\section{The GNS Construction}
\label{section:gns}

\begin{definition}[Cyclic Representation]
\label{definition:cyclic-representation}
    Let $A$ be a $C^*$-algebra, $(H, \pi)$ be a representation of $A$, and $\xi \in H$, then $\xi$ is a \textbf{cyclic vector} for $(H, \pi)$ if $\bracsn{\pi(x)(\xi)|x \in A}$ is dense in $H$. The representation $(H, \pi)$ is \textbf{cyclic} if it admits a cyclic vector. 
\end{definition}


\begin{lemma}
\label{lemma:cstar-state-kernel}
    Let $A$ be a $C^*$-algebra, $\phi \in S(A)$, and
    \[
    N_\phi = \bracsn{x \in A| \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = 0}
    \]

    then:
    \begin{enumerate}
        \item For any $x, y \in A$ with $x \in N_\phi$ or $y \in N_\phi$, $\dpn{x, y}{A} = 0$. 
        \item $N_\phi$ is a closed left ideal of $A$. 
    \end{enumerate}
\end{lemma}
\begin{proof}
    (1): By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, for any $x, y \in A$, 
    \[
    |\dpn{x, y}{\phi}|^2 \le \dpn{x, x}{\phi} \cdot \dpn{y, y}{\phi}
    \]

    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 $x, y \in N_\phi$, 
    \begin{align*}
        \dpn{x + y, x + y}{\phi} &= \dpn{x, x}{\phi} + \dpn{x, y}{\phi} + \dpn{y, x}{\phi} + \dpn{y, y}{\phi} \\
        &= \dpn{x, y}{\phi} + \dpn{y, x}{\phi} 
    \end{align*}

    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 $y \in A$, 
    \[
        \dpn{yx, yx}{\phi} = \dpn{x^*y^*yx, \phi}{A} = \dpn{x^*(y^*yx), \phi}{A} = \dpn{y^*yx, x}{\phi} =0 
    \]

    by (1). 
\end{proof}


\begin{definition}[GNS Triple]
\label{definition:gns-triple}
    Let $A$ be a unital $C^*$-algebra, $\phi \in S(A)$, and
    \[
    N_\phi = \bracsn{x \in A| \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = 0}
    \]

    Let $H_\phi^0 = A/N_\phi$, $H_\phi$ be its completion with respect to $\dpn{\cdot, \cdot}{\phi}$, and
    \[
    \pi_\phi^0: A \to B(H_\phi^0) \quad \pi_\phi^0(x)(y + N_\phi) = xy + N_\phi
    \]

    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:
    \begin{enumerate}
        \item $(H_\phi, \dpn{\cdot, \cdot}{\phi})$ is a Hibert space.
        \item $(H_\phi, \pi_\phi)$ is a well-defined representation of $A$. 
        \item $\xi_\phi = 1_A + N_\phi$ 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 $x, y \in A$,
        \[
        \dpn{x, y}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}
        \]
    \end{enumerate}
    
    The representation $(H_\phi, \pi_\phi)$ is the \textbf{cyclic representation of $A$ induced by $\phi$}, and the triple $(H_\phi, \pi_\phi, \xi_\phi)$ is the \textbf{Gelfand-Naimark-Segal (GNS) triple associated with $\phi$}. 
\end{definition}
\begin{proof}[Proof, {{\cite[Proposition 14.2]{Zhu}}}. ]
    (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 \autoref{lemma:cstar-state-kernel}, 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 $y \in A$, 
    \[
        \dpn{y, y}{\phi} - \dpn{xy, xy}{\phi} = \dpn{y^*y, \phi}{A} - \dpn{y^*x^*xy, \phi}{A} = \dpn{y^*(1 - x^*x)y, \phi}{A}
    \]

    Since $\sigma_A(x^*x) \subset [0, 1]$, $\sigma_A(1 - x^*x) \subset [0, 1]$ and is positive by \autoref{corollary:spectrum-characterisation-iff}. Thus there exists $z \in A$ positive such that $(1 - x^*x) = z^*z$, so
    \[
        \dpn{y, y}{\phi} - \dpn{xy, xy}{\phi} = \dpn{y^*z^*zy, \phi}{A} = \dpn{zy, zy}{\phi} \ge 0
    \]

    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 \hyperref[linear extension theorem]{theorem:linear-extension-theorem-normed}. 
    

    Now, let $x, y, z \in A$, then
    \[
    \pi_\phi^0(x)[\pi_\phi^0(y)(z + N_\phi)] = \pi_\phi^0(x)(yz + N_\phi) = xyz + N_\phi = \pi_\phi^0(xy)(z + N_\phi)
    \]

    and by uniqueness of continuous extensions, $\pi_\phi(x)\pi_\phi(y) = \pi_\phi(xy)$, so $\pi_\phi$ is a homomorphism. 

    Finally, 
    \[
        \dpn{\pi_\phi^0(x^*)y, z}{\phi} = \dpn{z^*x^*y, \phi}{A} = \dpn{y, xz}{\phi} = \dpn{y, \pi_\phi^0(x)z}{\phi}
    \]

    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 $H_\phi$.
\end{proof}

\begin{theorem}[Gelfand-Naimark-Segal]
\label{theorem:gns}
    Let $A$ be a unital $C^*$-algebra, then:
    \begin{enumerate}
        \item 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
        \[
        \dpn{x, y}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}
        \]
        \item 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 $A$. Moreover, if $(H_\phi, \pi_\phi, \xi_\phi)$ is the GNS triple associated with $\phi$, then there exists a unitary equivalence $U: H \to H_\phi$ such that $U\xi = \xi_\phi$. 
        \item 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 $B([l^2(P(A)); H_\phi])$. 
    \end{enumerate}
\end{theorem}
\begin{proof}
    (1): By the \hyperref[GNS construction]{definition:gns-triple}. 

    (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
    \[
    U: H^0 \to H_\phi^0 \quad \pi(x)\xi \mapsto \pi_\phi(x)\xi_\phi
    \]

    then for each $x, y \in A$ with $\pi(x - y)\xi = 0$,
    \begin{align*}
    0 &= \dpn{\pi(x - y)\xi, \pi(x - y)\xi}{H} = \dpn{(x - y)^*(x - y), \phi}{A} \\
    &= \dpn{x - y, x- y}{\phi} = \dpn{\pi_\phi(x - y)\xi_\phi, \pi_\phi(x - y)\xi_\phi}{H_\phi}
    \end{align*}
    

    and $\pi_\phi(x - y)\xi_\phi = 0$ as well. Thus $U$ is well-defined. Moreover, for each $x \in A$,
    \[
        \dpn{\pi(x)\xi, \pi(x)\xi}{H} = \dpn{x^*x, \phi}{A} = \dpn{x^*x, 1_A}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(x) \xi_\phi}{H_\phi}
    \]

    so $U$ is an isometry. For each $x, y \in A$,
    \begin{align*}
    U(\pi(x)[\pi(y)\xi]) &= U(\pi(xy)\xi) = \pi_\phi(xy)\xi_\phi \\
    &= \pi_\phi(x)[\pi_\phi(y)\xi_\phi] = \pi_\phi(x)[U(\pi(y)\xi)]
    \end{align*}
    

    so $U$ \hyperref[extends continuously]{theorem:linear-extension-theorem-normed} 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,
    \[
    0 \ne \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(x)\xi_\phi}{H_\phi}
    \]

    so $\pi_\phi(x) \ne 0$, and $\pi_{\mathcal{S}}(x) \ne 0$ as well. 

    By \autoref{corollary:cstar-positive-weakstar-dense}, 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 \autoref{theorem:continuity-of-homomorphism-c-star}, $\pi_{P(A)}(A)$ is closed in $B([l^2(P(A)); H_\phi])$. 
    
\end{proof}



