04_14.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 36}
\date{4/14/21} % Can also use \today

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

\par\noindent
Recall from last time: Let $M$ be a Riemannian manifold. We defined the differential operators $\del$ (gradient) and $\div$ (divergence), and we have
\[
	\begin{tikzcd}
		C^{\infty}(M) \arrow[r,shift left, "\del"] \arrow[d, symbol={=}] & \mf{X}(M) \arrow[l,shift left, "\scalebox{0.75}{$\begin{smallmatrix}\div\\ \ptxt{or}\\ -\div\end{smallmatrix}$}"] \arrow[d,symbol=\cong,"\,\,\ptxt{(metric dual)}"]\\
		C^{\infty}(M) \arrow[r, shift left, "d"] & \Omega^{1}(M) \arrow[l, shift left, "\scalebox{0.75}{$\begin{smallmatrix}-\delta\\ \ptxt{or}\\ \delta\end{smallmatrix}$}"]
	\end{tikzcd}
\]
``$\delta=-\div$ on the differential form side''\n

\par\noindent
We also defined the Laplacian on functions $\Laplacian:C^{\infty}(M)\to{}C^{\infty}(M)$ by $\Laplacian=\delta\of{}d$ iff $\Laplacian{}f=-\div(\del{}f)$.\n

\subsection*{$l^{2}$ Inner Products}

\defn{
	Assume $M$ is oriented. $\forall{}f,g\in{}C^{\infty}(M)$, we define the \u{$l^{2}$ inner product} by
	\[
		\iprod{f,g}_{l^{2}}=\int_{M}fg\,\,d\Vol
	\]
}

\par\noindent
We can extend this to sections of real vector bundles over $M$, $\mc{E}\overset{\pi}{\to}M$. Put a Euclidean structure on the fibers of $\mc{E}$: $\forall{}p\in{}M$, $\giprod_{p}$ is a Euclidean inner product on $\mc{E}_{p}=\pi\inv(p)$, varying smoothly with $p$.\n

\defn{
	$\forall{}s,t\in\Gamma_{0}(\mc{E})$ (compactly supported sections). Then we define the \u{$l^{2}$ inner product} by
	\[
		\iprod{s,t}_{l^{2}}=\int_{M}\underbrace{\iprod{s(p),t(p)}_{p}}_{\ptxt{function of $p$}}d\Vol
	\]
}

\par\noindent
Consider the case $\mc{E}=\bigwedge^{k}(T^{*}M)$. Then the Euclidean structure on $\bigwedge^{k}(T^{*}M)$ is induced by the Riemannian metric. For $k=1$, we simply have $T^{*}M\cong{}TM$ by the metric dual. For general $k$, $\forall{}p\in{}M$, let $V=T_{p}^{*}M$. We define an inner product on $\bigwedge^{k}V$:
\[
	(v_{1}\wedge\cdots\wedge{}v_{k},w_{1}\wedge\cdots\wedge{}w_{k})\eqdef\det\paren{\iprod{v_{i},w_{j}}}_{ij}
\]
Check: If $(e_{1},\ldots,e_{n})$ is an orthonormal basis of $V$, then $\set{e_{i_{1}}\wedge\cdots\wedge{}e_{i_{k}}\mid{}i_{1}<\cdots<i_{k}}$ is an orthonormal basis of $\bigwedge^{k}V$.\n

\par\noindent
In this way, we get the notion of an $l^{2}$ inner product of any two $k$-forms $\alpha,\beta\in\Omega^{k}(M)$ by
\[
	\iprod{\alpha,\beta}_{l^{2}}=\int_{M}\iprod{\alpha_{p},\beta_{p}}_{p}d\Vol
\]

\prop{
	$\forall{}f\in{}C^{\infty}(M),X\in\mf{X}(M)$, one has
	\[
		\iprod{\del{}f,X}_{l^{2}}=-\iprod{f,\div{}X}_{l^{2}}
	\]
	That is, $\forall{}f\in\Omega^{0}(M),\alpha\in\Omega^{1}(M)$,
	\[
		\iprod{df,\alpha}_{l^{2}}=\iprod{f,\delta\alpha}_{l^{2}}
	\]
	That is, $\delta=d^{*}$, the adjoint of $d$, so $\Laplacian=d^{*}d$.\nn
	Proof: Start with $\mc{L}_{fX}(d\Vol)=f\mc{L}_{X}(d\Vol)+Xf$. Now integrate:
	\[
		\int_{M}\mc{L}_{fX}(d\Vol)=\int_{M}\div(fX)d\Vol=0
	\]
	because $\d{}M=\emptyset$. So we have
	\[
		0=\int_{M}f\div(X)d\Vol+\int_{M}\underbrace{\iprod{X,\del{}F}}_{\hspace{-35pt}\mathrlap{X(f)=df(X)=\iprod{\del{}f,X}}}d\Vol
	\]
	So $0=\iprod{f,\div{}X}_{l^{2}}+\iprod{X,\del{}f}_{l^{2}}$.\proven
}

\cor{
	$\iprod{\Laplacian{}f,g}_{l^{2}}=\iprod{f,\Laplacian{}g}_{l^{2}}$.\n
}

\par\noindent
Now, generalize to $\Omega^{k}$. (The previous discussion was for $k=0$.)
\[
	\begin{tikzcd}
		\Omega^{k} \arrow[r, shift left, "d"] & \Omega^{k+1} \arrow[l, shift left, "\delta=d^{*}=?"]
	\end{tikzcd}
\]
Is there a $\delta$? What is it?\n

\par\noindent
In local coordinates, $\delta$ is \i{also} a differential operator of degree $1$. Try integrating by parts!\n

\par\noindent
Preliminary linear algebra: the Hodge star operator. Let $V$ be an $n$-dimensional vector space, oriented, with an inner product. We claim that $\forall{}k$, there exists $\star:\bigwedge^{k}V\to\bigwedge^{n-k}V$ linear such that for any positive oriented basis $(e_{1},\ldots,e_{n})$ of $V$, $\star(e_{1}\wedge\cdots\wedge{}e_{k})=e_{k+1}\wedge\cdots\wedge{}e_{n}$.\n

\ex{
	For $V=\R^{3}$ with the standard orientation,
	\[
		\map{\star:\bigwedge^{2}V}{\bigwedge^{1}V}{dx^{1}\wedge{}dx^{2}}{dx^{3}}
	\]
	(Now do it cyclically.)\n
}

\par\noindent
Note: $\dim\bigwedge^{k}=\binom{n}{k}=\binom{n}{n-k}=\dim\bigwedge^{n-k}$.\n

\par\noindent
Observe: On $\R^{3}$ in the calc 3 context, for $X\in\mf{X}(\R^{3})$, we define
\[
	\curl{}X=\del\times{}X\in\mf{X}(M)
\]
What is this object? Well,
\[
	\begin{tikzcd}
		\mf{X}(\R^{3}) \arrow[r,symbol=\cong] &[-15pt] \Omega^{1}(\R^{3}) \arrow[r,"d"] \arrow[r, bend right=30, ""{coordinate, name=Z1}, phantom] & \Omega^{2}(\R^{3}) \arrow[r,"\star"] \arrow[r, bend right=30, ""{coordinate, name=Z2}, phantom] & \Omega^{1}(\R^{3}) \arrow[r,symbol=\cong] &[-15pt] \mf{X}(\R^{3})
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			-- (\tikztostart.south)
			|- (Z1)
			-- (Z2)\tikztonodes
			-| (\tikztotarget.south)
			-- (\tikztotarget.south)
		}, "\curl"']
	\end{tikzcd}
\]
Note that this only works for $\dim{}=3$.\n

\par\noindent
Some properties of $\star$:
\begin{enumerate}
	\item We have
	\[
		\begin{tikzcd}
			\bigwedge^{k} \arrow[r,"\star"] \arrow[r, bend right=45, ""{coordinate, name=Z1}, phantom] & \bigwedge^{n-k} \arrow[r,"\star"] \arrow[r,bend right=45, ""{coordinate, name=Z2}, phantom] & \bigwedge^{k}
			\arrow[from=1-1, to=1-3, to path={
				-- (\tikztostart.south)
				|- (Z1)
				-- (Z2)\tikztonodes
				-| (\tikztotarget.south)
				-- (\tikztotarget.south)
			}, "(-1)^{k(n-k)}\Id"']
		\end{tikzcd}
	\]
	because
	\[
		e_{1}\wedge\cdots\wedge{}e_{k}\overset{\star}{\mapsto}e_{k+1}\wedge\cdots\wedge{}e_{n}\overset{\star}{\mapsto}(-1)^{\sigma}e_{1}\wedge\cdots\wedge{}e_{k}
	\]
	``$n-k$ signs, $k$ times''.
	\item $\star:\bigwedge^{n}V\to\bigwedge^{0}V=\R$ has $\star(\Vol)=1$.
	\item $\forall\alpha,\beta\in\bigwedge^{k}V$, $\iprod{\alpha,\beta}=\star(\alpha\wedge(\star\beta))\in\R$.
\end{enumerate}

\cor{
	Apply/extend $\star$ to forms on a compact, oriented, Riemannian manifold $M$ (with $\dim{}M=n$), $\Omega^{k}(M)$, by acting pointwise: $\star:\Omega^{k}(M)\to\Omega^{n-k}(M)$. Note: $\forall\alpha,\beta\in\Omega^{k}(M)$, $\iprod{\alpha,\beta}_{l^{2}}=\int_{M}\alpha\wedge(\star\beta)$.\n
}

\par\noindent
Back to our main question:\n

\prop{
	The adjoint $\delta$ of $d:\Omega^{k}\to\Omega^{k+1}$ is $\delta=(-1)^{nk+1}\star{}d\star$.\n
}

\par\noindent
Note: If $\beta\in\Omega^{k+1}$, $\star\beta\in\Omega^{n-k-1}$, so $d\star\beta\in\Omega^{n-k}$, so $\star{}d\star\beta\in\Omega^{k}$. Superficially, $\delta=\star{}d\star:\Omega^{k+1}\to\Omega^{k}$. Now, we prove it:\n

\par\noindent
Proof: Let $\alpha\in\Omega^{k},\beta\in\Omega^{k+1}$. We want to show $\iprod{d\alpha,\beta}_{l^{2}}=\iprod{\alpha,\delta\beta}_{l^{2}}$. We'll use integration by parts. Starting with the fact that $0=\int_{M}d(\alpha\wedge\star\beta)$, because $\alpha\wedge\star\beta$ is a $n-1$ form, so $d(\alpha\wedge\star\beta)$ is a top-degree form. By Stokes' theorem, since we have an empty boundary, this integral is $0$. Well,
\[
	0=\int_{M}d(\alpha\wedge\star\beta)=\underbrace{\int_{M}d\alpha\wedge\star\beta}_{=\iprod{d\alpha,\beta}_{l^{2}}}+(-1)^{k}\int_{M}\alpha\wedge(d\star\beta)
\]
So
\[
	\iprod{d\alpha,\beta}_{l^{2}}=(-1)^{?}\int_{M}\alpha\wedge(d\star\beta)=(-1)^{?}\int_{M}\alpha\wedge(\star\star)d\star\beta=(-1)^{?}\iprod{\alpha,\star{}d\star\beta}_{l^{2}}=(-1)^{?}\iprod{\alpha,\delta\beta}
\]
(We didn't do the sign computations, but they do work out.)\proven

\defn{
	The Laplacian on forms $\Laplacian:\Omega^{k}(M)\to\Omega^{k}(M)$ is $\Laplacian=\delta{}d+d\delta$.\n
}

\end{document}