Let $A$ be a C$^*$-algebra, concretely acting on a Hilbert space $H$. Suppose that $\xi_0\in H$ is cyclic and separating for $A$ (that is, the map $A\rightarrow H, a\mapsto a(\xi_0)$ is injective with dense range). Let $M=A''$ the von Neumann algebra generated by $A$.
Need $\xi_0$ still be separating for $M$? That is, $x\in M, x(\xi_0)=0 \implies x=0$?
Actually, I think I can prove this. We turn $\mathfrak A = \{ a(\xi_0) : a\in A \}$ into a left Hilbert algebra algebra in the obvious way. Then run the Tomita-Takesaki machinery (actually not needed in full generality as we start with a state, not a weight). Then the von Neumann algebra generated by $\mathfrak A$ is nothing but $M$, and so the general theory tells us that $\varphi(x) = \|x\xi_0\|$ will be a faithful weight on $M$, which is what we need.
Is this right? Surely this argument is far, far more complicated then necessary?
No, I don't think this is right: there seems no reason why the Tomita operator $S:a(\xi_0)\mapsto a^*(\xi_0)$ is preclosed.
