04_19.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 38}
\date{4/19/21} % Can also use \today

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

\par\noindent
Recall: The Laplace-Beltrami operator on forms: $\Laplacian$ on $\Omega^{k}$, where $\Laplacian=d\delta+\delta{}d$.\n

\ex{
	On $\R^{n}$, $\Laplacian$ on functions (i.e. $\Omega^{0}=C^{\infty}(\R^{n},\R)$). Then $\Laplacian=-\sum_{i=1}^{n}\frac{\d^{2}}{\d{}x_{i}^{2}}$. (This is using our sign convention.)\nn
	For $\omega\in\R^{n}$, $\Laplacian(e^{i\omega\cdot{}x})=\norm{\omega}^{2}e^{i\omega\cdot{}x}$.\n
}

\par\noindent
For forms $f_{I}dx^{I}$, where $I=\set{i_{1}<\cdots<i_{k}}$ is a multi-index, the Laplacian is $\Laplacian(f_{I}dx^{I})=(\Laplacian{}f_{I})dx^{I}$.\n

\par\noindent
Throughout our discussion today, we'll assume $M$ is compact and oriented.\n

\ex{
	$\mb{T}=\R^{n}/\Lambda$, where $\Lambda=\set{\sum{}k_{i}e_{i},k_{i}\in\Z,e_{1},\ldots,e_{n}\;\ptxt{linearly independent}}$. $\Laplacian_{\R^{n}}$ makes sense on $C^{\infty}(\mb{T})$. If $\omega\in\R^{n}$ is such that $\forall{}k\in\Lambda$, $\omega\cdot{}k\in{}2\pi\Z$, then $x\mapsto{}e^{i\omega\cdot{}x}$ is periodic w.r.t. $\Lambda$, so it's $C^{\infty}(\mb{T})$.\n
}

\thm{
	The eigenvalues of $\Lambda_{\mb{T}}$ on functions are $\norm{\omega}^{2}$ for $\omega$ satisfying the above condition. Such $\omega$ constitute the \i{dual lattice} $\Lambda^{*}$.\n
}

\thm{
	For $S^{2}$, distinct eigenvalues are $k(k+1)$ for $k\in\set{0,1,2,\ldots}$, with corresponding muliplicities $2k+1$. That is, listed, with multiplicities, we have eigenvalues
	\[
		\underbrace{0\vphantom{,}}_{k=0},\underbrace{2,2,2}_{k=1},\underbrace{6,6,6,6,6}_{k=2},\ldots
	\]
}

\par\noindent
In the general case, we have\n

\thm{
	(Spectral Theorem for the Laplacian on Forms) Let $M$ be a compact, oriented manifold. There exists an orthonormal basis (in the $l^{2}$ sense) of $\Omega^{k}$, $\set{\alpha_{j}:j=0,1,2,\ldots}$ an real eigenvalues $\lambda_{0}\le\lambda_{1}\le\lambda_{2}\le\cdots\to+\infty$ such that $\Laplacian\alpha_{j}=\lambda_{j}\alpha_{j}$. Moreover, $\lim_{j\to\infty}\lambda_{j}=+\infty$, i.e., the multiplicities are all finite.\n
}

\par\noindent
This yields the spectral decomposition of $\alpha\in\Omega^{k}$:
\[
	\alpha=\alpha_{H}+\sum_{\lambda>0}\alpha_{\lambda}\qquad\Laplacian\alpha_{\lambda}=\lambda\alpha_{\lambda},\Laplacian\alpha_{H}=0
\]
$\alpha_{H}$ is the ``harmonic part'', and the $\lambda$'s are all distinct eigenvalues. This decomposition is unique and orthogonal.\n

\par\noindent
Note: Because this is an infinite-dimensional space, there's a lot of analysis happening behind the scenes.\n

\par\noindent
Observe: $\alpha$ is smooth iff $\norm{\alpha_{\lambda}}_{l^{2}}=0$ (really $\lambda^{-\infty}$). In other words, it decays rapidly as $\lambda$ grows.\n

\defn{
	$G:\Omega^{k}\to\Omega^{k}$ is defined by $G(\alpha)=\sum_{\lambda>0}\frac{1}{\lambda}\alpha_{\lambda}$.\n
}

\par\noindent
Observe: $G$ is smooth.\n

\par\noindent
By definition,
\[
	\Laplacian{}G(\alpha)=\sum_{\lambda>0}\frac{1}{\lambda}\Laplacian\alpha_{\lambda}=\sum_{\lambda>0}\frac{1}{\lambda}\lambda\alpha_{\lambda}=\sum_{\lambda>0}\alpha_{\lambda}=\alpha-\alpha_{H}
\]

\lemma{
	$\comm{G,d}=0$ and $\comm{G,\delta}=0$.\nn
	Why should this be true? We've seen that $\comm{\Laplacian,d}=0$ and $\comm{\Laplacian,\delta}=0$. So $\forall\alpha_{\lambda}$, $\Laplacian\alpha_{\lambda}=\lambda\alpha_{\lambda}$. $d\alpha_{\lambda}$ is either $0$ or an eigenform of $\Laplacian$. So we get the commutative diagram
	\[
		\begin{tikzcd}[ampersand replacement=\&]
			\Omega^{k} \arrow[r, "d"] \arrow[d, "G"'] \& \Omega^{k+1} \arrow[d, "G"]\\
			\Omega^{k} \arrow[r, "d"] \& \Omega^{k+1}
		\end{tikzcd}
	\]
}

\cor{
	(Hodge Decomposition Theorem) $\Omega^{k}=\mc{H}^{k}\oplus{}d(\Omega^{k-1})\oplus\delta(\Omega^{k+1})$, where $\mc{H}^{k}=\ker(\Laplacian:\Omega^{k}\to\Omega^{k})$ is the set of harmonic forms.\n
}

\par\noindent
This decomposition is an orthogonal direct sum with respect to the $l^{2}$ inner prouct. In fact, if $\alpha\in\Omega^{k}$,
\[
	\alpha=\alpha_{H}+\Laplacian{}G(\alpha)=\alpha_{H}+\underbrace{d(\delta{}G(\alpha))}_{\in{}d(\Omega^{k-1})}+\underbrace{\delta(\overbrace{d(G(\alpha))}^{G(d\alpha)})}_{\in{}\delta(\Omega^{k+1})}
\]

\cor{
	If $\alpha\in\Omega^{k}$ is such that $d\alpha=0$, then $[\alpha]=[\alpha_{H}]$ (where $[\,\cdot\,]$ is the equivalence class in deRham cohomology).\nn
	Proof:
	\[
		\alpha=\alpha_{H}+d(\delta{}G(\alpha))+\delta(\underbrace{dG(\alpha)}_{G(d\alpha)=0})=\alpha_{H}+d(\delta{}G(\alpha))
	\]
	\proven
}

\par\noindent
Recall: $\Laplacian\alpha=0$ iff $d\alpha=0$ and $\delta\alpha=0$. In fact, $\forall{}c\in{}H^{k}_{\dR}(M)$, $\exists\unique\alpha_{H}\in[c]$ such that $\Laplacian\alpha_{H}=0$.\n

\cor{
	The Betti numbers of a manifold are finite.\nn
	Proof: $b_{k}=\dim{}H^{k}$. This is the multiplicity of $k$, which we know to be finite.\proven
}

\cor{
	(Poincar\'e Duality) The $\star$ operator induces $H^{k}(M)\cong{}H^{n-k}(M)$, where $n=\dim{}M$.\nn
	Proof: Follows from the fact that $\comm{\Laplacian,\star}=0$.\proven
}

\subsection*{Applications of Hodge Theory}

\par\noindent
Recall that $\forall{}p\in{}M$, $\forall{}u,v\in{}T_{p}M$, $\Ric_{p}^{B}(u,v)=\tr(T_{p}M\ni{}x\mapsto{}R_{p}(x,u)v\in{}T_{p}M)$. We saw that $\Ric^{B}$ is symmetric, and $\Ric(u)=\Ric^{B}(u,u)$.\n

\thm{
	(Bochner's Theorem) Assume $M$ is connected, compact, and oriented.
	\begin{enumerate}[label=(\alph*), leftmargin=4\parindent]
		\item If $\Ric\ge{}0$, $b_{1}\le\dim{}M$.
		\item If $\Ric\ge{}0$ and $\exists{}p_{0}\in{}M$ such that $\Ric_{p_{0}}^{B}$ is positive definite, then $b_{1}=0$.
	\end{enumerate}\up\nn
	Proof: Introduce a ``rough Laplacian'' associated to $\nabla$. (Recall that $T^{*}M\cong{}TM$, and think of the connection as $\nabla:\gamma(T^{*}M)\to\Gamma(T^{*}M\otimes{}T^{*}M)$.) The rough Laplacian is $\nabla^{*}\nabla:\Omega^{1}\to\Omega^{1}$. This works because we have $\nabla^{*}:\Gamma(T^{*}M\otimes{}T^{*}M)\to\Gamma(T^{*}M)$, using $l^{2}$ inner products associated with the Euclidean structures on the bundles.\nn
	It turns out this rough Laplacian has the same ymbol as the ordinary Laplacian. In fact, hey only differ by a term of order $0$! A calculation show that $\forall\alpha\in\Omega^{1}$, $\Laplacian\alpha=\nabla^{*}\nabla\alpha+\Ric^{B}(\alpha^{\sharp},\cdot)$, where $\alpha^{\sharp}$ is the metric dual to $\alpha$. Well,
	\begin{align*}
		\Laplacian\alpha=0 & \Rightarrow{}0=\int_{M}\iprod{\Laplacian\alpha,\alpha}d\Vol\\
		& \Rightarrow{}0=\underbrace{\norm{\Laplacian\alpha}^{2}}_{\mathclap{=\iprod{\nabla^{*}\nabla\alpha,\alpha}_{l^{2}}}}+\int_{M}\Ric^{B}(\alpha^{\sharp},\alpha^{\sharp})d\Vol\\
		& \Rightarrow\nabla\alpha=0\;\ptxt{and}\;\Ric(\alpha^{\sharp})=0\\
		& \Rightarrow\alpha\;\ptxt{is parallel-determined by any value}\;\alpha_{p}\in{}T_{p}^{*}M
	\end{align*}
}

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