02_10.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 10}
\date{2/10/21} % Can also use \today

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

\par\noindent
Recall from last time:
\defn{
	Given a vector bundle $\mathcal{E}\to{}M$ with connection $\nabla$, $X,Y\in\mf{X}(M)$, the \u{curvature operator} $\mathcal{R}$ of $\nabla$, evlaulated on $(X,Y)$, is
	\[
		\map{\mathcal{R}(X,Y):\Gamma(\mathcal{E})}{\Gamma(\mathcal{E})}{(X,Y)}{\brack{\nabla_{X},\nabla_{Y}}-\nabla_{\brack{X,Y}}}
	\]
}

\par\noindent
Observe: When Do Carmo defines the curvature operator (Chapter 4, Definition 2.1, in the case where $\mathcal{E}=TM$), they use the opposite sign.\n

\par\noindent
Observe: $\mathcal{R}$ is given by a tensor! What does that mean? Last time, using the second approach, we computed locally in a moving frame $(E_{1},\ldots,E_{r})$ (with associated connection matrix $\vartheta$) that $\brack{\nabla_{X},\nabla_{Y}}=\nabla_{\brack{X,Y}}+d\vartheta(X,Y)+\brack{\vartheta(X),\vartheta(Y)}$. So $\mathcal{R}$ has, for its components in the given frame, the components of the vector
\[
	(d\vartheta(X,Y)+\brack{\vartheta(X),\vartheta(Y)})\vec{f}\qquad{}s=f^{i}E_{i},\vec{f}=\begin{pmatrix}
		f^{1}\\
		\vdots\\
		f^{r}
	\end{pmatrix}
\]

\defn{
	$\Omega\eqdef{}d\vartheta+\vartheta\wedge\vartheta$ is the \u{curvature matrix} of $\nabla$ with respect to the moving frame $(E_{1},\ldots,E_{r})$.\n
}

\par\noindent
(This is true because we observed $(\vartheta\wedge\vartheta)(X,Y)=\brack{\vartheta(X),\vartheta(Y)}$.)\n

\par\noindent
In fact, $\forall{}p\in{}U=\dom(E_{i})$, $\mathcal{R}(X,Y)(s)(p)\in\mathcal{E}_{p}$ is the image of $s(p)$ by the linear transformation $\mathcal{E}_{p}\to\mathcal{E}_{p}$ whose matrix (in the basis $(E_{1}(p),\ldots,E_{r}(p))$) is $\Omega_{p}(X_{p},Y_{p})$.\n

\par\noindent
The virtue of this definition is that it's a well-defined global object! But it turns out to be a differential operator of order $0$, meaning there's no derivatives, so it's just multiplication. At each point it's given by a linear transformation of the fibers, with the matrix determined by $X_{p}$ and $Y_{p}$.\n

\par\noindent
Observe: The dependence on $X$ and $Y$ is punctual! $\forall{}p\in{}M$, $\mathcal{R}(X,Y)(s)(p)$ depends only on $X_{p},Y_{p}\in{}T_{p}M$ and $s(p)\in\mathcal{E}_{p}$.\n

\par\noindent
$\mathcal{R}$, as an object, is ``an $\End$-$\mathcal{E}$ valued $2$-form on $M$''. That is, $\forall{}p\in{}M$, $\forall{}u,v\in{}T_{p}M$, $\mathcal{R}_{p}(u,v):\mathcal{E}_{p}\to\mathcal{E}_{p}$ is a linear map, and $\mathcal{R}(\cdot,\cdot)$ is bilinear and skew-symmetric.\n

\par\noindent
Intuition: $\mathcal{R}$ is given by infinitesimal holonomy. Given a tiny loop at $p$ below, the holonomy of the path is approximately $\exp(\mathcal{R}_{p}(u,v))$ (using the matrix exponential).
\[
	\begin{tikzpicture}
		\def\arlen{0.1};
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		\draw [->] (0,0) -- (0,1);
		\draw[blue] ({0.5-\arlen},{0+\arlen}) -- (0.5,0) -- ({0.5-\arlen},{0-\arlen});
		\draw[blue] ({1+\arlen},{0.5-\arlen}) -- (1,0.5) -- ({1-\arlen},{0.5-\arlen});
		\draw[blue] ({0.5+\arlen},{1+\arlen}) -- (0.5,1) -- ({0.5+\arlen},{1-\arlen});
		\draw[blue] ({0+\arlen},{0.5+\arlen}) -- (0,0.5) -- ({0-\arlen},{0.5+\arlen});
		%
		\draw (0,-0.25) node {$p$};
		\draw (1.25,0) node {$\vec{u}$};
		\draw (0,1.25) node {$\vec{v}$};
	\end{tikzpicture}
\]

\par\noindent
Even though we're taling about an operator, it's given by a tensor. $\mathcal{R}$ itself is a section of
\[
	\underbrace{T^{*}M\vphantom{\mathcal{E}_{p}}\otimes{}T^{*}M}_{\ptxt{$2$-form part}}\otimes\underbrace{\mathcal{E}_{p}\otimes\mathcal{E}_{p}}_{\hspace{-35pt}\mathrlap{\ptxt{Endomorphism part}}}
\]

\par\noindent
We're well on our way to defining the Levi-Civita connection!\n

\par\noindent
Consider a vector bundle $\mathcal{E}\to{}M$, now with a positive definite inner product on each fiber. (In the case where $\mathcal{E}=TM$, this exactly is a Riemannian metric.)\n

\defn{
	A connection $\nabla$ on $\mathcal{E}$ (with $\giprod$) is said to preserve $\giprod$ iff $\forall\gamma:[a,b]\to{}M$, parallel transport $\mathcal{P}_{\gamma}:\mathcal{E}_{\gamma(a)}\to\mathcal{E}_{\gamma(b)}$ is an isometry, i.e., $\forall{}u,v\in\mathcal{E}_{\gamma(a)}$, $\iprod{\mathcal{P}_{\gamma}(u),\mathcal{P}_{\gamma}(v)}_{\gamma(B)}=\iprod{u,v}_{\gamma(a)}$.\n
}

\prop{
	Given a vector bundle $\mathcal{E}\to{}M$, inner product $\giprod$ on each fiber, and a connection $\nabla$, the following are equivalent:
	\begin{enumerate}[label=(\alph*), leftmargin=4\parindent]
		\item $\nabla$ preserves $\giprod$.
		\item $\forall{}s,t\in\Gamma(\mathcal{E})$, $\forall{}X\in\mf{X}(M)$, $X(\iprod{s,t})=\iprod{\nabla_{X}s,t}+\iprod{s,\nabla_{X}t}$. Note that $\iprod{s,t}$ is a function on $M$, which we can differentiate with respect to $X$. We can think of this as a sort of ``product rule''.
		\item $\forall(E_{1},\ldots,E_{r})$ local orthonormal frame (which exists by Gram-Schmidt), the connection matrix $\vartheta$ is skew symmetric, i.e., $\forall{}i,j$, $\theta_{j}^{i}=-\theta_{i}^{j}$.
	\end{enumerate}\up\n
	Proof: First, we show that (b) $\Leftrightarrow$ (c). Let $(E_{1},\ldots,E_{r})$ be our local orthonormal frame. Then there are functions $f^{i},g^{j}$ such that $s=f^{i}E_{i}$ and $t=g^{j}E_{j}$. Thus, we can form $\vec{f},\vec{g}$, and by orthonormality of the frame
	\[
		\iprod{s,t}=\sum_{i=1}^{r}\sum_{j=1}^{r}f^{i}g^{j}\underbrace{\iprod{E_{i},E_{j}}}_{=\delta_{ij}}=\sum_{i=1}^{r}f^{i}g^{i}=\vec{f}\cdot\vec{g}
	\]
	Thus, with a slight abuse of notation,
	\[
		\iprod{\nabla_{X}s,t}\ptxt{``=''}(\nabla_{X}\vec{f})\cdot\vec{g}=(X(\vec{f})+\vartheta(X)\vec{f})\cdot\vec{g}
	\]
	And
	\[
		\iprod{\nabla_{X}s,t}+\iprod{s,\nabla_{X}t}=\underbrace{X(\vec{f})\cdot\vec{g}+\vec{f}\cdot{}X(\vec{g})}_{=X(\vec{f}\cdot\vec{g})=X(\iprod{s,t})}+(\vartheta(X)\vec{f})\cdot\vec{g}+\vec{f}\cdot(\vartheta(X)\vec{g})
	\]
	So the product rule holds iff $\forall{}s,t$ / $\forall\vec{f},\vec{g}$, $(\vartheta(X)\vec{f})\cdot\vec{g}+\vec{f}\cdot(\vartheta(X)\vec{g})=0$, which is true iff $\vartheta(X)$ is skew-symmetric.\nn
	In order to show (a), we just change the setting a bit. Let $\gamma:[a,b]\to{}M$ be a smooth curve. Take $V,W\in\Gamma_{\gamma}(\mathcal{E})$. We claim that, just as above, we get
	\[
		\underbrace{\iprod{\frac{DV}{dt},W}}_{\mathclap{\ptxt{a function of $t$}}}+\iprod{V,\frac{DW}{dt}}-\frac{d}{dt}\iprod{V,W}=(\vartheta(\dot\gamma)\vec{f})\cdot\vec{g}+\vec{f}\cdot(\vartheta(\dot\gamma)\vec{g})
	\]
	Assume $V$ and $W$ are parallel along $\gamma$. By definition, this means $\frac{DV}{dt}=\frac{DW}{dt}=0$. Then
	\[
		-\frac{d}{dt}\iprod{V,W}=(\vartheta(\dot\gamma)\vec{f})\cdot\vec{g}+\vec{f}\cdot(\vartheta(\dot\gamma)\vec{g})
	\]
	Well,
	\begin{align*}
		\ptxt{$\nabla$ preserves $\giprod$} & \Leftrightarrow\frac{d}{dt}\iprod{V,W}=0,\;\forall{}V,W\;\ptxt{parallel}\\
		& \Leftrightarrow(\vartheta(\dot\gamma)\vec{f})\cdot\vec{g}+\vec{f}\cdot(\vartheta(\dot\gamma)\vec{g})=0\;\ptxt{in all instances}\\
		& \Leftrightarrow\vartheta\;\ptxt{is skew symmetric}
	\end{align*}
	\proven
}

\thm{
	Let $M$ be a Riemannian manifold. Then $\exists\unique\nabla$ on $\mathcal{E}=TM\to{}M$ such that
	\begin{enumerate}[label=(\alph*),leftmargin=4\parindent]
		\item $\nabla$ preserves the Riemannian metric. \i{(This depends on the choice of Riemannian metric.)}
		\item $\forall{}X,Y\in\mf{X}(M)$, $\nabla_{X}Y-\nabla_{Y}X=\brack{X,Y}$. \i{(This does not depend on the choice of Riemannian metric.)}
	\end{enumerate}\up\n
}

\defn{
	This $\nabla$ is called the \u{Levi-Civita connection} on $M$.\n
}

\end{document}