Uniqueness of the Riemann minimal exampleshera.ugr.es/doi/15000011.pdf · 2005-02-24 · minimal...

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Uniqueness of the Riemann minimal examples William H. Meeks III 1,* , Joaquı´n Pe´rez 2,** , Antonio Ros 2,** 1 Mathematics Department, University of Massachusetts, Amherst, MA 01003, USA 2 Departamento de Geometrı´a y Topologı´a, Universidad de Granada, E-Fuentenueva, 18071 Granada, Spain (FAX: 34 58 243281) Oblatum 30-V-1997 & 5-VIII-1997 Abstract. We prove that a properly embedded minimal surface in R 3 of genus zero with infinite symmetry group is a plane, a catenoid, a helicoid or a Riemann minimal example. We introduce the language of Hurwitz schemes to understand the underlying moduli space of surfaces in our setting. 1 Introduction In a posthumously published paper, B. Riemann [20] (see also [4], pp. 461– 464) used elliptic functions to classify all minimal surfaces in R 3 that are foliated by circles and straight lines in horizontal planes. He showed that these examples are the plane, the helicoid, the catenoid and a one-parameter family fR t g t>0 with infinite topology. The new surfaces R t , called Riemann minimal examples, intersect horizontal planes in lines at precisely integer heights, with the radius of the circles going to infinity near the lines. Each surface R t is also invariant under reflection in the x 1 ; x 3 -plane and by the translation by t; 0; 2. Furthermore, in the complement of a solid open cylinder in the direction of t; 0; 2, the surface R t consists topologically of disjoint punctured disks, each one close and asymptotic to a plane at integer height. Each R t is naturally conformally dieomorphic to a vertical cylinder in R 3 , punctured at integer heights corresponding to the ends. In particular, R t is conformally equivalent to C C f0g punctured in a discrete set of points with 0 and 1 being limit points of the planar ends. In topological Invent. math. 131, 107–132 (1998) * The research described in this paper was supported by research grants NSF DMS 9505101 and NSF DMS 9312087 ** Research partially supported by a DGYCYT Grant No. PB94-0796

Transcript of Uniqueness of the Riemann minimal exampleshera.ugr.es/doi/15000011.pdf · 2005-02-24 · minimal...

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Uniqueness of the Riemann minimal examples

William H. Meeks III1,*, Joaquõ n Pe rez2,**, Antonio Ros2,**

1 Mathematics Department, University of Massachusetts, Amherst, MA 01003, USA2 Departamento de GeometrõÂ a y TopologõÂ a, Universidad de Granada, E-Fuentenueva,

18071 Granada, Spain (FAX: 34 58 243281)

Oblatum 30-V-1997 & 5-VIII-1997

Abstract. We prove that a properly embedded minimal surface in R3 ofgenus zero with in®nite symmetry group is a plane, a catenoid, a helicoid ora Riemann minimal example. We introduce the language of Hurwitzschemes to understand the underlying moduli space of surfaces in oursetting.

1 Introduction

In a posthumously published paper, B. Riemann [20] (see also [4], pp. 461±464) used elliptic functions to classify all minimal surfaces in R3 that arefoliated by circles and straight lines in horizontal planes. He showed thatthese examples are the plane, the helicoid, the catenoid and a one-parameterfamily fRtgt>0 with in®nite topology. The new surfaces Rt, called Riemannminimal examples, intersect horizontal planes in lines at precisely integerheights, with the radius of the circles going to in®nity near the lines. Eachsurface Rt is also invariant under re¯ection in the �x1; x3�-plane and by thetranslation by �t; 0; 2�. Furthermore, in the complement of a solid opencylinder in the direction of �t; 0; 2�, the surface Rt consists topologically ofdisjoint punctured disks, each one close and asymptotic to a plane at integerheight.

Each Rt is naturally conformally di�eomorphic to a vertical cylinder inR3, punctured at integer heights corresponding to the ends. In particular, Rt

is conformally equivalent to C� � Cÿ f0g punctured in a discrete set ofpoints with 0 and 1 being limit points of the planar ends. In topological

Invent. math. 131, 107±132 (1998)

* The research described in this paper was supported by research grants NSF DMS 9505101 and

NSF DMS 9312087

**Research partially supported by a DGYCYT Grant No. PB94-0796

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terms, this means that Rt is a properly embedded periodic minimal surface inR3 of genus zero, with exactly two limit ends.

Our main theorem is that the Riemann examples are unique in thissetting:

Theorem 1 If M is a properly embedded periodic minimal planar domain in R3

with two limit ends, then M is one of the Riemann examples.

The above theorem is equivalent, after taking quotients by a translation orscrew motion symmetry, to proving that a genus one properly embeddedminimal surface with planar ends in a nonsimply connected ¯at three-manifold is the quotient of some Rt. We will demonstrate Theorem 1 byproving the uniqueness of Riemann examples in the quotient space. Thisresult follows from an exhaustive study of the global and local properties ofthe Flux map, which associates to each surface in our setting its ¯ux along ahorizontal section. In our analysis, we introduce Hurwitz schemes, whichseem to be the ideal setting for analyzing the type of moduli space and/oruniqueness questions that often arise in the global theory of minimal sur-faces.

Recent work of Kusner et al. [6] shows that an embedded periodic genuszero minimal surface that is not the plane or the helicoid must have exactlytwo limit ends. Their result, together with our uniqueness theorem, showsthat the following more general classi®cation holds:

Theorem 2 Suppose that M is a properly embedded minimal surface in R3 withgenus zero. If the symmetry group of M is in®nite, then M is a plane, acatenoid, a helicoid or a Riemann example. In particular, a properly embeddedminimal surface with genus zero has in®nite symmetry group if and only if it isfoliated by circles and/or lines in parallel planes.

For previous partial results on Theorem 1, see [5, 10, 12, 13, 17±19, 22].See also [13] for an up to date complete discussion and further references tothe history behind our Theorem 1. More results about singly-periodicminimal surfaces with planar ends can be found in [1, 2].

2 Proof of the Main Theorem

We now prove Theorem 1, assuming the results in later sections. Let M be aperiodic properly embedded minimal surface in R3 with genus zero and twolimit ends. Since every periodic minimal surface with more than one end hasa top and a bottom limit end, it follows that all the middle ends are simpleends, which in the case of ®nite genus means annular ends (see [7] for acomplete discussion of the ordering of the ends of a properly embeddedminimal surface). The structure theorem in Callahan et al. [3] implies thatthere exists a nontrivial screw motion or a translation in R3 which preservesthe surface and such that the quotient surface M=K, K being the cyclic group

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generated by the above rigid motion, has genus one and ®nitely many planarparallel ends. In particular, M=K has ®nite total curvature. For the re-mainder of the proof, we will suppose that the ends of M=K are horizontal.By a result of Pe rez and Ros [19], we have that K must be generated by atranslation. Hence, Theorem 1 follows from the next one:

Theorem 3 Let M be a properly embedded minimal surface in R3=T of genusone and a ®nite number of planar ends, T being a nontrivial translation. Then,M is a quotient of a Riemann example.

We will letS denote the space of properly embedded minimal (oriented) toriin a quotient of R3 by a translation T , that depends on the surface, with 2nhorizontal planar (ordered) ends. By the maximum principle at in®nity, see[11] and also [16], the ends are separated by a positive distance. Moreover,embeddedness insures that consecutive ends have reversed (vertical) limitnormal vectors. The allowed orders for the ends with normal vector�0; 0;ÿ1�, p1; . . . ; pn, and for the ends with normal vector �0; 0; 1�, q1; . . . ; qn,will be those in which the list �p1; q1; . . . ; pn; qn� corresponds to consecutiveends in the quotient space. We will identify in our space S two surfaceswhich di�er by a translation that preserves both orientation and the order ofthe above list of ends, or equivalently, that maps the ®rst end in the list ofone of the surfaces into the ®rst end of the list of the second surface.

A surface M 2S cuts transversally any horizontal plane nonasymptoticto its ends in a compact Jordan curve c. We will orient c in such a way thatthe ¯ux vector v along c has positive third coordinate. As the ¯ux is ahomological invariant and vanishes around a planar end, it follows that vdoes not depend on the height of the intersecting plane. From the results in[19] we know that v cannot be vertical. We will rescale our surfaces so thatthis ¯ux has always third coordinate equal to one. We de®ne the Flux map

F : S ÿ! R2 ÿ f0g

by letting F �M� be the horizontal part of v.Denote by R the subset of S consisting of the Riemann examples and

their rotations around the vertical x3-axis, see more details in Sect. 3. Weclaim R is open and closed in S: it is proved in Pe rez [18] that any smalldeformation in S of a Riemann example must be another Riemann ex-ample. This gives the openness of R. For closedness, observe that Riemannexamples are characterized by the fact of being foliated by circles or lines inhorizontal planes, see [20]. So, if a sequence of Riemann examples convergesto a surface M 2S in the uniform topology on compact subsets, thenhorizontal sections on M will be circles or lines and, thus, M 2 R.

Proof of Theorem 3. The result follows from a combination of the followingthree statements:� The Flux map is proper (which will be proved in Theorem 5).� The Flux map is open (see Theorem 6).

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� There exists a positive number e such that if M 2S satis®esjF �M�j < e, then M 2 R (Theorem 7).

Indeed, suppose S0 :�SÿR 6� ;. As S0 is open and closed in S and theFlux map F : S ÿ! R2 ÿ f0g is proper (Theorem 5) and open (Theorem 6),it follows that its restriction to S0 is also a proper and open map. In par-ticular, F �S0� � R2 ÿ f0g. This contradicts Theorem 7, which insures thatF �S0� omits a punctured neighborhood of zero in R2. Now the proof iscomplete. (

3 Background

We now recall some general properties to be satis®ed by any surface in thesetting of Theorem 1. With this aim, take an orientable surface M � R3=Tin the space S of properly embedded minimal tori in quotients of R3 bynontrivial translations, with 2n horizontal planar ends. Note that M has®nite total curvature and well-de®ned Gauss map, that we will identify withthe meromorphic function g obtained after stereographic projection fromthe North Pole of the sphere S2. We will also denote by / � @x3

@z dz the heightdi�erential, where z denotes a holomorphic coordinate. Both g;/ extendmeromorphically to the conformal compacti®cation R of M through the 2npunctures corresponding to the ends. As the ends are planar and horizontal,/ has neither poles or zeroes on R, g has double zeroes at n of the ends,p1; . . . ; pn, double poles at the remaining n ends, q1; . . . ; qn, and M has noother points with vertical normal vector. In particular, the minimal surfaceintersects transversally in a compact Jordan curve each horizontal plane inR3=T whose height does not coincide with the height of an end. Denote by cone of these horizontal compact sections. Then, from Weierstrass repre-sentation one easily deduces that the condition to close the real period of thesurface along c is equivalent toZ

c

1

g/ �

Zc

g/ : �1�

and in this case, the horizontal ¯ux along c is given by

F �M ; c� � iZc

g/ � iZc

1

g/ :

At an end pi with g�pi� � 0�2�, both / and g/ have no singularities, and 1g /

has a double pole without residue. Similarly, the period problem at an endqi with g�qi� � 1�2� can be solved if and only if the residue of g/ at qi

vanishes. Finally, note that as the sum of the residues of g/ and 1g / on R is

zero, to solve the period problem at the ends it su�ces to insure that

Respi

1

g/

� �� Resqi g/� � � 0; 1 � i � nÿ 1 : �2�

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Reciprocally, consider a genus one surface R and a pair �g; �c�� where g is ameromorphic function with degree 2n and divisor �g� � p21 . . . p2nqÿ21 . . . qÿ2n ,for distinct points pi; qj 2 R, and �c� is a nonzero homology class in H1�R;Z�.Denote by / the unique meromorphic di�erential on R satisfying

Rc / � 2pi.

Then, �g;/� de®nes a complete immersed minimal surface in a quotientof R3 by a translation if and only if (1) and (2) hold simultaneously.Furthermore, when it exists, the surface above has 2n embedded planarhorizontal ends, although the surface itself may not be embedded.

We now describe brie¯y the Riemann minimal examples, which will becharacterized in this paper. They form a one-parameter family fRkgk>0 ofproperly embedded singly periodic minimal surfaces, each one foliated bycircles and lines in parallel planes, which we will think about as horizontal.In fact, a surface satisfying such hypotheses must be a Riemann example.Each surface Rk is invariant by a cyclic group of translations, such that thequotient with a smaller number of ends is a twice punctured torus that canbe conformally parametrized by f�z;w� 2 C�C

��w2 � z�zÿ k��kz� 1�gÿf�0; 0�; �1;1�g, with Weierstrass data

g�z;w� � z; / � Akdzw

;

where Ak is a nonzero complex number satisfying A2k 2 R. Rk has a re¯ection

symmetry in a vertical plane (which in the parametrization above is the�x1; x3�-plane), and a 180 degree rotation around straight lines orthogonalto the symmetry plane, that are situated at half distance between each pairof consecutive ends and cut the surface orthogonally. If we consider anymultiple of the generator of the translation group leaving Rk invariant, wewill ®nd a quotient surface with the topology of a torus minus 2n points,n 2 N (here we have a di�erent parametrization of the Riemann minimalexamples from the one we used in the introduction).

We will use some standard results in several complex variables theory.For convenience of the reader, we list here these results as stated in the bookof Gri�ths and Harris [9]. A subset V of a complex manifold M is called ananalytic subvariety if for any p 2 M , there exists a neighborhood U of p in Mand a ®nite collection of holomorphic functions f1; . . . ; fk on U such thatU \ V � fq 2 U

��f1�q� � . . . � fk�q� � 0g, see [9], p. 20. In particular, V is aclosed subset. Analytic subvarieties of C are C itself and its discrete subsets.It follows directly from the de®nition that if V is an analytic subvatiety of Mand V 0 is an open and closed subset of V , then also V 0 is an analytic sub-variety of M .

Now we state some important theorems concerning analytic subvarieties.

Riemann's extension theorem ([9] p. 9) Let V be an analytic subvariety of anopen connected subset U of Cn, V 6� U , and f : U ÿ V ÿ! C a boundedholomorphic function. Then, f extends holomorphically to U.

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Proper mapping theorem ([9] p. 395) Let f : M ÿ! N be a holomorphic mapbetween complex manifolds and V � M an analytic subvariety. If f jV isproper, then f �V � is an analytic subvariety of N.

We will apply the result above only in when V is compact. In this case,properness is clear. If moreover N � C, then we conclude that f �V � is a®nite set of points.

In general, nonconstant holomorphic maps between manifolds with thesame dimension are not open maps. However, openness holds for the sub-class of ®nite maps, i.e. maps such that the inverse image of any point is®nite. In fact, we have the following theorem:

Openness theorem ([9] p. 667) Let U be an open subset of Cn with 0 2 U andf : U ÿ! Cn a holomorphic map such that fÿ1�f0g� � f0g. Then, there existsa neighborhood W of 0 in U such that f jW is an open map.

4 The properness of the Flux map

In this section we will prove the properness of the Flux mapF : S ÿ! R2 ÿ f0g on the space of properly embedded minimal genus oneexamples in R3=T for varying T . This is equivalent to proving that iffM�i���i 2 Ng is a sequence of examples whose horizontal ¯uxes (i.e., the two®rst components of the ¯ux vectors) in norm lie in some interval �1e ; e�; e > 0,then a subsequence of the fM�i�gi converges smoothly to another suchexample. Also, some of the results proved here will be useful in under-standing the limit of a sequence fM�i�gi whose horizontal ¯uxes converge tozero.

Lemma 1 Suppose that R � R3=T is a properly embedded non¯at orientableminimal surface with ®nite topology and horizontal planar ends. Also supposethat the absolute value of the Gaussian curvature of R is bounded from aboveby some positive constant c. Let R�t� be a parallel surface at distance t from R,0 � jtj < ���

cp

. Then, R�t� is embedded and has nonnegative mean curvature.Furthermore, the family fR�t���0 � jtj < ���

cp g gives rise to a foliation of a

regular neighborhood of R. In particular, the distance between the ends of R atin®nity is greater than or equal to 2

���cp

and the vertical component of T isgreater than or equal to 4n

���cp

, where 2n is the number of ends of R.

Proof. The proof of this assertion is standard in the case of a closedorientable minimal surface in a ¯at three dimensional torus, by a simpleapplication of the maximum principle for surfaces of nonnegative meancurvature. In the present noncompact setting with the ends of R asymptoticto planes, a simple application of the maximum principle for oppositlyoriented graphs of nonnegative mean curvature shows that the minimaldistance between two disjoint surfaces of nonpositive curvature (and op-posite unit normals), whose ends are asymptotic to planes, cannot go to zero

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at in®nity (see [16] for similar arguments). With this in mind, the proof ofthe Lemma is straightforward and details are left to the reader (see alsoLemma 4 in [21] for a detailed proof of a similar assertion in the case of anembedded minimal surface of ®nite total curvature in R3). (

Lemma 2 For any positive number k, there exists a positive constant e�k� suchthat the following statement holds. Suppose M is a complete minimal surfaceand P is a horizontal plane that intersects M orthogonally at a point x0, anda � M \ P is the intersection curve. If the vertical ¯ux of a is less than k, thenin the intrinsic neighborhood of distance k from x0, there is a point of M wherethe absolute Gaussian curvature is greater than e�k�.

Proof. Suppose that the lemma fails for some positive number k. In thiscase, there exists a complete minimal surface M�i� � R3, passing throughthe origin, with the �x1; x2�-plane P orthogonal to M�i� at the origin, andsuch that the intrinsic neighborhood of the origin, N�i�, of radius k, hasmaximum absolute curvature less than 1

i. Since the N�i� converge to verticalplanar disks as i!1, the vertical ¯ux of N�i� \ P is close to 2k for i large.This implies that the ¯ux of the M�i� is greater than k, which is a con-tradiction. This ®nishes the proof. h

Proposition 1 Suppose that f eM�i���i 2 Ng is a sequence of embedded Riemann-type minimal surfaces in R3 invariant under translation by T �i�, and such thateMi=T �i� � R3=T �i� has 2n horizontal planar ends. Suppose further that thepoint of maximal absolute Gaussian curvature of each eM�i� occurs at the originand equals to one. Then, after choosing a subsequence, we may assume that theeM�i� converge to a properly embedded minimal surface eM1, which must be oneof the following examples.

i) eM1 is a vertical catenoid. In this case, the ¯ux vectors of the eM�i�converge to �0; 0; 2p� and fkT �i�kgi is unbounded.

ii) eM1 is a vertical helicoid. In this case, we may assume after choosing asubsequence that fT �i�gi ! �0; 0; 2pk� for some ®xed positive integer k.Furthermore, the vertical components of the ¯uxes of the eM�i� go to in®nity asi!1.

iii) eM1 is another Riemann-type example. In this case, we may assumeafter choosing a subsequence that T �i� ! T 2 R3, eM1=T � R3=T has 2nhorizontal ends and the ¯uxes of eM�i� converge to the ¯ux of eM1.Proof. Lemma 1 implies that there are uniform local area bounds for thefamily f eM�i�gi. As curvature is also uniformly bounded, it follows, afterchoosing a subsequence, that f eM�i�gi converges smoothly on compactsubsets of R3 to a properly embedded minimal surface eM1 with absoluteGaussian curvature 1 at the origin.

By Lemma 1, the vertical component of T �i� is bounded from below by4n. Suppose that fkT �i�kgi is not bounded from above. Then, eM1 is a limitof surfaces contained in fundamental regions of eM�i�, which would imply

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that eM1 has ®nite total curvature. Since eM�i� has genus zero, the same holdsfor eM1, and by the uniqueness theorem of Lo pez and Ros [14], eM1 is acatenoid. As the eM�i� have no vertical normals, eM1 has the same propertyand thus it is a vertical catenoid. The maximum absolute Gaussian curva-ture of eM1 equals one, so its ¯ux vector equals �0; 0; 2p�. Since f eM�i�giconverges smoothly to eM1, there must be simple closed curves of eM�i�whose ¯uxes converge to �0; 0; 2p� as i!1. Thus, in the case of fkT �i�kgi isnot uniformly bounded, case i) of the Proposition must hold.

Suppose now that eM1 is not as in description i) above. After passing to asubsequence, we will assume that eM�i� converges to eM1. Since fkT �i�kgi isbounded and kT �i�k � 4n, a subsequence of the T �i� converges to somenontrivial vector T 2 R3. Thus eM1 is invariant under translation by T .Suppose that eM1 is simply connected. Since eM1 is periodic and nonplanar,Theorem 2 in [15] states that it must be a helicoid. Since the normal vectorson eM�i� are never vertical, the same holds for eM1. This is su�cient to showthat eM1 is a vertical helicoid and T is vertical. Since the maximum absoluteGaussian curvature of eM1 is one, T � �0; 0; 2pk� for some positive integer k.Since the eM�i� converge uniformly on compact subsets of R3 to eM1 and thevertical helicoid has in®nite vertical ¯ux, the last statement in ii) holds.

Suppose that T �i� ! T as i!1 and eM1 is not simply connected. Weclaim that eM1 is another Riemann-type example. First note that in a fun-damental slab region, eM1 has ®nite absolute total curvature at most 8pn.The reason for this bound on total curvature is that such a fundamental slabregion for eM1 is obtained as limit of parallel slab fundamental regions forthe eM�i�, and in each one of these regions the corresponding surface hasabsolute total curvature 8pn. This implies that the quotient surfaceM1 � eM1=T � R3=T has absolute total curvature at most 8np. Hence, M1has ®nite topology and its ends are annular. By Theorem 1 in [3], the ends ofM1 are planar. Since every such surface has Gauss map that misses onlypossibly two values corresponding to the normals of the ends and the Gaussmap of M1 misses the values f0;1g, then the ends of M1 are horizontal.We now show that M1 has exactly 2n ends, which are the limits of the 2nends of the M�i� � eM�i�=T �i�. By Lemma 1, the spacing between the ends ofthe eM�i� is bounded away from zero and hence, the relative heights of theends of the M�i� converge to distinct heights in R3=T . It is easy to check thatthese limit heights in R3=T correspond to ends of M1, since the level sets ofthese heights are noncompact (because they are smooth limits of noncom-pact curves on the M�i�). As the ®nite total curvature of M1 cannot begreater than the total curvature of the M�i�, it follows that M1 cannot havemore than 2n ends. This ®nishes the proof of the proposition. (

We now prove the main technical result of this section.

Theorem 4 Let fM�i���i 2 Ng �S be a sequence of examples with 2n ends inR3=T �i�, normalized so that the vertical component of the ¯ux of M�i� is one.Suppose that the horizontal ¯uxes of the family fM�i�gi are bounded. Then,the Gaussian curvature of the family fM�i�gi is uniformly bounded.

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Proof. Let ki :� maxM�i��������jKjp

(here K denotes Gaussian curvature), andR�i� :� kiM�i� � R3=kiT �i�. We will denote by eR�i� the lifted surface of R�i�to R3 (and similarly with eM�i� for M�i�). Assume that the theorem fails. Wewill derive a contradiction by proving the existence of almost helicoids inM�i� for i large and then, use them for calculating the horizontal part of the¯ux, which will turn out to be unbounded.

By replacing M�i� by a subsequence, we can suppose that ki is strictlyincreasing. After a translation, we will also assume that a point of maximalabsolute Gaussian curvature of eR�i� occurs at the origin. By Proposition 1, asubsequence of the feR�i�gi converges smoothly on compact sets of R3 toa properly embedded minimal surface H1, which must be a vertical catenoid,a vertical helicoid or another Riemann-type example. Since the vertical¯uxes of the eR�i� go to in®nity as i!1, Proposition 1 also implies that H1

is a vertical helicoid with vertical translation vector T , and a subsequence ofthe kiT �i� converges to T . Again after replacing the M�i� by this subse-quence, we may assume that the surfaces eR�i� converge to H1.

Now let pi : R3=kiT �i� ÿ! R2 denote the linear projection along thedirection of kiT �i� onto the �x1; x2�-plane R2 � R3=kiT �i�. From now on,we will only consider very large values of i, so that inside the solid cylinderpÿ1�D�, D � fx 2 R2jkxk < 1g, the surface R�i� is connected and extremelyclose to a piece of a vertical helicoid containing its axis and kiT �i� is almostvertical. On the compacti®ed genus one surface R�i� obtained by attachingto R�i� its 2n ends, H1�i� :� R�i� \ pÿ1i �D� is a open annulus that does notseparate R�i�, hence its complement is a compact annulus F �i� that containsthe 2n ends of R�i�. If the Gauss map gi of F �i� were never horizontal, thenthe extended projection pi : F �i� ÿ! �R2 [ f1g� ÿ D, sending the endpoints to1, would be a proper submersion and hence a covering map (notethat F �i� has no points above D by de®nition of H1�i�). This is impossiblesince �R2 [ f1g� ÿ D is simply connected but F �i� is not. Denote byS1 � fz 2 S2 � C

��kzk � 1g. Since H1�i� is almost a vertical helicoid inside ofpÿ1i �D�, �gijH1�i��ÿ1�S1� is one simple closed curve on H1�i� that covers evenly(with ®nite multiplicity) S1 through gi. Since the Gauss map is horizontalsomewhere on F �i�, the multiplicity of the above covering is less that thedegree of gi. Now it follows that for every h 2 S1 there exists at least onepoint in F �i� with normal vector h.

Recall that the maps gi : R�i� ÿ! C have at most 2n distinct branchvalues. Hence, there is a ®xed small e > 0 such that for every i, there exists ah�i� 2 S1 � S2 with the disk of radius e in S2 around h�i� disjoint from thebranch locus of gi. Renormalize eR�i� by translating a point p�i� with normalvector h�i� to the origin. As before, after choosing a subsequence we mayassume that the R�i� converge to a properly embedded minimal surface H2 inR3=T . If the curvature of H2 is not identically zero, then our previousarguments show that H2 must be a vertical helicoid.

We now show that the curvature of H2 is not identically zero, whichwill imply that H2 is a vertical helicoid. Consider the neighborhoodN�i� � fx 2 R�i���dist�x; p�i�� < 3

2 kTkg of p�i�. If the curvature of H2 were

Uniqueness of the Riemann minimal examples 115

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identically zero, then H2 would contain a component that is a ¯at verticalannulus in R3=T passing through the origin. This means that for i large,N�i� intersects the �x1; x2�-plane in R3=kiT �i� in either 1; 2 or 3 arcs near theorigin, which are C1-close (and converging as i gets large) to straight linesegments. The ®rst part of Lemma 1 shows that there can be just one arc fori large. The one arc intersection implies that for i large, N�i� contains a loopai of R�i� giving rise to the period vector kiT �i�. However, by constructiongi�ai� lies in the disk around h�i� with radius going to zero as i goes to1. Inparticular, gi�ai� eventually lies in the disk on S2 of radius e around h�i�.Since this disk is disjoint from the branch values of gi, the inverse image ofthis disk via gi consists of disks on R�i�, one of which contains ai. But thiscontradicts the fact that ai is homotopically nontrivial on R�i�. This con-tradiction completes the proof that H2 is a vertical helicoid in R3=T .

So far we have shown that for i large, the surfaces R�i� have two largeregions in which R�i� is close to two helicoids, that can occur in the limit. InR2 � R3=kiT �i�, choose disjoint open round disks D1�i�;D2�i� of the sameradius such that the solid cylinders pÿ1i �Dj�i�� capture the forming helicoidsH1;H2 minus neighborhoods of their ends. Furthermore, these disks can bechosen such that Hj�i� :� pÿ1i �Dj�i�� \ R�i� is a connected annulus ``closeto'' Hj, j � 1; 2. After replacing R�i� by a subsequence, we may assume thefollowing properties:

1. The radius r�i� of D1�i� goes to in®nity as i goes to 1.

2. limi!1r�i�ki� 0.

3. The normal line on the boundary ``helices'' of Hj�i� makes an angle ofless than 1

i with �0; 0; 1�.We claim that the extended Gauss map gi of R�i� ÿ H1�i� \ H2�i�� � is con-tained in the two disks of radius 1

i around the North and the South Poles inS2. By the open mapping property of gi on the compacti®cation R�i�, this isequivalent to proving that gi has no horizontal normal vectors onR�i� ÿ H1�i� \ H2�i�� �. If gi has such a horizontal vector, our previousargument used in the construction of H2 proves that for large i we haveanother vertical helicoid forming. Repeating this argument a ®nite numberof times yields a sequence H1�i�;H2�i�; . . . ;Hk�i� of almost helicoids on R�i�for i large. Since the absolute total curvature of R�i� is 8np and each Hj�i�has approximately some positive multiple of 4p for absolute total curvature,this process of construction cannot produce more that 2n such helicoids.The complement of the k helicoids in R�i� gives k closed annular stripsF1�i�; . . . ; Fk�i� (see Fig. 1). This means that one eventually arrives at asequence H1�i�; H2�i�; . . . ;Hk�i� and gi

ÿR�i� ÿ [k

j�1Hj�i��is contained in

some small neighborhood of f0;1g in S2. Consider the projection pi : F1�i�ÿ! R2 [ f1gÿ �ÿ [k

j�1Dj�i�. This projection is a proper submersion andtherefore a covering space. Since @F1�i� has two components, the boundaryof the base space can only have two components, which implies k � 2.

Now consider the implications that these results, already proven for R�i�,have for the original surfaces M�i� in R3=T �i�. In this case we will use some

116 W. H. Meeks et al.

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of the previous notation from the case of R�i�. Let pi : R3=T �i� ÿ! R2

denote the projection on the �x1; x2�-plane in the direction of T �i�, and1ki

H1�i�; 1ki

H2�i� the forming helicoids, contracted by ki. The cylindrical re-gions containing the contracted almost helicoids project onto the disks1ki

D1�i�; 1ki

D2�i�, both with the same radius going to zero as i goes to in®nity.Since a homothety does not change the Gauss map, the Gauss mapgi : M�i� ÿ! S2 (we are working with the same notation as in the case ofR�i�) restricted to M�i� ÿ ÿ 1

kiH1�i� [ 1

kiH2�i�

�is contained in the disks of

radius 1i around the points 0;1 2 S2.

Assume that F1�i�; F2�i� are labeled so that the normal vectors of M�i�are upward pointing on F1�i� and downward pointing on F2�i�. Let L�i�denote the line segment joining @D1�i� to @D2�i� that minimizes the distance.Recall that the compacti®ed torus M�i� is divided into four annuli, two ofthem corresponding to 1

kiH1�i�; 1

kiH2�i� and the other two annuli, that we

denote by F1�i�; F2�i�, corresponding to the two connected components ofthe covering (see Fig. 1 and Fig. 2)

pi : M�i� ÿ 1

kiH1�i� [ 1

kiH2�i�

� �ÿ! R2 [ f1gÿ �ÿ 1

kiD1�i� [ 1

kiD2�i�

� �:

Assume that F1�i�; F2�i� are labeled so that the normal vectors of M�i� areupward pointing on F1�i� and downward pointing on F2�i�. Let L1�i� denotea lift of L�i� to the component F1�i� of the above covering. Since M�i� isembedded, the lifts of L�i� to F1�i� [ F2�i� are cyclicly ordered by height, sowe can chose another lift, L2�i�, to lie on the sheet directly above L1�i�. Notethat L2�i� � F2�i�. We now join the end point of L1�i� in 1

kiH1�i� to the end

point of L2�i� in 1ki

H1�i� by a curve contained in 1ki

H1�i�, and similarly join upthe corresponding end points on 1

kiH2�i� (see Fig. 2).

Note that the joined up curve ci represents in homology the curve forwhich one calculates ¯uxes for M�i�. We choose the orientation of ci so thatthe vertical component of the ¯ux along ci is one. The orientation of ciinduces an orientation on L1�i� and L2�i�. We orient L�i� with the orientationof L1�i�.

Fig. 1

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We now wish to show that the length kL�i�k of L�i� goes to in®nity asi!1. Initially, we can assume that kci \

ÿ1ki

H1�i� [ 1ki

H2�i��k ! 0 as

i!1. We now give an estimate for the vertical ¯ux of ci. Since the normalvector to M�i� along ci \ F1�i� [ F2�i�� � makes an angle of less that 1i with thevertical, the vertical ¯ux along Lj�i� is less than or equal to sin 1

i

ÿ �kL�i�k.Hence the total vertical ¯ux, which is one, is bounded from above by ci \

1

kiH1�i� [ 1

kiH2�i�

� � � 2 sin1

i

� �kL�i�k

� ci \

1

kiH1�i� [ 1

kiH2�i�

� � � 2

ikL�i�k :

Thus kL�i�k > i3 for i large.

We now estimate the horizontal ¯ux of ci. First note that while theorientation of pi L1�i�� � and L�i� agree, the orientation of pi L2�i�� � and L�i�are opposite. Since pijF1 preserves orientation but pijF2 reverses orientation,the projection of the conormal ®eld along L1�i� and L2�i� projects to apositive function times the conormal ®eld along L�i�. Since the lengths of thepart of ci inside

1ki

H1�i� [ 1ki

H2�i� converge to zero as i!1, we know thatthe horizontal ¯ux of the limit can be calculated as the limit of the horizontal¯uxes along L1�i� and L2�i�. Recall that the sheets in F1�i� [ F2�i� converge tohorizontal sheets as i!1, and so the horizontal ¯ux of L1�i� [ L2�i� di-vided by the length of L�i� converges to 2 as i!1. As the length of L�i� isgreater than i

3, it follows that the horizontal ¯uxes of the M�i� are un-bounded, contrary to our hypotheses. This contradicts our earlier assump-tion that the curvature of the family fM�i�gi is unbounded, and completesthe proof of the theorem. (

We now prove the properness of the Flux map F : S ÿ! R2 ÿ f0g.Theorem 5 The Flux map is proper.

Proof. Consider a sequence fM�i���i 2 Ng of examples in S whose normed¯ux, kF M�i�� �k, lies in �e; 1e� for some e > 0. It is su�cient to show that asubsequence of the fM�i�gi converges to another surface in S with vertical¯ux one and horizontal ¯ux in �e; 1e�.

Fig. 2

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By Theorem 4, there is an upper bound on the Gaussian curvatures ofthe family fM�i�gi. For simplicity, we consider the lifts

eM�i� � R3 of M�i�,invariant under translation by T �i�. Translate each eM�i� so that a point ofmaximal absolute Gaussian curvature of eM�i� occurs at the origin. As beforewe may assume after choosing a subsequence that f eM�i�gi converges to asurface eM1. Since eM�i� has points with horizontal normal vector, Lemma 2implies that the maximums of the absolute Gaussian curvatures of thefamily f eM�i�g are bounded from below by a positive constant. Hence, themaximal absolute value of Gaussian of eM1, which occurs at the origin, issome positive number c2.

We now apply Proposition 1 to conclude that eM1 is another example inS. First, replace the surfaces eM�i� by 1

c�i� eM�i�, where c�i� is the maximalGaussian curvature of M�i�. Since c�i� ! c, the surfaces 1

c�i� eM�i� converge to1ceM1. Since the vertical ¯uxes of these surfaces are bounded, 1c eM1 is not a

helicoid. As the horizontal components of the ¯uxes are uniformly boundedaway from zero, 1c

eM1 cannot be a catenoid. Hence 1ceM1 is a Riemann-type

example with the ¯uxes of the 1c�i� eM�i� converging to the ¯ux of 1

ceM1. It

follows that the quotient surface M1 obtained from eM1 is an element of Sand the horizontal ¯ux of M1 lies in �e; 1e�. This completes the proof ofproperness. (

The last result of this section will be useful for proving the uniqueness ofthe Riemann examples among surfaces M 2S with very small horizontal¯ux, which will be demonstrated in Sect. 6.

Lemma 3 Suppose fM�i���i 2 Ng �S is a sequence of minimal surfaces suchthat the norm of the horizontal ¯uxes of M�i� goes to zero as i goes to in®nity.Then, for i large, M�i� is close to n translates of larger and larger compactregions of the vertical catenoid with vertical ¯ux one, together with very ¯atregions that are graphs and contain the planar ends of M�i�.

Proof. Since the sequence in the Lemma is arbitrary, it su�ces to provethere is some subsequence that satis®es the conclusions in the statement ofthe Lemma. We now prove this assertion.

For each i, choose an angle h�i� 2 S1 � C which is a regular value of theGauss map gi : M�i� ÿ! C. Let p�i; 1�; . . . ; p�i; 2n� denote the 2n distinctpreimages of h�i� on M�i�. Let eM�i� denote the lift of M�i� to R3 and ep�i; j�,1 � j � 2n, the lifts of the p�i; j� to a fundamental region of eM�i� in R3. LeteM�i; j� be the translate of eM�i� which has ep�i; j� at the origin. By Lemma 1and Theorem 4, there are uniform curvature estimates and local area boundsfor this family. Therefore, up to passing to a subsequence, f eM�i; 1�gi con-verges to a properly embedded surface eM�1; 1�.

By Lemma 2, eM�1; 1� is not a plane. Proposition 1 implies that eM�1; 1�is a vertical catenoid, a vertical helicoid or corresponds to the lift of anotherRiemann-type example (actually to apply Proposition 1 directly one needsto normalize the curvature as was done in the proof of the previous

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Theorem). Since the vertical ¯uxes of the eM�i; 1� are one, Proposition 1shows that eM�1; 1� is not a helicoid. Since a Riemann example has nonzerohorizontal ¯ux, Proposition 1 shows that eM�1; 1� must be a verticalcatenoid with vertical ¯ux one.

For i large, eM�i; 1� has a large compact region close to a catenoid, so thisregion can be chosen with injective Gauss map. It follows that the pointsp�i; j�, j � 2, are far from the origin on the surface. The previous argumentshows that a subsequence of f eM�i; 2�gi converges to a vertical catenoid withvertical ¯ux one, passing through the origin at ep�i; 2�. Thus, for each large®rst index and passing again to a subsequence, each surface has two regionswhich are close to translates of a ®xed large compact region of a verticalcatenoid with vertical ¯ux one. Continuing in this manner one can show thatpassing to a subsequence, around each ep�i; j�, the surface eM�i� is close to atranslation of larger and larger compact pieces of a ®xed vertical catenoidwith vertical ¯ux one. In particular, the vertical components of the trans-lation vectors of the eM�i� must be going to in®nity, since the large compactregions of the 2n catenoids embed simultaneously in the quotient spaces. Asthe level sets of any Riemann-type example are connected, these compactregions in the quotient space can be chosen as the intersection of M�i� withhorizontal slabs whose widths go to in®nity as i increases.

We now claim that in each one of the complementary slabs found above,the surface M�i� contains one of its planar ends and is a ¯at graph over ahorizontal plane minus two disjoint disks. Let A�i� denote one of thesecomplementary slabs. M�i� \ A�i� has two boundary components, each oneclose to a large circle and with almost vertical normal vector. Since the totalcurvature of the complement of the 2n catenoidal regions is arbitrarily small,a simple application of the openness property of the Gauss map shows thatthe Gauss map of M�i� \ A�i� is contained in a small neighborhood of onevertical direction, hence the projection on the horizontal plane is a propersubmersion that is one-to-one on each boundary component. Since thesurface M�i� \ A�i� is connected, elementary topology arguments imply thatthe projection is one-to-one. Now we prove that M�i� \ A�i� is di�eomor-phic to R2 minus two open disks. First, note that the projection is not acompact annulus, since M�i� \ A�i� lies locally on the outside of eachboundary curve. Since M�i� intersects each of the 2n complementary slabs innoncompact sets, and there are 2n ends on the surface, each such intersec-tion must be di�eomorphic to R2 minus two open disks. This proves thatM�i� \ A�i� is a ¯at graph over the horizontal plane minus two large disks, asdesired. Now the proof of the Lemma is complete. (

Remark 1 Suppose that fM�i�gi �S is a sequence whose horizontal ¯uxesconverge to zero. By Lemma 3, for i large, the branch points of the Gaussmap of the M�i� lie in the very ¯at regions which are graphs and contain theends. Hence, the branch values of the Gauss maps of these surfaces convergeto 0 and 1 in C. By carefully choosing the ¯at regions, each ¯at regiontogether with the point at in®nity is a degree two branched cover of a disk

120 W. H. Meeks et al.

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neighbourhood of 0 or1. The Riemann-Hurwitz formula implies that thereare exactly two branch points, one of them being a ®nite point in the surface.This remark will be used in the proof of Theorem 7.

5 The openness of the Flux map

Let n be a positive integer. We will denote by W the space of all lists�R; g; p1; . . . ; pn; q1; . . . ; qn; �c�� where R is a closed genus one Riemann sur-face and g : R ÿ! C is a meromorphic function of degree 2n, having ndistinct double zeroes p1; . . . ; pn and n distinct double poles q1; . . . ; qn on R.So, the divisor of g is given by �g� � p21 . . . p2nqÿ21 . . . qÿ2n . Note that we areassuming an ordering on the points pi and qi, and we will think of di�erentorderings as di�erent elements of W. The last component �c� in the listabove is a homology class in H1 Rÿ fp1; . . . ; pn; q1; . . . ; qng;Z� � whichinduces a nontrivial homology class in the torus R, �c� 2 H1�R;Z� ÿ f0g.Elements ofW will be called marked meromorphic maps, and will be denotedsimply by g. Riemann-Hurwitz formula gives that each g 2W has exactly2n branch points, not necessarily distinct, other than its zeroes and poles.The branch values corresponding to these points will be called C�-branchvalues.

Natural families of meromorphic functions with prescribed degree andgenus can be endowed with structures of ®nite-dimensional complex man-ifolds, called Hurwitz schemes, see for instance [8]. We will explain someaspects of this construction in our particular case.

Let g 2W with C�-branch values a1; . . . ; ak, k � 2n, without countingmultiplicity. A neighborhood, U�g�, of g inW can be described as follows.Take k � 2 small pairwise disjoint disks Di in C, centered at the points ai,1 � i � k, Dk�1 and Dk�2 centered at 0 and 1, respectively. Denote byX � CÿSDi, the complementary domain. The restrictiong : gÿ1�X� ÿ! X is an unbranched 2n-sheeted covering map with connectedtotal space. Each component of gÿ1�Di�, i � 1; . . . ; k � 2, is a disk andcontains at most one branch point of g, possibly of high multiplicity. Wheni � k � 1 or k � 2; gÿ1�Di� has just n components and g has just one simplebranch point, a double zero or a double pole, in each of these components.A marked meromorphic map f 2W lies in U�g� if it satis®es the followingconditions:

(i) f has no branch values on X and the covering maps f : fÿ1�X� ÿ! Xand g : gÿ1�X� ÿ! X are isomorphic. Thus we can identify, from the con-formal point of view, fÿ1�X� and gÿ1�X�.

(ii) Each component of fÿ1�Di� is a disk in fÿ1�C�; i � 1; . . . ; k � 2. Theidenti®cation in (i) determines a bijective correspondence between thecomponents of fÿ1�Di� and those of gÿ1�Di� and the total branch order of gand f at the corresponding components necessarily coincides. In particular,if g has a simple branch point in a component, the same holds for f at thecorresponding component.

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(iii) The correspondence in (ii) determines, when i � k � 1 and k � 2, abijection between the zeroes (resp. poles) of g and those of f .We impose thatthe ordering in the set of zeroes and poles of f coincides with the one in the zeroand pole set of g through this bijection.We will denote the zeroes and poles off again by p1; . . . ; pn and q1; . . . ; qn, respectively.

(iv) The homology class of g, �c� 2 H1 gÿ1�C� ÿ fp1; . . . ; pn;ÿ

q1; . . . ;qng;Z� can be represented by a closed curve c immersed in gÿ1�X�, which bymeans of the identi®cation in (i) induces a homology class in the puncturedsurface determinated by the meromorphic map f , also denoted by�c� 2 H1 fÿ1�C� ÿ fp1; . . . pn; q1; . . . ; qng;Z

ÿ �. Using (i) and (ii) we can check

that �c� 6� 0 in H1 fÿ1�C�;Zÿ �. We will impose on the marked meromorphic

map f that its associated homology class coincides with �c�.If an element g 2W has 2n distinct C�-branch values, then the same holdsfor any f 2W near g. Furthermore, the map which applied to f yields itslist of C�-branch values is a local chart for W around g. In particular,dimW � 2n. If g has multiple branch points or di�erent branch points withthe same C�-branch value, coordinates around g are a bit more complicatedto describe, see [8], but they follow the essential idea that elements inW arelocally parametrized by the list of their branch values in C�.

On W we have some globally de®ned holomorphic functionsri : W ÿ! C; 1 � i � 2n. For any g 2W, ri�g� is the symmetric elementarypolynomial of degree i in the variables a1; . . . ; a2n, where fa1; . . . ; a2ng is theset of C�-branching values of g, each one repeated as many times as itsmultiplicity indicates. Note that the ai themselves are not globally well-de®ned functions because fa1; . . . ; a2ng is an unordered set. Recall that thenumbers r1�g�; . . . ; r2n�g� are the coe�cients of the polynomial

Qi�zÿ ai�,

and so, they determine the unordered set fa1; . . . ; a2ng. There are only®nitely many (unmarked) meromorphic maps g with the same C�-branchvalues or, equivalently, with the same image under all the functionsri; i � 1; . . . ; 2n. As a consequence of the description above of the topologyin W, we have that given a1; . . . ; a2n 2 C, then the set fg 2W��r1�g� �a1; . . . ; r2n�g� � a2ng is a discrete subset ofW. Note that when nonvoid, theabove subset is in®nite because of the in®nitely many possible choices for �c�in H1 gÿ1�C� ÿ fp1; . . . ; pn; q1; . . . ; qng;Z

ÿ �.

For any g 2W, we have an unique holomorphic di�erential / � /�g� onthe complex torus gÿ1�C� satisfyingZ

c

/ � 2pi : �3�

Moreover, / depends holomorphically on g in the sense that the map fromU�g� � gÿ1�X� into C� given by

�f ; z�# /�f �/�g� �z� �4�

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is holomorphic. As �g;/� are potential Weierstrass data for surfaces in thesetting of our theorem, we will call W the space of Weierstrass represen-tations.

Later, we will need the following property for the analytic subvarieties ofW.

Lemma 4 The only compact analytic subvarieties of W are the ®nite subsets.

Proof. Let V be a compact analytic subvariety of W. From the propermapping theorem (see Sect. 3), it follows that ri�V� is a compact analyticsubvariety of C. This implies that ri�V� is a ®nite subset, for 1 � i � 2nand, thus,V itself is a discrete subset ofW. From this we obtain directly the®niteness of V. (

We consider the Period map P : W ÿ! C2n that associates to eachg � �gÿ1�C�; g; p1; . . . ; pn; q1; . . . ; qn; �c�� of W the 2n-tuple

P �g� �Zc

1

g/;Zc

g/; Resp11

g/

� �; . . . ;Respnÿ1

1

g/

� �; Resq1�g/�; . . . ;Resqnÿ1 g/� �

0B@1CA ;

where Resp means the residue of the corresponding di�erential at the pointp. We claim that P is a holomorphic map. Take f in a neighborhood U�g� ofg inW as above. Holomorphicity is clear for the ®rst two coordinates of Pbecause of the analytic dependence of (4). For the remaining coordinates,note that the residues appearing in P �f � can be computed as integrals alongcurves contained in the boundary of X.

We consider the subsetM ofW given by the marked meromorphic mapsg such that P �g� � z; z; 0; . . . ; 0� �, for some z 2 C. As we pointed out inSect. 3, for each g 2M the pair �g;/�g�� is the Weierstrass representation ofa complete minimal surface M immersed in R3=T , for some nonhorizontalvector T 2 R3, with 2n planar ends at the points p1; . . . ; pn; q1; . . . ; qn.Moreover, each compact horizontal section of this surface consists ofexactly one Jordan curve in M . Note that a geometric minimal surfaceproduces in®nitely many elements when considered in M, by taking all thepossible orderings of its ends and all the di�erent choices of the homologyclass �c� in the punctured torus that give the same class in the compacti®edsurface. Nevertheless, all those in®nitely many elements of M form a dis-crete subset in W. We will refer to M as the space of immersed minimalsurfaces.

Let beS denote the space of properly embedded minimal (oriented) toriin a quotient of R3 by a translation T , that depends on the surface, with 2nhorizontal planar (ordered) ends. Recall (see Sect. 2) that surfaces in S arede®ned up to translations. We embed S in M as follows. Each M 2 S

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determines a marked meromorphic map g � �gÿ1 C�; g; p1; . . . ; pn;ÿ

q1; . . . ; qn; �c��, where g is the (extended) Gauss map of M ; �p1; q1 . . . ; pn; qn�is the list of its ends and c is the compact horizontal section of M by a planewhich lies between the asymptotic planes at the ends p1 and q1, oriented insuch a way that the third coordinate of the conormal vector is alwayspositive.

The following assertion can be demonstrated by using the argumentsthat appear in the proof of Lemma 6.

Lemma 5 OnS, both topologies, the smooth convergence of the lifted surfaceson compact subsets of R3 and the one induced by the inclusion S �W,coincide.

Remark 2 The Flux map F : S ÿ! R2 ÿ f0g de®ned in Sect. 2 applied toM 2S is the horizontal part z 2 R2 ÿ f0g � Cÿ f0g of the ¯ux of M alongc. The map P applied to M is �iz;ÿiz; 0; . . . ; 0� 2 C2n. Thus, the Period maprestricted toS identi®es with the Flux map. We will use this identi®cation inthe proof of the next Lemma.

Lemma 6S is an open and closed subset inM. In particular, for any z 2 R2ÿf0g, the space S�z� � fM 2S��F �M� � zg is an analytic subvariety of W.

Proof. If fM�i���i 2 Ng �S is a sequence, thought of as maps, which con-verges to a surface M 2M, then it follows from Lemma 1 that the mappingM is an embedding. So, M 2S and the closedness is proved.

To prove the openness part, take a sequence fM�j�gj �M which con-verges to M 2S in the topology of W. If we denote by g the (extended)Gauss map of M and by fj the one of M�j�, then, with the notation in thebeginning of this Section, it follows that fj 2 U�g� for j large, and largecompact pieces of M�j� and M are parametrized by mapswj : gÿ1�X� ÿ! R3=Tj and w : gÿ1�X� ÿ! R3=T such that the wj convergeto w. In particular, wj is an embedding for j large. Moreover, we can as-sume, by taking the neighborhood U�g� small enough, that the connectedcomponents of M ÿ X consist either of small graphs D1; . . . ;Dr over convexplanar domains, or bounded graphs D�1; . . . ;D�2n over the complements oflarge convex domains in a horizontal plane.

If Dk�j� is the component of M�j� ÿ gÿ1�X� that corresponds to Dk, thenDk�j� is a disk whose boundary values are arbitrarily close, depending on j,to those of Dk, and whose Gauss map image (that coincides with the one ofDk) is small. In this situation, it is a standard fact that Dk�j� is a graph whichconverges to Dk. Therefore, for j large, the domain M�j� ÿ gÿ1�X�� �[D1�j� [ . . . [ Dr�j� is embedded in R3=Tj.

Consider now the component D�k�j� in M�j� ÿ gÿ1�X� that correspondsto D�k . It is a representative of a planar end of M�j�, with almost verticalGauss map and whose boundary values converge to those of D�k . So, for jlarge enough, each D�k�j� is a graph over the exterior of a Jordan curve in the

124 W. H. Meeks et al.

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�x1; x2�-plane, because its horizontal projection is a proper local di�eo-morphism that maps injectively its boundary onto a convex horizontalcurve. From the maximum principle at in®nity [11], the distance betweendistinct ends of M , D�1; . . . ;D�2n is positive and, so, we can assume that eachtwo of these ends are strictly separated by a horizontal plane. For j large,this plane will also separate the closed curves @D�k�j�; k � 1; . . . ; s. As the x3-coordinate on D�k�j� is a harmonic function which extends to the puncture, itfollows from the maximum principle that the planes above separate theentire ends D�1�j�; . . . ;D�2n�j�. Therefore, M�j� is embedded and we concludethat S is open in M.

For any z 2 C, the subset M�z� � fg 2W��P�g� � �iz;ÿiz; 0; . . . ; 0�g �M is clearly an analytic subvariety of W. Therefore, S�z� �M�z� \S,which is open and closed in M�z�, is also an analytic subvariety of W. (

Theorem 6 The Flux map F : S ÿ! R2 ÿ f0g is open.

Proof. We show that F is open in a neighborhood of any surface M in thespace S. Let z � F �M� 2 R2 ÿ f0g. Lemma 6 and the properness of F(Theorem 5) imply that S�z� � F ÿ1�z� is a compact analytic subvariety ofW. From Lemma 4, we obtain that S�z� is ®nite. Therefore, using theopenness Theorem in Sect. 3, we get that there exists a neighborhood U ofM in W such that P jU is open. Using Lemma 6, we can assume thatU \M � U \S. This gives directly the openness of F : U \S ÿ!R2 ÿ f0g, as we claimed. (

6 Surfaces with almost vertical ¯ux

Consider e 2 �0; 1�, and denote by D�e� � fz 2 C��kzk < eg and D��e� �

D�e� ÿ f0g. Given a � �a1; . . . ; a2n� 2 D��e�� �2n, a marked meromorphic mapassociated to a can be constructed as follows:

Consider 2n copies of C, labelled by C1; . . . ;C2n. For i odd, cut Ci alongtwo arcs ai; ai�1 joining 0 with ai and 1 with 1

ai�1, respectively. When i is

even, cut Ci along two arcs ai; ai�1 joining 1 with 1aiand 0 with ai�1, res-

pectively. Choose these arcs in such a way that consecutive copies Ci, Ci�1share the arc ai�1, so we can glue together Ci with Ci�1 (in the same way,glue C2n with C1 along the common arc a1), forming a genus one surface Rwith the 2n copies of C. The z map on each copy Ci is now well-de®ned on Ras a meromorphic map g with degree 2n, which has a double zero (resp.double pole) at 0 in each copy Ci (resp. at 1 in each Ci) and C�-branchvalues a1; 1a2 ; . . . ; a2nÿ1; 1

a2n, each one being shared by two of the copies of C.

The ordered list of zeroes and poles associated to our marked meromor-phic map is de®ned to be 01;11; . . . ; 0n;1n� �, where A1 � f01;11g,A2 � f11; 02g, A3 � f02;12g; . . . ;A2n � f1n; 01g are the sets of zeroes andpoles of the z-map on the copies C1; . . . ;C2n, respectively. Finally, thenontrivial homology class �c� 2 H1 Rÿ f01;11; . . . ; 0n;1ng;Z� � is de®ned tobe the class induced by the loop fkzk � 1g in the ®rst copy C1, see Fig. 3.

Uniqueness of the Riemann minimal examples 125

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In fact, this correspondence between lists �a1; . . . ; a2n� 2 D��e�� �2n andmarked meromorphic maps is a local parametrization in the complexmanifoldW. Also note that by choosing di�erent orderings for the zero andpole set and di�erent homology classes in the construction above, we con-struct an in®nite number of other elements inW, with the same underlyingmeromorphic function. However, these elements are far away, in thetopology of W, from the one constructed above.

If we consider the restriction of the Period map P to the open subset ofW where the above local chart is de®ned, then P can be considered as aholomorphic map from D��e�� �2n into C2n. Next we will prove that P extendsholomorphically to �D�e��2n. With this aim, we need the following result.

Lemma 7 Let S be a Riemann surface and ffj : S ÿ! C�gj a sequence ofinjective holomorphic functions. Assume that there exists a closed curve C � Ssuch that fj�C� is not nulhomotopic in C�, for any j 2 N. Then, there exists asubsequence of ffjgj, denoted in the same way, and a sequence fkjgj � C� suchthat fkjfjgj converges on compact subsets of S to a injective holomorphicfunction f : S ÿ! C�.

Proof. Take a point p 2 S and consider the functions efj � 1fj�p� fj. Viewed as

functions on S ÿ fpg, the family fefjgj omits the values 0; 1;1 in C. HenceMontel-Caratheodory Theorem says that a fefjgj is a normal family andthus, it has a subsequence, denoted in the same way, that converges to ameromorphic function f : S ÿ fpg ÿ! C.

Take a complex coordinate z, kzk < 2e, centered at p. From the meanvalue property of holomorphic maps we have that

1

2pe

Zfkzk�eg

efj�z�jdzj � 1 :

As the efj converge on fkzk � eg, we conclude that, if f were constant, then itwould be identically one. In particular, efj�C� would be contained in a smallneighborhood of 1 in C� for j large enough, which is contrary to theassumption on fj�C�. Therefore, f is nonconstant. The maximum modulus

Fig. 3

126 W. H. Meeks et al.

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principle implies that the efj converge to f on the whole surface S. In thissituation, a well-known theorem of Hurwitz implies that f is injective andhas no zeroes or poles on S. (

Lemma 8 The Period map P extends holomorphically to D�e�� �2n� Cn.

Proof. Following the notation at the beginning of this Section, note thatgiven �a1; . . . ; a2n� 2 D��e�� �2n with associated marked meromorphic mapg 2W, we can rewrite the Period map P in a more homogeneus way asfollows. Consider the 2n closed curves c � c1; c2; . . . ; c2n corresponding tothe loops fkzk � 1g on C1; . . . ;C2n, respectively. Note that all the ci arehomologous in R and each residue at a pole (resp. a zero) shared by con-secutive copies Cj, Cj�1 of the meromorphic di�erential g/ (resp. of 1g /) canbe computed as the expression 1

2pi

ÿ Rcj�1g/ÿ Rcj

g/� (resp. 12pi

ÿ Rcj�1

1g /

ÿ Rcj1g /��. Also note that if the copies Cj;Cj�1 have a zero along the

common arc aj�1, thenR

cjg/ � Rcj�1

g/, and the same holds for 1g / if aj�1

joins a pole with 1aj�1

. Now an elementary exercise in linear algebra showsthat the Period map can be expressed as the composition of a regular lineartransformation in C2n with the holomorphic map G : D��e�� �2n ÿ! C2n

de®ned by

G�a1; . . . ; a2n� � G�g� �Zc1

g/;Zc2

1

g/;Zc3

g/;Zc4

1

g/; . . . ;

Zc2nÿ1

g/;Zc2n

1

g/

0B@1CA :

As consequence, it su�ces to prove that G extends holomorphically toD�e�� �2n.As D�e�� �2nÿ D��e�� �2n is an analytic subvariety of D�e�� �2n, by the Riem-

ann extension theorem (see Sect. 3) G will extend holomorphically providedthat it is bounded. With this aim, consider a sequence fa�j� ��a1�j�; . . . ; a2n�j��gj � D��e�� �2n that converges to a � �a1; . . . ; a2n� 2 D�e�� �2n,such that at least one of the components of a is zero. By the construction atthe beginning of this Section, each list a�j� determines an elementg�j� � R�j�; g�j�; p1�j�; . . . ; pn�j�; q1�j�; . . . ; qn�j�; �c�j��� � in W and a holo-morphic one-form /�j� such that

Rc�j� /�j� � 2pi. Denote by b1; . . . ; bs the

distinct elements in the set of limit branch values fa1; 1a2 ; . . . ; a2nÿ1;1

a2n; 0;1g. For a � 1; . . . ; s, take pairwise disjoint small disks Da centered at

the ba, and let X � Cÿ [sa�1Da. For j large enough, g�j� : g�j�ÿ1�X� ÿ! X

is an unbranched 2n-sheeted covering, any two of these being isomorphic.This isomophism allows us to identify each component B1�j�; . . . ;Bk�j� ofg�j�ÿ1�X� with open subsets B1; . . . ;Bk of a collection of genus zero surfacesS1; . . . ; Sk, in such a way that g�j� restricted to Bm�j� identi®es with gm

restricted to Bm, for certain nonconstant meromorphic maps gm : Sm ÿ! C,m � 1 . . . ; k satisfying that the branch values of gm are contained infb1; . . . ; bsg and deg�g1� � . . .� deg�gk� � 2n. When we shrink the disks Da,the domains Bm increase until in the limit they ®ll the whole sphere Sm minusa ®nite subset Zm. Moreover, when Bm is viewed as a subdomain of the torus

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R�j�, each component of @Bm bounds a disk in R�j� except in two of them,which are homologous to the curve c�j�. These two boundary componentscorrespond to two elements in the list a�j� that converge to zero, anddetermine two distinct points Pm;Qm 2 Zm, where gm takes the value 0 and/or 1. On the other hand, the curves c1; . . . ; c2n in R�j� induce 2n curves(denoted again by c1; . . . ; c2n�, distributed in B1; . . . ;Bk such that each oneof them, when viewed inside the corresponding annulus Sm ÿ fPm;Qmg,generates its homology group.

Fix m 2 f1; . . . ; kg and consider the inclusion map from Bm into R�j�, forj large. Let C be one of the curves in fc1; . . . ; c2ng, which lies in Bm. Thus, Cis a generator of the ®rst homology group of Sm ÿ fPm;Qmg. Consider thecyclic covering P of R�j� with image subgroup the cyclic group generated bythe loop C in p1�R�j��, where C is viewed as a nontrivial homotopy class inR�j�. The total space of this covering can be conformally parametrized byC� so that C lifts to the homotopy class of fkzk � 1g in C�. Hence theinclusion of Bm into R�j� lifts to a holomorphic injective functionfj : Bm ÿ! C�. Lemma 7 implies that, after passing to a subsequence,fkjfjgj converges on compact subsets of Bm to a injective holomorphicfunction f : Bm ÿ! C�, where fkjgj is a sequence of nonzero complexnumbers. The domain Bm can be thought as an element in a increasingexhaustion of Sm ÿ Zm, so we can assume, again after choosing a subse-quence, that neither the sequence fkjgj nor the limit function f depend onBm, hence f : Sm ÿ Zm ÿ! C� is a injective holomorphic function and fkjfjgjconverges uniformly on compact subsets of Sm ÿ Zm to f . Therefore, f ex-tends to a biholomorphism between Sm and C. On the other hand, ournormalization on /�j� implies that its lifting to C� is P�/�j� � dz

z . Thus, /�j�restricted to Bm coincides with f �j

dzz � dfj

fj� d�kjfj�

kjfjand, so, when j goes to

in®nity, /�j� converges to / � dff on Sm ÿ Zm. Observe that / is a mero-

morphic di�erential on Sm with exactly two simple poles. As the integral of/�j� along the boundary components of Bm � Sm that enclose the pointsPm;Qm is �2pi, we conclude that the same holds for the one-form /, orequivalently, that the poles of / occur at Pm and Qm.

In summary, we have shown that after passing to a subsequence, whenj!1 the torus R�j� disconnects into k genus zero surfaces S1; . . . ; Sk, themeromorphic maps g�j� converge uniformly on compact subsets of each Sm

minus a ®nite subset Zm to a nonconstant meromorphic map gm : Sm ÿ! C,1 � m � k, in such a way that the total degree of g1; . . . ; gk is equal to 2n.Moreover, the holomorphic di�erentials /�j� converge uniformly on com-pact subsets of Sm ÿ Zm to the (unique) meromorphic di�erential / on Sm

which has just two simple poles at certain points Pm; Qm 2 Zm with residues1;ÿ1 at these points.

Now we conclude easily that fG�a�j��gj is bounded in C2n, because eachcomponent of G�a�j�� is written as an integral along a ®xed closed curvecontained in Sm ÿ Zm for some m, of certain holomorphic di�erentials thatconverge uniformly on the closed curve to another holomorphic di�erential.

(

128 W. H. Meeks et al.

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Remark 3 From the proof above, it follows that if a 2 �D�e��2n ÿ �D��e��2n

determines the spheres S1; . . . ; Sk, then the value of G at a is given by

G�a� �Zc1

g/;Zc2

1

g/;Zc3

g/;Zc4

1

g/; . . . ;

Zc2nÿ1

g/;Zc2n

1

g/

0B@1CA :

where g is taken as gm in the r-th component, provided that the curve cr liesin the sphere Sm, and / is the holomorphic di�erential on the annulusSm ÿ fPm;Qmg that has simple poles at Pm and Qm, and period 2pi along thegenerator of its homology.

Lemma 9 There exists a neighborhood U of 0 2 C2n such that P jU is biholo-morphic.

Proof.Using the arguments in the proof of Lemma 8, it su�ces to check thatthe di�erential at zero of G : D�e�� �2n ÿ! C2n is regular.

Fix j 2 f1; . . . ; 2ng and compute@G@aj�0�. As we showed in the proof of

Lemma 8, the 2n-tuple �0; . . . ; aj; . . . ; 0�; aj 2 D��e�, determines 2n±1 genuszero surfaces S1; . . . ; Sjÿ1; Sj�1; . . . ; S2n and meromorphic maps g1; . . . ; gjÿ1;gj�1; . . . ; g2n on these surfaces, satisfying

1. For m � 1; . . . ; jÿ 2; j� 1; . . . ; 2n, gm : Sm ÿ! C has degree one. If weparametrize Sm conformally by C, then gm�z� � z, and the curve cm is ho-mologous to fkzk � 1g in C�.

2. gjÿ1 : Sjÿ1 ÿ! C has degree two. When j is odd, gjÿ1 has a double zeroand aj is a C�-branch value with rami®cation order one. If j is even, itsbranch values are1 and 1

aj. Moreover, Sjÿ1 contains the curves cjÿ1; cj that

appear in the de®nition of G.

Take m 6� jÿ 1; j. The m-th component of G is eitherR

cmg/ or

Rcm

1g /,

depending on the parity of m, where cm lies in one of the surfaces in case 1.Then, as our normalization of / on Sm � C gives / � dz

z , it follows thatZcm

g/ �Zkzk�1

dz � 0;

Zcm

1

g/ �

Zkzk�1

dzz2� 0 :

This implies that the m-th component of G is identically zero along the curve�0; . . . ; aj; 0; . . . ; 0�, hence the same holds for its derivative.

We now consider the �jÿ 1�-th, j-th components of G. Assume ®rstlythat j is odd. If we parametrize Sjÿ1 by S � f�z;w� 2 C�C

��w2 � z�zÿ aj�g,then gjÿ1�z;w� � z, and the curves cjÿ1; cj are homologous inS ÿ f�0; 0�; �aj; 0�; �1;1�1; �1;1�2g to the two connected lifts offkzk � 1g to S by the z-projection. To compute G�0; . . . ; 0; aj; 0; . . . ; 0�,according to Remark 3, we need to determine a meromorphic di�erential /on S by the conditions

(1.a) / has no zeroes or poles in S ÿ f�1;1�1; �1;1�2g, and

Uniqueness of the Riemann minimal examples 129

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(1.b)Rkzk�1 / � 2pi (here fkzk � 1g denotes either of the two liftings of

the unit circle to S by the z-projection).

Conditions (1.a) and (1.b) imply that / � c�aj� dzw, where c�aj� 2 C� satis®es

the equation

c�aj�Zkzk�1

dz������������������z�zÿ aj�

p � 2pi :

Hence c�aj� is a holomorphic function of aj that extends to zero as c�0� � 1.Now by direct computation one has

ddaj

�����0

Zkzk�1

1

g/ � d

daj

�����0

c�aj�Zkzk�1

dz

z������������������z�zÿ aj�

p� c0�0�

Zkzk�1

dzz2� 1

2

Zkzk�1

dzz3� 0 :

ddaj

�����0

Zkzk�1

g/ � ddaj

�����0

c�aj�Zkzk�1

zdz������������������z�zÿ aj�

p� c0�0�

Zkzk�1

dz� 1

2

Zkzk�1

dzz� pi :

Similarly, if j is even, we can write Sjÿ1 � S � f�z;w� 2 C�C��w2

� ajzÿ 1g, g�z;w� � z, and the curves cjÿ1; cj are the liftings by g offkzk � 1g. Therefore, / � bc�aj� dz

zw, where bc�aj� is holomorphic in aj withbc�0� � i, and we obtain

ddaj

�����0

Zkzk�1

1

g/ � ÿpi;

ddaj

�����0

Zkzk�1

g/ � 0 :

In conclusion, the partial derivative of G at the origin is

@G@aj�0� � 0; . . . ; 0; �ÿ1�j�1pi; 0 . . . ; 0

� �;

and the di�erential of G at the origin is bijective, as desired. (

Now we can state the main result in this Section.

Theorem 7 There exists e > 0 such that if M 2 S satis®es jF �M�j < e, then Mis a Riemann minimal example.

130 W. H. Meeks et al.

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Proof. Consider the neighborhood U of 0 2 C2n where P jU is bijective givenby Lemma 9, and note that P extends by P �0� � 0.

By applying the properness Theorem 5 and the openness Theorem 6 toany connected component C in W corresponding to the space of Riemannexamples (which is open and closed in S), we deduce that in C we can ®ndRiemann examples with almost vertical ¯ux. By Remark 1, these surfaces liein U if the horizontal part of their ¯uxes are small enough. On the otherhand, if M 2S lies in U, then its horizontal ¯ux coincides with the hori-zontal ¯ux of a Riemann example R, suitably rotated around the verticalaxis. Thus P takes the same value at M and at R, which implies these surfacescoincide.

Now the theorem follows from the relation between the two ®rst com-ponents of P jS and the Flux map F (see Remark 2 in Sect. 5). (

References

1. M. Callahan, D. Ho�man, H. Karcher, A family of singly-periodic minimal surfacesinvariant under a screw-motion, Experiment. Math. 3(2), 157±182 (1993)

2. M. Callahan, D. Ho�man, W.H. Meeks III, Embedded minimal surfaces with an in®nitenumber of ends, Invent. Math. 96, 459±505 (1989)

3. M. Callahan, D. Ho�man, W.H. Meeks III, The structure of singly periodic minimalsurfaces, Invent. Math. 99, 455±481 (1990)

4. G. Darboux, The orie Ge ne rale des Surfaces, Vol. I, Chelsea Pub. Co. Bronx, New York(1972)

5. Y. Fang, F. Wei, On Uniqueness of Riemann's examples, to appear in Proceedings of theA.M.S.

6. R. Kusner, W.H. Meeks III, H. Rosenberg, personal communication7. C. Frohman, W.H. Meeks III, The ordering theorem for the ends of properly embedded

minimal surfaces, Topology 36(3), 605±617 (1997)8. W. Fulton, Hurwitz schemes and irreducibility of moduli of algebraic curves, Ann. of

Math. (2) 90, 542±575 (1969)9. P. Gri�ths, J. Harris, Principles of Algebraic Geometry (Pure and Applied Mathematics),

Wiley-Interscience, 197810. D. Ho�man, H. Karcher, H. Rosenberg, Embedded minimal annuli in R3 bounded by a

pair of straight lines, Comment. Math. Helv. 66(4), 599±617 (1991)11. R. Langevin, H. Rosenberg, A maximum principle at in®nity for minimal surfaces and

applications, Duke Math. J. 57, 819±828 (1988)12. F.J. Lo pez, M. Ritore , F. Wei, A characterization of Riemann's minimal surfaces, to

appear in J. Di�erential Geometry13. F.J. Lo pez, D. Rodrõ guez, Properly immersed singly periodic minimal cylinders in R3,

preprint14. F.J. Lo pez, A. Ros, On embedded complete minimal surfaces of genus zero, J. Di�erential

Geometry 33, 293±300 (1991)15. W.H. Meeks III, H. Rosenberg, The geometry of periodic minimal surfaces, Comm. Math.

Helv. 68, 538±578 (1993)16. W.H. Meeks III, H. Rosenberg, The maximum principle at in®nity for minimal surfaces in

¯at three manifolds, Comm. Math. Helv. 65, 2 255±270 (1990)17. J. Pe rez, Riemann bilinear relations on minimal surfaces, to appear in Math. Ann.18. J. Pe rez, On singly-periodic minimal surfaces with planar ends, Trans. of the A.M.S. 349(6),

2371±2389 (1997)19. J. Pe rez, A. Ros, Some uniqueness and nonexistence theorems for embedded minimal

surfaces, Math. Ann. 295(3), 513±525 (1993)

Uniqueness of the Riemann minimal examples 131

Page 26: Uniqueness of the Riemann minimal exampleshera.ugr.es/doi/15000011.pdf · 2005-02-24 · minimal surfaces with planar ends can be found in [1, 2]. 2 Proof of the Main Theorem We now

20. B. Riemann, UÈ ber die FlaÈ che vom kleinsten Inhalt bei gegebener Begrenzung, Abh. KoÈ nigl.d. Wiss. GoÈ ttingen, Mathem. Cl., 13, 3±52 (1867)

21. A. Ros, Embedded minimal surfaces: forces, topology and symmetries, Calc. Var. 4, 469±496 (1996)

22. E. Toubiana, On the minimal surfaces of Riemann, Comment. Math. Helv. 67(4), 546±570(1992)

132 W. H. Meeks et al.