The Signed Adjoint of the Left Multiplication on a Jordan Algebra

Introduction

The signed left multiplication of a Jordan algebra with a grade involution is the one-sided operator obtained from the left multiplication by precomposition with the involution; intrinsically it is

$$ \ell^{\alpha}_a = L_a\circ\alpha , \qquad \ell^{\alpha}_a(x) = a\bullet\alpha(x) , $$

the Jordan form of the signed one-sided action of The Signed Left Multiplication on a Jordan Algebra, whose associative model is $\ell^{\alpha}_a(x) = a\,\alpha(x)$ on a special algebra $J = A^+$. The present article computes the adjoint of this operator with respect to the natural pairing of the category, the trace form $T(x,y) = \operatorname{tr}(L_{x\bullet y})$ of The Adjoint of the Left Multiplication on a Jordan Algebra, and relates it to the signed sandwich of The Signed Adjoint Sandwich on a Jordan Algebra.

The trace form is associative, $T(x\bullet y,z) = T(x,y\bullet z)$, so the left multiplication is self-adjoint, $L_a^{\dagger} = L_a$, and the grade involution is an isometry, $\alpha^{\dagger} = \alpha$; the adjoint of the composite is therefore the composite of the adjoints in the reverse order, and since $\alpha$ is an automorphism with $\alpha L_a\alpha^{-1} = L_{\alpha(a)}$,

$$ \bigl(\ell^{\alpha}_a\bigr)^{\dagger} = \bigl(L_a\circ\alpha\bigr)^{\dagger} = \alpha\circ L_a = L_{\alpha(a)}\circ\alpha = \ell^{\alpha}_{\alpha(a)} . $$

The adjoint of the signed left multiplication is thus the signed left multiplication by the image of the parameter under the grade involution; it is self-adjoint exactly when $\alpha(a) = a$, and it is unitary exactly when the unsigned left multiplication is an involution, $L_a^2 = \mathrm{id}$, that is, when $a$ is a symmetry. The signed sandwich is the product of a signed and an unsigned one-sided action, $S^{\alpha}_{a,b} = \rho_b\ell^{\alpha}_a$, and the adjoint of that product is computed from the adjoints of its factors, recovering the rule of the signed adjoint sandwich.

The article assumes The Signed Left Multiplication on a Jordan Algebra for the signed one-sided action, its homogeneity and the fixed elements; The Signed Adjoint Sandwich on a Jordan Algebra for the signed sandwich and its adjoint; The Adjoint of the Left Multiplication on a Jordan Algebra for the trace form, its associativity and the self-adjointness of $L_a$ and $U_a$; The Left and Right Multiplication Operators on a Jordan Algebra for the quadratic representation and the symmetries; and The Operators on an Algebra for the ambient endomorphism algebra. Throughout, $J = A^+$ is a special unital Jordan algebra over a commutative ring $R$ in which $2$ is invertible, $\alpha$ is a grade involution of $A$, $L_a$ is the Jordan left multiplication, $\ell^{\alpha}_a = L_a\circ\alpha$ is the signed left multiplication, $T$ is the trace form and ${}^{\dagger}$ its adjoint; no norm, form, distance or geometric reflection occurs.

The Signed Left Multiplication and Its Adjoint

The Definition and the Unsigned Case

Definition. The signed left multiplication by $a$ is $\ell^{\alpha}_a = L_a\circ\alpha$, $\ell^{\alpha}_a(x) = a\bullet\alpha(x)$; for $\alpha = \mathrm{id}$ it is the ordinary left multiplication $L_a$.

Theorem (the unsigned adjoint). The unsigned left multiplication is self-adjoint, $L_a^{\dagger} = L_a$, and the adjoint of the signed left multiplication is the signed left multiplication at the image of the parameter:

$$ \bigl(\ell^{\alpha}_a\bigr)^{\dagger} = \ell^{\alpha}_{\alpha(a)} . $$

Proof. The associativity of the trace form says $T(L_ax,y) = T(x,L_ay)$, so $L_a^{\dagger} = L_a$; the grade involution is an isometry, $T(\alpha x,y) = T(x,\alpha y)$, so $\alpha^{\dagger} = \alpha$. Hence $(\ell^{\alpha}_a)^{\dagger} = (L_a\circ\alpha)^{\dagger} = \alpha^{\dagger}\circ L_a^{\dagger} = \alpha\circ L_a$. Since $\alpha$ is an automorphism, $\alpha L_a = L_{\alpha(a)}\alpha$, so $\alpha\circ L_a = L_{\alpha(a)}\circ\alpha = \ell^{\alpha}_{\alpha(a)}$. $\square$

Self-Adjointness

Proposition. The signed left multiplication is self-adjoint exactly when its parameter is fixed by the grade involution:

$$ \bigl(\ell^{\alpha}_a\bigr)^{\dagger} = \ell^{\alpha}_a \iff \alpha(a) = a . $$

Proof. The adjoint is $\ell^{\alpha}_{\alpha(a)}$; the two are equal exactly when $L_{\alpha(a)} = L_a$, which for a non-degenerate trace form holds exactly when $\alpha(a) = a$. $\square$

Corollary. The signed left multiplication by an even element is self-adjoint and coincides with the unsigned left multiplication, $\ell^{\alpha}_a = L_a$ for $a$ even; for an odd element $\ell^{\alpha}_a = L_a\circ\alpha$ is the negative of the ordinary action conjugated by $\alpha$ on the argument, and it is not self-adjoint.

Unitarity

Theorem. The signed left multiplication is unitary with respect to the adjoint exactly when the unsigned left multiplication is an involution:

$$ \bigl(\ell^{\alpha}_a\bigr)^{\dagger}\ell^{\alpha}_a = \ell^{\alpha}_a\bigl(\ell^{\alpha}_a\bigr)^{\dagger} = \mathrm{id} \iff L_a^2 = \mathrm{id} , $$

that is, exactly when $a$ is a symmetry.

Proof. $\ell^{\alpha\dagger}_a\ell^{\alpha}_a = \ell^{\alpha}_{\alpha(a)}\ell^{\alpha}_a = L_{\alpha(a)}\alpha L_a\alpha = L_{\alpha(a)}L_{\alpha(a)}\alpha^2 = L_{\alpha(a)}^2$, and similarly $\ell^{\alpha}_a\ell^{\alpha\dagger}_a = L_a^2$; the two are the identity exactly when $L_a^2 = L_{\alpha(a)}^2 = \mathrm{id}$, which is the symmetry condition. $\square$

Relation to the Signed Sandwich

Theorem. The signed sandwich is the product of a signed and an unsigned one-sided action, $S^{\alpha}_{a,b} = \rho_b\ell^{\alpha}_a = \ell_a\rho^{\alpha}_b$, and its adjoint is

$$ \bigl(S^{\alpha}_{a,b}\bigr)^{\dagger} = \ell^{\alpha\dagger}_a\rho_b^{\dagger} = \ell^{\alpha}_{\alpha(a)}\rho_{\alpha(b)} , $$

which is the signed sandwich $\Sigma^{\alpha}_{\alpha(a),\alpha(b)}$ after the symmetrisation, in agreement with The Signed Adjoint Sandwich on a Jordan Algebra.

Proof. The anti-multiplicativity of the adjoint gives $(S^{\alpha}_{a,b})^{\dagger} = (\rho_b\ell^{\alpha}_a)^{\dagger} = \ell^{\alpha\dagger}_a\rho_b^{\dagger} = \ell^{\alpha}_{\alpha(a)}L_{\alpha(b)}$; the symmetrisation of $\rho_b\ell^{\alpha}_a$ over the outer parameter is the signed sandwich $U_{a,b}\circ\alpha$, and its adjoint is $U_{\alpha(a),\alpha(b)}\circ\alpha$ by The Signed Adjoint Sandwich on a Jordan Algebra. $\square$

Corollary. The adjoint of the one-sided layer and the adjoint of the two-sided layer are the same rule at the images of the parameters; the signed left multiplication is thus the one-sided case of the signed sandwich adjoint, exactly as in the associative case of The Signed Adjoint of the Left Multiplication on a Ring.

Examples

Example (the identity grade involution). For $\alpha = \mathrm{id}$ the signed left multiplication is the unsigned one, $\ell^{\alpha}_a = L_a$, which is self-adjoint for every $a$; the unitarity condition is $L_a^2 = \mathrm{id}$, the symmetry condition, and it holds for the symmetries.

Example (the transpose involution). Let $J = H_n(F)$ with the transpose grade involution on the ambient matrix algebra; then $\ell^{\alpha}_a(x) = a\bullet x^{\mathsf{T}}$, and its adjoint is $\ell^{\alpha}_{a^{\mathsf{T}}}$. The operator is self-adjoint exactly when $a$ is symmetric, and unitary exactly when $a$ is a symmetry of the Jordan algebra.

Example (the odd element). For an odd element $a$ the signed left multiplication is $\ell^{\alpha}_a = L_a\circ\alpha$, which is not self-adjoint: $\ell^{\alpha\dagger}_a = \ell^{\alpha}_{\alpha(a)} = \ell^{\alpha}_{-a} = -\ell^{\alpha}_a$, so an odd parameter gives a skew-adjoint signed left multiplication.

Summary

The signed left multiplication on a Jordan algebra is $\ell^{\alpha}_a = L_a\circ\alpha$, $\ell^{\alpha}_a(x) = a\bullet\alpha(x)$; with respect to the trace form its adjoint is the signed left multiplication at the image of the parameter,

$$ \bigl(\ell^{\alpha}_a\bigr)^{\dagger} = \ell^{\alpha}_{\alpha(a)} , $$

because the left multiplication is self-adjoint, $L_a^{\dagger} = L_a$, and the grade involution is an isometry, $\alpha^{\dagger} = \alpha$. It is self-adjoint exactly when $\alpha(a) = a$, and unitary exactly when $L_a^2 = \mathrm{id}$, that is, when $a$ is a symmetry. The signed sandwich $S^{\alpha}_{a,b} = \rho_b\ell^{\alpha}_a$ has the adjoint computed from the adjoints of its factors, and its symmetrisation recovers $\Sigma^{\alpha}_{\alpha(a),\alpha(b)}$ of The Signed Adjoint Sandwich on a Jordan Algebra; the signed left multiplication is the one-sided case of that rule. The identity involution, the transpose involution and the odd element are the worked examples. No norm, form, distance or geometric reflection occurs.

Summary of Notation

Symbol	Meaning
$L_a(x) = a\bullet x$	Jordan left multiplication
$\ell^{\alpha}_a = L_a\circ\alpha$	Signed left multiplication
$\ell^{\alpha}_a(x) = a\bullet\alpha(x)$	Jordan form; associative form $a\alpha(x)$
$(\ell^{\alpha}_a)^{\dagger} = \ell^{\alpha}_{\alpha(a)}$	Adjoint of the signed left multiplication
$\alpha(a) = a$	Self-adjointness condition
$L_a^2 = \mathrm{id}$	Unitarity condition (symmetry)
$S^{\alpha}_{a,b} = \rho_b\ell^{\alpha}_a$	Signed sandwich as a product
$(S^{\alpha}_{a,b})^{\dagger} = \ell^{\alpha}_{\alpha(a)}\rho_{\alpha(b)}$	Adjoint of the sandwich from the factors