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03_29.tex
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03_29.tex
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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 29}
\date{3/29/21} % Can also use \today
\begin{document}
\maketitle
\setlength\RaggedRightParindent{\parindent}
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\par\noindent
Today's goal is to prove Cartan-Hadamard: If $M$ is a complete, connected Riemannian manifold with $K\le{}0$, then $\forall{}p\in{}M$, $\exp_{p}:T_{p}M\to{}M$ is a smooth covering map.\n
\defn{
A $C^{\infty}$ map $F:\tilde{M}\to{}M$ is a \u{smooth covering map} iff $\forall{}p\in{}M$, there's a neighborhood $V$ of $p$ such that\n $F\inv(V)=\bigcup_{\alpha}U_{\alpha}$, where $\alpha\ne\beta\Rightarrow{}U_{\alpha}\cap{}U_{\beta}=\emptyset$, and $\forall\alpha$, $\restr{F}_{U_{\alpha}}^{V}$ is a diffeomorphism. We say that $V$ is \u{evenly covered}.\n
}
\par\noindent
Observe: A smooth covering map $F:\tilde{M}\to{}M$ is always a local diffeomorphism, as the definition of local diffeomorphism is $\forall\tilde{p}\in\tilde{M}$, there's a neighborhood $U$ of $\tilde{p}$ and $V$ of $F(\tilde{p})$ such that $\restr{F}_{U}^{V}$ is a diffeomorphism.\n
\defn{
A smooth map $F:\tilde{M}\to{}M$ between Riemannian manifolds is a \u{local isometry} iff $\forall\tilde{p}\in{}M$, there's a neighborhood $U$ of $\tilde{p}$ and $V$ of $F(\tilde{p})$, $\restr{F}_{U}^{V}$ is an isometry.\n
}
\par\noindent
Some properties of a local isometry $F:\tilde{M}\to{}M$:
\begin{itemize}
\item $F$ is a local diffeomorphism.
\item If $\tilde\gamma:I\to\tilde{M}$ is a geodesic on $\tilde{M}$, then $\gamma=F\of\tilde\gamma$ is a geodesic on $M$.
\item If $c:[a,b]\to\tilde{M}$ is any path, $l(F\of{}c)=l(c)$.
\end{itemize}
\lemma{
Let $F:\tilde{M}\to{}M$ be a local isometry, where $\tilde{M}$ and $M$ are connected, complete Riemannian manifolds. Then $F$ is a surjective covering map.\nn
Proof: The main property of $F$ is $\forall{}p\in{}M$, $\tilde{p}\in{}F\inv(p)$, $\forall{}v\in{}T_{p}M$, $\exists\unique\tilde{v}\in{}T_{\tilde{p}}\tilde{M}$ such that $F_{*,\tilde{p}}(\tilde{v})=v$, and also $F\of(\exp_{\tilde{p}}(t\tilde{v}))=\exp_{p}(tv)$. We obtain the existence and uniqueness of $\tilde{v}$ because $dF_{\tilde{p}}$ is a bijection. And the equality with the exponential map is true because both sides are geodesics on $M$ with the same initial conditions. We'll say that we can ``lift'' geodesics from $M$ to $\tilde{M}$: choose $\tilde{p}\in{}F\inv(p)$. Then $\exists\unique\tilde\gamma$ geodesic on $\tilde{M}$ such that $(F\of\gamma)(t)=\exp_{p}(tv)$ and $\tilde\gamma(0)=\tilde{p}$.\nn
To show the map is surjective, let $\tilde{p}\in\tilde{M}$, and define $p=F(\tilde{p})$. Let $q\in{}M$. Then by completeness, there is a geodesic $\gamma$ on $M$ joining $p$ to $q$ -- $\gamma(0)=p$ and $\gamma(T)=q$. Let $\tilde\gamma$ be the lift of $\gamma$ to $\tilde{M}$ such that $\tilde\gamma(0)=\tilde{p}$. $\forall{}t$, $(F\of\tilde\gamma)(t)=\gamma(t)$. So $F(\tilde\gamma(T))=\gamma(T)=q$, so $q\in\im{}F$.\nn
Next, we show the map is a covering map. Let $p\in{}M$. Since $F$ is a local isometry, $F_{*}$ is always bijective, so $p$ is a regular value. Thus, $F\inv(p)=\bigsqcup_{\alpha}\set{\tilde{p}_{\alpha}}$ is the disjoint union of (at most) countably many points. Let $\varepsilon>0$ such that there's an open geodesic all $B_{\varepsilon}(p)\subset{}M$, centered at $p$ with radius $\varepsilon$. $\forall\alpha$, define the open metric ball $U_{\alpha}=\set{\tilde{q}\in\tilde{M}\mid{}\tilde{d}(\tilde{p}_{\alpha},\tilde{q})<\varepsilon}$. We claim that $F\inv(B_{\varepsilon}(p))=\bigcup_{\alpha}U_{\alpha}$, and the conditions of being a covering map are satisfied by the $U_{\alpha}$.\nn
Claim 1: $\forall\alpha$, $F$ maps $U_{\alpha}$ into $B_{\varepsilon}(p)$, and $\restr{F}_{U_{\alpha}}$ is a bijection (so as a result, the restriction of $F$ is a diffeomorphism). Proof: Pick $\tilde{q}\in{}U_{\alpha}$. Let $\tilde\gamma$ be a geodesic segment in $\tilde{M}$ joining $\tilde{p}_{\alpha}$ to $\tilde{q}$. Then $l(\tilde\gamma)<\varepsilon$. Consider $\gamma=F\of\tilde\gamma$, a geodesic of the same length, $l(\gamma)=l(\tilde\gamma)<\varepsilon$. Then $\im\gamma\subseteq{}B_{\varepsilon}(p)$, so $F(\tilde{q})\in{}B_{\varepsilon}(p)$. Now, we construct the inverse of $\restr{F}_{U_{\alpha}}^{B_{\varepsilon}(p)}$. Start with some $q\in{}B_{\varepsilon}(p)$. Lift the radial geodesic from $p$ to $q$ u to $\tilde{}$, starting at $\tilde{p}_{\alpha}$. Then its endpoint is the inverse of $q\in{}U_{\alpha}$.\nn
Claim 2: $\alpha\ne\beta\Rightarrow{}U_{\alpha}\cap{}U_{\beta}=\emptyset$. Proof: We will show $\tilde{d}(\tilde{p}_{\alpha},\tilde{b}_{\beta})>2\varepsilon$. By the triangle inequality, this suffices. Let $\tilde\gamma$ be he minimizing geodesic from $\tilde{p}_{\alpha}$ to $\tilde{p}_{\beta}$. Consider $\gamma=F\of\tilde\gamma$. We claim that $\gamma$ must exit $B_{\varepsilon}(p)$, because any geodesic contained in $B_{\varepsilon}(p)$ and passing through $p$ is a radial geodesic, so it must be minimizing. It's not, so thus, $l(\gamma)>2\varepsilon$.\nn
Claim 3: $F\inv(B_{\varepsilon})=\bigcup_{\alpha}U_{\alpha}$. Proof: $\supseteq$ is part of claim 1. For $\subseteq$, let $\tilde{q}\in{}F\inv(B_{\varepsilon}(p))$, so $F(q)\in{}B_{\varepsilon}(p)$. Then let $\gamma$ be the radial geodesic from $F(\tilde{q})$ back to $p$. Let $\tilde\gamma$ be the lift of $\gamma$, starting at $\tilde{q}$. $\tilde\gamma$ ends at $\tilde{}$ such that $F(\tilde{p})=$, so $\tilde{p}\in{}F\inv(p)$. Thus, $\exists\alpha$ \st{} $\tilde{p}=\tilde{p}_{\alpha}$, and $l(\tilde\gamma)=l(\gamma)<\varepsilon$.\proven
}
\lemma{
If $M$ is such that $K\le{}0$ everywhere, then there are no conjugate points.\nn
Proof: HW\n
}
\thm{
(Cartan-Hadamard) Let $M$ be a complete Riemannian manifold, with $K\le{}0$ everywhere, then $\forall{}p\in{}M$, $\exp_{p}:T_{p}M\to{}M$ is a smooth covering map.\n
}
\par\noindent
Note that the second lemma implies $\exp_{p}$ has no critical points. The idea of the proof is we put a metric on $T_{p}M$ that makes $\exp_{p}$ a local isometry. Then we have to check that this metric is complete. It is, because rays $t\mapsto{}tv$ are geodesics in this (crazy) metric, and they exist $\forall{}t$.\n
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