GREEN FUNCTION, PAINLEVÉ VI EQUATION, AND EISENSTEIN SERIES OF WEIGHT ONE

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1 GREEN FUNCTION, PAINLEVÉ VI EQUATION, AND EISENSTEIN SERIES OF WEIGHT ONE ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG ABSTRACT. We study the problem: How many singular points of a solution λt to the Painlevé VI equation with parameter 8, 8, 8, 3 8 might have in C \ {0, }? Here t 0 C \ {0, } is called a singular point of λt if λt 0 {0,, t 0, }. Based on Hitchin s formula, we explore the connection of this problem with Green function and the Eisenstein series of weight one. Among other things, we prove: i There are only three solutions which have no singular points in C \ {0, }. ii For a special type of solutions called real solutions here, any branch of a solution has at most two singular points in particular, at most one pole in C \ {0, }. iii Any Riccati solution has singular points in C \ {0, }. iv For each N 5 and N = 6, we calculate the number of the real j-values of zeros of the Eisenstein series E Nτ; k, k of weight one, where k, k runs over [0, N ] with gcdk, k, N =. The geometry of the critical points of the Green function on a flat torus E τ, as τ varies in the moduli M, plays a fundamental role in our analysis of the Painlevé IV equation. In particular, the conjectures raised in [] on the shape of the domain Ω 5 M, which consists of tori whose Green function has extra pair of critical points, are completely solved here. CONTENTS. Introduction. Painlevé VI: Overviews 3 3. Riccati solutions 6 4. Completely reducible solutions 5. Geometry of Ω Geometry of Ω Algebraic solutions 4 8. Further discussion 48 Appendix A. Picard solution and Hitchin s solution 50 Appendix B. Asymptotics of real solutions at {0,, } 5 References 56

2 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG. INTRODUCTION.. Painlevé property. In the literature, a nonlinear differential equation in one complex variable is said to possess the Painlevé property if its solutions have neither movable branch points nor movable essential singularities. For the class of second order differential equations. λ t = Ft, λ, λ, t CP, where Ft, λ, λ is meromorphic in t and rational in both λ and λ, Painlevé later completed by Gambier, [, 9] obtained the classification of those nonlinear ODEs which possess the Painlevé property. They showed that there were fifty canonical equations of the form. with this property, up to Möbius transformations. Furthermore, of these fifty equations, fortyfour are either integrable in terms of previously known functions such as elliptic functions, equivalent to linear equations, or are reduced to one of six new nonlinear ODEs which define new transcendental functions see eg. [7]. These six nonlinear ODEs are called Painlevé equations. Among them, Painlevé VI is often considered to be the master equation, because others can be obtained from Painlevé VI by the confluence. Due to its connection with many different disciplines in mathematics and physics, Painlevé VI has been extensively studied in the past several decades. See [, 3, 8, 0,, 3, 5, 0, 4, 5, 7, 8, 35] and the references therein. Painlevé VI PVI is written as. d λ dt = + λ + λ + λ λ λ t t t λ t [ dλ dt t + t + dλ λ t dt ] α + β t λ + γ t t t + δ λ λ t where α, β, γ, δ are four complex constants. From., the Painlevé property says that any solution λt is a multi-valued meromorphic function in C\{0, }. To avoid the multi-valueness of λt, it is better to lift solutions of. to its universal covering. It is known that the universal covering of C\{0, } is the upper half plane H = {τ Im τ > 0}. Then t and the solution λt can be lifted through the covering map τ t by.3 tτ = e 3τ e τ e τ e τ, λt = pτ τ e τ, e τ e τ where z = z τ is the Weierstrass elliptic function defined by.4 z τ := z + z ω ω Λ τ \{0} ω, and Λ τ := {m + nτ m, n Z} is the lattice generated by ω = and ω = τ. Also ω 3 = + τ and e i τ = ω i τ for i =,, 3. Consequently,,

3 GREEN FUNCTION AND PAINLEVÉ VI EQUATION 3 pτ satisfies the following elliptic form of PVI:.5 where ω 0 = 0 and d p τ dτ = 3 4π α i p τ + ω i τ, i=0.6 α 0, α, α, α 3 = α, β, γ, δ. This elliptic form was first discovered by Painlevé [30]. For more recent derivations of it, see [, 5]... Hitchin solutions. In this paper, we consider the special case α i = 8 for 0 i 3, i.e., d p τ.7 dτ = 3 3π p τ + ω i τ, i=0 which is the elliptic form of PVI 8, 8, 8, 3 8. Equation.7 has connections with some geometric problems. The well-known example is related to the construction of Einstein metrics in four dimension; see [5]. In the seminal work [5], Hitchin obtained his famous formula to express a solution pτ of.7 with some complex parameters r, s:.8 pτ τ = r + sτ τ + r + sτ τ ζ r + sτ τ rη τ + sη τ. Here η i τ = ζ ω i τ, i =,, are quasi-periods of the Weierstrass zeta function ζz τ = z ξ τdξ. By.8, he could construct an Einstein metric with positive curvature if r R and s ir, and an Einstein metric with negative curvature if r ir and s R. He also obtained an Einstein metric with zero curvature, but the corresponding solution of.7 is given by another formula other than.8. Indeed, this corresponds to the Riccati solutions of.7; see 3. For simplicity, we denote p r,s τ equivalently, λ r,s t via.3 to be the solution of.7 with the expression.8. It is obvious that if r, s Z := { 0, 0, 0,,, 0,, } + Z, then either ζ r + sτ τ rη τ + sη τ or ζ r + sτ τ rη τ + sη τ 0 in H. Hence for any complex pair r, s Z, p r,s τ is always a solution to.7, or equivalently, λ r,s t is a multi-valued solution to PVI 8, 8, 8, 3. We say that two solutions λ r,st and λ r,s t give or belong to 8 the same solution if λ r,s t is the analytic continuation of λ r,st along some closed loop in C \ {0, }. In 4, we will prove that λ r,s and λ r,s give the same solution to PVI 8, 8, 8, 3 8 if and only if s, r s, r γ mod Z for some matrix γ Γ = {γ SL, Z γ I mod }. In this paper, we are mainly concerned with the question of smoothness of solutions to PVI 8, 8, 8, 3 8, and the location of its singular points. Notice

4 4 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG that for a solution λt and a point t 0 C\{0, }, the RHS of Painlevé VI. has a singularity at t 0, λt 0 provided that λt 0 {0,, t 0, }. Therefore, in this paper, we say λt is smooth at t 0 if λt 0 {0,, t 0, }. Furthermore, a singular point t 0 C\{0, } is called of type 0,, 3 respectively if λt 0 = λt 0 = 0,, t 0 respectively. We take PVI 0,0,0, as an initial example for our discussion, because it can be transformed to PVI 8, 8, 8, 3 8 of our concern by a Bäcklund transformation cf. [8]. In the literature, the Bäcklund transformation plays a very useful role in the study of Painlevé VI; for example, for finding the algebraic solutions, see [0, 6, 4]. Conventionally, solutions of PVI 0,0,0,, the socalled Picard solutions, can be expressed in terms of Gauss hypergeometric functions. It was first found by Picard [3]. Let.9 ω t = iπf,, ; t, ω t = πf,, ; t be two linearly independent solutions of the Gauss hypergeometric equation.0 t tω t + tω t 4ωt = 0. Then Picard solution of PVI 0,0,0, can be expressed as. ˆλ ν,ν t = ν ω t + ν ω t ω t, ω t + + t 3, for some ν, ν Z, where ω t, ω t is the Weierstrass elliptic function with periods ω t and ω t. See [] for a proof. Obviously, ˆλ ν,ν t is smooth for all t C\{0, } if and only if ν, ν R \ Z see A.. The lifting of ˆλ ν,ν t by.3 is given by. ˆp ν,ν τ = ν + ν τ, which of course is a solution of the elliptic form of PVI 0,0,0, : d pτ = 0. dτ Then the Bäcklund transformation takes ˆλ ν,ν t into solution λ r,s t of PVI 8, 8, 8, 3 8 with r, s = ν, ν. Thus, in the elliptic form the Bäcklund transformation seems comparably simple. This fact and. are well known to experts. But it is difficult to find references for the proof. For the reader s convenience, we present a rigorous proof in Appendix A. It is surprising to us that after the Bäcklund transformation from PVI 0,0,0, to PVI 8, 8, 8, 3, PVI 8 8, 8, 8, 3 8 has only three solutions which are smooth in C\{0, }. Theorem.. There are only three solutions λt to PVI 8, 8, 8, 3 8 such that λt is smooth for all t C\{0, }. They are precisely λ 4,0 t, λ 0, t and λ 4 4, t. 4 Theorem. shows that the Bäcklund transformation does not preserve the smoothness of solutions. Thus, Theorem. can not be proved by applying Picard solutions and the Bäcklund transformation. We remark that

5 GREEN FUNCTION AND PAINLEVÉ VI EQUATION 5 the Bäcklund transformation is complicated due to not only the complicated form of birational maps between solutions but also the fact that it transforms a pair of solutions of the Hamiltonian system equivalently, the pair λt, λ t, but not the solution λt only. To prove Theorem., we start from the formula.8. Of course,.8 does not give the complete set of solutions to.7. The missing ones are solutions obtained from Riccati equations. For such Riccati solutions, we have some expressions like.8. By employing these expressions, we will prove in 6 that any Riccati solution has singularities in C\{0, }. Hence our strategy for the proof of Theorem. is to study the smoothness of λ r,s t for any complex pair r, s Z. From.8, it is easy to see that if r, s is not a real pair, then λ r,s t always possesses a singularity t 0 {0,, } indeed, infinitely many singularities, because there always exist infinitely many τ 0 H such that r + sτ 0 is a lattice point of the torus E τ0 := C/Λ τ0. So for the proof of Theorem. we could restrict ourselves to consider only r, s R \ Z. In this case, we introduce the Green function and the Hecke form to study it..3. Green function and Hecke form. Let Gz τ be the Green function on the torus E τ : { G z τ = δ0 z.3 E τ in E τ, E τ G z τ dz = 0, where δ 0 is the Dirac measure at 0 and E τ is the area of the torus E τ. We recall the analytic description of Gz τ in []. Recall the theta function ϑ := ϑ, where ϑ z; τ = i n= Then the Green function is given by n e n+ πiτ e n+πiz..4 Gz τ = Im z log ϑz; τ + π Im τ + Cτ, where Cτ is a constant so that E τ G = 0. Recall that η i τ = ζ ω i τ, i =,, are quasi-periods of ζz τ. Using log ϑ z = ζz η z and the Legendre relation η ω η ω = πi, we have.5 4πG z z τ = ζz τ rη τ sη τ, where z = r + sτ with r, s R. As mentioned before, ζ r + sτ τ rη τ sη τ 0 in H whenever r, s Z \ Z. Thus for r, s R \ Z,.5 shows that r + sτ is a non half-period critical point of Gz τ we call such critical point a nontrivial critical point if and only if ζr + sτ τ rη τ sη τ = 0. Naturally, we ask the question: How many nontrivial critical points might Gz τ have? Note that the nontrivial critical

6 6 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG points must appear in pair because Gz τ is an even function in z. This question was answered in the following surprising result: Theorem A. [] For any torus E τ, Gz τ has at most one pair of nontrivial critical points. Theorem B. [3] Suppose that Gz τ has one pair of nontrivial critical points. Then the three half-periods are all saddle points of Gz τ, i.e., the Hessian satisfies det D G ω k τ 0 for k =,, 3. For any r, s R \ Z, we define Z = Z r,s by.6 Z r,s τ := ζr + sτ τ rη τ sη τ, τ H. Clearly Z r,s is a holomorphic function in H. If r, s is an N-torsion point, i.e., r, s = k N, k N with 0 k, k < N and gcdk, k, N =, it was proved by Hecke in [4] that Z r,s τ is a modular form of weight with respect to ΓN = { A SL, Z A I mod N }. This modular form is called the Hecke form in []. Indeed, it is the Eisenstein series of weight with characteric r, s if r, s is an N-torsion point. Following [3, p.59], the Eisenstein series of weight is defined by E N τ, s; k, k := Im τ s mτ + n mτ + n s, m,n where m, n runs over Z under the condition 0 = m, n k, k mod N. It is known that E N τ, s; k, k is a meromorphic function in the s-plane and holomorphic at s = 0. Set E N τ; k, k := E N τ, 0; k, k. By using the Fourier expansions of both Z r,s τ and E N τ; k, k see [3, p.59] and [9, p.39], we have.7 Z r,s τ = NE N τ; k, k, if r, s k N, k N mod. Hence,.5 yields that Gz τ 0 has a critical N-torsion point k N + k N τ 0 with N 3 if and only if E Nτ 0; k, k = 0. Now we see the connection of the Hecke form with the solution p r,s τ or λ r,s t of.7: Z r,s τ appears in the denominator of the RHS of.8. When r, s R \ Z, the formula.8 implies that t 0 is a type 0 singularity, i.e., λ r,s t 0 = if and only if Z r,s τ 0 = 0, t 0 = tτ 0, or equivalently, the Green function Gz τ 0 has a nontrivial critical point r + sτ 0. By Theorem A, it means that Gz τ 0 have exactly five critical points in the torus E τ0 : ω ω, ω 3 and ±r + sτ 0. This connection and.8 together with the Painlevé property say that the Eisenstein series E Nτ; k, k of weight has only simple zeros; see Theorem 4.. The simplicity of zeros was also proved by Dahmen [7] as a consequence of his counting formula of algebraic integral Theorem B is used in the proof of Theorem. ii to be stated later. After establishing Theorem., we have a stronger version of Theorem B: det D G ω k τ < 0 for k =,, 3 if Gz τ has one pair of nontrivial critical points; see Proposition 6. in 6.,

7 GREEN FUNCTION AND PAINLEVÉ VI EQUATION 7 Lamé equations by the method of dessins d enfants. In 7 we will discuss the position and the number of those zeros of E Nτ 0; k, k. Recall the group action of SL, Z on the upper half plane H: τ = γ τ = aτ + b cτ + d, γ = a b SL, Z. c d Then we have the transformation law see 4.4 in 4:.8 Z r,s τ = cτ + dz r,s τ where s, r = s, r γ. From here, we see that Gz τ has five critical points whenever Gz τ has five critical points. Let M := H/SL, Z and Ω 5 := {τ M Gz τ has five critical points}, Ω 3 := {τ M Gz τ has three critical points}. Then we have Ω 3 Ω 5 = M by Theorem A. Moreover, from the proof of Theorem A in [], we know that Ω 5 M is open and Ω 3 is closed. In this paper we determine the geometry of Ω 3 and Ω 5 as conjectured in []: Theorem. Geometry of Ω 3 and Ω 5. i Both Ω 5 and Ω 3 = Ω 3 { } are simply connected in M = S. ii C = Ω 5 = Ω 3 = S. C\{ } = R is smooth. It consists of points τ so that some half-period is a degenerate critical point of Gz τ. The proof is given in 5 and 6. We actually prove a stronger result on Ω 5 : For any τ Ω 5, there is only one half period whose Hessian det D G vanishes. Theorem. is clearly closely related to the following question: What is the set of pairs r, s such that Z r,s τ has no zeros? We should write an alternative form of i in Theorem. to answer this question. We note that the following two statements hold:.9 Z r,s τ = ±Z r,s τ r, s ±r, s mod Z,.0 λ r,s τ = λ r,s τ r, s ±r, s mod Z. The statement.9 is trivial while.0 was proved in [6]. From both.9 and.0, we could assume r, s [0, ] [0, ]\ Z. Then i of Theorem. can be stated more precisely. For this purpose, we consider. F 0 = { τ H 0 Re τ, τ }. It is elementary to prove that F 0 is a fundamental domain for Γ 0 c.f. Remark 5.. Notice that F 0 is one half of the fundamental domain of Γ. The following theorem will imply i of Theorem.. Theorem.3. Let r, s [0, ] [0, ]\ Z. Then Z r,s τ = 0 has a solution τ F 0 if and only if r, s 0 := {r, s 0 < r, s <, r + s > }.

8 8 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG FIGURE. The lifted domain Ω 5 F 0 of Ω 5 is the domain bounded by the 3 curves corresponding to the loci of degenerate critical points. Moreover, the solution τ F 0 is unique for any r, s 0. We will see that Theorem. is a consequence of the non-existence part of Theorem.3 in 5. Indeed, the existence part of Theorem.3 has applications as well; see the next subsection, where we will discuss the singular points of a real solution λt..4. Real solution. It is well known that Painlevé VI governs the isomonodromic deformations of some linear ODE. In the elliptic form it is convenient to choose the ODE to be a generalized Lamé equation c.f..4. A solution λt of PVI 8, 8, 8, 3 8 is called a real solution if its associated monodromy of the generalized Lamé equation is unitary. In [6] it was proved that a solution λt is a real solution if and only if λt = λ r,s t for some r, s R \ Z. We call such a solution of PVI 8, 8, 8, 3 8 real because any solution with unitary monodromy representation must come from blowup

9 GREEN FUNCTION AND PAINLEVÉ VI EQUATION 9 solutions of the mean field equation; see [6]. We remark that real solutions do not mean real-valued solutions along the real-axis of t. Indeed, for.7 there are no real-valued solutions; see the discussion in Appendix B. The reasons we are studying real solutions are: i any algebraic solution is a real solution; ii any real solution is smooth for t R\{0, } see [6]; iii any real solution has no essential singularity even at 0, and see Appendix B. It is known see that tτ = e 3τ e τ e τ e τ maps any fundamental domain of Γ one-to-one and onto C \ {0, }. Then by the transformation.3, we see that any solution λtτ is single-valued and meromorphic whenever τ is restricted on a fundamental domain of Γ. In this paper, a branch of a solution λt to. means a solution λtτ defined for τ in a fundamental domain of Γ e.g. F given by.. Recall a singular point t 0 {0,, } of λt means λt 0 {0,, t 0, }. Denote C ± = {t Im t 0}. Then for real solutions we have: Theorem.4. Suppose λt is a real solution. Then any branch of λt has at most two singular points in C \ {0, }, and they must be different type singular points if the branch has exactly two singular points. Furthermore, the set. Ω 0 := {t C t is a type 0 singular point of some real solution} is open and simply connected and Ω 0 consists of three smooth curves connecting 0,, respectively. Remark.5. Theorem.4 shows that for each k {0,,, 3}, any branch of a real solution has at most one type k singular point in C \ {0, }. Theorem.4 will be proved in 6, where we will see that, the curve of Ω 0 connecting and 0 resp. connecting and, connecting 0 and is the image of the degenerate curve of ω resp. ω, ω 3 of Green function Gz τ in F 0 under the map tτ. Similarly, we can define.3 Ω k ± := {t C ± t is a type k singular point of some real solution}. Then Ω k = Ω 0 and any real solution is smooth in C \ Ω 0 and Ωk + = Ω 0 + = {t t Ω 0 } for k {,, 3}, Ω 0 + {0, }, which consists of three connected components that contain, 0, 0, and, + respectively. See the proof in Algebraic solution. A solution λt to PVI is called an algebraic solution if there is a polynomial h C[t, x] such that ht, λt 0. It is equivalent to that λt has only a finite number of branches. By our classification theorem for.7, λt is an algebraic solution of PVI 8, 8, 8, 3 8 if and only if λt = λ r,s t for some r, s Q N with N 3, where.4 Q N := { k N, k N gcdk, k, N =, 0 k, k N }

10 0 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG is the set of N-torsion points of exact order N. The classification of the algebraic solutions for PVI 8, 8, 8, 3 8 could be deduced from the Bäcklund transformation and Picard solutions, as shown in [6]. It is therefore natural to ask the following question: Is any singular point t 0 of an algebraic solution λt an algebraic number? Is the lifting τ 0 of t 0 a transcendental number? The first question is equivalent to asking whether the j-value of any zero of E Nτ; k, k is an algebraic number. Here jτ is the classical modular function, the j-invariant of τ, under the action by SL, Z; see.5 below. This question can be answered easily from the aspect of Painlevé VI or from the q-expansion principle in the theory of modular forms c.f. [9]. It is well known from the addition theorem of function that there is a polynomial Ψ N Z[x, y, g, g 3 ] such that if x, y is an N-torsion point of the elliptic curve y = 4x 3 g x g 3, then Ψ N x, y = 0. The degree of Ψ N is N, and y appears only with odd powers in Ψ N x, y if N is even; y appears only with even powers in Ψ N x, y if N is odd. See [6, p.7]. Now we come back to.9 and.. Suppose that ˆλt = ˆλ ν,ν t is a solution of PVI 0,0,0,, where ν, ν is an N-torsion point. Then by the above result and the formulae for ẽ k := ω kt ω t, ω t here ω 3 = ω + ω, see [6]: ẽ = + t 3, ẽ = + t 3, ẽ 3 = t t + 3, we see that there is a polynomial ˆP Q[t, x] such that ˆPt, ˆλt 0. This polynomial seems too complicated to be computed in general. By the Bäcklund transformation, we conclude that for any algebraic solution λt, there is a polynomial P Q[t, x] such that Pt, λt 0. Hence any singular point t of λ must be a root of a polynomial with integral coefficients, which implies that t is an algebraic number. Let t = tτ. Recall the classical modular function jτ of SL, Z: g τ.5 jτ := 3 78 g τ 3 7g 3 τ = 78 g τ 3 τ, where g τ and g 3 τ are the coefficients of the elliptic curve E τ : y = 4x 3 g τx g 3 τ, and the relation between tτ and jτ is.6 j = 56 t t + 3 t t. So if tτ is algebraic, then jτ is algebraic. Another way to see it is to use a general principle from the theory of modular forms. Since all the coefficients of the Fourier expansion of Z r,s τ are algebraic numbers, {jτ Z r,s τ = 0} are algebraic by the so-called

11 GREEN FUNCTION AND PAINLEVÉ VI EQUATION q-expansion principle c.f. [9]. However, we can prove more. Let us consider Z N τ := Z r,s τ. r,s Q N This is a modular form of weight Q N := #Q N with respect to SL, Z. For N 5, m := Q N 4 N and Z Nτ is invariant under SL, Z. Observe τ m that Z r,0 τ if s = 0,.7 Z r,s τ = Z 0, s τ if r = 0, Z r, s τ if r = 0, s = 0, which implies that any zero of Z Nτ τ m poles in H, we conclude that.8 must be doubled. Since Z Nτ τ m Z N τ τ m = C m l N j has no for some monic polynomial l N of j and nonzero constant C m. If N is odd, then Z N = 0. Hence Z Nτ has poles of order m at τ =, τ m equivalently, l N j is a polynomial of degree m = Q N 4. If N is even, then l N := deg l N < m. In any case, we have Z N τ = C m τ m l n HG 4 τ 3, τ, where HX, Y is a homogeneous polynomial of X, Y and G 4 τ = g τ/60 is the classical Eisenstein series of weight 4. By using the q-expansion of Z r,s τ, we can prove that l N j has rational coefficients. Theorem.6. For any N 5 with N = 6, the monic polynomial l N j determined by.8 has rational coefficients and satisfies i for any zero j 0 of l N j, there is an algebraic solution λ r,s t, r, s Q N, such that j 0 = jτ 0, where t 0 = tτ 0 satisfies λ r,s t 0 =. Conversely, for any algebraic solution λ r,s t, r, s Q N, if λ r,s t 0 = for some t 0 = tτ 0, then j 0 = jτ 0 is a zero of l N j. ii l N j has distinct roots. iii for any N = N, l N j and l N j have no common zeros. iv.9 deg l N = { QN where ϕ is the Euler function. Recall the elementary formulae.30 Q N = N p N, p prime 4 if N is odd, Q N 4 ϕ N if N is even, p, ϕn = N p N, p prime. p

12 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG Denote the j-value set of zeros of Z N τ by JN := {jτ Z r,s τ = 0 for some r, s Q N }. If N = 3, then Q N = 8 and Z N τ =const G 4 τ. Thus the zero τ = ρ := e πi 3 and J3 = {0}. By Theorem. and Lemma 5., we see that J4 = J6 =. For N 5, JN is just the zero set of l N j. Note that formula.9 also holds for N = 6, which gives deg l 6 = 0, so l 6 j is a non-zero constant. This also proves J6 =. The computation of l N seems to be difficult in general. However, by applying PVI, it is considerably easier for small N. Here are some examples:.3 J3 = {0}, J5 =.3 J7 = For N = 9, the polynomial is { }, J8 = { }, { ± 7559 }. 4 l 9 j =j j j So J9 = {a, b, b}, where a R and b R. Numerically,.33 a 86.3, b It seems that except for N = 3, all elements in JN are not algebraic integers. If this would be true, then by a classical result of Siegal and Schneider, all τ such that λtτ = for an algebraic solution λt are transcendental. A natural question is how to determine their location in the fundamental domain F of SL, Z, where.34 F = {τ H 0 Re τ <, τ, τ > } {ρ = e πi 3 }. The above examples show that there is at least one zero of l N j of where the corresponding τ is on the circular arc {τ H τ = }. Define J N = {r, s Q N r + s = and 3 < s < }, J + N = {r, s Q N r + s = and 0 < s < 3 }. Then we have the following interesting result. Theorem.7. For any N 5 with N = 6, l N j has exactly #J + N real zeros in 0, 78 and exactly #J N real zeros in, 0. Furthermore, l Nj has no zeros in {0} [78, +. Notice that in the fundamental domain F of SL, Z, the corresponding τ of any positive zero of l N j is on the circular arc {τ F τ = }; while the corresponding τ of any negative zero of l N j is on the line {τ F

13 GREEN FUNCTION AND PAINLEVÉ VI EQUATION 3 Re τ = }. We can use to check the validity of Theorem.7 for small values of N. For example, J + 5 = { 5, 5 }, J 5 = ; J+ 7 = { 3 7, 7 }, J 7 = { 7, 3 7 }; J + 8 = { 3 8, 8 }, J 8 = ; J+ 9 = { 4 9, 9 }, J 9 =. The proof of Theorems.6 and.7 will be given in 7. In 8 we will give some further remarks about Theorems.4 and.7. The explicit relation between Picard solutions and Hitchin s solutions will be given in Appendix A. Finally, Appendix B is denoted to the asymptotics of real solutions at {0,, }, which are needed in the computation of l N j.. PAINLEVÉ VI: OVERVIEWS In this section, we start with the discussion of Painlevé VI.: d λ dt = λ + λ + dλ λ t dt t + t + dλ λ t dt [ λ λ λ t + t t α + β t λ + γ t ] t t + δ λ λ t. It is well-known that. possesses the Painlevé property, which says that any solution λt has no branch points and no essential singularities at any t C\{0, }... Multi-valueness via single-valueness. The Painlevé property implies that although a solution λt is multi-valued in C, λt is a single-valued meromorphic function if t is restricted in C ± = {z = x + iy y 0}. That means if λt is analytically continued along a closed curve t = tɛ, t0 = t, in C + or C, then λt0 = λt. Due to the multi-valueness of a solution of., it is convenient to lift solutions and the equation to the universal covering. The universal covering space of C\{0, } is the upper half plane H. The covering map tτ is given in.3, by which, Painlevé VI. is transformed into the elliptic form.5. It is elementary that tτ is invariant under the action of γ Γ, where Γ = {A SL, Z A I mod }. That is tτ = tτ if and only if τ = γ τ = aτ+b a b cτ+d for some γ = c d Γ. Indeed tτ is the principal modular function of Γ. Let H = H Q, where Q is the set of rational numbers. Then it is well known that H /Γ = CP with three cusp point, 0, which are mapped to, 0, by tτ respectively. As the consequence of the isomorphism, we have t τ = dt τ = 0, τ H, dτ

14 4 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG namely the transformation tτ is locally one-to-one. Therefore, tτ maps any fundamental domain of Γ one-to-one onto C \ {0, }, and any solution λtτ is single-valued and meromorphic whenever τ is restricted in a fundamental domain of Γ. As pointed out in, throughout this article, a branch of a solution λt always means a solution λtτ defined for τ in a fundamental domain of Γ. The fundamental domain F of Γ is. F = { τ 0 Re τ <, τ, τ 3 > }. When τ ir +, e k τ are real-valued and satisfies e τ < e 3 τ < e τ see e.g. [6]. From here, it is easy to see that tir + = 0,, where ti = and ti0 = 0. Here we used lim τ i e τ = lim τ i e 3 τ = π 3 see 6. Furthermore, we could deduce from above that for any τ F, tτ R if and only if τ ir + {τ H τ = } {τ H Re τ = }. By the formula.8, we see that pτ τ is always a single-valued meromorphic function defined in H. However, as a solution of.7, pτ has a branch point at those τ such that pτ E τ [], where E τ [] := { ω k 0 k 3} is the set of -torsion points in E τ. The single-valueness of pτ τ is one of the advantages of the elliptic form. Recalling. and., F 0 is a half part of F. Then it is not difficult to prove that the transformation tτ maps the interior of F 0 onto the lower half plane C, and tτ maps F \F 0 onto C + ; see 6. Hence it is convenient to use τ F when a branch of solution λt with t C \ {0, } is discussed. Different branches of λt can be obtained from.8 by considering τ in another fundamental domain of Γ... Isomonodromic deformation. It is well known that Painlevé VI governs the isomonodromic deformation of some linear ODE. See [8] in this aspect. For the elliptic form.5, it was shown in [6] that it is convenient to use the so-called generalized Lamé equation GLE:. y = [ 3 j=0 n j nj + z + ω j +A ζ z + p ζ z p + B z + p + z p Suppose n j + Z. Then pτ is a solution of.5 if and only if there exist Aτ and Bτ such that GLE. preserves the monodromy as τ deforms. The formula to connect parameters of.5 and. is:.3 α j = nj +, j = 0,,, 3. See [6] for the proof. The advantage to employ GLE. is that for some cases, the monodromy representation is easier to describe. For example, let us consider n j = 0 for all j. Then the elliptic form of PVI is.7, and GLE is.4 y = [ 3 4 z + p + z p + A ζ z + p ζ z p + B] y. For any p E τ [], ±p are the singular points of.4 with local exponents and 3. We always assume that ±p are apparent singularities. If r, s ] y.

15 GREEN FUNCTION AND PAINLEVÉ VI EQUATION 5 C \ Z and pτ = p r,s τ is the solution given by.8, then we proved in [6] that the monodromy representation ρ : π E τ \{±p}, q 0 SL, C of GLE.4 is generated by ργ ± = I, ρl = e πis 0 0 e πis e πir 0, ρl = 0 e πir where q 0 is a base point, γ ± π E τ \{±p}, q 0 encircles ±p once and l, π E τ \{±p}, q 0 are two fundamental circles of the torus E τ such that γ + γ = l l l l. In particular, the monodromy representation ρ is completely reducible..3. Bäcklund transformation. In [8], Okamoto constructed the so-called Bäcklund transformations between solutions of Painlevé VI with different parameters. Indeed, this transformation is a birational transformation between the solutions of the corresponding Hamiltonian system, or equivalently, a birational transformation of λt, λ t together. Since λt and λ t are algebraically independent generally otherwise, Painlevé equation would be reduced to a first order ODE, the Bäcklund transformation is not a birational transformation of the solution λt only. For example, it is known that a solution λt of PVI 8, 8, 8, 3 8 can be obtained from a solution ˆλt of PVI 0,0,0, by the following Bäcklund transformation cf. [34, transformation s in p.73]:.5 λt = ˆλt + ˆµt, ˆµt = tt ˆλ ˆλ ˆλ. ˆλ ˆλ ˆλ t As mentioned in the Introduction, for PVI 0,0,0,, all its solutions are Picard solutions:.6 ˆλt = ˆλ ν,ν t = ν ω t + ν ω t ω t, ω t + t + 3, where ν, ν C \ Z and ω, t are given by.9. See [6, ]. In principle, Hitchin s formula.8 could be obtained from Picard solution.6 via.5, as mentioned by [6] and some other references. However, the computation of ˆλ t via.6 is actually very difficult, and in practice, it is not easy at all to obtain Hitchin s formula from Picard solution This is why we can not find a rigorous derivation of Hitchin s formula from Picard solution in the literature. In Appendix A, we will give a rigorous derivation from Hitchin s formula to Picard solution. In the literature, researchers often restrict the study of Painlevé VI to special parameters via Bäcklund transformations. This leaves the impression that the theory for different parameters may be much the same. However, this turns out not to be completely true in general. For example, it is easy to see from the expression A. that ˆλt is smooth for all t C\{0, } if and only if ν, ν R \ Z. But this assertion is obviously false for PVI 8, 8, 8, 3 8.,

16 6 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG However, the Bäcklund transformation is useful to discuss branch points and essential singularities, which are preserved under the Bäcklund transformation. For example, if t = 0 is a branch point for a solution, then after the Bäcklund transformation, t = 0 is still a branch point for the new solution. Another example is that t = is an essential singularity for λ r,s t if r R and s ir. Thus, t = is also an essential singularity for Picard solution ˆλ ν,ν t if ν R and ν ir. For the discussion of branch points for real solutions, please see Appendix B. 3. RICCATI SOLUTIONS First we review the classification theorem of solutions to the elliptic form.7 due to the associated monodromy representation of GLE.4. It was shown that solutions expressed in.8 does not contain all the solutions. Indeed, we have the following classification theorem proved in [6]. In this article, when we talk about the monodromy representation, we always mean the one of GLE.4. Theorem C. [6, Theorem 4.] Suppose pτ is a solution to the elliptic form.7. Then the followings hold: i The monodromy representation is completely reducible if and only if there exists r, s C \ Z such that pτ τ is given by.8. ii The monodromy representation is not completely reducible if and only if 3. λ t = p τ τ e τ, t = e 3 τ e τ e τ e τ e τ e τ 3. satisfies one of the following four Riccati equations: dλ dt = tt λ tλ + t, µ 0, 3.3 dλ dt = tt λ λ + t, µ λ, 3.4 dλ dt = tt λ t, µ λ, 3.5 dλ dt = tt λ + t λ t, µ λ t. Here µt is defined by the second formula in A., i.e., λt, µt satisfies the well-known Hamiltonian system of Painlevé VI. It is known that Riccati equations can be transformed into second order linear equations such as the Gauss hypergeometric equation. Hence, this classification shows that once the associated monodromy representation is not completely reducible, then solution λt can be expressed in terms of previously known functions, i.e., it does not define new transcendental functions.

17 GREEN FUNCTION AND PAINLEVÉ VI EQUATION 7 Now we discuss the solutions of which the associated monodromy representation is not completely reducible, and the results in this section will be used to prove Theorem. in 5. For GLE.4 in E τ with p E τ [], we proved in [6] that there is always a solution which is expressed by: 3.6 y a z := exp zζa + p + ζa p σ z a σ z + p σ z p, where the pair ±a E τ is uniquely determined by 3.7 A = [ζ p + a + ζ p a ζ p]. If a a mod Λ τ, then y a z and y a z are linearly independent solutions to.4. In this case, the monodromy representation associated to.4 is completely reducible. In fact, we proved in [6, Lemma.3] that the monodromy representation for.4 is not completely reducible if and only if a E τ []. Now we assume a = ω k E τ []. Let y z = y a z and y z = χzy z be a linearly independent solution of.4 to y z. Clearly it is equivalent to χz const and 3.8 χ z χ z + y z y z = 0, i.e., χ z = const y z. On the other hand, by using ζz ζz = 3.9 A = p ω k 4 p ω k. When a = 0, we have y z = σ z + p σ z p σz z, 3.7 is equivalent to z = c z p, and then 3.8 yields χz = cζz + pz. So for any cτ = 0, cτy, y is a fundamental system of solutions to GLE.4, where y z = ζz + pzy z. In particular, cτy 3.0 l j cτy = y 0 η j + pω j cτ y. Proposition 3.. The solutions of the Riccati equation 3. can be parametrized by C CP : 3. λ C t = p Cτ τ e τ e τ e τ Proof. We separate the proof into two steps., p C τ τ = η τ Cη τ. C τ Step. We prove that for any constant C CP, λ C t given by 3. solves the Riccati equation 3..

18 8 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG Fix any C CP and let pτ = p C τ, Aτ = pτ 4 in the generalized Lamé equation.4. pτ If C =, then pτ = η τ. Choose cτ = η τ + pττ. By the Legendre relation τη τ η τ = πi, cτ = πi. Thus by 3.0, we have l cτy = y 0 cτy, l 0 y cτy = y 0 cτy. y That is, GLE.4 is monodromy preserving as τ deforms, so p τ is a solution of the elliptic form.7 see Subsection.. If C =, then 3. gives η τ + pτ 0 and C = η τ+ pττ η τ+ pτ. Choose cτ = η τ + pτ. Clearly except a set of discrete points in H, cτ = 0 and so 3. l cτy = y 0 cτy, l y cτy = y 0 cτy. C y As before, we conclude that p C τ is a solution of the elliptic form.7. We remark that the second formula in 3. was previously obtained in [5, 33], where there does not contain the relation between λ C t and Riccati equations. Here, together with our result in [6], we conclude that λ C t actually satisfies the Ricatti equation 3.. Step. Let λt be any solution of the Riccati equation 3.. We prove the existence of C CP such that λt = λ C t. Define ±pτ by λt via 3. and Aτ = pτ 4. Then pτ is a pτ solution of the elliptic form.7, which implies that.4 is monodromy preserving as τ deforms. Therefore, there exists a fundamental system of solutions ỹ z; τ, ỹ z; τ to.4 such that the monodromy matrices M, M, which are defined by l j ỹ ỹ = M ỹ j, j =,, ỹ are independent of τ. We may assume pτ τ p τ τ, otherwise we are done. Then cτ := η τ + pτ 0. For any τ such that cτ = 0, cτy, y is also a fundamental system of solutions, so there is a b ỹ cτy an invertible matrix γ = such that = γ. Clearly c d ỹ y the monodromy matrices of cτy, y is given by 3., where 3.3 C := η τ + pτ ττ η τ + pτ τ may depend on τ. Then 0 M = γ γ = + bd ad bc b ad bc d ad bc bd ad bc,

19 M = γ GREEN FUNCTION AND PAINLEVÉ VI EQUATION 9 0 γ C = + bd ad bc C b ad bc C d ad bc C bd ad bc C Since M, M are independent of τ and bd 0, we conclude that C is a constant independent of τ. Consequently, 3.3 implies pτ τ = p C τ τ and so λt = λ C t. The proof is complete. Remark 3.. It is easy to see that if Im C > 0, then λ C t has singularities at least a pole in C\{0, }. However, it is not so obvious to see whether λ C t has singularities or not if Im C 0. In 6, we will exploit formulae 3. and 3.8 below to prove that any solution of the four Riccati equations has singularities in C\{0, }. Another observation is that C = gives that 3.4 λ t = η τ + e τ e τ e τ is a solution of PVI 8, 8, 8, 3 8. Since λ t, λ t and λ t t can have only simple zeros cf. [8, Proposition.4.], a direct consequence is Theorem 3.3. For fixed k {,, 3}, the followings hold: i Any zero of η τ + e k τ must be simple. ii 3.5 iii η τ+τe k τ η τ+e k τ Proof. Recall d dτ η τ + e k τ = for any τ H. πi is a locally one-to-one map from H to C { }. t = tτ = e 3τ e τ e τ e τ. Since t τ = 0 for all τ H, the assertion i follows readily from the fact that λ t for k =, λ t for k = and λ t t for k = 3 can have only simple zeros. For the assertion ii, we note from the Legendre relation and 3. that λ C t = η τ + e τ πi τ C. e τ e τ Fix any τ 0 H. If τ 0 is a zero of η + e, then d dτ η + e τ=τ0 =. So it suffices to consider the case η τ 0 + e τ 0 = 0. Then by letting C = τ 0 πi η τ 0 + e τ 0, we see that t 0 = tτ 0 is a zero of λ C t. Since λ C t has only simple zeros, we have d η + e πi τ=τ0 = 0. dτ τ C.

20 0 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG This, together with η τ 0 + e τ 0 πi d τ 0 C = 0, easily implies dτ η + e τ=τ0 = πi. This proves 3.5 for k =. Similarly, by considering λ C t and λ C t t, we can prove 3.5 for k =, 3. This proves the assertion ii. Finally, using the Legendre relation leads to Therefore, η τ+τe k τ η τ+e k τ η τ + τe k τ η τ + e k τ = τ πi η τ + e k τ. is locally one-to-one. This completes the proof. Remark 3.4. In 6, we will see that the Hessian of the Green function Gz τ at z = ω = : det D G τ = Cτ Im η τ + τe τ η τ + e τ for some Cτ > 0, provided that η τ + e τ = 0. The local one-to-one of the map η τ+τe τ is important for studying the curve in H where the η τ+e τ half-period ω is a degenerate critical point of Gz τ. See 6. Furthermore, we will prove a stronger result that η τ + e τ has only one zero in any fundamental domain of Γ; see Theorem 6.6. Similarly, we can prove that all solutions of the other three Riccati equations can be parametrized by CP. The calculation is as follows. Fix k {,, 3}. When a = ω k, by 3.6 it is easy to see that p e χz = k e k e i e k e j ζz ω k p e + e k k z e k e i e k e j satisfies 3.8, where {i, j} = {,, 3}\{k}. As before, for any cτ = 0, cτy z, y z is a fundamental system of solutions to.4, where y z = χzy z. In particular, cτy 3.6 l cτy 0 j = y Dη j+ω j +De k, y cτ where 3.7 D := p e k e k e i e k e j. Proposition 3.5. For C CP, we let λ C t = p Cτ τ e τ, where e τ e τ 3.8 p C τ τ = e kcη τ η τ + g 4 e k C τ. Cη τ η τ + e k C τ Then λ C t satisfies the Ricatti equation 3.3 if k =, 3.4 if k =, 3.5 if k = 3. Furthermore, such λ C t give all the solutions of these three Riccati equations.

21 GREEN FUNCTION AND PAINLEVÉ VI EQUATION Proof. We sketch the proof for fixed k {,, 3}. For any C CP, we let pτ ω k pτ ω k pτ = p C τ, Aτ = 4 in.4. If C =, i.e., Dη + + πi De k 0, then we choose cτ = Dη + τ + De k = η 0. By τ+e k τ 3.6, l cτy 0 cτy =, l y 0 y cτy 0 cτy =. y y If C =, then 3.8 gives Dη + + De k 0 and C = Dη +τ+de k Dη ++De k. Choose cτ = Dη + + De k. Then l cτy 0 cτy =, l y y cτy 0 cτy =. y C y Similarly as in Proposition 3., we see that p C τ is a solution of the elliptic form.7. Again the formula in 3.8 was first obtained in [33]. Here, together with our result in [6], we conclude that λ C t actually satisfies the Ricatti equation 3.3 if k =, 3.4 if k =, 3.5 if k = 3. The rest of the proof is similar to that of Proposition 3.. For solution p C τ of the Riccati equations given in Propositions 3. and 3.5, we let τ = γ τ and C = γ C for γ SL, Z. By using see 4 and the formula of p C τ τ, it is easy to prove 3.9 p C τ τ = cτ + d p C τ τ. Then we have the following result, which can be proved by the same argument of Proposition 4.4 in 4, so we omit the details of the proof here. Proposition 3.6. Let λ C t and λ C t solve the same one of the four Riccati equations Then they give the same solution to PVI 8, 8, 8, 3 8 if and only if C = γ C for some γ Γ. We conclude this section by a remark. In [6], Mazzocco classified solutions of PVI µ /,0,0, write PVI µ for convenience for µ + Z. Notice that PVI is precisely PVI 0,0,0, and PVI µ can be transformed to PVI via Bäcklund transformations. Mazzocco proved for µ + Z and µ =, say µ = for instance, PVI has two types of solutions: one is socalled Picard type solutions, which is obtained from Picard solutions.6 via Bäcklund transformations; the other one is so-called Chazy solutions, such as λt = 8 {[ω + νω + tω + νω ] 4tω + νω } ω + νω ω + νω [t ω + νω + ω + νω ][ω + νω + xω + νω ] where ν C, which will turn to be the singular solutions λ 0 t 0,, t or of PVI via Bäcklund transformations. Here together with Theorem C and our argument in, in principle, solutions of the four Riccati equations

22 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG could be obtained from Chazy solutions of PVI via Bäcklund transformations, but the process would be too complicated to be computed. 4. COMPLETELY REDUCIBLE SOLUTIONS 4.. Simple zeros of Hecke form. By Theorem C in 3, any solution λt of PVI 8, 8, 8, 3 8 with a completely reducible monodromy representation can be expressed by 3.: λt = pτ τ e τ, t = e 3 τ e τ e τ e τ e τ e τ, where pτ τ is given by.8 with some r, s C \ Z. From.8, we have the following application of the Painlevé property. Theorem 4.. Suppose r, s C \ Z is a pair of complex constants. Then the Hecke form Z r,s τ = ζr + sτ τ rη τ + sη τ has only simple zeros. Proof. First, we note that the situations r + sτ E τ [] and Z r,s τ = 0 can not occur simultaneously. If not, then there are τ 0 and m, n Z such that r + sτ 0 = m + nτ 0 + ω, where ω is any lattice points {0, ω, ω, ω 3 = ω + ω }, and also ζr + sτ 0 = rη τ 0 + sη τ 0. Without loss of generality, we might assume ω = ω. The other cases can be proved similarly. The second identity also implies Therefore, we have η τ 0 = ζ ω = ζ r m + s nτ 0 = ζ r + sτ 0 mη τ 0 nη τ 0 = r mη τ 0 + s nη τ 0. r m + s nτ 0 = 0, r m η τ 0 + s nη τ 0 = 0, which implies r m τ = 0 and s = n because the matrix η τ η τ is non-degenerate for any τ due to the Legendre relation. Obviously it contradicts to the assumption r, s Z. Now suppose Z r,s τ 0 = 0, which implies pτ 0 = by.8 because r + sτ 0 = 0. Consider the transformation τ 0 t 0 via 3.. Then by the Painlevé property, we know that λt has a pole at t = t 0 {0,, }. By substituting the local expansion of λt at t = t 0 into., it is easy to prove that the order of pole at t = t 0 is, which implies the zero of Z r,s at τ = τ 0 is simple. Remark 4.. If r, s is an N-torison point, i.e., r, s = k N, k N for positive integers k i, N 3 and gcdk, k, N =, then the function Z r,s τ is a modular form of weight with respect to the modular group ΓN. In this

23 GREEN FUNCTION AND PAINLEVÉ VI EQUATION 3 case, Theorem 4. was proved in [7], where the method of dessins d enfants was used. For a real pair of r, s, we will give an alternative proof in 5. Since α i = 8 for 0 i 3, it is easy to see that for k 3, pτ + ω k is also a solution of the elliptic form.7 provided that pτ is a solution of.7. Then we have the following result, which will be used in 5. Proposition 4.3. Given r, s C \ Z, we define r, s if k =, 4. r k, s k = r, s if k =, r, s if k = 3. Then p r,s τ + ω k = ±p r k,s k τ in E τ. Proof. It was proved in [6] that.8 is equivalent to ζr + sτ + p r,s τ + ζr + sτ p r,s τ rη τ + sη τ = 0. Form here, we easily obtain and so ζr k + s k τ + p r,s τ + ω k + ζr k + s k τ p r,s τ + ω k r k η τ + s k η τ = 0, p r,s τ + ω k τ = r k + s k τ τ + = p rk,s k τ τ. This completes the proof. r k + s k τ τ ζ r k + s k τ τ r k η τ + s k η τ We call a solution to.5 a real solution if the monodromy group of its associated GLE. is contained in SU. For the case α j = 8, pτ is a real solution if and only if it is given by.8 for some real pair r, s R \ Z. 4.. Modularity. In this subsection we study the modularity property of solutions to PVI 8, 8, 8, 3 8. Consider the pair z, τ C H and z = r + sτ. a b For any γ = SL, Z, conventionally γ can act on C H by c d z γz, τ := cτ+d, γ τ = z cτ+d, aτ+b cτ+d. Then z cτ + d = r + sτ cτ + d = r + s τ, where τ = γ τ and s, r = s, r γ. Using z 4. cτ + d τ = cτ + d z τ, τ = aτ + b cτ + d,

24 4 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG we derive and so 4.3 z ζ cτ + d τ = cτ + d ζ z τ, η τ η τ = cτ + dγ η τ. η τ Set r, s η τ, η τ T = rη τ + sη τ. Then r, s η τ, η τ T = cτ + dr, s η τ, η τ T and so 4.4 Z r,s τ = cτ + dz r,s τ. Together 4. and 4.4, we obtain 4.5 p r,s τ τ = cτ + d p r,s τ τ = pr,s τ cτ+d τ, where r + s τ, τ = γr + sτ, τ. Indeed, by a direct calculation, we could prove that p r,sτ cτ+d as a function of τ is a solution of the elliptic form.7 since p r,s τ is a solution of.7. Particularly, p r,sτ cτ+d = ±p r,s τ mod Λ τ. Recall that λ r,s t is the corresponding solution of., namely 4.6 λ r,s t = p r,sτ τ e τ. e τ e τ Then the above argument yields the following result. Proposition 4.4. λ r,s t and λ r,s t belong to the same solution of PVI 8, 8, 8, 3 8 if and only if s, r s, r γ mod Z by some γ Γ. Proof. For the sufficient part, assume s, r s, r γ mod Z by some γ Γ. Recall from [6, Lemma 4.] that 4.7 p r,s τ τ = p r, s τ τ r, s ± r, s mod Z, which implies that all elements in ±r, s + Z give precisely the same solution λ r,s t. Hence we may assume s, r = s, r γ by replacing s, r with some element in s, r + Z if necessary. Let l 0 H be a path starting from any fixed point τ 0 to τ 0 = γ τ 0. Then l := tl 0 π CP \{0,, }, t 0, where t τ = e 3τ e τ e τ e τ and t 0 = tτ 0. Let U H be a small neighborhood of τ 0 and denote V = tu. Since λ r,s t = p r,s τ τ e τ, τ U, e τ e τ so the analytic continuation l λ r,s t of λ r,s t along l satisfies l λ r,s t = p r,s γ τ γ τ e γ τ, τ U. e γ τ e γ τ

25 GREEN FUNCTION AND PAINLEVÉ VI EQUATION 5 On the other hand, s, r = s, r γ gives r + s τ, τ = γr + sτ, τ, where τ a b = γ τ. Moreover, γ = Γ gives c d 4.8 e j γ τ = cτ + d e j τ, j =,, 3. Then it follows from 4.5 and 4.8 that 4.9 namely λ r,s tγ τ = p r,s γ τ γ τ e γ τ e γ τ e γ τ = p r,sτ τ e τ = λ r,s tτ, τ U, e τ e τ 4.0 λ r,s t = l λ r,s t, t V. Conversely, assume that λ r,s t and λ r,s t represent different branches of the same solution in a small neighborhood V of t 0 CP \{0,, }. Then there is l π CP \{0,, }, t 0 such that 4.0 holds. Fix any τ 0 H such that t 0 = t τ 0 and let t l H denote the lifting path of l under the map t τ = e 3τ e τ e τ e τ such that its starting point is τ 0. Denote its ending point by τ 0. Then tτ 0 = t 0 = tτ 0, which implies τ 0 = γ τ 0 a b for some γ = Γ. Let U be a neighborhood of τ c d 0 such that tu V. Then 4.6 and 4.0 give 4.9. Define s, r := s, r γ, then r + s τ, τ = γ r + sτ, τ, where τ = γ τ, and so 4.5 gives 4. p r,s γ τ γ τ = cτ + d p r, s τ τ. Substituting 4. and 4.8 into 4.9 leads to p r,s τ τ = p r, s τ τ, τ U. Again by 4.7 we obtain r, s ± r, s mod Z, namely s, r s, r ±γ mod Z where ±γ Γ. Define for any N-torsion point r, s = k N, k N Q N, Γ r,s := { γ SL, Z s, r γ ±s, r mod Z }. Then p r,s τ τ is a modular form of weight with respect to Γ r,s in the sense p r,s τ τ = cτ + d p r,s τ τ, γ Γ r,s. For example, if r = 0, then { } a b Γ r,s = γ = SL, Z c d b 0, a ± mod N.

26 6 ZHIJIE CHEN, TING-JUNG KUO, CHANG-SHOU LIN, AND CHIN-LUNG WANG 5. GEOMETRY OF Ω 5 In this and the next sections, our main purpose is to prove Theorems.-.4. In these two sections, we mainly consider τ F 0, where F 0 H is the fundamental domain for Γ 0 defined by 5. F 0 := {τ H 0 Re τ, τ }. Remark 5.. Recall that a b Γ 0 N = {γ = SL, Z c 0 c d mod N}. It is well known that the modular curve X 0 N = H/Γ 0 N parametrizes the pair E, C of an elliptic curve E together with a cyclic subgroup C E with C = N. For N = p being a prime, [SL, Z : Γ 0 p] = p + and a fundamental domain for Γ 0 p is given by F = F SF STF ST p F, 0 where S =, T = and F is any fundamental domain for 0 0 SL, Z. For N = p =, X 0 parametrizes E, q with q a half period. An alternative choice of fundamental domain is F 0 = F TSF TS F notice that TS 3 = Id and TS fixes ρ = e πi/3. Recall the Hecke form Z r,s τ := ζr + sτ τ rη τ + sη τ, which is doubly periodic in r, s R. It is related to the Green function on E τ via Z r,s τ = 4π z Gr + sτ τ. Recall also the q expansion for ζ with q := e πiτ : 5. ζz τ = η τz πi + eπiz e πiz πi n= e πiz q n e πiz q n e πiz q n e πiz q n This can be deduced form the Jacobi triple product formula for theta function ϑ and the relation between ϑ and σ, see e.g. [36]. We use the Legendre relation η τ η = πi and the above q expansion to compute the q expansion for Z: 5.3 Z r,s τ = πis πi + eπiz e πiz πi n= where z = r + sτ. See also [9, 4]. e πiz q n e πiz q n e πiz q n e πiz q n.,