From 44d8bbd4da2e4610ca28a5d0f30715567424b537 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 2 Jun 2026 15:40:18 -0400 Subject: [PATCH] Added basic facts on maximal ideals. --- src/op/banach/definitions.tex | 5 ++- src/op/banach/ideal.tex | 59 ++++++++++++++++++++++++++++++++ src/op/banach/index.tex | 2 ++ src/op/banach/multiplicative.tex | 50 +++++++++++++++++++++++++++ src/op/notation.tex | 4 ++- 5 files changed, 118 insertions(+), 2 deletions(-) create mode 100644 src/op/banach/ideal.tex create mode 100644 src/op/banach/multiplicative.tex diff --git a/src/op/banach/definitions.tex b/src/op/banach/definitions.tex index 0c6c79c..bc3b4fa 100644 --- a/src/op/banach/definitions.tex +++ b/src/op/banach/definitions.tex @@ -12,8 +12,11 @@ \begin{definition}[Unital Banach Algebra] \label{definition:unital-banach-algebra} - Let $A$ be a Banach algebra, then $A$ is \textbf{unital} if there exists $1 \in A$ such that for any $x \in A$, $x1 = 1x = x$. In which case, $1$ is the unique \textbf{multiplicative identity} of $A$. + Let $(A, \norm{\cdot}_A)$ be a Banach algebra, then $A$ is \textbf{unital} if there exists $1 \in A$ such that for any $x \in A$, $x1 = 1x = x$. In which case, there exists an equivalent norm $\norm{\cdot}_{1}: A \to [0, \inftu)$ such that $\norm{1}_1 = 1$, and $A$ is always assumed to be equipped with this norm. \end{definition} +\begin{proof}[Proof, {{\cite[Proposition I.1.3]{Takesaki1}}}. ] + For each $x \in A$, let $L_x \in L(A; A)$ be defined by $y \mapsto xy$, and let $\norm{x}_1 = \norm{L_x}_{L(A; A)}$, then $\norm{x}_1 \le \norm{x}_A$ and $\norm{1}_1 = 1$. On the other hand, $\frac{\norm{x}_A}{\norm{1}_A} \le \norm{x}_1$, so $\norm{\cdot}_1$ is equivalent to $\norm{\cdot}_A$. +\end{proof} \begin{definition}[Homomorphism] \label{definition:banach-algebra-homomorphism} diff --git a/src/op/banach/ideal.tex b/src/op/banach/ideal.tex new file mode 100644 index 0000000..b00d1e3 --- /dev/null +++ b/src/op/banach/ideal.tex @@ -0,0 +1,59 @@ +\section{Ideals} +\label{section:banach-algebra-ideals} + +\begin{definition}[Ideal] +\label{definition:banach-algebra-ideal} + Let $A$ be a Banach algebra and $I \subset A$, then $I$ is a \textbf{left ideal} if: + \begin{enumerate} + \item $I$ is a subspace of $A$. + \item For each $x \in A$, $aI \subset I$. + \end{enumerate} + + and $I$ is a \textbf{two-sided ideal} if in addition to the above, + \begin{enumerate}[start=2] + \item For each $x \in A$, $Ia \subset I$. + \end{enumerate} + + \textit{In this document, the "sidedness" of ideals are omitted. Within each block, the statement should hold as long as the interpretation is consistent. } +\end{definition} + +\begin{definition}[Proper Ideal] +\label{definition:banach-algebra-proper-ideal} + Let $A$ be a Banach algebra and $I \subset A$ be an ideal, then $I$ is \textbf{proper} if $I \subsetneq A$. +\end{definition} + +\begin{definition}[Maximal Ideal] +\label{definition:maximal-ideal} + Let $A$ be a Banach algebra and $I \subset A$ be an ideal, then $I$ is \textbf{maximal} if: + \begin{enumerate} + \item $I$ is proper. + \item For any proper ideal $J \subsetneq A$ with $J \supset I$, $I = J$. + \end{enumerate} + + The set $\cm(A)$ of maximal two-sided ideals of $A$ is the \textbf{maximal ideal space} of $A$. +\end{definition} + +\begin{proposition} +\label{proposition:ideal-gymnastics} + Let $A$ be a unital Banach algebra and $I \subset A$ be a proper ideal, then: + \begin{enumerate} + \item $I$ contains no invertible elements. + \item $\ol I$ is also a proper ideal. + \item $I$ is contained in a maximal ideal. + \item If $I$ is maximal, then $I$ is closed. + \end{enumerate} +\end{proposition} +\begin{proof}[Proof, {{\cite{FollandHarmonic}}}. ] + (1): If $I$ contains an invertible element, then $1 \in I$ and thus $A = I$. + + (2): By \autoref{proposition:banach-algebra-inverse}, $G(A)$ is open. Since $G(A) \cap I = \emptyset$, $G(A) \cap \ol I = \emptyset$ as well. + + (3): By Zorn's lemma. + + (4): By (2). +\end{proof} + + + + + diff --git a/src/op/banach/index.tex b/src/op/banach/index.tex index 341ac20..aa071c7 100644 --- a/src/op/banach/index.tex +++ b/src/op/banach/index.tex @@ -4,5 +4,7 @@ \input{./definitions.tex} \input{./invertible.tex} \input{./igroup.tex} +\input{./ideal.tex} \input{./spectrum.tex} \input{./fc.tex} +\input{./multiplicative.tex} diff --git a/src/op/banach/multiplicative.tex b/src/op/banach/multiplicative.tex new file mode 100644 index 0000000..d39cb00 --- /dev/null +++ b/src/op/banach/multiplicative.tex @@ -0,0 +1,50 @@ +\section{Multiplicative Functionals} +\label{section:multiplicative-functional} + +\begin{definition}[Multiplicative Functional] +\label{definition:multiplicative-functional} + Let $A$ be a unital Banach algebra and $\phi \in A^*$, then $\phi$ is \textbf{multiplicative} if $\phi \ne 0$ and for each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$. +\end{definition} + +\begin{proposition} +\label{proposition:multiplicative-unit} + Let $A$ be a unital Banach algebra and $\phi \in A^*$ be a multiplicative functional, then + \begin{enumerate} + \item For each $x \in A$, $|\phi(x)| \le [x]_{sp} \le \norm{x}_A$. + \item $\norm{\phi}_{A^*} = 1$. + \item $\phi(G(A)) \subset \complex \setminus \bracs{0}$. + \end{enumerate} + +\end{proposition} +\begin{proof} + (3): For each $x \in G(A)$, $1 = \phi(xx^{-1}) = \phi(x)\phi(x^{-1}) \ne 0$. + + (1): By (3), for every $\lambda \in \complex$ with $|\lambda| > [x]_{sp}$, $\lambda x \in G(A)$ and $\lambda - \phi(x) \ne 0$. Therefore $\phi(x) \in \ol{B(0, [x]_{sp})}$. + + (2): For each $\lambda \in \complex$, $\phi(\lambda 1) = \lambda$, so $\norm{\phi}_{A^*} \le 1$. +\end{proof} + +\begin{definition}[Space of Multiplicative Linear Functionals] +\label{definition:multiplicative-linear-functional-space} + Let $A$ be a unital Banach algebra, then $\Omega(A)$ is the \textbf{space of multiplicative linear functionals}, which is a compact Hausdorff space under the weak-* topology. +\end{definition} +\begin{proof} + By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. +\end{proof} + +\begin{proposition} +\label{proposition:commutative-maximal-ideal-space} + Let $A$ be a commutative unital Banach algebra, then the mapping + \[ + \Omega(A) \to \cm(A) \quad \phi \mapsto \ker(\phi) + \] + + is a bijection. +\end{proposition} +\begin{proof} + For each $\phi \in \cm(A)$, $\ker(\phi)$ is an ideal of codimension $1$, and must be maximal. + + On the other hand, for each $I \in \Omega(A)$, the quotient $A/I$ is a field, so by the \hyperref[Gelfand-Mazur Theorem]{theorem:gelfand-mazur}, it is isomorphic to $\complex$. Thus the canonical projection $A \to A/I \iso \complex$ induces a multiplicative functional on $A$. +\end{proof} + + diff --git a/src/op/notation.tex b/src/op/notation.tex index 75ea846..0749488 100644 --- a/src/op/notation.tex +++ b/src/op/notation.tex @@ -10,5 +10,7 @@ $I(A)$ & The index group of $A$. & \autoref{definition:index-group} \\ $\sigma_A(x) = \sigma(x)$ & The spectrum of $x$ in $A$. & \autoref{definition:spectrum} \\ $R_x(\lambda)$ & The resolvent of $x$. & \autoref{definition:resolvent} \\ - $[x]_{sp}$ & The spectral radius of $x$. & \autoref{definition:spectral-radius} + $[x]_{sp}$ & The spectral radius of $x$. & \autoref{definition:spectral-radius} \\ + $\Omega(A)$ & Space of multiplicative functionals on $A$. & \autoref{definition:multiplicative-functional} \\ + $\cm(A)$ & Maximal ideal space of $A$. & \autoref{definition:maximal-ideal} \end{tabular} \ No newline at end of file