\section{Order Structures of $C^*$-Algebras}
\label{section:order-c-star-algebra}

\begin{definition}[Positive]
\label{definition:positive-c-star-algebra}
    Let $A$ be a $C^*$-algebra and $x \in A$, then $x$ is \textbf{positive} if there exists $y \in A$ such that $x = y^*y$. 
\end{definition}

\begin{proposition}
\label{proposition:positive-spectrum}
    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then $x$ is positive if and only if $\sigma_A(x) \subset [0, \infty)$. 
\end{proposition}
\begin{proof}
    Using the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $x$ is positive if and only if $\Gamma_{A[x]}(x) = \text{Id}$ is positive in $C(\sigma_A(x); \complex)$, if and only if $\sigma_A(x) = \Gamma_{A[x]}(x)(\Omega(A[x])) \subset [0, \infty)$ by \autoref{proposition:gelfand-transform-gymnastics}. 
\end{proof}

\begin{proposition}
\label{proposition:positive-norm-inequality}
    Let $A$ be a unital $C^*$-algebra and $x \in A_{sa}$, then the following are equivalent:
    \begin{enumerate}
        \item $x$ is positive.
        \item $\sigma_A(x) \subset [0, \infty)$. 
        \item There exists $\lambda \ge \norm{x}_A$ such that $\norm{\lambda - x}_A \le \lambda$. 
    \end{enumerate}
\end{proposition}
\begin{proof}[Proof, {{\cite[Lemma 11.3]{Zhu}}}. ]
    (1) $\Leftrightarrow$ (2): \autoref{proposition:positive-spectrum}. 

    (2) $\Leftrightarrow$ (3): By assumption, $\sigma_A(x) \subset \real$, so \autoref{theorem:c-star-normal-spectral-radius} implies that
    \[
    \norm{\lambda - x}_A = [\lambda - x]_{sp} = \sup\bracsn{\lambda - \mu|\mu\in \sigma_A(x)}
    \]

    which is bounded above by $\lambda$ if and only if $\sigma_A(x) \subset [0, \infty)$. 
\end{proof}

\begin{corollary}
\label{corollary:positive-ordering}
    Let $A$ be a unital $C^*$-algebra. For each $x, y \in A$, denote $x \ge y$ if $x - y$ is positive, then $(A, \le)$ is an ordered vector space. 
\end{corollary}
\begin{proof}
    By definition, the ordering is reflexive, antisymmetric, translation-invariant, and invariant under scaling by positive constants. It remains to show that $\le$ is transitive, or equivalently, the sum of two positive elements is positive. 

    Let $x, y \in A$ be positive, then $x + y$ is self-adjoint. Thus there exists $\lambda \ge \norm{x}_A$ and $\mu \ge \norm{y}_A$ such that $\norm{\lambda - x}_A \le \lambda$ and $\norm{\mu - y}_A \le \mu$, so $\norm{(\lambda + \mu) - (x + y)}_A \le \lambda + \mu$, and $x + y$ is positive by \autoref{proposition:positive-norm-inequality}. 
\end{proof}


\begin{definition}[Positive Square Root]
\label{definition:positive-square-root}
    Let $A$ be a $C^*$-algebra and $x \in A$ be positive, then there exists a unique positive element $y \in A$ such that $y^2 = x$. The element $y$ is the \textbf{positive square root} of $x$, denoted $\sqrt{x}$. 
\end{definition}
\begin{proof}
    Since $x$ is positive, $\sigma_A(x) \subset [0, \infty)$ by \autoref{proposition:positive-norm-inequality}. Therefore the square root function $f(t) = \sqrt{t}$ is defined and continuous on $\sigma_A(x)$. Using the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $f(x)$ is a positive element of $A$ such that $f(x)^2 = x$. 

    Let $y \in A$ such that $y^2 = x$, then by the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, $y = f(y^2) = f(x)$, so the square root is unique. 
\end{proof}


\begin{definition}[Absolute Value]
\label{definition:absolute-value-c-star}
    Let $A$ be a unital $C^*$-algebra and $x \in A$, then $|x| = \sqrt{x^*x}$ is the \textbf{absolute value} of $x$. 
\end{definition}

\begin{definition}[Positive and Negative Parts]
\label{definition:positive-negative-cstar-algebra}
    Let $A$ be a unital $C^*$-algebra and $x \in A$ be self-adjoint, then there exists unique positive elements $x^+, x^- \in A$ such that
    \begin{enumerate}
        \item $x = x^+ - x^-$.
        \item $x^+x^- = x^-x^+ = 0$. 
    \end{enumerate}
    
    The pair $(x^+, x^-)$ are the \textbf{positive and negative parts} of $x$.
\end{definition}
\begin{proof}
    Since $x$ is self-adjoint, $\sigma_A(x) \subset \real$ by \autoref{proposition:self-adjoint-spectrum}. Using the continuous functional calculus, existence is given by the functions $f^+(\lambda) = \lambda \vee 0$ and $f^-(\lambda) = \lambda \wedge 0$ and \autoref{proposition:positive-norm-inequality}. 

    On the other hand, for each $p \in \real[z]$ with $p(0) = 0$, (2) implies that $p(x) = p(x^+) + p(-x^-)$. By the \hyperref[Stone-Weierstrass Theorem]{theorem:stone-weierstrass}, $f(x) = f(x^+) + f(-x^-)$ for all $f \in C(\real; \real)$ with $f(0) = 0$. In particular, (1) then implies that $f^+(x) = f+(x^+) + f^+(-x^-) = f^+(x^+) = x^+$, and likewise $f^-(x) = x^-$. Therefore the decomposition is given uniquely by the continuous functional calculus. 
\end{proof}

\begin{remark}
\label{remark:positive-negative-cstar-algebra}
    The condition in the sign decomposition that $x^+x^- = x^-x^+ = 0$ is essential. Otherwise I may use silly decompositions like $0 = 1 - 1$. 
\end{remark}

\begin{lemma}
\label{lemma:cstar-inversion-order-reversing}
    Let $A$ be a unital $C^*$-algebra, $x, y \in G(A)$ be positive elements with $x \le y$, then $x^{-1} \ge y^{-1}$.
\end{lemma}
\begin{proof}
    Since $y - x \ge 0$ and $x$ is invertible, $y^{-1/2}(y - x)y^{-1/2} \ge 0$ as well. As such, 
    \[
    y^{-1/2}xy^{-1/2} \le y^{-1/2}yy^{-1/2} = 1
    \]

    Thus $\sigma_A(y^{-1/2}xy^{-1/2}) \subset \ol{B_\complex(0, 1)}$, and $\sigma_A(y^{1/2}x^{-1}y^{1/2}) \subset \complex \setminus B_\complex(0, 1)$. Hence $y^{1/2}x^{-1}y^{1/2} \ge 1$, and $x^{-1} \ge y^{-1}$. 
\end{proof}