04_09.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Leture 34}
\date{4/9/21} % Can also use \today

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

\section*{Pontryagin Classes}

\par\noindent
Observe: $\restr{\Pf}_{\so(n)}:\so(n)\to\R$ (for $n$ even) is $\Ad_{\SO(n)}$-invariant. So $\Pf(\Omega)$ is independent of choice of frame, where $\Omega$ is the curvature matrix with respect to that frame. We also have the Chern-Weil morphism:
\[
	\underbrace{\set{\ptxt{$\Ad_{\SO(n)}$invariant polynomials on $\so(n)$}}}_{I(\so(n))}\to\Omega^{*}(M)
\]
where $p\mapsto{}p(\Omega)$. An amazing fact is that $p(\Omega)$ is always closed! So we get $I(\so(n))\to{}H^{*}_{dR}(M)$, the deRham cohomology.\n

\ex{
	Elements of $I(\so(r))$ are Pontryagin polynomials. Let $A\in\so(r)$ be skew-symmetric (as usual, with $r$ even). We claim that $A\tpose=-A$ implies the characteristic polynomial is even.
	\[
		\det(\lambda{}I-A)=\sum_{k=0}^{r/2}\lambda^{r-2k}P_{k}(A)
	\]
	where $P_{k}(A)$ is homogeneous, and of degree $2r$.\n
}

\par\noindent
We can apply this idea to a rank-$r$ vector bundle $\mc{E}\to{}M$. The idea is to use a metric on each fiber of $\mc{E}$ to get an orthonormal frame, and then the connection to get curvature forms. We get $P_{r}(\Omega)$, a differential form of degree $4r$. ($\Omega$ is the curvature matrix w.r.t. the orthonormal moving frame.) $[P_{r}(\Omega)]\in{}H^{4r}(M)$.\n

\thm{
	The cohomology classes are independent of the connection chosen -- they're purely topological, and associated to $\mc{E}$.\n
}

\par\noindent
So Gauss-Bonnet implies that $[\ms{K}dV]\in{}H^{n}(M)$ is independent of the metric.\n

\par\noindent
Now, back to Gauss-Bonnet. We want to show
\[
	\int_{M}\ms{K}dV=\frac{\Vol(S^{n})}{2}\chi(M)\qquad\ptxt{using}\qquad\int_{M}\ms{K}dV=\int_{M}N^{*}(dV_{S^{n}})
\]
where $N$ is the Gauss spherical map.\n

\par\noindent
Degree Theory: \i{What happens when you pullback a top-degree form.} (See Lee Differentiable Manifolds page 457.)\n

\par\noindent
Preliminary (but still important) result:\n

\thm{
	Let $M$ be a compact, connected, oriented manifold. (Note: It must have empty boundary.) Then the integration map
	\[
		\map{\int_{M}:H^{n}(M)}{\R}{[\omega]}{\int_{M}\omega}
	\]
	is an isomorphism! (We know it's well-defined by Stokes' theorem.) As a result, $\dim{}H^{n}(M)=1$.\nn
	Proof: We'll work with compactly-supported forms in open sets. Observe that $\int_{M}$ is nonzero -- $\int_{M}d\Vol>0$. We know $\int_{M}$ is a linear map. So we need to show $\forall\omega\in\Omega^{n}(M)$ such that $\int_{M}\omega=0$, $\exists\eta\in\Omega^{n-1}(M)$ such that $d\eta=\omega$.\nn
	Step 1: Assume $\omega\in\Omega_{0}^{n}(\R^{n})$ such that $\int_{\R^{n}}\omega=0$. Then we claim $\exists\eta\in\Omega_{0}^{n-1}(\R^{n})$ such that $d\eta=\omega$. Observe that the homotopy axiom implies $H^{k}(\R^{n})=0$ for $k>0$, so such an $\eta$ exists, and the claim is that $\eta$ can be chosen to have compact support. For this, see Lemma 17.27 in Lee.\nn
	Now, back to the manifold case. Let $\set{U_{i}}$ be a finite cover $M$ (possible by compactness) such that $\forall{}i$, $U_{i}\cong\R^{n}$ diffeomorphically. WOLOG if $M_{k}=U_{1}\cup\cdots\cup{}U_{k}$, then $M_{k}\cap{}U_{k+1}=\emptyset$. (Use $M$'s connectedness, and renumber the $U_{i}$ if necessary. If no such $U_{i}$ existed, then union all of them, and we would have two disjoint open sets that cover $M$, making it disconnected. Oops!)\nn
	Introduction: If $\omega\in\Omega_{0}^{n}(M_{k})$ is such that $\int_{M_{k}}\omega=0$, then $\exists\eta\in\Omega_{0}^{n-1}(M_{k})$ such that $d\eta=\omega$. For $k=1$, see the previous claim. Then use induction and a partition of unity to complete the proof. (See Lee for the full details.)\proven
}

\defn{
	Let $F:M_{1}\to{}M_{2}$ be smooth, where $M_{1}$ and $M_{2}$ are compact, connected, oriented manifolds of the same dimension, $\dim{}M_{1}=\dim{}M_{2}=n$. Consider $F^{*}:H^{n}(M_{2})\to{}H^{n}(M_{1})$. By the previous result, we know that\n $\dim{}H^{n}(M_{2})=\dim{}H^{n}(M_{1})=1$. Thus, $F^{*}$ is multiplication by a scalar, and that number is called the \u{degree} of $F$.\n
}

\par\noindent
$\forall{}c\in{}H^{n}(M_{2})$, $\int_{M_{1}}F^{*}(c)=\deg{}F\int_{M_{2}}c$. That is,
\[
	\begin{tikzcd}
		H^{n}(M_{2}) \arrow[r,"F^{*}"] \arrow[d,"\int_{M_{2}}"] & H^{n}(M_{1}) \arrow[d,"\int_{M_{1}}"]\\
		\R \arrow[r,"\ptxt{mult. by}\;\deg{}F"'] & \R
	\end{tikzcd}
\]

\thm{
	Let $q\in{}M_{2}$ be a regular value of $F$. Write $F\inv(q)=\bigcup_{i=1}^{N}\set{p_{i}}$. This is a zero-manifold and compact, so it's the finite disjoint union of points. Define
	\[
		(-1)^{p_{i}}=\left\{\begin{array}{ll}
			1 & dF_{p}\;\ptxt{preserves orientation}\\
			-1 & \ptxt{otherwise}
		\end{array}\right.
	\]
	Then
	\[
		\deg(F)=\sum_{i=1}^{n}(-1)^{p_{i}}\in\Z
	\]
	Proof: $F$ is a local diffeomorphism at each $p_{i}$. So we can argue that $\exists{}V$ a neighborhood of $q$ and $U_{i}$ a neighborhood of each $p_{i}$ such that $\restr{F}_{U_{i}}^{V}$ is a diffeomorphism. That is, $F$ is evenly covered at $q$. Let $\omega\in\Omega_{0}^{n}(V)$ be a bump function such that $\int_{V}\omega=\int_{M_{2}}\omega=1$ (by extending $\omega$ to $0$ on $M_{2}$ outside of $V$). What is $\int_{M_{1}}F^{*}\omega$? Well, it's equal to $\deg(F)\cdot{}1=\deg(F)$. But $F\inv(V)$ is the union of the $U_{i}$'s, so
	\[
		\int_{M_{1}}F^{*}(\omega)=\sum_{i=1}^{N}\underbrace{\int_{U_{i}}(\restr{F}_{U_{i}})^{*}\omega}_{\hspace{-55pt}\mathrlap{=\pm{}1\;\ptxt{by diffeo invariance of integrals}}}=\sum_{i=1}^{N}(-1)^{p_{i}}
	\]
	\proven
}

\cor{
	Gauss-Bonnet reduces to the (purely topological) statement $\deg(N)=\frac{1}{2}\chi(M)$ ($N:M\to{}S^{n}$ is the Gauss map).\n
}

\par\noindent
Observe:
\begin{enumerate}
	\item If $F,F':M_{1}\to{}M_{2}$ are homotopic, then $\deg(F)=\deg(F')$, because $F^{*}=(F')^{*}$.
	\item If $M_{1}=\d{}W$, and $F:M_{1}\to{}M_{2}$ extends to $\tilde{F}:W\to{}M_{2}$, hen the degree of $F$ is $0$.
	\[
		\begin{tikzcd}
			M_{1} \arrow[r,"F"] \arrow[d, hookrightarrow] & M_{2}\\
			W \arrow[ur,"\tilde{F}"]
		\end{tikzcd}
	\]
	Prove this by using Stokes theorem to show $F^{*}=0$.
\end{enumerate}

\end{document}