Let $A$ be a $C^*$-algebra and $\rho$ be a pure state on $A$ i.e. $\rho\in PS(A)$. Let $\tilde{A}$ be the unitisation of $A$ and $u$ be a unitary element in $\tilde{A}$. If we define $\rho^u:A\rightarrow \mathbb{C}$ such that $\rho^u(a)=\rho(uau^*)$. I have to show that $\rho^u$ is also a pure state on $A$.

I know $A$ is a hereditary $C^*$-algebra in it's unitisation $\tilde{A}$, hence the map is well defined. Now for proving that it's a state I have to show somehow that $\operatorname{lim}_{\lambda}\rho^u(u_\lambda)=1$ where $\{u_{\lambda}\}_{\lambda\in \Lambda}$ is an approximate unit of $A$. I do not know how to proceed further.

Any form of help is highly appreciated. Thanks.

  • 1
    $\begingroup$ $uu_\lambda u^*$ is another approximate unit. $\endgroup$
    – Ruy
    Oct 29, 2020 at 15:42
  • $\begingroup$ @Ruy Can you please give some hint on how to prove this fact? $\endgroup$ Oct 29, 2020 at 16:20
  • 1
    $\begingroup$ $\|(uu_\lambda u^*)a - a\| = \|u^*(uu_\lambda u^*a - a)u\| = \|u_\lambda (u^*au) - u^*au\|.$ $\endgroup$
    – Ruy
    Oct 29, 2020 at 16:27


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