11_09.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 29}
\date{11/9/20} % Can also use \today

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

\subsection*{Motivation: Integration}

\par\noindent
How do we integrate on a manifold? Start with calc 3. Let $U\subseteq\R^{n}$ be an open set. Then we can change variables with
\[
	\int_{U}f(x)\,dx=\int_{V}f(x(y))\underbrace{\det \begin{pmatrix}
		\pd{x}{y}
	\end{pmatrix}}_{\ptxt{if positive}}\,dy
\]
for $x=x(y)$.\n

\par\noindent
Now, for the general interpretation: let $\map{F:V}{U}{y}{x=x(y)}$ be a diffeomorphism. Then $\begin{pmatrix}\pd{x}{y}\end{pmatrix}$ is the Jacobian of $F$. We'll interpret this as the pullback of the RHS integral by $F$ -- $F^{*}(dx^{1},\ldots,dx^{n})=?$. But what does $\det$ mean in general? It's an alternating, multilinear function, so let's work with that.\n

\defn{
	Let $V$ be a $n$-dimensional vector space, and $k\in\N$. A \u{$k$-covector} or \u{$k$-form} on $V$ is a multilinear map $\alpha:\underbrace{V\times\cdots\times{}V}_{k}\to\R$ that is alternating (i.e. if you swap two elements, the sign flips).\n
}

\par\noindent
Observe: Let $\sigma\in{}S_{k}$ (the symmetric group). Given $v_{1},\ldots,v_{k}$, $\alpha$ alternating, we have
\[
	\alpha(v_{\sigma(1)},\ldots,v_{\sigma(k)})=\underbrace{(-1)^{\sigma}}_{\mathclap{=\sgn(\sigma)}}\alpha(v_{1},\ldots,v_{k})
\]

\par\noindent
Observe: If $\set{v_{1},\ldots,v_{k}}$ is linearly dependent, and $\alpha$ is alternating, then $\alpha(v_{1},\ldots,v_{k})=0$.\n
Proof: Say $v_{1}=\lambda_{2}v_{2}+\cdots+\lambda_{k}v_{k}$. Then
\[
	\alpha(v_{1},\ldots,v_{k})=\alpha\paren{\sum_{i=2}^{k}\lambda_{i}v_{i},v_{2},\ldots,v_{k}}=\sum_{i=2}^{k}\lambda_{i}\alpha(v_{i},v_{2},\ldots,v_{i},\ldots,v_{k})=0
\]

\defn{
	$\bigwedge^{k}V^{*}$ is the set of alternating $\R$-multilinear functions on $V^{k}$. We say that $k$ is the \u{degree}.\n
}

\par\noindent
Observe: $k$-forms can be pulled back by linear maps.\n

\defn{
	If $F:V\to{}W$ is linear, and $\alpha\in{}\bigwedge^{k}W^{*}$, $v_{1},\ldots,v_{k}\in{}V$, then
	\[
		(F^{*}\alpha)(v_{1},\ldots,v_{k})=\alpha(F(v_{1}),\ldots,F(v_{k}))
	\]
	and $F^{*}\alpha\in\bigwedge^{k}V^{*}$.\n
}

\defn{
	Let $(\mathcal{E}^{1},\ldots,\mathcal{E}^{n})$ be an ordered basis of $V^{*}$. Let $A=\set{a_{1}<\cdots<a_{k}}\subset\set{1,\ldots,n}$. Define $\mathcal{E}^{A}:V\times\cdots\times{}V\to\R$ by
	\[
		\mathcal{E}^{A}(v_{1},\ldots,v_{k})=\det \begin{pmatrix}
			\mathcal{E}^{a_{i}}(v_{j})
		\end{pmatrix}_{(i,j)}
	\]
}

\ex{
	Say $V=\R^{3}$ with the standard basis. Then $\mathcal{E}^{13}((x_{1},x_{2},x_{3}),(y_{1},y_{2},y_{3}))=\det \begin{pmatrix}
		x_{1} & y_{1}\\
		x_{3} & y_{3}
	\end{pmatrix}$.\n
}

\prop{
	For a given $V$ and $k$, and an ordered basis $(\mathcal{E}^{1},\ldots,\mathcal{E}^{n})$ of $V^{*}$, the set $\set{\mathcal{E}^{I}:I\subset\set{1,\ldots,n},\#I=k}$ is a basis of $\bigwedge^{k}V^{*}$. In particular, $\dim\bigwedge^{k}V^{*}=\binom{n}{k}$.\n
}

\par\noindent
Observe: If $k>n$, $\bigwedge^{k}V^{*}=\set{0}$. If $k=n$, $\dim\bigwedge^{k}V^{*}=1$. If $k=1$, $\bigwedge^{1}V^{*}=V^{*}$.\n

\newpage
\par\noindent
As a warmup, let $k=2$> Let $\set{e_{1},\ldots,e_{n}}$ be a basis of $V$, and $\set{\mathcal{E}^{1},\ldots,\mathcal{E}^{n}}$ the corresonding dual basis of $V^{*}$. Note that $\mathcal{E}^{i}(e_{j})=\delta_{ij}$.\n
Say $\alpha\in\bigwedge^{2}V^{*}$, $v_{1}=\sum_{a=1}^{n}v_{1}^{a}e_{a}$, $v_{2}=\sum_{b=1}^{n}v_{2}^{b}e_{b}$. Then
\begin{align*}
	\alpha(v_{1},v_{2}) & =\alpha\paren{\sum_{a=1}^{n}v_{1}^{a}e_{a},\sum_{b=1}^{n}v_{2}^{b}e_{b}}\\
	& =\sum_{a=1}^{n}v_{1}^{a}\alpha\paren{e_{a},\sum_{b=1}^{n}v_{2}^{b}e_{b}}\\
	& =\sum_{a=1}^{n}\sum_{b=1}^{n}v_{1}^{a}v_{2}^{b}\alpha(e_{a},e_{b})\\
	& =\smashoperator[r]{\sum_{1\le{}a<b\le{}n}}v_{1}^{a}v_{2}^{b}\alpha(e_{a},e_{b})+v_{1}^{b}v_{2}^{a}\alpha(e_{b},e_{a})\\
	& =\sum_{1\le{}a<b\le{}n}\underbrace{(v_{1}^{a}v_{2}^{b}-v_{1}^{b}v_{2}^{a})}_{\displaystyle\hspace{-6em}\mathrlap{=\left|\begin{matrix}
			v_{1}^{a} & v_{2}^{a}\\ v_{1}^{b} & v_{2}^{b})
		\end{matrix}\right|=\mathcal{E}^{ab}(v_{1},v_{2})}}\alpha(e_{a},e_{b})\\
	& =\sum_{I=\set{a<b}}\alpha(e_{a},e_{b})\mathcal{E}^{I}(v_{1},v_{2})
\end{align*}
In general, $\displaystyle\alpha=\smashoperator{\sum_{1\le{}a<b\le{}n}}\alpha(e_{a},e_{b})\mathcal{E}^{ab}$.\n

\par\noindent
Now, for general $k\in\N$, fix $\alpha\in\bigwedge^{k}V^{*}$. For $i=1,\ldots,k$, let $v_{i}=\sum_{a=1}^{n}v_{i}^{a}e_{a}$. Then
\begin{align*}
	\alpha(v_{1},\ldots,v_{k}) & =\alpha\paren{\sum_{a_{1}=1}^{n}v_{1}^{a_{1}}e_{a_{1}},\ldots,\sum_{a_{k}}^{n}v_{k}^{a_{k}}e_{a_{k}}}\\
	& =\sum_{a_{1},\ldots,a_{k}=1}^{n}\paren{\prod_{j=1}^{k}v_{j}^{a_{j}}}\alpha(e_{a_{1}},\ldots,e_{a_{k}})\\
	& =\sum_{1\le{}a_{1}<\cdots<a_{k}\le{}n}\underbrace{\paren{\sum_{\sigma\in{}S_{k}}\paren{\prod_{j=1}^{k}v_{j}^{\sigma(a_{i})}}(-1)^{\sigma}}}_{\displaystyle\hspace{-12em}\mathrlap{=\det\begin{pmatrix}v_{j}^{a_{i}}\end{pmatrix}_{(i,j)}=\mathcal{E}^{A}(v_{1},\ldots,v_{k}),A=\set{a_{1},\ldots,a_{k}}}}\alpha(e_{a_{1}},\ldots,e_{a_{k}})\\
	& =\smashoperator[r]{\sum_{A=\set{a_{1}<\cdots<a_{k}}}}\alpha(e_{a_{1}},\ldots,e_{a_{k}})\mathcal{E}^{A}(v_{1},\ldots,v_{k})
\end{align*}
So in general, $\displaystyle\alpha=\smashoperator{\sum_{A=\set{a_{1},\cdots<a_{k}}}}\alpha(e_{a_{1}},\ldots,e_{a_{k}})\mathcal{E}^{A}$.\n

\section*{The Wedge Product}

\par\noindent
We want to take a $k$-form and an $l$-form and make a $k+l$-form.\n

\defn{
	For $\alpha\in\bigwedge^{k}V^{*}$, $\beta\in\bigwedge^{l}V^{*}$, define the \u{wedge product} of $\alpha$ and $\beta$, $\alpha\wedge\beta\in\bigwedge^{k+l}V^{*}$, by
	\begin{align*}
		(\alpha\wedge\beta)(v_{1},\ldots,v_{k+l}) & \eqdef\frac{1}{k!l!}\sum_{\sigma\in{}S_{k+l}}(-1)^{\sigma}\alpha(v_{\sigma(1)},\ldots,v_{\sigma(k)})\beta(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)})\\
		& =\sum_{\sigma\in\Sh(k,l)}(-1)^{\sigma}\alpha(v_{\sigma(1)},\ldots,v_{\sigma(k)})\beta(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)})\\
	\end{align*}
}

\defn{
	$\Sh(k,l)$ is the set of \u{$k-l$ shuffles}, which are permutations $\sigma\in{}S(k+l)$ such that $\sigma(1)<\cdots<\sigma(k)$ and\break
	$\sigma_{k+1}<\cdots<\sigma_{k+l}$.\n
}

\subsection*{The Skew-Symmetrizer}

\par\noindent
Given $\alpha\in\bigwedge^{k}V^{*}$ and $\beta\in\bigwedge^{l}V^{*}$, we can define
\[
	(\alpha\otimes\beta)(v_{1},\ldots,v_{k+l})=\alpha(v_{1},\ldots,v_{k})\beta(v_{k+1},\ldots,v_{k+l})
\]
As a map, $\alpha\otimes\beta:V^{k+l}\to\R$ is $k+l$-multilinear, but not alternating/skew-symmetric.\n

\defn{
	The \u{skew-symmetrizer} of a multilinear map $f:V^{m}\to\R$ is defined by
	\[
		A(f)(v_{1},\ldots,v_{m})=\frac{1}{m!}\sum_{\sigma\in{}S_{m}}(-1)^{\sigma}f(v_{\sigma(1)},\ldots,v_{\sigma(m)})
	\]
	\n
}

\lemma{
	$A(f)$ is skew-symmetric/alternating, and if $f$ is already skew-symmetric, $A(f)=f$.\nn
	Proof: Let $\tau\in{}S_{m}$. Then
	\begin{align*}
		A(f)(v_{\tau(1)},\ldots,v_{\tau(m)}) & =\frac{1}{m!}\sum_{\sigma\in{}S_{m}}(-1)^{\sigma}f(v_{\sigma(\tau(1))},\ldots,v_{\sigma(\tau(m))})\\
		& \labeledeq{(1)}\frac{1}{m!}\sum_{\mu\in{}S_{m}}\underbrace{(-1)^{\mu\tau}}_{\hspace{-3em}\mathrlap{=(-1)^{\tau}(-1)^{\mu}}}f(v_{\mu(1)},\ldots,v_{\mu(m)})\\
		& =(-1)^{\tau}A(f)
	\end{align*}
	with (1) true because if $\mu=\sigma\tau$, then $\sigma=\mu\tau\inv$.\proven
}

\par\noindent
Thus, we have $\alpha\wedge\beta=\frac{(k+l)!}{k!l!}A(\alpha\otimes\beta)$.\n

\ex{
	$k=l=1$. Then $\Sh(1,1)=S_{2}$, so
	\[
		(\alpha\wedge\beta)(v_{1},v_{2})=\alpha(v_{1})\beta(v_{2})-\alpha(v_{2})\beta(v_{1}).
	\]
}

\ex{
	$k=1$, $l=2$. Then the elements of $S_{3}$ are
	\[
		\begin{array}{ccc|c|c}
			\sigma(1) & \sigma(2) & \sigma(3) & \in\Sh(1,2)? & \sgn\\ \hline
			1 & 2 & 3 & \ptxt{Yes} & +\\
			1 & 3 & 2 & \ptxt{No}\\
			2 & 1 & 3 & \ptxt{Yes} & -\\
			2 & 3 & 1 & \ptxt{No}\\
			3 & 1 & 2 & \ptxt{Yes} & +\\
			3 & 2 & 1 & \ptxt{No}
		\end{array}
	\]
	so $\Sh(1,2)=\set{\cycle{1 & 2 & 3},\cycle{2 & 1 & 3},\cycle{3 & 1 & 2}}$, so
	\[
		(\alpha\wedge\beta)(v_{1},v_{2},v_{3})=\alpha(v_{1})\beta(v_{2},v_{3})-\alpha(v_{2})\beta(v_{1},v_{3})+\alpha(v_{3})\beta(v_{1},v_{2})
	\]
	If $\beta=\gamma\wedge\delta$, then we get
	\begin{align*}
		(\alpha\wedge(\gamma\wedge\delta))(v_{1},v_{2},v_{3})= & \alpha(v_{1})(\gamma(v_{2})\delta(v_{3})-\gamma(v_{3})\delta(v_{2}))\\
		& -\alpha(v_{2})(\gamma(v_{1})\delta(v_{3})-\gamma(v_{3})\delta(v_{1}))\\
		& +\alpha(v_{3})(\gamma(v_{1})\delta(v_{2})-\gamma(v_{2})\delta(v_{1}))\\
		= & \left|\begin{matrix}
			\alpha(v_{1}) & \alpha(v_{2}) & \alpha(v_{3})\\
			\gamma(v_{1}) & \gamma(v_{2}) & \gamma(v_{3})\\
			\delta(v_{1}) & \delta(v_{2}) & \delta(v_{3})
		\end{matrix}\right|
	\end{align*}
}

\lemma{
	$\mathcal{E}^{I}\wedge\mathcal{E}^{J}=\mathcal{E}^{IJ}$.\n
}

\lemma{
	$\mathcal{E}^{I}=\mathcal{E}^{I_{1}}\wedge\cdots\wedge\mathcal{E}^{I_{k}}$ ($\wedge$ is associative).\n
}

\prop{
	The wedge product is
	\begin{itemize}[leftmargin=4\parindent]
		\item Bilinear
		\item Associative
		\item Anti-commutative: $\beta\wedge\alpha=(-1)^{kl}\alpha\wedge\beta$, for $\alpha\in\bigwedge^{k}$, $\beta\in\bigwedge^{l}$
		\item If $F:V\to{}W$, $\alpha\in\bigwedge^{k}W^{*}$, $\beta\in\bigwedge^{l}W^{*}$, then $F^{*}(\alpha\wedge\beta)=(F^{\alpha})\wedge(F^{*}\beta)$
	\end{itemize}
}

\end{document}