88 lines
5.0 KiB
TeX
88 lines
5.0 KiB
TeX
\section{Order Structures of $C^*$-Algebras}
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\label{section:order-c-star-algebra}
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\begin{definition}[Positive]
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\label{definition:positive-c-star-algebra}
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Let $A$ be a $C^*$-algebra and $x \in A$, then $x$ is \textbf{positive} if there exists $y \in A$ such that $x = y^*y$.
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\end{definition}
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\begin{proposition}
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\label{proposition:positive-spectrum}
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Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then $x$ is positive if and only if $\sigma_A(x) \subset [0, \infty)$.
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\end{proposition}
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\begin{proof}
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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}.
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\end{proof}
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\begin{proposition}
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\label{proposition:positive-norm-inequality}
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Let $A$ be a unital $C^*$-algebra and $x \in A_{sa}$, then the following are equivalent:
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\begin{enumerate}
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\item $x$ is positive.
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\item $\sigma_A(x) \subset [0, \infty)$.
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\item There exists $\lambda \ge \norm{x}_A$ such that $\norm{\lambda - x}_A \le \lambda$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Lemma 11.3]{Zhu}}}. ]
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(1) $\Leftrightarrow$ (2): \autoref{proposition:positive-spectrum}.
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(2) $\Leftrightarrow$ (3): By assumption, $\sigma_A(x) \subset \real$, so \autoref{theorem:c-star-normal-spectral-radius} implies that
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\[
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\norm{\lambda - x}_A = [\lambda - x]_{sp} = \sup\bracsn{\lambda - \mu|\mu\in \sigma_A(x)}
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\]
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which is bounded above by $\lambda$ if and only if $\sigma_A(x) \subset [0, \infty)$.
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\end{proof}
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\begin{corollary}
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\label{corollary:positive-ordering}
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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.
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\end{corollary}
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\begin{proof}
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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.
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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}.
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\end{proof}
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\begin{definition}[Positive Square Root]
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\label{definition:positive-square-root}
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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 $\sqrt{x}$.
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\end{definition}
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\begin{proof}
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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 $f(x)^2 = x$.
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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.
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\end{proof}
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\begin{definition}[Absolute Value]
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\label{definition:absolute-value-c-star}
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Let $A$ be a unital $C^*$-algebra and $x \in A$, then $|x| = \sqrt{x^*x}$ is the \textbf{absolute value} of $x$.
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\end{definition}
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\begin{definition}[Positive and Negative Parts]
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\label{definition:positive-negative-cstar-algebra}
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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
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\begin{enumerate}
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\item $x = x^+ - x^-$.
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\item $x^+x^- = x^-x^+ = 0$.
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\end{enumerate}
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The pair $(x^+, x^-)$ are the \textbf{positive and negative parts} of $x$.
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\end{definition}
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\begin{proof}
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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}.
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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.
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\end{proof}
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\begin{remark}
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\label{remark:positive-negative-cstar-algebra}
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The condition in the sign decomposition that $x^+x^- = x^-x^+ = 0$ is essential. Otherwise I may use silly decompositions like $0 = 1 - 1$.
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\end{remark}
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