\section{The Gelfand-Naimark Theorem}
\label{section:gelfand-naimark}

\begin{theorem}[Gelfand-Naimark]
\label{theorem:gelfand-naimark}
    Let $A$ be a commutative unital $C^*$-algebra, then the Gelfand transform
    \[
    \Gamma_A: A \to C(\Omega(A); \complex) \quad \Gamma_A(x)(\phi) = \phi(x)
    \]

    is a unital $C^*$-isomorphism. 
\end{theorem}
\begin{proof}[Proof, {{\cite[Theorem 9.4]{Zhu}}}. ]
    By construction $\Gamma_A$ is a unital algebra homomorphism. 

    To see that $\Gamma_A$ preserves involutions, let $y \in A$ be self-adjoint. By \autoref{proposition:gelfand-transform-gymnastics} and \autoref{proposition:self-adjoint-spectrum}, $\Gamma_A(y)(\Omega(A)) = \sigma_A(y) \subset \real$, so $\Gamma_A(y) \in C(\Omega(A); \real)$. For any $x \in A$, write $x = \text{Re}(x) + i\text{Im}(x)$, where $\text{Re}(x)$ and $\text{Im}(x)$ are both self-adjoint, then
    \begin{align*}
        \Gamma_A(x^*) &= \Gamma_A(\text{Re}(x) - i\text{Im}(x)) \\
        &= \Gamma_A(\text{Re}(x)) - i\Gamma_A(\text{Im}(x)) \\
        &= \overline{\Gamma_A(\text{Re}(x)) + i\Gamma_A(\text{Im}(x))} = \overline{\Gamma_A(x)}
    \end{align*}

    so $\Gamma_A(x^*) = \Gamma_A(x)^*$. 

    Now, for each $x \in A$, \autoref{corollary:c-star-unique-norm} and \autoref{proposition:gelfand-transform-gymnastics} imply that
    \begin{align*}
        \norm{x}_A^2 &= \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)} \\
        &= \sup\bracs{|\Gamma_A(x^*x)(\phi)|\ | \phi \in \Omega(A)} \\
        &= \sup\bracs{|\Gamma_A(x)(\phi)|^2\ | \phi \in \Omega(A)} \\
        \norm{x}_A &= \norm{\Gamma_A(x)}_u
    \end{align*}

    Thus $\Gamma_A$ is an isometry, and $\Gamma_A(A)$ is a closed subalgebra of $C(\Omega(A))$. 

    Since $\Gamma_A(1_A) = 1$, $\Gamma_A(A)$ contains constants. As $\Gamma_A(A)$ separates points and is closed under complex conjugation, $\Gamma_A(A) = C(\Omega(A))$ by the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}.
\end{proof}

\begin{corollary}
\label{corollary:gelfand-naimark-converse}
    Let $A$ be a unital $C^*$-algebra, then the following are equivalent:
    \begin{enumerate}
        \item $A$ is commutative. 
        \item $\Gamma_A$ is a *-isomorphism. 
        \item $\Gamma_A$ is injective.
    \end{enumerate}
\end{corollary}

\begin{corollary}
\label{corollary:spectrum-characterisation-iff}
    Let $A$ be a commutative unital $C^*$-algebra and $x \in A$ be normal, then:
    \begin{enumerate}
        \item $x$ is self-adjoint if and only if $\sigma_A(x) \subset \real$. 
        \item $x$ is unitary if and only if $\sigma_A(x) \subset \partial B_\complex(0, 1)$.
        \item $x$ is positive if and only if $\sigma_A(x) \subset [0, \infty)$. 
        \item $x$ is a projection if and only if $\sigma_A(x) \subset \bracs{0,1}$. 
    \end{enumerate}
\end{corollary}
% NEEDS WORK

\begin{corollary}
\label{corollary:stonean-commutative-algebra}
    Let $A$ be a unital $C^*$-algebra, then $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected. 
\end{corollary}
\begin{proof}
    By \autoref{theorem:gelfand-naimark}, $A$ and $C(\Omega(A); \complex)$ are isomorphic as $C^*$-algebras. In particular, $A_{sa}$ and $C(\Omega(A); \real)$ are isomorphic as ordered vector spaces, so $A_{sa}$ is order complete if and only if $C(\Omega(A); \real)$ is order complete. Thus the \hyperref[Stone-Nakano Theorem]{theorem:stone-nakano-extremely-disconnected} implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected.
\end{proof}

\begin{corollary}
\label{corollary:linfinity-extremely-disconnected}
    Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremely disconnected. 
\end{corollary}
\begin{proof}
    By \autoref{corollary:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremely disconnected. 
\end{proof}