The Signed Sandwich on a Bimodule over an Algebra

Introduction

A two-sided operator on a module dresses one element between a left and a right factor: for $a, b \in A$ the unsigned sandwich is $S_{a,b}(x)=a x b$. When the algebra and the module carry a grade involution $\alpha$ — an order-two automorphism that preserves the products — the sandwich can be twisted in the middle, and the signed sandwich is $S^{\alpha}_{a,b}(x)=a\,\alpha(x)\,b$. The twist is a single substitution of $\alpha(x)$ for $x$, and it is the operator form of the parity sign the graded structures carry: an element of odd degree changes sign under $\alpha$, and the sandwich records that sign in the middle factor.

The article defines the two sandwiches, computes how each is built from the one-sided multiplications of Left and Right Multiplication of a Module, proves the exact relation between them, $S^{\alpha}_{a,b}=S_{a,b}\circ\alpha$, and works out the composition rule. The rule is the graded one: a product of two signed sandwiches is unsigned, and the whole family of sandwiches is $\mathbb{Z}/2$-graded by the parity of the twist. The article then specialises to the reflections, the signed sandwiches $r_u=S^{\alpha}_{u,u^{-1}}$ by a unit, and determines when such an operator is an involution.

The article assumes the module theory of Modules over an Algebra and the operator layer of Left and Right Multiplication of a Module and Module Endomorphisms. The reflected operators are developed further in Reflections as Signed Two-Sided Operators on a Bimodule over an Algebra, and the involution of the elements $\sigma$, the dagger built from it and the adjoint belong to the * Theory and * Operator Theory groups: none of them occurs here. The article stays inside Part I — no distance, norm, form, topology or limit — and the word reflection names an operator of order two, with no geometric reading. Throughout, $R$ is a commutative ring with $1 \neq 0$; $A$ is a unital associative $R$-algebra, not assumed commutative; $\alpha$ is a grade involution of $A$; ${}_A M_A$ is an $(A,A)$-bimodule carrying the compatible grade involution $\alpha$; and $L_a$, $R_b$ are the one-sided multiplications.

The Unsigned Sandwich

Definition

Definition. For $a, b \in A$ the unsigned sandwich with parameters $a$ and $b$ is the additive map

$$ S_{a,b} : M \to M, \qquad S_{a,b}(x) = a\,x\,b . $$

It is the composite $S_{a,b}=L_a\circ R_b=R_b\circ L_a$ of a left and a right multiplication by elements of $A$, and it is quadratic in its parameters: it is linear in neither $a$ nor $b$ alone when $A$ is noncommutative, since $S_{aa',b}(x)=a\bigl(a'xb\bigr)=S_{a,b}(a'x)$.

Proposition. For all $a,b,c,d \in A$,

$$ S_{a,b}\circ S_{c,d} = S_{ac,\,db}, \qquad S_{a,b}(1)=ab \quad (\text{on } {}_A A_A). $$

Proof. $S_{a,b}(S_{c,d}(x))=a(cxd)b=(ac)x(db)=S_{ac,db}(x)$; the value at the unit is the definition. $\square$

The operator monoid

Proposition. The unsigned sandwiches form a submonoid of $\operatorname{End}_R(M)$ under composition, with identity $S_{1,1}=\mathrm{id}_M$; if $a$ and $b$ are units then $S_{a,b}$ is invertible with

$$ S_{a,b}^{-1}=S_{a^{-1}\!,\,b^{-1}} . $$

Proof. The composition law is closure, and $S_{1,1}=\mathrm{id}_M$. If $a,b$ are units then $S_{a,b}S_{a^{-1},b^{-1}}=S_{1,1}=S_{a^{-1},b^{-1}}S_{a,b}$ by the law. $\square$

The converse is not claimed: an element of $A$ that is not a unit may act invertibly on a small module, and then the sandwich it defines is invertible even though its parameter is not. For the regular bimodule $M={}_A A_A$ the converse is a statement about units of $A$ and is not needed here.

The kernel of the operator is the additive subgroup $\ker S_{a,b}=\{x : axb=0\}$, and the operator vanishes exactly when $aMb=0$; in a noncommutative algebra the single product $ab$ does not control the operator, the reason being the same $2\times2$ example as in Left and Right Multiplication in a Ring.

The diagonal case

Proposition. For a unit $u \in A^{\times}$ the diagonal sandwich is the inner conjugation

$$ S_{u,u^{-1}}(x)=uxu^{-1}, $$

and $S_{u,u^{-1}}S_{v,v^{-1}}=S_{uv,(uv)^{-1}}$, so the diagonal sandwiches reproduce the group $A^{\times}$ acting by conjugation.

Proof. Immediate from the definitions and the composition law. $\square$

The Grade Involution

Definition on the algebra and on the module

Definition. A grade involution of $A$ is an algebra automorphism $\alpha$ with $\alpha^{2}=\mathrm{id}$; it is trivial when $\alpha=\mathrm{id}$. A graded bimodule over $(A,\alpha)$ is an $(A,A)$-bimodule $M$ together with an additive map, again written $\alpha : M \to M$, such that

$$ \alpha^{2}=\mathrm{id}, \qquad \alpha(amb)=\alpha(a)\,\alpha(m)\,\alpha(b) \quad (a,b \in A,\ m \in M). $$

When $2$ is invertible in $A$ and in the action on $M$ the pair splits,

$$ M=M_{\bar0}\oplus M_{\bar1}, \qquad M_{\bar0}=\{x : \alpha(x)=x\}, \qquad M_{\bar1}=\{x : \alpha(x)=-x\}, $$

the even and the odd parts, and $A_i M_j \subseteq M_{i+j}$; the bimodule is then graded in the sense of the $\mathbb{Z}/2$-grading. The grading is that of Superalgebras and Graded Structures, which owns it; only its order-two operator is used here.

Proposition. The even part $M_{\bar0}$ is an $(A_{\bar0},A_{\bar0})$-bimodule and the odd part $M_{\bar1}$ is an $(A_{\bar0},A_{\bar0})$-bimodule with products into $M_{\bar0}$; the products satisfy $A_i M_j \subseteq M_{i+j}$.

Proof. Read the compatibility $\alpha(amb)=\alpha(a)\alpha(m)\alpha(b)$: if $a,b$ are even and $m$ is even, $amb$ is even; if one factor is odd the sign flips accordingly. $\square$

Conjugation by the grade involution

Proposition. For all $a,b \in A$,

$$ \alpha\,L_a\,\alpha^{-1}=L_{\alpha(a)}, \qquad \alpha\,R_b\,\alpha^{-1}=R_{\alpha(b)}, \qquad \alpha\,S_{a,b}\,\alpha^{-1}=S_{\alpha(a),\alpha(b)} . $$

Proof. $\alpha(L_a(\alpha^{-1}(x)))=\alpha(a\,\alpha^{-1}(x))=\alpha(a)\,x=L_{\alpha(a)}(x)$ by the compatibility; the right case is the mirror, and the sandwich case is the two together. $\square$

Thus the grade involution acts on the monoid of sandwiches by the simultaneous substitution $a\mapsto\alpha(a)$, $b\mapsto\alpha(b)$, and it fixes $S_{a,b}$ exactly when both parameters are even.

The Signed Sandwich

Definition and the relation to the unsigned one

Definition. For $a,b \in A$ the signed sandwich is

$$ S^{\alpha}_{a,b} : M \to M, \qquad S^{\alpha}_{a,b}(x)=a\,\alpha(x)\,b . $$

It is the composite $S^{\alpha}_{a,b}=L_a\circ R_b\circ\alpha=S_{a,b}\circ\alpha$ of the unsigned sandwich with the grade involution, and it is additive.

Theorem. For all $a,b \in A$,

$$ S^{\alpha}_{a,b}=S_{a,b}\circ\alpha, \qquad\text{and}\qquad S^{\alpha}_{a,b}=\alpha\circ S_{\alpha^{-1}(a),\,\alpha^{-1}(b)} . $$

In particular the signed and the unsigned sandwiches coincide exactly when $\alpha=\mathrm{id}$, and $S^{\alpha}_{a,b}(x)=S_{a,b}(\alpha(x))$ for every $x$.

Proof. $S_{a,b}(\alpha(x))=a\alpha(x)b=S^{\alpha}_{a,b}(x)$, which is the first identity; for the second, $\alpha(S_{\alpha^{-1}(a),\alpha^{-1}(b)}(x))=\alpha(\alpha^{-1}(a)\,x\,\alpha^{-1}(b))=a\,\alpha(x)\,b$. $\square$

The signed sandwich is linear in each parameter, $S^{\alpha}_{a+a',b}=S^{\alpha}_{a,b}+S^{\alpha}_{a',b}$, and it is the middle argument that carries the twist; the parameters do not.

The graded composition rule

The compositions of signed and unsigned sandwiches follow the parity of the twist. This is the sign rule of the grading in operator form.

Theorem. For all $a,b,c,d \in A$,

$$ S_{a,b}\circ S_{c,d}=S_{ac,\,db}, \qquad S^{\alpha}_{a,b}\circ S_{c,d}=S^{\alpha}_{a\alpha(c),\,\alpha(d)b}, $$

$$ S_{a,b}\circ S^{\alpha}_{c,d}=S^{\alpha}_{ac,\,db}, \qquad S^{\alpha}_{a,b}\circ S^{\alpha}_{c,d}=S_{a\alpha(c),\,\alpha(d)b}. $$

Thus the product of two sandwiches of the same parity is unsigned and the product of two of opposite parity is signed: the family $\{S_{a,b}\}\cup\{S^{\alpha}_{a,b}\}$ is a monoid graded by $\mathbb{Z}/2$, the unsigned sandwiches even and the signed ones odd.

Proof. The first line is the unsigned law. For the second, $S^{\alpha}_{a,b}S_{c,d}=S_{a,b}\alpha S_{c,d}=S_{a,b}S_{\alpha(c),\alpha(d)}\alpha=S^{\alpha}_{a\alpha(c),\alpha(d)b}$, using the conjugation identity and $S_{a,b}S_{c',d'}=S_{ac',d'b}$. The third is $S_{a,b}S_{c,d}\alpha=S_{ac,db}\alpha=S^{\alpha}_{ac,db}$. The fourth is $S_{a,b}\alpha S_{c,d}\alpha=S_{a,b}S_{\alpha(c),\alpha(d)}=S_{a\alpha(c),\alpha(d)b}$, the two grade involutions cancelling. $\square$

Corollary. If the parameters are units then the signed sandwich is invertible, and

$$ \bigl(S^{\alpha}_{a,b}\bigr)^{-1}=S^{\alpha}_{\alpha(a)^{-1}\!,\,\alpha(b)^{-1}} \qquad (a,b \in A^{\times}). $$

Proof. $S^{\alpha}_{a,b}=S_{a,b}\alpha$ and $\alpha$ is invertible with $\alpha^{-1}=\alpha$; the inverse is $\alpha\,S_{a,b}^{-1}=\alpha\,S_{a^{-1},b^{-1}}=S^{\alpha}_{\alpha(a)^{-1},\alpha(b)^{-1}}$ by the conjugation identity. $\square$

Proposition (the diagonal case). For a unit $u$,

$$ r_u=S^{\alpha}_{u,u^{-1}}, \qquad r_u(x)=u\,\alpha(x)\,u^{-1}, $$

the signed inner conjugation by $u$, and

$$ r_u^{2}=S_{u\alpha(u),\,(u\alpha(u))^{-1}}=\operatorname{conj}_{u\alpha(u)} . $$

Proof. The first display is the definition. For the square, the fourth composition rule gives $r_u^{2}=S^{\alpha}_{u\alpha(u),\,\alpha(u^{-1})u^{-1}}$; here $\alpha(u^{-1})=\alpha(u)^{-1}$, so the second parameter is $\alpha(u)^{-1}u^{-1}=(u\alpha(u))^{-1}$, and the composite is the unsigned diagonal sandwich $S_{w,w^{-1}}$ with $w=u\alpha(u)$, that is, the inner conjugation by $w$. $\square$

The square lands in the unsigned sandwiches even though the reflection is signed, which is the parity rule in its simplest instance.

Reflections Realised by the Signed Sandwich

Definition and the involutive case

Definition. A reflection of the graded bimodule $(M,\alpha)$ is a signed two-sided operator $r_u=S^{\alpha}_{u,u^{-1}}$ by a unit $u$; it is a genuine reflection, that is, an operator of order two, exactly when $r_u^{2}=\mathrm{id}_M$.

Definition. The centralizer of the bimodule is

$$ C_A(M)=\{w \in A : wx=xw \text{ for all } x \in M\}, $$

the elements that act the same way from either side; it is a subalgebra of $A$ containing $Z(A)$, and it equals $Z(A)$ when $M$ is faithful.

Theorem. Let $u \in A^{\times}$ and put $w=u\alpha(u)$. The signed inner conjugation $r_u$ is a reflection if and only if $w \in C_A(M)$, and then it is the inner automorphism $\operatorname{conj}_w$ of order two,

$$ r_u^{2}=\mathrm{id}_M \iff u\,\alpha(u) \in C_A(M). $$

Proof. By the proposition above, $r_u^{2}=\operatorname{conj}_w$, and an inner conjugation is the identity exactly when the conjugating element acts centrally on the module, which is the definition of $C_A(M)$; when $w \in C_A(M)$ the map is the identity. Conversely $r_u^{2}=\mathrm{id}$ says $wx=xw$ for all $x$, so $w \in C_A(M)$. $\square$

Corollary. If $u$ is fixed by $\alpha$ then $w=u^{2}$; if $u$ is negated by $\alpha$ and $2$ is invertible then $w=-u^{2}$; in either case $w$ is central as soon as $u^{2}$ is, and then $r_u$ is a reflection.

Proof. Substitute $\alpha(u)=\pm u$. If $\alpha(u)=u$ then $u\alpha(u)=u^{2}$; if $\alpha(u)=-u$ then $u\alpha(u)=-u^{2}$. A unit and its negative are central together. $\square$

Examples

(a) The trivial grade involution. If $\alpha=\mathrm{id}$ then $S^{\alpha}_{a,b}=S_{a,b}$, every sandwich is even, the composition rule is the unsigned one, and $r_u=\operatorname{conj}_u$ with $r_u^{2}=\operatorname{conj}_{u^{2}}$: the criterion becomes $u^{2} \in C_A(M)$.

(b) The matrix algebra with an even/odd grading. Let $A=M_2(k)$ and $\alpha(X)=JXJ^{-1}$ with $J=\operatorname{diag}(1,-1)$, so $\alpha^{2}=\mathrm{id}$, and let $M=A$ with $\alpha$ acting entrywise; the even part is the diagonal matrices and the odd part the off-diagonal ones. For $u=E_{12}+E_{21}$ one has $\alpha(u)=-u$ and $u^{2}=I$, so $u\alpha(u)=-I \in Z(A)$ and $r_u$ is a reflection; explicitly $r_u(x)=u\alpha(x)u^{-1}$ and $r_u^{2}(x)=(-I)x(-I)^{-1}=x$.

(c) The degenerate case. Let $A=M_2(k)$, $\alpha=\mathrm{id}$ and $u=E_{12}+I$. Then $u^{2}=I+2E_{12}$ is not central, so $r_u=\operatorname{conj}_u$ has $r_u^{2}=\operatorname{conj}_{u^{2}}\neq\mathrm{id}$: the operator is an automorphism of infinite order, not a reflection. The failure is exactly the non-centrality of $u\alpha(u)$.

(d) A trivial action on the module. If the grade involution fixes every element of $M$, then $S^{\alpha}_{a,b}=S_{a,b}$ for all $a,b$, whatever $\alpha$ does on $A$: the grading is invisible to the operators, and the criterion reduces to the unsigned one, $u^{2} \in C_A(M)$.

Summary

Let $A$ be an $R$-algebra with a grade involution $\alpha$ and ${}_A M_A$ a bimodule carrying the compatible action of $\alpha$. The unsigned sandwich is $S_{a,b}(x)=axb=L_aR_b$, with $S_{a,b}S_{c,d}=S_{ac,db}$ and inverses $S_{a^{-1},b^{-1}}$ on the units, and the signed sandwich is $S^{\alpha}_{a,b}(x)=a\alpha(x)b=S_{a,b}\alpha$. The grade involution conjugates one-sided multiplications and sandwiches as $\alpha S_{a,b}\alpha^{-1}=S_{\alpha(a),\alpha(b)}$. The composition rule is graded: $S_{a,b}S_{c,d}=S_{ac,db}$ and $S^{\alpha}_{a,b}S^{\alpha}_{c,d}=S_{a\alpha(c),\alpha(d)b}$ are unsigned, while $S^{\alpha}_{a,b}S_{c,d}=S^{\alpha}_{a\alpha(c),\alpha(d)b}$ and $S_{a,b}S^{\alpha}_{c,d}=S^{\alpha}_{ac,db}$ are signed. The diagonal signed sandwich $r_u=S^{\alpha}_{u,u^{-1}}$ is the signed inner conjugation $x\mapsto u\alpha(x)u^{-1}$, and its square is the inner conjugation by $u\alpha(u)$; it is a reflection exactly when $u\alpha(u)$ lies in the centralizer $C_A(M)$ of the bimodule, which is $Z(A)$ for a faithful module. The case $\alpha=\mathrm{id}$ reduces everything to the unsigned theory, and the failure in the degenerate case is the non-centrality of $u\alpha(u)$. No involution of the elements, no dagger and no adjoint occurs; those belong to the * Theory and * Operator Theory groups of the category.

Summary of Notation

Symbol	Meaning
$A$	unital associative $R$-algebra, not assumed commutative
${}_A M_A$	$(A,A)$-bimodule
$\alpha$	grade involution, an automorphism of $A$ and of $M$ with $\alpha^{2}=\mathrm{id}$
$M_{\bar0}, M_{\bar1}$	even and odd parts of the graded bimodule
$S_{a,b}(x)=axb$	unsigned sandwich, the two-sided operator $L_aR_b$
$S^{\alpha}_{a,b}(x)=a\alpha(x)b$	signed sandwich
$S^{\alpha}_{a,b}=S_{a,b}\circ\alpha$	the signed sandwich is the unsigned one at $\alpha$
$S_{a,b}S_{c,d}=S_{ac,db}$	composition of unsigned sandwiches
$S^{\alpha}_{a,b}S^{\alpha}_{c,d}=S_{a\alpha(c),\alpha(d)b}$	composition of two signed sandwiches is unsigned
$r_u=S^{\alpha}_{u,u^{-1}}$	reflection, the signed inner conjugation by a unit
$r_u^{2}=\operatorname{conj}_{u\alpha(u)}$	a reflection is an involution iff $u\alpha(u) \in C_A(M)$
$C_A(M)$	centralizer of the bimodule, $\{w : wx=xw \text{ for all } x\}$
$L_a$, $R_b$	left and right multiplications of the module