Two-Sided Operators on the General Quaternionic Algebra

Introduction

The biquaternion algebra carries two families of one-sided operators, the left and the right multiplications, and their products are the two-sided operators $L_aR_b(y)=a\,y\,b$. Among them one member is singled out by the general quaternionic bilinear form and the natural conjugation: the choice $b=a^{\natural}$, giving the twisted two-sided operator

$$ \Theta_a=L_aR_{a^{\natural}},\qquad \Theta_a(y)=a\,y\,a^{\natural}. $$

This is the conjugation sandwich of the biquaternion algebra, the member of the general two-sided family whose second slot carries the natural conjugation in place of the inverse. The corpus also calls it the signed inner conjugation, a name it shares with the different operator $\tilde Q_1\mapsto\alpha(\tilde Q_6)\tilde Q_1\tilde Q_6^{-1}$; the remark which member of the general family this operator is below separates the two. Placing the conjugation where the inverse belongs is what makes the operator defined on every element and not only on the units, and it leaves the norm in front of the inner conjugation: on a vector the operator is the reflection exactly on the shell $\langle\tilde Q_6\rangle_{\natural}=-1$, and a scaled copy of it elsewhere.

This article treats the operator: its factors, its multiplication law, its injectivity up to sign, its adjoint for the general quaternionic bilinear form, the criterion that singles it out among the products $L_aR_b$, its automorphisms, and its dagger adjoint. The general Clifford theory of the family is Two-Sided Operators on a Clifford Algebra with Signed Inner Conjugation, to which this is the biquaternion instance; the one-sided factors are One-Sided Operators on the General Quaternionic Algebra; the groups cut out by the norm are The Pin and Spin Groups of the General Quaternionic Algebra; and the corresponding family over the Hermitian form is Mixed Inner Conjugation on the General Plain Sesqualgebra. The form, its matrix and its adjoint are The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra; nothing of the general theory is re-derived here.

The Operator and Its Factors

Definition. For $\tilde Q_6\in\mathbb{B}$ the twisted two-sided operator of $\tilde Q_6$, or the conjugation sandwich by $\tilde Q_6$, is the $\mathbb{C}$-linear map

$$ \Theta_{\tilde Q_6}:\mathbb{B}\longrightarrow\mathbb{B},\qquad \Theta_{\tilde Q_6}(\tilde Q_1)=\tilde Q_6\,\tilde Q_1\,\tilde Q_6^{\natural}. $$

Proposition (the factors, and the quadraticity in the parameter). With $L$ and $R$ the left and right multiplications,

$$ \Theta_{\tilde Q_6}=L_{\tilde Q_6}\circ R_{\tilde Q_6^{\natural}}=R_{\tilde Q_6^{\natural}}\circ L_{\tilde Q_6}, $$

the two factors commuting; and the parameter map is quadratic, not linear,

$$ \Theta_{\lambda\tilde Q_6}=\lambda^{2}\,\Theta_{\tilde Q_6},\qquad \Theta_{\tilde Q_6+\tilde Q_2}=\Theta_{\tilde Q_6}+\Theta_{\tilde Q_2}+\tilde Q_6\,[\,]\tilde Q_2^{\natural}+\tilde Q_2\,[\,]\tilde Q_6^{\natural}, $$

the last two terms being the cross term; in particular $\Theta_{-\tilde Q_6}=\Theta_{\tilde Q_6}$ and $\Theta_{i\tilde Q_6}=-\Theta_{\tilde Q_6}$.

Proof. The first identity is the definition, the second is $L_aR_b=R_bL_a$. For the quadraticity, $(\lambda\tilde Q_6)\,\tilde Q_1\,(\lambda\tilde Q_6)^{\natural}=\lambda^{2}\tilde Q_6\tilde Q_1\tilde Q_6^{\natural}$ because the natural conjugation is $\mathbb{C}$-linear; expanding the product $(\tilde Q_6+\tilde Q_2)\tilde Q_1(\tilde Q_6+\tilde Q_2)^{\natural}$ gives the displayed cross term.

Remark (why the parameter map cannot be linear). A linear map of the parameter would give $\Theta_{0}=0$ and $\Theta_{\lambda\tilde Q_6}=\lambda\Theta_{\tilde Q_6}$; but $\Theta_{\tilde Q_6}$ is built from two factors each linear in $\tilde Q_6$, so the scalar appears twice. The operator is linear in $\tilde Q_1$, which is what the applications use, and quadratic in $\tilde Q_6$; the sign $\Theta_{-\tilde Q_6}=\Theta_{\tilde Q_6}$ is the same statement at $\lambda=-1$.

Proposition (the sign and the inverse). For every element, $\Theta_{\tilde Q_6}=\Theta_{-\tilde Q_6}$; for an invertible $\tilde Q_6$,

$$ \Theta_{\tilde Q_6}=\langle\tilde Q_6\rangle_{\natural}\ \mathrm{Ad}_{\tilde Q_6},\qquad \mathrm{Ad}_{\tilde Q_6}(\tilde Q_1)=\tilde Q_6\,\tilde Q_1\,\tilde Q_6^{-1}, $$

and $\mathrm{Ad}_{\tilde Q_6}=\Theta_{\tilde Q_6}$ exactly on the norm-one set $\langle\cdot\rangle_{\natural}=1$.

Proof. The first identity is $\tilde Q_6^{\natural}=\langle\tilde Q_6\rangle_{\natural}\tilde Q_6^{-1}$ for a unit, and $\langle\tilde Q_6\rangle_{\natural}$ is central; the second follows from $\langle\cdot\rangle_{\natural}=1$.

Remark (why the sign is needed). The plain inner conjugation $\mathrm{Ad}_{\tilde Q_6}$ is an algebra automorphism and needs the inverse, so it is defined only on the units and is blind to the norm; the twisted form $\Theta_{\tilde Q_6}$ is defined for every element, is linear in its argument and quadratic in its parameter, and carries the norm as the scalar in front of the inner conjugation. On the norm-one set the two agree, and off it they differ by the scalar $\langle\tilde Q_6\rangle_{\natural}$ – which is the whole of the sign in the name.

Remark (which member of the general family this operator is, and the two senses of the name). The general two-sided family of Two-Sided Operators on a Clifford Algebra attaches to a pair (an automorphism $\theta$ of the left factor, an anti-automorphism $c$ of the right one) the operator $\Phi^{\theta,c}_x(y)=\theta(x)\,y\,c(x)$, and its five members are told apart by $c$. The member that the general layer calls signed inner conjugation is $\theta=\alpha$, $c=(\ )^{-1}$,

$$ \alpha(\tilde Q_6)\,\tilde Q_1\,\tilde Q_6^{-1}=\varepsilon_{\tilde Q_6}\,\mathrm{Ad}_{\tilde Q_6}, $$

with $\varepsilon_{\tilde Q_6}=(-1)^{k}$ the parity sign of a homogeneous $\tilde Q_6$. The operator of this article, $\Theta_{\tilde Q_6}=\tilde Q_6\,\tilde Q_1\,\tilde Q_6^{\natural}$, is the member $\theta=\mathrm{id}$, $c={}^{\natural}$, the conjugation sandwich, whose value at the unit is the Clifford norm; and the two are not the same operator, for on a unit

$$ \Theta_{\tilde Q_6}=\langle\tilde Q_6\rangle_{\natural}\,\mathrm{Ad}_{\tilde Q_6}, \qquad \alpha(\tilde Q_6)\,\tilde Q_1\,\tilde Q_6^{-1}=\varepsilon_{\tilde Q_6}\,\mathrm{Ad}_{\tilde Q_6}, $$

so they differ by the scalar $\langle\tilde Q_6\rangle_{\natural}\varepsilon_{\tilde Q_6}^{-1}$ and coincide exactly when

$$ \langle\tilde Q_6\rangle_{\natural}=\varepsilon_{\tilde Q_6}, $$

that is on the homogeneous elements of $\mathrm{Pin}$ — odd with $\langle\cdot\rangle_{\natural}=-1$ and even with $\langle\cdot\rangle_{\natural}=+1$. The difference is the normalisation of the sign: the general member puts the inverse on the right factor and the sign on the left, while the conjugation sandwich puts the conjugate in place of the inverse. Both normalisations are used in the corpus under the name "signed inner conjugation", and this article is about the second. On the vector subspace the difference is visible without computation: for a vector $\tilde Q_6$ the general member is $-\mathrm{Ad}_{\tilde Q_6}=\rho_{\tilde Q_6}$ for every odd parameter, whereas $\Theta_{\tilde Q_6}$ restricts to $-\langle\tilde Q_6\rangle_{\natural}\rho_{\tilde Q_6}$, which is the reflection exactly on the shell $\langle\cdot\rangle_{\natural}=-1$.

The Multiplication Law

Proposition (composition). For all $\tilde Q_6,\tilde Q_2$,

$$ \boxed{\; \Theta_{\tilde Q_6}\circ\Theta_{\tilde Q_2}=\Theta_{\tilde Q_6\tilde Q_2} \;}. $$

Proof. $\Theta_{\tilde Q_6}\bigl(\Theta_{\tilde Q_2}(\tilde Q_1)\bigr)=\tilde Q_6\tilde Q_2\tilde Q_1\tilde Q_2^{\natural}\tilde Q_6^{\natural}=\tilde Q_6\tilde Q_2\tilde Q_1(\tilde Q_6\tilde Q_2)^{\natural}$ because ${}^{\natural}$ is an anti-automorphism.

Proposition (multiplicativity up to the norm). For all $\tilde Q_6$ and all $\tilde Q_1,\tilde Q_3$,

$$ \Theta_{\tilde Q_6}(\tilde Q_1)\,\Theta_{\tilde Q_6}(\tilde Q_3)=\langle\tilde Q_6\rangle_{\natural}\,\Theta_{\tilde Q_6}(\tilde Q_1\tilde Q_3). $$

Proof. $\tilde Q_6\tilde Q_1\tilde Q_6^{\natural}\tilde Q_6\tilde Q_3\tilde Q_6^{\natural}=\tilde Q_6\tilde Q_1\langle\tilde Q_6\rangle_{\natural}\tilde Q_3\tilde Q_6^{\natural}=\langle\tilde Q_6\rangle_{\natural}\tilde Q_6\tilde Q_1\tilde Q_3\tilde Q_6^{\natural}$, moving the central scalar out. In particular $\Theta_{\tilde Q_6}$ is an algebra automorphism exactly when $\langle\tilde Q_6\rangle_{\natural}=1$, while for $\langle\tilde Q_6\rangle_{\natural}=-1$ the operator is multiplicative up to the sign $-1$, which is the algebraic meaning of the word signed; read in the other order the same identity is $\Theta_{\tilde Q_6}(\tilde Q_3)\Theta_{\tilde Q_6}(\tilde Q_1)=\langle\tilde Q_6\rangle_{\natural}\Theta_{\tilde Q_6}(\tilde Q_3\tilde Q_1)$, the statement with the two factors interchanged.

Proposition (injectivity up to sign, and the kernel). For units $\tilde Q_6,\tilde Q_2$, the assignment $\tilde Q_6\mapsto\Theta_{\tilde Q_6}$ is injective up to the sign of the parameter: $\Theta_{\tilde Q_6}=\Theta_{\tilde Q_2}$ if and only if $\tilde Q_2=\tilde Q_6$ or $\tilde Q_2=-\tilde Q_6$. Moreover $\Theta_{\tilde Q_6}=0$ if and only if $\tilde Q_6=0$. On the unit group the kernel of the assignment is therefore $\{\pm e_t\}$, and $\Theta_{\tilde Q_6}$ is invertible exactly when $\tilde Q_6$ is, with inverse $\Theta_{\tilde Q_6^{-1}}$.

Proof. If $\tilde Q_6=0$ then $\Theta_{\tilde Q_6}=0$; otherwise apply both sides to $e_t$: $\Theta_{\tilde Q_6}(e_t)=\tilde Q_6\tilde Q_6^{\natural}=\langle\tilde Q_6\rangle_{\natural}e_t$ and $\Theta_{\tilde Q_2}(e_t)=\langle\tilde Q_2\rangle_{\natural}e_t$, so $\langle\tilde Q_6\rangle_{\natural}=\langle\tilde Q_2\rangle_{\natural}$; then apply to a general $\tilde Q_1$: $\tilde Q_6\tilde Q_1\tilde Q_6^{\natural}=\tilde Q_2\tilde Q_1\tilde Q_2^{\natural}$; multiplying the two sides of $\tilde Q_6\tilde Q_1\tilde Q_6^{\natural}=\tilde Q_2\tilde Q_1\tilde Q_2^{\natural}$ on the left by $\tilde Q_6^{-1}$ and on the right by $\tilde Q_6^{-\natural}$ gives $\tilde Q_1=\tilde Q_6^{-1}\tilde Q_2\,\tilde Q_1\,\tilde Q_2^{\natural}\tilde Q_6^{-\natural}$, so the element $\tilde Q_8=\tilde Q_6^{-1}\tilde Q_2$ satisfies $\tilde Q_8\tilde Q_1\tilde Q_8^{\natural}=\tilde Q_1$ for all $\tilde Q_1$; taking $\tilde Q_1=e_t$ gives $\langle\tilde Q_8\rangle_{\natural}=1$ and then $\tilde Q_1\tilde Q_8=\tilde Q_8\tilde Q_1$ for all $\tilde Q_1$, so $\tilde Q_8$ is central, hence $\tilde Q_8=\lambda e_t$ with $\lambda^{2}=1$, that is $\tilde Q_8=\pm e_t$, so $\tilde Q_2=\pm\tilde Q_6$. The inverse statement is the composition law with $\tilde Q_6^{-1}$.

The Adjoint

Proposition (the adjoint for the general quaternionic bilinear form). The adjoint of $\Theta_{\tilde Q_6}$ for the general quaternionic bilinear form is $\Theta_{\tilde Q_6^{\natural}}$,

$$ \boxed{\; \bigl(\Theta_{\tilde Q_6}\bigr)^{N}=\Theta_{\tilde Q_6^{\natural}},\qquad\text{that is}\qquad \bigl\langle\Theta_{\tilde Q_6}\tilde Q_1,\tilde Q_3\bigr\rangle_{\natural}=\bigl\langle\tilde Q_1,\Theta_{\tilde Q_6^{\natural}}\tilde Q_3\bigr\rangle_{\natural} \;}. $$

Proof. The adjoint of a product of one-sided operators is the product of the adjoints in the reverse order, and the adjoints are $(L_{\tilde Q_6})^{N}=L_{\tilde Q_6^{\natural}}$ and $(R_{\tilde Q_2})^{N}=R_{\tilde Q_2^{\natural}}$ (One-Sided Operators on the General Quaternionic Algebra); hence $(\Theta_{\tilde Q_6})^{N}=(L_{\tilde Q_6}R_{\tilde Q_6^{\natural}})^{N}=(R_{\tilde Q_6^{\natural}})^{N}(L_{\tilde Q_6})^{N}=R_{\tilde Q_6^{\natural\,\natural}}L_{\tilde Q_6^{\natural}}=R_{\tilde Q_6}L_{\tilde Q_6^{\natural}}=\Theta_{\tilde Q_6^{\natural}}$, the factors commuting.

Corollary (the self-adjoint and the skew-adjoint operators). The operator $\Theta_{\tilde Q_6}$ is $\langle\cdot\rangle_{\natural}$-self-adjoint exactly when $\tilde Q_6^{\natural}=\pm\tilde Q_6$, that is exactly when the parameter lies in one of the two eigenspaces of the natural conjugation – the centre $\mathbb{C}_{\mathbb{B}}$ or the vector subspace $\mathrm{Vect}(\mathbb{B})$; and no nonzero operator $\Theta_{\tilde Q_6}$ is $\langle\cdot\rangle_{\natural}$-skew-adjoint.

Proof. Self-adjointness is $\Theta_{\tilde Q_6^{\natural}}=\Theta_{\tilde Q_6}$, that is $\tilde Q_6^{\natural}\tilde Q_1\tilde Q_6=\tilde Q_6\tilde Q_1\tilde Q_6^{\natural}$ for all $\tilde Q_1$, which after multiplication on both sides reads $2Q_{6t}(\tilde Q_1\vec Q_6-\vec Q_6\tilde Q_1)=0$ when $\tilde Q_6=Q_{6t}e_t+\vec Q_6$ is split into its scalar and pure parts and $\tilde Q_1$ is taken pure; hence $Q_{6t}=0$, so that $\tilde Q_6$ is pure and $\tilde Q_6^{\natural}=-\tilde Q_6$, or $\vec Q_6$ commutes with every pure element, so that $\vec Q_6=0$ and $\tilde Q_6$ is central with $\tilde Q_6^{\natural}=\tilde Q_6$. The two solutions are exactly the centre and the vector subspace. For skew-adjointness one would need $(\Theta_{\tilde Q_6})^{N}=-\Theta_{\tilde Q_6}$, that is $\Theta_{\tilde Q_6^{\natural}}=\Theta_{i\tilde Q_6}$. Evaluating at $e_t$ gives $\langle\tilde Q_6\rangle_{\natural}=-\langle\tilde Q_6\rangle_{\natural}$, so $\tilde Q_6$ is null; evaluating at a pure $\tilde Q_1$ and comparing the pure parts then gives $Q_{6t}^{2}\tilde Q_1$ against $\langle \vec Q_6\rangle_{\natural}\tilde Q_1$ on the orthogonal complement of $\vec Q_6$, and with $\langle\tilde Q_6\rangle_{\natural}=Q_{6t}^{2}+\langle \vec Q_6\rangle_{\natural}=0$ this forces $Q_{6t}=0$ and $\langle \vec Q_6\rangle_{\natural}=0$, leaving a pure null parameter, for which $\Theta_{\tilde Q_6^{\natural}}=\Theta_{\tilde Q_6}$ rather than $-\Theta_{\tilde Q_6}$ unless $\tilde Q_6=0$. So there is no nonzero skew-adjoint operator.

Remark (the sign and the symmetry). The two self-adjoint families are the two eigenspaces of the sign character of the algebra, which is why the criterion is a sign and not an inequality. The comparison with the Hermitian form is the sharpest illustration: in the dagger layer the adjoint pairs $\tilde Q_6$ with $\tilde Q_6^{*}$, self-adjointness cuts out $\mathbb{M}_+$ and skew-adjointness cuts out $\mathbb{M}_-$, both populated; here the adjoint pairs $\tilde Q_6$ with $\tilde Q_6^{\natural}$, and because $\natural$ is the sign on the vector subspace, the two eigenspaces of the bilinear case are the centre and the vector subspace while the skew-adjoint sector collapses (Mixed Inner Conjugation on the General Plain Sesqualgebra).

When a Product $L_aR_b$ Is Two-Sided

Proposition (the criterion). For all $\tilde Q_1,\tilde Q_3$ the product satisfies

$$ L_{\tilde Q_6}R_{\tilde Q_2}(\tilde Q_1)\,L_{\tilde Q_6}R_{\tilde Q_2}(\tilde Q_3)=L_{\tilde Q_6}R_{\tilde Q_2}\bigl(\tilde Q_1\,(\tilde Q_2\tilde Q_6)\,\tilde Q_3\bigr), $$

so it is multiplicative up to the scalar $\tilde Q_2\tilde Q_6$ – that is $L(\tilde Q_1)L(\tilde Q_3)=\mu L(\tilde Q_1\tilde Q_3)$ with $\mu=\tilde Q_2\tilde Q_6$ – exactly when the product $\tilde Q_2\tilde Q_6$ is central, $\tilde Q_2\tilde Q_6\in\mathbb{C}e_t$; and it is strictly multiplicative, $L(\tilde Q_1)L(\tilde Q_3)=L(\tilde Q_1\tilde Q_3)$, exactly when $\tilde Q_2\tilde Q_6=e_t$, that is $\tilde Q_2=\tilde Q_6^{-1}$ for a unit $\tilde Q_6$. The strictly multiplicative products are therefore the inner automorphisms $\mathrm{Ad}_{\tilde Q_6}$, and up to scalar multiples they are exactly the twisted two-sided operators, $\mathrm{Ad}_{\tilde Q_6}=\langle\tilde Q_6\rangle_{\natural}^{-1}\Theta_{\tilde Q_6}$.

Proof. The left side is $\tilde Q_6\tilde Q_1\tilde Q_2\tilde Q_6\tilde Q_3\tilde Q_2=\tilde Q_6\tilde Q_1(\tilde Q_2\tilde Q_6)\tilde Q_3\tilde Q_2$, which is $L_{\tilde Q_6}R_{\tilde Q_2}$ applied to $\tilde Q_1(\tilde Q_2\tilde Q_6)\tilde Q_3$. When $\tilde Q_2\tilde Q_6=\mu e_t$ the inserted scalar is central and may be moved out, giving $L(\tilde Q_1)L(\tilde Q_3)=\mu L(\tilde Q_1\tilde Q_3)$; conversely, for units $\tilde Q_6,\tilde Q_2$ the identity for all $\tilde Q_1,\tilde Q_3$ forces $\tilde Q_1(\tilde Q_2\tilde Q_6)\tilde Q_3=\mu\tilde Q_1\tilde Q_3$ after cancellation, hence $\tilde Q_2\tilde Q_6=\mu e_t$, and the strict case is $\mu=1$, giving $\tilde Q_2=\tilde Q_6^{-1}=\tilde Q_6^{\natural}/\langle\tilde Q_6\rangle_{\natural}$.

Corollary (the canonical two-sided member). With $\tilde Q_2=\lambda\tilde Q_6^{\natural}$, the product $\tilde Q_2\tilde Q_6=\lambda\tilde Q_6^{\natural}\tilde Q_6=\lambda \langle\tilde Q_6\rangle_{\natural}e_t$ is central, so $L_{\tilde Q_6}R_{\tilde Q_6^{\natural}}=\Theta_{\tilde Q_6}$ is multiplicative up to the norm; and since the strictly multiplicative products are the inner automorphisms $\mathrm{Ad}_{\tilde Q_6}=\langle\tilde Q_6\rangle_{\natural}^{-1}\Theta_{\tilde Q_6}$, the products $L_{\tilde Q_6}R_{\tilde Q_2}$ that are multiplicative up to a scalar are exactly the scalar multiples of the twisted two-sided operators $\Theta_{\tilde Q_6}$, and this is the sense in which the signed inner conjugation is the canonical two-sided operator of the algebra.

Proof. The centrality is the computation of the criterion; the converse is the criterion read backwards, $\tilde Q_2\tilde Q_6=\mu e_t$ giving $\tilde Q_2=\mu\tilde Q_6^{-1}$ for a unit $\tilde Q_6$, and $\tilde Q_6^{-1}=\tilde Q_6^{\natural}/\langle\tilde Q_6\rangle_{\natural}$ gives $\tilde Q_2$ proportional to $\tilde Q_6^{\natural}$.

Automorphisms

Proposition (the norm of the image and the automorphisms). For all $\tilde Q_6,\tilde Q_4$,

$$ \boxed{\; \langle\Theta_{\tilde Q_6}(\tilde Q_4)\rangle_{\natural}=\langle\tilde Q_6\rangle_{\natural}^{2}\langle\tilde Q_4\rangle_{\natural} \;}, $$

so that $\Theta_{\tilde Q_6}$ is an automorphism of the general quaternionic bilinear form exactly when $\langle\tilde Q_6\rangle_{\natural}=\pm1$; its determinant is $\det\Theta_{\tilde Q_6}=\langle\tilde Q_6\rangle_{\natural}^{4}$ in the coefficient basis.

Proof. $\langle\tilde Q_6\tilde Q_4\tilde Q_6^{\natural}\rangle_{\natural}=\langle\tilde Q_6\rangle_{\natural}\langle\tilde Q_4\rangle_{\natural}\langle\tilde Q_6^{\natural}\rangle_{\natural}=\langle\tilde Q_6\rangle_{\natural}^{2}\langle\tilde Q_4\rangle_{\natural}$ by multiplicativity, and $\langle\tilde Q_6^{\natural}\rangle_{\natural}=\langle\tilde Q_6\rangle_{\natural}$; the determinant follows because $\Theta_{\tilde Q_6}$ is the product of the left and right multiplications by $\tilde Q_6$ and $\tilde Q_6^{\natural}$ and each of those has determinant $\langle\tilde Q_6\rangle_{\natural}^{2}$ in the coefficient basis.

Corollary (the group of operators). The set $\{\Theta_{\tilde Q_6}:\langle\tilde Q_6\rangle_{\natural}=\pm1\}$ is a group under composition, isomorphic to $\{\tilde Q_6\in\mathbb{B}^\times:\langle\tilde Q_6\rangle_{\natural}=\pm1\}/\{\pm e_t\}$, and it lies in the orthogonal group of the form; the subset with $\langle\cdot,\cdot\rangle_{\natural}=1$ lies in the special orthogonal group and consists of the inner automorphisms.

Proof. The composition law and the injectivity up to sign make the assignment a group homomorphism with kernel $\{\pm e_t\}$; the preservation statement is the proposition, and the determinant is $+1$ when $\langle\cdot\rangle_{\natural}=1$; the restriction to the norm-one set is the pair $\Theta_{\tilde Q_6}=\mathrm{Ad}_{\tilde Q_6}$ of an automorphism. The group is the biquaternion pin group, developed in The Pin and Spin Groups of the General Quaternionic Algebra.

The Dagger Adjoint

Proposition (the adjoint for the Hermitian form). For the general plain sesquilinear form $\langle\tilde Q_1,\tilde Q_2\rangle_{*}=\mathrm{Sc}(\tilde Q_1\tilde Q_2^{*})$,

$$ \bigl\langle\Theta_{\tilde Q_6}\tilde Q_1,\tilde Q_3\bigr\rangle_{*}=\bigl\langle\tilde Q_1,\Theta_{\tilde Q_6^{*}}\tilde Q_3\bigr\rangle_{*}, $$

so that the adjoint is $\bigl(\Theta_{\tilde Q_6}\bigr)^{*}=\Theta_{\tilde Q_6^{*}}$; the operator is self-adjoint exactly for $\tilde Q_6^{*}=\pm\tilde Q_6$, that is for $\tilde Q_6$ in $\mathbb{M}_+$ or in $\mathbb{M}_-$, and it is unitary for the Hermitian form exactly when $\tilde Q_6^{*}\tilde Q_6=e_t$, that is on the slice $U$ of the unitary elements of norm one.

Proof. Multiplying out, $\langle\Theta_{\tilde Q_6}\tilde Q_1,\tilde Q_3\rangle_{*}=\mathrm{Sc}(\tilde Q_6\tilde Q_1\tilde Q_6^{\natural}\tilde Q_3^{*})$ and $\langle\tilde Q_1,\Theta_{\tilde Q_6^{*}}\tilde Q_3\rangle_{*}=\mathrm{Sc}(\tilde Q_1\tilde Q_6^{\natural}\tilde Q_3^{*}\tilde Q_6)$, and the two agree by the cyclic property of the scalar part together with $(\tilde Q_6^{*})^{\natural}=\tilde Q_6^{\natural*}$, which follows from ${}^{\natural}$ and ${}^{*}$ commuting. Self-adjointness is $(\Theta_{\tilde Q_6})^{*}=\Theta_{\tilde Q_6}$, that is $\Theta_{\tilde Q_6^{*}}=\Theta_{\tilde Q_6}$, that is $\tilde Q_6^{*}=\pm\tilde Q_6$; the plus is the Hermitian sector $\mathbb{M}_+$ and the minus the anti-Hermitian sector $\mathbb{M}_-$, both populated, since the operator is blind to the overall sign of its parameter. Unitarity is $(\Theta_{\tilde Q_6})^{*}\Theta_{\tilde Q_6}=\mathrm{id}$, that is $\Theta_{\tilde Q_6^{*}\tilde Q_6}=\mathrm{id}$, that is $\tilde Q_6^{*}\tilde Q_6=\pm e_t$, and the minus is impossible because $\langle\tilde Q_6^{*}\tilde Q_6\rangle_{\natural}=\langle\tilde Q_6^{*}\rangle_{\natural}\langle\tilde Q_6\rangle_{\natural}=\overline{\langle\tilde Q_6\rangle_{\natural}}\langle\tilde Q_6\rangle_{\natural}=\lvert \langle\tilde Q_6\rangle_{\natural}\rvert^{2}$ is a non-negative real while $\langle -e_t\rangle_{\natural}=-1$. The dagger layer, with its two sectors and its mixed operators, is Mixed Inner Conjugation on the General Plain Sesqualgebra.

Worked Examples

A unit of norm one. For $\tilde Q_6=e_t+\vec e_x$, $\langle\tilde Q_6\rangle_{\natural}=2$, so $\Theta_{\tilde Q_6}=2\,\mathrm{Ad}_{\tilde Q_6}$; the operator preserves the form up to $\langle\tilde Q_6\rangle_{\natural}^{2}=4$ and is not a form-preserving map. Scaling to $\tilde Q_7=(e_t+\vec e_x)/\sqrt2$ gives $\langle\cdot,\cdot\rangle_{\natural}=1$ and $\Theta_{\tilde Q_7}=\mathrm{Ad}_{\tilde Q_7}$, an inner automorphism and a form-preserving map.

A null parameter. For $\tilde Q_6=\vec e_x+i\vec e_y$, $\langle\tilde Q_6\rangle_{\natural}=0$ and $\Theta_{\tilde Q_6}(\tilde Q_1)=\tilde Q_6\tilde Q_1\tilde Q_6^{\natural}$; the operator is nonzero but has $\Theta_{\tilde Q_6}(e_t)=0$ and is not invertible, in agreement with the criterion that $\Theta$ is invertible exactly on the units.

A vector parameter. For $\tilde Q_5=\vec e_x$ one has $\tilde Q_5^{\natural}=-\vec e_x$, so $\Theta_{\tilde Q_5}(\tilde Q_1)=-\vec e_x\tilde Q_1 \vec e_x$, with $\langle\tilde Q_5\rangle_{\natural}=1$: the operator is a form-preserving map, $\Theta_{\tilde Q_5}(e_t)=e_t$ and $\Theta_{\tilde Q_5}(\vec e_y)=-\vec e_y$. Its restriction to the vector subspace is the reflection, up to sign, described in The Pin and Spin Groups of the General Quaternionic Algebra.

A parameter of norm minus one. For $\tilde Q_6=ie_t$, $\langle\tilde Q_6\rangle_{\natural}=-1$ and $\tilde Q_6^{\natural}=ie_t$, so $\Theta_{\tilde Q_6}(\tilde Q_1)=-\tilde Q_1$: the operator is a scalar, of norm $\pm1$ and orthogonal, of determinant $+1$; the kernel $\{\pm e_t\}$ of the assignment contains both $\tilde Q_6$ and $-\tilde Q_6$, and the two give the same operator.

Summary

The twisted two-sided operator of the biquaternion algebra is $\Theta_{\tilde Q_6}=L_{\tilde Q_6}R_{\tilde Q_6^{\natural}}$, the conjugation sandwich $\tilde Q_1\mapsto\tilde Q_6\tilde Q_1\tilde Q_6^{\natural}$: quadratic in the parameter, with $\Theta_{-\tilde Q_6}=\Theta_{\tilde Q_6}$ and $\Theta_{i\tilde Q_6}=-\Theta_{\tilde Q_6}$, defined for every element, and equal to $\langle\tilde Q_6\rangle_{\natural}$ times the inner conjugation $\mathrm{Ad}_{\tilde Q_6}$ on the units. It satisfies $\Theta_{\tilde Q_6}\circ\Theta_{\tilde Q_2}=\Theta_{\tilde Q_6\tilde Q_2}$ and $\Theta_{\tilde Q_6}(\tilde Q_1)\Theta_{\tilde Q_6}(\tilde Q_3)=\langle\tilde Q_6\rangle_{\natural}\Theta_{\tilde Q_6}(\tilde Q_1\tilde Q_3)$, so that it is an algebra automorphism exactly on the norm-one set and is multiplicative up to the sign $-1$ for norm minus one. The assignment is injective up to sign on the units, with kernel $\{\pm e_t\}$, and $\Theta_{\tilde Q_6}=0$ only for $\tilde Q_6=0$; its adjoint for the general quaternionic bilinear form is $\Theta_{\tilde Q_6^{\natural}}$, the self-adjoint operators being exactly the parameters lying in the centre or the vector subspace and the skew-adjoint ones nonexistent; and its adjoint for the Hermitian form is $\Theta_{\tilde Q_6^{*}}$, with self-adjointness on $\mathbb{M}_+\cup\mathbb{M}_-$ and unitarity on the slice $U$. Among the products $L_{\tilde Q_6}R_{\tilde Q_2}$, those that are multiplicative up to a scalar are exactly the ones with $\tilde Q_2\tilde Q_6$ central, that is, up to scalars, $\tilde Q_2$ proportional to $\tilde Q_6^{\natural}$, the strictly multiplicative ones being the inner automorphisms $\mathrm{Ad}_{\tilde Q_6}$; so it is $\Theta$ that is canonical. The operator is an automorphism of the general quaternionic bilinear form exactly when $\langle\tilde Q_6\rangle_{\natural}=\pm1$, and it is that condition that cuts out the pin and spin groups.

Summary of Notation

Further Reading

• The Four Adjoints of the Biquaternion Algebra in Examples, for the signed inner conjugation $\tilde Q_6^{\natural}\tilde Q_4\tilde Q_6^{-1}$ against the unsigned $\tilde Q_6\tilde Q_4\tilde Q_6^{-1}$, their difference $-2\vec Q_6\tilde Q_4\tilde Q_6^{-1}$ and the two matrices of the sign
• One-Sided Operators on the General Quaternionic Algebra, for the factors
• The Pin and Spin Groups of the General Quaternionic Algebra, for the groups cut out by the norm
• The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, for the general quaternionic bilinear form and its Gram matrix
• Mixed Inner Conjugation on the General Plain Sesqualgebra, for the corresponding family over the dagger
• Two-Sided Operators on a Clifford Algebra with Signed Inner Conjugation, for the general theory