I'm reading the paper Homogeneity of the State Space of Factors of Type $\rm{III}_1$ by Connes and Størmer. Homogeneity of the state space means that all normal states are approximately unitarily equivalent in the sense that for normal states $\psi,\phi$ on $M$ and $\epsilon>0$, there exists a unitary $u\in M$ such that $$ \| u\psi u^* - \phi\|<\epsilon, $$ where $u\psi u^* = \psi\circ\mathrm{Ad}_{u^*}$.

The proof of the implication "homogeneity implies type $\rm{III}_1$" confuses me. The proof seems to be reduced to Connes' Une classification des facteurs de type $\rm{III}$ paper (which is in french). Since type $\rm{III}_1$ is defined via the Connes invariant $S(M)$, which is the intersection of all modular spectra $\mathrm{Sp}\Delta_\psi$, they have to show that the modular spectrum is always $\mathbb R_+$. My question is:

Can one prove directly that approximate unitary equivalence of two normal states $\phi,\psi$ on a factor $M$ implies that they have the same modular spectrum, i.e., $\mathrm{Sp}\Delta_\psi=\mathrm{Sp}\Delta_\phi$?

If this were true, one could prove $\mathrm{Sp}\Delta_\psi=\mathbb R_+$ for all normal states $\psi$ using $M \cong M\otimes M_2(\mathbb C)$. Homogeneity would imply that $\psi\otimes \rho$ and $\psi$ have the same modular spectrum for all states $\rho$ on $M_2(\mathbb C)$ which can only hold if the spectrum is $\mathbb R_+$.

Edit: I think I might have an argument using ultrapower techniques but I'd still be very interested in a more direct argument.

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    $\begingroup$ One can indeed prove this using ultrapowers. Assume that $\psi$ and $\phi$ are faithful normal states on a von Neumann algebra $M$ and that $\psi$ and $\phi$ are approximately unitarily conjugate. Then the ultrapower states $\psi^\omega$ and $\phi^\omega$ are unitarily conjugate so that their modular operators have the same spectrum. By Corollary 4.8.(3) in doi.org/10.1016/j.jfa.2014.03.013 also the modular operators of $\psi$ and $\phi$ have the same spectrum. $\endgroup$ Dec 18, 2023 at 8:11


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