Norm and Invertibility in the Split-Complex Algebra

Introduction

This article studies the algebraic norm that the general plain bilinear product, the multiplication $Q_2Q_1$ of the split-complex algebra, carries, and the invertibility of the elements of that product.

A norm is relative to a product, and the algebraic norm of a product is the diagonal of its general plain bilinear form, the definition of The Two Forms and the Two Algebraic Norms of the Split-Complexoid Space: for the plain product the form is $\langle Q_2,Q_1\rangle=Q_2Q_1$ and its diagonal is

$$ \langle Q\rangle=Q^{2}=(q^{2}+q'^{2})+2qq'j, $$

the first of the two algebraic norms of the space and not the second, $\langle Q\rangle_{*}=Q\bar Q=q^{2}-q'^{2}$. The article defines it, records its value, its Gram matrix, its anisotropy and its vanishing set, and proves that it is multiplicative as an element of the ring and not a real norm. That is the whole of the algebraic norm of the plain product.

The invertibility is where the article records an absence. The plain product is associative and unital, so its invertible elements and its group of units are defined; but the algebraic norm of the product does not decide them. The element $1+j$ is a zero divisor and satisfies $\langle 1+j\rangle=2(1+j)\neq0$: the non-vanishing of the algebraic norm is not sufficient for invertibility. The converse failure cannot occur, because $\langle Q\rangle=0$ forces $Q=0$ over the reals; the norm is anisotropic and sees only the origin. The article states that absence rather than a criterion. The criterion of the plain product is furnished by the other algebraic norm, $\langle Q\rangle_{*}=Q\bar Q$, which is not the algebraic norm of the plain product and is studied in Norm and Invertibility in the General Split-Complex Algebra; the zero divisors are Zero Divisors of the Split-Complex Algebra.

The results. The plain product carries one algebraic norm, the diagonal of its general plain bilinear form, $\langle Q\rangle=Q^{2}=(q^{2}+q'^{2})+2qq'j$, $\mathbb{D}$-valued, anisotropic, of realified Gram matrix $\operatorname{diag}(1,1)$ in the basis $1,j$. It is multiplicative, $\langle QB\rangle=\langle Q\rangle\langle B\rangle$, by associativity of the ring. It is not a real norm: its value is not real in general, it defines no order and no modulus, and its vanishing set is the single point $\{0\}$. It does not decide invertibility: $\langle Q\rangle\neq0$ does not force $Q$ to be a unit, the witness being the zero divisor $1+j$. Each section below states its result first and then proves it.

The Algebraic Norm of the Plain Product

The general plain bilinear product carries exactly one algebraic norm, the diagonal of its general plain bilinear form, $\langle Q\rangle=Q^{2}=(q^{2}+q'^{2})+2qq'j$; it is $\mathbb{D}$-valued and anisotropic, of realified Gram matrix $\operatorname{diag}(1,1)$, and its vanishing set is the origin.

Definition

The algebraic norm of the plain product is the diagonal of the general plain bilinear form of the product,

$$ \langle Q\rangle=\langle Q,Q\rangle=Q^{2}, $$

for the form

$$ \langle Q_2,Q_1\rangle=Q_2Q_1, $$

the general plain bilinear form of The Two Forms and the Two Algebraic Norms of the Split-Complexoid Space. The form is symmetric, because the algebra is commutative. In coordinates,

$$ \langle Q\rangle=(q+jq')^{2}=q^{2}+q'^{2}+2qq'j, $$

so the algebraic norm takes its value in the whole algebra $\mathbb{D}$ and not in the base field.

Basic properties. Each is a direct reading of the definition.

• $\langle Q\rangle$ is $\mathbb{D}$-valued; it is real exactly when $qq'=0$, that is on the real line $\mathbb{R}_{\mathbb{D}}$ or on the split imaginary line $j\mathbb{R}_{\mathbb{D}}$.
• $\langle\cdot\rangle$ is homogeneous of degree two, $\langle\lambda Q\rangle=\lambda^{2}\langle Q\rangle$ for $\lambda\in\mathbb{R}$, and its polarisation is the form of the same letter, $\langle Q+B\rangle-\langle Q\rangle-\langle B\rangle=2\langle Q,B\rangle=2QB$.
• $\langle\cdot\rangle$ is anisotropic: it vanishes only at $Q=0$. It is therefore not indefinite, but it is not a real norm either, because its value is not real.
• Its realified Gram matrix on the basis $1,j$ is the identity $\operatorname{diag}(1,1)$, since the real part of $Q_2Q_1$ is $q_2q_1+q'_2q'_1$. The full $\mathbb{D}$-valued form has the symmetric Gram matrix $\begin{pmatrix}1&j\\j&1\end{pmatrix}$.
• Its vanishing set is $\{Q:\langle Q\rangle=0\}=\{0\}$.

The Polarisation

Proposition. The polar form of the algebraic norm is the symmetrised product,

$$ \boxed{\; \langle Q_2,Q_1\rangle=\tfrac12\bigl(\langle Q_2+Q_1\rangle-\langle Q_2\rangle-\langle Q_1\rangle\bigr)=Q_2Q_1 \;}, $$

and it is $\mathbb{D}$-valued and symmetric.

Proof. Expand $(Q_2+Q_1)^{2}=Q_2^{2}+Q_2Q_1+Q_1Q_2+Q_1^{2}$; the algebra is commutative, so the two middle terms are equal and the half-difference is $Q_2Q_1$.

Multiplicativity

Theorem. The algebraic norm of the plain product is multiplicative:

$$ \boxed{\;\langle QB\rangle=\langle Q\rangle\langle B\rangle,\qquad Q,B\in\mathbb{D}\;}. $$

Proof. By associativity and commutativity of the ring,

$$ \langle QB\rangle=(QB)^{2}=QBQB=Q^{2}B^{2}=\langle Q\rangle\langle B\rangle. $$

The multiplicativity is the multiplicativity of the square in any commutative associative algebra; it does not make $\langle\cdot\rangle$ a scalar norm, because the product on the right is the product of the ring and not the product of real numbers.

Corollary. The algebraic norm is invariant under the conjugation, $\langle\bar Q\rangle=\overline{\langle Q\rangle}$.

Proof. $\bar Q^{\,2}=\overline{Q^{2}}$, the bar being an automorphism of the algebra.

The Algebraic Norm Is Not a Real Norm

The algebraic norm is $\mathbb{D}$-valued, not real; its real part is the Euclidean form $q^{2}+q'^{2}$, positive definite, and its imaginary part is $2qq'$. It is therefore not a norm of the real vector space, although it is multiplicative.

The Real and the Imaginary Parts

Write $Q^{2}=B_{0}+jB_{1}$ with

$$ B_{0}=q^{2}+q'^{2},\qquad B_{1}=2qq'. $$

The real part $B_{0}$ is the Euclidean quadratic form of the plane, positive definite, of Gram matrix $\operatorname{diag}(1,1)$; the imaginary part $B_{1}$ is the hyperbolic cross term. The real part is not multiplicative: for $Q=B=1+j$ one has $B_{0}(Q)=B_{0}(B)=2$, while $QB=2+2j$ and $B_{0}(QB)=8$, not $4$. The full algebraic norm is multiplicative because it keeps both parts and multiplies in the ring; the real part alone does not.

The Vanishing Set and Anisotropy

Proposition. The algebraic norm vanishes only at the origin:

$$ \boxed{\;Q^{2}=0\iff Q=0\;}. $$

Proof. $Q^{2}=0$ gives $q^{2}+q'^{2}=0$ and $2qq'=0$. The first equation forces $q=q'=0$ over $\mathbb{R}$, since both squares are non-negative and their sum is zero.

So the algebraic norm is anisotropic. This is the sharp difference from the second algebraic norm $\langle Q\rangle_{*}=q^{2}-q'^{2}$, which vanishes on the two null lines and therefore on every zero divisor. The plain algebraic norm sees the zero-divisor set as a set of nonzero values and cannot separate it from the units.

The Shell of the Algebraic Norm

The shell of the algebraic norm at a central value $\lambda\in\mathbb{R}$ is the set

$$ \{Q:\langle Q\rangle=\lambda\}. $$

For $\lambda=0$ it is $\{0\}$. For $\lambda=1$ it is the four-element set $\{\pm1,\pm j\}$, the roots of $+1$ of Square Roots of One and the Idempotents of the Split-Complex Algebra. For $\lambda=-1$ it is empty, because the real part $q^{2}+q'^{2}$ is non-negative. The shell is not a level set of a real norm: for a real central value it is a finite set, of four points when $\lambda>0$, one point when $\lambda=0$, and empty when $\lambda<0$.

The Algebraic Norm Does Not Decide Invertibility

The algebraic norm does not decide invertibility: $\langle Q\rangle=0$ forces $Q=0$, which is not a unit, but $\langle Q\rangle\neq0$ does not force $Q$ to be a unit.

The algebraic norm of a product would decide invertibility if the invertible elements were exactly the elements of non-vanishing algebraic norm. For the plain product that statement is false in the direction that matters, and the article records the absence rather than a criterion.

Theorem (the algebraic norm does not decide invertibility). The implication

$$ \langle Q\rangle=0\Longrightarrow Q\text{ non-invertible} $$

holds, but the converse implication

$$ \langle Q\rangle\neq0\Longrightarrow Q\text{ invertible} $$

fails.

Proof. The first statement is the anisotropy of the norm: $\langle Q\rangle=0$ gives $Q=0$, and $0$ has no inverse. For the second, take $Q=1+j$. Its algebraic norm is

$$ \langle 1+j\rangle=(1+j)^{2}=1+2j+j^{2}=2+2j\neq0, $$

while $Q$ is a zero divisor, $(1+j)(1-j)=1-j^{2}=0$, hence not invertible.

Remark (the witness). The element $1+j$ is the whole of the failure. It is a nonzero zero divisor, so it lies on the null cone of the other algebraic norm, $\langle 1+j\rangle_{*}=1^{2}-1^{2}=0$; its plain algebraic norm is $2+2j$, a nonzero element of the ring. The vanishing set of $\langle\cdot\rangle$ is the origin and the vanishing set of $\langle\cdot\rangle_{*}$ is the two null lines; the zero-divisor set is the second vanishing set with the origin removed, and the first norm cannot see it.

Remark (what replaces the criterion). The plain product is not left without an invertibility theory: it is associative and unital, every element of its group of units has an inverse, and the zero divisors are characterised in Zero Divisors of the Split-Complex Algebra. What the article records is that this theory is not read from the algebraic norm of the plain product, which is the one algebraic norm the product carries; the criterion that does exist is carried by the other algebraic norm, $\langle Q\rangle_{*}=Q\bar Q$, and belongs to Norm and Invertibility in the General Split-Complex Algebra.

The Unit and the Group of Units

The plain product is associative and unital, with the two-sided unit $1$, and its invertible elements form the group of units $\mathbb{D}^{\times}$, a proper subset of $\mathbb{D}\setminus\{0\}$ because the product has zero divisors.

The unit of the plain product is $1$: it is a two-sided unit, $Q1=1Q=Q$, and it is the only one. The product is associative and commutative, so the inverse of an element is unique when it exists and the invertible elements form the group of units of the plain product, an abelian group.

The plain product has zero divisors: the identity $(1+j)(1-j)=0$ exhibits a product of two nonzero elements that is zero. The plain algebra is therefore not a division algebra, and the group of units is a proper subset of $\mathbb{D}\setminus\{0\}$. The zero divisors, their two families and their geometry are Zero Divisors of the Split-Complex Algebra, and they are cited here, not read.

The group is described by the other algebraic norm: an element is a unit exactly when $\langle Q\rangle_{*}=Q\bar Q\neq0$, and then $Q^{-1}=\bar Q/\langle Q\rangle_{*}$. That criterion, the four components of the group and the distribution of the units among the distinguished subspaces are the subject of Norm and Invertibility in the General Split-Complex Algebra; the present article only records that the group exists and that its boundary is not the vanishing set of $\langle\cdot\rangle$.

Summary

The general plain bilinear product $Q_2Q_1$ carries one algebraic norm, the diagonal of its general plain bilinear form,

$$ \langle Q\rangle=Q^{2}=(q^{2}+q'^{2})+2qq'j, $$

$\mathbb{D}$-valued, anisotropic, of realified Gram matrix $\operatorname{diag}(1,1)$ and vanishing set the origin. It is multiplicative, $\langle QB\rangle=\langle Q\rangle\langle B\rangle$, by associativity of the ring, and invariant under the conjugation, $\langle\bar Q\rangle=\overline{\langle Q\rangle}$.

It is not a real norm: its value is not real in general, its real part is the Euclidean form $q^{2}+q'^{2}$ and its imaginary part is $2qq'$, and it defines no order and no modulus. It does not decide invertibility: the zero divisor $1+j$ has $\langle 1+j\rangle=2+2j\neq0$, so non-vanishing is not sufficient; the implication $\langle Q\rangle=0\Rightarrow Q=0$ holds, but it only says that the origin is not a unit. The plain product is associative and unital, with unit $1$ and a genuine abelian group of units; the criterion is carried by the other algebraic norm $\langle Q\rangle_{*}=Q\bar Q$, which is not the algebraic norm of the plain product.

Summary of Notation

Further Reading

• The Two Forms and the Two Algebraic Norms of the Split-Complexoid Space, for the definition of the algebraic norm as the form at the two equal arguments, the two algebraic norms and their separation.
• Norm and Invertibility in the General Split-Complex Algebra, for the other algebraic norm, the criterion of invertibility and the group of units.
• Split-Complex Algebra, for the product, its associativity, its unit and its element theory.
• Zero Divisors of the Split-Complex Algebra, for the zero divisors, which the algebraic norm of the plain product does not detect.
• Square Roots of One and the Idempotents of the Split-Complex Algebra, for the shell at $1$ and the four roots of $+1$.