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Nuttall decomposition of a three-sheeted torus S. R. Nasyrov Kazan (Volga Region) Federal University
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2024, Volume 88, Issue 5, DOI: https://doi.org/10.4213/im9561 
Bibliographic databases:
  § 1. IntroductionWe first recall the definition of type II Hermite–Padé diagonal approximations. Consider some functions $f_j$, $1\leqslant j\leqslant m$, of complex variable $\tau$ holomorphic in a neighbourhood of infinity. Their type II Hermite–Padé diagonal approximants are rational functions
$$
\begin{equation}
\frac{Q_{nj}(\tau)}{P_n(\tau)},\qquad 1\leqslant j\leqslant m,
\end{equation}
\tag{1.1}
$$
such that the polynomials $Q_{nj}(\tau)$ and $P_n(\tau)$, $\operatorname{deg} P_n\leqslant mn$, satisfy
$$
\begin{equation*}
P_n(\tau)f_j(\tau)-Q_{nj}(\tau)=O(\tau^{-(n+1)}),\qquad \tau\to\infty, \quad 1\leqslant j\leqslant m.
\end{equation*}
\notag
$$
Assume that the functions $f_j$ can by analytically continued from infinity along every curve lying outside a fixed compact $\mathcal{E}$. An important problem is to find maximal convergence domains of the approximants to the given functions. In the case of a single function ($m=1$), the problem was solved by Stahl [1], [2]. He proved that such maximal domain is the exterior of a compact set $\mathcal{K}$. The compact set $\mathcal{K}$ is described via orthogonal critical trajectories of a quadratic differential related to the Green function of the exterior of $\mathcal{K}$. In the case of multiple functions, the problem is open. It is of interest to study it for finite sets $\mathcal{E}$. So, in what follows, $\mathcal{E}$ is a finite set. Nuttall [3] conjectured that the asymptotics of Padé–Hermite approximants is related to an $(m+1)$-sheeted compact Riemann surface $S$ over the Riemann sphere with branch points over the set $\mathcal{E}\subset \mathbb{C}$. Let $P_0,P_1,\dots,P_m$ be the points of $S$ lying over infinity. Denote by $p$ the projection, $p\colon S\to \overline{\mathbb{C}}$. Let $p_j$ be the restriction of $p$ on a sufficiently small neighbourhood of the point $P_j$ such that $p_j$ is injective. Consider an Abelian integral $G$ on $S$ regular at each point of $S$, except for the points $P_j$, $0\leqslant j\leqslant m$. At these points, $G$ behaves as
$$
\begin{equation}
G(p_j^{-1}(\tau))\sim \begin{cases} m\ln \tau, &\tau\to P_0, \\ -\ln \tau, &\tau\to P_j,\, 1\leqslant j \leqslant m, \\ \end{cases}
\end{equation}
\tag{1.2}
$$
as $\tau\to\infty$. In addition, we require that all the periods of $G$ are pure imaginary. Note that such differential is defined uniquely, up to an additive constant, and $g=\operatorname{Re} G$ is a single-valued harmonic function on $S$. For every $\tau\in \mathbb{C}\setminus \mathcal{E}$, there are exactly $(m+1)$ points of $S$ lying over $\tau$. Denote them by $\tau^{(0)},\tau^{(1)},\dots,\tau^{(m)}$. We also set $g_j(\tau):=g(\tau^{(j)})$. It is clear that we can fix a numeration of $\tau^{(0)},\tau^{(1)},\dots,\tau^{(m)}$ such that $g_0(\tau)\geqslant g_1(\tau)\geqslant\dots\geqslant g_m(\tau)$, $\tau\in \mathbb{C}\setminus \mathcal{E}$. We note that, at points where the values of some $g_j$ coincide, such a numeration is not unique. Consider the set of points $\tau$ on the complex plane at which the values of $g_j(\tau)$ are pairwise distinct. For such $\tau$, we have Consider the sets
$$
\begin{equation*}
S_j:=\{P\in S\colon P=\tau^{(j)}\ \ \text{for }\ \tau =p(P)\}.
\end{equation*}
\notag
$$
We will call $S_j$ the $j$th sheet of $S$. Note that we can obtain $S$ from the sheets $S_j$, $0\leqslant j\leqslant m$, by gluing them along some piecewise analytic arcs, and the projection of every boundary $\partial S_j$ of the leaf $S_j$ on the Riemann sphere consists of a finite set of analytic arcs. A decomposition of a compact Riemann surface into such Nuttall sheets is a very complicated problem with important applications in approximation theory (see, in this regard, some recent papers [4]–[8]). Even for $m=2$ and if the number of points of the set $\mathcal{E}$ is not large, the structure of the sheets is not completely studied. Consider, for example, the Hermite–Padé approximation problem for a pair of functions $f_1=f$, $f_2=f^2$, where all the points $a_j$ are distinct, the numbers $2\alpha_j$ are not integer, $\sum_{j=1}^3{\alpha_j}=0$. This problem was suggested by S. P. Suetin (see p. 998 in [9]).Consider the Riemann surface of the function and construct for it the functions $g_j$, $j=0,1,2$. Aptekatev and Tulyakov [9] fully studied the geometric structure of the set
$$
\begin{equation*}
\{\tau\in \mathbb{C} \mid \exists\, j,k\in \{0,1,2\}\colon j\neq k \ \text{ and }\ g_j(\tau)=g_k(\tau)\}
\end{equation*}
\notag
$$
in the case where the triangle $\Delta(a_1,a_2,a_3)$ with vertices $a_1$, $a_2$, $a_3$ is sufficiently close to a regular one. This set and its subsets play an important role in investigations of convergence of rational approximations. By the Nuttall conjecture, the set $\gamma_{01}:=\{\tau\in \mathbb{C} \mid g_0(\tau)=g_1(\tau)\}$ attracts the poles of the Hermite–Padé approximants. The set $\gamma_{12}:=\{\tau\in \mathbb{C} \mid g_1(\tau)=g_2(\tau)\}$ is also of interest. In this paper, we investigate the above for an arbitrary location of the points $a_1$, $a_2$, $a_3$. Let us describe the main results of the paper. 1. As is known, the Riemann surface of function (1.3) is a three-sheeted complex torus, that is, a parabolic Riemann surface; therefore, we can take the complex plane $\mathbb{C}$ as its universal covering space. With the help of the Weierstrass elliptic functions, we construct, in an explicit form, a function that performs this universal covering and also an Abelian integral on the universal covering (that is, on the complex plane $\mathbb{C}$) which is the lift of Nuttall’s Abelian integral defined on the torus. Such a covering has the property that rotations of the complex plane about the origin through $\pm 2\pi/3$ induce conformal automorphisms of the torus which leaves fixed the projections to the plane $\mathbb{C}$. 2. The Nuttall decomposition of the complex torus into sheets induces a decomposition of $\mathbb{C}$ into three sets; we will call them sheets on the universal covering space. We show that the boundaries of sheets are in the union of the null set $L(u,0)$ of a doubly periodic harmonic function $u$ with two logarithmic singularities in every parallelogram of periods, and whose sets are obtained by rotations of $L(u,0)$ about the origin through $\pm 2\pi/3$. 3. We fully describe the differential-geometric structure of the set $L(u,0)$. In particular, we show that the gradient $\nabla u$ has exactly two zeros (taking account of multiplicity) in every parallelogram of periods, and these zeros are in the set $L(u,0)$ if and only if either the points $a_1$, $a_2$ and $a_3$ are the vertices of an isosceles triangle with apex angle greater than $\pi/3$ or these three points lie on the same straight line. 4. In the case where the points $a_1$, $a_2$ and $a_3$ are the vertices of an isosceles triangle with apex angle $<\pi/3$, we completely describe the structure of the Nuttall sheets. In other cases, the structure is described under an additional assumption (Conjecture 11.1) that, on the universal covering space, in every parallelogram of periods, the intersection of the set $L(u,0)$ with the sets $e^{\pm 2\pi i/3}L(u,0)$, except for the points lying over $a_1$, $a_2$ and $a_3$, contains only three points, which are obtained from each other by a rotation about the origin through $\pm 2\pi/3$, in addition, every two of the three sets intersect transversely. These points correspond to three points on the torus with the same projection. The projections of the lines separating the sheets contain $\infty$; this fact takes place regardless of the validity of Conjecture 11.1. It is also shown that, in the case where $a_1$, $a_2$ and $a_3$ are on the same straight line, the zero Nuttall sheet is disconnected. This fact holds regardless of Conjecture 11.1. The paper is organized as follows. In § 2, we recall some properties of the Weierstrass elliptic $\mathfrak{P}$-, $\zeta$-, and $\sigma$-functions with arbitrary periods $\omega_1$ and $\omega_2$. In § 3, we consider the case where the lattice of periods is generated by two vectors equal in length and such that the angle between them is $\pi/3$. Some geometric properties of the Weierstrass functions are established and some auxiliary inequalities are proved. In § 4, we introduce a real-valued doubly-periodic function $V(z)$ playing an important role in the study of Nuttall’s Abelian integral on the three-sheeted torus, and investigate properties of the function and its gradient. In § 5, we describe uniformization of the Riemann surface of function (1.3). The universal covering is carried out by a function constructed with the help of the Weierstrass sigma function; the inverse is locally expressed via the Schwarz–Christoffel mapping. In § 6, we construct the induced Abelian integral on the universal covering space and, with the help of the Weierstrass sigma function, find equations for describing the boundary of the Nuttall sheets. These equations have the form $u(z)=0$, $u(e^{2\pi i/3}z)=0$ and $u(e^{-2\pi i/3}z)=0$, where $u(z)$ is a definite doubly-periodic harmonic function with logarithmic singularities at some points. These points are defined by a complex parameter $\alpha$, which has a transparent geometric meaning: a point $\alpha$ on the universal covering space corresponds to the infinite point lying on one of sheets of the Riemann surface of function (1.3). Moreover, $\alpha$ lies in the regular triangle $ABC$ with vertices $A=0$, $B=e^{\pi i/6}$ and $C=e^{-\pi i/6}$ and can coincide with any of its points except for its vertices. We also note that the points $A$, $B$ and $C$ on the universal covering space correspond to branch points of the Riemann surface of function (1.3). We next study the null set of $u$. Its structure depends essentially on whether it contains zeros of the gradient of $u$ (there are exactly two such zeros taking account of multiplicities). The main result of § 7 is Lemma 7.2, which states that if the zeros of the gradient are in the null set of $u$, then $\alpha$ is either on the boundary of the triangle $ABC$ or on some of its symmetry axis. The first case means that the points $a_1$, $a_2$ and $a_3$ are on the same straight line; in the second case, the triangle with vertices at $a_1$, $a_2$ and $a_3$ is isosceles. In § 8, it is proved that, in the case of isosceles triangle with vertices at $a_1$, $a_2$ and $a_3$, the zeros of the gradient of $u$ are in its null set if and only if the apex angle of this triangle is $\geqslant \pi/3$ (Theorem 8.1). In the case of angle $\pi/3$ (a regular triangle), the zero point of the gradient is multiple. In § 9, we consider the case where $a_1$, $a_2$ and $a_3$ are on the same straight line; in this case, the zeros of the gradient are in the null set of $u$. In § 10, we introduce a quadratic differential such that the function $u$ is constant on its trajectories. We study the structure of singular trajectories of this quadratic differential. In § 11, we describe the structure of the Nuttall sheets and the corresponding sets on the universal covering space depending on the geometry of the triangle $a_1a_2a_3$, that is, for various locations of the parameter $\alpha$ in the triangle $ABC$; we obtain this description under Conjecture 11.1 on the points of intersection of the null set $L(u,0)$ of $u$ and the sets obtained from this set by rotations about the origin through $\pm2\pi/3$. There are four essentially different cases to consider: 1) the triangle $a_1a_2a_3$ is isosceles with apex angle $<\pi/3$; 2) the triangle $a_1a_2a_3$ is not isosceles; 3) the triangle $a_1a_2a_3$ is isosceles with apex angle $>\pi/3$; 4) the points $a_1$, $a_2$ and $a_3$ are on the same straight line. In addition, the cases, where either the triangle $a_1a_2a_3$ is regular or $a_1$, $a_2$ and $a_3$ lie on the same straight line and one of the points is the mid-point of the segment connecting the other two, require a separate consideration. In § 12, we prove Conjecture 11.1 for the case 1). To this end, we consider the sets of points which are critical for the restrictions of the function $u$ to the straight lines forming the angle $\pi/3$ (or $-\pi/3$) with the real axis. We show that such sets are the preimages of some circles under the Weierstrass $\mathfrak{P}$-function, and study the range of the inclination angle of the tangents to this set. § 2. Properties of Weierstrass elliptic functionsThe main tool in the paper is the machinery of Weierstrass elliptic functions. Here, we recall some their properties. Next, in § 3 will give some additional properties if the lattice of periods $\Omega$ is generated by the vectors $\omega_1=\sqrt{3}$ and $\omega_2=\sqrt{3}\,e^{i\pi/3}$; in this case the nodes of the lattice $\Omega$ are at vertices of regular triangles of side $\sqrt{3}$ that form a triangulation of the plane. We first consider the Weierstrass elliptic functions of arbitrary periods1 $\omega_1$ and $\omega_2$. The Weierstrass $\mathfrak{P}$-function is a doubly-periodic function defined as the sum of the series
$$
\begin{equation}
\mathfrak{P}(z)=\frac{1}{z^2} +\mathop{{\sum}'} \biggl[\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\biggr],
\end{equation}
\tag{2.1}
$$
where the sum $\mathop{{\sum}'}$ is taken over all non-zero periods $\omega=m_1\omega_1+m_2\omega_2$, $m_1,m_2\in \mathbb{Z}$, of the lattice $\Omega$. We note some transparent and important properties of the $\mathfrak{P}$-function. The function $\mathfrak{P}$ is even. In the parallelogram of periods, it has a unique pole of the second order, and assumes each value two times (with multiplicity). Therefore, the Weierstrass $\mathfrak{P}$-function induces a two-sheeted branched covering of the extended complex plane by the torus $\mathbb{C}/\Omega$. The function $\mathfrak{P}$ satisfies the differential equation
$$
\begin{equation}
(\mathfrak{P}'(z))^2=4(\mathfrak{P}(z)-e_1)(\mathfrak{P}(z)-e_2)(\mathfrak{P}(z) -e_3),
\end{equation}
\tag{2.2}
$$
where
$$
\begin{equation*}
e_1=\mathfrak{P}\biggl(\frac{\omega_1}2\biggr),\qquad e_2=\mathfrak{P}\biggl(\frac{\omega_2}2\biggr),\qquad e_3=\mathfrak{P}\biggl(\frac{\omega_1+\omega_2}2\biggr).
\end{equation*}
\notag
$$
The Weierstrass zeta function is introduced by
$$
\begin{equation}
\zeta(z)=\frac{1}{z}+\mathop{{\sum}'}\biggl[\frac{1}{z-\omega} +\frac{1}{\omega}+\frac{z}{\omega^2}\biggr].
\end{equation}
\tag{2.3}
$$
This function is odd, and, in every parallelogram of periods, it has a unique simple pole with residue $1$, and $\zeta'(z)=-\mathfrak{P}(z)$. The values of this function change by an additive constant if its argument changes by a period, where $\eta_k=2\zeta(\omega_k/2)$. In addition, The next assertion is obtained with the help of (2.4) and in view of the well-known property of elliptic functions (see [10], p. 18, § $4^\circ$). Theorem 2.1. For all points $a$, $b$ not comparable modulo the lattice $\Omega$, the function $\zeta(z-a)-\zeta(z-b)$ is elliptic, it has two simple poles at points comparable with $a$ and $b$ modulo $\Omega$, and two zeros (taking into account multiplicities) $z_1$ and $z_2$ in every parallelogram of periods. In addition, that is, the zeros of this functions are symmetric with respect to the point $(a+b)/2$ modulo $\Omega$. The Weierstrass sigma function is defined by
$$
\begin{equation}
\sigma(z)=z\mathop{{\prod}'}\biggl(1-\frac{z}{\omega}\biggr)e^{z/\omega+z^2/(2\omega^2)}.
\end{equation}
\tag{2.6}
$$
It is odd, has simple zeros at points of $\Omega$, and satisfies In addition,
$$
\begin{equation}
\sigma(z+\omega_k)=-\sigma(z) e^{\eta_k(z+\omega_k/2)}, \qquad k=1,2.
\end{equation}
\tag{2.8}
$$
Elliptic functions can be represented as ratios of two products of sigma functions. Namely, the following result holds (see, for example, [10], Chap. III, § 14). Theorem 2.2. Let a non-constant elliptic function $g$, in the parallelogram of periods, has $n$ zeros at points $a_1,\dots,a_n$ and $n$ poles at points $b_1,\dots,b_n$ (the zeros and poles are counted with their multiplicities). Let $\omega=\sum_{k=1}^n(b_k-a_k)$. Then $\omega$ is an element of $\Omega$, and there exists a non-zero constant $C$ such that where $b_k^*=b_k$, $1\leqslant k\leqslant n-1$, and $b_n^*=b_n-\omega$. § 3. Elliptic functions in the case of a regular latticeIn what follows, we will assume that the lattice $\Omega$ is generated by the vectors
$$
\begin{equation}
\omega_1=\sqrt{3}, \qquad \omega_2=\sqrt{3}\, e^{i\pi/3}.
\end{equation}
\tag{3.1}
$$
In this case, the parallelogram of periods consists of two regular triangles $AIJ$ and $JIM$ of side $\sqrt{3}$ (see Fig. 1). Because of the symmetry of the lattice with respect to the real axis and straight lines that pass through the origin and form the angles $\pm 2\pi/3$ with the positive part of the real axis, and since $\Omega$ is invariant under rotations about points of the lattice by multiples of $2\pi/3$, we conclude that
$$
\begin{equation}
\mathfrak{P}(\overline{z})=\overline{\mathfrak{P}(z)},\qquad \zeta(\overline{z})=\overline{\zeta(z)},\qquad \sigma(\overline{z})=\overline{\sigma(z)},
\end{equation}
\tag{3.2}
$$
$$
\begin{equation}
\mathfrak{P}(e^{i\pi/3}z)=e^{-i2\pi/3}\mathfrak{P}(z),\quad \zeta(e^{i\pi/3}z)=e^{-i\pi/3}\zeta(z), \quad \sigma(e^{i\pi/3}z)=e^{i\pi/3}\sigma(z).
\end{equation}
\tag{3.3}
$$
We will show that
$$
\begin{equation}
\eta_1=\frac{2\pi}3,\qquad \eta_2=\frac{2\pi}3\, e^{-\pi i/3}.
\end{equation}
\tag{3.4}
$$
Indeed, $\omega_1=\sqrt{3}$, $\omega_2=\sqrt{3}\, e^{\pi i/3}$. In addition, by the second equality in (3.3), we have $\eta_2=\eta_1e^{-\pi i/3}$. Now an appeal to (2.5) shows that $\eta_1\sqrt{3}\, e^{\pi i/3}-\eta_1e^{-\pi i/3}\sqrt{3}=3\eta_1 i= 2\pi i$, which implies (3.4). In view of (3.3) we can write (2.2) as Differentiating, we obtainNow let us study the behaviour of the Weierstrass functions in the triangle $ABD$ with vertices $A=0$, $B=\sqrt{3}/2+i/2$ and $D=\sqrt{3}/2$ (see Fig. 1). Using the geometric meaning of the argument of the derivative of a conformal mapping, we obtain the following result. Theorem 3.1. 1) The image of the triangle $ABD$ under the function $\mathfrak{P}$ is the angle $\{-\pi/3 < \arg w < 0\}$ with the correspondence of points indicated in Fig. 1. 2) The image of the same triangle under the Weierstrass function $\zeta$ is the unbounded triangle $\{\operatorname{Re} w >\pi/3, -\pi/6<\arg w<0\}$. 3) The function $\mathfrak{P}'$ maps conformally the triangle $ABD$ onto the second quarter of the $w$-plane. 4) The function $\mathfrak{P}'/\mathfrak{P}$ maps conformally the triangle $ABD$ onto the second quarter of the $w$-plane with excluded ray $R$, which is a part of the ray2 $\{\arg w=5\pi/6\}$. Applying the Riemann–Schwarz symmetry principle a few times, we see that the image of the triangle $AIJ$ is the plane cut along three rays with vertices at the points
$$
\begin{equation*}
e_1>0,\qquad e_2=e^{-2\pi i/3}e_1,\qquad e_3=e^{2\pi i/3}e_1,
\end{equation*}
\notag
$$
and the origin lies at the continuations of these rays (see Fig. 1). It can be shown that where $B(\,{\cdot}, {\cdot}\,)$ is the Euler beta function. Consider the regular hexagon $\mathfrak{F}$ of side $1$ with centre at the origin, and whose one vertex is at the point $B$. This hexagon contains precisely one point from each of the equivalence classes modulo $\Omega$ (with evident identification of points from the opposite sides). We will call this hexagon fundamental, in analogy with the fundamental parallelogram. The following results can be obtained using the Riemann–Schwarz symmetry principle. Theorem 3.2. The Weierstrass $\mathfrak{P}$-function maps the interior of the fundamental hexagon $\mathfrak{F}$ with periods $(3.1)$ onto the two-sheeted Riemann surface obtained from the compact two-sheeted Riemann surface with branch points lying over $e_1$, $e_2$, $e_3$ and $\infty$ by making cuts on both sheets along the radial rays going from these points. In addition, the segments connecting the origin with the vertices and mid-points of the sides of $\mathfrak{F}$ and which lie on the rays $\{\arg z=k\pi/6\}$, $k\in \mathbb{Z}$, are mapped one-to-one onto the rays satisfying $\arg w=-k\pi/3$. Naturally, if the opposite sides of the fundamental hexagon $\mathfrak{F}$ are identified, then the image is a two-sheeted compact Riemann surface of genus $1$ (a complex torus) with branch points over the points $e_1$, $e_2$ and $e_3$. Let
$$
\begin{equation}
c=\frac{\eta_1}{\omega_1}=\frac{2\pi}{3\sqrt{3}}=1.2092\dots\,.
\end{equation}
\tag{3.7}
$$
Theorem 3.3. In the case of the lattice with periods (3.1), the Weierstrass zeta function is a univalent conformal mapping of the interior of $\mathfrak{F}$ onto the exterior of the regular hexagon centred at the origin such that the mid-point of one of its side is the point $\eta_1/2=\pi/3$, which is the image of the point $D$. In addition, each vertex and mid-point of every side of $\mathfrak{F}$ with affix $z$ is mapped to the point $c\overline{z}$, and the segment connecting $z$ with the origin is mapped to the radial ray with vertex at the point $c\overline{z}$. Now we consider the function The next result is immediate from Theorem 3.3. Corollary 3.1. The function $\widetilde{\zeta}(z)$ maps conformally the regular hexagon $\mathfrak{F}^*$ obtained from $\mathfrak{F}$ by the rotation about the origin through $\pi/6$ onto the exterior of a regular hexagon homothetic to $\mathfrak{F}^*$. In addition, each vertex and mid-point of a side of $\mathfrak{F}^*$ with affix $z$ is mapped to the point $c\overline{z}$, and the segment connecting $z$ with the origin is mapped to the radial ray with vertex at the point $c\overline{z}$. Lemma 3.1. In the interior of the triangle $ABD$:3 Proof. Assertion 1) is immediate from assertion 1) of Theorem 3.1.
2) The function $\beta(z):=z^2\mathfrak{P}(z)$ has a removable singularity at the origin and $\lim_{z\to0}\beta(z)=1$. Therefore, $\beta$ can be extended , as a continuous function, to the closure of the triangle $ABD$, and the points $A$, $B$ and $D$ are mapped to $1$, $0$ and $(\sqrt{3}/2)^2\mathfrak{P}(\sqrt{3}/2)=3e_1/4= 1.47459\dots$ . From assertion 1) of Theorem 3.1, it follows that $\beta$ maps the sides $AB$ and $AD$ of the triangle $ABD$ to some subsets of the positive part of the real axis. From the same theorem it also follows that, on the side $BD$, the function $\mathfrak{P}(z)$ takes positive values (except at the point $B$, which is mapped to the origin), therefore, on this side the function $ \arg \beta(z)=2\arg z+\arg \mathfrak{P}(z)=2\arg z $ strictly increases from $0$ to $\pi/3$ if $z$ varies over $BD$ from $D$ to $B$. Hence $BD$ is mapped to a simple curve $L_1$ which connects the point $D'=3e_1/4$ and the origin $B'$ (see Fig. 2). A simple reasoning based on the argument principle (see, for example, [11], Chap. 1, § 1) shows that the function $\beta(z)$ maps univalently the triangle $ABD$ onto the domain bounded by the segment of the real axis that connects the points $B'=0$ and $D'=3e_1/4$, and the curve $L_1$. Since, as noted above, for the points of $L_1$ the argument changes from $0$ to $\pi/3$, and on the rest of the boundary it is zero, we conclude that the domain lies in the angle $0<\arg w<\pi/3$, and so (3.10) holds. 3) Proceeding as in the proof of (3.10) and employing Theorem 3.1, we verify that the function $\gamma(z):=z\zeta(z)$ assumes positive values on the sides $AB$ and $AD$ of the triangle $ABD$. We will show that if $z$ is on the segment $BD$ and does not coincide with its end-points, then To this end, we note that the antiholomorphic function $w=c\overline{z}$ maps $ABD$ onto the triangle $\Delta_1$ with vertices at the points $\widetilde{A}_1=0$, $\widetilde{D}=\eta_1/2$, $\widetilde{B}=(\eta_1/2)(1-i\sqrt{3}/3)$, and the holomorphic function $\zeta(z)$ maps $ABD$ onto the domain $\Delta_2$, which is the complement of the triangle $\Delta_1$ with respect to the angle $-\pi/3<\arg w<0$ (see Fig. 3). Consider an inner point $z$ of the segment $BD$ and denote by $\widetilde{E}$ its image $\zeta(z)$ under the mapping $\zeta$.We will show that the point $\widetilde{E}$ lies on the segment $\widetilde{B}\widetilde{D}$ lower than the point $E$ with affix $c\overline{z}$. For this, we compare the conformal moduli of two quadrilaterals:4 $\Delta_1(\widetilde{A}_1, \widetilde{B},\widetilde{E}, \widetilde{D})$ and $\Delta_2(\widetilde{A}{}_2, \widetilde{B}, \widetilde{E}, \widetilde{D})$, where $\widetilde{A}{}_2$ is infinity. Let the triangle $\Delta_3$ be obtained from $\Delta_1$ by reflection with respect to the straight line $\operatorname{Re} w=\eta_1/2$. This triangle has vertices $\widetilde{A}_3$, $\widetilde{B}$ and $\widetilde{D}$, where $\widetilde{A}_3$ is the point symmetric to $\widetilde{A}_1$ with respect to the straight line $\operatorname{Re} w=\eta_1/2$. By the symmetry,
$$
\begin{equation*}
\operatorname{mod} \Delta_1(\widetilde{A}_1, \widetilde{B}, \widetilde{E},\widetilde{D}) = \operatorname{mod} \Delta_3(\widetilde{A}_3, \widetilde{B}, \widetilde{E},\widetilde{D}).
\end{equation*}
\notag
$$
In addition, since the modulus increases with extension of the domain, we have
$$
\begin{equation*}
\operatorname{mod} \Delta_3(\widetilde{A}_3, \widetilde{B}, \widetilde{E},\widetilde{D}) \leqslant \operatorname{mod} \Delta_2(\widetilde{A}_3, \widetilde{B}, \widetilde{E},\widetilde{D}).
\end{equation*}
\notag
$$
Now by Theorem 2.3.4 in [12],
$$
\begin{equation*}
\operatorname{mod} \Delta_2(\widetilde{A}_3, \widetilde{B}, \widetilde{E},\widetilde{D}) < \operatorname{mod} \Delta_2(\widetilde{A}_2, \widetilde{B}, \widetilde{E},\widetilde{D}).
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation}
\operatorname{mod}\Delta_1(\widetilde{A}_1, \widetilde{B}, \widetilde{E}, \widetilde{D}) < \operatorname{mod} \Delta_2(\widetilde{A}_2, \widetilde{B}, \widetilde{E},\widetilde{D}).
\end{equation}
\tag{3.16}
$$
Consider the quadrilateral $\Delta_1(\widetilde{A}_1, \widetilde{B}, \widetilde{E},\widetilde{D})$ and investigate the dependence of its conformal modulus from the location of its vertex $\widetilde{E}$ on the side $\widetilde{B}\widetilde{D}$. Another appeal to Theorem 2.3.4 in [12] shows that the conformal modulus is strictly increasing as $\widetilde{E}$ moves upward along $\widetilde{A}\widetilde{B}$. Now consider the quadrilateral $\Delta_1(\widetilde{A}_1, \widetilde{B}, {E},\widetilde{D})$. By the invariance of the conformal modulus under conformal and anticonformal mappings, we conclude that
$$
\begin{equation}
\operatorname{mod} \Delta_1(\widetilde{A}_1, \widetilde{B}, E,\widetilde{D}) = \operatorname{mod} \Delta_2(\widetilde{A}{}_2, \widetilde{B}, \widetilde{E}, \widetilde{D}).
\end{equation}
\tag{3.17}
$$
From (3.16) and (3.17) it follows that $\operatorname{mod}\Delta_1(\widetilde{A}_1, \widetilde{B},E, \widetilde{D}) > \operatorname{mod} \Delta_1(\widetilde{A}_1, \widetilde{B}, \widetilde{E},\widetilde{D})$. Therefore, on the segment $\widetilde{B}\widetilde{D}$, the point $\widetilde{E}$ with affix $\zeta(z)$ lies lower than the point $E$ with affix $c\overline{z}$. Hence $|{\arg \zeta(z)}|<|{\arg c\overline{z}}|=|{\arg z}|$, and (3.15) is proved. The moduli of the complex numbers $z$ and $\zeta(z)$ are strictly increasing as $z$ moves upward along the segment $BD$. Hence the function $|\gamma(z)|$ is strictly increases on the segment $BD$. From (3.15) we see that, for inner points of $BD$,
$$
\begin{equation*}
\arg \gamma(z)=\arg z+\arg \zeta(z)=|{\arg z}|-|{\arg \zeta(z)}|< 0,
\end{equation*}
\notag
$$
therefore, the image of this segment under the mapping $\beta$ is a simple curve $L_2$ lying in the lower half-plane (see Fig. 4). By the argument principle, the function $z\zeta(z)$ is univalent in the triangle $ABD$ and maps it onto the domain bounded by the segment on the positive part of the real axis with end-points $D''=\pi\sqrt{3}/6$ and $B''=2\pi\sqrt{3}/9$ and the curve $L_2$ lying in the lower half-plane. Now (3.11) follows. 4) As noted in the proof of (3.11), the function $c\overline{z}$ maps the triangle $ABD$ onto $\Delta_1$, and the function $\zeta(z)$ maps it onto the domain $\Delta_2$ (see Fig. 3). Thus, in the triangle, we have $\operatorname{Re}\zeta(z)>\eta_1/2>\operatorname{Re}[c\overline{z}]$, that is, $\operatorname{Re}[\zeta(z)-c\overline{z}]>0$. Besides, from (3.15) it follows that, for inner points of the triangle $ABD$, we have $|{\arg[\zeta(z)]}|\geqslant|{\arg (c\overline{z})}|$, in addition, by the maximum principle, this inequality is strong. Hence
$$
\begin{equation*}
|{\operatorname{Im}\zeta(z)}| =\operatorname{Re}\zeta(z)\tan |{\arg[\zeta(z)]}| >\operatorname{Re} (c\overline{z})\tan |{\arg(c\overline{z})}| =|{\operatorname{Im}(c\overline{z})}|.
\end{equation*}
\notag
$$
This shows that $\operatorname{Im}\zeta(z)<\operatorname{Im}(c\overline{z})$ in the triangle $ABD$, that is, $\operatorname{Im}[\zeta(z)-c\overline{z}] < 0$. Therefore, the point $\zeta(z)-c\overline{z}$ is in the fourth quarter, and (3.12) is proved. 5) By (3.11), we have $\operatorname{Im}(z(\zeta(z)-c\overline{z})) =\operatorname{Im}(z\zeta(z)-c|z|^2)=\operatorname{Im}(z\zeta(z))<0$, which implies the right-hand inequality in (3.13). The left-hand side is secured by (3.12), since
$$
\begin{equation*}
\arg z(\zeta(z)-c\overline{z}) =\arg z+\arg(\zeta(z)-c\overline{z})\geqslant \arg(\zeta(z)-c\overline{z}).
\end{equation*}
\notag
$$
6) By (3.10) and (3.13), we have
$$
\begin{equation*}
\arg\frac{\mathfrak{P}(z)}{(\zeta(z)-c\overline{z})^2} =\arg\mathfrak{P}(z)-2\arg(\zeta(z)-c\overline{z})<\pi.
\end{equation*}
\notag
$$
Besides, from (3.9) and (3.12) we obtain
$$
\begin{equation*}
\arg\frac{\mathfrak{P}(z)}{(\zeta(z)-c\overline{z})^2} =\arg (z^2\mathfrak{P}(z))-2\arg(z(\zeta(z)-c\overline{z}))>0.
\end{equation*}
\notag
$$
From these inequalities we deduce (3.14). This proves Lemma 3.1. Theorem 3.3, inequalities (3.15), and the symmetry of the zeta function imply the following lemma. Lemma 3.2. The function $w=(1/c)\overline{\zeta(z)}$ maps homeomorphically the boundary of the fundamental hexagon $\mathfrak{F}$ onto itself and the vertices and mid-points of sides of $\partial\mathfrak{F}$ are fixed points of this mapping. In addition, each vertex of $\mathfrak{F}$ is an attracting point for this mapping and the mid-point of every side is repelling in the sense that if $z$ is on a side of $\partial\mathfrak{F}$ between a vertex $\mathcal{A}$ and a mid-point $\mathcal{B}$ of the side, and so its image is on the same segment $\mathcal{AB}$ and lies closer to the vertex $\mathcal{A}$ than $z$. Remark 3.1. The same result also holds for the mapping which is complex conjugate to $w=(1/c)\widetilde{\zeta}(z)$. The next result follows from Lemma 3.2. Lemma 3.3. Let $l_k=\partial\mathfrak{F}\cap \{(k-1)\pi/6<\arg z<k\pi/6\}$, $k\in \mathbb{Z}$. Then the function $\operatorname{Im}[\zeta(z)-c\overline{z}]$ is positive on $l_k$ for $k=2, 5, 7, 9, 10, 12$ and negative for $k=1, 3, 4, 6, 8, 11$. We also note the following property of the function $\widetilde{\zeta}(z)$ defined by (3.8). Lemma 3.4. Let either $z=e^{2\pi i/3}r$ or $z=1+e^{2\pi i/3}r$, $1/2<r<1$. Then $\operatorname{Im} (\widetilde{\zeta}(z)-c\overline{z})<0$. Proof. Let $z=e^{2\pi i/3}r$, $1/2<r<1$. Then, using Theorem 3.3 and Remark 3.1, we conclude that $\arg\widetilde{\zeta}(z)=\arg \overline{z}=-2\pi/3$, $|\widetilde{\zeta}(z)|>c>c|z|$, and from this we obtain $\operatorname{Im} \widetilde{\zeta}(z) =\cos(2\pi/3)|\widetilde{\zeta}(z)|<\cos(2\pi/3)c|z| =\operatorname{Im} (c\overline{z})$.
In the case $z=1+e^{2\pi i/3}r$, $1/2<r<1$, we note that the point $z^*=e^{-\pi i/6}z$ is on the side $BD$ of the triangle $ABD$. Then, by (3.15), we obtain $|{\arg \zeta(e^{-\pi i/6}z)}| >|{\arg(e^{\pi i/6}\overline{z})}|$, and from this, taking into account that $-\pi/3<\arg \zeta(z^*)$, $\arg \overline{z^*}<-\pi/6$, $\operatorname{Re} \zeta(z^*)=\operatorname{Re} c \overline{z^*}=\eta_1>0$ on the side $BD$, we conclude that $-\pi/2<\arg \widetilde{\zeta}(z)<\arg (c\overline{z})<0$, $|\widetilde{\zeta}(z)|> |c\overline{z}|$. Therefore, $\operatorname{Im} \widetilde{\zeta}(z)<\operatorname{Im} (c\overline{z})<0$. This proves the lemma. Now we will establish some properties of the function $\arg(\zeta(z)-c\overline{z})$ for points lying on segments of length $1/2$ that pass through the point $\omega_1/2$ and form the angles $\pm \pi/6$ and $\pi/2$ with the real axis. Lemma 3.5. Let $t\in(0,1/2)$. Then Proof. The last assertion of the lemma is immediate from Theorem 3.3 and assertion 4 of Lemma 3.1. For the inequalities, it suffices to prove only the first inequality, because the second one is obtained from by symmetry. So, let us prove the first inequality. We will use the following geometric reasoning: if a smooth curve $z=z(t)$, $a\leqslant t \leqslant b$, is convex, does not contain $0$, and its tangent at each of its point, except, possibly, its end-point, does not contain the origin, then the function $\arg z(t)$ is monotone with respect to the parameter $t$ on $(a,b)$.
Consider the curve $\gamma(t)=\zeta(\omega_1/2+e^{i\pi/6}t) -c\overline{(\omega_1/2+e^{i\pi/6}t)}$, $t\in(0,1/2)$. We will show that this curve is convex. We have
$$
\begin{equation*}
\gamma_1(t):=-\gamma'(t) =e^{i\pi/6}\mathfrak{P}\biggl(\frac{\omega_1}2+e^{i\pi/6}t\biggr)+ce^{-i\pi/6}.
\end{equation*}
\notag
$$
Let us first verify that the curve $\gamma_1(t)$, $t\in(0,1/2)$ is convex. Using (3.6), we find the derivatives
$$
\begin{equation*}
\gamma_1'(t)=e^{i\pi/3}\mathfrak{P}'\biggl(\frac{\omega_1}2+e^{i\pi/6}t\biggr), \quad \gamma_1''(t)=i\mathfrak{P}''\biggl(\frac{\omega_1}2+e^{i\pi/6}t\biggr) =6i\mathfrak{P}^2\biggl(\frac{\omega_1}2 +e^{i\pi/6}t\biggr),
\end{equation*}
\notag
$$
$\arg \gamma_1''(t)=\pi/2+2 \arg \mathfrak{P}(\omega_1/2+e^{i\pi/6}t)$. Hence
$$
\begin{equation*}
\frac{d}{dt}\biggl[\arg \mathfrak{P}\biggr(\frac{\omega_1}2+e^{i\pi/6}t\biggr)\biggr] =\operatorname{Im} \biggl[e^{i\pi/6} \frac{\mathfrak{P}'(\omega_1/2+e^{i\pi/6}t)}{\mathfrak{P}(\omega_1/2+e^{i\pi/6}t)}\biggr]>0,
\end{equation*}
\notag
$$
since from assertion 4 of Theorem 3.1 we have, by the symmetry principle,
$$
\begin{equation*}
\arg\frac{\mathfrak{P}'(\omega_1/2+e^{i\pi/6}t)}{\mathfrak{P}(\omega_1/2+e^{i\pi/6}t)} \in \biggl(0,\frac{\pi}2\biggr).
\end{equation*}
\notag
$$
Therefore, $\arg \gamma_1''(t)$ strictly increases on $[0,1/2]$ from $\pi/2$ to $\pi/2+2 \arg \mathfrak{P}(\omega_1/2+(1/2)e^{i\pi/6}) =\pi/2+2\arg(2 e_1 e^{i\pi/3})=7\pi/6$. The curve $\gamma_1'(t)$ joins the origin with the point $e^{i\pi/3}\mathfrak{P}'(\omega_1/2+(1/2)e^{i\pi/6})=6 e_1^{3/2}e^{5\pi i/6}$. This implies that $\pi/2<\arg \gamma_1'(t)<5\pi/6$ on the curve $\gamma_1'(t)$, and since $\arg \gamma_1'(t)$ increases, the curve $\gamma_1(t)$ is convex.
The curve $\gamma_1(t)$ joins the points $e_1e^{i\pi/6}+c e^{-i\pi/6}=2.7499\ldots + 0.378458\dots i$ and $2e_1 i+c e^{-i\pi/6}=1.0472\ldots + 3.32763\dots i$. Since the origin is lower than every tangent to this curve, the function $\arg\gamma_1(t)$ is monotone on it. Hence the curve $\gamma(t)$ is convex. The curve $\gamma(t)$ connects the point with the origin. The angles of inclination of tangents to the curve at its end-points are equal to $\arg \gamma_1(0.5)=0.136767\dots$ and $\arg \gamma_1(0)=1.26591\dots$ . Hence the function $\arg \gamma(t)$ strictly increases and has extremal values at the points $t=0$ and $t=0.5$. We have $\arg \gamma(0.5)=7\pi/6$. To find $\lim_{t\to 0+}\arg \gamma(t)$, we note that $\arg \gamma(t)=\arg (\gamma(t)/t)$ and In addition, $\arg(e^{i\pi/6}e_1+ce^{-i\pi/6})=\delta_0$. This proves the lemma.§ 4. Function $V(z)$ and its propertiesConsider the function where the constant $c$ is defined by (3.7).The gradient of $V$ in the complex form5 is Lemma 4.1. The function $V$ has the following properties. 1) The function $V(z)$ has periods $\omega_1$ and $\omega_2$. 2) $V(\overline{z})=V(z)$. 3) $V(e^{i\pi/3}z)=V(z)$. 4) In the closure of the triangle $ABD$, the gradient $\nabla V(z)$ vanishes only at its vertices $D$ and $B$. In addition, $B$ is a point of maximum of $V$, and $D$ is its saddle point. 5) If we consider only the points $z=re^{i\phi}$ from the interior of the triangle $ABD$, then the function $V(re^{i\phi})$ increases with respect to $r$ for a fixed $\phi$ and increases with respect to $\phi$ for a fixed $r$. Proof. Assertion 1) follows from (2.8). The properties 2) and 3) are consequences of (3.2) and (3.3).
Let us prove 4). By (4.2), the equality $\nabla V=0$ is equivalent to As noted in the proof of Lemma 3.1, the function $w=\zeta(z)$ maps the triangle $ABD$ onto the domain $\Delta_2$, and $w=c\overline{z}$ maps it onto the triangle $\Delta_1$; in addition, the angular points are mapped to the angular points (see Fig. 3). Since the intersection of the closures of the domains $\Delta_1$ and $\Delta_2$ is the segment $\widetilde{B}\widetilde{D}$, the values of the functions $\zeta(z)$ and $\overline{z}$ in the closure of $ABD$ can only coincide if $z$ is on the side $BD$ of the triangle. It is evident that at the points $B$ and $D$ the gradient $\overline{\zeta(z)}-cz$ of $V(z)$ vanishes. At other points of the segment $BD$, this equality is impossible because of (3.15).The matrix of the second order partial derivatives of $V(z)$ (the Hessian matrix) is as follows:
$$
\begin{equation*}
\begin{pmatrix} -\operatorname{Re} \mathfrak{P}(z)-c & \operatorname{Im}\mathfrak{P}(z) \\ \operatorname{Im}\mathfrak{P}(z) & \operatorname{Re} \mathfrak{P}(z)-c \end{pmatrix}.
\end{equation*}
\notag
$$
At the point $B=\sqrt{3}/2+i/2$, we have $\mathfrak{P}(\sqrt{3}/2+i/2)=0$, therefore, the leading main minors of the matrix are $I_1=-c<0$ and $I_2=c^2>0$. Therefore, at the point $B$, the second differential of $V$ is a positive definite quadratic form, and $B$ is a point of maximum of $V$.
At the point $D=\sqrt{3}/2$ we have $\operatorname{Re}\mathfrak{P}(\sqrt{3}/2)=e_1$, $\operatorname{Im}\mathfrak{P}(\sqrt{3}/2)=0$, thus, $I_1=-e_1-c<0$, $I_2=c^2-e_1^2<0$, since $c=\eta_1/\sqrt{3}=1.2092\ldots<e_1=1.96611\dots$ . Consequently, at the point $D$, the second differential is an indefinite quadratic form, and $D$ is a saddle point of $V$. 5) Let a point $z=re^{i\phi}$ be in the interior of the triangle $ABD$. By (3.13),
$$
\begin{equation*}
\frac{\partial V(re^{i\phi})}{\partial r} =\operatorname{Re}(\zeta(re^{i\phi})e^{i\phi}-c r) =\frac1{r}\operatorname{Re}[z(\zeta(z)-c\overline{z})]>0,
\end{equation*}
\notag
$$
consequently, in $ABD$ the function $V$ is strictly monotone with respect to $r$ for a fixed $\phi$.
Now we will investigate the monotonicity of $V(re^{i\phi})$ with respect to $\phi$ in the triangle $ABD$. By (3.11),
$$
\begin{equation*}
\frac{\partial V(re^{i\phi})}{\partial \phi} =\operatorname{Re}(\zeta(re^{i\phi})ire^{i\phi}) =-\operatorname{Im} (z\zeta(z))>0.
\end{equation*}
\notag
$$
Lemma 4.1 is proved. We set $ L(V,\epsilon):=\{z\in \mathbb{C}\colon V(z)=\epsilon\}$; we will call it the $\epsilon$-level set of the function $V$. Let $d_0:=V(\sqrt{3}/2+i/2)=-0.569079\dots$ be the maximal value of $V$, and $c_0:=V(\sqrt{3}/2)=-0.612683\dots$ is its value at the saddle point. Let us investigate the behaviour of the gradient $\nabla V$ on $\epsilon$-level sets of $V$. For this, we first describe the structure of these sets for $-\infty<\epsilon<d_0$ (see Fig. 5). Lemma 4.2. 1) The set $L(V,c_0)$ is connected. Its intersection with the fundamental hexagon $\mathfrak{F}$ is a starlike (with respect to the origin) curve consisting of six smooth arcs, each of which connects the points $(\sqrt{3}/2)e^{(k-1)\pi i/3}$ and $(\sqrt{3}/2)e^{k\pi i/3}$, $0\leqslant k \leqslant 6$. This curve is symmetric with respect to the real axis and does not change under rotations through multiples of $\pi/3$ about the origin, the points $B$ and $C$, and also about all points equivalent to them modulo $\Omega$. 2) If $\epsilon<c_0$, then the intersection of the set $L(V,\epsilon)$ with the fundamental hexagon $\mathfrak{F}$ is a smooth starlike (with respect to the origin) curve symmetric with respect to the real axis and invariant under rotations through multiples of $\pi/3$ about the origin, the points $B$ and $C$, and all points equivalent to them modulo $\Omega$. 3) If $c_0<\epsilon<d_0$, then the intersection of the set $L(V,\epsilon)$ with $\mathfrak{F}$ is the union of six disjoint smooth arcs. One of the arcs is the intersection of $L(V,\epsilon)$ with the triangle $ABD$; denote it by $\kappa$. The arc $\kappa$ has end-points on the segments $BD$ and $BD_1$, where $D_1$ is the point symmetric to $D$ with respect to the ray $\ell\colon \arg z=\pi/6$, it is in the quadrangle $ADBD_1$. This arc is symmetric with respect to the ray $\ell$. The remaining arcs are obtained from $\kappa$ by a rotation about the origin through multiples of $\pi/3$. The connected component of the set $L(V,\epsilon)$ containing $\kappa$ is a smooth Jordan curve which is the union of this arc and two arcs which are obtained from $\kappa$ by a rotation about $B$ through multiples of $\pm 2\pi/3$. Proof. We first note that, because of the symmetry and periodicity of $V$, the set $L(V,\epsilon)$ is invariant with respect to rotations about the origin by multiples of $2\pi/3$. It is also symmetric with respect to all straight lines passing through the origin and forming angles $\pi k/6$, $k\in \mathbb{Z}$, with the positive part of the real axis and also with respect to straight lines which are their shifts by vectors of the lattice $\Omega$. Consequently, it is invariant under rotations about the points $B$ and $C$ by multiples of $2\pi/3$.
From the proof of assertion 4) of Lemma 4.1 it follows that, at the point $B$, the second differential of $V$ has the form $-(e_1+c)\, dx^2+(e_1-c)\, dy^2$. Thus, at this point, the graph of $V$ has two asymptotical directions, which are horizontal, and their projections to the plane $XOY$ are defined by the vectors Hence, in a neighbourhood of $B$, the set $L(V,c_0)$ consists of two smooth arcs with tangent vectors (4.3).Now consider the part of $L(V,c_0)$ lying in the interior of $ABD$. The function $V(x,y)$ increases with respect to $y$ for $x=\sqrt{3}/2$, since, by (3.15)
$$
\begin{equation*}
\frac{\partial V(z)}{\partial y}=-\operatorname{Im} \zeta(z)-cy =|{\operatorname{Im}\zeta(z)}|-c|{\operatorname{Im} z}|>0,
\end{equation*}
\notag
$$
for $z$ lying in the interior of the segment $BD$. Consider any point $Q$ from $BD$. By assertion 5) of Lemma 4.1, on the segment with end-points $A$ and $Q$, the function $V$ strictly increases from $(-\infty)$ to the value of $V$ at the point $Q$, which is greater than $c_0$. By continuity of $V$, on every such segment there is exactly one point of the set $L(V,c_0)$. Consequently, the intersection of $L(V,c_0)$ with the triangle $ABD$ is a smooth arc which is starlike with respect to the origin. Now assertion 1) of Lemma 4.2 follows by the symmetry of the function $V$ (Lemma 4.1, assertions 2) and 3)).
2) Let $\epsilon<c_0$. As in the proof of assertion 1), we make sure that, because of the monotonicity of $V(re^{i\phi})$ with respect to $r$ for a fixed $\phi$, the intersection of $L(V,\epsilon)$ with the triangle $ABD$ is a smooth arc starlike with respect to the origin. It connects some inner points of the segments $AB$ and $AD$. Now assertion 2) follows by the symmetry of $V$ (see Lemma 4.1). 3) Let $c_0<\epsilon<d_0$. On the segment $BD$, there exists a unique point $Q_\epsilon$ at which $V$ is equal to $\epsilon$. Again, using the monotonicity of $V(re^{i\phi})$ with respect to $r$, we make sure that if $Q$ is an inner point of the segment $BQ_\epsilon$, then the segment $AQ$ does not contain points of the set $L(V,\epsilon)$. If $Q$ is on the segment $BQ_\epsilon$, then on the segment $AQ$ there exists precisely one point of this set. This proves 3), and, therefore, Lemma 4.2. Let $\epsilon<d_0$ and let $\xi$ be a smooth arc which is the intersection of the triangle $ABD$ and the set $L(V,\epsilon)$. By Lemma 4.2, this arc is starlike with respect to the origin, consequently, it can be represented in the form $z=z(\phi)=r(\phi)e^{i\phi}$, $\phi_0\leqslant \phi\leqslant\pi/6$. Here, $\phi_0=0$ if $\epsilon\leqslant c_0$, and $\phi_0>0$ if $c_0<\epsilon< d_0$. Proof. We have $V(z(\phi))=\epsilon$, $\phi_0\leqslant \phi\leqslant\pi/6$. Differentiating this equality with respect to $\phi$, we obtain $V_x \dot{x}+V_y\dot{y}\equiv 0$ (here, the dot above a letter means differentiation with respect to $\phi$), that is, the vector of gradient $\nabla V$ at the point $z=z(\phi)$ is orthogonal to the tangent vector $\dot{z}$. We rewrite the orthogonality condition as
Now consider the function $\ln(\zeta(z)-c\overline{z})$ on the arc $\xi$. At $z=z(\phi)$, we have
$$
\begin{equation*}
\frac{d}{d\phi}\ln(\zeta(z)-c\overline{z}) =-\frac{\mathfrak{P}(z)\dot{z}+c\overline{\dot{z}}}{\zeta(z)-c\overline{z}}.
\end{equation*}
\notag
$$
By (4.4), $\overline{c\dot{z}}/[\zeta(z)-c\overline{z}]$ is a pure imaginary number, and hence
$$
\begin{equation*}
\begin{aligned} \, \frac{d}{d\phi}\ln|\zeta(z)-c\overline{z}\,| &=-\operatorname{Re}\frac{\mathfrak{P}(z)\dot{z}}{\zeta(z)-c\overline{z}} =-\operatorname{Re}\frac{\mathfrak{P}(z)}{(\zeta(z)-c\overline{z})^2}\, \dot{z}(\zeta(z)-c\overline{z}) \\ &=|\dot{z}(\zeta(z)-c\overline{z})|\operatorname{Im} \frac{\mathfrak{P}(z)}{(\zeta(z)-c\overline{z})^2}. \end{aligned}
\end{equation*}
\notag
$$
By assertion 5) of Lemma 3.1, the function $|\nabla V(z(\phi))| =|\zeta(z)-c\overline{z})|$ strictly increases on $[\phi_0,\pi/6]$, whence the lemma. Lemma 4.4. Let $z_1$ and $z_2$ be two points not belonging to the lattice $\Omega$ and non-equivalent modulo $\Omega$. If $ V(z_1)=V(z_2)$, $\nabla V(z_1)=\nabla V(z_2)$, then there exist $r>0$ and $k\in \mathbb{Z}$ such that Proof. Let $\epsilon=V(z_1)=V(z_2)$. Then the points $z_1, z_2\in L(V,\epsilon)$.
Since the difference $z_1-z_2$ in (4.5) is comparable with $re^{i\pi k/6}$ modulo $\Omega$, and since both the function $V$ and its gradient do not change under shifts by elements of $\Omega$, we can assume that $z_1$ and $z_2$ are in the fundamental hexagon $\mathfrak{F}$. Moreover, because of the symmetry of $V$, we can assume that $z_1$ is in the triangle $ABD$. If either $\arg z_1=0$ or $\arg z_1=\pi/6$, then from Lemma 4.3 and since $V$ is symmetric, it follows that $\mathfrak{F}$ contains precisely six points $z$, including $z_1$ itself, at which $|\nabla V(z)|=|\nabla V(z_1)|$; these points are obtained from $z_1$ by a rotation about the origin through multiples of $\pi/3$, and the point $z_2$ is one of them. Since, under the rotation of $z$ by $\pi/3$ about the origin, the gradient $\nabla V(z)$ is also rotated by $\pi/3$, we conclude that, for these points the equality $\nabla V(z)=\nabla V(z_1)$ holds only if $z=z_1$. Therefore, $z_2=z_1$, which is impossible. If $\phi =\arg z_1\neq 0$, $\pi/6$, then, by Lemma 4.3, in $\mathfrak{F}$ there are twelve points which are solutions of the equation $|\nabla V(z)|=|\nabla V(z_1)|$. They are obtained from the points $z_1$ and $\overline{z_1}$ by rotations about the origin through multiples of $\pi/3$. As in the above case, we verify that $z_2$ cannot be obtained from $z_1$ by a rotation about the origin through a multiple of $\pi/3$. Therefore, $z_2=\overline{z_1}e^{i m\pi/3}$, $m\in \mathbb{Z}$. Hence $ z_1-z_2=z_1-\overline{z_1}e^{i m\pi/3} =e^{i m\pi/6}(z_1e^{-i m\pi/6}-\overline{z_1}e^{i m\pi/6}) $, and, consequently, $\arg(z_1-z_2)=m\pi/6\pm\pi/2$, since the number $z_1e^{-im\pi/6}-\overline{z_1}e^{i m\pi/6}=2i\operatorname{Im}(z_1e^{-im\pi/6})$ is purely imaginary. This completes the proof of Lemma 4.4. § 5. Uniformization of the three-sheeted surfaceConsider the three-sheeted Riemann surface $R(a_1,a_2,a_3)$ of function (1.3) over the extended complex plane (the Riemann sphere) $\overline{\mathbb{C}}$ of complex variable $\tau$. Here, $a_1$, $a_2$ and $a_3$ are three pairwise distinct points of the complex plane. This surface is of genus $g=1$. If the points $a_1$, $a_2$ and $a_3$ do not lie on the same straight line, then we draw a circle $\mathcal{O}$ through these points and numerate them such that they follow in the order: $a_1$, $a_3$, $a_2$ as $\mathcal{O}$ is traversed counterclockwise. Consider the Moebius transform $T$ in the $\tau$-plane, which maps the points $a_1$, $a_2$ and $a_3$ to $\widetilde{a}_1=1$, $\widetilde{a}_2=e^{2\pi i/3}$, $\widetilde{a}_3=e^{-2\pi i/3}$, which are the cubic roots of unity; these points lie on the unit circle and are the vertices of a regular triangle inscribed in the unit circle. This transform maps the exterior of $\mathcal{O}$ onto the interior of the unit circle. In the case where $a_1$, $a_2$ and $a_3$ are on the same straight line, we map the half-plane bounded by this line onto the unit disc with the same correspondence of points: $a_j\mapsto \widetilde{a}_j$, $1\leqslant j\leqslant 3$. The surface $R(a_1,a_2,a_3)$ is mapped by $T$ onto the surface $R(\widetilde{a}_1,\widetilde{a}_2,\widetilde{a}_3)$. Note also that $T$ maps infinity to the point $ w_0=(\gamma-e^{-\pi i/3})/(\gamma-e^{\pi i/3})$, which lies in the unit disc; here, $\gamma=(a_3-a_2)/(a_3-a_1)$ is the unharmonic ratio of the points $a_1$, $a_2$, $a_3$ and $\infty$ (see Fig. 6). We also note that if $a_1$, $a_2$ and $a_3$ are on the same straight line, then $w_0$ is on the boundary of the unit disc. Therefore, the problem of uniformization of the surface $R(a_1,a_2,a_3)$ over the extended complex $\tau$-plane is reduced to a similar problem for the surface $R_0:=R(\widetilde{a}_1,\widetilde{a}_2,\widetilde{a}_3)$ over the extended complex $w$-plane, this surface corresponds to the function $\sqrt[3]{w^3-1}$. Let $\widetilde{A}_1$, $\widetilde{A}_2$ and $\widetilde{A}_3$ be the points of the surface $R_0$ lying over $\widetilde{a}_1$, $\widetilde{a}_2$ and $\widetilde{a}_3$ (see Fig. 6). Since the genus of $R_0$ is $1$ (the parabolic case), we can take the finite complex plane $\mathbb{C}$ as a universal covering space of $R_0$. The universal covering $\pi\colon \mathbb{C}\to R_0$ is carried out by a doubly periodic function $\pi$ with periods $\omega_1$, $\omega_2$. We can assume that $\pi(0)=\widetilde{a}_1$; this can be achieved by a shift of the plane. Denote by $p$ the projection of $R_0$ on $\overline{\mathbb{C}}$. The Riemann surface $R_0$ possesses a non-trivial group of (holomorphic) automorphisms $f$ that keep the projections of points, that is, $f\colon R_0\to R_0$ are such that ${p\circ \! f}=p$. In particular, any continuous mappings $f$ permuting its sheets is such an automorphism. Denote by $\mathrm{Aut}(R_0)$ the group of such mappings (we include in elements of $\mathrm{Aut}(R_0)$ the identical mapping, though it does not permute the sheets). Note that the points $\widetilde{A}_1$, $\widetilde{A}_2$ and $\widetilde{A}_3$ are fixed for every $f\in \mathrm{Aut}(R_0)$. Every automorphism $f\in \mathrm{Aut}(R_0)$ has a lift $\widetilde{f}$ on the universal covering space. Thus, there exists $\widetilde{f}\colon\mathbb{C}\to \mathbb{C}$ such that $f\circ\pi=\pi\circ\widetilde{f}$, that is, the diagram is commutative. Since $\pi(0)=\widetilde{a}_1$ and $f(\widetilde{a}_1)=\widetilde{a}_1$, we can assume that $\widetilde{f}(0)=0$. The last equality defines $\widetilde{f}$ in a unique way.The holomorphic automorphism $\widetilde{f}$ of $\mathbb{C}$ is a linear transform, consequently, $\widetilde{f}(z)=bz$, $z\in \mathbb{C}$, for some $b\neq 0$. A simple analysis of the behaviour of automorphisms in neighbourhoods of the fixed points $\widetilde{A}_1$, $\widetilde{A}_2$ and $\widetilde{A}_3$ shows that the subgroup of automorphisms $\widetilde{f}$ of the complex plane corresponding to the group of automorphisms rearranging the sheets of $R_0$ and satisfying $\widetilde{f}(0)=0$ is a cyclic group generated by a rotation about the origin through $2\pi/3$. Now consider the set $\pi^{-1}(\widetilde{A}_1)$. It is easy to show that this is a lattice $\Omega$ on the plane generated by two linear independent (over $\mathbb{R}$) vectors $\omega_1$ and $\omega_2$. From the definition of the lattice it follows that $\omega$ is invariant with respect to rotations about the origin by multiples of $2\pi/3$. Similarly, we show that the lattice is invariant with respect to rotations about every point of the plane corresponding to $\widetilde{A}_1$, $\widetilde{A}_2$ and $\widetilde{A}_3$. This establishes the following result. Lemma 5.1. The lattice $\Omega$ is contained in the set of vertices of some triangulation of the plane by regular triangles. In addition, if the lattice is invariant with respect to the rotation by $2\pi/3$ about some point, then the point is either a vertex or the centre of some triangle of the triangulation. Since the mapping $\pi$ is defined up to a linear automorphism of the plane, we can assume that the generators of the lattice $\Omega$ are the vectors
$$
\begin{equation}
\omega_1=\sqrt{3},\qquad \omega_2=\sqrt{3}\, e^{\pi i/3}.
\end{equation}
\tag{5.1}
$$
The point $\widetilde{A}_1$ corresponds to points of the lattice $\Omega$ under the mapping $\pi$. From Lemma 5.1 it follows that the centre of the triangle $T_1$ with vertices $0$, $\sqrt{3}$, $\sqrt{3}\, e^{\pi i/3}$ and points, equivalent to it, corresponds, under the mapping $\pi$, to one of the points $\widetilde{A}_2$, $\widetilde{A}_3$; in addition, the centre of the triangle $T_2$ with vertices $\sqrt{3}$, $\sqrt{3}\, e^{\pi i/3}$, $\sqrt{3}(1-e^{\pi i/3})$ and points equivalent to it corresponds to another one. Let, for definiteness, the centre of $T_1$ correspond to $\widetilde{A}_2$, and the centre of $T_2$, to $\widetilde{A}_3$. The triangles $T_1$ and $T_2$ form a parallelogram constructed by the vectors $\omega_1-\omega_2$ and $\omega_2$, it is a parallelogram of periods for the function $\pi$ that uniformizes $R_0$. Denote by $B$ the centre of $T_1$, and by $C$, the centre of $T_2$. Taking into account the doubly periodicity of the uniformizing function $\pi$, we will denote by $B_j$ and $C_k$ (with integer $j$ and $k$) the points equivalent to $B$ and $C$ modulo $\Omega$ (see Fig. 7). Now we will describe the universal covering in an explicit form. Consider the function conformally mapping the triangle $\Delta=ABC$ with vertices at the points $0$, $e^{-i\pi/6}$, $e^{i\pi/6}$ onto the unit disc with the correspondence of points:
$$
\begin{equation*}
0\mapsto 1, \qquad e^{-i\pi/6}\mapsto e^{i2\pi/3}, \qquad e^{i\pi/6}\mapsto e^{-i2\pi/3}.
\end{equation*}
\notag
$$
Continuing the function by the symmetry principle to the whole plane, we obtain a mapping, which covers $R_0$ by the complex plane. We note that the centre $z_0=\sqrt{3}/3$ of the triangle $\Delta$ is mapped to the origin, any point equivalent to $z_0$ modulo the lattice is also mapped to the origin. The point $(-z_0)$ is obtained from $z_0$ by applying an odd number of reflections with respect to sides of triangles of the triangulation, therefore, $(-z_0)$ and all points equivalent to it are mapped to infinity. Using Theorem 2.2, we can write the superposition of the universal covering $\pi$ and the projection $p$ via its zeros and poles:
$$
\begin{equation*}
p\circ \pi(z)=-\frac{\sigma(z-z_0)\sigma(z-e^{2\pi i/3}z_0)\sigma(z-e^{-2\pi i/3}z_0)} {\sigma(z+z_0)\sigma(z+e^{2\pi i/3}z_0)\sigma(z+e^{-2\pi i/3}z_0)}.
\end{equation*}
\notag
$$
Here, $\sigma(z)$ is the Weierstrass sigma function with periods $\omega_1=\sqrt{3}$ and $\omega_2=\sqrt{3}\, e^{i\pi/3}$; here, we used the fact that $\pi(0)=1$. The inverse can be expressed via the Schwarz–Christoffel integral
$$
\begin{equation}
z=\pi^{-1}(w)=C_1\int_0^w\frac{dw}{(w^3-1)^{2/3}}+z_0,\qquad C_1=-\frac{2\pi}{(\Gamma(1/3))^3}\,=-0.326807\dots,
\end{equation}
\tag{5.2}
$$
where $\Gamma$ is the Euler gamma function; here, we take into account that the integrand in (5.2) is a multi-valued function and its different branches correspond to different sheets of the surface $R_0$.
§ 6. Description of the Abelian integralNow we will discuss how to construct the Abelian integral $G$ on the Riemann surface $R(a_1,a_2,a_3)$ with the desired asymptotics (1.2) for $m=2$. For this, it is sufficient to describe the Abelian integral $G_0=G\circ T^{-1}$ on the Riemann surface $R_0$. Here, $T$ denotes the homeomorphism of the three-sheeted Riemann surfaces induced by the Moebius transform $T$; for simplicity of notation, we will denote it also by $T$. Note that the Moebius transform $T$ maps infinite points on $R(a_1,a_2,a_3)$ to finite points on $R_0$ lying over $w_0$. In a neighbourhood of $w_0$, we have where $a$ and $b\neq 0$ are some constants. Therefore, $G_0$ in a neighbourhood of the point lying over $w_0$ on the zero sheet has the asymptotics In neighbourhoods of the other two points lying over $w_0$, we haveThe Abelian integral $G_0$ induces the Abelian integral $\widetilde{G}:=G_0\circ\pi$ on the universal covering space (the complex plane) of $R_0$. In the triangle $ABC$ (see Fig. 7) there is exactly one point $\alpha$ that is mapped to $w_0$ under the function $p\circ\pi$. Hence the points $e^{2\pi i/3}\alpha$ and $e^{-2\pi i/3}\alpha$, which are obtained by a rotation of $\alpha$ about the origin by multiples of $2\pi/3$, corresponds to the other two points of $R_0$ which lie over $w_0$. Remark 6.1. Note that, under cyclic permutations of the points $a_1$, $a_2$ and $a_3$, the points $\widetilde{a}_1$, $\widetilde{a}_2$ and $\widetilde{a}_3$ are also permuted cyclically. At the same time, the point $w_0$ is rotated about the origin by multiples of $2\pi/3$ (see Fig. 6). Because of the symmetry of function (5.2), the point $\alpha$ on the universal covering space is rotated about the centre of the triangle $ABC$ by multiples of $2\pi/3$. Therefore, changing the numbering, if necessary, we can achieve that the point $\alpha$ is in one of the triangles into which the segments connecting the centre of the triangle $ABC$ with its vertices separate this triangle. If we consider, instead of $a_1$, $a_2$ and $a_3$, their complex conjugate points $\overline{a}_1$, $\overline{a}_2$ and $\overline{a}_3$, then, instead of $w_0$, we obtain $\overline{w}_0$, and $\alpha$ will change to $\overline{\alpha}$. A decomposition for the points $\overline{a}_1$, $\overline{a}_2$ and $\overline{a}_3$ will be obtained from the similar structure for the points $a_1$, $a_2$ and $a_3$ by the mirror symmetry. Therefore, we can assume that $\alpha$ is in one of the six triangles into which the triangle $ABC$ is separated by its bisectors. As a rule, we will use the triangle which contains the vertex $A$ and lies above the real axis; we will denote it by $\mathfrak{S}$. We also note that sometimes it is convenient to use other triangles. For example, we can study the problem for real $\alpha\in (\sqrt{3}/3,\sqrt{3}/2)$, rather than $\alpha$ from the intervals obtained from this one by a rotation about the centre of $ABC$ through multiples of $2\pi/3$. Using (6.1) and (6.2) and since the projection $\pi$ is conformal, we see that
$$
\begin{equation}
\widetilde{G}(z)\sim -2\ln (z-\alpha), \qquad z\to\alpha.
\end{equation}
\tag{6.3}
$$
In neighbourhoods of the other two points lying over $w_0$, we have
$$
\begin{equation}
\widetilde{G}(z)\sim \ln (z-e^{\pm 2\pi i/3}\alpha), \qquad z\to e^{\pm 2\pi i/3}\alpha.
\end{equation}
\tag{6.4}
$$
Consider the function
$$
\begin{equation}
\Phi(z):=-2\ln\sigma(z-\alpha)+\ln\sigma(z-e^{2\pi i/3}\alpha) +\ln\sigma(z-e^{-2\pi i/3}\alpha),
\end{equation}
\tag{6.5}
$$
where $\sigma(z)$ is the Weierstrass sigma function with periods $\omega_1=\sqrt{3}$ and $\omega_2=\sqrt{3}\,e^{i\pi/3}$. In neighbourhoods of the points $\alpha$, $e^{2\pi i/3}\alpha$ and $e^{-2\pi i/3}\alpha$, this function behaves as
$$
\begin{equation*}
\Phi(z)\sim \begin{cases} -2\ln (z-\alpha), &z\to \alpha, \\ \ln (z-e^{\pm2\pi i/3}\alpha), &z\to e^{\pm2\pi i/3}\alpha. \end{cases}
\end{equation*}
\notag
$$
Using (2.8), we obtain
$$
\begin{equation}
\ln\sigma(z+\omega_k)-\ln\sigma(z)=\eta_k(z+\omega_k/2)\pm \pi i,\qquad k=1,2,
\end{equation}
\tag{6.6}
$$
where $\eta_k=2\zeta(\omega_k/2)$ and the sign of the last expression is defined by the choice of the branch of logarithm. From (6.5) and (6.6), we have $\Phi(z+\omega_k)-\Phi(z)=3\alpha\eta_k \ (\operatorname{mod} 2\pi i)$, $k=1$, $2$, since
$$
\begin{equation*}
\begin{aligned} \, &\Phi(z+\omega_k)-\Phi(z) =-2\bigl(\ln\sigma(z-\alpha+\omega_k)-\ln\sigma(z-\alpha)\bigr) +\bigl(\ln\sigma(z-e^{2\pi i/3}\alpha+\omega_k) \\ &\qquad\qquad-\ln\sigma(z-e^{2\pi i/3}\alpha)\bigr) +\bigl(\ln\sigma(z-e^{-2\pi i/3}\alpha+\omega_k)-\ln\sigma(z-e^{-2\pi i/3}\alpha)\bigr) \\ &\qquad=-2\eta_k\biggl(z-\alpha+\frac{\omega_k}2\biggr) + \eta_k\biggl(z-\alpha e^{2\pi i/3} +\frac{\omega_k}2\biggr) +\eta_k\biggl(z-\alpha e^{-2\pi i/3}+\frac{\omega_k}2\biggr) \\ &\qquad=\eta_k\alpha(2-e^{2\pi i/3}-e^{-2\pi i/3})=3\alpha\eta_k \pmod{2\pi i}. \end{aligned}
\end{equation*}
\notag
$$
Next, using the known values of periods (5.1) and $\eta_k$ (see (3.4)), we conclude that the real part
$$
\begin{equation*}
\widetilde{g}(z):=-2\ln|\sigma(z-\alpha)| +|{\ln\sigma(z-e^{2\pi i/3}\alpha)}| +\ln\sigma|(z-e^{-2\pi i/3}\alpha)|-\sqrt{3}\, \eta_1\operatorname{Re}(\overline{\alpha}z),
\end{equation*}
\notag
$$
of the function $\widetilde{G}(z):=\Phi(z)-\sqrt{3}\, \eta_1\overline{\alpha}z$ is a doubly periodic harmonic function with periods $\omega_1$ and $\omega_2$. Therefore, we constructed the required Abelian integral $\widetilde{G}(z)$. Let us formulate this result as a theorem. Theorem 6.1. The Nuttall Abelian integral $G$ has the form $G=\widetilde{G}\circ \pi^{-1}\circ T$, where If a point $z\in\mathbb{C}$ and $w=\pi(z)$, then the preimage $\pi^{-1}(w)$ consists of the points $z$, $e^{2\pi i/3}z$, $e^{-2\pi i/3}z$ and of the points equivalent to them modulo the lattice $\Omega$. This implies that the zero sheet of $R_0$ corresponds to the set $S_0$ of points $z$ on the universal covering space such that
$$
\begin{equation*}
\widetilde{g}(z)>\widetilde{g}(ze^{2\pi i/3}) \quad \text{and} \quad \widetilde{g}(z)>\widetilde{g}(ze^{-2\pi i/3}).
\end{equation*}
\notag
$$
For the set $S_2$ corresponding to the second sheet, we have
$$
\begin{equation*}
\widetilde{g}(z)<\widetilde{g}(ze^{2\pi i/3}) \quad \text{and} \quad \widetilde{g}(z)<\widetilde{g}(ze^{-2\pi i/3}).
\end{equation*}
\notag
$$
Further, for the set $S_1$ corresponding to the first sheet, we have
$$
\begin{equation*}
\widetilde{g}(ze^{-2\pi i/3})<\widetilde{g}(z) <\widetilde{g}(ze^{2\pi i/3}) \quad \text{or} \quad \widetilde{g}(ze^{2\pi i/3})<\widetilde{g}(z)<\widetilde{g}(ze^{-2\pi i/3}).
\end{equation*}
\notag
$$
Let $0\leqslant j$, $k\leqslant2$, $j\neq k$. Denote
$$
\begin{equation*}
L_{jk}:=\{z\colon \widetilde{g}(ze^{2\pi j i/3})>\widetilde{g}(ze^{2\pi k i/3})\},\qquad \Gamma_{jk}=\{z\colon \widetilde{g}(ze^{2\pi j i/3})=\widetilde{g}(ze^{2\pi k i/3})\}.
\end{equation*}
\notag
$$
Since the group of rotations about the origin by multiples of $2\pi k /3$ is isomorphic to $\mathbb{Z}/3\mathbb{Z}$, we can define $L_{jk}$ and $\Gamma_{jk}$ for arbitrary integers $j$ and $k$ not comparable modulo $3$; for this, it is sufficient to take, instead of $j$ and $k$, their remainders of the division by $3$. In this notation,
$$
\begin{equation}
S_0=L_{01}\cap L_{02},\qquad S_2=L_{10}\cap L_{20},\qquad S_1=(L_{01}\cap L_{20})\cup (L_{02}\cap L_{10}).
\end{equation}
\tag{6.7}
$$
Note that, for every $j\neq k$, the three sets $S_{jk}$, $S_{kj}$ and $\Gamma_{jk}=\Gamma_{kj}$ are disjoint and their union is the whole complex plane. Remark 6.2. By (6.7), the boundaries of the “sheets” $S_j$ lie in the sets $\Gamma_{01}$ and $\Gamma_{02}$, that is, the arcs of $\Gamma_{12}$ are not in $\partial S_j$, $0\leqslant j\leqslant 2$. It is evident that, under the rotation by the $2\pi/3$, the sets $L_{jk}$ and $\Gamma_{jk}$ are mapped to the sets $L_{j-1,k-1}$ and $\Gamma_{j-1,k-1}$, respectively. Hence, for every $n\in \mathbb{Z}$, Using (6.8), we can write (6.7), for example, in the form
$$
\begin{equation}
\begin{gathered} \, S_0=L_{01}\cap e^{2\pi i/3} L_{10},\qquad S_2=L_{10}\cap e^{2\pi i/3}L_{01}, \\ S_1=(L_{01}e^{2\pi i/3}\cap L_{01})\cup (e^{2\pi i/3}L_{10}\cap L_{10}) \end{gathered}
\end{equation}
\tag{6.9}
$$
or
$$
\begin{equation}
\begin{gathered} \, S_0=e^{2\pi i/3} L_{12}\cap e^{-2\pi i/3} L_{21},\qquad S_2=e^{2\pi i/3} L_{21}\cap e^{-2\pi i/3}L_{12}, \\ S_1=e^{2\pi i/3} (L_{12}\cap e^{-2\pi i/3} L_{12}) \cup (e^{2\pi i/3}L_{21}\cap e^{-2\pi i/3} L_{21}). \end{gathered}
\end{equation}
\tag{6.10}
$$
Now we will study the sets $L_{jk}$. Because of the invariance of the lattice with respect to rotations about the origin by multiples of $2\pi/3$ and the homogeneity of $\sigma(z)=\sigma(z;\omega_1,\omega_2)$ qua function of $z$ and periods $\omega_1$ and $\omega_2$, we have $\ln|\sigma(ze^{\pm 2\pi i/3})|=\ln|\sigma(z)|$. Consequently,
$$
\begin{equation*}
\begin{aligned} \, \widetilde{g}(ze^{2\pi i/3}) &=-2\ln|\sigma(ze^{2\pi i/3}-\alpha)| +\ln|\sigma(ze^{2\pi i/3}-e^{2\pi i/3}\alpha)| \\ &\qquad+\ln|\sigma(ze^{2\pi i/3}-e^{4\pi i/3}\alpha)| -\sqrt{3}\, \eta_1\operatorname{Re}(\overline{\alpha}e^{2\pi i/3}z) \\ &=-2\ln|\sigma(z-e^{-2\pi i/3}\alpha)| +\ln|\sigma(z-\alpha)| +\ln|\sigma(z-e^{2\pi i/3}\alpha)| \\ &\qquad -\sqrt{3}\, \eta_1\operatorname{Re}(\overline{\alpha}e^{2\pi i/3}z). \end{aligned}
\end{equation*}
\notag
$$
Therefore, the inequality $\widetilde{g}(z)>\widetilde{g}(ze^{2\pi i/3})$ is equivalent to
$$
\begin{equation}
\ln|\sigma(z-\alpha)|-\ln|\sigma(z-e^{-2\pi i/3}\alpha)| <\frac{\sqrt{3}}3 \, \eta_1\operatorname{Re}\bigl(\overline{\alpha}(e^{2\pi i/3}-1)z\bigr).
\end{equation}
\tag{6.11}
$$
Similarly, the inequality $\widetilde{g}(ze^{-2\pi i/3})>\widetilde{g}(z)$ is equivalent to
$$
\begin{equation}
\ln|\sigma(z-\alpha)|-\ln|\sigma(z-e^{2\pi i/3}\alpha)| <\frac{\sqrt{3}}3 \, \eta_1\operatorname{Re}\bigl(\overline{\alpha}(e^{-2\pi i/3}-1)z\bigr),
\end{equation}
\tag{6.12}
$$
an the inequality $\widetilde{g}(ze^{2\pi i/3})>\widetilde{g}(ze^{-2\pi i/3})$ is equivalent to
$$
\begin{equation}
\ln|\sigma(z-e^{-2\pi i/3}\alpha)|-\ln|\sigma(z-e^{2\pi i/3}\alpha)| <\frac{\sqrt{3}}3 \, \eta_1\operatorname{Re} \bigl(\overline{\alpha}(e^{-2\pi i/3}-e^{2\pi i/3})z\bigr).
\end{equation}
\tag{6.13}
$$
Consider the function
$$
\begin{equation}
u(z):=\ln|\sigma(z-e^{-2\pi i/3}\alpha)| -\ln|\sigma(z-e^{2\pi i/3}\alpha)|-\eta_1\operatorname{Im} (\overline{\alpha} z).
\end{equation}
\tag{6.14}
$$
We note that we can express $u(z)$ in terms of the theta-function $\theta_{11}(z)=\theta_{11}(z;\tau)$, where $\tau=\omega_2/\omega_1=e^{\pi i/3}$:
$$
\begin{equation*}
u(z)=\ln|\theta_{11}(z-e^{-2\pi i/3}\alpha)|-\ln|\theta_{11}(z-e^{2\pi i/3}\alpha)|.
\end{equation*}
\notag
$$
With the use of $u(z)$, inequalities (6.11)–(6.13) can be written in the form $u(e^{-2\pi i/3}z)<0$, $u(e^{2\pi i/3}z)<0$ and $u(z)<0$. Now we will formulate the result thus obtained as a theorem. Theorem 6.2. Let the function $u$ be defined by equality (6.14), and let $L_{12}=\{z\in \mathbb{C} \colon u(z)<0\}$, $L_{21}=\{z\in \mathbb{C} \colon u(z)>0\}$, $L_{01}=e^{2\pi i/3}L_{12}$, $L_{10}=e^{2\pi i/3}L_{21}$, $L_{20}=e^{-2\pi i/3}L_{12}$, $L_{02}=e^{-2\pi i/3}L_{21}$. Then the sets $S_j$, $0\leqslant j\leqslant 2$, on the universal covering space corresponding to the Nuttall sheets has the form (6.7). Further, for short, we will call the sets $S_j$ on the universal covering space also sheets. Remark 6.3. Since the sets $L_{jk}$ can be obtained from each other by rotations about the origin through multiples of $2\pi/3$ and by taking the complement, it suffices, for their description, to study one of them, namely, $L_{12}=\{z\colon u(z)<0\}$. Hence the sets $S_{j}$ can be found with the help of (6.10). The common boundary of the sets $L_{12}$ and $L_{21}$ is the set $\Gamma_{12}=L(u,0):=\{z\colon u(z)=0\}$, therefore, the main problem is the description of this set. § 7. Null-set of the function $u$ and its critical points. The general caseConsider the $0$-set $L(u,0)$ of the function $u(z)$ defined by (6.14). The structure of the $0$-set of every harmonic function depends essentially on whether it contains zeros of the gradient $\nabla u$. Lemma 7.1. The equation $\nabla u=0$ has exactly two (taking account of multiplicity) solutions in the fundamental hexagon $\mathfrak{F}$. If $\alpha$ does not coincide with the centre $\sqrt{3}/3$ of the triangle $ABC$, then two of these solutions are distinct, otherwise, there is one multiple zero of $\nabla u$ also located at the point $\sqrt{3}/3$. Proof. The critical points of $u$, that is, the points at which its gradient vanishes, satisfy
By Theorem 2.1, the function $\zeta(z-e^{-2\pi i/3}\alpha)-\zeta(z-e^{2\pi i/3}\alpha)$ is doubly periodic and has two simple poles in $\mathfrak{F}$. Therefore, it is two-valent and assumes the value $(-\eta_1 \overline{\alpha} i)$ exactly at two points (taking into account multiplicities). Consequently, equation (7.1) has two solutions in $\mathfrak{F}$, that is, there are two critical points of $g$ in $\mathfrak{F}$ (taking account of multiplicities). We will investigate where $\nabla u$ has a multiple zero. We set $t=z+\alpha/2$ and $a=i\alpha\sqrt{3}/2$. Now we can write (7.1) as
$$
\begin{equation}
\zeta(t+a)-\zeta(t-a)-2c\overline{a}=0, \qquad c=\frac{\eta_1}{\omega_1}.
\end{equation}
\tag{7.2}
$$
If a solution $t$ of (7.2) is multiple, then the derivative of the function in its left-hand side vanishes at this point, that is, $\mathfrak{P}(t+a)-\mathfrak{P}(t-a)=0$. Taking into account the fact that $\mathfrak{P}$ is doubly periodic and even, and since, in every fundamental domain, it takes every value two times (with account of multiplicities), we conclude that $t+a\equiv \pm(t-a) \ (\operatorname{mod} \Omega)$. Consequently, either $2a$ or $2t$ belongs to $\Omega$.
By Remark 6.1, we can assume that $\alpha$ is in the triangle $\mathfrak{S}$. Hence $2a=i\alpha\sqrt{3}$ satisfies $0<\operatorname{Im} (2a)\leqslant 1$, $\pi/2\leqslant\arg(2a)\leqslant 2\pi/3$, therefore, this point cannot coincide with any of points from $\Omega$. Now consider the case $2t\in\Omega$. By periodicity, it is sufficient to consider the cases $t=0$, $t=\omega_1/2$, $t=\omega_2/2$ and $t=(\omega_1-\omega_2)/2$. If $t=0$, then from (7.2) it follows that $\zeta(a)-c\overline{a}=0$, that is, $\overline{a}=(1/c)\zeta(a)$. The point $a$ belongs to the interior of $\mathfrak{F}$. From Theorem 3.3 it follows that the point $-(1/c)\zeta(a)$ belongs to the exterior of $\mathfrak{F}$. Therefore, the case $t=0$ is impossible. If $t=\omega_1/2$, then by (2.4) we obtain From (7.2) and (7.3) we conclude that $\zeta(t+a)=c(t+\overline{a})$. By Lemma 3.2, the point $t+a$ coincides with a vertex of the fundamental hexagon. It is easy to see that, under the condition that $\alpha$ is in the triangle $ABC$, this is possible only if $a=i/2$, and then $\alpha=\sqrt{3}/3$.The cases $t=\omega_2/2$ and $t=(\omega_1+\omega_2)/2$ are considered in a similar way. This proves the lemma. Now we will study the problem when the set of zeros of $\nabla u$ has a non-empty intersection with $L(u,0)$. The following result holds. Lemma 7.2. Let $\nabla u=0$ at some point $z\in L(u,0)$. Then $\alpha$ is either on the boundary of $ABC$ or on one of its bisectors. Proof. Let $u(z)=0$ and $\nabla u(z)=0$. Denote
$$
\begin{equation}
z_1=z-e^{-2\pi i/3}\alpha,\qquad z_2=z-e^{2\pi i/3}\alpha.
\end{equation}
\tag{7.4}
$$
Then we can write the equality $u(z)=0$ as where the function $V$ is defined by (4.1).
Vanishing of the gradient of $u$ can be written in the form
$$
\begin{equation*}
\zeta(z-e^{-2\pi i/3}\alpha)-\zeta(z-e^{2\pi i/3}\alpha)+\eta_1 \overline{\alpha} i=0
\end{equation*}
\notag
$$
or $ \zeta(z_1)-c \overline{z}_1=\zeta(z_2)-c\overline{z}_2$, which is equivalent to From (7.5) and (7.6), by Lemma 4.4, for some $r>0$ and $k\in \mathbb{Z}$, we have $z_1-z_2\equiv re^{i\pi k/6} \ (\operatorname{mod} \Omega)$. Now by (7.4), we have, for some integer $m$ and $n$,
$$
\begin{equation}
\sqrt{3}\, i\alpha = re^{i\pi k/6}+m\omega_1+in\omega_2.
\end{equation}
\tag{7.7}
$$
Now we consider the triangulation of the plane by regular triangles with vertices at points of the lattice. Drawing bisectors in each triangle, we obtain a new triangulation $TR_1$, whose vertices are points of the lattice and centres of triangles of the first triangulation. Let $TR$ be the union of sides of triangles from the new triangulation. The set of points of the form $re^{i\pi k/6}+m\omega_1+in\omega_2$ is obtained from the union of rays emanating from the origin at angles which are multiples of $\pi /6$ with shifts by elements of the lattice $\Omega$. It is easy to see that this set coincides with $TR$. From (7.7) we conclude that the point $\sqrt{3}i\alpha$ lies in the regular triangle with vertices at the points $0$, $\omega_2$ and $\omega_2-\omega_1$, which is obtained from $ABC$ by the transform $z\mapsto \sqrt{3}\, i z$. Therefore, the point $\sqrt{3}\,i\alpha$ belongs to $TR$, that is, it lies either on the boundary of triangle or on some of its bisectors. This proves the lemma. Now we consider in detail the cases where $\alpha$ lies either on one of the bisectors or on the boundary of $ABC$. In Theorems 8.1 and 9.1 that follows, we will completely describe when in such cases the set of zeros of the gradient of $u$ intersects the null set of $u$. In particular, from these theorems it will follow that if the intersection is non-empty, then all zeros of the gradient lie on the set $L(u,0)$. § 8. The case of real $\alpha$Let $\alpha$ be on a bisector of $ABC$; without loss of generality we can assume that the bisector goes from the vertex $A$. Therefore, $\alpha$ is a real number, $\alpha\in (0,\sqrt{3}/2)$; this fact means that, in the initial configuration, three points $a_1$, $a_2$ and $a_3$ in the $\tau$-plane form an isosceles triangle: $|a_1-a_2|=|a_1-a_3|$. As noted above, by (6.8), for description of the sets $L_{jk}$, it is only sufficient to study the geometric structure of the set $L_{12}$, since other sets $L_{jk}$ are obtained from it by rotations through multiples of $2\pi /3$ and the complement operation (with deletion of the boundary). For real $\alpha$, the set $\Gamma_{12}$, which is the boundary of $L_{12}$, is symmetric with respect to the real axis, contains this axis, therefore, it is easier to describe this set. So, we will begin with description of the set $\Gamma_{12}=\{u(z)=0\}=L(u,0)$. For real $\alpha$, the equation $u(z)=0$ is equivalent to the equation
$$
\begin{equation}
\ln|\sigma(z-e^{-2\pi i/3}\alpha)| -\ln|\sigma(z-e^{2\pi i/3}\alpha)|-\eta_1\alpha\operatorname{Im} z=0,
\end{equation}
\tag{8.1}
$$
whose solution set contains the real axis. We will investigate, for which $\alpha$ the critical points of $u$ are located on the real axis. If $z=x$ is real, then, because of the symmetry of the function $\zeta(z)$ (see (3.2)), (7.1) is equivalent to
$$
\begin{equation}
\varphi(x):=\operatorname{Im} \zeta(x-e^{2\pi i/3}\alpha)-\frac{\eta_1}2\,\alpha=0.
\end{equation}
\tag{8.2}
$$
Lemma 8.1. The function $\varphi$ is periodic on the real axis with period $\sqrt{3}$. This function takes its maximal (positive) value at the points $x=-\alpha/2+n\sqrt{3}$, $n\in \mathbb{Z}$, and the minimal value at the points $x=-\alpha/2+\sqrt{3}/2+n\sqrt{3}$, $n\in \mathbb{Z}$. If $\alpha\in(0,\sqrt{3}/3)$, then the minimal value of $\varphi$ is positive, and if $\alpha\in(\sqrt{3}/3,\sqrt{3}/2]$, then it is negative. If $\alpha=\sqrt{3}/3$, then the minimal value is zero. Proof. Because of (2.4), the function $\varphi$ is periodic on the real axis with period $\sqrt{3}$. Since $\varphi'(x)=-\operatorname{Im} \mathfrak{P}(x-e^{2\pi i/3}\alpha)$, we see that $\varphi'(x)=0$ at the points where the function $\mathfrak{P}(x-e^{2\pi i/3}\alpha)=\mathfrak{P}(x+\alpha/2-i\alpha \sqrt{3}/2)$ takes real values. Therefore, we have either $x+\alpha/2\equiv0$ or $x+\alpha/2\equiv\sqrt{3}/2 \ (\operatorname{mod} \Omega)$.
Consider the function $\varphi$ on the segment $\Sigma:=[-\alpha/2,-\alpha/2+\sqrt{3}]$ of length $\omega_1$. It is easy to see that at the end-points of the segment $\Sigma$ the function $\varphi$ takes positive values, since
$$
\begin{equation*}
\varphi\biggl(-\frac{\alpha}2\biggr) =\operatorname{Im} \zeta\biggl(-\frac{i\alpha\sqrt{3}}2\biggr) -\frac{\eta_1}2\alpha>\operatorname{Im} \zeta(-i)-\frac{\eta_1}2\,\alpha =\frac{\eta_1}2\biggl(\frac2{\sqrt{3}}-\alpha\biggr)>0.
\end{equation*}
\notag
$$
Here, we used the fact that, by Theorem 3.3, the function $\zeta(z)$ maps the segment of the imaginary axis with end-points $0$ and $(-i)$ to the ray emanating from the point $\zeta(-i)=i\eta_1/\sqrt{3}$ upward. The function $\varphi'(x)=-\operatorname{Im} \mathfrak{P}(x-e^{2\pi i/3}\alpha)$ changes sign from “$+$” to “$-$” at the point $x=-\alpha/2+\sqrt{3}/2$, therefore, at the end-points of $\Sigma$, the function $\varphi$ takes maximal values, and $x=-\alpha/2+\sqrt{3}/2$ is its point of minimum. The value of $\varphi$ at this point is
$$
\begin{equation*}
\psi(\alpha):=\varphi\biggl(-\frac{\alpha}2+\frac{\sqrt{3}}2\biggr) =\operatorname{Im} \zeta\biggl(\frac{\sqrt{3}}2\,(1-i\alpha)\biggr)-\frac{\eta_1}2\,\alpha.
\end{equation*}
\notag
$$
An analysis of the behaviour of the function $\mathfrak{P}$ on the segment with end-points $\sqrt{3}/2$ and $\sqrt{3}/2-i3/2$ (Theorem 3.2) shows that the derivative
$$
\begin{equation*}
\psi'(\alpha)=\frac{\sqrt{3}}2\operatorname{Re}\mathfrak{P}\biggl(\frac{\sqrt{3}}2\, (1-i\alpha)\biggr) -\frac{\eta_1}2
\end{equation*}
\notag
$$
is strictly decreasing on $[0,\sqrt{3}]$ from to $\psi'(\sqrt{3})=-\infty$. Consequently, on the segment $[0,\sqrt{3}]$, the function $\psi$ first strictly increases and then strictly decreases. Since $\psi(0)=\operatorname{Im}\zeta(\sqrt{3}/2)=0$, we conclude that on the first part of the segment it is positive and on the second one it has at most one zero. Since we conclude that on the segment $[0,\sqrt{3}/2]$ the function $\psi$ has a unique zero at $\sqrt{3}/3$. This proves the lemma. Theorem 8.1. Let $\alpha$ be real, $0<\alpha\leqslant \sqrt{3}/2$. Then the set $\Gamma_{12}=L(u,0)$ contains the real axis. Moreover, the following is valid. 1) If $0<\alpha<\sqrt{3}/3$, then there are no critical points of the function $u(z)$ on the real axis. This case corresponds to the case of isosceles triangle $\Delta(a_1,a_2,a_3)$ with apex angle less than $\pi/3$. 2) If $\sqrt{3}/3<\alpha<\sqrt{3}/2$, then there is exactly two critical points of $u(z)$ on every segment on the real axis of length $\omega_1= \sqrt{3}$. Moreover, its gradient has one zero $\xi_1=\xi_1(\alpha)$ on $(0,\alpha)$ and one zero $\xi_2=\xi_2(\alpha)$ on $(\alpha,\sqrt{3}/2)$. Besides, $\xi_1+\xi_2=\sqrt{3}-\alpha$. This case corresponds to the case of isosceles triangle $\Delta(a_1,a_2,a_3)$ with apex angle greater than $\pi/3$. 3) At last, if $\alpha=\sqrt{3}/3$, then critical points of $u(z)$ are real and double. The segment $[0,\sqrt{3}]$ contains precisely one (double) critical point $\xi=\sqrt{3}/3$. This is the case of the regular triangle $\Delta(a_1,a_2,a_3)$. Proof. By Lemma 8.1, it is sufficient to establish that in the second case $\xi_1(\alpha)\in(0,\alpha)$ and $\xi_2(\alpha)\in(\alpha,\sqrt{3}/2)$. For this, we will show that the function $\varphi$, as defined by (8.2), is such that $\varphi(\alpha)<0$, $\varphi(0)>0$, $\varphi(\sqrt{3/2})>0$.
The value of $\varphi(x)=\operatorname{Im} \zeta(x-\alpha e^{2\pi i/3})-\eta_1x/2$ at the point $\alpha$ is
$$
\begin{equation*}
\varphi(\alpha)=\operatorname{Im} \zeta(\alpha-\alpha e^{2\pi i/3}) -\frac{\eta_1\alpha}2=\operatorname{Im} \zeta (\beta e^{-\pi i/6}) -\frac{c\beta}2=\operatorname{Im}(\zeta(z)-c\overline{z}),
\end{equation*}
\notag
$$
where $\beta=\sqrt{3}\, \alpha \in [1,1.5]$, $z=\beta e^{-\pi i/6}$. Because of periodicity of the function $\zeta(z)-c\overline{z}$, we have $\varphi(\alpha)= \operatorname{Im}(\zeta(\widetilde{z})-c\overline{\widetilde{z}})$, where $\widetilde{z}=z-\omega_1$. The point $\widetilde{z}$ is on the boundary of the fundamental hexagon $\mathfrak{F}$, and, for $\alpha=\sqrt{3}/3$ and $\alpha=\sqrt{3}/2$, this point coincides, respectively, with a vertex and the mid-point of a side of the hexagon. By Theorem 3.3, the function $\varphi$ vanishes at these points. By Lemma 3.3, if $\alpha$ is an inner point of $[\sqrt{3}/3,\sqrt{3}/2]$, then $\widetilde{z}$ is on the boundary arc $l_8$ of the fundamental hexagon, consequently, by this lemma, $\varphi(\alpha)<0$.
Now let us evaluate the sign of $\varphi(0)=-\operatorname{Im} \zeta(\alpha e^{2\pi i/3})-\eta_1\alpha/2$. By Theorem 3.1, the function $\zeta$ maps points of the interval $z=te^{2\pi i/3}$, $0<t<\sqrt{3}/2$, to points of the ray $z=te^{-2\pi i/3}$, $t>\eta_1\sqrt{3}/2$, therefore,
$$
\begin{equation*}
\varphi(0)=|{\operatorname{Im} \zeta(\alpha e^{2\pi i/3})}| -\frac{\eta_1\alpha}2> \biggl|\operatorname{Im} \zeta\biggl(\frac{\sqrt{3}}2 \, e^{2\pi i/3}\biggr)\biggr| -\frac{\eta_1\sqrt{3}}4=0.
\end{equation*}
\notag
$$
Now we consider
$$
\begin{equation*}
\varphi\biggl(\frac{\sqrt{3}}2\biggr)=\operatorname{Im} \zeta\biggl(\frac{\sqrt{3}}2 -\alpha e^{2\pi i/3}\biggr)-\frac{\eta_1\alpha}2 =-\operatorname{Im} \zeta\biggl(\frac{\sqrt{3}}2+\alpha e^{2\pi i/3}\biggr) -\frac{\eta_1\alpha}2.
\end{equation*}
\notag
$$
Let $\psi_2(\alpha):=-\operatorname{Im} \zeta(\sqrt{3}/2+\alpha e^{2\pi i/3})-\eta_1\alpha/2$, $\alpha\in[\sqrt{3}/3,\sqrt{3}/2]$. Then
$$
\begin{equation*}
\begin{aligned} \, \psi'_2(\alpha) &=\operatorname{Im} \biggl[e^{2\pi i/3} \mathfrak{P}\biggl(\frac{\sqrt{3}}2+\alpha e^{2\pi i/3}\biggr)\biggr] -\frac{\eta_1}2, \\ \psi''_2(\alpha) &=\operatorname{Im} \biggl[e^{4\pi i/3} \mathfrak{P}' \biggl(\frac{\sqrt{3}}2+\alpha e^{2\pi i/3}\biggr)\biggr]. \end{aligned}
\end{equation*}
\notag
$$
For $\alpha\in[\sqrt{3}/3,\sqrt{3}/2]$, the point $\sqrt{3}/2+\alpha e^{2\pi i/3}$ lies in the interior of the fundamental hexagon, moreover, it lies in its triangular part, which satisfies $\pi/6 < \arg z<\pi/3$. By Theorem 3.1 and the symmetry principle, the function $\mathfrak{P}'$ maps this triangle onto the first quarter, and hence, $-5\pi/6< \arg[e^{4\pi i/3} \mathfrak{P}'(\sqrt{3}/2+\alpha e^{2\pi i/3})]<-\pi/6$. As a result, $\psi''_2(\alpha)<0$.
Therefore, $\psi'_2(\alpha)$ strictly decreases on $[\sqrt{3}/3,\sqrt{3}/2]$. Since at $\alpha=\sqrt{3}/3$ we have $\psi'_2(\alpha)=-0.344663\ldots<0$, the derivative $\psi'_2(\alpha)$ is negative, that is, $\psi_2(\alpha)$ strictly decreases on $[\sqrt{3}/3,\sqrt{3}/2]$. For $\alpha=\sqrt{3}/3$, the point $\sqrt{3}/2+\alpha e^{2\pi i/3}$ coincides with the mid-point of a side of $\mathfrak{F}$, therefore, at this point Hence $\psi_2(\alpha)>0$ on $[\sqrt{3}/3,\sqrt{3}/2]$. This proves the theoremRemark 8.1. It is clear that, in the case 2) of Theorem 8.1, the critical points $\xi_1(\alpha)$ and $\xi_2(\alpha)$ depend continuously on $\alpha$. It can be shown that $\xi_2(\alpha)$ is strictly increasing and $\xi_1(\alpha)$ is strictly decreasing on $(\sqrt{3}/3, \sqrt{3}/2)$. As $\alpha\to \sqrt{3}/2$, we have $\xi_1(\alpha)\to 0$, $\xi_2(\alpha)\to \sqrt{3}/2$, and as $\alpha\to \sqrt{3}/3$, the values $\xi_1(\alpha)$ and $\xi_2(\alpha)$ also tend to $\sqrt{3}/3$. § 9. Location of the image of the pole on the side $AB$ of the triangle $ABC$Let $\alpha$ be on the side $AB$ of the triangle $ABC$. This means that in the initial configuration three points $a_1$, $a_2$ and $a_3$ are on the same straight line. Theorem 9.1. If $\alpha$ is on the side $AB$ of the triangle $ABC$, then the set $\Gamma_{12}=L(u,0)$ contains the straight line $L$ passing through the points $A$ and $B$. Besides, on this straight line, each segment of length $3$ contains two critical points of $u(z)$ non equivalent modulo $\Omega$. Proof. Let $\alpha=a e^{\pi i/6}$, $z=t e^{\pi i/6}$, where $a\in (0,1)$. Then the equation $u(z)=0$, where $u$ is defined by (6.14), can be written as
$$
\begin{equation}
\ln|\sigma(e^{\pi i/6}(t-e^{-2\pi i/3}a))| -\ln|\sigma(e^{\pi i/6}(t-e^{2\pi i/3}a))| -\eta_1 a\operatorname{Im} t=0.
\end{equation}
\tag{9.1}
$$
The function $\sigma(e^{\pi i/6}z)$ is such that
$$
\begin{equation*}
\sigma(e^{\pi i/6}z)=\overline{\sigma(e^{-\pi i/6}\overline{z})} =\overline{\sigma(e^{-\pi i/3}e^{\pi i/6}\overline{z})} =\overline{e^{-\pi i/3}\sigma(e^{\pi i/6}\overline{z})} =e^{\pi i/3}\overline{\sigma(e^{\pi i/6}\overline{z})},
\end{equation*}
\notag
$$
therefore, $|\sigma(e^{\pi i/6}z)|=|\sigma(e^{\pi i/6}\overline{z})|$. This implies that if $t$ is real, then (9.1) holds. Thus, the zero level set of $u$ contains the straight line $L=\{t e^{\pi i/6},\,t\in \mathbb{R}\}$.
Now we will show that critical points also lie on this straight line, and, in addition, each segment of length $3$ contains precisely two critical points. We write the equation for critical points as
$$
\begin{equation*}
e^{\pi i/6}\zeta\bigl(e^{\pi i/6}(t-e^{-2\pi i/3}a)\bigr) -e^{\pi i/6}\zeta\bigl(e^{\pi i/6}(t-e^{2\pi i/3}a)\bigr)+\eta_1 a i=0
\end{equation*}
\notag
$$
or, what is the same, with the use of notation (3.8),
$$
\begin{equation}
\widetilde{\zeta}(t-e^{-2\pi i/3}a)-\widetilde{\zeta}(t-e^{2\pi i/3}a)+\eta_1 a i=0.
\end{equation}
\tag{9.2}
$$
We have
$$
\begin{equation*}
\zeta(e^{\pi i/6}z)=\overline{\zeta(e^{-\pi i/6}\overline{z})} =\overline{\zeta(e^{-\pi i/3}e^{\pi i/6}\overline{z})} =\overline{e^{\pi i/3}\zeta(e^{\pi i/6}\overline{z})} =e^{-\pi i/3}\overline{\zeta(e^{\pi i/6}\overline{z})},
\end{equation*}
\notag
$$
and hence $\widetilde{\zeta}(z)=\overline{\widetilde{\zeta}(\overline{z})}$. If $t$ is real, then (9.2) can be written as
$$
\begin{equation*}
\psi(t):=\operatorname{Im} [\widetilde{\zeta}(t-e^{2\pi i/3}a)]-\frac{\eta_1}2\, a=0.
\end{equation*}
\notag
$$
The function $\psi(t)$, qua function of a complex variable $t$, has periods
$$
\begin{equation*}
m\sqrt{3}\, e^{i\pi/6}+n\sqrt{3}\, e^{-i\pi/6}, \qquad m, n\in\mathbb{Z}.
\end{equation*}
\notag
$$
In particular, letting $m=1$, $n=-1$, we find that $\psi$ has period $3$. Hence it has period $3$ qua function of a real argument. Now let us study the monotonicity of $\psi$. We have
$$
\begin{equation*}
\psi'(t)=-\operatorname{Im}\bigl[e^{\pi i/3} \mathfrak{P}\bigl(e^{\pi i/6}(t-e^{2\pi i/3}a)\bigr)\bigr].
\end{equation*}
\notag
$$
Let us now show that the derivative $\psi'(t)$ vanishes at the points $t_1=-a/2$ and $t_2=-a/2+3/2$.
Indeed, $\psi'(t)=-\operatorname{Im}[e^{\pi i/3} \mathfrak{P}(e^{\pi i/6}(t+a/2-i(\sqrt{3}/2)a))]$, Therefore, since $\mathfrak{P}$ is even and $a$ is real, we have, by Theorem 3.2,
$$
\begin{equation*}
\psi'(t_1)=-\operatorname{Im}\biggl[e^{\pi i/3} \mathfrak{P}\biggl(e^{\pi i/6} \biggl(i\, \frac{\sqrt{3}}2\, a\biggr)\biggr)\biggr] =-\operatorname{Im}\biggl[e^{\pi i/3} \mathfrak{P} \biggl(e^{2\pi i/3}\biggl(\frac{\sqrt{3}}2\biggr)a\biggr)\biggr]=0.
\end{equation*}
\notag
$$
Now consider $\psi'(t_2)=-\operatorname{Im}[e^{\pi i/3} \mathfrak{P}(e^{\pi i/6}(3/2+i\sqrt{3}\,a))]$. Since the period $\omega_1$ of $\mathfrak{P}$ is $\sqrt{3}$, we obtain
$$
\begin{equation*}
e^{\pi i/6}\biggl(\frac32+i\, \frac{\sqrt{3}}2\, a\biggr)-\omega_1 =e^{\pi i/6} \biggl(\frac32+i\, \frac{\sqrt{3}}2\, a-e^{-\pi i/6}\omega_1\biggr) =e^{2\pi i/3}(a+1)\frac{\sqrt{3}}2,
\end{equation*}
\notag
$$
and, as in the case of the point $t_1$, we conclude that $\psi'(t_2)=0$.
Now we will show that $\psi(t_1)>0$, $\psi(t_2)<0$. We have
$$
\begin{equation*}
\psi(t_1)=-\operatorname{Im}\biggl[e^{\pi i/6} \zeta\biggl(e^{2\pi i/3}\, \frac{\sqrt{3}}2 \, a\biggr)\biggr]-\frac{\eta_1}2\, a=-\operatorname{Im} \widetilde{\zeta}\biggl(\frac{\sqrt{3}}2\, ai\biggr) -\frac{\eta_1}2\, a.
\end{equation*}
\notag
$$
Since, by Corollary 3.1, the points $\widetilde{\zeta}((\sqrt{3}/2)ai)$ for $a\in(0,1)$ belong to the negative part of the imaginary axis and lie lower than the point $\widetilde{\zeta}((\sqrt{3}/2)i)$, we obtain
$$
\begin{equation*}
-\operatorname{Im}\widetilde{\zeta}\biggl(\frac{\sqrt{3}}2\, ai\biggr) > -\operatorname{Im}\widetilde{\zeta}\biggl(\frac{\sqrt{3}}2\, i\biggr) =c\, \frac{\sqrt{3}}2\, a>c\, \frac{\sqrt{3}}2\, a=\frac{\eta_1}2\, a.
\end{equation*}
\notag
$$
Thus, we have shown that $\psi(t_1)>0$. A similarly argument proves that $\psi(t_2)< 0$. By the intermediate value theorem, the equation $\psi(t)=0$ has a unique root on $[t_1,t_2]$. A similar assertion also holds for $[t_2,t_1+3]$. Theorem 9.1 is proved. Remark 9.1. A more detailed analysis shows that if $\alpha=a e^{\pi i/6}$, $a\in (0,0.5]$, then there exists precisely two critical points of the function $u(z)$ of the form $z_{1,2}=t_{1,2} e^{\pi i/6}$, $t_1$, $t_2\in[-1,1]$. Moreover, $t_1<0<t_2$, $t_1+t_2=-a$. As functions of $a$, the values $t_1$ and $t_2$ are decreasing and $t_1=-1$, $t_2=0.5$ for $a=0.5$. Besides, where $\xi=0.7865\dots$ is a unique root of the equation $|\mathfrak{P}(e^{i\pi/6}x)|=c$ on $[0,1]$; here, the constant $c$ is defined by (3.7). Let us prove, for example, that $-1\leqslant t_1<0<t_2<1$. By Corollary 3.1,
$$
\begin{equation*}
\psi(0)=-\operatorname{Im}\widetilde{\zeta}(e^{2\pi i/3}a)-\frac{\eta_1}2\, a>-\operatorname{Im}\widetilde{\zeta}(e^{2\pi i/3})-\frac{\eta_1}2=0.
\end{equation*}
\notag
$$
In addition,
$$
\begin{equation*}
\psi(-1)=\operatorname{Im}\widetilde{\zeta}(-1-e^{2\pi i/3}a) -\frac{\eta_1}2\, a= \operatorname{Im}[\widetilde{\zeta}(z)-c\overline{z}],
\end{equation*}
\notag
$$
where $z=-1-e^{2\pi i/3}a$ is on the half $W$ of the side of the regular hexagon $\mathfrak{F}^*$ located inside the angle $\pi< \arg z<7\pi/6$ for $0<a<1/2$. Since $W$ is mapped to itself by the complex conjugate function of $(1/c)\widetilde{\zeta}(z)$, and the fact that each vertex is attracting and every mid-point of the side of $\mathfrak{F}^*$ is repelling (see Remark 3.1 after Lemma 3.2), we obtain $0<\operatorname{Im}[\widetilde{\zeta}(z)-c\overline{z}]$, and this implies that $\psi(-1)<0$. Therefore, $-1\leqslant t_1<0$, $t_2>0$, and since $t_2=-t_1-\alpha$, we obtain $t_2<1$. We also note that asymptotics (9.3) can be established by the method of expansion in a small parameter. § 10. Trajectories of the quadratic differential related to the function $u$Here, we introduce the quadratic differential,6 associated with the function $u$, and describe the structure of its trajectories and the geometry of the null set of $u$ depending on the value of $\alpha$. Consider the function
$$
\begin{equation}
F(z):=\ln\sigma(z-e^{-2\pi i/3}\alpha)-\ln\sigma(z-e^{2\pi i/3}\alpha) +i \eta_1\overline{\alpha}z.
\end{equation}
\tag{10.1}
$$
Its real part is $u$ defined by (6.14). We have
$$
\begin{equation}
F'(z)=\zeta(z-e^{-2\pi i/3}\alpha)-\zeta(z-e^{2\pi i/3}\alpha) +i \eta_1\overline{\alpha}.
\end{equation}
\tag{10.2}
$$
Consider the quadratic differential on the plane7 Since $F'(z)$ is a doubly periodic function, this quadratic differential induces a quadratic differential on the torus $\mathbb{C}/\Omega$.Denote by $L(u,\gamma)=\{z\colon u(z)=\gamma\}$ the level set of $u$. Note that the sets $L(u,\gamma)$ can be disconnected, and their connected components are the union of trajectories of the quadratic differential $Q(z)\, dz^2$. Our task is to describe the differential-topological structure of $Q(z)\, dz^2$, in particular, the trajectories which contain critical points. Theorem 10.1. In each fundamental domain, the quadratic differential $Q(z)\, dz^2$ has two poles of the second order at the points ${p_1\equiv e^{2\pi i/3}\alpha}$ and $p_2\equiv e^{-2\pi i/3}\alpha \ (\operatorname{mod}\Omega)$. If $\alpha\neq \sqrt{3}/3$, then, in each fundamental domain, the differential has two distinct zeros of the second order at the points $z_1$, $z_2$; in addition, $z_1$ and $z_2$ are symmetric with respect to the point $-\alpha/2$ $(\operatorname{mod}\Omega)$. If $\alpha= \sqrt{3}/3$, then $Q(z)\, dz^2$ has in $\mathfrak{F}$ one zero of the fourth order at the point $\sqrt{3}/3\ (\operatorname{mod}\Omega)$. This result is immediate from (10.2), (10.3), Lemma 7.1, and Theorem 2.1. The function $u$ is doubly periodic and harmonic in the plane, except at the points $p_1$ and $p_2\ (\operatorname{mod} \Omega)$. Therefore, it induces on the torus $\mathbb{C}/\Omega$ a function $U$ harmonic except at the two points corresponding to $p_1$ and $p_2$. The following result is obtained via the maximum modulus principle. Theorem 10.2. Every level set $L(u,\gamma)$ is invariant with respect to shifts by vectors of the period lattice and induces a partition of the torus into two connected components, in one of which the inequality $U>\gamma$ holds and it contains the point corresponding to $p_1$, and, in the second one, $U<\gamma$ and this components contains the point corresponding to $p_2$. The image of every connected subset of $L(u,\gamma)$ (in particular, of a trajectory of the differential or the union of intersecting trajectories) under the factorizing mapping $\pi$ on the torus either does not split it, or splits it into two parts. In neighbourhoods of the poles $p_j$, we have $-Q(z)\, dz^2\sim -(z-p_j)^{-2}$, therefore, the trajectories of the differential are closed curves (see [15]). In a neighbourhood of a zero $z_j$, there is a finite number of arcs of trajectories going from it (four, if $z_j$ is a zero of the first order of the function $F'(z)$, and six, if it is a zero of the second order). This implies that each non-singular trajectory on the torus is a closed curve and a singular one connects two zeros, which possibly coincide. The following result is almost clear. Lemma 10.1. 1) Every non-singular trajectory $\varrho$ of the considered quadratic differential $Q(z)\, dz^2$ is a smooth curve on the plane, either closed or non-closed, and invariant under shifts by elements of $\Omega$. 2) Every singular trajectory $\varrho$ connects two distinct points $P_1$ and $P_2$ of the plane, which are zeros of the differential $Q(z)\,dz^2$. Now we will describe the singular trajectories of $Q(z)\,dz^2$, that is, the trajectories containing its zeros. Three cases are possible. 1) The parameter $\alpha=\sqrt{3}/3$. This case, which we call degenerate, was investigated in [9]. 2) The parameter $\alpha$ is either on a perpendicular drawn from the centre of the triangle $ABC$ to one of its side (not coinciding with the centre of $ABC$) or on one of the sides of $ABC$ (not coinciding with the vertices of $ABC$). Hence, by assertion 2) of Theorems 8.1 and 9.1, the null level set $L(u,0)$ contains both the points which are zeros of the quadratic differential. We will call this case singular. If $\alpha$ is on a perpendicular drawn from its centre, we will call this case singular of type I. If $\alpha$ is on a side of $ABC$, then we call it a singular case of type II. 3) The parameter $\alpha$ does not satisfy the conditions described in assertions 1) and 2). So, we will call the case non-singular. If $\alpha$ is on one of the segments connecting the centre of $ABC$ with its vertex we well say that it is non-singular symmetric case. In the opposite case, we have the non-singular symmetric case. Theorem 10.3. In case 1), the singular trajectories of $Q(z)\, dz^2$ are segments of length $\sqrt{3}$ that connect points of the form $\sqrt{3}/3+m\omega_1+n\omega_2$, $m,n\in \mathbb{Z}$. The union of the segments induces a partition of the plane into regular triangles with vertices at the indicated points. Every two adjacent triangles form a fundamental parallelogram. On the torus, this configuration corresponds to the union of three smooth curves with one common point and not intersecting at other points. Each of these curves corresponds to a segment which is a singular trajectory of the quadratic differential. At the common point, the curves intersect each other at angles $2\pi/3$. The union of all singular trajectories coincides with $L(u,0)$. In case 2), the singular trajectories form a connected set that coincides with the set $L(u,0)$. In case 3), the singular trajectories do not intersect the set $L(u,0)$, which consists of a countable number of connected components. These components are unbounded smooth curves $L_n$, $n\in\mathbb{Z}$, invariant under shifts by the period $\omega_1$. In addition, for every $n$, the curve $L_{n+2}$ is obtained from $L_n$ by a shift by $\omega_2$. These curves can be numerate such that the curves $L_{-1}$ and $L_1$ are symmetric to each other with respect to the point $-\alpha/2$, and $L_0$ contains the point $-\alpha/2$. Proof. As noted above, case 1) was studied in [9]. In case 2), the result follows from assertion 2) of Theorems 8.1 and 9.1. The conclusion of the theorem in case 3) can be easily proved with the help of theorems on the properties of quadratic differentials on compact Riemann surfaces [13]. Now, for completeness, we will describe singular trajectories of the quadratic differential $Q(z)\, dz^2$. Theorem 10.4. Let $\alpha\neq \sqrt{3}/3$ and a singular trajectory $\varrho$ connect two distinct points $P_1$ and $P_2$ which are zeros of the quadratic differential $Q(z)\, dz^2$. 1) In the singular case, the points $P_1$ and $P_2$ are not equivalent to each other modulo the lattice. In addition, there exist points $P_3$ and $P_4$ such that $P_1\sim P_3$, $P_2\sim P_4$ $(\operatorname{mod}\Omega)$ and the curvilinear Jordan quadrilateral $D$, the boundary of which consists of four smooth arcs $P_1P_2$, $P_2P_3$, $P_3P_4$ and $P_4P_1$, which are trajectories of the differential intersecting orthogonally at the points $P_j$, $1\leqslant j\leqslant 4$, is a circular domain for $Q(z)\,dz^2$. The union of all arcs obtaining from the described above four arcs, sides of the quadrilateral, by shifts by all vectors of the lattice $\Omega$, form a level set of $u$. This set separates the plane into a countable number of Jordan quadrilaterals bounded by four smooth arcs intersecting each other orthogonally at some points of $\Omega$. Every boundary arc is a trajectory connecting two points, one of which is equivalent to $P_1$ and the other is equivalent to $P_2$. The quadrilaterals adjacent with $D$, are equivalent modulo the lattice. If we denote one of them by $D_1$, then we can assert that it is a circular domain of the differential and all shifts of domains $D$ and $D_1$ by elements of the lattice induce a tiling of the complex plane by quadrilaterals (see Fig. 8, a). 2) In the non-singular case, the points $P_1$ and $P_2$ are equivalent modulo the lattice, they differ by the vector $\omega_1$ and there is one more trajectory $\varrho_1$, distinct from $\varrho$ and connecting these points. The angle between $\varrho$ and $\varrho_1$ at points of their intersection is $\pi/2$. The component of the level set, containing $\varrho$ and $\varrho_1$, is obtained from $\varrho\cup \varrho_1$ by the union of its shifts by multiples of $\omega_1$ (see Fig. 8, b). It is easy to prove Theorem 10.4 using standard topological arguments and properties of quadratic differentials [13]. At last, we will describe dynamics of the set $L(u,0)$ if the parameter $\alpha$ moves along the real axis from $0$ to $\sqrt{3}/2$, and then upward to the vertex $B$. If $\alpha$ is on $(0,\sqrt{3}/3)$, the intersection of $L(u,0)$ with the strip $0<\operatorname{Im} z<3 /2$ is an unbounded smooth curve $\Gamma=\Gamma(\alpha)$ invariant under shifts by $\omega_1$ and symmetric with respect to straight lines $\operatorname{Re} z=-\alpha/2+(k/2)\omega_1$, $k\in \mathbb{Z}$ (see Fig. 9, a). From Theorem 12.1 (see below) it follows that the curve is the graph of a smooth periodic function $y=h_\alpha(x)$, $x\in \mathbb{R}$. Numerical calculations show the function $h_\alpha(x)$ attains its maximum at points of the form $-\alpha/2+k \omega_1$ and minimum at the points $-\alpha/2\,{+}\,(k+1/2)\omega_1$ (it seems this fact can be proved rigorously). As $\alpha\to \sqrt{3}/3$, the maximal value of $h_\alpha(x)$ tends to $(3/2)$, and the minimal values tends to zero. In the limit, $\Gamma(\alpha)$ tends to an infinite polygonal line, which has vertices at the points $\sqrt{3}/3\,{+}\,k \omega_1$, $\sqrt{3}/3\,{+}\,k \omega_1\,{+}\,\omega_2$. This polygonal line separates the strip $0<\operatorname{Im} z<3 /2$ into regular triangles of side $\sqrt{3}$ (see Fig. 9, b). The “touch” by $\Gamma(\alpha)$ of the real axis in the limit, as $\alpha\to\sqrt{3}/3$, means that the two critical points of the function $u$ tend to the point $\sqrt{3}/3$ and forming at it a double critical point. We also note that as $\alpha\to 0+$, the curves $\Gamma(\alpha)$ tend to the limit position – this is the curve $\Gamma(0)$, which is the null set of the function $\operatorname{Im} (\zeta(z)-c \overline{z})$ in the considered strip. In other strips, the structure is obtained by shifts by multiples of $\omega_2$. Now we assume that $\alpha$ moves on $[\sqrt{3}/3,\sqrt{3}/2]$. As $\alpha$ grows from $\sqrt{3}/3$, the double critical point breaks up into two simple ones, $\xi_1(\alpha)$ and $\xi_2(\alpha)$. The set $L(u,0)$ is connected, and is the union of the horizontal straight lines $\operatorname{Im} z= 3k/2$, $k\in \mathbb{Z}$, and infinitely smooth curves, which intersect orthogonally the straight lines (see Fig. 8, a). These curves can be represented as the graphs of the smooth functions $x=f(y)$, $y\in \mathbb{R}$; they all are obtained from one of them by the reflection with respect to the straight line $\operatorname{Re} z=-\alpha/2$ and shifts by multiples of $\omega_1$. The set $L(u,0)$ induces a tiling of the complex plane, with rich symmetry properties. As $\alpha\to \sqrt{3}/2-$, the curves straighten up, and, for $\alpha=\sqrt{3}/2$, we obtain the tiling of the plane by rectangles. As $\alpha$ goes upward from $\alpha=\sqrt{3}/2$, the vertical straight lines remains unchanged, and the horizontal ones bend (see Fig. 10, a). As $\alpha\to \sqrt{3}/2+(1/2)i$, these lines have a limiting position — this is the null set of the function $\operatorname{Re} [\zeta(z-e^{\pi i/6})-c(\overline{z}-e^{-\pi i/6})]$. In Fig. 10, b), we also give the structure of the set $L(u,0)$ in the non-singular symmetric case. We note that if the parameter $\alpha$, corresponding to this case, tends to some $\alpha_0$ corresponding to the singular case, the adjacent components of the set $L(u,0)$, which are infinite smooth periodic curves, in the limit, close up at points which are zeros of the gradient of the function $u$ corresponding to $\alpha_0$. § 11. The structure of Nuttall sheets for various locations of points $a_1$, $a_2$ and $a_3$We first establish the following auxiliary assertion. Lemma 11.1. The vertices of the fundamental hexagon $\mathfrak{F}$, the point $\alpha$, and the points of the form $-\alpha/2+(l \omega_1+k\omega_2)/2$, $l,k\in \mathbb{Z}$, belong to the set $L(u,0)$. Proof. Let us first prove that the vertices of $\mathfrak{F}$ belong to $L(u,0)$. Consider, for example, the point $B$ with affix $e^{\pi i/6}$. We set $\beta=\alpha e^{-2\pi i/3}$. In view of (3.3), we have
$$
\begin{equation*}
\begin{aligned} \, u(e^{\pi i/6}) &=\ln|\sigma(e^{\pi i/6}-e^{-2\pi i/3}\alpha)| -\ln|\sigma(e^{\pi i/6}-e^{2\pi i/3}\alpha)| -\eta_1\operatorname{Im} (\overline{\alpha} e^{\pi i/6}) \\ &=\ln|\sigma(e^{\pi i/6}-\beta)|-\ln|\sigma(e^{\pi i/6} -e^{4\pi i/3}\beta)|-\eta_1\operatorname{Im} (\overline{\beta} e^{-\pi i/2}) \\ &=\ln|\sigma(e^{\pi i/6}-\beta)|-\ln|\sigma(e^{5\pi i/6}-\beta)| +\eta_1\operatorname{Re} {\beta}. \end{aligned}
\end{equation*}
\notag
$$
Since, by (2.8)
$$
\begin{equation*}
\begin{aligned} \, |\sigma(\beta-e^{5\pi i/6})| &=|\sigma(\beta-e^{\pi i/6}+\omega_1)| =|\sigma(\beta-e^{\pi i/6})e^{\eta_1(\beta-e^{\pi i/6}+\omega_1/2)}| \\ &=|\sigma(\beta-e^{\pi i/6})e^{\eta_1\beta}| =|\sigma(\beta-e^{\pi i/6})|e^{\eta_1\operatorname{Re}\beta}, \end{aligned}
\end{equation*}
\notag
$$
we have $u(e^{\pi i/6})=0$.
Now let us prove that a point of type $z=-\alpha/2+(l\omega_1+k\omega_2)/2$, $l,k\in \mathbb{Z}$, lies in $L(u,0)$. Since, under the symmetry with respect to the point $-\alpha/2$, the function $u$ changes its sign, we conclude that at the point $z^*=-\alpha/2-(l\omega_1+k\omega_2)/2$ we have $u(z^*)=-u(z)$. But the points $z^*$ and $z$ differ from each other by an element of the lattice, therefore, $u(z^*)=u(z)$. Consequently, $u(z)=0$. The equality $u(\alpha)=0$ is evident. This proves the lemma. Corollary 11.1. One of arcs in the $\tau$-plane, which are the projections of sheet separation arcs, goes through the point at infinity. This assertion is immediate from the fact that $\alpha\in L(u,0)$ and the description of the structure of the set $L(u,0)$ (see § 10). Degenerate caseIn the case $\alpha=\sqrt{3}/3$, the set $\Gamma_{12}$ is described in [9], and also in case 1) of Theorem 10.3. The corresponding set $L_{12}$ is shown in Fig. 11, a. Using (6.8) and (6.9), we conclude that the sheets $S_j$, $0\leqslant j\leqslant 2$, on the universal covering space (in the fundamental hexagon $\mathfrak{F}$) has the form shown in Fig. 11, b. Moreover, the sheet $S_0$ is connected, and the sheets $S_1$ and $S_2$ consist of six and three connected components. In the initial $\tau$-plane, the zero sheet is obtained from the plane by drawing cuts along three segments connecting the vertices of the regular triangle $\Delta(a_1,a_2,a_3)$ with its centre. The first sheet consists of six angles of size $\pi/3$ with common vertex at the centre of the triangle $\Delta(a_1,a_2,a_3)$ bounded by the rays, passing through the vertices of the triangle, and bisectors. At last, the second sheet consists of three angles of size $2\pi/3$ cut along the rays with vertices at $a_j$ such that the rays are bisectors of the angles. Non-degenerate caseNow we will study the Nuttall decomposition into sheets on the universal covering space in the non-degenerate cases. Consider $\alpha$ lying in the triangle $\mathfrak{S}$ bounded by the straight lines $y=0$, $y=(\sqrt{3}/3)x$ and $y=\sqrt{3}/3-\sqrt{3}x$, and not coinciding with its vertices. One of the vertices of the triangle is $A(0)$, and two other ones are at the points $M_1=\sqrt{3}/3$ and $N_1=\sqrt{3}/4+i/4$. By Remark 6.1, the set $L(u,0)$, for $\alpha$ lying on $M_1N_1$, is obtained from the corresponding set for $e^{-2\pi i/3}\alpha$ lying in the segment $M_1D$ of the real axis by a rotation through $2\pi/3$. From the results of §§ 8–10 we see that the structure of the set $L(u,0)$ is as follows. 1) If $\alpha$ is on the side $AM$ (the non-singular symmetric case), $L(u,0)$ consists of horizontal straight lines and simple smooth curves not intersecting the straight lines and invariant under the shift by $\omega_1$; in addition, the curves are obtained from each other by shifts by multiples of $\omega_2$. 2) If $\alpha$ is inside the triangle $\mathfrak{S}$ (the non-singular non-symmetric case), $L(u,0)$ consists of two families of simple curves. Curves from different families are disjoint, all the curves are invariant under the shift by $\omega_1$. Different curves from the same family are obtained from each other by shifts by multiples of $\omega_2$. 3) If $\alpha$ is on the side $M_1N_1$ (the singular case of type I), the set $L(u,0)$ consists of straight lines forming the angle $2\pi/3$ with the positive part of the real axis and simple smooth curves which are invariant with respect to the shift by $\omega_1+\omega_2$, intersect the straight lines orthogonally, and are obtained from each other by shifts by multiples of $\omega_1$ (and also of $\omega_2$). 4) If $\alpha$ is on the side $AN_1$ (the singular case of type II), the set $L(u,0)$ consists of straight lines forming the angle $\pi/6$ with the positive part of the real axis and simple smooth curves which are invariant with respect to the shift by the vector $\omega_1-\omega_2$, intersect these straight lines orthogonally, and are obtained from each other by shifts by multiples of $\omega_1$ (and also of $\omega_2$). Note that, in cases 1) and 3), the set $L(u,0)$ is mirror symmetric about the real axis, and, in case 4), it is symmetric about the straight line passing through the points $A$ and $N_1$. As $\alpha\to\sqrt{3}/3$, the curves described in properties 1)–3) either tend to the real axis or to broken lines with equal links and angles $0$, $\pm \pi/3$ of inclination to the real axis. The set $\Gamma:=\Gamma_{01}\cup\Gamma_{02}\cup \Gamma_{12}$ is very important for a description of the structure of the Nuttall sheets. We recall that $\Gamma_{01}$ and $\Gamma_{02}$ are obtained from $\Gamma_{12}$ by a rotation through $\pm 2\pi/3$, therefore, their structure is known. The set $\Gamma$ separates the plane into a countable number of parts. It has some symmetry. Lemma 11.2. The set $\Gamma$ is invariant with respect to shifts by vectors of $\Omega$ and rotations through $\pm 2\pi/3$ about the points $m_1\omega_1+m_2\omega_2$, $m_1, m_2\in \mathbb{Z}$, and vertices of the fundamental hexagon. In the case of real $\alpha$, it is also mirror symmetric with respect to the straight lines $y=(3/2)m$, $m\in \mathbb{Z}$, and also straight lines obtained from them by a rotation through $\pm 2\pi/3$ about the origin. Most of the assertions of Lemma 11.2 are immediate from the results of § 6. We will only show the invariance of the set under rotations about vertices of the fundamental hexagon. Due to periodicity and symmetry arguments, it is sufficient to consider the rotation about the point $z=i$. This rotation has the form $w=e^{2\pi i/3}(z-i)+i=e^{2\pi i/3}z+\sqrt{3}\, e^{\pi i/3}=e^{2\pi i/3}z+\omega_2$, whence the required invariance. We can consider the set $\Gamma$ as the graph with vertices at points which are the points of intersection of various lines from the set $\Gamma_{12}$ (it is non-empty only in singular cases), and their rotations through multiples of $2\pi/3$, and also of lines from various sets $\Gamma_{jk}$. The arcs which are sides of the graph are analytic. By the definition of the sets $\Gamma_{jk}$, we have $\Gamma_{12}=L(u,0)$, $\Gamma_{01}=e^{2\pi i/3}L(u,0)$ is the null set of the function $u(e^{-2\pi i/3}z)$, and $\Gamma_{02}=e^{-2\pi i/3}L(u,0)$ is the null set of the function $u(e^{2\pi i/3}z)$. Since, by (6.14), for every $z$ we have $u(z)+u(e^{2\pi i/3}z)+u(e^{-2\pi i/3}z)=0$, we conclude that
$$
\begin{equation*}
\mathfrak{V}:=\Gamma_{12}\cap \Gamma_{01} =\Gamma_{12}\cap \Gamma_{02}=\Gamma_{01}\cap \Gamma_{02} =\Gamma_{12}\cap \Gamma_{01}\cap \Gamma_{02}.
\end{equation*}
\notag
$$
By Lemma 11.1, the set $\mathfrak{V}$ contains all points equivalent to the points $A$, $B$ and $C$ modulo $\Omega$. We also note that the set of vertices of $\Gamma$ consists of points of the set $\mathfrak{V}$, and, in the singular cases, also of points equivalent to points $F$ and $G$, which are critical points of $u$ in the fundamental hexagon (instead of the hexagon, we can take any fundamental domain for the lattice $\Omega$). Unfortunately, we could not strictly describe the set $\mathfrak{V}$ for arbitrary $\alpha$. Numerous numerical calculations suggest the following conjecture. Conjecture 11.1. In the fundamental hexagon $\mathfrak{F}$, the set $\mathfrak{V}$, except for the centre and vertices of $\mathfrak{F}$, contains precisely three points $E$, $E_1$ and $E_2$, which are obtained from each other by rotations about the origin through multiples of $2\pi/3$. Through each of the three points, there pass three smooth arcs of the sets $\Gamma_{12}$, $\Gamma_{01}$ and $\Gamma_{12}$, and each two of them intersect each other transversally. The following result will be proved below in § 12. Theorem 11.1. Conjecture 11.1 is true in the non-singular symmetric case. This result allows us to completely describe the differential-topological structure of the Nuttall sheets in the non-singular symmetric case. Now we will give a description of the set $\Gamma$ and the structure of Nuttall sheets in the above four cases under the assumption that Conjecture 11.1 is valid in the cases 2)–4). Non-singular symmetric caseLet $\alpha\in (0,\sqrt{3}/3)$. As noted in Theorem 10.3, the set $\Gamma_{12}$ consists of an infinite number of connected components which are straight lines parallel to the real axis, and analytic curves invariant under the shift by $\omega_1=\sqrt{3}$. Consider the connected component $K$ of the set $\Gamma_{12}$ lying in the strip $0<\operatorname{Im} z<3/2$. Lemma 11.3. The curve $K$ contains an arc $\gamma_1$ connecting the points $B=e^{\pi i/6}$ and $C_1=i$, and an arc $\gamma_2$ connecting the points $B_1=e^{5\pi i/6}$ and $C_1=i$ such that $\gamma_1$ and $\gamma_2$ do not contain other points equivalent to $B$ and $C$ modulo $\Omega$. Proof. The simple smooth curve $K$ contains all the points equivalent to $B$ and $C$ and lying in the strip $0<\operatorname{Im} z<3/2$, and hence there exist two arcs $\gamma_1$ and $\gamma_2$ connecting in $K$ the point $C_1$ with points $B'$ and $B''$ equivalent to either $B$ or $C$, and not containing other points equivalent to them. We will show that $B'$ and $B''$ are equivalent to $B$. Indeed, if, for example, ${B'}\sim C$, then ${B'}$ differs from $C$ by $n\omega_1$, $n\in \mathbb{Z}\setminus \{0\}$. Combining $\gamma_1$ with its shifts by multiples of $\omega_1$, we obtain an infinite curve $\widetilde{\gamma}$ lying in $K$. But then $\widetilde{\gamma}$ should coincide with $K$, which is impossible, since $\gamma_1$, and therefore, $\widetilde{\gamma}$, do not contain any point equivalent to $B$.
Now we will show that one of the points $B'$, $B''$ coincides with $B$, and the other one is $B_1$. Consider, for example, $\gamma_1$ which connects $C_1$ and $B'=e^{\pi i/6}+k\omega_1$, $k\in \mathbb{Z}$. Let $\widetilde{\gamma}_1$ be obtained from $\gamma_1$ by reflection with respect to the real axis and by shift by a vector connecting the points $C'=e^{-\pi i/6}+k\omega_1$ and $C_1$. Then $\widetilde{\gamma}_1$ is contained in $K$. The smooth arc ${\gamma}_1\cup\widetilde{\gamma}_1$ has end-points $B'$ and $\widetilde{B}'=-(e^{-\pi i/6}+k\omega_1)$, which differ by $(2k+1)\omega_1$. Augmenting ${\gamma}_1\cup\widetilde{\gamma}_1$ with all its shifts by vectors $(2k+1)\omega_1$, we obtain all the set $K$. Since $K$ contains all points of the form $e^{i\pi/6}+ m\omega_1$, $m\in \mathbb{Z}$, we conclude that $2k+1=\pm 1$, that is, either $k=0$ or $k=-1$. Consequently, $B'$ coincides with either $B$ or $B_1=e^{i\pi/6}-\omega_1$. The same concerns the point $B''$. Therefore, one of the points $B'$, $B''$ is $B$, and the other one is $B_1$. This proves the lemma. It is easy to find the point $E$ which is in $\mathfrak{V}$ and which is not equivalent to $A$, $B$, and $C$. This is the point of intersection of $\gamma_1$ and the ray $\arg z=\pi/3$. Denote by $E_1$ and $E_2$ the points obtained from it by a rotation about the origin through $\pm 2\pi/3$. Let the affix of $E$ be $re^{\pi i/3}$ (see Fig. 12). Rotating $\gamma_2$ through $2\pi/3$ about the point $C_1(i)$ we obtain a subarc $\gamma^*_1$ of $\Gamma$, which connects the points $B$ and $C_1$ (as in the case of $\gamma_1$). In addition, $\gamma^*_1$ cannot coincide with $\gamma_1$, since both the arcs are smooth and form the angle $2\pi/3$ at the point $C_1$. Consequently, $\gamma^*_1$ intersect the ray $\arg z=\pi/3$ at some point, which, by Conjecture 11.1, coincides with $E$. As noted above, Conjecture 11.1 is valid in the non-singular symmetric case under consideration (see Theorem 11.1. We also see that, by the symmetry of $\Gamma$ with respect to the ray $\arg z=\pi/3$, the arcs $\gamma_1$ and $\gamma^*_1$ are symmetric with respect to the ray. Now, rotating $\gamma_1\cup\gamma^*_1$ about the point $C_1$ through $(-2\pi/3)$, we obtain $\gamma_2\cup \gamma^*_2$, where $\gamma^*_2$ is symmetric to $\gamma_2$ with respect to the ray $\arg z=2\pi/3$; they intersect each other at the point $E_2^*=(\sqrt{3}-r)e^{2\pi i/3}$. Further, rotating $\gamma_2\cup\gamma^*_2$ about $B_1$, we obtain the arcs $\gamma_3$ and $\gamma^*_3$, which intersect at $E_1=-r$. Continuing this process, we obtain the arcs $\gamma_k$, $\gamma^*_k$ intersecting each other at the points $E_1$, $E^*$, $E_2$, $E^*_1$, where $E^*$ and $E^*_1$ are obtained from $E^*_2$ by a rotation about the origin through $\pm 2\pi/3$. Moreover, the points $E$ and $E^*$, $E_1$ and $E_1^*$, and also $E_2$ and $E_2^*$, are pairwise equivalent modulo $\Omega$. Now we divide each of the arcs $\gamma_1$ and $\gamma_1^*$ into two arcs $\alpha_1$ and $\beta_1$, $\alpha^*_1$ and $\beta^*_1$ lying in the angles $0<\arg z<\pi/3$ and $\pi/3<\arg z<2\pi/3$. A simple analysis shows that the zero sheet $S_0$ is a hexagon bounded by $\alpha_1^*$, the arc that is obtained from $\alpha_1^*$ by the rotation through $2\pi/3$ about the point $B$, the segment connecting the points $A$ and $E$, and also three arcs, symmetric to the above ones with respect to the real axis. The sheet $S_2$ is connected, it is a curvilinear hexagon $AE_2^*B_1E_1C_2E^*$. The sheet $S_1$ consists of four connected components: the bigons $B_1E_1$, $E_1C_2$ and the curvilinear quadrangles $AEC_1E_2^*$, $AE^*B_2E_2$. To the boundary arcs of the hexagons forming the sheets $S_0$ and $S_2$ there correspond the arcs in the $\tau$-plane connecting the image of $e$ of the point $E$ with $a_1$, $a_2$, $a_3$ on the Riemann sphere (see Fig. 13). Note that to the $\alpha_1$ and to the arc symmetric to it with respect to the real axis there correspond the arcs $\mu_1$ and $\mu_2$ in the $\tau$-plane connecting the points $a_2$ and $a_3$ with the point $e$ corresponding to $E$. Similarly, to $\alpha_1^*$ and to the arc symmetric to it there correspond the arcs $\mu_1^*$ and $\mu_2^*$ connecting the points $a_2$ and $a_3$ with $e$. The arcs $\mu_1$ and $\mu_1^*$ (as well as the arcs $\mu_2$ and $\mu_2^*$ symmetric to them) form the boundary of a wing-like Jordan domain. Remark 11.1. Numerical calculations show that the angle of inclination $\beta=\beta(\alpha)$ of the tangent to the line $\Gamma_{12}$ at the point of intersection $E$ decreased on $(0,\sqrt{3}/3)$ qua function of $\alpha$, and $\lim_{\alpha\to \sqrt{3}/3}\beta(\alpha)= -\pi/3$, $\lim_{\alpha\to 0}\beta(\alpha)=-0.660365\ldots<-\pi/6=-0.523599$. In addition, the inner angles of the “wings” at the angular point $e$ change from the limit value $2\arg(c-e^{2\pi i/3}e_1)-\pi/3=0.27353\dots$ to $\pi/3$. Non-singular non-symmetric caseWe will show that, in the non-singular non-symmetric case, the picture is qualitatively the same as in the non-singular symmetric case; naturally, in contrast to it, here, there is no the symmetry with respect to the rays $\arg z=\pi k/3$, $k\in\mathbb Z$. Let $\alpha=\alpha_1$ be an arbitrary point satisfying the conditions of case 2) and not lying on the real axis. Note that, by Remark 6.1, we can assume that there is a curve $\Gamma=\{\alpha(t),\,0\leqslant t\leqslant 1\}$ in the triangle $ABC$ connecting $\alpha$ with some point $\alpha_0\in(0,\sqrt{3}/3)$ not intersecting (excluding the point $\alpha_0$ itself) the boundary of the triangle and perpendiculars from its centre to the sides of the triangle; we can assume that $\alpha(0)=\alpha_0$, $\alpha(1)=\alpha_1$. Since, by Lemma 7.2 and assertion 1) of Theorem 8.1, for $\beta = \alpha(t)\,{\in}\,\Gamma$, the zeros of the gradient $\nabla u(z) = \nabla u(z;\beta)$ are not on the null level $L(u(\,{\cdot}\,;\beta),0)$, with the help of the implicit function theorem, we conclude that there is a smooth isotopy $f(z,t)\colon\mathbb{C}\times[0,1]\to\mathbb{C}$ such that, for every $t\in [0,1]$, the mapping $f_t(z):=f(z,t)$ is a diffeomorphism of $\mathbb{C}$ onto $\mathbb{C}$ and maps $L(u(\,{\cdot}\,;\alpha_0),0)$ on $L(u(\,{\cdot}\,;\alpha(t)),0)$. Consequently, the differential-topological structure of the set $L(u(\,{\cdot}\,;\beta),0)$ is the same for all $\beta\in\Gamma$, in particular, the structure of $L(u(\,{\cdot}\,;\alpha_1),0)$ in the non-symmetric case is similar to that of $L(u(\,{\cdot}\,;\alpha_0),0)$ in the non-singular symmetric case. Recalling that $G_{12}=L(u,0)$, we obtain the sets $\Gamma_{12}(\alpha_0)$ and $\Gamma_{12}(\alpha_1)$ corresponding to the values $\alpha_0$ and $\alpha_1$ of the parameter $\alpha$, are obtained from each other with the help of this isotopy. The same concerns the sets $\Gamma_{02}(\alpha_0)$ and $\Gamma_{02}(\alpha_1)$, and the sets $\Gamma_{01}(\alpha_0)$ and $\Gamma_{01}(\alpha_1)$. Under Conjecture 11.1, we see that the number of points of intersection does not change under the isotopy, and the intersection points change continuously depending on $t$. This implies that the picture in the non-singular case is similar to that in the symmetric one. To construct the graph $\Gamma$, we consider two connected components, $K_1$ and $K_2$, of $\Gamma_{12}$ such that $K_1$ contains the point $A$, and $K_2$ contains the points $B$ and $C$. Next, we apply the rotation of $K_2$ through $2\pi/3$ about the origin. The result of the rotation intersects $K_1$ at the point $E$, which generates additional vertices in the graph $\Gamma$. And further, we proceed as in the non-singular symmetric case. Singular case of type IIn this case, because of the symmetry of $\Gamma_{12}$ with respect to the real axis for real $\alpha$, it is convenient to consider $\alpha$ lying not on the boundary of the triangle $\mathfrak{S}$, but on the interval $(\sqrt{3}/3, \sqrt{3}/2)$ (see Remark 6.1). Consider the symmetric arc of the set $\Gamma_{12}$ connecting the points $B$ and $C$, the mid-point of this arc is the critical point $G$ (see Fig. 14). Let us rotate the arc about the point $B$ through the angle $(-2\pi/3)$, and then rotate the arc thus obtained about the point $C_1$ through $(-2\pi/3)$. We next continue the process and rotate each obtained arc through the same angle about the points $B_1$, $C_2$ and $B_2$. As a result, we obtain a curve, which is the boundary of a curvilinear hexagon $CBC_1B_1C_2B_2$ (this hexagon is a fundamental domain). There is another arc in $\Gamma_{12}$ symmetric with respect to the real axis; this arc connects the points $C_1$ and $B_2$, and its centre is another critical point $F$. This arc intersects the rays $\arg z=\pm \pi/3$ at the points $E$ and $E_2$ symmetric with respect to the real axis. In addition, $E$ is obtained from $E_2$ by the rotation about the origin through $2\pi/3$. Denote by $E_1$ the point obtained from $E$ by the rotation about the origin through $2\pi/3$. According to Conjecture 11.1, there are no other points which are vertices of the hexagon $\Gamma$, except for $A$, $B$, $C$, $G$, $F$, and the point in this hexagon equivalent to them. Now the structure of the graph is fully defined (see Fig. 14). Let us now describe the structure of sheets. The zero sheet $S_0$ is connected; it is a curvilinear hexagon $AE_2C\widetilde{E}_1BE$, where $\widetilde{E}_1$ is a point obtained by a shift of the point $E_1$ by $\omega_1$. This sheet corresponds to the exterior of three Jordan arcs in the $\tau$-plane connecting the point $e$ (which is the image of the point $E$) with the points $a_1$, $a_2$ and $a_3$ (see Fig. 15). Note that the curvilinear triangles $B_1C_2E_1$ and $A E_2 E$ correspond to the Jordan domains $\Omega_2$ and $\Omega_1$ with the cut along the segment $a_1e$, which is the image of $AE$ (coinciding with the image of $AE_2$). Let $\Omega_3$ be the complement of $\Omega_1\cup\Omega_2$. The second sheet consists of three components. On the universal covering space they correspond to the curvilinear quadrangle $AF_1E_1F_2$ and two symmetric curvilinear pentagons $F_1EG_2^*C_1G_1$ and $F_2G_2B_2G_1^*E_2$. In the $\tau$-plane, the corresponding domains are $\Omega_1$ with cut along the segment $a_1f$, which is the image of $AF_1$ (and $AF_2$), and also the upper and lower halves of $\mathbb{C}\setminus \Omega_1$ with cuts along the arcs $ga_2$ and $ga_3$, which are the images of the arcs $GC$ and $GB$ and lie on the boundary of $\Omega_2$. The first sheet consists of six components. They are the upper and lower halves of the domains $\Omega_k$, $1\leqslant k\leqslant 3$. As in the non-singular symmetric case, the angle between boundary arcs of domains $\Omega_1$ and $\Omega_2$ at the point $e$ depends on the value of $\alpha$. Singular case of the type IIBy Remark 6.1, we can consider the case $\alpha=\sqrt{3}/2+i y_0$, $0<y_0<0.5$. In this case, the set $\Gamma_{12}$ consists of vertical straight lines $x=(\sqrt{3}/2)k$, $k\in \mathbb{Z}$, and a countable number of unbounded curves invariant with respect to shifts by multiples of $\omega_1$; we will call them horizontal lines or trajectories. The horizontal lines are obtained from each other by shifts by multiples of $\omega_2$. Note that the horizontal curves tend, as $y_0\to 0$, to the real axis, since, for $y_0=0$, the horizontal curves turn into straight lines and induce, together with vertical straight lines, a tiling of the plane by rectangles. We note that the boundary of the fundamental hexagon $\mathfrak{F}$ is contained in the graph $\Gamma$. Let first $y_0>0$. By Remark 9.1, the critical point $F$ lies on the imaginary axis lower than the origin and the point $G$ is located on the segment $BC$ below the real axis (see Fig. 16). Then the arc $FG$ of the horizontal trajectory intersects the ray $\arg z=-\pi/6$ at some point $E$. By Conjecture 11.1, such a point is unique. The set $\Gamma_{12}$ is symmetric with respect to the imaginary axis, therefore, it contains the arc $FG^*$ symmetric to $FG$ with respect to this axis. Rotating the arc $G^*FG$ and the segment $B_2C_1$ about the origin by multiples of $2\pi/3$, we obtain a part of the graph $\Gamma$ on the universal covering space located in the interior of the fundamental hexagon. Now we will describe the structure of sheets on the universal covering space. In contrast to the previous cases, now the zero sheet is disconnected and consists of two components. One of them is obtained from the two curvilinear quadrangles $CBF_1E$ and $B_1C_2E_2F_2$ by identifying the equivalent points of the sides $BC$ and $B_1C_2$ that differ by $\omega_1$; the second one is the curvilinear quadrangle $AF_1E_1F_2$. The second sheet also consists of two components. One of them is the curvilinear hexagon $B_2G_1^*EAE_2G_2$, and the second is the curvilinear quadrangle $E_1 G_2^*C_1G_1$. The first sheet consists of four components. Now we will describe the structure of sheets in the $\tau$-plane. The image of $EF$ and the arc $FE_2$ symmetric to it bound a Jordan domain $\Omega_1$ in the $\tau$-plane containing in its interior the point $a_1$, which is the image of $A$. The image of $EG$ and the arc $E_2G^*$ symmetric to it bound a Jordan domain $\Omega_2$, which contains the point $a_2$ (the image of $C$) in its interior. The zero sheet consists of the domain $\Omega_1$ with cut along the segment $a_1f$ and the exterior of the union of $\Omega_1$ and the segment $[a_2,e]$. The second sheet consists of the domain $\Omega_2$ cut along the segment $ga_2$ and the exterior of the union of the domain $\Omega_2$ and the rays $(-\infty, a_1)$, $(a_3,\infty)$. At last, the first sheet consists of the upper and lower halves of the domain $\Omega_2$ and its complement (see Fig. 17). Now let $y_0=0$. We note that as $y_0\to 0$ the “horizontal” trajectories straighten and tend to the straight lines $y=(3/2)k$, $k\in \mathbb{Z}$, and the curvilinear triangle $EE_1E_2$ contracts to a point. The part of the graph $\Gamma$ lying in the fundamental hexagon, coincides with the union of the boundary of the hexagon, the six segments connecting its opposite vertices and the six segments connecting mid-points of its opposite sides. The zero sheet consists of one component obtained from the triangles $ABC$ and $AB_1C_2$ by identifying equivalent points of the sides $BC$ and $B_1C_2$. The second sheet consists of two components, which are the parts of the fundamental hexagon lying in the angles $\pi/3+\pi k<\arg z<2\pi/3+\pi k$, $k=0,1$. The first sheet consists of four components lying in the angles $\pi/6+\pi k/2<\arg z<\pi/3+\pi k/2$, $0\leqslant k\leqslant 5$. In the $\tau$-plane, the sheets are arranged as follows. We can assume that the segment connecting the points $a_2$ and $a_3$, is on the real axis, $a_2=-a_3$, $a_3>0$. Since the point $a_1$ is its mid-point, $a_1=0$. Hence the zero sheet is the exterior of the segment $a_2a_3$. The second sheet consists of the right and left half-planes cut along the rays $(-\infty,a_2)$ and $(a_3,\infty)$. At last, the first sheet is the union of four components, which are four quarters of the plane. Numerical calculations show that as $y_0\to 0$, the domain $\Omega_1$ contracts to a point and the domain $\Omega_2$ grows and tends to the half-plane. § 12. Proof of Conjecture 11.1 in the non-singular symmetric caseNow we will prove Theorem 11.1 on the validity of Conjecture 11.1 in the non-singular symmetric case, that is, in the case where the parameter $\alpha$ is on the interval $(0,\sqrt{3}/3)$. To this end, we need the following result. Theorem 12.1. The angle of inclination to the real axis of the tangent to each line from $L(u,0)$ at each of its point lies in the interval $(-\pi/3,\pi/3)$. Now our problem is to study the intersection of $L(u,0)$ with the straight lines $z(t)=z_0+e^{i\phi}t$, $t\in \mathbb{R}$. Theorem 12.1 will be proved once we show that, for $|\phi-\pi/2|\leqslant\pi/6$, these straight lines intersect each component of $L(u,0)$ at a single point. It suffices to consider the case where $|\phi-\pi/2|=\pi/6$, that is, $\phi=\pi/3$ and $\phi=2\pi/3$; in addition, because of the symmetry of $L(u,0)$ with respect to the point $-\alpha/2$, we only need to investigate the case $\phi= \pi/3$. Therefore, let $\phi=\pi/3$. We have
$$
\begin{equation*}
u(z):=\ln\biggl|\sigma\biggl(z+\frac{\alpha}2 +i\, \frac{\sqrt{3}}2\ \alpha\biggr) \biggr| -\ln\biggl|\sigma\biggl(z+\frac{\alpha}2 -i \, \frac{\sqrt{3}}2\, \alpha\biggr) \biggr| -\eta_1\operatorname{Im} (\overline{\alpha} z).
\end{equation*}
\notag
$$
Setting $u_1(z)=u(z-\alpha/2)$, we have $L(u,0)=-\alpha/2+L(u_1,0)$, where $L(u_1,0)=\{z\colon u_1(z)=0\}$. Since $L(u,0)$ is obtained from $L(u_1,0)$ by a horizontal shift, we can replace $u$ by $u_1$. We have $u_1(z)=\ln|\sigma(z+i(\sqrt{3}/2)\alpha)|-\ln|\sigma(z-i(\sqrt{3}/2)\alpha)| -\eta_1\operatorname{Im} (\overline{\alpha} z)$. Setting $\beta=(\sqrt{3}/2)\alpha$, $\beta\in(0,1/2)$, we have
$$
\begin{equation*}
\begin{aligned} \, u_1(z) &=\ln|\sigma(z+i\beta)|-\ln|\sigma(z-i\beta)| - \frac2{\sqrt{3}}\, \eta_1\operatorname{Im} (\overline{\beta} z) \\ &=\ln|\sigma(z+i\beta)|-\ln|\sigma(z-i\beta)|-2c \operatorname{Im} (\overline{\beta} z). \end{aligned}
\end{equation*}
\notag
$$
Consider the function where $x_0$ is a fixed point on the real axis; we will study its dependence on $t\in \mathbb{R}$. We set
$$
\begin{equation}
A=A(\beta)=\mathfrak{P}'(i\beta),\quad B=B(\beta)=\mathfrak{P}(i\beta),\quad C=C(\beta)=2(\zeta(i\beta)+ci\overline{\beta}).
\end{equation}
\tag{12.2}
$$
Lemma 12.1. The set of extreme points of the function $f(t)=u_1(z(t))$ lies in the set of points $z$ such that Proof. We have $f'(t)=\operatorname{Re}\{e^{i\pi/3}[\zeta(z(t)+i\beta) -\zeta(z(t)-i\beta)+2ci\overline{\beta}\,]\}$. From the properties of elliptic functions, it follows that
Therefore, local extremes of $f(t)$ lie in the set of $z$ satisfying
$$
\begin{equation}
\arg[\zeta(z+i\beta)-\zeta(z-i\beta)\,+2ci\overline{\beta}\,]=-\frac{\pi}3\pm \frac{\pi}2
\end{equation}
\tag{12.5}
$$
or
$$
\begin{equation}
\arg\biggl[-\frac{\mathfrak{P}'(i\beta)}{\mathfrak{P}(z)- \mathfrak{P}(i\beta)} +2(\zeta(i\beta)+ci\overline{\beta}\,)\biggr]=-\frac{\pi}3\pm \frac{\pi}2.
\end{equation}
\tag{12.6}
$$
Taking into account notation (12.2), we can write (12.6) as
$$
\begin{equation}
\arg\biggl[-\frac{A}{\mathfrak{P}(z)-B}+C\biggr]=-\frac{\pi}3\pm \frac{\pi}2.
\end{equation}
\tag{12.7}
$$
But (12.7) is equivalent to (12.3). Lemma 12.1 is proved. We also need the following result. Lemma 12.2. The disc contains the points $e_2=\mathfrak{P}(\omega_2/2)$ and $e_3=\mathfrak{P}(\omega_3/2)$, where $\omega_3=\omega_1e^{2\pi i/3}$l; the point $e_1=\mathfrak{P}(\omega_1/2)$ lies in the exterior of this disc. Proof. 1) Consider the case of the point $e_1$. Conditions (12.5) and (12.3) are equivalent (see the proof of Lemma 12.1), and hence the inequality $|e_1-w_0|>R$ is equivalent to saying that
$$
\begin{equation*}
\operatorname{Re}\biggl\{ e^{i\pi/3}\biggl[\zeta\biggl(\frac{\omega_1}2+i\beta\biggr) -\zeta\biggl(\frac{\omega_1}2-i\beta\biggr)+2ci\overline{\beta}\,\biggr]\biggr\}>0.
\end{equation*}
\notag
$$
We have
$$
\begin{equation*}
\begin{aligned} \, &\zeta\biggl(\frac{\omega_1}2+i\beta\biggr)-\zeta\biggl(\frac{\omega_1}2-i\beta\biggr) +2ci\overline{\beta}=\zeta\biggl(\frac{\omega_1}2+i\beta\biggr) +\zeta\biggl(i\beta-\frac{\omega_1}2\biggr)+2ci\overline{\beta} \\ &\qquad=2\zeta\biggl(\frac{\omega_1}2+i\beta\biggr)-\eta_1+2ci\overline{\beta} = 2\biggl(\zeta\biggl(\frac{\omega_1}2+i\beta\biggr) -c\overline{\biggl(\frac{\omega_1}2+i\beta\biggr)}\, \biggr). \end{aligned}
\end{equation*}
\notag
$$
So, we only need to prove that
$$
\begin{equation*}
-\frac{5\pi}6<\arg\biggl[\biggl\{\zeta\biggl(\frac{\omega_1}2+i\beta\biggr) -c\overline{\biggl(\frac{\omega_1}2+i\beta\biggr)}\,\biggr\}\biggr]<\frac{\pi}6.
\end{equation*}
\notag
$$
But this inequality is true, because $\arg[\{\zeta(\omega_1/2+i\beta)-c\overline{(\omega_1/2+i \beta)}\}]=-\pi/2$ by Lemma 3.5.
2) Now consider the case of the point $e_2$. Proceeding as in the proof of assertion 1), taking into account the equality $\omega_2=e^{i\pi/3}\omega_1$ and (3.3), we obtain
$$
\begin{equation*}
\begin{aligned} \, &\zeta\biggl(\frac{\omega_2}2+i\beta\biggr) -\zeta\biggl(\frac{\omega_2}2-i\beta\biggr) +2ci\overline{\beta} \\ &\qquad= 2e^{-i\pi/3}\biggl(\zeta\biggl(\frac{\omega_1}2 +ie^{-i\pi/3}\beta\biggr)-c\overline{\biggl(\frac{\omega_1}2 +ie^{-i\pi/3}\beta\biggr)}\, \biggr). \end{aligned}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\begin{aligned} \, &\arg\biggl\{e^{i\pi/3}\biggl[\zeta\biggl(\frac{\omega_2}2+i\beta\biggr) -\zeta\biggl(\frac{\omega_2}2-i\beta\biggr) +2ci\overline{\beta}\,\biggr]\biggr\} \\ &\qquad=\arg\biggl[\zeta\biggl(\frac{\omega_1}2+e^{i\pi/6}\beta\biggr) -c\overline{\biggl(\frac{\omega_1}2+e^{i\pi/6}\beta\biggr)}\,\biggr] \end{aligned}
\end{equation*}
\notag
$$
and the inequality $|w-e_2|<R$ is equivalent to
$$
\begin{equation*}
\operatorname{Re} e^{i\pi/3}\biggl[\zeta\biggl(\frac{\omega_2}2+i\beta\biggr) -\zeta\biggl(\frac{\omega_2}2-i\beta\biggr)+2ci\overline{\beta}\,\biggr]<0,
\end{equation*}
\notag
$$
or, what is the same,
$$
\begin{equation*}
\frac{\pi}2<\arg\biggl[\frac{\zeta(\omega_1/2+e^{i\pi/6}\beta) -c\overline{(\omega_1/2+e^{i\pi/6}\beta)}}{z}\biggr]<\frac{3\pi}2,
\end{equation*}
\notag
$$
which is valid because of Lemma 3.5.
3) Now consider the case of $e_3$. We have $\omega_3=\omega_2-\omega_1=\omega_1e^{i2\pi/3}$, and, therefore,
$$
\begin{equation*}
\begin{gathered} \, \begin{aligned} \, &\zeta\biggl(\frac{\omega_3}2+i\beta\biggr) -\zeta\biggl(\frac{\omega_3}2-i\beta\biggr) +2ci\overline{\beta} \\ &\qquad=2e^{-i2\pi/3}\biggl(\zeta\biggl(\frac{\omega_1}2 +ie^{-i2\pi/3}\beta\biggr) -c\overline{\biggl(\frac{\omega_1}2+ie^{-i2\pi/3}\beta\biggr)}\,\biggr), \end{aligned} \\ \begin{aligned} \, &\arg\biggl\{ e^{i\pi/3}\biggl[\zeta\biggl(\frac{\omega_3}2+i\beta\biggr) -\zeta\biggl(\frac{\omega_3}2-i\beta\biggr) +2ci\overline{\beta}\,\biggr]\biggr\} \\ &\qquad=-\frac{\pi}3+\arg\biggl[\zeta\biggl(\frac{\omega_1}2+e^{i\pi/6}\beta\biggr) -c\overline{\biggl(\frac{\omega_1}2+e^{i\pi/6}\beta\biggr)}\,\biggr]. \end{aligned} \end{gathered}
\end{equation*}
\notag
$$
It is easy to see that the condition $|w-e_2|<R$ is equivalent to
$$
\begin{equation*}
\frac{5\pi}6<\arg\biggl[\zeta\biggl(\frac{\omega_1}2+e^{-i\pi/6}\beta\biggr) -c\overline{\biggl(\frac{\omega_1}2+e^{-i\pi/6}\beta\biggr)}\,\biggr]<\frac{11\pi}6,
\end{equation*}
\notag
$$
which also holds by Lemma 3.5. This proves the lemma. From Lemma 12.1 it follows that the set of extreme points for all the functions $f(t)=f(t,x_0)$ is the preimage of the circle $w=w_0+Re^{i\theta}$, $\theta\in [0,2\pi]$ under the mapping realized by the Weierstrass $\mathfrak{P}$-function. By (3.5), this function satisfies the differential equation8 $\mathfrak{P}'(z)=-2\sqrt{\mathfrak{P}^3(z)-e_1^3}$, therefore, its inverse, up to a sign and shifts by elements of the lattice $\Omega$, has the form
$$
\begin{equation*}
z=\mathfrak{P}^{-1}(w)=-\frac{1}{2}\int_{\omega_1/2}^w\frac{dw}{\sqrt{w^3-e_1^3}}.
\end{equation*}
\notag
$$
Consider the curve, which is the image of the circle (12.3) under $\mathfrak{P}^{-1}(w)$; this curve has the form
$$
\begin{equation*}
z=z(\theta)=-\frac{1}{2}\int_{\omega_1/2}^{w_0+Re^{i\theta}}\frac{dw}{\sqrt{w^3-e_1^3}}.
\end{equation*}
\notag
$$
The angle of inclination of its tangent to the real axis is
$$
\begin{equation*}
\gamma=\gamma(\theta)=\pi+\arg\biggl[\frac{iRe^{i\theta}} {\sqrt{(w_0+Re^{i\theta})^3-e_1^3}}\biggr] =\frac{3\pi}{2}-\frac{1}{2}\arg\frac{{(w_0/R+e^{i\theta})^3-(e_1/R)^3}}{e^{i2\theta}}.
\end{equation*}
\notag
$$
Setting $w_0/R=a$, $e_1/R=r$, we have
$$
\begin{equation}
\gamma(\theta)=\frac{3\pi}{2}-\frac{1}{2}\arg\frac{(e^{i\theta}+a)^3}{e^{i2\theta}}.
\end{equation}
\tag{12.9}
$$
Our problem is to describe the range of the function $\gamma(\theta)$. Note that, by Lemma 12.2,
$$
\begin{equation}
|a-r|>1,\qquad |a-re^{2\pi i/3}|<1,\qquad |a-re^{-2\pi i/3}|<1.
\end{equation}
\tag{12.10}
$$
In addition, from (12.4) we see that
$$
\begin{equation*}
a=\frac{B}{R}+ e^{i\phi}, \quad\text{where} \quad \phi=\arg \frac{Ae^{i\pi/3}}{2\operatorname{Re}(Ce^{i\pi/3})}.
\end{equation*}
\notag
$$
We have $R>0$, $B=\mathfrak{P}(i\beta)$ is a (negative) real number, $\arg A=\arg C=-\pi/2$, and hence $\operatorname{Re}(Ce^{i\pi/3})>0$ and $ \phi=\arg(Ae^{i\pi/3})=-\pi/6$. This implies that $a=B/R+\sqrt{3}/2-i/2$, that is, $\operatorname{Im} a=-1/2$. Next, (12.10) implies that the regular triangle with vertices at the point $-a+r$, $-a+re^{2\pi i/3}$ and $-a+re^{-2\pi i/3}$ is such that its vertex $-a+r$ is outside the unit circle, and the two other vertices are inside the circle. This implies $\chi=-\operatorname{Re} a>0$, in addition, $\sqrt{3}/6<\chi<\sqrt{3}/2$. We next require the following lemma, which gives us information about the value of $\gamma(\theta)$. Lemma 12.3. Let a complex number $a=-\chi-i/2$ and a real number $r$ be such that $\sqrt{3}/6<\chi<\sqrt{3}/2$, $r>0$, and inequalities (12.10) hold. Then we can fix a continuous branch of the argument of the function $[(z+a)^3-r^3]/z^2$ on the unit circle $|z|=1$ such that Remark 12.1. We first note that if the origin is in the interior of a circle $\Gamma$ and $\gamma$ is a non-closed subarc of $\Gamma$ on which a continuous branch of argument is fixed, then the argument changes on $\gamma$ monotonically. If the origin is outside of $\Gamma$, then there are two tangents to $\Gamma$ passing through it. If $\gamma$ does not contain the points of tangency, then the argument on $\gamma$ changes monotonically. If $\gamma$ contains only a tangency point, then, at this point, the argument has a global extremum, and the other extremum is attained at one of the end-points of $\gamma$. We will use this fact below. Proof of Lemma 12.3. Consider the function This function has three zeros. By (12.10), the zero $z_1=-a+r$ is outside the unit circle, and the zeros $z_2=-a+re^{2\pi i /3}$ and $z_3=-a+re^{4\pi i/3}$ are inside it. Hence
$$
\begin{equation*}
\chi+r>\frac{\sqrt{3}}{2},\qquad \biggl(\chi-\frac{r}{2}\biggr)^2 + \biggl(\frac{1}{2}+\frac{\sqrt{3}}{2}r\biggr)^2<1,
\end{equation*}
\notag
$$
that is, Moreover, $0<r<\sqrt{3}/3$. For a fixed $\chi$, $\sqrt{3}/6<\chi<\sqrt{3}/2$, we have On the interval $\sqrt{3}/6<\chi<\sqrt{3}/2$, the difference between $r_2(\chi)$ and $r_1(\chi)$ attains its maximum at $\chi=1-\sqrt{3}/6$; the maximal value is $2-\sqrt{3}=0.2679\dots$ .
Denote $g(z;\chi,r)=f(z;-\chi-i/2,r)$, $h(z;\chi,r)=g(z;\chi,r)/z^2$. We need to study the function
$$
\begin{equation}
\arg h(e^{i\theta};\chi(r),r)=\arg g(e^{i\theta};\chi(r),r)-2\theta
\end{equation}
\tag{12.13}
$$
and to prove that, under some choice of branch of the argument, we have which is equivalent to (12.11).
I. We first consider the case $\chi=\chi_1(r)$. We fix $\theta$ and investigate the behaviour of the function $\arg g(e^{i\theta};\chi_1(r),r)$ for $0<r<\sqrt{3}/3$. In this case,
$$
\begin{equation*}
\begin{aligned} \, z_1 &=\chi_1(r)+r+\frac{i}2=\frac{\sqrt{3}}2+\frac{i}2=e^{i\pi/6}, \\ z_2 &=\chi_1(r)-\frac{r}2+\frac{ir\sqrt{3}}2+\frac{i}2=e^{i\pi/6}-\rho e^{-i\pi/6}, \\ z_3 &=\chi_1(r)-\frac{r}2-\frac{ir\sqrt{3}}2+\frac{i}2=e^{i\pi/6}-\rho e^{i\pi/6}, \end{aligned}
\end{equation*}
\notag
$$
where $0<\rho=r\sqrt{3}<1$. Since, for $\chi=\chi_1(r)$, the zero $z_1$ of the function $f(z;a,r)$ is on the unit circle, we obtain
$$
\begin{equation*}
g(e^{i\theta};\chi_1(r),r)=(e^{i\theta}-e^{i\pi/6})(e^{i\theta}-e^{i\pi/6}+\rho e^{-i\pi/6})(e^{i\theta}-e^{i\pi/6}+\rho e^{i\pi/6}).
\end{equation*}
\notag
$$
Therefore, the function $\arg g(e^{i\theta};\chi_1(r),r)$ has discontinuity at $\theta=\pi/6$. We set
$$
\begin{equation}
b=e^{i\theta}-e^{i\pi/6} =2e^{i(\theta/2+7\pi/12)}\sin\biggl(\frac{\theta}{2}-\frac{\pi}{12}\biggr).
\end{equation}
\tag{12.15}
$$
We can assume that $\pi/6<\theta<13\pi/6$. In this case, $\sin(\theta/2-\pi/12)>0$ and $\arg b=\theta/2+7\pi/12$ changes continuously on $(\pi/6,\,13\pi/6)$. On this interval, we can fix a continuous branch of the argument of the function $\arg g(e^{i\theta};\chi_1(r),r)$ such that
$$
\begin{equation}
\arg g(e^{i\theta};\chi_1(r),r)= \frac{\theta}{2}+\frac{7\pi}{12} +\arg\frac{e^{i\theta}-\tau e^{i\pi/6}}{e^{-i\theta}+i-\tau e^{i\pi/6}},
\end{equation}
\tag{12.16}
$$
where $\tau=1-\rho$. Consider the Moebius transform for $z=(1-\rho)e^{i\pi/6}$, $0<\rho<1$. We have
$$
\begin{equation}
T(z)=\frac{e^{i\theta}-e^{i\pi/6}+e^{i\pi/6}-z}{e^{-i\theta} -e^{-i\pi/6}+e^{i\pi/6}-z} =S(\zeta):=\frac{b-\zeta}{\overline{b}-\zeta},
\end{equation}
\tag{12.18}
$$
where $\zeta=z-e^{i\pi/6}$, and $b$ is defined by (12.15). For $z=(1-\rho)e^{i\pi/6}$, we have
$$
\begin{equation*}
\zeta=t e^{i\pi/6},\qquad -1<t=-\rho=-\sqrt{3}\, r<0.
\end{equation*}
\notag
$$
The set of these $\zeta$ describes the segment $\Delta$ with end-points at $0$ and $\zeta_1=-e^{i\pi/6}$.
From (12.13), (12.16)–(12.18) we have
$$
\begin{equation}
\arg h(e^{i\theta};\chi_1(r),r) =\arg S(t e^{i\pi/6})-\frac{3\theta}{2} + \frac{7\pi}{12}.
\end{equation}
\tag{12.19}
$$
The function $S(\zeta)$ maps the real axis to the unit circle. Under the mapping $S(\zeta)$, the ray $\zeta=-\rho e^{i\pi/6}$, $\rho>0$, forming the angle $\pi/6$ with the real axis is sent to the arc $\gamma=\gamma^\theta$ of the circle $\Gamma = \Gamma^\theta$, which forms the angle $\pi/6$ with the unit circle and has end-points $S(\infty) = 1$ and Under the mapping $S$, the point $\zeta_1$ goes to the point In addition,Remark 12.2. It is easy to see that each circle that contains the arcs $\gamma=\gamma^\theta$ for various $\theta$ passes through the point $w=1$ and intersects it at the angle $\pi/6$. This implies that these arcs form a pencil of circles passing through $w=1$ and tangent to the straight line $L$, which passes through the points $w=1$ and $w=e^{i5\pi/3}$. Naturally, the straight line $L$ itself is an element of this pencil, and this line passes through infinity. It is evident that each two elements of the pencil intersect each other only at the point $w=1$. Let $\gamma_1=\gamma_1^\theta$ be the part of the arc $\gamma=\gamma^\theta$ corresponding to the segment $\Delta$ with end-points $0$ and $\zeta_1=-e^{i\pi/6}$ under the mapping $S$. In Fig. 18, we draw the unit circle and typical arcs $\gamma_1^\theta$ for various values of $\theta$. It is clear that with increasing $\theta$ the end-point $w_0$ of the arc $\gamma_1$ which lies on the unit circle moves in the counterclockwise direction. It is easily checked that, for $\pi/6<\theta<\pi/2$, the arc $\gamma_1$ is outside the unit disc. The arc $\gamma_1$ tends, as $\theta\to \pi/6+0$, to the arc connecting $1$ and $e^{i4\pi/3}$. The arc $\gamma_1$ tends, as $\theta\to \pi/2-0$, to the ray $L_0$, which is a part of the straight line $L$, which goes from the point $e^{i5\pi/3}$ and contains the point $1$ on its extension. For $\pi/2<\theta<5\pi/6$, the arc $\gamma_1$ is also outside the unit disc, but its direction of convexity changes. The arc $\gamma_1$ tends, as $\theta\to \pi/2+0$, to the same ray $L_0$, and, as $\theta\to 5\pi/6-0$, its end-points get infinitesimally closer to each other and it contracts to the point $1$. For $5\pi/6<\theta<13\pi/6$ the arc $\gamma_1$ is inside the unit disc. As $\theta\to 5\pi/6+0$, it contracts to a point and as $\theta\to 13\pi/6-0$ it tends to the arc connecting the points $1$ and $e^{i4\pi/3}$, but lying not outside the unit circle, as in the case $\pi/6<\theta<\pi/2$, but inside it. Now we will consider the indicated three cases in more detail. 1) Let $\pi/6<\theta<\pi/2$. In this case, the arc $\gamma$ is outside the unit disc. Since the point $b$ has argument $\theta/2+7\pi/12$ lying between $2\pi/3$ and $5\pi/6$, the point $b$ is above the straight line $\zeta=t e^{i\pi/6}$, $t\in \mathbb{R}$, therefore, the point $0=S(b)$ is inside the disc bounded by the circle $\Gamma$ containing $\gamma$. Consequently, the argument of the points $S(t e^{i\pi/6})$ changes monotonically with respect to $t$. It is easy to show that the argument also increases with $t$. Therefore, it attains its extremes at the end-points of the segment $[0,1]$. Hence
$$
\begin{equation}
\frac{3\theta}{2}-\frac{\pi}{4}<\arg S(t e^{i\pi/6}) <\theta+\frac{7\pi}{6}, \qquad -1<t<0.
\end{equation}
\tag{12.23}
$$
Now, since $\pi/6<\theta<\pi/2$, we have, by (12.19),
$$
\begin{equation}
\frac{\pi}{3}<\arg h(e^{i\theta};\chi_1(r),r) <-\frac{\theta}{2}+\frac{7\pi}{4}<\frac{5\pi}{3}, \qquad 0<r<\frac{\sqrt{3}}{3}.
\end{equation}
\tag{12.24}
$$
2) Let $\pi/2<\theta<5\pi/6$. As in case 1), $\gamma$ is outside the unit disc. Let us calculate the arguments of the end-points of $w_0$ and $w_1$ of $\gamma_1$. We have
$$
\begin{equation}
\arg w_0=\theta+\frac{7\pi}{6}\in \biggl(\frac{5\pi}{3},2\pi\biggr),\qquad \arg w_1=\frac{3\theta}{2}+\frac{3\pi}{4}\in \biggl(\frac{11\pi}{12},2\pi\biggr).
\end{equation}
\tag{12.25}
$$
This implies that the arc $\gamma_1$ is in the lower half-plane. In addition, $\gamma_1$ is in the half-plane bounded by the straight line $L$ and lies below it. Consequently, $\gamma_1$ is lower than the straight line passing through the origin and forming the angle $\pi/3$ with the positive direction of the real axis. Hence
$$
\begin{equation}
\frac{4\pi}3<\arg S(t e^{i\pi/6})<2\pi, \qquad -1<t<0.
\end{equation}
\tag{12.26}
$$
Consequently, taking into account the range of $\theta$ and using (12.19), we have
$$
\begin{equation}
\frac{2\pi}{3}< \frac{23\pi}{12}-\frac{3\theta}{2} <\arg h(e^{i\theta};\chi_1(r),r) <\frac{31\pi}{12}-\frac{3\theta}{2}<\frac{11\pi}{6}, \qquad 0<r<\frac{\sqrt{3}}{3}.
\end{equation}
\tag{12.27}
$$
3) At last, let $5\pi/6<\theta<13\pi/6$. Then the arc $\gamma$ is inside the unit disc. Its subarc $\gamma_1$ connects the points $w_1$ and $w_0$. By (12.20) and (12.22), we have
$$
\begin{equation*}
\arg w_1=3\frac{\theta}2+\frac{3\pi}4>\theta+\frac{7\pi}6=\arg w_0.
\end{equation*}
\notag
$$
It is easy to show that, for $7\pi/6<\theta<13\pi/6$, the origin is inside the circle $\Gamma$, therefore, the function $\arg S(t e^{i\pi/6})$ changes monotonically on $\gamma_1$. If $5\pi/6<\theta<7\pi/6$, then the origin is outside the circle $\Gamma$ which contains $\gamma_1$. Consider the tangent to $\Gamma$ passing through the origin. It is easy to show that the point of tangency is on the distance
$$
\begin{equation*}
d=\frac{\sqrt{1-4\sin^2(\theta/2+7\pi/12)}}{|2\sin(\pi/4-\theta/2)|}
\end{equation*}
\notag
$$
from the origin. By (12.21), we obtain $d<|w_0|,|w_1|<1$. This means that the point of tangency is not on the arc $\gamma_1$. Therefore, the function $\arg S(t e^{i\pi/6})$ changes monotonically on $\gamma_1$. Hence
$$
\begin{equation}
\theta+\frac{7\pi}{6} <\arg S(t e^{i\pi/6}) < \frac{3\theta}{2}+\frac{3\pi}{4}, \qquad -1<t<0.
\end{equation}
\tag{12.28}
$$
Therefore,
$$
\begin{equation}
\frac{2\pi}{3}<\frac{7\pi}{4}-\frac{\theta}{2}<\arg h(e^{i\theta};\chi_1(r),r) < \frac{4\pi}{3}, \qquad 0<r<\frac{\sqrt{3}}{3}.
\end{equation}
\tag{12.29}
$$
Thus, we have shown that inequalities (12.24), (12.27) and (12.29) hold on the three possible intervals which may contain $\theta$. This implies (12.14). II. Now we consider the case $\chi_1(r)<\chi<\chi_2(r)$. We claim that if $\chi$ is fixed and $r$ increases on $r_1(\chi)<r<r_2(\chi)$, then the inequalities are preserved. We first need the following auxiliary assertion. Let $\varphi_1<\arg z < \varphi_2$. We set $\Gamma(\varphi_1,\varphi_2):=\{z\,{\in}\, \mathbb{C}\mid \varphi_1 < \arg z < \varphi_2\}$. Lemma 12.4. Let $s\colon z\mapsto z-r$ be the shift by $r>0$ to the left of points in the complex plane. 1) If $0\leqslant\varphi_1\leqslant\pi\leqslant\varphi_2\leqslant 2\pi$, then $s(\Gamma(\varphi_1,\varphi_2))\subset \Gamma(\varphi_1,\varphi_2)$. 2) If $0\leqslant\varphi_1<\varphi_2\leqslant \pi$, then $s(\Gamma(\varphi_1,\varphi_2))\subset \Gamma(\varphi_1,\pi)$. 3) If $\pi\leqslant\varphi_1<\varphi_2\leqslant 2\pi$, then $s(\Gamma(\varphi_1,\varphi_2))\subset \Gamma(\pi,\varphi_2)$. The proof of Lemma 12.4 is clear. Of course, this lemma can be applied in case where $2\pi k \leqslant\varphi_1\leqslant\pi\leqslant\varphi_2\leqslant 2\pi(k+1)$ for some $k\in \mathbb{Z}$. In this case, $\varphi_1$ and $\varphi_2$ can be replaced by $\varphi_1-2\pi k$ and $\varphi_2-2\pi k$. There are several cases to consider. 1) Let $\pi/6<\theta<\pi/2$. Then (12.23) holds. Therefore, $S(t e^{i\pi/6})$ lies in the angle $\Gamma(\phi_1(\theta),\phi_2(\theta))$, where $\phi_1(\theta)=3\theta/2-\pi/4$, $\phi_2(\theta)=\theta+7\pi/6$. By the constraints on $\theta$, we have $\phi_1(\theta)\in (0,\pi)$, $\phi_2(\theta)\in (4\pi/3,5\pi/3)$. With increasing $r$ the value $g(e^{i\theta};\chi,r)=f(e^{i\theta};-\chi-i/2,r)=(e^{i\theta}-\chi-i/2)^3-r^3$ obtains a negative increment. In view of assertion 1) of Lemma 12.4, estimate (12.23) holds for the increased $r$. This verifies (12.24). 2) Let $\pi/2<\theta<5\pi/6$. Then (12.26) holds. Applying assertion 3) of Lemma 12.4 for the increased $r$, we have the inequality which differs from (12.26) with the change of $4\pi/3$ by $\pi$ in the lower estimate. Now, instead of (12.27), we have
$$
\begin{equation}
\frac{\pi}{3}< \arg h(e^{i\theta};\chi_1(r),r)<\frac{11\pi}{6}, \qquad -1<t<0.
\end{equation}
\tag{12.31}
$$
3) Let $5\pi/6<\theta<13\pi/6$. Then, according to (12.28), the value $S(t e^{i\pi/6})$ is in the angle $\Gamma(\phi_1(\theta),\phi_2(\theta))$, where $\phi_1(\theta)=\theta+7\pi/6$, $\phi_2(\theta)=3\theta/2+3\pi/4$. There are three cases to consider. a) Let $5\pi/6<\theta<3\pi/2$. Then $2\pi<\phi_1(\theta)<8\pi/3$, $2\pi<\phi_2(\theta)<3\pi$. Applying assertion 2) of Lemma 12.4, we conclude that $\phi_1(\theta)<\arg S(te^{i\pi/6})<3\pi$ for the changed $r$, and, therefore, instead of (12.24), we have
$$
\begin{equation*}
\frac{2\pi}3<\arg h(e^{i\theta};\chi_1(r),r) <3\pi+\frac{7\pi}{12}-\frac{3\theta}{2} <2\pi+\frac{\pi}{3}.
\end{equation*}
\notag
$$
Note that the upper estimate in this inequality is not attained, since the equality is only possible for $\theta=5\pi/6$, but, for this value, the arc $\gamma_1$ contracts to the point $w=1$, and $\arg S(te^{i\pi/6})$ is $2\pi<3\pi$ on this arc. b) Let $3\pi/2<\theta<11\pi/6$. Then $8\pi/3<\phi_1(\theta)<3\pi$, $3\pi<\phi_2(\theta)<7\pi/2$. In this case, we apply assertion 1) of Lemma 12.4, which shows that estimate (12.29) does not change. c) Let $11\pi/6<\theta<13\pi/6$. Then $3\pi<\phi_1(\theta)<10\pi/3$, $7\pi/2<\phi_2(\theta)<4\pi$. Therefore, the inequality $3\pi<\arg S(te^{i\pi/6})<\phi_2(\theta)$ holds for the increased $r$, and now from this inequality, we obtain, in lieu of (12.29),
$$
\begin{equation*}
\frac{\pi}{3}<3\pi+\frac{7\pi}{12}-\frac{3\theta}{2} <\arg h(e^{i\theta};\chi_1(r),r)<\frac{11\pi}{6}.
\end{equation*}
\notag
$$
This completes the proof of Lemma 12.3. Now let us return to the proof of Theorem 12.1. From Lemma 12.3, by (12.9), it follows that the angle of inclination of tangents to curves which are the loci of the extremes of functions (12.1) does not assume the value $\pi/3$. Therefore, each straight line of the form $z(t)=x_0+e^{i \pi/3}t$, $x_0\in \mathbb{R}$, intersects each of these curves exactly at one point. Hence each such a straight line intersects each components of the set $L(u_1,0)$ at one point. Therefore, the angle of inclination of every tangent to every component of the set $L(u_1,0)$, and consequently of $L(u,0)$, is never equal to $\pi/3$. This completes the proof of Theorem 12.1. Proof of Theorem 11.1. Our problem is to describe the set $\mathfrak{V}$, which is the intersection of the sets $\Gamma_{12}$, $\Gamma_{01}=e^{2\pi i/3}\Gamma_{12}$ and $\Gamma_{02}=e^{-2\pi i/3}\Gamma_{12}$. According to the above, the set $\mathfrak{V}$ coincides with the intersection of each two of these three sets. Besides, $\mathfrak{V}$ contains the points $A$, $B$, $C$, and all points of the plane equivalent to them modulo $\Omega$.
From Theorems 8.1 and assertion 3) of Theorem 10.3, taking into account Theorem 12.1, we conclude that the set $\Gamma_{12}=L(u,0)$ consists of an infinite number of straight lines parallel to the real axis and obtained from it by shifts by vectors $(3/2)n$, $n\in \mathbb{Z}$, and also of an infinite number of curves obtained from each other by shifts by vectors $\omega_2n$, $n\in \mathbb{Z}$. The curve which lies in the horizontal strip $0<\operatorname{Im} z<3/2$ is the graph of the single-valued $\omega_1$-periodic function $y=f(x)$, $x\in \mathbb{R}$. By rectilinear components we will mean the straight lines incoming to the set $\Gamma_{12}$, the described above curves are curvilinear components. The sets $\Gamma_{01}$ and $\Gamma_{02}$ obtained from $\Gamma_{12}$ by the rotation through $\pm 2\pi/3$, also contain rectilinear and curvilinear components. It is evident that the points equivalent to $A$ belong to rectilinear components of all three sets $\Gamma_{12}$, $\Gamma_{01}$ and $\Gamma_{02}$, and the points equivalent to $B$ and $C$ are from their curvilinear components. We will show that, for every point of the set $\mathfrak{V}$ not equivalent to $B$ and $C$, there is at least one rectilinear component of one of the sets $\Gamma_{12}$, $\Gamma_{01}$ and $\Gamma_{02}$ containing it. Assume the contrary. Then three curvilinear components intersect at some point $P$. Consider the tiling (triangulation) of the plane by regular triangles which have vertices at all points of the lattice $\Omega$. Now $P$ belongs to some triangle of this triangulation. But the centre $Q$ of this triangle is equivalent to either $B$ or $C$. Therefore, the points $P$ and $Q$ belong to all three sets $\Gamma_{jk}$. Since every triangle of the triangulation is contained in some horizontal strip $(3/2)(n-1)<\operatorname{Im} z<(3/2)n$, $n\in \mathbb{Z}$, and in this strip there is a unique curvilinear component of the set $\Gamma_{12}$, it follows that the points $P$ and $Q$ belong to the same curvilinear component of $\Gamma_{12}$. Since the sets $\Gamma_{01}$ and $\Gamma_{02}$ are obtained from $\Gamma_{12}$ by rotations through $\pm 2\pi/3$, and since these rotations transform triangles of the triangulation into each other, we conclude that the points $P$ and $Q$ belong to the same curvilinear component of the set $\Gamma_{01}$, and also to the same component of $\Gamma_{02}$. Consider the vector $\overrightarrow{PQ}$. This vector is parallel to some tangent to the curve, which the curvilinear component of $\Gamma_{12}$, at some point lying between $P$ and $Q$. By Theorem 12.1, the angle between this vector and the real axis is $<\pi/3$. Similarly, considering the curvilinear components of $\Gamma_{01}$ and $\Gamma_{02}$ containing both the points $P$ and $Q$, we see that the angles between the vector $\overrightarrow{PQ}$ and the straight lines obtained from the real axis by the rotations through $\pm \pi/3$ is $<\pi/3$. But this is evidently impossible. Therefore, each point of the set $\mathfrak{V}$ not equivalent to $B$ and $C$ is contained in at least one rectilinear component. Now consider the point of intersection of the curvilinear component of the set $\Gamma_{12}=L(u,0)$ lying in the strip $0<\operatorname{Im} z<3/2$, with the straight lines $y=\pm \sqrt{3}\, x$. By Theorem 12.1, its intersection with each of the straight lines consists of one point, in addition, this intersection is transversal. Therefore, we have two points, $z_1$ and $z_2$, of intersection of this component with the straight lines $y=\pm \sqrt{3}\,x$. Let $\arg z_1=\pi/3$, $\arg z_2=2\pi/3$. The curvilinear components of the sets $\Gamma_{01}$ and $\Gamma_{02}$ are obtained from those of the set $\Gamma_{12}$ by rotations through $\pm 2\pi/3$. Therefore, to describe, up to shifts by elements of the lattice, the points of the set $\mathfrak{V}$ non-equivalent to $B$ and $C$, it is sufficient to study the set of six points $z_k$, $1\leqslant k\leqslant 6$, such that $z_3= e^{2\pi i/3} z_1$, $z_4= e^{2\pi i/3} z_2$, $z_5= e^{4\pi i/3} z_1$, $z_6= e^{4\pi i/3} z_2$. We claim that the points $z_4$, $z_5$ and $z_6$ differ from $z_1$, $z_2$ and $z_3$ by elements of $\Omega$. Consider, for example, the points $z_3$ and $z_6$. We will show that $z_6-z_3=\omega_1$. Indeed, because of the invariance of the set $\Gamma$ with respect to rotations through $\pm 2\pi/3$ about the vertices of the fundamental hexagon and the origin, the point $z_1$ is obtained from $z_6$ by a rotation through $-2\pi/3$ about the point $B(e^{\pi i/6})$. Since $z_3=e^{2\pi i/3}z_1$, we obtain $z_3=e^{2\pi i/3}(e^{-2\pi i/3}(z_6-e^{\pi i/6})+e^{\pi i/6})=z_6-\omega_1$. Thus, up to equivalence modulo the lattice, we obtain three points $z_1$, $z_3$ and $z_5$ lying in the same fundamental domain, and they are obtained from each other by rotations through $\pm 2\pi/3$. At last, let us verify that, at each point of intersection, the arcs which are components of the three sets $\Gamma_{jk}$ intersect transversally. Consider, for example, the point of intersection of the curvilinear component of the set $\Gamma_{12}=L(u,0)$ lying in the strip $0<\operatorname{Im} z<3/2$, with the straight line $y=\sqrt{3}x$ lying in $\Gamma_{02}$. From Remark 11.1 it follows that the angle between them at the point of their intersection is distinct from $\pi/2$. Since $\Gamma_{01}$ is symmetric to $\Gamma_{02}$ with respect to the straight line $y=\sqrt{3}\,x$, each two of these three sets intersect each other transversally at this point. Theorem 11.1 is proved. In conclusion, the author express his gratitude to A. I. Aptekarev for drawing his attention to the problem considered here and for valuable remarks. The author is also grateful to an anonymous referee for careful reading of the manuscript, remarks, and suggestions that improved the text. 
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\Bibitem{Nas24} |
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