\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{mathrsfs}
\usepackage{latexsym}
\usepackage[all]{xy}
\setlength{\parindent}{0pt}
\setlength{\parskip}{2ex}
\addtolength{\hoffset}{-1cm}
\addtolength{\textwidth}{2cm}
\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{\transversal}{\pitchfork}
\newcommand{\wh}{[unintelligible]}
\newcommand{\aA}{\mathbb{A}}
\newcommand{\cC}{\mathbb{C}}
\newcommand{\Dd}{\mathscr{D}}
\newcommand{\eE}{\mathbb{E}}
\newcommand{\Ff}{\mathscr{F}}
\newcommand{\hH}{\mathbb{H}}
\newcommand{\Ll}{\mathscr{L}}
\newcommand{\Mm}{\mathscr{M}}
\newcommand{\Nn}{\mathscr{N}}
\newcommand{\nN}{\mathbb{N}}
\newcommand{\Oo}{\mathscr{O}}
\newcommand{\pP}{\mathbb{P}}
\newcommand{\qQ}{\mathbb{Q}}
\newcommand{\rR}{\mathbb{R}}
\newcommand{\Ss}{\mathscr{S}}
\newcommand{\Uu}{\mathscr{U}}
\newcommand{\Vv}{\mathscr{V}}
\newcommand{\Xx}{\mathscr{X}}
\newcommand{\zZ}{\mathbb{Z}}
\newcommand{\la}[1]{\mathfrak{#1}}
\newcommand{\injects}{\hookrightarrow}
\newtheorem{conj}{Conjecture}

\title{January 13, 2007}
\begin{document}
\maketitle

Dennis won't be here.  Dinner tonight is at Ming's.  You guys don't seem to be very good at signing up.  If you don't we'll do it at the restaurant.  Yasha, you're good at computing, will you do it?  Very good.  We're pleased to start with Edward Looijenga.

Thank you.  Thank you for inviting me.  Thank you for organizing this, bringing together an interesting group of people.  This is what I like about the topic, it brings together people from different fields.

When I gave this talk the title it now has, it soon dawned on me that I had given myself too tall an order.  There were limitations on time and my competence, so I'll concentrate on one specific topic.  I'm referring to the mapping class group.  Why is this an interesting thing to have a presentation of?  It appears elsewhere, in topological quantum field theory and in number theory.  Let's start off by recalling a few things.  

Start with a compact connected oriented surface with a single boundary.  So let $S_{g,1}=S,$ and then $\pi_0(Diff^+(S,\boundary S))=\Gamma_{g,1}$ and $\pi_0(Diff^+(S))=\Gamma^1_g$ and there is a surjection of the first onto the second.  Dehn twists are in this mapping class group.  There is the familiar idea that if $\alpha$ and $\beta$ are disjoint then the Dehn twists commute, while if they intersect in one point they satisfy the braid relation $D_aD_bD_a=D_bD_aD_b.$  There is a hyperelliptic counterpart of this, which requires a hyperelliptic involution (one whose quotient space is a disk).  So look at the diffeomorphisms which commute with this, take the connected components, and that is the hyperelliptic mapping class group $H_{g,1}.$  So by folklore
\begin{enumerate}
\item $H_{g,1}\to \Gamma_{g,1}$ is injective and has image the elements that commute with this involution.
\item $H_{g,1}$ is generated by the Dehn twists subject to the braid relations.
\item the square of the product of the generators is [somehow related to] the involution
\end{enumerate}
[I can't see the board.]
\section{Fukaya}
[I do not take lectures for overhead talks.]

\section{McMullen}
There's an interesting comparison, it's a theme, between moduli spaces and symmetric spaces.  Consider $\varphi$ an isometry from $\hH^1\to \hH^n/\Gamma.$  So here $\bar{\varphi}(\hH^1)$ can be wild.  But if you look at a hyperbolic two-plane, a result says that $\bar{\varphi}(\hH^2)$ is always geometric.   It's $N^k$ isometrically immersed in $M^n$ with finite volume.  What is remarkable is that every two-plane, not just most, have geometric structure.

The way one goes about proving this is y passing to the frame bundle, and looking at an orbit of $SL_2(\rR).$  The same thing is true in the case of moduli space.  If we look at the bundle of one-forms, we can get an $SL_2(\rR)$ action on the moduli space.  If you have a holomorphic one-form, you can think of $\cC$ as $\rR^2,$ take a map that takes the square to a parallelogram.  The composition $d(A\circ f)$ still gives a complex structure.  We define that to be $A$ applied to $(X,\omega).$  These are area preserving so it's useful for these to act on unit things in moduli space.

So underlying this is the following, in heuristic form.  Orbit closures and invariant ergodic measures for the action of $SL_2(\rR)$ on $M_2$ are algebraic.  They could be, well, let $SL(X,\omega)$ be the stablizing (for $(X,\omega)$) subgroup of $SL_2(\rR).$  If $Z$ is the closure, it might be a bundle over a curve, if $Z$ is a lattice, it could be $\Omega_1E_D, \Omega_1 M_2(Z),$ or it could be all of moduli space.

So what I want to spend most of this talk describing is the Teichmueller curves in genus two.  There are curves totally geodesic for the hyperbolic metric.  I want to find all of the pairs so that the stabilizer is a lattice.

If you want insight into this question, rather than talking to an algebraic geometer you should talk to a billiard theorist.

Everyone is familiar with the idea of a billiard ball bouncing off of the walls of a square.  You can reduce to the geodesic flow on a flat torus.  You could have rational slope, so that the geodesic closes and you get a periodic trajectory.  Or it could be irrational so that you get uniform distribution.  We call things dynamically optimal if these are the only two possibilities.  Look at an $L$-shaped billiard table, that has trajectories like this one which are neither closed nor uniformly distributed.  To analyze this on an $L$-shaped table, we can look at reflections of this polygon in its sides.
\section{}
I'll talk about algebraic cobordisms, work from 2006 with Levine.  I'll work with algebraic varieties, in particular over $\cC.$ Think of them in projective space as zero loci of polynomials, here nonsingular and quasiprojective.

For motivation I'll talk about topological bordism.  If I have $\boundary W-M_1\sqcup M_2$ then I want to set $\boundary W$ to zero and say something like $M_1\sim M_2.$  So complex bordism is well-known.  I can say Quillen, Milnor, Novikov.  This is for a weakly complex manifold.  If $M$ is a smooth real manifold equipped with $E\to M$ a complex vector bundle which as a real bundle has the underlying structure of $TM\otimes \rR^n.$  This is what a weakly complex manifold is.  Take $MU_\star(X)$ for $X$ a topological space.  I want $MU_i(X)$ to be the free Abelian group generated by $[M\to X]$ where $M$ is weakly complex and the map is continuous, and then quotient by the bordism relation $[\boundary W\to X].$  If you've never thought of this before, the peculiar thing is the weakly complex objects.  Do it because it's well-behaved.  It's a generalized homology theory, meaning it satisfies every axiom other than the dimension axiom.  It's a beautiful theory, if $X$ is a point and I look with $\qQ$ coefficients, then $MU_\star(\bullet)$ is a ring over $\qQ$ with classes determined by projective space $\qQ[\cC\pP^1,\cC\pP^2,\cC\pP^3,\ldots].$

This is the picture we want to think about in algebraic geometry.  How can we think about this?  The most naive idea will be the first one we'll consider.  You look at an algebraic variety $Y$ which is nonsingular, and have a map $\pi$ to some curve, say $\pP^1.$  I look at the fibers over $0$ and $\infty,$ and I assume that these fibers $Y_0,\ Y_\infty$ are smooth.  I could interpret the bordism relation $\boundary Y=0$ means that $[Y_0]-[Y_\infty]=0.$  This has no flavor of bordism theory.  We know that the topology doesn't change at all.  So this turns out to be not such a good idea.  So even though we're at an early stage we come to a fork.  How could we develop this theory?  The left fork, which is the first one which was taken by Levine and Morel pretty recently, to use as a guide Quillen's axiomatic formalism for complex cobordism.  One follows this and comes to an algebraic cobordism theory.  The right fork, the main path of this lecture, is to look harder for some kind of geometric interpretation.  In fact the approaches are the same, give the same theory.

So the first thing I want to say, I want to talk about the axiomatic approach and then about the geometric approach.  Then there's the theorem that they're the same, then I'll say some applications.

So, um, some first discussion now about the axiomatic approach.  So, I, you should ask questions if I get stuck.  There will be not just no moduli space, but no specific variety unless you ask for it.  Look at the category of schemes over $\cC.$  I want to consider the simplest possible schemes so let them be quasiprojective.  This algebraic bordism, I want to associate to every $X\in Sch$ an Abelian group $\Omega_\star(X),$ and the way this axiomatic approach goes, fully motivated by Quillen's view of complex cobordism, is to define $\Omega_\star$ as the universal oriented bordism homology theory on $Sch.$  The outcome is that they prove that there is a universal theory that satisfies these axioms.  They are all ones thta you would not deny.

\begin{itemize}
\item It should be additive under $\sqcup.$
\item There should be pushforward under projective maps.  If $f:X\to Y$ then $f_\star:\Omega_\star(X)\to \Omega_\star(Y).$\
\item There should be pullback for locally complete intersection morphism, so that if $f:X\to Y$ there is a pullback.
\item for a bilinear pair, we have $\Omega_\star(X)\otimes\Omega_\star(Y)\to \Omega_\star(X\times Y)
\item for $X=*$ then there is a $1$ in $\Omega_\star(Y)$.
\item Fol $f:E\to X$ a vectr bundle, we have $\Omega^*(X)\cong \Omega^*(Y).$
\end{itemize}
The theorem is that there exists $\Omega_\star, have the universal such theory.

Now I want to go down the other fork, look for a geometric approach.  That idea is that if we want a bordism theory, we need to allow something to happen, we can't allow just smooth fibers.

What does it mean in algebraic geometry when something happens?  I want to look at a map $Y\to $\pP^1$ but now I want to look at the fibers, what is the simplest thing that can happen when you allow things to happen?  You get smoth double points.  For instance this is two planes stuck together transversally.

This is the kind of picture I want so I want to write down a relation.  Now how do I write $\boundary Y=0?$  I could write $[A]-[B]-[C]=0$ but that's just wrong.

So what about this picture, with an elliptic curve.  There should be a correction for the intersection.  The right way to correct this, well, the intersection is the wrong dimension, there's a nice way to do it.  If you only remember one thing from this talk, this is the thing to remember.  This is the third try, and this will be correct.  The equation you want to write down is 
$$
[A]-[B]-[C]+[\pP_D]=0.$$  
You look at $D$ here.  It sits inside $B$ and $C.$  We can think of the normal bundles $N_{B/D}$ and $N_{C/D}.$  So $\pP_D$ over $D$ is $\pP(N_{B/D}\oplus \Oo_D)$  The transverse properties mean that $N_{B/D}\otimes N_{C/D}=\O_D.$  So they're just opposite of one another.

The main theorem is that
\begin{thm}(Levine, P.)
Double point relation, we only allow double point degeneracy.  The double point relations define $\Omega_\star.$
\end{thm}
So by this we mean we take the free Abelian group on $[M\to X]$ where $M$ is nonsingular quasiprojective and the map is projective, modded out by the relation $[A\to X]-[B\to X]-[C\to X]+[\pP_D\to X].$

Suppose I have a family with some $A$ and three pieces coming together.  On every intersection there would be a vector bundle and on the lower degeneracies.  But that's taken care of.

The first issue is what about the proof?  I have the algebraic formulation.  If I look at the theory $$\Omega_\star^{geom}\to_{easy} \omega^{axiom}_\star.$$

So $f_\star(\pi^\star)[1\in\Omega_\star(\star)].$
So I said, this is the easy way.  It's hard to show that it goes the other way.  To construct the arrow backwards you want to show that the definition satisfies the axioms of bordism homology theory.

What is $\Omega_\star(\bullet)?$ We'll come to \wh in a second.  Over a point this is an incredibly concrete thing.  You get $M$ a variety modding out by the unique map every time.  There is a map to $MU_\star(\bullet).$  We know this is generated by projective spaces, and each one is hit by the projective spaces so this is surjective.  One wants to know whether we are getting all the algebraic bordism of a point.  It's an isomorphism, as it turns out, and this is a nontrivial statement due to Levine and Morel.  
$$\xymatrix{
	0\\
	\Omega_\star(\bullet)\ar[u]\ar[r]^\sim&MU_\star(\bullet)\ar[r]\ar[d]_\cong&0
	\lL\ar[u]^\cong\ar[r]^\sim&\lL}$$

What is the main nontrivial input?  The new thing that is required is the weak factorization theorem, which says that if I have two nonsingular birational projective varieties, and if I can factor it by blowing up and down with smooth centers, this is due to Wlodarczyk.  

If I want to connect this to Chow groups, the answer is that the Chow and K theory live inside this.  So a formal group law for a ring $R$ is $F(u,v)\in R[[u,v]]$ such that
\begin{enumerate}
\item $F(u,0)=F(0,u)=u$
\item $F(u,v)=F(v,u)$
\item $F(F(u,v),w)=F(u,F(v,w))$
\end{enumerate}
So $\lL$ is defined as $\zZ[A_{ij}]_{i,j\ge 1}$ modulo the formal group law relations.  Here $F=u+v+\sum A_{ij}u^iv^j.$
So a formal group law on $R$ is the same as a map $\lL\to R.$  Then $\Omega_\star$ has Chern class operators $c(L):Z^\star Z_\star \Omega_\star(X)\to \Omega_\star (X)$ with $c(L_1\otimes L_2)=c(L_1)+C(L_2)+\ldots$  

[missed a bit]

If you want to see the connection to the Chow ring, another theorem is that the Chow rings arise by taking $\Omega_\star(X)\otimes_\lL \zZ,$ killing all the generators.  That gives you $A_\star(X).$  For Chow rings, with the formal group law, we know the rule is $c(L_1\otimes L_2)=c(L_1)+c(L_2).$

[That maps into $MU_\star(X)?$  Why is it not an isomorphism when it is not a point?]

It's not because this contains all the Chow data.

[What if you did something in the geometric definition where you did bundles over any connected algebraic curve instead of over $\pP^1$ times your thing?]

It's the same thing.  Oh wait, what do I want to say here?  

[So if, if you divide it couldn't contain all of Chow.]

So this would be algebraic equivalence?

[You didn't talk about a relative version or a long exact sequence?]

In algebraic geometry you often instead get a short right exact sequence.

[Do you have time to say anything about applications?]

These theories involving integration over moduli spaces, relative theory or symplectic field theory, if you understand what happens when varieties break once, these theories should be determined by what they do to projective spaces.

[With cobordism of manifolds, it's obvious this is transitive by gluing.  Is that obvious in this case?]

You can try to do the following thing.  Look at $\pP^1,$ if you have two $\pP^1$s in families, you can try to do this thing, and sometimes you can do this.  You can think about doing it.

[Let's start again in a half hour.]

 




\end{document}


dAwdA (t) + dA(t)w *dA