Proof of The Tracial W*-Closure of a Cyclic Tracial Operator Algebra: the Commutant of the Right Action is a Tracial W*-Probability Space Equal to the Double Commutant

Each result cited below is universally quantified over the data in its own statement. We prove the claims in the order 1 (containment), 3 (right-bounded vectors), 2 (tracial W*-probability space), 4 (commutant), 5 (double commutant); each uses only the claims proved before it.

Preliminaries. By The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §conjugation, $J$ is a conjugation of $H$ in the sense of Conjugation of a Complex Hilbert Space §conjugation, with $J(S\Omega)=S^{*}\Omega$ for $S\in\mathcal{A}$ and $J\Omega=\Omega$. By Basic Properties of Commutants: Unital Algebras Closed under Weak Limits, Order Reversal, the Triple Commutant, and Conjugation §conjugation, $JAJ\in\mathcal{L}(H)$ for every $A\in\mathcal{L}(H)$, so $J\mathcal{A}J\subseteq\mathcal{L}(H)$ and $M=(J\mathcal{A}J)'$ is defined; that clause also gives $(JAJ)^{*}=JA^{*}J$, $(JAJ)(JBJ)=J(AB)J$, $J(JAJ)J=A$, $J(A+B)J=JAJ+JBJ$ and $J(cA)J=\overline{c}\,JAJ$ for $A,B\in\mathcal{L}(H)$ and $c\in\mathbb{C}$, which we use without further mention. Since $\lVert J\xi\rVert^{2}=\langle J\xi,J\xi\rangle=\langle\xi,\xi\rangle$, $J$ preserves norms, and $J\xi-J\eta=J(\xi-\eta)$ by additivity. By The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §right-action, for all $S,U\in\mathcal{A}$,

Two consequences. (P1) $J\mathcal{A}J$ is closed under adjoints: $(JSJ)^{*}=JS^{*}J$ with $S^{*}\in\mathcal{A}$ by Cyclic Tracial Operator Algebras and Their Traces §star-algebra. (P2) $J\mathcal{A}J\Omega=\mathcal{A}\Omega$: $JSJ\Omega=S^{*}\Omega\in\mathcal{A}\Omega$, and conversely $S\Omega=(S^{*})^{*}\Omega=J S^{*}J\Omega$ with $S^{*}\in\mathcal{A}$, using $(S^{*})^{*}=S$ from 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. Let $D=\mathcal{A}\Omega$. By Cyclic Tracial Operator Algebras and Their Traces §star-algebra, $D$ is a linear subspace ($S\Omega+cU\Omega=(S+cU)\Omega$ and $0=(0\cdot I)\Omega$), and by Cyclic Tracial Operator Algebras and Their Traces §cyclic it is dense in $H$. (P3) A vector $v\in H$ with $\langle\eta,v\rangle=0$ for every $\eta\in D$ is zero: for every $\eta\in D$, $\lVert v\rVert^{2}=\langle v-\eta,v\rangle+\langle\eta,v\rangle=\langle v-\eta,v\rangle\le\lVert v-\eta\rVert\,\lVert v\rVert$ by Cauchy-Schwarz Inequality in a Complex Inner Product Space, and $\lVert v-\eta\rVert$ can be made smaller than any real $\varepsilon>0$ by density of $D$; so $\lVert v\rVert^{2}\le\varepsilon\lVert v\rVert$ for every $\varepsilon>0$, which forces $\lVert v\rVert=0$, i.e. $v=0$.

Claim 1. Let $S\in\mathcal{A}$. For every $U\in\mathcal{A}$ we have $JUJ\in\mathcal{A}'$, so $S(JUJ)=(JUJ)S$. Thus $S\in\mathcal{L}(H)$ commutes with every element of $J\mathcal{A}J$, i.e. $S\in(J\mathcal{A}J)'=M$ by The Commutant of a Set of Bounded Operators on a Complex Hilbert Space §commutant. Hence $\mathcal{A}\subseteq M$. By Basic Properties of Commutants: Unital Algebras Closed under Weak Limits, Order Reversal, the Triple Commutant, and Conjugation §order, $J\mathcal{A}J\subseteq(J\mathcal{A}J)''=\bigl((J\mathcal{A}J)'\bigr)'=M'$.

Claim 3. Let $\zeta$ and $C$ be as in the claim. Define $T_{0}:D\to H$ by $T_{0}(S\Omega)=JS^{*}J\zeta$ for $S\in\mathcal{A}$. It is well defined: if $S,S'\in\mathcal{A}$ with $S\Omega=S'\Omega$, put $R=S-S'\in\mathcal{A}$, so $R\Omega=0$ and $R^{*}=S^{*}-S'^{*}$ 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; then $R^{*}\in\mathcal{A}$ and, by the hypothesis applied to $R^{*}$ and by The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §trace,

It is linear: for $S,U\in\mathcal{A}$ and $c\in\mathbb{C}$, $(S+cU)^{*}=S^{*}+\overline{c}\,U^{*}$ 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, so $T_{0}(S\Omega+cU\Omega)=J(S^{*}+\overline{c}\,U^{*})J\zeta=JS^{*}J\zeta+c\,JU^{*}J\zeta=T_{0}(S\Omega)+c\,T_{0}(U\Omega)$. It is bounded by $C$: $\lVert T_{0}(S\Omega)\rVert=\lVert JS^{*}J\zeta\rVert\le C\lVert S^{*}\Omega\rVert=C\lVert S\Omega\rVert$, again by the hypothesis (for $S^{*}\in\mathcal{A}$) and The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §trace. By Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §linear there is $T\in\mathcal{L}(H)$ extending $T_{0}$ with $\lVert T\rVert_{\mathrm{op}}\le C$. Since $I^{*}=I$ 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 $JIJ\xi=J(J\xi)=\xi$, we get $T\Omega=T_{0}(I\Omega)=JIJ\zeta=\zeta$.

$T\in M$: let $U\in\mathcal{A}$. For $S\in\mathcal{A}$, $JUJ(S\Omega)=SU^{*}\Omega$ with $SU^{*}\in\mathcal{A}$, and $(SU^{*})^{*}=US^{*}$ 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, so

The operators $T\,JUJ$ and $JUJ\,T$ in $\mathcal{L}(H)$ agree on $D$, so they are equal by Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §equality. Hence $T$ commutes with $J\mathcal{A}J$, i.e. $T\in M$.

Uniqueness: let $T,T'\in M$ with $T\Omega=T'\Omega=\zeta$. Then $R=T-T'\in M$ by Basic Properties of Commutants: Unital Algebras Closed under Weak Limits, Order Reversal, the Triple Commutant, and Conjugation §algebra and $R\Omega=0$. For $S\in\mathcal{A}$, $R(JSJ\Omega)=JSJ(R\Omega)=0$, so $R$ vanishes on $J\mathcal{A}J\Omega=D$ (P2), and $R=0$ by Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely §equality.

Claim 2. We check the three conditions of Cyclic Tracial Operator Algebras and Their Traces §triple for $(H,M,\Omega)$. Condition (a): by Basic Properties of Commutants: Unital Algebras Closed under Weak Limits, Order Reversal, the Triple Commutant, and Conjugation §algebra applied to $\mathcal{S}=J\mathcal{A}J$, $I\in M$ and $M$ is closed under sums, scalar multiples and products; by (P1), $(J\mathcal{A}J)^{*}\subseteq J\mathcal{A}J$, so the same clause gives $T^{*}\in M$ for $T\in M$. Condition (b): $\lVert\Omega\rVert=1$ by Cyclic Tracial Operator Algebras and Their Traces §cyclic for $\mathcal{A}$, and $M\Omega\supseteq\mathcal{A}\Omega=D$ by Claim 1, so $M\Omega$ is dense, since every open ball of $H$ meets $D$.

Next, $J(T\Omega)=T^{*}\Omega$ for every $T\in M$. Let $S\in\mathcal{A}$. Using $\langle J\xi,J\eta\rangle=\langle\eta,\xi\rangle$ and $JJ\xi=\xi$, then $S^{*}\Omega=JSJ\Omega$, then $(JSJ)^{*}=JS^{*}J$, then $T(JS^{*}J)=(JS^{*}J)T$ (as $T\in M$) and $JS^{*}J\Omega=(S^{*})^{*}\Omega=S\Omega$; the two steps that move an operator across the inner product are the adjoint relation, by Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint (in the last step applied to $T^{*}$, whose adjoint is $T$ 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):

So $v=JT\Omega-T^{*}\Omega$ satisfies $\langle\eta,v\rangle=0$ for all $\eta\in D$, and $v=0$ by (P3).

Condition (c): for $S,T\in M$, using the adjoint relation (by Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint), $J(S\Omega)=S^{*}\Omega$, $J(T\Omega)=T^{*}\Omega$ and $\langle J\xi,J\eta\rangle=\langle\eta,\xi\rangle$ with $\xi=S\Omega$, $\eta=JT\Omega$,

So $(H,M,\Omega)$ is a cyclic tracial operator algebra. By Basic Properties of Commutants: Unital Algebras Closed under Weak Limits, Order Reversal, the Triple Commutant, and Conjugation §order, $M''=(J\mathcal{A}J)'''=(J\mathcal{A}J)'=M$, so $(H,M,\Omega)$ is a tracial W*-probability space by Tracial W*-Probability Spaces §space. Its trace is $\tau_{M}(T)=\langle\Omega,T\Omega\rangle$ by Tracial W*-Probability Spaces §trace and Cyclic Tracial Operator Algebras and Their Traces §trace, which equals $\tau_{\mathcal{A}}(T)$ for $T\in\mathcal{A}\subseteq M$. Finally $J$ is a conjugation of $H$ with $J(T\Omega)=T^{*}\Omega$ for every $T\in M$, so by the uniqueness in The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §conjugation, applied to $(H,M,\Omega)$, $J$ is the conjugation of $(H,M,\Omega)$.

Claim 4. First $JMJ\subseteq M'$: for $T\in M$, The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §right-action applied to the cyclic tracial operator algebra $(H,M,\Omega)$, whose conjugation is $J$ by Claim 2, gives $JTJ\in M'$. Conversely let $y\in M'$ and put $\zeta=Jy\Omega$. For $S\in\mathcal{A}$ we have $S\in M$ by Claim 1, so $Sy=yS$, and

by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound. By Claim 3 with $C=\lVert y\rVert_{\mathrm{op}}$ there is $T\in M$ with $T\Omega=Jy\Omega$. Then $JTJ\in M'$, and $JTJ\Omega=JT\Omega=JJy\Omega=y\Omega$ since $J\Omega=\Omega$. Thus $JTJ-y\in M'$ (by Basic Properties of Commutants: Unital Algebras Closed under Weak Limits, Order Reversal, the Triple Commutant, and Conjugation §algebra) vanishes at $\Omega$, and The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §separating, applied to $(H,M,\Omega)$, gives $y=JTJ\in JMJ$. Hence $M'=JMJ$.

Claim 5. Since $\mathcal{A}\subseteq M$ (Claim 1), two applications of Basic Properties of Commutants: Unital Algebras Closed under Weak Limits, Order Reversal, the Triple Commutant, and Conjugation §order give $M'\subseteq\mathcal{A}'$ and $\mathcal{A}''\subseteq M''=M$ (Claim 2). Conversely let $y\in\mathcal{A}'$. Then $JyJ\in\mathcal{L}(H)$ and, for $S\in\mathcal{A}$, $(JyJ)(JSJ)=J(yS)J=J(Sy)J=(JSJ)(JyJ)$, so $JyJ\in(J\mathcal{A}J)'=M$. Hence $y=J(JyJ)J\in JMJ=M'$ by Claim 4. So $\mathcal{A}'\subseteq M'$, and Basic Properties of Commutants: Unital Algebras Closed under Weak Limits, Order Reversal, the Triple Commutant, and Conjugation §order gives $M=M''\subseteq\mathcal{A}''$. Therefore $M=\mathcal{A}''$.