inclusion of von Neumann algebras implies reversing inequality of its modular operators

16 Views Asked by At

I'm working with von Neumann algebras and I stumbled with this statement in a work of Borchers (1999)

enter image description here

where we must think $\mathcal{N}$ and $\mathcal{M}$ as concrete von Neumann algebras over the same Hilbert space, and $\Delta_\mathcal{M}$ is the modular operator associated to $\mathcal{M}$ and similarly for $\Delta_\mathcal{N}$.

I suppose there is something I'm not catching because assuming $\mathcal{N}\subseteq\mathcal{M}$, I could only arrive to $S_\mathcal{M}|_{Dom(\Delta_\mathcal{N})}=S_\mathcal{N}$ (to do that I assume there is a cyclic and separating vector for $\mathcal{N}$) where $S_R$ denotes the Tomita-Takesaki operator of the algebra $R$, impliying $\Delta_\mathcal{M}|_{Dom(\Delta_\mathcal{N})}=\Delta_\mathcal{N}$.

However, I couldn't find the proof of Borcher's assertion.

Thanks in advance to any help.