10_14.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 19}
\date{10/14/20} % Can also use \today

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

\thm{
	The image of an embedding $F:M\to{}N$ is a regular submanifold of $N$.\nn
	Proof: Let $q\in{}F(M)$. We need to show that there are coordinates of $N$ near $q$ adapted to $F(M)$. Let $p\in{}M$ \st{} $F(p)=q$. By the immersion theorem, there are corodinates $(U,\phi=(x^{1},\ldots,x^{m}))$ of $M$ with $p\in{}U$, and $(V,\psi=(y^{1},\ldots,y^{n}))$ of $N$ with $q\in{}V$, with $U\subseteq{}F\inv(V)$, such that $\tilde{F}(I)=(I,0)$ (with $m-n$ zeros).
	\[
		\begin{tikzcd}[ampersand replacement=\&]
			\&[-15pt] U \arrow[r, "F"] \arrow[d, "\phi"] \& V \arrow[d, "\psi"] \&[-15pt]\\
			\R^{m} \arrow[r, symbol=\supseteq] \& \phi(U) \arrow[r, "\tilde{F}"] \& \psi(V) \arrow[r, symbol=\subseteq] \& \R^{n}
		\end{tikzcd}
	\]
	Well, $F(U)=\set{w\in{}V\mid{}y^{j}(w)=0,\forall{}j\in\set{1,\ldots,n}}$. The point is that because $F$ is an embedding (i.e. $F|^{F(M)}:M\to{}F(M)$ is a homeomorphism), then $F(U)$ is a relatively open set of $F(M)$. So there is an open set $W\subseteq{}N$ such that $F(U)=F(M)\cap{}W$. Therefore, $F(M)\cap{}W\set{w\inv{}V\mid{}y^{j}(w)=0,\forall{}j\in\set{1,\ldots,n}}$. So $\psi$ on $V\cap{}W$ is adapted to $F(M)$ at $q$.\proven
}

\par\noindent
We want a stronger statement for when an injective immersion is an embedding.\n

\prop{
	Let $F:M\to{}N$ be a continuous proper map between manifolds. Then $F$ is closed.\nn
	Proof: Let $C\subseteq{}M$ be a closed set, and $q\in\closure{F(C)}$. We need to show $q\in{}F(C)$. Well, let $V\subseteq{}N$ be an open neighborhood of $q$ such that $\closure{V}$ is compact. (We can do this because Euclidean spaces are locally compact.) Observe then that $q\in\closure{F(C)\cap\closure{W}}$. $F\inv(\closure{V})$ is compact, since $F$ is proper, so $C\cap{}F\inv(\closure{V})$ is compact. Therefore, $F(C\cap{}F\inv(\closure{V}))$ is compact, so it's closed. But $F(C\cap{}F\inv(\closure{V}))=F(C)\cap\closure{V}$ (this is a set-theoretic fact). SO $F(C)\cap\closure{V}$ is closed. Since $q\in\closure{F(C)\cap\closure{V}}$, $q\in{}F(C)\cap\closure{V}$, so $q\in{}F(C)$. Thus, $F$ is closed.\proven
}

\subsection*{Vector Fields}

\defn{
	Let $M$ be a smooth manifold of dimension $n$. A \u{vector field on $M$} is a secion of $TM\overset{\pi}{\to}M$. That is, it's a map $X:M\to{}TM$ \st{} $\forall{}p\in{}M$, $X_{p}\in{}T_{p}M$.\n
}

\par\noindent
This is a slight abuse of notation. Update: Write $X(p)=(p,X_{p})$. The $p$ is the same because $X$ is a section $\pi\of{}X=I_{M}$.\n

\par\noindent
Basically, a vector field on a manifold assigns a tangent vector to every point in a manifold.\n

\defn{
	A vector field $X$ on $M$ is \u{smooth} ($C^{\infty}$) iff it is smooth as a map between manifolds $X:M\to{}TM$.\n
}

\par\noindent
What does this mean (concretely)?\n

\par\noindent
Let's think about smooth sections of vector bundles in general (e.g. $TM$ and $T^{*}M$). Let
\[
	\begin{tikzcd}
		\mathcal{E} \arrow[d,"\pi"]\\
		M
	\end{tikzcd}
\]
be a $C^{\infty}$ vector bundle of rank $\rho\in\N$. This means $\mathcal{E}$ is a manifold, $\pi$ is $C^{\infty}$ and onto, $\forall{}p\in{}M$, $\pi\inv(p)$ has the structure of a vector space over $\R$ of dimension $\rho$, and there exists a family $\set{(U_{\alpha},\chi_{\alpha})}$ such that $\set{U_{\alpha}}$ is an open cover of $M$ and $\forall\alpha$, $\chi_{\alpha}$ is a diffeomorphism, and we have the vector bundle trivialization
\[
	\begin{tikzcd}
		\pi\inv(U_{\alpha}) \arrow[rr, "\chi_{\alpha}"] \arrow[dr, "\pi"] &[-10pt] &[-10pt] U_{\alpha}\times\R^{\rho} \arrow[dl]\\
		& U_{\alpha}
	\end{tikzcd}
\]
\st{} $\chi_{\alpha}$ is linear on fibers, i.e., $\forall{}p\in{}U_{\alpha}$, we have
\[
	\restr{\chi_{\alpha}}{\pi\inv(p)}:\pi\inv(p)\overset{\cong}{\to}\set{p}\times\R^{\rho}\cong\R^{\rho}
\]
The mapping from $\pi\inv(p)$ to $\R^{\rho}$ is a linear isomorphism.\n

\ex{
	Let $\mathcal{E}=TM$. The trivializations are induced by coordinate charts. Let $U\subseteq{}M$, with $\phi=(x^{1},\ldots,x^{n}):U\to\R^{n}$ be a coordinate chart. Then $\forall{}p\in{}U,i\in\set{1,\ldots,n}$, $\pdat{}{x^{i}}{p}\in{}T_{p}U=T_{p}M$. We have
	\[
		\begin{tikzcd}[ampersand replacement=\&]
			\pi\inv(U) \arrow[rr,"\chi"] \arrow[dr, "\pi"] \&[-10pt] \&[-10pt] U\times\R^{n} \arrow[dl]\\
			\& U
		\end{tikzcd}
	\]
	defined by
	\[
		\chi\inv(p,v)=\paren{p,\sum_{i=1}^{n}v^{i}\pdat{}{x^{i}}{p}}
	\]
	with $p\in{}U$ and $v=(v^{1},\ldots,v^{n})\in\R^{n}$.\n
	We proved that $\forall{}p\in{}M$, $\set{\pdat{}{x^{i}}{p}:i=1,\ldots,m}$ is a basis of $T_{p}M$. So $\chi\inv$ is invertible and a linear isomorphism of the fibers.\n
	(The definition of the smooth structure on $TM$ is such that $\chi$ as above is a diffeomorphism.)\n
}

\ex{
	$\displaystyle T^{*}M=\bigcup_{p\in{}M}\set{p}\times{}T_{p}^{*}M$\n
	Again, a coordinate chart on $M$ induces a trivialization.\n
	Use: $\forall\alpha\in{}T_{p}^{*}M$, $\alpha=\sum_{i=1}^{n}\alpha_{i}\restr{dx^{i}}{p}$, where $\restr{dx^{i}}{p}$ is dual to $\pdat{}{x^{i}}{p}$, and $\set{\restr{dx^{i}}{p}\mid{}i=1,\ldots,n}$ is a basis of $T_{p}^{*}M$.\n
}

\par\noindent
Back to
\[
	\begin{tikzcd}
		E \arrow[d,"\pi"]\\
		M \arrow[u, bend right=60, "s", swap]
	\end{tikzcd}
\]
Let $s$ be a section (i.e. $\pi\of{}s=I_{M}$). When is $s$ smooth? We will answer this question using trivializations. Given a trivialization $\chi$,
\[
	\begin{tikzcd}
		\pi\inv(U) \arrow[rr, "\chi"] \arrow[dr, "\pi"] & & U\times\R^{p} \arrow[dl, "\pi"]\\
		& U \arrow[ul, bend right=30, "\restr{s}{U}", swap, pos=0.3] \arrow[ur, bend right=30, "\hspace{-10pt}\chi\of(\restr{s}{U})\eqdef{}s_{\chi}", swap]
	\end{tikzcd}
\]
We have $\forall{}p\in{}U$, $s_{\chi}(p)=(p,\star)$, where $\star$ is given by a function $F:U\to\R^{\rho}$. $F$ is just the projection onto $\R^{\rho}$ composed with $s_{\chi}$. So $\forall{}p\in{}U$, $s_{\chi}(p)=(p,F(p))$.\n

\par\noindent
Claim: $\restr{s}{U}$ is smooth iff $F:U\to\R^{\rho}$ is smooth.\nn
Proof: Next time...

\end{document}