The Specificities of the Symmetric Plain Algebra

Introduction

The symmetric plain product is the commutative half of the plain product of $\mathbb{B}=\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$,

$$ \tilde Q_1\bullet\tilde Q_2=\tfrac12(\tilde Q_1\tilde Q_2+\tilde Q_2\tilde Q_1), $$

the multiplication of Introduction to the Symmetric Plain Algebra. It makes the complex space a commutative unital Jordan algebra, and of the twelve structures of The Algebraic Structures over the Biquaternion $\mathbb{C}$ Space it is the only one that is a Jordan product. What is specific to the block is the structure a Jordan product carries and an arbitrary commutative product does not: a square that is the plain square, a multiplication operator that is the symmetric half of the two plain multiplications, a quadratic identity of degree two with a generic trace, a generic norm and an inverse, a quadratic representation, and a Peirce theory at the idempotents. Each is an explicit computation on the four coordinates, and each is developed below in the block's own terms.

The block is exact in the opposite sense as well. The Jordan law is weaker than associativity, so the block does not inherit the parent's operator calculus: no non-trivial idempotent gives a projector, a square-zero element gives an operator of index three rather than a square-zero operator, the Peirce decomposition is cubic rather than quadratic, and the invertible elements form no group. Those absences are stated with the theorems they accompany, and §The Isolation among the Twelve names the one law from which the theorems and the absences both follow.

The Product and Its Square

Proposition (the shape of the product). For $\tilde Q_1=Q_{1t}e_t+\vec Q_1$ and $\tilde Q_2=Q_{2t}e_t+\vec Q_2$,

$$ \tilde Q_1\bullet\tilde Q_2=\bigl[Q_{1t}Q_{2t}-(\vec Q_1,\vec Q_2)\bigr]e_t+Q_{1t}\vec Q_2+Q_{2t}\vec Q_1. $$

Proof. The plain product is $\tilde Q_1\tilde Q_2=[Q_{1t}Q_{2t}-(\vec Q_1,\vec Q_2)]e_t+Q_{1t}\vec Q_2+Q_{2t}\vec Q_1+\vec Q_1\times\vec Q_2$, with $\vec Q_1\times\vec Q_2$ the cross product of the vector parts; the reverse product carries $-\vec Q_1\times\vec Q_2$ in its place, and the half-sum is the display.

The block is therefore the plain product with the cross term dropped, and the cross term is exactly the antisymmetric part of the plain product.

Proposition (square, commutativity, unit). The block is commutative, unital with unit $e_t$, and

$$ \tilde Q\bullet\tilde Q=\tilde Q^2=\bigl[Q_t^2-(\vec Q,\vec Q)\bigr]e_t+2Q_t\vec Q. $$

Proof. Commutativity is read on the shape of the product, and $e_t\bullet\tilde Q=\tfrac12(e_t\tilde Q+\tilde Qe_t)=\tilde Q$. The cross product cancels on the diagonal, $\vec Q\times\vec Q=0$, so $\tilde Q\bullet\tilde Q=\tilde Q\tilde Q$, and the displayed form is the plain square.

Corollary (the square of the block is the square of the algebra). Because the antisymmetric part of the plain product is alternating, it vanishes on the diagonal and carries none of the square. The square of the block and the square of the algebra are the same map, so the idempotents, the nilpotents, the square roots and the spectrum of the block are those of the algebra, and the generic norm of §The Degree-Two Structure is the algebraic norm. The block and the algebra part company at the product and not at the square.

Proposition (the block is not associative).

$$ \vec e_x\bullet(\vec e_x\bullet \vec e_y)=0,\qquad (\vec e_x\bullet \vec e_x)\bullet \vec e_y=-\vec e_y . $$

Proof. $\vec e_x\bullet \vec e_x=\vec e_x^2=-e_t$ and $\vec e_x\bullet \vec e_y=\tfrac12(\vec e_x\vec e_y+\vec e_y\vec e_x)=\tfrac12(\vec e_z-\vec e_z)=0$. The left side is then $\vec e_x\bullet0=0$ and the right side is $(-e_t)\bullet \vec e_y=-\vec e_y$.

Proposition (polarisation).

$$ \tilde Q_1\bullet\tilde Q_2=\tfrac12\bigl((\tilde Q_1+\tilde Q_2)^{\bullet2}-\tilde Q_1^{\bullet2}-\tilde Q_2^{\bullet2}\bigr). $$

Proof. The product is $\mathbb{C}$-bilinear and commutative, so the square of a sum expands as the sum of the two squares and twice the product.

So the quadratic map $\tilde Q\mapsto\tilde Q^{\bullet2}$, together with the linear structure, determines the product; the square is the primitive object and the product is its polarisation.

The Multiplication Operator

Proposition (the operator is the symmetric half). For every $\tilde Q_3$,

$$ L^{\bullet}_{\tilde Q_3}=\tfrac12\bigl(L_{\tilde Q_3}+R_{\tilde Q_3}\bigr). $$

Proof. $L^{\bullet}_{\tilde Q_3}\tilde Q_2=\tilde Q_3\bullet\tilde Q_2=\tfrac12(\tilde Q_3\tilde Q_2+\tilde Q_2\tilde Q_3)=\tfrac12(L_{\tilde Q_3}+R_{\tilde Q_3})\tilde Q_2$, and the identity of maps follows.

The operator of the block is thus the mean of a plain left and a plain right multiplication, and the two compose with cross terms; this is the exact sense in which the block is a symmetrisation of the parent's operator theory and not a restriction of it.

Theorem (the operator identity of the block). The block satisfies

$$ \boxed{\; L^{\bullet}_{\tilde Q_1}L^{\bullet}_{\tilde Q_1^{\bullet2}}=L^{\bullet}_{\tilde Q_1^{\bullet2}}L^{\bullet}_{\tilde Q_1}\qquad\text{for every }\tilde Q_1 \;}, $$

equivalently the Jordan identity $(\tilde Q_1^{\bullet2}\bullet\tilde Q_2)\bullet\tilde Q_1=\tilde Q_1^{\bullet2}\bullet(\tilde Q_2\bullet\tilde Q_1)$.

Proof. The block is the symmetrisation of the associative plain product, so it is Jordan-admissible by The Symmetric and Antisymmetric Parts of an Algebra Product; the Jordan identity is that admissibility, and for a commutative product it is equivalent to the operator identity, the two sides being $L^{\bullet}_{\tilde Q_1}L^{\bullet}_{\tilde Q_1^{\bullet2}}\tilde Q_2$ and $L^{\bullet}_{\tilde Q_1^{\bullet2}}L^{\bullet}_{\tilde Q_1}\tilde Q_2$ after commutativity is used twice.

Proposition (the block fails left alternativity).

$$ \vec e_x\bullet(\vec e_x\bullet \vec e_y)=0\neq-\vec e_y=(\vec e_x\bullet \vec e_x)\bullet \vec e_y . $$

Proof. From the non-associativity display above. Left alternativity is the demand $L^{\bullet}_{\tilde Q_1}{}^{2}=L^{\bullet}_{\tilde Q_1^{\bullet2}}$ at every element, and this is its failure.

Corollary (no projector from a non-trivial idempotent). A non-trivial idempotent $\tilde\Pi$ satisfies $L^{\bullet}_{\tilde\Pi}{}^{2}\neq L^{\bullet}_{\tilde\Pi}$. The unit alone gives a projector, $L^{\bullet}_{e_t}=\mathrm{id}$, and the origin gives the zero operator.

Proof. Write the idempotent as $\tilde\Pi=\tfrac12(e_t+\vec q)$ with $\vec q$ a pure root of $+1$, so $\vec q^2=e_t$ and the scalar coordinate of $\vec q$ vanishes; then $L^{\bullet}_{\tilde\Pi}=\tfrac12(\mathrm{id}+L^{\bullet}_{\vec q})$. If $L^{\bullet}_{\tilde\Pi}$ were a projector, squaring and comparing would give $(L^{\bullet}_{\vec q})^{2}=\mathrm{id}$. But for a pure $\vec q$ the square of its operator is $L^{\bullet}_{\vec q}{}^{2}\tilde Q_2=Q_{2t}-(\vec q,\vec W)\vec q$, which vanishes at every vector $\vec W$ of the two-dimensional complex subspace orthogonal to $\vec q$; such a $\vec W$ is nonzero, so $(L^{\bullet}_{\vec q})^{2}\neq\mathrm{id}$ and $L^{\bullet}_{\tilde\Pi}$ is not a projector.

Proposition (a square-zero element gives an operator of index three). For a pure isotropic $\tilde\Upsilon$, that is $Q_t=0$ and $(\vec Q,\vec Q)=0$, the operator $L^{\bullet}_{\tilde\Upsilon}$ satisfies $L^{\bullet}_{\tilde\Upsilon}{}^{2}\neq0$ and $L^{\bullet}_{\tilde\Upsilon}{}^{3}=0$, with image not contained in the kernel; the witness is $\tilde\Upsilon=\vec e_x+i\vec e_y$.

Proof. Square Roots of Zero and the Nilpotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space: the elements of square zero are exactly the pure isotropic vectors, and their left multiplication has the stated index, the image $\mathrm{span}\{e_t,\tilde\Upsilon\}$ not lying in the kernel.

So the block has neither of the two operator constructions of the parent: an idempotent produces no projector and a square-zero element produces no flag. Both failures are the failure of left alternativity, and neither is repaired by any choice of element.

The Idempotents and the Cubic Peirce Theory

Proposition (the idempotents of the block). The idempotents $\tilde\Pi\bullet\tilde\Pi=\tilde\Pi$ are the idempotents of the plain product: the origin, the unit $e_t$, and the elements

$$ \tilde\Pi=\tfrac12(e_t+\vec q),\qquad \vec q\ \text{pure},\ \vec q^2=e_t, $$

They are read off the quadratic identity at integral trace, $(T,N)=(0,0)$, $(2,1)$ and $(1,0)$, so that the non-trivial idempotents are exactly the isotropic ones.

Proof. The square of the block is the plain square, so the equation is $\tilde\Pi^2=\tilde\Pi$, classified in Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space; the quadratic identity of §The Degree-Two Structure is then evaluated at the three traces.

Corollary (the cubic Peirce identity). For every idempotent $\tilde\Pi$,

$$ L^{\bullet}_{\tilde\Pi}\bigl(2L^{\bullet}_{\tilde\Pi}-\mathrm{id}\bigr)\bigl(L^{\bullet}_{\tilde\Pi}-\mathrm{id}\bigr)=0, $$

equivalently $2L^{\bullet}_{\tilde\Pi}{}^{3}-3L^{\bullet}_{\tilde\Pi}{}^{2}+L^{\bullet}_{\tilde\Pi}=0$; the eigenvalues of $L^{\bullet}_{\tilde\Pi}$ lie in $\{0,\tfrac12,1\}$.

Proof. Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, verified at the witness $\tilde\Pi=\tfrac12(e_t+i\vec e_x)$, whose operator has the eigenvalues $0$, $\tfrac12$ and $1$.

Corollary (the Jordan Peirce decomposition). For a non-trivial idempotent,

$$ \mathbb{B}=\mathbb{B}_{1}\oplus\mathbb{B}_{1/2}\oplus\mathbb{B}_{0},\qquad \mathbb{B}_{\lambda}=\{\tilde Q_2:\tilde\Pi\bullet\tilde Q_2=\lambda\tilde Q_2\}, $$

with $\mathbb{B}_1=\mathbb{C}\tilde\Pi$ of dimension one, $\mathbb{B}_0=\mathbb{C}(e_t-\tilde\Pi)$ of dimension one and $\dim\mathbb{B}_{1/2}=2$. At $\tilde\Pi=\tfrac12(e_t+i\vec e_x)$ the three dimensions are $1,2,1$.

Proof. Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space; the dimensions are the multiplicities computed there.

This is the Peirce theory of the block, and it is cubic: the decomposition is into one middle space, not into a pair of off-diagonal corners. There are no corner subalgebras, the complementary idempotent $e_t-\tilde\Pi$ contributes the line $\mathbb{B}_0$ and not a second central piece, and the middle space is a single two-dimensional space carrying no product of its own. The quadratic Peirce calculus is the parent's, and it is not available here.

The Degree-Two Structure

Theorem (the quadratic identity). Every element satisfies

$$ \boxed{\; \tilde Q^{\bullet2}-T(\tilde Q)\tilde Q+\langle\tilde Q\rangle_{\natural}e_t=0,\qquad T(\tilde Q)=2Q_t,\qquad \langle\tilde Q\rangle_{\natural}=\sum_{\mu\in\{t,x,y,z\}}Q_\mu^2 \;}. $$

Proof. On the square of §The Product and Its Square and $T(\tilde Q)\tilde Q=2Q_t^2e_t+2Q_t\vec Q$, the difference is $\bigl(Q_t^2-(\vec Q,\vec Q)-2Q_t^2+\langle\tilde Q\rangle_{\natural}\bigr)e_t$, and the coefficient vanishes because $\langle\tilde Q\rangle_{\natural}=Q_t^2+(\vec Q,\vec Q)$.

The functional $T$ is the generic trace of the block and $\langle\cdot\rangle_{\natural}$ its generic norm; the identity is the generic minimal polynomial, of degree two, so the block is a Jordan algebra of degree two. The discriminant is

$$ T(\tilde Q)^2-4\langle\tilde Q\rangle_{\natural}=-4(\vec Q,\vec Q), $$

which vanishes exactly on the isotropic cone of the vector part; there the two generic roots coincide, and the element $\vec e_x+i\vec e_y$ is the witness with a nonzero vector part.

Theorem (the inverse). An element with $\langle\tilde Q\rangle_{\natural}\neq0$ is invertible in the Jordan sense, with

$$ \boxed{\; \tilde Q^{-1}=\frac{2Q_te_t-\tilde Q}{\langle\tilde Q\rangle_{\natural}}=\frac{\tilde Q^{\natural}}{\langle\tilde Q\rangle_{\natural}} \;}, $$

which is the inverse of the algebra; the invertibility criterion is $\langle\tilde Q\rangle_{\natural}\neq0$.

Proof. Dividing the quadratic identity by $\langle\tilde Q\rangle_{\natural}$ exhibits $(T(\tilde Q)e_t-\tilde Q)/\langle\tilde Q\rangle_{\natural}$ as a Jordan inverse, and $T(\tilde Q)e_t-\tilde Q=2Q_te_t-\tilde Q=\tilde Q^{\natural}$; the formula of the associative inverse is Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra.

Theorem (the quadratic representation). For $U_{\tilde Q_1}\tilde Q_2=2(\tilde Q_1\bullet\tilde Q_2)\bullet\tilde Q_1-\tilde Q_1^{\bullet2}\bullet\tilde Q_2$,

$$ \langle U_{\tilde Q_1}\tilde Q_2\rangle_{\natural}=\langle\tilde Q_1\rangle_{\natural}^2\langle\tilde Q_2\rangle_{\natural}, $$

while $\langle\cdot\rangle_{\natural}$ itself is not multiplicative, $\langle\tilde Q_1\bullet\tilde Q_2\rangle_{\natural}\neq \langle\tilde Q_1\rangle_{\natural}\langle\tilde Q_2\rangle_{\natural}$ in general.

Proof. Norm and Invertibility in the Symmetric Plain Algebra, where the failure of multiplicativity and its replacement by the quadratic representation are proved on the coordinates.

Corollary (the block is not a division Jordan algebra). The isotropic cone $\{\langle\tilde Q\rangle_{\natural}=0\}$ is the union of the origin and the zero-divisor cone of the algebra, so the norm form is non-degenerate but isotropic, of realified signature $(4,4)$; it is the Hermitian form, and not the norm form, that is definite. The elements of square zero are the pure isotropic vectors, a proper subset of the cone, and $e_t+i\vec e_x$ separates the two notions: $\langle e_t+i\vec e_x\rangle_{\natural}=0$ while $(e_t+i\vec e_x)^{\bullet2}=2e_t+2i\vec e_x\neq0$.

Proof. The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra, where the cone, the signature and the separation are computed.

What the Block Does Not Inherit

Three things the parent product has are absent, and the absences belong to the block's description as much as its theorems.

• No projector calculus. A non-trivial idempotent does not give a projector (§The Multiplication Operator), so there is no image–kernel splitting of the type $\mathbb{B}=\tilde\Pi\mathbb{B}\oplus(e_t-\tilde\Pi)\mathbb{B}$ and no ideal generated by an idempotent. The unit gives the identity operator and the origin the zero operator, and those are the only projectors produced by idempotents.
• No associative Peirce calculus. The Peirce theory is the cubic one of §The Idempotents and the Cubic Peirce Theory: one middle space, no corners, no matrix units, and no module theory over the algebra. The element theory of the block — the generic trace and norm, the inverse, the quadratic representation — is the substitute, and it is a different subject.
• No group of units. The invertible elements are not closed under $\bullet$: the pair $\vec e_x,\vec e_y$ is invertible, $\langle \vec e_x\rangle_{\natural}=\langle \vec e_y\rangle_{\natural}=1$, with product $\vec e_x\bullet \vec e_y=0$. The block also has its own zero-divisor set, strictly larger than the algebra's isotropic cone, and it is not the complement of the invertible elements: $\vec e_x$ is a zero divisor, $\vec e_x\bullet \vec e_y=0$, and yet is invertible. The criterion $\det L^{\bullet}_{\tilde Q}=Q_t^2\langle\tilde Q\rangle_{\natural}=0$ is Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, and it separates the two notions.

The Isolation among the Twelve

One law separates the block from the other eleven structures, and it is stated here to close the article, not to enumerate them. The block is commutative and satisfies the Jordan identity, equivalently the operator identity of §The Multiplication Operator; a commutative product with the Jordan identity is a Jordan product, and the block is the only Jordan product of the twelve.

Theorem. Of the twelve structures of The Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, the symmetric plain product is the only one that is commutative and satisfies the Jordan identity.

Proof. The twelve are generated from the plain product by twisting a slot with a conjugation or by (anti)symmetrising, and each of the three operations destroys at least one clause of the criterion. The four general products are non-commutative, the four antisymmetric parts are non-commutative, the symmetric quaternionic product is commutative but fails the identity, and the two symmetric sesquilinear products are only conjugate-commutative and fail the identity. The block is commutative by construction and satisfies the identity by Jordan-admissibility, so it alone meets both clauses.

Two fine points, kept because they are about the law and not about the list. The Jordan identity alone does not isolate the block: it holds at every element of the associative general plain product, at every element of the general quaternionic product, whose square is central, and vacuously at every element of the two alternating antisymmetrisations. What excludes those four is the commutativity clause, and it is the conjunction that the definition names. So the correct exclusive statement is that no other structure is a Jordan product, and not that no other structure satisfies the identity.

Summary

The symmetric plain product $\tilde Q_1\bullet\tilde Q_2=\tfrac12(\tilde Q_1\tilde Q_2+\tilde Q_2\tilde Q_1)$ is the plain product with the cross term dropped, $\tilde Q_1\bullet\tilde Q_2=[Q_{1t}Q_{2t}-(\vec Q_1,\vec Q_2)]e_t+Q_{1t}\vec Q_2+Q_{2t}\vec Q_1$, commutative and unital, with the plain square, $\tilde Q_2^{\bullet2}=[Q_{2t}^2-(\vec Q_2,\vec Q_2)]e_t+2Q_{2t}\vec Q_2$. Its multiplication operator is the symmetric half $L^{\bullet}_{\tilde Q_3}=\tfrac12(L_{\tilde Q_3}+R_{\tilde Q_3})$ of the two plain multiplications, and it satisfies the operator identity $L^{\bullet}_{\tilde Q_1}L^{\bullet}_{\tilde Q_1^{\bullet2}}=L^{\bullet}_{\tilde Q_1^{\bullet2}}L^{\bullet}_{\tilde Q_1}$, equivalent to the Jordan identity. The block fails left alternativity at $\vec e_x,\vec e_y$, and therefore yields no projector from a non-trivial idempotent and no flag from a square-zero element, whose operator has index three instead. Its idempotents are $0$, $e_t$ and the isotropic $\tfrac12(e_t+\vec q)$ with $\vec q$ a pure root of $+1$; at a non-trivial idempotent the operator obeys the cubic Peirce identity with eigenvalues $0,\tfrac12,1$ and the space splits as $\mathbb{C}\tilde\Pi\oplus\mathbb{B}_{1/2}\oplus\mathbb{C}(e_t-\tilde\Pi)$ with dimensions $1,2,1$. The block is of degree two: every element satisfies $\tilde Q_2^{\bullet2}-T(\tilde Q_2)\tilde Q_2+\langle\tilde Q_2\rangle_{\natural}e_t=0$ with $T=2Q_{2t}$ and $\langle\cdot\rangle_{\natural}=\sum_\mu Q_{2,\mu}^2$, its inverse is $\tilde Q_2^{\natural}/\langle\tilde Q_2\rangle_{\natural}$, its norm obeys $\langle U_{\tilde Q_1}\tilde Q_2\rangle_{\natural}=\langle\tilde Q_1\rangle_{\natural}^2\langle\tilde Q_2\rangle_{\natural}$ though it is not multiplicative, and the isotropic cone makes it a non-division Jordan algebra with pure isotropic elements of square zero proper inside the cone. The block inherits none of the parent's projector calculus, associative Peirce calculus or group of units, and its zero-divisor set is larger than the non-invertible elements. Of the twelve structures it is the only Jordan product, the four general and four antisymmetric products failing by non-commutativity, the symmetric quaternionic product by the identity, and the two symmetric sesquilinear products by both clauses read in the conjugate sense.

Summary of Notation

Further Reading

• Kevin McCrimmon, A Taste of Jordan Algebras (Springer, 2004), for the Jordan identity, the operator identity, the cubic Peirce identity and the degree-two theory.
• Richard D. Schafer, An Introduction to Nonassociative Algebras (Academic Press, 1966), for the left and the right multiplications and the alternative laws.
• The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra, for the element theory the law delivers.
• The Multiplication Operators of the Symmetric Plain Algebra, for the operators, the derivations and the structure algebra.
• The Symmetric and Antisymmetric Parts of an Algebra Product, for the Jordan-admissibility of an associative algebra.
• The Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the twelve operations and the isolation table.