MODULI SPACES OF EMBEDDED CONSTANT MEAN CURVATURE SURFACES WITH FEW ENDS AND SPECIAL SYMMETRY

KARSTEN GROSSE-BRAUCKMANN AND ROBERT B. KUSNER

Abstract. We give necessary conditions on complete embedded cmc surfaces with three or four ends subject to reflection symmetries. The respective submoduli spaces are twodimensional varieties corresponding to singular orbifold points in the moduli spaces of general cmc surfaces. We characterize fundamental domains of our cmc surfaces by associated great circle polygons in the three-sphere.

We are interested in explicitly parametrizing the moduli space M_{g,k} of complete, connected, properly embedded surfaces in ℝ^{3} with finite genus g and a finite number k of labeled ends having nonzero constant mean curvature. By rescaling we may assume this constant is 1, the mean curvature of the unit sphere. Two surfaces in ℝ^{3} are identified as points in M_{g,k} if there is a rigid motion of ℝ^{3} carrying one surface to the other. Moreover, we shall include in M_{g,k} a somewhat larger class of constant mean curvature (cmc) surfaces, the almost (or Alexandrov) embedded surfaces, which are immersed surfaces bounding immersions of handlebodies into ℝ^{3}. The space M_{g,k} is a finite dimensional real analytic variety, and in a neighborhood of a surface with no L^{2}-Jacobi fields it is a 3k − 6-dimensional manifold [KMP] for each k ≥ 3.

A few of the cmc moduli spaces M_{g,k} are known explicitly: the only embedded compact cmc surface is a round sphere [A], so M_{g,0} is either a point (g = 0) or empty (g > 0); M_{g,1} is empty, since there are no one-ended examples [M]; two-ended examples are necessarily the Delaunay unduloids [KKS], which are simply-periodic surfaces of revolution whose minimal radius or neckradius ρ ∈ (0, \frac{1}{2}] parametrizes M_{0,2}, and M_{g,2} is empty for g > 0. The Kapouleas construction [Kp] shows that M_{g,k} is not empty for every k ≥ 3 and every g. Furthermore, an embedded end of a cmc surface is asymptotically a Delaunay unduloid [KKS], and this defines in particular the neckradii and axes of the ends, which are related via a balancing formula. We call the surfaces in M_{g,k} the k-unduloids of genus g, or simply k-unduloids if their genus is 0.

In the present paper we focus on cmc surfaces with few ends and special symmetries: these give two-dimensional submoduli spaces of M_{g,3} and M_{g,4}, which can be thought of as certain sets of singular orbifold points within the larger cmc-moduli space. The genus g triunduloids comprising M_{g,3} are special in that a priori each has a plane of reflection symmetry [KKS], which we will think of as a horizontal plane. Their moduli space is

Date: October 1996.

Supported by SFB 256 at Universität Bonn, and NSF grant DMS 94-04278 at UMassAmherst.

[figure: ^{ρ_{2}  ρ_{2} ρ_{1} ρ_{1}}; ρ^{S}; ρ^{A}; ρ^{A}; α; ρ; ρ; ρ; ρ; α; ρ; ρ; ρ; ρ]

Figure 1. Axes and neckradius of rectangular, isosceles, and rhombic surfaces, as well as for a dihedrally symmetric 4-unduloid of genus 1.

3-dimensional at non-degenerate points [KMP]. Here we study the Y-shaped isosceles triunduloids in M_{0,3} which have an additional symmetry in the form of a vertical plane of reflection as depicted in Figure 1.2. (We will deal with the general triunduloid case elsewhere [GKS].) We also study a submoduli space of the 6-dimensional space M_{0,4} of 4unduloids, namely the rectangular 4-unduloids with the two extra vertical symmetry planes depicted in Figure 1.1.

Our results are motivated in part by some observations in [G], where a one-parameter family of dihedrally symmetric k-unduloids is constructed assuming their k asymptotic axes are arranged in a plane with equal angles 2π/k. More precisely, two dihedrally symmetric k-unduloids were shown to exist for neckradii in the interval (0, \frac{1}{k}) and one surface was found corresponding to the right endpoint. This suggested there might be some general constraints limiting the allowed neckradius of the ends, and restricting the number of cmc surfaces within these limits.

Indeed, our main result here is a bound on the neckradii for the two symmetry types we consider (Theorems 14 and 20). These bounds follow from an analysis of the fundamental domains and their associated boundary contours. Using Lawson’s theorem [L] [Ka], if there is a cmc surface with the assumed reflection symmetries, then there is an associated minimal surface in the three-sphere 𝕊^{3} whose boundary consists of great circle arcs. The boundary polygons exist only with certain lengths, and this restricts the range of neckradii the Delaunay ends can have. We need to assume enough symmetry to guarantee that a fundamental domain is simply-connected, not only so that Lawson’s theorem is applicable, but also to have the number of boundary arcs low enough to handle the spherical trigonometry explicitly.

We also prove that for given asymptotic axes there are exactly 2 associated spherical boundary contours for each neckradius below some maximal value, and exactly one at the maximum. We believe (but do not prove) that each contour bounds a unique minimal surface in 𝕊^{3}, so that our characterization of the family of spherical polygons would lead to a characterization of the family of cmc surfaces of the given symmetry type.

To deduce the existence of a complete cmc surface from a fundamental domain may require solving a period problem. For rectangular 4-unduloids, the period problem is solved

SYMMETRIC CMC SURFACES WITH FEW ENDS 3

by symmetry alone, while for isosceles triunduloids the known asymptotics of each end together with the balancing formula suffice. This is no longer true for some closely related symmetry types of surfaces, like the rhombic 4-unduloids (see Fig. 1.3) and dihedrally symmetric k-unduloids of genus 1 (Fig. 1.4); their period problem can only be solved experimentally [GP]. The rhombic case is particularly interesting since it allows for a midsection, on which bubbles can be added or deleted continuously. A similar property already holds for the isosceles family (Subsection 5.5.4): as one follows a loop in the isosceles moduli space winding around the point corresponding to a surface with a cylindrical “stem” end, bubbles are generated (or deleted) on the stem. These observations suggest that the moduli spaces M_{0,k} may be connected.

1. Preliminaries

1.1. Lawson’s theorem. Our main tool is this consequence of Bonnet’s fundamental theorem, as formulated by Lawson [L, p.364]:

Theorem 1. Let M be a simply connected immersed minimal surface in 𝕊^{3}. Then there exists an isometric immersed cmc surface M̃ ⊂ ℝ^{3}, and vice versa. Furthermore, a planar arc of (Schwarz) reflection in M̃ corresponds to a great circle arc in M.

We call M and M̃ conjugate cousin surfaces or simply associated surfaces.

We will consider fundamental domains of cmc surfaces with respect to a group of reflections, whose entire boundary consists in planar arcs of reflection symmetry (geodesic curvature arcs).

Corollary 2. A cmc surface M̃ bounded by geodesic curvature arcs is associated to a spherical minimal domain M bounded by great circle arcs.

The fundamental domains of the finite topology cmc surfaces we study are never compact, and some bounding geodesic curvature arcs are infinitely long. We call such an arc a ray or a line if it extends in one or both directions to infinity, respectively. We use the same terminology for the boundary of the associated minimal domains in 𝕊^{3}: each arc of the boundary is a parameterized great circle arc, which in the case of rays and lines covers the great circle infinitely many times.

1.2. Karcher’s Hopf fields. A Hopf field is defined by the oriented unit tangent vectors to the great circle fibres of a fixed, oriented Hopf fibration of 𝕊^{3}. Each oriented great circle is tangent to a unique Hopf field. Since there is an 𝕊^{2} worth of fibres in a Hopf fibration, and since the space of oriented great circles in 𝕊^{3} is 𝕊^{2} × 𝕊^{2}, there is an 𝕊^{2} worth of such fibrations. Each fibration can be identified with the tangent directions at some point p in 𝕊^{3}. In the following A, B, C will always denote a fixed positively oriented orthonormal basis of the tangent space T_{p}𝕊^{3} = ℝ^{3}. With these choices, a Hopf field is determined by a linear combination aA + bB + cC with a^{2} + b^{2} + c^{2} = 1.

4 GROSSE-BRAUCKMANN AND KUSNER

The spherical boundary contours associated to a cmc fundamental domain bounded by geodesic curvature arcs have a property independent of the length of the arcs: each arc of the contour determines a Hopf field. More precisely the Hopf fields for associated great circle arcs are unique with the following properties: (i) fields determined by consecutive great circle arcs make an angle equal to the angle enclosed by the geodesic curvature arcs at a vertex (this is the dihedral angle of the two symmetry planes containing the arcs); (ii) when measured in terms of Hopf fields, the tangent plane along the boundary of the spherical minimal surface rotates from one vertex to the next as much as the tangent plane rotates between two vertices on the cmc domain.

Hopf fields were added to Lawson’s conjugate surface construction by Karcher. See [Ka] or [G] for proofs and details.

1.3. Ends. Suppose we divide a cmc surface by its reflection symmetry group. If a symmetry plane contains the axis of an end, we call the portion to one side of this plane a half end. If another symmetry plane contains the axis and is orthogonal to the first plane, then the portion contained in the wedge between these planes is a quarter end.

We want to characterize the pair of rays on the boundary of quarter or half ends. To do this we look at the case of a quarter or half Delaunay surface first. A quarter end of a Delaunay unduloid with neckradius ρ has an associated boundary polygon with two great circle rays; the associated spherical minimal surface domain, which is part of a spherical helicoid, contains perpendicular great circle arcs of length

\frac{πρ}{2} and \frac{π(1 − ρ)}{2}. (1)

Similarly a half end has perpendiculars of length

πρ and π(1 − ρ). (2)

Since associated surfaces are isometric, this follows from the fact that there are symmetry circles of radius ρ and 1 − ρ running around a neck or bulge of the unduloid, respectively. These circles alternate at distance π/2 for ρ ≠ 1/2; in case ρ = 1/2 the helicoid associated to the cylinder is foliated by these perpendiculars.

An end of an almost embedded cmc surface with finite topology is asymptotically a Delaunay unduloid [KKS], so its neckradius and axis are defined. It turns out that the great circle rays bounding quarter and half ends coincide with the rays on the boundary associated to the limiting quarter and half Delaunay ends.

Lemma 3. Let M̃ be a cmc fundamental domain which contains a quarter end of neckradius ρ. The associated minimal domain M in 𝕊^{3} is then bounded by two great circle rays. The asymptotic limit of the surface contains perpendicular great circle arcs of lengths given as in (1). Similarly, a half end of neckradius ρ is bounded by two great circle rays, and its asymptotic limit contains perpendiculars of lengths given as in (2).

Proof. By Lawson’s theorem M has great circle boundary rays. Each embedded end of M̃ is exponentially asymptotic (in the C^{∞} topology — see Subsection 1.4 below) to a Delaunay unduloid [KKS]. Thus M is asymptotic to a spherical helicoid. A great circle ray asymptotic to a great circle must actually cover the great circle. Hence the boundary rays of M are the great circle rays bounding the associated limiting spherical helicoid.

1.4. Continuity of families. We define a topology on the set of great circle k-gons with fixed Hopf fields by taking the set of arclengths in ℝ_{+}^{k} as continuous coordinates. We will also need a topology on families of polygonal contours containing a consecutive pair of rays. The pair will always have a shortest perpendicular tangent to a given Hopf direction. In this case we can truncate the rays at the endpoints of this perpendicular, and include their truncated lengths (modulo 2π), as well as the length of this perpendicular into the coordinates.

For families of cmc surfaces there are various choices for a topology. If we are given a cmc surface we can write sufficiently C^{0}-close cmc surfaces as graphs over compact subsets and use the C^{1}-norm of these graphs. By elliptic regularity theory, the resulting topology is equivalent to the C^{∞}-topology. Thus from Lawson’s theorem we obtain the following fact.

Lemma 4. Consider a family of cmc surfaces invariant under a fixed group of reflections, such that their fundamental domains are simply connected and bounded by geodesic curvature arcs. The map which assigns the great circle polygons bounding the associated minimal domains to these surfaces is continuous.

1.5. Balancing and Kapouleas’ existence result. Let us associate a force vector

f := 2πρ(1 − ρ)a ∈ ℝ^{3} (3)

to a Delaunay unduloid with neckradius ρ, whose axis points in the direction of the unit vector a. Similarly, for each end of a surface in M_{g,k}, the asymptotic Delaunay limit defines neckradius and axes, and hence forces f_{1}, …, f_{k}. The balancing formula

∑_{i=1}^{k} f_{i} = 0, (4)

is a necessary condition, which in fact holds in much larger generality [KKS].

It is a deep fact that balancing is also, in a certain sense, a sufficient condition for the existence of cmc surfaces: among Kapouleas’ results [Kp] is that k-unduloids exist for a dense set of asymptotic axis directions whose force vectors f_{1}, …, f_{k} are balanced and small. This smallness condition means that the neckradii are also small. Here our aim is to study phenomena which occur for large neckradius.

2. Reflection symmetry

2.1. Alexandrov reflection. A. D. Alexandrov [A] invented a reflection method which was first applied to show that the round sphere is the only compact (almost) embedded cmc surface. This reflection principle was generalized to noncompact cmc surfaces in ℝ^{3} as follows [KKS]:

Theorem 5. If an (almost) embedded, complete cmc surface M lies in a planar slab of finite thickness, then there is a plane P in this slab which is a symmetry plane for M. Furthermore, each symmetric open half M^{+} and M^{−} of M is locally a graph lying to one side of P. In particular the normal takes values in open hemispheres on each open half of M. More precisely, viewing M as the isometric immersion I of an abstract surface Σ, we can take Σ = Σ̄^{+}∪Σ̄^{−} with M^{±} = I(Σ^{±}). The restriction of I to Σ^{±} can then be written in the form I = (ι, ±u) where u is a positive function over some surface domain U, and ι is an isometric immersion of U into P. In case M is embedded, we may take M = Σ and U ⊂ P, so that each half is globally a graph.

Let us prove that each half M^{±} lies to one side of P; all other statements are proved in Remark 2.13 of [KKS]. The boundary ∂M^{±} is contained in P, that is, u = 0 on ∂U. By the asymptotics u actually takes a minimum. Suppose that this minimum value is negative. Since M^{+} is locally a graph its mean curvature vector points downward. Using a plane parallel to P as a comparison leads to a contradiction.

Since each end of an almost embedded cmc surface is asymptotically Delaunay, the only assumption we need — in order to apply the previous theorem — involves the configuration of the asymptotic axes.

Corollary 6. (i) If the axes of the ends of an (almost) embedded, complete cmc surface M are contained in a slab parallel to a plane P then P is a symmetry plane. (ii) If an (almost) embedded, complete cmc surface M has all its axes parallel, then M is a Delaunay unduloid.

For part (ii) the above argument now shows that M must be cylindrically bounded, that is, within a bounded distance of a line. This means we can apply the reflection argument of the theorem in any direction about this line, and conclude that M is a surface of revolution about an axis parallel to this line.

2.2. Schwarz reflection. If we have a cmc surface with reflectional symmetries then we can recover it by Schwarz reflection from a fundamental domain bounded by geodesic curvature arcs. In the situation of the above theorem, because M^{±} is to one side of the horizontal symmetry plane P, the intersection M ∩ P consists of geodesic curvature arcs. We are interested in what happens when further symmetry planes are present. In particular we consider a vertical symmetry plane V, which is a plane orthogonal to the horizontal symmetry plane P.

SYMMETRIC CMC SURFACES WITH FEW ENDS 7

The vertical reflection acts isometrically on the abstract surface Σ with a fixed point set F, and again I(F) ⊂ V ∩ M consists of geodesic curvature arcs. In the embedded case I(F) = V ∩ M.

When M has genus 0 and k ends we can represent Σ^{+} (or Σ^{−}, or even U) by a closed disk D with k points removed from ∂D corresponding to the ends of M. Furthermore, if we take a conformal representation, then an isometry of Σ^{+} acts by a hyperbolic isometry of D. Thus the fixed point set F^{+} = F ∩ Σ^{+} of a reflection is a hyperbolic geodesic. This geodesic joins either a pair of boundary punctures, or a boundary point and a puncture, or a pair of boundary points. The corresponding cases on M^{+} are either a geodesic curvature line, or a ray, or an arc, respectively.

For genus 1 we can replace the disk representing M^{+} with an annulus, whereas for genus g > 1 we must allow for h handles attached to the disk as well as c open disks removed from the interior of D, where g = c + 2h.

We summarize our discussion above as follows.

Lemma 7. For almost embedded k-unduloids M of genus 0, under the assumption of (horizontal) coplanar ends, the fixed point set F^{+} under reflection in a vertical mirror plane V is connected and is either an arc in M^{+} ∩ V from the horizontal plane P to itself, a ray from P to an end, or a line from one end to another.

3. The trigonometry of Lawson quadrilaterals

[figure: β; l; t; r; s; p̃; β; \frac{π}{2} − l; r; l; s; t; \frac{π}{2} −t; \frac{π}{2} −s; C; A; −A; sinβA−cosβC; −B; p]

Figure 2. Quadrilateral bounding a euclidean cmc fundamental domain and an isometric spherical minimal domain (righthand sketch is meant as a stereographic projection).

3.1. Existence. Our analysis rests on spherical trigonometry, namely the trigonometry of the boundary polygons of the spherical minimal surfaces associated to the fundamental domains for the cmc surfaces with few ends. The basic polygon for our analysis is a quadrilateral in 𝕊^{3} with Hopf fields −A, C, sin β A − cos β C, −B. Such a quadrilateral has three right angles, and one oriented angle β ∈ [0, 2π) enclosed by the C- and (sin β A− cos β C)-arcs at some point p.

8 GROSSE-BRAUCKMANN AND KUSNER

We will use the letters l, t, r, s to denote the lengths of a quadrilateral as in Figure 2. We call quadrilaterals with the described Hopf fields, positive lengths, and 0 < l ≤ π/4 Lawson quadrilaterals. We denote them Γ(l, t, r, s; β). They arise from the truncation of associated quarter ends with an asymptotic perpendicular of length l given by Lemma 3; if we choose the shorter perpendicular then indeed 0 < l ≤ π/4. The Lawson quadrilaterals are the associated boundaries of cmc fundamental domains coming from doubly periodic cmc surfaces in case β = π/3, π/4, or π/6 [L] [G].

In this section we classify all Lawson quadrilaterals. We first construct two basic families of quadrilaterals with edge lengths at most π and then extend them to obtain all quadrilaterals. The family (i) in the following lemma was already described in Lemma 3.1 of [G].

Lemma 8. For each β ∈ (0, π/2)∪(π/2, π) there exists a continuous one-parameter family of Lawson quadrilaterals Γ(l, t, r, s; β) as follows:

(i) If 0 < β < π/2 the lengths l, t, r, s range from 0, 0, 0, 0 past β/2, π/4, π/4, β/2 to 0, π/2, π/2, β. Here r, s, t are monotonic, whereas l monotonically increases on the first part of the family and monotonically decreases on the second half. (ii) If π/2 < β < π the lengths l, t, r, s range from 0, π, 0, π past β/2, 3π/4, π/4, π − β/2, to 0, π/2, π/2, π − β. Again r, s, t are monotonic, whereas l monotonically increases on the first part of the family and monotonically decreases on the second half.

Proof. A Lawson quadrilateral with 0 < l, t, s < π/2 and 0 < r, β < π, is uniquely characterized by the four formulas

(5)  cos s cos r = cos l cos t,
(6)  sin s cos r = sin l sin t,
(7)  cos s cos l = cos r cos t + cos β sin r sin t,
(8)  sin s sin l = cos r sin t − cos β sin r cos t.

These four formulas can be obtained using the spherical cosine law, and each is implied by the remaining three, so that one parameter is free. Indeed in [G] it is shown that for 0 < l, t, r, s, β < π/2 the three equations

(9)  tan s = tan l tan t,
(10)  tan 2t = cos β tan 2r,
(11)  cos 2l = \sqrt{cos^{2}β sin^{2} 2r + cos^{2} 2r},

are equivalent to (5) – (8) except in case r = π/4 ⇔ t = π/4 ⇔ l = β/2 ⇔ s = β/2. For all other 0 < r < π/2 we get a length 0 < l ≤ β/2 by (11), 0 < t < π/2 by (10) and 0 < s < π/2 by (9). This gives a continuous family of quadrilaterals for which (5) – (8) hold, so that part (i) is proved.

SYMMETRIC CMC SURFACES WITH FEW ENDS 9

To prove (ii) note that 0 < l, t, r, s, β < π/2 solve (5) – (8) if and only if Γ(l, π − t, r, π − s; π − β) is a Lawson quadrilateral. To see this we extend Figure 2.2 to the left up to the next perpendicular of length l. Then the extended part plus the portion to the left of the bold quadrilateral form together a large quadrilateral. Up to the motion (A, B, C) ↦ (A, −B, −C) this quadrilateral has exactly the Hopf fields of the bold quadrilateral, but π/2 < β < π. Thus it is a Lawson quadrilateral, and the claims for (ii) follow from (i).

The case β = π/2 is somewhat different. The equations (5) – (8) can be checked directly to confirm the existence of the following Lawson quadrilaterals with lengths at most π.

Lemma 9. There exist right-angled Lawson quadrilaterals Γ(l, t, r, s; π/2)

(i) with 0 < l ≤ π/4, r = π/2 − l, and constant lengths s = t = π/2, (ii) with 0 < l = r ≤ π/4, and s = t = π, and (iii) with l = r = π/4, and 0 < s = t.

These right-angled quadrilaterals bifurcate, for instance at Γ(π/4, π/2, π/4, π/2; π/2), which is of type (i) and (ii). As Figure 2 indicates, the right-angled Lawson quadrilaterals are the associated boundaries of fundamental domains for the Delaunay unduloids. Therefore these bifurcations can be given the following geometrical interpretation. For the non-cylindrical unduloids the planar curvature circles are spaced at distance π/2 along the meridians. Small and large circles alternate, so that at distance π/2 there are two different circles (case (i) of the Lemma), and at distance π two equal ones (case (ii)). On the cylinder, there is a continuous family of planar symmetry circles having any distance s = t > 0 (case (iii)).

3.2. Uniqueness. We now want to characterize all Lawson quadrilaterals. In the general case l, t, r, s ∈ ℝ_{+} the ‘quadratic’ equations (9) – (11) imply (5) – (8) only up to sign. Checking cases for the latter set of equations gives the uniqueness result:

Lemma 10. All Lawson quadrilaterals with 0 ≤ β < π are generated from those listed in Lemma 8 and 9 by the following substitutions:

(i) adding 2πn for n ∈ ℕ to r, s, or t; (ii) replacing (s, t) by (s + π, t + π); (iii) replacing (r, t) by (r + π, t + π); (iv) replacing (r, s) by (r + π, s + π).

Whereas there are no Lawson quadrilaterals with β = π, the quadrilaterals with π < β < 2π are obtained by another substitution from Lemma 10:

Lemma 11. Lawson quadrilaterals with π < β < 2π are obtained from Lemma 8 and 9 using the following substitutions:

(i) replacing (r, β) by (2π − r, β + π), or (ii) replacing (r, s, β) by (π − r, s + π, β + π). All such quadrilaterals are obtained from these by the substitutions listed in Lemma 10.

4. Rectangular surfaces

Consider 4-unduloids of arbitrary genus with three orthogonal symmetry planes. For combinatorial reasons the axes of the ends must all be contained in one of the planes, which we call horizontal. On the other hand, by Corollary 6(ii) no two axes coincide, and there are only two ways for a 4-unduloid to have this symmetry:

• The axes form a right-angled cross, so that they are contained in the intersection of the horizontal with the vertical symmetry planes as in Figure 1.1. Opposite ends are congruent and thus there are only two neckradii ρ_{1} and ρ_{2}. Since the axes are at right angles to each other we call such 4-unduloids rectangular.

• All ends are congruent and their axes form an angle α or π/2−α with the intersection of the two vertical symmetry planes as in Figure 1.3. There can be a finite length midpiece in between the intersection of two pairs of the four axes. We call this case rhombic.

The two cases have the dihedrally symmetric 4-unduloids [G] in common.

We assume our surfaces have genus 0 for the remainder of this section. In our conformal model given in Subsection 2.2 we represent each half of a four-ended surface by a closed disk with four boundary punctures. One vertical reflection can either (i) fix a pair of opposite punctures and interchange the other, or (ii) induce a transposition of two pairs of punctures. The fixed point set under the vertical reflection joins the fixed punctures in case (i), and by Lemma 7 there is a geodesic curvature line on each half of the cmc surface. In case (ii) the fixed point set is a geodesic curvature arc which under the horizontal reflection completes to a closed loop.

Any other vertical reflection must be of the same type, unless the surface is dihedrally symmetric. Thus case (i) corresponds to the rectangular surfaces, while (ii) is the rhombic case. In particular, a rectangular surface has a fundamental domain bounded by one line and two rays meeting in a right angle at a point p̃ as in Fig. 3.1. On the other hand, a rhombic surface has fundamental domain bounded by two arcs and two rays, all meeting at right angles. The rectangular case, which we consider here, does not pose a period problem; the period problem posed by the rhombic surfaces can be dealt with numerically [GP].

The rectangular fundamental domain contains two quarter ends. To simplify we use asymptotic perpendiculars provided by Lemma 3 to truncate the spherical contour. We truncate at the necks, and by (1) the truncated arcs have lengths

0 < l_{1}, l_{2} ≤ π/4. (12)

The resulting pentagon, with the truncated line of length s and two truncated rays of lengths t_{1}, t_{2}, is right-angled. We denote it by Γ(t_{1}, l_{1}, s, l_{2}, t_{2}).

[figure: s_{1} + s_{2}; t_{1}; t_{2}; l_{1}; l_{2}; p̃; β; r; l_{2}; s_{2}; t_{2}; \frac{π}{2} −β; C; −A; sin βA−cos βC; −B; A; t_{1}; l_{1}; −C; s_{1}; p]

Figure 3. Generating cmc fundamental domain for a rectangular surface, and spherical boundary polygon of associated spherical minimal surface. The spherical contour is truncated by two asymptotic perpendiculars which are associated to the two dotted symmetry lines of length l_{1}, l_{2} contained in the asymptotic cmc limit of the end.

To reduce the spherical trigonometry of these pentagons to the quadrilaterals of Section 3 we determine the Hopf fields of the pentagons (Figure 3). We start at the intersection point p of the geodesic rays. The Hopf fields are then A for one ray, −C for its asymptotic perpendicular, −B for the line, −A for the other asymptotic perpendicular, and C for the returning ray. We refer to a right-angled pentagon with these Hopf fields and positive lengths satisfying (12) as a truncated rectangular contour. The main geometric observation is a decomposition of these pentagons into quadrilaterals as indicated in Figure 3.2:

Lemma 12. The set of truncated rectangular contours Γ(t_{1}, l_{1}, s, l_{2}, t_{2}) is in 1-1 correspondence to pairs of Lawson quadrilaterals Γ_{1}(l_{1}, t_{1}, r, s_{1}; β) and Γ_{2}(l_{2}, t_{2}, r, s−s_{1}; π/2−β) with 0 < r, β < π/2.

Proof. If the two quadrilaterals are glued along their arcs of length r in such a way that the two complementary angles face each other and are contained in the same tangent plane, a right-angled pentagon is formed. Moreover, this pentagon has the desired Hopf fields.

On the other hand, suppose the pentagon is given. We consider a great circle through p which meets the geodesic containing the opposite −B arc orthogonally in a point q. Extending s and t_{1}, t_{2} further if necessary, two quadrilaterals Γ_{1} and Γ_{2} result. By orthogonality, the Hopf field of the diagonal is of the form sin β A − cos β C with β ∈ (0, 2π). We claim that in fact 0 < β, r < π/2.

We use the following fact, which is immediate from Lemmas 8, 10, and 11: a Lawson quadrilateral with β ∈ (0, π) and β ≠ π/2 has 0 < r mod π < π/2, whereas a quadrilateral with β ∈ (π, 2π) and β ≠ 3π/2 has π/2 < r mod π < π.

Suppose that the quadrilateral Γ_{2} has β ∈ (π/2, π). This means that the quadrilateral Γ_{1} has angle in (3π/2, 2π). By the preceding fact there is no consistent choice of r and this case is impossible. The same argument applies to β ∈ (3π/2, 2π). The case β ∈ (π, 3π/2)

12 GROSSE-BRAUCKMANN AND KUSNER

leads to a consistent choice r ∈ (π/2, π). However, in this case we can replace the diagonal by an arc in the opposite direction; the arc in the opposite direction has β ∈ (0, π/2) and meets the antipodal point of the former endpoint at a length π − r ∈ (0, π/2). Finally, since l_{1}, l_{2} > 0 the angle β can not be an integer multiple of π/2, and hence the claim is proved.

As β ∈ (0, π/2) the length of r can be taken to be in (0, π/2) by Lemma 8(i) and Lemma 10. Furthermore, again by Lemma 10 we see that q is contained in the pentagon and the extension is not necessary.

We now analyze the pentagons via the two quadrilaterals of the lemma. Note that the family of dihedrally symmetric 4-unduloids constructed in [G] satisfies 0 < ρ ≤ 1/4. These surfaces lead to symmetric pentagons Γ(l, t, s, t, l) with 0 < l ≤ π/8: they decompose into two equal Lawson quadrilaterals Γ(l, t, r, s/2; π/4) as given by Lemma 8(i). For the general case the two quadrilaterals are different but l_{1} + l_{2} ≤ π/4 still holds.

Lemma 13. There exists a continuous two-parameter family of truncated rectangular pentagon contours

G_{trunc} = {Γ(t_{1}, l_{1}, s, l_{2}, t_{2}) | 0 < l_{1}, l_{2} and l_{1} + l_{2} ≤ π/4, 0 < t_{1}, t_{2}, s < π/2}.

For each pair l_{1}, l_{2} > 0 with l_{1} + l_{2} < π/4 the family contains exactly two distinct contours, while for l_{1} + l_{2} = π/4 there is only one.

Proof. We determine a quadrilateral Γ_{1}(l_{1}, t_{1}, r, s_{1}; β) with angle 0 < β < π/2 and all edgelengths in (0, π/2) as follows. We take 0 < l_{1} < π/4 and r with l_{1} < r < π/2 − l_{1} as parameters. Then (11) (with l = l_{1}) gives

(13) cos^{2}β = \frac{cos^{2}2l_{1} − cos^{2}2r}{sin^{2}2r}

and we obtain a β ≥ 2l_{1}. Since l_{1} < r the numerator of (13) is positive, and the fraction is at most 1, so that we can choose β < π/2. Then Lemma 8(i) gives existence of a family of quadrilaterals Γ_{1}(l_{1}, t_{1}, r, s; β) with 0 < t_{1} = t < π/2 and 0 < s < β < π/2 as in (10) and (9). These quadrilaterals depend continuously on the parameters r and l_{1}.

With r and β fixed, we now construct another quadrilateral Γ_{2}(l_{2}, t_{2}, r, s − s_{1}; π/2 − β). According to (11) and Lemma 8(i) there is such a quadrilateral with 0 < l_{2} < π/4 determined by

(14) cos^{2}2l_{2} = cos^{2}(π/2 − β) sin^{2}2r + cos^{2}2r,

and 0 < t_{2} < π/2 and s − s_{1} < π/2 − β given by (10) and (9). Consequently, the quadrilaterals Γ_{2} form a family which is continuous in (l_{1}, r). By Lemma 12 this gives a truncated rectangular pentagon contour.

We now want to show that all pairs l_{1}, l_{2} with l_{1} + l_{2} ≤ π/4 are attained, such that there are two different pentagons for l_{1} + l_{2} < π/4, and one if equality holds. For this we

�

[figure: [Kp]; ρ_{1}; ρ_{2}; \frac{1}{2}; \frac{1}{2}; [B]; [B]; [G]]

Figure 4. The moduli space of rectangular boundary contours can be represented by a two-sheeted covering of the triangle defined by ρ_{1}, ρ_{2} > 0 and ρ_{1}+ρ_{2} < 1/2; the two sheets are glued along the dashed line ρ_{1}+ρ_{2} = 1/2.

consider the extremal choices of r. If r ↘ l_{1} then by (13) β ↗ π/2, and thus by (14) l_{2} ↘ 0. On the other hand, for r = π/4, (13) implies β = 2l_{1}, and from (14) we conclude 2l_{2} = π/2 − 2l_{1}.

For given l_{1}, it follows from continuity that a choice of r in the lower interval (l_{1}, π/4] gives all l_{2} ∈ (0, π/4 − l_{1}]. Taking r-derivatives of (13) and (14) it is elementary to see that the function l_{2}(r) is strictly monotonic. Thus each l_{2} in (0, π/4 − l_{1}] is taken exactly once, and in particular l_{1} + l_{2} ≤ π/4.

The upper r-interval [π/4, π/2 − l_{1}) also yields quadrilaterals Γ_{2} with l_{2} ∈ (0, π/4 − l_{1}]. This follows with the same arguments since the limit r ↗ π/2−l_{1} is analogous to r ↘ l_{1}.

Fixing the pentagon at p we extend the lengths t_{1}, t_{2} and s to infinity. We denote by G the family of extended contours which results this way from G_{trunc}. Although there are many other choices of pentagons for rectangular surfaces besides the family G_{trunc}, the extended contours are unique. This follows from the decomposition of the pentagon into quadrilaterals (Lemma 12), with the Uniqueness Lemma 10. Furthermore, on the boundary of G_{trunc} the lengths l_{1} or l_{2} vanish. Thus the family G is maximal:

Theorem 14. The neckradii of a rectangular 4-unduloid satisfy

0 < ρ_{1} + ρ_{2} ≤ \frac{1}{2}. (15)

Furthermore, all rectangular 4-unduloids have a fundamental domain associated to a spherical minimal surface with boundary in the continuous two-parameter family G; there are two contours which satisfy (15) with strict inequality, and one for equality.

The boundary of the moduli space is given by the contours with vanishing ρ_{1} or ρ_{2}; the degenerate limiting contours bound surfaces associated to Delaunay surfaces with an orthogonal string of spheres.

We believe that our two-parameter family G is in 1-1 correspondence to a continuous family of rectangular cmc surfaces. One sheet of the boundary contours with r > π/4

14 GROSSE-BRAUCKMANN AND KUSNER

corresponds to surfaces with more spherical centers, while on the other sheet with r < π/4 the centers are smaller. In particular a a rescaled sequence of surfaces in the latter sheet with ρ_{1} = λρ_{2} → 0 converges to a minimal surface with four catenoid ends of alternating logarithmic growth.

To prove existence, one would have to solve Plateau’s problem for the spherical contours in G with minimal surfaces. We believe these minimal surfaces are unique, at least in the class with almost embedded associate surfaces. So far existence is known for the following cases (see Figure 4):

• On an infinite subset in the neighborhood of the origin ρ_{1} = ρ_{2} = 0 by Kapouleas [Kp]. This corresponds to the sheet of contours with spherical centers. It is conceivable that these surfaces form in fact an open continuous family (cf. Remark 4.6 of [Kp]).

• On the diagonal of G where the corresponding dihedrally symmetric 4-unduloids attain the neckradii 0 < ρ_{1} = ρ_{2} ≤ 1/4 [G].

• Existence of the maximal neckradius family with ρ_{1} + ρ_{2} = 1/2 for either ρ_{1} or ρ_{2} small was shown by Berglund [B]. Asymptotically, these surfaces are almost cylinders in one direction, and almost spherical unduloids in the other.

It is known that there are only finitely many components in the moduli space of rectangular surfaces with neckradius greater than any fixed ε > 0; this follows from the curvature and area bounds of [KK].

Doubly periodic surfaces of rectangular type. Using Kapouleas’ method [Kp], doubly periodic surfaces with rectangular lattice can be found. Here we want to let the lattice size vary, and require only that the two generating vectors are orthogonal (we still fix H = 1). Under this assumption Kapouleas has countably many families F_{m,n} in a neighborhood of the following degenerate situation: on one set of parallel axes there are m ∈ ℕ spheres in between the junction spheres and likewise n ∈ ℕ spheres in the other direction. For these surfaces the two small neckradii ρ_{1} and ρ_{2} are parameters. The boundary of the associated spherical fundamental domains form a family of contours contained in a family G_{m,n}. This family is obtained from G_{trunc} as follows: for a contour in G_{trunc} the arc t_{1} is extended by nπ/2, t_{2} by mπ/2, and s by (m + n)π/2.

Since a continuous family of cmc surfaces gives rise to a continuous family of associated spherical boundary polygons (Lemma 4) the maximality of G_{trunc} gives the maximality of G_{m,n}:

Theorem 15. For each m, n ∈ ℕ × ℕ, Kapouleas’ doubly periodic rectangular surfaces F_{m,n} are contained in a different connected component of the moduli space of doubly periodic surfaces with a set of two orthogonal lattice vectors.

Again we believe that the Plateau problem for the contours G_{m,n} can be solved with a continuous family of surfaces extending those of Kapouleas. The numerical results of [GP]

SYMMETRIC CMC SURFACES WITH FEW ENDS 15

make us doubt that the theorem continues to hold in the class of all (not necessarily rectangular) doubly periodic surfaces. Rather, it seems that only the imposition of a sufficiently strong symmetry group separates the moduli space into different connected components.

5. Isosceles triunduloids

The class M_{g,3} of triunduloids of any genus is special in that a (horizontal) symmetry plane is present [KKS]. Indeed by balancing (4) the three force vectors of the ends must be contained in a plane, which is a symmetry plane by Corollary 6(ii). We assume an additional orthogonal (vertical) symmetry plane and call the triunduloids of this type isosceles. The symmetry implies the axes for the ends of an isosceles triunduloid resemble a Y, and in particular, the axes all meet at a point (this is also true for a general triunduloid and can be proven using torques [K1]). The intersection of the two symmetry planes contains the axis of one end, which we call the stem. The other two ends are congruent, and their axes, the arms, enclose a well-defined angle

α ∈ (0, π/2)

with the intersection of the symmetry planes (see Figure 1.2). For α = 0 the surfaces would be I-shaped which is impossible by Corollary 6(i). For α = π/2 the surfaces are T-shaped; this and any larger angle are excluded by the balancing formula (4). We remark that isosceles triunduloids with π/2 < α < π exist outside the almost embedded class, for instance with nodoid ends on either stem or arms [Kp].

As with the symmetric 4-unduloids in the previous section, we can characterize a fundamental domain for any isosceles triunduloid of genus 0. The conformal model in Subsection 2.2 lets us represent either half of a triunduloid as a closed disk with three boundary punctures, and the vertical reflection must fix one of the punctures and interchange the other two. It follows from Lemma 7 that the fixed point set of the vertical reflection consists of a geodesic curvature line symmetric about the horizontal mirror plane P, and whose two ends run out the stem end. Each symmetric half is then a geodesic curvature ray which meets P only at its endpoint p̃. A fundamental domain for any isosceles triunduloid is then bounded by this ray in the vertical plane, together with a geodesic curvature ray and line in the horizontal plane (see Figure 5.1).

Again the known asymptotics of the ends allows us to truncate the associated infinite contour with its asymptotic perpendiculars. Starting with the point p associated to p̃ (Figure 5) the Hopf fields of the truncated contour can be seen to be −B (ray of half end), cos α A − sin α C (asymptotic perpendicular for half end), −B (line), −A (asymptotic perpendicular of quarter end), C (ray). We can require

0 < l ≤ \frac{π}{4} and 0 < r ≤ \frac{π}{2} , (16)

16 GROSSE-BRAUCKMANN AND KUSNER

if we truncate the contour with the shorter geodesics, i.e. we cap the ends at the necks not at the bubbles. We call a pentagon with the above Hopf fields which satisfies (16) a truncated isosceles contour and denote it with its five lengths Γ(b, r, s − b, l, t).

5.1. Trigonometry. Like the rectangular case, the main idea is to reduce the trigonometry of the pentagons to quadrilaterals. In this case one of the two quadrilaterals in the decomposition is what we call a Clifford rectangle Γ_{C}(b, r). This is a right-angled quadrilateral of edgelength b and r which is subset of a Clifford torus. Opposite arcs are Clifford parallel, and the Hopf fields can be taken as A, −B, −A, cos(2b)B − sin(2b)C. Thus in the special case b = π/4 a Clifford rectangle is also a rectangular Lawson quadrilateral of the type in Lemma 9(iii). In the general case the Hopf fields are easily derived from those of a Clifford torus (see [G]).

Lemma 16. (i) A Lawson quadrilateral Γ_{L}(l, t, r, s; π/2 − α + 2b) and a Clifford rectangle Γ_{C}(b, r) can be combined to form a truncated isosceles contour Γ(b, r, s − b, l, t) if condition (16) and the following hold: s > b, and 0 < π/2 − α + 2b < π. (ii) Conversely, a pentagon satisfying (16) and kπ < s−b, t < (k+1)π for sufficiently large k ∈ ℤ can be decomposed into a Lawson quadrilateral and a Clifford rectangle as in (i).

Proof. (i) If the Clifford rectangle is glued to the Lawson quadrilateral such that the b-arc of the rectangle is subset of the s-arc of the quadrilateral, and the r-arcs agree (as indicated in Figure 5), then a pentagon for isosceles surfaces results. (ii) We construct a great circle arc through the point p which meets the opposite −B geodesic line orthogonally. Since both rays of the half end have the same −B Hopf field, they have a continuous set of perpendiculars which are all Clifford parallel to the r-arc. Thus a Clifford parallel to the r-arc (in the orientation indicated in Figure 5) at distance b gives a diagonal with Hopf field cos(α − 2b) A − sin(α − 2b) C. We have to go the same distance b on both −B arcs to obtain the Clifford rectangle of the r-arc and the diagonal; this means that the diagonal only meets an extension of the opposite −B arc.

Writing the Hopf field of the diagonal in the form sin(π/2−α+2b) A−cos(π/2−α+2b) C we see that the other quadrilateral formed has only one non-right angle

β := \frac{π}{2} − α + 2b

at p and hence is a Lawson quadrilateral.

The problem to determine the possible neckradii amounts to investigating how the length l of the asymptotic perpendicular truncating the quarter end (stem) relates to the length r of the perpendicular truncating the half end (arms). For the Lawson quadrilateral Γ_{L}(l, t, r, s; π/2 − α + 2b) we obtain from (11) an expression involving the length b,

(17) cos^{2} 2l = sin^{2}(α − 2b) sin^{2} 2r + cos^{2} 2r.

[figure: l; t; α; b; r; s − b; p̃; β; l; s − b; t; cosαA−sinαC; sinβA−cosβC; C; −A; −B; b; b; r; −B; p]

Figure 5. Generating cmc fundamental domain for an isosceles surface with dotted asymptotic curvature lines, and spherical boundary polygon of associated spherical minimal surface truncated with the asymptotic perpendiculars.

5.2. Balancing and the period problem. There is a period problem for isosceles surfaces: the associated cmc domain must have the two rays bounding the half end in the same plane, not just in parallel planes. Boundary contours which can bound surfaces with vanishing periods form a codimension one family within all truncated isosceles contours. Although we do not give an existence proof, we can select such a family using two facts: the balancing formula, and the asymptotics of the ends. Technically the condition we obtain is a further necessary condition on the boundary contours.

The balancing formula (4) relates the forces f^{A} of the arms to the force f^{S} of the stem

|f^{S}| = 2 cos α |f^{A}|.

Thus (3) yields for the minimal neckradii ρ^{S} and ρ^{A}

2πρ^{S}(1 − ρ^{S}) = 2 cos α 2πρ^{A}(1 − ρ^{A}). (18)

These neckradii are the asymptotic Delaunay radii, and hence can be expressed

2πρ^{S} = 4l and 2πρ^{A} = 2r

in terms of the arc lengths of the truncated contour. We obtain

l(π − 2l) = cos α r(π − r). (19)

Solving this quadratic equation for those l admissible by (16) we arrive at the following statement on the pentagon lengths, which, by (18) is in fact a result on the neckradii.

Lemma 17. Let Γ be the contour bounding a minimal surface in 𝕊^{3} associated to the fundamental domain of a (balanced) isosceles surface. If the great circle rays of the half end in Γ have a shortest perpendicular with length r, then the great circle rays of the quarter end have shortest perpendiculars of length

0 < l = \frac{π}{4} − \sqrt{\frac{π^{2}}{16} − cos α \frac{r(π − r)}{2}} ≤ \frac{π}{4} . (20)

In particular, (20) has only a solution for those 0 ≤ r ≤ π/2 which are not greater than

r_{max}(α) := \{ \begin{matrix}\frac{π}{2} (1 − \sqrt{1 − \frac{1}{2 cos α}}) & for & 0 < α ≤ \frac{π}{3} \\ \frac{π}{2} & for & \frac{π}{3} ≤ α < \frac{π}{2} . \end{matrix}

5.3. Truncated associated boundary contours. We now study the truncated isosceles contours via the Decomposition Lemma 16. The first step is to describe a family of quadrilaterals satisfying (17) and (20). Eliminating l from these equations means that (α, r, b) is a zero of the function

f(α, r, b) := sin^{2}(α − 2b) sin^{2}2r + cos^{2}2r − sin^{2} \sqrt{\frac{π^{2}}{4} − 2 cos α r(π − r)}.

Conversely, a zero (α, r, b) of f and the length l defined by (20) solve (17). We now want to determine all zeros of f. By the periodicity of f in b it will be sufficient to confine b to some interval of length π. Specifically, we set

D := \{(α, r, b) | 0 < α < π/2, 0 < r ≤ r_{max}(α), \frac{α}{2} − \frac{π}{4} ≤ b ≤ \frac{α}{2} + \frac{π}{4} \} .

Lemma 18. On the set D, all zeros of f form a continuous two-parameter family homeomorphic to a disk

Z := {(α, r, b) ∈ D | 0 < r ≤ R(α), b = b_{i}(α, r) for i = 1, 2},

where

R(α) := π \frac{1 − cos α}{2 − cos α}.

Here b_{1}(α, r) ∈ (α/2 − π/4, α/2] and b_{2}(α, r) ∈ [α/2, α/2 + π/4) are two continuous functions with b_{1}(α, R(α)) = b_{2}(α, R(α)) = α/2.

Figure 6. The function R(α) and the projection of Z.

[figure: α; R(α); \frac{π}{2}; \frac{π}{3}; arccos \frac{2}{3}; \frac{π}{2}; \frac{π}{3}; \frac{π}{4}; σ; Z_{0}^{−}; Z_{0}^{+}; Z_{1,2}]

SYMMETRIC CMC SURFACES WITH FEW ENDS 19

Proof. It is straightforward to show that f(α, r, b) is defined on D, i.e. R(α) ≤ r_{max}(α). We want to find zeros of f on D. We establish b_{1}(α, r) first for r in the open interval 0 < r < R(α). An explicit formula for b_{1}(α, r) involves many square roots, so it is more straightforward to apply the implicit function theorem. To do so we claim:

(i) \frac{∂f}{∂b} < 0 on D, (ii) f(α, r, α/2 − π/4) > 0 for 0 < α < π/2 and 0 < r ≤ R(α), and (iii) f(α, r, α/2) < 0 for 0 < α < π/2 and 0 < r < R(α).

The derivative

\frac{∂f}{∂b} = −2 sin(2α − 4b) sin^{2}2r (21)

is clearly negative on D which gives claim (i). To prove (ii) observe sin^{2}(α−2(α/2−π/4)) =

1, so we have f(α, r, α/2−π/4) = sin^{2}r+cos^{2}r−sin^{2} (2\sqrt{\frac{π^{2}}{16} − cos α\frac{r(π−r)}{2}}) = cos^{2}(2\sqrt{…}).

This is non-negative and vanishes exactly at r = 0.

Proving (iii) amounts to a further elementary calculation: f(α, r, α/2) < 0 is equivalent to

sin^{2} (\frac{π}{2} − 2r) < sin^{2} \sqrt{\frac{π^{2}}{4} − 2 cos αr(π − r)}.

Since 0 < \sqrt{…} < π/2 and −π/2 < π/2 − 2r < π/2 this is equivalent to

|\frac{π}{2} − 2r| < \sqrt{\frac{π^{2}}{4} − 2 cos αr(π − r)}.

Squaring yields

\frac{π^{2}}{4} − 2r(π − 2r) < \frac{π^{2}}{4} − 2 cos αr(π − r),

which is equivalent to π − 2r > cos α(π − r), π(1 − cos α) > (2 − cos α)r or, finally, 0 < r < R(α).

For each α the implicit function theorem gives a differentiable function b_{1}(α, r) where 0 < r < R(α). The function b_{1} is unique in the subset of D with π/4 − α/2 < b < α/2 by (i).

It is a direct consequence of the previous calculation that (α, R(α), α/2) is also a zero of f. Moreover from (21) it follows that this is the only zero in D with r = R(α) and b ≤ α/2. Thus we can continuously extend b_{1} by setting b_{1}(α, R(α)) = α/2. Furthermore, if r > R(α) it follows from the calculation that 2\sqrt{π^{2}/4 − 2 cos αr(π − r)} < |π/2 − 2r|; thus f(α, r, b) > sin^{2}(α − 2b) sin^{2}2r > 0, and there are no zeros of f for r > R(α).

To obtain b_{2} ∈ [α/2, π/4) we set b_{2}(α, r) := α − b_{1}(α, r). Then α − 2b_{1} = 2b_{2} − α and this leaves f is invariant. All properties claimed for b_{2} follow from b_{1}.

To determine the Lawson quadrilaterals in the form for Lemma 16, we have to select values of l, t, and s satisfying (5) – (8). Whereas there is a continuous way to choose l,

20 GROSSE-BRAUCKMANN AND KUSNER

the two lengths t and s which reflect the position of the truncation by the asymptotic perpendiculars must have a discontinuity on Z. In the following lemma we first establish the quadrilaterals on Z_{i} = {(α, b, r) ∈ Z | b = b_{i}(α, r), r ≠ R(α)} for i = 1, 2, and Z_{0} := {(α, α/2, R(α))}. Then we discuss the continuity problem. It will be useful to decompose Z_{0} further, namely into the two sets Z_{0}^{−} := {(α, b, r) ∈ Z_{0} | α < arccos(2/3)} and Z_{0}^{+} := {(α, b, r) ∈ Z_{0} | α > arccos(2/3)}, and the point σ := (arccos(2/3), arccos(2/3)/2, π/4). We can now state the main technical lemma for the isosceles case.

Lemma 19. All Lawson quadrilaterals with l, r, α, b ∈ D satisfying (17), (20), and (16), and 0 < s, t ≤ π are of the following form:

(i) There are two disk-type continuous two-parameter families of Lawson quadrilaterals G_{i} := {Γ_{L}(l, t, r, s; π/2 − α + 2b) | (α, r, b) ∈ Z_{i}} for i = 1, 2. (ii) There are three continuous one-parameter families of Lawson quadrilaterals G_{0}^{±} := {Γ_{L}(l, t, r, s; π/2 − α + 2b) | (α, r, b) ∈ Z_{0}^{±}} and G_{0}^{σ} := {Γ_{L}(l, t, r, s; π/2 − α + 2b) | (α, r, b) = σ}. Furthermore, the length l of all these quadrilaterals is a continuous function on Z which vanishes on ∂Z. The lengths t, s are continuous on Z_{1} ∪ Z_{0}^{+} ∪ Z_{2}. When (α, r, b) ∈ Z_{2} approaches a point in Z_{0}^{−}, we have lim s = lim t = π; whereas for (α, r, b) ∈ Z_{1} approaching such a point, lim s = lim t = 0.

Proof. A unique value 0 < l ≤ π/4 on Z is determined by (20). Since r and α are continuous on Z so is l. On the portion of ∂Z with r = 0 it follows that l ≡ 0. The remaining boundary of Z satisfies α = π/2. Again, it follows directly from (20) that l ↘ 0 when this boundary arc is approached.

(i) We now want to define s, t on Z_{i}. For the points in Z_{1} with b_{1} < α/2 we have 0 < β < π/2. In this case (10) and (9) define 0 < s, t < π/2 continuously, and uniquely within (0, π]. This gives G_{1}.

In the proof of Lemma 8 we pointed out that the following substitution is a bijection of Lawson quadrilaterals:

(l, r, s, t; β) ↦ (l, r, π − s, π − t; π − β). (22)

When we take the quadrilaterals G_{1}, this substitution gives the quadrilaterals G_{2} parameterized by Z_{2} with π/2 < β < π and unique lengths π/2 < s, t < π. (ii) We now discuss the case b = α/2 or β = π/2. On Z_{0}^{−} we have R(α) < π/4. From (17) follows cos^{2}2l = cos^{2}2R(α), so that l = R(α) by (16). This satisfies (20) too, thus the Lawson quadrilaterals are of the form Γ(R(α), t, R(α), s; π/2). Lemma 9(ii) gives the oneparameter family Γ(R(α), π, R(α), π; π/2) defining G_{0}^{−}. By Lemma 10 these are all such Lawson quadrilaterals with 0 < s, t ≤ π.

On Z_{0}^{+} we have R(α) > π/4 and l = π/2 − R(α) satisfies (20). Now Lemma 9(i) gives the quadrilaterals Γ(π/2 − R(α), π/2, R(α), π/2; π/2) which are again unique. These give the family G_{0}^{+}.

[figure: α; \frac{π}{3}; arccos \frac{2}{3}; \frac{π}{4}; maximal; neckradius family; \frac{π}{2}; 0; spheroidal; boundary; ρ^{S} = ρ^{A} = 0; noidal; boundary; ρ^{S} = ρ^{A} = 0; Delaunay plus string boundary, ρ^{S} = 0; string at neck; string at bulge; cyl.; arms; cyl.; stem; σ; symmetric; family; Z_{0}^{−}; Z_{0}^{+}; \} ∑_{i=1}^{3} ρ_{i} ≤ 1; \} ∑_{i=1}^{3} ρ_{i} ≤ 4\frac{1−cos α}{2−cos α}]

Figure 7. The moduli space of the isosceles surfaces represented schematically as the interior of a triangle. We indicate some curves of constant neckradius with solid lines for the arms, and dotted lines for the stem. Existence is only proved for the symmetric family [G] and in an infinite set in a neighborhood of the spheroidal boundary [Kp].

Finally, at the point σ with R(α) = l = π/4 there is an entire one-parameter family G_{0}^{σ} of unique quadrilaterals Γ(π/4, s, π/4, s; π/2) with 0 < s < π given by Lemma 9(iii). This proves (ii).

We have to compare the quadrilaterals of (ii) with the one-sided limits taken in Z_{1} and Z_{2}. The points of Z_{1} converging to Z_{0}^{−} are of the form (α, r, b_{1}(α, r)) with α < arccos(2/3). Thus for r ↗ R(α) we find t ↘ 0 by (10) and s ↘ 0 by (9). On the other hand the limits from Z_{2} (given by the b_{2} branch through the substitution (22) from the b_{1} values) have s, t ↗ π.

However, for the limits converging to Z_{0}^{+}, we obtain continuity: the limiting quadrilaterals with (α, r, b_{1}) for r ↗ R(α) > π/4 have t ↗ π/2 by (10), and s ↗ π/2 by (9). The same lengths are obtained for the similar limit with (α, r, b_{2}) via (22).

5.4. All isosceles boundary contours. The truncated pentagons obtained by applying Lemma 16 to the family of quadrilaterals constructed in Lemma 19 have an arc with (formal) length s−b, which may be negative. This is avoided by taking s sufficiently large.

22 GROSSE-BRAUCKMANN AND KUSNER

In particular, there is no problem if we extend the arcs with lengths b, t, s − b to infinity. This way we obtain the boundary contours for isosceles surfaces directly. Also, the infinite contours no longer have the discontinuity along Z_{0}^{−}.

Theorem 20. An isosceles surface of angle 0 < α < π/2 and with neckradii (ρ^{S}, ρ^{A}) satisfies

(23) ρ^{A} ≤ ρ_{max}^{A}(α) := \frac{1−cos α}{2−cos α},

(24)  ρ^{S} ≤ ρ_{max}^{S}(α) := min \{\frac{cos α}{2−cos α}, 1 − \frac{cos α}{2−cos α}\} .

Furthermore, there is a maximal family G of contours which bound a minimal domain in 𝕊^{3} associated to a fundamental domain of an isosceles surface. This family is homeomorphic to a two-dimensional disk. On ∂G the contours degenerate with ρ^{S} = 0. Each pair of neckradii satisfying the strict inequalities in (23) and (24) arises for two different boundary contours, and equality is satisfied with one contour.

In particular the neckradius sum over all three ends satisfies

(25) 2ρ^{A} + ρ^{S} ≤ min (1, 4 \frac{1 − cos α}{2 − cos α}) .

Proof. Extending s and t of the quadrilaterals in G_{1} ∪ G_{2} ∪ G_{0}^{±} ∪ G_{0}^{σ} by π and applying Lemma 16 yields non-degenerate pentagons. Then we extend the pentagon arcs of length t, b, and s − b to infinity. This extension leads to a unique boundary contour for G_{0}^{σ}, and thus all isosceles contours arising from Lemma 19 are parameterized by Z. Furthermore, they are continuous in Z since l, r are continuous by Lemma 19. These are also all isosceles boundary contours, since truncation of a given isosceles contour leads to a pentagon for isosceles surfaces for which we can assume 0 < s, t ≤ π if we move the perpendiculars; for such quadrilaterals, uniqueness is proven in Lemma 19.

Finally, on ∂Z we have l ≡ 0 so that ρ^{S} = 0. Thus G is a maximal family.

Again we expect a continuous family of isosceles cmc surfaces to be in a 1-1 relation to our boundary contours. Existence is known in two cases only:

• For an infinite set in a neighborhood of the spheroidal boundary with ρ^{S} = ρ^{A} = 0 and spherical center (see 5.5.1 below) [Kp], and

• for the one-parameter subfamily with of dihedrally symmetric triunduloids with α = π/3 by [G].

5.5. The geometry of the isosceles family. On the assumption that our family of contours is bijective to a family of cmc surfaces, we discuss the geometrical properties of this family. Using a numerical existence scheme, we found such a two parameter family of isosceles surfaces, and give images in the paper [GP].

SYMMETRIC CMC SURFACES WITH FEW ENDS 23

5.5.1. The moduli space boundary. There are three geometrically distinguished arcs on the boundary of isosceles contours, i.e. the contours parameterized with ∂Z. Whereas for a non-degenerate isosceles surface the position of necks is only known asymptotically, in the limiting case (where an end has neckradius 0) there are self-intersection points of the boundary rays. In particular, the length of the curvature line from p̃ to a first pinched neck on the end is well-defined, and given by t mod π and b mod π.

On the spheroidal boundary, parameterized with (α, 0, α) ∈ Z we have ρ^{S} = ρ^{A} = 0, so that the limiting configuration consists of strings of spheres with isosceles symmetry. The limiting lengths of the curvature arcs are t = π and b = α; these are exactly the values to be expected for the degenerate surfaces with junction spheres in the center, when the necks pinch off. Isosceles surfaces obtained with Kapouleas’ method [Kp] are located in a neighborhood of this boundary component, and their limit set gives the boundary arc.

On the noidal boundary, parameterized with (α, 0, 0), the ends are also spherical (ρ^{S} = ρ^{A} = 0) but the contours have b = t = 0. This is the expected value for trinoidal junctions, i.e. the three strings of spheres are attached to a point. A blow-up in the center of the cmc surfaces such that the necks are scaled to constant radius gives in the limit a minimal trinoid with isosceles symmetry. The existence of k-unduloids close to this noidal boundary component might follow from analogous work in the constant scalar curvature setting [MP].

The noidal and spheroidal boundary agree in a limiting contour with α = 0. In this degenerate case the two arms coincide and form a double string of spheres.

Finally there is an arc which we call Delaunay plus string. Here only ρ^{S} = 0, but ρ^{A} ∈ (0, π/2]. For the half with b < α/2, there is a string of spheres attached to a Delaunay neck. When ρ^{A} → 0 the Delaunay surface itself tends to a string of spheres, and the limit agrees with the noidal boundary case for α = π/2. At the other endpoint, ρ^{A} → π/2 the Delaunay surface is a cylinder. This leads over to the other half of the family with b > α/2 which is a similar family with the string of spheres attached at a Delaunay bubble. Its limiting case is spheroidal.

5.5.2. A triunduloid with one cylindrical end. The contour at σ ∈ Z has arm neckradius ρ^{A} = 1/4 and the stem has the cylindrical neckradius ρ^{S} = 1/2. The arms enclose an angle of 2 arccos(2/3) ≈ 96.4^{∘}. This is the only contour in the family with a cylindrical neckradius. The arms can attain the cylindrical neckradius only on the moduli space boundary.

5.5.3. The maximal neckradius family. This one-parameter subfamily of contours in G is parameterized by Z_{0} and realizes the maximal neckradii for a given α. It includes the contour with cylindrical stem. The right hand side of (24) gives the minimal and maximal asymptotic radius of the stem; on Z_{0}^{−} the first radius is minimal, while on Z_{0}^{+} the second one is. At the cylindrical contour (at parameter σ) the minimum flips and the consequence is the non-differentiability of the maximal neckradius sum (25) at this point: the neckradius

24 GROSSE-BRAUCKMANN AND KUSNER

sum is 1 for arccos(2/3) ≤ α < π/2, but smaller than 1 for 0 < α < arccos(2/3). If instead of the neckradius we replace ρ^{S} by the asymptotic bulgeradius ρ_{0}^{S} = cos α/(2 − cos α) then 2ρ^{A} + ρ_{0}^{S} = 1 on the maximal family.

It is interesting to observe that the asymptotic position of the minimum neckradius shifts by π/2 at σ, so that the selection of an asymptotic planar curvature circle of radius ρ^{S} can not be analytic. Indeed, our values s mod π, t mod π given by Lemma 19(ii) are discontinuous on the maximal neckradius family at σ, when they jump by π/2, half a translational period of the end.

5.5.4. Marked bubbles. It may appear that selecting a family without requiring (16) would avoid the discontinuity of t and s in our description, so that a perpendicular of the boundary rays could be marked in a continuous way over the family. However, we want to show that a discontinuity in the marking must be present around σ.

Let γ ⊂ Z − {σ} be a loop winding once about the point σ. The isosceles boundary contours are continuous on γ. Nevertheless, according to Lemma 19, the corresponding path of quadrilaterals Γ_{L}(γ) has a discontinuity in s, t of ±π, when Z_{0}^{−} is crossed. Thus the asymptotic perpendicular moves by one full period along the stem. This means that on the closed loop γ a bubble is created or deleted. Note also that the discontinuity in t, s can be located on any curve joining ∂Z to σ.

As a consequence, a closed curve winding n times about σ generates or deletes n bubbles on the stem. For all isosceles surfaces, except for the one with cylindrical stem, we can mark a bubble on the stem. The corresponding “marked” moduli space is then the universal covering of the annulus Z − {σ}.

References

[A] A.D. Alexandrov: Uniqueness theorems for surfaces in the large, Vestnik Leningrad University, 19(13): 5–8 (1958)

[B] J. Berglund: Dissertation, University of Massachusetts, Amherst, 1996

[G] K. Große-Brauckmann: New surfaces of constant mean curvature, Math. Zeit. 214, 527–565 (1993)

[GP] K. Große-Brauckmann, K. Polthier: Constant mean curvature surfaces derived from Delaunay’s and Wente’s surfaces, SFB 288 preprint, TU-Berlin (1996); and in: Visualization and Mathematics (H.-C. Hege and K. Polthier, eds.), Springer Berlin, to appear

[GKS] K. Große-Brauckmann, R. Kusner, J.M. Sullivan: work in progress

[Ka] H. Karcher: The triply periodic minimal surfaces of A. Schoen and their constant mean curvature companions, Man. math. 64, 291–357 (1989)

[Kp] N. Kapouleas: Complete constant mean curvature surfaces in Euclidean three-space, Ann. of Math. 131, 239–330 (1990)

[KK] N. Korevaar, R. Kusner: The global structure of constant mean curvature surfaces, Invent. Math. 114, 311–332 (1993)

[KKS] N. Korevaar, R. Kusner, B. Solomon: The structure of complete embedded surfaces with constant mean curvature, J. Diff. Geom. 30, 465–503 (1989)

[K1] R. Kusner: Bubbles, conservation laws, and balanced diagrams. In: Geometric analysis and computer graphics (P. Concus, R. Finn, D. Hoffman, eds.), 103–108, Springer New York 1990 [K2] R. Kusner: Natural Lagrangian and symplectic structures for the moduli spaces of constant mean

curvature surfaces and related geometric variational problems. Preprint, 1996 [KMP] R. Kusner, R. Mazzeo, D. Pollack: The moduli space of complete embedded constant mean cur-

vature surfaces, Geom. Funct. Anal. 6, 120–137 (1996)

[L] H.B. Lawson: Complete minimal surfaces in 𝕊^{3}. Ann. of Math. 92, 335–374 (1970)

[MP] R. Mazzeo, F. Pacard: Constant scalar curvature metrics with isolated singularities, preprint 1996 [M] W.H. Meeks: The topology and geometry of embedded surfaces of constant mean curvature, J.

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Universität Bonn, Mathematisches Institut, Beringstr. 1, 53115 Bonn, Germany

  E-mail address: kgb@math.uni-bonn.de

Mathematics Department, University of Massachusetts, Amherst MA 01003, USA

  E-mail address: kusner@gang.umass.edu