10_26.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 23}
\date{10/26/20} % Can also use \today

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

\subsection*{Uniqueness of Integral Curves}

\par\noindent
Last time, given $X\in\mathfrak{X}(M)$, we defined integral curves of $X$, and proved \i{local} existence and uniqueness by reducing to the Euclidean case and using theory from ordinary differential equations. Today, we'll start with global uniqueness.\n

\lemma{
	Assume that $c_{1},c_{2}:(\alpha,\beta)\to{}M$ (with $\alpha<0<\beta$) are integral curves of $X$, and $c_{1}(0)=c_{2}(0)=p$. Then $\forall{}t\in(\alpha,\beta)$, $c_{1}(t)=c_{2}(t)$.\nn
	Proof: Assume not. Then $S\eqdef\set{t\in(\alpha,\beta)\mid{}t>0,c_{1}(t)\ne{}c_{2}(t)}\ne\emptyset$. Well, this set is bounded below and nonempty, so let $\tau=\inf{}S$.\nn
	We claim that $c_{1}(\tau)=c_{2}(\tau)$ -- assume not. Then because $M$ is Hasudorff, there exist neighborhoods $U_{1}$ around $c_{1}(\tau)$ and $U_{2}$ around $c_{2}(\tau)$ such that $U_{1}\cap{}U_{2}=\emptyset$. By the continuity of $c_{1}$ and $c_{2}$, $\exists{}t<\tau$ such that $c_{1}(t)\ne{}c_{2}(t)$. Thus, $\tau$ is not the infimum of $S$. Oops! This is a contradiction, so we must have $c_{1}(\tau)=c_{2}(\tau)$.\nn
	Now, we use local uniqueness of integral curves of $X$. Let $c$ be an integral curve with initial condition\break $c(\tau)=c_{1}(\tau)=c_{2}(\tau)$. By local uniqueness, $c$ must agree with $c_{1}$ and $c_{2}$ on a neighborhood of $\tau$, so $\tau<\inf{}S$. Oops!\nn
	Therefore, we must have $c_{1}(t)=c_{2}(t)$, $\forall{}t\in(\alpha,\beta)$.\proven
}

\cor{
	Given $X\in\mathfrak{X}(M)$ and $p\in{}M$, there is an interval $(\alpha(p),\beta(p))$ containing $0$ (possibly with $\alpha(p)=-\infty$ and/or $\beta(p)=+\infty$), and an integral curve $c:(\alpha(p),\beta(p))\to{}M$ of $X$ with $c(0)=p$, such that for any other integral curve $\tilde{c}:I\to{}M$ (with $I$ and open interval) with $\tilde{c}(0)=p$, one has $I\subset(\alpha(p),\beta(p))$ and on $I$, $\restr{c}{I}=\tilde{c}$.\nn
	Proof: Let $\mathcal{I}$ be the set of intervals which are domains for some integral curve $c$ of $X$ with $c(0)=p$. Then we have $(\alpha(p),\beta(p))=\bigcup_{I\in\mathcal{I}}I$. Any wo $c_{1}:I_{1}\to{}M$, $c_{2}:I_{2}\to{}M$ agree on their overlap, $I_{1}\cap{}I_{2}$, so they define an integral curve on their union, $I_{1}\cup{}I_{2}$. Doing this for all $I\in\mathcal{I}$ gives us the desired integral curve.\proven
}

\defn{
	Such an integral curve is the unique \u{maximal integral curve} of $X$ through $p$ at $t=0$.\n
}

\par\noindent
We can refer to a chapter on vector fields in the book by Boothby, but it's a little too detailed.\n

\defn{
	Given $X\in\mathfrak{X}(M)$, define $\mathcal{W}=\set{(t,p)\in\R\times{}M\mid{}t\in(\alpha(p),\beta(p))}\subseteq\R\times{}M$. This is the domain of a map $\phi:\mathcal{W}\to{}M$, which takes a pair $(t,p)$ to he unique maximal integral curve of $X$ with initial condition $p$ at time $t$. In other words, $\forall(t,p)\in\mathcal{W}$, $\phi(0,p)=p$ and $\pd{\phi}{t}(t,p)=X_{\phi(t,p)}$. $\phi$ is called the \u{flow} of $X$.\n
}

\thm{
	$\mathcal{W}\subseteq\R\times{}M$ is open, and $\phi$ is a smooth map (of $t,p$).\n
}

\par\noindent
This general theorem is rather challenging to prove, but the local version (which is included in our textbook) is sufficient for our purposes.\n

\thm{
	Let $X\in\mathfrak{X}(M)$, $p\in{}M$. Then there exists a neighborhood $V$ of $p$, $\varepsilon>0$, and a smooth map $\phi:(-\varepsilon,\varepsilon)\times{}V\to{}M$ such that
	\begin{itemize}[topsep=0pt, itemsep=0pt, leftmargin=4\parindent]
		\item $\phi(0,q)=q$, $\forall{}q\in{}V$
		\item $\pd{\phi}{t}(t,q)=X_{\phi(t,q)}$. (Note: This is the velocity of the curve $t\mapsto\phi(t,q)$ at time $t$. That is, $t\mapsto\phi(t,q)$ is an integral curve of $X$ with initial condition $q$.)
	\end{itemize}\up\nn
	Proof: Just quote Calc IV/diffeq. In case $M=\R^{n}$, this is a theorem. Then just use local coordinates to reduce any manifold to the Euclidean case.\proven
}

\newpage
\par\noindent
Note: This isn't as fancy as the previous (global) version, but it's enough for our purposes. $(-\varepsilon,\varepsilon)\times{}V$ is sometimes referred to as a ``flow box''. $\mathcal{W}$ may be quite complicated, but the flow boxes are always easy to work with.\n

\par\noindent
Main points:
\begin{enumerate}[topsep=0pt, itemsep=0pt]
	\item $\phi$ is $C^{\infty}$ in $(t,p)$. We refer to this as ``smooth dependence on initial conditions''.
	\item $\varepsilon>0$ can be uniform on $V$.
\end{enumerate}\up\n

\par\noindent
Notation: It is standard to write $\phi(t,p)=\phi_{t}(p)$. This emphasizes, in the local flow theorem, that $\forall{}t\in(-\varepsilon,\varepsilon)$, we can think of the map $\phi_{t}:V\to{}M$. This is called the ``time $t$ map''. Think about it as moving every point in $V$ by time $t$ along their respective integral curves. In other words, it takes the blob $V$ to a new blob $\phi_{t}(V)$.\n

\thm{
	Given $X\in\mathfrak{X}(M)$, $p\in{}M$. If $t,s,t+s\in(\alpha(p),\beta(p))$, then
	\[
		\phi_{t}(\phi_{s}(p))=\phi_{t+s}(p)=\phi_{s+t}(p)=\phi_{s}(\phi_{t}(p)).
	\]
	This is known as the ``$1$-parameter group''.\nn
	Proof: Fix $s$, and consider the curves $t\mapsto\phi_{t+s}(p)$ and $t\mapsto\phi_{t}(\phi_{s}(p))$. Both are integral curves of $X$, with the same initial conditions (check that this is true), so by uniqueness, they're the same curve. The rest follows from commutativity of addition.\proven
}

\par\noindent
So altogether, if there are no domain issues,
\[
	\phi_{t}\of\phi_{s}=\phi_{t+s}=\phi_{s+t}=\phi_{s}\of\phi_{t}.
\]
For complete fields, this shows that $t\mapsto\phi_{t}$ is a map $\R\to\set{\ptxt{All diffeomorphisms }M\to{}M}\eqdef\Diff(M)$, and this map is a group morphism from $(\R,+)\to\Diff(M)$!

\end{document}