# Question about the homogeneity of the state space of a type $\rm{III}_1$ factor

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.

• 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. Dec 18, 2023 at 8:11