09_04.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 3}
\date{9/4/20} % Can also use \today

\begin{document}
\maketitle
\setlength\RaggedRightParindent{\parindent}
\RaggedRight

\subsection*{Group Actions}

\defn{
	Let $G$ be a group, $X$ a set. A \underline{left action} of $G$ on $X$ is a map
	\[
		\map{G\times{}X}{X}{(g,x)}{g\cdot{}x}
	\]
	such that
	\begin{enumerate}[topsep=0pt, itemsep=0pt, label=\alph*)]
		\item if $e\in{}G$ is the identity, $\forall{}x\in{}X$, $e\cdot{}x=x$
		\item $\forall{}g_{1},g_{2}\in{}G$, $\forall{}x\in{}X$, $(g_{1}g_{2})\cdot{}x=g_{1}\cdot(g_{2}\cdot{}x)$.
	\end{enumerate}\up\n
}

\par\noindent
In other words, if $\forall{}g\in{}G$, we define the map
\[
	\map{L_{g}:X}{X}{x}{g\cdot{}x}
\]
then $L_{e}=I_{X}$ and $L_{g_{1}g_{2}}=L_{g_{1}}\of{}L_{g_{2}}$.\n

\defn{
	Given a group action, if $x\in{}X$, the \underline{orbit} of $x$ is the set $G\cdot{}x=\set{y\in{}X\mid{}\exists{}g\in{}G\ptxt{ \st }g\cdot{}x=y}$.\n
}

\lemma{
	The orbits partition $X$, i.e., $x\sim{}y$ iff $G\cdot{}x=G\cdot{}y$ is an equivalence relation.\n
}

\par\noindent
Notation: $X/G$ and $G\backslash{}X$ are both valid. We'll stick with $G\backslash{}X$. (This is the quotient space whose points are the orbits of points in $X$.)\n

\defn{
	Assume $X$ is a topological space, and the group $G$ acts on $X$ (on the left). The action is \underline{by continuous maps} iff $\forall{}G\in{}G$, $L_{g}:X\to{}X$ is continuous.\n
}

\par\noindent
Observe that $\forall{}g,L_{g}$ is a homeomorphism, because $\exists{}g\inv\in{}G$, so $L_{g\inv}$ is continuous, and $L_{g}\of{}L_{g\inv}=I_{X}=L_{g\inv}\of{}L_{g}$.\n

\lemma{
	If $G$ acts by continuous maps, the orbit relation is open.\n
	Proof: Let $U\subseteq{}X$ be open. We need to show that saturation $\hat U$ of $U$ is open.\n
	\begin{align*}
		\hat U & =\set{x\in X\mid \exists y\in{}U\ptxt{ \st{} }x\sim y}\ptxt{ ($\sim$ being the orbit relation)}\\
		& =\set{x\in{}X\mid{}\exists{}y\in{}U,g\in{}G\ptxt{ \st{} }y=g\cdot{}x}
	\end{align*}
	Thus,
	\[
		\hat{U}=\bigcup_{g\in{}G}g\cdot{}U=\bigcup_{g\in{}G}\set{g\cdot{}x\mid{}x\in{}U}=\bigcup_{g\in{}G}L_{g}(U)
	\]
	$L_{g}$ is a homeomorphism, so it is an open map, so each $L_{g}(U)$ is open, so $\hat{U}$ is open.\proven
}

\defn{
	A \underline{topological group} is a group $G$ with a topology \st{} the maps
	\[
		\map{G\times{}G}{G}{(g,k)}{gk}\qquad\ptxt{and}\qquad\map{G}{G}{g}{g\inv}
	\]
	are continuous.\n
}

\par\noindent
Aside: Later on, when we have a manifold, and these maps are smooth, then this is a Lie group.\n

\ex{
	$\GL(n,\R)\subseteq\R^{n\times{}n}\cong\R^{n^{2}}$, the set of invertible $n\times{}n$ matrices.\n
	In fact, this is an open subset, since it's described by $\GL(n,\R)=\set{M\in\R^{n\times{}n}\mid{}\det M\ne{}0}$, i.e.,\break $\GL(n,\R)=\det\inv(\R\cut\set{0})$. Because $\det$ is a continuous map from $\R^{n\times{}n}$ to $\R$ and $\R\cut\set{0}$ is open, we get that $\GL(n,\R)$ is open.\n
	Note that $\GL(n,\R)$ is a topological group, with the induced topology. In fact, any subgroup of a topological group is naturally a topological group with respect to the subspace topology.\n
}

\ex{
	$O(n,\R)=\set{g\in\GL(n,\R)\mid{}g\inv=g\tpose}$.\n
	$\GL(n,\C)\subseteq\C^{n^{2}}\cong\R^{2n^{2}}$. Note that $\GL(n,\C)\subseteq\GL(2n,\R)$, since $\C\cong\R^{2}$.\n
	$U(n)=\set{g\in\GL(n,\C)\mid{}g\inv=\conj{g}\tpose}$.\n
}

\defn{
	If $G$ is a topological group acting on a topological space $X$, the action is \underline{continuous} iff $G\times{}X\to{}X$ is a continuous map.\n
}

\lemma{
	A continuous action is an action by continuous maps. (I.e. $\forall{}g\in{}G$, $L_{g}:X\to{}X$ is continuous.)\n
}

\ex{
	$G=S^{1}=\set{z\in\C:\abs{z}=1}=U(1)$ acts on $S^{2n+1}\subseteq\C^{n+1}$ by $\lambda\in{}S^{1}$, $(z_{1},\ldots,z_{n+1})\in{}S^{2n+1}$,\break $\lambda\cdot(z_{1},\ldots,z_{n+1})=(\lambda{}z_{1},\ldots,\lambda{}z_{n+1})$. This is a continuous action.\n
}

\par\noindent
Question: Suppose $G$ is a topological group acting on $X$. (So the orbit relation is open.) When is $G\backslash{}X$ Hausdorff?\n
Well, this is true iff the graph of the orbit relation is closed.\n

\par\noindent
Define
\[
	\star\quad\map{G\times{}X}{X\times{}X}{(g,x)}{(x,g\cdot{}x)}
\]
This is a continuous map, whose image is the graph of the orbit relation.\n

\prop{
	If $G$ and $X$ are both compact, and $X$ is Hausdorff, then $G\backslash{}X$ is Hausdorff.\n
	Proof: The image of $\star$ is compact, and compact subsets of Hausdorff spaces are closed, so the orbit relation is closed.\proven
}

\ex{
	$S^{1}\times{}S^{2n+1}\to{}S^{2n+1}$ as above.\n
	Then the proposition implies $\C\P^{n}=S^{1}\backslash{}S^{2n+1}$ is Hausdorff and second-countable.\n
	Note: $\C\P^{n}\cong\set{\ptxt{$1$-dimensional subspaces of $\C^{n+1}$}}$.\n
}

\end{document}