(I asked this on math.stackexchange, without response).

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, $x\in M, x(\xi_0)=0 \implies x=0$?

It is standard (and easy to prove) that this is equivalent to $\xi_0$ be cyclic for $M'$. However, the usual proof breaks down, and does not show this to be equivalent to $\xi_0$ being separating for $A$.

I think I can prove this using left Hilbert algebras. 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. Actually, it's not at all clear to me that this is correct-- I don't see why the map $S:\mathfrak A \rightarrow \mathfrak A; a\xi_0 \mapsto a^*\xi_0$ is preclosed. So now I suspect there might be a counter-example...


The answer is yes for trace vectors, but no in general. 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 $$\phi(f)=\int_C f(x)_{11}\, dx + \int_{[0,1]\setminus C}\mathrm{tr}f(x)\,dx.$$ 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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  • $\begingroup$ I took the liberty of correcting the LaTeX. This looks good for me! Many thanks... $\endgroup$ – Matthew Daws Apr 7 '12 at 6:33

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