04_07.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 33}
\date{4/7/21} % Can also use \today

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

\par\noindent
Today, we'll take the first step towards proving Gauss-Bonnet. Let $M\subseteq\R^{n+1}$ be a compact, oriented manifold of even dimension $\dim{}M=n=2m$, with $N:M\to{}S^{n}$ given by the orientation. Define the Gaussian curvature $\ms{K}:M\to\R$ by $\ms{K}dV_{M}=N^{*}dV_{S^{n}}$.\n

\thm{
	$\ms{K}$ is intrinsic to the Riemannian metric of $N$.\n
}

\par\noindent
Recall the Weingarten formula: $\forall{}p\in{}M$, we have the commutative diagram
\[
	\begin{tikzcd}[ampersand replacement=\&]
		T_{p}M \arrow[r,"-dN"] \arrow[dr,"S_{N(p)}"'] \& T_{N(p)}S^{n} \arrow[d,symbol=\cong]\\[-10pt]
		\& T_{p}M
	\end{tikzcd}
\]
where $T_{N(p)}S^{n}\cong{}T_{p}M$ isometrically by translation. $S_{N_{p}}$ is the shape operator by the Weingarten formula, so
\[
	\restr{\ms{K}dV_{M}}_{p}=\restr{S_{N(p)}^{*}dV_{M}}_{p}\qquad\Rightarrow\qquad\ms{K}=\det{}S_{N(p)}=\prod_{i=1}^{n}\kappa_{i}
\]
To prove this, we use orthonormal moving frames on $M$. Let $(E_{1},\ldots,E_{n})$ be a positive orthonormal moving frame. We get the curvature matrix $(\Omega_{j}^{i})$, a matrix of $2$-forms.
\[
	\forall{}X,Y\in\mf{X}(M),\quad{}R(X,Y)(E_{j})=\Omega_{j}^{i}(X,Y)E_{i}\quad\Rightarrow\quad\Omega_{j}^{i}(X,Y)=R(X,Y,E_{j},E_{i})
\]
Also, we have Gauss' formula:
\[
	0=R(W,X,Y,Z)+\iprod{B(W,Y),B(X,Z)}-\iprod{B(W,Z),B(X,Y)}
\]
So
\[
	\Omega_{j}^{i}(E_{k},E_{l})=\iprod{B(E_{i},E_{k}),B(E_{j},E_{l})}-\iprod{B(E_{i},E_{l}),B(E_{j},E_{k})}
\]
Recall: $S_{ki}=S_{ik}\eqdef\iprod{S(E_{i}),E_{k}}=\iprod{B(E_{i},E_{k}),N}$ is the $N$-component of $B(E_{i},E_{k})$ $(S_{ij})$ is the matrix of $S$. Written all together, we have
\[
	\Omega_{j}^{i}(E_{k},E_{l})=S_{ik}S_{jl}-S_{il}S_{jk}
\]

\prop{
	Let $n=2m$. Then with $\sigma_{n}$ the symmetric group,
	\[
		\ms{K}dV=\frac{1}{n!}\sum_{\alpha\in\sigma_{n}}(-1)^{\alpha}\bigwedge_{i=1}^{m}\Omega_{\alpha(2i)}^{\alpha(2i-1)}\eqdef\Pf(\Omega)
	\]
}

\defn{
	$\Pf(\Omega)$ is called the \u{Pfaffian} of $\Omega$.\n
}

\par\noindent
We'll prove that the Pfaffian is independent of choice of moving frame. It's enough to show $\Pf(\Omega)(E_{1},\ldots,E_{n})=\ms{K}$. Well, we introduce
\[
	Q=\set{\varphi\in\sigma_{n}\mid{}\forall{}i\in\set{1,\ldots,n},\varphi(2i-1),\varphi(2i)\in\set{2i-1,2i}}
\]
We then begin to compute
\begin{align*}
	\Pf(\Omega)(E_{1},\ldots,E_{n}) & =\frac{1}{n!2^{m}}\sum_{\alpha,\beta\in\sigma_{n}}(-1)^{\alpha}(-1)^{\beta}\prod_{i=1}^{m}\Omega_{\alpha(2i)}^{\alpha(2i-1)}(E_{\beta(2i-1)},E_{\beta(2i)})\\
	& =\frac{1}{n!2^{m}}\sum_{\alpha,\beta\in\sigma_{n}}(-1)^{\alpha}(-1)^{\beta}\sum_{\varphi\in{}Q}(-1)^{\varphi}\prod_{i=1}^{m}S_{\alpha(2i-1)\beta\varphi(2i-1)}S_{\alpha(2i)\beta\varphi(2i)}\\
	& =\frac{1}{n!2^{m}}\sum_{\varphi\in{}Q}\sum_{\alpha,\beta\in\sigma_{n}}(-1)^{\alpha}(-1)^{\beta}(-1)^{\varphi}\prod_{i=1}^{m}S_{\alpha(2i-1)\beta\varphi(2i-1)}S_{\alpha(2i)\beta\varphi(2i)}
\end{align*}
Eventualy, this computation will give us $\det{}S$.\n

\defn{
	(Official definition) For $X=(x_{j}^{i})$ an $n\times{}n$ matrix of commuting variables (with $n=2m$ even), we define the \u{Pfaffian}
	\[
		\Pf(X)=\frac{1}{n!}\sum_{\alpha\in\sigma_{n}}(-1)^{\alpha}\prod_{i=1}^{m}X_{\alpha(2i)}^{\alpha(2i-1)}
	\]
}

\lemma{
	$\forall{}X,Y$, $\Pf(Y\tpose{}XY)=\det(Y)\Pf(X)$.\nn
	Proof: Just another direct computation.\n
}

\par\noindent
Now, go back to moving frames and curvature matrices. Considering local frames, write $E_{i}=a_{i}^{j}F_{j}$ on $U\subseteq{}M$. Then $A(p)=(a_{i}^{j}(p))\in\SO(n)$, and we konw that $\Omega_{F}=A\inv\Omega_{E}A$. So $\Pf(\Omega_{F})=\det(A)\Pf(\Omega_{E})=\Pf(\Omega_{E})$.\n

\par\noindent
Thus, $\Pf(\Omega_{F})=\Pf(\Omega_{E})$. So by the usual arguments, there's a unique global top-degree form on $M$ such that for any moving frame on $U$, it agrees with $\Pf(\Omega)$. Therefore, by our proposition, $\ms{K}dV$ is of that form.\proven

\par\noindent
Question: Are there other combinations of the $\Omega_{j}^{i}$'s that give global forms on $M$? We need some polynomial $P:\so(n)\to\R$ such that $\forall{}A\in\SO(n)$, $\forall{}X\in\so(n)$, $P(A\inv{}XA)=P(X)$. Given such a $P$, repeat the previous argument to show that there's a global form $\varpi$ such that on any $U$ with a moving frame $E_{1},\ldots,E_{n}$, $\restr{\varpi}_{U}=P(\Omega)$.\n

\par\noindent
$P(\Omega_{E})=P(\Omega_{F})$, so $dP(\Omega)=0$ always ($\forall{}P$ invariant). We get the Chern-Weil morphism:
\[
	\set{\Ad\ptxt{-invariant polynomials on}\;\so(n)}\to{}H^{*}M
\]

\end{document}