Proof of The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra

Each result cited below is universally quantified over the data in its own statement. The following facts are used throughout.

(Alg) By Cyclic Tracial Operator Algebras and Their Traces §star-algebra, $\mathcal{A}\subseteq\mathcal{L}(H)$ contains $I$ and is closed under sums, scalar multiples, products and adjoints.

(Adj) Let $S,T\in\mathcal{L}(H)$ and $c\in\mathbb{C}$. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint, $S$ has an adjoint $S^{*}\in\mathcal{L}(H)$, with $\langle S^{*}\eta,\xi\rangle=\langle\eta,S\xi\rangle$ (Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint). By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus and Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique, $(S^{*})^{*}=S$, so $\langle S\eta,\xi\rangle=\langle\eta,S^{*}\xi\rangle$; moreover $(S+T)^{*}=S^{*}+T^{*}$, $(cS)^{*}=\overline{c}\,S^{*}$, $(ST)^{*}=T^{*}S^{*}$, $I^{*}=I$, and $\langle\xi,S^{*}S\xi\rangle=\lVert S\xi\rVert^{2}$.

(Dense) Put $D=\mathcal{A}\Omega$. We have $S\Omega+T\Omega=(S+T)\Omega$, $c(S\Omega)=(cS)\Omega$ and $0=(0I)\Omega$. By (Alg), $D$ is therefore a linear subspace of $H$, and it is dense in $H$ by Cyclic Tracial Operator Algebras and Their Traces §cyclic.

Proof of clause 1 (Trace). Recall $\tau(S)=\langle\Omega,S\Omega\rangle$ (Cyclic Tracial Operator Algebras and Their Traces §trace). Linearity of $\tau$ follows from conditions 2 and 3 of Complex Inner Product Space: $\tau(S+T)=\langle\Omega,S\Omega+T\Omega\rangle=\tau(S)+\tau(T)$ and $\tau(cS)=c\,\tau(S)$. Next, $\tau(I)=\langle\Omega,\Omega\rangle=\lVert\Omega\rVert^{2}=1$ by Cyclic Tracial Operator Algebras and Their Traces §cyclic. The identity $\tau(ST)=\tau(TS)$ is Cyclic Tracial Operator Algebras and Their Traces §tracial. By condition 1 of Complex Inner Product Space and (Adj), $\tau(S^{*})=\langle\Omega,S^{*}\Omega\rangle=\overline{\langle S^{*}\Omega,\Omega\rangle}=\overline{\langle\Omega,S\Omega\rangle}=\overline{\tau(S)}$. By (Adj), $\tau(S^{*}S)=\langle\Omega,S^{*}S\Omega\rangle=\lVert S\Omega\rVert^{2}$. Since $S^{*}\in\mathcal{A}$, the last identity applied to $S^{*}$, together with $(S^{*})^{*}=S$ and traciality, gives

Both norms are nonnegative, so $\lVert S^{*}\Omega\rVert=\lVert S\Omega\rVert$ by Existence and Uniqueness of the Nonnegative Square Root.

Proof of clause 2 (Separation). Let $S\in\mathcal{A}$ with $S\Omega=0$, and let $T,U\in\mathcal{A}$. By (Adj), (Alg) and traciality,

since the linear map $TU^{*}$ sends $0$ to $0$. Thus $\langle\eta,S\xi\rangle=\langle\eta,0\xi\rangle$ for all $\xi,\eta$ in the dense linear subspace $D$. By Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §equality, applied with $K=H$ and $E=D$, $S=0$.

Now let $T\in\mathcal{A}'$ with $T\Omega=0$. For $S\in\mathcal{A}$, $T(S\Omega)=S(T\Omega)=S0=0$ by The Commutant of a Set of Bounded Operators on a Complex Hilbert Space §commutant. So $T$ and $0$, both in $\mathcal{L}(H)$, agree on $D$, and $T=0$ by the final sentence of Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §equality.

Finally, if $S\in\mathcal{A}$ and $\tau(S^{*}S)=0$, then $\lVert S\Omega\rVert^{2}=0$ by clause 1. Hence $S\Omega=0$ by condition 4 of Complex Inner Product Space, and $S=0$ by the first part.

Proof of clause 3 (Conjugation). The map on $D$. Define $J_0:D\to H$ by $J_0(S\Omega)=S^{*}\Omega$ for $S\in\mathcal{A}$. This is well defined. If $S,S'\in\mathcal{A}$ satisfy $S\Omega=S'\Omega$, then $(S-S')\Omega=0$ with $S-S'\in\mathcal{A}$, so clause 1 gives $\lVert(S-S')^{*}\Omega\rVert=\lVert(S-S')\Omega\rVert=0$. By (Adj), $(S-S')^{*}=S^{*}-S'^{*}$, hence $S^{*}\Omega=S'^{*}\Omega$. For $S,T\in\mathcal{A}$ and $c\in\mathbb{C}$, (Adj) gives

and $\lVert J_0(S\Omega)\rVert=\lVert S\Omega\rVert$ by clause 1. By Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §conjugate-linear, applied with $K=H$ and $C=1$, there is a unique continuous $J:H\to H$ extending $J_0$. It is additive, satisfies $J(c\xi)=\overline{c}\,J\xi$, and satisfies $\lVert J\xi\rVert=\lVert\xi\rVert$ for all $\xi\in H$. By construction $J(S\Omega)=S^{*}\Omega$ for $S\in\mathcal{A}$.

Involution. For $\xi,\eta\in H$ and $c\in\mathbb{C}$, $J(J(\xi+\eta))=J(J\xi+J\eta)=J(J\xi)+J(J\eta)$ and $J(J(c\xi))=J(\overline{c}\,J\xi)=c\,J(J\xi)$, by claim 1 of Properties of Complex Conjugation and Modulus. Also $\lVert J(J\xi)\rVert=\lVert\xi\rVert$. So $J\circ J$ is linear with bound $1$ and lies in $\mathcal{L}(H)$ (Bounded Linear Maps between Complex Normed Spaces and the Operator Norm §bounded). For $S\in\mathcal{A}$ we have $S^{*}\in\mathcal{A}$, so $J(J(S\Omega))=J(S^{*}\Omega)=(S^{*})^{*}\Omega=S\Omega$. So $J\circ J$ and $I$ agree on $D$, and $J\circ J=I$ by Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §equality.

The inner product identity. For $S,T\in\mathcal{A}$, (Adj), (Alg) and traciality give

We extend this identity to all of $H$ by continuity. First, if $a_k\to a$ and $b_k\to b$ in $H$, then $\langle a_k,b_k\rangle\to\langle a,b\rangle$ in $\mathbb{C}$. Indeed, sesquilinearity (conditions 2 and 3 of Complex Inner Product Space, claims 1 and 2 of Elementary Properties of a Complex Inner Product) gives $\langle a_k,b_k\rangle-\langle a,b\rangle=\langle a_k-a,b_k-b\rangle+\langle a_k-a,b\rangle+\langle a,b_k-b\rangle$. So by claim 1 of The Induced Norm is a Norm, and Induces a Metric and claim 7 of Properties of Complex Conjugation and Modulus,

Given $\varepsilon>0$, put $M=\lVert a\rVert+\lVert b\rVert+1$. Once $\lVert a_k-a\rVert$ and $\lVert b_k-b\rVert$ are below both $1$ and $\varepsilon/(3M)$, each term is below $\varepsilon/3$.

Now let $\xi,\eta\in H$, and use (Dense), Dense Subset of a Topological Space and Sequential Characterization of the Closure in a Metric Space to choose sequences $(\xi_k),(\eta_k)$ in $D$ converging to $\xi,\eta$. Then $\lVert J\xi_k-J\xi\rVert=\lVert J(\xi_k-\xi)\rVert=\lVert\xi_k-\xi\rVert$, so $J\xi_k\to J\xi$, and likewise $J\eta_k\to J\eta$. By the continuity just shown, $\langle J\xi_k,J\eta_k\rangle\to\langle J\xi,J\eta\rangle$ and $\langle\eta_k,\xi_k\rangle\to\langle\eta,\xi\rangle$. These are the same sequence by the identity on $D$. So $\langle J\xi,J\eta\rangle=\langle\eta,\xi\rangle$ by Uniqueness of Limits in a Metric Space. Hence $J$ is a conjugation of $H$ (Conjugation of a Complex Hilbert Space §conjugation) with $J(S\Omega)=S^{*}\Omega$ for $S\in\mathcal{A}$.

Uniqueness. Let $J'$ be any conjugation of $H$ with $J'(S\Omega)=S^{*}\Omega$ for $S\in\mathcal{A}$. Then $\lVert J'\xi\rVert^{2}=\langle J'\xi,J'\xi\rangle=\langle\xi,\xi\rangle=\lVert\xi\rVert^{2}$, so $\lVert J'\xi\rVert=\lVert\xi\rVert$ (Norm Induced by a Complex Inner Product). Additivity and conjugate homogeneity with $\overline{-1}=-1$ give $J'\xi-J'\eta=J'(\xi-\eta)$, hence $\lVert J'\xi-J'\eta\rVert=\lVert\xi-\eta\rVert$. So $\delta=\varepsilon$ meets Continuous Map Between Metric Spaces, and $J'$ is continuous on $H$. It extends $J_0$, so $J'=J$ by the uniqueness in Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §conjugate-linear. Finally $J\Omega=J(I\Omega)=I^{*}\Omega=\Omega$ by (Adj).

Proof of clause 4 (Right action). Let $S\in\mathcal{A}$. Since $S\in\mathcal{L}(H)$ and $J$ is a conjugation, $JSJ\in\mathcal{L}(H)$ by Basic Properties of Commutants: Unital Algebras Closed under Weak Limits, Order Reversal, the Triple Commutant, and Conjugation §conjugation. For $T\in\mathcal{A}$ we have $T^{*},ST^{*}\in\mathcal{A}$ by (Alg), so clause 3 and (Adj) give

Now let $U\in\mathcal{A}$. For every $T\in\mathcal{A}$, $UT\in\mathcal{A}$, so by the formula just proved

Thus $(JSJ)U$ and $U(JSJ)$, both in $\mathcal{L}(H)$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations, agree on $D$. By Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §equality, $(JSJ)U=U(JSJ)$. As $U\in\mathcal{A}$ was arbitrary, $JSJ\in\mathcal{A}'$ by The Commutant of a Set of Bounded Operators on a Complex Hilbert Space §commutant.