02_17.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 13}
\date{2/17/21} % Can also use \today

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

\par\noindent
Review: Given $(M,g)$ a Riemannian manifold, $\exists\unique{}G\in\mf{X}(TM)$ \st{} the integral curves of $G$ are the lifts of geodesics.\n

\par\noindent
Notation: $\forall{}(q,v)\in{}TM$, $t\mapsto\gamma(t,q,m)$ is the geodesic with initial condition $(q,v)$. It is the projection onto $M$ of the integral curve of $G$, starting at $(q,v)$. For given $(q,v)$, $\gamma(t,q,v)$ has a maximal domain of definition, an interval in $t$, that depends on $(q,v)$.\n

\lemma{
	(Lemma 1 from Last Time) $\forall{}p\in{}M$, $\exists{}V\subseteq{}M$, a neighborhood of $p$, and $\exists\varepsilon,\delta>0$ \st{} $\gamma$ is defined on the set $(-\delta,\delta)\times\set{(q,v)\in{}TV:\norm{v}<\varepsilon}$. That is, $\gamma(t,q,v)$ is defined $\forall{}q\in{}V$, $v\in{}T_{q}V$, $\norm{v}<\varepsilon$, $\abs{t}<\delta$, and $\gamma$ is smooth as a map.\n
}

\defn{
	Given $V\subseteq{}M$, $\varepsilon>0$, we define the \u{$\varepsilon$-ball tangent bundle}, by $B_{\varepsilon}(TV)=\set{(q,v)\in{}TV:\norm{v}<\varepsilon}$. This is a fiber bundle $B_{\varepsilon}(TV)\to{}V$, whose fibers are open balls of radius $\varepsilon$, centered at $0$.\n
}

\defn{
	We also define the \u{unit tangent bundle} of $M$, $S_{1}(TM)$, by $S_{\varepsilon}(TV)=\set{(q,v)\in{}TV:\norm{v}=\varepsilon}$. This is a fiber bundle $S_{1}(TM)\to{}M$, whose fibers are $S^{n-1}$.\n
}

\ex{
	The unit tangent bundle of $S^{2}$ is isomorphic to
	\[
		\set{(\vec{q},\vec{v})\in\R^{3}\times\R^{3}:\norm{\vec{q}}=1,\norm{\vec{v}}=1,\vec{q}\cdot\vec{v}=0}
	\]
	(The final condition is a tangency condition). In turn, this is diffeomorphic to $\SO(3)$ as manifolds, by $(\vec{q},\vec{v})\mapsto(\vec{q},\vec{v},\vec{q}\times\vec{v})$. Treating the three output vectors as columns of a matrix yields an orthogonal matrix with determinant $1$.\n
}

\par\noindent
Observe: $G$ is tangent to $S_{\varepsilon}(TM)$, $\forall\varepsilon>0$. This is a fancy way to say that, along a geodesic, speed is constant. Because $\frac{d}{dt}\iprod{\dot\gamma,\dot\gamma}=0$, we know $\norm{\dot\gamma}$ is constant, so the integral curves of $G$ are fully contained in $S_{\varepsilon}(TM)$.\n

\cor{
	If $M$ is compact, every geodesic is defined $\forall{}t\in\R$, i.e., $G$ is complete.\nn
	Proof: $M$ is compact, so $\forall\varepsilon>0$, $S_{\varepsilon}(TM)$ is compact, and any field on a compact manifold is complete.\proven
}

\lemma{
	(Lemma 2 from Last Time) (Homogeneity of Geodesic Flow) Let $(q,v)\in{}TM$, $a>0$. If $\gamma(t,q,v)$ is defined for $\abs{t}<\delta$, then $\gamma(t,q,av)$ is defined for $\abs{t}<\frac{\delta}{a}$, and $\gamma(t,q,av)=\gamma(at,q,v)$.\nn
	Proof: Check that both sides satisfy the geodesic equation $\nabla_{\dot\gamma}\dot\gamma=0$, and have the same initial conditions (namely, $(q,av)$).\proven
}

\prop{
	(Do Carmo 2.7) $\forall{}p\in{}M$, $\exists{}V\subseteq{}M$ a neighborhood of $p$, and $\varepsilon>0$ \st{} $\forall(q,v)\in{}B_{\varepsilon}(TV)$, $\gamma(t,q,v)$ is defined for $\abs{t}<43$. (Note: we really just need it to be defined for $t=1$, so we can get our exponential map. But $43$ is such a nice number.)\nn
	Proof: Let $V,\delta,\varepsilon_{1}$ be as in Lemma 1, so that $\forall(q,v)\in{}B_{\varepsilon_{1}}(TV)$, $\gamma(t,q,v)$ is defined for $\abs{t}<\delta$. Choose $a>0$ \st{} $\abs{t}<\delta\Leftarrow{}a\abs{t}=\abs{at}<43$ -- specifically, choose $a=\frac{\delta}{43}$. Now, by Lemma 2, $\gamma(t,q,\frac{\delta}{43}v)$ is defined for $\abs{t}<43$ if $\norm{v}<\varepsilon_{1}$. Now define $\varepsilon=\varepsilon_{1}\cdot\frac{\delta}{43}$. Thus, $\frac{\delta}{43}\norm{v}<\varepsilon_{1}\Leftrightarrow\norm{v}<\varepsilon$.\proven
}

\defn{
	Let $p\in{}M$, $V\subset{}M$ a neighborhood of $p$, and $\varepsilon$ as in the previous proposition. The we define
	\begin{enumerate}[leftmargin=4\parindent]
		\item $\map{\exp:B_{\varepsilon}(TV)}{M}{(q,v)}{\gamma(1,q,v)}$
		\item $\map{\exp_{p}:B_{\varepsilon}(0)}{M}{v}{\gamma(1,p,v)}$
	\end{enumerate}\up\n
}

\par\noindent
Observe: Both $\exp$ and $\exp_{p}$ are differentiable.\n

\lemma{
	$\forall{}p\in{}M$, $d(\exp_{p})_{v=0}$ is the identity.
	\[
		\begin{tikzcd}[ampersand replacement=\&]
			T_{0}(T_{p}M) \arrow[r] \arrow[d,symbol=\cong] \& T_{p}M\\[-10pt]
			T_{p}M \arrow[ur,"\Id\;\ptxt{(claimed)}", swap]
		\end{tikzcd}
	\]
	Proof: Use curves to compute $d(\exp_{p})_{0}$. Take a curve in $T_{p}M$, starting at $0\in{}T_{p}M$, e.g., $t\mapsto{}tw$ for some $w\in{}T_{p}M$. Then
	\[
		d(\exp_{p})_{0}(w)=\restr{\frac{d}{dt}\exp_{p}(tw)}_{t=0}=\restr{\frac{d}{dt}\gamma(1,p,tw)}_{t=0}=\restr{\frac{d}{dt}\gamma(t,p,w)}_{t=0}=w
	\]
	\proven
}

\cor{
	$\exp_{p}$ is a local diffeomorphism near $0$, i.e., $\exists\mathscr{N}\subset{}T_{p}M$, a neighborhood of $0$ such that $\restr{\exp_{p}}_{\mathscr{N}}:\mathscr{N}\overset{\sim}{\to}U\osubset{}M$, for some $U$.
}

\end{document}