Association and the Transpose on the Quaternion Algebra

Introduction

The quaternion product pairs with the scalar functional into one bilinear form, the product form $\langle\tilde q_1,\tilde q_2\rangle=\mathrm{Sc}(\tilde q_1\tilde q_2)$, and that form has a transpose calculus: every real-linear operator $T$ of $\mathbb{H}$ has a transpose $T^{\sim}$ defined by $\langle T\tilde q_1,\tilde q_2\rangle=\langle\tilde q_1,T^{\sim}\tilde q_2\rangle$, and the transposes of the one- and two-sided multiplications are computable in closed form. The form of the quaternion product is read in this article; the operators themselves are Two-Sided Operators on the Quaternion Algebra, below this article in this category, and One-Sided Operators on the Quaternion Algebra, below this article in this category; the enveloping algebra is The Enveloping Operator Algebra of the Quaternion Algebra and the Bi-module Structure, below this article in this category.

The form is symmetric and indefinite of signature $(1,3)$. It is not the positive definite coordinate pairing, and the whole of the calculus below is the bookkeeping of that gap: the Gram matrix is the diagonal matrix of the signature, and the transpose of an operator is the operator's matrix conjugated by it.

The Plain Form

Definition. The product form is $\langle\tilde q_1,\tilde q_2\rangle=\mathrm{Sc}(\tilde q_1\tilde q_2)$, the scalar part of the quaternion product, a bilinear form on $\mathbb{H}$.

Proposition. $\mathrm B$ is symmetric, non-degenerate, and indefinite of signature $(1,3)$; in the basis $e_t,\vec e_x,\vec e_y,\vec e_z$ its Gram matrix is

$$ D=\operatorname{diag}(1,-1,-1,-1). $$

Proof. Symmetry is the identity $\mathrm{Sc}(\tilde q_1\tilde q_2)=\mathrm{Sc}(\tilde q_2\tilde q_1)$, read from the coordinate formula: the scalar part of a product is $q_{1t}q_{2t}-q_{1x}q_{2x}-q_{1y}q_{2y}-q_{1z}q_{2z}$, which is unchanged when the two factors are exchanged. Non-degeneracy and the Gram matrix are the table of the scalar parts of the sixteen products: $\mathrm B(e_t,e_t)=1$, $\mathrm B(\vec e_\mu,\vec e_\mu)=-1$, and $\mathrm B(e_\mu,e_\nu)=0$ for $\mu\neq\nu$.

Corollary. $\mathrm B$ is the polar form of the quadratic form $\mathfrak{Q}(\tilde q)=\mathrm{Sc}(\tilde q^2)=q_t^2-q_x^2-q_y^2-q_z^2$, and the associated quadratic form is indefinite.

Proof. $\langle\tilde q_1,\tilde q_2\rangle=\tfrac12(\mathfrak{Q}(\tilde q_1+\tilde q_2)-\mathfrak{Q}(\tilde q_1)-\mathfrak{Q}(\tilde q_2))$ by bilinearity, and $D$ has one positive and three negative diagonal entries.

The Association Functional

Definition. The association functional is the trilinear form $A(\tilde q_1,\tilde q_2,\tilde q_3)=\mathrm{Sc}(\tilde q_1\tilde q_2\tilde q_3)$.

Theorem (cyclic invariance). $A(\tilde q_1,\tilde q_2,\tilde q_3)=A(\tilde q_2,\tilde q_3,\tilde q_1)=A(\tilde q_3,\tilde q_1,\tilde q_2)$.

Proof. Apply the symmetry of the form to the pair $\tilde q_1\tilde q_2$ and $\tilde q_3$: $\mathrm{Sc}(\tilde q_1\tilde q_2\tilde q_3)=\mathrm{Sc}(\tilde q_3\tilde q_1\tilde q_2)$, which is the cyclic shift; the second is the shift of the first.

Corollary (the association is immaterial). $A$ is unchanged when the two multiplications are associated in the other order, $A(\tilde q_1,\tilde q_2,\tilde q_3)=\mathrm{Sc}((\tilde q_1\tilde q_2)\tilde q_3)=\mathrm{Sc}(\tilde q_1(\tilde q_2\tilde q_3))$, by associativity alone; cyclic invariance is the additional statement that the two arguments can be moved without changing the value.

Definition. The association functional defines the sandwich operators $T_{\tilde q_4,\tilde q_5}=L_{\tilde q_4}R_{\tilde q_5}$, $\tilde q\mapsto\tilde q_4\tilde q\tilde q_5$, and $A(\tilde q_1,\tilde q_2,\tilde q_3)=\mathrm B(L_{\tilde q_1}R_{\tilde q_3}\tilde q_2,e_t)$ reads the functional through one of them.

The Transposes of the One-Sided Operators

Definition. For a real-linear operator $T$ of $\mathbb{H}$ the transpose $T^{\sim}$ is the operator with $\langle T\tilde q_1,\tilde q_2\rangle=\langle\tilde q_1,T^{\sim}\tilde q_2\rangle$ for all $\tilde q_1,\tilde q_2$. Non-degeneracy of $\mathrm B$ makes $T^{\sim}$ unique, and $T\mapsto T^{\sim}$ is an anti-automorphism of the operator algebra.

Theorem. $L_{\tilde q_4}^{\sim}=R_{\tilde q_4}$ and $R_{\tilde q_5}^{\sim}=L_{\tilde q_5}$.

Proof. $\langle L_{\tilde q_4}\tilde q_1,\tilde q_2\rangle=\mathrm{Sc}(\tilde q_4\tilde q_1\tilde q_2)$ and $\langle\tilde q_1,R_{\tilde q_4}\tilde q_2\rangle=\mathrm{Sc}(\tilde q_1\tilde q_2\tilde q_4)$, equal by cyclic invariance of the association functional; hence $L_{\tilde q_4}^{\sim}=R_{\tilde q_4}$, and the second is the mirror with the two sides swapped.

Corollary. $L_{\tilde q_4}$ is self-transpose exactly when $\tilde q_4$ is central; the self-transpose left multiplications are the scalar ones.

Proof. $L_{\tilde q_4}=R_{\tilde q_4}$ exactly when $\tilde q_4$ commutes with every element, that is, by the centre theorem, when $\tilde q_4\in\mathbb{R}e_t$.

The Transpose of a Two-Sided Operator

Theorem. For the sandwich $T_{\tilde q_4,\tilde q_5}=L_{\tilde q_4}R_{\tilde q_5}$, $\tilde q\mapsto\tilde q_4\tilde q\tilde q_5$,

$$ \boxed{\; T_{\tilde q_4,\tilde q_5}^{\sim}=L_{\tilde q_5}R_{\tilde q_4}=T_{\tilde q_5,\tilde q_4} \;} .$$

Proof. The transpose is an anti-automorphism, $(\tilde q_1\tilde q_2)^{\sim}=\tilde q_2^{\sim}\tilde q_1^{\sim}$, so $T_{\tilde q_4,\tilde q_5}^{\sim}=(L_{\tilde q_4}R_{\tilde q_5})^{\sim}=R_{\tilde q_5}^{\sim}L_{\tilde q_4}^{\sim}=L_{\tilde q_5}R_{\tilde q_4}$, which is $T_{\tilde q_5,\tilde q_4}$.

Corollary. The sandwich is self-transpose exactly when $\tilde q_5\in\mathbb{R}\tilde q_4$ or $\tilde q_4\in\mathbb{R}\tilde q_5$.

Proof. The equality $L_{\tilde q_4}R_{\tilde q_5}=L_{\tilde q_5}R_{\tilde q_4}$ says $\tilde q_4\tilde q\tilde q_5=\tilde q_5\tilde q\tilde q_4$ for every $\tilde q$. At $\tilde q=e_t$ it gives $\tilde q_4\tilde q_5=\tilde q_5\tilde q_4$, so $\tilde q_5$ commutes with $\tilde q_4$; if $\tilde q_5=\mu\tilde q_4$ for some real $\mu$ the two sides are equal, which is the sufficiency. Conversely, either $\tilde q_4$ is central, and then the identity reads $\tilde q\tilde q_4\tilde q_5=\tilde q_5\tilde q\tilde q_4$, that is $\tilde q\tilde q_4\tilde q_5=\tilde q_4\tilde q_5\tilde q$ with the product $\tilde q_4\tilde q_5$ central because $\tilde q_4$ is and $\tilde q_5$ commutes with it, so $\tilde q_5$ is central too and $\tilde q_5\in\mathbb{R}\tilde q_4$; or $\tilde q_4$ is not central, its commutant is the plane $\mathbb{R}e_t+\mathbb{R}\tilde q_4$, and $\tilde q_5=\lambda+\mu\tilde q_4$ gives $\tilde q_4\tilde q\tilde q_5-\tilde q_5\tilde q\tilde q_4=\lambda(\tilde q_4\tilde q-\tilde q\tilde q_4)$, which vanishes for all $\tilde q$ only if $\lambda=0$, that is $\tilde q_5\in\mathbb{R}\tilde q_4$. The degenerate cases $\tilde q_4=0$ or $\tilde q_5=0$ are included, both sides being $0$.

Theorem (the natural sign is self-transpose). The operator $\natural:\tilde q\mapsto\tilde q^{\natural}$ satisfies $\natural^{\sim}=\natural$.

Proof. $\langle\tilde q_1^{\natural},\tilde q_2\rangle=\mathrm{Sc}(\tilde q_1^{\natural}\tilde q_2)=\sum_\mu q_{1\mu} q_{2\mu}$ and $\langle\tilde q_1,\tilde q_2^{\natural}\rangle=\mathrm{Sc}(\tilde q_1\tilde q_2^{\natural})=\sum_\mu q_{1\mu} q_{2\mu}$; the two agree for all $\tilde q_1,\tilde q_2$, so the operator is self-transpose. The formula displays the coordinate pairing as the product form read through the natural sign in either slot.

The Matrix Formula for the Transpose

Theorem. Let $T$ be real-linear with matrix $G$ in the basis $e_t,\vec e_x,\vec e_y,\vec e_z$ and let $D=\operatorname{diag}(1,-1,-1,-1)$. Then $T^{\sim}$ has matrix

$$ \boxed{\; G^{\sim}=D\,G^{\mathsf T}D \;} .$$

Proof. For all $\tilde q_1,\tilde q_2$, $\langle T\tilde q_1,\tilde q_2\rangle=(Gq_1)^{\mathsf T}Dq_2=q_1^{\mathsf T}G^{\mathsf T}Dq_2$ and $\langle\tilde q_1,T^{\sim}\tilde q_2\rangle=q_1^{\mathsf T}D(G^{\sim}q_2)=q_1^{\mathsf T}DG^{\sim}q_2$, where $q_1,q_2$ are the coordinate columns; equating the two for all $q_1,q_2$ gives $G^{\mathsf T}D=DG^{\sim}$, that is $G^{\sim}=D^{-1}G^{\mathsf T}D=D\,G^{\mathsf T}D$ because $D^{-1}=D= D^{\mathsf T}$.

Corollary. The transpose matrix is the matrix transpose conjugated by the signature matrix; in particular the transposes of $L_{\tilde q_4}$ and $R_{\tilde q_4}$ are obtained by $L_{\tilde q_4}^{\sim}=D\,L_{\tilde q_4}^{\mathsf T}D=R_{\tilde q_4}$ and $R_{\tilde q_4}^{\sim}=D\,R_{\tilde q_4}^{\mathsf T}D=L_{\tilde q_4}$.

Proof. The formula applied to the matrix of each operator, with the identifications of the theorem above.

Corollary (the operator table). The transposes of the elementary operators are

Operator	Transpose
$L_{\tilde q_4}$	$R_{\tilde q_4}$
$R_{\tilde q_5}$	$L_{\tilde q_5}$
$L_{\tilde q_4}R_{\tilde q_5}$	$L_{\tilde q_5}R_{\tilde q_4}$
$\natural$	$\natural$

Proof. The four statements above.

The Identity against the Quaternion Adjoint

Theorem. The product form read through the natural sign in either slot is the positive definite coordinate pairing:

$$ \boxed{\; \langle\tilde q_1^{\natural},\tilde q_2\rangle=\langle\tilde q_1,\tilde q_2^{\natural}\rangle=\sum_{\mu}q_{1\mu} q_{2\mu}=\mathrm{Sc}(\tilde q_1^{\natural}\tilde q_2) \;} .$$

Proof. $\tilde q_1^{\natural}\tilde q_2=\sum_\mu q_{1\mu} q_{2\mu}+\text{(a vector part)}$ and $\tilde q_1\tilde q_2^{\natural}$ likewise; the scalar part is the sum of the coordinate products, the vector parts being sums of the basis vectors without a scalar component.

Corollary. The natural sign is the adjoint of the identity of the product form: it carries the indefinite form to the positive definite coordinate one, and its self-transposeness is the statement that the two pairings agree.

Summary

The product form $\langle\tilde q_1,\tilde q_2\rangle=\mathrm{Sc}(\tilde q_1\tilde q_2)$ is symmetric, non-degenerate and indefinite of signature $(1,3)$, with Gram matrix $D=\operatorname{diag}(1,-1,-1,-1)$ and polar quadratic form $\mathrm{Sc}(\tilde q^2)=q_t^2-q_x^2-q_y^2-q_z^2$. The association functional $\mathrm{Sc}(\tilde q_1\tilde q_2\tilde q_3)$ is cyclically invariant, and the association of the two multiplications does not change it. The transpose of a real-linear operator is $G^{\sim}=D\,G^{\mathsf T}D$ in matrix form, and on the elementary operators it is $L_{\tilde q_4}^{\sim}=R_{\tilde q_4}$, $R_{\tilde q_5}^{\sim}=L_{\tilde q_5}$, $(L_{\tilde q_4}R_{\tilde q_5})^{\sim}=L_{\tilde q_5}R_{\tilde q_4}$ and $\natural^{\sim}=\natural$.

Summary of Notation

Symbol	Meaning
$\langle\tilde q_1,\tilde q_2\rangle=\mathrm{Sc}(\tilde q_1\tilde q_2)$	The product form
$A(\tilde q_1,\tilde q_2,\tilde q_3)=\mathrm{Sc}(\tilde q_1\tilde q_2\tilde q_3)$	The association functional
$D=\operatorname{diag}(1,-1,-1,-1)$	Gram matrix of $\mathrm B$
$T^{\sim}$	Transpose of the operator $T$
$L_{\tilde q_4},R_{\tilde q_5}$	Left and right multiplications
$\natural$	The natural sign, the quaternion conjugation