*}$ is positive definite, so $\langle\tilde Q\rangle_{*}=0$ exactly at $\tilde Q=0$, the block has no nonzero isotropic vector and no invertible element, and its element classification is two-way, the origin and everything else. The article states the absence, then the algebraic norm, then the failure of multiplicativity and the polarisation.
The results. Every value of the product $\tilde Q_1\star\tilde Q_2 = \langle\tilde Q_1,\tilde Q_2\rangle_{*}e_t$ is central, so the block has no unit, no inverse and no group of units, and no element is invertible. The block carries one algebraic norm, $\langle\tilde{Q}\rangle_{*} = \sum_{\mu\in\{t,x,y,z\}}\lvert Q_\mu\rvert^{2}$, real, non-negative and positive definite of signature $(8,0)$ on the real space underlying $\mathbb{B}$, vanishing only at $\tilde{Q} = 0$, with image of its square map the real ray $\mathbb{R}_{\geq0}e_t$. The algebraic norm is not multiplicative: $\langle\tilde Q_1\star\tilde Q_2\rangle_{*} = \lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2}$, with defect the discriminant $\langle\tilde Q_1\rangle_{*}\langle\tilde Q_2\rangle_{*} - \lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2}\geq0$ vanishing exactly on linearly dependent pairs, and it polarises to the Hermitian form. The definiteness is why it cannot decide invertibility: $\langle\tilde{Q}\rangle_{*} = 0$ exactly at the origin, so the block has no nonzero isotropic vector and its element classification is two-way.
The Relative Notions
The word unit, the word inverse and the word group of units name a product, so the definitions are stated once for a general product and then read on the products of this space; they are in Definitions for the Study of the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space.
Definition (recalled). Let $\star$ be a product on $\mathbb{B}$. An element $e$ is a left unit of $\star$ when $e\star\tilde Q_1 = \tilde Q_1$ for every $\tilde Q_1\in\mathbb{B}$, a right unit when $\tilde Q_1\star e = \tilde Q_1$ for every $\tilde Q_1\in\mathbb{B}$, and a two-sided unit, or an identity, when it is both. When a two-sided unit $e$ exists, an element $\tilde Q_1$ is invertible of $\star$ when there is a $\tilde Q_2$ with $\tilde Q_1\star\tilde Q_2 = \tilde Q_2\star\tilde Q_1 = e$, the element $\tilde Q_2$ being the inverse of $\tilde Q_1$ of $\star$; the group of units of $\star$ is the set of the invertible elements, a group under $\star$. When no two-sided unit exists, the three notions are empty and the group of units is not defined.
Throughout this article the product is the symmetric plain sesqualgebra $\star$, and another product is named where it is meant.
The Absence of a Unit
The block has no unit of any kind: there is no left, no right and no two-sided unit $e$, because every value of the product is central and the block has non-central elements.
Theorem (the block has no unit). The block has no left unit and no right unit: there is no element $e$ with $e\star\tilde{Q} = \tilde{Q}$ for every $\tilde{Q}\in\mathbb{B}$, and none with $\tilde{Q}\star e = \tilde{Q}$ for every $\tilde{Q}\in\mathbb{B}$. Hence there is no two-sided unit either.
Proof. Every value of the product is central, $e\star\tilde{Q} = \langle e,\tilde{Q}\rangle_{*}e_t \in \mathbb{C}e_t$. The element $\vec e_x$ is not central, so $e\star \vec e_x \neq \vec e_x$ for every $e$, and $\vec e_x\star e \neq \vec e_x$ for every $e$; there is therefore no left unit, no right unit and no two-sided unit. In particular the idempotent $e_t$ is not a unit: $e_t\star \vec e_x = \langle e_t,\vec e_x\rangle_{*}e_t = 0$, since the basis is orthonormal for $\langle\cdot,\cdot\rangle_{*}$.
Remark (the reason is the image). The failure is structural: a product whose image is the central line cannot reproduce an element with a vector part, and the block has non-central elements. The same reason deprives the symmetric quaternionic product of a unit, while the symmetric plain product, whose image is the whole space, has the unit $e_t$.
The Absence of an Inverse and of the Group of Units
No element of the block is invertible of $\star$, and the group of units of $\star$ is not defined, both notions being empty for want of a two-sided unit.
Theorem (no element is invertible, and there is no group of units). No element of $\mathbb{B}$ is invertible of $\star$, and the group of units of $\star$ is not defined.
Proof. The notions of §The Relative Notions presuppose a two-sided unit, and §The Absence of a Unit shows that the block has none; the definition is therefore empty. There is no invertible element and no inverse formula.
Remark (what the block does not have). The block has no unit, no inverse, no group of units, no division structure and no three-way classification of the elements into the zero, the invertible and the zero divisors; its element classification is two-way and is read in §The Absence of a Criterion from the Algebraic Norm. The block neither inherits the invertible elements of another product nor carries a substitute group of its own elements.
The Algebraic Norm of the Block
The block carries one algebraic norm, $\langle\tilde{Q}\rangle_{*} = \mathrm{Sc}(\tilde Q\star\tilde Q) = \sum_{\mu\in\{t,x,y,z\}}\lvert Q_\mu\rvert^{2}$, real, non-negative and positive definite; its Gram matrix is the identity on the basis $e_t,\vec e_x,\vec e_y,\vec e_z$ and its square map has image the real ray $\mathbb{R}_{\geq0}e_t$.
Definition. The algebraic norm of the block is the diagonal of the scalar-part form of the product,
$$ \langle\tilde{Q}\rangle_{*} = \mathrm{Sc}\bigl(\tilde Q\star\tilde Q\bigr) = \langle\tilde{Q},\tilde{Q}\rangle_{*} = \sum_{\mu\in\{t,x,y,z\}}\lvert Q_\mu\rvert^{2} \in \mathbb{R}_{\geq 0}, $$
a real non-negative number, and it is the coefficient of the square of the element,
$$ \tilde{Q}\star\tilde{Q} = \langle\tilde{Q}\rangle_{*}e_t . $$
The equation names the product $\star$: the algebraic norm is relative to the product, and the algebraic norms of the other three rows are different quadratic maps on the same space (The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra).
Basic properties. Each is a direct reading of the definition.
- $\langle\tilde{Q}\rangle_{*}$ is a real number, $\overline{\langle\tilde{Q}\rangle_{*}} = \langle\tilde{Q}\rangle_{*}$, because each $\lvert Q_\mu\rvert^{2}$ is real.
- $\langle\tilde{Q}\rangle_{*}$ is non-negative, and $\langle\tilde{Q}\rangle_{*} = 0$ exactly at $\tilde{Q} = 0$; the quadratic form is positive definite, of signature $(8,0)$ on the real space underlying $\mathbb{B}$.
- $\langle\cdot,\cdot\rangle_{*}$ is invariant under the three conjugations, $\langle\tilde{Q}^{\natural}\rangle_{*} = \langle\overline{\tilde{Q}}\rangle_{*} = \langle\tilde{Q}^{*}\rangle_{*} = \langle\tilde{Q}\rangle_{*}$, because each conjugation preserves the moduli $\lvert Q_\mu\rvert$ of the coordinates.
- $\langle\cdot,\cdot\rangle_{*}$ is homogeneous of degree two over the reals but not over $\mathbb{C}$: $\langle\lambda\tilde{Q}\rangle_{*} = \lvert\lambda\rvert^{2}\langle\tilde{Q}\rangle_{*}$ for $\lambda\in\mathbb{C}$.
- $\langle\cdot,\cdot\rangle_{*}$ is the polarisation source of the form $\langle\tilde Q_1,\tilde Q_2\rangle_{*}$, recovered from the diagonal in §The Polarisation of the Algebraic Norm.
Proposition (the Gram matrix and the square map). On the basis $e_t,\vec e_x,\vec e_y,\vec e_z$ the form has Gram matrix the identity, $\langle e_\mu,e_\nu\rangle_{*} = \delta_{\mu\nu}$, so $\langle e_\mu\rangle_{*} = 1$; the square map $\tilde{Q}\mapsto\tilde{Q}\star\tilde{Q}$ has image the real ray $\mathbb{R}_{\geq 0}e_t$ and is surjective onto it.
Proof. The Gram statement is $\langle e_\mu,e_\nu\rangle_{*} = \sum_\sigma (e_\mu)_\sigma\overline{(e_\nu)_\sigma} = \delta_{\mu\nu}$, and the square display gives the image. The ray is attained since $\tilde{Q} = \sqrt{\lambda}e_t$ gives $\langle\tilde{Q}\rangle_{*} = \lambda$ for $\lambda\geq 0$.
The Failure of Multiplicativity
The algebraic norm of the block is not multiplicative: $\langle\tilde Q_1\star\tilde Q_2\rangle_{*} = \lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2}$, the defect $\langle\tilde Q_1\rangle_{*}\langle\tilde Q_2\rangle_{*} - \lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2}\geq0$ vanishes exactly on linearly dependent pairs, and multiplicativity holds on the centre alone.
Theorem (the algebraic norm of a product). For all $\tilde Q_1,\tilde Q_2$,
$$ \boxed{\; \langle\tilde Q_1\star\tilde Q_2\rangle_{*} = \lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2} \;}, $$
the square of the modulus of the form on the two arguments; consequently $\langle\tilde Q_1\star\tilde Q_2\rangle_{*} = \langle\tilde Q_1\rangle_{*}\langle\tilde Q_2\rangle_{*}$ exactly when the two arguments are linearly dependent over $\mathbb{C}$.
Proof. The product is central, $\tilde Q_1\star\tilde Q_2 = \langle\tilde Q_1,\tilde Q_2\rangle_{*}e_t$, and $\langle \lambda e_t\rangle_{*} = \lvert\lambda\rvert^{2}$. So the algebraic norm of the product is $\lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2}$. For the equality case, the quadratic form $t\mapsto \langle\tilde Q_1 - t\tilde Q_2\rangle_{*}\geq 0$ in the complex variable $t$ has discriminant $\langle\tilde Q_1\rangle_{*}\langle\tilde Q_2\rangle_{*} - \lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2}$, so this quantity is non-negative, and the vanishing of the discriminant is the vanishing of the form at its minimising argument, that is the proportionality of $\tilde Q_1$ and $\tilde Q_2$.
Corollary (the defect is the discriminant). For all $\tilde Q_1,\tilde Q_2$,
$$ \boxed{\; \langle\tilde Q_1\rangle_{*}\langle\tilde Q_2\rangle_{*} - \langle\tilde Q_1\star\tilde Q_2\rangle_{*} = \langle\tilde Q_1\rangle_{*}\langle\tilde Q_2\rangle_{*} - \lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2} \geq 0 \;}, $$
with equality exactly when $\tilde Q_1$ and $\tilde Q_2$ are linearly dependent. The algebraic norm is therefore submultiplicative and multiplicative exactly on linearly dependent pairs.
Remark (the identity that replaces multiplicativity). The block has one exact multiplicativity, read on the form rather than on the algebraic norm: for all $\tilde Q_1,\tilde Q_2,\tilde Q_3$,
$$ \langle\tilde Q_1\star\tilde Q_2,\tilde Q_3\rangle_{*} = \langle\tilde Q_1,\tilde Q_2\rangle_{*}\,\langle e_t,\tilde Q_3\rangle_{*}, $$
and dually $\langle\tilde Q_3,\tilde Q_1\star\tilde Q_2\rangle_{*} = \langle\tilde Q_3,e_t\rangle_{*}\overline{\langle\tilde Q_1,\tilde Q_2\rangle_{*}}$. The first factor $\langle\tilde Q_1,\tilde Q_2\rangle_{*}$ is the coefficient of the product, the second is the evaluation of the form on the idempotent $e_t$; the product is central, so the form evaluates it only through that one coefficient.
Remark (the algebraic norm is multiplicative on the centre). On the central line the block is the field rule $(\lambda e_t)\star(\mu e_t) = \lambda\overline{\mu}e_t$ and the algebraic norm is the squared modulus, $\langle \lambda e_t\rangle_{*} = \lvert\lambda\rvert^{2}$; on central elements it is multiplicative, $H\bigl((\lambda e_t)\star(\mu e_t)\bigr) = \langle \lambda e_t\rangle_{*}\langle \mu e_t\rangle_{*}$, because $\mathbb{C}e_t$ is a field. Multiplicativity holds on the centre and fails off it, and the failure is measured by the corollary.
The Polarisation of the Algebraic Norm
The polar form of $\langle\cdot,\cdot\rangle_{*}$ is twice the real part of the sesquilinear form, $\langle\tilde Q_1 + \tilde Q_2\rangle_{*} - \langle\tilde Q_1\rangle_{*} - \langle\tilde Q_2\rangle_{*} = 2\Re \langle\tilde Q_1,\tilde Q_2\rangle_{*}$, and the full Hermitian form is recovered from the algebraic norm by the four-term polarisation $\langle\tilde Q_1,\tilde Q_2\rangle_{*} = \tfrac{1}{4}\sum_{k=0}^{3}i^{k}\langle\tilde Q_1 + i^{k}\tilde Q_2\rangle_{*}$.
Theorem (the polarisation). The polar form of the quadratic map $\langle\cdot,\cdot\rangle_{*}$ is twice the real part of the sesquilinear form,
$$ \langle\tilde Q_1 + \tilde Q_2\rangle_{*} - \langle\tilde Q_1\rangle_{*} - \langle\tilde Q_2\rangle_{*} = 2\Re \langle\tilde Q_1,\tilde Q_2\rangle_{*}, $$
and the full Hermitian form is recovered from the algebraic norm by
$$ \langle\tilde Q_1,\tilde Q_2\rangle_{*} = \tfrac{1}{4}\sum_{k=0}^{3}i^{k}\,\langle\tilde Q_1 + i^{k}\tilde Q_2\rangle_{*}. $$
Proof. The first display expands $\langle\tilde Q_1+\tilde Q_2\rangle_{*} = \langle\tilde Q_1\rangle_{*}+\langle\tilde Q_2\rangle_{*}+\langle\tilde Q_1,\tilde Q_2\rangle_{*}+\overline{\langle\tilde Q_1,\tilde Q_2\rangle_{*}}$. For the second, insert the four values $k = 0,1,2,3$ and use $\langle\tilde Q_1 + i^k\tilde Q_2\rangle_{*} = \langle\tilde Q_1\rangle_{*}+\langle\tilde Q_2\rangle_{*}+2\Re\bigl(i^{-k}\langle\tilde Q_1,\tilde Q_2\rangle_{*}\bigr)$; the four coefficients $1,i,-1,-i$ average to the real and imaginary parts of $\langle\cdot,\cdot\rangle_{*}$, giving $\langle\cdot,\cdot\rangle_{*}$.
Remark (the polarisation is the Hermitian one). The second display is the polarisation of a sesquilinear form and not of a bilinear one: the four coefficients are the fourth roots of unity, not the pair $\pm1$ of the symmetric polarisation, which is the form of the Hermitian symmetry $\langle\tilde Q_2,\tilde Q_1\rangle_{*} = \overline{\langle\tilde Q_1,\tilde Q_2\rangle_{*}}$. The first display extracts only the real part, because the diagonal $\langle\cdot,\cdot\rangle_{*}$ of a Hermitian form is real; the imaginary part is carried by the four-term display.
The Absence of a Criterion from the Algebraic Norm
The block has no invertible element, so no criterion can name one, and its algebraic norm decides nothing: $\langle\tilde{Q}\rangle_{*} = 0$ holds exactly at $\tilde{Q} = 0$, so the block has no nonzero element of zero algebraic norm and no nonzero isotropic vector.
Theorem (the algebraic norm does not decide invertibility). The block has no invertible element, so no criterion can name one; and its algebraic norm has no nonzero zero:
$$ \boxed{\; \langle\tilde{Q}\rangle_{*} = 0 \quad\Longleftrightarrow\quad \tilde{Q} = 0 \;}. $$
Proof. The first clause is the theorem on the absence of the inverse. The equivalence is the positive definiteness: $\langle\tilde{Q}\rangle_{*} = \sum_\mu\lvert Q_\mu\rvert^{2}$ is a sum of non-negative reals, vanishing exactly when every $Q_\mu$ vanishes.
Remark (what replaces the criterion). A criterion would equate the invertible elements with the complement of a hypersurface of the algebraic norm. For the block the algebraic norm is definite, it vanishes only at the origin, and it therefore separates nothing: the block has no nonzero element of zero algebraic norm, or, in the vocabulary of forms, no nonzero isotropic vector. Its element classification is two-way, the zero element and the non-zero elements, and the non-zero elements are all of one kind, the zero divisors of Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space. The definiteness is the exact opposite of the general plain product, whose algebraic norm $\langle\cdot\rangle$ is indefinite and does not decide invertibility either (Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra).
Corollary (the substitute for the group of units). The block has no group of units; the set that replaces it is the group of the linear maps of $\mathbb{B}$ that preserve the form $\langle\cdot,\cdot\rangle_{*}$, together with its algebra, and it is read in The Multiplication Operators of the Symmetric Plain Sesqualgebra. The block carries no subgroup of its own elements to play that role, since it has no invertible element.
The Comparison with the General Plain Algebra
The parent is unital and the block is not; the two products carry different algebraic norms $\langle\cdot\rangle$ and $\langle\cdot\rangle_{*}$; neither algebraic norm decides invertibility, the block having no invertible element at all; and the block's zero divisors are every nonzero element while the parent's are a proper subset.
| clause | general plain algebra | symmetric plain sesqualgebra |
|---|---|---|
| algebraic norm | $\langle\tilde{Q}\rangle = \mathrm{Sc}(\tilde Q\tilde Q) = Q_t^2-(\vec Q,\vec Q)$, complex, indefinite | $\langle\tilde{Q}\rangle_{*} = \mathrm{Sc}(\tilde Q\star\tilde Q) = \sum_\mu\lvert Q_\mu\rvert^{2}$, real, definite |
| multiplicativity of the algebraic norm | $\langle\cdot\rangle$ not multiplicative | $\langle\cdot\rangle_{*}$ submultiplicative, $\langle\tilde Q_1\star\tilde Q_2\rangle_{*} = \lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2}$ |
| isotropic vectors | the complex cone of $B$ | none but the origin |
| unit | $e_t$, two-sided | none |
| invertible elements | defined; not read from $B$ | none |
| zero divisors | a proper subset | every non-zero element |
The parent is unital and the block is not; the two products carry different algebraic norms, $\langle\cdot\rangle$ and $\langle\cdot\rangle_{*}$, and neither algebraic norm decides invertibility: the parent has invertible elements that its algebraic norm $\langle\cdot\rangle$ does not see, and the block has no invertible element at all. The last three rows are the absence and the shape of a rank-one product.
Summary
The symmetric plain sesqualgebra $\tilde Q_1\star\tilde Q_2 = \langle\tilde Q_1,\tilde Q_2\rangle_{*}e_t$ has no unit, because every value of the product is central and $\mathbb{B}$ has non-central elements; consequently it has no inverse, no group of units and no division structure, and the relative definitions are empty. It carries one algebraic norm, the diagonal of its scalar-part form,
$$ \langle\tilde{Q}\rangle_{*} = \mathrm{Sc}\bigl(\tilde Q\star\tilde Q\bigr) = \sum_{\mu\in\{t,x,y,z\}}\lvert Q_\mu\rvert^{2}, $$
a real non-negative definite quadratic map of signature $(8,0)$, invariant under the three conjugations. The algebraic norm is not multiplicative: $\langle\tilde Q_1\star\tilde Q_2\rangle_{*} = \lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2}$, the defect is the discriminant $\langle\tilde Q_1\rangle_{*}\langle\tilde Q_2\rangle_{*} - \lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2}\geq 0$, vanishing exactly on linearly dependent pairs, and multiplicativity holds on the centre alone. What replaces it is the exact identity $\langle\tilde Q_1\star\tilde Q_2,\tilde Q_3\rangle_{*} = \langle\tilde Q_1,\tilde Q_2\rangle_{*}\langle e_t,\tilde Q_3\rangle_{*}$, and the algebraic norm polarises to the Hermitian form, $\Re H$ at the symmetric step and $\langle\cdot,\cdot\rangle_{*}$ at the four-term display. The absence of a criterion is the definiteness, $\langle\tilde{Q}\rangle_{*} = 0\Leftrightarrow\tilde{Q} = 0$: the block has no nonzero isotropic vector and no invertible element, and its element classification is two-way.
Summary of Notation
| Symbol | Meaning |
|---|---|
| $\tilde Q_1\star\tilde Q_2 = \langle\tilde Q_1,\tilde Q_2\rangle_{*}e_t$ | the product of the block |
| $\langle\tilde Q_1,\tilde Q_2\rangle_{*} = \sum_\mu Q_{1,\mu}\overline{Q_{2,\mu}}$ | the Hermitian form defining the block, the scalar-part form of the product |
| $\langle\tilde{Q}\rangle_{*} = \langle\tilde{Q},\tilde{Q}\rangle_{*} = \sum_\mu\lvert Q_\mu\rvert^{2}$ | the algebraic norm of the block, one of the four algebraic norms of the space |
| $\tilde{Q}\star\tilde{Q} = \langle\tilde{Q}\rangle_{*}e_t$ | the square map, image $\mathbb{R}_{\geq0}e_t$ |
| $\langle\tilde Q_1\star\tilde Q_2\rangle_{*} = \lvert \langle\tilde Q_1,\tilde Q_2\rangle_{*}\rvert^{2}$ | the algebraic norm of a product |
| $\langle\tilde Q_1\star\tilde Q_2,\tilde Q_3\rangle_{*} = \langle\tilde Q_1,\tilde Q_2\rangle_{*}\langle e_t,\tilde Q_3\rangle_{*}$ | the identity that replaces multiplicativity |
| $\langle\tilde Q_1,\tilde Q_2\rangle_{*} = \tfrac14\sum_k i^{k}\langle\tilde Q_1 + i^{k}\tilde Q_2\rangle_{*}$ | the Hermitian polarisation of the algebraic norm |
| $\langle\tilde{Q}\rangle_{*} = 0\Leftrightarrow\tilde{Q} = 0$ | the definiteness, the substitute for the criterion |
| no unit, no inverse, no group of units | the absence carried by the block |
Further Reading
- The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, for the definition of the algebraic norm as the form at the two equal arguments and the four algebraic norms of the space.
- The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, for the Hermitian form $\langle\cdot,\cdot\rangle_{*}$ and the table of the four forms.
- Introduction to the Symmetric Plain Sesqualgebra, for the product, its class, its table and its placement among the twelve structures.
- Definitions for the Study of the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, §The Unit and the Inverse, for the relative notions of the unit, the inverse and the group of units.
- Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, for the algebraic norm $\langle\cdot\rangle$ of the plain product, which is not the form $\langle\cdot,\cdot\rangle_{*}$ of this article.
- The Hermitian Form as a Product on the Symmetric Plain Sesqualgebra, for the form $\langle\cdot,\cdot\rangle_{*}$ read as the product, its Gram matrix, its positivity and its restriction to the remarkable subspaces.
- Positivity and the Hermitian Cone of the Biquaternion Algebra with Hermitian Adjoint, for the positivity and the cone of $\langle\cdot,\cdot\rangle_{*}$.
- Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the zero-divisor set that is the complement of the origin.
- The Multiplication Operators of the Symmetric Plain Sesqualgebra, for the group of linear maps that preserve $\langle\cdot,\cdot\rangle_{*}$, the substitute for the group of units.