One-Sided Operators on the General Quaternionic Algebra

Introduction

The biquaternion algebra acts on itself in two ways, on the left and on the right, and the two actions are the one-sided operators

$$ L_{\tilde Q_6}(\tilde Q_1)=\tilde Q_6\tilde Q_1,\qquad R_{\tilde Q_2}(\tilde Q_1)=\tilde Q_1\tilde Q_2 . $$

Every two-sided operator is a product of one left and one right multiplication, and the twisted two-sided operator of the algebra, the conjugation sandwich $\Theta_{\tilde Q_6}=L_{\tilde Q_6}R_{\tilde Q_6^{\natural}}$, is the member of that family selected by the natural conjugation. This article treats the pair: the two composition laws, the commutation of the left and the right action, the two adjoints, the norm of a multiplication, and the criterion that decides which two-sided products are multiplicative. The two-sided operator itself is Two-Sided Operators on the General Quaternionic Algebra; the general Clifford theory of the one-sided families is One-Sided Operators on a Clifford Algebra with Signed Inner Conjugation, of which this is the biquaternion instance; the corresponding one-sided operators over the dagger are One-Sided Operators on the General Plain Sesqualgebra; and the form that supplies the adjoints is The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra.

The Two Multiplications

Definition. For $\tilde Q_6,\tilde Q_2\in\mathbb{B}$ the left multiplication by $\tilde Q_6$ and the right multiplication by $\tilde Q_2$ are the $\mathbb{C}$-linear maps

$$ L_{\tilde Q_6}:\tilde Q_1\mapsto\tilde Q_6\tilde Q_1,\qquad R_{\tilde Q_2}:\tilde Q_1\mapsto\tilde Q_1\tilde Q_2 . $$

Proposition (linearity, the identity, injectivity). The maps $\tilde Q_6\mapsto L_{\tilde Q_6}$ and $\tilde Q_2\mapsto R_{\tilde Q_2}$ are $\mathbb{C}$-linear and injective; $L_{e_t}=R_{e_t}=\mathrm{id}$; and $L_{\tilde Q_6}$ and $R_{\tilde Q_2}$ are bijections exactly when $\tilde Q_6$ and $\tilde Q_2$ are units, with inverses $L_{\tilde Q_6}^{-1}=L_{\tilde Q_6^{-1}}$ and $R_{\tilde Q_2}^{-1}=R_{\tilde Q_2^{-1}}$.

Proof. Linearity is linearity of the product in each factor. For injectivity, $L_{\tilde Q_6}=0$ forces $\tilde Q_6=L_{\tilde Q_6}(e_t)=0$, and $R_{\tilde Q_2}=0$ forces $\tilde Q_2=R_{\tilde Q_2}(e_t)=0$; the identity element gives $L_{e_t}=R_{e_t}=\mathrm{id}$. For the inverses, $L_{\tilde Q_6}L_{\tilde Q_6^{-1}}=L_{\tilde Q_6\tilde Q_6^{-1}}=L_{e_t}=\mathrm{id}$, and only a unit has a two-sided inverse; the injectivity of the parameter map shows that no non-unit can be invertible, since a nonzero non-unit is a zero divisor.

Remark (the parameter map is linear here). Unlike the two-sided operator, which is quadratic in its parameter, each one-sided operator is linear in its parameter: $L_{\lambda\tilde Q_6+\mu\tilde Q_2}=\lambda L_{\tilde Q_6}+\mu L_{\tilde Q_2}$, and likewise for $R$. This is the formal difference between the one-sided and the two-sided layers, and it is the reason a one-sided operator alone cannot carry the norm: the norm appears only in the composition of two.

The Composition Laws and the Two Actions

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

$$ \boxed{\; L_{\tilde Q_6}\circ L_{\tilde Q_2}=L_{\tilde Q_6\tilde Q_2},\qquad R_{\tilde Q_6}\circ R_{\tilde Q_2}=R_{\tilde Q_2\tilde Q_6},\qquad L_{\tilde Q_6}\circ R_{\tilde Q_2}=R_{\tilde Q_2}\circ L_{\tilde Q_6} \;}. $$

Proof. $L_{\tilde Q_6}L_{\tilde Q_2}(\tilde Q_1)=\tilde Q_6\tilde Q_2\tilde Q_1=L_{\tilde Q_6\tilde Q_2}(\tilde Q_1)$; $R_{\tilde Q_6}R_{\tilde Q_2}(\tilde Q_1)=\tilde Q_1\tilde Q_2\tilde Q_6=R_{\tilde Q_2\tilde Q_6}(\tilde Q_1)$; and the left and right multiplications commute by associativity, $\tilde Q_6(\tilde Q_1\tilde Q_2)=(\tilde Q_6\tilde Q_1)\tilde Q_2$.

Remark (the two laws are opposite). The left law preserves the order of the parameters and the right law reverses it: $L$ is a representation of the algebra and $R$ is a representation of the opposite algebra, $R_{\tilde Q_6}R_{\tilde Q_2}=R_{\tilde Q_2\tilde Q_6}$. This is the formal reason the two-sided product $L_{\tilde Q_6}R_{\tilde Q_2}$ is not multiplicative in general, and the reason the natural conjugation, which is an anti-automorphism, is the right partner for the left multiplication.

Proposition (the double centraliser). The operators commuting with every left multiplication are exactly the right multiplications: $\{T:TL_{\tilde Q_6}=L_{\tilde Q_6}T\text{ for all }\tilde Q_6\}=\{R_{\tilde Q_2}\}$. Dually, the operators commuting with every right multiplication are the left multiplications; and the operators commuting with both families are the scalars, $T=\lambda\,\mathrm{id}$.

Proof. A right multiplication commutes with every left multiplication by associativity, which is one inclusion. For the converse, let $T$ commute with all $L_{\tilde Q_6}$, and put $\tilde Q_2=T(e_t)$; then $T(\tilde Q_6)=T(L_{\tilde Q_6}e_t)=L_{\tilde Q_6}T(e_t)=\tilde Q_6\tilde Q_2=R_{\tilde Q_2}(\tilde Q_6)$, so $T=R_{\tilde Q_2}$. An operator commuting with both families commutes with all products $L_{\tilde Q_6}R_{\tilde Q_2}$, hence is multiplication by a central element, that is by a scalar.

The Two-Sided Products and the Sign

Definition. A two-sided operator is a product $L_{\tilde Q_6}R_{\tilde Q_2}$; the twisted two-sided operator is the member $\Theta_{\tilde Q_6}=L_{\tilde Q_6}R_{\tilde Q_6^{\natural}}$.

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 $\tilde Q_2\tilde Q_6$ exactly when $\tilde Q_2\tilde Q_6$ is central, and strictly multiplicative exactly when $\tilde Q_2\tilde Q_6=e_t$; up to scalar multiples, the products that are multiplicative in this sense are exactly the twisted two-sided operators.

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 $\tilde Q_2\tilde Q_6=\mu e_t$ is scalar it may be moved out, giving $L(\tilde Q_1)L(\tilde Q_3)=\mu L(\tilde Q_1\tilde Q_3)$, and the strict case is $\mu=1$. Conversely, for units $\tilde Q_6,\tilde Q_2$ the identity for all $\tilde Q_1,\tilde Q_3$ forces the factor $\tilde Q_2\tilde Q_6$ to be scalar. With $\tilde Q_2\tilde Q_6=\mu e_t$ and $\tilde Q_6$ a unit, $\tilde Q_2=\mu\tilde Q_6^{-1}=\mu\tilde Q_6^{\natural}/\langle\tilde Q_6\rangle_{\natural}$, proportional to $\tilde Q_6^{\natural}$.

Proposition (the sign, and the factorisation). For every element,

$$ \Theta_{\tilde Q_6}=L_{\tilde Q_6}R_{\tilde Q_6^{\natural}}=\langle\tilde Q_6\rangle_{\natural}\,\mathrm{Ad}_{\tilde Q_6}\quad(\tilde Q_6\text{ a unit}),\qquad \Theta_{\tilde Q_6}(\tilde Q_1)\Theta_{\tilde Q_6}(\tilde Q_3)=\langle\tilde Q_6\rangle_{\natural}\,\Theta_{\tilde Q_6}(\tilde Q_1\tilde Q_3), $$

and $\Theta_{\tilde Q_6}\Theta_{\tilde Q_2}=\Theta_{\tilde Q_6\tilde Q_2}$. The right factor is the inverse right multiplication up to the norm, $R_{\tilde Q_6^{\natural}}=\langle\tilde Q_6\rangle_{\natural}R_{\tilde Q_6^{-1}}$ on a unit, so it is the left factor $L_{\tilde Q_6}$ that carries the conjugation, and the norm scalar that appears in the multiplication law is the price of that choice.

Proof. Each identity is the corresponding identity of Two-Sided Operators on the General Quaternionic Algebra, read in the one-sided notation; the factorisation $R_{\tilde Q_6^{\natural}}=\langle\tilde Q_6\rangle_{\natural}R_{\tilde Q_6^{-1}}$ is $\tilde Q_6^{\natural}=\langle\tilde Q_6\rangle_{\natural}\tilde Q_6^{-1}$ for a unit.

Remark (why the left factor carries the sign). A one-sided operator does not act on the vector subspace: no left multiplication and no right multiplication carries $\mathbb{C}\{\vec e_x,\vec e_y,\vec e_z\}$ into itself except the scalar ones, so the sign cannot be detected one-sidedly. It is the product of the two factors that acts on the vector subspace as a form-preserving map, and the sign sits in the left factor as the reflection of the parameter.

Remark (the two normalisations of the sign). The general two-sided family $\Phi^{\theta,c}_x(y)=\theta(x)y\,c(x)$ of Two-Sided Operators on a Clifford Algebra has five members, and the name "signed inner conjugation" is used in the corpus for two of them: the member $\theta=\alpha$, $c=(\ )^{-1}$, which is $\alpha(\tilde Q_6)\tilde Q_1\tilde Q_6^{-1}=\varepsilon_{\tilde Q_6}\mathrm{Ad}_{\tilde Q_6}$ with the parity sign $\varepsilon$, and the member $\theta=\mathrm{id}$, $c={}^{\natural}$, the conjugation sandwich $\tilde Q_6\tilde Q_1\tilde Q_6^{\natural}=\langle\tilde Q_6\rangle_{\natural}\mathrm{Ad}_{\tilde Q_6}$, which is the operator of this article. They coincide exactly on the homogeneous parameters with $\langle\tilde Q_6\rangle_{\natural}=\varepsilon_{\tilde Q_6}$. The one-sided factors below are those of the second normalisation, the one whose right factor is $R_{\tilde Q_6^{\natural}}=\langle\tilde Q_6\rangle_{\natural}R_{\tilde Q_6^{-1}}$; the factors of the first differ by the same scalar and are the ones used in Biquaternion Versors and the Orthogonal Group.

The Adjoints for the Bilinear Form

Proposition (the two adjoints). For the general quaternionic bilinear form,

$$ \bigl(L_{\tilde Q_6}\bigr)^{N}=L_{\tilde Q_6^{\natural}},\qquad \bigl(R_{\tilde Q_2}\bigr)^{N}=R_{\tilde Q_2^{\natural}} . $$

Proof. $\langle L_{\tilde Q_6}\tilde Q_4,\tilde Q_1\rangle_{\natural}=\mathrm{Sc}(\tilde Q_6\tilde Q_4\tilde Q_1^{\natural})$, and by the cyclicity of the scalar part this equals $\mathrm{Sc}(\tilde Q_4\tilde Q_1^{\natural}\tilde Q_6)=\mathrm{Sc}(\tilde Q_4(\tilde Q_6^{\natural}\tilde Q_1)^{\natural})=\langle\tilde Q_4,L_{\tilde Q_6^{\natural}}\tilde Q_1\rangle_{\natural}$, since ${}^{\natural}$ is an anti-automorphism; the right case is the same computation with the factors in the other order.

Corollary (the self-adjoint, the skew-adjoint and the congruent multiplications). The left multiplication $L_{\tilde Q_6}$ is $N$-self-adjoint exactly for $\tilde Q_6$ central, $N$-skew-adjoint exactly for $\tilde Q_6$ pure, and an automorphism of the form exactly for $\langle\tilde Q_6\rangle_{\natural}=1$; for the right multiplication the same three criteria hold with $\tilde Q_2$ in place of $\tilde Q_6$.

Proof. By the injectivity of the parameter map, $L_{\tilde Q_6^{\natural}}=L_{\tilde Q_6}$ is $\tilde Q_6^{\natural}=\tilde Q_6$, that is $\tilde Q_6$ central; $L_{\tilde Q_6^{\natural}}=-L_{\tilde Q_6}=L_{-\tilde Q_6}$ is $\tilde Q_6^{\natural}=-\tilde Q_6$, that is $\tilde Q_6$ pure; and the preservation condition is read from the norm of the image, $\langle L_{\tilde Q_6}\tilde Q_4\rangle_{\natural}=\langle\tilde Q_6\rangle_{\natural}\langle\tilde Q_4\rangle_{\natural}$, which is $1$ for all $\tilde Q_4$ exactly when $\langle\tilde Q_6\rangle_{\natural}=1$.

Remark (the two sectors are populated here). Both the self-adjoint and the skew-adjoint sector are non-empty for the one-sided operators, the centre $\mathbb{C}_{\mathbb{B}}$ and the vector subspace $\mathrm{Vect}(\mathbb{B})$: the sign character of the algebra splits the parameters, and the two sectors are its two eigenspaces. The contrast with the two-sided operator, where the skew-adjoint sector collapses, is one more sign that the two-sided operator is the twisted object and the one-sided ones are not.

The Norm and the Determinant of a Multiplication

Proposition (the conformal factor and the determinant). For all $\tilde Q_6,\tilde Q_4$,

$$ \langle L_{\tilde Q_6}\tilde Q_4\rangle_{\natural}=\langle\tilde Q_6\rangle_{\natural}\,\langle\tilde Q_4\rangle_{\natural},\qquad \langle R_{\tilde Q_6}\tilde Q_4\rangle_{\natural}=\langle\tilde Q_6\rangle_{\natural}\,\langle\tilde Q_4\rangle_{\natural}, $$

so that each multiplication scales the form by the norm of its parameter; in the coefficient basis,

$$ \boxed{\; \det L_{\tilde Q_6}=\det R_{\tilde Q_6}=\langle\tilde Q_6\rangle_{\natural}^{2} \;}. $$

Proof. $\langle\tilde Q_6\tilde Q_4\rangle_{\natural}=\langle\tilde Q_6\rangle_{\natural}\langle\tilde Q_4\rangle_{\natural}$ is the multiplicativity of the norm, and the right case is the same with the factors interchanged. For the determinant, the matrix realisation $\mathsf{M}_2$ of the algebra sends $L_{\tilde Q_6}$ to the map $B\mapsto\mathsf{M}_2(\tilde Q_6)B$ on $M_2(\mathbb{C})$, which is $\mathsf{M}_2(\tilde Q_6)\otimes I_2$; the determinant of a Kronecker product is $\det(\mathsf{M}_2(\tilde Q_6))^{2}\det(I_2)^{2}=\det(\mathsf{M}_2(\tilde Q_6))^{2}=\langle\tilde Q_6\rangle_{\natural}^{2}$, since $\det\mathsf{M}_2(\tilde Q_6)=\langle\tilde Q_6\rangle_{\natural}$.

Corollary (the form-preserving multiplications). The multiplications that preserve the form are exactly those by an element of norm one, and their group is the norm-one slice of $\mathbb{B}^\times$; the multiplications by an element of norm $-1$ reverse the sign of the form.

Proposition (the matrix models of the multiplications). The two multiplications are the two Kronecker products of the matrix $\mathsf{M}_2(\tilde Q_6)$ with the identity:

$$ \mathsf{M}_2\bigl(L_{\tilde Q_6}\tilde Q_4\bigr)=\mathsf{M}_2(\tilde Q_6)\,\mathsf{M}_2(\tilde Q_4),\qquad \mathsf{M}_2\bigl(R_{\tilde Q_6}\tilde Q_4\bigr)=\mathsf{M}_2(\tilde Q_4)\,\mathsf{M}_2(\tilde Q_6), $$

so that $L_{\tilde Q_6}$ is the left factor and $R_{\tilde Q_6}$ the right factor of the regular representation: each is the Kronecker product of $\mathsf{M}_2(\tilde Q_6)$ with the identity $I_2$, the left one carrying $\mathsf{M}_2(\tilde Q_6)$ itself and the right one carrying its transpose, the transpose entering because the two factors of a Kronecker product act on the two sides of the matrix. Independently of which slot the identity occupies, $\mathsf{M}_2(\tilde Q_6)$ acts on the left factor for $L_{\tilde Q_6}$ and $\mathsf{M}_2(\tilde Q_6)^{\mathsf T}$ on the right factor for $R_{\tilde Q_6}$, so the two determinants are $\det\mathsf{M}_2(\tilde Q_6)^{2}=\langle\tilde Q_6\rangle_{\natural}^{2}$ each and the two traces are $2\,\mathrm{tr}\,\mathsf{M}_2(\tilde Q_6)=4Q_{6t}$ each; the two commute for every pair of parameters, which is the associativity of the algebra read in the regular representation.

Proof. The multiplication map $B\mapsto\mathsf{M}_2(\tilde Q_6)B$ on $M_2(\mathbb{C})\cong\mathbb{C}^{4}$ is a Kronecker product of $\mathsf{M}_2(\tilde Q_6)$ with $I_2$, of determinant $\det\mathsf{M}_2(\tilde Q_6)^{2}=\langle\tilde Q_6\rangle_{\natural}^{2}$ by the multiplicativity of the determinant of a Kronecker product, and $B\mapsto B\mathsf{M}_2(\tilde Q_6)$ is the same product with $\mathsf{M}_2(\tilde Q_6)$ replaced by its transpose, of the same determinant because the determinant is invariant under transposition. The trace of either product is $\mathrm{tr}\,\mathsf{M}_2(\tilde Q_6)\cdot\mathrm{tr}\,I_2=2Q_{6t}\cdot2=4Q_{6t}$ because $\mathrm{tr}\,\mathsf{M}_2(\tilde Q_6)=2Q_{6t}$; the commutation is the mixed-product rule for Kronecker products, $(A\otimes I_2)(I_2\otimes A^{\mathsf T})=(I_2\otimes A^{\mathsf T})(A\otimes I_2)$.

The Hermitian Layer

Proposition (the dagger adjoints). For the general plain sesquilinear form,

$$ \boxed{\; \bigl(L_{\tilde Q_6}\bigr)^{*}=L_{\tilde Q_6^{*}},\qquad \bigl(R_{\tilde Q_2}\bigr)^{*}=R_{\tilde Q_2^{*}} \;}. $$

Proof. $\langle L_{\tilde Q_6}\tilde Q_4,\tilde Q_1\rangle_{*}=\mathrm{Sc}(\tilde Q_6\tilde Q_4\tilde Q_1^{*})$ and $\langle\tilde Q_4,L_{\tilde Q_6^{*}}\tilde Q_1\rangle_{*}=\mathrm{Sc}(\tilde Q_4(\tilde Q_6^{*}\tilde Q_1)^{*})=\mathrm{Sc}(\tilde Q_4\tilde Q_1^{*}\tilde Q_6)$; the two agree by the cyclicity of the scalar part, since $(\tilde Q_6^{*}\tilde Q_1)^{*}=\tilde Q_1^{*}\tilde Q_6$.

Corollary (the sectors for the dagger). For the Hermitian form, $L_{\tilde Q_6}$ is self-adjoint exactly for $\tilde Q_6\in\mathbb{M}_+$, skew-adjoint exactly for $\tilde Q_6\in\mathbb{M}_-$, and unitary exactly on the slice $a^{*}a=e_t$; the two sectors are the two halves of the Hermitian splitting, and the dagger layer is where the pair of sectors is symmetric (One-Sided Operators on the General Plain Sesqualgebra).

Worked Examples

A scalar parameter. For $\tilde Q_6=\lambda e_t$ the multiplications are scalars, $L_{\tilde Q_6}=R_{\tilde Q_6}=\lambda\,\mathrm{id}$; they are self-adjoint for every $\lambda$, since a scalar parameter is central, and a form-preserving map exactly for $\lambda^{2}=1$, that is $\lambda=\pm1$.

A vector parameter. For $\tilde Q_5=\vec e_x$, $\langle\tilde Q_5\rangle_{\natural}=1$ and $\tilde Q_5^{\natural}=-\vec e_x$, so $L_{\tilde Q_5}$ is skew-adjoint and an automorphism of the form; the right multiplication $R_{\tilde Q_5}$ has the same two properties, and $L_{\tilde Q_5}R_{\tilde Q_5}=\Theta_{\tilde Q_5}$ is the reflection-like operator of the two-sided layer.

A null parameter. For $\tilde Q_6=\vec e_x+i\vec e_y$, $\langle\tilde Q_6\rangle_{\natural}=0$: both multiplications are nonzero, neither is invertible, and both collapse the form, $\langle L_{\tilde Q_6}\tilde Q_4\rangle_{\natural}=0$ for every $\tilde Q_4$.

A unit of norm one. For $\tilde Q_6=(e_t+\vec e_x)/\sqrt2$ the multiplications are automorphisms of the form, and the two-sided product $L_{\tilde Q_6}R_{\tilde Q_6^{\natural}}$ is an inner automorphism.

Summary

The one-sided operators of the biquaternion algebra are the left and right multiplications $L_{\tilde Q_6}(\tilde Q_1)=\tilde Q_6\tilde Q_1$ and $R_{\tilde Q_2}(\tilde Q_1)=\tilde Q_1\tilde Q_2$: linear and injective in the parameter, invertible exactly for units, with the composition laws $L_{\tilde Q_6}L_{\tilde Q_2}=L_{\tilde Q_6\tilde Q_2}$, $R_{\tilde Q_6}R_{\tilde Q_2}=R_{\tilde Q_2\tilde Q_6}$ and $L_{\tilde Q_6}R_{\tilde Q_2}=R_{\tilde Q_2}L_{\tilde Q_6}$. The operators commuting with all left multiplications are exactly the right multiplications, and the operators commuting with both families are the scalars. For the general quaternionic bilinear form the adjoints are $(L_{\tilde Q_6})^{N}=L_{\tilde Q_6^{\natural}}$ and $(R_{\tilde Q_2})^{N}=R_{\tilde Q_2^{\natural}}$; a left multiplication is self-adjoint exactly for a central parameter, skew-adjoint exactly for a pure one, and a form-preserving map exactly for norm one, and each multiplication scales the form by the norm of its parameter. For the Hermitian form the adjoints are the same with ${}^{*}$, with the two sectors $\mathbb{M}_+$ and $\mathbb{M}_-$ and unitarity on the slice $a^{*}a=e_t$. The two-sided products $L_{\tilde Q_6}R_{\tilde Q_2}$ are multiplicative up to the scalar $\tilde Q_2\tilde Q_6$ exactly when $\tilde Q_2\tilde Q_6$ is central and strictly multiplicative exactly when $\tilde Q_2\tilde Q_6=e_t$, that is, up to scalars, exactly for the twisted two-sided operators $\Theta_{\tilde Q_6}=L_{\tilde Q_6}R_{\tilde Q_6^{\natural}}$, whose multiplicativity holds up to the norm.

Summary of Notation

Further Reading

• Two-Sided Operators on the General Quaternionic Algebra, for the product of the two factors
• One-Sided Operators on the General Plain Sesqualgebra, for the dagger layer
• The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, for the general quaternionic bilinear form and its Gram matrix
• One-Sided Operators on a Clifford Algebra with Signed Inner Conjugation, for the general theory