Structure Theorems for Constant Mean Curvature Surfaces Bounded

by a Planar Curve

Ricardo Sá Earp^{∗} and Fabiano Brito^{†}

Mathematics Department Pontificia Universidade Catolica 22453, Rio de Janeiro, Brazil

William H. Meeks, III^{‡} Mathematics Department University of Massachusetts

Amherst, MA 01003

Harold Rosenberg Mathematics Department Université de Paris 7, 75251 Paris, France

December 17, 1990

∗Research supported by grant from CAPES.

†Research supported by grant from CNPQ.

‡Research supported by grant DE-FG02-86ER250125 of the Applied Mathematical Science subprogram of the Office of Energy Research, U.S. Department of Energy, and National Science Foundation grant DMS-8900285.

1 Introduction

A circle C in ^{3} is the boundary of two spherical caps of constant mean curvature H for any positive number H, which is at most the radius of C. It is natural to ask whether spherical caps are the only possible examples. Some examples of constant mean curvature immersed tori by Wente [7] indicate that there are compact genus-one immersed constant mean curvature surfaces with boundary C that are approximated by compact domains in Wente tori; however, this has not been proved. Still one has the conjecture:

Conjecture 1 A compact constant mean curvature surface bounded by a circle is a spherical cap if either of the following conditions hold:

1. The surface has genus 0 and is immersed;

2. The surface is embedded.

If M is a compact embedded constant mean curvature surface in ^{3} with boundary C and M is contained in one of the two halfspaces determined by the plane containing C, the Alexander reflection method [1] immediately proves M has the planar reflectional symmetries of C; hence, M is a surface of revolution. Since the only compact constant mean curvature surfaces of revolution are spherical caps (by Delaunay’s classification [3] of constant mean curvature surfaces of revolution), Conjecture 1 holds for the subclass of surfaces that are embedded and contained in a halfspace. It is therefore of interest to obtain natural geometric conditions that force a compact embedded constant mean curvature surface to be contained in a halfspace.

One result of our paper is to give the following sufficient condition for a compact constant mean curvature surface to be contained in a halfspace.

Theorem 1 Let C be a convex curve in a plane P and let M be a compact connected surface with boundary C. Assume M is embedded, of constant mean curvature, and transverse to P along C. Then M is contained in one of the halfspaces of ^{3} determined by P.

Theorem 1 can be generalized to the case where the planar curve C is not necessarily convex.

Theorem 2 Let M, C and P be as in Theorem 1, except that C is not necessarily convex. Let H^{+} and H^{−} denote the closed halfspaces determined by P and indexed so that there is a connected component M^{+} of M ∩ H^{+} with C ⊂ ∂M_{+}. Let M^{−} = M ∩ H^{−}. Let D ⊂ P be the planar disk with ∂D = C. Then

1. M ∪ D is an embedded surface;

2. M^{+} = M ∩ H^{+};

3. M^{−} is a graph over the compact domain in P bounded by ∂M^{+} − C;

4. No component of ∂M^{+} lies completely outside of the convex hull of C.

Acknowledgements. We wish to thank Remi Langevin for helpful suggestions related to the proof of Theorem 1.

2 The proof of Theorem 1.

The proof of Theorem 1 uses the balancing formula [5]:

∫_{D} Y · η_{D} = \frac{1}{2H} ∫_{∂M} Y · ν .

Here Y is a constant vector field in ^{3}, H is the positive mean curvature of M, ν is the inward pointing conormal to M along ∂M, D ⊂ P is the planar disk with boundary C, and η_{D} is the unit normal vector field to D. The orientation of η_{D} is that which orients the cycle M ∪ D when M is oriented by its mean curvature vector.

Proof of the balancing formula. The flux of Y across the oriented cycle M ∪ D is zero; hence

∫_{M} Y · η_{M} + ∫_{D} Y · η_{D} = 0 ,

where the unit normal η_{M} of M is oriented by the mean curvature vector of M.

Let X denote the position vector field of M, so that ΔX = 2Hη_{M}. Then

∫_{M} Y · η_{M} = \frac{1}{2H} ∫_{M} Y · ΔX = \frac{1}{2H} ∫_{M} Δ(Y · X)

= −\frac{1}{2H} ∫_{∂M} Y · \frac{∂X}{∂ν} = −\frac{1}{2H} ∫_{∂M} Y · ν ,

since \frac{∂X}{∂ν} = ν. □

We remark that the balancing formula still holds when M is not transverse to P along C.

Proof of Theorem 1. Assume the contrary, so that M meets P elsewhere than C. We can assume M meets P transversally by making a small vertical translation (we think of P as horizontal).

Let C_{1}, …, C_{n} be the simple closed curves of M in Int(D) (if there are any). For each j, 1 ≤ j ≤ n, let C_{j}^{+}(ε) be the planar curve on M, near C_{j}, obtained by intersecting M with the horizontal plane P (ε), at height ε. Similarly, let C_{j}^{−}(ε) be the curve in M ∩ P (−ε) that is near C_{j}.

We form an embedded surface N by removing from M the annuli bounded by the C_{j}^{+}(ε) ∪ C_{j}^{−}(ε) and attaching the horizontal planar disks D_{j}^{+} ∪ D_{j}^{−}, bounded by the C_{j}^{+}(ε) ∪ C_{j}^{−}(ε). Also we attach D to M along C. To ensure that N is embedded, one uses different values of ε when several C_{j} are concentric; see Figure 1.

Figure 1:

Let M̃ be the connected component of N that contains C. M̃ separates ^{3} into two components; let A be the closure of the bounded component. The nonsmooth points of M̃ are on C and the C_{j}^{±}(ε) and the mean curvature vector of M̃ points into A. Notice that this is possible since M̃ is obtained by attaching disks to a smooth connected submanifold of M (where the mean curvature vector is never zero) and the mean curvature vector extends across these disks (cf. Figure 1). We orient M̃ by the mean curvature vector. Denote the set P − D by Ext(D) and the interior of D by Int(D). We will see that M is disjoint from Ext(D) and Int(D), thus proving the theorem.

We first observe that M ∩ Int(D) ≠ ∅ and M ∩ Ext(D) = ∅ leads to a contradiction. Assume now that M ∩ Ext(D) = ∅. Let M_{1} be the connected component of M ∩ _{+}^{3} that contains C, where _{+}^{3} = {(x_{1}, x_{2}, x_{3}) | x_{3} ≥ 0}. M_{1} together with a proper submanifold D_{1} of D bound an embedded threemanifold W_{1} ⊂ _{+}^{3}; see Figure 2.

We apply the balancing formula to M_{1} ∪ D_{1}:

∫_{∂M_{1}} Y · ν = 2H ∫_{D_{1}} Y · η_{D_{1}} ,

with Y = e_{3} = (0, 0, 1). Notice that the conormal ν along ∂M_{1} points into M_{1}, hence, ν · e_{3} > 0 along ∂M_{1} and ∫ _{∂M_{1}}Y · ν > ∫ _{C}Y · ν, when M_{1} has other components in Int(D).

The right side of the balancing formula is 2H(area D_{1}) < 2H(area D). Now apply the balancing formula to M ∪ D: ∫ _{C}Y · ν = 2H∫ _{D}Y · η_{D},

again with Y = e_{3}. Since the right side is 2H(area D), this is incompatible with the previous inequalities. Thus we can assume M ∩ Ext(D) ≠ ∅.

We will show that M̃ ∩ Ext(D) ≠ ∅ leads to a contradiction by using the Alexandrov reflection principle with vertical planes coming from infinity towards M̃.

Recall that M̃ separates ^{3} into two connected components (A is the bounded component) and M̃ is smooth away from C and the disks D_{j}^{+} and D_{j}^{−} contained in it. First we observe that M̃ ∩ Ext(D) has no components that are null homotopic in Ext(D). To see this, suppose α were such a component. Let ℓ be an infinite line segment, starting at a point of D and intersecting α in at least two points. Consider a family Q(t), t < ∞, of parallel vertical planes coming from infinity and orthogonal to ℓ. Suppose the family Q(t) intersects M̃∩Ext(D) for the first time at t = t_{0}. Continuing the movement of Q(t_{0}) by parallel translation towards C, would produce for some t_{1} < t_{0} a point of tangential contact of M̃ with the reflection of M̃ ∩ (⋃_{t_{1}≤t≤t_{0}} Q(t) ∩ M̃) in Q(t_{1}), before reaching C. This would yield a plane of symmetry of M̃ with D on one side of the plane; a contradiction. We remark that the Alexandrov reflection principle applies here because M̃ bounds the compact region A and the first point of contact of the symmetry with M̃ occurs before D, hence at a smooth point of M̃. We used the fact that C is convex here to ensure that a symmetry of α touches α before the plane reaches C.

So we can assume Q(t) touches M̃ ∩ Ext(D) for the first time along a curve α that is homotopic to C in Ext(D). Let E be the annulus in Ext(D) bounded by α ∪ C. Observe first that Int(E) contains no components α_{1} of M̃ ∩ Ext(D) that are homotopic to C in E. This follows by using the reflection principle with vertical planes Q(t) as above: a symmetry of M̃ would intersect M̃ for a first time before arriving at D; see Figure 3.

Hence the mean curvature vector along α ∪ C points into E; in particular, along C, it points towards Ext(D). But this contradicts the balancing formula:

∫_{C} Y · ν = 2H ∫_{D} Y · η_{D} ,

where Y = e_{3}, so Y · ν > 0 along C. Since H⃗ points towards Ext(D) along C, η_{D} = −e_{3}; so the right side is −2H(area D). □

The proof of Theorem 1 generalizes immediately to constant mean curvature hypersurfaces in ^{n+1} to give:

Theorem 3 Let C be a convex (n−1)-sphere in a hyperplane P ⊂ ^{n+1} and let M be a compact submanifold with boundary C. Assume M is embedded of constant mean curvature, and transverse to P along C. Then M is contained in one of the halfspaces of ^{3} determined by P.

We make one further conjecture directly related to Theorem 1.

Conjecture 2 A compact, embedded, constant mean curvature surface in _{+}^{3} bounded by a convex planar curve in the x_{1}x_{2}-plane must have genus zero. (This conjecture is even unknown for soap bubbles bounding a convex planar curve.)

3 The proof of Theorem 2.

We now prove Theorem 2. The main idea of the proof is to construct an abstract flat three-manifold that isometrically submerses into ^{3} and then to apply Alexandrov reflection in this abstract manifold.

Let M, M^{+}, M^{−}, C, D and P be as defined in the statement of Theorem 2. Assume for convenience that P is the (x_{1}, x_{2})-plane, H^{+} is the upper halfspace in ^{3} and H^{−} is the lower halfspace. Let C_{1}, …, C_{n} be the simple closed curves in M ∩ Int(D) (if there are any). As in the proof of Theorem 1 define the related curves C_{1}^{+}(ε), C_{1}^{−}(ε), …, C_{n}^{+}(ε), C_{n}^{−}(ε) bounded by the planar disks D_{1}^{+}, D_{1}^{−}, …, D_{n}^{+}, D_{n}^{−}. Similarly, define the embedded surface N by removing from M the annuli bounded by the C_{j}^{+}(ε) ∪ C_{j}^{−}(ε) and attaching the disks D_{j}^{+} ∪ D_{j}^{−}. Also we attach D to M along C. As in the proof of Theorem 1, to ensure that N is embedded, one uses different values of ε when several of the C_{j} are concentric. We may choose these values of ε sufficiently small so that the image of the orthogonal projection of ∪∂D_{j}^{+} onto P is an embedded 1-manifold, as is the projection of ∪∂D_{j}^{−}.

Note that there is a natural partial ordering of the disks D_{1}^{+}, …, D_{n}^{+} as well as for D_{1}^{−}, …, D_{n}^{−}. Namely, we say that D_{i}^{+} ≥ D_{j}^{+} if the third coordinate of D_{i}^{+} is greater than the third coordinate of D_{i}^{+} and if C_{i} ⊂ P lies inside C_{j} ⊂ P. Of course D_{i}^{+} ≥ D_{j}^{+} geometrically just means that D_{i}^{+} lies over D_{j}^{+}. Similarly we can partially order the disks D_{1}^{−}, …, D_{n}^{−} and it makes sense to make the statement that D_{i}^{−} lies over D_{j}^{−}.

Let M̃ denote the component of N that contains D. M̃ separates ^{3} into two components; let A be the closure of the bounded component. Note that we orient M̃ by its mean curvature vector and this vector points into A.

Assertion 1 M̃ ∩ (∪D_{i}^{−}) = ∅.

Assume for the moment that Assertion 1 holds and we will show how the theorem follows. First note that M̃∩(∪D_{i}^{+}) = ∅ by applying the flux formula to the cycle M̃. (See Figure 2 and the part of the proof of Theorem 1 where it is shown that M̃ ∩ Ext(D) = ∅ implies M̃ ∩ Int(D) = ∅.) This proves M̃ = M ∪ D and therefore part 1 of Theorem 2 is proved. It follows that A is nonsmooth only along C. Also note by the flux formula applied to M̃ and Y = e_{3}, the mean curvature vector of M̃ along C points into the region above D. Therefore A lies above D.

The usual application of the Alexandrov reflection principle, applied using horizontal planes with negative third coordinate, shows that M^{−} = M̃ ∩ H^{−} is a graph over its projection onto P. This proves part 3 holds.

We now prove the part of M above P is connected; this will establish part 2 of Theorem 2. Suppose, on the contrary, that the part of M above P is not connected. Let Σ be the union of all components of M above P except the component M^{+} that contains C. Σ together with a planar domain in P bounds a compact region R in H^{+}. We claim that M^{+} is disjoint from R. Otherwise M^{+} is contained in a connected component R_{1} of R. Then take a path Γ in R_{1} joining a highest point of ∂R_{1} to a highest point of M^{+} (this is easy since M^{+} is connected). Γ meets M̃ only at its end points. At the end point on ∂R_{1}, the mean curvature points down (and into A), hence Γ ⊂ A. But at the end point of Γ on M^{+} the mean curvature vector is also pointing into A. Since it’s also pointing down there, this is impossible. This proves M^{+} is disjoint from R.

Note that every path in Int(R) is contained completely inside A or completely outside A. The reason for this is that such a path is disjoint from M since M ∩ H^{+} = M^{+} ∪ Σ. However every component of R intersects Int(A) near a highest point of the component. Hence, R ⊂ A.

Now the Alexandrov reflection principle, using horizontal planes above P, shows Σ is a graph over a domain in P. It is possible to apply Alexandrov reflection in this case because M^{+}, and hence D, is disjoint from R. We know that M^{−} is also a graph over a domain in P. Let x ∈ M^{−} ∩ Σ. Σ is not vertical at x since P would be a plane of symmetry of M if this were so. (Σ is a graph so Alexandrov reflection would show that P is a plane of symmetry of M if Σ were vertical at x.) It follows that M^{−}∪Σ is a graph of a function over a smooth domain in P with zero boundary values and on some component domain the function changes sign. Hence on this component the function has both local maxima and local minima and the sign of the mean curvature of the graph at these points must be opposite. Thus M^{−} ∩ Σ = ∅ and Σ = ∅ as desired. This completes the proof of parts 1, 2 and 3. The proof of part 4 now proceeds as in the proof of Theorem 1; one does Alexandrov reflection using vertical planes. It remains to prove Assertion 1.

Proof of Assertion 1. Suppose M̃ ∩ (∪D_{i}^{−}) ≠ ∅. After reindexing we may assume that M̃ ∩ (⋃_{i=1}^{n} D_{i}^{−}) = ⋃_{i=1}^{k} D_{i}^{−}. Let M_{1}, …, M_{p} denote the components of N, other than M̃, that have at least one of the disks D_{1}^{+}, …, D_{k}^{+} in their boundary. Each M_{i} separates ^{3} into two components; let A_{i} denote the closure of the bounded component. Note that the mean curvature of M_{i} points into A_{i}. We will orient all planar disks in ∂A or in ∂A_{i} by the inward pointing normal.

Choose an i ≤ k. We know that D_{i}^{−} ⊂ ∂A. Let j be such that D_{i}^{+} ⊂ ∂A_{j}. Since the mean curvature vector of the annulus on M joining ∂D_{i}^{−} to ∂D_{i}^{+} is continuous, D_{i}^{−} is oppositely oriented from D_{i}^{+}. Let Q_{i} denote the solid cylinder bounded by D_{i}^{−} ∪ D_{i}^{+} together with the annulus on M bounded by ∂D_{i}^{+} ∪ ∂D_{i}^{−}.

Form a space X as follows. First take the disjoint union of A, A_{1}, …, A_{k}. For each i, i ≤ k, such that the outward orientation of Q_{i} agrees with the orientation of D_{i}^{−} ∪ D_{i}^{+}, we attach Q_{i} to the disjoint union of A, A_{1}, …, A_{k}. Note that the space X is a flat compact three-manifold that submerses isometrically into ^{3}.

Suppose for the moment that there exists a disk D_{t_{1}}^{−} in ∂X such that D_{t_{1}}^{−} is also in ∂A, i.e., 1 ≤ t_{1} ≤ k. In particular D_{t_{1}}^{−} is oriented by e_{3}. Notice that there exists a disk D_{j}^{−} ⊂ ∂A that lies directly above D_{t_{1}}^{−}. (A lies above D and A lies above D_{t_{1}}^{−}.)

Claim 1 Suppose E_{1}^{+} is the component in {A, A_{1}, …, A_{p}} such that D_{j}^{+} ⊂ ∂E_{1}^{+}. Then there does not exist a disk of E_{1}^{+} that lies between D_{j}^{+} and D_{t_{1}}^{+} and furthermore D_{t_{1}}^{+} ⊂ ∂E_{1}^{+}.

Claim 1 will allow us to modify X as follows. By Claim 1, for every disk D_{t_{1}}^{−} in ∂X such that D_{t_{1}}^{−} is also in ∂A, the component E_{1}^{+} in {A, A_{1}, …, A_{s}} such that D_{j}^{+} ⊂ ∂E_{1}^{+} has the property that D_{t_{1}}^{+} ⊂ ∂E_{1}^{+} and there no disks of E_{1}^{+} between D_{j}^{+} and D_{t_{1}}^{+}. We next modify X by lifting the cylinder Q_{t_{1}} into X and then form \overline{X − Q_{t_{1}}}. After removing all such solid cylinders in X, we obtain a new manifold X̃ with the property that every disk D_{s}^{−} ⊂ ∂X satisfies D_{s}^{−} ∩ ∂A = ∅.

Proof of Claim 1. Suppose that there is a disk D_{t_{2}}^{+} ⊂ ∂E_{1}^{+} that lies between D_{j}^{+} and D_{t_{1}}^{+}. We choose D_{t_{2}}^{+} to be the lowest such disk, i.e., there are no disks of ∂E_{1}^{+} between D_{j}^{+} and D_{t_{2}}^{+}. Then D_{t_{2}}^{+} is oriented by −e_{3}. Now the disk D_{t_{2}}^{−} is between D_{t_{1}}^{−} and D_{j}^{−}. D_{t_{2}}^{−} is not in ∂A (since D_{j}^{−} is the disk of ∂A directly above D_{t_{1}}^{−}) and D_{t_{2}}^{−} ⊂ A, therefore the connected component of N containing D_{t_{2}}^{−} is contained in A. Let E_{2}^{−} be the compact region bounded by this connected component; E_{2}^{−} ⊂ A. Since D_{t_{2}}^{+} is oriented by −e_{3}, D_{t_{2}}^{−} is oriented by e_{3}. Hence there is a disk D_{t_{3}}^{−} of ∂E_{2}^{−} directly above D_{t_{2}}^{−}. Clearly D_{t_{3}}^{+} lies between D_{t_{2}}^{+} and D_{j}^{+}. Since D_{t_{3}}^{+} ⊄ ∂E_{1}^{+} and D_{t_{3}}^{+} ⊂ E_{1}^{+}, the connected component of N that contains D_{t_{3}}^{+} is contained in E_{1}^{+}. Therefore the compact region of ^{3} bounded by this component is contained in E_{1}^{+}. We label this region E_{3}^{+}. Since D_{t_{3}}^{+} is oriented by e_{3}, there exists a disk D_{t_{4}}^{+} of ∂E_{3}^{+} directly above D_{t_{3}}^{+}. The connected component of N that contains D_{t_{4}}^{−} is contained in E_{2}^{−} so the compact region E_{4}^{−}, that it bounds, is also contained in E_{2}^{−}. Clearly this process yields an infinite sequence of compact regions E_{2}^{−}, E_{4}^{−}, etc., which is impossible. This proves that there is no disk of ∂E_{1}^{+} between D_{t_{1}}^{+} and D_{j}^{+}.

Next we prove D_{t_{1}}^{+} ⊂ ∂E_{1}^{+}. If not, then the connected component of N that contains D_{t_{1}}^{+} is contained in E_{1}^{+}, since D_{t_{1}}^{+} ⊂ E_{1}^{+}. Let E_{2}^{+} be the compact region bounded by this connected component. We know D_{t_{1}}^{+} is oriented by −e_{3} so there is a disk D_{t_{2}}^{+} of ∂E_{2}^{+} directly below D_{t_{1}}^{+} and above D_{j}^{+}. Now D_{t_{2}}^{−} is between D_{t_{1}}^{−} and D_{j}^{−} and D_{t_{2}}^{−} is not in ∂A. Hence the connected component of N that contains D_{t_{2}}^{−} is contained in A. Let E_{3}^{−} be the compact region bounded by this connected component. Clearly this process (just as in the last paragraph) yields an infinite sequence of compact regions E_{3}^{−}, E_{5}^{−}, …, which is impossible. This proves Claim 1. □

Now consider the disks of ∂X̃ in H^{+}. We claim there are no disks of ∂X̃ that are directly above the D_{i}^{+} that came from the D_{i}^{−} of ∂A oriented by −e_{3}. One proves this exactly as the proof of Claim 1: just turn ^{3} over (x_{3} → −x_{3}) and call such a disk of ∂X̃ (if it existed) D_{t_{1}}^{−}. Then proceed exactly as in the proof of Claim 1.

Now consider the surface M̃^{−} = M̃∩H^{−}. Let A_{−t} = {p ∈ A | x_{3}(p) ≤ −t} for any −t ≤ 0. Let P_{t} = P + (0, 0, t) and let R_{t}: ^{3} → ^{3} be reflection in P_{t}.

Consider A_{−t} ⊂ X̃ and the largest value −t_{0} such that the induced reflection R̃_{t}: A_{−t} → X̃ is defined. Since there are no disks in ∂X̃ directly above the D_{i}^{+} that came from the D_{i}^{−} of ∂A oriented by −e_{3}, the Alexandrov reflection principle shows that t_{0} = 0 and M̃^{−} is a graph over its projection Δ onto P. (It is helpful to think of M̃^{−} as having its boundary on P by letting ε go to zero.) Notice that Δ ⊂ X̃ and Δ ∩ C = ∅. On the other hand, a component of Δ that has a boundary curve in D must also have another boundary curve in P −D and so such a component must contain the curve C. Since Δ ∩ C = ∅, we conclude that Δ ∩ D = ∅. This establishes Assertion 1 and completes the proof of Theorem 2. □

Remark 1 Brito and Sá Earp [2] recently gave a somewhat simpler proof of Theorem 2, again using the balancing formula and a reflection principle related to the Alexandrov reflection principle.

References

[1] A. D. Alexandrov. Uniqueness theorems for surfaces in the large I. Vestnik Leningrad Univ. Math., 11(19):5–17, 1956.

[2] F. Brito and R. Sá Earp. Geometric configurations of constant mean curvature surfaces with planar boundary. to appear in Anais da Academia Brasileira de Ciências.

[3] G. Darboux. Leçons sur la théorie générale des surfaces et les applications géometriques du calcul infinitésimal. Gauthier-Villars, Paris, 1st part, 2nd edition, 1914.

[4] N. Korevaar, R. Kusner, W. H. Meeks III, and B. Solomon. Constant mean curvature surfaces in hyperbolic space. to appear in American Journal of Math.

[5] N. Korevaar, R. Kusner, and B. Solomon. The structure of complete embedded surfaces with constant mean curvature. Journal of Differential Geometry, 30:465–503, 1989.

[6] R. Kusner. Global geometry of extremal surfaces in three-space. PhD thesis, University of California, Berkeley, 1988.

[7] H. Wente. Counterexample to a conjecture of H. Hopf. Pacific Journal of Math., 121:193–243, 1986.