[proofplan] We work in the GNS representation of the faithful normal state $\tau$ and compute the Tomita operator on the dense cyclic subspace $M\Omega_\tau$. Traciality shows that this Tomita operator preserves norms and extends to an antiunitary involution whose modular operator is the identity. Since the modular automorphism group is implemented by the unitary group $\Delta_\tau^{it}$, the identity modular operator forces every modular automorphism to be the identity. [/proofplan]

[step:Represent $M$ on the GNS Hilbert space of $\tau$] By [citetheorem:9308], applied to the normal state $\tau$, there exist a [Hilbert space](/page/Hilbert%20Space) $H_\tau$, a normal unital representation \begin{align*} \pi_\tau:M\to\mathcal{L}(H_\tau), \end{align*} and a cyclic vector $\Omega_\tau\in H_\tau$ such that \begin{align*} \tau(a)=(\pi_\tau(a)\Omega_\tau,\Omega_\tau)_{H_\tau} \end{align*} for every $a\in M$. The [inner product](/page/Inner%20Product) is linear in the first variable. Because $\tau$ is faithful, $\pi_\tau$ is faithful. Indeed, if $a\in M$ satisfies $\pi_\tau(a)=0$, then \begin{align*} \tau(a^*a)=(\pi_\tau(a^*a)\Omega_\tau,\Omega_\tau)_{H_\tau}=0, \end{align*} and faithfulness of $\tau$ gives $a^*a=0$, hence $a=0$. We therefore identify $M$ with $\pi_\tau(M)$ and write $a\Omega_\tau$ for $\pi_\tau(a)\Omega_\tau$. The dense subspace \begin{align*} D_0:=M\Omega_\tau=\{a\Omega_\tau:a\in M\}\subseteq H_\tau \end{align*} has inner product \begin{align*} (a\Omega_\tau,b\Omega_\tau)_{H_\tau}=\tau(b^*a) \end{align*} for all $a,b\in M$. [/step]

[step:Compute the Tomita operator on the cyclic subspace]Define the Tomita operator on $D_0$ by the conjugate-[linear map](/page/Linear%20Map) \begin{align*} S_0:D_0\to D_0,\qquad S_0(a\Omega_\tau)=a^*\Omega_\tau. \end{align*} This is well-defined: if $a\Omega_\tau=0$, then \begin{align*} 0=\|a\Omega_\tau\|_{H_\tau}^2=\tau(a^*a), \end{align*} so faithfulness of $\tau$ gives $a=0$, and hence $a^*\Omega_\tau=0$. For $a,b\in M$, using the GNS inner product formula and then traciality of $\tau$, we obtain \begin{align*} (S_0(a\Omega_\tau),S_0(b\Omega_\tau))_{H_\tau}=(a^*\Omega_\tau,b^*\Omega_\tau)_{H_\tau}=\tau(ba^*). \end{align*} Since $\tau$ is tracial, \begin{align*} \tau(ba^*)=\tau(a^*b). \end{align*} Using the GNS inner product formula again, \begin{align*} \tau(a^*b)=(b\Omega_\tau,a\Omega_\tau)_{H_\tau}. \end{align*} Thus \begin{align*} (S_0(a\Omega_\tau),S_0(b\Omega_\tau))_{H_\tau}=(b\Omega_\tau,a\Omega_\tau)_{H_\tau}. \end{align*}[/step]

[guided]The point of this step is to locate exactly where traciality enters. The Tomita operator is initially defined only on the dense subspace $D_0=M\Omega_\tau$, and it reverses the algebra element: \begin{align*} S_0(a\Omega_\tau)=a^*\Omega_\tau. \end{align*} Because $S_0$ is conjugate-linear, the correct isometry identity is not preservation of the inner product in the usual linear sense, but rather the flipped identity \begin{align*} (S_0\xi,S_0\eta)_{H_\tau}=(\eta,\xi)_{H_\tau}. \end{align*} We verify this on vectors $\xi=a\Omega_\tau$ and $\eta=b\Omega_\tau$ with $a,b\in M$. The GNS inner product gives \begin{align*} (S_0(a\Omega_\tau),S_0(b\Omega_\tau))_{H_\tau}=(a^*\Omega_\tau,b^*\Omega_\tau)_{H_\tau}=\tau((b^*)^*a^*). \end{align*} Since $(b^*)^*=b$, this becomes \begin{align*} (S_0(a\Omega_\tau),S_0(b\Omega_\tau))_{H_\tau}=\tau(ba^*). \end{align*} Now traciality says $\tau(xy)=\tau(yx)$ for all $x,y\in M$. Applying this with $x=b$ and $y=a^*$ gives \begin{align*} \tau(ba^*)=\tau(a^*b). \end{align*} Finally, the GNS inner product formula gives \begin{align*} (b\Omega_\tau,a\Omega_\tau)_{H_\tau}=\tau(a^*b). \end{align*} Combining the displayed equalities yields \begin{align*} (S_0(a\Omega_\tau),S_0(b\Omega_\tau))_{H_\tau}=(b\Omega_\tau,a\Omega_\tau)_{H_\tau}. \end{align*} This is exactly the antiunitary isometry identity on the dense domain $D_0$.[/guided]

[step:Conclude that the modular operator is the identity] Taking $b=a$ in the identity just proved gives \begin{align*} \|S_0(a\Omega_\tau)\|_{H_\tau}^2=\|a\Omega_\tau\|_{H_\tau}^2 \end{align*} for every $a\in M$. Hence $S_0$ is an isometry on the dense subspace $D_0$. Since $S_0^2(a\Omega_\tau)=a\Omega_\tau$ for every $a\in M$, the range of $S_0$ is also $D_0$. Therefore the closure $S_\tau$ of $S_0$ is an antiunitary involution on all of $H_\tau$. For an antiunitary involution one has \begin{align*} S_\tau^*S_\tau=I_{H_\tau}. \end{align*} By definition of the modular operator in the Tomita construction, \begin{align*} \Delta_\tau=S_\tau^*S_\tau. \end{align*} Consequently \begin{align*} \Delta_\tau=I_{H_\tau}. \end{align*} [/step]

[step:Use the modular implementation to identify every automorphism] By the definition of the modular automorphism group associated to the faithful normal state $\tau$, the automorphism $\sigma_t^\tau:M\to M$ is implemented in the GNS representation by \begin{align*} \pi_\tau(\sigma_t^\tau(x))=\Delta_\tau^{it}\pi_\tau(x)\Delta_\tau^{-it} \end{align*} for every $x\in M$ and every $t\in\mathbb R$. Since $\Delta_\tau=I_{H_\tau}$, functional calculus gives \begin{align*} \Delta_\tau^{it}=I_{H_\tau}. \end{align*} Thus \begin{align*} \pi_\tau(\sigma_t^\tau(x))=\pi_\tau(x) \end{align*} for every $x\in M$ and every $t\in\mathbb R$. Because $\pi_\tau$ is faithful, this implies \begin{align*} \sigma_t^\tau(x)=x. \end{align*} Therefore $\sigma_t^\tau=\operatorname{id}_M$ for every $t\in\mathbb R$. [/step]