Compactness of the space of embedded minimal surfaces with free boundary in threemanifolds with nonnegative Ricci curvature and convex boundary
Abstract.
We prove a lower bound for the first Steklov eigenvalue of embedded minimal hypersurfaces with free boundary in a compact dimensional Riemannian manifold which has nonnegative Ricci curvature and strictly convex boundary. When , this implies an apriori curvature estimate for these minimal surfaces in terms of the geometry of the ambient manifold and the topology of the minimal surface. An important consequence of the estimate is a smooth compactness theorem for embedded minimal surfaces with free boundary when the topological type of these minimal surfaces is fixed.
1. Introduction
In a series of recent papers [10], [11] and [12], Fraser and Schoen studied an extremal problem for the first Steklov eigenvalue on compact surfaces with boundary and proved that minimal surfaces in Euclidean balls with free boundary on the boundary of the ball realize the extrema. The equatorial disk [37] and the critical catenoid [10] [11] [12] uniquely maximize among metrics on the disk and annulus respectively. In the recent preprint [10], they were able to prove existence of extremal metrics for genus zero orientable surfaces with any number of boundary components, and the extrema are achieved by properly embedded minimal surfaces in the unit ball in with free boundary. The nonorientable case of a Möbius band was also studied in detail. This motivates the question of finding more examples of properly embedded minimal surfaces in the unit ball. In the case without boundary, Lawson [22] proved that a closed orientable surface of any genus can be realized as an embedded minimal surface in the standard round sphere (while any nonorientable closed surface except can be realized as a minimal immersion into ). A central question in this direction is the following:
Question 1: Which compact orientable surfaces with boundary can be realized as properly embedded minimal surfaces in the unit ball with free boundary?
Since is simply connected, any properly embedded surface in must be orientable. By the results of Fraser and Schoen [10] [11], we know any genus zero orientable surface can be minimally embedded into as a free boundary solution. Therefore, it remains to look for surfaces of higher genus. Another related question one could ask is the following:
Question 2: Given a compact orientable surface with boundary, in how many ways can it be realized as a properly embedded minimal surface in the unit ball with free boundary?
Note that any such minimal surface comes in a continuous family because of the isometry group of the unit ball (which is a compact group). Therefore, we should look at minimal embeddings up to congruences. Using the holomorphicity of Hopf differential, Nitsche [28] proved that the equatorial totally geodesic disk is the only immersed minimal disk in with free boundary (up to congruences). For the annulus, we have the following conjecture:
Conjecture 1.1.
The critical catenoid is the unique properly embedded minimal annulus in with free boundary up to congruences.
One should compare this conjecture with the longstanding conjecture of Lawson [23], which asserts that the Clifford torus is the unique embedded minimal torus in up to congruences. Various partial results were obtained in this direction with additional assumptions ([27] [32] [36]). Very recently, Lawson’s conjecture was proved in full generality by Brendle [4] (see [5] for a nice survey of this conjecture).
In this paper, we prove that the space of properly embedded minimal surfaces in with free boundary is compact in the topology, if we fixed the topological type of the surface. In fact, we prove that the compactness result holds in any compact manifold with nonnegative Ricci curvature and strictly convex boundary . This result is similar to the classical compactness of minimal surfaces in closed manifolds with positive Ricci curvature of Choi and Schoen [7]. Note that proper embeddedness is an essential assumption in our theorem.
Theorem 1.2.
Let be a compact dimensional Riemannian manifold with nonempty boundary . Suppose has nonnegative Ricci curvature and the boundary is strictly convex with respect to the inward unit normal. Then the space of compact properly embedded minimal surfaces of fixed topological type in with free boundary on is compact in the topology for any .
The key ingredient in the proof is a lower bound on the first Steklov eigenvalue for properly embedded minimal surfaces with free boundary in terms of the boundary convexity of . Combining with an equality from [12], this gives an apriori upper bound on the length of the boundary in terms of the topology of the minimal surface and the boundary convexity of . By an isoperimetric inequality of White [38], we get an upper bound on the area of the minimal surface as well. These together give an apriori bound on the norm of the second fundamental form of the minimal surface. By a removable singularity theorem and curvature estimates similar to the ones in [7], we obtain the smooth compactness theorem above.
The outline of this paper is as follows. In section 2, we prove some general facts about minimal hypersurfaces with free boundary in a Riemannian manifold with nonnegative Ricci curvature and convex boundary. The key results are the isoperimetric inequality (Lemma 2.2) and the connectedness principle (Corollary 2.5). When , we prove that any such manifold is diffeomorphic to the unit ball (Theorem 2.11). In section 3, we prove a lower bound for the first Steklov eigenvalue of a properly embedded minimal hypersurface in terms of the boundary convexity of the ambient manifold. Then, we specialize to dimension three and prove a removable singularity theorem and curvature estimates at the free boundary in sections 4 and 5. In section 6, we give a proof of our main compactness theorem (Theorem 1.2).
Acknowledgements. The authors would like to thank Professor Richard Schoen for many useful discussions and his interest in this work. They would also like to express their gratitude to the anonymous referee for all the useful comments.
2. Minimal hypersurfaces with free boundary
Let be a compact dimensional Riemannian manifold with nonempty boundary . Let be the metric on and be the Riemannian connection on . The second fundamental form of , with respect to the inner unit normal , is given by , where are tangent to . The mean curvature of is then defined as the trace of ; i.e., where is any orthonormal basis for the tangent bundle . All manifolds are assumed to be smooth up to the boundary unless otherwise stated.
Let be a compact hypersurface (possibly with boundary) properly immersed in ; that is, . We say that is a minimal hypersurface with free boundary if is a minimal hypersurface (i.e., the mean curvature vanishes) and meets orthogonally along . If is an embedding, we treat as a submanifold of and take to be the inclusion map . Suppose is twosided; that is, there exists a globally defined unit normal vector field on . Any normal vector field on has the form for some and the second variation (see [34] for example) of the volume functional with respect to is
(2.1)  
where is the gradient operator on , is the Ricci curvature of , and is the second fundamental form of with respect to the unit normal . Here, and are the volume forms on and respectively. Note that is tangent to along since meets orthogonally along . We say that is stable if for any smooth function on . Otherwise, is unstable. The following lemma is an immediate consequence of formula (2.1).
Lemma 2.1.
Let be an dimensional compact Riemannian manifold with nonempty boundary . Suppose has nonnegative Ricci curvature and the boundary is strictly convex with respect to the inward unit normal; i.e., there exists a constant such that for any unit vector tangent to .
Then, any twosided, properly immersed, smooth minimal hypersurface with nonempty free boundary must be unstable. Moreover, if is orientable, then the th relative integral homology group vanishes.
Proof.
Taking in (2.1), the curvature assumptions imply that (since ). Therefore, is unstable. To prove the second assertion, suppose . Let be a nontrivial primitive class in . We claim that one can choose a compact embedded minimal hypersurface with free boundary (which may be empty) such that in and minimizes volume among all hypersurfaces homologous to relative to (see, for example, Corollary 9.9 in [9]). This can be seen as follows. By PoincareLefschetz duality and noting , we have
So corresponds to a map , which we can assume to be smooth by the Whitney approximation theorem. Then, for any regular value of , the preimage is a properly embedded compact orientable hypersurface which represents in . Since is stable, the curvature estimates for stable minimal hypersurfaces ( [14] [33] [35]) imply that is smooth up to the boundary (which may be empty), except possibly along a singular set of Hausdorff dimension at most . Assume first the singular set is empty. If , then we have a contradiction with the above statement since is stable and twosided. If , then it follows from Lemma 2.2 below that this is impossible. In case is nonempty, a cutoff argument near the singular set gives the same conclusion since has codimension greater than . ∎
The next lemma is an isoperimetric inequality for minimal hypersurfaces in , which holds under slightly weaker curvature assumptions than those in Lemma 2.1.
Lemma 2.2 (Isoperimetric inequality).
Let be a compact dimensional Riemannian manifold with nonempty boundary . Suppose has nonnegative Ricci curvature and the boundary is strictly mean convex with respect to the inward unit normal.
Then, contains no smooth, closed, embedded minimal hypersurface. Furthermore, if , then there exists a constant , depending only on , such that
(2.2) 
for any smooth immersed minimal hypersurface in .
Proof.
It suffices to show that if contains no smooth, closed, embedded minimal hypersurface, then the isoperimetric inequality (2.2) follows from Theorem 2.1 of [38]. Suppose not, and let be a smooth, closed embedded minimal hypersurface in . Since is strictly mean convex, we have by the strong maximum principle (see [38]). Therefore, and are a positive distance apart; i.e., . Since both and are compact, there exists a minimizing geodesic (parametrized by arc length) that realizes the distance between and . Since is minimizing, lies in the interior of except at . Moreover, is orthogonal to and at the end points. Pick any orthonormal basis for and let be their parallel extensions to normal vector fields along . If we look at the second variation of with respect to the normal variation fields , and sum over , we get
since , is minimal, and is strictly mean convex. Therefore, for some and hence cannot be stable. This contradicts that is a minimizing geodesic from to . This proves the lemma. ∎
Remark 2.3.
We will later apply (2.2) to properly embedded minimal surfaces with free boundary. However, the isoperimetric inequality (2.2) applies in general to any minimal hypersurface without any assumptions on the boundary .
The next lemma shows that under the curvature assumptions in Lemma 2.1, any two properly embedded minimal hypersurfaces must intersect each other.
Lemma 2.4.
Let be an dimensional compact orientable Riemannian manifold with nonempty boundary . Suppose has nonnegative Ricci curvature and the boundary is strictly convex with respect to the inward unit normal. Then, any two properly embedded orientable minimal hypersurfaces and in with free boundaries on must intersect; i.e., .
Proof.
We argue by contradiction. Suppose there are two disjoint properly embedded minimal hypersurfaces , with free boundaries on . Without loss of generality, we assume that both and are connected. Note that and are nonempty by Lemma 2.2. Since by Lemma 2.1, there exists a compact connected domain such that , where is a smooth domain in . On , let and be the distance functions from and respectively. Since has nonnegative Ricci curvature and , are minimal, we have and on in the barrier sense (see Definition 1 in [6]) away from and . Notice that we have used the fact that these minimal hypersurfaces meet orthogonally so that for any point in , is realized by a geodesic from to an interior point on . Hence, the same calculation as in the Laplace comparison theorem for the case without boundary applies here. Therefore, in the barrier sense on . We claim that is constant on .
If attains an interior minimum in , the generalized Hopf maximum principle in [6] implies that is constant. Since is continuous, the global minimum must be achieved by some point . Since is strictly convex, the outward normal derivative of is strictly positive on . Indeed the outward normal derivative of each of and is positive on . Therefore, the minimum lies on or . Assume, without loss of generality, that . Observe that on , and is a point on that is closest to . Since is convex and is connected, there exists a minimizing geodesic that connects to , and such a geodesic is disjoint from and meets and orthogonally at the end points. On the other hand, as on , actually realizes the distance between and in . This implies that is constant on , but then has an interior minimum, which implies that is constant in . However, since the outward normal derivative of is strictly positive on and is nonempty, this is a contradiction. Therefore, and cannot be disjoint. ∎
Corollary 2.5 (Connectedness principle).
Under the curvature assumptions on and in Lemma 2.4, any properly embedded minimal hypersurface in with free boundary is connected.
One can prove Lemma 2.4 using a form of Reilly’s formula [31]. In this paper, we will look at compact manifolds with piecewise smooth boundary. We first observe that Reilly’s formula holds for such manifolds provided that the function is smooth enough away from the singular set of the boundary. Note that there is a sign difference in the formula from that in [8], since we are using the inward unit normal instead of the outward unit normal.
Lemma 2.6 (Reilly’s formula).
Let be a compact manifold with piecewise smooth boundary . Suppose is a continuous function on where , and is the singular set. Assume that there exists some , depending only on , such that for all . Then, Reilly’s formula holds:
(2.3)  
Here, is the Ricci tensor of ; and are the Laplacian, Hessian, and gradient operators on respectively; and are the intrinsic Laplacian and gradient operators on each ; is the inward unit normal of ; and are the mean curvature and second fundamental form of in with respect to the inward unit normal respectively.
Proof.
Using Lemma 2.6, we give an alternative proof of Lemma 2.4.
Alternative Proof of Lemma 2.4.
We will prove Lemma 2.4 by contradiction. Suppose and are connected and disjoint. Let be the connected domain bounded by and modulo as before. Note that is a compact manifold with piecewise smooth boundary , where . Let int denote the interior of . Consider the following mixed DirichletNeumann boundary value problem on :
(2.4) 
where is the outward unit normal on . Since and meet orthogonally along their boundaries, there exists a function such that
Letting , the mixed boundary value problem (2.4) is equivalent to the following mixed boundary value problem with zero boundary data:
(2.5) 
Since , classical results for elliptic equations with homogeneous boundary data ([1] and [25]) imply that a solution to (2.5) exists in the classical sense and the solution , where , and therefore is a solution to (2.4), with uniform estimates away from the singular set . Applying Reilly’s formula in Lemma 2.6 to and , as , we obtain
(2.6) 
The boundary terms for and vanish since and are minimal and is constant on and . Since and , (2.6) implies that and hence is locally constant on , which is impossible since and on but on . Thus, we have a contradiction. ∎
Remark 2.7.

Note that the free boundary condition comes in the proof in a rather subtle way that gives enough regularity to the mixed boundary value problem (2.4) to apply Reilly’s formula in Lemma 2.6. If the free boundary condition is dropped, the theorem is no longer true. For example, there are disjoint flat disks in the unit ball in .

The theorem does not hold if we only assume that and is only weakly convex. For example, let be the dimensional unit ball in and be the product manifold smoothly capped off by two unit half balls at the two ends. Then all the slices , , are mutually disjoint embedded minimal hypersurfaces with free boundary on . However, there are some rigidity results in this case (see [29]).
We now prove a result about the connectedness of for with nonnegative curvature and strictly mean convex boundary.
Proposition 2.8.
Let be an dimensional compact Riemannian manifold with nonempty boundary . Suppose has nonnegative Ricci curvature and the boundary is strictly mean convex with respect to the inward unit normal. Then is connected, and the homomorphism
induced by the inclusion map is surjective.
Proof.
The proof is similar to the one in Lemma 2.2. If is not connected, there exists a minimizing geodesic from one component of to another component that realizes the distance between them. However, the second variation formula and the curvature assumptions on and imply that is unstable, which is a contradiction. Therefore, is connected. (One can also give a different proof using Reilly’s formula (2.3). Suppose is not connected. Let be one of its components. Take to be a harmonic function that is equal to one on and is equal to zero on , which is nonempty. Then, Reilly’s formula implies that is constant, which is a contradiction.) The same argument applies to the universal cover of . Thus is connected and this implies the surjectivity of the homomorphism as in [23]. ∎
Remark 2.9.
As noted in [23], Proposition 2.8 remains true if is assumed to be only piecewise smooth and if the interior angle between two smooth boundary pieces is always less than .
From the alternative proof of Lemma 2.4, we actually proved that if is a connected manifold with nonnegative Ricci curvature and piecewise smooth boundary , where we can decompose with nonempty and strictly convex with respect to the inward unit normal, and minimal, then must be connected. The following corollary is an immediate consequence following the arguments in the proof of Theorem 2 in [23].
Corollary 2.10.
Let be a compact dimensional Riemannian manifold with nonempty boundary . Suppose has nonnegative Ricci curvature and the boundary is strictly convex with respect to the inward unit normal. Let be a properly embedded minimal hypersurface in with free boundary on . If both and are orientable, then divides into two connected components and .
When , we get much stronger topological restrictions on from the curvature and boundary convexity assumptions.
Theorem 2.11.
Let be a compact Riemannian 3manifold with nonempty boundary . Assume has nonnegative Ricci curvature.

If is orientable and is strictly mean convex with respect to the inner unit normal, then is diffeomorphic to a dimensional handlebody.

If is strictly convex with respect to the inner unit normal, then is diffeormorphic to the ball .
Remark 2.12.
Note that we do not need to assume is orientable in case (b); it follows as a consequence. For higher dimensions , we conjecture that in case (b), has finite fundamental group (see Conjecture 1.3 in [24]).
Proof.
First, we assume that is orientable. If is strictly mean convex and nonempty, then it is connected, by Proposition 2.8. Using Theorem 5 in [26], we know that is a handlebody. If is strictly convex, then we also have from Lemma 2.1, which implies that is diffeomorphic to the ball .
Suppose is nonorientable and is strictly convex. Then the orientable double cover is the ball . Therefore, is a double cover of ; thus is homeomorphic to . However, since is the boundary of a compact manifold, by a theorem of Pontrjagin [30], all the StiefelWhitney numbers of vanish. However . This is a contradiction. So must be orientable. ∎
3. Steklov Eigenvalue Estimate
In this section, we prove a lower bound for the first Steklov eigenvalue of a compact properly embedded minimal hypersurface satisfying the free boundary condition in a compact orientable manifold with boundary, where has nonnegative Ricci curvature and is strictly convex. We refer the reader to section 2 of [12] for a brief description of the DirichlettoNeumann map and Steklov eigenvalues.
Theorem 3.1.
Let be an dimensional compact orientable Riemannian manifold with nonempty boundary . Suppose has nonnegative Ricci curvature and the boundary is strictly convex with respect to the inward unit normal. Let be a constant such that for any unit vector tangent to .
Let be a properly embedded minimal hypersurface in with free boundary on . Supose one of the following holds,

is orientable; or

is finite;
then we have the following eigenvalue estimate
where is the first nonzero Steklov eigenvalue of the DirichlettoNeumann map on .
Proof.
We first assume that is orientable. By Corollary 2.5 and Corollary 2.10, is connected and divides into two connected components and . Take . Let where . Thus, . Note that is not necessarily connected, but each component of must intersect along some component of . Otherwise, would have more than one component, which would contradict Proposition 2.8.
Let be a first eigenfunction of the DirichlettoNeumann map on ; i.e., there exists such that
(3.1) 
where is the outward conormal vector of with respect to , and . Recall that . Let be the harmonic extension of to :
Next, we consider the Dirichlet boundary value problem on the compact manifold with piecewise smooth boundary :
(3.2) 
Note that the Dirichlet boundary data is continuous. Standard results on elliptic boundary problems ([1], [2], [3]) imply that a classical solution for (3.2) exists and , for every , together with uniform estimates away from the singular set . Applying Reilly’s formula (2.3), we have
where and are the inward unit normals of and respectively, with respect to . Without loss of generality, we can assume that the integral . Otherwise, we choose instead. Using the fact that and , integrating by parts gives
(3.3) 
where and are the outward conormal vectors of with respect to and respectively. Since meets orthogonally along , we have and along the common boundary . Since , the gradient is continuous on up to the singular set . Therefore,
Putting this back into (3.3) and using the boundary condition in (3.1), we get
Since , another integration by parts on implies that . As is nonconstant on (since is nonconstant on ), we get . This proves the theorem when is orientable.
Now, suppose is not orientable but is finite. Let be the universal cover of . Then satisfies the same curvature assumptions as . Since is finite, is compact and is a finite covering. Let be the lifting of ; i.e., . Since is simply connected and is properly embedded, both and are orientable. By the result above, . But the pullback by of the first Steklov eigenfunction of into is again an eigenfunction of . Therefore, . The proof of Theorem 3.1 is completed. ∎
Since is simply connected, we have the following corollary.
Corollary 3.2.
Let be a compact properly embedded minimal hypersurface in , the Euclidean unit ball, with free boundary on . Then .
It is known ([12]) that for a minimal submanifold properly immersed in the unit ball in with free boundary on the unit sphere, the coordinate functions are Steklov eigenfunctions with eigenvalue . It is natural to ask if this is the first Steklov eigenvalue when the minimal submanifold is properly embedded and has codimension one.
Conjecture 3.3.
Let be a compact properly embedded minimal hypersurface in , the Euclidean unit ball, with free boundary on . Then .
From now on, we will assume that . In [12], Fraser and Schoen proved that if is a compact orientable surface of genus with boundary components of total length , then . Combining this with a bound of Kokarev [20] and Theorem 3.1, we get the following estimate on the boundary length of a minimal surface with free boundary in terms of its topology.
Proposition 3.4.
Let and be the same as in Theorem 3.1. Assume that dim . Then,
Remark 3.5.
By Theorem 2.11, we know that is diffeomorphic to the unit ball , which is simply connected, and hence any properly embedded surface in is automatically orientable.
Corollary 3.6.
Let be a compact properly embedded minimal hypersurface in , the Euclidean unit ball, with free boundary on . Then .
4. Removable Singularity Theorem
In this section, we prove a removable singularity result at the free boundary for properly embedded minimal surfaces with free boundary in a compact Riemannian 3manifold with boundary.
Theorem 4.1.
Let be a Riemannian 3manifold with boundary and let be a point on . Suppose is a (possibly nonorientable) minimal surface with smooth boundary and finite Euler characteristic which is properly embedded in . Assume that meets orthogonally along . If lies in the closure of as a point set, then is a smooth properly embedded minimal surface in .
Proof.
Assume first that is orientable. Let denote the Riemann surface (with boundary) determined by the induced metric on , and let be a conformal harmonic embedding with . Since the Euler characteristic of is finite, is conformally equivalent to a compact Riemann surface (with boundary) with a finite number of disks and points removed. Therefore, there exist (open or closed) arcs or points such that is a compact Riemann surface with boundary. Since lies in the closure of as a point set, this implies that we can extend continuously to . We claim that all are points on the boundary. Note that for each , . Suppose is an arc. Since is continuous up to and is harmonic with a constant value along , by a result in [15], is up to . Since is conformal, along and therefore we can extend past to take the constant value and this extension is still . This results in a weakly harmonic map which is , and therefore a classical harmonic map ([18]). However, since is constant on an open set, must be identically constant on , which is a contradiction. Therefore, each is a point. Moreover, we see that is not an interior point, since is a proper embedding and lies in the closure of . Therefore, extends smoothly ([19]) across to a harmonic map from to . If were a boundary branch point of , then by the asymptotic expansion near a branch point at the free boundary (Lemma 1 of [17]), there would be a line of selfintersection emanating from , which contradicts that is embedded. Therefore, extends as a proper minimal immersion from . Since is properly embedded, the maximum principle for minimal surfaces with free boundary implies that (otherwise, there would be two minimal halfdisks with free boundary that touch at one point at , which would violate the maximum principe). Hence, is a smooth properly embedded compact minimal surface in with smooth free boundary on .
Now suppose is not orientable. Let be the orientable double cover of and let be the Riemann surface determined by . The same argument as above gives a proper minimal immersion from . Choose a sufficiently small such that is a disjoint union of open sets with . Since is properly embedded, by the maximum principle, we have for all . Therefore, is a smooth properly embedded compact minimal surface in with smooth free boundary on . This proves Theorem 4.1.
∎
5. Curvature Estimates
In this section, we extend the wellknown “small total curvature” estimate of Choi and Schoen [7] to the free boundary case.
Theorem 5.1.
Let be a compact Riemannian 3manifold with boundary. Then there exists small enough (depending only on and ) such that the following holds: let , and suppose is a compact properly immersed minimal surface in with free boundary on such that . Then, there exists depending only on the geometry of in such that if
then we have
where is a constant depending only on the geometry of in .
Proof.
Choose such that
Let be chosen so that
Therefore, as , by the choice of and , we have
We rescale the metric on by setting . Then is still a minimal surface with respect to with free boundary on . In the rescaled metric, we have
(5.1) 
where . We claim that we can choose sufficiently small enough so that . In this case,
Suppose that , then . Therefore,
If , then
(5.2) 
By the monotonicity formula for minimal surfaces with free boundary ([14]), when is sufficiently small, there is a constant , depending on the geometry of , such that the area with respect to satisfies . Thus, (5.2) implies that
Together with (5.2), this implies that
(5.3) 
for any . On the other hand, using (5.1), for each , the connected component of containing is a graph over some open set of of a function with uniformly bounded gradient and Hessian. Note that is minimal, hence satisfies a uniformly elliptic equation. If , we can apply the interior Schauder estimate (Corollary 6.3 of [13]) to get a uniform estimate for . If , using the free boundary condition, satisfies a homogeneous boundary condition in the Fermi coordinates, hence the Schauder estimate for uniformly elliptic equations with homogeneous boundary conditions ([1]) again implies a uniform estimate for . Therefore, in any case, we have
for some constant depending only on the geometry of . This contradicts (5.3) above when is sufficiently small (depending on the geometry of ). As a result, when is chosen small enough, then . So we are done.
∎
6. The Smooth Compactness Theorem
We prove our main compactness result in this section.
Theorem 6.1.
Let be a compact dimensional Riemannian manifold with nonempty boundary . Suppose has nonnegative Ricci curvature and the boundary is strictly convex with respect to the inward unit normal. Then the space of compact properly embedded minimal surfaces of fixed topological type in with free boundary on is compact in the topology for any .
Proof.
Note that by Theorem 2.11, is diffeomorphic to the unit ball , hence is simply connected. Let be a compact properly embedded minimal surface with free boundary on . Then is orientable. Suppose has genus with boundary components. From the Gauss equation and the minimality of , for any , we have
where and are the sectional curvatures of the plane with respect to and respectively. We can Integrate the equality above over and apply the GaussBonnet theorem to obtain
where is the geodesic curvature of with respect to and is the Euler characteristic of . Since meets orthogonally along , is equal to , where is the unit tangent vector for . Therefore, there exists a constant depending only on the upper bound of the sectional curvature of and the principal curvatures of so that
Using the isoperimetric inequality (2.2) and the apriori length bound in Proposition 3.4, we obtain
where is a constant depending only on the geometry of the ambient manifold .
Let be a sequence of compact properly embedded minimal surfaces of fixed topological type. Using the same covering argument on [7, P.390391], we can extract a subsequence of , which we still call , and a finite number of points such that converges in the topology to some in for any sufficiently small . Here, is a properly embedded minimal surface (possibly with multiplicity) in with free boundary on . Note that some may lie on . By the removable singularity theorem (Theorem 4.1), is a compact properly embedded minimal surface with free boundary. The only thing left to prove is that has multiplicity as the limit of .
Recall that is orientable. As converges to in for any sufficiently small , there exists a large enough such that is locally a union of graphs over by the curvature estimate in Theorem 5.1. We claim that there is only one sheet. Suppose not, then since is orientable,we can order the sheets , where . We claim that in this case, we would have as , which would contradict the eigenvalue estimate in Theorem 3.1.
To prove that as , we define a Lipschitz function on such that