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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 26}
\date{3/22/21} % Can also use \today

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

\par\noindent
Continuing from last time, we need to compute $\iprod{J^{(3)},J'}$.\n

\lemma{
	$J^{(3)}=-\frac{D}{dt}\mc{R}(J,\dot\gamma)\dot\gamma$. (Prove using properties of Jacobi fields.)\n
}

\par\noindent
Claim: $\restr{\frac{D}{dt}\mc{R}(J,\dot\gamma)\dot\gamma}_{t=0}=\restr{\mc{R}(J',\dot\gamma)\dot\gamma}_{t=0}$.\n
Proof: Let $W\in\Gamma_{\gamma}(TM)$. Compute
\[
	\frac{d}{dt}\iprod{\mc{R}(J,\dot\gamma)\dot\gamma,W}=\iprod{\frac{D}{dt}\mc{R}(J,\dot\gamma)\dot\gamma,W}+\iprod{\mc{R}(J,\dot\gamma)\dot\gamma,\cancelto{\mathrlap{0\;\ptxt{at}\;t=0}}{\frac{Dw}{dt}}}
\]
and
\[
	\frac{d}{dt}\iprod{\mc{R}(J,\dot\gamma)\dot\gamma,W}=\frac{d}{dt}(J,\dot\gamma,\dot\gamma,W)=\frac{d}{dt}(W,\dot\gamma,\dot\gamma,J)=\iprod{\frac{D}{dt}\cancelto{0}{\mc{R}(W,\dot\gamma)}\dot\gamma,J}+\underbrace{\iprod{\mc{R}(W,\dot\gamma)\dot\gamma},\frac{DJ}{dt}}_{\hspace{-5pt}\mathrlap{=(W,\dot\gamma,\dot\gamma,J')=(J',\dot\gamma,\dot\gamma,W)}}
\]
Thus, at $t=0$, we have $\restr{(J',\dot\gamma,\dot\gamma,W)}_{t=0}=\restr{\iprod{\frac{D}{dt}\mc{R}(J,\dot\gamma)\dot\gamma,W}}_{t=0}=\iprod{\mc{R}(J',\dot\gamma)\dot\gamma,W}$.\n
We conclude that $\mc{R}(J',\dot\gamma)\dot\gamma=\frac{D}{dt}\mc{R}(J,\dot\gamma)\dot\gamma$.\n

\par\noindent
Thus, $\iprod{J^{(3)},J'}=\iprod{\mc{R}(J',\dot\gamma)\dot\gamma,J'}=K_{0}(J',\dot\gamma)$. This proves the lemma, and completes the proof we began last time.\proven

\section*{Conjugate Points}

\defn{
	Let $\gamma$ be a geodesic, $p,q\in\im\gamma$ distinct. Then $p,q$ are \u{conjugate} along $\gamma$ iff there exists a nonzero Jacobi field of $\gamma$ that vanishes at $p$ and $q$.\n
}

\ex{
	With constant negative curvature $K<0$, suppose $\gamma(0)=p$, $\gamma(t_{1})=q$. Then $J(t)=A\sinh(\sqrt{-K}t)W(t)$. So there are no conjugate points.\nn
	If $K>0$, then $J(t)=A\sin(\sqrt{K}t)$, so $t=\frac{\pi}{\sqrt{k}}$ yields a conjugate point.\n
}

\defn{
	The \u{multiplicity} of $q$ as a conjugate point of $p$ is $\dim\set{J\mid{}J(q)=0,J(p)=0}$.\n
}

\par\noindent
Claim: Multiplicity is at most $n-1$, and $n-1$ is achieved by the sphere $S^{n}$.\n

\par\noindent
Note: ``Being conjugate'' is a symmetric relation, but it's not transitive!\n

\prop{
	Let $q=\exp_{p}(tv_{q})$, $v_{q}\in{}T_{p}M$. Then $q$ is conjugate to $p$ iff $v_{q}$ is a critical point of $\exp_{p}$, iff $d(\exp_{p})_{v_{q}}$ has a nontrivial kernel.\nn
	Proof: Simply recall how to compute $d(\exp_{p})_{v_{q}}$: $J(1)=d(\exp_{p})_{v_{q}}(w)$ when $J(0)=0$ and $J'(0)=w$ (by parameterizing $\gamma$ by arc length, and rescaling so that $t_{1}=1$).\proven
}

\par\noindent
Moreover, $\dim\ker{}d(\exp_{p})_{v_{q}}$ is the multiplicity of $q$, and it is at most $n-1$, because $v_{q}\not\in\ker{}d(\exp_{p})_{v_{q}}$, because $d(\exp_{p})_{v_{q}}(v_{q})=\dot\gamma(t_{1})\ne{}0$, because $\gamma$ is nontrivial, because $p\ne{}q$.\n

\thm{
	(I) Let $q=\exp_{p}(t_{1}v)$, $\norm{v}=1$ be a conjugate point of $p$. Then $\forall{}t_{2}>t_{1}$, $t\mapsto\exp_{p}(tv)$ is \i{not} minimizing on $[0,t_{2}]$.\nn
	Proof: This is simply a nice application of the second variation form.\proven
}

\par\noindent
Think of the sphere -- if we go from the north pole to past the south pole, the curve isn't minimizing.\n

\par\noindent
Note: There are no conjugate points on a cylinder, but not all geodesics are minimizing.\n

\par\noindent
Note: The converse is not globally true.\n

\thm{
	(II) Let $p,q\in\im\gamma$, for $\gamma$ a geodesic, and assume $p,q$ are not conjguate, and there are no conjugate points between $p$ and $q$. Then any proper variation of the arc $\overarc{pq}$ is such that $\forall{}s$ sufficiently small, the $t$-curves are longer than $\overarc{pq}$.\n
}

\thm{
	\i{Second Variation Formula, with one jump in $V'$.} Let $\gamma$ be a geodesic, defined for $t\in[0,t_{2}]$. Suppose $V$ is a proper, continuous, infinitesimal variation (with respect to $s$), and is smooth on $[0,t_{1}]$ and $[t_{1},t_{2}]$ (assume $1$-sided derivatives exist at $t_{1}$). Then
	\[
		E''(0)=-\int_{0}^{t_{2}}\iprod{V,V''+\mc{R}(V,\dot\gamma)\dot\gamma}\,dt-\iprod{V(t_{1}),\Delta{}V'(t_{1})}
	\]
	where $\Delta{}V'(t_{1})=V'(t_{1}^{+})-V'(t_{1}^{-})$ (using the one-sided derivatives).\nn
	Proof: See Do Carmo, page 197. (They prove this result for any number of jumps.)\n
}

\par\noindent
Now, back to the philosophy that this is a sort of Hessian of the energy functional $\mc{E}:\set{\ptxt{paths}\;p\rightsquigarrow{}q}\to\R$. In the above formula, $V$ is a tangent vector, so the right hand side should be a quadratic form on $V$. But what is the associated bilinear form?\n

\prop{
	Given the same assumptions as above, with any number of jumps, we have
	\[
		E''(0)=\int_{0}^{t_{2}}\iprod{V',V'}-\iprod{\mc{R}(V,\dot\gamma)\dot\gamma,V}\,dt
	\]
	and moreover, the symmetric bilinear form is
	\[
		I(V,W)=\int_{0}^{t}\iprod{V',W'}-\underbrace{\iprod{\mc{R}(V,\dot\gamma)\dot\gamma,W}}_{\ptxt{pairwise symmetric in}\;V,W}\,dt
	\]
	so $E''(0)=I(V,V)$ (and clearly, $I$ is symmetric).\n
}

\end{document}