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Unitary monodromy of Lamé differential operators F.Beukers

Unitary monodromy of Lam′e di?erential operators

F.Beukers

May17,2007

Abstract

The classical second order Lam′e equation contains a so-called accessory parameter B.In this paper we study for which values of B the Lam′e equation has a monodromy group which is conjugate to a subgroup of SL(2,R)(unitary monodromy with inde?nite hermitian form).We refomulate the problem as a spectral problem and give an asymptotic expansion for the spectrum.

1Introduction

Consider the di?erential equation in the complex plane

P(z)y +1

P (z)y+(

?B)y=0,P(z)=4z3?g2z?g3(1)

and where P(z)has three distinct zeros z1,z2,z3.This is the Lam′e equation with parameter n=?1/2.See[WW].This equation is Fuchsian and has four singular points,z1,z2,z3,∞and local exponents0,1/2at the?nite singularities and exponents1/4,1/4at∞.Conversely any linear di?erential equation with these four singularities and local exponents is necessarily of the form given above.Only the parameter B is not determined by the location of the singularities and their exponents.This parameter is known as the accessory parameter.Let G be the monodromy group of equation(1).

We shall be interested in the following question.

Question1.1(Accessory parameter problem)We call G unitary if it admits a non-trivial G-invariant hermitian form on C2(not necessarily positive de?nite).Given P(z),for which complex values of B is G unitary?

For di?erential equations in the p-adic domain there is the similar question for which values of B there exist solutions with p-adic radius of convergence one(see[Dwork]).This problem is studied in a separate paper[Be2002].We like to consider the present problem as the∞-adic version of it.

There is a very picturesque interpretation of the condition”unitary”on the monodromy group G in the case when z1,z2,z3are real with ordering z1

the four segments(?∞,z1),(z1,z2),(z2,z3),(z3,∞)onto four segments of circles,which we denote by I,II,III,IV.Moreover,the segments I and IV are tangent,due to the exponent di?erences zero at∞.The pairs(I,II),(II,III),(III,IV)intersect at right angles because of the local exponent di?erences1/2at the?nite singularities.In particular,let us choose y1 such that it is the holomorphic solution around z2whose powerseries has constant term1and let y2be the unique solution starting with(z?z2)1/2(1+···).Then a possible image of the Schwarz map looks

However,one should be careful with pictures like this.The above image corresponds to the situation when y1has no zeros on the intervals(z1,z2)and(z2,z3).When y1does have zeros on these intervals,the image of the Schwarz map may overlap itself several times.

Let S be the group generated by the complex re?ections in the four circles.Let S?S be the subgroup of index two of automorphisms of P1.Then it also well-known that the monodromy group G modulo scalars is precisely S.

Clearly there is a circle C with0as origin,which passes through the point of tangency between I,IV.The following Lemma is more or less obvious.

Lemma1.2The group G is unitary if and only if C is orthogonal to the circle segments I,IV.

The question of the orthogonality of circles like C was asked by F.Klein in[Kl1907].Based on the so-called oscillation theorems of Hilbert it can be established that there exists an in?nite, but discrete,set of real values of B for which C is indeed orthogonal.At that time this result was considered a possible step towards the solution of the uniformisation problem.

Very soon afterwards,E.Hilb,[Hi1909]established a similar theorem for di?erential equation with four singularities and general exponents.

Now that the uniformisation problem has been solved by methods from analysis,we can reverse the situation and show the existence of at least one B for which(1)has unitary monodromy. Let D be the hyperbolic disc and j:D→C\{z1,z2,z3}be the universal cover of P1minus the points z=z1,z2,z3,∞.Apply the Schwarzian derivative

S:η(z)→η

2 2

Then S (j ?1)is a rational function Q of z regular outside of z 1,z 2,z 3,∞and poles of order at most two.It is known that u +Qu =0is a Fuchsian equation of order two such that there exist two solutions u 1,u 2with the property that j (u 1/u 2)=z .Replacing u by 4z 3?g 2z ?g 3y ,the equation changes into an equation of type (1).The projectivised monodromy of this equation is precisely the covering group of our universal cover.

It is the purpose of this paper to rewrite the unitarity problem of G as a spectral problem in the space of real-analytic functions on C \{z 1,z 2,z 3}with the values of B as spectrum.Moreover,an asymptotic analysis of the spectrum indicates that eigenvalues to the problem occur in abundance.See Conjecture 5.1and Theorem 7.1.In particular we see that most eigenvalues B are complex numbers.

2Unitary groups

Let H be a Hermitean matrix with det(H )=0.We de?ne the corresponding unitary group U (H )by

U (H )={g ∈GL (2,C )|g t Hg =H }.

Notice that for any h ∈GL (2,C )we have

U (h t

Hh )=h ?1U (H )h.

We can conjugate the group U (H )in such a way that the corresponding Hermitian matrix is

either 1001 or 100?1

.In the ?rst case we call H ,and its conjugated versions,positive de?nite,in the second case inde?nite.

In this paper we shall deal with inde?nite Hermitean forms.In particular we take the form

H 0= 0i ?i 0

as standard Hermitean form.

Proposition 2.1The unitary group U (H 0)is the group generated by SL (2,R )and the diag-onal matrices λI 2with λ∈C ,|λ|=1.

Proof.Suppose g = a b c d

∈U (H 0).Then,from g t H 0g =H 0it follows that ac ?ac ad ?bc bc ?ad bd ?bd = 01?10

.The ?rst case we look at is when a =0.In that case we get ?bc =1and bd ∈R .Note that b =0.Choose β∈R and b ?∈C with |b ?|=1such that b =βb ?.From bd ∈R it follows that d is a real multiple of b .So d =δb ?where δ∈R .Similarly it follows from the ?rst equation that c =γb ?for some γ∈R .More particularly,?bc =1implies that ?βγ=1.So we see that g =b ? 0βγδ

with |b ?|=1and ?γβ=1.3

This proves our assertion when a=0.

Similarly we deal with the case d=0.Let us now assume that a,d=0.Put a=αa?with α∈R and|a?|=1.From ac?ac=0it follows that ac∈R.Hence c is a real multiple of a. Put c=γa?for someγ∈R.Similarly it follows from bd?bd=0that b is a real multiple of d,say b=λd withλ∈https://www.sodocs.net/doc/c05117240.html,stly it follows from ad?bc=1that ad(1?γλ)=1.We now conclude that d is a real multiple of a.Let us now put d=δa?.Then ad?bc=1implies that

1=(αδ?βγ)|a?|2=αδ?βγ.

Hence

g=a?

αβ

γδ

αδ?γβ=1

as asserted.qed

3Monodromy groups

In this section we gather some information on the monodromy representation corresponding to(1).Fix a base point z0in the complex plane and letΓ1,Γ2,Γ3be simple closed loops beginning and ending in z0which encircle respectively the points z1,z2,z3counter clockwise. Let M1,M2,M3be the corresponding local monodromy matrices.The local monodromy ma-trix M∞around in?nity is given by the relation M1M2M3M∞=Id.Note that the?nite local monodromies have eigenvalues1,?1,hence M21=M22=M23=Id and M1,M2,M3are re?ections.The matrix M∞has coinciding eigenvalues i,i or?i,?i and,consequently,trace ±2i.It cannot be a scalar since we always have logarithmic solutions around z=∞.So we conclude that M∞is a parabolic element(in this paper a scalar element is not considered to be parabolic).Let G be the group generated by these local monodromies.

A?rst remark we like to make is that G acts irreducibly on the space of solutions.Suppose on the contrary that G acts reducibly.Then M1,M2,M3,M∞have a common eigenvector v.Letλ1,λ2,λ3,λ∞be the corresponding eigenvalues.Their product should be one.But this is impossible sinceλ∞=i and the other eigenvalues are±1.So we conclude that G acts irreducibly.The irreducibility of G also implies that any G-invariant hermitian form is non-degenerate and uniquely determined up to scalars.

In the following Proposition we give necessary and su?cient conditions for the unitarity of the group generated by three involutions whose product is parabolic.

Proposition3.1Let P,Q,R∈GL(2,C)be re?ections(eigenvalues1,?1)and suppose that P QR is parabolic with trace±2i.Let G be the group generated by P,Q,R and denote by t M the trace of a2×2-matrix M.Then the following statements are equivalent

1.G is unitary

2.t P Q,t QR,t P R are real.

3.t P Q and t QR are real and satisfy(t2

P Q ?4)(t2

?4)≥16.

In the proof the following Lemma is useful.

Lemma3.2Let P,Q,R∈GL(2,C)be re?ections(eigenvalues1,?1).Then

t2P Q+t2QR+t2P R?t P Q t QR t P R=2+t(P QR)2.

Suppose in addition that P QR is parabolic with trace±2i and t P Q,t QR,t P R∈R.Then |t P Q|,|t QR|,|t P R|>2.

Proof.The identity

t2P Q+t2QR+t2P R?t P Q t QR t P R=2+t(P QR)2

can be proven by a straightforward computation.Suppose P QR is parabolic with trace±2i, the matrix(P QR)2is parabolic with trace?2.Hence

t2P Q+t2QR+t2P R?t P Q t QR t P R=0.

Supppose al three traces in this equation are real.Consider the equation as a quadratic

equation in t P Q.Then its discriminant should be≥0.This means,(t P R t QR)2?4(t2

P R +t2

)≥

0and hence

(t2P R?4)(t2QR?4)≥16.

Similar inequalities hold for any other pair of traces.The inequalities imply that either the absolute values of all traces are>2,as asserted,or that all traces are zero.

In the latter case we consider the group G generated by P,Q,R in more detail.Since the trace of P Q is zero and determinant1,the eigenvalues of P Q are±i.Hence(P Q)2=?Id. Using this and P2=Q2=Id we get P Q=?QP.Similarly for the other pairs.From this we conclude that G modulo scalars is an abelian group of order4.Hence G is a?nite group and P QR cannot be parabolic,since parabolic elements have in?nite order.qed Proof of Proposition3.1.From the proof of Lemma3.2we see that t P Q,t QR,t RP∈R

is equivalent to t P Q,t QR∈R and(t2

P Q ?4)(t2

?4)≥16.Hence it remains to show the

equivalence of(1)and(2).

First we prove(1)?(2).So suppose that G is unitary.Since G contains the parabolic element P QR the signature of the hermitian form should be(1,1).Without loss of generality we may then assume that the hermitian form is given by H0as de?ned in the previous section.Since P,Q,R are determinant?1matrices,it follows from Proposition2.1that they should be of the form iN,where N∈SL(2,R).Hence P Q,QR,P R∈SL(2,R)and so their traces are real.

Proof of(2)?(1).Suppose that t P Q,t QR,t P R are real.Since P QR is parabolic with trace ±2i,Lemma3.2tells us that all traces have absolute value>2.By conjugation we can see

to it that

?i0

Suppose that Q=

p q

r?p

.Choose a,b∈C such that a2+b2=1and(a2?b2)p+ab(q+r)=

0.Then conjugation by M=

a b

?b a

leaves P?xed and changes Q into

MQM?1=

0?2abp+a2q?b2r

?2abp?b2q+a2r0

Adopting this conjugation we get a new Q =

0q r 0

with q r =1.From the fact that t P Q ∈R it follows that i (q ?r )∈R .Put q =iλ.Substitute this in q r =1to ?nd that r =

?i/λ.Now i (q ?r )∈R implies that λ+1/λ∈R .We also know that |λ+1/λ|=|t P Q |>2.Hence λ∈R ,and we conclude

Q =i 0λ?1/λ0

∈iSL (2,R ).Now put R = p q r ?p

for some suitable p,q,r ∈C .Notice that t P R =i (q ?r )and t QR =i (pλ?r/λ).Solving for q,r gives

q =i t P R λ?t QR

λ?1/λr =i t P R /λ?t QR λ?1/λ.In particular we see that q,r are purely imaginary.The determinant of R is ?1.In other words,?p 2?qr =?1.Hence we ?nd

?p 2=qr ?1=?(t P R λ?t QR )(t P R /λ?t QR )(λ?1/λ)2

?1=?t 2P R +t 2QR ?t P Q t P R t QR (λ?1/λ)2

?1In the last line we used λ+1/λ=t P Q .We now use the trace identity of Lemma 3.2to ?nd

?p 2=

t 2P Q (λ?1/λ)2?1=(λ+1/λ)2(λ?1/λ)2?1=4(λ?1/λ)2

Hence p =

2i (λ?1/λ)=2i t 2P Q ?4.In particular p is purely imaginary.So we conclude that R is i times a matrix from SL (2,R ).Now we see that our normalised P,Q,R have the standard form H 0as common invariant form.

qed

4A spectral problem

We remind the reader of the concept of Wronskian determinant of a second order equation y +py +qy =0where p,q are analytic functions in a simply connected domain U ?C .Let

y 1,y 2be two independent solutions and consider W =y 1y 2?y 2y 1.It is easy to see that W satis?es the di?erential equation W =?pW ,hence W =αexp ? pdz for some non-zero

constant α.In particular W has no zeros or poles in U .

In what follows we shall also take into account real analytic solutions of our di?erential equa-tion.Since such functions are not holomorphic we must replace the complex di?erentiation

operator d

dz by its real counterpart.Let us therefore rewrite our second order equation in the

form

?2z y+p?z y+qy=0(2)

where

?z=1

In the following Proposition we show that the real solution space is a four dimensional R-vector space.

Proposition4.1Let U be as above.Let y1,y2be two independent complex analytic solutions of(2)in U.Then the R-vector space of real C2solutions in U is four dimensional and spanned by y1y1,Re(y1y2),Im(y1y2),y2y2.

We denote the real valued C2-functions on U by C2(U,R)and the complex-valued ones by C2(U,C).

Proof.Let u be any real C2-solution on U of(2).Choose A,B∈C2(U,C)such that

u=Ay1+By2

?z u=Ay 1+By 2

We do this by solving for A,B.We?nd

A=(y 2u?y2?z u)/W

B=(?y 1u+y1?z u)/W

In particular,A,B∈C2(U)since the Wronskian W has no zeros in U.A simple calculation shows that?z A=?z B=0,hence A,B are anti-holomorphic functions,i.e.A,B are holomor-phic.Let us rewrite(2)in the form My=0,where M=?2z+p?z+q.Note that Mu=0 since u is real.After substitution of u=Ay1+By2this yields

0=(MA)y1+(MB)y2.(3)

Since?z commutes with any holomorphic di?erential operator we get M?z u=?z Mu=0. Hence

0=(MA) 1+(MB) 2(4) Solving equations(3)and(4)yields MA=MB=0.So A and B are holomorphic functions satisfying My=0.Hence there exist complex constantsα,β,γ,δsuch that A=αy1+βy2 and B=γy1+δy2on U.We conclude that

u=αy1y1+βy1y2+γy2y1+δy2y2.

Hence we see that our solution space is spanned by the functions of our Proposition.

It remains to show that the four functions y i y j are C -linear independent.Suppose that there exist α,β,γ,δsuch that

0=αy 1y 1+βy 1y 2+γy 2y 1+δy 2y 2.

Apply the operator ?z ,

0=αy 1y 1+βy 1y 2+γy 2y 1+δy 2y 2.

These two inequalities together yield αy 1+βy 2=0and γy 1+δy 2=0.Since y 1,y 2are independent functions this implies α=β=γ=δ=0.Hence our functions are indeed independent.qed Proposition 4.2Let G be the monodromy group of the linear second order di?erential equa-tion y +py +qy =0where p,q ∈C (z ).Let S be the set of poles of p and q .Then G is unitary if and only if there exists a non-trivial real,C 2function f on C \S which is a solution of (2).Moreover,this function f is uniquely determined up to a constant factor.

Proof.Suppose ?rst that G is unitary.Hence there exists a non-trivial 2×2-matrix H so that H t =H and g t Hg =H for all g ∈G .The bar denotes complex conjugation and the superscript t denotes transposition of a matrix.Let y 0,y 1be any two independent solutions around a non-singular point z 0.Then we see that f =(y 1,y 2)H y 1y 2

is invariant under monodromy.Hence f can be extended globally throughout C \S .Moreover it is real-valued and real-analytic.Furthermore,f cannot be identically zero since the functions y i y j are linearly independent according to Proposition 4.1.Finally,since ?z y 1=?z y 2=0we see that f satis?es our equation (2).

Suppose conversely that we have a global real,C 2-function f satisfying equation (2).Choose a simply connected domain U ∈C \S and two independent complex analytic solutions y 1,y 2of (2),de?ned on U .In Proposition 4.1we have seen that there exist unique numbers α,β,γ,δsuch that

f =αy 1y 1+βy 1y 2+γy 2y 1+δy 2y 2.

Hence

f =(y 1,y 2)H

y 1y 2 where H = αβγδ

.Since f is real-valued,conjugation and transposition show that H t =H .Furthermore,f is globally de?ned,hence g t Hg =H for any g ∈G .

Finally the existence two independent global real solutions to 2would imply the existence of two independent G -invariant Hermitian forms.However this is impossible in view of the irreducible action of G .Therefore f is uniquely determined up to scalars.qed Let us now return to our equation (1)and its monodromy group G .Note that Proposition

4.2has now turned the unitary problem of G into the problem

Lf =Bf,

f ∈C 2(C \{z 1,z 2,z 3},R )(5)

where

L =P ?2z +(P /2)?z +z/48

The interesting point here is that we now restated our unitarity problem as an eigenvalue problem on C2(C\{z1,z2,z3}).In particular,the eigenspace for any dimension is at most one by the uniqueness(up to scalars)of f.It would be of great interest to determine the complete spectrum.

To initiate this study we rewrite equation(1)using elliptic functions.As is well known the elliptic curve E:y2=4x3?g2x?g3can be parametrised by a suitable Weierstrass?-function as follows,x=?(z),y=? (z).Replace z by?(z)in(1)to get

d2y dz2+

?(z)

y=0.(6)

We denote the period lattice of E byΛ.The Lam′e equation can thus be considered either as a di?erential equation in C with doubly periodic coe?cients and singularities inΛ,or a di?erential equation on the elliptic curve E with a single singularity at the point∞of E.In both cases the local exponents at the singularity read1/2,1/2.The fundamental group of E minus∞is the free group on two generators,where the generators are formed by a basisγ1,γ2 of the closed paths on E.In the covering space C→C/Λ=E these closed loops correspond to two periodsω1,ω2∈Λsay.The commutatorγ1γ2γ?11γ?12is the closed simple path around the point∞on E.Since the local exponents there are equal and1/2,the local monodromy matrix at∞is parabolic with trace?2.The monodromy group H of(6)is generated by the subgroup of G consisting of the determinant1elements.So,with the above notations H= M1M2,M2M3 .for example.In particular H has index two in G.

The spectral problem(5)can now be lifted to

?2z u+?(z)

u=Bu,u∈C2(C\Λ),u real?valued and even(7)

The extra condition that u is even(i.e.u(?z)=u(z))is a remnant of the fact that u is a pullback via the covering map of E from the original Lam′e spectral problem(5).

5Asymptotic analysis

Instead of looking at the spectral problem(7)we consider the more general version ?2z u?n(n+1)?(z)u=Bu,u∈C2(C\Λ),u real?valued and even(8) which corresponds to the general Lam′e equation with parameter n.We take n∈R.The problem(7)corresponds to the choice n=?1/2.

LetΛbe the lattice corresponding to the elliptic curve y2=4x3?g2x?g3and letω1,ω2be a Z-basis satisfying Im(ω2/ω1)>0.

Byη1,η2we denote the quasi-periods corresponding to the latticeΛ.They are de?ned by ηi=ζ(z+ωi)?ζ(z)whereζ(z)is the Weierstrassζ-function de?ned byζ (z)=??(z)and ζ(z)=?ζ(?z).We let?=(ω2ω1?ω1ω2)/2i.Note that?is real and positive because Im(ω2/ω1)>0.Moreover,?is the area of a fundamental paralellogram ofΛ.We also de?ne a=η1ω2?η2ω1.

In this section we provide evidence in the form of a perturbation calculation for the validity of the following conjecture.

Conjecture5.1Let notations be as above.Up to order1/|l0|the values of B in problem(7)

are given by

B=l20?n(n+1)

2i?

(9)

where l0∈Λ,the lattice generated byω2,ω1.

Let us?rst solve(8)when n(n+1)is replaced by0and B is replaced by l2for convenience. The real-valued solutions of?2z u=l2u are given by

Ae lz+lz+B(e lz?lz+e?lz+lz)+C

(e lz?lz?e?lz+lz)+De?lz?lz

with A,B,C,D∈R.The condition that u is even sees to it that C=0and A=D,so we get

u=A(e lz+lz+e?lz?lz)+B(e lz?lz+e?lz+lz).

If A=0we see that u cannot be periodic with respect toΛ.So A=0and periodicity of u now implies the existence of m1,m2∈Z such that

lω1?lω1=?2πim2

lω2?lω2=2πim1

Hence l=π(m2ω2+m1ω1)/?.

Let us now turn to an asymptotic study of the original problem(8)with B=l2when|l|→∞. We?rst?nd an approximate solution of the complex di?erential equation u ?n(n+1)?u=l2u. Put u=e lz+β(z)for someβ(z).We?nd,β +(β )2+2lβ ?n(n+1)?(z)=0.Now consider

the asymptotic expansion

β(z)=β1(z)

β2(z)

+···

After substitution into the di?erential equation and comparison of equal powers of1/l we can ?nd theβk recursively as follows,

2β 1?n(n+1)?=0

2β 2+β 1=0

2β 3+β 1β 2+β 2=0

···

We shall consider the second order approximation ofβ.Denote byζ(z)the Weierstrassζ-function,i.e.ζ (z)=??(z).Then,β1(z)=?n(n+1)ζ(z)/2.From the second equation we inferβ2(z)=?β 1(z)/2=?n(n+1)?(z)/4.

Let us?rst compute the spectrum up to order1/|l|.We perturb the solution e lz?lz+c.c.to

exp

lz?n(n+1)ζ(z)/2l?lz+/2l

+c.c.

Here c.c stands for complex conjugate.The periodicity conditions now imply the existence of m 1,m 2∈Z such that

lω1?lω1?n (n +1)η12l +n (n +1)η12l =?2πim 2+O (1/|l |2)(10)lω2?lω2?n (n +1)η22l +n (n +1)η22l =2πim 1+O (1/|l |2)(11)

where ηi =ζ(z +ωi )?ζ(z )for i =1,2.Put l 0=π(m 2ω2+m 1ω1)/?and l =l 0+ .Then,up to second order in 1/l 0,

ω1? ω1=n (n +1)η1/2l 0?c.c

ω2? ω2=n (n +1)η2/2l 0?c.c

Solution of yields

l =l 0?n (n +1)2i ? a l 0+b l 0 +O 1|l 0|2

where a =(η1ω2?η2ω1)and b =(η2ω1?η1ω2).Notice that b equals 2πi because of Legendre’s relation.We conclude,

B =l 2=l 20?n (n +1)2i ? a +2πi l 0l 0 +O 1|l 0| ,as conjectured.

In the Section 6we shall see that the agreement with numerical data in the case n =?1/2is remarkably well.In Section 7we give a proof with convergent series in the case when n =1.6Numerical data

Now let us take n =?1/2,which is the original problem.The generators of the monodromy of equation (1)can be computed numerically as follows.Fix a non-singular point a 0and construct three simple loops,each enclosing one of the ?nite singularities exactly once.Let γbe such a loop.We discretize γby choosing N points a 0,a 1,a 2,...,a N =a 0on γwhich are regularly spaced and whose increasing indices follow the orientation of γ.Now write equation

(1)as a ?rst order system

y =v

v =?P

2P v ?z/4?B 2P

y and solve this system numerically with a Runge-Kutta method (we used fourth order)using the points a i .As initial column vectors we chose (1,0)t and (0,1)t .After numerical integration

we obtain two column vectors (a,c )t and (b,d )t .The monodromy matrix corresponding to γ

is then approximated by a b c d

.We can now compute M 1,M 2,M 3to any accuracy we like,by increasing the number of in-terpolation points a i .In order to determine B such that the monodromy group is unitary

we use Proposition3.1and interpolation.For any choice of B we can compute the traces

t12:=t M

1M2and t23:=t M

2M3

.These are complex analytic functions of B.Choose a value

B0of B and a nearby value B1.The derivative of t12,t23at B0can be approximated by

λ=t12(B1)?t12(B0)

B1?B0

andμ=

t23(B1)?t23(B0)

B1?B0

The linear approximations of t12,t23at B=B0now read t12(B)=t12(B0)+λ(B?B0)and t23(B)=t23(B0)+μ(B?B0).We now solve x∈C from the system of equations

Im(t12(B0)+λx)=0,Im(t23(B0)+μx)=0.

A brief computation gives us

x=μIm(t12(B0))?λIm(t23(B0))

Im(λμ)

Supposedly the value B0+x should be a closer approximation to a spectral value of B than B0.We then repeat the argument with the new value of B.In practice this turns out to work very well.After we computed a spectral value to high enough order of precision it remains to

check that the third trace t M

1M3is also real.

In the following we carry out the computations described above for a number of values of g2,g3,m,n and compare them with the asymptotic approximations given by(9).

First we take g2=4,g3=0.This corresponds to the elliptic curve y2=4x3?4x which has a square lattice.In this case equation(1)has an extra symmetry with respect to x→?x,B→?B.Hence the spectrum of(7)has the symmetry B→?B.An obvious spectral value is B= 0.In that case we can write down an explicit solution for(1),namely2F1(1/8,1/8,3/4;z2). It is well-known that second order di?erential equations for hypergeometric functions have triangle groups as monodromy groups,and triangle groups are unitary.

The periods read

ω1=2

√

x3?x

=2.62206···

andω2=iω1=2.62206i.The quasi-periods read

η1=?2

xdx

√

x3?x

=1.19814···

andη2=?iη1=?1.19814i.We check thatη1ω2?η2ω1=0andπ/4?=0.114237.Hence (9)yields the approximated eigenvalues

1.43554(m+ni)2+0.114237m+ni m?ni

Here is a table with some asymptotic and numerical eigenvalues for the eigenvalue problem (7)with g2=4,g3=0.In addition we list the traces of A=M1M2,B=M1M3,C=M2M3.

m n numerical asymptotic t A t B t C

value value

10 1.5526 1.5497-2.004629.5424-29.6103

20 5.8568 5.8564 2.00001606.12606.123

11 2.9823i 2.9853i-26.191-26.191683.914

3013.034313.0341-213463.8-13463.8

21 4.3752+5.8326i 4.3751+5.8335i24.3597-591.325-14380.1

12-4.3752+5.8326i-4.3751+5.8335i-591.32524.3597-14380.1

3111.5758+8.6814i11.5757+8.6817i-23.7666-13352.8316787

2211.5978i11.5986i569.847569.847324724

13-11.5758+8.6814i-11.5757+8.6817i-13352.8-23.7665316787

327.22167+17.3315i7.22164+17.3319i-556.42713125.2-7303200

23-7.22167+17.3315i-7.22164+17.3319i13125.2-556.426-7303200

3325.9536i25.954i-12917.7-12917.7166867000

7The case n=1

In the previous section we have considered the parameter choice n=?1/2in our Lam′e equation.We shall now concentrate on the case n=1.In that case we can write down explicit solutions and thus provide evidence for the correctness of the asymptotic analysis we carried out for general n and in particular n=?1/2.

So we consider the equation

d2y

?(2?(z)+B)y=0.

Choose a such that B=?(a)and let us suppose that2a is not in the period lattice.In that case two independent solutions of the Lam′e equation are given by f(z),f(?z)where

f(z)=eζ(a)z σ(z?a)σ(z)

andσ(z)is the Weierstrassσ-function de?ned by

σ(z)=z

ω∈Λ

exp

2ω2

It is a holomorphic function on C with zeros in the lattice points ofΛ.Furthermore,for any ω∈Λ,

σ(z+ω)=±eη(ω)(z+ω/2)σ(z)

where we use+ifω/2∈Λand?if not.It follows from this functional equation that

f(z+ω)=eζ(a)ω?aηf(z).

Notice by the way that f(z)is independent of the choice of a moduloΛ.We now study for which values of a,and hence B,the Lam′e equation allows a non-trivial real valuedΛ-periodic solution,symmetric in z.

A basis for the real valued symmetric solutions is given by

U(z)=f(z)f(?z)+f(?z)f(z),V(z)=|f(z)|2+|f(?z)|2

Suppose that f(z+ω)=λf(z).Then f(?z?ω)=λ?1f(?z).TheΛ-periodic linear combi-nations of U(z),V(z)are either U(z)itself ifλ=λfor allω∈Λor V(z)if|λ|=1for allω. Suppose f(z+ωj)=λj f(z)for j=1,2.Then the?rst case corresponds toλ1,λ2∈R and the second to|λ1|=|λ2|=1.

Let us consider the second case?rst.We have necessarily

ζ(a)ω1?η1a=?2πir2,ζ(a)ω2?η2a=2πir1

with r1,r2∈R.Solving this for a andζ(a)we get

ζ(a)=r1η1+r2η2,a=r1ω1+r2ω2.

Hence we must solve

ζ(r1ω1+r2ω2)=r1η1+r2η2

in r1,r2∈R.Notice thatζ(x1ω1+x2ω2)?x1η1?x2η2is periodic in x1,x2with period1, so we can restrict ourselves to?nding solutions with0≤r1,2<1.We expect at most?nitely many solutions.

Let us now consider the case whenλ1,λ2∈R.This implies that there exist integers m1,m2 such that

ζ(a)ω1?ζ(a)ω1?aη1+aη1=?2πm2

ζ(a)ω2?ζ(a)ω2?aη2+aη2=2πm1

We solve this equation recursively in a for large values of|m1|+|m2|.The observation is that a should be close to a lattice point,which we can take to be0.So let us put l=ζ(a)and notice that a=1/l+O(1/|l|2).So our equation can be rewritten as

lω1?lω1?η1

η1

=?2πim2+O(|l|?2)

lω2?lω2?η2

η2

=2πim1+O(|l|?2)

Notice that this precisely the problem(10)for n=1.The di?erence is now that we have convergence instead of asymptotic approximation.Our conclusion follows by noticing that B=?(a)=ζ(a)2+O(1/|ζ(a)|)=l2+O(1/|l|)for a very close to0.So we have,

Theorem7.1Conjecture5.1is true when n=1.

8References

[Be2002]Beukers,On Dwork’s accessory parameter problem,Math.Z.241(2002),425-444. [Hi1908]E.Hilb,¨Uber Kleinsche Theoreme in der Theorie der linearen Di?erentialgleichungen, Math.Ann.66(1909),215-257.

[Kl1907]F.Klein,Bemerkungen zur Theorie der linearen Di?erentialgleichungen zweiter Ord-nung,Math.Ann.64(1907),175-196.

[KRV1979]L.Keen,H.E.Rauch,A.T.Vasquez,Moduli of punctured tori and the accessory param-eter of Lam′e’s equation,Trans.AMS255(1979),201-230.

[MSW2002]D.Mumford,C.Series,D.Wright,Indra’s Pearls,The vision of Felix Klein,Cambridge University Press2002.

[Ne1949]Z.Nehari,On the accessory problem of a Fuchsian di?erential equation,Amer.J.Math.

71(1949),24-39.

[WW1980]E.T.Whittaker,G.N.Watson,A course on Modern Analysis,4th edition,Cambridge University Press,1980.

Unitary monodromy of Lamé differential operators F.Beukers

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