The Signed Adjoint of the Left Multiplication on a Module over an Algebra

Introduction

The signed left multiplication is the one-sided operator $L^{\alpha}_a(m)=a\,\alpha(m)$ that carries the parity sign of the grading. This article computes its adjoint with respect to an $\alpha$-invariant pairing; the answer is the signed left multiplication by the parameter transformed by the composite $\beta=\alpha\sigma$, $(L^{\alpha}_a)^{*}=L^{\alpha}_{\beta(a)}$. It is the one-sided specialisation of the signed adjoint sandwich and the companion of the unsigned rule $(L_a)^{*}=L_{\sigma(a)}$.

The article is the sixth of the * Operator Theory group of this category. It assumes the signed left multiplication and its elementary properties from The Signed Left Multiplication on a Module over an Algebra, the signed adjoint sandwich from The Signed Adjoint of the Sandwich on a Bimodule over an Algebra, and the pairing and adjoint of The Adjoint of a Module Homomorphism; the unsigned one-sided adjoint is The Adjoint of a Module Homomorphism, and the reflection case is The Signed Adjoint of the Reflection on a Bimodule over an Algebra. The article stays inside Part I: no distance, norm, form with a norm, positivity, topology or limit. Throughout, $R$ is a commutative ring with $1 \neq 0$ in which $2$ is invertible, $(A,\sigma)$ is an involutive $R$-algebra, $\alpha$ is a grade involution of $A$ commuting with $\sigma$, $\beta=\alpha\sigma$, $M$ is a graded left $A$-module with an $\alpha$-invariant balanced $\sigma$-sesquilinear pairing, $L_a(m)=am$ is the unsigned left multiplication, and $L^{\alpha}_a=L_a\circ\alpha$, $L^{\alpha}_a(m)=a\,\alpha(m)$.

The Signed Left Multiplication and Its Adjoint

The one-sided adjoint

Theorem. For every $a \in A$ the adjoint of the signed left multiplication is the signed left multiplication by the transformed parameter:

$$ \bigl(L^{\alpha}_a\bigr)^{*}=L^{\alpha}_{\beta(a)}, \qquad \beta=\alpha\sigma . $$

Proof. The signed left multiplication is the signed sandwich with right parameter $1$, $L^{\alpha}_a=S^{\alpha}_{a,1}$. By The Signed Adjoint of the Sandwich on a Bimodule over an Algebra, $\bigl(S^{\alpha}_{a,1}\bigr)^{*}=S^{\alpha}_{\beta(a),\beta(1)}$, and $\beta(1)=1$ because $\beta$ is an involution of the algebra. Hence the adjoint is $S^{\alpha}_{\beta(a),1}=L^{\alpha}_{\beta(a)}$. $\square$

Proof (direct). Write $L^{\alpha}_a=L_a\circ\alpha$, so by anti-multiplicativity $\bigl(L^{\alpha}_a\bigr)^{*}=\alpha^{*}(L_a)^{*}$. The pairing is $\alpha$-invariant, so $\alpha^{*}=\alpha$, and The Adjoint of a Sandwich on a Bimodule over an Algebra gives $(L_a)^{*}=L_{\sigma(a)}$. Hence $\bigl(L^{\alpha}_a\bigr)^{*}=\alpha L_{\sigma(a)}=L^{\alpha}_{\alpha\sigma(a)}$, using the identity $\alpha L_b=L^{\alpha}_{\alpha(b)}$ of The Signed Left Multiplication on a Module over an Algebra. $\square$

The parameter transforms by the composite $\beta=\alpha\sigma$, exactly as for the general signed sandwich: the grading and the involution both act, and only their product is visible.

Relation to the unsigned adjoint and to the grading

Proposition. The adjoint of the signed left multiplication and the adjoint of the unsigned left multiplication are related by the grade involution:

$$ \bigl(L^{\alpha}_a\bigr)^{*}=\alpha\,(L_{\sigma(a)})^{*}=\alpha\,L_{\sigma(a)}, \qquad \bigl(L_a\bigr)^{*}=L_{\sigma(a)} . $$

Proof. The first is the direct computation in the proof of the theorem; the second is the unsigned rule. $\square$

Corollary (parity). The signed left multiplications are odd for the $\mathbb{Z}/2$-grading of the sandwich monoid, and the adjoint sends an odd operator to an odd operator: $\bigl(L^{\alpha}_a\bigr)^{*}=L^{\alpha}_{\beta(a)}$. The adjoint commutes with the grading, as it does for the two-sided operators.

Proof. Immediate from the theorem and the parity statement of The Signed Adjoint of the Sandwich on a Bimodule over an Algebra. $\square$

Relation to the Signed Sandwich

The one-sided operators inside the sandwich

Proposition. For all $a,b \in A$ the signed sandwich factors through the signed left multiplication,

$$ S^{\alpha}_{a,b}=L^{\alpha}_a\circ R_{\alpha(b)}, \qquad L^{\alpha}_a=S^{\alpha}_{a,1}, \qquad R^{\alpha}_b:=S^{\alpha}_{1,b} . $$

Proof. $L^{\alpha}_a(R_{\alpha(b)}(x))=a\,\alpha(x\alpha(b))=a\,\alpha(x)\,\alpha(\alpha(b))=a\,\alpha(x)\,b=S^{\alpha}_{a,b}(x)$, and the other identities are the parameter choices. $\square$

Corollary. The adjoint of the signed sandwich is the composite of the adjoints of its factors in the reverse order, and the one-sided adjoints assemble it:

$$ \bigl(S^{\alpha}_{a,b}\bigr)^{*}=\bigl(R_{\alpha(b)}\bigr)^{*}\bigl(L^{\alpha}_a\bigr)^{*}=R_{\sigma(\alpha(b))}L^{\alpha}_{\beta(a)}=R_{\beta(b)}L^{\alpha}_{\beta(a)}=S^{\alpha}_{\beta(a),\beta(b)} . $$

Proof. Anti-multiplicativity and the theorem, with the right-handed adjoint $R_c^{*}=R_{\sigma(c)}$ obtained from the opposite algebra as in The Adjoint of the Sandwich on a Bimodule over an Algebra. For the middle equality, $\sigma(\alpha(b))=\sigma\alpha(b)=\alpha\sigma(b)=\beta(b)$ because $\alpha$ and $\sigma$ commute. For the last, $R_dL^{\alpha}_c=S^{\alpha}_{c,d}$ directly, since $R_dL^{\alpha}_c(x)=c\,\alpha(x)\,d$. $\square$

The one-sided case is therefore not an analogy but a factorisation: the sandwich is a signed left multiplication composed with a right multiplication, and its adjoint is the composite of the one-sided adjoints.

Self-Adjointness and Unitarity

Self-adjoint and skew-adjoint signed left multiplications

Proposition. The kernel of the parameter map $a \mapsto L^{\alpha}_a$ is the annihilator $\operatorname{Ann}_A(M)=\{a : aM=0\}$. Consequently

$$ \bigl(L^{\alpha}_a\bigr)^{*}=L^{\alpha}_a \iff \beta(a)-a \in \operatorname{Ann}_A(M), \qquad \bigl(L^{\alpha}_a\bigr)^{*}=-L^{\alpha}_a \iff \beta(a)+a \in \operatorname{Ann}_A(M). $$

For a faithful module these read $\beta(a)=a$ and $\beta(a)=-a$.

Proof. $L^{\alpha}_a=L^{\alpha}_{a'}$ is $L^{\alpha}_{a-a'}=0$, that is $(a-a')M=0$, by the kernel statement of The Signed Left Multiplication on a Module over an Algebra; applying this to the theorem gives the criteria. $\square$

So the self-adjoint signed left multiplications are those with $\beta$-symmetric parameter (modulo the annihilator), the analogue of the symmetric operators among the unsigned left multiplications.

Unitary signed left multiplications

Proposition. A signed left multiplication is unitary exactly when its parameter is a unitary element of $A$:

$$ \bigl(L^{\alpha}_a\bigr)^{*}L^{\alpha}_a=L^{\alpha}_a\bigl(L^{\alpha}_a\bigr)^{*}=\mathrm{id} \iff \sigma(a)a=a\sigma(a)=1 . $$

Proof. This is the unitarity criterion of The Signed Adjoint of the Sandwich on a Bimodule over an Algebra with $(a,b)=(a,1)$. $\square$

Corollary. The unitary signed left multiplications form a group isomorphic to the unitary group of $A$ modulo the annihilator: the map $a \mapsto L^{\alpha}_a$ restricted to the unitary elements of $A$ is multiplicative on unitary parameters and has kernel the unitary elements of $\operatorname{Ann}_A(M)$.

Proof. For unitary $a,b$ the composition rule gives $L^{\alpha}_aL^{\alpha}_b=L_{a\alpha(b)}$, and $a\alpha(b)$ is unitary when $a$ and $b$ are; the kernel is computed as above. $\square$

Kernel, Image and the Paired Complement

The complement identities

Proposition. For every $a$,

$$ \ker\bigl(L^{\alpha}_a\bigr)^{*}=\bigl(\operatorname{im}L^{\alpha}_a\bigr)^{\mathrm{c}}, \qquad \operatorname{im}\bigl(L^{\alpha}_a\bigr)^{*}={}^{\mathrm{c}}\bigl(\ker L^{\alpha}_a\bigr) . $$

Proof. These are the general identities of The Adjoint of a Module Homomorphism for the operator $L^{\alpha}_a$. $\square$

The explicit complements

Proposition. With $\operatorname{im}L^{\alpha}_a=aM$ and $\ker L^{\alpha}_a=\alpha(\ker L_a)$,

$$ \ker\bigl(L^{\alpha}_a\bigr)^{*}=\ker L_{\sigma(a)}, \qquad \operatorname{im}\bigl(L^{\alpha}_a\bigr)^{*}=\beta(a)M . $$

Proof. The first is $(aM)^{\mathrm{c}}=\{n : \langle am,n\rangle=0\ \forall m\}=\{n : \langle m,\sigma(a)n\rangle=0\ \forall m\}=\ker L_{\sigma(a)}$ by non-degeneracy. The second is the image of $L^{\alpha}_{\beta(a)}$, namely $\beta(a)M$. $\square$

Corollary. The two descriptions agree: $\alpha(\ker L_{\beta(a)})=\ker L_{\sigma(a)}$, which is the identity $\alpha(\ker L_c)=\ker L_{\alpha(c)}$ for $c=\beta(a)$.

Proof. $\alpha(\ker L_{\beta(a)})=\{m : \beta(a)\,\alpha(m)=0\}=\{m : \alpha(\sigma(a)m)=0\}=\ker L_{\sigma(a)}$ using $\alpha(\beta(a))=\sigma(a)$. $\square$

Examples

(a) The trivial grading. For $\alpha=\mathrm{id}$ the theorem reduces to $(L_a)^{*}=L_{\sigma(a)}$, the unsigned rule.

(b) The trivial involution. For $\sigma=\mathrm{id}$ the adjoint is $(L^{\alpha}_a)^{*}=L^{\alpha}_{\alpha(a)}$, so a signed left multiplication is self-adjoint exactly when its parameter is even.

(c) The regular module. For $M={}_A A$ with the regular pairing, $\operatorname{Ann}_A(M)=0$, so the criteria are the exact equalities $\beta(a)=a$ and $\beta(a)=-a$, and the unitary signed left multiplications are the unitary elements of $A$.

(d) A module with annihilator. For $A=k[x]/(x^{2})$ acting on $M=k[x]/(x)$ with $x\cdot m=0$, the annihilator is the maximal ideal $(x)$ and every parameter with the same image in $k$ gives the same signed left multiplication: the self-adjointness criterion degenerates to congruence modulo $(x)$, and $\ker L^{\alpha}_a$ is all of $M$ for $a \in (x)$.

(e) The Clifford algebra. For $A$ a Clifford algebra with canonical involution $\sigma$ and grade involution $\alpha$, the signed left multiplication $L^{\alpha}_v$ by a vector $v$ has adjoint $L^{\alpha}_{\beta(v)}$ with $\beta=\alpha\sigma$; it is self-adjoint exactly for $\beta(v)=v$, which for a vector holds exactly when the vector's parity is even, and it is unitary exactly for a unit vector $\sigma(v)v=1$.

Summary

For a graded module with an $\alpha$-invariant balanced $\sigma$-sesquilinear pairing, the adjoint of the signed left multiplication is the signed left multiplication by the composite $\beta=\alpha\sigma$, $\bigl(L^{\alpha}_a\bigr)^{*}=L^{\alpha}_{\beta(a)}$; this is the one-sided case of the signed adjoint sandwich, obtained directly from $\alpha^{*}=\alpha$, $(L_a)^{*}=L_{\sigma(a)}$ and the identity $\alpha L_b=L^{\alpha}_{\alpha(b)}$. The adjoint preserves the grading parity. Because the signed left multiplication is the signed sandwich with right parameter $1$, the sandwich adjoint factors through the one-sided adjoints. Self-adjointness holds exactly when $\beta(a)-a$ annihilates the module, skew-adjointness when $\beta(a)+a$ does, and unitarity exactly when the parameter is unitary in $A$; for a faithful module the criteria are the equalities $\beta(a)=a$, $\beta(a)=-a$ and $\sigma(a)a=a\sigma(a)=1$. The kernel and the image of the adjoint are described by the paired complements: $\ker(L^{\alpha}_a)^{*}=\ker L_{\sigma(a)}$ and $\operatorname{im}(L^{\alpha}_a)^{*}=\beta(a)M$, in agreement with the general theory.

Summary of Notation

Symbol	Meaning
$R$, $(A,\sigma)$	base ring, involutive $R$-algebra
$\alpha$, $\beta=\alpha\sigma$	grade involution, and the composite involution
$M$, $\langle\cdot,\cdot\rangle$	graded left module with α-invariant balanced σ-sesquilinear pairing
$L_a$	unsigned left multiplication, $L_a(m)=am$
$L^{\alpha}_a=L_a\alpha$	signed left multiplication, $L^{\alpha}_a(m)=a\alpha(m)$
$(L^{\alpha}_a)^{*}=L^{\alpha}_{\beta(a)}$	the adjoint of the signed left multiplication
$\alpha L_b=L^{\alpha}_{\alpha(b)}$	the conjugation identity used in the proof
$S^{\alpha}_{a,b}=L^{\alpha}_a R_{\alpha(b)}$	the signed sandwich through the one-sided operators
$\operatorname{Ann}_A(M)$	the annihilator, source of the degenerate criteria
$\ker(L^{\alpha}_a)^{*}=\ker L_{\sigma(a)}$	the kernel of the adjoint