We investigate the phase diagram of the complex cubic unitary ensemble of random matrices with the potential $V(M)=−13M3+tM$, where t is a complex parameter. As proven in our previous paper [Bleher et al., J. Stat. Phys. 166, 784–827 (2017)], the whole phase space of the model, $t∈C$, is partitioned into two phase regions, $Oone−cut$ and $Otwo−cut$, such that in $Oone−cut$ the equilibrium measure is supported by one Jordan arc (cut) and in $Otwo−cut$ by two cuts. The regions $Oone−cut$ and $Otwo−cut$ are separated by critical curves, which can be calculated in terms of critical trajectories of an auxiliary quadratic differential. In Bleher et al. [J. Stat. Phys. 166, 784–827 (2017)], the one-cut phase region was investigated in detail. In the present paper, we investigate the two-cut region. We prove that in the two-cut region, the endpoints of the cuts are analytic functions of the real and imaginary parts of the parameter t, but not of the parameter t itself (so that the Cauchy–Riemann equations are violated for the endpoints). We also obtain the semiclassical asymptotics of the orthogonal polynomials associated with the ensemble of random matrices and their recurrence coefficients. The proofs are based on the Riemann–Hilbert approach to semiclassical asymptotics of the orthogonal polynomials and the theory of S-curves and quadratic differentials.

This work is a continuation of the works of Bleher and Deaño1,2 and Bleher, Deaño, and Yattselev.3 The main goal of the whole project is to analyze the large N asymptotics of the partition function and correlation functions in the unitary ensemble of random matrices with the cubic potential $V(M)=−13M3+tM$, where t is a complex parameter. The partition function of this ensemble is formally defined by the matrix integral over the space of N × N Hermitian matrices,

$ZNRM(t)=∫HNe−NTr−13M3+tMdM.$

The integral is formal because it is divergent, and it needs a regularization. To define such a regularization, consider the (formal) identity

$ZNRM(t)=πN(N−1)/2∏k=1Nk!∫R…∫R∏1≤j

{see Ref. 4 [formula (1.2.18)]}. The random matrix partition function $ZNRM(t)$ can now be regularized by changing the contour of integration from the real line $R$ to the contour Γ depicted in Fig. 1, which goes from (−) to (eπi/3). Hence, we are interested in studying the regularized eigenvalue partition function defined as

$ZN(t)≔∫Γ…∫Γ∏1≤j
(1.1)

which is well defined (the integrals are convergent) for all values $t∈C$.

As shown in Ref. 2, the partition function ZN(t) can be used to study the topological expansion in the cubic ensemble of random matrices and enumeration of regular graphs of degree 3 on Riemannian surfaces. To be more precise, let us consider the cubic potential in the form M2/2 − uM3, where u > 0, and the corresponding eigenvalue matrix integral is given by

$ΞN(u)=∫Γ…∫Γ∏1≤j
(1.2)

We would like to consider all the possible values, including the complex ones, of the parameter u. Formula (1.2) is not very convenient for this purpose because the contour of integration Γ should be rotated to secure the convergence of the integral depending on the argument of u. Instead, let us make the change of variables

$ζj=(3u)−1/3zj+16u,$

which yields that

$ζj22−uζj3−1108u2=−zj33+tzj,$

where

$t=14(3u)4/3.$
(1.3)

This implies the relation between the partition functions,

$ΞN(u)=(4t)N2/4e−2t3/2N/3ZN(t).$

As proven in Ref. 2, the free energy of the cubic model $FN(u)$ admits an asymptotic expansion as N in powers of N−2, that is,

$FN(u)≔1N2lnΞN(u)ΞN(0)∼∑g=0∞F(2g)(u)N2g$
(1.4)

for any u in the interval 0 ≤ u < uc, where uc ≔ 31/418−1 is a critical point. In addition, the functions F(2g)(u) admit an analytic continuation to the disk |u| < uc on the complex plane, and if we expand them in powers of u,

$F(2g)(u)=∑j=1∞f2j(2g)u2j(2j)!,$
(1.5)

then the coefficient $f2j(2g)$ is a positive integer that counts the number of 3-valent connected graphs with 2j vertices on a Riemann surface of genus g. Asymptotic expansion (1.4) is called a topological expansion. For more details on this aspect of the theory, we refer the reader to the classical papers;5,6 the monograph of Forrester7 (Sec. 1.6); and the works of Mulase,8 Di Francesco,9 Ercolani and McLaughlin,10,11 Eynard,12 and references therein, or a very readable introduction by Zvonkin.13 It is noteworthy that the general idea of a topological expansion goes back to the classical work of t’ Hooft.14

As shown in Ref. 2, the coefficients $f2j(2g)/(2j)!$ of power series (1.5) behave, when j, as

$f2j(2g)(2j)!=K2gj5g−72uc2j1+Oj−1/2,K2g>0.$

This implies that uc is the radius of convergence of each power series (1.5). In fact, u = uc is a singular point of each of the functions (1.5). The topological expansion in a neighborhood of the critical point uc has been obtained in the work of Bleher and Deaño,1 and it is closely related to the Painlevé I equation as follows: consider a formal series

$∑g=0∞Γ5g−12ucgK2g6⋅31/4λ1−5g2,$
(1.6)

and it can be shown that there exists a one-parameter family of solutions {yα(λ)}, $α∈R$, of the Painlevé I differential equation

$y″(λ)=a0y2(λ)−a1λ,$

with $a0=252394$ and $a1=2323−54$, that admits an asymptotic expansion (1.6) as λ → −. This family consists of the Boutroux tronquée solutions, which means that they are asymptotically free of poles in one of the five canonical sectors of angle 2π/5 in the complex plane that appear naturally when considering rotational symmetries of the Painlevé I equation. The parameter α appears in the Painlevé I Riemann–Hilbert problem, as described by Kapaev in Ref. 15, and in the context of the cubic model, it is related to the choice of the original contour of integration Γ (see Refs. 1 and 2). In the present case, we have α = 1, which makes the solution of Painlevé I tritronquée, that is, asymptotically free of poles in four of the five canonical sectors of angle 2π/5 in the complex plane. We refer to the reader to Ref. 1 (Sec. 2.1) and also the work of Joshi and Kitaev16 for more details.

It is noteworthy that the key ingredient in the proof of topological expansion (1.4) in Ref. 8 is the derivation of semiclassical asymptotic formulas for the recurrence coefficients $γn2$ and βn [see (1.13), where the parameter t is related to u via (1.3)] of the corresponding monic orthogonal polynomials Pn(z) = zn + ⋯ defined via orthogonality relations

$∫ΓPn(z)zke−N12z2−uz3dz=0,k∈{0,1,…,n−1}.$

The idea of using the orthogonal polynomials in ensembles of random matrices goes back to the classical works of Dyson and Mehta (see Refs. 17 and 18 and references therein). In particular, as proven in Ref. 2, for any u ∈ [0, uc), there exists ɛ > 0 such that as N, n with $1−ε≤nN≤1+ε$, the recurrence coefficients $γn2$ and βn admit asymptotic expansions in powers of $1N2$,

$γn2∼∑k=0∞1N2kg2knN,u,βn∼∑k=0∞1N2kb2knN+12N,u,$
(1.7)

where the functions g2k(s, u), b2k(s, u) do not depend on n and N and are analytic in s at s = 1.

In Ref. 1, this asymptotic expansion is extended to the double scaling asymptotic expansion of the recurrence coefficients at the critical point uc. In the double scaling regime, we set

$nN=1+vN−4/5,$

where $v∈R$ is a scaling variable. Then, as proven in Ref. 1, at u = uc, the recurrence coefficients $γn2$ and βn admit asymptotic expansions in powers of N−2/5 as N,

$γn2∼γc2+∑k=1∞1N2k/5p2k(v),βn∼βc+∑k=1∞1N2k/5q2k(ṽ),$
(1.8)

where the functions p2k(v), $q2k(ṽ)$ are expressed in terms of the Boutroux tritronquée solution to the Painlevé I equation, $γc2,βc$ are known explicitly, and $ṽ=v+N−1/5/2$. As shown in Ref. 1, expansions (1.7) and (1.8) can be extended for large N to u in overlapping intervals: [0, ucN−0.79] for (1.7) and [ucN−0.65, uc] for (1.8), and this can be used to obtain the double scaling asymptotic formula for the partition function at uc. In particular, let

$u−uc=2−1253−74λN−45,$

where λ is a complex scaling variable in the double scaling regime. Then, for λ outside of a neighborhood of the poles of y(λ), the Boutroux tritronquée solution to Painlevé I equation, the partition function ΞN(u) can be written as

$ΞN(u)=ΞNreg(u)ΞNsing(λ)1+O(N−ε),ε>0,$
(1.9)

where the regular and singular factors are given by

$ΞNreg(u)=eN2[a+b(u−uc)+c(u−uc)2]+dandΞNsing(λ)=e−Y(λ),$

where a, b, c, d are some explicit constants and Y(λ) is a solution of the differential equation Y″(λ) = y(λ) with the boundary condition

$Y(λ)=2645(−λ)5/2−148log(−λ)+O((−λ)−5/2),λ→−∞.$

Asymptotic formula (1.9) is used in Ref. 1 to prove the conjecture of David19,20 that the poles of the tritronquée solution y(λ) give rise to zeros of ΞN(u).

As we have stressed before, the topological expansions from Refs. 1–3 were obtained by first analyzing the associated orthogonal polynomials defined by

$∫ΓzkPn(z;t,N)e−NV(z;t)dz=0,k∈{0,…,n−1},$
(1.10)

where

$V(z;t)≔−z33+tz,t∈C.$
(1.11)

Due to the non-Hermitian character of the above relations, it might happen that polynomial satisfying (1.10) is non-unique. In this case, we understand by Pn(zt, N) the monic non-identically zero polynomial of the smallest degree (such a polynomial is always unique). One way of connecting the partition function to orthogonal polynomials is via three term recurrence relations. More precisely, let

$hn(t,N)≔∫ΓPn2(z;t,N)e−NV(z;t)dz=Dn(t,N)Dn−1(t,N),$
(1.12)

where D−1(t, N) ≡ 1 and $Dn(t,N)≔det∫Γzi+je−NV(z;t)dzi,j=0n$ is the Hankel determinant of the moments of the measure of integration in (1.10). It easy to see that each Dn(t, N) is an entire function of t, and therefore, each hn(t, N) is meromorphic in $C$. Hence, given n, the set of the values t for which there exists k ∈ {0, …, n} such that hk(t, N) = 0 is countable with no limit points in the finite plane. Outside of this set, the standard argument using orthogonality (1.10) shows that

$zPn(z;t,N)=Pn+1(z;t,N)+βn(t,N)Pn(z;t,N)+γn2(t,N)Pn−1(z;t,N),$
(1.13)

and by analytic continuation, (1.13) extends to those values of t for which Dn−1(t, N)Dn(t, N) ≠ 0 [that is, n + 1-th and nth polynomials appearing (1.13) have the prescribed degrees], where

$γn2(t,N)=hn(t,N)/hn−1(t,N).$
(1.14)

Note that (1.13) remains meaningful even if Dn−2(t, N) = 0. In this case, hn−2(t, N) = 0, hn−1(t, N) = , and hn(t, N) is finite, which means that Pn−2(zt, N) is orthogonal to itself and, therefore, is equal to Pn−1(zt, N) but not to Pn(zt, N). Thus, $γn2(t,N)=0$, and the last term in (1.13) is zero. Note also that if hn−1(t, N) ≠ 0 while hn(t, N) = 0 [that is, Dn−1(t, N) ≠ 0 and Dn(t, N) = 0], then we simply have that $γN2(t,N)=0$ by analytic continuation. However, this means that the polynomial Pn(zt, N) is orthogonal to itself and, therefore, Pn+1(zt, N) = Pn(zt, N). Then, the analytic continuation argument necessitates that βn(t, N) = , which also can be seen from the determinantal representation of βn(t, N) as it is the difference of the subleading coefficients of Pn(zt, N) and Pn+1(zt, N). It is further known that the recurrence coefficients $γN2(t,N)$ satisfy the Toda equation,

$∂2FN(t)∂t2=γN2(t,N),FN(t)=1N2logZN(t).$

Another way of connecting the partition function to orthogonal polynomials is through the formula

$ZN(t)=N!∏n=0N−1hn(t,N)=N!DN−1(t,N).$

As it turns out, there are two different regions in the complex t-plane for which the behavior of the polynomials Pn(zt, N), and therefore of the partition function, is different. Colloquially, we call them one-cut and two-cut regions, a distinction that will become clear later. In Refs. 1–3, the partition function has been analyzed in and on the boundary of the one-cut region. The goal of this work is to start the analysis in the two-cut region. More precisely, here we only consider the asymptotics of the orthogonal polynomials and their recurrence coefficients and postpone the analysis of the partition function for the future project. The structure of this paper is as follows:

• In Secs. II and III, we describe equilibrium measures and corresponding S-curves for the cubic model under consideration (those describe asymptotic behavior of the normalized counting measures of zeros of the orthogonal polynomials in the weak * sense). This leads us to a precise description of the phase diagram (i.e., of the one- and two-cut regions) of the cubic model on the complex t-plane.

• In Sec. IV, we present the main results of this paper: asymptotic formulas for the orthogonal polynomials and their recurrence coefficients in the two-cut phase region.

• Sections VIX are devoted to the proof of our main results; in them, we obtain various results about the detailed structure of the S-curves, derive main properties of the dominant terms of the expansions, apply the Riemann–Hilbert approach to asymptotics of the orthogonal polynomials, and finally prove the expansions themselves.

It is well understood that the zeros of polynomials satisfying (1.10) asymptotically distribute as a certain weighted equilibrium measure on an S-contour corresponding to the weight function (1.11). In this section, we discuss these notions in greater detail. Our consideration will use the works of Huybrechs, Kuijlaars, and Lejon21 and Kuijlaars and Silva.22 Let us start with some definitions.

Definition 2.1.
LetVbe an entire function. The logarithmic energy in the external field ReVof a measureνin the complex plane is defined as
$EV(ν)=∬log1|s−t|dν(s)dν(t)+∫ReV(s)dν(s).$
The equilibrium energy of a contour Γ in the external field ReVis equal to
$EV(Γ)=infν∈M(Γ)EV(ν),$
(2.1)
where$M(Γ)$denotes the space of Borel probability measures on Γ.

When ReV(s) − log|s| → + as Γ ∋ s, there exists a unique minimizing measure for (2.1), which is called the weighted equilibrium measure of Γ in the external field ReV, say, μΓ [see Ref. 23 (Theorem I.1.3) or Ref. 21]. The support of μΓ, say, JΓ, is a compact subset of Γ. The equilibrium measure μ = μΓ is characterized by the Euler–Lagrange variational conditions,

$2Uμ(z)+ReV(z)=ℓ,z∈JΓ,≥ℓ,z∈Γ\JΓ,$
(2.2)

where = Γ is a constant, the Lagrange multiplier, and Uμ(z)≔ −log|zs|dμ(s) is the logarithmic potential of μ [see Ref. 23 (Theorem I.3.3)].

Observe that due to the analyticity of the integrand in (1.1), Γ can be varied without changing the value of ZN(t). Henceforth, we suppose that partition function (1.1) and orthogonal polynomials (1.10) are defined with $Γ∈T$, where $T$ is the following class of contours:

Definition 2.2.
We shall denote by$T$the collection of all piecewise smooth contours that extend to infinity in both directions, each admiting a parameterizationz(s),$s∈R$, for which there existsɛ ∈ (0, π/6) ands0 > 0 such that
$|arg(z(s))−π/3|≤ε,s≥s0,|arg(z(s))−π|≤ε,s≤−s0,$
(2.3)
where arg(z(s)) ∈ [0, 2π).

Despite the above flexibility, it is well understood in the theory of non-Hermitian orthogonal polynomials, starting with the works of Stahl24–26 and Gonchar and Rakhmanov,27 that one should use the contour whose equilibrium measure has support symmetric (with the S-property) in the external field ReV. We make this idea precise in the following definition:

Definition 2.3.
The supportJΓhas the S-property in the external field ReVif it consists of a finite number of open analytic arcs and their endpoints, and on each arc, it holds that
$∂∂n+2UμΓ+ReV=∂∂n−2UμΓ+ReV,$
(2.4)
where$∂∂n+$and$∂∂n−$are the normal derivatives from the (+)- and (−)-sides of Γ. We shall say that a curve$Γ∈T$is an S-curve in the field ReVifJΓhas the S-property in this field.

It is also understood that geometrically JΓ is comprised of critical trajectories of a certain quadratic differential. Recall that if Q is a meromorphic function, a trajectory (respectively, orthogonal trajectory) of a quadratic differential −Q(z)dz2 is a maximal regular arc on which

$−Q(z(s))z′(s)2>0respectively,−Q(z(s))z′(s)2<0$

for any local uniformizing parameter. A trajectory is called critical if it is incident with a finite critical point [a zero or a simple pole of −Q(z)dz2] and it is called short if it is incident only with finite critical points. We designate the expression critical (orthogonal) graph of −Q(z)dz2 for the totality of the critical (orthogonal) trajectories −Q(z)dz2.

The following theorem is a specialization to V(zt) of Ref. 22 (Theorems 2.3 and 2.4).

Theorem 2.1.

LetV(zt) be given by(1.11).

1. There exists a contour$Γt∈T$such that
$EV(Γt)=supΓ∈TEV(Γ).$
(2.5)
2. The supportJtof the equilibrium measure$μt≔μΓt$has the S-property in the external field ReV(zt), and the measureμtis uniquely determined by(2.4). Therefore,μtand its supportJtare the same for every Γtsatisfying(2.5).

3. The function
$Q(z;t)=V′(z;t)2−∫dμt(s)z−s2,z∈C\Jt,$
(2.6)
is a polynomial of degree 4.
4. The supportJtconsists of some short critical trajectories of the quadratic differentialQ(zt)dz2, and the equation
$dμt(z)=−1πiQ+1/2(z;t)dz,z∈Jt,$
(2.7)
holds on each such critical trajectory, where$Q1/2(z;t)=12z2+O(z)$asz[in what follows,Q1/2(zt) will always stand for such a branch].

Much information on the structure of the critical graphs of a quadratic differential can be found in the excellent monographs.28–30 Since deg Q = 4, Jt consists of one or two arcs, corresponding (respectively) to the cases where Q(zt) has two simple zeros and one double zero and the case where it has four simple zeros. Away from Jt, one has freedom in choosing Γt. In particular, let

$U(z;t)≔Re2∫ezQ1/2(s;t)ds=ℓΓt−Re(V(z;t))−2Uμt(z),$
(2.8)

where eJt is any and the second equality follows from (2.6) [since the constant $ℓΓt$ in (2.2) is the same for both connected components of Jt and the integrand is purely imaginary on Jt, the choice of e is indeed not important]. Clearly, $U(z;t)$ is a subharmonic function (harmonic away from Jt), which is equal to zero Jt by (2.2). The trajectories of −Q(zt)dz2 emanating out of the endpoints of Jt belong to the set ${z:U(z;t)=0}$, and it follows from the variational condition (2.2) that $Γt\Jt⊂{z:U(z;t)<0}$. However, within the region ${z:U(z;t)<0}$, the set Γt\Jt can be varied freely. The geometry of the set ${z:U(z;t)<0}$ is described further below in Theorems 3.1 and 3.2.

The partition of the phase space into one- and two-cut regions as well as the structure of Γt and its dependence on t has been heuristically described in Refs. 31 and 32. Even before these works, the phase diagram in Fig. 3 was heuristically described in Ref. 19. A mathematically rigorous description was provided in Ref. 3 but only in the one-cut region. Let us quickly recall the important notions from Ref. 3.

Denote by $C$ the critical graph of an auxiliary quadratic differential

$−(1+1/s)3ds2$
(3.1)

[see Fig. 2(a)]. It was shown in Ref. 3 (Sec. 5) that $C$ consists of five critical trajectories emanating from −1 at the angles 2πk/5, k ∈ {0, 1, 2, 3, 4}, one of them being (−1, 0), other two forming a loop crossing the real line approximately at 0.635, and the last two approaching infinity along the imaginary axis without changing the half-plane (upper or lower). Given $C$, define

$Δ≔x:2x3∈C.$

Furthermore, put $Ωone−cut$ to be the shaded region on Fig. 2(b) and set

$∂Ωone−cut=Δbirthb∪−2−1/3∪Δsplit∪eπi/32−1/3∪Δbirtha,$

where $Δsplit$ connects −2−1/3 and eπi/32−1/3, $Δbirthb$ extends to infinity in the direction of the angle 7π/6, while $Δbirtha$ extends to infinity in the direction of the angle π/6. Let

$t(x)≔(x3−1)/x$

and set

$tcr≔3⋅2−2/3=t−2−1/3,Oone−cut≔t(Ωone−cut),Csplit≔tΔsplit,Cbirthb≔tΔbirthb,Cbirtha≔tΔbirtha,S≔(tcr,∞),e2πi/3S≔z:e−2πi/3z∈S$
(3.2)

(see Fig. 3). The function t(x) is holomorphic in $Ωone−cut$ with a non-vanishing derivative there. It maps $Ωone−cut$ onto $Oone−cut$ in a one-to-one fashion. Hence, the inverse map x(t) exists and is holomorphic.

FIG. 2.

Schematic representation of (a) the critical graph $C$; (b) the set Δ (solid lines) and the domain $Ωone−cut$ (shaded region).

FIG. 2.

Schematic representation of (a) the critical graph $C$; (b) the set Δ (solid lines) and the domain $Ωone−cut$ (shaded region).

Close modal
FIG. 3.

Domain $Oone−cut$ (shaded region); $∂Oone−cut$ consisting of the open bounded arc $Csplit$, two open semi-unbounded arcs $Cbirtha$ and $Cbirthb$, and two points $tcr$ and $e2πi/3tcr$; the semi-unbounded open horizontal rays S and e2πi/3S (dashed lines).

FIG. 3.

Domain $Oone−cut$ (shaded region); $∂Oone−cut$ consisting of the open bounded arc $Csplit$, two open semi-unbounded arcs $Cbirtha$ and $Cbirthb$, and two points $tcr$ and $e2πi/3tcr$; the semi-unbounded open horizontal rays S and e2πi/3S (dashed lines).

Close modal

Convention 1.

Below, we adopt the following convention: Γ(z1, z2) (respectively, Γ[z1, z2]) stands for the trajectory or orthogonal trajectory (respectively, the closure of) of the differentialQ(zt)dz2connectingz1andz2, oriented fromz1toz2, and$Γz,eiθ∞$[respectively,$Γeiθ∞,z$] stands for the orthogonal trajectory ending atz, approaching infinity at the angleθ, and oriented away fromz(respectively, oriented towardz).33

The following theorem has been proven in Ref. 3 (Theorem 3.2), and it describes the geometry of Γt when $t∈Ōone−cut$.

Theorem 3.1.
LetμtandQ(zt) be as in Theorem 2.1, and let$Jt=supp(μt)$. When$t∈Ōone−cut$, the polynomialQ(zt) is of the form
$Q(z;t)=14(z−a(t))(z−b(t))(z−c(t))2,$
(3.3)
witha(t),b(t), andc(t) given by
$a(t)≔x(t)−i2/x(t),b(t)≔x(t)+i2/x(t),c(t)≔−x(t),$
(3.4)
where$x(t)$is the branch holomorphic in$Oone−cut$satisfying$x(0)=eπi/3$. The setJtconsists of a single arc and the following:
• If$t∈Oone−cut$, thenJt = Γ[a, b] and an S-curve$Γt∈T$can be chosen as

• $Γeπi∞,a∪Jt∪Γb,eπi/3∞$whentbelongs to the connected component bounded by$S∪Csplit∪e2πi/3S$[seeFigs. 4(a)4(e) ];

• $Γeπi∞,a∪Jt∪Γ(b,c)∪Γc,eπi/3∞$whentS[seeFig. 4(f) ];

• $Γeπi∞,c∪Γ(c,a)∪Jt∪Γb,eπi/3∞$whente2πi/3S;

• $Γeπi∞,a∪Jt∪Γb,e−πi/3∞∪Γe−πi/3∞,c∪Γc,eπi/3∞$whentbelongs to the connected component bounded by$S∪Cbirthb$[seeFig. 4(g) ];

• $Γeπi∞,c∪Γc,e−πi/3∪Γe−πi/3∞,a∪Jt∪Γb,eπi/3∞$whentbelongs to the connected component bounded by$e2πi/3S∪Cbirtha$.

• If$t=tcr$(respectively,$t=e2πi/3tcr$), thenJt = Γ[a, b],ccoincides withb(respectively,a), and an S-curve$Γt∈T$can be chosen as in Case I(a) [seeFig. 5(a) ].

• If$t∈Csplit$, thenJt = Γ[a, c] ∪ Γ[c, b] and an S-curve$Γt∈T$can be chosen as in Case I(a) [seeFig. 5(b) ].

• If$t∈Cbirthb$(respectively,$t∈Cbirtha$), thenJt = Γ[a, b] and an S-curve$Γt∈T$can be chosen as in Case I(d) [respectively, Case I(e)] [seeFig. 5(c) ].

FIG. 4.

Schematic representation of the critical (solid) and critical orthogonal (dashed) graphs of −Q(zt)dz2 when $t∈Oone−cut$. The bold curves represent the preferred S-curve Γt. Shaded region is the set ${U(z;t)<0}$.

FIG. 4.

Schematic representation of the critical (solid) and critical orthogonal (dashed) graphs of −Q(zt)dz2 when $t∈Oone−cut$. The bold curves represent the preferred S-curve Γt. Shaded region is the set ${U(z;t)<0}$.

Close modal

Now, let $Otwo−cut≔C\Ōone−cut$. Then, the following theorem holds:

Theorem 3.2.
LetμtandQ(zt) be as in Theorem 2.1,$Jt=supp(μt)$. When$t∈Otwo−cut$, the polynomialQ(zt) is of the form
$Q(z;t)=14(z−a1(t))(z−b1(t))(z−a2(t))(z−b2(t)),$
(3.5)
witha1(t),b1(t),a2(t), andb2(t) all distinct. The real and imaginary parts ofai(t), bi(t) are real analytic functions of Re(t) and Im(t) when$t∈Otwo−cut$; however, at no point of$Otwo−cut$any of the functionsai(t), bi(t) is analytic. The S-curve Γtcan be chosen as
$Γeπi∞,a1(t)∪Jt,1∪Γb1(t),e−πi/3∞∪Γe−πi/3∞,a2(t)∪Jt,2∪Γb2(t),eπi/3∞,$
whereJt = Jt,1Jt,2and$Jt,i≔Γai(t),bi(t)$,i ∈ {1, 2}; seeFig. 6[this also explains how we choose the labeling of the zeros ofQ(zt) in the considered case]. Moreover, it holds that
$a1(t),b1(t)→a(t*),b1(t),a2(t)→c(t*),anda2(t),b2(t)→b(t*)$
(3.6)
astt*with$t*∈Cbirtha∪e2πi/3tcr$,$t*∈Csplit$, and$t*∈Cbirthb∪tcr$, respectively.

FIG. 6.

The schematic representation of the critical and critical orthogonal graphs of −Q(zt)dz2 when $t∈Otwo−cut$. The bold curves represent the preferred S-curve Γt. Shaded region is the set ${U(z;t)<0}$.

FIG. 6.

The schematic representation of the critical and critical orthogonal graphs of −Q(zt)dz2 when $t∈Otwo−cut$. The bold curves represent the preferred S-curve Γt. Shaded region is the set ${U(z;t)<0}$.

Close modal

We prove Theorem 3.2 in Sec. V.

In this section, we assume that $t∈Otwo−cut$ and that Q(zt) and Jt = Jt,1Jt,2 are as in Theorem 3.2. We also put

$Et≔a1(t),b1(t),a2(t),b2(t),Jt,i°≔Jt,i\Et,andJt°≔Jt,1°∪Jt,2°.$

Convention 2.

When it comes to the definition of the S-contour Γt, it will be more practical for us to change the choice of Γtmade in Theorem 3.2. Rather than including unbounded trajectories$Γb1(t),e−πi/3∞$and$Γe−πi/3∞,a2(t)$, we shall replace them by a smooth Jordan arc, say,It, connectingb1(t) anda2(t) such thatIt° ≔ It\Etlies entirely within the region${U(z;t)<0}$while there exist$s1(t)∈Γb1(t),e−πi/3∞$and$s2(t)∈Γe−πi/3∞,a2(t)$for which$Γb1(t),s1(t)∪Γs2(t),a2(t)⊂It°$(seeFig. 14 ).

To describe the asymptotics of the orthogonal polynomials themselves, we need to construct the Szegő function of eV(z;t).

Proposition 4.1.
Let constantsς(t) anddi(t) be given by
$ς(t)≔2t3d0(t)anddi(t)≔∫ItsidsQ1/2(s;t),$
(4.1)
where, as usual, we use the branch$Q1/2(z;t)=12z2+O(z)$asz. Then, the Szegő function
$D(z;t)≔exp12V(z;t)+13z+∫It3ς(t)s−zdsQ1/2(s;t)Q1/2(z;t)$
(4.2)
is holomorphic and non-vanishing in$C̄\Γ[a1(t),b2(t)]$with continuous traces onJt° ∪ It° that satisfy
$D+(s;t)D−(s;t)=eV(s;t),s∈Jt°,D+(s;t)=D−(s;t)e2πiς(t),s∈It°.$
(4.3)
Denote byμ1(t) ≔ ∫sdμt(s) the first moment of the equilibrium measureμt. Then,
$D(z;t)=exp13−(ςd1)(t)21+t2+μ1(t)−3(ςd2)(t)/23z+O1z2$
(4.4)
asz. In particular, it holds that$D(∞;t)=exp1−t(d1/d0)(t)3$.

We shall also denote by $D(z;t)≔D(z;t)/D(∞;t)$ the normalized Szegő function. We prove Proposition 4.1 in Sec. VI A.

To describe the geometric growth of the orthogonal polynomials, let us define

$Q(z;t)≔∫b2(t)zQ1/2(s;t)ds,z∈C\Γeπi∞,b2(t)$
(4.5)

(where we deviate slightly from Convention 1 and denote by Γ[s1, s2] a subarc of Γt connecting s1 to s2; for example, Γ(eπi, a2] includes It rather than the short trajectories connecting b1 to a2). Observe that $U(z;t)=2Re(Q(z;t))$ as defined in (2.8). This function has the following properties:

Proposition 4.2.
Let constantsτ(t), ω(t) be given by
$τ(t)≔−1πi∫ItQ1/2(s;t)dsandω(t)≔−1πi∫Jt,1Q+1/2(s;t)ds.$
(4.6)
These constants are necessarily real [in fact,ω(t) = μt(Jt,1); see(2.7)]. The function$eQ(z;t)$is holomorphic in$C\Γ[a1(t),b2(t)]$, and there exists a constant*(t) such that
$expV(z;t)−ℓ*(t)2+Q(z;t)=z+O(1)asz→∞.$
(4.7)
Moreover,$Q(z;t)$possesses continuous traces onJt° ∪ It° that are purely imaginary onJt° and satisfy
$eQ+(s;t)+Q−(s;t)=1,s∈Jt,2°,eQ+(s;t)+Q−(s;t)=e2πiτ(t),s∈Jt,1°,eQ+(s;t)=eQ−(s;t)−2πiω(t),s∈It°.$
(4.8)

We prove Proposition 4.2 in Sec. VI B. Observe that it follows from Theorem 3.2 that $|eQ(z;t)|$ is less than 1 when $U(z;t)<0$ (the shaded areas of Fig. 6), is equal to 1 on the critical trajectories (black curves), and otherwise is greater than 1. The constant *(t) is such that $Re(ℓ*(t))=ℓΓt$ [see (2.2)]. Moreover, the following identity holds:

Proposition 4.3.
Set
$B(t)≔−∫Jt,1dsQ+1/2(s;t)/∫ItdsQ1/2(s;t).$
(4.9)
Let$θ(u;B(t))$be the Riemann theta function associated with$B(t)$, i.e.,
$θ(u;B(t))=∑k∈ZexpπiB(t)k2+2πiuk,u∈C.$
(4.10)
Then, it holds that
$eℓ*(t)=4D(∞;t)eπiτ(t)(ς(t)+ω(t)+B(t)τ(t))b1(t)−a1(t)+b2(t)−a2(t)θ(ς(t)+ω(t)+B(t)τ(t))θ(0)2.$
(4.11)

We prove this proposition in Sec. VII F. We also explain right after the statement of Proposition 7.1 further below that b1(t) − a1(t) + b2(t) − a2(t) ≠ 0. Another auxiliary function we need is given by

$A(z;t)≔12z−b2(t)z−a2(t)z−b1(t)z−a1(t)1/4+z−b2(t)z−a2(t)z−b1(t)z−a1(t)−1/4$
(4.12)

for $z∈C̄\Jt$, where the branches are chosen so that the summands are holomorphic in $C̄\Jt$ and have value 1 at infinity. As explained in Sec. VII B, this function can be analytically continued through each side of Jt° and is non-vanishing in the domain of the definition.

Given a sequence ${Nn}n∈N$, we define further below in (7.23) functions Θn(zt), which are ratios of Riemann theta functions (4.10) with various arguments. To shorten the presentation of the main results, we only discuss main features of the functions Θn(zt) and defer the detailed construction and description of further properties to Sec. VII.

Proposition 4.4.
Functions Θn(zt) are holomorphic in$C̄\Γ[a1(t),b2(t)]$with at most one zero there. They have continuous traces onIt° that satisfy
$Θn+(s;t)=Θn−(s;t)e2πi(nω(t)+(n−Nn)ς(t)).$
Assume that there exists a constantN*such that |nNn| ≤ N*for all$n∈N$. Then, for anyδ > 0, there exists a constantC(t, δ, N*) such that
$|A(z;t)Θn(z;t)|≤C(t,δ,N*),z∈C̄\⋃e∈Et|z−e|<δ,$
that is, including the traces on Γ[a1(t), b2(t)]. Givenɛ > 0, let$N(t,ε)$be a subsequence of indicesnsuch that Θn(zt) is non-vanishing in {|z| ≥ 1/ɛ}. Then, there exists a constantc(t, ɛ) > 0 such that
$|Θn(∞;t)|≥c(t,ε),n∈N(t,ε).$

As in the case of the Szegő functions $D(z;t)$, it would be convenient for us to renormalize Θn(zt) at infinity. Thus, we set ϑn(zt) ≔ Θn(zt)/Θn(t). Observe that $ϑn(z;t)DNn−n(z;t)enQ(z;t)$ are functions holomorphic in $C\Jt$.

Proposition 4.4 has substance only if the sets $N(t,ε)$ have infinite cardinality. Recall $B(t)$ from (4.9). It follows from the general theory of Riemann surfaces (see Sec. VII A) that $Im(B(t))>0$. In particular, any $s∈C$ can be uniquely written as $x+B(t)y$ for some $x,y∈R$.

Proposition 4.5.

Write$ς(t)=:x(t)+B(t)y(t)$,$x(t),y(t)∈R$. It holds that

• if$x(t),y(t)∈Z$, then the functions Θn(zt) do not depend on the choice ofNnand at least one of the integersn, n + 1 belongs to$N(t,ε)$;34

• otherwise, for any natural numberN* > 0, there exists a choice ofNnsuch that |nNn| ≤ N*and$N(t,ε)=N$for allɛsmall enough.

On the other hand, if the sequence ${Nn}n∈N$ is fixed, we can claim the following proposition:

Proposition 4.6.
Let${Nn}n∈N$be a sequence such that |nNn| ≤ N*for someN* ≥ 0. The sequence$N(t,ε)$is infinite for allɛsmall enough unless there exist integersd > 0, k, i1, i2, m1, m2such that
$ς(t)=(i1+B(t)i2)/d,ω(t)d=(k−1)i1+m1d,andτ(t)d=(k−1)i2+m2d,$
(4.13)
where at least one of the fractionsi1/d,i2/dis irreducible,dis even,i1andi2are odd, and the sequence {Nn} is such that all but finitely many numbersnkNnare divisible byd/2 but not byd. Moreover, the following special cases take place:
• If there exists an infinite subsequence {nl} such that$Nnl+1−Nnl∈{0,1}$(in particular, this happens whenNn = n, in which case the differences are always equal to 1), then at least one of the integersnl, nl + 1 belongs to$N(t,ε)$.

• If one of the triplesω(t), x(t), 1 orτ(t), y(t), 1 is rationally independent, then at least one of the integersn, n + 1 belongs to$N(t,ε)$.

We prove Propositions 4.4–4.6 in Secs. VII CVII E, respectively.

Theorem 4.1.
Let$t∈Otwo−cut$and${Nn}n=1∞$be a sequence such that |nNn| ≤ N*for someN*fixed. LetPn(zt, N) be defined by (1.10) and (1.11) and
$ψn(z;t)≔Pn(z;t,Nn)e−n(V(z;t)−ℓ*(t))/2.$
Givenɛ > 0, let$N(t,ε)$be as in Proposition 4.4. Then, for all$n∈N(t,ε)$large enough, it holds that35
$ψn(z;t)=AϑnDNn−n(z;t)+Oεn−1enQ(z;t)$
(4.14)
locally uniformly in$C\Jt$; moreover,
$ψn(s;t)=AϑnDNn−n+(s;t)enQ+(s;t)+AϑnDNn−n−(s;t)enQ−(s;t)+Oεn−1$
(4.15)
locally uniformly onJt°.

Recall that each ϑn(zt) might have a single zero in $C\Jt$. If these zeros accumulate at a point z* along a subsequence of $N(t,ε)$, then the polynomials Pn(zt, Nn) will have a single zero approaching z* along this subsequence by (4.14) and Rouche’s theorem. With this exception, it also follows from (4.14) that Pn(zt, Nn) are eventually zero free on compact subsets $C\Jt$. The main part of the Proof of Theorem 4.1 is carried out in Sec. VIII with auxiliary details relegated to Secs. VI and VII.

The functions ϑn(zt) were defined as pull-backs from one of the sheets (a copy of $C̄\Jt$) of certain ratios of Riemann theta functions defined on the (two-sheeted) Riemann surface of Q1/2(zt). To describe asymptotics of the recurrence coefficients, we shall need the pull-backs from the other sheet as well, which, for the purpose of the next theorem, we denote by $ϑn*(z;t)$. To complicate matters further, our analysis requires another related family of ratios of theta functions, which, in the next theorem, we will denote by $ϑ̃n(z;t)$ [see (7.24) for the rigorous definition of these functions]. The Proof of Proposition 4.4, in fact, shows that $ϑn*(∞;t)=Oε(1)$ and $ϑ̃n(∞;t)=Oε(1)$ for $n∈N(t,ε)$.

Theorem 4.2.
Let$Sk(t)≔∑i=12bik(t)−aik(t)$fork ∈ {1, 2}. In the setting of Theorem 4.1, it holds for all$n∈N(t,ε)$large enough that
$hn(t,Nn)=π2e−nℓ*(t)S1(t)D2(n−Nn)(∞;t)ϑn*(∞;t)+Oεn−1,γn2(t,Nn)=116S12(t)ϑn*(∞;t)ϑ̃n(∞;t)+Oεn−1,$
(4.16)
where$D(∞;t)$is as in Proposition 4.1,*(t) as in(4.7), and$Oε(⋅)$is interpreted in the same way as in Theorem 4.1. Furthermore, it holds that
$βn(t,Nn)=ϑn*(∞;t)S2(t)2S1(t)+ddzlogϑn(1/z;t)+logϑn*(1/z;t)z=0+Oεn−1ϑn*(∞;t)+Oεn−1,$
(4.17)
where we agree that$ϑn*(∞;t)ddzlogϑn*(1/z;t)z=0=ddzϑn*(1/z;t)z=0$even when$ϑn*(∞;t)=0$.

Of course, if there exists a subsequence of $N(t,ε)$ along which the values $ϑn*(∞;t)$ are separated away from zero, then formula (4.17) can be further simplified. However, such a subsequence might not exist [in particular, one can deduce from (7.7) and (7.20) that it does not exist when Nn = n and ω(t) = τ(t) = 1/2]. Note that formulas (4.16) are not immediately reflective of the equality in (1.14). However, as we point out further in (9.12), the leading term of the asymptotics of $γn2(t,Nn)$ can be rewritten to make this connection more transparent. Theorem 4.2 is proven in Sec. IX, but its proof heavily relies on the material of Sec. VIII.

In this section, we prove Theorem 3.2. We do it in several steps. In Sec. V A, we gather results about quadratic differentials that will be important to us throughout the proof. In Sec. V B, we show the validity of formula (3.5); that is, we prove that we are indeed in the two-cut case when $t∈Otwo−cut$. In Sec. V C, we show that the critical and critical orthogonal graphs of

$ϖt(z)≔−Q(z;t)dz2$

do look like as depicted in Fig. 6. In Sec. V D, we describe the dependence of the zeros of Q(zt) on t by showing that the variational condition (2.2) and the S-property (2.6) yield that the zeros satisfy a certain system of real equations with non-zero Jacobian [see (5.17) and (5.18)] and that this system is, in fact, uniquely solved by them. Finally, in Sec. V E, we establish the limits in (3.6).

To start, let us also recall the following important result, known as Teichmüller’s lemma [see Ref. 30 (Theorem 14.1)]. Let P be a geodesic polygon of a quadratic differential, that is, a Jordan curve in $C̄$ that consists of a finite number of trajectories and orthogonal trajectories of this differential. Then, it holds that

$∑z∈P1−θ(z)2+ord(z)2π=2+∑z∈int(P)ord(z),$
(5.1)

where ord(z) is the order of z with respect to the considered differential and θ(z) ∈ [0, 2π], zP, is the interior angle of P at z. Both sums in (5.1) are finite since only critical points of the differential have a non-zero contribution.

Let us briefly recall the main properties of the differential ϖt(z). The only critical points of ϖt(z) are the zeros of Q(zt) and the point at infinity. Regular points have order 0, the order of a zero of Q(zt) is equal to its multiplicity, and infinity is a critical point of order −8. Through each regular point passes exactly one trajectory and one orthogonal trajectory of ϖt(z), which are orthogonal to each other at the point. Two distinct (orthogonal) trajectories meet only at critical points [see Ref. 30 (Theorem 5.5)]. As Q(zt) is a polynomial, no finite union of (orthogonal) trajectories can form a closed Jordan curve, while a trajectory and an orthogonal trajectory can intersect at most once [see Ref. 29 (Lemma 8.3)]. Furthermore, (orthogonal) trajectories of ϖt(z) cannot be recurrent (dense in two-dimensional regions) [see Ref. 28 (Theorem 3.6)]. From each critical point of order m > 0, there emanate m + 2 critical trajectories whose consecutive tangent lines at the critical point form an angle 2π/(m + 2). Furthermore, since infinity is a pole of order 8, the critical trajectories can approach infinity only in six distinguished directions, namely, asymptotically to the lines Lπ/6, Lπ/6, and Lπ/2, where Lθ = {z: z = reiθ, r ∈ (−, )}. In fact, there exists a neighborhood of infinity such that any trajectory entering it necessarily tends to infinity [see Ref. 30 (Theorem 7.4)]. This discussion also applies to orthogonal trajectories. In particular, they can approach infinity asymptotically to the lines L0, Lπ/3, and L2π/3 only.

Denote by $G$ the critical graph of ϖt(z), that is, the totality of all the critical trajectories of ϖt(z). Then [see Ref. 28 (Theorem 3.5)], the complement of $G$ can be written as a disjoint union of either half-plane or strip domains. Recall that a half-plane (or end) domain is swept by trajectories unbounded in both directions that approach infinity along consecutive critical directions. Its boundary is connected and consists of a union of two unbounded critical trajectories and a finite number (possibly zero) of short trajectories of ϖt(z). The map $z↦∫z−ϖt$ maps end domains conformally onto half planes ${z∈C|Re(z)>c}$ for some $c∈R$ that depends on the domain and extends continuously to the boundary. Similarly, a strip domain is again swept by trajectories unbounded in both directions, but its boundary consists of two disjoint ϖt(z)-paths, each of which is comprised of two unbounded critical trajectories and a finite number (possibly zero) of short trajectories. The map $z↦∫z−ϖt$ maps strip domains conformally onto vertical strips ${w∈C|c1 for some $c1,c2∈R$ depending on the domain and extends continuously to their boundaries. The number c2c1 is known as the width of a strip domain and can be calculated in terms of ϖt(z) as

$Re∫pq−ϖt,$
(5.2)

where p, q belong to different components of the boundary of the domain.

Assume to the contrary that for a given $t∈Otwo−cut$, we are in one-cut case. That is, there exists a choice of a, b, and c such that the polynomial Q(zt) from (2.6) has the form (3.3). It follows from (2.6) in conjunction with (1.11) that

$Q(z;t)=−z2+t2−1z+Oz−22=(z2−t)24+z+C$
(5.3)

for some constant C. Then, by equating the coefficients in (3.3) and (5.3), we obtain a system of equations

$a+b+2c=0,ab+c2+2(a+b)c=−2t,2abc+(a+b)c2=−4.$
(5.4)

Setting x ≔ (a + b)/2 and eliminating the product ab from the second and third relations in (5.4) yield

$x3−tx−1=0,$
(5.5)

which is exactly the equation appearing before (3.2). Given any solution of (5.5), say, x(t), then a(t), b(t), and c(t) are necessarily expressed via (3.4). Theorem 2.1 and the variational condition (2.2) imply that there must exist a contour $Γt∈T$ [this class of contours was defined right after (2.3)] such that

$U(z;t)≤0for allz∈Γt$
(5.6)

[see (2.8)]. In what follows, we shall show that no such contour exists in $T$ for any of the three possible choices of x(t) solving (5.5) when $t∈Otwo−cut$ and Q(zt) is given by (3.3) and (3.4).

In accordance with the above strategy, observe that the solutions of (5.5) can be written as

$xk(t)=uk(t)+t3uk(t),uk(t)≔12−14−t3271/3e2kπi/3,$
(5.7)

k ∈ {0, 1, 2}, with all branches being principal. It can be readily verified that x1(t) is analytic in $C\(e2πi/3S̄,S̄)$ (here, $⋅̄$ means topological closure) [see (3.2) for the definition of the ray S] and

$x0(t)=e4πi/3x1te−2πi/3,x1(t)=e4πi/3x1t̄e2πi/3̄,x2(t)=x1t̄̄.$
(5.8)

Furthermore, noting that the function x(t), defined after (3.2), maps $Ωone−cut$ onto $Oone−cut$, it can be easily checked that x(t) is evaluated as shown on Fig. 7, where the dashed lines are the chosen branch cuts of x1(t). In particular, x(t) can be analytically continued across $Csplit$, $Cbirtha$, and $Cbirthb$ [see (3.2) and Fig. 7]. In what follows, we consider what happens in the case of each of these continuations.

Continue x(t) into $Otwo−cut$ by either x2(t) or x0(t), that is, analytically across either $Cbirthb$ or $Cbirtha$ (see Fig. 7). The first and the last symmetries in (5.8) then yield that ϖt(z) is either equal to

$ϖt̄(z̄)̄orϖte−2πi/3ze−4πi/3.$

Since the set $Otwo−cut$ is symmetric with respect to the line Lπ/3, its rotation by −2π/3 is equal to its reflection across the real axis. Thus, the critical graph ϖt(z) when $t∈Otwo−cut$ and x(t) is continued by either x2(t) or x0(t) being equal to the reflection across the real axis or the rotation by 4π/3 of the critical graph of $ϖt*(z)$ for some t* such that $t̄*∈Otwo−cut$. These graphs were studied in Ref. 3 (Theorems 3.2 and 3.4) and determined to have the structure as depicted on Figs. 4(a)4(c) (or the reflection of these three panels across the line L2π/3). A direct examination shows that none of these critical graphs form a curve in $T$ for which (5.6) holds (such a curve must belong to the closure of the gray regions on Fig. 4).

Suppose now that we continue x(t) by x1(t), that is, analytically across $Csplit$. For such a choice of x(t), the critical graph of ϖt(z) was studied in Ref. 21 when tLπ/3 (in which case the critical graph is symmetric with respect to L2π/3). In particular, it was shown that x(t) ∈ L2π/3, no union of critical trajectories join a and b, and no critical trajectory of ϖt(z) crosses the line L2π/3 when $t∈Lπ/3∩Otwo−cut$ [see Ref. 21 (Lemma 3.2)]. Since critical trajectories cannot intersect, can approach infinity asymptotically to the lines Lπ/6, Lπ/2, and Lπ/6 only, and must obey Teichmüller’s lemma (5.1), the critical graph of ϖt(z) must be as on Fig. 8. Clearly, this critical graph does not yield a curve in $T$ for which (5.6) holds. Thus, to complete the proof, we need to argue that the structure of the critical trajectories of ϖt(z) remains the same for all $t∈Otwo−cut$ in the considered case. Observe that it is enough to show that all the trajectories out of b approach infinity.

Recall that the trajectories emanating out of b are part of the level set ${U(z;t)=0}$ [see (2.8)]. When $t∈Lπ/3∩Otwo−cut$, these trajectories approach infinity at the angles −π/6, π/6, and π/2 (see Fig. 8). Since the values of $U(z;t)$ analytically depend on t, the same must be true in a neighborhood of each such t. This will remain so until one of the trajectories hits a critical point different from the one at infinity. As can be seen from Fig. 8, this critical point must necessarily be c = −x [see (3.4)]. That is, as long as $U(−x;t)≠0$, the trajectories out of b will asymptotically behave as on Fig. 8. When x(t) is continued into $Otwo−cut$ by x1(t), its values lie within the gray region on Fig. 9(b) [see also Fig. 2(b)]. The values 2x3(t) lie within the gray region on Fig. 9(a) [see also Fig. 2(a)]. It was verified in Ref. 3 (Sec. 5.3) that

$U(−x;t)=Re23∫−12x31+1s3/2ds,$

where the path of integration lies within the shaded domain on Fig. 9(a). Hence, $U(−x;t)=0$ if and only if 2x3 belongs to a trajectory of −(1 + 1/s)3ds2 emanating from −1. These trajectories are drawn on Figs. 2 and 9(a) (black lines). Thus, $U(−x;t)≠0$ in the considered case as claimed.

FIG. 9.

Shaded regions represent the domains within which $2x13(t)$ [panel (a)] and x1(t) [panel (b)] change when $t∈Otwo−cut$.

FIG. 9.

Shaded regions represent the domains within which $2x13(t)$ [panel (a)] and x1(t) [panel (b)] change when $t∈Otwo−cut$.

Close modal

Let, as usual, Q(zt) be the polynomial guaranteed by Theorem 2.1. According to what precedes, it has the form (3.5) when $t∈Otwo−cut$. Recall the properties of the differential ϖt(z) = −Q(zt)dz2 described at the beginning of Sec. V A. In particular, it has four critical points of order 1, which, for a moment, we label as z1(t), z2(t), z3(t), z4(t) [these are the zeros of Q(zt)], a critical point of order −8 at infinity, and no other critical points. It follows from Theorem 2.1(4) and (2.2) with (2.8) that

$Re∫Γt[zi(t),zj(t)]Q+1/2(z;t)dz=0,$
(5.9)

where Γt[zi(t), zj(t)] is the subarc of Γt with endpoints zi(t), zj(t) and $Q+1/2(z;t)$ is the trace of Q1/2(zt) on the positive side of Γt. Equation (5.9) implies the existence of three short critical trajectories of ϖt(z). Indeed, if all three critical trajectories out of a zero zi(t) approach infinity, then zi(t) must belong to a boundary of at least one strip domain. Let zj(t) be a different zero of Q(zt) belonging to the other component of the boundary of this strip domain. Then, it follows from (5.2) and (5.9) that the width of this strip domain is 0, which is impossible. Thus, each zero of Q(zt) must be coincident with at least one short trajectory. Therefore, either there is a zero, say, z4(t), connected by short trajectories to the remaining three zeros or there are at least two short trajectories connecting two pairs of zeros. In the latter case, label these zeros by a1(t), b1(t) and a2(t), b2(t). If the other two trajectories out of both a1(t) and b1(t) approach infinity, one of these zeros again must belong to the boundary of a strip domain with either a2(t) or b2(t) belonging to the other component of the boundary. As before, (5.9) yields that the width of this strip domain is 0, which, again, is impossible. Thus, in this case, there also exists a third short critical trajectory. Then, we choose a labeling of the zeros so that b1(t) and a2(t) are connected by this trajectory.

Since short critical trajectories cannot form closed curves, there cannot be any more of them. That is, the remaining critical trajectories are unbounded. Consider the two unbounded critical trajectories out of z1(t) in the case where short ones form a threefold [see Fig. 10(c)]. Since critical trajectories cannot intersect and the remaining zeros are connected to z1(t) by short critical trajectories, the unbounded critical trajectories out of z1(t) delimit a half-plane domain and, in particular, must approach infinity along consecutive critical directions [those are given by the angles (2k + 1)π/6, k ∈ {0, …, 5}; see Sec. V A]. Clearly, the same is true for the unbounded critical trajectories out of z2(t) and z3(t) as well as for the unbounded critical trajectories out of a1(t), b2(t), and the union of the unbounded critical trajectory out of b1(t), the short critical trajectory connecting b1(t) to a2(t), and the unbounded critical trajectory out of a2(t) in the case where short critical trajectories form a Jordan arc (see Fig. 10).

FIG. 10.

Geometries of the critical graph of ϖt(z). Shaded regions represent the open set ${U(z;t)<0}$, the white regions represent the open ${U(z;t)>0}$, and the red dashed arcs form Γt\Jt.

FIG. 10.

Geometries of the critical graph of ϖt(z). Shaded regions represent the open set ${U(z;t)<0}$, the white regions represent the open ${U(z;t)>0}$, and the red dashed arcs form Γt\Jt.

Close modal

Now, let $U(z;t)$ be given by (2.8). Clearly, $U(z;t)$ is a subharmonic function that is equal to zero on Jt [see (2.2)]. Since $U(z;t)$ must have the same sign on both sides of each subarc of Jt by (2.4), it follows from the maximum principle for subharmonic functions that it is positive there. Furthermore, since trajectories of ϖt(z) cannot form a closed Jordan curve, all the connected components of the open set ${U(z;t)<0}$ must necessarily extend to infinity. Since Re(V(zt)) is the dominant term of $U(z;t)$ around infinity [see (2.8)], for any δ > 0, there exists R > 0 sufficiently large so that

$Sπ/3,δ∪Sπ,δ∪S−π/3,δ∩{|z|>R}⊂{U(z;t)<0},S0,δ∪S2π/3,δ∪S−2π/3,δ∩{|z|>R}⊂{U(z;t)>0},$
(5.10)

where Sθ,δ ≔ {|arg(z) − θ| < π/6 − δ}. Altogether, the critical graph of ϖt(z) must look like either on Fig. 6 or on Fig. 10.

It remains to show that ϖt(z) cannot have the critical graph as on any of the panels of Fig. 10. To this end, recall that the contour Γt must contain Jt and two unbounded arcs extending to infinity in the directions π/3 and π (red dashed unbounded arcs on Fig. 10). Let Γ* be obtained from Γt by dropping the short trajectory that is a part of Jt and whose removal keeps Γ* connected (this can be done for any of the panels on Fig. 10). Observe that Γ* also belongs to $T$. Let μ* be the weighted equilibrium distribution on Γ* as defined in Definition 2.1. Since Γ* ⊂ Γt, it holds that $μ*∈M(Γt)$. Moreover, since μ*μt, $EV(Γ*)=EV(μ*)>EV(μt)=EV(Γt)$ [see (2.1)]. However, the last inequality clearly contradicts (2.5).

We have shown that the critical graph of ϖt(z) must look like on Fig. 6. As the critical orthogonal and critical trajectories cannot intersect, the structure of the critical orthogonal graph is uniquely determined by structure of the critical graph. Now, we can completely fix the labeling of the zeros of Q(zt) by given the label a1(t) to one that is incident with the orthogonal critical trajectory extending to infinity asymptotically to the ray arg(z) = π.

We start with some general considerations. Let f(z) and g(z) be analytic functions of z = x + iy. Consider a determinant of the form

$D=∂xRe(f)∂yRe(f)*∂xIm(f)∂yIm(f)*∂xRe(g)∂yRe(g)*,$

where the entries of the third column are not important for the forthcoming computation. Due to Cauchy–Riemann relations, it holds that f′ = x Re(f) + ix Im(f) = y Im(f) − iy Re(f). Therefore,

$D=Re(f′)−Im(f′)*Im(f′)Re(f′)*Re(g′)−Im(g′)*=f′if′*Im(f′)Re(f′)*Re(g′)−Im(g′)*=i2f′if′*f′̄−if′̄*Re(g′)−Im(g′)*$

by adding the second row times i to the first one and then multiplying the second row by −2i and adding the first row to it. It further holds that

$D=i22f′if′*0−if′̄*g′−Im(g′)*=i22f′0*0−if′̄*g′−ig′̄/2*=12f′0*0f′̄*g′g′̄*,$

where we added the second column times −i to the first one, then added the first column times −i/2 to the second one, and then factored 2 from the first column, −i from the second one, and 1/2 from the third row.

Now, let fj(z1, z2, z3, z4), j ∈ {1, 2, 3, 4, 5}, be analytic functions in each variable zi = xi + iyi. We would like to compute the Jacobian of the following system of real-valued functions of x1, y1, …, x4, y4:

$Re(f1),Im(f1),Re(f2),Im(f2),Re(f3),Im(f3),Re(f4),Re(f5).$
(5.11)

That is, we are interested in

$Jac=Re(f11)−Im(f11)Re(f12)−Im(f12)Re(f13)−Im(f13)Re(f14)−Im(f14)Im(f11)Re(f11)Im(f12)Re(f12)Im(f13)Re(f13)Im(f14)Re(f14)Re(f21)−Im(f21)Re(f22)−Im(f22)Re(f23)−Im(f23)Re(f24)−Im(f24)Im(f21)Re(f21)Im(f22)Re(f22)Im(f23)Re(f23)Im(f24)Re(f24)Re(f31)−Im(f31)Re(f32)−Im(f32)Re(f33)−Im(f33)Re(f34)−Im(f34)Im(f31)Re(f31)Im(f32)Re(f32)Im(f33)Re(f33)Im(f34)Re(f34)Re(f41)−Im(f41)Re(f42)−Im(f42)Re(f43)−Im(f43)Re(f44)−Im(f44)Re(f51)−Im(f51)Re(f52)−Im(f52)Re(f53)−Im(f53)Re(f54)−Im(f54),$

where $fji≔∂zifj$. By performing the same row and column operations as for the determinant D above, we get that

$Jac=12if110f120f130f1400f̄110f̄120f̄130f̄14f210f220f230f2400f̄210f̄220f̄230f̄24f310f320f330f3400f̄310f̄320f̄330f̄34f41f̄41f42f̄42f43f̄43f44f̄44f51f̄51f52f̄52f53f̄53f54f̄54$

with the constant in front of the determinant coming from (i/2)3 × 24 × (−i)4 × (1/2)2, where the first factor is due to multiplications of the second, fourth, and sixth rows by −2i, the second is due to factoring 2 from the odd columns, the third one comes from factoring −i from the even columns, and the fourth factor is due to factoring 1/2 from the last two rows. Assume further that

$f1(z1,z2,z3,z4)=z1+z2+z3+z4,f2(z1,z2,z3,z4)=z1z2+z1z3+z1z4+z2z3+z2z4+z3z4,f3(z1,z2,z3,z4)=z2z3z4+z1z3z4+z1z2z4+z1z2z3.$
(5.12)

Then, by using the above explicit expressions and subtracting the first (respectively, second) column from the third, fifth, and seventh (respectively, fourth, sixth, and eighth), we get that

$Jac=12i1000000001000000**z1−z20z1−z30z1−z40**0z̄1−z̄20z̄1−z̄30z̄1−z̄4**z1−z2(z3+z4)−10z1−z3(z2+z4)−10z1−z4(z2+z3)−10**0z̄1−z̄2(z̄3+z̄4)−10z̄1−z̄3(z̄2+z̄4)−10z̄1−z̄4(z̄2+z̄3)−1**(z1−z2)g42(z̄1−z̄2)ḡ42(z1−z3)g43(z̄1−z̄3)ḡ43(z1−z4)g44(z̄1−z̄4)ḡ44**(z1−z2)g52(z̄1−z̄2)ḡ52(z1−z3)g53(z̄1−z̄3)ḡ53(z1−z4)g54(z̄1−z̄4)ḡ54,$

where $gji(z1,z2,z3,z4)≔(z1−zi)−1(fji−fj1)(z1,z2,z3,z4)$, j ∈ {4, 5} and i ∈ {2, 3, 4}. Hence,

$Jac=|z1−z2|2|z1−z3|2|z1−z4|22i101010010101z3+z40z2+z40z2+z300z̄3+z̄40z̄2+z̄40z̄2+z̄3g42ḡ42g43ḡ43g44ḡ44g52ḡ52g53ḡ53g54ḡ54.$

Absolutely analogous computation now implies that

$Jac=|z1−z2|2|z1−z3|2|z1−z4|2|z2−z3|2|z2−z4|22i10100101h43h̄43h44h̄44h53h̄53h54h̄54,$

where $hji(z1,z2,z3,z4)≔(z2−zi)−1(gji−gj2)(z1,z2,z3,z4)$, j ∈ {4, 5} and i ∈ {3, 4}. The above expression immediately yields that

$Jac=∏i
(5.13)

where $kj(z1,z2,z3,z4)≔(z3−z4)−1(gj4−gj3)(z1,z2,z3,z4)$, j ∈ {4, 5}. Finally, let

$w(z)≔(z−z1)(z−z2)(z−z3)(z−z4)$

be a branch such that $w(z)=z2+O(z)$ as z with branch cuts γ12 and γ34 that are bounded, disjoint, and smooth and where γij connects zi to zj. Furthermore, select a smooth arc γ32 disjoint (except for the endpoints) from the previous two. Set

$f4(z1,z2,z3,z4)≔4∫γ12w(z)dzandf5(z1,z2,z3,z4)≔4∫γ32w(z)dz,$
(5.14)

where we integrate w(z) on the positive side of γ12. Let O ⊂ {zizj, ij, i, j ∈ {1, 2, 3, 4}} be a domain such that there exist arcs γij(z1, z2, z3, z4) with the above properties for each (z1, z2, z3, z4) ∈ O, which, in addition, possess parameterizations that depend continuously on each variable z1, z2, z3, z4. Then, the functions fj(z1, z2, z3, z4), j ∈ {4, 5}, are analytic in each variable zi for (z1, z2, z3, z4) ∈ O. Furthermore, it can be readily computed that

$gji=2∫w(z)dz(z−z1)(z−zi),hji=−2∫w(z)dz(z−z1)(z−z2)(z−zi),andkj=2∫dzw(z),$

where the integrals are taken over γ12 when j = 4 and γ32 when j = 5. Trivially, (5.13) can be rewritten as

$Jac=4∏i
(5.15)

Now, consider the Riemann surface $S≔z≔(z,w):w2=(z−z1)(z−z2)(z−z3)(z−z4)$. Denote by $π:S→C̄$ the natural projection π(z) = z and write w(z) for a rational function on $S$ such that z = (z, w(z)). Let βπ−1(γ12) and απ−1(γ32). Orient these cycles so that

$2∫γ12dzw(z)=∮βdzw(z)and2∫γ32dzw(z)=∮αdzw(z).$

Observe that the cycles α, β form the right pair at the point of their intersection and that $S\{α∪β}$ is simply connected. Hence, the cycles α, β form a homology basis on $S$. Since the genus of $S$ is 1, it has a unique (up to multiplication by a constant) holomorphic differential. It is quite easy to check that this differential is dz/w(z). Hence, we get from (5.15) that

$Jac=∏i0$
(5.16)

when (z1, z2, z3, z4) ∈ O, where the last inequality was shown by Riemann.

Now, let $Q(z;t)=14(z−a1(t))(z−b1(t))(z−a2(t))(z−b2(t))$ be the polynomial from Theorem 2.1. It can be easily deduced from (2.6) that

$f1(a1,b1,a2,b2)=0,f2(a1,b1,a2,b2)=−2t,f3(a1,b1,a2,b2)=−4,$
(5.17)

where the functions fi(z1, z2, z3, z4), i ∈ {1, 2, 3}, are given by (5.12).

Fix $t*∈Otwo−cut$ and let δ* > 0 be small enough so that all four disks ${|z−zi*|≤δ*}$ are disjoint, where $z1*=a1(t*)$, $z2*=b1(t*)$, $z3*=a2(t*)$, and $z4*=b2(t*)$. Let $O≔{(z1,z2,z3,z4):|zi−zi*|<δ*}$, $si*$ be the point of intersection of ${|z−zi*|=δ*}$ and $Jt*$, i ∈ {1, 2, 3, 4}, and $ui*$ be the point of intersection of ${|z−zi*|=δ*}$ and Γ(b1(t*), a2(t*)), i ∈ {2, 3}, where, as usual, Γ(a, b) is the subarc of the trajectories of ϖt(z) connecting a and b. Then, we can choose $γ12=[z1,s1*]∪Γ(s1*,s2*)∪[s2*,z2]$, $γ32=[z3,u3*]∪Γ(u3*,u2*)∪[u2*,z2]$, and $γ34=[z3,s3*]∪Γ(s3*,s4*)∪[s4*,z4]$, where [a, b] is the line segment connecting a and b in $C$ (see Fig. 11). Clearly, the arcs γij continuously depend on (z1, z2, z3, z4) ∈ O. Now, the relations (5.9) can be rewritten as

$Ref4a1,b1,a2,b2=0andRef5a1,b1,a2,b2=0$
(5.18)

with fj(z1, z2, z3, z4), j ∈ {4, 5}, given by (5.14), where we set w(z) ≔ Q1/2(zt). It follows from (5.16) and the implicit function theorem that there exists a neighborhood of t* in which system (5.17) and (5.18) is uniquely solvable and the solution, say, $(a1*(t),b1*(t),a2*(t),b2*(t))$, is such that the real and imaginary parts of $ai*(t),bi*(t)$ are real analytic functions of Re(t) and Im(t) for t in this neighborhood. These local solutions are unique only locally around the point (a1(t*), b1(t*), a2(t*), b2(t*)), and we still need to argue that they do coincide with the zeros ai(t), bi(t) of Q(zt) [of course, it holds that $ai*(t*)=ai(t*)$ and $bi*(t*)=bi(t*)$].

In what follows, we always assume that t belongs to a disk centered at t* of small enough radius so that the functions $ai*(t),bi*(t)$ are defined and continuous in this disk. Let

$Q*(z;t)≔14(z−a1*(t))(z−b1*(t))(z−a2*(t))(z−b2*(t))andϖt*(z)≔−Q*(z;t)dz2.$
(5.19)

Furthermore, let $U*(z;t)$ be defined as in (2.8) with Q(zt) replaced by Q*(zt). For the moment, choose the branch cut for Q*(z;t)1/2 as in the paragraph between (5.17) and (5.18). Each function $U*(z;t)$ is harmonic off the chosen branch cut and can be continued harmonically across it by $−U*(z;t)$. Moreover, it follows immediately from their definition that the functions $U*(z;t)$ are uniformly bounded above and below on any compact set for all considered values of the parameter t. Thus, they converge to $U(z;t*)$ locally uniformly in $C\{a1(t*),b1(t*),a2(t*),b2(t*)}$ as tt*. Since the critical graph of $ϖt*(z)$ is the zero-level set of $U*(z;t)$, it converges to the critical graph of $ϖt*(z)$ in any disk {|z| < R}. Due to relations (5.18), the argument at the beginning of Sec. V C also shows that the critical graph of $ϖt*(z)$ has three short critical trajectories, which, due to uniform convergence, necessarily connect $a1*(t)$ to $b1*(t)$, $b1*(t)$ to $a2*(t)$, and $a2*(t)$ to $b2*(t)$ (the disk around t* can be decreased if necessary). Thus, arguing as in Sec. V C and using uniform convergence, we can show that Fig. 6 also schematically represents the critical and critical orthogonal graphs of $ϖt*(z)$. Moreover, let us now take the branch cut for Q*(z;t)1/2, say, $Jt*$, along the short critical trajectories of $ϖt*(z)$ connecting $ai*(t)$ to $bi*(t)$. Then, the shading on Fig. 6 corresponds to regions where $U*(z;t)$ is positive (white) and negative (gray).

Define $Γt*$ to be the union of the critical orthogonal trajectory of $ϖt*(z)$ that connects infinity to $a1*(t)$, its short critical trajectories, and the critical orthogonal trajectory that connects $b2*(t)$ to infinity. Orient it so that the positive direction proceeds from $a1*(t)$ to $b2*(t)$. Let the measures $μt*$ be given by (2.7) with Q(zt) replaced by Q*(zt) and Jt replaced by $Jt*$. Clearly, each $μt*$ is a positive measure. Moreover, it has a unit mass by the Cauchy theorem and since $Q*(z;t)1/2=(z2−t)/2+1/z+O(1/z2)$ due to (5.17) [see also (5.3)]. Thus, it holds that

$F*(z;t)≔Q*(z;t)1/2+V′(z;t)2−∫dμt*(s)z−s=Oz−2$

as z and F*(zt) is holomorphic in $C̄\Jt*$. It follows from the well-known behavior of Cauchy integrals of smooth densities [see Ref. 36 (Sec. 1.8)] that the traces of F*(zt) on $Jt*$ are bounded. It further follows from the Sokhotski–Plemelj formulas [see Ref. 36 (Sec. 1.4)] that

$F+*(s;t)−F−*(s;t)=Q+*(s;t)1/2−Q−*(s;t)1/2−∫Q+*(w;t)1/2w−zdwπi++∫Q+*(w;t)1/2w−zdwπi−=Q+*(s;t)1/2−Q−*(s;t)1/2−2Q+*(s;t)1/2≡0$

for $s∈Jt*$. Hence, F*(zt) is an entire function and, therefore, is identically zero. This observation, in particular, yields that

$U*(z;t)≔Re2∫b2*(t)zQ*(s;t)1/2ds=ℓt*−Re(V(z;t))−2Uμt*(z)$

for some constant $ℓt*$ [see also (2.8)]. Since $U*(z;t)$ can be harmonically continued across $Jt*$ by $−U*(z;t)$, we get that $μt*$ satisfies (2.4); that is, $Jt*$ has the S-property in the field Re(V(zt)). Since $Γt*∈T$, it follows from the uniqueness part of Theorem 2.1(2) that $μt*=μt$. In particular, $ai*(t)=ai(t)$ and $bi*(t)=bi(t)$, i ∈ {1, 2}.

Since any compact subset of $Otwo−cut$ can be covered by finitely many disks where the above considerations hold, the functions ai(t), bi(t) continuously depend on $t∈Otwo−cut$, and moreover, their real and imaginary parts are real analytic functions of Re(t) and Im(t).

Let us now show that at no point in $Otwo−cut$ are the functions ai(t), bi(t) analytic in t. Suppose, for the sake of contradiction, that one of these functions, say, b2(t), is analytic in t at some point $t0∈Otwo−cut$. Then, all the other endpoint functions, a1(t), b1(t), and a2(t), are analytic in t at t0 as well. Indeed, from (5.12) and (5.17), we have that

$a1+b1+a2=−b2,a1b1+a1a2+b1a2=−2t−b2(a1+b1+a2)=−2t+b22,a1b1a2=−4−b2(a1b1+a1a2+b1a2)=−4−b2(−2t+b22)=−4+2tb2−b23.$

Hence, a1(t), b1(t), a2(t) are roots of the cubic equation

$z3+b2z2+(−2t+b22)z+(4−2tb2+b23)=0.$

Since a1(t), b1(t), a2(t) are pairwise different, they analytically depend on the coefficients of the cubic equation. That is, they are analytic functions of t and b2. This implies that a1(t), b1(t), and a2(t) are analytic in t at t0. The analyticity of a1(t), b1(t), a2(t), and b2(t) yields that the integral

$1πi∫Γt[a1(t),b1(t)]Q+1/2(s;t)ds$

is also an analytic function of t at t0. However, the integral above is equal to −ω(t) [see (4.6)], which is a real number [see (2.7)]. Hence, ω(t) must be constant in U. On the other hand, it follows from (3.6), the proof of which in Subsection V E is independent of the current considerations, that ω(t) is not constant in $Otwo−cut$. We also have already shown that ω(t) is real analytic in Re(t) and Im(t) and, therefore, cannot be locally constant. This contradiction proves that the endpoints a1(t), b1(t), a2(t), and b2(t) are not analytic functions of t at any point in $Otwo−cut$.

It follows from Ref. 37 (Theorem 5.11) that if t remains in a bounded set, so do the zeros of Q(zt). Fix $t*∈∂Otwo−cut$, and let {tm} be a sequence such that tmt* as m. Restricting to a subsequence if necessary, we see that there exist $ai*,bi*$ such that e(tm) → e* as m, where e ∈ {a1, b1, a2, b2}. Clearly, polynomials Q(ztm) converge uniformly on compact subsets of $C$ to Q*(z), the polynomial with zeros $ai*,bi*$ and leading coefficient 1/4. Let ϖ*(z) ≔ −Q*(z)dz2. Then, repeating the argument after (5.19), we get that the critical graphs of $ϖtm(z)$ converge to the critical graph of ϖ*(z) in any disk |z| < R. Furthermore, as stated in Ref. 28 (Theorem 3.3), there exists Rt > 0 such that every trajectory of ϖt(z) entering {|z| > Rt} necessarily remains in {|z| > Rt} and tends to infinity. Moreover, examination of the proof of Ref. 28 (Theorem 3.3) also shows that one can take $Rt=2−1/2maxi|ai(t)|,|bi(t)|$ and that if a trajectory enters S(2k−1)π/3,δ ∩ {|z| > Rt}, k ∈ {0, 1, 2}, it stays in this sector [see (5.10)]. Put $R*≔supmRtm$. As all the differentials $ϖtm(z)$ have structurally “the same” critical graph (see Fig. 6) and all the trajectories of $ϖtm(z)$ and ϖ*(z) entering S(2k−1)π/3,δ ∩{|z| > R*} must remain there, tend to infinity, and not intersect, the unbounded critical trajectories of ϖ*(z) behave like on Fig. 6.

Observe that the first and the third equations of (5.17) [see also (5.12)] must remain true for $a1*,b1*,a2*,b2*$ as well. Hence, we cannot simultaneously have that $a1*=b1*$ and $a2*=b2*$. As short trajectories cannot form loops and their tangent vectors cannot become parallel at each critical point, the above considerations yield that the critical and, therefore, critical orthogonal graphs of ϖ*(t) look like either on Fig. 6 or on Fig. 5 or as the graphs obtained by reflection across the line L2π/3 of the graphs on Fig. 5. Repeating the arguments at the end of Sec. V D, we see that ϖ*(t) gives rise to an S-contour in $T$. The limits in (3.6) now follow from the uniqueness of such a S-contour and Theorem 3.1 [in particular, the critical graph ϖ*(t) cannot look like as on Fig. 6].

In this section, we prove Propositions 4.1 and 4.2 as well as discuss other properties of $Q(z;t)$. We consider the parameter $t∈Otwo−cut$ to be fixed and stop indicating the dependence on t of the various quantities appearing below whenever this does not introduce ambiguity and is convenient.

It follows from (1.11) and (2.6) and the choice of the branch of Q1/2(z) that

$Q1/2(z)=z2−t2+1z+μ1z2+O1z3.$
(6.1)

It further follows from the choice of the constant ς, d1 in (4.1) that

$z+∫It3ςs−zdsQ1/2(s)=z−2tz−3ςd1z2−3ςd2z3+O1z4$

as z. Thus, the analyticity properties of $D(z)$ as well as (4.4) now follow from the fact that the product of the above functions behaves like

$−3V(z)2+1−3ςd12+t2+μ1−3ςd2/2z+O1z2$

as z. Moreover, since $Q+1/2(s)=−Q−1/2(s)$ for sJt, we get the first relation in (4.3). The second relation in (4.3) follows from the Plemelj–Sokhotski formula

$∫Itςs−zdsQ1/2(s)+−∫Itςs−zdsQ1/2(s)−=2πiςQ1/2(z),z∈It°.$

Since the arc It is homologous to the short critical trajectory of −Q(z)dz2 connecting b1 and a2 and Jt,1 = Γ[a1, b1] is such a trajectory (see Fig. 6 and Conventions 1 and 2), the constants τ, ω are indeed real. Let

$g(z)≔∫log(z−s)dμt(s),z∈C\Γeπi∞,b2,$
(6.2)

where we take the principal branch of log(· − s) holomorphic outside of $Γeπi∞,s$ and μt is the equilibrium measure defined in (2.7). It follows directly from definition (6.2) that

$∂zg(z)=∫dμt(s)z−s,$

where, as usual, z ≔ (x − iy)/2. Therefore, it can be deduced from (2.2) and (2.6) that

$g(z)=V(z)−ℓ*2+∫b2zQ1/2(s)ds=V(z)−ℓ*2+Q(z),$
(6.3)

where, as usual, we take the branch $Q1/2(z)=12z2+O(z)$, * is a constant such that the equality holds at b2 [note that Re(*) = ; see (2.2)], and $Q(z)$ is given by (4.5). Property (4.7) clearly follows from (6.2) and (6.3). In view of (6.3), let us define

$ϕe(z)≔2∫ezQ1/2(s)ds,e∈{a1,b1,a2,b2},$
(6.4)

holomorphically in $C\Γ(eπi∞,b2]$ when e = b2, in $C\Γ[a1,eπi/3∞)$ when e = a1, and in $C\Γ(eπi∞,b1)∪Γ(a2,eπi/3∞)$ when e ∈ {b1, a2}. Clearly, $ϕb2(z)=2Q(z)$. One can readily check that

$ϕb2(z)=ϕa2(z)±2πi(1−ω),ϕb1(z)±2πi(1−ω)+2πiτ,ϕa1(z)±2πi+2πiτ,z∈C\Γ,$
(6.5)

where the plus sign is used if z lies to the left of Γ and the minus sign if z lies to the right of it, and

$ϕb2±(s)=±2πiμtΓ[s,b2],s∈Γ(a2,b2),±2πiμtΓ[s,b2]+2πiτ,s∈Γ(a1,b1).$
(6.6)

The jump relations in (4.8) now easily follow from (6.5) and (6.6).

For future use, let us record that (6.3), (6.5), and (6.6) imply that

$g+(s)−g−(s)=0,s∈Γ(b2,eπi/3∞),±ϕb2±(s),s∈Γ(a2,b2),2πi(1−ω),s∈Γ(b1,a2),±ϕb2±(s)−2πiτ,s∈Γ(a1,b1),2πi,s∈Γ(eπi∞,a1),$
(6.7)

and that

$g+(s)+g−(s)−V(s)+ℓ*=ϕb2(s),s∈Γ(b2,eπi/3∞),0,s∈Γ(a2,b2),ϕa2(s),s∈Γ(b1,a2),2πiτ,s∈Γ(a1,b1),ϕa1(s)+2πiτ,s∈Γ(eπi∞,a1).$
(6.8)

Given e ∈ {a1, b1, a2, b2}, let

$Ue≔z:|z−e|<δeρ(t)/3,$
(6.9)

where δe ∈ (0, 1] to be adjusted later, and we shall specify the function ρ(t) at the end of this subsection. Set

$Je≔Ue∩JtandIe≔Ue∩(Γt\Jt),$
(6.10)

where the arcs Je and Ie inherit their orientation from Γt, and we assume that the value of ρ(t) is small enough so that these arcs are connected and that Ie is a subarc of the orthogonal critical trajectory of −Q(z)dz2 emanating from e (see the remarks about Γt at the very beginning of Sec. IV). The latter fact and Theorem 3.2 yield that

$ϕe(s)<0,s∈Ie$
(6.11)

(see Fig. 6). In fact, the same reasoning shows that (6.11) holds not only on Ie, but on Γ(eπi, a1) when e = a1, on Γ(b2, eπi/3) when e = b2, and for Re(ϕe(z)) on Γ(b1, a2) when e ∈ {b1, a2} (observe that these functions are also monotone on the respective arcs). Furthermore, each function ϕe(z) is analytic in Ue\Je, and its traces on Je satisfy

$ϕe±(s)=±2πiνeμ(Js,e)=2πe±3πiνe/2μ(Js,e),$
(6.12)

where Js,e is the subarc of Je with endpoints e and s,

$νe≔1,e∈{b1,b2},−1,e∈{a1,a2},$
(6.13)

and the second equality follows from (2.7) and (6.4). Since |ϕe(z)| ∼ |ze|3/2 as ze, it follows from (6.11) and (6.12) that we can define an analytic branch of $(−ϕe)2/3(z)$ in Ue that is positive on Ie and satisfies $(−ϕe)2/3(s)=−2πμt(Js,e)2/3$, sJe. Since $(−ϕe)2/3(z)$ has a simple zero at e, it is conformal in Ue for all radii small enough. Altogether, $(−ϕe)2/3(z)$ maps e into the origin, is conformal in Ue, and satisfies

$(−ϕe)2/3(Je)⊂(−∞,0),(−ϕe)2/3(Ie)⊂(0,∞).$
(6.14)

Furthermore, if we define $(−ϕe)1/6(z)$ to be holomorphic in Ue\Je and positive on Ie, then

$(−ϕe)+1/6(s)=νei(−ϕe)−1/6(s),s∈Je.$
(6.15)

To specify ρ(t), let ρe(t) be the radius of the largest disk around e for which Je, Ie are connected and in which $(−ϕe)2/3(z)$ is conformal. Observe that the disk around e of radius ρe(t) cannot contain other endpoints of Jt besides e. We set ρ(t) ≔ mine{ρe(t)}. Then, the disks Ue in (6.9) are necessarily disjoint. Observe also that ρ(t) is non-zero for all $t∈Otwo−cut$ and continuously depends on t due to continuous dependence on t of ϕe(z), which in itself follows from Theorem 3.2 and (6.4).

In this section, we prove Propositions 4.3–4.6 as well as discuss some related results. As in Sec. VI, we omit indicating the explicit dependence on t whenever convenient.

In this subsection, we discuss properties of the Riemann surface $S$ that has already appeared between (5.15) and (5.16). Once again, set

$S≔z≔(z,w):w2=Q(z).$
(7.1)

We denote by $π:S→C̄$ the natural projection π(z) = z and by ·* a holomorphic involution on $S$ acting according to the rule z* = (z, −w). We use notation z, s, a for points on $S$ with natural projections z, s, a.

The function w(z), defined by w2(z) = Q(z), is meromorphic on $S$ with simple zeros at the ramification points Et = {a1, b1, a2, b2} = π−1(Et), double poles at the points on top of infinity, and is otherwise non-vanishing and finite. Put

$Δ≔π−1(Jt)andS=D(0)∪Δ∪D(1),$

where the domains D(k) project onto $C̄\Jt$ with labels chosen so that $2w(z)=(−1)kz2+O(z)$ as z approaches the point on top of infinity within D(k). For $z∈C̄\Jt$, we let z(k) stand for zD(k) with, as agreed, π(z) = z. We define a homology basis on $S$ in the following way: we let

$α≔π−1(It)andβ≔π−1(Jt,1),$

where α is oriented toward b1 within D(0) and β is oriented so that α, β form the right pair at b1 (see Figs. 12 and 13). It also will be convenient to put $Sα,β≔S\{α∪β}$, which is simply connected.

The surface $S$ has genus one. Thus, there exists a unique holomorphic differential on $S$ normalized to have a unit period on α, say, $H$. In fact, it can be explicitly expressed as

$H(z)=∮αdsw(s)−1dzw(z).$
(7.2)

With this notation, the choice of the homology basis, and the definition of w(z), (4.9) becomes

$B=∮βH,Im(B)>0,$
(7.3)

where the last inequality is a classical result of Riemann (see Ref. 38).

Given the normalized holomorphic differential, we define

$a(z)≔∫b2zH,$
(7.4)

where we restrict z as well as the path of integration to $Sα,β$. The function $a(z)$ is a holomorphic function in $Sα,β$ with continuous traces on α, β (away from the point of their intersection) that satisfy

$a+(s)−a−(s)=−B,s∈α\{b1},1,s∈β\{b1},$
(7.5)

by the normalization of $H$ and the definition of $B$. Moreover, it continuously extends to $∂Sα,β$, the topological boundary of $Sα,β$ (the rectangle on Fig. 13). Note that

$τ=12πi∮αw(s)dsandω=−12πi∮βw(s)ds,$
(7.6)

where τ, ω are the constants from (4.6). It readily follows from (7.5) and (7.6) that

$∮∂Sα,β(wa)(s)ds=∮βw(s)(a+−a−)(s)ds+∮αw(s)(a+−a−)(s)ds=−2πi(ω+Bτ),$

where $∂Sα,β$ is oriented so that $Sα,β$ remains on the left when $∂Sα,β$ is traversed in the positive direction. On the other hand, the function $(wa)(z)$ is meromorphic in $Sα,β$ with only two singularities, both polar, at (0) and (1). Moreover, since w(z*) = −w(z) and $a(z*)=−a(z)$, the residues at those poles coincide. Therefore, it holds that

$ω+Bτ=−12πi∮∂Sα,β(wa)(s)ds=−2resz=∞(0)(wa)(z).$

Recall that $wz(0)=Q1/2(z)$, and therefore, it has expansion (6.1) as z(0)(0). This and (7.4) also allow us to deduce that

$az(0)=a∞(0)+∫∞(0)z(0)H=a∞(0)−∮αdsw(s)−12z+2t3z3+O1z4$

as z(0)(0). Therefore, taking into account (4.1) with the definition of α and the symmetry $a(z*)=−a(z)$, we get that

$ς+ω+Bτ=ς+2a∞(0)+4t3∮αdsw(s)−1=2∫b2∞(0)H=∫∞(1)∞(0)H,$
(7.7)

where the path of integration lies entirely in $Sα,β$ and is involution-symmetric.

Denote by $Jac(S)≔C/{Z+BZ}$ the Jacobi variety of $S$. We shall represent elements of $Jac(S)$ as equivalence classes $[s]={s+j+mB:j,m∈Z}$, where $s∈C$. One can readily deduce from (7.5) that $a(z)$, defined in (7.4), is essentially a holomorphic bijection from $S$ onto $Jac(S)$. Formally, we define this bijection, known as Abel’s map, by

$z∈S↦∫b2zH∈Jac(S).$
(7.8)

Given $s∈C$, we denote by z[s] the unique point in $S$ such that $∫b2z[s]H=[s]$.

Proposition 7.1.
Letτ, ωbe given by(4.6)andςby(4.1). Furthermore, let {Nn} be a sequence as in Theorem 4.1. Denote byzn,k = zn,k(t) the unique solution$z[sn,k(t)]$of the Jacobi inversion problem with
$sn,k(t)≔∫b2p(k)H+(n−Nn)ς+(ω+Bτ)n,p=p(t)≔b1b2−a1a2(b2−a2)+(b1−a1),$
(7.9)
k ∈ {0, 1}. Then, for any subsequence$N*$, the point(0)is a topological limit point of${zn,1}n∈N*$if and only if(1)is a topological limit point of${zn,0}n∈N*$. Moreover, it holds that [sn,0(tNn)] = [sn−1,1(tNn)]; that is,zn,0(Nn) = zn−1,1(Nn).

The behavior of the points zn,k with respect to n can be extremely chaotic. Assuming that nNn is constant, it is known that if the numbers ω and τ are rational, then there exist only finitely many distinct points zn,k; when ω and τ are irrational, all the points zn,k are distinct, lie on a Jordan curve if 1, ω, and τ are rationally dependent, and are dense on the whole surface $S$ otherwise.39

Let us also point out that the point p is always finite. Indeed, it follows from (1.11) and (2.6) that b1 + b2 + a1 + a2 = 0 [see also (5.12) and (5.17)]. Thus, if b2 + b1a1a2 = 0 were true, then we would have b1 = −b2 and a1 = −a2. In this case, a1b1a2 + a1b1b2 + a1a2b2 + b1a2b2 = 0. However, this expression must be equal to −4 according to the above references.

Proof of Proposition 7.1.
Define
$γ(z)≔z−b2z−a2z−b1z−a11/4,z∈C̄\Jt,$
(7.10)
where γ(z) is holomorphic off Jt = supp(μt) and the branch is chosen so that γ() = 1. Furthermore, set
$A(z)=γ(z)+γ−1(z)2andB(z)≔γ(z)−γ−1(z)−2i.$
(7.11)
Observe that the function A(z) was already defined in (4.12). The functions A(z) and B(z) are holomorphic in $C̄\Jt$ and satisfy
$A(∞)=1,B(∞)=0,andA±(s)=±B∓(s),s∈Jt°.$
(7.12)
Note that the equation (AB)(z) = 0 can be rewritten as γ4(z) = 1 and has two solutions, namely, and the point p from (7.9). In fact, unless pJt°, it a zero of B(z). Indeed, it is enough to show that γ(p) = 1 in the latter case. Let Liγ4(Jt,i), i ∈ {1, 2}, which are unbounded arcs connecting the origin to the point at infinity. Let $L⊂C̄\Jt$ be an arc connecting the point at infinity and p. Then, γ4(L) is a closed curve that contains 1 and does not intersect the arcs Li and, therefore, does not wind around the origin. Thus, analytic continuation of the principal branch of the 1/4-root from 1 along γ4(L) leads back to the value 1 at the point 1. However, this continuation is exactly the continuation of γ(z) from the point at infinity to p along L, which does imply that γ(p) = 1 as claimed.

It follows from (7.12) that

$(B/A)(z),z∈D(0),−(A/B)(z),z∈D(1),$
(7.13)

is a rational function on $S$ with two simple zeros (0) and p(0) and two simple poles (1) and p(1) [if it happens that pJt°, then we choose $p(0)∈S$ precisely in such a way that it is a zero of (7.13) and so p(1) is a pole of (7.13); it is, of course, still true that these points are distinct and $πp(k)=p$]. Therefore, Abel’s theorem yields that

$∫p(0)∞(1)H=∫p(1)∞(0)H,$
(7.14)

while relations (7.9), in particular, imply that

$∫p(0)zn,0H=∫p(1)zn,1H.$
(7.15)

Let zk be a topological limit of a subsequence ${zni,k}$. Holomorphy of the differential $H$ implies that

$∫p(k)zni,kH=∫p(k)zkH+∫zkzni,kH→∫p(k)zkH$

as i, where the integral from zk to $zni,k$ is taken along the path that projects into a segment joining zk and $zni,k$. The first claim of the proposition now follows from (7.14) and (7.15) and the unique solvability of the Jacobi inversion problem on $S$. Observe that

$∫p(0)p(1)H=∫∞(1)∞(0)H=ς+ω+Bτ$
(7.16)

by (7.7) and (7.14). Since

$sn,0(Nn)=sn−1,1(Nn)+∫p(1)p(0)H+ς+ω+Bτ,$

the second conclusion of the proposition easily follows.

As we shall show further below, the functions Θn(zt) from Proposition 4.4 vanish at zn,1 when it belongs to D(0) and do not vanish at all when zn,1 does not belong to D(0). Hence, the subsequences $N(ε)=N(t,ε)$ from Proposition 4.4 can be equivalently defined as

$N(ε)≔n∈N:zn,1∉D(0)∩π−1|z|≥1/ε.$

Denote by r[s] the distance from 0 to [s], viewed as a lattice in $C$. That is,

$r[s]≔minj,m∈Zs+j+Bm,$

where s is any representative of [s]. Furthermore, for each r[s] > 0, choose ɛs > 0 small enough so that

$a(z)−a(∞(0))1/ε,$

where we also assume that ɛs is small enough so that always $Uεs⊂Sα,β$. Then,

$a(z1)−a(z2)=[s]≠⇒either z1∉Uεs or z2∉Uεs$
(7.17)

as otherwise it must hold $0.

Assume that either x or y is not an integer. Let $zn,1−,zn,1,zn,1+$ be solutions of the Jacobi inversion problem (7.9) with N equal to n − 1, n, n + 1, respectively. Then, it follows from (7.9) that

$a(zn,1±)−a(zn,1)=∓ς.$

As [±ς] ≠  by assumption, the second assertion of the proposition follows from (7.17).

Assume now that both x and y are integers. It readily follows from (7.9) that the definition of zn,k does not depend on the choice of Nn in this case. The first assertion of the proposition holds since

$a(zn+1,1)−a(zn,1)=ω+Bτ≠0,$

where the last conclusion follows from the fact ω = μt(Jt,1) ∈ (0, 1) by the very definition in (4.6).

Let k = Nn+1Nn. It follows from (7.9) that

$a(zn+1,1)−a(zn,1)=(1−k)ς+ω+Bτ.$
(7.18)

Set $K≔{k∈Z:r[(1−k)ς+ω+Bτ]=0}$. We have already shown that $r[ω+Bτ]>0$. It also holds that $r[ς+ω+Bτ]>0$ since $[ς+ω+Bτ]≠$ by (7.7) and the unique solvability of the Jacobi inversion problem. The following lemma easily follows from (7.17).

Lemma 7.2.

Assume that |nNn| ≤ N*, and let$εk≔ε(1−k)ς+ω+Bτ$.

1. If$K=∅$, then at least one of the integersn, n + 1 belongs to$N(ε)$for all$ε≤min|k|≤2N*+1εk$.

2. If$K={k′}$, there exists an infinite subsequence {nl} such that$Nnl+1−Nnl≠k′$, and therefore, at least one of the integersnl, nl + 1 belongs to$N(ε)$for all$ε≤min|k|≤2N*+1,k≠k′εk$.

3. If there exists an infinite subsequence {nl} such that$Nnl+1−Nnl∈{0,1}$, then at least one of the integersnl, nl + 1 belongs to$N(ε)$for allɛ ≤ min{ɛ0, ɛ1}.

Inclusion $k∈K$ means that $[(1−k)ς+ω+Bτ]=$. Hence, both triples ω, x, 1 and τ, y, 1 are rationally dependent, $ς=x+By$. Assume that $k′,k′′∈K$, k′ ≠ k″. Then, it follows from (7.18) that $[ω+Bτ]=[(k′−1)ς]$ and [(k″ − k′)ς] = . The latter relation implies the first representation in (4.13), while the former gives the other two. It is easy to see in this case that $K=k′+dZ$. That is, if $K$ has at least two elements, then it is an arithmetic progression, ω, τ are rational numbers, ς has rational coordinates in the basis $1,B$, and the second and third relations of (4.13) must be satisfied. Thus, we can claim the following lemma:

Lemma 7.3.

If one of the triplesω, x, 1 orτ, y, 1 is rationally independent, then$K=∅$. If not all numbersω, τ, x, yare rational or they all rational but the second and third relations of(4.13)do not hold, then either$K=∅$or$K={k′}$.

Assume now that all three relations of (4.13) take place. That is, $[ω+Bτ]=[(k−1)ς]$ and [] =  for some integers k, d. It follows from (7.14) that

$∫b2p(1)H=12∫p(0)p(1)H=12∫∞(1)∞(0)H+j+Bm=∫b2∞(0)H+j+Bm2$

for some $j,m∈Z$, where we use involution-symmetric paths of integration. Note that j, m cannot be simultaneously even as this would contradict unique solvability of the Jacobi inversion problem. In fact, it holds that

$∫p(1)∞(0)H=∫b2∞(0)H+∫b2p(0)H=∫b2b1H=1+B2,$
(7.19)

i.e., both j, m must be odd. Indeed, the first equality follows from the symmetry $a(z*)=−a(z)$. Next, observe that the function $(−1)kγ−2z(k)−1$ is meromorphic on $S$ with zeros at (0) and p(0) and poles at b1 and b2, all simple. Hence, the second equality is a consequence of the same symmetry and Abel’s theorem. Furthermore, consider an involution-symmetric cycle represented by the anti-diagonal on Fig. 13, oriented so that it proceeds toward b1 in D(0). It is clearly homologous to α + β. Then, the symmetry of $a(z)$ and the normalization in (7.2) and (7.3) yield the final equality in (7.19). Note that the conclusion of (7.19) a posteriori holds with the bounds of integration in the left most integral of (7.19) in flipped. Therefore, adding $∫∞(0)b2H$ to both sides of (7.9) gives us

$∫∞(0)zn,1H=1+B2+(n−Nn)ς+(ω+Bτ)n=1+B2+(nk−Nn)ς.$
(7.20)

Since ς has rational coordinates in the basis $1,B$ with denominator d, the right-hand side of (7.20) has at most d distinct values that depend only on ρ ∈ {0, …, d − 1}, the remainder of the division of nkNn by d. Let zρ, ρ ∈ {0, …, d − 1}, be such that

$∫∞(0)zρH=1+B2+ρς.$

Clearly, ${zn,1}n∈N⊆{zρ}ρ=0d−1$. Thus, it only remains to investigate when zρ = (0) or, equivalently, when $[(1+B)/2+ρς]=$. From the representation of ς in (4.13), it must simultaneously hold that ρ = d(2lj − 1)/(2ij), j ∈ {1, 2}, for some $l1,l2∈Z$. Since one of the pairs (ij, d) is co-prime and ρ ∈ {0, …, d − 1}, this is possible only if ρ = 0 or ρ = d/2 (in the second case, of course, d must be even and ij = 2lj − 1 must be odd), and the former is ruled out since $[(1+B)/2]≠$ (or equivalently, since 0 ≠ 2lj − 1).

Lemma 7.4.

If all three relations of(4.13)take place, then the Jacobi inversion problem(7.9)forzn,1has only finitely many distinct solutions and(0)is one of them if and only ifdis even,i1, i2are odd, andnkNnis divisible byd/2 but not byd.

Recall that Abel’s map (7.8) is essentially carried out by the function $a(z)$ defined in (7.4). We shall consider the extension $ã(z)$ of $a(z)$ to the whole surface $S$ defined by setting $ã(s)≔a+(s)$ for sα and sβ\{b1}. Given such an extension and (7.9), there exist unique integers jn,k, mn,k such that

$ã(zn,k)=ãp(k)+(n−Nn)ς+(ω+Bτ)n+jn,k+Bmn,k,k∈{0,1}.$
(7.21)

Recall the definition of θ(u) in (4.10). The function θ(u) is holomorphic in $C$ and enjoys the following periodicity properties:

$θ(u+j+Bm)=exp−πiBm2−2πiumθ(u),j,m∈Z.$
(7.22)

It is also known that θ(u) vanishes only at the points of the lattice $B+12$. Let

$Θn,k(z)=exp−2πimn,k+τna(z)θa(z)−ã(zn,k)−B+12θa(z)−ãp(k)−B+12.$
(7.23)

Then, we define Θn(zt) from Proposition 4.4 as well as the other functions appearing in the statement of Theorem 4.2 by

$Θn(z;t)≔Θn,1(0)(z),ϑn(z)≔Θn,1(0)(z)Θn,1(0)(∞),ϑn*(z)≔Θn,1(1)(z)Θn,1(0)(∞),andϑ̃n(z)≔Θn,0(0)(z)Θn,0(1)(∞),$
(7.24)

where F(i)(z), i ∈ {0, 1}, stands for the pull-back under π(z) of a function F(z) from D(i) into $C̄\Jt$.

The functions Θn,k(z) are meromorphic on $Sα,β$ with exactly one pole, which is simple and located at p(k), and exactly one zero, which is also simple and located at zn,k (observe that these functions can be analytically continued as multiplicatively multivalued functions on the whole surface $S$; thus, we can talk about simplicity of a pole or zero regardless whether it belongs to the cycles of a homology basis or not). Moreover, according to (7.5), (7.21), and (7.22), they possess continuous traces on α, β away from b1 that satisfy

$Θn,k+(s)=Θn,k−(s)exp−2πi(ωn+(n−Nn)ς),s∈α\{b1},exp−2πiτn,s∈β\{b1}.$
(7.25)

Recall functions A(z), B(z) from (7.11) and (7.12). To discuss boundedness properties of Θn,k(z) and for the asymptotic analysis in Sec. VIII, it will be convenient to define

$Mn,0(z)=Θn,0(z)B(z),z∈D(0),A(z),z∈D(1),andMn,1(z)=Θn,1(z)A(z),z∈D(0),−B(z),z∈D(1).$
(7.26)

These functions are holomorphic on $S\{α∪β∪Δ}$ since the pole of Θn,k(z) is canceled by the zero of B(z). Each function Mn,k(z) has exactly two zeros, namely, zn,k and (k). It follows from (7.12) and (7.25) that

$Mn,k±(0)(s)=∓Mn,k∓(1)(s),s∈Jt,2°,Mn,k±(0)(s)=∓e−2πiτnMn,k∓(1)(s),s∈Jt,1°,Mn,k±(i)(s)=e(−1)i2πi(nω+(n−Nn)ς)Mn,k∓(i)(s),s∈It°.$
(7.27)

It further follows from (7.10) and (7.11) that |Mn,k(z)| ∼ |ze|−1/4 as zeE unless zn,k coincides with e in which case the exponent becomes 1/4. Assume now that there exists N* ≥ 0 such that |nNn| ≤ N* for all $n∈N$. Then, for each δ > 0 there exists C(δ, N*) independent of n such that

$|Mn,k(z)|≤C(δ,N*),z∈Oδ≔S\∪e∈Eπ−1{|z−e|<δ}.$
(7.28)

Indeed, observe that {log|Mn,k(z)|} is a family of subharmonic functions in Oδ\α [the jump of Mn,k(z) is unimodular on β]. By the maximum principle for subharmonic functions, log|Mn,k(z)| reaches its maximum on (Oδ\α), where the maximum is clearly finite. Since the sequence {nNn} is bounded by assumption and the range of $ã(z)$ is bounded by construction, so are the sequences {mn,k + τn} and {jn,k + ωn} [see (7.21)], and recall that jn,k + ωn are real and $Im(B)>0$. Thus, any limit point of {log|Mn,k(z)|} is obtained by taking simultaneous limit points of {nNn}, {mn,k + τn}, and {jn,k + ωn}, computing the corresponding solution zk of the Jacobi inversion problem (7.21) and plugging all of these quantities into the right-hand side of (7.23). Hence, all these limit functions are also bounded above on the closure of Oδ\α, which proves (7.28). Finally, it holds that

$Mn,k∞(1−k)≥cε,n∈N(ε),$
(7.29)

for some constant cɛ > 0 by a similar compactness argument combined with the definition of $N(ε)$ in Proposition 4.4, the observation that Mn,k(z) is non-zero at (1−k) when (1−k)zn,k, and the last conclusion of Proposition 7.1.

Define a function F(z) on $S\{α∪β}$ by setting $Fz(1)≔1/Fz(0)$ and

$Fz(0)≔D−1(z)expV(z)/2+Q(z)$

[see (4.2) and (4.5)]. It follows from (4.3) and (4.8) that

$F+(s)=F−(s)exp2πi(ω+ς),s∈α\{b1},exp2πiτ,s∈β\{b1},$
(7.30)

where one needs to recall that It is oriented toward a2, while $α∩S(0)$ is oriented toward b1. Moreover, it follows from (4.7) that

$Fz(0)=eℓ*/2D−1(∞)z+O(1)$
(7.31)

as z. On the other hand, we get from (7.16) that there exist integers j, m such that

$ãp(1)−ãp(0)=ς+ω+Bτ+j+Bm.$
(7.32)

Thus, we get similarly to (7.23) and (7.25) that the function

$Θ(z)≔exp−2πi(m+τ)a(z)θa(z)−ãp(1)−B+12θa(z)−ãp(0)−B+12$
(7.33)

is meromorphic in $S\{α∪β}$, with a simple zero at p(1), a simple pole at p(0), and otherwise non-vanishing and finite, whose traces on the cycles of the homology basis satisfy

$Θ+(s)=Θ−(s)exp−2πi(ω+ς),s∈α\{b1},exp−2πiτ,s∈β\{b1}.$
(7.34)

Combining (7.30), (7.31), and (7.34) yields that the function (FΘ)(z) is rational on $S$ with the divisor p(1) + (1)p(0)(0). That is, this function is a constant multiple of the reciprocal of (7.13). Recall the definition of S1 in Theorem 4.2. Observe that

$A(∞)=1andB(z)=S14i1z+Oz−2$

as z. Then, we get from (7.13) and (7.31) and the symmetry F(z)F(z*) ≡ 1 that

$eℓ*=D2(∞)16S12Θ∞(1)Θ∞(0).$
(7.35)

Thus, it only remains to obtain the expression for the last fraction. We get from (7.19) that

$a∞(0)−ãp(1)=1+B2+u+Bv$

for some integers u, v. Hence, we get from (7.7), (7.32), and (7.22) that

$Θ∞(0)=e−πi(m+τ)(ς+ω+Bτ)θ(Bv)θ(ς+ω+Bτ+B(m+v))=eπi(2v+m−τ)(ς+ω+Bτ)+πiB(2v+m)θ(0)θ(ς+ω+Bτ)$

and similarly

$1Θ∞(1)=e−πi(m+τ)(ς+ω+Bτ)θ(B(m+v))θ(Bv−ς−ω−Bτ)=e−πi(2v+m+τ)(ς+ω+Bτ)−πiB(2v+m)θ(0)θ(−ς−ω−Bτ).$

Since θ(·) is an even function, this finishes the proof of the proposition.

As agreed before, we omit the dependence on t whenever it does not cause ambiguity. In what follows, it will be convenient to set

$I≔1001,σ1≔0110,andσ3≔100−1.$

We are seeking solutions of the following sequence of Riemann–Hilbert problems for 2 × 2 matrix functions (RHP-Y):

• Y(z) is analytic in $C\Γ$ and $limC\Γ∋z→∞Y(z)z−nσ3=I$;

• Y(z) has continuous traces on Γ\{a1, b1, a2, b2} that satisfy
$Y+(s)=Y−(s)1e−NnV(s)01,$
where, as before, V(z) is given by (1.11) and the sequence {Nn} is such that |nNn| ≤ N* for some N* > 0.
The connection of RHP-Y to orthogonal polynomials was first demonstrated by Fokas, Its, and Kitaev in Ref. 40 and lies in the following. If the solution of RHP-Y exists, then it is necessarily of the form
$Y(z)=Pn(z)CPne−NnV(z)−2πihn−1Pn−1(z)−2πihn−1CPn−1e−NnV(z),$
(8.1)
where Pn(z) = Pn(zt, Nn) are the polynomial satisfying orthogonality relations (1.10), hn = hn(t, Nn) are the constants defined in (1.12), and $Cf(z)$ is the Cauchy transform of a function f given on Γ, i.e.,
$(Cf)(z)≔12πi∫Γf(s)s−zds.$

Below, we show the solvability of RHP-Y for all $n∈N(t,ε)$ large enough following the framework of the steepest descent analysis introduced by Deift and Zhou.41 The latter lies in a series of transformations that reduce RHP-Y to a problem with jumps asymptotically close to identity.

Suppose that Y(z) is a solution of RHP-Y. Put

$T(z)≔enℓ*σ3/2Y(z)e−n(g(z)+ℓ*/2)σ3,$
(8.2)

where the function g(z) is defined by (6.2) and * was introduced in (6.3). Then,

$T+(s)=T−(s)e−n(g+(s)−g−(s))en(g+(s)+g−(s)−V(s)+ℓ*)+(n−Nn)V(s)0e−n(g−(s)−g+(s)),$

s ∈ Γ, and therefore, we deduce from (4.7) and (6.6)–(6.8) that T(z) solves RHP-T.

• T(z) is analytic in $C\Γ$ and $limC\Γ∋z→∞T(z)=I$;

• T(z) has continuous traces on Γ\{a1, b1, a2, b2} that satisfy
$T+(s)=T−(s)1en(2πiτ+ϕa1(s))+(n−Nn)V(s)01,s∈Γ(eiπ∞,a1),1enϕb2(s)+(n−Nn)V(s)01,s∈Γ(b2,eπi/3∞),e2πiωnenϕa2(s)+(n−Nn)V(s)0e−2πiωn,s∈Γ(b1,a2),$
and
$T+(s)=T−(s)e−nϕb2+(s)e(n−Nn)V(s)0e−nϕb2−(s),s∈Γ(a2,b2),e−n(ϕb2+(s)−2πiτ)e2πiτn+(n−Nn)V(s)0e−n(ϕb2−(s)−2πiτ),s∈Γ(a1,b1).$

Clearly, if RHP-T is solvable and T(z) is the solution, then by inverting (8.2), one obtains a matrix Y(z) that solves RHP-Y.

As usual in the steepest descent analysis of matrix Riemann–Hilbert problems for orthogonal polynomials, the next step is based on the identity

$e−n(ϕb2+(s)−C)enC+(n−Nn)V(s)0e−n(ϕb2−(s)−C)=10e−nϕb2−(s)−(n−Nn)V(s)1×0enC+(n−Nn)V(s)−e−nC−(n−Nn)V(s)010e−nϕb2+(s)−(n−Nn)V(s)1$

that follows from (6.6), where C = 2πiτ when s ∈ Γ(a1, b1) and C = 0 when s ∈ Γ(a2, b2). To carry it out, we shall introduce two additional systems of arcs.

Denote by J± smooth homotopic deformations of Jt within the region $Re(ϕb2(z))>0$ such that J+ lies to the left and J to the right of Jt (see Fig. 14) [both connected subarcs of J+ (respectively, J) are oriented from ai to bi, i ∈ {1, 2}]. We shall fix the way these arcs emanate from e ∈ {a1, b1, a2, b2}. In particular, let Ue be given by (6.9) and $(−ϕe)2/3(z)$ be as in (6.14). Then, we require that

$arg(−ϕe)2/3(z)=±νe(2π/3),z∈Ue∩J±,$
(8.3)

where νe is defined by (6.13). This requirement always can be fulfilled due to conformality $(−ϕe)2/3(z)$ in Ue and the choice of the branch in (6.14).

FIG. 14.

The thick curves represent Γ and thinner black curves represent J±. The shaded part represents regions where Re(ϕe(z)) < 0 (they are the same for each e ∈ {a1, b1, a2, b2}).

FIG. 14.

The thick curves represent Γ and thinner black curves represent J±. The shaded part represents regions where Re(ϕe(z)) < 0 (they are the same for each e ∈ {a1, b1, a2, b2}).

Close modal

Denote by O± the open sets delimited by J± and Jt. Set

$S(z)≔T(z)10∓e−nϕb2(z)−(n−Nn)V(z)1,z∈O±,I,otherwise.$
(8.4)

Then, if T(z) solves RHP-T, S(z) solves RHP-S:

• S(z) is analytic in $C\(Γ∪J+∪J−)$ and $limC\Γ∋z→∞S(z)=I$;

• S(z) has continuous traces on Γ\{a1, b1, a2, b2} that satisfy RHP-T (b) on Γ(eπi, a1), Γ(b1, a2), and Γ(b2, eπi/3), as well as
$S+(s)=S−(s)0e(n−Nn)V(s)−e−(n−Nn)V(s)0,s∈Γ(a2,b2),0e2πiτn+(n−Nn)V(s)−e−2πiτn−(n−Nn)V(s)0,s∈Γ(a1,b1),10e−nϕb2(s)−(n−Nn)V(s)1,s∈J±.$

As before, since transformation (8.4) is invertible, a solution of RHP-S yields a solution of RHP-T.

The Riemann–Hilbert problem for the global parametrix is obtained from RHP-S by removing the quantities that are asymptotically zero from the jump matrices in RHP-S (b). The latter can be easily identified with the help of (6.5) and by recalling that the constant τ is real. Thus, we are seeking the solution of RHP-N:

• N(z) is analytic in $C̄\Γ[a1,b2]$ and N() = I;

• N(z) has continuous traces on Γ(a1, b2)\{b1, a2} that satisfy
$N+(s)=N−(s)0e(n−Nn)V(s)−e−(n−Nn)V(s)0,s∈Γ(a2,b2),0e2πiτn+(n−Nn)V(s)−e−2πiτn−(n−Nn)V(s)0,s∈Γ(a1,b1),e2πiωn00e−2πiωn,s∈Γ(b1,a2).$

We shall solve this problem only for $n∈N(ε)=N(t,ε)$ from Proposition 4.4. In fact, to solve RHP-N, we only need to exclude indices n for which zn,1 = (0) (in this case zn,0 = (1); see Proposition 7.1) as should be clear from (8.5). However, for further analysis, we shall need estimate (7.29), where it is crucial that $n∈N(ε)$ for some ɛ > 0.

Let the functions Mn,k(z) be given by (7.26) and $D(z)=D(z;t)$ be defined by (4.2). With the notation introduced right after (7.23), a solution of RHP-N is given by

$N(z)=M−1(∞)M(z),M(z)≔Mn,1(0)(z)Mn,1(1)(z)Mn,0(0)(z)Mn,0(1)(z)D(Nn−n)σ3(z).$
(8.5)

Indeed, RHP-N (a) follows from holomorphy of $D(z)$ and Mn,k(z) discussed in Proposition 4.1 and right after (7.26). Fulfillment of RHP-N (b) can be checked by using (4.3) and (7.27). Observe also that det(N(z)) ≡ 1. Indeed, as the jump matrices in RHP-N (b) have unit determinants, det(N(z)) is holomorphic through Γ(a1, b1), Γ(b1, a2), and Γ(a2, b2). It also has at most square root singularities at {a1, b1, a2, b2} as explained right after (7.27). Thus, it is holomorphic throughout $C̄$ and, therefore, is a constant. The normalization at infinity implies that this constant is 1.

The jumps discarded in RHP-N are not uniformly close to the identity around each point e ∈ {a1, b1, a2, b2}. The goal of this section is to solve RHP-S in the disks Ue [see (6.9)] with a certain matching condition on the boundary of the disks. More precisely, we are looking for a matrix functions Pe(z) that solves RHP-P:

• Pe(z) has the same analyticity properties as S(z) restricted to Ue [see RHP-S (a)];

• Pe(z) satisfies the same jump relations as S(z) restricted to Ue [see RHP-S (b)];

• $Pe(z)=N(z)I+O(n−1)$ holds uniformly on ∂Ue as n.

Again, we shall solve RHP-P only for $n∈N(ε)$ with $O(⋅)=Oε(⋅)$ in RHP-P(c).

Let Ue, Je, and Ie, e ∈ {a1, b1, a2, b2}, be as in (6.9) and (6.10). Furthermore, let A(ζ) be the Airy matrix.42,43 That is, it is analytic in $C\(−∞,∞)∪L−∪L+$, $L±≔ζ:arg(ζ)=±2π/3$, and satisfies

$A+(s)=A−(s)01−10,s∈(−∞,0),1011,s∈L±,1101,s∈(0,∞),$

where the real line is oriented from − to and the rays L± are oriented toward the origin. It is known that A(ζ) has the following asymptotic expansion44 at infinity:

$A(ζ)e23ζ3/2σ3∼ζ−σ3/42∑k=0∞sk00tk(−1)ki(−1)ki123ζ3/2−k,$
(8.6)

where the expansion holds uniformly in $C\(−∞,∞)∪L−∪L+$, and

$s0=t0=1,sk=Γ(3k+1/2)54kk!Γ(k+1/2),tk=−6k+16k−1sk,k≥1.$

Let us write AeA if e ∈ {b1, b2} and Aeσ3Aσ3 if e ∈ {a1, a2}. It can be easily checked that σ3Aσ3 has the same jumps as A only with the reversed orientation of the rays. Moreover, one needs to replace i by −i in (8.6) when describing the behavior of σ3Aσ3 at infinity. Let $ζe(z)≔−n(3/4)ϕe(z)2/3$, which is conformal in Ue [see (6.14)]. Furthermore, put

$Je(z)≔e(Nn−n)V(z)σ3/2I,e=b2,eπi(±ω)nσ3,e=a2,eπi(±ω−τ)nσ3,e=b1,e−πiτnσ3,e=a1,$

where we use ω if z lies to the left of Γ and use −ω if z lies to the right of Γ. Then, it can be readily verified by using (6.5) that

$Pe(z)≔Ee(z)Aeζe(z)e(2/3)ζe3/2(z)σ3Je(z)$
(8.7)

satisfies RHP-P (a) and (b) for any matrix Ee(z) holomorphic in Ue. It follows immediately from (8.6) that RHP-P (c) will be satisfied if

$Ee(z)≔NJe−1(z)1−νei−νei1ζeσ3/4(z)2,$
(8.8)

provided this matrix function is holomorphic in Ue, where νe was defined in (6.13). By using RHP-N (b) and (6.15), one can readily check that Ee(z) is holomorphic in Ue\{e}. Since ζe(z) has a simple zero at e, it also follows from (8.5) and the claim after (7.27) that Ee(z) can have at most square root singularity at e and, therefore, is, in fact, holomorphic in the entire disk Ue as needed.

In fact, it follows from (8.6)–(8.8) that

$Pe(z)∼N(z)I+1n∑k=0∞Pe,k(z)nk,$
(8.9)

where the expansion inside the parentheses holds uniformly on ∂Ue and locally uniformly for $t∈Otwo−cut$ and

$Pe,k−1(z)=Je−1(z)1−νei−νei1sk00tk(−1)kνeiνe(−1)ki1Je(z)−ϕe(z)2−k,k≥1.$
(8.10)

Set $Σ≔(Γ\Jt)∪J+∪J−∩D∪∪e∂Ue$, $D≔C\∪eŪe$ We shall show that for all $n∈N(ε)$ large enough there exists a matrix function R(z) that solves the following Riemann–Hilbert problem (RHP-R):

• R(z) is holomorphic in $C\Σ$ and $limC\Γ∋z→∞R(z)=I$;

• R(z) has continuous traces on Σ° (points with the well-defined tangent) that satisfy
$R+(s)=R−(s)Pe(s)N−1(s),s∈∂Ue,N(s)10e−nϕb2(s)−(n−Nn)V(s)1N−1(s),s∈J±∩D,$
where ∂Ue is oriented clockwise, and
$R+(s)=R−(s)N(s)1en(2πiτ+ϕa1(s))+(n−Nn)V(s)01N−1(s),s∈Γ(eiπ∞,a1)∩D,N−(s)e2πiωnenϕa2(s)+(n−Nn)V(s)0e−2πiωnN+−1(s),s∈Γ(b1,a2)∩D,N(s)1enϕb2(s)+(n−Nn)V(s)01N−1(s),s∈Γ(b2,eπi/3∞)∩D.$

Observe that RHP-R is a well posed problem as det(N(z)) ≡ 1, as explained after (8.5), and therefore, the matrix is invertible. Recall also that the entries of N(z) and N−1(z) are uniformly bounded on Σ for $n∈N(ε)$ according to (7.28) and (7.29).

To prove solvability of RHP-R, let us show that the jump matrices in RHP-R (b) are close to the identity. To this end, set

$Δ(s)≔(R−−1R+)(s)−I,s∈Σ.$
(8.11)

Since the entries of N(z) are uniformly bounded on each ∂Ue with respect to $n∈N(ε)$, it holds by RHP-P (c) and (8.9) that

$Δ(s)∼1n∑k=0∞NPe,kN−1(s)nk,$
(8.12)

where this pseudo-expansion45 is valid uniformly on ∂Ue. Thus, it holds that

$‖Δ‖L∞(∪e∂Ue)=Oεn−$