Image of single GNS representation and kernel vs nullspace

Question

I have a very basic question about GNS construction and null-spaces vs kernels of linear functionals.

Let $A$ be a unital $\mathrm{C}^*$-algebra and $\rho$ a state on $A$. Consider the GNS construction $(\pi,H,\xi)$.

Where $\rho_0\in B(H)^*$ is given by: $$\rho_0(b)=\langle\xi,b\xi\rangle$$ $\rho$ factors through $\pi$, $\rho=\rho_0\circ\pi$. Now consider $\pi(A)\subseteq B(H)$.

Is there any sense in saying that $\pi(A)$ is "an image of the part of $A$ that $\rho$ can see?" This is surely related to the relationship between $\ker \rho$ and $N_\rho=\{a\in A\mid\rho(a^*a)=0\}$.

Answer 1

I think the phrase

$\pi(A)$ is "an image of the part of $A$ that $\rho$ can see?"

is wrong. It implies that if $\rho$ doesn't see some part of $A$, the same is true for $\pi$, and that's not always the case. When $\rho$ is faithful then $\pi$ is faithful, but the converse is not true; it is possible for $\pi$ to be faithful even if $\rho$ is not.

Consider for instance $A=M_2(\mathbb C)$ and $\rho(a)=a_{11}$. Then $$N_\rho=\{a\in A:\ a_{11}=a_{21}=0\}.$$ So $H=A/N_\rho$ is $2$-dimensional. And $\pi:A\to B(H)$ is faithful, since $A$ is simple. Thus in this case $\rho$ only "sees" the $1,1$ entry of $a$, but $\pi$ sees it all.