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\author{Gabriel C. Drummond-Cole}
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\title{Dennis Seminar\\January 31, 2007}
\begin{document}
\maketitle

The idea of algebraic topology is that you have topology as in topological spaces and maps and you also have geometry as in things that are locally Euclidean, like manifolds, and metrics on them.  Topology is sort of like geometry except you let the distances stretch.  Then there's algebra and there's a transition from topology and geometry to algebra, and that's algebraic topology, and that gives a photograph of topology.  Depending on how much algebra you put in, you can get more finely grained photography.

For example, the fundamental group might go from a space with a basepoint to a group.  This is algebra pure and simple.  This algebraic object, the group, is problematical.  It's a strong invariant.  If you take a knot complement, the fundamental group along with the subgroup from the meridian and the longitude completely determines the knot.

We talked last semester about how a three dimensional manifold tells you a lot by its fundamental group.  On the other hand, it's a famous theorem in logic that the problem of group recognition is not computable.

There is something better, more amenable to analysis, but weaker, because it doesn't control the fundamental group, and that is the chain complex.  What I want to do now, I want to think of this dictionary as like a metaphor.  On one side you have topology and geometry, and on the other side algebra, but in the metaphor, space will be represented by a chain complex.  This is a sequence of things and maps between them:
$$
\xymatrix{\ldots\ar[r]&C_{i+1}\ar[r]^{\delta}&C_i\ar[r]^{\delta}&C_{i-1}\ar[r]^{\delta}&\ldots}
$$
For $R$ any ring you let these $C_i$ be $R$-modules and assume that $\delta^2=0.$  Usually we assume that these are free modules.  Some of the arguments don't require that.

Usually in algebraic topology they sort of downplay this and traditionally you can, well, the kernel of $\delta_i$ contains the image of $\delta_{i+1}.$  They are not always equal, so you define $H_i(C,\delta)=\ker \delta_i/\im \delta_{i+1}.$  These are the famous homology groups.  If $R$ is $\zZ$ then these are Abelian groups.  In general these are the $i$th homology $R$-module.

More recently, one has learned not to do this passage to homology.  You think of passing to homology as giving you something canonical but you're throwing away some geometric information.  Many invariant statements about the chain complex can be stated in terms of the homology, and it's useful, but there is not a perfect correspondence between them.

All right.  Well, there are a lot of interesting operatinos on chain complexes.  A lot of them have immediate cognates on spaces, although we may have to sharpen things.

If we have two chain complexes, we can take $C_.\otimes D_.,$ which has in its $n$th component $\oplus_{i+j=n} C_i\otimes D_j$ and $$\delta_\otimes (c\otimes d)=\delta c \otimes d +(-1)^c c\otimes \delta d.$$
\begin{exercise}
Show that $\delta_\otimes\circ\delta_\otimes=0.$
\end{exercise}
If you defined it without the sign, you would get $\delta'_\otimes(c\otimes d)=2(\boundary c\otimes \boundary d)\ne 0.$

The way this exercise works, you're supposed to do homework, you have four handins of homework.  I don't like anyone who doesn't do their homework.  It tells me immediately whether you understand anything.   There's this famous sentence due to Kozsul, repeated by Quillen in the 1969 ``Rational Homotopy Theory,'' when an object of degree $p$ passes over an object of degree $q,$ the sign $(-1)^{pq}$ accrues.  It gets added to the story.  Every object in algebraic topology has a dimension, and this map has a dimension.  The sign, it's only the parity of $p$ and $q$ that count.  In this formula, you pass $\delta$ over $c$ and $\otimes.$  That rule works very well, it's not completely perfect, but I don't want to talk about that.  If you use this you'll be fine until you get to a point where a sign mistake won't hurt you anyway.

This corresponds to Cartesian product.

[Why doesn't this have similar properties, projection maps?]

If you want to project onto the $C$ factor.  Suppose you have a map $D_.\to R$ which projects onto one factor, usually in degree zero.  You can take this map $\otimes id$ and that takes $D_.\times C_.\to C_.$  The chain complexes that come from spaces have augmentation.

\begin{exercise}
Check whether augmentation projections are categorical.
\end{exercise}

A mapping between chain complexes $\pi:C_.\to D_.$ means a map of $R$-modules $\pi_i:C_i\to D_i$ and the commutative diagram
$$\xymatrix{
	C_i\ar[d]_\delta\ar[r]^\pi&D_i\ar[d]^\delta\\C_{i-1}\ar[r]_\pi&D}$$
An augmentation map is a map $C_.\to R,$ a split surjection.  Split means there is a map back whose composition gives the identity.

Actually, chain complexes have a life of their own.  In some sense it's true that you can model a space by a chain complex, but they have a life of their own, these are not unmodified chain complexes, they have augmentations, et cetera.

This is one of the best ones, this is a great one here: there is the notion of deformation of a chain mapping.  A mapping of chain complexes is called a chain mapping.  Later we'll consider maps between the two pieces where the maps don't commute.  There's a definition
\begin{defi}
$\pi$ is deformable to $\pi'$ (both chain mappings) if and only if there exists a degree 1 mapping $h:C_.\to D_{.+1}$ such that $$\pi-\pi'=\boundary h-(-1)^{|\delta h}|}h\boundary=\boundary h + h\boundary.$$
\end{defi}
There's a cute little aside here.  If you have an operator $D$ with $D^2=0,$ and if you take $DA+AD,$ composing with $D$ on the right and on the left are both equal to $DAD.$  Any expression like $\boundary h + h\boundary$ trivially commutes with $\boundary.$  This is a trivial way of commuting.  There's kind of a picture that goes with this.  You can imagine I have the some space.  Under some mapping $\pi$ you can go to a space like that, or another map $\pi'$ could go to a space like this.  You could deform from one to the other via a sort of carpetting $h.$  If you put in $(\pi-\pi')(x)$ is equal to $\boundary h(x)+h(\boundary x).$  [Pictures.]

That explains the sign formula.  This is an interesting definition that mirrors the interesting geometry well, even with the signs.

Now, algebraic topology was kind of a gift of topology to mathematics.  Having objects with this kind of differential and the ability to do deformations was a new level of algebra.  This algebraic model of deformation is one of the best ports of this discussion.

Now we have a new notion of isomorphism using deformation.  Other words for this, you also have chain homotopy.  If you're a topologist and you're doing this you call this a chain homotopy.  You say $\pi$ is chain homotopic to $\pi'.$  Then using deformation or chain homotopy, you say $C_.$ and $D_.$ are quasiisomorphic if and only if, well, let me do this in steps.  A map $\pi$ is called a quasiisomorphism if and only if there exists a homotopy inverse $\pi':D_.\to C_.$ so that $\pi\circ \pi'$ is deformable to the identity, and $\pi'\pi$ is deformable to the identity.

It turns out that everything that you do, well, with any mathematical object anything you do is an invariant of strict isomorphism.  It's pretty clear what strict isomorphism means.  This is a weaker notion of equivalence.  Anything you do to a chain complex is an isomorphism invariant.  If you do something to a chain complex, if it's not a quasiisomorphism invariant, you should throw it away.

There is a consequence of quasiisomorphism which is used to make a variant definition.  Suppose you have a quasiisomorphism.  Oh, I didn't do this in the right order.  Let me go back to chain mappings and make two remarks.  \begin{itemize}\item First, a chain mapping induces a mapping from the homology of $C$ to the homology of $D.$  When you have chain complexes you can form these graded $R$-modules in homology, remember.  Why is that true?  A chain mapping, if an element is in the kernel, then its image under $\pi$ is in the kernel of $\boundary.$  So it maps the kernel into the kernel, what are called the cycles to the cycles.  If you have something whose preimage is a boundary, you can pull back along the other direction to see that it is a boundary as well.  This is functorial, it respects the identity morphism and composition.  
\item So $\pi_.=\pi_.'$ if they are deformable to one another.  Suppose you have $\eta$ a cycle, you want $\pi(z)-\pi'(z).$  These are supposed to represent homology classes, this is $\boundary (hz).$
\item A quasiisomorphism induces an isomorphism on homology.  
\end{itemize}
So homology is a good construction because it's a quasiisomorphism invariant.  Usually you make choices in geometry, and you get things that are quasiisomorphic.  This is kind of a new idea that was introduced in about 1920.  These ideas contain the first examples of more elaborate objects called categories.   

The question that is raised, what about the converse.  Suppose you have a map one way which induces an isomorphism on homology, can you find a map back to make this a quasiisomorphism?  Not always, but here's a more general notion of quasiisomorphism.  Let's call this $H$-quasiisomorphism.  Make the converse true in the new definition.  $H$-quasiisomorphism is the equivalence relation induced by maps which induce homology isomorphisms.

These notions are not equivalent in general.  But if you make an assumption, there's a theorem.  I think the theorem is
\begin{thm}Assume we are in the category $BF$ of free chain complexes, so that $C_i$ is a free $R$-module.  The proof that I know needs them to be bounded from below.  \\
\begin{enumerate}  Inside $BF$ homology is also bounded from below. It's in $B.$ 
\item If $C_.$ is a chain complex whose homology is bounded from below, then there exists a chain complex $D$ and a chain mapping $D\to C$ inducing an isomorphism or $C_.$  So, up to that notion.
\item Inside $BF$ $H$ quasiisomorphism implies quasiisomorpism. 
\end{enumerate}
\end{thm}

As a corollary you get the foundations of homological algebra.  the fundamental objects are rings and $R$-modules.  You can make an $R$-module into a chain complex, putting it ($A$) in degree zero, with everything else zero.  I write positive integers off to the left.  You choose a free module $F$ that maps onto $A$ and then map it in.  This will have some kernel, so you choose another free module mapping onto the kernel, and then you can extend into an even longer string of $F_i.$

[Missed some.]

\begin{exercise}
Prove part one of the theorem, that there exists a free resolution of a free bounded below complex.
\end{exercise}

In the remaining few minutes, let me go back and discuss how we get chain complexes from topology.

Chain complexes in topology were almost introduced by Poincar\'e.  It's pretty cool to read ``Complement Analysis Situs.'' He proved Poincar\'e duality, and then Heegaard rebutted him, and he almost introduced this in order to rebut the rebuttal.  He had pieces and their boundaries, and started taking linear combinations of their chunks.  Then he defined his Betti numbers that way.  He was only working with the elements of the complexes and never formed the groups.  He would count the number of linearly independent cycles.  Betti used integer coefficients, so it included torsion.  Apparently, Emmy Noether emphasized that there was a group, an Abelian group here.  It would be interesting to find where the first chain complex is written down.  There were some geometric chain complexes.  Brouwer made one, Alexander, Cech, this one is pretty cool, and then the whopper, Eilenberg.

Given any topological space $X$, let $C_i(X)$ be the free $R$-module (say $\zZ$) generated by all maps $(f:\Delta_i\to X)$ where $\Delta_i$ is the standard simplex of dimension $i.$  So $C_0$ is the free $R$-module geneerated by points.  For every continuous map of the interval into the space, you could attach an interval to any two points in the same component.  Then you have boundary maps for each of these.

This is functorial and you have homology right here.  You have a homotopy functor on all spaces to Abelian groups.  I don't like teaching about homology this way, but there it all is.  One idea is that if you could break your space up into nice pieces, you cauld make a finite dimensional madel  You can read of the boundary from the [Picture].

In the last five minutes, there's no advantage to taking anything besides $\zZ,$ because the construction for $R$ is functorial.

\section{Terilla}
[I got here late.]

We're going to have two deformation directions, the $t$ and the $\hbar$ directions.  Let $(V,d,[\,,\,])$ be a differential graded Lie algebra.  You get $V=\oplus_i V^i, d:V^i\to V^{i+1}$ satisfying $d^2=0,$ the bracket $[\,,\,]:V^k\otimes V^\ell\to V^{k+\ell}$ which is skew symmetric and Jacobi (graded) and satisfies $d[a,b]=[da,b]\pm [a,db].$  So we deform the differential $d$ in the direction of an inner derivation, $$d\leadsto d+[\gamma,\bullet]\equiv d_\gamma,$$ so that $d_\gamma(a)=da+[\gamma,a],$ such that $(V,d_\gamma,[\,,\,])$ is a differential graded Lie algebra.  We need $\gamma\in V^1$ for this to be true, $d_\gamma$ will be a derivation of $[\,,\,]$ by Jacobi for $V,$ and then the hard one is $d_\gamma^2=0.$

So This is implied by $d\gammma+\frac{1}{2}[\gamma,\gamma]=0.$  This is because $d_\gamma(x)=dx+[\gamma,x]$ so that 
$$d_\gamma^2(x)=d(dx+[\gamma,x])+[\gamma,dx+[\gamma,x]]$$$$
=[d\gamma,x]+[\gamma,dx]+[\gamma,dx]+[\gamma,[\gamma,x]]=[d\gamma+\frac{1}{2}[\gamma,\gamma],x].$$

This led to this equation, called the Maurer-Cartan equation for $\gamma\in V^1,\ d\gamma+\frac{1}{2}[\gamma,\gamma]=0.$

[Why inner deformations?]

You have a deformation problem of a structure $\Ss.$  Then $\Ss$ leads to a differential graded Lie algebra with deformations in the direction of an inner derivation.

A derivation $f$ of a product (in this case the Lie bracket) is a linear map which satisfies the derivation (Liebnitz) property with respect to the multiplication.  An inner derivation is one of the form $f=[\gamma,\bullet].$ This is the infinitessimal version of an inner automorphism.

If we were talking about derivations of $A$ an associative algebra, then the Lie algebra would be the Hochschild complex.  Deformations of a complex structure gives the Lie algebra of sections of $\Omega^{\cdot,p}\otimes T,\bar{\boundary},[\,,\,]$

What do I mean by the $t$-direction?  What I often want to do is allow $\gamma$ to have parameters.  For example, if you have a differential graded Lie algebra, and you have a commutative associative algebra $(A,m),$ then you can form the tensor product and you get  $(V\otimes_k A, d, [\,,\,]\otimes m),$ a differential graded Lie algebra.

If $A$ were to be something like $A=k[s,t],$ then a typical element of $V\otimes A$ would look like $s\gamma+t^2\alpha,$ and another would look like $st\beta.$  Then the result is $$[s\gamma+t^2\alpha,st\beta]=\pm s^2t[\gamma,\beta]\pm st^3[\alpha,\beta].$$

Involving parameters can be illuminating.  If $\gamma\in V^1\otimes k[[t]],$ then $d\gamma+\frac{1}{2}[\gamma,\gamma]=0$ breaks up into a bunch of different equations.  I even want to assume that there is no $\gamma_0,$ so that $\gamma\in V^1\otimes tk[[t]].$  In general you don't have a nice map from $V\otimes R\to V.$  But there is a nice map from $V\otimes tk[[t]]\to V$ by taking $t$ to zero.  In order for this term to go away, I need $\gamma$ to start with the $t$ term and not the constant term.  I want to use $\hat{\otimes}$  which means that I'm using $tV^1[[t]].$  The inverse limit of $V$ tensor the polynomials is what we call the completed tensor product of $V$ tensor the inverse limit of the polynomials.  You have $tk[t]/t^2\gets tk[t]/t^3\gets\cdots,$ so you have the inverse limit.  You tensor first and then take inverse limits.  There is a tidy effective abstract language to describe this, in terms of [Artin functors?]

Okay, so I have $d\gamma_1=0.$  Next you have $d\gamma_2+[\gamma_1,\gamma_1]=0.$  The next equation will be $d\gamma_3+[\gamma_1,\gamma_2]+[\gamma_2,\gamma_1]=0.$

So you want to find the solution starting with $\gamma$ and moving in the $t$ direction.  You want to build this $\gamma$ piece by piece.  You start with $\gamma_1,$ and then look for a $\gamma_2$ to solve the second equation, and then continue onward if you can.  You find a solution mod $t^2$ and then one mod $t^3,$ and so on.  If $d\gamma_1=0,$ then $\gamma=t\gamma_1$ solves the Maurer Cartan equation modulo $t^2.$  But does there exist a $\gamma_2$ such that $\gamma=t\gamma_1+t^2\gamma_2$ solves the Maurer Cartan equation modulo $t^3?$  This is the same as saying, does there exist $\gamma_2$ such that $d\gamma_2=-[\gamma_1,\gamma_1]?.$  Well, $d$ of the right hand side is zero, so the right hand side is closed, but is it exact?  So you look at the set of cohomology classes which are zero when bracketed with themselves in homology.  That's a quadratic locus in the homology.

Look at the next one $[\gamma_1,\gamma_2]$ is closed by Jacobi, but only because $d\gamma_2=-[\gamma_1,\gamma_1].$

One idea in these lectures is to discuss differential graded Lie algebras and associative algebras, and how $L_\infty$ minimal models or $A_\infty$ minimal models are like solving the Maurer Cartan equations with noncommutative parameter rings.  This is somehow related, as well, to Massey products in homology.

\begin{defi}
Suppose $H=H^1(V)$ is finite dimensional.  An element $$\Gamma=\sum t_i\gamma_i + t_it_j\gamma_{ij}+\ldots\in V^1\otimes k[[H^*]]$$ where $\{\gamma_i\}$ are a basis for $H$ and the $t_i$ are the dual basis, satisfying $d\Gamma+\frac{1}{2}[\Gamma,\Gamma]=0$ is called a versal solution to the Maurer Cartan equation.  There should be a map of parameter rings to take $\Gamma$ to any solution.
\end{defi}
This kind of summarizes the $t$ direction.  Let me give one example of a situation in which you always have a universal solution.  I'll put some sentence on the board.

There exists a versal solution to the Maurer Cartan equation if and only if the bracket and all higher Lie Massey products vanish in homology, which is true if and only if the $L_\infty$ minimal model for the Lie algebra is trivial, if and only if there exists a formality map $V,d,[\,,\,],0\to H,0,0,$ the homology with trivial differential and bracket.

What is a BV-algebra?  $(V,\Delta,\cdot).$  Here $V=\oplus V_i,$ and $(V,\cdot)$ is a commutative associative algebra, and $\Delta:V^k\to V^{k+1},$ where $$[a,b]=\Delta(ab)\pm a\Delta b\pm (\Delta a) b.$$
where $[\,,\,]$ is a derivation of $\cdot$ in each variable, and $\Delta^2=0,$ which implies $V,\Delta,[\,,\,]$ is a differential graded Lie algebra.

The theorem is that for any $BV$ algebra, there exists a versal solution to the Maurer Cartan equation in $(V,\Delta,[\,,\,])$ (unique up to gauge).

For culture, deformations of a complex structure motivated all of this in the 50s.  The spaces are complicated, singularities, so on.  If it's a Calabi-Yau K\:ahler manifold, it's a BV algebra and you have this versal solution.

There was a theorem proven by Baranikov and Kontsevich which said, if your manifold is Calabi-Yau and K\:ahler, you have the solution.  They used the $d-\bar{d}$ lemma, which is unnecessary.  You just need to notice the $BV$ algebra.

[Some discussion of whether their solution would coincide.]

I will make three short discussions.  \begin{enumerate}\item Suppose $\Delta\gamma_1=0.$  Consider $[\gamma_1, \gamma_1],$ well, by the definition of the bracket, this is $\Delta(\gamma_1^2).$  So you can choose a $\gamma_2.$  This or something like it works all the way down.
\item Because of the interaction between $\Delta$ and $\cdot,$ it turns out $\Delta(e^x)=0$ if and only if $\Delta x + \frac{1}{2}[x,x]=0.$  This gives you another way of finding a universal solution.  Take $\sum t_i\gamma_i,$ with $\gamma_i$ a basis for $H,$ then this satisfies $\Delta\gamma=0.$  Then let $\Gamma=\log \gamma.$  So $\Delta(e^\Gamma)=\Delta(e^{\log\gamma})=\Delta(\gamma)=0.$
\item You can look at $(V,\Delta)$ as an $L_\infty$ algebra with no bracket or anything.  Then you can look at, there's a gauge action on coderivations.  I'm saying that $(V,\Delta+[\,,\,])$ is like a gauge transformation by $\cdot.$  You can think of $\cdot,$ $[\,,\,]$ and $\Delta$ all as coderivations.  You can get rid of the bracket by changing gauge by the product.  You should be able to gauge transform a solution in the trivial $L_\infty$ setting to the desired setting.
\end{enumerate}
So BV algebras are important, they arise in the situation of BV quantization.  I have $(V,d,\Delta, \cdot)$ is a BV algebra.  But $(V,d,\cdot)$ is a differential graded algebra and $d\Delta+\Delta d=0.$  You can replace $\Delta$ with $d+\Delta.$  This generates the exact same bracket since $d$ is a derivation.  As a differential graded Lie algebra you don't get anything new.  The homology might be different.

Here is the key step.  Introduce a new variable $\hbar.$  Now look at $(V[[\hbar]],K=d+\hbar \Delta, [\,,\,]).$  This is a differential graded Lie algebra over $k[[\hbar]].$  This is not necessarily formal.  The logarithm of $\Gamma$ will have $\frac{1}{\hbar}.$  So $dx+\hbar \Delta x + \frac{1}{2} [x,x]=0$ if and only if $K\frac{e^{\frac{x}{\hbar}}.$

If $K(\gamma_1)=0,$ then $K(\gamma_1^2)=\hbar[\gamma_1,\gamma_1].$  So it's not closed anymore, unless $\gamma_1^2$ is divisible by $\hbar.$  Typically everything is defined up to homotopy in the $t$ direction.  Now when you have this other differential, you can fold it in and forget about it, so you want to incorporate it with an $\hbar.$

Let me state a precise theorem.
\begin{thm}
Suppose that $\{x_\alpha\}, where x_\alpha=x_\alpha^0+x_\alpha^1\hbar+\cdots$ is a basis of representatives for $K$-cohomology of $V[[\hbar]],$ such that $\{x_\alpha^0\}$ is a basis for the $d$-cohomology of $V.$  Then there exists a versal solution in $(V[[\hbar]],K,[\,,\,]).$   There is also a connection on the moduli space.  This will have an $\hbar$ dependence but in this case there will be no $\hbar$ dependence.
\end{thm}

What is this real secret?  I'm being recorded.  Deformations of anything lead you naturally to $L_\infty$ algebras.  An $L_\infty$ algebra $D$ is a map $SV\to SV$ which is a coderivation of square zero.  It turns out with coderivations you can bracket them.  There exists a very interesting and completely natural $\bigstar$ product $Hom(SV,SV)[[\hbar]]^{\otimes 2}\to Hom(SV,SV)[[\hbar]]$ which is associative and noncommutative.  When $\hbar$ is zero, it recovers the pre-Lie product on the coderivations.  This sits in an $\hbar$-enriched structure, and the product can be understood in terms of gluing operations of Riemann surfaces.  These can be naturally enriched in ways that correspond to Riemann surfaces.  You do the same thing as here, but keep track of the $\hbar.$

\section{Nathalie Wahl}

I want to talk about various forms of homological stability.  I'll eventually talk about three-manifolds.  It's easier to start with surfaces, which are easier to understand.  I'll be talking about orientable and nonorientable surfaces, both at once.  $S_{g,k}$ is a surface of genus $g$ with $k$ boundary components.  Also nonorientable ones, and by such a picture, $F_{n,k}$ removes $n$ disks and glues in $n$ Mobius bands. I will also call $n$ the genus.

This is one type of map you can look at, I want to look at all sorts of maps you have between these surfaces.  You can include $S_{g,k}\to S_{g,k+1}$ by gluing on a punctured torus, or glue on a pair of pants to go to $S_{g,k+1}$ for $k\ge 1.$  You can also go to $S_{g,k-1}$ by gluing on disks.  You have the same ones on the nonorientable case.  $F_{n,k}\to F_{n+1,k},\ F_{n,k}\to F_{n,k+1},F_{n,k}\to F_{n,k-1}.$  Call these maps $\alpha,\beta,\gamma.$

So now the mapping class groups, $\Gamma(S_{g,k})=\pi_0(Diff^+(S_{g,k}, \boundary)).$  There is a short exact sequence
$$\zZ\to \Gamma(S_{g,k})\to \Gamma(S_{g,k-1})$$
For a nonorientable surface or $k\ne 0$, of course you don't need the $+$ for orientation.  So now $\alpha,\beta,\gamma$ induce maps.
\begin{thm}
Harer, Ivanov
$$\alpha_*:H_i(\Gamma_{g,k},\zZ)\to H_i(\Gamma_{g+1,k},\zZ),
\beta_*:H_i(\Gamma_{g,k},\zZ)\to H_i(\Gamma_{g,k+1},\zZ),
\gamma_*:H_i(\Gamma_{g,k},\zZ)\to H_i(\Gamma_{g,k-1},\zZ)$$
are isomorphisms for $i\le \frac{g-1}{2}$ or $\frac{g-2}{2}$ for $\gamma.$\end{thm}
The homology of $\Gamma$ is the same as the homology of the classifying space of the group, and there is a theorem which says that the components of the diffeomorphism group of a surface are contractible.  This is true except for unstable surfaces.

As a consequence, $B\Gamma\cong B\text{Diff}$ which is the classifying space for surface bundles.  So the homology of the group is isomorphic to the homology, this gives you characteristic classes of surface bundles.  If there is boundary then the moduli space is also a model for this, so this is about the homology of moduli space.

The same holds for nonorientale surfaces
\begin{thm}Wahl\\
$$\alpha_*:H_i(M(F_{n,k}),\zZ)\to H_i(M(F_{n+1,k}),\zZ),
\beta_*:H_i(M(F_{n,k}),\zZ)\to H_i(M(F_{n,k+1}),\zZ),
\gamma_*:H_i(M(F_{n,k}),\zZ)\to H_i(M(F_{n,k-1}),\zZ)$$
are isomorphisms for $i\le \frac{n-2}{4}$ or $\frac{n-4}{4}$ for $\gamma.$\end{thm}
Now you can push things farther and calculate the stable homology.

So $\Gamma_\infty$ is the colimit (direct limit) of $\Gamma_{g,1}\to \Gamma_{g+1,1}\to\ldots.$  Similarly $M_\infty=\cup M_{n,1}.$

[This would be the moduli space of Riemann surfaces where you only change the complex structure on a compact part.]

The bigger my genus, the more this map becomes an isomorphism.  So $H_i(\Gamma_\infty)\cong H_i(\Gamma_{g,k})$ as long as $i\le \frac{g-2}{2}.$
\begin{thm}Madsen, Weiss\\
$H_*(\Gamma_\infty)\cong H_*(\Omega^\infty \cC\pP^\infty_{-1})$ which we also talk about as $\Omega^\infty \gG^+_2.$  Rationally this is very easy, the homology of $\Gamma_\infty$ is a polynomial algebra $\qQ[k_1,k_2,\ldots]$ where $|k_i|=2i.$\end{thm}
\begin{thm}
$H_*(\M_\infty)\cong H_*(\Omega^\infty \gG_{-2})$ so that rationally $H^*(M_\infty,\qQ)\cong \qQ[\eta_1,\eta_2,\ldots]$ where $|\eta_i|=4.$  We even have $$H^*(M_\infty, \zZ[\frac{1}{2}])\cong H^*(\Omega^\infty S^\infty(BO(2)+),\zZ[\frac{1}{2}])$$\end{thm}
The cohomology of the mapping class groups are the same as those of the diffeomorphism group so these give classes of surface bundles.  If I start with a surface bundle $S\to E\to B$ I can get the vertical tangent bundle $T_pE\to E.$  Above $e$ I put the tangent plane of $S$ at the point $e.$  These are classified by maps to $\cC\pP^\infty$ if $E$ is orientable, $Gr^+(2,\rR^\infty),$ and in the nonoriented case I get a map to the nonoriented Grassmanians $BO(2)\cong Gr(2,\rR^\infty)$.

So I get classes $e\in H^2$ or $p\in H^4.$  I want a class in the cohomology of $B,$ and there is a transfer which goes from the cohomology of $E$ to the cohomology of $B.$

[At this point I stopped taking notes.]


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