34.11 The GNS Construction
Definition 34.11.1 (Cyclic Representation).label Let $A$ be a $C^{*}$-algebra, $(H, \pi)$ be a representation of $A$, and $\xi \in H$, then $\xi$ is a cyclic vector for $(H, \pi)$ if $\bracsn{\pi(x)(\xi)|x \in A}$ is dense in $H$. The representation $(H, \pi)$ is cyclic if it admits a cyclic vector.
Lemma 34.11.2.label Let $A$ be a $C^{*}$-algebra, $\phi \in S(A)$, and
then:
- (1)
For any $x, y \in A$ with $x \in N_{\phi}$ or $y \in N_{\phi}$, $\dpn{x, y}{A}= 0$.
- (2)
$N_{\phi}$ is a closed left ideal of $A$.
Proof. (1): By the Cauchy-Schwarz inequality, for any $x, y \in A$,
If $x \in N_{\phi}$ or $y \in N_{\phi}$, then the above inequality shows that $\dpn{x, y}{\phi}= 0$.
(2): As the zero set of a continuous function on $A$, $N_{\phi}$ is closed.
For any $x, y \in N_{\phi}$,
By (1), $\dpn{x, y}{\phi}= \dpn{y, x}{\phi}= 0$. Therefore $x + y \in N_{\phi}$.
Finally, for each $x \in N_{\phi}$ and $y \in A$,
by (1).$\square$
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 $H_{\phi}$.$\square$
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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