01_29.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 5}
\date{1/29/21} % Can also use \today

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

\par\noindent
Some stuff we can do with Riemannian metrics:
\begin{itemize}
	\item A Riemannian metric on $M$ allows us to define the length of a tangent vector, and the angle between two tangent vectors. For $v,w\in{}T_{p}M$, $\norm{v}=\sqrt{\iprod{v,v}_{p}}$, and $\iprod{v,w}_{p}=\norm{v}\norm{w}\cos\theta$ defines $\theta$, the angle between $v$ and $w$.
	\item A Riemannian metric on $M$ allows us to define the length of a curve in $M$. Say $\gamma:[a,b]\to{}M$. We define its length
	\[
		L(\gamma)=\int_{a}^{b}\norm{\dot\gamma(t)}\,dt=\int_{a}^{b}\sqrt{\iprod{\dot\gamma(t),\dot\gamma(t)}_{\gamma(t)}}\,dt
	\]
	If $\im\gamma\subseteq{}U$, a coordinate patch, with coordinates $(x^{1}(t),\ldots,x^{k}(t))$, then
	\[
		L(\gamma)=\int_{a}^{b}\sqrt{g_{ij}(\gamma(t))\dot{x}^{i}(t)\dot{x}^{j}(t)}\,dt\qquad{}g_{ij}=\iprod{\pd{}{x^{i}},\pd{}{x^{j}}}\in{}C^{\infty}(U)
	\]
\end{itemize}

\lemma{
	$L(\gamma)$ is invariant under re-parameterizations of $\gamma$. Consider $[\alpha,\beta]\overset{\sim}{\underset{t}{\to}}[a,b]\overset{\gamma}{\to}M$. Then
	\[
		\int_{\alpha}^{\beta}\norm{\frac{d\gamma}{ds}(t(s))}\,ds=\int_{a}^{b}\norm{\frac{d\gamma}{dt}(t)}\,dt
	\]
	Proof: The chain rule says $\frac{d\gamma}{ds}=\frac{dt}{ds}\frac{d\gamma}{dt}$. So $\norm{\frac{d\gamma}{ds}}=\norm{\frac{dt}{ds}}\norm{\frac{d\gamma}{dt}}$. Now, use the change of variables formula for integrals.\proven
}

\defn{
	Let $M$ be a connected Riemannian manifold. We define the \u{Riemannian distance function} by, $\forall{}p,q\in{}M$,\n $d(p,q)=\inf\set{L(\gamma)\mid{}\gamma\ptxt{ is a continuous curve, or ``path'', joining $p$ and $q$ that is piecewise smooth}}$.\n
}

\par\noindent
Note: We can replace this definition with just ``smooth'' -- the definitions are equivalent. But that's harder to prove, and this definition will be useful later on.\n

\par\noindent
Note that because $M$ is connected, it's path connected. This means the set of lengths of curves connecting pairs of points is nonempty, so the infimum exists. And because $L(\gamma)\ge{}0$, $d(p,q)\ge{}0$.\n

\thm{
	$d$ is a metric, or distance function, i.e., $\forall{}p,q,r\in{}M$,
	\begin{enumerate}[topsep=0pt, itemsep=0pt, leftmargin=4\parindent, label=(\roman*)]
		\item $d(p,q)\ge{}0$, with $d(p,q)=0\Leftrightarrow{}p=q$
		\item $d(p,q)=d(q,p)$
		\item $d(p,q)+d(q,r)\ge{}d(p,r)$
	\end{enumerate}\up\nn
	Proof: (This is only a partial proof)
	\begin{enumerate}[topsep=0pt, itemsep=0pt, leftmargin=4\parindent, label=(\roman*)]
		\item If $p=q$, take a trivial/constant path. $\dot\gamma\equiv{}0$, so $L(\gamma)=0$. The converse remains to be shown: that $d(p,q)=0$ implies $p=q$. This will be a corollary of the ``Gauss lemma'', which we'll do later on.
		\item We never assumed reparameterizations couldn't reverse the direction of the curve. They can, which directly proves $d(p,q)=d(q,p)$.
		\item Among the paths joining $p$ and $r$ are paths that travel through $q$. Specifically, given any path from $p$ to $q$ and any path from $q$ to $r$, we can concatenate them to get a path from $p$ to $r$.
	\end{enumerate}\up\n
	\proven
}

\par\noindent
Observe: The topology defined by $d$ is the same as the given topology on $M$.\n

\par\noindent
Sometimes, but not always, the infimum is attained, i.e., there exists a minimizing path. In fact, if such a path exists, it's always smooth.\n

\ex{
	Minimizing paths on the sphere are arcs of great circles -- intersections of the sphere with hyperplanes through the origin.\n
}

\ex{
	Minimizing paths don't always exist! Consider $M=\R^{2}\cut\set{0}$. For $p=(43,0)$ and $q=(-43,0)$, $d(p,q)=86$, but there's no path of that length between them (since you can't go through the origin).\n
}

\par\noindent
Refer to Do Carmo, Chapter 1, \sectionSymbol{}2 for more details.\n

\subsection*{Volume Element of an Oriented Riemannian Manifold}

\par\noindent
Reminder: An orientation on an orientable manifold $M$ is determined by a class of top-degree differential forms, $\nu$, with the defining property that $\forall{}p\in{}M$ and any $(v_{1},\ldots,v_{n})$, a positive basis of $T_{p}M$, $\nu_{p}(v_{1},\ldots,v_{n})>0$.\n

\par\noindent
In particular, $\nu$ is nowhere-vanishing. Conversely, a nowhere-vanishing top-degree form can be used to define positive bases, and in turn, an orientation.\n

\defn{
	If $M$ is a orientable Riemannian manifold, its \u{volume form} $\nu$ is defined by the property that $\forall{}p\in{}M$, $\forall(v_{1},\ldots,v_{n})$, a positive, orthonormal basis of $T_{p}M$, one has $\nu_{p}(v_{1},\ldots,v_{n})=1$.\n
}

\par\noindent
Observe: If this condition holds for some positive orthonormal basis, it holds for all positive orthonormal bases. The important calculation is as follows: Fix $p\in{}M$, $(v_{1},\ldots,v_{n})$ a positive orthonormal basis of $T_{p}M$, and $(e_{1},\ldots,e_{n})$ any other ordered basis of $T_{p}M$. Then we can write each $e_{i}=\sum_{l=1}^{n}a_{i}^{l}v_{l}$. For any top-degree form $\nu$, $\nu_{p}(e_{1},\ldots,e_{n})=\det(a_{i}^{l})\nu(v_{1},\ldots,v_{n})$, so if $(e_{1},\ldots,e_{n})$ is also positive and orthonormal, then $\det(a_{i}^{l})=1$.\proven

\subsection*{Computation of the Volume Form in Coordinates}

\par\noindent
Start with $(x_{1},\ldots,x_{n})$, a positive coordinate system with domain $U$. (Recall that this means $\forall{}p\in{}U$, $(\pdat{}{x^{1}}{p},\ldots,\pdat{}{x^{n}}{p})$ is a positive basis of $T_{p}M$.) Apply Gram-Schmidt to each basis (pointwise). We obtain vector fields $v_{1},\ldots,v_{n}$ on $U$ which are orthonormal at each point. And Gram-Schmidt shows the $v$'s are related to the partial derivatives by a smooth matrix, so $\forall{}j$, $v_{j}\in\mathfrak{X}(U)$. And, by possibly permuting the $v_{i}$'s, we can ensure it's a positive basis at every point.\n

\par\noindent
In fact, let's write $\pd{}{x^{i}}=\sum_{l}a_{i}^{l}v_{l}$. Then
\[
	g_{ij}=\iprod{\pdat{}{x^{i}}{p},\pdat{}{x^{j}}{p}}=\sum_{k,l}a_{i}^{l}a_{j}^{k}\underbrace{\iprod{v_{k},v_{l}}}_{=\delta_{k,l}}=\sum_{k}a_{i}^{k}a_{j}^{k}=AA\tpose\quad\ptxt{for}\quad{}A=(a_{i}^{k})
\]
So $\det(g_{ij})=\det(A)^{2}>0$. On the other hand, with our Riemannian volume form $\nu$,
\[
	\nu\paren{\pd{}{x^{1}},\ldots,\pd{}{x^{n}}}=\underbrace{\det(A)}_{\mathclap{=\sqrt{\det(g_{ij})}}}\underbrace{\nu(v_{1},\ldots,v_{n})}_{=1}=\sqrt{\det(g_{ij})}
\]
So in coordinates, $\nu=\sqrt{\det(g_{ij})}\,dx^{1}\wedge\cdots\wedge{}dx^{n}$.\n

\defn{
	The volume of a subset $U$ of a Riemannian manifold is $\Vol(U)=\int_{U}\nu$, where $\nu$ is the Riemannian volume form, if this integral is finite.\n
}

\lemma{
	For any coordinate system $(y^{1},\ldots,y^{n})$ on $U$, positive or not, the Riemannian integral
	\[
		\int_{U}\sqrt{\det\paren{\iprod{\pd{}{y^{i}},\pd{}{y^{j}}}}}\,\underbrace{dy^{1}\cdots{}dy^{n}}_{\mathclap{\ptxt{Riemann integral}}}
	\]
	is equal to $\Vol(U)$.\n
}

\par\noindent
This is true because the change of variables formula for a Riemann integral involves the absolute value of the Jacobian. So in the end, orientation is \i{not} needed to compute volumes of manifolds. In fact, we can even compute volumes of non-orientable manifolds! We generalize by using partitions of unity.

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