Involution and polar identities for the Tomita operator

If $x\mathrm{\Omega }=y\mathrm{\Omega }$ for $x,y\in M$, then $(x-y)\mathrm{\Omega }=0$ and separatingness gives $x=y$. Thus $S_0$ is well defined; it is antilinear by $(ax+by{)}^{\ast }=\bar{a}{x}^{\ast }+\bar{b}{y}^{\ast }$, and its domain is dense by cyclicity. The same reasoning makes $F_0$ well defined; [F4] makes $M^{\mathrm{\prime }}\mathrm{\Omega }$ dense. Also $S_0^2(x\mathrm{\Omega })=x\mathrm{\Omega }$ and $F_0^2(x^{\mathrm{\prime }}\mathrm{\Omega })=x^{\mathrm{\prime }}\mathrm{\Omega }$.

By [F4], $\mathrm{\Omega }$ is cyclic for $M^{\mathrm{\prime }}$ because it is separating for $M$, and separating for $M^{\mathrm{\prime }}$ because it is cyclic for $M$. Applying [F2] to the Tomita operator $F_0$ of $M^{\mathrm{\prime }}$ therefore proves directly that $F_0$ is closable; applying [F2] to $M$ proves that $S_0$ is closable. For $x\in M$ and $x^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$, the adjoint identity and commutation give $(x^{\mathrm{\prime }}\mathrm{\Omega }\mathrm{\mid }{S}_{0}x\mathrm{\Omega })=(x^{\mathrm{\prime }}\mathrm{\Omega }\mathrm{\mid }{x}^{\ast }\mathrm{\Omega })=(x\mathrm{\Omega }\mathrm{\mid }{x}^{\mathrm{\prime }\ast }\mathrm{\Omega })=(x\mathrm{\Omega }\mathrm{\mid }{F}_0{x}^{\mathrm{\prime }}\mathrm{\Omega })\mathrm{.}$ By [F3] this says $F_0\subseteq S_0^{\ast }$ and $S_0\subseteq F_0^{\ast }$.

Let $\xi \in D(S)$ and choose ${\xi }_n\in M\mathrm{\Omega }$ with ${\xi }_n\to \xi$ and $S_0{\xi }_n\to S\xi$, using the definition of graph closure. Since $S_0^2{\xi }_n={\xi }_n$, the graph points $(S{\xi }_n,S(S{\xi }_n))=(S_0{\xi }_n,{\xi }_n)$ converge to $(S\xi ,\xi )$. Closedness of $S$ implies $S\xi \in D(S)$ and $S^2\xi =\xi$. Hence $S$ is an involution, so it is injective and its range is the dense domain $D(S)$.

Apply the closed-antilinear polar result [F2] to $S$. Injectivity and dense range make its partial anti-isometry $J$ antiunitary. The inverse of $S=J{\mathrm{\Delta }}^{1\mathrm{/}2}$ is ${\mathrm{\Delta }}^{-1\mathrm{/}2}{J}^{\ast }=J^{\ast }(J{\mathrm{\Delta }}^{-1\mathrm{/}2}{J}^{\ast })$, whose second factor is positive self-adjoint. Since $S^{-1}=S$, uniqueness of the polar decomposition gives $J=J^{\ast }$ and ${\mathrm{\Delta }}^{1\mathrm{/}2}=J{\mathrm{\Delta }}^{-1\mathrm{/}2}J$. Thus $J^2=I$ and $J\mathrm{\Delta }J={\mathrm{\Delta }}^{-1}$. For antiunitary conjugation the functional-calculus rule is $Jf(\mathrm{\Delta })J=\bar{f}(J\mathrm{\Delta }J)$. Taking $f(s)=s^{it}$ gives $J{\mathrm{\Delta }}^{it}J=({\mathrm{\Delta }}^{-1}{)}^{-it}={\mathrm{\Delta }}^{it}$, so $J$ commutes with every ${\mathrm{\Delta }}^{it}$.

Because $S$ is the graph closure of $S_0$, their adjoints agree: the adjoint identity for $S_0$ extends along convergent graph pairs to $S$, and restriction gives the converse. Hence $F=S^{\ast }=S_0^{\ast }$ and $F_0\subseteq F$. From $S=J{\mathrm{\Delta }}^{1\mathrm{/}2}$, the antilinear adjoint identity [F3] and self-adjointness of ${\mathrm{\Delta }}^{1\mathrm{/}2}$ give $F={\mathrm{\Delta }}^{1\mathrm{/}2}J$. Functional calculus applied to $J\mathrm{\Delta }J={\mathrm{\Delta }}^{-1}$ gives ${\mathrm{\Delta }}^{1\mathrm{/}2}J=J{\mathrm{\Delta }}^{-1\mathrm{/}2}$ on their common domain.

Since $S_0\mathrm{\Omega }=F_0\mathrm{\Omega }=\mathrm{\Omega }$ and $F_0\subseteq F$, steps 3.1 and 4.1 give ${\mathrm{\Delta }}^{1\mathrm{/}2}\mathrm{\Omega }=J\mathrm{\Omega }={\mathrm{\Delta }}^{-1\mathrm{/}2}\mathrm{\Omega }$. For the spectral measure of $\mathrm{\Delta }$, this equality has squared norm difference ${\int }_{(0,\mathrm{\infty })}\mathrm{\mid }{\lambda }^{1\mathrm{/}2}-{\lambda }^{-1\mathrm{/}2}{\mathrm{\mid }}^{2}d{\mu }_{\mathrm{\Omega }}(\lambda )=0$. The integrand vanishes only at $\lambda =1$, so $\mathrm{\Delta }\mathrm{\Omega }=\mathrm{\Omega }$ and then $J\mathrm{\Omega }=\mathrm{\Omega }$.