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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.

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I've just asked this at MO: mathoverflow.net/questions/93295/… –  Matthew Daws Apr 6 '12 at 9:32
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You might wish to apply the strikethrough here, as well –  user16299 Apr 7 '12 at 3:03
    
@Yemon: Thanks! I should mentioned that Narutaka Ozawa has provided a nice counter-example over at MO. –  Matthew Daws Apr 7 '12 at 6:35

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A counter-example was given on mathoverflow by Narutaka Ozawa:


Take a closed nowhere dense subset $C\subset [0,1]$ with positive measure and consider the state $\phi$ on $C([0,1],M_2)$ defined by $$\phi(f)=\int_C f(x)_{11}dx+\int_{[0,1]\setminus C} \operatorname{tr}f(x)\,dx$$ Here $f(x)_{11}$ is the $(1,1)$-entry of $f(x)\in M_2$. Then $\phi$ is a faithful state, and the GNS vector is a cyclic separating vector. However it is not separating for the double commutant, which is $L^\infty ([0,1],M_2)$.

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