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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 28}
\date{3/26/21} % Can also use \today

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

\par\noindent
We're currently trying to prove:
\thm{
	Let $M$ be connected, $p\in{}M$ such that $\exp_{p}$ is defined on all of $T_{p}M$. Let $q\in{}M$. Then there's a geodesic $\gamma$ from $p$ to $q$ such that $d(p,q)=l(\gamma)$.\nn
}

\par\noindent
Continuing from last time, we want to show that $\sup\mathscr{T}=d(p,q)$, i.e., $d(p,q)\in\mathscr{T}$, i.e., $d(p,q)=d(p,q)+d(\gamma(d(p,q)),q)$, i.e., $d(\gamma(d(p,q)),q)=0$. Because $d$ is a distance function, it's enough to show $\gamma(d(p,q))=q$.\n

\par\noindent
Let $t_{1}=\sup\mathscr{T}$. Because $\mathscr{T}$ is closed, we know $t_{1}\in\mathscr{T}$. Assume that $t_{1}<d(p,q)$ (we will show a contradiction). Then $\exists\delta>0$ \st{} $t_{1}+\delta<d(p,q)$, and there exists a geodesic sphere $S_{\delta}$ centered at $\gamma(t_{1})$ with radius $\delta$. Let $y\in{}S_{\delta}$ minimizing the map
\[
	\map{S_{\delta}}{\R}{x}{d(x,q)}
\]
We want to show $t_{1}+\delta\in{}\mathscr{T}$.\n

\par\noindent
Claim 1: $d(y,q)=d(p,q)-(t_{1}+\delta)$.\n
Proof: The lemma from last time implies that $d(p,q)-t_{1}=d(\gamma(t_{1}),q)=\delta+d(y,q)$.\proven

\par\noindent
Claim 2: $d(p,y)=t_{1}+\delta$.\n
Proof: $\le$ follows directly from the triangle inequality. $\ge$ is true because $d(p,q)\le{}d(p,y)+d(y,q)$, so
\[
	d(p,y)\ge{}d(p,q)-(d(p,q)-(t_{1}+\delta))\ge{}t_{1}+\delta
\]
by claim 1.\proven

\par\noindent
Claim 3: $y=\gamma(t_{1}+\delta)$.\n
Proof: Consider the path $\gamma(t)$ for $0\le{}t\le{}t_{1}$, followed by a radial geodesic from $\gamma(t_{1})$ to $y$. So the path is overall from $p$ to $y$. By claim 2, this path is length minimizing, so it's a geodesic with the same initial conditions as $\gamma$. This implies it must be $\gamma$, with the domain $0\le{}t\le{}t_{1}+\delta$.\n

\par\noindent
Claims 1 and 3 together imply that $t_{1}+\delta\in\mathscr{T}$, a contradiction. This completes the proof.\proven

\thm{
	(Hopf-Rinow) Let $M$ be connected. Assume $\exists{}p\in{}M$ \st{} $\exp_{p}$ is defined on all of $T_{p}M$. Then
	\begin{enumerate}[label=(\alph*), leftmargin=4\parindent, topsep=0pt, itemsep=0pt]
		\item Every closed and bounded set is compact.
		\item $M$ is complete as a metric space.
		\item $M$ is geodesically complete, i.e., $\forall{}q\in{}M$, $\exp_{q}$ is defined on all of $T_{q}M$.
	\end{enumerate}\up\n
	In fact, these are all equivalent. We will show this by: assumption $\Rightarrow$ (a) $\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ assumption.\nn
	Proof: assumption $\Rightarrow$ (a): Let $S\subset{}M$ be closed and bounded. Then $\forall{}y\in{}S$, $\exists\gamma_{y}$ a minimizing geodesic from $p$ to $y$. $S$ is bounded, so $\exists{}R>0$ \st{} $\forall{}y\in{}S$, $l(\gamma_{y})<R$. So $S\subset\exp_{p}\paren{\overline{B_{R}(0)}}$, where $B_{R}(0)\subset{}T_{p}M$. Because $\overline{B_{R}(0)}$ is compact and $\exp_{p}$ is continuous, $\exp_{p}\paren{\overline{B_{R}(0)}}$ is also compact, so $S$ is contained in a compact set, and is thus compact.\nn
	(a) $\Rightarrow$ (b): This only requires facts of point-set topology. If $(x_{n})$ is Cauchy, then it is bounded. So its image is compact. Thus, there's a convergent subsequence, so $(x_{n})$ converges to its limit point.\nn
	(b) $\Rightarrow$ (c): We prove this by contradiction. Assume $\exists{}q\in{}M$, $T>0$, and $v\in{}T_{q}M$ with $\norm{v}=1$, such that $\gamma(t)=\exp_{q}(tv)$ exists $\forall{}t\in[0,)$, but not beyond $T$. Take $t_{1}<t_{2}<$ in $[0,T)$ so that $(t_{n})$ converges to $T$, and let $x_{n}=\gamma(t_{n})$. Note that $d(x_{n},x_{m})\le\abs{t_{n}-t_{m}}$, $\forall{}n,m$. Since $(t_{n})$ converges, $(x_{n})$ is Cauchy, so $\exists{}w\in{}M$ such that $\lim_{n\to\infty}x_{n}=w$. Let $W$ be a totally normal neighborhood of $w$, and $\delta>0$ such that $\forall{}x\in{}W$, the geodesic ball $B(x,\delta)\supseteq{}W$. Then $\exists{}N\in\N$, such that $\forall{}n,m>N$, $\abs{t_{n}-t_{m}}<\delta/2$, and $\gamma(t_{m})\in{}W$. Pick $n>N$. Use $\exp_{x_{n}}$ to ``relaunch'' the geodesic. $s\mapsto\exp_{x_{n}}(s\dot\gamma(t_{n}))$ exists for $\abs{s}<\delta$. This extends $\gamma$ past $T$, since $t_{n}+\delta>T$. Oops!\nn
	(c) $\Rightarrow$ assumption: Trivial.\proven
}

\defn{
	If $M$ satisfies these properties, we call $M$ a \u{complete Riemannian manifold}.\n
}

\par\noindent
We obtain yet another version of Bonnet-Myer as a corollary:
\thm{
	(Bonnet-Myer III) Let $M$ be a compact, connected Riemannian manifold, and assume $\Ric>\paren{\frac{\pi}{l}}^{2}$. Then $M$ is compact and the diameter of $M$ is no more than $l$.\nn
	Proof: We already know that any geodesic with a length of at least $l$ is not minimizing. Also, any pair of points can be joined by a minimizing geodesic. Thus, the diameter of $M$ is at most $L$, so $M$ is bounded and compact.\proven
}

\end{document}