The Signed Adjoint Sandwich on an Algebra

Introduction

The signed sandwich is the two-sided operator $S_{a,b}(x)=a\,\alpha(x)\,b$ built from a left multiplication, a right multiplication and a grade involution $\alpha$; it is the algebraic home of the reflections and of the conjugation by a unit, and the two-sided companion of the signed left multiplication. Its signed adjoint with respect to the twisted pairing $\{x,y\}=\tau(x\sigma(y))$ is again a signed sandwich, $S_{a,b}^{*_\sigma}=S_{\delta(a),\delta(b)}$ with $\delta=\sigma\alpha$, and the parameters are not reversed: the twisted adjoint of the signed sandwich is the signed sandwich of the $\delta$-images, in the same order. The sandwich is unitary, $S^{*_\sigma}S=SS^{*_\sigma}=\mathrm{id}$, exactly when the two products $\delta(a)\alpha(a)$ and $\alpha(b)\delta(b)$ are central and mutually inverse; for the reflection $b=\alpha(a)^{-1}$ this is the single condition that $\sigma(a)a$ be central and fixed by $\alpha$, and for the one-sided operator $b=1$ it is $\delta(a)\alpha(a)=1$.

This article computes the signed adjoint, the inverse and the composition of the sandwiched operators, derives the unitarity condition $u^{*}u=uu^{*}=1$ it defines, and prepares the reflection of The Signed Adjoint of the Reflection on an Algebra. It assumes The Signed Sandwich on an Algebra for the operator $S_{a,b}$, the signed conjugation $\rho_u=S_{u,u^{-1}}$ and the composition and inverse rules, The Adjoint of the Left Multiplication on an Algebra for the one-sided case, Involutions of the Operator Algebra and The Adjoint in an Involutive Algebra for the adjoint operation and the twisted pairing, Frobenius Algebras for the trace and the pairing, Involutive Linear Algebras for the involution and the grade involution, and Unitary Elements of an Involutive Algebra for the unitary elements and the group they form. The graded version of the same computation is The Graded Adjoint Action on a Module over an Algebra; the analytic reading of the unitary operators is Part II. This article stays inside Part I: no distance, norm, form with a norm, topology or limit.

Throughout, $k$ is a field of characteristic not two, $A$ is a finite-dimensional unital associative $k$-algebra, $\tau$ is a trace whose pairing $\langle x,y\rangle=\tau(xy)$ is nondegenerate, $\sigma$ is an involution of $A$ with $\tau(\sigma(x))=\tau(x)$, and $\alpha$ is an involutive automorphism of $A$ with $\alpha^{2}=\mathrm{id}$ commuting with $\sigma$ and preserving the trace, $\tau(\alpha(x))=\tau(x)$. The twist is $\delta=\sigma\alpha$, an anti-automorphism of order two; the twisted pairing is $\{x,y\}=\tau(x\sigma(y))$, the unsigned sandwich is $T_{a,b}(x)=axb$, the signed sandwich is $S_{a,b}(x)=a\alpha(x)b$, and the adjoints for the two pairings are $T^{*_\sigma}$ and $T^{*}$.

The Signed Adjoint

Theorem. With respect to the twisted pairing,

$$ \{S_{a,b}x,y\}=\{x,\,S_{\delta(a),\delta(b)}y\}, \qquad \text{that is} \qquad S_{a,b}^{*_\sigma}=S_{\delta(a),\delta(b)} . $$

The signed adjoint of a sandwich is the sandwich of the two $\delta$-images, with the order of the parameters not reversed, and the assignment $S_{a,b}\mapsto S_{\delta(a),\delta(b)}$ is an involution of the signed sandwich space.

Proof. Compute the left side, $\{S_{a,b}x,y\}=\tau(a\alpha(x)b\,\sigma(y))=\tau(\sigma(y)a\alpha(x)b)$ by cyclicity, and use the $\alpha$-invariance of the trace to write it as $\tau(\alpha(\sigma(y)a\alpha(x)b))=\tau(\alpha\sigma(y)\alpha(a)x\alpha(b))$; by the commutation $\alpha\sigma=\sigma\alpha$ this is $\tau(\sigma(\alpha(y))\alpha(a)x\alpha(b))=\tau(x\alpha(b)\sigma(\alpha(y))\alpha(a))$, again by cyclicity. On the other side, $\{x,S_{\delta(a),\delta(b)}y\}=\tau(x\,\sigma(\delta(a)\alpha(y)\delta(b)))=\tau(x\,\sigma(\delta(b))\sigma(\alpha(y))\sigma(\delta(a)))$, and $\sigma\delta=\sigma\sigma\alpha=\alpha$ gives $\sigma(\delta(a))=\alpha(a)$ and $\sigma(\delta(b))=\alpha(b)$, so this is the same expression. Hence the identity; the order-two property is $\delta^{2}=\mathrm{id}$, and the image is again a signed sandwich, so the signed sandwich space is stable.

Proposition (the plain pairing). With respect to the plain pairing,

$$ S_{a,b}^{*}=S_{\alpha(b),\alpha(a)}, $$

so the plain adjoint of a signed sandwich is the signed sandwich of the two $\alpha$-images with the order reversed; and the two adjoints are related by the conjugation of the operator algebra,

$$ S_{a,b}^{*_\sigma}=c_{\sigma}\bigl(S_{a,b}^{*}\bigr), \qquad c_{\sigma}(T)=\sigma T\sigma . $$

Proof. For the plain pairing, $\langle S_{a,b}x,y\rangle=\tau(a\alpha(x)by)=\tau(bya\alpha(x))=\tau(\alpha(bya)x)$ by the $\alpha$-invariance and cyclicity, which is $\langle x,S_{\alpha(b),\alpha(a)}y\rangle$, since $\alpha(bya)=\alpha(b)\alpha(y)\alpha(a)$. The relation is the dictionary $T^{*_\sigma}=\sigma T^{*}\sigma$ of The Adjoint in an Involutive Algebra, applied to $T=S_{a,b}$: $\sigma S_{\alpha(b),\alpha(a)}\sigma(x)=\sigma(\alpha(b)\alpha(\sigma(x))\alpha(a))=\sigma(\alpha(a))\,\sigma(\alpha(\sigma(x)))\,\sigma(\alpha(b))=\delta(a)\alpha(x)\delta(b)$, using $\alpha\sigma=\sigma\alpha$ and $\sigma^{2}=\mathrm{id}$.

Corollary (the grade involution is self-adjoint). The grade involution itself is the sandwich $S_{1,1}=\alpha$, and it is self-adjoint for both pairings, $\alpha^{*}=\alpha$ and $\alpha^{*_\sigma}=\alpha$.

Proof. For the twisted pairing the theorem gives $S_{1,1}^{*_\sigma}=S_{\delta(1),\delta(1)}=S_{1,1}$, because $\delta(1)=1$; for the plain pairing the proposition gives $S_{1,1}^{*}=S_{\alpha(1),\alpha(1)}=S_{1,1}$, because $\alpha(1)=1$.

The Inverse and the Composition

Theorem. The signed sandwich satisfies the composition rules of The Signed Sandwich on an Algebra,

$$ S_{a,b}S_{c,d}=T_{a\alpha(c),\,\alpha(d)b}, \qquad S_{a,b}^{-1}=S_{\alpha(a)^{-1},\,\alpha(b)^{-1}}, $$

and the unsigned sandwich $T_{c,d}$ is the identity exactly when $c$ is a central unit and $d=c^{-1}$.

Proof. The composition and the inverse are read from The Signed Sandwich on an Algebra: the product of two signed sandwiches is an unsigned sandwich, and the sandwich $S_{a,b}=T_{a,b}\alpha$ is invertible exactly when $a$ and $b$ are units. For the identity criterion, $T_{c,d}(x)=cxd=x$ for all $x$ gives $cd=1$ at $x=1$ and then $cx=xc$ for all $x$, that is $c$ central with $d=c^{-1}$; conversely a central $c$ with $d=c^{-1}$ gives $cxc^{-1}=x$.

The identity criterion is the reason a unitarity condition on a sandwich carries a central factor and not only $1$: the parameters $(a,b)$ and $(\lambda a,\lambda^{-1}b)$ with $\lambda$ central determine the same sandwich, so the unitarity is a condition modulo this scaling.

The Unitarity Condition

Theorem. The signed sandwich is unitary for the twisted pairing, $S_{a,b}^{*_\sigma}S_{a,b}=S_{a,b}S_{a,b}^{*_\sigma}=\mathrm{id}$, exactly when

$$ c=\delta(a)\alpha(a) \text{ is central}, \qquad \text{and} \qquad d=\alpha(b)\delta(b)=c^{-1} . $$

Equivalently, $S_{a,b}^{*_\sigma}=S_{a,b}^{-1}$, and the two products agree because the operators are square matrices over $k$.

Proof. By the theorem and the composition rule, $S_{a,b}^{*_\sigma}S_{a,b}=S_{\delta(a),\delta(b)}S_{a,b}=T_{c,d}$ with $c=\delta(a)\alpha(a)$ and $d=\alpha(b)\delta(b)$. The identity criterion of the preceding section says that $T_{c,d}=\mathrm{id}$ exactly when $c$ is central and $d=c^{-1}$, which is the displayed condition; the equivalence with $S^{*_\sigma}=S^{-1}$ is the inverse rule, and one of the two products being the identity suffices for the other because a one-sided inverse of a square matrix is two-sided.

Corollary (the reflection). For the signed conjugation $\rho_u=S_{u,u^{-1}}$ the unitarity condition is the single condition

$$ \rho_u \text{ is unitary} \iff \sigma(u)u \text{ is central and } \alpha\bigl(\sigma(u)u\bigr)=\sigma(u)u . $$

so the reflection is a unitary operator exactly when $\sigma(u)u$ is an element of the even centre of the graded algebra.

Proof. With $b=\alpha(a)^{-1}$ and $a=u$ one computes $\alpha(b)\delta(b)=\alpha(\alpha(u)^{-1})\delta(\alpha(u)^{-1})=u^{-1}\delta(\alpha(u))^{-1}$ and $\delta(\alpha(u))=\sigma(u)$, so $d=(\sigma(u)u)^{-1}$; and $c=\delta(u)\alpha(u)=\alpha(\sigma(u)u)$ because $\alpha\sigma=\sigma\alpha$. The general condition $d=c^{-1}$ is then $\sigma(u)u=\alpha(\sigma(u)u)$, and the centrality of $c=\alpha(\sigma(u)u)$ is the centrality of $\sigma(u)u$.

Corollary (the one-sided case). For the signed left multiplication $S_{a,1}=\Lambda(a)$ the unitarity condition is

$$ S_{a,1} \text{ is unitary} \iff \delta(a)\alpha(a)=1 \iff \alpha(a) \text{ is a unitary element of } A , $$

since $\delta(a)=\sigma(\alpha(a))$ and the equation $\sigma(\alpha(a))\alpha(a)=1$ is the unitarity of $\alpha(a)$ for $\sigma$.

Proof. In the general condition $d=\alpha(1)\delta(1)=1$, so $c^{-1}=1$ and $c=1$, that is $\delta(a)\alpha(a)=1$; with $\delta=\sigma\alpha$ and $v=\alpha(a)$ this is $\sigma(v)v=1$, the unitarity of $v$. The full development of the one-sided operator is The Signed Adjoint of the Left Multiplication on an Algebra.

Corollary (the unitary elements). The unitary elements of the algebra for the anti-automorphism $\delta$ are the units $u$ with

$$ u^{*_\sigma}u=uu^{*_\sigma}=1, \qquad u^{*_\sigma}=\delta(u), $$

and they form a subgroup of $A^{\times}$; the one-sided signed operators $\Lambda(u)$ with $u$ of this form are unitary, and the signed sandwiches with $a$ and $b$ satisfying the two conditions above are the unitary sandwiches.

Proof. The group statement is Unitary Elements of an Involutive Algebra applied to the involutive algebra $(A,\delta)$; the identification of the unitary one-sided operators is the corollary above with $\delta(u)=u^{-1}$, and the two-sided statement is the theorem.

Examples

(a) The matrix sandwich. $A=M_n(k)$ with the transpose $\sigma$ and the grade involution $\alpha$ of a $\mathbb{Z}/2$-grading; the signed adjoint of $S_{a,b}(X)=a\alpha(X)b$ is $S_{\delta(a),\delta(b)}$ with $\delta=\sigma\alpha$, and for $\alpha=\mathrm{id}$ the adjoint of $X\mapsto aXb$ is $X\mapsto a^{\mathsf{T}}Xb^{\mathsf{T}}$, the transpose of the matrix parameters applied to $X$. The centrality condition is the scalarity of the parameter products.

(b) The signed conjugation. For $b=a^{-1}$ and $\alpha=\mathrm{id}$ the sandwich is the inner automorphism $x\mapsto axa^{-1}$, and its twisted adjoint is $x\mapsto\sigma(a)x\sigma(a)^{-1}$; it is self-adjoint only when $\sigma(a)$ is a central multiple of $a$, and unitary only when $\sigma(a)a$ is central, which for $\alpha=\mathrm{id}$ is the condition $\sigma(a)a\in Z(A)$.

(c) The reflection and the odd element. For a graded algebra and an odd element $u$ with $u^{2}$ central, the signed conjugation $\rho_u$ is a reflection with $\rho_u^{2}=\iota_{-u^{2}}$; it is unitary exactly when $\sigma(u)u$ is central and even. The Clifford algebra supplies the standard instance, and its metric reading is Part II and Quadratic Forms and Clifford Algebras.

(d) The unit case. $a=b=1$ gives the grade involution $S_{1,1}=\alpha$, whose signed adjoint is itself; it is unitary because $\alpha^{2}=\mathrm{id}$ and $\alpha$ is self-adjoint, which is the case $c=d=1$ of the general condition.

(e) The central scaling. The parameters $(a,b)$ and $(\lambda a,\lambda^{-1}b)$ with $\lambda$ central give the same signed sandwich when $\lambda$ is fixed by $\alpha$; the unitarity condition is invariant under the scaling, since $\delta(\lambda a)\alpha(\lambda a)=\lambda^{2}\delta(a)\alpha(a)$ and $\alpha(\lambda^{-1}b)\delta(\lambda^{-1}b)=\lambda^{-2}\alpha(b)\delta(b)$, the two products squaring back into inverse positions.

Summary

For the twisted pairing the signed sandwich $S_{a,b}(x)=a\alpha(x)b$ has signed adjoint $S_{a,b}^{*_\sigma}=S_{\delta(a),\delta(b)}$ with $\delta=\sigma\alpha$, the parameters not reversed, while for the plain pairing the adjoint is $S_{\alpha(b),\alpha(a)}$, the parameters reversed and twisted by $\alpha$; the two are related by the conjugation $c_{\sigma}(T)=\sigma T\sigma$, and the grade involution $\alpha=S_{1,1}$ is self-adjoint for both. The sandwich is unitary, $S^{*_\sigma}S=SS^{*_\sigma}=\mathrm{id}$, exactly when the central elements $c=\delta(a)\alpha(a)$ and $d=\alpha(b)\delta(b)$ are inverse to one another; for the reflection $b=\alpha(a)^{-1}$ this reduces to the centrality and the $\alpha$-fixedness of $\sigma(a)a$, and for the one-sided operator $b=1$ to $\delta(a)\alpha(a)=1$, which is the unitarity of $\alpha(a)$. The unitary elements for $\delta$ satisfy $u^{*_\sigma}u=uu^{*_\sigma}=1$ with $u^{*_\sigma}=\delta(u)$ and form a subgroup of the units; the signed sandwich space is stable under the signed adjoint, and the identity is reached only up to the central scaling $(a,b)\sim(\lambda a,\lambda^{-1}b)$ of the parameters.

Summary of Notation

Symbol	Meaning
$S_{a,b}(x)=a\alpha(x)b$	the signed sandwich
$T_{a,b}(x)=axb$	the unsigned sandwich
$\alpha$, $\sigma$, $\delta=\sigma\alpha$	the grade involution, the involution, and their twist
$\{x,y\}=\tau(x\sigma(y))$	the twisted pairing
$S_{a,b}^{*_\sigma}=S_{\delta(a),\delta(b)}$	the signed adjoint; no reversal
$S_{a,b}^{*}=S_{\alpha(b),\alpha(a)}$	the plain adjoint; reversal
$S_{a,b}S_{c,d}=T_{a\alpha(c),\alpha(d)b}$	the composition rule
$S_{a,b}^{-1}=S_{\alpha(a)^{-1},\alpha(b)^{-1}}$	the inverse
$T_{c,d}=\mathrm{id}\iff c\in Z(A),\ d=c^{-1}$	the identity criterion
$c=\delta(a)\alpha(a)$, $d=\alpha(b)\delta(b)$	the unitarity parameters
unitary $\iff c\in Z(A),\ d=c^{-1}$	the unitarity condition
$\sigma(u)u\in Z(A)$, $\alpha$-fixed	the unitary reflections $\rho_u$
$u^{_\sigma}u=uu^{_\sigma}=1$, $u^{*_\sigma}=\delta(u)$	the unitary elements for $\delta$