03_17.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 24}
\date{3/17/21} % Can also use \today

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

\par\noindent
Continuing with Jacobi fields from last time. Recall: For $\gamma$ a geodesic, $J\in\Gamma_{\gamma}(TM)$ is a Jacobi field iff $\frac{D^{2}}{dt^{2}}J+\mc{R}(J,\dot\gamma)\dot\gamma=0$. It's a fact that $J$ is a Jacobi field iff $V=\restr{\d_{s}f}_{s=0}$ for some variation $f$ of $\gamma$ by geodesics. We proved part of this in class, and will prove the rest in a homework problem.\n

\par\noindent
Let $E_{1},\ldots,E_{n}\in\Gamma_{\gamma}(TM)$ be an orthonormal parallel frame. We can write any $J\in\Gamma_{\gamma}(TM)$ as $J(t)=f^{i}(t)E_{i}(t)$ for some smooth functions $f^{i}$. Using the fact that $\frac{D}{dt}E_{i}\equiv{}0$, we get $\frac{D^{2}}{dt^{2}}J=\ddot{f}^{i}E_{i}$, and by linearity, the Jacobi equation becomes a system of ODEs. $\forall{}i$,
\[
	\ddot{f}^{i}+\underbrace{(E_{i},\dot\gamma,\dot\gamma,E_{j})}_{a_{ij}=a_{ji}}f^{j}\equiv{}0\qquad\ptxt{and}\qquad(E_{i},\dot\gamma,\dot\gamma,E_{j})f^{j}=\mc{R}(J,\dot\gamma)\dot\gamma=\mc{R}(f^{j}E_{j},\dot\gamma)\dot\gamma=f^{j}\mc{R}(E_{j},\dot\gamma)\dot\gamma
\]
The $i$th component of $\mc{R}(J,\dot\gamma)\dot\gamma$ is
\[
	a_{ij}=f^{i}(E_{j},\dot\gamma,\dot\gamma,E_{i})=f^{i}(\dot\gamma,E_{i},E_{j},\dot\gamma)=a_{ji}
\]
So our system of equations is
\[
	\ddot{f}^{i}(t)+a_{ij}f^{j}(t)=0\quad{}1\le{}i\le{}n
\]

\cor{
	A Jacobi field is uniquely determined by $J(0)$ and $\frac{DJ}{dt}(0)$. In fact, we have an isomorphism
	\[
		\arraycolsep=2pt
		\begin{array}{rcl}
			\set{\ptxt{Jacobi fields along}\;\gamma} & \cong & T_{\gamma(0)}M\oplus{}T_{\gamma(0)}M\\
			J & \mapsto & (J(0),\frac{DJ}{dt}(0))
		\end{array}
	\]
	This tells us that the space of Jacobi fields along $\gamma$ has dimension $2n$.\n
}

\lemma{
	Let $J$ be a Jacobi field. Then $\exists{}a,b\in\R$ such that $\iprod{J(t),\dot\gamma(t)}=a+bt$.\nn
	Proof: It's enough to show $\frac{d}{dt}\iprod{J,\dot\gamma}$ is constant. Well,
	\[
		\frac{d}{dt}\paren{\frac{d}{dt}\iprod{J(t),\dot\gamma(t)}}=\frac{d}{dt}\iprod{\frac{DJ}{dt}(t),\dot\gamma(t)}=\iprod{\frac{D^{2}J}{dt^{2}}(t),\dot\gamma(t)}=-\iprod{\mc{R}(J,\dot\gamma)\dot\gamma(t),\dot\gamma(t)}=0
	\]
	with the last equality because curvature is skew-symmetric.\proven
}

\cor{
	If $J(0)$ and $\frac{DJ}{dt}(0)$ are orthogonal to $\dot\gamma(0)$, then they remain orthogonal for all $t$.\n
}

\defn{
	A Jacobi field satisfying the above condition is called a \u{normal Jacobi field}. The set of normal Jacobi fields forms a dimension $2(n-1)$ subspace.\n
}

\lemma{
	$(\dot\gamma,t\dot\gamma)$ span the space of tangential Jacobi fields.\n
}

\lemma{
	The space of Jacobi fields has a natural symplectic structure. $\forall{}J_{1},J_{2}$ Jacobi fields, the quantity
	\[
		\Omega(J_{1},J_{2})=\iprod{J_{1},\frac{DJ_{2}}{dt}}-\iprod{\frac{DJ_{1}}{dt},J_{2}}
	\]
	is constant w.r.t. $t$. We take $\Omega$ to be the symplectic form.\n
}

\par\noindent
Observe: The space of normal Jacobi fields is a symplectic subspace (i.e. the restriction of $\Omega$ is still non-degenerate). It corresponds to a certain subspace of $T_{(\dot\gamma(0),\dot\gamma(0))}(T^{*}M)$.\n

\par\noindent
We now check the lemma above. All we need to do is show $\Omega(J_{1},J_{2})$ is constant w.r.t. $t$. So we differentiate:
\[
	\frac{d}{dt}\Omega(J_{1},J_{2})=\frac{d}{dt}\paren{\iprod{J_{1},\frac{DJ_{2}}{dt}}-\iprod{\frac{DJ_{1}}{dt},J_{2}}}=\cancel{\iprod{\frac{DJ_{1}}{dt},\frac{DJ_{2}}{dt}}}+\iprod{J_{1},\frac{D^{2}J_{2}}{dt^{2}}}-\cancel{\iprod{\frac{DJ_{1}}{dt},\frac{DJ_{2}}{dt}}}-\iprod{\frac{D^{2}J_{1}}{dt^{2}},J_{2}}
\]
We can then use the Jacobi equation to cancel out the remaining terms.\proven

\ex{
	Let $M$ be an oriented surface, and take $\norm{\dot\gamma}\equiv{}1$. Then $(E_{1},E_{2})=(\dot\gamma,\dot\gamma^{\perp})$ is an orthonormal frame. We write down the Jacobi equations:
	\[
		(E_{i},\dot\gamma,\dot\gamma,E_{j})=\left\{\begin{array}{ll}
			0 & i=1\;\ptxt{or}\;j=1\\
			k & i=j=2
		\end{array}\right.
	\]
	where $k$ is the Gaussian curvature. Write $J=f^{1}\dot\gamma+f^{2}\dot\gamma^{\perp}$. Then $\ddot{f}^{1}=0$ iff $f^{1}=a+bt$, for $a,b\in\R$. And $\ddot{f}^{2}=kf^{2}=0$ (assume $k$ is constant for this problem). Then $f^{2}(t)=Ae^{i\sqrt{k}}+Be^{-i\sqrt{k}}$. $f^{2}$ must be real, so we solve this well-known type of differential equation, and if $k>0$, we get the following:
	\[
		J(t)=(A\cos\sqrt{k}t+B\sin\sqrt{k}t)\dot\gamma^{\perp}
	\]
	If $J(0)=0$, then $J(t)=B\sin(\sqrt{k}t)\dot\gamma^{\perp}$ Observe that $J(\frac{\pi}{\sqrt{k}})=0$. We say that ``$\gamma(0)$ and $\gamma(\frac{\pi}{\sqrt{k}})$ are conjugate''.\n
	If $k<0$, then we replace $\sin$ and $\cos$ with $\sinh$ and $\cosh$. Then $J(0)=0$ implies $J(t)=B\sinh(\sqrt{k}t)\dot\gamma^{\perp}$. In this case, $J(t)\ne{}0$, $\forall{}t\ne{}0$.\n
}

\par\noindent
One application of this is computing $d(\exp_{p})_{v}$ for $v\ne{}0$.\n

\prop{
	Given $v,w\in{}T_{p}M$, $d(\exp_{p})_{v}(w)=J(1)$, where $J$ is the Jacobi field such that $J(0)=0$ and $\frac{DJ}{dt}(0)=w$.\nn
	Proof: $d(\exp_{p})_{v}(w)=\restr{\frac{d}{dt}\exp_{p}(v+sw)}_{s=0}$. Define $f(s,t)=\exp_{p}(t(v+sw))$. $\forall{}s$, $t\mapsto{}f(s,t)$ is a geodesic. Define $J(t)=\restr{\d_{s}f}_{s=0}$. We know this is a Jacobi field, and claim that $\frac{DJ}{dt}(0)=w$.\n
}

\end{document}