SU(3) Toda system - Waseda UniversitySU(3) Toda system Chang-Shou Lin National Taiwan University...

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SU(3) Toda system Chang-Shou Lin National Taiwan University December 11-15,2017 Waseda University Chang-Shou Lin (TIMS) SU(3) Toda system December 11-15,2017 1 / 43

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Page 1: SU(3) Toda system - Waseda UniversitySU(3) Toda system Chang-Shou Lin National Taiwan University December 11-15,2017 Waseda University Chang-Shou Lin (TIMS) SU(3) Toda system December

SU(3) Toda system

Chang-Shou Lin

National Taiwan University

December 11-15,2017

Waseda University

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Introduction

In these two lectures, I am going to discuss how to study the integrableelliptic equations such as

∆u + eu = 8πnδ0 in Eτ, (0.1)

(the Liouville equation) ∆u + 2eu − ev = 4πn1δ0

∆v + 2ev − eu = 4πn2δ0in Eτ, (0.2)

(SU(3) Toda system)

where Eτ = C/Z ⊕ Zτ, Im τ > 0 and δ0 is the Dirac measure at 0. Asolution u of (0.1) means: (a)u ∈ C2(Eτ \ {0}) satisfies ∆u + eu = 0 inEτ \ {0}; (b)u(z) = 4n log |z|+ O(1), as z → 0.

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Contents

Part I

§1. Picard Potential.

§2. Liouville equation.

§3. Proof of Theorem 1.2.

§4. Application of premodular forms

Part II SU(3) Toda System

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§ 1 Picard potential

Let q(z) be an elliptic function defined on Eτ. If the linear 2nd ODE:

y′′

= (q(z) + B)y in C, for any B ∈ C, (1.1)

has a fundamental solution which is meromorphic in C, q(z) is called aPicard potential.⇒ The order of any pole of q ≤ 2.

Theorem (Gesztesy-Weikard)Suppose q(z) is an elliptic function. Then q(z) is a Picard potential iff q(z)is an algebra-geometric solution of KdV hierarchy equation.

Spectral polynomial Qq(B) and the algebraic curveΓq = {(B ,W)|W2 = Qq(B)} is the spectral curve.

Monodromy representation of (1.1). A singular point p of q is calledapparent if any solution of (1.1) is meromorphic at p.

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Since any singular point of q(z) is apparent, the local monodromy at anysingular point is the identity matrix I. Thus the monodromy representationis reduced to a homomorphism ρB : π1(Eτ)→ SL2(C). That is, ify1(z), y2(z) are two solutions of (1.1), then ∃Si , i = 1, 2:

(y1(z + 1), y2(z + 1)) = (y1(z), y2(z))S1

(y1(z + τ), y2(z + τ)) = (y1(z), y2(z))S2

such that S1 · S2 = S2 · S1 (Note Si depends on B).There are two cases:Case 1: There are two common eigenfunctions of S1 and S2. Then S1 andS2 can be diagonalized simultaneously:

S1 =

(e iθ1 00 e−iθ1

), S2 =

(e iθ2 00 e−iθ2

), θi ∈ C.

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In this case, θ1 and θ2 are monodromic data. Case 1 is called completelyreducible.Case 2: There is only one common eigenfunction of S1 and S2. After anormalization, S1 and S2 are

S1 =

(1 01 1

), S2 =

(1 0c 1

), c ∈ C ∪ {∞}.

If c = ∞, then S1 =

(1 00 1

)and S2 =

(1 01 1

). In this case, c is the only

monodromic data. Case 2 is called not completely reducible.

If q(z) is even elliptic, i.e., q(−z) = q(z). By letting x = ℘(z)

(Weierstrass elliptic function of order 2), (1.1) can be projected to anODE (still Fuchsian) of second order defined on C. Then themonodromy representation of this new ODE would be irreducible⇔case 1.

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Go back to (1.1). Then it can be proved that

Case 1 happens⇔ Qq(B) = 0.

The classic theory developed in 19th century (by Hermite...etc) can beapplied to prove this statement.The first natural question is the following uniqueness question: Under whatkind of condition on q(z), the Riemann-Hilbert correspondence B → ρB is1-1 ?Let

q(z) =3∑

i=0

ni(ni + 1)℘(z + ωi/2),

where ni ∈ Z≥0, ω0 = 0, ω1 = 1, ω2 = τ and ω3 = 1 + τ. Then q(z) is aPicard potential (it was first proved by Treibich-Verdier, thus it is called theTreibich-Verdier potential).

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Theorem 1.1

Let qn(z) =∑3

i=0 ni(ni + 1)℘(z + ωi/2) and LB ,n(y) = y′′

− (qn(z) + B)y.Then B → ρB is 1-1.

Note that this one-one correspondence is not the same as the one-onecorrespondence of the classic direct monodromy problem, because theconnection matrix is not specified here. See the book ”PainleveTranscendent” by Fokas etc. There are more than one method to proveTheorem 1.1. One of proofs is to apply the Painleve VI equation andOkamoto transformation, and this method yields stronger results: Fixn = (n0, n1, n2, n3), we define

n± = {(m0,m1,m2,m3)|∃i ∈ {0, 1, 2, 3},mi = ni ± 1,mj = nj , j , i}.

Then {y′′

= (qm(z) + B)y |m ∈ n±} → {the monodromy group} is one-one.

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The spectral polynomial Qn(B) is very difficult to compute explicitly. Thereis a simple fact about Qn(B):

If τ = ib , b > 0, then Qn(B) is a polynomial with real coefficients.

We post the following question: are all roots of Qn(B) real and distinct ?Recently Z.J. Chen and I completely solved this question. First, we call(n0, n1, n2, n3) satisfies the (∗) condition, if

either n0 + n3 − (n1 + n2) ≥ 2, n0 ≥ 1 and n3 ≥ 1

or n1 + n2 − (n0 + n3) ≥ 2, n1 ≥ 1 and n2 ≥ 1

are satisfied.

Theorem 1.2Let τ = ib , b > 0. All roots of the spectral polynomial Qn(B) are real anddistinct iff (∗) does not hold.

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Interestingly, the proof is not algebraic, but analytic in nature. We willpresent a sketch of the proof later. Indeed, the proof is based on theconnection of the PDE (0.1) and the ODE (1.1) with q = Treibich-Verdierpotential. This connection is part of our general theory, developed byChai-Lin-Wang, Lin-Wang, Chen-Kuo-Lin etc. I will discuss it below.We first discuss how to characterize the monodromic data for aTreibich-Verdier potential. Recall the hyperelliptic curve

Γn + Γqn = {(B ,W)|W2 = Qn(B)} ∪ {(∞,∞)}.Step 1: Construct an embedding πn : Γn → SymNEτ,N =

∑3i=0 ni . This

embedding is constructed from the zeros of a common eigenfunction ofthe monodromy group. We would not go to discuss it. Here SymNEτ is thesymmetric space of N copies of Eτ .After the embedding is constructed, we define the addition map σn by:

σn(B ,W) =∑N

i=0 ai , where (a1, · · ·, aN) = πn(B ,W).

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Theorem 1.3 (Chen-Kuo-Lin)

The degree of σn =∑3

i=0 ni(ni + 1) = N∗

By Theorem 1.3, the function field K(Γn)/K(Eτ) is a finite extension withthe degree = N∗. Now we want to find a primitive generator of thisextension. Put

zn(B ,W) = ζ(N∑

i=1

ai −

3∑k=1

ηkωk

2) −

N∑i=1

ζ(ai) +3∑

k=1

ηkωk

2

Here ζ(z) is an odd function and ζ′

(z) = −℘(z). The three constant ηk isthe quasi-periods of ζ:

ζ(z + ωk ) = ζ(z) + ηk , k = 1, 2, 3.

We can prove zn(B ,W) is a meromorphic function on the spectral curveΓn. Moreover, we can prove:

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Step 2:

Theorem 1.4 (Chen-Kuo-Lin)zn(B ,W) is a primitive generator of the extension of K(Γn) over K(Eτ).Furthermore, there exist a polynomialW(X ;℘(σn), ℘

(σn), τ) ∈ Q[ei(τ), ℘(σn), ℘′

(σn)][x] such that

W(zn(B ,W);℘(σn(B ,W)), ℘′

(σn)(B ,W); τ) ≡ 0 in Γn.

RemarkIndeed, zn(B ,W) is related to the computation of the monodromy data ofLB ,n. The proof of Theorem 1.3 and Theorem 1.4 rely on the explicit formof solutions of this ODE. We skip the proofs. But I want to remark that thecrucial step of Theorem 1.4 is to apply Theorem 1.1.

ei(τ) = ℘(ωi2 ), i = 1, 2, 3.

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Step 3: Construct the premodular form.Let Zr ,s(τ) = ζ(r + sτ) − η1(τ)r − η2(τ)s, for any (r , s) ∈ C2 \ 1

2Z2. (If

(r , s) ∈ 12Z

2, then either Zr ,s(τ) is not well-defined or Zr ,s(τ) ≡ 0). ThisZr ,s(τ) actually is the Eisenstain series of weight 1 with the characteristic(r , s) if (r , s) ∈ Q2 \ 1

2Z2. Then we put

Z (n)r ,s (τ) = W(Zr ,s(τ);℘(r + sτ), ℘

(r + sτ); τ).

Z (n)r ,s (τ) is a modular form of weight

∑3k=0

nk (nk+1)2 w.r.t Γ(N) if (r , s) is

N-torsion point. The important property of Z (n)r ,s is

Z (n)r ,s (τ0) = 0 for some τ0 ∈ H (the upper half plane) iff the ODE (1.1) withτ = τ0 and some B ∈ C has monodromic matrices:

S1 =

(e is 00 e−is

), S2 =

(e−ir 00 e ir

).

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§ 2 Liouville equation

We consider (0.1) with four singular points

∆u + eu = 8π3∑

k=0

nkδωk /2, nk ∈ Z≥0. (1.2)

Equation (1.2) is an integrable equation which is described by the classicalLiouville theorem.

Theorem (Liouville)For any solution u(z) of (2.1), there exists a meromorphic function f(z) onC such that

u(z) = log8|f

(z)|2

(1 + |f(z)|2)2 . (1.3)

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This function f(z) is called a developing function for the solution u. We canrecover f through the following formula

S{f } = uzz −12

u2z , (1.4)

where S{f } = ( f′′

f ′)′

− 12 ( f

′′

f ′)2: the Schwarz derivative of f . Now we assume

u(z) = u(−z) is an even solution. Since the RHS of (1.4) is elliptic and hassingularities only at 0 and half periods, there is B(u) ∈ C

S{f } = −2{∑

nk (nk + 1)℘(z +ωk

2) + B}. (1.5)

Furthermore,

f(z) =y1(z)

y2(z), z ∈ C, (1.6)

where y1(z), y2(z) are

y′′

= (3∑

k=0

(nk (nk + 1))℘(z + ωk/2) + B)y. (1.7)

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By a direct computation, (1.3) implies the projective monodromyrepresentation of f(z) is unitary. Indeed we have

Theorem 1.5The Liouville equation (1.2) at Eτ0 has a even solution iff there is B0 ∈ C

such that the monodromy group of ODE (1.1) with qn(z) + B0 is unitary.

Together with the premodualr form Z (n)r ,s (τ) introduced before, Theorem 1.5

is equivalent to

Theorem 1.6Equation (1.2) (the PDE) has a even solution in Eτ0 iff there is a real pair(r , s) < 1

2Z2, such that Z (n)

r ,s (τ0) = 0

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§3 Proof of Theorem 1.2

In the followings, we always assume τ = ib , b > 0.Step 1: If (∗) condition does not hold, then all roots of the spectralpolynomial Qqn(B , τ) are real and distinct.There is a classical method to prove this kind of results: (1) Sturmsequence: g0, g1, · · ·, gn, s.t ., gi(B0) = 0⇒ gi−1(B0) · gi+1(B0) < 0 andg0(z) has real and distinct roots⇒gn has real and distinct roots. Forexample, this method works for the case n1 = n2 = n3 = 0 (this ODE iscalled Lame). But this method alone can not prove step 1. We have toimprove this classical method.Step 2: If the spectral polynomial Qqn(B , ib) has real roots only then theLiouville equation (1.2) has no solution in Eib .

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We prove (a) if Qqn(B , ib) has real roots only then all the roots aredistinct;(b) the monodromy representation of ODE (1.1) is not unitary; (c)by Theorem 1, the PDE (2.1) has no even solutions. We use the KdVtheory to prove (a).Step 3: If (∗) condition holds then there is τ0 = ib0 such that the PDE (1.2)has a solution. By step 2 Qqn(B) has some complex roots.The existence in Step 3 was proved by Eremenko and Gabrielov.

Corollary 1.7The PDE (1.2) has no solution for all τ = ib , b > 0 iff (∗) does not hold.

In particular,∆u + eu = 8πnδ0 in Eτ

has no solution for any n ∈ N if τ = ib , b > 0.

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§4. Application of premodular forms

We want to mention some applications of the theory. As we know, ℘(z|τ)satisfies the equation

℘′2

= 4℘3 − g2(τ)℘ − g3(τ)

where g2(τ), g3(τ) are constant multiple of the Eisenstein series of weight4,6 respectively. They are modular form of weight 4 and 6. But theirderivatives are not. Surprisingly, we can apply our premodular formZ (3,0,0,0)

r ,s (τ) to find all the critical points of g2(τ).

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Consider F0 = {τ ∈ H|0 ≤ τ ≤ 1, |τ − 12 | ≥

12 }. F0 is the fundamental domain

of Γ0(2) = {

(a bc d

)∈ SL2(Z)|c is even integers}.

Theorem 1.8 (Chen-Lin)

Let γ =

(a bc d

)∈ Γ0(2). Then

(i) g′

2(τ) = 0 has a solution in γ(F0) if and only c , 0. In this case,g′

2(τ) = 0 has only one solution.

(ii) C = {τ ∈ F0|g′

2(γτ) = 0, γ =

(a bc d

), c , 0}. C ∩ F0 = a disjoint union

of three smooth curve.

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Contents

Part II

§1. Main Results.

§2. Integrability.

§3. Fundamental Lemmas.

§4. Proof of Theorem 2.2.

§5. Proof of Theorem 2.3.

§6. Conclusion.

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§1 Main results

The SU(3) Toda system on Eτ: ∆u + 2eu − ev = 4πn1δ0

∆v + 2ev − eu = 4πn2δ0on Eτ. (2.1)

Here we assume ni ∈ Z≥0. Applying a recent result byLin-Wei-Yang-Zhang, we can prove

Theorem 2.1If n1 , n2 mod 3, then (2.1) has one even solution for any torus Eτ.

Theorem 2.1 is a highly non-trivial result. The proof is based on sharpestimates for a sequence of bubbling solutions if such a sequence ofsolutions exist. By the hypothesis n1 , n2 mod 3, we then exclude thepossibility of bubbling of solutions, and the existence follows from this apriori estimates by applying a general theory from the variationalapproach.

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So in this talk, I assume n1 = n2 mod 3. The difficulty of (2.1) is thepossibility of the existence of bubbling solutions. However, (2.1) is alsointegrable elliptic PDE. Like part I, we hope to apply this integrability tostudy (2.1). From this direction, we prove two results

Theorem 2.2Assume n2 = n1 + 3l, l ≥ 0. If n1 is odd and n2 is even, then (2.1) has noeven solutions for any Eτ.

We will present a proof of Theorem 2.2 later.

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The second result is

Theorem 2.3Assume n1 even and n2 = n1 + 3. Then the Toda system (2.1) ∆u + 2eu − ev = 4πnδ0

∆v + 2ev − eu = 4π(n + 3)δ0on Eτ. (2.2)

has an even solution iff the Liouville equation

∆u + eu = 8π(n + 2)δ0 on Eτ (2.3)

has a solution.

Corollary 2.4If τ = ib, then (2.2) has no even solutions.

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§2 Integrability

Next we explain why (2.1) is integrable. Put

U =2u + v

3, V =

2v + u3

.

Then (U,V) satisfies ∆U + e2U−V = 4πγ1δ0,

∆V + e2V−U = 4πγ2δ0,(2.4)

whereγ1 =

2n1 + n2

3, γ2 =

2n2 + n1

3, γi ∈ Z≥0.

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Now consider the following linear ODE:

L + (∂z − Vz)(∂z + Vz − Uz)(∂z + Uz) = ∂3z + W2∂z + W3, (2.5)

where

W2 = Uzz − U2z + Vzz − V2

z + UzVz ,

W3 = Uzzz − 2UzUzz + UzVzz + VzU2z − UzVz .

From (2.5), we have Lf = 0, f = e−U.

Theorem (Lin-Wei-Ye)(W2)z = (W3)z = 0.ConsequentlyW2 and W3 are elliptic functions withdouble periods 1 and τ.

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By applying

U(z) = 2γ1 ln |z − pi |+ O(1), V(z) = 2γ2 ln |z − pi |+ O(1) near pi ,

we have

W2(z) = −[γ1(γ1 + 1) + γ2(γ2 + 1) − γ1γ2]1

(z − pi)2 + O(1

z − pi)

+−α

(z − pi)2 + O(1

z − pi),

W3(z) = [2γ1(γ1 + 1) + γ1γ2(γ1 − γ2 − 1)]1

(z − pi)3 + O(1

(z − pi)2 )

(z − pi)3 + O(1

(z − pi)2 ).

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Now suppose U(z) and V(z) are even. Then

W2(z) = α℘(z) + B , W3 = β℘′

(z), B ∈ C.

So,Ly = y

′′′

− (α℘(z) + B)y′

+ β℘′

(z)y

is a Fushsian ODE of third order.Let yi(z) be a system of fundamental solution of the ODE Ly = 0. Since fis real-valued, f can be written as

f =∑i,j

Mi,jyi(z)yj(z), Mi,j ∈ C.

Lin-Wei-Ye proved that (Mi,j) > 0 and then

e−U = f =∑

i

|yi |2, where Lyi = 0. (2.6)

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Then a direct computation shows that the local exponents of L at thelattice point are

−γ1,−γ1 + (n1 + 1),−γ1 + (n1 + 1) + n2 + 1. (2.7)

Since the difference of two exponents is an integer, So (2.3) might havesolution with logarithmic singularity. However if L is derived from a solutionof SU(3) Toda system, then L obviously have no logarithmic singularity,that is, 0 is an apparent singularity.Because (2.6), the monodromy group of L is unitary. Conversely, if themonodromy group is unitary, then the RHS of (2.6) is invariant under thetransformation z → z + ωi , in particular,

U(z) = U(z + 1) = U(z + τ).

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We summarize: putα = γ1(γ1 + 1) + γ2(γ2 + 1) − γ1γ2,

β = 2γ1(γ1 + 1) + γ1γ2(γ1 − γ2 − 1).(2.8)

Theorem 2.5Equation (2.1) has a even solution on Eτ0 iff ∃B0 such that the monodromygroup of

L =d3

dz3 − (α℘(z) + B0)ddz

+ β℘′

(z) on Eτ0

is unitary.

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§3 Fundamental Lemmas

We let Ly =

d3

dz3 y − α℘(z)dydz

+ β℘′

(z)y,

LBy = Ly − Bddz

y, B ∈ C

where (α, β) is given by (2.8). Set n = n1 and n2 = n + 3l. Then the localexponents of L are

−n − l, 1 − l, n + 2l + 2. (2.9)

Since the difference of any two exponents is an integer, LBy = 0 mighthave solutions with logatithmic singularity at 0. The singularity point 0 iscalled apparent if there are no such solutions. Suppose for a fixed B ∈ C,0 is an apparent singularity of LB . Then the monodromy representation ρcan be reduced to a homomorphism of π1(Eτ)→ SL(3,C), which image iscalled the monodromy group of LB .

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Obviously, the monodromy group is Abelian. Denote by Sj the monodromymatrix (with respect to a fundamental solution) under the transfomationz → z + ωj . Then S1 · S2 = S2 · S1.The following results are simple, but fundamental (and surprising!).

Lemma 2.6Let l be odd. Then for any B ∈ C, LBy = 0 has an unique even ellipticsolution y1 of the form

y1(z) =m∑

j=0

Cj(B)℘(z)j ,m =l − 1

2(2.10)

where Cm(B) = 1,Cj(B) ∈ Q[g2, g3][B] with degree Cj(B) = m − j.Furthermore, Cj(B) is homogenous of weight m − j, where the weightB , g2, g3 are 1,2,3 respectively.

Note that the order of pole at 0 of y1(z) is l − 1.

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Lemma 2.7Let n and l be even. Then for any B ∈ C, LB has an unique even ellipticsolution Y1 of the form

Y1(z) =k∑

j=0

Cj(B)℘(z)j , k =n + l

2(2.11)

where Ck (B) = 1,Cj(B) ∈ Q[g2, g3][B] with degree k − j. Furthermore,Cj(B) is homogenous of weight k − j, where the weight B , g2, g3 are 1,2,3respectively.

The proof of both lemmas are straightforward. Substituting the forms of(2.10) or (2.11), we get a 4-terms recursive formula of Cj , for 0 ≤ j ≤ k .Starting from Ck , we can solve all Cj , j = 0, · · ·, k − 1. The recursiverelation holds because the hypothesis on n and l.

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Now there is only one missing, i.e., (n.l) = (even,odd). For this case, weconsider the dual ODE of LB

L∗(y) =d3

dz3 y − α℘(z)y − (α + β)℘′

(z)y. (2.12)

Then L∗ has local exponent

−n − 2l, l + 1, n + l + 2 (2.13)

Note that there are two positive exponents, and if n is even, then n + 2l isalways even. The proof of Lemma 2.6 yields

Lemma 2.8Let n be even. Then for any B ∈ C, the dual equation L∗B(y) = 0 has anunique even elliptic solution of the form

Y1(z) =k ∗∑

j=0

Dj(B)℘(z)j , k ∗ =n + 2l

2

where Dk ∗(B) = 1,Dj(B) ∈ Q[g2, g3][B] with degree k ∗ − j. Furthermore,Dj(B) is homogenous of weight k ∗ − j.

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Note: We could prove LB is apparent iff L∗B is also apparent.

RemarkThe existence of one elliptic solution does not imply LB has only the

apparent singularities. That is S1 =

1 0 00 ∗ ∗

0 ∗ ∗

and S2 =

1 0 00 ∗ ∗

0 ∗ ∗

which

means that LB can be reduced to a second order ODE by the help of itsdual. Unfortunately, this second order ODE has more complicatedpotentials by increasing the number of singular points (all are apparent!).In any case, by using that dual operator of L, we could prove if n is even,then for any B ∈ C, LB has apparent singularities only.

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§4 Proof of Theorem 2.2

Recall the local exponents of L are

−n − l, 1 − l, n + 2l + 2.

Since both n and l are odd (n2 = n + 3l is even), by Lemma 2.6 wealready have an elliptic solution y1(z) whose order of pole at z = 0 is l − 1.In this case, we do not expect LB could have apparent singularity only forany B ∈ C.

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Lemma 2.9

Set k = n+l2 . Then there exists a (monic) polynomial P(B) ∈ Q[g2, g3][B]

with degree k − l−12 = n+1

2 and homogenous weight k − l−12 such that 0 is

an apparent singularity of (1.6) iff P(B) = 0

The proof is standard: we substitute y(z) =∑∞

j=0 cjz2j−2k , c0 = 1 into theequation. Then we get a recursive formula which solves cj from c0, · · ·, cj−1.The coefficient φj of cj in this formula vanishes iff j ∈ {0, k − l−1

2 }. Startingfrom c0 = 1 we could uniquely solve cj(B) from j = 0, 1, · · ·, k − l−1

2 − 1. Atj = k − l−1

2 , we have 0 = φjcj = a sum fromc0, · · ·, cj−1 + P(B). SoP(B) = 0 iff cj(B) can be solved. ⇔ 0 is apparent.

Since both n and l are odd, the local exponent −(n + l), where n + l iseven. So we expect there should be a similar result as Lemma 2.9 for evenelliptic solutions.

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Lemma 2.10

Let n, l, k as before. Then there exists a polynomial P(B) ∈ Q[g2, g3][B] ofdegree k − l−1

2 such that LBy = 0 has an even elliptic solution y2 of degreek in x = ℘(z) iff P(B) = 0.

The proof is very similar to Lemmas in §3. Now if P(B) = 0, the solutiony2(z) has a singularity at 0 with order 2k = n + l, however, the oneobtained in Lemma 2.6 has the pole with order l − 1. They are different.Let y3(z) be the solution LBy = 0 with the local exponent n + 2l + 2 (this isthe largest exponent). Then y3(z) can not have logarithmic singularity at 0.Hence all the solution of LBy = 0 are meromorphic⇒P(B)=0. If P(B) hasdistinct root and degP(B) = degP(B), then P(B) = P(B).

So we have to prove P(B) has distinct roots.

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Lemma 2.11

The polynomial P(B; τ) has distinct roots expect for finitely many SL(2,Z)orbits of τ.

Let D(τ) be the discriminate of the polynomial P(B; τ). To prove Lemma2.11, it suffices to show that D(τ) , 0 (since D(τ) is holomorphic in τ,indeed D(τ) is a modular form w.r.t SL2(Z)). Indeed we proveD(i) , 0, i.e., P(B; i) has real and distinct roots. The result that D(i) , 0 isnot obvious. But we omit the proof here.

Proof of Theorem 2.2. We want to prove for any B, the monodromy groupis not unitary. Otherwise, if the monodromy is unitary, by Lemma 2.6 andLemma 2.11(they imply there are two elliptic solution), the monodromymatrix Si (under z → z + ωi) has eigenvalue with multiplity ≥ 2. Butdet Si = 1⇒ all the eigenvalues of Si = 1⇒ Si = I3, i = 1, 2⇒ All thesolutions of LBy = 0 is (even) elliptic. But the solution y3(z) with the localexponent n + 2l + 2 is holomorphic⇒ y3(z) could not be elliptic, acontradiction.

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§5 Proof of Theorem 2.3

The integer l = 1 for Theorem 2.3. Theorem 2.3 is only a special case forour project wich tries to study the monodromy of LB for even n ≥ 0 and anyl ≥ 1. The case l = 1⇒ α = (n + 2)(n + 3) and β = 0. So

LBy = y′′′

− ((n + 2)(n + 3)℘(z) + B)y′

. (2.14)

We will compare (2.14) with the Lame equation

LBy = y′′

− ((n + 2)(n + 3)℘(z) + B)y. (2.15)

So Y(z) =∫

y(z) is a solution of (2.14)⇔ y(z) is a solution of (2.15). Aswe discussed in Part I, (2.15) is related to

∆u + eu = 8π(n + 2)δ0 in Eτ. (2.16)

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Then Theorem 2.3 is an immediate consequence of the following result.

Theorem 2.12Let n ≥ 2 be even. Then the monodromy of

f′′′

(z) = (n(n + 1)℘(z) + B)f′

, for some B ∈ C (2.17)

is unitary iff the monodromy

y′′

= (n(n + 1)℘(z) + B)y′

(2.18)

is unitary.

RemarkIf n is odd, then Theorem 2.12 does not hold true.

The proof is not easy at all. So we skip it.

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§6 Conclusion

(1) Suppose n is even. We could prove ( in another paper) that LBy = 0has apparent singularity only (the proof is first to prove its dual is apparent,and secondly, we prove LB is apparent iff its dual is apparent). By Lemma2.8, the dual of 3rd order ODE can be decomposed into a product of asecond order ODE and the trivial first order (both are Fuchsian) We couldprove this second order ODE is Picard. Thus there exists a polynomialQ(n,l)(B) s.t. the monodromy representation of LB is completely reducibleiff Q(n,l)(B) , 0. However the premodular form of this second order ODE isdifficult to compute. We are still trying to understand it better.

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(2) The Gelfand-Dicky flows for third order ODE is related to the eigenvalueproblem LBy = Ly −By. It is known that L is an algebro-geometric solutionof the Boussinesq hierarchy iff Ly + By = 0 has only apparent singularityfor any B ∈ C.In spite the similar phenomena (i.e., LB = L − B∂z has only apparentsingularity), we do not know whether the eigenvalue problemLBy = Ly − By

is related to an integrable flow or not. This issue should beinvestigated further.Our goal is to study whether the theory in §1 of Part I can be extendedto a third order ODE on Eτ or not. This is the main concern of our futureproject.

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