TITLE: Circle passing through intersection points of two bigger circles
QUESTION [1 upvotes]: Suppose the equations of two intersecting circles are given.Now how to find the equation of circle passing through the points of intersection of the larger circles?
Now please dont tell me that i got to solve the two equations for point of intersections :-P!! I guess there must be a shorter method..any ideas?
P.S. i MEANT THE SMALLEST POSSIBLE CIRCLE.

REPLY [0 votes]: Let the equations of two intersecting circles be $C_1=0$ and $C_2=0$.
Then the equation of family of circles passing through the intersection points can be given by $C_1 + tC_2 = 0$, $t\ne -1 $. It is easy to see that this equation satisfies the points that are common to both the circles.


TITLE: Examples of dual bases. Intuition behind the dual basis.
QUESTION [2 upvotes]: So we have our definition of the dual basis:
We call the ordered basis $\beta^*=\{\textbf{f}_1,\dots ,\textbf{f}_n\}$ of $V^*$ that satisfies $\textbf{f}_i(x_j)=\delta_{ij}$ the dual basis of $\beta$.
Take this example:
The space of square $n \times n$ matrices. We have an example of a linear functional $\textbf{f}:V \to \mathbb{R}$ that maps a matrix $A$ to $tr(A)$. (an example functional just to demonstrate my understanding).
What would be the dual basis to the standard basis for this space?
Another example $\mathbb{R}^2$:
Let $\beta =\{(2,1),(1,3)\}$. We wish to find the dual basis $\beta^*=\{\textbf{f}_1,\textbf{f}_2\}$. To explicitly determine a formula for $\textbf{f}_1$ we consider
$1=\textbf{f}_1(2,1)=\textbf{f}_1(2e_1+e_2)=2\textbf{f}_1(e_1)+\textbf{f}_1(e_2)$
$0=\textbf{f}_1(1,3)=\textbf{f}_1(e_1+3e_2)=\textbf{f}_1(e_1)+3\textbf{f}_1(e_2)$
Why do we use the standard basis here instead of the given basis explicitly?
After all this is done, does the dual basis always induce a natural inner product? If so, what does that really mean intuitively and what does this do for us?

REPLY [1 votes]: If you have a basis $\mathcal B$ of a vector space $V$, the dual basis $\mathcal B^*$ of $V^*$ simply consists in the (parallel) projections on the vectors of $\mathcal B$.
Thus, in the example you mention, the standard basis is the set of matrices $E_{ij}=(e_{kl})$, where $e_{kl}=0$ if $(k,l)\ne (i,j)$, $e_{ij}=1$. Its dual basis is made up of the maps
$$A=(a_{kl})\longmapsto a_{ij}.$$
Second question: the standard basis is used just because all cordinates are given in the standard basis… Note that since $f_1=2e_1+e_2$, we have $f_1^*=2e_1^*+e_2^*$.
Last question: no, the dual basis does not induce an inner product, because this notion is valid for vector spaces over abstract fields, while inner products are deined for real (or complex) vector spaces. However, for real vector spaces the notions are linked through a natural pairing:
\begin{align*}
V\times V^*&\to \mathbf R,\\
(x,u)&\mapsto u(x).
\end{align*}


TITLE: How does the nature of bonding between individual atoms in a molecule affect the van der Waal interactions between molecules?
QUESTION [0 upvotes]: I am currently studying certain materials. Some of these materials have ionic bonds, while some others have covalent bonds. Moreover, each unit of these materials interacts with the adjacent unit via van der Waals interaction. I have observed that the strength of the van der Waals interaction is different in case of materials with different type of bonds.
Is there any correlation between the nature of bonds and the strength of van der Waals interactions between such sub-units of the material?

REPLY [0 votes]: The term van der Waals force is a catchall for several different kinds of forces. However all of the forces are due to electric dipoles, and the strength of the force is related to the dipole moments of the interacting molecules.
The strongest interaction is when both interacting molecules have a permanent electric dipole. For example water molecules have a large permanent dipole and the dipole-dipole interactions produce a relatively large force between them.
An example of the other extreme would be the interaction between two noble gas atoms. These do not have a permanent dipole, but they have a (very small) oscillating dipole due to random thermal fluctuations in their electron density. These random ocillating dipoles in two atoms interact to produce a relative weak force called the London dispersion force.
And finally we the the case in between where one of the interacting molecules as a permanent dipole and the other doesn't.
So as a general rule a permanent dipole produces a strong VdW force, and we get dipoles when the bonds in the molecules are polar. If there is no permanent dipole the higher the polarisibility of the molecule the larger the London dispersion force.


TITLE: Prove that the integral operator has no eigenvalues
QUESTION [6 upvotes]: Let $V$ be the vector space of all real valued continuous functions. Prove that the linear operator $\displaystyle\int_{0}^{x}f(t)dt$ has no eigenvalues.

REPLY [4 votes]: Hint: Suppose you have $\int_{0}^{x} f(t)dt = \lambda \cdot f(t)$. Differentiate both sides - you should easily be able to solve the resultant differential equation. Is your solution truly an eigenvector if it is nontrivial?


TITLE: How can a laser pointer have range of several kilometers in atmosphere?
QUESTION [2 upvotes]: Laser pointers manufacturers claim that some pointers have range of several kilometers.
Okay, they use a powerful laser, but that powerful laser usually has power less than one watt. Okay, the laser beam is very focused.
But what about atmosphere particles like dust and vapor? Why don't these particles diffuse the beam and make it lose energy?
How does it happen that a less than one watt laser can have several kilometers range in atmosphere?

REPLY [0 votes]: Upshot: the range of a laser is greater because the light is concentrated in a very narrow beam.
Without going into much detail of the power of the laser, I think the problem here has to do with the definition of 'intensity'. Wikipedia lists several different units for intensity, the one that most closely matches 'brightness' as perceived by the eye is probably radiance.
The units of radiance are
$$
L = W·sr^{−1}·m^{−2}
$$
or watt per steradian per square metre, a 'steradian' being a two-dimensional angle on the sky.
Now given a fixed power for a certain light source, the difference between ordinary lighting and a laser are two-fold:

The surface ($m^{2}$) of a laser is much smaller (probably a factor of 10)
The divergence ($sr$) of the laser beam is also much smaller (factor ~100 or more).

Both these factors mean that the radiance for the same power ($W$) a laser beam is at least 1000 times as 'bright' as a regular light source. Since the brightness is higher, the beam will be visible up to a longer distance.
It is important to note that this result is regardless of the absorbance or scattering that occurs, which in principle happens for both laser and non-laser beams.


TITLE: Can we have a one-one function from [0,1] to the set of irrational numbers?
QUESTION [17 upvotes]: Since both of them are uncountable sets, we should be able to construct such a map. Am I correct?
If so, then what is the map?

REPLY [1 votes]: There are lots of good answers already, but I thought I might add one more using a flexible method that shows up a lot in descriptive set theory.
Note that $\mathbb{R} \setminus \mathbb{Q}$ is the intersection of countably many dense open sets $(G_n : n \in \mathbb{N})$; just take $G_n = \mathbb{R} \setminus \{q_i\}$ for some enumeration $(q_i : i \in \mathbb{N})$ of the rationals.
Note also that there is an injection from $[0,1]$ to the Cantor space $2^\mathbb{N}$ given by binary expansions.  If there is a choice between a binary expansion that is eventually zeroes and another that is eventually ones, take the one that is eventually zeroes. (This injection is not continuous.)
So it suffices to prove the following general fact:

Let $(G_n : n \in \mathbb{N})$ be a sequence of dense open subsets of $\mathbb{R}$.  Then there is an injection from $2^\mathbb{N}$ to $\bigcap_{n \in \mathbb{N}} G_n$. (As we will see, there is a continuous injection.)

By recursion, we can define for each finite sequence $s$ of natural numbers a closed interval $I_s$ satisfying the following properties:

$I_s$ has length between zero and $2^{-n}$ where $n$ is the length of the finite sequence $s$,
$I_t \subset I_s$ whenever $t$ extends $s$,
$I_s \cap I_t = \emptyset$ whenever $s$ and $t$ are incompatible, and
$I_s \subset G_n$ where $n$ is the length of $s$.

This binary tree of intervals is called a Cantor scheme.  If desired, we can define $I_s$ explicitly as $[q_i,q_j]$ where $(i,j)$ is the lexicographically least pair of natural numbers making this interval obey the rules.
Every element $(d_i : i \in \mathbb{N})$ of the Cantor space $2^\mathbb{N}$ determines a branch in this binary tree.  Our continuous injection is obtained by mapping $(d_i : i \in \mathbb{N})$ to the unique point in the intersection of the corresponding sequence of closed intervals
$$I_{d_0} \supset I_{d_0,d_1} \supset I_{d_0,d_1,d_2} \supset \cdots.$$


TITLE: Amplification of magnetic field
QUESTION [2 upvotes]: can we by any means amplify magnetic signal as we can with electric signal. As both electric and magnetic field can be represented in the form of a wave the analogy seems to be natural.
I want the input and output as magnetic signal.

REPLY [1 votes]: I presume your idea revolves around permanent magnets, if so stacking decreasing sized magnets in a pyramid configuration grants you a powerful localized field at the smallest point. Halbach configurations allow for combined field manipulation which comes with the downside of slowly degrading the total field strength over time.
Please do let me know what you find as I am researching a similer querry.


TITLE: Extending a collection of objects by the repeated application of a collection of functions
QUESTION [2 upvotes]: I am trying to find the correct terminology for a process. The following description is handwavy but the goal is to give the idea of the process I am trying to describe mathematically.
We start with a collection (/set/category/...?) $\mathcal{C_{0}}$ of objects $c_{i}$ and a collection $\mathcal{F}$ of functions $f_{i}$ (/maps/morphisms/...?) that can be applied to every object $c_{i}$ in $\mathcal{C}$. The arity of the functions can be anything (some can take one argument, some others two, some others three, etc...). Now we apply the following process: at each step $n$, all the functions $f_{i}$ are applied to all the objects in $\mathcal{C_{n}}$ and all the combinations of objects (for the functions that takes more than one argument) to produce $\mathcal{C_{n+1}}$ as the union of $\mathcal{C_{n}}$ and the results of the application of the functions. We call $\mathcal{C}$ the collection of objects obtained at infinity.
Is there anything in set theory or category theory (or other domains of maths) that would resemble something along these lines and where (ideally) $\mathcal{C_{0}}$, $\mathcal{F}$, and $\mathcal{C}$ would have an associated terminology. I am not necessarily looking for an exact correspondence with what I described (which is very handwavy anyway), but I am looking for mathematical notions/entry points to investigate around the idea.
Illustration: if $\mathcal{C}_{0} = \{2\}$ and $\mathcal{F} = \{f, g\}$ where $f$ is the function that gives the next prime number if given a prime number, and $0$ otherwise and where $g$ is the product of two numbers, ultimately $\mathcal{C} = \mathbb{N} \setminus \{1\}$ (if I'm not mistaken).

REPLY [3 votes]: For me, given a set of operators $\mathcal F$ on a set $X$, and given a subset $Y\subseteq X$, it would seem natural to say something like "the $\mathcal F$-closure of $X$" to denote the new subset of $X$ obtained by applying things in $\mathcal F$ until nothing new is obtained.
I'm not sure if there is an impediment to relaxing this to include operators on a proper class rather than a set, but perhaps there is, depending on what the operators do.
As a special case, if you are viewing a group $G$ acting on a set $X$, the $G$-closure of a single element is called an orbit.


text:
\begin{document}
\date{}
\subjclass[2010]{37K10, 05C30}
\keywords{Grothendieck's ``dessins d'enfants'', ribbon graphs, Virasoro constraints, topological recursion}

\author{M.~Kazarian,~P.~Zograf}
\address{Steklov Mathematical Institute\\8 Gubkin St.\\Moscow 119991 Russia}
\email{kazarian{\char'100}mccme.ru}
\address{St.Petersburg Department of the Steklov Mathematical Institute\\
Fontanka 27\\ St. Petersburg 191023, and
Chebyshev Laboratory of St. Petersburg State University\\
14th Line V.O. 29B\\
St.Petersburg 199178 Russia}
\email{zograf{\char'100}pdmi.ras.ru}

\title[Virasoro constraints and topological recursion]{Virasoro constraints and topological recursion for Grothendieck's dessin counting}

\begin{abstract}
We compute the number of coverings of ${\C}P^1\setminus\{0, 1, \infty\}$ with a given monodromy type over~$\infty$ and given numbers of preimages of 0 and 1. We show that the generating function for these numbers enjoys several remarkable integrability properties: it obeys the Virasoro constraints, an evolution equation, the KP (Kadomtsev-Petviashvili) hierarchy and satisfies a topological recursion in the sense of Eynard-Orantin.
\end{abstract}

\maketitle

\tableofcontents

\section{Introduction and preliminaries}
Enumerative problems arising in various fields of mathematics, from combinatorics and representation theory to algebraic geometry and low-dimensional topology, often bear much in common. In many cases the generating functions associated with these problems exhibit similar behavior -- in particular, they may satisfy
\begin{itemize}
\item
Virasoro constraints,
\item
Evolution equations of the ``cut-and-join'' type,
\item
Integrable hierarchy (such as Kadomtsev-Petviashvili (KP), Korteveg-\linebreak DeVries (KdV) or Toda equations),
\item
Topological recursion (also known as Eynard-Orantin recursion).
\end{itemize}
Simple Hurwitz numbers provide one of the best studied examples of such an enumerative problem -- indeed, their generating function satisfies the celebrated cut-and-join equation \cite{GJ1}, the Virasoro constraints (via the ELSV theorem \cite{ELSV}
and the famous Mumford's Grothendieck-Riemann-Roch formula \cite{M} it reduces to the Witten-Kontsevich potential), the KP hierarchy \cite{O}, \cite{KL} or \cite{K}, and the topological recursion \cite{EMS}. Other examples include the Witten-Kontsevich theory, Mirzakhani's Weil-Petersson volumes, Gromov-Witten invariants of the complex projective line, invariants of knots, etc. (see \cite{EO1}, \cite{EO2} for a review).

These remarkable integrability properties of generating functions usually result from matrix model reformulations of the corresponding counting problems. However, in this paper we show that for the enumeration of Grothendieck's {\em dessins d'enfants} all these properties follow from pure combinatorics in a rather straightforward way.

The origin of Grothendieck's theory of dessins d'enfants~\cite{G} lies in the famous result by Belyi:

\begin{theorem} {\rm (Belyi,~\cite{B})}
A smooth complex algebraic curve~$C$ is defined over the field of algebraic numbers~$\Qb$ if and only if there exist a non-constant meromorphic function~$f$ on~$C$ (or a holomorphic branched cover~$f:C\to\pl$) that is ramified only over the points $0,1,\infty\in\pl$.
\end{theorem}

We call~$(C,f)$, where~$C$ is a smooth complex algebraic curve and~$f$ is a meromorphic function on~$C$ unramified over~$\pl\setminus\{0,1,\infty\}$, a {\em Belyi pair}. For a Belyi pair~$(C,f)$ denote by~$g$ the genus of~$C$ and by~$d$ the degree of~$f$. Consider the inverse image~$f^{-1}([0,1])\subset C$ of the real line segment $[0,1]\subset\pl$. This is a connected bicolored graph with~$d$ edges, whose vertices of two colors are the preimages of 0 and 1 respectively, and the ribbon graph structure is induced by the embedding $f^{-1}([0,1])\hookrightarrow C$. (Recall that a ribbon graph structure is given by prescribing a cyclic order of half-edges at each vertex of the graph.)

The following is straightforward (cf. also~\cite{LZ}):

\begin{lemma}\label{Gr}{\rm (Grothendieck,~\cite{G})}
There is a one-to-one correspondence between the isomorphism classes of Belyi pairs and connected bicolored ribbon graphs.
\end{lemma}

\begin{definition}
A connected bicolored ribbon graph representing a Belyi pair is called Grothendieck's {\em dessin d'enfant}.\footnote{An important observation of Grothendieck that, by Belyi's theorem, the absolute Galois group ${\rm Gal}(\Qb/\Q)$ naturally acts on dessins,
lies beyond the scope of this paper; we refer the reader to \cite{LZ} for details.}
\end{definition}

Let~$(C,f)$ be a Belyi pair of genus~$g$ and degree~$d$, and let~$\G=f^{-1}([0,1])\hookrightarrow C$ be the corresponding dessin. Put $k=|f^{-1}(0)|,\;l=|f^{-1}(1)|$ and $m=|f^{-1}(\infty)|$, then we have $2g-2=d-(k+l+m)$. We assume that the poles of $f$ are labeled and denote the set of their orders by $\mu=(\mu_1,\ldots,\mu_m)$, so that $d=\sum_{i\geq 1}\mu_i$. The triple~$(k,l,\mu)$ will be called here the {\em type} of the dessin~$\G$, and the set of all dessins of type~$(k,l,\mu)$ will be denoted by~$\Dc_{k,l;\mu}$.

Actually, instead of the dessin $\G=f^{-1}([0,1])$ corresponding to a Belyi pair~$(C,f)$ it is more convenient to consider the graph $\G^*=\overline{f^{-1}(1/2+\sqrt{-1}\R)}$ dual to $\Gamma$ (where the bar denotes the closure in $C$), see Fig.~\ref{dual}. The graph $\G^*$ is connected, has $m$ ordered vertices of even degrees $2\mu_1,\ldots,2\mu_m$ at the poles of $f$ and inherits a natural ribbon graph structure. Moreover, the boundary components (faces) of $\G^*$ are naturally colored: a face is colored in white (resp. in gray) if it contains a preimage of 0 (resp. 1), and every edge of $\G^*$ belongs to precisely two boundary components of different color.

\begin{figure}[hbt]
 \begin{center}
 \includegraphics[width=5cm]{dual}
 \caption{Decomposition of $\pl$ into two 1-gons.}
 \label{dual}
\end{center}
\end{figure}

In this paper we are interested in the weighted count of labeled dessins d'enfants of a given type. Namely, define
\begin{align*}
N_{k,l}(\mu)=N_{k,l}(\mu_1,\ldots,\mu_m)=\sum_{\G\in\Dc_{k,l,\mu}}\frac{1}{|{\rm Aut}_b \G|}\;,
\end{align*}
where ${\rm Aut}_b \G$ denotes the group of automorphisms of $\G$ that preserve the boundary componentwise.\footnote{Equivalently, we
can put $N_{k,l}(\mu)=\sum_{\G\in\Dc_{k,l,\mu}}\frac{1}{|{\rm Aut}_v \G^*|}\;,$
where ${\rm Aut}_v \G^*$ is the group of automorphisms of the dual graph $\G^*$ preserving each vertex pointwise. A closely related problem of the weighted count of unlabeled dessins $\G$ with weights $\frac{1}{|{\rm Aut}\,\G|}$ is equivalent to the above one.
If one treats $\mu$ as the unordered partition $[1^{m_1}2^{m_2}\ldots]$, where $m_j=\#\{\mu_i=j\}$, then the corresponding number of dessins of type $(k,l,\mu)$ is equal to $\frac{1}{|{\rm Aut}\,\mu|}\,N_{k,l}(\mu)$ with
$|{\rm Aut}\,\mu|=m_1!m_2!\ldots\;.$}
Consider the total generating function
\begin{align}\label{gf}
F(s,u,v,p_1,p_2,\dots) = \sum_{k,l,m\geq 1}\frac{1}{m!}\sum_{\mu\in\Z_+^m} N_{k,l}(\mu) s^{d} u^k v^l\, p_{\mu_1}\ldots p_{\mu_m}\;,
\end{align}
where the second sum is taken over all ordered sets $\mu=(\mu_1,\ldots,\mu_m)$ of positive integers, and
$d=\sum_{i=1}^m \mu_i$.

The objective of this paper is to show that the generating function $F$ satisfies all four integrability properties listed at the beginning of this section -- namely, Virasoro constraints, an evolution equation, the KP (Kadomtsev-Petviashvili) hierarchy and a topological recursion. We prove the Virasoro constraints by a bijective combinatorial argument and derive from them all other properties of $F$.\footnote{While this paper was in preparation, similar results were independently obtained by matrix integration methods in \cite{AC} and generalized further in \cite{AMMN}.}  As a result, we obtain a simpler version of the topological recursion in terms of homogeneous components of $F$. We also revisit the problem of enumeration of the ribbon graphs with a prescribed boundary type. Topological recursion for this problem was first established in \cite{EO2} (cf. also \cite{DMSS}). In this paper we give a different, more streamlined proof of it based on the Virasoro constraints and show that the corresponding generating function satisfies an evolution equation and the KP hierarchy as well. These (and other) examples convincingly demonstrate that Virasoro constraints imply topological recursion and are in fact equivalent to it.

Additionally, we show how our results can be applied to effectively enumerate orientable maps and hypermaps regardless of the boundary type. In particular, we present a very straightforward derivation of the famous Harer-Zagier recursion \cite{HZ} for the numbers of genus $g$ polygon gluings from the Walsh-Lehman formula \cite{WL} (a higher genus generalization of Tutte's recursion \cite{T}).\footnote{Recently we came across the paper \cite{CC} where this result was proven along similar lines. However, the authors of \cite{CC} do not explicitly use Virasoro constraints that considerably simplify and clarify the proof.}

\section{Virasoro constraints}

\subsection{Virasoro constraints for the numbers of dessins}

For any integer~$n\geq 0$ consider the differential operator
\begin{multline}
L_n=-\frac{n+1}{s}\frac{\partial}{\partial p_{n+1}}+(u+v)n\frac{\partial}{\partial p_{n}}
+\sum_{j=1}^\infty p_j(n+j)\frac{\partial}{\partial p_{n+j}}\\
{}+\sum_{i+j=n}ij\frac{\partial^2}{\partial p_{i} \partial p_{j}}+\delta_{0,n}uv\;.\label{V}
\end{multline}

A straightforward check shows that for any integer~$m,n\geq 0$
\begin{align*}
[L_m, L_n]=(m-n)L_{m+n}\,.
\end{align*}
In other words, the operators~$L_n$ form (a half of) a representation of the Virasoro (or, rather, Witt) algebra.

The main technical statement of this section is the following

\begin{theorem}\label{Virasoro}
The partition function $e^F=e^{F(s,u,v,p_1,p_2,\dots)}$ satisfies the infinite system of non-linear differential equations (Virasoro constraints)
\begin{align}\label{cons}
L_ne^F=0\;.
\end{align}
The equations~\eqref{cons} determine the partition function~$e^F$ uniquely.
\end{theorem}

\begin{proof}
The Virasoro constraints~\eqref{cons} can be re-written as follows:
\begin{align}
\frac{n+1}{s}\frac{\partial F}{\partial p_{n+1}}&=
\sum_{j=1}^\infty p_j(n+j)\frac{\partial F}{\partial p_{n+j}}+(u+v)n\frac{\partial F}{\partial p_{n}}\nonumber\\
&+\sum_{i+j=n}ij\left(\frac{\partial^2 F}{\partial p_{i} \partial p_{j}}+
\frac{\partial F}{\partial p_{i}} \frac{\partial F}{\partial p_{j}}\right) + \delta_{0,n}uv\;.\label{vc}
\end{align}
Eq.~(\ref{vc}) for $n+1=\mu_1$ can be further re-written as a recursion relation for the coefficients $N_{k,l}(\mu)$ of $F$:
\begin{align}
\mu_1\,N_{k,l}(\mu_1,&\ldots,\mu_m)
=\sum_{j=2}^m (\mu_1+\mu_j-1)N_{k,l}(\mu_1+\mu_j-1,\mu_2,\ldots,\widehat{\mu}_j,\ldots,\mu_m)\nonumber\\
&{}+(\mu_1-1)(N_{k-1,l}(\mu_1-1,\mu_2,\ldots,\mu_m)+N_{k,l-1}(\mu_1-1,\mu_2,\ldots,\mu_m))\nonumber\\
&{}+\sum_{i+j=\mu_1-1}ij \bigg(N_{k,l}(i,j,\mu_2,\ldots,\mu_m)\nonumber\\
&\qquad{}+\mathop{\sum_{k_1+k_2=k}}_{l_1+l_2=l}\quad\sum_{I\sqcup J=\{2,\ldots,m\}}
N_{k_1,l_1}(i,\mu_I)N_{k_2,l_2}(j,\mu_J)\bigg)\;,\label{vt}
\end{align}
where $\mu_I=\mu_{i_1},\ldots,\mu_{i_k},\;I=\{i_1,\ldots,i_k\}$, and the hat means that the corresponding term is omitted.\footnote{Formula (\ref{vt}) is a ``bicolored" analogue of Tutte's recursion, cf. \cite{T}, Eq.~2.1, for $g=0$ and \cite{WL}, Eq.~(6), for any $g\geq 0$ (a more general form of Tutte's recursion one can find, e.~g., in \cite{EO2}).} This recursion is valid for $\sum_{i=1}^m\mu_i>1$ and expresses the numbers~$N_{k,l}(\mu)$ recursively in terms of~$N_{1,1}(1)=1$.

We prove this recursion similar to \cite{WL} (cf. also \cite{DMSS}, \cite{EO2}, \cite{N}) by establishing a direct bijection between dessins counted in the left and right hand sides of~\eqref{vt}. Here it is more convenient to deal with the dual graphs instead.
Let~$\G^*$ be the ribbon graph dual to a dessin $\G$ of type $(k,l,\mu)$. There are $2\mu_1$ ways to pick a half-edge incident to the first vertex of $\G^*$. Following \cite{DMSS} we label this half-edge with an arrow (labeling of half-edges allow us to forget about nontrivial automorphisms). When $\G$ varies over the set $\Dc_{k,l,\mu}$, this gives twice the number in the l.h.s. of~\eqref{vt}.

Let us now express the same number in terms of dessins with one edge less. This can be done by contracting (or expanding) the labeled edges in the dual graphs in a way that preserves the proper coloring of faces. The following possibilities can occur:

\begin{enumerate}[(i)]
\item The labeled edge connects the first vertex with the $j$-th vertex, $j\neq 1$. Contracting this edge we get a ribbon graph with properly bicolored faces of type $(k,l,\mu_1+\mu_j-1,\mu_2,\ldots,\widehat{\mu}_j,\ldots,\mu_m)$, see~Fig.~\ref{contract}.
Conversely, given a graph of type $(k,l,\mu_1+\mu_j-1,\mu_2,\ldots,\widehat{\mu}_j,\ldots,\mu_m)$, there are $2(\mu_1+\mu_j-1)$ ways to split its first vertex into two ones of degrees $2\mu_1$ and $2\mu_j$. Since $j$ can vary from 2 to $m$, this gives twice the first sum in the r.h.s. of ~\eqref{vt}.

\begin{figure}[hbt]
 \begin{center}
 \includegraphics[width=9cm]{contract}
 \caption{Contracting an edge with different endpoints.}
 \label{contract}
\end{center}
\end{figure}

\item The labeled edge forms a loop that bounds a white 1-gon, see~Fig.~\ref{loop}. Contracting such a loop we reduce both $k$ and $\mu_1$ by 1, leaving $l$ and $\mu_j,\; j=2,\ldots,m,$ unchanged. Conversely, if we have a graph of type $(k-1,l,\mu_1-1,\mu_2,\ldots,\mu_m$, we can insert a loop into any of the $\mu_1-1$ gray sectors at the first vertex in order to get a graph of type $(k,l,\mu_1,\ldots,\mu_m)$, and 2 ways to label one of its half-edges. The case of a loop bounding a gray 1-gon can be treated verbatim, giving twice the second term in the r.h.s. of~\eqref{vt}.

\begin{figure}[hbt]
 \begin{center}
 \includegraphics[width=8cm]{loop}
 \caption{Contracting a loop that bounds a 1-gon.}
 \label{loop}
\end{center}
\end{figure}

\item The labeled edge forms a loop whose half-edges are not adjacent relative to the cyclic order of half-edges at the first vertex. Contracting such a loop we split the first vertex into two ones, say, of degrees $2i$ and $2j$, where $i+j=\mu_1-1$, see Fig.~\ref{split}. Under this operation the graph may remain connected, or may split into two connected components. In the former case we get a graph of type $(k,l,i,j,\mu_2,\ldots,\mu_m)$. Reversing this operation, we join the first two vertices and add a loop. We can place the labeled half-edge of the loop in any of the $2i$ sectors at the first vertex, but its other half-edge can be placed only in one of $j$ sectors of different color at the second vertex (otherwise it will not be compatible with the face coloring). This gives us twice the third term in the r.h.s. of~\eqref{vt}. The latter case when the graph becomes disconnected can be treated similarly.

\begin{figure}[hbt]
 \begin{center}
 \includegraphics[width=8cm]{split}
 \caption{Contracting a loop that splits the vertex into two ones of degrees $2i$ and $2j$ with $i+j=\mu_1-1$.}
 \label{split}
\end{center}
\end{figure}
\end{enumerate}

The operations (i)--(iii) are reversible and compatible with the face coloring, thus establishing a required bijection. This proves the Virasoro constraints~\eqref{vc}. It is also not hard to see that the Virasoro constraints determine the partition function~$e^F$ uniquely, since they are equivalent to the recursion~\eqref{vt}.
\end{proof}

\begin{corollary}\label{evo}
Put
\begin{align*}
\Lambda_1&=\sum_{i=2}^\infty (i-1)p_i\,\frac{\partial}{\partial p_{i-1}}\;,\nonumber\\
M_1&=\sum_{i=2}^\infty \sum_{j=1}^{i-1} \left((i-1)p_j p_{i-j}\,\frac{\partial}{\partial p_{i-1}}
+ j(i-j) p_{i+1}\,\frac{\partial^2}{\partial p_j \partial p_{i-j}}\right)\;.
\end{align*}
Then the partition function~$e^F$ satisfies the evolution equation
\begin{align}
\frac{\partial e^F}{\partial s}=((u+v)\Lambda_1+M_1+uvp_1)e^F\;,\label{eveq}
\end{align}
and is uniquely determined by the initial condition $F\left|_{s=0}\right.=0$. In other words,~$e^F$ is explicitly given by the formula
\begin{align*}
e^F=e^{s((u+v)\Lambda_1+M_1+uvp_1)}\,1
\end{align*}
were ``1'' stands for the constant function identically equal to~1.
\end{corollary}
\begin{proof}
Multiply the both sides of~\eqref{V} by~$p_{n+1}$ and sum over~$n$. We get
\begin{align*}
\sum_{n=0}^\infty p_{n+1}L_n
&=\sum_{n=0}^\infty p_{n+1}\left(-\frac{n+1}{s}\frac{\partial}{\partial p_{n+1}}+(u+v)n\frac{\partial}{\partial p_{n}}\right.\nonumber\\
&\hspace{0.4in}+\sum_{j=1}^\infty p_j(n+j)\frac{\partial}{\partial p_{n+j}}
+\sum_{i+j=n}\left. ij\frac{\partial^2}{\partial p_{i} \partial p_{j}}\right)+uvp_1\nonumber\\
&=-\frac{\partial}{\partial s}+(u+v)\Lambda_1+M_1+uvp_1\;,
\end{align*}
and the required statement immediately follows from~\eqref{cons}.
\end{proof}

\begin{remark}
A different proof of Corollary \ref{evo} has recently appeared in \cite{Z}.
\end{remark}

\begin{remark}\label{ps}
Denote by $\psi$ the principal specialization of the partition function $e^F$:
\begin{align*}
\psi=\psi(s,t,u,v)=e^{F(s,u,v,p_1,p_2,\ldots)}\left|_{p_i=t^i}\right.\;,
\end{align*}
where $t$ is a new formal variable. It is not hard to check that
\begin{align*}
\Lambda_1 e^F\left|_{p_i=t^i}\right.=t\left(t\frac{d}{dt}\right)\psi\,, \qquad M_1 e^F\left|_{p_i=t^i}\right.=t\left(t\frac{d}{dt}\right)^2\psi\,.
\end{align*}
Then, with the help of the obvious identity $s\frac{\D\psi}{\D s}=t\frac{\D\psi}{\D t}$, the evolution equation \eqref{eveq} translates into the following equation for the wave function $\psi$:
\begin{align*}
\frac{1}{st}\left(t\frac{d}{dt}\right)\psi=(u+v)\left(t\frac{d}{dt}\right)\psi+\left(t\frac{d}{dt}\right)^2\psi+uv\psi\;.
\end{align*}
It can be further rewritten as the Schr\"odinger equation
\begin{align}\label{qc}
t^2\,\frac{d^2\psi}{dt^2}+\left((u+v+1)\,t-\frac{1}{s}\right)\,\frac{d\psi}{dt}+uv\psi=0\;
\end{align}
Eq.~\eqref{qc} is often referred to as the {\em quantum curve equation} in the literature on topological recursions.
Note that the coefficients of $\log\psi$ enumerate dessins with given numbers of white and black vertices, and a given number of edges regardless of genus (or the number of boundary components).
\end{remark}

Another observation is that the generating function $F=F(s,u,v,p_1,p_2,\dots)$ satisfies an infinite system of non-linear partial differential equations called the KP (Kadomtsev-Petviashvili) hierarchy (this means that the numbers~$N_{k,l}(\mu)$ additionally obey an infinite system of recursions). The KP hierarchy is one of the best studied completely integrable systems in mathematical physics. Below are the first few equations of the hierarchy:
\begin{equation}\label{KP}
\begin{aligned}
&F_{22}=-\frac12\,F_{11}^2+F_{31}-\frac1{12}\,F_{1111}\;,\\
&F_{32}=-F_{11}F_{21}+F_{41}-\frac16F_{2111}\;,\\
&F_{42}=-\frac12\,F_{21}^2-F_{11}F_{31}+F_{51}+\frac18\,F_{111}^2
+\frac1{12}\,F_{11}F_{1111}-\frac14\,F_{3111}+\frac1{120}\,F_{111111}\;,\\
&F_{33}=\frac13\,F_{11}^3-F_{21}^2-F_{11}F_{31}+F_{51}
+\frac14\,F_{111}^2+\frac13\,F_{11}F_{1111}-\frac13\,F_{3111}\\
&\hspace{3.5in}+\frac1{45}\,F_{111111}\;,
\end{aligned}
\end{equation}
where the subscript~$i$ stands for the partial derivative with respect to~$p_i$.

The exponential~$Z=e^F$ of any solution is called a {\em tau function} of the hierarchy. The space of solutions (or the space of tau functions) has a nice geometric interpretation as an infinite-dimensional Grassmannian (called the {\em Sato Grassmannian}), see~\cite{MJD} or~\cite{K} for details. In particular, the space of solutions is homogeneous: there is a Lie algebra $\widehat{\mathfrak{gl}(\infty)}$ that acts infinitesimally on the space of solutions, and the action of the corresponding Lie group is transitive.

\begin{corollary}\label{tau}
The generating function $F=F(s,u,v,p_1,p_2,\dots)$ satisfies the infinite system of KP equations~\eqref{KP} with respect to $p_1,p_2,\dots$ for any parameters~$s,u,v$. Equivalently, the partition function~$Z=e^F$ is a 3-parameter family of KP tau functions.
\end{corollary}
\begin{proof}
To begin with, we notice that~$1$ is obviously a KP tau function. Then, since
$p_1, \Lambda_1, M_1\in\widehat{\mathfrak{gl}(\infty)}$
(cf.~\cite{K}), the linear combination $s((u+v)\Lambda_1+M_1 +uvp_1)$ also belongs to $\widehat{\mathfrak{gl}(\infty)}$ for any~$s,u,v$. The exponential $e^{s((u+v)\Lambda_1+M_1+uvp_1)}$ therefore preserves the Sato Grassmannian and maps KP tau functions to KP tau functions. Thus,~$e^F$ is a KP tau function as well, and~$F$ is a solution to KP hierarchy.
\end{proof}

\begin{remark}
Corollary~\ref{tau} was earlier proven in~\cite{GJ2} by a different method.
However,~\cite{GJ2} contains no analogs of the Virasoro constraints or the evolution equation.
\end{remark}

At the end of this subsection we will sketch how to enumerate dessins with $k$ white vertices, $l$ black vertices, $d$ edges and $m$ boundary components regardless of the partition $\mu=(\mu_1,\ldots,\mu_m)$.
To these ends, consider the specialization operator
\begin{align}
\theta: F\mapsto F|_{p_i=t, i=1,2,\ldots}\label{sp}
\end{align}
and put $f=\theta(F)$. This specialization is more subtle than the one considered in Remark \ref{ps}, and the coefficients of $f$ do not mix dessins of different genera since by Euler's formula $(k+l)-d+m=2-2g$. Expanding $f$ into a series in the variables $s$ and $t$, we recompose it as
\begin{align}
f(s,t,u,v)=\sum_{g=0}^\infty\sum_{d=2g+1}^\infty \frac{f_{g,d}(t,u,v)}{d}\,s^d\;,
\end{align}
where each coefficient $f_{g,d}(t,u,v)$ is a homogeneous polynomial in $t,u,v$ of degree $d+2-2g$ with integer coefficients.

Furthermore, using the Virasoro constraints (\ref{V}) with $n=0,1,2$ and the homgeneity equation
$$s\frac{\partial F}{\partial s}=\sum_{i=1}^\infty ip_i\frac{\partial F}{\partial p_i}\;,$$
we can express the specializations of partial derivatives of $F$ with respect to the variables $p_1,p_2,p_3$ in terms of $s$-derivatives of $f$. More precisely, a straightforward computation yields
\begin{lemma}
We have
\begin{align*}
&\theta(F_{1})=s^2f'+suv, \\
&\theta(F_{11})=s^2(s^2f'+suv)', \\
&\theta(F_{1111})=s^2(s^2(s^2(s^2f'+suv)')')',\\
&2\theta(F_{2})=(1+s(u+v-t))(s^2f'+suv)-suv, \\
&3\theta(F_{3})=2s(u+v-t)\theta(F_{2})+(1-st)\theta(F_{1})+s\theta(F_{11})+s\theta(F_{1})^2-suv, \\
&\theta(F_{12})=s^2\theta(F_{2})', \\
&\theta(F_{13})=s^2\theta(F_{3})', \\
&2\theta(F_{22})=(1+s(u+v-t))\theta(F_{12})+3s\theta(F_{3})-2s\theta(F_{2}),
\end{align*}
where the subscript $i$ stands for the partial derivative with respect to $p_i$, and the prime $'$ denotes the derivative in $s$.
\end{lemma}
Applying the specialization operator $\theta$ to the first KP equation
$$\theta(F_{22})=-\frac12\,\theta(F_{11})^2+\theta(F_{31})-\frac1{12}\,\theta(F_{1111})\,,$$
cf. (\ref{KP}), and using the above formulas, we get an ordinary differential equation for $f$ as a function of $s$ that translates into the following quadratic recursion:
\begin{align}
(d+1)f_{g,d}&=(2d-1)\,a\,f_{g,d-1}+(d-2)\,b\,f_{g,d-2}+(d-1)^2(d-2)f_{g-1,d-2}\nonumber\\
            &+\sum_{i=0}^g\sum_{j=1}^{d-3}(4+6j)(d-2-j)f_{i,j}f_{g-i,d-2-j}\;,\label{ad}
\end{align}
where $a=t+u+v,\; b=4(tu+tv+uv)-a^2.$ Starting with $f_{0,1}=tuv$, one can recursively compute the polynomials $f_{g,d}$ for all $g$ and $d$.

Finally, let us restrict ourselves to the case of dessins with one boundary component (or bicolored polygon gluings). Denote by $h_{g,d}$ the linear term in $f_{g,d}$ with respect to $t$. Then the recursion (\ref{ad}) takes the form
\begin{align*}
(d+1)h_{g,d}&=(2d-1)(u+v)h_{g,d-1}-(d-2)(u-v)^2 h_{g,d-2}\\
            &+(d-1)^2(d-2)h_{g-1,d-2}\;,
\end{align*}
and we reproduce the well-known result of \cite{A} on the enumeration of genus $g$ gluings of a bicolored $2d$-gon with given numbers of white and black vertices (cf. also \cite{J}).

\subsection{Virasoro constraints for the numbers of ribbon graphs}

A closely related, but somewhat different enumerative problem was considered in~\cite{WL}. Recall that a dessin d'enfant $f^{-1}([0,1])$ is a bicolored connected ribbon graph with vertices ``colored" by either 0 or 1 depending on whether~$f$ maps the vertex to 0 or 1 in~${\C}P^1$. One can similarly try to enumerate all (not necessarily bicolored) connected ribbon graphs, and this is the problem that was addressed in~\cite{WL}. To make it precise, let us label the boundary components of a ribbon graph~$\G$ (or, equivalently, the vertices of the dual graph~$\G^*$) by integers from 1 to~$m$, and let $\mu_1,\ldots,\mu_m$ be the lengths of the boundary components of~$\G$ (or the degrees of vertices of~$\G^*$).

Ribbon graphs can naturally be represented by dessins of a special type called {\em clean dessins} in \cite{DMSS}. Namely, color each vertex of a ribbon graph in white and place black vertices at the midpoints of edges. Such a dessin corresponds to a covering of $\pl$ of even degree $d$ with ramification of type $[2^{d/2}]$ over 1 and arbitrary ramification over 0 and $\infty$.
As before, we put $k=|f^{-1}(0)|$ (the number of vertices of the ribbon graph~$\G$), $l=|f^{-1}(1)|=d/2$ (the number of edges of $\G$), and $m=|f^{-1}(\infty)|$ (the number of boundary components of~$\G$). Clearly, we have $k-d/2+m=2-2g$.

Denote by $D_{g,m}(\mu)=D_{g,m}(\mu_1,\ldots,\mu_m)$ the number of genus~$g$ ribbon graphs with~$m$ labeled vertices of degrees $\mu_1,\ldots,\mu_m$ counted with weights $\frac{1}{|{\rm Aut}_v\,\G|}$, where the automorphisms preserve each vertex of~$\G$ pointwise. Apparently, the same numbers enumerate pure dessins with $m$ labeled boundary components of lengths $(2\mu_1,\ldots,2\mu_m)$. The following recursion for $D_{g,m}(\mu)$ was derived in~\cite{WL}, Eq.~(6):\footnote{\label{M} This is (a specialization of) Tutte's recursion for arbitrary $g$, cf.~\cite{EO2}. This formula, undeservedly forgotten, was recently reproduced in \cite{DMSS}, Eq.~(3.15). Note that the second term in the r.h.s. of~\eqref{wl}, corresponding to a loop bounding a 1-gon, was inadvertently omitted there. This required some ``modification'' of the numbers $D_{g,m}(\mu_1,\ldots,\mu_m)$ in \cite{DMSS}.}
\begin{align}
\mu_1 D_{g,m}(\mu_1,\ldots,&\mu_m)
=\sum_{j=2}^m (\mu_1+\mu_j-2)D_{g,m-1}(\mu_1+\mu_j-2,\mu_2,\ldots,\widehat{\mu}_j,\ldots,\mu_m)\nonumber\\
&{}+2(\mu_1-2)D_{g,m}(\mu_1-2,\mu_2,\ldots,\mu_m)\nonumber\\
&{}+\sum_{i+j=\mu_1-2}ij \bigg(D_{g-1,m+1}(i,j,\mu_2,\ldots,\mu_m)\nonumber\\
&{}+\sum_{g_1+g_2=g}\quad\sum_{I\sqcup J=\{2,\ldots,m\}}
D_{g_1,|I|+1}(i,\mu_I)D_{g_2,|J|+1}(j,\mu_J)\bigg)\;.\label{wl}
\end{align}
Recursion~\eqref{wl} is valid for all $g\geq 0,\;m\geq 1$, and~$\mu$ such that $d=\sum_{i=1}^m\mu_i>2$, whereas for~$d=2$ the only nonzero numbers are $D_{0,1}(2)=1/2,\;D_{0,2}(1,1)=1$.
Below we give a convenient interpretation of this recursion in terms of PDEs.

Similar to (\ref{gf}), introduce the generating function
\begin{align}\label{gfM}
\tF(s,u,p_1,p_2,\ldots)
&=\sum_{g=0}^\infty\sum_{m=1}^\infty\frac{1}{m!} \sum_{\mu\in{\Z}_+^m} D_{g,m}(\mu)s^d u^k p_{\mu_1}\ldots p_{\mu_m}\;,
\end{align}
where $d=\sum_{i=1}^m\mu_i$, $k=2-2g-m+d/2$, and $\mu=(\mu_1,\ldots,\mu_m)$ (compared to (\ref{gf}), we omit here the trivial factor $v^{d/2}$ that carries no additional information in this case).

\begin{theorem}\label{rg} The generating function $\tF$ enjoys the following integrability properties:
\begin{enumerate}[(i)]
\item Let
\begin{align*}
\tL_n=-\frac{n+2}{s^2}\frac{\partial}{\partial p_{n+2}}
+2\,u\,n\frac{\partial}{\partial p_{n}}
+\sum_{j=1}^\infty p_j(n+j)\frac{\partial}{\partial p_{n+j}}\\
{}+\sum_{i+j=n}ij\frac{\partial^2}{\partial p_i \partial p_j}+\delta_{-1,n}u p_1+\delta_{0,n}u^2\;,
\end{align*}
where $n\geq -1$. Then the partition function $e^\tF$ satisfies the infinite system of PDE's (``Virasoro constraints'')
\begin{align*}
\tL_n e^\tF=0
\end{align*}
that determine $\tF$ uniquely.

\item Put
\begin{align*}
&\Lambda_2=\sum_{i=3}^\infty (i-2)p_i\,\frac{\partial}{\partial p_{i-2}}+\frac12p_1^2\;,\nonumber\\
&M_2=\sum_{i=2}^\infty \sum_{j=1}^{i-1} \left((i-2)p_j p_{i-j}\,\frac{\partial}{\partial p_{i-2}}
+ j(i-j) p_{i+2}\,\frac{\partial^2}{\partial p_j \partial p_{i-j}}\right)\;.
\end{align*}
Then $e^\tF$ satisfies the evolution equation
\begin{align*}
\frac{1}{s}\frac{\partial e^\tF}{\partial s}=(2\,u\,\Lambda_2+M_2+u^2p_2)e^\tF\,,
\end{align*}
that, together with the initial condition~$\tF|_{s=0}=0$,
determines~$\tF$ uniquely. In other words,~$e^\tF$ is explicitly given by the formula
\begin{align*}
e^\tF=e^{\frac{s^2}{2}(2\,u\,\Lambda_2+M_2+u^2p_2)}\,1\;.
\end{align*}

\item The partition function $e^\tF$ is a tau function of the KP hierarchy, i.e. its logarithm $\tF(s,u,p_1,p_2,\ldots)$ satisfies~\eqref{KP} for any $s$ and $u$.

\end{enumerate}
\end{theorem}

\begin{proof}
Part (i) of the theorem is just a reformulation of the recursions (\ref{wl}) for
$\mu_1=n+2$. Note that the operators~$\tL_n$ obey the commutation relations
$[\tL_m,\tL_n]=(m-n)\tL_{m+n}$
for $m>n\geq -1$.

To prove (ii) we multiply $\tL_n$ by~$p_{n+2}$ and sum over~$n$:
\begin{align}
\sum_{n=-1}^\infty p_{n+2}\tL_n
&=\sum_{n=-1}^\infty p_{n+2}\bigg(-\frac{n+2}{s}\frac{\partial}{\partial p_{n+2}}+2\,u\,n\frac{\partial}{\partial p_{n}}\nonumber\\
&\hspace{0.6in}
{}+\sum_{j=1}^\infty p_j(n+j)\frac{\partial}{\partial p_{n+j}}
+\sum_{i+j=n}ij\frac{\partial^2}{\partial p_{i} \partial p_{j}}\bigg)+u\,p_1^2+u^2p_2\nonumber\\
&=-\frac{1}{s}\frac{\partial}{\partial s}+2\,u\,\Lambda_2+M_2+u^2 p_2\;.\label{evom}
\end{align}

Part (iii) follows from the fact that $\Lambda_2, M_2$ and $p_2$ belong to $\widehat{\mathfrak{gl}(\infty)}$,
cf. Corollary~\ref{tau}.
\end{proof}

\begin{remark}
Similar to Eq.~\eqref{qc} we can write the quantum curve equation for the wave function
$$
\widetilde{\psi}=\widetilde{\psi}(s,t,u)=e^{\tF(s,u,p_1,p_2,\ldots)}\left|_{p_i=t^i}\right.\;,
$$
that reads in this case as follows:
$$
t^2\,\frac{d^2\widetilde{\psi}}{dt^2}+\left(2(u+1)\,t-\frac{1}{s^2\,t}\right)\,\frac{d\widetilde{\psi}}{dt}+(u+u^2)\,\widetilde{\psi}=0\;.
$$
Note that this equation differs from the one obtained in \cite{MS}. The reason for that is explained in Footnote \ref{M} above. To fix that, put $Z(x,\hbar)=e^{-\frac{\log x}{\hbar}}\,\widetilde{\psi}|_{s=\hbar/x,t=1,u=1/\hbar}$. Then one has
$$\left(\hbar^2\frac{d^2}{dx^2}+\hbar\frac{d}{dx}+1\right)Z(x,\hbar)=0$$
precisely like in \cite{MS}.
\end{remark}

To complete this section, we will show that the Walsh-Lehman formula (\ref{wl}) implies the Harer-Zagier \cite{HZ} recursion for the numbers of orientable polygon gluings. We will follow the same lines as at the end of the previous subsection. To begin with, put $\tf=\theta(\tF)=\tF|_{p_i=t, i=1,2,\ldots}$, where the specialization operator $\theta$ is defined by Eq.~(\ref{sp}).
The coefficients of $\tf$ enumerate ribbon graphs with given numbers of vertices, edges and boundary components and, therefore, do not mix graphs of different genera.
Rearrange the series $f$ as follows:
\begin{align}
\tf(s,t,u)=\sum_{g=0}^\infty\sum_{l=2g}^\infty \frac{\tf_{g,l}(t,u)}{2l}\,s^{2l}\;,
\end{align}
where $l=d/2$, and each coefficient $\tf_{g,l}(t,u)$ is a homogeneous polynomial in $t,u$ of degree $l+2-2g$ with integer coefficients.
Like in the case of dessins, using the Virasoro constraints of Theorem \ref{rg} (i) with $n=-1,0,1$, we can express the specializations of partial derivatives of $\tF$ with respect to the variables $p_1,p_2,p_3$ in terms of $s$-derivatives of $\tf$. A straightforward computation yields
\begin{lemma}
We have
\begin{align*}
&\theta(\tF_{1})=s^3\tf'+s^2tu, \\
&\theta(\tF_{11})=s^3\theta(\tF_{1})'-s^2\theta(\tF_{1})+s^2u, \\
&\theta(\tF_{111})=s^3\theta(\tF_{11})'-2s^2\theta(\tF_{11}),\\
&\theta(\tF_{1111})=s^3\theta(\tF_{111})'-3s^2\theta(\tF_{111}),\\
&2\theta(\tF_{2})=s^3\tf'+s^2u^2, \\
&4\theta(\tF_{22})=s^6\tf''+3s^5\tf'+2s^4u^2, \\
&\theta(\tF_{13})=s^5((2s^2u+1-s^2t)(s\tf')'+(2s^2u+2-s^2t)\tf')+s^6(2tu^2-t^2u)+3s^4u^2,
\end{align*}
where, as before, the subscript $i$ stands for the partial derivative with respect to $p_i$, and the prime $'$ denotes the derivative in $s$.
\end{lemma}
Applying the specialization operator $\theta$ to the first KP equation (\ref{KP})
and using the above formulas, we get an ordinary differential equation for $\tf$ as a function of $s$ that translates into the following quadratic recursion:
\begin{align}
(l+1)\tf_{g,l}&=2(2l-1)(t+u)\tf_{g,l-1}+(2l-1)(2l-3)(l-1)\tf_{g-1,l-2}\nonumber\\
              &+3\sum_{i=0}^g\sum_{j=0}^{l-2}(2j+1)(2(l-2-j)+1)\tf_{i,j}\tf_{g-i,l-2-j}\;,\label{hz}
\end{align}
where we put by definition $\tf_{0,0}=u$.	This is essentially the formula from \cite{CC}, but derived in a more straightforward way. Note that starting with $\tf_{0,1}=t^2u+tu^2$, one can recursively compute the polynomials $\tf_{g,l}$ for all $g$ and $l$.

To enumerate genus $g$ ribbon graphs with one boundary component (or $2l$-gon gluings), it is sufficient to consider the linear terms $\epsilon_{g,l}$ in $\tf_{g,l}$ with respect to $t$. Then Eq.~(\ref{hz}) turns into the famous Harer-Zagier recursion
\begin{align*}
(l+1)\epsilon_{g,l}=2(2l-1)u\epsilon_{g,l-1}+(2l-1)(2l-3)(l-1)\epsilon_{g-1,l-2}\;,
\end{align*}
cf. \cite{HZ}.

\section{Topological recursion}

The generating function $F=\sum_{g,m}F_{g,m}$ of \eqref{gf} enumerating Belyi pairs~$(C,f)$ (or Grothendieck's dessins) can be naturally decomposed into components with fixed~$g$ and $m$, where $g$ is the genus of~$C$ and~$m$ is the number of poles of~$f$:
\begin{align}
F_{g,m}(s,u,v,p_1,p_2,\dots)
= \frac{1}{m!}\sum_{\mu_1,\dots,\mu_m}\sum_{k+l=d-m+2-2g}
   N_{k,l}(\mu) s^{d} u^k v^l\, p_{\mu_1}\ldots p_{\mu_m}\label{gm}
	\end{align}
(here, as usual, $d=\sum\mu_i$). Another way to collect these numbers into a generating series is to use the
\emph{$m$-point correlation functions}
$$W_{g,m}(x_1,\dots,x_m)
 =\sum_{\mu_1,\dots,\mu_m}\sum_{k+l=d-m+2-2g} N_{k,l}(\mu)\,x_1^{\mu_1}\dots x_m^{\mu_m}.$$

\emph{Topological recursion} (cf. \cite{EO2}) is an ``ansatz'' that allows to reconstruct the coefficients of certain generating series recursively in~$g$ and~$m$. Traditionally, it is formulated in terms of correlation functions or, rather,
\emph{differentials}

\begin{align*}
w_{g,m}(x_1,\dots,x_m)&=\frac{\D^mW_{g,m}}{\D x_1\ldots \D x_m} \,dx_1\ldots dx_m\\
&=\sum_{\mu}\sum_{\substack{k,\,l\\k+l=d-m+2-2g}} N_{k,l}(\mu)\prod_{i=1}^m \mu_i x_i^{\mu_i-1} dx_i.
\end{align*}

We will present the topological recursion in terms of components~$F_{g,m}$ of the generating function $F$. The advantage of this approach is that we need only one set of variables~$p_i$ for all~$g$ and~$n$. The two approaches being equivalent, it proves out, however, that many properties of the recursion become more clear in terms of~$p$-variables.

One of the nice features of topological recursion is that the generating functions~$F_{g,m}$ become polynomilas under a linear change of variables $p_i$. The components~$F_{g,m}$ of the total generating function $F$ are infinite formal series in $p_i$, and their polynomiality is far from being an immediate consequence of the Virasoro constraints for $F$. On the other hand, this polynomiality automatically follows from the equations of topological recursion. Another advantage of topological recursion is its universality. For a variety of enumerative problems it takes the same form, differing only in initial conditions.

Introduce the formal variables~$x$ and~$z$ related by
\begin{equation}
z(x)=\sqrt{\frac{1-\beta  s  x}{1-\alpha  s  x}},\qquad
  x(z)=\frac{z^2-1}{s  \left(\alpha  z^2-\beta \right)},
\label{eqzx}\end{equation}
where
\begin{equation}
\a=(\sqrt{u}-\sqrt{v})^2,\qquad
 \b=(\sqrt{u}+\sqrt{v})^2.
\label{eqabs}
\end{equation}
We consider~\eqref{eqzx} as a formal change of coordinates on the complex projective line $\pl$ near the point $x=0$ (resp., $z=1$)
depending on the parameters $\a,\b,s$ (or $u,v,s$).

Put
\begin{equation}
T_j(p)=\sum_{i=1}^\infty c_j^{(i)}p_i\,,
\label{eqT}\end{equation}
where the coefficients $c_j^{(i)}$ are defined by the relation
$$\sum_{i=1}^\infty c_j^{(i)}x^i=(z(x))^{j}-1\,.$$

\begin{theorem}\label{th2}
Let $F_{g,m}$ be the infinite series defined by~\eqref{gm}. Then
\begin{enumerate}[(i)]
\item For each~$g,m$ with~$2g-2+m>0$ there exists a polynomial~$G_{g,m}$ of the variables~$t_j$, $j\in\Z_\odd$, such that
$$F_{g,m}(p)=G_{g,m}(t)\bigm|_{t_j=T_j(p)}$$
(i.e. each~$F_{g,m}$ is a polynomial in the linear functions $T_{\pm1},T_{\pm3},T_{\pm5},\dots$).
\item The polynomials~$G_{g,m}$ can be recursively computed starting from $G_{0,3}$ and $G_{1,1}$, cf. Eqs.~\eqref{eq1}--\eqref{eq4} below.
\end{enumerate}
\end{theorem}

Let us now formulate the recursion for the polynomials~$G_{g,m}$ precisely. This can be done in terms of the so-called \emph{spectral curve}. In our case the spectral curve is the projective line~$\pl$ equipped with the globally defined holomorphic involution $z\mapsto -z$ with respect to some affine coordinate~$z$.

By a {\em Laurent form} we understand here a globally defined meromorphic 1-form on $\pl$ with poles only at~$0$ and~$\infty$. Denote by~$L$ the space of odd Laurent forms relative to the involution $z\mapsto -z$. The forms $d(z^j)=j\,z^{j-1}\,dz$, $j\in\Z_\odd$, provide a convenient basis in~$L$. Let~$P_L$ denote the projector to the space~$L$ in the space of all Laurent forms. For a Laurent form~$\phi$ its projection $\psi=P_L(\phi)$ to $L$ is uniquely determined by the requirement that the form
$\psi-\phi_\odd$ is regular at both~$0$ and~$\infty$, where $\phi_\odd(z)=\frac12(\phi(z)-\phi(-z))$ is the odd part
of~$\phi$. More explicitly, the action of~$P_L$ is given by the formula
\begin{align}
(P_L\phi)(z)&=\sum_{i=0}^\infty\res_{w=0}(\phi(w)w^{2i+1})\;z^{-2i-2}dz-
\sum_{i=0}^\infty\res_{w=\infty}(\phi(w)w^{-2i-1})\;z^{2i}dz\nonumber\\
&=\res_{w=0}\left(\phi(w)\frac{w\;dz}{z^2-w^2}\right)+\res_{w=\infty}\left(\phi(w)\frac{w\;dz}{z^2-w^2}\right)\;.
\label{res}\end{align}

Note that for the validity of this definition it will suffice to assume that~$\phi$ is defined in a neighborhood of the
points~$0$ and~$\infty$, or even that~$\phi$ is  a formal Laurent series at these points. On the other hand,
the form~$P_L\phi\in L$ is always globally defined on~$\pl$.

In fact, the recursion applies not to the polynomials~$G_{g,m}$ themselves, but to certain $1$-forms~$U_{g,m}$. For the set of variables $z$ and $t=(t_{\pm 1},t_{\pm 3},\ldots)$ introduce the differential operator
$$\d_{z,t}=\sum_{j\in\Z_\odd}jz^{j-1}\pd{}{t_j}\,.$$
For $2g-2+m>0$ put
\begin{align*}
U_{g,m}(z)&=\d_{z,t} G_{g,m}(z)=\sum_{j\in\Z_\odd}jz^{j-1}\pd{G_{g,m}}{t_j}\,.
\end{align*}
As we will see later, $U_{g,m}=U_{g,m}(z)dz$ is an odd Laurent form on~$\pl$ that is polynomial in~$t_j$.

\begin{remark}\label{remdelta}
The operator~$\d_{z,t}$ written in terms of $x$ and $p$-variables becomes
$$\d_{x,p}=\frac{dz}{dx}\sum_{i=1}^\infty ix^{i-1}\,\pd{}{p_i}\,,$$
where $x$ is related to $z$ by (\ref{eqzx}) and $t_j=T_j(p)$, see~\eqref{eqT}.
(For brevity we will omit the subscripts `$p$' and `$t$' by
$\d$, always associating $p$- and $t$-variables with $x$ and $z$ respectively.)
More precisely, assume that a function~$f$ of $p$-variables can be expressed as a composition  $f=h\circ T$, where $h$ is a function of $t$-variables and~$T$ is the linear change~\eqref{eqT}. Then we have $\d_xf\,dx=(\d_zh)|_{t_k=T_k(p)}\,dz$.

In particular, if~$g$ is a polynomial in $t$-variables, then~$\d_xfdx$ is a Laurent form in~$z$ (with coefficients depending on $p_i$'s). Indeed, the operators on both sides satisfy the Leibnitz rule, and therefore it is sufficient to prove the equality for the case~$h=t_k$, that is,
$$\d_x T_k(p)dx=k z^{k-1} dz,\qquad z=z(x)\,,$$
which is essentially the definition of the linear functions~$T_k(p)$, cf.~(\ref{eqT}).
\end{remark}

In the unstable cases (i.e. when $2g-2+m\leq 0$) the definition of~$U_{g,m}$ should be modified. Namely, we set $U_{0,1}=0$ and
define~$U_{0,2}$ by the following formal expansions
\begin{equation}
\begin{aligned}
U_{0,2}(z)dz&=-\sum_{i=0}^\infty t_{-2i-1}z^{2i}dz=-(t_{-1}+t_{-3}z^2+\dots)\,dz,\quad z\to 0,\\
U_{0,2}(z)dz&=-\sum_{i=0}^\infty t_{2i+1}z^{-2i}d(z^{-1})=(t_1z^{-2}+t_3z^{-4}+\dots)\,dz,\quad z\to\infty,\\
\end{aligned}
\label{U02}
\end{equation}

In general, the homogeneous degree~$m$ polynomial~$G_{g,m}$ can be recovered form the form~$U_{g,m}$ by the Euler formula
\begin{equation}
G_{g,m}=\frac{1}{m}\sum t_k\pd{G_{g,m}}{t_k}=\frac1m\,\O(U_{g,m},U_{0,2})\;,
\label{eq1}\end{equation}
where for odd forms~$\phi$ and~$\psi$ we set
$$\O(\phi,\psi)=-\O(\psi,\phi)=\res_{z=0}\left(\phi\int\psi\right)+\res_{z=\infty}\left(\phi\int\psi\right)\;.$$

The last ingredient needed to write down the topological recursion is the form
\begin{equation}\label{eta}
\eta=\eta(z)dz=\frac{\s\,z^2dz}{(z^2-1)^2 (\a\,z^2 -\b)},\qquad
\s=(\a-\b)^2=16\,u\,v.
\end{equation}
This form is odd and has the property that the dual vector field
$$\frac{1}{\eta}=\frac{1}{\eta(z)}\,\frac{d}{dz}=\frac{\a\,z^4-(2\,\a+\b)\,z^2+\a +2\,\b -\b\,z^{-2}}{\s}\,\frac{d}{dz}$$
is meromorphic with poles of order~$2$ at~$z=0$ and $z=\infty$ and regular elsewhere in~$\pl$.

The main recursive relation of this paper is
\begin{equation}
U_{g,m}=P_L\Biggl(\frac{1}{2\eta}\Biggl(\d U_{g-1,m+1}+\sum_{\substack{g_1+g_2=g,\\m_1+m_2=m+1}}U_{g_1,m_1}U_{g_2,m_2}\Biggr)\Biggr)
\label{eq2}\end{equation}
(here and below we tacitly assume that $\d f=\d_z f\,dz$).
Note that almost all terms in the sum on the right hand side of (\ref{eq2}) belong to~$L$, so that~$P_L$ is identical on these terms. Therefore, (\ref{eq2}) can equivalently be rewritten as
$$
U_{g,m}=\frac{1}{2\eta}\Biggl(\d U_{g-1,m+1}+\mathop{\sum{}^*}_{\substack{g_1+g_2=g,\\m_1+m_2=m+1}}
U_{g_1,m_1}U_{g_2,m_2}\Biggr)+P_L\left(\frac1\eta\,U_{g,m-1}U_{0,2}\right)\;,
$$
where the star $^*$ by the summation sign means that the terms involving~$U_{0,2}$ are excluded (recall that $U_{0,1}=0$ by assumption). This recursion relation is valid for all~$g$ and~$n$ with~$2g-2+m>1$. Moreover, it applies for~$(g,m)=(0,3)$ as well:
\begin{equation}
U_{0,3}=P_L\left(\frac{U_{0,2}^2}{2\eta}\right).
\label{eq3}\end{equation}
In the case $(g,m)=(1,1)$ the formula is not applicable since~$\d U_{0,2}$ is not defined. This is why we set by definition
\begin{equation}
U_{1,1}=U_{1,1}(z)\,dz=P_L\left(\frac{1}{2\eta}\left(\frac{dz}{2z}\right)^2\right)=\frac{1}{2\eta}\left(\frac{dz}{2z}\right)^2
=\frac{1}{8z^2\eta(z)}\,dz\,.
\label{eq4}\end{equation}

Eqs.~\eqref{eq1}--\eqref{eq4} concretize the second statement of Theorem~\ref{th2}.

Below we list the polynomials $U_{g,m}$ and $G_{g,m}$ for small~$g$ and~$m$:

\begin{align*}
\s\,U_{0,3}(z)&=\frac{\a}{2}\,t_1^2-\frac{\b}{2}\,t_{-1}^2\,z^{-2},\\[-12pt]\\
\s\,G_{0,3}&=\frac{\a}{6}\,t_1^3+\frac{\b}{6}\,t_{-1}^3,\\[-12pt]\\
\s\,U_{1,1}(z)&=\frac{\a}{8}\,z^2 -\frac{2\a+\b}{8}
  +\frac{\a+2\b}{8}\,z^{-2}-\frac{\b}{8}\,z^{-4},\\[-12pt]\\
\s\,G_{1,1}&=\frac{\a}{24}\,t_3-\frac{2\a+\b}{8}\,t_1-\frac{\a+2\b}{8}\,t_{-1}+\frac{\b}{24}\,t_{-3},\\[-12pt]\\
\s^2\,U_{0,4}(z)&=\frac{\a^2}{2}t_1^3\,z^2
 +\Bigl(\frac{\a^2}{2}\,t_3\, t_1^2-\frac{\a(2\a+\b)}{2}\,t_1^3-\frac{\a\b}{2}\,t_{-1}^2\, t_1\Bigr)\\
 &+\Bigl(\frac{\b(\a+2\b)}{2}\,t_{-1}^3+\frac{\a\b}{2}\,t_1^2\, t_{-1}-\frac{\b^2}{2}\,t_{-3}\, t_{-1}^2\Bigr)\,z^{-2}
 -\frac{\b^2}{2}\,t_{-1}^3\,z^{-4},\\[-12pt]\\
\s^2\,G_{0,4}&=\frac{\a^2}{6}\,t_1^3\, t_3-\frac{\a(2\a+\b)}{8}\,t_1^4-\frac{\a\b}{4}\,t_1^2\, t_{-1}^2
     -\frac{\b(\a+2\b)}{8}\,t_{-1}^4+\frac{\b^2}{6}\,t_{-3}\,t_{-1}^3,\\[-12pt]\\
\s^2\,U_{1,2}(z)&=\frac{5\a^2}{8}\,t_1\,z^4+\Bigl(\frac{\a^2}{8}\,t_3-\frac{3\a(2\a+\b)}{4}\,t_1\Bigr)\,z^2\\
&+\Bigl(\frac{10\a^2+16\a\b+\b^2}{8}\,t_1+\frac{\a^2}{8}\,t_5+\frac{\a\b}{2}\,t_{-1}-\frac{\a(2\a+\b)}{4}\,t_3\Bigr)\\
&+\Bigl(-\frac{\a^2+16\a\b+10\b^21}{8}\,t_{-1}+\frac{\b(\a+2\b)}{4}\,t_{-3}-\frac{\a\b}{2}\,t_1-\frac{\b^2}{8}\,t_{-5}\Bigr)\,z^{-2}\\
&+\Bigl(\frac{3\b(\a+2\b)}{4}\,t_{-1}-\frac{\b^2}{8}\,t_{-3}\Bigr)\,z^{-4}
  -\frac{5\b^2}{8}\,t_{-1}\,z^{-6},\\[-12pt]\\
\s^2\,G_{1,2}&=\frac{\a^2}{8}\,t_1\,t_5+\frac{\a^2}{48}\,t_3^2-\frac{\a(2\a+\b)}{4}\,t_1\,t_3+\frac{\a^2+16\a\b+10\b^2}{16}\,t_{-1}^2\\
&+\frac{\a\b}{2}\,t_{-1}\,t_1+\frac{10\a^2+16\a\b+\b^2}{16}\,t_1^2-\frac{\b(\a+2\b)}{4}\,t_{-3}\,t_{-1}\\
&+\frac{\b^2}{48}\,t_{-3}^2+\frac{\b^2}{8}\,t_{-5}\,t_{-1}\;,
\end{align*}
where $\a,\b,\s$ are given by~\eqref{eqabs},~\eqref{eta}. This list can be continued further on.

\begin{remark}
Here we compare our form of the topological recursion~\eqref{eq2} with the one that can be found in the literature, see, e.g.~\cite{EO1}, \cite{EO2}.
\begin{enumerate}[(i)]
\item Traditionally, the spectral curve comes with an embedding to (a compactification of)~$\C^2$ by means of certain meromorphic functions~$X,Y$ on~$\pl$. These functions are chosen so that~$X$ is even with respect to the involution $z\mapsto -z$, and~$\eta=Y\,dX$. In our case we could have set, for example,
$$X=\frac1x=s\,\frac{\a z^2-\b}{z^2-1},\qquad Y=-\frac{\a-\b}{2s}\,\frac{z}{\alpha z^2-\beta}.$$
The formulas of the topological recursion, however, involve the coordinates~$X$ and~$Y$ only in the combination~$\eta=Y\,dX$.
\item The topological recursion is usually formulated in terms of $m$-point correlators. They are related to the homogeneous components~$F_{g,m}$ of the generating function~$F$ by the formulas
\begin{align}
w_{g,m}(x_1,\dots,x_m)&=\d_{x_1}\ldots\d_{x_m} F_{g,m}\,dx_1\ldots dx_m\nonumber\\
&=\d_{x_2}\ldots\d_{x_m} U_{g,m}(x_1)\,dx_1\ldots dx_m\;,
\label{eqwgn}
\end{align}
where $\d_{x_j}=\sum_{i=1}^\infty i\,x_j^{i-1}\,\pd{}{p_i}$. Via the change of variables~\eqref{eqzx}, $w_{g,m}$ can be viewed as a meromorphic $m$-differential on $\left(\pl\right)^m$ that is a Laurent form with respect to each of its arguments provided~$2g-2+m>0$.
\item A version of~\eqref{eqwgn} holds also for~$(g,m)=(0,2)$. Namely, $w_{0,2}(z_1,z_2)=\d_{z_2}U_{0,2}\,dz_2$ is the odd part of the \emph{Bergman kernel} $B(z_1,z_2)=\frac{dz_1\,dz_2}{(z_1-z_2)^2}$:
$$w_{0,2}(z_1,z_2)=\d_{z_2}U_{0,2}\,dz_2=\frac12\left(B(z_1,z_2)-B(z_1,-z_2)\right).$$
This can be interpreted as an equality of asymptotic expansions of the left and right hand sides at $z_1\to 0$ and~$z_1\to\infty$,
cf.~\eqref{U02}:
$$w_{0,2}(z_1,z_2)=
\begin{cases}-\sum_{i=0}^\infty d(z_2^{-2i-1})\;z_1^{2i}dz_1,&z_1\to 0\;,\medskip\\
-\sum_{i=0}^\infty d(z_2^{2i+1})\;z_1^{-2i}d(z_1^{-1}),& z_1\to\infty\;.\end{cases}$$
\item The projector~$P_L$, see \eqref{res}, is given by the contour integral
\begin{align*}
(P_L\phi)(z)\,dz=\frac{1}{2\pi\sqrt{-1}}\left(\int_{|w|=\epsilon}\frac{\phi(w)\,w\,dw}{z^2-w^2}
+\int_{|w|=1/\epsilon}\frac{\phi(w)\,w\,dw}{z^2-w^2}\right)\,dz\;,
\end{align*}
for small $\epsilon > 0$, where
$$\frac{w\;dz}{z^2-w^2}=\frac12\int\limits_{-w}^w B(z,\cdot).$$
This explains the appearance of the Bergman kernel in the majority of expositions of the topological recursion.
\item The above items (i)--(iv) demonstrate that the traditional form of the topological recursion in terms of the correlators~$w_{g,n}$ is obtainable from~\eqref{eq2} by applying $\d_{z_2}\ldots\d_{z_n}$ to the both sides of it.
\end{enumerate}
\end{remark}

\section{Proof of the topological recursion}

\subsection{Master Virasoro equation}

As we will see below, the topological recursion relations are just the equivalently reformulated Virasoro constraints.
To begin with, let us collect the Virasoro constraints~\eqref{vc} into a single equation by multiplying the~$n$th equation by~$x^n$ (where~$x$ is a formal variable) and summing them up:
\begin{multline*}
\sum_{n=0}^\infty x^n\Biggl(-\frac1s(n+1)\pd{F}{p_{n+1}}+(u+v)\,n\pd{F}{p_n}+
\sum_{j=1}^\infty p_j(n+j)\pd{F}{p_{n+j}}\Biggr.\\\Biggl.+
\sum_{i+j=n}ij\left(\pd{^2F}{p_i\D p_j}+\pd{F}{p_i}\pd{F}{p_j}\right)\Biggr)+uv=0.
\end{multline*}
This equation can be simplified. Notice that
\begin{align*}
\sum_{n=0}^\infty&\; x^n\left(-\frac1s(n+1)\pd{F}{p_{n+1}}+(u+v)\,n\pd{F}{p_n}\right)
  =\left(-\frac1s+(u+v)\,x\right)\d_x F\;,\\
\sum_{n=0}^\infty&\; x^n\sum_{i+j=n}ij\left(\pd{^2F}{p_i\D p_j}+\pd{F}{p_i}\pd{F}{p_j}\right)=
 x^2\left(\d_x^2 F+(\d_x F)^2\right)\;,
\end{align*}
where, as in the previous section, $\d_x=\sum_{n=1}^\infty nx^{n-1}\pd{}{p_n}$ (cf. Remark \ref{remdelta}).
As for the third term in the sum, we use the identity $p_j=\d_y^{-1}(j\,y^{j-1})=\d_y^{-1}d_y(y^j)$, where~$y$ is a new independent formal variable and $d_y=\frac{d}{dy}$, to re-write it as follows:
\begin{align*}
\sum_{n=0}^\infty\; x^n\sum_{j=1}^\infty& p_j(n+j)\pd{F}{p_{n+j}}\\
   &=\sum_{n=0}^\infty\;\sum_{i+j=n} x^{i}\,p_j\,n\,\pd{F}{p_n}
   =\d_y^{-1} d_y\left(\sum_{n=0}^\infty\;\sum_{i+j=n} x^{i}\,y^j\,n\,\pd{F}{p_n}\right)\\
   &=\d_y^{-1} d_y\left(\sum_{k=0}^\infty \frac{x^{k+1}-y^{k+1}}{x-y}\, k\,\pd{F}{p_k}\right)
   =\d_y^{-1} d_y\left(\frac{x^2{\d_x F}-y^2\d_y F}{x-y}\right)
\end{align*}
This yields the following \emph{master Virasoro equation} that unifies all Eqs.~\eqref{vc}:
\begin{align}
\Bigl(-\frac1s+(u+v)\,x\Bigr)\d_x F&+x^2\left(\d_x^2 F+(\d_x F)^2\right)\nonumber\\
&+\d_y^{-1} d_y\left(\frac{x^2\d_x F-y^2\d_y F}{x-y}\right)+uv=0\;.
\label{mve}
\end{align}

\subsection{Unstable terms and the spectral curve}

Our immediate goal is to extract the homogeneous terms in~\eqref{mve} contributing to $\d_x F_{g,n}$ for fixed~$g$ and~$n$. We start with the unstable cases. For~$g=0$ and~$n=1$ we get
\begin{equation}
x^2\,(\d_x F_{0,1})^2+\Bigl(-\frac1{s}+(u+v)\,x\Bigr)\d_x F_{0,1}+u v=0\,.\label{eqspectr}
\end{equation}
Solving this equation for $x\,\d_x F_{0,1}$, choosing the proper root and expanding it into the Taylor series at~$x=0$ we get
\begin{align*}
x\,\d_x F_{0,1}&=\frac{1}{2} \left(\frac{1}{s x}-u-v-\sqrt{\left(\frac{1}{s x}-u-v\right)^2-4 u v}\right)\\
&=u\,v\,s\,x+u\,v\,(u+v)\,s^2x^2+ u\,v\,\left(u^2+3\,u\,v+v^2\right)\,s^3x^3 +\dots
\end{align*}
and
$$F_{0,1}=u\,v\,s\,p_1+\frac12u\,v\,(u+v)\,s^2p_2+\frac13u\,v\,\left(u^2+3\,u\,v+v^2\right)\,s^3p_3 +\dots.$$

\begin{definition} The \emph{spectral curve} is the Riemann surface of the algebraic function $x\,\d_x F_{0,1}$ in the $x$-variable.
\end{definition}

In other words, the spectral curve is an algebraic curve such that~$x$ and~$x\d_x F_{0,1}$ are globally defined univalued meromorphic functions on it. In our case the spectral curve is given by~\eqref{eqspectr}. It is rational (admits a rational parametrization). Let~$z$ be an affine coordinate on~$\pl$. Its choice is not important, but, for convenience, we choose it in such a way that the two critical points of the function~$x$ on~$\pl$ are~$z=0$ (with the critical value~$x=\frac{1}{s\b}$) and~$z=\infty$ (with the critical value~$x=\infty$). The corresponding rational parametrization has the following form:
\begin{align*}
x&=\frac{z^2-1}{s\,(\a  z^2-\b)},&
z&=\sqrt{\frac{1-\b  s x}{1-\a s x}},\\
x\,\d_x F_{0,1}&=\sqrt{uv}\,\frac{1-z}{1+z},&\d_x F_{0,1}\frac{dx}{dz}&=\frac{8\,u\,v\,z}{(z+1)^2 \left(\alpha z^2-\beta \right)},
\end{align*}
where~$\a,\b$ are related to~$u,v$ by~\eqref{eqabs}.
All functions entering these equalities can be regarded either as rational functions in $z$-variable or as formal power expansions of these functions at~$z=1$.

We continue with the terms with~$g=0$ and~$n=2$ in~\eqref{mve}. We have
$$\Bigl(-\frac1s+(u+v)x\Bigr)\d_x F_{0,2}+2\,x^2\,\d_x F_{0,1}\,\d_x F_{0,2}
+\d_y^{-1} d_y\left(\frac{x^2\,\d_x F_{0,1}-y^2\,\d_y F_{0,1}}{x-y}\right)
=0\,,$$
from where we get
$$\d_y\,\d_x\,F_{0,2}=\frac{d_y\left(\frac{x^2\,\d_x\, F_{0,1}-y^2\,\d_y\, F_{0,1}}{x-y}\right)}
     {\frac1s-(u+v)x-2\,x^2\,\d_x F_{0,1}}\;.$$
This equality uniquely determines~$\d_y\d_x F_{0,2}dxdy$ as a meromorphic bidifferential on~$\pl\times\pl$. Substituting the obtained above expressions for~$x,\;dx/dz$ and $\d_x F_{0,1}$ into the last formula, after some miraculous cancellations we finally get
\begin{equation}
\d_y\d_x F_{0,2}\,dx\,dy=\frac{dz\,dw}{(z+w)^2}=-B(z,-w),
\label{ddF02}\end{equation}
where $B(z,w)=\frac{dz\,dw}{(z-w)^2}$ is the Bergman kernel, and $w$ is related to $y$ by the same formulas \eqref{eqzx} that relate $z$ to $x$.

\subsection{Rational recursion formula}

Now we look at the homogeneous terms of genus $g$ and degree $m$ with $2g-2+m>1$ in \eqref{mve}. To begin with, let us extract the unstable terms from the expression~$(\d F)^2$ in~\eqref{mve} in order to re-group them with the other summands. Multiplying~\eqref{mve} by $x^{-2}$ we get
\begin{align*}
&2\left(-\frac{1}{2\,s\,x^2}+\frac{u+v}{2\,x}+\d_x F_{0,1}\right)\,\d_x(F-F_{0,1}-F_{0,2})+\d_x^2 F\\
&\quad +\bigl(\d_x(F-F_{0,1}-F_{0,2})\bigr)^2+x^{-2}\d_y^{-1}\,d_y\,\left(\frac{x^2\,\d_x(F-F_{0,1})-y^2\,\d_y(F-F_{0,1})}{x-y}\right)\\
&\hspace{2.4in}+2\d_x(F-F_{0,1})\,\d_x F_{0,2}-(\d_x F_{0,2})^2=0.
\end{align*}
Let us rewrite the coefficients of this equation in the $z$-coordinate. It is convenient to put
$$
\eta=\eta(x)\,dx=-\left(-\frac{1}{2\,s\,x^2}+\frac{u+v}{2\,x}+\d_x F_{0,1}\right)\,dx\;,
$$
so that in terms of the coordinate $z$
$$
\eta=\eta(z)\,dz=\frac{(\a-\b)^2\,z^2}{(z^2-1)^2 (\a\,z^2 -\b)}\,dz\,.
$$
Then, using the already known expressions for~$x$, $\frac{dx}{dz}$,~$\d_x F_{0,1}$, and~$\d_y\d_x F_{0,2}$, we find that
\begin{align}
\d_x F_{0,2}=-\d_y^{-1}\,d_y\left(\frac{1}{z+w}\right).
\label{delf}
\end{align}
From the above identities we obtain the following equation for~$\d_x F$:
\begin{align}
\d_x(F-F_{0,1}-F_{0,2})&=\frac{1}{2\,\eta(x)}\Bigl(\d_x^2 F+(\d_x(F-F_{0,1}-F_{0,2}))^2-(\d_xF_{0,2})^2\Bigr)\nonumber\\
+\d_y^{-1}\,d_y&\left(\frac{w}{z^2-w^2}\left(\frac{\d_x(F-F_{0,1})}{\eta(x)}-\frac{\d_y (F-F_{0,1})}{\eta(y)}\right)\right)\,
\frac{dz}{dx}\;.\label{eqvirz}
\end{align}
In particular, for the homogeneous components $U_{g,m}=\d_x F_{g,m}$ with~$2g-2+m>1$ this equation reads
\begin{multline}
U_{g,m}(z)=\frac{1}{2\,\eta(z)}\Biggl(\d_z U_{g-1,m+1}(z)+\mathop{\sum{}^{\mathrlap{*}}}_{\substack{g_1+g_2=g,\\m_1+m_2=m+1}}
U_{g_1,m_1}(z)\,U_{g_2,m_2}(z)\Biggr)\\
 {}+\d_w^{-1}\,d_w\left(\frac{w}{z^2-w^2}\left(\frac{U_{g,m-1}(z)}{\eta(z)}-\frac{U_{g,m-1}(w)}{\eta(w)}\right)\right)\;,
 \label{eqrat}\end{multline}
where the star $^*$ by the summation sign means that the unstable terms with $2g-2+m\leq 0$ are omitted.
We refer to this equation as the \emph{rational recursion formula} for the forms~$U_{g,m}=U_{g,m}\,(z)dz$.

This equation allows to prove the polynomiality property for the forms~$U_{g,m}$ by induction in~$g$ and~$m$. Indeed, assume that~$U_{g',m'}$ is a Laurent form in~$z$ with coefficients polynomially depending on~$t_k=T_k(p)$, $k\in Z_\odd$, for all~$(g',m')$ with $0<2g'-2+m'<2g-2+m$. Then the first summand in~\eqref{eqrat} obviously also has this form. Let us examine the second summand. Note that the operator $\d_w^{-1}\,d_w$ is well defined on the space of odd Laurent polynomials in variable $w$, so let us check that this condition always holds. Indeed,
the function~$\frac{U_{g,m-1}(z)}{\eta(z)}$ is an even Laurent polynomial in~$z$, therefore, it can be represented as a Laurent polynomial in~$z^2$. Therefore, $\frac1{z^2-w^2}\left(\frac{U_{g,m-1}(z)}{\eta(z)}-\frac{U_{g,m-1}(w)}{\eta(w)}\right)$ is a Laurent polynomial in~$z^2$ and~$w^2$ regular at~$z=\pm w$.  Multiplying it by~$w$ and applying~$d_w$ we obtain an odd Laurent form in~$w$. The polynomiality property for the forms~$U_{g,m}$ follows now from Remark~\ref{remdelta}.

Utilising Remark~\ref{remdelta} once again, we obtain the polynomiality property for~$F_{g,m}$ with~$2g-2+m>0$ as well. This proves the main assertion of Theorem~\ref{th2} (under the assumption that it is valid for the initial terms~$F_{0,3}$
and~$F_{1,1}$; this is checked below).

\subsection{The residual formalism}

Since the both sides of~\eqref{eqrat} belong to~$L$, it is convenient to equate the projections of the terms entering this equality by applying~$P_L$ to each of them. The terms of the first summand on the right hand side already belong to~$L$, so~$P_L$ is identical on them. Compute the image of the second summand on the right hand side under the projection~$P_L$. The key observation is that~$P_L$ can be applied to the two terms of this summand separately. In particular, the form~$\frac{dz}{z^2-w^2}$ is regular both at~$z=0$ and at~$z=\infty$ so that it does not contribute to the image. It remains to compute the term
\begin{align*}
&P_L\!\left(\d_w^{-1}d_w\!\left(\frac{w}{z^2-w^2}\frac{U_{g,m-1}(z)}{\eta(z)}\right)dz\right)\\
&\hspace{2in}=P_L\!\left(\frac{U_{g,m-1}(z)}{\eta(z)}\,\d_w^{-1}\!\left(\frac{(z^2+w^2)}{(z^2-w^2)^2}dz\,dw\right)\right).
\end{align*}
The form $\frac{(z^2+w^2)}{(z^2-w^2)^2}\,dz\,dw$ is not Laurent so that~$\d_w^{-1}$ is not applicable to it directly. However, what we actually need in order to apply~$P_L$ is the expansion of this form at~$z=0$ and~$z=\infty$. The coefficients of these expansions are Laurent with respect to~$w$:
\begin{align*}
&\frac{(z^2+w^2)}{(z^2-w^2)^2}dz\,dw=-\sum_{i=0}^\infty d(w^{-2i-1})z^{2i}dz=
  -\d_w\sum_{i=0}^\infty t_{-2i-1}z^{2i}dz,\quad z\to 0,\\
&\frac{(z^2+w^2)}{(z^2-w^2)^2}dz\,dw=-\sum_{i=0}^\infty d(w^{2i+1})z^{-2i}d(z^{-1})\\
&\hspace{2.2in}=-\d_w\sum_{i=0}^\infty t_{2i+1}z^{-2i}d(z^{-1}),\quad z\to \infty.
\end{align*}
In other words, the form $\d_w^{-1}\Bigl(\frac{(z^2+w^2)}{(z^2-w^2)^2}\,dz\,dw\Bigr)$ is well defined in some neighborhoods of the points~$z=0$ and~$z=\infty$ and coincides with the form~$U_{0,2}$ defined by~\eqref{U02}. This proves the equality~\eqref{eq2} of the topological recursion.

\subsection{Initial terms of the recursion}

In order to finish the proof of Theorem~\ref{th2}, it remains to check it for the initial terms of the recursion, that is, for~$U_{1,1}=\d_x F_{1,1}\,dx$ and~$U_{0,3}=\d_x F_{0,3}dx$.

For the case~$g=m=1$, equating the corresponding terms in~\eqref{eqvirz} we get
$$U_{1,1}=\frac{1}{2\eta(x)}\d_x^2F_{0,2}\,dx\,,$$
and using the explicit formula~\eqref{ddF02} for~$\d^2F_{0,2}$, we obtain the required formula~\eqref{eq4} for~$U_{1,1}$.

For the case~$g=0$,~$m=3$ the computations are slightly more involved. Eq.~\eqref{eqvirz} uniquely determines the form~$U_{0,3}$ as
$$U_{0,3}=
 \d_w^{-1}\, d_w\left(\frac{w}{z^2-w^2}\left(\frac{\d_x F_{0,2}(x)}{\eta(x)}-\frac{\d_y F_{0,2}(y)}{\eta(y)}\right)\right)\,dz
 -\frac{(\d_x F_{0,2})^2}{2\,\eta(x)}\,dx\;,
$$
where, as before, $y$ is related to $w$ by the same formulas \eqref{eqzx} that relate $x$ to $z$.

The form~$\d_x F_{0,2}\,dx$ being known, cf.~\eqref{delf}, we directly compute
$$U_{0,3}=\frac{1}{32 u v}\left(\a\,t_1^2dz-\b t_{-1}^2\frac{dz}{z^2} \right)=P_L\left(\frac{(U_{0,2}(z))^2}{2\eta(z)}\,dz\right)$$
which agrees with~\eqref{eq3}. This completes the proof of Theorem~\ref{th2}.

\section{Topological recursion for ribbon graphs}

Here we discuss the toplogical recursion for the numbers $D_{g,m}$ of genus $g$ ribbon graphs with $m$ labeled boundary components of lengths $\mu_1,\ldots,\mu_m$, see Section 2. A topological recursion for these numbers was first obtained in~\cite{EO1},
Theorem 7.3 (it was later rediscovered in \cite{DMSS}). We present a simple proof of this recursion in terms of the homogeneous components~$\tF_{g,m}$ of the generating function~\eqref{gfM}
that follows directly from the Virasoro constraints, cf. Theorem~\ref{rg},~(i). In fact, the argument is quite parallel to that of the case of dessins d'enfants. This is why we skip the details paying attention only to the differences between these two enumerative problems. More specifically, we have the same spectral curve $\C P^1$ equipped with an affine coordinate~$z$ and the
involution~$z\mapsto-z$. What is different, is the choice of the local coordinate~$x$ at the point~$z=1$ and the form~$\eta$.

Consider the linear functions $\tT_k(p)$ given by~\eqref{eqT} with
\begin{equation}\label{zxM}
z(x)=\sqrt{\frac{1+2\sqrt{u}\; s\, x}{1-2\sqrt{u}\; s\, x}}
\end{equation}

\begin{theorem}\label{th3}
Let $\tF_{g,m}$ be the infinite series defined by
\begin{align*}
\tF_{g,m}(s,u,p_1,p_2,\ldots)
&=\frac{1}{m!} \sum_{\mu\in{\Z}_+^m} D_{g,m}(\mu)s^d u^k p_{\mu_1}\ldots p_{\mu_m}\;.
\end{align*}
Then
\begin{enumerate}[(i)]
\item For each~$g,m$ with~$2g-2+m>0$ there exist a polynomial~$\tG_{g,m}$ of the variables~$t_j$, $j\in\Z_\odd$, such that
$$\tF_{g,m}(p)=\tG_{g,m}(t)\bigm|_{t_j=\tT_j(p)}$$
(i.e. each~$\tF_{g,m}$ is a polynomial in the linear functions $\tT_{\pm1},\tT_{\pm3},\tT_{\pm5},\ldots$).
\item The polynomials~$\tG_{g,m}$ can be recursively computed, starting from $\tG_{0,3}$ and $\tG_{1,1}$, by the same Eqs.~\eqref{eq2}--\eqref{eq4} with $\eta$ given by the formula
$$
\eta=\eta(z)\,dz=-\frac{16uz^2dz}{(1-z^2)^3}\;.
$$
\end{enumerate}
\end{theorem}

\begin{proof}
Like in the case of dessins, we start with the master Virasoro equation that readily follows from Theorem~\ref{th2},~(i):
\begin{align*}
\Bigl(-\frac1{s^2}+2\,u\,x^2\Bigr)\,\d_x \tF&+x^3\Bigl(\d_x^2 \tF+(\d_x \tF)^2\Bigr)\\
&+\d_y^{-1}\,d_y\left(\frac{x^3\,\d_x \tF-y^3\,d_y \tF}{x-y}\right)+u^2\,x+u\,p_1=0\;.
\end{align*}
The spectral curve in this case (an analog of \eqref{eqspectr} above) is given by the equation
$$x^2\,(\d_x\tF_{0,1})^2-x\,\d_x\tF_{0,1}\Bigl(\frac{1}{s^2 x^2}-2 u\Bigr)+u^2=0\,.$$
Solving this equation for $x\,\d_x\tF_{0,1}$ we get the following rational parametrization of the spectral curve
\begin{align*}
&x(z)=\frac{z^2-1}{2s\sqrt{u}\;(z^2+1)}\;,\\
&x\,\d_x\tF_{0,1}=\frac{1-2\,s^2\,u\,x^2-\sqrt{1-4\,s^2\,u\,x^2}}{2s^2\,x^2}=u\left(\frac{1-z}{1+z}\right)^2\;,
\end{align*}
with the inverse change $z(x)$ given by~\eqref{zxM}.
The genus $0$ two point correlator is the same as in the case of dessins:
$$\d_y\d_x\tF_{0,2}\,dx\,dy=\frac{dz\,dw}{(z+w)^2},$$
and instead of the form $\eta$ we have
$$\tilde\eta=\tilde\eta(x)dx=\left(-\d_x \tF_{0,1}-\frac{u}{x}+\frac{1}{2\,s^2\,x^3}\right)\,dx=-\frac{16\,u\,z^2}{(1-z^2)^3}\,dz\;.$$
Thus, the master Virasoro equation in $z$-coordinate acquires the same form~\eqref{eqvirz} and implies the topological recursion~\eqref{eqrat} with $\eta$ replaced by $\tilde\eta$.
\end{proof}

Below are the first few terms of the recursion:
\begin{align*}
U_{0,3}&=\frac{1}{32\,u}\left(t_1^2-t_{-1}^2z^{-2}\right)\,,\\[-12pt]\\
G_{0,3}&=\frac{1}{96\,u} \left(t_{1}^3+t_{-1}^3\right)\,,\\[-12pt]\\
U_{1,1}&=\frac{1}{128\,u}\left(z^2-3+3\,z^{-2}-z^{-4}\right)\,,\\[-12pt]\\
G_{1,1}&=\frac{1}{384\,u}\left(t_{3}-9 t_{1}-9 t_{-1}+t_{-3}\right)\,,\\[-12pt]\\
U_{0,4}&=\frac{1}{512\,u^2}\left(t_1^3 z^2+(t_3 t_1^2-3 t_1^3-t_{-1}^2 t_1)\right.\\[-12pt]\\
&\hspace{.9in}\left.+(t_1^2 t_{-1}+3 t_{-1}^3-t_{-3} t_{-1}^2)\,z^{-2}-t_{-1}^3\,z^{-4}\right)\,,\\[-12pt]\\
G_{0,4}&=\frac{1}{6144\,u^2}\left(4 t_1^3 t_3-9 t_1^4-6 t_1^2 t_{-1}^2-9 t_{-1}^4+4 t_{-3} t_{-1}^3\right)\,,\\[-12pt]\\
U_{1,2}&=\frac{1}{2048\,u^2}\left(5\,t_1\,z^4+(t_3-18 t_1)\,z^2+(4 t_{-1}+27 t_1-6 t_3+t_5)\right.\\[-12pt]\\
 &\hspace{.3in}\left.+(-t_{-5}+6 t_{-3}-27 t_{-1}-4 t_1)\,z^{-2}+(18 t_{-1}-t_{-3})\,z^{-4}-5\,t_{-1}\,z^{-6}\right)\,,\\[-12pt]\\
G_{1,2}&=\frac{1}{12288\,u^2}\left(6 t_1 t_5+t_3^2-36 t_1 t_3+81 t_1^2+24 t_{-1} t_1\right.\\[-12pt]\\
 &\hspace{2in}\left.+81 t_{-1}^2-36 t_{-1} t_{-3}+t_{-3}^2+6 t_{-5} t_{-1}\right)\,.
\end{align*}

\medskip
\noindent
{\bf Acknowledgments.}
The main results of the paper, Theorems 4 and 5, were obtained under support of the Russian Science Foundation grant 14-21-00035. The work of MK was additionally supported by the President of Russian Federation grant NSh-5138.2014.1 and by the RFBR grant 13-01-00383. PZ acknowledges hospitality of the Center for Quantum Geometry of Moduli Spaces at Aarhus University. We thank JSC ``Gazprom Neft" for funding short-term visits of MK to the Chebyshev Laboratory at SPbSU. We are grateful to L.~Chekhov, B.~Eynard, P.~Norbury and G.~Schaeffer for useful discussions, and to the anonymous referee for correcting a few typos and suggesting several improvements in the text.

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