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\title{Stringy Topology Notes\\January 17, 2006}
\author{Gabriel C. Drummond-Cole}
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\begin{document}
\maketitle

We need to know if people are interested in the excursion today.  We have to have the bus company, a quick poll, are many people interested in the excursion. 

All the travel reimbursements are by MSRI.  For trainees only, that were offered money, we have your money here.  Everybody else you have to get it from MSRI.

\section{Teichner}
Good morning, I'll try my best to warm you up.  I can walk around and stretch, I have a slight advantage here.

I'm going to first recall some things I said yesterday.  I'll start with a reminder of the EFT version of $K$-theory and then do an analogy of a CFT version for ``elliptic cohomology.''  I'll try to draw analogues of everything.
\subsection{Reminder of the EFT version of $K$-theory}
Let me remind you of the space $\mathrm{EFT}_m=\{D$ on $\mathscr{H}\},$ where $\mathscr{H}$ is a graded $Cl_n$-algebra.  These were odd self-adjoint $Cl_n$-linear and $e^{-D^2}$ is compact.  Up to homotopy type you can put finite rank or anything in between.  

Because of the condition here this is a very easy space.  The eigenvalues are real, the eigenvalues are discrete and the eigenspaces are finite dimensional.  Because of linearity the eigenspaces are Clifford modules.  The eigenspace at $-\lambda$ by oddness is the grading involution of the eigenspace at $\lambda,$ so the $0$-eigenspace is a $\mathbb{Z}_2$-graded Clifford module.

There's a nice map into finite dimensional graded Clifford modules by $D\to \ker\ D.$  If I mod out by those that extend to Clifford $n+1$-modules (graded) then I get an isomorphism.

[What is the supersymmetry here?]

It is just that $D$ is odd.  This is the easy version.  So this map, if you have a configuration, you can push anything that's not in the kernel to infinity.  Then if there's a configuration, I should say, the topology comes from, if two points move together, the eigenspaces add, or eigenspaces can move to infinity and vanish.  They're all orthogonal and their sum is not $\mathscr{H}.$  If two parts come together at $0,$ they have to come in on both sides simultaneously.  If you allow such homotopies you get the quotient by the $n+1$ modules.  You probably remember that this is the Atiyah-Bott-Shapiro version of $KO_n(pt).$

[What is $\infty?$]

$D$ is very unbounded, if $D=\infty$ then $e^{-D^2}$ is compact in the usual sense.  Instead of $\pi_0(EFT_n)$ you take homotopy classes of \wh.  

What we did, we made this simple space complicated by introducing EFTs, we did this because people tried to generalize the simple version and failed.

The other definition I wanted to remind you of is that this space $\mathrm{EFT}_n$ is also the space of all $1$-dimensional supersymmetric Euclidean Field theories of defree $n.$  If you have a field theory $E$ then $E(pt)=\mathscr{H}.$  Then degree $n$ means you get graded $Cl_n$-modules.  You can also study $E([0,t]).$  You get $e^{-tD^2}.$  You get this if you're not supersymmetric, you will also get some $D$ if you are supersymmetric.

One nice thing is, you can take a field theory and evaluate it also on a circle of length $t,\ S^1_t.$  So $E(S^1_t)\in\mathbb{R}.$  The beauty is that if you evaluate an EFT on $S^1_t$ it will not depend on $t$ and it's going to be an integer.  This real number is just the supertrace $str(E([0,t])).$  This follows from the axioms of a field theory.  The very first argument in a physics book on this will say that that this is independent of $t$ and an integer.  It's the index from $KO_n\ n\equiv 0\pmod 4$ or the superdimension.

[Strictly speaking this only works for $n=0.$  Also the circle needs a spin structure.]

Now let me enlarge my diagram a little bit.

$$
\xymatrix{
M^n\in\{\mathrm{closed\ Riemannian\ spin\ manifolds,\ dim} n\}\ar[d]\ar@{->>}[rr]\ar_{\hat{A}\ \mathrm{genus,\ analytic\ genus}}[ddr] &&\Omega_n^{Spin}\ar[d]_{\mathrm{Atiyah's}\ \alpha}\\
 \Dirac_M\in \pi_0(\mathrm{EFT}_n)\ar[dr]_{\mathrm{evaluate\ on}\ S^1_t}\ar[rr]_\cong&&KO_n(pt)\ar[dl]_{\mathrm{index}}\\&\mathbb{Z}
}$$
\subsection{}
So now say we have 
$$\xymatrix{
\mathrm{closed\ Riemannian\ string\ manifolds, dim}\ n, w_1=w_2=0=\frac{p_1}{2}\ar@{.>}[d] \ar[ddr]^{\mathrm{Witten\ genus}} \ar@{->>}[rr] &&\Omega_n^{string}\ar[d]_{\mathrm{Ando-Hopkins-Strickland}}\\
\pi_0(CFT_n)\ar[dr]^{\mathrm{evaluate\ on\ tori}}\ar@{.>}[rr]&&TMF_n(pt)\ar[dl]\\
&MF_n
}$$
This theory of topological modular forms maps to every elliptic cohomology theory uniquely.   It's not elliptic and is called this because it maps to modular forms.

The Witten genus is the index of the Dirac operator on the loop space, $\mathrm{index}^{S^1}(\Dirac_{LM}).$  The way he got to this is, he, and this is a pleasure to write this down again, he used the Atiyah-Bott fixed point theorem to calculate the index, the fixed point is the manifold.  That formula makes perfect sense and gives you the genus.  The refinement from $\Omega_n^{string}$ is a lifting of the Witten genus to something much more interesting.  One of the points of trying to make the space $\pi_0(CFT_n)$ is making a lot of new index theorems, we can't do this yet because the arrows are dotted.  Just as you can think of the index as information about the EFT, you can get it from information about the CFT.  So Witten had this whole quantum field theory in mind.  This is just one part of that.  It's  the supertrace associated to the annulus of a certain length.  There are other operators corresponding to other surfaces.
This is how you think of the Dirac operator, restrict to this surface and take the infinitessimal generators.  The infinitessimal generator for $E(\mathrm{annulus})=$``$\mathrm{index}^{S^1}(\Dirac_{LM})$''$\subset\sigma$-model of $M.$

Before the amazing thing was that the evaluation was an integer, here it is an integral modular form.

I think I made the analogy pretty well, I hate to erase this.

Now $MF_*:=$ integral (weak) modular forms.  A modular form of weight $k$ is a holomorphic section of $Det^{\otimes k}\to \mathscr{M}_{tori}\cong S^2,$ which contains a cusp $\tau=i\infty,$ we write $q=e^{2\pi i \tau}.$

Integral means you have integer coefficients in your $q$-expansion and weak means it's only meromorphic at the cusp.

So you get $MF_*\cong \mathbb{Z}[c_4,c_6,\Delta^{\pm 1}]$ with $c_4^3=c_6^2=1728\Delta.$

I should say the degree $n$ is defined as $2k,$ twice the weight.  This is because $Det^{\otimes k}\stackrel{\mathrm{spin\ str.}}{=}Pf^{\otimes 2k}.$  This is again only for dimensions divisible by four.

The $TMF\to MF$ is a rational isomorphism.

This was the easy part.  Instead of working over arbitrary rings, just use supersymmetry and you get the solid arrow, that's the punchline.
\subsection{}
\begin{defi}
$\mathrm{CFT}_n$ is the space of all $2$-dimensional extended supersymmetric CFTs of degree $n.$
\end{defi}
So a regular CFT is a functor from the $2$-dimensional bordism category of conformal structures to $Vect,$ so that is everything but three words.

Supersymmetry is what gives you holomorphic and the map by evaluating on tori.  The definition will be tomorrow.  

Now this business about extended.  First the definition and then the why.  We really have a functor from a cobordism $2$-category $CB^2\to Vect^2.$  You have $0$-morphisms, $1$-morphisms, and $2$-morphisms, I'm not going to take you through the definition, but the objects are $0$-dimensional conformal manifolds, the morphisms $1$-dimensional conformal manifolds, and $2$-morphisms $2$-dimensional conformal manifolds.  These have corners.  If you just do the gluings you can do, you'll rediscover the definition of a $2$-category.

In $Vect$ we also want to extend.  We'll take von Neumann algebras as objects, bimodules as morphisms and intertwiners as $2$-morphisms.  I don't want to say too much about this aspect but the reason we want this is
\begin{conj}
$\Omega \mathrm{CFT}_n\cong\mathrm{CFT}_{n+1}$\end{conj}
So we think we can define an $\Omega$-spectrum, but only if we use the extended version of a $CFT.$  What I really want to concentrate on is the degree $n.$  This will build in the correct weight of a modular form and be in analogy to the Clifford module structure in the $K$-theory. 

[You can make a space?]

Just as we fixed a Hilbert space we'll fix a von Neumann algebra and a bimodule.

$E$ has degree $n$ if \begin{itemize}
\item $E(X^1)$ is a module, $\mathbb{Z}_2$-graded, over $C(X)^{\otimes n},$ where $C(x):=\mathbb{C}l(\mathrm{spinors\ of}\ X).$  

[What's the inner product on the spinors?  Oh, they happen to have one on a $1$-manifold.]  

So the analogy is that $C(pt)=Cl_1$ and $C(pt)^{\otimes n}=Cl_n.$  This is cheating a little, it's really a superpoint, a good way to excuse yourself.

Now the other thing about the degree is, well, if $\Sigma^2$ is a conformal surface, there is a canonical graded irreducible module over the Clifford algebra over $C(\boundary \Sigma)$ namely $\mathrm{Fock}(\Sigma).$  Even for $\Sigma$ closed you get $\mathbb{C}$ and this is $Pf(\Sigma)$ which is still interesting.  You have to take restrictions of the \wh of spinors to the boundary, that's a Lagrangian, \wh.

I should have explained things better, I'm rushing.  The degree, it's a module of this Clifford algebra tensor $n.$
\item Really only $E(\Sigma, \psi)\in E(\boundary \Sigma)$ is defined for $\psi\in \mathrm{Fock}(\Sigma)^{\otimes n}$ and the association $(\Sigma,\psi)\mapsto E(\Sigma,\psi)$ is $C(X)^{\otimes n}$-linear in $\psi.$

Maybe I should point out that the Fock space has a canonical element, the vacuum element.  You can reduce to that element but somehow that's not quite right.

Note that on the torus we really get a degree $n$ modular form.  You get a number, and an element of something, you get a section of the dual power of the Pfaffian line.  
\end{itemize}
\begin{exercise}
This is the punchline, I will explain it to anyone who asks.  For $K$-theory this is equivalent to$E([0,t])\in \mathscr{K}_{Cl_n}(\mathscr{H})\subset \mathscr{K}(\mathscr{H}).$
\end{exercise}

Sorry for going over.

[Yesterday an important role was played by EFTs being a monoidal functor.  On the left, well, these are symmetric monoidal bicategories, so I should read \wh's 200-page definition of this.]

Just like in $K$-theory it came down to understanding intervals.  Now it comes down to triangles, but they have an interesting moduli space.  This whole machinery will reduce in the end, you only have one bicategory to understand, that has the right structures.

\section{Sullivan, String topology}

I happened to look at my abstract somewhere and it's totally unfamiliar to me.  Is Ralph Cohen here?  There's a longer blurb somewhere.  So I'm going to try to address myself mainly to the younger people in the audience, but occasionally make advanced comments about them.  So first let's, let's take a space and divide it into cell, you have the boundaries of the cells, you can get the homology, what is the structure on this.

You have the diagonal map $X\to X\times X,$ which is not cellular but can be homotoped to something cellular.  You get approximately a coalgebra structure.  With coefficients in a field you get a coassociative cocommutative coalgebra on homology $V\to V\otimes V$
 by the K\:unneth theorem.
 
 So that makes the homology into a coalgebra.  Now you have this structure.  There's another operation if $X$ is a manifold.  Then the cell decomposition, if you look at the dual cells, these dual constructions you get a construction like this, the pieces are cones over spheres.  They're cones in general but not necessarily cones on spheres.  This is the dual cell decomposition.  Then you can make a deformation of one cell composition into the other.  On various levels you construct a homotopy from one decomposition to another, then you can take two cycles, realize one in the dual decomposition, they're always transversal, so you get a product structure.  It turns out when you pass to homology this is independent of choices and you get a map $V\otimes V\to V.$
 
 Now, what is the relationship between them, they satisfy a compatibility, which is the following.  The coproduct, we just found an algebra structure by intersection.  Because of this structure, $V\otimes V$ is a module over $V$ on the left and right.  So compatibility is that the coproduct is a map of (two-sided) modules.  So the structure of the homology of a manifold is that you have two operations and they satisfy this compatibility.
 
 Any questions about that?  Let me prove this.  The proof is just a picture.  You have to look at the diagonal, that's hard, even for two dimensions. I'll draw the two manifold, imagine this is the diagonal.  You want to show that, let me just do the picture.  You take two cycles that have been made transversal.  Let me draw them on the diagonal, maybe just one.  It turns out the unit in the intersection algebra is the whole manifold.  So cross the, $a\otimes 1,$ I want to change it's name from $X$ to $M,$ this is $a\times M.$  Now we draw the $b$-cycle here on the diagonal.  Now notice that if you did $a\cap b$ and apply the diagonal, here's the intersection.  That is also the intersection of $b$ with $a\times m.$
 
The point is that the one factor $M$ is transversal to the diagonal.  Transversality of $b$ with $a$ to transversality of $a\times M$ with $\Delta b.$  

Just draw the picture and look at it.  That shows the diagonal map is a map of modules, bimodules.

This argument works even for an open manifold.  The homology of an open manifold has a certain kind of structure with a compatibility between them.

This is an example of a symmetric algebraic structure.  What that means is that you have a vector space or something, you have an underlying space and a lot of maps like $V^2\to V, V\to V^2,$ or like $V^3\to V^2.$ 

Then various compositions of these will be subject to some relations.  So the associative law is an identity between two ways of composition.

These things like quantum field theories and so on seem to come supplied with various structures, so it's interesting to come up with a language to describe these things.

I gave you one minor example, with two operations and an identity.  You can then talk about the dual structure, take the same structure and turn it around.  An algebra gives you a coalgebra.  An algebraic structure is called symmetric if when you reverse it you get the same structure.

A Lie algebra is not a symmetric structure, you reverse the arrows and get a coLie algebra.  The module condition in terms of diagrams, the module condition with the arrows reversed is the same relation.  A Hopf algebra is another example of an algebraic structure.

One imagines that a lot of these structures are going to be symmetric, these string things.  The next sentence is, if $X$ has a structure then its dual has that structure.

[If they're graded do you call them super symmetric?]

No, but I want everything to be graded.  If you reverse all the arrows you get the same structure on the dual space. There are nuances if it's infinite dimensional.

[You said Hopf.]

That's true.  When you have a symmetric structure, you have the dual.  You have two of these structures, one on homology and the other on the dual space (cohomology, but I'm not getting into that yet).

So you can ask whether these two structures are isomorphic.  Is there an isomorphism between the space and the dual space carrying one structure to another.

If the manifold is a closed manifold then this is true, $V\sim V^*,$ and then some people would call this a commutative Frobenius algebra.  A topologist would call it a Poincar\'e duality algebra.  So here we have, if so, then we can say that we have a self-dual structure.  We have Poincar\'e duality.

So there's this distinction, sort of like an open manifold has a symmetric structure, but it's not neccesarily self dual.

That's elementary, here's an advanced version.  There are important notions of $\infty$-algebras.  As long as $k$ and $l$ are nonzero you can always define the $\infty$ version, then you can symmetrize and talk about a quasiisomorphism between these two structures, an $\infty$-version of a commutative frobenius algebra.

The structure will be free, replacing vector spaces by chain complexes, with more generators, and when you take homology you get your original structure.

So this is a way to include this concept into the concept of Poincar\'e duality.  

Now why are these $\infty$ structures important.  At the chain level, I guess it was Ezra, who was talking about it, they can be moved from one model to another.  The underlying chain complexes, all you need to know is they have the same homology, you can move the structure over, but it becomes $\infty.$  That's the workhorse use of this concept.  The meaning is independent of the triangulation and the choice of representatives.

If you take the homology, go back to the space, the homology coalgebra structure, it's commutative and associative, if you try to do it on the chain level, you can symmetrize but then it won't be associative, you can fill in the associator with a homotopy, so due to Quillen, the infinity version of this will give the rational homotopy groups of a space.

So that's, and also historically, if you work over $\mathbb{Z}$ you can't make it commutative, Steenrod has a famous paper where he sort of does this $\infty$ and gets the Steenrod squares.

So this gives a bunch of important information in topology.  There's the recent work of Mandell, that says, if you take all the homotopies, the $E_\infty$-structure, that determines the homotopy type of $X.$  I conjecture that if you do the same thing for a manifold with the pairing, that there's an infinity version that would determine a lot, for instance the homeomorphism type of the manifold.

This intersection theory will lead to operations on the free loop space of a manifold, this is really based on the intersection.  Before I get there I want to say more about the intersection theory.  So um, I talked some to Fukaya, he's developed this point quite far, I'll only use a little, enough for what I need later.  So let's think about, for example, Poincar\'e duality, the statement that these two spaces are isomorphic.  So we have, I'm going to introduce some notation.  So $C_.$ will be the chains, and the cochains will be $C^.$  This is cohomology.  With field coefficients this will be dual.  We want maps $C^k \stackrel{?}{\to} C_{d-k}$ and $C_.\stackrel{?}{\to} C^.$

These won't be isomorphisms, what we want is a chain map in each direction so that the composition is chain homotopic to the identity.  So the map $C^k\to C_{d-k}$ is an element of $C_k\otimes C_{d-k}.$  So putting them together I get something in $\oplus C_k\otimes C_{d-k}.$  So I need a chain, this $\alpha$ is going to be the diagonal, but I have to deform it.   This map is $\Delta\cap U.$  The thing in the opposite direction should be in $C^k\otimes C^{d-k}.$  One way is cap product with the diagonal.  For this second one the idea is a cocycle that lives near the diagonal.  On a manifold this is like the Thom class of the bundle over the diagonal.  So the picture is, you take a cycle, and you cap with the cycle cross $M$ and project into the factor, that gives you a cocycle.

Anyway, there's a few simple properties that imply each composition is the identity.

I'm going very slow here.  The map the other way, the smoothing, that's more subtle.  If you were working with all chains, form are dual to chains, that would be a form that represents the Thom class, this is a cocycle or a differential form, now I'll make an advanced remark.  There's a strong feeling that a lot of the information about the structure of the manifold is contained in this element here.  To build this concretely takes a lot of work.  The way I'm going to use this is, by using this diffuse diagonal I can always intersect things without worrying about transversality.  I want to do computations relative to some graphs that come up in string topology.  I need a concrete picture of this.

[Why are you calling this diffuse?]

A cocycle is diffuse.  A cocycle dual to something gives one at each place.  It's made of the cycle and nearby things.

The Thom class construction will create a diffuse class like that.

You can always form a cap product with a chain, that means there's some kind of transversality working somewhere.

The next idea is what I'll call a real chain.  I'm going to go to real coefficients to make this concrete construction.  Here, going back to this advanced remark, if you have a space possibly with singularities, you can draw this picture and if you have a fundamental class, if you can find an object like this with support near a manifold, it has the homology properties of a manifold.  Then you can define the Pontryagin classes, so by logic their value is contained completely in the existence of this cohomology class

Now I want to consider cochains as forms.  A finite dimensinal family of chains with a measure on the parameter space is the dual space to forms.  Now what I want to do is build this cocycle as a differential form, which also determines a dual element, it's sort of like a real chain as well.  If you have $\omega,$ you can define $T_\omega(\eta)=\int_M \omega\wedge\eta.$  I want to build a real chain that will live near the diagonal and hav this form.  Its linear function on forms by integrating will be the same as a form and as a real chain.

So what is this thing?  Drop the triangulation and just think of intersecting objects in a manifold.  If you want to intersect things you look at $X\times Y\cap \Delta.$  You want this to be transversal to the diagonal.  It would be just as good if it were transversal to some other version of the diagonal.  This is a discussion from about page 80 in deRham's book.  I discuss this in 1976 in Inventiones (Serre) in ``Cycles for dynamical study for...''

You cover the manifold by charts and think of one ball as being Euclidean space contracted radially.  The translation group extends to be the identity outside the ball, $C^\infty$ up to the boundary.  You can take a Gaussion measure.  $\mathbb{R}^n,$ you, not Gaussian, a bump function with small support.  For each translation in the ball, the map will be the identity autside a ball and then you have a little bump.

Then you take the family, order the balls in some way and then there are various things you can do.  Take the cartesian product, that's your parameter space, take the product measure, then compose all of these little maps.  So the final composition is a small perterbation of the diagonal.

Then take them together with an ordering, if you compose two with disjoint support they commute, otherwise they overlap.  So the parameter space has no toplogical significance, I have a weight somewhere.  This is exactly what's in deRham's book.

The linear function it determines is actually given by, there's a differential form in here, the Thom class, $\int_{\mathrm{real\ chain}}=\int_M U\wedge \eta.$

The claim now, I'll use this to extend partially defined string topology operations to make them globally defined over the reals.  Notice right away, $c\times c',$ these being currents, take any two things like that and the formula $U\wedge $``c,'' that works, then you get the perturbed section, this is discussed in deRham's book.  So then $c\pitchfork c'.$  This is $\pi_1(U\wedge (c\times c')).$

The intersection will be defined by $\int_\lambda (c\times c'\pitchfork \lambda.$  This is a geometric version of this chain.  $\lambda$ is the perturbed diagonal.  Now we have the perturbed intersection theory.

[Why is it independent of the choice of the measure?]

It's not.  It's like, when did you stop beating your wife, it assumes, it's not independent.  It's the same homology class, homologically it's the same, it's a local perturbation.  That's the whole point of going into $\infty$-algebras, chain homotopies, that's the greatest gift of topology to the rest of math.  That's another advanced statement.

[Does this intersection commute with boundary?]

Yes, that's a good question.  Think of, well, if I may give my own tribute, I heard that when Bott gave a course on differential topology, he would say the most important thing about a smooth compact manifold is this smoothing operator.  Any object on a manifold can be transported and averaged, and this commutes with $d.$  This shows the deRham cohomology is finite dimensional.  The homology, as a Banach space, the identity map is compact so it's finite dimensional.

I've been wondering for thirty years, how do you read off the Pontryagin classes out of this $\mu,$ off and on, at least once a week.

\section{Nathalie Wahl, Mapping Class Groups and nonorientable}
We had to cancel the trip to Patzcuoro tomorrow.  Those of you who already paid, we thank you for your contribution.

So let me, I'll be talking about mapping class groups of nonorientable surfaces.

Recall:\begin{thm}
(Madsen-Weiss)\\
$\mathbb{Z}\times B\Gamma_\infty\cong_{H_*}\Omega^\infty\mathbb{CP}^\infty_{-1}=\Omega^\infty\mathbb{G}_{-2}^+=MTSO(2).$
\end{thm}
$B\cC_2^+\cong \Omega^\infty\Sigma\GG_{-2}^+$ (G-M-T-W) \\and then $B\cC\cong B\cC_{2,b}^+$ \\and then $\Omega B\cC_{2,b}^+\cong_{H_*}\mathbb{Z}\times B\Gamma_\infty$ (Tillman)

For the third you need Harer's stability theorem.

You can play with oriented cobordism, but your ground structure doesn't have this.  Can we use this nice theorem to say something about a nice group?  We need a version of Harer's stability.

This is also the same pattern as S\o rin's talk, using graphs for the first part and so on.

Fix notation: $S_{n,k}=\#_n\mathbb{RP}^2-(\amalg_k D^2).$  A crash course in topology, if you take $T^2\#\mathbb{RP}^2\cong \#_3\mathbb{RP}^2.$
Now $M_{n,k}=\pi_0\mathrm{Diff}(S_{n,k},\boundary),$ the mapping class group.
\\
$M_\infty=\mathrm{colim}(M_{n,1}\to M_{n+1,1}\to\cdots).$

We have maps
$$\xymatrix{
M_{n,k}\ar[d]_\gamma\ar[r]_\alpha\ar[dr]^\beta&M_{n+1,k}\\
M_{n,k-1}&M_{n,k+1}}$$
\begin{thm}
[W] $\alpha_i:H_i(M_{n,k},\mathbb{Z})\to H_i(M_{n+1,k},\mathbb{Z})$ is surjective for $n\ge 4i-1,$ an isomorphism for $n\ge 4i+2.$
\\ $\beta_i:H_i(M_{n,k},\mathbb{Z})\to H_i(M_{n,k+1},\mathbb{Z})$ is an isomorphism for $n\ge 4i+2.$
\\ $\gamma_i:H_i(M_{n,1},\mathbb{Z})\to H_i(M_{n,0},\mathbb{Z})$ is an isomorphism for $n\ge 4i+4.$
\end{thm}
\begin{cor}
$\mathbb{Z}\times BM_\infty\cong_{H_*}\Omega^\infty\GG_{-2}=MTO(2).$
\end{cor}
\begin{cor}
$H_*(M_\infty,\mathbb{Z}[1/2])\cong H_*(\Omega^\infty S^\infty(BO(2)_+),\mathbb{Z}[1/2]).$
\end{cor}
[What is $\GG_{-2}?]$

These are the orthogonal bundle to the canonical bundle for the Grassmannian of non-oriented $2$-planes, and then you pull back the big guy-

[I remember, thank you.]

$H^*(M_{\infty},\mathbb{Q})=\mathbb{Q}[\xi_1,\xi_2,\ldots]$ with $|\xi_i|=4i.$

Note that $\Omega B\cC_2\cong_{H_*}\mathbb{Z}\times BM_\infty.$  The right side is nonorientable, the left non-oriented.

The classes $x_i$ you have
$$\xymatrix{S\ar[r]&E\ar[d]\\&B}\leadsto \xymatrix{TE\ar[d]\\E}\leadsto p_1\in H^4(E).$$
Here this is the non-oriented surface bundle leading to the vertical tg bundle, a two-plane bundle.

Here $\xi_i=\tau^*(p_1^i)\in H^{4i}(B)$ where $\tau^*$ is the Becker Gottlieb transfer.

[What's that?]

[Some explanation.]


This which I just erased tells you something about the stable homotopy of the mapping class group.  It was known before that $H(M_{n,0})=\mathbb{Z}/2$ for $n\ge 7.$

Let me move on to giving you an idea of the proof of the theorem.
\subsection{Proof of the stability theorem}
If you have $G_n,n\ge 0,$ and you want stability for the homology, you want to find a  simplicial object, maybe a simplicial complex $X_n$ with $G_n$ acting on $X_n$ and then \begin{itemize}\item$X_n$ highly connected
\item $St(p-\mathrm{simplex})\sim G_{n-p-1}$
\item Action as transitive as possible (and transitive on vertices).
\end{itemize}
[This lets you do something.]

The more connected, the better the stability, the stabilizer helps with the spectral sequence, and the third condition makes the spectral sequence simpler.

This tells you $d^1:H_q(St(0-\mathrm{simplex}))\to H_qG_n.$

I'll give you a couple of examples.

Let me look at orientable case (Harer).

He looks at two complexes and picks two points on two different boundaries, and looks at arcs between them inside the surface.  You can also take the points on the same boundary.

To prove, the hard part of the stability theorem is not to build the complexes, but to get something highly connected.

Let $\mathscr{A}(S,\Delta_0,\Delta_1)$ has $\Delta_i$ sets of points in $\boundary S$ so this is a simplicial complex with vertices as isotopy classes of embedded arcs from $\Delta_0$ to $\Delta_1.$

Then $\langle I_0,I_p\rangle$ is a $p$-simplex if $I_j$ are disjoint and not pairwise isotopic.

This is a subcomplex of $\langle I_0,I_1\rangle$ such that $S\backslash\{I_0,I_p\}$ connected.  We get either $MCG(S_g,{k-1})\to MCG(S_{g,k})$ or $MCG(S_{g-1,k+1})\to MCG(S_{g,k}).$
\begin{thm}(Harer)\\
$\mathscr{A}(S,\Delta_0,\Delta_1)$ is $(4g+k+k'+m+\ell-6)$-connected.
\\
$\mathscr{A}_0(S,\Delta_0,\Delta_1)$ is $(2g+k'-3)$-connected.
\end{thm}

In the nonorientable case we have a transitive action on the vertices.

We can distinguish two paths, one of which goes through a nonorientable thing.  One way is to keep track of what happens to the normal bundle.  Then $\vec{Delta}$ is an ``oriented set of points''
\begin{defi} An arc in $(S,\vec{\Delta})$ is one-sided if its normal bundle identifies the orientation of its endpoints.
\end{defi}
If you look at a one-sided arc from a point to itself, it's exactly what you would expect.

So you can define the same kind of complexes as before, so $\mathscr{D}(S,\vec{\Delta})$ is the complex of one-sided arcs in $(S,\vec{\Delta})$ and $\mathscr{D}_0(S,\vec{\Delta})$ is the subcomplex of $\langle I_0,\ldots, I_p\rangle$ such that $S\backslash(I_0,I_p),$ connected an nonorientable.

\begin{thm}$\mathscr{D}(S,\vec{\Delta})$ is $(2n+k+k'+m+\ell-6)$-connected,$\mathscr{D}_0(S,\vec{\Delta})$ is $Ch{n}{2}-2$-connected.
If $S$ is oriented then \wh
\end{thm}
I'm out of time.
\section{Godin}
I did not take notes.
\section{Ramirez}
I do not take notes for talks with transparencies.


\end{document}
[[[a]]]
d^3=0

[[[a]b]] [[[a]]b] [[[b]a]] [[[b]]a] [[[ab]]]
d^2(ab)=d(da*b)+d(db*a)+a*d^2(b)+b*d^2(a)

