11_18.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 33}
\date{11/18/20} % Can also use \today

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

\section*{Global formulae for $d$}

\par\noindent
First, we want to consider the pairing of forms and vector fields. If $\alpha\in\Omega^{k}$ and $X_{1},\ldots,X_{k}\in\mathfrak{X}(M)$, then $\alpha(X_{1},\ldots,X_{k}):M\to\R$ is a function on $M$.\n

\prop{
	(1) If $\alpha\in\Omega^{1}(M)$, $Y_{0},Y_{1}\in\mathfrak{X}(M)$, then
	\[
		(d\alpha)(Y_{0},Y_{1})=Y_{0}(\alpha(Y_{1}))-Y_{1}(\alpha(Y_{0}))-\alpha(\brack{Y_{0},Y_{1}})
	\]
	Proof: WOLOG assume $\alpha=f\,dg$. Then $d\alpha=df\wedge{}dg$. Thus,
	\[
		(d\alpha)(Y_{0},Y_{1})=df(Y_{0})dg(Y_{1})-df(Y_{1})dg(Y_{0})=Y_{0}(f)Y_{1}(g)-Y_{1}(f)Y_{0}(g)
	\]
	Well, the RHS of this is
	\[
		Y_{0}(fY_{1}(g))-Y_{1}(fY_{0}(g))-f\brack{Y_{0},Y_{1}}g=(Y_{0}f)(Y_{1}g)+\cancel{fY_{0}Y_{1}g}-(Y_{1}fY_{0}g+\cancel{fY_{1}Y_{0}g})-f(\cancel{Y_{0}Y_{1}g}-\cancel{Y_{1}Y_{0}g})
	\]
	\proven
}

\prop{
	(2) If $\alpha\in\Omega^{2}(M)$, $Y_{0},Y_{1},Y_{2}\in\mathfrak{X}(M)$, then
	\[
		(d\alpha)(Y_{0},Y_{1},Y_{2})=Y_{0}\alpha(Y_{1},Y_{2})-Y_{1}\alpha(Y_{0},Y_{2})+Y_{2}\alpha(Y_{0},Y_{1})-\alpha(\brack{Y_{0},Y_{1}},Y_{2})+\alpha(\brack{Y_{0},Y_{2}},Y_{1})-\alpha(\brack{Y_{1},Y_{2}},Y_{0})
	\]
	In general, if $\alpha\in\Omega^{k}(M)$, $Y_{0},\ldots,Y_{k}\in\mathfrak{X}(M)$, then
	\[
		(d\alpha)(Y_{0},\ldots,Y_{k})=\sum_{j=0}^{k}(-1)^{j}Y_{j}\alpha(Y_{0},\ldots,\widehat{Y_{j}},\ldots,Y_{k})+\smashoperator{\sum_{0\le{}i<j\le{}k}}(-1)^{i+j}\alpha(\brack{Y_{i},Y_{j}},Y_{0},\ldots,\widehat{Y_{i}},\ldots,\widehat{Y_{j}},\ldots,Y_{k})
	\]
	where $\widehat{\;\,\cdot\;\,}$ indicates that we omit that term.\n
}

\par\noindent
Refer to Proposition 20.14 in Tu/Proposition 14.32 in Lee.\n

\par\noindent
Another way to think of this is, gven $\alpha\in\Omega^{k}(M)$, $X_{1},\ldots,X_{k}\in\mathfrak{X}(M)$, and $p\in{}M$, we have $(X_{i})_{p}\in{}T_{p}M$, so
\[
	\map{\alpha_{p}:\smash{\underbrace{T_{p}M\times\cdots\times{}T_{p}M}_{k}}=(T_{p}M)^{k}}{\R}{p}{\alpha_{p}((X_{1})_{p},\ldots,(X_{k})_{p})\in\R}
\]

\section*{Lie Derivatives of Forms}

\defn{
	Let $X\in\mathfrak{X}(M)$, $\alpha\in\Omega^{k}(M)$, and $p\in{}M$. Let $\varphi$ be the local flow of $X$. Form the curve $(-\varepsilon,\varepsilon)\ni{}t\mapsto(\varphi_{t}^{*}\alpha)_{p}\in\bigwedge^{k}(T_{p}^{*}M)$. (Note that $\bigwedge^{k}(T_{p}^{*}M)$ is independent of $t$.) Then we defined
	\[
		(\mathcal{L}_{X}\alpha)_{p}=\restr{\frac{d}{dt}(\varphi_{t}^{*}\alpha)_{p}}{t=0}
	\]
}

\par\noindent
Observe: If $k=0$, then $\alpha$ is a function on $M$, so
\[
	(\mathcal{L}_{X}\alpha)_{p}=\restr{\frac{d}{dt}(\alpha\of\varphi_{t})_{p}}{t=0}=\restr{\frac{d}{dt}(\alpha(\varphi_{t}(p)))}{t=0}=\restr{\dot\varphi_{t}\vphantom{\frac{d}{dt}}}{\mathrlap{t=0}}(\alpha)
\]
Thus, $(\mathcal{L}_{X}\alpha)_{p}=X(\alpha)(p)$ if $\alpha$ is a function.\n

\subsection*{First Properties}

\prop{
	\begin{enumerate}[leftmargin=4\parindent, topsep=0pt, itemsep=0pt, label=\alph*)]
		\item The expression $\mathcal{L}_{X}(\alpha)$ is $\R$-linear w.r.t $X$ and $\alpha$.
		\item $\mathcal{L}_{X}(\alpha\wedge\beta)=\mathcal{L}_{X}(\alpha)\wedge\beta+\alpha\wedge\mathcal{L}_{X}(\beta)$ -- $\mathcal{L}_{X}$ is a derivation.
		\item $\mathcal{L}_{X}(d\alpha)=d\mathcal{L}_{X}(\alpha)$.
		\item If $F:M\to{}M$ is a diffeomorphism, then $\mathcal{L}_{X}(F^{*}\alpha)=F^{*}\paren{\mathcal{L}_{F_{*}(X)}\alpha}$.
	\end{enumerate}\up\nn
	Proof: (b) holds because $\varphi_{t}^{*}(\alpha\wedge\beta)=(\varphi_{t}^{*}\alpha)\wedge(\varphi_{t}^{*}\beta)$. We just need to show the right hand side satisfies he usual product rule with respect to $\frac{d}{dt}$.\n
	(c) holds because $\varphi_{t}^{*}(d\alpha)=d(\varphi_{t}^{*}\alpha)$, $\forall{}t$. This uses the fact that $d$ and $\frac{d}{dt}$ commute, i.e., $\frac{\d^{2}}{\d{}t\d{}x^{j}}=\frac{\d^{2}}{\d{}x^{j}\d{}t}$.\n
	(d) holds because for $p\in{}M$, $v_{1},\ldots,v_{k}\in{}T_{p}M$,
	\[
		\mathcal{L}_{X}(F^{*}\alpha)_{p}(v_{1},\ldots,v_{k})=\restr{\frac{d}{dt}\smash{\underbrace{\varphi_{t}^{*}(F^{*}\alpha)_{p}\vphantom{\frac{d}{dt}}}_{\mathclap{(\varphi_{t}^{*}\of{}F^{*})(\alpha)=(F\of\varphi_{t})^{*}(\alpha)}}}(v_{1},\ldots,v_{k})}{t=0}=\restr{\frac{d}{dt}\alpha((F\of\varphi_{t})_{*,p})(v_{1}),\ldots,(F\of\varphi_{t})_{*,p}(v_{k})}{t=0}
	\]
	Observe: $t\mapsto{}F\of\varphi_{t}$ is the integral curve of $F_{*}(X)$ at $F(p)$. So
	\[
		\mathcal{L}_{X}(F^{*}\alpha)_{p}(v_{1},\ldots,v_{k})=\cdots=(\mathcal{L}_{F_{*}X}\alpha)(F_{*,p}(v_{1}),\ldots,F_{*,p}(v_{k}))=F^{*}(\mathcal{L}_{F_{*}(X)}\alpha)(v_{1},\ldots,v_{k})
	\]
	\proven
}

\cor{
	If $\mathcal{L}_{X}\alpha=0$ (everywhere), then $\varphi_{t}^{*}\alpha=\alpha$, $\forall{}t$.\nn
	Proof: We will show $\forall{}p\in{}M$, the curve $t\mapsto{}(\varphi_{t}^{*}\alpha)_{p}\in\bigwedge^{k}T_{p}^{*}M$ is constant.\nn
	Proof: $\forall{}s$,
	\[
		\restr{\frac{d}{dt}\varphi_{t}^{*}\alpha}{t=s}=\restr{\frac{d}{dt}(\varphi_{t+s}^{*}\alpha)}{t=0}=\restr{\frac{d}{dt}\varphi_{t}^{*}(\varphi_{s}^{*}(\alpha))}{t=0}=\mathcal{L}_{X}(\varphi_{s}^{*}\alpha)=\varphi_{s}^{*}\mathcal{L}_{\underbrace{(\varphi_{s})_{*}(X)}_{=X}}\alpha=\varphi_{s}^{*}\underbrace{\mathcal{L}_{X}\alpha}_{=0}=0
	\]
	\proven
}

\subsection*{Towards Cartan's Formula}

\par\noindent
We need interior multiplication/contraction.\n

\defn{
	Given $\alpha\in\Omega^{k}$, $X\in\mathfrak{X}(M)$, we define $\iota_{X}\alpha\in\Omega^{k-1}$ by, $\forall{}p\in{}M$ and $v_{1},\ldots,v_{k-1}\in{}T_{p}M$,
	\[
		(\iota_{X}\alpha)_{p}(v_{1},\ldots,v_{k-1})=\alpha_{p}(X_{p},v_{1},\ldots,v_{k})
	\]
}

\thm{
	(Cartan's Magic Formula) $\mathcal{L}_{X}=d\of\iota_{X}+\iota_{X}\of{}d:\Omega^{k}\to\Omega^{k}$.\n
}

\par\noindent
Picture:
\[
	\begin{tikzcd}
		\cdots \arrow[r,"d"] & \Omega^{k-1} \arrow[r,"d"] & \Omega^{k} \arrow[r,"d"] \arrow[dl,"\iota_{X}",swap] \arrow[d,"\mathcal{L}_{X}", Rightarrow] & \Omega^{k+1} \arrow[r,"d"] \arrow[dl,"\iota_{X}"] & \cdots\\
		\cdots \arrow[r,"d"] & \Omega^{k-1} \arrow[r,"d"] & \Omega^{k} \arrow[r,"d"] & \Omega^{k+1} \arrow[r,"d"] & \cdots
	\end{tikzcd}
\]
where $\mathcal{L}_{X}$ is the ``sum of two paths''.\n

\par\noindent
On Friday, we'll show that when $\alpha=dx\wedge{}dy$ and $M=\R^{2}$, then $\mathcal{L}_{X}(\alpha)=(\div{}X)\alpha$, in the calc III sense.

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