04_12.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 35}
\date{4/12/21} % Can also use \today

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

\section*{Degree Theory}

\par\noindent
Given $F:M_{1}\to{}M_{2}$, where $M_{1}$ and $M_{2}$ are compact, connected, oriented manifolds of dimension $n=\dim{}M_{1}=\dim{}M_{2}$. Then $\exists\deg(F)\in\Z$ such that
\begin{enumerate}[label=\alph*)]
	\item $F^{*}:H^{n}(M_{2})\to{}H^{n}(M_{1})$ is ``multiplication by $\deg{}F$''.
	\item $\forall{}a\in\Omega^{n}(M_{2})$, $\int_{M_{1}}F^{*}\alpha=\deg(F)\int_{M_{2}}\alpha$.
	\item If $q\in{}M_{2}$ is a regular value of $F$, so $F\inv(q)=\bigcup_{i=1}^{N}\set{p_{i}}$, then $\deg(F)=\sum_{i=1}^{N}(-1)^{p_{i}}$, where
	\[
		(-1)^{p_{i}}=\left\{\begin{array}{ll}
			1 & dF_{p}\;\ptxt{preserves orientation}\\
			-1 & \ptxt{otherwise}
		\end{array}\right.
	\]
\end{enumerate}

\par\noindent
Application:\n
\thm{
	If $M_{1}=\d{}W$, and $f:M_{1}\to{}M_{2}$ extends to a smooth function $\tilde{F}:W\to{}M_{2}$, then $\deg(F)=0$.
	\[
		\begin{tikzcd}[ampersand replacement=\&]
			M_{1} \arrow[r,"F"] \arrow[d,hookrightarrow,"\iota"'] \& M_{2}\\
			W \arrow[ur,"\tilde{F}"']
		\end{tikzcd}
	\]
	Proof: Let $\alpha\in\Omega^{p}(M_{2})$. Then
	\[
		\int_{M_{1}}F^{*}\alpha=\int_{M_{1}}\iota^{*}\tilde{F}^{*}\alpha\labeledeq{(1)}\int_{W}d\tilde{F}^{*}\alpha=\int_{W}\tilde{F}^{*}d\alpha=0
	\]
	because $\alpha$ is a top-degree form, and (1) because of Stokes' theorem.\proven
}

\cor{
	There is no $C^{\infty}$ map $\closure{B}^{n}\to{}S^{n}$ (where $\closure{B}^{n}$ is the closed unit ball in $\R^{n}$) which is the identity on $S^{n}=\d\closure{B}^{n}$.\nn
	Proof: $\deg(I_{S^{n}})=1\ne{}0$.\proven
}

\par\noindent
(We presented some examples of Gauss-Bonnet and degree theory, but I'm not yet talented enough with TikZ to reproduce them.)\n

\section*{The Laplacian and Hodge Theory}

\par\noindent
Let $(M,g)$ be a Riemannian manifold.\n

\defn{
	The \u{gradient} is
	\[
		\map{\del:C^{\infty}(M)}{\mf{X}(M)}{f}{\del{}f}
	\]
	where $\del{}f$ is metric-dual to $df$. That is, $\forall{}v\in{}T_{p}M$, $\iprod{\del{}f(p),v}=df_{p}(v)$.\n
}

\par\noindent
$\del$ has a product rule: $\del(fg)=f\del{}g+g\del{}f$.\n

\defn{
	The \u{divergence} $\div:\mf{X}(M)\to{}C^{\infty}(M)$ is defined by $\forall{}X\in\mf{X}(M)$, $\mc{L}_{X}\Vol=(\div{}X)\Vol$, where $\Vol\in\Omega^{n}(M)$ is a volume form.\n
}

\par\noindent
Observe: This is really a local definition, and $\div$ is independent of orientation.\n

\par\noindent
Properties:
\begin{enumerate}[label=(\alph*)]
	\item In coordinate, $X=f^{i}\pd{}{x^{i}}$, then $\div{}X=\frac{1}{\sqrt{g}}\pd{}{x^{i}}(\sqrt{g}f^{i})$. (Note: we're using the shorthand $\sqrt{g}=\sqrt{\det(g_{ij})}$).
	\item $\div(fX)=f\div{}X+X(f)$.
	\item $(\div{}X)(p)=\tr(T_{p}M\ni{}v\mapsto(\nabla_{v}X)(p)\in{}T_{p}M)$.
\end{enumerate}

\thm{
	(Divergence Theorem) Let $M$ be a manifold-with-boundary, with the boundary $\d{}M$ oriented by $\nu$, an outward point unit normal vector field so that $\Vol_{\d{}M}=\iota_{\nu}\Vol_{M}$. Then $\forall{}X\in\mf{X}(M)$,
	\[
		\int_{M}(\div{}X)\Vol_{M}=\int_{\d{}M}\iprod{X,\nu}\Vol_{\d{}M}
	\]
	The right-hand side is the flux of $X$ out of $M$.\n
}

\cor{
	$\displaystyle(\div{}X)(p)=\lim_{\varepsilon\searrow{}0}\frac{1}{\Vol{}B_{\varepsilon}(p)}\int_{B_{\varepsilon}(p)}(\div{}X)d\Vol_{M}\labeledeq{Thm}\,\lim_{\varepsilon\searrow{}0}\frac{1}{\Vol{}B_{\varepsilon}(p)}\int_{S_{\varepsilon}(p)}\iprod{X,\nu}\Vol_{S_{\varepsilon}(p)}$\n
	``$(\div{}X)(p)$ is the infinitesimal flux per unit volume at $p$.''\n
}

\par\noindent
Observe: If $\d{}M=\emptyset$, then we get $\int_{M}(\div{}X)\Vol_{M}=0$. Proof: The left-hand side is simply
\[
	\int_{M}\mc{L}_{X}\Vol_{M}=\int_{M}d(\iota_{X}\Vol_{M})=\int_{\d{}M}j^{*}(\iota_{X}\Vol_{M})
\]
where $j:\d{}M\hookrightarrow{}M$. Then, we check that $j^{*}(\iota_{X}\Vol_{M})=\iprod{X,\nu}\Vol_{\d{}M}$. Use that $X=\iprod{X,\nu}+\eta$, where $\eta$ is tangent to $\d{}M$.\proven

\defn{
	The \u{laplacian} on functions is
	\[
		\map{\Laplacian:C^{\infty}(M)}{C^{\infty}(M)}{f}{\Laplacian{}f=-\div(\del{}f)}
	\]
	Note that some people don't use the negative sign.\n
}

\par\noindent
Combining the previous formulas, in coordinates, we get
\begin{align*}
	\Laplacian{}f & =-\frac{1}{\sqrt{g}}\pd{}{x^{i}}\paren{\sqrt{g}g^{ij}\pd{f}{x^{j}}}\\
	& =-\frac{1}{\sqrt{g}}\paren{\sqrt{g}g^{ij}\frac{\d^{2}f}{\d{}x^{i}\d{}x^{j}}+\ptxt{lower order terms}}\\
	& =-g^{ij}\frac{\d^{2}f}{\d{}x^{i}\d{}x^{j}}+\ptxt{lower order terms}
\end{align*}
So the highest order term only depend son the metric, not derivatives of the metric.\n

\par\noindent
Next time, we'll consider the following commutative diagram:
\[
	\begin{tikzcd}
		C^{\infty}(M) \arrow[r, shift left, "\del"] \arrow[d,symbol={=}] & \mf{X}(M) \arrow[l, shift left, "-\div"] \arrow[d,symbol=\cong,"\quad\ptxt{metric dual}"]\\
		C^{\infty}(M) \arrow[r, shift left, "d"] & \Omega^{1}(M) \arrow[l, shift left, "\delta"]
	\end{tikzcd}
\]
We'll see that $\delta=d^{*}$ (the adjoint) if we use $l^{2}$ inner products on $C^{\infty}(M)$ and $\Omega^{1}(M)$. On functions, $\Laplacian=\delta{}d=d^{*}d$.

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