11_16.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 32}
\date{11/16/20} % Can also use \today

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

\subsection*{The Exterior Differential}

\thm{
	Let $M$ be a manifold. Then there exists a unique operator $d$
	\[
		\Omega^{0}\overset{d}{\xrightarrow{\hspace{0.5cm}}}\Omega^{1}\overset{d}{\xrightarrow{\hspace{0.5cm}}}\Omega^{2}\overset{d}{\xrightarrow{\hspace{0.5cm}}}\cdots\overset{d}{\xrightarrow{\hspace{0.5cm}}}\Omega^{k-1}\overset{d}{\xrightarrow{\hspace{0.5cm}}}\Omega^{k}
	\]
	\st{}
	\begin{enumerate}[itemsep=0pt, topsep=0pt, leftmargin=4\parindent, label=(\arabic*)]
		\item $d$ is $\R$-linear.
		\item $d:\Omega^{0}\to\Omega^{1}$ is the usual differential of a function.
		\item $d^{2}=0$.
		\item The anti-derivation property: $d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^{\deg\alpha}\alpha\wedge{}d\beta$. (This is also known as the super-symmetric version of Leibniz' rule.)
		\item If $F:M\to{}N$ is smooth, then $\forall\alpha\in\Omega^{k}N$, $d(F^{*}\alpha)=F^{*}(d\alpha)$. (This property is called ``Naturality''.)
	\end{enumerate}\up\n
}

\par\noindent
Note: In theory, the $d$ is a different operator for each $\Omega^{i}$. But it would be inconvenient to call them different names.\n

\par\noindent
Proof: We'll begin by proving this for $U\osubseteq\R^{n}$, and then use coordinates. This works because the anti-derivation property implies locality.\n

\par\noindent
For $\R^{n}$: If $\displaystyle\alpha=\sum_{I}a_{I}dx^{I}\in\Omega^{k}(\R^{n})$, then the properties imply that we must have $\displaystyle{}d\alpha=\sum_{I}\sum_{j=1}^{n}\pd{a_{I}}{x^{j}}dx^{j}\wedge{}dx^{I}$.\n
Use this as our definition of $d$ on $\R^{n}$. Now, we must check the $5$ properties. $1$ and $4$ follow directly from its definition. By linearlity, we can check this solely on ``monomials''.\n

\par\noindent
\b{Anti-derivation:} Let $\alpha=a\,dx^{I}$, $\beta=b\,dx^{J}$, $a,b\in{}C^{\infty}$. Then
\begin{align*}
	d(\alpha\wedge\beta) & =d(a\,dx^{I}\wedge{}b\,dx^{J})\\
	& =d(ab\,dx^{I}\wedge{}dx^{J})\\
	& =d(ab\,dx^{IJ})\\
	& =\sum_{j=1}^{n}\pd{(ab)}{x^{j}}\,dx^{j}\wedge{}dx^{IJ}\\
	& =\sum_{j=1}^{n}\brack{b\pd{a}{x^{j}}+a\pd{b}{x^{j}}}\,dx^{j}\wedge{}dx^{IJ}\\
	& =\sum_{j=1}^{n}\pd{a}{x^{j}}\,dx^{j}\wedge{}dx^{I}\wedge(b\,dx^{J})+a\sum_{j=1}^{n}\paren{\pd{b}{x^{j}}\,dx^{j}\wedge{}dx^{I}}\wedge{}dx^{J}\\
	& =\sum_{j=1}^{n}\pd{a}{x^{j}}\,dx^{j}\wedge{}dx^{I}\wedge(b\,dx^{J})+(-1)^{k}a\,dx^{I}\sum_{j=1}^{n}\pd{b}{x^{j}}\,dx^{j}\wedge{}dx^{I}\\
	& =d\alpha\wedge\beta+(-1)^{k}\alpha\wedge{}d\beta
\end{align*}
\n

\newpage
\par\noindent
\b{$\mathbf{d^{2}=0}$:}
\begin{align*}
	d\brack{d(a\,dx^{I})} & =d\brack{\sum_{j=1}^{n}\pd{a}{x^{j}}\,dx^{j}\wedge{}dx^{I}}\\
	& \labeledeq{(1)}\sum_{j=1}^{n}d\paren{\pd{a}{x^{j}}}\wedge{}dx^{j}\wedge{}dx^{I}\\
	& =\underbrace{\paren{\sum_{j=1}^{n}\sum_{l=1}^{n}\frac{\d^{2}a}{\d{}x^{l}\d{}x^{j}}\,dx^{l}\wedge{}dx^{j}}}_{\mathclap{\ptxt{Observe: Terms with $j=l$ vanish.}}}\wedge{}dx^{I}\\
	& =\paren{\sum_{1\le{}j<l\le{}n}\frac{\d^{2}a}{\d{}x^{l}\d{}x^{j}}\,dx^{l}\wedge{}dx^{j}+\frac{\d^{2}a}{\d{}x^{j}\d{}x^{l}}\,dx^{j}\wedge{}dx^{l}}\wedge{}dx^{I}\\
	& =\paren{\sum_{1\le{}j<l\le{}n}\paren{\frac{\d^{2}a}{\d{}x^{l}\d{}x^{j}}-\frac{\d^{2}}{\d{}x^{j}\d{}x^{l}}}\,dx^{l}\wedge{}dx^{j}}\wedge{}dx^{I}\\
	& \labeledeq{(2)}0
\end{align*}
with (1) because $dx^{j}\wedge{}dx^{I}$ is constant, so $d(dx^{j}\wedge{}dx^{I})=0$, and (2) by Clairaut's Theorem, which tells us that\break $\frac{\d^{2}a}{\d{}x^{l}\d{}x^{j}}=\frac{\d^{2}a}{\d{}x^{j}\d{}x^{l}}$, so their difference is $0$.\n

\par\noindent
\b{Naturality:} Say $F:V\to{}U$ is $C^{\infty}$, $V\subseteq\R^{m}$, $U\subseteq\R^{n}$. Let $\alpha\in\Omega^{k}(U)$ of the form $\alpha=a\,dx^{I}$. We need to prove that $d(F^{*}\alpha)=F^{*}(d\alpha)$. Say $F=(F^{1},\ldots,F^{n})$ with $F^{j}:V\to\R$. We proved that
\[
	F^{*}(dx^{i_{1}})\wedge\cdots\wedge{}F^{*}(dx^{i_{k}})=F^{*}(dx^{I})=dF^{i_{1}}\wedge\cdots\wedge{}dF^{i_{k}}\eqdef{}dF^{I}
\]
So $F^{*}\alpha=(\alpha\of{}F)F^{*}(dx^{I})=(\alpha\of{}F)dF^{I}$, and thus,
\[
	d(F^{*}\alpha)=d(\alpha\of{}F)\wedge{}dF^{I}+(\alpha\of{}F)\underbrace{d(dF^{I})}_{=0}=d(\alpha\of{}F)\wedge{}dF^{I}
\]
\[
	F^{*}(d\alpha)=F^{*}\brack{\sum_{j=1}^{n}\pd{a}{x^{j}}\,dx^{j}\wedge{}dx^{I}}=\sum_{j=1}^{n}F^{*}\paren{\pd{a}{x^{j}}}\,dx^{j}\wedge{}dx^{I}=\sum_{j=1}^{n}\paren{\pd{a}{x^{j}}\of{}F}\underbrace{\paren{F^{*}(dx^{j})}}_{dF^{j}}\wedge{}dF^{I}=d(\alpha\of{}F)\wedge{}dF^{I}
\]
So we conclude that $d(F^{*}\alpha)=F^{*}(d\alpha)$.\n

\par\noindent
Now, we use coordinates to do it on a manifold.\n

\defn{
	Given $\alpha\in\Omega^{k}(M)$, $\phi:U\to\R^{n}$ a coordinate system, define $d\alpha\in\Omega^{k+1}(U)$ by $\phi^{*}\brack{d(\phi\inv)^{*}\alpha}$, where $d$ is the Euclidean version, and $(\phi\inv)^{*}\alpha\in\Omega^{k}(\phi(U))$.\n
}

\par\noindent
Now, we need to prove that $d\alpha$ on $U$ is independent of choice of coordinates. Suppose $\psi:V\to\R^{n}$ is another coordinate chart (WOLOG $U\cap{}V\ne\emptyset$). Call $U$ the intersection of their domains. Write $\psi=F\of\phi$
\[
	\begin{tikzcd}
		&[-15pt] U \arrow[dl, "\phi", swap] \arrow[dr, "\psi"] &[-15pt]\\
		\phi(U) \arrow[rr, "F", swap] & & \psi(U)
	\end{tikzcd}
\]
Then
\begin{align*}
	\psi^{*}\brack{d(\psi\inv)^{*}\alpha} & =\phi^{*}\of{}F^{*}\brack{d(\phi\inv\of{}F\inv)^{*}\alpha}\\
	& =\phi^{*}\of{}F^{*}\brack{d((F\inv)^{*}\of(\phi\inv)^{*})\alpha}\\
	& =\phi^{*}\of{}F^{*}\of(F\inv)^{*}\brack{d(\phi\inv)^{*}\alpha}\\
	& =\phi^{*}\brack{d(\phi\inv)^{*}\alpha}
\end{align*}
\n

\par\noindent
As for showing uniqueness on manifolds, use that anti-derivation implies locality, as with vector fields.\proven

\ex{
	$\R^{3}$
	\begin{align*}
		d(f\,dx+g\,dy+h\,dz) & =df\wedge{}dx+dg\wedge{}dy+dh\wedge{}dz\\
		& =(\d_{x}f\,dx+\d_{y}f\,dy+\d_{z}f\,dz)+\cdots\\
		& =\d_{y}f\,dy\wedge{}dx+\d_{z}f\,dz\wedge{}dx+\cdots\\
		& =(\d_{x}g-\d_{y}f)\,dx\wedge{}dy+(\d_{x}h-\d_{z}f)\,dx\wedge{}dz+\cdots
	\end{align*}
	We get the components of the curl of the vector field $<f,g,h>$.\n
}

\defn{
	$\alpha\in\Omega^{k}(M)$ is \u{closed} if $d\alpha=0$, and \u{exact} if $\exists\beta\in\Omega^{k-1}(M)$ such that $\alpha=d\beta$.\n
}

\par\noindent
Note: $d^{2}=0$ means exact implies closed.\n

\defn{
	The \u{$k$th DeRahm Cohomology} $H^{k}(M)=\ker(d:\Omega^{k}\to\Omega^{k+1})/\im(d:\Omega^{k-1}\to\Omega^{k})$ (with the vector space quotient).\n
}

\ex{
	If $M=S^{1}$, $d\theta$ dual to $\pd{}{\theta}$ is closed. (Any top-degree form is closed.)\n
	Claim: $d\theta$ is not exact, and in fact, $H^{1}(S^{1})=\R$.\n
}

\end{document}