Annihilators of the General Quaternionic Algebra
Introduction
The zero divisor of a product is a property of one element alone: it says that some nonzero partner exists that the element kills. The annihilator says which partners those are, and it is the finer object. It is the set of elements killed by a fixed element, $$ A_\star(\tilde Q_1)=\{\tilde Q_2:\tilde Q_1\star\tilde Q_2=0\},\qquad B_\star(\tilde Q_1)=\{\tilde Q_2:\tilde Q_2\star\tilde Q_1=0\}, $$ the left and the right annihilator (Definitions for the Study of the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, §The Zero Divisor and the Annihilator). When the product is carried by a multiplication operator the annihilator is the kernel of that operator, so it is a linear subspace.
The block of this article is the general quaternionic multiplication $$ \tilde Q_1\star\tilde Q_2=\tilde Q_1^{\natural}\tilde Q_2 =\bigl[Q_{1t}Q_{2t}+(\vec Q_1,\vec Q_2)\bigr]+Q_{1t}\vec Q_2-Q_{2t}\vec Q_1-\vec Q_1\times\vec Q_2, $$ the product of the natural conjugate of the first factor with the second (Introduction to the General Quaternionic Algebra of Biquaternions). It is neither commutative nor alternating, and yet a two-sided involution ties the two sides: the natural conjugation is a $\mathbb{C}$-linear anti-automorphism of order two, so it carries the right annihilator $\{\tilde Q_2:\tilde Q_2^{\natural}\tilde Q_1=0\}$ onto the left annihilator $\{\tilde Q_1^{\natural}\tilde Q_2=0\}$, and the two annihilators of an element coincide. The annihilator vanishes off the zero-divisor cone $C(\tilde Q_1)=0$ and is a complex plane on it. The article owns that reading, the operator with its rank, and the shape of the annihilator on the remarkable subspaces; the zero-divisor set and its comparison with the other three general products are the cited article's.
The results. The left and the right annihilator of an element coincide for every element of $\mathbb{B}$, unit or zero divisor alike, because the natural conjugation is a $\mathbb{C}$-linear anti-automorphism of order two; the quaternionic product is the one general product whose annihilators are not exchanged but equal. The multiplication operator $L_{\tilde Q_1}$ has trace $4Q_{1t}$ and determinant $C(\tilde Q_1)^{2}$ with $C(\tilde Q_1)=\sum_\mu Q_{1\mu}^{2}$: it is invertible off the zero-divisor cone, while on the cone it has rank two and a kernel of complex dimension two. The common annihilator is therefore $\{0\}$ off the cone and a complex plane on it, and the plane contains the element itself. Each section below states its result first and then proves it.
The Annihilator of a Product
The annihilator of an element is the set of partners that it kills; the left annihilator is the kernel of the left multiplication and the right annihilator the kernel of the right multiplication, and both are nonzero exactly when the element is a zero divisor of the corresponding side.
Definition (annihilator). Let $\star$ be a product on $\mathbb{B}$ and let $\tilde Q_1\in\mathbb{B}$. The left annihilator and the right annihilator of $\tilde Q_1$ in $\star$ are $$ A_\star(\tilde Q_1)=\{\tilde Q_2:\tilde Q_1\star\tilde Q_2=0\},\qquad B_\star(\tilde Q_1)=\{\tilde Q_2:\tilde Q_2\star\tilde Q_1=0\}. $$ They are the kernels of the left and the right multiplication, and they are nonzero exactly when $\tilde Q_1$ is a zero divisor of the corresponding side.
Throughout this article the product is the general quaternionic multiplication $\star$, and unqualified, annihilator means the annihilator of $\star$. The left and the right multiplication are the operator $$ L_{\tilde Q_1}\tilde Q_2=\tilde Q_1\star\tilde Q_2=\tilde Q_1^{\natural}\tilde Q_2,\qquad R_{\tilde Q_1}\tilde Q_2=\tilde Q_2\star\tilde Q_1=\tilde Q_2^{\natural}\tilde Q_1, $$ a multiplication by the natural conjugate. The block is the sibling of the general plain algebra, and the two share the cone and its plane but differ in the coincidence of the two sides.
The Two Annihilators Coincide
For every $\tilde Q_1\in\mathbb{B}$, unit or zero divisor alike, the two annihilators coincide, $B_\star(\tilde Q_1)=A_\star(\tilde Q_1)$, because the natural conjugation is a $\mathbb{C}$-linear anti-automorphism of order two.
Theorem (the coincidence). For every $\tilde Q_1\in\mathbb{B}$, unit or zero divisor alike, $$ \boxed{\; B_\star(\tilde Q_1)=A_\star(\tilde Q_1) \;}. $$
Proof. Let $\tilde Q_2\in\mathbb{B}$. Applying the natural conjugation ${}^{\natural}$ to the equation $\tilde Q_1^{\natural}\tilde Q_2=0$ gives $(\tilde Q_1^{\natural}\tilde Q_2)^{\natural}=0$, which is $\tilde Q_2^{\natural}\tilde Q_1=0$ because ${}^{\natural}$ is an anti-automorphism of order two; the argument reverses because ${}^{\natural}$ is a bijection of $\mathbb{B}$ onto itself. Hence $\tilde Q_1\star\tilde Q_2=0$ exactly when $\tilde Q_2\star\tilde Q_1=0$, and the two annihilators contain exactly the same elements.
The coincidence is a property of this product: it holds because ${}^{\natural}$ is a $\mathbb{C}$-linear anti-automorphism of order two, so the first factor can be conjugated away without changing the annihilation. The general plain algebra has no such involution and its two annihilators are generally distinct.
The Multiplication Operator and the Rank
The operator $L_{\tilde Q_1}$ has trace $4Q_{1t}$ and determinant $C(\tilde Q_1)^{2}$ with $C(\tilde Q_1)=\sum_\mu Q_{1\mu}^{2}$; it is invertible with trivial kernel when $C(\tilde Q_1)\neq0$, and of rank two with kernel of complex dimension two when $\tilde Q_1\neq0$ and $C(\tilde Q_1)=0$.
Proposition (the operator and its determinant). In the basis $e_t,\vec e_x,\vec e_y,\vec e_z$ the operator $L_{\tilde Q_1}$ of an element $\tilde Q_1=Q_{1t}e_t+\vec Q_1$ has the matrix of multiplication by $\tilde Q_1^{\natural}$, of trace $4Q_{1t}$ and determinant $$ \boxed{\; \det L_{\tilde Q_1}=C(\tilde Q_1)^{2},\qquad C(\tilde Q_1)=\sum_{\mu\in\{t,x,y,z\}}Q_{1\mu}^{2} \;}. $$
Proof. The matrix is the left multiplication matrix of the quaternion $\tilde Q_1^{\natural}$, whose determinant is the square of the quaternion norm $C(\tilde Q_1^{\natural})=C(\tilde Q_1)$; the natural conjugation keeps the four squares, so the determinant is $C(\tilde Q_1)^{2}$. The trace is the sum of the four diagonal entries, $4Q_{1t}$.
Corollary (the rank and the annihilator). The rank of $L_{\tilde Q_1}$ and the annihilator of $\tilde Q_1$ are
| $\tilde Q_1$ | $C(\tilde Q_1)$ | $\operatorname{rank}L_{\tilde Q_1}$ | $\dim\ker L_{\tilde Q_1}$ |
|---|---|---|---|
| $0$ | $0$ | $0$ | $4$ |
| $\tilde Q_1\neq0$, $C(\tilde Q_1)\neq0$ | $\neq0$ | $4$ | $0$ |
| $\tilde Q_1\neq0$, $C(\tilde Q_1)=0$ | $0$ | $2$ | $2$ |
Proof. The determinant $C(\tilde Q_1)^{2}$ is nonzero exactly when $C(\tilde Q_1)\neq0$, and in that case the operator is invertible and the annihilator is $\{0\}$. When $C(\tilde Q_1)=0$ the determinant vanishes; the multiplication by the nonzero $\tilde Q_1^{\natural}$ in the plain algebra is then of rank two, because the plain algebra is the matrix algebra $M_2(\mathbb{C})$ and a nonzero element of the cone is a matrix of rank one (Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors), whence the kernel is of complex dimension two. The origin gives the zero operator.
The Annihilator of an Element
A plain unit has trivial annihilator, while a nonzero zero divisor has a common annihilator of complex dimension two that contains the element itself.
Theorem (the dimension of the annihilator). Let $\tilde Q_1\in\mathbb{B}$.
- If $\tilde Q_1$ is a unit of the plain product, $C(\tilde Q_1)\neq0$, then $A_\star(\tilde Q_1)=B_\star(\tilde Q_1)=\{0\}$.
- If $\tilde Q_1\neq0$ is a zero divisor, $C(\tilde Q_1)=0$, then $A_\star(\tilde Q_1)$ and $B_\star(\tilde Q_1)$ are equal, of complex dimension two, and the common annihilator contains $\tilde Q_1$.
Proof. The first clause is the rank table. For the second, the annihilator is the kernel of the multiplication by $\tilde Q_1^{\natural}$, of rank two on the cone; and $\tilde Q_1^{\natural}\tilde Q_1=C(\tilde Q_1)e_t=0$, so $\tilde Q_1$ lies in the annihilator.
Example. For the pure zero divisor $\tilde Q_1=\vec e_x+i\vec e_y$, whose vector part has $(\vec Q_1,\vec Q_1)=1-1=0$, the common annihilator is $$ \boxed{\; A_\star(\vec e_x+i\vec e_y)=\mathbb{C}\{\vec e_x+i\vec e_y,\ \vec e_z-ie_t\} \;}, $$ and for the non-pure zero divisor $\tilde Q_1=e_t+i\vec e_x$, of algebraic norm $1-1=0$, $$ A_\star(e_t+i\vec e_x)=\mathbb{C}\{e_t+i\vec e_x,\ \vec e_y+i\vec e_z\}. $$ In both cases the element lies in its own annihilator and the annihilator is a complex plane.
The Dichotomy by the Null Cone
The annihilator reads the cone twice: off the cone it is $\{0\}$ and the element is a plain unit, on the cone it is a complex plane and the element is a rank-one matrix.
The zero-divisor set of the block is the punctured cone $\{\tilde Q_1\neq0:C(\tilde Q_1)=0\}$, a single equation, and the annihilator is the finer reading of the same equation: off the cone the annihilator is $\{0\}$ and the element is a unit of the plain product, and on the cone the annihilator is a complex plane and the element is a rank-one matrix. The block is the sibling of the general plain algebra and the twin of the symmetric quaternionic algebra: the first has the same cone and the same plane but its two sides are distinct, and the second has the same coincidence of the two sides but its annihilator is a hyperplane for every nonzero element rather than a plane on a cone. The annihilator is the invariant that separates the three.
The Annihilator on the Remarkable Subspaces
On a nonzero central element the annihilator is $\{0\}$; on the vector subspace it is $\{0\}$ unless the vector part is isotropic, and a complex plane on the isotropic locus; and on the Hermitian and anti-Hermitian subspaces it is again a complex plane for a nonzero zero divisor.
The centre. A nonzero central element $\tilde Q_1=\lambda e_t$ has $L_{\lambda e_t}\tilde Q_2=\lambda\tilde Q_2$, the scalar operator, so it is a unit of the plain product and its annihilator is $\{0\}$.
The vector subspace. A nonzero pure element $\tilde Q_1=\vec Q_1$ has $L_{\vec Q_1}\tilde Q_2=-\vec Q_1\tilde Q_2$ and $C(\tilde Q_1)=(\vec Q_1,\vec Q_1)$; its annihilator is $\{0\}$ unless the vector part is isotropic, and on the isotropic locus it is the complex plane of the elements the pure quaternion $\vec Q_1$ kills by left multiplication.
The Hermitian and anti-Hermitian subspaces. On the Hermitian subspace the natural conjugation is the coefficientwise conjugation of the vector part, and the annihilator of a nonzero Hermitian zero divisor is again a complex plane; the elements of the annihilator are those killed by the Hermitian quaternion, a left ideal of the plain algebra.
Summary
The annihilator of an element in a product is the set of partners the element kills, the left and the right annihilator being the kernels of the left and the right multiplication. The general quaternionic multiplication $\tilde Q_1\star\tilde Q_2=\tilde Q_1^{\natural}\tilde Q_2$ carries a $\mathbb{C}$-linear anti-automorphism of order two, so its two annihilators of an element always coincide. The operator $L_{\tilde Q_1}$ has determinant $C(\tilde Q_1)^{2}$ and trace $4Q_{1t}$: it is invertible off the cone and of rank two and kernel a complex plane on the cone, and the plane contains the element itself. The block shares the cone and the plane with the general plain algebra and the coincidence of the two sides with the symmetric quaternionic algebra, and it is the only one of the three with both properties.
Summary of Notation
| Symbol | Meaning |
|---|---|
| $\tilde Q_1\star\tilde Q_2=\tilde Q_1^{\natural}\tilde Q_2$ | the general quaternionic multiplication, the operation $\mathrm{GQA}$ |
| $A_\star(\tilde Q_1),B_\star(\tilde Q_1)$ | the left and the right annihilator of $\tilde Q_1$ in the block, equal by the anti-automorphism ${}^{\natural}$ |
| $L_{\tilde Q_1}$ | the left multiplication operator, $L_{\tilde Q_1}\tilde Q_2=\tilde Q_1^{\natural}\tilde Q_2$, whose kernel is the annihilator |
| $\det L_{\tilde Q_1}=C(\tilde Q_1)^{2}$, $\operatorname{tr}L_{\tilde Q_1}=4Q_{1t}$ | the determinant and the trace of the operator |
| $\{0\}$ off the cone | the annihilator of an element of nonzero central square, a unit of the plain product |
| the complex plane on the cone | the annihilator of a nonzero zero divisor, containing the element itself |
| $C(\tilde Q_1)=\sum_\mu Q_{1\mu}^{2}$ | the central square, the criterion of the cone |
Further Reading
- Definitions for the Study of the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, §The Zero Divisor and the Annihilator, for the zero divisor and the annihilator of a general product.
- Introduction to the General Quaternionic Algebra, for the operation, its table and its anti-automorphism.
- Zero Divisors of the General Quaternionic Algebra, above this article in this category, for the criterion, the two families and the comparison with the other general products.
- Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, for the plain algebra, the cone and the rank-one matrix reading.
- Annihilators in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, the block whose annihilator is a hyperplane for every nonzero element.