03_01.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 17}
\date{3/1/21} % Can also use \today

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

\par\noindent
Recall the setup from last time:

\begin{tikzpicture}
	\draw[pattern=north west lines, pattern color=gray] (0,-1) rectangle (2,1);
	\draw[thick, ->] (-0.5,0) -- (4,0) node[anchor=west]{$t$};
	\draw[thick, <->] (0,-2) -- (0,2) node[anchor=east]{$s$};
	\draw (0,1) node[anchor=east]{$+\delta$} -- (2,1) -- (2,-1) -- (0,-1) node[anchor=east]{$-\delta$};
	\draw[-to, thick, red] (0,0) -- (2,0);
	\draw[-to, red] (0,0.4) -- (2,0.4);
	\draw[-to, red] (0,0.8) -- (2,0.8);
	\draw[-to, red] (0,-0.4) -- (2,-0.4);
	\draw[-to, red] (0,-0.8) -- (2,-0.8);
	\draw[thick, blue] (0.5,-0.9) -- (0.5,0.9);
	\draw[thick, blue] (1,-0.9) -- (1,0.9);
	\draw[thick, blue] (1.5,-0.9) -- (1.5,0.9);
	%
	\draw[-to] (5,0) -- (6,0);
	%
	\draw (7,1) arc (45:-15:3) node (M) {} arc (120:60:6);
	\node (p) at (9,0) {};
	\fill (p) circle [radius=0.05cm];
	\draw (p) + ({1.25*0.87}, {1.25*0.5}) node (v) {};
	\draw[-to, thick] (p) -- (v);
	\draw[-to, red] (p) arc (105:93.75:4) node (s11) {} arc (93.75:82.5:4) node (s21) {} arc (82.5:71.25:4) node (s31) {} arc (71.25:60:4);
	\draw[-to, red, thick] (p) arc (120:75:4) node (exppv) {};
	\draw[-to, red] (p) arc (135:90:4);
	\draw[-to, red] (p) arc (150:105:4) node (tcurve) {};
	\draw[-to, red] (p) arc (165:153.75:4) node (s12) {} arc (153.75:142.5:4) node (s22) {} arc (142.5:131.25:4) node (s32) {} arc (131.25:120:4);
	\fill (exppv) circle [radius=0.05cm];
	\draw[-to, red] (p) arc (120:75:4);
	\draw[blue, thick, bend right=30] (s11.center) to (s12.center);
	\draw[blue, thick, bend right=30] (s21.center) to (s22.center);
	\draw[blue, thick, bend right=30] (s31.center) to (s32.center);
	%
	\draw (5.5,0) node[anchor=south]{$f$};
	\draw (p) node[anchor=north east]{$p$};
	\draw (v) + (0,0) node {$v$};
	\draw (exppv) node[anchor=west]{$\exp_{p}(v)$};
	\draw[red] (tcurve) + (0.5,0) node[anchor=west]{$t$-curves};
	\draw[red] (2.5,-0.5) node[anchor=north west] {Original geodesic};
	\draw[red, dashed] (2.05,-0.05) -- (2.6,-0.6);
	\draw[blue] (1,1.5) node {$s$-curves};
	\draw (2,0) node[anchor=south west]{$t=1$};
	\draw (M) + (0.5,0.75) node{$M$};
\end{tikzpicture}

\defn{
	A \u{vector field along $f$} is a lift $\tilde{f}$ of $f$ to $TM$. I.e., $\tilde{f}$ is defined such that the following diagram commutes:
	\[
		\begin{tikzcd}[ampersand replacement=\&]
			\& TM\arrow[d,"\pi"] \\[-10pt]
			D \arrow[ur,"\tilde{f}"] \arrow[r,"f"'] \& M
		\end{tikzcd}
	\]
}

\par\noindent
Note that such a lift isn't unique!\n

\ex{
	One such lift is $\displaystyle\tilde{f}=\left\{\begin{array}{l}f_{t}\\ f_{s}\end{array}\right.$. For such a $\tilde{f}$, we can define $\frac{D}{dt}\tilde{f}$ and $\frac{D}{ds}\tilde{f}$ by restricting $\tilde{f}$ to $t$ and $s$ curves, respectively.\n
}

\prop{
	$\frac{D}{dt}f_{s}=\frac{D}{ds}f_{t}$ at each $(t,s)$.\nn
	Proof: We will compute in local coordinates $(x^{1},\ldots,x^{n})$. Let $X_{i}=\pd{}{x^{i}}$, $\forall{}i$. We write $f(t,s)=(x^{1}(t,s),\ldots,x^{n}(t,s))$, where $x^{i}(t,s):\dom(f)\to\R$. Note that we can write $f_{s}=\pd{x^{i}}{s}X_{i}(f(t,s))$, and likewise for $f_{t}$. We now compute
	\[
		\frac{D}{dt}f_{s}=\frac{\d^{2}x^{i}}{\d{}t\d{}s}X_{i}+\pd{x^{i}}{s}\frac{D}{dt}X_{i}
	\]
	We know $f_{t}=\pd{x^{j}}{t}X_{j}$, so because $\frac{D}{dt}$ is the covariant derivative with respect to $f_{t}$,
	\[
		\frac{D}{dt}X_{i}=\pd{x^{j}}{t}\nabla_{X_{j}}X_{i}\qquad\frac{D}{dt}f_{s}=\frac{\d^{2}x^{i}}{\d{}t\d{}s}X_{i}+\pd{x^{i}}{s}\pd{x^{j}}{t}\nabla_{X_{j}}X_{i}
	\]
	Computing similarly, we also get
	\[
		\frac{D}{ds}f_{t}=\frac{\d^{2}x^{i}}{\d{}s\d{}t}X_{i}+\pd{x^{i}}{t}\pd{x^{j}}{s}\nabla_{X_{j}}X_{i}
	\]
	By Clairaut's theorem, $\frac{\d^{2}x^{i}}{\d{}t\d{}s}=\frac{\d^{2}x^{i}}{\d{}s\d{}t}$, so the first term of $\frac{D}{dt}f_{s}$ and $\frac{D}{ds}f_{t}$ are equal. Furthermore, because the Levi-Civita connection is torsion-free, $\comm{X_{i},X_{j}}=0$, so $\nabla_{X_{j}}X_{i}=\nabla_{X_{i}}X_{j}$. This means we can swap the coefficients in the second term to show equality. We conclude that $\frac{D}{dt}f_{s}=\frac{D}{ds}f_{t}$.\proven
}

\par\noindent
Observe: We can ask if $\frac{D}{ds}$ and $\frac{D}{dt}$ commute. We'll see on Friday that the answer is no, because curvature comes into play.\n

\par\noindent
We're now ready to prove Gauss' lemma...\newpage

\lemma{
	(Gauss' Lemma) In a normal neighborhood of $p$, radial geodesics are orthogonal to geodesic spheres.\nn
	Proof: Let $p\in{}M$ and $\varepsilon>0$ such that $\exp_{p}:B_{\varepsilon}(0)\overset{\sim}{\to}\exp_{p}(B_{\varepsilon}(0))$ (with $B_{\varepsilon}(0)\subseteq{}T_{p}M$ and $\exp_{p}(B_{\varepsilon}(0))\subseteq{}M$) is a diffeomorphism onto its image. Take $r\in(0,\varepsilon)$, so $S_{r}\subseteq{}T_{p}M$ is the sphere of radius $r$. Then choose any curve $s\mapsto{}c(s)\in{}S_{r}$, for an arbtirarily small domain $s\in(-\delta,\delta)$. Define $f(t,s)=\exp_{p}(tc(s))$. Illustration:\n
	\begin{tikzpicture}
		\draw[thick, <->] (-3,0) -- (3,0);
		\draw[thick, <->] (0,-3) -- (0,3);
		\draw (2,0) arc (0:20:2) node (c1) {} arc (20:45:2) node (c2) {} arc (45:70:2) node (c3) {} arc (70:360:2);
		\draw[pattern=north west lines, pattern color=gray] (0,0) -- (c1.center) arc (20:70:2) -- cycle;
		\draw[red, -to] (0,0) -- (c1.center);
		\draw[red, thick, -to] (0,0) -- (c2.center);
		\draw[red, -to] (0,0) -- (c3.center);
		\draw[to-to, thick, blue] (c1.center) arc (20:70:2);
		%
		\draw[-to] (5,0) -- (6,0);
		%
		\draw (7,1) arc (45:-15:3) node (M) {} arc (120:60:6);
		\node (p) at (9,0) {};
		\fill (p) circle [radius=0.05cm];
		\draw[-to, red] (p) arc (105:93.75:4) node (s11) {} arc (93.75:82.5:4) node (s21) {} arc (82.5:71.25:4) node (s31) {};
		\draw[-to, red, thick] (p) arc (135:101.25:4);
		\draw[-to, red] (p) arc (165:153.75:4) node (s12) {} arc (153.75:142.5:4) node (s22) {} arc (142.5:131.25:4) node (s32) {};
		\draw[pattern=north west lines, pattern color=gray] (p.center) arc (105:71.25:4) to[bend right=30] (s32.center) arc (131.25:165:4);
		\draw[-to, red] (p) arc (105:93.75:4) arc (93.75:82.5:4) arc (82.5:71.25:4);
		\draw[-to, red] (p) arc (165:153.75:4) arc (153.75:142.5:4) arc (142.5:131.25:4);
		\draw[blue, thick, bend right=30, to-to] (s31.center) to (s32.center);
		%
		\draw (5.5,0) node[anchor=south]{$\exp_{p}$};
		\draw (p) node[anchor=north east]{$p$};
		\draw (M) + (0.5,0.75) node{$M$};
		\draw (2.5,-2.5) node[anchor=south east] {$T_{p}M$};
		\draw[blue] (c3) node[anchor=south] {$c(s)$};
		\draw[red] (1,0.75) node {$t$};
		\draw (-1.5,1.5) node[anchor=north west] {$S_{r}$};
		\draw (0,0) node[anchor=north east] {$p$};
	\end{tikzpicture}\n
	The key calculation we'll perform is $\frac{d}{dt}\iprod{f_{t},f_{s}}$; we want to show it's equal to $0$. Well,
	\begin{align*}
		\frac{d}{dt}\iprod{f_{t},f_{s}} & =\iprod{\frac{D}{dt}f_{t},f_{s}}+\iprod{f_{t},\frac{D}{dt}f_{s}}\\
		& \labeledeq{(1)}\iprod{0,f_{s}}+\iprod{f_{t},\frac{D}{dt}f_{s}}\\
		& =\iprod{f_{t},\frac{D}{dt}f_{s}}\\
		& \labeledeq{(2)}\iprod{f_{t},\frac{D}{ds}f_{t}}\\
		& =\frac{1}{2}\iprod{\frac{D}{ds}f_{t},f_{t}}+\frac{1}{2}\iprod{f_{t},\frac{D}{ds}f_{t}}\\
		& =\frac{1}{2}\frac{d}{ds}\iprod{f_{t},f_{t}}\\
		& \labeledeq{(3)}\frac{1}{2}\frac{d}{ds}\norm{c(s)^{2}}\\
		& =\frac{1}{2}\frac{d}{ds}r^{2}\\
		& =0
	\end{align*}
	with (1) because $t\mapsto\exp_{p}(tv)=G(1,p,tv)=G(t,p,v)$ is a geodesic, (2) because of the proposition from earlier, and (3) because $\iprod{f_{t},f_{t}}$ is constant with repsect to $t$, so we can choose to evaluate it at $t=0$. Now, we can evaluate $\restr{\iprod{f_{t},f_{s}}}_{t=0}=\iprod{c(s),0}=0$, so we get that, for all $t,s$, $\iprod{f_{t},f_{s}}=0$.\proven
}

\par\noindent
Why are we done? Well, we can find $f_{t}(t)$ by $f(t,s=0)$ WOLOG, so $f(t,0)$ is the velocity of the radial geodeisc $t\mapsto\exp_{p}(tv(0))$, and $f(t,0)$ is an arbitrary tangent vector to the geodesic sphere $\exp_{p}(S_{r})$. So we conclude that the tangent space of a point $q$ on the geodesic sphere is perpendicular to the geodesic $\exp_{p}(tv)$, where $\exp_{p}(v)=q$.\newpage

\cor{
	If $U=\exp_{p}(B_{\varepsilon}(0))$ is a normal neighborhood of $p$, and $q\in{}U$, then the shortest path from $p$ to $q$ is $t\mapsto\exp_{p}(tv)$ ($0\le{}t\le{}1$), where $\exp_{p}(v)=q$. (By path, we mean a continuous, piecewise $C^{1}$ function.)\nn
	Proof: Assume $c:[0,1]\to{}U$, with $c(0)=p$ and $c(1)=q$, is a smooth path, and its image is contained in $U$. Write $(\exp_{p})\inv(c(t))=r(t)w(t)$, where $r(t)\ge{}0$ and $\norm{w(t)}\equiv{}1$. Consider the family $f(r,t)=\exp_{p}(rw(t))$, so that $c(t)=f(r(t),t)$. Then $\frac{dc}{dt}=\frac{dr}{dt}f_{r}+f_{t}$, and $f_{r}$ and $f_{t}$ are perpendicular for all $t$, so we can use the Pythagorean theorem to find
	\[
		\norm{\frac{dc}{dt}}^{2}=\norm{\frac{dr}{dt}f_{r}}^{2}+\norm{f_{t}}^{2}=\abs{\frac{dr}{dt}}^{2}\norm{f_{r}}^{2}+\norm{f_{t}}^{2}=\abs{\frac{dr}{dt}}^{2}+\norm{f_{t}}^{2}\ge\abs{\frac{dr}{dt}}^{2}
	\]
	Using this inequality, we can bound the length of $c$:
	\[
		l(c)=\int_{0}^{1}\norm{\frac{dc}{dt}}\,dt\ge\int_{0}^{1}\abs{\frac{dr}{dt}}\,dt\ge\int_{0}^{1}\frac{dr}{dt}\,dt=r(1)-r(0)=r(1)
	\]
	But $r(t)$ is the length of the radial geodesic $r\mapsto\exp_{p}(rw(1))$, joining $p$ to $q$. (Note that equality holds iff $\norm{f_{t}}^{2}\equiv{}0$, which is true iff $c$ is the radial geodesic.)\nn
	Now, we must consider the case where the image of $c$ is not contained in $U$. Well, there must be some $t_{0}\in(0,1)$ \st{} $c(t_{0})$ is on the geodesic sphere psasing through $q$. We know the length of $c:[0,t_{0}]$ is no smaller than the length of a radial geodesic directly to $q$, so the inequality still holds. See the illustration below:\n
	\begin{tikzpicture}
		\draw (0,0) node (p) {};
		\fill (p) circle (0.05);
		\draw (p) ellipse (3 and 1);
		\draw (p) ellipse (4 and 2);
		\node (q) at ($(0,0) + (30:3 and 1)$) {};
		\node (m) at ($(0,0) + (60:3 and 1)$) {};
		\fill (q) circle (0.05);
		\fill (m) circle (0.05);
		\draw[red, thick, -to] (0,0) -- (q.center) node[near start,anchor=north west] {Radial Geodesic};
		\draw[black!60!green, thick, -to] (0,0) to[bend right=20] (m.center) to[bend left=80, looseness=4] (q.center);
		%
		\draw (p) node[anchor=east] {$p$};
		\draw (0,-1) node[anchor=south] {Geodesic sphere};
		\draw (0,-2) node[anchor=south] {$U$};
		\draw (q) + (0.1,0) node[anchor=west] {$q$};
		\draw[black!60!green] (m) +(0.1,0.4) node[anchor=south] {$c$};
	\end{tikzpicture}\n
}

\cor{
	$d(p,q)=\inf(l(c))$, over the set of all $c$ joining $p$ and $q$, is actually a distance function.\n
}

\par\noindent
We showed all the other parts earlier -- the only thing left to check is that $d(p,q)=0\;\Rightarrow\;p=q$. We'll prove this by contraposition next time, but the idea is to assume that $p$ and $q$ are distinct, and then construct a normal neighborhood of $p$ that doesn't contain $q$. Then we know that $d(p,q)$ must be larger than the radius of the geodesic sphere, which is nonzero.\n

\end{document}