Norm and Invertibility in the Symmetric Plain Algebra
Introduction
This article studies the algebraic norm that the symmetric plain algebra, the biquaternion space under the Jordan product $\tilde Q_1\bullet\tilde Q_2=\tfrac12(\tilde Q_1\tilde Q_2+\tilde Q_2\tilde Q_1)$ of Introduction to the Symmetric Plain Algebra, carries, and the invertibility of its elements.
A norm is relative to 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. The symmetrisation keeps the scalar part of a product, so the block carries the algebraic norm of the general plain bilinear product, the general plain bilinear algebraic norm
$$ \langle\tilde Q\rangle=\mathrm{Sc}\bigl(\tilde Q\bullet\tilde Q\bigr)=\sum_{\mu\in\{t,x,y,z\}}\varepsilon_\mu Q_\mu^{2}=Q_t^{2}-Q_x^{2}-Q_y^{2}-Q_z^{2}, $$
complex-valued, indefinite and not multiplicative. The article defines it, records its Gram matrix, its signature and its vanishing set, and proves the failure of multiplicativity; the four general products carry four algebraic norms in all, and $\langle\cdot\rangle$ is the one of this row.
The invertibility is where the article records an absence. The block is commutative and unital, with the unit $e_t$, but it is not associative, and its invertibility is governed by a quadratic form that is not its algebraic norm. The element $e_t+\vec e_x$ is invertible and satisfies $\langle e_t+\vec e_x\rangle=0$, and the element $e_t+i\vec e_x$ is a zero divisor and satisfies $\langle e_t+i\vec e_x\rangle=2$: the non-vanishing of the algebraic norm is neither necessary nor sufficient for invertibility, and the article states that absence rather than a criterion. The invertibility the block does have is the invertibility of the Jordan theory, carried by a different quadratic form and studied in The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra.
The results. The block carries one algebraic norm, the diagonal of its scalar-part form, $\langle\tilde Q\rangle=Q_t^{2}-Q_x^{2}-Q_y^{2}-Q_z^{2}$, complex-valued and indefinite, of Gram matrix $\mathrm E=\operatorname{diag}(1,-1,-1,-1)$ and realification of signature $(4,4)$, with the complex cone as its vanishing set. It is not multiplicative, the pair $(\vec e_x,\vec e_y)$ separating the two sides. It does not decide invertibility: the unit $e_t+\vec e_x$ has $\langle\cdot\rangle=0$ and the zero divisor $e_t+i\vec e_x$ has $\langle\cdot\rangle=2$, so the non-vanishing of the algebraic norm is neither necessary nor sufficient; the invertibility the block does have is that of the Jordan theory, carried by a different quadratic form. Each section below states its result first and then proves it.
The Algebraic Norm of the Block
The block carries exactly one algebraic norm, the diagonal of its scalar-part form, $\langle\tilde Q\rangle=Q_t^{2}-Q_x^{2}-Q_y^{2}-Q_z^{2}$; it is complex-valued and indefinite, of Gram matrix $\mathrm E=\operatorname{diag}(1,-1,-1,-1)$ and realification of signature $(4,4)$, and its vanishing set is the complex cone.
Definition. The algebraic norm of the block is the diagonal of the scalar-part form of the product,
$$ \langle\tilde Q\rangle=\mathrm{Sc}\bigl(\tilde Q\bullet\tilde Q\bigr)=\sum_{\mu\in\{t,x,y,z\}}\varepsilon_\mu Q_\mu^{2}=Q_t^{2}-Q_x^{2}-Q_y^{2}-Q_z^{2}, $$
equivalently $\langle\tilde Q\rangle=\langle\tilde Q,\tilde Q\rangle$ for the form
$$ \langle\tilde Q_1,\tilde Q_2\rangle=\mathrm{Sc}\bigl(\tilde Q_1\bullet\tilde Q_2\bigr)=\mathrm{Sc}\bigl(\tilde Q_1\tilde Q_2\bigr)=Q_{1t}Q_{2t}-(\vec Q_1,\vec Q_2), $$
which is the general plain bilinear form of The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra. The symmetrisation does not change the scalar part, so the block carries the algebraic norm of the general plain bilinear product; the form and its diagonal are of the same letter.
Basic properties. Each is a direct reading of the definition.
- $\langle\tilde Q\rangle$ is complex-valued in general; it is real exactly on the union of the quaternion subspace $\mathbb{H}_{\mathbb{B}}$ and its imaginary translate $i\mathbb{H}_{\mathbb{B}}$.
- $\langle\cdot\rangle$ is homogeneous of degree two, $\langle\lambda\tilde Q\rangle=\lambda^{2}\langle\tilde Q\rangle$ for $\lambda\in\mathbb{C}$, and its polarisation is the form of the same letter, $\langle\tilde Q_1+\tilde Q_2\rangle-\langle\tilde Q_1\rangle-\langle\tilde Q_2\rangle=2\langle\tilde Q_1,\tilde Q_2\rangle$.
- $\langle\cdot\rangle$ is not positive definite and not definite.
- Its Gram matrix on the basis is the sign matrix $\mathrm E=\operatorname{diag}(1,-1,-1,-1)$, and the realification of $\langle\cdot\rangle$ is non-degenerate of signature $(4,4)$ on $\mathbb{B}\cong\mathbb{R}^{8}$.
- Its vanishing set is the complex cone $\{\langle\tilde Q\rangle=0\}$, the complex quadric of The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, and not the zero-divisor set of the block.
The Algebraic Norm Is Not Multiplicative
The algebraic norm of the block is not multiplicative: $\langle\tilde Q_1\bullet\tilde Q_2\rangle\neq \langle\tilde Q_1\rangle\langle\tilde Q_2\rangle$ in general, the pair $(\vec e_x,\vec e_y)$ separating the two sides.
Proposition. The algebraic norm of the block is not multiplicative:
$$ \boxed{\; \langle\tilde Q_1\bullet\tilde Q_2\rangle\neq \langle\tilde Q_1\rangle\,\langle\tilde Q_2\rangle\quad\text{in general.} \;} $$
Proof. The pair $\vec e_x,\vec e_y$ separates the two sides: $\vec e_x\bullet \vec e_y=\tfrac12(\vec e_x\vec e_y+\vec e_y\vec e_x)=\tfrac12(\vec e_z-\vec e_z)=0$, so $\langle \vec e_x\bullet \vec e_y\rangle=\langle 0\rangle=0$, while $\langle \vec e_x\rangle\langle \vec e_y\rangle=(-1)(-1)=1$. The two values differ, so no multiplicativity holds.
Remark (the multiplicative algebraic norm is another form). Of the four algebraic norms of the space the multiplicative one is $\langle\cdot\rangle_{\natural}$, the algebraic norm of the quaternionic product, $\langle\tilde Q\rangle_{\natural}=\sum_\mu Q_\mu^{2}$; it is a different quadratic form on the same space, and the block, whose product is not associative and has zero divisors, fails multiplicativity for its own algebraic norm $\langle\cdot\rangle$. The multiplicativity that the block does retain is the one of the quadratic representation of the Jordan theory, a statement about a different quadratic form and not about $\langle\cdot,\cdot\rangle$.
The Unit and the Degeneracy of the Naive Inverse
The block is commutative and unital with unit $e_t$, but it is not associative, and its naive inverse equation $\tilde Q_1\bullet\tilde Q_2=e_t$ is degenerate: the element $\tilde\Upsilon=\vec e_x+i\vec e_y$ satisfies it for a two-dimensional affine family of $\tilde Q_2$.
The unit of the block is $e_t$: the product is commutative, so its one-sided inverse equations coincide, and $e_t$ is the two-sided unit (Introduction to the Symmetric Plain Algebra). The block is not associative: $(\tilde Q_1\bullet\tilde Q_2)\bullet\tilde Q_3\neq\tilde Q_1\bullet(\tilde Q_2\bullet\tilde Q_3)$ in general, and the single equation $\tilde Q_1\bullet\tilde Q_2=e_t$ is degenerate: the element $\tilde\Upsilon=\vec e_x+i\vec e_y$ satisfies $\tilde\Upsilon\bullet\tilde Q_2=e_t$ for a two-dimensional affine family of $\tilde Q_2$, because its left multiplication has a two-dimensional kernel. The naive inverse equation therefore does not define an inverse, and the block is treated with the inverse of the Jordan theory in The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra.
The Absence of a Criterion from the Algebraic Norm
The algebraic norm does not decide invertibility: $\langle\tilde Q\rangle=0$ does not force $\tilde Q$ non-invertible, and $\langle\tilde Q\rangle\neq0$ does not force $\tilde Q$ invertible.
The algebraic norm of a product would decide invertibility if the invertible elements were exactly the elements of non-vanishing algebraic norm. For the block that statement is false in both directions, and the article records the absence rather than a criterion.
Theorem (the algebraic norm does not decide invertibility). Neither implication holds:
$$ \boxed{\; \langle\tilde Q\rangle=0\ \not\Longrightarrow\ \tilde Q\text{ non-invertible}, \qquad \langle\tilde Q\rangle\neq0\ \not\Longrightarrow\ \tilde Q\text{ invertible} \;}. $$
Proof. For the first implication, the element $\tilde Q=e_t+\vec e_x$ has $\langle\tilde Q\rangle=1^{2}-1^{2}=0$, and it is invertible: $(e_t+\vec e_x)\bullet\tfrac12(e_t-\vec e_x)=\tfrac12(e_t+\vec e_x)(e_t-\vec e_x)=\tfrac12\cdot2e_t=e_t$ and $\tilde Q^{\bullet2}\bullet\tfrac12(e_t-\vec e_x)=\tilde Q$, so $\tfrac12(e_t-\vec e_x)$ is its inverse of the block. For the second implication, the element $\tilde Q=e_t+i\vec e_x$ has $\langle\tilde Q\rangle=1^{2}-(i)^{2}=1+1=2\neq0$, and it is a zero divisor, $(e_t+i\vec e_x)(e_t-i\vec e_x)=e_t^{2}-(i\vec e_x)^{2}=e_t-e_t=0$, hence non-invertible while its algebraic norm does not vanish.
Remark (the two witnesses). The element $e_t+\vec e_x$ is invertible with vanishing algebraic norm, so the algebraic norm is not necessary for invertibility; the element $e_t+i\vec e_x$ is non-invertible with non-vanishing algebraic norm, so it is not sufficient either. The vanishing set of $\langle\cdot\rangle$ is the complex cone and the zero-divisor set is the cone of a different quadratic form; the two cones meet but neither contains the other.
Remark (what replaces the criterion). The block is not left without an invertibility theory: it is commutative and unital, the Jordan inverse is defined by its two equations, and the zero divisors are characterised in Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space. That theory is not read from the algebraic norm of the block, which is the one algebraic norm the block carries; the criterion it uses is carried by a different quadratic form and belongs to the article that owns that form.
Summary
The symmetric plain product $\tilde Q_1\bullet\tilde Q_2=\tfrac12(\tilde Q_1\tilde Q_2+\tilde Q_2\tilde Q_1)$ carries one algebraic norm, the diagonal of its scalar-part form, the general plain bilinear algebraic norm
$$ \langle\tilde Q\rangle=\mathrm{Sc}\bigl(\tilde Q\bullet\tilde Q\bigr)=\sum_{\mu\in\{t,x,y,z\}}\varepsilon_\mu Q_\mu^{2}=Q_t^{2}-Q_x^{2}-Q_y^{2}-Q_z^{2}, $$
complex-valued, of Gram matrix $\mathrm E=\operatorname{diag}(1,-1,-1,-1)$ and realification of signature $(4,4)$, with the complex cone as its vanishing set, and it is not multiplicative, the pair $(\vec e_x,\vec e_y)$ separating the two sides, $\langle \vec e_x\bullet \vec e_y\rangle=\langle 0\rangle=0$ against $\langle \vec e_x\rangle\langle \vec e_y\rangle=1$.
The block is commutative and unital, with the unit $e_t$, and it is not associative, the naive inverse equation $\tilde Q_1\bullet\tilde Q_2=e_t$ being degenerate at $\tilde\Upsilon=\vec e_x+i\vec e_y$. Its algebraic norm does not decide invertibility: the invertible element $e_t+\vec e_x$ has $\langle\cdot\rangle=0$ and the zero divisor $e_t+i\vec e_x$ has $\langle\cdot\rangle=2$, so the non-vanishing of the algebraic norm is neither necessary nor sufficient. The article records that absence; the invertibility the block does have is the invertibility of the Jordan theory, carried by a different quadratic form, which is not the algebraic norm of the block.
Summary of Notation
| Symbol | Meaning |
|---|---|
| $\tilde Q_1\bullet\tilde Q_2=\tfrac12(\tilde Q_1\tilde Q_2+\tilde Q_2\tilde Q_1)$ | the symmetric plain product, the multiplication of the block |
| $\langle\tilde Q_1,\tilde Q_2\rangle=\mathrm{Sc}(\tilde Q_1\bullet\tilde Q_2)=\sum_\mu\varepsilon_\mu Q_{1\mu} Q_{2\mu}$ | the general plain bilinear form, the scalar-part form of the product |
| $\langle\tilde Q\rangle=\langle\tilde Q,\tilde Q\rangle=\sum_\mu\varepsilon_\mu Q_\mu^{2}=Q_t^{2}-Q_x^{2}-Q_y^{2}-Q_z^{2}$ | the algebraic norm of the block, one of the four algebraic norms of the space |
| $\mathrm E=\operatorname{diag}(1,-1,-1,-1)$ | the Gram matrix of $\langle\cdot\rangle$; the signature of its realification is $(4,4)$ |
| $\{\langle\tilde Q\rangle=0\}$ | the complex cone, the vanishing set of the algebraic norm |
| $e_t$ | the two-sided unit of the block |
| $\tilde\Upsilon=\vec e_x+i\vec e_y$ | the witness of the degeneracy of the naive inverse equation |
| $e_t+\vec e_x$, $e_t+i\vec e_x$ | the witnesses: invertible with $\langle\cdot\rangle=0$, and zero divisor with $\langle\cdot\rangle=2$ |
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 general plain bilinear form $\langle\cdot,\cdot\rangle$ and the table of the four forms.
- Introduction to the Symmetric Plain Algebra, for the product, its unit, its commutativity and its non-associativity.
- The Trace Form and the Invariance of the Symmetric Plain Algebra, for the form the algebraic norm polarises.
- The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra, for the invertibility the block does have, carried by a different quadratic form.
- Definitions for the Study of the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, §The Unit and the Inverse, for the relative definitions of the unit and the inverse.
- Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the zero divisors, which the algebraic norm of the block does not detect.
- Remarkable Subspaces and the Four General Products, Forms and Norms, for the restriction of $\langle\cdot\rangle$ to the remarkable subspaces with its signature.