The Involution on the Operator Algebra

Introduction

The algebra of bounded operators of a Banach algebra is an algebra without an involution, but a form on the algebra turns it into one: the adjoint of a bounded operator $T$ is the operator $T^\dagger$ determined by $\{Tx,y\} = \{x,T^\dagger y\}$, and the map $T \mapsto T^\dagger$ is additive, involutive and anti-multiplicative, so it is an anti-automorphism of the operator algebra of order two. The form of the category is the trace pairing $\{x,y\} = \tau(x\sigma(y))$ attached to a continuous $\sigma$-invariant trace $\tau$; when it is nondegenerate the adjoint exists and is unique, the adjointable operators form a closed subalgebra, and the left regular representation $a \mapsto L_a$ becomes a *-representation, $(L_a)^\dagger = L_{\sigma(a)}$. This article fixes the form of the category, defines the adjoint and proves that it is an order-two anti-automorphism of the operator algebra.

The article assumes the Banach algebra, the bounded operators, the operator norm and the left and right multiplications from Operators on a Banach Algebra; the involution, its continuity, the fixed and skew elements and the isometry of the involution from Involutive Banach Algebras and the Gelfand–Naimark Theorem; the self-adjoint and unitary elements from Adjoints in a Banach Algebra; the completeness and closed subalgebras of the operator algebra from The Operator Algebra of a Banach Space; and the trace, the pairing and the operator involution of the finite-dimensional case from Frobenius Algebras and The Adjoint in an Involutive Algebra. The signed and reflection adjoints are the articles that follow; the grade involution $\alpha$ appears there.

Throughout, $A$ is a unital Banach algebra over $\mathbb{C}$ with a continuous involution $\sigma$, $a^* = \sigma(a)$; $\tau : A \to \mathbb{C}$ is a continuous $\sigma$-invariant trace, $\tau(xy) = \tau(yx)$ and $\tau(\sigma(x)) = \tau(x)$, whose trace pairing

$$ \{x,y\} = \tau\bigl(x\,\sigma(y)\bigr) , \qquad x,y \in A , $$

is nondegenerate — this is the form of the category; $B(A)$ is the unital Banach algebra of bounded operators; $L_a(x) = ax$ and $R_b(x) = xb$; an operator $T \in B(A)$ is adjointable when there is $T^\dagger$ with $\{Tx,y\} = \{x,T^\dagger y\}$ for all $x,y$; and $\mathcal{A}(A)\subseteq B(A)$ is the set of adjointable operators.

The Form of the Category

Proposition (the form is sesquilinear and symmetric). The trace pairing is conjugate-linear in the second variable, linear in the first, and

$$ \{x,y\} = \overline{\{y,x\}} , \qquad \{\sigma(x),\sigma(y)\} = \overline{\{x,y\}} , $$

and it is nondegenerate by hypothesis; it is the twisted pairing $\{x,y\} = \langle x,\sigma(y)\rangle$ of the plain pairing $\langle x,y\rangle = \tau(xy)$.

Proof. Conjugate-linearity in the second variable is the conjugate-linearity of $\sigma$ and the linearity of $\tau$; for symmetry, $\overline{\{y,x\}} = \overline{\tau(y\sigma(x))} = \tau(\sigma(y\sigma(x))) = \tau(x\sigma(y)) = \{x,y\}$, using the $\sigma$-invariance of $\tau$ and $\sigma(y\sigma(x)) = x\sigma(y)$; the last identity is the definition of the twisted pairing. $\square$

Remark (the $\mathrm{C}^*$-specialisation). When $A$ is a $\mathrm{C}^*$-algebra of operators on a Hilbert space $H$ with the trace class and $\tau$ the Hilbert–Schmidt trace, the form of the category is the Hilbert–Schmidt inner product $\{x,y\} = \operatorname{tr}(x y^*)$, and the adjoint of an operator is the usual Hilbert-space adjoint; the trace pairing is then positive definite and the theory of this article is the operator theory of Operator Algebras read through the form.

The Adjoint

Theorem (existence, uniqueness and the elementary laws). Let $T \in B(A)$ be adjointable. Then $T^\dagger$ is unique, and the assignment $T \mapsto T^\dagger$ is additive, conjugate-linear when the scalar twist is present, involutive and anti-multiplicative:

$$ (S + T)^\dagger = S^\dagger + T^\dagger , \qquad (ST)^\dagger = T^\dagger S^\dagger , \qquad (T^\dagger)^\dagger = T , \qquad (\lambda T)^\dagger = \bar\lambda\,T^\dagger . $$

Hence the adjoint is an anti-automorphism of the algebra $\mathcal{A}(A)$ of order two, an isomorphism $\mathcal{A}(A) \to \mathcal{A}(A)^{\mathrm{op}}$, and the adjointable operators form a subalgebra of $B(A)$.

Proof. Uniqueness and additivity are the nondegeneracy of the form: $\{x,(S+T)^\dagger y\} = \{(S+T)x,y\} = \{Sx,y\} + \{Tx,y\} = \{x,S^\dagger y\} + \{x,T^\dagger y\}$, and the form is conjugate-linear in the second argument. For the product, $\{x,(ST)^\dagger y\} = \{STx,y\} = \{Tx,S^\dagger y\} = \{x,T^\dagger S^\dagger y\}$. The order two is $\{x,Ty\} = \{T^\dagger x,y\} = \{x,(T^\dagger)^\dagger y\}$, and the compatibility with the scalars is the conjugation. $\square$

Theorem (self-adjoint and skew operators). An operator $T$ is self-adjoint when $T^\dagger = T$ and skew when $T^\dagger = -T$; the self-adjoint operators form a real linear subspace and the skew operators the real subspace $i\mathcal{A}(A)^+$; every adjointable $T$ decomposes as

$$ T = \tfrac12(T + T^\dagger) + \tfrac12(T - T^\dagger) , $$

and the self-adjoint and the skew operators are closed when the adjoint is continuous. The self-adjoint operators are exactly the operators fixed by the involution, and they are the analogue in $\mathcal{A}(A)$ of the self-adjoint elements of $A$.

Proof. The decomposition is formal and the parts are self-adjoint and skew by the involutivity and conjugate-linearity; closedness is the closedness of the fixed set of a continuous map, the adjoint being continuous for the topology of bounded convergence when $\sigma$ and $\tau$ are. $\square$

Theorem (the left regular representation is a *-representation). The left and right multiplications are adjointable, with

$$ (L_a)^\dagger = L_{\sigma(a)} , \qquad (R_b)^\dagger = R_{\sigma(b)} , $$

and the map $a \mapsto L_a$ is a *-representation of $(A,\sigma)$ in $\mathcal{A}(A)$: $L_{ab} = L_aL_b$, $L_{a^*} = (L_a)^\dagger$. Hence the involution of the elements and the adjoint of the operators agree on the image of the regular representation, an agreement proved and not assumed.

Proof. $\{L_ax,y\} = \tau(ax\sigma(y)) = \tau(x\sigma(y)a) = \tau(x\sigma(\sigma(a)y)) = \{x,L_{\sigma(a)}y\}$ by the cyclicity of $\tau$ and $\sigma^2 = \mathrm{id}$; the right-handed computation is the mirror. Multiplicativity is associativity and the adjoint formula is the computation. $\square$

The Adjointable Operators

Proposition (closure and completion). The adjointable operators form a subalgebra $\mathcal{A}(A)$ of $B(A)$, closed under the adjoint; it is a closed subalgebra when the adjoint is continuous, and it is complete when $B(A)$ is, so the involution descends to the completion of $\mathcal{A}(A)$. The adjoint is continuous for the topology of bounded convergence.

Proof. Closure under products, sums and the adjoint is the theorem; the adjoint map $T \mapsto T^\dagger$ is the composite of the transpose with the form, continuous when the form is continuous and nondegenerate, and a closed $\dagger$-stable subalgebra is complete after completion with its involution extended by continuity. $\square$

Example (the matrix algebra). For $A = M_n(\mathbb{C})$ with $\sigma$ the conjugate transpose and $\tau$ the trace, the trace pairing is the Hilbert–Schmidt inner product, the adjoint of an operator $T$ on $M_n$ is the Hilbert–Schmidt adjoint, and the self-adjoint operators are the operators fixed by it; the left regular representation is the *-representation of $M_n$ on itself.

Example (the group algebra). For the group algebra of a finite group with $\tau(\sum a_g u_g) = a_1$ and the involution $u_g^* = u_{g^{-1}}$, the trace pairing is nondegenerate and the adjoint of the left multiplication by a group element is the left multiplication by its inverse. The trace pairing is the base of the operator theory of the group algebra, Group Algebras.

The Transpose, the Form and the Matrix Model

Definition. An operator $T$ has a transpose $T^{\mathsf t}$ relative to the plain pairing $\langle x,y\rangle = \tau(xy)$ when $\langle Tx,y\rangle = \langle x,T^{\mathsf t}y\rangle$; the adjoint is then the twist of the transpose by the involution.

Proposition (transpose and adjoint). If $T$ has a transpose $T^{\mathsf t}$ then $T$ is adjointable and

$$ T^\dagger = \sigma\,T^{\mathsf t}\,\sigma , $$

so the adjoint is the transpose conjugated by the involution; the transpose is linear and multiplicative, $(ST)^{\mathsf t} = T^{\mathsf t}S^{\mathsf t}$, and the adjoint inherits its anti-multiplicativity through the twist.

Proof. $\{Tx,y\} = \tau(Tx\,\sigma(y)) = \langle Tx,\sigma(y)\rangle = \langle x,T^{\mathsf t}\sigma(y)\rangle = \tau(x\,T^{\mathsf t}\sigma(y))$; and $\{x,\sigma(T^{\mathsf t}\sigma(y))\} = \tau(x\,\sigma(\sigma(T^{\mathsf t}\sigma(y)))) = \tau(x\,T^{\mathsf t}\sigma(y))$, so $T^\dagger = \sigma T^{\mathsf t}\sigma$. The multiplicativity of the transpose is the associativity of the product, and the twist by the anti-automorphism $\sigma$ reverses the order, giving the anti-multiplicativity of the adjoint. $\square$

Example (the matrix model). For $A = M_n(\mathbb{C})$ with $\sigma$ the conjugate transpose and $\tau$ the trace, the plain pairing is $\langle X,Y\rangle = \operatorname{tr}(XY)$ and the transpose of an operator is the ordinary matrix transpose of the operator on the matrix space; the adjoint is its conjugate transpose, and the two coincide for real matrices.

Summary

The form of the category of the operator theory on a Banach algebra $A$ with a continuous involution $\sigma$ and a continuous $\sigma$-invariant trace $\tau$ is the trace pairing $\{x,y\} = \tau(x\sigma(y))$, nondegenerate by hypothesis and sesquilinear; it defines the adjoint $T^\dagger$ by $\{Tx,y\} = \{x,T^\dagger y\}$. The adjoint is unique, additive, conjugate-linear and anti-multiplicative, $(ST)^\dagger = T^\dagger S^\dagger$ and $(T^\dagger)^\dagger = T$, so it is an order-two anti-automorphism of the algebra of adjointable operators, an isomorphism onto its opposite; the self-adjoint ($T^\dagger = T$) and skew ($T^\dagger = -T$) operators are real subspaces of the adjointable operators, closed when the adjoint is continuous, and every adjointable operator is their average. The left and right multiplications are adjointable with $(L_a)^\dagger = L_{\sigma(a)}$ and $(R_b)^\dagger = R_{\sigma(b)}$, so the left regular representation is a *-representation, and the adjointable operators form a closed subalgebra that is complete with $B(A)$. The signed, reflection and graded adjoints are the articles that follow.

Summary of Notation

Symbol	Meaning
$A$, $\sigma$, $a^* = \sigma(a)$	Banach algebra with a continuous involution
$\tau$, $\tau(xy) = \tau(yx) = \tau(\sigma(x))$	Continuous $\sigma$-invariant trace
$\{x,y\} = \tau(x\sigma(y))$	The form of the category (trace pairing), nondegenerate
$T^\dagger$, $\{Tx,y\} = \{x,T^\dagger y\}$	The adjoint of an operator
$(ST)^\dagger = T^\dagger S^\dagger$, $(T^\dagger)^\dagger = T$	Anti-multiplicativity and order two
$T^\dagger = T$, $T^\dagger = -T$	Self-adjoint and skew operators
$\mathcal{A}(A)$	The adjointable operators, a closed subalgebra
$(L_a)^\dagger = L_{\sigma(a)}$	The regular representation is a `*`-representation
$T^{\mathsf t}$, $T^\dagger = \sigma T^{\mathsf t}\sigma$	Transpose for the plain pairing; the adjoint