$J\Delta^{it}J = \Delta^{it}$ for modular operators associated to left Hilbert algebra

Question

Let $\mathcal{A}$ be a (full) left Hilbert algebra with Hilbert space completion $H$. Let $J: H \to H$ be the modular conjugation and $\Delta$ be the associated modular operator. Is it true that $$J \Delta^{it}J = \Delta^{it}$$ for $t\in \mathbb{R}?$

Answer 1

And because it was so much fun, here is a second answer, proving something more general: If $A$ is a self-adjoint operator on a Hilbert space $H$ and $U$ an anti-unitary operator on $H$ (i.e. a surjective anti-linear operator such that $\langle U\xi,U\eta\rangle=\langle\eta,\xi\rangle$ for all $\xi,\eta\in H$), then $f(UAU^\ast)=Uf(A)^\ast U^\ast$ for all Borel functions $f\colon \mathbb R\to\mathbb C$:

Let $E$ denote the spectral measure of $A$. A direct computation shows that $U E(\cdot)U^\ast$ is a spectral measure. If $f\colon \mathbb R\to\mathbb C$ is a Borel function and $\xi\in U(\mathrm{dom}(f(A)))$, then $$ \int_{\mathbb R}\lvert f(\lambda)\rvert^2\,d\langle \xi,UE(\lambda)U^\ast\xi\rangle=\lVert f(A)U^\ast\xi\rVert^2<\infty. $$ Moreover, if $\xi,\eta\in U(\mathrm{dom}(f(A)))$, then $$ \int_{\mathbb R}f(\lambda)\,d\langle \xi,UE(\lambda)U^\ast\eta\rangle=\int_{\mathbb R}f(\lambda)\,d\langle U^\ast\eta,E(\lambda)U^\ast\xi\rangle=\langle U^\ast \eta,f(A)U^\ast\xi\rangle=\langle \xi,U f(A)^\ast U^\ast\eta\rangle. $$ In particular, taking $f=\mathrm{id}_{\mathbb R}$, we see that $UE(\cdot)U^\ast$ is the spectral measure of $UA U^\ast$. Given this fact, the last displayed equation implies $$ \langle \xi,f(UAU^\ast)\eta\rangle=\langle\xi,Uf(A)^\ast U^\ast\eta\rangle, $$ hence $f(UAU^\ast)=U f(A)^\ast U^\ast$.

Now you can apply this result to $A=\Delta$, $U=J$ and $f(\lambda)=\lambda^{-it}$ with $t\in\mathbb R$.

Answer 2

I presume you already know that $J\Delta J=\Delta^{-1}$ (that's Lemma VI.1.5. (v) (d) in Takesaki II). Taking square roots, we get by induction $J\Delta^{2^{-n}} J=\Delta^{-2^{-n}}$ for all $n\in\mathbb Z$. This already implies $J\Delta^{-\bar z}J=\Delta^z$ for all $z\in\mathbb C$ by a complex analysis argument:

For $\xi\in \bigcap_{n\in\mathbb Z}\mathrm{dom}(\Delta^n)$ the functions $z\mapsto \Delta^z \xi$ and $z\mapsto J\Delta^{-\bar z}J \xi$ are analytic. This may not be entirely obvious for the second function because of the $\bar z$, but the anti-linearity of $J$ comes to our rescue: If $\Delta^zJ\xi=\sum_k z^k\eta_k$, then $$ J\Delta^{-\bar z}J\xi=J\sum_{k=0}^\infty(-\bar z)^k\eta_k=\sum_{k=0}^\infty (-z)^kJ\eta_k. $$ By the identity theorem for analytic functions, we obtain $\Delta^z\xi=J\Delta^{-\bar z}J\xi$. As $\bigcap_{n\in\mathbb Z}\mathrm{dom}(\Delta^n)$ is a core for $\Delta^z$ for all $z\in\mathbb C$, we conclude $\Delta^z=J\Delta^{-\bar z}J$. In particular, $\Delta^{it}=J\Delta^{it}J$ for all $t\in\mathbb R$.