From 60c0f80a93d659dcdfa39092741eba120953eac9 Mon Sep 17 00:00:00 2001 From: Derived Cat Date: Thu, 19 Sep 2024 16:05:25 -0400 Subject: [PATCH] update Group Action --- group.tex | 71 +++++++++++----- set_theory.tex | 23 ++++- topological_group.tex | 191 ------------------------------------------ 3 files changed, 71 insertions(+), 214 deletions(-) diff --git a/group.tex b/group.tex index 765b3b9..9db456a 100644 --- a/group.tex +++ b/group.tex @@ -276,11 +276,8 @@ \subsection{Definitions} If $G$ acts on $X$ by $\sigma$, we say $(X,\sigma)$ is a \textbf{$G$-set}. If there is no ambiguity, we simply say $X$ is a $G$-set. \end{definition} -The $G$-sets and $G$-maps form a category $G\text{-}\mathsf{Set}$ and we have category isomorphism -\begin{align*} - G\text{-}\mathsf{Set} & \stackrel{\sim}{\longrightarrow}[\mathsf{B}G, \mathsf{Set}] \\ - \sigma & \longmapsto \sigma(-) -\end{align*} + + \begin{proposition}{Equivalent Definition of Group Actions}{} Let $G$ be a group and $X$ be a set. A group action of $G$ on $X$ can be alternatively defined as a map \begin{align*} @@ -299,23 +296,13 @@ \subsection{Definitions} \end{proposition} We say $X$ is a right $G$-set if $X$ is a left $G^{\mathrm{op}}$-set. -\begin{example}{Trivial Group Action}{} - Let $G$ be a group and $X$ be a set. The \textbf{trivial group action} of $G$ on $X$ is defined as $\sigma_g=\mathrm{id}_X$ for all $g\in G$. -\end{example} -\begin{example}{Actions on $X$ Induce Actions on $2^X$}{acting_on_power_set} - If a group $G$ acts on a set $X$, then $G$ acts on the power set $2^X$ by - \[ - g\cdot A=\{ g\cdot x\mid x\in A\} . - \] -\end{example} - -\begin{definition}{Equivariant Map}{} - Let $G$ be a group and $X,Y$ be $G$-sets. A map $f:X\to Y$ is called \textbf{equivariant} if for all $g\in G$ and $x\in X$, we have +\begin{definition}{$G$-equivariant Map}{} + Let $G$ be a group and $(X,\sigma)$, $(Y,\sigma')$ be $G$-sets. A map $f:X\to Y$ is called \textbf{$G$-equivariant} if for all $g\in G$ and $x\in X$, we have \[ f(g\cdot x)=g\cdot f(x) . \] - Equivalently, $f$ is equivariant if it is a natural transformation $f:\sigma(-)\implies \sigma'(-)$ + Equivalently, $f$ is $G$-equivariant if it is a natural transformation $f:\sigma(-)\implies \sigma'(-)$ such that for any $g\in G$, the following naturality diagram commutes \[ \begin{tikzcd}[ampersand replacement=\&, column sep=1.7em, row sep=small] \mathsf{B}G \& \bullet \arrow[rr, "g"] \& \& \bullet \\ @@ -326,6 +313,35 @@ \subsection{Definitions} \] \end{definition} +\begin{definition}{Category of $G$-sets}{} + The categories of left $G$-sets, denoted by $G\text{-}\mathsf{Set}$, are defined as follows: + \begin{itemize} + \item Objects: $G$-sets. + \item Morphisms: $G$-equivariant maps. + \item Composition of morphisms is the composition of functions. + \end{itemize} + $G\text{-}\mathsf{Set}$ can be identified with the functor category $[\mathsf{B}G,\mathsf{Set}]$, given by the following isomorphism of categories +\begin{align*} + G\text{-}\mathsf{Set} & \stackrel{\sim}{\longrightarrow}[\mathsf{B}G, \mathsf{Set}] \\ + (X,\sigma ) & \longmapsto \left(\bullet \rightarrow X,\;\, \sigma:G\to \mathrm{Aut}_{\mathsf{Set}}(X)\right) +\end{align*} + +\end{definition} + + +\begin{example}{Trivial Group Action}{} + Let $G$ be a group and $X$ be a set. The \textbf{trivial group action} of $G$ on $X$ is defined as $\sigma_g=\mathrm{id}_X$ for all $g\in G$. +\end{example} + + +\begin{example}{Actions on $X$ Induce Actions on $2^X$}{acting_on_power_set} + If a group $G$ acts on a set $X$, then $G$ acts on the power set $2^X$ by + \[ + g\cdot A=\{ g\cdot x\mid x\in A\} . + \] +\end{example} + + \begin{definition}{Product of $G$-Sets}{} The \textbf{product} of two $G$-sets $X$ and $Y$ is defined as the set $X\times Y$ with the $G$-action \[ @@ -467,9 +483,26 @@ \subsection{Definitions} \begin{enumerate}[(i)] \item $x\in X^G\iff \mathrm{Stab}_G(x)=G$. \item $\ker \left(G\to \mathrm{Aut}_{\mathsf{Set}}(X)\right)=\bigcap\limits_{x\in X}\mathrm{Stab}_G(x)$. + \item $\mathrm{Stab}_G(gx)=g\mathrm{Stab}_G(x)g^{-1}$ for any $g\in G$. Hence, $\{\mathrm{Stab}_G(x)\mid x\in X\}$ is a conjugacy class in $G$. \end{enumerate} + If $X \curvearrowleft G$ is a right action, then we have $\mathrm{Stab}_G(xg)=g^{-1}\mathrm{Stab}_G(x)g$ for any $g\in G$. \end{proposition} - +\begin{proof} + \begin{enumerate}[(i)] + \item + $$ + x \in X^G \iff \forall g \in G, gx = x \iff \mathrm{Stab}_G(x) = G. + $$ + \item + $$ + g \in \ker \left(G \to \mathrm{Aut}_{\mathsf{Set}}(X) \right) \iff \forall x \in X ,\; gx = x \iff g \in \bigcap_{x \in X} \mathrm{Stab}_G(x). + $$ + \item Let $h \in \mathrm{Stab}_G(gx)$, meaning $h(gx) = gx$. + Applying $g^{-1}$ to both sides, we get $g^{-1}h(gx) = g^{-1}(gx)=x$. Thus, $g^{-1}hg \in \mathrm{Stab}_G(x)$, meaning $h \in g \mathrm{Stab}_G(x) g^{-1}$. + Conversely, if $h \in g \mathrm{Stab}_G(x) g^{-1}$, then $h = g k g^{-1}$ for some $k \in \mathrm{Stab}_G(x)$. Therefore, $h(gx) = g(kx) = gx$, so $h \in \mathrm{Stab}_G(gx)$. + Thus, $\mathrm{Stab}_G(gx) = g \mathrm{Stab}_G(x) g^{-1}$. + \end{enumerate} +\end{proof} \begin{definition}{Faithful Group Action}{} diff --git a/set_theory.tex b/set_theory.tex index 284d5b4..b52a0dd 100644 --- a/set_theory.tex +++ b/set_theory.tex @@ -161,14 +161,29 @@ \subsection{Relation} \section{Function} \begin{proposition}{}{} Let $f:X\to Y$ be a map. Suppose that $A_\alpha,A,E\subseteq X$ and $B_\alpha,B,F\subseteq Y$. We have -\begin{itemize} - - +\begin{enumerate}[(i)] \item $f\left(\bigcup\limits_{\alpha\in I}A_\alpha\right)=\bigcup\limits_{\alpha\in I}f\left(A_\alpha\right)\ $, $f\left(\bigcap\limits_{\alpha\in I}A_\alpha\right)\subseteq\bigcap\limits_{\alpha\in I}f\left(A_\alpha\right)\ $, $f(E-A)\supseteq f(E)-f(A)$. \item $f^{-1}\left(\bigcup\limits_{\alpha\in I}B_\alpha\right)=\bigcup\limits_{\alpha\in I}f^{-1}\left(B_\alpha\right)\ $, $f^{-1}\left(\bigcap\limits_{\alpha\in I}B_\alpha\right)=\bigcap\limits_{\alpha\in I}f^{-1}\left(B_\alpha\right)\ $, $f^{-1}(F-B)=f^{-1}(F)-f^{-1}(B)$. \item $A\subseteq f^{-1}(f(A))\ $, $B\supseteq f(f^{-1}(B))$. -\end{itemize} + \item If $f$ is surjective, then $f\left(f^{-1}(B)\right)=B$. + \item If $f$ is injective, then $f\left(\bigcap\limits_{\alpha\in I}A_\alpha\right)=\bigcap\limits_{\alpha\in I}f\left(A_\alpha\right)$, where the indexed set $I$ is nonempty. + \item $f\left(A\cap f^{-1}(B)\right)=f(A)\cap B$. +\end{enumerate} \end{proposition} +\begin{prf} + \begin{enumerate}[(i)] + \item Omited. + \item Omited. + \item Omited. + \item Omited. + \item Omited. + \item On the one hand, + \[ + f\left(A\cap f^{-1}(B)\right)\subseteq f(A)\cap f\left(f^{-1}(B)\right)\subseteq f(A)\cap B. + \] + On the other hand, if $y\in f(A)\cap B$, then there exists $x\in A$ such that $f(x)=y$. Since $f(x)\in B$, we have $x\in f^{-1}(B)$, which implies $x\in A\cap f^{-1}(B)$. Therefore, $y\in f\left(A\cap f^{-1}(B)\right)$. Hence, $f(A)\cap B\subseteq f\left(A\cap f^{-1}(B)\right)$. + \end{enumerate} +\end{prf} \begin{proposition}{Equivalent Characterization of Injections}{} Let $f:X\to Y$ be a map. The following are equivalent: diff --git a/topological_group.tex b/topological_group.tex index 2f81335..8b13789 100644 --- a/topological_group.tex +++ b/topological_group.tex @@ -1,192 +1 @@ -\chapter{Topological Group} -\section{Topological Group} -\begin{definition}{Topological Group}{} - A \textbf{topological group} is a group $G$ equipped with a topology $\tau$ such that the group multiplication map - \begin{align*} - \mu:G\times G&\longrightarrow G\\ - (g,h)&\longmapsto gh - \end{align*} - and the inversion map - \begin{align*} - \sigma:G&\longrightarrow G\\ - g&\longmapsto g^{-1} - \end{align*} - are continuous maps. -\end{definition} - -\noindent Topological groups are the group objects in the category $\mathsf{Top}$. -\begin{definition}{Topological Group Category}{} - Topological groups form a category $\mathsf{TopGrp}$, where the morphisms are continuous group homomorphisms. -\end{definition} - -\noindent An isomorphism of topological groups is a group isomorphism that is also a homeomorphism of the underlying topological spaces. -\begin{proposition}{}{} - The category $\mathsf{TopGrp}$ is complete. Limits in $\mathsf{TopGrp}$ commute with - \begin{itemize} - \item forgetful functor $\mathsf{TopGrp}\to\mathsf{Top}$ - \item forgetful functor $\mathsf{TopGrp}\to\mathsf{Grp}$ - \end{itemize} -\end{proposition} - -\begin{prf} - It is enough to prove the existence and commutation for products and equalizers. Let $R_i, i \in I$ be a collection of topological rings. Take the usual product $R=\prod R_i$ with the product topology. Since $R \times R=\prod\left(R_i \times R_i\right)$ as a topological space (because products commutes with products in any category), we see that addition and multiplication on $R$ are continuous. Let $a, b: R \rightarrow R^{\prime}$ be two homomorphisms of topological rings. Then as the equalizer we can simply take the equalizer of $a$ and $b$ as maps of topological spaces, which is the same thing as the equalizer as maps of rings endowed with the induced topology. -\end{prf} - - -\begin{proposition}{Subgroups of Topological Group are Topological Groups}{} - Let $G$ be a topological group and $H$ be a subgroup of $G$. Then $H$ is a topological group with the subspace topology induced by $G$. -\end{proposition} - -\begin{prf} - Since $H$ is a subgroup of $G$, the group multiplication map $\mu:G\times G\to G$ restricts to a map $\mu|_{H\times H}:H\times H\to H$. Since the inclusion $i:H\hookrightarrow G$ is continuous, - \[ - \mu|_{H\times H}:H \times H \xrightarrow{i\times i} G\times G\xrightarrow{\mu}\mu(G) \xrightarrow{i'} H - \] - is also continuous. Similarly, the inversion map $\sigma:G\to G$ restricts to a map $\sigma|_H:H\to H$ and $\sigma|_H$ is continuous. Hence $H$ is a topological group. - -\end{prf} - - -\begin{proposition}{Translation Invariance}{} - For any $a \in G$, left or right multiplication by $a$ yields a homeomorphism $G \rightarrow G$. -\end{proposition} - -\begin{prf} - Given any $a\in G$, let - \begin{align*} - L_a=\mu(a,\cdot):G&\longrightarrow G\\ - g&\longmapsto ag - \end{align*} - be the left multiplication map and - \begin{align*} - R_a=\mu(\cdot,a):G&\longrightarrow G\\ - g&\longmapsto ga - \end{align*} - be the right multiplication map. Since the group multiplication map $\mu:G\times G\to G$ is continuous, $L_a$ and $R_a$ must be continuous. Note that $L_a^{-1}=L_{a^{-1}}$ and $R_a^{-1}=R_{a^{-1}}$. Then we see $L_a^{-1}$ and $R_a^{-1}$ are also continuous maps. Hence $L_a$ and $R_a$ are homeomorphisms. -\end{prf} - - -\begin{corollary}{}{translation_invariance_cor} - Given any $a \in G$ and $S \subseteq G$, let's denote $a S:=\{a s: s \in S\}$ and $S a:=\{s a: s \in S\}$. Then - \begin{itemize} - \item $S$ is open $\iff$ $a S$ is open $\iff$ $a S$ is open. - \item $S$ is closed $\iff$ $a S$ is closed $\iff$ $a S$ is closed. - \end{itemize} -\end{corollary} - -\begin{proposition}{Neighborhood Basis at $1_G$ Determines the Topology of $G$}{} - Given a topological group $G$, if $\mathcal{N}$ is a neighborhood basis of the identity element $1_G$, then for all $x \in X$, - \[ - x \mathcal{N}:=\{x N: N \in \mathcal{N}\} - \] - is a neighborhood basis of $x$ in $G$. In particular, the topology on $G$ is completely determined by any neighborhood basis at the identity element. -\end{proposition} - -\begin{prf} - Let $x \in G$ and $V$ be any neighborhood of $x$. There exists an open set $U$ such that $x\in U\subseteq V$. By \Cref{th:translation_invariance_cor}, $x^{-1} U$ is an open neighborhood of $1_G$. Since $\mathcal{N}$ is a neighborhood basis of $1_G$, there exists $N \in \mathcal{N}$ such that $N \subseteq x^{-1} U$. Then there exists $x N \in x \mathcal{N}$ such that $x N \subseteq U$. Hence $x \mathcal{N}$ is a neighborhood basis of $x$. -\end{prf} - - - -\begin{definition}{Inverse Limit in $\mathsf{TopGrp}$}{} - Let $\mathsf{I}$ be a \hyperref[th:filtered_category]{filtered} \hyperref[th:thin_category]{thin category} and $F:\mathsf{I}^{\mathrm{op}}\to \mathsf{TopGrp}$ be a functor. Similar to the \hyperref[th:inverse_limit_of_groups]{inverse limit in \textsf{Grp}}, we can unpack the information of $F$ into an inverse system $\left(\left(G_i\right)_{i \in I},\left(f_{i j}\right)_{i \leq j \in I}\right)$. The inverse limit of this inverse system is $\varprojlim F$, also denoted by $\varprojlim_{i\in I}G_i$.\\ - To give a concrete construction of $\varprojlim_{i\in I}G_i$, we can take the inverse limit of the underlying group and endow it with the subspace topology induced by the product topology on $\prod_{i\in I}G_i$. -\end{definition} - - -\section{Continuous Topological Group Action} -\begin{definition}{Compact Open Topology}{} - Let $X$ be a topological space and $K$ be a compact subset of $X$. The \textbf{compact open topology} on $\mathrm{Hom}_{\mathsf{Top}}(X,Y)$ is the topology generated by the subbasis - \[ - \mathcal{S}:=\left\{f\in \mathrm{Hom}_{\mathsf{Top}}(X,Y)\midv K\text{ is compact in }X,\;V\text{ is open in }Y,\;f(K)\subseteq V\right\}. - \] -\end{definition} - - -\begin{definition}{Group Action on Topological Space by Homeomorphisms}{} - A \textbf{group action} on a topological space $X$ is a group homomorphism $\rho:G\to \mathrm{Aut}_{\mathsf{Top}}(X)$, where $\mathrm{Aut}_{\mathsf{Top}}(X)$ is the group of all homeomorphisms from $X$ to itself. -\end{definition} - - -\begin{definition}{Continuous Topological Group Action on Topological Space}{} - A \textbf{continuous topological group action} on a topological space $X$ is a group homomorphism $\rho:G\to \mathrm{Aut}_{\mathsf{Set}}(X)$, where $G$ is a topological group, such that the following map induced by $\rho$ - \begin{align*} - \varrho:G\times X&\longrightarrow X\\ - (g,x)&\longmapsto \rho(g)(x) - \end{align*} - is continuous. In this case, we have $\mathrm{im}\rho \subseteq \mathrm{Aut}_{\mathsf{Top}}(X)$. -\end{definition} - -\begin{prf} - For any $g\in G$, - \begin{align*} - \rho(g): X &\longrightarrow X\\ - x &\longmapsto \varrho(g,x) - \end{align*} - is continuous and has a continuous inverse $\rho(g^{-1})$. Hence $\rho(g)\in \mathrm{Aut}_{\mathsf{Top}}(X)$. -\end{prf} - -From the definition, we see if $varpho:G\times X\to X$ is continuous topological group action on topological space $X$, it is also a group action on $X$ by homeomorphisms. If $G$ is discrete, then the converse holds. - -\begin{proposition}{Discrete Group Acts Continuously on Topological Space $\iff$ Acts by Homeomorphisms}{} - Let $G$ be a group acting on the underlying set of a topological space $X$ through a group homomorphism $\rho:G\to \mathrm{Aut}_{\mathsf{Set}}(X)$. Then the following are equivalent: - \begin{enumerate}[(i)] - \item $G$ equipped with discrete topology acts continuously on $X$ - \item $G$ acts by homeomorphisms on $X$, i.e., - $\mathrm{im}\rho \subseteq \mathrm{Aut}_{\mathsf{Top}}(X)$ - \end{enumerate} -\end{proposition} -\begin{prf} - We only need to prove (ii)$\implies$ (i). For any open set $U\subseteq X$, we have - \begin{align*} - \varrho^{-1}(U)&=\{(g,x)\in G\times X\mid \varrho(g,x)\in U\}\\ - &=\{(g,x)\in G\times X\mid \rho(g)(x)\in U\}\\ - &=\{(g,x)\in G\times X\mid x\in \rho(g)^{-1}(U)\}\\ - &=\bigcup_{g\in G}\left(\left\{g\right\}\times \rho(g)^{-1}(U) \right) - \end{align*} - Since $\rho(g)$ is a homeomorphism, $\rho(g)^{-1}(U)$ is open for any open set $U$. Since $G$ is discrete, each $\left\{g\right\}\times \rho(g)^{-1}(U)$ is open in $G\times X$. Hence $\varrho^{-1}(U)$ as a union of open sets is open in $G\times X$, which implies $\varrho$ is continuous. -\end{prf} - - -\begin{definition}{Orbit Space}{} - Let $G$ be a group acting on a topological space $X$. The \textbf{orbit space} of $X$ under the action of $G$ is the quotient space $G\backslash X $ obtained by identifying all points in $X$ that are in the same orbit. $G\backslash X $ is equipped with the quotient topology: a subset $U\subseteq G\backslash X $ is open if and only if $\pi^{-1}(U)$ is open in $X$, where $\pi:X\to G\backslash X$ is the quotient map. -\end{definition} - - - -\begin{proposition}{}{} - For any continuous action of a topological group $G$ on a topological space $E$, the quotient map $p: E \rightarrow G\backslash E$ is an open map. -\end{proposition} - -\begin{prf} - For any $g \in G$ and any subset $U \subseteq M$, we define a set $g \cdot U \subseteq M$ by -$$ -g \cdot U=\{g \cdot x: x \in U\} . -$$ -If $U \subseteq M$ is open, then $\pi^{-1}(\pi(U))$ is equal to the union of all sets of the form $g \cdot U$ as $g$ ranges over $G$. Since $p \mapsto g \cdot p$ is a homeomorphism, each such set is open, and therefore $\pi^{-1}(\pi(U))$ is open in $M$. Becaues $\pi$ is a quotient map, this implies that $\pi(U)$ is open in $G\backslash M$, and therefore $\pi$ is an open map. -\end{prf} - - - - -\section{Topological Ring} -\begin{definition}{Topological Ring}{} - A \textbf{topological ring} is a ring $R$ equipped with a topology $\tau$ such that the ring addition map - \begin{align*} - +:R\times R&\longrightarrow R\\ - (a,b)&\longmapsto a+b - \end{align*} - the ring multiplication map - \begin{align*} - \cdot:R\times R&\longrightarrow R\\ - (a,b)&\longmapsto a\cdot b - \end{align*} - and the addition inverse map - \begin{align*} - -:R&\longrightarrow R\\ - a&\longmapsto -a - \end{align*} - are continuous maps. -\end{definition} -