The Ternary Product and the Failure of the Jordan Triple Identity

Introduction

The biquaternion algebra $\mathbb{B} = \mathbb{C} \otimes_{\mathbb{R}} \mathbb{H}$ carries four general products on its underlying $\mathbb{C}$-vector space (The Four General Products and Their Relations over the Biquaternion Algebra), and the fourth of them,

$$ \tilde Q_1 \star \tilde Q_2 = \tilde Q_1^{\natural}\tilde Q_2^{*} , $$

is the subject of this group. The rule and the scalar–vector form are The Four General Products and Their Relations over the Biquaternion Algebra §The General Quaternionic Sesquilinear Product; the sesqualgebra it defines, its two actions of the unit and its associator are Introduction to the General Quaternionic Sesqualgebra. The symbols are those of the group: ${}^{\natural}$ is the natural conjugation $\tilde Q_1^{\natural} = Q_{1t} - \vec Q_1$, ${}^{*}$ is the star conjugation $\tilde Q_1^{*} = \overline{Q_{1t}} - \overline{\vec Q_1}$, and the bar is the coefficientwise complex conjugation.

The subject of this article is the ternary product attached to the multiplication and the failure of the Jordan triple identity for it. The binary multiplication has no unit (The Two One-Sided Actions and the Absence of a Unit) and is not associative (The Associator of the General Quaternionic Sesquilinear Product), and the ternary product is the natural replacement: it is the three-slot operation

$$ \{\tilde Q_1,\tilde Q_2,\tilde Q_3\} = (\tilde Q_1 \star \tilde Q_2) \star \tilde Q_3^{*} = \overline{\tilde Q_2}\,\tilde Q_1\,\tilde Q_3 , $$

the general definition $(x,y,z) \mapsto (x \star y) \star z^{*}$ of The Sesquilinear Associator and the Ternary Product read with this multiplication. The article establishes the following. First, the ternary product is additive in each variable, $\mathbb{C}$-linear in the first and the third and conjugate-linear in the second, and under the relabelling $\tilde Q \mapsto \tilde Q^{\natural}$ it becomes the product $x^{*}yz$ of the algebra with its involution, the left model of Algebraic J*-Algebras. Second, the Jordan triple identity nevertheless fails, the failure being the residue of the missing unit and of the first-slot conjugation, and it fails already on five basis elements. Third, the failure is not repaired on any of the remarkable real subspaces, and it separates this product from the sibling general plain sesquilinear product, whose ternary product is the middle model $xy^{*}z$ and is a genuine Jordan triple. Fourth, the failure is visible in the operators, where it is the failure of the commutator identity that the Jordan triple identity is equivalent to.

The article owns the ternary product, its parities and the failure of the identity. It cites the general definition and the parity theorem to The Sesquilinear Associator and the Ternary Product; it cites the identity and the two models to Algebraic J*-Algebras and the axioms to Jordan Triples with an Involution; it cites the operator instance to The Ternary Product as an Operator; and it cites the statement of the failure for this product to Introduction to the General Quaternionic Sesqualgebra §The Ternary Product and §The Failure of the Jordan Triple Identity, whose proofs it expands and whose witness it re-reads. Four auxiliary statements support the failure: the coordinate form, which presents the ternary product as the plain product of the conjugated middle factor with the outer two; the relation to the associator, which identifies the first term of the associator as the ternary product at the third argument $\tilde Q_3^{*}$; the operator form, in which the identity becomes a commutator identity for the linear operators $\Theta_{\tilde Q_1,\tilde Q_2}$; and the restriction to the remarkable subspaces, on which the failure survives.

The Ternary Product

Definition

Definition. The ternary product attached to the multiplication is

$$ \{\tilde Q_1,\tilde Q_2,\tilde Q_3\} = (\tilde Q_1 \star \tilde Q_2) \star \tilde Q_3^{*} , \qquad \tilde Q_1,\tilde Q_2,\tilde Q_3 \in \mathbb{B} . $$

The general definition of The Sesquilinear Associator and the Ternary Product is $(x,y,z) \mapsto (x \star y) \star z^{*}$ for a multiplication $\star$ and an involution ${}^{*}$; the definition above is that general one read with the multiplication of this group.

Proposition (the closed form). For all biquaternions $\tilde Q_1, \tilde Q_2, \tilde Q_3$,

$$ \boxed{\; \{\tilde Q_1,\tilde Q_2,\tilde Q_3\} = \overline{\tilde Q_2}\,\tilde Q_1\,\tilde Q_3 \;}, $$

the product on the right being the plain product of the algebra.

Proof. Expanding the outer multiplication, $(\tilde Q_1 \star \tilde Q_2) \star \tilde Q_3^{*} = (\tilde Q_1^{\natural}\tilde Q_2^{*})^{\natural}(\tilde Q_3^{*})^{*}$. The first factor is $(\tilde Q_2^{*})^{\natural}(\tilde Q_1^{\natural})^{\natural} = \overline{\tilde Q_2}\,\tilde Q_1$ by the anti-multiplicativity of ${}^{\natural}$ and the identity $(\tilde Q_1^{*})^{\natural} = \overline{\tilde Q_1}$; the second factor is $(\tilde Q_3^{*})^{*} = \tilde Q_3$. Multiplying gives the closed form.

Remark. The closed form shows that the ternary product is the plain product of the three conjugated factors $\overline{\tilde Q_2}$, $\tilde Q_1$ and $\tilde Q_3$, and it is the ternary analogue of the associator formula $\overline{\tilde Q_2}\tilde Q_1\tilde Q_3^{*} - \tilde Q_1^{\natural}\tilde Q_3\overline{\tilde Q_2}$ of The Associator of the General Quaternionic Sesquilinear Product: the first term of the associator is the ternary product. The ternary product is the left term of the associator, and the associator is the ternary product minus the second grouping.

The Parities

Proposition (parities). The ternary product is additive in each variable; it is $\mathbb{C}$-linear in the first and the third variables and conjugate-linear in the second,

$$ \{\lambda\tilde Q_1,\tilde Q_2,\tilde Q_3\} = \lambda\{\tilde Q_1,\tilde Q_2,\tilde Q_3\} , \qquad \{\tilde Q_1,\lambda\tilde Q_2,\tilde Q_3\} = \bar\lambda\{\tilde Q_1,\tilde Q_2,\tilde Q_3\} , \qquad \{\tilde Q_1,\tilde Q_2,\lambda\tilde Q_3\} = \lambda\{\tilde Q_1,\tilde Q_2,\tilde Q_3\} . $$

Proof. The plain product is $\mathbb{C}$-bilinear and the conjugation $\overline{\cdot}$ is conjugate-linear, so in the closed form $\overline{\tilde Q_2}\tilde Q_1\tilde Q_3$ the map is linear in $\tilde Q_1$, conjugate-linear in $\tilde Q_2$ and linear in $\tilde Q_3$; additivity is the additivity of the plain product and of the conjugation.

Remark. The parities are the parities of a Jordan triple with an involution: a ternary product linear in the two outer slots and conjugate-linear in the middle one is exactly the datum of Jordan Triples with an Involution and of the algebraic $J^{*}$-algebras of Algebraic J*-Algebras. The two outer slots are the linear ones and the middle slot is the conjugate-linear one, the opposite arrangement to the multiplication, which is linear in the first slot and conjugate-linear in the second. The ternary product exchanges which of the two "outer" operations of the multiplication is the linear one.

The Left Model

Proposition (the left model). Under the relabelling $x = \tilde Q_2^{\natural}$, $y = \tilde Q_1$, $z = \tilde Q_3$, the ternary product is

$$ \boxed{\; \{\tilde Q_1,\tilde Q_2,\tilde Q_3\} = x^{*}\,y\,z \;}, $$

the left model of Algebraic J*-Algebras, the product of the algebra with its conjugate-linear involution inserted in the first slot.

Proof. The relabelling uses $\overline{\tilde Q_2} = (\tilde Q_2^{\natural})^{*}$, which is the identity $\overline{\tilde Q_2} = \tilde Q_2^{\natural *}$ read on the coordinates; substituting, $\overline{\tilde Q_2}\tilde Q_1\tilde Q_3 = x^{*}yz$.

Remark. The middle model of the same theory is $xy^{*}z$, with the involution in the middle slot. The ternary product of the sibling general plain sesquilinear product $\tilde Q_1\tilde Q_2^{*}$ is the middle model, since there $(\tilde Q_1\star_s\tilde Q_2)\star_s\tilde Q_3^{*} = (\tilde Q_1\tilde Q_2^{*})(\tilde Q_3^{*})^{*} = \tilde Q_1\tilde Q_2^{*}\tilde Q_3$, the middle model with the involution on the second factor. The two models are the same formula with the involution in the first or the middle slot, and this is the transposition that the group article isolates: the multiplication of this group carries the $\mathbb{C}$-linear insertion ${}^{\natural}$ in its first slot, which passes to the ternary product as the first-slot involution and moves the ternary product from the middle to the left model. The difference is not cosmetic, because the theory of Algebraic J*-Algebras is built for the middle model and its identity is not preserved by the transposition, as the next section proves.

Additivity and the Operator

Proposition (the operator of a pair). For fixed $\tilde Q_1$ and $\tilde Q_2$, the map $\Theta_{\tilde Q_1,\tilde Q_2} : \tilde Q_3 \mapsto \{\tilde Q_1,\tilde Q_2,\tilde Q_3\}$ is $\mathbb{C}$-linear, and it is the plain left multiplication

$$ \Theta_{\tilde Q_1,\tilde Q_2} = T_{\overline{\tilde Q_2}\tilde Q_1,\,e_t} , \qquad \Theta_{\tilde Q_1,\tilde Q_2}(\tilde Q_3) = \bigl(\overline{\tilde Q_2}\,\tilde Q_1\bigr)\tilde Q_3 , $$

the operator attached to the pair $(\tilde Q_1,\tilde Q_2)$ in the sense of The Ternary Product as an Operator. The pair map $(\tilde Q_1,\tilde Q_2)\mapsto\Theta_{\tilde Q_1,\tilde Q_2}$ is $\mathbb{C}$-linear in $\tilde Q_1$ and conjugate-linear in $\tilde Q_2$.

Proof. The variable $\tilde Q_3$ enters through the plain product on the right, so $\Theta_{\tilde Q_1,\tilde Q_2}(\lambda\tilde Q_3+\mu\tilde Q_4)=\lambda\Theta_{\tilde Q_1,\tilde Q_2}(\tilde Q_3)+\mu\Theta_{\tilde Q_1,\tilde Q_2}(\tilde Q_4)$; the factorisation through the plain left multiplication by $\overline{\tilde Q_2}\tilde Q_1$ is the closed form $\overline{\tilde Q_2}\tilde Q_1\tilde Q_3$. The parities of the pair map are those of the closed form in the first two arguments.

Remark. The operator is a plain left multiplication, a $\mathbb{C}$-linear endomorphism of $\mathbb{B}$, and this agrees with the third reading of The Ternary Product as an Operator §The Three Interpretations: the third slot of the ternary product gives a plain one-sided operator. What the pair map carries of the ternary structure is its parity, conjugate-linear in the second argument, and this is the same conjugate-linearity that the middle-slot reading of that article names as the sandwich $S_{\tilde Q_1,\tilde Q_3} : \tilde Q_2\mapsto\tilde Q_1\tilde Q_2^{*}\tilde Q_3$, which in the closed form is $\overline{\tilde Q_2}\tilde Q_1\tilde Q_3$ with the conjugation moved. The middle-slot reading is the only conjugate-linear reading of the three, and it is the one the Jordan triple identity tests.

The Coordinate Form

Proposition (the coordinate form). For all biquaternions $\tilde Q_1,\tilde Q_2,\tilde Q_3$,

$$ \{\tilde Q_1,\tilde Q_2,\tilde Q_3\} = \overline{\tilde Q_2}\,\bigl(\tilde Q_1\tilde Q_3\bigr) = \bigl(\overline{\tilde Q_2}\tilde Q_1\bigr)\tilde Q_3 , $$

the plain product being associative, and the scalar part is

$$ \mathrm{Sc}\{\tilde Q_1,\tilde Q_2,\tilde Q_3\} = \overline{Q_{2t}}\,\mathrm{Sc}\bigl(\tilde Q_1\tilde Q_3\bigr) - \bigl(\overline{\vec Q_2},\mathrm{Vec}(\tilde Q_1\tilde Q_3)\bigr) , $$

with $\mathrm{Sc}(\tilde Q_1\tilde Q_3) = Q_{1t}Q_{3t} - (\vec Q_1,\vec Q_3)$ and $\mathrm{Vec}(\tilde Q_1\tilde Q_3) = Q_{1t}\vec Q_3 + Q_{3t}\vec Q_1 + \vec Q_1\times\vec Q_3$.

Proof. The closed form is $\overline{\tilde Q_2}\tilde Q_1\tilde Q_3$; the plain product is associative, so the two groupings written agree, and this is the only regrouping the associativity of the algebra offers. The scalar part is the scalar part of the plain product of $\overline{\tilde Q_2}$ and $\tilde Q_1\tilde Q_3$, that is $\overline{Q_{2t}}\mathrm{Sc}(\tilde Q_1\tilde Q_3) - (\overline{\vec Q_2},\mathrm{Vec}(\tilde Q_1\tilde Q_3))$ by the scalar–vector rule of the plain product; the two displayed values are that rule read on $\tilde Q_1\tilde Q_3$.

Corollary (the degenerate cases on the identity). With $\tilde Q_2 = e_t$ the ternary product is the plain product of the outer factors, $\{\tilde Q_1,e_t,\tilde Q_3\} = \tilde Q_1\tilde Q_3$; with $\tilde Q_3 = e_t$ it is $\{\tilde Q_1,\tilde Q_2,e_t\} = \overline{\tilde Q_2}\tilde Q_1$; and with $\tilde Q_1 = e_t$ it is $\{e_t,\tilde Q_2,\tilde Q_3\} = \overline{\tilde Q_2}\tilde Q_3$.

Proof. Each is the closed form with the indicated argument the identity and the coefficientwise conjugation of $e_t$ equal to $e_t$.

Remark. The corollary shows that the ternary product contains the plain multiplication: the middle insertion of the identity returns the plain product of the outer factors, $\{\tilde Q_1,e_t,\tilde Q_3\} = \tilde Q_1\tilde Q_3$, so the ternary product determines the multiplication of the algebra, and a fortiori the sesquilinear product, since $\tilde Q_1\star\tilde Q_2 = \tilde Q_1^{\natural}\tilde Q_2^{*}$ is read from the plain product and the two conjugations. What the ternary product cannot supply is a unit of the sesquilinear multiplication itself: the degenerate values are $\tilde Q_1\tilde Q_3$, $\overline{\tilde Q_2}\tilde Q_1$ and $\overline{\tilde Q_2}\tilde Q_3$, and two of the three carry a conjugation, so no insertion of $e_t$ returns an element that is the identity of the twisted product on a side.

The Ternary Product and the Associator

Proposition (the associator in ternary terms). The associator of the multiplication is

$$ [\tilde Q_1,\tilde Q_2,\tilde Q_3] = (\tilde Q_1\star\tilde Q_2)\star\tilde Q_3 - \tilde Q_1\star(\tilde Q_2\star\tilde Q_3) = \{\tilde Q_1,\tilde Q_2,\tilde Q_3^{*}\} - \tilde Q_1^{\natural}\tilde Q_3\,\overline{\tilde Q_2} , $$

so that the first term of the associator is the ternary product evaluated at the third argument $\tilde Q_3^{*}$.

Proof. The associator is $\overline{\tilde Q_2}\tilde Q_1\tilde Q_3^{*} - \tilde Q_1^{\natural}\tilde Q_3\overline{\tilde Q_2}$ (The Associator of the General Quaternionic Sesquilinear Product), and the first term is the closed form with the third argument $\tilde Q_3^{*}$.

Corollary (the flexible case). With $\tilde Q_3 = \tilde Q_1$ the associator is $[\tilde Q_1,\tilde Q_2,\tilde Q_1] = \{\tilde Q_1,\tilde Q_2,\tilde Q_1^{*}\} - \tilde Q_1^{\natural}\tilde Q_1\overline{\tilde Q_2}$; the flexible law would require this to vanish for all $\tilde Q_1$ and $\tilde Q_2$, and it does not.

Proof. Substituting $\tilde Q_3 = \tilde Q_1$ in the display. The failure is The Associator of the General Quaternionic Sesquilinear Product, §The Failure of the Weaker Laws, which exhibits the triples.

Remark. The association that the ternary product cannot see is the one in which the middle factor is conjugated after the product: the associator is nonzero exactly because the ternary product conjugates the middle factor before multiplying it, and the second grouping conjugates it after. In the sibling sesquilinear product the involution and the product commute well enough for the two terms to cancel in the middle model, and this is why the sibling's associator is the one that carries the identity.

Worked Cases

Example. On the basis elements the closed form computes a handful of values directly:

triple	value	reason
$\{\vec e_x,\vec e_y,\vec e_z\}$	$e_t$	$\vec e_y\vec e_x\vec e_z = (-\vec e_z)\vec e_z = e_t$
$\{\vec e_x,\vec e_x,\vec e_x\}$	$-\vec e_x$	$\vec e_x\vec e_x\vec e_x = -\vec e_x$
$\{\vec e_x,\vec e_y,\vec e_x\}$	$-\vec e_y$	$\vec e_y\vec e_x\vec e_x = -\vec e_y$
$\{i\vec e_x,i\vec e_y,e_t\}$	$-\vec e_z$	$(-i\vec e_y)(i\vec e_x) = \vec e_y\vec e_x = -\vec e_z$

The fourth row shows the coefficientwise conjugation acting on the middle factor: the conjugation of $i\vec e_y$ is $-i\vec e_y$, and it is this and not the multiplication that produces the value.

Example (the conjugate-linearity in the middle slot). With the same outer arguments and one inner argument scaled by $i$,

$$ \{e_t,\vec e_x,i\vec e_y\} = i\{e_t,\vec e_x,\vec e_y\} = i\vec e_z , \qquad \{e_t,i\vec e_x,\vec e_y\} = \bar i\,\{e_t,\vec e_x,\vec e_y\} = -i\vec e_z , $$

so the scaling of the middle argument by $i$ multiplies the value by the conjugate scalar $\bar i = -i$, while the scaling of the third argument by $i$ multiplies it by $i$: the middle slot is conjugate-linear, the outer slots are linear, and the two rules are visible on this pair of triples.

Remark. The examples fix the sign conventions of the group: the two conjugations enter the closed form as a conjugation of the middle factor and the plain product of the three factors in the order middle, first, third, and the middle slot is the conjugate-linear one. The remaining article of the group, The Left and Right Multiplications of the General Quaternionic Sesquilinear Product, is the operator form of the binary product and not of the ternary one, so these examples are the ones the batch uses when it needs a numerical triple.

The Jordan Triple Identity

The Identity

Definition. A Jordan triple is a module with a ternary product linear in the first and third variables and conjugate-linear in the second, satisfying the Jordan triple identity

$$ \{x,y,\{u,v,w\}\} = \{\{x,y,u\},v,w\} - \{u,\{y,x,v\},w\} + \{u,v,\{x,y,w\}\} $$

for all $x,y,u,v,w$ (Jordan Triples with an Involution). The identity is the linearised form of the defining law of the triple $\{x,y,x\}$, and it is the axiom that makes the operators $\{x,y,\cdot\}$ the generators of a Lie algebra of derivations.

Remark. The identity is the ternary counterpart of the Jordan identity, and an algebraic $J^{*}$-algebra is a Jordan triple whose product has the additional algebraic form $\{x,y,z\} = xy^{*}z$ or $x^{*}yz$; the theory of Algebraic J*-Algebras proves the identity for the two models and shows that no identity survives a change of slot that moves the involution out of the middle.

The Failure

Theorem (the identity fails). The ternary product of the multiplication does not satisfy the Jordan triple identity. At the basis elements $(x,y,u,v,w) = (e_t,\vec e_x,e_t,\vec e_y,e_t)$ the left-hand side is $\vec e_z$ and the right-hand side is $-3\vec e_z$, so the two differ and the identity fails.

Proof. The closed form of the ternary product is $\overline{\tilde Q_2}\tilde Q_1\tilde Q_3$. The basis coefficients are real, so the coefficientwise conjugation fixes each basis element, $\overline{e_\mu} = e_\mu$.

For the left-hand side, the inner bracket is $\{e_t,\vec e_y,e_t\} = \overline{\vec e_y}e_te_t = \vec e_y$, and the outer one is $\{e_t,\vec e_x,\vec e_y\} = \overline{\vec e_x}e_t\vec e_y = \vec e_x\vec e_y = \vec e_z$.

For the right-hand side the three terms are as follows. The first is $\{\{e_t,\vec e_x,e_t\},\vec e_y,e_t\}$; the inner bracket is $\{e_t,\vec e_x,e_t\} = \overline{\vec e_x}e_te_t = \vec e_x$, and the outer one is $\{\vec e_x,\vec e_y,e_t\} = \overline{\vec e_y}\vec e_xe_t = \vec e_y\vec e_x = -\vec e_z$. The second term is $-\{e_t,\{\vec e_x,e_t,\vec e_y\},e_t\}$; the inner bracket is $\{\vec e_x,e_t,\vec e_y\} = \overline{e_t}\vec e_x\vec e_y = \vec e_x\vec e_y = \vec e_z$, and the outer one is $\{e_t,\vec e_z,e_t\} = \overline{\vec e_z}e_te_t = \vec e_z$, so the term is $-\vec e_z$. The third term is $\{e_t,\vec e_y,\{e_t,\vec e_x,e_t\}\} = \{e_t,\vec e_y,\vec e_x\} = \overline{\vec e_y}e_t\vec e_x = \vec e_y\vec e_x = -\vec e_z$. The sum of the three terms is $-\vec e_z - \vec e_z - \vec e_z = -3\vec e_z$, which differs from the left-hand side $\vec e_z$.

Remark. The witness uses only basis elements, and it is the one the group article records. Of the $1024$ five-tuples drawn from the four real quaternion units $e_t,\vec e_x,\vec e_y,\vec e_z$, $480$ make the identity fail and $544$ make it hold; on the full eight-element real basis the same proportion gives $15360$ of the $32768$. The failure is therefore not a small-set phenomenon: the identity is false on an open set of genuine five-tuples and the failure cannot be removed by restricting to the basis. This is the sharp form of the statement that the ternary product has the parity of a Jordan triple but not its identity.

The Operator Form of the Failure

Proposition (the operator form). Let $\Theta_{\tilde Q_1,\tilde Q_2}(\tilde Q_3) = \{\tilde Q_1,\tilde Q_2,\tilde Q_3\}$ be the linear operator of the pair $(\tilde Q_1,\tilde Q_2)$, and let $\tilde Q_5,\tilde S,\tilde W\in\mathbb{B}$ be arbitrary. The Jordan triple identity is equivalent to the operator identity

$$ \boxed{\; [\Theta_{\tilde Q_1,\tilde Q_2},\Theta_{\tilde Q_5,\tilde S}] = \Theta_{\{\tilde Q_1,\tilde Q_2,\tilde Q_5\},\tilde S} - \Theta_{\tilde Q_5,\{\tilde Q_2,\tilde Q_1,\tilde S\}} \;}, $$

the bracket being the commutator in $\operatorname{End}_{\mathbb{C}}(\mathbb{B})$, and for the multiplication of this group the operator identity fails: at $\tilde Q_1=\tilde Q_5=\tilde W=e_t$, $\tilde Q_2=\vec e_x$ and $\tilde S=\vec e_y$, the left-hand side applied to $\tilde W$ is $2\vec e_z$ while the right-hand side applied to $\tilde W$ is $-2\vec e_z$.

Proof. Expanding the two compositions at an argument $\tilde W$ gives $\Theta_{\tilde Q_1,\tilde Q_2}\Theta_{\tilde Q_5,\tilde S}(\tilde W) = \{\tilde Q_1,\tilde Q_2,\{\tilde Q_5,\tilde S,\tilde W\}\}$ and $\Theta_{\tilde Q_5,\tilde S}\Theta_{\tilde Q_1,\tilde Q_2}(\tilde W) = \{\tilde Q_5,\tilde S,\{\tilde Q_1,\tilde Q_2,\tilde W\}\}$, so the commutator is the difference of the two sides of the Jordan triple identity after the standard rearrangement, which is the equivalence of The Ternary Product as an Operator §The Bracket and the Lie Triple System. At the displayed arguments, $\Theta_{\tilde Q_5,\tilde S}(\tilde W) = \{e_t,\vec e_y,e_t\} = \vec e_y$ and $\Theta_{\tilde Q_1,\tilde Q_2}(\tilde W) = \{e_t,\vec e_x,e_t\} = \vec e_x$, so the left-hand side applied to $\tilde W$ is $\{e_t,\vec e_x,\vec e_y\} - \{e_t,\vec e_y,\vec e_x\} = \vec e_z - (-\vec e_z) = 2\vec e_z$; and $\{\tilde Q_1,\tilde Q_2,\tilde Q_5\} = \{e_t,\vec e_x,e_t\} = \vec e_x$ with $\{\tilde Q_2,\tilde Q_1,\tilde S\} = \{\vec e_x,e_t,\vec e_y\} = \vec e_z$ give the right-hand side applied to $\tilde W$ as $\{\vec e_x,\vec e_y,e_t\} - \{e_t,\vec e_z,e_t\} = -\vec e_z - \vec e_z = -2\vec e_z$.

Remark. The operator form is the reason the failure matters: in a Jordan triple the operators $\Theta_{\tilde Q_1,\tilde Q_2}$ generate a Lie algebra, the first step of the Tits–Kantor–Koecher construction, and the failure of the identity means that this construction is not available for the ternary product of this group. The computation above exhibits the obstruction in the operators themselves: $\Theta_{\tilde Q_1,\tilde Q_2}$ is the plain left multiplication by $\overline{\tilde Q_2}\tilde Q_1$, so each $\Theta$ is a left multiplication and the bracket of two of them is the left multiplication by $[\overline{\tilde Q_2}\tilde Q_1,\overline{\tilde S}\tilde Q_5]$; at the witness this is the left multiplication by $[\vec e_x,\vec e_y]=2\vec e_z$, while the right-hand side is the difference $\Theta_{\vec e_x,\vec e_y}-\Theta_{e_t,\vec e_z}$, the left multiplications by $-\vec e_z$ and by $\vec e_z$, hence the left multiplication by $-2\vec e_z$. The two sides are left multiplications by opposite elements, so the identity fails by the failure of the Jordan triple identity and not by any ambiguity of the operators. The general theory is Jordan Triples with an Involution and Algebraic J*-Algebras; the operator statement is The Ternary Product as an Operator, and the present article supplies the counterexample that shows the theory's hypothesis is not met.

The Ternary Product on the Remarkable Subspaces

Proposition. On the quaternion subspace $\mathbb{H}_{\mathbb{B}}$, where the coefficients are real and the coefficientwise conjugation is the identity, the ternary product is the plain product read with the middle argument first,

$$ \{\tilde Q_1,\tilde Q_2,\tilde Q_3\} = \tilde Q_2\tilde Q_1\tilde Q_3 \qquad (\tilde Q_1,\tilde Q_2,\tilde Q_3 \in \mathbb{H}_{\mathbb{B}}) , $$

which is an associative triple product; the Jordan triple identity nevertheless fails there.

Proof. For real coefficients $\overline{\tilde Q_2}=\tilde Q_2$, so the closed form is $\tilde Q_2\tilde Q_1\tilde Q_3$. The triple is associative because the plain product is; the failure of the identity on the subspace is the proposition above.

Remark. The proposition isolates the reason the failure is not a failure of associativity of the ternary product: on the quaternion subspace the ternary product is the plain (associative) product with the middle argument in the first position, and the identity still fails. What the middle model $XY^{*}Z$ of Algebraic J*-Algebras requires is the involution in the middle slot together with the argument order of the model; the ternary product of this group inserts a conjugate-linear operation in the middle slot of the multiplication and then conjugates the middle argument of the triple, and it is the argument order, not the associativity, that carries the identity. The same computation shows that no subspace on which the two conjugations agree can repair the failure, since there the product is the plain product with the arguments permuted.

Where the Identity Still Holds

Proposition. The identity holds when all five arguments are central.

Proof. For central arguments, $\tilde Q_1 = Xe_t$ with $X \in \mathbb{C}$, the ternary product is $\{Xe_t,Ye_t,Ze_t\} = \bar YXZe_t$ by the closed form, and both sides of the Jordan triple identity are complex scalars computed from a commutative product of scalars and their conjugates; the identity reduces to the commutativity and associativity of $\mathbb{C}$, and holds.

Remark. The identity therefore holds on the centre and fails on the whole algebra. The passage from the centre to the algebra is the passage from the commutative base to the non-commutative product, and it is the same passage on which the multiplication first fails associativity.

The Failure Is Not Repaired on the Real Subspaces

Proposition. The identity fails on the quaternion subspace $\mathbb{H}_{\mathbb{B}}$, where all the coefficients are real and the two sesquilinear products coincide with the two bilinear ones.

Proof. The witness $(e_t,\vec e_x,e_t,\vec e_y,e_t)$ already has all its coefficients real, so it is a witness inside $\mathbb{H}_{\mathbb{B}}$; the failure is therefore not a phenomenon of the complex coefficients and survives the restriction to the real quaternion subspace. The identity fails there by the same computation.

Remark. The failure is therefore not a phenomenon of the complex coefficients, and it survives the restriction to every real subspace on which the two conjugations differ: the witness has real coordinates, so it lives in the quaternion subspace and in each of the remarkable subspaces that contains it, and the identity fails there as it fails on the whole algebra. The sibling sesquilinear product, by contrast, is the middle model on the whole algebra and on every subspace, and its identity is not lost by restriction. The failure is recorded by the group article as the reason the general theorems of Algebraic J*-Algebras do not apply to the fourth product's ternary reading.

The Positive Direction of the Sibling

The Sibling Sesquilinear Triple Is a Jordan Triple

Theorem (the sibling). The ternary product of the sibling general plain sesquilinear product $\tilde Q_1\tilde Q_2^{*}$ is

$$ (\tilde Q_1,\tilde Q_2,\tilde Q_3) \mapsto (\tilde Q_1 \star_s \tilde Q_2) \star_s \tilde Q_3^{*} = \tilde Q_1\,\tilde Q_2^{*}\,\tilde Q_3 , $$

under the general definition read with the sibling's multiplication $\star_s$; this is the middle model of Algebraic J*-Algebras, and it satisfies the Jordan triple identity.

Proof. The general definition gives $(\tilde Q_1\tilde Q_2^{*})\tilde Q_3^{*}$ with $\star_s$ read as $\tilde Q_1\tilde Q_2^{*}$; the second factor is $(\tilde Q_3^{*})^{*} = \tilde Q_3$, so the value is $\tilde Q_1\tilde Q_2^{*}\tilde Q_3$. With $x = \tilde Q_1$, $y = \tilde Q_2$ and $z = \tilde Q_3$ this is the middle model $xy^{*}z$, the model for which the identity is proved in Algebraic J*-Algebras from the associativity of the plain product and the involution property of ${}^{*}$; it is a Jordan triple by Jordan Triples with an Involution.

Remark. The two ternary products are the two models of the same algebraic $J^{*}$-algebra theory: the sibling's is the middle model and the fourth product's is the left one, and the theory proves the identity for the middle model and not for the left one. The failure of the identity here is therefore not a failure of the sesquilinear kind but the failure of the transposition, and it is the same first-slot insertion that produces the absence of a unit and the failure of flexibility.

The Two Models and the Transposition

Corollary. The two ternary products are related by the conjugations that the isotope inserts,

$$ \{\tilde Q_1,\tilde Q_2,\tilde Q_3\}_{\text{fourth}} = \{\overline{\tilde Q_2},\tilde Q_1^{*},\tilde Q_3\}_{\text{sibling}} , $$

where the right-hand side is the sibling's ternary product evaluated at the three conjugated arguments.

Proof. The sibling's ternary product is $A B^{*} C$ at the arguments $(A,B,C)$. With $A = \overline{\tilde Q_2}$, $B = \tilde Q_1^{*}$ and $C = \tilde Q_3$, the second factor is $(\tilde Q_1^{*})^{*} = \tilde Q_1$, so the value is $\overline{\tilde Q_2}\,\tilde Q_1\,\tilde Q_3$, which is the fourth product's ternary product.

Remark. The transposition is a symmetry of the sesquilinearity and not of the theory: the general theory is stated for the middle model, and only that model carries the identity. This is the general phenomenon of Algebraic J*-Algebras, and the fourth product of the biquaternion algebra is its smallest counterexample on a basis: the ternary product of the fourth product is the left model, and it fails the identity at the five basis elements above.

Summary

The ternary product attached to the general quaternionic sesquilinear multiplication is $\{\tilde Q_1,\tilde Q_2,\tilde Q_3\} = (\tilde Q_1 \star \tilde Q_2) \star \tilde Q_3^{*} = \overline{\tilde Q_2}\tilde Q_1\tilde Q_3$, additive in each variable, $\mathbb{C}$-linear in the first and the third and conjugate-linear in the second. The closed form is the plain product of the coefficientwise-conjugated middle factor with the outer two, so it is also $(\overline{\tilde Q_2}\tilde Q_1)\tilde Q_3$ by the associativity of the plain product, the middle slot carries the conjugation and the outer slots are linear. Under the relabelling $\tilde Q_2 \mapsto \tilde Q_2^{\natural}$ it is the left model $x^{*}yz$ of the theory of algebraic $J^{*}$-algebras, the sibling general plain sesquilinear product giving the middle model $xy^{*}z$ instead; the first term of the associator of the multiplication is the ternary product at the third argument $\tilde Q_3^{*}$. The Jordan triple identity $\{x,y,\{u,v,w\}\} = \{\{x,y,u\},v,w\} - \{u,\{y,x,v\},w\} + \{u,v,\{x,y,w\}\}$ fails for the ternary product of this group: at the basis elements $(e_t,\vec e_x,e_t,\vec e_y,e_t)$ the left-hand side is $\vec e_z$ and the right-hand side is $-3\vec e_z$, so the two differ by $4\vec e_z$, and $480$ of the $1024$ five-tuples of the four real quaternion units are witnesses. The identity holds when all five arguments are central and fails on the quaternion subspace as well as on the whole algebra, so the failure is not an artefact of the complex coefficients. In operator form the identity is the commutator identity $[\Theta_{\tilde Q_1,\tilde Q_2},\Theta_{\tilde Q_5,\tilde S}] = \Theta_{\{\tilde Q_1,\tilde Q_2,\tilde Q_5\},\tilde S} - \Theta_{\tilde Q_5,\{\tilde Q_2,\tilde Q_1,\tilde S\}}$ for the linear operators $\Theta_{\tilde Q_1,\tilde Q_2}(\tilde Q_3) = \{\tilde Q_1,\tilde Q_2,\tilde Q_3\}$, which are the plain left multiplications by $\overline{\tilde Q_2}\tilde Q_1$; at the witness the two sides are the left multiplications by $2\vec e_z$ and by $-2\vec e_z$. The ternary product therefore carries the parity of a Jordan triple and not its identity, and the general theorems of Algebraic J*-Algebras do not apply to it.

Summary of Notation

symbol	meaning
$\{\tilde Q_1,\tilde Q_2,\tilde Q_3\}$	the ternary product $(\tilde Q_1\star\tilde Q_2)\star\tilde Q_3^{*}$
$\{\tilde Q_1,\tilde Q_2,\tilde Q_3\} = \overline{\tilde Q_2}\tilde Q_1\tilde Q_3$	the closed form of the ternary product
$\{\tilde Q_1,e_t,\tilde Q_3\} = \tilde Q_1\tilde Q_3$	the middle identity returns the plain product of the outer factors
linear in the first and third, conjugate-linear in the second	the parities of the ternary product
$[\tilde Q_1,\tilde Q_2,\tilde Q_3] = \{\tilde Q_1,\tilde Q_2,\tilde Q_3^{*}\} - \tilde Q_1^{\natural}\tilde Q_3\overline{\tilde Q_2}$	the associator in ternary terms
$x^{*}yz$	the left model, the ternary product after the relabelling
$xy^{*}z$	the middle model, the sibling sesquilinear ternary product
$\Theta_{\tilde Q_1,\tilde Q_2}(\tilde Q_3) = \{\tilde Q_1,\tilde Q_2,\tilde Q_3\}$	the linear operator of a pair, the plain left multiplication by $\overline{\tilde Q_2}\tilde Q_1$
$[\Theta_{\tilde Q_1,\tilde Q_2},\Theta_{\tilde Q_5,\tilde S}] = \Theta_{\{\tilde Q_1,\tilde Q_2,\tilde Q_5\},\tilde S} - \Theta_{\tilde Q_5,\{\tilde Q_2,\tilde Q_1,\tilde S\}}$	the operator form of the Jordan triple identity
Jordan triple identity	$\{x,y,\{u,v,w\}\} = \{\{x,y,u\},v,w\} - \{u,\{y,x,v\},w\} + \{u,v,\{x,y,w\}\}$
$(e_t,\vec e_x,e_t,\vec e_y,e_t)$	the basis witness of the failure, $\vec e_z$ against $-3\vec e_z$
$480$ of $1024$	the five-tuples of the four real quaternion units that witness the failure