# Stokes graphs of the Rabi problem with real parameters

René Langøen  Department of Mathematics, University of Bergen, Norway
<EMAIL_ADDRESS>,  Irina Markina  Department of Mathematics, University of
Bergen, Norway<EMAIL_ADDRESS>and  Alexander Solynin  Department of
Mathematics and Statistics, Texas Tech University, Lubbock, TX, USA
<EMAIL_ADDRESS>
###### Abstract.

The goal of this paper is to study the geometry of the Stokes graphs
associated with the problem, which was introduced by Isidor Rabi in 1937 to
model reactions of atoms to the harmonic electric field with frequency close
to the natural frequency of the atoms. In the standard Garnier form, the Rabi
model is a matrix linear differential equation with three physical parameters,
which are: the level of separation of the fermion mode $\Delta$, the boson-
fermion coupling $g$, and the eigenvalue $E$ of the Hamiltonian relevant to
this model. The qualitative behavior of solutions of this type of problems is
often described in terms of the Stokes graphs of associated quadratic
differential, which in the case of Rabi problem can be represented in the form
$Q_{0}(z)\,dz^{2}=-\frac{z^{4}+c_{3}z^{3}+c_{2}z^{2}+c_{1}z+c_{0}}{(z-1)^{2}(z+1)^{2}}\,dz^{2}$
with the coefficients $c_{k}$, $k=0,1,2,3$, depending on the parameters
$\Delta$, $g$, and $E$. In this paper, we first give a complete classification
of possible generic topological types of domain configurations and Stokes
graphs of this quadratic differential assuming that its coefficients $c_{k}$
are real and the zeros of its numerator are distinct from its poles. Then we
identify the set of coefficients $(c_{3},c_{2},c_{1},c_{0})\in\mathbb{R}^{4}$,
which correspond to particular choices of the physical parameters $\Delta$,
$g$, and $E$. The structure of Stokes graphs and domain configurations of
quadratic differentials, which appear as asymptotic cases when the parameters
of the Rabi problem tend to infinity, also will be discussed.

###### Key words and phrases:

Rabi problem, Stokes graph, quadratic differential, critical trajectory

###### 2020 Mathematics Subject Classification:

34M40, 34M65, 30C10

###### Contents

  1. 1 Introduction
  2. 2 Rabi model and associated quadratic differential
  3. 3 Basics on quadratic differentials
  4. 4 Stokes graphs and domain configurations of $Q_{0}(z)\,dz^{2}$
  5. 5 Domain configurations for the Rabi model
  6. 6 Asymptotic behavior for rescaled Rabi problem
  7. A List of notation
  8. B Zoo of Stokes graphs.

##  1\. Introduction

In this work, we study geometry of Stokes graphs and domain configurations of
quadratic differentials associated with the Rabi problem. These graphs and
configurations provide important information on the qualitative behavior of
solutions to this problem. The problem was introduced by Isidor Issak Rabi
[12] as a model describing how a rapidly varying weak magnetic field affects
an oriented atom possessing nuclear spin. Thus, the Rabi problem deals with
reactions of the atom to the harmonic electric field with frequency close to
the atom’s natural frequency. Despite its simplicity, the quantum Rabi model
is not exactly solvable. The problem, originated in mathematical physics as a
model describing a simple harmonic oscillator or two level quantum system
[22], has obtained numerous applications; in particular, in the theory of
quantum computing and other areas of quantum mechanics under different
coupling regimes, see [21].

The Rabi model in the standard Garnier form [8] is a system of linear matrix
equations on the Riemann sphere, where the coefficient matrix has two simple
poles at $p_{1},p_{2}\in\mathbb{C}$ and a pole of order two at $\infty$. It
has three physical parameters: $\Delta\in\mathbb{R}$ is the level of
separation of the fermion mode, $g\in\mathbb{C}$ is the boson-fermion
coupling, and $E\in\mathbb{C}$ is the eigenvalue of the Hamiltonian defined by
the physical problem. The detailed description of the Rabi problem will be
given in Section 2. The governing equation of this problem is a second order
linear ODE, which after linear change of variables can be written in the form

(1.1) $\frac{d^{2}}{dz^{2}}y(z)+Q(\Delta,E,g,z)y(z)=0,$

where
$Q(\Delta,E,g,z)=-(1/4)(z^{4}+a_{3}z^{3}+a_{2}z^{2}+a_{1}z+a_{0})z^{-2}(z+4g^{2})^{-2}$
is a rational function with coefficients $a_{k}=a_{k}(\Delta,E,g)$,
$k=0,1,2,3$, depending on the parameters of the Rabi problem.

The general scheme to study the equation of type (1.1) includes the following
3 steps.

  * 1.

Construction of the Stokes graph embedded in $\overline{\mathbb{C}}$ and
identification of its faces. We recall that the Stokes graph of equation (1.1)
is the graph consisting of the critical trajectories of the quadratic
differential

(1.2)
$Q(\Delta,E,g,z)\,dz^{2}=-\frac{1}{4}\frac{z^{4}+a_{3}z^{3}+a_{2}z^{2}+a_{1}z+a_{0}}{z^{2}(z+4g^{2})^{2}}\,dz^{2},\quad
z\in\mathbb{C}\cup\\{\infty\\}.$

  * 2.

Finding the fundamental solution in each domain that is a face of the Stokes
graph.

  * 3.

Identifying the matrices relating the fundamental solutions defined in
different domains in order to get the “global” fundamental solution.

In this paper, we focus on the first step of this general scheme assuming that
the parameters $\Delta$, $E$, and $g^{2}\not=0$ of the Rabi problem are real
numbers. Thus, in our study we assume that the boson-fermion coupling
$g\not=0$ is either real or pure imaginary nonzero number. These our
assumptions imply, in turn, that the coefficients $a_{k}$ of the quadratic
differential (1.2) are real numbers.

The relations between properties of solutions of certain ODE’s and the
structure of critical trajectories of related quadratic differentials were
explored by many authors working with differential equations. One of the
primary references here is the monograph [4] by M. V. Fedoryuk. For more
recent results on applications of quadratic differentials to specific
differential equations, we refer to the papers [11], [13], and [1].

We structure our paper as follows. In Section 2, we describe the Rabi model
and remind useful facts from the theory of linear ODE’s. Furthermore, we use
linear change of variables to rewrite the quadratic differential (1.2) in a
more symmetric form (2.8), which is convenient to work with. We also give
explicit expressions for the coefficients $c_{k}$ of the quadratic
differential (2.8) as functions of the parameters of the Rabi problem. In
Section 3, we collect definitions and results from the theory of quadratic
differentials needed for the study of the Stokes graphs of equation (1.1).

Section 4 contains classification of possible critical graphs and domain
configurations of the quadratic differential (2.8) assuming that its
coefficients $c_{k}$ are real numbers not necessarily related to the
parameters $\Delta$, $E$, and $g$ of the Rabi problem. To avoid having too
many cases and figures in just one paper, we also assume in what follows that
the quadratic differential $Q_{0}(z)\,dz^{2}$ defined by (2.8) has _full set
of critical points_ , which means that the zeros of the numerator in (2.8) are
distinct from the poles $\pm 1$. The description of the Stokes graphs and
domain configurations of the so-called _depressed quadratic differential of
the Rabi problem_ , when cancellation of zeros and poles happens and
$Q_{0}(z)\,dz^{2}$ has at most simple pole at least one of the points $z=\pm
1$, will be postponed for a sequel paper.

Our classification of Stokes graphs and domain configurations of
$Q_{0}(z)\,dz^{2}$ is based on the number of real zeros, on types of domains
present in the domain configuration of this quadratic differential, and on the
positions of zeros on boundaries of these domains.

As it is expected, the classification includes many cases. We want to mention
here the analogy between classifications of critical graphs of quadratic
differentials with prescribed number of critical points and classifications of
real algebraic curves of a given degree. As it is well known, the
classification of cubic curves, first suggested by I. Newton in the
seventeenth century and completed later, contains 78 types of curves. A
different, more topological classification of cubics, was discussed in [20].
Interestingly enough, the modern approach to the classification problem for
real algebraic curves, initiated by J.C. Langer and D.A. Singer in [10] and
then used in [16], reveals that shapes of these curves can be identified as
critical graphs of appropriate quadratic differentials defined on compact
Riemann surfaces.

In Section 5, we describe the set of coefficients
$(c_{3},c_{2},c_{1},c_{0})\in\mathbb{R}^{4}$ of the quadratic differential
$Q_{0}(z)\,dz^{2}$, which correspond to the real values of the Rabi parameters
$\Delta$, $E$, and $g^{2}$. In our last Section 6, we study the limit domain
configurations, when the boson-fermion coupling $g$ tends to $\infty$,
assuming that the parameters $\Delta$ and $E$ are certain functions of $g$.
The latter study is motivated by the isomonodromic problem associated to the
Rabi model [2], when one tries to relate the parameters of the model providing
the same monodromy data for the ODE. This problem is closely related to the
tau-function of the Painlevé V equation and it could be helpful in the study
of the quantized spectrum of the Rabi model.

Appendix A contains notations consistently used throughout the paper. Appendix
B, that is the “Zoo” of Stokes graphs and domain configurations, contains
examples of possible Stokes graphs and domain configurations of the quadratic
differential $Q_{0}(z)\,dz^{2}$ that is the symmetrized form of the quadratic
differential (1.2). These Stokes graphs and domain configurations are
described in details in Section 4.

##  2\. Rabi model and associated quadratic differential

Before describing the Rabi model, we present general facts from the theory of
linear ODE needed for our study. Consider the matrix linear differential
equation

(2.1) $\frac{d\Psi(z)}{dz}=A(z)\Psi(z)$

in a domain $\Omega\subset\overline{\mathbb{C}}$. Here

$A(z)=\begin{pmatrix}a_{11}&a_{21}\\\
a_{12}&a_{22}\end{pmatrix}\quad{\mbox{and}}\quad\Psi(z)=\begin{pmatrix}f_{11}&f_{21}\\\
f_{12}&f_{22}\end{pmatrix}$

are $2\times 2$ matrices with coefficients, depending on $z\in\Omega$. Each of
the columns $f_{1}=\begin{pmatrix}f_{11}\\\ f_{12}\end{pmatrix}$ and
$f_{2}=\begin{pmatrix}f_{21}\\\ f_{22}\end{pmatrix}$ of the unknown matrix
valued function $\Psi(z)$ is a solution to the equation

$\frac{df_{k}(z)}{dz}=A(z)f_{k}(z),\quad k=1,2.$

If $A(z)$ is holomorphic in $\Omega$, then for every $z_{0}\in\Omega$ there is
a unique fundamental solution $\Psi(z)$ to (2.1) holomorphic at $z_{0}$,
satisfying the normalization condition $\Psi(z_{0})=I$, where $I$ is the
identity matrix. Any other fundamental solution $\widetilde{\Psi}(z)$, which
is holomorphic at $z_{0}$, has the form $\widetilde{\Psi}(z)=\Psi(z)C$, where
$C$ is a constant matrix. If $A(z)$ is meromorphic in $\Omega$, then the
behavior of solutions to (2.1) near singular points of $A(z)$ is more
complicated; see, for instance, [5].

The functions $f_{11}$ and $f_{21}$ are solutions to the second order ODE

(2.2) $\frac{d^{2}f}{dz^{2}}+p(z)\frac{df}{dz}+q(z)f(z)=0$

with

$p(z)=-{\rm Tr}A-\frac{1}{a_{12}}\frac{da_{12}(z)}{dz}=-{\rm
Tr}A-\frac{d}{dz}\log a_{12},$

and

$q(z)=\frac{a_{11}}{a_{12}}\frac{da_{12}(z)}{dz}-\frac{da_{11}(z)}{dz}+\det
A=a_{11}\frac{d}{dz}\log a_{12}-\frac{da_{11}(z)}{dz}+\det A.$

Note that if $f_{11}(z)$ and $f_{21}(z)$ are linearly independent solutions of
(2.2), then

$\begin{pmatrix}f_{11}(z)&f_{21}(z)\\\
f^{\prime}_{11}(z)&f^{\prime}_{12}(z)\end{pmatrix}$

is the fundamental solution to (2.1).

Furthermore, changing variable in (2.2) via $f(z)=\psi(z)y(z)$ with

(2.3) $\psi(z)={\rm
exp}\left(-\frac{1}{2}\,\int_{z_{0}}^{z}p(\tau)\,d\tau\right),$

we rewrite (2.2) in the following equivalent form:

(2.4) $y^{\prime\prime}(z)+Q(z)y(z)=0,$

where

(2.5) $Q(z)=q(z)-\frac{1}{4}p^{2}(z)-\frac{1}{2}p^{\prime}(z).$

Now we turn to the Rabi model that is a physical model describing a simple
harmonic oscillator, or two level quantum system [2, 22]. The Rabi model in
the standard Garnier form [8] is governed by the differential equation (2.1)
with the matrix

(2.6) $A(z)=\frac{\sigma_{3}}{2}+\frac{A_{0}}{z}+\frac{A_{t}}{z-t},$

where $t=-4g^{2}$, $z\in\overline{\mathbb{C}}$,

$\sigma_{3}=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix},\quad
A_{0}=\begin{pmatrix}E+g^{2}&-\Delta\\\ 0&0\end{pmatrix},\quad
A_{t}=\begin{pmatrix}0&0\\\ -\Delta&E+g^{2}\end{pmatrix}.$

Here $\sigma_{3}$ is one of the Pauli matrices, $2\Delta$ is the energy
difference between the two fermion levels, $g$ is the boson-fermion coupling,
and $E$ is the eigenvalue of a Hamiltonian defined by the physical problem. In
a general setting of the Rabi problem, the energy difference $\Delta$ is real
but the boson-fermion coupling $g$ and the eigenvalue $E$ may be complex
numbers. As we already mentioned in the Introduction, below in this paper we
always assume that the parameters $\Delta$ and $E$ are real numbers and that
$g\not=0$ is either real or pure imaginary nonzero number. In this case, the
parameter $t=-4g^{2}\not=0$ is real and therefore the function $A(z)$ defined
by (2.6) has a pole at the point $z=t$, $t\not=0$, of the real axis.

With the matrix $A(z)$ defined in (2.6), system (2.1) has two regular singular
points $z=0$, $z=t$, and one irregular singular point at $z\in\infty$ of
Poisson rank 1; see [5].

Reducing the matrix differential equation to the second order ODE, as it was
described earlier in this section, we obtain the following equation:

(2.7) $f^{\prime\prime}(z)+p(z)f^{\prime}(z)+q(z)f(z)=0,$

where

$p(z,t)=\frac{1-\theta}{z}-\frac{\theta}{z-t},$
$q(z)=-\frac{1}{4}+\frac{1}{z}\Big{(}-\frac{1}{2}+\frac{\Delta^{2}}{t}-\frac{\theta^{2}}{t}-\frac{\theta}{2}\Big{)}+\frac{1}{z-t}\Big{(}-\frac{\Delta^{2}}{t}+\frac{\theta^{2}}{t}+\frac{\theta}{2}\Big{)},$

and

$\theta={\rm Tr}A_{0}={\rm Tr}A_{t}=E+g^{2}=E-\frac{t}{4}.$

Changing variable in (2.7) via (2.3), we obtain equation (2.4) with the
function $Q(z)=Q(z,t)$, where

$Q(z,t)=-\frac{1}{4}\frac{z^{4}+a_{3}z^{3}+a_{2}z^{2}+a_{1}z+a_{0}}{z^{2}(z-t)^{2}},$

with the coefficients $a_{k}$, $k=0,1,2,3$, given by

$\displaystyle a_{3}$ $\displaystyle=$ $\displaystyle-2t+2,\quad\quad
a_{2}=t^{2}-t(2\theta+4)+4\Delta^{2}-1,$ $\displaystyle a_{1}$
$\displaystyle=$ $\displaystyle
t^{2}(2\theta+2)-t(4\Delta^{2}-2\theta-2),\quad\quad
a_{0}=t^{2}(\theta^{2}-1).$

The goal now is to describe possible Stokes graphs of the equation (2.4) or,
equivalently, describe possible structures of the critical trajectories of the
quadratic differential $Q(z,t)\,dz^{2}$. Changing variables in the quadratic
differential $Q(z,t)\,dz^{2}$ via the linear transformation
$z\to\frac{t}{2}(1-z)$ and then multiplying the resulting quadratic
differential by $4$, we obtain the following more symmetric form of this
quadratic differential, which is easier to work with:

(2.8)
$Q_{0}(z)\,dz^{2}=-\frac{P_{0}(z)}{(z-1)^{2}(z+1)^{2}}\,dz^{2}=-\frac{z^{4}+c_{3}z^{3}+c_{2}z^{2}+c_{1}z+c_{0}}{(z-1)^{2}(z+1)^{2}}\,dz^{2},$

where the coefficients of the numerator
$P_{0}(z)=z^{4}+c_{3}z^{3}+c_{2}z^{2}+c_{1}z+c_{0}$ are given by

(2.9) $\displaystyle c_{3}$ $\displaystyle=g^{-2},\quad
c_{2}=\frac{1}{4g^{4}}(8Eg^{2}+4\Delta^{2}+4g^{2}-1),$ (2.10) $\displaystyle
c_{1}$ $\displaystyle=-\frac{1}{2g^{4}}(4g^{2}+2E+1),\quad
c_{0}=-\frac{1}{4g^{4}}(4\Delta^{2}-4E^{2}-4E+1).$

We note here that critical trajectories and domain configuration of the
resulting quadratic differential $Q_{0}(z)\,dz^{2}$ coincide with those of
$Q(z,t)\,dz^{2}$ up to scaling, reflection with respect to the imaginary axis,
and translation in the horizontal direction. Thus, in what follows we work
with the rescaled quadratic differential $Q_{0}(z)\,dz^{2}$ given by equation
(2.8) assuming that it has full set of critical points, i.e. we assume that
$P_{0}(\pm 1)\not=0$ and therefore in the cases under consideration
$Q_{0}(z)\,dz^{2}$ has double poles at the points $\pm 1$.

##  3\. Basics on quadratic differentials

In this section, we first recall definitions and some basic facts about
quadratic differentials. Some notations, needed for future use, will also be
introduced. After that we discuss basic characteristics of the quadratic
differential (2.8), assuming that all its coefficients $c_{k}$ are real. In
particular, few basic properties of the Stokes graph of this quadratic
differential and its domain configuration, which do not depend on the
positions of the zeros of $Q_{0}(z)\,dz^{2}$, will be mentioned. The detailed
description of properties of the Stokes graphs and domain configurations,
which depend on the positions of these zeros, will be given in the next
section.

In this paper, we deal with quadratic differentials defined on the Riemann
sphere $\overline{\mathbb{C}}$. For more general theory of quadratic
differentials defined on Riemann surfaces the interested reader may consult
classical monographs by J. Jenkins [6] and K. Strebel [17].

A quadratic differential on a domain $D\subset\overline{\mathbb{C}}$ is a
differential form $Q(z)\,dz^{2}$ with meromorphic $Q(z)$ and with the
conformal transformation rule

(3.1)
$Q_{1}(\zeta)\,d\zeta^{2}=Q(\varphi(z))\left(\varphi^{\prime}(z)\right)^{2}\,dz^{2},$

where $\zeta=\varphi(z)$ is a conformal map from $D$ onto a domain $G$ in the
extended plane of the parameter $\zeta$.

The zeros and poles of $Q(z)$ are critical points of $Q(z)\,dz^{2}$, in
particular, zeros and simple poles are finite critical points and poles of
order greater than $1$ are infinite critical points of $Q(z)\,dz^{2}$.

A trajectory (respectively, orthogonal trajectory) of $Q(z)\,dz^{2}$ is a
closed analytic Jordan curve or maximal open analytic arc $\gamma\subset D$
such that

$Q(z)\,dz^{2}>0\quad{\mbox{along
$\gamma$}}\quad\quad({\mbox{respectively}},Q(z)\,dz^{2}<0\quad{\mbox{along
$\gamma$}}).$

A trajectory $\gamma$ is called _critical_ if at least one of its end points
is a finite critical point of $Q(z)\,dz^{2}$.

In the theory of ODE’s, critical trajectories of the quadratic differential
$Q(z)\,dz^{2}$ associated with equation (2.4) are known as the _Stokes lines_
of this equation. Accordingly, a finite critical point of $Q(z)\,dz^{2}$, that
is an end point of a Stokes line, is called _turning point_ of the equation
(2.4). We have to stress here, that terminology concerning quadratic
differentials used in this paper may differ from the one used by some other
authors. For instance, in the Fedoryuk’s monograph [4] the Stokes lines are
defined as our orthogonal critical trajectories. In this work, we will stick
with definitions used consistently in the classical publications of J. Jenkins
[6, 7], K. Strebel [17], and others.

Let $\Phi_{Q}$, $\overline{\Phi}_{Q}$, and $G_{Q}=\partial\overline{\Phi}_{Q}$
denote the union of points of critical trajectories of a quadratic
differential $Q(z)\,dz^{2}$ defined on $\overline{\mathbb{C}}$, the closure of
this union of points, and the boundary of this closure, respectively. It is
known, see [6, Theorem 3.5], that the set $G_{Q}$ is either empty or it
consists of a finite number of critical trajectories of $Q(z)\,dz^{2}$.
Furthermore, each of these critical trajectories in each direction has an end
point situated at some point in $\overline{\mathbb{C}}$ and at least one of
these end points is a finite critical point of $Q(z)\,dz^{2}$. If
$G_{Q}\not=\emptyset$, it is called the _critical graph_ of $Q(z)\,dz^{2}$.
Interestingly enough, the study initiated in [14] and continued in [15] shows
that every weighted planar graph $G$ with simply connected and doubly
connected faces can be realized as the critical graph of a certain quadratic
differential defined on $\overline{\mathbb{C}}$.

In relation with ODE’s, the critical graph $G_{Q}$ is also known as the
_Stokes graph_ of equation (2.4), see [4]. As usual, the critical trajectories
that are components of $G_{Q}$ are edges of $G_{Q}$, their end points are
vertices of $G_{Q}$, and the connected components $\Omega_{k}$ of
$\overline{\mathbb{C}}\setminus G_{Q}$ are faces of $G_{Q}$.

The set $\\{\Omega_{k}\\}$ of all faces of $Q(z)\,dz^{2}$ is called the domain
configuration of $Q(z)\,dz^{2}$. The domains $\Omega_{k}$, which are bounded
by the Stokes lines, play an essential role in the asymptotic analysis of
ODE’s. According to the Basic Structure Theorem of J. Jenkins, [6, Theorem
3.5], the set $\\{\Omega_{k}\\}$ consists of a finite number of domains
$\Omega_{k}$, each of which belongs to one of the following 5 types:

  1. $\bullet$

$\Omega_{k}$ is called a circle domain of $Q(z)\,dz^{2}$, if it is a simply
connected domain bounded by a finite number of critical trajectories, which
end points are finite critical points of $Q(z)\,dz^{2}$, and such that
$\Omega_{k}$ contains exactly one critical point of $Q(z)\,dz^{2}$, called the
_center_ of $\Omega_{k}$, which is a pole of order $2$.

  2. $\bullet$

$\Omega_{k}$ is called a ring domain of $Q(z)\,dz^{2}$, if it is a doubly
connected domain, free of critical points and critical trajectories, such that
each of two boundary components of $\Omega_{k}$ consists of a finite number of
critical trajectories with end points in the set of finite critical points of
$Q(z)\,dz^{2}$.

  3. $\bullet$

$\Omega_{k}$ is called an end domain of $Q(z)\,dz^{2}$, if it is a simply
connected domain, free of critical points and critical trajectories, which
boundary consists of a finite number of critical trajectories, such that two
of them have a common end point at a pole of order $\geq 2$, called the
_vertex_ of $\Omega_{k}$, while all other end points of these critical
trajectories are finite critical points of $Q(z)\,dz^{2}$.

  4. $\bullet$

$\Omega_{k}$ is called a strip domain of $Q(z)\,dz^{2}$, if it is a simply
connected domain, free of critical points and critical trajectories, which
boundary, consisting of four or more critical trajectories, has exactly two
distinct boundary points, called _vertices_ , which belong to the set of
infinite critical points of $Q(z)\,dz^{2}$. The boundary arcs joining vertices
of a strip domain are called the _sides_ of this domain. It is also required
that each side consists of two or more critical trajectories.

  5. $\bullet$

$\Omega_{k}$ is called a density domain of $Q(z)\,dz^{2}$ if every trajectory,
which crosses $\Omega_{k}$, is dense in $\Omega_{k}$.

Below in this paper, we use the following notations. By $(a,b)$ and $[a,b]$ we
denote, respectively, open and closed intervals having end points at $z=a$ and
$z=b$. Notations $(-\infty,a)$, $(b,\infty)$, etc, will be used to denote
infinite intervals on the real axis. If $\gamma$ is a rectifiable arc in a
domain $D$, where a quadratic differential $Q(z)\,dz^{2}$ is defined, then its
$Q$-length is defined as $|\gamma|_{Q}=\int_{\gamma}|Q(z)|^{1/2}\,|dz|$. Also,
if $a,b\in\overline{\mathbb{C}}$ are not infinite critical points of
$Q(z)\,dz^{2}$ and the open interval $(a,b)$ does not contain critical points
of $Q(z)\,dz^{2}$, then we define $[a,b]_{Q}$ as follows:

(3.2) $[a,b]_{Q}=\int_{a}^{b}\sqrt{Q(z)}\,dz,$

where the integration is taken along the interval $[a,b]$. In what follows, we
mostly work with the real or imaginary parts of the integrals defined as in
(3.2). In such cases, we assume that the branch of the square root in (3.2) is
chosen such that these real or imaginary parts are non-negative.

An important property of quadratic differentials is that the transformation
rule (3.1) respects trajectories, orthogonal trajectories, and their
$Q$-lengths, as well as it respects domain configurations and critical points
together with their multiplicities and trajectory structure nearby.

In the classification of domain configurations of $Q_{0}(z)\,dz^{2}$,
presented in Section 4, we routinely use a few simple properties of critical
graphs of quadratic differentials, which, for convenience of the reader, are
collected in the following lemma.

######  Lemma 3.1.

Let $Q(z)\,dz^{2}$ be a quadratic differential on $\overline{\mathbb{C}}$,
which does not have density domains among its faces. Then the following
properties hold:

  1. 1.

Let $D\subset\mathbb{C}$ be a simply connected domain, which boundary consists
of a finite number of critical trajectories of $Q(z)\,dz^{2}$ and their end
points. Let $P$ and $N$ denote the number of poles and zeros of $Q(z)\,dz^{2}$
on $\overline{D}$, where poles and zeros in $D$ are counted with their
multiplicities and poles and zeros on $\partial D$, considered as boundary
critical points of $Q(z)\,dz^{2}$, are counted with half of their
multiplicities. Then

$P-N=2.$

  2. 2.

Let $\Gamma$ be a connected boundary component of a face $\Omega$. If the
connected component of $\overline{\mathbb{C}}\setminus\Gamma$, containing
$\Omega$, also contains at least one zero of $Q(z)\,dz^{2}$, then $\Omega$ is
a ring domain.

  3. 3.

Let $\Gamma$ be a Jordan arc consisting of a finite number of critical
trajectories of $Q(z)\,dz^{2}$, which end points are infinite critical points
of $Q(z)\,dz^{2}$, and there is no other infinite critical point on $\Gamma$.
If $\Gamma$ is a proper boundary arc on the boundary of a face $\Omega$ of the
quadratic differential $Q(z)\,dz^{2}$, then $\Omega$ is a strip domain.

  4. 4.

If $Q(z)\,dz^{2}$ has $n\geq 1$ zeros, counting multiplicity, in a Jordan
domain $D\subset\mathbb{C}$ and does not have other critical points on
$\overline{D}$, then there are at least $n+2$ critical trajectories of
$Q(z)\,dz^{2}$ crossing the boundary of $D$.

_Proof._ The formula in part 1 of this lemma is just a simplest special case
of the relation between zeros and poles given in [6, Lemma 3.2]. Parts 2 and 3
immediately follow from the Basic Structure Theorem, see [6, Theorem 3.5], and
from the definitions of end, circle, ring, and strip domains given above.

Part 4 is a property known in the Graph theory without relation to quadratic
differentials. It can be proved as follows. Consider the graph $G_{Q}(D)$ that
is the restriction of the critical graph $G_{Q}$ onto the domain $D$. Since
$Q(z)\,dz^{2}$ does not have poles in a simply connected domain $D$, it
follows from part 1 of Lemma 3.1 that $G_{Q}(D)$ does not have cycles.

It is enough to prove the property required in this part assuming that the
graph $G_{Q}(D)$ is connected. Indeed, if the result holds for every connected
component of $G_{Q}(D)$, then adding the numbers of trajectories exiting $D$
for each connected subgraph of $G_{Q}(D)$, we obtain the required result for
$G_{Q}(D)$. Thus, we assume that $G_{Q}(D)$ is connected and without cycles;
in the Graph theory such graphs are known as _trees_. We may start with the
case when all zeros of $Q(z)\,dz^{2}$, which are in the domain $D$, are
simple. Then the degree of each vertex of $G_{Q}(D)$ is $3$. Thus, the sum of
all these degrees is $3n$. If the vertices $v_{1}$ and $v_{2}$ of $G_{Q}(D)$
are connected by an edge of $G_{Q}(D)$, we merge these vertices along this
edge to obtain a graph $G^{1}$ with $n-1$ vertices, with total degree of
vertices equal $3n-2$, and with the same number of edges having end points on
$\partial D$ as for the graph $G_{Q}(D)$. This merging procedure can be done
$n-1$ times until we obtain a graph $G^{n-1}$ with a single vertex of degree
$n+2$ such that every edge of $G^{n-1}$ has its end point on $\partial D$.
Since the number of the end points on $\partial D$ for the graph $G^{n-1}$ is
the same as for the graph $G_{Q}(D)$, the required result is proved. $\Box$

Next, we present a list of basic properties of the quadratic differential
$Q_{0}(z)\,dz^{2}$ defined in equation (2.8), which always hold when all its
coefficients $c_{k}$ are real.

  1. (1)

The quadratic differential $Q_{0}(z)\,dz^{2}$ has at most three distinct poles
and therefore it follows from Jenkins’s _Three poles lemma_ that the domain
configuration of $Q_{0}(z)\,dz^{2}$ does not contain density domains [6].

  2. (2)

Since all the coefficients of the rational function $Q_{0}(z)$ are real, the
complex zeros of $Q_{0}(z)\,dz^{2}$ are in conjugate pairs, the number of real
zeros (counting multiplicity) is even, and the critical graph and domain
configuration of $Q_{0}(z)\,dz^{2}$ are symmetric with respect to the real
line.

  3. (3)

Since $Q_{0}(z)\,dz^{2}=-(1+o(1))\,dz^{2}$ as $z\to\infty$, it follows that
$Q_{0}(z)\,dz^{2}$ has a pole of order four at $z=\infty$ with two critical
directions defined by condition $-1\cdot\,dz^{2}>0$. Thus, $d_{1}=i$ and
$d_{2}=-i$ are the critical directions of $Q_{0}(z)\,dz^{2}$ at $z=\infty$.
Furthermore, the domain configuration of $Q_{0}(z)\,dz^{2}$ always includes
exactly two end domains, the “left domain” $\Omega^{l}_{e}$ and the “right
domain” $\Omega^{r}_{e}$, such that $\Omega^{l}_{e}\supset(-\infty,-a)$ and
$\Omega^{r}_{e}\supset(a,+\infty)$, for all $a>0$ big enough. This, together
with the symmetry property, imply that, if $a>0$ is big enough, then the
intervals $(-\infty,-a)$ and $(a,+\infty)$ lie on orthogonal trajectories of
$Q_{0}(z)\,dz^{2}$.

  4. (4)

Let $e_{k}$, $k=1,2,3,4$, denote the zeros of the numerator $P_{0}(z)$ of the
quadratic differential (2.8). In the case $e_{k}\not=\pm 1$, $k=1,2,3,4$, the
quadratic differential $Q_{0}(z)\,dz^{2}$ has two second order poles and
therefore it may have at most two circle domains centered at the poles $z=-1$
and $z=1$. If such circle domains exist, we denote them as
$\Omega_{c}(-1)\ni-1$ and $\Omega_{c}(1)\ni 1$. Furthermore,
$Q_{0}(z)\,dz^{2}$ may have at most one ring domain $\Omega_{r}$, which, if
exists, must separate the poles $z=-1$ and $z=1$ from the pole $z=\infty$.

For the long classification of possible domain configurations of the quadratic
differential (2.8) presented in the next section, it is convenient to
introduce necessary notations and fix some terminology.

Everywhere below, $\gamma_{a,b}$ stands for a critical trajectory, including
its end points, which starts at $a$ and ends at $b$. Thus, $\gamma_{b,a}$ is
the same critical trajectory as $\gamma_{a,b}$ but with opposite orientation.
If a critical trajectory $\gamma_{a,b}$ is symmetric with respect to the real
axis, then we add superscripts “l”, “c”, and “r”, like $\gamma_{a,b}^{l}$,
$\gamma_{a,b}^{c}$, $\gamma_{a,b}^{r}$, to indicate when $\gamma_{a,b}$
crosses the real axis to the left of the pole $z=-1$, in between the poles
$z=-1$ and $z=1$, or to the right of the pole $z=1$. In the case, when $a=b$,
we use a shorter notation $\gamma_{a}^{l}=\gamma_{a,a}^{l}$, etc, assuming
counter clock-wise orientation of $\gamma_{a}^{l}$, etc. An additional
superscript “$-$”, like $\gamma_{a}^{l-}$, etc, will be used to indicate that
this critical trajectory is oriented clock-wise in the case under
consideration.

By $\Gamma_{e}^{l}$, $\Gamma_{e}^{r}$, $\Gamma_{c}(-1)$ and $\Gamma_{c}(1)$,
we denote the boundaries of the domains $\Omega_{e}^{l}$, $\Omega_{e}^{r}$,
$\Omega_{c}(-1)$ and $\Omega_{c}(1)$ assuming positive orientation of these
boundaries, with respect to corresponding domains. Also, $\Gamma_{r}^{(out)}$
and $\Gamma_{r}^{(inn)}$ will denote the outer and inner boundary components
of the ring domain $\Omega_{r}$, where we assume that $\Gamma_{r}^{(out)}$ is
oriented in the positive direction with respect to $\Omega_{r}$ and
$\Gamma_{r}^{(inn)}$ is oriented in the negative direction with respect to
$\Omega_{r}$.

By $\Omega_{s}(a,b)$ we denote a strip domain with vertices $a$ and $b$. This
notation with $a=-i\infty$ and/or $b=i\infty$ means that $\Omega_{s}(a,b)$
approaches its vertex $a=\infty$ along the negative direction of the imaginary
axis and it approaches its vertex $b=\infty$ along the positive direction of
the imaginary axis. When $a=-i\infty$, and/or $b=i\infty$, by
$\Gamma_{s}^{l}(a,b)$ and $\Gamma_{s}^{r}(a,b)$ we denote the left and right
sides of $\Omega_{s}(a,b)$, respectively. In the case when $a=b$, which can
happen when $a=\pm 1$, by $\Gamma_{s}^{(out)}(a,b)$ and
$\Gamma_{s}^{(inn)}(a,b)$ we denote the outer and inner sides of
$\Omega_{s}(a,b)$, respectively. Furthermore, $\Gamma_{s}^{+}(-1,1)$ and
$\Gamma_{s}^{-}(-1,1)$ will denote the sides of $\Omega_{s}(-1,1)$ lying in
the upper half-plane and in the lower half-plane, respectively.

To characterize the behavior of $Q_{0}(z)\,dz^{2}$ near its poles $z=\pm 1$,
we introduce the following notations. Let

$\alpha_{-1}=-\frac{1}{4}(1+c_{3}+c_{2}+c_{1}+c_{0}),\quad\quad\alpha_{1}=-\frac{1}{4}(1-c_{3}+c_{2}-c_{1}+c_{0}).$

If $\alpha_{k}\not=0$, $k=-1,1$, then $\alpha_{k}$ is the leading coefficient
of the Laurent expansion of $Q_{0}(z)$ at the pole $z=k$. Therefore,
$Q_{0}(z)\,dz^{2}$ has second order pole at $z=k$ if $\alpha_{k}\not=0$. Since
$c_{k}$, $k=0,1,2,3$, are real, it follows that, if $\alpha_{k}\not=0$, then
there is a trajectory or orthogonal trajectory of $Q_{0}(z)\,dz^{2}$
surrounding the point $z=k$, which $Q_{0}$-length will be denoted by
$\delta_{k}>0$; i.e.

(3.3)
$\delta_{k}=\left|\int_{|z-k|=\epsilon}\sqrt{Q_{0}(z)}\,dz\right|=2\pi\,|\alpha_{k}|^{1/2},\quad
k=-1,1,$

where $\epsilon>0$ is small enough.

We already mentioned that all the domain configurations of the quadratic
differential $Q_{0}(z)\,dz^{2}$ with real coefficients are symmetric with
respect to the real axis. Also, in many cases, we will have pairs of domain
configurations, which are symmetric to each other with respect to the
imaginary axis. This happens, for instance, when positions of zeros in one
configuration are symmetric with respect to the imaginary axis to positions of
zeros in the other configuration. In cases like this, we will describe with
details one of these configurations and then mention that the other one is the
_mirror configuration_ of the configuration described above.

##  4\. Stokes graphs and domain configurations of $Q_{0}(z)\,dz^{2}$

In this section, we assume that the coefficients $c_{k}$, $k=0,1,2,3$, of the
numerator $P_{0}(z)$ of the quadratic differential $Q_{0}(z)\,dz^{2}$ are real
and that $Q_{0}(z)\,dz^{2}$ has full set of critical points. Under these
assumptions, $P_{0}(z)$ has four zeros (counting multiplicity), which are
either real, not equal to $\pm 1$, or in conjugate pairs. Below, we describe
how the Stokes graphs and domain configurations of $Q_{0}(z)\,dz^{2}$ depend
on the positions of these zeros. In particular, how they depend on the number
of real zeros of $P_{0}(z)$. Thus, we will distinguish between three main
cases: case I, when there are no real zeros, case II with two real zeros, and
case III with four real zeros. In what follows, we describe in details
possible domain configurations in the _generic cases_ ; i.e. when the zeros
$e_{k}$, $k=1,2,3,4$, of $P_{0}(z)$ are distinct. The remaining cases, which
we call _degenerate cases_ , will be also mentioned but most details will be
left to the interested reader.

I. Suppose that there are no real zeros. Let $e_{1}=\alpha_{1}+i\beta_{1}$ and
$e_{2}=\alpha_{2}+i\beta_{2}$, with $\beta_{1}>0$, and $\beta_{2}>0$, be zeros
of $Q_{0}(z)\,dz^{2}$. Then their complex conjugates $e_{3}=\overline{e_{1}}$
and $e_{4}=\overline{e_{2}}$ are also zeros of $Q_{0}(z)\,dz^{2}$. In this
case, the intervals $(-\infty,-1)$, $(-1,1)$, and $(1,\infty)$ are orthogonal
trajectories of $Q_{0}(z)\,dz^{2}$. This implies that $Q_{0}(z)\,dz^{2}$ has
two circle domains $\Omega_{c}(-1)$ and $\Omega_{c}(1)$ with respective
boundaries $\Gamma_{c}(-1)$ and $\Gamma_{c}(1)$, each of which contains at
least one of the pairs $\\{e_{1},e_{3}\\}$ and $\\{e_{2},e_{4}\\}$ of the
zeros of $Q_{0}(z)\,dz^{2}$. Changing numeration, if necessary, we may assume
that $e_{1},e_{3}\in\Gamma_{c}(-1)$.

Depending on the positions of $e_{1}$ and $e_{2}$ on the boundaries, we have
the following subcases.

1\. Suppose that $e_{1}\in\Gamma_{c}(-1)$ and
$e_{2}\not\in\Gamma_{c}(-1)\cup\Gamma_{c}(1)$. Then we also must have
$e_{1}\in\Gamma_{c}(1)$. This happens if and only if the following
inequalities hold:

(4.1)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[-1+\varepsilon,e_{1}]_{Q_{0}}-\operatorname{Im}[-1+\varepsilon,e_{2}]_{Q_{0}}\right)<0,$
(4.2)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[1+\varepsilon,e_{1}]_{Q_{0}}-\operatorname{Im}[1+\varepsilon,e_{2}]_{Q_{0}}\right)<0.$

Roughly speaking, inequalities (4.1) and (4.2) mean that the zeros $e_{1}$ and
$e_{2}$ are not connected by critical trajectories having end points at the
finite critical points of $Q_{0}(z)\,dz^{2}$ and that the zero $e_{1}$ is
closer, in terms of the $Q_{0}$-metric, to the poles $\pm 1$ than the zero
$e_{2}$.

Since $e_{2}\not\in\Gamma_{c}(-1)\cup\Gamma_{c}(1)$, it follows that each of
the boundaries $\Gamma_{c}(-1)$ and $\Gamma_{c}(1)$ consists of two critical
trajectories joining $e_{1}$ and $e_{3}$. Precisely,
$\Gamma_{c}(-1)=\gamma_{e_{1},e_{3}}^{l}\cup\gamma_{e_{3},e_{1}}^{c}$, where
$\gamma_{e_{1},e_{3}}^{l}$ intersects the interval $(-\infty,-1)$ and
$\gamma_{e_{3},e_{1}}^{c}$ intersects the interval $(-1,1)$ and
$\Gamma_{c}(1)=\gamma_{e_{1},e_{3}}^{c}\cup\gamma_{e_{3},e_{1}}^{r}$, where
$\gamma_{e_{1},e_{3}}^{c}$ coincides up to orientation with
$\gamma_{e_{3},e_{1}}^{c}$, and $\gamma_{e_{3},e_{1}}^{r}$ intersects the
interval $(1,\infty)$.

It follows also that the set
$\Gamma_{r}^{(inn)}=\gamma_{e_{1},e_{3}}^{l}\cup\gamma_{e_{3},e_{1}}^{r}$ is a
closed Jordan curve that is a boundary component of one of the faces of
$Q_{0}(z)\,dz^{2}$. By part 2 of Lemma 3.1, in the case under consideration,
this face must be a ring domain $\Omega_{r}$. Furthermore, the second boundary
component $\Gamma_{r}^{(out)}$ of the ring domain $\Omega_{r}$ must consist of
two critical trajectories joining $e_{2}$ and $e_{4}$. Thus,
$\Gamma_{r}^{(out)}=\gamma_{e_{2},e_{4}}^{l}\cup\gamma_{e_{4},e_{2}}^{r}$,
where $\gamma_{e_{2},e_{4}}^{l}$ intersects the interval $(-\infty,-1)$ and
$\gamma_{e_{4},e_{2}}^{r}$ intersects the interval $(1,\infty)$.

The two remaining critical trajectories
$\gamma_{e_{2},i\infty}\subset\mathbb{H}_{+}$ and
$\gamma_{-i\infty,e_{4}}\subset\mathbb{H}_{-}$ are arcs on the boundaries
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{2}}^{l}\cup\gamma_{e_{2},i\infty}$
and
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{2}}\cup\gamma_{e_{2},e_{4}}^{r}\cup\gamma_{e_{4},-i\infty}$
of the end domains $\Omega^{l}_{e}$ and $\Omega^{r}_{e}$, respectively. Figure
3 Case I-1 gives an example of the domain configuration.

2\. Suppose that $e_{1}\in\Gamma_{c}(-1)$ but $e_{1}\not\in\Gamma_{c}(1)$ and
$e_{2}\in\Gamma_{c}(1)$ but $e_{2}\not\in\Gamma_{c}(-1)$. This position of
zeros happens if and only if the limit in inequality (4.1) is negative as in
the case 1, but the limit in inequality (4.2) is positive. The latter
conditions imply that the zeros $e_{1}$ and $e_{2}$ are not connected by
critical trajectories having end points at the finite critical points of
$Q_{0}(z)\,dz^{2}$ and that $e_{1}$ is, in terms of the $Q_{0}$-metric, closer
to the pole $-1$ than $e_{2}$, and $e_{2}$ is closer to the pole $1$ than
$e_{1}$.

Same argument as in part I-1 above, implies that there are critical
trajectories $\gamma_{e_{1},e_{3}}^{c}$, crossing the interval $(-1,1)$ at
$x_{1}$, and $\gamma_{e_{2},e_{4}}^{c}$, crossing the interval $(x_{1},1)$
such that
$\Gamma_{c}(-1)=\gamma_{e_{1},e_{3}}^{l}\cup\gamma_{e_{3},e_{1}}^{c}$ is the
boundary of the circle domain $\Omega_{c}(-1)$ and
$\Gamma_{c}(1)=\gamma_{e_{2},e_{4}}^{c}\cup\gamma_{e_{4},e_{2}}^{r}$ is the
boundary of $\Omega_{c}(1)$.

Since $e_{1}\in\Gamma_{c}(-1)\setminus\Gamma_{c}(1)$ and
$e_{2}\in\Gamma_{c}(1)\setminus\Gamma_{c}(-1)$, it follows that there is a
certain “gap” between circle domains $\Omega_{c}(-1)$ and $\Omega_{c}(1)$,
which must be a strip domain $\Omega_{s}(-i\infty,i\infty)$ with the vertices
at $-i\infty$ and $i\infty$ and with sides
$\Gamma_{s}^{l}=\gamma_{-i\infty,e_{3}}\cup\gamma_{e_{3},e_{1}}^{c}\cup\gamma_{e_{1},i\infty}$
and
$\Gamma_{s}^{r}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{2}}^{c}\cup\gamma_{e_{2},i\infty}$.

The boundaries of the end domains $\Omega^{l}_{e}$ and $\Omega^{r}_{e}$ are
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{3}}\cup\gamma_{e_{3},e_{1}}^{l}\cup\gamma_{e_{1},i\infty}$
and
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{2}}\cup\gamma_{e_{2},e_{4}}^{r}\cup\gamma_{e_{4},-i\infty}$,
respectively. Figure 3 Case I-2 gives an example of the domain configuration.

3\. Suppose now that the boundary of one of the domains $\Omega_{c}(-1)$ and
$\Omega_{c}(1)$ contains both zeros $e_{1}$ and $e_{2}$, but the boundary of
the other domain contains only one of these zeros. Assume, without loss of
generality, that $e_{1},e_{2}\in\Gamma_{c}(-1)$ and $e_{2}\in\Gamma_{c}(1)$
but $e_{1}\not\in\Gamma_{c}(1)$. This implies that $e_{1}\not=e_{2}$ and that
$e_{1}$ and $e_{2}$ are connected by the critical trajectory
$\gamma_{e_{1},e_{2}}\subset\mathbb{H}_{+}$ of $Q_{0}(z)\,dz^{2}$. By
symmetry, $e_{3}$ and $e_{4}$ are connected by the critical trajectory
$\gamma_{e_{3},e_{4}}\subset\mathbb{H}_{-}$. Furthermore, it implies that
$\Omega_{c}(-1)$ has a common boundary arc with each of the domains
$\Omega_{c}(1)$, $\Omega_{e}^{l}$, and $\Omega_{e}^{r}$. The latter implies,
in turn, that $\Omega_{c}(-1)$, $\Omega_{c}(1)$, $\Omega_{e}^{l}$, and
$\Omega_{e}^{r}$ are the only domains in the domain configuration of
$Q_{0}(z)\,dz^{2}$. This configuration occurs if and only if the limit in each
of inequalities (4.1) and (4.2) is zero and, additionally, the following
inequality holds:

(4.3)
$\operatorname{Re}\,[b,e_{1}]_{Q_{0}}+\operatorname{Re}\,[a,e_{1}]_{Q_{0}}>\operatorname{Re}\,[b,e_{2}]_{Q_{0}}+\operatorname{Re}\,[a,e_{2}]_{Q_{0}},$

where the points $-1<a<1$ and $b>1$ are chosen so that the points
$e_{1},e_{2},a$ and the points $e_{1},e_{2},b$ do not lie on a straight line.

In this case, the boundary of the circle domain $\Omega_{c}(1)$ is
$\Gamma_{c}(1)=\gamma_{e_{2},e_{4}}^{c}\cup\gamma_{e_{4},e_{2}}^{r}$, and the
boundary of $\Omega_{c}(-1)$ is
$\Gamma_{c}(-1)=\gamma_{e_{1},e_{3}}^{l}\cup\gamma_{e_{3},e_{4}}\cup\gamma_{e_{4},e_{2}}^{c}\cup\gamma_{e_{2},e_{1}}$.

The boundaries of the end domains $\Omega_{e}^{l}$ and $\Omega_{e}^{l}$ are
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{3}}\cup\gamma_{e_{3},e_{1}}^{l}\cup\gamma_{e_{1},i\infty}$
and
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup\gamma_{e_{1},e_{2}}\cup\gamma_{e_{2},e_{4}}^{r}\cup\gamma_{e_{4},e_{3}}\cup\gamma_{e_{3},-i\infty}$,
respectively. See Figure 4 Case I-3.

In the case when $e_{1},e_{2}\in\Gamma_{c}(1)$ and $e_{2}\in\Gamma_{c}(-1)$
but $e_{1}\not\in\Gamma_{c}(-1)$, the domain configuration is the mirror
configuration to the configuration described above as it is shown in Figure 4
Case I-3-m.

The remaining case, when the boundary of each of the domains $\Omega_{c}(-1)$
and $\Omega_{c}(1)$ contains both zeros $e_{1}$ and $e_{2}$, is our first
degenerate case. Indeed, the argument used earlier in part I-3 shows that, if
$e_{1}\not=e_{2}$, then there is a critical trajectory $\gamma_{e_{1},e_{2}}$,
which belongs to the boundaries of both $\Omega_{c}(-1)$ and $\Omega_{c}(1)$.
The latter easily leads to a contradiction. Hence, this case happens if and
only if $e_{1}=e_{2}$. Thus, $Q_{0}(z)\,dz^{2}$ has only two zeros $e_{1,2}$
and $e_{3,4}$ of order two each. The Stokes graph and domain configuration for
this degenerate case are shown in Figure 5.

II. Suppose that there are two real zeros $e_{1}$ and $e_{2}$ and two complex
conjugate zeros $e_{3}=\alpha_{3}+i\beta_{3}$ with $\beta_{3}>0$ and
$e_{4}=\overline{e_{3}}$. Below, we classify possible domain configurations,
first, depending on positions of the real zeros with respect to the poles
$z=-1$ and $z=1$, and then by positions of zeros on the boundaries of domains
present in the domain configurations of $Q_{0}(z)\,dz^{2}$.

In the generic cases discussed in this part and illustrated in Figures 6–24 in
Appenix B, we assume that all these zeros are distinct and that $e_{k}\not=\pm
1$, $k=1,2$.

1\. Let $e_{1}<e_{2}<-1$. In this case, the interval $(e_{1},e_{2})$ is a
trajectory of $Q_{0}(z)\,dz^{2}$ and the intervals $(-\infty,e_{1})$,
$(e_{2},-1)$, $(-1,1)$, and $(1,\infty)$ are orthogonal trajectories of
$Q_{0}(z)\,dz^{2}$. This implies that the quadratic differential
$Q_{0}(z)\,dz^{2}$ has two circle domains $\Omega_{c}(-1)$, $\Omega_{c}(1)$,
two end domains $\Omega_{e}^{l}$, $\Omega_{e}^{r}$, and, possibly, some other
domains.

One more conclusion, which easily follows from the assumption $e_{1}<e_{2}<-1$
is that none of the points of the interval $(e_{1},e_{2}]$ belongs to the
boundary of $\Omega_{e}^{l}$. Indeed, if $x_{0}\in\Gamma_{e}^{l}$,
$e_{1}<x_{0}\leq e_{2}$, then there is a Jordan arc
$\gamma\subset\Omega_{e}^{l}$ symmetric with respect to the real axis, which
has both its end points at $x_{0}$ and crosses the real axis at some point
$x_{1}<e_{1}$. Thus, the zero $e_{1}$ is the only critical point of
$Q_{0}(z)\,dz^{2}$ that is inside the domain bounded by $\gamma$. Now, the
required result follows from part 4 of Lemma 3.1 or, alternatively, it can be
proved as follows. Since the critical trajectories, different from the
interval $(e_{1},e_{2})$, each of which has one end point at $e_{1}$, must
have a second end point at some critical point of $Q_{0}(z)\,dz^{2}$, these
trajectories must intersect the curve $\gamma$. Since $\gamma$ lies in the end
domain $\Omega_{e}^{l}$, the latter is not possible. Similar argument shows
that none of the points of the interval $[e_{1},e_{2})$ belongs to the
boundary of $\Omega_{c}(-1)$.

Thus, each of the boundaries $\Gamma_{e}^{l}$ and $\Gamma_{c}(-1)$ may contain
$1$, $2$, or $3$ zeros of $Q_{0}(z)\,dz^{2}$. Accordingly, we have the
following subcases.

(a) Suppose that $e_{1}$ is the only zero on the boundary of $\Omega_{e}^{l}$
and $e_{2}$ is the only zero on the boundary of $\Omega_{c}(-1)$. In this
case, the boundary of the end domain $\Omega^{l}_{e}$ is
$\Gamma^{l}_{e}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$ and the
boundary of the circle domain $\Omega_{c}(-1)$ is
$\Gamma_{c}(-1)=\gamma_{e_{2}}^{c}$, where the critical trajectory
$\gamma_{e_{2}}^{c}$ has both its end points at $e_{2}$ and intersects the
interval $(-1,1)$ at some point $x_{1}$. The latter also implies that
$e_{3},e_{4}\in\Gamma_{c}(1)$ but $e_{1}\not\in\Gamma_{c}(1)$. Thus, in terms
of the $Q_{0}$-metric, the zero $e_{2}$ is closer to the pole $-1$ than
$e_{3}$ and the zero $e_{3}$ is closer to the pole $1$ than $e_{2}$. This
means that the assumptions in this part hold if and only if the following
inequalities hold:

(4.4)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[-1-\varepsilon,e_{2}]_{Q_{0}}-\operatorname{Im}[-1-\varepsilon,e_{3}]_{Q_{0}}\right)<0,$
(4.5)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[1+i\varepsilon,e_{2}]_{Q_{0}}-\operatorname{Im}[1+i\varepsilon,e_{3}]_{Q_{0}}\right)>0,$

Since $e_{3},e_{4}$ are the only zeros on $\Gamma_{c}(1)$ it follows that
$\Gamma_{c}(1)=\gamma_{e_{3},e_{4}}^{c}\cup\gamma_{e_{4},e_{3}}^{r}$, where
$\gamma_{e_{3},e_{4}}^{c}$ intersects $(-1,1)$ at some point $x_{2}$,
$x_{1}<x_{2}<1$. The remaining critical trajectories are
$\gamma_{e_{3},i\infty}$ and $\gamma_{e_{4},-i\infty}$. Under these
circumstances, there is one more face of the Stokes graph of
$Q_{0}(z)\,dz^{2}$ that is a strip domain $\Omega_{s}(-i\infty,i\infty)$ with
sides
$\Gamma_{s}^{l}(-i\infty,i\infty)=\gamma_{-i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{c}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},i\infty}$
and
$\Gamma_{s}^{r}(-i\infty,i\infty)=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{3}}^{c}\cup\gamma_{e_{3},i\infty}$.
Figure 6 Case II-1-a gives an example of the domain configuration.

The mirror configuration, shown in Figure 6 Case II-1-a-m, occurs when
$1<e_{2}<e_{1}$, $e_{1}$ is the only zero on $\Gamma_{e}^{r}$, and $e_{2}$ is
the only zero on $\Gamma_{c}(1)$. This case happens if and only if

(4.6)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[1+\varepsilon,e_{2}]_{Q_{0}}-\operatorname{Im}[1+\varepsilon,e_{3}]_{Q_{0}}\right)<0,$
(4.7)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[-1+i\varepsilon,e_{2}]_{Q_{0}}-\operatorname{Im}[-1+i\varepsilon,e_{3}]_{Q_{0}}\right)>0.$

(b) Suppose that $e_{1}$ is the only zero on $\Gamma_{e}^{l}$ and that
$e_{3},e_{4}\in\Gamma_{c}(-1)$ but $e_{1},e_{2}\not\in\Gamma_{c}(-1)$. This
assumption holds if and only if the limits in inequalities (4.4) and (4.5) are
positive. It is immediate from the latter assumption that
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$,
$\Gamma_{c}(-1)=\gamma_{e_{3},e_{4}}^{l}\cup\gamma_{e_{4},e_{3}}^{c}$ and
$\Gamma_{c}(1)=\gamma_{e_{3},e_{4}}^{c}\cup\gamma_{e_{4},e_{3}}^{r}$.

Furthermore, the set
$\Gamma_{r}^{(inn)}=\gamma_{e_{3},e_{4}}^{l}\cup\gamma_{e_{4},e_{3}}^{r}$ is a
boundary component of a face of the Stokes graph of $Q_{0}(z)\,dz^{2}$, which
in this case must be a ring domain $\Omega_{r}$ of $Q_{0}(z)\,dz^{2}$, see
Lemma 3.1. Under these circumstances, the only possibility for the outer
boundary component of $\Omega_{r}$ is that
$\Gamma_{r}^{(out)}=\gamma_{e_{2}}^{r}$, where $\gamma_{e_{2}}^{r}$ has both
end points at $e_{2}$ and crosses the interval $(1,\infty)$.

The boundary of the end domain $\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$.
Figure 7 Case II-1-b gives an example of the domain configuration.

The mirror configuration shown in Figure 7 Case II-1-b-m, occurs when
$1<e_{2}<e_{1}$, $e_{1}$ is the only zero on $\Gamma_{e}^{r}$,
$e_{3},e_{4}\in\Gamma_{c}(1)$ but $e_{1},e_{2}\not\in\Gamma_{c}(1)$. This case
happens if and only if the limits in inequalities (4.6) and (4.7) are
positive.

(c) Let $e_{1}$ be the only zero on $\Gamma_{e}^{l}$ and
$e_{2},e_{3},e_{4}\in\Gamma_{c}(-1)$. As we have mentioned earlier,
$e_{1}\not\in\Gamma_{c}(-1)$. This happens if and only if the limits in
inequalities (4.4) and (4.5) are zero. Since
$e_{2},e_{3},e_{4}\in\Gamma_{c}(-1)$, it follows that there are critical
trajectories $\gamma_{e_{2},e_{3}}\subset\mathbb{H}_{+}$ and
$\gamma_{e_{2},e_{4}}\subset\mathbb{H}_{-}$, which belong to the boundary of
$\Omega_{c}(-1)$. Therefore in this case,
$\Gamma_{c}(-1)=\gamma_{e_{2},e_{4}}\cup\gamma_{e_{4},e_{3}}^{c}\cup\gamma_{e_{3},e_{2}}$
and $\Gamma_{c}(1)=\gamma_{e_{3},e_{4}}^{c}\cup\gamma_{e_{4},e_{3}}^{r}$.
Under these circumstances, the only possibilities for the boundaries of the
end domains are the following:
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$ and
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}e_{3}}\cup\gamma_{e_{3},e_{4}}^{r}\cup\gamma_{e_{4},e_{2}}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$.
This also implies that $\Omega_{c}(-1)$, $\Omega_{c}(1)$, $\Omega_{e}^{l}$,
and $\Omega_{e}^{r}$ are the only domains in the domain configuration of
$Q_{0}(z)\,dz^{2}$. The case is illustrated in Figure 8 Case II-1-c.

The mirror configuration, shown in Figure 8 Case II-1-c-m, occurs when
$1<e_{2}<e_{1}$, $e_{1}$ is the only zero on $\Gamma_{e}^{r}$,
$e_{2},e_{3},e_{4}\in\Gamma_{c}(1)$. This case happens if and only if the
limits in inequalities (4.6) and (4.7) are zero.

(d) Suppose that $\Gamma_{e}^{l}$ contains two zeros, which in this case are
$e_{3}$ and $e_{4}$. In terms of the $Q_{0}$-metric, the latter means that the
zeros $e_{1}$ and $e_{2}$ are closer to the poles $z=\pm 1$ than the zeros
$e_{3}$ and $e_{4}$. This assumption holds if and only if the limits in
inequalities (4.4) and (4.5) are negative. In this case, there are critical
trajectories $\gamma_{e_{3},e_{4}}^{l}$, crossing the interval
$(-\infty,e_{1})$, and $\gamma_{e_{3},e_{4}}^{r}$, crossing the interval
$(1,\infty)$ at some point $x_{2}$, and therefore we also have that
$e_{3},e_{4}\in\Gamma_{e}^{r}$, but $e_{1},e_{2}\not\in\Gamma_{e}^{r}$. The
boundaries of the end domains $\Omega_{e}^{l}$ and $\Omega_{e}^{r}$ are
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{3}}^{l}\cup\gamma_{e_{3},i\infty}$
and
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{4}}^{r}\cup\gamma_{e_{4},-i\infty}$.
It follows also that the set
$\Gamma_{r}^{(out)}=\gamma_{e_{3},e_{4}}^{l}\cup\gamma_{e_{4},e_{3}}^{r}$ is a
boundary component of one of the faces of $Q_{0}(z)\,dz^{2}$. Since the
interior of $\Gamma_{r}^{(out)}$ contains more than one critical points, by
Lemma 3.1, this face must be a ring domain $\Omega_{r}$. The latter implies
that there is a critical trajectory $\gamma_{e_{1}}^{r}$ having both end
points at $e_{1}$, which crosses the interval $(1,\infty)$ at some point
$x_{1}$, $1<x_{1}<x_{2}$. Then, $\Gamma_{r}^{(inn)}=\gamma_{e_{1}}^{r}$.

Under these conditions, we must have one more critical trajectory
$\gamma_{e_{2}}^{c}$, which has both its end points at $e_{2}$ and crosses the
interval $(-1,1)$. Therefore, the boundary of the circle domain
$\Omega_{c}(-1)$ is $\Gamma_{c}(-1)=\gamma_{e_{2}}^{c}$ and the boundary of
the circle domain $\Omega_{c}(1)$ is
$\Gamma_{c}(1)=\gamma_{e_{1}}^{r}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{c}\cup[e_{2},e_{1}]$.
Figure 9 Case II-1-d gives an example of the domain configuration described
above.

The mirror configuration, shown in Figure 9 Case II-1-d-m, occurs when
$1<e_{2}<e_{1}$, and when $e_{3}$ and $e_{4}$ are the only zero on
$\Gamma_{e}^{r}$. This case happens if and only if the limits in (4.6) and
(4.7) are negative.

(e) Suppose now that $e_{1},e_{3},e_{4}\in\Gamma_{e}^{l}$. This case happens
if and only if the limits in (4.4) and (4.5) are zero. Then the boundary of
the end domain $\Omega_{e}^{l}$ must contain critical trajectories
$\gamma_{e_{1},e_{3}}$ and $\gamma_{e_{1},e_{4}}$ and therefore
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{1}}\cap\gamma_{e_{1},e_{3}}\cup\gamma_{e_{3},i\infty}$.
The latter implies, in turn, that
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{4}}^{r}\cup\gamma_{e_{4},-i\infty}$.
Under these circumstances the remaining possibility is that there is a
critical trajectory $\gamma_{e_{2}}^{c}$, which has both its end points at
$e_{2}$ and crosses the interval $(-1,1)$. The latter implies that
$\Gamma_{c}(-1)=\gamma_{e_{2}}^{c}$,
$\Gamma_{c}(1)=\gamma_{e_{1},e_{4}}\cup\gamma_{e_{4},e_{3}}^{r}\cup\gamma_{e_{3},e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{1}}^{c}\cup[e_{2},e_{1}]$
and that $\Omega_{c}(-1)$, $\Omega_{c}(1)$, $\Omega_{e}^{l}$, and
$\Omega_{e}^{r}$ are the only domains in the domain configuration of
$Q_{0}(z)\,dz^{2}$ in the case under consideration. Figure 10 Case II-1-e
illustrates the domain configuration.

The mirror configuration, shown in Figure 10 Case II-1-e-m, occurs when
$1<e_{2}<e_{1}$, and when $e_{1},e_{3},e_{4}\in\Gamma_{e}^{r}$. This case
happens if and only if the limits in (4.6) and (4.7) are zero.

2\. Let $-1<e_{1}<e_{2}<1$. Then, the interval $(e_{1},e_{2})$ is a trajectory
of $Q_{0}(z)\,dz^{2}$ and the intervals $(-\infty,-1)$, $(-1,e_{1})$,
$(e_{2},1)$, and $(1,\infty)$ are orthogonal trajectories of
$Q_{0}(z)\,dz^{2}$. As in the case 1 above, this implies that the quadratic
differential $Q_{0}(z)\,dz^{2}$ has two circle domains $\Omega_{c}(-1)$,
$\Omega_{c}(1)$, two end domains $\Omega_{e}^{l}$, $\Omega_{e}^{r}$, and,
possibly, some other domains. Also, it is not difficult to see that either
$[e_{1},e_{2}]\subset\Gamma_{e}^{l}$ or
$[e_{1},e_{2}]\cap\Gamma_{e}^{l}=\emptyset$. Similarly, we have that either
$[e_{1},e_{2}]\subset\Gamma_{e}^{r}$ or
$[e_{1},e_{2}]\cap\Gamma_{e}^{r}=\emptyset$.

The latter implies that each of the boundaries $\Gamma_{e}^{l}$ and
$\Gamma_{e}^{r}$ may contain $2$ or $4$ zeros of $Q_{0}(z)\,dz^{2}$. Thus, we
have the following subcases.

(a) Suppose that $e_{3},e_{4}$ are the only zeros on each of the boundaries
$\Gamma_{e}^{l}$ and $\Gamma_{e}^{r}$. This means that the zeros $e_{1}$ and
$e_{2}$ are closer, in terms of the $Q_{0}$-metric, to the poles $z=\pm 1$
than the zeros $e_{3}$ and $e_{4}$. The latter happens if and only the
following inequalities hold:

(4.8)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[-1+\varepsilon,e_{1}]_{Q_{0}}-\operatorname{Im}[-1+\varepsilon,e_{3}]_{Q_{0}}\right)<0$
(4.9)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[1-\varepsilon,e_{2}]_{Q_{0}}-\operatorname{Im}[1-\varepsilon,e_{3}]_{Q_{0}}\right)<0.$

Since $e_{3},e_{4}$ are the only zeros on $\Gamma_{e}^{l}$ and on
$\Gamma_{e}^{r}$, it follows that there are critical trajectories
$\gamma_{e_{3},e_{4}}^{l}$ that crosses the interval $(-\infty,-1)$ at some
point $x_{1}$ and $\gamma_{e_{3},e_{4}}^{r}$ crossing the interval
$(1,\infty)$ an some point $x_{2}$. The boundaries of the end domains
$\Omega_{e}^{l}$ and $\Omega_{e}^{r}$ are
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{3}}^{l}\cup\gamma_{e_{3},i\infty}$
and
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{4}}^{r}\cup\gamma_{e_{4},-i\infty}$.

The set
$\Gamma_{r}^{(out)}=\gamma_{e_{3},e_{4}}^{l}\cup\gamma_{e_{4},e_{3}}^{r}$ is a
Jordan curve that is an outer boundary component of a face of the Stokes graph
of $Q_{0}(z)\,dz^{2}$, which in this case must be a ring domain $\Omega_{r}$
by Lemma 3.1. Under these circumstances, there are critical trajectories
$\gamma_{e_{1}}^{l}$ that crosses the interval $(x_{1},-1)$ and
$\gamma_{e_{2}}^{r}$ crossing the interval $(1,x_{2})$. Then the inner
boundary component of $\Omega_{r}$ is
$\Gamma^{(inn)}=\gamma_{e_{1}}^{l}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r}\cup[e_{2},e_{1}]$.
The boundaries of the circle domains $\Omega_{c}(-1)$ and $\Omega_{c}(1)$ are
$\Gamma_{c}(-1)=\gamma_{e_{1}}^{l}$ and $\Gamma_{c}(1)=\gamma_{e_{2}}^{r}$.
Figure 11 Case II-2-a gives an example of the domain configuration.

(b) Suppose that $e_{3},e_{4}$ are the only zeros on the boundary
$\Gamma_{e}^{l}$ and $e_{1},e_{2}$ are the only zeros on the boundary
$\Gamma_{e}^{r}$. The latter happens if and only the limit in (4.8) is
positive and the limit in (4.9) is negative. The boundary of the end domain
$\Omega_{e}^{l}$ is
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{3}}^{l}\cup\gamma_{e_{3},i\infty}$.
Under these assumptions, the domain configuration must contain a strip domain
$\Omega_{s}(-i\infty,i\infty)$ with the vertices at $\pm i\infty$, which
separates the end domains $\Omega_{e}^{l}$ and $\Omega_{e}^{r}$. It has sides
$\Gamma_{s}^{l}(-i\infty,i\infty)=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{3}}^{c}\cup\gamma_{e_{3},i\infty}$
and
$\Gamma_{s}^{r}(-i\infty,i\infty)=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$.

The circle domains have the boundaries
$\Gamma_{c}(-1)=\gamma_{e_{3},e_{4}}^{l}\cup\gamma_{e_{4},e_{3}}^{c}$,
$\Gamma_{c}(1)=\gamma_{e_{2}}^{r}$.

In this case, the boundary of the end domain $\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$.
Figure 12 Case II-2-b gives an example of such a domain configuration.

The mirror configuration, shown in Figure 12 Case II-2-b-m, occurs when
$-1<e_{2}<e_{1}<1$, and when $e_{3},e_{4}$ are the only zeros on the boundary
$\Gamma_{e}^{r}$ and $e_{1},e_{2}$ are the only zeros on the boundary
$\Gamma_{e}^{l}$. This case happens if and only if the limit in (4.8) is
negative and the limit in (4.9) is positive.

(c) Suppose that $e_{3},e_{4}$ are the only zeros on $\Gamma_{e}^{l}$ and all
zeros $e_{1},e_{2},e_{3},e_{4}$ belong to the boundary $\Gamma_{e}^{r}$. This
happens if and only if the limit in (4.8) is zero (then the limit in (4.9) is
also zero) and
$\operatorname{Re}[e_{1},e_{3}]_{Q_{0}}<\operatorname{Re}[e_{2},e_{3}]_{Q_{0}}$.
The boundary of the end domain $\Omega_{e}^{l}$ is the same as in the cases
(a) and (b) above,
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{3}}^{l}\cup\gamma_{e_{3},i\infty}$.
Since $\Gamma_{e}^{r}$ contains all zeros, it follows that there are critical
trajectories $\gamma_{e_{1},e_{3}}\subset\mathbb{H}_{+}$ and
$\gamma_{e_{1},e_{4}}\subset\mathbb{H}_{-}$. In this case,
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},e_{4}}\cup\gamma_{e_{4},-i\infty}$.
The boundary of the circle domain $\Omega_{c}(-1)$ is
$\Gamma_{c}(-1)=\gamma_{e_{3},e_{4}}^{l}\cup\gamma_{e_{4},e_{1}}\cup\gamma_{e_{1},e_{3}}$
and the boundary of the circle domain $\Omega_{c}(1)$ is
$\Gamma_{c}(1)=\gamma_{e_{2}}^{r}$. Figure 13 Case II-2-c shows the Stokes
graph and domain configuration.

The mirror configuration, shown in Figure 13 Case II-2-c-m, occurs when
$-1<e_{2}<e_{1}<1$, and when $e_{3},e_{4}$ are the only zeros on the boundary
$\Gamma_{e}^{r}$ and all zeros $e_{1},e_{2},e_{3},e_{4}$ belong to the
boundary $\Gamma_{e}^{l}$. This case happens if and only if the limits in
(4.8) and 4.9 are zero and
$\operatorname{Re}[e_{1},e_{3}]_{Q_{0}}<\operatorname{Re}[e_{2},e_{3}]_{Q_{0}}$.

3\. Let $e_{1}<-1<e_{2}<1$. The intervals $(-\infty,e_{1})$, $(e_{2},1)$ and
$(1,\infty)$ are orthogonal trajectories of $Q_{0}(z)\,dz^{2}$ and the
intervals $(e_{1},-1)$, $(-1,e_{2})$ are trajectories of $Q_{0}(z)\,dz^{2}$.
Therefore, there is only one circle domain $\Omega_{c}(1)$ and there is at
least one strip domain having one or both its vertices at $z=-1$. Notice that
the intervals $[e_{1},-1]$ and $[-1,e_{2}]$ can not lie on the boundary of the
end domain $\Omega_{e}^{r}$. The latter implies that $e_{3},e_{4}$ are the
only zeros of $Q_{0}(z)\,dz^{2}$, which belong to $\Gamma_{e}^{r}$. Therefore,
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{4}}^{r}\cup\gamma_{e_{4},-i\infty}$
in all subcases considered below. Also, as in the previous cases, it is not
difficult to see that $e_{2}$ can not belong to the boundary of the end domain
$\Omega_{e}^{l}$. Thus, $\Gamma_{e}^{l}$ may contain $1$, $2$, or $3$ zeros of
$Q_{0}(z)\,dz^{2}$. Accordingly, we have the following subcases.

(a) Let $e_{1}$ be the only zero on $\Gamma_{e}^{l}$. Then
$\Gamma^{l}_{e}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$. This
together with part 3 of Lemma 3.1 imply that there are strip domains
$\Omega_{s}(-1,i\infty)$ and $\Omega_{s}(-i\infty,-1)$, symmetric to each
other with respect to the real axis, which left sides are
$\Gamma_{s}^{l}(-1,i\infty)=[-1,e_{1}]\cup\gamma_{e_{1},i\infty}$ and
$\Gamma_{s}^{l}(-i\infty,-1)=\gamma_{-i\infty,e_{1}}\cup[e_{1},-1]$,
respectively. The right side of each of the strip domains
$\Omega_{s}(-1,i\infty)$ and $\Omega_{s}(-1,-i\infty)$ must contain at least
one zero of $Q_{0}(z)\,dz^{2}$. Thus, we have three possibilities.

($\alpha$) Let $e_{3}$ be the only zero on $\Gamma_{s}^{r}(-1,i\infty)$. Then,
by symmetry, $e_{4}$ is the only zero on $\Gamma_{s}^{r}(-i\infty,-1)$. Under
the assumptions of this case, the latter happens if and only if the following
holds:

(4.10)
$2\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}+2\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}=\delta_{-1},$

where $\delta_{-1}$ is defined in (3.3). Under these conditions, there are
critical trajectories $\gamma_{-1,e_{3}}\subset\mathbb{H}_{+}$, which joins
the pole $-1$ and $e_{3}$, and $\gamma_{-1,e_{4}}\subset\mathbb{H}_{-}$, which
joins $-1$ and $e_{4}$. Therefore,
$\Gamma_{s}^{r}(-1,i\infty)=\gamma_{-1,e_{3}}\cup\gamma_{e_{3},i\infty}$ and
$\Gamma_{s}^{r}(-i\infty,-1)=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},-1}$.

Under these circumstances, the remaining critical trajectory is
$\gamma_{e_{2}}^{r}$ and the boundary of the circle domain $\Omega_{c}(1)$ is
$\Gamma_{c}(1)=\gamma_{e_{2}}^{r}$.

It follows from our discussion above that there is one more face of the Stokes
graph of $Q_{0}(z)\,dz^{2}$, which in this case must be a strip domain
$\Omega_{s}(-1-1)$ with both its vertices at $-1$ having
$\Gamma_{s}^{(out)}(-1,-1)=\gamma_{-1,e_{4}}\cup\gamma_{e_{4},e_{3}}^{r}\cup\gamma_{e_{3},-1}$
as its outer side and
$\Gamma_{s}^{(inn)}(-1,-1)=[-1,e_{2}]\cup\gamma_{e_{2}}^{r}\cup[e_{2},-1]$ as
its inner side. Figure 14 Case II-3-a-$\alpha$ shows an example of the Stokes
graph and domain configuration.

The mirror configuration, shown in Figure 14 Case II-3-a-$\alpha$-m, occurs
when $-1<e_{2}<1<e_{1}$, $e_{1}$ is the only zero on $\Gamma_{e}^{r}$, and
$e_{3}$ is the only zero on $\Gamma_{s}^{l}(1,i\infty)$. This case happens if
and only if the following holds:

(4.11)
$2\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}+2\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}=\delta_{1}.$

($\beta$) Let $e_{2}$ be the only zero on each of the sides
$\Gamma_{s}^{r}(-1,i\infty)$ and $\Gamma_{s}^{r}(-i\infty,-1)$. This case
happens if and only if

(4.12)
$2\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}=2\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}+\delta_{-1}.$

Under these assumptions, there are critical trajectories
$\gamma_{e_{2},i\infty}$ and $\gamma_{e_{2},-i\infty}$ and therefore
$\Gamma_{s}^{r}(-1,i\infty)=[-1,e_{2}]\cup\gamma_{e_{2},i\infty}$ and
$\Gamma_{s}^{r}(-i\infty,e_{2})=\gamma_{-i\infty,e_{2}}\cup[e_{2},-1]$.

Hence, there is one more face of the Stokes graph of $Q_{0}(z)\,dz^{2}$, which
is a strip domain $\Omega_{s}(-i\infty,i\infty)$ with its vertices at
$i\infty$ and $-i\infty$. The sides of this strip domain are
$\Gamma_{s}^{l}(-i\infty,i\infty)=\gamma_{-i\infty,e_{2}}\cup\gamma_{e_{2},i\infty}$
and
$\Gamma_{s}^{r}(-i\infty,i\infty)=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{3}}^{c}\cup\gamma_{e_{3},i\infty}$.
In this case, the boundary of the circle domain $\Omega_{c}(1)$ is
$\Gamma_{c}(1)=\gamma_{e_{3},e_{4}}^{c}\cup\gamma_{e_{4},e_{3}}^{r}$, see
Figure 15 Case II-3-a-$\beta$.

The mirror configuration, shown in Figure 15 Case II-3-a-$\beta$-m, occurs
when $-1<e_{2}<1<e_{1}$, $e_{1}$ is the only zero on $\Gamma_{e}^{r}$, and
$e_{2}$ is the only zero on each of the sides $\Gamma_{s}^{l}(1,i\infty)$ and
$\Gamma_{s}^{l}(-i\infty,1)$. This case happens if and only if

(4.13)
$2\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}=2\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}+\delta_{1}.$

($\gamma$) Suppose now that $e_{2},e_{3}\in\Gamma_{s}^{r}(-1,i\infty)$ and
$e_{2},e_{4}\in\Gamma_{s}^{r}(-1,-i\infty)$. This case happens if and only if
$\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}=\delta_{-1}$.

Since zeros $e_{2}$ and $e_{3}$ belong to the same side of the strip domain
$\Omega_{s}(-1,i\infty)$, we have that
$\Gamma_{s}^{r}(-1,i\infty)=[-1,e_{2}]\cup\gamma_{e_{2},e_{3}}\cup\gamma_{e_{3},i\infty}$.
By symmetry,
$\Gamma_{s}^{r}(-i\infty,-1)=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{2}}\cup[e_{2},-1]$.
The latter implies, in turn, that the boundary of the circle domain
$\Omega_{c}(1)$ is
$\Gamma_{c}(1)=\gamma_{e_{2},e_{4}}\cup\gamma_{e_{4},e_{3}}^{r}\cup\gamma_{e_{3},e_{2}}$.
The Stokes graph and the domain configuration are shown in Figure 16 Case
II-3-a-$\gamma$.

The mirror configuration, shown in Figure 16 Case II-3-a-$\gamma$-m, occurs
when $-1<e_{2}<1<e_{1}$, $e_{1}$ is the only zero on $\Gamma_{e}^{r}$, and
$e_{2},e_{3}\in\Gamma_{s}^{l}(1,i\infty)$. This case happens if and only if
$\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}=\delta_{1}$.

(b) Suppose that $\Gamma_{e}^{l}$ contains two zeros, which are $e_{3},e_{4}$.
This case happens if and only if

$2\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}=2\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}+\delta_{-1}.$

Then
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{3}}^{l}\cup\gamma_{e_{3},i\infty}$.
This also implies that $e_{3},e_{4}\in\Gamma_{e}^{r}$ and, therefore,
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{4}}^{r}\cup\gamma_{e_{4},-i\infty}$,
where $\gamma_{e_{3},e_{4}}^{r}$ intersects $(1,\infty)$ at some point
$x_{3}$.

Then, the set
$\Gamma_{r}^{(out)}=\gamma_{e_{3},e_{4}}^{r}\cup\gamma_{e_{4},e_{3}}^{l}$ is
an outer boundary component of a face of the Stokes graph of
$Q_{0}(z)\,dz^{2}$, which must be a ring domain $\Omega_{r}$, see Lemma 3.1.
The inner boundary component of this ring domain contains just one zero
$e_{1}$ and therefore $\Gamma_{r}^{(inn)}=\gamma_{e_{1}}^{r}$, where
$\gamma_{e_{1}}^{r}$ intersects the real axis at some point $x_{2}$,
$1<x_{2}<x_{3}$. Since $\Omega_{r}$ separates $e_{1},e_{2}$ from
$e_{3},e_{4}$, this case happens if and only if
$2\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}>\delta_{-1}$.

Under these circumstances, the only possibility for the boundary of the circle
domain $\Omega_{c}(1)$ is that $\Gamma_{c}(1)=\gamma_{e_{2}}^{r}$, where
$\gamma_{e_{2}}^{r}$ intersects the real axis at some point $x_{1}$,
$1<x_{1}<x_{2}$.

The latter implies that the remaining face of the Stokes graph is the strip
domain $\Omega_{s}(-1,-1)$ with both vertices at $-1$, which outer and inner
sides are
$\Gamma_{s}^{(out)}(-1,-1)=[-1,e_{1}]\cup\gamma_{e_{1}}^{r}\cup[e_{1},-1]$ and
$\Gamma_{s}^{(inn)}(-1,-1)=[-1,e_{2}]\cup\gamma_{c_{2}}^{r}\cup[e_{2},-1]$,
see Figure 17 Case II-3-b.

The mirror configuration, shown in Figure 17 Case II-3-b-m, occurs when
$-1<e_{2}<1<e_{1}$ and when $e_{3}$ and $e_{4}$ are the only zeros on
$\Gamma_{e}^{r}$. This case happens if and only if

$2\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}=2\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}+\delta_{1}.$

(c) Suppose now that $\Gamma_{e}^{l}$ contains zeros $e_{1},e_{3},e_{4}$. This
case happens if and only if
$2\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}=\delta_{-1}.$ This implies that
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{1}}\cup\gamma_{e_{1},e_{3}}\cup\gamma_{e_{3},i\infty}$.
This also implies that $e_{3},e_{4}\in\Gamma_{e}^{r}$ and, therefore,
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{4}}^{r}\cup\gamma_{e_{4},-i\infty}$,
where $\gamma_{e_{3},e_{4}}^{r}$ intersects $(1,\infty)$ at some point
$x_{2}$.

In this case, the set
$[-1,e_{1}]\cup\gamma_{e_{1},e_{4}}\cup\gamma_{e_{4},e_{3}}^{r}\cup\gamma_{e_{3},e_{1}}\cup[e_{1},-1]$
is connected and therefore, by Lemma 3.1, it is an outer side
$\Gamma_{s}^{(out)}(-1,-1)$ of the strip domain $\Omega_{s}(-1,-1)$. The inner
side of this strip domain is
$\Gamma_{s}^{(inn)}(-1,-1)=[-1,e_{2}]\cup\gamma_{e_{2}}^{r}\cup[e_{2},-1]$,
where $\gamma_{e_{2}}^{r}$ intersects the real axis at some point $x_{1}$,
$1<x_{1}<x_{2}$.

Under the made assumptions, the remaining possibility for the boundary of the
circle domain $\Omega_{c}(1)$ is that $\Gamma_{c}(1)=\gamma_{e_{2}}^{r}$,
where $\gamma_{e_{2}}^{r}$. The Stokes graph has four faces only, as it is
shown in Figure 18 Case II-3-c.

The mirror configuration, shown in Figure 18 Case II-3-c-m, occurs when
$-1<e_{2}<1<e_{1}$ and $\Gamma_{e}^{r}$ contains zeros $e_{1},e_{3},e_{4}$.
This case happens if and only if
$2\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}=\delta_{1}$.

4\. Consider the case when $e_{1}<-1$ and $e_{2}>1$. In this case the
intervals $(-\infty,e_{1})$ and $(e_{2},\infty)$ are orthogonal trajectories
of $Q_{0}(z)\,dz^{2}$ and the intervals $(e_{1},-1)$, $(-1,1)$ and $(1,e_{2})$
are trajectories of $Q_{0}(z)\,dz^{2}$. Therefore, there are no circle domains
in this case and, in all possible subcases, there is a strip domain
$\Omega_{s}(-1,1)$ with vertices at the poles $-1$ and $1$, which is symmetric
with respect to the real axis.

As in the previous cases, each of the boundaries $\Gamma_{e}^{l}$ and
$\Gamma_{e}^{r}$ of the end domains may contain $1$, $2$, or $3$ zeros and
$e_{2}\not\in\Gamma_{e}^{l}$, $e_{1}\not\in\Gamma_{e}^{r}$. Thus, we consider
the following subcases.

(a) Let $e_{1}$ be the only zero on $\Gamma_{e}^{l}$. Then
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$. This
combined with Lemma 3.1 imply that there are strip domains
$\Omega_{s}(-1,i\infty)$ and $\Omega_{s}(-i\infty,-1)$, symmetric to each
other with respect to the real axis, which left and right sides are
$\Gamma_{s}^{l}(-1,i\infty)=[-1,e_{1}]\cup\gamma_{e_{1},i\infty}$ and
$\Gamma_{s}^{r}(-i\infty,-1)=\gamma_{-i\infty,e_{1}}\cup[e_{1},-1]$,
respectively.

In its turn, the boundary of $\Omega_{e}^{r}$ also may have $1$, $2$, or $3$
zeros. Thus, we have three possibilities.

($\alpha$) Let $e_{2}$ be the only zero on $\Gamma_{e}^{r}$. This happens if
and only if $2\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}<\delta_{-1}$ and
$2\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}<\delta_{1}$. Then,
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{2}}\cup\gamma_{e_{2},-i\infty}$. This also
implies that there are strip domains $\Omega_{s}(1,i\infty)$ and
$\Omega_{s}(-i\infty,1)$, symmetric to each other with respect to the real
axis, which right sides are
$\Gamma_{s}^{r}(1,i\infty)=[1,e_{2}]\cup\gamma_{e_{2},i\infty}$ and
$\Gamma_{s}^{r}(-i\infty,1)=\gamma_{-i\infty,e_{2}}\cup[e_{2},1]$,
respectively. The right sides of each of the strip domains
$\Omega_{s}(-1,i\infty)$ and $\Omega_{s}(-i\infty,-1)$ and the left sides of
each of the strip domains $\Omega_{s}(1,i\infty)$ and $\Omega_{s}(-i\infty,1)$
must contain at least one zero of $Q_{0}(z)\,dz^{2}$. This implies that
$\Gamma_{s}^{r}(-1,i\infty)=\gamma_{-1,e_{3}}\cup\gamma_{e_{3},i\infty}$,
$\Gamma_{s}^{r}(-i\infty,-1)=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},-1}$,
$\Gamma_{s}^{l}(1,i\infty)=\gamma_{1,e_{3}}\cup\gamma_{e_{3},i\infty}$, and
$\Gamma_{s}^{l}(-i\infty,1)=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},1}$.
Under these circumstances, the sides of the strip domain $\Omega_{s}(-1,1)$
are $\Gamma_{s}^{+}(-1,1)=\gamma_{-1,e_{3}}\cup\gamma_{e_{3},1}$ and
$\Gamma_{s}^{-}(-1,1)=\gamma_{-1,e_{4}}\cup\gamma_{e_{4},1}$. The
corresponding domain configuration is shown in Figure 19 Case II-4-a-$\alpha$.

($\beta$) Suppose that $\Gamma_{e}^{r}$ contains two zeros, which in this case
are $e_{3}$ and $e_{4}$. This case happens if and only if
$2\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}+2\operatorname{Im}[e_{2},e_{3}]+\delta_{1}=\delta_{-1}$.
Then,
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{4}}^{r}\cup\gamma_{e_{4},-i\infty}$.
This also implies that the right sides of the strip domains
$\Omega_{s}(-1,i\infty)$ and $\Omega_{s}(-i\infty,-1)$ are
$\Gamma_{s}^{r}(-1,i\infty)=\gamma_{-1,e_{3}}\cup\gamma_{e_{3},i\infty}$ and
$\Gamma_{s}^{r}(-i\infty,-1)=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},-1}$,
respectively. Furthermore, the assumptions imply that $e_{2}$ is the only zero
on the boundary of the strip domain $\Omega_{s}(-1,1)$. Therefore, there are
critical trajectories $\gamma_{-1,e_{2}}^{+}\subset\mathbb{H}_{+}$ and
$\gamma_{-1,e_{2}}^{-}\subset\mathbb{H}_{-}$ and the sides of
$\Omega_{s}(-1,1)$ are
$\Gamma_{s}^{+}(-1,1)=\gamma_{-1,e_{2}}^{+}\cup[e_{2},1]$ and
$\Gamma_{s}^{-}(-1,1)=\gamma_{-1,e_{2}}^{-}\cup[e_{2},1]$. The remaining face
of $Q_{0}(z)\,dz^{2}$ in this case is the strip domain $\Omega_{s}(-1,-1)$,
symmetric with respect to the real axis, with both its vertices at $-1$ and
sides
$\Gamma_{s}^{(out)}(-1,-1)=\gamma_{-1,e_{4}}\cup\gamma_{e_{4},e_{3}}^{r}\cup\gamma_{e_{3},-1}$
and
$\Gamma_{s}^{(inn)}(-1,-1)=\gamma_{-1,e_{2}}^{-}\cup\gamma_{e_{2},-1}^{+}$.
Figure 20 Case II-4-a-$\beta$ gives an example of a domain configuration.

The mirror configuration, shown in Figure 20 Case II-4-a-$\beta$-m, occurs
when $e_{1}>1$, $e_{2}<-1$ and when $e_{3}$, $e_{4}$ are the only zeros of
$\Gamma_{e}^{l}$. This case happens if and only if
$2\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}+2\operatorname{Im}[e_{2},e_{3}]+\delta_{-1}=\delta_{1}$.

($\gamma$) Suppose now that $e_{2},e_{3},e_{4}\in\Gamma_{e}^{r}$. This happens
if and only if
$2\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}+\delta_{1}=\delta_{-1}$. Then,
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{2}}\cup\gamma_{e_{2},e_{4}}\cup\gamma_{e_{4},-i\infty}$.
This also implies that the right sides of the strip domains
$\Omega_{s}(-1,i\infty)$ and $\Omega_{s}(-i\infty,-1)$ are
$\Gamma_{s}^{r}(-1,i\infty)=\gamma_{-1,e_{3}}\cup\gamma_{e_{3},i\infty}$ and
$\Gamma_{s}^{r}(-i\infty,-1)=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},-1}$,
respectively.

Furthermore, our assumptions in this case imply that $e_{2}$, $e_{3}$, and
$e_{4}$ belong to the boundary of the strip domain $\Omega_{s}(-1,1)$.
Therefore,
$\Gamma_{s}^{+}(-1,1)=\gamma_{-1,e_{3}}\cup\gamma_{e_{3},e_{2}}\cup[e_{2},1]$
and
$\Gamma_{s}^{-}(-1,1)=\gamma_{-1,e_{4}}\cup\gamma_{e_{4},e_{2}}\cup[e_{2},1]$.
There are no other domains in the domain configuration of $Q_{0}(z)\,dz^{2}$,
see Figure 21 Case II-4-a-$\gamma$.

The mirror configuration, shown in Figure 21 Case II-4-a-$\gamma$-m, occurs
when $e_{1}>1$, $e_{2}<-1$ and when $e_{2},e_{3},e_{4}\in\Gamma_{e}^{l}$. This
case happens if and only if
$2\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}+\delta_{-1}=\delta_{1}$.

(b) Suppose that each of the boundaries $\Gamma_{e}^{l}$ and $\Gamma_{e}^{r}$
contains two zeros, which are $e_{3}$ and $e_{4}$ for each of these
boundaries. Then there are critical trajectories $\gamma_{e_{3},e_{4}}^{l}$,
which crosses the real axis at some point $x_{1}<e_{1}$ and
$\gamma_{e_{3},e_{4}}^{r}$, which crosses the real axis at some point
$x_{2}>e_{2}$. This immediately implies that
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{3}}^{l}\cup\gamma_{e_{3},i\infty}$
and
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{4}}^{r}\cup\gamma_{e_{4},-i\infty}$.
In this case, the closed Jordan curve
$\Gamma_{r}^{(out)}=\gamma_{e_{3},e_{4}}^{l}\cup\gamma_{e_{4},e_{3}}^{r}$ is
an outer boundary component of a face of the quadratic differential
$Q_{0}(z)\,dz^{2}$, which must be a ring domain $\Omega_{r}$. The inner
boundary component $\Gamma_{r}^{(inn)}$ of $\Omega_{r}$ may contain one or
both of the zeros $e_{1}$, $e_{2}$. Thus, we have the following subcases.

($\alpha$) Let $e_{1}\in\Gamma_{r}^{(inn)}$, but
$e_{2}\not\in\Gamma_{r}^{(inn)}$. This case happens if and only if
$\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}<\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}$
and $\delta_{-1}>\delta_{1}$. Then, $\Gamma_{r}^{(inn)}=\gamma_{e_{1}}^{r}$,
where $\gamma_{e_{1}}^{r}$ is the critical trajectory with both end points at
$e_{1}$, which crosses the real axis at some point $x_{3}$,
$e_{2}<x_{3}<x_{2}$. Under the assumptions, there are critical trajectories
$\gamma_{e_{2},-1}^{+}\subset\mathbb{H}_{+}$ and
$\gamma_{e_{2},-1}^{-}\subset\mathbb{H}_{-}$ having their end points at the
points $-1$ and $e_{2}$. This implies that
$\Gamma_{s}^{+}(-1,1)=\cup\gamma_{-1,e_{2}}^{+}\cup[e_{2},1]$ and
$\Gamma_{s}^{-}(-1,1)=\gamma_{-1,e_{2}}^{-}\cup[e_{1},1]$.

Under these circumstances, there is one more face of the Stokes graph of
$Q_{0}(z)\,dz^{2}$, which is a strip domain $\Omega_{s}(-1,-1)$ symmetric with
respect to the real axis having both its vertices at the pole $-1$. In this
case,
$\Gamma_{s}^{(out)}(-1,-1)=[-1,e_{1}]\cup\gamma_{e_{1}}^{r}\cup[e_{1},-1]$ and
$\Gamma_{s}^{(inn)}(-1,-1)=\gamma_{-1,e_{2}}^{+}\cup\gamma_{e_{2},-1}^{-}$.
Figure 22 Case II-4-b-$\alpha$ gives an example of the Stokes graph and domain
configuration.

The mirror configuration, shown in Figure 22 Case II-4-b-$\alpha$-m, occurs
when $e_{1}>1$, $e_{2}<-1$, when $e_{3},e_{4}$ are the only zeros on
$\Gamma_{e}^{l}$ and on $\Gamma_{e}^{r}$, and when
$e_{1}\in\Gamma_{r}^{(inn)}$ but $e_{2}\not\in\Gamma_{r}^{(inn)}$. This case
happens if and only if
$\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}<\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}$
and $\delta_{-1}<\delta_{1}$.

($\beta$) The points $e_{1},e_{2}$ belong to $\Gamma_{r}^{(inn)}$ if and only
if
$\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}=\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}$
and $\delta_{-1}=\delta_{1}$. In this case, there are critical trajectories
$\gamma_{e_{1},e_{2}}^{+}\subset\mathbb{H}_{+}$ and
$\gamma_{e_{1},e_{2}}^{-}\subset\mathbb{H}_{-}$ having their end points at
$e_{1}$ and $e_{2}$ and, therefore,
$\Gamma_{r}^{(inn)}=\gamma_{e_{1},e_{2}}^{-}\cup\gamma_{e_{2},e_{1}}^{+}$.
This also implies that
$\Gamma_{s}^{+}(-1,1)=[-1,e_{1}]\cup\gamma_{e_{1},e_{2}}^{+}\cup[e_{2},1]$ and
$\Gamma_{s}^{-}(-1,1)=[-1,e_{1}]\cup\gamma_{e_{1},e_{2}}^{-}\cup[e_{2},1]$,
see Figure 24 Case II-4-b-$\beta$.

(c) Suppose now that $e_{1},e_{3},e_{4}\in\Gamma_{e}^{l}$. Then
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},e_{1}}\cup\gamma_{e_{1},e_{3}}\cup\gamma_{e_{3},i\infty}$.
In its turn, the boundary of $\Omega_{e}^{r}$ may have $1$, $2$, or $3$ zeros
and $e_{1}\not\in\Gamma_{s}^{r}$. In the case, when $e_{2}$ is the only zero
on $\Gamma_{e}^{r}$, the domain configuration is the mirror configuration
mentioned in part II-4-a-$\gamma$ above. The remaining subcases are the
following.

($\alpha$) Suppose that $e_{3}$ and $e_{4}$ are the only zeros on
$\Gamma_{e}^{r}$. This case happens if and only if
$\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}=0$ and $\delta_{-1}>\delta_{1}$. In
this case there are critical trajectories
$\gamma_{e_{1},e_{3}}\subset\mathbb{H}_{+}$,
$\gamma_{e_{1},e_{4}}\subset\mathbb{H}_{-}$,
$\gamma_{-1,e_{2}}^{+}\subset\mathbb{H}_{+}$ and
$\gamma_{-1,e_{2}}^{-}\subset\mathbb{H}_{-}$. This implies that
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{4}}^{r}\cup\gamma_{e_{4},-i\infty}$
and that $\Gamma_{s}^{+}(-1,1)=\gamma_{-1,e_{2}}^{+}\cup[e_{2},1]$ and
$\Gamma_{s}^{-}(-1,1)=\gamma_{-1,e_{2}}^{-}\cup[e_{2},1]$. The remaining face
of $Q_{0}(z)\,dz^{2}$ in this case is the strip domain $\Omega_{s}(1,1)$
symmetric with respect to the real axis with both its vertices at $1$.
Therefore,
$\Gamma_{s}^{(out)}(1,1)=[-1,e_{1}]\cup\gamma_{e_{1},e_{4}}\cup\gamma_{e_{4},e_{3}}^{r}\cup\gamma_{e_{3},e_{1}}\cup[e_{1},-1]$
and $\Gamma_{s}^{(inn)}(1,1)=\gamma_{-1,e_{2}}^{-}\cup\gamma_{e_{2},-1}^{+}$.
Figure 23 Case II-4-c-$\alpha$ gives an example of a domain configuration.

The mirror configuration for this case, shown in Figure 23 Case
II-4-c-$\alpha$-m, occurs when $e_{1}>1$, $e_{2}<-1$, when
$e_{1},e_{3},e_{4}\in\Gamma_{e}^{r}$ and when $e_{3}$ and $e_{4}$ are the only
zeros on $\Gamma_{e}^{l}$. This case happens if and only if
$\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}=0$ and $\delta_{-1}<\delta_{1}$.

($\beta$) Suppose that $\Gamma_{e}^{r}$ contains three zeros, which, in this
case, are $e_{2}$, $e_{3}$ and $e_{4}$. This case is self-mirrored, it happens
if and only if
$\operatorname{Im}[e_{1},e_{3}]_{Q_{0}}=\operatorname{Im}[e_{2},e_{3}]_{Q_{0}}=0$.
In this case,
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup\gamma_{e_{3},e_{2}}\cup\gamma_{e_{2},e_{4}}\cup\gamma_{e_{4},-i\infty}$.
The latter implies that $\Omega_{e}^{l}$, $\Omega_{e}^{r}$ and
$\Omega_{s}(-1,1)$ are the only domains in the domain configuration of
$Q_{0}(z)\,dz^{2}$ and that the sides of $\Omega_{s}(-1,1)$ are
$\Gamma_{s}^{+}(-1,1)=[-1,e_{1}]\cup\gamma_{e_{1},e_{3}}\cup\gamma_{e_{3},e_{2}}\cup[e_{2},1]$
and
$\Gamma_{s}^{-}(-1,1)=[-1,e_{1}]\cup\gamma_{e_{1},e_{4}}\cup\gamma_{e_{4},e_{2}}\cup[e_{2},1]$,
see Figure 24 Case II-4-c-$\beta$.

III. Suppose that all zeros of $P_{0}(z)$ are real. In the generic cases,
illustrated in Figures 25–45 in Appenix B, we assume that all these zeros are
distinct and that $e_{k}\not=\pm 1$, $k=1,2,3,4$. Possible degenerate cases
appeared from this part are shown in Figures 46–50 in Appendex B. Depending on
the number of zeroes on each of the intervals $(-\infty,-1)$, $(-1,1)$ and
$(1,\infty)$, we consider the following subcases.

1\. Suppose that $e_{1}<e_{2}<e_{3}<e_{4}<-1$. In this case, the intervals
$(-\infty,e_{1})$, $(e_{2},e_{3})$, $(e_{4},-1)$, $(-1,1)$, and $(1,\infty)$
are orthogonal trajectories of $Q_{0}(z)\,dz^{2}$ and the intervals
$(e_{1},e_{2})$ and $(e_{3},e_{4})$ are trajectories of $Q_{0}(z)\,dz^{2}$.
This implies that there are two circle domains $\Omega_{c}(-1)$ and
$\Omega_{c}(1)$. Furthermore, the topological argument based on Lemma 3.1
shows that $e_{1}$ is the only zero on the boundary of $\Omega_{e}^{l}$ and
$e_{4}$ is the only zero on the boundary of $\Omega_{c}(-1)$. Therefore,
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$ and
$\Gamma_{c}(-1)=\gamma_{e_{4}}^{c}$. Under the assumptions of this case, there
are closed critical trajectories $\gamma_{e_{3}}^{r}$ intersecting
$(1,\infty)$ at $x_{1}$ and $\gamma_{e_{2}}^{r}$ intersecting
$(x_{1},\infty)$. This implies that the circle domain $\Omega_{c}(1)$ has the
boundary
$\Gamma_{c}(1)=\gamma_{e_{3}}^{r}\cup[e_{3},e_{4}]\cup\gamma_{e_{4}}^{r-}\cup[e_{4},e_{3}]$
and the end domain $\Omega_{e}^{r}$ has the boundary
$\Gamma_{e}^{r}=\gamma_{-i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},i\infty}$.
Under these circumstances, there is one more face of $Q_{0}(z)\,dz^{2}$, which
is the ring domain $\Omega_{r}$ with the boundary components
$\Gamma_{r}^{(out)}=\gamma_{e_{2}}^{r}$ and
$\Gamma_{r}^{(inn)}=\gamma_{e_{3}}^{r}$.

The Stokes graph and domain configuration are shown in Figure 25 Case III-1.

The mirror configuration, shown in Figure 25 Case III-1-m, occurs if
$1<e_{4}<e_{3}<e_{2}<e_{1}$.

2\. Suppose that $-1<e_{1}<e_{2}<e_{3}<e_{4}<1$. In this case, the intervals
$(-\infty,-1)$, $(-1,e_{1})$, $(e_{2},e_{3})$, $(e_{4},-1)$, and $(1,\infty)$
are orthogonal trajectories of $Q_{0}(z)\,dz^{2}$ and the intervals
$(e_{1},e_{2})$ and $(e_{3},e_{4})$ are trajectories of $Q_{0}(z)\,dz^{2}$. As
before, this implies that there are two circle domains $\Omega_{c}(-1)$ and
$\Omega_{c}(1)$. Furthermore, using Lemma 3.1 we conclude that $e_{1}$ is the
only zero on the boundary of $\Omega_{c}(-1)$ and $e_{4}$ is the only zero on
the boundary of $\Omega_{c}(1)$. Therefore,
$\Gamma_{c}(-1)=\gamma_{e_{1}}^{l}$ and $\Gamma_{c}(1)=\gamma_{e_{4}}^{r}$.
Under the assumptions of this case, there are critical trajectories
$\gamma_{e_{2},i\infty}$, $\gamma_{e_{2},-i\infty}$, $\gamma_{e_{3},i\infty}$,
and $\gamma_{e_{3},-i\infty}$. This implies that in this case there is a strip
domain $\Omega_{s}(-i\infty,i\infty)$ with sides
$\Gamma_{s}^{l}(-i\infty,i\infty)=\gamma_{-i\infty,e_{2}}\cup\gamma_{e_{2},i\infty}$
and
$\Gamma_{s}^{r}(-i\infty,i\infty)=\gamma_{-i\infty,e_{3}}\cup\gamma_{e_{3},i\infty}$.

The domain configuration in this case is shown in Figure 26 Case III-2.

3\. Suppose that $e_{1}<e_{2}<e_{3}<-1<e_{4}<1$. In this case, the intervals
$(-\infty,e_{1})$, $(e_{2},e_{3})$, $(e_{4},1)$, and $(1,\infty)$ are
orthogonal trajectories of $Q_{0}(z)\,dz^{2}$ and the intervals
$(e_{1},e_{2})$, $(e_{3},-1)$, and $(-1,e_{4})$ are trajectories of
$Q_{0}(z)\,dz^{2}$. Thus, there is only one circle domain $\Omega_{c}(1)$ in
this case.

A topological argument similar to the one used in the proof of Lemma 3.1,
which is based on the information obtained from the Basic Structure Theorem
[6, Theorem 3.5], shows that, in the case under consideration, there are three
critical trajectories $\gamma_{e_{k}}^{r}$, $k=2,3,4$, such that
$\gamma_{e_{k}}^{r}$ crosses the interval $(1,\infty)$ at the point $x_{k}$,
$1<x_{4}<x_{3}<x_{2}$. Moreover, $\gamma_{e_{3}}^{r}$ is contained in the
Jordan domain bounded by $\gamma_{e_{2}}^{r}$, and $\gamma_{e_{4}}^{r}$ is
contained in a Jordan domain bounded by $\gamma_{e_{3}}^{r}$. This implies
that there exist critical trajectories $\gamma_{e_{1},i\infty}$ and
$\gamma_{e_{1},-i\infty}$. Now, when the Stokes graph of $Q_{0}(z)\,dz^{2}$ is
identified, one can easily see that the domain configuration of
$Q_{0}(z)\,dz^{2}$ consists of end domains $\Omega_{e}^{l}$ and
$\Omega_{e}^{r}$, circle domain $\Omega_{c}(1)$, ring domain $\Omega_{r}$, and
strip domain $\Omega_{s}(-1,-1)$. The corresponding boundaries, boundary
components, and sides are the following:
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$,
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$,
$\Gamma_{c}(1)=\gamma_{e_{4}}^{r}$, $\Gamma_{r}^{(out)}=\gamma_{e_{2}}^{r}$,
$\Gamma_{r}^{(inn)}=\gamma_{e_{3}}^{r}$,
$\Gamma_{s}^{+}(-1,1)=[-1,e_{3}]\cup\gamma_{e_{3}}^{r}\cup[e_{3},-1]$,
$\Gamma_{s}^{-}(-1,1)=[-1,e_{4}]\gamma_{e_{4}}^{r}\cup[e_{4},-1]$.

The Stokes graph and domain configuration are shown in Figure 27 Case III-3.

The mirror configuration, shown in Figure 27 Case III-3-m, occurs when
$-1<e_{4}<1<e_{3}<e_{2}<e_{1}$.

4\. Suppose that $e_{1}<e_{2}<e_{3}<-1<1<e_{4}$. In this case, the intervals
$(-\infty,e_{1})$, $(e_{2},e_{3})$, and $(e_{4},\infty)$ are orthogonal
trajectories of $Q_{0}(z)\,dz^{2}$ and the intervals $(e_{1},e_{2})$,
$(e_{3},-1)$, $(-1,1)$, and $(1,e_{4})$ are trajectories of
$Q_{0}(z)\,dz^{2}$. Thus, there are no circle domains in this case and there
is a strip domain $\Omega_{s}(-1,1)$, which is symmetric with respect to the
real axis.

Using the topological argument based on the Basic Structure Theorem [6,
Theorem 3.5], we conclude that there are critical trajectories
$\gamma_{e_{1},i\infty}\subset\overline{\mathbb{H}}_{+}$ and
$\gamma_{e_{1},-i\infty}\subset\overline{\mathbb{H}}_{-}$. The latter implies
that $\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$.

Since the strip domain $\Omega_{s}(-1,1)$ is symmetric with respect to the
real axis, its boundary $\partial\Omega_{s}(-1,1)$ must contain at least one
of the zeros $e_{3}$, $e_{4}$. Thus, we consider the following subcases.

(a) Let $e_{3}$ be the only zero on $\partial\Omega_{s}(-1,1)$. This case
happens if and only if $\delta_{-1}<\delta_{1}$. This inequality implies that
there are critical trajectories
$\gamma_{e_{3},1}^{+}\subset\overline{\mathbb{H}}_{+}$ and
$\gamma_{e_{3},1}^{-}\subset\overline{\mathbb{H}}_{-}$. Therefore, the upper
and lower sides of the strip domain $\Omega_{s}(-1,1)$ are, respectively,
$\Gamma_{s}^{+}(-1,1)=[-1,e_{3}]\cup\gamma_{e_{3},1}^{+}$ and
$\Gamma_{s}^{-}(-1,1)=[-1,e_{3}]\cup\gamma_{e_{3},1}^{-}$.

As concerns critical trajectories, different from the interval
$(e_{1},e_{2})$, which have at least one end point at $e_{2}$, there are three
possibilities.

($\alpha$) There are critical trajectories
$\gamma_{e_{2},1}^{+}\subset\overline{\mathbb{H}}_{+}$ and
$\gamma_{e_{2},1}^{-}\subset\overline{\mathbb{H}}_{-}$. This subcase happens
if and only if $\delta_{-1}+2[e_{2},e_{3}]_{Q_{0}}<\delta_{1}$. Under these
assumptions, there is a strip domain $\Omega_{s}(1,1)$, the inner and the
outer sides of which are
$\Gamma_{s}^{(inn)}=\gamma_{1,e_{3}}^{-}\cup\gamma_{e_{3},1}^{+}$ and
$\Gamma_{s}^{(out)}=\gamma_{1,e_{2}}^{-}\cup\gamma_{e_{2},1}^{+}$. Under these
circumstances, the set
$\Gamma_{s}^{l}(1,i\infty)=\gamma_{1,e_{2}}^{+}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},i\infty}$
is a boundary arc of one of the faces of the Stokes graph of
$Q_{0}(z)\,dz^{2}$, which in this case must be a strip domain
$\Omega_{s}(1,i\infty)$. Thus, $\Gamma_{s}^{l}(1,i\infty)$ is the left side of
$\Omega_{s}(1,i\infty)$. Since the right side of $\Omega_{s}(1,i\infty)$ must
contain at least one zero of $Q_{0}(z)\,dz^{2}$, the only possibility is that
$\Gamma_{s}^{r}(1,i\infty)=[1,e_{4}]\cup\gamma_{e_{4},i\infty}$, where
$\gamma_{e_{4},i\infty}\subset\overline{\mathbb{H}}_{+}$ is the critical
trajectory of $Q_{0}(z)\,dz^{2}$ joining $e_{4}$ and $\infty$. Since the
trajectory structure of $Q_{0}(z)\,dz^{2}$ is symmetric with respect to the
real axis, it follows that there is a strip domain $\Omega_{s}(-i\infty,1)$
with left and right sides
$\Gamma_{s}^{l}(-i\infty,1)=\gamma_{-i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2},1}^{-}$
and $\Gamma_{s}^{r}(-i\infty,1)=\gamma_{-i\infty,e_{4}}\cup[e_{4},1]$,
respectively. The latter also implies that the boundary of the end domain
$\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{4}}\cup\gamma_{e_{4},-i\infty}$. There are
no other domains in the domain configuration of $Q_{0}(z)\,dz^{2}$ in this
case. The Stokes graph and domain configuration are shown in Figure 28 Case
III-4-a-$\alpha$.

The mirror configuration, shown in Figure 28 Case III-4-a-$\alpha$-m, occurs
when $e_{4}<-1<1<e_{3}<e_{2}<e_{1}$ and it occurs if and only if
$\delta_{1}+2[e_{2},e_{3}]_{Q_{0}}<\delta_{-1}$.

($\beta$) There is a critical trajectory $\gamma_{e_{2}}^{r}$, which
intersects the interval $(e_{4},\infty)$. This subcase happens if and only if
$\delta_{-1}<\delta_{1}<\delta_{-1}+2[e_{2},e_{3}]_{Q_{0}}$. In this case the
boundary of the end domain $\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r-}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$.

Furthermore, the curve $\gamma_{e_{2}}^{r}$ is an outer boundary component of
one of the faces of the Stokes graph of $Q_{0}(z)\,dz^{2}$, which must be a
ring domain $\Omega_{r}$. Therefore, $\Gamma_{r}^{(out)}=\gamma_{e_{2}}^{r}$.
Under these circumstances there is a critical trajectory $\gamma_{e_{4}}^{l}$
intersecting the interval $(e_{2},e_{3})$, which is the inner boundary
component of $\Omega_{r}$, i.e. $\Gamma_{r}^{(inn)}=\gamma_{e_{4}}^{l}$. The
remaining face of the Stokes graph in this case is a strip domain
$\Omega_{s}(1,1)$ symmetric with respect to the real axis, the inner and outer
sides of which are
$\Gamma_{s}^{(inn)}(1,1)=\gamma_{1,e_{3}}^{+}\cup\gamma_{e_{3},1}^{-}$ and
$\Gamma_{s}^{(out)}(1,1)=[1,e_{4}]\cup\gamma_{e_{4}}^{l}\cup[e_{4},1]$. The
Stokes graph and domain configuration are shown in Figure 29 Case
III-4-a-$\beta$.

The mirror configuration for this, shown in Figure 29 Case III-4-a-$\beta$-m,
occurs when $e_{4}<-1<1<e_{3}<e_{2}<e_{1}$ and it occurs if and only if
$\delta_{1}<\delta_{-1}<\delta_{1}+2[e_{2},e_{3}]_{Q_{0}}$.

($\gamma$) There are critical trajectories $\gamma_{e_{2},e_{4}}^{+}$ and
$\gamma_{e_{2},e_{4}}^{-}$. This subcase happens if and only if
$\delta_{1}=\delta_{-1}+2[e_{2},e_{3}]_{Q_{0}}$. The boundary of the end
domain $\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2},e_{4}}^{+}\cup\gamma_{e_{4},e_{2}}^{-}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$.
The remaining face of the Stokes graph in this case is the strip domain
$\Omega_{s}(1,1)$ symmetric with respect to the real axis, the inner and outer
sides of which are
$\Gamma_{s}^{(inn)}(1,1)=\gamma_{1,e_{3}}^{+}\cup\gamma_{e_{3},1}^{-}$ and
$\Gamma_{s}^{(out)}(1,1)=[1,e_{4}]\cup\gamma_{e_{4},e_{2}}^{+}\cup\gamma_{e_{2},e_{4}}^{-}\cup[e_{4},1]$.
The Stokes graph and domain configuration are shown in Figure 30 Case
III-4-a-$\gamma$.

The mirror configuration, shown in Figure 30 Case III-4-a-$\gamma$-m, occurs
when $e_{4}<-1<1<e_{3}<e_{2}<e_{1}$ and it occurs if and only if
$\delta_{-1}=\delta_{1}+2[e_{2},e_{3}]_{Q_{0}}$.

(b) Let $e_{4}$ be the only zero on $\partial\Omega_{s}(-1,1)$. This case
happens if and only if $\delta_{1}<\delta_{-1}$. This inequality implies that
there are critical trajectories
$\gamma_{-1,e_{4}}^{+}\subset\overline{\mathbb{H}}_{+}$ and
$\gamma_{-1,e_{4}}^{-}\subset\overline{\mathbb{H}}_{-}$. Therefore, the upper
and lower sides of the strip domain $\Omega_{s}(-1,1)$ are
$\Gamma_{s}^{+}(-1,1)=\gamma_{-1,e_{4}}^{+}\cup[e_{4},1]$ and
$\Gamma_{s}^{-}(-1,1)=\gamma_{-1,e_{4}}^{-}\cup[e_{4},1]$, respectively. In
this case, the set
$\Gamma_{s}^{(inn)}(-1,-1)=\gamma_{-1,e_{4}}^{-}\cup\gamma_{e_{4},-1}^{+}$
must be an inner side of the strip domain $\Omega_{s}(-1,-1)$. Under these
circumstances, there is a critical trajectory $\gamma_{e_{3}}^{r}$
intersecting the interval $(e_{4},\infty)$ at some point $x_{1}$. The outer
side of $\Omega_{s}(-1,-1)$ in this case is
$\Gamma_{s}^{(out)}(-1,-1)=[-1,e_{3}]\cup\gamma_{e_{3}}^{r}\cup[e_{3},-1]$.

Furthermore, the curve $\Gamma_{r}^{(inn)}=\gamma_{e_{3}}^{r}$ is an inner
component of some face of the quadratic differential $Q_{0}(z)\,dz^{2}$, which
must be a ring domain $\Omega_{r}$. The latter implies that there is a
critical trajectory $\gamma_{e_{2}}^{r}$ intersecting the interval
$(x_{1},\infty)$. In this case, $\Gamma_{r}^{(out)}=\gamma_{e_{2}}^{r}$. Now,
when all critical trajectories of $Q_{0}(z)\,dz^{2}$ are identified, the
boundary of the end domain $\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r-}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$.
The Stokes graph and domain configuration are shown in Figure 31 Case III-4-b.

The mirror configuration for this case, shown in Figure 31 Case III-4-b-m,
occurs when $e_{4}<-1<1<e_{3}<e_{2}<e_{1}$ and it occurs if and only if
$\delta_{-1}<\delta_{1}$.

(c) Let $e_{3},e_{4}\in\partial\Omega_{s}(-1,1)$. This case happens if and
only if $\delta_{-1}=\delta_{1}$. This equality implies that there are
critical trajectories
$\gamma_{e_{3},e_{4}}^{+}\subset\overline{\mathbb{H}}_{+}$ and
$\gamma_{e_{3},e_{4}}^{-}\subset\overline{\mathbb{H}}_{-}$. Therefore, the
upper and lower sides of the strip domain $\Omega_{s}(-1,1)$ are
$\Gamma_{s}^{+}(-1,1)=[-1,e_{3}]\cup\gamma_{e_{3},e_{4}}^{+}\cup[e_{4},1]$ and
$\Gamma_{s}^{-}(-1,1)=[-1,e_{3}]\cup\gamma_{e_{3},e_{4}}^{-}\cup[e_{4},1]$,
respectively. In this case, the set
$\Gamma_{r}^{(inn)}=\gamma_{e_{3},e_{4}}^{-}\cup\gamma_{e_{4},e_{3}}^{+}$ must
be an inner boundary component of the ring domain $\Omega_{r}$. Under these
circumstances, there is a critical trajectory $\gamma_{e_{2}}^{r}$
intersection the interval $(e_{4},\infty)$. The outer side of $\Omega_{r}$ in
this case is $\Gamma_{r}^{(out)}=\gamma_{e_{2}}^{r}$.

Now, when all critical trajectories of $Q_{0}(z)\,dz^{2}$ are identified, the
boundary of the end domain $\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r-}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$.
The Stokes graph and domain configuration are shown in Figure 32 Case III-4-c.

The mirror configuration, shown in Figure 32 Case III-4-c-m, occurs when
$e_{4}<-1<1<e_{3}<e_{2}<e_{1}$ and it occurs if and only if
$\delta_{-1}=\delta_{1}$.

5\. Suppose that $e_{1}<-1<e_{2}<e_{3}<e_{4}<1$. In this case, the intervals
$(-\infty,e_{1})$, $(e_{2},e_{3})$, $(e_{4},1)$, and $(1,\infty)$ are
orthogonal trajectories of $Q_{0}(z)\,dz^{2}$ and the intervals $(e_{1},-1)$,
$(-1,e_{2})$ and $(e_{3},e_{4})$ are trajectories of $Q_{0}(z)\,dz^{2}$. Thus,
in this case there is only one circle domain $\Omega_{c}(1)$.

The topological restrictions from the Basic Structure Theorem [6, Theorem 3.5]
imply that there exist critical trajectories $\gamma_{e_{1},i\infty}$,
$\gamma_{e_{1},-i\infty}$, $\gamma_{e_{2},i\infty}$,
$\gamma_{e_{2},-i\infty}$, $\gamma_{e_{3},i\infty}$, and
$\gamma_{e_{3},-i\infty}$. Moreover, the latter implies that there is a
critical trajectory $\gamma_{e_{4}}^{r}$ that is the boundary of the circle
domain $\Omega_{c}(1)$; i.e. $\Gamma_{c}(1)=\gamma_{e_{4}}^{r}$. In this case,
the boundaries of the end domains $\Omega_{e}^{l}$ and $\Omega_{e}^{r}$ are
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$ and
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{3}}\cup[e_{3},e_{4}]\cup\gamma_{e_{4}}^{r}\cup[e_{4},e_{3}]\cup\gamma_{e_{3},-i\infty}$.

Furthermore, there exist three strip domains $\Omega_{s}(-1,i\infty)$,
$\Omega_{s}(-1,-i\infty)$ and $\Omega_{s}(-i\infty,i\infty)$. The
corresponding sides of these strip domains are the following:
$\Gamma_{s}^{l}(-1,i\infty)=[-1,e_{1}]\cup\gamma_{e_{1},i\infty}$ and
$\Gamma_{s}^{r}(-1,i\infty)=[-1,e_{2}]\cup\gamma_{e_{2},i\infty}$,
$\Gamma_{s}^{l}(-i\infty,-1)=\gamma_{-i\infty,e_{1}}\cup[e_{1},-1]$ and
$\Gamma_{s}^{r}(-i\infty,-1)=\gamma_{-i\infty,e_{2}}\cup[e_{2},-1]$,
$\Gamma_{s}^{l}(-i\infty,i\infty)=\gamma_{-i\infty,e_{2}}\cup\gamma_{e_{2},i\infty}$
and
$\Gamma_{s}^{r}(-i\infty,i\infty)=\gamma_{-i\infty,e_{3}}\cup\gamma_{e_{3},i\infty}$.
The Stokes graph and domain configuration are shown in Figure 33 Case III-5.

The mirror configuration, shown in Figure 33 Case III-5-m, occurs when
$-1<e_{4}<e_{3}<e_{2}<1<e_{1}$.

6\. Suppose that $e_{1}<e_{2}<-1<e_{3}<e_{4}<1$. In this case, the intervals
$(-\infty,e_{1})$, $(e_{2},-1)$, $(-1,e_{3})$, $(e_{4},1)$, and $(1,\infty)$
are orthogonal trajectories of $Q_{0}(z)\,dz^{2}$ and the intervals
$(e_{1},e_{2})$ and $(e_{3},e_{4})$ are trajectories of $Q_{0}(z)\,dz^{2}$.
This implies that there are two circle domains $\Omega_{c}(-1)$ and
$\Omega_{c}(1)$. As in the previous cases, the topological argument based on
the Basic Structure Theorem [6, Theorem 3.5] and Lemma 3.1 implies that
$e_{1}$ is the only zero on the boundary of $\Omega_{e}^{l}$ and $e_{4}$ is
the only zero on the boundary of $\Omega_{c}(1)$. Therefore,
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}^{l}\cup\gamma_{e_{1},i\infty}$ and
$\Gamma_{c}(1)=\gamma_{e_{4}}^{r}$, where $\gamma_{e_{4}}^{r}$ crosses
$(1,\infty)$ at some point $x_{1}$.

Under the assumptions of this case, the boundary of $\Omega_{c}(-1)$ may
contain 1 or 2 zeros. Thus, to identify the remaining domains, we consider
three subcases.

(a) Let $e_{2}$ be the only zero on $\Gamma_{c}(-1)$. Under the assumptions of
Case III-6, the latter happens if and only if

(4.14)
$\lim_{\varepsilon\to+0}\left([e_{2},-1-\varepsilon]_{Q_{0}}-[-1+\varepsilon,e_{3}]_{Q_{0}}\right)<0.$

In this case, $\Gamma_{c}(-1)=\gamma_{e_{2}}^{c}$, where $\gamma_{e_{2}}^{c}$
crosses the interval $(-1,e_{3})$. Thus, there is one more face of
$Q_{0}(z)\,dz^{2}$ that is the strip domain $\Omega_{s}(-i\infty,i\infty)$
with the sides
$\Gamma_{s}^{l}(-i\infty,i\infty)=\gamma_{-i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{c}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},i\infty}$
and $\Gamma_{s}^{r}=\gamma_{-i\infty,e_{3}}\cup\gamma_{e_{3},i\infty}$.
Finally the boundary of the end domain $\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{-i\infty,e_{3}}\cup[e_{3},e_{4}]\cup\gamma_{e_{4}}^{r}\cup[e_{4},e_{3}]\cup\gamma_{e_{3},i\infty}$.
The Stokes graph and domain configuration are shown in Figure 34 Case III-6-a.

The mirror configuration, shown in Figure 34 Case III-6-a-m, occurs when
$-1<e_{4}<e_{3}<1<e_{2}<e_{1}$ and with these assumptions it happens if and
only if

(4.15)
$\lim_{\varepsilon\to+0}\left([e_{2},1+\varepsilon]_{Q_{0}}-[1-\varepsilon,e_{3}]_{Q_{0}}\right)<0.$

(b) Let $e_{3}$ be the only zero on $\Gamma_{c}(-1)$. Under the assumptions of
the case III-6, the latter happens if and only if the limit in (4.14) is
positive. In this case, $\Gamma_{c}(-1)=\gamma_{e_{3}}^{l}$, where
$\gamma_{e_{3}}^{l}$ crosses the interval $(e_{2},-1)$. Under these
circumstances, the set
$\Gamma_{r}^{(inn)}=\gamma_{e_{3}}^{l}\cup[e_{3},e_{4}]\cup\gamma_{e_{4}}^{r}\cup[e_{4},e_{3}]$
is a boundary component of a face of the Stokes graph of $Q_{0}(z)\,dz^{2}$,
which must be a ring domain $\Omega_{r}$ by Lemma 3.1. This implies that there
is a critical trajectory $\gamma_{e_{2}}^{r}$, which crosses the interval
$(x_{4},\infty)$, where $x_{4}$ is defined earlier in the case 3, and
therefore $\Gamma_{r}^{(out)}=\gamma_{e_{2}}^{r}$.

Finally the boundary of the end domain $\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$.
The Stokes graph and domain configuration are shown in Figure 35 Case III-6-b.

The mirror configuration, shown in Figure 35 Case III-6-b-m, occurs when
$1<e_{4}<e_{3}<1<e_{2}<e_{1}$ and if and only if the limit in (4.15) is
positive.

(c) Suppose now that $e_{2},e_{3}\in\Gamma_{c}(-1)$. Under the assumptions of
Case III-6, the latter happens if and only if the limit in (4.14) is zero. In
this case,
$\Gamma_{c}(-1)=\gamma_{e_{2},e_{3}}^{+}\cup\gamma_{e_{3},e_{2}}^{-}$ and
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2},e_{3}}^{+}\cup[e_{3},e_{4}]\cup\gamma_{e_{4}}^{r}\cup[e_{4},e_{3}]\cup\gamma_{e_{3},e_{2}}^{-}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$.
The Stokes graph and domain configuration are shown in Figure 36 Case III-6-c.

The mirror configuration, shown in Figure 36 Case III-6-c-m, occurs when
$-1<e_{4}<e_{3}<1<e_{2}<e_{1}$ and if and only if the limit in (4.15) is zero.

7\. Suppose that $e_{1}<e_{2}<-1<1<e_{3}<e_{4}$. In this case, the intervals
$(-\infty,e_{1})$, $(e_{2},-1)$, $(-1,1)$, $(1,e_{3})$, and $(e_{4},\infty)$
are orthogonal trajectories of $Q_{0}(z)\,dz^{2}$ and the intervals
$(e_{1},e_{2})$ and $(e_{3},e_{4})$ are trajectories of $Q_{0}(z)\,dz^{2}$.
Thus, there are two circle domains $\Omega_{c}(-1)$ and $\Omega_{c}(1)$.

First, we mention few topological obstructions for the critical trajectories
starting at zeros $e_{2}$ and $e_{3}$. There is no trajectories with one end
point at one of these zeros and second end point at $\infty$. Indeed, if such
a trajectory, say $\gamma_{e_{2},i\infty}$, exists (then
$\gamma_{e_{2},-i\infty}$ exists as well), then the zero $e_{1}$ is the only
critical point of $Q_{0}(z)\,dz^{2}$ in the simply connected domain $D\ni
e_{1}$ bounded by the curve
$\gamma_{-i\infty,e_{2}}\cup\gamma_{e_{2},i\infty}$, which is impossible, see
part 1 of Lemma 3.1. Similar argument shows that in the case under
consideration there are no trajectories joining the zeros $e_{2}$ and $e_{3}$,
and there are no trajectories with both end points at $e_{2}$ or at $e_{3}$,
which cross the interval $(1,e_{3})$ or the interval $(e_{2},-1)$,
respectively. Thus, we are left with the following subcases.

(a) Suppose that there are critical trajectories $\gamma_{e_{2}}^{c}$ and
$\gamma_{e_{3}}^{c}$ crossing $(-1,1)$ at some points $x_{1}$ and $x_{2}$,
respectively, such that $x_{1}<x_{2}$. This case happens if and only if the
following inequalities hold:

(4.16)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[e_{2},-1+i\varepsilon]_{Q_{0}}-\operatorname{Im}[e_{3},-1+i\varepsilon]_{Q_{0}}\right)<0,$
(4.17)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[e_{3},1+i\epsilon]_{Q_{0}}-\operatorname{Im}[e_{2},1+i\epsilon]_{Q_{0}}\right)<0.$

Under these circumstances, $\Gamma_{c}(-1)=\gamma_{e_{2}}^{c}$,
$\Gamma_{c}(1)=\gamma_{e_{3}}^{c}$ and there are critical trajectories
$\gamma_{e_{1},i\infty}$, $\gamma_{e_{1},-i\infty}$, $\gamma_{e_{4},i\infty}$,
and $\gamma_{e_{4},-i\infty}$. This implies that
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$ and
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{4}}\cup\gamma_{e_{4},-i\infty}$.
Furthermore, this implies that there is one more face of the Stokes graph of
$Q_{0}(z)\,dz^{2}$, which is the strip domain $\Omega_{s}(-i\infty,i\infty)$
with the sides
$\Gamma_{s}^{l}(-i\infty,i\infty)=\gamma_{-i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{c}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},i\infty}$
and
$\Gamma_{s}^{r}(-i\infty,i\infty)=\gamma_{-i\infty,e_{4}}\cup[e_{4},e_{3}]\cup\gamma_{e_{3}}^{c}\cup[e_{3},e_{4}]\cup\gamma_{e_{4},i\infty}$,
see Figure 37 Case III-7-a.

(b) Suppose that there are critical trajectories $\gamma_{e_{2}}^{r}$ crossing
$(e_{4},\infty)$ and $\gamma_{e_{3}}^{c}$ crossing $(-1,1)$. This case happens
if and only if the limit in (4.16) is positive and the limit in (4.17) is
negative. These conditions imply that there is a critical trajectory
$\gamma_{e_{4}}^{l}$, which crosses the interval $(e_{2},-1)$. The remaining
two critical trajectories in this case are $\gamma_{e_{1},i\infty}$ and
$\gamma_{e_{1},-i\infty}$. Now, when all critical trajectories are identified,
the domain configuration of $Q_{0}(z)\,dz^{2}$ consists of end domains
$\Omega_{e}^{l}$, $\Omega_{e}^{r}$, circle domains $\Omega_{c}(-1)$,
$\Omega_{c}(1)$, and a ring domain $\Omega_{r}$. The corresponding boundaries
are the following:
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$,
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$,
$\Gamma_{c}(-1)=\gamma_{e_{4}}^{l}\cup[e_{4},e_{3}]\cup\gamma_{e_{3}}^{c}\cup[e_{3},e_{4}]$,
$\Gamma_{c}(1)=\gamma_{e_{3}}^{c}$, $\Gamma_{r}^{(out)}=\gamma_{e_{2}}^{r}$,
and $\Gamma_{r}^{(inn)}=\gamma_{e_{4}}^{l}$. The Stokes graph and domain
configuration are shown in Figure 38 Case III-7-b.

The mirror configuration, shown in Figure 38 Case III-7-b-m, occurs when
$e_{4}<e_{3}<-1<1<e_{2}<e_{1}$ and if and only if the following inequalities
hold:

(4.18)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[e_{2},1+i\varepsilon]_{Q_{0}}-\operatorname{Im}[e_{3},1+i\varepsilon]_{Q_{0}}\right)>0,$
(4.19)
$\lim_{\varepsilon\to+0}\left(\operatorname{Im}[e_{3},-1+i\epsilon]_{Q_{0}}-\operatorname{Im}[e_{2},-1+i\epsilon]_{Q_{0}}\right)<0.$

(c) Suppose that there are critical trajectories $\gamma_{e_{2},e_{4}}^{+}$
and $\gamma_{e_{2},e_{4}}^{-}$. This case happens if and only if the limits in
(4.16) and (4.17) are zero. The remaining critical trajectories in this case
are $\gamma_{e_{1},i\infty}$, $\gamma_{e_{1},-i\infty}$, and
$\gamma_{e_{3}}^{c}$. Under these conditions, the domain configuration of
$Q_{0}(z)\,dz^{2}$ consists of end domains $\Omega_{e}^{l}$, $\Omega_{e}^{r}$
and circle domains $\Omega_{c}(-1)$, $\Omega_{c}(1)$. The corresponding
boundaries are the following:
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$,
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2},e_{4}}^{+}\cup\gamma_{e_{4},e_{2}}^{-}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},-i\infty}$,
$\Gamma_{c}(-1)=\gamma_{e_{2},e_{4}}^{+}\cup[e_{4},e_{3}]\cup\gamma_{e_{3}}^{c}\cup[e_{3},e_{4}]\cup\gamma_{e_{4},e_{2}}^{-}$,
and $\Gamma_{c}(1)=\gamma_{e_{3}}^{c}$, see Figure 39 Case III-7-c.

The mirror configuration, shown in Figure 39 Case III-7-c-m, occurs when
$e_{4}<e_{3}<-1<1<e_{2}<e_{1}$ and if and only if the limits in (4.18) and
(4.19) are zero.

8\. Suppose that $e_{1}<e_{2}<-1<e_{3}<1<e_{4}$. In this case, the intervals
$(-\infty,e_{1})$, $(e_{2},-1)$, $(-1,e_{3})$, $(e_{4},\infty)$ are orthogonal
trajectories of $Q_{0}(z)\,dz^{2}$ and the intervals $(e_{1},e_{2})$,
$(e_{3},1)$ and $(1,e_{4})$ are trajectories of $Q_{0}(z)\,dz^{2}$. Thus,
there is only one circle domain $\Omega_{c}(-1)$.

Applying the topological argument based on the Basic Structure Theorem [6,
Theorem 3.5] and Lemma 3.1 once more, we conclude that $e_{1}$ is the only
zero on the boundary of $\Omega_{e}^{l}$ and therefore,
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}^{l}\cup\gamma_{e_{1},i\infty}$.
Furthermore, similar topological argument implies that
$e_{1},e_{4}\not\in\Gamma_{c}(-1)$. Therefore, $\Gamma_{c}(-1)$ may contain
one of the zeros $e_{2}$, $e_{3}$ or both these zeros. Thus, we have the
following subcases.

(a) Suppose that $e_{2}\in\Gamma_{c}(-1)$ but $e_{3}\not\in\Gamma_{c}(-1)$.
This happens if and only if the limit in (4.16) is negative. Then
$\Gamma_{c}(-1)=\gamma_{e_{2}}^{c}$, where $\gamma_{e_{2}}^{c}$ intersects the
interval $(-1,e_{3})$. In this case,
$\Gamma_{s}^{l}(-i\infty,i\infty)=\gamma_{-i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{c}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},i\infty}$
is a boundary arc of one of the faces of the Stokes graph of
$Q_{0}(z)\,dz^{2}$, which in this case must be a strip domain
$\Omega_{s}(-i\infty,i\infty)$ having $\Gamma_{s}^{l}(-i\infty,i\infty)$ as
its left side. The right side of $\Omega_{s}(-i\infty,i\infty)$ is
$\Gamma_{s}^{r}(-i\infty,i\infty)=\gamma_{-i\infty,e_{3}}\cup\gamma_{e_{3},i\infty}$.
Under these circumstances, there are critical trajectories
$\gamma_{e_{4},i\infty}\subset\mathbb{H}_{+}$ and
$\gamma_{e_{4},-i\infty}\subset\mathbb{H}_{-}$ and, hence, the boundary of the
end domain $\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{4}}\cup\gamma_{e_{4},-i\infty}$. This also
implies that there are strip domains
$\Omega_{s}(1,i\infty)\subset\mathbb{H}_{+}$ and
$\Omega_{s}(-i\infty,1)\subset\mathbb{H}_{-}$, which sides are
$\Gamma_{s}^{l}(1,i\infty)=[1,e_{3}]\cup\gamma_{e_{3},i\infty}$,
$\Gamma_{s}^{r}(1,i\infty)=[1,e_{4}]\cup\gamma_{e_{4},i\infty}$ and
$\Gamma_{s}^{l}(-i\infty,1)=\gamma_{-i\infty,e_{3}}\cup[e_{3},1]$,
$\Gamma_{s}^{r}(-i\infty,1)=\gamma_{-i\infty,e_{4}}\cup[e_{4},1]$,
respectively. The Stokes graph and domain configuration are shown in Figure 40
Case III-8-a.

The mirror configuration for this case, shown in Figure 40 Case III-8-a-m,
occurs if and only if $e_{4}<-1<e_{3}<1<e_{2}<e_{1}$ and the limit in (4.16)
is negative.

(b) Suppose that $e_{3}\in\Gamma_{c}(-1)$ but $e_{2}\not\in\Gamma_{c}(-1)$.
This happens if and only if the limit in (4.16) is positive. Then
$\Gamma_{c}(-1)=\gamma_{e_{3}}^{l}$, where $\gamma_{e_{3}}^{l}$ intersects the
interval $(e_{2},-1)$ at some point $x_{1}$. In this case,
$\Gamma_{s}^{(inn)}(1,1)=[1,e_{3}]\cup\gamma_{e_{3}}^{l}\cup[e_{1},1]$ is a
boundary arc of one of the faces of the Stokes graph of $Q_{0}(z)\,dz^{2}$,
which in this case must be a strip domain $\Omega_{s}(1,1)$ having
$\Gamma_{s}^{(inn)}(1,1)$ as its inner side. The outer side
$\Gamma_{s}^{(out)}(1,1)$ of $\Omega_{s}(1,1)$ must contain at least one of
the zeros $e_{2}$, $e_{4}$. Therefore, we have the following three subcases.

($\alpha$) Let $e_{2}\in\Gamma_{s}^{(out)}(1,1)$ but
$e_{4}\not\in\Gamma_{s}^{(out)}(1,1)$. This subcase happens if and only if
$2\operatorname{Im}\,\left([e_{2},i]_{Q_{0}}+[i,e_{3}]_{Q_{0}}\right)<\delta_{1}$.
In this subcase, there are critical trajectories
$\gamma_{e_{2},1}^{+}\subset\overline{\mathbb{H}}_{+}$ and
$\gamma_{e_{2},1}^{-}\subset\overline{\mathbb{H}}_{-}$ such that
$\Gamma_{s}^{(out)}(1,1)=\gamma_{1,e_{2}}^{+}\cup\gamma_{e_{2},1}^{-}$.

Under these circumstances, there are critical trajectories
$\gamma_{e_{4},i\infty}\subset\mathbb{H}_{+}$ and
$\gamma_{e_{4},-i\infty}\subset\mathbb{H}_{-}$ and, hence, the boundary of the
end domain $\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{4}}\cup\gamma_{e_{4},-i\infty}$. This also
implies that there are strip domains
$\Omega_{s}(1,i\infty)\subset\mathbb{H}_{+}$ and
$\Omega_{s}(-i\infty,1)\subset\mathbb{H}_{-}$, which sides are
$\Gamma_{s}^{l}(1,i\infty)=\gamma_{1,e_{2}}^{+}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},i\infty}$,
$\Gamma_{s}^{r}(1,i\infty)=[1,e_{4}]\cup\gamma_{e_{4},i\infty}$ and
$\Gamma_{s}^{l}(-i\infty,1)=\gamma_{-i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2},1}^{-}$,
$\Gamma_{s}^{r}(-i\infty,1)=\gamma_{-i\infty,e_{4}}\cup[e_{4},1]$,
respectively. The Stokes graph and domain configuration are shown in Figure 41
Case III-8-b-$\alpha$.

The mirror configuration, shown in Figure 41 Case III-8-b-$\alpha$-m, occurs
if and only if $e_{4}<-1<e_{3}<1<e_{2}<e_{1}$ and
$2\operatorname{Im}\,\left([e_{2},i]_{Q_{0}}+[i,e_{3}]_{Q_{0}}\right)<\delta_{-1}$.

($\beta$) Let $e_{4}\in\Gamma_{s}^{(out)}(1,1)$ but
$e_{2}\not\in\Gamma_{s}^{(out)}(1,1)$. This subcase happens if and only if
$2\operatorname{Im}\,\left([e_{2},i]_{Q_{0}}+[i,e_{3}]_{Q_{0}}\right)>\delta_{1}$.
These assumptions imply that
$\Gamma_{s}^{(out)}(1,1)=[1,e_{4}]\cup\gamma_{e_{4}}^{l}\cup[e_{4},1]$, where
$\gamma_{e_{4}}^{l}$ is a critical trajectory of $Q_{0}(z)\,dz^{2}$, which
intersect the interval $(e_{2},-1)$ at some point $x_{2}<x_{1}$.

Under conditions of this subcase, the critical trajectory $\gamma_{e_{4}}^{l}$
is an inner boundary component of a face of the Stokes graph of
$Q_{0}(z)\,dz^{2}$, which must be a ring domain $\Omega_{r}$. Hence,
$\Gamma_{r}^{(inn)}=\gamma_{e_{4}}^{l}$. Under these circumstances, the outer
boundary component $\Gamma_{r}^{(out)}$ of $\Omega_{r}$ must contain the zero
$e_{4}$. Therefore, in this case there is a critical trajectory
$\gamma_{e_{2}}^{r}$ intersecting the interval $(e_{4},\infty)$ and
$\Gamma_{r}^{(out)}=\gamma_{e_{2}}^{r}$. There are no other critical
trajectories in this case, which implies that
$\Gamma_{e}^{r}=\gamma_{-i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2}}^{r}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},i\infty}$.
The Stokes graph and domain configuration are shown in Figure 42 Case
III-8-b-$\beta$.

The mirror configuration, shown in Figure 42 Case III-8-b-$\beta$-m, occurs if
and only if $e_{4}<-1<e_{3}<1<e_{2}<e_{1}$ and
$2\operatorname{Im}\,\left([e_{2},i]_{Q_{0}}+[i,e_{3}]_{Q_{0}}\right)>\delta_{-1}$.

($\gamma$) Let $e_{2},e_{4}\in\Gamma_{s}^{(out)}(1,1)$. This subcase happens
if and only if the following equality holds:
$2\operatorname{Im}\,\left([e_{2},i]_{Q_{0}}+[i,e_{3}]_{Q_{0}}\right)=\delta_{1}$.
Since the zeros $e_{2}$ and $e_{4}$ both belong to $\Gamma_{s}^{(out)}(1,1)$
it follows that there are critical trajectories
$\gamma_{e_{2},e_{4}}^{+}\subset\overline{\mathbb{H}}_{+}$ and
$\gamma_{e_{2},e_{4}}^{-}\subset\overline{\mathbb{H}}_{-}$. In this case,
$\Gamma_{s}^{(out)}(1,1)=[1,e_{4}]\cup\gamma_{e_{4},e_{2}}^{+}\cup\gamma_{e_{2},e_{4}}^{-}\cup[e_{4},1]$.
Now, when all critical trajectories are identified, the boundary of the end
domain $\Omega_{e}^{r}$ is
$\Gamma_{e}^{r}=\gamma_{-i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2},e_{4}}^{-}\cup\gamma_{e_{4},e_{2}}^{+}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},i\infty}$.
There are no other domains in the domain configuration of $Q_{0}(z)\,dz^{2}$.
The Stokes graph and domain configuration are shown in Figure 43 Case
III-8-b-$\gamma$.

The mirror configuration, shown in Figure 43 Case III-8-b-$\gamma$-m, occurs
if and only if $e_{4}<-1<e_{3}<1<e_{2}<e_{1}$ and
$2\operatorname{Im}\,\left([e_{2},i]_{Q_{0}}+[i,e_{3}]_{Q_{0}}\right)=\delta_{-1}$.

(c) Suppose that $e_{2},e_{3}\in\Gamma_{c}(-1)$. This happens if and only if
the limit in (4.16) is zero. Since the zeros $e_{2}$ and $e_{3}$ both belong
to $\Gamma_{c}(-1)$ it follows that there are critical trajectories
$\gamma_{e_{2},e_{3}}^{+}\subset\overline{\mathbb{H}}_{+}$ and
$\gamma_{e_{2},e_{3}}^{-}\subset\overline{\mathbb{H}}_{-}$. In this case,
$\Gamma_{c}(-1)=\gamma_{e_{2},e_{3}}^{-}\cup\gamma_{e_{3},e_{2}}^{+}$.

Furthermore,
$\Gamma_{s}^{l}(1,i\infty)=[1,e_{3}]\cup\gamma_{e_{3},e_{2}}^{+}\cup[e_{2},e_{1}]\cup\gamma_{e_{1},i\infty}$
is a boundary arc of one of the faces of the Stokes graph of
$Q_{0}(z)\,dz^{2}$, which must be a strip domain $\Omega_{s}(1,i\infty)$
having $\Gamma_{s}^{l}(1,i\infty)$ as its left side. The only possibility for
the right side of $\Omega_{s}(1,i\infty)$ is that
$\Gamma_{s}^{r}(1,i\infty)=[1,e_{4}]\cup\gamma_{e_{4},i\infty}$. Similarly, we
conclude that there is a strip domain $\Omega_{s}(-i\infty,1)$ with sides
$\Gamma_{s}^{l}(-i\infty,1)=\gamma_{-i\infty,e_{1}}\cup[e_{1},e_{2}]\cup\gamma_{e_{2},e_{3}}^{-}\cup[e_{3},1]$
and $\Gamma_{s}^{r}(-i\infty,1)=\gamma_{-i\infty,e_{4}}\cup[e_{4},1]$.

The latter also implies that the boundary of the end domain $\Omega_{e}^{r}$
is $\Gamma_{e}^{r}=\gamma_{-i\infty,e_{4}}\cup\gamma_{e_{4},i\infty}$. There
are no other domains in the domain configuration of $Q_{0}(z)\,dz^{2}$. The
Stokes graph and domain configuration are shown in Figure 44 Case III-8-c.

The mirror configuration, shown in Figure 44 Case III-8-c-m, occurs if and
only if $e_{4}<-1<e_{3}<1<e_{2}<e_{1}$ and the limit in (4.17) is zero.

9\. Suppose that $e_{1}<-1<e_{2}<e_{3}<1<e_{4}$. In this case, the intervals
$(-\infty,e_{1})$, $(e_{2},e_{3})$, $(e_{4},\infty)$ are orthogonal
trajectories of $Q_{0}(z)\,dz^{2}$ and the intervals $(e_{1},-1)$,
$(-1,e_{2})$ and $(e_{3},1)$, $(1,e_{4})$ are trajectories of
$Q_{0}(z)\,dz^{2}$. There are no circle domains.

As in the previous cases, the topological constrains related to the Basic
Structure Theorem [6, Theorem 3.5] and Lemma 3.1 imply that there exist eight
critical trajectories, each having one of its end points at $i\infty$ or
$-i\infty$. These critical trajectories are: $\gamma_{e_{k},i\infty}$,
$k=1,2,3,4$ and $\gamma_{e_{k},-i\infty}$, $k=1,2,3,4$. Now, when the Stokes
graph of $Q_{0}(z)\,dz^{2}$ is identified, one can easily see that the domain
configuration of $Q_{0}(z)\,dz^{2}$ consists of end domains $\Omega_{e}^{l}$,
$\Omega_{e}^{r}$ and five strip domains, which are $\Omega_{s}(-1,i\infty)$,
$\Omega_{s}(-1,-i\infty)$, $\Omega_{s}(1,i\infty)$, $\Omega_{s}(1,-i\infty)$,
and $\Omega_{s}(-i\infty,i\infty)$. The boundaries of the end domains
$\Omega_{e}^{l}$ and $\Omega_{e}^{r}$ are
$\Gamma_{e}^{l}=\gamma_{-i\infty,e_{1}}\cup\gamma_{e_{1},i\infty}$ and
$\Gamma_{e}^{r}=\gamma_{i\infty,e_{4}}\cup\gamma_{e_{4},-i\infty}$,
respectively. The sides of the strip domains $\Omega_{s}(-1,i\infty)$,
$\Omega_{s}(-1,-i\infty)$, $\Omega_{s}(1,i\infty)$, $\Omega_{s}(1,-i\infty)$,
$\Omega_{s}(-i\infty,i\infty)$ are
$\Gamma_{s}^{l}(-1,i\infty)=[-1,e_{1}]\cup\gamma_{e_{1},i\infty}$ and
$\Gamma_{s}^{r}(-1,i\infty)=[-1,e_{2}]\cup\gamma_{e_{2},i\infty}$,
$\Gamma_{s}^{l}(-i\infty,-1)=\gamma_{-i\infty,e_{1}}\cup[e_{1},-1]$ and
$\Gamma_{s}^{r}(-i\infty,-1)=\gamma_{-i\infty,e_{2}}\cup[e_{2},-1]$,
$\Gamma_{s}^{l}(1,i\infty)=[1,e_{3}]\cup\gamma_{e_{3},i\infty}$ and
$\Gamma_{s}^{r}(1,i\infty)=[1,e_{4}]\cup\gamma_{e_{4},i\infty}$,
$\Gamma_{s}^{l}(-i\infty,1)=\gamma_{-i\infty,e_{3}}\cup[e_{3},1]$ and
$\Gamma_{s}^{r}(-i\infty,1)=\gamma_{-i\infty,e_{4}}\cup[e_{4},1]$,
$\Gamma_{s}^{l}(-i\infty,i\infty)=\gamma_{-i\infty,e_{2}}\cup\gamma_{e_{2},i\infty}$
and
$\Gamma_{s}^{r}(-i\infty,i\infty)=\gamma_{-i\infty,e_{3}}\cup\gamma_{e_{3},i\infty}$,
respectively. The Stokes graph and domain configuration in this case are shown
in Figure 45 Case III-9.

IV. In this part, we discuss possible Stokes graphs and domain configurations
for the degenerate cases. Precisely, we describe changes, which occur in the
Stokes graphs and domain configurations when two or more zeros merge. In the
case without real zeros, there is only one degenerate configuration described
in Case I-3 and illustrated in Figure 5. In the cases with two or four real
zeros, we have the following possibilities.

1\. In the case with two real zeros $e_{1}$ and $e_{2}$, these zeros can merge
if and only if both belong to one of the intervals $(-\infty,-1)$, $(-1,1)$,
or $(1,\infty)$. In all these cases an interval $(e_{1},e_{2})$ is one of the
critical trajectories of $Q_{0}(z)\,dz^{2}$ and therefore
$\gamma_{e_{1},e_{2}}=[e_{1},e_{2}]$. Thus, when $e_{1}$ and $e_{2}$ merge,
the arc $\gamma_{e_{1},e_{2}}$ shrinks to a point $e_{1,2}$, while the
structure of domain configuration remains the same as in the corresponding
generic case. Therefore, the Stokes graphs and domain configurations in the
degenerate cases mentioned above are the same as in the generic cases II-1 and
II-2 shown in Figures 6-13 except that the interval $[e_{1},e_{2}]$ shrinks to
a single point.

2\. Similar situation occurs in the case with four real zeros when the
intervals $(e_{1},e_{2})$ and/or $(e_{3},e_{4})$ are critical trajectories of
$Q_{0}(z)\,dz^{3}$ as it is illustrated in Figures 25-44. In these cases, if
$e_{1}$ merges with $e_{2}$ and/or $e_{3}$ merges with $e_{4}$, then the
domain configuration remains the same as in the generic case shown in the
corresponding figure except that the interval $[e_{1},e_{2}]$ and/or the
interval $[e_{3},e_{4}]$ shrinks to a single point.

3\. In case III with four real zeros there are situations when a ring domain
and/or strip domain collapses if two or more zeros merge. Precisely, in cases
III-1, III-3, III-4-b, III-4-c and in their mirror cases the ring domain
$\Omega_{r}$ collapses, when the zeros $e_{2}$ and $e_{3}$ merge forming a
double zero $e_{2,3}$. The corresponding degenerate domain configurations for
these cases are shown in Figures 46, 48, 50, and 51.

Furthermore, in cases III-2, III-5, III-9 and in their mirror cases the strip
domain $\Omega_{s}(-i\infty,i\infty)$ collapses when $e_{2}$ and $e_{3}$ merge
forming a double zero $e_{2,3}$ as it is shown in Figures 47, 52, and 53.
Similarly, in cases III-4-a-$\alpha$ and III-4-a-$\gamma$ the strip domain
$\Omega_{s}(1,1)$ collapses and in the corresponding mirror cases
III-4-a-$\alpha$-m and III-4-a-$\gamma$-m the strip domain $\Omega_{s}(-1,-1)$
collapses when the zeros $e_{2}$ and $e_{3}$ merge to a double zero $e_{2,3}$;
see Figures 50, and 51.

In the remaining case, that is III-4-a-$\beta$, the ring domain $\Omega_{r}$
and the strip domain $\Omega_{s}(1,1)$ both collapse when the zeros $e_{2}$
and $e_{3}$ merge to a double zero $e_{2,3}$. Similarly, in the mirror case
III-4-a-$\beta$-m the domains $\Omega_{r}$ and $\Omega_{s}(-1,-1)$ both
collapse when $e_{2}$ and $e_{3}$ merge forming a double zero $e_{2,3}$. The
resulting degenerate domain configurations in these cases are those shown in
Figures 50, and 51.

Further merging of zeros, when a double zero merges with one or two single
zeros, does not change the domain configurations. Possible cases are the
following. In each of the cases III-1-deg, III-1-deg-m and III-2-deg shown in
Figures 46 and 47, the double zero $e_{2,3}$ can merge with $e_{1}$, then the
edge $[e_{1},e_{2,3}]$ shrinks to a point forming a triple zero, or $e_{2,3}$
can merge with $e_{4}$, then the edge $[e_{4},e_{2,3}]$ shrinks to a point
again forming a triple zero, or $e_{2,3}$ can merge with with both $e_{1}$ and
$e_{4}$ forming a zero of order four, then both edges $[e_{1},e_{2,3}]$ and
$[e_{2,3}],e_{4}]$ shrink to this zero of order four.

In all cases shown in Figures 48-52, the double zero $e_{2,3}$ can merge with
$e_{1}$ forming a triple zero. In all these cases with triple zero, the domain
configurations contain the same domains as shown in Figures 48-52 and the
Stokes graphs consists of the same edges as in these figures, except that the
edge $[e_{1},e_{2,3}]$ shrinks to a point forming a triple zero of the
corresponding quadratic differential.

##  5\. Domain configurations for the Rabi model

The existence and properties of solutions to Rabi problem depend on the values
of physical parameters $\Delta$, $E$, and $g$, while our classification of the
Stokes graphs for this problem is given in terms of the number of real zeros
and some other characteristics of associated quadratic differential
$Q_{0}(z)\,dz^{2}$. Thus, to apply the results presented in Section 4 to Rabi
problem, we have to identify which of the types I, II, or III of Stokes graphs
and domain configurations of $Q_{0}(z)\,dz^{2}$ correspond to a particular
choice of the Rabi parameters $\Delta$, $E$, and $g$.

For given $\Delta$, $E$, and $g$, the coefficients $c_{k}$ of the quartic
polynomial $P_{0}(z)$, that is the numerator of $Q_{0}(z)$ in formula (2.8),
are expressed explicitly by formulas (2.9) and (2.10) as functions of
$\Delta$, $E$, and $g$. Thus, to use our classification of Stokes graphs to
study the Rabi problem, we have to determine if the polynomial $P_{0}(z)$ with
these coefficients has no real zeros, has two real zeros, or it has four real
zeros.

The theory of quartic equations, which origin goes back to the work of
Lodovico Ferrari in the $16^{th}$ century, is well known and contains all
information on distribution of zeros of such equations, which we need for our
study. Precisely, in Propositions 5.1 and 5.2 below, we present classical
results of J. L. Lagrange [9, Chapitre V, Article III, Section 39, p. 67]
(see, also, Chapter IV, Section 7 in [3]) interpreted in terms of the
parameters $\Delta$, $E$, and $g$ of the Rabi problem. As it was shown by J.
L. Lagrange, the number of real roots of $P_{0}(z)$ depends on the signs of
the discriminant $\mathcal{D}_{0}$ and two additional characteristics,
$\mathcal{P}_{0}$ and $\mathcal{Q}_{0}$, of the polynomial $P_{0}$, which are
defined as follows:

$\displaystyle\mathcal{D}_{0}$ $\displaystyle=$
$\displaystyle-27c_{3}^{4}c_{0}^{2}+18c_{3}^{3}c_{2}c_{1}c_{0}-4c_{3}^{3}c_{1}^{3}-4c_{3}^{2}c_{2}^{3}c_{0}+c_{3}^{2}c_{2}^{2}c_{1}^{2}+144c_{3}^{2}c_{2}c_{0}^{2}$
$\displaystyle-$ $\displaystyle
6c_{3}^{2}c_{1}^{2}c_{0}-80c_{3}c_{2}^{2}c_{1}c_{0}+18c_{3}c_{2}c_{1}^{3}+16c_{2}^{4}c_{0}-4c_{2}^{3}c_{1}^{2}-192c_{3}c_{1}c_{0}^{2}$
$\displaystyle-$ $\displaystyle
128c_{2}^{2}c_{0}^{2}+144c_{2}c_{1}^{2}c_{0}-27c_{1}^{4}+256c_{0}^{3},$
$\mathcal{P}_{0}=8c_{2}-3c_{3}^{2},\quad\mathcal{Q}_{0}=64c_{0}-16c_{2}^{2}+16c_{3}^{2}c_{2}-16c_{3}c_{1}-3c_{3}^{4}.$

Using equations (2.9) and (2.10), we express $\mathcal{D}_{0}$,
$\mathcal{P}_{0}$, and $\mathcal{Q}_{0}$, as the following functions of the
physical parameters $\Delta$, $E$, and $g$:

$\displaystyle\mathcal{D}_{0}$ $\displaystyle=$
$\displaystyle-\frac{1}{4g^{20}}\left(1024\Delta^{6}g^{8}-1024\Delta^{4}E^{2}g^{8}+2048\Delta^{6}Eg^{6}-2048\Delta^{4}E^{3}g^{6}\right.$
$\displaystyle-$ $\displaystyle
1024\Delta^{4}Eg^{8}+4608\Delta^{2}Eg^{10}-4096E^{3}g^{10}+512\Delta^{8}g^{4}+512\Delta^{6}E^{2}g^{4}$
$\displaystyle+$ $\displaystyle
1024\Delta^{6}g^{6}-1024\Delta^{4}E^{4}g^{4}-3072\Delta^{4}E^{2}g^{6}+2048\Delta^{4}g^{8}+7680\Delta^{2}E^{2}g^{8}$
$\displaystyle+$ $\displaystyle
2304\Delta^{2}g^{10}-8192E^{4}g^{8}-6144E^{2}g^{10}+1728g^{12}+512\Delta^{8}Eg^{2}$
$\displaystyle-$ $\displaystyle
512\Delta^{6}E^{3}g^{2}+512\Delta^{6}Eg^{4}-2048\Delta^{4}E^{3}g^{4}+3328\Delta^{4}Eg^{6}+1536\Delta^{2}E^{3}g^{6}$
$\displaystyle+$ $\displaystyle