The Conjugate-Commutative Law and the Idempotents of the Symmetric Plain Sesqualgebra
Introduction
The symmetric plain sesqualgebra is the operation $\tilde Q_1\star\tilde Q_2 = \mathrm{Sc}(\tilde Q_1\tilde Q_2^{*})e_t = \langle\tilde Q_1,\tilde Q_2\rangle_{*}e_t$ on the biquaternion space, the Hermitian form $\langle\cdot,\cdot\rangle_{*}$ read as a multiplication (Introduction to the Symmetric Plain Sesqualgebra). This article develops the law that the operation forces, which is not commutativity but conjugate-commutativity, and it reads the law through the two involutions of the sesqualgebra and against the law of the sibling symmetric blocks. The objects the law determines — the idempotents, the annihilator and the square-zero elements — are stated here only as consequences and are proved in their own articles of the block: Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, Zero Divisors 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.
The result that organises the article is that the law and the symmetry of the form are one statement. Because every value of the product is central, the law can only be conjugate-commutativity — a commutativity read through the conjugation of the central value — and it reads on the form as the Hermitian symmetry $\langle\tilde Q_1,\tilde Q_2\rangle_{*} = \overline{\langle\tilde Q_2,\tilde Q_1\rangle_{*}}$. On the centre the product is $(\lambda e_t)\star(\mu e_t) = \lambda\overline{\mu}e_t$, the multiplication of the field by its conjugate, and the idempotent equation there is $\lvert\lambda\rvert^{2} = \lambda$, whose roots are $0$ and $1$; the reduction is Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space.
Two further consequences are drawn, both of them consequences of the definiteness of the form. The two-sided annihilator of the block is zero, because $\langle\cdot,\cdot\rangle_{*}$ is non-degenerate; and there is no nonzero element of square zero, because the form is strictly positive at the two equal arguments off the origin. Both statements are the exact opposites of the quaternionic row, where the form is indefinite and the quadratic map kills nonzero elements. They are proved in Zero Divisors 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. The article closes with the comparison of the law in the sibling symmetric blocks and with the criterion that comparison feeds, in The Specificities of the Symmetric Plain Sesqualgebra.
The Conjugate-Commutative Law
Theorem (the law). For all $\tilde Q_1,\tilde Q_2$,
$$ \boxed{\; \tilde Q_1\star\tilde Q_2 = \overline{\tilde Q_2\star\tilde Q_1} \;}, $$
the bar acting on the central coefficient; the operation is conjugate-commutative and not commutative.
Proof. $\langle\tilde Q_2,\tilde Q_1\rangle_{*} = \sum_\mu Q_{2,\mu}\overline{Q_{1,\mu}} = \overline{\sum_\mu Q_{1,\mu}\overline{Q_{2,\mu}}} = \overline{\langle\tilde Q_1,\tilde Q_2\rangle_{*}}$, and the value is $\overline{\langle\tilde Q_1,\tilde Q_2\rangle_{*}}e_t$, the coefficientwise conjugate of $\tilde Q_1\star\tilde Q_2$.
Proposition (the two involutions act, one as an automorphism and one as the reversal). The natural conjugation preserves the product and the Hermitian conjugation reverses it up to the coefficientwise conjugate:
$$ \boxed{\; \tilde Q_1^{\natural}\star\tilde Q_2^{\natural} = \tilde Q_1\star\tilde Q_2 , \qquad \tilde Q_1^{*}\star\tilde Q_2^{*} = \overline{\tilde Q_1\star\tilde Q_2} = \tilde Q_2\star\tilde Q_1 \;}. $$
Proof. On the coordinates, $(\tilde Q_1^{\natural})_\mu = \varepsilon_\mu Q_{1,\mu}$ and $(\tilde Q_1^{*})_\mu = \varepsilon_\mu\overline{Q_{1,\mu}}$ with $\varepsilon = (1,-1,-1,-1)$. Hence $\langle\tilde Q_1^{\natural},\tilde Q_2^{\natural}\rangle_{*} = \sum_\mu \varepsilon_\mu Q_{1,\mu}\,\overline{\varepsilon_\mu Q_{2,\mu}} = \sum_\mu \varepsilon_\mu^{2}Q_{1,\mu}\overline{Q_{2,\mu}} = \langle\tilde Q_1,\tilde Q_2\rangle_{*}$, since $\varepsilon_\mu = \pm1$ is real; and $\langle\tilde Q_1^{*},\tilde Q_2^{*}\rangle_{*} = \sum_\mu \varepsilon_\mu\overline{Q_{1,\mu}}\,\overline{\varepsilon_\mu\overline{Q_{2,\mu}}} = \sum_\mu \varepsilon_\mu^{2}\overline{Q_{1,\mu}}\,Q_{2,\mu} = \overline{\langle\tilde Q_1,\tilde Q_2\rangle_{*}}$. The second equality of the display is the conjugate-commutative law.
Remark (why the two involutions differ). The natural conjugation is $\mathbb{C}$-linear and acts as an automorphism of the block; the Hermitian conjugation carries the conjugate-linear slot into the linear one, so it reverses the two arguments and conjugates the central value. In the plain bilinear product the roles are the other way round: there the natural conjugation alone is the antiautomorphism, and the product carries no conjugation of the base field.
Remark (why the law is conjugate and not plain). A plain commutativity $\tilde Q_1\star\tilde Q_2 = \tilde Q_2\star\tilde Q_1$ reads $\langle\tilde Q_1,\tilde Q_2\rangle_{*} = \langle\tilde Q_2,\tilde Q_1\rangle_{*}$, a form that is symmetric rather than Hermitian; the block instead satisfies $\langle\tilde Q_1,\tilde Q_2\rangle_{*} = \overline{\langle\tilde Q_2,\tilde Q_1\rangle_{*}}$, which is exactly the Hermitian symmetry of the form. The law of the product and the symmetry of the form are one statement. This is the reason the block is conjugate-commutative while the symmetric plain block, whose form $Q_{1t}Q_{2t} - (\vec Q_1,\vec Q_2)$ is symmetric, is commutative.
The Idempotents
The idempotents of the block, the elements with $\tilde Q\star\tilde Q = \tilde Q$, are exactly $0$ and $e_t$, and the reduction is the one the law forces. The square is central, so an idempotent lies in the centre, where the product is the field rule $(\lambda e_t)\star(\mu e_t) = \lambda\overline{\mu}e_t$; the equation there reads $\lvert\lambda\rvert^{2} = \lambda$, whose roots are $0$ and $1$, and $e_t$ is an idempotent and not a unit, since $\tilde Q\star e_t = Q_te_t$ and $e_t\star\tilde Q = \overline{Q_t}e_t$ reproduce no element with a vector part. The classification, its proof and the comparison with the two central siblings are Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space; the present article records the reduction because the law is read through the field rule of the centre, and cites the rest.
The Annihilator and the Square-Zero Elements
The two-sided annihilator of the block is $\{0\}$, because the form $\langle\cdot,\cdot\rangle_{*}$ is non-degenerate, and the block has no nonzero square-zero element, because the diagonal of $\langle\cdot,\cdot\rangle_{*}$ is strictly positive off the origin; it does have vanishing products of nonzero elements, $e_\mu\star e_\nu = 0$ for $\mu\neq\nu$, the value vanishing exactly on the orthogonal pairs. The two statements are proved in Zero Divisors 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, where the index-three nilpotents are also read; this article cites them and does not restate them, and it records only that both are consequences of the definiteness of $\langle\cdot,\cdot\rangle_{*}$, the property the law shares with the form.
The Comparison with the Sibling Symmetric Blocks
The law. The law together with the unit separates the block from the two bilinear parts $\mathrm{SPA}$ and $\mathrm{SQA}$. The two bilinear blocks are both commutative, and the sesquilinear one is conjugate-commutative; among the two commutative blocks only the plain one has a unit.
| block | product | field of scalars | law | unit | form |
|---|---|---|---|---|---|
| $\mathrm{SPA}$ | $\tilde Q_1\bullet\tilde Q_2$ | $\mathbb{C}$-bilinear | commutative | $e_t$ two-sided | $Q_{1t}Q_{2t} - (\vec Q_1,\vec Q_2)$ |
| $\mathrm{SQA}$ | $\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}e_t$ | $\mathbb{C}$-bilinear | commutative | none | $Q_{1t}Q_{2t} + (\vec Q_1,\vec Q_2)$ |
| $\mathrm{SPS}$ | $\langle\tilde Q_1,\tilde Q_2\rangle_{*}e_t$ | sesquilinear | conjugate-commutative | none | $Q_{1t}\overline{Q_{2t}} + (\vec Q_1,\overline{\vec Q_2})$ |
The symmetric plain block is the one with a unit and the one whose product has a vector part; the two central blocks have neither, because the identity of a central-valued product cannot reproduce a vector part. The laws are commutative, conjugate-commutative, and the idempotents follow: the plain block has the two families $\tfrac12(e_t + \tilde\Xi i)$ attached to the square roots of $-1$ (The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra), the two central blocks have only $0$ and $e_t$ (Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, The Radical and the Isotropic Elements of the Symmetric Quaternionic Algebra).
The two conjugate-commutative blocks. The law does not separate the block from the other sesquilinear symmetric part of the twelve, the symmetric quaternionic sesquilinear block $\mathrm{SQS}$, which is conjugate-commutative as well; the two are separated by the image and by the form. There the value is $$ \boxed{\; \mathrm{SQS}(\tilde Q_1,\tilde Q_2) = \langle\tilde Q_1,\tilde Q_2\rangle_{\natural*}e_t - Q_{1t}\overline{\vec Q_2} - \overline{Q_{2t}}\,\vec Q_1 , \qquad \langle\tilde Q_1,\tilde Q_2\rangle_{\natural*} = Q_{1t}\overline{Q_{2t}} - (\vec Q_1,\overline{\vec Q_2}) \;}, $$ with a non-central value that lies in no subspace of the remarkable subspaces and an indefinite scalar form $\langle\cdot,\cdot\rangle_{\natural*}$, while here the value is the central $\langle\tilde Q_1,\tilde Q_2\rangle_{*}e_t$ and the form $\langle\cdot,\cdot\rangle_{*}$ is definite. The two blocks share everything else at the level of the law: both are sesquilinear over $(\mathbb{C},\bar{\cdot}\,)$, both satisfy $\tilde Q_1\star\tilde Q_2 = \overline{\tilde Q_2\star\tilde Q_1}$, and neither has a unit. The difference is the sign carried by the vector part of the parent pairing, which the plain block cancels and the quaternionic block keeps; the same sign swaps the definite diagonal $\langle\tilde Q_2,\tilde Q_2\rangle_{*} = \sum_\mu\lvert Q_{2,\mu}\rvert^{2}$ for the indefinite $\langle\tilde Q_2,\tilde Q_2\rangle_{\natural*} = \lvert Q_{2t}\rvert^{2} - \lvert\vec Q_2\rvert^{2}$ (Introduction to the Symmetric Quaternionic Sesqualgebra).
The readings through the involutions. The plain block is $\mathbb{C}$-bilinear, and the natural conjugation is an antiautomorphism of the plain product, hence an automorphism of the symmetric block, $(\tilde Q_1\bullet\tilde Q_2)^{\natural} = \tilde Q_1^{\natural}\bullet\tilde Q_2^{\natural}$; there is no conjugation of the base field, and so no second reading. The sesquilinear-central block carries that conjugation, and its law has the two readings, $\tilde Q_1\star\tilde Q_2 = \tilde Q_1^{\natural}\star\tilde Q_2^{\natural}$, an automorphism, and $\tilde Q_1^{*}\star\tilde Q_2^{*} = \overline{\tilde Q_1\star\tilde Q_2}$, the reversal. The pair of readings is the mark that the product carries the conjugation of the base field.
The idempotents and their projectors. The idempotents of the block are the two central ones, and their multiplications are also its projectors; the classification, the failure of the bijection with the roots of $-1$ and the comparison with the sibling blocks are Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, and the idempotent theory of the general plain sesquilinear product, of which the two are the restriction, is Idempotents of the General Plain Sesqualgebra. The element $e_t$ is an idempotent of the block and of the plain product at once, being the nontrivial idempotent of the one and the identity of the other, and it is a unit of the plain product and not of the block, which is the whole difference.
Worked Examples
The centre. On the centre the product is the field rule $(\lambda e_t)\star(\mu e_t) = \lambda\overline{\mu}e_t$, the multiplication of $\mathbb{C}$ by its conjugate. Its idempotents are $0$ and $1$, its involutive automorphisms are the two, and the conjugate-commutative law is the statement that $z\overline{w} = \overline{w\overline{z}}$.
A Hermitian element. For $\tilde Q = e_t + i\vec e_z$ one has $\tilde Q\star\tilde Q = \langle\tilde Q,\tilde Q\rangle_{*}e_t = 2e_t$, so $\tilde Q$ is not an idempotent: the square is central while $\tilde Q$ is not, and idempotence in the block is a condition on the central elements alone.
An element of square zero searched in vain. Take $\tilde Q = e_t + i\vec e_x$, the zero divisor of the plain product. Its diagonal is $\langle\tilde Q,\tilde Q\rangle_{*} = 2$, so $\tilde Q\star\tilde Q = 2e_t \neq 0$: an element that is a zero divisor of the algebra is a unit-like element of the block, in the sense that its square is nonzero. The block and the algebra disagree on the element.
Summary
The symmetric plain sesqualgebra satisfies the conjugate-commutative law $\tilde Q_1\star\tilde Q_2 = \overline{\tilde Q_2\star\tilde Q_1}$, read through the coefficientwise conjugation of the central value; the law and the Hermitian symmetry of the form $\langle\cdot,\cdot\rangle_{*}$ are one statement. The natural conjugation acts as an automorphism of the block, $\tilde Q_1^{\natural}\star\tilde Q_2^{\natural} = \tilde Q_1\star\tilde Q_2$, and the Hermitian conjugation as the reversal, $\tilde Q_1^{*}\star\tilde Q_2^{*} = \overline{\tilde Q_1\star\tilde Q_2} = \tilde Q_2\star\tilde Q_1$. On the centre the law is the field rule $(\lambda e_t)\star(\mu e_t) = \lambda\overline{\mu}e_t$, the multiplication of $\mathbb{C}$ by its conjugate. In the comparison of the sibling blocks the law is commutative, commutative, conjugate-commutative, and the plain block alone has a unit: the plain product carries a vector part, the two central products do not. The other conjugate-commutative block, $\mathrm{SQS}$, is not separated by the law but by the image and the form: its value carries a vector part and its scalar form $\langle\cdot,\cdot\rangle_{\natural*}$ is indefinite, where here the value is central and $\langle\cdot,\cdot\rangle_{*}$ is definite, and the separation of the twelve by the law and the image is the criterion of The Specificities of the Symmetric Plain Sesqualgebra. The idempotents, the annihilator and the square-zero elements that the law forces, and the index-three nilpotents beyond the square, are Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, Zero Divisors 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.
Summary of Notation
| Symbol | Meaning |
|---|---|
| $\tilde Q_1\star\tilde Q_2 = \langle\tilde Q_1,\tilde Q_2\rangle_{*}e_t$ | the product of the block |
| $\langle\tilde Q_1,\tilde Q_2\rangle_{*} = \sum_\mu Q_{1,\mu}\overline{Q_{2,\mu}}$ | the Hermitian form of the block |
| $\tilde Q_1\star\tilde Q_2 = \overline{\tilde Q_2\star\tilde Q_1}$ | the conjugate-commutative law |
| $\tilde Q_1^{\natural}\star\tilde Q_2^{\natural} = \tilde Q_1\star\tilde Q_2$, $\tilde Q_1^{*}\star\tilde Q_2^{*} = \overline{\tilde Q_1\star\tilde Q_2}$ | the two involutions: automorphism and reversal |
| $(\lambda e_t)\star(\mu e_t) = \lambda\overline{\mu}e_t$ | the product on the centre, the field rule the law is read through |
| $\{0, e_t\}$ | the idempotents, read in Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space |
| $\{0\}$ | the two-sided annihilator, read in Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space |
| $\mathrm{Vect}(\mathbb{B})$ | the index-three nilpotents, read in Square Roots of Zero and the Nilpotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space |
Further Reading
- Introduction to the Symmetric Plain Sesqualgebra, for the product, its class, its table and the failure of the Jordan identity.
- The Specificities of the Symmetric Plain Sesqualgebra, for the criterion of the law and the centrality against which the eleven other structures are tested.
- The Hermitian Form as a Product on the Symmetric Plain Sesqualgebra, for the form read as the product, its Gram matrix, its positivity and its restriction to the remarkable subspaces.
- Norm and Invertibility in the Symmetric Plain Sesqualgebra, for the norm, the definiteness and the absence of a unit.
- Square Roots of One and the Idempotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the idempotents $0$ and $e_t$, the field rule of the centre and the halving map from the roots of one.
- Zero Divisors in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the punctured-space zero-divisor set and the annihilators.
- Square Roots of Zero and the Nilpotents in the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the index-three nilpotents of the vector subspace and the absence of square-zero elements.
- Positivity and the Hermitian Cone of the Biquaternion Algebra with Hermitian Adjoint, for the cone generated by the squares of the block.
- The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra, the commutative unital sibling and its two idempotent families.
- The Radical and the Isotropic Elements of the Symmetric Quaternionic Algebra, the central sibling with an indefinite form, whose isotropic elements are the zero divisors.
- Introduction to the Symmetric Quaternionic Sesqualgebra, the other conjugate-commutative block of the twelve, whose law is compared with this one's and whose value carries a vector part.