10_19.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 21}
\date{10/19/20} % Can also use \today

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

\par\noindent
Last time, we showed that a smooth vector field $X\in\mathfrak{X}(M)$ defines a derivation
\[
	\map{X:C^{\infty}(M)}{C^{\infty}(M)}{f}{(p\mapsto{}X_{p}([f]))}
\]
Here, we're thinking of $X$ as an operator, i.e., $f\mapsto{}X(f)$. Note that $\forall{}p\in{}M$, $X_{p}\in{}T_{p}M$.\n

\prop{
	The commutator of any two derivations $C^{\infty}(M)\to{}C^{\infty}(M)$ is a derivation.\n
	Proof: This is just an algebraic calculation.\n
}

\par\noindent
Today, we'll prove the converse -- that for any derivation $D$, there is a unique vector field $X\in\mathfrak{X}(M)$ such that $D=X$ (as an operator). So overall, we will have showed a one-to-one correspondence between derivations and vector fields. To do this, we need ``bump functions''.\n

\prop{
	Let $U\subseteq{}M$ open, $p\in{}U$. Then $\exists\chi\in{}C^{\infty}(M)$ \st
	\begin{enumerate}[topsep=5pt, itemsep=0pt, leftmargin=4\parindent, label=(\arabic*)]
		\item $\supp(\chi)=\closure{\set{q\in{}M:\chi(q)\ne{}0}}\subseteq{}U$ (and it is compact)
		\item $\exists{}V$ open with $p\in{}V$ such that $\restr{\chi}{V}\equiv{}1$.\n
		Note: (1) implies that $\closure{V}\subseteq{}U$.
	\end{enumerate}\up\n
	\b{Defn:} Such a $\chi$ is called a \u{bump function} at $p$.\nn
	Proof: It's enough to consider the case where $p=0\in\R^{n}$, as we can use a chart near $p$ to define $\chi$ in some neighborhood of $p$, and then extend $\chi$ to be $0$ outside that neighborhood.\n
	Start with the case where $n=1$ (i.e. $\R$). (See also \sectionSymbol{}13 in the book.) Start with
	\[
		f(x)=\left\{\begin{array}{ll}
			e^{-1/x} & x>0\\
			0 & x\le{}0
		\end{array}\right.
	\]
	We claim that $f$ is $C^{\infty}$ on $\R$. (This is because $\forall{}k\in\N$, $f^{(k)}(0)$ is defined.)\nn
	\hspace*{15pt}Note: $f$ is a famous example of a non-analytic function.\nn
	Next, let $g(x)=\frac{f(x)}{f(x)+f(1-x)}$. Note: $\forall{}x\in\R$, $f(x)+f(1-x)\ne{}0$, so $g$ is well-defined, and $C^{\infty}$. If $x\ge{}1$, then $f(1-x)=0$, so $g(x)=1$. If $x\le{}0$, $f(x)=0$, so $g(x)=0$.\nn
	Next, choose some $a,b\in\R_{>0}$ with $0<a^{2}<b^{2}$, and define $h(x)=g(\frac{x-a^{2}}{b^{2}-a^{2}})$. Then finally, take $\rho(x)=1-h(x^{2})$. Then we have $\restr{\rho}{[-a,a]}\equiv{}1$, and $\restr{\rho}{(-\infty,-b]\cup[b,\infty)}\equiv{}0$, and $\rho$ is $C^{\infty}$.\nn
	For $\R^{n}$, let $\chi(x)=\rho(\norm{x}^{2})$. Then $\supp\chi$ is a subset of a ball around the origin, and $\chi$ restricted to a smaller ball is always $1$.\proven
}

\defn{
	$D:C^{\infty}(M)\to{}C^{\infty}(M)$ is a \u{local operator} if $\forall{}f,g\in{}C^{\infty}(M)$, $\forall{}U\osubseteq{}M$, if $\restr{f}{U}=\restr{g}{U}$, then $\restr{D(f)}{U}=\restr{D(g)}{U}$.\n
}

\prop{
	A derivation $D:C^{\infty}(M)\to{}C^{\infty}(M)$ is a local operator.\nn
	Proof: By linearity of $D$, WOLOG $g\equiv{}0$. Assume that $\restr{f}{U}\equiv{}0$, and let $p\in{}U$. Let $\chi\in{}C^{\infty}(M)$ be a bump function at $p$ with $\supp(\chi)\subset{}U$. Note: $\chi\cdot{}f\equiv{}0$ on $M$, so $D(\chi{}f)=0$. Well, by the chain rule, $D(\chi{}f)=\chi{}D(f)+fD(\chi)$. If we evaluate at $p$, we have $f(p)=0$ and $\chi(p)=1$, so $0=0+D(f)(p)$, so $D(f)(p)=0$. Thus, $\restr{D(f)}{U}\equiv{}0$.\proven
}

\par\noindent
Note: One can show that every local (linear) operator is a differential operator.\n

\newpage
\thm{
	Let $D:C^{\infty}(M)\to{}C^{\infty}(M)$ be a derivation. Then $\exists{}X\in\mathfrak{X}(M)$ such that $D=X$ (as an operator).\nn
	Proof: Let $p\in{}M$. To define $X_{p}\in{}T_{p}M$, pick some $[f]\in{}C^{\infty}_{p}(M)$. Let $f:U\to\R$ represent this germ. Let $\chi$ be a bump function at $p$ with $\supp(\chi)\subseteq{}U$. Define $\tilde{f}:M\to\R$ where $\tilde{f}=\chi{}f$, i.e.,
	\[
		\tilde{f}(p)=\left\{\begin{array}{ll}
			\chi(p)f(p) & p\in{}U\\
			0 & p\in{}M\cut{}U
		\end{array}\right.
	\]
	Observe that $\tilde{}f\in{}C^{\infty}(M)$, and since $\tilde{f}$ agrees with $f$ in some open neighborhood $V$ of $p$, it's an extension of $\restr{f}{V}$. Define $X_{p}([f])=D(\tilde{f})(p)$. We need to justify that this is well-defined -- what if we changed our representation of $[f]$, or chose a different $\chi$? Is the number $D(\tilde{f})(p)$ invariant with respect to these changes? Yes! Under the above changes, there's no effect on the germ $[\chi{}f]\in{}C^{\infty}_{p}(M)$, and we just proved that $D$ is local.\nn
	Next, we need to show that $X$, as it's defined above, is smooth. Let $\phi=(x^{1},\ldots,x^{n})$ be any coordinate system on $U\subset{}M$. Then
	\[
		\restr{X}{U}=\sum_{j=1}^{n}X(x^{j})\pd{}{x^{j}}
	\]
	where $X(x^{j})$ is a function on $U$. We need to check that each $X(x^{j})$ is smooth. Again, we will use a bump function at $p\in{}U$. By definition, $X(x^{j})(p)=D(\tilde{x}^{j})(p)$, where $\tilde{x}^{j}=\chi\cdot{}x^{j}$ (extended by $0$ outside of $U$). And by our assumption, $D(\tilde{x}^{j})\in{}C^{\infty}(M)$.\nn
	We conclude that $X(x^{j})\in{}C^{\infty}(M)$, so $X$ is smooth.\proven
}

\cor{
	If $X,Y\in\mathfrak{X}(M)$, then $[X,Y]$ (treating $X$ and $Y$ as operators) is itself a vector field.\n
}

\end{document}