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

\begin{document}\maketitle
\section{Geometry of Teichmueller Spaces}
[I do not take notes on slide talks.]
\section{Penner}
I'd like to tell you today and tomorrow about a relationship of geometry and combinatorics.  The geometry I'm referring to is of the Teichmueller space $\Tt_g^s$ of surfaces $F_g^s$ with $s\ge 1$ punctured surfaces, crossed with $R_+^s.$  

The combinatorics is fat or ribbon graphs, where each vertex has a cyclic ordering on its edges.  A fat graph is data for a skinny surface.  I'l explain a relationship between the geometry and the combinatorics.

We know that Teichmueller theory can be treated either conformally or in the hyperbolic setting, and the relationship between the combinatorics and the geometry depends on the point of view.

There is a process $JS$ which takes a fat graph to a conformal surface.  Put positive numbers on the edges, and run around the boundary of the associated skinny surface.  Take an annulus of height one and circumfrence given by the lengths of the edges, and glue the annulus in.  The resulting object doesn't have a Riemannian metric, but you can get one, and then you put in a concentric conformal structure.  The $R_+^s$ is the perimeters.

The incredible theorem is that in the appropriate sense this is an isomorphism.  As you vary over homotopy types of embeddings of fat graphs into the surface and over all fat graphs and lengths, you get a one to one onto map to conformal surfaces.  

For the hyperbolic structure setting, you have the convex hull construction, which goes in the other direction.  Suppose I have a hyperbolic surface.  I want to use the upper sheet of the hyperboloid and Minkowski isometries and a subgroup $\Gamma< SO_+(2,1).$  I'll interpret the real numbers as sizes of horocycles, to give me finite cutoffs.  

Consider the open positive light cone.  This is naturally identified with the set of all horocycles in $\hH.$  You let $h(u)=w\in \hH$ such that $uw=\frac{-1}{\sqrt{2}}.$  Any negative constant would do, I'm taking this number for convenience.

An $s$-tuple of horocycles gives rise to $s\ \Gamma$ orbits of points in $L^+.$  This is a $\Gamma$-invariant subset of the light cone.

{\it Fact.}The extreme edges of the closed convex hull of these $s$ orbits project to arcs in $F_g^s$ decomposing it into ideal polygons.  These are discrete points in the light cone.  Take the closed convex hull which has extreme edges.  Project these back to the surface and you get arcs on the surface decomposing it into polygons $\Delta$

Where is the fat graph?  It's the Poincar\'e dual to $\Delta.$  I'm going to freely pass back and forth between the two formalisms.

I haven't told you the numbers on the edges of the fat graph.  I'll tell you presently the numbers you put on the edges that come from the hyperbolic information.

These two constructions are not inverses but I conjecture that $JS\crc chc$ is bounded distortion in the Teichmueller metric.  My goal is to tell you how the combinatorics responds when the geometry degenerates.  If the conjecture is true then whatever I say in the hyperbolic setting is true in the conformal setting.

While $(JS)^{-1}$ is highly transcendental (involves solving the Beltrami equation), the map $(chc)^{-1}$ is algebraic.  

The hyperbolic treatment depends on global affine coordinates in this bundle over Teichmueller space which do not exist in the conformal setting.  The basic invariants of the story are:

Given a hyperbolic surface and a horocycle around each hole, $\tilde{\Gamma}\in \Tt_g^s\times \rR^s_+$ and given a homotopy class of an arc $\alpha$ connecting punctures, you can look at the length along the geodesic $\alpha$ between the two horocycles.  The horocycles give us a natural way to truncate this path between punctures.

My treatment is very 19th century, it depends on the coordinates.  Define the $\lambda$-length $\lambda(\alpha;\hat{\Gamma})=\sqrt{e^\delta}.$  I get a negative sign on $\delta$ if the horocycles intersect.

The $\lambda$-lengths satisfy many equations.
\begin{lemma}
\begin{enumerate}
\item[a.] $\lambda(h(u),h(v))=\sqrt{u\cdot v}$ for $u,v$ in the light cone with $\alpha$ the unique geodesics connecting their centers, orthogonal to them.
\item[b.] Suppose I have a decorated triangle with $\lambda$-lengths $a,b,c.$  Then \wh is $c/ab.$
\item[c.] Suppose I have a decorated quadrilateral, triangulated.  What is the length of the other diagonal?  Well, $\underbrace{ef}_{\text{diagonals}}=ac+bd,$ which is called a Ptolemy relation because of Ptolemy's theorem about inscribed quadrilaterals.
\item[d.] Under the conformal map sending $u,v,w$ to $(0,1,\infty),$ then $\zeta\mapsto {-bd}{ac}$ if the square is
$$\xymatrix{
\zeta\ar@{-}[r]^d\ar@{-}[d]_c&u\\w\ar@{-}[ur]^e\ar@{-}[r]_b&v\ar@{-}[u]_a}$$
\item[e.] The volume of the tetrahedron  determined, well, the simplicial coordinate of $e$ is $$E=\frac{a^2+b^2-e^2}{abe}+\frac{c^2+d^2-e^2}{cde}$$ 
Then $abcdE=2\sqrt{2}$ is the volume of the tetrahedron in Minkowski space.
\end{enumerate}
\end{lemma}
That's the background, and here's the theorem of today:
\begin{thm}(Global coordinates)\\
Fix an ideal triangulation that is a decomposition of the surface into triangles by arcs connecting punctures.  Then there is a natural map $\Tt_g^s\times \rR^s_+\to \rR^\Delta_+$ which maps 
$$\tilde{\Gamma}\mapsto (\alpha\mapsto \lambda(\alpha,\hat{\Gamma}))$$
and this is a real analytic homeomorphism onto.  Furthermore, the Weil-Peterson $2$-form pulls back to 
$$
\sum_{\text{triangles labeled by }a,b,c} d\log a\wedge d\log b +d\log b\wedge d\log c + d\log c\wedge d\log a
$$
The action of the mapping class group in coordinates is given by a permutation of these coordinates followed by these Ptolemy translations.  These are the only coordinates where the MCG action is tractable and so is the WP form.
\end{thm}
In the remaining time I would like to give you a fairly complete proof, in fact two.  Let me give a sketch.  I start with two little claims that I won't prove.  First, given a triple of rays in the light cone and a triple of numbers, there is a unique triple of points in the light cone realizing the triple of rays as $\lambda$-lengths (inner products).  This can just be done explicitly in Minkowski space.

The second claim is that given $x,y$ in the light cone, and positive reals $a,b,$ there exists a unique point on either side of the plane of $0,x,y$ realizing $a,b$ as $\lambda$ lengths.   This you can prove with a calculation or synthetically.

I've described to you via this calculation of $\lambda$ lengths what the map is.  I have to calculate an inverse.  I have to build a hyperbolic structure given a combinatorial skeleton.

Consider the topological universal cover $\tilde{F}$ of $F_g^s,$ and lift $\Delta$ to a squiggly ideal triangulation.  We can lift numbers from downstairs to upstairs, the putative $\lambda$-lengths.  

I will produce a Fuchsian group by straightening this to a tesselation of the Poincar\'e disk, and lift these ideal points to the open positive light cone.  Choose three rays in the light cone and a triangle in the universal cover.  Apply the first claim to get the lengths on this triangle realized on the rays.  On either side of the plane I can add the next triangle.  I can avoid what I've already put down by choosing a side.  This is bounded geometry.  There are only finitely many cross-ratios.  Every time I add a triangle I move some definite distance.  Whenever I do a left and right turn I move a definite distance.  That shows the surjection property.  

I have to produce the Fuchsian group.  To construct it, choose a fundamental domain, and suppose that $x$ and $y$ are identified with $x',y'.$  Extend one triangle too many, and there's a unique transformation taking $xyz$ to $x'y'z'.$  The uniqueness statement gives you some invariance under $\pi_1.$  The group up to conjugacy is independent of the choice of fundamental domain.

I want to give an alternative proof that I will need tomorrow for the asymptotics of the cell decomposition.  This is a construction of Fock-Thurston.  There's a little trick.  Take the fat graph $G$ dual to $\Delta$ and blow up each vertex of the fat graph into a small triangle.  Define matrices, the right and left turn matrices:
$$\left(\begin{array}{cc}-1&-1 \\1&0 \end{array}\right),\left(\begin{array}{cc} 0&1\\-1&-1 \end{array}\right)$$
and 
$$X_\zeta=\left(\begin{array}{cc}0&-\zeta^{-1/2} \\\zeta^{1/2}&0 \end{array}\right)$$
where $\zeta$ is the crossratio $-\frac{bd}{ac}.$

Now you can get the Fuchsian group immediately, calculating the monodromy by moving along with these matrices; The ordered products of these matrices give monodromies of the representations.  You use $X_\zeta$ to move across from one small triangle to the next.

As to the third part of the theorem, I never spoke of marking, so that if $\varphi\in MCG_g^s,$ then $\lambda(\alpha,\hat{\Gamma})=\lambda(\varphi\alpha,\varphi\hat{\Gamma}),$ so $\lambda(\alpha,\varphi\hat{\Gamma})=\lambda(\varphi^{-1}\alpha,\hat{\Gamma}).$  The fact, which I'll prove tomorrow, follows from the existence of the cell decomposition, and is that Whitehead moves, removing diagonals and putting in the other one (contracting an edge and expanding it the other way) act transitively on ideal triangulations.  We want to know the numbers, we apply the inverse homeomorphism.  Do the Whitehead moves to get back to the original triangulation, and you can get all the $\lambda$ lengths by these Ptolemy moves.

Short of a proof of the Weil-Peterson part, I've sketched a proof.  I want to explain tomorrow what happens to fat graphs as you approach nodal curves.  Thanks for your attention.








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