Let $\varphi$ be a state of the $C^*$-algebra $A$, $B\subset A$ a hereditary subalgebra and $K_\varphi:=\{x\in B : 0\le x \le 1, \varphi(x)=1\}$. Let $\pi_\varphi:A\rightarrow \mathcal{B}(H_\varphi)$ be the GNS representation corresponding to $\varphi$. We can write $\varphi(x)=\langle\pi_\varphi(x)\xi_\varphi,\xi_\varphi\rangle$ for a cyclic vector $\xi_\varphi\in H_\varphi$ belonging to $\pi_\varphi$.

I already showed, that for every state $f\in S(A)$, $f\neq\varphi$, there exists $x\in K_\varphi$ such that $f(x)<1$. I now want to show, that for all $\eta\in H_\varphi$ with $\eta \perp \xi_\varphi$, there exists $x\in K_\varphi$ sucht that $\pi_\varphi(x)\eta=0$.

I have no idea how so solve this. I only know that $\langle\pi_\varphi(x)\eta,\zeta\rangle<1$ for any $\zeta\in H_\varphi$. I don't know what I'm missing.

Thank you in advance!

Edit: I found that $\varphi$ is a pure state. With $\varphi$ a pure state, this simply follows from Kadison's transitivity theorem.

  • 2
    $\begingroup$ Something is off. If you take $f=\varphi$, you say you proved that there exists $x\in K_\varphi$ with $\varphi(x)<1$, a contradiction. $\endgroup$ – Martin Argerami Sep 12 '18 at 22:02
  • $\begingroup$ I forgot to mention that $f$ can't be $\varphi$. Sorry, edited it in! $\endgroup$ – T.Gel Sep 13 '18 at 7:30
  • $\begingroup$ I found that $\varphi$ is a pure state and then it simply follows from Kadison's transitivity theorem. $\endgroup$ – T.Gel Sep 13 '18 at 11:41
  • $\begingroup$ At least you require that $f\ne\varphi$ on $B$; in many examples it is easy to construct $f\ne\varphi$ but they agree on $B$. And now you say that $\varphi$ is pure; that's nowhere in your question, where you say "let $\varphi$ be a state". $\endgroup$ – Martin Argerami Sep 13 '18 at 18:06

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