Suppose that $\mathcal{A}$ is a unital $\mathrm{C}$-algebra with a faithful state $\varphi:\mathcal{A}\to \mathbb{C}$, that is, for all $f\in \mathcal{A}$,
$$\varphi(f^*f)=0\implies f=0.$$
I understand that $\varphi$ has a normal extension to a state $\omega_\varphi$ on the bidual $\mathcal{A}^{**}$. I want to say that $\varphi$ faithful does not in general imply that $\omega_\varphi$ is faithful.
I want to use this counterexample, but I find the bidual a very slippery beast. Let $\mathcal{A}=C([0,1])$ and $\varphi:=\int_{[0,1]}\cdot d\lambda(t)$ integration against the Lesbesgue measure. I want to say that there is something like a Kronecker delta in $\mathcal{A}^{**}$:
$$\delta_{1/2}(x)=\begin{cases}1 & \text{ if }x=1/2 \\ 0 & \text{otherwise}\end{cases}.$$
I know that $\varphi$ is faithful on $C([0,1])$ but:
$$\omega_\varphi(\delta_{1/2})=\int_{[0,1]}\delta_{1/2}(t)\,d\lambda(t)=0,$$
and so $\omega_\varphi$ is not faithful on the bidual. Does this basically check out?
Best Answer
I think your idea is fine. To make things more concrete, consider the state $\delta\in A^*$ given by $\delta(f)=f(0)$ (any other point would do).
The state $\varphi$ extends, as you say, to a normal state on $A^{**}$. This is done canonically by using the view of $A^{**}$ as the enveloping von Neumann algebra, which we can see as $$\tag1 A''=\overline{\bigoplus_\rho\rho(A)}^{\rm sot}, $$ where $\rho$ runs over the unitary equivalence classes of representations of $A$. One of the $\rho$ used is $\pi_\varphi$, the GNS representation of $\varphi$. Since we then have $$ \varphi(a)=\langle\pi_\varphi(a)\Omega_\varphi,\Omega_\varphi\rangle, $$ the obvious extension $\omega_\varphi(T)=\langle T\Omega_\varphi,\Omega_\varphi\rangle$ is normal, being a point-state. Because $\pi_\varphi$ and $\pi_\delta$ are not unitarily equivalent (one is faithful, the other one isn't), they correspond to different direct summands in $(1)$. Let $P\in A''$ the projection onto the $\delta$ component (I haven't thought about enveloping von Neumann algebras in a long time, but I think it's standard that $P$ is in the center of $A''$, by using the fact that all representations of $A$ are quotients of the universal representation). Then $\omega_\varphi$ is orthogonal to the range of $P$, and thus $\omega_\varphi(P)=0$.