03_10.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 21}
\date{3/10/21} % Can also use \today

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

\par\noindent
Review: second variation formula. If $\gamma$ is a geodesic, and $f$ a proper variation of $\gamma$, $V=\restr{\d_{s}}_{s=0}$, then
\[
	E''(0)=-\int_{0}^{a}\iprod{V,\frac{D^{2}V}{dt^{2}}+\mc{R}(V,\dot\gamma)(\dot\gamma)}\,dt
\]

\par\noindent
Next, we're going to take a closer look at the curvature, $\mc{R}$. Recall its definition: If $X,Y\in\mf{X}(M)$, then the curavture is $\mc{R}(X,Y):\mf{X}(M)\to\mf{X}(M)$, defined by $\mc{R}(X,Y)=\comm{\nabla_{X},\nabla_{Y}}-\nabla_{\comm{X,Y}}$. It turns out that $\mc{R}(X,Y)$ is given by the action of a tensor $\mc{R}$ of the form $\forall{}p\in{}M;u,v\in{}T_{p}M$, $\mc{R}_{p}(u,v):T_{p}M\to{}T_{p}M$, a linear map.\n

\par\noindent
As an operator, $\mc{R}(X,Y)(Z)_{p}=\mc{R}_{p}(X_{p},Y_{p})(Z_{p})\in{}T_{p}M$. Also, $\mc{R}$ shows up as an obstruction to finding a covariant-constant frame $\nabla_{E_{i}}E_{j}\equiv{}0$.\n

\section*{Curvature Identities}

\par\noindent
(Covered in Do Carmo, Chapter 4, \sectionSymbol{}2)\n

\par\noindent
The first identity we consider is the Bianchi identity: $\forall{}X,Y,Z\in\mf{X}(M)$, $\mc{R}(X,Y)Z+\R(Z,X)Y+\R(Y,Z)X=0$, due to $\nabla$ being torsion-free. The proof is simply a computation, and can be found on page 91 of Do Carmo.\n

\prop{
	Introduce $X,Y,Z,T\in\mf{X}(M)$, and define $(X,Y,Z,T)\eqdef\iprod{\mc{R}(X,Y)Z,T}$.
	\begin{enumerate}[itemsep=0pt, leftmargin=4\parindent, label=(\alph*)]
		\item $(X,Y,Z,T)+(Y,Z,X,T)+(Z,X,Y,T)=0$ (Proved via the Bianchi identity)
		\item $(X,Y,Z,T)=-(Y,X,Z,T)$ (Because $\mc{R}(X,Y)=-\mc{R}(Y,X)$)
		\item $(X,Y,Z,T)=-(X,Y,T,Z)$ (Because $\nabla$ perserves $\giprod$)
		\item $(X,Y,Z,T)=(Z,T,X,Y)$ (Follow from the Bianchi identity and some algebra)
	\end{enumerate}\up\n
}

\par\noindent
In coordinates $(x^{1},\ldots,x^{n})$, $X_{i}=\pd{}{x^{i}}$, then $\mc{R}$ has components in the coordinate system, $\mc{R}_{ijk}^{l}\in{}C^{\infty}(U)$ (where $U$ is the domain of the coordinate chart) such that
\[
	\underbrace{\mc{R}(X_{i},X_{j})X_{k}}_{\in\mf{X}(U)}=\mc{R}_{ijk}^{l}X_{l}
\]
But how do we compute $\mc{R}_{ijk}^{l}$?\n

\lemma{
	$\mc{R}_{ijk}^{l}=\Gamma_{jk}^{l}\Gamma_{il}^{s}-\Gamma_{ik}^{l}\Gamma_{jl}^{s}+\d_{i}\Gamma_{jk}^{s}-\d_{j}\Gamma_{ik}^{s}$. (Recall that Do Carmo uses the opposite sign for $\mc{R}$.)\nn
	Proof: This is a messy computation. The complete details can be found on pages 92-93 of Do Carmo.\n
}

\par\noindent
Observe that $\Gamma$ depends on the first derivatives of $g_{ij}$, so $\mc{R}$ depends on the second derivatives of $g_{ij}$.\n

\defn{
	$(X_{i},X_{j},X_{k},X_{s})\eqdef\mc{R}_{ijks}$. $X_{ijks}\eqdef\mc{R}_{ijk}^{l}g_{ls}$.\n
}

\par\noindent
This allows us to rephrase the identities much more concisely:
\begin{enumerate}[label=(\alph*), itemsep=0pt]
	\item $\mc{R}_{ijks}+\mc{R}_{jkis}+\mc{R}_{kijs}=0$
	\item $\mc{R}_{ijks}=-\mc{R}_{jiks}$
	\item $\mc{R}_{ijks}=-\mc{R}_{ijsk}$
	\item $\mc{R}_{ijks}=\mc{R}_{ksij}$
\end{enumerate}
Nobody \i{really} understands this whole tensor. The whole thing is a monster. But we can understand parts of it.\n

\par\noindent
One part which we can understand is the sectional curvature, which shows up in the second variation formula.\n

\defn{
	$\forall{}u,v\in{}T_{p}M$,
	\[
		\abs{u\wedge{}v}=\sqrt{\norm{u}^{2}\norm{v}^{2}-\iprod{u,v}^{2}}
	\]
	is the area of the parallelogram spanned by $u$ and $v$.\n
}

\lemma{
	If $u$ and $v$ are linearly independent, then
	\[
		K(u,v)\eqdef\frac{\mc{R}(u,v,v,u)}{\abs{u\wedge{}v}^{2}}=\frac{\iprod{\mc{R}(u,v)v,u}}{\abs{u\wedge{}v}^{2}}
	\]
	$K$ depends only on the plane $\pi(u,v)=\vspan(u,v)$.\nn
	Proof: Check that the RHS is invariant under each of the following ``moves'':
	\begin{itemize}[leftmargin=4\parindent, itemsep=0pt, topsep=0pt]
		\item $(u,v)\rightsquigarrow(\lambda{}u,v)$ (for $\lambda\ne{}0$)
		\item $(u,v)\rightsquigarrow(v,u)$
		\item $(u,v)\rightsquigarrow(u+\lambda{}v,v)$
	\end{itemize}\up\n
	We will finish proving this next time.\n
}

\defn{
	$K(\pi)=K(u,v)$ is called the \u{sectional curvature} of $p$ at $\pi$.\n
}

\end{document}