The Radical and the Isotropic Elements of the Symmetric Quaternionic Algebra
Introduction
The symmetric quaternionic multiplication $\tilde Q_1\star\tilde Q_2=\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}e_t$ (Introduction to the Symmetric Quaternionic Algebra) collapses every product onto the central line, so the element theory of the operation is read entirely from the coefficient $\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}=Q_{1t}Q_{2t}+(\vec Q_1,\vec Q_2)$ and its diagonal. This article reads that coefficient as a pairing and develops the two objects the article title names: the radical, which is the set of elements annihilated by the whole algebra on both sides, and the isotropic elements, the elements whose square vanishes. It then records the derived objects the collapse produces — the derived law and the ternary law — and identifies the isotropic elements with the zero divisors of the parent plain algebra.
The Product as a Pairing with Values in the Centre
Because the operation is $\mathbb{C}$-bilinear, it is the same thing as a family of linear maps, and because its value is central it is the same thing as a bilinear form.
Definition. For $\tilde Q_4\in\mathbb{B}$ the left multiplication by $\tilde Q_4$ is the $\mathbb{C}$-linear map
$$ L^{\star}_{\tilde Q_4}: \mathbb{B}\longrightarrow\mathbb{C}e_t,\qquad L^{\star}_{\tilde Q_4}\tilde Q_1 = \tilde Q_4\star\tilde Q_1 = \langle\tilde Q_4,\tilde Q_1\rangle_{\natural}\,e_t . $$
Because the product is commutative the right multiplication coincides with it, $R^{\star}_{\tilde Q_4}=L^{\star}_{\tilde Q_4}$; the operator is examined in The Multiplication Operators of the Symmetric Quaternionic Algebra.
Proposition (the operation is a form read on the central line). The operation is determined by $\langle\cdot\rangle_{\natural}$ and $\langle\cdot\rangle_{\natural}$ is recovered from it: for every pair,
$$ \boxed{\; \langle\tilde Q_1,\tilde Q_2\rangle_{\natural} = \operatorname{Sc}\bigl(\tilde Q_1\star\tilde Q_2\bigr) \;}, $$
the coefficient of $e_t$ in the value. In particular $\star$ is the quaternion form placed on the central line, and the only datum of the operation is the symmetric $\mathbb{C}$-bilinear form $\langle\cdot\rangle_{\natural}$.
Proof. The value is $\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}e_t$ by the theorem of Introduction to the Symmetric Quaternionic Algebra of Biquaternions, and its scalar part is $\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}$. Conversely a bilinear operation given by a form is the form.
Remark (an algebraic form). The coefficient $\langle\cdot\rangle_{\natural}$ is a $\mathbb{C}$-bilinear form, symmetric, with values in $\mathbb{C}$; its values are complex and not real in general, as at $\tilde Q_1=e_t$, $\tilde Q_2=ie_t$, where $N=i$. The diagonal $\langle\cdot\rangle_{\natural}$ is the algebraic norm of Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, and $\langle\cdot\rangle_{\natural}$ is multiplicative on the algebra. The reader must keep the two readings apart: the diagonal $\langle\tilde Q_1\rangle_{\natural}$ is multiplicative and central-valued, while the form $\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}$ is only bilinear.
The Radical
Definition. The radical or two-sided annihilator of $\star$ is the set of elements annihilated by every element on both sides,
$$ \operatorname{Rad}(\star) = \{\tilde Q_4\in\mathbb{B}: \tilde Q_4\star\tilde Q_1=0 \text{ and } \tilde Q_1\star\tilde Q_4=0 \text{ for every } \tilde Q_1\in\mathbb{B}\} = \{\tilde Q_4\in\mathbb{B}: L^{\star}_{\tilde Q_4}=0\}. $$
Because the operation is commutative the two-sided and the one-sided definitions agree, and the radical is the common kernel of the left multiplications, $\bigcap_{\tilde Q_1}\ker L^{\star}_{\tilde Q_1}=\{\tilde Q_4: \langle\tilde Q_4,\tilde Q_1\rangle_{\natural}=0 \text{ for all } \tilde Q_1\}$.
Theorem (the radical is trivial). $\operatorname{Rad}(\star)=\{0\}$: the only element annihilated by the whole algebra is $0$.
Proof. $\langle\cdot\rangle_{\natural}$ is the standard symmetric form of $\mathbb{C}^4$ in the basis $e_t,\vec e_x,\vec e_y,\vec e_z$, whose Gram matrix is $I_4$; a form with invertible Gram matrix has no nonzero element paired to zero with the whole space. Concretely $\tilde Q_4\star e_t=Q_{4t}e_t$ and $\tilde Q_4\star \vec e_\mu=Q_{4\mu} e_t$ for $\mu\in\{x,y,z\}$, so $\tilde Q_4\star\tilde Q_1=0$ for all $\tilde Q_1$ forces $Q_{4t}=Q_{4x}=Q_{4y}=Q_{4z}=0$.
Corollary (the operation is not the zero operation, and it is non-degenerate as a form). The operation is nonzero — $\langle e_t,e_t\rangle_{\natural}=1$ — and the form $\langle\cdot\rangle_{\natural}$ is non-degenerate, of rank $4$ over $\mathbb{C}$ and of Gram matrix $I_4$ in the basis. The image of each nonzero left multiplication is the whole central line $\mathbb{C}e_t$: for $\tilde Q_4\ne0$ some coordinate $Q_{4\mu}\ne0$, so $L^{\star}_{\tilde Q_4}e_\mu=Q_{4\mu} e_t\ne0$ and no nonzero element acts by zero.
Remark (a trivial radical and a poor element theory are not the same fact). A non-degenerate pairing is what one expects of a form, and it does not prevent the operation from being element-theoretically poor: the radical is trivial, yet the operation has no unit, its idempotents reduce to $0$ and $e_t$, and its square-zero set is a cone of complex dimension $3$. The triviality of the radical is a statement about the pairing; the poverty is a statement about the product. The two must not be confused.
The Isotropic Elements
Definition. An element $\tilde Q$ is isotropic when its square under $\star$ vanishes, $\tilde Q\star\tilde Q=0$; equivalently when $\langle\tilde Q\rangle_{\natural}=0$. The isotropic cone of the operation is the set of isotropic elements.
Theorem (the isotropic elements are the zero divisors of the parent). An element is isotropic for $\star$ exactly when $\langle\tilde Q\rangle_{\natural}=\sum_\mu Q_\mu^2=0$, and the nonzero isotropic elements are exactly the zero divisors of the parent plain product, $\{N=0\}\setminus\{0\}=\mathcal{Z}_{\tilde Q_1\tilde Q_2}$; the zero-divisor set of the block itself is strictly larger, being the whole punctured space (Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space). The cone is the affine cone over the quadric $\{[Q_t:Q_x:Q_y:Q_z]:\sum_\mu Q_\mu^2=0\}$ of $\mathbb{P}^3(\mathbb{C})$, of complex dimension $2$ in the projective space and $3$ in the algebra, and it is the isotropic cone of the form $\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}$ and of the norm $\langle\tilde Q_1\rangle_{\natural}$ alike (Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors, Norm and Invertibility in the Symmetric Quaternionic Algebra). It splits into the pure family, the elements with $Q_t=0$, a cone of complex dimension $2$, and the non-pure family, the elements with $Q_t\ne0$, of complex dimension $3$; the two are disjoint and their union is the cone (Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors).
Proof. $\tilde Q\star\tilde Q=\langle\tilde Q\rangle_{\natural}e_t$ by the square formula of Introduction to the Symmetric Quaternionic Algebra of Biquaternions, and this vanishes exactly when $\langle\tilde Q\rangle_{\natural}=0$. The identification of the set $N=0$ with the zero divisors and its description as a quadric cone are Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors.
Example (the standard isotropic elements). For $\tilde Q=e_t+i\vec e_x$ the norm is $N=1+i^2=0$, so the element is isotropic; it lies on the Hermitian subspace $\mathbb{M}_+$ and is a non-pure zero divisor. For $\tilde Q=\vec e_x+i\vec e_y$ the norm is $N=1+i^2=0$ as well, and the element is a pure zero divisor, lying in the vector subspace. For $\tilde Q=\vec e_x+\vec e_y+\vec e_z$ the norm is $N=1+1+1=3\ne0$, and the element is not isotropic.
Remark (isotropy is not the vanishing of the value). An element with $\langle\tilde Q_1\rangle_{\natural}=0$ still multiplies nonzero elements to nonzero values: the product $\tilde Q_1\star\tilde Q_2=\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}e_t$ can be nonzero even when $\langle\tilde Q_1,\tilde Q_1\rangle_{\natural}=0$, because the form is indefinite and not positive. The isotropy of $\langle\cdot\rangle_{\natural}$ is the vanishing of the form on one direction, and it is the diagonal that reads it; the radical, by contrast, is the vanishing on all directions, and it is trivial. Isotropy is a property of each element, and the radical is a property of the space.
Remark (the idempotents and the square-zero elements). The idempotents of the operation are $0$ and $e_t$ alone, and the square-zero elements are exactly the isotropic ones, that is the punctured cone $\{N=0\}\setminus\{0\}$. Both are Common Theory slots of the block — Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space and Square Roots of Zero and the Nilpotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space — and they are cited here rather than restated; the point kept here is only that the idempotents and the nilpotents meet only at the origin, that the idempotents meet the square-zero elements nowhere, and that the square-zero elements are not the square-zero operators of The Specificities of the Symmetric Quaternionic Algebra, which are the vector subspace $Q_t=0$.
The Derived Objects and the Collapse of the Laws
Proposition (the derived law collapses). The commutator of the operation vanishes identically,
$$ \boxed{\; [\tilde Q_1,\tilde Q_2]_{\star} = \tilde Q_1\star\tilde Q_2-\tilde Q_2\star\tilde Q_1 = 0 \;}, $$
so the operation has no derived Lie structure: the derived operation of $\mathrm{SQA}$ is the zero operation on the whole space. The derived object of the operation is therefore the one-dimensional algebra $\mathbb{C}e_t=\mathbb{C}_{\mathbb{B}}$, with the multiplication of $\mathbb{C}$.
Proof. The operation is commutative. The image is the central line, on which the operation is the multiplication of $\mathbb{C}$ by the proposition of Introduction to the Symmetric Quaternionic Algebra of Biquaternions.
Proposition (the associative law does not collapse). The operation is not associative: at $\tilde Q_1=\tilde Q_2=\vec e_x$, $\tilde Q_3=e_t$ the two bracketings are
$$ (\tilde Q_1\star\tilde Q_2)\star\tilde Q_3 = e_t, \qquad \tilde Q_1\star(\tilde Q_2\star\tilde Q_3) = 0 , $$
so the associator is the nonzero element $e_t$. The general associator is
$$ (\tilde Q_1\star\tilde Q_2)\star\tilde Q_3-\tilde Q_1\star(\tilde Q_2\star\tilde Q_3) = \Bigl(\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}Q_{3t} - Q_{1t}\langle\tilde Q_2,\tilde Q_3\rangle_{\natural}\Bigr)e_t . $$
Proof. $(\tilde Q_1\star\tilde Q_2)\star\tilde Q_3=\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}e_t\star\tilde Q_3=\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}Q_{3t}e_t$, and $\tilde Q_1\star(\tilde Q_2\star\tilde Q_3)=\langle\tilde Q_2,\tilde Q_3\rangle_{\natural}\,L^{\star}_{\tilde Q_1}e_t=\langle\tilde Q_2,\tilde Q_3\rangle_{\natural}Q_te_t$. At the witness the first is $\langle \vec e_x,\vec e_x\rangle_{\natural}\cdot 1\cdot e_t=e_t$ and the second is $\langle \vec e_x,e_t\rangle_{\natural}\cdot 0=0$.
Remark (the ternary law). The ternary object the operation carries is the one obtained by composing the quaternionic bracket with the product,
$$ \boxed{\; T(\tilde Q_1,\tilde Q_2,\tilde Q_3) = [\tilde Q_1,\tilde Q_3]_{\natural}\star\tilde Q_2 = \langle[\tilde Q_1,\tilde Q_3]_{\natural},\tilde Q_2\rangle_{\natural}e_t \;}, $$
which is linear in each of its three arguments and alternating in the first and the third, the bracket $[\tilde Q_1,\tilde Q_3]_{\natural}=2\bigl(Q_{1t}\vec Q_3-Q_{3t}\vec Q_1-\vec Q_1\times\vec Q_3\bigr)$ being alternating. It does not vanish in general: at $\tilde Q_1=\vec e_x$, $\tilde Q_2=\vec e_z$, $\tilde Q_3=\vec e_y$ the bracket is $[\tilde Q_1,\tilde Q_3]_{\natural}=-2\vec e_z$ and $\langle -2\vec e_z,\vec e_z\rangle_{\natural}=-2$, so $T=-2e_t$, and the sum $[\tilde Q_1,\tilde Q_3]_{\natural}\star\tilde Q_2+\tilde Q_2\star[\tilde Q_1,\tilde Q_3]_{\natural}=2\langle[\tilde Q_1,\tilde Q_3]_{\natural},\tilde Q_2\rangle_{\natural}e_t$ is twice this value and is not $0$. The operation therefore does not kill the bracket by symmetrisation; what is zero is the commutator of $\star$ itself, not its composition with the bracket.
The Relation to the Zero Divisors
The identification of the isotropic elements with the zero divisors makes the element theory of the operation a small chapter of the zero-divisor geometry of $\mathbb{B}$, which is algebraic: the cone $N=0$ is a quadric cone, cut by the remarkable subspaces into the pieces recorded in Remarkable Subspaces under the Symmetric Quaternionic Algebra of Biquaternions. On the two definite rows $\mathbb{H}_{\mathbb{B}}$ and $i\mathbb{H}_{\mathbb{B}}$ the norm is strictly signed and the cone meets the row only at the origin; on the vector subspace the cone is the pure zero-divisor cone; on the two Hermitian subspaces the cone is their real null cone.
Proposition (the operation kills nothing on the definite rows). No nonzero element of $\mathbb{H}_{\mathbb{B}}$ or of $i\mathbb{H}_{\mathbb{B}}$ is isotropic for $\star$.
Proof. On $\mathbb{H}_{\mathbb{B}}$ the form is positive definite, $N=\sum_\mu q_\mu^2\ge0$, vanishing only at the origin; on $i\mathbb{H}_{\mathbb{B}}$ it is negative definite, $N=-\sum_\mu(q'_\mu)^2\le0$, vanishing only at the origin (Remarkable Subspaces under the General Quaternionic Algebra).
Remark (the radical again, from the subspaces). The triviality of the radical is compatible with a large isotropic cone, and the two are read on the Gram data: the form has Gram matrix $I_4$, which is invertible and carries no radical; the cone is the set where the diagonal vanishes, which is the isotropic set of an indefinite form on a complex space. A well-defined form with an isotropic cone of positive dimension is the generic situation.
Summary
The symmetric quaternionic multiplication is a pairing: $\tilde Q_1\star\tilde Q_2=\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}e_t$, the quaternion form placed on the central line, and the form is recovered from the product as the coefficient of $e_t$. Its radical — the two-sided annihilator — is trivial, the Gram matrix of $N$ being the unit matrix; the operation is therefore not the zero operation and its form is non-degenerate. Its isotropic elements are the zero divisors, $\langle\tilde Q\rangle_{\natural}=\sum_\mu Q_\mu^2=0$, forming the affine cone over the quadric $\sum_\mu Q_\mu^2=0$ of $\mathbb{P}^3(\mathbb{C})$, of complex dimension $3$ in the algebra; the standard witnesses are $e_t+i\vec e_x$ on the Hermitian subspace and $\vec e_x+i\vec e_y$ on the vector subspace. The square-zero elements are exactly the isotropic ones; the idempotents are $0$ and $e_t$ alone; the derived law collapses, the commutator of $\star$ being identically zero; the associative law does not collapse, the associator being $\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}Q_{3t}-Q_{1t}\langle\tilde Q_2,\tilde Q_3\rangle_{\natural}$ times $e_t$, nonzero at $\vec e_x,\vec e_x,e_t$; and the ternary object $\langle[\tilde Q_1,\tilde Q_3]_{\natural},\tilde Q_2\rangle_{\natural}e_t$ does not vanish. The two definite rows of the remarkable subspaces contain no nonzero isotropic element.
Summary of Notation
| Symbol | Meaning |
|---|---|
| $\tilde Q_1\star\tilde Q_2=\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}e_t$ | the symmetric quaternionic multiplication |
| $\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}=Q_{1t}Q_{2t}+(\vec Q_1,\vec Q_2)$ | the quaternion form, the coefficient of the product |
| $\langle\tilde Q\rangle_{\natural}=\langle\tilde Q,\tilde Q\rangle_{\natural}=\sum_\mu Q_\mu^2$ | the norm form, the diagonal of $\langle\cdot\rangle_{\natural}$ |
| $L^{\star}_{\tilde Q_4}$ | the left (and right) multiplication, $L^{\star}_{\tilde Q_4}\tilde Q_1=\langle\tilde Q_4,\tilde Q_1\rangle_{\natural}e_t$ |
| $\operatorname{Rad}(\star)$ | the two-sided annihilator, $\{0\}$ |
| $[\tilde Q_1,\tilde Q_2]_{\natural}$ | the quaternionic bracket $\tilde Q_1^{\natural}\tilde Q_2-\tilde Q_2^{\natural}\tilde Q_1$ |
| $[\tilde Q_1,\tilde Q_2]_{\star}$ | the commutator of the operation, identically zero |
| $T(\tilde Q_1,\tilde Q_2,\tilde Q_3)=\langle[\tilde Q_1,\tilde Q_3]_{\natural},\tilde Q_2\rangle_{\natural}e_t$ | the ternary object of the operation |
| $\mathbb{C}_{\mathbb{B}},\mathrm{Vect}(\mathbb{B}),\mathbb{H}_{\mathbb{B}},i\mathbb{H}_{\mathbb{B}},\mathbb{M}_+,\mathbb{M}_-$ | the remarkable subspaces |
Further Reading
- Introduction to the Symmetric Quaternionic Algebra, for the operation and its square.
- Norm and Invertibility in the Symmetric Quaternionic Algebra and Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the diagonal, the non-degenerate form and the maximal zero-divisor set of the block.
- Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space and Square Roots of Zero and the Nilpotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the two idempotents and for the square-zero elements.
- The Symmetrised Quaternionic Product and the Hermitian Subspace, for the centrality and the coefficient in full.
- Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra and Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, for the norm, its polarisation and the isotropic cone.
- Square Roots of One and the Idempotents of the Biquaternion Algebra, for the idempotents of the algebra and their contrast with the idempotents of $\mathrm{SQA}$.
- Remarkable Subspaces under the Symmetric Quaternionic Algebra, for the isotropic elements on each subspace.
- The Multiplication Operators of the Symmetric Quaternionic Algebra, for the operators $L^{\star}_{\tilde Q_4}$ in full.