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\begin{document}
\title[Masur--Veech volumes]{Masur--Veech volumes of quadratic differentials and their asymptotics}
\author{Di Yang, Don Zagier, Youjin Zhang}
\dedicatory{Dedicated to the memory of Boris Anatol'evich Dubrovin}
\maketitle
\begin{abstract}
Based on the Chen--M\"oller--Sauvaget formula, we apply the theory of integrable systems 
to derive three equations for the generating series of the 
Masur--Veech volumes ${\rm Vol} \, \Q_{g,n}$ associated with the principal strata 
of the moduli spaces of quadratic differentials, 
and propose refinements of the 
conjectural formulas given in~\cite{DGZZ,ADGZZ}
 for the large genus asymptotics
 of ${\rm Vol} \, \Q_{g,n}$ and of the associated area Siegel--Veech constants.
\end{abstract}

\medskip

\setcounter{tocdepth}{1}
\tableofcontents

\section{Statements of the results}
Let $\M_{g,n}$ denote the moduli space of complex 
algebraic curves of genus~$g$ with~$n$ distinct marked points, 
and $\Q_{g,n}$ the moduli space of pairs $(\C,q)$, 
where $\C \in \M_{g,n}$ is a smooth algebraic curve and $q$ is a 
meromorphic quadratic differential on~$\C$ with only simple poles at the marked points. 
This moduli space of quadratic differentials~$\Q_{g,n}$ 
is endowed with the canonical symplectic structure. 
The induced volume element on $\Q_{g,n}$ is called 
the Masur--Veech (MV) volume element. 
Denote by~$\MV_{g,n}$ the volume of~$\Q_{g,n}$; see e.g.~\cite{DGZZ} for its meaning.
Recently,  Chen--M\"oller--Sauvaget~\cite{CMS} proved that  
the volumes ${\rm Vol} \, \Q_{g,n}$ with $2g-2+n>0$ 
can be expressed in terms of linear Hodge integrals as follows:
\beq\label{CMSformula}
{\rm Vol} \, \Q_{g,n} \= 2^{2g+1} \frac{\pi^{6g-6+2n} (4g-4+n)!}{(6g-7+2n)!} \,
\sum_{j=0}^g \int_{\overline{\mathcal{M}}_{g,3g-3+2n-j}}  
\frac{\lambda_j  \psi_{n+1}^2 \cdots \psi_{3g-3+2n-j}^2}{(3g-3+n-j)!} \,,
\eeq
where $\overline{\mathcal{M}}_{g,k}$ denotes the Deligne--Mumford 
compactification of~$\mathcal{M}_{g,k}$,  
$\psi_i$ denotes the first Chern class of the $i_{\rm th}$ tautological line 
bundle on $\overline{\mathcal{M}}_{g,k}$, and $\lambda_j$ denotes the 
$j_{\rm th}$ Chern class of the rank~$g$ Hodge bundle $\mathbb{E}_{g,k}$ 
on $\overline{\mathcal{M}}_{g,k}$.  
The goal of the present paper is to study the numbers 
${\rm Vol} \, \Q_{g,n}$ by using the Chen--M\"oller--Sauvaget (CMS) formula.

For $g,n\geq 0$, we define 
\beq\label{definisionagn}
a_{g,n} \= \begin{cases}
\sum_{j=0}^g \frac1{(3g-3+n-j)!}\int_{\overline{\mathcal{M}}_{g,3g-3+2n-j}} 
\lambda_j  \psi_{n+1}^2 \cdots \psi_{3g-3+2n-j}^2 \,, & 2g-2+n>0\,, \\
0\,, & {\rm otherwise.}\\
\end{cases}
\eeq
Note that the $a_{g,n}$ are rational numbers, and differ from ${\rm Vol} \, \Q_{g,n}$ only by 
some simple factors. Define a generating series $\H(x,\e)$ for the numbers $a_{g,n}$, called the MV free energy, by 
\beq\label{defintionofHxe}
\H(x,\e) \:= \sum_{g,n\geq0} \e^{2g-2} \frac{x^n}{n!} a_{g,n} \,.
\eeq
The first result of this paper is then given by the following theorem.
\begin{theorem} \label{thmequation}
The series $\H(x,\e)$ satisfies the following two equations:
\begin{align}
& \bigl[\p_x(\H_+\,-\, \H_-)\bigr]^2 \+ \p_x^2 \bigl( \H_+ + \H_-\bigr) \= \frac{2x}{\e^2} \,,  \label{theequation} \\
& \biggl(\e  \p_\e + \frac12x\p_x - \frac{\e^2}{24} \p_x^3\biggr)  \bigl(\H_{+} - \H_{-}\bigr)    \+ 
\frac{\e^2}{12}  \bigl[\p_x(\H_{+}-\H_{-})\bigr]^3   \=  0  \,, \label{anotherequation} 
\end{align} 
where 
$\H_{\pm} := \H\bigl(x\pm\frac{i \e}{2},\e\bigr)$.
\end{theorem} 

A statement equivalent to equation~\eqref{theequation} is given by the following corollary. 
\begin{cor}\label{thmrecursion}
For all $g\ge0$ and $n\geq 2$, the numbers $a_{g,n}$ %defined in~\eqref{definisionagn}
can be uniquely determined by the following recursion relation
\begin{align}
&  a_{g,q+2} \= \frac{q!}2 \sum_{g_1+g_2+j_1+j_2=g \atop n_1+n_2=q+4+2(j_1+j_2)} 
\frac{(-1)^{j_1+j_2}  a_{g_1,n_1} a_{g_2,n_2} }{4^{j_1+j_2} (2j_1+1)!(2j_2+1)! (n_1-2j_1-2)!(n_2-2j_2-2)!}  \nn\\
& \qquad \qquad \,-\, \sum_{j=1}^g \frac{(-1)^j a_{g-j,q+2j+2}}{4^j (2j)!}   \+ \delta_{q,1} \delta_{g,0}   
 \label{recursionagn}
\end{align}
along with the boundary condition $a_{0,2}=0$ (cf.~\eqref{definisionagn}), where $q\geq 0$.
\end{cor}

Another corollary of Theorem~\ref{thmequation} 
is the following non-linear differential equation for the series~$\H$.
\begin{cor}\label{Kazdeduced} 
The series $\H=\H(x,\e)$ satisfies the following equation:
\beq \label{deducedfrom}
 \e \p_\e \p_x (\H) \+ x  \p_x^2 ( \H) \+ \frac12  \p_x(\H) \,-\, \frac{\e^2}4  \bigl[\p_x^2(\H)\bigr]^2 \,-\, \frac{\e^2}{24}  \p_x^4(\H)  \= 0 \,.
\eeq
\end{cor}
\noindent The proof will be given in Section~\ref{section3}. 
We also show there that equation~\eqref{deducedfrom} implies a recursion given by Kazarian in~\cite{Kazarian} for the Hodge integrals   
\[\frac{(5g-3-j)(5g-5-j)}{(3g-3-j)!}\int_{\overline{\mathcal{M}}_{g,3g-3-j}} 
\lambda_j  \psi_{1}^2 \cdots \psi_{3g-3-j}^2\,,\quad 0\le j\le g \, .\]

A third corollary of Theorem~\ref{thmequation} (which apart from the boundary conditions 
is in fact equivalent to equation~\eqref{deducedfrom}) is the following recursion for the numbers~$a_{g,n}$. 
\begin{cor}\label{thmrecursion}
For all $g\geq 0$ and $n\geq 1$, the numbers $a_{g,n}$ are given recursively by
\begin{align}
&  a_{g, n} \= 
\frac{1}{2} \sum_{g_1,g_2\ge 0 \atop g_1+g_2=g} \sum_{ n_i\geq 2,  (g_i,n_i)\neq (0,3), i=1,2 \atop n_1+n_2=n+3}  
\binom{n-1}{n_1-2}
\, \frac{a_{g_1,n_1} a_{g_2,n_2}}{4g-4+n}   \+  \frac1{12} \, \frac{a_{g-1,n+3}}{4g-4+n}  
 \label{recursionagntwo} 
\end{align}
if $2g-2+n > 0$, $(g,n) \not\in \{ (0,3) , \,  (0,4) \}$, $a_{0,3}=a_{0,4}=1$ and $a_{0,1}=a_{0,2}=a_{-1,n}=0$.
\end{cor}
The recursion relations~\eqref{recursionagn} or~\eqref{recursionagntwo} both give rapid (polynomial-time) 
algorithms for computing $a_{g,n}$ for $n\geq 2$ or $n\geq 1$, respectively. The first few values $a_{g,n}$ 
are given by the following table. 
\begin{table}[H]
%[!htbp]
\begin{center}\tiny 
    \begin{tabular} {|c|c|c|c|c|c|c|c |N}
    \hline
    & $n=0$ & $n=1$ & $n=2$ & $n=3$ & $n=4$ & $n=5$ & $n=6$ &   \\ [8pt]
    \hline
    $g=0$ &  0 & 0 & 0 & 1 & 1 &  3  & 15  & \\ [8pt]
    \hline
    $g=1$ & 0 & $\frac1{12}$ & $\frac{1}{8}$ & $\frac{11}{24}$ & $\frac{21}{8}$ & $\frac{163}{8}$ & $\frac{1595}{8}$  &  \\[8pt]
    \hline
    $g=2$ & $\frac1{96}$ & $\frac{29}{640}$ & $\frac{337}{1152}$ & $\frac{319}{128}$ & $\frac{10109}{384}$ & $\frac{42445}{128}$ & 
    $\frac{620641}{128}$  & \\[8pt]
    \hline
    $g=3$ & $\frac{575}{21504}$ & $\frac{20555}{82944} $ & $\frac{77633}{27648}$ & 
    $\frac{1038595}{27648}$ & $\frac{16011391}{27648}$ & $\frac{31040465}{3072}$ & 
    $\frac{201498115}{1024}$ &  \\[8pt]
    \hline
    $g=4$ & $\frac{2106241}{7962624}$ & $\frac{1103729}{294912}$ & $\frac{160909109}{2654208}$ & 
    $\frac{14674841399}{13271040}$ & $\frac{99177888029}{4423680}$ &
    $\frac{442442475179}{884736}$ & $\frac{10765584400823}{884736}$ &  \\[8pt]
  \hline
    \end{tabular}
\end{center}
\caption{The numbers $a_{g,n}$ with $0\leq g\leq 4$ and $0\leq n\leq 6$.} \label{table1}
\end{table}

The following proposition describes the property of ${\rm Vol} \, \Q_{g,n}$,
which will enable us to determine also $a_{g,0}$ and $a_{g,1}$ from~\eqref{theequation}, 
and $a_{g,0}$ from~\eqref{anotherequation} or~\eqref{deducedfrom}.
\begin{prop}[\cite{AEZ,ABCDGLW,CMS}] \label{prp1lem}
The following properties of the MV volumes hold:
\begin{align}
&{\rm Vol} \, \Q_{0,n}
\= \frac{\pi^{2n-6}}{2^{n-5}}\,,\qquad \forall \, n\ge 3\,;\label{thm0n}\\
&{\rm Vol} \, \Q_{1,n}
\= \frac{\pi^{2n}}3 \biggl(\frac{n! }{ (2 n-1)!!} \+ \frac{2n}{(2 n-1) 2^n}\biggr)\,, \qquad \forall \,n\ge 1\,;\label{thm1n}\\
&{\rm Vol} \, \Q_{g,n} \= 2^{2g+1+n} \frac{\pi^{6g-6+2n} (4g-4+n)!}{(6g-7+2n)!} 
\sum_{j=0}^{g}  \frac{\langle \lambda_j \tau_2^{3g-3-j} \rangle_g}{(3g-3-j)!} \biggl(\frac{5g-5-j}{2}\biggr)_n\,,\label{thmgn}
\end{align}
where $g\geq 2$, $n\geq 0$, $(b)_n:=b(b+1)\cdots (b+n-1)$ denotes the increasing Pochhammer symbol, and 
we used Witten's notation: for a cohomology class $\gamma\in H^*(\overline{\M}_{g,n};\CC)$, 
\[\langle \gamma \tau_{i_1} \cdots \tau_{i_n} \rangle_g \:=  
\int_{\overline{\mathcal M}_{g,n}} 
\gamma \, \psi_1^{i_1} \cdots \psi_n^{i_n} \,,\quad i_1,\dots,i_n\geq 0 \,. \]
\end{prop}
\noindent The explicit expression for ${\rm Vol} \, \Q_{0,n}$, $n\geq 3$ was conjectured 
by Kontsevich, and was proved by Athreya-Eskin-Zorich in~\cite{AEZ}. 
The formula~\eqref{thm1n} was conjecturally given by Andersen et. al.~\cite{ABCDGLW}, and
the formula~\eqref{thmgn} is equivalent to the 
Conjecture~5.4 of~\cite{ABCDGLW} (to see the equivalence, cf.~\cite{CMS}). 
A proof of Proposition~\ref{prp1lem} was given in~\cite{CMS}. 
In this paper we give a different proof of this proposition based on the following lemma. 

\begin{lemma} \label{Txstructure} 
Let $T=\sqrt{1-2x}$. Define 
the power series $\H_g(x)$, $g\geq 0$ by   
\beq\label{Hgenusexpansion}
\H(x,\e) \;=:\; \sum_{g\geq0} \e^{2g-2} \H_g(x) \,.
\eeq
Then we have
\begin{align}
\HHod_0(x) &\= 
\frac1{40}  \,-\, \frac{T^2}{12} \+ \frac{T^4}{8} \,-\, \frac{T^5}{15}\,, \label{h0hodgeT} \\
\HHod_1(x) & \= \frac1{24} \log \frac1T \+ \frac{1}{24}  \,(1-T) \,, \label{h1hodgeT} \\
\HHod_2(x) & \= \frac{7}{1440} \, \frac{1}{T^5} \+ \frac{5}{1152} \, \frac{1}{T^4} \+ \frac{7}{5760} \, \frac{1}{T^3}  \,. \label{h2hodgeT} 
\end{align}
In general, we have the following expression for $\HHod_g(x)$:
\begin{align}
\HHod_g(x) & \= 
\sum_{j=0}^{g}  \frac{\langle \lambda_j \tau_2^{3g-3-j} \rangle_g}{(3g-3-j)!} \frac{1}{T^{5g-5-j}}\,, \quad g\geq 2\,. \label{ggeq2hodgeT}
\end{align}
\end{lemma}
\noindent We give in Section~\ref{section2} a proof of Lemma~\ref{Txstructure} 
by using the CMS formula~\eqref{CMSformula} and the Dubrovin-Zhang formalism~\cite{DZ-norm, DLYZ1, DuY} on  
Hodge integrals.
Substituting the expansion~\eqref{Hgenusexpansion} into~\eqref{deducedfrom} we find
\beq\label{hgrecursion6}
x \H_{g}'' \+ \Bigl(2g-\frac32\Bigr) \H_{g}' \,-\, 
\frac14 \sum_{g_1,g_2\geq 0 \atop g_1+g_2=g} \H_{g_1}'' \H_{g_2}'' \,-\, \frac1{24} \H_{g-1}'''' \=0 \,.
\eeq
Here, prime, ``\,$'$\," denotes $d/dx$. It turns out that this formula together with Lemma~\ref{Txstructure} 
determines $\H_g$, $g\geq 0$, and therefore the $a_{g,n}$, uniquely for all $g,n\geq0$.

Recently, Aggarwal, Delecroix, Goujard, Zograf and Zorich~\cite{ADGZZ} 
proposed a conjectural formula for the large~$g$ leading asymptotics 
of ${\rm Vol} \, \Q_{g,n}$ (the conjectural formula was given originally in~\cite{DGZZ} for $n=0$). 
The ADGZZ conjecture was very recently proved in~\cite{A3}.
Our next result is a refinement of the ADGZZ conjecture 
to the following more precise asymptotic statement.
\begin{conj}\label{conjasymg}
For any fixed $n\geq 0$, we have the asymptotic formula:
\beq
{\rm Vol} \, \Q_{g,n} \;\sim\; \frac{2^{12g+4n-10}}{3^{4g+n-4}\pi}  \, \sum_{k=0}^\infty \frac{m_k(n)}{g^k}\,,\quad g\to\infty\,,  \label{mvlargeg} 
\eeq
where each $m_k(n)$ is a polynomial in~$n$ with coefficients in $\QQ[\pi^2]$, with the first four values 
(with $M=-\pi^2/144$  for convenience)
given by 
\begin{align}
m_0(n) &\= 1\,, \quad m_1 (n) \= M\,, \nn\\
 m_2 (n) & \= \frac M{24}\,n^3\,-\,\frac{3M}8\,n^2 \+\frac{4M-27M^2}{6}\,n \+\frac{M+19M^2}2\,,  \nn \\ 
 m_3 (n) &\= - \frac{8M+27M^2}{288}n^4 \+ \frac{17M+65M^2}{48} n^3 
  \,-\,\frac{860M+1890M^2-14256M^3}{576} n^2 \nn \\  
 &\qquad \+ \frac{104M \,-\, 373M^2 \,-\, 6156 M^3 }{48}  n 
 \nn\\
 &\qquad \, - \, \frac{ 55M - 3615M^2 - 28650M^3 + 126846M^4 }{180} \,.\nn 
\end{align}  
\end{conj}

The asymptotic formula~\eqref{mvlargeg} with $\sum_{k=0}^\infty m_k(n)/g^k$ 
replaced by~$1$ is the ADGZZ conjecture. We refer to~\cite{A1,A2,CMSZ,CMZ,EZ,S2,S3} for 
the analogues of the ADGZZ conjecture and Conjecture~\ref{conjasymg} (cf. also Conjecture~\ref{conjecture2} 
in Section~\ref{section4} below) for the MV volumes and for the related area Siegel--Veech constants associated with 
the moduli spaces of abelian differentials, and the proofs of these analogues via different approaches.
Conjecture~\ref{conjasymg} can also be stated in terms of the numbers $a_{g,n}$ defined in~\eqref{definisionagn} as
\beq a_{g,n} \;\sim\; \frac{(6g-7+2n)!}{(4g-4+n)! }\frac{2^{10g+4n-11}}{3^{4g+n-4}\pi^{6g-5+2n}} \, 
  \sum_{k=0}^\infty \frac{m_k(n)}{g^k} \,, \quad g\rightarrow \infty\,.
\eeq

Conjecture~\ref{conjasymg}, like the related Conjecture~\ref{conjecture2} which will be stated in Section~\ref{section4} below, 
is completely empirical. Specifically, we computed the 
values of $a_{g,n}$ numerically for $g\le 100$ and a number of small values of~$n$, then interpolated by the 
numerical method explained in~\cite{Zagier}, \cite[Section~5]{GM}
and elsewhere to get an asymptotic power series in~$1/g$ with coefficients 
known to high precision, and then used polynomial interpolation and the LLL (Lenstra--Lenstra--Lovasz) method to recognize the coefficients 
as polynomials in~$n$ with coefficients in $\QQ[\pi^2]$.

\begin{remark}
It would be interesting to investigate the following generating series:
\beq
C_n(\e)\:= \sum_{g\geq 0} \e^{2g-2} a_{g,n}\,, \quad n\geq 0\,.
\eeq
In other words, $\H(x,\e)=\sum_{n\geq 0}\frac{x^n}{n!} C_n(\e)$. 
Equation~\eqref{deducedfrom} then implies the following relations for $C_n(\e)$:
\beq
C_{n+4} \= \frac{24}\e C_{n+1}' \+ 12\frac{2n+1}{\e^2} C_{n+1} \,-\, 6n! \sum_{n_1+n_2=n} \frac{C_{n_1+2} C_{n_2+2}}{n_1! \,n_2!} \,, \quad n\geq 0\,. 
\eeq
Similarly, equation~\eqref{anotherequation} implies relations for the analogue of~$C_n(\e)$ for $\H_+-\H_-$.
Understanding of $C_n(\e)$ or its analogue might be useful for proving the above Conjecture~\ref{conjasymg}. 
\end{remark}

The paper is organized as follows. 
In Section~\ref{section2}, we review the Dubrovin-Zhang theory and give a proof of Lemma~\ref{Txstructure}.
In Section~\ref{section3}, we prove Theorem~\ref{thmequation}.
In Section~\ref{section4}, we extend a conjectural formula for the large genus asymptotics of the 
area Siegel--Veech constants.

\medskip

\noindent {\bf Acknowledgements.} 
We would like to thank Dawei Chen, Martin M\"oller, and Motohico Mulase
for helpful suggestions. Part of the work of D.Y. was done during his visit in MPIM; 
he thanks MPIM for excellent working conditions and financial support. 
This work was partially supported by NSFC No.\,11771238.

\section{The Hodge free energy}\label{section2}
In this section we first give a short review of the Dubrovin-Zhang approach to Hodge integrals~\cite{DLYZ1,Buryak,DZ-norm,DuY}, 
and then specialize our discussions to linear Hodge integrals 
and prove Lemma~\ref{Txstructure}.
Recall that the genus~$g$ Hodge free energy $\HHod_g(\bt;\bs)$ is defined by
\begin{align}
&\HHod_g(\bt;\bs) \= \sum_{k\geq 0} \sum_{\,i_1, \dots, i_k\geq 0} \frac{t_{i_1}\cdots t_{i_k}}{k!} \int_{\overline{\mathcal M}_{g,k}} 
 \Omega_{g,k}(\bs) \, \psi_1^{i_1} \cdots \psi_k^{i_k}\,,\\
&\Omega_{g,k}(\bs) \:= \exp \biggl(\,\sum_{j\geq 0} s_{2j-1} \, {\rm ch}_{2j-1}(\mathbb{E}_{g,k})\biggr) \,.
\end{align}
Here $g\geq 0$, $\bt = (t_0,t_1,\dots)$, $\bs=(s_1,s_3,\dots)$, 
$t_0,t_1,t_2,\dots$, $s_1,s_3,\dots$ are indeterminates, 
and ${\rm ch}_1, {\rm ch}_3, {\rm ch}_5$, $\dots$ denote components of the Chern character of~$\mathbb{E}_{g,k}$. 
Define the total Hodge free energy $\HHod$ by
$$
 \HHod\=\HHod(\bt;\bs;\epsilon) \= \sum_{g\geq0} \HHod_g(\bt;\bs)\,\epsilon^{2g-2} \,.
$$

Let $\vv \in \CC[[\bt]]$ be the unique power series solution to the following equation:
\beq\label{hodo}
\sum_{i\geq 0} \frac{t_i}{i!} v^i \= v \,. 
\eeq
It is well known that this unique power series $\vv=\vv(\bt)$ has the explicit expression
\beq\label{genus0solution}
\vv(\bt) \=  
\sum_{k\geq 1} \frac1k \sum_{p_1,\dots,p_k\geq 0\atop p_1+\dots+p_k=k-1} \frac{t_{p_1}}{p_1!} \cdots \frac{t_{p_k}}{p_k!}\;.
\eeq
Denote 
\beq
v_m(\bt) \= \p_{t_0}^m(v(\bt))\,, \quad m\geq 0\,. 
\eeq

\medskip

\noindent {\bf Theorem A} (\cite{DLYZ1})  
\textit{The genus~0 and~1 Hodge free energies have the expressions
\begin{align}
& \HHod_0(\bt; \bs) \= 
\frac{\vv(\bt)^3}6 \m \sum_{i\geq 0} t_i  \frac{\vv(\bt)^{i+2}}{i! (i+2)}  
\+ \frac12 \sum_{i,j\geq 0} t_i t_j \frac{\vv(\bt)^{i+j+1}}{(i+j+1) i! j!}\,,\label{Hodgegenus0}\\
& \HHod_1(\bt; \bs) \= \frac1{24} \log\vv_1(\bt) \+ \frac{s_1}{24} \vv(\bt) \,.\label{Hodgegenus1}
\end{align}
For $g\geq 2,$ there exist elements 
\[\Hod_g(z_1,\dots,z_{3g-2}; s_1,s_3,\dots,s_{2g-1})\in \CC\bigl[z_1,\dots,z_{3g-2},z_1^{-1}; s_1,s_3,\dots,s_{2g-1}\bigr]\]
satisfing the conditions
\begin{align} 
& \sum_{m=1}^{3g-2} m \, z_m \, \frac{\p \Hod_g}{\p z_m} \= (2g-2)\,\Hod_g \, , \label{homog1}\\
& \sum_{m=2}^{3g-2} (m-1) \, z_m \, \frac{\p \Hod_g}{\p z_m} \+ \sum_{j=1}^g (2j-1) \, s_{2j-1} \, \frac{\p \Hod_g}{\p s_{2j-1}} 
  \= (3g-3)\, \Hod_g \, , \label{homog2}
\end{align}
such that 
\begin{equation}
\HHod_g(\bt; \bs) \= \Hod_g \bigl( \vv_1(\bt),\dots,\vv_{3g-2}(\bt);s_1,s_3,\dots,s_{2g-1}\bigr) \,. \label{generalhodge}
\end{equation}}

\noindent This theorem was proved in~\cite{DLYZ1}; see also~\cite{DuY} for a straightforward proof.

Define 
\beq\label{defineu}
u \= u(\bt;\bs;\e) \:= \e^2 \frac{\p^2 \H(\bt;\bs;\e)}{\p t_0^2} \,,
\eeq
then according to~\cite{DLYZ1}, $u$ satisfies an integrable hierarchy of tau-symmetric Hamiltonian evolutionary 
PDEs, called the Hodge hierarchy,  
which is a deformation of the KdV hierarchy~\cite{Witten,Kontsevich} and has the form
\beq
\frac{\p u}{\p t_k} \= P \frac{\delta \bar h_k}{\delta u(x)} \,, \quad k\geq 0.
\eeq
Here $P=\p_x \+ \cdots$ is a Hamiltonian operator, $\bar h_k$, $k\geq 0$ are Hamiltonians.

In~\cite{DYZ} Theorem~A was applied under a particular specialization of $\bt,\bs$, which  
gives the classical Hurwitz numbers according to the ELSV formula. In this paper, we consider a {\it different} specialization.
Firstly, we specialize~${\bf s}$ to ${\bf s}={\bf s}^*$ as follows:
\beq
s_{2k-1}^* \:= (2k-2)! \, s^{2k-1}, \quad k\geq 1 \,.
\eeq
Denote by $\Lambda_{g,k}(s):=\sum_{j=0}^g \lambda_j \, s^j$ the Chern polynomial of~$\mathbb{E}_{g,k}$. Applying the relationship between 
the Chern classes and the Chern character, and 
using Mumford's relations~\cite{Mumford} \[{\rm ch}_{2m}(\mathbb{E}_{g,k})\=0\,, \quad m\geq 1\,,\] 
we obtain $\Omega_{g,k}(\bs=\bs^*) =  \Lambda_{g,k}(s)$. So we have 
\beq\label{linearhodgepotential}
\HHod_g(\bt;\bs^*) \= \sum_{n\geq 0} \sum_{\,i_1, \dots, i_n\geq 0} \frac{t_{i_1}\cdots t_{i_n}}{n!} \int_{\overline{\mathcal M}_{g,n}} 
 \Lambda_{g,n}(s) \, \psi_1^{i_1} \cdots \psi_n^{i_n} \,.
\eeq
Secondly, we specialize $\bt$ to $\bt=\bt^*$ given by
\beq\label{specialvalue}
t_0^* \= x \,, \quad t_1^*  \= 0 \,,  \quad t_2^* \= 1\,, \quad t_i^* \= 0~(i\geq 3) \,.
\eeq
Substituting~\eqref{specialvalue} into~\eqref{linearhodgepotential} we arrive at
\beq\label{Hodgextj}
\HHod_g(\bt^*;\bs^*) \=   \sum_{n_0\geq0} 
\frac{x^{n_0}}{n_0!} \, \sum_{j=0}^g s^j
 \frac{\langle \lambda_j  \, \tau_0^{n_0} \, \tau_2^{3g-3+n_0-j} \rangle_g}{ (3g-3+n_0-j)!} \,.
\eeq
From the definition of~$a_{g,n}$ given in~\eqref{definisionagn}, it follows that 
the MV free energy is a specialized linear Hodge free energy. More precisely, we have the following lemma.
\begin{lemma} \label{twofreeenergiesequal}
For any $g\geq 0$, the following identities hold:
\beq
\H_g(x) \= \HHod_g(\bt^*;\bs^*)|_{s=1} \,,
\eeq
where $\H_g(x)$ is the $g^{\rm th}$ part of the MV free energy~\eqref{Hgenusexpansion}.
Equivalently, we have
\beq\label{MVhodge}
{\rm Vol} \, \Q_{g,n} 
\= 2^{2g+1} \frac{\pi^{6g-6+2n} (4g-4+n)!}{(6g-7+2n)!} \,\p_x^n\bigl(\HHod_g(\bt^*;\bs^*)\bigr)\big|_{x=0,s=1} \,.
\eeq

\end{lemma}

Let us now apply Theorem~A to the computation of~$\HHod_g(\bt^*;\bs^*)$, which, due to~\eqref{MVhodge}, 
gives rise to~${\rm Vol} \, \Q_{g,n}$.
Substituting~\eqref{specialvalue} into~\eqref{hodo} we find that $v=v(\bt^*)$ satisfies the following quadratic equation
\beq
x \+ \frac{v^2}2  \= v\,.
\eeq
By solving this and observing that the power series~$v$ starts with~$x$, 
we obtain 
\[v(\bt^*)= 1-\sqrt{1-2 x}\,.\]
Denote 
\beq
T \:= \sqrt{1-2x} \, .
\eeq
Then by noticing $ \p_x \= -\frac1T \p_T  $
we find 
\beq\label{vmam}
v_m (\bt^*) \=  \frac{(2m-3)!!}{T^{2m-1}} \+ \delta_{m,0} \,, \quad m\geq 0\,.
\eeq

\begin{lemma} \label{jetsstructure} 
The power series $\HHod_g(\bt^*;\bs^*)$ of $x,t$ 
are given explicitly for $g=0,1,2$ by
\begin{align}
\HHod_0(\bt^*;\bs^*) &\= 
\frac1{40}  \,-\, \frac{T^2}{12} \+ \frac{T^4}{8} \,-\, \frac{T^5}{15} \,, \label{h0hodges} \\
\HHod_1(\bt^*;\bs^*) & \= \frac1{24} \log \frac1T \+ \frac{s}{24}  \,(1-T) \,, \label{h1hodges} \\
\HHod_2(\bt^*;\bs^*) & \= \frac{7}{1440} \, \frac{1}{T^5} \+ \frac{5}{1152} \, \frac{s}{T^4} \+ \frac{7}{5760} \, \frac{s^2}{T^3}  \,. \label{h2hodges}
\end{align}
In general, for $g\geq 2$, $\HHod_g(\bt^*;\bs^*)$ has the following expression:
\beq\label{ggeq2hodges}
\HHod_g(\bt^*;\bs^*) \= 
\sum_{j=0}^{g}  \frac{\langle \lambda_j \tau_2^{3g-3-j} \rangle_g}{(3g-3-j)!} \frac{s^j}{T^{5g-5-j}}\,, \quad g\geq 2\,.
\eeq
\end{lemma}

\begin{proof}
By substituting~\eqref{vmam} into~\eqref{Hodgegenus0} and \eqref{Hodgegenus1}, we arrive at the
formulas for $\HHod_0(\bt^*;\bs^*)$ and $\HHod_1(\bt^*;\bs^*)$, respectively.
The formula for $\HHod_2(\bt^*;\bs^*)$ can be obtained by using the algorithm of~\cite{DLYZ1} with $v_m(\bt^*)$ given by~\eqref{vmam}.
To show the validity of the formula for $\HHod_g(\bt^*;\bs^*)$, $g\geq 2$, we first observe that, according to \eqref{generalhodge}, \eqref{vmam} 
and the homogeneity conditions \eqref{homog1}, \eqref{homog2}, the function $\HHod_g(\bt^*;\bs^*)$ 
can be written in the form
\beq\label{ggeq2hodgeform}
\HHod_g(\bt^*;\bs^*) \= \sum_{j=0}^{g}  \frac{C_{g,j} s^j}{T^{5g-5-j}} \,, \quad g\geq 2\,,
\eeq
where $C_{g,j}\in\QQ$.
Therefore,
\[
\HHod_g(\bt^*;\bs^*)|_{x=0} \= \sum_{j=0}^{g}  C_{g,j} s^j \, , \quad g\geq 2 \,.
\]
On the other hand, it follows from~\eqref{Hodgextj} that 
\[\HHod_g(\bt^*;\bs^*)|_{x=0} \= \sum_{j=0}^g  
\frac{\langle \lambda_j \tau_2^{3g-3-j} \rangle_g}{(3g-3-j)!} s^j \,.\]
By comparing the coefficients of $s^j$ in the two formulas given above we arrive at
\beq
C_{g,j} \= \frac{\langle \lambda_j \tau_2^{3g-3-j} \rangle_g}{(3g-3-j)!}\,,\quad j=0,\dots,g\,,
\eeq
where $g\ge 2$. The lemma is proved.
\end{proof}

\smallskip

\begin{proof}[Proof of Lemma~\ref{Txstructure}]
By putting $s=1$ in Lemma~\ref{jetsstructure}, we arrive at the result of Lemma~\ref{Txstructure}.
\end{proof}

Now let us give a proof of Proposition~\ref{prp1lem} based on Lemma~\ref{Txstructure}.

\begin{proof}[Proof of Proposition~\ref{prp1lem}]
By using~\eqref{h0hodgeT} and the fact that $\frac{d}{dx} = -\frac1T \frac{d}{dT}$ we have
\begin{align}
& \HHod_0'(x) \= \frac16 - \frac{T^2}2  + \frac{T^3}3 \,, 
\quad \HHod_0''(x) \= v(\bt^*) \,,\\
& \frac{d^n\HHod_0(x)}{dx^n}  \= v_{n-2}(\bt^*) \=  \frac{ (2n-7)!!}{T^{2n-5}} \,, \quad n\geq 3\,.
\end{align}
Therefore,
$
\frac{d^n \HHod_0(x) }{dx^n}\big|_{x=0} = 
(2n-7)!! \, \delta_{n\geq 3}$.
Due to the definition~\eqref{defintionofHxe} and the CMS formula this gives~\eqref{thm0n}. 
Similarly, by using~\eqref{h1hodgeT} we obtain
\beq
\frac{d^n\HHod_1(x)}{dx^n} \= 
\frac{\delta_{n\geq1}}{24} \frac{2^{n-1}(n-1)!}{T^{2n}} \+\frac{\delta_{n,0}}{24} \log \frac1T \+ \frac{1}{24} \frac{(2n-3)!! }{T^{2n-1}} \+ \frac{\delta_{n,0}}{24} \,,
\eeq
from which we arrive at~\eqref{thm1n}.
Finally, by using~\eqref{ggeq2hodgeT} we have for $g\geq 2$,
\beq
\frac{d^n\HHod_g(x)}{dx^n} \= 
\sum_{j=0}^{g}\frac{\langle \lambda_j \tau_2^{3g-3-j} \rangle_g}{(3g-3-j)!} \frac{\prod_{i=0}^{n-1}(5g-5-j+2i)}{T^{5g-5-j+2n}} \, ,
\eeq
which yields formula~\eqref{thmgn}. 
Proposition~\ref{prp1lem} is proved. 
\end{proof}

\begin{remark}
The explicit expressions of the numbers 
$\langle \lambda_g \tau_2^{2g-3} \rangle_g$ that appear in~\eqref{thmgn} of Proposition~\ref{prp1lem} are given by the following
$\lambda_g$-conjecture proven in~\cite{FP,ELSV}:
\beq
\frac{\langle \lambda_g \tau_2^{2g-3} \rangle_g}{(2g-3)!} \= 
\frac{2^{2g-1}-1}{2^{2g-1}} (4g-7)!! \, \frac{|B_{2g}|}{(2g)!} \,, \qquad g\geq 2\,,
\eeq
where $B_k$ denotes the $k_{\rm th}$-Bernoulli number. The number 
$\langle \tau_2^{3g-3} \rangle_g$ for $g\geq 2$
has the expression~\cite{IZ}:
\beq\label{IZformula}
\frac{\langle \tau_2^{3g-3} \rangle_g}{(3g-3)!} \= \frac{24^{-g} c_g}{(5g-3)(5g-5)} \,,
\eeq
where  $c_g$ are given by the recursion
\beq\label{cgdef}
c_g \=  50\,(g-1)^2\,c_{g-1} + \frac12 \sum_{h=2}^{g-2} c_h \, c_{g-h},\quad  g\geq 3
\eeq
together with 
$c_0=-1, c_1=2, c_2=98$.
\end{remark}

Proposition~\ref{prp1lem} and formula~\eqref{IZformula} imply immediately 
the following corollary.
\begin{cor}\label{cor3}
For any fixed $g\geq 0$, the following asymptotic formula is true:
\beq\label{largen}
{\rm Vol} \, \Q_{g,n} \;\sim\;  \kappa_g \, \frac{n^{\frac{g}2} \pi^{2n}}{2^n} \quad (n\rightarrow \infty) \,,
\eeq
where 
\beq\label{kappag}
 \kappa_g \= \frac{64\,\pi^{6g-\frac{11}2}}{384^g\,\Gamma(\frac{5g-1}2)} \, c_g \,,  
\eeq
and $c_g$ are defined by \eqref{cgdef}.
\end{cor}
\noindent The reader may notice that certain universality 
found in~\cite{DYZ} about asymptotics of enumerations 
related to~$\overline{\mathcal{M}}_{g,n}$ 
reappears in~\eqref{largen}, \eqref{kappag}. 
The first few~$\kappa_g$ are given by 
$\kappa_0 = 32/\pi^6$, $\kappa_1 = \pi^{\frac12}/3$,  $\kappa_2 = 7 \pi^6/1080$, 
$\kappa_3 = 245 \pi^{25/2}/7962624$.

\section{Relations for the MV volumes}\label{section3}
The goal of this section is to prove Theorem~\ref{thmequation} and Corollary~\ref{Kazdeduced}. 

\begin{proof}[Proof of Theorem~\ref{thmequation}]
It was shown by Buryak~\cite{Buryak} that the Hodge hierarchy associated with~$\Lambda(s)$ 
 is normal Miura equivalent \cite{DZ-norm,DLYZ1} 
to the intermediate long wave (ILW) hierarchy. 
To be precise, define~$\tilde u=\tilde u(\bt;s;\e)$ by 
\beq\label{Miuratimes}
\tilde u(\bt;s;\e) \:= \sum_{g=0}^\infty \e^{2g}  \frac{(-1)^g s^g}{2^{2g} (2g+1)! } \frac{\p^{2g} u}{\p t_0^{2g}},
\eeq
where $u$ is defined in~\eqref{defineu} with the specialization $\bs=\bs^*$; then $\tilde u$
satisfies~\cite{Buryak} the ILW hierarchy, which has the first two flows
\begin{align}
& \tilde u_{t_1} \= \tilde u \frac{\p \tilde u}{\p t_0}  
\+ \sum_{g\geq1} \frac{|B_{2g}|}{(2g)!} \e^{2g} s^{g-1} \frac{\p^{2g+1} \tilde u}{\p t_0^{2g+1}}\,, \label{ilw1}\\
& \tilde u_{t_2}\= \frac 1 2 \tilde u^2 \frac{\p \tilde u}{\p t_0} \+ \sum_{g\geq 1} \frac{|B_{2g}|}{(2g)!} \e^{2g} 
\frac{s^{g-1}}{4}\biggl(2 \p_{t_0} \Bigl(\tilde u\frac{\p^{2g} \tilde u}{\p t_0^{2g}}\Bigr) + \frac{\p^{2g+1} (\tilde u^2)}{\p t_0^{2g+1}} \biggr) \nn\\
& \qquad \qquad \+ \sum_{g\geq 2} \frac{|B_{2g}|}{(2g)!} (g+1) \e^{2g} s^{g-2} \frac{\p^{2g+1} \tilde u}{\p t_0^{2g+1}} \, .\label{ilw2-zh}
\end{align}

Let us now do the specialization~\eqref{specialvalue} with $s=1$, and denote the series $u(\bt^*;\bs^*;\e)|_{s=1}$, $\tilde u(\bt^*;s;\e)|_{s=1}$ by
$u=u(x,\e)$, $\tilde u=\tilde u(x,\e)$, 
respectively. Then $u(x,\e) =\e^2\p_x^2(\H(x,\e))$, and 
from~\eqref{Miuratimes} it follows that 
$\tilde u(x,\e)$ and $u(x,\e)$ are related by
\beq\label{utu}
 \tilde u \= \sum_{g=0}^\infty \e^{2g}  \frac{(-1)^g}{2^{2g} (2g+1)! } \frac{\p^{2g} u}{\p x^{2g}} \,.
\eeq

\begin{prop} \label{BuryakILW}
The series~$\tilde u = \tilde u(x,\e)$ satisfies the following non-linear equation:
\beq\label{EL}
x \+ \frac{\tilde u^2}{2} \+ \sum_{g=1}^\infty \e^{2g} \frac{|B_{2g}|}{(2g)!} \frac{\p^{2g} \tilde u}{\p x^{2g}} \= \tilde u \,.
\eeq
\end{prop}
\begin{proof}
Recall that the Hodge partition function $Z=Z(\bt;\bs;\e):=e^{\H(\bt;\bs;\e)}$ satisfies the 
string equation (cf.~e.g.~\cite{DLYZ1,DuY}), that is,
\beq\label{stringeqZ}
\sum_{i=0} t_{i+1} \frac{\p Z}{\p t_i} \+ \frac{t_0^2}{2\epsilon^2} Z \+ \frac{s_1}{24} Z \= \frac{\p Z}{\p t_0} \,. 
\eeq
Dividing both sides of~\eqref{stringeqZ} by~$Z$ 
and differentiating with respect to~$x$ we obtain 
\beq\label{stringOmega}
\sum_{i=0}^\infty t_{i+1} \frac{\p^2 \H(\bt;\bs;\e)}{\p t_i\p x} \+ \frac{x}{\epsilon^2}  \= \frac{\p^2 \H(\bt;\bs;\e)}{\p x^2} \,. 
\eeq
We recall that 
\beq\label{twopointfunctions}
\e^2\frac{\p^2 \H(\bt;\bs;\e)}{\p t_i\p x} \= \Omega_{i,0}\bigl(u(\bt;\bs;\e),u_x(\bt;\bs;\e),\dots\bigr)\,,\quad i\geq0\,, 
\eeq
where $\Omega_{i,0}$ are certain differential polynomials~\cite{Buryak,DLYZ1} of~$u$. 
Then by using the Miura transformation~\eqref{Miuratimes} we obtain 
\beq\label{stringOmegatilde}
\sum_{i=0}^\infty t_{i+1} \widetilde \Omega_{i,0}\bigl(\tilde u(\bt;\bs;\e), \tilde u_x(\bt;\bs;\e),\dots\bigr) \+ x  
\= \tilde u(\bt;\bs;\e)\,. 
\eeq
Here $\widetilde\Omega_{i,0}$, $i\geq 0$ are differential polynomials of~$\tilde u$. 
Buryak~\cite{Buryak} showed that the Miura transformation~\eqref{Miuratimes} 
transforms the Hamiltonian structure~$P$
of the linear Hodge hierarchy to $\p_x$, in particular, $\tilde u(\bt;\bs;\e)$ satisfies the Hamiltonian system
\beq
 \frac{\p \tilde u}{\p t_1} \= \p_x \frac{\delta \tilde {\bar h}_1}{\delta \tilde u(x)}   \,,
\eeq 
where
\[ \tilde {\bar h}_1 \=  \int \biggl(\frac{\tilde{u}^3}{6} \+ \sum_{g=1}^\infty \frac{|B_{2g}|}{2(2g)!} \tilde u \tilde u_{2g} \biggr) dx\,. \]
Therefore, according to~\cite{DLYZ1} we know that 
\beq\label{Omega10}
\widetilde \Omega_{1,0} \= \frac{\delta \tilde {\bar h}_1}{\delta \tilde u(x)} \=  \frac{\tilde u^2}{2} \+ \sum_{g=1}^\infty \e^{2g} \frac{|B_{2g}|}{(2g)!} \frac{\p^{2g} \tilde u}{\p x^{2g}}  \,.
\eeq
Thus equation~\eqref{stringOmegatilde} with the specialization $s=1$
leads to~\eqref{EL}. The proposition is proved.
\end{proof}

We are in a position of proving equation~\eqref{theequation}.
Indeed, observe that 
\beq
\sum_{g\geq 1} \e^{2g} \frac{|B_{2g}|}{(2g)!} \p_x^{2g}  \= 1 \,-\,\frac{ i}2 \e \, \p_x  \,-\, \frac{i\e\p_x}{e^{i \e \p_x}-1},
\eeq
so it follows from ~\eqref{EL} that
\beq
x \+ \frac{\tilde u^2}{2} \,-\,\frac{ i}2 \e \, \p_x (\tilde u)  \,-\, \frac{i\e\p_x}{e^{i \e \p_x}-1} (\tilde u)  \= 0 \,.
\eeq
By using the fact that $\tilde u = -i \e \p_x (\H_+-\H_-)$ we arrive at equation~\eqref{theequation}. 

We will now prove equation~\eqref{anotherequation}.
We first switch on the $t_2$-dependence and denote it by~$t$ 
in the specialization~\eqref{specialvalue}. 
More precisely, we consider 
\begin{align}
& \H \= \H(x,t,\e) \:=  \sum_{g,n\geq 0} 
\sum_{j=0}^g \frac{\langle\lambda_j \tau_0^n \tau_2^{3g-3+n-j}\rangle}{(3g-3+n-j)!} \, \e^{2g-2} \frac{x^n}{n!} t^{3g-3+n-j} \, ,
\end{align}
and denote $\H_{\pm} := \H\bigl(x\pm\frac{i \e}{2},t,\e\bigr)$. 
Then by using equation~\eqref{ilw1} and an argument like the one we used above to derive  
equation~\eqref{theequation}, we find that~$\H$ satisfies the following equation:
\begin{align}
&t\frac{\e^2}2 \Bigl[\p_x(\H_+\,-\, \H_-)\Bigr]^2 \+ t\frac{\e^2}2 \p_x^2 \bigl( \H_+ + \H_-\bigr) \,-\, (1-t) \, i \,\e \, \p_x(\H_+\,-\, \H_-) \= x \,.\label{H21} 
\end{align}
Then by using equations~\eqref{ilw2-zh} and~\eqref{H21} we obtain the following equation for~$\H$:
\begin{align}
& -i \,\e\, \p_t \, (\H_+\,-\, \H_-) \= \frac16 \tu^3 + \frac34 \tu^2 + \tu - \frac{i\e}2 \tu \tu_x -\frac{3i\e}4 \tu_x - \frac{\e^2}6 \tu_{xx} 
 \+ \frac{x}{2t} - \frac{i\e}{4t}  \nn\\
& \qquad \qquad \qquad \qquad \qquad \quad - \frac{1+2t}{2t} \e^2 \p_x^2(\H_-) + \frac{i\e^3}{4} \p_x^3(\H_-) - \frac{\e^2}2  \tu \, \p_x^2(\H_-) \,. \label{H22}
\end{align}
Here we recall that $\tilde u=-i\e\p_x(\H_+\,-\, \H_-)$, and we also used Theorem~A to get the constant in~$x$ term $- i\e/4t $.
It is not difficult to deduce from Theorem~A the following homogeneity property for~$\H$:
\beq\label{homogenH}
t \frac{\p \H}{\p t} \+ \Bigl(x-\frac1t\Bigr) \frac{\p \H}{\p x} \+ \e \frac{\p \H}{\p \e} \= - \frac1{24} - \frac1{24 t} -\frac{x^2}{2 \e^2 t} \,.
\eeq
From the above equations \eqref{H21}--\eqref{homogenH} we arrive at equation~\eqref{anotherequation}. 
The theorem is proved.
\end{proof}

Let us proceed to prove Corollary~\ref{Kazdeduced}.
\begin{proof}[Proof of Corollary~\ref{Kazdeduced}] 
Differentiating equation~\eqref{anotherequation} with respect to~$x$ we obtain 
\[
\biggl(\e \p_x \p_\e  + \frac12 \p_x + \frac12x\p_x^2 - \frac{\e^2}{24} \p_x^4\biggr)  \bigl(\H_{+} - \H_{-}\bigr)    \+ 
\frac{\e^2}{4} \bigl[\p_x(\H_{+}-\H_{-})\bigr]^2  \bigl[\p_{x}^2(\H_{+}-\H_{-})\bigr]  \=  0 \,,\]
so from equation~\eqref{theequation} it follows that 
\[
\biggl(\e \p_\e  + \frac12  + x\p_x - \frac{\e^2}{24} \p_x^3\biggr) \circ \p_x \bigl(\H_{+} - \H_{-}\bigr)    \,-\,
\frac{\e^2}{4} \Bigl[ \bigl(\p_x^2 ( \H_+)\bigr)^2 - \bigl(\p_{x}^2(\H_{-})\bigr)^2\Bigr]  \=  0   \,. \]
Observing that $\bigl[ x  \p_x,e^{\pm i\e\p_x/2}\bigr]= \mp \frac{i\e}2 e^{\pm i\e\p_x/2} \p_x $ one can simplify this equation and find
\beq\label{equiv1kaz}
\Bigl(e^{\frac{i\e\p_x}2}-e^{-\frac{i\e\p_x}2}\Bigr)  
\Biggl[\biggl(\e \p_\e  + \frac12  + x\p_x - \frac{\e^2}{24} \p_x^3\biggr) \circ \p_x \bigl(\H \bigr)    \,-\,
\frac{\e^2}{4} \bigl[\p_x^2 ( \H )\bigr]^2 \Biggr]  \=  0   \,. 
\eeq
Since the operator $(e^{\frac{i\e\p_x}2}-e^{-\frac{i\e\p_x}2}) / \p_x$ 
is invertible on power series of~$x$, 
we find that equation~\eqref{equiv1kaz} is equivalent to
\beq\label{equiv2kaz}
\p_x  \Biggl[\biggl(\e \p_\e  + \frac12  + x\p_x - \frac{\e^2}{24} \p_x^3\biggr) \circ \p_x \bigl(\H \bigr)    \,-\,
\frac{\e^2}{4} \bigl[\p_x^2 ( \H )\bigr]^2 \Biggr]  \=  0   \,. 
\eeq
It follows that 
\[\biggl(\e \p_\e  + \frac12  + x\p_x - \frac{\e^2}{24} \p_x^3\biggr) \circ \p_x \bigl(\H \bigr)    \,-\,
\frac{\e^2}{4} \bigl[\p_x^2 ( \H )\bigr]^2 \= C(\e)\,,\]
where $C(\e)=\sum_{g\geq0} \e^{2g-2} C_g$ with $C_g$ being constants.  It remains to show that $C_g$ all vanish. Indeed, 
for $g=0$ and $g=1$, this can be verified directly with the explicit expressions of~$\H_0$ and~$\H_1$ given in Lemma~\ref{Txstructure}.
For $g\geq2$, by using Lemma~\ref{Txstructure} and the fact that $\p_x = -\frac{1}{T}\p_T$ we arrive at $C_g=0$.
The  corollary is proved.
\end{proof}

Let us now show that Corollary~\ref{Kazdeduced} implies Kazarian's recursion 
on the linear Hodge integrals    
  $\frac{(5g-3-j)(5g-5-j)}{(3g-3-j)!}\int_{\overline{\mathcal{M}}_{g,3g-3-j}} 
\lambda_j  \psi_{1}^2 \cdots \psi_{3g-3-j}^2$.
Indeed, differentiating~\eqref{deducedfrom} with respect to~$x$ we find that 
the series~$u=\e^2 \p_x^2(\H)$ satisfies the equation
\beq\label{equationforumay1}
2\e u_{\e} \+ 2x u_x \,-\, u  \=  \p_x \Bigl(\frac12 u^2\Bigr)\+\frac1{12}\e^2 u_{xxx} \,.
\eeq
Denote 
\beq\label{genusexpansionsuandutilde}
 u(x,\e) =: \; \sum_{g\geq 0}\e^{2g} u^{[g]}(x)\,.
 \eeq
Then we can write~\eqref{equationforumay1} equivalently as follows:
\beq\label{newrecursionforu}
\Bigl(4g-1+2x\frac{d}{dx}\Bigr) \bigl(u^{[g]}\bigr) \=  \frac12 \frac{d}{dx} \biggl(\sum_{g_1+g_2=g} u^{[g_1]}u^{[g_2]}\biggr)
 \+ \frac1{12} \frac{d^3}{dx^3} \bigl(u^{[g-1]}\bigr) \,,\quad g\geq 0\,.
\eeq
To proceed we note that it follows easily from Lemma~\ref{Txstructure} that $u^{[g]}(x)$ has the expression 
\begin{align}
& u^{[0]} = 1 - T \,, \quad  
u^{[1]} \= \frac{1}{12} \frac{1}{T^{4}} \+ \frac{1}{24} \frac{1}{T^{3}}  \,, \label{u0u1}\\
& u^{[g]} \= \sum_{j=0}^{g} \frac{\langle \lambda_j \tau_2^{3g-3-j} \rangle_g}{(3g-3-j)!} 
\frac{\prod_{i=0}^{1}(5g-5-j+2i)}{T^{5g-1-j}} \,, \quad g\geq 2\,. \label{expandug}
\end{align}
Thus using the 
fact that $\frac{d}{dx}= -\frac1T \frac{d}{dT}=:D_T$ we find that~\eqref{newrecursionforu} is equivalent to 
\beq\label{dtuu}
\bigl(4g-1+(1-T^2)D_T\bigr) \bigl(u^{[g]}\bigr) \=  \frac12 D_T \biggl(\sum_{g_1,g_2\geq 0 \atop g_1+g_2=g} u^{[g_1]}u^{[g_2]}\biggr) \+ \frac1{12} D_T^3 \bigl(u^{[g-1]}\bigr)\,.
\eeq
Substituting~\eqref{u0u1}, \eqref{expandug} into~\eqref{dtuu} we find 
\begin{align}
c_{g,j} \= & \frac{g+1-k}{5g-2-j}\, c_{g,j-1} \+ \frac{(5g-6-j)(5g-4-j)}{12} \, c_{g-1,j}\notag \\ 
& \+ \frac12 \sum_{g_1,g_2\geq 1, j_1,j_2\geq 0 \atop g_1+g_2=g,  j_1+j_2=j} c_{g_1,j_1} c_{g_2,j_2} \,,\qquad g\geq1,\, 0\leq j\leq g\,,\label{kaz}
\end{align}
where the numbers $c_{g,j}$ are defined by 
\beq
c_{g,j} \:=  \frac{\langle \lambda_j \tau_2^{3g-3-j} \rangle_g}{(3g-3-j)!}  \prod_{i=0}^{1}(5g-5-j+2i) \,.
\eeq
The recursion relations~\eqref{kaz} for $c_{g,j}$ were obtained by Kazarian~\cite{Kazarian} 
from the KP hierarchy~\cite{Kazarian1} satisfied by the linear Hodge integrals.

It is not clear at the moment whether Corollary~\ref{Kazdeduced} and Lemma~\ref{Txstructure} imply Theorem~\ref{thmequation}.

We end this section with two remarks on the computational aspects.
Firstly, as a consequence of equation~\eqref{theequation} and Lemma~\ref{Txstructure}, 
the~$u^{[g]}$ can be computed from the recursion 
\begin{align}
& u^{[0]} \= 1-T \,, \nn\\
& u^{[g]} \= \frac1{2T} \sum_{0\leq g_1,g_2\leq g-1 \atop g_1+g_2+j_1+j_2=g} \Bigl(-\frac14\Bigr)^{j_1+j_2}
\frac{D_T^{2j_1}\bigl(u^{[g_1]}\bigr) D_T^{2j_2}\bigl(u^{[g_2]}\bigr)}{(2j_1+1)!(2j_2+1)!} 
 \,-\, \frac1T \sum_{j=1}^g \Bigl(-\frac14\Bigr)^{j} \frac{D_T^{2j}\bigl(u^{[g-j]}\bigr)}{(2j)!}  \,, \nn
\end{align}
where $g\geq 1$. 
Then one can further compute $\H_g$, $g\geq 2$ from $u^{[g]}$ via
\beq
\HHod_g \= \sum_{j=0}^{g}  \frac{C_{g,j}}{T^{5g-5-j}} \,, \qquad 
C_{g,j} \= \frac{{\rm coefficient~of}~1/T^{5g-1-j}~{\rm in}~ u^{[g]}}{(5g-3-j)(5g-5-j)}  \quad (0\leq j\leq g)\,.
\eeq
Secondly, the series $\tilde u$ (see~\eqref{utu}) also presents good properties. Denote 
\beq\label{genusexpansionsutilde} 
\tilde u(x,\e) =: \; \sum_{g\geq 0}\e^{2g} \tilde u^{[g]}(x) \,. 
\eeq
If then follows from \eqref{u0u1}, \eqref{expandug}
that $\tilde u^{[g]}$ has the expression
\begin{align}
& \tilde u^{[0]} = 1 - T \,, \quad \tilde u^{[1]} \= \frac{1}{12 T^4} \,, \quad \tilde u^{[g]} \=  \sum_{j=0}^g  \frac{d_{g,j}}{T^{5g-1-j}} ~ (g\geq2)\,, \label{expandtu}
\end{align}
where $d_{g,j}\in \QQ$ are constants. In terms of intersection numbers we have for $g\ge 2$,
\begin{align}
& \tilde u^{[g]} \= 
\sum_{g_1=0}^{g-2}  \frac{(-1)^{g_1}}{2^{2g_1} (2g_1+1)! }  \sum_{j=0}^{g-g_1} \frac{\langle \lambda_j \tau_2^{3g-3g_1-3-j} \rangle_{g-g_1}}{(3g-3g_1-3-j)!} \frac{\prod_{i=0}^{1+2g_1}(5g-5g_1-5-j+2i)}{T^{5g-g_1-1-j}} \nn\\
&\qquad \quad   \+\frac{(-1)^{g-1}2^{2g}}{12}   \frac1{T^{4g}} \+ \frac{(-1)^{g} (4g-3)!!}{2^{2g} (2g+1)! }  \, \frac{5-2g}{6}\, \frac{1}{T^{4g-1}} \,.\nn
\end{align}
Substituting~\eqref{genusexpansionsutilde} 
into~\eqref{EL} we find that  $\tilde u^{[g]}$, $g \geq 0$ satisfy the following recursion
\begin{align}
& \tilde u^{[0]} \= 1 - T \,, \label{polyalg1}\\
& \tilde u^{[g]} \= \frac1{2T} \sum_{g_1=1}^{g-1}  \tilde u^{[g_1]}\tilde u^{[g-g_1]} \+ 
\frac1{T}\sum_{g_1=1}^{g} \frac{|B_{2g_1}|}{(2g_1)!} D_T^{2g_1} \bigl(\tilde u^{[g-g_1]}\bigr) \,, \quad g\geq 1\,. \label{polyalg2}
\end{align}
This recursion gives 
an algorithm for computing~$\tilde u$. 
From~\eqref{utu} we know that 
\[
u \=  \tilde u \+ \sum_{g\geq 1} \e^{2g} \frac{2^{2g-1}-1}{2^{2g-1}}  \frac{|B_{2g}|}{(2g)!}  D_T^{2g} (\tilde u) \,.
\]
Therefore, for $g\geq 0$, 
\[u^{[g]} \=\tilde u^{[g]} \+ \sum_{g_1=1}^g \frac{2^{2g_1-1}-1}{2^{2g_1-1}} \frac{|B_{2g_1}|}{(2g_1)!} 
D_T^{2g_1} \bigl( \tilde u^{[g-g_1]}\bigr) \,. \]
So this gives rise to another algorithm for computing the MV volumes. 
One could also use~\eqref{anotherequation} to study~$\tilde u$.

\section{Asymptotics of the area Siegel--Veech constants}\label{section4}
In this section we use Goujard's formula to compute the area Siegel--Veech (SV) constants 
associated with principal strata of moduli spaces of quadratic differetials. 
Indeed, according to Goujard~\cite{Goujard} the area 
SV constants can be expressed explicitly in terms of 
the number~$a_{g,n}$ as follows:
\beq\label{Goujardformula}
C_{\rm area} (\mathcal{Q}_{g,n}) \= \frac{\pi^{-2}}{4a_{g,n}} 
\Biggl(n(n-1)a_{g,n-1} \+  a_{g-1,n+2}  \; + \!\! \sum_{g_1,g_2\ge 0, \, n_1,n_2 \ge 1 \atop {g_1+g_2=g, \, n_1+n_2=n+2\atop 3g_i-3+n_i > 0 \, (i=1,2) }} 
\binom{n}{n_1-1} a_{g_1,n_1}a_{g_2,n_2} 
\Biggr) \,.
\eeq 
The result in this section is a refinement 
of the conjectural formula for the large~$g$ asymptotics 
of $C_{\rm area} (\mathcal{Q}_{g,n})$ given in~\cite{DGZZ,ADGZZ} to the following more precise asymptotic statement.

\begin{conj}\label{conjecture2}
For any fixed $n\geq 0$, we have the asymptotic formula
\beq
C_{\rm area} (\mathcal{Q}_{g,n}) \;\sim\;  \sum_{k=0}^\infty \frac{C_k(n)}{g^k} \,,\qquad g\to\infty \,,
\label{asvclargeg} 
\eeq
where each $C_k(n)$ is a polynomial with rational coefficients in~$n$ and $M=-\pi^2/144$, 
with the first four of them being 
\begin{align}
C_0(n) &\= \frac14\,, \quad C_1 (n) \= \frac1{48}n^2\,-\, \frac3{16} n \+ \frac{1-2 M}{4} \,, \nn\\
C_2 (n) & \= - \frac{5+12 M}{576} \,n^3 
\+ \frac{59+180 M}{576}\,n^2 
 \,-\, \frac{11+24 M-72 M^2}{32}  \,n 
 \+ \frac{23+15 M-648 M^2}{72} \,,  \nn \\ 
C_3 (n) &\= 
\frac{4+17M+54 M^2}{1152}\, n^4 \,-\, 
\frac{179+978 M+3564 M^2}{3456} \, n^3 \nn\\
 & \qquad \+ \frac{929+5169 M+13554 M^2-42768 M^3}{3456} \, n^2  \nn\\ 
 & \qquad \,-\, \frac{989+4851 M -4428 M^2 - 192456 M^3}{1728} \,  n \nn\\
 & \qquad \+  
 \frac{295+1165 M-16140 M^2-105300 M^3+253692 M^4}{720} \,.\nn 
\end{align}
\end{conj}

The asymptotic formula~\eqref{asvclargeg} with $\sum_{k=0}^\infty C_k(n)/g^k$ replaced by~$1/4$ 
becomes the ADGZZ conjecture for the area SV constants.  As we mentioned in the Introduction, the 
above Conjecture~\ref{conjecture2} is also not based on theoretical reasoning but on numerical computations. 
Very recently Aggarwal~\cite{A3} proved the ADGZZ conjecture for the area SV constants by showing that   
the leading term asymptotics in~\eqref{mvlargeg} implies the leading term asymptotics in~\eqref{asvclargeg} 
with the knowledge of Goujard's formula~\eqref{Goujardformula}. 
However, we do not know whether
Conjecture~\ref{conjasymg} implies Conjecture~\ref{conjecture2} in the same way. This would be an interesting
point to investigate next. 


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\medskip
\medskip


\noindent Di Yang

\noindent School of Mathematical Sciences, University of Science and Technology of China\\
Hefei 230026, P.R.~China \\
diyang@ustc.edu.cn

\medskip

\noindent Don Zagier

\noindent Max-Planck-Institut f\"ur Mathematik, Bonn 53111, Germany \\
and International Centre for Theoretical Physics, Trieste 34014, Italy \\
\noindent dbz@mpim-bonn.mpg.de

\medskip

\noindent Youjin Zhang

\noindent Department of Mathematical Sciences, Tsinghua University \\ 
Beijing 100084, P.R.~China\\
youjin@mail.tsinghua.edu.cn



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