A criterion for uniqueness of tangent cones at infinity for minimal surfaces
AA CRITERION FOR UNIQUENESS OF TANGENTCONES AT INFINITY FOR MINIMAL SURFACES
PAUL GALLAGHER
Abstract.
We partially resolve a conjecture of Meeks on the as-ymptotic behavior of minimal surfaces in R with quadratic areagrowth. Introduction
Let Σ be an embedded minimal surface in R . By the monotonicityformula, the area density Θ( r ) := A (Σ ∩ B r ) πr is nondecreasing. If lim r →∞ Θ( r ) = Θ( ∞ ) = k < ∞ , we say that Σ has quadratic area growth, or the area growth of k planes. Date : March 20, 2017.
Figure 1. ∼ minimal) a r X i v : . [ m a t h . DG ] M a r INIMAL SURFACE TANGENT CONES AT INFINITY 2
Figure 2. ∼ minimal)For surfaces with the growth of 2 planes, there are two canonicalexamples: the catenoid (Fig 1), and Scherk’s singly periodic surfaces,which occur in a one parameter family (Fig 2 and Fig 3), where theparameter is the angle betwee the two leaves. As the angle goes tozero, the Scherk surfaces approach a catenoid on compact sets after anappropriate rescaling. In 2005, Meeks and Wolf proved the followingtheorem: Theorem 1.1. [MW] Suppose that Σ is an embedded minimal surfacein R which has infinite symmetry group and Θ( ∞ ) <
3. Then Σ iseither a catenoid or a Scherk example.Meeks has conjectured that the symmetry condition in the abovemay be removed:
Conjecture 1.2. [M] Let Σ be an embedded minimal surface in R with area growth of 2 planes. Then Σ is either a catenoid or a Scherkexample. INIMAL SURFACE TANGENT CONES AT INFINITY 3
Figure 3. ∼ minimal)However, an initial difficulty with the above is that it is not yetknown that a minimal surface with quadratic growth even needs tobe asymptotic to a catenoid or a Scherk example. By the compactnesstheory from GMT, it is known that if Σ is an embedded minimal surfacewith quadratic area growth, then for any sequence r i → ∞ , there existsa subsequence ρ i such that Σ /ρ i ∩ B converges to a minimal cone C in the varifold topology. Such a cone C is called a tangent cone atinfinity . A priori, there may be many tangent cones at infinity.This leads to the following conjecture, also due to Meeks: Conjecture 1.3. [M] Let Σ be an embedded minimal surface in R with quadratic area growth. Then Σ has a unique tangent cone atinfinity. INIMAL SURFACE TANGENT CONES AT INFINITY 4
In the case of finite genus, this had already been resolved by Collin[C], who proved that any minimal surface with finite genus and qua-dratic area growth must be asymptotic to a single multiplicity k plane.In particular, when combined with a result of Schoen [S], this resolvesMeeks’ full conjecture in the case of finite genus - that is, the onlyminimal surface with the area growth of two planes and finite genus isthe catenoid.In this paper, we prove that Meeks’ conjecture holds true underadditional assumptions: Theorem 1.4.
Let Σ be an embedded minimal surface with the areagrowth of k planes. Suppose that there exists α < R sufficiently large, there exists a line l R Σ ∩ B R ∩ { d ( x, l R ) > R α } is a union of at least 2 k disks Σ i and such that ∂ Σ i is homotopicallynontrivial in ∂ ( B R ∩ { d ( x, l R ) > R α } ). Then Σ has a unique tangentcone at infinity.This leads to the following: Corollary 1.5.
Let Σ be an embedded minimal surface with quadraticarea growth. Let C α = { x + x ≤ R α } . Then if for some R , Σ \ ( B R ∪ C α ) is a union of 2 k topological disksΣ i each with finitely many boundary components, then Σ has a uniquetangent cone at infinity.Note that the corollary substitutes the homotopy requirement fromthe theorem for the existence of a single line around which we can baseour sublinearly growing set.1.1. Acknowledgments.
The author would like to thank his advisor,William Minicozzi, as well as Jonathan Zhu, Frank Morgan, Ao Sun,and Nick Strehlke for their comments and suggestions throughout thewriting of this paper.2.
Proof of Theorem 1.4
The proof of this begins with the following:
INIMAL SURFACE TANGENT CONES AT INFINITY 5
Lemma 2.1 (Lower Area Bound) . Suppose that Σ satisfies the condi-tions of Theorem 1.4. Then for some C = C (Σ)Area( B R ∩ Σ) > kπR − CR α +1 . Proof.
We will work on each leaf Σ i separately, and the lemma willcome from adding the area of all the leaves together.First note that B R ∩{ d ( x, l R ) > R α } = T R is a rotationally symmetricsolid torus and (since Σ i is a disk), ∂ Σ i is contractible in T R . However,since T R is rotationally symmetric, the smallest spanning disk for anysuch curve has area at least that of a vertical cross section C . Anysuch vertical cross section consists of a half-circle of radius R minus astrip of length 2 R and width CR α . Thus, we have A (Σ i ) ≥ A ( C ) ≥ π R − CR α +1 (cid:3) Remark 2.2.
Note that Lemma 2.1 implies that there are in factexactly 2 k disks in the statement of Theorem 1.4.We make a definition: Definition 2.3.
The error at scale r of a minimal surfaces with areagrowth of k planes is defined as e ( r ) = k − Area(Σ ∩ B r ) πr Thus, the Lemma 2.1 is equivalent to the statement:(1) e ( r ) ≤ Cr α − We now apply an argument of Brian White [W] to prove uniquenessof the tangent cone.
Lemma 2.4.
Let Σ satisfy the following: ∃ R , α < R < r < ∞ ,(2) e ( r ) < Cr − α Then Σ has a unique tangent cone at infinity.
INIMAL SURFACE TANGENT CONES AT INFINITY 6
Proof.
Define F ( z ) = z/ | z | . Then note that A ( F (Σ ∩ ( B r \ B s ))) is equalto the area of the projection of Σ ∩ ( B r \ B s )) onto the unit sphere. Wewill bound this area. We have: A ( F (Σ ∩ ( B r \ B s ))) = (cid:90) Σ ∩ B r \ B s | x N || x | d Σ ≤ (cid:20)(cid:90) Σ ∩ B r \ B s | x N | | x | d Σ (cid:21) / (cid:20)(cid:90) Σ ∩ B r \ B s | x | d Σ (cid:21) / By monotonicity, the term inside the first bracket is smaller than e ( s ).Also, the term in the second bracket can be bounded by distance andarea. Thus, we get that the above is smaller than e ( s ) / ( s − A ( B r ∩ Σ)) / Now, by equation (2), along with the fact that A ( B r ∩ Σ) < kπr , wehave that this is bounded by Cs ( α − / (cid:20)(cid:16) rs (cid:17) ( r − A ( B r ∩ Σ)) (cid:21) / ≤ C rs (1 − α ) / Pick s and r such that s ≤ r ≤ s . Then A ( F (Σ ∩ ( B r \ B s ))) ≤ Cs ( α − / We then sum the above bound to see A ( F (Σ ∩ ( B n r \ B r ))) = n (cid:88) k =1 A ( F (Σ ∩ ( B k r \ B k − r ))) ≤ C n (cid:88) k =1 (2 k r ) ( α − / ≤ Cr (1 − α ) / − (1 − α ) / As r → ∞ , this term goes to zero. Thus, the area of the projection ofΣ \ B r approaches zero as r gets large, which means that the tangentcone must be unique. (cid:3) Proof of Corollary 1.5
Let Σ i be one of the components of Σ \ ( C α ∪ B R ). Then note that theclosure of Σ i in R must be conformally equivalent to D with finitely INIMAL SURFACE TANGENT CONES AT INFINITY 7 many boundary points removed. Take a neighborhood N of one ofthese missing boundary points which does not come close to any othermissing boundary points. Then N ⊂ Σ i has exactly one boundarycomponent. There are two options for the shape of ∂N .(1) The function x | ∂N is unbounded in both directions.(2) x | ∂N is bounded in one direction.Note that x cannot be bounded in both directions, as then ∂N would be compact, which it is not.We temporarily assume that Option 1 occurs. Let γ be the portionof ∂N which is not on the boundary of C α ∪ B R . Take R larger so that γ ⊂ B R . Let R >> R . Then some component of N ∩ B R ∩ C cα willsatisfy the conditions of Theorem 1.4. This implies that it is possibleto prove the Lower Area Bound lemma for this component, and inparticular, the area must be asymptotic to πR / Lemma 3.1.
Under our assumptions, Option 2 is not possible.
Proof.
Suppose that Option 2 occurs. WLOG, let x | ∂N be boundedbelow by 0, and let ( x , x , ∈ ∂N be the point at which that min-imum is achieved. Let ρ = ( x + x ) / . Let C be a catenoid wherethe radius of the center geodesic is strictly larger than 2 ρ . Then by asimple application of the maximum principle, N must intersect C . Inparticular, this implies that inf ∂B R x | N < C + log R .Now, consider a sequence of R i such that Σ ∩ B R i converges to atangent cone at infinity. By compactness, R − i N ∩ ∂B R i must eitherconverge to a union of geodesics on B or must disappear at infinity.However, due to the discussion of the previous paragraph, N cannotdisappear at infinty, and so must converge to a nontrivial union ofgeodesics Γ j , possibly with endpoints at the north or south poles.Let p be a nonsmooth point on ∪ Γ j . Then there must exist a neigh-borhood S of p such that | A | restricted to S ∩ R − i N is unboundedas i → ∞ . However, since N is a minimal disk with quadratic areagrowth bounds, | A | ( x ) must be bounded by C/d ( x ), where d ( x ) is thedistance of x from the boundary of N . INIMAL SURFACE TANGENT CONES AT INFINITY 8
Suppose that p is not equal to the south pole. Then we can chooseour neighborhood S of p to stay away from the x axis, so we will havethat | A | < C uniformly on S ∩ R − i N . Suppose that p is equal to thesouth pole. Then by the assumption of Option 2, ∂N is only containedin the region x ≥
0. So, we can choose S = B / ( p ), and this impliesthe same uniform | A | bound.Therefore, there will be no nonsmooth points of ∪ Γ j , which impliesthat Γ j consists of a single great circle passing through the north pole.In particular, this implies that there are some (cid:15) ( R i ) → R − i N ∩ B is greater than π − (cid:15) ( R i ), where (cid:15) → R i → ∞ . Thus, we have at least 2 k components of Σ \C α , each ofwhich has area growth at least πR / kπR , no component can havegrowth πR . (cid:3) Future Directions
There are several potential extensions of the work above. Theorem1.4 and Corollary 1.5 effectively assume that all tangent cones of Σare unions of planes with a common axis . It is likely not significantlymore difficult to show that the same result holds in the case when theone-dimensional singular set is more complicated, as long as away froma sublinearly growing neighborhood, Σ is a union of disks. That is, wehave the following as another potential step towards the resolution ofMeeks’ Conjecture:
Conjecture 4.1.
Let Σ have the area growth of k planes, and supposethat there exists a uniform α < R > R >>
1, thefollowing is true: There exist line segments L i ( R ), 1 ≤ i ≤ m ( R ) < M such that outside of an α − sublinearly growing neighborhood of ∪ L i ( R ),Σ ∩ B R is a union of disks. Then Σ has a unique tangent cone at infinity. References [C] P. Collin,
Topologie et courbure des surfaces minimales proprement plonges de R , Ann. of Math., (145) Global Problems in Classical Minimal Surface Theory ,Global theory of minimal surfaces, Clay Math. Proc., vol. 2, Amer. Math. Soc.,Providence, RI, 2005, p. 453-469.
INIMAL SURFACE TANGENT CONES AT INFINITY 9 [MW] William H. Meeks, III and Michael Wolf,
Minimal surfaces with the areagrowth of two planes: The case of infinite symmetry , Journal of the AMS, (20)
Uniqueness, Symmetry, and Embeddedness of Minimal Sur-faces , Journal of Differential Geometry, (18) (1983), 791-809.[W] Brian White,
Tangent Cones to Two-Dimensional Area-Minimizing IntegralCurrents are Unique , Duke Mathematics Journal, (50)(50)