I study the paper "Doplicher and Longo: Standard and split inclusions of von Neumann algebras" and have a question about the proof of Lemma 3.2.
Let $\omega$ be a faithful normal state on a type I factor $N$ on a Hilbert space $H$ (e.g. $N=B(H)$, the bounded operators on $H$), and $\rho \in B(H)$ the trace class Radon--Nikodym derivative of $\omega$ (i.e. $\omega(A) = \mathrm{tr}(\rho A)$ for all $A\in N$). By the spectral theorem, $$\rho = \bigoplus_{\lambda \in \sigma(\rho)\backslash \{0\}} \lambda E_{\lambda}, $$ where $E_{\lambda}$ is the spectral projection on the eigenspace corresponding to the eigenvalue $\lambda$. The eigenspaces $E_\lambda H$ are finite-dimensional for $\lambda \neq 0$.
I want to understand why the group $\mathcal{U}_\omega$, the group of all unitaries of $N$ that commute with $\rho$, is compact in the strong operator topology. If $U \in \mathcal{U}_\omega$, then $U = \bigoplus_{\lambda \in \sigma(\rho)} U_\lambda$, where $U_\lambda$ is unitary and keeps $E_\lambda H$ invariant.
The argument given in the paper is as follows: Because the spaces $E_\lambda H$, $\lambda \neq 0$, are finite-dimensional, the group of all unitaries on $E_\lambda H$ is compact. Thus, $\mathcal{U}_\omega$ is a direct product of compact groups, and therefore compact in the strong operator topology by Tychonoff's theorem. However, this argument seems to ignore the space $E_0 H$, which is possibly infinite-dimensional.
My first thought was that the faithulness of $\omega$ may imply that $0$ is not an eigenvalue of $\rho$. However, as a counterexample, take a normalised vector $x \in H$ that is separating for $N$. Then $\rho_x=\langle x, \cdot \rangle x$ induces a faithful state on $N$, and $0$ is an eigenvalue of $\rho$ with infinite multiplicity.
If $\omega$ is faithful, then $E_0=0$. Indeed, you have $$ \omega(E_0)=\omega(I-\sum_{\lambda\ne0}E_\lambda)=\operatorname{Tr}(\rho(I-\sum_{\lambda\ne0}E_\lambda))=0, $$ so $E_0=0$.
Doplicher and Longo carry the proof in the case where $N=B(H)$, acting on $H$. In that case there is no separating vector and $E_0\in N$, which makes the above argument work.