\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathrsfs}
\usepackage[all]{xy}
\setlength{\parindent}{0pt}
\setlength{\parskip}{2ex}
\addtolength{\hoffset}{-1cm}
\addtolength{\textwidth}{2cm}
\title{MAT :Differential Topology\\August 31, 2004}
\author{Gabriel C. Drummond-Cole}
\newtheorem{lemma}{Lemma}
\newtheorem{exercise}{Exercise}
\newtheorem{prop}{Proposition}
\newtheorem{example}{Example}
\newtheorem{defi}{Definition}
\newtheorem{thm}{Theorem}
\newtheorem{cor}{Corollary}
\newtheorem{remark}{Remark}
\newcommand{\boundary}{\delta}
\newcommand{\interior}{\ ^\circ}
\newcommand{\percent}{\%}
\newcommand{\transpose}{t}
\newcommand{\crossing}[4]{\xymatrix{#1\ar[dr]|\hole&#2\\#3\ar[ur]&#4}}
%\newcommand{\prop}{\begin{prop}}
%\newcommand{\endprop}{\end{prop}}
%\newcommand{\exercise}{\begin{exercise}}
%\newcommand{\endexercise}{\end{exercise}}
\begin{document}
\noindent
\maketitle
This is a basic class in differential topology.  We hope to do at least the basic concept of differential topology.  As you may have already checked on the previous syllabus (and hopefully I'll have current information on the webpage later today or tomorrow), we'll start with topological manifolds.  Then smooth manifolds, submanifolds, imbeddings, and then vector bundles starting with the tangent bundle, and then Sard's theorem, transversality.  Sard's theorem is a very important theorem.  It comes up where you wouldn't expect it to resolve everything, where you just say that the set of principal values has zero measure.

The field of differential topology has to do with global invariants of manifolds.  What else? Maybe some homotopy stability, equivalences of when two maps are homotopic.  It's not easy to see that two maps with the same degree must be homotopic, although the converse is quite easy.  So we'll talk about degree theory, and something else, intersection theory hopefully, and if we have time Morse theory or some other theory that everyone will be interested in.

There will probably be homework, a midterm and possibly a final.  The exams will be take-home.  It's good for learning.  At the end maybe each of you would be presenting a topic.  And then give a little presentation to the class.  It might be twenty minutes so you'd just be giving a basic idea.  That would be in the last weeks of the class.  When we have homework and everything it will be on my webpage.

What I want to do is, if you can write your emails on a piece of paper here, so that if I need to communicate with you, I can.  Any questions?

Myself I'm not going to stick to any book, but the main reference books will be listed on the webpage.  Most of the stuff we do will be in Hirsch (?).  As a reference book it's good.  I didn't put anything on reserve yet, but I probably should, right?  Other questions?  Feel free to email me if you want to talk to me.  

So maybe we can start.  Today there are no scary things.  We'll talk about topological manifolds.  The usual notation is $M^n$ for an $n$-dimensional manifold.  For us this will be a Hausdorff topological space which is locally homeomorphic to $\mathbb{R}^n.$

The local condition says that for every point $x$ in $M,$ there is a neighborhood (which you can assume to be open) of $x$ homeomorphic to $\mathbb{R}^n.$  We'll also usually want a countable basis, and sometimes you can relax the Hausdorff assumption.

Now it might seem that the Hausdorff condition is implied by the local condition, but that is not the case, e.g., the line with two origins, which you may have seen in a topology class.  That is, $M=\mathbb{R}\cup\{p\}.$  The topology is that $U$ is open if and only if $U\subset \mathbb{R}$ is open or $p\in U$ and $(U\cap\mathbb{R})\cup\{0\}$ is a neighborhood of 0 in $\mathbb{R}.$  So this is a 1-dimensional non-Hausdorff "manifold."

Since we're dealing with homeomorphism as our equivalence, local invariants such as curvature don't even exist.  One more remark: sometimes people start by defining this to be a metric space.  With our other assumptions, second countability implies metrizability.
\begin{itemize}
\item The basic example is $\mathbb{R}^n.$
\item We can look at open subsets of $\mathbb{R}^n,$ for instance an open ball $B^n.$
\item Open subsets of any manifold are also manifolds themselves.
\end{itemize}
We don't want to think of these as sitting in $\mathbb{R}^n,$ although that's always possible, but as things by themselves.

Now I should have said above that "there exists a neighborhood and an integer $n$ such that the neighborhood is homeomorphic to $\mathbb{R}^n.$"  So here we can start talking about dimension
\subsection{dimension}
Let $U,V$ be two neighborhoods of $x\in M.$  Is it possible for $U\cong \mathbb{R}^n,\ V\cong\mathbb{R}^m,$ and $n\ne m?$  No, it is not.  It relies on the fact that $\mathbb{R}^n$ and $\mathbb{R}^m$ are not homeomorphic.  You've probably seen before that $\mathbb{R}$ and $\mathbb{R}^2$ are not homeomorphic, since removing a point from each yields one connected and one disconnected space.
\begin{remark}
In higher dimensions, it is nontrivial to see the failure of the homeomorphism.
\end{remark}
Now when your manifold is disconnected, dimension can differ on different components.

\begin{example}
$S^1,$ the circle.  This is the only connected compact 1-dimensional manifold.  We can view this in 3 ways.
\begin{itemize}
\item $\{(x,y)\in\mathbb{R}^2|x^2+y^2-1=0\}=f^{-1}(0)$ where $f:\mathbb{R}^2\to\mathbb{R}$ is $f(x,y)=x^2+y^2+1.$  Now $0$ is a regular value of $f$ so the rank of the Jacobian is 1.  Then the inverse function theorem tells us that $S^1$ is a manifold.  Many manifolds arise this way, and it's a great way to prove that something is a manifold.
\item $S^1=\mathbb{R}/\mathbb{Z}$ (identification as a group quotient) with the quotient topology.  This readily lends itself to the definition of the torus.
\item $S^1=[0,1]/\{0\equiv 1\}.$   This is again with the quotient topology.  This idea of identifying boundary components is useful when gluing surfaces or attaching handles.
\end{itemize}
\end{example}
Note: $\mathbb{R}$ and $S^1$ are the only connected $1$-dimensional manifolds.

Now let's say you wanted to show that $S^1$ was a manifold via the definition.  Then you could use the usual one $\phi:(0,1)\to S^1$ by $t\to (\cos 2\pi t,\sin 2\pi t)$ and a complementary function $\phi:(-1/2,1/2)\to S^1$ by the ame formula.  Another thing is the stereographic projection.  This is important because it generalizes to show that the higher dimensional spheres are manifolds as well.

Particularly, the "sphere," i.e., the 2-sphere, $S^2,$ is our first compact $2$-manifold.  We'll get another manifold but first let me remind you that $M_1^{n_1}\times M_2^{n_2}$ is an $n_1+n_2$-dimesnsional manifold.  So $S^1\times S^1$ is the torus $T^2.$

We can also get tori of higher dimensions by looking at the products of more than two circles, of course.  We can also obtain the torus as $\mathbb{R}^2/\mathbb{Z}^2$ or by identifying the opposite sides of a square.  You can get surfaces of higher genus this way as well.  You can also obtain the 2-holed handlebody by attaching handles, i.e., as the connected sum of two tori.

Now we want to talk about the mobius band and the projective plane.  These are familiar spaces.  We will obtain the mobius band as an identification of the square.  For the real projective plane, we can identify as $\mathbb{R}P^2=\{$ linesin $\mathbb{R}^3$ through the origin$\},$ or as $S^2$ with antipodes identified, i.e., as $\mathbb{R}^3\backslash\{0\}/\{x\equiv \lambda x|\lambda\ne 0\}.$  You can't visualize this in $\mathbb{R}^3.$

So this generalizes immediately to $\mathbb{R}P^n.$  These differ according as whether $n$ is even or odd.  For $n$ odd we get orientable spaces, while for $n$ even you get nonorientable spaces; for instance $\mathbb{R}P^1$ is the circle.

Another thing I want to mention today are manifolds with boundary.  You can think of, say, the disk or the mobius band.  These are not manifolds as previously defined.  Instead we allow a manifold-with-boundary to have points with neighborhoods homeomorphic to $\mathbb{H}^n=\{(x_1,\ldots,x_n)\in\mathbb{R}^n:x_1\ge 0\}.$

Note that if $M$ sits in $\mathbb{R}^n$ then $\boundary M$  is different from the old sense of boundary.  E.g., $(0,1)$ in $\mathbb{R}$ has $\boundary (0,1)=\emptyset,$ where as the topological boundary is $\{0,1\}.$  A compact manifold with no boundary is called closed.

Next time we'll start with smooth manifolds and discuss smooth manifolds.  That's all I wanted to say today.

\end{document}
