The Jordan Multiplication Operators

Introduction

A Jordan algebra is a commutative algebra $J$ whose product, written $\bullet$, satisfies the Jordan identity in place of associativity. As in every algebra, each element $a$ carries the operator of multiplication by it,

$$ L_a : J \longrightarrow J, \qquad L_a(x) = a \bullet x , $$

and this family of operators is the subject of the present article. The family is linear in $a$ and contains the identity when $J$ is unital, but it is not multiplicative in $a$: the failure of associativity shows up as the non-commutation of $L_a$ and $L_b$, and the Jordan identity is exactly the statement that each $L_a$ commutes with the multiplication by its own square,

$$ [L_a, L_{a^2}] = 0 . $$

The article assembles the operators generated by the $L_a$, the multiplication algebra $\operatorname{Mult}(J)$, and inside it the quadratic representation $U_a = 2L_a^2 - L_{a^2}$ and its polarisation $U_{a,b}$. The quadratic representation is the two-sided analogue of a one-sided multiplication, and it satisfies the fundamental formula $U_{U_a b} = U_a U_b U_a$, which is the reason the multiplication algebra is the natural home of the structure group of $J$. The commutators $[L_a, L_b]$ are the inner derivations, and the article records their place in the derivation algebra of Jordan Algebras. The trace form $T(x,y) = \operatorname{tr}(L_{x \bullet y})$, defined for a Jordan algebra free of finite rank, is shown to be symmetric and associative, so that every multiplication operator is self-adjoint for it.

The article assumes the theory of Jordan Algebras — the axioms, the symmetrisation $A^+$ of an associative algebra, power associativity, the Peirce decomposition and the trace form — and adds nothing to it. The non-associative product is treated nowhere else here; the Lie counterpart, where the multiplication operators are the adjoint operators $\operatorname{ad}_a$, is The Adjoint Action of a Lie Algebra, and the analogue for an associative algebra is Left and Right Multiplication in a Ring. Throughout, $R$ is a commutative ring with identity $1 \neq 0$ in which $2$ is invertible whenever a coefficient $\tfrac12$ is written, and $(J, \bullet)$ is a Jordan $R$-algebra; the unital cases are marked.

The Multiplication Operators

Definition and linearity

Definition. For $a \in J$ the multiplication operator by $a$ is the $R$-linear map

$$ L_a : J \longrightarrow J, \qquad L_a(x) = a \bullet x . $$

The assignment is $R$-linear in the subscript, and it is injective when $J$ is unital, because $L_a = 0$ forces $a = L_a(1) = 0$. The set $L(J) = \{L_a : a \in J\}$ is an $R$-submodule of $\operatorname{End}_R(J)$ isomorphic to $J$, but it is not a subalgebra: the product of two multiplication operators is generally not a multiplication operator.

Proposition. For all $a, b \in J$ and $r \in R$,

$$ L_{a+b} = L_a + L_b, \qquad L_{ra} = rL_a, \qquad L_1 = \mathrm{id}_J \text{ when } J \text{ is unital}. $$

If $J = A^+$ is the symmetrisation of an associative algebra $A$ with product $x \bullet y = \tfrac12(xy+yx)$, then

$$ L_a = \tfrac12\bigl(\lambda_a + \rho_a\bigr), $$

where $\lambda_a(x) = ax$ and $\rho_a(x) = xa$ are the one-sided multiplications of $A$.

Proof. The first two identities are the bilinearity of $\bullet$; the unital case is $1 \bullet x = x$. For the special case, $L_a(x) = \tfrac12(ax+xa) = \tfrac12(\lambda_a+\rho_a)(x)$. $\square$

Proposition (the reflection of a non-multiplicative family). The identity $L_aL_b = L_{a\bullet b}$ holds for all $a, b$ exactly when $J$ is associative; for a Jordan algebra that is not associative it fails, and the failure is measured by the operator

$$ [L_a, L_b] = L_aL_b - L_bL_a . $$

Proof. $L_aL_b(x) = a \bullet (b \bullet x)$ and $L_{a\bullet b}(x) = (a\bullet b)\bullet x$; these agree for all $x$ exactly when $\bullet$ is associative, and a power-associative algebra is associative when $L_aL_b = L_{a\bullet b}$ for all pairs. The definition of the commutator is immediate. $\square$

Thus the family $L(J)$ is closed under addition and under scalar multiplication but not under composition, and the obstruction to composition is the commutator, which is the subject of the next section.

The Jordan Identity in Operator Form

Proposition. For every $a \in J$, the Jordan identity $a^2 \bullet (a \bullet x) = (a^2 \bullet x) \bullet a$ is equivalent to

$$ [L_a, L_{a^2}] = 0 . $$

Polarising gives the linearised form, and, when $2$ and $3$ are invertible, the cyclic form

$$ [L_a, L_{b \bullet c}] + [L_b, L_{c \bullet a}] + [L_c, L_{a \bullet b}] = 0 \qquad \text{for all } a, b, c \in J . $$

Proof. The equivalence is the identity $(a^2 \bullet (a \bullet x)) - ((a^2 \bullet x)\bullet a) = 0$ for all $x$. The polarisations are those of Jordan Algebras: expand $[L_{a+tb}, L_{(a+tb)^2}] = 0$ over $R[t]$ and collect the coefficient of $t$, then expand with three parameters. The cyclic form at $c = a$ is the linearised form, and conversely the linearised form at $b = a$ reads $3[L_a, L_{a^2}] = 0$. $\square$

The content of the Jordan identity is therefore: multiplication by $a$ commutes with multiplication by $a^2$, and this survives linearisation. It does not make the family commutative — $[L_a,L_b]$ is generally nonzero — but it makes the subalgebra generated by the powers of one element commutative and associative, which is power associativity.

The Multiplication Algebra

Definition

Definition. The multiplication algebra of $J$ is the $R$-subalgebra

$$ \operatorname{Mult}(J) = \langle L_a : a \in J \rangle \subseteq \operatorname{End}_R(J) $$

generated by the multiplication operators and the identity. Its elements are the multiplication operators of $J$, and a Jordan algebra is called associative when $\operatorname{Mult}(J) = L(J)$, that is, when every generated operator is again a single multiplication.

The multiplication algebra is the smallest associative algebra in which the Jordan product can be read off: $a \bullet b = L_a(b) = L_aL_b(1)$ when $J$ is unital, so the product is recovered from $\operatorname{Mult}(J)$ by evaluation at the unit. It is generally non-commutative. For a special Jordan algebra $J = A^+$ the multiplication algebra is the $R$-span of the two-sided operators $x \mapsto pxq$, since $\lambda_p\rho_q$ generates it together with the one-sided multiplications; the identification is made below.

The Quadratic Representation

Definition. For $a, b \in J$ the quadratic representation is the operator

$$ U_{a,b} = L_aL_b + L_bL_a - L_{a\bullet b} \in \operatorname{Mult}(J), $$

and the quadratic representation of an element $a$ is $U_a = U_{a,a}$; with the coefficient $\tfrac12$ the polarised form is

$$ U_{a,b}(x) = \tfrac12\bigl(2a\bullet(b\bullet x) + 2b\bullet(a\bullet x) - 2(a\bullet b)\bullet x\bigr) = a\bullet(b\bullet x) + b\bullet(a\bullet x) - (a\bullet b)\bullet x . $$

The operator $U_{a,b}$ is symmetric in its parameters, $U_{a,b} = U_{b,a}$, and $R$-bilinear; $U_a$ is quadratic in $a$, and polarisation recovers $U_{a,b}$ from the diagonal,

$$ U_{a,b} = \tfrac12\bigl(U_{a+b} - U_a - U_b\bigr), \qquad U_a = U_{a,a} = 2L_a^2 - L_{a^2} . $$

Remark. The operators $U_{a,b}$ and $U_a$ are the quadratic representations. They are the two-sided operators generated by the one-sided multiplications, and they are defined here only as elements of the multiplication algebra. Their theory — the special form $U_{a,b}(x) = \tfrac12(axb+bxa)$, $U_a(x) = axa$, the fundamental formula $U_{U_ab} = U_aU_bU_a$, invertibility, the action on the Peirce spaces and the inner structure group — is the subject of The Left and Right Multiplication Operators on a Jordan Algebra, next in this category, and is not developed here.

The Symmetrised Multiplication Algebra

$\operatorname{Mult}(J)$ is an associative algebra, and as such it carries the symmetrised product $\{S, T\} = ST + TS$; the symmetrised algebra $\operatorname{Mult}(J)^+$ is a Jordan algebra, and it is the Jordan algebra generated by the multiplication operators $L_a$. Its elements include the $L_a$, the $U_{a,b} = \{L_a, L_b\} - L_{a\bullet b}$, and every symmetric product of the generators. The $L_a$ themselves generate $\operatorname{Mult}(J)^+$ under $\{$ , $\}$ together with $R$-linear combinations, and the identity is the unit. That the product closes inside a Jordan algebra is automatic: the symmetrisation of an associative algebra is always a Jordan algebra, by Jordan Algebras.

Proposition. The map $a \mapsto L_a$ is an injective $R$-linear map $J \to \operatorname{Mult}(J)^+$ whose image generates $\operatorname{Mult}(J)^+$; it is a Jordan homomorphism exactly when $J$ is associative.

Inner Derivations

The Commutator of Two Multiplications

Proposition. For all $a, b \in J$ the operator $[L_a, L_b] = L_aL_b - L_bL_a$ is a derivation of $J$:

$$ [L_a,L_b](x \bullet y) = [L_a,L_b](x) \bullet y + x \bullet [L_a,L_b](y) . $$

Proof. The linearised Jordan identity, with $c = x$ and the roles permuted, gives

$$ [L_a, L_b]L_x = L_{[L_a,L_b]x}, $$

because $[L_a,L_{b\bullet x}] + [L_b,L_{x\bullet a}] + [L_x,L_{a\bullet b}] = 0$ rearranges to $L_{[L_a,L_b]x}$. Applying both sides to $y$ gives the derivation identity. $\square$

The derivations of $J$ form a Lie algebra $\operatorname{Der}(J)$ under the commutator, as in Jordan Algebras. The span of the operators $[L_a,L_b]$ is the ideal of inner derivations, written $\operatorname{inn}(J)$; it is an ideal of $\operatorname{Der}(J)$ because $[\delta,[L_a,L_b]] = [[\delta,L_a],L_b] + [L_a,[\delta,L_b]]$ and $[\delta,L_a] = L_{\delta a}$ is again a multiplication operator. The inner derivations are the infinitesimal automorphisms of $J$ that are visible inside $\operatorname{Mult}(J)$, and the quotient $\operatorname{Der}(J)/\operatorname{inn}(J)$ is the Lie algebra of the outer automorphism group of $J$.

Example. For $J = A^+$ and the halved product, $[L_a,L_b] = \tfrac14[\lambda_a+\rho_a, \lambda_b+\rho_b] = \tfrac14(\lambda_{[a,b]} - \rho_{[a,b]})$; on $A$ it acts by $x \mapsto \tfrac14([a,b]x - x[a,b])$, the inner derivation of $A$ by $\tfrac14[a,b]$, symmetrised. Every inner derivation of the special Jordan algebra therefore comes from an inner derivation of $A$.

The Trace Form

Definition. Let $J$ be free of finite rank as an $R$-module. The trace form is

$$ T : J \times J \longrightarrow R, \qquad T(x, y) = \operatorname{tr}(L_{x \bullet y}) . $$

By Jordan Algebras it is symmetric and $R$-bilinear, and every derivation of $J$ is skew for it, $T(\delta x, y) + T(x, \delta y) = 0$.

Theorem. The trace form is associative,

$$ T(x \bullet y, z) = T(x, y \bullet z) \qquad \text{for all } x, y, z \in J , $$

and consequently every multiplication operator is self-adjoint for it,

$$ T(L_a x, y) = T(x, L_a y). $$

Proof (special case). For $J = A^+$ with the halved product one computes $\operatorname{tr}(L_c) = n\operatorname{tr}(c)$ for every $c \in A$, where $n = \dim_R A$ and $\operatorname{tr}$ on the right is the matrix trace; hence

$$ T(x,y) = \operatorname{tr}(L_{x\bullet y}) = n\operatorname{tr}(x \bullet y) = \tfrac n2\operatorname{tr}(xy + yx) = n\operatorname{tr}(xy). $$

Then $T(x\bullet y, z) = \tfrac n2\operatorname{tr}\bigl((xy+yx)z\bigr)$ and $T(x, y\bullet z) = \tfrac n2\operatorname{tr}\bigl(x(yz+zy)\bigr)$, and the two agree because $\operatorname{tr}(yxz) = \operatorname{tr}(xzy)$. The self-adjointness follows from associativity and symmetry:

$$ T(L_a x, y) = T(a \bullet x, y) = T(x, a \bullet y) = T(x, L_a y). \qquad \square $$

Remark. The associativity of the trace form is the statement that the trace form of a Jordan algebra is associative in the sense of Jordan theory; for a general Jordan algebra it follows from the linearised Jordan identity exactly as the special case above follows from associativity of $A$. The identity $T(\delta x, y) = -T(x, \delta y)$ for derivations and the identity $T(L_a x, y) = T(x, L_a y)$ are then two readings of the same associativity, with $\delta$ skew and $L_a$ symmetric because a derivation reverses one factor and a multiplication does not.

Corollary. The trace form is invariant under the inner structure group in the sense $T(U_a x, U_a y) = T(x, U_a U_a y)$, which follows from the self-adjointness of $U_a$; since $U_a$ is self-adjoint and $U_a^2 = U_{a^2}$ up to the fundamental formula, this is a statement about the quadratic representations and not about the operator $U_a$ being orthogonal. An automorphism of $J$ that lies in the multiplication algebra therefore preserves $T$ exactly when its own action on the coefficients does, and the invariance statements are algebraic: no positivity, no signature and no norm of the form occurs.

Worked Examples

The Simplest Jordan Algebra

Let $J = R^2$ with the coordinatewise product and the two standard idempotents $e_1 = (1,0)$, $e_2 = (0,1)$. Then $\bullet$ is associative, $L_a$ is the diagonal matrix of the coordinates of $a$,

$$ L_{(u,v)} = \begin{pmatrix} u & 0 \\ 0 & v \end{pmatrix}, \qquad U_{(u,v)} = \begin{pmatrix} u^2 & 0 \\ 0 & v^2 \end{pmatrix} = L_{(u^2,v^2)} , $$

and $\operatorname{Mult}(J) = L(J)$ is the diagonal algebra, of dimension two. The commutator $[L_a,L_b]$ vanishes: the algebra is associative, and $U_a = L_{a^2}$ is itself a multiplication. The trace form is not the standard form: $T((u,v),(u',v')) = \operatorname{tr}(L_{uu'+vv'}) = 2(uu'+vv')$, the factor $2$ being the rank.

A Spin Factor

Let $J = JSpin_2 = R \oplus R^2$ with the product $(\alpha, v) \bullet (\beta, w) = (\alpha\beta + B(v,w),\ \alpha w + \beta v)$ of Jordan Algebras, where $B$ is the standard form on $R^2$ with orthonormal basis $e_1, e_2$. In the basis $1, e_1, e_2$ the multiplication operators by the two basis vectors are

$$ L_{e_1} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad L_{e_2} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}, $$

both symmetric, and the quadratic representation of $e_1$, which satisfies $e_1 \bullet e_1 = 1$, is

$$ U_{e_1} = 2L_{e_1}^2 - L_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}, $$

the reflection in the line $Re_1$ of the vector plane together with the identity on the scalar line; the same computation for $e_2$ gives the reflection in $Re_2$. This example shows the general feature that $L_a$ need not be invertible or diagonalisable, and that $U_a$, not $L_a$, is the operator of order two attached to a symmetry $a$ with $a \bullet a = 1$.

Summary

For a Jordan algebra $J$, the multiplication operators $L_a(x) = a\bullet x$ are linear in $a$ and injective when $J$ is unital, but they are not multiplicative: $L_aL_b = L_{a\bullet b}$ holds exactly for the associative algebras. The Jordan identity is $[L_a, L_{a^2}] = 0$, with the linearised and cyclic forms obtained by polarisation. The operators generated by the $L_a$ form the multiplication algebra $\operatorname{Mult}(J) \subseteq \operatorname{End}_R(J)$, in which the quadratic representation $U_{a,b} = L_aL_b + L_bL_a - L_{a\bullet b}$ and its diagonal $U_a = 2L_a^2 - L_{a^2}$ live; in the special case $J = A^+$ they are $U_{a,b}(x) = \tfrac12(axb+bxa)$ and $U_a(x) = axa$, with the theory developed in The Left and Right Multiplication Operators on a Jordan Algebra. The commutators $[L_a,L_b]$ are the inner derivations, forming the ideal $\operatorname{inn}(J)$ of $\operatorname{Der}(J)$. The symmetrisation $\operatorname{Mult}(J)^+$ is the Jordan algebra generated by the multiplication operators, and the trace form $T(x,y) = \operatorname{tr}(L_{x\bullet y})$ is symmetric, associative, and makes every multiplication operator self-adjoint.

Summary of Notation

Symbol	Meaning
$R$	Commutative ring with identity, $2$ invertible where $\tfrac12$ is used
$(J, \bullet)$	Jordan algebra
$L_a$, $L_a(x) = a \bullet x$	Multiplication operator by $a$
$L(J)$	Image of $L$, an $R$-submodule isomorphic to $J$
$[L_a, L_b]$	Commutator of two multiplications; an inner derivation
$\operatorname{Mult}(J) = \langle L_a \rangle$	Multiplication algebra, generated by the $L_a$
$U_{a,b} = L_aL_b + L_bL_a - L_{a\bullet b}$	Quadratic representation (polarised)
$U_a = 2L_a^2 - L_{a^2}$	Quadratic representation of $a$
$U_{U_a b} = U_aU_bU_a$	Fundamental formula
$\operatorname{Mult}(J)^+$	Jordan algebra generated by the multiplication operators
$\operatorname{Der}(J)$, $\operatorname{inn}(J)$	Derivations; inner derivations $\operatorname{span}\{[L_a,L_b]\}$
$T(x,y) = \operatorname{tr}(L_{x\bullet y})$	Trace form; symmetric, associative
$\lambda_a$, $\rho_a$	Left and right multiplication in an associative algebra
$A^+$	Symmetrisation of an associative algebra