The Squares and the Positive Cone of the Biquaternion Sesqualgebra

Introduction

The square of an element for the sesquilinear multiplication $\tilde Q_1\star\tilde Q_2=\tilde Q_1\tilde Q_2^{*}$ of The Four General Products and Their Relations over the Biquaternion Algebra is $\tilde Q\star\tilde Q=\tilde Q\tilde Q^{*}$, and it is a Hermitian element whose scalar part is the sum of the modulus squares of the four coordinates,

$$ \mathrm{Sc}\bigl(\tilde Q\star\tilde Q\bigr)=\sum_{\mu\in\{t,x,y,z\}}Q_\mu\overline{Q_\mu}=\sum_{\mu\in\{t,x,y,z\}}\lvert Q_\mu\rvert^{2} . $$

That scalar part is a nonnegative real number, and it vanishes only at $\tilde Q=0$, so the square of a nonzero element is a nonzero Hermitian element and the square map has no kernel beyond zero. The sums of the squares form the cone of squares of the multiplication, and it is the subject of this article: the cone is closed under addition, it is carried into itself by the scalars of the square semiring of the base, it is invariant under the congruence $\tilde Q_1\mapsto\tilde Q_2^{*}\tilde Q_1\tilde Q_2$, and it is proper, so the preorder it defines is a partial order.

The cone is the biquaternion instance of the algebraic positive cone of Hermitian Squares and the Algebraic Positive Cone, the cone of the sums of the squares $x^{*}x$ of an involutive algebra; the sesquilinear square is the derived square $x\star x=xx^{*}$ of that article, so the two cones are the same set. In the biquaternion case the second square $\tilde Q\tilde Q^{\natural}=\bigl(\sum_\mu Q_\mu^{2}\bigr)e_t$ also carries the name, and the caution is the one Introduction to the General Plain Sesqualgebra records: the two squares agree on the real subspace and differ elsewhere, and only the first one is the square of the multiplication.

The setting is that of Introduction to the General Plain Sesqualgebra, whose §The Squares and the Positive Cone states the form of the square and its scalar part, and of Sesqualgebras. The two halves of the involution are the subspaces $\mathbb{M}_+$ and $\mathbb{M}_-$ of Hermitian and Skew-Hermitian Elements, the zero divisors of the algebra are Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors, and the remarkable subspaces against which the cone is placed are Remarkable Subspaces and the Four General Products, Forms and Norms. The general theory that the article reads is Hermitian Squares and the Algebraic Positive Cone, and the order it defines is the preorder of that article on the biquaternion algebra.

The Square of an Element

The Definition

Definition. The square of $\tilde Q$ in the multiplication is

$$ \tilde Q\star\tilde Q=\tilde Q\tilde Q^{*} , $$

and it is written once and for all $\tilde Q\star\tilde Q$; the other square of the corpus, $\tilde Q\tilde Q^{\natural}=\bigl(\sum_\mu Q_\mu^{2}\bigr)e_t$, is called the natural square and is not the square of the multiplication.

Remark. The two squares are the two central readings of the product with one factor conjugated, and Introduction to the General Plain Sesqualgebra keeps them apart by the caution that both are called the square in different places of the corpus. The present article uses only the first, and the second appears once, in §The Two Squares, where the contrast is computed.

The Square Is Hermitian

Theorem. For every $\tilde Q$ the square $\tilde Q\star\tilde Q$ is Hermitian,

$$ \boxed{\;\bigl(\tilde Q\star\tilde Q\bigr)^{*}=\tilde Q\star\tilde Q\;}, $$

and it is the product of $\tilde Q$ with its own conjugate.

Proof. The square is $\tilde Q\tilde Q^{*}$, and the involution reverses the order and is of order two, so $(\tilde Q\tilde Q^{*})^{*}=(\tilde Q^{*})^{*}\tilde Q^{*}=\tilde Q\tilde Q^{*}$.

Remark. Self-adjointness is the first structural fact of the square and the reason the idempotents of the multiplication reduce to the Hermitian idempotents of the algebra, in Square Roots of One and the Idempotents of the General Plain Sesqualgebra. It also says that the square takes its values in the subspace $\mathbb{M}_+$, so the cone generated by the squares is a cone inside $\mathbb{M}_+$.

The Scalar Part

Theorem (the scalar part of the square). For every $\tilde Q_1,\tilde Q_2$ the scalar part of the product is sesquilinear in the two arguments,

$$ \mathrm{Sc}\bigl(\tilde Q_1\star\tilde Q_2\bigr)=\sum_{\mu\in\{t,x,y,z\}}Q_{1,\mu}\overline{Q_{2,\mu}} , $$

and the square of $\tilde Q$ has the scalar part

$$ \mathrm{Sc}\bigl(\tilde Q\star\tilde Q\bigr)=\sum_{\mu\in\{t,x,y,z\}}Q_\mu\overline{Q_\mu}=\sum_{\mu\in\{t,x,y,z\}}\lvert Q_\mu\rvert^{2} , $$

a nonnegative real number that vanishes if and only if $\tilde Q=0$.

Proof. The scalar part of $\tilde Q_1\tilde Q_2^{*}$ is $Q_{1t}\overline{Q_{2t}}-(\vec Q_1,-\overline{\vec Q_2})=Q_{1t}\overline{Q_{2t}}+(\vec Q_1,\overline{\vec Q_2})=\sum_\mu Q_{1,\mu}\overline{Q_{2,\mu}}$, by the scalar–vector form of the product of Introduction to the General Plain Sesqualgebra, §The Scalar Part of the Multiplication. At $\tilde Q_1=\tilde Q_2$ the sum is $\sum_\mu\lvert Q_{2,\mu}\rvert^{2}$, a sum of modulus squares; it is nonnegative, and it vanishes only when every coordinate vanishes.

Remark. The display is the reason the square is non-vanishing: a nonzero square has a strictly positive scalar part, so it is not zero and its scalar part fixes a genuine sign. It is also the reason the general plain bilinear sum $\sum_\mu Q_\mu^{2}$ and the scalar part of the square are not the same function: the first is a general plain bilinear sum, the second a sum of modulus squares, and they agree exactly on the real subspace.

The Coordinates of the Square

Proposition. In the scalar–vector form, with $\tilde Q=Q_te_t+\vec Q$,

$$ \tilde Q\star\tilde Q=\Bigl(\sum_{\mu\in\{t,x,y,z\}}\lvert Q_\mu\rvert^{2}\Bigr)e_t+\bar Q_t\vec Q-Q_t\overline{\vec Q}-\vec Q\times\overline{\vec Q} , $$

and the vector part is the sum of the three terms displayed.

Proof. Multiply $\tilde Q=Q_t+\vec Q$ by $\tilde Q^{*}=\bar Q_t-\overline{\vec Q}$ with the multiplication of the basis; the scalar part is the sum of the modulus squares by the theorem above, and the vector part collects the terms $\bar Q_t\vec Q$ and $-Q_t\overline{\vec Q}$ from the two scalars and the cross term $-\vec Q\times\overline{\vec Q}$.

Remark. The vector part need not vanish, so the square of an element is a genuinely biquaternionic Hermitian element and not a real scalar. The real scalar part is the only part whose sign is fixed, and it is the part the properness of the cone is read from.

The Squares of the Basis

Example. Every basis element squares to the unit,

$$ e_\mu\star e_\mu=e_t , \qquad \mu\in\{t,x,y,z\} , $$

and likewise $i\vec e_\mu\star i\vec e_\mu=e_t$ for the three vector directions. Indeed $\vec e_\mu\star \vec e_\mu=\vec e_\mu \vec e_\mu^{*}=\vec e_\mu(-\vec e_\mu)=-\vec e_\mu^{2}=e_t$, and $e_t\star e_t=e_te_t^{*}=e_t$; for $i\vec e_\mu$ one has $(i\vec e_\mu)^{*}=i\vec e_\mu$ and $(i\vec e_\mu)^{2}=i^{2}\vec e_\mu^{2}=(-1)(-e_t)=e_t$, so $i\vec e_\mu\star i\vec e_\mu=(i\vec e_\mu)^{2}=e_t$.

Remark. The basis elements are therefore not small squares but unit-sized ones, and the square of a Hermitian element of square $e_t$ is $e_t$ throughout. The family of Hermitian elements with $\tilde Q^{2}=e_t$ is the family that the matrix model sends to the Hermitian involutions, and it is the family that parametrises the projections of Square Roots of One and the Idempotents of the General Plain Sesqualgebra, each projection being $\tfrac12(e_t+\tilde\Xi)$ with $\tilde\Xi$ of square $-e_t$.

The Two Squares

Proposition. The square of the multiplication and the natural square agree exactly on the elements whose four coordinates are real,

$$ \tilde Q\star\tilde Q=\tilde Q\tilde Q^{\natural} \quad\Longleftrightarrow\quad Q_\mu\in\mathbb{R}\ \text{ for }\ \mu\in\{t,x,y,z\} , $$

so they agree on the real subspace $\mathbb{R}^{4}$ and differ off it. The simplest witness is $\tilde Q=ie_t$, where the square of the multiplication is $e_t$ and the natural square is $-e_t$, a difference of $2e_t$.

Proof. Write the coordinates in the real form $\tilde Q=\tilde Q_1+i\tilde Q_2$ with $\tilde Q_1,\tilde Q_2\in\mathbb{H}_{\mathbb{B}}$ real quaternions. The star and the natural conjugation act on that form by $\tilde Q^{*}=\tilde Q_1^{\natural}-i\tilde Q_2^{\natural}$ and $\tilde Q^{\natural}=\tilde Q_1^{\natural}+i\tilde Q_2^{\natural}$, because both maps are $\mathbb{R}$-linear and the star conjugates the coefficient $i$; hence the difference of the two squares is $\tilde Q\tilde Q^{*}-\tilde Q\tilde Q^{\natural}=-2i\,\tilde Q\tilde Q_2^{\natural}$. If the two squares are equal then $\tilde Q\tilde Q_2^{\natural}=0$, that is $\tilde Q_1\tilde Q_2^{\natural}+i\tilde Q_2\tilde Q_2^{\natural}=0$; the first term is real and the second is the purely imaginary scalar $i\bigl(Q_{2t}^{2}+\lVert\vec Q_2\rVert^{2}\bigr)$, so the equation forces $\tilde Q_2\tilde Q_2^{\natural}=0$, whence $\tilde Q_2=0$ and the four coordinates of $\tilde Q$ are real. Conversely, for real coordinates one has $\tilde Q^{*}=\tilde Q^{\natural}$ and the two squares coincide. At $\tilde Q=ie_t$ one has $\tilde Q_1=0$ and $\tilde Q_2=e_t$, so the difference is $-2i\,(ie_t)e_t=2e_t$, while $\tilde Q\star\tilde Q=i\bar i\,e_t=e_t$ and $\tilde Q\tilde Q^{\natural}=i^{2}e_t=-e_t$.

Remark. The distinction matters for the vanishing of the square. The natural square $\bigl(\sum_\mu Q_\mu^{2}\bigr)e_t$ vanishes exactly on the zero divisors of the algebra, a six-real-dimensional set, by Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors; the square of the multiplication vanishes exactly at $\tilde Q=0$. The two statements are the two halves of the corpus's treatment of squares, and the cone of the multiplication is built from the first and not the second.

The Cone of Squares

The Definition

Definition. The cone of squares of the sesquilinear multiplication is the set of the finite sums of squares,

$$ C(\mathbb{B})=\Bigl\{\sum_{i=1}^{n}\tilde Q_i\star\tilde Q_i : n\ge0,\ \tilde Q_i\in\mathbb{B}\Bigr\} , $$

the empty sum being $0$.

Remark. The cone is the biquaternion instance of the algebraic positive cone of Hermitian Squares and the Algebraic Positive Cone, and the square of the multiplication is the derived square $x\star x=xx^{*}$ of Sesqualgebras; the two definitions therefore coincide and $C(\mathbb{B})$ is the algebraic positive cone of the algebra with the involution ${}^{*}$.

Elementary Closure

Proposition. The cone contains $0$, it is closed under addition, and it is contained in the Hermitian subspace:

$$ 0\in C(\mathbb{B}) , \qquad C(\mathbb{B})+C(\mathbb{B})\subseteq C(\mathbb{B}) , \qquad C(\mathbb{B})\subseteq\mathbb{M}_+ . $$

Proof. The element $0$ is the empty sum. Two finite sums add by concatenation. Each square is Hermitian by §The Square of an Element, and a sum of Hermitian elements is Hermitian.

Remark. The closure under addition is what distinguishes the cone from the set of the squares, which is not closed: the sum of two squares is a sum of two squares but need not be a square. The cone is the smallest set that contains every square and is closed under addition, and it imposes no further relation.

Proposition. The cone is generated by the squares of the involution as well: $C(\mathbb{B})=\bigl\{\sum_i\tilde Q_i^{*}\star\tilde Q_i^{*}\bigr\}$, and $x\star x=xx^{*}$ while $x^{*}\star x^{*}=x^{*}x$ for every $x$.

Proof. Replacing each $\tilde Q_i$ by $\tilde Q_i^{*}$ in the definition and using $*^{2}=\mathrm{id}$ exchanges $\tilde Q\tilde Q^{*}$ and $\tilde Q^{*}\tilde Q$. The identifications are $\tilde Q\star\tilde Q=\tilde Q\tilde Q^{*}$ and $\tilde Q^{*}\star\tilde Q^{*}=\tilde Q^{*}\tilde Q$.

Remark. The cone can therefore be read from the two sides of the involution, and the two readings agree set-theoretically even though the individual squares differ.

The Scalars of the Cone

Proposition. The square semiring of the base is the set of the nonnegative real numbers,

$$ Q(\mathbb{C})=\Bigl\{\sum_j\mu_j\bar\mu_j : \mu_j\in\mathbb{C}\Bigr\}=\mathbb{R}_{\ge0} , $$

and for every $\lambda\in\mathbb{R}_{\ge0}$ the cone is preserved, $\lambda\,C(\mathbb{B})\subseteq C(\mathbb{B})$.

Proof. The square semiring is that of Hermitian Squares and the Algebraic Positive Cone, §The Scalar Action; over $\mathbb{C}$ with the conjugation each term $\mu\bar\mu$ is a nonnegative real number and every nonnegative real number occurs, and the general element of the semiring is a finite sum of such. The invariance is the proposition of that article, applied to the base $\mathbb{C}$ and the involution ${}^{*}$.

Remark. The scalars that preserve the cone are the nonnegative real numbers and no more, and they are the fixed ring $\mathbb{R}$ cut down to the nonnegative half. The cone is therefore not a complex cone but a real one, which is the algebraic shadow of the sesquilinearity of the square.

Congruence

Proposition. Let $\tilde Q_1\in C(\mathbb{B})$ and $\tilde Q_2\in\mathbb{B}$. Then the congruence of $\tilde Q_1$ by $\tilde Q_2$,

$$ \tilde Q_1\longmapsto\tilde Q_2^{*}\,\tilde Q_1\,\tilde Q_2 , $$

is again an element of the cone, so the congruence of every element carries the cone into itself.

Proof. This is the invariance of the algebraic positive cone under the congruence, Hermitian Squares and the Algebraic Positive Cone, §Congruence: it suffices to treat $\tilde Q_1=\tilde Q_3^{*}\tilde Q_3$ and to use the associativity of the algebra, $\tilde Q_2^{*}(\tilde Q_3^{*}\tilde Q_3)\tilde Q_2=(\tilde Q_3\tilde Q_2)^{*}(\tilde Q_3\tilde Q_2)$. The general element is a sum of such.

Remark. The congruence is built from the algebra product, not from the multiplication alone, and that is why the proposition needs the associativity of the algebra. It is the algebraic content of the statement that the positive elements stay positive under a change of frame; for an invertible $\tilde Q_2$ the congruence is a bijection of the cone.

Properness

Theorem. The cone of squares is proper,

$$ \boxed{\;C(\mathbb{B})\cap\bigl(-C(\mathbb{B})\bigr)=\{0\}\;}, $$

so the preorder $\tilde Q_2\preceq\tilde Q_3$ defined by $\tilde Q_3-\tilde Q_2\in C(\mathbb{B})$ is a partial order.

Proof. Take the additive map $f=\mathrm{Re}\circ\mathrm{Sc}:\mathbb{B}\to\mathbb{R}$, which is additive because the scalar part is additive and the real part is additive. On a square, $f(\tilde Q\star\tilde Q)=\mathrm{Sc}(\tilde Q\star\tilde Q)=\sum_\mu\lvert Q_\mu\rvert^{2}$ by §The Square of an Element, which is a nonnegative real number and is strictly positive for $\tilde Q\neq0$. The sufficient criterion for properness of Hermitian Squares and the Algebraic Positive Cone, §The Order, therefore applies with $K=\mathbb{R}$, and gives $C(\mathbb{B})\cap(-C(\mathbb{B}))=\{0\}$ and $f(\tilde Q_4)>0$ for every nonzero $\tilde Q_4\in C(\mathbb{B})$. The preorder is a partial order by the same article.

Remark. Properness is the point at which the biquaternion cone differs from the cones of the two bilinear products: for the products $\tilde Q_1\tilde Q_2$ and $\tilde Q_1^{\natural}\tilde Q_2$ the involution is trivial on the scalars and the cone of the squares is not proper, by the non-proper examples of Hermitian Squares and the Algebraic Positive Cone. The conjugation in the involution is exactly what makes the square a sum of modulus squares and the cone proper, and the next section reads the criterion in the coordinates.

Remark (the sign of a positive element). The map $f=\mathrm{Re}\circ\mathrm{Sc}$ is the strict positive functional that the criterion supplies, and it is the only scalar datum used. It is not an isomorphism-invariant of the cone alone, but the properness it proves is a property of the cone, so the choice does not enter the conclusion.

The Cone and the Hermitian Subspace

The Cone Inside the Subspace

Proposition. The cone is a subset of the Hermitian subspace, it is closed under addition, and its additive hull is the whole Hermitian subspace,

$$ C(\mathbb{B})\subseteq\mathbb{M}_+ , \qquad \mathbb{M}_+=C(\mathbb{B})-C(\mathbb{B}) . $$

Proof. The inclusion is §Elementary Closure. For the hull, that an additive submonoid generates its group is general; the content is that the group generated by $C(\mathbb{B})$ is all of $\mathbb{M}_+$ and not a proper subgroup. Under the matrix model $\mathsf{M}_2$ of Biquaternion $2\times2$ Matrix Element Representation, the cone is the set of the sums of the Hermitian squares $\mathsf{M}_2(\tilde Q)\mathsf{M}_2(\tilde Q)^{\dagger}$, that is the set of the Hermitian elements of $M_2(\mathbb{C})$ that are sums of the products $XX^{\dagger}$, which are exactly the elements with nonnegative eigenvalues; every Hermitian matrix is the difference of two such, by its spectral decomposition. The transport by $\mathsf{M}_2$ gives the statement.

Remark. The two displays are the algebraic statement that the Hermitian subspace is the linear span of the cone, and they are the reason the cone is the right positive object of the Hermitian half: the half is exactly the set of the differences of positive elements, and the positive elements are those the cone contains.

The Two Halves and the Cone

Proposition. The cone is disjoint from the anti-Hermitian subspace except at $0$,

$$ C(\mathbb{B})\cap\mathbb{M}_-=\{0\} , $$

and the skew-Hermitian elements are exactly the elements of the form $\tilde Q-\tilde Q^{*}$.

Proof. An element of the intersection is Hermitian and skew-Hermitian, hence $0$; the second statement is the decomposition $\tilde Q=\tfrac12(\tilde Q+\tilde Q^{*})+\tfrac12(\tilde Q-\tilde Q^{*})$ into the Hermitian and skew-Hermitian parts of Hermitian and Skew-Hermitian Elements.

Remark. The cone is a cone of the positive half and crosses the negative half only at the origin, which is the element-level form of the properness proved above.

The Cone Among the Remarkable Subspaces

The remarkable subspaces of Remarkable Subspaces and the Four General Products, Forms and Norms place the cone as follows: it lies in the Hermitian subspace $\mathbb{M}_+$, its intersection with the centre is the nonnegative real multiples of the unit, and it meets the anti-Hermitian subspace only at the origin. The scalar part and the trace are the two readings of the same projection, and the cone is the part of $\mathbb{M}_+$ on which they are nonnegative.

Remark. The table is the placement and not a new computation: it reads the inclusion $C(\mathbb{B})\subseteq\mathbb{M}_+$ against the other five subspaces. The entry for the real subspace is that $\tilde Q\star\tilde Q=\bigl(\sum_\mu Q_\mu^{2}\bigr)e_t$ for a real $\tilde Q$, a nonnegative multiple of the unit, so the real half meets the cone in the ray of the unit and in nothing else.

The Vanishing Square and the Zero Divisors

Theorem. The square of the multiplication vanishes only at the origin,

$$ \boxed{\;\tilde Q\star\tilde Q=0 \quad\Longleftrightarrow\quad \tilde Q=0\;}, $$

whereas the natural square vanishes exactly on the zero divisors of the algebra,

$$ \tilde Q\tilde Q^{\natural}=0 \quad\Longleftrightarrow\quad \sum_{\mu\in\{t,x,y,z\}}Q_\mu^{2}=0 , $$

a six-real-dimensional set.

Proof. The first is the theorem on the scalar part: $\lvert\mathrm{Sc}(\tilde Q\star\tilde Q)\rvert=\sum_\mu\lvert Q_\mu\rvert^{2}$ vanishes only at $\tilde Q=0$. The second is $\tilde Q\tilde Q^{\natural}=\bigl(\sum_\mu Q_\mu^{2}\bigr)e_t$ by the definition of the natural square, and the classification of the zero divisors of $\mathbb{B}$ is Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors.

Remark. The two vanishing sets are the two faces of the square: the square of the multiplication has the smallest possible vanishing set, the origin, and the natural square has the largest possible nontrivial one, the zero divisor set. The idempotents of the multiplication are the nonzero Hermitian elements with $\tilde Q\star\tilde Q=\tilde Q$, and they are the projections of Square Roots of One and the Idempotents of the General Plain Sesqualgebra, not the zero divisors.

The Matrix Model

Proposition. Under the isomorphism $\mathsf{M}_2:\mathbb{B}\to M_2(\mathbb{C})$ of Biquaternion $2\times2$ Matrix Element Representation, which sends the involution to the conjugate transpose, the square becomes

$$ \mathsf{M}_2\bigl(\tilde Q\star\tilde Q\bigr)=\mathsf{M}_2(\tilde Q)\mathsf{M}_2(\tilde Q)^{\dagger} , $$

the cone of squares becomes the set of the Hermitian elements of $M_2(\mathbb{C})$ that are sums of the products $XX^{\dagger}$, and the additive map of the properness theorem becomes the trace,

$$ \mathrm{tr}\bigl(\mathsf{M}_2(\tilde Q)\mathsf{M}_2(\tilde Q)^{\dagger}\bigr)=\sum_{i,j=1}^{2}\lvert X_{ij}\rvert^{2}=2\sum_{\mu\in\{t,x,y,z\}}\lvert Q_\mu\rvert^{2} . $$

Proof. The first identity is the multiplicativity of $\mathsf{M}_2$ on the sesquilinear product, $\mathsf{M}_2(\tilde Q_1\star\tilde Q_2)=\mathsf{M}_2(\tilde Q_1)\mathsf{M}_2(\tilde Q_2)^{\dagger}$, together with $\mathsf{M}_2(\tilde Q^{*})=\mathsf{M}_2(\tilde Q)^{\dagger}$. The trace identity is $\sum_{i,j}\lvert X_{ij}\rvert^{2}$ for the matrix $\mathsf{M}_2(\tilde Q)$, and the comparison with the scalar part is the factor two from $\mathrm{tr}(I)=2$ and the identity $\mathrm{tr}(\mathsf{M}_2(\tilde Q))=2Q_t$ of Biquaternion $2\times2$ Matrix Element Representation together with the fact that the modulus is computed on the four coordinates.

Remark. The trace is the matrix reading of the map $\mathrm{Sc}$, and the factor $2$ is the only difference between the two. The properness of the cone is therefore the properness of the cone of the Hermitian squares of $M_2(\mathbb{C})$, which is the standard proper cone of the matrix algebra.

Summary

The square of the sesquilinear multiplication is $\tilde Q\star\tilde Q=\tilde Q\tilde Q^{*}$; it is Hermitian, its scalar part is the sum of the modulus squares of the four coordinates, and it vanishes only at the origin. The sums of the squares form the cone $C(\mathbb{B})$, which contains $0$, is closed under addition, is contained in the Hermitian subspace $\mathbb{M}_+$ whose additive hull it generates, is preserved by the nonnegative real scalars, is invariant under the congruence $\tilde Q_1\mapsto\tilde Q_2^{*}\tilde Q_1\tilde Q_2$, and is proper, so the preorder it defines is a partial order. The cone is the algebraic positive cone of the algebra with the involution ${}^{*}$, the biquaternion instance of Hermitian Squares and the Algebraic Positive Cone, and it is proper where the cones of the two bilinear products are not. The natural square $\tilde Q\tilde Q^{\natural}=\bigl(\sum_\mu Q_\mu^{2}\bigr)e_t$ vanishes on the zero divisors, and the square of the multiplication vanishes at the origin alone; the cone is built from the first of the two squares and not the second.

Summary of Notation

Further Reading

• I. N. Herstein, Rings with Involution (University of Chicago Press, 1976), for the positive cone of an involutive ring and the Hermitian squares.
• Nathan Jacobson, Structure and Representations of Jordan Algebras (American Mathematical Society, 1968), for the cone of squares and the order it defines in a Jordan algebra.
• Max Koecher, The Minnesota Notes on Jordan Algebras and Their Applications (Springer Lecture Notes in Mathematics 1710, 1999), for the algebraic positive cone and the order of a Jordan algebra.
• Kevin McCrimmon, A Taste of Jordan Algebras (Springer, 2004), for the quadratic form of an algebra with involution and the cone of the Hermitian squares.
• Richard S. Pierce, Associative Algebras (Springer, 1982), for the Hermitian elements of a finite-dimensional algebra and the products $xx^{*}$.