The Multiplication Operators of the Symmetric Quaternionic Algebra

Introduction

The symmetric quaternionic multiplication is commutative, so its left multiplications and its right multiplications coincide; the operator of the block is the single family

$$ L^{\star}_{\tilde Q_5}: \mathbb{B}\longrightarrow\mathbb{C}e_t,\qquad L^{\star}_{\tilde Q_5}\tilde Q_4 = \tilde Q_5\star\tilde Q_4 = \langle\tilde Q_5,\tilde Q_4\rangle_{\natural}e_t , $$

of The Radical and the Isotropic Elements of the Symmetric Quaternionic Algebra. This article reads the family as the operator layer of the block: the matrices, the ranks and the traces of the operators, their composition and the algebra they generate, their adjoints with respect to the form $\langle\cdot\rangle_{\natural}$, the derivations the block produces, and the two groups of linear maps attached to it — the group that preserves the form and the group that preserves the product. It follows the pattern of the operator articles of the parent row, The Left Multiplications of the Quaternionic Product and the Opposite Monoid and One-Sided Operators on the General Quaternionic Algebra of Biquaternions, and it cites rather than restates the general theory of the left multiplications of an algebra.

The operation and its formula are Introduction to the Symmetric Quaternionic Algebra; the form $\langle\cdot\rangle_{\natural}$ and its role as the coefficient are The Quaternion Form as a Product on the Symmetric Quaternionic Algebra; the derivation layer of a general algebra is Automorphisms and Derivations of Algebras; the matrix models are Introduction to the Symmetric Quaternionic Algebra. Nothing of the enriched layer of Part II is used.

The Operators, Their Matrices and Their Ranks

Proposition (the matrix of a multiplication). In the basis $e_t,\vec e_x,\vec e_y,\vec e_z$ the matrix of $L^{\star}_{\tilde Q_5}$ has its first row equal to the coordinates of $\tilde Q_5$ and every other entry zero:

$$ \bigl[L^{\star}_{\tilde Q_5}\bigr]_{t\mu} = Q_{5\mu}, \qquad \bigl[L^{\star}_{\tilde Q_5}\bigr]_{\nu\mu}=0 \ (\nu\in\{x,y,z\}), $$

that is, the four columns are the four multiples $Q_{5t}e_t,Q_{5x}e_t,Q_{5y}e_t,Q_{5z}e_t$.

Proof. $L^{\star}_{\tilde Q_5}e_\mu=\langle\tilde Q_5,e_\mu\rangle_{\natural}e_t=Q_{5\mu} e_t$, so the $\mu$-th column is $Q_{5\mu} e_t$, with coordinates $Q_{5\mu}$ at the first place and zero elsewhere.

Corollary (rank and image). For $\tilde Q_5\ne0$ the rank of $L^{\star}_{\tilde Q_5}$ is $1$ and its image is the whole central line $\mathbb{C}e_t$; for $\tilde Q_5=0$ it is the zero operator. The kernel is the hyperplane

$$ \ker L^{\star}_{\tilde Q_5} = \{\tilde Q_4\in\mathbb{B}: \langle\tilde Q_5,\tilde Q_4\rangle_{\natural}=0\}, $$

of complex dimension $3$ for $\tilde Q_5\ne0$; it is the hyperplane of the elements paired to zero with $\tilde Q_5$ by the form $\langle\cdot,\cdot\rangle_{\natural}$, and it contains $\tilde Q_5$ exactly when $\tilde Q_5$ is isotropic.

Proof. The columns span $\mathbb{C}e_t$ if some $Q_{5\mu}\ne0$, and the image is central; the kernel is the kernel of the linear form $\langle\tilde Q_5,\cdot\rangle_{\natural}$, whose rank is $1$ because the form is non-degenerate. The last statement is $\langle\tilde Q_5,\tilde Q_5\rangle_{\natural}=\langle\tilde Q_5\rangle_{\natural}=0$.

Proposition (trace and determinant). For every $\tilde Q_5$,

$$ \boxed{\; \operatorname{Tr}\bigl(L^{\star}_{\tilde Q_5}\bigr) = Q_{5t}, \qquad \det\bigl(L^{\star}_{\tilde Q_5}\bigr) = 0 \;}. $$

Proof. The trace is the sum of the diagonal entries of the matrix above, and only the entry $(t,t)=Q_{5t}$ is nonzero. The determinant vanishes because the rank is at most $1$ on a space of dimension $4$; for $\tilde Q_5=0$ the operator is zero.

Remark (a rank-one layer). The operator layer of the block is a rank-one layer: every nonzero multiplication has rank one, so the family is as far from the identity operator as a nonzero family can be, and no element of the algebra is a unit. The trace detects the scalar part of the multiplier and nothing else: the vector part of the multiplier is invisible to the trace, and the operator family separates the elements by the kernel and not by the trace.

Composition and the Algebra of Operators

Proposition (the composition law). The multiplication operators compose by

$$ \boxed{\; L^{\star}_{\tilde Q_5}\circ L^{\star}_{\tilde Q_1} = Q_{5t}\,L^{\star}_{\tilde Q_1} \;}, $$

and the family $\{L^{\star}_{\tilde Q_5}:\tilde Q_5\in\mathbb{B}\}$ is a complex vector space of dimension $4$, the monoid of multiplications of the block.

Proof. $L^{\star}_{\tilde Q_5}(L^{\star}_{\tilde Q_1}\tilde Q_4)=L^{\star}_{\tilde Q_5}\bigl(\langle\tilde Q_1,\tilde Q_4\rangle_{\natural}e_t\bigr)=\langle\tilde Q_1,\tilde Q_4\rangle_{\natural}\langle\tilde Q_5,e_t\rangle_{\natural}e_t=Q_{5t}\langle\tilde Q_1,\tilde Q_4\rangle_{\natural}e_t=Q_{5t}L^{\star}_{\tilde Q_1}\tilde Q_4$.

Corollary (the map is faithful but not multiplicative). The map $\tilde Q_5\mapsto L^{\star}_{\tilde Q_5}$ is injective, its kernel being the radical $\{0\}$ of the operation, so the operator family represents the algebra faithfully as a vector space. It is not a representation of the operation: the composition law compares $L^{\star}_{\tilde Q_5}\circ L^{\star}_{\tilde Q_1}=Q_{5t}L^{\star}_{\tilde Q_1}$ with $L^{\star}_{\tilde Q_5\star\tilde Q_1}=\langle\tilde Q_5,\tilde Q_1\rangle_{\natural}L^{\star}_{e_t}$, and the two differ, as at $\tilde Q_5=\vec e_x$, $\tilde Q_1=\vec e_x$, where the first is $0$ and the second is $L^{\star}_{e_t}\ne0$. The monoid of multiplications is not commutative either, $L^{\star}_{e_t}\circ L^{\star}_{\vec e_x}=L^{\star}_{\vec e_x}$ while $L^{\star}_{\vec e_x}\circ L^{\star}_{e_t}=0$, although the operation is commutative.

Remark (the operator algebra in the basis). The four operators $L^{\star}_{e_\mu}$ are the matrices with a single nonzero row: $L^{\star}_{e_\mu}$ has the entry $1$ at position $(t,\mu)$ and zeros elsewhere. Their products are $L^{\star}_{e_t}\circ L^{\star}_{e_\nu}=L^{\star}_{e_\nu}$ and $L^{\star}_{\vec e_k}\circ L^{\star}_{e_\nu}=0$ for $k\in\{x,y,z\}$, so the algebra generated by the family is the four dimensional algebra of the matrices whose only possibly nonzero row is the first, and its multiplication is the product of such row matrices. This is the shadow of the collapse: the operator family carries the four coordinates of the multiplier in a single row, and no product of two multiplications can recover the vector part of a multiplier.

The Adjoint with Respect to the Form

Definition. The adjoint of a complex-linear map $T$ with respect to $\langle\cdot\rangle_{\natural}$ is the linear map $T^{\dagger}$ determined by

$$ \langle T\tilde Q_2,\tilde Q_3\rangle_{\natural} = \langle\tilde Q_2,T^{\dagger}\tilde Q_3\rangle_{\natural} \qquad \text{for all } \tilde Q_2,\tilde Q_3\in\mathbb{B}, $$

which exists and is unique because $\langle\cdot\rangle_{\natural}$ is non-degenerate.

Proposition (the adjoint of a multiplication). For every $\tilde Q_5$,

$$ \boxed{\; \bigl(L^{\star}_{\tilde Q_5}\bigr)^{\dagger}\tilde Q_3 = \sum_\mu Q_{5\mu} Q_{3t}\,e_\mu , \qquad\text{and}\qquad L^{\star}_{\tilde Q_5}\ \text{is self-adjoint}\quad\Longleftrightarrow\quad \tilde Q_5\in\mathbb{C}e_t \;}. $$

Proof. $\langle L^{\star}_{\tilde Q_5}\tilde Q_2,\tilde Q_3\rangle_{\natural}=\langle\langle\tilde Q_5,\tilde Q_2\rangle_{\natural}e_t,\tilde Q_3\rangle_{\natural}=\langle\tilde Q_5,\tilde Q_2\rangle_{\natural}Q_{3t}=\sum_\mu Q_{5\mu} Q_{2\mu} Q_{3t}$, and $\langle\tilde Q_2,\sum_\mu Q_{5\mu} Q_{3t}e_\mu\rangle_{\natural}=\sum_\mu Q_{2\mu} Q_{5\mu} Q_{3t}$; the two agree for all $\tilde Q_2$, which gives the displayed adjoint. For self-adjointness, compare the adjoint with $L^{\star}_{\tilde Q_5}\tilde Q_3=\langle\tilde Q_5,\tilde Q_3\rangle_{\natural}e_t$: the identity $\sum_\mu Q_{5\mu} Q_{3t}e_\mu=\bigl(\sum_\mu Q_{5\mu} Q_{3\mu}\bigr)e_t$ for all $\tilde Q_3$ forces $Q_{5\mu}=0$ for $\mu\in\{x,y,z\}$.

Remark (the adjoint is the operator of a linear form). The adjoint $\tilde Q_3\mapsto\sum_\mu Q_{5\mu} Q_{3t}e_\mu$ scales the whole element $\tilde Q_5$ by the scalar part of the argument, so its value lies in the line $\mathbb{C}\tilde Q_5$ and carries a vector part whenever $\tilde Q_5$ is not central; the multiplication, by contrast, has a central value for every argument. The two also read different data: the adjoint depends on the argument through $Q_{3t}$ alone, while the multiplication reads it through the full pairing $\langle\tilde Q_5,\cdot\rangle_{\natural}$, so the multiplication is self-adjoint exactly for a central multiplier, the case $L^{\star}_{Q_{5t}e_t}\tilde Q_2=Q_{5t}Q_{2t}e_t$, in which it too reads only the scalar part.

The Two Groups of Linear Maps

Definition. The group of linear maps preserving the form is

$$ G_N = \{T\in GL(\mathbb{B}): \langle T\tilde Q_2,T\tilde Q_3\rangle_{\natural}=\langle\tilde Q_2,\tilde Q_3\rangle_{\natural} \text{ for all } \tilde Q_2,\tilde Q_3\}, $$

and the group of transformations preserving the product is

$$ G_{\star} = \{T\in GL(\mathbb{B}): T(\tilde Q_2\star\tilde Q_3)=(T\tilde Q_2)\star(T\tilde Q_3) \text{ for all } \tilde Q_2,\tilde Q_3\}. $$

Proposition (the two matrix conditions). In the basis, with $G=I_4$ the Gram matrix of $N$,

$$ T\in G_N \quad\Longleftrightarrow\quad T^{\mathsf T}T=I_4, \qquad T\in G_{\star} \quad\Longleftrightarrow\quad T^{\mathsf T}T=I_4 \ \text{and}\ T e_t=e_t . $$

Proof. The first equivalence is the invariant form written on the coordinates: $\langle T\tilde Q_2,T\tilde Q_3\rangle_{\natural}=(T\tilde Q_2)^{\mathsf T}(T\tilde Q_3)=\tilde Q_2^{\mathsf T}T^{\mathsf T}T\tilde Q_3$, and this is $\langle\tilde Q_2,\tilde Q_3\rangle_{\natural}$ for all $\tilde Q_2,\tilde Q_3$ exactly when $T^{\mathsf T}T=I_4$. For the second, $T(\tilde Q_2\star\tilde Q_3)=\langle\tilde Q_2,\tilde Q_3\rangle_{\natural}Te_t$ must equal $(T\tilde Q_2)\star(T\tilde Q_3)=\langle T\tilde Q_2,T\tilde Q_3\rangle_{\natural}e_t$; taking $\tilde Q_2=\tilde Q_3=e_t$ gives $Te_t=\langle Te_t,Te_t\rangle_{\natural}e_t$, so $Te_t=\lambda e_t$ with $\lambda=\lambda^2$ and, $T$ being invertible, $\lambda=1$; then $\langle T\tilde Q_2,T\tilde Q_3\rangle_{\natural}=\langle\tilde Q_2,\tilde Q_3\rangle_{\natural}$, which is $T^{\mathsf T}T=I_4$.

Corollary (dimensions and Lie algebras). Over $\mathbb{C}$, the group $G_N$ has complex dimension $6$ and its Lie algebra is $\{X:X^{\mathsf T}+X=0\}$, the space of antisymmetric matrices, of complex dimension $6$. The group $G_{\star}$ has complex dimension $3$ and its Lie algebra is $\{X:X^{\mathsf T}+X=0,\ Xe_t=0\}$, the antisymmetric matrices killing $e_t$, of complex dimension $3$. The group $G_{\star}$ contains the maps $\operatorname{diag}(1,u_x,u_y,u_z)$ with $u_k=\pm1$, whose action negates the chosen vector directions and fixes the unit.

Proof. A complex-linear $T$ preserving $\langle\cdot\rangle_{\natural}$ is a matrix with $T^{\mathsf T}T=I_4$; the Lie algebra is the first-order form of this condition, $X^{\mathsf T}+X=0$, of dimension $4\cdot3/2=6$. Adding $Te_t=e_t$ to the conditions of $G_{\star}$ kills the first column of $X$, which removes $3$ parameters and gives complex dimension $3$. The diagonal sign maps satisfy $T^{\mathsf T}T=I_4$ and fix $e_t$.

Remark (the structure group). The structure group of the block is the group of invertible linear maps that transport the product to a nonzero scalar multiple of itself,

$$ G^{\mathrm{str}}_{\star} = \{T\in GL(\mathbb{B}): T(\tilde Q_2\star\tilde Q_3)=\lambda\,(T\tilde Q_2)\star(T\tilde Q_3) \text{ for some } \lambda\in\mathbb{C}^{\times}\}. $$

The computation of the proposition gives the conditions $Te_t\in\mathbb{C}e_t$ and $\langle T\tilde Q_2,T\tilde Q_3\rangle_{\natural}=\mu \langle\tilde Q_2,\tilde Q_3\rangle_{\natural}$ for a nonzero scalar $\mu$, that is $T^{\mathsf T}T=\mu I_4$ together with the invariance of the central line. The form-preserving group is the case $\mu=1$ of the rescaling condition, with no requirement on $Te_t$; the group preserving the product is the intersection with $Te_t=e_t$, and there the value $\mu=1$ is forced back, since $\langle Te_t,Te_t\rangle_{\natural}=\mu \langle e_t,e_t\rangle_{\natural}$ reads $1=\mu$. The three groups are related by

$$ G_{\star} \subset G^{\mathrm{str}}_{\star}, \qquad G_{\star} = G^{\mathrm{str}}_{\star}\cap\{Te_t=e_t\}. $$

The structure group contains maps that rescale the form, and the product-preserving group does not: rescaling the product is not the same as preserving it, exactly as for a quadratic form.

The Derivations of the Block

Definition. A derivation of the block is a complex-linear map $D$ with

$$ D(\tilde Q_2\star\tilde Q_3) = (D\tilde Q_2)\star\tilde Q_3 + \tilde Q_2\star(D\tilde Q_3) \qquad\text{for all } \tilde Q_2,\tilde Q_3\in\mathbb{B}. $$

Proposition (the weight, and its vanishing). A derivation is the same thing as a complex-linear map $D$ satisfying $D e_t=\delta e_t$ and

$$ \langle D\tilde Q_2,\tilde Q_3\rangle_{\natural} + \langle\tilde Q_2,D\tilde Q_3\rangle_{\natural} = \delta\,\langle\tilde Q_2,\tilde Q_3\rangle_{\natural} \qquad\text{for all } \tilde Q_2,\tilde Q_3 , $$

for one complex number $\delta$. The weight is forced to be zero: at $\tilde Q_2=\tilde Q_3=e_t$ the identity reads $De_t=2(De_t)_te_t$, which with $De_t=\delta e_t$ is $\delta=2\delta$. Every derivation therefore has $De_t=0$ and $D^{\mathsf T}+D=0$, and these maps, the Lie algebra of $G_N$ cut by $De_t=0$, form the whole derivation algebra, of complex dimension $3$.

Proof. Writing the definition on the value, $D(\tilde Q_2\star\tilde Q_3)=\langle\tilde Q_2,\tilde Q_3\rangle_{\natural}De_t$ and $(D\tilde Q_2)\star\tilde Q_3+\tilde Q_2\star(D\tilde Q_3)=\bigl(\langle D\tilde Q_2,\tilde Q_3\rangle_{\natural}+\langle\tilde Q_2,D\tilde Q_3\rangle_{\natural}\bigr)e_t$. The vector part of the equality forces $De_t$ to be central, $De_t=\delta e_t$, and the central part is the displayed identity. At $\tilde Q_2=\tilde Q_3=e_t$ that identity gives $De_t=\langle De_t,e_t\rangle_{\natural}e_t+\langle e_t,De_t\rangle_{\natural}e_t=2(De_t)_te_t$, and $(De_t)_t=\delta$, so $\delta=0$. With $\delta=0$ the identity is $\langle D\tilde Q_2,\tilde Q_3\rangle_{\natural}+\langle\tilde Q_2,D\tilde Q_3\rangle_{\natural}=0$, that is $D^{\mathsf T}+D=0$ in the basis, together with $De_t=0$; antisymmetric matrices killing $e_t$ form the complex dimension $3$ space displayed.

Remark (the inner derivations vanish). The inner derivations of the operation are the maps $\tilde Q_2\mapsto \tilde Q_5\star\tilde Q_2-\tilde Q_2\star\tilde Q_5$, and they vanish identically because the operation is commutative; the derivation algebra of the block is therefore entirely made of outer derivations. The bracket of two multiplication operators, $[L^{\star}_{\tilde Q_5},L^{\star}_{\tilde Q_1}]=Q_{5t}L^{\star}_{\tilde Q_1}-Q_{1t}L^{\star}_{\tilde Q_5}$, is not a derivation: its value on $\tilde Q_2\star\tilde Q_3$ is zero, while the sum $(D\tilde Q_2)\star\tilde Q_3+\tilde Q_2\star(D\tilde Q_3)$ is $\bigl(Q_{5t}\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}-Q_{1t}\langle\tilde Q_5,\tilde Q_2\rangle_{\natural}\bigr)Q_{3t}e_t+\bigl(Q_{5t}\langle\tilde Q_1,\tilde Q_3\rangle_{\natural}-Q_{1t}\langle\tilde Q_5,\tilde Q_3\rangle_{\natural}\bigr)Q_{2t}e_t$, which does not vanish in general; the operator layer is not a Lie algebra of derivations here, in contrast with the associative case.

Summary

The operator layer of the symmetric quaternionic algebra is the single commutative family $L^{\star}_{\tilde Q_5}\tilde Q_4=\langle\tilde Q_5,\tilde Q_4\rangle_{\natural}e_t$, with matrix in the basis equal to the row of the coordinates of $\tilde Q_5$ in the first row and zero elsewhere. Every nonzero multiplication has rank $1$, image the central line $\mathbb{C}e_t$ and kernel the hyperplane $\langle\tilde Q_5,\cdot\rangle_{\natural}=0$ of complex dimension $3$; its trace is $Q_{5t}$ and its determinant is $0$. The multiplications compose by $L^{\star}_{\tilde Q_5}\circ L^{\star}_{\tilde Q_1}=Q_{5t}L^{\star}_{\tilde Q_1}$: the family is a faithful $4$-dimensional vector space of operators, its operators are the matrices with a single nonzero row, and it is neither commutative nor multiplicative. The adjoint of $L^{\star}_{\tilde Q_5}$ with respect to $\langle\cdot\rangle_{\natural}$ is the map $\tilde Q_3\mapsto\sum_\mu Q_{5\mu} Q_{3t}e_\mu$, and the multiplication is self-adjoint exactly when the multiplier is central. The group preserving the form is $\{T:T^{\mathsf T}T=I_4\}$, of complex dimension $6$, and the group preserving the product is its subgroup with $Te_t=e_t$, of complex dimension $3$, containing the diagonal sign maps; the structure group rescales the form and the product. A derivation is a map with $De_t=\delta e_t$ and $\langle D\tilde Q_2,\tilde Q_3\rangle_{\natural}+\langle\tilde Q_2,D\tilde Q_3\rangle_{\natural}=\delta \langle\tilde Q_2,\tilde Q_3\rangle_{\natural}$, where the weight $\delta$ is forced to be $0$; the derivation algebra is the $3$-dimensional space $\{De_t=0,\ D^{\mathsf T}+D=0\}$, entirely outer.

Summary of Notation

Further Reading

• The Radical and the Isotropic Elements of the Symmetric Quaternionic Algebra, for the multiplications and the radical
• The Quaternion Form as a Product on the Symmetric Quaternionic Algebra, for the form $\langle\cdot\rangle_{\natural}$, its Gram matrix and its non-degeneracy
• The Left Multiplications of the Quaternionic Product and the Opposite Monoid and One-Sided Operators on the General Quaternionic Algebra, for the left multiplications of the parent product
• Automorphisms and Derivations of Algebras, for the general theory of a derivation and its inner part
• Introduction to the Symmetric Quaternionic Algebra, for the same operators in the realization
• Introduction to the Symmetric Quaternionic Algebra, for the same operators in the regular model
• Remarkable Subspaces under the Symmetric Quaternionic Algebra, for the product on the remarkable subspaces