Tomita modular automorphisms and the upper-strip KMS condition

By [F2], ${\sigma }_t(M)=M$. Each ${\sigma }_t$ is conjugation by the unitary ${\mathrm{\Delta }}^{it}$, so it is complex-linear, multiplicative, adjoint-preserving and bijective, with inverse ${\sigma }_{-t}$. The spectral calculus gives ${\mathrm{\Delta }}^{i(s+t)}={\mathrm{\Delta }}^{is}{\mathrm{\Delta }}^{it}$, hence ${\sigma }_s{\sigma }_t={\sigma }_{s+t}$ and ${\sigma }_0=I$. To check normality, let $0\le a_i\uparrow a$ be a bounded increasing net in $M$. Unitary conjugation is an order isomorphism of $\mathcal{B}(H)$: if $b$ bounds every ${\mathrm{\Delta }}^{it}{a}_i{\mathrm{\Delta }}^{-it}$, then ${\mathrm{\Delta }}^{-it}b{\mathrm{\Delta }}^{it}$ bounds every $a_i$, so $a\le {\mathrm{\Delta }}^{-it}b{\mathrm{\Delta }}^{it}$ and ${\mathrm{\Delta }}^{it}a{\mathrm{\Delta }}^{-it}\le b$. Thus ${\mathrm{\Delta }}^{it}{a}_i{\mathrm{\Delta }}^{-it}\uparrow {\mathrm{\Delta }}^{it}a{\mathrm{\Delta }}^{-it}$ in $\mathcal{B}(H)$. By [F2] the limit is in $M$, hence is also the supremum in $M$. By the definition in [F7], each restriction is a normal $\ast$-automorphism.

For $A,B\in M$, both $A^{\ast }\mathrm{\Omega }$ and $B\mathrm{\Omega }$ lie in $D({\mathrm{\Delta }}^{1\mathrm{/}2})$, because $Sx\mathrm{\Omega }=x^{\ast }\mathrm{\Omega }$ for $x\in M$ and $\mathrm{\parallel }Sx\mathrm{\Omega }\mathrm{\parallel }=\mathrm{\parallel }{\mathrm{\Delta }}^{1\mathrm{/}2}x\mathrm{\Omega }\mathrm{\parallel }$. Define $F_{A,B}(z)=({\mathrm{\Delta }}^{i\bar{z}\mathrm{/}2}{A}^{\ast }\mathrm{\Omega }\mid {\mathrm{\Delta }}^{-iz\mathrm{/}2}B\mathrm{\Omega }),$ where $(u\mathrm{\mid }v)=⟨v,u⟩$ is linear in its second argument. By [F3], both vectors $A^{\ast }\mathrm{\Omega }$ and $B\mathrm{\Omega }$ belong to $D({\mathrm{\Delta }}^{1\mathrm{/}2})$; [F4] then makes each vector factor bounded and continuous on the closed strip. The first is anti-holomorphic as a vector-valued function of $z$ and enters the first, conjugate-linear slot, while the second is holomorphic in the second slot. Hence $F_{A,B}$ is bounded and continuous on the strip and holomorphic inside.

By [F8], $t\mapsto {\mathrm{\Delta }}^{it}\xi$ is norm-continuous for every $\xi \in H$. For fixed $x\in M$ and $\xi ,\eta \in H$, the coefficient $⟨{\sigma }_t(x)\xi ,\eta ⟩$ is therefore continuous. If $T$ is finite rank, $\mathrm{Tr}(T{\sigma }_t(x))$ is a finite sum of such continuous coefficients. For arbitrary trace-class $T$, choose finite-rank $T_n\to T$ in trace norm by [F6]; uniformly in $t$, $\mathrm{\mid }\mathrm{Tr}((T-T_n){\sigma }_t(x))\mathrm{\mid }\le \mathrm{\parallel }T-T_n{\mathrm{\parallel }}_1\mathrm{\parallel }x\mathrm{\parallel }$. Thus every element of the predual $Q$ evaluates continuously on ${\sigma }_t(x)$, which is point-ultraweak continuity.

Since ${\mathrm{\Delta }}^{it}\mathrm{\Omega }=\mathrm{\Omega }$ and ${\mathrm{\Delta }}^{it}$ is unitary, $\omega ({\sigma }_t(x))=⟨{\mathrm{\Delta }}^{it}x\mathrm{\Omega },\mathrm{\Omega }⟩=⟨x\mathrm{\Omega },{\mathrm{\Delta }}^{-it}\mathrm{\Omega }⟩=\omega (x)\mathrm{.}$ So $\omega$ is invariant under ${\sigma }_t$ and under ${\alpha }_t={\sigma }_{-t}$. Also $\omega (I)=\mathrm{\parallel }\mathrm{\Omega }{\mathrm{\parallel }}^2=1$ and $\omega (x^{\ast }x)=\mathrm{\parallel }x\mathrm{\Omega }{\mathrm{\parallel }}^2\ge 0$, so $\omega$ is a state; because $\mathrm{\Omega }$ is separating for $M$, this same identity shows that $\omega$ is faithful. By [F5], $\omega$ is a normal vector functional on $M$. Hence it is a faithful normal state as asserted.

For real $t$, unitarity of ${\mathrm{\Delta }}^{it\mathrm{/}2}$ and the identity ${\mathrm{\Delta }}^{-it}B\mathrm{\Omega }={\sigma }_{-t}(B)\mathrm{\Omega }$ give $F_{A,B}(t)=(A^{\ast }\mathrm{\Omega }\mathrm{\mid }{\mathrm{\Delta }}^{-it}B\mathrm{\Omega })=(\mathrm{\Omega }\mathrm{\mid }A{\mathrm{\Delta }}^{-it}B\mathrm{\Omega })=\omega (A{\alpha }_t(B))\mathrm{.}$ This is the lower boundary in [F7].

At $z=t+i$, ${\mathrm{\Delta }}^{i\bar{z}\mathrm{/}2}{A}^{\ast }\mathrm{\Omega }={\mathrm{\Delta }}^{it\mathrm{/}2}JA\mathrm{\Omega }$ and ${\mathrm{\Delta }}^{-iz\mathrm{/}2}B\mathrm{\Omega }={\mathrm{\Delta }}^{-it\mathrm{/}2}J{B}^{\ast }\mathrm{\Omega }$, because $Sx\mathrm{\Omega }=x^{\ast }\mathrm{\Omega }$ and $S=J{\mathrm{\Delta }}^{1\mathrm{/}2}$. By [F2], $J$ commutes with ${\mathrm{\Delta }}^{it\mathrm{/}2}$; antiunitarity gives $(Ju\mathrm{\mid }Jv)=(v\mathrm{\mid }u)$. Therefore $F_{A,B}(t+i)=({\mathrm{\Delta }}^{-it\mathrm{/}2}{B}^{\ast }\mathrm{\Omega }\mathrm{\mid }{\mathrm{\Delta }}^{it\mathrm{/}2}A\mathrm{\Omega })=(B^{\ast }\mathrm{\Omega }\mathrm{\mid }{\mathrm{\Delta }}^{it}A\mathrm{\Omega })=\omega (B{\sigma }_t(A))\mathrm{.}$ Invariance from step 2.2 changes the last expression to $\omega ({\sigma }_{-t}(B)A)=\omega ({\alpha }_t(B)A)$, the upper boundary in [F7].

Steps 2.2–3.1 provide the bounded continuous strip function with both required boundary values for each $A,B\in M$. By [F7], $\omega$ is upper-strip $(\alpha ,1)$ KMS. Step 1.1 gives the asserted point-ultraweakly continuous automorphism group, and [F2] gives $JMJ=M^{\mathrm{\prime }}$. No norm-continuity assertion on all of $M$ is made.