12_07.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 38}
\date{12/7/20} % Can also use \today

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

\section*{Sard's Theorem (with an Application)}

\par\noindent
As a preliminary, we have to talk about sets of measure $0$.\n

\defn{
	Informally speaking, $S\subset\R^{n}$ has \u{measure zero} iff $\forall\varepsilon>0$, $S$ can be covered by countably many $n$-cubes of total volume less than $\varepsilon$.\n
}

\prop{
	If $S$ has measure $0$, and $F:S\to\R^{m}$ is smooth, then $F(S)$ has measure $0$.\nn
	Proof: Based on the fact that $C^{\infty}$ functions are Lipschitz on compact sets. I.e., $\norm{F(p)-F(q)}<C\norm{p-q}$ for some constant $C\in\R_{>0}$.\n
}

\defn{
	A subset $S\subset{}M$ has \u{measure zero} iff $\forall(U,\phi)$, a coordinate chart, the set $\phi(U\cap{}S)\subseteq\R^{n}$ has measure zero.\n
}

\prop{
	Equivalently, $S$ can be covered by countably many charts $\set{(U_{i},\phi_{i})}$ \st{} $\forall{}i$, $\phi_{i}(U_{i}\cap{}S)$ has measure zero.\n
}

\thm{
	(Sard's Theorem) If $F:M\to{}N$ is smooth, the set of crtical values of $F$ has measure $0$.\n
}

\par\noindent
Reminder: $q\in{}N$ is a regular value iff $\forall{}q\in{}F\inv(p)$, $F_{*,p}$ is surjective.\n
\phantom{Reminder: }$q\in{}N$ is a critical value iff $q$ is not a regular value.\n

\par\noindent
Note: If $q\not\in\Im(F)$, then $q$ is a regular value.\n

\cor{
	The set of regular values of $F$ is dense in $N$. (It's the complement of a set of measure zero.) In particular, if $F:M\to{}N$ is smooth, and $\dim{}M<\dim{}N$, then the only regular values are $N\cut\Im(F)$, so we conclude that $N\cut\Im(F)$ is dense, and $\Im(F)$ has measure zero. In particular, submanifolds of nonzero codimension have measure zero.\n
}

\par\noindent
(Recall: A set $S$ is dense if $\forall{}U$ open, $U\cap{}S\ne\emptyset$.)\n

\section*{The Embedding Theorem}

\thm{
	(Whitney Embedding Theorem) Let $M$ be an $n$-dimensional manifold. Then $M$ can be embedded in $\R^{2n+1}$ and immersed in $\R^{2n}$. (This is the weak version.)\n
}

\thm{
	$M$ can be embedded in $\R^{2n}$. (This is the strong version.)\n
}

\par\noindent
Proof of the weak version: We start by embedding $M$ into some $\R^{N}$. Then we successively project $M\subset\R^{N}$ onto ``good hyperplanes''. The first step is to cover $M$ with an atlas $\set{(U_{i},\phi_{i})}_{i=1,\ldots,k}$. Let $\set{\chi_{i}}_{i=1,\ldots,k}$ be a subordinate partition of unity: $\forall{}i$, $\supp\chi_{i}\subset{}U_{i}$.\n

\par\noindent
Now, define
\[
	\forall{}i,\;\map{\psi_{i}:M}{\R^{n}}{p}{\left\{\begin{array}{ll}
		\chi_{i}(p)\phi_{i}(p) & p\in{}U_{i}\\
		0 & p\not\in{}U_{i}
	\end{array}\right.}
	\qquad
	\map{F:M}{\R^{kn+k\eqdef{}N}}{p}{(\psi_{1}(p),\ldots,\psi_{k}(p),\chi_{1}(p),\ldots,\chi_{k}(p))}
\]
We claim that $F$ is injective, and an immersion. Assume $p,q\in{}M$ \st{} $F(p)=F(q)$.\n
Then $\exists{}i$ \st{} $\chi_{i}(p)=\chi_{i}(q)\ne{}0$. Thus, $p,q\in{}U_{i}$. So $F(p)=F(q)\Rightarrow{}\psi_{i}(p)=\psi_{i}(q)\Rightarrow\phi_{i}(p)=\phi_{i}(q)\Rightarrow{}p=q$.\n

\par\noindent
Now, to show that $F$ is an immersion. Assume $v\in{}T_{p}M$ \st{} $F_{*,p}(v)=0$. Again, choose $i$ \st{} $\chi_{i}(p)\ne{}0$, so $p\in{}U_{i}$. Then
\[
	0=(\psi_{i})_{*,p}(v)=d\chi_{i}(v)\phi_{i}(p)+\chi_{i}(p)d\phi_{p}(v)
\]
since $\psi_{i}=\chi_{i}\cdot\phi_{i}$. But also $d\chi_{i}(v)=0$, so $d(\phi_{i})_{p}(v)=0$, so we must have $v=0$ (as $\phi_{i}$ is a diffeomorphism).\proven

\par\noindent
The next step is lowering the dimension. Let $\P=\set{l\subseteq\R^{N}\ptxt{ subspaces of dimension $1$}}$. $\forall{}l\in\P$, let $\pi_{l}:\R^{N}\to{}l^{\perp}$, where $l^{\perp}$ is the orthogonal complement (and a hyperplane). We claim that $\exists{}l\in\P$ \st{} $\restr{\pi_{l}}{M}:M\to{}l^{\perp}$ is an embedding, provided that $N>2n+1$.\n

\par\noindent
Well, we need to find an $l$ \st{} $\restr{\pi_{l}}{M}$ is injective and an immersion. Consider
\[
	\map{G:M\times{}M\cut\set{(p,p)\mid{}p\in{}M}}{\P}{\underbrace{(p,q)}_{p,q\in\R^{N}}}{\R\underbrace{(p-q)}_{\ne{}0}}
\]
$G$ is smooth. If $\dim(M\times{}M)<\dim(P)$, then $\P\cut\Im(G)$ is dense, so if we pick $l\in\P\cut\Im(G)$, then $\restr{\pi_{l}}{M}$ is injective. Because $dim(M\times{}M)=2n$, and we need $2n<N-1$, we require $N>2n+1$.\n

\par\noindent
Now, to ensure $\restr{\pi_{l}}{M}$ is an immersion, let
\[
	\map{H:SM\eqdef\set{(p,v)\in{}TM\mid{}\norm{v}=1}}{\P}{(p,v)}{\R{}v}
\]
We claim that if $l\in\P\cut\Im(H)$, then $\restr{\pi_{l}}{M}$ is an immersion. We perform a similar dimension count as before, and we get to $N>2n$.\proven

\end{document}