Added basics of Banach algebras.

refs.bib

@@ -122,3 +122,13 @@
   year={1978},
   publisher={Springer}
 }
+@book{Zhu,
+  title={An Introduction to Operator Algebras},
+  author={Zhu, K.},
+  isbn={9780849378751},
+  lccn={93007172},
+  series={Studies in Advanced Mathematics},
+  url={https://books.google.ca/books?id=XHLj7bz8hOIC},
+  year={1993},
+  publisher={Taylor \& Francis}
+}

src/op/banach/igroup.tex

@@ -0,0 +1,34 @@
+\section{The Index Group}
+\label{section:index-group}
+
+\begin{definition}[Identity Component]
+\label{definition:identity-component}
+    Let $A$ be a unital Banach algebra, then the connected component $G_0(A)$ of $G(A)$ containing $1$ is the \textbf{identity component} of $G(A)$, and:
+    \begin{enumerate}
+        \item $G_0(A)$ is an open, closed, and normal subgroup of $G(A)$. 
+        \item The cosets of $G_0(A)$ are the connected components of $G(A)$. 
+    \end{enumerate}
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem 2.4]{Zhu}}}. ]
+    (1): Since $G_0(A)$ is connected, it is open and closed. By \autoref{proposition:locally-path-connected-properties}, $G_0(A)$ coincides with the path component of $G(A)$ containing $1$. 
+
+    Let $x, y \in G_0(A)$, then there exists paths $f, g \in C([0, 1]; G_0(A))$ such that $f(0) = g(0) = 1$, $f(1) = x$, and $g(1) = y$. The concatenation of $f$ and $xg$ then is a path from $1$ to $xy$, so $xy \in G_0(A)$. In addition, $t \mapsto f(t)^{-1}$ is a path from $1$ to $x^{-1}$, so $x^{-1} \in G_0(A)$ as well. Therefore $G_0(A)$ is a subgroup of $G(A)$. 
+
+    Finally, let $x \in G(A)$, then $x^{-1}G_0(A)x$ is a connected subset of $G(A)$ containing $1$, so $x^{-1}G_0(A)x \subset G_0(A)$ and $G_0(A) \subset xG_0(A)x^{-1}$. Since the above holds for all $x \in G(A)$, $x^{-1}G_0(A)x = G_0(A)$. 
+
+    (2): For each $x \in G(A)$, $xG_0(A)$ is connected, closed, and open, so it is a connected component by \autoref{lemma:union-connected-components}. 
+\end{proof}
+
+% Note: this setup appears to work in general topological groups. There does not seem to be any payoffs as of now.
+% However, should there be any, the proof should be improved to avoid the path argument. 
+
+
+\begin{definition}[Index Group]
+\label{definition:index-group}
+    Let $A$ be a unital Banach algebra, then the \textbf{index group} $I(A)$ is the quotient $G(A)/G_0(A)$, which is a discrete group. 
+\end{definition}
+\begin{proof}
+    By \autoref{definition:identity-component}, the components of $G(A)$ are the cosets of $G_0(A)$, so the quotient group is discrete. 
+\end{proof}
+
+

src/op/banach/index.tex

@@ -1,5 +1,7 @@
-\chapter{$C*$-Algebras}
+\chapter{Banach Algebras}
 \label{chap:banach-algebras}
 
 \input{./definitions.tex}
-\input{./invertible.tex}
+\input{./invertible.tex}
+\input{./igroup.tex}
+\input{./spectrum.tex}

src/op/banach/invertible.tex

@@ -4,15 +4,17 @@
 
 \begin{definition}[Invertible]
 \label{definition:banach-algebra-invertible}
-    Let $A$ be a unital Banach algebra and $x \in A$, then $x$ is \textbf{invertible} if there exists $x^{-1} \in A$ such that $xx^{-1} = x^{-1}x = 1$. The set $G(A)$ denotes the collection of all invertible elements in $A$. 
+    Let $A$ be a unital Banach algebra and $x \in A$, then $x$ is \textbf{invertible} if there exists $x^{-1} \in A$ such that $xx^{-1} = x^{-1}x = 1$. The set $G(A)$ denotes the group of all invertible elements in $A$. 
 \end{definition}
 
-\begin{lemma}
+\begin{lemma}[Neumann Series]
 \label{lemma:neumann-series}
     Let $A$ be a unital banach algebra and $x \in B_A(1, 1)$, then $x \in G(A)$ with
     \[
     x^{-1} = \sum_{n = 0}^\infty (1 - x)^n
     \]
+
+    and $\normn{x^{-1}}_A \le (1 - \norm{x})_A^{-1}$. 
 \end{lemma}
 \begin{proof}
     Since $\norm{1 - x}_A < 1$, the series converges absolutely. Let $y = \sum_{n = 0}^\infty (1 - x)^n$, then
@@ -50,3 +52,27 @@
     (3): Since the inversion map is locally a power series, it is $C^\infty$ by \autoref{theorem:termwise-differentiation}. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:swap-invertible}
+    Let $A$ be a unital Banach algebra and $x, y \in A$, then $1 - xy \in G(A)$ if and only if $1 - yx \in G(A)$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition I.3.4]{Zhu}}}. ]
+    If $1 - xy \in G(A)$, then
+    \begin{align*}
+        (1 - yx)[y(1 - xy)^{-1}x + 1] &= y(1 - xy)^{-1}x + 1 \\
+        &-yxy(1 - xy)^{-1}x - yx \\
+        &= y[(1 - xy)^{-1} - xy(1 - xy)^{-1}]x + 1 - yx \\
+        &= yx + 1 - yx = 1
+    \end{align*}
+    
+    Similarly, 
+    \begin{align*}
+        [y(1 - xy)^{-1}x + 1](1 - yx) &=  y(1 - xy)^{-1}x + 1 \\
+        &-y(1 - xy)^{-1}xyx - yx \\
+        &= y[(1 - xy)^{-1} - (1 - xy)^{-1}xy]y + 1 - yx \\
+        &= yx + 1 - yx = 1
+    \end{align*}
+    
+\end{proof}
+
+

src/op/banach/spectrum.tex

@@ -0,0 +1,104 @@
+\section{The Spectrum}
+\label{section:spectrum}
+
+\begin{definition}[Spectrum]
+\label{definition:spectrum}
+    Let $A$ be a unital Banach algebra and $x \in A$, then
+    \[
+    \sigma(x) = \sigma_A(x) = \bracs{\lambda \in \complex| \lambda - x \not\in G(A)}
+    \]
+
+    is the \textbf{spectrum} of $x$ in $A$, and its complement is the \textbf{resolvent set} of $x$. 
+\end{definition}
+
+\begin{definition}[Spectral Radius]
+\label{definition:spectral-radius}
+    Let $A$ be a unital Banach algebra and $x \in A$, then
+    \[
+    [x]_{sp} = \sup\bracs{|\lambda|: \lambda \in \sigma_A(x)}
+    \]
+
+    is the \textbf{spectral radius} of $x$. 
+\end{definition}
+
+\begin{definition}[Resolvent]
+\label{definition:resolvent}
+    Let $A$ be a unital Banach algebra and $x \in A$, then
+    \[
+    R_x: \complex \setminus \sigma_A(x) \to A \quad \lambda \mapsto \frac{1}{\lambda - x}
+    \]
+
+    is the \textbf{resolvent} function of $x$, which is holomorphic on $\complex \setminus \sigma_A(x)$.
+\end{definition}
+\begin{proof}
+    By \autoref{proposition:banach-algebra-inverse}, the mapping $y \mapsto y^{-1}$ is smooth on $G(A)$. Hence $R_x: \complex \setminus \sigma_A(x) \to A$ is holomorphic. 
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:spectrum-non-empty}
+    Let $A$ be a unital Banach algebra and $x \in A$, then $\sigma_A(x) \ne \emptyset$. 
+\end{proposition}
+\begin{proof}
+    Assume for contradiction that $\sigma_A(x) = \emptyset$, then by \autoref{definition:resolvent}, the resolvent
+    \[
+    R_x: \complex \setminus \sigma_A(x) \to A \quad \lambda \mapsto \frac{1}{\lambda - x}
+    \]
+
+    is an entire function. Since
+    \[
+    R_x(\lambda) = \frac{1}{\lambda - x} = \frac{\lambda^{-1}}{1 - \lambda^{-1}x}
+    \]
+
+    which tends to $0$ as $|\lambda| \to \infty$, $R_x \in H(\complex; A) \cap C_0(\complex; A)$. By \hyperref[Liouville's Theorem]{theorem:liouville}, $R_x = 0$, which is impossible. 
+\end{proof}
+
+\begin{theorem}[Gelfand-Naimark]
+\label{theorem:gelfand-naimark}
+    Let $A$ be a unital Banach algebra. If every non-zero element of $A$ is invertible, then $A$ is isometrically isomorphic to $\complex$. 
+\end{theorem}
+\begin{proof}
+    Let $x \in A$. By \autoref{proposition:spectrum-non-empty}, there exists $\lambda \in \sigma_A(x)$. Since $\lambda \cdot 1 - x$ is not invertible and every non-zero element of $A$ is invertible, $x = \lambda \cdot 1$. Therefore the mapping $\complex \to A$ defined by $\lambda \mapsto \lambda \cdot 1$ is an isometric isomorphism. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:spectral-radius-hadamard}
+    Let $A$ be a unital Banach algebra and $x \in A$, then $[\cdot]_{sp} = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$. 
+\end{proposition}
+\begin{proof}
+    Let $r = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$, then for any $\lambda \in \complex$ with $|\lambda| > r$, the series $\sum_{n = 0}^\infty \lambda^{-n-1}x^n$ converges absolutely, and to the inverse of $(x - \lambda)$. Therefore $r \ge [\cdot]_{sp}$. 
+
+    Let $D = \bracs{\lambda \in \complex|\ |\lambda| > [x]_{sp}}$, then since the series $\sum_{n = 0}^\infty \lambda^{-n-1}x^n$ is the expansion of $R_x$ at infinity, which is defined on $D$, it must converge on $D$. Let $\phi \in A^*$, then by the preceding discussion, the series $\sum_{n = 0}^\infty \lambda^{-n-1}\dpn{x^n, \phi}{A}$ converges on $D$. Thus for any $\lambda \in D$, 
+    \[
+    \sup_{n \in \natz}|\lambda^{-n-1} \dpn{x^n, \phi}{A}|^{1/n} < \infty
+    \]
+
+    By the \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness}, 
+    \[
+    \sup_{n \in \natz}|\lambda|^{-n-1} \cdot \normn{x^n}_A < \infty
+    \]
+
+    Therefore $\limsup_{n \to \infty}\norm{x^n}_A^{1/n} \le [x]_{sp}$. 
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:spectrum-compact}
+    Let $A$ be a unital Banach algebra and $x \in A$, then $\sigma_A(x)$ is a compact subset of $B_\complex(0, \norm{x}_A)$. 
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:banach-algebra-inverse}, $G(A)$ and hence the resolvent set of $x$ is open, so $\sigma_A(x)$ is closed. By \autoref{proposition:spectral-radius-hadamard}, $[x]_{sp} \le \norm{x}_A$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:spectrum-product-gymnastics}
+    Let $A$ be a unital Banach algebra and $x, y \in A$, then:
+    \begin{enumerate}
+        \item $\sigma(xy) \cup \bracs{0} = \sigma(yx) \cup \bracs{0}$. 
+        \item $[xy]_{sp} = [yx]_{sp}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\lambda \in \complex \setminus \bracs{0}$, then by \autoref{proposition:swap-invertible}, $\lambda - xy \in G(A)$ if and only if $\lambda - yx \in G(A)$. 
+\end{proof}
+

src/op/notation.tex

@@ -5,4 +5,10 @@
     \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
     \hline
     $1$ & Identity element of a unital algebra. & \autoref{definition:unital-banach-algebra} \\
+    $G(A)$ & Invertible group of a unital algebra. & \autoref{definition:banach-algebra-invertible} \\
+    $G_0(A)$ & The identity component of $G(A)$. & \autoref{definition:identity-component} \\
+    $I(A)$ & 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}
 \end{tabular}