## Separating vectors for C$^*$-algebras

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

(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...

-

## 1 Answer

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 $L^\infty([0,1],M_2)$).

-

I took the liberty of correcting the LaTeX.  This looks good for me!  Many thanks... – Matthew Daws Apr 7 2012 at 6:33
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9329742789268494, "perplexity_flag": "head"}