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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 34}
\date{11/20/20} % Can also use \today

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

\section*{Lie Derivatives of Forms}

\ex{
	Let $X=\ang{F^{1},\ldots,F^{n}}\in\mathfrak{X}(\R^{n})$, $\mu=dx^{1}\wedge\cdots\wedge{}dx^{n}$, and $\phi$ the flow of $X$. Then
	\[
		\mathcal{L}_{X}(\mu)=\restr{\frac{d}{dt}\phi_{t}^{*}\mu}{t=0}
	\]
	Well, $\phi_{t}^{*}\mu=\det(J(\phi_{t})_{*})\mu$, so
	\[
		\restr{\det(J(\phi_{t})_{*})}{t=0}=\tr\paren{\restr{\vphantom{\frac{d}{dt}}\smash{\underbrace{\frac{d}{dt\vphantom{a_{b_{c}}}}J(\phi_{t})_{*}}_{\star}}}{t=0}}\underbrace{\det(\phi_{t=0})}_{=1}
	\]
	For $\star$, do $\frac{d}{dt}$ first, and then $\pd{}{x^{i}}$. And $\det(\phi_{t=0})=1$, because $\phi_{t}$ is the flow of $X$, so $\frac{d}{dt}$ is just $X$. Thus, we have
	\[
		\tr \begin{pmatrix}
			\horizontalMatrixLine & \del{}F_{1} & \horizontalMatrixLine\\
			& \vdots\\
			\horizontalMatrixLine & \del{}F_{n} & \horizontalMatrixLine
		\end{pmatrix}=\sum_{j=1}^{n}\pd{F_{j}}{x^{j}}=\div(X)
	\]
	We conclude that $\mathcal{L}_{X}\mu=(\div{}X)\mu$.\proven
}

\thm{
	(Cartan's Magic Formula) $\mathcal{L}_{X}=\iota_{X}\of{}d+d\of\iota_{X}:\Omega^{k}\to\Omega^{k}$.\nn
	Proof: Let $P_{X}=\iota_{X}\of{}d-d\of\iota_{X}$. Then $P_{X}$ has the following properties:
	\begin{enumerate}[leftmargin=4\parindent, topsep=0pt, itemsep=0pt, label=\arabic*)]
		\item $\R$-linearity
		\item It's a derivation w.r.t. $\wedge$: $P_{X}(\alpha\wedge\beta)=P_{X}(\alpha)\wedge\beta+\alpha\wedge{}P_{X}(\beta)$
		\item It commutes with $d$
		\item $f\in{}C^{\infty}(M)\Rightarrow{}P_{X}(f)=X(f)$
		\item $P_{X}$ is local
	\end{enumerate}\up\n
	These properties belong to a unique operator. (Compute in coordinates, and by linearity, just use monomials.)
	\[
		P_{X}(a\,dx^{I})=P_{X}(a)\wedge{}dx^{I}+a\wedge{}P_{X}(dx^{I})=X(a)\,dx^{I}+aP_{X}(dx^{I})
	\]
	We then expand $P_{X}(dx^{I})$ with induction.\proven
}

\ex{
	For $k=2$:
	\[
		P_{X}(dx^{1}\wedge{}dx^{2})=P_{X}(dx^{1})\wedge{}dx^{2}+dx^{1}\wedge{}P(dx^{2})=dX(x^{1})\wedge{}dx^{2}+dx^{1}\wedge{}dX(x^{2})
	\]
	This is unique, and it just uses the five properties.\n
}

\subsection*{Applications of Cartan's Formula}

\defn{
	A \u{symplectic manifold} is a pair $(M,\omega)$, with $M$ a manifold, and $\omega\in\Omega^{2}(M)$ such that
	\begin{enumerate}[leftmargin=4\parindent, topsep=0pt, itemsep=0pt, label=\arabic*)]
		\item $\forall{}p\in{}M,v\in{}T_{p}M\cut\set{0}$, $\iota_{v}(\omega_{p})=\omega_{p}(v,\cdot):T_{p}M\to\R$ is nonzero.
		\item $d\omega=0$.
	\end{enumerate}\up\n
}

\par\noindent
Question: What are the symmetries of a symplectic manifold $(M,\omega)$? Specifically, are there one-parameter groups $\varphi_{t}:M\to{}M$ such that $\forall{}t$, $\varphi_{t}^{*}\omega=\omega$?\n
Use the Lie derivative: If $X\in\mathfrak{X}(M)$ is the generator of $\varphi_{t}$, then
\[
	\varphi_{t}^{*}\omega=\omega\Leftrightarrow\mathcal{L}_{X}\omega=0\Leftrightarrow\iota_{X}\underbrace{d\omega}_{=0}+d\iota_{X}\omega=0\Leftrightarrow{}d\iota_{X}\omega=0
\]

\defn{
	One particular class of such $X$'s comes from the following: Take any function $H\in{}C^{\infty}(M)$, and define $X$ by $\iota_{X}\omega=-dH$. By non-degeneracy of $\omega$, $X$ is unique! And,
	\[
		\map{T_{p}M}{T_{p}^{*}M}{v}{\iota_{v}\omega}
	\]
	has no kernel, so it's a bijection. This $X$ is called the \u{Hamilton field} of $H$.\n
}

\exer{
	Take $M=\R^{2n}$ with coordinates $(x,p)$, where $x$ and $p$ are the standard coordinates in $\R^{n}$. Let
	\[
		\omega=\sum_{i=1}^{n}dp^{i}\wedge{}dx^{i}\qquad{}H=\frac{1}{2}\norm{p}^{2}+V(x)
	\]
	Compute the Hamilton field, and show that the integral curves satisfy $\ddot{x}(t)=-\del{}V(x(t))$. This is better known as Newton's second law!\n
}

\par\noindent
We're now done with Lie derivatives.

\section*{Integration of Forms on Oriented Manifolds}

\par\noindent
First, we have to define orientation. Let $V$ be an $n$-dimensional vector space. Let $\mathscr{B}(V)$ be the set of all ordered bases of $V$. For $e\in\mathscr{B}$, $e=(e_{1},\ldots,e_{n})$ is an ordered basis of $V$. Observe: $\forall{}e,f\in\mathscr{B}$, $\exists\unique{}M\in\GL(n,\R)$ such that $\forall{}i$, $e_{i}=Mf_{i}$.\n

\defn{
	$e\sim{}f\Leftrightarrow\det{}M>0$. We say $e$ and $f$ define the same orientation of $V$.\n
}

\par\noindent
Check:
\begin{enumerate}[topsep=0pt, itemsep=0pt, label=\arabic*)]
	\item $\sim$ is an equivalence relation
	\item $\mathscr{B}/\sim$ has two elements
\end{enumerate}\up\n

\defn{
	An \u{orientation} of $V$ is a choice of an equivalence class in $\mathscr{B}/\sim$. Bases in that equivalence class are said to be \u{positive}.\n
}

\par\noindent
Alternatively, consider the set of nonzero top-degree forms, $\underbrace{\paren{{\bigwedge}^{n}V}}_{\dim=1}\cut\set{0}$. When we take away $0$, any $\mu\in\paren{{\bigwedge}^{n}V}\cut\set{0}$ defines an orientation by: a basis $e$ is positive iff $\mu(e)>0$. The orientation defined by $\mu$ only depends on which connected component of $\paren{\bigwedge^{n}V}\cut\set{0}$ contains $\mu$. Conversely, an orientation singles out one of the two components of $\paren{\bigwedge^{n}V}\cut\set{0}$.\n

\par\noindent
Conclusion: An orientation is a choice of a connected component of $\paren{\bigwedge^{n}V}\cut\set{0}$.\n

\par\noindent
Now, we move on to manifolds! Note: Not all manifolds are orientable (e.g. the Mobius band).\n

\defn{
	An \u{orientation} on $M$ (if it exists) is a continuous choice of orientation of each tangent space.\n
}

\par\noindent
Continuity means $\forall{}p\in{}M$, there exists a continuous moving frame $(E_{1},\ldots,E_{n})$ such that at every point $q$ in the domain of $E$, $(E_{1}(q),\ldots,E_{n}(q))$ is a positive basis of $T_{q}M$.\n

\lemma{
	A connected manifold can have either two orientations, or it's non-orientable.\nn
	Proof: The idea is if the manifold is orientable, then consider orientations $\mathscr{O}_{1}$ and $\mathscr{O}_{2}$. Define $F:M\to\R$ by $f(p)=1$ if $\mathscr{O}_{1}(p)=\mathscr{O}_{2}(p)$, and $0$ otherwise. (Note: we don't really ever use this notation.) Then by the continuity of $\mathscr{O}_{1}$ and $\mathscr{O}_{2}$, $f$ is continuous, so $f$ is locally constant, so if $M$ is connected, then $f$ is constant.\proven
}

\defn{
	Let $M$ be an oriented manifold. Then a \u{positive atlas} on $M$ is an atlas $\set{(U_{\alpha},\phi_{\alpha})}$ of $M$ such that $\forall\alpha$, the moving frame $\paren{\pd{}{x^{1}_{\alpha}},\ldots,\pd{}{x^{n}_{\alpha}}}$ is a positive frame.\n
}

\newpage
\lemma{
	\begin{enumerate}[leftmargin=4\parindent, topsep=0pt, itemsep=0pt, label=\arabic*)]
		\item The transition functions $F$ and $G$ of any two elements in a positive atlas satisfy $\det(J(F))=1=\det(J(G))$.
		\item An oriented manifold always has a positive atlas.
	\end{enumerate}\up\nn
	Proof: This is very tedious! Idea:\n
	1) Recall that $J(F)$ is actually the change of basis matrix between $\paren{\pd{}{x^{1}_{\alpha}},\ldots,\pd{}{x^{n}_{\alpha}}}$.
	2) Start with any atlas. $\forall\alpha$, $\set{\pd{}{x^{1}_{\alpha}},\ldots,\pd{}{x^{n}_{\alpha}}}$ is either positive, or not. If it is positive, do nothing, and keep $\phi_{\alpha}$. If it's not positive, relable (switch) $x^{1}$ and $x^{2}$. Now it's positive!\n
	\proven
}

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