From 6d84f0ea47997c6623777cefdf16e7cf7e02ac9a Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Wed, 17 Jun 2026 17:17:37 -0400 Subject: [PATCH] Performed housekeeping for topological groups. --- src/fa/tvs/definition.tex | 97 +------------------- src/topology/groups/definition.tex | 131 ++++++++++++++++++++++++++++ src/topology/groups/index.tex | 5 ++ src/topology/index.tex | 1 + src/topology/uniform/definition.tex | 2 +- 5 files changed, 142 insertions(+), 94 deletions(-) create mode 100644 src/topology/groups/definition.tex create mode 100644 src/topology/groups/index.tex diff --git a/src/fa/tvs/definition.tex b/src/fa/tvs/definition.tex index 9302d12..4889fa2 100644 --- a/src/fa/tvs/definition.tex +++ b/src/fa/tvs/definition.tex @@ -6,18 +6,13 @@ \label{definition:tvs} Let $E$ be a vector space over $K \in \bracs{\real, \complex}$ and $\topo \subset 2^E$ be a topology. If \begin{enumerate} - \item[(TVS1)] $E \times E \to E$ with $(x, y) \mapsto x + y$ is continuous. - \item[(TVS2)] $K \times E \to E$ with $(\lambda, x) \mapsto \lambda x$ is continuous. + \item[(TVS1)] The addition map $E \times E \to E$ with $(x, y) \mapsto x + y$ is continuous. + \item[(TVS2)] The scalar multiplication map $K \times E \to E$ with $(\lambda, x) \mapsto \lambda x$ is continuous. \end{enumerate} then the pair $(E, \topo)$ is a \textbf{topological vector space}. \end{definition} -\begin{definition}[Translation-Invariant Topology] -\label{definition:translation-invariant-topology} - Let $E$ be a vector space and $\topo$ be a topology on $E$, then $\topo$ is \textbf{translation-invariant} if for any $U \in \topo$ and $y \in E$, $U + y \in \topo$. -\end{definition} - \begin{lemma} \label{lemma:tvs-translation-invariant} Let $E$ be a TVS over $K \in \RC$, then the topology of $E$ is translation-invariant. @@ -26,26 +21,8 @@ Let $U \subset E$ open and $y \in E$, then $U + y$ is the preimage of $U$ by the map $x \mapsto x - y$. By (TVS1), $U + y$ is open. \end{proof} -\begin{definition}[Translation-Invariant Uniformity] -\label{definition:translation-invariant-uniformity} - Let $E$ be a vector space, $\fU$ be a uniformity on $E$, and $U \in \fU$, then $U$ is \textbf{translation-invariant} if for every $z \in E$, - \[ - U = \bracs{(x + z, y + z)|(x, y) \in U} - \] - and $\fU$ is \textbf{translation-invariant} if there exists a fundamental system of translation-invariant entourages. -\end{definition} - -\begin{lemma} -\label{lemma:translation-invariant-symmetric} - Let $E$ be a vector space and $\fU$ be a translation-invariant uniformity, then $\fU$ admits a fundamental system of symmetric, translation-invariant entourages. -\end{lemma} -\begin{proof} - Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By \autoref{lemma:symmetricfundamentalentourage}, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages. -\end{proof} - - -\begin{proposition}[{{\cite[I.1.4]{SchaeferWolff}}}] +\begin{proposition} \label{proposition:tvs-uniform} Let $E$ be a TVS over $K \in \bracs{\real, \complex}$, then: \begin{enumerate} @@ -56,73 +33,7 @@ The space $E$ will always be assumed to be equipped with its translation-invariant uniformity. \end{proposition} \begin{proof} - (2): Firstly, for any $V \in \fB_0$, $U_V$ is translation-invariant. - \begin{enumerate} - \item[(FB1)] For any $V, V' \in \fB_0$, there exists $W \in \fB_0$ such that $W \subset V \cap V'$. In which case, $U_V \cap U_{V'} \supset U_W \in \fB$. - \item[(UB1)] For any $V \in \fB_0$, $0 \in V$, so $\Delta \subset U_V$. - \item[(UB2)] Let $V \in \fB_0$, then by (TVS1), there exists $W \in \fB_0$ such that $W + W \subset V$. For any $(x, y), (y, z) \in U_W$, $(x - y), (y - z) \in W$, so $(x - z) \in V$. Thus $U_W \circ U_W \subset U_V$. - \end{enumerate} - - By \autoref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$. - - (1): Let $\mathfrak{V}$ be a translation-invariant uniformity on $E$ inducing the topology. For any symmetric, translation-invariant entourage $V \in \mathfrak{V}$, $V(x) = V(0) + x$ for all $x \in E$, and $(x, y) \in V$ if and only if $y - x \in V$, if and only if $x - y \in V$. Thus $V = U_{V(0)}$. - - Let $W \in \cn(0)$, then by \autoref{lemma:translation-invariant-symmetric}, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_W$. Thus $\mathfrak{V} \supset \fU$. - - Let $V \in \mathfrak{V}$. Using \autoref{lemma:translation-invariant-symmetric}, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_0$ with $W \subset V(0)$. In which case, $U_W \subset V$, and $\fU \supset \mathfrak{V}$. -\end{proof} - -\begin{proposition} -\label{proposition:tvs-closure} - Let $E$ be a TVS over $K \in \RC$, $A \subset E$, and $\fB \subset \cn(0)$ be a fundamental system of neighbourhoods, then - \[ - \ol{A} = \bigcap_{U \in \fB}\bracs{A + U| U \in \fB} - \] - -\end{proposition} -\begin{proof} - Let $V \in \cn(0)$ be balanced and $U_V = \bracs{(x, y) \in E \times E| x - y \in V}$, then $y \in U_V(A)$ if and only if there exists $x \in A$ such that $(x, y) \in U_V$. This is equivalent to $x - y \in V$ and $y - x \in V$, so $U_V(A) = A + V$. - - Assume without loss of generality that $\fB$ consists of symmetric entourages. By \autoref{proposition:tvs-uniform}, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and \autoref{proposition:uniformclosure} implies that - \[ - \ol{A} = \bigcap_{V \in \fB}\bracs{U_V(A)| U \in \fB} = \bigcap_{V \in \fB}U + A - \] - -\end{proof} - -\begin{proposition}[{{\cite[I.1.1]{SchaeferWolff}}}] -\label{proposition:tvs-set-operations} - Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$, then: - \begin{enumerate} - \item If $A$ is open, then $A + B$ is open. - \item If $A$ is closed and $B$ is compact, then $A + B$ is closed. - \end{enumerate} -\end{proposition} -\begin{proof} - $(1)$: For every $x \in B$, $A + x$ is open by \autoref{definition:translation-invariant-topology}, so - \[ - A + B = \bigcup_{x \in B}(A + x) - \] - - is open. - - $(2)$: Let $x \in \overline{A + B}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap (A + B) \ne \emptyset$, so $(U - B) \cap A \ne \emptyset$, and $\fB = \bracs{U - B| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that - \[ - y \in \bigcap_{U \in \fF}\overline{U - B} = \bigcap_{U \in \fF}\overline{(-B) + U} - \] - - - By \autoref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so - \[ - y \in \bigcap_{U \in \fF}\overline{(-B) + U} \subset \bigcap_{U \in \fF}[(-B) + U + U] - \] - - Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus - \[ - y \in \bigcap_{U \in \fF}[(-B) + U + U] \subset \bigcap_{V \in \cn(0)}[(x-B) + V] = \overline{x - B} = x - B - \] - - so $x \in y + B \subset A + B$. + By \autoref{definition:group-translation-invariant-uniformity}. \end{proof} \begin{definition}[Balanced/Circled] diff --git a/src/topology/groups/definition.tex b/src/topology/groups/definition.tex new file mode 100644 index 0000000..f607161 --- /dev/null +++ b/src/topology/groups/definition.tex @@ -0,0 +1,131 @@ +\section{Group Topologies} +\label{section:group-topologies} + + +\begin{definition}[Topological Group] +\label{definition:topological-group} + Let $G$ be a group and $\mathcal{T} \subset 2^G$ be a topology. If + \begin{enumerate}[label=(TG\arabic*)] + \item The composition map $G \times G \to G$ with $(g, h) \mapsto gh$ is continuous. + \item The inversion map $G \to G$ with $g \mapsto g^{-1}$ is continuous. + \end{enumerate} + + then the pair $(E, \mathcal{T})$ is a \textbf{topological group}. +\end{definition} + +\begin{definition}[Translation-Invariant Topology] +\label{definition:translation-invariant-topology} + Let $G$ be a group and $\topo \subset 2^G$ be a topology, then $\topo$ is \textbf{left translation-invariant} if for every $U \in \topo$ and $g \in G$, $gU \in \topo$, and \textbf{right translation-invariant} if for every $U \in \topo$ and $g \in G$, $Ug \in \topo$. If $\topo$ is both left and right translation-invariant, then $\topo$ is \textbf{translation-invariant}. +\end{definition} + +\begin{definition}[Translation-Invariant Uniformity] +\label{definition:translation-invariant-uniformity} + Let $G$ be a group, $\fU$ be a uniformity on $G$, and $U \in \fU$, then $U$ is \textbf{left translation-invariant} if for every $z \in G$, + \[ + U = zU = \bracs{(zx, zy)|(x, y) \in U} + \] + + and \textbf{right translation-invariant} if for every $z \in G$, + \[ + U = Uz = \bracs{(xz, yz)|(x, y) \in U} + \] + + The uniformity $\fU$ is \textbf{left/right translation-invariant} if it admits a fundamental system of left/right translation-invariant entourages. +\end{definition} + +\begin{lemma} +\label{lemma:translation-invariant-symmetric} + Let $G$ be a group and $\fU \subset 2^{G \times G}$ be a left/right translation-invariant uniformity, then $\fU$ admits a fundamental system of symmetric, left/right translation-invariant entourages. +\end{lemma} +\begin{proof} + Let $z \in G$, then the map $(x, y) \mapsto (zx, zy)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) = zU \cap zV$, and $U \cap V$ is left translation-invariant. By \autoref{lemma:symmetricfundamentalentourage}, $\fU$ admits a left fundamental system of symmetric, translation-invariant entourages. +\end{proof} + +\begin{definition} +\label{definition:group-translation-invariant-uniformity} + Let $G$ be a topological group, then: + \begin{enumerate}[label=(L\arabic*)] + \item There exists a unique left translation-invariant uniformity $\fU_L$ on $G$ that induces its topology. + \item For each $U \in \cn_G(1)$, let $U_{L, V} = \bracsn{(x, y) \in G^2|x^{-1}y \in V}$, then $\fB_L = \bracs{U_{L, V}|V \in \cn_G(1)}$ is a fundamental system of entourages for $\fU_L$. + \end{enumerate} + + and + \begin{enumerate}[label=(R\arabic*)] + \item There exists a unique right translation-invariant uniformity $\fU_R$ on $G$ that induces its topology. + \item For each $U \in \cn_G(1)$, let $U_{R, V} = \bracsn{(x, y) \in G^2|yx^{-1} \in V}$, then $\fB_R = \bracs{U_{R, V}|V \in \cn_G(1)}$ is a fundamental system of entourages for $\fU_R$. + \end{enumerate} + + The uniformities $\fU_L$ and $\fU_R$ are the \textbf{left} and \textbf{right uniformities} of $G$, respectively. The inversion map $G \to G$ with $g \mapsto g^{-1}$ is an isomorphism of the left and right uniformities. +\end{definition} +\begin{proof}[Proof, {{\cite[I.1.4]{SchaeferWolff}}}. ] + (L2): For each $V \in \cn_G(0)$, $U_{L, V}$ is left translation-invariant. + \begin{enumerate} + \item[(FB1)] For each $V, V' \in \cn_G(1)$, $V \cap V' \in \cn_G(0)$, so $U_{L, V \cap V'} = U_{L, V} \cap U_{L, V'}$. + \item[(UB1)] For each $V \in \cn_G(1)$, $1 \in V$, so $\Delta \subset U_{L, V}$. + \item[(UB2)] For each $V \in \cn_G(1)$, by (TG1), there exists $W \in \cn_G(1)$ such that $WW \subset V$. In which case, $U_{L, W} \circ U_{L, W} \subset U_{L, V}$. + \end{enumerate} + + By \autoref{proposition:fundamental-entourage-criterion}, $\fB_L$ forms a fundamental system of entourages for a left translation-invariant uniformity $\fU_L$ on $G$. + + (L1): Let $\fV$ be a left translation-invariant uniformity on $G$. For each symmetric, left translation-invariant entourage $V \in \fV$, and $g \in G$, + \[ + V(g) = \bracsn{h \in G| (g, h) \in V} = g V(1) = \bracs{h \in G| g^{-1}h \in V(1)} + \] + + so $V = U_{L, V(1)}$. Therefore $\fV \subset \fU_L$ by \autoref{lemma:translation-invariant-symmetric}. + + On the other hand, for each $W_0 \in \cn_G(1)$, there exists a symmetric, left translation-invariant entourage $W \in \fV$ such that $W(1) \subset W_0$. In which case, $W = U_{L, W(1)} \subset U_{L, W_0(1)}$, and $\fV \supset \fU_L$. +\end{proof} + +\begin{definition}[Left/Right Continuous] +\label{definition:left-right-continuous} + Let $G$ be a topological group, $Y$ be a uniform space, and $f: G \to Y$, then $f$ is \textbf{left/right uniformly continuous} if it is uniformly continuous with respect to the left/right uniformity of $G$. +\end{definition} + +\begin{proposition} +\label{proposition:tvs-closure} + Let $G$ be a topological group, $A \subset G$, and $\fB \subset \cn_G(1)$ be a fundamental system of neighbourhoods, then + \[ + \ol{A} = \bigcap_{U \in \fB}\bracs{AU| U \in \fB} = \bigcap_{U \in \fB}\bracs{UA|U \in \fB} + \] + +\end{proposition} +\begin{proof} + By \autoref{proposition:uniformclosure}. +\end{proof} + + +\begin{proposition}[{{\cite[I.1.1]{SchaeferWolff}}}] +\label{proposition:tvs-set-operations} + Let $G$ be a topological group and $A, B \subset G$, then + \begin{enumerate} + \item If $A$ is open, then $AB$ is open. + \item If $A$ is closed and $B$ is compact, then $AB$ is closed. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): For every $x \in B$, $Ab$ is open by translation invariance, so + \[ + AB = \bigcup_{x \in B}(Ab) + \] + + is open. + + (2): Let $x \in \overline{AB}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap AB \ne \emptyset$, so $UB^{-1} \cap A \ne \emptyset$, and $\fB = \bracsn{UB^{-1}| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that + \[ + y \in \bigcap_{U \in \fF}\overline{UB^{-1}} = \bigcap_{U \in \fF}\overline{UB^{-1}} + \] + + + By \autoref{proposition:tvs-closure}, $\overline{UB^{-1}} \subset UUB^{-1}$, so + \[ + y \in \bigcap_{U \in \fF}\overline{UB^{-1}} \subset \bigcap_{U \in \fF}[UUB^{-1}] + \] + + Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus + \[ + y \in \bigcap_{U \in \fF}[UUB^{-1}] \subset \bigcap_{V \in \cn_G(1)}[xVB^{-1}] = \overline{xB^{-1}} = xB^{-1} + \] + + so $x \in yB \subset AB$. +\end{proof} \ No newline at end of file diff --git a/src/topology/groups/index.tex b/src/topology/groups/index.tex new file mode 100644 index 0000000..c7567be --- /dev/null +++ b/src/topology/groups/index.tex @@ -0,0 +1,5 @@ +\chapter{Topological Groups} +\label{chap:topological-group} + +\input{./definitions.tex} + diff --git a/src/topology/index.tex b/src/topology/index.tex index c12bc4e..b82bad1 100644 --- a/src/topology/index.tex +++ b/src/topology/index.tex @@ -5,4 +5,5 @@ \input{./uniform/index.tex} \input{./functions/index.tex} \input{./metric/index.tex} +\input{./groups/index.tex} \input{./notation.tex} diff --git a/src/topology/uniform/definition.tex b/src/topology/uniform/definition.tex index 03914c2..3e59f7a 100644 --- a/src/topology/uniform/definition.tex +++ b/src/topology/uniform/definition.tex @@ -25,7 +25,7 @@ \label{definition:slice} Let $X, Y$ be sets, $U \subset X \times Y$, and $A \subset X$, then \[ - U(A) = \bracs{y \in Y: (x, y) \in U, x \in A} + U(A) = \bracs{y \in Y| (x, y) \in U, x \in A} \] is the \textbf{slice} of $U$ at $A$.