Subset of GNS Hilbert space generated by von Neumann algebra $\mathcal{M}$ and its commutant $\mathcal{M}^\prime$ Let $\mathcal{M}$ be a von Neumann algebra with a faithful normal state $\varphi$. Let $L^2(\mathcal{M}, \varphi)$ be the GNS-Hilbert space associated with $\varphi$ and $\Omega_\mathcal{M}$ the corresponding canonical cyclic vector. We can then view $\mathcal{M}$ as a von Neumann subalgebra of $\mathcal{B}(L^2(\mathcal{M},\varphi))$. Since $\varphi$ is faithful, the vector $\Omega_M$ is both cyclic (i.e. $\mathcal{M}\Omega_\mathcal{M} \subseteq L^2(M,\varphi)$ is dense) and separating (i.e. $x\Omega_\mathcal{M}=0$ implies $x=0$ for every $x \in \mathcal{M}$).This implies that $\Omega_{\mathcal{M}}$ is also cyclic and separating for the commutant $\mathcal{M}^\prime \subseteq \mathcal{B}(L^2 (M,\varphi))$ of $\mathcal{M}$.
I was wondering: for $x \in \mathcal{M}$ consider $x\Omega_{\mathcal{M}} \in L^2 (M,\varphi)$. Is it possible to find $y \in \mathcal{M}^\prime$ such that $x\Omega_\mathcal{M}=y\Omega_{\mathcal{M}}$? Or equivalently, is $\mathcal{M}\Omega_\mathcal{M}=\mathcal{M}^\prime \Omega_\mathcal{M}$?
 A: To answer your first question, I believe this can be solved using standard Tomita-Takesaki theory. In particular, recall that the Tomita operator is the unbounded antilinear operator $S_0$ defined as $S_0x\Omega_M=x^*\Omega_M$ for all $x\in M$, where $\Omega_M$ is the cyclic vector on $M$. Note that $S_0^2=I$ and so $S_0=S_0^{-1}$. Furthermore, it is well-known that $S_0$ is a closable operator often denoted by $S$.
Now for any $x\in M$, I claim that $S_0xS_0^{-1}=S_0xS_0\in M'$. Let $z,w\in M$, and observe that $$(S_0xS_0)zw\Omega_M=S_0xw^*z^*\Omega=zwx^*\Omega_M$$ similarly, $$z(S_0xS_0)w\Omega_M=zS_0xw^*\Omega_M=zwx^*\Omega_M$$ and so we can conclude that $S_0xS_0\in M'$. So given an $x\in M$, you can conjugate it with the Tomita operator to get a $y=S_0xS_0\in M'$ or vice versa.
I'm fairly new to Tomita-Takasaki theory and von Neumann algebras in general, so I apologized in advance if I've made any egregious errors.
Edit: As noted in the comments I am unable to show that the operator $S_0xS_0$ is actually bounded. In the case that $M$ is semi-finite and $\varphi$ is a tracial state it follows that $S$ is actually a unitary and so clearly $SxS$ is bounded but is this true only when $\varphi$ is tracial?
