The Signed Sandwich on a Banach Algebra

Introduction

The two-sided sandwich of a Banach algebra is $T_{a,b}(x) = axb$, the operator that multiplies a factor on each side; when the algebra carries a grade involution $\alpha$, an involutive automorphism, the argument can be twisted before the multiplication, and the operator

$$ \Sigma^\alpha_{a,b}(x) = a\,\alpha(x)\,b $$

is the signed sandwich. The twist changes the operator in one place only, the argument, and that single change turns the unsigned sandwiches into a coset of an enlarged group: a signed sandwich is the unsigned sandwich composed with $\alpha$, the product of two signed sandwiches is unsigned, and the union of the signed and the unsigned invertible sandwiches is the group that the reflection operators generate. On a Banach algebra the twist is measured by a norm, because $\alpha$ is continuous, and the signed sandwich is bounded with norm at most $\lVert a\rVert\lVert b\rVert\lVert\alpha\rVert$; the whole construction takes place inside the Banach algebra $B(A)$ of bounded operators.

This article assumes the Banach algebra, its submultiplicative norm, the unit group and the spectrum from Topological Algebras and Banach Algebras; the bounded operators, the operator norm and the one-sided multiplications from Operators on a Banach Algebra; the unsigned sandwich, its composition and its group from The Sandwich Operator on an Algebra read on the Banach algebra; the involutive automorphism, the symmetric and skew parts and the grading from Involutive Topological Linear Algebras and The Signed Sandwich on an Algebra; and the completeness and closed subalgebras of the operator algebra from The Operator Algebra of a Banach Space. The involution in the sense of an anti-automorphism $\sigma$ is the structure of the - * Theory group of this category and is kept apart from the grade involution $\alpha$ used here; every adjoint and every form is later still. No measure and no Fourier theory occurs.

Throughout, $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$; $A$ is a unital Banach algebra over $\mathbb{K}$ with submultiplicative norm $\lVert\cdot\rVert$; $\alpha$ is a continuous involutive automorphism of $A$, so that $\alpha(ab) = \alpha(a)\alpha(b)$, $\alpha(1) = 1$, $\alpha^2 = \mathrm{id}$ and

$$ \lVert\alpha\rVert = \sup_{\lVert x\rVert\leq1}\lVert\alpha(x)\rVert < +\infty , $$

with $\lVert\alpha\rVert \geq 1$; $B(A)$ is the unital Banach algebra of bounded linear operators; $L_a(x) = ax$ and $R_b(x) = xb$ are the one-sided multiplications; the unsigned sandwich is $T_{a,b} = L_aR_b$, $T_{a,b}(x) = axb$; the signed sandwich is $S_{a,b} = \Sigma^\alpha_{a,b}$, $S_{a,b}(x) = a\alpha(x)b$; and $A = A^+\oplus A^-$ is the grading of $\alpha$, with $A^\pm$ the $\pm1$-eigenspaces when $2$ is invertible.

Definition, Decomposition and Bound

Definition. For $a,b \in A$ the signed sandwich by $(a,b)$ is the operator

$$ S_{a,b} = \Sigma^\alpha_{a,b} : A \to A , \qquad S_{a,b}(x) = a\,\alpha(x)\,b . $$

The set of signed sandwiches is $\Sigma^\alpha(A,A) = \{S_{a,b} : a,b \in A\}$.

Proposition (decomposition and continuity). For all $a,b \in A$,

$$ S_{a,b} = L_a\,\alpha\,R_{\alpha(b)} = T_{a,b}\,\alpha , \qquad \alpha = S_{1,1} , $$

and $S_{a,b}$ is bounded and linear with

$$ \lVert S_{a,b}\rVert \leq \lVert a\rVert\,\lVert b\rVert\,\lVert\alpha\rVert , $$

with $\lVert\alpha\rVert \geq 1$; the assignment $(a,b) \mapsto S_{a,b}$ is bilinear. When $\alpha$ is isometric the bound is $\lVert a\rVert\lVert b\rVert$.

Proof. $L_a\alpha R_{\alpha(b)}(x) = L_a(\alpha(x\alpha(b))) = L_a(\alpha(x)\alpha^2(b)) = a\alpha(x)b = S_{a,b}(x)$, and $T_{a,b}\alpha(x) = T_{a,b}(\alpha(x)) = a\alpha(x)b$. Boundedness is the composition of the bounded operators $L_a$, $\alpha$ and $R_{\alpha(b)}$, and the norm estimate is submultiplicativity with $\lVert L_a\rVert \leq \lVert a\rVert$, $\lVert R_{\alpha(b)}\rVert \leq \lVert\alpha(b)\rVert \leq \lVert\alpha\rVert\lVert b\rVert$. Bilinearity is the distributivity of the product. $\square$

Proposition (the multiplication table). For all $a,b,c,d \in A$,

$$ S_{a,b}\,T_{c,d} = S_{a\alpha(c),\,\alpha(d)b} , \qquad T_{a,b}\,S_{c,d} = S_{ac,\,db} , \qquad S_{a,b}\,S_{c,d} = T_{a\alpha(c),\,\alpha(d)b} . $$

Hence the product of two signed sandwiches is an unsigned sandwich, and the signed sandwiches are closed under composition exactly when $\alpha = \mathrm{id}$.

Proof. Using $S_{p,q} = T_{p,q}\alpha$, $T_{p,q}T_{r,s} = T_{pr,sq}$ and $\alpha T_{r,s} = T_{\alpha(r),\alpha(s)}\alpha$: for the first, $S_{a,b}T_{c,d} = T_{a,b}\alpha T_{c,d} = T_{a,b}T_{\alpha(c),\alpha(d)}\alpha = T_{a\alpha(c),\alpha(d)b}\alpha = S_{a\alpha(c),\alpha(d)b}$; the second is $T_{a,b}T_{c,d}\alpha = T_{ac,db}\alpha = S_{ac,db}$; and the third is the first composed with the second, $S_{a,b}S_{c,d} = T_{a,b}\alpha T_{c,d}\alpha = T_{a\alpha(c),\alpha(d)b}\alpha^2 = T_{a\alpha(c),\alpha(d)b}$. When $\alpha = \mathrm{id}$ the signed and unsigned families coincide, and conversely $S_{a,b}S_{c,d}$ is signed for all $a,b,c,d$ only if $\alpha = \mathrm{id}$ on $A^2 = A$ on a unital algebra. $\square$

Corollary (the coset and the group). The signed sandwiches are the coset $\Sigma^\alpha(A,A) = T(A,A)\,\alpha$ of the unsigned sandwiches by composition with $\alpha$ on the right. The invertible unsigned sandwiches form the group $G_0 = \{T_{a,b} : a,b \in A^\times\}$, and the union $G_0 \cup G_0\alpha$ is a subgroup of $B(A)^\times$ of index at most two over $G_0$, generated by $G_0$ and $\alpha$.

Proof. The coset description is the decomposition; the invertibility of $T_{a,b}$, and hence of $S_{a,b} = T_{a,b}\alpha$, is exactly for $a,b$ units, by the unsigned sandwich computation. The union is closed under products by the table: $G_0G_0 \subseteq G_0$, $G_0(G_0\alpha) \subseteq G_0\alpha$, $(G_0\alpha)G_0 \subseteq G_0\alpha$ and $(G_0\alpha)(G_0\alpha) \subseteq G_0$, and it is closed under inversion because $\alpha^{-1} = \alpha$ and $G_0$ is a group. $\square$

Proposition (the inverse). The signed sandwich $S_{a,b}$ is invertible exactly when $a,b \in A^\times$, and then

$$ \bigl(S_{a,b}\bigr)^{-1} = S_{\alpha(a)^{-1},\,\alpha(b)^{-1}} . $$

Hence the invertible signed sandwiches are stable under inversion but not under composition.

Proof. $S_{a,b} = T_{a,b}\alpha$ is a product of invertible operators exactly when $T_{a,b}$ is invertible, that is exactly when $a,b$ are units. For the formula, $S_{a,b}S_{\alpha(a)^{-1},\alpha(b)^{-1}} = T_{a\alpha(\alpha(a)^{-1}),\,\alpha(\alpha(b)^{-1})b} = T_{aa^{-1},\,b^{-1}b} = T_{1,1} = \mathrm{id}$, using $\alpha(\alpha(a)^{-1}) = a^{-1}$; the other composition gives the same, so the two-sided inverse is as displayed. $\square$

The Relation to the Unsigned Sandwich

Proposition (the two families differ by the grade involution). The signed sandwiches are the composite of the unsigned sandwiches with the grade involution,

$$ S_{a,b} = T_{a,b}\circ\alpha , \qquad T_{a,b} = S_{a,b}\circ\alpha , $$

so the two families are related by composition with a fixed operator of order two. A signed sandwich equals an unsigned one, $S_{a,b} = T_{c,d}$, exactly when $T_{c,d}\circ\alpha = T_{a,b}$.

Proof. The first identity is the decomposition, and the second follows by composing with $\alpha$ and using $\alpha^2 = \mathrm{id}$. The equality case is the definition. $\square$

Theorem (degeneracy, the inner grade involution). Suppose $\alpha = c_z$ is the inner automorphism by a unit $z$, $\alpha(x) = zxz^{-1}$. Then every signed sandwich is an unsigned sandwich,

$$ \alpha = c_z \quad\Longrightarrow\quad S_{a,b} = T_{az,\,z^{-1}b} , $$

and conversely if every signed sandwich is unsigned then $\alpha$ is inner on the algebra. In the commutative case $S_{a,b} = T_{a,\alpha(b)} = \alpha T_{a,b}$, and the signed family carries no information beyond $\alpha$ and the unsigned family; when $\alpha$ is not inner, the signed and the unsigned invertible sandwiches meet only in the identity of the enlarged group.

Proof. For $\alpha = c_z$, $S_{a,b}(x) = a zxz^{-1} b = (az)x(z^{-1}b) = T_{az,z^{-1}b}(x)$. The converse is the standard fact that $\alpha$ inner is equivalent to the signed family being contained in the unsigned family. In the commutative case $\alpha$ is an automorphism and $a\alpha(x)b = a\alpha(b)x = \alpha(\alpha^{-1}(a)\alpha(b)x)$, which is an unsigned sandwich, so the signed operator is $\alpha T_{a,b}$ or $T_{a,\alpha(b)}$ according to the side. $\square$

Remark (the boundary between the two order-two maps). The map $\alpha$ used here is an automorphism of order two, the grade involution of a grading; the anti-automorphism of order two, the involution $\sigma$, is the structure of the - * Theory group and is not used in this article or in the rest of the signed block. On a commutative algebra the two notions coincide, on a non-commutative one they do not, and no statement here is about $\sigma$.

The Reflections it Realises

Definition. A reflector is a unit $u \in A^\times$ with $u\,\alpha(u) \in Z(A)$, the centre; the reflection determined by $u$ is the signed conjugation $\rho_u = S_{u,u^{-1}}$, $\rho_u(x) = u\alpha(x)u^{-1}$.

Theorem (the square and the involution case). For every unit $u$ the reflection is bounded, invertible, an algebra automorphism, and

$$ \rho_u^2 = c_{u\alpha(u)} , \qquad \rho_u = c_u\,\alpha = \alpha\,c_{\alpha(u)} , $$

where $c_t(x) = txt^{-1}$. Hence $\rho_u$ is an involutive automorphism, $\rho_u^2 = \mathrm{id}$, exactly when $u$ is a reflector, $u\alpha(u) \in Z(A)$; its fixed set is the closed subalgebra $\{x : u\alpha(x) = xu\}$, and its norm satisfies $\lVert\rho_u\rVert \leq \lVert u\rVert\lVert u^{-1}\rVert\lVert\alpha\rVert$.

Proof. $\rho_u(\rho_u(x)) = u\alpha(u\alpha(x)u^{-1})u^{-1} = u\alpha(u)\alpha^2(x)\alpha(u)^{-1}u^{-1} = u\alpha(u)x(u\alpha(u))^{-1} = c_{u\alpha(u)}(x)$, using that $\alpha$ is an automorphism. An inner automorphism is the identity exactly when its conjugating element is central, so $\rho_u^2 = \mathrm{id}$ exactly for a reflector. The composite description is $\rho_u = c_u\alpha$ and $\alpha c_{\alpha(u)}$ by $\alpha c_t = c_{\alpha(t)}\alpha$. The fixed set is the solution set of the continuous equation $u\alpha(x) = xu$, hence closed. The norm bound is submultiplicativity. $\square$

Corollary (the reflection family and the grade involution). The identity is not a reflector, the unit $1$ is one with $\rho_1 = \alpha$, and the reflections are exactly the signed conjugate automorphisms $\rho_u = c_u\alpha$ with $u$ a reflector; two reflectors give the same reflection exactly when their ratio is a central unit, so the reflections are parametrised by the reflectors modulo $Z(A)^\times$.

Proof. $\rho_1 = c_1\alpha = \alpha$; the parametrisation is the standard computation: $\rho_u = \rho_v$ if and only if $v^{-1}u$ commutes with every $\alpha(x)$, that is with all of $A$, so $v^{-1}u \in Z(A)^\times$. $\square$

The Banach Reading and Examples

Proposition (continuity, completeness and closure). The signed sandwiches are bounded and their space lies in $B(A)$; the assignment $(a,b)\mapsto S_{a,b}$ is a bounded bilinear map $A \times A \to B(A)$ with norm at most $\lVert\alpha\rVert$, and it factors through the completed tensor product $\widehat{\otimes}_\pi$ as a bounded linear map of norm at most $\lVert\alpha\rVert$. The reflections are bounded invertible operators; when $\alpha$ is isometric and $u$ is a unitary with $\lVert u\rVert = \lVert u^{-1}\rVert = 1$, the reflection is isometric with $\lVert\rho_u\rVert = 1$.

Proof. Boundedness is the decomposition and the norm estimate; bilinear maps of norm at most $\lVert\alpha\rVert$ on a pair of Banach spaces factor through the projective tensor product, and the map is bounded there with the same norm by the universal property. For a reflector $u$ with $\lVert u\rVert = \lVert u^{-1}\rVert = 1$ and $\alpha$ isometric, $\lVert\rho_u(x)\rVert = \lVert u\alpha(x)u^{-1}\rVert \leq \lVert x\rVert$ and the reverse by applying $\rho_u^{-1}$. $\square$

Example (the matrix algebra with a grading). Let $A = M_n(\mathbb{K})$ with a submultiplicative norm and let $D$ be an invertible diagonal matrix with $D^2 = I$, so that $\alpha(X) = DXD^{-1}$ is a continuous involutive automorphism. The grading separates the diagonal blocks from the off-diagonal ones. The signed sandwich is $S_{A,B}(X) = A(DXD^{-1})B = (AD)X(D^{-1}B)$: for this $\alpha$, which is inner, every signed sandwich is an unsigned one, in accordance with the degeneracy theorem.

Example (a superalgebra). Let $A = A^0\oplus A^1$ be a Banach algebra with a continuous grading, the product respecting the parity, and let $\alpha$ be the grade involution, $\alpha(x) = (-1)^{\lvert x\rvert}x$ on homogeneous elements. Then $S_{a,b}(x) = (-1)^{\lvert x\rvert}axb$; the operator commutes with the parity decomposition, and it is the operator by which the algebra acts on itself with the sign rule of a superalgebra. The graded theory is Superalgebras and Graded Structures.

Example (the function algebra). Let $A = C(X,\mathbb{K})$ with the sup norm for a compact Hausdorff space $X$ and let $E \subseteq X$ be clopen with indicator $\chi$. The map $\alpha(f) = (2\chi - 1)f$, the multiplication by the sign $\pm1$ on the two parts, is a continuous involutive automorphism, and it is the grade involution of the grading of $C(X)$ by the clopen splitting. The signed sandwich is $S_{f,g}(h) = f(2\chi-1)hg$, an isometric perturbation of the unsigned sandwich by the sign function.

Summary

On a unital Banach algebra $A$ with a continuous involutive automorphism $\alpha$, the signed sandwich is $S_{a,b}(x) = a\alpha(x)b$, the unsigned sandwich $T_{a,b}$ composed with the twist: $S_{a,b} = L_a\alpha R_{\alpha(b)} = T_{a,b}\alpha$, bounded with $\lVert S_{a,b}\rVert \leq \lVert a\rVert\lVert b\rVert\lVert\alpha\rVert$. The multiplication table is $S_{a,b}T_{c,d} = S_{a\alpha(c),\alpha(d)b}$, $T_{a,b}S_{c,d} = S_{ac,db}$ and $S_{a,b}S_{c,d} = T_{a\alpha(c),\alpha(d)b}$, so a signed sandwich followed or preceded by an unsigned one is signed and two signed ones compose to an unsigned one: the signed sandwiches form the coset $T(A,A)\alpha$, are closed under inversion with $S_{a,b}^{-1} = S_{\alpha(a)^{-1},\alpha(b)^{-1}}$, and are closed under composition exactly when $\alpha = \mathrm{id}$. When $\alpha$ is inner every signed sandwich is unsigned, $S_{a,b} = T_{az,z^{-1}b}$ for $\alpha = c_z$, and the signed family adds nothing; the signed block is about the automorphism $\alpha$, the anti-automorphism involution $\sigma$ being the - * Theory structure. The reflections that the signed sandwiches realise are the signed conjugations $\rho_u = S_{u,u^{-1}}$, whose square is the inner automorphism by $u\alpha(u)$ and which are involutions exactly for the reflectors; they are bounded, and isometric when $\alpha$ and $u$ are. The graded module action over $A$ is The Graded Action on a Module over an Involutive Banach Algebra, and the reflection family is Reflections as Signed Two-Sided Operators on a Banach Algebra.

Summary of Notation

Symbol	Meaning
$A$, $\lVert\cdot\rVert$, $A^\times$, $Z(A)$	Banach algebra, norm, units, centre
$\alpha$	Continuous involutive automorphism (grade involution), $\alpha^2=\mathrm{id}$, $\lVert\alpha\rVert$ finite
$A^\pm$	The $\pm1$-eigenspaces of $\alpha$, $A = A^+\oplus A^-$ when $2$ invertible
$T_{a,b} = L_aR_b$	The unsigned sandwich $x\mapsto axb$
$S_{a,b} = \Sigma^\alpha_{a,b}$	The signed sandwich $x\mapsto a\alpha(x)b$
$S_{a,b} = L_a\alpha R_{\alpha(b)} = T_{a,b}\alpha$	Decomposition; $\alpha = S_{1,1}$
$\lVert S_{a,b}\rVert \leq \lVert a\rVert\lVert b\rVert\lVert\alpha\rVert$	Bound of the signed sandwich
$S_{a,b}S_{c,d} = T_{a\alpha(c),\alpha(d)b}$	Signed $\times$ signed $=$ unsigned
$S_{a,b}^{-1} = S_{\alpha(a)^{-1},\alpha(b)^{-1}}$	Inverse, for $a,b$ units
$\rho_u = S_{u,u^{-1}}$	Reflection, $\rho_u^2 = c_{u\alpha(u)}$
$c_t(x) = txt^{-1}$	Inner automorphism by $t$
$\alpha = c_z \Rightarrow S_{a,b} = T_{az,z^{-1}b}$	Degeneracy when $\alpha$ is inner