GREEN FUNCTION, PAINLEV ´E 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
λ(t)to the Painlev´e VI equation with parameter(18,−18 ,18,38) might have in C\ {0, 1}? Here t0∈C\ {0, 1}is called a singular point of λ(t)if λ(t0)∈
{0, 1, t0,∞}. 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, 1}. (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) inC\ {0, 1}. (iii) Any Riccati solution has singular points inC\ {0, 1}. (iv) For each N≥5 and N6=6, we calculate the number of the real j-values of zeros of the Eisenstein series E1N(τ; k1, k2)of weight
one, where(k1, k2)runs over[0, N−1]2with gcd(k1, k2, N) =1.
The geometry of the critical points of the Green function on a flat torus Eτ, as τ varies in the moduliM1, plays a fundamental role in our analysis
of the Painlev´e IV equation. In particular, the conjectures raised in [22] on the shape of the domainΩ5 ⊂ M1, which consists of tori whose Green
function has extra pair of critical points, are completely solved here.
CONTENTS
1. Introduction 2
2. Painlev´e VI: Overviews 13
3. Riccati solutions 16
4. Completely reducible solutions 22
5. Geometry ofΩ5 26
6. Geometry of ∂Ω5 31
7. Algebraic solutions 41
8. Further discussion 48
Appendix A. Picard solution and Hitchin’s solution 50
Appendix B. Asymptotics of real solutions at{0, 1,∞} 51
References 56
1. INTRODUCTION
1.1. Painlev´e property. In the literature, a nonlinear differential equation in one complex variable is said to possess the Painlev´e property if its solu-tions have neither movable branch points nor movable essential singulari-ties. For the class of second order differential equations
(1.1) λ00(t) =F(t, λ, λ0), t∈CP1,
where F(t, λ, λ0)is meromorphic in t and rational in both λ and λ0, Painlev´e (later completed by Gambier, [11, 29]) obtained the classification of those nonlinear ODEs which possess the Painlev´e property. They showed that there were fifty canonical equations of the form (1.1) with this property, up to M ¨obius transformations. Furthermore, of these fifty equations, forty-four 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. [17]). These six nonlinear ODEs are called Painlev´e equations. Among them, Painlev´e VI is often considered to be the master equation, because oth-ers can be obtained from Painlev´e VI by the confluence. Due to its connec-tion with many different disciplines in mathematics and physics, Painlev´e VI has been extensively studied in the past several decades. See [1, 3, 8, 10, 12, 13, 15, 20, 24, 25, 27, 28, 35] and the references therein.
Painlev´e VI (PVI) is written as d2λ dt2 = 1 2 1 λ+ 1 λ−1+ 1 λ−t dλ dt 2 − 1 t + 1 t−1+ 1 λ−t dλ dt (1.2) + λ(λ−1) (λ−t) t2(t−1)2 " α+β t λ2 +γ t−1 (λ−1)2 +δ t(t−1) (λ−t)2 # , where α, β, γ, δ are four complex constants. From (1.2), the Painlev´e
prop-erty says that any solution λ(t)is a multi-valued meromorphic function in
C\{0, 1}. To avoid the multi-valueness of λ(t), it is better to lift solutions of (1.2) to its universal covering. It is known that the universal covering
of C\{0, 1}is the upper half plane H = {τ | Im τ > 0}. Then t and the
solution λ(t)can be lifted through the covering map τ 7→t by
(1.3) t(τ) = e3(τ)−e1(τ)
e2(τ)−e1(τ)
, λ(t) = ℘(p(τ)|τ)−e1(τ)
e2(τ)−e1(τ) ,
where℘(z) = ℘(z|τ)is the Weierstrass elliptic function defined by
(1.4) ℘(z|τ):= 1 z2 +
∑
ω∈Λτ\{0} 1 (z−ω)2 − 1 ω2 ,and Λτ := {m+nτ|m, n ∈ Z}is the lattice generated by ω1 = 1 and
p(τ)satisfies the following elliptic form of PVI: (1.5) d 2p(τ) dτ2 = − 1 4π2 3
∑
i=0 αi℘0 p(τ) +ω2i|τ , where ω0=0 and (1.6) (α0, α1, α2, α3) = α,−β, γ,12 −δ .This elliptic form was first discovered by Painlev´e [30]. For more recent derivations of it, see [1, 25].
1.2. Hitchin solutions. In this paper, we consider the special case αi = 18
for 0≤i≤3, i.e., (1.7) d 2p(τ) dτ2 = − 1 32π2 3
∑
i=0 ℘0 p(τ) + ωi 2|τ ,which is the elliptic form of PVI(1
8,−81,18,38). Equation (1.7) has connections with some geometric problems. The well-known example is related to the construction of Einstein metrics in four dimension; see [15]. In the seminal
work [15], Hitchin obtained his famous formula to express a solution p(τ)
of (1.7) with some complex parameters r, s:
(1.8) ℘(p(τ)|τ) = ℘ (r+sτ|τ) + ℘0(r+sτ|τ)
2(ζ(r+sτ|τ)− (rη1(τ) +sη2(τ))) . Here ηi(τ) = 2ζ(ω2i|τ), i = 1, 2, are quasi-periods of the Weierstrass zeta function ζ(z|τ) =−Rz
℘(ξ|τ)dξ.
By (1.8), 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 (1.7) is given by another formula other than
(1.8). Indeed, this corresponds to the Riccati solutions of (1.7); see§3.
For simplicity, we denote pr,s(τ)(equivalently, λr,s(t)via (1.3)) to be the solution of (1.7) with the expression (1.8). It is obvious that if
(r, s)∈ 1 2Z2:= (0, 0),(0,12),(21, 0),(12,12) +Z2, then either ζ(r+sτ|τ)− (rη1(τ) +sη2(τ))≡∞ or ζ(r+sτ|τ)− (rη1(τ) + sη2(τ))≡0 inH. Hence for any complex pair(r, s)6∈ 12Z2, pr,s(τ)is always
a solution to (1.7), or equivalently, λr,s(t) is a (multi-valued) solution to
PVI(1
8,−81,18,38). We say that two solutions λr,s(t)andλr0,s0(t)give (or belong to) the same solution if λr0,s0(t) is the analytic continuation of λr,s(t) along some closed loop inC\ {0, 1}. In§4, we will prove that λr,sand λr0,s0give the same solution to PVI(1
8,−81,18,38) if and only if (s
0, r0) ≡ (s, r)·γ modZ2 for some matrix
γ∈ Γ(2) ={γ∈SL(2,Z)|γ≡ I2 mod 2}.
In this paper, we are mainly concerned with the question of smoothness of solutions to PVI(1
that for a solution λ(t) and a point t0 ∈ C\{0, 1}, the RHS of Painlev´e VI (1.2) has a singularity at (t0, λ(t0))provided that λ(t0) ∈ {0, 1, t0,∞}. Therefore, in this paper, we say λ(t)is smooth at t0if λ(t0) 6∈ {0, 1, t0,∞}. Furthermore, a singular point t0∈ C\{0, 1}is called of type 0 (1, 2, 3 respec-tively) ifλ(t0) =∞ (λ(t0) =0, 1, t0respectively).
We take PVI(0,0,0,1
2) as an initial example for our discussion, because it
can be transformed to PVI(1
8,−81,18,38)of our concern by a B¨acklund transfor-mation (cf. [28]). In the literature, the B¨acklund transfortransfor-mation plays a very useful role in the study of Painlev´e VI; for example, for finding the algebraic
solutions, see [10, 26, 24]. Conventionally, solutions of PVI(0,0,0,1
2), the so-called Picard solutions, can be expressed in terms of Gauss hypergeometric functions. It was first found by Picard [31]. Let
(1.9) ω1(t) =−iπF(12,12, 1; 1−t), ω2(t) =πF(12,12, 1; t)
be two linearly independent solutions of the Gauss hypergeometric equa-tion
(1.10) t(1−t)ω00(t) + (1−2t)ω0(t)−1
4ω(t) =0. Then Picard solution of PVI(0,0,0,1
2)can be expressed as
(1.11) ˆλν1,ν2(t) = ℘(ν1ω1(t) +ν2ω2(t)|ω1(t), ω2(t)) +
1+t
3 ,
for some(ν1, ν2) 6∈ 12Z2, where℘(·|ω1(t), ω2(t))is the Weierstrass elliptic
function with periods ω1(t) and ω2(t). See [12] for a proof. Obviously,
ˆλν1,ν2(t)is smooth for all t∈ C\{0, 1}if and only if(ν1, ν2)∈ R
2\1
2Z2(see (A.1)). The lifting of ˆλν1,ν2(t)by (1.3) is given by
(1.12) ˆpν1,ν2(τ) =ν1+ν2τ,
which of course is a solution of the elliptic form of PVI(0,0,0,1
2):
d2p(τ)
dτ2 = 0. Then the B¨acklund transformation takes ˆλν1,ν2(t) into solution λr,s(t) of PVI(1
8,−81,18,38) with (r, s) = (ν1, ν2). Thus, in the elliptic form the B¨acklund transformation seems comparably simple. This fact and (1.12) 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 B¨acklund transformation from PVI(0,0,0,1 2) to PVI(1
8,−81,18,38), PVI(18,−81,18,38) has only three solutions which are smooth in
C\{0, 1}.
Theorem 1.1. There are only three solutionsλ(t)to PVI(1
8,−81,18,38)such thatλ(t) is smooth for all t∈C\{0, 1}. They are preciselyλ1
4,0(t),λ0,14(t)andλ14,14(t). Theorem 1.1 shows that the B¨acklund transformation does not preserve the smoothness of solutions. Thus, Theorem 1.1 can not be proved by ap-plying Picard solutions and the B¨acklund transformation. We remark that
the B¨acklund transformation is complicated due to not only the compli-cated form of birational maps between solutions but also the fact that it transforms a pair of solutions of the Hamiltonian system (equivalently, the pair
(λ(t), λ0(t))), but not the solution λ(t)only.
To prove Theorem 1.1, we start from the formula (1.8). Of course, (1.8) does not give the complete set of solutions to (1.7). The missing ones are solutions obtained from Riccati equations. For such Riccati solutions, we have some expressions like (1.8). By employing these expressions, we will
prove in§6 that any Riccati solution has singularities in C\{0, 1}. Hence
our strategy for the proof of Theorem 1.1 is to study the smoothness of
λr,s(t)for any complex pair(r, s)6∈ 12Z2.
From (1.8), it is easy to see that if (r, s)is not a real pair, then λr,s(t)
al-ways possesses a singularity t0 6∈ {0, 1,∞}(indeed, infinitely many
singu-larities), because there always exist infinitely many τ0 ∈H such that r+sτ0
is a lattice point of the torus Eτ0 := C/Λτ0. So for the proof of Theorem 1.1
we could restrict ourselves to consider only(r, s) ∈ R2\1
2Z2. In this case, we introduce the Green function and the Hecke form to study it.
1.3. Green function and Hecke form. Let G(z|τ)be the Green function on
the torus Eτ: (1.13) ( −∆G(z|τ) =δ0(z)−|E1τ| in Eτ, R EτG(z|τ)dz=0,
where δ0 is the Dirac measure at 0 and|Eτ|is the area of the torus Eτ. We
recall the analytic description of G(z|τ)in [22]. Recall the theta function
ϑ :=ϑ1, where ϑ1(z; τ) =−i ∞
∑
n=−∞ (−1)ne(n+1 2)2πiτe(2n+1)πiz.Then the Green function is given by
(1.14) G(z|τ) =− 1
2πlog|ϑ(z; τ)| +
(Im z)2
2 Im τ +C(τ),
where C(τ)is a constant so that RE
τG = 0. Recall that ηi(τ) = 2ζ(
ωi
2 |τ), i = 1, 2, are quasi-periods of ζ(z|τ). Using (log ϑ)z = ζ(z)−η1z and the
Legendre relation η1ω2−η2ω1=2πi, we have
(1.15) −4πGz(z|τ) =ζ(z|τ)−rη1(τ)−sη2(τ),
where z = r+sτ with r, s ∈ R. As mentioned before, ζ(r+sτ|τ)−
rη1(τ)−sη2(τ) ≡ 0 in H whenever(r, s) ∈ 12Z2\Z2. Thus for (r, s) ∈
R2\1
2Z2, (1.15) shows that r+sτ is a non half-period critical point of G(z|τ)
(we call such critical point a nontrivial critical point) if and only if ζ(r+
sτ|τ)−rη1(τ)−sη2(τ) = 0. Naturally, we ask the question: How many
points must appear in pair because G(z|τ) is an even function in z. This question was answered in the following surprising result:
Theorem A. [22] For any torus Eτ, G(z|τ)has at most one pair of nontrivial
critical points.
Theorem B.[23] Suppose that G(z|τ)has one pair of nontrivial critical points. Then the three half-periods are all saddle points of G(z|τ), i.e., the Hessian satisfies det D2G(ωk 2 |τ)≤0 for k=1, 2, 3. 1 For any(r, s)∈R2\1 2Z2, we define Z =Zr,sby (1.16) Zr,s(τ):= ζ(r+sτ|τ)−rη1(τ)−sη2(τ), ∀τ∈H.
Clearly Zr,sis a holomorphic function inH. If(r, s)is an N-torsion point,
i.e., (r, s) = (k1
N,kN2) with 0 ≤ k1, k2 < N and gcd(k1, k2, N) = 1, it was
proved by Hecke in [14] that Zr,s(τ)is a modular form of weight 1 with
respect toΓ(N) ={A∈SL(2,Z)| A≡ I2 (mod N)}. This modular form
is called the Hecke form in [21]. Indeed, it is the Eisenstein series of weight 1 with characteric(r, s)if(r, s)is an N-torsion point. Following [32, p.59], the Eisenstein series of weight 1 is defined by
EN1(τ, s; k1, k2):= (Im τ)s
∑
(m,n)(mτ+n)−1|mτ+n|−2s,
where(m, n)runs overZ2under the condition 06= (m, n)≡ (k1, k2)mod N.
It is known that E1N(τ, s; k1, k2) is a meromorphic function in the s-plane
and holomorphic at s=0. Set EN1(τ; k1, k2):=E1N(τ, 0; k1, k2). By using the Fourier expansions of both Zr,s(τ)and E1N(τ; k1, k2)(see [32, p.59] and [9, p.139]), we have
(1.17) Zr,s(τ) =NE1N(τ; k1, k2), if(r, s)≡ (kN1,kN2) mod 1.
Hence, (1.15) yields that G(z|τ0)has a critical N-torsion point kN1 +kN2τ0with N≥3 if and only if EN1(τ0; k1, k2) =0.
Now we see the connection of the Hecke form with the solution pr,s(τ)
(or λr,s(t)) of (1.7): Zr,s(τ)appears in the denominator of the RHS of (1.8). When(r, s)∈R2\1
2Z2, the formula (1.8) implies that t0is a type 0 singular-ity, i.e., λr,s(t0) = ∞ if and only if Zr,s(τ0) = 0, t0 = t(τ0), or equivalently, the Green function G(z|τ0)has a nontrivial critical point r+sτ0. By Theorem A, it means that G(z|τ0)have exactly five critical points in the torus Eτ0:
ω1
2 ,
ω2
2 ,
ω3
2 and±(r+sτ0). This connection and (1.8) together with the Painlev´e
property say that the Eisenstein series E1N(τ; k1, k2) of weight 1 has only
simple zeros; see Theorem 4.1. The simplicity of zeros was also proved by Dahmen [7] as a consequence of his counting formula of algebraic integral 1Theorem B is used in the proof of Theorem 1.2 (ii) to be stated later. After establishing
Theorem 1.2, we have a stronger version of Theorem B: det D2G(ωk
2 |τ) <0 for k=1, 2, 3 if
Lam´e equations by the method of dessins d’enfants. In§7 we will discuss the position and the number of those zeros of E1N(τ0; k1, k2).
Recall the group action of SL(2,Z)on the upper half planeH:
τ0 =γ·τ= aτ+b cτ+d, γ= a b c d ∈ SL(2,Z).
Then we have the transformation law (see (4.4) in§4):
(1.18) Zr0,s0(τ0) = (cτ+d)Zr,s(τ) where(s0, r0) = (s, r)·γ−1.
From here, we see that G(z|τ0)has five critical points whenever G(z|τ)has
five critical points. LetM1:=H/SL(2,Z)and
Ω5 :={τ∈ M1|G(z|τ)has five critical points}, Ω3 :={τ∈ M1|G(z|τ)has three critical points}.
Then we haveΩ3∪Ω5 =M1by Theorem A. Moreover, from the proof of
Theorem A in [22], we know thatΩ5 ⊂ M1 is open andΩ3 is closed. In
this paper we determine the geometry ofΩ3andΩ5as conjectured in [22]:
Theorem 1.2(Geometry ofΩ3andΩ5).
(i) BothΩ5and ¯Ω3 =Ω3∪ {∞}are simply connected inM1 ∼=S2. (ii) C=∂Ω5 =∂ ¯Ω3 ∼=S1. C\{∞} ∼=R is smooth. It consists of points τ so
that some half-period is a degenerate critical point of G(z|τ).
The proof is given in§5 and §6. We actually prove a stronger result on
∂Ω5: For any τ ∈∂Ω5, there is only one half period whose Hessian det D2G
vanishes.
Theorem 1.1 is clearly closely related to the following question: What
is the set of pairs (r, s) such that Zr,s(τ) has no zeros? We should write an
alternative form of (i) in Theorem 1.2 to answer this question. We note that the following two statements hold:
(1.19) Zr,s(τ) =±Zr0,s0(τ)⇐⇒ (r, s)≡ ±(r0, s0) (mod Z2), (1.20) λr,s(τ) =λr0,s0(τ)⇐⇒ (r, s)≡ ±(r0, s0) (mod Z2).
The statement (1.19) is trivial while (1.20) was proved in [6]. From both (1.19) and (1.20), we could assume (r, s) ∈ [0, 1]× [0,12]\1
2Z2. Then (i) of Theorem 1.2 can be stated more precisely. For this purpose, we consider
(1.21) F0={τ∈H|0≤Re τ ≤1, |τ− 12| ≥ 12}.
It is elementary to prove that F0is a fundamental domain forΓ0(2)(c.f.
Re-mark 5.1). Notice that F0 is one half of the fundamental domain of Γ(2).
The following theorem will imply (i) of Theorem 1.2.
Theorem 1.3. Let(r, s)∈ [0, 1]× [0,12]\1
2Z2. Then Zr,s(τ) = 0 has a solution τ∈ F0if and only if
FIGURE1. The lifted domain ˜Ω5 ⊂ F0 of Ω5 is the domain bounded by the 3 curves corresponding to the loci of degen-erate critical points.
Moreover, the solutionτ∈ F0is unique for any(r, s)∈ 40.
We will see that Theorem 1.1 is a consequence of the non-existence part
of Theorem 1.3 in§5. Indeed, the existence part of Theorem 1.3 has
appli-cations as well; see the next subsection, where we will discuss the singular points of a real solution λ(t).
1.4. Real solution. It is well known that Painlev´e VI governs the isomon-odromic deformations of some linear ODE. In the elliptic form it is conve-nient to choose the ODE to be a generalized Lam´e equation (c.f. (2.4)). A solution λ(t)of PVI(1
8,−81,18,38) is called a real solution if its associated mon-odromy of the generalized Lam´e equation is unitary. In [6] it was proved that a solution λ(t)is a real solution if and only if λ(t) = λr,s(t)for some
(r, s) ∈ R2\1
2Z2. We call such a solution of PVI(1
8,−81,18,38) real because any solution with unitary monodromy representation must come from blowup
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 (1.7) 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, 1}(see [6]);
(iii) any real solution has no essential singularity even at 0, 1 and ∞ (see
Appendix B).
It is known (see §2) that t(τ) = e3(τ)−e1(τ)
e2(τ)−e1(τ) maps any fundamental
do-main ofΓ(2)one-to-one and ontoC\ {0, 1}. Then by the transformation
(1.3), we see that any solution λ(t(τ)) is single-valued and meromorphic
whenever τ is restricted on a fundamental domain ofΓ(2). In this paper, a
branch of a solutionλ(t)to (1.2) means a solution λ(t(τ))defined for τ in a
fundamental domain ofΓ(2)(e.g. F given by (2.1)).
Recall a singular point t0 6∈ {0, 1,∞}of λ(t)means λ(t0)∈ {0, 1, t0,∞}.
DenoteC± ={t| Im t≷0}. Then for real solutions we have:
Theorem 1.4. Suppose λ(t)is a real solution. Then any branch of λ(t)has at
most two singular points inC\ {0, 1}, and they must be different type singular
points if the branch has exactly two singular points. Furthermore, the set (1.22) 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, 1,∞ respectively.
Remark 1.5. Theorem 1.4 shows that for each k∈ {0, 1, 2, 3}, any branch of a
real solution has at most one type k singular point inC\ {0, 1}. Theorem 1.4
will be proved in§6, where we will see that, the curve of ∂Ω(−0)connecting
∞ and 0 (resp. connecting 1 and ∞, connecting 0 and 1) is the image of the
degenerate curve of ω1
2 (resp.
ω2
2 ,
ω3
2 ) of Green function G(z|τ)in F0under
the map t(τ). Similarly, we can define
(1.23) Ω(k)
± :={t ∈C±|t is a type k singular point of some real solution}.
Then Ω(−k) = Ω−(0) and Ω(+k) = Ω (0)
+ = {t|t−1 ∈ Ω
(0)
− } for k ∈ {1, 2, 3},
and any real solution is smooth inC\ (Ω(0)
− ∪Ω
(0)
+ ∪ {0, 1}), which consists
of three connected components that contain(−∞, 0),(0, 1)and(1,+∞)
re-spectively. See the proof in§6.
1.5. Algebraic solution. A solution λ(t) to PVI is called an algebraic
so-lution if there is a polynomial h ∈ C[t, x] such that h(t, λ(t)) ≡ 0. It is
equivalent to that λ(t)has only a finite number of branches. By our
classi-fication theorem for (1.7), λ(t)is an algebraic solution of PVI(1
8,−81,18,38)if and only if λ(t) =λr,s(t)for some(r, s)∈QN with N≥3, where
(1.24) QN := n k1 N, k2 N gcd(k1, k2, N) =1, 0≤ k1, k2≤ N−1 o
is the set of N-torsion points of exact order N. The classification of the
algebraic solutions for PVI(1
8,−81,18,38) could be deduced from the B¨acklund transformation and Picard solutions, as shown in [26]. It is therefore natural to ask the following question:
Is any singular point t0 of an algebraic solutionλ(t)an algebraic number? Is the liftingτ0of t0a transcendental number?
The first question is equivalent to asking whether the j-value of any zero of E1N(τ; k1, k2)is an algebraic number. Here j(τ)is the classical modular
function, the j-invariant of τ, under the action by SL(2,Z); see (1.25) below.
This question can be answered easily from the aspect of Painlev´e VI or from the q-expansion principle in the theory of modular forms (c.f. [19]).
It is well known from the addition theorem of ℘ function that there is a
polynomialΨN ∈ Z[x, y, g2, g3]such that if (x, y)is an N-torsion point of
the elliptic curve y2 = 4x3−g
2x−g3, then ΨN(x, y) = 0. The degree of
ΨN is N
2−1
2 , and y appears only with odd powers inΨN(x, y)if N is even;
y appears only with even powers inΨN(x, y)if N is odd. See [16, p.272].
Now we come back to (1.9) and (1.11). Suppose that ˆλ(t) = ˆλν1,ν2(t)is a solution of PVI(0,0,0,1
2), where (ν1, ν2)is an N-torsion point. Then by the above result and the formulae for ˜ek := ℘(ωk(t)
2 |ω1(t), ω2(t))(here ω3 = ω1+ω2, see [26]): ˜e1= − 1+t 3 , ˜e2=1− 1+t 3 , ˜e3=t− t+1 3 ,
we see that there is a polynomial ˆP∈ Q[t, x]such that
ˆ
P(t, ˆλ(t))≡0.
This polynomial seems too complicated to be computed in general. By the
B¨acklund transformation, we conclude that for any algebraic solution λ(t),
there is a polynomial P ∈ Q[t, x]such that P(t, λ(t)) ≡ 0. Hence any
sin-gular point t of λ must be a root of a polynomial with integral coefficients, which implies that t is an algebraic number.
Let t=t(τ). Recall the classical modular function j(τ)of SL(2,Z):
(1.25) j(τ):=1728 g2(τ) 3 g2(τ)3−27g3(τ)2 =1728g2(τ) 3 ∆(τ) ,
where g2(τ) and g3(τ) are the coefficients of the elliptic curve Eτ: y2 =
4x3−g
2(τ)x−g3(τ), and the relation between t(τ)and j(τ)is
(1.26) j=256(t
2−t+1)3 t2(t−1)2 . So if t(τ)is algebraic, then j(τ)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 Zr,s(τ)
q-expansion principle (c.f. [19]). However, we can prove more. Let us con-sider
Z(N)(τ):=
∏
(r,s)∈QNZr,s(τ).
This is a modular form of weight |QN| := #QN with respect to SL(2,Z).
For N≥5, m := |QN|
24 ∈N and Z(N)(τ)
∆(τ)2m is invariant under SL(2,Z). Observe that (1.27) Zr,s(τ) = −Z1−r,0(τ) if s=0, −Z0,1−s(τ) if r=0, −Z1−r,1−s(τ) if r6=0, s6=0, which implies that any zero ofZ∆((Nτ))(2mτ)must be doubled. Since
Z(N)(τ)
∆(τ)2m has no
poles inH, we conclude that
(1.28) Z(N)(τ)
∆(τ)2m =C2m(`N(j)) 2
for some monic polynomial `N of j and nonzero constant C2m. If N is
odd, then Z(N)(∞) 6= 0. Hence
Z(N)(τ)
∆(τ)2m has poles of order 2m at τ = ∞,
equivalently,`N(j)is a polynomial of degree m = |QN|
24 . If N is even, then
lN :=deg`N < m. In any case, we have
Z(N)(τ) =C2m∆(τ)2m−2lnH(G4(τ)3,∆(τ))2,
where H(X, Y)is a homogeneous polynomial of X, Y and G4(τ) =g2(τ)/60
is the classical Eisenstein series of weight 4. By using the q-expansion of Zr,s(τ), we can prove that`N(j)has rational coefficients.
Theorem 1.6. For any N ≥ 5 with N 6= 6, the monic polynomial`N(j) deter-mined by (1.28) has rational coefficients and satisfies
(i) for any zero j0of`N(j), there is an algebraic solutionλr,s(t),(r, s)∈ QN, such that j0= j(τ0), where t0=t(τ0)satisfiesλr,s(t0) =∞. Conversely, for any algebraic solutionλr,s(t), (r, s) ∈ QN, if λr,s(t0) = ∞ for some t0=t(τ0), then j0= j(τ0)is a zero of`N(j).
(ii) `N(j)has distinct roots.
(iii) for any N1 6=N2,`N1(j)and`N2(j)have no common zeros. (iv) (1.29) deg`N = ( |QN| 24 if N is odd, |QN| 24 − 12ϕ(N2) if N is even, whereϕ(·)is the Euler function.
Recall the elementary formulae
(1.30) |QN| = N2
∏
p|N, p prime 1− 1 p2 , ϕ(N) =N∏
p|N, p prime 1− 1 p .Denote the j-value set of zeros of Z(N)(τ)by
J(N):={j(τ)|Zr,s(τ) =0 for some(r, s)∈ QN}.
If N = 3, then|QN| = 8 and Z(N)(τ) =const·G4(τ)2. Thus the zero τ =
ρ := eπi3 and J(3) = {0}. By Theorem 1.1 and Lemma 5.2, we see that
J(4) = J(6) = ∅. For N ≥ 5, J(N)is just the zero set of`N(j). Note that
formula (1.29) also holds for N = 6, which gives deg`6 = 0, so`6(j)is a
non-zero constant. This also proves J(6) =∅.
The computation of`Nseems to be difficult in general. However, by
ap-plying PVI, it is considerably easier for small N. Here are some examples: (1.31) J(3) ={0}, J(5) =n5·212 35 o , J(8) =207646 38 , (1.32) J(7) =n 211 57·34(−333009±175519 √ 21)o.
For N =9, the polynomial is
`9(j) =j3+86191391040000000 78815638671875 j 2+19885648112869441536 78815638671875 j − 7205712225604271603712 78815638671875 .
So J(9) ={a, b, ¯b}, where a∈R and b6∈R. Numerically,
(1.33) a≈186.3, b≈ −639.9+285.0×√−1.
It seems that except for N =3, all elements in J(N)are not algebraic integers.
If this would be true, then by a classical result of Siegal and Schneider, all
τ such that λ(t(τ)) = ∞ for an algebraic solution λ(t)are transcendental. A natural question is how to determine their location in the fundamental
domain F of SL(2,Z), where
(1.34) F ={τ∈H|0≤Re τ <1, |τ| ≥1, |τ−1| >1} ∪ {ρ= eπi3 }.
The above examples show that there is at least one zero of`N(j)of where
the corresponding τ is on the circular arc{τ∈H| |τ| =1}. Define
JN−={(r, s)∈ QN |2r+s=1 and 13 <s< 12}, JN+={(r, s)∈ QN |2r+s=1 and 0< s< 13}. Then we have the following interesting result.
Theorem 1.7. For any N ≥ 5 with N 6= 6,`N(j)has exactly #JN+real zeros in
(0, 1728)and exactly #JN−real zeros in(−∞, 0). Furthermore,`N(j)has no zeros in{0} ∪ [1728,+∞).
Notice that in the fundamental domain F of SL(2,Z), the corresponding
τ of any positive zero of`N(j)is on the circular arc{τ∈ F| |τ| =1}; while
Re τ = 1
2}. We can use (1.31)-(1.33) to check the validity of Theorem 1.7 for
small values of N. For example, J5+= {(2 5,15)}, J5−=∅; J7+ ={(37,17)}, J7−= {(27,37)}; J8+={(3 8, 2 8)}, J8− =∅; J + 9 = {(49, 1 9)}, J9−=∅.
The proof of Theorems 1.6 and 1.7 will be given in§7. In §8 we will give
some further remarks about Theorems 1.4 and 1.7. 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, 1,∞}, which are needed in the computation of`N(j). 2. PAINLEVE´ VI: OVERVIEWS
In this section, we start with the discussion of Painlev´e VI (1.2): d2λ dt2 = 1 2 1 λ+ 1 λ−1+ 1 λ−t dλ dt 2 − 1 t + 1 t−1+ 1 λ−t dλ dt + λ(λ−1) (λ−t) t2(t−1)2 " α+β t λ2 +γ t−1 (λ−1)2 +δ t(t−1) (λ−t)2 # . It is well-known that (1.2) possesses the Painlev´e property, which says that
any solution λ(t)has no branch points and no essential singularities at any
t ∈C\{0, 1}.
2.1. Multi-valueness via single-valueness. The Painlev´e property implies
that although a solution λ(t)is multi-valued inC, λ(t)is a single-valued
meromorphic function if t is restricted inC± = {z = x+iy|y ≷ 0}. That
means if λ(t)is analytically continued along a closed curve t=t(e), t(0) =
t(1), inC+(orC−), then λ(t(0)) =λ(t(1)).
Due to the multi-valueness of a solution of (1.2), it is convenient to lift solutions and the equation to the universal covering. The universal
cover-ing space ofC\{0, 1}is the upper half planeH. The covering map t(τ)is
given in (1.3), by which, Painlev´e VI (1.2) is transformed into the elliptic form (1.5).
It is elementary that t(τ)is invariant under the action of γ∈ Γ(2), where Γ(2) ={A∈SL(2,Z)|A≡ I2mod 2}.
That is t(τ) =t(τ0)if and only if τ0 = γ·τ= aτ+b
cτ+d for some γ= a b
c d
∈
Γ(2). Indeed t(τ) is the principal modular function of Γ(2). Let H∗ =
H∪Q, where Q is the set of rational numbers. Then it is well known that
H∗/Γ(2) ∼= CP1with three cusp point∞, 0, 1 which are mapped to 1, 0, ∞
by t(τ)respectively. As the consequence of the isomorphism, we have
t0(τ) = dt
namely the transformation t(τ)is locally one-to-one. Therefore, t(τ)maps
any fundamental domain of Γ(2) one-to-one ontoC\ {0, 1}, and any
so-lution λ(t(τ))is single-valued and meromorphic whenever τ is restricted
in a fundamental domain of Γ(2). As pointed out in§1, throughout this
article, a branch of a solution λ(t)always means a solution λ(t(τ))defined
for τ in a fundamental domain ofΓ(2).
The fundamental domain F2ofΓ(2)is
(2.1) F2= τ|0≤Re τ <2, |τ−12| ≥ 12, |τ−32| > 12 .
When τ ∈ iR+, ek(τ)are real-valued and satisfies e2(τ) < e3(τ) < e1(τ) (see e.g. [6]). From here, it is easy to see that t(iR+) = (0, 1), where t(i∞) =
1 and t(i0) = 0. Here we used limτ→i∞e2(τ) = limτ→i∞e3(τ) = −π
2
3 (see
§6). Furthermore, we could deduce from above that for any τ ∈ F2, t(τ)∈
R if and only if τ∈iR+∪ {τ∈H| |τ−1
2| = 12} ∪ {τ∈H| Re τ=1}.
By the formula (1.8), we see that ℘(p(τ)|τ) is always a single-valued
meromorphic function defined inH. However, as a solution of (1.7), p(τ)
has a branch point at those τ such that p(τ) ∈ Eτ[2], where Eτ[2] :=
{ωk
2 |0≤ k ≤3}is the set of 2-torsion points in Eτ. The single-valueness of
℘(p(τ)|τ)is one of the advantages of the elliptic form.
Recalling (1.21) and (2.1), F0is a half part of F2. Then it is not difficult to
prove that the transformation t(τ)maps the interior of F0 onto the lower
half planeC−, and t(τ)maps F2\F0ontoC+; see§6. Hence it is convenient to use τ∈ F2when a branch of solution λ(t)with t∈ C\ {0, 1}is discussed.
Different branches of λ(t)can be obtained from (1.8) by considering τ in
another fundamental domain ofΓ(2).
2.2. Isomonodromic deformation. It is well known that Painlev´e VI gov-erns the isomonodromic deformation of some linear ODE. See [18] in this aspect. For the elliptic form (1.5), it was shown in [6] that it is convenient to use the so-called generalized Lam´e equation (GLE):
(2.2) y00= " ∑3 j=0nj nj+1 ℘z+ωj 2 + 3 4(℘ (z+p) + ℘ (z−p)) +A(ζ(z+p)−ζ(z−p)) +B # y. Suppose nj 6∈ 12+Z. Then p(τ)is a solution of (1.5) if and only if there exist
A(τ)and B(τ)such that GLE (2.2) preserves the monodromy asτ deforms. The
formula to connect parameters of (1.5) and (2.2) is:
(2.3) αj = 12 nj+ 12
2
, j=0, 1, 2, 3.
See [6] for the proof. The advantage to employ GLE (2.2) is that for some cases, the monodromy representation is easier to describe. For example, let
us consider nj =0 for all j. Then the elliptic form of PVI is (1.7), and GLE is
(2.4) y00= 3
4(℘ (z+p) + ℘ (z−p)) +A(ζ(z+p)−ζ(z−p)) +B y.
For any p6∈ Eτ[2],±p are the singular points of (2.4) with local exponents
−1
C2\1
2Z2and p(τ) = pr,s(τ)is the solution given by (1.8), then we proved
in [6] that the monodromy representation ρ : π1(Eτ\{±p}, q0)→ SL(2,C) of GLE (2.4) is generated by ρ(γ±) =−I2, ρ(`1) = e−2πis 0 0 e2πis , ρ(`2) = e2πir 0 0 e−2πir ,
where q0 is a base point, γ± ∈ π1(Eτ\{±p}, q0) encircles ±p once and
`1,2 ∈ π1(Eτ\{±p}, q0) are two fundamental circles of the torus Eτ such
that γ+γ− = `−21`−11`2`1. In particular, the monodromy representation ρ is completely reducible.
2.3. B¨acklund transformation. In [28], Okamoto constructed the so-called B¨acklund transformations between solutions of Painlev´e VI with different parameters. Indeed, this transformation is a birational transformation be-tween the solutions of the corresponding Hamiltonian system, or equiva-lently, a birational transformation of(λ(t), λ0(t))together. Since λ(t)and
λ0(t)are algebraically independent generally (otherwise, Painlev´e equation
would be reduced to a first order ODE), the B¨acklund transformation is not
a birational transformation of the solution λ(t)only.
For example, it is known that a solution λ(t)of PVI(1
8,−81,18,38) can be ob-tained from a solution ˆλ(t)of PVI(0,0,0,1
2)by the following B¨acklund
trans-formation (cf. [34, transtrans-formation s2in p.723]):
(2.5) λ(t) = ˆλ(t) + 1
2 ˆµ(t), µˆ(t) =
t(t−1)ˆλ0−ˆλ(ˆλ−1)
2 ˆλ(ˆλ−1)(ˆλ−t) .
As mentioned in the Introduction, for PVI(0,0,0,1
2), all its solutions are Picard solutions:
(2.6) ˆλ(t) = ˆλν1,ν2(t) = ℘(ν1ω1(t) +ν2ω2(t)|ω1(t), ω2(t)) +
t+1
3 ,
where (ν1, ν2) ∈ C2\12Z2 and ω1,2(t) are given by (1.9). See [26, 12]. In principle, Hitchin’s formula (1.8) could be obtained from Picard solution (2.6) via (2.5), as mentioned by [26] and some other references. However, the computation of ˆλ0(t)via (2.6) 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 Painlev´e VI to spe-cial parameters via B¨acklund 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.1) that ˆλ(t)is smooth for all t∈C\{0, 1}if and only if(ν1, ν2)∈R2\12Z2. But this assertion is obviously false for PVI(1
However, the B¨acklund transformation is useful to discuss branch points and essential singularities, which are preserved under the B¨acklund
trans-formation. For example, if t= 0 is a branch point for a solution, then after
the B¨acklund transformation, t = 0 is still a branch point for the new
so-lution. Another example is that t = 1 is an essential singularity for λr,s(t)
if r ∈ R and s ∈ iR. Thus, t = 1 is also an essential singularity for Picard solution ˆλν1,ν2(t)if ν1 ∈R and ν2∈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 (1.7) due to the associated monodromy representation of GLE (2.4). It was shown that solutions expressed in (1.8) 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 (2.4).
Theorem C. ([6, Theorem 4.2]) Suppose p(τ)is a solution to the elliptic form (1.7). Then the followings hold:
(i) The monodromy representation is completely reducible if and only if there exists(r, s)∈C2\1
2Z2such that℘(p(τ)|τ)is given by (1.8).
(ii) The monodromy representation is not completely reducible if and only if
(3.1) λ(t) = ℘(p(τ)|τ)−e1(τ)
e2(τ)−e1(τ)
, t= e3(τ)−e1(τ)
e2(τ)−e1(τ) satisfies one of the following four Riccati equations:
(3.2) dλ dt =− 1 2t(t−1)(λ 2−2tλ+t), µ≡0, (3.3) dλ dt = 1 2t(t−1)(λ 2−2λ+t), µ≡ 1 2λ, (3.4) dλ dt = 1 2t(t−1)(λ 2−t), µ≡ 1 2(λ−1), (3.5) dλ dt = 1 2t(t−1)(λ 2+2(t−1)λ−t), µ≡ 1 2(λ−t).
Hereµ(t)is defined by the second formula in (A.2), i.e.,(λ(t), µ(t)) sat-isfies the well-known Hamiltonian system of Painlev´e 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 λ(t)can be expressed in terms
of previously known functions, i.e., it does not define new transcendental functions.
Now we discuss the solutions of which the associated monodromy rep-resentation is not completely reducible, and the results in this section will
be used to prove Theorem 1.1 in§5. For GLE (2.4) in Eτwith p6∈Eτ[2], we
proved in [6] that there is always a solution which is expressed by: (3.6) ya1(z):=exp
1
2z(ζ(a1+p) +ζ(a1−p))
σ(z−a1)
pσ(z+p)σ(z−p),
where the pair±a1∈ Eτis uniquely determined by
(3.7) A= 1
2[ζ(p+a1) +ζ(p−a1)−ζ(2p)].
If a1 6≡ −a1 mod Λτ, then y−a1(z)and ya1(z)are linearly independent so-lutions to (2.4). In this case, the monodromy representation associated to (2.4) is completely reducible. In fact, we proved in [6, Lemma 2.3] that the monodromy representation for (2.4) is not completely reducible if and only if a1 ∈Eτ[2].
Now we assume a1 = ω2k ∈ Eτ[2]. Let y1(z) = ya1(z) and y2(z) =
χ(z)y1(z)be a linearly independent solution of (2.4) to y1(z). Clearly it is equivalent to χ(z)6≡const and
(3.8) χ00(z)
χ0(z) +2
y10(z)
y1(z)
=0, i.e., χ0(z) =const·y1(z)−2.
On the other hand, by using 2ζ(z)−ζ(2z) =−1
2 ℘00(z) ℘0(z), (3.7) is equivalent to (3.9) A=−1 4 ℘00(p−ωk 2 ) ℘0(p−ωk 2 ) . When a1 =0, we have y1(z)−2 = σ (z+p)σ(z−p) σ(z)2 =c(℘(z)− ℘(p)),
and then (3.8) yields χ(z) =c(ζ(z) + ℘(p)z). So for any c(τ)6=0,(c(τ)y1, y2)
is a fundamental system of solutions to GLE (2.4), where y2(z) = (ζ(z) +
℘(p)z)y1(z). In particular, (3.10) `∗j c(τ)y1 y2 = ηj+℘(1p)ωj 0 c(τ) 1 ! c(τ)y1 y2 .
Proposition 3.1. The solutions of the Riccati equation (3.2) can be parametrized
by C∈CP1 : (3.11) λC(t) = ℘(pC(τ)|τ)−e1(τ) e2(τ)−e1(τ) , ℘(pC(τ)|τ) = η2(τ)−Cη1(τ) C−τ .
Proof. We separate the proof into two steps.
Step 1. We prove that for any constant C ∈ CP1, λ
C(t)given by (3.11) solves the Riccati equation (3.2).
Fix any C∈ CP1and let p(τ) = pC(τ), A(τ) =−14℘ 00(p(τ))
℘0(p(τ)) in the gener-alized Lam´e equation (2.4).
If C =∞, then℘(p(τ)) = −η1(τ). Choose c(τ) = η2(τ) + ℘(p(τ))τ. By the Legendre relation τη1(τ)−η2(τ) =2πi, c(τ) = −2πi. Thus by (3.10), we have `∗1c(τ)y1 y2 =1 0 0 1 c(τ)y1 y2 ,`∗2c(τ)y1 y2 = 1 0 1 1 c(τ)y1 y2 .
That is, GLE (2.4) is monodromy preserving as τ deforms, so p∞(τ)is a
solution of the elliptic form (1.7) (see Subsection 2.2).
If C 6= ∞, then (3.11) gives η1(τ) + ℘(p(τ)) 6≡0 and C = ηη21((ττ)+℘()+℘(pp((ττ))))τ. Choose c(τ) = η1(τ) + ℘(p(τ)). Clearly except a set of discrete points in
H, c(τ)6=0 and so (3.12) `∗1c(τ)y1 y2 =1 0 1 1 c(τ)y1 y2 ,`∗2c(τ)y1 y2 = 1 0 C 1 c(τ)y1 y2 .
As before, we conclude that pC(τ)is a solution of the elliptic form (1.7).
We remark that the second formula in (3.11) was previously obtained
in [15, 33], where there does not contain the relation between λC(t) and
Riccati equations. Here, together with our result in [6], we conclude that
λC(t)actually satisfies the Ricatti equation (3.2).
Step 2. Let λ(t)be any solution of the Riccati equation (3.2). We prove
the existence of C∈CP1such that λ(t) =λ
C(t). Define±p(τ)by λ(t)via (3.1) and A(τ) = −1
4
℘00(p(τ))
℘0(p(τ)). Then p(τ)is a solution of the elliptic form (1.7), which implies that (2.4) is monodromy preserving as τ deforms. Therefore, there exists a fundamental system of solutions(˜y1(z; τ), ˜y2(z; τ))to (2.4) such that the monodromy matrices M1,
M2, which are defined by
`∗j ˜y1 ˜y2 =Mj ˜y1 ˜y2 , j =1, 2,
are independent of τ. We may assume℘(p(τ)|τ)6≡ ℘(p∞(τ)|τ), otherwise
we are done. Then c(τ) := η1(τ) + ℘(p(τ)) 6≡ 0. For any τ such that
c(τ)6=0,(c(τ)y1, y2)is also a fundamental system of solutions, so there is
an invertible matrix γ = a b
c d
such that ˜y1 ˜y2 = γc(τ)y1 y2 . Clearly the monodromy matrices of(c(τ)y1, y2)is given by (3.12), where
(3.13) C := η2(τ) + ℘(p(τ)|τ)τ
η1(τ) + ℘(p(τ)|τ) may depend on τ. Then
M1 =γ1 01 1 γ−1 = 1+ bd ad−bc − b2 ad−bc d2 ad−bc 1−adbd−bc ! ,
M2 =γ 1 0C 1 γ−1= 1+ bd ad−bcC −b 2 ad−bcC d2 ad−bcC 1− bd ad−bcC ! .
Since M1, M2 are independent of τ and bd 6≡ 0, we conclude that C is
a constant independent of τ. Consequently, (3.13) implies ℘(p(τ)|τ) =
℘(pC(τ)|τ)and so λ(t) =λC(t).
The proof is complete.
Remark 3.2. It is easy to see that if Im C> 0, then λC(t)has singularities (at least a pole) inC\{0, 1}. However, it is not so obvious to see whether λC(t)
has singularities or not if Im C ≤ 0. In§6, we will exploit formulae (3.11)
and (3.18) (below) to prove that any solution of the four Riccati equations has singularities inC\{0, 1}.
Another observation is that C=∞ gives that
(3.14) λ∞(t) =−η1(τ) +e1(τ)
e2(τ)−e1(τ) is a solution of PVI(1
8,−81,18,38). Since λ∞(t), λ∞(t)−1 and λ∞(t)−t can have only simple zeros (cf. [18, Proposition 1.4.1]), a direct consequence is
Theorem 3.3. For fixed k∈ {1, 2, 3}, the followings hold: (i) Any zero of η1(τ) +ek(τ)must be simple.
(ii)
(3.15) d
dτ((η1(τ) +ek(τ))
−1)6= 1
2πi for anyτ∈H.
(iii) η2(τ)+τek(τ)
η1(τ)+ek(τ) is a locally one-to-one map fromH to C∪ {∞}. Proof. Recall
t =t(τ) = e3(τ)−e1(τ)
e2(τ)−e1(τ) .
Since t0(τ)6=0 for all τ ∈H, the assertion (i) follows readily from the fact
that λ∞(t)(for k = 1), λ∞(t)−1 (for k = 2) and λ∞(t)−t (for k = 3) can have only simple zeros.
For the assertion (ii), we note from the Legendre relation and (3.11) that
λC(t) =−
η1(τ) +e1(τ)−τ2πi−C e2(τ)−e1(τ)
.
Fix any τ0 ∈H. If τ0is a zero of η1+e1, then ddτ((η1+e1)−1)|τ=τ0 = ∞. So it suffices to consider the case η1(τ0) +e1(τ0)6=0. Then by letting
C=τ0− 2πi
η1(τ0) +e1(τ0) ,
we see that t0 = t(τ0)is a zero of λC(t). Since λC(t)has only simple zeros, we have d dτ η1+e1− 2πi τ−C |τ=τ0 6=0.
This, together with η1(τ0) +e1(τ0)− τ2πi0−C = 0, easily implies ddτ((η1 + e1)−1)|τ=τ0 6=
1
2πi. This proves (3.15) for k = 1. Similarly, by
consider-ing λC(t)−1 and λC(t)−t, we can prove (3.15) for k = 2, 3. This proves
the assertion (ii).
Finally, using the Legendre relation leads to
η2(τ) +τek(τ) η1(τ) +ek(τ) = τ− η 2πi 1(τ) +ek(τ) . Therefore, η2(τ)+τek(τ)
η1(τ)+ek(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 G(z|τ)
at z= ω1
2 = 12:
det D2G(12|τ) =−C(τ)·Im η2(τ) +τe1(τ)
η1(τ) +e1(τ)
for some C(τ) > 0, provided that η1(τ) +e1(τ)6= 0. The local one-to-one of the map η2(τ)+τe1(τ)
η1(τ)+e1(τ) is important for studying the curve inH where the half-period ω1
2 is a degenerate critical point of G(z|τ). See§6. Furthermore,
we will prove a stronger result that η1(τ) +e1(τ)has only one zero in any
fundamental domain ofΓ(2); see Theorem 6.6.
Similarly, we can prove that all solutions of the other three Riccati
equa-tions can be parametrized by CP1. The calculation is as follows. Fix k ∈
{1, 2, 3}. When a1 = ω2k, by (3.6) it is easy to see that χ(z) =−( ℘(p)−ek ek−ei)(ek−ej) ζ(z− ωk 2 )− 1+ek ℘(p)−ek (ek−ei)(ek−ej) z
satisfies (3.8), where {i, j} = {1, 2, 3}\{k}. As before, for any c(τ) 6=
0, (c(τ)y1(z), y2(z)) is a fundamental system of solutions to (2.4), where y2(z) =χ(z)y1(z). In particular, (3.16) `∗j c(τ)y1 y2 = 1 0 −Dηj+ωj(1+Dek) c(τ) 1 ! c(τ)y1 y2 , where (3.17) D := ℘(p)−ek (ek−ei)(ek−ej) .
Proposition 3.5. For C∈CP1, we letλ
C(t) = ℘(peC2((ττ))−|τe)1−(τe1)(τ), where
(3.18) ℘(pC(τ)|τ) =
ek(Cη1(τ)−η2(τ)) + (g42 −2e2k)(C−τ) Cη1(τ)−η2(τ) +ek(C−τ)
.
Then λC(t) satisfies the Ricatti equation (3.3) if k = 1, (3.4) if k = 2, (3.5)
if k = 3. Furthermore, such λC(t) give all the solutions of these three Riccati
Proof. We sketch the proof for fixed k ∈ {1, 2, 3}. For any C ∈ CP1, we let p(τ) = pC(τ), A(τ) = −14
℘00(p(τ)−ωk2 )
℘0(p(τ)−ωk2 ) in (2.4). If C = ∞, i.e., Dη1+ (1+ Dek) ≡ 0, then we choose c(τ) = Dη2+τ(1+Dek) = η1(τ2πi)+ek(τ) 6≡0. By (3.16), `∗1c(τ)y1 y2 =1 0 0 1 c(τ)y1 y2 ,`∗2c(τ)y1 y2 = 1 0 1 1 c(τ)y1 y2 . If C 6= ∞, then (3.18) gives Dη1+ (1+Dek) 6≡ 0 and C = DDηη21++(τ(11++DeDekk)). Choose c(τ) =Dη1+ (1+Dek). Then `∗1c(τ)y1 y2 =1 0 1 1 c(τ)y1 y2 ,`∗2c(τ)y1 y2 = 1 0 C 1 c(τ)y1 y2 . Similarly as in Proposition 3.1, we see that pC(τ)is a solution of the elliptic form (1.7). Again the formula in (3.18) was first obtained in [33]. Here,
together with our result in [6], we conclude that λC(t)actually satisfies the
Ricatti equation (3.3) if k = 1, (3.4) if k = 2, (3.5) if k = 3. The rest of the
proof is similar to that of Proposition 3.1.
For solution pC(τ)of the Riccati equations given in Propositions 3.1 and
3.5, we let τ0 = γ·τ and C0 = γ·C for γ ∈ SL(2,Z). By using (4.2)-(4.4)
(see§4) and the formula of℘(pC(τ)|τ), it is easy to prove (3.19) ℘(pC0(τ0)|τ0) = (cτ+d)2℘(pC(τ)|τ).
Then we have the following result, which can be proved by the same
argu-ment of Proposition 4.4 in§4, so we omit the details of the proof here.
Proposition 3.6. Let λC(t) and λC0(t) solve the same one of the four Riccati equations (3.2)-(3.5). Then they give the same solution to PVI(1
8,−81,18,38)if and only if C0 =γ·C for someγ∈Γ(2).
We conclude this section by a remark. In [26], Mazzocco classified so-lutions of PVI((2µ−1)2/2,0,0,1
2) (write PVIµ for convenience) for µ ∈
1 2 +Z.
Notice that PVI1
2 is precisely PVI(0,0,0,12) and PVIµ can be transformed to PVI1
2 via B¨acklund transformations. Mazzocco proved for µ ∈
1
2+Z and µ6= 1
2, say µ= −21 for instance, PVI−21 has two types of solutions: one is so-called Picard type solutions, which is obtained from Picard solutions (2.6) via B¨acklund transformations; the other one is so-called Chazy solutions, such as e λ(t) = 1 8{[ω2+νω1+2t(ω02+νω10)]2−4t(ω20+νω10)2}2 (ω2+νω1)(ω02+νω01)[2(t−1)(ω02+νω01) +ω2+νω1][ω2+νω1+2x(ω20+νω01)]
(where ν ∈ C), which will turn to be the singular solutions λ0(t) ≡ 0, 1, t
or∞ of PVI1
2 via B¨acklund transformations. Here together with Theorem C
could be obtained from Chazy solutions of PVI−1
2 via B¨acklund
transforma-tions, but the process would be too complicated to be computed. 4. COMPLETELY REDUCIBLE SOLUTIONS
4.1. Simple zeros of Hecke form. By Theorem C in§3, any solution λ(t)
of PVI(1
8,−81,18,38)with a completely reducible monodromy representation can be expressed by (3.1): λ(t) = ℘(p(τ)|τ)−e1(τ) e2(τ)−e1(τ) , t= e3(τ)−e1(τ) e2(τ)−e1(τ) , where℘(p(τ)|τ)is given by (1.8) with some(r, s) ∈ C2\1
2Z2. From (1.8), we have the following application of the Painlev´e property.
Theorem 4.1. Suppose(r, s)∈C2\1
2Z2is a pair of complex constants. Then the Hecke form Zr,s(τ) =ζ(r+sτ|τ)− (rη1(τ) +sη2(τ))has only simple zeros. Proof. First, we note that the situations r+sτ ∈ Eτ[2]and Zr,s(τ) = 0 can
not occur simultaneously. If not, then there are τ0and m, n ∈ Z such that
r+sτ0 = m+nτ0+ω2, where ω is any lattice points{0, ω1, ω2, ω3 = ω1+ ω2}, and also ζ(r+sτ0) =rη1(τ0) +sη2(τ0). Without loss of generality, we
might assume ω=ω1. The other cases can be proved similarly.
The second identity also implies 1 2η1(τ0) =ζ( ω1 2 ) =ζ((r−m) + (s−n)τ0) = ζ(r+sτ0)−mη1(τ0)−nη2(τ0) = (r−m)η1(τ0) + (s−n)η2(τ0). Therefore, we have (r−m−12) + (s−n)τ0 =0, (r−m−12)η1(τ0) + (s−n)η2(τ0) =0, which implies r−m−1
2 =0 and s=n because the matrix
1 τ
η1(τ) η2(τ)
is non-degenerate for any τ due to the Legendre relation. Obviously it
contradicts to the assumption(r, s)6∈ 1
2Z2.
Now suppose Zr,s(τ0) =0, which implies℘(p(τ0)) =∞ by (1.8) because
℘0(r+sτ0) 6= 0. Consider the transformation τ0 7→ t0 via (3.1). Then by
the Painlev´e property, we know that λ(t)has a pole at t = t0 6∈ {0, 1,∞}.
By substituting the local expansion of λ(t)at t = t0 into (1.2), it is easy to
prove that the order of pole at t = t0is 1, which implies the zero of Zr,sat
τ=τ0is simple.
Remark 4.2. If(r, s)is an N-torison point, i.e., (r, s) = (k1
N,kN2)for positive integers ki, N ≥ 3 and gcd(k1, k2, N) = 1, then the function Zr,s(τ) is a
case, Theorem 4.1 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 αi = 18 for 0 ≤i ≤3, it is easy to see that for 1 ≤ k ≤3, p(τ) +ω2k is also a solution of the elliptic form (1.7) provided that p(τ)is a solution of
(1.7). Then we have the following result, which will be used in§5.
Proposition 4.3. Given(r, s)∈C2\1 2Z2, we define (4.1) (rk, sk) = (r−12, s) if k=1, (r, s− 12) if k=2, (r−1 2, s−12) if k=3. Then pr,s(τ) + ω2k =±prk,sk(τ)in Eτ.
Proof. It was proved in [6] that (1.8) is equivalent to
ζ(r+sτ+pr,s(τ)) +ζ(r+sτ−pr,s(τ))−2(rη1(τ) +sη2(τ)) =0. Form here, we easily obtain
ζ(rk+skτ+ (pr,s(τ) + ω2k)) +ζ(rk+skτ− (pr,s(τ) + ω2k)) −2(rkη1(τ) +skη2(τ)) =0, and so ℘(pr,s(τ) +ω2k|τ) =℘ (rk+skτ|τ) + ℘0(rk+skτ|τ) 2(ζ(rk+skτ|τ)− (rkη1(τ) +skη2(τ))) =℘(prk,sk(τ)|τ).
This completes the proof.
We call a solution to (1.5) a real solution if the monodromy group of its associated GLE (2.2) is contained in SU(2). For the case αj = 18, p(τ)is a real solution if and only if it is given by (1.8) for some real pair(r, s)∈R2\1
2Z2. 4.2. Modularity. In this subsection we study the modularity property of solutions to PVI(1
8,−81,18,38). Consider the pair(z, τ)∈ C×H and z= r+sτ.
For any γ = a b
c d
∈ SL(2,Z), conventionally γ can act onC×H by
γ(z, τ):= (cτz+d, γ·τ) = (cτz+d,acττ++db). Then z cτ+d = r+sτ cτ+d =r 0+s0τ0, where τ0 =γ·τ and(s0, r0) = (s, r)·γ−1. Using (4.2) ℘ z cτ+d τ 0 = (cτ+d)2℘(z|τ), τ0 = aτ+b cτ+d,
we derive ζ z cτ+d τ 0= (cτ+d)ζ(z|τ), and so (4.3) η2(τ0) η1(τ0) = (cτ+d)γ·ηη2(τ) 1(τ) . Set(r, s)· (η1(τ), η2(τ))T =rη1(τ) +sη2(τ). Then(r0, s0)· (η1(τ0), η2(τ0))T = (cτ+d)(r, s)· (η1(τ), η2(τ))T and so (4.4) Zr0,s0(τ0) = (cτ+d)Zr,s(τ).
Together (4.2) and (4.4), we obtain
(4.5) ℘ pr0,s0(τ0)|τ0= (cτ+d)2℘(pr,s(τ)|τ) = ℘ pr,s(τ) cτ+d τ 0,
where (r0 +s0τ0, τ0) = γ(r+sτ, τ). Indeed, by a direct calculation, we could prove that pr,s(τ)
cτ+d as a function of τ0 is a solution of the elliptic form
(1.7) since pr,s(τ)is a solution of (1.7). Particularly, pcr,sτ+(τd) =±pr0,s0(τ0)mod Λτ0. Recall that λr,s(t)is the corresponding solution of (1.2), namely
(4.6) λr,s(t) =
℘(pr,s(τ)|τ)−e1(τ) e2(τ)−e1(τ)
. Then the above argument yields the following result.
Proposition 4.4. λr,s(t)andλr0,s0(t)belong to the same solution of PVI(1 8,−81,18,38) if and only if(s, r)≡ (s0, r0)·γ modZ2by someγ∈Γ(2).
Proof. For the sufficient part, assume (s, r) ≡ (s0, r0)·γ modZ2 by some
γ∈Γ(2). Recall from [6, Lemma 4.2] that
(4.7) ℘(pr,s(τ)|τ) = ℘ (p˜r,˜s(τ)|τ)⇐⇒ (r, s)≡ ±(˜r, ˜s) (mod Z2),
which implies that all elements in±(r, s) +Z2give precisely the same
solu-tion λr,s(t). Hence we may assume(s, r) = (s0, r0)·γ by replacing(s, r)with
some element in(s, r) +Z2if necessary. Let`
0 ⊂H be a path starting from any fixed point τ0to τ00 = γ·τ0. Then` := t(`0)∈ π1(CP1\{0, 1,∞}, t0), where t(τ) = e3(τ)−e1(τ)
e2(τ)−e1(τ) and t0 = t(τ0). Let U ⊂ H be a small
neighbor-hood of τ0and denote V =t(U). Since
λr0,s0(t) = ℘(pr0,s0(τ)|τ)−e1(τ) e2(τ)−e1(τ)
, τ∈U,
so the analytic continuation`∗λr0,s0(t)of λr0,s0(t)along`satisfies
`∗λr0,s0(t) = ℘(pr
0,s0(γ·τ)|γ·τ)−e1(γ·τ) e2(γ·τ)−e1(γ·τ)
On the other hand, (s, r) = (s0, r0)·γ gives (r0+s0τ0, τ0) = γ(r+sτ, τ), where τ0 =γ·τ. Moreover, γ=a b c d ∈ Γ(2)gives (4.8) ej(γ·τ) = (cτ+d)2ej(τ), j =1, 2, 3.
Then it follows from (4.5) and (4.8) that
λr0,s0(t(γ·τ)) = ℘(pr 0,s0(γ·τ)|γ·τ)−e1(γ·τ) e2(γ·τ)−e1(γ·τ) (4.9) = ℘(pr,s(τ)|τ)−e1(τ) e2(τ)−e1(τ) =λr,s(t(τ)), τ ∈U, namely (4.10) λr,s(t) = `∗λr0,s0(t), t∈V.
Conversely, assume that λr,s(t)and λr0,s0(t)represent different branches
of the same solution in a small neighborhood V of t0 ∈ CP1\{0, 1,∞}.
Then there is` ∈π1(CP1\{0, 1,∞}, t0)such that (4.10) holds. Fix any τ0∈
H such that t0 = t(τ0) and let t−1(`) ⊂ H denote the lifting path of ` under the map t(τ) = e3(τ)−e1(τ)
e2(τ)−e1(τ) such that its starting point is τ0. Denote its ending point by τ00. Then t(τ00) = t0 = t(τ0), which implies τ00 = γ·τ0
for some γ = a b
c d
∈ Γ(2). Let U be a neighborhood of τ0 such that
t(U) ⊂ V. Then (4.6) and (4.10) give (4.9). Define(˜s, ˜r):= (s0, r0)·γ, then
(r0+s0τ0, τ0) =γ(˜r+ ˜sτ, τ), where τ0 = γ·τ, and so (4.5) gives
(4.11) ℘(pr0,s0(γ·τ)|γ·τ) = (cτ+d)2℘(p˜r,˜s(τ)|τ). Substituting (4.11) and (4.8) into (4.9) leads to
℘(pr,s(τ)|τ) = ℘ (p˜r,˜s(τ)|τ), τ∈U.
Again by (4.7) we obtain(r, s) ≡ ±(˜r, ˜s)modZ2, namely (s, r) ≡ (s0, r0)·
(±γ)modZ2where±γ∈ Γ(2).
Define for any N-torsion point(r, s) = (k1
N, k2
N)∈QN,
Γ(r,s):=γ∈SL(2,Z)| (s, r)·γ≡ ±(s, r)modZ2 .
Then℘(pr,s(τ)|τ)is a modular form of weight 2 with respect toΓ(r,s)in the sense
℘(pr,s(τ0)|τ0) = (cτ+d)2℘(pr,s(τ)|τ), ∀γ∈ Γ(r,s).
For example, if r=0, then
Γ(r,s) = γ=a b c d ∈SL(2,Z) b≡0, a≡ ±1 mod N .
