The Specificities of the Symmetric Plain Algebra

Introduction

The symmetric plain algebra of $\mathbb{H}$ carries 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. This article gathers the properties that characterise it — commutativity, the unit, the Jordan identity and the definiteness of its generic norm — and the classes that degenerate in it: the idempotents reduce to $0$ and $e_t$, the nilpotents reduce to $0$, the Peirce decomposition is trivial, and the formal reality of its underlying product fails. It closes with the isolation of the block among the companion products of the chapter.

This is not the algebra $\mathbb{H}$. The two algebras share the space, the addition, the real scalars and the unit, but $\bullet$ is a different multiplication and the projection onto the scalar line is not multiplicative. Throughout, the natural sign ${}^{\natural}$ is the quaternion conjugation, the bar is the identity because $\mathbb{R}$ has no nontrivial involution, and the star coincides with the natural sign and gives nothing new.

The Properties of the Block

Commutativity, the Unit and the Jordan Identity

Theorem. The product $\bullet$ is commutative, unital with two-sided unit $e_t$, satisfies the Jordan identity

$$ \boxed{\;(\tilde q_1\bullet\tilde q_2)\bullet(\tilde q_1\bullet\tilde q_1)=\tilde q_1\bullet\bigl(\tilde q_2\bullet(\tilde q_1\bullet\tilde q_1)\bigr)\;}, $$

and is not associative; it is power-associative, all bracketings of a power of one element agreeing.

Proof. Commutativity and the unit are Introduction to the Symmetric Plain Algebra, and so is the Jordan identity, which follows because $\bullet$ is the symmetrisation of the associative plain product. Associativity fails at $\tilde q_1=\tilde q_2=\vec e_x$, $\tilde q_3=\vec e_y$: $(\vec e_x\bullet \vec e_x)\bullet \vec e_y=-\vec e_y$ against $\vec e_x\bullet(\vec e_x\bullet \vec e_y)=0$. Power-associativity holds because the subalgebra generated by one element is the commutative associative span of its powers, the block being the symmetrisation of an associative algebra.

So the block is a commutative unital Jordan algebra over $\mathbb{R}$, and it is a special Jordan algebra, being the symmetrisation of an associative one.

The Formal Reality and the Definiteness of the Norm

Definition. A Jordan algebra over $\mathbb{R}$ is formally real when a sum of squares vanishes only when every term vanishes:

$$ x_1^2+x_2^2+\cdots+x_n^2=0\implies x_1=x_2=\cdots=x_n=0, $$

where the square is the square of the Jordan product.

Theorem (the definiteness of the norm). The generic norm $\langle\tilde q\rangle_{\natural}=\tilde q\tilde q^{\natural}=q_t^2+|\vec q|^2$ is positive definite and vanishes only at the origin; consequently the block has no nonzero nilpotent, every nonzero element is invertible, and the block is a degree-two Jordan algebra with an anisotropic generic norm.

Proof. The norm is a sum of squares of the coordinates, so it is non-negative and vanishes only when $\tilde q=0$; no nonzero nilpotent follows from the quadratic identity $\tilde q^2-2q_t\tilde q+\langle\tilde q\rangle_{\natural}e_t=0$, and invertibility is Norm and Invertibility in the Symmetric Plain Algebra.

Theorem (the product with the identity involution is not formally real). With the identity involution the block is not formally real in the sense of the definition:

$$ \vec e_x\bullet \vec e_x+e_t\bullet e_t=-e_t+e_t=0, $$

with both terms nonzero.

Proof. $\vec e_x\bullet \vec e_x=\vec e_x^2=-e_t$ and $e_t\bullet e_t=e_t^2=e_t$. In general the square of a pure element is $\vec q_a\bullet\vec q_a=-|\vec q_a|^2e_t$, negative, while the square of $e_t$ is $+e_t$, so the square of a pure unit element cancels the square of the unit. Equivalently, the trace form $\operatorname{Tr}(L^{\bullet}_{\tilde q_1}L^{\bullet}_{\tilde q_2})=4q_{1t}q_{2t}-2\,\vec q_1\cdot\vec q_2$ is indefinite of signature $(1,3)$, so the Euclidean criterion fails too.

Theorem (the natural involution is formally real). The quaternion conjugation $\natural$ is an involution of the block, $(\tilde q_1\bullet\tilde q_2)^{\natural}=\tilde q_1^{\natural}\bullet\tilde q_2^{\natural}$, its self-adjoint part is the scalar line $\mathbb{R}e_t$, and a sum of squares of self-adjoint elements is a sum of positive scalars, vanishing only when every term vanishes. The natural involution of the block is formally real.

Proof. $(\tilde q_1\bullet\tilde q_2)^{\natural}=\tfrac12(\tilde q_2^{\natural}\tilde q_1^{\natural}+\tilde q_1^{\natural}\tilde q_2^{\natural})=\tilde q_1^{\natural}\bullet\tilde q_2^{\natural}$; the fixed set of $\natural$ is $\mathbb{R}e_t$, and $(a e_t)^2=a^2e_t$ with $a^2\geq0$, so a sum $\sum a_k^2e_t$ vanishes only when every $a_k$ vanishes.

Remark (the two readings of the menu's formal reality). The outline of the block puts "the formal reality of $\mathbb{H}$ as a Jordan algebra" among the properties. The statement is correct read on the natural involution — the algebra $\mathbb{H}$ with its conjugation is formally real, its self-adjoint part being the scalar line — and it is the definiteness of the generic norm that it produces; the statement is false read on the product with the identity involution, where the witness $\vec e_x^2+e_t^2=0$ stands. The two readings differ, and the article states both: the block is a special Jordan algebra with a formally real natural involution and a definite generic norm, but its underlying product is not formally real. The self-adjoint part being one-dimensional, the involution formal reality is the least informative of the two, and the definiteness of the norm is the operative property.

Remark (which operation carries the sum-of-squares condition of the product). The sum-of-squares condition for the product is met by the general quaternionic product, whose square is $|\tilde q|^2e_t$ and is non-negative; but it is not a commutative Jordan product, $\mathrm{GQA}$ being non-commutative, so it does not carry the formal reality of a Jordan algebra. The Euclidean spin factor $JSpin_4\cong H_2(\mathbb{C})$, the degree-two formally real Jordan algebra on this space, is $\mathbb{R}\oplus\mathbb{R}^3$ with the product

$$ (a\cdot e_t+\vec q_a)\bullet'(b\cdot e_t+\vec v)=(ab+\vec q_a\cdot\vec v)e_t+(a\vec v+b\vec q_a), $$

whose vector-vector sign is opposite to the block's; it is the spin factor of Jordan Algebras, §Spin Factors, and it is not one of the companion products of the menu. So the block is the only commutative unital Jordan algebra among the companion products, and the owner of the definite norm, but its product is not formally real, the sum-of-squares condition for the product being carried by the general quaternionic product, which is not a commutative Jordan product.

The Degenerate Classes

The Idempotents

Theorem. The idempotents of the block are exactly $0$ and $e_t$:

$$ \boxed{\;\tilde\pi\bullet\tilde\pi=\tilde\pi\iff\tilde\pi\in\{0,e_t\}\;}. $$

Proof. The square of the block is the plain square, so the equation is $\tilde\pi^2=\tilde\pi$. The vector part gives $(2\pi_t-1)\vec\pi=0$, so either $\vec\pi=0$, and then $\pi_t^2=\pi_t$, giving $\tilde\pi\in\{0,e_t\}$, or $\pi_t=\tfrac12$, and then the scalar part gives $|\vec\pi|^2=-\tfrac14$, which has no real solution.

Corollary (the class is two-point). The idempotent class is a two-point class, the two trivial idempotents; the block has no nontrivial idempotent, and therefore no projector calculus.

The Nilpotents and the Square-Zero Elements

Theorem. The block has no nonzero nilpotent; its square-zero elements are exactly the origin:

$$ \boxed{\;\tilde q\bullet\tilde q=0\iff\tilde q=0\;}. $$

Proof. The equation is $\tilde q^2=0$, whose scalar and vector equations are $q_t^2-|\vec q|^2=0$ and $2q_t\vec q=0$. If $\vec q=0$ then $q_t=0$; if $q_t=0$ then $|\vec q|^2=0$ and $\vec q=0$.

Remark (zero divisors are not nilpotents). The block has a large zero-divisor set, the nonzero pure elements, but no nonzero nilpotent: the two classes are distinct, a zero divisor being an element with a nonzero partner and a nilpotent an element whose own square vanishes, and they coincide neither here nor in general.

The Degree-Two Structure

The Quadratic Identity and the Inverse

Theorem. Every element satisfies the quadratic identity $\tilde q^2-T(\tilde q)\tilde q+\langle\tilde q\rangle_{\natural}e_t=0$, and it is invertible exactly when $\langle\tilde q\rangle_{\natural}\neq0$, that is, when $\tilde q\neq0$, with Jordan inverse

$$ \boxed{\;\tilde q^{-1}=\frac{T(\tilde q)e_t-\tilde q}{\langle\tilde q\rangle_{\natural}}=\frac{\tilde q^{\natural}}{\langle\tilde q\rangle_{\natural}},\qquad \det L^{\bullet}_{\tilde q}=q_t^2\,\langle\tilde q\rangle_{\natural}\;}. $$

Proof. These are Norm and Invertibility in the Symmetric Plain Algebra, and the determinant is The Multiplication Operators of the Symmetric Plain Algebra, below this article in this category. The block is of degree two, so the whole degree-two structure is carried by the trace $T$, the norm $\langle\cdot\rangle_{\natural}$ and the unit.

The Peirce Decomposition

Theorem. For the two idempotents the Peirce decomposition is trivial:

$$ J_1(e_t)=\mathbb{H},\quad J_{1/2}(e_t)=J_0(e_t)=0; \qquad J_0(0)=\mathbb{H},\quad J_1(0)=J_{1/2}(0)=0, $$

where $J_\lambda(c)=\{\tilde q : c\bullet\tilde q=\lambda\tilde q\}$.

Proof. The multiplication operator of $e_t$ is the identity, $L^{\bullet}_{e_t}=\mathrm{id}$, so every element is in the eigenvalue $1$ space and the other two spaces vanish; the multiplication operator of $0$ is the zero operator, so every element is in the eigenvalue $0$ space.

Corollary. A nontrivial idempotent would decompose the block into three Peirce subspaces of dimensions $1,\ 2,\ 1$; since the block has none, its Peirce decomposition is degenerate and carries no projector calculus. The decomposition is recorded here and nowhere else in the chapter.

The Isotropic Cone

Proposition. The generic norm has no isotropic element but the origin, while the quadratic trace $\mathrm{Sc}(\tilde q\bullet\tilde q)=q_t^2-|\vec q|^2$ has the isotropic cone

$$ q_t^2=|\vec q|^2, $$

of real dimension three in the algebra.

Proof. The norm is positive definite; the quadratic trace is the plain quadratic form $\mathrm{Sc}(\tilde q^2)$, whose zero set is the displayed cone. The cone of the quadratic trace is that of the trace form of The Trace Form and the Invariance of the Symmetric Plain Algebra, below this article in this category, and it must not be confused with the vanishing of the generic norm.

The Isolation of the Block among the Companion Products

Theorem. Among the companion products of The 6 Products of the Quaternioid Real Space, the symmetric plain product is the unique one that is commutative, unital and Jordan, and it is the unique one whose algebra is a commutative unital Jordan algebra.

Proof. The algebra product $\tilde q_1\tilde q_2$ is associative and unital but not commutative; the antisymmetric plain product $\mathrm{APA}$ is alternating, has no unit and is not commutative; the general quaternionic product $\mathrm{GQA}$ is not commutative and only left unital. So $\mathrm{SPA}$ is the only one of the companion products with all three properties. The three properties single out the block, but they do not make its product formally real: the block is a Jordan algebra whose underlying product is not formally real, and the operation whose square is non-negative, $\mathrm{GQA}$, is not a commutative Jordan product.

Remark (the three generating operations). The isolation is read off three operations and no more: commutativity separates the block from the algebra product and the general quaternionic product; the unit separates it from the structures without a two-sided unit; and the Jordan identity completes the criterion. The three together single out $\mathrm{SPA}$; each alone fails to.

Summary

The symmetric plain algebra of $\mathbb{H}$ is a commutative unital special Jordan algebra over $\mathbb{R}$, not associative but power-associative, and with a positive definite generic norm $\langle\tilde q\rangle_{\natural}=q_t^2+|\vec q|^2$: sums of squares of the coordinates vanish only at the origin, the block has no nonzero nilpotent, and every nonzero element is invertible. Its underlying product is nevertheless not formally real with the identity involution: $\vec e_x\bullet \vec e_x+e_t\bullet e_t=-e_t+e_t=0$ with both terms nonzero, because the square of a pure element is negative, and its trace form is indefinite of signature $(1,3)$; the natural involution $\natural$ of the block is formally real, with the scalar line as self-adjoint part, and the definiteness of the generic norm is the operative property the outline calls the formal reality of $\mathbb{H}$ as a Jordan algebra. Its idempotents are exactly $0$ and $e_t$, so its class is two-point and it has no nontrivial projector; its nilpotents are only $0$, so its zero divisors, the nonzero pure elements, are not nilpotents; its Peirce decomposition is degenerate, for the same reason; and its generic norm has no isotropic element but the origin, while its quadratic trace has the indefinite cone $q_t^2=|\vec q|^2$. The degree-two structure is carried by the trace $T(\tilde q)=2q_t$, the norm $\langle\tilde q\rangle_{\natural}=q_t^2+|\vec q|^2$, the quadratic identity and the Jordan inverse. Among the companion products the block is the unique one that is commutative, unital and Jordan, and the criterion is read off those three operations.

Summary of Notation

Further Reading

• Introduction to the Symmetric Plain Algebra, for the product, the unit and the Jordan identity.
• Norm and Invertibility in the Symmetric Plain Algebra, for the norm, the trace and the inverse.
• Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the zero divisors that are not nilpotents.
• The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra, below this article in this category, for the square and the inverse in full.
• Jordan Algebras, for the special Jordan algebras, the formal reality and the classification of the formally real ones.
• The 6 Products of the Quaternioid Real Space and The Algebraic Structures over the Quaternioid $\mathbb{R}$ Space, for the companion products and their laws.