10_28.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 24}
\date{10/28/20} % Can also use \today

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

\par\noindent
Recall: If $X\in\mathfrak{X}(M)$, $p\in{}M$, then there exists a neighborhood $V$ of $p$, $\varepsilon>0$, and a function $\phi:(-\varepsilon,\varepsilon)\times{}V\to{}M$ such that $\forall{}q\in{}V$, $t\mapsto\phi(t,q)$ is an integral curve of $X$ with $\phi(0,p)=p$.\n

\par\noindent
Notation: $\phi(t,q)=\phi_{t}(q)$. So $\forall{}t\in(-\varepsilon,\varepsilon)$, we can think of $\phi_{t}:V\to{}M$ (a ``time $t$ map'').\n

\par\noindent
Notation: Fix $X\in\mathfrak{X}(M)$. Then $\forall{}p\in{}M$, $(\alpha(p),\beta(p))$ is the domain of the (unique) maximal integral curve of $X$ through $p$.\n

\par\noindent
$\forall{}t\in\R$, let $M_{t}=\set{p\in{}M\mid{}t\in(\alpha(p),\beta(p))}$. $M_{t}$ is the set of points whose integral curves are defined at time $t$. This is a little fussy for our purposes, since most of the vector fields we care about are complete.\n

\par\noindent
From last time, $\mathcal{W}=\set{(p,t)\in{}M\times\R\mid{}p\in{}M_{t}}$. Recall that $\mathcal{W}$ is open in $M\times\R$. The map $\phi:\mathcal{W}\to{}M$ is the global flow of $X$.\n

\par\noindent
Observe: For our purposes, we don't need the global theory as much. We'll concentrate on:
\begin{enumerate}[topsep=0pt, itemsep=0pt, label=\alph*)]
	\item Local flows $\phi:(-\varepsilon,\varepsilon)\times{}V\to{}M$ (uniform time)
	\item Complete fields, i.e., those $X\in\mathfrak{X}(M)$ for which $\forall{}p\in{}M$, $(\alpha(p),\beta(p))=\R$.
\end{enumerate}\up\n

\par\noindent
Recall the example from last friday.
\ex{
	$X=yx^{2}\pd{}{x}$. Then $x(t)=\frac{x(0)}{1-x(0)y(0)t}$ and $y(t)=y(0)$. This is not a complete vector field. $\phi_{t}(x,y)=(\frac{x}{1-xyt},y)$.\n
}

\ex{
	(From physics) Let $M=\R^{2}$ with coordinates $(x,p)$. Let $\dot{x}=p$ and $\dot{p}=0$. Then the integral curves of $x=p\pd{}{x}+0\pd{}{p}$ are of the form $x(t)=tp+x(0)$, $p(t)=p(0)$. Thus,
	\[
		\phi_{t}(x,p)=(x+tp,p)=\begin{pmatrix}
			1 & t\\
			0 & 1
		\end{pmatrix}\begin{pmatrix}
			x\\
			p
		\end{pmatrix}
	\]
	This is a linear shear. Note that the integral curves are just horizontal lines.\n
}

\prop{
	If $t,s,t+s\in(\alpha(p),\beta(p))$, then $\phi_{t+s}(p)=\phi_{t}(\phi_{s}(p))$.\n
}

\par\noindent
We call this our group law. This means, where defined, $\phi_{t+s}=\phi_{t}\of\phi_{s}$, $\forall{}t,s\in\R$.\n

\par\noindent
Let's amplify this idea. Suppose the vector field is complete. From the group law, $\forall{}t\in\R$, $\phi_{t}\of\phi_{-t}=I$, i.e., $\phi_{t}:M\to{}M$ is a diffeomorphism with inverse $(\phi_{t})\inv=\phi_{-t}$.\n

\par\noindent
Also, the mapping $\R\to\Diff(M)$ is a group morphism from $(\R,+)$ to $(\Diff(M),\of)$. In this case, $\phi$ is called a one-parameter group of diffeomorphisms.\n

\par\noindent
More generally, one can consider smooth maps $\phi:\R\times{}M\to{}M$, and define $\phi_{t}(p)=\phi(t,p)$. Then $\set{\phi_{t}}_{t\in\R}$ is a smooth one-parameter family of maps $M\to{}M$. The ones which satisfy the group law $\phi_{t+s}=\phi_{t}\of\phi_{s}$ correspond precisely to vector fields. Specifically, let $X_{p}$ be the velocity at $t=0$ of the integral curve $t\mapsto\phi_{t}(p)$.\n

\defn{
	$X$ is the \u{infinitesimal generator} of the $1$-parameter subgroup $\phi_{t}$.\n
}

\lemma{
	(Translation Lemma) Let $\phi$ be the $1$-parameter group (flow) generated by $X\in\mathfrak{X}(M)$. Then $\forall{}s\in\R,p\in{}M$, $t\mapsto\phi_{t+s}(p)$ is the integral curve of $X$ through $\phi_{s}(p)$.\nn
	Proof: Use the calc 1 chain rule and the group law. $\phi_{t+s}(p)=\phi_{t}(\phi_{s}(p))$. Now differentiate both sides with respect to $t$.\proven
}

\thm{
	If $M$ is compact, any $X\in\mathfrak{X}(M)$ is complete, i.e., all maximal integral curves of $X$ have domain $\R$.\n
}

\par\noindent
We're not quite ready to prove this yet, but we will soon.\n

\lemma{
	(Uniform Time Lemma) For any $M$, for any $X\in\mathfrak{X}(M)$, if $\exists\varepsilon>0$ such that all maximal integral curves' domains contain $(-\varepsilon,\varepsilon)$, then $X$ is complete.\nn
	Proof: Assume $X$ is not complete. THen $\exists{}p\in{}M$ \st{} $\beta(p)<\infty$ (the argument would follow identically if instead $\alpha(p)>-\infty$). Let $t_{0}\in\R$ \st{} $\beta(p)-\varepsilon<t_{0}<\beta(p)$, and consider the curve
	\[
		c(t)=\left\{\begin{array}{ll}
			\phi_{t}(p) & \alpha(p)<t<t_{0}\\
			\phi_{t-t_{0}}(\phi_{t_{0}}(p)) & -\varepsilon<t-t_{0}<\varepsilon
		\end{array}\right.
	\]
	Then this is an integral curve of $X$, with $c(0)=p$, and it is defined for all $t$ such that $t_{0}-\varepsilon<t<\varepsilon+t_{0}$. Note that $\varepsilon+t_{0}>\beta(p)$. Since it is defined for all $t\in(\alpha(p),\varepsilon+t_{0})$, this is a contradiction, as $(\alpha(p),\beta(p))$ is the maximal domain of an integral curve through $p$.\proven
}

\end{document}