Square Roots of One and the Idempotents of the General Plain Sesqualgebra

Introduction

The general plain sesqualgebra is the operation $\tilde Q_1\star\tilde Q_2=\tilde Q_1\tilde Q_2^{*}$ on the biquaternion space $\mathbb{B}=\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$ (Introduction to the General Plain Sesqualgebra, The Four General Products and Their Relations over the Biquaternion Algebra). This article treats two subjects, and in this order. The first is the square roots of the central value $+1$, the elements $\tilde\Xi$ with

$$ \tilde\Xi\star\tilde\Xi=e_t . $$

The second is the idempotents of the block, the elements $\tilde\Pi$ with $\tilde\Pi\star\tilde\Pi=\tilde\Pi$. The two subjects are one subject, because the halving map

$$ \tilde\Xi\longmapsto\tilde\Pi=\tfrac12\bigl(e_t+\tilde\Xi\bigr) $$

is a bijection from the Hermitian roots of $+1$ onto the idempotents, with inverse $\tilde\Xi=2\tilde\Pi-e_t$. The computation is two lines: if $\tilde\Xi$ is Hermitian with $\tilde\Xi\star\tilde\Xi=e_t$ then $\tilde\Xi^{*}=\tilde\Xi$ and $\tilde\Xi^{2}=\tilde\Xi\tilde\Xi^{*}=e_t$, so $\tilde\Pi$ is Hermitian with $\tilde\Pi\star\tilde\Pi=\tilde\Pi^{2}=\tfrac14(e_t+2\tilde\Xi+e_t)=\tilde\Pi$; and conversely every idempotent of the block is Hermitian, and $\tilde\Xi=2\tilde\Pi-e_t$ is Hermitian with $\tilde\Xi^{2}=4\tilde\Pi-4\tilde\Pi+e_t=e_t$. Every idempotent of the block is read through that map, on the Hermitian roots of $+1$ alone. The non-Hermitian roots of $+1$ do not give idempotents: the halved element is then not Hermitian, and the block detects it. Each Hermitian root carries two idempotents and not one, the complete orthogonal pair $\tilde\Pi_+(\tilde\Xi)=\tfrac12(e_t+\tilde\Xi)$ and $\tilde\Pi_-(\tilde\Xi)=\tfrac12(e_t-\tilde\Xi)$, whose addition, subtraction and multiplication are computed in §The Two Idempotents of a Root and §The Arithmetic of the Pair of Idempotents.

The article is algebraic. The roots of the general quadratic $\tilde\Xi\star\tilde\Xi=\tilde Q_1$ are the companion article Square Roots of a General Element in the General Plain Sesqualgebra; the roots of $0$ and the nilpotents are Square Roots of Zero and the Nilpotents of the General Plain Sesqualgebra; the zero-divisor set is Zero Divisors of the General Plain Sesqualgebra; and the ideals the idempotents generate, with the Peirce decomposition the pair of a root indexes, are Ideals and Peirce Decomposition of the General Plain Sesqualgebra. The roots of $-1$ are a short remark at the end of the roots part, they are used nowhere, and they open nothing.

The results. The block-square is the Hermitian element $\tilde\Xi\star\tilde\Xi=\tilde\Xi\tilde\Xi^{*}=(\lvert q_r\rvert^{2}+\lvert q_i\rvert^{2})e_t+2i(q_{r0}\vec v-q_{i0}\vec q+\vec q\times\vec v)$ in the real halves $\tilde\Xi=q_r+iq_i$, whose scalar part is the definite $\sum_\mu\lvert\Xi_\mu\rvert^{2}\geq0$. The scalar part decides the roots of $+1$: a central value has a root only when it is a non-negative real, the value $0$ forcing the origin, and the roots of $+1$ are exactly the single family $\tilde\Xi=Aw$ over the unit complex scalars $A$ and the real unit quaternions $w$, that is $\mathrm{U}(1)\cdot S^{3}$, of real dimension four, containing $e_t$, $-e_t$ and the pure imaginary unit vectors $\tilde\Xi$. The Hermitian roots of $+1$ are $\pm e_t$ and the roots $\tilde\Xi=i\vec u$, and the halving map $\tilde\Xi\mapsto\tfrac12(e_t+\tilde\Xi)$ is a bijection from them onto the idempotents of the block, which are $0$, $e_t$ and the $\tfrac12(e_t+\tilde\Xi)$ over the real unit vectors $\vec u$: a real two-parameter family of rank-one Hermitian idempotents. Each Hermitian root carries the complete pair $\tilde\Pi_\pm(\tilde\Xi)=\tfrac12(e_t\pm\tilde\Xi)$, the two members orthogonal in the block and in the algebra at once, summing to $e_t$, differing by the root, and their two subscripts the two eigenvalues of multiplication by the root.

The equations name a product, so the roots and the idempotents are relative to it. Throughout this article the square is the one of the general plain sesquilinear product, the Hermitian square $\tilde\Xi\star\tilde\Xi=\tilde\Xi\tilde\Xi^{*}$; the unqualified word idempotent means an idempotent of that product, and an idempotent of the associative multiplication $\tilde Q_1\tilde Q_2$ is called an idempotent of the algebra.

The Roots of One

The roots of $+1$ are exactly the elements $\tilde\Xi=Aw$ over the unit complex scalars $A$ and the real unit quaternions $w$, that is the single family $\mathrm{U}(1)\cdot S^{3}$ of real dimension four, containing $e_t$, $-e_t$ and the pure imaginary unit vectors $\tilde\Xi$; no root of $+1$ is a zero divisor.

A root of $+1$ is an element $\tilde\Xi\in\mathbb{B}$ with $\tilde\Xi\star\tilde\Xi=e_t$, the square being the one of the block.

The Reduction

The block-square is Hermitian with scalar part $\mathrm{Sc}(\tilde\Xi\star\tilde\Xi)=\sum_\mu\lvert\Xi_\mu\rvert^{2}\geq0$, and equating it to a central value $\lambda e_t$ requires $\lambda$ to be a non-negative real number, so the negative and the non-real scalars are excluded, and so is the value $0$.

Theorem (the Hermitian square and its scalar part). For $\tilde\Xi=q_r+iq_i$ with $q_r,q_i$ real quaternions,

$$ \boxed{\;\tilde\Xi\star\tilde\Xi=\tilde\Xi\tilde\Xi^{*}=\bigl(\lvert q_r\rvert^{2}+\lvert q_i\rvert^{2}\bigr)e_t+2i\Bigl(q_{r0}\vec v-q_{i0}\vec q+\vec q\times\vec v\Bigr)\;}, $$

where $q_{r0},q_{i0}$ are the real scalar parts of $q_r,q_i$ and $\vec q,\vec v$ their vector parts. The square is Hermitian, its scalar part is

$$ \mathrm{Sc}\bigl(\tilde\Xi\star\tilde\Xi\bigr)=\lvert q_r\rvert^{2}+\lvert q_i\rvert^{2}=\sum_{\mu\in\{t,x,y,z\}}\lvert\Xi_\mu\rvert^{2}\geq0, $$

a real non-negative number vanishing only at $\tilde\Xi=0$.

Proof. The involution is $(\ )^{*}=\overline{\ }\,\circ\,{}^{\natural}$, and $q_r,q_i$ are real, so $\tilde\Xi^{*}=q_r^{\natural}-iq_i^{\natural}$. Expanding $\tilde\Xi\tilde\Xi^{*}=(q_r+iq_i)(q_r^{\natural}-iq_i^{\natural})$ and using $q_rq_r^{\natural}=\lvert q_r\rvert^{2}$, $q_iq_i^{\natural}=\lvert q_i\rvert^{2}$, and the product rule of the real quaternions for $q_iq_r^{\natural}-q_rq_i^{\natural}$ — whose scalar part vanishes and whose vector part is $-2q_{i0}\vec q+2q_{r0}\vec v+2\vec q\times\vec v$, giving the display after a sign change — yields the formula. The Hermitianness is $(\tilde\Xi\tilde\Xi^{*})^{*}=\tilde\Xi\tilde\Xi^{*}$, and the scalar part is read off the display.

Corollary (the scalar trichotomy). The equation $\tilde\Xi\star\tilde\Xi=\lambda e_t$ requires, for a complex scalar $\lambda$,

$$ \sum_{\mu\in\{t,x,y,z\}}\lvert\Xi_\mu\rvert^{2}=\lambda\qquad\text{and}\qquad q_{r0}\vec v-q_{i0}\vec q+\vec q\times\vec v=0 . $$

Consequently a central value $\lambda$ has a root only when $\lambda$ is a non-negative real number: the negative and the non-real scalars are excluded, and the value $0$ forces $\tilde\Xi=0$, which the definition of a root excludes.

Proof. The two sides are equal exactly when the scalar and the vector parts are equal, and the scalar part of $\lambda e_t$ is $\lambda$. The scalar equation forces $\lambda=\sum_\mu\lvert\Xi_\mu\rvert^{2}$, a real number $\geq0$, so a negative or a non-real $\lambda$ is excluded and $\lambda=0$ gives $\tilde\Xi=0$. The value $0$ is treated in Square Roots of Zero and the Nilpotents of the General Plain Sesqualgebra.

The Roots of Plus One

Theorem (the roots of $+1$). The roots of $+1$ are exactly the elements

$$ \boxed{\;\tilde\Xi=Aw,\qquad A\in\mathbb{C}\ \text{with}\ \lvert A\rvert=1,\ w\in\mathbb{H}\ \text{real with}\ \lvert w\rvert=1\;}, $$

that is, the family $\{Aw\}=\mathrm{U}(1)\cdot S^{3}$, of real dimension four, containing $e_t$, $-e_t$ and the pure imaginary unit vectors, the roots $\tilde\Xi=i\vec u$ over the real unit vectors $\vec u$.

Proof. By the reduction the square is $e_t$ exactly when $\lvert q_r\rvert^{2}+\lvert q_i\rvert^{2}=1$ and $q_{r0}\vec v-q_{i0}\vec q+\vec q\times\vec v=0$. Dot the vector equation with $\vec q\times\vec v$: the first two terms are orthogonal to $\vec q\times\vec v$ and the third is $\lvert\vec q\times\vec v\rvert^{2}$, so $\vec q\times\vec v=0$; hence $\vec q$ and $\vec v$ are parallel, and the vector equation reads $q_{r0}\vec v=q_{i0}\vec q$. If $\vec q\neq0$ then $\vec v=\lambda\vec q$ for a real $\lambda$, the equation gives $q_{i0}=\lambda q_{r0}$ and therefore $q_i=q_{i0}+\vec v=\lambda(q_{r0}+\vec q)=\lambda q_r$: the halves are parallel as full quaternions, $\tilde\Xi=q_r+iq_i=(1+i\lambda)q_r$, and writing $A=(1+i\lambda)/\lvert 1+i\lambda\rvert$, a unit complex scalar, and $w=\lvert 1+i\lambda\rvert\,q_r$, a real quaternion since $q_r$ is one, gives $\tilde\Xi=Aw$ with $\lvert w\rvert=1$ from the scalar equation. If $\vec q=0$ then $q_r=q_{r0}$ is a real scalar and the equation reads $q_{r0}\vec v=0$: either $\vec v=0$, and $\tilde\Xi=(q_{r0}+iq_{i0})e_t=Ae_t$ with $A=q_{r0}+iq_{i0}$, a unit complex scalar because $q_{r0}^{2}+q_{i0}^{2}=1$, is already of the required form with $w=e_t$, or $q_{r0}=0$, and $\tilde\Xi=iq_i=i(q_{i0}+\vec v)$ is of the required form with the unit complex scalar $A=i$ and the real unit $w=q_{i0}+\vec v$. In both cases $\tilde\Xi=Aw$ with $A$ a unit complex scalar and $w$ a real unit quaternion. Conversely every $Aw$ with $A$ a unit complex scalar and $w$ real unit has $q_r=(\mathrm{Re}A)w$, $q_i=(\mathrm{Im}A)w$, hence $\lvert q_r\rvert^{2}+\lvert q_i\rvert^{2}=\lvert A\rvert^{2}\lvert w\rvert^{2}=1$ and $q_{r0}\vec v-q_{i0}\vec q+\vec q\times\vec v=\mathrm{Re}A\,\mathrm{Im}A\,\vec w\times\vec w=0$, the first two terms cancelling and the cross product of a vector with itself vanishing, so it is a root. The parametrisation $(A,w)\mapsto Aw$ has kernel $\{(1,e_t),(-1,-e_t)\}$, of order two, so the root set is a single real four-dimensional family, and the named elements are the cases $A=1$, $A=-1$ and $A=i$ with $w=\vec u$.

Corollary (the roots of $+1$ are never zero divisors). Every root of $+1$ has $\mathrm{Sc}(\tilde\Xi\star\tilde\Xi)=1$, hence $\langle\tilde\Xi\rangle_{*}=1$, and its central square is a complex number of modulus one,

$$ C(Aw)=A^{2},\qquad\text{in particular}\qquad C(\pm e_t)=1,\qquad C(i\vec u)=-1 . $$

Every root of $+1$ is therefore one-sidedly invertible, and the root set of $+1$ is disjoint from the zero-divisor cone.

Proof. The normalisation is the scalar equation $\mathrm{Sc}=1$ and the algebraic norm $\langle\cdot\rangle_{*}$ of the reduction. For the central square, $C(\tilde\Xi)=\sum_\mu\Xi_\mu^{2}$ with $\Xi_\mu=Aw_\mu$ and $w_\mu$ real, so $C(\tilde\Xi)=A^{2}\sum_\mu w_\mu^{2}=A^{2}$, since $\sum_\mu w_\mu^{2}=\lvert w\rvert^{2}=1$, of modulus one and never zero; the named values are the cases $A=\pm1$ and $A=i$. The zero-divisor criterion is $C=0$ (Zero Divisors of the General Plain Sesqualgebra), so the root set misses the cone; the one-sided invertibility of the roots is Norm and Invertibility in the General Plain Sesqualgebra.

Remark (the shape of the root set). The parent's root set of $+1$ is several families together with the isolated elements $\pm i$; the block has its roots of $+1$ in the single family $\mathrm{U}(1)\cdot S^{3}$, of real dimension four, and no root of the other two values. The block's set is a single orbit and the parent's is not, and the parent's two isolated roots have no counterpart in the block.

The Hermitian Roots of One

Definition (the Hermitian roots of $+1$). A root of $+1$ is Hermitian when $\tilde\Xi^{*}=\tilde\Xi$.

Theorem (the Hermitian roots of $+1$). The Hermitian roots of $+1$ are the two central scalars $\pm e_t$ together with the pure imaginary unit vectors, the roots $\tilde\Xi=i\vec u$ over the real unit vectors $\vec u$, a real two-parameter family.

Proof. A Hermitian element is $\xi_te_t+i\vec\xi'$ with $\xi_t\in\mathbb{R}$ and $\vec\xi'$ a real vector, and its square in the algebra is $\xi_t^{2}e_t+\lvert\vec\xi'\rvert^{2}e_t+2\xi_ti\vec\xi'$ (since $(i\vec\xi')^{2}=-(\vec\xi')^{2}=\lvert\vec\xi'\rvert^{2}e_t$). Since the element is Hermitian its block-square is its algebra square, and the equation is $(\xi_t^{2}+\lvert\vec\xi'\rvert^{2})e_t+2\xi_ti\vec\xi'=e_t$, whose solution is $\xi_t=\pm1$ with $\vec\xi'=0$, or $\xi_t=0$ with $\lvert\vec\xi'\rvert=1$. The last is $\tilde\Xi$.

Remark (the Hermitian roots are the involutions of the block). A Hermitian root of $+1$ is the element form of an involution of the Hermitian subspace, and the three kinds are the two central involutions $\pm e_t$ and the family $\tilde\Xi=i\vec u$ over the real unit vectors $\vec u$. These are exactly the roots that the halving map turns into idempotents, and the non-Hermitian roots of $+1$ are dropped by the idempotent theory.

The roots of minus one. In a product that is two-sidedly linear the central imaginary carries the roots of $+1$ to the roots of $-1$, because $(i\tilde\Xi)\star(i\tilde\Xi)=i^{2}(\tilde\Xi\star\tilde\Xi)=-\tilde\Xi\star\tilde\Xi$; the mechanism is not universal. The block is sesquilinear, so the second slot conjugates and $(i\tilde\Xi)\star(i\tilde\Xi)=i\,\tilde\Xi\star\tilde\Xi\,\overline{i}=\tilde\Xi\star\tilde\Xi$: the scalar multiplication changes nothing, the roots of $-1$ are not read off the roots of $+1$ by such a map, and the root set of $-1$ is empty, as §The Reduction already shows.

The Idempotents

Every idempotent of the block is a Hermitian idempotent of the algebra, and the idempotents of the block are $0$, $e_t$ and the elements $\tilde\Pi_+(\tilde\Xi)=\tfrac12(e_t+\tilde\Xi)$ over the non-central Hermitian roots $\tilde\Xi$ of $+1$, a real two-parameter family; each is the member attached to $+1$ of the complete pair $\tilde\Pi_\pm(\tilde\Xi)$ of a Hermitian root, and the halving map $\tilde\Xi\mapsto\tfrac12(e_t+\tilde\Xi)$ is a bijection from the Hermitian roots of $+1$ onto them.

The Definition

Definition. An element $\tilde\Pi$ is idempotent for the product $\star$, or a $\star$-idempotent, when

$$ \tilde\Pi\star\tilde\Pi=\tilde\Pi . $$

The equation names a product, so the notion is relative to it; the general definition, with the orthogonal, complete and primitive families, the Hermitian idempotent and the Peirce decomposition, is Definitions for the Study of the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, §The Idempotent, the Hermitian Idempotent and the Peirce Decomposition, and throughout this article the product is the general plain sesquilinear product $\tilde Q_1\star\tilde Q_2=\tilde Q_1\tilde Q_2^{*}$ of the block. The idempotents of the associative product $\tilde Q_1\tilde Q_2$ are called the idempotents of the algebra, and they are written $\tilde\Pi$ with $\tilde\Pi^{2}=\tilde\Pi$.

Idempotents of an associative product encode direct summands: an idempotent $\tilde\Pi$ of the algebra and its complement $e_t-\tilde\Pi$ split the underlying $\mathbb{C}$-module,

$$ \mathbb{B}=\mathbb{B}\tilde\Pi\oplus\mathbb{B}(e_t-\tilde\Pi),\qquad \mathbb{B}=\tilde\Pi\mathbb{B}\oplus(e_t-\tilde\Pi)\mathbb{B}, $$

and every such decomposition arises from an idempotent. A primitive idempotent, one that is not a sum of two nonzero orthogonal idempotents, generates a minimal left ideal; the general statements are Definitions for the Study of the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, and the decomposition carried by a plain idempotent is Ideals and Peirce Decompositions of the Biquaternion Algebra.

Every Idempotent of the Block Is Hermitian

Theorem. Let $\tilde\Pi$ satisfy $\tilde\Pi\star\tilde\Pi=\tilde\Pi$. Then $\tilde\Pi$ is Hermitian,

$$ \tilde\Pi^{*}=\tilde\Pi . $$

Proof. The equation reads $\tilde\Pi\tilde\Pi^{*}=\tilde\Pi$. The square $\tilde\Pi\tilde\Pi^{*}$ is fixed by the involution for every element, $(\tilde\Pi\tilde\Pi^{*})^{*}=\tilde\Pi\tilde\Pi^{*}$ (Introduction to the General Plain Sesqualgebra, §The Squares and the Positive Cone). Applying ${}^{*}$ to the equation therefore gives $\tilde\Pi\tilde\Pi^{*}=\tilde\Pi^{*}$, and the left-hand sides of the equation and of its conjugate are the same, so the right-hand sides are equal, $\tilde\Pi=\tilde\Pi^{*}$.

Remark. The proof uses two things and no more: the square is Hermitian, and the involution is injective. The self-adjointness is therefore forced on every idempotent of the block, and it is the reason the equation reduces to the algebra's: a Hermitian element has $\tilde\Pi\star\tilde\Pi=\tilde\Pi\tilde\Pi^{*}=\tilde\Pi\tilde\Pi=\tilde\Pi^{2}$, so its two squares coincide.

The Reduction

Corollary. An element is idempotent for the multiplication $\star$ if and only if it is a Hermitian idempotent of the algebra:

$$ \tilde\Pi\star\tilde\Pi=\tilde\Pi \quad\Longleftrightarrow\quad \tilde\Pi=\tilde\Pi^{*}\ \text{ and }\ \tilde\Pi^{2}=\tilde\Pi . $$

Proof. If $\tilde\Pi\star\tilde\Pi=\tilde\Pi$ then $\tilde\Pi$ is Hermitian by the theorem, and $\tilde\Pi\star\tilde\Pi=\tilde\Pi\tilde\Pi^{*}=\tilde\Pi^{2}$, so $\tilde\Pi^{2}=\tilde\Pi$. Conversely a Hermitian idempotent has $\tilde\Pi\star\tilde\Pi=\tilde\Pi\tilde\Pi^{*}=\tilde\Pi\tilde\Pi=\tilde\Pi^{2}=\tilde\Pi$.

Proposition (the scalar part of the equation). Let $\tilde\Pi$ satisfy $\tilde\Pi\star\tilde\Pi=\tilde\Pi$. Then the scalar part of $\tilde\Pi$ is the sum of the modulus squares of its coordinates,

$$ \Pi_t=\mathrm{Sc}\bigl(\tilde\Pi\star\tilde\Pi\bigr)=\sum_{\mu\in\{t,x,y,z\}}\Pi_\mu\overline{\Pi_\mu}=\sum_{\mu\in\{t,x,y,z\}}\lvert\Pi_\mu\rvert^{2}, $$

a real number, and it is nonzero unless $\tilde\Pi=0$.

Proof. The scalar part of the square is $\sum_\mu\lvert\Pi_\mu\rvert^{2}$ by the reduction of the roots part; taking the scalar part of $\tilde\Pi\star\tilde\Pi=\tilde\Pi$ gives the display. The sum of the modulus squares is a nonnegative real number, and it vanishes only when every coordinate vanishes.

Remark. The display is a constraint and not a definition: it rules out at once the algebra idempotents with a non-real scalar part, and it is the idempotent reading of the nonnegativity of the sesquilinear square that The Squares and the Positive Cone of the Biquaternion Sesqualgebra develops.

The Classification

Theorem (the idempotents of the block). The idempotents of the sesquilinear multiplication $\star$ are $0$, $e_t$ and the elements

$$ \boxed{\;\tilde\Pi_+(\tilde\Xi)=\tfrac12\bigl(e_t+\tilde\Xi\bigr) , \qquad \tilde\Xi=i\vec u , \qquad \vec u\in\mathbb{R}^{3},\ (\vec u,\vec u)=1\;}, $$

where $\vec u=\sum_{\mu\in\{x,y,z\}}u_\mu\vec e_\mu$ ranges over the real vectors of the vector subspace with $(\vec u,\vec u)=\sum_\mu u_\mu^{2}=1$. There are no others.

Proof. By the reduction the idempotents of the block are the Hermitian elements with $\tilde\Pi^{2}=\tilde\Pi$. A Hermitian element is $\tilde\Pi=\alpha e_t+i\vec w$ with $\alpha\in\mathbb{R}$ and $\vec w$ a real vector, and its square in the algebra is

$$ \tilde\Pi^{2}=\alpha^{2}e_t+2\alpha\,i\vec w+(i\vec w)^{2}=\bigl(\alpha^{2}+\lvert\vec w\rvert^{2}\bigr)e_t+2\alpha\,i\vec w , $$

since $\vec w^{2}=-\lvert\vec w\rvert^{2}e_t$ and $(i\vec w)^{2}=i^{2}\vec w^{2}=\lvert\vec w\rvert^{2}e_t$. Equating this to $\tilde\Pi=\alpha e_t+i\vec w$ splits the equation into the vector part $(2\alpha-1)\vec w=0$ and the scalar part $\alpha^{2}+\lvert\vec w\rvert^{2}=\alpha$. If $\vec w=0$ the scalar equation is $\alpha^{2}=\alpha$, so $\alpha=0$ or $\alpha=1$, giving the trivial idempotents $0$ and $e_t$. If $\vec w\neq0$ then $\alpha=\tfrac12$, the scalar equation gives $\lvert\vec w\rvert^{2}=\tfrac14$, and writing $\vec w=\tfrac12\vec u$ with $(\vec u,\vec u)=1$ gives $\tilde\Pi=\tfrac12(e_t+\tilde\Xi)$. No other case occurs, and every listed element is a Hermitian idempotent of the algebra and therefore an idempotent of the block.

Remark. The idempotents of the block are exactly the Hermitian idempotents of $\mathbb{B}$, and the word projection is reserved for the idempotent operators they generate by multiplication (§The Operator Interpretation); the name pure state of Idempotents of the General Plain Algebra is the same element read in the Hermitian subspace. The two families of algebra idempotents that are not Hermitian are the ones the block discards, and they are treated in §The Contrast With the Algebra.

The Coordinates

Proposition (the coordinate form). For a real vector $\vec u$ with $(\vec u,\vec u)=1$ and coordinates $u_x,u_y,u_z\in\mathbb{R}$,

$$ \tilde\Pi_+(\tilde\Xi)=\tfrac12e_t+\tfrac{i}{2}\sum_{\mu\in\{x,y,z\}}u_\mu\vec e_\mu , $$

so its scalar part is $\tfrac12$, its vector part is $\tfrac{i}{2}\vec u$, and its four coordinates are $\bigl(\tfrac12,\tfrac{i}{2}u_x,\tfrac{i}{2}u_y,\tfrac{i}{2}u_z\bigr)$. Its scalar part is the sum of the modulus squares of its coordinates,

$$ \tfrac12=\tfrac14+\tfrac14\sum_\mu u_\mu^{2}, $$

which is the scalar-part identity of the reduction read on this family.

Proof. The coordinates are read from the display, and $\sum_\mu u_\mu^{2}=1$ gives the scalar-part identity $\Pi_t=\sum_\mu\lvert\Pi_\mu\rvert^{2}$.

Remark. The scalar part $\tfrac12$ is forced by the normalisation $(\vec u,\vec u)=1$ and is the same for the whole family; the family is the whole set of the Hermitian idempotents and not a proper subfamily of it.

The Complement and the Pairing of the Directions

Proposition (the complementary idempotent, and the antipode). For every non-central Hermitian root $\tilde\Xi$ of $+1$,

$$ e_t-\tilde\Pi_+(\tilde\Xi)=\tilde\Pi_-(\tilde\Xi)=\tilde\Pi_+(-\tilde\Xi), $$

and the two are idempotents of the algebra, orthogonal and summing to the unit,

$$ \tilde\Pi_+(\tilde\Xi)\tilde\Pi_-(\tilde\Xi)=0 , \qquad \tilde\Pi_+(\tilde\Xi)+\tilde\Pi_-(\tilde\Xi)=e_t . $$

Proof. The first identity is $e_t-\tfrac12(e_t+\tilde\Xi)=\tfrac12(e_t-\tilde\Xi)=\tilde\Pi_-(\tilde\Xi)$, and $\tilde\Pi_-(\tilde\Xi)=\tilde\Pi_+(-\tilde\Xi)$. For the orthogonality, multiply out, using $\tilde\Xi^{2}=e_t$:

$$ \tilde\Pi_+(\tilde\Xi)\tilde\Pi_-(\tilde\Xi)=\tfrac14\bigl(e_t+\tilde\Xi\bigr)\bigl(e_t-\tilde\Xi\bigr)=\tfrac14\bigl(e_t-\tilde\Xi^{2}\bigr)=\tfrac14\bigl(e_t-e_t\bigr)=0 , $$

and the sum is immediate. The pair $\tilde\Pi_+(\tilde\Xi),\tilde\Pi_-(\tilde\Xi)$ is the pair of the Hermitian root $\tilde\Xi$ in the sense of §The Two Idempotents of a Root.

Proposition (when two idempotents are orthogonal). For two non-central Hermitian roots $\tilde\Xi_1,\tilde\Xi_2$ of $+1$, of axes the real unit vectors $\vec u$ and $\vec u_{m}$,

$$ \tilde\Pi_+(\tilde\Xi_1)\tilde\Pi_+(\tilde\Xi_2)=0 \quad\Longleftrightarrow\quad \tilde\Xi_2=-\tilde\Xi_1 . $$

Proof. Multiplying out and using $\vec u\vec u_{m}=-(\vec u,\vec u_{m})e_t+\vec u\times\vec u_{m}$ for the product of two pure real vectors gives

$$ \tilde\Pi_+(\tilde\Xi_1)\tilde\Pi_+(\tilde\Xi_2)=\tfrac14\Bigl(\bigl(1+(\vec u,\vec u_{m})\bigr)e_t-\vec u\times\vec u_{m}+i\bigl(\vec u+\vec u_{m}\bigr)\Bigr). $$

The scalar part vanishes when $(\vec u,\vec u_{m})=-1$, and for two real unit vectors this forces $\vec u_{m}=-\vec u$, that is $\tilde\Xi_2=-\tilde\Xi_1$, because then $(\vec u+\vec u_{m},\vec u+\vec u_{m})=(\vec u,\vec u)+2(\vec u,\vec u_{m})+(\vec u_{m},\vec u_{m})=0$ and a real vector whose square vanishes is zero. With $\vec u_{m}=-\vec u$ the vector part is $-\vec u\times(-\vec u)+i(\vec u-\vec u)=0$ as well. Conversely $\vec u_{m}=-\vec u$ gives $0$ by the previous proposition.

Remark. Orthogonality is therefore as sparse as it can be: an idempotent is orthogonal, as an algebra idempotent, to one other idempotent of the family and to no third one, so the family is a family of complementary pairs and not an orthogonal set. The Peirce decomposition of the algebra relative to the standard pair is Ideals and Peirce Decomposition of the General Plain Sesqualgebra.

The Standard Pair

Example. The two idempotents at the directions $\pm \vec e_z$ are the standard idempotents of the algebra,

$$ \tilde\Pi_1(\vec u_z)=\tfrac12(e_t+i\vec e_z)=\tilde\Pi_+ , \qquad \tilde\Pi_2(\vec u_z)=\tfrac12(e_t-i\vec e_z)=\tilde\Pi_- , $$

and they are the images under $\mathsf{M}_2$ of the two diagonal matrix units of the matrix model: $\tilde\Pi_1(\vec u_z)$ corresponds to $E_{11}$ and $\tilde\Pi_2(\vec u_z)$ to $E_{22}$, as the transport of The General Plain Sesqualgebra in the $2\times2$ Matrix Element Representation $M_2(\mathbb{C})$ records. Since the equalities are equalities in the algebra, they hold for the block as well, a Hermitian element having the same square in the two products. $i\vec e_z$ is a root of $+1$, $(i\vec e_z)^2 = e_t$, and $\tilde\Pi_+$ and $\tilde\Pi_-$ are its two associated idempotents.

The Two Idempotents of a Root

A Hermitian root $\tilde\Xi$ of $+1$ carries the two idempotents $\tilde\Pi_+(\tilde\Xi)=\tfrac12(e_t+\tilde\Xi)$ and $\tilde\Pi_-(\tilde\Xi)=\tfrac12(e_t-\tilde\Xi)=e_t-\tilde\Pi_+(\tilde\Xi)$; each is an idempotent of the block, the two are complementary, and the root is their difference, $\tilde\Xi=\tilde\Pi_+(\tilde\Xi)-\tilde\Pi_-(\tilde\Xi)$.

Let $\tilde\Xi$ be a Hermitian root of $+1$, so that $\tilde\Xi^{*}=\tilde\Xi$ and $\tilde\Xi^{2}=\tilde\Xi\tilde\Xi^{*}=e_t$ (§The Hermitian Roots of One). The two idempotents of the root are the elements

$$ \tilde\Pi_+(\tilde\Xi)=\tfrac12\bigl(e_t+\tilde\Xi\bigr),\qquad \tilde\Pi_-(\tilde\Xi)=\tfrac12\bigl(e_t-\tilde\Xi\bigr). $$

The pair carries the two signs rather than two numbers, as §The Idempotent Convention of Conventions in Mathematics provides for a complete pair: the subscript $+$ names the member attached to $+1$ and the subscript $-$ the member attached to $-1$, an attachment §The Arithmetic of the Pair of Idempotents computes.

Each member is Hermitian, $\tilde\Pi_\pm^{*}=\tilde\Pi_\pm$, since $\tilde\Xi$ is, and a Hermitian element has the same square in the two products (§The Reduction), so each is idempotent for the block as soon as it is idempotent for the algebra:

$$ \tilde\Pi_+(\tilde\Xi)^{2}=\tfrac14\bigl(e_t+2\tilde\Xi+\tilde\Xi^{2}\bigr)=\tfrac14\bigl(2e_t+2\tilde\Xi\bigr)=\tilde\Pi_+(\tilde\Xi), $$

$$ \tilde\Pi_-(\tilde\Xi)^{2}=\tfrac14\bigl(e_t-2\tilde\Xi+\tilde\Xi^{2}\bigr)=\tfrac14\bigl(2e_t-2\tilde\Xi\bigr)=\tilde\Pi_-(\tilde\Xi). $$

Both are therefore idempotents of the block, and both are idempotents in the classification of §The Classification. The two are complementary, $\tilde\Pi_-(\tilde\Xi)=e_t-\tilde\Pi_+(\tilde\Xi)$, and their difference recovers the root,

$$ \tilde\Pi_+(\tilde\Xi)-\tilde\Pi_-(\tilde\Xi)=\tilde\Xi . $$

The trivial roots give the trivial pairs, the pair of $+e_t$ being $(e_t,0)$ and the pair of $-e_t$ being $(0,e_t)$.

Remark (opposite roots, the same pair). The two Hermitian roots $\tilde\Xi$ and $-\tilde\Xi$ carry the same unordered pair with its members exchanged,

$$ \tilde\Pi_+(-\tilde\Xi)=\tilde\Pi_-(\tilde\Xi),\qquad \tilde\Pi_-(-\tilde\Xi)=\tilde\Pi_+(\tilde\Xi), $$

so the passage from a root to its pair is two-to-one. Conversely every idempotent $\tilde\Pi$ of the block other than $0$ and $e_t$ arises in this way, being the member attached to $+1$ of the Hermitian root $\tilde\Xi=2\tilde\Pi-e_t$. Read on the idempotents, the pair of the root $\tilde\Xi$ is the pair $\tilde\Pi_+(\tilde\Xi),\tilde\Pi_-(\tilde\Xi)$ of §The Complement and the Pairing of the Directions, and the pair of the root $-\tilde\Xi$ is the same two elements exchanged.

The Arithmetic of the Pair of Idempotents

The pair of a Hermitian root is complete, $\tilde\Pi_++\tilde\Pi_-=e_t$; the root is its difference, $\tilde\Pi_+-\tilde\Pi_-=\tilde\Xi$; the pair is orthogonal in the block and in the algebra at once, $\tilde\Pi_+\star\tilde\Pi_-=\tilde\Pi_-\star\tilde\Pi_+=0$; and multiplication by the root is the sign change of the pair on both sides, $+1$ on $\tilde\Pi_+$ and $-1$ on $\tilde\Pi_-$.

Throughout this section $\tilde\Xi$ is a Hermitian root of $+1$ and $\tilde\Pi_+=\tilde\Pi_+(\tilde\Xi)$, $\tilde\Pi_-=\tilde\Pi_-(\tilde\Xi)$ its two idempotents, as in §The Two Idempotents of a Root.

Addition. The pair is complete: the two idempotents sum to the unit,

$$ \tilde\Pi_+ + \tilde\Pi_-=\tfrac12\bigl(e_t+\tilde\Xi\bigr)+\tfrac12\bigl(e_t-\tilde\Xi\bigr)=e_t . $$

Subtraction. The two idempotents differ by the root,

$$ \tilde\Pi_+ - \tilde\Pi_-=\tfrac12\bigl(e_t+\tilde\Xi\bigr)-\tfrac12\bigl(e_t-\tilde\Xi\bigr)=\tilde\Xi, $$

so that the root is read on the pair as $\tilde\Xi=\tilde\Pi_+-\tilde\Pi_-$, equivalently $\tilde\Xi=2\tilde\Pi_+-e_t$. Each member is the complement of the other, and replacing the root by $-\tilde\Xi$ interchanges them.

Multiplication. Both members are Hermitian, so their block products coincide with their algebra products and

$$ \tilde\Pi_+\star\tilde\Pi_-=\tilde\Pi_+\tilde\Pi_-^{*}=\tilde\Pi_+\tilde\Pi_-=\tfrac14\bigl(e_t+\tilde\Xi\bigr)\bigl(e_t-\tilde\Xi\bigr)=\tfrac14\bigl(e_t-\tilde\Xi^{2}\bigr)=0, $$

and equally $\tilde\Pi_-\star\tilde\Pi_+=0$: the pair is orthogonal, in the block and in the algebra at once. With the idempotence of the two members the four products are

so the $\mathbb{C}$-span $\mathbb{C}\tilde\Pi_+\oplus\mathbb{C}\tilde\Pi_-$ is carried by the block product into itself, componentwise up to the conjugation of the second scalar,

$$ \bigl(s\tilde\Pi_++t\tilde\Pi_-\bigr)\star\bigl(s'\tilde\Pi_++t'\tilde\Pi_-\bigr)=s\overline{s'}\,\tilde\Pi_++t\overline{t'}\,\tilde\Pi_- , $$

while the algebra product on the same span is the componentwise product $ss'\tilde\Pi_++tt'\tilde\Pi_-$, so that the block product differs from the algebra product of the span by the conjugate on the second factor alone.

Multiplication by the root. The root acts on the pair by its two signs, and on the left and on the right alike. On the left, using $\tilde\Xi^{*}=\tilde\Xi$,

$$ \tilde\Xi\star\tilde\Pi_+=\tilde\Xi\tilde\Pi_+^{*}=\tilde\Xi\tilde\Pi_+=\tfrac12\bigl(\tilde\Xi+\tilde\Xi^{2}\bigr)=\tfrac12\bigl(e_t+\tilde\Xi\bigr)=\tilde\Pi_+, $$

$$ \tilde\Xi\star\tilde\Pi_-=\tilde\Xi\tilde\Pi_-=\tfrac12\bigl(\tilde\Xi-\tilde\Xi^{2}\bigr)=-\tfrac12\bigl(e_t-\tilde\Xi\bigr)=-\tilde\Pi_- , $$

and on the right $\tilde\Pi_+\star\tilde\Xi=\tilde\Pi_+$ and $\tilde\Pi_-\star\tilde\Xi=-\tilde\Pi_-$ by the same computation. The subscripts of the pair are therefore the eigenvalues of multiplication by the root: the member attached to $+1$ is the one the root fixes, and the member attached to $-1$ the one the root negates. Since the pair is complete the two members split the block, $\mathbb{B}=\mathbb{B}\tilde\Pi_+\oplus\mathbb{B}\tilde\Pi_-=\tilde\Pi_+\mathbb{B}\oplus\tilde\Pi_-\mathbb{B}$, along the two signs of the root; the decomposition and its Peirce refinement are Ideals and Peirce Decomposition of the General Plain Sesqualgebra.

The Halving Map and the Bijection With the Roots of One

Theorem (the idempotents of the block are the halved Hermitian roots of $+1$). The map $\tilde\Xi\mapsto\tfrac12(e_t+\tilde\Xi)$ is a bijection from the Hermitian roots of $+1$ onto the idempotents of the block, with inverse $\tilde\Xi=2\tilde\Pi-e_t$,

$$ 0=\tfrac12(e_t-e_t),\qquad e_t=\tfrac12(e_t+e_t),\qquad \tfrac12(e_t+\tilde\Xi),\ \ \tilde\Xi=i\vec u,\ \ \vec u\in\mathbb{R}^{3},\ (\vec u,\vec u)=1 . $$

Proof. A Hermitian root $\tilde\Xi$ of $+1$ satisfies $\tilde\Xi^{*}=\tilde\Xi$ and $\tilde\Xi^{2}=\tilde\Xi\tilde\Xi^{*}=e_t$, so $\tilde\Pi=\tfrac12(e_t+\tilde\Xi)$ is Hermitian and has square $\tfrac14(e_t+2\tilde\Xi+\tilde\Xi^{2})=\tfrac14(2e_t+2\tilde\Xi)=\tilde\Pi$: it is a Hermitian idempotent of the algebra, hence an idempotent of the block by the reduction. Conversely an idempotent $\tilde\Pi$ of the block is a Hermitian idempotent of the algebra, and $\tilde\Xi=2\tilde\Pi-e_t$ is Hermitian with square $(2\tilde\Pi-e_t)^{2}=4\tilde\Pi-4\tilde\Pi+e_t=e_t$, a Hermitian root of $+1$. The two maps are inverse, and the list is the classification of §The Hermitian Roots of One and §The Classification.

Remark (the family and the pair). The halving map carries the two trivial Hermitian roots $\pm e_t$ to the trivial idempotents $e_t$ and $0$, and the family $\tilde\Xi=i\vec u$ to the Hermitian idempotents; the pair $\{\tilde\Xi,-\tilde\Xi\}$ of Hermitian roots becomes the complementary pair $\{\tilde\Pi,e_t-\tilde\Pi\}$ of idempotents, which is the pair of the root of §The Two Idempotents of a Root, read on the idempotents of §The Complement and the Pairing of the Directions. The Hermitian roots of $+1$ are the spine of the idempotent theory.

The Contrast With the Algebra

The idempotents of the algebra are the halving map applied to its own roots of $+1$, a real four-parameter family together with the trivial pair; the idempotents of the block are the Hermitian ones, so the two sets agree exactly on the trivial pair and the Hermitian idempotents, and every non-Hermitian idempotent of the algebra is dropped by the block.

The idempotents of the associative product are the elements $\tilde\Pi=\tfrac12(e_t+\tilde\Xi)$ over the roots $\tilde\Xi$ of $+1$ of the algebra, $\tilde\Xi^{2}=e_t$, by the halving map of Square Roots of One and the Idempotents of the Biquaternion Algebra, where those roots are classified; the map $\tilde\Xi\mapsto\tilde\Pi$ is a bijection from the algebra's roots of $+1$ onto its idempotents, and the non-trivial roots give the non-trivial idempotents, a real four-parameter family. The block's idempotents are the Hermitian idempotents of the algebra, by the reduction, and they are the subfamily with $\tilde\Xi$ Hermitian.

Example (the simplest non-Hermitian idempotent). Take the root of $+1$ of the algebra $\tilde\Xi=\sqrt2\,i\vec e_x-\vec e_y$, so that $\tilde\Xi^{2}=e_t$ and

$$ \tilde\Pi=\tfrac12(e_t+\tilde\Xi)=\tfrac12e_t+\tfrac12\bigl(\sqrt2\,i\vec e_x-\vec e_y\bigr) $$

is an idempotent of the algebra, not Hermitian: $\tilde\Xi^{*}=\sqrt2\,i\vec e_x+\vec e_y\neq\tilde\Xi$, so $\tilde\Pi^{*}\neq\tilde\Pi$. Its square in the block is

$$ \tilde\Pi\star\tilde\Pi=\tilde\Pi\tilde\Pi^{*}=e_t+\tfrac{\sqrt2}{2}\,i\vec e_x+\tfrac{\sqrt2}{2}\,i\vec e_z\neq\tilde\Pi . $$

Proof. The idempotence for the algebra is the halving map at the root $\tilde\Xi$ of $+1$; the failure of Hermitianness is read from $\tilde\Xi^{*}=\sqrt2\,i\vec e_x+\vec e_y$. For the square, $\tilde\Pi\tilde\Pi^{*}=\tfrac14(e_t+\tilde\Xi)(e_t+\tilde\Xi^{*})=\tfrac14(e_t+\tilde\Xi+\tilde\Xi^{*}+\tilde\Xi\tilde\Xi^{*})$. Here $\tilde\Xi+\tilde\Xi^{*}=2\sqrt2\,i\vec e_x$ and

$$ \tilde\Xi\tilde\Xi^{*}=\bigl(\sqrt2\,i\vec e_x-\vec e_y\bigr)\bigl(\sqrt2\,i\vec e_x+\vec e_y\bigr)=3e_t+2\sqrt2\,i\vec e_z , $$

by the multiplication of the basis, so the sum is $\tfrac14(4e_t+2\sqrt2\,i\vec e_x+2\sqrt2\,i\vec e_z)=e_t+\tfrac{\sqrt2}{2}i\vec e_x+\tfrac{\sqrt2}{2}i\vec e_z$, which differs from $\tilde\Pi$.

Remark. The element is the distinguishing case of Introduction to the General Plain Sesqualgebra: it is idempotent for the bilinear product and not for the sesquilinear one. Its scalar part is $\tfrac12$, but $\mathrm{Sc}(\tilde\Pi\star\tilde\Pi)=1$, so the scalar-part identity of §The Reduction fails at the same element, which is a second reading of the same failure.

The Matrix Model

Under the isomorphism $\mathsf{M}_2$ the idempotents correspond to the rank-one Hermitian idempotents of $M_2(\mathbb{C})$, $\mathsf{M}_2(\tilde\Pi_+(\tilde\Xi))=\tfrac12(I+i\mathsf{M}_2(\vec u))$, and the family is exactly the set of those idempotents.

Proposition. Under the isomorphism $\mathsf{M}_2:\mathbb{B}\to M_2(\mathbb{C})$ of The General Plain Sesqualgebra in the $2\times2$ Matrix Element Representation $M_2(\mathbb{C})$, which satisfies $\mathsf{M}_2(\tilde Q_2^{*})=\mathsf{M}_2(\tilde Q_2)^{\dagger}$ and $\mathsf{M}_2(\tilde Q_1\tilde Q_2^{*})=\mathsf{M}_2(\tilde Q_1)\mathsf{M}_2(\tilde Q_2)^{\dagger}$, the idempotents correspond to the rank-one Hermitian idempotents of the matrix algebra,

$$ \mathsf{M}_2\bigl(\tilde\Pi_+(\tilde\Xi)\bigr)=\tfrac12\bigl(I+i\mathsf{M}_2(\vec u)\bigr), $$

and the family is the set of the rank-one Hermitian idempotents of $M_2(\mathbb{C})$.

Proof. The image of a Hermitian idempotent is Hermitian, $\mathsf{M}_2(\tilde\Pi)^{\dagger}=\mathsf{M}_2(\tilde\Pi^{*})=\mathsf{M}_2(\tilde\Pi)$, and idempotent, $\mathsf{M}_2(\tilde\Pi)^{2}=\mathsf{M}_2(\tilde\Pi^{2})=\mathsf{M}_2(\tilde\Pi)$. The image of $\tilde\Pi_+(\tilde\Xi)$ is computed from $\mathsf{M}_2(\tilde Q)=\sum_\mu Q_\mu\mathsf{M}_2(e_\mu)$ on the coordinates of §The Coordinates, and $\mathsf{M}_2(\tilde\Xi)=i\mathsf{M}_2(\vec u)$ because $\mathsf{M}_2$ is $\mathbb{C}$-linear. The matrices $i\mathsf{M}_2(\vec e_x),i\mathsf{M}_2(\vec e_y),i\mathsf{M}_2(\vec e_z)$ are Hermitian, traceless and $i\mathsf{M}_2(\vec e_\mu)^{2}=I$, and they form a real basis of the traceless Hermitian matrices; hence $i\mathsf{M}_2(\vec u)=\sum_\mu u_\mu\,i\mathsf{M}_2(\vec e_\mu)$ is traceless Hermitian with square $(\vec u,\vec u)I$, and for $(\vec u,\vec u)=1$ the element $\tfrac12(I+i\mathsf{M}_2(\vec u))$ is a Hermitian idempotent of rank one because its trace is one. Conversely every rank-one Hermitian idempotent of $M_2(\mathbb{C})$ is $\tfrac12(I+H)$ with $H$ traceless Hermitian and $H^{2}=I$, and every such $H$ is $i\mathsf{M}_2(\vec u)$ for a real vector $\vec u$ with $(\vec u,\vec u)=1$.

Remark. The matrix image is the standard rank-one projection of a two-dimensional space, and the parametrisation by the real unit vectors $\vec u$ is the parametrisation of those projections by the traceless Hermitian involutions. The standard pair is $\mathsf{M}_2(\tilde\Pi_+)=E_{11}$ and $\mathsf{M}_2(\tilde\Pi_-)=E_{22}$.

The Operator Interpretation

An element idempotent for $\star$ has its right multiplication $R^{\star}_{\tilde\Pi}$ a projector, while its left multiplication is a projector only for $\tilde\Pi=0$; the elements whose right multiplication is a projector are the idempotents of the plain product, a family of complex dimension two that strictly contains the idempotents of the block.

The idempotents of the block are a class of elements, and they must not be confused with the projectors of the block, which are a class of operators. An element idempotent for the block gives a projector under its right multiplication, and the converse fails, because the right projectors are the idempotents of the plain product; on the left the only projector is the zero operator.

Corollary (idempotent elements and their operators). An element $\tilde\Pi$ idempotent for $\star$ has $R^{\star}_{\tilde\Pi}$ a projector, because $\tilde\Pi\star\tilde\Pi=\tilde\Pi$ gives $(R^{\star}_{\tilde\Pi})^{2}=R^{\star}_{\tilde\Pi}$. The left multiplication is not a projector unless $\tilde\Pi=0$. The elements whose right multiplication is a projector are the idempotents of the plain product, a family of complex dimension two, and they strictly contain the idempotents of the block.

Proof. The operator statement on the right is the right-composition law $R^{\star}_{\tilde Q_1}R^{\star}_{\tilde Q_2}=R^{\star}_{\tilde Q_1\tilde Q_2}$ read at $\tilde Q_1=\tilde Q_2=\tilde\Pi$ (The Specificities of the General Plain Sesqualgebra). The class of the projectors is the projector criterion of the same article: $R^{\star}_{\tilde Q_1}$ is a projector exactly when $\tilde Q_1$ is a plain idempotent, and the plain idempotents are the halved plain roots of $+1$, a complex quadric of which the block's idempotents are the Hermitian subfamily of real dimension two. On the left there is no composition law, and a conjugate-linear projector is the zero operator, so $L^{\star}_{\tilde\Pi}$ is a projector only for $\tilde\Pi=0$.

Remark (the half that the block keeps). On the element level the block keeps the Hermitian idempotents and drops the rest; on the operator level the right multiplications recover the whole plain idempotent set, read at the conjugated index. The gap between the two is the gap between the block's element theory and its right-operator theory, and it is the mirror of the gap that the general quaternionic product shows on its left side. The projectors themselves, their images and their kernels, are treated in the operator articles of the block.

Summary

The square of an element of the general plain sesqualgebra is the Hermitian element $\tilde\Xi\star\tilde\Xi=\tilde\Xi\tilde\Xi^{*}=(\lvert q_r\rvert^{2}+\lvert q_i\rvert^{2})e_t+2i(q_{r0}\vec v-q_{i0}\vec q+\vec q\times\vec v)$ in the real halves $\tilde\Xi=q_r+iq_i$, whose scalar part is the definite quantity $\sum_\mu\lvert\Xi_\mu\rvert^{2}$. The scalar part decides the central values: no negative real or non-real scalar has a root, the scalar part being non-negative, and plus one has a root, the root set being exactly the family $Aw$ over the unit complex scalars $A$ and the real unit quaternions $w$, that is $\mathrm{U}(1)\cdot S^{3}$, a single real four-parameter family containing $e_t$, $-e_t$ and the pure imaginary unit vectors $\tilde\Xi$. The Hermitian roots of $+1$ are $\pm e_t$ and the roots $\tilde\Xi=i\vec u$, and the halving map $\tilde\Xi\mapsto\tfrac12(e_t+\tilde\Xi)$ is a bijection from them onto the idempotents of the block, which are $0$, $e_t$ and the $\tfrac12(e_t+\tilde\Xi)$: an element is idempotent for the block exactly when it is a Hermitian idempotent of the algebra, each Hermitian root carries the complete orthogonal pair $\tilde\Pi_\pm(\tilde\Xi)=\tfrac12(e_t\pm\tilde\Xi)$, whose sum is $e_t$, whose difference is the root, whose two members multiply to $0$ in the block and in the algebra at once, and whose subscripts are the two eigenvalues of multiplication by the root, the complement of an idempotent is the idempotent $\tilde\Pi_-(\tilde\Xi)=\tilde\Pi_+(-\tilde\Xi)$ at the antipodal direction, and under the matrix model the family is the set of the rank-one Hermitian idempotents.

Summary of Notation

Further Reading

• Introduction to the General Plain Sesqualgebra, for the product, its class, its two slot involutions and the positive cone of the square that the reduction uses.
• Definitions for the Study of the Algebraic Structures over the Biquaternion $\mathbb{C}$ Space, for the relative definition of the root, the idempotent, the Hermitian idempotent and the Peirce decomposition.
• Norm and Invertibility in the General Plain Sesqualgebra, for the algebraic norm, the square and the definiteness of its scalar part.
• The Squares and the Positive Cone of the Biquaternion Sesqualgebra, for the positive cone that the square of the block generates.
• Square Roots of a General Element in the General Plain Sesqualgebra, for the general reduction of the root equation of which the roots of $+1$ are one central case.
• Square Roots of Zero and the Nilpotents of the General Plain Sesqualgebra, for the roots of $0$, which the same square excludes, and the nilpotents of index three.
• Square Roots of One and the Idempotents of the Biquaternion Algebra, for the parent's roots of $+1$ and the halving map, whose Hermitian half is the set computed here.
• Square Roots of One and the Idempotents of the Quaternion Algebra, for the idempotents of the associative product, whose Hermitian half the block keeps.
• Zero Divisors of the General Plain Sesqualgebra, for the criterion $C=0$ that no root of $+1$ satisfies and among which the idempotents sit.
• The Specificities of the General Plain Sesqualgebra, for the right-composition law and the projector criterion whose projector family is the whole plain idempotent set.
• The General Plain Sesqualgebra in the $2\times2$ Matrix Element Representation $M_2(\mathbb{C})$, for the model in which the idempotents correspond to the rank-one Hermitian idempotents $E_{11},E_{22}$ and their conjugates.