Definition 34.11.3 (GNS Triple).label Let $A$ be a unital $C^{*}$-algebra, $\phi \in S(A)$, and
Let $H_{\phi}^{0} = A/N_{\phi}$, $H_{\phi}$ be its completion with respect to $\dpn{\cdot, \cdot}{\phi}$, and
For each $x \in A$, let $\pi_{\phi}(x)$ be the continuous extension of $\pi_{\phi}^{0}(x)$ to an element of $B(H_{\phi})$, then:
- (1)
$(H_{\phi}, \dpn{\cdot, \cdot}{\phi})$ is a Hibert space.
- (2)
$(H_{\phi}, \pi_{\phi})$ is a well-defined representation of $A$.
- (3)
$\xi_{\phi} = 1_{A} + N_{\phi}$ is a unit vector in $H_{\phi}$, and $\bracsn{\pi_\phi(x)\xi_\phi| x \in A}$ is dense in $H_{\phi}$. Moreover, for each $x, y \in A$,
\[\dpn{x, y}{\phi}= \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}\]
The representation $(H_{\phi}, \pi_{\phi})$ is the cyclic representation of $A$ induced by $\phi$, and the triple $(H_{\phi}, \pi_{\phi}, \xi_{\phi})$ is the Gelfand-Naimark-Segal (GNS) triple associated with $\phi$.
Proof, [Proposition 14.2, Zhu93]. (2): Fix $x \in A$, then for each $y_{1}, y_{2} \in A$ with $y_{1} - y_{2} \in N_{\phi}$, $x(y_{1} - y_{2}) \in N_{\phi}$ by Lemma 34.11.2, so $\pi_{\phi}^{0}(x)$ is well-defined on $A/N_{\phi}$.
By rescaling, assume without loss of generality that $\norm{x}_{A} \le 1$. In which case, for each $y \in A$,
Since $\sigma_{A}(x^{*}x) \subset [0, 1]$, $\sigma_{A}(1 - x^{*}x) \subset [0, 1]$ and is positive by Corollary 34.6.3. Thus there exists $z \in A$ positive such that $(1 - x^{*}x) = z^{*}z$, so
and $\dpn{y, y}{\phi}\ge \dpn{xy, xy}{\phi}$. Therefore $\pi_{\phi}^{0}(x)$ extends continuously into an element of $B(H_{\phi})$ by the linear extension theorem.
Now, let $x, y, z \in A$, then
and by uniqueness of continuous extensions, $\pi_{\phi}(x)\pi_{\phi}(y) = \pi_{\phi}(xy)$, so $\pi_{\phi}$ is a homomorphism.
Finally,
By uniqueness of continuous extensions, $\pi_{\phi}(x^{*}) = \pi_{\phi}(x)^{*}$. Therefore $\pi_{\phi}$ is a *-homomorphism, and $(H_{\phi}, \pi_{\phi})$ is a representation of $A$.
(3): Since $\phi$ is a state, $\dpn{1_A, 1_A}{\phi}= 1$, and $1_{A}$ is a unit vector. As $H_{\phi}$ is the completion of $A/N_{\phi}$ and $A/N_{\phi} = \bracsn{\pi_\phi(x)(1_A + N_\phi)| x \in A}$, $\bracsn{\pi_\phi(x)(1_A + N_\phi)| x \in A}$ is dense in $H_{\phi}$.
For each $x, y \in A$, $\dpn{x, y}{\phi}= \dpn{\pi_\phi(x)(1_A + N_\phi), \pi_\phi(y)(1_A + N_\phi)}{H_\phi}$ by well-definedness of the inner product on $H_{\phi}$.$\square$
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