On cyclic and separating vector

Question

Can we always have atleast two orthogonal joint cyclic and separating vectors when vN algebra $M$ is in standard form acting on GNS?

Answer 1

Yes. Let $A$ be any non-trivial unital C$^*$-algebra and $\varphi$ a faithful state. Assume first that $p\in A$, where $p$ is a proper projection; then $0<\varphi(p)<1$. Let $b=2p-1$. Then $\sigma(b)=\{-1,1\}$. We have $\varphi(b)=2\varphi(p)-1$, so $\varphi(b)\not\in\sigma(b)$. This tells us that $c=b-\varphi(b)1$ is invertible.

Because $c$ is invertible, it is cyclic and separating: indeed, if $\hat a c=0$, this means $ac=0$ so $a=0$. And $\hat a=(ac^{-1})\hat c$. So $\hat c$ is cyclic and separating. Also, $$ \langle \hat c,\hat 1\rangle=\varphi(c)=0, $$ so $c$ and $\hat 1$ are orthogonal.

When there are no projections, I cannot make the above work (you can always get elements "close" to be a projection, but I cannot get the right estimates). Anyway, here is a way to construct an element $b$ and a state $\varphi$ with $b-\varphi(b) 1$ invertible.

Start with $X\subset\mathbb R$, compact. If $X$ is not connected, write $X=X_1\cup X_2$ with $X_1,X_2$ compact and disjoint. Then $p=1_{X_1}\in C(X)$, so taking $\varphi(f)=\tfrac1{m(X_1)}\,\int f$ we get $p$ and $\varphi$ as above. When $X$ is connected, it is an interval. So assume $X=[0,1]$ (it's easy to generalize to $[a,b]$), let $f(t)= e^{2\pi i t}$; the spectrum of $f$, equal to its range, is the unit circle. Let $\varphi$ be the state $\varphi(g)=\tfrac1{2\pi^2}\int_0^{2\pi} t\, g(t) dt$. Then $$ \varphi(f)=\frac1{2\pi^2}\int_0^1\,t\,e^{2\pi i t}\,dt=\frac{-1+(1-4i\pi^2)e^{4i\pi^2}}{8\pi^4}, $$ which is not in the unit circle. So $f-\varphi(f)$ is invertible.

Now assume that $A$ is any nontrivial unital C$^*$-algebra (that is, $A$ is not $\mathbb C$ so that there exists a selfadjoint element with spectrum with at least two points). Let $a\in A$ be selfadjoint, with $\sigma(a)$ not a point. Via the Gelfand transform, we have $C^*(a)\simeq C(X)$, with $X\subset\mathbb R$ compact. By the above, there exists $b\in C^*(a)$ and a state $\varphi$ on $C^*(a)$ with $b-\varphi(b)$ invertible. Since $\varphi$ is a state, $\varphi(1)=1$. Now extend via Hahn-Banach to all of $A$. Since $\|\varphi\|=\varphi(1)$ even after the extension, the extension is positive. So now we have a state $\varphi$ on $A$ and $b\in A$ with $b-\varphi(b)$ invertible.