02_01.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 6}
\date{2/1/21} % Can also use \today

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

\par\noindent
A brief roadmap for the next few weeks:
\[
	\begin{tikzcd}
		\ptxt{Riemannian Metrics} \arrow[ddr] &[-40pt] &[-40pt] \ptxt{Connections} \arrow[d]\\
		& & \begin{array}{c}
			\ptxt{Parallel Transport}\\
			\ptxt{and Curvature}
		\end{array} \arrow[dl]\\
		& \begin{array}{c}
			\ptxt{If $M$ is Riemannian, then}\\
			\ptxt{$TM$ has a special connection}
		\end{array} \arrow[d]\\
		& \ptxt{Etc.}
	\end{tikzcd}
\]

\par\noindent
First, some motivation... (See also Lee Riemannian Geometry, Chapter 4)\n

\par\noindent
Start with $\gamma:(a,b)\to\R^{n}$. Let $Y=\sum_{i}f^{i}\pd{}{x^{i}}\in\mathfrak{X}(\R^{n})$, a smooth vector field in $\R^{n}$. We can compute $\restr{\frac{d}{dt}Y(\gamma(t))}{t=t_{0}}$ in $\R^{n}$. But what's really happening? Well, we're computing
\[
	\restr{\frac{d}{dt}Y(\gamma(t))}{t=t_{0}}=\lim_{h\to{}0}\frac{Y(\gamma(t_{0}+h))-Y(\gamma(t_{0}))}{h}
\]
But we can only do this in $\R^{n}$, not on manifolds in general! Strictly speaking, $Y(\gamma(t_{0}+h))\in{}T_{\gamma(t_{0}+h)}\R^{n}$ and $Y_{\gamma(t_{0})}\in{}T_{\gamma(t_{0})}\R^{n}$. We can take their difference because we're identifying all tangent spaces of $\R^{n}$ with each other, using translations of $\R^{n}$. And $\set{\ptxt{translations of $\R^{n}$}}\cong\R^{n}$ as a vector space. In other words, we can translate vectors in $\R^{n}$ ``parallel to themselves''.\n

\par\noindent
For $Y=\sum_{i}f^{i}\pd{}{x^{i}}$, we get a formula:
\[
	\restr{\frac{d}{dt}Y(\gamma(t))}{t=t_{0}}=\sum_{i=1}^{n}df_{\gamma(t_{0})}^{i}(\dot\gamma(t_{0}))\pd{}{x^{i}}\eqdef\paren{\bar\nabla_{\dot\gamma(t_{0})}Y}(\gamma(t_{0}))
\]
(Note that $\pd{}{x^{i}}$ is a constant frame on $\R^{n}$, so we can use $\pd{}{x^{i}}$ and $\pdat{}{x^{i}}{p}$ interchangeably).\n

\defn{
	If $Y=\sum_{i=1}^{n}f^{i}\pd{}{x^{i}}\in\mathfrak{X}(\R^{n})$, $p\in\R^{n}$, and $v\in{}T_{p}M$, we define
	\[
		\paren{\bar\nabla_{v}Y}(p)\eqdef\sum_{i=1}^{n}df^{i}_{p}(v)\pd{}{x^{i}}
	\]
}

\par\noindent
We can think of $\paren{\bar\nabla_{v}Y}(p)$ as a vector, which only depends on the values of $Y$ along a curve $\gamma$ as above.\n

\par\noindent
The question remains: Is there something analogous to this on manifolds? It may look a bit like a Lie derivative, but note that $\bar\nabla$ is \b{not} a Lie derivative!\n

\par\noindent
Recall: Given $X,Y\in\mathfrak{X}(M)$, we can define $\mathcal{L}_{X}Y$ using the flow $\varphi$ of $X$:
\[
	(\mathcal{L}_{X}Y)(p)=\lim_{t\to{}0}\frac{(\varphi_{-t})_{*,\varphi_{t}(p)}(X_{\varphi_{t}(p)})-X_{p}}{t}
\]
In this case, we need $X$ as a vector field, whereas above, we just need a vector. And $\mathcal{L}_{X}Y$ is dependent on $X$, but there are infinitely many vector fields $X$ \st{} $X_{p}=v$ (with $v$ as above).\n

\par\noindent
In fact, we need some additional structure on the manifold, because we cannot natural identify $T_{p}M$ with $T_{q}N$, when $p\ne{}q$. This additional structure is called a connection.\n

\defn{
	Let $\mathcal{E}\to{}M$ be a vector bundle. A \u{connection} on $\mathcal{E}$ is an operator
	\[
		\map{\del:\mathfrak{X}(M)\times\Gamma(\mathcal{E})}{\Gamma(\mathcal{E})}{(X,s)}{\nabla_{X}s}
	\]
	that satisfies:
	\begin{enumerate}[topsep=0pt, itemsep=0pt, leftmargin=4\parindent, label=\arabic*)]
		\item $\forall{}X,Y\in\mathfrak{X}(M)$, $\forall{}s\in\Gamma(\mathcal{E})$, $\nabla_{X+Y}s=\nabla_{X}s+\nabla_{Y}s$
		\item $\forall{}f\in{}C^{\infty}(M)$, $\nabla_{fX}s=f\nabla_{X}s$
		\item $\nabla_{X}(fs)=f\nabla_{X}s+X(f)s=f\nabla_{X}s+df(X)s$
	\end{enumerate}\up\n
	Because of properties 1 and 2, we say that a connection is ``linear in $X$ over $C^{\infty}(M)$''.\n
}

\par\noindent
Note that although our definition above uses vector fields, we will show that this dependence is pointwise.\n

\ex{
	$\nabla=\bar\nabla$ on $\mathcal{E}=T\R^{n}$.\n
}

\prop{
	If $\nabla$ is a connection on $\mathcal{E}\to{}M$, then $\forall{}X\in\mathfrak{X}(M),s\in\Gamma(\mathcal{E}),p\in{}M$, $(\nabla_{X}s)(p)\in\mathcal{E}_{p}=\pi\inv(p)$ only depends on $X_{p}$ and the values of $s$ in an arbitrarily small open neighborhood of $p$.\nn
	Proof: Let $U$ be a neighborhood of $p$; $\chi$ a bump function supported on $U$, with $\chi\equiv{}1$ on $V\osubseteq{}U$, a smaller open neighborhood of $p$. Consider $\nabla_{X}(\chi{}s)=X(\chi)s+\chi\nabla_{X}s$. Evaluate at $p$: the right hand side is just $(\nabla_{X}s)(p)$ because $\chi\equiv{}1$ on $V$ and $X(\chi)\equiv{}0$ on $V$ (because $\chi$ is constant on $V$, and $X$ is a derivation). The computation of $(\nabla_{X}s)(p)$ can be localized to, say, a coordinate neighborhoood of $p$.\nn
	Let $X=\sum_{i}a^{i}\pd{}{x^{i}}$ in local coordinates. Then, by $C^{\infty}$ linearity of $\nabla$,
	\[
		(\nabla_{X}s)(p)=\sum_{i}a^{i}(p)(\nabla_{\pd{}{x^{i}}}s)(p)
	\]
	If $X(p)=0$ (which is true iff $\forall{}j$, $a^{j}(p)=0$), then $(\nabla_{X}s)(p)=0$. So if $X(p)=\tilde{X}(p)$, then $(\nabla_{X}s)(p)=(\nabla_{\tilde{X}}s)(p)$.\proven
}

\par\noindent
Observe: Given $\nabla$ and $s$, $(\nabla_{X}s)(p)\in\mathcal{E}_{p}$ depends only on $X(p)$, and does so linearly! So $\nabla$ and $s$ define a map
\[
	\map{T_{p}M}{\mathcal{E}_{p}}{v}{(\nabla_{v}s)(p)}
\]
which is itself an element of $T_{p}^{*}M\otimes\mathcal{E}_{p}$. Therefore, $\nabla$ can be thought of as an operator $\nabla:\Gamma(\mathcal{E})\to\mathcal{E}(T^{*}M\otimes\mathcal{E})$, whose image is ``$\mathcal{E}$-valued differential forms''.\n

\subsection*{Local Expression of a $\nabla$}

\par\noindent
Let $\mathcal{E}\to{}M$ be a vector bundle, with connection $\nabla$, $U\osubseteq{}M$, and $(E_{1},\ldots,E_{r})$ a moving frame of $\mathcal{E}$ over $U$. That is, $\forall{}j$, $E_{j}\in\Gamma(\restr{\mathcal{E}}{U})$, and at each $p\in{}U$, $(E_{1}(p),\ldots,E_{r}(p))$ is a basis of $\mathcal{E}_{p}$. So if $s\in\Gamma(\restr{\mathcal{E}}{U})$, then $\exists{}f^{i}\in{}C^{\infty}(U)$ \st{} $s=\sum_{j}f^{j}E_{j}$. So $\forall{}X\in\mathfrak{X}(U)$, we get
\[
	\nabla_{X}s=\sum_{j=1}^{r}f^{j}\nabla_{X}E_{j}+X(f^{j})E_{j}
\]
But
\[
	\nabla_{X}E_{j}=\sum_{i=1}^{r}\theta_{j}^{i}(X)E_{i}
\]
By the discussion above, $\forall{}i,j$, $\theta_{j}^{i}\in\Omega^{1}(U)$, a one-form. So we can define $\vartheta=(\theta_{j}^{i})$, an $r\times{}r$ matrix of one-forms on $U$, depending on the moving frame $(E_{1},\ldots,E_{r})$. In fact, this $\vartheta$ determines $\nabla$ on $U$!

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