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

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

Since $$\mathcal N \subseteq \mathcal M$$, it follows by standard arguments that $$\Delta_{\mathcal N} \ge \Delta_{\mathcal M}$$

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 $$S_{\mathcal{M}}(x)=S_\mathcal{N}(x)$$ for $$x\in \operatorname{Dom}(S_\mathcal{N})$$ (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 $$R$$. This implies $$\Delta_\mathcal{M}(x)=\Delta_\mathcal{N}(x)$$, for $$x\in \operatorname{Dom}(\Delta_\mathcal{N})$$.

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

• What is the work of Borchers? Apr 17, 2023 at 20:36

It can be shown by using the "characteristics matrix" of unbounded operators (see here). For an unbounded operator $$X$$, we write the projection $$\mathcal{P}$$ onto the closed subspace $$\mathcal{H} \oplus \mathcal{H}$$ generated by the graph using a two-by-two matrix $$p_{i,k}$$ $$p^*_{i,k} = p_{j,k}$$ where the following map holds $$X \colon p_{1,1}\psi + p_{1,2} \phi \rightarrow p_{2,1}\psi + p_{2,2} \phi,\quad \psi,\phi \in \mathcal{H}$$ so one finds that the inverse of $$p_{1,1}$$ is given by $$p_{1,1} = (1+XX^*)^{-1}$$ and for an extension of $$X$$, which we call $$Y$$, we have $$\mathcal{P}(Y) \geq \mathcal{P}(X)$$. Then one sees that $$(1+YY^*)^{-1} \geq (1+XX^*)^{-1}$$
Now for the inclusion of von Neumann algebra $$\mathcal{N} \subset \mathcal{M}$$, we do the same thing. Since now $$X$$ is an anti-linear operator (and thus $$p_{1,2}$$ and $$p_{2,1}$$ are anti-linear), the second Hilbert space must be replaced by complex conjugate Hilbert space. We get the following $$(1+\Delta_{\mathcal{M}}^*)^{-1} \geq (1+\Delta_{\mathcal{N}}^*)^{-1}$$ which is $$\Delta_{\mathcal{N}} \geq \Delta_{\mathcal{M}}$$