\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathrsfs}
\usepackage[all]{xy}
\setlength{\parindent}{0pt}
\setlength{\parskip}{2ex}
\addtolength{\hoffset}{-1cm}
\addtolength{\textwidth}{2cm}
\title{Dennis Sullivan Course Notes\\March 4, 2005}
\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}
\newtheorem{asdefi}{Analysis Situs Definition}
\newcommand{\boundary}{\delta}
\newcommand{\interior}{\ ^\circ}
\newcommand{\percent}{\%}
\newcommand{\transpose}{t}
\newcommand{\wh}{[unintelligible]}
\newcommand{\crossing}[4]{\xymatrix{#1\ar[dr]|\hole&#2\\#3\ar[ur]&#4}}
\begin{document}
\maketitle

I used to be at this place called IHES and physicists and mathematicians worked together.  Once I saw them arguing with one another, for like twenty minutes.  They didn't understand what a quotient vector space was.  They had a Hilbert space, and they wanted a basis in terms of the original functions.  They could have just passed to the quotient.

I remember this one student, this engineering student.  She was smart, I remember the weekend that she learned what it was.  Only mathematicians do that.  This, I made that longer so he'd get out of the room.  He's having a problem with having a concept that's sort of abstract.  It's very simple.

Anyway, uh, so, uh, what did you want to ask me?

[Students aren't coming to my office hours.  Make them.]

[How?]

[Throttle them.]

[Withdraw points.]

[Why are you looking for trouble?]

[I just want them to learn.]

Right.

[It's a big day in math when you realize that you can do whatever you want in math.  Let whatever.]

I could suppose my way out of a paper bag.  

So, you're my grader for this course but you're not here on Monday.  So I need, I'm going to be away March 11, Friday, and the following Monday too.  So, I want, I'm going to have a review and quiz for my undergraduate class.  You can do the review but you can't do the quiz.  You can't do the Monday but you can do the Friday.

[By definition of Friday?]

Right.  You can tell them about your office hours.  I need a volunteer to give the quiz on Monday.

This problem Travis gave me, I feel like I'm the graduate student and he's the aloof advisor.  I need a volunteer to survey Ren\'e Thom's cobordism theory on Friday and then Monday.  It's great stuff.

All right.  So I'm struggling with this problem here.

[Can you classify two by two symmetric matrices?]

Over what ring?  I got into it, got to a Diophantine problem and then felt was it was like, and got tired of pursuing that line.  I remembered the original question was, ``which ones are realized by manifolds?''  This question actually helps you understand the classification.

So you've got $T$ and the pairing $\langle,\rangle.$  So either we're in $4l+1$ and this is skew symmetric with $T=H_{2i}$ or it's $4l-1$ and it's symmetric with $T=H_{2i-1}.$

There's going to be a construction from $L$ to $T.$  It's a functor.  Let's do the symmetric one first.  Suppose I have a free abelian group of finite rank and some pairing $L\otimes L\to \mathbb{Z},$ symmetric but not necessarily perfect.  So you have a symmetric matrix $a_{ib}$ where $a_{ij}=\langle x_i,x_j\rangle.$  So the determinant is nonzero but it need not be $\pm 1.$  A perfect pairing it would be a unit.

In the rationals, consider tensoring with the rationals, then it is a perfect pairing over the rationals.  Take $V=L\otimes \mathbb{Q}.$

Then you define the dual lattice $L^*$ as $\{all\ v\in V:\langle v,x\rangle\in\mathbb{Z}$ for all $x\in L\}.$  Notice that $L\subset L^*,$ but $L^*$ is bigger precisely when $L$ is not a perfect pairing.

We have this map $L\to Hom(L,\mathbb{Z}).$  This is injective but not necessarily onto.  Now we consider the vectors in the rational version.  $L^*$ is in fact $Hom(L,\mathbb{Z}).$

You then get a short exact sequence $0\to L\to L^*\to T\to 0,$ where $T=L^*/L$ is torsion, and $|T|=\det(a_{ij}).$ 

Okay, so now notice that we have, if we take $\langle L^*,L^*\rangle\to \mathbb{Q}.$  This implies that $T\otimes T\to\mathbb{Q}/\mathbb{Z}.$  Take two elements of $T,$ lift to $L^*,$ and take the inner product.  Since this changes by an integer if you change by $L,$ you get the quotient by the integers.

This torsion pairing is perfect.  It arises geometrically.  Suppose, I'll say that $(T,\langle,\rangle)$ is an algebraic boundary if there exists $L,\langle$ giving it.


Say you have a manifold $W$ with boundary$M$  with torsion in the middle and zero elsewhere.  Now you have $L=H_{2l}W,\ L^*=H_{2l}(W,M),\ T=H_{2l-1}M.$

You get a long exact sequence of homology associated to a short exact sequence of chain complexes.

Then there's a duality between $H_i$ of the manifold and $H_{n-i}$ of the relative homology.  Let's pretend that there's no torsion.  Modulo torsion it says the intersection pairing is a perfect pairing.  Your cycles have intersection matrix, the interior homology is not affected by adding a disk; then the paiting is perfect.  Suppose $H_{2l}(W,M)$ is free and $H_{2l}(W,M)$ is torsion.  If the map $H_{2l}(W,M)\to H_{2l-1}M$ is $\alpha$ then $\alpha^*:H_{2l}M\to H_{2k}(W,M)$ is relative.

Now what?  By dualizing and playing around we get a  theory.  So, so we deduce, this is going to be half the proof of a theorem.


\end{document}