\documentclass[11pt]{article}

  \usepackage{latexsym}
  \usepackage{comment}
  %\usepackage{showlabels}
  \usepackage{amsmath,amsthm,amsfonts,amssymb,graphicx,epsfig,latexsym,color,amssymb,mathabx,graphicx}
  %\usepackage{showkeys}
  \usepackage{epsf,enumerate,bbm,stmaryrd}
  \usepackage[shortlabels]{enumitem}
  \usepackage{tikz}
  \def\rank{\operatorname{rank}}

  \newtheorem*{thm}{Theorem}
  \newtheorem*{lemma}{Lemma}
  
  \newcommand{\mb}[1]{\marginpar{\tiny #1 --mb}}

  \bibliographystyle{plain}   %plain

  \def\R{{\mathbb R}}
  \def\Q{{\mathbb Q}}

  \def\Z{{\mathbb Z}}
  \def\N{{\mathbb N}}
  \def\H{{\mathcal H}}  
  \def\G{{\Gamma}}
  \def\S{{\mathcal S}}
  \def\s{{\sigma}}
  \def\t{{\tau}}
  \def\e{{\varepsilon}}
  \def\F{{\mathcal F}}
  \def\X{{\mathcal X}}
  \def\P{{\mathcal P}}
  \def\pa{pseudo-Anosov\ }
  \def\diam{\operatorname{diam}}
  \def\spec{\operatorname{spec}}
  \def\stab{\operatorname{Stab}}
  \def\diag{\operatorname{diag}}
  \def\SL{\operatorname{SL}}
  \def\GL{\operatorname{GL}}
  \def\Sym{\operatorname{Sym}}
  \def\FF{{\mathbb F}}
  \def\A{{\mathcal A}}
  \def\l{{\lambda}}
  \def\C{{\mathbb C}}

  \newcommand{\<}{\langle}
  \renewcommand{\>}{\rangle}
  \newcommand{\intprod}{\mathbin{\raisebox{\depth}{\scalebox{1}[-1]{$\lnot$}}}}

  \begin{document}


  
  \vskip 1cm

  \centerline{\Large {de Rham cohomology}}
\vskip 1cm

\subsection*{Computations}

The main tools are Mayer-Vietoris and homotopy invariance (Poincar\'e lemma).

\begin{enumerate}[1.]

\item Compute the de Rham cohomology of the M\"obius strip.

\item Compute by induction on $n$ the de Rham cohomology of $\R^2$ with
  $n$ points removed. You can assume that the points are on the
  $x$-axis, although that doesn't affect the answer.

\item Let $M$ be a manifold. A {\it good cover} of $M$ is a finite open
  cover $\{U_1,U_2,\cdots,U_m\}$ such that for any subcollection the
  intersection
  $$U_{i_1}\cap U_{i_2}\cap\cdots\cap U_{i_k}$$
  is either empty or contractible. Show that de Rham cohomology
  $H^p(M)$ of $M$
  is finite dimensional for all $p$. Hint: Induct on $m$.

  Remark: Compact smooth manifolds admit a good cover. E.g. fix a
  Riemannian metric, and then it turns out that all sufficiently small
  balls are convex, in the sense that any two points are connected by
  a unique geodesic (shortest path), which implies that any finite
  cover by such balls is good.

\end{enumerate}

\subsection*{Real projective space}

\begin{enumerate}[resume]

\item Show that $H^p(\R P^n)=0$ when $0<p<n$. Hint: 
Given a closed $p$-form $\omega$ on $\R P^n$, pull it back to $S^n$ by
the projection $\pi:S^n\to \R P^n$ and use your knowledge of
$H^p(S^n)$ to find a $(p-1)$-form $\eta$ on $S^n$ with
$d\eta=\pi^*\omega$. Then arrange by averaging that $a^*\eta=\eta$ for
the antipodal map $a:S^n\to S^n$ and show that $\eta$ induces a form on
$\R P^n$.

\item Finish the computation of $H^p(\R P^n)$ by showing that $H^n(\R
  P^n)$ is 0 when $n$ is even and it is $\R$ when $n$ is odd. (Recall
  that $\R P^n$ is orientable iff $n$ is odd.) Hint: Use
  Mayer-Vietoris for open sets $U,V$ analogous to the sets we used for
  $\C P^n$ in class, and show that it gives an exact sequence
  $$0\to H^{n-1}(\R P^{n-1})\to \R\to H^n(\R P^n)\to 0$$
  and then induct on $n$ starting with $\R P^1=S^1$.
  
\end{enumerate}

\subsection*{The $n$-torus}

\begin{enumerate}[resume]

\item \label{2d}
  Think of the 2-torus $T^2$ as the quotient $\R^2/\Z^2$. the
  forms $dx,dy$ on $\R^2$ are translation invariant and descend to
  $T^2$, and they are closed. Show that their cohomology classes are
  independent in $H^1(T^2)$ by finding a closed submanifold along
  which the integral of one is 0 and of the other it is
  nonzero. (Perhaps a good warmup is to argue that the class of $dx$
  generates $H^1(S^1)$.)

\item Generalize Problem \ref{2d} to the $n$-torus and show that $\dim
  H^k(T^n)\geq \binom{n}{k}$.

\item Show that in fact $\dim
  H^k(T^n)= \binom{n}{k}$ using the Mayer-Vietoris sequence and
  induction on $n$ as
  follows. Write $S^1=U\cup V$ where $U,V$ are open intervals and
  $U\cap V$ consists of two intervals. Thus $T^n=T^{n-1}\times U\cup
  T^{n-1}\times V$. In the M-V sequence we have a map
  $$H^k(T^{n-1}\times U) \oplus H^k(T^{n-1}\times V)\to
  H^k(T^{n-1}\times (U\cap V))$$ which can be identified with
  $$H^k(T^{n-1})\oplus H^k(T^{n-1})\to H^k(T^{n-1})\oplus
  H^k(T^{n-1})$$
  The image of this map is the diagonal (this follows from the
  Poincar\'e lemma). Show that this implies that we have an exact
  sequence
  $$0\to H^{k-1}(T^{n-1})\to H^k(T^n)\to H^k(T^n)\to 0$$ and then induct.
  
\end{enumerate}

\subsection*{Miscellaneous}



\begin{enumerate}[resume]

\item Let $M$ be a closed (compact, no boundary) orientable manifold
  of dimension $n$ and let $\omega$ be an $(n-1)$-form on $M$. Show
  that $d\omega$ vanishes at some point of $M$.


\item Let $M\subset \R^n$ be a closed $(n-1)$-manifold. Show that $M$
  separates $\R^n$ into at least 2 components. Hint: If
  $\R^n\smallsetminus M$ is connected, show that there is a map
  $S^1\to\R^n$ with $I_2(f,M)=1 \mod 2$. Why is that a contradiction?
  You can use the fact that connected manifolds are smooth-path
  connected (can you prove it?)

\item If $M$ is also connected show that it separates $\R^n$ into exactly 2
  components and it is the boundary of the closure of each, so in
  particular $M$ is orientable. So for example $\R P^2$ cannot be
  embedded in $\R^3$. Hint: Tubular neighborhood theorem.


\end{enumerate}



\end{document}
