Modular Ordinary Differential Equations on $\mathrm{SL}(2,\mathbb{Z})$ of
Third Order

Modular Ordinary Differential Equations on
$\boldsymbol{\mathrm{SL}(2,\mathbb{Z})}$
of Third Order and Applications

Zhijie CHEN a, Chang-Shou LIN b and Yifan YANG c

Z. Chen, C.-S. Lin and Y. Yang

a) Department of Mathematical Sciences, Yau Mathematical Sciences Center,
a) Tsinghua University, Beijing, 100084, China<EMAIL_ADDRESS>
b) Center for Advanced Study in Theoretical Sciences, National Taiwan
University,
b) Taipei 10617, Taiwan<EMAIL_ADDRESS>
c) Department of Mathematics, National Taiwan University
c) and National Center for Theoretical Sciences, Taipei 10617, Taiwan
<EMAIL_ADDRESS>
Received June 24, 2021, in final form February 13, 2022; Published online
February 22, 2022

In this paper, we study third-order modular ordinary differential equations
(MODE for short) of the following form
$y^{\prime\prime\prime}+Q_{2}(z)y^{\prime}+Q_{3}(z)y=0$,
$z\in\mathbb{H}=\\{z\in\mathbb{C}\,|\,\operatorname{Im}z>0\\}$, where
$Q_{2}(z)$ and $Q_{3}(z)-\frac{1}{2}Q_{2}^{\prime}(z)$ are meromorphic modular
forms on ${\rm SL}(2,\mathbb{Z})$ of weight $4$ and $6$, respectively. We show
that any quasimodular form of depth $2$ on ${\rm SL}(2,\mathbb{Z})$ leads to
such a MODE. Conversely, we introduce the so-called Bol representation
$\hat{\rho}\colon{\rm SL}(2,\mathbb{Z})\to{\rm SL}(3,\mathbb{C})$ for this
MODE and give the necessary and sufficient condition for the irreducibility
(resp. reducibility) of the representation. We show that the irreducibility
yields the quasimodularity of some solution of this MODE, while the
reducibility yields the modularity of all solutions and leads to solutions of
certain ${\rm SU}(3)$ Toda systems. Note that the ${\rm SU}(N+1)$ Toda systems
are the classical Plücker infinitesimal formulas for holomorphic maps from a
Riemann surface to $\mathbb{CP}^{N}$.

modular differential equations; quasimodular forms; Toda system

11F11; 34M03

##  1 Introduction

Let $Ly=0$ be a Fuchsian ordinary differential equation of third order defined
on the upper half plane
$\mathbb{H}=\\{z\in\mathbb{C}\,|\,\operatorname{Im}z>0\\}$:

$Ly:=y^{\prime\prime\prime}+Q_{2}(z)y^{\prime}+Q_{3}(z)y=0,\qquad
z\in\mathbb{H},$ (1.1)

where ${}^{\prime}:=\frac{{\rm d}}{{\rm d}z}$. Near a regular point $z_{0}$ of
$Ly=0$, a local solution $y(z)$ can be obtained by giving the initial values
$y^{(k)}(z_{0})$, $k=0,1,2$, and then $y(z)$ could be globally defined through
analytic continuation. However, globally $y(z)$ might be multi-valued. If
$Ly=0$ is defined on $\mathbb{C}$, then the monodromy representation from
$\pi_{1}(\mathbb{C}\setminus\\{\text{singular points}\\})$ to
$\mathrm{SL}(3,\mathbb{C})$ is introduced to characterize the multi-valueness
of solutions. If the potentials $Q_{2}(z)$ and $Q_{3}(z)$ are elliptic
functions with periods $1$ and $\tau$ ($\operatorname{Im}\tau>0$) and any
solution of $Ly=0$ is single-valued and meromorphic, then the monodromy
representation reduces to a homomorphism from $\pi_{1}(E_{\tau})$ to
$\mathrm{SL}(3,\mathbb{C})$, where
$E_{\tau}:=\mathbb{C}/(\mathbb{Z}+\mathbb{Z}\tau)$ is the elliptic curve. The
well-known examples are the integral Lamé equations and its generalizations;
see, e.g., [4, 5, 6] for some recent developments of this subject. Note that
$\pi_{1}(E_{\tau})$ is abelian. In this paper, we consider the case that
$Ly=0$ is defined on $\mathbb{H}$ and instead of the monodromy representation
defined on $\pi_{1}(\mathbb{H}\setminus\\{\text{singular points}\\})$, we
study a new representation defined on a discrete non-abelian group $\Gamma$
that is related to the modular property of $\Gamma$ acting on $\mathbb{H}$. It
is called the Bol representation in this paper as in [25] where the Bol
representation was first introduced for second order differential equations.

Let $\Gamma$ be a discrete subgroup of $\mathrm{SL}(2,\mathbb{R})$ that is
commensurable with $\mathrm{SL}(2,\mathbb{Z})$. Equation (1.1) is called a
_modular ordinary differential equation_ (MODE for short) on $\Gamma$ if
$Q_{2}(z)$ and $Q_{3}(z)-\frac{1}{2}Q_{2}^{\prime}(z)$ are meromorphic modular
forms on $\Gamma$ of weight $4$ and $6$, respectively. Modular ordinary
differential equations (of general order) appear prominently in the study of
rational conformal field theories (see, e.g., [1, 10, 12, 14, 19, 21, 27,
34]). They provide a practical tool for classifying rational conformal field
theories. As an object in the theory of modular forms, modular differential
equations have also been studied by mathematicians. See, for example, [9, 11,
13, 17, 18, 29].

Given a MODE (1.1), it is natural to ask whether there are solutions
satisfying some modular property. The main goal of this paper is to study when
MODE (1.1) has solutions that lead to modular forms or quasimodular forms. The
approach is to calculate the Bol representation, which will be explained
below.

First we recall some basic notions from the ODE aspect. Equation (1.1) is
called _Fuchsian_ if the order of any pole of $Q_{j}(z)$ is at most $j$,
$j=2,3$. At the cusp $\infty$, we let $q_{N}={\rm e}^{2\pi{\rm i}z/N}$, where
$N$ is the width of $\infty$ in $\Gamma$. Then $\frac{{\rm d}}{{\rm
d}z}=\frac{2\pi{\rm i}}{N}q_{N}\frac{{\rm d}}{{\rm d}q_{N}}$ and so (1.1)
becomes

$\left(q_{N}\frac{{\rm d}}{{\rm d}q_{N}}\right)^{3}y+\left(\frac{N}{2\pi{\rm
i}}\right)^{2}Q_{2}(z)q_{N}\frac{{\rm d}}{{\rm
d}q_{N}}y+\left(\frac{N}{2\pi{\rm i}}\right)^{3}Q_{3}(z)y=0.$ (1.2)

From here we see that

$\text{\it\eqref{eq-01} is Fuchsian at $\infty$ if and only if $Q_{2}(z)$ and
$Q_{3}(z)$ are holomorphic at $\infty$},$

and similar conclusions hold for other cusps of $\Gamma$. By (1.2), the
indicial equation at the cusp $\infty$ is given by

$\kappa^{3}+\Big{(}\frac{N}{2\pi{\rm
i}}\Big{)}^{2}Q_{2}(\infty)\kappa+\left(\frac{N}{2\pi{\rm
i}}\right)^{3}Q_{3}(\infty)=0,$

the roots of which are called the local exponents of (1.1) at $\infty$,
denoted by $\kappa_{\infty}^{(1)}$, $\kappa_{\infty}^{(2)}$ and
$\kappa_{\infty}^{(3)}$, satisfying $\sum_{j}\kappa_{\infty}^{(j)}=0$. In this
paper, we always assume that the exponent differences
$\kappa_{\infty}^{(j)}-\kappa_{\infty}^{(1)}$ are _integers_ for $j=2,3$, Then
$\sum_{j}\kappa_{\infty}^{(j)}=0$ implies
$\kappa_{\infty}^{(j)}\in\frac{1}{3}\mathbb{Z}$ for all $j$ and so we may
assume
$\kappa_{\infty}^{(1)}\leq\kappa_{\infty}^{(2)}\leq\kappa_{\infty}^{(3)}$.
Similar assumptions are made for other cusps.

On the other hand, let $z_{0}\in\mathbb{H}$ be a singular point of (1.1) and
write

$Q_{j}(z)=A_{j}(z-z_{0})^{-j}+O\big{(}(z-z_{0})^{-j+1}\big{)}\qquad\text{\it
at }z_{0},$

then the indicial equation at $z_{0}$ is given by

$\kappa(\kappa-1)(\kappa-2)+A_{2}\kappa+A_{3}=0,$

the roots of which are the local exponents of $(\ref{eq-01})$ at $z_{0}$,
denoted by $\kappa_{z_{0}}^{(1)}$, $\kappa_{z_{0}}^{(2)}$ and
$\kappa_{z_{0}}^{(3)}$, satisfying $\sum_{j}\kappa_{z_{0}}^{(j)}=3$. In this
paper, we always assume that the exponent differences
$\kappa_{z_{0}}^{(j)}-\kappa_{z_{0}}^{(1)}$ are _integers_ for $j=2,3$. Then
$\sum_{j}\kappa_{z_{0}}^{(j)}=3$ implies
$\kappa_{z_{0}}^{(j)}\in\frac{1}{3}\mathbb{Z}$ for all $j$ and so we may
assume $\kappa_{z_{0}}^{(1)}\leq\kappa_{z_{0}}^{(2)}\leq\kappa_{z_{0}}^{(3)}$.
Since the exponent differences are integers, (1.1) might have solutions with
logarithmic singularities at $z_{0}$. See Appendix A for all possibilities of
the solution structure of (1.1) at $z_{0}$. The singularity $z_{0}$ is called
_apparent_ if (1.1) has no solutions with logarithmic singularities at
$z_{0}$. In this case, the three local exponents must be distinct, i.e.,
$\kappa_{z_{0}}^{(1)}<\kappa_{z_{0}}^{(2)}<\kappa_{z_{0}}^{(3)}$; see, e.g.,
Appendix A. In this paper, we always assume that $L$ is _apparent at any
singularity $z_{0}\in\mathbb{H}$_. More precisely, we assume that the MODE
(1.1) satisfies

  * (H1)

The MODE (1.1) is Fuchsian on $\mathbb{H}\cup\\{\text{cusps}\\}$;

  * (H2)

At any singular point $z_{0}\in\mathbb{H}$,
$\kappa_{z_{0}}^{(1)}<\kappa_{z_{0}}^{(2)}<\kappa_{z_{0}}^{(3)}$ satisfy
$\kappa_{z_{0}}^{(1)}\in\frac{1}{3}\mathbb{Z}_{\leq 0}$ and
$\kappa_{z_{0}}^{(j)}-\kappa_{z_{0}}^{(1)}\in\mathbb{Z}$ for $j=2,3$.
Furthermore, $z_{0}$ is apparent.

  * (H3)

At any cusp $s$ of $\Gamma$,
$\kappa_{s}^{(1)}\leq\kappa_{s}^{(2)}\leq\kappa_{s}^{(3)}$ satisfies
$\kappa_{s}^{(1)}\in\frac{1}{3}\mathbb{Z}_{\leq 0}$ and
$\kappa_{s}^{(j)}-\kappa_{s}^{(1)}\in\mathbb{Z}$ for $j=2,3$.

The motivation of all these assumptions will be clear from Theorem 1.1 below.

Note $\ln q_{N}=\frac{2\pi{\rm i}}{N}z$. Under our assumption (H3), (1.1)
might have solutions containing $(\ln q_{N})^{2}=-\frac{4\pi^{2}}{N^{2}}z^{2}$
terms; see Remark A.9. In this case, we call the cusp $\infty$ to be
_completely not apparent_ or _maximally unipotent_ , because under the Bol
representation $\hat{\rho}$ that will be introduced below, the corresponding
matrix $\hat{\rho}(T)\in\mathrm{SL}(3,\mathbb{C})$ of
$T=\left(\begin{smallmatrix}1&N\\\ 0&1\end{smallmatrix}\right)\in\Gamma$ has
eigenvalues $\\{1,1,1\\}$ but $\operatorname{rank}(\hat{\rho}(T)-I_{3})=2$,
i.e., $\hat{\rho}(T)$ is maximally unipotent. Here $I_{3}$ denotes the
$3\times 3$ identity matrix.

One class of the MODEs can be derived from quasimodular forms of depth $2$.
The notion of quasimodular forms was first introduced by Kaneko and Zagier
[20]. See Section 2 for a brief overview of basic properties of quasimodular
forms. In particular, given a holomorphic function $\phi(z)$ satisfying

$(\phi\big{|}_{2}\gamma)(z):=(cz+d)^{-2}\phi(\gamma z)=\phi(z)+\frac{\alpha
c}{cz+d}$

for all $\gamma=\left(\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix}\right)$
for some nonzero complex number $\alpha$, any quasimodular form $f(z)$ of
weight $k$ and depth $2$ with character $\chi$ can be expressed as

$f(z)=f_{0}(z)+f_{1}(z)\phi(z)+f_{2}(z)\phi(z)^{2},$

where $f_{j}(z)$ is a modular form on $\Gamma$ of weight $k-2j$ with character
$\chi$ and $f_{2}\neq 0$. Consider

$\displaystyle\begin{pmatrix}h_{1}(z)\\\ h_{2}(z)\\\
h_{3}(z)\end{pmatrix}:=\begin{pmatrix}z^{2}f(z)+\alpha
z(f_{1}(z)+2f_{2}(z)\phi(z))+\alpha^{2}f_{2}(z)\\\
2zf(z)+\alpha(f_{1}(z)+2f_{2}(z)\phi(z))\\\ f(z)\end{pmatrix}$ (1.3)

and define

$W_{f}(z):=\det\begin{pmatrix}h_{1}&h_{1}^{\prime}&h_{1}^{\prime\prime}\\\
h_{2}&h_{2}^{\prime}&h_{2}^{\prime\prime}\\\
h_{3}&h_{3}^{\prime}&h_{3}^{\prime\prime}\end{pmatrix}$ (1.4)

to be the Wronskian associated to $f$. Then $W_{f}(z)$ is a modular form on
$\Gamma$ of weight $3k$ with character $\chi^{3}$; see Lemma 2.1 for a proof.
This $W_{f}(z)$ was first introduced by Pellarin [28] for
$\Gamma=\mathrm{SL}(2,\mathbb{Z})$ and $\phi(z)=E_{2}(z)$.

Now we define $g_{j}(z):=\frac{h_{j}(z)}{\sqrt[3]{W_{f}(z)}}$, then

$\det\begin{pmatrix}g_{1}&g_{1}^{\prime}&g_{1}^{\prime\prime}\\\
g_{2}&g_{2}^{\prime}&g_{2}^{\prime\prime}\\\
g_{3}&g_{3}^{\prime}&g_{3}^{\prime\prime}\end{pmatrix}=1,$

and a further differentiation leads to

$\det\begin{pmatrix}g_{1}&g_{1}^{\prime}&g_{1}^{\prime\prime\prime}\\\
g_{2}&g_{2}^{\prime}&g_{2}^{\prime\prime\prime}\\\
g_{3}&g_{3}^{\prime}&g_{3}^{\prime\prime\prime}\end{pmatrix}=0,$ (1.5)

so $g_{3}(z)$ is a solution of (1.1) with

$Q_{2}(z):=\frac{g_{1}^{\prime\prime\prime}g_{2}-g_{1}g_{2}^{\prime\prime\prime}}{g_{1}g_{2}^{\prime}-g_{1}^{\prime}g_{2}},\qquad
Q_{3}(z):=\frac{g_{1}^{\prime}g_{2}^{\prime\prime\prime}-g_{2}^{\prime}g_{1}^{\prime\prime\prime}}{g_{1}g_{2}^{\prime}-g_{1}^{\prime}g_{2}}.$
(1.6)

It is easy to see that $Q_{2}(z)$ and $Q_{3}(z)$ are single-valued, and
$g_{1}$, $g_{2}$ are also solutions of (1.1). Our first result reads as
follows.

######  Theorem 1.1.

Let $Q_{2}(z)$ and $Q_{3}(z)$ be given by (1.6). Then

  * $(1)$

(1.1) is a MODE, i.e., $Q_{2}(z)$ and $Q_{3}(z)-\frac{1}{2}Q_{2}^{\prime}(z)$
are meromorphic modular forms on $\Gamma$ $($with trivial character$)$ of
weight $4$ and $6$, respectively.

  * $(2)$

$(H1)$–$(H3)$ hold for (1.1).

Furthermore, for $\Gamma=\mathrm{SL}(2,\mathbb{Z})$, we have that

  * $(3)$

At the elliptic point ${\rm i}$, $\big{\\{}3\kappa_{\rm i}^{(1)},3\kappa_{\rm
i}^{(2)},3\kappa_{\rm i}^{(3)}\big{\\}}\equiv\\{0,0,1\\}\ \mathrm{mod}\ 2$.

  * $(4)$

At the elliptic point $\rho=\frac{-1+\sqrt{3}i}{2}$,
$\kappa_{\rho}^{(j)}\in\mathbb{Z}$ for all $j$ and
$\big{\\{}\kappa_{\rho}^{(1)},\kappa_{\rho}^{(2)},\kappa_{\rho}^{(3)}\big{\\}}\equiv\\{0,1,2\\}\
\mathrm{mod}\ 3$.

We emphasize that (1) and (2) in Theorem 1.1 hold for any $\Gamma$, not only
for $\Gamma=\mathrm{SL}(2,\mathbb{Z})$. They will lay the ground for our
future study of general MODEs on other congruence subgroups.

As an example, in Section 6, we will work out the MODE in the case $f(z)$ is
an extremal quasimodular form on $\mathrm{SL}(2,\mathbb{Z})$, introduced first
in [18]; see Theorem 6.2.

Now we introduce the notion of the Bol representation of $\Gamma$ associated
to the MODE (1.1), which was first introduced in [25] for second order MODEs.
It is well known that any (local) solution $y(z)$ of (1.1) can be extended to
a multi-valued function in $\mathbb{H}$ through analytic continuation. Fix a
point $z_{0}\in\mathbb{H}$ that is not a singular point of (1.1) and let $U$
be a simply-connected neighborhood of $z_{0}$ that contains no singularities
of (1.1). For $\gamma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\Gamma$, choose a path $\sigma$ from $z_{0}$ to
$\gamma z_{0}$ and consider the analytic continuation of $y(z)$, $z\in U$,
along the path. Then $y(\gamma z)$ is well-defined in $U$. Define

$(y|_{-2}\gamma)(z):=(cz+d)^{2}y(\gamma z),\qquad z\in U,$

then by a direct computation or by using Bol’s identity [3], we see that
$(y|_{-2}\gamma)(z)$ is also a solution of (1.1). Thus, given a fundamental
system of solutions $Y(z)=(y_{1}(z),y_{2}(z),y_{3}(z))^{t}$, there is
$\hat{\gamma}\in\mathrm{SL}(3,\mathbb{C})$ such that

$(Y\big{|}_{-2}\gamma)(z)=\hat{\gamma}Y(z),$

where the fact $\det\hat{\gamma}=1$ follows from that the Wronskians of $Y$
and $(Y\big{|}_{-2}\gamma)$ are the same. Obviously, this matrix
$\hat{\gamma}$ depends on the choice of the path $\sigma$. However, under the
above assumptions, all local monodromy matrices are $\varepsilon I_{3}$ with
$\varepsilon^{3}=1$, so different choices of $\sigma$ will only possibly
change $\hat{\gamma}$ to ${\rm e}^{\pm\frac{2\pi{\rm i}}{3}}\hat{\gamma}$.
From here, we see that there is a well-defined homomorphism
$\rho\colon\Gamma\to\mathrm{PSL}(3,\mathbb{C})$ such that

$(Y\big{|}_{-2}\gamma)(z)={\rm e}^{\frac{2\pi{\rm
i}k}{3}}\rho(\gamma)Y(z),\qquad k\in\\{0,\pm 1\\},$

where $y_{j}(\gamma z)$ are always understood to take analytic continuation
along the same path for $j=1,2,3$. This homomorphism $\rho$ will be called the
_Bol representation_ as in [25]. For the convenience of computations, it is
better to lift $\rho$ to a homomorphism $\hat{\rho}\colon\Gamma\to{\rm
GL}(3,\mathbb{C})$ as follows. Suppose that we can find a multi-valued
meromorphic function $F(z)$ such that: (i) The analytic continuation of
$\hat{y}(z):=F(z)y(z)$, where $y(z)$ is any solution of (1.1), gives rise to a
single-valued holomorphic function on $\mathbb{H}$, and (ii) $F(z)^{3}$ is a
modular form on $\Gamma$ of weight $3k$ with some character, where
$k\in\mathbb{N}$. Such $F(z)$ can be constructed explicitly when $\Gamma$ is a
triangle group. Then by letting $\hat{Y}(z):=F(z)Y(z)$, there is
$\hat{\rho}(\gamma)\in{\rm GL}(3,\mathbb{C})$ such that

$\big{(}\hat{Y}\big{|}_{\ell}\gamma\big{)}(z)=\hat{\rho}(\gamma)\hat{Y}(z),\qquad\text{where
}\ell=k-2.$

This homomorphism $\hat{\rho}\colon\Gamma\to{\rm GL}(3,\mathbb{C})$, as a lift
of $\rho$, will also be called the _Bol representation_ since there is no
confusion arising. Naturally we consider the following problem:

######  Question.

Can we characterize, in terms of local exponents, the MODEs (1.1) whose Bol
representations are irreducible?

One purpose of this paper is to answer this question for the case
$\Gamma=\mathrm{SL}(2,\mathbb{Z})$. For $\Gamma=\mathrm{SL}(2,\mathbb{Z})$,
the above $F(z)$ can be taken to be

$F(z):=\Delta(z)^{-\kappa_{\infty}^{(1)}}E_{4}(z)^{-\kappa_{\rho}^{(1)}}E_{6}(z)^{-\kappa_{\rm
i}^{(1)}}\prod_{j=1}^{m}F_{j}(z)^{-\kappa_{z_{j}}^{(1)}},$ (1.7)

where

$E_{4}(z)=1+240\sum_{n=1}^{\infty}\frac{n^{3}q^{n}}{1-q^{n}},\qquad
E_{6}(z)=1-504\sum_{n=1}^{\infty}\frac{n^{5}q^{n}}{1-q^{n}},\qquad q={\rm
e}^{2\pi{\rm i}z},$ (1.8)

are the Eisenstein series of weight $4$ and $6$, respectively,

$\Delta(z)=\frac{E_{4}(z)^{3}-E_{6}(z)^{2}}{1728}=q-24q^{2}+252q^{3}-1472q^{4}+\cdots,$

${\rm i}=\sqrt{-1}$ and $\rho=\big{(}{-}1+\sqrt{3}{\rm i}\big{)}/2$ are the
elliptic points of $\mathrm{SL}(2,\mathbb{Z})$, $\\{z_{1},\dots,z_{m}\\}$
$\sqcup\\{{\rm i},\rho,\infty\\}$ denotes the set of singular points of the
MODE (1.1) mod $\mathrm{SL}(2,\mathbb{Z})$,
$t_{j}:=E_{4}(z_{j})^{3}/E_{6}(z_{j})^{2}$ and
$F_{j}(z):=E_{4}(z)^{3}-t_{j}E_{6}(z)^{2}$. Then $F(z)^{3}$ is a modular form
of weight $3(\ell+2)$, where the integer $\ell=k-2$ is given by

$\ell:=-2-12\kappa_{\infty}^{(1)}-4\kappa_{\rho}^{(1)}-6\kappa_{\rm
i}^{(1)}-12\sum_{j=1}^{m}\kappa_{z_{j}}^{(1)}.$

In other words, besides the assumptions (H1)–(H3), we need to assume further
that $\kappa_{\rho}^{(1)}\in\mathbb{Z}$ such that $\ell\in\mathbb{Z}$.
Consequently, we will see from Lemma 3.2 that the Bol representation
$\hat{\rho}$ is indeed a group homomorphism from $\mathrm{SL}(2,\mathbb{Z})$
to $\mathrm{SL}(3,\mathbb{C})$.

######  Remark 1.2.

The choice of $F(z)$ is not unique since we can multiply $F(z)$ by a
holomorphic modular form to obtain a new one. Different choices of $F(z)$’s
may give different weights $k$ (and so $\ell$) but keeping
$\hat{\rho}(\gamma)$ invariant. For example, when the MODE (1.1) comes from a
quasimodular form $f(z)$ of depth $2$ on $\Gamma$ as shown in Theorem 1.1,
then one choice is to take $\sqrt[3]{W_{f}(z)}$ as $F(z)$, i.e.,
$\hat{Y}=(h_{1},h_{2},h_{3})^{t}$ defined in (1.3). Note that for
$\Gamma=\mathrm{SL}(2,\mathbb{Z})$, $\sqrt[3]{W_{f}(z)}$ might be different
from the $F(z)$ given by (1.7). To obtain that $\sqrt[3]{W_{f}(z)}$ equals to
the $F(z)$ given by (1.7), we need to assume that $f_{0}$, $f_{1}$, $f_{2}$
have no common zeros.

Note from $\sum_{j}\kappa_{\rm i}^{(j)}=3$ that we have either
$\big{\\{}3\kappa_{\rm i}^{(1)},3\kappa_{\rm i}^{(2)},3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{0,0,1\\}\ \mathrm{mod}\ 2$ or $\big{\\{}3\kappa_{\rm
i}^{(1)},\allowbreak 3\kappa_{\rm i}^{(2)},3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{1,1,1\\}\ \mathrm{mod}\ 2$. Our second result of
this paper is

######  Theorem 1.3.

Let $\Gamma=\mathrm{SL}(2,\mathbb{Z})$ and suppose that the MODE (1.1)
satisfies $(H1)$–$(H3)$ and $\kappa_{\rho}^{(1)}\in\mathbb{Z}$. Then the Bol
representation $\hat{\rho}$ is irreducible if and only if
$\big{\\{}3\kappa_{\rm i}^{(1)},3\kappa_{\rm i}^{(2)}$, $3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{0,0,1\\}\ \mathrm{mod}\ 2$.

Note that all irreducible representations of $\mathrm{SL}(2,\mathbb{Z})$ of
rank up to $5$ have been classified by Tuba and Wenzl [31].111We thank the
referee for providing the reference. One may use their results, the work of
Westbury [32], and Lemma 3.10 below to give another proof of Theorem 1.3
different from that given in Section 3. See Remark 3.12.

As an application of Theorem 1.3, we can show that the converse statement of
Theorem 1.1 holds. More precisely, we have

######  Theorem 1.4.

Let $\Gamma=\mathrm{SL}(2,\mathbb{Z})$ and suppose that the MODE (1.1)
satisfies $(H1)$–$(H3)$ and $\kappa_{\rho}^{(1)}\in\mathbb{Z}$. Let $y_{+}(z)$
be the solution of (1.1) of the form
$y_{+}(z)=q^{\kappa_{\infty}^{(3)}}\sum_{j=0}^{\infty}c_{j}q^{j}$, $c_{0}=1$,
and $F(z)$ be defined by (1.7).

  * $(1)$

If $\big{\\{}3\kappa_{\rm i}^{(1)},3\kappa_{\rm i}^{(2)},3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{0,0,1\\}\ \mathrm{mod}\ 2$, then
$\hat{y}_{+}(z):=F(z)y_{+}(z)$ is a quasimodular form of weight $\ell+2$ and
depth $2$.

  * $(2)$

If $\big{\\{}3\kappa_{\rm i}^{(1)},3\kappa_{\rm i}^{(2)},3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{1,1,1\\}\ \mathrm{mod}\ 2$, then the Bol
representation $\hat{\rho}$ is trivial, i.e., $\hat{\rho}(\gamma)=I_{3}$ for
all $\gamma\in\mathrm{SL}(2,\mathbb{Z})$. In particular, $12|\ell$ and
$\hat{y}(z):=F(z)y(z)$ is a modular form of weight $\ell$ for any solution
$y(z)$ of (1.1).

Together with Theorems 1.4 and 1.1, we can obtain

######  Corollary 1.5.

Let $\Gamma=\mathrm{SL}(2,\mathbb{Z})$ and suppose $(H1)$–$(H3)$ and
$\kappa_{\rho}^{(1)}\\!\in\mathbb{Z}$ hold for the MODE (1.1). Then $(3)$ and
$(4)$ in Theorem 1.1 are equivalent.

In the reducible case, Theorem 1.4(2) can be applied to construct solutions of
the $\mathrm{SU}(3)$ Toda system. See Section 4 for the precise statement. The
Toda system is an important integrable system in mathematical physics. In
algebraic geometry, the $\mathrm{SU}(N+1)$ Toda system is exactly the
classical infinitesimal Plücker formula associated with holomorphic maps from
Riemann surfaces to $\mathbb{CP}^{N}$; see, e.g., [7, 23, 24] and references
therein for the recent development of the Toda system.

The rest of this paper is organized as follows. In Section 2, we give the
proof of Theorem 1.1, namely we will prove that every quasimodular form of
depth $2$ leads to a MODE (1.1) satisfying the conditions (H1)–(H3). We focus
on the case $\Gamma=\mathrm{SL}(2,\mathbb{Z})$ from Section 3. Theorems
1.3–1.4 and Corollary 1.5 will be proved in Section 3. In Section 4, we
discuss the reducible case and prove the converse statement of Theorem 1.4(2).
We also give an application to the $\mathrm{SU}(3)$ Toda system. In Section 5,
we discuss the criterion on the existence of the MODE (1.1) which is Fuchsian
and apparent throughout $\mathbb{H}$ with prescribed local exponents at
singularities and at cusps. In Section 6, as examples of MODEs, we will work
out the explicit expressions of $Q_{j}(z)$’s for an extremal quasimodular form
$f(z)$. Finally in Appendix A, we recall the theory of the solution structure
of third order ODEs at a regular singular point.

##  2 Quasimodular forms of depth 2
and its associated 3rd order MODE

The main purpose of this section is to prove Theorem 1.1. Let $\Gamma$ be a
discrete subgroup of $\mathrm{SL}(2,\mathbb{R})$ that is commensurable with
$\mathrm{SL}(2,\mathbb{Z})$ and $\chi\colon\Gamma\to\mathbb{C}^{\times}$ be a
character of $\Gamma$ of finite order. A holomorphic function $f(z)$ defined
on the upper half plane $\mathbb{H}$ is a modular form of weight $k$ with
character $\chi$ if the following conditions hold:

  * (1)

$(f\big{|}_{k}\gamma)(z):=(cz+d)^{-k}f(\gamma z)=\chi(\gamma)f(z)$ for any
$\gamma\in\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\Gamma$;

  * (2)

$f$ is holomorphic at any cusp $s$ of $\Gamma$.

For example, the Eisenstein series $E_{4}(z)$ and $E_{6}(z)$ in (1.8) are
modular forms of weight $4$ and $6$ on $\mathrm{SL}(2,\mathbb{Z})$,
respectively. We let $\mathfrak{M}_{k}(\Gamma,\chi)$ denote the space of
modular forms of weight $k$ with character $\chi$ on $\Gamma$. For example,
for $\Gamma=\mathrm{SL}(2,\mathbb{Z})$, $\infty$ is the only cusp and we
assume that the character is trivial, i.e., $\chi\equiv 1$. Then condition (1)
implies $f(z+1)=f(z)$, which implies that $f(z)$ can be viewed as a function
of $q={\rm e}^{2\pi{\rm i}z}$, and condition (2) just means that $f$ is
holomorphic at $q=0$.

The notion of quasimodular forms was introduced by Kaneko and Zagier [20].
Originally, they are defined as the holomorphic parts of nearly holomorphic
modular forms. For our purpose, it suffices to know that a holomorphic
function $f(z)$ is a quasimodular form of weight $k$ and depth $r$ with
character $\chi$ on $\Gamma$ if and only if $f(z)$ can be expressed as

$f(z)=\sum_{j=0}^{r}f_{j}(z)\phi(z)^{j},$

where $f_{j}(z)\in\mathfrak{M}_{k-2j}(\Gamma,\chi)$ with $f_{r}\not\equiv 0$
and $\phi(z)$ is a holomorphic function satisfying that

$(\phi\big{|}_{2}\gamma)(z):=(cz+d)^{-2}\phi(\gamma z)=\phi(z)+\frac{\alpha
c}{cz+d}$ (2.1)

for all $\gamma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\Gamma$ for some nonzero complex number $\alpha$
and $\phi(z)$ is holomorphic at cusps of $\Gamma$. This $\phi(z)$ is called a
quasimodular form of weight $2$ and depth $1$ on $\Gamma$. For example, if
$\Gamma$ is a subgroup of $\mathrm{SL}(2,\mathbb{Z})$, we can always let

$\phi(z)=E_{2}(z):=\frac{1}{2\pi{\rm
i}}\frac{\Delta^{\prime}(z)}{\Delta(z)}=1-24\sum_{n=1}^{\infty}\frac{nq^{n}}{1-q^{n}},$

and so $\alpha=\frac{6}{\pi{\rm i}}$. We let
$\widetilde{\mathfrak{M}}_{k}^{\leq r}(\Gamma,\chi)$ denote the space of
quasimodular forms of weight $k$ and depth $\leq r$ with character $\chi$. One
basic property is that the quasi-modularity is invariant under the
differentiation, namely if $f(z)\in\widetilde{\mathfrak{M}}_{k}^{\leq
r}(\Gamma,\chi)$, then $f^{\prime}(z)\in\widetilde{\mathfrak{M}}_{k+2}^{\leq
r+1}(\Gamma,\chi)$; see [33, Proposition 20]. We refer the reader to [8, 20,
33] for the general theory of quasimodular forms.

Now we consider the Wronskian $W_{f}(z)$ (see [28]) associated to

$f(z)=f_{0}(z)+f_{1}(z)\phi(z)+f_{2}(z)\phi(z)^{2}\in\widetilde{\mathfrak{M}}_{k}^{\leq
2}(\Gamma,\chi),$

where $f_{j}(z)\in\mathfrak{M}_{k-2j}(\Gamma,\chi)$ with $f_{2}\neq 0$. Then

$\displaystyle\big{(}f\big{|}_{k}\gamma\big{)}(z):={}$
$\displaystyle(cz+d)^{-k}f(\gamma z)$ (2.2) $\displaystyle={}$
$\displaystyle\chi(\gamma)\sum_{j=0}^{2}f_{j}(z)\left(\phi(z)+\frac{\alpha
c}{cz+d}\right)^{j},\qquad\gamma=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in\Gamma.$

As in [28], we set

$\displaystyle P_{f}(t):=\sum_{j=0}^{2}f_{j}(z)\left(\phi(z)+\alpha
t\right)^{j},$ $\displaystyle
Q_{f}(t):=t^{2}P_{f}(1/t)=f(z)t^{2}+\alpha(f_{1}(z)+2f_{2}(z)\phi(z))t+\alpha^{2}f_{2}(z),$
$\displaystyle F_{f}(z):=\begin{pmatrix}Q_{f}(t)\\\ \dfrac{\partial}{\partial
t}Q_{f}(t)\vspace{1mm}\\\ \dfrac{1}{2}\dfrac{\partial^{2}}{\partial
t^{2}}Q_{f}(t)\end{pmatrix}_{|{t=z}}$
$\displaystyle\hphantom{F_{f}(z)~{}}{}=\begin{pmatrix}z^{2}f(z)+\alpha
z(f_{1}(z)+2f_{2}(z)\phi(z))+\alpha^{2}f_{2}(z)\\\
2zf(z)+\alpha(f_{1}(z)+2f_{2}(z)\phi(z))\\\
f(z)\end{pmatrix}=:\begin{pmatrix}h_{1}(z)\\\ h_{2}(z)\\\
h_{3}(z)\end{pmatrix}.$ (2.3)

and define $W_{f}(z)$ as in (1.4) to be the Wronskian associated to $f$.

######  Lemma 2.1.

$W_{f}(z)$ is a modular form on $\Gamma$ of weight $3k$ with character
$\chi^{3}$.

###### Proof.

For $\Gamma=\mathrm{SL}(2,\mathbb{Z})$, this result is proved in [28] and can
be also derived from Mason [26, Lemma 3.1], and the approaches in [26, 28] can
be easily applied to general discrete subgroups $\Gamma$. Here we provide an
elementary proof for general $\Gamma$ for completeness. By (2.2) and $ad-bc=1$
we have

$\displaystyle f(\gamma z)=\chi(\gamma)(cz+d)^{k}\left[f+\frac{\alpha
c}{cz+d}(f_{1}+2f_{2}\phi)+\frac{\alpha^{2}c^{2}}{(cz+d)^{2}}f_{2}\right]$
$\displaystyle\hphantom{f(\gamma
z)}{}=\chi(\gamma)(cz+d)^{k-2}\big{(}c^{2}h_{1}+cdh_{2}+d^{2}h_{3}\big{)},$
$\displaystyle h_{2}(\gamma z)=2\frac{az+b}{cz+d}f(\gamma
z)+\alpha(f_{1}(\gamma z)+2f_{2}(\gamma z)\phi(\gamma z))$
$\displaystyle\hphantom{h_{2}(\gamma
z)}{}=\chi(\gamma)(cz+d)^{k-2}\bigg{[}2\frac{az+b}{cz+d}((cz+d)^{2}f+\alpha
c(cz+d)(f_{1}+2f_{2}\phi)$ $\displaystyle\hphantom{h_{2}(\gamma
z)=}{}+\alpha^{2}c^{2}f_{2})+\alpha(f_{1}+2f_{2}\phi)+\frac{2\alpha^{2}c}{cz+d}f_{2}\bigg{]}$
$\displaystyle\hphantom{h_{2}(\gamma
z)}{}=\chi(\gamma)(cz+d)^{k-2}[2ach_{1}+(ad+bc)h_{2}+2bdh_{3}],$
$\displaystyle h_{1}(\gamma z)=\frac{(az+b)^{2}}{(cz+d)^{2}}f(\gamma
z)+\alpha\frac{az+b}{cz+d}(f_{1}(\gamma z)+2f_{2}(\gamma z)\phi(\gamma
z))+\alpha^{2}f_{2}(\gamma z)$ $\displaystyle\hphantom{h_{1}(\gamma
z)}{}=\chi(\gamma)(cz+d)^{k-2}\bigg{[}\frac{(az+b)^{2}}{(cz+d)^{2}}\big{(}(cz+d)^{2}f+\alpha
c(cz+d)(f_{1}+2f_{2}\phi)$ $\displaystyle\hphantom{h_{1}(\gamma
z)=}{}+\alpha^{2}c^{2}f_{2}\big{)}+\frac{az+b}{cz+d}\Big{[}\alpha(f_{1}+2f_{2}\phi)+\frac{2\alpha^{2}c}{cz+d}f_{2}\Big{]}+\frac{\alpha^{2}}{(cz+d)^{2}}f_{2}\bigg{]}$
$\displaystyle\hphantom{h_{1}(\gamma
z)}{}=\chi(\gamma)(cz+d)^{k-2}\big{[}a^{2}h_{1}+abh_{2}+b^{2}h_{3}\big{]}.$

Thus

$F_{f}(\gamma z)=\chi(\gamma)(cz+d)^{k-2}AF_{f}(z),\qquad\text{where}\quad
A:=\begin{pmatrix}a^{2}&ab&b^{2}\\\ 2ac&ad+bc&2bd\\\
c^{2}&cd&d^{2}\end{pmatrix}.$

Together with the fact $\det A=(ad-bc)^{3}=1$, it is easy to see that

$\displaystyle W_{f}(\gamma z)=\det(F_{f}(\gamma z),F_{f}^{\prime}(\gamma
z),F_{f}^{\prime\prime}(\gamma z))=\chi(\gamma)^{3}(cz+d)^{3k}W_{f}(z).$

In particular, for the case $\gamma=\left(\begin{smallmatrix}1&N\\\
0&1\end{smallmatrix}\right)$, where $N$ is the width of the cusp $\infty$, the
transformation law shows that $W_{f}(z)$ is a polynomial only in $f_{0}$,
$f_{1}$, $f_{2}$, $\phi$, and their derivatives. Thus, by the computation
above and the fact that the ring of quasimodular forms is invariant under
differentiation, $W_{f}(z)$ is a quasimodular form that is actually modular.
In other words, $W_{f}(z)\in\mathfrak{M}_{3k}\big{(}\Gamma,\chi^{3}\big{)}$.
This completes the proof. ∎

Now we define $g_{j}(z):=h_{j}(z)/\sqrt[3]{W_{f}(z)}$. Then (1.5) holds, so
$g_{3}(z)$ is a solution of

$Ly:=y^{\prime\prime\prime}+Q_{2}(z)y^{\prime}+Q_{3}(z)y=0,$ (2.4)

where

$Q_{2}(z):=\frac{g_{1}^{\prime\prime\prime}g_{2}-g_{1}g_{2}^{\prime\prime\prime}}{g_{1}g_{2}^{\prime}-g_{1}^{\prime}g_{2}},\qquad
Q_{3}(z):=\frac{g_{1}^{\prime}g_{2}^{\prime\prime\prime}-g_{2}^{\prime}g_{1}^{\prime\prime\prime}}{g_{1}g_{2}^{\prime}-g_{1}^{\prime}g_{2}}.$
(2.5)

Note that

  * (i)

$g_{1}(z)$ and $g_{2}(z)$ are also solutions of (2.4) and $g_{1}$, $g_{2}$,
$g_{3}$ are linearly independent.

  * (ii)

Although $g_{j}$ might be multi-valued, $g_{j}^{\prime}/g_{j}$ is single-
valued. Thus $Q_{j}(z)$ is single-valued and meromorphic on $\mathbb{H}$ for
$j=1,2$. We will see from Theorem 2.2 below that any pole of $Q_{j}(z)$ comes
from zeros of $W_{f}(z)$.

For $z_{0}\in\mathbb{H}$, we denote
$\kappa_{z_{0}}^{(1)}\leq\kappa_{z_{0}}^{(2)}\leq\kappa_{z_{0}}^{(3)}$ to be
the local exponents of (2.4) at $z_{0}$.

######  Theorem 2.2.

Under the above notations, $Q_{2}(z)$ and
$Q_{3}(z)-\frac{1}{2}Q_{2}^{\prime}(z)$ are meromorphic modular forms on
$\Gamma$ $($with trivial character$)$ of weight $4$ and $6$, respectively.
Furthermore, all singular points of (2.4) on $\mathbb{H}$ comes from the zeros
of $W_{f}(z)$, $(H1)$–$(H3)$ hold, and every cusp of $\Gamma$ is completely
not apparent for (2.4).

###### Proof.

To prove the modularity of $Q_{j}$, we consider

$\tilde{y}(z):=\big{(}y\big{|}_{-2}\gamma\big{)}(z)=(cz+d)^{2}y(\gamma
z),\qquad\gamma\in\Gamma.$

Then

$\tilde{y}^{\prime}(z)=y^{\prime}(\gamma z)+2c(cz+d)y(\gamma
z),\qquad\tilde{y}^{\prime\prime\prime}(z)=(cz+d)^{-4}y^{\prime\prime\prime}(\gamma
z),$

so

$\displaystyle L\tilde{y}=(cz+d)^{-4}\big{\\{}y^{\prime\prime\prime}(\gamma
z)+(cz+d)^{4}Q_{2}(z)y^{\prime}(\gamma z)$
$\displaystyle\hphantom{L\tilde{y}=}{}+\big{[}(cz+d)^{6}Q_{3}(z)+2c(cz+d)^{5}Q_{2}(z)\big{]}y(\gamma
z)\big{\\}}.$ (2.6)

Recalling $g_{3}(z)=\frac{f(z)}{\sqrt[3]{W_{f}(z)}}$, we have

$\displaystyle\big{(}g_{3}\big{|}_{-2}\gamma\big{)}(z)$
$\displaystyle=\bigg{(}\frac{f}{\sqrt[3]{W_{f}}}\Big{|}_{-2}\gamma\bigg{)}(z)=\frac{(cz+d)^{2}f(\gamma
z)}{\sqrt[3]{W_{f}(\gamma
z)}}=\frac{\varepsilon}{\sqrt[3]{W_{f}(z)}}\big{(}c^{2}h_{1}+cdh_{2}+d^{2}h_{3}\big{)}$
$\displaystyle=\varepsilon\big{(}c^{2}g_{1}(z)+cdg_{2}(z)+d^{2}g_{3}(z)\big{)},$

where $\varepsilon^{3}=1$. Thus $\big{(}g_{3}\big{|}_{-2}\gamma\big{)}(z)$ is
also a solution of (2.4). From this and (2.6), we have

$\displaystyle Q_{2}(\gamma z)=(cz+d)^{4}Q_{2}(z),$ $\displaystyle
Q_{3}(\gamma z)=(cz+d)^{6}Q_{3}(z)+2c(cz+d)^{5}Q_{2}(z),$

so $\big{(}Q_{2}\big{|}_{4}\gamma\big{)}=Q_{2}$ and
$\big{(}(Q_{3}-\frac{1}{2}Q_{2}^{\prime})\big{|}_{6}\gamma\big{)}=Q_{3}-\frac{1}{2}Q_{2}^{\prime}$.
This proves the modularity of $Q_{2}$ and $Q_{3}$.

To prove (H1)–(H3), we let $z_{0}$ be any pole of $Q_{j}(z)$ for some $j=1,2$.
Clearly $g_{j}(z)=(z-z_{0})^{\alpha_{j}}(c_{j}+O(z-z_{0})^{j})$ near $z_{0}$
for some $\alpha_{j}\in\frac{1}{3}\mathbb{Z}$ and $c_{j}\neq 0$. By replacing
$g_{2}$ by $g_{2}-\frac{c_{2}}{c_{1}}g_{1}$ if necessary, we may assume
$\alpha_{1}\neq\alpha_{2}$. Then we easily deduce from (2.5) that

$\displaystyle
Q_{2}(z)=\frac{\alpha_{1}(\alpha_{1}-1)(\alpha_{1}-2)-\alpha_{2}(\alpha_{2}-1)(\alpha_{2}-2)}{(\alpha_{2}-\alpha_{1})(z-z_{0})^{2}}+O\big{(}(z-z_{0})^{-1}\big{)},$
$\displaystyle
Q_{3}(z)=\frac{\alpha_{1}\alpha_{2}[(\alpha_{2}-1)(\alpha_{2}-2)-(\alpha_{1}-1)(\alpha_{1}-2)]}{(\alpha_{2}-\alpha_{1})(z-z_{0})^{3}}+O\big{(}(z-z_{0})^{-2}\big{)},$

so $z_{0}$ is a regular singular point of (2.4). Thus, (2.4) is Fuchsian on
$\mathbb{H}$.

Let $z_{0}$ be any singular point. It follows from
$g_{j}(z)=\frac{h_{j}(z)}{\sqrt[3]{W_{f}(z)}}$ that

$g_{j}(z)=(z-z_{0})^{\frac{-\text{ord}_{z_{0}}W_{f}}{3}}\sum_{l\geq
0}d_{j}(z-z_{0})^{j},$ (2.7)

where $\text{ord}_{z_{0}}W_{f}$ denotes the zero order of $W_{f}(z)$ at
$z_{0}$. Since $(g_{1},g_{2},g_{3})$ is a fundamental system of solutions of
(2.4) and $g_{j}$’s have no logarithmic singularities at $z_{0}$, we conclude
from (2.7) and Remark A.8 that (1) the local exponents
$\kappa_{z_{0}}^{(j)}\in\frac{1}{3}\mathbb{Z}$ and are all distinct; (2) the
exponent differences are all nonzero integers, namely
$m_{z_{0}}^{(j)}:=\kappa_{z_{0}}^{(j+1)}-\kappa_{z_{0}}^{(j)}-1$ are
nonnegative integers for $j=1,2$; (3) $z_{0}$ is an apparent singularity of
(2.4).

Since

$Q_{j}(z)=A_{j}(z-z_{0})^{-j}+O\big{(}(z-z_{0})^{-j+1}\big{)},\qquad j=2,3,$
(2.8)

then the indicial equation of (2.4) at $z_{0}$ is

$\kappa(\kappa-1)(\kappa-2)+A_{2}\kappa+A_{3}=0,$ (2.9)

which implies $\sum_{j=1}^{3}\kappa_{z_{0}}^{(j)}=3$ and so
$\kappa_{z_{0}}^{(1)}=-\frac{2m_{z_{0}}^{(1)}+m_{z_{0}}^{(2)}}{3}\in\frac{1}{3}\mathbb{Z}_{\leq
0}$. This proves (H2). Remark that if $z_{0}$ is not a zero of $W_{f}(z)$,
i.e., $\text{ord}_{z_{0}}W_{f}=0$, then it follows from (2.7) that
$\kappa_{z_{0}}^{(1)}\in\mathbb{Z}_{\geq 0}$ and so
$\kappa_{z_{0}}^{(1)}=m_{z_{0}}^{(1)}=m_{z_{0}}^{(2)}=0$, i.e., the local
exponents at $z_{0}$ are $\\{0,1,2\\}$. This already implies $A_{2}=A_{3}=0$.
Together with the fact that $z_{0}$ is apparent, we easily deduce from the
Frobenius method that both $Q_{2}(z)$ and $Q_{3}(z)$ are holomorphic at
$z_{0}$, a contradiction with that $z_{0}$ is a singular point. Thus $z_{0}$
is a zero of $W_{f}(z)$. This proves that all singular points of (2.4) on
$\mathbb{H}$ come from the zeros of $W_{f}(z)$.

Let $N$ be the width of the cusp $\infty$ and $q_{N}={\rm e}^{2\pi{\rm
i}z/N}$. Since modular forms $f_{j}(z)$, $W_{f}(z)$ are holomorphic in terms
of $q_{N}$ and $z=\frac{N}{2\pi{\rm i}}\ln q_{N}$, we see from (2.3) that

$g_{j}(z)=\frac{h_{j}(z)}{\sqrt[3]{W_{f}(z)}}=\sum_{k=0}^{3-j}(\ln
q_{N})^{k}q_{N}^{-\frac{\operatorname{ord}_{\infty}W_{f}}{3}}\sum_{t=0}^{\infty}c_{j,k,t}q_{N}^{t},\qquad
j=1,2,3,$

so $\infty$ is also a regular singular point of (2.4), i.e., (2.4) is Fuchsian
at $\infty$ and so $Q_{j}(z)$ is holomorphic at $\infty$ for $j=2,3$. Since
$(\ln q_{N})^{2}$ appears in the expression of $g_{1}(z)$, we see from Remark
A.9 that $\infty$ is completely not apparent and the local exponents
$\kappa_{\infty}^{(1)}\leq\kappa_{\infty}^{(2)}\leq\kappa_{\infty}^{(3)}$
satisfy $\kappa_{\infty}^{(j)}\in\frac{1}{3}\mathbb{Z}$ and

$m_{\infty}^{(1)}:=\kappa_{\infty}^{(2)}-\kappa_{\infty}^{(1)}\in\mathbb{Z}_{\geq
0},\qquad
m_{\infty}^{(2)}:=\kappa_{\infty}^{(3)}-\kappa_{\infty}^{(2)}\in\mathbb{Z}_{\geq
0}.$

Note that the indicial equation at $\infty$ is

$\kappa^{3}+\left(\frac{N}{2\pi{\rm
i}}\right)^{2}Q_{2}(\infty)\kappa+\left(\frac{N}{2\pi{\rm
i}}\right)^{3}Q_{3}(\infty)=0,$

which implies $\sum\kappa_{\infty}^{(j)}=0$ and so
$\kappa_{\infty}^{(1)}=-\frac{2m_{\infty}^{(1)}+m_{\infty}^{(2)}}{3}$. We now
consider other cusps.

Assume that $s$ is another cusp of $\Gamma$ different from $\infty$. Let
$\sigma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\mathrm{SL}(2,\mathbb{Z})$ be a matrix such that
$\sigma\infty=s$. Regarding $f(z)$ as a quasimodular form on
$\Gamma^{\prime}=\ker\chi\cap\mathrm{SL}(2,\mathbb{Z})$, we can express $f(z)$
as
$f(z)=\widetilde{f}_{0}(z)+\widetilde{f}_{1}(z)E_{2}(z)+\widetilde{f}_{2}(z)E_{2}(z)^{2}$
for some $\widetilde{f}_{j}(z)\in\mathfrak{M}_{k-2j}(\Gamma^{\prime})$. We
check that

$\displaystyle\left(g_{3}|_{-2}\sigma\right)(z)$
$\displaystyle=\frac{(cz+d)^{2}f(\sigma z)}{\sqrt[3]{W_{f}(\sigma
z)}}=\epsilon\frac{(cz+d)^{2}(f|_{k}\sigma)(z)}{\sqrt[3]{(W_{f}|_{3k}\sigma)(z)}}$
$\displaystyle=\epsilon\frac{(cz+d)^{2}p_{1}(z)+\alpha
c(cz+d)p_{2}(z)+\alpha^{2}c^{2}p_{3}(z)}{\sqrt[3]{(W_{f}|_{3k}\sigma)(z)}},$

where

$\displaystyle
p_{1}(z)=\big{(}\widetilde{f}_{0}|_{k}\sigma\big{)}(z)+\big{(}\widetilde{f}_{1}|_{k-2}\sigma\big{)}(z)E_{2}(z)+\big{(}\widetilde{f}_{2}|_{k-4}\sigma\big{)}(z),$
$\displaystyle
p_{2}(z)=\big{(}\widetilde{f}_{1}|_{k-2}\sigma\big{)}(z)+2\big{(}\widetilde{f}_{2}|_{k-4}\sigma\big{)}(z)E_{2}(z),$
$\displaystyle p_{3}(z)=\big{(}\widetilde{f}_{2}|_{k-4}\sigma\big{)}(z),$

$\alpha=6/\pi{\rm i}$ and $\epsilon$ is a third root of unity. Except for
$cz+d$, every term in the expression has a $q_{M}$-expansion, where $M$ is the
width of the cusp $\sigma$ and $q_{M}={\rm e}^{2\pi{\rm i}z/M}$. Since
$s\neq\infty$, we have $c\neq 0$. This shows that there is a local solution at
the cusp $s$ having a factor $z^{2}$. According to the solution structure
discussed in the appendix, the point $s$ must be completely not apparent and
we have
$\kappa_{s}^{(2)}-\kappa_{s}^{(1)},\kappa_{s}^{(3)}-\kappa_{s}^{(1)}\in\mathbb{Z}$.
By the same reasoning as in the case of the cusp $\infty$, the sum
$\kappa_{s}^{(1)}+\kappa_{s}^{(2)}+\kappa_{s}^{(3)}$ is equal to $0$ and hence
$\kappa_{s}^{(j)}\in\frac{1}{3}\mathbb{Z}$ for all $j$. This proves (H1),
(H3), and that every cusp is completely not apparent. ∎

For a MODE, the local exponents are invariant under $z_{0}\to\gamma z_{0}$ for
any $\gamma\in\Gamma$.

######  Proposition 2.3.

Let $z_{0}$ be a singular point of (2.4). Then the local exponents of (2.4) at
$\gamma z_{0}$ are the same for all $\gamma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\Gamma$.

###### Proof.

Let

$Q_{j}(z)=\tilde{A}_{j}(z-\gamma z_{0})^{-j}+O\big{(}(z-\gamma
z_{0})^{-j+1}\big{)},\qquad j=2,3.$

Recalling (2.8) and (2.9). we only need to prove
$\big{(}\tilde{A}_{2},\tilde{A}_{3}\big{)}=(A_{2},A_{3})$.

Since

$\gamma z-\gamma
z_{0}=\frac{z-z_{0}}{(cz+d)(cz_{0}+d)}=\frac{z-z_{0}}{(cz_{0}+d)^{2}}(1+O(z-z_{0}))\qquad\text{as
$z\to z_{0}$},$

we have for $z\to z_{0}$ that

$\displaystyle Q_{2}(z)$ $\displaystyle=(cz+d)^{-4}Q_{2}(\gamma z)$
$\displaystyle=(cz_{0}+d)^{-4}(1+O(z-z_{0}))\big{[}\tilde{A}_{2}(\gamma
z-\gamma z_{0})^{-2}+O\big{(}(\gamma z-\gamma z_{0})^{-1}\big{)}\big{]}$
$\displaystyle=\tilde{A}_{2}(z-z_{0})^{-2}+O\big{(}(z-z_{0})^{-1}\big{)},$

so $\tilde{A}_{2}=A_{2}$. Since $Q_{3}-\frac{1}{2}Q_{2}^{\prime}$ is a modular
form of weight $6$, a similar argument implies
$\tilde{A}_{3}+\tilde{A}_{2}=A_{3}+A_{2}$ and so $\tilde{A}_{3}=A_{3}$. ∎

Now we consider $\Gamma=\mathrm{SL}(2,\mathbb{Z})$ and discuss the local
exponents of (2.4) at the elliptic points ${\rm i}$ and $\rho$, namely to
complete the proof of Theorem 1.1. For this purpose, we note that the remark
below could be used to simplify some computations.

######  Remark 2.4.

If $f_{0}$, $f_{1}$, $f_{2}$ have a common zero
$z_{0}\in\mathbb{H}\cup\\{\infty\\}$, we take $M(z)$ to be a holomorphic
modular form such that it has only one simple zero at $z_{0}$. Then
$f_{j}(z)/M(z)$ are holomorphic modular forms and $f(z)/M(z)$ is a
quasimodular form. Since $W_{f}(z)=M(z)^{3}W_{f/M}(z)$ and so

$g_{j}(z)=\frac{h_{j}(z)}{\sqrt[3]{W_{f}(z)}}=\frac{h_{j}(z)/M(z)}{\sqrt[3]{W_{f/M}(z)}},$

namely all $g_{j}(z)$’s are invariant by replacing $f(z)$ by $f(z)/M(z)$, so
the differential equation (2.4) derived from $f(z)$ and that from $f(z)/M(z)$
are the same (Note that $g_{j}(z)$ are solutions of (2.4) but neither $f(z)$
nor $f(z)/M(z)$ are solutions of (2.4), so we do not mean that $f(z)/M(z)$
satisfies the same differential equation as $f(z)$). Therefore, without loss
of generality, we assume throughout the section that $f_{0}$, $f_{1}$, $f_{2}$
have no common zeros on $\mathbb{H}\cup\\{\infty\\}$.

By applying this remark, we have

######  Lemma 2.5.

Let $\Gamma_{\infty}=\pm\langle T\rangle$ be the stabilizer subgroup of
$\infty$ in $\Gamma$. Then there are at most two right cosets
$\Gamma_{\infty}\gamma$ in $\Gamma_{\infty}\backslash\Gamma$ $($the set of
right cosets of $\Gamma_{\infty}$ in $\Gamma)$ such that $f(\gamma z_{0})=0$.

###### Proof.

Suppose that there are three distinct right cosets $\Gamma_{\infty}\gamma$ in
$\Gamma_{\infty}\backslash\Gamma$ such that $f(\gamma z_{0})=0$. Without loss
of generality, we assume that one of them is the coset of $I$, i.e.,
$f(z_{0})=0$. Now we have

$\left(f|_{k}\gamma\right)(z)=\chi(\gamma)\left(f(z)+\frac{\alpha
c}{cz+d}(f_{1}(z)+2f_{2}(z)\phi(z))+\left(\frac{\alpha
c}{cz+d}\right)^{2}f_{2}(z)\right)$

for $\gamma\in\Gamma$. If $f(\gamma_{1}z_{0})=f(\gamma_{2}z_{0})=0$ for
$\gamma_{1}$ and $\gamma_{2}$ in two different cosets in
$\Gamma_{\infty}\backslash\Gamma$, then we have
$f_{1}(z_{0})+2f_{2}(z_{0})\phi(z_{0})=f_{2}(z_{0})=0$. However, this implies
that $f_{0}(z_{0})\neq 0$ since $f_{j}(z)$ are assumed to have no common zeros
on $\mathbb{H}$, and hence $f(z_{0})\neq 0$, a contradiction. ∎

Now we let $\Gamma={\rm SL}(2,\mathbb{Z})$. Given a character $\chi$, we have
$\chi(T)={\rm e}^{2\pi{\rm i}m/24}$ for some integer $m\in[0,23]$. Recall the
Dedekind eta function

$\eta(z)={\rm e}^{2\pi{\rm i}z/24}\prod_{n=1}^{\infty}\big{(}1-{\rm
e}^{2n\pi{\rm i}z}\big{)}.$

Since the MODE associated to $f$ is the same as that associated to
$f/\eta^{m}$, by considering $f/\eta^{m}$ if necessary, we can always assume
$\chi(T)=1$ and so $\chi(S)=\chi(R)=1$, i.e., we can always assume that the
character $\chi$ is trivial for $\Gamma={\rm SL}(2,\mathbb{Z})$.

###### Proof of Theorem 1.1.

The conclusions (1) and (2) are proved in Theorem 2.2. It suffices to consider
$\Gamma=\mathrm{SL}(2,\mathbb{Z})$ and prove (3) and (4).

Let $z_{0}\in\\{{\rm i},\rho\\}$. By Lemma 2.5, we have $f(\gamma z_{0})\neq
0$ for some $\gamma\in\mathrm{SL}(2,\mathbb{Z})$. Then it follows from
Proposition 2.3 that

$\kappa_{z_{0}}^{(1)}=\kappa_{\gamma z_{0}}^{(1)}=\operatorname{ord}_{\gamma
z_{0}}\frac{f(z)}{\sqrt[3]{W_{f}(z)}}=-\frac{1}{3}\operatorname{ord}_{z_{0}}W_{f}(z).$

Since $W_{f}(z)$ is a modular form of weight $3k$ on
$\mathrm{SL}(2,\mathbb{Z})$, it follows from the valence formula for modular
forms (see, e.g., [30]) that

$\frac{\operatorname{ord}_{\rm
i}W_{f}}{2}+\frac{\operatorname{ord}_{\rho}W_{f}}{3}\equiv\frac{k}{4}\quad\
\mathrm{mod}\ 1.$

This implies
$\kappa_{\rho}^{(1)}=-\frac{\operatorname{ord}_{\rho}W_{f}}{3}\in\mathbb{Z}_{\leq
0}$ and $3\kappa_{\rm i}^{(1)}=-\operatorname{ord}_{\rm i}W_{f}\equiv k/2\
\mathrm{mod}\ 2$.

Recall that we may assume that $f_{0}$, $f_{1}$, and $f_{2}$ have no common
zeros. When $k\equiv 0\ \mathrm{mod}\ 4$, we have

$m_{\rm i}^{(2)}\equiv 3\kappa_{\rm i}^{(1)}\equiv\frac{k}{2}\equiv 0\quad\
\mathrm{mod}\ 2,$

and we are done. When $k\equiv 2\ \mathrm{mod}\ 4$, the weights of $f_{0}$ and
$f_{2}$ are congruent to $2$ modulo $4$ and their expansions in $w=(z-{\rm
i})/(z+{\rm i})$ are of the form

$f_{0}(z)=(1-w)^{k}\sum_{n=0}^{\infty}a_{2n+1}w^{2n+1},\qquad
f_{2}(z)=(1-w)^{k-4}\sum_{n=0}^{\infty}c_{2n+1}w^{2n+1},$

while the expansion of $f_{1}(z)$ is of the form

$f_{1}(z)=(1-w)^{k-2}\sum_{n=0}^{\infty}b_{2n}w^{2n}.$

(See Proposition 5.1 and Remark 5.2 below.)

Let $h_{j}(z)$, $j=1,2,3$, be given by (2.3). Then the local exponent of
$ah_{1}(z)+bh_{2}(z)+ch_{3}(z)$ at $z={\rm i}$ must be one of

$\big{\\{}0,\kappa_{\rm i}^{(2)}-\kappa_{\rm i}^{(1)},\kappa_{\rm
i}^{(3)}-\kappa_{\rm i}^{(1)}\big{\\}}$ (2.10)

for any $(a,b,c)\in\mathbb{C}^{3}$. Consider the function

$\displaystyle h_{1}+{\rm
i}h_{2}-h_{3}=z^{2}\big{(}f_{0}+f_{1}E_{2}+f_{2}E_{2}^{2}\big{)}+\alpha
z(f_{1}+2f_{2}E_{2})+\alpha^{2}f_{2}$ $\displaystyle\hphantom{h_{1}+{\rm
i}h_{2}-h_{3}=}{}+2{\rm i}z\big{(}f_{0}+f_{1}E_{2}+f_{2}E_{2}^{2}\big{)}+{\rm
i}\alpha(f_{1}+2f_{2}E_{2})-\big{(}f_{0}+f_{1}E_{2}+f_{2}E_{2}^{2}\big{)}$
$\displaystyle\hphantom{h_{1}+{\rm i}h_{2}-h_{3}}{}=(z+{\rm
i})^{2}f_{0}+(z+{\rm i})((z+{\rm i})E_{2}+\alpha)f_{1}+((z+{\rm
i})E_{2}+\alpha)^{2}f_{2}.$

We compute that $z={\rm i}(1+w)/(1-w)$ and hence

$z+{\rm i}=\frac{2{\rm i}}{1-w},\qquad\frac{{\rm d}w}{{\rm d}z}=\frac{2{\rm
i}}{(z+{\rm i})^{2}}=\frac{(1-w)^{2}}{2{\rm i}}.$ (2.11)

Also, since $E_{2}=\frac{1}{2\pi{\rm i}}{\rm d}\log\Delta(z)/{\rm d}z$ and the
expansion of $\Delta(z)$ is of the form

$\Delta(z)=(1-w)^{12}\sum_{n=0}^{\infty}d_{2n}w^{2n},$

we find that

$\displaystyle E_{2}(z)$
$\displaystyle=-\frac{(1-w)^{2}}{4\pi}\left(-\frac{12}{1-w}+\frac{\sum
2nd_{2n}w^{2n-1}}{\sum d_{2n}w^{2n}}\right)$
$\displaystyle=\frac{3}{\pi}(1-w)-\frac{(1-w)^{2}}{4\pi}\sum_{n=0}^{\infty}d_{2n+1}^{\prime}w^{2n+1}$
(2.12)

for some power series $\sum d_{2n+1}^{\prime}w^{2n+1}$. It follows that, by
(2.11),

$(z+{\rm i})E_{2}(z)+\frac{6}{\pi{\rm i}}=\frac{1-w}{2\pi{\rm
i}}\sum_{n=0}^{\infty}d_{2n+1}^{\prime}w^{2n+1}.$

From this, we see that

$h_{1}+{\rm i}h_{2}-h_{3}=(1-w)^{k-2}\sum_{n=0}^{\infty}e_{2n+1}w^{2n+1}$

for some $e_{j}$. This, together with (2.10), implies that either $\kappa_{\rm
i}^{(2)}-\kappa_{\rm i}^{(1)}$ or $\kappa_{\rm i}^{(3)}-\kappa_{\rm i}^{(1)}$
is odd. This proves the assertion (3) that $\big{\\{}3\kappa_{\rm
i}^{(1)},3\kappa_{\rm i}^{(2)},3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{0,0,1\\}$ $\ \mathrm{mod}\ 2$.

The proof of (4) is similar. We consider the function

$h_{1}-\overline{\rho}h_{2}+\overline{\rho}^{2}h_{3}=(z-\overline{\rho})^{2}f_{0}+(z-\overline{\rho})((z-\overline{\rho})E_{2}+\alpha)f_{1}+((z-\overline{\rho})E_{2}+\alpha)^{2}f_{2}.$

Setting $w=(z-\rho)/(z-\overline{\rho})$, we have
$z=(\rho-\overline{\rho}w)/(1-w)$,

$z-\overline{\rho}=\frac{\sqrt{3}{\rm i}}{1-w},\qquad\frac{{\rm d}w}{{\rm
d}z}=\frac{(1-w)^{2}}{\sqrt{3}{\rm i}},$

and

$(z-\overline{\rho})E_{2}(z)+\frac{6}{\pi{\rm i}}=\frac{1-w}{2\pi{\rm
i}}\sum_{n=0}^{\infty}d_{3n+2}^{\prime}w^{3n+2}$ (2.13)

(since the expansion of $\Delta(z)$ is $(1-w)^{12}\sum d_{3n}w^{3n}$ for some
$d_{3n}$). When $k\equiv 1\ \mathrm{mod}\ 3$, the expansions of $f_{j}$ are of
the form $f_{0}=(1-w)^{k}\sum a_{3n+1}w^{3n+1}$, $f_{1}=(1-w)^{k-2}\sum
b_{3n+2}w^{3n+2}$, and $f_{2}=(1-w)^{k-4}\sum c_{3n}w^{3n}$. Therefore, the
expansion of $h_{1}-\overline{\rho}h_{2}+\overline{\rho}^{2}h_{3}$ is of the
form

$h_{1}-\overline{\rho}h_{2}+\overline{\rho}^{2}h_{3}=(1-w)^{k-2}\sum_{n=0}^{\infty}e_{3n+1}w^{3n+1},$

which implies that either $\kappa_{\rho}^{(2)}-\kappa_{\rho}^{(1)}$ or
$\kappa_{\rho}^{(3)}-\kappa_{\rho}^{(1)}$ is congruent to $1$ modulo $3$.
Since the sum of $\kappa_{\rho}^{(j)}$ is $3$, we deduce that the set
$\big{\\{}\kappa_{\rho}^{(1)},\kappa_{\rho}^{(2)},\kappa_{\rho}^{(3)}\big{\\}}$
is congruent to $\\{0,1,2\\}$ modulo $3$. Likewise, when $k\equiv 2\
\mathrm{mod}\ 3$, we can show that
$h_{1}-\overline{\rho}h_{2}+\overline{\rho}^{2}h_{3}=(1-w)^{k-2}\sum
e_{3n+2}w^{3n+2}$ and obtain the same conclusion.

For the case $k\equiv 0\ \mathrm{mod}\ 3$, we need to make the computation
more precise. Let

$\displaystyle f_{0}(z)=(1-w)^{k}\sum_{n=0}^{\infty}a_{3n}w^{3n},$
$\displaystyle f_{1}(z)=(1-w)^{k-2}\sum_{n=0}^{\infty}b_{3n+1}w^{3n+1},$
$\displaystyle f_{2}(z)=(1-w)^{k-4}\sum_{n=0}^{\infty}c_{3n+2}w^{3n+2}$

be the expansions of $f_{j}$. A computation similar to (2.12) yields

$E_{2}(z)=\frac{2\sqrt{3}}{\pi}(1-w)-\frac{(1-w)^{2}}{2\pi\sqrt{3}}\sum_{n=0}^{\infty}d_{3n+2}^{\prime}w^{3n+2}$

and hence

$\displaystyle f(z)=(1-w)^{k}\sum_{n=0}^{\infty}a_{3n}w^{3n}$
$\displaystyle\hphantom{f(z)=}{}+(1-w)^{k-1}\left(\frac{2\sqrt{3}}{\pi}-\frac{1-w}{2\pi\sqrt{3}}\sum_{n=0}^{\infty}d_{3n+2}^{\prime}w^{3n+2}\right)\left(\sum_{n=0}^{\infty}b_{3n+1}w^{3n+1}\right)$
$\displaystyle\hphantom{f(z)=}{}+(1-w)^{k-2}\left(\frac{2\sqrt{3}}{\pi}-\frac{1-w}{2\pi\sqrt{3}}\sum_{n=0}^{\infty}d_{3n+2}^{\prime}w^{3n+2}\right)^{2}\left(\sum_{n=0}^{\infty}c_{3n+2}w^{3n+2}\right).$

On the other hand, by (2.13), we have

$\displaystyle h_{1}-\overline{\rho}h_{2}+\overline{\rho}^{2}h_{3}$
$\displaystyle\qquad=(1-w)^{k-2}\left(-3\sum_{n=0}^{\infty}a_{3n}w^{3n}+\frac{\sqrt{3}}{2\pi}\left(\sum_{n=0}^{\infty}d_{3n+2}^{\prime}w^{3n+2}\right)\left(\sum_{n=0}^{\infty}b_{3n+1}w^{3n+1}\right)\right.$
$\displaystyle\left.\qquad\quad{}-\frac{1}{4\pi^{2}}\left(\sum_{n=0}^{\infty}d_{3n+2}^{\prime}w^{3n+2}\right)^{2}\left(\sum_{n=0}^{\infty}c_{3n+2}w^{3n+2}\right)\right).$

We then check that the expansion of
$h_{1}-\overline{\rho}h_{2}+\overline{\rho}^{2}h_{3}+3f$ is of the form

$(1-w)^{k-2}\sum_{n=0}^{\infty}\big{(}e_{3n+1}w^{3n+1}+e_{3n+2}w^{3n+2}\big{)},$

which again implies that either $\kappa_{\rho}^{(2)}-\kappa_{\rho}^{(1)}$ or
$\kappa_{\rho}^{(3)}-\kappa_{\rho}^{(1)}$ is not congruent to $0$ modulo $3$
and hence
$\big{\\{}\kappa_{\rho}^{(1)},\kappa_{\rho}^{(2)},\kappa_{\rho}^{(3)}\big{\\}}\equiv\\{0,1,2\\}\
\mathrm{mod}\ 3$. This completes the proof. ∎

##  3 The MODE on $\boldsymbol{\mathrm{SL}(2,\mathbb{Z})}$

The purpose of this section is to prove Theorems 1.3 and 1.4, and Corollary
1.5. Let $Q_{2}(z)$ and $Q_{3}(z)-\frac{1}{2}Q_{2}(z)$ be meromorphic modular
forms on $\mathrm{SL}(2,\mathbb{Z})$ of weight $4$ and $6$ respectively, i.e.,

$Ly:=y^{\prime\prime\prime}(z)+Q_{2}(z)y^{\prime}(z)+Q_{3}(z)y(z)=0,\qquad
z\in\mathbb{H}$ (3.1)

is a MODE. In this section, we use the notations
$S=\bigl{(}\begin{smallmatrix}0&-1\\\ 1&0\end{smallmatrix}\bigr{)}$,
$T=\bigl{(}\begin{smallmatrix}1&1\\\ 0&1\end{smallmatrix}\bigr{)}$ and
$R=ST=\bigl{(}\begin{smallmatrix}0&-1\\\ 1&1\end{smallmatrix}\bigr{)}$ such
that $S^{2}=R^{3}=-I_{2}$. Suppose the MODE (3.1) has regular singularities at
$\\{z_{1},\dots,z_{m}\\}\sqcup\\{{\rm i},\rho,\infty\\}$ mod
$\mathrm{SL}(2,\mathbb{Z})$, where $\rho=\big{(}{-}1+\sqrt{3}{\rm i}\big{)}/2$
is the fixed point of $R$.

###  3.1 Proof of Theorem 1.4(1)

First we want to give the proof of Theorem 1.4(1), which is long and will be
separated into several lemmas. For this purpose, throughout this section we
always assume that

  * (S1)

$\kappa_{\infty}^{(1)},\kappa_{\rm
i}^{(1)},\kappa_{z_{j}}^{(1)}\in\frac{1}{3}\mathbb{Z}_{\leq 0}$ and
$\kappa_{\rho}^{(1)}\in\mathbb{Z}_{\leq 0}$ such that

$\ell:=-2-12\kappa_{\infty}^{(1)}-4\kappa_{\rho}^{(1)}-6\kappa_{\rm
i}^{(1)}-12\sum_{j=1}^{m}\kappa_{z_{j}}^{(1)}\in\mathbb{Z}.$

(Note $\ell\in\mathbb{Z}$ implies $\ell\in 2\mathbb{Z}$.) Furthermore,
$m_{z}^{(j)}:=\kappa_{z}^{(j+1)}-\kappa_{z}^{(j)}-1\in\mathbb{Z}_{\geq 0}$ for
$z\in\\{z_{1},\dots,z_{m}\\}\sqcup\\{{\rm i},\rho\\}$ and $j=1,2$, and
$m_{\infty}^{(j)}:=\kappa_{\infty}^{(j+1)}-\kappa_{\infty}^{(j)}\in\mathbb{Z}_{\geq
0}$ for $j=1,2$.

  * (S2)

ODE (3.1) is apparent at any singular point
$z\in\\{z_{1},\dots,z_{m}\\}\sqcup\\{{\rm i},\rho\\}$.

  * (S3)

$\big{\\{}3\kappa_{\rm i}^{(1)},3\kappa_{\rm i}^{(2)},3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{0,0,1\\}\ \mathrm{mod}\ 2$.

######  Remark 3.1.

Note that (S1) and (S2) are equivalent to the assumptions (H1)–(H3) and
$\kappa_{\rho}^{(1)}\in\mathbb{Z}_{\leq 0}$ in Theorem 1.3, while (S3) is
needed to obtain quasimodular forms as stated in Theorem 1.3(1). In view of
Theorem 1.1, the above assumptions (S1)–(S3) are necessary for the validity of
Theorem 1.4(1). We will prove below that they are also sufficient.

Set $t_{j}:=E_{4}(z_{j})^{3}/E_{6}(z_{j})^{2}\not\in\\{0,1,\infty\\}$ and
$F_{j}(z):=E_{4}(z)^{3}-t_{j}E_{6}(z)^{2}$. Then this modular form $F_{j}(z)$
has only one simple zero at $z_{j}$, up to
$\mathrm{SL}(2,\mathbb{Z})$-equivalence. Define

$F(z):=\Delta(z)^{-\kappa_{\infty}^{(1)}}E_{4}(z)^{-\kappa_{\rho}^{(1)}}E_{6}(z)^{-\kappa_{\rm
i}^{(1)}}\prod_{j=1}^{m}F_{j}(z)^{-\kappa_{z_{j}}^{(1)}}.$

Then $F(z)^{3}$ is a modular form of weight $3(\ell+2)$.

Clearly for any solution $y(z)$ of (3.1),

$\hat{y}(z):=F(z)y(z)$

is single-valued and holomorphic on $\mathbb{H}$. Furthermore, its order at
$z\in\\{z_{1},\dots,z_{m}\\}\sqcup\\{{\rm i},\rho\\}$ is one of
$\big{\\{}0,m_{z}^{(1)}+1,m_{z}^{(1)}+m_{z}^{(2)}+2\big{\\}}$, and at $\infty$
is one of
$\big{\\{}0,m_{\infty}^{(1)},m_{\infty}^{(1)}+m_{\infty}^{(2)}\big{\\}}$.

Fix a fundamental system of solutions $Y(z)=(y_{1}(z)$,
$y_{2}(z),y_{3}(z))^{t}$ of (3.1) and let $\hat{Y}(z):=F(z)Y(z)$. Then for any
$\gamma\in\mathrm{SL}(2,\mathbb{Z})$, there is a matrix
$\hat{\rho}(\gamma)\in{\rm GL}(3,\mathbb{C})$ such that

$\big{(}\hat{Y}\big{|}_{\ell}\gamma\big{)}(z)=\hat{\rho}(\gamma)\hat{Y}(z).$
(3.2)

This is a lifting of the Bol representation. We will use freely the notation
$\hat{\gamma}=\hat{\rho}(\gamma)$ just for convenience.

######  Lemma 3.2.

There holds $\det\hat{\rho}(\gamma)=1$ for any
$\gamma\in\mathrm{SL}(2,\mathbb{Z})$. That is, $\hat{\rho}$ is a group
homomorphism from $\mathrm{SL}(2,\mathbb{Z})$ to $\mathrm{SL}(3,\mathbb{C})$.

###### Proof.

The proof is similar to that of [25, Lemma 4.2], where the second-order MODE
was studied. Let

$W(z)=\det\begin{pmatrix}y_{1}&y_{1}^{\prime}&y_{1}^{\prime\prime}\\\
y_{2}&y_{2}^{\prime}&y_{2}^{\prime\prime}\\\
y_{3}&y_{3}^{\prime}&y_{3}^{\prime\prime}\end{pmatrix},\qquad\hat{W}(z)=\det\begin{pmatrix}\hat{y}_{1}&\hat{y}_{1}^{\prime}&\hat{y}_{1}^{\prime\prime}\\\
\hat{y}_{2}&\hat{y}_{2}^{\prime}&y_{2}^{\prime\prime}\\\
\hat{y}_{3}&\hat{y}_{3}^{\prime}&\hat{y}_{3}^{\prime\prime}\end{pmatrix}.$

Then $W(z)\equiv C$ is a nonzero constant. By (3.2) we have

$\big{(}\hat{W}\big{|}_{3(\ell+2)}\gamma\big{)}(z)=\det\hat{\rho}(\gamma)\hat{W}(z).$

Since $\hat{W}(z)=F(z)^{3}W(z)=CF(z)^{3}$, we also have

$\displaystyle\hat{W}\big{|}_{3(\ell+2)}\gamma=C\big{(}F^{3}\big{|}_{3(\ell+2)}\gamma\big{)}=\frac{\big{(}F^{3}\big{|}_{3(\ell+2)}\gamma\big{)}(z)}{F(z)^{3}}\hat{W}(z).$

Thus

$\det\hat{\rho}(\gamma)=\frac{\big{(}F^{3}\big{|}_{3(\ell+2)}\gamma\big{)}(z)}{F(z)^{3}}.$

This proves $\det\hat{\rho}(\gamma)=1$ because $F(z)^{3}$ is a modular form of
weight $3(\ell+2)$ on $\mathrm{SL}(2,\mathbb{Z})$. ∎

Remark that under our assumption $\kappa_{\infty}^{(1)},\kappa_{\rm
i}^{(1)},\kappa_{z_{j}}^{(1)}\in\frac{1}{3}\mathbb{Z}_{\leq 0}$ and
$\kappa_{\rho}^{(1)}\in\mathbb{Z}_{\leq 0}$, $y(z)^{3}$ is a single-valued and
meromorphic function on $\mathbb{H}$ for any solution $y(z)$ of (3.1).

######  Lemma 3.3.

Under the assumptions $(S1)$–$(S3)$, there is at least one solution $y(z)$ of
(3.1) such that $y(z)^{3}$ is not a meromorphic modular form of weight $-6$.

###### Proof.

Suppose the conclusion is not true, namely $y(z)^{3}$ is a meromorphic modular
form of weight $-6$ for any solution $y(z)$. Then by the well-known valence
formula for modular forms (see, e.g., [30]), we obtain

$\frac{\operatorname{ord}_{\rm
i}\big{(}y^{3}\big{)}}{2}+\frac{\operatorname{ord}_{\rho}\big{(}y^{3}\big{)}}{3}\equiv\frac{1}{2}\quad\
\mathrm{mod}\ \mathbb{Z},$

so $\operatorname{ord}_{\rm i}\big{(}y^{3}\big{)}$ is odd. Since
$\operatorname{ord}_{\rm i}\big{(}y^{3}\big{)}$ can be chosen as any one of
$3\kappa_{\rm i}^{(1)}$, $3\kappa_{\rm i}^{(2)}$, $3\kappa_{\rm i}^{(3)}$,
these three numbers are all odd, clearly a contradiction with our assumption
(S3). ∎

######  Lemma 3.4.

Under the assumptions $(S1)$–$(S3)$, (3.1) is completely not apparent at
$\infty$.

To prove Lemma 3.4, we need the following well-known lemma due to Beukers and
Heckman [2].

######  Lemma 3.5 ([2]).

Let $n\geq 2$ and $H\subset{\rm GL}(n,\mathbb{C})$ be a subgroup generated by
two matrices $A$, $B$ such that $\operatorname{rank}(A-B)\leq 1$. Then $H$
acts irreducibly on $\mathbb{C}^{n}$ if and only if $A$ and $B$ have distinct
eigenvalues.

Let $\\{a_{1},\dots,a_{n}\\}$ and $\\{b_{1},\dots,b_{n}\\}$ be the eigenvalues
of $A$ and $B$ respectively. The following lemma, due to Levelt, is to recover
$A$ and $B$ by their eigenvalues. See [2] for a proof.

######  Lemma 3.6 (cf. [2]).

Suppose that $\operatorname{rank}(A-B)=1$ and $a_{1},\dots,a_{n}$,
$b_{1},\dots,b_{n}$ are all nonzero complex numbers with $a_{i}\neq b_{j}$ for
any $i$, $j$. Then up to a common conjugation in ${\rm GL}(n,\mathbb{C})$, $A$
and $B$ can be uniquely determined by

$A=\begin{pmatrix}0&0&\cdots&0&-A_{n}\\\ 1&0&\cdots&0&-A_{n-1}\\\
0&1&\cdots&0&-A_{n-2}\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&\cdots&1&-A_{1}\end{pmatrix},\qquad
B=\begin{pmatrix}0&0&\cdots&0&-B_{n}\\\ 1&0&\cdots&0&-B_{n-1}\\\
0&1&\cdots&0&-B_{n-2}\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&\cdots&1&-B_{1}\end{pmatrix},$

where $A_{j}$’s and $B_{j}$’s are given by

$\displaystyle\prod_{j=1}^{n}(t-a_{j})=t^{n}+A_{1}t^{n-1}+\cdots+A_{n},$
$\displaystyle\prod_{j=1}^{n}(t-b_{j})=t^{n}+B_{1}t^{n-1}+\cdots+B_{n}.$

######  Corollary 3.7.

Let $A$, $B$ be $3\times 3$ matrices such that $\operatorname{rank}(A-B)\leq
1$. Suppose that $A^{2}=I_{3}$ and the eigenvalues of $A$ are $\\{1,-1,-1\\}$.
Then $A$, $B$ have common eigenvalues and so the group $H$ generated by $A$
and $B$ acts on $\mathbb{C}^{3}$ reducibly.

###### Proof.

This corollary is trivial if $\operatorname{rank}(A-B)=0$, i.e., $A=B$. So we
may assume $\operatorname{rank}(A-B)\allowbreak=1$. Assume by contradiction
that $A$ and $B$ have no common eigenvalues, then by
$(t-1)(t+1)^{2}=t^{3}+t^{2}-t-1$, we see from Lemma 3.6 that $A$ is conjugate
to

$\begin{pmatrix}0&0&1\\\ 1&0&1\\\ 0&1&-1\end{pmatrix}$

and so $A^{2}\neq I_{3}$, a contradiction with the assumption $A^{2}=I_{3}$.

Thus $A$ and $B$ have common eigenvalues, and it follows from Lemma 3.5 that
$H$ acts on $\mathbb{C}^{3}$ reducibly. ∎

###### Proof of Lemma 3.4.

Suppose that the statement is not true. Then
$\operatorname{rank}\big{(}\hat{T}-I_{3}\big{)}\leq 1$ (note $\hat{T}=I_{3}$
if and only if $\infty$ is apparent). Then $\hat{R}=\hat{S}\hat{T}$ implies
$\operatorname{rank}\big{(}\hat{S}-\hat{R}\big{)}\leq 1$.

By
$\hat{R}^{3}=\hat{\rho}\big{(}R^{3}\big{)}=\hat{\rho}(-I_{2})=(-1)^{\ell}I_{3}=I_{3}$
and $\det\hat{R}=1$, we have either $\hat{R}=\lambda I_{3}$ for some
$\lambda^{3}=1$ or $\hat{R}$ is conjugate to
$\operatorname{diag}\big{(}1,\varepsilon,\varepsilon^{2}\big{)}$ where
$\varepsilon={\rm e}^{2\pi{\rm i}/3}$. Similarly, by
$\hat{S}^{2}=\hat{\rho}\big{(}S^{2}\big{)}=\hat{\rho}(-I_{2})\allowbreak=I_{3}$
and $\det\hat{S}=1$, we have either $\hat{S}=I_{3}$ or $\hat{S}$ is a
conjugate of $\operatorname{diag}(1,-1,-1)$.

If $\hat{S}=I_{3}$, then by
$\operatorname{rank}\big{(}\hat{S}-\hat{R}\big{)}\leq 1$ we obtain
$\hat{R}=I_{3}$. This implies that for any solution $y(z)$ of (3.1),
$\hat{y}(z)^{3}=F(z)^{3}y(z)^{3}$ is a modular form of weight $3\ell$ and so
$y(z)^{3}$ is a meromorphic modular form of weight $-6$, a contradiction with
Lemma 3.3.

Thus $\hat{S}$ is a conjugate of $\operatorname{diag}(1,-1,-1)$. If
$\hat{R}=\lambda I_{3}$ for some $\lambda^{3}=1$, then by $\lambda\neq-1$ we
obtain

$1\geq\operatorname{rank}\big{(}\hat{S}-\hat{R}\big{)}=\operatorname{rank}(\operatorname{diag}(1-\lambda,-1-\lambda,-1-\lambda))\geq
2,$

a contradiction.

So $\hat{R}$ is conjugate to
$\operatorname{diag}\big{(}1,\varepsilon,\varepsilon^{2}\big{)}$. By Corollary
3.7, there is a subspace $V\subsetneqq\mathbb{C}^{3}$ which is invariant under
the actions $\hat{S}$ and $\hat{R}$. If $\dim V=2$, then there is an
invertible matrix $P$ such that

$P\hat{S}P^{-1}=\begin{pmatrix}A_{1}&0\\\ *&a_{1}\end{pmatrix},\qquad
P\hat{R}P^{-1}=\begin{pmatrix}B_{1}&0\\\ *&b_{1}\end{pmatrix},$

where $A_{1}$ and $B_{1}$ are $2\times 2$ matrices. This implies

$\operatorname{rank}\begin{pmatrix}A_{1}-B_{1}&0\\\
*&a_{1}-b_{1}\end{pmatrix}=\operatorname{rank}\big{(}\hat{R}-\hat{S}\big{)}\leq
1.$ (3.3)

Note $a_{1}\in\\{1,-1\\}$ and
$b_{1}\in\big{\\{}1,\varepsilon,\varepsilon^{2}\big{\\}}$. If $a_{1}\neq
b_{1}$, then (3.3) implies $A_{1}=B_{1}$, namely $\hat{S}$ and $\hat{R}$ have
two common eigenvalues, a contradiction. So $a_{1}=b_{1}=1$. Then the
eigenvalues of $A_{1}$ are $\\{-1,-1\\}$, so $A_{1}=-I_{2}$. Similarly,
$B_{1}$ is conjugate to
$\operatorname{diag}\big{(}\varepsilon,\varepsilon^{2}\big{)}$. Thus

$1\geq\operatorname{rank}(A_{1}-B_{1})=\operatorname{rank}\big{(}\operatorname{diag}\big{(}{-}1-\varepsilon,-1-\varepsilon^{2}\big{)}\big{)}=2,$

a contradiction. So $\dim V=1$, which implies the existence of an invertible
matrix $P$ such that

$P\hat{S}P^{-1}=\begin{pmatrix}a_{1}&0\\\ *&A_{1}\end{pmatrix},\qquad
P\hat{R}P^{-1}=\begin{pmatrix}b_{1}&0\\\ *&B_{1}\end{pmatrix},$

where $A_{1}$ and $B_{1}$ are $2\times 2$ matrices. Clearly the same argument
as (3.3) also yields a contradiction. This completes the proof. ∎

Let $\hat{y}_{3}(z):=\hat{y}_{+}(z)=F(z)y_{+}(z)$,
$\hat{y}_{1}(z):=(\hat{y}_{3}\big{|}_{\ell}S)(z)$ and
$\hat{y}_{2}(z):=(\hat{y}_{3}\big{|}_{\ell}R)(z)$, and
$y_{j}(z):=\hat{y}_{j}(z)/F(z)$ for $j=1,2,3$.

######  Lemma 3.8.

Under the assumptions $(S1)$–$(S3)$, $\hat{y}_{1}(z)$, $\hat{y}_{2}(z)$ and
$\hat{y}_{3}(z)$ are linearly independent and $\hat{y}_{1}(z)$ can be written
as

$\hat{y}_{1}(z)=\beta z^{2}\hat{y}_{3}(z)+z\hat{m}_{1}^{*}(z)+\hat{m}_{2}(z),$
(3.4)

where $\beta\neq 0$ is a constant and $\hat{m}_{1}^{*}(z)$, $\hat{m}_{2}(z)$
are of the form

$\frac{\hat{m}_{1}^{*}(z)}{F(z)}=m_{1}^{*}(z)=q^{\kappa_{\infty}^{(2)}}\sum_{j\geq
0}c_{j,1}q^{j},\qquad\frac{\hat{m}_{2}(z)}{F(z)}=m_{2}(z)=q^{\kappa_{\infty}^{(1)}}\sum_{j\geq
0}c_{j,2}q^{j}.$ (3.5)

###### Proof.

Under our assumption, Lemma 3.4 says that $\infty$ is completely not apparent,
so it follows from Remark A.9 that (3.1) has a basis of solutions of the form
$(y_{-},y_{\perp},y_{+})$, where $y_{+}$, $y_{\perp}$, and $y_{-}$ are given
by (A.19) and (A.20).

Step 1. We show that $\hat{y}_{1}(z)$ is linearly independent with
$\hat{y}_{3}(z)$.

Suppose not, i.e., there is some constant $\alpha\neq 0$ such that

$(\hat{y}_{3}\big{|}_{\ell}S)(z)=\hat{y}_{1}(z)=\alpha\hat{y}_{3}(z).$

Then with respect to $(Fy_{-},Fy_{\perp},\hat{y}_{3}(z))^{t}$, we have

$\hat{\rho}(S)=\begin{pmatrix}S_{1}&*\\\ 0&\alpha\end{pmatrix},$

where $S_{1}$ is a $2\times 2$ matrix. Then $\hat{\rho}(S)^{2}=I_{3}$ implies
$S_{1}^{2}=I_{2}$ and $\alpha^{2}=1$, i.e., $\alpha=\pm 1$.

On the other hand, it follows from the expressions of
$(y_{-},y_{\perp},y_{+})$ in Remark A.9 that with respect to
$(Fy_{-},Fy_{\perp},\hat{y}_{3}(z))^{t}$,

$\hat{\rho}(T)=\begin{pmatrix}1&2&*\\\ 0&1&1\\\ 0&0&1\end{pmatrix},$

so

$\hat{\rho}(R)=\hat{\rho}(S)\hat{\rho}(T)=\begin{pmatrix}S_{1}&*\\\
0&\alpha\end{pmatrix}\begin{pmatrix}1&2&*\\\ 0&1&1\\\ 0&0&1\end{pmatrix}.$

From this and $\hat{\rho}(R)^{3}=I_{3}$, we obtain

$\left(S_{1}R_{1}\right)^{3}=I_{2},\qquad\text{where}\quad
R_{1}:=\begin{pmatrix}1&2\\\ 0&1\end{pmatrix},$ (3.6)

and $\alpha^{3}=1$, so $\alpha=1$. Then $\det S_{1}=\det\hat{\rho}(S)=1$,
which together with $S_{1}^{2}=I_{2}$ easily implies $S_{1}=\pm I_{2}$, and
then (3.6) yields $\pm\bigl{(}\begin{smallmatrix}1&6\\\
0&1\end{smallmatrix}\bigr{)}=I_{2}$, a contradiction.

Step 2. By Step 1, there exists $(\beta,\delta)\neq(0,0)$ and $\epsilon$ such
that

$\hat{y}_{1}=\beta Fy_{-}+\delta Fy_{\perp}+\epsilon\hat{y}_{3}.$ (3.7)

We claim that $\beta\neq 0$ and (3.4) holds.

Assume by contradiction that $\beta=0$, i.e., $\hat{y}_{1}=\delta
Fy_{\perp}+\epsilon\hat{y}_{3}$ with $\delta\neq 0$. Then with respect to
$(Fy_{-},\hat{y}_{1},\hat{y}_{3})$ we have

$\hat{\rho}(T)=\begin{pmatrix}1&\frac{2}{\delta}&a\\\ 0&1&\delta\\\
0&0&1\end{pmatrix}\qquad\text{for some }a\in\mathbb{C},$

and

$\hat{\rho}(S)=\begin{pmatrix}-1&b&b\\\ 0&0&1\\\
0&1&0\end{pmatrix}\qquad\text{for some }b\in\mathbb{C},$

where we used $\hat{\rho}(S)^{2}=I_{3}$, $\det\hat{\rho}(S)=1$,
$\hat{y}_{1}(z)=(\hat{y}_{3}\big{|}_{\ell}S)(z)$ and

$(\hat{y}_{1}\big{|}_{\ell}S)(z)=(\hat{y}_{3}\big{|}_{\ell}(-I_{2}))(z)=(-1)^{-\ell}\hat{y}_{3}(z)=\hat{y}_{3}(z).$
(3.8)

Thus

$\hat{\rho}(R)=\hat{\rho}(S)\hat{\rho}(T)=\begin{pmatrix}-1&b-\frac{2}{\delta}&b+b\delta-a\\\
0&0&1\\\ 0&1&\delta\end{pmatrix}.$

But this implies that $-1$ is an eigenvalue of $\hat{\rho}(R)$, a
contradiction with $\hat{\rho}(R)^{3}=I_{3}$. This proves $\beta\neq 0$ and so
it follows from the expression of $(y_{-},y_{\perp},y_{+})$ in Remark A.9 that
(3.4) and (3.5) hold.

Step 3. We show that $\hat{y}_{1}(z)$, $\hat{y}_{2}(z)$ and $\hat{y}_{3}(z)$
are linearly independent.

In fact, we see from $R=ST$ that
$\hat{y}_{2}(z)=(\hat{y}_{+}\big{|}_{\ell}R)(z)=(\hat{y}_{1}\big{|}_{\ell}T)(z)=\hat{y}_{1}(z+1)$,
from which and (3.4) we obtain

$\hat{y}_{2}=\hat{y}_{1}+2\beta
z\hat{y}_{3}+\hat{m}_{1}^{*}+\beta\hat{y}_{3}.$ (3.9)

By (3.7) and (3.9) we have

$\begin{pmatrix}\hat{y}_{1}\\\ \hat{y}_{2}\\\
\hat{y}_{3}\end{pmatrix}=A\begin{pmatrix}Fy_{-}\\\ Fy_{\perp}\\\
\hat{y}_{3}\end{pmatrix}\qquad\text{with}\quad
A=\begin{pmatrix}\beta&\delta&\epsilon\\\ \beta&2\beta+\delta&*\\\
0&0&1\end{pmatrix}.$

Since $\det A=2\beta^{2}\neq 0$, we obtain that $\hat{y}_{1}(z)$,
$\hat{y}_{2}(z)$ and $\hat{y}_{3}(z)$ are also linearly independent. This
completes the proof. ∎

Recalling (3.4), we set $\hat{m}_{1}(z)$ and $\hat{m}_{0}(z)$ to be

$\displaystyle\hat{m}_{1}^{*}(z)=\hat{m}_{1}(z)+\frac{\pi{\rm
i}}{3}\hat{m}_{2}(z)E_{2}(z),$ (3.10)
$\displaystyle\hat{y}_{3}(z)=\left(\frac{\pi{\rm
i}}{6}\right)^{2}\hat{m}_{2}(z)E_{2}(z)^{2}+\frac{\pi{\rm
i}}{6}\hat{m}_{1}(z)E_{2}(z)+\hat{m}_{0}(z).$ (3.11)

The following result can be seen as the converse statement of Theorem 2.2.

######  Theorem 3.9.

Under the assumptions $(S1)$–$(S3)$, the following hold.

  * $(a)$

$\beta=1$.

  * $(b)$

$\hat{m}_{j}(z)$ are meromorphic modular forms of weight $\ell+2-2j$ for
$j=0,1,2$, that is, $\hat{y}_{3}(z)$ is a quasimodular form of weight $\ell+2$
with depth $2$.

###### Proof.

(a) By Lemma 3.8, we can take
$\hat{Y}(z)=(\hat{y}_{1}(z),\hat{y}_{2}(z),\hat{y}_{3}(z))^{t}$ to be a basis
and let $\hat{T}$, $\hat{S}$, $\hat{R}$ denote the associated matrices
$\hat{\rho}(T)$, $\hat{\rho}(S)$ and $\hat{\rho}(R)$ of the Bol
representation. Recalling (3.8) that
$(\hat{y}_{3}\big{|}_{\ell}S)=\hat{y}_{1}(z)$ and
$(\hat{y}_{1}\big{|}_{\ell}S)=\hat{y}_{3}(z)$, we have

$\hat{S}=\begin{pmatrix}0&0&1\\\ \lambda&-1&\lambda\\\
1&0&0\end{pmatrix},\qquad\text{for some }\lambda\in\mathbb{C},$ (3.12)

where $\hat{S}^{2}=I_{3}$ is used. Note from $SR=S^{2}T=-T$ that

$\hat{y}_{1}\big{|}_{\ell}R=\hat{y}_{3}\big{|}_{\ell}SR=\hat{y}_{3}\big{|}_{\ell}(-T)=\hat{y}_{3},$

and from $R^{2}=T^{-1}S$ that

$\hat{y}_{2}\big{|}_{\ell}R=\hat{y}_{3}\big{|}_{\ell}R^{2}=\hat{y}_{3}\big{|}_{\ell}\big{(}T^{-1}S\big{)}=\hat{y}_{3}\big{|}_{\ell}S=\hat{y}_{1},$

so

$\hat{R}=\begin{pmatrix}0&0&1\\\ 1&0&0\\\ 0&1&0\end{pmatrix}.$ (3.13)

Therefore,

$\displaystyle\hat{T}=\hat{S}^{-1}\hat{R}=\hat{S}\hat{R}=\begin{pmatrix}0&0&1\\\
\lambda&-1&\lambda\\\ 1&0&0\end{pmatrix}\begin{pmatrix}0&0&1\\\ 1&0&0\\\
0&1&0\end{pmatrix}=\begin{pmatrix}0&1&0\\\ -1&\lambda&\lambda\\\
0&0&1\end{pmatrix}.$ (3.14)

On the other hand, by (3.9) we have

$\displaystyle\hat{y}_{2}\big{|}_{\ell}T=\hat{y}_{1}\big{|}_{\ell}T+2\beta
z\hat{y}_{3}+\hat{m}_{1}^{*}+3\beta\hat{y}_{3}=-\hat{y}_{1}+2\hat{y}_{2}+2\beta\hat{y}_{3}.$

Hence (3.14) yields $\lambda=2$ and $\beta=1$. This proves (a). In particular,
it is easy to see from (3.12)–(3.14) that the representation $\hat{\rho}$ is
irreducible, i.e., there is no proper nontrivial subspace of
$\mathbb{C}\hat{y}_{1}+\mathbb{C}\hat{y}_{2}+\mathbb{C}\hat{y}_{3}$ that is
invariant under $\hat{\rho}(\mathrm{SL}(2,\mathbb{Z}))$.

(b) To prove (b), we note that (3.4) and (3.9) become

$\displaystyle\hat{y}_{1}(z)=z^{2}\hat{y}_{3}(z)+z\hat{m}_{1}^{*}(z)+\hat{m}_{2}(z),$
(3.15)
$\displaystyle\hat{y}_{2}=\hat{y}_{1}+2z\hat{y}_{3}+\hat{m}_{1}^{*}+\hat{y}_{3},$
(3.16)

which implies that

$y^{*}(z):=2zy_{3}(z)+m_{1}^{*}(z)$

is also a solution of (3.1) and

$\hat{y}_{2}=\hat{y}_{1}+\hat{y}^{*}+\hat{y}_{3},\qquad\text{where}\quad\hat{y}^{*}=Fy^{*}=2z\hat{y}_{3}+\hat{m}_{1}^{*}.$

By (3.12) and $\lambda=2$, we have

$2\big{(}\tfrac{-1}{z}\big{)}\hat{y}_{1}+\hat{m}_{1}^{*}\big{|}_{\ell}S=\hat{y}^{*}\big{|}_{\ell}S=(\hat{y}_{2}-\hat{y}_{1}-\hat{y}_{3})\big{|}_{\ell}S=\hat{y}_{1}-\hat{y}_{2}+\hat{y}_{3}=-\hat{y}^{*}=-2z\hat{y}_{3}-\hat{m}_{1}^{*}.$

From here and (3.15), we obtain

$z(\hat{m}_{1}^{*}\big{|}_{\ell}S)(z)=z\hat{m}_{1}^{*}(z)+2\hat{m}_{2}(z).$
(3.17)

On the other hand, by (3.15),

$\hat{y}_{3}=\hat{y}_{1}\big{|}_{\ell}S=\big{(}\tfrac{-1}{z}\big{)}^{2}\hat{y}_{1}+\big{(}\tfrac{-1}{z}\big{)}(\hat{m}_{1}^{*}\big{|}_{\ell}S)+\hat{m}_{2}\big{|}_{\ell}S,$

which implies

$\hat{y}_{1}=z^{2}\hat{y}_{3}+z(\hat{m}_{1}^{*}\big{|}_{\ell}S)-z^{2}(\hat{m}_{2}\big{|}_{\ell}S).$

Again by (3.15), we have

$z(\hat{m}_{1}^{*}\big{|}_{\ell}S)(z)=z\hat{m}_{1}^{*}(z)+\hat{m}_{2}(z)+z^{2}(\hat{m}_{2}\big{|}_{\ell}S)(z).$
(3.18)

Thus (3.17) and (3.18) imply
$\hat{m}_{2}\big{|}_{\ell-2}S=z^{2}(\hat{m}_{2}\big{|}_{\ell}S)=\hat{m}_{2}$.
This proves that $\hat{m}_{2}(z)$ is a modular form of weight $\ell-2$.

Recalling $E_{2}\big{|}_{2}S=E_{2}+6/\pi{\rm i}z$, it follows from (3.10) and
(3.17) that

$\displaystyle z\left(\hat{m}_{1}+\frac{\pi{\rm
i}}{3}\hat{m}_{2}E_{2}\right)+2\hat{m}_{2}$
$\displaystyle=z(\hat{m}_{1}^{*}\big{|}_{\ell}S)=z\left\\{\hat{m}_{1}\big{|}_{\ell}S+\frac{\pi{\rm
i}}{3}(\hat{m}_{2}\big{|}_{\ell-2}S)\left(E_{2}+\frac{6}{\pi{\rm
i}z}\right)\right\\}$
$\displaystyle=z(\hat{m}_{1}\big{|}_{\ell}S)+\frac{\pi{\rm
i}z}{3}\hat{m}_{2}E_{2}+2\hat{m}_{2},$

which yields $\hat{m}_{1}\big{|}_{\ell}S=\hat{m}_{1}$. This proves that
$\hat{m}_{1}(z)$ is a modular form of weight $\ell$.

Finally, to prove the modularity of $\hat{m}_{0}(z)$, we use (3.15) and (3.11)
to obtain

$\displaystyle z^{2}\hat{y}_{3}+z\left(\hat{m}_{1}+\frac{\pi{\rm
i}}{3}\hat{m}_{2}E_{2}\right)+\hat{m}_{2}=\hat{y}_{1}=\hat{y}_{3}\big{|}_{\ell}S$
$\displaystyle\qquad{}=\left(\frac{\pi{\rm
i}z}{6}\right)^{2}(\hat{m}_{2}\big{|}_{\ell-2}S)\Big{(}E_{2}+\frac{6}{\pi{\rm
i}z}\Big{)}^{2}+\frac{\pi{\rm
i}z^{2}}{6}(\hat{m}_{1}\big{|}_{\ell}S)\Big{(}E_{2}+\frac{6}{\pi{\rm
i}z}\Big{)}+\hat{m}_{0}\big{|}_{\ell}S$
$\displaystyle\qquad{}=z^{2}\hat{y}_{3}+z\left(\hat{m}_{1}+\frac{\pi{\rm
i}}{3}\hat{m}_{2}E_{2}\right)+\hat{m}_{2}+\hat{m}_{0}\big{|}_{\ell}S-z^{2}\hat{m}_{0},$

which implies
$\hat{m}_{0}\big{|}_{\ell+2}S=z^{-2}(\hat{m}_{0}\big{|}_{\ell}S)=\hat{m}_{0}$.
This proves that $\hat{m}_{0}(z)$ is a modular form of weight $\ell+2$. The
proof is complete. ∎

Clearly the above arguments imply Theorem 1.4(1).

###  3.2 Proofs of Theorems 1.3 and 1.4, and Corollary 1.5

In this section, we complete the proof of Theorems 1.3 and 1.4, and Corollary
1.5. First we need the following general observation.

######  Lemma 3.10.

Let $(S1)$–$(S2)$ hold. Then the eigenvalues of $\hat{\rho}(R)$ are precisely

${\rm e}^{-\frac{\pi{\rm
i}}{3}(\ell+2\kappa)},\qquad\kappa\in\big{\\{}0,\kappa_{\rho}^{(2)}-\kappa_{\rho}^{(1)},\kappa_{\rho}^{(3)}-\kappa_{\rho}^{(1)}\big{\\}},$
(3.19)

and the eigenvalues of $\hat{\rho}(S)$ are precisely

$i^{-\ell-2\kappa},\qquad\kappa\in\big{\\{}0,\kappa_{\rm i}^{(2)}-\kappa_{\rm
i}^{(1)},\kappa_{\rm i}^{(3)}-\kappa_{\rm i}^{(1)}\big{\\}}.$ (3.20)

###### Proof.

Under our assumption (S2) that $\rho$ is apparent, it follows from Remark 5.2
and Lemma 5.5 below that any solution $y(z)$ of (3.1) has an expansion of the
form

$\frac{1}{(1-w)^{2}}\sum_{n=0}^{\infty}a_{n}w^{n+\kappa},\qquad
w=w(z)=\frac{z-\rho}{z-\overline{\rho}}$

at $\rho$, where $a_{0}\neq 0$ and
$\kappa\in\big{\\{}\kappa_{\rho}^{(j)}\colon j=1,2,3\big{\\}}$. Hence,
$\hat{y}(z)$ has an expansion of the form

$(1-w)^{\ell}\sum_{n=0}^{\infty}b_{n}w^{n+\kappa},\qquad\kappa\in\big{\\{}0,\kappa_{\rho}^{(2)}-\kappa_{\rho}^{(1)},\kappa_{\rho}^{(3)}-\kappa_{\rho}^{(1)}\big{\\}}$

with $b_{0}\neq 0$. Recalling that $\rho=\big{(}{-}1+\sqrt{3}{\rm i}\big{)}/2$
is a fixed point of $R=\bigl{(}\begin{smallmatrix}0&-1\\\
1&1\end{smallmatrix}\bigr{)}$, we denote

$\zeta:=c\rho+d=\rho+1={\rm e}^{\pi{\rm i}/3}.$

Then a direct computation gives

$w(Rz)=\zeta^{-2}w(z),\qquad 1-w(Rz)=\zeta^{-1}(z+1)(1-w(z)),$

so

$\left(\hat{y}\big{|}_{\ell}R\right)(z)=\zeta^{-\ell}(1-w)^{\ell}\sum_{n=0}^{\infty}b_{n}\big{(}\zeta^{-2}w\big{)}^{n+k}.$

Therefore, $\hat{y}(z)$ is an eigenfunction of $\hat{\rho}(R)$ if and only if
the series expansion of $\hat{y}(z)$ is of the form

$(1-w)^{\ell}\sum_{n=0}^{\infty}b_{3n}w^{3n+\kappa},\qquad\kappa\in\big{\\{}0,\kappa_{\rho}^{(2)}-\kappa_{\rho}^{(1)},\kappa_{\rho}^{(3)}-\kappa_{\rho}^{(1)}\big{\\}}$

and the corresponding eigenvalue is $\zeta^{-\ell-2\kappa}={\rm
e}^{-\frac{\pi{\rm i}}{3}(\ell+2\kappa)}$. Note from
$\hat{\rho}(R)^{3}=(-1)^{\ell}I_{3}=I_{3}$ that $\hat{\rho}(R)$ can be
diagonizalied. We claim that the eigenvalues of $\hat{\rho}(R)$ are precisely
those in (3.19).

Indeed, for any
$\kappa\in\big{\\{}0,\kappa_{\rho}^{(2)}-\kappa_{\rho}^{(1)},\kappa_{\rho}^{(3)}-\kappa_{\rho}^{(1)}\big{\\}}$,
we define

$N_{\kappa}:=\\#\big{\\{}\tilde{\kappa}\in\big{\\{}0,\kappa_{\rho}^{(2)}-\kappa_{\rho}^{(1)},\kappa_{\rho}^{(3)}-\kappa_{\rho}^{(1)}\big{\\}}\,\big{|}\,\zeta^{-\ell-2\tilde{\kappa}}=\zeta^{-\ell-2\kappa}\big{\\}}.$

Clearly (3.19) holds if $N_{\kappa}=3$ for some $\kappa$ (and so for all
$\kappa$). So we only consider the case $N_{\kappa}\in\\{1,2\\}$ for all
$\kappa$. Assume by contradiction that there are $N_{\kappa}+1\in\\{2,3\\}$
linearly independent eigenfunctions

$\hat{y}_{j}=(1-w)^{\ell}\sum_{n=0}^{\infty}b_{j,3n}w^{3n+\kappa},\qquad
b_{1,0}=b_{2,0}=b_{N_{\kappa}+1,0}\neq 0,\qquad 1\leq j\leq N_{\kappa}+1,$

corresponding to the same eigenvalue $\zeta^{-\ell-2\kappa}$. Then

$\hat{y}_{1}-\hat{y}_{2}=(1-w)^{\ell}\sum_{n=n_{0}}^{\infty}(b_{1,3n}-b_{2,3n})w^{3n+\kappa}$

is also an eigenfunction of $\zeta^{-\ell-2\kappa}$, where $n_{0}\geq 1$ is
the smallest integer such that $b_{1,3n_{0}}-b_{2,3n_{0}}\allowbreak\neq 0$.
This implies
$\kappa,\kappa+3n_{0}\in\big{\\{}0,\kappa_{\rho}^{(2)}-\kappa_{\rho}^{(1)},\kappa_{\rho}^{(3)}-\kappa_{\rho}^{(1)}\big{\\}}$,
already a contradiction if $N_{\kappa}=1$. If $N_{\kappa}=2$, then by using
the linear combination of $\hat{y}_{1}$, $\hat{y}_{2}$, $\hat{y}_{3}$, there
is another $n_{1}\geq 1$ satisfying $n_{1}\neq n_{0}$ such that
$\kappa,\kappa+3n_{0},\kappa+3n_{1}\in\big{\\{}0,\kappa_{\rho}^{(2)}-\kappa_{\rho}^{(1)},\kappa_{\rho}^{(3)}-\kappa_{\rho}^{(1)}\big{\\}}$,
again a contradiction with $N_{\kappa}=2$. Thus, for any $\kappa$, the
dimension of eigenfunctions of $\zeta^{-\ell-2\kappa}$ is at most
$N_{\kappa}$. This implies the assertion (3.19).

The proof of (3.20) is similar and is omitted here. ∎

Note that if (S3) does not hold, it follows from $\kappa_{\rm
i}^{(1)}+\kappa_{\rm i}^{(2)}+\kappa_{\rm i}^{(3)}=3$ that
$\big{\\{}3\kappa_{\rm i}^{(1)},3\kappa_{\rm i}^{(2)},3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{1,1,1\\}\ \mathrm{mod}\ 2$. We have

######  Theorem 3.11.

Let $(S1)$–$(S2)$ hold and suppose $\big{\\{}3\kappa_{\rm
i}^{(1)},3\kappa_{\rm i}^{(2)},3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{1,1,1\\}\ \mathrm{mod}\ 2$. Then $12|\ell$ and for
any solution $y(z)$ of (3.1), $\hat{y}(z)$ is a modular form of weight $\ell$.
In particular, the representation $\hat{\rho}$ is trivial.

###### Proof.

By (3.20) and $\big{\\{}3\kappa_{\rm i}^{(1)},3\kappa_{\rm
i}^{(2)},3\kappa_{\rm i}^{(3)}\big{\\}}\equiv\\{1,1,1\\}$ $\ \mathrm{mod}\ 2$,
we see that the eigenvalues of $\hat{S}$ are all the same, so we see from
$\hat{S}^{2}=I_{3}$ that $\hat{S}=I_{3}$. Consequently, $\hat{T}=\hat{R}$ and
then $\hat{T}^{3}=\hat{R}^{3}=I_{3}$. Since the eigenvalues of $\hat{T}$ are
$\\{1,1,1\\}$, we obtain $\hat{R}=\hat{T}=I_{3}$, i.e., the representation
$\hat{\rho}$ is trivial and $\hat{y}(z)$ is a modular form of weight $\ell$
for any solution $y(z)$. Furthermore, it follows from Lemma 3.10 that ${\rm
e}^{-\frac{\pi{\rm i}}{3}\ell}={\rm i}^{-\ell}=1$, so $\ell\equiv 0\
\mathrm{mod}\ 12$. ∎

###### Proof of Theorems 1.3 and 1.4.

Theorem 1.3 follows from Theorems 3.9 and 3.11. ∎

###### Proof of Corollary 1.5.

Under the assumptions (H1)–(H3) and $\kappa_{\rho}^{(1)}\in\mathbb{Z}$, we
have $\kappa_{\rho}^{(j)}\in\mathbb{Z}$ for all $j$. Together with
$\kappa_{\rho}^{(1)}+\kappa_{\rho}^{(2)}+\kappa_{\rho}^{(3)}=3$, we have
either $\kappa_{\rho}^{(1)}\equiv\kappa_{\rho}^{(2)}\equiv\kappa_{\rho}^{(3)}\
\mathrm{mod}\ 3$ or
$\big{\\{}\kappa_{\rho}^{(1)},\kappa_{\rho}^{(2)},\kappa_{\rho}^{(3)}\big{\\}}\equiv\\{0,1,2\\}\
\mathrm{mod}\ 3$.

First suppose $\big{\\{}3\kappa_{\rm i}^{(1)},3\kappa_{\rm
i}^{(2)},3\kappa_{\rm i}^{(3)}\big{\\}}\equiv\\{0,0,1\\}\ \mathrm{mod}\ 2$.
Then Theorem 3.9 holds, in particular, $\hat{R}\neq I_{3}$ and the eigenvalues
can not be all the same. This together with (3.19) imply that
$\kappa_{\rho}^{(1)}\equiv\kappa_{\rho}^{(2)}\equiv\kappa_{\rho}^{(3)}\
\mathrm{mod}\ 3$ is impossible, so

$\big{\\{}\kappa_{\rho}^{(1)},\kappa_{\rho}^{(2)},\kappa_{\rho}^{(3)}\big{\\}}\equiv\\{0,1,2\\}\quad\
\mathrm{mod}\ 3.$ (3.21)

Conversely, suppose (3.21) holds. If $\big{\\{}3\kappa_{\rm
i}^{(1)},3\kappa_{\rm i}^{(2)},3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{1,1,1\\}\ \mathrm{mod}\ 2$, then Theorem 3.11
implies $\hat{R}=I_{3}$, which together with (3.19) imply
$\kappa_{\rho}^{(1)}\equiv\kappa_{\rho}^{(2)}\equiv\kappa_{\rho}^{(3)}\
\mathrm{mod}\ 3$, a contradiction with (3.21). Thus $\big{\\{}3\kappa_{\rm
i}^{(1)},3\kappa_{\rm i}^{(2)},3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{0,0,1\\}\ \mathrm{mod}\ 2$. ∎

######  Remark 3.12.

We note the under the assumption that the eigenvalues of $\hat{\rho}(T)$ are
all $1$, Proposition 2.5 and Corollary to Theorem 2.9 of [31] and results of
[32] imply that $\hat{\rho}$ is irreducible if and only if the eigenvalues of
$\hat{\rho}(S)$ and $\hat{\rho}(R)$ are $1$, $-1$, $-1$ and $1$, ${\rm
e}^{2\pi{\rm i}/3}$, ${\rm e}^{-2\pi{\rm i}/3}$, respectively. Our Theorem 1.3
shows that the irreducibility property of $\hat{\rho}$ is solely determined by
the local exponents at ${\rm i}$. This link between the results of [31, 32]
and Theorem 1.3 is provided by Lemma 3.10. In other words, one may also use
results of [31, 32] and Lemma 3.10 to give an alternative proof of Theorem
1.3. Our approach has the advantage that it directly shows that
$\hat{y}_{+}(z)$ is a quasimodular form of depth $2$. (Note that Westbury’s
paper [32] does not seem to be easily available. We refer the reader to the
introduction section of [22] for a quick review of Westbury’s results.)

##  4 Reducibility and $\boldsymbol{\mathrm{SU}(3)}$ Toda systems on
$\boldsymbol{\mathrm{SL}(2,\mathbb{Z})}$

In view of Theorem 3.11 or equivalently Theorem 1.4(2), it is natural to ask
whether the converse statement holds or not. The purpose of this section is to
establish such a converse statement and apply it to the $\mathrm{SU}(3)$ Toda
system. Let $\Gamma$ be a discrete subgroup of $\mathrm{SL}(2,\mathbb{R})$
commensurable with $\mathrm{SL}(2,\mathbb{Z})$. In general, there are at least
three sources of modular forms and quasimodular forms that will give rise to
third-order MODEs on $\Gamma$:

  1. (i)

If $f(z)\in\widetilde{\mathfrak{M}}_{k}^{\leq 2}(\Gamma,\chi)$, then
$f(z)/\sqrt[3]{W_{f}(z)}$ satisfies a third-order MODE on $\Gamma$. This case
has been studied in Section 2.

  2. (ii)

If $f(z)=f_{1}(z)\phi(z)+f_{0}(z)\in\widetilde{\mathfrak{M}}_{k}^{\leq
1}(\Gamma,\chi_{1})$ and $g(z)\in\mathfrak{M}_{k-1}(\Gamma,\chi_{2})$ with
$\chi_{1}(-I_{2})\chi_{2}(-I_{2})\allowbreak=-1$, then a similar argument as
Theorem 2.2 shows that

$f(z)/\sqrt[3]{W_{f,g}(z)},\qquad(zf+\alpha
f_{1})/\sqrt[3]{W_{f,g}(z)}\qquad\text{and}\qquad g(z)/\sqrt[3]{W_{f,g}(z)}$

are solutions of some third-order MODE on $\Gamma$. Here

$W_{f,g}(z)=\det\begin{pmatrix}f&f^{\prime}&f^{\prime\prime}\\\ zf+\alpha
f_{1}&(zf+\alpha f_{1})^{\prime}&(zf+\alpha f_{1})^{\prime\prime}\\\
g&g^{\prime}&g^{\prime\prime}\end{pmatrix}.$

  3. (iii)

If $f(z)\in\mathfrak{M}_{k}(\Gamma,\chi_{1})$,
$g(z)\in\mathfrak{M}_{k}(\Gamma,\chi_{2})$, and
$h(z)\in\mathfrak{M}_{k}(\Gamma,\chi_{3})$ for some characters $\chi_{j}$ of
$\Gamma$, then $f(z)/\sqrt[3]{W_{f,g,h}(z)}$, $g(z)/\sqrt[3]{W_{f,g,h}(z)}$,
$h(z)/\sqrt[3]{W_{f,g,h}(z)}$ are solutions of some third-order MODE on
$\Gamma$. Here

$W_{f,g,h}(z)=\det\begin{pmatrix}f&f^{\prime}&f^{\prime\prime}\\\
g&g^{\prime}&g^{\prime\prime}\\\ h&h^{\prime}&h^{\prime\prime}\end{pmatrix}.$
(4.1)

To simplify the situation, we impose the condition that the values of
$\chi_{j}(T)$ in the second and the third cases are all the same (so that
$\rho(T)$ has only one eigenvalue with multiplicity $3$). In the case
$\Gamma=\mathrm{SL}(2,\mathbb{Z})$, this condition implies that $\chi_{j}$ are
all the same, say, $\chi_{j}=\chi$ for all $j$, so Case (ii) will not occur.
Moreover, in Case (iii), we can divide $f$, $g$, $h$ by an eta-power
$\eta(z)^{m}$ satisfying ${\rm e}^{2\pi{\rm i}m/24}=\chi(T)$. The differential
equation corresponding to $f(z)/\eta(z)^{m}$ is the same as that corresponding
to $f(z)$. Thus, in the case $\Gamma=\mathrm{SL}(2,\mathbb{Z})$, we may assume
that $\chi$ is trivial.

######  Lemma 4.1.

Let $f,g,h\in\mathfrak{M}_{k}(\mathrm{SL}(2,\mathbb{Z}))$ be three linearly
independent modular forms of weight $k$ on $\mathrm{SL}(2,\mathbb{Z})$. Define
$W_{f,g,h}(z)$ by (4.1). Then $W_{f,g,h}$ is a modular form of weight $3(k+2)$
on $\mathrm{SL}(2,\mathbb{Z})$.

###### Proof.

Let $F(z)=(f(z),g(z),h(z))^{t}$, which satisfies $F(\gamma z)=(cz+d)^{k}F(z)$
for all $\gamma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\mathrm{SL}(2,\mathbb{Z})$. Then the assertion
follows from the basic properties of the determinant function. ∎

######  Theorem 4.2.

Let $f,g,h\in\mathfrak{M}_{k}(\mathrm{SL}(2,\mathbb{Z}))$ be linearly
independent modular forms and $\mathcal{L}y=0$ be the differential equation
satisfied by $f/\sqrt[3]{W_{f,g,h}}$, $g/\sqrt[3]{W_{f,g,h}}$, and
$h/\sqrt[3]{W_{f,g,h}}$. Then $Q_{2}(z)$ and
$Q_{3}(z)-\frac{1}{2}Q_{2}^{\prime}(z)$ are meromorphic modular forms of
weight $4$ and $6$ respectively. Furthermore, $(H1)$–$(H3)$ hold and $\infty$
is also apparent.

###### Proof.

The proof is similar as that of Theorem 2.2. The only difference is that
$\infty$ is also apparent because $f/\sqrt[3]{W_{f,g,h}}$,
$g/\sqrt[3]{W_{f,g,h}}$ and $h/\sqrt[3]{W_{f,g,h}}$ are linearly independent
solutions. ∎

######  Proposition 4.3.

Let $f,g,h\in\mathfrak{M}_{k}(\mathrm{SL}(2,\mathbb{Z}))$ be linearly
independent modular forms and $\mathcal{L}y=0$ be the differential equation
satisfied by $f/\sqrt[3]{W_{f,g,h}}$, $g/\sqrt[3]{W_{f,g,h}}$, and
$h/\sqrt[3]{W_{f,g,h}}$. Then the local exponents of $\mathcal{L}y=0$ at the
elliptic points ${\rm i}$ and $\rho=\big{(}{-}1+\sqrt{3}{\rm i}\big{)}/2$
satisfy

  1. $1)$

$\kappa_{\rm i}^{(2)}-\kappa_{\rm i}^{(1)},\kappa_{\rm i}^{(3)}-\kappa_{\rm
i}^{(1)}\equiv 0\ \mathrm{mod}\ 2$, and

  2. $2)$

$\kappa_{\rho}^{(j)}\in\mathbb{Z}$ for all $j$, and
$\kappa_{\rho}^{(1)}\equiv\kappa_{\rho}^{(2)}\equiv\kappa_{\rho}^{(3)}\
\mathrm{mod}\ 3$.

###### Proof.

Note that every solution $y(z)$ of $\mathcal{L}y=0$ can be written as

$(af(z)+bg(z)+ch(z))/\sqrt[3]{W_{f,g,h}(z)}$

for some $a,b,c\in\mathbb{C}$. As $a$, $b$, and $c$ vary, the order of $y(z)$
at ${\rm i}$ (respectively, $\rho$) will go through all possible local
exponents of $\mathcal{L}y=0$ at ${\rm i}$ (respectively, $\rho$). Since

$\frac{\operatorname{ord}_{\rm
i}(af+bg+cz)}{2}+\frac{\operatorname{ord}_{\rho}(af+bg+cz)}{3}\equiv\frac{\operatorname{ord}_{\rm
i}(f)}{2}+\frac{\operatorname{ord}_{\rho}(f)}{3}\quad\ \mathrm{mod}\
\mathbb{Z},$

so

$\displaystyle\operatorname{ord}_{\rm i}(af+bg+cz)-\operatorname{ord}_{\rm
i}(f)\equiv 0\quad\ \mathrm{mod}\ 2,$
$\displaystyle\operatorname{ord}_{\rho}(af+bg+cz)-\operatorname{ord}_{\rho}(f)\equiv
0\quad\ \mathrm{mod}\ 3$

hold for any $(a,b,c)\neq(0,0,0)$. From here and

$\displaystyle\kappa_{\rm i}^{(j)}+\tfrac{1}{3}\operatorname{ord}_{\rm
i}W_{f,g,h}(z)\in\\{\operatorname{ord}_{\rm
i}(af+bg+cz)\,|\,(a,b,c)\neq(0,0,0)\\},$
$\displaystyle\kappa_{\rho}^{(j)}+\tfrac{1}{3}\operatorname{ord}_{\rho}W_{f,g,h}(z)\in\\{\operatorname{ord}_{\rho}(af+bg+cz)\,|\,(a,b,c)\neq(0,0,0)\\}$

for all $j$, we obtain the assertion (1) and
$\kappa_{\rho}^{(2)}-\kappa_{\rho}^{(1)},\kappa_{\rho}^{(3)}-\kappa_{\rho}^{(1)}\equiv
0\ \mathrm{mod}\ 3$. This together with
$\kappa_{\rho}^{(1)}+\kappa_{\rho}^{(2)}+\kappa_{\rho}^{(3)}=3$ imply
$\kappa_{\rho}^{(j)}\in\mathbb{Z}$ for all $j$ and so the assertion (2) holds.
∎

The above result is precisely the converse statement of Theorem 1.4(2).

######  Example 4.4.

Recall that the smallest weight $k$ such that
$\dim\mathfrak{M}_{k}(\mathrm{SL}(2,\mathbb{Z}))=3$ is $24$. Let
$f(z)=E_{4}(z)^{6}$, $g(z)=E_{4}(z)^{3}\Delta(z)$, and $h(z)=\Delta(z)^{2}$,
which form a basis for $\mathfrak{M}_{24}(\mathrm{SL}(2,\mathbb{Z}))$. To
determine the differential equation

$\mathcal{L}y:=D_{q}^{3}y(z)+Q(z)D_{q}y(z)+\left(\frac{1}{2}D_{q}Q(z)+R(z)\right)y(z)=0$

satisfied by $f/\sqrt[3]{W_{f,g,h}}$, $g/\sqrt[3]{W_{f,g,h}}$, and
$h/\sqrt[3]{W_{f,g,h}}$, we use Ramanujan’s identities

$D_{q}E_{2}=\frac{E_{2}^{2}-E_{4}}{12},\qquad
D_{q}E_{4}=\frac{E_{2}E_{4}-E_{6}}{3},\qquad
D_{q}E_{6}=\frac{E_{2}E_{6}-E_{4}^{2}}{2},$

and compute that $W_{f,g,h}(z)=cE_{4}(z)^{6}E_{6}(z)^{3}\Delta(z)^{3}$ for
some nonzero number $c$. Noticing that $W_{f,g,h}(z)$ has a zero of order $3$
at ${\rm i}$ and a zero of order $6$ at $\rho$, we know that $\kappa_{\rm
i}^{(1)}=-1$, $\kappa_{\rho}^{(1)}=-2$, which, by Proposition 4.3, implies
that $\kappa_{\rm i}^{(2)}=1$, $\kappa_{\rm i}^{(3)}=3$,
$\kappa_{\rho}^{(2)}=1$, and $\kappa_{\rho}^{(3)}=4$. In other words, the
indicial equations at ${\rm i}$ and at $\rho$ are $(x+1)(x-1)(x-3)=0$ and
$(x+2)(x-1)(x-4)=0$, respectively. Also, we have
$\operatorname{ord}_{\infty}f-\frac{1}{3}\operatorname{ord}_{\infty}W_{f,g,h}=-1$,
$\operatorname{ord}_{\infty}g-\frac{1}{3}\operatorname{ord}_{\infty}W_{f,g,h}=0$,
and
$\operatorname{ord}_{\infty}h-\frac{1}{3}\operatorname{ord}_{\infty}W_{f,g,h}=1$,
which implies that the indicial equation at $\infty$ is $(x+1)x(x-1)$.
Therefore, according to Lemmas 5.7, 5.8, and 5.9, the meromorphic modular
forms $Q(z)$ and $R(z)$ in $\mathcal{L}y$ are

$Q(z)=-E_{4}(z)-\frac{3}{4}\frac{E_{4}(z)(E_{4}(z)^{3}-E_{6}(z)^{2})}{E_{6}(z)^{2}}+\frac{8}{9}\frac{E_{4}(z)^{3}-E_{6}(z)^{2}}{E_{4}(z)^{2}}$

(note that this can also be computed directly using (2.5)) and

$R(z)=s_{\rm i}^{(1)}\frac{E_{4}(z)^{3}-E_{6}(z)^{2}}{E_{6}(z)}$

for some complex number $s_{\rm i}^{(1)}$. Using the apparentness condition at
${\rm i}$, we can show that $s_{\rm i}^{(1)}=0$. In other words, the
differential equation is
$D_{q}^{3}y(z)+Q(z)D_{q}y(z)+\frac{1}{2}D_{q}Q(z)y(z)=0$. We remark that the
reason why the differential equation is of this special form is due to the
fact that it is the symmetric square of some second order MODE.

It is worth to point out that Theorem 4.2 can be applied to construct
solutions of the $\mathrm{SU}(3)$ Toda system

$\begin{cases}\displaystyle\Delta v_{1}+2{\rm e}^{v_{1}}-{\rm
e}^{v_{2}}=4\pi\sum_{k=1}^{N}n_{1,k}\delta_{p_{k}},\\\ \displaystyle\Delta
v_{2}+2{\rm e}^{v_{2}}-{\rm
e}^{v_{1}}=4\pi\sum_{k=1}^{N}n_{2,k}\delta_{p_{k}}\end{cases}\qquad\text{in}\
\mathbb{R}^{2},$

where $\Delta=\frac{\partial^{2}}{\partial
x_{1}^{2}}+\frac{\partial^{2}}{\partial x_{2}^{2}}$ is the Laplace operator
and $\delta_{p}$ denotes the Dirac measure at $p$. We always use the complex
variable $w=x_{1}+{\rm i}x_{2}$. Then the Laplace operator
$\Delta=4\partial_{w\bar{w}}$.

As in Theorem 4.2, we let
$f,g,h\in\mathfrak{M}_{k}(\mathrm{SL}(2,\mathbb{Z}))$ be linearly independent
modular forms and

$\displaystyle
y^{\prime\prime\prime}(z)+Q_{2}(z)y^{\prime}(z)+Q_{3}(z)y(z)=0,\qquad
z\in\mathbb{H}$ (4.2)

be the MODE satisfied by $y_{1}(z):=f(z)/\sqrt[3]{W_{f,g,h}}$,
$y_{2}(z):=g(z)/\sqrt[3]{W_{f,g,h}}$, and
$y_{3}(z):=h(z)/\sqrt[3]{W_{f,g,h}}$. Denote the set of regular singular
points of (4.2) modulo $\mathrm{SL}(2,\mathbb{Z})$ on $\mathbb{H}$ by
$\mathcal{S}=\\{z_{1},\dots,z_{m},{\rm i},\rho\\}$. For each
$z\in\mathcal{S}$, it follows from Theorem 4.2 that there are
$m_{z}^{(1)},m_{z}^{(2)}\in\mathbb{Z}_{\geq 0}$ such that the local exponents
of (4.2) at $\gamma z$ are the same as those at $z$ and are given by

$\kappa_{z}^{(1)}=-\frac{2m_{z}^{(1)}+m_{z}^{(2)}}{3},\qquad\kappa_{z}^{(2)}=\kappa_{z}^{(1)}+m_{z}^{(1)}+1,\qquad\kappa_{z}^{(3)}=\kappa_{z}^{(2)}+m_{z}^{(2)}+1$

for any $\gamma\in\mathrm{SL}(2,\mathbb{Z})$. Similarly, there are
$m_{\infty}^{(1)},m_{\infty}^{(2)}\in\mathbb{N}$ such that the local exponents
of (4.2) at the cusp $\infty$ are given by

$\kappa_{\infty}^{(1)}=-\frac{2m_{\infty}^{(1)}+m_{\infty}^{(2)}}{3},\qquad\kappa_{\infty}^{(2)}=\kappa_{\infty}^{(1)}+m_{\infty}^{(1)},\qquad\kappa_{\infty}^{(3)}=\kappa_{\infty}^{(2)}+m_{\infty}^{(2)}.$

Given any $\lambda,\mu>0$, we define

$\displaystyle{\rm
e}^{-U_{1;\lambda,\mu}(z)}:=\frac{1}{4}\big{(}\lambda^{2}\mu^{-1}|y_{1}|^{2}+\mu^{2}\lambda^{-1}|y_{2}|^{2}+\lambda^{-1}\mu^{-1}|y_{3}|^{2}\big{)},$
$\displaystyle{\rm
e}^{-U_{2;\lambda,\mu}(z)}:=\frac{1}{4}\big{[}\lambda\mu|W(y_{1},y_{2})|^{2}+\lambda^{-2}\mu|W(y_{2},y_{3})|^{2}+\lambda\mu^{-2}|W(y_{3},y_{1})|^{2}\big{]},$

where $W(y_{i},y_{j}):=y_{i}^{\prime}y_{j}-y_{j}^{\prime}y_{i}$. Note that
${\rm e}^{-U_{k;\lambda,\mu}(z)}$ is single-valued for any $z\in\mathbb{H}$
and $0<{\rm e}^{-U_{k;\lambda,\mu}(z)}<\infty$ as long as
$z\notin\mathrm{SL}(2,\mathbb{Z})\mathcal{S}$. We have

######  Lemma 4.5.

Given any $\lambda,\mu>0$, there holds

$\begin{cases}\Delta U_{1;\lambda,\mu}+{\rm
e}^{2U_{1;\lambda,\mu}-U_{2;\lambda,\mu}}=0,\\\ \Delta U_{2;\lambda,\mu}+{\rm
e}^{2U_{2;\lambda,\mu}-U_{1;\lambda,\mu}}=0\end{cases}\qquad\text{in}\
\mathbb{H}\setminus(\mathrm{SL}(2,\mathbb{Z})\mathcal{S}).$ (4.3)

###### Proof.

The proof can be easily adopted from [7, 23, 24]; we sketch the proof here for
the reader’s convenience. Given any $\lambda,\mu>0$, we define

$\mathcal{W}_{\lambda,\mu}:=\begin{pmatrix}\lambda^{\frac{3}{2}}y_{1}&\mu^{\frac{3}{2}}y_{2}&y_{3}\\\
\lambda^{\frac{3}{2}}y_{1}^{\prime}&\mu^{\frac{3}{2}}y_{2}^{\prime}&y_{3}^{\prime}\\\
\lambda^{\frac{3}{2}}y_{1}^{\prime\prime}&\mu^{\frac{3}{2}}y_{2}^{\prime\prime}&y_{3}^{\prime\prime}\end{pmatrix}.$

Since the Wroksian $W(y_{1},y_{2},y_{3})=1$, we have
$\det\mathcal{W}_{\lambda,\mu}=(\lambda\mu)^{\frac{3}{2}}$. Define a positive
definite matrix

$R_{\lambda,\mu}:=(\lambda\mu)^{-1}\mathcal{W}_{\lambda,\mu}\overline{\mathcal{W}_{\lambda,\mu}}^{T}.$

then $\det R_{\lambda,\mu}=1$. For $1\leq m\leq 3$, we let $R_{\lambda,\mu;m}$
denote the leading principal minor of $R_{\lambda,\mu}$ of dimension $m$.
Since $y_{j}(z)$ is holomorphic in
$\mathbb{H}\setminus(\mathrm{SL}(2,\mathbb{Z})\mathcal{S})$, a direct
computation leads to (see, e.g., [24])

$R_{\lambda,\mu;m}(\partial_{z\bar{z}}R_{\lambda,\mu;m})-(\partial_{z}R_{\lambda,\mu;m})(\partial_{\bar{z}}R_{\lambda,\mu;m})=R_{{\lambda,\mu;m-1}}R_{\lambda,\mu;m+1},\qquad
m=1,2,$ (4.4)

for $z\in\mathbb{H}\setminus(\mathrm{SL}(2,\mathbb{Z})\mathcal{S})$, where
$R_{\lambda,\mu;0}:=1$.

On the other hand,

$\displaystyle\frac{1}{4}R_{\lambda,\mu;1}=\frac{1}{4}(\lambda\mu)^{-1}\big{(}\lambda^{3}|y_{1}|^{2}+\mu^{3}|y_{2}|^{2}+|y_{3}|^{2}\big{)}={\rm
e}^{-U_{1;\lambda,\mu}(z)}.$

Define ${\rm e}^{-V_{\lambda,\mu}(z)}:=\frac{1}{4}R_{\lambda,\mu;2}$. we will
prove that ${\rm e}^{-V_{\lambda,\mu}(z)}={\rm e}^{-U_{2;\lambda,\mu}(z)}$.

Note that $R_{\lambda,\mu;3}=\det R_{\lambda,\mu}=1$. Letting $m=1$ in (4.4)
leads to (note $0<{\rm e}^{-U_{1;\lambda,\mu}},{\rm
e}^{-V_{\lambda,\mu}}<+\infty$ in
$\mathbb{H}\setminus(\mathrm{SL}(2,\mathbb{Z})\mathcal{S})$)

$\displaystyle 4{\rm e}^{-V_{\lambda,\mu}}$
$\displaystyle=R_{\lambda,\mu;2}=16\big{[}{\rm
e}^{-U_{1;\lambda,\mu}}\big{(}\partial_{z\bar{z}}{\rm
e}^{-U_{1;\lambda,\mu}}\big{)}-\big{(}\partial_{z}{\rm
e}^{-U_{1;\lambda,\mu}}\big{)}\big{(}\partial_{\bar{z}}{\rm
e}^{-U_{1;\lambda,\mu}}\big{)}\big{]}$ $\displaystyle=-16{\rm
e}^{-2U_{1;\lambda,\mu}}\partial_{z\bar{z}}U_{1;\lambda,\mu}=-4{\rm
e}^{-2U_{1;\lambda,\mu}}\Delta U_{1;\lambda,\mu}\quad\text{in
}\mathbb{H}\setminus(\mathrm{SL}(2,\mathbb{Z})\mathcal{S}),$ (4.5)

and letting $m=2$ in (4.4) leads to

$\displaystyle 4{\rm e}^{-U_{1;\lambda,\mu}}$
$\displaystyle=R_{\lambda,\mu;1}=16\big{[}{\rm
e}^{-V_{\lambda,\mu}}(\partial_{z\bar{z}}{\rm
e}^{-V_{\lambda,,\mu}})-\big{(}\partial_{z}{\rm
e}^{-V_{\lambda,\mu}}\big{)}\big{(}\partial_{\bar{z}}{\rm
e}^{-V_{\lambda,\mu}}\big{)}\big{]}$ $\displaystyle=-16{\rm
e}^{-2V_{\lambda,\mu}}\partial_{z\bar{z}}V_{\lambda,\mu}=-4{\rm
e}^{-2V_{\lambda,\mu}}\Delta V_{\lambda,\mu}\quad\text{in
}\mathbb{H}\setminus(\mathrm{SL}(2,\mathbb{Z})\mathcal{S}).$

Furthermore, we insert ${\rm
e}^{-U_{1;\lambda,\mu}}=\frac{1}{4}\sum|a_{j}y_{j}|^{2}$ (where
$a_{1}=\lambda\mu^{-1/2}$, $a_{2}=\mu\lambda^{-1/2}$ and
$a_{3}=(\lambda\mu)^{-1/2}$) into (4.5), which leads to

$\displaystyle\frac{1}{4}{\rm e}^{-V_{\lambda,\mu}}$ $\displaystyle={\rm
e}^{-U_{1;\lambda,\mu}}\big{(}\partial_{z\bar{z}}{\rm
e}^{-U_{1;\lambda,\mu}}\big{)}-\big{(}\partial_{z}{\rm
e}^{-U_{1;\lambda,\mu}}\big{)}\big{(}\partial_{\bar{z}}{\rm
e}^{-U_{1;\lambda,\mu}}\big{)}$
$\displaystyle=\frac{1}{16}\left[\left(\sum|a_{j}y_{j}|^{2}\right)\left(\sum|a_{j}y_{j}^{\prime}|^{2}\right)-\left(\sum
a_{j}^{2}y_{j}^{\prime}\overline{y_{j}}\right)\left(\sum
a_{j}^{2}y_{j}\overline{y_{j}^{\prime}}\right)\right]$
$\displaystyle=\frac{1}{16}\big{[}|W(a_{1}y_{1},a_{2}y_{2})|^{2}+|W(a_{2}y_{2},a_{3}y_{3})|^{2}+|W(a_{3}y_{3},a_{1}y_{1})|^{2}\big{]},$

so ${\rm e}^{-V_{\lambda,\mu}(z)}={\rm e}^{-U_{2;\lambda,\mu}(z)}$. This
proves that $(U_{1;\lambda,\mu}(z),U_{2;\lambda,\mu}(z))$ solves the Toda
system (4.3). ∎

Now any $\tilde{z}\in\mathcal{S}$, it follows from the local behavior of
$y_{j}$’s that near $\gamma\tilde{z}$,

$\displaystyle
U_{1;\lambda,\mu}(z)=-2\kappa_{\tilde{z}}^{(1)}\ln|z-\gamma\tilde{z}|+O(1),$
$\displaystyle
U_{2;\lambda,\mu}(z)=-2\big{(}2-\kappa_{\tilde{z}}^{(3)}\big{)}\ln|z-\gamma\tilde{z}|+O(1).$

Similarly, at the cusp $\infty$, we have

$\displaystyle U_{1;\lambda,\mu}(z)=-2\kappa_{\infty}^{(1)}\ln|q|+O(1),$
$\displaystyle U_{2;\lambda,\mu}(z)=2\kappa_{\infty}^{(3)}\ln|q|+O(1),$

where $q={\rm e}^{2\pi{\rm i}z}$. Since $f$, $g$, $h$, $W_{f,g,h}$ are modular
forms of weights $k$, $k$, $k$ and $3(k+2)$ respectively, we easily obtain

$|y_{j}(\gamma
z)|^{2}=\frac{|y_{j}(z)|^{2}}{|cz+d|^{4}},\qquad|W(y_{i},y_{j})(\gamma
z)|^{2}=\frac{|W(y_{i},y_{j})(z)|^{2}}{|cz+d|^{4}}$

for any $\gamma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\mathrm{SL}(2,\mathbb{Z})$, so

$U_{j;\lambda,\mu}(\gamma z)=U_{j;\lambda,\mu}(z)+4\ln|cz+d|,\qquad j=1,2.$

Now we define

$(u_{1;\lambda,\mu},u_{2;\lambda,\mu}):=(2U_{1;\lambda,\mu}-U_{2;\lambda,\mu},2U_{2;\lambda,\mu}-U_{1;\lambda,\mu}).$

Then we have

$\begin{cases}\Delta u_{1;\lambda,\mu}+2{\rm e}^{u_{1;\lambda,\mu}}-{\rm
e}^{u_{2;\lambda,\mu}}=0,\\\ \Delta u_{2;\lambda,\mu}+2{\rm
e}^{u_{2;\lambda,\mu}}-{\rm
e}^{u_{1;\lambda,\mu}}=0\end{cases}\qquad\text{in}\
\mathbb{H}\setminus(\mathrm{SL}(2,\mathbb{Z})\mathcal{S}),$

and near $\gamma\tilde{z}$,

$u_{1;\lambda,\mu}(z)=2m_{\tilde{z}}^{(1)}\ln|z-\gamma\tilde{z}|+O(1),\qquad
u_{2;\lambda,\mu}(z)=2m_{\tilde{z}}^{(2)}\ln|z-\gamma\tilde{z}|+O(1),$

while at the cusp $\infty$,

$u_{1;\lambda,\mu}(z)=2m_{\infty}^{(1)}\ln|q|+O(1),\qquad
u_{2;\lambda,\mu}(z)=2m_{\infty}^{(2)}\ln|q|+O(1).$

Therefore, $(u_{1;\lambda,\mu},u_{2;\lambda,\mu})$ is a solution of the
following $\mathrm{SU}(3)$ Toda system

$\begin{cases}\displaystyle\Delta u_{1}+2{\rm e}^{u_{1}}-{\rm
e}^{u_{2}}=4\pi\sum_{\gamma}\bigg{(}m_{\rm i}^{(1)}\delta_{\gamma
i}+m_{\rho}^{(1)}\delta_{\gamma\rho}+\sum_{j=1}^{m}m_{z_{j}}^{(1)}\delta_{\gamma
z_{j}}\bigg{)}\quad\text{on}\ \mathbb{H},\\\ \displaystyle\Delta u_{2}+2{\rm
e}^{u_{2}}-{\rm e}^{u_{1}}=4\pi\sum_{\gamma}\bigg{(}m_{\rm
i}^{(2)}\delta_{\gamma
i}+m_{\rho}^{(2)}\delta_{\gamma\rho}+\sum_{j=1}^{m}m_{z_{j}}^{(2)}\delta_{\gamma
z_{j}}\bigg{)}\quad\text{on}\ \mathbb{H},\\\ \displaystyle
u_{k}(z)=2m_{\infty}^{(k)}\ln|q|+O(1)\quad\text{as}\
\operatorname{Im}z\to\infty,\\\ u_{j}(\gamma
z)=u_{j}(z)+4\ln|cz+d|,\quad\forall\gamma\in\mathrm{SL}(2,\mathbb{Z}).\end{cases}$
(4.6)

Now consider the modular function $w\colon\mathbb{H}\to\mathbb{C}$ defined by

$w=w(z):=\frac{E_{4}(z)^{3}}{E_{4}(z)^{3}-E_{6}(z)^{2}}.$

It is well known that $w(z)$ is holomorphic, surjective and

$w({\rm i})=1,\qquad w(\rho)=0,\qquad w(\infty)=\infty.$

A direct computation gives

$w^{\prime}(z)=-2\pi{\rm
i}\frac{E_{4}(z)^{2}E_{6}(z)}{E_{4}(z)^{3}-E_{6}(z)^{2}}.$

Denote $p_{j}:=w(\gamma z_{j})$. Then all points of
$\\{p_{1},\dots,p_{m},1,0\\}\subset\mathbb{C}$ are distinct. Now we define
$(v_{1}(w),v_{2}(w))$ for $w\in\mathbb{C}$ by

$u_{k}(z)=v_{k}(w(z))+2\ln|w^{\prime}(z)|,\qquad z\in\mathbb{H}.$

Since $w(\gamma z)=w(z)$ gives $w^{\prime}(\gamma z)=(cz+d)^{2}w^{\prime}(z)$,
it follows from $u_{k}(\gamma z)=u_{k}(z)+4\ln|cz+d|$ that $v_{k}(w)$ is well-
defined for $w\in\mathbb{C}$. Now outside $\\{p_{1},\dots,p_{m},1,0\\}$, we
have

$\displaystyle\Delta u_{k}(z)$
$\displaystyle=4\partial_{z\bar{z}}u_{k}(z)=4\partial_{w\bar{w}}v_{k}(w)|w^{\prime}(z)|^{2}=|w^{\prime}(z)|^{2}\Delta
v_{k}(w)$ $\displaystyle={\rm e}^{u_{k^{\prime}}(z)}-2{\rm
e}^{u_{k}(z)}=|w^{\prime}(z)|^{2}\big{(}{\rm e}^{v_{k^{\prime}}(w)}-2{\rm
e}^{v_{k}(w)}\big{)},$

so

$\Delta v_{k}(w)+2{\rm e}^{v_{k}(w)}-{\rm e}^{v_{k^{\prime}}(w)}=0,$

where $\\{k,k^{\prime}\\}=\\{1,2\\}$. Furthermore, since
$w^{\prime}(z_{j})\neq 0$, we have that at $w=p_{j}=w(z_{j})$,

$\displaystyle
v_{k}(w)=u_{k}(z)-2\ln|w^{\prime}(z)|=2m_{z_{j}}^{(k)}\ln|z-z_{j}|+O(1)=2m_{z_{j}}^{(k)}\ln|w-p_{j}|+O(1).$

At $w=w({\rm i})=1$, since $\operatorname{ord}_{\rm i}(w-1)=2$ and
$\operatorname{ord}_{\rm i}w^{\prime}=1$, we have

$\displaystyle v_{k}(w)$
$\displaystyle=u_{k}(z)-2\ln|w^{\prime}(z)|=2\big{(}m_{{\rm
i}}^{(k)}-1\big{)}\ln|z-{\rm i}|+O(1)$ $\displaystyle=\big{(}m_{{\rm
i}}^{(k)}-1\big{)}\ln|w-1|+O(1).$

At $w=w(\rho)=0$, since $\operatorname{ord}_{\rho}w=3$ and
$\operatorname{ord}_{\rho}w^{\prime}=2$, we have

$\displaystyle v_{k}(w)$
$\displaystyle=u_{k}(z)-2\ln|w^{\prime}(z)|=2\big{(}m_{\rho}^{(k)}-2\big{)}\ln|z-\rho|+O(1)$
$\displaystyle=\frac{2\big{(}m_{\rho}^{(k)}-2\big{)}}{3}\ln|w|+O(1).$

At $w=w(\infty)=\infty$, since

$w(z)=Cq^{-1}(1+O(q)),\qquad w^{\prime}(z)=-2\pi{\rm i}Cq^{-1}(1+O(q)),$

where $q={\rm e}^{2\pi{\rm i}z}$ and $C\neq 0$ is a constant, so

$\displaystyle v_{k}(w)$
$\displaystyle=u_{k}(z)-2\ln|w^{\prime}(z)|=2\big{(}m_{\infty}^{(k)}+1\big{)}\ln|q|+O(1)$
$\displaystyle=-2\big{(}m_{\infty}^{(k)}+1\big{)}\ln|w|+O(1).$

Therefore, (4.6) is equivalent to

$\begin{cases}\displaystyle\Delta v_{1}+2{\rm e}^{v_{1}}-{\rm
e}^{v_{2}}=4\pi\frac{m_{{\rm
i}}^{(1)}-1}{2}\delta_{1}+4\pi\frac{m_{\rho}^{(1)}-2}{3}\delta_{0}+4\pi\sum_{j=1}^{m}m_{z_{j}}^{(1)}\delta_{p_{j}}\
\text{in}\ \mathbb{R}^{2},\\\ \displaystyle\Delta v_{2}+2{\rm e}^{v_{2}}-{\rm
e}^{v_{1}}=4\pi\frac{m_{{\rm
i}}^{(2)}-1}{2}\delta_{1}+4\pi\frac{m_{\rho}^{(2)}-2}{3}\delta_{0}+4\pi\sum_{j=1}^{m}m_{z_{j}}^{(2)}\delta_{p_{j}}\
\text{in}\ \mathbb{R}^{2},\\\
v_{k}(w)=-2\big{(}m_{\infty}^{(k)}+1\big{)}\ln|w|+O(1)\quad\text{as}\
|w|\to\infty.\end{cases}$ (4.7)

Note from Proposition 4.3 that

$m_{{\rm i}}^{(k)}+1=\kappa_{{\rm i}}^{(k+1)}-\kappa_{{\rm i}}^{(k)}\equiv 0\
\mathrm{mod}\ 2\qquad\text{and}\qquad
m_{\rho}^{(k)}+1=\kappa_{\rho}^{(k+1)}-\kappa_{\rho}^{(k)}\equiv 0\
\mathrm{mod}\ 3,$

so $\frac{m_{{\rm i}}^{(k)}-1}{2}\in\mathbb{Z}_{\geq 0}$ and
$\frac{m_{\rho}^{(2)}-2}{3}\in\mathbb{Z}_{\geq 0}$ for $k=1,2$. In conclusion,
starting from any given linearly independent modular forms
$f,g,h\in\mathfrak{M}_{k}(\mathrm{SL}(2,\mathbb{Z}))$, we can construct a two-
parametric family of solutions to certain Toda system (4.7):

######  Theorem 4.6.

$(v_{1;\lambda,\mu}(w),v_{2;\lambda,\mu}(w))$ defined by

$v_{k;\lambda,\mu}(w(z))=u_{k;\lambda,\mu}(z)-2\ln|w^{\prime}(z)|,\qquad
z\in\mathbb{H}$

are a two-parametric family of solutions of the $\mathrm{SU}(3)$ Toda system
(4.7), where $\lambda,\mu>0$ can be arbitrary.

##  5 Polynomial systems derived from the conditions (H1)–(H3)

In view of Theorem 1.3 proved in Section 3, a natural question is whether
given a prescribed set of singular points and the local exponents at
singularities and at the cusps, there exist MODEs (1.1) satisfying the
conditions (H1)–(H3). We will see in this section that this problem of
existence is equivalent to that of solving a certain system of polynomial
equations. Note that in view of Theorems 1.1 and 4.2 such a MODE (1.1) exists
for certain sets of data.

###  5.1 Solution expansions for MODEs

Let $\Gamma$ be a discrete subgroup of $\mathrm{SL}(2,\mathbb{R})$
commensurable with $\mathrm{SL}(2,\mathbb{Z})$ and (1.1) be a MODE on
$\Gamma$. To verify the apparentness of a singular point $z_{0}$ of (1.1), we
use the classical Frobenius method. However, since (1.1) is modular, it will
be more convenient that all functions are expanded in terms of $\widetilde{w}$
as introduced in [25] rather than $z-z_{0}$.

Fix $z_{0}\in\mathbb{H}$, we let $w=(z-z_{0})/(z-\overline{z}_{0})$ and

$\widetilde{w}=\frac{w}{1+Aw},\qquad\text{with}\quad
A=\frac{4\pi\phi^{\ast}(z_{0})\operatorname{Im}z_{0}}{\alpha},$ (5.1)

where $\phi(z)$ is the quasimodular form of weight $2$ and depth $1$ on
$\Gamma$, i.e.,

$\displaystyle\big{(}\phi\big{|}_{2}\gamma\big{)}(z)=\phi(z)+\frac{\alpha_{0}c}{2\pi{\rm
i}(cz+d)},\qquad\gamma=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in\Gamma,$

for some nonzero complex number $\alpha_{0}$ (note $\alpha_{0}=2\pi{\rm
i}\alpha$ for $\alpha$ given in (2.1)), and
$\phi^{*}(z):=\phi(z)+\frac{\alpha_{0}}{2\pi{\rm i}(z-\bar{z})}$. Clearly
$\phi^{*}(z)$ satisfies

$\big{(}\phi^{\ast}\big{|}_{2}\gamma\big{)}(z)=\phi^{\ast}(z),\qquad\gamma\in\Gamma.$

The advantage of the expansion in terms of $\widetilde{w}$ is the following
result.

######  Proposition 5.1 ([25, Propositions A.4 and A.7]).

Let $f(z)$ be a meromorphic modular form of weight $k$ on $\Gamma$. Then
$f(z)$ admits an expansion of the form

$f(z)=(1-(1+A)\widetilde{w})^{k}\sum_{n=n_{0}}^{\infty}a_{n}(-4\pi(\operatorname{Im}z_{0})\widetilde{w})^{n}.$

Furthermore, if $z_{0}$ is an elliptic point with the stabilizer subgroup
$\Gamma_{z_{0}}$ of order $N$, then $a_{n}=0$ whenever $k+2n\not\equiv 0\
\mathrm{mod}\ N$.

######  Remark 5.2.

Note that when $f(z)$ is a holomorphic modular form, the coefficients $a_{n}$
in the series can be expressed in terms of the Serre derivatives of $f(z)$.
See [25, Proposition A.4] for the precise statement.

Also note that when $z_{0}$ is an elliptic point, we have $A=0$ and hence
$\widetilde{w}=w$. This is because $\phi^{\ast}$ transforms like a modular
form of weight $2$, but any modular form of weight $2$ will vanish at every
elliptic point.

By Proposition 5.1, we can write

$\displaystyle Q(z):=\frac{Q_{2}(z)}{(2\pi{\rm
i})^{2}}=(1-(1+A)\widetilde{w})^{4}\sum_{n=-2}^{\infty}a_{n}(-4\pi(\operatorname{Im}z_{0})\widetilde{w})^{n},$
$\displaystyle R(z):=\frac{Q_{3}(z)-\frac{1}{2}Q_{2}^{\prime}(z)}{(2\pi{\rm
i})^{3}}=(1-(1+A)\widetilde{w})^{6}\sum_{n=-3}^{\infty}b_{n}(-4\pi(\operatorname{Im}z_{0})\widetilde{w})^{n}.$

Then (1.1) is equivalent to

$D_{q}^{3}y(z)+Q(z)D_{q}y(z)+\left(\frac{1}{2}D_{q}Q(z)+R(z)\right)y(z)=0,$
(5.2)

where $D_{q}:=q\frac{{\rm d}}{{\rm d}q}=\frac{1}{2\pi{\rm i}}\frac{{\rm
d}}{{\rm d}z}$. In particular, Proposition 5.1 yields

######  Corollary 5.3.

Suppose $-I_{2}\in\Gamma$ and let $z_{0}$ be the elliptic point of order $e$
of $\Gamma$. Then $N=2e$ and so $a_{n}=0$ if $n\not\equiv e-2\ \mathrm{mod}\
e$ and $b_{n}=0$ if $n\not\equiv e-3\ \mathrm{mod}\ e$.

For later usage, we recall Bol’s identity in the following form.

######  Lemma 5.4.

Let $\gamma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\mathrm{GL}(2,\mathbb{C})$ and $r$ be a positive
integer. Set

$w=\gamma z=(az+b)/(cz+d).$

Then for a $(r+1)$-th differentiable function $g(z)$, we have

$\frac{{\rm d}^{r+1}}{{\rm
d}z^{r+1}}\left(\frac{(\det\gamma)^{r/2}}{(a-cw)^{r}}g(w)\right)=\frac{(a-cw)^{r+2}}{(\det\gamma)^{r/2+1}}\frac{{\rm
d}^{r+1}}{{\rm d}w^{r+1}}g(w).$

###### Proof.

Bol’s identity states that

$\big{(}y\big{|}_{-r}\gamma\big{)}^{(r+1)}(z)=\big{(}y^{(r+1)}\big{|}_{r+2}\gamma\big{)}(z).$

Noticing that

$a-cw=a-c\frac{az+b}{cz+d}=\frac{\det\gamma}{cz+d},$

we find that the factor $(\det\gamma)^{1/2}/(cz+d)$ appearing in the slash
operator can be written as

$\frac{(\det\gamma)^{1/2}}{cz+d}=\frac{a-cw}{(\det\gamma)^{1/2}},$ (5.3)

which yields the version of Bol’s identity stated in the lemma. ∎

To apply Frobenius’ method by using the expansion in Proposition 5.1, we need
the following result.

######  Lemma 5.5.

Let $Q(z)$ and $R(z)$ be meromorphic modular forms of weight $4$ and $6$,
respectively, on $\Gamma$. Assume that $\widetilde{Q}(x)=\sum_{n\geq
n_{0}}a_{n}x^{n}$ and $\widetilde{R}(x)=\sum_{n\geq n_{0}}b_{n}x^{n}$ are the
power series such that

$\displaystyle
Q(z)=(1-(1+A)\widetilde{w})^{4}\widetilde{Q}(-4\pi(\operatorname{Im}z_{0})\widetilde{w}),$
$\displaystyle
R(z)=(1-(1+A)\widetilde{w})^{6}\widetilde{R}(-4\pi(\operatorname{Im}z_{0})\widetilde{w}).$
(5.4)

Then

$y(z)=\frac{1}{(1-(1+A)\widetilde{w})^{2}}\sum_{n=0}^{\infty}c_{n}(-4\pi(\operatorname{Im}z_{0})\widetilde{w})^{n+\alpha}$
(5.5)

is a solution of (5.2) if and only if the series
$\widetilde{y}(x)=\sum_{n=0}^{\infty}c_{n}x^{n+\alpha}$ satisfies

$\frac{{\rm d}^{3}}{{\rm d}x^{3}}\widetilde{y}(x)+\widetilde{Q}(x)\frac{{\rm
d}}{{\rm d}x}\widetilde{y}(x)+\left(\frac{1}{2}\frac{{\rm d}}{{\rm
d}x}\widetilde{Q}(x)+\widetilde{R}(x)\right)\widetilde{y}(x)=0.$ (5.6)

###### Proof.

Let

$\gamma=\begin{pmatrix}-4\pi\operatorname{Im}z_{0}&0\\\
0&1\end{pmatrix}\begin{pmatrix}1&0\\\
A&1\end{pmatrix}\begin{pmatrix}1&-z_{0}\\\
1&-\overline{z}_{0}\end{pmatrix}=\begin{pmatrix}-4\pi\operatorname{Im}z_{0}&(4\pi\operatorname{Im}z_{0})z_{0}\\\
1+A&-Az_{0}-\overline{z}_{0}\end{pmatrix}$

with $\det\gamma=-4\pi(\operatorname{Im}z_{0})(z_{0}-\overline{z}_{0})$. Let
$x=\gamma z=-4\pi(\operatorname{Im}z_{0})\widetilde{w}$. Note that if we write
$\gamma$ as $\gamma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)$, then

$\frac{(a-cx)^{2}}{\det\gamma}=2\pi{\rm i}(1-(1+A)\widetilde{w})^{2}.$

Thus, by (5.3), $y(z)$ and $\widetilde{y}(x)$ are related by

$y(z)=2\pi{\rm i}\big{(}\widetilde{y}\big{|}_{-2}\gamma\big{)}(z).$

Hence, applying Lemma 5.4 with $r=2$, we obtain

$D_{q}^{3}y(z)=\frac{1}{(2\pi{\rm i})^{2}}\frac{{\rm d}^{3}}{{\rm
d}z^{3}}\big{(}\widetilde{y}\big{|}_{-2}\gamma\big{)}(z)=(1-(1+A)\widetilde{w})^{4}\frac{{\rm
d}^{3}}{{\rm d}x^{3}}\widetilde{y}(x).$

Also, a direct computation yields, by (5.3),

$\frac{{\rm d}x}{{\rm d}z}=\frac{{\rm d}\gamma z}{{\rm
d}z}=\frac{\det\gamma}{(cz+d)^{2}}=\frac{(a-cx)^{2}}{\det\gamma}=2\pi{\rm
i}(1-(1+A)\widetilde{w})^{2},$

and

$\frac{{\rm d}\widetilde{w}}{{\rm
d}z}=-\frac{1}{4\pi\operatorname{Im}z_{0}}\frac{{\rm d}x}{{\rm
d}z}=\frac{(1-(1+A)\widetilde{w})^{2}}{2{\rm i}\operatorname{Im}z_{0}}.$

Hence,

$\displaystyle D_{q}y(z)$ $\displaystyle=\frac{1}{2\pi{\rm i}}\frac{{\rm
d}}{{\rm
d}z}\left(\frac{1}{(1-(1+A)\widetilde{w})^{2}}\widetilde{y}(x)\right)$
$\displaystyle=-\frac{1+A}{2\pi(\operatorname{Im}z_{0})(1-(1+A)\widetilde{w})}\widetilde{y}(x)+\frac{{\rm
d}}{{\rm d}x}\widetilde{y}(x),$

and

$\displaystyle D_{q}Q(z)$ $\displaystyle=\frac{1}{2\pi{\rm i}}\frac{{\rm
d}}{{\rm d}z}\left((1-(1+A)\widetilde{w})^{4}\widetilde{Q}(x)\right)$
$\displaystyle=(1-(1+A)\widetilde{w})^{6}\left(\frac{1+A}{\pi(\operatorname{Im}z_{0})(1-(1+A)\widetilde{w})}\widetilde{Q}(x)+\frac{{\rm
d}}{{\rm d}x}\widetilde{Q}(x)\right).$

Putting everything together, we find that

$\displaystyle
D_{q}^{3}y(z)+Q(z)D_{q}y(z)+\left(\frac{1}{2}D_{q}(z)+R(z)\right)y(z)$
$\displaystyle\qquad{}=(1-(1+A)\widetilde{w})^{4}\left(\frac{{\rm d}^{3}}{{\rm
d}x^{3}}\widetilde{y}(x)+\widetilde{Q}(x)\frac{{\rm d}}{{\rm
d}x}\widetilde{y}(x)+\left(\frac{1}{2}\frac{{\rm d}}{{\rm
d}x}\widetilde{Q}(x)+\widetilde{R}(x)\right)\widetilde{y}(x)\right).$

Thus, $y(z)$ is a solution of (5.2) if and only if $\widetilde{y}(x)$ is a
solution of (5.6). ∎

We will see from Lemma 5.10 below that Corollary 5.3 and Lemma 5.5 can be
applied to prove that (1.1) or equivalently (5.2) is apparent at elliptic
points of order $e\geq 3$. This is a great advantage of using expansions in
terms of $\widetilde{w}$ rather than $z-z_{0}$.

######  Remark 5.6.

In practice, the power series $\widetilde{Q}(x)$ and $\widetilde{R}(x)$ can be
computed using Proposition A.4 of [25]. For example, for the Eisenstein series
$E_{4}(z)$ and $E_{6}(z)$ on $\mathrm{SL}(2,\mathbb{Z})$, we find that

$\displaystyle
E_{4}(z)=(1-(1+A)\widetilde{w})^{4}\left(B-\frac{1}{3}Cu+\frac{5}{72}B^{2}u^{2}-\frac{5}{432}BCu^{3}+\cdots\right),$
$\displaystyle
E_{6}(z)=(1-(1+A)\widetilde{w})^{6}\left(C-\frac{1}{2}B^{2}u+\frac{7}{48}BCu^{2}+\cdots\right),$
(5.7)

where $u=-4\pi(\operatorname{Im}z_{0})\widetilde{w}$, $B=E_{4}(z_{0})$, and
$C=E_{6}(z_{0})$.

###  5.2 Existence of $\boldsymbol{Q(z)}$ and $\boldsymbol{R(z)}$

In this section we shall discuss the criterion of the existence of meromorphic
modular forms $Q(z)$ and $R(z)$ of weight $4$ and $6$, respectively, on
$\mathrm{SL}(2,\mathbb{Z})$, such that the differential equation (5.2) is
Fuchsian and apparent throughout $\mathbb{H}$ with prescribed local exponents
at singularities and at cusps.

Throughout the section, we let $z_{j}$, $j=1,\dots,m$, be
$\mathrm{SL}(2,\mathbb{Z})$-inequivalent point on $\mathbb{H}$, none of which
is an elliptic point. We let ${\rm i}=\sqrt{-1}$ and
$\rho=\big{(}{-}1+\sqrt{3}{\rm i}\big{)}/2$ be the unique elliptic point of
order $2$ and $3$ of $\mathrm{SL}(2,\mathbb{Z})$, respectively. For $z=z_{j}$,
${\rm i}$, or $\rho$, we assume that
$\kappa_{z}^{(1)}<\kappa_{z}^{(2)}<\kappa_{z}^{(3)}$, are rational numbers in
$\frac{1}{3}\mathbb{Z}$ such that
$\kappa_{z}^{(1)}+\kappa_{z}^{(2)}+\kappa_{z}^{(3)}=3$ and
$\kappa_{z}^{(2)}-\kappa_{z}^{(1)}\in\mathbb{Z}$ (and hence
$\kappa_{z}^{(3)}-\kappa_{z}^{(1)}\in\mathbb{Z})$. When $z={\rm i}$, we
further assume that

$\big{\\{}3\kappa_{\rm i}^{(1)},3\kappa_{\rm i}^{(2)},3\kappa_{\rm
i}^{(3)}\big{\\}}\equiv\\{0,0,1\\}\ \mathrm{mod}\ 2.$ (5.8)

Also, when $z=\rho$, we note that the assumptions on $\kappa_{\rho}^{(j)}$
above imply that $\kappa_{\rho}^{(j)}\in\mathbb{Z}$ for all $j$. We further
assume that

$\big{\\{}\kappa_{\rho}^{(1)},\kappa_{\rho}^{(2)},\kappa_{\rho}^{(3)}\big{\\}}\equiv\\{0,1,2\\}\quad\
\mathrm{mod}\ 3.$ (5.9)

Finally, for the cusp $\infty$ of $\mathrm{SL}(2,\mathbb{Z})$, we let
$\kappa_{\infty}^{(1)}\leq\kappa_{\infty}^{(2)}\leq\kappa_{\infty}^{(3)}$ be
three rational numbers in $\frac{1}{3}\mathbb{Z}$ such that
$\kappa_{\infty}^{(1)}+\kappa_{\infty}^{(2)}+\kappa_{\infty}^{(3)}=0$ and
$\kappa_{\infty}^{(2)}-\kappa_{\infty}^{(1)},\kappa_{\infty}^{(3)}-\kappa_{\infty}^{(2)}\in\mathbb{Z}$.
We shall consider the problem whether there exist meromorphic modular forms
$Q(z)$ and $R(z)$ of weight $4$ and $6$, respectively, on
$\mathrm{SL}(2,\mathbb{Z})$, such that (3.1) is Fuchsian and apparent
throughout $\mathbb{H}$ and the local exponents $\kappa$’s are given as above.

######  Lemma 5.7.

Let notations ${\rm i}$ and $\rho$ be as above. Then meromorphic modular forms
$Q(z)$ and $R(z)$ of weight $4$ and $6$, respectively, on
$\mathrm{SL}(2,\mathbb{Z})$ that have poles of order at most $2$ and $3$,
respectively, at points $\mathrm{SL}(2,\mathbb{Z})$-equivalent to $z_{j}$,
${\rm i}$, or $\rho$ and are holomorphic at other points and cusps are of the
form

$\displaystyle Q(z)=r_{\infty}E_{4}(z)+r_{{\rm
i}}^{(2)}\frac{E_{4}(z)\Delta_{0}(z)}{E_{6}(z)^{2}}+r_{\rho}^{(2)}\frac{\Delta_{0}(z)}{E_{4}(z)^{2}}$
$\displaystyle\hphantom{Q(z)=}{}+\sum_{j=1}^{m}\left(r_{z_{j}}^{(2)}\frac{E_{4}(z)\Delta_{0}(z)^{2}}{F_{j}(z)^{2}}+r_{z_{j}}^{(1)}\frac{E_{4}(z)\Delta_{0}(z)}{F_{j}(z)}\right),$
$\displaystyle R(z)=s_{\infty}E_{6}(z)+s_{{\rm
i}}^{(3)}\frac{\Delta_{0}(z)^{2}}{E_{6}(z)^{3}}+s_{{\rm
i}}^{(1)}\frac{\Delta_{0}(z)}{E_{6}(z)}$
$\displaystyle\hphantom{R(z)=}{}+s_{\rho}^{(3)}\frac{E_{6}(z)\Delta_{0}(z)}{E_{4}(z)^{3}}+\sum_{j=1}^{m}\sum_{k=1}^{3}s_{z_{j}}^{(k)}\frac{E_{6}(z)\Delta_{0}(z)^{k}}{F_{j}(z)^{k}},$
(5.10)

where $\Delta_{0}(z)=1728\Delta(z)=(E_{4}(z)^{3}-E_{6}(z)^{2})$ and
$F_{j}(z)=E_{4}(z)^{3}-t_{j}E_{6}(z)^{2}$ with
$t_{j}=E_{4}(z_{j})^{3}/E_{6}(z_{j})^{2}$.

###### Proof.

By Corollary 5.3, there are no meromorphic modular forms of weight $4$ having
a pole at ${\rm i}$ or $\rho$ with a nonzero residue. Also, the order of a
meromorphic modular form of weight $6$ on $\mathrm{SL}(2,\mathbb{Z})$ at ${\rm
i}$ is necessarily odd, while that at $\rho$ is congruent to $0$ modulo $3$.
Thus, we can take $r_{\rm i}^{(2)}$, $r_{\rho}^{(2)}$, $r_{z_{j}}^{(2)}$,
$r_{z_{j}}^{(1)}$ such that

$\displaystyle Q(z)-r_{{\rm
i}}^{(2)}\frac{E_{4}(z)\Delta_{0}(z)}{E_{6}(z)^{2}}-r_{\rho}^{(2)}\frac{\Delta_{0}(z)}{E_{4}(z)^{2}}-\sum_{j=1}^{m}\left(r_{z_{j}}^{(2)}\frac{E_{4}(z)\Delta_{0}(z)^{2}}{F_{j}(z)^{2}}+r_{z_{j}}^{(1)}\frac{E_{4}(z)\Delta_{0}(z)}{F_{j}(z)}\right)$

is a holomorphic modular form of weight $4$ on $\mathrm{SL}(2,\mathbb{Z})$, so
it must be a multiple of $E_{4}(z)$. The proof for $R(z)$ is similar. ∎

We first determine the indicial equations of (5.2) at $\infty$, ${\rm i}$,
$\rho$, and $z_{j}$, $j=1,\dots,m$.

######  Lemma 5.8.

Suppose that $Q(z)$ and $R(z)$ are meromorphic modular forms given by (5.10).
Then the indicial equation of (5.2) at the cusp $\infty$ is

$x^{3}+r_{\infty}x+s_{\infty}=0.$

###### Proof.

It is clear that $Q(z)=r_{\infty}+O(q)$ and $R(z)=s_{\infty}+O(q)$. Assume
that there is a solution of (5.2) of the form $y(z)=q^{\alpha}(1+O(q))$,
$\alpha\in\mathbb{R}$. We compute that

$\displaystyle
D_{q}^{3}y(z)+Q(z)D_{q}y(z)+\left(\frac{1}{2}D_{q}Q(z)+R(z)\right)y(z)$
$\displaystyle\qquad{}=\alpha^{3}q^{\alpha}+r_{\infty}\alpha
q^{\alpha}+s_{\infty}q^{\alpha}+O\big{(}q^{\alpha+1}\big{)},$

from which we see that the indicial equation at $\infty$ is
$x^{3}+r_{\infty}x+s_{\infty}=0$. ∎

We now consider the indicial equation of (5.2) at a point in $\mathbb{H}$. Let
$z_{0}$ be one of $z_{j}$, ${\rm i}$, or $\rho$ and $\widetilde{w}$ be defined
by (5.1) with $\phi^{\ast}(z)=E_{2}^{\ast}(z):=E_{2}(z)+6/(\pi{\rm
i}(z-\overline{z}))$. Recall that in Lemma 5.5 we have proved that if
$\widetilde{Q}(x)$ and $\widetilde{R}(x)$ are the Laurent series in $x$ such
that (5.4) holds. Then

$y(z;\alpha)=\frac{1}{(1-(1+A)\widetilde{w})^{2}}\sum_{n=0}^{\infty}c_{n}(\alpha)(-4\pi(\operatorname{Im}z_{0})\widetilde{w})^{n+\alpha},\qquad
c_{0}(\alpha)=1,$

is a solution of (5.2) if and only if the series
$\widetilde{y}(x;\alpha)=\sum_{n=0}^{\infty}c_{n}(\alpha)x^{n+\alpha}$
satisfies

$\frac{{\rm d}^{3}}{{\rm
d}x^{3}}\widetilde{y}(x;\alpha)+\widetilde{Q}(x)\frac{{\rm d}}{{\rm
d}x}\widetilde{y}(x;\alpha)+\left(\frac{1}{2}\frac{{\rm d}}{{\rm
d}x}\widetilde{Q}(x)+\widetilde{R}(x)\right)\widetilde{y}(x;\alpha)=0.$ (5.11)

Let

$\widetilde{Q}(x)=\sum_{n=-2}^{\infty}a_{n}x^{n},\qquad\frac{1}{2}\frac{{\rm
d}}{{\rm
d}x}\widetilde{Q}(x)+\widetilde{R}(x)=\sum_{n=-3}^{\infty}b_{n}x^{n},$ (5.12)

where each $a_{n}$ and $b_{n}$ is linear in the parameters $r$’s and $s$’s.
Then following the computation in Appendix A, we see that (5.11) is equivalent
to

$R_{n}(\alpha):=f(\alpha+n)c_{n}(\alpha)+\sum_{k=0}^{n-1}[(\alpha+k)a_{n-k-2}+b_{n-k-3}]c_{k}(\alpha)=0$
(5.13)

for all $n\geq 0$, where $f(t):=t(t-1)(t-2)+a_{-2}t+b_{-3}$. In particular,
$R_{0}(\alpha)\colon f(\alpha)=0$ is the indicial equation at $z_{0}$, i.e.,

$f(t)=\big{(}t-\kappa_{z_{0}}^{(1)}\big{)}\big{(}t-\kappa_{z_{0}}^{(2)}\big{)}\big{(}t-\kappa_{z_{0}}^{(3)}\big{)}.$

Using (5.7), we can work out the coefficients $a_{-2}$ and $b_{-3}$.

######  Lemma 5.9.

Suppose that $Q(z)$ and $R(z)$ are meromorphic modular forms given by (5.10).
Then the indicial equations of (5.2) at ${\rm i}$, $\rho$, and $z_{j}$,
$j=1,\dots,m$, are

$\displaystyle x(x-1)(x-2)+4r_{{\rm i}}^{(2)}x-4r_{{\rm i}}^{(2)}-8s_{{\rm
i}}^{(3)}=0,$ $\displaystyle
x(x-1)(x-2)-9r_{\rho}^{(2)}x+9r_{\rho}^{(2)}+27s_{\rho}^{(3)}=0,$
$\displaystyle
x(x-1)(x-2)+\frac{r_{z_{j}}^{(2)}}{t_{j}}x-\frac{r_{z_{j}}^{(2)}}{t_{j}}+\frac{s_{z_{j}}^{(3)}}{t_{j}^{2}}=0,$

respectively.

###### Proof.

For the elliptic point ${\rm i}$, we know that $E_{6}({\rm i})=0$. Hence,
using (5.7), we find that

$\displaystyle\widetilde{Q}(x)=\frac{4r_{{\rm
i}}^{(2)}}{x^{2}}+\dots,\qquad\widetilde{R}(x)=-\frac{8s_{{\rm
i}}^{(2)}}{x^{3}}+\cdots.$

Therefore, we have $a_{-2}=4r_{{\rm i}}^{(2)}$ and $b_{-3}=-4r_{{\rm
i}}^{(2)}-8s_{{\rm i}}^{(2)}$. Similarly, for the elliptic point $\rho$, using
$E_{4}(\rho)=0$ and (5.7) again, we find that $a_{-2}=-9r_{\rho}^{(2)}$ and
$b_{-3}=9r_{\rho}^{(2)}+27s_{\rho}^{(3)}$. For the point $z_{j}$, letting
$B=E_{4}(z_{j})$ and $C=E_{6}(z_{j})$, we compute that

$\displaystyle\frac{\Delta_{0}(z)}{(1-(1+A)\widetilde{w})^{12}}=\big{(}B^{3}-C^{2}\big{)}+O(\widetilde{w})=C^{2}(t_{j}-1)+O(\widetilde{w}),$
$\displaystyle\frac{F_{j}(z)}{(1-(1+A)\widetilde{w})^{12}}=\left(B-\frac{1}{3}Cu+O\big{(}\widetilde{w}^{2}\big{)}\right)^{3}-t_{j}\left(C-\frac{1}{2}B^{2}u+O\big{(}\widetilde{w}^{2}\big{)}\right)^{2}$
$\displaystyle\hphantom{\frac{F_{j}(z)}{(1-(1+A)\widetilde{w})^{12}}}{}=(t_{j}-1)B^{2}Cu+O\big{(}\widetilde{w}^{2}\big{)},$

where $u=-4\pi(\operatorname{Im}z_{j})\widetilde{w}$. It follows that

$\widetilde{Q}(x)=r_{z_{j}}^{(2)}\frac{(t_{j}-1)^{2}BC^{4}}{(t_{j}-1)^{2}B^{4}C^{2}x^{2}}+\cdots=\frac{r_{z_{j}}^{(2)}}{t_{j}x^{2}}+\dots,$

and

$\widetilde{R}(x)=s_{z_{j}}^{(3)}\frac{(t_{j}-1)^{3}C^{7}}{(t_{j}-1)^{3}B^{6}C^{3}}+\cdots=\frac{s_{z_{j}}^{(3)}}{t_{j}^{2}x^{3}}+\cdots.$

Therefore, $a_{-2}=r_{z_{j}}^{(2)}/t_{j}$ and
$b_{-3}=-r_{z_{j}}^{(2)}/t_{j}+s_{z_{j}}^{(3)}/t_{j}^{2}$. This proves the
lemma. ∎

The two lemmas show that the parameters $r_{\infty}$, $s_{\infty}$, $r_{{\rm
i}}^{(2)}$, $s_{{\rm i}}^{(3)}$, $r_{\rho}^{(2)}$, $s_{\rho}^{(3)}$,
$r_{z_{j}}^{(2)}$, and $s_{z_{j}}^{(3)}$, $j=1,\dots,m$, solely depend on the
local exponents $\kappa$’s. The remaining parameters are

$\begin{cases}r_{z_{j}}^{(1)},s_{z_{j}}^{(2)},s_{z_{j}}^{(1)},\quad\text{for}\
j=1,\dots,m,\\\ s_{{\rm i}}^{(1)}.\end{cases}$

That is, the number of remaining parameters is $3m+1$. We now show that the
apparentness condition will impose $3m+1$ polynomial constraints on the
remaining parameters. For convenience, we let $\mathbf{r}$ and $\mathbf{s}$
denote $r_{z_{1}}^{(1)},\dots,r_{z_{m}}^{(1)}$ and $s_{{\rm
i}}^{(1)},s_{z_{1}}^{(2)},s_{z_{1}}^{(1)},\dots,s_{z_{m}}^{(2)},s_{z_{m}}^{(1)}$,
respectively.

Let $z_{0}\in\\{{\rm i},\rho,z_{1},\dots,z_{m}\\}$. Assume that $\alpha$ is
one of the local exponents $\kappa_{z_{0}}^{(k)}$. We observe that one can
recursively determine $c_{n}(\alpha)$ using (5.13) as long as $f(\alpha+n)\neq
0$. Thus, there always exists a solution
$\widetilde{y}\big{(}x;\kappa_{z_{0}}^{(3)}\big{)}$ with local exponent
$\kappa_{z_{0}}^{(3)}$; see, e.g., Lemma A.2. For the exponent
$\alpha=\kappa_{z_{0}}^{(2)}$, because

$f\big{(}\alpha+\kappa_{z_{0}}^{(3)}-\kappa_{z_{0}}^{(2)}\big{)}=f\big{(}\kappa_{z_{0}}^{(3)}\big{)}=0,$

we can only solve for $c_{n}(\alpha)$ up to
$n=\kappa_{z_{0}}^{(3)}-\kappa_{z_{0}}^{(2)}-1$. At
$n^{\prime}=\kappa_{z_{0}}^{(3)}-\kappa_{z_{0}}^{(2)}$, the relation
$R_{n^{\prime}}(\alpha)$ becomes

$\sum_{k=0}^{n^{\prime}-1}[(\alpha+k)a_{n^{\prime}-k-2}+b_{n^{\prime}-k-3}]c_{k}(\alpha)=0.$
(5.14)

If $c_{k}(\alpha)$, $k=0,\dots,n^{\prime}-1$, do not satisfy this relation,
then there is no solution with exponent $\kappa_{z_{0}}^{(2)}$. If
$c_{k}(\alpha)$, $k=0,\dots,n^{\prime}-1$, satisfy this relation, we then can
choose $c_{n^{\prime}}(\alpha)$ to be any number and solve for $c_{n}(\alpha)$
recursively and obtain a solution
$\widetilde{y}\big{(}x;\kappa_{z_{0}}^{(2)}\big{)}$ with local exponent
$\kappa_{z_{0}}^{(2)}$. Likewise, for the local exponent
$\alpha=\kappa_{z_{0}}^{(1)}$, there will be two conditions (5.14)
corresponding to $n^{\prime}=\kappa_{z_{0}}^{(2)}-\kappa_{z_{0}}^{(1)}$ and
$n^{\prime}=\kappa_{z_{0}}^{(3)}-\kappa_{z_{0}}^{(1)}$ that $c_{k}(\alpha)$
must satisfy. Thus, there are three polynomial equations

$P_{z_{0},k_{1},k_{2}}(\mathbf{r},\mathbf{s})=0,\qquad(k_{1},k_{2})\in\\{(1,2),(1,3),(2,3)\\}$

that $\mathbf{r}$ and $\mathbf{s}$ need to satisfy. However, we will see in a
moment that when $z_{0}$ is an elliptic point, some or all of the three
polynomial equations hold trivially.

######  Lemma 5.10.

Assume that $z_{0}$ is an elliptic point of order $e$ of
$\mathrm{SL}(2,\mathbb{Z})$. Let
$(k_{1},k_{2})\in\\{(1,2),\allowbreak(1,3),(2,3)\\}$. If
$\kappa_{z_{0}}^{(k_{2})}-\kappa_{z_{0}}^{(k_{1})}\not\equiv 0\ \mathrm{mod}\
e$, then the polynomial $P_{z_{0},k_{1},k_{2}}(\mathbf{r},\mathbf{s})$ is
identically zero. In particular, under our assumptions (5.8) and (5.9), the
differential equation (5.2) is apparent at $\rho$ for any $\mathbf{r}$ and
$\mathbf{s}$, and also there is at most one pair $(k_{1},k_{2})$ such that
$P_{{\rm i},k_{1},k_{2}}(\mathbf{r},\mathbf{s})\neq 0$.

###### Proof.

By Corollary 5.3, the coefficients $a_{n}$ in the expansion of
$\widetilde{Q}(x)$ vanish whenever $n\not\equiv-2\allowbreak\ \mathrm{mod}\
e$. Likewise, the coefficients $b_{n}$ in $\widetilde{R}(x)$ vanish whenever
$n\not\equiv-3\ \mathrm{mod}\ e$. Using these facts, we find that the
condition (5.13) reduces to

$f(\alpha+n)c_{n}(\alpha)+\sum_{k\equiv n\ \mathrm{mod}\ e,\,k\leq
n-1}[(\alpha+k)a_{n-k-2}+b_{n-k-3}]c_{k}(\alpha)=0.$ (5.15)

Then we can easily prove by induction up to $n=n^{\prime}-1$ that
$c_{n}(\alpha)=0$ whenever $n\not\equiv 0\ \mathrm{mod}\ e$. Now if
$n^{\prime}\not\equiv 0\ \mathrm{mod}\ e$, then (5.14) automatically holds
because every summand is $0$ due to the facts that $a_{n^{\prime}-k-2}$ and
$b_{n^{\prime}-k-3}$ are nonzero only when $k\equiv n^{\prime}\ \mathrm{mod}\
e$ and $c_{k}(\alpha)$ is nonzero only when $k\equiv 0\ \mathrm{mod}\ e$, but
$k$ cannot be congruent to $n^{\prime}$ and $0$ at the same time. This proves
the lemma. ∎

In the following, we let $P_{{\rm i}}(\mathbf{r},\mathbf{s})$ denote the only
nonzero polynomial $P_{{\rm i},k_{1},k_{2}}(\mathbf{r},\mathbf{s})$ in the
lemma. The discussion above shows that there are $3m+1$ polynomial equations
$P_{{\rm i}}(\mathbf{r},\mathbf{s})=0$,
$P_{z_{j},k_{1},k_{2}}(\mathbf{r},\mathbf{s})=0$, $j=1,\dots,m$,
$(k_{1},k_{2})\in\\{(1,2),(1,3),(2,3)\\}$ in $3m+1$ variables such that (5.2)
is Fuchsian and apparent throughout $\mathbb{H}$ and all
$\mathrm{SL}(2,\mathbb{Z})$-inequivalent singularities belong to $\\{{\rm
i},\rho,z_{1},\dots,z_{m}\\}$ with the given local exponents if and only if
the parameters $\mathbf{r}$ and $\mathbf{s}$ are common roots of the
polynomials. We now consider the degree of these polynomials.

######  Proposition 5.11.

We have

$\deg
P_{z_{j},k_{1},k_{2}}(\mathbf{r},\mathbf{s})=\begin{cases}\kappa_{z_{j}}^{(k_{2})}-\kappa_{z_{j}}^{(k_{1})},&\text{if
}(k_{1},k_{2})=(1,2)\text{ or }(2,3),\\\
\kappa_{z_{j}}^{(3)}-\kappa_{z_{j}}^{(1)}-1,&\text{if
}(k_{1},k_{2})=(1,3),\end{cases}$

and

$\deg P_{{\rm i}}(\mathbf{r},\mathbf{s})=\big{(}\kappa_{\rm
i}^{(k_{2})}-\kappa_{\rm i}^{(k_{1})}\big{)}/2,$

where $(k_{1},k_{2})$ is the unique pair such that $\kappa_{\rm
i}^{(k_{2})}-\kappa_{\rm i}^{(k_{1})}\equiv 0\ \mathrm{mod}\ 2$. To be more
precise, for a polynomial $P(\mathbf{r},\mathbf{s})$ in $\mathbf{r}$ and
$\mathbf{s}$, we let $\operatorname{LT}(P)$ denote the sum of the terms of
highest degree in $P$. Then, up to nonzero scalars, we have

$\operatorname{LT}(P_{z_{j},k_{1},k_{2}}(\mathbf{r},\mathbf{s}))=\prod_{k=1}^{\kappa_{z_{j}}^{(k_{2})}-\kappa_{z_{j}}^{(k_{1})}}\big{(}\big{(}\kappa_{z_{j}}^{(k_{1})}+k-3/2\big{)}d_{1}r_{z_{j}}^{(1)}+d_{2}s_{z_{j}}^{(2)}\big{)}$

for $(k_{1},k_{2})=(1,2)$ or $(2,3)$, and

$\displaystyle\operatorname{LT}(P_{z_{j},1,3}(\mathbf{r},\mathbf{s}))=\big{(}d_{3}s_{z_{j}}^{(1)}+d_{1}^{\prime}r_{z_{1}}^{(1)}+\cdots+d_{m}^{\prime}r_{z_{m}}^{(1)}\big{)}$
$\displaystyle\hphantom{\operatorname{LT}(P_{z_{j},1,3}(\mathbf{r},\mathbf{s}))=}{}\times\prod_{k=1,\,k\neq\kappa_{z_{j}}^{(2)}-\kappa_{z_{j}}^{(1)},\kappa_{z_{j}}^{(2)}-\kappa_{z_{j}}^{(1)}+1}\big{(}\big{(}\kappa_{z_{j}}^{(k_{1})}+k-3/2\big{)}d_{1}r_{z_{j}}^{(1)}+d_{2}s_{z_{j}}^{(2)}\big{)},$

where $d_{1},d_{2},d_{3},d_{1}^{\prime},\dots,d_{m}^{\prime}$ are complex
numbers with $d_{1},d_{2},d_{3}\neq 0$.

Also, $\operatorname{LT}(P_{\rm i}(\mathbf{r},\mathbf{s}))$ is a product of
$\big{(}\kappa_{\rm i}^{(k_{2})}-\kappa_{\rm i}^{(k_{1})}\big{)}/2$ linear
sums in $s_{{\rm i}}^{(1)}$ and $\mathbf{r}$ with the coefficients of $s_{{\rm
i}}^{(1)}$ being nonzero.

###### Proof.

By Lemma 5.7, each of $a_{n}$ and $b_{n}$ in (5.12) is a linear combination of
$r$’s and $s$’s. The key observation here is that only $r_{z_{j}}^{(2)}$
(which has been determined by the local exponents and is regarded as a
constant) appears in $a_{-2}$, only $r_{z_{j}}^{(2)}$ and $r_{z_{j}}^{(1)}$
appear in $a_{-1}$, only $r_{z_{j}}^{(2)}$ and $s_{z_{j}}^{(3)}$ (which has
been determined by the local exponents and is regarded as a constant) appear
in $b_{-3}$, only $r_{z_{j}}^{(2)}$, $r_{z_{j}}^{(1)}$, $s_{z_{j}}^{(3)}$, and
$s_{z_{j}}^{(2)}$ appear in $b_{-2}$, and only $r_{z_{j}}^{(2)}$,
$r_{z_{j}}^{(1)}$, $s_{z_{j}}^{(3)}$, $s_{z_{j}}^{(2)}$, and $s_{z_{j}}^{(1)}$
appear in $b_{-1}$. In particular, we have

$\operatorname{LT}(a_{-1})=d_{1}r_{z_{j}}^{(1)},\qquad\operatorname{LT}(b_{-2})=-\frac{1}{2}d_{1}r_{z_{j}}^{(1)}+d_{2}s_{z_{j}}^{(2)},$
(5.16)

where $d_{1}$ and $d_{2}$ are the leading coefficients in the series
expansions of $E_{4}(z)\Delta(z)/F_{j}(z)$ and
$E_{6}(z)\Delta(z)^{2}/F_{j}(z)^{2}$ in $\widetilde{w}$ (and hence are
nonzero). Consider (5.14) for the cases
$(\alpha,n^{\prime})=\big{(}\kappa_{z_{j}}^{(2)},\kappa_{z_{j}}^{(3)}-\kappa_{z_{j}}^{(2)}\big{)}$
and
$(\alpha,n^{\prime})=\big{(}\kappa_{z_{j}}^{(1)},\kappa_{z_{j}}^{(2)}-\kappa_{z_{j}}^{(1)}\big{)}$
first. From (5.13) and (5.16), we can easily show inductively that, up to
$n=n^{\prime}-1$,

$\displaystyle\operatorname{LT}(c_{n}(\alpha))$
$\displaystyle=\prod_{k=1}^{n}\left(-\frac{\operatorname{LT}((\alpha+k-1)a_{-1}+b_{-2})}{f(\alpha+k)}\right)$
$\displaystyle=\prod_{k=1}^{n}\left(-\frac{(\alpha+k-3/2)d_{1}r_{z_{j}}^{(1)}+d_{2}s_{z_{j}}^{(2)}}{f(\alpha+k)}\right),$
(5.17)

where $d_{1}$ and $d_{2}$ are the two nonzero complex numbers in (5.16), and
hence

$\operatorname{LT}(P_{z_{j},k_{1},k_{2}}(\mathbf{r},\mathbf{s}))=\prod_{k=1}^{\kappa_{z_{0}}^{(k_{2})}-\kappa_{z_{0}}^{(k_{1})}}\left(-\frac{(\alpha+k-3/2)d_{1}r_{z_{j}}^{(1)}+d_{2}s_{z_{j}}^{(2)}}{f(\alpha+k)}\right)$

for $(k_{1},k_{2})=(1,2)$ or $(2,3)$.

We now consider (5.14) for the remaining case
$(\alpha,n^{\prime})=\big{(}\kappa_{z_{j}}^{(1)},\kappa_{z_{j}}^{(3)}-\kappa_{z_{j}}^{(1)}\big{)}$.
Let $n^{\prime\prime}=\kappa_{z_{j}}^{(2)}-\kappa_{z_{j}}^{(1)}$. Up to
$n=n^{\prime\prime}-1$, the terms of highest degree in $c_{n}(\alpha)$ is
given by (5.17). Since $P_{z_{j},1,2}(\mathbf{r},\mathbf{s})=0$ is assumed to
hold, (5.13) holds for $n=n^{\prime\prime}$ for arbitrary
$c_{n^{\prime\prime}}(\alpha)$. Here, we simply choose
$c_{n^{\prime\prime}}(\alpha)$ to be $0$. Then we have, by (5.13)