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\author{Gabriel C. Drummond-Cole}
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\title{Moduli Space of Riemann Surfaces}

\begin{document}\maketitle
\section{Ravi Vakil\\ Introduction to the Moduli Space}
So let me begin by recalling some notions from yesterday.  We're dealing with the moduli space of Riemann surfaces of genus at least two $\Mm_g,$ and I mentioned two variations on this, $\mgnbar,$ where you have stable nodal curves.  Somewhere between these lies $\mgn^{rt}$ where all the genus occurs in only one component.

I introduced cohomology classes on each of these spaces.  We had $\kappa$ classes $\kappa_i\in H^{2i}(\Mm).$  More important are the $\psi_i\in H^2(\mgnbar^{rt}).$  The line bundle defining this is the cotangent bundle at a marked point.  It's complex codimension one.

There are also the $\lambda_i\in H^{2i}(\Mm),$ which is the first Chern class of the vector space (bundle) of differentials.

Faber's conjecture says that to multiply a bunch of $\kappa$ classes, if the total codimension is greater than $g-2$ you get zero, and $\kappa_{a_1}\cdots \kappa_{a_n}=\alpha \kappa_{g-2}$ if the $a_i$ add to $g-2.$  Faber's conjecture tells us what $\alpha$ is.

We want to say that we know all the intersection products if we know how to multiply all the $\psi$ classes together.

Let me hold off on this, but this will be the content of the next ten or fifteen minutes.  We move from the question about $\mgnbar$ to the analogous question in $\mgnbar^{rt},$ and we multiply $\psi_i$ together to get a dimension $g-2+n$ class.  Then push them down to $\Mm_g$ and what multiple of $\kappa_{g-2}$ do we get?

The answer to one version of this was given by Witten's conjecture, around 1990.  You take these numbers, you put them in a generating function, and the generating function satisfies a differential equation which gives you a recursion to allow you to calculate these numbers.  There are around seven proofs.  Each one has given a lot of additional insight.  It was proven first by Kontsevich, then by Okounkov and Pandharipande, then Mirzakhani, then Kazarian-Lando, Kim-Ki Liu, and Kazarian.  I wouldn't be surprised if I've forgotten one.

So these proofs, how is such a proof to begin?  You need to get ahold of the top intersections.  This leads me to the ELSV formula, which all but Kontsevich and Mirzakhani's pass through.  It's an incredibly important result.  It has aged extremely well.  It answered an important question and the idea behind it has been used since then in a number of ways.  This is what gives us an entrance.  This deals with Hurwitz numbers, which count branched covers of $\cC\pP^1,$ and this will depend on a degree $d,$ a genus $g$ of the cover, and a fixed branching over $\infty$ corresponding to a fixed partition $\alpha.$  The remaining branch points will be simple, meaning two sheets coming together.  How many tdo we need?  It's $r=d+n+2g-2,$ and you count things appropriately, counting things rationally if there are automorphisms.

$H_\alpha^g$ is the number of ways of choosing $r$ transpositions which generate $S_d$ and their product is $\alpha,$ and then divided by $d!.$  There's a notion that will come up later of double Hurwitz numbers, which is counting branched covers of the projective line with two partitions $\alpha$ and $\beta.$  

Why is this true?  On explanation is that each $\sigma_i$ corresponds to the monodromy around  a branch point, and then $\alpha$ corresponds to the monodromy around $\infty.$  A second explanation, strictly worse, is by degeneration.  Pinch the target to a chain of hot dogs with a branch point in each piece.  Then what can happen to the cover?  You can check with a local calculation that the covers are a bunch of trivial spheres like $x\to x^n,$ except for one which is not a trivial cover which accounts for the single branch point.  You can count these degenerated covers instead, which gives this formula.

Any Hurwitz number that you want can be computed by hand.  The structure in terms of character theory is also easy to get your hands on.  
\begin{thm}
$$H^g_\alpha=r!(\prod \frac{\alpha_i^{\alpha_i}}{\alpha_i!})\int_{\mgnbar}\frac{1-\lambda_1\ldots \pm \lambda_g}{(1-\alpha_1\psi_1)\cdots (1-\alpha_n\psi_n)}$$\end{thm}
We will focus on the integral.  When you expand it out, you get a bunch of terms
$$
\sum_{a_1+\ldots+a_n+k=3g-3+n}(-1)^k \left(\int_{\mgnbar}\lambda_k\psi_1^{a_1}\cdots\psi_n^{a_n}\right)\alpha_1^{a_1}\cdots\alpha_n^{a_n}.$$
I want to compute a whole lot of Hurwitz numbers with fixed $g$ and $n.$  You know now the numbers are equal to a combinatorial factor times a polynomial of degree $3g-3+n.$  Knowing enough Hurwitz numbers, you can compute the coefficients of the polynomial.  A computer program can do this.  These are packaged in a nice way.  You could formalize this to pick off a coefficient.  It gives the whole package, in a way that has to do with the representation and character theory of the symmetric group.

This takes these mysterious numbers and packages them in a way we can work with.

Let me give you an overview of a proof of the ELSV formula.  I want to just give an overview because it's hard to do in a reasonable amount of time.

How does this relate to the original proof?  I'm not sure.  This is joint with Graber.  We're trying to count branched covers of $\pP^1.$  We want to use localization and equivariant cohomology.  We take the space of branched covers, and we'll fix branching over $\infty$ to be $\alpha.$  This is not compact.  It's so badly noncompact it has no fixed points.  We want a compactification.  

This comes from Gromov Witten theory and at first blush is hideous, it's the space of (relative) stable maps to $\pP^1.$  If you want some sort of metaphor, it's like compactifying $\mgn$ to $\mgnbar.$  Take maps of nodal curves to $\pP^1.$  I want the preimage of infinity (in a first approximation) to correspond to the partition $\alpha.$  I have fixed $g,\alpha.$  Now I allow nodal curves, and a curve component could be crushed to a point.  I again want every genus zero curve to have three special points and isolated genus one curves to have one special point.  This gives something Hausdorff, and you want no continuous families of automorphisms.

You need to allow one other thing that is slightly worse.  What happens when you try to crash junk into $\infty?$  You pop off a sphere, our original $\pP^1$ breaks off a new component near $\infty.$  Things line up nicely over the breaking.  You can do this more than once, have a chain of $\pP^1$s.  You'll have to take my word that this gives you a moduli space.  It is indeed compact, which is probably the only nice thing I can say about it.  It's not the same dimension, it's not smooth.  Now we're sad again.

It turns out, and this is the magic of Gromov-Witten, it wants to be the right dimension, and the space of branched covers has a known dimension with a fundamental class.  This compactification has a virtual fundamental class of the expected dimension.  In the algebro-geometric part of Gromov-Witten theory, a lot of things are not hard, but this is hard.  If you want to prove something hard you're forced to struggle with this.

Now, sadly, I'm in the process of erasing what we're trying to prove, but given how much I'll be waving my hands, I'm not too upset.

Now localization works on this space.  You have a problem, a cohomology calculation you want to do.  Localization turns this into a calculation on the fixed point loci.  Consider these branched covers with junk happening over here, and I want $r$ fixed branch points.  That's the calculation I want to do.  I want to choose a linearization.  That's like forcing all the branch points to zero.  I look for all the maps I can write down which are fixed by the torus.  I can't have any branching added to $\infty.$  Nothing is happening between $0$ and $\infty.$  The rest of it, a single connected curve of genus $g$ with $n$ marked points, is over $0.$  Then you apply the recipe, write down the fixed points, and then looking at various pieces of this picture gives you ingredients of this formula half visible under erasure on the board.  The recipe spits out the ELSV formula.  If you see $(1-\alpha_n\psi_n)$ in the denominator, you'll get this sort of thing.  The group is $\cC^*$ acting algebraically on $\cC\pP^1.$

Now let's take this lesson to heart when dealing with Faber's conjecture.  There are various things that might make you flinch, like that one of these is about numbers and the other about cohomology classes.  There are not many tools in Gromov-Witten theory.  People basically just use localization and degeneration, and that's what we'll use here.

Let me start with facts of geometry, the Riemann-Hurwitz formula.  Say you have a map from a curve to $\pP^1.$  The curve need not be connected.  Say that there's no branching between $0$ and $\infty.$  Then it has to be a union of spheres mapping like $x\to x^\alpha.$  What if you allow yourself one branching?  Then you can have something which is a bunch of spheres and then one genus zero interesting cover.  Let me wrench that up and say we have a map from a curve with rational tails, of genus $g.$  You've got to see a genus $g$ curve there.  Let's say you have a map like this.  Then you need at least $2g$ branching away from $0$ and $\infty.$  If I have precisely $2g$ branch points, then I have one sheet with all the branching of genus $g$ and then the others are spheres with no branching.

We want to intersect $\psi$ classes now, restricting attention to maps with rational tails.  We look for relative stable maps to $\pP^1$ but now the open set which just has rational tails.  Now let me draw a picture.  You have a genus $g$ curve and then some spheres.  These are the maps we're considering.  We're looking for a class of dimension $2g-1.$  Now, what's the dimension of this space of branched covers with $r$ branch boints.  Fix all but $2g-1$ branched points.  Now I have a geometric object in which I see a genus $g$ curve, $\alpha$ points, and all the ingredients are coming together.  Call this thing a Hurwitz class.  I want to compute it, and so I need to identify it with something involving $\psi$ classes.  Let me first degenerate.  So break my target so each branch points lies on its own chain, the fixed branch points.  The other $2g-1$ can vary, but this is a curve with rational tails.  I need at least $2g$ branching in the middle of one of these components.  The only way I can get $2g$ in one component is if all of them are in the same component, and now you know what this looks like.  You have one genus $g$ part and then simple spheres with no branching.

You get a tree with no loops because you have rational tails.  You can write the Hurwitz class in a comprehensible way as a multiple of something simple.

Now you use localization.  What does the picture of the fixed points look like?  A reasonably typical fixed point, you can have some spreading in the target and you've got all but $2g-1$ branch points.  The ones I had my hands on go to zero and I have $2g-1$ left.  Where could they be?  They could be in the component of $\infty.$  We know this has rational tails, and through a chain of reasoning the genus is over zero.  Now the recipe gives you a calculation which will give you the $\psi$ classes.  Everything else is genus zero which is combinatorial and discrete.  Then from the component of $\infty$ you get double Hurwitz numbers, thankfully genus zero ones.

This ends the head-hurting part of the story.  Let me summarize and extract the morals that you get from the recipe and the flavor.   The first moral is that these classes in fact vanish in codimension bigger than $g-2+n.$  This proves the vanishing part of Faber's conjecture.

Secondly, you can compute by hand any top intersection.  Perhaps a better thing to say is that they are combinatorially tractable.  More precisely, if you take Faber's intersection numbers, after massaging things very cleverly, you get these from genus zero double Hurwitz numbers.

As an example of that, these are understood in the case where you have at most three parts.  We think it could be done for four or five parts. 

Let me just promise that when Ian talks, you can forget most of what I said, and the only thing to remember is the localization.  You have a bipartite graph with vertices over $0$ and $\infty.$

This is the second of my pair of lectures and I want to sum up what I think I said.  First of all I discussed the moduli spaces and discussed the tautological subring.  All natural classes lie in this subring.  Understanding this ring modulo the perfect pairing conjecture involves understanding top intersections.  A key idea for the moduli space of stable curves is ELSV, which gets you from cohomology classes to factorization of a permutation into transpositions.  To approach Faber's conjecture you allow marked points and allow bubbling and get $\psi$ classes, and so then you localize and get something tractable involving double Hurwitz numbers.  

That's all I wanted to say, thank you very much for your time.
\section{Zogrof}
Remember $\Hh_n(X)$ was the space of generic meromophic functions on $X$ of degree $n.$  As I said last time, the tangent space to $\Hh_n(X)$ at $f$ can be identified with $H^0(X,f^*T\bar{\cC}).$  Let me say a few words about why this is so.  Let's take our surface, pick up some points on it, then on the sphere the set of values as a curve, this is the image of $x$ and the infinitessimal deformation of this family is just the tangent vector.  It assigns to each point on the Riemann surface a tangent vector in the sphere to its image.  You get a pullback of the tangent line bundle to the sphere.  Let me draw the corresponding commutative diagram:
$$\xymatrix{
f^*T\bar{\cC}\ar[r]^f_{T\bar{\cC}}\ar[d]&T\bar{\cC}\ar[d]\\X\ar[r]^f&\bar{\cC}}$$
Now I will fix canonical metrics on $X$ and the Riemann sphere.  On $X$ we take the hyperbolic metric and on $\bar{\cC}$ the metric of constant positive curvature.  If we have two holomorphic sections of the line bundle $f^*T\bar{\cC}$ called $\xi,\eta,$ then $$
\langle \xi,\eta\rangle = \int_X (f_{T\bar{\cC}}(\xi),f_{T\bar{\cC}}(\eta))_{\bar{\cC}}dp.
$$
I will represent $f=p/q$ in a canonical way.  $p$ and $q$ should be holomorphic sections.  Let me remind you that there is a projection $\cC^2\backslash\{0\}\to \bar{\cC}$ that takes $(z_1,z_2)\to z_1/z_2.$  Denote by $L$ the tautological line bundle on the complex projective line $\cC.$  This line bundle is a subbundle of the trivial two-dimensional bundle on the complex projective line.  Now we have the meromorphic function and we can consider the pullback of the tautological line bundle
$$\xymatrix{
\Oo_X\oplus \Oo_X\ar@/_/[dd]\ar[r]&\Oo_{\bar{\cC}}\oplus \Oo_{\bar{\cC}}\ar@/_/[dd]\\
f^*L\ar[d]\ar[r]\ar[u]& L\ar[d]\ar[u]\\X\ar[r]&\bar{\cC}}
$$
Now
$$\xymatrix{
&&\Oo_X\\
f^*L\ar[urr]^p\ar[drr]_q\ar[r]&\Oo_X\oplus \Oo_X \ar[ur]\ar[dr]\\
&&\Oo_X}
$$
and $p,q\in H^0(X,f^*L^*)$ of degree $n.$
Here $n\ge 2g-1.$  Okay, so $f$ is represented as a ratio of two holomorphic sections.  This scalar product can be written in a simpler form now:
$$
(f_{T\bar{\cC}}(\xi),f_{T\bar{\cC}}(\eta))_{\bar{\cC}}=\frac{\xi\bar{\eta}}{(|p|^2+|q|^2)^2}
$$
And this ratio we integrate over the sphere.  We are using the constant positive curvature metric $\frac{|dz|^2}{(1+|z|^2)^2}$

Okay, now so there is a scalar product in each fiber of the tangent bundle to the space of generic meromorphic functions that gives rise to a Hermitian metric.  An interesting observation is that this metric is scalar, which is where determinants of Laplacians come into play.

Denote this Hermitian scalar product by $\langle\,,\,\rangle_X$ and then define $\omega_X=-\frac{1}{2} Im\langle\ ,\ \rangle_X$ and I want to show that this is closed.
Now
$$
C^0(X,f^*T\bar{\cC})\stackrel{{\bar{\partial}}}{\to} C^{0,1}(X,f^*T\bar{\cC})$$ and there is the adjoint $\bar{\partial}^*$ in the other direction.  Then the Laplace operator $\Delta=\bar{\partial}^*\bar{\partial}.$

Let $\lambda_X=\wedge^{2n-g+1}T\Hh_n(X);$ this is a holomorphic line bundle.  Slightly change the Hodge norm of this holomorphic line bundle, and consider the equivalent norm $||\ ||_Q=\frac{||\ ||}{(\det\ \Delta)^{\frac{1}{2}}}.$  We can calculate the Chern form.

Denote by $c_1$ the Chern form $c_1(\lambda_X,||\ ||_Q).$  \begin{thm}This is precisely $\frac{-1}{\pi^2}\omega_X.$\end{thm}
One of the consequences is that $\langle,\ \rangle_X$ is K\:ahler.

This ratio is in fact a local K\:ahler potential for this metric on the space of generic meromorphic functions on the Riemann surface.

Now I want to gereralize this to the case where the complex structure on the space also varies.  I want to generalize to branched covers of the projective line.  This is joint with Leon Takhtajan.  Denote by $\Hh_{g,n}$ the space of generic meromorphic functions of degree $n,$ or better to say, the space of isomorphism classes of $(X,f)$ where $X$ is a Riemann surface of genus $g$ and $f$ a meromorphic function of degree $n.$

To begin with we need to describe $T_{(X,f)}\Hh_{g,n}.$  We will describe this as a space of sections which are no longer holomorphic.  Let me draw one more commutative diagram:
$$
\xymatrix{
TX\ar[dr]^\alpha\ar[drr]^{df}\ar[ddr]\\
&f^*T\bar{\cC}\ar[d]\ar[r]^{f_{T\bar{\cC}}}&T\bar{\cC}\ar[d]
\\
&X\ar[r]^f&\bar{\cC}
}$$
So the differential factors through the pullback of the tangent bundle.  It induces a map 
$$\xymatrix{C^0(X,TX)\ar@/^/[r]^{\alpha_*}&C^0(X,f^*T\bar{\cC})}$$
Formally $\alpha=f_{T\bar{\cC}}^{-1}\circ df$ and $\alpha_*(\nu)=(p'q-pq')\nu.$  This is a well-defined one-form on $X$ with coefficients in $f^*T\bar{\cC}.$  Well, going back to the tangent space to the space of meromorphic functions, I can identify it as follows:
$$
T_{(X,f)}\Hh_{g,n}\cong C^0(X,f^*T\bar{\cC})/\alpha_*C^0(X,TX).$$
One can interpret this in terms of sheaf cohomology and then the problem is to identify certain cochains with smooth actions.  After we do that we get this isomorphism.  The codimension is the right one, $2g-2n-2,$ so we get the right dimension of the tangent space.

Now let's look at $\Hh_n(X)/Aut\ X\injects \Hh_{g,n}\to \Mm_g.$  There is a short exact sequence of tangent spaces
$$
0\to T_f\Hh_n(X)\to T_(X,f)\Hh_{g,n}\to T\Mm_g\to 0.$$
We can replace these with isomorphic things to get
$$
0\to H^0(X,f^*T\bar{\cC})\to \frac{C^0(X,f^*T\bar{C})}{\alpha_*C^0(X,TX)}\stackrel{\alpha_*^{-1}\bar{\partial}}{\to} H^{0,1}_{\bar{\partial}}(X,TX)\to 0.$$
Now we need a splitting $\beta$ to invert the last map, and there is such a splitting.  This is also natural.  Let me describe it.

We need $\varphi=\beta(\mu)$ and $\bar{\partial}\varphi=\alpha_*\mu$ along with $\phi\perp H^0(X,f^*T\bar{\cC}).$  This gives
$$
T_{(X,f)}\Hh_{g,n}\cong H^0(X,f^*T\bar{\cC})\oplus \beta H^{0,1}_{\bar{\partial}}(X,TX).$$
Now
$$
\langle \xi,\eta\rangle_{g,n}=\int_X (f_{T\bar{\cC}}\xi,f_{T\bar{\cC}}\eta)_{\bar{\cC}}d\rho$$ and consider $\omega_{g,n}=-\frac{1}{2}Im\ \langle\ ,\ \rangle.$  
Let me form the analogue of the K\:ahler theorem to finish the talk.

The final statement, we need to consider a Dirac type operator $D=\left(\begin{array}{cc}\bar{\partial} & -\alpha_*\\0&\partial \rho\end{array}\right)$ where this operator acts on two-component sections $\psi=\left(\begin{array}{c}\varphi\\\mu\end{array}\right).$ Now $D\psi=0$ if and only if $\bar{\partial}\psi=\alpha_*(\mu)$ and $\partial \rho \mu=0.$

Again consider the determinant of the tangent bundle
$$
\lambda_{g,n} = \wedge^{2n+2g-2}T\Hh_{g,n}$$ with $||\ ||_Q=\frac{||\ ||}{(det\ D^*D)^{\frac{1}{2}}}$
\begin{thm}
$$c_1(\lambda_{g,n},||\ ||_Q)=\frac{-1}{\pi^2}\omega_{g,n}.$$\end{thm}

\section{Ian Morrison, birational geometry of moduli spaces}
I've been asked to do an introductory survey.  I will include classical results and talk about some recent work.  I'm trying to balance telling you what's true and why it's true to keep prerequisites to a minimum.  There are notes I'm working on that have several things that I'm not including in these talks.  That's at projectivepress.com

So the theme that I'd like to convince you of, this goes back to Mumford and Joe Harris is the greatest proponent, is that any question you ask about varieties you should ask about the moduli space.  In the case of birational geometry there is a beautiful interplay between things you know about a single curve, about a family of curves, and the geometry of the moduli space.  Tomorrow I'll talk about divisors and cones, and \wh on Thursday.

Let me start with the space $\Mm_g,$ and over this space we have the universal curve $\Cc_g.$  On that curve we have the relative dualizing sheaf $\omega_{\Cc_g/\Mm_g}.$  I'm going to ignore that this only exists off the curves with automorphisms, but that's high codimension in general.  We can take the direct image of this dualizing sheaf and we get a bundle that I call $\Lambda$ because I want to define $\lambda=c_1(\Lambda).$  We could start with the first Chern class of $\omega,$ square it, and then push it down with $\pi.$  It's good to say that for $g\ge 3, pic(\Mm_g)=\qQ\lambda.$

The next space we want to work with is $\bar{\Mm}_g$ with a singularity condition, that there should be three nodes on a rational component be a numerical condition to say that the relative dualizing sheaf should be an ample curve on this base.

The class I want to focus on is in  $\Delta= \bar{\Mm}_g\backslash \Mm_g.$  The main fact from the deformation theory of nodal curves is that near the origin in the versal deformation space, we're looking at a neighborhood of the origin in $\cC^{3g-3}.$  A collection of the hyperplanes where things are nodal meet transversally.  What is the normal line to that hyperplane at the origin?  It is the tensor product of the two tangent lines to the node.  That is the key to doing enumerative calculations with the boundary.

One consequence of this description is that $\Delta$ is the closure of the locus of stable curves with a single node.
One componont, $\Delta_0$ or $\Delta_{irreducible},$ I want te pinch off a node, called irreducible because pinching off gives an irreducible surface.  Reducible would separate into genus one and genus two.

Now $$Pic(\bar{\Mm}_g)=\qQ\lambda\oplus\qQ\Delta_{ir}\oplus^{\lfloor g/2\rfloor}\Delta_i$$

After $\Mm_g$ and $\bar{\Mm}_g$ comes $\mgnbar.$  My condition is that the dualizing sheaf twisted by the points gives $\Mm.$  Here we can have $\bar{\Mm}_{0,3},$ a point.  $\bar{\Mm}_{0,4}$ is a sphere with three marked points.  What does $\bar{Mm}_{0,5}$ look like?
$$
\xymatrix{
&&&&\\
\infty\ar@{-}[rrrr]&&&&
\\1\ar@{-}[rrrr]&&&&
\\0\ar@{-}[rrrr]&&&&
\ar[uuuurrrr]
}
$$ but we need to blow up at the intersections.  This has three special fibers, each of which is the nodal intersection of two $\pP^1$ with two marked points on it.

I have the canonical maps.  For example I have the forgetful maps $\bar{\Mm}_{g,P}\to \bar{\Mm}_{g,N}$ for $N\subset P.$
Let's say I forget a point $q$ from $\bar{\Mm}_{g,N\cup\{q\}}.$  Then for each $j\le N$ I can look at the fiber above $j,$ where I break off a $\pP^1$ where $j$ is and mark that sphere with $j$ and $q$ (there are no moduli).

Then I can get classes $\sigma_j^*(c_1(\omega_\pi))=\omega_\pi|_{\Sigma_j}.$

There is something different about genus zero, you can't have an irreducible node and in any genus the classes $\Delta_i$ are no longer irreducble.  I need to specify a partition $P$ and I have to pay attention in genus zero to make sure there are two markings on a genus zero part.

Now gluing maps.  I can glue two specified points $q,r$ together, and this moves to the component of irreducible nodes, and this is a surjective map $\xi:\bar{\Mm}_{g-1,N\cup\{q,r\}}\to \Delta_{ir}\subset \bar{\Mm}_{g,N}$

There is the other one
$$\eta:\bar{\Mm}_{i,P\cup\{q\}}\times \bar{\Mm}_{g-i,N\backslash p\cup \{r\}}\to \Delta_{ir}\subset \bar{\Mm}_{g,n}.$$
For a fixed curve on the right we call this map $\theta.$

To use these we want some information.  So as an example, $\theta^*(\lambda)=\lambda.$  Here $H^0(c_1\omega_C)=H^0(B,\omega_B)\oplus H^0(B,\omega_B)$ and then I get
$\Lambda_C=\Lambda_B\oplus H^0(H,\omega_B)\otimes\Oo$

What happens when I pull back $\Delta_i,Q?$  It's boringly simple, it pulls back to itself with one exception.  If I pull back $\Delta_{i,P}$ then the image lies inside $\Delta_{i,P}.$

[fatigued.]
\section{Generating function for modular graphs}
Modular graphs are simply the graphs that arise as dual graphs of curves.  The generating functions for such graphs appear in many different topics, so such generating functions can be discussed in different languages.  Maybe I'll start with the language of Gaussian integrals.

Suppose you have a real $r$-dimensional space and a Gaussian measure with mean $X$ and covariance measure $\Sigma.$  This simply means that we have the density normal distribution $Nd\xi=\frac{1}{2\pi}^{r/2}\sqrt{\det\ \Sigma}}exp(-\frac{1}{2}(x-\xi)^{-1}\Sigma^{-1}(x-\xi)d\xi.$

Then for a function $U$ I define 
$$
\Psi(X,\Sigma)=\int_\rR^r exp(U(\xi))Nd\xi
$$
Let me talk about the $U$ I want to deal with, which is traditionally denoted $U(y,\hbar)$ which is
$$
\frac{U_0(y)}{\hbar}+U_1(y)+U_2(y)+\ldots\in \frac{1}{\hbar}\cC[(y,\hbar)].
$$
Now I add an $\hbar$ to $N$ to make it  
$$
Nd\xi=\frac{1}{2\pi\hbar}^{r/2}\sqrt{\det\ \Sigma}}exp(-\frac{1}{2\hbar}(x-\xi)^{-1}\Sigma^{-1}(x-\xi)d\xi
$$
We're mainly interested in $\Psi=\log\tilde{\Psi}.$
This function $\Psi$ satisfies very simple differential equations if $U$ is well-behaved.  Let $\sigma$ be the components of the metric $\Sigma,$ then 
$$\left\{ \begin{array}{l}\frac{\partial \tilde{\Psi}}{\partial \sigma_{ij}}=\frac{\hbar}{\{ij\}!}\frac{\partial \tilde{\Psi}}{\partial x_i\partial x_j}\\\tilde{\Psi}(y,0)=e^U\end{array}
$$
Where $\{ij\}$ has $1$ in the $i$ and $j$ place and zero elsewhere.

If I take $\Sigma=\sigma E$ then instead of this system of equations I have the heat equation $\frac{\partial\tilde{\Psi}{\partial \sigma}=\frac{\hbar}{2}\Delta\tilde{\Psi}.$

What do we get for $\Psi?$
$$
\frac{\partial \Psi}{\partial \sigma_{ij}}=\frac{\hbar}{\{ij\}!}\left[\frac{\partial^2 \psi}{\partial x_i\partial x_j}+\frac{\partial \psi}{\partial x_i}\frac{\partial \psi}{\partial x_j}, \psi(x,0)=U.$$

Then for $g=0$ we get $\frac{\partial \psi_0}{\partial \sigma_{ij}}=\frac{1}{\{ij\}!}\frac{\partial \psi_0}{\partial x_i}\frac{\partial \psi_0}{\partial x_j}.$  Then the solution is the following functional equation, with $F=\nabla_X U$ and $H=\nabla_X F:$
$$
\Phi(X,\Sigma)=F(X+\Sigma\Phi(X,\Sigma))
$$
where $\Phi=\nabla_X \psi.$
We write this thusly because it makes things easier to see, namely
$A(X)=X-\Sigma F(X)$ and $B(X)=X+\Sigma\Phi(X,\Sigma).$  Then $A(B(X))=X$ and $B(A(X))=X.$

For $g=1,\ \psi_1=U_1(X+\Sigma\Phi(X,\Sigma))-\frac{1}{2}\tr\ \log (E-\Sigma H(X+\Sigma \Phi)).$
This is a long calculation but is relatively trivial.

Right now I'm telling you everything without graphs.

For $g>1$ we have the theorem.  There are stable graph polynomials $P_g$ depending on $\Sigma$ and another parameter.  I must return to $U_g(X),$ and I want to write it as $\sum \frac{a_{g,N}X^N}{N!}.$  Then $P_g$ depends on $a_{g,N}$ for stable situations, $|N|+2g-2>0.$  These satisfy recurrence and homogeneity relations so that the solution can be written $$\Psi_g(X,\Sigma)=P_g(\frac{\partial^N U_g}{\partial X^N}(X+\Sigma\Phi), \Sigma(E-H(X+\Sigma Phi)\Sigma)^{-1})
$$
The inverse is $\Sigma+\Sigma H\Sigma+ \Sigma H\Sigma H\Sigma+\ldots$

Now I want to explain how to get all this story starting from graphs.  I still gave no proofs but since we have a system of equations, theoretically we can prove these without appealing to graphs.

What are these graphs?  On each vertex I want to write a nonnegative integer called the genus, and I want to call the half-edges coming out from $1$ to $r.$  If I want to join them, then I can do that and the edge gets the label $ij.$  I don't want to deal with stability.  Now let me introduce some notation.  For each vertex I will discuss the multiindex of its indexes.  Then $N(\Gamma)$ is the multiindex of the external half-edges.
The genus of the graph is
$g(\Gamma)=\sum_v g(V)+b_2(\Gamma)-b_0(\Gamma)+1.$

Then $\tilde{\Psi}(a_{g,N},X,\Sigma,\hbar)=\sum_\Gamma \frac{1}{Aut\ \Gamma}\prod_v a_{g(v)U(v)}\prod_e \sigma_{ij}\frac{X^{N(\Gamma)}}{N(\Gamma)!\hbar^{g-1}.$

If we don't order tails we should omit the factorial.

Then I want to define another function $\Psi$ which is the same but for all connected graphs.  A standard combinatorial statement is that $\tilde{\Psi}=exp(\Psi).$  For $X=0$ we kill all the summands corresponding to graphs with edges.  The sum for those is $\Psi(0,\Sigma,t)=U.$  The last thingi is that this function $\psi$ \wh.  It is not hard to understand that the same generating function for graphs with one fixed edge is the derivative by $\sigma_{ij}.$  Removing the connection gives a graph with two marked half-edges.  The genus of this graph is \wh.  Either we disconnect a cycle or the graph deconnects.  This explains $H$ here.  If $i$ and $j$ are equal we need two positive orderengs.

$\psi_0$ is the generating function for trees.  This is given in the generating functions, it's hard to see.  They all worked with $\sigma=Id.$

Now I want to speak about more interesting things to our theme about moduli spaces of graphs.

[Too fatigued.]




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