The Signed Adjoint Sandwich on a Bimodule over an Algebra

Introduction

The signed sandwich $S^{\alpha}_{a,b}(x)=a\,\alpha(x)\,b$ twists the two-sided action by the grade involution $\alpha$. When the bimodule carries also an involution $\sigma$ and a pairing compatible with both, the adjoint of a signed sandwich can be computed, and the answer is a signed sandwich again, with parameters transformed by the composite $\alpha\sigma$. The article proves the formula, derives the unitarity condition $u^{*}u=uu^{*}=1$ it defines, and shows that the adjoint respects the $\mathbb{Z}/2$-grading of the sandwich monoid.

The article is the fourth of the * Operator Theory group of this category. It assumes the signed sandwich, its relation to the unsigned one and its composition rules from The Signed Sandwich on a Bimodule over an Algebra; the unsigned adjoint from The Adjoint of the Sandwich on a Bimodule over an Algebra; the pairing and the adjoint of The Adjoint of a Module Homomorphism; and the involution of The Involution on the Endomorphism Ring of a Module. The self-adjointness of the reflections, and its failure in the degenerate case, is The Signed Adjoint of the Reflection on a Bimodule over an Algebra, and the one-sided case is The Signed Adjoint of the Left Multiplication on a Module 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 $\sigma$, ${}_A M_A$ is a graded $(A,A)$-bimodule with grade involution $\alpha$ commuting with the actions in the graded sense, and the pairing is $\sigma$-sesquilinear, balanced and $\alpha$-invariant.

The Graded Pairing

The compatibility of the pairing with the grading

Definition. The pairing $\langle\cdot,\cdot\rangle$ on the graded bimodule $(M,\alpha)$ is $\alpha$-invariant when

$$ \langle\alpha(x),\alpha(y)\rangle=\langle x,y\rangle \qquad (x,y \in M). $$

Equivalently, because $\alpha^{2}=\mathrm{id}$,

$$ \langle\alpha(x),y\rangle=\langle x,\alpha(y)\rangle \qquad (x,y \in M). $$

The pairing is balanced for the involution $\sigma$ when it satisfies the two identities of The Adjoint of the Sandwich on a Bimodule over an Algebra, $\langle ax,y\rangle=\langle x,\sigma(a)y\rangle$ and $\langle xb,y\rangle=\langle x,y\sigma(b)\rangle$.

The two compatibility conditions are the two halves of "the pairing sees both structures": $\sigma$ governs the algebra and $\alpha$ the grading, and the identities say that moving an element across the pairing applies the appropriate structure on the other side.

The grade involution is self-adjoint and unitary

Proposition. For an $\alpha$-invariant pairing the grade involution $\alpha$ is self-adjoint and unitary with respect to the adjoint ${}^{*}$:

$$ \alpha^{*}=\alpha, \qquad \alpha^{*}\alpha=\alpha\alpha^{*}=\mathrm{id} . $$

Proof. The invariance identities give $\langle\alpha(x),y\rangle=\langle x,\alpha(y)\rangle$, which is the defining identity of the adjoint $\alpha^{*}=\alpha$. Then $\alpha^{*}\alpha=\alpha^{2}=\mathrm{id}$ and likewise. $\square$

So the grading and the involution of the module are not independent: an $\alpha$-invariant pairing makes the grade involution an isometry, and the adjoint of the grade involution is the grade involution.

The Adjoint of the Signed Sandwich

The main computation

Theorem. Let $\beta=\alpha\sigma$ be the composite of the grade involution and the involution of the algebra, an involution of $A$ because $\alpha$ and $\sigma$ commute. Then for all $a,b \in A$,

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

Proof. Since $S^{\alpha}_{a,b}=S_{a,b}\circ\alpha$ and the adjoint is anti-multiplicative, $(S^{\alpha}_{a,b})^{*}=\alpha^{*}(S_{a,b})^{*}$. By the proposition above $\alpha^{*}=\alpha$, and by The Adjoint of the Sandwich on a Bimodule over an Algebra $(S_{a,b})^{*}=S_{\sigma(a),\sigma(b)}$. Hence $(S^{\alpha}_{a,b})^{*}=\alpha\,S_{\sigma(a),\sigma(b)}$. Finally, the conjugation identity $S^{\alpha}_{c,d}=\alpha S_{\alpha^{-1}(c),\alpha^{-1}(d)}$ of The Signed Sandwich on a Bimodule over an Algebra gives $\alpha S_{\sigma(a),\sigma(b)}=S^{\alpha}_{\alpha^{-1}\sigma(a),\,\alpha^{-1}\sigma(b)}$, and $\alpha^{-1}\sigma=(\alpha^{-1}\sigma)=\alpha\sigma$ because $\alpha^{-1}=\alpha$ commutes with $\sigma$. $\square$

The parameters are transformed by $\beta=\alpha\sigma$, the product of the two involutions, not by $\sigma$ alone: the twist in the middle of the sandwich and the twist of the algebra both act, and they compose.

Parity and the adjoint

Corollary (the adjoint respects the grading). The adjoint of a signed sandwich is signed and the adjoint of an unsigned sandwich is unsigned:

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

Consequently the adjoint ${}^{*}$ preserves the $\mathbb{Z}/2$-grading of the monoid $\{S_{a,b}\}\cup\{S^{\alpha}_{a,b}\}$: a sandwich of parity zero is sent to a sandwich of parity zero, and a sandwich of parity one to a sandwich of parity one.

Proof. The first display is the theorem; the second is its $\alpha=\mathrm{id}$ case; the parity statement follows because the adjoint of an unsigned sandwich is unsigned and the adjoint of a signed sandwich is signed, and it is additive. $\square$

Specialisations

Proposition. In the two extreme cases the formula reduces to the known ones:

(i) for $\alpha=\mathrm{id}$, $(S_{a,b})^{*}=S_{\sigma(a),\sigma(b)}$, the unsigned adjoint;

(ii) for $\sigma=\mathrm{id}$, $(S^{\alpha}_{a,b})^{*}=S^{\alpha}_{\alpha(a),\alpha(b)}$, the grading alone.

Proof. Immediate from $\beta=\alpha\sigma$. $\square$

The Unitarity Condition

Unitary signed sandwiches

Definition. A signed sandwich is unitary when

$$ (S^{\alpha}_{a,b})^{*}S^{\alpha}_{a,b}=S^{\alpha}_{a,b}(S^{\alpha}_{a,b})^{*}=\mathrm{id}, $$

in the sense of the unitary elements of The Involution on the Endomorphism Ring of a Module.

Theorem. Suppose the sandwich map is injective. Then $S^{\alpha}_{a,b}$ is unitary if and only if $a$ and $b$ are unitary elements of $A$:

$$ (S^{\alpha}_{a,b})^{*}S^{\alpha}_{a,b}=S^{\alpha}_{a,b}(S^{\alpha}_{a,b})^{*}=\mathrm{id} \iff \sigma(a)a=a\sigma(a)=1,\quad \sigma(b)b=b\sigma(b)=1 . $$

Proof. By the theorem and the composition rules of the signed sandwiches,

$$ (S^{\alpha}_{a,b})^{*}S^{\alpha}_{a,b}=S^{\alpha}_{\beta(a),\beta(b)}S^{\alpha}_{a,b}=S_{\beta(a)\alpha(a),\,\alpha(b)\beta(b)}, \qquad S^{\alpha}_{a,b}(S^{\alpha}_{a,b})^{*}=S^{\alpha}_{a,b}S^{\alpha}_{\beta(a),\beta(b)}=S_{a\alpha(\beta(a)),\,\alpha(\beta(b))b}. $$

With $\beta=\alpha\sigma$ and $\alpha^{2}=\mathrm{id}$, the parameters are $\beta(a)\alpha(a)=\alpha(\sigma(a)a)$, $\alpha(b)\beta(b)=\alpha(b\sigma(b))$ in the first product, and $a\alpha(\beta(a))=a\sigma(a)$, $\alpha(\beta(b))b=\sigma(b)b$ in the second. Both products equal $\mathrm{id}=S_{1,1}$ exactly when $\sigma(a)a=a\sigma(a)=1$ and $\sigma(b)b=b\sigma(b)=1$, by injectivity of the sandwich map. $\square$

The parameter criterion is independent of the grading: the grade involution enters the adjoint formula but cancels from the unitarity condition, so the unitary signed sandwiches are those with unitary parameters, exactly as in the unsigned case.

The unitary signed sandwiches form a group

Proposition. The unitary signed sandwiches with a fixed right parameter, or with both parameters fixed as unitary elements, form a group under composition; more precisely, if $a,a'$ and $b,b'$ are unitary in $A$ then $S^{\alpha}_{a,b}S^{\alpha}_{a',b'}$ is the unsigned sandwich $S_{a\alpha(a'),\,\alpha(b')b}$ and it is unitary.

Proof. The composition rule $S^{\alpha}_{a,b}S^{\alpha}_{a',b'}=S_{a\alpha(a'),\alpha(b')b}$ is that of The Signed Sandwich on a Bimodule over an Algebra; the product of unitary parameters $\alpha(a')$ and $b'$ is unitary, and $\alpha$ preserves unitarity because it is an involution of the algebra, so the parameters $a\alpha(a')$ and $\alpha(b')b$ are unitary and the product is unitary by the previous theorem. $\square$

Hence the unitary signed sandwiches generate a group of isometries of the pairing, and its elements are the products of a unitary parameter on the left, the grade involution, and a unitary parameter on the right.

Reflections

Corollary. For a unit $u$ the reflection $r_u=S^{\alpha}_{u,u^{-1}}$ has a reflection as its adjoint,

$$ (r_u)^{*}=S^{\alpha}_{\beta(u),\,\beta(u)^{-1}}=r_{\beta(u)}, $$

and $r_u$ is unitary exactly when $u$ is a unitary element of $A$.

Proof. The adjoint formula with $S^{\alpha}_{u,u^{-1}}$ and the relation $\beta(u^{-1})=\beta(u)^{-1}$ give the first display; the unitarity criterion with $(a,b)=(u,u^{-1})$ gives the second. $\square$

The self-adjointness of the reflection — the case $r_{\beta(u)}=r_u$ — and its failure are the subject of The Signed Adjoint of the Reflection on a Bimodule over an Algebra.

Compatibility with the Involution

The two involutions of the algebra

Proposition. The adjoint of a signed sandwich is a signed sandwich whose parameters carry the composite involution $\beta=\alpha\sigma$; the involution $\sigma$ and the grade involution $\alpha$ therefore enter asymmetrically: $\sigma$ acts on the parameters of the unsigned adjoint, and $\alpha$ acts through the conjugation identity of the signed sandwich, so only the product $\alpha\sigma$ is visible in the answer.

Proof. The theorem gives $\beta=\alpha\sigma$; the roles of the two factors are as described in its proof. $\square$

Corollary. The adjoint operation and the grade involution commute on the sandwiches,

$$ (S^{\alpha}_{a,b})^{*}=\alpha\,(S_{\alpha\sigma(a),\alpha\sigma(b)})\quad\text{and}\quad \alpha\,S_{a,b}=S^{\alpha}_{\alpha(a),\alpha(b)}, $$

so applying the adjoint and then the grade involution is the same as applying the grade involution and then the adjoint of the corresponding unsigned sandwich.

Proof. Both displays are the identities already established, combined. $\square$

Degenerate Cases

A pairing that is not $\alpha$-invariant

Proposition. If the pairing is not $\alpha$-invariant then $\alpha$ need not be self-adjoint and the adjoint of the grade involution, $\alpha^{*}$, is an operator different from $\alpha$; the formula for the adjoint of a signed sandwich acquires the extra factor $\alpha^{*}$ and reads $(S^{\alpha}_{a,b})^{*}=\alpha^{*}S_{\sigma(a),\sigma(b)}$, which is a signed sandwich only when $\alpha^{*}$ maps signed sandwiches to signed sandwiches.

Proof. The proof of the theorem used only $\alpha^{*}=\alpha$ from invariance; without it the factor $\alpha^{*}$ remains. $\square$

The composite is not an involution

Proposition. If $\alpha$ and $\sigma$ do not commute then $\beta=\alpha\sigma$ is not an involution and $\beta^{k}$ cycles; the adjoint of a signed sandwich is still a signed sandwich with parameters $\beta(a),\beta(b)$, but the map on parameters is no longer an involution, and applying the adjoint twice returns $S^{\alpha}_{\beta^{2}(a),\beta^{2}(b)}$, which differs from $S^{\alpha}_{a,b}$ unless $\beta^{2}=\mathrm{id}$.

Proof. $(S^{\alpha}_{a,b})^{**}=S^{\alpha}_{\beta^{2}(a),\beta^{2}(b)}$ from the theorem; $\beta^{2}=\mathrm{id}$ exactly when $\alpha$ and $\sigma$ commute. $\square$

A non-injective sandwich map

Proposition. If the sandwich map is not injective then the unitarity criterion is a criterion on the operator, and the parameters of a unitary signed sandwich are determined only up to the kernel of the sandwich map.

Proof. The adjoint depends only on the operator and $S^{\alpha}_{a,b}$ depends only on the image of $(a,b)$ modulo the kernel. $\square$

Examples

(a) The trivial grading. For $\alpha=\mathrm{id}$ the signed sandwich is the unsigned sandwich and the theorem reduces to $(S_{a,b})^{*}=S_{\sigma(a),\sigma(b)}$.

(b) The trivial involution. For $\sigma=\mathrm{id}$ the adjoint of a signed sandwich is $S^{\alpha}_{\alpha(a),\alpha(b)}$, so the adjoint acts on the parameters by the grade involution alone.

(c) The group algebra of a finite group. For $A=R[G]$ with $\sigma(g)=g^{-1}$, $M=A$ the regular bimodule, and $\alpha$ the grading by $G/G^{2}$, the adjoint of $S^{\alpha}_{g,h}(x)=g\alpha(x)h$ is $S^{\alpha}_{\alpha\sigma(g),\alpha\sigma(h)}=S^{\alpha}_{\alpha(g)^{-1},\alpha(h)^{-1}}$ for group elements; the unitarity condition is $g,h$ of order dividing the exponent forced by $\sigma$ and the grading, and it fails for elements of order not two when the grading is trivial.

(d) The Clifford algebra. For $A$ a Clifford algebra with its canonical involution $\sigma$ and grade involution $\alpha$, the adjoint of the signed sandwich gives the standard formula for conjugation by a versor composed with the reflection of the sandwich, and the unitarity condition is the versor condition $\sigma(v)v=1$.

Summary

On a graded bimodule with a balanced $\sigma$-sesquilinear pairing that is $\alpha$-invariant, the grade involution is self-adjoint and unitary, $\alpha^{*}=\alpha$, and the adjoint of the signed sandwich is $\bigl(S^{\alpha}_{a,b}\bigr)^{*}=S^{\alpha}_{\beta(a),\beta(b)}$ with $\beta=\alpha\sigma$ the composite of the grade involution and the algebra involution. The adjoint preserves the $\mathbb{Z}/2$-grading: unsigned sandwiches go to unsigned sandwiches and signed to signed. The unitarity condition $u^{*}u=uu^{*}=\mathrm{id}$ for a signed sandwich holds exactly when its two parameters are unitary elements of $A$, independently of the grading, and the unitary signed sandwiches compose to unsigned sandwiches with unitary parameters, generating a group of isometries. The reflection $r_u=S^{\alpha}_{u,u^{-1}}$ has adjoint $r_{\beta(u)}$ and is unitary exactly when $u$ is unitary in $A$; its self-adjointness is treated separately. When the pairing is not $\alpha$-invariant the formula acquires the factor $\alpha^{*}$, and when $\alpha$ and $\sigma$ do not commute the parameter map is no longer an involution.

Summary of Notation

Symbol	Meaning
$R$, $(A,\sigma)$	base ring, involutive $R$-algebra
$\alpha$	grade involution of $A$, commuting with $\sigma$
${}_A M_A$	graded $(A,A)$-bimodule with grade involution $\alpha$
$\langle\cdot,\cdot\rangle$	balanced σ-sesquilinear, α-invariant pairing
$\alpha^{*}=\alpha$	the grade involution is self-adjoint and unitary
$S_{a,b}$, $S^{\alpha}_{a,b}$	unsigned and signed sandwiches
$\beta=\alpha\sigma$	the composite involution
$(S^{\alpha}_{a,b})^{*}=S^{\alpha}_{\beta(a),\beta(b)}$	the adjoint of the signed sandwich
$\sigma(a)a=a\sigma(a)=1$	unitarity of the parameter
$r_u=S^{\alpha}_{u,u^{-1}}$	the reflection, $(r_u)^{*}=r_{\beta(u)}$