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

\begin{document}\maketitle
We're talking about systems of evolutionary PDEs
$$u_t^i=A_j^i(u)u^j_x+\sum_{k\ge 1}\epsilon^k Q_k^i(u,u_x,\ldots, u^{(k+1)})$$
This is called a hierarchy if we have many times $t$ becomes $t_1,t_2,\ldots$ and the disprsionless limit takes $\epsilon$ to $0.$ 

Semisimplicity says the eigenvalues, that is, the solutions to $\det A^i_j(x)-\lambda \delta^i_j =0$ are distinct.

I will always assume this, and indeed the stronger condition of hyperbolicity.  

So starting from an integrable hierarchy and moving to the dispersionless limit yields a simplified dispersionless hierarchy.  This procedure preserves integrability, that is, the operators all still commute.  The question I want to address is the question of reconstruction.  This question has never been addressed before the advent of Gromov Witten, to figure out how to arise at a given dispersionless hierarchy, to classify all integrable hierarchies having this as its dispersionless limit.  This is inspired by trying to find the higher genus Gromov Witten invariants starting from genus zero.  To give a simple example, take the case $n=1$ and the hierarchy will be the hierarchy of equations $\frac{\partial V}{\partial t_k}=\frac{V^k}{k!}\frac{\partial V}{\partial x}.$  It is not hard to check that these commute
$\frac{\partial }{\partial t_k}\frac{\partial }{\partial t_r}=\frac{\partial}{\partial t_r}\frac{\partial}{\partial t_k}.$  This can be checked directly or with the Poisson bracket, $\{V(x),H_k\}$ with
$$H_j=\frac{1}{2\pi}\int_0^{2\pi}\frac{v^{k+2}}{(k+2)!}dx$$
and $\{V(x),V(y)\}_1=\delta'(x-y)$;
$\{V(x),V(y)\}_2=v(x)\delta'(x-y)+\frac{1}{2}v_x\delta(x-y)$.
So given two functionals $H$ and $F$ on $V,$ you compute 
$$
\{H,F\}_2=\frac{1}{2\hbar}^2\int_0\int_0 dxdy\frac{\partial H}{\partial v(x)}\{v(x),v(y)\}_2 \frac{\partial F}{\partial v(x)}$$and I obtain
$$\frac{1}{2\pi}\int_0^{2\pi}dx\frac{\partial H}{\partial v(x)}\left[v(x)\frac{\partial F}{\partial v(x)}'+\frac{1}{2}v_x \frac{\partial F}{\partial v(x)}\right]$$
This is the standard dual of the Lie algebra of vector fields on $S^1.$

Then we want to deform, well, these two formulas look very different, but do the following.  

{\it Remark.} Do the substitution $v=\frac{1}{4}\tilde{v}^2.$  The second Poisson bracket is $\{\tilde{v})k),\tilde{v}(y)\}_2=\delta'(x-y).$  From the point of view of differential geometry this is just a change of basis, of dependent variables.  There is no difference between the two of them with respect to the group of local diffeomorphisms.

This is a very simple exercise, but following this observation, let me give some more general setting beyond $n=1.$  The Poisson bracket on $\Ll(M^n)$ comes from a flat metric on $M^n$ $ds^2=\eta_{ij}dx^idv^j.$  Take the inverse matrix $\eta^{ij}.$  Then the bracket $\{v^i(x),v^j(x)\}=\eta^{ij}\delta'(x-y).$  This is odd and the right hand side does not depend on the coordinate so we get Jacobi for free.  This constant bracket was introduced almost 25 years ago as a Poisson bracket of hydrodynamic type.

To have a biHamiltonian structure on the loop space we must have a pair of flat metrics both flat in different coordinates.  Such a pair determines a flat pencil of metrics, and under mild assumptions these correspond to Frobenius manifolds.  I'm arriving at one of the main points.  I want to introduce an equivalence relation.  For the dispersionless limit, this is clear, it is just the group of local diffeomorphisms.  I now want to extend this to the full class of PDEs.  This will be the class of change of dependent variables $u^i\to \tilde{u}^i$ which in the first term is a local diffeomorphism $F_0^i(u)$ but then adds 
$$\sum_{k\ge 1}\epsilon^k \underbrace{F^i_k(,u_k,\ldots,u^{(k)}}_{\text{degree }k}$$
These transformations form a group and we call two things equivalent if they are taken to one another by an element of the group.

\begin{defi}
An integrable hierarchy is trivial if it is equivalent to the dispersionless limit.
\end{defi}

The main technical trick to proceed to the classification.  We have a suitable infinite dimensional manifold (the loop space) we have ``changes of coordinates'' on the loop space, we have vector fields.  So now we can define the various homologies, the biHamiltonian and so on.
\begin{lemma}
Getzler\\
If $M^n$ is a ball then $H^{>0}(\Ll(M^n),\eta^{ij}\delta'(x-y))=0.$ \end{lemma}
So the deformation theory is trivial.

The proof is not that complicated.  From the finite dimensional intuition this is a little bit surprising.  In that case we can rotate the leaves of the finite dimensional foliation.  This is the basis for going ahead and we can look at biHamiltonian cohomology.

An important technical step is the definition of quasitriviality.  Consider a bigger group.  Again the leading term will be a local diffeomorphism, but this time $F^i_k$ is a rational function (of degree $k$) in the jets.  The description of the infinitessimal deformation space is completely different.
\begin{thm}
Assume $M^n$ is a ball.  Then $H^2_{quasi}(\pi_1,\pi_2)$ is always trivial.
\end{thm}
In particular that means as soon as we allow rational transformation functions, all tails can be killed.  This is a quite technical theorem, but let me show how it works in an example, the KdV hierarchy.  The dispersionless version is
$v_{t_1}=vv_x, v_{t_2}=\frac{1}{2}v^2 v_x.$  Do the substitution
$$u=v+\frac{\epsilon^2}{24}(\log v_x)_xx+\ldots$$ so that if we do the substitution you obtain $u_{t_1}=uu_x+\frac{\epsilon^2}{12}u_{xxx}$ and so on, the hierarchy before taking the limit.  Doing the calculations there are a lot of cancellations.  What is important is that necessarily unexpected higher order derivatives are involved in the formulae for the quasitriviality.  An important lemma was the $3g-2$ lemma, that $m_k\le \frac{3k}{2}.$
\begin{thm} Now I want to compute $H^2(\pi_1,\pi_2)$ without the quasi equivalence relation, with polynomial jets.

This, is isomorphic to $\bigoplus^n C^\infty(\rR)$.
\end{thm}
My time is almost over.  Next time I will give a more explicit formula for how to establish such an isomorphism.  To give you some idea of how to establish this, consider the example $n=1,$ when I recall the dispersionless pencil was $\delta'(x-y)$ and the second one was $u(x)\delta'+\frac{1}{2}u_x\delta(x-y).$  The infinitessimal deformation is obtained adding the term $3\epsilon^2 c(u)\delta'''(x-y)+\ldots$
Given $c(u)$ one determines uniquely the deformation class of this structure.  In particular, if this coefficient is zero then the pencil is trivial.

For choosing $c(u)=c,$ a constant, one obtains the central extension of the Lie algebra of vector fields on the circle, the Virosora with central charge $c.$  So we call these $c_i(u_k)$ the central invariants.

One more example, I will finish.  One example is, given a simple Lie algebra $\la{g},$ then one can associate to that the Drinfel'd construction, an integrable hierarchy.  You do something with the twisted Kac-Moody Lie algebra $\hat{\la{g}}^{(1)}.$  The dispersionless Poisson pencil comes from the Frobenius structure on $\Ll^2/W(\la{g})$ and we want to see what happens here.  The theorem for the $A_n,B_n,C_n,D_n,G_2$ is that $c_i=\frac{1}{48}\langle \alpha_i,\alpha_i\rangle_{\la{g}}$.

Now about existence.  Can you extend a given deformation to a \wh.  We cannot do this, even for one component it is very hard.  We can prove existence up to order eight.  So 
\begin{thm}
This is a partial result.  Given a semisimple Frobenius manifold, there exists a unique hierarchy with $c_1=c_2=\ldots=\frac{1}{24}$ so that the dispersionless limit is described in terms of the semisimple structure in the standard way.
\end{thm}
\begin{enumerate}
\item The first step is the quasitriviality theorem.  We need a suitable transformation, which is rational.  
\item Then we use the $3g-2$ lemma, which says that the quasitriviality can be looked for in a particular case.  \item Then we have to determine the quasitriviality transformation from the Virosora constraints.  Then this is unique from the $3g-2$ lemma.  \item Then step four is to plug this into the dispersionless hierarchy.  Then we prove cancellation of the denominators.  This is because Virosora acts by symmetries described in terms of linear differential operators.  This is basically the idea of proving the vanishing of the denominators.
\item This we cannot do.  To prove polynomiality in the jets of the biHamiltonian structure, that would mean we could start with a standard one, and move to a rational one, but to prove the cancellation of the denominators, well, I'm going to stop now.
\end{enumerate}

\end{document}
