# On a class of elliptic orthogonal polynomials and their integrability

Harini Desiraju1<EMAIL_ADDRESS>,  Tomas Lasic Latimer1
<EMAIL_ADDRESS>and  Pieter Roffelsen1
<EMAIL_ADDRESS>1School of Mathematics and Statistics,
University of Sydney, Camperdown, NSW 2006

###### Abstract.

Building upon the recent works of Bertola; Fasondini, Olver and Xu, we define
a class of orthogonal polynomials on elliptic curves and establish a
corresponding Riemann-Hilbert framework. We then focus on the special case,
defined by a constant weight function, and use the Riemann-Hilbert problem to
derive recurrence relations and differential equations for the orthogonal
polynomials. We further show that the sub-class of even polynomials is
associated to the elliptic form of Painlevé VI, with the tau function given by
the Hankel determinant of even moments, up to a scaling factor. The first
iteration of these even polynomials relates to the special case of Painlevé VI
studied by Hitchin in relation to self-dual Einstein metrics.

###### Contents

  1. 1 Introduction
  2. 2 Riemann-Hilbert problems and moments
  3. 3 Linear problems for the general case and recurrence relations
  4. 4 Linear problems for the even case and the elliptic form of Painlevé VI
  5. 5 The Painlevé VI tau-function and Hankel determinants
  6. A Elliptic functions and their periodicity properties
  7. B List of polynomials
  8. C Structure of the moment matrix

##  1\. Introduction

Orthogonal polynomials constitute a fundamental class of special functions
with important applications to a wide array of topics, from combinatorics to
signal processing. Particularly, they provide useful tools to understand
universality of random matrix ensembles [Dei99] and large $N$ limits of matrix
models [DGZ95], and describe special solutions to integrable systems such as
Painlevé equations [Van17].

Traditionally, orthogonal polynomials define a basis of real polynomials,
orthogonal with respect to an inner product defined by integrating against a
weight function on a subset of the real line. For example, Hermite polynomials
are orthogonal with respect to the weight $e^{-x^{2}}$ on the real line.
Moreover, it is known that generalisations of classical Chebyshev and Jacobi
polynomials can be described by elliptic functions, see for instance [Akh90,
Car60, IVY01, Ree45, VZ09].

In the past couple of years there were notable breakthroughs in defining
orthogonal polynomials directly on elliptic curves, with modern techniques
facilitating a systematic analysis of their properties [Ber21, Ber22, BGK22,
FOX22, FOX23]. Inspired by these works, we consider families of meromorphic
functions built out of the the Weierstrass $\wp$-function and its derivative,
which are orthogonal with respect to a given weight function, and call them
elliptic orthogonal polynomials (EOPs) due to their proximity in construction
of those introduced by Heine [Hei78] and Rees [Ree45].

We establish a general framework to analyse such polynomials using their
moments and the Riemann-Hilbert method. Furthermore, when the weight is
constant, we show that the even EOPs, indexed by $k$, are related to the
elliptic form of Painlevé VI. For $k=1$, the parameters of the elliptic
Painlevé VI equation are
$\left(\frac{1}{8},\frac{1}{8},-\frac{1}{8},\frac{3}{8}\right)$, for which the
general solution is known to be described by elliptic functions [Hit95,
Man98].

The notion of EOPs we use in this paper is as follows: let $\tau$ be an
element of the upper half-plane $\mathbb{H}$ and $\pi(z)$ be an elliptic
function with periods $1$ and $\tau$. We call $\pi(z)$ an elliptic polynomial
if all of its poles are located on the lattice
$\mathbb{Z}+\mathbb{Z}\cdot\tau$. Its degree is $n$ if the pole at $z=0$ is of
order $n$, and we call it monic if, for $n\geq 0$,

$\pi(z)=z^{-n}(1+\mathcal{O}(z)),\qquad\textrm{as}\quad z\to 0.$

Note, in particular, that there exists no elliptic polynomial of degree one.
We consider sequences of elliptic polynomials $(\pi_{n})_{n\geq 0,n\neq 1}$,
with $\pi_{n}$ monic of degree $n$ for $n\in\mathbb{N}_{\neq 1}$, which
satisfy an orthogonality condition of the form

$\int_{\frac{\tau}{2}}^{\frac{\tau}{2}+1}\pi_{m}(z)\pi_{n}(z){\sf
w}(z)dz=\delta_{mn}h_{n},$ (1.1)

for some $h_{n}\in\mathbb{C}$, where $\delta_{mn}$ is the Kronecker delta
function and $w(z)$ is an $L^{1}$ function on the interval
$\gamma:=[\tfrac{\tau}{2},\tfrac{\tau}{2}+1]$, called the weight function, for
all $m,n\in\mathbb{N}_{\neq 1}$. In this case, we call $(\pi_{n})_{n\geq
0,n\neq 1}$ a sequence of elliptic orthogonal polynomials (EOPs). The choice
of support $\gamma$ is motivated by the fact that it is invariant under
negation and complex conjugation on the torus

$\mathbb{T}:=\mathbb{C}/(\mathbb{Z}+\mathbb{Z}\cdot\tau),$

and does not contain $[0]\in\mathbb{T}$.

We note the following analogy between elliptic polynomials and traditional
complex polynomials. A complex polynomial can be characterised as a
meromorphic function on $\mathbb{CP}^{1}$, with at most one pole, at $\infty$.
Analogously, an elliptic polynomial, as defined above, can be characterised as
a meromorphic function on the torus $\mathbb{T}$, with only at most one pole,
at $[0]\in\mathbb{T}$.

Figure 1. The orthogonality interval in the fundamental domain $\mathbb{T}$.

A useful basis for elliptic polynomials can be constructed in terms of the
Weierstrass $\wp$-function and its $z$-derivative,

$\displaystyle\mathcal{B}=\\{\mathcal{E}_{n}\\}_{{n\geq 0,n\neq 1}},$
$\displaystyle\mathcal{E}_{2k}=\wp(z)^{k},$
$\displaystyle\mathcal{E}_{2k+3}=-\tfrac{1}{2}\wp^{\prime}(z)\wp(z)^{k},$
$\displaystyle k\geq 0,$ (1.2)

chosen such that $\mathcal{E}_{n}$ is monic of degree $n$, for
$n\in\mathbb{N}_{\neq 1}$. Recall the isomorphism

$\displaystyle x=\wp(z),$ $\displaystyle y=\wp^{\prime}(z),$ (1.3)

from $\mathbb{T}$ to the cubic curve

$y^{2}=4x^{3}-g_{2}x-g_{3},$ (1.4)

where $g_{2,3}$ are the elliptic invariants. Under this map, the EOP $\pi_{n}$
becomes a bivariate polynomial in $\\{x,y\\}$ of weighted degree $n$, where

$\operatorname{deg}(x)=2,\quad\operatorname{deg}(y)=3.$

Under this identification, our definition of elliptic polynomials then
coincides with that used in [FOX22], and the transformation (1.3) results in
an equivalent definition of the bases of elliptic polynomials to that given in
(1.2).

When ${\sf w}(z)$ is even in $z$ around the midpoint of the contour $\gamma$,
that is,

${\sf w}(\tfrac{1}{2}(1+\tau)+z)={\sf w}(\tfrac{1}{2}(1+\tau)-z)\quad
z\in[0,1],$ (1.5)

the monic elliptic-polynomials naturally split into even and odd polynomials:

$\displaystyle\pi_{2k}(z,\tau)$
$\displaystyle=\sum_{i=0}^{k}\widehat{a}_{i,2k}(\tau)\wp(z)^{k-i},\qquad\widehat{a}_{0,2k}=1,$
(1.6) $\displaystyle\pi_{2k+3}(z,\tau)$
$\displaystyle=-\sum_{i=0}^{k}\widehat{a}_{i,2k+3}(\tau)\wp(z)^{\prime}\wp(z)^{k-i}/2,\qquad\widehat{a}_{0,2k+3}=1,$
(1.7)

where,

$\displaystyle\pi_{2k}(-z)=\pi_{2k}(z),$
$\displaystyle\pi_{2k+3}(-z)=-\pi_{2k+3}(z).$ (1.8)

Henceforth, we refer to $\pi_{2k}$ and $\pi_{2k+3}$ as the respective even and
odd EOPs. Note that $\pi_{2k}(z)$ is purely a function of $x=\wp(z)$ according
to (1.2), and such polynomials are similar in nature to the Akheizer
polynomials in [IC08], and generalised Jacobi polynomials studied in the
literature by [Car60, Hei78, IVY01, Ree45] among others.

In this paper, we show how the EOPs, for any choice of weight function, can be
written in terms of determinants of moments and characterised as the $(1,1)$
entry of the unique solution of a corresponding $2\times 2$ Riemann-Hilbert
problem (RHP).

The connection between orthogonal polynomials and the RHP was established in
the 1990’s by Fokas, Its, and Kitaev [FIK91]. Since then it has been
instrumental in a number of settings to prove a variety of results for
different classes of orthogonal polynomials
Deift1999strong,qRHP,etnavol25pp369-392, most often finding application in
determining the asymptotics of different classes of orthogonal polynomials,
and universality of random matrix ensembles. RHPs were also used in [Ber22] to
determine large degree asymptotic behaviours within an analytically distinct
class of orthogonal polynomials on elliptic curves, allowing for an additional
pole in the basis.

A family of orthogonal polynomials corresponding to a given weight function
are completely determined by the moments of the weight function [Sze39]. Such
a representation plays an important role in several aspects of orthogonal
polynomials, as well as their applications in areas such as combinatorics and
probability theory, see the survey articles [CKS16, Kön05] for example.
Specifically, the Hankel determinants of moments were exploited to study the
partition functions of ensembles of random matrices and to study the
corresponding tau-functions or, equivalently, the solutions of integrable
equations. See Forrester and Witte [FW06] for this approach in the case of
Painlevé VI and Bertola [Ber08] for other classes of isomonodromic systems. In
the present case, the relevant moments and the determinant are defined as

$\displaystyle\mu_{i,j}:=\int_{\gamma}\mathcal{E}_{i}(z)\mathcal{E}_{j}(z){\sf
w}(z)dz,$ $\displaystyle D_{n}:=\det(\mu_{i,j})_{i,j=0}^{n-1}.$

As is the case for classical orthogonal polynomials [FIK91], the
aforementioned RHP for EOPs is uniquely solvable if and only if the
determinant $D_{n}\neq 0$. Furthermore, the moment matrix is a block matrix
consisting of a checkerboard pattern of odd and even moments. Consequently its
determinant factorises into two Hankel determinants, consisting of even and
odd moments respectively, both of which define Painlevé tau-functions. A
similar result connecting Hankel determinants constructed out of certain
elliptic functions and the tau-function of Painlevé VI was obtained by [BCH15]
from the study of generalized Jacobi polynomials.

###  1.1. Outlook

We list a few possible future directions following the results presented here:

  1. (1)

with the construction in Section 2, one can systematically increase the
complexity of the weight function. For example, a weight function that is
algebraic in $\wp(z)$, which may lead to solutions of other integrable
equations,

  2. (2)

the large degree asymptotics of EOPs can be studied using the RHPs described
here.

  3. (3)

Hankel determinants are related to partition functions of random matrix
ensembles. It would be interesting to see the elliptic extensions of such
relations.

  4. (4)

Another natural question would be the extension of the Riemann-Hilbert setup
developed here to study orthogonal polynomials on higher genus surfaces
[FOX22].

###  1.2. Outline

In Section 2 we begin by describing the general and even polynomials in terms
of the determinants of their moments, and the construct the solutions $Y_{n}$
and $Y_{2k}$ of the Riemann-Hilbert problems associated to each class of
polynomials respectively. Theorem 2.1 proves the existence and uniqueness of
the solution of the general RHP. For the remainder of the paper we set the
weight ${\sf w}(z)\equiv 1$.

In Section 3, we obtain differential and discrete linear systems satisfied by
the solution of the general RHP $Y_{n}$ in Propositions 3.1 and 3.2
respectively. In Theorems 3.1 and 3.2 we derive recurrence relations for the
polynomials $\pi_{n}$ and the coefficients.

In Section 4, Theorem 4.1 shows that the linear problems satisfied by the
solution of the even RHP $Y_{2k}$ describe the Lax pair of the elliptic form
of Painlevé VI. For $k=1$, the monodromy exponents assume a special form which
is related to the Hitchin case of Painlevé VI as shown in Proposition 4.3.

In Section 5, we derive explicit solutions of Painlevé VI built out of Hankel
determinants of even moments, see Theorem 5.1. Furthermore, we show that these
Hankel determinants are the corresponding Painlevé VI tau-functions, in
Theorem 5.2.

###  1.3. Acknowledgements

We thank Nalini Joshi and Milena Radnović for valuable discussions and
suggestions. H.D. acknowledges the support of Australian Research Council
Discovery Project #DP200100210. P.R. acknowledges the support of Australian
Research Council Discovery Project #DP210100129. TL’s research was supported
the Australian Government Research Training Program and by the University of
Sydney Postgraduate Research Supplementary Scholarship in Integrable Systems.

A part of the work was done during the authors’ residence at the Isaac Newton
Institute during the Fall 2022 semester and they thank the Møller institute
and organisers of the program ”Applicable resurgent asymptotics: towards a
universal theory” for their hospitality. HD thanks the Simons foundation for
supporting her INI visit.

##  2\. Riemann-Hilbert problems and moments

In this section, we begin by detailing the representation of the general and
even EOPs in terms of determinants of moments. We then show show that the
consecutive EOPs, and their suitable Cauchy transforms, form a unique solution
to a corresponding Riemann-Hilbert problem (RHP). We relate the existence and
uniqueness of the polynomials to the unique solvability of the RHP and find
explicit expressions for the determinants of the solutions. This will in turn
allow us to derive differential and difference linear systems satisfied by the
polynomials for the case ${\sf w}(z)\equiv 1$.

###  2.1. Moments in the general case

We start by assuming $\tau\in i\mathbb{R}$, so that the Weierstrass
$\wp$-function is real on $\gamma$, i.e the basis $\mathcal{B}$ of elliptic
polynomials consists of real functions on $\gamma$. If, in addition, the
weight function ${\sf w}(z)$ is strictly positive, then

$\langle f,g\rangle=\int_{\gamma}{f(z)g(z){\sf w}(z)dz}$

defines an inner product on the space of real elliptic polynomials and the
Gram–Schmidt process shows that the corresponding EOPs exist and are unique,
with $\pi_{n}$ given by

$\pi_{n}(z)=D_{n}^{-1}\begin{vmatrix}\mu_{0,0}&\mu_{0,2}&\mu_{0,3}&\ldots&\mu_{0,n}\\\
\mu_{2,0}&\mu_{2,2}&\mu_{2,3}&\ldots&\mu_{2,n}\\\
\mu_{3,0}&\mu_{3,2}&\mu_{3,3}&\ldots&\mu_{3,n}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
\mu_{n-1,0}&\mu_{n-1,1}&\mu_{n-1,2}&\ldots&\mu_{n-1,n}\\\
\mathcal{E}_{0}(z)&\mathcal{E}_{2}(z)&\mathcal{E}_{3}(z)&...&\mathcal{E}_{n}(z)\end{vmatrix},$
(2.1)

where

$\mu_{i,j}=\int_{\gamma}\mathcal{E}_{i}(z)\mathcal{E}_{j}(z){\sf
w}(z)dz\qquad(i,j\in\mathbb{N}_{\neq 1}),$ (2.2)

and

$D_{n}=|S_{n}|,\quad
S_{n}:=\begin{vmatrix}\mu_{0,0}&\mu_{0,2}&\mu_{0,3}&\ldots&\mu_{0,n-1}\\\
\mu_{2,0}&\mu_{2,2}&\mu_{2,3}&\ldots&\mu_{2,n-1}\\\
\mu_{3,0}&\mu_{3,2}&\mu_{3,3}&\ldots&\mu_{3,n-1}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
\mu_{n-1,0}&\mu_{n-1,1}&\mu_{n-1,2}&\ldots&\mu_{n-1,n-1}\end{vmatrix}.$ (2.3)

See Appendix C for an illustration. The following identity then follows from
the above equation,

$h_{n}:=\int_{\frac{1}{2}\tau+0}^{\frac{1}{2}\tau+1}\pi_{n}(z)^{2}{\sf
w}(z)dz=\int_{\frac{1}{2}\tau+0}^{\frac{1}{2}\tau+1}\mathcal{E}_{n}\pi_{n}(z){\sf
w}(z)dz=\frac{D_{n+1}}{D_{n}}.$ (2.4)

For $\tau\notin i\mathbb{R}$, or ${\sf w}(z)$ is not strictly positive, the
polynomial $\pi_{n}(z)$ exists and is unique if and only if the determinant
$D_{n}$ is nonzero. Similarly, we will find that the associated RHP has a
unique solution if and only if $D_{n}\neq 0$.

In Figures 2, 3 and 4, the first couple of EOP’s are plotted on $\gamma$, with
$\tau=i$ and ${\sf w}(z)\equiv 1$.

Figure 2. Graph displaying function values of the even EOPs $\pi_{0}$,
$\pi_{2}$, $\pi_{4}$ and $\pi_{6}$, on the interval $\tfrac{1}{2}\tau+[0,1]$,
in red, blue, green and purple respectively, with $\tau=i$. Figure 3. Graph
displaying function values of the odd EOPs $\pi_{3}$, $\pi_{5}$, $\pi_{7}$ and
$\pi_{9}$, on the interval $\tfrac{1}{2}\tau+[0,1]$, in red, blue, green and
purple respectively, with $\tau=i$. Figure 4. Graph displaying function values
of the EOPs $\pi_{3}$, $\pi_{4}$, $\pi_{5}$ and $\pi_{6}$, on the interval
$\tfrac{1}{2}\tau+[0,1]$, in red, blue, green and purple respectively, with
$\tau=i$.

###  2.2. Riemann-Hilbert problem for the general case

The first step in defining the RHP is to analyse the Cauchy transform of the
polynomials

$\displaystyle\mathcal{C}(\pi_{n})(z):=\int_{\gamma}\pi_{n}(w)C(w,z){\sf
w}(w)\frac{dw}{2\pi i}.$ (2.5)

Generally, the Cauchy kernel is a meromorphic function in $z$ and a one form
in $w$, with residues $\pm 1$ at its poles, i.e it is an Abelian differential
of the third kind. The precise form of such a kernel on a torus is not unique
and several examples can be found in the literature [BI88, Rod87, DDG20,
Ber22]. For the present case, we consider the scalar kernel

$\displaystyle C(w,z)=\zeta(w-z)-\zeta(w),$ (2.6)

where $\zeta(.)$ denotes the Weierstrass $\zeta$-function (see A). This Cauchy
kernel has the following properties.

  1. (1)

The periodicity of $\zeta$-function, see equation (A.6), implies that

$\displaystyle C(w,z+1)=C(w,z)-\eta_{1}(\tau),$ $\displaystyle
C(w,z+\tau)=C(w,z)-\eta_{2}(\tau).$ (2.7)

  2. (2)

The kernel $C(w,z)$ has poles at $w=z,w=0$ with residues $\pm 1$ respectively,

  3. (3)

In the limit $w\to z$,

$\displaystyle C(w,z)=\frac{1}{w-z}+\mathcal{O}(w-z),$ (2.8)

  4. (4)

and in the limit $z\to 0$,

$\displaystyle
C(w,z)=\sum_{k=1}^{\infty}(-1)^{k}\frac{z^{k}}{k!}\zeta^{(k)}(w),$ (2.9)

where $\zeta^{(k)}(\cdot)$ denotes the $k$th derivative of $\zeta$.

The following statements then hold true for (2.5).

######  Lemma 2.1.

The Cauchy transform of $\pi_{n}$ is a piecewise analytic, doubly periodic
function for $n\geq 2$.

###### Proof.

It is immediate from the definition, that $\mathcal{C}(\pi_{n})(z)$ is
analytic away from $\gamma$. From the relations (2.7), we see that

$\displaystyle\mathcal{C}(\pi_{n})(z+1)=\mathcal{C}(\pi_{n})(z)-\int_{\gamma}\eta_{1}(\tau)\pi_{n}(w){\sf
w}(w)\frac{dw}{2\pi i},$ (2.10)

and due to orthogonality (1.1), the second term vanishes for $n\neq 0$.
Similar computation holds for $\mathcal{C}(\pi_{n})(z+\tau)$. ∎

######  Lemma 2.2.

In the asymptotic limit $z\to 0$, the polynomials have an asymptotic expansion
of the form

$\displaystyle\pi_{n}(z)=\sum_{j=0}^{\infty}\frac{c_{j,n}(\tau)}{z^{n-j}},\qquad
c_{0,n}=1,$ (2.11)

and, similarly, their Cauchy transform have an asymptotic expansion of the
form,111all the coefficients are $\tau$-dependent unless stated otherwise. We
often omit the $\tau$\- dependence of the coefficients for ease of notation.

$\displaystyle\mathcal{C}(\pi_{n})(z)=\frac{h_{n}(\tau)}{2\pi
i}\sum_{j=0}^{\infty}\widetilde{c}_{j,n}(\tau)z^{n+j-1},$
$\displaystyle\widetilde{c}_{0,n}=1,$
$\displaystyle\widetilde{c}_{1,n}=-c_{1,n+1}.$ (2.12)

###### Proof.

Using the Taylor expansion around $z=0$ of the Cauchy kernel in equation
(2.9), which holds uniformly in $w\in\gamma$, we obtain the following
asymptotic expansion as $z\rightarrow 0$,

$\displaystyle\mathcal{C}(\pi_{n})(z)$ $\displaystyle=\frac{1}{2\pi
i}\int_{\gamma}\sum_{k=1}^{\infty}(-1)^{k}\frac{z^{k}}{k!}\zeta^{(k)}(w)\pi_{n}(w){\sf
w}(w)\frac{dw}{2\pi i}$ (2.13) $\displaystyle=\frac{1}{2\pi
i}\sum_{k=1}^{\infty}(-1)^{k}\frac{z^{k}}{k!}\int_{\gamma}\zeta^{(k)}(w)\pi_{n}(w){\sf
w}(w)\frac{dw}{2\pi i}$ (2.14) $\displaystyle=\frac{1}{2\pi
i}\sum_{k=n-1}^{\infty}(-1)^{k}\frac{z^{k}}{k!}\int_{\gamma}\zeta^{(k)}(w)\pi_{n}(w){\sf
w}(w)\frac{dw}{2\pi i}$ (2.15) $\displaystyle=\frac{h_{n}(\tau)}{2\pi
i}\sum_{j=0}^{\infty}\widetilde{c}_{j,n}(\tau)z^{n+j-1},$ (2.16)

where, in the third equality we used orthogonality and the fact that
$\zeta^{(k)}(w)$ is an elliptic polynomial of degree $k$, for $k\geq 1$, with
coefficients given by

$\frac{h_{n}(\tau)}{2\pi
i}\widetilde{c}_{j,n}(\tau)=\frac{(-1)^{n-1+j}}{(n-1+j)!}\int_{\gamma}\zeta^{(n-1+j)}(w)\pi_{n}(w){\sf
w}(w)\frac{dw}{2\pi i},$

for $j\geq 0$. To compute the leading order coefficients, note that

$\frac{(-1)^{n-1+j}}{(n-1+j)!}\zeta^{(n-1+j)}(w)=w^{-(n+j)}(1+\mathcal{O}(w^{2})),$

as $w\rightarrow 0$, so that

$\frac{(-1)^{n-1}}{(n-1)!}\int_{\gamma}\zeta^{(n-1)}(w)\pi_{n}(w){\sf
w}(w)dw=\int_{\gamma}\pi_{n}(w)^{2}dw=h_{n},$

and

$\frac{(-1)^{n}}{n!}\int_{\gamma}\zeta^{(n)}(w)\pi_{n}(w){\sf
w}(w)dw=\int_{\gamma}(\pi_{n+1}(w)-c_{1,n+1}\pi_{n}(z))\pi_{n}(w){\sf
w}(w)dw=-c_{1,n+1}h_{n}.$

Therefore, $\widetilde{c}_{0,n}=1$ and $\widetilde{c}_{1,n}=-c_{1,n+1}$ and
the lemma follows. ∎

######  Remark 2.1.

For ${\sf w}(z)\equiv 1$, the value of $h_{n}$ specialises and all the odd-
indexed coefficients vanish, that is, $c_{j,n}=\widetilde{c}_{j,n}=0$ for
$j\geq 1$ odd.

Let us now define a RHP such that the 11 entry of its solution is the
polynomial $\pi_{n}(z)$.

Riemann-Hilbert problem 1.  The Riemann-Hilbert Problem comprises of finding a
$2\times 2$ matrix valued function $Y_{n}(z,\tau)$ with the following
properties:

  * •

$Y_{n}(z,\tau)$ is analytic in $z\in\mathbb{T}\setminus(\gamma\cup\\{0\\})$.

  * •

The following jump condition hold for $z\in\gamma$:

$\displaystyle Y_{n,+}(z,\tau)=Y_{n,-}(z,\tau)\left(\begin{array}[]{cc}1&{\sf
w}(z)\\\ 0&1\end{array}\right),$ (2.19)

where, following the standard notation, $\pm$ indicate the piece-wise analytic
functions to the left and right side respectively of $\gamma$ w.r.t its
orientation, see Figure 1.

  * •

In the limit $z\to 0$:

$\displaystyle
Y_{n}(z,\tau)=\left(\mathbb{1}+\mathcal{O}(z)\right)\left(\begin{array}[]{cc}z^{-n}&0\\\
0&z^{n-2}\end{array}\right).$ (2.22)

######  Theorem 2.1.

Let $n\geq 3$, then RHP 2.1 is uniquely solvable if and only if the
determinant $D_{n}\neq 0$, in which case the solution is given by

$\displaystyle
Y_{n}(z,\tau)=\left(\begin{array}[]{cc}\pi_{n}(z)&\mathcal{C}(\pi_{n})(z)\\\
\frac{2\pi i}{h_{n-1}}\pi_{n-1}(z)&\frac{2\pi
i}{h_{n-1}}\mathcal{C}(\pi_{n-1})(z)\end{array}\right),\qquad(n\geq 3).$
(2.25)

###### Proof.

To prove the theorem, we analyse the two rows of a solution to the RHP
separately.

We start with the first row, $(Y_{11}(z),Y_{12}(z))$. The conditions imposed
in the RHP translate to

  * •

$Y_{11}(z)$ and $Y_{12}(z)$ are analytic and doubly periodic on
$\mathbb{T}\setminus(\gamma\cup\\{0\\})$,

  * •

$Y_{11}(z)$ is analytic on $\gamma$ and $Y_{12}(z)$ satisfies the jump
condition

$(Y_{12})_{+}(x)=(Y_{12})_{-}(x)+Y_{11}(x){\sf w}(x)\qquad(x\in\gamma).$

  * •

asymptotic conditions

$Y_{11}(z)=z^{-n}(1+\mathcal{O}(z)),\quad
Y_{12}(z)=\mathcal{O}(z^{n-1});\qquad z\rightarrow 0.$

We refer to this as the row one RHP.

Note that, $Y_{11}(z)$ is necessarily a monic elliptic polynomial of degree
$n$. In order to study $Y_{12}(z)$, we first consider the following function,

$r(z):=Y_{12}(z)-\mathcal{C}(Y_{11})(z)+\zeta(z)\int_{\frac{\tau}{2}}^{\frac{\tau}{2}+1}Y_{11}(x){\sf{\sf
w}}(x)dx.$

Note that $r(z)$ has no jump on $\gamma$ and it is periodic with respect to
$1$ and $\tau$, so it must be an elliptic function. Furthermore, by the
asymptotic conditions,

$r(z):=z^{-1}\int_{\frac{\tau}{2}}^{\frac{\tau}{2}+1}Y_{11}(x){\sf
w}(x)dx+\mathcal{O}(1)\qquad(z\rightarrow 0).$

Since $r(z)$ has no poles away from the lattice
$\mathbb{Z}+\mathbb{Z}\cdot\tau$ and there exists no elliptic function of
degree one, it follows that $r(z)$ must be a constant and

$\int_{\frac{\tau}{2}}^{\frac{\tau}{2}+1}Y_{11}(x){\sf w}(x)dx=0.$ (2.26)

But, by the last equality, and the asymptotics of $Y_{11}(z)$ and $Y_{12}(z)$,
we now have

$r(z)=Y_{12}(z)-\mathcal{C}(Y_{11})(z)=\mathcal{O}(z^{n-1})$

as $z\rightarrow 0$. Thus, $r(z)\equiv 0$ and it follows that

$Y_{12}(z)=\mathcal{C}(Y_{11})(z).$

In particular, by the asymptotics of $Y_{12}(z)$

$\mathcal{C}(Y_{11})(z)=\mathcal{O}(z^{n-1})\quad(z\rightarrow 0).$

In other words,

$\int_{\frac{\tau}{2}}^{\frac{\tau}{2}+1}Y_{11}(z)\zeta^{(k)}(z){\sf
w}(z)dz=0$

for $1\leq k\leq n-2$, implying that

$\int_{\frac{\tau}{2}}^{\frac{\tau}{2}+1}Y_{11}(z)\mathcal{E}_{m}(z){\sf
w}(z)dz=0$ (2.27)

for $2\leq m\leq n-1$. We have already seen that this equality also holds for
$m=0$, see equation (2.26).

Now, let us write

$Y_{11}(z)=c_{0}\mathcal{E}_{0}+c_{2}\mathcal{E}_{2}+c_{3}\mathcal{E}_{3}+\ldots+c_{n-1}\mathcal{E}_{n-1}+\mathcal{E}_{n}.$

Then equation (2.27) is equivalent to

$S_{n}\begin{pmatrix}c_{0}\\\ c_{2}\\\ \vdots\\\
c_{n-1}\end{pmatrix}=\begin{pmatrix}-\mu_{n,0}\\\ -\mu_{n,2}\\\ \vdots\\\
-\mu_{n,n-1}\end{pmatrix},$ (2.28)

where $S_{n}$ is the matrix defined in equation (2.3).

All in all, we find that a solution $(Y_{11}(z),Y_{12}(z))$ to the row one RHP
exists and is unique if and only if equation (2.28) has a unique solution,
which in turn is true if and only if $D_{n}\neq 0$. In particular, in that
case, $\pi_{n}(z)$ exists and

$Y_{11}(z)=\pi_{n}(z),\quad Y_{12}(z)=\mathcal{C}(\pi_{n})(z).$

The second row of $Y$ is analysed in much the same way. We note that the
second row $(Y_{21}(z),Y_{22}(z))$ must satisfy

  * •

$Y_{21}(z)$ and $Y_{22}(z)$ are analytic and doubly periodic on
$\mathbb{T}\setminus(\gamma\cup\\{0\\})$,

  * •

$Y_{21}(z)$ is analytic on $\gamma$ and $Y_{22}(z)$ satisfies the jump
condition

$(Y_{22})_{+}(x)=(Y_{22})_{-}(x)+Y_{21}(x){\sf w}(x)\qquad(x\in\gamma).$

  * •

asymptotic conditions

$Y_{21}(z)=\mathcal{O}(z^{-(n-1)})),\quad
Y_{22}(z)=z^{n-2}(1+\mathcal{O}(z))\qquad(z\rightarrow 0).$

We refer to this as the row two RHP.

It follows that $Y_{21}(z)$ is required to be an elliptic polynomial of degree
less or equal to $n-1$,

$Y_{22}(z)=\mathcal{C}(Y_{21})(z),$

and we find that

$\int_{\frac{\tau}{2}}^{\frac{\tau}{2}+1}Y_{21}(z)\mathcal{E}_{m}(z){\sf
w}(z)dz=\delta_{m,n-1}2\pi i,$ (2.29)

for $m=0,2,3,\ldots,n-1$. Upon writing

$Y_{21}(z)=c_{0}\mathcal{E}_{0}+c_{2}\mathcal{E}_{2}+c_{3}\mathcal{E}_{3}+\ldots+c_{n-1}\mathcal{E}_{n-1},$

this is equivalent to

$S_{n}\begin{pmatrix}c_{0}\\\ c_{2}\\\ \vdots\\\ c_{n-2}\\\
c_{n-1}\end{pmatrix}=\begin{pmatrix}0\\\ 0\\\ \vdots\\\ 0\\\ 2\pi
i\end{pmatrix},$ (2.30)

where $S_{n}$ is the matrix defined in equation (2.3).

All in all, we find that a solution $(Y_{21}(z),Y_{22}(z))$ to row two RHP
exists and is unique if and only if equation (2.30) has a unique solution,
which in turn is true if and only if $D_{n}\neq 0$.

Furthermore, if $D_{n}\neq 0$, then

$\frac{2\pi i}{h_{n-1}}\pi_{n-1}(z)=\frac{2\pi
i}{D_{n}}\;\widetilde{\pi}_{n-1}(z),$

is well-defined and necessarily

$Y_{21}(z)=\frac{2\pi i}{h_{n-1}}\pi_{n-1}(z),\quad Y_{22}(z)=\frac{2\pi
i}{h_{n-1}}\mathcal{C}(\pi_{n-1})(z).$

We have now shown that both rows of a solution to the RHP exist if and only if
$D_{n}\neq 0$, and, in such case, we proven that formula (2.1) holds. The
theorem follows. ∎

The asymptotic expansions in Lemma 2.2 now help us obtain the precise form of
the determinant of the solution of the RHP: 2.1 $Y_{n}$.

######  Lemma 2.3.

For ${\sf w}(z)\equiv 1$, the determinant of the solution takes the form

$\displaystyle\det Y_{n}(z,\tau)=\wp(z,\tau)+\alpha_{n}(\tau)=:f_{n}(z),$
(2.31)

where $\alpha_{n}(\tau)$ is given by

$\displaystyle\alpha_{n}:=c_{2,n}+\widetilde{c}_{2,n-1}-\beta_{n},$
$\displaystyle\beta_{n}:=\frac{h_{n}}{h_{n-1}}.$ (2.32)

###### Proof.

Setting ${\sf w}(z)\equiv 1$, let us begin by noting that $\det Y_{n}$

  1. (1)

is a doubly-periodic function

$\displaystyle\det Y_{n}(z+\tau)=\det Y_{n}(z+1)=\det Y_{n}(z),$ (2.33)

  2. (2)

and has no jump on $\gamma$,

$\displaystyle\det Y_{n,+}=\det Y_{n,-}.$ (2.34)

With the asymptotic behaviour of $\pi_{n}(z)$ (2.11), and
$\mathcal{C}(\pi_{n})(z)$ (2.12) in the determinant of $Y_{n}$ (2.25), we see
that in the limit $z\to 0$

$\displaystyle\det
Y_{n}(z,\tau)=\frac{1}{z^{2}}+c_{2,n}+\widetilde{c}_{2,n-1}-\frac{h_{n}}{h_{n-1}}+\mathcal{O}(z^{2}).$
(2.35)

Since $\det Y_{n}$ is an elliptic function due to properties (2.33)-(2.34), it
only has one pole, which is of order 2 at $z=0$, and therefore must be equal
to the Weierstrass $\wp(z)$-function plus a constant $\alpha_{n}$ defined in
(2.25), (2.32). ∎

Let us now restrict to even polynomials and repeat the methods presented
above.

###  2.3. Hankel determinants and the even case

Consider the vector space of even elliptic polynomials,

$\mathcal{P}_{\text{even}}=\mathcal{P}_{\text{even}}(\tau)=\\{\text{even
elliptic functions with only a pole at $0$}\\},$

with corresponding basis

$\\{1,\wp(z),\wp(z)^{2},\wp(z)^{3},\ldots\\}.$

Let ${\sf w}$ be an even weight function on the interval
$\tfrac{1}{2}\tau+[0,1]$, that is,

${\sf w}(\tfrac{1}{2}+\tfrac{1}{2}\tau+x)={\sf
w}(\tfrac{1}{2}+\tfrac{1}{2}\tau-x),\qquad x\in\tfrac{1}{2}\tau+[0,1].$

For any natural number $k\in\mathbb{N}$, we define the $k$th orthogonal
polynomial $\pi_{2k}(z)$ with respect to ${\sf w}$, if it exists, by the
conditions

$\displaystyle\int_{\frac{1}{2}\tau+0}^{\frac{1}{2}\tau+1}\wp(z)^{m}\pi_{2k}(z){\sf
w}(z)dz=0,\quad(0\leq m<2k),$
$\displaystyle\pi_{2k}(z)=\wp(z)^{k}(1+\mathcal{O}(\wp(z)^{-1}))\quad(z\rightarrow
0).$

For $i+j=k\in\mathbb{N}$,

$\displaystyle\nu_{2i,2j}\equiv\nu_{2(i+j)}=\nu_{2k}:=\int_{\frac{1}{2}\tau+0}^{\frac{1}{2}\tau+1}\wp(z)^{k}{\sf
w}(z)dz.$ (2.36)

Then, $\pi_{2k}(z)$ exists if and only if the Hankel determinant of moments

$\Delta_{2k}:=\begin{vmatrix}\nu_{0}&\nu_{2}&\ldots&\nu_{2k-2}\\\
\nu_{2}&\nu_{4}&\ldots&\nu_{2k}\\\ \vdots&\vdots&\ddots&\vdots\\\
\nu_{2k-2}&\nu_{2k}&\ldots&\nu_{4k-2}\end{vmatrix},$ (2.37)

is nonzero, with $\Delta_{0}=1$. In turn, $\pi_{2k}$ is explicitly given by

$\pi_{2k}(z)=\frac{1}{\Delta_{2k}}\begin{vmatrix}\nu_{0}&\nu_{2}&\ldots&\nu_{2k}\\\
\nu_{2}&\nu_{4}&\ldots&\nu_{2k+2}\\\ \vdots&\vdots&\ddots&\vdots\\\
\nu_{2k-2}&\nu_{2k}&\ldots&\nu_{4k-2}\\\
1&\wp(z)&\ldots&\wp(z)^{k}\end{vmatrix},$

further implying that

$h_{2k}:=\int_{\frac{1}{2}\tau+0}^{\frac{1}{2}\tau+1}\pi_{2k}(z)^{2}{\sf
w}(z)dz=\int_{\frac{1}{2}\tau+0}^{\frac{1}{2}\tau+1}\wp(z)^{k}\pi_{2k}(z){\sf
w}(z)dz=\frac{\Delta_{2k+2}}{\Delta_{2k}}.$ (2.38)

Another direct consequence is that the even polynomial can now be expanded as

$\pi_{2k}(z)=\wp(z)^{k}-\frac{\Gamma_{2k}}{\Delta_{2k}}\wp(z)^{k-1}+\frac{\Lambda_{2k}}{\Delta_{2k}}\wp(z)^{k-2}+\mathcal{O}(\wp(z)^{k-3}),$
(2.39)

where

$\displaystyle\Gamma_{2k}$
$\displaystyle:=\begin{vmatrix}\nu_{0}&\nu_{2}&\ldots&\nu_{2k-4}&\nu_{2k}\\\
\nu_{2}&\nu_{4}&\ldots&\nu_{2k-2}&\nu_{2k+2}\\\ \
\vdots&\vdots&\ddots&\vdots&\vdots\\\
\nu_{2k-2}&\nu_{2k}&\ldots&\nu_{4k-6}&\nu_{4k-2}\end{vmatrix},$
$\displaystyle\Lambda_{2k}$
$\displaystyle:=\begin{vmatrix}\nu_{0}&\nu_{2}&\ldots&\nu_{2k-6}&\nu_{2k-2}&\nu_{2k}\\\
\nu_{2}&\nu_{4}&\ldots&\nu_{2k-4}&\nu_{2k}&\nu_{2k+2}\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\
\nu_{2k-2}&\nu_{2k}&\ldots&\nu_{4k-8}&\nu_{4k-4}&\nu_{4k-2}\end{vmatrix},$

for $k\geq 1$ with $\Gamma_{2}=\mu_{2}$ and
$\Gamma_{0}=\Lambda_{2}=\Lambda_{0}=0$.

######  Proposition 2.1.

For ${\sf w}(z)\equiv 1$, the moments satisfy the following recursion

$\displaystyle(8k+12)\nu_{2k+4}=(2k+1)g_{2}\nu_{2k}+2kg_{3}\nu_{2k-2}.$ (2.40)

###### Proof.

With the prescribed weight function, the moment (2.36) reads

$\nu_{2k}=\int_{\frac{1}{2}\tau+0}^{\frac{1}{2}\tau+1}\wp(z)^{k}dz=\oint_{\text{cycle}}\frac{x^{k}}{y}dx,$
(2.41)

where $y^{2}=4x^{3}-g_{2}x-g_{3}$, as can be seen by a change of variables
$x=\wp(z)$. Note that expression

$\frac{d}{dx}x^{k}y=kx^{k-1}y+\frac{1}{2y}(12x^{k+2}-g_{2}x^{k}),$

from which we obtain the following identity

$k\oint_{cycle}x^{k-1}y\,dx=\tfrac{g_{2}}{2}\nu_{2k}-6\nu_{2k+4},$ (2.42)

for $k\geq 1$. Therefore,

$\displaystyle 4\nu_{2k+4}=$ $\displaystyle
4\oint_{\text{cycle}}\frac{4x^{k+2}}{y}dx,$ (2.43) $\displaystyle=$
$\displaystyle\oint_{\text{cycle}}\frac{x^{k-1}y^{2}+g_{2}x^{k}+g_{3}x^{k-1}}{y}dx$
(2.44) $\displaystyle=$
$\displaystyle\oint_{\text{cycle}}x^{k-1}ydx+g_{2}\nu_{2k}+g_{3}\nu_{2k-2}$
(2.45) $\displaystyle=$
$\displaystyle\frac{1}{k}(\tfrac{g_{2}}{2}\nu_{2k}-6\nu_{2k+4})+g_{2}\nu_{2k}+g_{3}\nu_{2k-2},$
(2.46)

where we used identity (2.42) in the last equality. This gives the recursion
in the proposition. ∎

The first values are

$\displaystyle\nu_{0}=1,$ $\displaystyle\nu_{1}=-2\eta_{1},$
$\displaystyle\nu_{2}=\frac{1}{12}g_{2},$
$\displaystyle\nu_{3}=\frac{1}{10}(g_{3}-3g_{2}\eta_{1}),$

where $\eta_{1}=\zeta(\frac{1}{2})$ is the first period of the second kind.
Correspondingly, the first few even Hankel determinants, defined in equation
(2.37), are given by

$\displaystyle\Delta_{0}=\Delta_{2}=1,$
$\displaystyle\Delta_{4}=\frac{1}{12}(g_{2}-48\eta_{1}^{2}),$
$\displaystyle\Delta_{6}=\frac{1}{37800}(25g_{2}^{3}-378g_{3}^{2}+108g_{2}g_{3}\eta_{1}-1872g_{2}^{2}\eta_{1}^{2}+43200g_{3}\eta_{1}^{3}).$

The Hankel determinants of moments are functions of the modular parameter
$\tau\in\mathbb{H}$, and they can be written as explicit functions in $\tau$
using the equations for $g_{2},g_{3}$ and $\eta_{1}$, as functions of $\tau$,
in Appendix A. They satisfy the following symmetries,

$\Delta_{2k}(\tau)^{\diamond}=\Delta_{2k}(\tau^{\diamond}),\quad\Delta_{2k}(\tau+2)=\Delta_{2k}(\tau)\quad(\tau\in\mathbb{H}),$

where $(\alpha+\beta i)^{\diamond}=-\alpha+\beta i$ for
$\alpha,\beta\in\mathbb{R}$.

In Figure 5, the zero distribution of $\Delta_{2k}(\tau)$ is displayed in
blue, for $k=2,4,6,8$, in the upper-half plane cut off by

$\Im\;\tau\geq\tfrac{1}{\pi}\log(\tfrac{5}{4}).$

The reason for this cut, is that the numerics become unstable near the real
line. In particular, even though only finitely many zeros are shown in the
plots, there might in fact be an infinite number of zeros accumulating at
points $\tau\in\mathbb{Z}$, for the even Hankel determinants $\Delta_{2k}$,
$k\geq 2$.

(a) $k=2$

(b) $k=3$

(c) $k=4$

(d) $k=5$

Figure 5. In these plots the zero distributions of $\Delta_{2k-1}(\tau)$ and
$\Delta_{2k}(\tau)$ are displayed in respectively red and blue, for
$k=2,3,4,5$, in the domain defined by $-1\leq\Re\;\tau\leq 1$ and
$\Im\;\tau\geq\tfrac{1}{\pi}\log(\tfrac{5}{4})\approx 0.071$. Furthermore, in
each plot the dashed line is the line
$\Im\;\tau=\tfrac{1}{\pi}\log(\tfrac{5}{4})$.

######  Remark 2.2.

A similar computation follows for the odd case starting from (2.2):

$\displaystyle\nu_{2i+3,2j+3}=\int_{\frac{\tau}{2}}^{1+\frac{\tau}{2}}(\wp^{\prime}(z))^{2}\wp(z)^{i+j}dz=\oint_{\text{cycle}}yx^{k}dx$
(2.47)

with the change of variables as in (2.41). After a direct manipulation, we get

$\displaystyle\nu_{2k+3}=4\nu_{2k+6}-g_{2}\nu_{2k+4}-g_{3}\nu_{2k}.$ (2.48)

The determinant of the odd moments is

$\Delta_{2k+3}:=\begin{vmatrix}\nu_{3}&\nu_{5}&\ldots&\nu_{2k+1}\\\
\nu_{5}&\nu_{7}&\ldots&\nu_{2k+3}\\\ \vdots&\vdots&\ddots&\vdots\\\
\nu_{2k+1}&\nu_{2k+3}&\ldots&\nu_{4k-1}\end{vmatrix}.$

We have plotted the zero distributions of the first couple of odd Hankel
determinants in red in Figure 5.

Note that $\Delta_{2k}$ and $\Delta_{2k+1}$ are Hankel determinants, for
$k\geq 0$. However, the determinant $D_{n}$ is generally not Hankel, but has a
checkerboard pattern of even and odd moments leading to the factorisation

$\displaystyle D_{n}=\Delta_{n}\Delta_{n+1},$ $\displaystyle n\geq 2.$

###  2.4. Riemann-Hilbert problem for the even case

If the weight function is even on $\gamma$, i.e. equation (1.5) holds, the
sequence of orthogonal polynomials splits into two sequences of even and odd
EOPs and we can define a RHP corresponding to the even polynomials $\pi_{2k}$,
$k\geq 0$. The analysis for the odd case mirrors the even one and so we
restrict our analysis to the even case.

Analogous to RHP 2.1, the even polynomials $\pi_{2k}$ appear as the 1,1
elements of the solution

$\displaystyle Y_{2k}(z,\tau)$
$\displaystyle=\left(\begin{array}[]{cc}\pi_{2k}(z)&\mathcal{C}(\pi_{2k}{\sf
w})(z)\\\ \frac{2\pi i}{h_{2k-2}}\pi_{2k-2}(z)&\frac{2\pi
i}{h_{2k-2}}\mathcal{C}(\pi_{2k-2}{\sf w})(z)\end{array}\right),$
$\displaystyle k\geq 1,$ (2.51)

to the following RHP.

Riemann-Hilbert problem 2.

  * •

The function $Y_{2k}(z,\tau)$ is piece-wise analytic on
$\mathbb{T}\backslash\left(\gamma\cup\\{0\\}\right)$,

  * •

for $z\in\gamma$, the following jump condition holds

$\displaystyle
Y_{2k,+}(z,\tau)=Y_{2k,-}(z,\tau)\left(\begin{array}[]{cc}1&{\sf w}(z)\\\
0&1\end{array}\right),$ (2.54)

  * •

and in the limit $z\to 0$,

$\displaystyle Y_{2k}(z,\tau)$
$\displaystyle=\left(\mathbb{1}+\mathcal{O}(z)\right)\left(\begin{array}[]{cc}z^{-2k}&0\\\
0&z^{2k-3}\end{array}\right).$ (2.57)

Analogously to Theorem 2.1, RHP is 2.4 is solvable if and only if the Hankel
determinant of even moments $\Delta_{2k}(\tau)\neq 0$.

Mimicking the analysis for the general polynomials, we begin with the
asymptotic behaviour of the even polynomials and their Cauchy transform.

######  Lemma 2.4.

In the limit $z\to 0$, the polynomials

$\displaystyle\pi_{2k}(z)=\sum_{i=0}^{\infty}\frac{a_{i,2k}}{z^{2(k-i)}},\qquad
a_{0,2k}=1,$ (2.58)

and their Cauchy transform

$\displaystyle\mathcal{C}(\pi_{2k})(z)=\frac{h_{2k}}{2\pi
i}\sum_{i=0}^{\infty}\widetilde{a}_{i,2k}z^{2(k+i)-1},\qquad\widetilde{a}_{0,2k}=1,$
$\displaystyle\widetilde{a}_{1,2k}=-a_{1,2k+2}.$ (2.59)

###### Proof.

The proof is the same as for Lemma 2.2. ∎

######  Lemma 2.5.

The determinant of $Y_{2k}$ is

$\displaystyle\det Y_{2k}(z,\tau)=-\frac{\wp^{\prime}(z)}{2}.$ (2.60)

###### Proof.

Consider the determinant $\det Y_{2k}(z)$ of the solution to RHP 2.4. Note
that $\det Y_{2k}(z)$ has a trivial jump along $\gamma$ and thus extends to an
elliptic function with only a pole at $z=0$. Furthermore, from the asymptotic
behaviour of $Y_{2k}(z)$, it follows that

$\det Y_{2k}(z)=z^{-3}(1+\mathcal{O}(z))\qquad(z\rightarrow 0).$

Finally, note that $Y_{2k}(-z)\sigma_{3}$ also solves RHP 2.4, thus
$Y_{2k}(z)=Y_{2k}(-z)\sigma_{3}$, and $\det Y_{2k}(-z)=-\det Y_{2k}(z)$. We
have now shown that $\det Y_{2k}(z)$ is an odd, monic, elliptic polynomial of
degree $3$. There exists only one such polynomial,
$-\frac{1}{2}\wp^{\prime}(z)$. The lemma follows. ∎

##  3\. Linear problems for the general case and recurrence relations

In this section we show that the solution of the RHP 2.1 for the general
polynomials $\pi_{n}$ satisfies a system of linear differential and difference
equations, and their compatibility condition leads to the recurrence relation
for the polynomials and the recursion for their coefficients. Henceforth, we
use the notation

${}^{\prime}=\frac{d}{dz},$ $\displaystyle\dot{}=\frac{d}{d\tau}$

######  Proposition 3.1.

The solution of the RHP: 2.1 solves the linear difference equation

$\displaystyle Y_{n+1}=R_{n}Y_{n},$ $\displaystyle
R_{n}=\frac{1}{f_{n}}\left(\begin{array}[]{cc}-\wp^{\prime}(z)/2&-\frac{h_{n}}{2\pi
i}f_{n+1}\\\ \frac{2\pi i}{h_{n}}f_{n}&0\end{array}\right),$ $\displaystyle
n\geq 3,$ (3.3)

where $f_{n}$ is the determinant of $Y_{n}$ (2.31).

###### Proof.

We begin with the $(n+1)^{th}$ solution (2.25)

$\displaystyle
Y_{n+1}(z,\tau)=\left(\begin{array}[]{cc}\pi_{n+1}(z)&\mathcal{C}(\pi_{n+1})(z)\\\
\frac{2\pi i}{h_{n}}\pi_{n}(z)&\frac{2\pi
i}{h_{n}}\mathcal{C}(\pi_{n})(z)\end{array}\right),$ (3.6)

and observe that

$\displaystyle Y_{n+1}Y_{n}^{-1}\det Y_{n}$
$\displaystyle=\left(\begin{array}[]{cc}\pi_{n+1}(z)&\mathcal{C}(\pi_{n+1})(z)\\\
\frac{2\pi i}{h_{n}}\pi_{n}(z)&\frac{2\pi
i}{h_{n}}\mathcal{C}(\pi_{n})(z)\end{array}\right)\left(\begin{array}[]{cc}\frac{2\pi
i}{h_{n-1}}\mathcal{C}(\pi_{n-1})(z)&-\mathcal{C}(\pi_{n})(z)\\\ -\frac{2\pi
i}{h_{n-1}}\pi_{n-1}(z)&\pi_{n}(z)\end{array}\right)$ (3.11)
$\displaystyle=\left(\begin{array}[]{cc}\frac{2\pi
i}{h_{n-1}}\left(\pi_{n+1}\mathcal{C}(\pi_{n-1})-\pi_{n-1}\mathcal{C}(\pi_{n+1})\right)&-\frac{h_{n}}{2\pi
i}\det(Y_{n+1})\\\ \frac{2\pi i}{h_{n}}\det(Y_{n})&0\end{array}\right).$
(3.14)

The 11 element in the above expression is determined as follows. In the limit
$z\to 0$,

$\displaystyle\frac{2\pi
i}{h_{n-1}}\left(\pi_{n+1}(z)\mathcal{C}(\pi_{n-1})(z)-\pi_{n-1}(z)\mathcal{C}(\pi_{n+1})(z)\right)=\frac{1}{z^{3}}\left(1+\mathcal{O}(z)\right),$
(3.15)

and following from the periodicity properties of $Y_{n}$, $\det Y_{n}$, the
LHS of (3.14) is a matrix valued elliptic function. Therefore,

$\displaystyle\left(Y_{n+1}Y_{n}^{-1}\det
Y_{n}\right)_{11}=-\frac{\wp^{\prime}(z)}{2}.$ (3.16)

The above expression along with (2.31) gives (3.3). ∎

######  Proposition 3.2.

The solution of the RHP: 2.1 $Y_{n}$ solves the following linear differential
equation for $n\geq 3$:

$\displaystyle Y_{n}^{\prime}=L_{n}Y_{n},$ $\displaystyle
L_{n}=\frac{1}{f_{n}}\left(\begin{array}[]{cc}n\wp^{\prime}(z)/2&\frac{h_{n}}{2\pi
i}\left((n-1)f_{n}+nf_{n+1}\right)\\\ \frac{2\pi
i}{h_{n-1}}\left((2-n)f_{n-1}+(1-n)f_{n}\right)&(2-n)\wp^{\prime}(z)/2\end{array}\right).$
(3.19)

###### Proof.

With the derivative222We drop the $z$, $\tau$ dependence in favour of brevity.
of $Y_{n}$ in (2.25), we get

$\displaystyle Y_{n}^{\prime}Y_{n}^{-1}\det Y_{n}$
$\displaystyle=\left(\begin{array}[]{cc}\pi_{n}^{\prime}(z)&\partial_{z}\mathcal{C}(\pi_{n})(z)\\\
\frac{2\pi i}{h_{n}}\pi_{n}^{\prime}(z)&\frac{2\pi
i}{h_{n}}\partial_{z}\mathcal{C}(\pi_{n})(z)\end{array}\right)\left(\begin{array}[]{cc}\frac{2\pi
i}{h_{n-1}}\mathcal{C}(\pi_{n-1})(z)&-\mathcal{C}(\pi_{n})(z)\\\ -\frac{2\pi
i}{h_{n-1}}\pi_{n-1}(z)&\pi_{n}(z)\end{array}\right).$ (3.24)

Recalling the asymptotic behaviour of the following entities near $z\to 0$ in
Lemma 2.2 now paying attention to the fact that with our weight the
polynomials split into odd and even parts,

$\displaystyle\pi_{n}(z)$
$\displaystyle=\frac{1}{z^{n}}+\frac{c_{2,n}}{z^{n-2}}+\dots,$ (3.25)
$\displaystyle\mathcal{C}(\pi_{n})(z)$ $\displaystyle=\frac{h_{n}}{2\pi
i}\left(z^{n-1}+\widetilde{c}_{2,n}z^{n+1}+\dots\right),$ (3.26)

the asymptotic behaviour of the 11 element of (3.24) is

$\displaystyle\left(Y_{n}^{\prime}Y_{n}^{-1}\det Y_{n}\right)_{11}$
$\displaystyle=-\frac{n}{z^{3}}-\frac{1}{z}\left(n\widetilde{c}_{2,n-1}+(n-2)c_{2,n}+(n-1)\beta_{n}\right)+\mathcal{O}(z),$
(3.27)

and because the LHS is an elliptic function,

$\displaystyle\left(Y_{n}^{\prime}Y_{n}^{-1}\det
Y_{n}\right)_{11}=n\frac{\wp^{\prime}(z)}{2},$ (3.28)

with the constraint

$\displaystyle n\widetilde{c}_{2,n-1}+(n-2)c_{2,n}+(n-1)\beta_{n}=0.$ (3.29)

Also note that

$\displaystyle c_{2,n}+\widetilde{c}_{2,n-2}=0.$ (3.30)

The 12 element in the limit $z\to 0$ is

$\displaystyle\left(Y_{n}^{\prime}Y_{n}^{-1}\det
Y_{n}\right)_{12}=\frac{h_{n}}{2\pi
i}\left(\frac{n}{z^{2}}+\frac{(n-1)}{z^{2}}+(2n+1)\widetilde{c}_{2,n}+(2n-3)c_{2,n}+\mathcal{O}(z^{2})\right),$
(3.31)

and we can check that

$\displaystyle(n-1)f_{n}+nf_{n+1}$
$\displaystyle=(n-1)\left(\wp(z)+c_{2,n}+\widetilde{c}_{2,n-1}-\beta_{n}\right)+n\left(\wp(z)+c_{2,n+1}+\widetilde{c}_{2,n}-\beta_{n+1}\right)$
$\displaystyle\mathop{=}^{\eqref{eq:betan_c}}\frac{(n-1)}{z^{2}}+\frac{n}{z^{2}}+(n-1)c_{2,n}+(n-1)\widetilde{c}_{2,n-1}+n\widetilde{c}_{2,n-1}+(n-2)c_{2,n}$
$\displaystyle+nc_{2,n+1}+n\widetilde{c}_{2,n}+(n+1)\widetilde{c}_{2,n}+(n-1)c_{2,n+1}$
$\displaystyle\mathop{=}^{\eqref{detcons1}}\frac{(n-1)}{z^{2}}+\frac{n}{z^{2}}+(2n-3)c_{2,n}+(2n+1)\widetilde{c}_{2,n}.$
(3.32)

Therefore,

$\displaystyle\left(Y_{n}^{\prime}Y_{n}^{-1}\det
Y_{n}\right)_{12}=\frac{h_{n}}{2\pi i}\left((n-1)f_{n}+nf_{n+1}\right).$
(3.33)

Similarly, the 21 element

$\displaystyle\left(Y_{n}^{\prime}Y_{n}^{-1}\det Y_{n}\right)_{21}=\frac{2\pi
i}{h_{n-1}}\left(\frac{1-n}{z^{2}}+\frac{2-n}{z^{2}}+(5-2n)c_{2,n-1}+(1-2n)\widetilde{c}_{2,n-1}+\mathcal{O}(z^{2})\right),$
(3.34)

and as before, we can see that

$\displaystyle(2-n)f_{n-1}+(1-n)f_{n}$
$\displaystyle=(2-n)\left(\wp(z)+c_{2,n-1}+\widetilde{c}_{2,n-2}-\beta_{n-1}\right)$
$\displaystyle+(1-n)\left(\wp(z)+c_{2,n}+\widetilde{c}_{2,n-1}-\beta_{n}\right)$
$\displaystyle=\frac{(2-n)}{z^{2}}+\frac{(1-n)}{z^{2}}+(2-n)c_{2,n-1}+(2-n)\widetilde{c}_{2,n-2}-(2-n)\beta_{n-1}$
$\displaystyle+(1-n)c_{2,n}+(1-n)\widetilde{c}_{2,n-1}-(1-n)\beta_{n}$
$\displaystyle\mathop{=}^{\eqref{eq:betan_c}}\frac{(2-n)}{z^{2}}+\frac{(1-n)}{z^{2}}+(2-n)c_{2,n-1}+(2-n)\widetilde{c}_{2,n-2}+(1-n)\widetilde{c}_{2,n-2}$
$\displaystyle+(3-n)c_{2,n-1}+(1-n)c_{2,n}+(1-n)\widetilde{c}_{2,n-1}-n\widetilde{c}_{2,n-1}+(2-n)c_{2,n}$
$\displaystyle\mathop{=}^{\eqref{detcons1}}\frac{(2-n)}{z^{2}}+\frac{(1-n)}{z^{2}}+(5-2n)c_{2,n-1}+(1-2n)\widetilde{c}_{2,n-1}.$
(3.35)

Therefore,

$\displaystyle\left(Y_{n}^{\prime}Y_{n}^{-1}\det Y_{n}\right)_{21}=\frac{2\pi
i}{h_{n-1}}\left((2-n)f_{n-1}+(1-n)f_{n}\right).$ (3.36)

Finally, we see that

$\displaystyle\left(Y_{n}^{\prime}Y_{n}^{-1}\det
Y_{n}\right)_{22}=(2-n)\frac{\wp^{\prime}(z)}{2},$ (3.37)

which finishes the proof of the proposition. ∎

An immediate consequence of the above proposition is that the 11 entry of the
linear equation (3.19) gives an ODE for the polynomials.

######  Corollary 3.1.

The polynomials $\pi_{n}$ solve the following second order differential
equation

$\displaystyle\pi_{n}^{\prime\prime}(z)$
$\displaystyle=\left(\frac{\wp^{\prime}(z)}{f_{n}}+n\left(\frac{f_{n+1}}{f_{n}}\right)^{\prime}\left((n-1)+n\frac{f_{n+1}}{f_{n}}\right)^{-1}\right)\pi_{n}^{\prime}$
$\displaystyle+\left(\left(\frac{n\wp^{\prime}(z)}{2f_{n}}\right)^{\prime}-n\left(\frac{f_{n+1}}{f_{n}}\right)^{\prime}\frac{n\wp^{\prime}(z)}{2\left((n-1)f_{n}+nf_{n+1}\right)}-\det
L_{n}\right)\pi_{n}.$

###  3.1. Recurrence relations

With the above linear system, as first step, we obtain the recurrence relation
for the elliptic-polynomials. We find

$\pi_{n+1}=-\frac{\wp^{\prime}\pi_{n}}{2f_{n}}-\frac{\beta_{n}f_{n+1}}{f_{n}}\pi_{n-1},$
(3.38)

which follows from the first entry of (3.3). Note that $f_{n}$ as defined in
(2.31) depends on $z$. Iterating the above equation once again yields the
following theorem.

######  Theorem 3.1.

The polynomials $\\{\pi_{n}\\}_{n=4}^{\infty}$, satisfy the relation

$\pi_{n+2}=\left(\wp-B_{n}\right)\pi_{n}-\beta_{n}\beta_{n-1}\pi_{n-2},$
(3.39)

where,

$B_{n}=\beta_{n+1}+\beta_{n}+\alpha_{n}+\alpha_{n+1}.$

Furthermore,

$-\alpha_{n}^{3}+\frac{g_{2}}{4}\alpha_{n}-\frac{g_{3}}{4}=\beta_{n}(\alpha_{n-1}-\alpha_{n})(\alpha_{n+1}-\alpha_{n}).$
(3.40)

###### Proof.

Iterating (3.38) by $n\pm 1$ we find that

$\pi_{n+2}=\left(\frac{(\wp^{\prime})^{2}}{4(\wp+\alpha_{n})(\wp+\alpha_{n+1})}-\beta_{n+1}\frac{\wp+\alpha_{n+2}}{\wp+\alpha_{n+1}}-\beta_{n}\frac{\wp+\alpha_{n-1}}{\wp+\alpha_{n}}\right)\pi_{n}-\beta_{n}\beta_{n-1}\pi_{n-2}.$
(3.41)

However, from the difference equation of the even case (4.56), we see that
$\pi_{n+2}$ can be written in terms of $\pi_{n}$ and $\pi_{n-2}$ using the
equation

$\pi_{n+2}=\left(\wp-B_{n}\right)\pi_{n}-\beta_{n}\beta_{n-1}\pi_{n-2},$
(3.42)

where $B_{n}$ is a constant. Moreover, the behaviour of the first term in
(3.41) near $z=0$ gives the expression

$B_{n}=\beta_{n+1}+\beta_{n}+\alpha_{n}+\alpha_{n+1}.$

Considering the behaviour of (3.41) near the poles of $\wp+\alpha_{n}$ we
further determine that

$-\alpha_{n}^{3}+\frac{g_{2}}{4}\alpha_{n}-\frac{g_{3}}{4}=\beta_{n}(\alpha_{n-1}-\alpha_{n})(\alpha_{n+1}-\alpha_{n}),$
(3.43)

and the theorem follows. ∎

We now use the compatibility condition of the linear system (3.3), (3.19) to
obtain the recurrence relation for the coefficients of the elliptic
polynomials.

######  Theorem 3.2.

The compatibility condition

$\displaystyle R_{n}^{\prime}-L_{n+1}R_{n}+R_{n}L_{n}=0,$ (3.44)

gives the following recurrence relations:

$\displaystyle\beta_{n}$
$\displaystyle=\frac{g_{3}-g_{2}\alpha_{n}+4\alpha_{n}^{3}}{4(\alpha_{n-1}-\alpha_{n})(\alpha_{n}-\alpha_{n+1})},$
(3.45) $\displaystyle\alpha_{n+1}$
$\displaystyle=\frac{(1-n)\alpha_{n}\left(4\alpha_{n}^{3}-3g_{2}\alpha_{n}+4g_{3}\right)-\alpha_{n-1}\left(4(n-2)\alpha_{n}^{3}+ng_{2}\alpha_{n}-(2n-1)g_{3}\right)}{4n\alpha_{n}^{3}+(n-1)\alpha_{n-1}\left(g_{2}-12\alpha_{n}^{2}\right)+g_{2}(n-2)\alpha_{n}-g_{2}(2n-3)}.$
(3.46)

###### Proof.

Substituting (3.3) and (3.19) into Equation (3.44) we find that the (1,1)
entry is the only non-zero term, and is cubic in $\wp(z)$. Equating the
coefficients of the powers of order 3, 2, 1, 0 to zero we find the following
set of equations respectively

$\displaystyle
0=(n-2)\alpha_{n}-(n+1)\alpha_{n+1}+(2n-3)\beta_{n}-(2n+1)\beta_{n+1},$ (3.47)
$\displaystyle
3\alpha_{n}\alpha_{n+1}+\frac{{g_{2}}}{2}=((n-2)\alpha_{n-1}+(n-1)\alpha_{n}+2(2n-3)\alpha_{n+1})\beta_{n}$
$\displaystyle\hskip
85.35826pt-((4n+2)\alpha_{n}+n\alpha_{n+1}+(n+1)\alpha_{n+2})\beta_{n+1},$
(3.48) $\displaystyle
3g_{3}+g_{2}n\alpha_{n}-g_{2}(n-1)\alpha_{n+1}=4\alpha_{n+1}(2(n-2)\alpha_{n-1}+2(n-1)\alpha_{n}+(2n-3)\alpha_{n+1})\beta_{n}$
$\displaystyle\hskip
156.49014pt-4\alpha_{n}((2n+1)\alpha_{n}+2n\alpha_{n+1}+2(n+1)\alpha_{n+2})\beta_{n+1},$
(3.49)
$\displaystyle-\alpha_{n+1}(g_{2}\alpha_{n}+g_{3}(n-2))+g_{3}(n+1)\alpha_{n}=-4(n\alpha_{n+1}+(n+1)\alpha_{n+2})\alpha_{n}^{2}\beta_{n+1}$
$\displaystyle\hskip
213.39566pt+4((n-2)\alpha_{n-1}+(n-1)\alpha_{n})\alpha_{n+1}^{2}\beta_{n}.$
(3.50)

The recurrence for $\beta_{n}$ (3.45) follows from (3.43). Substituting (3.45)
in (3.47), we obtain a constraint that is cubic in $\alpha_{n}$,
$\alpha_{n+1}$ and linear in $\alpha_{n-1}$, $\alpha_{n+2}$:

$\displaystyle\alpha_{n+2}=\frac{X}{Y}$

with

$\displaystyle X$
$\displaystyle=\alpha_{n+1}\left(2g_{2}(1-2n)\alpha_{n}+\alpha_{n-1}\left(4(n-2)\alpha_{n}^{2}+2g_{2}n+g_{2}\right)+4(n-1)\alpha_{n}^{3}+g_{3}(2n-3)\right)$
$\displaystyle-
g_{3}(2n+1)(\alpha_{n-1}-\alpha_{n})+4n(\alpha_{n}-\alpha_{n-1})\alpha_{n+1}^{3}-4(2n-1)(\alpha_{n-1}-\alpha_{n})\alpha_{n}\alpha_{n+1}^{2},$
$\displaystyle Y$
$\displaystyle=\alpha_{n}(4\alpha_{n}((n-2)\alpha_{n-1}+(n-1)\alpha_{n})+g_{2}(3-2n))+4(n+1)(\alpha_{n-1}-\alpha_{n})\alpha_{n+1}^{2}$
$\displaystyle+4(2n-1)\alpha_{n}(\alpha_{n}-\alpha_{n-1})\alpha_{n+1}+g_{3}(2n-3).$

Manipulating the set of equations (3.47)-(3.50) using the above expression
gives (3.46). ∎

Initial conditions for the recurrence (3.46) of the $\alpha_{n}$ in Theorem
3.2 can be computed directly, yielding

$\displaystyle\alpha_{3}$
$\displaystyle=\frac{3g_{3}-4g_{2}\eta_{1}}{g_{2}-48\eta_{1}^{2}},$
$\displaystyle\alpha_{4}$
$\displaystyle=\frac{5g_{2}^{3}-108g_{3}^{2}+108g_{2}g_{3}\eta_{1}-432g_{2}^{2}\eta_{1}^{2}+8640g_{3}\eta_{1}^{3}}{18(3g_{3}-4g_{2}\eta_{1})(g_{2}-48\eta_{1}^{2})},$

where we remark that explicit formulas for $g_{2},g_{3}$ and $\eta_{1}$ are
given in Appendix A.

##  4\. Linear problems for the even case and the elliptic form of Painlevé VI

In this section, we restrict to the case of the even elliptic polynomials.
Firstly, we express the polynomials in terms of Hankel determinants. Secondly,
the solution of the RHP yields the Lax pair of the elliptic form of Painlevé
VI where the modular parameter $\tau$ assumes the role of the isomonodromic
time. Moreover, the recurrence relation coming from the corresponding discrete
linear system provides a formulation of the solution of the elliptic form of
the Painlevé VI equation.

###  4.1. Lax pair of the elliptic form of Painlevé VI

######  Theorem 4.1.

For $k\geq 1$, the solution (2.51) solves the following pair of linear
equations that correspond to the elliptic form of Painlevé VI

$\displaystyle L_{2k}(z,\tau)$
$\displaystyle:=Y^{\prime}_{2k}(z,\tau)Y_{2k}(z,\tau)^{-1}=\sum_{i=1}^{3}\wp^{\prime}(z)\frac{{L}_{2k}^{(i)}}{(\wp(z)-e_{i})},$
(4.1) $\displaystyle M_{2k}(z,\tau)$
$\displaystyle:=\frac{d}{d\tau}{Y}_{2k}(z,\tau)Y_{2k}(z,\tau)^{-1}=\sum_{i=1}^{3}\frac{{L}_{2k}^{(i)}(\dot{\wp}(z)-\dot{e}_{i})}{2(\wp(z)-e_{i})},$
(4.2)

where the matrices $L_{2k}^{(i)}$ are given in (4.34).

We begin with the $z$-derivative and then compute the $\tau$-derivative.

###### Proof.

Let us the following asymptotics of the even polynomials and their Cauchy
transform Lemma 2.4333we drop the $z$, $\tau$ dependence for the remainder of
the proof.

$\displaystyle\pi_{2n}$
$\displaystyle=\frac{1}{z^{2n}}+\frac{a_{1,2n}}{z^{2n-2}}+\frac{a_{2,2n}}{z^{2n-4}}+\mathcal{O}(z^{6-2n}),$
(4.3) $\displaystyle\mathcal{C}(\pi_{2n})(z)$
$\displaystyle=\frac{h_{2k}}{2\pi
i}\left(z^{2n-1}+\widetilde{a}_{1,2n}z^{2n+1}+\widetilde{a}_{2,2n}z^{2n+3}+\mathcal{O}(z^{2n+5})\right).$
(4.4)

The leading behaviour of solution $Y_{2k}$ (2.51) near $z\to 0$ is then

$\displaystyle
Y_{2k}=\left(\mathbb{1}+z^{2}U+z^{4}V+\mathcal{O}(z^{6})\right)\left(\begin{array}[]{cc}z^{-2k}&0\\\
0&z^{2k-3}\end{array}\right),$ (4.7)

where

$\displaystyle U=\left(\begin{array}[]{cc}a_{1,2k}&\frac{h_{2k}}{2\pi i}\\\
\frac{2\pi i}{h_{2k-2}}&\widetilde{a}_{1,2k-2}\end{array}\right),$
$\displaystyle V=\left(\begin{array}[]{cc}a_{2,2k}&\frac{h_{2k}}{2\pi
i}\widetilde{a}_{1,2k}\\\ \frac{2\pi
i}{h_{2k-2}}a_{1,2k-2}&\widetilde{a}_{2,2k-2}\end{array}\right),$ (4.12)

and due to (2.59), $U$, $V$ are traceless. Then, in the same limit,

$\displaystyle Y_{2k}^{\prime}Y_{2k}^{-1}\det Y_{2k}$
$\displaystyle=\frac{1}{z^{4}}\left(\begin{array}[]{cc}-2k&0\\\
0&2k-3\end{array}\right)+\frac{1}{z^{2}}\left(2U+\left[U,\left(\begin{array}[]{cc}-2k&0\\\
0&2k-3\end{array}\right)\right]\right)$ (4.17)
$\displaystyle+4V-2U^{2}+\left[V,\left(\begin{array}[]{cc}-2k&0\\\
0&2k-3\end{array}\right)\right]+\mathcal{O}(z^{2}).$ (4.20)

Note that the LHS of the above expression is an elliptic function

$\displaystyle
Y_{2k}^{\prime}(z+\tau)Y_{2k}(z+\tau)^{-1}=Y_{2k}^{\prime}(z+1)Y_{2k}(z+1)^{-1}=Y_{2k}^{\prime}(z)Y_{2k}(z)^{-1}.$
(4.21)

Therefore using (2.60) we have

$\displaystyle
Y_{2k}^{\prime}Y_{2k}^{-1}=\frac{1}{\wp^{\prime}(z)}\left(\widetilde{L}_{2k}^{(2)}\wp^{2}(z)+\widetilde{L}_{2k}^{(1)}\wp(z)+\widetilde{L}_{2k}^{(0)}\right),$
(4.22)

where

$\displaystyle\widetilde{L}_{2k}^{(2)}=-2\left(\begin{array}[]{cc}-2k&0\\\
0&2k-3\end{array}\right),$
$\displaystyle\widetilde{L}_{2k}^{(1)}=-2\left(2U+\left[U,\left(\begin{array}[]{cc}-2k&0\\\
0&2k-3\end{array}\right)\right]\right)$ (4.27)
$\displaystyle\widetilde{L}_{2k}^{(0)}=-2\left(4V-2U^{2}+\left[V,\left(\begin{array}[]{cc}-2k&0\\\
0&2k-3\end{array}\right)\right]\right),$ (4.30)

and

$\displaystyle\operatorname{tr}\widetilde{L}_{2k}^{(1)}=0$
$\displaystyle\operatorname{tr}\widetilde{L}_{2k}^{(0)}=-\frac{2h_{2k-2}}{i\pi}\left(\beta_{2k}+a_{1,2k}^{2}\right).$
(4.31)

Furthermore, using the cubic equation

$\displaystyle\left(\wp^{\prime}(z)\right)^{2}=4\left(\wp-
e_{1}\right)\left(\wp-e_{2}\right)\left(\wp-e_{3}\right),$ (4.32)

(4.22) can be re-written as

$\displaystyle
L_{2k}(z,\tau):=Y_{2k}^{\prime}Y_{2k}^{-1}=\wp^{\prime}(z)\left(\frac{{L}_{2k}^{(1)}}{\wp(z)-e_{1}}+\frac{{L}_{2k}^{(2)}}{\wp(z)-e_{2}}+\frac{{L}_{2k}^{(3)}}{\wp(z)-e_{3}}\right),$
(4.33)

with the relations

$\begin{split}{L}_{2k}^{(1)}&=\frac{e_{1}(1-e_{2})\widetilde{L}_{2k}^{(2)}+\widetilde{L}_{2k}^{(0)}-e_{1}e_{3}\widetilde{L}_{2k}^{(1)}}{4(e_{1}-e_{2})(e_{1}-e_{3})},\\\
{L}_{2k}^{(2)}&=\frac{e_{2}(e_{1}-1)\widetilde{L}_{2k}^{(2)}+e_{2}e_{3}\widetilde{L}_{2k}^{(1)}-\widetilde{L}_{2k}^{(0)}}{4(e_{1}-e_{2})(e_{2}-e_{3})},\\\
{L}_{2k}^{(3)}&=\frac{\widetilde{L}_{2k}^{(0)}+e_{3}^{2}\widetilde{L}_{2k}^{(1)}+e_{3}(1+2e_{3})\widetilde{L}_{2k}^{(2)}}{4(e_{1}-e_{3})(e_{2}-e_{3})}.\end{split}$
(4.34)

Now we derive the $\tau$-derivative. From (4.1) we have

$\displaystyle\partial_{z}\log
Y_{2k}(z,\tau)=\sum_{i=1}^{3}{L}_{2k}^{(i)}\partial_{z}\log\left(\wp(z)-e_{i}\right),$
(4.35)

implying that $Y_{2k}$ has the following local behaviour around
$z-\frac{\omega_{i}}{2}\to 0$:

$\displaystyle Y_{2k}(z,\tau)\sim
G_{i}\left(\wp(z)-e_{i}\right)^{\Lambda_{i}}\left(\mathbb{1}+\mathcal{O}(z^{2})\right)C_{i},$
(4.36)

where

$\displaystyle\Lambda_{i}=G_{i}^{-1}{L}_{2k}^{(i)}G_{i}.$ (4.37)

The $\tau$-derivative in the vicinity of $z=\frac{w_{i}}{2}$ is

$\displaystyle\frac{dY_{2k}}{d\tau}=\left(\frac{dG_{i}}{d\tau}\left(\wp(z)-e_{i}\right)^{\Lambda_{i}}+G_{i}\Lambda_{i}\left(\wp(z)-e_{i}\right)^{\Lambda_{i}-\mathbb{1}}\left(\dot{\wp(z)}-\dot{e}_{i}\right)\right)\left(\mathbb{1}+\mathcal{O}(z^{2})\right)C_{i},$
$\displaystyle\Rightarrow$
$\displaystyle\frac{dY_{2k}}{d\tau}Y_{2k}^{-1}=\frac{dG_{i}}{d\tau}G_{i}^{-1}+\frac{{L}_{2k}^{(i)}\left(\dot{\wp(z)}-\dot{e}_{i}\right)}{(\wp(z)-e_{i})}+\mathcal{O}(z^{4}).$
(4.38)

Moreover, near $z=0$,

$\displaystyle\dot{Y}_{2k}Y_{2k}^{-1}\sim
z^{2}\left(U+\dot{U}\right)+\mathcal{O}(z^{4}).$ (4.39)

Therefore, using Liouville theorem, we can uniquely determine that

$\displaystyle\frac{dY_{2k}}{d\tau}Y_{2k}^{-1}=:M_{2k}(z,\tau)=\sum_{i=1}^{3}\frac{{L}_{2k}^{(i)}\left(\dot{\wp}(z)-\dot{e}_{i}\right)}{(\wp(z)-e_{i})},$
(4.40)

and the following periodicity relations hold

$\displaystyle M_{2k}(z+1,\tau)=M_{2k}(z+,\tau),$ $\displaystyle
M_{2k}(z+\tau,\tau)=M_{2k}(z,\tau)+L_{2k}(z,\tau).$ (4.41)

∎

The matrices $L_{2k}$, $M_{2k}$ are in fact the Lax pair of the elliptic form
of the Painlevé VI [Tak01], which, with a change of variables reduces to the
usual Lax pair, as will be shown in Proposition 4.3. Furthermore, we can
compute the eigenvalues of the residue matrices of $L_{2k}$, which turn out to
have the following values owing to the specific form of the determinant of
$Y_{2k}$ (2.60).

For what follows it will be useful to make the linear system (4.1) traceless.
To do this, we start by noting that $L_{2k}(z)$ (4.1) is an elliptic matrix
function, and its trace is obtained using Jacobi’s formula,

$\operatorname{tr}L_{2k}(z)=\frac{d}{dz}\log\det
Y_{2k}(z)=\frac{\wp^{\prime\prime}(z)}{\wp^{\prime}(z)},$

where used that the determinant (2.60)

$\displaystyle\det Y_{2k}(z)=-\tfrac{1}{2}\wp^{\prime}(z)=f_{2k}.$ (4.42)

We now use the following gauge transformation to obtain a traceless linear
system:

$\mathcal{Y}_{2k}(z)=f_{2k}^{-\frac{1}{2}}Y_{2k}(z),$ (4.43)

so that

$\mathcal{Y}^{\prime}_{2k}(z)=\mathcal{L}_{2k}(z)\mathcal{Y}_{2k}(z),\qquad\mathcal{L}_{2k}(z)=L_{2k}(z)-\frac{\wp^{\prime\prime}(z)}{2\wp^{\prime}(z)}\mathbb{1},$
(4.44)

######  Proposition 4.1.

The monodromy exponents $\pm\theta_{i}$ around the singularities $\omega_{i}$,
and $\pm\theta_{0}$ around the singularity $z=0$ of the linear system (4.44),
are given by

$\displaystyle\theta_{0}=\frac{1}{2}\left(4k-{3}\right),$
$\displaystyle\theta_{i}=\frac{1}{2},$ $\displaystyle i=1,2,3,$ (4.45)

and $k=1$ is the case related to self-dual Einstein metrics [Tod94, Hit95].

###### Proof.

We start with the behaviour of $\mathcal{Y}_{2k}$ at $z\to 0$:

$\displaystyle\mathcal{Y}_{2k}=(\mathbb{1}+\mathcal{O}(z))\left(\begin{array}[]{cc}z^{-2k}&0\\\
0&z^{2k-3}\end{array}\right)\left(\begin{array}[]{cc}z^{3/2}&0\\\
0&z^{3/2}\end{array}\right),$ (4.50)

therefore the monodromy exponents around $z=0$ are
$\pm\frac{1}{2}\left(4k-{3}\right)$, which can be read off from the
asymptotics (4.7). The coefficient matrix has simple poles at
$z=w_{1},w_{2},w_{1}+w_{2},0$ coming from the term $\wp^{\prime}(z)$. We now
determine the eigenvalues of the residue matrices at the singularities. The
asymptotic behaviour of $\mathcal{Y}_{2k}(z)$ near $z=w_{i}$, $i=1,2,3$ is

$\mathcal{Y}_{2k}(z)=c^{-\frac{1}{2}}(z-w_{1})^{-\frac{1}{2}}Y_{2k}(w_{1})(I+\mathcal{O}(z-w_{1})),$

as $f_{2k}=(z-w_{i})(c+\mathcal{O}(z-w_{i}))$ for some $c\neq 0$. Moreover,
(4.42) implies that $\det Y_{2k}(w_{i})$ is zero and therefore $Y_{2k}(w_{i})$
is a rank one matrix and can be expressed as

$Y_{2k}(w_{i})C=\begin{pmatrix}0&*\\\ 0&*\end{pmatrix},$

where the second column is nonzero with $C$ being a constant matrix. We then
obtain that for $z\to w_{i}$,

$\displaystyle\mathcal{Y}_{2k}(z)C=U_{0}(\mathbb{1}+\mathcal{O}(z-w_{i}))(z-w_{i})^{\frac{1}{2}\sigma_{3}},$
$\displaystyle U_{0}\in SL_{2}(\mathbb{C})$

as $z\rightarrow w_{i}$, for a matrix. Therefore the monodromy exponents
around $z=w_{i}$ are $\pm\frac{1}{2}$. ∎

We will see in what follows (from (4.71)) that the unique zero of the $(1,2)$
entry of $L_{2k}(z)$ will be $z=Q(\tau)$, where $Q(\tau)$ satisfies the
elliptic form of Painlevé VI [Tak01] in the present case reads

$\displaystyle(2\pi
i)^{2}\frac{d^{2}Q(\tau)}{d\tau^{2}}=\sum_{i=0}^{3}\alpha_{i}\wp^{\prime}(Q(\tau)+w_{i}),$
(4.51)

where $w_{0}=0$, and

$\displaystyle\alpha_{0}=\frac{(\theta_{0}-1)^{2}}{2}=\frac{(4k-5)^{2}}{8},$
$\displaystyle\alpha_{1}=-\frac{\theta_{1}^{2}}{2}=-\frac{1}{8},$
$\displaystyle\alpha_{2}=\frac{\theta_{3}^{2}}{2}=\frac{1}{8},$
$\displaystyle\alpha_{3}=\frac{\left(1-\theta_{2}^{2}\right)}{2}=\frac{3}{8}.$
(4.52)

For $k=1$ this is the Hitchin case [Hit95, Man98], and the corresponding
solution $Q=Q(\tau)$, is given by

$\wp(Q(\tau))=e_{1}+e_{2}+2\frac{16\eta_{1}^{3}+g_{2}\eta_{1}-g_{3}}{48\eta_{1}^{2}-g_{2}}.$
(4.53)

This formula follows from equation (5.3).

We finish this subsection with a recursive formula for the $Y_{2k}$.

######  Proposition 4.2.

The matrix function $Y=Y_{2k}$ satisfies the following discrete evolution with
respect to $k$,

$\displaystyle Y_{2k+2}=R_{2k}Y_{2k},$ $\displaystyle
R_{2k}:=\left(\begin{array}[]{cc}\wp(z)+a_{1,2k+2}-a_{1,2k}&-\frac{h_{2k}}{2\pi
i}\\\ \frac{2\pi i}{h_{2k}}&0\end{array}\right),$ (4.56)

for $k\geq 1$.

###### Proof.

The proof is similar to that of Proposition 3.1 and follows from a direct
substitution of (2.51) with the expressions of Lemmas 2.4 and 2.5. ∎

From the above proposition, we obtain the following three-term recurrence
relation for the orthogonal polynomials,

$\displaystyle\wp(z)\pi_{2k}=\pi_{2k+2}+\widetilde{\alpha}_{2k}\pi_{2k}+\beta_{2k}\pi_{2k-2},$
(4.57)

which coincides with (3.39), as can be seen by using the relations (2.32),
(2.12), and (2.59).

###  4.2. Relation to Painlevé VI

Under a simultaneous transformation of the dependent and independent
variables,

$\displaystyle t=\frac{e_{3}-e_{1}}{e_{2}-e_{1}},$ $\displaystyle
u(t)=\frac{\wp(Q(\tau))-e_{1}}{e_{2}-e_{1}},$ (4.58)

the elliptic form of Painlevé VI (4.51) reduces to the usual form [Tak01]

$\displaystyle\ddot{u}=$
$\displaystyle\left(\frac{1}{u}+\frac{1}{u-1}+\frac{1}{u-t}\right)\frac{\dot{u}^{2}}{2}-\left(\frac{1}{t}+\frac{1}{t-1}+\frac{1}{u-t}\right)\dot{u}$
$\displaystyle+\frac{u(u-1)(u-t)}{2t^{2}(t-1)^{2}}\left((2k-\tfrac{1}{2})^{2}-\frac{t}{4u^{2}}+\frac{(t-1)}{4(u-1)^{2}}+\frac{3t(t-1)}{4(u-t)^{2}}\right),$

where $u\equiv u_{k}(t)$, and $\dot{}$ denotes derivative with respect to $t$.

In this section, we derive a corresponding transformation for the Lax pair.

######  Proposition 4.3.

The change of variables

$\displaystyle x=\frac{\wp(z)-e_{1}}{e_{2}-e_{1}},$ $\displaystyle
t=\frac{e_{3}-e_{1}}{e_{2}-e_{1}},$
$\displaystyle\widetilde{Y}_{2k}(x,t)=\mathcal{Y}_{2k}(z,\tau)\left(e_{1}-e_{2}\right)^{-(k-\frac{3}{4})\;\sigma_{3}},$
(4.59)

transforms the linear system in Theorem 4.1 into the following $4$-point
Fuchsian system, with the singularities at $w_{1},w_{2},w_{3},0$ mapped to
$0,1,t,\infty$ respectively, and corresponding deformation equation,

$\frac{d\widetilde{Y}_{2k}}{dx}=A_{2k}\widetilde{Y}_{2k},\quad\frac{d\widetilde{Y}_{2k}}{dt}=B_{2k}\widetilde{Y}_{2k},$
(4.60)

with

$\displaystyle
A_{2k}(x,t)=\frac{A_{2k}^{(1)}}{x}+\frac{A_{2k}^{(2)}}{x-1}+\frac{A_{2k}^{(3)}}{x-t},\qquad
B_{2k}(x,t)=-\frac{A_{2k}^{(3)}}{x-t},$

and coefficient matrices above are related to $\mathcal{L}_{2k}$ as

$\displaystyle
A^{(i)}_{2k}=\left(e_{2}-e_{1}\right)^{(4k-3)\sigma_{3}/4}\mathcal{L}_{2k}^{(i)}\left(e_{2}-e_{1}\right)^{-(4k-3)\sigma_{3}/4},\qquad
i=1,2,3.$ (4.61)

In particular, the monodromy of $\widetilde{Y}_{2k}(x,t)$, with respect to
$x$, is constant in $t$.

###### Proof.

The linear system (4.44):

$\displaystyle\frac{\partial}{\partial
z}\mathcal{Y}_{2k}(z,\tau)=\wp^{\prime}(z)\left(\frac{\mathcal{L}_{2k}^{(1)}(\tau)}{\wp(z)-e_{1}}+\frac{\mathcal{L}_{2k}^{(2)}(\tau)}{\wp(z)-e_{2}}+\frac{\mathcal{L}_{2k}^{(3)}(\tau)}{\wp(z)-e_{3}}\right)\mathcal{Y}_{2k}(z,\tau)$

under the change of variables (4.59) reads

$\displaystyle\frac{\partial}{\partial
x}\mathcal{Y}_{2k}(x,t)=\left(\frac{\mathcal{L}_{2k}^{(1)}(t)}{x}+\frac{\mathcal{L}_{2k}^{(2)}(t)}{x-1}+\frac{\mathcal{L}_{2k}^{(3)}(t)}{x-t}\right)\mathcal{Y}_{2k}(x,t).$
(4.62)

Let us now understand the change of variables for the equation (4.2) $\tau$
after the gauge transformation (4.43):

$\displaystyle\frac{d}{d\tau}\mathcal{Y}_{2k}(z,\tau)$
$\displaystyle=\left(\sum_{i=1}^{3}\frac{(\dot{\wp}(z)-\dot{e}_{i})\mathcal{L}_{2k}^{(i)}}{(\wp(z)-e_{i})}\right)\mathcal{Y}_{2k}(z,\tau).$
(4.63)

We begin by simplifying the sum on the right-hand side. To this end, we note
that the definition of $x$ in equation (4.59), and the fact that
$\frac{dx}{d\tau}=0$, imply

$\displaystyle\dot{\wp}(z)$
$\displaystyle=\frac{\left(\wp(z)-e_{1}\right)\dot{e}_{2}-\left(\wp(z)-e_{2}\right)\dot{e}_{1}}{e_{2}-e_{1}}.$

From the above equation, and the identity $e_{1}+e_{2}+e_{3}=0$, we obtain the
following relations

$\displaystyle\frac{\dot{\wp}(z)-\dot{e}_{1}}{\wp(z)-e_{1}}$
$\displaystyle=\frac{\dot{e}_{2}-\dot{e}_{1}}{e_{2}-e_{1}},$
$\displaystyle\frac{\dot{\wp}(z)-\dot{e}_{2}}{\wp(z)-e_{2}}$
$\displaystyle=\frac{\dot{e}_{2}-\dot{e}_{1}}{e_{2}-e_{1}},$
$\displaystyle\frac{\dot{\wp}(z)-\dot{e}_{3}}{\wp(z)-e_{3}}$
$\displaystyle=\frac{\dot{e}_{2}-\dot{e}_{1}}{e_{2}-e_{1}}-3\frac{e_{1}\dot{e}_{2}-\dot{e}_{1}e_{2}}{(e_{2}-e_{1})(\wp(z)-e_{3})}.$

Using the above identities, we can express the sum on the right-hand side of
equation (4.63) rationally in $x$, giving

$\displaystyle\frac{d}{d\tau}\mathcal{Y}_{2k}(x,\tau)\mathcal{Y}_{2k}(x,\tau)^{-1}=\left(\mathcal{L}_{2k}^{(1)}+\mathcal{L}_{2k}^{(2)}+\mathcal{L}_{2k}^{(3)}\right)\frac{(\dot{e}_{2}-\dot{e}_{1})}{e_{2}-e_{1}}-\frac{3\mathcal{L}_{2k}^{(3)}\left(e_{1}\dot{e}_{2}-e_{2}\dot{e}_{1}\right)}{\left(x-\left(\frac{e_{3}-e_{1}}{e_{2}-e_{1}}\right)\right)(e_{2}-e_{1})^{2}},$
(4.64)

where

$\displaystyle\mathcal{L}_{2k}^{(1)}+\mathcal{L}_{2k}^{(2)}+\mathcal{L}_{2k}^{(3)}=\frac{1}{4}(4k-3)\sigma_{3},$
(4.65)

as can be seen from (4.34), along with the gauge transformation (4.44). We now
observe that, the gauge transformation

$\displaystyle\widetilde{{Y}}:=\mathcal{Y}_{2k}\left(e_{1}-e_{2}\right)^{-(4k-3)\;\sigma_{3}/4},$
(4.66)

removes the constant term in (4.64), giving

$\displaystyle\frac{d}{d\tau}\widetilde{Y}(x,\tau)\widetilde{Y}(x,\tau)^{-1}=-\frac{3\left(e_{1}\dot{e}_{2}-e_{2}\dot{e}_{1}\right)}{(e_{2}-e_{1})^{2}}\frac{A^{(3)}_{2k}}{\left(x-\left(\frac{e_{3}-e_{1}}{e_{2}-e_{1}}\right)\right)}.$
(4.67)

Using the identity below coming from the change of variable $\tau\to t$

$\displaystyle\frac{dt}{d\tau}=3\frac{(\dot{e}_{2}e_{1}-\dot{e}_{1}e_{2})}{(e_{2}-e_{1})^{2}}$
(4.68)

implies that

$\displaystyle\frac{d}{dt}\widetilde{Y}(x,t)\widetilde{Y}(x,t)^{-1}=-\frac{A^{(3)}_{2k}(t)}{\left(x-t\right)}.$
(4.69)

Furthermore, (4.62) with the gauge transformation (4.66) and (4.61) is

$\displaystyle\frac{d}{dx}\widetilde{Y}(x,t)\widetilde{Y}(x,t)^{-1}=\frac{A^{(1)}_{2k}}{x}+\frac{A^{(2)}_{2k}}{x-1}+\frac{A^{(3)}_{2k}}{x-t}.$
(4.70)

which finishes the proof of the proposition. ∎

We note that the following rational matrix is traceless,

$A^{(1)}_{2k}+A^{(2)}_{2k}+A^{(3)}_{2k}=-\frac{\theta_{0}}{2}\sigma_{3},\quad\theta_{0}:=\tfrac{3}{2}-2k,$

and

$|A^{(i)}_{2k}|=-\frac{\theta_{i}^{2}}{4},\quad\theta_{i}\equiv\theta=\frac{1}{2}\qquad(i=1,2,3),$

as can be obtained from Proposition 4.1 and (4.61). Introducing standard
coordinates $(u_{k},v_{k},g_{k})$, see e.g. [JM81], through

$\displaystyle(A_{2k})_{12}(x,t)=-\theta_{0}g_{k}\frac{x-u_{k}}{2x(x-t)(x-1)},$
(4.71)
$\displaystyle(A_{2k})_{11}(x,t)=v_{k}-\frac{\theta}{2}\left(\frac{1}{u_{k}}+\frac{1}{u_{k}-t}+\frac{1}{u_{k}-1}\right),$

we obtain that $u\equiv u_{k}$ and $v\equiv v_{k}$ satisfy the equations below

$\displaystyle t(t-1)\dot{u}$
$\displaystyle=-\tfrac{1}{2}t+u\left(t-\tfrac{1}{2}u+2p(u-t)(u-1)\right),$
(4.72) $\displaystyle t(t-1)\dot{v}$
$\displaystyle=\tfrac{3}{16}+\frac{\theta_{0}}{4}(\theta_{0}-2)-tv(1+v)+vu+2(1+t)v^{2}u-3v^{2}u^{2},$
(4.73)

and $g(t)=g_{k}(t)$ solves

$\frac{g_{k}^{\prime}(t)}{g_{k}(t)}=(1-4k)\frac{u(t)-t}{2t(t-1)}.$ (4.74)

Note that the zero of $\left(\mathcal{L}_{2k}\right)_{12}$ being $z=Q(\tau)$
follows from (4.71) under the change of variable (4.58).

##  5\. The Painlevé VI tau-function and Hankel determinants

In this section, we find explicit formulas for the solution $u_{k}$ of
Painlevé VI, introduced in (4.71), in terms of the Hankel determinants of
moments defined in equation (2.37). To this end, we compute the first couple
of matrix coefficients in the the expansion around $z=\infty$ of the solution
to RHP 2.4.

######  Lemma 5.1.

The asymptotic expansion of the solution of RHP 2.4 around $z=0$ can be
written as

$Y_{2k}(z)=\left(\mathbb{1}+\wp(z)^{-1}U+\wp(z)^{-2}V+\mathcal{O}(\wp(z)^{-3})\right)\left(\begin{array}[]{cc}\wp(z)^{k}&0\\\
0&\wp(z)^{-k}\end{array}\right)\left(\begin{array}[]{cc}1&0\\\
0&-\frac{1}{2}\wp^{\prime}(z)\end{array}\right),$ (5.1)

where the matrices $U$ and $V$ are given explicitly by

$U=\begin{pmatrix}-\frac{\Gamma_{2k}}{\Delta_{2k}}&\frac{\Delta_{2k+2}}{2\pi
i\Delta_{2k}}&\\\ \frac{2\pi
i\Delta_{2k-2}}{\Delta_{2k}}&+\frac{\Gamma_{2k}}{\Delta_{2k}}\end{pmatrix},\qquad
V=\begin{pmatrix}\vspace{1mm}\frac{\Lambda_{2k}}{\Delta_{2k}}&\frac{\Gamma_{2k+2}}{2\pi
i\Delta_{2k}}&\\\ -\frac{2\pi i\Gamma_{2k-2}}{\Delta_{2k}}&{\sf
v}_{22}\end{pmatrix},$

with

${\sf
v}_{22}=\frac{\Gamma_{2k}^{2}}{\Delta_{2k}^{2}}+\frac{\Delta_{2k-2}\Delta_{2k+2}}{\Delta_{2k}}-\frac{\Lambda_{2k}}{\Delta_{2k}}.$

###### Proof.

Recall the explicit, and unique, solution $Y_{2k}(z)$ of RHP 2.4, defined in
equation (2.51). Note that $\widehat{Y}(z)=Y_{2k}(-z)\sigma_{3}$, also
satisfies all the conditions in the RHP, and thus

$Y_{2k}(z)=Y_{2k}(-z)\sigma_{3}.$

It follows from this symmetry that $Y_{2k}(z)$ admits an expansion in powers
of $\wp(z)$ as given in the lemma.

We proceed to compute the coefficient matrices $U$ and $V$. The expressions
for ${\sf u}_{11}$, ${\sf v}_{11}$, ${\sf u}_{21}$ and ${\sf v}_{21}$ follow
directly from the expansions of the corresponding orthogonal polynomials in
equation (2.39). Next, by equation (2.12), we have

$\mathcal{C}(\pi_{2k})(z)=\frac{h_{2k}}{2\pi
i}z^{2k-1}(1+\mathcal{O}(z^{2})),$

from which the expression for $u_{12}$ follows. Finally, note that
$|Y_{2k}(z)|=-\tfrac{1}{2}\wp^{\prime}(z)$ implies

$|\mathbb{1}+\wp(z)^{-1}U+\wp(z)^{-2}V|=1+\mathcal{O}(\wp(z)^{-3})$

as $z\rightarrow 0$, which is equivalent to

$\operatorname{Tr}U=0,\quad\operatorname{Tr}V+|U|=0,$

as can also be seen from (4.7). (4.12). The expressions for ${\sf u}_{22}$ and
${\sf v}_{22}$ are obtained from the above two equations. ∎

It follows from the asymptotic expansion (5.1) for $Y_{2k}(z)$, that
$\widetilde{Y}_{2k}(x)$ has an expansion around $x=\infty$ of the form

$\displaystyle\widetilde{Y}_{2k}(x)$ $\displaystyle=\Psi_{2k}(x)G(x),$
$\displaystyle\Psi_{2k}(x)$
$\displaystyle=I+x^{-1}\widetilde{U}+x^{-2}\widetilde{V}+\mathcal{O}(x^{-3}),$
$\displaystyle G(x):$
$\displaystyle=x^{(k-\frac{3}{4})\sigma_{3}}\bigg{(}1+\frac{e_{2}}{(e_{1}-e_{2})x}\bigg{)}^{k\;\sigma_{3}}((1-t/x)(1-1/x))^{-\frac{1}{4}\sigma_{3}},$

where

$\widetilde{U}=\frac{U}{e_{1}-e_{2}},\qquad\widetilde{V}=\frac{V-e_{2}\;U}{(e_{1}-e_{2})^{2}}.$

Now, by equation (4.70), we can express the coefficient matrix $A_{2k}(x)$, in
terms of $\Psi_{2k}(x)$, as follows,

$\widetilde{A}_{2k}(x)=\Psi_{2k}^{\prime}(x)\Psi_{2k}(x)^{-1}+\Psi_{2k}(x)G^{\prime}(x)G(x)^{-1}\Psi_{2k}(x)^{-1}.$
(5.2)

This expression allows us to compute the coordinates $(u_{k},v_{k},g_{k})$
introduced in equation (4.71). Indeed, by expanding the right-hand side of
equation (5.2) around $x=\infty$, we get the following expansion for its
$(1,2)$-entry,

$\displaystyle(\widetilde{A}_{2k})_{12}(x)$
$\displaystyle=-\tfrac{1}{2}(4k-1)\widetilde{\sf
u}_{12}x^{-2}+\tfrac{1}{2}s\;x^{-3}+\mathcal{O}(x^{-4}),$ $\displaystyle s:$
$\displaystyle=-(4k+1)t-(4k-3)\widetilde{\sf u}_{11}\widetilde{\sf
u}_{12}+\left(1+t+4k\frac{e_{2}}{e_{1}-e_{2}}\right)\widetilde{\sf u}_{12}.$

Comparing this asymptotic expansion with equation (4.71), and recalling that

$e_{1}-e_{2}=4\mathcal{K}(t)^{2},$

where $\mathcal{K}(t)$ is the complete elliptic integral of the first kind
(see Appendix A), we obtain

$\displaystyle u_{k}(t)$
$\displaystyle=\frac{1}{4\mathcal{K}(t)^{2}}\left(\frac{2k-3}{2k-1}\frac{\Gamma_{2k}}{\Delta_{2k}}-\frac{2k+1}{2k-1}\frac{\Gamma_{2k+1}}{\Delta_{2k+1}}\right)+\frac{1+t}{3},$
(5.3) $\displaystyle g_{k}(t)$ $\displaystyle=-\frac{(2k-1)}{4\pi
i(2k-3)}\mathcal{K}(t)^{1-2k}h_{2k}(t).$ (5.4)

We are now in a position to prove the following theorem.

######  Theorem 5.1.

For $k\geq 0$,

$u_{k}(t)=\frac{2t(t-1)}{2k-1}\left(\frac{\dot{\Delta}_{2k}(t)}{{\Delta}_{2k}(t)}-\frac{\dot{\Delta}_{2k+2}(t)}{{\Delta}_{2k+2}(t)}\right)+1-\frac{\mathcal{E}(t)}{\mathcal{K}(t)},$
(5.5)

solves $P_{\text{VI}}$, where $\mathcal{K}(t)$ and $\mathcal{E}(t)$ denote the
complete elliptic integrals of the first and second kind respectively (see
Appendix A), with parameter values

$\theta_{1}=\theta_{2}=\theta_{3}=\tfrac{1}{2},\quad\theta_{0}=\tfrac{1}{2}(3-2k).$

###### Proof.

Recalling that $\Delta_{0}(t)=\Delta_{2}(t)=1$ and

$\Delta_{4}(t)=\tfrac{16}{3}\mathcal{K}(t)^{2}((t-1)\mathcal{K}(t)^{2}-2(t-2)\mathcal{K}(t)\mathcal{E}(t)-3\mathcal{E}(t)^{2}),$

it can be check by direct calculation that

$\displaystyle u_{0}(t)$
$\displaystyle=1-\frac{\mathcal{E}(t)}{\mathcal{K}(t)},$ (5.6a) $\displaystyle
u_{1}(t)$
$\displaystyle=1-\frac{\mathcal{E}(t)}{\mathcal{K}(t)}-\frac{2t(t-1)}{3}\frac{\dot{\Delta}_{4}(t)}{\Delta_{4}(t)},$
(5.6b)

solve $P_{\text{VI}}$ for the parameter values indicated in the theorem.

Now, assume $k\geq 2$. By combining the differential equation for the gauge
factor $g_{k}(t)$, equation (4.74), with the explicit expression for
$g_{k}(t)$ in terms of $h_{2k}(t)$, equation (5.4), we obtain

$\frac{\dot{h}_{2k}(t)}{h_{2k}(t)}=-\frac{(2k-1)}{2t(t-1)}\left(\frac{\mathcal{E}(t)}{\mathcal{K}(t)}+u_{k}(t)-1\right).$

Solving this equation for $u_{k}(t)$ and using equation (2.38), we obtain the
expression for $u_{k}(t)$ given in the theorem. Since we already know that
$u_{k}(t)$ solves Painlevé VI, the theorem follows. ∎

###  5.1. The Painlevé VI $\tau$-function

The Painlevé VI tau function $\mathcal{T}_{k}(t)$, corresponding to the linear
system (4.60), can be defined by

$\zeta_{k}(t)=t(t-1)\frac{d}{dt}\log{\mathcal{T}_{k}(t)},$ (5.7)

up to a multiplicative constant, where

$\displaystyle\zeta_{k}=$
$\displaystyle(t-1)\operatorname{Tr}{{A}_{2}{A}_{3}}+t\operatorname{Tr}{{A}_{1}{A}_{3}}$
$\displaystyle=$ $\displaystyle
u_{k}(u_{k}-t)(u_{k}-1)v_{k}^{2}-\tfrac{1}{2}(t-2(1+t)u_{k}+3u_{k}^{2})v_{k}$
(5.8)
$\displaystyle-\tfrac{1}{2}k(2k-3)u_{k}+\tfrac{1}{4}(4k^{2}-6k+1)t-\tfrac{1}{8}.$

It is an analytic function on the universal covering space of the punctured
sphere $\mathbb{CP}^{1}\setminus\\{0,1,\infty\\}$, and satisfies the ODE

$(t(t-1)\ddot{\zeta}(t))^{2}=-2\begin{vmatrix}\tfrac{1}{8}&t\dot{\zeta}-\zeta&\tfrac{3}{8}-\frac{1}{4}\theta_{0}^{2}+\dot{\zeta}\\\
t\dot{\zeta}-\zeta&\tfrac{1}{8}&(t-1)\dot{\zeta}-\zeta\\\
\tfrac{3}{8}-\frac{1}{4}\theta_{0}^{2}+\dot{\zeta}&(t-1)\dot{\zeta}-\zeta&\tfrac{1}{8}\end{vmatrix}.$

In the following theorem, we give an explicit expression for
$\mathcal{T}_{k}(t)$.

######  Theorem 5.2.

For $k\geq 0$,

$\mathcal{T}_{k}(t)=t^{\frac{1}{8}}(1-t)^{\frac{1}{8}}(2\mathcal{K}(t))^{-n(2k-3)}\Delta_{2k}(t).$
(5.9)

###### Proof.

To prove the theorem, it is enough to derive the following expression for
$\zeta_{k}(t)$,

$\zeta_{k}(t)=t(t-1)\frac{\dot{\Delta}_{2k}(t)}{\Delta_{2k}(t)}+\frac{1}{2}(2k)(2k-3)\left(\frac{\mathcal{E}(t)}{\mathcal{K}(t)}+t-1\right)+\frac{1}{8}(2t-1).$
(5.10)

for $k\geq 0$. We will prove this expression by induction.

We first deal with the cases $k=0$ and $k=1$. We recall the explicit
expressions for $u_{0}$ and $u_{1}$ in equations (5.6). Using equation (4.72),
we obtain the following corresponding expressions for $v_{0}$ and $v_{1}$,

$\displaystyle v_{0}(t)$ $\displaystyle=0,$ $\displaystyle v_{1}(t)$
$\displaystyle=-\frac{3\Delta_{4}(t)}{32\mathcal{K}(t)\mathcal{E}(t)(\mathcal{K}(t)-\mathcal{E}(t))((1-t)\mathcal{K}(t)-\mathcal{E}(t)},$

and consequently, using equation (5.8), we obtain the following expressions
for $\zeta_{0}(t)$ and $\zeta_{1}(t)$,

$\displaystyle\zeta_{0}(t)$ $\displaystyle=\tfrac{1}{8}(2t-1),$
$\displaystyle\zeta_{1}(t)$
$\displaystyle=\tfrac{1}{8}(2t-1)-\tfrac{1}{2}\left(\frac{\mathcal{E}(t)}{\mathcal{K}(t)}+t-1\right).$

This shows that equation (5.10) holds for $k=0,1$.

Next, we derive a recursive equation for $\zeta_{k}(t)$. To this end, we note
that the recurrence in Proposition 4.2, translates to the following recurrence
for $\widetilde{Y}_{2k}(x,t)$,

$\widetilde{Y}_{2k+2}(x,t)=R_{2k}(x,t)\widetilde{Y}_{2k}(x,t),$

where

$R_{2k}(x,t)=\begin{pmatrix}x+r_{11}&r_{12}\\\ r_{21}&0\end{pmatrix},$

with

$r_{12}=\frac{\theta_{0}k}{2(\theta_{0}-1)},\quad
r_{21}=-\frac{2(\theta_{0}-1)}{\theta_{0}k},$

and

$\displaystyle\frac{1}{4}\theta_{0}(\theta_{0}-2)r_{11}=$
$\displaystyle\tfrac{1}{2}u_{k}(u_{k}-t)(u_{k}-1)v_{k}^{2}-\tfrac{1}{4}\left(3u_{k}^{2}-2(t+1)u_{k}+t\right)v_{k}$
$\displaystyle+\tfrac{3}{32}(3u_{k}-t-1)+\tfrac{1}{8}\theta_{0}(\theta_{0}-2)(u_{k}-t-1).$

Here $\theta_{0}=\tfrac{3}{2}-k$, as before.

From this recursive formula, we obtain the following recurrence for the
coefficient matrix of the linear system in equation (4.60),

$\widetilde{A}_{2k+2}(x,t)=R_{2k}(x,t)\widetilde{A}_{2k}(x,t)R_{2k}(x,t)^{-1}+\left(\frac{\partial}{\partial
x}R_{2k}(x,t)\right)R_{2k}(x,t)^{-1}.$

Direct substitution now yields a very compact recursive formula for
$\zeta_{k}(t)$,

$\zeta_{k+1}(t)=\zeta_{k}(t)+(\theta_{0}-1)(u_{k}(t)-t).$ (5.11)

By combining this recursive formula with the equation for $u_{k}(t)$ in
Theorem 5.1, equation (5.10) follows by induction. This completes the proof of
the theorem. ∎

######  Corollary 5.1.

From equations (5.11), (5.5) and (5.7), we obtain the following recursion for
the Painlevé VI tau function $\mathcal{T}_{k}(t)$,

$\displaystyle s_{k}\mathcal{T}_{k-1}(t)\mathcal{T}_{k+1}(t)=$ $\displaystyle
4(4k-3)^{2}t^{2}(t-1)^{2}\mathcal{T}_{k}(t)\ddot{\mathcal{T}}_{k}(t)-4(4k-1)(4k-5)t^{2}(t-1)^{2}\dot{\mathcal{T}}_{k}(t)^{2}$
$\displaystyle+2((4k-3)^{2}+1)t(t-1)(2t-1)\mathcal{T}_{k}(t)\dot{\mathcal{T}}_{k}(t)$
$\displaystyle+\left[2(k-1)(2k-1)(4k^{2}-6k+1+t-t^{2})-\tfrac{1}{4}\right]\mathcal{T}_{k}(t)^{2},$

for $k\geq 1$. Here, the $s_{k}$ are some nonzero constants which are not
rigidly defined in general. However, using the exact formula (5.9), they
become numerical constants, and the first few are given by

$s_{1}=-3,\quad s_{2}=525,\quad s_{3}=6237,\quad s_{4}=27885,\quad
s_{5}=82365.$

##  Appendix A Elliptic functions and their periodicity properties

The Weierstrass cubic reads as

$\displaystyle\left(\wp^{\prime}(z)\right)^{2}=4\wp^{3}(z)-g_{2}\wp(z)-g_{3}=4\left(\wp(z)-e_{1}\right)\left(\wp(z)-e_{2}\right)\left(\wp(z)-e_{3}\right),$
(A.1)

where

$\displaystyle e_{1}=\wp\left(\frac{1}{2}\right),$ $\displaystyle
e_{2}=\wp\left(\frac{\tau}{2}\right),$ $\displaystyle
e_{3}=\wp\left(\frac{1+\tau}{2}\right).$ (A.2)

The Weierstrass $\wp$-function is doubly periodic

$\displaystyle\wp(z+1)=\wp(z),$ $\displaystyle\wp(z+\tau)=\wp(z),$ (A.3)

and has a double pole at zero

$\displaystyle\lim_{z\to 0}z^{2}\wp(z)=1.$ (A.4)

The Weierstrass $\zeta$-function is defined to be the anti-derivative of
$\wp(z)$ uniquely characterised by

$\displaystyle\zeta^{\prime}(z)=-\wp(z),\quad\zeta(z)=\frac{1}{z}+\mathcal{O}(z)\quad(z\rightarrow
0),$ (A.5)

and has the following periodic properties

$\displaystyle\zeta(z+1)=\zeta(z)+\eta_{1}(\tau),$
$\displaystyle\zeta(z+\tau)=\zeta(z)+\eta_{2}(\tau),$ (A.6)

which in turn define the Weierstrass $\eta$-functions.

The elliptic nome is defined by

$q=\exp{\frac{i\pi\omega_{3}}{\omega_{1}}}=\exp{i\pi(1+\tau)},$

and we define

$t=\frac{e_{3}-e_{2}}{e_{1}-e_{2}}=\lambda(\tau),$

where $\lambda(\cdot)$ is the modular lambda function.

We have the following explicit expressions for $(e_{1},e_{2},e_{3})$ in terms
of $q$ and $t$,

$\displaystyle(e_{1},e_{2},e_{3})=$
$\displaystyle\frac{\pi^{2}}{3}\left(\theta_{3}(0,q)^{4}+\theta_{4}(0,q)^{4},\theta_{2}(0,q)^{4}-\theta_{4}(0,q)^{4}),-\theta_{2}(0,q)^{4}-\theta_{3}(0,q)^{4}\right)$
$\displaystyle=$ $\displaystyle\frac{4}{3}\mathcal{K}(t)^{2}(2-t,-1-t,2t+1),$

where $\theta_{j}(z,q)$ denotes the $j$th Jacobi elliptic function for $1\leq
j\leq 4$. In particular

$\theta_{2}(0,q)=2q^{\frac{1}{4}}\sum_{n=0}^{\infty}q^{n(n+1)},\quad\theta_{3}(0,q)=1+2\sum_{n=0}^{\infty}q^{n^{2}},\quad\theta_{4}(0,q)=\theta_{3}(0,-q).$

This yields the following formulas for the invariants $\\{g_{2},g_{3}\\}$ in
terms of $q$ and $t$,

$\displaystyle g_{2}$
$\displaystyle=\frac{64}{3}(t^{2}-t+1)\mathcal{K}(t)^{4},$
$\displaystyle=\frac{4\pi^{4}}{3}(\theta_{2}(0,q)^{8}-\theta_{2}(0,q)^{4}\theta_{3}(0,q)^{4}+\theta_{3}(0,q)^{8}),$
$\displaystyle g_{3}$
$\displaystyle=\frac{256}{27}(2t-1)(t-2)(t+1)\mathcal{K}(t)^{6},$
$\displaystyle=\frac{8\pi^{6}}{27}\left(\theta_{2}(0,q)^{12}+\theta_{3}(0,q)^{12}-\frac{3}{2}\theta_{2}(0,q)^{4}\theta_{3}(0,q)^{4}(\theta_{2}(0,q)^{4}+\theta_{3}(0,q)^{4}\right).$

Finally, we note the following useful formula for $\eta_{1}$,

$\displaystyle\eta_{1}=$
$\displaystyle-\frac{\pi^{2}}{6}\frac{\theta_{1}^{(3)}(0,q)}{\theta_{1}^{(1)}(0,q)}$
$\displaystyle=$
$\displaystyle\frac{2}{3}\mathcal{K}(t)((t-2)\mathcal{K}(t)+3\mathcal{E}(t)),$

where $\theta_{1}^{(j)}(z,q)$ denotes the $j$th derivative of
$\theta_{1}(z,q)$ with respect to $z$.

##  Appendix B List of polynomials

$\displaystyle\pi_{0}$ $\displaystyle=1,$ $\displaystyle\pi_{2}$
$\displaystyle=\wp(z)+a_{1,1},$ $\displaystyle\pi_{3}$
$\displaystyle=-\tfrac{1}{2}\wp^{\prime}(z),$ $\displaystyle\pi_{4}$
$\displaystyle=\wp^{2}(z)+a_{1,2}\wp+a_{2,2},$ $\displaystyle\pi_{5}$
$\displaystyle=-\tfrac{1}{2}\wp^{\prime}(z)\wp+b_{2,2}\wp(z).$

We compute the first few coefficients

$\displaystyle
a_{1,1}=2\eta_{1}(\tau),\,\,a_{1,2}=2\left(\frac{4\eta_{1}(\tau)g_{2}-3g_{3}}{5g_{2}-240\eta_{1}^{2}(\tau)}\right),\,\,a_{2,2}=4\eta_{1}(\tau)\left(\frac{4\eta_{1}(\tau)g_{2}-3g_{3}}{5g_{2}-240\eta_{1}^{2}(\tau)}\right)-\frac{g_{2}}{12},$
(B.1)

and

$\displaystyle h_{1}(\tau)=1,$ $\displaystyle
h_{2}(\tau)=-4\eta_{1}^{2}(\tau)+\frac{g_{2}}{12},$ $\displaystyle
h_{3}(\tau)=\frac{1}{5}\left(g_{2}\eta_{1}^{2}(\tau)-3g_{3}\right).$ (B.2)

##  Appendix C Structure of the moment matrix

Let us elaboratte on the structure of the moment matrix $D_{n}$ (2.3). There
are three points to note about the moments (2.2):

  1. (1)

all the mixed moments vanish, i.e for all $i,j$

$\displaystyle\mu_{2i,2j+1}=0,$ (C.1)

  2. (2)

the following symmetry property holds

$\displaystyle\mu_{i,j}=\mu_{j,i},$ (C.2)

  3. (3)

and, generally

$\displaystyle\mu_{2i,2j}\neq\mu_{2i-1,2j+1}.$ (C.3)

With the above properties, following figure illustrates the moment matrix for
$n=9$.

Figure 6. Moment matrix $D_{9}$ (2.3) with the  odd and  even moments are
colour coded.

Note that the matrix above has a block structure with the checkerboard pattern
generated by the element

$\displaystyle
M_{i,j}=\left(\begin{array}[]{cc}{\color[rgb]{1,0.6484375,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0.6484375,0}\mu_{2i-1,2j-1}}&0\\\
0&{\color[rgb]{0.25,0.5,0.75}\definecolor[named]{pgfstrokecolor}{rgb}{0.25,0.5,0.75}\mu_{2i,2j}}\end{array}\right),\qquad
i,j\geq 2.$ (C.6)

The even bands are a trivial consequence of dimensions of the space of
meromorphic functions in the genus 1 case.

## References

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