From 26fdb527ce19c0cab680ad2a865d3d7ae6a255cb Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Fri, 3 Jul 2026 16:20:41 -0400 Subject: [PATCH] Added more facts about the positive square root. --- document.tex | 2 ++ src/op/c-star/cont.tex | 56 +++++++++++++++++++++++++++++++++++++++++ src/op/c-star/index.tex | 4 +-- src/op/c-star/order.tex | 49 ++++++++++++++++++++++++++++++++++++ 4 files changed, 109 insertions(+), 2 deletions(-) diff --git a/document.tex b/document.tex index 4d8831d..3e7f699 100644 --- a/document.tex +++ b/document.tex @@ -6,6 +6,8 @@ Hi, welcome to my digital garden, where I collect math results that I learn. +Despite being presented in a linear order, I will frequently reference things between chapters and sections. + Occasionally, I make up some definitions to play with. These definition blocks will always have a * at the end of its title to indicate that it lives mostly in my head. These terms will always be referenced with a link to their definition block. \input{./src/cat/index} diff --git a/src/op/c-star/cont.tex b/src/op/c-star/cont.tex index 184b06b..a9375a6 100644 --- a/src/op/c-star/cont.tex +++ b/src/op/c-star/cont.tex @@ -43,6 +43,12 @@ \begin{enumerate}[start=3] \item Under the identification $\Omega(A[x]) = \sigma_A(x)$, for each $f \in C(\sigma_A(x); \complex)$, $f(x) = \Gamma^{-1}_{A[x]}(f)$. \item The mapping $f \mapsto f(x)$ is a unital *-isomorphism. + \item For each $U \subset \complex$ open and $f \in C(U; \complex)$, the mapping + \[ + \bracs{x \in A|\sigma_A(x) \subset U} \to A \quad x \mapsto f(x) + \] + + is continuous. \end{enumerate} \end{definition} @@ -58,8 +64,58 @@ (3): The mapping $f \mapsto f(x)$ is a *-homomorphism. + (6): Fix $x \in A$ and let $K \in \cn_U(\sigma_A(x); \complex)$ be compact. By \autoref{proposition:spectrum-continuous}, there exists $r > 0$ such that $\sigma_A(y) \subset K$ for all $y \in B_A(x, r)$. + + Let $\eps > 0$. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $p \in \complex[z, \ol z]$ such that $\sup_{\lambda \in K}|p(\lambda) - f(\lambda)| < \eps$. In which case, + \begin{align*} + \norm{f(x) - f(y)}_A &\le \norm{f(x) - p(x)}_A + \norm{f(y) - p(y)}_A + \norm{p(x) - p(y)}_A \\ + &\le \norm{p(x) - p(y)}_A + 2\eps + \end{align*} + + for all $y \in B_A(x, r)$. Since $p$ is a continuous map on $A$, there exists $\delta \in (0, r)$ such that $\norm{p(x) - p(y)}_A < \eps$ for all $y \in B_A(x, \delta)$. Therefore $\norm{f(x) - f(y)}_A < 3\eps$ for all $y \in B_A(x, \delta)$. + (Uniqueness): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}. \end{proof} +\begin{theorem}[Spectral Mapping Theorem (Continuous)] +\label{theorem:spectral-mapping-continuous} + Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then: + \begin{enumerate} + \item For every $f \in C(\sigma_A(x); \complex)$, $\sigma_A(f(x)) = f(\sigma_A(x))$. + \item For every $f \in C(\sigma_A(x); \complex)$ and $g \in C(\sigma_A(f(x)); \complex)$, $(f \circ g)(x) = f(g(x))$. + \end{enumerate} +\end{theorem} +\begin{proof} + (1): Since $f(x) \in A[x]$, by \autoref{proposition:gelfand-transform-gymnastics} and definition of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, + \[ + \sigma_A(f(x)) = \sigma_{A[x]}(f(x)) = \Gamma_{A[x]}(f(x))(\sigma_A(x)) = f(\sigma_A(x)) + \] + + (2): Firstly, the theorem holds directly if $f, g \in \complex[z, \ol z]$ are polynomials in $z$ and $\ol z$. + + Suppose that $g \in \complex[z, \ol z]$ but $f$ is arbitrary. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $\seq{p_n} \subset \complex[z, \ol z]$ such that $p_n \to f$ uniformly on $\sigma_A(x)$. By property (6) of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, + \[ + (g \circ f)(x) = \limv{n}(g \circ f_n)(x) = \limv{n}g(f_n(x)) = \limv{n}g(f(x)) + \] + + Finally, suppose that both $f$ and $g$ are arbitary. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $\seq{p_n} \subset \complex[z, \ol z]$ such that $p_n \to g$ uniformly on $\sigma_A(f(x))$. By continuity of the continuous functional calculus, + \[ + (g \circ f)(x) = \limv{n}(g_n \circ f)(x) = \limv{n}g_n(f(x)) = \limv{n}g(f(x)) + \] +\end{proof} + + + +\begin{corollary} +\label{corollary:normal-spectrum-identity} + Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then: + \begin{enumerate} + \item $x$ is self-adjoint if and only if $\sigma(x) \subset \real$. + \item $x$ is unitary if and only if $\sigma(x) \subset \partial B_\complex(0, 1)$. + \item $x$ is a projection if and only if $\sigma(x) \subset \bracs{0, 1}$. + \end{enumerate} +\end{corollary} + + diff --git a/src/op/c-star/index.tex b/src/op/c-star/index.tex index bf7240d..d89067e 100644 --- a/src/op/c-star/index.tex +++ b/src/op/c-star/index.tex @@ -5,7 +5,7 @@ \input{./sub.tex} \input{./unitary.tex} \input{./sa.tex} -\input{./order.tex} \input{./homomorphism.tex} \input{./gelfand.tex} -\input{./cont.tex} \ No newline at end of file +\input{./cont.tex} +\input{./order.tex} \ No newline at end of file diff --git a/src/op/c-star/order.tex b/src/op/c-star/order.tex index 706670e..5c95954 100644 --- a/src/op/c-star/order.tex +++ b/src/op/c-star/order.tex @@ -6,5 +6,54 @@ 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 II.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} +