# Metrics with Positive constant curvature and modular differential equations

Jia-Wei Guo, Chang-Shou Lin, and Yifan Yang  Department of Mathematics,
National Taiwan University, Taipei, Taiwan 10617<EMAIL_ADDRESS>Center for Advanced Study in Theoretical Sciences (CASTS), National Taiwan
University, Taipei, Taiwan 10617<EMAIL_ADDRESS>Department of
Mathematics, National Taiwan University and National Center for Theoretical
Sciences, Taipei, Taiwan 10617<EMAIL_ADDRESS>
###### Abstract.

Let $\mathbb{H}^{\ast}=\mathbb{H}\cup\mathbb{Q}\cup\\{\infty\\}$, where
$\mathbb{H}$ is the complex upper half-plane, and $Q(z)$ be a meromorphic
modular form of weight $4$ on $\mathrm{SL}(2,\mathbb{Z})$ such that the
differential equation $\mathcal{L}:y^{\prime\prime}(z)=Q(z)y(z)$ is Fuchsian
on $\mathbb{H}^{\ast}$. In this paper, we consider the problem when
$\mathcal{L}$ is apparent on $\mathbb{H}$, i.e., the ratio of any two nonzero
solutions of $\mathcal{L}$ is single-valued and meromorphic on $\mathbb{H}$.
Such a modular differential equation is closely related to the existence of a
conformal metric $ds^{2}=e^{u}|dz|^{2}$ on $\mathbb{H}$ with curvature $1/2$
that is invariant under $z\mapsto\gamma\cdot z$ for all
$\gamma\in\mathrm{SL}(2,\mathbb{Z})$.

Let $\pm\kappa_{\infty}$ be the local exponents of $\mathcal{L}$ at $\infty$.
In the case $\kappa_{\infty}\in\frac{1}{2}\mathbb{Z}$, we obtain the following
results:

  1. (a)

a complete characterization of $Q(z)$ such that $\mathcal{L}$ is apparent on
$\mathbb{H}$ with only one singularity (up to
$\mathrm{SL}(2,\mathbb{Z})$-equivalence) at $i=\sqrt{-1}$ or
$\rho=(1+\sqrt{3}i)/2$, and

  2. (b)

a complete characterization of $Q(z)$ such that $\mathcal{L}$ is apparent on
$\mathbb{H}^{\ast}$ with singularities only at $i$ and $\rho$.

We provide two proofs of the results, one using Riemann’s existence theorem
and the other using Eremenko’s theorem on the existence of conformal metric on
the sphere.

In the case $\kappa_{\infty}\notin\frac{1}{2}\mathbb{Z}$, we let
$r_{\infty}\in(0,1/2)$ be defined by $r_{\infty}\equiv\pm\kappa_{\infty}\
\mathrm{mod}\ 1$. Assume that $r_{\infty}\notin\\{1/12,5/12\\}$. A special
case of an earlier result of Eremenko and Tarasov says that
$1/12<r_{\infty}<5/12$ is the necessary and sufficient condition for the
existence of the invariant metric. The threshold case
$r_{\infty}\in\\{1/12,5/12\\}$ is more delicate. We show that in the threshold
case, an invariant metric exists if and only if $\mathcal{L}$ has two linearly
independent solutions whose squares are meromorphic modular forms of weight
$-2$ with a pair of conjugate characters on $\mathrm{SL}(2,\mathbb{Z})$. In
the non-existence case, our example shows that the monodromy data of
$\mathcal{L}$ are related to periods of the elliptic curve $y^{2}=x^{3}-1728$.

We thank the referee for providing the reference [12] and for detailed
comments that improve the exposition of the paper.

##  1\. Introduction

A meromorphic function $Q$ on the upper half-plane $\mathbb{H}$ is called a
meromorphic modular form of weight $k\in\mathbb{Z}$ with respect to
$\mathrm{SL}(2,\mathbb{Z})$ if $Q$ satisfies

$Q(\gamma\cdot z)=(cz+d)^{k}Q(z),\quad\gamma=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in\mathrm{SL}(2,\mathbb{Z}),$

and $Q$ is also meromorphic at the cusp $\infty$. When $k=0$, a meromorphic
modular form is called a modular function. We refer to [1] and [19] for the
elementary theory of (holomorphic) modular forms. Given a meromorphic modular
form $Q$ of weight $4$ on $\mathrm{SL}(2,\mathbb{Z})$, we consider a Fuchsian
modular differential equation of second order on $\mathbb{H}$

(1.1) $y^{\prime\prime}=Q(z)y\quad\text{on}\ \mathbb{H},\qquad
y^{\prime}:=\frac{dy}{dz}.$

The differential equation (1.1) is called Fuchsian if the order of any pole of
$Q$ is less than or equal to $2$. At $\infty$, by using $q=e^{2\pi iz}$, (1.1)
can be written as

(1.2)
$\left(q\frac{d}{dq}\right)^{2}y=-\frac{1}{4\pi^{2}}y^{\prime\prime}=-\frac{Q(z)}{4\pi^{2}}y.$

So (1.1) is Fuchsian at $\infty$ if and only if $Q$ is holomorphic at
$\infty$.

Suppose that $z_{0}$ is a pole of $Q$. The local exponents of (1.1) are
$1/2\pm\kappa$, $\kappa\geq 0$. If the difference $2\kappa$ of the two local
exponents is an integer, then the ODE (1.1) might have a solution with a
logarithmic singularity at $z_{0}$. A singular point $z_{0}$ of (1.1) is
called _apparent_ if the local exponents are $1/2\pm\kappa$ with
$\kappa\in\frac{1}{2}\mathbb{Z}_{\geq 0}$ and any solution of (1.1) has no
logarithmic singularity near $z_{0}$. In such a case, it is necessary that
$\kappa>0$. The ODE (1.1) or $Q$ is called _apparent_ if (1.1) is apparent at
any pole of $Q$ on $\mathbb{H}$. Clearly, if (1.1) is apparent then the local
monodromy matrix at any pole is $\pm I$, where $I$ is the $2\times 2$ identity
matrix.

A solution $y(z)$ of (1.1) might be multi-valued. For
$\gamma\in\mathrm{SL}(2,\mathbb{Z})$, $y(\gamma\cdot z)$ is understood as the
analytic continuation of $y$ along a path connecting $z$ and $\gamma\cdot z$.
Even though $y(\gamma\cdot z)$ is not well-defined, the slash operator of
weight $k$ ($k\in\mathbb{Z}$) can be defined in the usual way by

(1.3) $\left(y\big{|}_{k}\gamma\right)(z):=(cz+d)^{-k}y(\gamma\cdot
z),\quad\gamma=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in\mathrm{SL}(2,\mathbb{Z}),$

where $\gamma\cdot z=(az+b)/(cz+d)$. We have the well-known Bol’s identity [2]

$\left(y\big{|}_{-1}\gamma\right)^{(2)}(z)=\left(y^{(2)}\big{|}_{3}\gamma\right)(z).$

Hence, if $y(z)$ is a solution of (1.1), then
$\left(y\big{|}_{-1}\gamma\right)(z)$ is also a solution of (1.1). Here
$f^{(k)}(z)$ is the $k$-th derivative of $f(z)$.

Suppose that (1.1) is _apparent_ and $y_{i}$, $i=1,2$, are two independent
solutions. Since the local monodromy matrix at any pole of $Q$ is $\pm I$, the
ratio $h(z)=y_{2}(z)/y_{1}(z)$ is well-defined and meromorphic on
$\mathbb{H}$. By Bol’s identity, both
$\left(y_{i}\big{|}_{-1}\gamma\right)(z)$ are solutions of (1.1), where
$y_{1}(\gamma\cdot z)$ and $y_{2}(\gamma\cdot z)$ are understood as the
analytic continuation of $y_{1}(z)$ and $y_{2}(z)$ along the same path
connecting $z$ and $\gamma\cdot z$. Note that since (1.1) is assumed to be
apparent, difference choices of paths from $z$ to $\gamma\cdot z$ only result
in sign changes in $y_{1}(\gamma\cdot z)$ and $y_{2}(\gamma\cdot z)$.
Therefore, there is a matrix $\rho(\gamma)$ in $\mathrm{GL}(2,\mathbb{C})$
such that

(1.4) $\begin{pmatrix}\left(y_{1}\big{|}_{-1}\gamma\right)(z)\\\
\left(y_{2}\big{|}_{-1}\gamma\right)(z)\end{pmatrix}=\pm\rho(\gamma)\begin{pmatrix}y_{1}(z)\\\
y_{2}(z)\end{pmatrix}.$

Note that $\det\rho(\gamma)=1$ because the two Wronskians of fundamental
solutions $\left(y_{1}\big{|}_{-1}\gamma,y_{2}\big{|}_{-1}\gamma\right)$ and
$(y_{1},y_{2})$ are equal. Hence $\rho$ is a homomorphism from
$\mathrm{SL}(2,\mathbb{Z})$ to $\mathrm{PSL}(2,\mathbb{C})$. In this paper, we
call the homomorphism $\gamma\mapsto\pm\rho(\gamma)$ _the Bol representation_
associated to (1.1).

There is an old problem in conformal geometry related to (1.1). The problem is
to find a metric $ds^{2}$ with curvature $1/2$ on $\mathbb{H}$ that is locally
conformal to the flat metric and invariant under the change
$z\mapsto\gamma\cdot z$, $\gamma\in\mathrm{SL}(2,\mathbb{Z})$. Write
$ds^{2}=e^{u}\left|dz\right|^{2}$. Below, we collect some basic results
concerning the metric which will be proved in Section $2$.

  1. (1)

The curvature condition is equivalent to saying that $u$ satisfies the
curvature equation (2.5). Then

(1.5) $Q(z)=-\frac{1}{2}\left(u_{zz}-\frac{1}{2}u^{2}_{z}\right)$

is a meromorphic function.

  2. (2)

The invariant condition ensures that $Q$ is a meromorphic modular form of
weight $4$ with respect to $\mathrm{SL}(2,\mathbb{Z})$ and holomorphic at
$\infty$. Moreover, $Q(\infty)\leq 0$.

  3. (3)

The metric might have conic singularity at some $p\in\mathbb{H}$ with a conic
angle $\theta_{p}$, and the metric is smooth at $p$ if and only
$\theta_{p}=1$. Thus $Q$ has a pole at $p$ if and only $ds^{2}$ has a conic
singularity at $p$ (i.e., $\theta_{p}\neq 1$), provided that
$p\not\in\left\\{\gamma\cdot
i,\gamma\cdot\rho:\gamma\in\mathrm{SL}(2,\mathbb{Z})\right\\}$, where
$i=\sqrt{-1}$ and $\rho=(1+\sqrt{-3})/2$.

  4. (4)

Let $1/2\pm\kappa_{p}$, $\kappa_{p}>0$ be the local exponents at $p$ of (1.1)
with this $Q$. Then $\theta_{p}=2\kappa_{p}/e_{p}$, where $e_{p}$ is the
elliptic order of $p$. Moreover, if $\kappa_{p}\in\frac{1}{2}\mathbb{Z}$ for
any $p$, then (1.1) is automatically apparent.

We say the solution $u$ or the metric $e^{u}\left|dz\right|^{2}$ _realizes_
$Q$ or the associated ODE (1.1) is realized by $u$. We note that for a given
$Q$, finding a metric $e^{u}\left|dz\right|^{2}$ realizing $Q$ is equivalent
to solving the curvature equation (2.5) in Section 2 with the RHS being
$4\pi\sum n_{p}\delta_{p}$, where $n_{p}=2\kappa_{p}-1$, $\delta_{p}$ is the
Dirac measure at $p\in\mathbb{H}$ and the summation runs over all poles of $Q$
on $\mathbb{H}$. In particular, $\kappa_{p}\in\frac{1}{2}\mathbb{N}$, if and
only if the coefficient $n_{p}\in\mathbb{N}$, the set of positive integers.

In view of this connection, throughout the paper, we assume that the ODE (1.1)
satisfy the following conditions ($\mathbf{H}_{1}$) or ($\mathbf{H}_{2}$).

($\mathbf{H_{1})}$ The ODE (1.1) is apparent with the local exponents
$1/2\pm\kappa_{p}$ at any pole $p$ of $Q$,
$\kappa_{p}\in\frac{1}{2}\mathbb{N}$, and $Q(\infty)\leq 0$. Denote the local
exponents at $\infty$ by $\pm\kappa_{\infty}$. Moreover, if
$p\not\in\left\\{i,\rho\right\\}$, then $\kappa_{p}>1/2$.

Note that $Q(z)$ is smooth at $p$ if and only if $\kappa_{p}=1/2$, so the
requirement $\kappa_{p}>1/2$ means that that $Q(z)$ has a pole at $p$. Note
that by (4), the angle $\theta_{\rho}$ at $\rho$ is $2\kappa_{\rho}/3$ and
$\theta_{i}$ at $i$ is $\kappa_{i}$.

($\mathbf{H_{2}}$) The angles $\theta_{\rho}$ and $\theta_{i}$ are not
integers.

Suppose $\kappa_{\infty}\not\in\frac{1}{2}\mathbb{N}$. Then there is
$r_{\infty}\in(0,1/2)$ such that

(1.6) $\text{either}\ \kappa_{\infty}\equiv r_{\infty}\ \mathrm{mod}\
1\quad\text{or}\quad\kappa_{\infty}\equiv-r_{\infty}\ \mathrm{mod}\ 1.$

######  Theorem 1.1.

Suppose that (1.1) satisfies ($\mathbf{H}_{1}$), ($\mathbf{H}_{2}$), and
$\kappa_{\infty}\not\in\frac{1}{2}\mathbb{N}$. If $1/12<r_{\infty}<5/12$, then
there is an invariant metric of curvature $1/2$ realizing $Q$. Moreover, the
metric is unique. Conversely, if $Q$ is realized then $1/12\leq r_{\infty}\leq
5/12$.

Furthermore, assume that $r_{\infty}=1/12$ or $r_{\infty}=5/12$. Let $\chi$ be
the character of $\mathrm{SL}(2,\mathbb{Z})$ determined by

$\chi(T)=e^{2\pi i/6},\qquad\chi(S)=-1,$

where $T=\left(\begin{smallmatrix}1&1\\\ 0&1\end{smallmatrix}\right)$ and
$S=\left(\begin{smallmatrix}0&-1\\\ 1&0\end{smallmatrix}\right)$. Then there
is an invariant metric of curvature $1/2$ realizing $Q$ if and only there are
two solutions $y_{1}(z)$ and $y_{2}(z)$ of (1.1) such that $y_{1}(z)^{2}$ and
$y_{2}(z)^{2}$ are meromorphic modular forms of weight $-2$ with character
$\chi$ and $\overline{\chi}$, respectively, on $\mathrm{SL}(2,\mathbb{Z})$.

######  Remark.

Let $\mathbb{H}^{*}=\mathbb{H}\cup\mathbb{Q}\cup\left\\{\infty\right\\}$.
Since $\mathrm{SL}(2,\mathbb{Z})\backslash\mathbb{H}^{\ast}$ is conformally
diffeomorphic to the standard sphere $S^{2}$, Theorem 1.1 can be formulated in
terms of the existence of metrics on $S^{2}$ with prescribed singularities at
poles of $Q$ and prescribed angle $\theta_{p}$ at each singular point $p$. In
this sense, Theorem 1.1 is a special case of a result of Eremenko and Tarasov
[12]111We thank the referee for pointing out this and providing the
reference., quoted as Theorem A.1 in the appendix. In the appendix, we give an
alternative and self-contained proof of their result in the form of Theorem
A.3 as it is elementary and involves only straightforward matrix computation.
(In the notation of Theorem A.3, Theorem 1.1 corresponds to the case
$\theta_{1}=1/2$, $\theta_{2}=1/3$, and $\theta_{3}=2r_{\infty}$ or
$\theta_{3}=1-2r_{\infty}$, depending on whether $2r_{\infty}\leq 1/2$ or
$2r_{\infty}>1/2$.)

The threshold case $r_{\infty}\in\\{1/12,5/12\\}$ is more delicate. In Section
6, we provides examples of existence and nonexistence of an invariant metric
with $r_{\infty}=1/12$. Our examples suggest that to each $Q(z)$ with
$r_{\infty}\in\\{1/12,5/12\\}$, one may associate a meromorphic differential
$1$-form $\omega$ of the second kind on a certain elliptic curve $E$, and
whether there exists an invariant metric realizing $Q$ hinges on whether
$\omega$ is exact, i.e., whether $\omega$ is the identity element in the first
de Rham cohomology group of $E$. Also, in the nonexistence example, we find
that the entries in the monodromy matrices can be expressed in terms of
periods or the central value of the $L$-function of the elliptic curve
$y^{2}=x^{3}-1728$. We plan to study the threshold case in more details in the
future.

Motivated by Theorem 1.1, we consider the datas given below.

(1.7) $\begin{split}&\text{A set of positive half-integers }\
\kappa_{\rho},\kappa_{i},\kappa_{j}\in\mathbb{N}/2,j=1,2,\ldots,m,\\\
&\text{such\ that}\ 2\kappa_{\rho}/3\not\in\mathbb{N},\
\kappa_{i}\not\in\mathbb{N};\ \text{a set of inequivalent points}\
p_{j}\in\mathbb{H},\\\ &j=1,2,\ldots,m;\text{ and a positve number}\
\kappa_{\infty}.\end{split}$

######  Definition 1.2.

We say $Q$ is _equipped with ( 1.7)_ if

  1. (i)

$\left\\{\rho,i,z_{j}:1\leq j\leq m\right\\}$ are the set of poles of $Q$;

  2. (ii)

The local exponents of $Q$ at $\rho,i,z_{j}$ are $1/2\pm\kappa_{\rho}$,
$1/2\pm\kappa_{i}$ and $1/2\pm\kappa_{j}$, respectively;

  3. (iii)

$Q$ is apparent on $\mathbb{H}$; and

  4. (iv)

The local exponents at $\infty$ are $\pm\kappa_{\infty}$.

######  Theorem 1.3.

Given (1.7), there are modular forms $Q$ of weight $4$ equipped with (1.7).
Moreover, the number of such $Q$ is at most $\prod^{m}_{j=1}(2\kappa_{j})$.

To prove the theorem, we first show that there is a finite set of polynomials
such that the set of $Q(z)$ equipped with (1.7) is in a one-to-one
correspondence with the set of common zeros of the polynomial. Then the
theorem follows immediately from the clasical Bézout theorem. Note that
Eremenko and Tarasov [12, Theorem 2.4] has proved a stronger result, which in
our setting states that for generic singular points $z_{1},\ldots,z_{m}$, the
number of $Q(z)$ is precisely $\prod_{j=1}^{m}(2\kappa_{j})$.

If the local exponents at $\infty$ are $\pm n/4$, $n$ is odd, then our second
result asserts that there is a modular form of weight $-4$ coming from the
equation. In the following, we use $T=\left(\begin{smallmatrix}1&1\\\
0&1\end{smallmatrix}\right)$ and $S=\left(\begin{smallmatrix}0&-1\\\
1&0\end{smallmatrix}\right)$.

######  Theorem 1.4.

Suppose that ($\mathbf{H}_{1}$) and ($\mathbf{H}_{2}$) hold and
$\kappa_{\infty}=n/4$, $n$ a positive odd integer. Then there is a constant
$c\in\mathbb{C}$ such that $F(z):=y_{-}(z)^{2}+cy_{+}(z)^{2}$ satisfies

$\left(F\big{|}_{-2}T\right)(z)=\left(F\big{|}_{-2}S\right)(z)=-F(z),$

where

$y_{\pm}(z)=q^{\pm n/4}\left(1+\sum_{j\geq 1}c_{j}^{\pm}q^{j}\right)$

are solutions of (1.1).

The constant $c$ is rational if all coefficients of $Q(z)/\pi^{2}$ in the
$q$-expansion are rational. We conjecture $c$ is positive, but it is not
proved yet. Obviously, $F(z)^{2}$ is a modular form of weight $-4$ with
respect to $\mathrm{SL}(2,\mathbb{Z})$. Let $\Gamma_{2}$ be the group
generated by $T^{2}=\left(\begin{smallmatrix}1&2\\\
0&1\end{smallmatrix}\right)$ and $ST=\left(\begin{smallmatrix}0&-1\\\
1&1\end{smallmatrix}\right)$, which is an index $2$ subgroup of
$\mathrm{SL}(2,\mathbb{Z})$. Then $F$ is a modular form of weight $-2$ on
$\Gamma_{2}$. This fact can help us to compute $c$ and $F(z)$ explicitly. For
example, if $Q(z)=-\pi^{2}n^{2}E_{4}(z)/4$, then $F(z)$ is holomorphic on
$\mathbb{H}$, but with a pole of order $n$ at $\infty$ ($\Gamma_{2}$ has only
one cusp $\infty$ and two elliptic points of order $3$). Thus it is not
difficult to prove

######  Corollary 1.5.

Let $Q(z)=-\pi^{2}(n/2)^{2}E_{4}(z)$, where $n$ is a positive odd integer.
Then there is a polynomial $P_{n-1}(x)\in\mathbb{Q}[x]$ of degree $(n-1)/2$
such that

$F(z)=\frac{E_{4}(z)}{\Delta(z)^{1/2}}P_{n-1}(j(z)).$

Here $E_{4}$ and $E_{6}$ are the Eisenstein series of weight $4$ and $6$ on
$\mathrm{SL}(2,\mathbb{Z})$ respectively:

$\displaystyle
E_{4}(z)=1+240\sum_{m=1}^{\infty}\frac{m^{3}q^{m}}{1-q^{m}}=1+240\sum^{\infty}_{m=1}\left(\sum_{d|n}d^{3}\right)q^{n},\quad
q=e^{2\pi iz},$ $\displaystyle
E_{6}(z)=1-504\sum_{m=1}^{\infty}\frac{m^{5}q^{m}}{1-q^{m}}=1-504\sum^{\infty}_{m=1}\left(\sum_{d|n}d^{5}\right)q^{n},$

$\Delta(z)=(E_{4}(z)^{3}-E_{6}(z)^{2})/1728=q-24q^{2}+\cdots$, and
$j(z)=E_{4}(z)^{3}/\Delta(z)$.

For small $n$, $P_{n-1}$ are shown in the following list.

$\begin{array}[]{c|ll}\hline\cr\hline\cr n&F&P_{n-1}\\\ \hline\cr
1&y_{-}^{2}+3(2^{3}y_{+})^{2}&1\\\ 3&y_{-}^{2}+3(2^{12}y_{+})^{2}&j-1536\\\
5&y_{-}^{2}+3(2^{18}7^{1}y_{+})^{2}&j^{2}-2240j+1146880\\\
7&y_{-}^{2}+3(2^{28}3^{1}y_{+})^{2}&j^{3}-3072j^{2}+2752512j-704643072\\\
9&(7y_{-})^{2}+3(2^{34}11^{1}13^{1}y_{+})^{2}&\begin{split}49j^{4}&-192192j^{3}+253034496j^{2}-\\\
&-125954949120j+19346680184832\end{split}\\\ \hline\cr\hline\cr\end{array}$

In practice, it seems not easy to verify the apparentness at a singular point
with local exponents $1/2\pm\kappa$, $\kappa\in\frac{1}{2}\mathbb{N}$. Take a
simple example

$\left(q\frac{d}{dq}\right)^{2}y=-\frac{1}{4\pi^{2}}y^{\prime\prime}=\left(\frac{n}{2}\right)^{2}E_{4}(z)y\quad\text{on}\
\mathbb{H}.$

The local exponents at $\infty$ are $\pm n/2$. The standard method to verify
the apparentness at $\infty$ is to show that there is a solution $y_{-}(z)$
having a $q$-expansion

$y_{-}(z)=q^{-n/2}\left(1+\sum_{j\geq 1}c_{j}q^{j}\right).$

Suppose $E_{4}(z)=\sum^{\infty}_{j\geq 0}b_{j}q^{j}$. Substituting the
$q$-expansion of $y_{-}$ and $E_{4}$ into the equation, then the coefficient
$c_{j}$ satisfies

(1.8)
$\left(\left(j-\frac{n}{2}\right)^{2}-\left(\frac{n}{2}\right)^{2}\right)c_{j}=\left(\frac{n}{2}\right)^{2}\sum_{k+\ell=j,\
\ell<j}b_{k}c_{\ell}.$

For $j=1,2,\ldots,n-1$, $c_{j}$ can be determined from $c_{0}=1$. However at
$j=n$, the LHS of (1.8) vanishes. Therefore, $\infty$ is apparent if and only
the RHS of (1.8) is $0$ at $j=n$. If $n$ is small, then it is easy to check
that the RHS of (1.8) is not $0$ at $j=n$. For a general $n$, nevertheless, it
seems not easy to see why it does not vanish from the recursive relation
(1.8). Thus for a modular ODE, the standard method is not efficient for this
purpose. We need other ideas.

We consider

(1.9)
$y^{\prime\prime}(z)=\pi^{2}\left(rE_{4}(z)+s\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}+t\frac{E_{4}(z)^{4}}{E_{6}(z)^{2}}\right)y(z),$

where $r,s$ and $t$ are constant parameters. For simplicity, we denote the
potential of (1.9) by $Q_{3}(z;r,s,t)$ or $Q_{3}(z)$ for short. Modulo
$\mathrm{SL}(2,\mathbb{Z})$, (1.9) has singularities only at $\rho$ and $i$
(recall that $E_{4}(z_{0})=0$ if and only if $z_{0}$ is equivalent to $\rho$
under $\mathrm{SL}(2,\mathbb{Z})$ and $E_{6}(z_{0})=0$ if and only if $z_{0}$
is equivalent to $i$). Assume the local exponents of (1.9) are
$1/2\pm\kappa_{i}$ at $i=\sqrt{-1}$ and $1/2\pm\kappa_{\rho}$ at
$\rho=(1+\sqrt{-3})/2$. Then it is easy to prove that $s=s_{\kappa_{\rho}}$,
$t=t_{\kappa_{i}}$, where

(1.10)
$s_{\kappa_{\rho}}=\frac{1-4\kappa^{2}_{\rho}}{9},\qquad\text{and}\qquad
t_{\kappa_{i}}=\frac{1-4\kappa^{2}_{i}}{4}.$

See Section 3 for the computation.

At $\infty$, the local exponents are $\pm\kappa_{\infty}$ if and only if

$r+s_{\kappa_{\rho}}+t_{\kappa_{i}}=-(2\kappa_{\infty})^{2}.$

In the following, we set the triple $\left(n_{i},\ n_{\rho},\
n_{\infty}\right)$ by

$(n_{i},n_{\rho},n_{\infty})=\left(\kappa_{i},\frac{2\kappa_{\rho}}{3},2\kappa_{\infty}\right).$

######  Theorem 1.6.

The modular differential equation (1.9) is apparent throughout
$\mathbb{H}\cup\\{\text{cusps}\\}$ if and only if the triple
$(n_{i},n_{\rho},n_{\infty})$ are positive integers satisfying (i) the sum of
these three integers is odd, and (ii) the sum of any two of these integers is
greater than the third. Moreover, In such a case, the ratio of any two
solutions is a modular function on $\mathrm{SL}(2,\mathbb{Z})$.

For example, if

$Q(z)=\pi^{2}\left(\frac{23}{36}E_{4}(z)-\frac{9n^{2}-1}{9}\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}-\frac{3}{4}\frac{E_{4}(z)^{4}}{E_{6}(z)^{2}}\right),\quad
n\in\mathbb{N},$

then we have $(n_{i},n_{\rho},n_{\infty})=(1,n,n)$. By Theorem 1.6, (1.9) is
apparent throughout $\mathbb{H}\cup\\{\text{cusps}\\}$. On the other hand,
$\infty$ is not apparent for the ODE

$y^{\prime\prime}(z)=-\pi^{2}n^{2}E_{4}(z)y(z).$

As discussed in (1.8), it seems very difficult to verify ($\mathbf{H}_{1}$).
So we would like to present some examples to show how to verify the condition
($\mathbf{H}_{1}$). The first example is

(1.11)
$y^{\prime\prime}(z)=\pi^{2}\left(rE_{4}(z)+s\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}\right)y(z),$

where $r,s$ are constant parameters. For simplicity, we denote the potential
of (1.11) by $Q_{1}(z;r,s)$ or $Q_{1}(z)$ for short. The singular points
modulo $\mathrm{SL}(2,\mathbb{Z})$ is $\rho$ only. If the local exponents are
$1/2\pm\kappa_{\rho}$, then a simple calculation in Section 3 shows
$s=s_{\kappa_{\rho}}$, where $s_{\kappa_{\rho}}$ is given in (1.10).

######  Theorem 1.7.

Let $\kappa_{\rho}\in\frac{1}{2}\mathbb{N}$.

  1. (a)

Assume $3\nmid 2\kappa_{\rho}$. Then $Q_{1}(z;r,s)$ is apparent if
$s=s_{\kappa_{\rho}}$ and any $r\in\mathbb{C}$.

  2. (b)

Assume $3|2\kappa_{\rho}$. Then there exists a polynomial
$P(x)\in\mathbb{Q}[x]$ of degree $2\kappa_{\rho}/3$ such that $Q_{1}(z;r,s)$
with $s=s_{\kappa_{\rho}}$ is apparent if and only if $r$ is a root of $P(x)$.
Moreover, $r$ satisfies

(1.12)
$r+s_{\kappa_{\rho}}=-\left(\ell+\frac{1}{2}\right)^{2},\quad\text{where}\
\ell=0,1,2,\ldots,\frac{2\kappa_{\rho}}{3}-1.$

Next, we consider the ODE

(1.13)
$y^{\prime\prime}(z)=\pi^{2}\left(rE_{4}(z)+t\frac{E_{4}(z)^{4}}{E_{6}(z)^{2}}\right)y(z)\quad\text{on}\
\mathbb{H},$

where $r$ and $t$ are constant parameters. For simplicity, the potential of
(1.13) is denoted by $Q_{2}(z;r,t)$ or $Q_{2}(z)$ for short. Similar to
(1.11), (1.13) has local exponents $1/2\pm\kappa_{i}$ at $i$ if and only if
$t=t_{\kappa_{i}},$ where $t_{\kappa_{i}}$ is given in (1.10).

######  Theorem 1.8.

Let $\kappa_{i}\in\frac{1}{2}\mathbb{N}$.

  1. (a)

Assume $\kappa_{i}\in\frac{1}{2}+\mathbb{Z}_{\geq 0}$. Then (1.13) is apparent
if and only if $t=t_{\kappa_{i}}$ and any $r\in\mathbb{C}$.

  2. (b)

Assume $\kappa_{i}\in\mathbb{N}$. Then there exists a polynomial
$P(x)\in\mathbb{Q}[x]$ of degree $\kappa_{i}$ such that (1.13) with
$t=t_{\kappa_{i}}$ is apparent if and only if $r$ is a root of $P(x)$.
Moreover, $r$ satisfies

(1.14)
$r+t_{\kappa_{i}}=-\left(\ell\pm\frac{1}{3}\right)^{2},\quad\begin{cases}\ell=0,2,4,\ldots,\kappa_{i}-1,\quad&\text{if}\
\kappa_{i}\ \text{is\ odd},\\\
\ell=1,3,5,\ldots,\kappa_{i}-1,\qquad&\text{if}\ \kappa_{i}\ \text{is\
even}.\end{cases}$

We use the Frobenius method to prove Part (a) of Theorem 1.7 and Theorem 1.8.
However, due to the modularity, our expansion of functions are expanded in
terms of powers of $w_{\rho}:=(z-\rho)/(z-\bar{\rho})$ and
$w_{i}:=(z-i)/(z+i)$, not powers of $z-\rho$ and $z-i$ as the standard method
does. This kind of expansion has been used in [20] and [22]. We will see in
Section 3 that this type of expansions not only simplifies computations
greatly, but also obtains the degree of $P(x)$ in Theorem 1.7(b) and Theorem
1.8(b) precisely.

We will present two proofs of (1.12) in Theorem 1.7(b) and (1.14) in Theorem
1.8(b) in Section 4 and Section 5. One is to apply Riemann’s existence theorem
on compact Riemann surfaces. The other is to apply the existence theorems of
the invariant metrics with curvature $1/2$. This geometric theorems are
obtained by Eremenko [10, 11]. Hopefully, these methods are useful for
treating this kind of problems in modular differential equations.

The paper is organized as follows. In Section 2, we will discuss the
connection between the invariant metric $ds^{2}=e^{u}\left|dz\right|^{2}$ of
curvature $1/2$ and modular ODEs, in particular, the relation among the
behavior of $u$ near cusps, angles and the local exponents of the realized
modular ODE by $u$. In Section 3, we will explain the expansion of modular
forms in terms of the natural coordinate $w=(z-z_{0})/(z-\bar{z}_{0})$, and
prove Theorem 1.7(a) and Theorem 1.8(a). Both Theorem 1.7(b) and Theorem
1.8(b) are proved in Section 4, and Theorem 1.6 is proved in Section 5.
Finally, we will prove Theorem 1.1 and Theorem 1.4 to complete the paper in
Section 6 and Section 7 respectively.

##  2\. Curvature equations and the modular ODEs

### 2.1.

Let $M$ be a compact Riemann surface, $p\in M$, and $z$ be a complex
coordinate in an open neighborhood $U$ of $p$ with $z(p)=0$. We consider the
following curvature equation:

(2.1) $4u_{z\bar{z}}+e^{u}=f\quad\text{on}\ U,$

where $f=4\pi\sum\alpha_{i}\delta_{p_{i}}$ is a sum of Dirac measures and
$0\neq\alpha_{i}>-1$. The assumption $\alpha_{i}>-1$ ensures that $e^{u}$ is
locally integrable in a neighborhood of $p_{i}$. The $L^{1}$-integrability
implies

(2.2) $u(z)=2\alpha_{i}\log\left|z-p_{i}\right|+O(1)\quad\text{near}\ p_{i}.$

This is a general result from the elliptic PDE theory, see [4, 5].

Let $w=w(z)$ be a coordinate change and set

(2.3) $\hat{u}(w)=u(z)-2\log\left|{\frac{dw}{dz}}\right|.$

Then $\hat{u}(w)$ also satisfies

$4\hat{u}_{w\bar{w}}+e^{\hat{u}}=\hat{f},\qquad
f=4\pi\sum\alpha_{i}\delta_{\hat{p}_{i}},$

where $\hat{p}_{i}=w(p_{i})$. In other words, $e^{u}\left|dz\right|^{2}$ is
invariant under a coordinate change. Since $u$ has singularities at $p_{i}$,
the metric $ds^{2}=e^{u}\left|dz\right|^{2}$ has a conic singularity at
$p_{i}$. If $u$ is a solution of (2.1), then the metric
$ds^{2}=e^{u}\left|dz\right|^{2}$ has curvature $1/2$ at any point
$p\not\in\left\\{p_{i}\right\\}$. Suppose that $M$ is covered by
$\left\\{U_{i}\right\\}$ and $z_{i}$ is a coordinate in $U_{i}$. We call the
collection $\left\\{u_{i}\right\\}$ to be a solution of (2.1) on $M$ if
$u_{i}$ is a solution of (2.1) on $U_{i}$ for each $i$ and satisfy the
transformation law $u_{j}=u_{i}-2\log\left|\frac{dz_{j}}{dz_{i}}\right|$ on
$U_{i}\cap U_{j}$.

Let $g$ be a metric of $M$ with the curvature $K$, and the equation (2.1) on
$M$ is equivalent to the curvature equation:

(2.4)
$\Delta_{g}\hat{u}+e^{\hat{u}}-K=4\pi\sum{\alpha}_{i}\delta_{p_{i}}\quad\text{on}\
M,$

where $\Delta_{g}$ is the Beltrami-Laplace operator of $(M,g)$. We could
normalize the metric $g$ such that the area of $M$ is equal to $1$. In the
case when $g$ has a constant curvature, (2.4) can be written as

$\Delta_{g}\hat{u}+\rho\left(\frac{e^{\hat{u}}}{\int
e^{\hat{u}}}-1\right)=4\pi\sum{\alpha}_{i}(\delta_{p_{i}}-1)\quad\text{on}\
M.$

This nonlinear PDE is often call _a mean field equation_ in analysis. See [3,
5, 4, 8, 7, 6] and [15, 16, 17] for the recent development of mean field
equations.

In this paper, we consider the compact Riemann surface that is the quotient of
$\mathbb{H}^{*}:=\mathbb{H}\cup\mathbb{Q}\cup\left\\{\infty\right\\}$ by a
finite index subgroup $\Gamma$ of $\mathrm{SL}(2,\mathbb{Z})$, and the
equation (2.1) is defined on the upper half space $\mathbb{H}$:

(2.5) $4u_{z\bar{z}}+e^{u}=4\pi\sum\alpha_{i}\delta_{p_{i}}\quad\text{on}\
\mathbb{H},$

where the RHS is invariant under the action of $\Gamma$, i.e., the set
$\left\\{p_{i}\right\\}$ is invariant under the action of $\Gamma$ and
$\alpha_{i}=\alpha_{j}$ if $p_{i}=\gamma\cdot p_{j}$ for some
$\gamma\in\Gamma$. The transformation law (2.3) for coordinate change is
equivalent to asking $u$ to satisfy

(2.6) $u(\gamma
z)=u(z)+4\log\left|cz+d\right|,\quad\forall\gamma=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in\Gamma.$

Let $s$ be a cusp of $\Gamma$ and $\gamma\in\mathrm{SL}(2,\mathbb{Z})$ be a
matrix such that $\gamma\cdot\infty=s$. Then we define $u_{\gamma}$ by

$u_{\gamma}(z):=u(\gamma\cdot z)-4\log\left|cz+d\right|.$

Thus, $u$ is required to satisfy the following behavior near $s$: there is
$\alpha_{s}>0$ such that

(2.7) $e^{u_{\gamma}(z)}=\left|q_{N}\right|^{4\alpha_{s}}(c+o(1)),\quad
q_{N}=e^{2\pi iz/N},\ c>0,$

where $N$ is the width of the cusp $s$ and $o(1)\rightarrow 0$ as
$q_{N}\rightarrow 0$. Given the RHS of (2.5) and a positive $\alpha_{s}$ at
the cusp $s$, we ask for a solution $u$ of (2.5) satisfying (2.6) and (2.7) at
any cusp.

The conic angle $\theta$, defined at a singularity $p_{i}$ or a cusp $s$, is
an important geometric quantity. Suppose that a metric $ds^{2}$, conformal to
the flat metric $\left|dz\right|^{2}$, has a conic singularity at $p$, and $w$
is a coordinate near $p$ with $w(p)=0$. If

(2.8) $ds^{2}=\left|w\right|^{2(\theta-1)}(c+o(1))\left|dw\right|^{2},\quad
c>0,$

then we call $\theta$ the _angle_ at $p$, and $2\pi\theta$ the _total angle_
at $p$. Since $ds^{2}$ is required to have a finite area, the angle $\theta$
is always _positive_. Note that $ds^{2}$ is smooth (as a metric) at $p$ if and
only if $\theta=1$.

Next, we want to calculate the angles of $ds^{2}=e^{u}\left|dz\right|^{2}$, if
$u$ is a solution of (2.5). Note that $z$ is not a coordinate of $M$ if
$p_{i}$ is an elliptic point of order $e_{i}>1$. Indeed, $w=(z-p_{i})^{e_{i}}$
is the local coordinate near $p_{i}$. For simplicity, we denote $z-p_{i}$ by
$z$ ($z(p_{i})=0$). By (2.2), we have
$u(z)=2\alpha_{i}\log\left|z\right|+O(1)$, i.e.,
$e^{u(z)}=\left|z\right|^{2\alpha_{i}}(c_{0}+o(1))$, $c_{0}>0$. Then

$e^{u(z)}\left|dz\right|^{2}=\left|w\right|^{(2\alpha_{i}+2)/e_{i}-2}(d+o(1))\left|dw\right|^{2},\quad
d>0.$

By (2.8), we have

(2.9) $\theta_{i}=\frac{\alpha_{i}+1}{e_{i}}.$

At a cusp $s$, the coordinate is $q_{N}=e^{2\pi iz/N}$, where $N$ is the width
of the cusp $s$. By (2.7),

$e^{u_{\gamma}(z)}\left|dz\right|^{2}=\left|q_{N}\right|^{4\alpha_{s}-2}(c+o(1))\left|dq_{N}\right|^{2},\quad
c>0.$

So the angle $\theta_{s}$ at $s$ is

(2.10) $\theta_{s}=2\alpha_{s}.$

### 2.2. Integrability and modular differential equations

Equation (2.5) is also known as an integrable system. There are two important
features related to the integrability. One is that

(2.11)
$Q(z):=-\frac{1}{2}\left(u_{zz}-\frac{1}{2}u^{2}_{z}\right)\quad\text{is\ a\
meromorphic\ function},$

because $Q(z)_{\bar{z}}=-\frac{1}{2}(u_{z\bar{z}z}-u_{z\bar{z}}u_{z})=0$ by
(2.5).

######  Lemma 2.1.

Each $p_{i}$ is a double pole of $Q(z)$ with the expansion
$\frac{\alpha_{i}}{2}\left(\frac{\alpha_{i}}{2}+1\right)(z-p_{i})^{-2}+O\left((z-p_{i})^{-1}\right)$.

###### Proof.

Since $u(z)=2\alpha_{i}\log\left|z-p_{i}\right|+O(1)$ near $p_{i}$, we have
$u_{z}(z)=\alpha_{i}(z-p_{i})^{-1}+O(1)$ and
$u_{zz}(z)=-\alpha_{i}(z-p_{i})^{-2}+O\left((z-p_{i})^{-1}\right)$. Then the
lemma follows immediately. ∎

On the other hand, the Liouville theorem asserts that locally any solution $u$
can be expressed as

(2.12)
$u(z)=\log\frac{8\left|h^{\prime}(z)\right|^{2}}{\left(1+\left|h(z)\right|^{2}\right)^{2}},$

where $h(z)$ is a meromorphic function. Recall the Schwarz derivative

(2.13)
$\left\\{h,z\right\\}=\left(\frac{h^{\prime\prime}}{h^{\prime}}\right)^{\prime}-\frac{1}{2}\left(\frac{h^{\prime\prime}}{h^{\prime}}\right)^{2}.$

Note that the Schwarz derivative can be used to recover $h$ from $u$. Indeed,
a direct computation from (2.12) yields that

(2.14) $\left\\{h,z\right\\}=-2Q(z).$

See [3, 15, 16, 17] for the detail of the proofs (2.12)–(2.14). The
meromorphic function $h$ is called a _developing map_ for the solution $u$.
Any two developing maps $h_{i}$, $i=1,2$, of $u$ have the same Schwarz
derivative by (2.14), thus they can be connected by a Möbius transformation,

(2.15) $h_{2}(z)=\frac{ah_{1}(z)+b}{ch_{1}(z)+d},\quad\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in\mathrm{SL}(2,\mathbb{C}).$

By (2.12), we obtain

(2.16)
$\frac{\left|h^{\prime}_{1}(z)\right|^{2}}{\left(1+\left|h_{1}(z)\right|^{2}\right)^{2}}=\frac{\left|h_{2}^{\prime}(z)\right|^{2}}{\left(1+\left|h_{2}(z)\right|^{2}\right)^{2}},$

which implies that the matrix $\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)$ is an unitary matrix.

Next, we recall the classical Hermite theorem, see [21].

######  Theorem A.

Let $y_{i}$, $i=1,2$, be two independent solutions of

$y^{\prime\prime}=Q(z)y.$

Then the ratio $h(z)=y_{2}(z)/y_{1}(z)$ satisfies
$\left\\{h,z\right\\}=-2Q(z)$.

Let $Q(z)$ be the meromorphic function (2.11) obtained from the solution $u$.
Consider the ODE

(2.17) $y^{\prime\prime}=Q(z)y.$

Then (2.11) and the Hermite theorem together imply that $h(z)$ is a ratio of
two solutions of (2.16).

######  Theorem 2.2.

Suppose $u$ is a solution of (2.5). Then (2.17) satisfies ($\mathbf{H}_{1}$)
and the followings hold.

  1. (a)

The function $Q(z)$ is a meromorphic modular form of weight $4$ with respect
to $\Gamma$ and holomorphic at any cusp. Moreover, at a cusp $s$, $Q(s)<0$.

  2. (b)

(2.17) is Fuchsian and the local exponents of (2.17) at $p_{i}$ are
$-\alpha_{i}/2$, $\alpha_{i}/2+1$, and $\pm\alpha_{s}$ at a cusp.

  3. (c)

If $\alpha_{i}\in\mathbb{N}$ for all $i$, then (2.17) is apparent.

###### Proof.

(a) By the chain rule, we have

$\displaystyle(u\kern 2.15277pt\vbox{\hbox{$\scriptstyle\circ$}}\kern
2.15277pt\gamma)_{z}(z)$ $\displaystyle=u_{z}(\gamma\cdot z)(cz+d)^{-2},$
$\displaystyle(u\kern 2.15277pt\vbox{\hbox{$\scriptstyle\circ$}}\kern
2.15277pt\gamma)_{zz}(z)$ $\displaystyle=u_{zz}(\gamma\cdot
z)(cz+d)^{-4}-u_{z}(\gamma\cdot z)\frac{2c}{(cz+d)^{3}}.$

Thus

$\displaystyle(u\kern 2.15277pt\vbox{\hbox{$\scriptstyle\circ$}}\kern
2.15277pt\gamma)_{zz}-\frac{1}{2}(u\kern
2.15277pt\vbox{\hbox{$\scriptstyle\circ$}}\kern 2.15277pt\gamma)^{2}_{z}$
$\displaystyle=\left(u_{zz}(\gamma\cdot z)-\frac{1}{2}u^{2}_{z}(\gamma\cdot
z)\right)$ $\displaystyle\qquad\times(cz+d)^{-4}-u_{z}(\gamma\cdot
z)\cdot\frac{2c}{(cz+d)^{3}}.$

On the other hand, the transformation law (2.6) yields

$\displaystyle(u\kern 2.15277pt\vbox{\hbox{$\scriptstyle\circ$}}\kern
2.15277pt\gamma)_{z}(z)$ $\displaystyle=u_{z}(z)+\frac{2c}{(cz+d)},$
$\displaystyle(u\kern 2.15277pt\vbox{\hbox{$\scriptstyle\circ$}}\kern
2.15277pt\gamma)_{zz}$ $\displaystyle=u_{zz}-\frac{2c^{2}}{(cz+d)^{2}}.$

Hence, we have

$\displaystyle(u\kern 2.15277pt\vbox{\hbox{$\scriptstyle\circ$}}\kern
2.15277pt\gamma)_{zz}-\frac{1}{2}(u\kern
2.15277pt\vbox{\hbox{$\scriptstyle\circ$}}\kern 2.15277pt\gamma)^{2}_{z}$
$\displaystyle=\left(u_{zz}-\frac{1}{2}u^{2}_{z}\right)-u_{z}(z)\cdot\frac{2c}{(cz+d)}-\frac{4c^{2}}{(cz+d)^{2}}$
$\displaystyle=\left(u_{zz}-\frac{1}{2}u^{2}_{z}\right)-\frac{2c}{(cz+d)}(u\kern
2.15277pt\vbox{\hbox{$\scriptstyle\circ$}}\kern 2.15277pt\gamma)_{z}(z).$

Since

$\frac{-2c}{(cz+d)}(u\kern 2.15277pt\vbox{\hbox{$\scriptstyle\circ$}}\kern
2.15277pt\gamma)_{z}=\frac{-2c}{(cz+d)^{3}}u_{z}(\gamma\cdot z),$

we find that $Q:=-\frac{1}{2}\left(u_{zz}-\frac{1}{2}u^{2}_{z}\right)$
satisfies

$Q(\gamma\cdot z)=Q(z)\cdot(cz+d)^{4}.$

This proves the modularity of $Q$.

To prove the holomorphy of $Q$ at cusps, without loss of generality, we may
assume that the cusp $s$ is $\infty$. Then $q_{N}=e^{2\pi iz/N}$ is the local
coordinate near $\infty$, where $N$ is the width of the cusp $\infty$. By the
transformation law of coordinate changes, the solution $\hat{u}$ in terms of
$q_{N}$ should be expressed by
$\hat{u}(q_{N})=u(z)-2\log\left|\frac{dq_{N}}{dz}\right|$. Thus,

$e^{\hat{u}(q_{N})}=\frac{8\left|h^{\prime}(z)\right|^{2}}{\left(1+\left|h(z)\right|^{2}\right)^{2}}\left|\frac{dq_{N}}{dz}\right|^{2}=8\left|\frac{d}{dq_{N}}h(z)\right|^{2}\left(1+\left|h(z)\right|^{2}\right)^{-2}.$

Hence the developing map $h(z)=\hat{h}(e^{2\pi iz/N})=\hat{h}(q_{N})$, where
$q_{N}=e^{2\pi iz/N}$. Note that

$\displaystyle\left\\{h,z\right\\}$
$\displaystyle=\left\\{\hat{h},q_{N}\right\\}\left(\frac{dq_{N}}{dz}\right)^{2}+\left\\{q_{N},z\right\\}$
$\displaystyle=\left\\{\hat{h},q_{N}\right\\}q^{2}_{N}\left(\frac{-4\pi^{2}}{N^{2}}\right)+\frac{2\pi^{2}}{N^{2}}.$

Since

$-\frac{1}{2}\left\\{\hat{h},q_{N}\right\\}=\hat{u}_{q_{N}q_{N}}-\frac{1}{2}\hat{u}_{q_{N}}^{2}=\frac{\alpha}{2}\left(\frac{\alpha}{2}+1\right)q^{-2}_{N}+O\left(q^{-1}_{N}\right),$

where $\alpha=\theta-1$, $\theta$ is the angle at $\infty$, we have

$\begin{split}\lim_{\mathrm{Im\,}z\rightarrow\infty}Q(z)&=-\frac{\pi^{2}}{N^{2}}\left(1+\frac{4\alpha}{2}\left(\frac{\alpha}{2}+1\right)\right)=-\frac{\pi^{2}}{N^{2}}(1+\alpha)^{2}<0,\end{split}$

because $\alpha>-1$. This proves Part (a).

Part (b) is a consequence of Lemma 2.1.

For Part (c), if $\alpha_{i}\in\mathbb{N}$ then the local exponents
$-\alpha_{i}/2$ and $\alpha_{i}/2+1$ can be written as $1/2\pm\kappa_{i}$,
$\kappa_{i}=(\alpha_{i}+1)/2\in\frac{1}{2}\mathbb{N}$ and by the Liouville
theorem (2.12), we see easily that $h(z)$ can not have a logarithmic
singularity at $p_{i}$. The fact that $h(z)$ is a ratio of two solutions of
(2.17) implies any solution of (2.17) has no logarithmic singularity. This
proves Part (c). ∎

Together with the Liouville theorem, we have

######  Proposition 2.3.

Suppose $Q$ is a meromorphic modular form of weight $4$ on
$\mathrm{SL}(2,\mathbb{Z})$. If there are two independent solutions $y_{1}$
and $y_{2}$ of (2.17) such that $h(z)=y_{2}(z)/y_{1}(z)$ satisfies $h(\gamma
z)=\frac{ah(z)+b}{ch(z)+d}$ for some unitary matrix
$\left(\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix}\right)$ depending on
$\gamma$, for any $\gamma\in\mathrm{SL}(2,\mathbb{Z})$, then $Q$ can be
realized.

###### Proof.

Let
$u(z)=\log\frac{8\left|h^{\prime}(z)\right|^{2}}{\left(1+\left|h(z)^{2}\right|\right)^{2}}$.
Since $h(z)$ is unitary, $u(z)$ is well-defined on $\mathbb{H}$ and satisfies
(2.6). Further, the Liouville theorem says that $u(z)$ satisfies (2.5). ∎

### 2.3. Examples

In this subsection, we will give some examples to indicate how to determine
$Q$ provided that the RHS of (2.5) is known and $\alpha_{\infty}$ is given at
$\infty$. Here, $\Gamma=\mathrm{SL}(2,\mathbb{Z})$.

Example 1. Assume that the RHS of (2.5) is equal to $0$. Then
$Q:=-\frac{1}{2}\left(u_{zz}-\frac{1}{2}u^{2}_{z}\right)$ is a holomorphic
modular form of weight $4$. Thus,

(2.18) $Q(z)=\pi^{2}rE_{4}(z).$

Since $\pm\alpha_{\infty}$ are the local exponents of (1.1) at $\infty$, we
have $r=-4\alpha^{2}_{\infty}$. Thus, $Q$ is uniquely determined. Note that at
$\infty$, the angle $\theta_{\infty}$ is equal to $2\alpha_{\infty}$.

Example 2. Assume that the RHS of (2.5) is $4\pi n\sum\delta_{p}$, where the
summation is over $\gamma\cdot\rho$ for every $\gamma\in\Gamma$. Then $Q$ is a
meromorphic modular form of weight $4$ whose poles occur at $\gamma\cdot\rho$
and the order is $2$. Thus, $E_{4}(z)^{2}Q(z)$ is holomorphic a modular form
of wright $12$, and then

$Q(z)=\pi^{2}\left(rE_{4}(z)+s\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}\right),$

where we recall that the graded ring of modular forms on
$\mathrm{SL}(2,\mathbb{Z})$ is generated by $E_{4}(z)$ and $E_{6}(z)$. By
Theorem 2.2, the local exponents at $\rho$ are $-n/2$ and $n/2+1$, which
implies $\kappa_{\rho}=(n+1)/2$, $s=(1-4\kappa^{2}_{\rho})/9$, and
$-(r+s)/4=\alpha^{2}_{\infty}$. Thus $Q$ is uniquely determined. Moreover, the
angles $\theta_{j}$ in this example are $\theta_{i}=1/2$,
$\theta_{\rho}=(n+1)/3$ and $\theta_{\infty}=2\alpha_{\infty}$.

Example 3. Assume that the RHS of (2.5) is equal to $4\pi n\sum\delta_{p}$,
where the summations is over $\gamma\cdot i$ for any $\gamma\in\Gamma$.
Reasoning as Example 2, we have

(2.19)
$Q(z)=\pi^{2}\left(rE_{4}(z)+t\frac{E_{4}(z)^{4}}{E_{6}(z)^{2}}\right).$

By Theorem 2.2, we have

$\kappa_{i}=\frac{n+1}{2},\quad
t=\frac{1-4\kappa^{2}_{i}}{4},\quad\text{and}\quad r+t=-4\alpha^{2}_{\infty}.$

Thus $Q$ is uniquely determined. Moreover, we have $\theta_{i}=(n+1)/2$,
$\theta_{\rho}=1/3$, and $\theta_{\infty}=2\alpha_{\infty}$.

Example 4. Assume the RHS of (2.5) is
$4\pi\left(n\sum_{p_{1}}\delta_{p_{1}}+m\sum_{p_{2}}\delta_{p_{2}}\right)$,
where $p_{1},p_{2}$ run over zeros of $E_{4}(z)$ and $E_{6}(z)$, respectively.
Then

(2.20)
$Q(z)=\pi^{2}\left(rE_{4}(z)+s\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}+t\frac{E_{4}(z)^{4}}{E_{6}(z)^{2}}\right).$

The conditions on the local exponents at $\rho$, $i$ and $\infty$ yield that

$\displaystyle
s=\frac{1-4\kappa_{\rho}^{2}}{9},\quad\kappa_{\rho}=\frac{n+1}{2};\quad
t=\frac{1-4\kappa^{2}_{i}}{4},\quad\kappa_{i}=\frac{m+1}{2};$ $\displaystyle
r+s+t=-4\alpha^{2}_{\infty}.$

Then $Q$ is uniquely determined. Moreover, $\theta_{1}=(m+1)/2$,
$\theta_{2}=(n+1)/3$ and $\theta_{\infty}=2\alpha_{\infty}$.

### 2.4 Eremenko’s theorem

A. Eremenko [10, 11] gave a necessary and sufficient conditions of the angles
$\theta_{i}$, $1\leq i\leq 3$, at the three singular points $i,\rho,\infty$
for the existence of $u$ of (2.5)-(2.7).

When one of angles is an integer, the following conditions are required.

(A) If only one (say $\theta_{1}$) of angles is an integer, then either
$\theta_{2}+\theta_{3}$ or $\left|\theta_{2}-\theta_{3}\right|$ is an integer
$m$ of opposite parity to $\theta_{1}$ with $m\leq\theta_{1}-1$. If all the
angles are integers, then (1) $\theta_{1}+\theta_{2}+\theta_{3}$ is odd, and
(2) $\theta_{i}<\theta_{j}+\theta_{k}$ for $i\neq j\neq k$.

Eremenko’s theorem. If one of $\theta_{j}$ is an integer, then a necessary and
sufficient condition for the existence of a conformal metric of positive
constant curvature on the sphere with three conic singularities of angles
$\theta_{1}$, $\theta_{2}$, $\theta_{3}$ ($\theta_{j}\neq 1$, $1\leq j\leq
3$), is that $\left\\{\theta_{1},\theta_{2},\theta_{3}\right\\}$ satisfies
(A). Moreover, if (A) holds and there is only one integral angle, then the
metric is unique.

##  3\. Expansions of Eisenstein series at $\rho$ and $i$

The $q$-expansion of a modular form $f(z)$, i.e., the expansion of $f(z)$ with
respect to the local parameter $q$ at the cusp $\infty$, is frequently studied
and is of great significance in many problems in number theory. Here we shall
review properties of series expansions of modular forms at a point
$z_{0}\in\mathbb{H}$, other than the cusp $\infty$.

######  Definition 3.1.

Let $\Gamma$ be a Fuchsian subgroup of the first kind of
$\mathrm{SL}(2,\mathbb{R})$. Let $f(z)$ be a meromorphic modular form of
weight $k$ on $\Gamma$. Given $z_{0}\in\mathbb{H}$, let

$w=w_{z_{0}}(z)=\frac{z-z_{0}}{z-\overline{z}_{0}}.$

The expansion of the form

(3.1) $f(z)=(1-w)^{k}\sum_{n\geq n_{0}}\frac{b_{n}}{n!}w^{n}$

is called the _power series expansion_ of $f$ at $z_{0}$.

One advantage of this expansion is that its coefficients $b_{n}$ have a simple
expression in terms of the Shimura-Maass derivatives of $f$. To state the
result, we recall that if $f:\mathbb{H}\to\mathbb{C}$ is said to be _nearly
holomorphic_ if it is of the form

$f(z)=\sum_{d=0}^{n}\frac{f_{d}(z)}{(z-\overline{z})^{d}}$

for some holomorphic functions $f_{d}$. If $k$ is an integer and
$f:\mathbb{H}\to\mathbb{C}$ is a nearly holomorphic function such that

$f\left(\frac{az+b}{cz+d}\right)=(cz+d)^{k}f(z)$

for all $\left(\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix}\right)\in\Gamma$
and each $f_{d}$ is holomorphic at every cusp, then we say $f$ is a _nearly
holomorphic modular form_ of weight $k$ on $\Gamma$.

For a nearly holomorphic function $f:\mathbb{H}\to\mathbb{C}$, we define its
_Shimura-Maass_ derivative of weight $k$ by

$(\partial_{k}f)(z):=\frac{1}{2\pi
i}\left(f^{\prime}(z)+\frac{kf(z)}{z-\overline{z}}\right).$

We have the following important properties of Shimura-Maass derivatives.

######  Lemma 3.2 ([20, Equations (1.5) and (1.8)]).

For any nearly holomorphic functions $f,g:\mathbb{H}\to\mathbb{C}$, any
integers $k$ and $\ell$, and any $\gamma\in\mathrm{GL}^{+}(2,\mathbb{R})$, we
have

$\partial_{k+\ell}(fg)=(\partial_{k}f)g+f(\partial_{\ell}g)$

and

$\partial_{k}\left(f\big{|}_{k}\gamma\right)=(\partial_{k}f)\big{|}_{k+2}\gamma.$

######  Remark.

The second property in the lemma implies that if $f$ is a nearly holomorphic
modular form of weight $k$ on $\Gamma$, then $\partial_{k}f$ is a nearly
holomorphic form of weight $k+2$ on $\Gamma$.

Set also

$\partial_{k}^{n}f=\partial_{k+2n-2}\ldots\partial_{k}f.$

Then the coefficients $b_{n}$ in (3.1) has the following expression.

######  Proposition 3.3 ([22, Proposition 17]).

If $f(z)$ is a holomorphic modular form of weight $k$ on $\Gamma$, then the
coefficients $b_{n}$ in (3.1) are

$b_{n}=(\partial^{n}_{k}f)(z_{0})(-4\pi\,\mathrm{Im\,}z_{0})^{n}$

for $n\geq 0$. That is, we have

$f(z)=(1-w)^{k}\sum_{n=0}^{\infty}\frac{(\partial^{n}_{k}f)(z_{0})(-4\pi\,\mathrm{Im\,}z_{0})^{n}}{n!}w^{n}.$

Note that there is a misprint in Proposition 17 [22]. The proof of the
proposition shows that
$b_{n}=(\partial^{n}f)(z_{0})(-4\pi\,\mathrm{Im\,}z_{0})^{n}$, but the
statement misses the minus sign.

We will use these properties of power series expansions of modular forms to
show that the apparentness of (1.1) at a point $z_{0}$ will imply the
apparentness at $\gamma z_{0}$ for all $\gamma\in\mathrm{SL}(2,\mathbb{Z})$.
We first prove two lemmas. The first lemma relates the power series expansion
of a meromorphic modular form at $z_{0}$ to that at $\gamma z_{0}$.

######  Lemma 3.4.

Assume that $f$ is a meromorphic modular form of weight $k$ on
$\mathrm{SL}(2,\mathbb{Z})$. Assume that the power series expansion of $f$ at
$z_{0}\in\mathbb{H}$ is

$f(z)=(1-w)^{k}\sum_{n=n_{0}}^{\infty}a_{n}w^{n},\qquad
w=w_{z_{0}}(z)=\frac{z-z_{0}}{z-\overline{z}_{0}}.$

For $\gamma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\mathrm{SL}(2,\mathbb{Z})$, let
$\widetilde{w}=w_{\gamma z_{0}}(z)=(z-\gamma
z_{0})/(z-\gamma\overline{z}_{0})$. Then the power series expansion of $f$ at
$\widetilde{z}_{0}$ is

$(cz_{0}+d)^{k}(1-\widetilde{w})^{k}\sum_{n=n_{0}}^{\infty}a_{n}\left(\frac{cz_{0}+d}{c\overline{z}_{0}+d}\right)^{n}\widetilde{w}^{n}.$

###### Proof.

Since every meromorphic modular form on $\mathrm{SL}(2,\mathbb{Z})$ can be
written as the quotient of two holomorphic modular forms on
$\mathrm{SL}(2,\mathbb{Z})$, it suffices to prove the lemma under the
assumption that $f$ is a holomorphic modular form.

According to Proposition 3.3, the power series expansions of $f$ at $z_{0}$
and at $\gamma z_{0}$ are

$(1-w)^{k}\sum_{n=0}^{\infty}\frac{(\partial^{n}_{k}f)(z_{0})(-4\pi\,\mathrm{Im\,}z_{0})^{n}}{n!}w^{n}$

are

$(1-\widetilde{w})^{k}\sum_{n=0}^{\infty}\frac{(\partial^{n}_{k}f)(\gamma
z_{0})(-4\pi\,\mathrm{Im\,}\gamma z_{0})^{n}}{n!}\widetilde{w}^{n},$

respectively. Since $\partial^{n}f(z)$ is modular of weight $k+2n$ (see the
remark following Lemma 3.2), we have

$(\partial^{n}f)(\gamma z_{0})=(cz_{0}+d)^{k+2n}(\partial^{n}f)(z_{0}).$

Also,

(3.2) $\mathrm{Im\,}\gamma z_{0}=\frac{\mathrm{Im\,}z_{0}}{|cz_{0}+d|^{2}}.$

Thus, if the power series expansion of $f$ at $z_{0}$ is

$(1-w)^{k}\sum_{n=0}^{\infty}\frac{b_{n}}{n!}w^{n},$

then that of $f$ at $\gamma z_{0}$ is

$\begin{split}&(1-\widetilde{w})^{k}\sum_{n=0}^{\infty}\frac{b_{n}}{n!}\frac{(cz_{0}+d)^{k+2n}}{|cz_{0}+d|^{2n}}\widetilde{w}^{n}\\\
&\qquad=(cz_{0}+d)^{k}(1-\widetilde{w})^{k}\sum_{n=0}^{\infty}\frac{b_{n}}{n!}\left(\frac{cz_{0}+d}{c\overline{z}_{0}+d}\right)^{n}\widetilde{w}^{n}.\end{split}$

This proves the lemma. ∎

The next lemma expresses $y^{\prime\prime}(z)$ in terms of $w$.

######  Lemma 3.5.

Let $z_{0}\in\mathbb{H}$ and set
$w=w_{z_{0}}(z)=(z-z_{0})/(z-\overline{z}_{0})$. If

$y(z)=\frac{1}{1-w}\sum_{n=0}^{\infty}a_{n}w^{\alpha+n}$

for some real number $\alpha$, then

$\frac{d^{2}}{dz^{2}}y(z)=\frac{(1-w)^{3}}{(z_{0}-\overline{z}_{0})^{2}}\sum_{n=0}^{\infty}a_{n}(\alpha+n)(\alpha+n-1)w^{\alpha+n-2}.$

###### Proof.

We first note that

$1-w=\frac{z_{0}-\overline{z}_{0}}{z-z_{0}}$

and hence

(3.3)
$\frac{dw}{dz}=\frac{z_{0}-\overline{z}_{0}}{(z-z_{0})^{2}}=\frac{(1-w)^{2}}{z_{0}-\overline{z}_{0}},\quad\frac{d^{2}w}{dz^{2}}=-2\frac{z_{0}-\overline{z}_{0}}{(z-z_{0})^{3}}=-\frac{2(1-w)^{3}}{(z_{0}-\overline{z}_{0})^{2}}.$

Let $g(w)=\sum a_{n}w^{\alpha+n}$. We compute that

$\frac{dy}{dz}=\left(\frac{1}{(1-w)^{2}}g(w)+\frac{1}{1-w}\frac{dg(w)}{dw}\right)\frac{dw}{dz}$

and

$\begin{split}\frac{d^{2}y}{dz^{2}}&=\left(\frac{2}{(1-w)^{3}}g(w)+\frac{2}{(1-w)^{2}}\frac{dg(w)}{dw}+\frac{1}{1-w}\frac{d^{2}g(w)}{dw^{2}}\right)\left(\frac{dw}{dz}\right)^{2}\\\
&\qquad+\left(\frac{1}{(1-w)^{2}}g(w)+\frac{1}{1-w}\frac{dg(w)}{dw}\right)\frac{d^{2}w}{dz^{2}}.\end{split}$

Using (3.3), we reduce this to

$\frac{d^{2}y}{dz^{2}}=\frac{(1-w)^{3}}{(z_{0}-\overline{z}_{0})^{2}}\frac{d^{2}g(w)}{dw^{2}}.$

This proves the lemma. ∎

######  Proposition 3.6.

Suppose that $Q$ is a meromorphic modular form of weight $4$ with respect to
$\mathrm{SL}(2,\mathbb{Z})$ such that (1.1) is Fuchsian. Let $z_{0}$ be a pole
of $Q$. Then the local exponents of (1.1) at $\gamma z_{0}$ are the same for
all $\gamma\in\mathrm{SL}(2,\mathbb{Z})$. Also, if (1.1) is apparent at
$z_{0}$, then it is apparent at $\gamma z_{0}$ for all
$\gamma\in\mathrm{SL}(2,\mathbb{Z})$.

###### Proof.

Let $\gamma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\mathrm{SL}(2,\mathbb{Z})$,
$w=(z-z_{0})/(z-\overline{z}_{0})$, and $\widetilde{w}=(z-\gamma
z_{0})/(z-\gamma\overline{z}_{0})$. It suffices to prove that if

$y(z)=\frac{1}{1-w}w^{\alpha}\sum_{n=0}^{\infty}c_{n}w^{n}$

is a solution of (1.1) near $z_{0}$, then

$\widetilde{y}(z)=\frac{1}{1-\widetilde{w}}\widetilde{w}^{\alpha}\sum_{n=0}^{\infty}c_{n}(C\widetilde{w})^{n},\qquad
C=\frac{cz_{0}+d}{c\overline{z}_{0}+d},$

is a solution of (1.1) near $\gamma z_{0}$.

Since (1.1) is assumed to be Fuchsian, the order of poles of $Q(z)$ at $z_{0}$
is at most $2$. We have

$\begin{split}Q(z)=(1-w)^{4}\sum_{n=-2}^{\infty}a_{n}w^{n}\end{split}$

for some complex numbers $a_{n}$. Then by Lemma 3.5, $y(z)$ being a solution
of (1.1) near $z_{0}$ means that

(3.4)
$\begin{split}&\frac{1}{(2i\,\mathrm{Im\,}z_{0})^{2}}\sum_{n=0}^{\infty}c_{n}(\alpha+n)(\alpha+n-1)w^{\alpha+n-2}\\\
&\qquad=\left(\sum_{n=-2}^{\infty}a_{n}w^{n}\right)\left(\sum_{n=0}^{\infty}c_{n}w^{\alpha+n}\right).\end{split}$

On the other hand, by Lemmas 3.5 and 3.4, we have

$\begin{split}Q(z)=(cz_{0}+d)^{4}(1-\widetilde{w})^{4}\sum_{n=-2}^{\infty}a_{n}(C\widetilde{w})^{n}\end{split}$

near $\gamma z_{0}$ and

$\begin{split}\widetilde{y}^{\prime\prime}(z)&=\frac{C^{2}(1-\widetilde{w})^{3}}{(2i\,\mathrm{Im\,}\gamma
z_{0})^{2}}\sum_{n=0}^{\infty}c_{n}(\alpha+n)(\alpha+n-1)C^{n}\widetilde{w}^{\alpha+n-2}\\\
&=(cz_{0}+d)^{4}\frac{(1-\widetilde{w})^{3}}{(2i\,\mathrm{Im\,}z_{0})^{2}}\sum_{n=0}^{\infty}c_{n}(\alpha+n)(\alpha+n-1)C^{n}\widetilde{w}^{\alpha+n-2},\end{split}$

where in the last step we have used (3.2) and
$C=(cz_{0}+d)/(c\overline{z}_{0}+d)$. From these two expressions and (3.4), we
see that if $y(z)$ is a solution of (1.1) near $z_{0}$, then
$\widetilde{y}(z)$ is a solution of (1.9) near $\gamma z_{0}$, and the proof
is completed. ∎

For our purpose, we need the following properties of power series expansions
of modular forms on $\mathrm{SL}(2,\mathbb{Z})$. These properties are well-
known to experts (see [13], for example). For convenience of the reader, we
reproduce the proofs here.

######  Lemma 3.7.

Let

$w_{i}(z)=\frac{z-i}{z+i}.$

Then

$w_{i}(-1/z)=-w_{i}(z),\qquad 1-w_{i}(-1/z)=-iz(1-w_{i}(z)).$

Also, let $\rho=(1+\sqrt{-3})/2$,

$w_{\rho}(z)=\frac{z-\rho}{z-\overline{\rho}}$

and $\gamma=\left(\begin{smallmatrix}0&-1\\\ 1&-1\end{smallmatrix}\right)$.
Then

$w_{\rho}(\gamma z)=e^{2\pi i/3}w_{\rho}(z),\qquad 1-w_{\rho}(\gamma
z)=e^{4\pi i/3}(z-1)(1-w_{\rho}(z)).$

###### Proof.

The proof is straightforward. Here we will only provide details for the case
of $w_{\rho}(z)$.

We have

$w_{\rho}(z)=\begin{pmatrix}1&-\rho\\\ 1&-\overline{\rho}\end{pmatrix}z.$

Hence,

$w_{\rho}(\gamma z)=\begin{pmatrix}1&-\rho\\\
1&-\overline{\rho}\end{pmatrix}\begin{pmatrix}0&-1\\\ 1&-1\end{pmatrix}z.$

We then compute that

$\begin{pmatrix}1&-\rho\\\
1&-\overline{\rho}\end{pmatrix}\begin{pmatrix}0&-1\\\
1&-1\end{pmatrix}\begin{pmatrix}1&-\rho\\\
1&-\overline{\rho}\end{pmatrix}^{-1}=\begin{pmatrix}(-1-\sqrt{-3})/2&0\\\
0&(-1+\sqrt{-3})/2\end{pmatrix}.$

It follows that

$w_{\rho}(\gamma z)=e^{2\pi i/3}w_{\rho}(z).$

Then we have

$1-w_{\rho}(\gamma
z)=1-\rho^{2}w_{\rho}(z)=1-\frac{\rho^{2}z+1}{z-\overline{\rho}}=\frac{(1-\rho^{2})(z-1)}{z-\overline{\rho}},$

while

$1-w_{\rho}(z)=\frac{\rho-\rho^{-1}}{z-\overline{\rho}}.$

Hence,

$1-w_{\rho}(\gamma z)=-\rho(z-1)(1-w_{\rho}(z))=e^{4\pi
i/3}(z-1)(1-w_{\rho}(z)).$

This proves the lemma. ∎

From the lemma, we deduce the following properties of expansions of modular
forms at $i$ and $\rho$. These properties will be crucial in the proofs of
Theorem 1.7(a) and Theorem 1.8(a).

######  Corollary 3.8.

Let $f(z)$ be a meromorphic modular form of even weight $k$ on
$\mathrm{SL}(2,\mathbb{Z})$. Suppose that the power series expansion of $f$ at
$i$ is

$f(z)=(1-w_{i}(z))^{k}\sum_{n=n_{0}}^{\infty}a_{n}w_{i}(z)^{n},\qquad
w_{i}(z)=\frac{z-i}{z+i}.$

Then $a_{n}=0$ whenever $n+k/2\not\equiv 0\ \mathrm{mod}\ 2$. Also, if the
power series expansion of $f$ at $\rho=(1+\sqrt{-3})/2$ is

$f(z)=(1-w_{\rho}(z))^{k}\sum_{n=n_{0}}^{\infty}b_{n}w_{\rho}(z)^{n},\qquad
w_{\rho}(z)=\frac{z-\rho}{z-\overline{\rho}},$

then $b_{n}=0$ whenever $n+k/2\not\equiv 0\ \mathrm{mod}\ 3$.

###### Proof.

Here we will only prove the case of $\rho$. Let
$\gamma=\left(\begin{smallmatrix}0&-1\\\ 1&-1\end{smallmatrix}\right)$. Since
$f(z)$ is a meromorphic modular form of weight $k$ on
$\mathrm{SL}(2,\mathbb{Z})$, we have

$f(\gamma
z)=(z-1)^{k}f(z)=(z-1)^{k}(1-w_{\rho}(z))^{k}\sum_{n=n_{0}}^{\infty}b_{n}w_{\rho}(z)^{n}$

On the other hand, by the lemma above, we have

$f(\gamma z)=e^{4\pi
ik/3}(z-1)^{k}(1-w_{\rho}(z))^{k}\sum_{n=n_{0}}^{\infty}b_{n}e^{2\pi
in/3}w_{\rho}(z)^{n}.$

Comparing the two expressions, we conclude that $b_{n}=0$ whenever
$n+k/2\not\equiv 0\ \mathrm{mod}\ 3$. ∎

To determine local exponents of modular differential equations at $\rho$ and
$i$, we need to know the leading terms of the expansions of
$E_{6}(z)^{2}/E_{4}(z)^{2}$ and $E_{4}(z)^{4}/E_{6}(z)^{2}$.

######  Lemma 3.9.

  1. (a)

Let

$w_{\rho}=w_{\rho}(z)=\frac{z-\rho}{z-\overline{\rho}}.$

Then we have

$\pi^{2}\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}=(1-w_{\rho}^{4})\left(\frac{3}{4}w_{\rho}^{-2}+\sum_{n=1}^{\infty}a_{n}w_{\rho}^{n}\right)$

for some complex numbers $a_{n}$ such that $a_{n}=0$ whenever $n\not\equiv 1\
\mathrm{mod}\ 3$.

  2. (b)

Let

$w_{i}=w_{i}(z)=\frac{z-i}{z+i}.$

Then

$\pi^{2}\frac{E_{4}(z)^{4}}{E_{6}(z)^{2}}=(1-w_{i})^{4}\left(\frac{1}{4}w_{i}^{-2}+\sum_{n=0}^{\infty}b_{n}w_{i}^{n}\right)$

for some complex numbers $b_{n}$ such that $a_{n}=0$ whenever $n\not\equiv 0\
\mathrm{mod}\ 2$.

###### Proof.

It is known that, as an analytic function on $\mathbb{H}$, $E_{4}(z)$ has a
simple zero at $\rho$. Also, $E_{6}(\rho)\neq 0$. Thus, by Corollary 3.8,

$\pi^{2}\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}=(1-w_{\rho})^{4}\left(a_{-2}w_{\rho}^{-2}+\sum_{n=1}^{\infty}a_{n}w_{\rho}^{n}\right)$

for some complex numbers $a_{n}$ such that $a_{n}=0$ whenever $n\not\equiv 1\
\mathrm{mod}\ 3$. To determine the leading coefficient $a_{-2}$, we use the
well-known Ramanujan’s identity

$\frac{1}{2\pi i}E_{4}^{\prime}(z)=\frac{E_{2}(z)E_{4}(z)-E_{6}(z)}{3},$

where $E_{2}(z)$ is the Eisenstein series of weight $2$ on
$\mathrm{SL}(2,\mathbb{Z})$ (see [22, Proposition 15]). Hence,

$\begin{split}\lim_{z\to\rho}w_{\rho}(z)\frac{E_{6}(z)}{E_{4}(z)}&=\frac{E_{6}(\rho)}{\rho-\overline{\rho}}\lim_{z\to\rho}\frac{z-\rho}{E_{4}(z)}=\frac{E_{6}(\rho)}{\sqrt{3}i}\frac{1}{E^{\prime}_{4}(\rho)}\\\
&=-\frac{E_{6}(\rho)}{2\pi\sqrt{3}}\frac{3}{E_{2}(\rho)E_{4}(\rho)-E_{6}(\rho)}=\frac{\sqrt{3}}{2\pi},\end{split}$

which implies that $a_{-2}=3/4$. This proves Part (a).

The proof of Part (b) is similar. We use another identity

$\frac{1}{2\pi i}E_{6}^{\prime}(z)=\frac{E_{2}(z)E_{6}(z)-E_{4}(z)^{2}}{2}$

of Ramanujan’s to conclude that the leading term of
$\pi^{2}E_{4}(z)^{4}/E_{6}(z)^{2}$ is $w_{i}^{-2}/4$. We omit the details. ∎

######  Corollary 3.10.

The local exponents of the modular differential equation (1.9) at $\rho$ and
at $i$ are roots of

$x^{2}-x+\frac{9}{4}s=0$

and

$x^{2}-x+t=0,$

respectively.

###### Proof.

Here we prove only the case of $\rho$; the proof of the case of $i$ is
similar.

Let $w=w_{\rho}(z)=(z-\rho)/(z-\overline{\rho})$. Assume that

$y(z)=\frac{1}{1-w}\sum_{n=0}^{\infty}a_{n}w^{\alpha+n},\quad a_{0}\neq 0,$

is a solution of (1.9). By Lemmas 3.9 and 3.5, we have

$y^{\prime\prime}(z)=-\frac{(1-w)^{3}}{3}\left(\alpha(\alpha-1)a_{0}w^{\alpha-2}+\cdots\right)$

while

$\begin{split}&\pi^{2}\left(rE_{4}(z)+s\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}+t\frac{E_{4}(z)^{4}}{E_{6}(z)^{2}}\right)y(z)\\\
&\qquad=(1-w)^{3}\left(\frac{3}{4}sa_{0}w^{\alpha-2}+\cdots\right).\end{split}$

Comparing the leading terms, we see that the exponent $\alpha$ satisfies
$\alpha^{2}-\alpha+9s/4=0$. ∎

We are now ready to prove Part (a) of Theorem 1.7.

###### Proof of Theorem 1.7(a).

By Proposition 3.6, we only need to determine when (1.11) is apparent at
$\rho$.

Let $\kappa_{\rho}\in\frac{1}{2}\mathbb{N}$ and set
$s=s_{\kappa_{\rho}}=(1-4\kappa_{\rho})/9$ so that the local exponents of the
modular differential equation (1.11) with $s=s_{\kappa_{\rho}}$, i.e.,

(3.5)
$y^{\prime\prime}(z)=\pi^{2}\left(rE_{4}(z)+s_{\kappa_{\rho}}\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}\right)y(z)$

at $\rho$ are $1/2\pm\kappa_{\rho}$, by Corollary 3.10.

Let $w=w_{\rho}(z)=(z-\rho)/(z-\overline{\rho})$. According to Corollary 3.8
and Lemma 3.9, we have

(3.6) $\pi^{2}E_{4}(z)=(1-w)^{4}\sum_{n=1}^{\infty}a_{n}w^{n},$

and

(3.7)
$\pi^{2}\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}=(1-w)^{4}\left(\frac{3}{4}w^{-2}+\sum_{n=1}^{\infty}b_{n}w^{n}\right),$

where $a_{n}$ and $b_{n}$ are complex numbers satisfying

(3.8) $a_{n}=b_{n}=0\quad\text{if}\quad n\not\equiv 1\ \mathrm{mod}\ 3.$

We also remark that $a_{1}\neq 0$ since the zero $\rho$ of $E_{4}(z)$, as a
holomorphic function on $\mathbb{H}$, is simple.

Now the differential equation (3.5) is apparent at $\rho$ if and only if it
has a solution of the form

$y(z)=\frac{1}{1-w}w^{1/2-\kappa_{\rho}}\sum_{n=0}^{\infty}c_{n}w^{n}\quad\text{with}\
c_{0}=1.$

Plugging this series into (3.5) and using Lemma 3.5, (3.6), and (3.7), we find
that the coefficients $c_{n}$ need to satisfy

(3.9)
$n\left(n-2\kappa_{\rho}\right)c_{n}=-3\sum_{j=0}^{n-2}c_{j}(ra_{n-j-2}+s_{\kappa_{\rho}}b_{n-j-2}).$

Due to (3.8) and (3.9), we can inductively prove that

(3.10) $c_{n}=0\quad\text{if}\ n\not\equiv 0\ \mathrm{mod}\ 3.$

Since the left-hand side of (3.9) vanishes when $n=2\kappa_{\rho}$, (3.5) is
apparent at $\rho$ if and only if

(3.11)
$\sum_{j=0}^{2\kappa_{\rho}-2}c_{j}(ra_{2\kappa_{\rho}-j-2}+s_{\kappa_{\rho}}b_{2\kappa_{\rho}-j-2})=0.$

Suppose $3\nmid 2\kappa_{\rho}$. Then, $j\equiv 0\ \mathrm{mod}\ 3$ and
$2\kappa_{\rho}-j-2\equiv 1\ \mathrm{mod}\ 3$ cannot hold simultaneously.
Hence, by (3.8) and (3.10), the condition (3.11) always holds for any $r$,
i.e., (3.5) is apparent at $\rho$ for any $r$. This proves (a).

For the case $3|2\kappa_{\rho}$, considering $r$ as an indeterminate and using
(3.9) to recursively express $c_{n}$ as polynomials in $r$, we find that
$c_{n}$ is a polynomial of degree exactly $n/3$ in $r$ when $3|n$ and
$n<2\kappa_{\rho}$. (Note that we use the fact that $a_{1}\neq 0$ to conclude
that the degree is $n/3$.) Thus, the left-hand side of (3.11) is a polynomial
$P(r)$ of degree $2\kappa_{\rho}/3$ in $r$ and (3.5) is apparent at $\rho$ if
and only if $r$ is a root of this polynomial $P(x)$. This proves Part (b)
except the identity (1.12). ∎

The proof of Theorem 1.8(a) except (1.14) is very similar to that of Theorem
1.7 and will be omitted.

##  4\.  Riemann’s existence theorem and its application.

In this section, we will use Riemann’s existence theorem to prove Theorems
1.6, 1.7(b), and 1.8(b). The basic idea is as follows.

Let $h(z)$ be a modular function on some subgroup $\Gamma$ of finite index of
$\mathrm{SL}(2,\mathbb{Z})$. A simple computation shows that both
$y_{1}(z)=1/\sqrt{h^{\prime}(z)}$ and $y_{2}(z)=h(z)/\sqrt{h^{\prime}(z)}$ are
solutions of

$y^{\prime\prime}(z)=Q(z)y(z),\qquad Q(z)=-\frac{1}{2}\\{h(z),z\\},$

where $\\{h(z),z\\}$ is the Schwarz derivative. Using either properties of
Schwarz derivatives or direct computation, we can verify that $\\{h(z),z\\}$
is a meromorphic modular form of weight $4$ on $\Gamma$. When $h(z)$ has
additional symmetry, $\\{h(z),z\\}$ can be modular on a larger group. Note
that, by construction, this differential equation
$y^{\prime\prime}(z)=Q(z)y(z)$ is apparent on $\mathbb{H}$. Thus, one way to
prove the theorems is simply to prove the existence of a modular function
$h(z)$ such that $-\\{h(z),z\\}/2=Q(z)$ for each $Q(z)$ appearing in the
theorems. To achieve this, we will use Riemann’s existence theorem.

Since some of the readers may not be familiar with Riemann’s existence
theorem, here we give a quick overview of this important result in the theory
of Riemann surfaces. The exposition follows [18, Chapter III].

Let $F:X\to Y$ be a (branched) covering of compact Riemann surfaces of degree
$d$. A point $y$ of $Y$ is a _branch point_ if the cardinality of $F^{-1}(y)$
is not $d$ and a point $x$ of $X$ is a _ramification point_ if $F$ is not
locally one-to-one near $x$. (In particular, $F(x)$ is a branch point.) Let
$B$ be the (finite) set of branch points on $Y$ under $F$. Pick a point
$y_{0}\in Y-B$ so that $F^{-1}(y_{0})$ has $d$ points, say
$x_{1},\ldots,x_{d}$. Every loop $\gamma$ in $Y-B$ based at $y_{0}$ can be
lifted to $d$ paths $\widetilde{\gamma}_{1},\ldots,\widetilde{\gamma}_{d}$
with $\widetilde{\gamma}_{j}(0)=x_{j}$ and
$\widetilde{\gamma}_{j}(1)=x_{j^{\prime}}$ for some $x_{j^{\prime}}$. The map
$j\mapsto j^{\prime}$ is then a permutation in $S_{d}$. The permutation
depends only on the homotopy class of $\gamma$. In this way, we get a
monodromy representation

$\rho:\pi_{1}(Y-B,y_{0})\to S_{d}.$

Note that since $F^{-1}(Y-B)$ is connected, the image of $\rho$ is a
transitive subgroup of $S_{d}$. Also, let $b\in B$ and $a_{1},\ldots,a_{k}$ be
the points in $F^{-1}(b)$ with ramification indices $m_{1},\ldots,m_{k}$,
respectively. We can show that if $\gamma$ is a small loop in $Y-B$ around $b$
based at $y_{0}$, then $\rho(\gamma)$ is a product of disjoint cycles of
lengths $m_{1},\ldots,m_{k}$.

To state the version of Riemann’s existence theorem used in the paper, let us
consider the case $Y=\mathbb{P}^{1}(\mathbb{C})$. Let
$B=\\{b_{1},\ldots,b_{n}\\}$ be the set of branch points of
$F:X\to\mathbb{P}^{1}(\mathbb{C})$. Let $\gamma_{j}$, $j=1,\ldots,n$, be loops
that circles $b_{j}$ once but no other branch points. Then
$\pi_{1}(\mathbb{P}^{1}(\mathbb{C})-B,y_{0})$ is generated by the homotopy
classes $[\gamma_{j}]$, subject to a single relation
$[\gamma_{1}]\ldots[\gamma_{n}]=1$ (with a suitable ordering of the points
$b_{j}$). Thus, the image of $\rho$ is generated by
$\sigma_{j}=\rho(\gamma_{j})$ satisfying the relation
$\sigma_{1}\cdots\sigma_{n}=1$. Then Riemann’s existence theorem states as
follows (see [18, Corollary 4.10]).

######  Theorem B (Riemann’s existence theorem).

Let $B=\\{b_{1},\ldots,b_{n}\\}$ be a finite subset of
$\mathbb{P}^{1}(\mathbb{C})$. Then there exists a one-to-one correspondence
between the set of isomorphism classes of coverings
$F:X\to\mathbb{P}^{1}(\mathbb{C})$ of compact Riemann surfaces of degree $d$
whose branch points lie in $B$ and the set of (simultaneous) conjugacy classes
of $n$-tuples $(\sigma_{1},\ldots,\sigma_{n})$ of permutations in $S_{d}$ such
that $\sigma_{1}\ldots\sigma_{n}=1$ and the group generated by the
$\sigma_{j}$’s is transitive.

Moreover, if the disjoint cycle decomposition of $\sigma_{j}$ is a product of
$k$ cycles of lengths $m_{1},\ldots,m_{k}$, then $F^{-1}(b_{j})$ has $k$
points with ramification indices $m_{1},\ldots,m_{k}$, respectively.

We now use this result to prove Theorems 1.6, 1.7(b), and 1.8(b). Since the
proofs are similar, we will provide details only for Theorem 1.7(b).

###### Proof of Theorem 1.7(b).

Assume that $3|2\kappa_{\rho}$. Let $\Gamma_{2}$ be the subgroup of index of
$2$ of $\mathrm{SL}(2,\mathbb{Z})$ generated by

$\gamma_{1}=\begin{pmatrix}1&-1\\\
1&0\end{pmatrix},\qquad\gamma_{2}=\begin{pmatrix}0&1\\\ -1&-1\end{pmatrix}.$

Note that

$\gamma_{1}\gamma_{2}=\begin{pmatrix}1&2\\\ 0&1\end{pmatrix}.$

The group $\Gamma_{2}$ has a cusp $\infty$ and two elliptic points
$\rho_{1}=(1+\sqrt{-3})/2$ and $\rho_{2}=(-1+\sqrt{-3})/2$ of order $3$, fixed
by $\gamma_{1}$ and $\gamma_{2}$, respectively. Let

$j_{2}(z)=\frac{E_{6}(z)}{\eta(z)^{12}},$

which is a Hauptmodul for $\Gamma_{2}$, and set

$J_{2}(z)=\frac{24}{j_{2}(z)}.$

We have $J_{2}(\infty)=0$, $J_{2}(\rho_{1})=1/\sqrt{-3}$, and
$J_{2}(\rho_{2})=-1/\sqrt{-3}$.

Set $\ell_{0}=2\kappa_{\rho}/3$. We first show that for each
$\ell\in\\{0,\ldots,\ell_{0}-1\\}$, there exists a modular function $h(z)$ on
$\Gamma_{2}$ such that the covering
$h:X(\Gamma_{2})\to\mathbb{P}^{1}(\mathbb{C})$ of compact Riemann surfaces is
ramified precisely at $\infty$, $\rho_{1}$, and $\rho_{2}$ with ramification
index $2\ell+1$, $\ell_{0}$, and $\ell_{0}$, respectively. Note that by the
Riemann-Hurwitz formula, such a covering has degree $\ell_{0}+\ell$, i.e.,
such a modular function $h(z)$ will be a rational function of degree
$\ell_{0}+\ell$ in $J_{2}(z)$.

Consider the two $\ell_{0}$-cycles

$\sigma_{1}=(1,\ldots,\ell_{0}),\qquad\sigma_{2}=(\ell_{0}+\ell,\ell_{0}+\ell-1,\ldots,\ell+1)$

in the symmetric group $S_{\ell_{0}+\ell}$. Since $\ell<\ell_{0}$, we have

$\sigma_{2}\sigma_{1}=(1,\ldots,\ell,\ell_{0}+\ell,\ell_{0}+\ell-1,\ldots,\ell_{0}),$

which is a $(2\ell+1)$-cycle. (Notice that if $\ell\geq\ell_{0}$, then
$\sigma_{1}$ and $\sigma_{2}$ are disjoint.) It is clear that when
$\ell<\ell_{0}$, the subgroup generated by $\sigma_{1}$ and $\sigma_{2}$ is a
transitive subgroup of $S_{\ell_{0}+\ell}$. Thus, by Riemann’s existence
theorem, there exists a covering of compact Riemann surfaces
$H:X\to\mathbb{P}^{1}(\mathbb{C})$ of degree $\ell_{0}+\ell$ ramified at three
points $\zeta_{1}$, $\zeta_{2}$, and $\zeta_{3}$ of
$\mathbb{P}^{1}(\mathbb{C})$ with corresponding monodromy $\sigma_{1}$,
$\sigma_{2}$, and $\sigma_{1}^{-1}\sigma_{2}^{-1}$, respectively. By the
Riemann-Hurwitz formula, the genus of $X$ is $0$, and $H$ is a rational
function from $\mathbb{P}^{1}(\mathbb{C})$ to $\mathbb{P}^{1}(\mathbb{C})$.
Furthermore, by applying a suitable linear fractional transformation on the
variable of $H$, we may assume that the three ramified points in
$H^{-1}(z_{j})$ are $0=J_{2}(\infty)$, $1/\sqrt{-3}=J_{2}(\rho_{1})$, and
$-1/\sqrt{-3}=J_{2}(\rho_{2})$, respectively. Set $h(z)=H(J_{2}(z))$. Then
$h(z)$ has the required properties that the only points of $X(\Gamma_{2})$
ramified under $h:X(\Gamma_{2})\to\mathbb{P}^{1}(\mathbb{C})$ are $\rho_{1}$,
$\rho_{2}$, and the cusp $\infty$ with ramified indices $\ell_{0}$,
$\ell_{0}$, and $2\ell+1$, respectively.

Now consider the Schwarz derivative $\\{h(z),z\\}$, which is a meromorphic
modular form of weight $4$ on $\Gamma_{2}$. We claim that it is in fact
modular on the bigger group $\mathrm{SL}(2,\mathbb{Z})$.

Indeed, to show $\\{h(z),z\\}$ is modular on $\mathrm{SL}(2,\mathbb{Z})$, it
suffices to prove that $\\{h(z),z\\}\big{|}T=\\{h(z),z\\}$, where
$T=\left(\begin{smallmatrix}1&1\\\ 0&1\end{smallmatrix}\right)$. Let
$\widetilde{h}(z)=h(z+1)$. Now the automorphism on $X(\Gamma_{2})$ induced by
$T$ interchanges $\rho_{1}$ and $\rho_{2}$. Thus, the ramification data of the
covering $\widetilde{h}:X(\Gamma_{2})\to\mathbb{P}^{1}(\mathbb{C})$ is the
same as that of $h$. By the Riemann’s existence theorem, $h$ and
$\widetilde{h}$ are related by a linear fractional transformation, i.e.,
$\widetilde{h}=(ah+b)/(ch+d)$ for some $a,b,c,d\in\mathbb{C}$ with $ad-bc\neq
0$. It follows that $\\{h(z),z\\}\big{|}T=\\{h(z),z\\}$ by the well-known
property $\\{(af(z)+b)/(cf(z)+d),z\\}=\\{f(z),z\\}$ of the Schwarz derivative.
This proves that $\\{h(z),z\\}$ is a meromorphic modular form of weight $4$ on
the larger group $\mathrm{SL}(2,\mathbb{Z})$.

Furthermore, since $\rho_{1}$ is an elliptic point of order $3$, a local
parameter for $\rho_{1}$ as a point on the compact Riemann surface
$X(\Gamma_{2})$ is $w^{3}$, where $w=(z-\rho)/(z-\overline{\rho})$. Therefore,
we have

$h(z)=d_{0}+\sum_{n=3\ell_{0}}^{\infty}d_{n}w^{n},$

for some complex numbers $d_{n}$ with $d_{3\ell_{0}}\neq 0$ and $d_{n}=0$
whenever $3\nmid n$. For convenience, set

$\begin{split}A&=\sum_{n=3\ell_{0}}^{\infty}nd_{n}w^{n-1},\\\
B&=\sum_{n=3\ell_{0}}^{\infty}n(n-1)d_{n}w^{n-2},\\\
C&=\sum_{n=3\ell_{0}}^{\infty}n(n-1)(n-2)d_{n}w^{n-3}.\end{split}$

Using (3.3), we compute that

$\begin{split}h^{\prime}(z)&=\frac{(1-w)^{2}}{\rho-\overline{\rho}}A,\\\
h^{\prime\prime}(z)&=\frac{(1-w)^{4}}{(\rho-\overline{\rho})^{2}}B-2\frac{(1-w)^{3}}{(\rho-\overline{\rho})^{2}}A,\\\
h^{\prime\prime\prime}(z)&=\frac{(1-w)^{6}}{(\rho-\overline{\rho})^{3}}C-6\frac{(1-w)^{5}}{(\rho-\overline{\rho})^{3}}B+6\frac{(1-w)^{4}}{(\rho-\overline{\rho})^{3}}A,\end{split}$

and hence

$\\{h(z),z\\}=\frac{(1-w)^{4}}{(\rho-\overline{\rho})^{2}}\left(\frac{C}{A}-\frac{3}{2}\frac{B^{2}}{A^{2}}\right)=-\frac{(1-w)^{4}}{3}\left(\frac{1-9\ell_{0}^{2}}{2w^{2}}+cw+\cdots\right)$

for some $c$. It follows that, by (3.7),

$\\{h(z),z\\}+2\pi^{2}s_{\kappa_{\rho}}\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}},\qquad
s_{\kappa_{\rho}}=\frac{1-4\kappa_{\rho}^{2}}{9}=\frac{1}{9}-\ell_{0}^{2},$

is a holomorphic modular form of weight $4$ on $\mathrm{SL}(2,\mathbb{Z})$. By
comparing the leading coefficients of the Fourier expansions at the cusp
$\infty$, we conclude that,

$\\{h(z),z\\}=-2\pi^{2}\left(rE_{4}(z)+s_{\kappa_{\rho}}\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}\right),$

where $r=-(2\ell+1)^{2}/4-s_{\kappa_{\rho}}=\ell_{0}^{2}-(2\ell+1)^{2}/4-1/9$.
Equivalently, $1/\sqrt{h^{\prime}(z)}$ and $h(z)/\sqrt{h^{\prime}(z)}$ are
solutions of (1.11) which also implies that the singularity of (1.11) at
$\rho$ is apparent.

Finally, since we have found $\ell_{0}$ different $r$ such that (1.11) has an
apparent singularity at $\rho$ for the given $s_{\kappa_{\rho}}$, by Part (a),
this proves the theorem. ∎

######  Example.

For small $\kappa_{\rho}$, the modular functions $h(z)$ appearing in the proof
are given by

$\begin{array}[]{cccc}\hline\cr\hline\cr\kappa_{\rho}&\ell&(r,s)&h(z)\\\
\hline\cr\displaystyle\frac{3}{2}&0&\displaystyle\left(\frac{23}{36},-\frac{8}{9}\right)\phantom{\Bigg{|}}&J_{2}\\\
3&0&\displaystyle\left(\frac{131}{36},-\frac{35}{9}\right)\phantom{\Bigg{|}}&\displaystyle\frac{J_{2}}{1-3J_{2}^{2}}\\\
3&1&\displaystyle\left(\frac{59}{36},-\frac{35}{9}\right)\phantom{\Bigg{|}}&\displaystyle\frac{J_{2}^{3}}{1+9J_{2}^{2}}\\\
\hline\cr\hline\cr\end{array}$

###### Proof of Theorem 1.8(b).

Assume that $\kappa_{i}\in\mathbb{N}$. Let $\Gamma_{3}$ be the subgroup of
index $3$ of $\mathrm{SL}(2,\mathbb{Z})$ generated by

$\gamma_{1}=\begin{pmatrix}1&-2\\\
1&-1\end{pmatrix},\qquad\gamma_{2}=\begin{pmatrix}1&-1\\\
2&-1\end{pmatrix},\qquad\gamma_{3}=\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}.$

We note that

$\gamma_{1}\gamma_{2}\gamma_{3}=\begin{pmatrix}1&3\\\ 0&1\end{pmatrix}.$

The group $\Gamma_{3}$ has one cusp and three elliptic points $z_{1}=1+i$,
$z_{2}=(1+i)/2$, and $z_{3}=i$ of order $2$, fixed by $\gamma_{j}$, $j=1,2,3$,
respectively. Let

$j_{3}(z)=\frac{E_{4}(z)}{\eta(z)^{8}}$

be a Hauptmodul for $\Gamma_{3}$ and set

$J_{3}(z)=12j_{3}(z)^{-1}.$

Note that $j_{3}(z)^{3}$ is equal to the elliptic $j$-function $j(z)$. Since
$j(i)=1728$ and $j(\rho)=0$, we have
$\\{J_{3}(z_{1}),J_{3}(z_{2}),J_{3}(z_{3})\\}=\\{1,e^{2\pi i/3},e^{4\pi
i/3}\\}$, $J_{3}(\rho)=\infty$, and $J_{3}(\infty)=0$.

Consider the case $r+t_{\kappa_{i}}=-(\ell+1/3)^{2}$ first. Our goal here is
to construct a modular function $h(z)$ on $\Gamma_{3}$, for each $\ell$ in the
range, such that the covering $h:X(\Gamma_{3})\to\mathbb{P}^{1}(\mathbb{C})$
has degree

$d=\frac{1}{2}(3\kappa_{i}+3\ell-1)$

and is ramified at precisely the cusp $\infty$ and the three elliptic points
$z_{1}$, $z_{2}$, and $z_{3}$ with ramification indices $3\ell+1$,
$\kappa_{i}$, $\kappa_{i}$, and $\kappa_{i}$, respectively. (Notice that
$\kappa_{i}$ and $\ell$ have opposite parities, so $d$ is an integer.) Since
the covering has four branch points, it is not easy to apply Riemann’s
existence theorem directly to get $h(z)$. Instead, we shall use the following
idea.

For convenience, set

(4.1) $m=\frac{1}{2}(\kappa_{i}+\ell-1),\qquad
m^{\prime}=\frac{1}{2}(\kappa_{i}-\ell-1).$

We claim that there exists a rational function $H(x)$ of degree $d$ in $x$ of
the form

$H(x)=\frac{x^{3\ell+1}G(x)^{3}}{F(x)^{3}},\quad\deg F(x)=m,\quad\deg
G(x)=m^{\prime},$

such that $xF(x)G(x)$ is squarefree and

$H(x)-1=\frac{(x-1)^{\kappa_{i}}L(x)}{F(x)^{3}}$

for some polynomial $L$ of degree $d-\kappa_{i}$ with no repeated roots. That
is, $H(x)$ is a rational function such that

  1. (i)

the covering $H:\mathbb{P}^{1}(\mathbb{C})\to\mathbb{P}^{1}(\mathbb{C})$
branches at precisely $\infty$, $0$, and $1$ (note that by the Riemann-Hurwitz
formula, $H$ cannot have other branch points),

  2. (ii)

the monodromy $\sigma_{\infty}$ around $\infty$ is a product of $m$ disjoint
$3$-cycles, the monodromy $\sigma_{0}$ around $0$ is a disjoint product of a
$(3\ell+1)$-cycle and $m^{\prime}$ $3$-cycles, and the monodromy $\sigma_{1}$
around $1$ is a $\kappa_{i}$-cycle,

  3. (iii)

the unique unramified point in $H^{-1}(\infty)$ is $\infty$, the unique point
of ramification index $3\ell+1$ in $H^{-1}(0)$ is $0$, and the unique ramified
point in $H^{-1}(1)$ is $1$.

Suppose that such a rational function $H(x)$ exists. We define
$h:X(\Gamma_{3})\to\mathbb{P}^{1}(\mathbb{C})$ by

$h(z)=H(J_{3}(z)^{3})^{1/3}=\frac{J_{3}(z)^{3\ell+1}G(J_{3}(z)^{3})}{F(J_{3}(z)^{3})}.$

From the construction, we see that $h$ ramifies only at $z_{1}=1+i$,
$z_{2}=(1+i)/2$, $z_{3}=i$, and $\infty$ with ramification indices
$\kappa_{i}$, $\kappa_{i}$, $\kappa_{i}$, and $3\ell+1$, respectively. Then
following the proof of Theorem 1.7(b), we can prove that the Schwarz
derivative $\\{h(z),z\\}$ is a meromorphic modular form on the larger group
$\mathrm{SL}(2,\mathbb{Z})$ and that

$\\{h(z),z\\}=-2\pi^{2}\left(rE_{4}(z)+t_{\kappa_{i}}\frac{E_{4}(z)^{4}}{E_{6}(z)^{2}}\right),\quad
r=-\left(\ell+\frac{1}{3}\right)^{2}-t_{\kappa_{i}},$

which is equivalent to the assertion that $1/\sqrt{h^{\prime}(z)}$ and
$h(z)/\sqrt{h^{\prime}(z)}$ are solutions of (1.13) with $t=t_{\kappa_{i}}$
and $r=-(\ell+1/3)^{2}-t_{\kappa_{i}}$ and hence implies that (1.13) is
apparent with these parameters.

It remains to prove that a rational function $H(x)$ with properties described
above exists. According to Riemann’s existence theorem, it suffices to find
$\sigma_{\infty}$ that is a product of $m$ disjoint $3$-cycles and
$\sigma_{1}$ that is a $\kappa_{i}$-cycle in $S_{d}$ such that
$\sigma_{1}\sigma_{\infty}$ is a disjoint product of a cycle of length
$3\ell+1$ and $m^{\prime}$ cycles of length $3$. Indeed, we find that we may
choose

$\sigma_{\infty}=(2,3,4)(5,6,7)\ldots(3m-1,3m,3m+1)$

and

$\sigma_{1}=(1,2,5,8,\ldots,3m-1,3m^{\prime}+1,3m^{\prime}-2,\ldots,7,4).$

Then

$\sigma_{1}\sigma_{\infty}=(1,2,3)(4,5,6)\ldots(3m^{\prime}-2,3m^{\prime}-1,3m^{\prime})(3m^{\prime}+1,3m^{\prime}+2,\ldots,d).$

This settles the case $r+t_{\kappa_{i}}=-(\ell+1/3)^{2}$.

The case $r+t_{\kappa_{i}}=-(\ell-1/3)^{2}$ can be dealt with in the same way.
The difference is that the rational function $H(x)$ in this case has degree

$d=\frac{3}{2}(\kappa_{i}+\ell-1)$

and is of the form

$H(x)=\frac{x^{3\ell-1}G(x)^{3}}{F(x)^{3}},\quad\deg F(x)=m,\quad\deg
G(x)=m^{\prime},$

where $m$ and $m^{\prime}$ are the same as those in (4.1), such that
$xF(x)G(x)$ is squarefree and

$H(x)-1=\frac{(x-1)^{\kappa_{i}}L(x)}{F(x)^{3}}$

for some polynomial $L(x)$ of degree $d-\kappa_{i}$ with no repeated roots.
I.e., $\sigma_{\infty}$ in this case is a disjoint product of $m$ $3$-cycles,
$\sigma_{0}$ is a a disjoint product of $(3\ell-1)$-cycle and $m^{\prime}$
$3$-cycles, and $\sigma_{1}$ is a $\kappa_{i}$-cycle. We choose

$\sigma_{\infty}=(1,2,3)(4,5,6)\ldots(3m-2,3m-1,3m)$

and

$\sigma_{1}=(1,4,7,\ldots,3m-2,3m,3m-3,\ldots,3\ell)$

with

$\sigma_{1}\sigma_{\infty}=(1,2,3,4,\ldots,3\ell-1)(3\ell,3\ell+1,3\ell+2)\ldots(3m-3,3m-2,3m-1).$

The rest of proof is the same as the case of
$r+t_{\kappa_{i}}=-(\ell+1/3)^{2}$. This completes the proof that (1.14) is
the complete list of parameters $r$ such that (1.13) with $t=t_{\kappa_{i}}$
is apparent. ∎

######  Example.

For small $\kappa_{i}$, the modular functions $h(z)$ in the proof are given by

$\begin{array}[]{cccc}\hline\cr\hline\cr\kappa_{i}&\ell\pm 1/3&(r,t)&h(z)\\\
\hline\cr
1&\displaystyle\frac{1}{3}&\displaystyle\left(\frac{23}{36},-\frac{3}{4}\right)\phantom{\Bigg{|}}&J_{3}\\\
2&\displaystyle\frac{2}{3}&\displaystyle\left(\frac{119}{36},-\frac{15}{4}\right)\phantom{\Bigg{|}}&\displaystyle\frac{J_{3}^{2}}{1+2J_{3}^{3}}\\\
2&\displaystyle\frac{4}{3}&\displaystyle\left(\frac{71}{36},-\frac{15}{4}\right)\phantom{\Bigg{|}}&\displaystyle\frac{J_{3}^{4}}{1-4J_{3}^{3}}\\\
\hline\cr\hline\cr\end{array}$

###### Proof of Theorem 1.6.

Assume that $n_{i}$, $n_{\rho}$, and $n_{\infty}$ are positive integers
satisfying the two conditions. We note that the parameters $r$, $s$, and $t$
in (1.9) are

(4.2) $r=-n_{\infty}^{2}+n_{\rho}^{2}+n_{i}^{2}-\frac{13}{36},\qquad
s=\frac{1}{9}-n_{\rho}^{2},\qquad t=\frac{1}{4}-n_{i}^{2}.$

Let

$d=\frac{1}{2}(n_{i}+n_{\rho}+n_{\infty}-1).$

By the second condition, we have

$d-n_{i}=\frac{1}{2}(n_{\rho}+n_{\infty}-n_{i}-1)\geq 0$

and similarly, $d-n_{\rho}\geq 0$. Thus, there are cycles of lengths $n_{i}$
and $n_{\rho}$ in the symmetric group $S_{d}$. Choose

$\sigma_{1}=(1,\ldots,n_{i}),\qquad\sigma_{2}=(d,d-1,\ldots,d-n_{\rho}+1)$

By the second condition again, we have

$n_{i}-(d-n_{\rho}+1)=\frac{1}{2}(n_{i}+n_{\rho}-n_{\infty}-1)\geq 0.$

In other words, the two cycles are not disjoint. We then compute that

$\sigma_{2}\sigma_{1}=(1,\ldots,d-n_{\rho},d,d-1,\ldots,n_{i}).$

This is a cycle of length

$d-n_{\rho}+(d-n_{i}+1)=2d-n_{\rho}-n_{i}+1=n_{\infty}.$

It is clear that the subgroup of $S_{d}$ generated by $\sigma_{1}$ and
$\sigma_{2}$ is transitive. Thus, by Riemann’s existence theorem, given three
distinct points $\zeta_{1}$, $\zeta_{2}$, and $\zeta_{3}$ on
$\mathbb{P}^{1}(\mathbb{C})$, there is a covering
$H:X\to\mathbb{P}^{1}(\mathbb{C})$ of compact Riemann surfaces of degree $d$
branched at $\zeta_{1}$, $\zeta_{2}$, and $\zeta_{3}$ with monodromy
$\sigma_{1}$, $\sigma_{2}$, and $\sigma_{3}=\sigma_{1}^{-1}\sigma_{2}^{-1}$,
respectively. By the Riemann-Hurwitz formula, the genus of $X$ is $0$ and we
may assume that $X=\mathbb{P}^{1}(\mathbb{C})$. Applying a suitable linear
fractional transformation (i.e., an automorphism of $X$) if necessary, we may
assume that the ramification points on $X$ are $1728=j(i)$, $0=j(\rho)$, and
$\infty=j(\infty)$ with ramification indices $n_{i}$, $n_{\rho}$, and
$n_{\infty}$, respectively. Let $h:X_{0}(1)\to\mathbb{P}^{1}(\mathbb{C})$ be
defined by $h(z)=H(j(z))$. Following the same computation as in the proof of
Theorem 1.7(b), we can show that

$\\{h(z),z\\}=-2\pi^{2}\left(rE_{4}(z)+s\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}+t\frac{E_{4}(z)^{4}}{E_{6}(z)^{2}}\right)$

with $r$, $s$, and $t$ given as (4.2) (details omitted). This implies that the
singularities of (1.9) are all apparent.

Conversely, assume that the differential equation (1.9) is apparent throughout
$\mathbb{H}\cup\\{\text{cusps}\\}$. Let $\pm n_{\infty}/2$ be the local
exponents at $\infty$. Then a fundamental pair of solutions near $\infty$ is

$y_{\pm}(z)=q^{\pm
n_{\infty}/2}\left(1+\sum_{n=1}^{\infty}c_{n}^{\pm}q^{n}\right).$

Let $h(z)=y_{+}(z)/y_{-}(z)$. Since (1.9) is apparent throughout $\mathbb{H}$,
$h(z)$ is a single-valued function on $\mathbb{H}$. Arguing as in the second
proof of Theorem 1.6, we see that $h(z)$ is a modular function on
$\mathrm{SL}(2,\mathbb{Z})$. Now since

$\\{h(z),z\\}=-2\pi^{2}\left(rE_{4}(z)+s\frac{E_{6}(z)^{2}}{E_{4}(z)^{2}}+t\frac{E_{4}(z)^{4}}{E_{6}(z)^{2}}\right)$

have poles only at points equivalent to $\rho$ or $i$ under
$\mathrm{SL}(2,\mathbb{Z})$, the covering
$X_{0}(1)\to\mathbb{P}^{1}(\mathbb{C})$ defined by $z\mapsto h(z)$ can only
ramify at $\rho$, $i$, or $\infty$. From the computation above, we see that
their ramification indices must be $n_{\rho}$, $n_{i}$, and $n_{\infty}$,
respectively. Then by the Riemann-Hurwitz formula, $n_{\rho}+n_{i}+n_{\infty}$
must be odd and the degree of the covering is
$(n_{\rho}+n_{i}+n_{\infty}-1)/2$. Since the ramification indices $n_{\rho}$,
$n_{i}$, and $n_{\infty}$ cannot exceed the degree of the covering, we
conclude that the sum of any two of $n_{\rho}$, $n_{i}$, and $n_{\infty}$ must
be greater than the remaining one. This completes the proof of the theorem. ∎

##  5\. Eremenko’s Theorem and its applications

###### Second proof of (1.12).

In Section 2.3, Example 2 shows that the angle of $Q_{1}$ at $i,\rho$ and
$\infty$ are

(5.1)
$\theta_{1}=\frac{1}{2},\quad\theta_{2}=\frac{2\kappa_{\rho}}{3},\quad\text{and}\quad\theta_{\infty}=\sqrt{-(r+s_{\kappa_{\rho}})}.$

First, we consider $\theta_{2}$ is even, say $\theta_{2}=2\ell_{0}$. By
Eremanko’s Theorem in Section 2, the curvature equation (2.5) has a solution
if and only if either $\left|\theta_{\infty}-\theta_{1}\right|=2\ell+1$ or
$\theta_{\infty}+\theta_{1}=2\ell+1$ for some $\ell\in\mathbb{Z}_{\geq 0}$ and
$\ell\leq\ell_{0}-1$. Since $\theta_{\infty}>0$, the condition
$\left|\theta_{\infty}-\theta_{1}\right|=2\ell+1\geq 1$ implies
$\theta_{\infty}-\theta_{1}>0$ and then $\theta_{\infty}-\theta_{1}=2\ell+1$.
This is equivalent to
$-(r+s_{\kappa_{\rho}})=\theta^{2}_{\infty}=(2\ell+1+1/2)^{2}$,
$\ell=0,\ldots,\ell_{0}-1$. The second condition
$\theta_{\infty}+\theta_{1}=2\ell+1$ is equivalent to
$-(r+s_{\kappa_{\rho}})=\theta^{2}_{\infty}=(2\ell+1/2)^{2}$,
$\ell=0,\ldots,\ell_{0}-1$. Therefore, there are exactly $2\ell_{0}$ different
$\theta_{\infty}$ such that the curvature equation (2.5) has a solution and
each of such a curvature equation is associated with the modular form
$Q_{1}(z;r,s_{\kappa_{\rho}})$ with $(r,s_{\kappa_{\rho}})$ where
$r+s_{\kappa_{\rho}}=-(\ell+1/2)^{2}$ for some
$\ell\in\left\\{0,\ldots,2\ell_{0}-1\right\\}$. By Theorem 2.2, for each
$(r,s_{\kappa_{\rho}})$, the ODE (1.11) is apparent. However, the first part
of Theorem 1.7(b) says that there exists a polynomial $P(x)$ of degree
$2\kappa_{\rho}/3$ such that (2.5) with $(r,s_{\kappa_{\rho}})$ is apparent if
and only if $P(r)=0$. Therefore, $P(r)$ has distinct roots and each root
satisfies $r+s_{\kappa_{\rho}}=-(\ell+1/2)^{2}$ for some integer $\ell$,
$0\leq\ell\leq 2\ell_{0}-1=\theta_{2}-1$. The proves (1.12) when $\theta_{2}$
is even.

For the case $\theta_{2}$ is odd, the idea of the proof is basically the same.
By noting $\theta_{1}=1/2$, the Eremenko theorem in Section 2 implies either
$\left|\theta_{\infty}-1/2\right|=\ell$ or $\theta_{\infty}+1/2=\ell$, where
$\ell$ is even because $\theta_{2}$ is odd. The first condition can be
replaced by $\theta_{\infty}-1/2=\ell$. Thus we have
$\theta_{\infty}=\ell+1/2$ or $\theta_{\infty}=\ell-1/2=(\ell-1)+1/2$, that is
$r+s=-(\ell+1/2)^{2}$, $\ell=0,1,2,\ldots,\theta_{2}-1$. The proof of (1.12)
is complete. ∎

###### Second proof of (1.14).

The angles for $Q_{2}(z)$ are $\theta_{1}=\kappa_{i}$, $\theta_{2}=1/3$, and
$\theta_{\infty}=\sqrt{-(r+t_{i})}$, where $\frac{1}{2}\pm\kappa_{i}$ are the
local exponents of (1.13). Hence

$\kappa_{i}-\frac{1}{2}+1=m+\frac{1}{2}$

i.e., $\theta_{1}=\kappa_{i}$ is an integer. Hence, there is a solution $u$ of
(2.5)-(2.7) with the RHS equals to $4\pi n\sum\delta_{p}$, where the summation
runs over $\gamma\cdot i$, $\gamma\in\mathrm{SL}(2,\mathbb{Z})$, if and only
if either
$\theta_{\infty}-\theta_{2}=\left|\theta_{\infty}-\theta_{2}\right|=\ell$ or
$\theta_{\infty}+\theta_{2}=\ell$ where $\ell\leq\kappa_{i}-1$ and $\ell$ has
the opposite parity of $\kappa_{i}$. Hence, $\theta_{\infty}=\ell\pm 1/3$ and
$r+t_{i}=-(\ell\pm 1/3)^{2}$. This proves (1.14). ∎

###### Second proof of Theorem 1.6.

Suppose that the ODE (1.9) has local exponents $\pm n_{\infty}$ at $\infty$,
$n_{\infty}\in\frac{1}{2}\mathbb{N}$. We claim that _( 1.9) is apparent
throughout $\mathbb{H}^{\ast}$ if and only if $Q_{3}(z)=Q_{3}(z;r,s,t)$ is
realized by a metric with curvature $1/2$_. It is clear that the second
statement implies the first statement. So it suffices to prove the other
direction.

Suppose that (1.9) is apparent throughout $\mathbb{H}^{\ast}$. Let
$y_{\pm}(z)=q^{\pm n_{\infty}/2}\left(1+O(q)\right)$ be two solutions of (1.9)
and set $h(z)=y_{+}(z)/y_{-}(z)$. Since (1.9) is apparent on $\mathbb{H}$,
$h(z)$ is a meromorphic single-valued function on $\mathbb{H}$ and its Schwarz
derivative is $-2Q_{3}(z)$. Recall Bol’s theorem that there is a homomorphism
$\rho:\mathrm{SL}(2,\mathbb{Z})\rightarrow\mathrm{PSL}(2,\mathbb{C})$ such
that

$\begin{pmatrix}\left(y_{1}\big{|}_{-1}\gamma\right)(z)\\\
\left(y_{2}\big{|}_{-1}\gamma\right)(z)\end{pmatrix}=\pm\rho(\gamma)\begin{pmatrix}y_{1}(z)\\\
y_{2}(z)\end{pmatrix},\quad\gamma\in\mathrm{SL}(2,\mathbb{C}).$

Clearly, $\rho(T)=\pm I$ because $\infty$ is apparent. Note that $\ker\rho$ is
a normal subgroup of $\mathrm{SL}(2,\mathbb{Z})$ and contains $\gamma
T\gamma^{-1}$ for any $\gamma\in\mathrm{SL}(2,\mathbb{Z})$. In particular,
$\ker\rho$ contains both $T=\left(\begin{smallmatrix}1&1\\\
0&1\end{smallmatrix}\right)$ and $STS^{-1}=\left(\begin{smallmatrix}1&0\\\
-1&1\end{smallmatrix}\right)$, where $S=\left(\begin{smallmatrix}0&-1\\\
1&0\end{smallmatrix}\right)$. Since $\left(\begin{smallmatrix}1&1\\\
0&1\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}1&0\\\
-1&1\end{smallmatrix}\right)$ generate $\mathrm{SL}(2,\mathbb{Z})$, we
conclude that $\ker\rho=\mathrm{SL}(2,\mathbb{Z})$. In other words,
$\rho(\gamma)=\pm I$ and $h(z)$ is a modular function on
$\mathrm{SL}(2,\mathbb{Z})$. Thus we have a solution
$u:=\log\frac{8\left|h^{\prime}(z)\right|^{2}}{\left(1+\left|h(z)\right|^{2}\right)^{2}}$
which realizes $Q_{3}$. This proves the claim.

Now, we apply the Eremenko theorem with the angles given by
$\theta_{1}=\kappa_{i}$, $\theta_{2}=2\kappa_{\rho}/3$ and
$\theta_{3}=n_{\infty}$. Our necessary and sufficient condition in Theorem 1.6
is identically the same as the condition of Eremenko’s theorem for the
existence of $u$ with three integral angles. This proves Theorem 1.6. ∎

######  Theorem 5.1.

Suppose $\kappa_{i}\in\mathbb{N}$ and
$\kappa_{\rho},\kappa_{\infty}\in\frac{1}{2}\mathbb{N}$ such that
$2\kappa_{\rho}/3\in\mathbb{N}$. If $Q_{3}(z;r,s,t)$ is apparent at $\rho$ and
$i$, then $Q$ can be realized.

###### Proof.

By the assumption, we have that $\theta_{i}$, $1\leq i\leq 3$, are all
integers. Now, given $\kappa_{i}$ and $\kappa_{\rho}$, $s$ and $t$ are
determined by the same formula in our paper. Further, there are polynomials
$P_{1}$ and $P_{2}$:

  1. $\bullet$

$Q_{3}(z;r,s,t)$ is apparent at $i$ if and only if $P_{1}(r)=0$, and $\deg
P_{1}(r)=\kappa_{i}$.

  2. $\bullet$

$Q_{3}(z;r,s,t)$ is apparent at $\rho$ if and only if $P_{2}(r)=0$, and $\deg
P_{2}=2\kappa_{\rho}/3$.

Therefore, $Q_{3}(z;r,s,t)$ is apparent at $i$ and $\rho$ if and only if

$r\in\left\\{r:P_{1}(r)=P_{2}(r)=0\right\\}.$

Now, we claim that under the assumption $\theta_{1}\in\mathbb{N}$,
$Q_{3}(z;r,s,t)$ is apparent if and only if the local exponents at $\infty$
are $\pm\kappa_{\infty}/2$, $\kappa_{\infty}\in\mathbb{N}$ and the curvature
equation has a solution.

By Eremenko’s Theorem (Section 2.4), (recall $\theta_{1}=\kappa_{i}$,
$\theta_{2}=2\kappa_{\rho}/3$, $\theta_{3}=2\kappa_{\infty}$) the curvature
equation has a solution if and only if $\theta_{1}+\theta_{2}+\theta_{3}$ is
odd and $\theta_{i}<\theta_{j}+\theta_{k}$, $i\neq j\neq k$. This condition is
equivalent to

  1. (a)

$\theta_{2}-\theta_{1}<\theta_{3}<\theta_{2}+\theta_{1},\quad\text{and}$

  2. (b)

$\theta_{1}-\theta_{2}<\theta_{3}<\theta_{1}+\theta_{2}.$

Since $\theta_{1}+\theta_{2}+\theta_{3}$ is odd, we have $\theta_{2}$
solutions of the curvature equation if $\theta_{1}>\theta_{2}$, $\theta_{1}$
solutions if $\theta_{2}>\theta_{1}$.

Now, $\deg P_{1}=\kappa_{i}=\theta_{1}$ and $\deg
P_{2}=2\kappa_{\rho}/3=\theta_{2}$. Then

$\displaystyle\min\left\\{\theta_{1},\theta_{2}\right\\}$
$\displaystyle\geq\left|\left\\{r:P_{1}(r)=P_{2}(r)=0\right\\}\right|$
$\displaystyle=\geq\\#\ \text{of\ curvature\
equations}\geq\min\left\\{\theta_{1},\theta_{2}\right\\}.$

Thus

$\left|\left\\{r:P_{1}(r)=P_{2}(r)=0\right\\}\right|=\\#\ \text{of\ curvature\
equations}.$

This proves the theorem. ∎

######  Remark.

In fact, the proof shows that if $\deg P_{i}\leq\deg P_{j}$, then $P_{i}$ is a
factor of $P_{j}$.

##  6\. proof of Theorem 1.1 and Theorem 1.4

###### Proof of Theorem 1.1.

Let $\rho$ be the Bol representation associated to (1.1), and set
$T=\left(\begin{smallmatrix}1&1\\\ 0&1\end{smallmatrix}\right)$,
$S=\left(\begin{smallmatrix}0&-1\\\ 1&0\end{smallmatrix}\right)$, and
$R=TS=\left(\begin{smallmatrix}1&-1\\\ 1&0\end{smallmatrix}\right)$. They
satisfy

(6.1) $S^{2}=-I,\quad\text{and}\quad R^{3}=-I.$

Assume that ($\mathbf{H}_{1}$) and ($\mathbf{H}_{2}$) hold. It follows from
either [12, Theorem 2.5], quoted as Theorem A.1 in the appendix, or Theorem
A.3 (with $\theta_{1}=1/2$, $\theta_{2}=1/3$, and $\theta_{3}=2r_{\infty}$ or
$\theta_{3}=1-2r_{\infty}$, depending on whether $2r_{\infty}\leq 1/2$ or
$2r_{\infty}>1/2$) in the appendix that if $1/12<r_{\infty}<5/12$, then an
invariant metric realizing $Q(z)$ exists, and if $0<r_{\infty}<1/12$ or
$5/12<r_{\infty}\leq 1/2$, then there does not exist an invariant metric
realizing $Q(z)$. So here we are concerned with the case $r_{\infty}=1/12$ or
$r_{\infty}=5/12$.

Assume that $r_{\infty}=1/12$. Then there exists a basis
$\\{y_{1}(z),y_{2}(z)\\}$ for the solution space of (1.1) such that

(6.2) $\rho(T)=\pm\begin{pmatrix}\epsilon&0\\\
0&\overline{\epsilon}\end{pmatrix},\qquad\epsilon=e^{2\pi i/12}.$

Since $S^{2}=-I$, we have $\rho(S)^{2}=\pm I$. The matrix $\rho(S)$ cannot be
equal to $\pm I$ as the relation $R=TS$ will imply that the eigenvalues of
$\rho(R)$ are $\pm e^{2\pi i/12}$ or $\pm e^{-2\pi i/12}$, which is absurd. It
follows that $\operatorname{tr}\rho(S)=0$ and we have

(6.3) $\rho(S)=\pm\begin{pmatrix}a&b\\\
c&-a\end{pmatrix},\qquad\rho(R)=\pm\rho(T)\rho(S)=\pm\begin{pmatrix}\epsilon
a&\epsilon b\\\ \bar{\epsilon}c&-a\bar{\epsilon}\end{pmatrix}$

for some $a,b,c\in\mathbb{C}$. Since $\rho(R)^{3}=\pm I$, $\det\rho(R)=1$, and
$\rho(R)\neq\pm I$ by a similar reason as above, the characteristic polynomial
of $\rho(R)$ has to be $x^{2}-x+1$ or $x^{2}+x+1$. In particular, we have
$\operatorname{tr}\rho(R)=\pm 1$, i.e., $a(\epsilon-\overline{\epsilon})=\pm
1$ and hence $a=\pm i$ and $bc=0$. Under the assumption that there is an
invariant metric realizing $Q(z)$, the matrices $\rho(S)$, $\rho(T)$, and
$\rho(R)$ must be unitary, after a simultaneous conjugation. (See the
discussion in Section 2.2.) If one of $b$ and $c$ is not $0$, this cannot
happen. Therefore, we have $b=c=0$. This implies that the function
$y_{1}(z)^{2}$, which is meromorphic throughout $\mathbb{H}$ since the local
exponents at every singularity are in $\frac{1}{2}\mathbb{Z}$, satisfies

$y_{1}(Tz)^{2}=e^{2\pi i/6}y_{1}(z)^{2},\qquad
y_{1}(Sz)^{2}=-z^{-2}y_{1}(z)^{2}.$

It follows that $y_{1}(z)^{2}$ is a meromorphic modular form of weight $-2$
with character $\chi$ on $\mathrm{SL}(2,\mathbb{Z})$. Likewise, we can show
that $y_{2}(z)^{2}$ is a meromorphic modular form of weight $-2$ with
character $\overline{\chi}$. This proves that if there is an invariant metric
realizing $Q(z)$, then there are solutions $y_{1}(z)$ and $y_{2}(z)$ with the
stated properties. The proof of the case $r_{\infty}=5/12$ is similar and is
omitted.

The proof of the converse statement is easy. If there exist solutions
$y_{1}(z)$ and $y_{2}(z)$ of (1.1) such that $y_{1}(z)^{2}$ and $y_{2}(z)^{2}$
are meromorphic modular forms of weight $-2$ with character $\chi$ and
$\overline{\chi}$, respectively, on $\mathrm{SL}(2,\mathbb{Z})$, then
$y_{1}(Tz)^{2}=e^{2\pi i/6}y_{1}(z)^{2}$ and $y_{2}(Tz)^{2}=e^{-2\pi
i/6}y_{2}(z)^{2}$, which implies that $y_{1}(z)^{2}$ and $y_{2}(z)^{2}$ are of
the form $y_{1}(z)^{2}=q^{1/6}\sum_{j\geq n_{0}}c_{j}q^{j}$ and
$y_{2}(z)^{2}=q^{-1/6}\sum_{j\geq n_{0}}d_{j}q^{j}$. It follows that
$r_{\infty}=1/12$ or $r_{\infty}=5/12$. It is clear that with respect to the
basis $\\{y_{1}(z),y_{2}(z)\\}$, the Bol representation is given by

$\rho(T)=\pm\begin{pmatrix}e^{2\pi i/12}&0\\\ 0&e^{-2\pi
i/12}\end{pmatrix},\qquad\rho(S)=\pm\begin{pmatrix}\pm i&0\\\
0&-i\end{pmatrix},$

and hence is unitary. It follows that there is an invariant metric of
curvature $1/2$ realizing $Q(z)$. This proves the theorem. ∎

We now give two examples with $r_{\infty}=1/12$, one of which can be realized
by some invariant metric of curvature $1/2$, while the other of which can not.
Note that Theorem 1 of [10] implies that when (1.1) does not have
$\mathrm{SL}(2,\mathbb{Z})$-inequivalent singularities outside $\\{i,\rho\\}$,
$1/12<r_{\infty}<5/12$ is the necessary and sufficient condition for the
existence of an invariant metric of curvature $1/2$ realizing $Q$. The
examples we provide below show that when (1.1) has
$\mathrm{SL}(2,\mathbb{Z})$-inequivalent singularities other than $i$ and
$\rho$, this condition is no longer a necessary condition.

######  Example.

Let $\eta(z)=q^{1/24}\prod_{n=1}^{\infty}(1-q^{n})=\Delta(z)^{1/24}$,

(6.4) $x(z)=\frac{E_{4}(z)}{\eta(z)^{8}}=q^{-1/3}+\cdots,\qquad
y(z)=\frac{E_{6}(z)}{\eta(z)^{12}}=q^{-1/2}+\cdots,$

and $h(z)=x(z)/y(z)=q^{1/6}+\cdots$. They are modular functions on the unique
normal subgroup $\Gamma$ of $\mathrm{SL}(2,\mathbb{Z})$ of index $6$ such that
$\mathrm{SL}(2,\mathbb{Z})/\Gamma$ is cyclic. (Another way to describe
$\Gamma$ is that $\Gamma=\ker\chi$, where $\chi$ is the character of
$\mathrm{SL}(2,\mathbb{Z})$ such that $\chi(S)=-1$ and $\chi(R)=e^{2\pi
i/3}$.) Using Ramanujan’s identities

$\begin{split}D_{q}E_{2}(z)&=\frac{E_{2}(z)^{2}-E_{4}(z)}{12},\\\
D_{q}E_{4}(z)&=\frac{E_{2}(z)E_{4}(z)-E_{6}(z)}{3},\\\
D_{q}E_{6}(z)&=\frac{E_{2}(z)E_{6}(z)-E_{4}(z)^{2}}{2},\end{split}$

where $D_{q}=qd/dq$ (see [22, Proposition 15]) and the relation
$\Delta(z)=(E_{4}(z)^{3}-E_{6}(z)^{2})/1728$, we can compute that

$\\{h(z),z\\}=(2\pi i)^{2}Q_{0}(z)$

where

$Q_{0}(z)=E_{4}(z)\left(-\frac{1}{72}-\frac{9(E_{4}(z)^{3}-E_{6}(z)^{2})^{2}}{(3E_{4}(z)^{3}-2E_{6}(z)^{2})^{2}}+\frac{5}{2}\frac{E_{4}(z)^{3}-E_{6}(z)^{2}}{3E_{4}(z)^{3}-2E_{6}(z)^{2}}\right).$

Thus,

$y_{+}(z)=\frac{h(z)}{\sqrt{D_{q}h(z)}}=q^{1/12}+\cdots,\quad
y_{-}(z)=\frac{1}{\sqrt{D_{q}h(z)}}=q^{-1/12}+\cdots$

are solutions of the differential equation $y^{\prime\prime}(z)=Q(z)y(z)$,
where $Q(z)=-(2\pi i)^{2}Q_{0}(z)/2$. The meromorphic modular form $Q(z)$ has
only one $\mathrm{SL}(2,\mathbb{Z})$-inequivalent singularity at the point
$z_{1}$ such that $3E_{4}(z_{1})^{3}-2E_{6}(z_{1})^{2}=0$ and is holomorphic
at the elliptic points $i$ and $\rho$. In the notation of Theorem 1.1, we have
$r_{\infty}=1/12$. This provides an example of an invariant metric of
curvature $1/2$ realizing a meromorphic modular form of weight $4$ with a
threshold $r_{\infty}$. Note that with respect to the basis
$\\{y_{+},y_{-}\\}$, the Bol representation is given by

$\rho(T)=\pm\begin{pmatrix}e^{2\pi i/12}&0\\\ 0&e^{-2\pi
i/12}\end{pmatrix},\qquad\rho(S)=\pm\begin{pmatrix}i&0\\\ 0&-i\end{pmatrix},$

both of which are unitary. (The information about $\rho(S)$ follows from the
transformation formula $\eta(-1/z)=\sqrt{z/i}\eta(z)$ and the fact that
$D_{q}h(z)=C\eta(z)^{4}(3E_{4}(z)^{3}-2E_{6}(z)^{2})/E_{6}(z)^{2}$ for some
constant $C$.)

######  Example.

Let $x(z)$ and $y(z)$ be defined by (6.4), and $\Gamma$ be the unique normal
subgroup of $\mathrm{SL}(2,\mathbb{Z})$ of index $6$ such that
$\mathrm{SL}(2,\mathbb{Z})/\Gamma$ is cyclic. The modular curve
$X(\Gamma):=\Gamma\backslash\mathbb{H}^{\ast}$ has one cusp of width $6$, no
elliptic points, and is of genus $1$. Since the modular functions $x(z)$ and
$y(z)$ on $\Gamma$ have only a pole of order $2$ and $3$, respectively, at the
cusp $\infty$ and are holomorphic elsewhere, they generate the function field
of $X(\Gamma)$. Then from the relation
$E_{4}(z)^{3}-E_{6}(z)^{2}=1728\eta(z)^{24}$, we see that $x(z)$ and $y(z)$
satisfies

$y^{2}=x^{3}-1728,$

which we may take as the defining equation for $X(\Gamma)$. Let $f(z)$ be a
meromorphic modular form of weight $2$ on $\Gamma$ such that all residues on
$\mathbb{H}$ are $0$. Equivalently, let $\omega=f(z)\,dz$ be a meromorphic
differential $1$-form of the second kind on $X(\Gamma)$. Consider

$y_{1}(z)=\frac{1}{\sqrt{f(z)}}\int_{z_{0}}^{z}f(u)\,du,\qquad
y_{2}(z)=\frac{1}{\sqrt{f(z)}},$

where $z_{0}$ is a fixed point in $\mathbb{C}$ that is not a pole of $f(z)$.
Under the assumption that all residues of $f(z)$ are $0$, the integral in the
definition of $y_{1}(z)$ does not depend on the choice of path of integration
from $z_{0}$ to $z$. A straightforward computation shows that the Wronskian of
$y_{1}$ and $y_{2}$ is a constant and hence $y_{1}(z)$ and $y_{2}(z)$ are
solutions of the differential equation $y^{\prime\prime}(z)=Q(z)y(z)$, where

$Q(z)=\frac{3f^{\prime}(z)^{2}-2f(z)f^{\prime\prime}(z)}{4f(z)^{2}}$

can be shown to be a meromorphic modular form of weight $4$ on $\Gamma$. (The
numerator of $Q(z)$ is a constant mulitple of the Rankin-Cohen bracket
$[f,f]_{2}$ and hence a mermomorphic modular form of weight $8$. See [9].) By
construction, this differential equation is apparent throughout $\mathbb{H}$.
Furthermore, if $f(z)$ is chosen in a way such that $f(\gamma
z)=\chi(\gamma)(cz+d)^{2}f(z)$ holds for all
$\gamma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\mathrm{SL}(2,\mathbb{Z})$ for some character
$\chi$ of $\mathrm{SL}(2,\mathbb{Z})$ with $\Gamma\subset\ker\chi$, then
$Q(z)$ is modular on $\mathrm{SL}(2,\mathbb{Z})$. We now utilize this
construction of modular differential equations to find $Q(z)$ that cannot be
realized, i.e., the monodromy group is not unitary.

We let $\omega_{1}=dx/y$ and $\omega_{2}=d(x/y^{3})$. Note that $\omega_{1}$
is a holomorphic $1$-form on the curve $y^{2}=x^{3}-1728$, while $\omega_{2}$
is an exact $1$-form and hence a meromorphic $1$-form of the second kind.
Using Ramanujan’s identities, we check that $\omega_{1}=f_{1}(z)\,dz$ and
$\omega_{2}=f_{2}(z)\,dz$ with

$f_{1}(z)=-\frac{2\pi i}{3}\eta(z)^{4},\quad f_{2}(z)=2\pi
i\frac{\eta(z)^{4}}{E_{6}(z)^{4}}\left(\frac{7}{6}E_{4}(z)^{3}\Delta(z)+576\Delta(z)^{2}\right).$

Now we choose, say,

$\omega=-\frac{3}{2\pi i}\left(\omega_{1}+\omega_{2}\right)$

and let $f(z)=q^{1/6}+\cdots$ be the meromorphic modular form of weight $2$
such that $\omega=f(z)\,dz$. Let $y^{\prime\prime}(z)=Q(z)y(z)$ be the
differential equation obtained from $f(z)$ using the construction described
above. Note that $f(z+1)=e^{2\pi i/6}f(z)$ and using
$\eta(-1/z)=\sqrt{z/i}\eta(z)$, we have $f(-1/z)=-z^{2}f(z)$. Thus, $f(\gamma
z)=\chi(\gamma)(cz+d)^{2}f(z)$ for all $\gamma=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)\in\mathrm{SL}(2,\mathbb{Z})$, where $\chi$ is the
character of $\mathrm{SL}(2,\mathbb{Z})$ such that $\chi(T)=e^{2\pi i/6}$ and
$\chi(S)=-1$. According the discussion above, the function $Q(z)$ is a
meromorphic modular form of weight $4$ with trivial character on
$\mathrm{SL}(2,\mathbb{Z})$. Note that $f(z)$ has zeros at points where
$6E_{6}(z)^{4}-7E_{4}(z)^{3}\Delta(z)-3456\Delta(z)^{2}=0$. Now let us compute
its Bol representation.

We choose $z_{0}=i\infty$ and find that

$y_{2}(z)=q^{-1/12}\left(1+\sum_{j=1}^{\infty}c_{j}q^{j}\right),\qquad
y_{1}(z)=q^{1/12}\sum_{j=0}^{\infty}d_{j}q^{j}$

for some $c_{j}$ and $d_{j}$ with $d_{0}\neq 0$. Therefore, the local
exponents at $\infty$ are $\pm 1/12$ and

$\rho(T)=\pm\begin{pmatrix}e^{2\pi i/12}&0\\\ 0&e^{-2\pi i/12}\end{pmatrix}.$

Also, since $f(-1/z)=-z^{2}f(z)$, we have

$\begin{split}\int_{i\infty}^{-1/z}f(u)\,du&=\int_{0}^{z}f(-1/u)\,\frac{du}{u^{2}}=-\int_{0}^{z}f(u)\,du\\\
&=-\int_{0}^{i\infty}f(u)\,du-\int_{i\infty}^{z}f(u)\,du.\end{split}$

Thus,

$\rho(S)=\pm\begin{pmatrix}i&C\\\ 0&-i\end{pmatrix},\qquad
C=i\int_{0}^{i\infty}f(u)\,du.$

Now recall that $\omega=f(z)\,dz$ is equal to $-3(\omega_{1}+\omega_{2})/(2\pi
i)$. Since $\omega_{2}=d(x/y^{3})$ is an exact $1$-form on $X(\Gamma)$ and the
modular curve $X(\Gamma)$ has only one cusp, which in particular says that
$\infty$ and $0$ are mapped to the same point on $X(\Gamma)$ under the natural
map $\mathbb{H}^{\ast}\to X(\Gamma)$, the integral
$\int_{0}^{i\infty}f_{2}(u)\,du$ is equal to $0$. Therefore, we have

$C=i\int_{0}^{i\infty}\eta(u)^{4}\,du.$

This constant $C$ can be expressed in terms of the central value of the
$L$-function of the elliptic curve $E:y^{2}=x^{3}-1728$, which is known to be
nonzero. From this, it is straightforward to check that there is no
simultaneous conjugation such that $\rho(T)$ and $\rho(S)$ both become
unitary.

###### Proof of Theorem 1.4.

We use the notations in the proof of Theorem 1.1. Since $\kappa_{\infty}=n/4$
for some odd integer $n$, with respect to the basis $\\{y_{+}(z),y_{-}(z)\\}$,
we have $\rho(T)=\pm\left(\begin{smallmatrix}i&0\\\
0&-i\end{smallmatrix}\right)$. If $\rho(S)=\pm I$, then
$\rho(R)=\pm\left(\begin{smallmatrix}i&0\\\ 0&-i\end{smallmatrix}\right)$,
which is a contradiction to $\rho(R)^{3}=\pm I$. Hence, $\rho(S)\neq\pm I$,
and we have $\operatorname{tr}\rho(S)=0$. Then, by a choosing a suitable
scalar $r$, the matrix of $\rho(S)$ with respect to $\\{ry_{+}(z),y_{-}(z)\\}$
will be of the form

$\rho(S)=\pm\begin{pmatrix}a&b\\\ b&-a\end{pmatrix}$

for some $a,b\in\mathbb{C}$ with $a^{2}+b^{2}=-1$, while $\rho(T)$ is still
$\pm\left(\begin{smallmatrix}i&0\\\ 0&-1\end{smallmatrix}\right)$. Set
$F(z)=r^{2}y_{+}(z)^{2}+y_{-}(z)^{2}$. We then compute that $F(Tz)=-F(z)$ and

$\begin{split}\left(F|_{-2}S\right)(z)&=(ary_{+}(z)+by_{-}(z))^{2}+(bry_{+}(z)-ay_{-}(z))^{2}\\\
&=-r^{2}y_{1}(z)^{2}-y_{2}(z)^{2}=-F(z).\end{split}$

This proves the theorem. ∎

##  7\. Existence of the curvature equation

In this section, we will prove Theorem 1.3 equipped with the data (1.7). The
main purpose of this section is to prove the existence and the number of such
$Q$ equipped with data (1.7). The discussion will be divided into several
cases depending on $\kappa_{\rho}$ and $\kappa_{i}$.

######  Lemma 7.1.

Suppose $F(z)$ is a modular form of weight $4$ with respect to
$\mathrm{SL}(2,\mathbb{Z})$, and is holomorphic except at $\rho$ and $i$. If
the pole order of $F(z)$ at $\rho$ or $i$ $\leq 1$, then $F(z)$ is
holomorphic.

###### Proof.

Let $n_{1}$ and $n_{2}$ be the orders of poles at $i$ and $\rho$ respectively.
The counting zero formula of meromorphic modular form (see [19]) says

$m-\frac{n_{1}}{2}-\frac{n_{2}}{3}=\frac{4}{12},\qquad m\ \text{is a non-
negative integer}.$

By the assumption, $n_{i}\leq 1$. From the identity, it is easy to see
$n_{1}\leq 0$ and $n_{2}\leq 0$. ∎

Let $t_{j}=E_{6}(z_{j})^{2}/E_{4}(z_{j})^{3}$ and define
$F_{j}(z)=E_{6}(z)^{2}-t_{j}E_{4}(z)^{3}$. By the theorem of counting zeros of
modular forms [19, p. 85, Theorem 3], $F_{j}(z)$ has a (simple) zero at
$z_{j}\in\mathbb{H}$.

######  Lemma 7.2.

Suppose that $Q$ satisfies the conditions (i) and (ii) in Definition 1.2. Then

(7.1)
$\begin{split}&Q=\pi^{2}\left(Q_{3}(z;r,s,t)+\sum^{m}_{j=1}\frac{r_{1}^{(j)}E_{4}(z)^{4}F_{j}(z)+r^{(j)}_{2}E_{4}(z)^{7}}{F_{j}(z)^{2}}\right),\end{split}$

where $r$, $r^{(j)}_{1}$ are free parameters and $s$, $t$, $r^{(j)}_{2}$ are
uniquely determined by

(7.2) $\begin{split}&s=s_{\kappa_{\rho}}:=(1-4\kappa^{2}_{\rho})/9,\quad
t=t_{\kappa_{i}}=(1-4\kappa^{2}_{i})/4,\quad\text{and}\\\
&r^{(j)}_{2}=r^{(j)}_{2,\kappa_{j}}=t_{j}(t_{j}-1)^{2}(1-4\kappa^{2}_{j})/4.\end{split}$

###### Proof.

Let $\hat{Q}$ denote the RHS of (7.1). Then it is a straightforward
computation to show that (ii) in Definition 1.2 holds at $p_{j}$ if and only
if $s=s_{\kappa_{\rho}}$ if $p_{j}=\rho$, $t=t_{\kappa_{i}}$ if $p_{j}=i$, and
$r^{(j)}_{2}=r^{(j)}_{2,\kappa_{j}}$ if $p_{j}=z_{j}$. By the choice of $s$,
$t$ and $r^{(j)}_{2}$, $Q-\hat{Q}$ might contain simple poles only. Further,
we can choose $r^{(j)}_{1}$ to make $Q-\hat{Q}$ holomorphic at $z_{j}$. By
Lemma 7.1, $Q-\hat{Q}$ is automatically smooth at $\rho$ and $i$. Therefore,
$Q-\hat{Q}$ is a holomorphic modular form of weight $4$, and the lemma follows
immediately because $E_{4}(z)$, up to a constant, is the only holomorphic
modular form of weight $4$. ∎

Now we are in the position to prove Theorem 1.3.

###### Proof of Theorem 1.3.

We first calculate the parameters $r,r^{(j)}_{1}$, $1\leq j\leq m$, such that
$Q$ is apparent at $z_{j}$. For simplicity, we assume $j=1$. From (7.1), we do
the Taylor expansion at $z=z_{1}$.

$\displaystyle Q(z)=a_{-2}(z-z_{1})^{-2}+(r_{1}b_{-1}+a_{-1})(z-z_{1})^{-1}$
$\displaystyle+\sum^{\infty}_{j=0}\left(a_{j}+r_{1}b_{j}+c_{j}\left(r,r^{(2)}_{1},\ldots,r^{(m)}_{1}\right)\right)\left(z-z_{1}\right)^{j}:=\sum^{\infty}_{j=-2}A_{j}\left(z-z_{1}\right)^{j},$

where $a_{j},b_{j}$ are independent of $r$, $r^{(j)}_{1}$ and
$c_{j}(r,r^{(2)}_{1},\ldots,r^{(m)}_{1})$ is linear in all variables, and also

$y(z)=(z-z_{1})^{1/2-\kappa_{1}}\left(1+\sum^{\infty}_{j=1}d_{j}(z-z_{1})^{j}\right).$

Then we derive the recursive formula by comparing both sides of (1.1) with $Q$
in (7.1),

(7.3) $j(j-2\kappa_{1})d_{j}=\sum_{k+\ell=j-2,\ k<j}d_{k}A_{\ell},\quad
A_{-1}=a_{-1}+r_{1}b_{-1},$

where $d_{0}=1$ and

$d_{1}=\frac{1}{1-2\kappa_{1}}d_{0}A_{-1}=\frac{b_{-1}}{1-2\kappa_{1}}r_{1}+\text{terms\
of\ lower\ orders}.$

By induction,

(7.4) $\displaystyle
j(j-2\kappa_{1})d_{j}=d_{j-1}A_{-1}+d_{j-2}A_{0}+d_{j-3}A_{1}+\cdots+d_{0}A_{j-2}$
$\displaystyle=\frac{b^{j-1}_{-1}}{(1-2\kappa_{1})\cdots\left((j-1)-2\kappa_{1}\right)}r^{j-1}_{1}+\
\text{terms\ of\ lower\ orders}.$

At $j=2k_{1}$, the RHS of (7.4) is

$P_{1}\left(r,r^{(1)}_{1},\ldots,r^{(m)}_{1}\right):=d_{2\kappa_{1}-1}A_{-1}+d_{2\kappa_{1}-2}A_{0}+\cdots+d_{0}A_{\kappa_{1}-2}.$

Clearly, $\deg P_{1}=2\kappa_{1}$ and

(7.5) $P_{1}=B_{0}r^{2\kappa_{1}}_{1}+\text{terms\ of\ lower\ orders},\quad
B_{0}\neq 0.$

We summarized what are known:

  1. $\bullet$

$\kappa_{i}\not\in\mathbb{N}$, then $Q$ is apparent at $i$ for any tuple
$\left(r,r^{(j)}_{1}\right)$.

  2. $\bullet$

$2\kappa_{p}/3\not\in\mathbb{N}$, then $Q$ is apparent at $\rho$ for any tuple
$\left(r,r^{(j)}_{1}\right)$,

  3. $\bullet$

$1/2\pm\kappa_{j}$, there is a polynomial
$P_{j}\left(r,r^{(1)}_{1},\ldots,r^{(m)}_{1}\right)$ of degree $2\kappa_{j}$
such that $Q$ is apparent at $z_{j}$ if and only if
$P_{j}\left(r,r^{(1)}_{1},\ldots,r^{(m)}_{1}\right)=0$.

Since $\kappa_{\infty}$ is given, we have
$\kappa_{\infty}=\sqrt{-Q(\infty)}/2$, and then

(7.6) $r+\sum^{m}_{j=1}\left(1-t_{j}\right)r^{(1)}_{j}+e=0,$

where $e$ is given. By Bezout’s theorem, we have
$N=\prod^{m}_{j=1}(2\kappa_{j})$ common roots with multiplicity of (7.5) and
(7.6) because by (7.5) it is easy to see that there are no solutions at
$\infty$. This proves the theorem.

∎

##  Appendix A Curvature equations on $S^{2}$ with multiple singularities

Let $\mathbb{H}^{\ast}=\mathbb{H}\cup\mathbb{Q}\cup\\{\infty\\}$. Since
$\mathrm{SL}(2,\mathbb{Z})\backslash\mathbb{H}^{\ast}\simeq\mathbb{C}\cup\\{\infty\\}$,
the equation (2.5) in the case $\Gamma=\mathrm{SL}(2,\mathbb{Z})$ can be
transformed into the mean field equations on $\mathbb{C}$:

(A.1) $\begin{cases}\Delta
u+e^{u}=4\pi\left(\alpha_{1}\delta_{0}+\alpha_{2}\delta_{1}+\sum^{m}_{j=1}n_{j}\delta_{p_{j}}\right)\quad\text{on}\
\mathbb{C},\\\ u(z)=-(4+2\alpha_{3})\log\left|z\right|+O(1)\quad\text{as}\
\left|z\right|\rightarrow\infty,\end{cases}$

where we assume that the isomorphism maps the points $i=\sqrt{-1}$,
$\rho=(1+\sqrt{-3})/2$, and $\infty$ of
$\mathrm{SL}(2,\mathbb{Z})\backslash\mathbb{H}^{\ast}$ to $0$, $1$, and
$\infty$, respectively, $\delta_{p}$ is the Dirac measure at $p\in\mathbb{C}$,
$\alpha_{k}>-1$ for $k=1,2,3$ and $n_{j}\in\mathbb{N}$. For any solution $u$
of (A.1), the conformal metric $e^{u}|dz|^{2}$ has the angles $\lambda_{1}$,
$\lambda_{2}$, and $\sigma_{j}$ at $0$, $1$, and $p_{j}$, respectively, where

(A.2)
$\lambda_{1}=\alpha_{1}+1,\qquad\lambda_{2}=\alpha_{2}+1,\qquad\sigma_{j}=n_{j}+1.$

Throughout the appendix, we assume that $\alpha_{k}$ are not integers for
$k=1,2,3$ and all $p_{j}$ are distinct. To find a solution for (A.1), we first
associate to (A.1) a second-order ODE

(A.3) $y^{\prime\prime}(z)+Q(z)y(z)=0,\qquad z\in\mathbb{C},$

where

(A.4) $\displaystyle Q(z)=$
$\displaystyle\left(\frac{\tfrac{\alpha_{1}}{2}(\tfrac{\alpha_{1}}{2}+1)}{z^{2}}+\frac{r_{1}}{z}\right)+\left(\frac{\tfrac{\alpha_{2}}{2}(\tfrac{\alpha_{2}}{2}+1)}{(z-1)^{2}}+\frac{r_{2}}{z-1}\right)$
$\displaystyle\qquad+\sum_{j=1}^{m}\frac{\tfrac{n_{j}}{2}(\tfrac{n_{j}}{2}+1)}{(z-p_{j})^{2}}+\frac{s_{j}}{z-p_{j}}$

for some free parameters $r_{0},r_{1},s_{j}$. It is known that (A.1) has a
solution if and only if the monodromy group of (A.3) is projectively unitary.

Note that the local exponents of (A.3) at $0$ and $1$ are
$\\{-\alpha_{1}/2,1+\alpha_{1}/2\\}$ and $\\{-\alpha_{2}/2,1+\alpha_{2}/2\\}$,
respectively. Since $\alpha_{1},\alpha_{2}\notin\mathbb{Z}$, the differences
of the local exponents are not integers. At each $p_{j}$, there is a
polynomial $P_{j}(r_{1},r_{2},s_{j})$ such that (A.3) is apparent if and only
if $P_{j}(r_{1},r_{2},s_{j})=0$. The derivation of the polynomials $P_{j}$ is
the same as Lemma 7.2. Moreover, the asymptotic behavior of $u$ at $\infty$
yields that (A.3) is Fuchsian at $\infty$ with local exponents $-\alpha_{3}/2$
and $1+\alpha_{3}/2$. Thus, we have

$r_{1}+r_{2}+\sum_{j}s_{j}=0$

and

$\begin{split}\frac{\alpha_{\infty}}{2}(\frac{\alpha_{\infty}}{2}+1)=&\lim_{z\to\infty}z^{2}Q(z)\\\
=&r_{1}+\sum_{j=1}^{m}s_{j}p_{j}+\sum_{k\in\\{0,1\\}}\frac{\alpha_{k}}{2}(\frac{\alpha_{k}}{2}+1)+\sum_{j=1}^{m}\frac{n_{j}}{2}(\frac{n_{j}}{2}+1).\end{split}$

Therefore, for given local exponent data for (A.1), the Bézout theorem implies
that there are at most $\prod_{j=1}^{m}(n_{j}+1)$ distinct $Q$ such that (A.3)
realizes the mean field equation (A.1) for given data. Theorem 2.5 of [12] is
to give a necessary and sufficient condition to ensure that the projective
monodromy group of (A.3) is unitary, i.e., that (A.1) has a solution.

######  Theorem A.1 ([12, Theorem 2.5]).

Suppose that $\alpha_{1},\alpha_{2},\alpha_{3}$ are not integers and all
combinations

(A.5) $\alpha_{1}\pm\alpha_{2}\pm\alpha_{3}\text{ are not integers}$

for any choice of signs. Then (A.1) has a solution if and only if

$\cos^{2}\pi\alpha_{1}+\cos^{2}\pi\alpha_{2}+\cos^{2}\pi\alpha_{3}+2(-1)^{\sigma+1}\cos\pi\alpha_{1}\cos\pi\alpha_{2}\cos\pi\alpha_{3}<1,$

where $\sigma=\sum_{j=1}^{m}n_{j}$. Moreover, the number of distinct solutions
of (A.1) is less than or equal to $\prod_{j=1}^{m}(n_{j}+1)$.

We remark that the notations $\alpha_{j}$ here differ from those used in [12]
by $1$.

Note that when (A.1) arises from the differential equation (1.1) considered in
Theorem 1.1, we have

$\alpha_{1}=\kappa_{i}-1,\quad\alpha_{2}=2\kappa_{\rho}/3-1,\quad\alpha_{3}=2\kappa_{\infty},\quad
n_{j}=2\kappa_{p_{j}}-1,$

where $\kappa_{i},\kappa_{\rho},\kappa_{p_{j}}\in\frac{1}{2}\mathbb{N}$ are
the local exponent data in ($\mathbf{H}_{1}$) and ($\mathbf{H}_{2}$). Hence,
$\alpha_{1}\in\frac{1}{2}+\mathbb{Z}$ and
$\alpha_{2}=\pm\frac{1}{3}+\mathbb{Z}$ and the condition (A.5) is equivalent
to $r_{\infty}\neq 1/12,5/12$. Thus, the first half of Theorem 1.1 is a
special case of Eremenko and Tarasov’s theorem. In the remainder of the
appendix, we provide an alternative and self-contained proof of Theorem A.1.

For $k=1,2,3$, let $\theta_{k}\in(0,1/2]$ be real numbers such that

(A.6) $\alpha_{k}\equiv\pm\theta_{k}\ \mathrm{mod}\
1,\quad\text{and}\quad\alpha_{k}=\ell_{k}\pm\theta_{k}.$

Let $S=\\{0,1,\infty,p_{1},\ldots,p_{m}\\}$ be the set of singular points of
(A.3). Choose a base point $z_{0}$ near $\infty$ and consider the monodromy
represenation $\rho:\pi_{1}(\mathbb{C}\setminus
S,z_{0})\to\mathrm{SL}(2,\mathbb{C})$ of (A.3). Let
$\beta_{j},\gamma_{k}\in\pi_{1}(\mathbb{C}\setminus S,z_{0})$ such that
$\beta_{j}$, $1\leq j\leq m$, (resp. $\gamma_{0}$, $\gamma_{1}$) is a simple
loop encircling $p_{j}$ (resp. $0$, $1$) counterclockwise, while
$\gamma_{\infty}$ is a simple loop around $\infty$ clockwise such that

$\gamma_{0}\gamma_{1}\prod_{j=1}^{m}\beta_{j}=\gamma_{\infty},\quad\text{in
}\pi_{1}(\mathbb{C}\setminus S,z_{0}).$

Since the local exponents at $\infty$ are $\\{-\alpha_{3}/2,1+\alpha_{3}/2\\}$
with $\alpha_{3}=\ell_{3}\pm\theta_{3}$ and any solution has no logarithmic
singularities, we can choose local solutions $y_{\infty,+}$, $y_{\infty,-}$
near $\infty$ such that with respect to $(y_{\infty,+},y_{\infty,-})$, the
monodromy matrix $\rho(\gamma_{\infty})$ is given by

(A.7) $\begin{split}\rho(\gamma_{\infty})&=\begin{pmatrix}e^{\pi
i(\theta_{3}\pm\ell_{3})}&0\\\ 0&e^{-\pi
i(\theta_{3}\pm\ell_{3})}\end{pmatrix}\\\
&=(-1)^{\ell_{3}}\begin{pmatrix}e^{\pi i\theta_{3}}&0\\\ 0&e^{-\pi
i\theta_{3}}\end{pmatrix}=:(-1)^{\ell_{3}}T.\end{split}$

For any $1\leq j\leq m$, since the local exponents at $p_{j}$ are
$\\{-n_{j}/2,1+n_{j}/2\\}$ with $n_{j}\in\mathbb{N}$, we see that the
monodromy matrix $\rho(\beta_{j})$ is $(-1)^{n_{j}}I_{2}$. Set

(A.8) $R:=(-1)^{\ell_{1}}\rho(\gamma_{0})^{-1},\qquad
S:=(-1)^{\ell_{2}}\rho(\gamma_{1}).$

We have

$(-1)^{\ell_{1}+\ell_{2}}R^{-1}S\prod_{j}^{m}(-1)^{n_{j}}I_{2}=(-1)^{\ell_{3}}T,$

i.e.,

(A.9) $S=(-1)^{\sum_{j}n_{j}+\sum_{k}\ell_{k}}RT.$

Let $R$, $S$, and $T$ be three matrices in $\mathrm{SL}(2,\mathbb{C})$ such
that

  1. (i)

the eigenvalues of $R$, $S$, and $T$ are $\delta_{1}^{\pm 1}$,
$\delta_{2}^{\pm 1}$ and $\delta_{3}^{\pm 1}$, respectively, where
$\delta_{j}=e^{\pm\pi i\theta_{j}}$ with $0<\theta_{j}<1$ and $i=\sqrt{-1}$,

  2. (ii)

the triple $(\theta_{1},\theta_{2},\theta_{3})$ satisfies

$0<\theta_{i}+\theta_{j}\leq 1,\quad\forall i\neq j,$

and

  3. (iii)

$\theta_{3}=\max_{1\leq j\leq 3}\theta_{j}$ and
$T=\operatorname{diag}(\delta_{3},\bar{\delta}_{3})=\left(\begin{smallmatrix}\delta_{3}&0\\\
0&\bar{\delta}_{3}\end{smallmatrix}\right)\in\mathrm{SU}(2,\mathbb{C})$.

######  Lemma A.2.

Suppose $R=\left(\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix}\right)$, $T$,
$S=RT\in\mathrm{SL}(2,\mathbb{C})$ satisfy (i)-(iii). Then the following hold.

  1. (a)

$\left|a\right|<1$ if and only if $\theta_{1}+\theta_{2}>\theta_{3}$.

  2. (b)

$\left|a\right|=1$ if and only if $\theta_{1}+\theta_{2}=\theta_{3}$.

###### Proof.

Note

$S=RT=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\begin{pmatrix}\delta_{3}&0\\\
0&\bar{\delta}_{3}\end{pmatrix}=\begin{pmatrix}\delta_{3}a&b\bar{\delta}_{3}\\\
\delta_{3}c&d\bar{\delta}_{3}\end{pmatrix}.$

Using the invariance of $\operatorname{tr}R$ and $\operatorname{tr}S$ under
conjugation, we have

$\begin{cases}a+d=\delta_{1}+\bar{\delta}_{1}\in\mathbb{R},\\\
\delta_{3}a+\bar{\delta}_{3}d=\delta_{2}+\bar{\delta}_{2}\in\mathbb{R}.\end{cases}$

Since $\delta_{3}\neq\pm 1$, we easily obtain

(A.10) $d=\bar{a},\quad
a=\frac{\delta_{2}+\bar{\delta}_{2}-\bar{\delta}_{3}(\delta_{1}+\bar{\delta}_{1})}{\delta_{3}-\bar{\delta}_{3}}.$

Consequently,

$\displaystyle
a=\frac{2\cos\pi\theta_{2}-2\bar{\delta}_{3}\cos\pi\theta_{1}}{\pm
2i\sin\pi\theta_{3}}=\pm\frac{i(\bar{\delta}_{3}\cos\pi\theta_{1}-\cos\pi\theta_{2})}{\sin\pi\theta_{3}}.$

Thus

(A.11)
$\begin{split}\left|a\right|^{2}&=\frac{(\bar{\delta}_{3}\cos\pi\theta_{1}-\cos\pi\theta_{2})(\delta_{3}\cos\pi\theta_{1}-\cos\pi\theta_{2})}{\sin^{2}\pi\theta_{3}}\\\
&=\frac{\cos^{2}\pi\theta_{1}-2\cos\pi\theta_{1}\cos\pi\theta_{3}\cos\pi\theta_{2}+\cos^{2}\pi\theta_{2}}{\sin^{2}\pi\theta_{3}}.\end{split}$

Let

$\displaystyle\Delta$
$\displaystyle:=\cos^{2}\pi\theta_{1}-2\cos\pi\theta_{1}\cos\pi\theta_{2}\cos\pi\theta_{3}+\cos^{2}\pi\theta_{2}-\sin^{2}\pi\theta_{3}$
$\displaystyle=\cos^{2}\pi\theta_{1}+\cos^{2}\pi\theta_{2}+\cos^{2}\pi\theta_{3}-(1+2\cos\pi\theta_{1}\cos\pi\theta_{2}\cos\pi\theta_{3}).$

Then (A.11) implies that $\Delta<0$ if and only if $\left|a\right|<1$.

Now using the formulas $\cos(x+y)=\cos x\cos y-\sin x\sin y$ and
$\cos^{2}x=(1+\cos(2x))/2$, we deduce that

$\begin{split}\Delta&=\cos^{2}\pi\theta_{3}-\cos\pi\theta_{3}(\cos\pi(\theta_{1}+\theta_{2})+\cos\pi(\theta_{1}-\theta_{2}))\\\
&\qquad+\frac{1}{2}(\cos(2\pi\theta_{1})+\cos(2\pi\theta_{2}))\\\
&=\cos^{2}\pi\theta_{3}-\cos\pi\theta_{3}(\cos\pi(\theta_{1}+\theta_{2})+\cos\pi(\theta_{1}-\theta_{2}))\\\
&\qquad+\cos\pi(\theta_{1}+\theta_{2})\cos\pi(\theta_{1}-\theta_{2}),\end{split}$

so

$\Delta=\left(\cos\pi\theta_{3}-\cos\pi(\theta_{1}+\theta_{2})\right)\left(\cos\pi\theta_{3}-\cos\pi(\theta_{1}-\theta_{2})\right).$

Since the assumptions (i)-(iii) give $1>\theta_{3}>|\theta_{1}-\theta_{2}|$,
we have $\cos\pi\theta_{3}-\cos\pi(\theta_{1}-\theta_{2})<0$, so the desired
results follow. The proof is complete. ∎

We now give an alternative proof of Theorem A.1, which is stated in the
following equivalent form.

######  Theorem A.3.

Assume that (A.5) holds.

  1. (a)

Suppose that $\sum_{k=1}^{3}\ell_{k}+\sum^{m}_{j=1}n_{j}$ is an even integer.
Then (A.1) has a solution if and only if $\theta_{i}+\theta_{j}>\theta_{k}$
for any $i\neq j\neq k$.

  2. (b)

Suppose that $\sum_{k=1}^{3}\ell_{k}+\sum^{m}_{j=1}n_{j}$ is an odd integer.
Then (A.1) has a solution if and only if $\theta_{1}+\theta_{2}+\theta_{3}>1$.

###### Proof.

Let $R$, $S$, and $T$ be defined by (A.7) and (A.8). We need to determine when
they are simultaneously conjugate to unitary matrices, under the assumption
that (A.5) holds.

Consider first the case $\sum\ell_{k}+\sum n_{j}$ is even. In such a case, we
have $S=RT$. Since for any permutation $\tau$ of the three points $0$, $1$,
and $\infty$, there is always a Möbius transformation $\gamma$ satisfying
$\gamma z=\tau(z)$ for all $z\in\\{0,1,\infty\\}$, without loss of generality,
we may assume that $\theta_{3}=\max_{k}\theta_{k}$. Then the condition
$\theta_{i}+\theta_{j}>\theta_{k}$ for any $i\neq j\neq k$ simply means
$\theta_{1}+\theta_{2}>\theta_{3}$, which we assume now. Moreover, we may
assume that $T=\left(\begin{smallmatrix}\delta_{3}&0\\\
0&\overline{\delta}_{3}\end{smallmatrix}\right)$ after a common conjugation,
where $\delta_{3}=e^{\pi i\theta_{3}}$.

Write $R=\left(\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix}\right)$. By
(A.10), we have $d=\overline{a}$. By Lemma A.2,
$\theta_{1}+\theta_{2}>\theta_{3}$ if and only if $|a|<1$ and hence
$bc=|a|^{2}-1<0$. Set $P=\left(\begin{smallmatrix}\mu&0\\\
0&1\end{smallmatrix}\right)$, where $\mu$ is a real number such that

$\mu^{2}=-\frac{bc}{|c|^{2}}=-\frac{b}{\overline{c}}.$

We have $P^{-1}TP=T$ and

$P^{-1}RP=\begin{pmatrix}a&\mu^{-1}b\\\ \mu c&d\end{pmatrix},$

which is unitary since $\mu^{-1}b=-\mu\overline{c}=-\overline{\mu c}$. This
proves that if $\theta_{1}+\theta_{2}>\theta_{3}$, then (A.1) has a solution.

Conversely, suppose that (A.1) has a solution. Then there exists a matrix $P$
such that $\hat{T}=P^{-1}TP$ and $\hat{R}=P^{-1}RP$ are both unitary. Now it
is known that every matrix in $\mathrm{SU}(2,\mathbb{C})$ is conjugate to a
diagonal matrix and the conjugation can be taken inside
$\mathrm{SU}(2,\mathbb{C})$. Hence, there exists a matrix $Q$ in
$\mathrm{SU}(2,\mathbb{C})$ such that $Q^{-1}\hat{T}Q=T$. Then
$Q^{-1}\hat{R}Q\in\mathrm{SU}(2,\mathbb{C})$. In particular, the $(1,1)$-entry
of $Q^{-1}\hat{R}Q$ has absolute value $\leq 1$. Since $PQ$ commutes with $T$
and $T$ is diagonal but not a scalar matrix, $PQ$ must be a diagonal matrix.
Therefore, the $(1,1)$-entry of $R$ also has absolute value $\leq 1$. It
follows that, by Lemma A.2, $\theta_{1}+\theta_{2}>\theta_{3}$ (as the case
$\theta_{1}+\theta_{2}=\theta_{3}$ is excluded from our consideration by
(A.5)). We conclude that under the assumptions that (A.5) holds and that
$\sum\ell_{k}+\sum n_{j}$ is even, (A.1) has a solution if and only if
$\theta_{i}+\theta_{j}>\theta_{k}$ for any $i\neq j\neq k$.

For the case $\sum\ell_{k}+\sum n_{j}$ is odd, we simply apply the result in
Part (a) to $\theta_{1},\theta_{2},1-\theta_{3}$ with $T$ replaced by $-T$ and
conclude that (A.1) has a solution if and only if
$\theta_{1}+\theta_{2}+\theta_{3}>1$. This completes the proof. ∎

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