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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.

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    $\begingroup$ What is the work of Borchers? $\endgroup$
    – LSpice
    Apr 17, 2023 at 20:36

1 Answer 1

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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 $$\Delta_{\mathcal{N}} \geq \Delta_{\mathcal{M}}$$

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