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.

  • 1
    $\begingroup$ What is the work of Borchers? $\endgroup$
    – LSpice
    Apr 17, 2023 at 20:36

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.