The Square, the Idempotents and the Jordan Inverse of the Symmetric Plain Algebra
Introduction
The symmetric plain algebra of $\mathbb{H}$ carries the Jordan product $\tilde q_1\bullet\tilde q_2=\tfrac12(\tilde q_1\tilde q_2+\tilde q_2\tilde q_1)$ of Introduction to the Symmetric Plain Algebra. It is the specialisation of the degree-two Common Theory of The Two General Products of an Algebra with Involution to the real quaternions.
This is not the algebra $\mathbb{H}$. The two algebras share the space, the addition, the real scalars and the unit, but $\bullet$ is a different multiplication and the projection onto the scalar line is not multiplicative. Throughout, the natural sign ${}^{\natural}$ is the quaternion conjugation, the bar is the identity because $\mathbb{R}$ has no nontrivial involution, and the star coincides with the natural sign and gives nothing new.
The Square Map
Definition. The square map of the block is
$$ Q:\tilde q\mapsto\tilde q\bullet\tilde q=\tilde q^2,\qquad \mathfrak{Q}(\tilde q)=\bigl(q_t^2-|\vec q|^2\bigr)e_t+2q_t\vec q . $$
It is the diagonal of the product; the polarisation $\mathfrak{Q}(\tilde q_1,\tilde q_2)=\tfrac12\bigl(\mathfrak{Q}(\tilde q_1+\tilde q_2)-\mathfrak{Q}(\tilde q_1)-\mathfrak{Q}(\tilde q_2)\bigr)$ is the product, $\mathfrak{Q}(\tilde q_1,\tilde q_2)=\tilde q_1\bullet\tilde q_2$.
Proposition (the relation to the norm). The square, the trace and the norm are tied by
$$ \mathrm{Sc}\,\tilde q^2=q_t^2-|\vec q|^2=2q_t^2-\langle\tilde q\rangle_{\natural},\qquad \langle\tilde q\rangle_{\natural}=q_t^{\,2}+|\vec q|^{\,2}, $$
and by the quadratic identity $\tilde q^2-T(\tilde q)\tilde q+\langle\tilde q\rangle_{\natural}e_t=0$. The square of the block is the plain square; it is not the quaternionic square, whose value is the central element $\tilde q^{\natural}\tilde q=\langle\tilde q\rangle_{\natural}e_t$, and the two agree only on the scalar line.
Proof. The square is displayed above; the scalar part is $q_t^2-|\vec q|^2$, the vector part $2q_t\vec q$. The plain square is quoted from Introduction to the Quaternion Algebra and the quaternionic square from Introduction to the General Quaternionic Algebra, below this article in this category; the quadratic identity is Norm and Invertibility in the Symmetric Plain Algebra.
Remark (the two general products on one square). The chapter carries two general products, the plain product $\tilde q_1\tilde q_2$ and the quaternionic product $\tilde q_1^{\natural}\tilde q_2$; the square of the block is the square of the first, and the symmetrisation of the second, The Symmetrised Quaternionic Product and the Hermitian Subspace, below this article in this category, has the central square $\langle\tilde q\rangle_{\natural}e_t$. The division of the real chapter is between these two squares, and the block's square is the non-central one.
The Idempotents and Their Degeneracy
Definition. An idempotent of the block is an element with $\tilde\pi\bullet\tilde\pi=\tilde\pi$.
Theorem. The idempotents of the block are exactly
$$ \boxed{\;\tilde\pi\in\{0,\ e_t\}\;}, $$
the two trivial idempotents; there is no nontrivial one.
Proof. The square is the plain square, so the equation is $\tilde\pi^2=\tilde\pi$. The vector part gives $(2\pi_t-1)\vec\pi=0$, so either $\vec\pi=0$ and the scalar part gives $\pi_t^2=\pi_t$, giving $\tilde\pi\in\{0,e_t\}$, or $\pi_t=\tfrac12$ and the scalar part gives $|\vec\pi|^2=-\tfrac14$, with no real solution.
Remark (the degeneracy in a division algebra). A nontrivial idempotent is built, in a split algebra, from a pure root of $+e_t$ by $\tilde\pi=\tfrac12(e_t+\vec q_a)$, and the reason the construction fails here is that the pure square is negative: $\vec q_a\bullet\vec q_a=-|\vec q_a|^2e_t$, so a pure element can be a root of $-e_t$ but never a root of $+e_t$. Over a division algebra with a negative definite pure part the construction never closes, and the idempotent class collapses to the two trivial elements; the collapse is the algebraic form of the absence of the split.
Corollary (the associated idempotents). The only idempotents associated with the zero divisors are the trivial ones: a nonzero pure element $\vec q_a$ is a zero divisor but the element $\tfrac12(e_t+\vec q_a)$ has square $\tfrac14(1-|\vec q_a|^2)e_t+\tfrac12\vec q_a$, which is not an idempotent for any real $\vec q_a$.
The Peirce Decomposition
Theorem (the cubic identity). The multiplication operator satisfies the cubic Peirce identity
$$ \boxed{\;2\,(L^{\bullet}_{\tilde\pi})^3-3\,(L^{\bullet}_{\tilde\pi})^2+L^{\bullet}_{\tilde\pi}=0\;} $$
at every idempotent $\tilde\pi$, so the spectrum is contained in $\{0,\tfrac12,1\}$.
Proof. The identity holds in every Jordan algebra for an idempotent, by the Peirce theory of Jordan Algebras; in the block it is checked directly from $L^{\bullet}_{e_t}=\mathrm{id}$ and $L^{\bullet}_{0}=0$.
Theorem (the Peirce decomposition). For the two idempotents the decomposition is trivial:
$$ J_1(e_t)=\mathbb{H},\quad J_{1/2}(e_t)=J_0(e_t)=0; \qquad J_0(0)=\mathbb{H},\quad J_1(0)=J_{1/2}(0)=0, $$
where $J_\lambda(\tilde\pi)=\{\tilde q : \tilde\pi\bullet\tilde q=\lambda\tilde q\}$.
Proof. $L^{\bullet}_{e_t}=\mathrm{id}$ has the single eigenvalue $1$ with multiplicity four; $L^{\bullet}_{0}=0$ has the single eigenvalue $0$ with multiplicity four.
Remark (the would-be shape). A nontrivial idempotent of a degree-two block of this kind would split the algebra into three Peirce spaces of dimensions $1,2,1$; since the block has only $0$ and $e_t$, the middle space never opens and the Peirce calculus is degenerate. This is the specific-theory reading of the degeneracy of the previous section, and it is the reason the block carries no projector calculus.
The Jordan Inverse
Theorem. An element is invertible exactly when $\langle\tilde q\rangle_{\natural}\neq0$, that is, when $\tilde q\neq0$, and the inverse is
$$ \boxed{\;\tilde q^{-1}=\frac{T(\tilde q)e_t-\tilde q}{\langle\tilde q\rangle_{\natural}}=\frac{2q_te_t-\tilde q}{\langle\tilde q\rangle_{\natural}}=\frac{\tilde q^{\natural}}{\langle\tilde q\rangle_{\natural}}\;}. $$
Proof. This is Norm and Invertibility in the Symmetric Plain Algebra; the inverse satisfies $\tilde q\bullet\tilde q^{-1}=e_t$ and $\tilde q^{\bullet2}\bullet\tilde q^{-1}=\tilde q$.
Proposition (the two-sidedness and the failure of multiplicativity). The inverse is two-sided and involutive, $(\tilde q^{-1})^{-1}=\tilde q$; but the inversion is not a homomorphism, and the invertible elements do not form a group, since $\vec e_x,\vec e_y$ are invertible and $\vec e_x\bullet \vec e_y=0$ has no inverse.
Proof. The two-sidedness is commutativity and the first defining equation; the failure of multiplicativity is the pair $\vec e_x,\vec e_y$.
The Quadratic Representation
Definition. The quadratic representation of the block is
$$ U_{\tilde q_1}=2\,(L^{\bullet}_{\tilde q_1})^2-L^{\bullet}_{\tilde q_1^{\,\bullet2}}, \qquad U_{\tilde q_1}(\tilde q_2)=2\,\tilde q_1\bullet(\tilde q_1\bullet\tilde q_2)-\tilde q_1^{\bullet2}\bullet\tilde q_2 . $$
Theorem. The quadratic representation is the plain sandwich operator and it is multiplicative in the norm:
$$ \boxed{\;U_{\tilde q_1}(\tilde q_2)=\tilde q_1\,\tilde q_2\,\tilde q_1,\qquad \langle U_{\tilde q_1}\tilde q_2\rangle_{\natural}=\langle\tilde q_1\rangle_{\natural}^2\langle\tilde q_2\rangle_{\natural}\;}. $$
Proof. Expanding $U_{\tilde q_1}(\tilde q_2)$ in the symmetrised plain product, $U_{\tilde q_1}(\tilde q_2)=\tfrac12\bigl(\tilde q_1(\tilde q_1\tilde q_2+\tilde q_2\tilde q_1)+(\tilde q_1\tilde q_2+\tilde q_2\tilde q_1)\tilde q_1\bigr)-\tfrac12(\tilde q_1^2\tilde q_2+\tilde q_2\tilde q_1^2)=\tilde q_1\tilde q_2\tilde q_1$ by the associativity of the plain product. The norm identity is the multiplicativity of $\langle\cdot\rangle_{\natural}$ applied twice, $\langle\tilde q_1\tilde q_2\tilde q_1\rangle_{\natural}=\langle\tilde q_1\rangle_{\natural}^2\langle\tilde q_2\rangle_{\natural}$.
Remark (the effect on the norm). For a nonzero $\tilde q$ the operator $U_{\tilde q}$ is invertible with inverse $U_{\tilde q^{-1}}$, and $U_{\tilde q^{-1}}=U_{\tilde q}^{-1}$; the norm identity is the degree-two form of the composite, and it is the quadratic representation statement of the degree-two Common Theory of The Two General Products of an Algebra with Involution read on the block.
Summary
The square map of the block is the plain square, $\mathfrak{Q}(\tilde q)=(q_t^2-|\vec q|^2)e_t+2q_t\vec q$, polarising to the product, with the scalar part $2q_t^2-\langle\tilde q\rangle_{\natural}$ and the quadratic identity $\tilde q^2-T(\tilde q)\tilde q+\langle\tilde q\rangle_{\natural}e_t=0$. The idempotents are exactly $0$ and $e_t$, because a nontrivial idempotent would need a pure root of $+e_t$ and the pure square is $-|\vec q_a|^2e_t$; the Peirce decomposition is trivial, $J_1(e_t)=\mathbb{H}$ and $J_0(0)=\mathbb{H}$ with the other spaces vanishing. The Jordan inverse is $\tilde q^{-1}=(T(\tilde q)e_t-\tilde q)/\langle\tilde q\rangle_{\natural}=\tilde q^{\natural}/\langle\tilde q\rangle_{\natural}$ for $\tilde q\neq0$, two-sided and involutive but not multiplicative, and the invertible elements do not form a group. The quadratic representation is $U_{\tilde q_1}=2(L^{\bullet}_{\tilde q_1})^2-L^{\bullet}_{\tilde q_1^{\,\bullet2}}$, equal to the sandwich $\tilde q_1\tilde q_2\tilde q_1$, and $\langle U_{\tilde q_1}\tilde q_2\rangle_{\natural}=\langle\tilde q_1\rangle_{\natural}^2\langle\tilde q_2\rangle_{\natural}$. Everything but the idempotents and the Peirce decomposition has the same shape over $\mathbb{C}$; there the idempotents are many and the Peirce calculus opens.
Summary of Notation
| Symbol | Meaning |
|---|---|
| $\mathfrak{Q}(\tilde q)=\tilde q\bullet\tilde q=\tilde q^2$ | the square map |
| $\vec q_a\bullet\vec q_a=-\lvert\vec q_a\rvert^2e_t$ | the square of a pure element |
| $\{0,e_t\}$ | the idempotents |
| $J_\lambda(\tilde\pi)$ | the Peirce spaces, trivial at $0$ and $e_t$ |
| $\tilde q^{-1}=\tilde q^{\natural}/\langle\tilde q\rangle_{\natural}$ | the Jordan inverse |
| $U_{\tilde q_1}=2(L^{\bullet}_{\tilde q_1})^2-L^{\bullet}_{\tilde q_1^{\,\bullet2}}$ | the quadratic representation, $=\tilde q_1\tilde q_2\tilde q_1$ |
| $\langle U_{\tilde q_1}\tilde q_2\rangle_{\natural}=\langle\tilde q_1\rangle_{\natural}^2\langle\tilde q_2\rangle_{\natural}$ | the norm identity |
Further Reading
- Introduction to the Symmetric Plain Algebra, for the product and the square.
- Norm and Invertibility in the Symmetric Plain Algebra, for the norm, the trace and the inverse.
- The Specificities of the Symmetric Plain Algebra, for the same objects read as the block's properties.
- The Multiplication Operators of the Symmetric Plain Algebra, below this article in this category, for the operator whose square and cube enter the representation.
- Jordan Algebras, for the Peirce decomposition and the cubic identity.
- The Two General Products of an Algebra with Involution, for the degree-two Common Theory specialised here.