I'm studying Conway's Functional Analysis.

In chapter 8.5, we learn about GNS representation. Theorem 8.5.17 states that

$\mathcal{A}$ is separable then, we can select representation $(\pi, \mathcal{H})$ that $\mathcal{H}$ is separable.

And next line, every separable C*-algebra has a cyclic representation that is isometric.

I want to prove this, and Conway suggests that $f_n$ is countable weak* dense subset of $S_\mathcal{A}$(which is state space of $\mathcal{A}$), then if we set $f = \sum 2^{-n}f_n$ then $\pi_f$ is an isometry.($\pi_f$ is from GNS construction)

Since GNS Construction suggest that if we make $\pi_f$, then it is cyclic representation, so I need to prove that $\pi_f$ is isometry. But I cannot get any way to get $\pi_f$'s norm.

if $e_f$ is cyclic vector of $\pi_f$, $$\|p_f(a) \|^2 \geq \|\pi_f(a) e_f\|^2 = \langle \pi_f(a^* a) e_f, e_f\rangle = f(a^* a) = \sum 2^{-n}f_n(a^* a)$$

and we know that $\|a\| = \sup\{f(a) : f \in S_\mathcal{A} \}$ for positive element $a$ but I don't know that $\sum 2^{-n}f_n(a^* a) \geq \|a^* a\| - \epsilon$. How can i deal with this problem?


1 Answer 1


So $f$ is a state. Lets first show that it is faithful, meaning that $f(a^*a)>0$ whenever $a\neq0$:

Suppose $a\in A$ and $a\neq0$, then there is a state $\varphi$ with $\varphi(a^*a)>0$. Since $f_n$ is weak* dense you have a sub-sequence with $f_{n_k}\to \varphi$ in the weak* topology, but this topology is just so that you can recover $f_{n_k}(a^*a)\to\varphi(a^*a)$, hence there is some $f_n$ with $f_n(a^*a)>0$. In particular $f(a^*a)>0$.

Now lets show a general statement. Namely that if $\pi_f$ is the GNS representation onto a faithful state then $\pi$ is an isometry. For this statement we use the non-trivial statement that injective C* morphisms are isometries. So all we check is that $\pi$ doesn't have a kernel:

Suppose that $\pi(a) =0$, then $\pi(a^*a)=\pi(a)^*\pi(a)=0$. It follows that for the ground-state $|f\rangle$ you get $\langle f,\pi(a^*a)f\rangle = f(a^*a) =0$, implying that $a=0$ by faithfulness of $f$.


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