Theorem 34.11.4 (Gelfand-Naimark-Segal).label Let $A$ be a unital $C^{*}$-algebra, then:
- (1)
For each $\phi \in S(A)$, there exists a triple $(H_{\phi}, \pi_{\phi}, \xi_{\phi})$ where $(H_{\phi}, \pi_{\phi})$ is a representation of $A$, $\xi_{\phi}$ is a cyclic unit vector of $(H_{\phi}, \pi_{\phi})$, and
\[\dpn{x, y}{\phi}= \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}\] - (2)
For each representation $(H, \pi)$ of $A$ with cyclic unit vector $\xi$, the mapping
\[\phi: A \to \complex \quad x \mapsto \dpn{\pi(x)\xi, \xi}{H}\]is a state on $A$. Let $(H_{\phi}, \pi_{\phi}, \xi_{\phi})$ be the GNS triple associated with $\phi$, then there exists a unitary equivalence $U: H \to H_{\phi}$ such that $U\xi = \xi_{\phi}$.
- (3)
For each $\mathcal{S}\subset S(A)$, the mapping
\[\pi_{\mathcal{S}}: A \to B([l^{2}(\mathcal{S}); H_{\phi}]) \quad \pi_{\mathcal{S}}(x)(\eta)_{\phi} = \pi_{\phi}(x)(\eta_{\phi})\]is a representation of $A$, which is injective if for every $x \in A$, there exists $\phi \in \mathcal{S}$ with $\dpn{x^*x, \phi}{A}\ne 0$.
In particular, $A$ is isomorphic to a closed subalgebra of $B([l^{2}(P(A)); H_{\phi}])$.
Proof. (1): By the GNS construction.
(2): For each $x \in A$, if $x$ is positive, then so is $\pi(x)$, so $\dpn{\pi(x)\xi, \xi}{H}\ge 0$. Since $\xi$ is a unit vector, $\dpn{\pi(1_A)\xi, \xi}{H}= \dpn{\xi, \xi}{H}= 1$, and $\phi$ is a state.
Let $H^{0} = \bracsn{\pi(x)\xi|x \in A}$ and $H_{\phi}^{0} = \bracsn{\pi_\phi(x)\xi_\phi|x \in A}$. Define
then for each $x, y \in A$ with $\pi(x - y)\xi = 0$,
and $\pi_{\phi}(x - y)\xi_{\phi} = 0$ as well. Thus $U$ is well-defined. Moreover, for each $x \in A$,
so $U$ is an isometry. For each $x, y \in A$,
so $U$ extends continuously to a unitary equivalence between $(H, \pi)$ and $(H_{\phi}, \pi_{\phi})$, with $U(\xi) = \xi_{\phi}$.
(3): Suppose that for each $x \in A$, there exists $\phi \in \mathcal{S}$ such that $\dpn{x^*x, \phi}{A}\ne 0$. In which case,
so $\pi_{\phi}(x) \ne 0$, and $\pi_{\mathcal{S}}(x) \ne 0$ as well.
By Corollary 34.10.9, for each $x \in A$, there exists $\phi \in P(A)$ with $\dpn{x^*x, \phi}{A}\ne 0$, so $\pi_{P(A)}$ is injective. By Theorem 34.5.3, $\pi_{P(A)}(A)$ is closed in $B([l^{2}(P(A)); H_{\phi}])$.$\square$
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