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update Group Action
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71 changes: 52 additions & 19 deletions group.tex
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Expand Up @@ -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*}
Expand All @@ -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
\[
\begin{tikzcd}[ampersand replacement=\&, column sep=1.7em, row sep=small]
\mathsf{B}G \& \bullet \arrow[rr, "g"] \& \& \bullet \\
Expand All @@ -326,6 +313,35 @@ \subsection{Definitions}
\]
\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
\[
Expand Down Expand Up @@ -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}{}
Expand Down
23 changes: 19 additions & 4 deletions set_theory.tex
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Expand Up @@ -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:
Expand Down
191 changes: 0 additions & 191 deletions topological_group.tex
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@@ -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}

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