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\title{Math 217 -- Geometry and Physics -- Lec02}
\author{UCLA, Fall 2014}
\date{\formatdate{06}{10}{2014}} % Activate to display a given date or no date (if empty),
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\begin{document}
\maketitle


\section{More on manifolds}


\subsection{Remarks}

\begin{enumerate}
\item Partition of unity: $M = \cup_{\alpha \in I} U_\alpha$ where $U_\alpha$ coordinate charts. There exists $\rho_\alpha : M \rightarrow \R$, $\rho_\alpha \geq 0$ smooth with $\opname{supp}(\rho_\alpha) \subset U_\alpha$ and $\sum_\alpha \rho_\alpha = 1$. $\omega = \sum_\alpha \rho_\alpha \omega$, $\opname{supp}(\rho_\alpha \omega) \subset U_\alpha$.
\item Sheaf theory. local $\implies$ global (Cohomology)
\item Characteristic classes. local computations $\implies$ global invariants.
\item $M ( = \cup_{\alpha \in I} U_\alpha) \overset{f}{\longrightarrow} N ( = \cup_\beta V_\beta)$ is smooth iff $\psi_\beta \circ f \circ \varphi_\alpha^{-1} : \varphi_\alpha(U_\alpha) \rightarrow \psi_\beta (V_\beta)$ [is smooth?]. Where $\varphi_\alpha : U_\alpha \rightarrow \R^n$, $\psi_\beta : V_\beta \rightarrow \R^n$. 
\item $\CP^n = \C^n \cup \CP^{n-1} = \dots = \C^n \cup \C^{n-1} \cup \dots \cup \C \cup \set{ \rho_+ }$ [rho sub what?]
\end{enumerate}

Given a tangent bundle $TM$ we have a vector field (smooth section) $X = \sum_i a_i \pd{}{x_i}$ and differential form $\omega = \sum_i a_i dx_i$, where $\langle dx_i, \pd{}{x_j} \rangle = \delta_{ij}$. Furthermore, a differential $p$-form $\eta = \sum_I a_I dx_I \in A^p (M)$, where $I = \set{ 1 \leq i_1 < \dots < i_p \leq n }$ and $1 \leq p \leq n$, is a section on $\Lambda^p T^* M$. $p = \deg \eta$.

The exterior product $\Lambda^p T^* M \overset{\pi}{\longrightarrow} M$. 

Define $A(M) = A^*(M) = \oplus_{p=0}^n A^p (M)$. Then $\eta \wedge \eta' = (-1)^{pq} \eta' \wedge \eta$. 

$M$ is orientable iff there exists a nowhere vanishing $n$-form $\omega$ on $M^n$. 

$d : A^p (M) \rightarrow A^{p+1}(M)$, $d\eta = \sum_I da_I \wedge dx_I$.

$da_I = \sum_{j=1}^n \pd{a_I}{x_j} dx_j$.

\begin{exercise}
As an exercise, show that
\begin{enumerate}
\item $d^2 = 0$. $A^p(M) \overset{d}{\longrightarrow} A^{p+1}(M) \overset{d}{\longrightarrow} A^{p+2}(M)$. $\deg\eta = p$.
\item Leibniz rule: $d(\eta \wedge \eta') = d\eta \wedge \eta' + (-1)^p \eta \wedge d\eta'$.
\item $\int_M \omega = \sum_{\alpha \in I} \int_{U_\alpha} (\rho_\alpha \omega) = \int_M (\sum_\alpha \rho_\alpha) \omega$. 

$\int_{U_\alpha} (\rho_\alpha \omega) = \int_{\varphi(U_\alpha) \subseteq \R^n} (\varphi_\alpha^{-1})^* (\rho_\alpha \omega)$
\end{enumerate}
\end{exercise}

\subsection{Stokes}

$\int_M d\eta = \int_{\partial M} \eta$ where $\eta \in A^{n-1}(M)$ implies that $\int_M d\eta = 0$ if $M$ closed compact. 

Consider: $f : M \rightarrow M'$ a smooth map. Then the pullback $f^* \eta'$ of $\eta' = \sum_I b_I(y) dy_I \in A(M')$ is given by 
\begin{equation}
f^* \eta'(x) = \sum_I b_I(f(x)) df_I.
\end{equation}

\begin{exercise}
Show that
\begin{enumerate}
\item Chain rule: $f^* \circ d_{M'} = d_M f^*$
\item $f^*(\eta_1 \wedge \eta_2) = (f^* \eta_1) \wedge (f^* \eta_2)$
\end{enumerate}
\end{exercise}

\subsection{de Rham Cohomology}

\begin{equation}
H_{\text{dR}}^p(M, \R) = \frac{\ker d}{\Im d} \Big|_{A^p(M)}
\end{equation}
where of course $\Im d \subseteq \ker d$. Then 
\begin{equation}
0 \rightarrow A^0(M) \overset{d}{\longrightarrow} A^1(M) \rightarrow \dots \overset{d}{\longrightarrow} A^p(M) \overset{d}{\longrightarrow} \dots \overset{d}{\longrightarrow} A^n(M) \rightarrow 0
\end{equation}

\subsection{Singular homology, Cohomology}

Famous theorem of de Rham: $H_{\text{dR}}^p(M, \R) \cong H_{\text{sing}}^p(M, \R)$ when $M$ is a compact, closed manifold. We will also see later that it is $\cong \ker \Delta_d$, where $\Delta_d$ is the elliptic operator. 

Cohomology ring: $H_\text{dR}^* (M) = \oplus_{p=0}^n H_\text{dR}^p(M)$. 

\begin{eqn}
\begin{matrix}
H_\text{dR}^p(M) & \times & H_\text{dR}^q(M) & \rightarrow & H_\text{dR}^{p+q}(M) \\
[\eta] = [\eta+d\alpha] & & [\eta'] = [\eta+d\alpha'] & \mapsto & [\eta \wedge \eta']
\end{matrix}
\end{eqn}

\subsection{Poincare duality}

\begin{eqn}
\begin{matrix}
f^* & : & H_\text{dR}^*(M') & \rightarrow & H_\text{dR}^*(M) \\
&& [\eta'] & \mapsto & [f^* \eta']
\end{matrix}
\end{eqn}

\begin{eqn}
\begin{matrix}
H_\text{dR}^p(M) & \times & H_\text{dR}^{n-p}(M) & \rightarrow \R \\
[\eta] = [\eta+d\alpha] &\quad &[\eta'] = [\eta'] &\mapsto \int_M \eta \wedge \eta'
\end{matrix}
\end{eqn}


\subsection{Betti numbers}
Since
\begin{equation}
H_\text{dR}^{n-}(M, \R) \cong H_\text{dR}^p(M, \R) \cong H_\text{dR}^p(M, \R)
\end{equation}
we have $b_p(M) = b_{n-p}(M)$, where
\begin{equation}
b_p(M) = \dim_\R H_\text{dR}^p(M, \R)
\end{equation}
is the $p$th Betti number of $M$.

Euler number: $\chi(M) = \sum_{p=0}^n (-1)^p b_p(M) = \int_M^{2n} e(TM)$ by Gauss-Bonnet, where $e(TM)$ is the Euler characteristic class.

\begin{corollary}
$\dim_\R M = \text{odd} \implies \chi(M) = 0$
\end{corollary}

\begin{conjecture}[Chern]
Let $M$ be an affine-flat compact closed manifold. Then $\chi(M) = 0$. 
\end{conjecture}








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