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FINITE TYPE MINIMAL ANNULI IN S2 ⇥ R

L. HAUSWIRTH, M. KILIAN, AND M. U. SCHMIDT

Abstract. We study minimal annuli in S2⇥R of finite type by relatingthem to harmonic maps C ! S2 of finite type. We rephrase an iterationby Pinkall-Sterling in terms of polynomial Killing fields. We discussspectral curves, spectral data and the geometry of the isospectral set.We consider polynomial Killing fields with zeroes and the correspondingsingular spectral curves, bubbletons and simple factors. We investigatethe di↵erentiable structure on the isospectral set of any finite type mini-mal annulus. We apply the theory to a 2-parameter family of embeddedminimal annuli foliated by horizontal circles.

Introduction

In the last decade there has been an interest in extending minimal surfacetheory to the target space S2 ⇥R [27, 20, 21, 12, 15]. A minimal surface thatis conformally immersed in S2⇥R is essentially described by a harmonic mapG : ⌦ ⇢ C ! S2. There is an important subclass of such harmonic maps thathave an algebraic description, the so-called harmonic maps of finite type. Forexample, all constant mean curvature (cmc) tori in R3, S3, H3 have harmonicGauss maps of finite type [14, 24]. The description of conformally immersedproper minimal annuli of finite type is analogous to the well known theory ofconstant mean curvature tori of finite type [14, 24, 2, 4, 9, 10, 7, 23]. Thusperiodic minimal immersions X : C ! S2 ⇥ R of finite type can be describedby algebraic data. Up to some finite dimensional and compact degrees offreedom the immersion is determined by the so-called spectral data (a, b).This consists of two polynomials of degree 2g respectively g + 1 for someg 2 N. The polynomial a(�) encodes a hyperelliptic Riemann surface calledthe spectral curve. The genus of the spectral curve is called spectral genus .The other polynomial b(�) encodes the extrinsic closing conditions. Thiscorrespondence is called the algebro-geometric correspondence. In particular,we characterize those algebraic curves which are the spectral curves of minimalannuli in S2 ⇥ R. The spectral data is the starting point for a deformationtheory to be used in [13].

1991 Mathematics Subject Classification. 53A10.L. Hauswirth was partially supported by the ANR-11-IS01-0002 grant.

1

2 L. HAUSWIRTH, M. KILIAN, AND M. U. SCHMIDT

Let us briefly outline this paper. In the first section we provide a shortintroduction to the local theory of minimal surfaces in S2⇥R and prove a Sym-Bobenko type formula. In the second section we discuss polynomial Killingfields, how these are related to the Pinkall-Sterling iteration and define thespectral curve. In the third section we invoke the Iwasawa factorization of theunderlying loop groups, mention the generalized Weierstrass representation,and briefly discuss the Symes method [29, 5, 6]. In the fourth section wetreat the spectral curve, the isospectral action, and prove some properties ofthe isospectral set. In particular we discuss singular spectral curves and howto remove singularities with the isospectral action. The isospectral set I(a)of a minimal annulus consists of all minimal annuli with the same spectraldata (a, b). We decompose I(a) with the help of a group action into orbitsand identify each orbit smoothly with a commutative Lie group. These orbitshave di↵erent dimensions. The lower dimensional orbits are in the closures ofthe higher dimensional ones. Our main results are

(1) The isospectral sets I(a) are compact.(2) If the 2g-roots of a are pairwise distinct, then there is one orbit dif-

feomorphic to

I(a) ⇠= (S1)g.

(3) If a has a double root ↵0 2 S1, then there is a di↵eomorphism

I(a) ⇠= I(a) with a(�) = ↵0(�� ↵0)2a(�).

(4) If a has a double root ↵0 /2 S1 there are two invariant submanifoldsdi↵eomorphic to

I(a) ⇠= I(a) [ G⇥ I(a) with G = C or C⇥

and a(�) = (�� ↵0)2(1� �↵0)

2a(�) .

In the fifth section we turn to periodic finite type harmonic maps, spectraldata and how the closing conditions are encoded in the spectral curve. Insection 6 we encounter bubbletons, simple factors and prove a factorizationtheorem for polynomial Killing fields with zeroes. In particular we show thatchanging the line in the simple factor preserves the period, and that for somechoices of lines the nodal singularity on the spectral curve disappears. In thelast section we compute the spectral data of minimal annuli in S2⇥R that arefoliated by circles. These low spectral genus examples will play an importantrole in a forthcoming paper [13]. This paper presents a mostly self-containedaccount of the integrable systems approach to minimal surfaces in S2 ⇥ R offinite type, but may also serve as a companion for other integrable surfacegeometries.

FINITE TYPE MINIMAL ANNULI IN S2 ⇥ R 3

1. Minimal surfaces in S2 ⇥ R

We study conformal minimal immersions X : ⌦ ⇢ C ! S2 ⇥ R where ⌦ isa simply connected domain of C. We write X = (G, h) for G : ⌦ ! S2 andh : ⌦ ! R, and call G the horizontal, and h the vertical components of X.If we denote by (C,�2(u)|du|2) the complex plane with the metric induced bythe stereographic projection of S2 (�2(u) = 4/(1+ |u|2)2), the map G : ⌦ ! Csatisfies

Gzz

+2G

1 + |G|2Gz

Gz

= 0 .

The holomorphic quadratic Hopf di↵erential associated to the harmonic mapG is given by

Q(G) = (� �G(z))2Gz

Gz

(dz)2 := �(z)(dz)2 .

The function � depends on z, whereas Q(G) does not.By conformality the induced metric is of the form ds2 = ⇢2(z)|dz|2, and

writing z = x+ y the partial derivatives satisfy |Xx

|2 = |Xy

|2 and Xx

? Xy

.Conformality reads

|Gx

|2�

+ (hx

)2 = |Gy

|2�

+ (hy

)2 and hGx

, Gy

i�

+ (hx

)(hy

) = 0 ,

hence (hz

)2(dz)2 = �Q(G). The zeroes of Q are double, and we can define ⌘as the holomorphic 1-form ⌘ = ±2i

pQ. The sign is chosen so that

h = Re

Z⌘ .

The unit normal vector n in S2 ⇥ R has third coordinate

hn, @

@t

i = n3 =|g|2 � 1

|g|2 + 1where g2 := �G

z

Gz

.

We define the real function ! : C ! R by

n3 := tanh! .

We express the di↵erential dG independently of z by

dG = Gz

dz +Gz

dz =1

2� �Gg�1⌘ � 1

2� �Gg ⌘ ,

and the metric ds2 is given in a local coordinate z by

ds2 = (|Gz

|�

+ |Gz

|�

)2|dz|2 =1

4(|g|�1 + |g|)2|⌘|2 = 4 cosh2 !|Q| .

We remark that the zeroes of Q correspond to the poles of !, so that theimmersion is well defined. Moreover the zeroes of Q are points where thetangent plane is horizontal. The Jacobi operator is

L =1

4|Q| cosh2 !�@2x

+ @2y

+Ric(n) + |dn|2�


and can be expressed in terms of Q and ! by

(1.1) L =1

4|Q| cosh2 !

✓@2x

+ @2y

+ 4|Q|+ 2|r!|2cosh2 !

◆.

Since n3 = tanh! is a Jacobi field obtained by vertical translation in S2 ⇥R,we have L tanh! = 0 and

�! + 4|Q| sinh(!) cosh(!) = 0 ,

where � = @2x

+ @2y

is the Laplacian of the flat metric.

Minimal annuli. Consider a minimal annulus A properly immersed in S2 ⇥R. If A is tangent to a horizontal plane {x3 = t}, the set A\ {x3 = t} defineson A a set of analytic curves with isolated singularities at points where thetangent plane of A is horizontal. Near such a singularity q, there are 2k + 2smooth branches meeting at equal angles, for some integer k � 1.

Annuli are transverse to every horizontal plane S2 ⇥ {t}. To see that weclaim that A \ {x3 = t} defines at least three connected components. Atmost two of them are non-compact because the immersion is proper. Hencethere is a compact disk in A with boundary in {x3 = t}, a contradiction to themaximum principle. To prove the claim consider A1, A2, ..., A2n distinct localcomponents at q of A \ {x3 = t}. We know that the A

i

alternate between{x3 � t} and {x3 t}. If A1 and A3 are not in the same component of{x3 � t}, then A2 yields a third component and A \ {x3 = t} has at leastthree connected components. If A1 and A3 are in the same component of{x3 � t} we can construct a cycle ↵13 in {x3 � t} which meets S2 ⇥ {t} onlyat q. We consider ↵0 in {x3 > 0} joining a point x of A1 and y of A2. Thenjoin x to y by a local path ↵1 in A going through q. Let ↵13 = ↵0 \ ↵1. IfA2 and A4 were in the same component we could find a cycle ↵24 on A whichmeets ↵13 in a single point, which is impossible since the genus of A is zero.If A2 and A4 are not in the same component of {x3 t}, then A1 yields athird component.

Hence properly immersed annuli are transverse to horizontal planes andthe third coordinate map h : A ! R is a proper harmonic function on eachend of A with dh 6= 0. Then each end of A is parabolic and the annuluscan be conformally parameterized by C/⌧Z. We will consider in the followingconformal minimal immersions X : C ! S2 ⇥ R with X(z + ⌧) = X(z).

Since dh 6= 0, the Hopf di↵erential Q has no zeroes. If h⇤ is the harmonicconjugate of h, we can use the holomorphic map (h + h⇤) : C ! C toparameterize the annulus by the conformal parameter z = x + y. In thisparametrization the period of the annulus is ⌧ 2 R and

X(z) = (G(z), y) with X(z + ⌧) = X(z) .


We say that we have parameterized the surface by its third component. Weremark that Q = 1

4 (dz)2 and ! satisfies the sinh-Gordon equation

(1.2) �! + sinh(!) cosh(!) = 0 .

Remark 1.1. We relax the condition ⌧ 2 R⇥ to ⌧ 2 C⇥, but we will param-eterize our annuli conformally such that Q = � dz2 is constant, independentof z and 4|�| = 1. This means that the third coordinate is linear.

In summary we have proven the following

Theorem 1.2. A proper minimal annulus is parabolic and X : C/⌧Z ! S2⇥Rhas conformal parametrization X(z) = (G(z), h(z)) with

(1) Harmonic map G : C/⌧Z ! S2, and h(z) = Re(� e ⇥/2z).(2) Constant Hopf di↵erential Q = 1

4 exp( ⇥) dz2.

(3) The metric of the immersion is ds2 = cosh2(!) dz ⌦ dz.(4) The third coordinate of the unit normal vector is n3 = tanh!.(5) The function ! : C/⌧Z ! R is a solution of (1.2).

Sym-Bobenko type formula. We use a description of harmonic maps intothe symmetric space S2 in terms of SU2-valued frames.

Identify su2⇠= R3,

�w u+ v

�u+ v � w

� ⇠= (u, v, w), so S2 ⇢ R3 consists of length

kXk =pdetX = 1 elements in su2 and hX1, X2i = � 1

2 tr�X1X2

�. Pick

�3 =�

00 �

� 2 S2 .

Let T denote the stabiliser of �3 under the adjoint action of SU2 on su2, sothat S2 ⇠= SU2/T. If ⇡ : SU2 ! SU2/T is the coset projection, then a mapF : ⌦ ⇢ C ! SU2 with G(z) = ⇡ � F (z) is called a frame of G, and we haveG(z) = F (z)�3F�1(z).

Harmonic maps come in S1-families (associated families), associate to thesame solution of the sinh-Gordon equation. Here � 2 S1 parameterizes anassociated family of harmonic maps, and the frame of such an associatedfamily is called an extended frame.

A method to obtain an extended frame is to write down an integrable1-form which integrates to an extended frame by solving the integrabilitycondition. We present a result which is similar to results in Bobenko [3]. Wenext specify for a real solution of the sinh-Gordon equation (1.2) a minimalsurface in S2 ⇥ R with a particularly simple vertical component.

Theorem 1.3. Let ! : C ! R be a solution of the sinh-Gordon equation.Let �, � 2 S1 be arbitrary but fixed, and F

�

(z) the solution of F�1�

dF�

=↵(!), F

�

(0) = where

(1.3) ↵(!) =1

4

✓2!

z

��1�e!

�e�! �2!z

◆dz +

1

4

✓�2!z

�e�!

�� e! 2!z

◆dz .


Then the map X�

(z) = (F�

(z)�3 F�

(z)�1, Re(�p��1z)) with � 2 S1

defines an associate family of conformal minimal immersions X�

: C ! S2⇥Rwith metric

ds2 = cosh2 ! dz ⌦ dz ,

and Hopf di↵erential

Q = 14 � � �

�1 dz2 .

Remark 1.4. Reality condition. For � 2 S1, ↵�

= ↵(!) in (1.3) takesvalues in su2, so F

�

takes values in SU2. For general � 2 C⇥ we have F�

2SL2(C). From the relation ↵1/�

t = �↵�

, the solution of F�1�

dF�

= ↵(!)satisfies the reality condition

(1.4) F1/�

t

= F�1�

.

Proof. Decomposing ↵(!) = ↵0 dz + ↵00 dz into (1, 0) and (0, 1) parts, wecompute

↵0z

=1

4

✓2!

zz

��1!z

�e!

� �!z

e�! �2!zz

◆,

↵00z

=1

4

✓ �2!zz

� �!z

e�!

�!z

�e! 2!zz

◆.

Using F�1�

dF�

= ↵ in the integrability condition (F�

)zz

= (F�

)zz

gives ↵0z

�↵00z

= [↵0, ↵00], and a direct computation shows this is equivalent to the sinh-Gordon equation (1.2). Hence we can integrate dF

�

= F�

↵ to obtain a mapF�

: C ! SL2(C) and define G�

= F�

�3 F�1�

. The vertical component ish = 2 (�(��)1/2��1/2z + (��)�1/2�1/2z. Its partial derivatives of are h

z

=

� 2 (��)1/2��1/2 and h

z

= 2 (��)�1/2�1/2. Then

h(X�

)z

, (X�

)z

i = � 12 tr

�[↵0, �3]

2�+ (h

z

)2 = 0 ,

h(X�

)z

, (X�

)z

i = � 12 tr

�[↵00, �3]

2�+ (h

z

)2 = 0 ,

and the conformal factor computes to

2h(X�

)z

, (X�

)z

i = 2h[↵0, �3], [↵00, �3]i+ 2 (h

z

)(hz

)

= 12 cosh(2!) +

12 = cosh2(!) .

As Q = h(G�

)z

, (G�

)z

i (dz)2 = � 12 tr

�[↵0, �3]2

�(dz)2 = 1

4��1 (dz)2 is holo-

morphic, we conclude that G is harmonic. ⇤

Remark 1.5. Isometric normalisation 1. By conformal parametrizationwe can choose 4|�| = 1 (the annulus is transverse to horizontal planes). Wehave constants �, � 2 S1 which are related to the Hopf di↵erential, namely4Q = ��1 dz2. We can normalize the parametrization with � = 1 and a


constant |�| = 1. For a given extended frame F�

which satisfies the equationF�1�

dF�

= ↵(!), F�

(0) = , we consider the U(1)-gauge

(1.5) g(�) =

✓�1/2 00 ��1/2

◆2 T .

Then F�

= g(�)�1F�

g(�) induces the immersion X which di↵ers from X by

a rotation in S2. The third coordinate h = h is preserved while G�

(z) =F�

�3F�1�

= g(�)�1G�

(z)g(�) and for ↵(!) as in (1.3) with � = 1 and � = ��have

F�1�

dF�

= g(�)�1↵(!)g(�) .

2. Polynomial Killing fields and spectral curves

We explain in this section how solutions of the sinh-Gordon equation giverise to polynomial Killing fields as solutions of a Lax equation. PolynomialKilling fields in turn define spectral curves, which are hyperelliptic Riemannsurfaces.

If ! is a solution of the sinh-Gordon equation, we consider a deformation!t

= ! + t u + O(t2). If !t

is a one parameter family of solutions of thesinh-Gordon equation, then the variational function u : C ! R satisfies thelinearized sinh-Gordon equation

(2.1) �u+ u cosh(2!) = 0 .

Definition 2.1. A solution ! of the sinh-Gordon equation is of finite type

if there exist g 2 N such that

(2.2) ��

(z) =��1

4

✓0 e!

0 0

◆+

gX

n=0

�n✓

un

(z) e!⌧n

(z)e!�

n

(z) �un

(z)

◆

is a solution of the Lax equation

(2.3) d��

= [��

, ↵(!)]

for some functions un

, ⌧n

,�n

: C ! C, and some � 2 S1, and � = 1 in ↵(!).

Proposition 2.2. Suppose ��

is of the form (2.2) for some arbitrary ! : C !R, and that �

�

solves the Lax equation (2.3) with ↵(!) as in (1.3), � = 1 and|�| = 1. Then:

(1) The function ! is a solution of the sinh-Gordon equation (1.2).(2) The functions u

n

are solutions of the linearized sinh-Gordon equation(2.1).

(3) The following iteration gives a formal solution of d��

= [��

, ↵(!)].Let u

n

, �n

, ⌧n�1, with u

n

solution of (2.1) be given. Now solve thesystem

⌧n;z = 1

2 � e�2!un

, ⌧n;z = 4 � !

z

un;z � 2 � u

n;zz


for ⌧n;z and ⌧

n;z. Then define un+1 and �

n+1 by

un+1 = �2 ⌧

n;z � 4 !z

⌧n

, �n+1 = � e2!⌧

n

+ 4 � un+1;z .

(4) Each ⌧n

is defined up to a complex constant cn

, so un+1 is defined up

to �4 cn

!z

.

(5) u0 = !z

, ug�1 = c!

z

for some c 2 C, and �g�1/�

t

also solves (2.3).

Proof. Inserting (2.2) into (2.3) and comparing coe�cients yields

4un;z + e2!�

n+1 � �⌧n

= 0 ,(2.4a)

4un;z + ��

n

� e2!⌧n�1 = 0 ,(2.4b)

4!z

⌧n

+ 2⌧n;z � u

n+1 = 0 ,(2.4c)

2e!⌧n;z � �e�!u

n

= 0 ,(2.4d)

2e!�n;z + �e�!u

n

= 0 ,(2.4e)

4!z

�n

+ 2�n,z

+ un�1 = 0 .(2.4f)

(1) Solving (2.4b) for �n+1, (2.4c) for u

n+1, (2.4d) for ⌧n;z, and inserting

these, and un+1;z and ⌧

n;zz into (2.4a) gives e2! �⌧n

(16!zz

�e�2!+e2!) = 0,which implies (1.2) if ⌧

n

6= 0.(2) @(2.4a)� 1

2 e2!(2.4f) + 12 �e

�!(2.4d) reads

4un;zz +

12un

�e2! + ��e�2!

�= 0 .

(3) The equation for ⌧n;z is (2.4d). Taking the z-derivative of (2.4a) and using

(2.4a), (2.4c) and (2.4e) gives ⌧n;z = 4 �!

z

un;z � 2 �u

n;zz. The equations forun+1, �n+1 are given by (2.4c) respectively (2.4b).

(4) In the iteration (3) the function ⌧n

is determined up to an integrationconstant. This gives an additional term !

z

in un+1.

(5) Left to the reader. ⇤

Pinkall-Sterling [24] constructed a series of special solutions of the inductionof Proposition 2.2 (3) via an auxiliary function � as follows: For a givensolution u

n

of the linearized sinh-Gordon equation (2.1), consider the function� : C ! C defined by

(2.5) �n;z = 4!

z

un;z , �

n;z = �un

sinh! cosh! .

Then ⌧n

= 2 ��12�n � u

n;z

�and u

n+1 := (un

)zz

� !z

�n

. This defines ahierarchy of solutions of (2.1). Applying this iteration to the trivial solutionu0 ⌘ 0 yields the sequence, whose first four terms are

u0 = 0 ,

u1 = !z

,

u2 = !zzz

� 2!3z

,

u3 = !zzzzz

� 10!zzz

!3z

� 10!2zz

!z

+ 6!5z

.


This infinite sequence produces solutions of the linearized sinh-Gordon equa-tion on C. These come from the iteration (2.5), and Pinkall-Sterling prove inProposition 3.1 [24], that �

n

depends only on ! and its k-th derivatives withk 2n + 1. The fact that we consider on C a uniformly bounded solutionof the sinh-Gordon equation ! : C ! R, implies by Schauder estimates thateach u

n

is uniformly bounded on C.

Proposition 2.3. A proper minimal annulus A immersed in S2 ⇥ R withbounded curvature and linear area growth has a metric ds2 = cosh2 !|ds|2,where ! : C/⌧C ! R is a finite type solution of the sinh-Gordon equation.We say that the annulus is of finite type.

Proof. A first step is to prove that the function ! : C/⌧C ! R is uniformlybounded. Consider a sequence of points p

n

in A such that !(pn

) is divergingto infinity and consider a sequence of translations t

n

e3 such that A + tn

e3is a sequence of annuli with p

n

+ tn

e3 points of S2 ⇥ {0}. Then by thebounded curvature hypothesis there is a sub-sequence converging locally toA0, a properly immersed minimal surface. The linear area growth assumptionassure that A0 is an annulus. But our hypothesis leads to a pole occurringat the height t = 0 since |!| ! 1. The limit normal vector n3(pn) =tanh!

n

(pn

) ! ±1 and the annulus A0 would be tangent to the height S2⇥{0},a contradiction to the maximum principle. Thus

supz2A

|!| C0 .

Now we apply Schauder estimates to the sinh-Gordon equation to obtain aCk,↵ estimate on the solution of the sinh-Gordon equation on C/⌧Z. Thereexists a constant C0 > 0 such that for any k 2 N

|!|A,k,↵

C0 .

Meeks-Perez-Ros [19] provide us with the following

Theorem 2.4. [19] An elliptic operator Lu = �u+ q u on a cylinder S1 ⇥Rhas for bounded and continuous q a finite dimensional kernel on the space ofuniformly bounded C2 -functions on S1 ⇥ R.

Since solutions u0, u1, u2... of the linearized sinh-Gordon equation are so-lutions depending only on ! and its higher derivatives, this family is a finitedimensional family by Theorem 2.4. Thus there is a g 2 N, and there existai

2 C such thatgX

i=1

ai

ui

= 0 .

This algebraic relation implies that ! is of finite type, and ensures the exis-tence of a polynomial Killing field �

�

of degree g. To achieve that, one hasto prescribe the right constants c0, c1, . . . , cg in the iteration procedure, andset the g + 1-coe�cient to zero. ⇤


Potentials and polynomial Killing fields. To parameterize real solutionsof the sinh-Gordon equation we make the following

Definition 2.5. The set of potentials is

Pg

= { ⇠�

=gX

d=�1

⇠d

�d | ⇠�1 2 �0 R+

0 0

�, tr(⇠�1⇠0) 6= 0,

⇠d

= �⇠g�1�d

t

2 sl2(C) for d = �1, . . . , g }.Remark 2.6. Isometric normalization 2. For � 2 S1, we denote by P

g

(�),potentials with residues

⇠�1 2 �0 �R+

0 0

�.

These correspond to the normalization of remark 1.4 and there is an isomor-phism P

g

(�) ! Pg

, given by

⇠�

7! g(�)�1⇠�

g(�) .

Each ⇠�

2 Pg

satisfies the reality condition

(2.6) �g�1⇠1/�t

= �⇠�

.

In other words, for ⇠�

2 Pg

, we have a map S1 ! su2,� 7! �1�g

2 ⇠�

. Thepolynomial

a(�) := �� det ⇠�

then satisfies the reality condition

(2.7) �2ga(1/�) = a(�) .

On su2 the determinant is the square of a norm, thus we have for � 2 S1 that

(2.8) ��g a(�) 0 for � 2 S1 .

When g is even, ⇠0, ..., ⇠ g

2�1 are independent 2⇥2 traceless complex matrices.

For odd g, ⇠0, ..., ⇠ g�32

are independent 2 ⇥ 2 traceless complex matrices and

⇠ g�12

2 su2. Thus the space of potentials Pg

of real finite type solutions of the

sinh-Gordon equation is an open subset of a 3g + 1 dimensional real vectorspace. The condition tr(⇠�1⇠0) 6= 0 implies that a(0) 6= 0 and by symmetry thehighest coe�cient of a is therefore non-zero. Thus � 7! a(�) is a polynomialof degree 2g with complex coe�cients, and we denote such by C2g[�]. Define

Mg

= {a 2 C2g[�] | a(�) = �� det ⇠�

with ⇠�

2 Pg

}= {a 2 C2g[�] | a(0) 6= 0,�2ga(1/�) = a(�)

and ��g a(�) 0 for � 2 S1} ,M0

g

= {a 2 Mg

| ��ga(�) < 0 for |�| = 1} .

(2.9)

Thus M0g

is an open subset of the 2g + 1 dimensional real vector space Mg

.


Definition 2.7. Polynomial Killing fields are maps ⇣�

: C ! Pg

(see Defini-tion 2.5) which solve the Lax equation

d⇣�

= [ ⇣�

, ↵(!) ] with ⇣�

(0) = ⇠�

2 Pg

.

We use solutions ��

of the Lax equation to construct polynomial Killingrelated to a finite type solution ! : C ! R of the sinh-Gordon equation.

Lemma 2.8. For a solution ��

of (2.3) with � = 1 in ↵(!) (see formula(1.3)), there exists constants � 2 S1 and k 2 R+ such that

⇣�

(z) = k��

(z)� k�g�1 �1/�(z)t

and ⇣�

(0) = ⇠�

is a polynomial Killing field.

Proof. The map ⇣�

satisfies the reality condition (2.6) so that

�g�1⇣1/�(z)t

= �⇣�

(z).

It remains to prove that the residues ⇣�1 and ⇠�1 are upper triangular withpurely imaginary non-zero coe�cient and trace(⇠�1⇠0) 6= 0. We use the fol-lowing

Remark 2.9. Isometric normalization 3. We write ↵�,�,�

(!) := ↵(!) forthe 1-form (1.3) and �

�,�,�

(z) := ��

(z) the associate solution (2.2) of the Laxequation (2.3). We use the unitary matrix g(�) defined in (1.5). Now �

��,1,�

solves (2.3) with ↵��,1,�(!) = ↵

�,�

�1,�

(!), so g(�)��,1,�g(�)�1 solves (2.3)

with ↵(!) = g(�)↵�,�

�1,�

g(�)�1 = ↵�,1,��1

�

. We conclude that

��,1,��1

�

= g(�)��,1,�g(�)

�1 .

In particular, if ⇠�

2 Pg

(�), then a solution of the Lax equation of definition2.7 satisfies ⇣

�

: C ! Pg

(�). This can be deduced from proposition 2.2, where⌧�1(z) = ⌧�1(0).

The remark 2.9 proves that changing � by ��1� changes the highest coef-ficient of �

�

by

�g

(��1�) = �g�1�g

(�) .

We compute the residue ⇠�1 at z = 0 with � = ��1�, and choose a unimodularnumber �, such that

14 e! � �

g

(�)e! = 14 e! � �1�g�

g

(�)e! 2 R+ .

We choose k to normalize the residue with k�1 = 1 + 4 �1�g�g

(�) 2 R+.There remains to compute with �0 = �e�2!/4,

tr(⇠�1⇠0) =�14 e!

� �14 �e

�!

�= ��/16 6= 0 .

Since ⇠�

2 Pg

, the Lax equation assures that ⇣�

(z) 2 Pg

(see remark 3.3). ⇤


Spectral curve. Suppose ⇠�

2 Pg

and ⇣�

is the polynomial Killing fieldwith ⇣

�

(0) = ⇠�

. Suppose further that the polynomial a(�) = �� det ⇠�

has2g-pairwise distinct roots. Define

⌃⇤ = {(⌫,�) 2 C2 | det(⌫ � ⇣�

) = 0}= {(⌫, �) 2 C2 | ⌫2 = � det ⇠

�

= ��1a(�)} .(2.10)

By construction we have a map � : ⌃⇤ ! C⇥ of degree 2, which is branched atthe 2g simple roots of the polynomial a. By declaring the points over � = 0, 1to be two further branch points, we then have 2g + 2 branch points. This2-point compactification ⌃ is called the spectral curve of the polynomialKilling field ⇣

�

.The Riemann-Hurwitz formula gives that the spectral curve ⌃ is a hyperel-

liptic Riemann surface of genus g, and its genus is called the spectral genus.It has three involutions

� : (�, ⌫) 7! (�,�⌫) ,% : (�, ⌫) 7! (��1,��1�g ⌫) ,

⌘ : (�, ⌫) 7! (��1, �1�g ⌫) .

(2.11)

The involution � is called the hyperelliptic involution. Note that ⌘ has nofixed points (a(1) 2 R�) and % fixes S1 pointwise. In particular, roots of aare symmetric with respect to inversion across the unit circle so that a(↵

i

) =0 , a(1/↵

i

) = 0.

3. Harmonic maps and Weierstrass representation

The generalized Weierstrass representation [8] gives a correspondence be-tween harmonic maps, extended frames and potentials. To formulate thegeneralized Weierstrass representation for harmonic maps into S2 we needvarious loop groups and a loop group factorization.

For real r 2 (0, 1], denote the circle Sr

= {� 2 C | |�| = r}, the diskIr

= {� 2 C | |�| < r} and the annulus Ar

= {� 2 C | r < |�| < 1/r}.The loop group of SL2(C) is the infinite dimensional Lie group ⇤

r

SL2(C) =O(S

r

, SL2(C)) of analytic maps Sr

! SL2(C).We need two subgroups. The first is

⇤r

SU2 = {F�

2 O(Ar

, SL2(C) | F�2S1 2 SU2} .

Thus

(3.1) F�

2 ⇤r

SU2 () F1/�

t

= F�1�

.

The second subgroup that participates in the Iwasawa decomposition is

⇤+r

SL2(C) = {B�

2 O(Ir

[ Sr

, SL2(C)) | B0 =�⇢ c

0 1/⇢

�

for ⇢ 2 R+ and c 2 C} .


The normalization that B0 is upper-triangular with real diagonals ensuresthat

⇤r

SU2 \ ⇤+r

SL2(C) = { } .The following important result is due to Pressley-Segal [26], and generalizedby McIntosh [22].

Theorem 3.1. Multiplication ⇤r

SU2 ⇥ ⇤+r

SL2(C) ! ⇤r

SL2(C) is a realanalytic di↵eomorphism onto. The unique splitting of �

�

2 ⇤r

SL2(C) into��

= F�

B�

with F�

2 ⇤r

SU2 and B�

2 ⇤+r

SL2(C) is called the r-Iwasawa

decomposition of ��

or just Iwasawa decomposition when r = 1.

Before specializing to the finite type theory of harmonic maps G : C ! S2,let us briefly recall the generalized Weierstrass representation [8]. Set

⇤1�1sl2(C) = {⇠

�

2 O(C⇥, sl2(C)) | (�⇠�

)|�=0 2 �

0 C⇥

0 0

�} .A potential is a holomorphic 1-form on C with values in ⇤1

�1sl2(C). Sup-pose that we have such a potential ⇠

�

dz with ⇠�

2 ⇤1�1sl2(C). To obtain a

corresponding extended frame F�

is a two step procedure:

(1) Solve the holomorphic ODE d��

= ��

⇠�

to obtain a map C !⇤r

SL2(C), z 7! ��

(z).(2) The r-Iwasawa factorization �

�

(z) = F�

(z)B�

(z) at each z 2 C givesan extended frame C ! ⇤

r

SU2, z 7! F�

(z).

Note that while ��

is holomorphic in z 2 C, the resulting extended frame isnot, since it also depends on z by the reality condition (1.4). It is provenin [8] that each extended frame can be obtained from a potential ⇠

�

by theIwasawa decomposition. Hence for any conformal minimal immersion X =(G, h) : C ! S2 ⇥ R there is a potential and corresponding extended framewhich frames G.

An extended frame is of (semi-simple) finite type, if there exists a g 2 N,and it has a corresponding potential ⇠

�

dz with ⇠�

2 Pg

⇢ ⇤1�1sl2(C). Hence

harmonic maps of finite type come from constant (1, 0)-forms with values inthe finite dimensional space P

g

, and thus have an algebraic description. Inthe finite type case the first step in the above two step procedure is explicit,since then �

�

= exp(z ⇠�

). Thus extended frames of finite type are obtainedby factorizing exp(z ⇠

�

) = F�

B�

with ⇠�

2 Pg

. This step can be made explicitin terms of theta functions on the spectral curve (see Bobenko [2]).

Expanding a polynomial Killing field ⇣�

: C ! Pg

as

(3.2) ⇣�

(z) =�0 ��1(z)0 0

��1 +

�↵0(z) �0(z)�0(z) �↵0(z)

�+ . . .+

�↵

g

(z) �

g

(z)�

g

(z) �↵

g

(z)

��g

we associate a matrix 1-form defined by

(3.3) ↵(⇣�

) =

✓↵0(z) ��1(z)��1

�0(z) �↵0(z)

◆dz �

✓↵0(z) �0(z)��1(z)� �↵0(z)

◆dz .


The following Proposition is well known, and the correspondence betweenpotentials, polynomial Killing fields and extended frames is known as ’Symesmethod’ [5, 6, 29].

Proposition 3.2. For each ⇠�

2 Pg

there is a unique polynomial Killingfield ⇣

�

: C ! Pg

solving d⇣�

= [ ⇣�

, ↵(⇣�

) ] with ⇣�

(0) = ⇠�

. The unitaryfactor F

�

: C ! ⇤SU2 of the Iwasawa decomposition exp(z ⇠�

) = F�

B�

isa solution of F�1

�

dF�

= ↵(⇣�

) with initial value F�

(0) = and ⇣�

(z) =B

�

(z) ⇠�

B�1�

(z) = F�1�

(z) ⇠�

F�

(z).

Proof. Clearly ⇣�

= B�

(z) ⇠�

B�1�

(z) = F�1�

(z) ⇠�

F�

(z) uniquely solves d⇣�

=[ ⇣

�

, F�1�

dF�

] with ⇣�

(0) = ⇠�

, so it remains to show F�1�

dF�

= ↵(⇣�

).Now F�1

�

dF�

= B�

⇠�

B�1�

� dB�

B�1�

, so by the reality condition (1.4) wehave F�1

�

dF�

= a�1��1 + a0 + a1�. If we decompose aj

= a0j

dz + a00j

dz into(1, 0) and (0, 1) parts, then we have in addition that

a000 = �a00t

, a001 = �a0�1

t

, a00�1 = �a01t

.

Now F�1�

(F�

)z

= ⇣�

� (B�

)z

B�1�

, and (B�

)z

B�1�

is holomorphic at � = 0.

Hence a0�1 = ⇣�1. Further, F�1�

(F�

)z

= �(B�

)z

B�1�

implies a00�1 = 0. Itremains to determine a0.

Expand B�

= B0 + B1� + . . .. Now a00 = ⇣0 � (B0)z B�10 . Since B0 is

upper-triangular, then so is (B0)z B�10 . Hence the lower-diagonal term of a00

is �0, and the upper-diagonal term of a000 is ��0. Also a000 = �(B0)z B�10 is

upper-triangular, so the lower-diagonal entry of a000 is zero, and consequentlyalso the upper-diagonal entry of a00 is zero.

Finally, writing B(0) = B0 =�⇢ c

0 1/⇢

�, and a00 =

�u v

w �u

�, then u = ⇢�1⇢

z

and u = ↵0 � ⇢�1⇢z

. These two equations, and since ⇢ is real analytic, give2u = ↵0. ⇤

Remark 3.3. With initial data ��1(0) 2 R+, the Lax equation gives ��1(z) 2R+. For a given potential ⇠

�

and polynomial Killing field ⇣�

: C ! Pg

we de-fine ! : C ! R by setting 4��1 := e!. The iteration implies that 2↵0 = !

z

.To express �0 in terms of ↵0 and ��1, we consider the Lax equation andfind (�0)z = �2↵0�0. Then �0 = qe�! where q is a holomorphic function.The term q is constant. The reason is that along the parameter z, we havea(�) = �� det ⇣

�

(z) = �� det ⇠�

and a(0) = ��1�0 = q/4. Coe�cients of ⇣�

depend only on higher derivatives of ! point wise in z, and ↵(⇣�

) = ↵(!).

Remark 3.4. Isometric normalization 4. With initial data ��1(0) 2�R+ (i.e. ⇠

�

2 Pg

(�)), the Iwasawa decomposition gives a solution of the Lax

equation ⇣�

: C ! Pg

(�) given by ⇣�

(z) = F�1�

(z)⇠�

F�

(z).


4. Isospectral set

The set I(a) consists of all initial conditions ⇠�

which give rise to the samespectral curve ⌃ and the same o↵-diagonal product a(0) = ��1�0.

Definition 4.1. Define for polynomial Killing field ⇣�

: C ! Pg

(�) as in(3.2), and a(�) = �� det ⇣

�

the set

I�

(a) := { ⇠�

2Pg

(�) | � det ⇠�

= �a(�) and

��1�0 = a(0) = � 116e

(1�g)✓ := � 116e

⇥ } .

When � = 1, we write I(a). The set I(a) is called the isospectral set of thepolynomial Killing field ⇣

�

.

We next define the isospectral action ⇡ : Cg ⇥ I(a) ! I(a).

Definition 4.2. Let ⇠�

2 I(a) and t = (t0, . . . , tg�1) 2 Cg, and

(4.1) exp�⇠�

g�1X

i=0

��iti

�= F

�

(t)B�

(t)

the Iwasawa factorization. Define the map ⇡(t) : I(a) ! I(a) by

⇡(t) ⇠�

= B�

(t) ⇠�

B�1�

(t) .

Since F�

(t)B�

(t) commutes with ⇠�

we have

⇡(t) ⇠�

= B�

(t) ⇠�

B�1�

(t) = F�1�

(t) ⇠�

F�

(t) .

Proposition 4.3. The map ⇡(t) : I(a) ! I(a) defines a commutative groupaction

⇡(t+ t0) = ⇡(t)⇡(t0) = ⇡(t0)⇡(t) .

Proof. For t, t0 2 Cg we have

F�

(t+ t0)B�

(t+ t0) = F�

(t)B�

(t)F�

(t0)B�

(t0) = F�

(t0)B�

(t0)F�

(t)B�

(t) .

Hence

exp(B�

(t)⇠�

B�1�

(t)⌃��it0i

) = B�

(t)F�

(t0)B�

(t0)B�1�

(t)

= F�1�

(t)F�

(t+ t0)B�

(t+ t0)B�1�

(t) ,

exp(B�

(t0)⇠�

B�1�

(t0)⌃��iti

) = B�

(t0)F�

(t)B�

(t)B�1�

(t0)

= F�1�

(t0)F�

(t+ t0)B�

(t+ t0)B�1�

(t0) .


Set B�

(t) = B�

(t+ t0)B�1�

(t0) and B�

(t0) = B�

(t+ t0)B�1�

(t). Then

⇡(t+ t0)⇠�

= B�

(t+ t0) ⇠�

B�1�

(t+ t0)

= B�

(t0)B�

(t) ⇠�

B�1�

(t)B�1�

(t0)

= B�

(t0)⇡(t)⇠�

B�1�

(t0) = ⇡(t0)⇡(t)⇠�

= B�

(t)B�

(t0) ⇠�

B�1�

(t0)B�1�

(t)

= B�

(t)⇡(t0)⇠�

B�1�

(t) = ⇡(t)⇡(t0)⇠�

.

⇤Removing singularities on ⌃. Elements of di↵erent isospectral sets maygive the same extended frame up to conjugation by g 2 T (see remark 1.4 andbelow). This is the case in particular if an initial value ⇠

�

has a root at some� = ↵0 2 C⇥. Then the corresponding polynomial Killing field ⇣

�

also has aroot at � = ↵0 for all z 2 C. In this case we may reduce the order of ⇠

�

and⇣�

without changing the immersion. This situation corresponds to a singularspectral curve because then the polynomial a(�) = �� det ⇠

�

has a root oforder at least two at ↵0. We can remove such a singularity by changing thesurface by an isometry. We describe this change below.

Proposition 4.4. Suppose a polynomial Killing field ⇣�

has roots in � 2C⇥. Then there is a polynomial p(�) such that ⇣

�

/p(�) has no roots in � 2C⇥. If F

�

and F�

are the extended frames of ⇣�

respectively ⇣�

/p(�), then

F�

(p(0) z) = F�

(z). Hence there is g(�) 2 T with ⇣�

:= g(�)�1(⇣�

/p(�)) g(�) :C ! P

g

and F (z) := g(�)�1F�

(p(0)z)g(�) induces a minimal surface X�

congruent to X�

in S2 ⇥ R.

Proof. Suppose the polynomial Killing field ⇣�

(z) = F�1�

(z)⇠�

F�

(z) has a rootat � = ↵0. Define

p(�) =

(p�↵0�+p�↵0 if ↵0 2 S1

(�� ↵0)(1� ↵0�) if ↵0 2 C⇥\S1 .If ⇣

�

has a simple root at � = ↵0 2 C⇥, then ⇣�

/p(�) : C ! ⇤g�deg p

�1 sl2(C)does not vanish at ↵0. Then there is a map C : C ! ⇤+SL2(C) such that

F�

(z)B�

(z) = exp(z ⇠�

) = exp(p(�) z ⇠�

) = exp(p(0) z ⇠�

)C(z)

= F�

(p(0)z)B�

(p(0)z)

and

F�

(z) = F�

(p(0) z) =

(F�

(p�↵0z) if ↵0 2 S1

F�

(�↵0z) if ↵0 2 C⇥\S1 .We consider � =

p�↵0 when |↵0| = 1 and � = �↵0/|↵0| in the other case.We conjugate ⇣

�

/p(�) by g(�) 2 T. Hence by remarks 1.5, 2.6, 2.9, 3.4, the


immersion X�

obtained from F�

(z) = g(�)�1F�

(p(0) z) g(�) is congruent tothe immersion X

�

. ⇤

Hence amongst all polynomial Killing fields that give rise to an extendedframe of finite type there is one of smallest possible degree. We say that apolynomial Killing field has minimal degree if and only if it has neither rootsnor poles in � 2 C⇥. We summarize two results by Burstall-Pedit [5, 6]:

Proposition 4.5. For an extended frame of finite type there exists a uniquepolynomial Killing field of minimal degree. There is a smooth 1-1 correspon-dence between the set of extended frames of finite type and the set of polynomialKilling fields without zeroes.

Proof. We briefly outline how to prove the existence and uniqueness of aminimal element. If the initial value ⇠

�

gives rise to an extended frame F�

,then the corresponding polynomial Killing field ⇣

�

can be modified accordingto Proposition 4.4 so that ⇣

�

is of minimal degree, and still giving rise to F�

.Hence there exists a polynomial Killing field ⇣

�

2 Pg

of least degree givingrise to F

�

.For the uniqueness, let ⇣

�

and ⇣�

both solve d⇣�

+ [↵(!), ⇣�

] = 0, withdeg⇣

�

� deg⇣�

. We can assume that ⇣�

, ⇣�

have no roots (if not, we simplifythe polynomial Killing field using Proposition 4.4). We use the iteration ofProposition 2.2. We prove that there is a polynomial q 2 Ck[�] such that

(4.2) ⇣�

(z) = q(�) ⇣�

(z) .

The polynomial q is constructed recursively by considering coe�cients un

,�n

, ⌧n�1 and u

n

, �n

, ⌧n�1. Since ⌧�1, ⌧�1, are constant, there is q0 with

⌧�1 = q0⌧�1. This implies that u0 = q0u0, �0 = q0�0 and there is q1 such that⌧0 = q0⌧0 + q1⌧�1. By the iteration, if there are constants q0, q1, ..., q` with

u0 = q0u0 , u1 = q0u1 + q1u0 , . . . u`

= q0u`

+ ...+ q`

u0 ,

�0 = q0�0 , �1 = q0�1 + q1�0 , . . . �`

= q0�` + ...+ q`

�0 ,

⌧�1 = q0⌧�1 , ⌧0 = q0⌧0 + q1⌧�1 , . . . ⌧`�1 = q0⌧`�1 + ...+ q

`

⌧�1 .

the iteration implies that there is q`+1 such that ⌧

`

= q0⌧` + ...+ q`+1⌧�1 and

this proves (4.2). Now, since ⇣�

, ⇣�

have no roots, the polynomial q(�) = q0is constant and since the residues coincide, we conclude that ⇣

�

= ⇣�

. ⇤

Remark 4.6. Since the Iwasawa factorization is a di↵eomorphism, and allother operations involved in obtaining an extended frame from a polynomialKilling field are smooth, the resulting minimal surface depends smoothly on theentries of the polynomial Killing field, and thus also smoothly on the entriesof its initial value.


For ⇠�

2 I(a) the polynomial a(�) = �� det ⇠�

has the form

(4.3) a(�) = bgY

i=1

(�� ↵i

)(1� �↵i

) , b 2 R� .

The condition (2.8) implies a(1) 0 so that b 2 R�.

Lemma 4.7. (1) If a has a double root ↵0 2 S1, then we have an isomorphism

I(a) ⇠= I(�↵0 (�� ↵0)�2 a) .

(2) If a has double root ↵0 /2 S1 then I(a) = {⇠�

2 I(a) | ⇠↵0 6= 0} [ {⇠

�

2I(a) | ⇠

↵0 = 0} and there is an isomorphism

{⇠�

2 I(a) | ⇠↵0 = 0} ⇠= I((�� ↵0)

�1(1� ↵0�)�1 a) .

Proof. (1) If a has a double root at ↵0 2 S1 then for any ⇠�

2 I(a), we have

⇠↵0 = 0. We can remove this root by Proposition 4.4, with � =

p�↵0, toobtain the isomorphism

⇠�

7! (p�↵0�+

p�↵0)�1g(�)�1⇠

�

g(�) 2 I(�↵0 (�� ↵0)�2 a) .

(2) Suppose a has a double root at ↵0 /2 S1. Then the isospectral set splitsinto a part which contains potentials with a zero at ↵0, which we can remove,and the set of potentials not zero at ↵0. But in this last case, this meansthat ⇠

↵0 is a nilpotent matrix. We use Proposition 4.4, with � = �↵0/|↵0|, toobtain the isomorphism

⇠�

7! 1��↵0

11�↵0�

g(�)�1⇠�

g(�) 2 I( 1��↵0

11�↵0�

a) .

⇤Theorem 4.8. Isospectral sets are compact. If the 2g-roots of the polynomiala(�) = �� det ⇠

�

for ⇠�

2 Pg

are pairwise distinct, then I(a) ⇠= (S1)g is aconnected smooth g-dimensional manifold di↵eomorphic to a g-dimensionalreal torus.

The proof of Theorem 4.8 follows in several steps. The compactness is aconsequence of the next Proposition 4.9. The second statement follows fromPropositions 4.10 and 4.11. We shall prove several properties of the map

(4.4) A : Pg

! Mg

, ⇠�

7! �� det ⇠�

.

Proposition 4.9. The map A in (4.4) is proper.

Proof. Since A is continuous it su�ces to show that pre-images of compactsets are bounded. Let K ⇢ M

g

be compact. Then the map S1 ⇥ K !R, (�, a) 7! ��ga(�) is bounded. For the compactness of the pre-image itsu�ces to show that all Laurent-coe�cients of a ⇠

�

2 Pg

are bounded, ifA(⇠

�

) 2 K. Fix a polynomial a 2 K and consider the isospectral set I(a) asa closed subset of the (3g+1)-dimensional vector space P

g

. For d = (1� g)/2


the map �d⇠�

is traceless and skew-hermitian for |�| = 1. The determinantof traceless skew-hermitian 2 ⇥ 2 matrices is the square of a norm k · k. TheLaurent-coe�cients of ⇠

�

=P

g

i=�1 �i ⇠

i

are

⇠i

=1

2⇡

Z

S1��i⇠

�

d�

�.

Using the norm gives

k⇠i

k 1

2⇡

Z

S1k�(1�g)/2⇠

�

k d�

� sup

�2S1

p��ga(�) .

Thus each entry of ⇠i

is bounded on S1, so A is proper and I(a) thereforecompact. ⇤Proposition 4.10. Suppose ⇠

�

2 Pg

has no roots in � 2 C⇥. Then the mapA in (4.4) has maximal rank 2g + 1. Let a(�) = �� det ⇠

�

. Then I(a) is ag-dimensional sub-manifold of P

g

.

Proof. Since det is the square of a norm on su2, at all roots of a on S1, thecorresponding ⇠

�

2 I(a) has to vanish. If ⇠�

is without roots, then a 2 M0g

.We show that the derivative of the map A has rank 2g+1 at a potential ⇠

�

without roots, and then invoke the implicit function theorem. Hence it su�cesto prove that for all roots ↵0 of a of order n, and all � 2 C there exists atangent vector ⇠

�

along Pg

at ⇠�

, such that the corresponding derivative of ais equal to

a(�) = �� det ⇠�

tr(⇠�1�

⇠�

) =�a(�)

(�� ↵0)m+

�m�a(�)

(1� ↵0�)m

with m = 1, . . . , n. Such a vector field fixes all roots of a except ↵0. Theset of these vector fields form a 2g-dimensional real vector space. Besides theroots of a(�) also the coe�cient b in (4.3) can be changed by this variation.We consider ⇠

�

= t ⇠�

with

a(�) = �� det ⇠�

tr(⇠�1�

⇠�

) = �2�t det ⇠�

.

This vector field preserves all roots of a(�), but changes a(1) with variationalfield a(1) = ta(1), and b 2 R� is changing non-trivially. This will prove thetheorem.

Now we construct vector fields ⇠�

. If ↵0 is a root of a of order n, thendet ⇠

↵0 vanishes, and ⇠↵0 is nilpotent. For a nonzero nilpotent 2 ⇥ 2-matrix

⇠↵0 there exists a matrix Q 2 su2 such that ⇠

↵0 = [Q, ⇠↵0 ]. To prove this

remark, observe that it holds if ⇠0 =�0 10 0

�, by setting Q0 =

�1 00 �1

�. The

general statement now follows since there exists g 2 SU2 with ⇠↵0 = g�1⇠0g,

and setting Q = g�1Q0g.We need the following basic fact: For A, B 2 sl2(C) and A 6= 0, we have

(4.5) tr(AB) = 0 () B = [C,A] with some C 2 sl2(C) .


For any ⇠�

2 Pg

, we have ⇠2�

= � det(⇠�

) and at a root ↵0 of a(�) of ordern, there exists for any m = 1, ..., n, a matrix

Q�

(m) = Q0 +Q1(�� ↵0) + ...+Qm�1(�� ↵0)

m�1

with Q0, . . . , Qn�1 2 sl2(C), such that ⇠�

� [Q�

(m), ⇠�

] has at ↵0 a rootof order m. The matrix Q

�

(m) is constructed inductively using (4.5). Weremark that at ↵0, the function � 7! tr(⇠2

�

) has a root of order n. Thenthere is Q0 such that ⇠

�

� [Q0, ⇠�] = (�� ↵0)⇠�,1 and tr(⇠�

(⇠�

� [Q0, ⇠�])) =(��↵0)tr(⇠�⇠�,1) has a root at ↵0 of order n. Then there is a matrix Q1 with⇠�,1 � [Q1, ⇠�] = (�� ↵0)Q2. Now we define for m = 1, ..., n

Q�

(m) = �tQ0 � tQ1(1� ↵0�)��1 � ...� tQ

m�1(1� ↵0�)m�1�m�1.

Then ⇠�

� [Q�

(m), ⇠�

] has at ↵�10 a root of order m. Now we define

qm

(�) = �

(��↵0)m+

�m�

(1� ↵0�)m,

Q�

(m) = �

(��↵0)mQ

�

(m) + �

m

�

(1�↵0�)mQ

�

(m) .

There exists some P 2 su2 such that

⇠�

= qm

(�)⇠�

� [Q�

(m), ⇠�

] + [P, ⇠�

] 2 T⇠

�

Pg

.

To see that we need to check ˙⇠�1 2 R+�0 10 0

�and ˙|a|(0) = 0. Note that if

A 2 su2 and B =�0 10 0

�, then [A, B] =

�↵ x

0 �↵

�with ↵ 2 C, x 2 R. Then we

can choose P 2 su2 such that

˙⇠�1 =�⇠�1 � �[Q0(m), ⇠�1]

(�↵0)n+ [P, ⇠�1] 2 R+

✓0 10 0

◆.

For the second condition we have (aa+ a¯a)(0) = |a|2(0)(�↵0)�n(� + �), andwe can choose � 2 C such that the variational field keeps |a(0)| unchangedalong the deformation. This proves that ⇠

�

2 T⇠

�

Pg

, and such vector fieldsspan a 2g+1 dimensional real vector space of vector fields in the complementof the kernel of the map. This proves that around ⇠

�

without zeroes, I(a) isa real g-dimensional manifold. ⇤

Proposition 4.11. For all ⇠�

2 Pg

without roots the vector fields of theisospectral group action generate at ⇠

�

a real g-dimensional subspace of thetangent space P

g

at ⇠�

.

Proof. The vector field (t0, . . . , tg�1) of the isospectral action at ⇠�

takes thevalues

⇠�

=⇥�g�1X

i=0

��iti

⇠�

�+, ⇠

�

⇤= �⇥�g�1X

i=0

��iti

⇠�

��, ⇠

�

⇤.


Hereg�1X

i=0

��iti

⇠�

=�g�1X

i=0

��iti

⇠�

�++

�g�1X

i=0

��iti

⇠�

��

is the Lie algebra decomposition of the Iwasawa decomposition. For A 2sl2(C) with A 6= 0 we have that {B 2 sl2(C) | [A, B] = 0} = {xA | x 2 C}.Hence the vector field corresponding to (t0, . . . , tg�1) vanishes at ⇠

�

if and

only if there exists a decomposition of the polynomialP

g�1i=0 �

�iti

= f+(�) +f�(�) into complex functions such that

�g�1X

i=0

��iti

⇠�

�+= f+⇠

�

and�g�1X

i=0

��iti

⇠�

��= f�⇠

�

.

Hence f+(��1) = �gf+(�) and f�(0) = 0.The polynomial

Pg�1i=0 �

�iti

is a linear combination of such functions if andonly if t

g�1�i

= ti

. The subspace of such (t0, . . . , tg�1) is a real g-dimensionalsubspace of Cg. This implies the proposition. ⇤

Recall Mg

from (2.9), and define

M1g

= { a 2 Mg

| a has 2g-pairwise distinct roots } ,P1g

= { ⇠�

2 Pg

| a(�) = �� det ⇠�

2 M1g

} .Proposition 4.12. For all a 2 M1

g

, the isospectral action ⇡ : Rg ⇥ I(a) !I(a) acts transitively on I(a) and the mapping A : P1

g

! M1g

is a principal

bundle with fibre I(a) = (S1)g.

Proof. At all roots of ⇠�

2 Pg

, the determinant det ⇠�

has a higher orderroot. Hence all ⇠

�

2 P1g

have no roots on C⇥. Proposition 4.10 implies thatA : P1

g

! M1g

has maximal rank 2g + 1 and induces a fibre bundle, whosefibres are real g-dimensional manifolds. The isospectral action preserves thedeterminant and thus the fibres. Proposition 4.11 implies that for all ⇠

�

2 P1g

the corresponding orbit of the isospectral action is an open submanifold of thecorresponding fibre. If ⇡(t

n

) ⇠�

converges to ⇠�

2 P1g

for a sequence (tn

)n2N

in Rg, then the orbit of ⇠�

is again an open submanifold of the correspondingfibre. Thus ⇡(t

n

) ⇠�

belongs to the orbit of ⇠�

for su�ciently large n 2 N.Then ⇡(t

n

) ⇠�

= ⇡(t0n

) ⇠�

and ⇠�

= ⇡(tn

� t0n

) ⇠�

is in the orbit of ⇠�

. Thisshows that the orbits of the isospectral set I(a) are open and closed subman-ifolds of the fibre. Due to Proposition 4.9 the fibers are compact. Thereforeall orbits are compact as well. Due to Proposition 4.11, for all ⇠

�

2 P1g

thestabilizer subgroup

(4.6) �⇠

�

= {t 2 Rg | ⇡(t) ⇠�

= ⇠�

}is discrete and Rg/�

⇠

�

is di↵eomorphic to the connected component of thefibre of ⇠

�

. We conclude that �⇠

�

is a lattice in Rg isomorphic to Zg. This


shows that A : P1g

! M1g

is a fibre bundle, whose fibres have connectedcomponents all isomorphic to (S1)g.

It remains to prove that I(a) has only one connected component. We firstshow that M0

g

(see (2.9)) is path connected: If a, a 2 M0g

satisfy

(4.7) t a(0) + (1� t) a(0) 6= 0 for all t 2 [0, 1]

then t a+ (1� t) a 2 M0g

for all t 2 [0, 1]. If a, a 2 M0g

do not satisfy (4.7),

then modifying � 7! a(�) by the rotation � 7! e� g✓a(e ✓�) for some suitable✓ 2 R we can ensure that (4.7) holds. Hence M0

g

is path connected.Since M1

g

is an open subset of M0g

, whose complement M0g

\ M1g

hascodimension at least 2, we conclude that M1

g

is connected and A : P1g

! M1g

has maximal rank 2g + 1 at any point. Hence it remains to show that thereexists at least one a 2 M1

g

for which I(a) has only one connected component.Denote the entries of ⇠

�

2 Pg

by polynomials ↵,�, � so that

⇠�

=

✓↵(�) �(�)�(�) �↵(�)

◆.

Then a(�) = �� det ⇠�

= �↵2 + ��. Here ↵ is a polynomial of degree at

most g � 1. For |�| = 1 the polynomial �1�g

2 ↵ 2 R and �1�g↵2 0, andtherefore �1�g�� 2 R.

Now we consider a potential ⇠�

with �� = �, and � has only roots on|�| = 1. We claim in this case that I(a) is connected. This condition on �

implies that a(�) = �↵2+�2 and �g�(��1) = ��(�) = ��(�). Now observethat ��g�2 0 and ��ga(�) �1�g↵2 0.

Let ⇠�

2 I(a) with entries ↵, �, �. We construct a family ⇠�,t

=�↵

t

�

t

�

t

�↵

t

�

with ↵t

= t↵+(1� t)↵. We prove that there exist polynomials �t

, �t

uniquelydefined such that ⇠

�,t

2 I(a) for all t 2 [0, 1], with �1 = �, �1 = � and �0 = �,�0 = �.

Since �1�g↵2 0 for |�| = 1 we have �1�g↵2t

0 for all t 2 [0, 1]. Nowwe consider the polynomial p

t

(�) = a(�) � �1�g↵2t

. If ↵0 is a root of pt

(�),then ↵�1

0 is a root of pt

(�). For t = 0 we know that p0 = �� where rootsof � are symmetric to the roots of �. At roots of p

t

we can define �t

and �t

with �1 2 R+. At t = 1, we have p = �2 and all the roots of p1 are doubleroots on the unit circle. Then � is defined uniquely, and I(a) is connected, ifwe can find such a path with a 2 M1

g

.Therefore we consider � = �g + 1 and ↵ = k(�g�1 + 1) with k 2 R+.

Then 0 = �g + 1 = �g�1 + 1 implies �g�1(1� �) = 0. Then the polynomials� and ↵ do not have common roots. Hence at k = 0 we have ��ga(�) =�1�g↵2 + ��g�2 0 for |�| = 1, and a has only double roots on the unitcircle. For k > 0 small enough, the roots change. But there are no roots onthe unit circle and the roots are simple and conjugate, so that a 2 M1

g

. ⇤


5. Periods

Suppose X�

= (G�

, h�

) is an associated family of minimal surfaces inS2 ⇥ R. For one member of this family to be periodic, say for � = 1 withperiod ⌧ 2 C⇥, this means that X1(z + ⌧) = X1(z) for all z 2 C, or equiv-alently G1(z + ⌧) = G1(z) and h1(z + ⌧) = h1(z) for all z 2 C. If G

�

(z) =F�

(z)�3 F�

(z)�1 and h�

(z) = Re(�4p��1�0��1 z) = Re(� e ⇥/2

p��1 z)

(where Q = 14e

⇥(dz)2) then periodicity reads⇥F�11 (z)F1(z + ⌧), �3

⇤= 0 and Re(� e ⇥/2 ⌧) = 0 .

The monodromy of an extended frame F�

with respect to the period ⌧ isthe matrix

M�

(⌧) = F�

(z)�1 F�

(z + ⌧) .

Thus periodicity of the horizontal part reads [M�

(⌧), �3] = 0. Due to (1.3),the monodromy C⇥ ! SL2(C), � 7! M

�

(⌧) is a holomorphic map with essen-tial singularities at � = 0, 1.

For a periodic immersion its conformal factor ! is periodic, and hence also↵(!) in (1.3) is periodic. This in turn implies that dM

�

(⌧) = 0 so thatM

�

(⌧) does not depend on z. Hence M�

(⌧) = F�

(0)�1 F�

(⌧) = F�

(⌧) sinceF�

(0) = .Let ⇣

�

be a periodic solution of the Lax equation (2.3) with initial value⇠�

2 Pg

, with period ⌧ so that ⇣�

(z + ⌧) = ⇣�

(z) for all z 2 C. Then alsothe corresponding ↵(⇣

�

) in (3.3) is ⌧ -periodic. Let dF�

= F�

↵(⇣�

), F�

(0) =and M

�

(⌧) = F�

(⌧) be the monodromy with respect to ⌧ . Then for z = 0 wehave ⇠

�

= ⇣�

(0) = ⇣�

(⌧) = F�1�

(⌧) ⇠�

F�

(⌧) = M�1�

(⌧) ⇠�

M�

(⌧) and thus

[M�

(⌧), ⇠�

] = 0 .

The monodromy takes values in SU2 for |�| = 1. The monodromy depends onthe choice of base point, but its conjugacy class and hence eigenvalues µ, µ�1

do not. The eigenspaces of M(�, ⌧) depend holomorphically on (⌫, �). Theeigenvalues of ⇠

�

and M�

(⌧) are di↵erent functions on the spectral curve ⌃.

Proposition 5.1. Let M�

=�A C

B D

� 2 ⇤SU2 and ⇠�

=�↵ �

� �↵

� 2 Pg

with

[M�

, ⇠�

] = 0. Assume ⌫ 6= 0 and µ2 6= 1. Then M�

and ⇠�

have sameeigenvectors + = (1, (⌫ � ↵)/�), � = (1, �(⌫ + ↵)/�) with

⇠�

+ = ⌫ + and M + = µ + ,

⇠�

� = �⌫ � and M � = µ�1 � .

The involution ⌘ : (�, ⌫) ! (��1, �1�g ⌫) acts on µ by ⌘⇤µ = µ.

Proof. From the reality condition �g�1⇠1/�t = �⇠

�

or equivalently ⇠1/� =

��1�g⇠�

t and with � = (1, �(⌫ + ↵)/�) we have

⇠1/� � = ��1�g

✓↵ �� ↵

◆ � = �1�g ⌫ � .


Since M1/�

t �1 = M�

, and M�

+ = µ +, and + = (1, (⌫ � ↵)/�) =(1, �/(⌫ + ↵)), we obtain the system

A+ C�

⌫ + ↵= µ and B +D

�

⌫ + ↵= µ

�

⌫ + ↵.

This implies by direct computation that M1/� � = M�1�

t

� = µ �. ⇤

At � = 0 and � = 1 a monodromy M�

(⌧) has essential singularities. Nextwe study the behavior of µ = µ(�, ⌧) at these two points.

Lemma 5.2. Let = (�, z) = (1, h(�, z)) be an eigenvector of ⇣�

(z) =F�

(z)�1 ⇠�

F�

(z). Then there exists a complex function f = f(�, z) such that

(5.1) f(�, z) (�, z) = F�1�

(z) (�, 0)

which satisfies

(1) f�1df = � 4��1/2 exp( ⇥/2) dz +O(1) in a neighborhood of � = 0 ,

(2) f�1df = � 4�1/2 exp(� ⇥/2) dz +O(1) in a neighborhood of � = 1.

Proof. Note that h(�, z) and f(�, z) are 2-valued in � because they dependon the choice of eigenvalue. By Proposition 5.1 an eigenvector of

⇣�

(z) =

✓↵(�, z) �(�, z)�(�, z) �↵(�, z)

◆

associate to the eigenvalue ⌫ =p

a(�)��1 is given by (�, z) = (1, (⌫ � ↵)/�).Now using ⇣

�

(z) = F�1�

(z) ⇠�

F�

(z) we see that F�1�

(z) (�, 0) is an eigenvec-tor of ⇣

�

(z) and it is collinear to (�, z). This defines the function f(�, z).Di↵erentiating (5.1) reads df + f d = �↵(⇣

�

)F�1�

|z=0, and then

(5.2) f�1df = �↵(⇣�

) � d .

In a neighborhood of � = 0, we have ⌫2 = � det ⇣�

= �↵

2+��

�

= ��1�0

�

+

O(1) = �e

⇥

16� +O(1). Expanding at � = 0

↵(⇣�

) =�↵0 ��1�

�1

�0 �↵0

�dz +O(1) ,

and considering the first entry of the vector equation (5.2) yields

f�1df = �↵0dz � ��1��1 ⌫ � ↵(z)

�(z)dz +O(1)

= �⌫dz +O(1) =� e ⇥/2

4p�

dz +O(1) .

In a neighborhood of � = 1, we have

↵(⇣�

) =� �↵0 ��0

��1� ↵0

�dz +O(1)


and we obtain (ii) by considering the eigenvector =⇣

�

⌫�↵

, 1⌘and looking

at the second entry gives

f�1df = ��1��

⌫ � ↵dz +O(1) =

��1�g�1p�a2g

p�dz +O(1)

=� e� ⇥/2

p�

4dz +O(1) .

Further, ⇠d

= � ¯⇠tg�1�d

implies �g�1 = ��0 and a2g = ��

g�1�g = ��0��1 =

e� ⇥/16. ⇤Using these properties we compute the local behavior of µ(�, ⌧) near � = 0

and � = 1.

Proposition 5.3. Let X : A ! S2 ⇥ R be an immersed finite type minimalcylinder with spectral curve ⌃. Then there exists a meromorphic di↵erentiald lnµ on ⌃ with second order poles without residues at � = 0, 1 so thatd lnµ� 1

4 ⌧ exp( ⇥/2) dp��1 extends holomorphically to � = 0, and d lnµ�

14 ⌧ exp(� ⇥/2) d

p� extends holomorphically to � = 1.

This di↵erential is the logarithmic derivative of a function µ : ⌃ ! Cwhich transforms under the involutions (2.11) as �⇤µ = µ�1, %⇤µ = µ�1 and⌘⇤µ = µ. Further µ = ±1 at each branch point of ⌃.

Proof. If ⌫ 6= 0, the eigenspace of ⇣�

is a complex 1-dimensional vector spaceand since [M

�

, ⇣�

] = 0, every eigenvector of ⇣�

with associated eigenvalue⌫ 6= 0 is an eigenvector of M

�

with eigenvalue µ which depends only on (⌫, �).If ⌫ = 0 then µ = ±1. Note that µ is a non-zero holomorphic function on⌃⇤. At � = 0 and � = 1, the monodromy has essential singularities and wethus need the local analysis of Lemma 5.2. If ⌧ is the period of the annulus,we have (⌧) = (0) and f(⌧) (0) = F�1

�

(0). Then µ = f�1(⌧). This

proves that at � = 0, d lnµ � 4 exp( ⇥/2) ⌧ dp��1

is holomorphic, and at

� = 1 the di↵erential d lnµ� 4 exp(� ⇥/2) ⌧ dp� extends holomorphically.

The di↵erential d lnµ has second order poles without residues at � = 0 andat � = 1.

If + and � are eigenvectors associated to eigenvalues ±⌫ of ⇣�

, the corre-sponding eigenvalues µ± of M

�

satisfy µ+µ� = 1. To see how the involution⌘ acts on µ, we note that since M

�

satisfies (1.4), we have that µ is thecorresponding eigenvalue of M1/� associated to µ by Proposition 5.1. Thus

⌘⇤µ = µ. Similarly %⇤µ = µ�1.If ⌫ = 0, then det ⇣

�

= 0, so let (e1, e2) 2 C2 such that ⇣�

(e1) = 0 and⇣�

(e2) = �e1. Since M�

2 SL2(C), let (e1, e2) be a basis of eigenvectors of M�

associate to µ and µ�1 and ⇣�

(e1) = 1�e1, ⇣�(e2) = 2�e1. Inserting this inM

�

⇣�

(ei

) = ⇣�

M�

(e1) proves that µ = µ�1. This proves that the holomorphicfunction µ takes values ±1 at each branch point of ⌃. ⇤


Next we relate the eigenvalues µ to the isospectral action. We prove inparticular that the existence of such a holomorphic function is a su�cientcondition to close the period of a polynomial Killing field with any initialpotential.

Proposition 5.4. The stabilizer �⇠

�

in (4.6) depends only on the orbit of ⇠�

.If � 2 �

⇠

�

satisfies F�

(�) = ± for some ⇠�

2 Pg

then the same is true forevery element in the orbit of ⇠

�

. The period ⌧ is related to t = (⌧, 0 , ..., 0) 2�⇠

�

.

Proof. This follows from the commuting property of the isospectral action,since

⇡(�)⇡(t) ⇠�

= ⇡(t)⇡(�) ⇠�

= ⇡(t) ⇠�

.

⇤

Proposition 5.5. Assume ⇠�

2 Pg

has no roots. Then � 2 �⇠

�

if and onlyif there exists on ⌃ a function µ which satisfies the following properties:

(1) µ is holomorphic on ⌃⇤ and there exist holomorphic functions f, g onC⇥ with µ = f ⌫ + g.

(2) �⇤µ = µ�1, %⇤µ = µ�1, ⌘⇤µ = µ and µ = ±1 at branch points of ⌃.(3) d lnµ is a meromorphic 1-form with d lnµ � d(

Pg�1i=0 �i�

�i⌫) holo-

morphic in a neighborhood of � = 0 and d lnµ + d(P

g�1i=0 �i�

i+1�g⌫)is holomorphic at � = 1.

Proof. For � 2 Rg, we write exp(P

g�1i=0 �i�

�i⇠�

) = F�

(�)B�

(�) for the Iwa-sawa decomposition. Then � 2 �

⇠

�

if and only if [F�

(�), ⇠�

] = [B�

(�), ⇠�

] =0. Hence F

�

(�) and B�

(�) act trivially on ⇠�

and ⇠�

= ⇣�

(0) = ⇣�

(�) =F�1�

(�) ⇠�

F�

(�) = B�

(�) ⇠�

B�1�

(�). Since F�

(�) and B�

(�) commute with⇠�

, we have by (4.5) that F�

(�) = f(�) ⇠�

+ 12 tr(F�

(�)) and B�

(�) =e(�) ⇠

�

+ 12 tr(B�

(�)) with functions e , f depending only on �. In this casewe define on the spectral curve ⌃ the eigenvalue function µ(�) of F

�

(�) by

µ(�) = f(�) ⌫ + g(�) = f(�) ⌫ + 12 tr(F�

(�)).

We prove that µ satisfies properties (1), (2) and (3). Since � 7! F�

(�) isholomorphic, the function � 7! tr(F

�

(�)) is holomorphic on ⌃⇤. Since F�

(�)and ⇠

�

commute, can write f ⇠�

= F�

(�) � 12 tr(F�

(�)) ), and since ⇠�

has

no zeroes conclude that � 7! f(�) is holomorphic. Hence µ : ⌃ ! C isholomorphic and f, g have no poles on C⇥, proving (1).

Property (2) follows since µ is the eigenvalue of F�

(�) and F�

satisfies (1.4).At � = 0, the eigenvalue of the matrix B

�

(�) = e(�)⇠�

+ 12 trace(B�

(�))is holomorphic, so e(0) = 0 (⇠

�

has a pole at � = 0). Hence B0(�) = . Since

exp(g�1X

i=0

�i

��i⌫)


is the product of eigenvalues of the product F�

(�)B�

(�), then at � = 0, thisis precisely the value of µ, which proves (3). Similarly, using %⇤d lnµ = d lnµallows to deal with the point � = 1.

Conversely, assume µ : ⌃ ! C satisfies conditions (1),(2) and (3). Assumethere are two holomorphic function f, g : C⇥ ! C with µ = f⌫ + g. HenceF = f ⇠

�

+ g is holomorphic on ⌃⇤ and belongs to the first factor of theIwasawa decomposition (a consequence of (2)). Due to (3), the matrix

B�

= F�1�

exp(g�1X

i=0

�i

��i⇠�

)

is holomorphic in a neighborhood of � = 0 and [B�

, ⇠�

] = 0. Since B�

�12 tr(B�

) is at � = 0 proportional to�0 10 0

�, B

�

belongs to the second fac-tor of the Iwasawa decomposition. This shows that F

�

B�

is the Iwasawadecomposition of

exp(g�1X

i=0

�i

��i⇠�

) = F�

B�

.

Since [F�

(�), ⇠�

] = 0, we conclude that � 2 �⇠

�

. ⇤

Remark 5.6. Holomorphic functions f, g : C⇥ ! C are given by

µ = 12

�µ��

⇤µ

⌫

�⌫ + 1

2 (µ+ �⇤µ) = f ⌫ + g.

Theorem 8.2 in Forster [11] assures that f, g extend to holomorphic functionson C⇥. At fixed points of the involution � (zeroes of ⌫), the function µ� �⇤µhas zeroes.

In case ⌫ has higher order roots, the function (µ��⇤µ)/⌫ may have a pole.Then the condition that µ = f⌫ + g with f, g holomorphic is stronger than µbeing holomorphic on ⌃⇤.

The 1-form d lnµ is meromorphic on ⌃, and changes sign under the hyper-elliptic involution.

The closing conditions for regular spectral curves. In the following werestrict to the case where a has only simple roots. Then the spectral curve ⌃is a hyperelliptic curve without singularities. The closing condition is simplestas we only need to check the existence of the holomorphic ⌫ by Remark 5.6.

Proposition 5.7. Let a 2 C2g[�] satisfy (2.7) and (2.8). Then on ⌃ thereexist for all ⌧ 2 C⇥ a unique meromorphic di↵erential � such that

(1) �⇤� = ��, %⇤ = ��, ⌘⇤� = �.(2) � has second order poles at 0 and 1 without residues, and no other

poles.

(3) � = 14 ⌧e ⇥/2d

p��1

+O(1) at � = 0 and � = 14 ⌧e� ⇥/2d

p�+O(1)

at � = 1.


(4)R 1/↵

i

↵

i

� = Re⇣R 1/↵

i

↵

i

�⌘= 0 for all roots ↵

i

of a where the integral is

computed along the line segment [↵i

, 1/↵i

].

In conclusion there exists a unique b 2 Cg+1[�] with � = b d�

⌫ �

2 which satisfies

�g+1b(��1) = �b(�).

Proof. We make the Ansatz that � = b d�

⌫ �

2 . The conditions (1), (2) and(3) fix the highest and lowest coe�cient of b, so there remain g real freecoe�cients of b. These coe�cients correspond to holomorphic di↵erentials

satisfying condition (4) Re(R 1/↵

i

↵

i

�) = 0. The first equality comes from thereality condition (1) and reads

Z↵

i

/|↵i

|

↵

i

� =

Z 1/↵i

↵

i

/|↵i

|� .

Concerning uniqueness, holomorphic di↵erentials whose integrals along cyclesare imaginary, are zero by Riemann’s bilinear relations. ⇤Definition 5.8. We define a compact Riemann surface with boundary by⌃ = ⌃ \ [�

i

where �i

are closed cycles over the straight lines connecting thebranch points ↵

i

and 1/↵i

.

Corollary 5.9. Let a 2 C2g[�] satisfy (2.7) and (2.8).

(1) If there is ⌧ 2 C and b which satisfies (1)-(4) of Proposition 5.7, then

there exists a unique meromorphic function h(�) : ⌃ ! C such that�⇤h(�) = �h(�) and dh = b d�

⌫ �

2 .

(2) I(a) corresponds to minimal annuli in S2⇥R if and only if there exists⌧ 2 C⇥ with ⌧e ⇥/2 2 R⇥, and such that the polynomial b definingthe function h(�) : ⌃ ! C, satisfies �⇤h(�) = �h(�) and dh = b d�

⌫�

2 .This function continuously extends to boundary segments connecting↵i

and 1/↵i

and then takes values on ⇡Z at all roots of (�� 1) a(�).

Proof. (1) In a small neighborhood of ⌃ over � = 0, the function h is uniquelydetermined by dh = b d�

⌫ �

2 up to some additive constant. This constant isdetermined by �⇤h(�) = �h(�) in this small neighborhood. By conditions onb we have

R�

i

dh = 0, so

Z

�

i

b d�

⌫ �2= 2

Z 1/↵i

↵

i

b d�

⌫ �2= 0 .

Now we can uniquely extend the function h to ⌃.(2) For an immersed annulus

X1(z) = (F1(z)�3 F�11 (z), Re (� e ⇥/2z))

in S2 ⇥R the extended frame F�

(z) admits a period ⌧ 2 C⇥ with ⌧e ⇥/2 2 R,and periodic Killing field ⇣

�

(z + ⌧) = ⇣�

(z). This implies that d lnµ is a


meromorphic di↵erential with second order pole at 0 and 1 without residuesand no other poles and satisfies condition (1)-(3) of Proposition 5.7. Theintegrals of d lnµ along closed cycles are integer multiples of 2⇡ , since thefunction µ is globally single-valued by condition (4). Since the extended frameF1(z) is periodic then µ(1) = ±1. Since µ2 = 1 at branch points ↵

i

, it isequivalent to the condition lnµ(↵

i

) 2 ⇡Z. Hence there is a polynomial bassociated to ⌧ 2 C such that d lnµ = b d�

⌫�

2 .Conversely, consider (a, b) such that on the spectral curve ⌃ the mero-

morphic di↵erential � = b d�

⌫ �

2 satisfiesR↵

i

1� 2 ⇡ Z and

R↵

j

↵

i

� 2 ⇡ Z. While

h(�) =R�

1� is a multiple-valued function on ⌃, the function eh : ⌃ ! C is

again holomorphic. It is described in Proposition 5.5 where ⌧ is given by theresidue of � at 0 and 1. Then ⇣

�

(z + ⌧) = ⇣�

(z), and we can integrate theextended frame F

�

. There remains to prove that F1(⌧) = ± . We remarkthat a solution of the characteristic equation of a solution of F�1

�

dF�

= ↵(⇣�

)defines a meromorphic di↵erential d lnµ which satisfies (1)–(3) of Proposition5.7. By uniqueness of such di↵erentials, d lnµ = � and µ(1) = eh(1) = ±1 sothat F1 is ⌧ -periodic. ⇤Definition 5.10. The spectral data of a minimal cylinder of finite type inS2 ⇥ R with sym point at � = 1 is a pair (a, b) 2 C2g[�]⇥ Cg+1[�] such that

(i) �2ga(��1) = a(�), ��ga(�) 0 for all � 2 S1 and a(0) = � 116e

⇥,

(ii) �g+1b(��1) = �b(�),

(iii) b(0) = ⌧e

⇥

32 2 e ⇥/2R (closing condition of the third coordinate).

(iv) Re⇣R 1/↵

i

↵

i

b d�

⌫�

2

⌘= 0 for all roots ↵

i

of a where the integral is computed

along the straight segment [↵i

, 1/↵i

].(v) The unique function h : ⌃ ! C, where ⌃ = ⌃\ [ �

i

and �i

are closedcycles over the straight lines connecting ↵

i

and 1/↵i

, satisfies

�⇤h(�) = �h(�) and dh =b d�

⌫�2.

This function continuously extends to boundary segments connecting↵i

and 1/↵i

and then takes values on ⇡Z at all roots of (�� 1) a(�).(vi) When a has higher order roots then eh = f ⌫ + g for holomorphic

f, g : C⇥ ! C with f(1) = 0.

6. Bubbletons

The term ’bubbleton’ is due to Sterling-Wente [28]. They are the solitonsof the theory, and finite type solutions of the sinh-Gordon equations withbubbletons have singular spectral curves. For more details on the relation-ship between bubbletons, Bianchi-Backlund transformations, simple factorsand cmc surfaces we refer to [17, 16, 18] and the references therein. ByProposition 4.4 we can discard roots of a potential ⇠

�

without changing the


surface, but it remains to discuss higher order roots of det ⇠�

where ⇠�

is notsemi-simple.

We now consider potentials ⇠�

for which the polynomial a(�) = �� det ⇠�

has higher order roots. Roots of ⇠�

come in symmetric pairs ↵0, 1/↵0 2 C⇥

and we set in this section � = �↵0/|↵0|. Because the polynomial � 7! a(�)is homogeneous of degree 2, such roots generate even roots of order at least 2in the polynomial a. Then there exists a polynomial a with

a(�) = (�� ↵0)2(1� ↵0�)

2a(�).

We study such ⇠�

and show that they can be factored into a product of simplefactors (see below) and a potential ⇠

�

2 I(a). If ⇠↵0 6= 0 but ord

↵0 det ⇠� � 2,then the matrix ⇠

↵0 is nilpotent and defines a complex line

L = ker ⇠↵0 = im ⇠

↵0 2 CP1.

Let v1 2 C2 be a unit vector which spans L, and complement v1 to an or-thonormal basis (v1, v2) of C2. Set

p(�) = (�� ↵0)(1� ↵0�)

and let QL

2 SU2 with QL

(e1) = v1. The entries of ⇠�

in the basis (v1, v2)are given by

(6.1) Q�1L

⇠(�)QL

:=

✓↵(�)p(�) �(�)(1� ↵0�)2

�(�)(�� ↵0)2 �↵(�)p(�)◆

where �↵(�)2 � ��(�)�(�) = a(�). If (u1, u2) 2 ker ⇠↵0 then (�u2, u1) 2

ker ⇠1/↵0.

Simple factors. Define for � 6= ↵0, ↵�10 the SL2(C)-valued maps

⇡↵0(�) :=

0

@

q��↵01�↵0 �

0

0q

1�↵0 �

��↵0

1

A , and ⇡L

:= QL

⇡↵0Q

�1L

.

At � = 0, we apply the QR-decomposition to get

⇡L

(0) = QL

⇡↵0(0)Q

�1L

= Q0,LR0,L = R1,LQ1,L

where Q0,L, Q1,L 2 SU2 and R0,L, R1,L are upper triangular matrices of theform

�⇢ r

0 ⇢

�1

�for ⇢ 2 R+ both depending on ↵0 and L.

Definition 6.1. Let L 2 CP1 and ↵0 2 C with r < min{|↵0|, 1/|↵0|}. A left

simple factor is a map

gL,↵0(�) = ⇡�1

L

(�)Q0,L .

Then H`

↵0= {g

L,↵0 | L 2 CP1} ⇠= CP1, and H`

↵0⇢ ⇤+

r

SL2(C).A right simple factor is a map

hL,↵0(�) = Q1,L⇡

�1L

(�) .

Then Hr

↵0= {h

L,↵0 | L 2 CP1} ⇠= CP1, and Hr

↵0⇢ ⇤+

r

SL2(C).


Lemma 6.2. There is a one-to-one correspondence between H`

↵0and Hr

↵0via

(6.2) gL,↵0 = ⇡�1

L

Q0,L = Q0,L⇡�1L

0 = hL

0,↵0

where

L0 = Q�10,LL and Q1,L0 = Q0,L .

Conversely, if L0 is given then we can define a QR-decomposition ⇡L

0 =R1,L0Q1,L0 with

L = Q1,L0L0.

Proof. For any A 2 SU2 and L0 = AL we have QAL

= AQL

. Hence ⇡L

0 =Q

AL

⇡↵0Q

�1AL

= A⇡L

A�1. Now we define L0 = Q�10,LL and we have g�1

L,↵0=

Q�10,L⇡L = ⇡

L

0Q�10,L. With ⇡

L

(0) = Q0,LR0,L, we observe that

⇡L

0(0) = Q�10,L⇡L(0)Q0,L = R0,LQ0,L = R1,L0Q1,L0

and hence Q1,L0 = Q0,L. Then g�1L,↵0

(�) = ⇡L

0Q�10,L = ⇡

L

0Q�11,L0 = h�1

L

0,↵0

(�). If

L0 = Q�10,LL, so that L = Q0,LL0 = Q1,L0L0. ⇤

Now we consider the matrix ⇠�

= p�1(�) g�1L,↵0

⇠�

gL,↵0 .

Proposition 6.3. Let ⇠�

2 I(a) with ⇠↵0 6= 0 and ord

↵0 det ⇠� � 2. Let� = �↵0/|↵0|. Then ⇠

�

uniquely factorizes as

⇠�

= p(�) gL,↵0 ⇠� g

�1L,↵0

with L = ker ⇠↵0 and ⇠

�

2 Pg

(�) with ⇠�

= g(�)�1⇠�

g(�) 2 I(a) (i.e. ⇠�

2I�

(a)).

Proof. The matrix ⇠↵0 uniquely defines the line L. Consider

⇠�

= p�1(�) g�1L,↵0

⇠�

gL,↵0 .

We need to prove that ⇠�

has no poles at ↵0, ↵�10 and ⇠

�

2 I�

(a). First define

Q�1L

⇠�

QL

:=

✓↵(�)p(�) �(�)(1� ↵0�)2

�(�)(�� ↵0)2 �↵(�)p(�)◆

.

The following matrix has neither pole nor zero at � = ↵0, ↵�10 :

Q�1L

Q0,L⇠�Q�10,LQL

=⇡↵0ppQ�1

L

⇠�

QL

⇡�1↵0pp

=

✓↵(�) �(�)�(�) �↵(�)

◆.

The residue of ⇠�

at � = 0 is R0,L�⇠�1R�10,L. As R0,L 2 ⇤+SL2(C), we conclude

that the residue takes values in �R+�0 10 0

�. Therefore ⇠

�

2 Pg�2(�). By

remark 2.6 on Isometric normalization we conclude that ⇠�

= g(�)�1⇠�

g(�) 2I(a). ⇤

Using the relation (6.2) between left and right factors immediately yields


Corollary 6.4. Let ⇠�

2 I(a) with ⇠↵0 6= 0 and ord

↵0a(�) � 2. Then ⇠�

uniquely factorizes as

⇠�

= p(�)hL

0,↵0 ⇠� h

�1L

0,↵0

with (L0, ⇠�

) 2 CP1 ⇥ I�

(a) where L0 = Q�10,LL and L = ker ⇠

↵0 .

Remark 6.5. The factorizations of Proposition 6.3 and Corollary 6.4 give

rise to pairs in (L0, ⇠�

) 2 CP1⇥I�

(a), and we say that we decompose such ⇠�

into (L, ⇠�

) or (L0, ⇠�

) or into (L, ⇠�

) or (L0, ⇠�

) where ⇠�

= g(�)�1⇠�

g(�) 2I(a) and L = g(�)�1L, L0 = g(�)�1L0 depending on the context.

A special situation occurs when L0? is an eigenline of ⇠↵0 .

Proposition 6.6. Suppose ⇠�

decomposes into (L0, ⇠�

) and ⇠↵0 L

0? = L0?.Then ⇠

�

has zeroes at � = ↵0, 1/↵0. Furthermore the singularity of the spec-tral curve is removable and up to a conformal change of coordinate the poten-

tials ⇠�

and ⇠�

induce the same extended frame F�

(p(0)z) = F�

(z).

Proof. We prove that

dimC ker ⇠↵0 = dimC ker p(↵0)⇡

�1L

0 (↵0)⇠↵0⇡L0(↵0) = 2 .

If L0? =< v2 > is an eigenline of ⇠↵0 then ⇡

L

0(�)v2 =q

1�↵0�

��↵0v2 and

⇡�1L

0 (↵0)⇠↵0⇡L0(↵0)v2 = µv2 and ⇠↵0v2 = 0 .

If L0 =< v1 >, we have ⇡L

0(�)v1 =q

��↵01�↵0�

v1 and

p(↵0)⇡�1L

0 (↵0)⇠↵0⇡L0(↵0)v1 = 0 .

Then ⇠�

has a zero at � = ↵0 and we can remove it without changing theextended frame by Proposition 4.4. ⇤Terng-Uhlenbeck formula. Let ⇠

�

2 Pg

with ⇠↵0 6= 0 , ord

↵0a(�) � 2 for

some ↵0 2 C⇥ \ S1. Suppose ⇠�


), and let 0 < r <min{|↵0, 1/|↵0|}. Now consider the unitary factor F

�

: R2 ! ⇤r

SU2(C) ofthe r-Iwasawa decomposition

exp(z ⇠�

) = F�

B�

and define F�

: R2 ! ⇤r

SU2(C) to be the unitary factor of the r-Iwasawadecomposition

exp(z p(�) ⇠�

) = F�

B�

.

Terng-Uhlenbeck [30] obtained a relationship between F�

and F�

and found

F�

(z) = hL

0,↵0 F�

(z)h�1L

0(z),↵0with L0(z) = tF

↵0(z)L0 .

We provide a proof of this in the appendix. We next show that closing condi-tions are preserved when changing the first factor in the factorization (L0, ⇠

�

).


Proposition 6.7. If there is (L00, ⇠�) 2 CP1 ⇥ I

�

(a), such that

⇠0,� = p(�)hL

00,↵0

⇠�

h�1L

00,↵0

induces a minimal annulus with period ⌧ , then for any L01 2 CP1, the extended

frame associate to ⇠1,� = p(�)hL

01,↵0

⇠�

h�1L

01,↵0

produces a ⌧ -periodic minimal

annulus.

Proof. The annulus induced by F0,� is periodic at � = 1, so F0,1(z + ⌧) =F0,1(z). In particular F0,1(⌧) = and the solution of sinh-Gordon equation isperiodic !0(z+ ⌧) = !0(z). This period condition on !0 implies ⇣0,�(z+ ⌧) =⇣0,�(z) for any � 2 C⇥ and z 2 C, since entries of ⇣0,� depend only on !0

and its derivatives. Now we remark that the associate polynomial Killingfield ⇣0,�(z) 2 I(a) decomposes uniquely as (L0

0(z), ⇣�(z)) 2 CP1 ⇥ I�

(a)where L0

0(z) = Q�10,L0(z)

L0(z) and L0(z) = ker ⇣0,↵0(z). Thus ⇣0,↵0(z) periodic

implies that L0(z) and L00(z) are periodic. The decomposition of ⇣0,�(z) is

given by

⇣0,�(z) = p(�)gL0(z),↵0

⇣�

(z)g�1L0(z),↵0

= p(�)hL

00(z),↵0

⇣�

(z)h�1L

00(z),↵0

and ⇣�

(z+ ⌧) = ⇣�

(z) for any � 2 C⇥. We can recover explicitly this relationby using the decomposition with L0 = Q�1

0,LL and the formula

F�

(z) = hL

00,↵0

F�

(z)h�1L

00(z),↵0

where L00(z) =

tF↵0(z)L

00. Then

⇣0,�(z) = F�

(z)�1⇠0,�F�

(z)

= hL

00(z),↵0

F�1�

(z)h�1L

00,↵0

⇠0,�hL

00,↵0

F�

(z)h�1L

00(z),↵0

.

Since L0(z + ⌧) = L0(z), we have L00(z + ⌧) = L0

0(z). Periodicity F1(⌧) =implies F1(⌧) = , and then there is a ⌧ -periodic extended frame without abubbleton associate to the polynomial Killing field ⇣

�

. Now we consider ⇠1,�associate to (L0

1, ⇠�) and we prove that its extended frame has the same period⌧ . To see that it remains to prove that F

↵0(⌧) = to conclude the periodicity

L01(⌧ + z) = tF

↵0(⌧)L01(z) = L0

1(z).

If L00(⌧) = L0

0, then L00 = L0

0(⌧) = t

¯F↵0(⌧)L

00 and L0

0 is an eigenvalue oftF

↵0(⌧) = F�11/↵0

(⌧), hence (L00)

? is an eigenline of F↵0(⌧).

Now ⇣�

(⌧) = ⇣�

(0) for all � 2 C⇥ implies [⇠0,↵0 , F↵0(⌧)] = 0. Thus (L00)

?

is not an eigenline of ⇠0,↵0 , since the polynomial Killing field ⇠0,↵0 would havea zero and we could remove the singularity of the spectral curve.

Recall that if A 2 SL2(C) and B 2 sl2(C), then [A,B] = 0 implies A =xB + y . This implies F

↵0(⌧) = x⇠0,↵0 + y . Now ⇠0,↵0L?0 6= L?

0 implies

x = 0 and thus F↵0(⌧) = ± . ⇤


Group action on bubbletons. We prove that there is a group action whichacts transitively on the first factor L0 of the decomposition ⇠

�

into (L0, ⇠�

).For ⇠

�

2 Pg

and � 2 C, define

m(�) ⇠�

=

8<

:

⇣�

��↵0+ ��

1�↵0�

⌘�

1�g

2 ⇠�

, when g = 2k + 1 ,⇣

�

��↵0+ ��

1�↵0�

⌘(�

�g

2 + �1�g

2 )⇠�

, when g = 2k .

Let exp(m(�)⇠�

) = F�

(�)B�

(�) be the r-Iwasawa factorization for r <min{|↵0|, 1/|↵0|}. We have a complex 1-dimensional isospectral group ac-tion ⇡ : C⇥ I(a) ! I(a) defined by

(6.3) ⇠�

(�) := ⇡(�) ⇠�

= F�

(�)�1 ⇠�

F�

(�) = B�

(�) ⇠�

B�1�

(�) .

The potential ⇠�

(�) has a decomposition (L0(�), ⇠�

(�)), and we prove that⇠�

(�) = ⇠�

is invariant under this action.

Theorem 6.8. Suppose a potential ⇠�


) 2 CP1⇥ I(a).Then the action (6.3) on ⇠

�

preserves the second term ⇠�

. If det ⇠↵0 6= 0, then

C acts on L0 2 CP1\{L01, L

02} transitively where (L0

1)?, (L0

2)? are eigenlines

of ⇠↵0 and fixed points of the action. If det ⇠

↵0 = 0, then C acts on L0 2CP1\{L0

3} transitively where (L03)

? is the eigenline of ⇠↵0 .

Proof. Using remark 6.5, we prove the theorem with L0 = g(�)L0 and ⇠�

=g(�)⇠

�

g(�)�1. Now L0 is an eigenline of ⇠↵0 if and only if L0 is an eigenline of

⇠�

. We consider for � 2 C the map

(�,�) =

8<

:

⇣�

��↵0+ ��

1�↵0�

⌘�

1�g

2 p(�) ⇠�

, when g = 2k + 1 ,⇣

�

��↵0+ ��

1�↵0�

⌘(�

�g

2 + �1�g

2 )p(�) ⇠�

, when g = 2k .

Then satisfies t (1/�,�) = � (�,�) and 2 ⇤r

su2. We have by the r-Iwasawa decomposition exp( (�,�)) = F

�

(�). Further, since B�

(�) = wehave

F�1�

(�) ⇠�

F�

(�) = B�

(�) ⇠�

B�1�

(�) = ⇠�

.

Suppose ⇠�

(�) has a decomposition (L0(�), ⇠�

(�)). We prove next that ⇠�

(�) =⇠�

is invariant by the group action m(�)⇠�

.From F

�

(�)B�

(�) = hL

0,↵0 exp( (�,�))h

�1L

0,↵0

= hL

0,↵0 F�

(�)h�1L

0,↵0

, andby the Terng-Uhlenbeck formula we obtain

F�

(�) = hL

0,↵0 F�

(�)h�1L

0(�),↵0where L0(�) = tF

↵0(�)L0 .

Applying the action and the invariance of the conjugation of ⇠�

by F�

(�) wehave

⇠�

(�) = ⇡(�)⇠�

= p(�)hL

0(�),↵0⇠�

h�1L

0(�),↵0.


This proves that ⇠�

is invariant while L0(�) changes under the group action.We consider now the map � 7! L0(�), and prove that this map spans CP1 (ifL0(0) 6= L0

0 and L0(0) /2 {L01, L

02, L

03} ).

First we assume det ⇠↵0 6= 0 (so also det ⇠1/↵0

6= 0). We denote by v1, v2the eigenvectors of ⇠1/↵0

, and by ⌫1, ⌫2 the corresponding eigenvalues. Since

trace ⇠�

= 0, we have ⌫2 = �⌫1. Now we have for

�0 =

(↵g/2�5/20 (1� |↵0|2) for g odd

↵g/2�30 (1� |↵0|2) for g even

that t (↵0,�) = � (1/↵0,�) = �� 0 ⇠1/↵0and tF

↵0(�) = exp(� (1/↵0,�)).With L0 =< xv1 + yv2 > we obtain

L0(�) = tF↵0(�)(xv1 + yv2) = exp(� (1/↵0,�))(xv1 + yv2)

= exp(��0⌫1)x v1 + exp(��0⌫2)yv2 = w .

HencetF

↵0(�)L0 =< w >

and the map L0 : CP1 ! CP1 given by L0(�) =< w > is surjective since

� : CP1 ! hexp(��0⌫1)xv1 + exp(��0⌫1)yv2i 2 CP1

is surjective.Now assume det ⇠

↵0 = det ⇠1/↵0= 0. There exists a basis (v1, v2) with

⇠1/↵0v1 = 0 and ⇠1/↵0

v2 = v1, and then

exp(� (1/↵0, �))(x v1 + y v2) = (x� ��0y) v1 + y v2.

For y 6= 0 the map � : CP1 ! h(x� ��0y)v1+ yv2i 2 CP1 is surjective. Wheny = 0 then L0 = v1 is an eigenvector of ⇠

↵0 , and there is no bubbleton. ⇤

7. Spectral curves of the Riemann family

The Riemann family consists of embedded minimal annuli in S2⇥R that arefoliated by horizontal constant curvature curves of S2. From [12] these annulican be conformally parameterized by their third coordinate with Q = 1

4 (dz)2

and the metric ds2 = cosh2 ! |dz|2 is obtained from real-analytic solutions ofthe Abresch system [1]

(7.1)

⇢�! + sinh! cosh! = 0 ,!xy

� !x

!y

tanh! = 0 .

The second equation is the condition that the curve x 7! (G(x, y), y) hasconstant curvature. This condition induces a separation of variables of thesinh-Gordon equation, and solutions can be described by two elliptic functions

(7.2) f(x) =�!

x

cosh!and g(y) =

�!y

cosh!


of real variables x and y respectively, and for c < 0, d < 0 solve the system

�(fx

)2 = f4 + (1 + c� d)f2 + c , � fxx

= 2f3 + (1 + c� d)f ,

�(gy

)2 = g4 + (1 + d� c)g2 + d , � gyy

= 2g3 + (1 + d� c)g .

We can then recover the function ! by

(7.3) sinh! = (1 + f2 + g2)�1(fx

+ gy

) .

The spectral curves of members of the Riemann family have spectral genus 0,1 or 2. The spectral genus zero case consists of flat annuli �⇥R, where � ⇢ S2is a great circle. In this case the solution of the sinh-Gordon equation is thetrivial solution ! ⌘ 0. The spectral genus 1 case consists of solutions of thesinh-Gordon equation that only depend on one real variable. The correspond-ing minimal annuli are analogous to associate family members of Delaunaysurfaces. In particular the spectral genus 1 case contains the rotational annuliand helicoids, and these are foliated by circles. Amongst the spectral genus 2surfaces, the corresponding minimal annuli are again foliated by circles, butno longer have rotational symmetry. This condition endows the spectral curvewith an additional symmetry.

Spectral genus 0. We first study annuli with spectral curves of genus 0.Inserting the trivial solution of the sinh-Gordon equation !0 ⌘ 0 into (1.3),and setting � = 1, gives

(7.4) ↵(!0) =1

4

✓0 ��1

0

◆dz +

1

4

✓0� 0

◆dz

The solution of F�1�

dF�

= ↵(!0), F�

(0) = is given by

F�

= exp

✓

4

✓0 ��1z + z

z + � z 0

◆◆

=

0

@ cos�14 (

zp�

+ zp�)� sin

�14 (

zp�

+z

p�)�

p�p

� sin�14 (

zp�

+ zp�)�

cos�14 (

zp�

+ zp�)�

1

A .

(7.5)

The horizontal part of (F�

�3 F�1�

, Re(� ��1/2z)) computes to

F�

�3 F�1�

=

✓cos

�Re (z ��1/2)

��1/2 sin

�Re (z ��1/2)

�

��1/2 sin�Re (z ��1/2)� � cos

�Re (z ��1/2)

�◆

.

Identifying su2⇠= R3,

�w u+ v

�u+ v � w

� ⇠= (u, v, w), evaluating the associatedfamily at � = 1, and writing z = x + y, we obtain the conformal minimalimmersion C ! S2 ⇥ R ⇢ R4 given by

X1(x, y) =�sinx, 0, cosx, y

�.

Restricting X1 to the strip (x, y) 2 [0, 2⇡]⇥ R then gives an embedded min-imal flat annulus in S2 ⇥ R. Evaluating the associated family at some other

point �0 2 S1, then ⌧ -periodicity requires that Re (⌧��1/20 ) 2 2⇡ Z.


We next compute the corresponding spectral data (a, b). Since

F�

= exp(z ⇠�

� z ⇠1/�t

) for ⇠�

= 4

�0 �

�1

1 0

�

coincides with the extended flat frame (7.5) computed above, we concludethat ⇠

�

is a potential for the flat surface. Hence a(�) = �� det ⇠�

= �1/16,and the spectral curve (2.10) is the 2-point compactification of {(⌫, �) | ⌫2 =��1/16}. The flat annulus has the simplest possible spectral curve. It isa genus zero hyperelliptic curve, so a double cover of CP1 with two branchpoints.

The eigenvalues of F�

in (7.5) are exp(± 4 (z��1/2 + z�1/2)). Therefore

the logarithmic eigenvalue (up to sign) of the monodromy with respect to thetranslation z 7! z + 2⇡ is

lnµ(�) =⇡

2

⇣��1/2 + �1/2

⌘.

Then

d lnµ =⇡ (�� 1)

4�3/2d� =

⇡ (1� �)

16�2⌫d�

since �3/2 = �4 �2⌫. Thus b(�) = ⇡

16 (1� �).

Spectral genus 1. We apply the Pinkall-Sterling iteration to the case where! satisfies ↵!

z

+�!z

= 0 for ↵,� 2 C. The relation implies that |↵| = |�| andup to a change of coordinate we assume without loss of generality that !

x

= 0.Then !

z

= �!z

, and we are exactly in the setting of Abresch’s system [1] andthere is a constant d < 0 with �b2

y

= (b2 + 1)(b2 + d) where

(7.6) b(y) =�!

y

cosh!and sinh! =

by

1 + b2.

We now use the Pinkall-Sterling iteration to compute the polynomial Killingfield. Starting with u�1 = ��1 = 0 and ⌧�1 = /4, and using 4!

zz

=� 1

2 sinh(2!), gives

u0 = �4 !z

⌧�1 = !z

, �0 = � e2!⌧�1 + 4 � u0;z = 14 � e�2! .

We use the function �0 to compute ⌧0 = 2 �( 12�0�u0;z). We have u0 = �!z

,and then

�0;z = �4!z

!zz

= 14 (cosh(2!))z ,

�0;z = �!z

sinh! cosh! = 14 (cosh(2!))z .

Then ⌧0 = 2 � (!zz

+ 18 cosh(2!)) =

14 � e

�2!.At the next step we find u1 = �2 ⌧

o;z � 4 !z

⌧0 = 0, �1 = � e2!⌧0 = 4 and⌧1 = 0. This gives the polynomial Killing field (2.2) as

⇣�

=4

✓ �2!y

e!��1 + � e�!

� e�! + e!� 2!y

◆.


Then a(�) = �� det ⇣�

= � 116

�� + (2 cosh(2!) + 4!2

y

)�+ ��2�. Using (7.6)

gives2!2

y

+ cosh 2! = 2b2 cosh2 ! + 1 + 2 sinh2 ! = 1� 2d .

so that a(�) = � 116

�� + 2(1� 2d)�+ ��2

�. Its roots are (2d�1±2

pd2 � d) �.

If � = ±1, then this polynomial satisfies the additional symmetry

(7.7) �2ga(1/�) = a(�) ,

for g = 1, and it has two real roots. The corresponding spectral curveis a double cover of CP1 branched at 4 points, so a hyperelliptic curve ofgenus 1. To close the surface, we have to close the third coordinate withQ = 1

4��10 (dz)2 = 1

4 (dz)2. Riemann annuli of spectral genus 1 satisfy !

x

= 0or !

y

= 0. This means that �0 = ±�. We can parameterize the annulus insuch a way that �0 = 1, and apply the iteration with � = 1. This correspondsto the case where y 7! !(y) depends only on its third coordinate and describesrotational examples. In the other case where �0 = 1 and � = �1, this corre-sponds to the helicoidal surfaces, where the surface is foliated by horizontalgeodesics. In this case the function ! depends only on the variable x.

Symmetric spectral genus 2. We next consider general real-analytic solu-tions of Abresch’s system. In this case we first prove that they correspond tospectral genus 2 surfaces.

Lemma 7.1. Every solution ! : C ! R of Abresch’s system (7.1) satisfies

(7.8) !zzz

� 2!3z

= � 14 !z

+ 12 (c� d)!

z

.

Proof. Di↵erentiating !z

= � 12 (f � g) cosh(!) gives

!zz

= 14 (f � g)2 sinh(!) cosh(!)� 1

2 (f � g)z

cosh(!) .

Now using the equation of the system, we have

2(f � g)z

= fx

� gy

= (1+f

2+g

2)(g2�f

2+d�c)f

x

+g

y

= g

2�f

2+d�c

sinh(!) .

Then !zz

= tanh(!) !2z

� 14 coth(!)

�g2 � f2 + d� c

�and thus

tanh(!) !zz

� tanh2(!)!2z

+ 14 (g

2 � f2) = 14 (c� d) 2 R

since the imaginary part is tanh(!)(!xy

� tanh(!)!x

!y

) = 0. Using theexpression for f

x

+ gy

and gy

� fx

we obtain

2fx

= (1 + f2 + g2) sinh(!)� (f2 � g2 + c� d) sinh�1(!) ,

2gy

= (1 + f2 + g2) sinh(!) + (f2 � g2 + c� d) sinh�1(!) .

To understand higher order derivative we write

�2(g2 � f2 + (d� c))z

= (2ffx

+ 2 ggy

)

= (f + g)(1 + f2 + g2) sinh(!)� (f � g)(f2 � g2 + c� d) sinh�1(!) .


We can check that

18 (f + g) cosh(!) = � 1

4!z

,

18 (f + g)(f2 + g2) cosh(!) = cosh(!)

8 (f � g)3 + cosh(!)2 fg(f � g)

= � !

3z

cosh2(!)� fg!

z

.

Now we compute

!zzz

� 2!3z

= �2!3z

+ !3z

sech2(!) + 2 tanh(!) !z

!zz

+ 14 !z

csch2(!)�g2 � f2 + d� c

�

� 14 coth(!)

�g2 � f2 + d� c

�z

= �2!3z

+ !3z

sech2(!)� !3z

sech2(!)� fg!z

+ 2!z

�tanh2(!) !2

z

+ 14 (f

2 � g2 + c� d)�

+ 14 !z

(g2 � f2 + d� c) csch2(!)� 14 !z

� 18 (f � g)(f2 � g2 + c� d) coth(!) csch(!)

= � 14 !z

+ !z

�2 tanh2(!) !2

z

� 2!2z

�

+ 12 (c� d)!

z

+ 12 (f � g)2!

z

= � 14 !z

+ 12 (c� d)!

z

.

⇤

Next we use the iteration of Pinkall-Sterling to compute the spectral curveassociated to the algebraic relation (7.8). It is a priori a one-parameter familyof algebraic relations but ! itself encodes other invariant quantities than theone found in the expression of !

zz

. The Pinkall-Sterling iteration gives

u�1 = u2 = ��1 = ⌧2 = 0 , u0 = !z

, u1 = �!z

,

�0 = 14 �e

�2! , �1 = 2 e�2!�!zz

+ !2z

+ 14 (c� d)

�, �2 = � 1

4 �,

⌧�1 = 14 , ⌧0 = 2 �

�!2z

� !zz

+ 14 (c� d)

�, ⌧1 = � 1

4 ��2e�2! .

This defines a solution of the Lax equation � by (2.2) of degree N = 2. Toobtain a polynomial Killing field we skew-symmetrize and define

⇣�

(z) =1

2�

�

(z)� �

2�1/�(z)

t

.

Then � ⇣1/�t

= �⇣�

and has ��1-coe�cient 18 e!(1 � �)

�0 10 0

�. For ⇣

�

to be

P2-valued we require that 18 e!(1 � �) 2 R+, which means that � = �1. In

this case, Q = 14��

�1(dz)2 implies that the Sym point is at � = �1 and the


entries of the polynomial Killing field are

↵ = (!z

� �!z

) ,

� = 4��1e! +

�e�!(!2

z

+ !zz

)� e!(!2z

+ !zz

)

+ 14 (c� d)(e�! � e!)

�� 4� e�! ,

� = � 4e�! + �

�e�!(!2

z

+ !zz

)� e!(!2z

+ !zz

)

+ 14 (c� d)(e�! � e!)

�+ 4�

2e! .

To compute the spectral curve, we have only to compute a(�) = �(↵2 + � �)at one point. We choose a point where !(x0, y0) = @

x

f(x0) = @y

g(y0) = 0.At this point !

zz

= !zz

= 0, thus

f(x0) = f0 = �@x

!(x0) =12

��1 + d� c+p��,

g(y0) = g0 = �@y

!(y0) =12

��1 + c� d+p��.

where � = (1 + c� d)2 � 4c = (1 + d� c)2 � 4d.Writing a(�) = � (↵2 + ��) = a0 + a1� + a2�2 + a3�3 + a4�4, and using

!z

(x0, y0) = � 12 (f0� g0) and !z

(x0, y0) = � 12 (f0+ g0), a computation gives

the real coe�cients

a0 = a4 = 116 ,

a1 = a3 = 12 (!

2z

+ !2z

) = 14 (f

20 � g20),

a2 = (!4z

+ !4z

)� 2 (!z

!z

+ !2z

!2z

)� 18 = � 1

8 � 12g

20 � 1

2f20 � f2

0 g20 .

The four real roots of a(�) are �1� 2f20 ± 2

pf20 + f4

0 , 1 + 2g20 ± 2p

g20 + g40 ,and a(�) satisfies the additional symmetry (7.7) for g = 2. In summary thespectral data of the 2-parameter family of the Riemann family are given by

Proposition 7.2. The genus 0 spectral data of an embedded annulus withSym point � = 1 is given by

1) a(�) = � 116 and b(�) = ⇡

16 (1� �)

The genus 1 spectral data of an embedded annulus with Sym point � = 1 isgiven by

2) a(�) = 116↵ (��↵)(↵��1) for ↵ 2 (0, 1) and b(�) = b(0)

�

(��)(��1),

with � 2 (↵, 1) and b(0) 2 R both determined by ↵.

3) a(�) = �116� (�+�)(��+1) for � 2 (0, 1) and b(�) = b(0)

�

(1��)(1+�)and b(0) 2 R determined by �.

The genus 2 spectral data of an embedded annulus with Sym point � = 1 isgiven by

4) a(�) = 116�↵ (� � ↵)(↵� � 1)(� + �)(�� + 1) for ↵,� 2 (0, 1) and

b(�) = b(0)�

(1 + �)(� � �)(�� 1) for � 2 (↵, 1) and b(0) 2 R both

determined by ↵ and �.


In conclusion, the polynomial a satisfies the additional symmetry �2ga(1/�) =a(�) and

a) �g+1b(1/�) = b(�) if a has a root ↵ 2 R+ and b(0) 2 R;b) �g+1b(1/�) = �b(�) if a has only roots in R� and b(0) 2 R.

Proof. We have seen above that spectral curves of the Riemann family havean additional involution (�, ⌫) ! (��1, �1�g⌫), since in all cases �2ga(1/�) =a(�). Now depending on a(�), we construct a function h which satisfies theclosing condition of the annulus. We prove that there are constants � andb(0) such that b satisfies the closing condition of Proposition 5.7.

First we remark that �g+1b(1/�) = �b(�) by construction. We look forh satisfying �⇤h = �h and dh = b d�

⌫�

2 . First we need to prove that h is well

defined on ⌃ (Definition 5.8).In cases 1) and 3), there is a root ↵ 2 (0, 1). Along the segment (↵, 1/↵),

the polynomial b(�) 2 R and ⌫ 2 R. Since b(�) has exactly one root in theinterval (↵, 1) at � 2 (↵, 1), there exists exactly one value of � which cancelsthe following integral for a given ↵ and �. Using the additional symmetry,there is a real � 2 (↵, 1) with

Z 1/↵

↵

b

⌫�2d� = 2

Z 1

↵

b

⌫�2d� = 0 .

Moreover, in cases 2) and 3), we have by the reality condition thatZ �1/�

��

b

⌫�2d� = 0 .

Now the function h with dh = b d�

⌫�

2 and �⇤h = �h is well defined on ⌃, thecurve with the two cycles around (�1/�, ��) and (↵, 1/↵) removed.

For : (�, ⌫) ! (�, ⌫) on ⌃ have ⇤h = �h, since b(�) = b(�).On the real axis between � = 0 and � = ↵ (or � = 1 in the case 2)), the

polynomial a takes real positive values. Then the segment (0, ↵] (or (0, 1] inthe case 2) is a set of fixed points for the involution . On this segment wededuce that h = ⇤h and the function h is purely imaginary on this segment.Since the integral

R 1

↵

d lnµ = 0, the function h is imaginary at � = 1 andh(↵) = h(1) 2 R.

The involution % in (2.11) leaves S1 invariant, and we have %⇤dh = �dh.Hence dh 2 R on S1. Thus on the unit circle h stays imaginary, so inparticular h 2 R at � = �1.

The segment (�1, ��) is a set of fixed points for and the function h 2 Ron this segment. Since on the real line the function a(�) changes sign andbecome real negative on (��, 0), the function h 2 R on this segment. We canthen deduce that h(��) = 0 at this point. Now we can choose the value ofb(0) to get a multiple value of ⇡ at the sym point � = 1. This proves theclosing condition and concludes the proof of the proposition. ⇤


Lemma 7.3. If � is a root of b, then the corresponding function |µ(�)| 6= 1.

Proof. For � 2 [↵, ↵�1], the function h = lnµ is real andR 1

↵

dh = 0. Then �is a root of dh and is contained in (↵, 1). Since Reh(↵) = Reh(1) = 0, thevalue � is the local critical point of h. Then Reh(�) 6= 0, and thus |µ| 6= 1. ⇤

Appendix A. Terng-Uhlenbeck Formula

Proposition A.1. Let hL

0,↵0 2 Hr

↵0the simple factor with ↵0 2 C⇥ \ S1,

with r < min{|↵0|, 1/|↵0|} and L0 2 CP1. Then

F�

(z) = hL

0,↵0 F�

(z)h�1L

0(z),↵0with L0(z) = tF

↵0(z)L0 .

Proof. By r-Iwasawa decomposition

F�

(z)B�

(z) = exp(z⇠�

) = exp(zp(�)hL

0,↵0 ⇠�h

�1L

0,↵0

)

= hL

0,↵0 exp(zp(�)⇠�)h

�1L

0,↵0

2 ⇤r

SL2(C)

and hL

0,↵0 2 ⇤+

r

SL2(C) (hL

0,↵0(0) = Q1,L0Q

L

0⇡�1↵0

(0)Q�1L

0 = R�11,L0 at � = 0

and ⇡ is holomorphic for r < |↵0|). Thenexp(zp(�)⇠

�

)h�1L

0,↵0

= F�

(z)B0�

(z) = F�

(z) B�

(z)h�1L

0,↵0

.

Now we have

F�

(z)B�

(z) = hL

0,↵0 exp(zp(�)⇠�)h

�1L

0,↵0

= hL

0,↵0 F�

(z)B0�

(z)

= (hL

0,↵0 F�

(z)H)(H�1B0�

(z)) .

By uniqueness of the r-Iwasawa decomposition we have only to prove that if

H = h�1L

0(z),↵02 Hr

↵0with L0(z) = t

¯F↵0(z)L

0 then hL

0,↵0 F�

(z)H 2 ⇤r

SU2.Clearly

hL

0,↵0 F�

(z)h�1L

0(z),↵02 ⇤

r

SL2(C)

is holomorphic on Ar

away from ↵0 and 1/↵0, and SU2-valued on S1. At theroots ↵0 and 1/↵0, we have simple poles and we have to study the residues of

G�

(z) = ⇡�1L

0 (�)F�

(z)⇡L

0(z)(�).

Now we consider the simple factor L0(z) = tF↵0(z)L

0. Let (L0(z), L0(z)?) be

an orthonormal basis of C2. Note that L0(z) = t

¯F↵0(z)L

0 = F�11/↵0

(z)L0 and

F↵0(z)

�1L0? = L0(z)?. When �! 1/↵0, we have

lim�!1/↵0

G�

L0(z) = lim�!1/↵0

q��↵01�↵0 �

QL

0⇡�1↵0

Q�1L

0 F�

¯F t

↵0L0(z) = L0 ,

lim�!↵

�10

(1�↵0�)G�

L0?(z) = lim�!↵

�10

(1�↵0�)q

1�↵0 �

��↵0Q

L

0⇡�1↵0

Q�1L

0 F�

L0?(z) = 0 .


When �! ↵0, we compute

lim�!↵0

(�� ↵0)G�

L0(z) = lim�!↵0

(�� ↵0)q

��↵01�↵0 �

QL

0⇡�1↵0

Q�1L

0 F�

L0(z) = 0 ,

lim�!↵0

G�

L0?(z) = lim�!↵0

q1�↵0 �

��↵0Q

L

0⇡�1↵0

Q�1L

0 F�

F�1↵0

L0?(z) = L0? .

This proves the proposition. ⇤

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Mannheim, Germany.E-mail address: [email protected]

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