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Index bounds and existence results for minimal surfaces and harmonic maps Sargent, Pam 2018

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Index bounds and existence results for minimalsurfaces and harmonic mapsbyPam SargentA thesis submitted in partial fulfilment of the requirements forthe degree ofDoctor of PhilosophyinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)The University of British Columbia(Vancouver)June 2018c© Pam Sargent 2018The following individuals certify that they have read, and recommend to the Facultyof Graduate and Postdoctoral Studies for acceptance, the dissertation entitled:Index bounds and existence results for minimal surfaces and harmonic mapssubmitted by Pamela Sargent in partial fulfillment of the requirements for the degreeof Doctor of Philosophy in Mathematics.Examining Committee:Ailana Fraser, MathematicsSupervisorJingyi Chen, MathematicsSupervisory Committee MemberAlbert Chau, MathematicsSupervisory Committee MemberTai-Peng Tsai, MathematicsUniversity ExaminerIngrid Stairs, AstronomyUniversity ExamineriiAbstractIn this work, we focus on three problems. First, we give a relationship betweenthe eigenvalues of the Hodge Laplacian and the eigenvalues of the Jacobi operatorfor a free boundary minimal hypersurface of a Euclidean convex body. We thenuse this relationship to obtain new index bounds for such minimal hypersurfaces interms of their topology. In particular, we show that the index of a free boundaryminimal surface in a convex domain in R3 tends to infinity as its genus or the numberof boundary components tends to infinity. Second, we consider the relationshipbetween the kth normalized eigenvalue of the Dirichlet-to-Neumann map (the kthSteklov eigenvalue) and the geometry of rotationally symmetric Mo¨bius bands. Morespecifically, we look at the problem of finding a metric that maximizes the kth Stekloveigenvalue among all rotationally symmetric metrics on the Mo¨bius band. We showthat such a metric can always be found and that it is realized by the induced metricon a free boundary minimal Mo¨bius band in B4. Third, we consider the existenceproblem for harmonic maps into CAT(1) spaces. If Σ is a compact Riemann surface,X is a compact locally CAT(1) space and ϕ : Σ → X is a continuous finite energymap, we use the technique of harmonic replacement to prove that either there existsa harmonic map u : Σ → X homotopic to ϕ or there exists a conformal harmonicmap v : S2 → X. To complete the argument, we prove compactness for energyminimizers and a removable singularity theorem for conformal harmonic maps.iiiLay SummaryA minimal surface is a surface that locally minimizes area. Free boundary minimalsurfaces of a ball are a special class of minimal surfaces that meet the boundary ofthe ball orthogonally. A minimal surface may not have the smallest area; the areacould decrease by perturbing the surface in certain directions. First, we relate thesurface’s topology to the number of directions in which perturbations yield decreasesin area.A Mo¨bius band is constructed by twisting one end of a strip of paper 180◦ and glu-ing the ends of the paper together. Second, we construct examples of free boundaryminimal Mo¨bius bands.The energy of a map between two spaces measures the amount a map stretchesthe original space. Third, we show that one can always find a smallest-energy mapbetween a surface and a space with a notion of distance whose curvature cannot betoo large.ivPrefaceThis thesis is based on three previous works, two of which have been published inacademic journals, and the other of which is in preparation.The material presented in Chapter 2 and Appendix A is based on the paper “Indexbounds for free boundary minimal hypersurfaces of convex bodies” [55] appearing inthe journal Proceedings of the American Mathematical Society, Volume 145 (2017),pages 2467-2480. I chose this problem under the guidance of my supervisor, AilanaFraser, and was responsible for all aspects of this work.The material in Chapter 3 is based on a recent project “Free boundary minimalMo¨bius bands in B4”, which is currently in preparation to be submitted to an aca-demic journal. Again, this problem was chosen under the guidance of my supervisor,and I was responsible for all aspects of this work.The material presented in Chapter 4 and Appendix B is based on the paper“Existence of harmonic maps into CAT(1) spaces” [6] which will appear in the journalCommunications in Analysis and Geometry. This was a joint work with ChristineBreiner, Ailana Fraser, Lan-Hsuan Huang, Chikako Mese and Yingying Zhang.vTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ixAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xi1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Index bounds for free boundary minimal hypersurfaces of convex bodies 21.2 Free boundary minimal Mo¨bius bands in B4 . . . . . . . . . . . . . . 41.3 Existence of harmonic maps into singular spaces . . . . . . . . . . . . 61.4 Layout . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 Index Bounds for Free Boundary Minimal Surfaces of ConvexBodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1.1 Free Boundary Minimal Hypersurfaces in Convex Bodies . . . 102.1.2 The Index of a Minimal Hypersurface . . . . . . . . . . . . . . 11vi2.1.3 Examples and Existence Results . . . . . . . . . . . . . . . . . 122.2 Notation and Conventions . . . . . . . . . . . . . . . . . . . . . . . . 152.3 Preliminary Calculations . . . . . . . . . . . . . . . . . . . . . . . . . 172.4 Proofs of Main Theorems . . . . . . . . . . . . . . . . . . . . . . . . . 272.4.1 Eigenvalue Relationship . . . . . . . . . . . . . . . . . . . . . 272.4.2 Index Bound . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 Constructing Free Boundary Minimal Mo¨bius Bands in B4 . . . . 333.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.1.1 The Dirichlet-to-Neumann Map and Steklov Eigenvalue Problem 343.1.2 Extremal Steklov Eigenvalue Problem and Free Boundary Min-imal Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.2 The Steklov eigenvalue problem for rotationally symmetric metrics onthe Mo¨bius band . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.3 Free boundary minimal Mo¨bius bands in B4 . . . . . . . . . . . . . . 484 Existence of Harmonic Maps into CAT(1) Spaces . . . . . . . . . . 514.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.1.1 Harmonic maps . . . . . . . . . . . . . . . . . . . . . . . . . . 534.1.2 CAT(1) Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 554.1.3 Main results and outline . . . . . . . . . . . . . . . . . . . . . 574.2 Preliminary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.3 Monotonicity and removable singularity theorem . . . . . . . . . . . . 644.4 Harmonic Replacement Construction . . . . . . . . . . . . . . . . . . 73Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87A The Dimension of the Space of Harmonic 1-Forms with DirichletBoundary Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94B Quadrilateral Estimates and Energy Convexity . . . . . . . . . . . . 97B.1 Quadrilateral estimates . . . . . . . . . . . . . . . . . . . . . . . . . . 97viiB.2 Energy Convexity, Existence, Uniqueness, and Subharmonicity . . . . 109viiiList of Figures1.1 An illustration of a geodesic triangle in M (left) and a comparisontriangle in the model space Mκ (right). Toponogov’s Theorem impliesthat d(Pt, Rs) ≤ d(P˜t, R˜s). . . . . . . . . . . . . . . . . . . . . . . . . 74.1 An illustration of a triangle in a geodesic space (X, d) (left) and acomparison triangle in S2 (right). If the geodesics connecting the sidesof the triangle in (X, d) are shorter than the corresponding geodesicsfor the comparison triangle in S2, then (X, d) is called a CAT(1) space. 56B.1 An illustration of the quadrilateral PQRS from Lemma B.1.2. . . . 99B.2 An illustration of the triangle ∆PQS, and the points Sη and Pη′ fromLemma B.1.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101B.3 An illustration of the quadrilateral PQRS, and the points Pη, P1−η,Qη′ and Q1−η′ from Lemma B.1.5. . . . . . . . . . . . . . . . . . . . . 104B.4 An illustration of the quadrilateral PQRS with P = u0(x()), Q =u0(x), R = u1(x) and S = u1(x()) used in the proof of Lemma B.2.1. 110ixAcknowledgementsAbove all others, I would like to thank my supervisor Ailana Fraser. This work wouldnot have been possible without her constant support and guidance. I am gratefulfor the direction and advice she has offered me at each stage of my academic career,and consider myself very lucky to have had such an understanding, generous anddedicated supervisor.I would also like to thank Fok-Shuen Leung, who has acted as my teaching mentorwhile I have been at UBC. He has played a huge role in my growth as an educator,and is a constant source of inspiration. I will be forever indebted to him for hisgenerosity.I would like to thank the members of my supervisory committee, Albert Chau,Jingyi Chen and Tai-Peng Tsai, for their advice and their time.I would like to thank my partner, my family and my friends for all of the sup-port and encouragement that they have given me throughout my graduate studentcareer. I would not have made it this far without the unwavering encouragementan emotional support of my partner, Dimitrios Karslidis. His humour, kindness andsupport have helped me persist through the roughest parts of graduate school. Iwould like to extend a special thanks to my officemates, Rob Fraser and Matt Coles,for all of the advice and assistance they have offered me over the last seven years.Finally, I would like to thank my parents, who have no idea what I do, but who havebeen immensely supportive and encouraging regardless.xDedicationTo my friends and family.xiChapter 1IntroductionThis work is devoted to problems related to free boundary minimal surfaces and har-monic maps. Minimal surfaces are critical points of the area functional and satisfycertain nonlinear partial differential equations. Examples of minimal surfaces includesoap films, helicoids (the geometric shape of DNA and double-spiral staircases), aswell as catenoids (minimal surfaces obtained by rotating catenaries about their di-rectrices). While important in geometry, they also have significant applications inother fields of mathematics and, in fact, played a crucial role in the celebrated proofof the Poincare´ conjecture. In addition, minimal surface theory also has physical ap-plications in fluid interface problems and deep connections to fundamental questionsin general relativity.In addition to problems in minimal surface theory, we look at problems concerningthe existence of harmonic maps into singular spaces. A natural notion of energy for amap between geometric spaces is defined by measuring the total stretch of the map ateach point of the domain and then integrating it over the domain. Harmonic maps arecritical points of the energy functional. They can be seen as both a generalizationof harmonic functions in complex analysis and a higher dimensional analogue ofparameterized geodesics in Riemannian geometry. In the absence of a totally geodesicmap, a harmonic map is perhaps the most natural way to map one given geometricspace into another. The theory of harmonic maps has proven to have importantapplications; for example, the existence theory for harmonic two-spheres played a1role in the proof of a generalization of the classical sphere theorem to pointwisequarter-pinching. Other important applications of harmonic maps include those inrigidity problems and in Teichmu¨ller theory amongst others.More specifically, we focus on three problems: finding index bounds for freeboundary minimal hypersurfaces of convex bodies, constructing free boundary min-imal Mo¨bius bands in the 4-dimensional Euclidean ball, and proving the existenceof harmonic maps from a compact Riemann surface into a compact locally CAT(1)space. Here, we outline the problems and state the main results, and outline thelayout of the thesis.1.1 Index bounds for free boundary minimal hy-persurfaces of convex bodiesIf M is a Riemannian manifold with ∂M 6= ∅ and Σ is a n-dimensional submanifoldwith nonempty boundary ∂Σ ⊂ ∂M , then Σ is a free boundary minimal submanifoldif it is a critical point for the volume functional among all n-dimensional submanifoldswhose boundary lie in the boundary of M . It is easy to show that Σ is a free boundaryminimal submanifold of M if and only if it has zero mean curvature and if it meetsthe boundary of M orthogonally. In the case when M = B3, the simplest exampleof a free boundary minimal surface is the equatorial disk.Despite their name, minimal submanifolds do not, in general, minimize volumeand instead are saddle points of the volume functional. Roughly speaking, the indexof a minimal submanifold measures the degree to which it does not minimize volumeand intuitively corresponds to the number of independent directions in which onecan perturb the submanifold and decrease its volume.In [55], we ask whether one can estimate the index of a free boundary minimalhypersurface of a convex body in terms of the hypersurface’s topology and dimension.This question, and the approach taken to answer it, was motivated by the work ofSavo who, in [56], gave a lower bound on the index of a minimal hypersurface ofSn+1 in terms of the hypersurface’s topology and dimension. In particular, if Mn is2a minimal hypersurface of Sn+1 with first Betti number β1(M), Savo showed thatInd(M) ≥ β1(M)(n+22) + n+ 2.To obtain this result, Savo first found a relationship between the eigenvalues of theJacobi operator J and the eigenvalues of the Hodge Laplacian ∆1 on one-forms.Namely, if λj(J) is the jth eigenvalue of the Jacobi operator and λj(∆1) is the jtheigenvalue of the Hodge Laplacian, then Savo showed thatλj(J) ≤ λm(j)(∆1)− 2(n− 1),where m(j) =(n+22)(j − 1) + 1.We prove the analogous result for free boundary minimal hypersurfaces of convexbodies by analyzing the Hodge Laplacian on one-forms. Unlike the case of minimalhypersurfaces of Sn+1, however, here our hypersurfaces have boundary which forcesus to analyze boundary value problems for the Hodge Laplacian. Specifically, weanalyze the relationship between the eigenvalues of the Jacobi operator and theeigenvalues of the Hodge Laplacian on one-forms with absolute boundary conditionsand obtain the following theorem.Theorem 1.1.1. Let Mn be an orientable free boundary minimal hypersurface of aconvex body in Rn+1 with Jacobi operator J . Then, for all positive integers j, onehas thatλj(J) ≤ λm(j)(∆1),where m(j) =(n+12)(j−1)+1 and λm(j)(∆1) is the m(j)th eigenvalue of the Laplacianeigenvalue problem with absolute boundary conditions.We are then able to use this to get a lower bound for the index of free boundaryminimal hypersurfaces of convex bodies in terms of the topology and dimension ofthe surface. In particular, if we let β1a = dimH1a(M) be the first absolute Bettinumber of M , we get the following index estimate.Theorem 1.1.2. (Index Bound) If M is an orientable free boundary minimal hy-3persurface of a convex body in Rn+1, thenInd(M) ≥⌊β1a +(n+12)− 1(n+12) ⌋ .In the special case of a free boundary minimal surface of a convex body in R3with genus g and k boundary components, this reduces toInd(M) ≥⌊2g + k + 13⌋.This result provides new index bounds for free boundary minimal surfaces of B3with large topology. In particular, it shows that Ind(M) ≥ 4 when 2g + k ≥ 11and Ind(M) tends to infinity as the genus or the number of boundary componentstends to infinity. This was obtained simultaneously, but independently and throughdifferent methods, by Ambrozio, Carlotto and Sharp [3].1.2 Free boundary minimal Mo¨bius bands in B4We are also interested finding explicit constructions of free boundary minimal sub-manifolds. Here, we focus our attention on constructing free boundary minimalMo¨bius bands in B4. Our construction is somewhat indirect and is motivated by theworks of Fraser and Schoen [24, 26] and Fan, Tam and Yu [21].In [26], Fraser and Schoen provide a connection between metrics that maximizethe kth Steklov eigenvalue on a surface and the geometry of that surface. Morespecifically, they show that a metric that maximizes the kth Steklov eigenvaluearises geometrically as the induced metric on a free boundary minimal surface ofa Euclidean ball by showing that one can construct a conformal minimal immersioninto a ball from the eigenfunctions corresponding to the Steklov eigenvalue. As aconsequence, if one can find the metrics that maximize the Steklov eigenvalues, thenone has existence of free boundary minimal surfaces of balls.This problem, however, is quite difficult to solve and in general does not yield4explicit solutions. As an alternative, in previous work Fraser and Schoen [24], andlater Fan, Tam and Yu [21], consider the more specialized problem of maximizingSteklov eigenvalues over all rotationally symmetric metrics on the annulus. In thismore specialized setting, the result of Fraser and Schoen [26] no longer guaranteesthat the maximizing metric, if it exists, arises as the metric on a free boundaryminimal surface of a Euclidean ball. However, restricting their attention to thissmaller class of metrics allowed them in [24] to solve the problem explicitly throughthe method of separation of variables. Fraser and Schoen showed that there is ametric that maximizes the first Steklov eigenvalue and that this metric is the inducedmetric on the critical catenoid, a free boundary minimal surface of B3. Fan, Tamand Yu considered the same problem for higher Steklov eigenvalues and showedthat, except in the case of the 2nd Steklov eigenvalue, whose supremum cannot beachieved, there is a metric that maximizes the kth Steklov eigenvalue and that itcorresponds to the induced metric on a free boundary minimal surface of B3 or B4.More specifically, they show that the metrics that maximize the Steklov eigenvaluesare the metrics induced on either the n-critical catenoid or the so-called n-criticalMo¨bius band. This provided explicit constructions of new free boundary minimalsurfaces in B3 and B4. Further, Fan, Tam and Yu conjectured that the supremum ofthe 2nd Steklov eigenvalue can never be achieved for any surface.Motivated by these works, we provide constructions of free boundary minimalMo¨bius bands in B4 by explicitly finding metrics that maximize Steklov eigenvaluesamong all rotationally symmetric metrics on the Mo¨bius band. In particular, weobtain the following theorem.Theorem 1.2.1. For all n ≥ 1, the maximum of the nth Steklov eigenvalue amongall rotationally symmetric metrics on the Mo¨bius band is achieved by the metric ona free boundary minimal Mo¨bius band in B4 given explicitly by the immersionΦ(t, θ) =1Rn(2n sinh(t) cos(θ), 2n sinh(t) sin(θ), cosh(2nt) cos(2nθ), cosh(2nt) sin(2nθ)),where Rn =√4n2 sinh2(Tn,1) + cosh2(2nTn,1) and (t, θ) ∈ [−Tn,1, Tn,1]× S1/ ∼.5In particular, Theorem 1.2.1 shows that the supremum of the 2nd Steklov eigen-value can be achieved and that the conjecture of Fan, Tam and Yu is false.1.3 Existence of harmonic maps into singular spacesAnother topic closely related to minimal surfaces is the theory of harmonic maps fromtwo-dimensional domains. The focus of our third problem is on obtaining existenceresults for harmonic maps into singular spaces. In the smooth setting, the celebratedwork of Sacks and Uhlenbeck [53] developed an existence theory for harmonic mapsfrom surfaces into compact Riemannian manifolds; see also the related works ofLemaire [44], Sacks-Uhlenbeck [54], and Schoen-Yau [58]. In chapter 4, we extendthe Sacks-Uhlenbeck existence theory to the case where the target is a metric spacewith an upper curvature bound.For some applications, it is important to consider harmonic maps when thesmooth Riemannian target is replaced by a singular space. The seminal works ofGromov-Schoen [28] and Korevaar-Schoen [40] consider harmonic maps from a Rie-mannian domain into a non-Riemannian target. In particular, they generalized theclassical notion of the energy of a map in order to define the notion of a harmonicmap. As one can not use variational methods to obtain an Euler-Lagrange equationfor the energy functional in the singular setting, here, a harmonic map is defined tobe a map that is locally energy minimizing. Further exploration of harmonic maptheory in the singular setting includes works of Jost [33], J. Chen [8], Eells-Fuglede[18] and Daskalopoulos-Mese [12].The classical notion of curvature also needs to be generalized in the singular set-ting. In the smooth setting, if M is a Riemannian manifold with sectional curvaturebounded above by κ and Mκ is the model space with constant sectional curvature κ,then Toponogov’s Theorem allows us to compare the lengths of geodesics in geodesictriangles in M and the corresponding geodesic triangles in Mκ (see Figure 1.1). Inthe singular setting, one uses this idea in reverse to define the notion of a metic spacewith curvature bounded above by κ.The above mentioned works all assume non-positivity of curvature (NPC) in the6P QRPtRsMP˜ Q˜R˜P˜tR˜sMκFigure 1.1: An illustration of a geodesic triangle inM (left) and a comparison trianglein the model space Mκ (right). Toponogov’s Theorem implies that d(Pt, Rs) ≤d(P˜t, R˜s).target space, and this curvature condition is heavily used. Without the assumptionof non-positive curvature, the existence problem for harmonic maps becomes morecomplicated and, in general, is not well understood even in the smooth setting.In chapter 4, we investigate the existence theory for harmonic maps in the casewhen the target curvature is bounded above by a constant that is not necessarily 0. Inthis direction, we mention the local existence result of Serbinowski [60] for harmonicmaps from Riemannian manifold domains. Our third problem specifically focuseson obtaining existence results for harmonic maps when the domain is a compactRiemann surface and the target is a compact locally CAT(1) space, that is, a completemetric space with curvature bounded above by 1 in the sense outlined above. Weobtain the following theorem.Theorem 1.3.1. Let Σ be a compact Riemann surface, X a compact locally CAT(1)space and ϕ ∈ C0∩W 1,2(Σ, X). Then either there exists a harmonic map u : Σ→ Xhomotopic to ϕ or a nontrivial conformal harmonic map v : S2 → X.This provides a generalization of the Sacks and Uhlenbeck existence result tothe case of metric space targets. The method used by Sacks and Uhlenbeck is notaccessible in the singular setting as it depends on a priori estimates stemming fromthe Euler-Lagrange equation of their perturbed energy functional. In the singular7setting, one can no longer use variational methods to obtain an Euler-Lagrangeequation. To circumnavigate this, we exploit the local convexity of the target CAT(1)space.1.4 LayoutThe focus of chapter 2 the proof of Theorem 1.1.2. We introduce the problemby providing an overview of free boundary minimal submanifolds, the Morse indexof a minimal submanifold and the Hodge Laplacian. We also provide all of thecalculations needed to prove Theorem 1.1.2.In chapter 3, we focus on proving Theorem 1.2.1. We introduce the Dirichlet-to-Neumann map and the Steklov eigenvalue problem, explicitly calculate the eigenval-ues and eigenfunctions for rotationally symmetric metrics on the Mo¨bius band, andprove a series of lemmas to determine which metric maximizes the k-th eigenvalue.We conclude the chapter by proving Theorem 1.2.1.Chapter 4 is devoted to proving Theorem 1.3.1. We outline both the definitionof energy and harmonicity for maps into metric spaces and CAT(1) spaces. Wethen prove compactness of energy minimizing maps into CAT(1) spaces, and prove aremovable singularity theorem. We then prove Theorem 1.3.1 using a local harmonicreplacement construction.Appendix A is an appendix to chapter 2. Here, we explicitly calculate the firstabsolute Betti number for a surface of genus g with k boundary components.Appendix B is an appendix to chapter 4. Here, we provide all of the details ofthe quadrilateral estimates in CAT(1) spaces, local energy convexity, and the localexistence and uniqueness results needed throughout chapter 4.8Chapter 2Index Bounds for Free BoundaryMinimal Surfaces of ConvexBodies2.1 IntroductionIn this chapter we look at the problem of obtaining lower bounds on the index offree boundary minimal surfaces of convex bodies in terms of their topology. Indexestimates for minimal surfaces are generally difficult to obtain, and there are fewminimal surfaces for which the index is explicitly known. However, index boundscan help in the classification of minimal surfaces, especially when the topology isexplicitly represented in the bounds, and have applications in understanding the re-lationships between the curvature and topology of Riemannian manifolds. Moreover,minimal surfaces whose index is known have proven to be useful in other problems;Urbano’s [68] index characterization of the Clifford torus as being the unique min-imal surface in S3 of index 5 was recently used by Marques and Neves [46] in theircelebrated proof of the longstanding Willmore Conjecture. In [56], Savo was ableto obtain index bounds for minimal hypersurfaces in Sn in terms of their topologymaking use of the Laplacian on 1-forms. His work has inspired the approach taken9here.2.1.1 Free Boundary Minimal Hypersurfaces in Convex Bod-iesA submanifold M of a compact Riemannian manifold M with boundary ∂M ⊂ ∂Mis said to be a free boundary minimal submanifold in M if it is a critical point forthe volume functional among submanifolds with boundary in ∂M . That is, M is afree boundary minimal submanifold of M if for every admissible variation Mt of M ,ddtVol(Mt)∣∣t=0= 0. The first variation formula for a variation Mt of M with variationfield V is given by,ddtVol(Mt)∣∣t=0= −∫M〈V,H〉dV +∫∂M〈V, η〉dA,where η is the outward unit conormal vector field. It follows thatM is a free boundaryminimal submanifold of M if and only if H ≡ 0 and η is orthogonal to T (∂M), i.e.,M meets ∂M orthogonally.Throughout, we will focus our attention on free boundary minimal hypersurfacesMn properly immersed in convex bodies Bn+1. Here, a convex body is a smooth,compact, connected (n + 1)-dimensional submanifold of Rn+1 for which the scalarsecond fundamental form of the boundary satisfies h∂B(U,U) < 0 (with respect tothe outward pointing normal vector) for all vectors U tangent to ∂B.We will also place some attention on the special case when B = B, the Euclideanball, as there are more existence results for free boundary minimal hypersurfacesof Euclidean balls. Free boundary minimal hypersurfaces of Euclidean balls havealso been shown to have an alternative characterization: in [24], Fraser and Schoenshowed that if Σk is a properly immersed submanifold of the Euclidean unit ballBn+1, then Σ is a free boundary minimal submanifold if and only if the coordinatefunctions of the immersion are (Steklov) eigenfunctions of the Dirichlet-to-Neumannmap with (Steklov) eigenvalue 1. Furthermore, free boundary minimal surfaces inBn+1 are extremal surfaces for the Steklov eigenvalue problem.102.1.2 The Index of a Minimal HypersurfaceSuppose that Mn ⊂ Bn+1 is a free boundary minimal hypersurface and that N is asmooth unit normal vector field. Then, for a normal variation with variation fielduN , the second variation formula isd2dt2Vol(Mt)∣∣t=0=∫M(‖∇u‖2 − ‖A‖2u2) dV + ∫∂Mh∂B(N,N)u2dA.Let J denote the Jacobi operator (also called the stability operator),J = ∆− ‖A‖2,and let Q denote the associated symmetric bilinear form,Q(u) =∫M[‖∇u‖2 − ‖A‖2u2] dV + ∫∂Mh∂B(N,N)u2 dA=∫Mu · Ju dV +∫∂M(∂u∂η+ h∂B(N,N)u)u dA.We say that λ(J) is an eigenvalue of J with eigenfunction u ∈ C∞(M) ifJu = λu on M,∂u∂η+ h∂B(N,N)u = 0 on ∂M.The (Morse) index of a minimal hypersurface is the maximal dimension of asubspace of C∞(M) on which the second variation is negative.A free boundary minimal hypersurface is said to be stable if it has index 0. Forfree boundary minimal hypersurfaces in Bn+1, there are none which are stable. Thisis easy to see since if we use the variation with variation field 1 ·N , then we get thatQ(1) = −∫M‖A‖2 dV +∫∂M(0 + h∂B(N,N)) · 1 dA < 0.Hence, any free boundary minimal hypersurface in Bn+1 has index at least 1.11It is well known that the equatorial disk has index 1. Moreover, it is the onlyfree boundary minimal hypersurface on Bn+1 to have index 1. To see this, note that,by the above argument, the index is at least 1. Now, suppose Ind(D) > 1. Thenthere is a two-dimensional subspace S of normal variations containing the variation1 · N on which the second variation of area is negative. Let V ∈ S be a normalvariation orthogonal to 1 · N , i.e.∫M〈V,N〉 dx1dx2 = 0. Then, V has zero averagein the sense that∫MV dx1dx2 = 0. Now, consider the constraint that the surfacemust divide the volume of the ball in half. Subject to this additional constraint, anyequatorial disk is area minimizing. However, a variation with zero average preservesthe constraint. Hence, the second variation of area for this variation must be non-negative, a contradiction.One can also show that, if M is not the equatorial disk, then Ind(M) ≥ 3 (seeTheorem 3.1 in [26])2.1.3 Examples and Existence ResultsFor general convex bodies different from Bn, little is known about the existence of freeboundary minimal submanifolds. In [64], Struwe showed the existence of a (possiblybranched) immersed free boundary minimal disk in any domain in R3 diffeomorphicto B3, and Gru¨ter and Jost [30] showed that there is an embedded free boundaryminimal disk in any convex body in R3. Jost [32] was also able to show that anyconvex body in R3 actually contains at least three embedded free boundary minimaldisks. More recently, Maximo, Nunes and Smith [47] showed that any convex bodyin R3 contains a minimal annulus. By the above argument, we know that any freeboundary minimal hypersurface of a convex body has index at least one. However,little else is known regarding the existence and index of minimal surfaces of greatertopological complexity in convex bodies.If we focus on free boundary minimal submanifolds of Euclidean balls, then moreis known. The simplest examples of free boundary minimal submanifolds in Bn+1are the equatorial k-planes Dk ⊂ Bn+1. By [50] and [27], any simply connected12free boundary minimal surface in Bn must be a flat equatorial disk, and it is wellknown that the equatorial disk has index 1 (see p. 3741 in [23]). In fact, it is theonly free boundary minimal surface of B3 to have index 1. However, there are nowmany examples of free boundary minimal surfaces of different topological type. Thecritical catenoid, a minimal surface with genus 0 and 2 boundary components, is anexplicit example of such a surface. In [26], Fraser and Schoen proved the existenceof free boundary minimal surfaces in B3 with genus 0 and k boundary componentsfor any k > 0. Using gluing techniques, in [22] Folha, Pacard and Zolotareva gavean independent construction of free boundary minimal surfaces in B3 of genus 0with k boundary components for k large. They were also able to use the sametechniques to construct a genus 1 free boundary minimal surface with k boundarycomponents for k large. Examples of free boundary minimal surfaces in B3 withany sufficiently large genus and 3 boundary components have also been constructed.Specifically, Kapouleas and M. Li [36] constructed such surfaces by using gluingtechniques to glue an equatorial disk to a critical catenoid, and Ketover [39] usedvariational methods to construct such surfaces. Furthermore, Kapouleas and Wiygul[37] used gluing techniques to construct free boundary minimal surfaces with oneboundary component and genus g for sufficiently large g. Less is known about theindex of such surfaces. By the above argument, the equatorial disk has index 1 andDevyver [15], Smith and Zhou [63] and Tran [67] have independently shown that thecritical catenoid has index 4. If M is not an equatorial disk, then Tran also showedthat Ind(M) ≥ 4.In this chapter, we give a relationship between the eigenvalues of the Jacobioperator and the eigenvalues of the Laplacian on 1-forms and, as a corollary, obtainnew index bounds for orientable free boundary minimal hypersurfaces of convexbodies. More specifically, our first main result is:Theorem. 1.1.1 Let Mn be an orientable free boundary minimal hypersurface of aconvex body in Rn+1 with Jacobi operator J . Then, for all positive integers j, onehas thatλj(J) ≤ λm(j)(∆1),13where m(j) =(n+12)(j−1)+1 and λm(j)(∆1) is the m(j)th eigenvalue of the Laplacianeigenvalue problem with absolute boundary conditions.Let β1a = dimH1a(M) be the first absolute Betti number of M . Our second mainresult is:Theorem. 1.1.2 (Index Bound) If M is an orientable free boundary minimal hy-persurface of a convex body in Rn+1, thenInd(M) ≥⌊β1a +(n+12)− 1(n+12) ⌋ .Corollary 2.1.1. If M is an orientable free boundary minimal surface of a convexbody in R3 with genus g and k boundary components, thenInd(M) ≥⌊2g + k + 13⌋.Corollary 2.1.1 provides new index bounds for free boundary minimal surfaces ofB3 with large topology. In particular, it shows that Ind(M) ≥ 4 when 2g + k ≥ 11and Ind(M) tends to infinity as the genus or the number of boundary componentstends to infinity. Corollary 2.1.1 was obtained simultaneously, but independently,by Ambrozio, Carlotto and Sharp [3]. In particular, they use different methods toobtain similar Morse index estimates for free boundary minimal hypersurfaces ofstrictly mean convex domains of Euclidean spaces.The remainder of the chapter is structured as follows: In the second section, weoutline the basic notation and conventions that we will use throughout the chapterand give a brief introduction to the Hodge Laplacian on p-forms. Here, we definethe Hodge Laplacian on p-forms and then focus on the special case when p = 1.We also introduce the two main boundary conditions for the eigenvalue problem ofthe Laplacian for 1-forms on a manifold with boundary. In the third section, weprovide several preliminary calculations that will ultimately allow us to see how theJacobi operator acts on specific test functions, which will be needed to prove ourmain results. We give the proofs of our two main results in the fourth section.142.2 Notation and ConventionsLet Mn be an orientable free boundary minimally immersed hypersurface in a convexEuclidean domain Bn+1 (∂M 6= ∅). Throughout, we will let N be a unit normalvector field on M .Let D denote the Levi-Civita connection on Rn+1 and ∇ the Levi-Civita connec-tion on M . We will let A denote the second fundamental form of M ⊂ B, and S theassociated shape operator. That is, for X, Y ∈ Γ(TM),A(X, Y ) = (DXY )N = 〈DXY,N〉 ·NS(X) = − (DXN)T ,so that 〈A(X, Y ), N〉 = 〈S(X), Y 〉In this setting, the Gauss equation tells us that, for any X, Y, Z,W ∈ Γ(TM),0 = RRn+1(X, Y, Z,W ) = 〈A(X,W ), A(Y, Z)〉 − 〈A(X,Z), A(Y,W )〉, (2.1)and the Codazzi equation tells us that, for any X, Y, Z ∈ Γ(TM),(∇XA)(Y, Z)− (∇YA)(X,Z) = (RRn+1(X, Y )Z)N = 0, (2.2)where(∇XA)(Y, Z) = (DXA(Y, Z))N − A(∇XY, Z)− A(Y,∇XZ).For any parallel vector field V in Rn+1, we have the orthogonal decompositionV = V + V N ,where V ∈ TM is the orthogonal projection of V onto M and V N = 〈V ,N〉·N ∈ NM .Since parallel vector fields on Rn+1 and their orthogonal projections onto M will be15used throughout, we introduce the following vector spaces:P = {parallel vector fields on Rn+1},P = {vector fields on M which are orthogonal projections of elements of P}.Throughout, we will let ∆p denote the Hodge Laplacian on p-forms (though thep will usually be dropped for convenience) and we will let ∇∗∇ denote the roughLaplacian on vector fields. So, if ω is a p-form on M and ξ is a vector field on M ,then∆pω = (dδ + δd)ω∇∗∇ξ = −n∑j=1(∇ej∇ejξ −∇∇ej ejξ),where d is the exterior derivative, δ is the codifferential, and {e1, . . . , en} is any localorthonormal frame of TM . Recall that a vector field X is dual to a 1-form θ if andonly if 〈X, Y 〉 = θ(Y ) for all Y ∈ Γ(TM). If ξ is the vector field dual to ω, then onecan also define the Hodge Laplacian of ξ, denoted ∆ξ, to be the vector field dual tothe 1-form ∆1ω. The Bochner formula relates the two Laplacians:∆ξ = ∇∗∇ξ + Ric(ξ),where Ric is seen as a symmetric endomorphism of TM .To get a bound on the index of M , we will consider the following eigenvalueproblem defined by the absolute boundary conditions:J1ω = λω,i∗ιηω = i∗ιηdω = 0,where i is the inclusion ∂M ↪→ M , ιη denotes interior multiplication by η and J1 isthe Jacobi operator on 1-forms defined by J1 = ∆1 − ‖A‖2. We will often drop thesubscripts for convenience. These absolute boundary conditions are a generalization16of Neumann boundary conditions for functions. We say that ω is tangential on ∂Mif i∗ιηω = 0, i.e., ω vanishes whenever its argument is normal to the boundary of M .So, if ω satisfies the absolute boundary conditions, then both ω and dω are tangential(dω is tangential whenever one of its arguments is normal to ∂M).We define the following space of harmonic 1-formsH1N(M) = {ω ∈ Ω1(M) | ∆ω = 0, ω satisfies the absolute boundary conditions},and note that β1a = dimH1a(M) = dimH1N(M), where H1a(M) is the first absolutecohomology group of M .2.3 Preliminary CalculationsThe calculations done here are analogous to those done by Savo in [56] for the caseof a minimal hypersurface in Sn+1. In Sn+1, a hypersurface has two normal vectors(one tangent to the sphere and one normal to both the sphere and the hypersurface)whereas a free boundary minimal hypersurface of a convex body Bn just has one. Theabsence of a second normal vector simplifies many of the preliminary calculations.However, a minimal hypersurface in Sn+1 has no boundary, so the main barrier inmodifying the approach of Savo to this free boundary setting is presence of boundaryterms. To deal with these boundary terms, we extend a result of Ros [52] to arbitrarydimensions.Lemma 2.3.1. Let V ∈ P and let V ∈ P be its orthogonal projection onto M . LetA and S be the second fundamental form and shape operator (respectively) of theimmersion φ : M → Bn. Then(a) ∇〈V ,N〉 = −S(V ).(b) ∆〈V ,N〉 = |S|2〈V ,N〉.17Proof. To show (a), take any X ∈ Γ(TM). Then we have that〈∇〈V , V 〉, X〉 = d (〈V ,N〉) (X)= X(〈V ,N〉)= 〈DXV ,N〉+ 〈V ,DXN〉= 〈V ,DXN〉,since V is parallel. Now, since 〈N,DXN〉 = 12X (‖N‖2) ≡ 0 and [X, V ] is tangent toM , we have that〈V ,DXN〉 = 〈V ,N〉 · 〈N,DXN〉+ 〈V,DXN〉= −〈DXV,N〉= −〈DVX + [X, V ], N〉= −〈DVX,N〉= 〈X, (DVN)T 〉.Hence, ∇〈V ,N〉 = −S(V ).For (b), let {e1, . . . en} denote normal coordinate vector fields centred at a pointp ∈M . Then (at p),−∆〈V ,N〉 =n∑i=1〈∇ei∇〈V ,N〉, ei〉=n∑i=1〈∇ei (DVN)T , ei〉=n∑i=1ei〈DVN, ei〉 − 〈(DVN)T ,∇eiei〉= −n∑i=1ei〈N,DV ei〉18= −n∑i=1ei〈N,A(V, ei)〉= −n∑i=1〈DeiN,A(V, ei)〉+ 〈N,DeiA(V, ei)〉.Since DeiN has no normal component, and A is symmetric, we have that∆〈V ,N〉 =n∑i=1〈N, (DeiA(ei, V ))N〉.Now,(DeiA(ei, V ))N = (∇eiA)(ei, V ) + A(∇eiV, ei) + A(V,∇eiei)= (∇eiA)(ei, V ) + A(∇eiV, ei),and from the Codazzi equation (2.2)(∇eiA)(ei, V ) = (∇VA)(ei, ei)= (DVA(ei, ei))N − A(∇V ei, ei)− A(ei,∇V ei).So,(DeiA(ei, V ))N = (DVA(ei, ei))N − 2A(ei,∇V ei) + A(∇eiV, ei).NowA(ei,∇eiV ) = (Dei∇V ei)N ,and, at p,〈Dei∇V ei, N〉 = −〈∇V ei, DeiN〉=n∑i=1〈V, ej〉〈∇ejei, DeiN〉 = 019Moreover, since M is minimal,n∑i=1(DVA(ei, ei))N =(DV(n∑i=1A(ei, ei)))= 0.Therefore,∆〈V ,N〉 =n∑i=1〈N, (DeiA(ei, V ))N〉=n∑i=1〈N,A(∇eiV, ei)〉=n∑i=1〈N,A(ei,∇eiV )〉=n∑i=1〈N,Dei∇eiV 〉= −n∑i=1〈(DeiN)T ,∇eiV 〉.Lemma 2.3.2. For any vector field ξ ∈ Γ(TM) and any V ∈ P with orthogonalprojection V ,(a) ∆ξ = ∇∗∇ξ − S2(ξ).(b) ∇∗∇V = S2(V ), ∆V = 0.Proof. Let {e1, . . . , en} be local normal coordinate vector fields centred at a point20p ∈M . Then, at p,Ric(ξ) =n∑i,j=1Ric(gijei, ξ)ej=n∑i=1Ric(ei, ξ)ei=n∑i,k=1RM(ek, ei, ξ, ek)eiUsing the minimality of M and the Gauss equation (2.1), we have thatn∑k=1RM(ek, ei, ξ, ek) =n∑k=1〈A(ek, ek), A(ei, ξ)〉 − 〈A(ek, ξ), A(ei, ek)〉= −n∑k=1〈A(ek, ξ), A(ei, ek)〉.Now−n∑k=1〈A(ek, ξ), A(ei, ek)〉 = −n∑k=1〈Dξek, N〉〈Deiek, N〉= −n∑k=1〈ek, DξN〉〈ek, DeiN〉= −〈(DξN)T , DeiN〉= 〈Dei(DξN)T , N〉= 〈D(DξN)T ei + [ei, (DξN)T ], N〉= −〈ei,(D(DξN)TN)T 〉= −〈ei, S2(ξ)〉.Therefore, Ric(ξ) = −S2(ξ), and (a) follows from the Bochner formula.To see that ∇∗∇V = S2(V ), we’ll first show that ∇∗∇N = 0 in the sense that if{e1, . . . , en} are again local normal coordinate vector fields centred at p ∈ M , then,21at p,n∑i=1(Dei(DeiN)T)T= 0.Since, DeiN is tangential,n∑i=1(Dei(DeiN)T)T=n∑i=1(DeiDeiN)T=n∑i,j=1〈DeiDeiN, ej〉ej=n∑i,j=1(ei〈DeiN, ej〉 − 〈DeiN,∇eiej〉)ej= −n∑i,j=1ei〈N,Deiej〉ej= −n∑i,j=1ei〈N,A(ej, ei)〉ej= −n∑i,j=1(〈DeiN,A(ej, ei)〉+ 〈N,Dei(A(ej, ei))〉) ej= −n∑i,j=1〈N,Dei(A(ej, ei))〉ej.Now, using the Codazzi equation (2.2), we have that(Dei(A(ej, ei)))N = (∇eiA) (ej, ei) + A(∇eiej, ei) + A(ej,∇eiei)= (∇eiA) (ej, ei)=(∇ejA) (ei, ei)=(Dej(A(ei, ei)))N − 2A(∇ejei, ei)=(Dej(A(ei, ei)))N.22Therefore,n∑i=1(Dei(DeiN)T)T= −n∑i,j=1〈N,Dei(A(ej, ei))〉ej= −n∑i,j=1〈N,Dej(A(ei, ei))〉ej= −n∑j=1〈N,Dej(n∑i=1A(ei, ei))〉ej = 0,again using the minimality of M .Now, if we write V = V −〈V ,N〉N , then we can use this calculation and the factthat V is parallel to help us calculate ∇∗∇V .∇∗∇V =n∑i=1(Dei(DeiV )T)T − (Dei(Dei(〈V ,N〉N))T )T= −n∑i=1(Dei(ei(〈V ,N〉)N + 〈V ,N〉DeiN)T)T= −n∑i=1(Dei(〈V ,N〉DeiN))T= −n∑i=1(ei(〈V ,N〉)DeiN + 〈V ,N〉DeiDeiN)T= −(n∑i=1ei(〈V ,N〉)DeiN)− 〈V ,N〉∇∗∇N= −n∑i=1〈V,DeiN〉DeiN=n∑i=1〈DeiV,N〉DeiN23=n∑i=1〈DV ei + [ei, V ], N〉DeiN= −n∑i=1〈ei, DVN〉DeiN= DDV NN = S2(V ).The fact that ∆V = 0 now follows from (a).Lemma 2.3.3. Let V ,W ∈ P and let V,W ∈ P be their orthogonal projections ontoM . Then, for any ξ ∈ Γ(TM),(a) ∆〈V, ξ〉 = 2〈S(V ), S(ξ)〉+ 〈V,∆ξ〉 − 2〈V ,N〉〈S,∇ξ〉.(b) 〈∇〈V ,N〉,∇〈W, ξ〉〉 = −〈W,N〉〈S(V ), S(ξ)〉 − 〈W,∇S(V )ξ〉.(c) ∆(〈V ,N〉〈W, ξ〉) = |S|2〈V ,N〉〈W, ξ〉+ 2(〈W,N〉〈S(V ), S(ξ)〉+ 〈W,∇S(V )ξ〉). + 〈V ,N〉(2〈S(W ), S(ξ)〉+ 〈W,∆ξ〉 − 2〈W,N〉〈S,∇ξ〉).Proof. Let {e1, . . . , en} be local normal coordinate vector fields centred at a pointp ∈M . Then, at p,∆〈V, ξ〉 = −n∑i=1eiei〈V, ξ〉= −n∑i=1ei (〈∇eiV, ξ〉+ 〈V,∇eiξ〉)= −n∑i=1〈∇ei∇eiV, ξ〉+ 2〈∇eiV,∇eiξ〉+ 〈V,∇ei∇eiξ〉= 〈∇∗∇V, ξ〉 − 2〈∇V,∇ξ〉+ 〈V,∇∗∇ξ〉,where 〈∇V,∇ξ〉 = ∑ni=1〈∇eiV,∇eiξ〉. From Lemma 2.3.2 we have that ∇∗∇V =24S2(V ) and ∇∗∇ξ = ∆ξ + S2(ξ). We also have that〈S2(V ), ξ〉 = 〈DDV NN, ξ〉= −〈N,DDV Nξ〉= −〈N,DξDVN + [DVN, ξ]〉= 〈DξN,DVN〉 = 〈S(ξ), S(V )〉,and, similarly, 〈V, S2(ξ)〉 = 〈S(V ), S(ξ)〉. Therefore,∆〈V, ξ〉 = 2〈S(V ), S(ξ)〉+ 〈V,∆ξ〉 − 2〈∇V,∇ξ〉.Finally,〈∇eiV,∇eiξ〉 = 〈DeiV,∇eiξ〉= 〈Dei(V − 〈V ,N〉N),∇eiξ〉= −〈V ,N〉〈DeiN,∇eiξ〉= 〈V ,N〉〈S(ei),∇eiξ〉.Hence, summing over i gives us that∆〈V, ξ〉 = 2〈S(V ), S(ξ)〉+ 〈V,∆ξ〉 − 2〈V ,N〉〈S,∇ξ〉.From Lemma 2.3.1(a) we know that ∇〈V ,N〉 = −S(V ), so we just need tocalculate ∇〈W, ξ〉. First, notice that for any vector field X on M , since W is parallel,∇XW =(DX(W − 〈W,N〉N))T= − (X(〈W,N〉)N + 〈W,N〉DXN)T = 〈W,N〉S(X)Hence,〈∇〈W, ξ〉, X〉 = X(〈W, ξ〉) = 〈∇XW, ξ〉+ 〈W,∇Xξ〉= 〈W,N〉〈S(X), ξ〉+ 〈W,∇Xξ〉.25So, for X = −S(V ) (= ∇〈V ,N〉), we have that〈∇〈V ,N〉,∇〈W, ξ〉〉 = −〈W,N〉〈S2(V ), ξ〉 − 〈W,∇S(V )ξ〉= −〈W,N〉〈S(V ), S(ξ)〉 − 〈W,∇S(V )ξ〉.Now (c) follows from (a) and (b) and Lemma 2.3.1(b).Let U = {V ∈ P | ‖V ‖ ≡ 1}. Then U can naturally be identified with Sn if weendow it with the measure µ = n+1Vol(Sn)dvSn .Lemma 2.3.4. For any X,Y ∈ Rn+1,∫U〈V ,X〉〈V , Y 〉 dV = 〈X,Y 〉.The proof of Lemma 2.3.4 follows from a direct, but tedious, calculation afterchanging to spherical coordinates and repeatedly applying the integral identity∫sinm x dx = − 1msinm−1 x cosx+m− 1m∫sinm−2 x dx.The following lemma was originally proved by Ros [52] for free boundary minimalsurfaces in a smooth domain in R3. Here, we extend his proof to obtain the analogousresult for free boundary minimal hypersurfaces in smooth domains in Rn.Lemma 2.3.5. Suppose ξ is a vector field on M dual to a 1-form ω which satisfiesthe absolute boundary conditions. Then, at a point p ∈ ∂M ,〈∇ηξ, ξ〉 = h∂B(N,N)‖ξ‖2.Proof. Let η be the (outward pointing) conormal vector along ∂M . Then, since ωsatisfies the absolute boundary conditions on ∂M , at p we have thatω(η) = 0,dω(η, t) = η(ω(t))− t(ω(η))− ω([η, t]) = 0,26for any vector t ∈ Tp(∂M). In particular, if ξ is the vector field dual to ω, then thefirst condition implies that ξp ∈ Tp(∂M), and so the second condition implies thatdω(η, ξ) = 0 at p. Now,〈ξ,∇ηξ〉 = η〈ξ, ξ〉 − 〈ξ,∇ηξ〉 = (∇ηω)(ξ),and we claim that (∇ηω)(ξ) = (∇ξω)(η). To see this, note that, by definition,(∇ξω)(η)− (∇ηω)(ξ) = ξ(ω(η))− ω(∇ξη)− η(ω(ξ)) + ω(∇ηξ).However,ω(∇ξη)− ω(∇ηξ) = ω(∇ξη −∇ηξ) = ω([ξ, η]),and, since dω(η, ξ) = 0, ω([η, ξ]) = η(ω(ξ))− ξ(ω(η)). Therefore(∇ξω)(η)− (∇ηω)(ξ) = ξ(ω(η))− η(ω(ξ)) + ω([η, ξ]) = 0.So,〈ξ,∇ηξ〉 = (∇ηω)(ξ) = (∇ξω)(η).Now, since ξ is tangent to ∂M and ω(η) = 0 on ∂M ,(∇ξω)(η) = ξ(ω(η))− ω(∇ξη) = 〈∇ξξ, η〉 = h∂B(ξ, ξ).Hence,〈∇ηξ, ξ〉 = 〈ξ,∇ηξ〉 = h∂B(ξ, ξ) = h∂B(N,N)‖ξ‖2.2.4 Proofs of Main Theorems2.4.1 Eigenvalue RelationshipTheorem. 1.1.1 Let Mn be an orientable free boundary minimal hypersurface of a27convex body in Rn+1 with Jacobi operator J . Then, for all positive integers j, onehas thatλj(J) ≤ λm(j)(∆1),where m(j) =(n+12)(j−1)+1 and λm(j)(∆1) is the m(j)th eigenvalue of the Laplacianeigenvalue problem with absolute boundary conditions.Lemma 2.4.1. For V ,W ∈ P, letXV,W = 〈V ,N〉W − 〈W,N〉V.Let ξ be any vector field on M and consider the function u = 〈XV,W , ξ〉. ThenJu = 〈XV,W ,∆ξ〉+ 2v,where v is the smooth functionv = 〈∇S(V )ξ,W 〉 − 〈∇S(W )ξ, V 〉.Proof of Lemma 2.4.1. Since u = 〈XV,W , ξ〉 = 〈V ,N〉〈W, ξ〉 − 〈W,N〉〈V, ξ〉, frompart (c) of Lemma 2.3.3, (after some cancellations) we get that∆u = |S|2u+ 〈XV,W ,∆ξ〉+ 2v,and so Ju = 〈XV,W ,∆ξ〉+ 2v.Proof of Theorem 1.1.1. Let {φ1, φ2, . . . , } be an orthonormal basis for L2(M) givenby eigenfunctions of J , where φi is an eigenfunction associated to λi(J). Let Vm(∆1) =⊕mi=1ENλi(∆1), where ENλi(∆1) is the space of eigenforms of ∆1 associated with λ1(∆1)with absolute boundary conditions. We want to find ω ∈ V m(∆1), ω 6≡ 0, for which∫M〈XV,W , ξ〉φidV = 0, (2.3)28for i = 1, . . . , j − 1 and for all V ,W ∈ P , where ξ is the vector field dual toω. Since XV,W is a skew-symmetric bilinear function of V ,W , and since dimP =dimRn+1 = n + 1, there are(n+12)equations that need to be satisfied in (2.3) foreach i, and therefore(n+12)(j − 1) homogeneous linear equations in total. So, ifm(j) =(n+12)(j−1)+1, then we’re guaranteed that there is a ω ∈ V m(j)(∆1), ω 6≡ 0,whose dual vector field satisfies (2.3) for all V,W and for i = 1, . . . j − 1. From themin-max principle and Lemma 2.4.1 we have that,λj(J)∫Mu2 dV ≤∫MuJu dV +∫∂M(∂u∂η+ h∂B(N,N)u)u dA=∫Mu〈XV,W ,∆ξ〉 dV + 2∫Muv dV +∫∂M(∂u∂η+ h∂B(N,N)u)u dA.(2.4)In addition,∂u∂η= η(〈V ,N〉〈W, ξ〉 − 〈W,N〉〈V , ξ〉)= 〈V ,DηN〉〈W, ξ〉+ 〈V ,N〉(〈DηW, ξ〉+ 〈W,Dηξ〉)− 〈W,DηN〉〈V, ξ〉+ 〈W,N〉(〈DηV , ξ〉+ 〈V ,Dηξ〉) .We’ll now use an integration technique that exploits Lemma 2.3.4 to help us sim-plify (2.4). We’ll then apply Lemma 2.3.5 to get the claimed eigenvalue relationship.Using the product metric on U × U , Lemma 2.3.4 implies that (pointwise)∫U×Uu2 dV dW = 2‖ξ‖2,∫U×Uu〈XV,W ,∆ξ〉 dV dW = 2〈ξ,∆ξ〉,∫U×Uuv dV dW = 0,29∫U×Uu〈V ,DηN〉〈W, ξ〉 dV dW = 0,∫U×Uu〈V ,N〉〈W,Dηξ〉 dV dW = 〈ξ,Dηξ〉 = 12η(‖ξ‖2).Therefore, integrating (2.4) over U × U yields2λj(J)∫M‖ξ‖2 dV ≤ 2∫M〈ξ,∆ξ〉 dV +∫∂M(η(‖ξ‖2) + 2h∂B(N,N)‖ξ‖2) dA.From Lemma 2.3.5 we know that η(‖ξ‖2) = 2h∂B(N,N)‖ξ‖2 on ∂M , since ξ is thedual vector field of a 1-form satisfying the absolute boundary conditions. Moreover,since ξ is the dual vector field to a linear combination of eigenforms of ∆1, it nowfollows that2λj(J)∫M‖ξ‖2 dV ≤ 2λm(j)(∆1)∫M‖ξ‖2 dV + 4∫∂Mh∂B(N,N)‖ξ‖2 dA.Since h∂B(U,U) < 0 for any vector tangent to ∂B, we get that2λj(J)∫M‖ξ‖2 dV ≤ 2λm(j)(∆1)∫M‖ξ‖2 dV.Now, since ω 6≡ 0, we can divide both sides by the L2(M)-norm of ξ to getλj(J) ≤ λm(j)(∆1).Remark 2.4.2. We note that when m(j) ≤ dimH1N(M), i.e. when ω is a linearcombination of harmonic forms and therefore a harmonic form itself, we actually getthe strict inequality λj(J) < λm(j)(∆1) = 0. This follows from the fact that ω 6≡ 0implies that ω|∂M 6≡ 0 (see Theorem 3.4.4 on p.131 of [59]), and so we get the strictinequality 4∫∂Mh∂M(N,N)‖ξ‖2 dA < 0.302.4.2 Index BoundTheorem. 1.1.2 (Index Bound) If M is an orientable free boundary minimal hy-persurface of a convex body in Rn+1, thenInd(M) ≥⌊β1a +(n+12)− 1(n+12) ⌋ .Proof. Suppose j is such that m(j) ≤ dimH1N(M) := β1a. Then λj(J) < λm(j)(∆) =0, so Ind(M) ≥ j. Now, m(j) = (n+12)(j − 1) + 1 ≤ β1a, so j ≤⌊β1a+(n+12 )−1(n+12 )⌋. Hence,Ind(M) ≥⌊β1a+(n+12 )−1(n+12 )⌋.Corollary. 2.1.1 If M is an orientable free boundary minimal surface in a convexbody in R3 with genus g and k boundary components, thenInd(M) ≥⌊2g + k + 13⌋.Proof. Since β1a = 2g + k − 1 for a surface (see Appendix A), this follows directlyfrom Theorem 1.1.2.Remark 2.4.3. We note that Corollary 2.1.1 can also be obtained by using the workof Ros. In [52], Ros shows that if ω is a harmonic 1-form and ξ is its dual vectorfield, then∆ξ + ‖A‖2ξ = 2〈∇ω,A〉N,and, if ω satisfies the absolute boundary conditions, then〈∇ηξ, ξ〉 = h∂B(N,N)‖ξ‖2.So, for ξ = (ξ1, ξ2, ξ3), if we use the notation Q(ξ, ξ) =∑3i=1Q(ξi, ξi) and assume31ω 6≡ 0,Q(ξ, ξ) = −∫M〈∆ξ + ‖A‖2ξ, ξ〉dV +∫∂M(〈∇ηξ, ξ〉+ h∂B(N,N)‖ξ‖2)dA= 2∫∂Mh∂B(N,N)‖ξ‖2dA < 0.Hence Q(X,X) < 0, and we get that dimH1N(M)− 3 · Ind(M) = (2g + k − 1)− 3 ·Ind(M) ≤ 0, or Ind(M) ≥ d (2g+k−1)3e = b2g+k+13c.32Chapter 3Constructing Free BoundaryMinimal Mo¨bius Bands in B43.1 IntroductionIn this chapter we look at the problem of constructing free boundary minimal Mo¨biusbands in B4 by solving an extremal eigenvalue problem. Though there are some ex-istence results (see 2.1.3 for an outline of currently known examples) for free bound-ary minimal surfaces in B3, explicit constructions are less common. As mentionedin 2.1.3, extremal eigenvalue techniques, gluing techniques and min-max techniqueshave been used to successfully construct free boundary minimal surfaces with specifictopology in B3. Here we take the approach inspired by the work of Fraser and Schoen[24, 26], and Fan, Tam and Yu [21] in which we use eigenfunctions that maximizethe Steklov eigenvalues for rotationally symmetric metrics to construct immersedfree boundary minimal Mo¨bius bands in B4.333.1.1 The Dirichlet-to-Neumann Map and Steklov Eigen-value ProblemIn [25], Fraser and Schoen showed that there is a connection between Steklov eigen-value problems on surfaces with boundary and free boundary minimal surfaces inthe unit ball. In particular, they showed that metrics that maximize the kth Stekloveigenvalue on surfaces with boundary arise from the metrics on free boundary min-imal surfaces in a Euclidean ball. If (Σ, g) is a compact Riemannian manifold withboundary, the Steklov eigenvalue problem is:∆gu = 0 on Σ∂u∂η= σu on ∂Σ,where η is the outward unit normal vector to ∂Σ, σ ∈ R, and u ∈ C∞(Σ). Stekloveigenvalues are eigenvalues of the Dirichlet-to-Neumann map, which sends a givensmooth function on the boundary to the normal derivative of its harmonic extensionto the interior. That is, if u ∈ C∞(∂Σ) and if u¯ ∈ C∞(Σ) is its harmonic extension,then the Dirichlet-to-Neumann map is the map L : C∞(∂Σ)→ C∞(Σ) defined byL(u) =∂u¯∂η.The Dirichlet-to-Neumann map is a non-negative, self-adjoint operator with discretespectrumσ0 = 0 < σ1 ≤ σ2 ≤ . . . ≤ σk ≤ . . .→∞.The first nonzero Steklov eigenvalue of L can be characterized variationally asσ1 = inf∫∂Σ u=0∫Σ|∇u|2dvΣ∫∂Σu2dv∂Σ34and in general,σk = inf{∫Σ|∇u|2dvΣ∫∂Σu2dv∂Σ:∫∂Σuφj = 0 for j = 0, 1, 2, . . . , k − 1.},where φj is an eigenfunction corresponding to the eigenvalue σj, for j = 1, 2, . . . , k−1.3.1.2 Extremal Steklov Eigenvalue Problem and Free Bound-ary Minimal SurfacesA classical result by Weinstock [69] shows that, on a simply-connected planar domain,the maximum of the first normalized Steklov eigenvalue is achieved by the rounddisk in the Euclidean plane. In [26] Fraser and Schoen proved the existence of ametric that maximizes the first normalized eigenvalue on any surface of genus zero,and showed that it is realized by the induced metric on a free boundary minimalsurface in B3. In the case of the annulus, the surface of genus zero with 2 boundarycomponents, they characterized the maximizing metric as the induced metric onthe critical catenoid. They also proved the existence of a maximizing metric on theMo¨bius band, and characterized it as the induced metric on the critical Mo¨bius band,the surface obtained by suitably scaling the embeddingφ(t, θ) = (2 sinh(t) cos(θ), 2 sinh(t) sin(θ), cosh(2t) cos(2θ), cosh(2t) sin(2θ))to lie in the unit ball, where (t, θ) ∈ [−T0, T0] × S1 and T0 is the unique positivesolution of coth(t) = 2 tanh(2t).In the special case of rotationally symmetric metrics on the annulus and Mo¨biusband, in [24, 26], Fraser and Schoen explicitly calculated the eigenvalues and eigen-functions of the Dirichlet-to-Neumann map and showed that the critical catenoid andcritical Mo¨bius band maximize the first normalized eigenvalue among all rotationallysymmetric metrics. Motivated by the work of Fraser and Schoen, in [21], Fan, Tamand Yu considered the problem of maximizing the kth normalized Steklov eigenvalueover all rotationally symmetric metrics on a cylinder. They showed that, except for35the 2nd normalized Steklov eigenvalue, the maximum is achieved by either the n-critical catenoid or the so-called n-Mo¨bius band. However, they also showed that thesupremum of the 2nd normalized Steklov eigenvalue can not be achieved. Girouardand Polterovich proved that for simply-connected planar domains, the supremum ofthe second normalized Steklov eigenvalue is 4pi and can not be achieved. This ledFan, Tam and Yu to conjecture that the supremum of the second normalized Stekloveigenvalue can never be achieved.In this chapter, we consider the problem of maximizing the kth normalized Stekloveigenvalue on the Mo¨bius band over all rotationally symmetric metrics. In particular,we show that this problem is solvable for all k, i.e. for each k, among all rotationallysymmetric metrics on the Mo¨bius band, there is a metric that maximizes the kthnormalized Steklov eigenvalue and it is achieved by a free boundary minimal Mo¨biusband in B4. Specifically, our main result is:Theorem. 1.2.1 For all n ≥ 1, the maximum of the nth Steklov eigenvalue amongall rotationally symmetric metrics on the Mo¨bius band is achieved by the metric ona free boundary minimal Mo¨bius band in B4 given explicitly by the immersionΦ(t, θ) =1Rn(n sinh(t) cos(θ), n sinh(t) sin(θ), cosh(nt) cos(nθ), cosh(nt) sin(nθ)),where Rn =√n2 sinh2(Tn,1) + cosh2(nTn,1) and (t, θ) ∈ [−Tn,1, Tn,1]× S1/ ∼.In particular, this provides a counterexample to Fan, Tam and Yu’s conjecture.In [38], Karpukhin et al. showed that, for k > 1, the supremum of the kth normal-ized eigenvalue of the Laplacian on a sphere cannot be achieved. This, together withthe result of Girouard and Polterovich, could suggest that, in general, the supremumof higher normalized Steklov eigenvalues might not be achievable. The results ofchapter 3, which show that the supremum of the kth normalized Steklov eigenvalueamong rotationally symmetric metrics on the Mo¨bius band is achievable, are inter-esting in that they could suggest that, for the Mo¨bius band, the supremum of thekth normalized Steklov eigenvalue among all metrics might actually be achievable.Based on the case when k = 1, one might expect that when maximizing metrics36exist, the maximizing metrics are rotationally symmetric.The remainder of the chapter is structured as follows: In the second section, weintroduce the Steklov eigenvalue problem for rotationally symmetric metrics on theMo¨bius band. Here, we prove a series of lemmas needed to find the rotationallysymmetric metric on the Mo¨bius band that maximizes the kth normalized Stekloveigenvalue. In the third section we prove that there is a metric that maximizes thekth normalized Steklov eigenvalue, and we use the corresponding eigenfunctions toconstruct a free boundary minimal surface and prove the main theorem.3.2 The Steklov eigenvalue problem for rotation-ally symmetric metrics on the Mo¨bius bandLet Σ be a Mo¨bius band, i.e. Σ = [−T, T ]× S1/ ∼, where (t, θ) ∼ (t′, θ′) if t′ = −tand θ′ = θ + pi. From [26] we know that the critical Mo¨bius band maximizes thefirst normalized Steklov eigenvalue over all smooth metrics on the Mo¨bius band.In general, from [25] we know that a metric on Σ that maximizes the kth Stekloveigenvalue among all smooth metrics on Σ arises as the induced metric on a freeboundary minimal Mo¨bius band in B4. However, solving this optimization problemis, in general, quite difficult. Here we investigate the simpler problem of finding arotationally symmetric metric on Σ that maximizes the kth Steklov eigenvalue amongall rotationally symmetric metrics on Σ. That is, we consider metrics of the formg = f(t)2(dt2 + dθ2),where f : [−T, T ]→ R is a smooth function satisfying f(t) = f(−t). Let η = 1f(T )∂∂tbe the outward unit conormal on ∂Σ. Our goal is to maximize the kth nonzeronormalized eigenvalues σ˜k(T ) = σk(T )Lg(∂Σ) = 2pif(T )σk(T ).If u(t, θ) is Steklov eigenfunction on Σ, u(t, θ) is a harmonic map (with respectto the flat metric) satisfying the boundary conditions u(t, θ) = u(−t, θ + pi) andis an eigenfunction of the Dirichlet-to-Neumann map. We may use the method of37separation of variables to get u(t, θ) = α(t)β(θ), with α(t) = α(−t) and β(θ) =β(θ + pi) andα′′(t)α(t)= −β′′(θ)β(θ)= k2.If k = 0, then α(t) = A + Bt and β(t) = C + Dθ. Since α(t) = α(−t) andβ(θ) = β(θ+pi), we must have that 2Bt = 0 and C = C+Dpi, so B = D = 0. Thus,α(t) = A and β(θ) = C, so u is constant. However, since u is an eigenfunction of theDirichlet-to-Neumann map, on ∂Σddηu = σu⇒ 1f(T )α′(T ) = σα(T )β(θ),so σ = 0.If k 6= 0, then it is easy to show thatα(t) = Ak sinh(kt) +Bk cosh(kt) and β(θ) = Ck sin(kθ) +Dk cos(kθ).So, since α(t)β(θ) = α(−t)β(θ + pi),(Ak sinh(kt) +Bk cosh(kt)) (Ck sin(kθ) +Dk cos(kθ))= (−Ak sinh(kt) +Bk cosh(kt)) (Ck sin(kθ) cos(kpi) +Dk cos(kθ) cos(kpi))= (−Ak sinh(kt) +Bk cosh(kt))(Ck sin(kθ)(−1)k +Dk cos(kθ)(−1)k)So,Ak sinh(kt)(1 + (−1)k) (Ck sin(kθ) +Dk cos(kθ))+Bk cosh(kt)(1− (−1)k) (Ck sin(kθ) +Dk cos(kθ)) = 0.If k is even, then we have that2Ak sinh(kt) (Ck sin(kθ) +Dk cos(kθ)) = 0,38so either Ak = 0 or Ck = Dk = 0. However, if Ck = Dk = 0, then β(θ) ≡ 0. Thismeans that u ≡ 0, which is not possible since u is an eigenfunction. Thus, Ak = 0and the eigenfunctions areu(t, θ) = cosh(kt) (Ck sin(kθ) +Dk cos(kθ)) .for some constants Ck, Dk. Now, since u is an eigenfunction of the Dirichlet-to-Neumann map, on ∂Σ, ∂∂ηu = σku, and so1f(T )k sinh(kT ) (Ck sin(kθ) +Dk cos(kθ)) = σk cosh(kt) (Ck sin(kθ) +Dk cos(kθ)) .Hence,σk(T ) =kf(T )tanh(kT ).If k is odd, then we have that2Bk cosh(kt) (Ck sin(kθ) +Dk cos(kθ)) = 0,so, similarly to the previous case, we conclude that Bk = 0. Thus, the eigenfunctionsareu(t, θ) = sinh(kt) (Ck sin(kθ) +Dk cos(kθ)) .for some constants Ck, Dk. Now, again, since u is an eigenfunction of the Dirichlet-to-Neumann map, on ∂Σ we get that1f(T )k cosh(kT ) = σk sinh(kT ),soσk(T ) =kf(T )coth(kT ).39Thus, the nonzero eigenvalues of the Dirichlet-to-Neumann map areλk(T ) =2kf(T )tanh(2kT ), and µk(T ) =(2k − 1)f(T )coth((2k − 1)T ),k = 1, 2, . . ., and the normalized eigenvalues areλ˜k(T ) = 4pik tanh(2kT ), and µ˜k(T ) = 2pi(2k − 1) coth((2k − 1)T ),k = 1, 2, . . ..Lemma 3.2.1. Let k, l ≥ 1. Then(i) λ˜k < λ˜k+1, µ˜l < µ˜l+1. Furthermore, λ˜n < µ˜n+1 for n ≥ 1, and each λ˜k and µ˜lhas multiplicity 2.(ii) λ˜k(T ) is monotone increasing in T and µ˜l(T ) is monotone decreasing in T .(iii) λ˜k(∞) := limT→∞λ˜k(T ) = 4pik and µ˜l(∞) := limT→∞µ˜l(T ) = 2pi(2l − 1).Proof. First, (i) and (iii) are clear by direct calculation. Now, (ii) follows from thefact thatdλ˜kdT= 8pik2sech2(2kT ) > 0 anddµ˜ldT= −2pi(2l − 1)2csch2((2l − 1)T ) < 0.Lemma 3.2.2. There exists T > 0 such that λ˜k(T ) = µ˜l(T ) if and only if l ≤ k.Moreover, T is unique if it exists.Proof. Let Fk,l(T ) = λ˜k(T )− µ˜l(T ) = 2pi (2k tanh(2kT )− (2l − 1) coth ((2l − 1)T )).Then Fk,l(T ) is continuous on (0,∞) andlimT→0Fk,l(T ) = −∞ and limT→∞Fk,l(T ) = 2pi(2k − (2l − 1)).Thus limT→∞Fk,l(T ) > 0 if and only if l ≤ k. Furthermore, Fk,l(T ) is monotone increas-ing on (0,∞) since λ˜k(T ) is monotone increasing and µ˜l(T ) is monotone decreasing.Hence there exists a unique T > 0 for which λ˜k(T ) = µ˜l(T ) if and only if l ≤ k.40Definition 3.2.3. For l ≤ k let Tk,l be the unique positive number such thatλ˜k(Tk,l) = µ˜l(Tk,l).Lemma 3.2.4. For l ≤ k, Tk,l is decreasing in k and increasing in l.Proof. Since λ˜k(T ) < λ˜k+1(T ), we have thatµ˜l(Tk,l) = λ˜k(Tk,l) < λ˜k+1(Tk,l).Hence, Fk+1,l(Tk,l) > 0, where Fk,l is as in the proof of Lemma 3.2.2, and, again,limT→0Fk+1,l(T ) = −∞. Hence Tk+1,l < Tk,l. Similarly, if l + 1 ≤ k,λ˜k(Tk,l) = µ˜l(Tk,l) < µ˜l+1(Tk,l),and so Fk,l+1(Tk,l) < 0. Since limt→∞Fk,l+1(T ) > 0, it follows that Tk,l < Tk,l+1.For fixed k > 0, let s = bk2c. By Lemma 3.2.4, if k ≥ 2, we see that we candecompose (0,∞) as(0,∞) = (0, Tk−1,1) ∪(s⋃j=2[Tk−j+1,j−1, Tk−j,j))∪ [Tk−s,s,∞).Note that if k = 1, then s = 0 and we do not decompose (0,∞).Lemma 3.2.5. For k ≥ 1,σ˜2k−1(T ) = σ˜2k(T ) ≤λ˜k(Tk,1) if T ∈ (0, Tk−1,1)λ˜k−j+1(Tk−j+1,j) if T ∈ [Tk−j+1,j−1, Tk−j,j), 2 ≤ j ≤ sλ˜k/2(∞) if T ∈ [Tk−s,s,∞), s = k2 , k evenλ˜(k+1)/2(T(k+1)/2,(k+1)/2) if T ∈ [Tk−s,s,∞), s = k−12 , k odd.Proof. First suppose T ∈ (0, Tk−1,1). Then, since λ˜k−1(T ) is increasing in T and41µ˜1(T ) is decreasing in T by Lemma 3.2.1, we have thatλ˜k−1(T ) < λ˜k−1(Tk−1,1) = µ˜1(Tk−1,1) < µ˜1(T ).Since each λ˜n(T ) and each µ˜n(T ) have multiplicity two, either σ˜2k−1(T ) = σ˜2k(T ) =λ˜k(T ) or σ2k−1(T ) = σ˜2k(T ) = µ˜1(T ). Now, since Tk,l is decreasing in k by Lemma3.2.4, 0 < Tk,1 < Tk−1,1. So, if T ≤ Tk,1,λ˜k(T ) ≤ λ˜k(Tk,1) = µ˜1(Tk,1) ≤ µ˜1(T )and so σ˜2k−1(T ) = σ˜2k(T ) = λ˜k(T ) ≤ λ˜k(Tk,1). Otherwise, Tk,1 < T < Tk−1,1, soλ˜k(T ) > λ˜k(Tk,1) = µ˜1(Tk,1) > µ˜1(T ),and σ˜2k−1(T ) = σ˜2k(T ) = µ˜1(T ) < µ˜1(Tk,1) = λ˜k(Tk,1). Hence, in either case,σ˜2k−1(T ) = σ˜2k(T ) ≤ λ˜k(Tk,1).Now, if 2 ≤ j ≤ s and T ∈ [Tk−j+1,j−1, Tk−j,j), thenλ˜k−j+1(T ) > λ˜k−j+1(Tk−j+1,j−1) = µ˜j−1(Tk−j+1,j−1) > µ˜j−1(T )andλ˜k−j(T ) < λ˜k−j(Tk−j,j) = µ˜j(Tk−j,j) < µ˜j(T ).So, either σ˜2k−1(T ) = σ˜2k(T ) = λ˜k−j+1(T ) or σ˜2k−1(T ) = σ˜2k(T ) = µ˜j(T ). Again,since Tk,l is decreasing in k and increasing in l by Lemma 3.2.4, Tk−j+1,j−1 <Tk−j+1,j < Tk−j,j. So, if Tk−j+1,j−1 ≤ T ≤ Tk−j+1,j, thenλ˜k−j+1(T ) ≤ λ˜k−j+1(Tk−j+1,j) = µ˜j(Tk−j+1,j) ≤ µ˜j(T )and so σ˜2k−1(T ) = σ˜2k(T ) = λ˜k−j+1(T ) ≤ λ˜k−j+1(Tk−j+1,j). Otherwise Tk−j+1,j <T < Tk−j,j, soλ˜k−j+1(T ) > λ˜k−j+1(Tk−j+1,j) = µ˜j(Tk−j+1,j) > µ˜j(T )42and σ˜2k−1(T ) = σ˜2k(T ) = µ˜j(T ) ≤ λ˜k−j+1(Tk−j+1,j). Hence, in either case, σ˜2k−1(T ) =σ˜2k(T ) ≤ λ˜k−j+1(Tk−j+1,j).If T ∈ [Tk−s,s,∞), s = k2 , k even (Tk−s,s = Tk/2,k/2), thenλ˜k/2(T ) ≥ λ˜k/2(Tk/2,k/2) = µ˜k/2(Tk/2,k/2) ≥ µ˜k/2(T ).Furthermore, Tk/2,k/2+1 is undefined by Lemma 3.2.2, soλ˜k/2(T ) < µ˜k/2+1(T ) ∀ T > 0.Hence, σ˜2k−1(T ) = σ˜2k(T ) = λ˜k/2(T ) < λ˜k/2(∞).Finally, if T ∈ [Tk−s,s,∞), s = k−12 , k odd (Tk−s,s = T(k+1)/2,(k−1)/2), thenλ˜(k+1)/2(T ) ≥ λ˜(k+1)/2(T(k+1)/2,(k−1)/2) = µ˜(k−1)/2(T(k+1)/2,(k−1)/2) ≥ µ˜(k−1)/2(T ).Furthermore, T(k−1)/2,(k+1)/2 is undefined by Lemma 3.2.2, soλ˜(k−1)/2(T ) < µ˜(k+1)/2(T ) ∀T > 0.So, either σ˜2k−1(T ) = σ˜2k(T ) = λ˜(k+1)/2(T ) or σ˜2k−1(T ) = σ˜2k(T ) = µ˜(k+1)/2(T ).Now, T(k+1)/2,(k+1)/2 is defined by Lemma 3.2.2 and T(k+1)/2,(k−1)/2 < T(k+1)/2,(k+1)/2by Lemma 3.2.4. If T(k+1)/2,(k−1)/2 ≤ T ≤ T(k+1)/2,(k+1)/2, thenλ˜k+1/2(T ) ≤ λ˜(k+1)/2(T(k+1)/2,(k+1)/2) = µ˜(k+1)/2(T(k+1)/2,(k+1)/2) ≤ µ˜(k+1)/2(T )and so σ˜2k−1(T ) = σ˜2k(T ) = λ˜(k+1)/2(T ) ≤ λ˜(k+1)/2(T(k+1)/2,(k+1)/2). Otherwise T >T(k+1)/2,(k+1)/2 andλ˜k+1/2(T ) > λ˜(k+1)/2(T(k+1)/2,(k+1)/2) = µ˜(k+1)/2(T(k+1)/2,(k+1)/2) > µ˜(k+1)/2(T ).So, σ˜2k−1(T ) = σ˜2k(T ) = µ˜(k+1)/2(T ) < µ˜(k+1)/2(T(k+1)/2,(k+1)/2) andµ˜(k+1)/2(T(k+1)/2,(k+1)/2) = λ˜(k+1)/2(T(k+1)/2,(k+1)/2). Thus, in either case, σ˜2k−1(T ) =σ˜2k(T ) ≤ µ˜(k+1)/2(T(k+1)/2,(k+1)/2).43Lemma 3.2.6. Letf(t) = sinh(t) cosh(t)− t and g(t) = sinh(t) cosh(t)− tt2.Then f(t) > 0 and g′(t) > 0 for all t > 0.Proof. We have thatf ′(t) = cosh2(t) + sinh2(t)− 1= 2 sinh2(t)which is positive for t > 0. Since f(0) = 0, it follows that f(t) > 0 for t > 0.Nowg′(t) =f ′(t)t2 − 2tf(t)t4=2t(sinh2(t) + 1)− 2 sinh(t) cosh(t)t3=2t cosh2(t)− 2 sinh(t) cosh(t)t3.Since cosh2(t) = 14(e2t + 2 + e−2t) and cosh(t) sinh(t) = 14(e2t − e−2t), we get thatg′(t) =(2t− 2)e2t + (2t+ 2)e−2t + 4t4t3=14t3[((2t− 2)∞∑k=0(2t)kk!)+((2t+ 2)∞∑k=0(−1)k (2t)kk!)+ 4t]=14t3[( ∞∑k=0(2t)k+1k!)− 2( ∞∑k=0(2t)kk!)+( ∞∑k=0(−1)k(2t)k+1k!)+ 2( ∞∑k=0(−1)k(2t)k(k!))+ 4t]=14t3[2( ∞∑k=0(2t)2k+1(2k)!)− 4( ∞∑k=0(2t)2k+1(2k + 1)!)+ 4t]44=14t3[ ∞∑k=1(2(2k)!− 4(2k + 1)!)(2t)2k+1]=14∞∑k=12(2k − 1)(2k + 1)!(2t)2k−2=14∞∑k=02(2k + 1)(2k + 3)!(2t)2kSince all of the coefficients are positive, it follows that g′(t) > 0 for t > 0.Lemma 3.2.7. Let x(a, b) be the unique positive solution ofa tanh(ax) = b coth(bx)for a ≥ b > 0. Letu(a, b) = a tanh(ax(a, b)).Then u(a, b) < u(a+ c, b− c) for a ≥ b > c > 0.Proof. Differentiating the first equation with respect to a yieldstanh(ax) + asech2(ax)(x+ a∂x∂a)= −b2csch2(bx) · ∂x∂aand so∂x∂a=− tanh(ax)− axsech2(ax)a2sech2(ax) + b2csch2(bx)< 0.Similarly,∂x∂b=coth(bx)− bxcsch2(bx)a2sech2(ax) + b2csch2(bx)=sinh(bx) cosh(bx)− bxsinh2(bx)(a2sech2(ax) + b2csch2(bx))> 0,where we have used Lemma 3.2.6 to conclude its sign.Now, since u(a, b) = b coth(bx(a, b)) and ∂x∂a< 0,∂u∂a= −b2csch2(bx) · ∂x∂a> 0.45Similarly,∂u∂b= a2sech2(ax) · ∂x∂b> 0.Hence, (∂u∂a)(∂u∂b) = b2(sinh(ax) cosh(ax) + ax)a2(sinh(bx) cosh(bx)− bx) >b2(sinh(ax) cosh(ax)− ax)a2(sinh(bx) cosh(bx)− bx) ≥ 1by Lemma 3.2.6 since a ≥ b. Note that the inequality is strict when a > b. Thus,for f(t) = u(a+ t, b− t),f ′(t) =∂u∂a− ∂u∂b,and so f ′(t) > 0 for t > 0. Hence u(a, b) < u(a+ c, b− c) for a ≥ b > c > 0.Corollary 3.2.1. For k ≥ l > c > 0 we have thatλ˜k(Tk,l) < λ˜k+c(Tk+c,l−c).Proof. By Lemma 3.2.7, for k ≥ l > c > 0 we have that u(2k, 2l − 1) < u(2k +2c, 2l − 1− 2c). Hence λ˜k(Tk,l) < λ˜k+c(Tk+c,l−c).In particular, this tells us thatλ˜k−j+1(Tk−j+1,j) < λ˜k(Tk,1),for 2 ≤ j < s and, when k is odd,λ˜(k+1)/2(T(k+1)/2,(k+1)/2) < λ˜k(Tk,1).So, when k is odd, by Lemma 3.2.5 we have that σ˜2k−1(T ) = σ˜2k(T ) ≤ λ˜k(Tk,1), andwhen k is even, σ˜2k−1(T ) = σ˜2k(T ) ≤ max(λ˜k/2(∞), λ˜k(Tk,1)).Lemma 3.2.8. For k ≥ 2 even,λ˜k/2(∞) < λ˜k(Tk,1)46Proof. First note that λ˜k/2(∞) = 2pik and λ˜k(Tk,1) = 4pik tanh(2kTk,1), so, if we letk = 2n with n ≥ 1, then we need to show that 8pin tanh(4nT2n,1) > 4pin. Nowtanh(4nT ) =e4nT − e−4nTe4nT + e−4nT=12⇔ T = log(3)8n.Since 2pi coth(T2n,1) = 8pin tanh(4nT2n,1), if coth(log(3)8n)> 2n = 4n tanh(4n log(3)8n),then it would follow that 2pi coth(T2n,1) = 8pin tanh(4nT2n,1) > 4pin. By directcalculation, we have thatcoth(log(3)8n)=31/8n + 131/8n31/8n − 131/8n=31/4n + 131/4n − 1 ,so coth(log(3)8n)> 2n is equivalent to31/4n + 131/4n − 1 > 2n⇔31/4n + 12n(31/4n − 1) > 1.Letf(n) =31/4n + 12n(31/4n − 1) .Then, thinking of n as a positive real number,f ′(n) =(− log(3)4n2· 31/4n)· (2n · (31/4n − 1))− (31/4n + 1) · (2 (31/4n − 1)+ 2n(− log(3)4n2· 31/4n))4n2 (31/4n − 1)2=− log(3)2n· 31/2n + log(3)2n· 31/4n − 2 · 31/2n + 2 + log(3)2n· 31/2n + log(3)2n· 31/4n4n2 (31/4n − 1)2=log(3)n· 31/4n + 2 · (1− 31/2n)4n2 (31/4n − 1)2 .Since the denominator is always positive, we will focus on the numerator. Using47Taylor series we have thatlog(3)n· 31/4n + 2 · (1− 31/2n) = log(3)n( ∞∑k=01k!(log(3)4n)k)− 2( ∞∑k=11k!(log(3)2n)k)=( ∞∑k=04k!(log(3)4n)k+1)−( ∞∑k=12k+1k!(log(3)4n)k)=( ∞∑k=14(k − 1)!(log(3)4n)k)−( ∞∑k=12k+1k!(log(3)4n)k)=∞∑k=1(4(k − 1)! −2k+1k!)·(log(3)4n)k.So, since 4(k−1)! =2k+1k!for k = 1, 2, and 4(k−1)! <2k+1k!for all k ≥ 3, it follows thatf ′(n) < 0 for all n > 0 and so f(n) is monotone decreasing.Now,limn→∞2n · (31/4n − 1) = limn→∞31/4n − 112n= limn→∞− log(3)4n2· 31/4n− 12n2=log(3)2,solimn→∞f(n) =4log(3)> 1.Thus, f(n) is bounded below by 1 and socoth(log(3)8n)=31/4n + 131/4n − 1 > 2n.Therefore, 8pin tanh(4nT2n,1) = 2pi coth(T2n,1) > 4pin, and so λ˜k/2(∞) < λ˜k(Tk,1).3.3 Free boundary minimal Mo¨bius bands in B4Here, using the results from the previous section, we first show that we can alwaysfind a rotationally symmetric metric that maximizes the kth Steklov eigenvalue. Wethen use the eigenfunctions corresponding to these maximal Steklov eigenvalues to48get constructions of free boundary minimal Mo¨bius bands in B4.Theorem 3.3.1. Let k ≥ 1 and Mk = supT>0(σ˜k(T )). Then M2k−1 = M2k = λ˜k(Tk,1),and is attained precisely when T = Tk,1.Proof. This follows directly from Lemma 3.2.1 and Lemma 3.2.8.Consider the immersed surface in R4 given byx(t, θ) = 2n sinh(t) cos(θ)y(t, θ) = 2n sinh(t) sin(θ)z(t, θ) = cosh(2nt) cos(2nθ)w(t, θ) = cosh(2nt) sin(2nθ)for t ∈ [−Tn,1, Tn,1]. Now, since the coordinate functions are Steklov eigenfunctions,they are harmonic extensions of their restriction to ∂Σ. Furthermore, if we letΦ(t, θ) = (x(t, θ), y(t, θ), z(t, θ), w(t, θ)), then∂Φ∂t= (n cosh(t) cos(θ), n cosh(t) sin(θ), n sinh(nt) cos(nθ), n sinh(nt) sin(nθ)) ,∂Φ∂θ= (−n sinh(t) sin(θ), n sinh(t) cos(θ),−n cosh(nt) sin(nθ), n cosh(nt) cos(nθ)) .It follows that∂Φ∂t· ∂Φ∂θ= 0 and∣∣∣∣∂Φ∂t∣∣∣∣ = ∣∣∣∣∂Φ∂θ∣∣∣∣ = n2 (sinh2(t) + cosh2(nt)). Hence,Φ is also conformal and so we see that the immersion defined by the coordinatefunctions is a minimal immersion. Moreover, since |Φ| is constant on ∂Σ, it followsfrom the maximum principle that Φ defines a surface contained in a ball centredat the origin of radius√4n2 sinh2(Tn,1) + cosh2(2nTn,1). To obtain a free boundaryminimal Mo¨bius band in B4, we scale the portion of this immersed surface insidethe ball centred at the origin of radius√4n2 sinh2(Tn,1) + cosh2(2nTn,1) to lie in B4.This yields the following:Theorem. 1.2.1 For all n ≥ 1, the maximum of the nth Steklov eigenvalue amongall rotationally symmetric metrics on the Mo¨bius band is achieved by the metric on49a free boundary minimal Mo¨bius band in B4 given explicitly by the immersionΦ(t, θ) =1Rn(2n sinh(t) cos(θ), 2n sinh(t) sin(θ), cosh(2nt) cos(2nθ), cosh(2nt) sin(2nθ)),where Rn =√4n2 sinh2(Tn,1) + cosh2(2nTn,1) and (t, θ) ∈ [−Tn,1, Tn,1]× S1/ ∼.50Chapter 4Existence of Harmonic Maps intoCAT(1) Spaces4.1 IntroductionIn this chapter we prove an existence result for harmonic maps from compact Rie-mann surfaces into complete metric spaces with an upper curvature bound. Thetheory of harmonic maps has proven to have important applications; for example,the existence theory for harmonic two-spheres of Sacks and Uhlenbeck [53] was ex-tended by Micallef and Moore [49] and used to prove a generalization of the classicalsphere theorem to pointwise quarter-pinching. Other important applications of har-monic maps include those in rigidity problems (for example, [62], [10], [28]) and inTeichmu¨ller theory (for example, [70], [16], [14]) amongst others.For some of the above mentioned applications, it has been necessary to considerharmonic maps when the smooth Riemannian target is replaced by a singular space.The seminal works of Gromov-Schoen [28] and Korevaar-Schoen [40] consider har-monic maps from a Riemannian domain into a non-Riemannian target. Furtherexploration of harmonic map theory in the singular setting includes works of Jost[33], J. Chen [8], Eells-Fuglede [18] and Daskalopoulos-Mese [12]. However, all ofthe above mentioned works assume non-positivity of curvature (NPC) in the target51space.When the curvature of the target space is allowed to be positive, the existenceproblem for harmonic maps becomes more complicated, and in many ways, moreinteresting. Although the general problem is not well understood, a breakthroughwas achieved in the case of two-dimensional domains by Sacks and Uhlenbeck [53].Indeed, they discovered a “bubbling phenomena” for harmonic maps; more specifi-cally, they prove the following dichotomy: given a finite energy map from a Riemannsurface into a compact Riemannian manifold, either there exists a harmonic maphomotopic to the given map or there exists a branched minimal immersion of the2-sphere. We also mention the related works of Lemaire [44], Sacks-Uhlenbeck [54],and Schoen-Yau [58].The goal of this chapter is to provide a generalization of the Sacks and Uhlenbeckexistence result to the case of metric space targets. We specifically look at the settingin which the target is a CAT(1) space, i.e. a complete metric space with curvaturebounded above by 1 in the sense of Alexandrov. The method used by Sacks andUhlenbeck is not accessible in the singular setting as it depends on a priori estimatesstemming from the Euler-Lagrange equation of their perturbed energy functionaland, in the singular setting, one can no longer use variational methods to obtainan Euler-Lagrange equation. Here, we develop an alternative method that insteadexploits the local convexity of the target CAT(1) space.Our original motivation for considering the existence problem in the singularsetting was to develop an approach to the non-smooth uniformization problem offinding a conformal (or more generally, a quasisymmetric) parameterization of ametric space homeomorphic to the 2-sphere, via harmonic map methods. We expectto able to use an application of our theorem to solve the non-smooth uniformizationproblem in the special case when the metric space in question has an additionalproperty that it is locally CAT(1).Before stating our result precisely, we first describe the setting of our problem inmore detail by outlining harmonic maps with singular targets and CAT(1) spaces.524.1.1 Harmonic mapsLet (Mm, g) and (Nn, h) be two Riemannian manifolds, and let u : M → N . Thenthere is a natural notion of the energy of u, E(u), which roughly measures the amountthe map u stretches M . More precisely, if {ei}mi=1 is a local orthonormal frame forTxM , then the energy density at x ∈M is|dux|2 = Trg(u∗h) =m∑i=1|dux(ei)|2= gαβ(x)hij(u(x))∂ui∂xα∂uj∂xβ,and the energy of u is defined to beE(u) =∫M|du|2dµg.A harmonic map is then a critical point of the energy functional.When the target space is no longer a smooth manifold but simply a completemetric space, Gromov and Schoen [28] and Korevaar and Schoen [40] developed aSobolev space theory for maps into metric spaces and harmonic maps theory intocomplete metric spaces with non-positive curvature in the sense of Alexandrov. If(Ω, g) is a Riemannian domain and (X, d) is a complete metric space, then a mapu : Ω → X is in L2(Ω, X) if u is a Borel measurable function with separable rangeand for some P ∈ X, ∫Ωd2(u(x), P )dµg <∞.To define the Sobolev space W 1,2(Ω, X) ⊂ L2(Ω, X), we need to define the energyof a map u : Ω → X when X is a complete metric space. We first define the-approximate energy density eu : Ω→ R byeu (x) =∫S(x,)d2(u(x), u(y))2· dσx,(y)n−1,where σx, is the induced measure on the -sphere S(x, ) centred at x. The -53approximating energy Eu : Cc(Ω)→ R is thendEu (φ) =∫Ωφeu dµg.We will often suppress the superscript d when the context is clear. An L2 mapu : Ω→ X is said to have finite energy ifE(u) = supφ∈Cc(Ω), 0≤φ≤1lim sup→0Eu (φ) <∞,and the Sobolev space W 1,2(Ω, X) is defined to the be the subset of L2(Ω, X) con-sisting of finite energy maps. In the case that u has finite energy, there is an energydensity function |∇u|2(x) such thateu (x)dµg ⇀ |∇u|2(x)dµg.For u ∈ W 1,2(Ω, X) and a smooth vector field V ∈ Γ(Ω), there is a directional energydensity function |u∗(V )|2(x) ∈ L1(Ω) such that|u∗(V )|2(x) = lim→0d2(u(x), u(expx(V )))2for a.e. x ∈ Ω,the energy density is given by|∇u|2(x) = 1ωn−1∫Sn−1⊂TxΩ|u∗(V )|2(x)dσ,and the energy of u isE(u) =∫Ω|∇u|2dµ.Given two finite energy maps u and v, the distance d(u, v) : Ω → R+ betweenthem belongs to the Sobolev space W 1,2(Ω). Therefore one can make the followingdefinition: u = v on ∂Ω if d(u, v) ∈ W 1,20 (Ω). A finite energy map u : Ω → X is54energy minimizing ifE(u) = inf{E(v) : v ∈ W 1,2(Ω, X), v = u on ∂Ω}.Given h ∈ W 1,2(Ω, X), we defineW 1,2h (Ω, X) = {f ∈ W 1,2(Ω, X) : h = f on ∂Ω}.Definition 4.1.1. We say that a map u : Ω→ X is harmonic if it is locally energyminimizing with locally finite energy; precisely, for every p ∈ Ω, there exist r > 0,ρ > 0 and P ∈ X such that h = u∣∣Br(p)has finite energy and minimizes energyamong all maps in W 1,2h (Br(p),Bρ(P )), where Br(p) is the geodesic ball in Ω ofradius r centred at p and Bρ(P ) is the geodesic ball in X of radius ρ centred at P .We refer the reader to [40] for further details and background.4.1.2 CAT(1) SpacesRoughly speaking, a CAT(1) space is a complete metric space with curvature boundedabove by 1 in the sense of triangle comparison.A complete metric space (X, d), is a geodesic space if for each P,Q ∈ X, thereexists a curve γPQ such that the length of γPQ is exactly d(P,Q). We call γPQ ageodesic between P and Q.Remark 4.1.2. For ease of notation, we will often denote d(P,Q) by dPQ.We determine a weak notion of an upper sectional curvature bound on X by usingcomparison triangles. Given any three points P,Q,R ∈ X such that dPQ + dQR +dRS < 2pi, the geodesic triangle 4PQR is the triangle in X with sides given by thegeodesics γPQ, γQR, γRS.Let 4P˜ Q˜R˜ denote a geodesic triangle on the standard sphere S2 such thatdPQ = dP˜ Q˜, dQR = dQ˜R˜ and dRP = dR˜P˜ . We call 4P˜ Q˜R˜ a comparison trianglefor the geodesic triangle 4PQR. Note that a comparison triangle is convex sincethe perimeter of the geodesic triangle is less than 2pi.55Given a geodesic space (X, d) and a geodesic γPQ with dPQ < pi, for τ ∈ [0, 1] let(1− τ)P + τQ denote the point on γPQ at distance τdPQ from P . That is,d((1− τ)P + τQ, P ) = τdPQ.Definition 4.1.3. Let (X, d) be a complete geodesic space. Then X is a CAT(1)space if, given any geodesic triangle 4PQR (with perimeter less than 2pi) and acomparison triangle 4P˜ Q˜R˜ in S2,dPtRs ≤ dP˜tR˜s (4.1)wherePt = (1− t)P + tQ, Rs = (1− s)R + sQ,P˜t = (1− t)P˜ + tQ˜, R˜s = (1− s)R˜ + sQ˜.P QRPtRsP˜ Q˜R˜P˜tR˜sFigure 4.1: An illustration of a triangle in a geodesic space (X, d) (left) and a com-parison triangle in S2 (right). If the geodesics connecting the sides of the triangle in(X, d) are shorter than the corresponding geodesics for the comparison triangle inS2, then (X, d) is called a CAT(1) space.The simplest examples of CAT(1) spaces are the complete Riemannian manifoldswith curvature bounded above by 1. In particular, S2 is a CAT(1) space. However,there are many examples of CAT(1) spaces other than Riemannian manifolds.A metric space (X, d) is said to be locally CAT(1) if every point of X has a56geodesically convex CAT(1) neighbourhood. Note that for a compact locally CAT(1)space, there exists a radius r(X) > 0 such that for all y ∈ X, Br(X)(y) is a compactCAT(1) space. We refer the reader to Section 2.2 of [7] for further background onCAT(1) spaces.4.1.3 Main results and outlineThe goal of this chapter is to prove a result analogous to the existence result of Sacksand Uhlenbeck when the target space is a compact CAT(1) space. More specifically,we obtain the following theorem.Theorem. 1.3.1 Let Σ be a compact Riemann surface, X a compact locally CAT(1)space and ϕ ∈ C0∩W 1,2(Σ, X). Then either there exists a harmonic map u : Σ→ Xhomotopic to ϕ or a nontrivial conformal harmonic map v : S2 → X.Sacks and Uhlenbeck used the perturbed energy method in the proof of Theorem1.3.1 for Riemannian manifolds. In doing so, they rely heavily on a priori estimatesprocured from the Euler-Lagrange equation of the perturbed energy functional. Oneof the difficulties in working in the singular setting is that, because of the lack oflocal coordinates, one does not have a P.D.E. derived from a variational principle(e.g. harmonic map equation). In order to prove results in the singular setting, wecannot rely on P.D.E. methods. To this end, we use a 2-dimensional generalizationof the Birkhoff curve shortening method [4], [5]. This local replacement processcan be thought of as a discrete gradient flow. This idea was used by Jost [33] togive an alternative proof of the Sacks-Uhlenbeck theorem in the smooth setting.More recently, in studying width and proving finite time extinction of the Ricci flow,Colding-Minicozzi [9] further developed the local replacement argument and proveda new convexity result for harmonic maps and continuity of harmonic replacement.However, even these arguments rely on the harmonic map equation and hence do nottranslate to our case. The main accomplishment of our method is to eliminate theneed for a P.D.E. by using the local convexity properties of the target CAT(1) space.(The necessary convexity properties of a CAT(1) space are given in Appendix B.)57For clarity, we provide a brief outline of the harmonic replacement construction.Given ϕ : Σ → X, we set ϕ = u00 and inductively construct a sequence of energydecreasing maps uln where n ∈ N∪{0}, l ∈ {0, . . . ,Λ}, and Λ depends on the geometryof Σ. The sequence is constructed inductively as follows. Given the map u0n, wedetermine the largest radius, rn, in the domain on which we can apply the existenceand regularity of Dirichlet solutions (see Lemma 4.2.1) for this map. Given a suitablecover of Σ by balls of this radius, we consider Λ subsets of this cover such that everysubset consists of non-intersecting balls. The maps uln : Σ → X, l ∈ {1, . . . ,Λ} aredetermined by replacing ul−1n by its Dirichlet solution on balls in the l-th subset of thecovering and leaving the remainder of the map unchanged. We then set u0n+1 := uΛn tocontinue by induction. There are now two possibilities, depending on lim inf rn = r.If r > 0, we demonstrate that the sequence we constructed is equicontinuous andhas a unique limit that is necessarily homotopic to ϕ. Compactness for minimizers(Lemma 4.2.2) then implies that the limit map is harmonic. If r = 0, then bubblingoccurs. That is, after an appropriate rescaling of the original sequence, the newsequence is an equicontinuous family of harmonic maps from domains exhausting C.As in the previous case, this sequence converges on compact sets to a limit harmonicmap from C to X. We extend this map to S2 by a removable singularity theoremdeveloped in section 4.3.We now give an outline of the chapter. In section 4.2, we introduce some notationand provide the results that are necessary in order to perform harmonic replacementand obtain a harmonic limit map. In particular, we state the existence and reg-ularity results for Dirichlet solutions and prove compactness of energy minimizingmaps into a CAT(1) space. In section 4.3, we prove our removable singularity theo-rem. Namely, in Theorem 4.3.6 we prove that any conformal harmonic map from apunctured surface into a CAT(1) space extends as a locally Lipschitz harmonic mapon the surface. This theorem extends to CAT(1) spaces the removable singularitytheorem of Sacks-Uhlenbeck [53] for a finite energy harmonic map into a Riemannianmanifold, provided the map is conformal. The proof relies on two key ideas. First, forharmonic maps u0 and u1 into a CAT(1) space, while d2(u0, u1) is not subharmonic, amore complicated weak differential inequality holds if the maps are into a sufficiently58small ball (Theorem B.2.4 in Appendix B.2, [60]). Using this inequality, we prove alocal removable singularity theorem for harmonic maps into a small ball. The secondkey idea, Theorem 4.3.4, is a monotonicity of the area in extrinsic balls in the targetspace, for conformal harmonic maps from a surface to a CAT(1) space. This theoremextends the classical monotonicity of area for minimal surfaces in Riemannian mani-folds to metric space targets. The proof relies on the fact that the distance functionfrom a point in a CAT(1) space is almost convex on a small ball. In application, themonotonicity is used to show that a conformal harmonic map defined on Σ\{p} iscontinuous across p. Then the local removable singularity theorem can be appliedat some small scale. Section 4.4 contains the harmonic replacement constructionoutlined above and the proof of the main theorem, Theorem 1.3.1. Note that we givecomplete proofs of several difficult estimates for quadrilaterals in a CAT(1) spacein Appendix B.1. The estimates are stated in the unpublished thesis [60] withoutproof. We apply these estimates in Appendix B.2 to give complete proofs of someenergy convexity, existence, uniqueness, and subharmonicity results (also stated in[60]) that are used throughout this chapter.4.2 Preliminary resultsThroughout this chapter we let (Ω, g) denote a Lipschitz Riemannian domain and(X, d) a locally CAT(1) space. We denote a geodesic ball in Ω of radius r centredat p ∈ Ω by Br(p) and a geodesic ball in X of radius ρ centred at P ∈ X by Bρ(P ).The following results will be used in the proof of the main theorem, Theorem 1.3.1.Lemma 4.2.1 (Existence, Uniqueness and Regularity of the Dirichlet solution). Forany finite energy map h : Ω→ Bρ(P ) ⊂ X, where ρ ∈ (0,min{r(X), pi4}), the Dirich-let solution exists. That is, there exists a unique element Dirh ∈ W 1,2h (Ω,Bρ(P ))that minimizes energy among all maps in W 1,2h (Ω,Bρ(P )). Moreover, if Dirh(∂Ω) ⊂Bσ(P ) for some σ ∈ (0, ρ), then Dirh(Ω) ⊂ Bσ(P ). Finally, the solution Dirh is lo-cally Lipschitz continuous with Lipschitz constant depending only on the total energyof the map and the metric on the domain.59For further details see Lemma B.2.2 in Appendix B.2, [60], and [7].Lemma 4.2.2 (Compactness for minimizers into CAT(1) space). Let (X, d) be aCAT(1) space and Br ⊂ Ω a geodesic (and topological) ball of radius r > 0 where(Ω, g) is a Riemannian manifold. Let ui : Br → X be a sequence of energy minimizerswith Eui [Br] ≤ Λ for some Λ > 0.Suppose that ui converges uniformly to u on Br and that there exists P ∈ X suchthat u(Br) ⊂ Bρ/2(P ) where ρ is as in Lemma 4.2.1. Then u is energy minimizingon Br/2.Proof. We will follow the ideas of the proof of Theorem 3.11 [41]. Rather than provethe bridge principle for CAT(1) spaces, we will modify the argument and appealdirectly to the bridge principle for NPC spaces (see Lemma 3.12 [41]).Since ui → u uniformly and u(Br) ⊂ Bρ/2(P ), there exists I large such that forall i ≥ I, ui(Br) ⊂ Bρ(P ). By Lemma 4.2.1, there exists c > 0 depending only onΛ and g such that for all i ≥ I, ui|B3r/4 is Lipschitz with Lipschitz constant c. Itfollows that for t > 0 small, there exists C > 0 depending on c and the dimension ofΩ such thatEui [Br/2\Br/2−t] ≤ Ct. (4.2)For ε > 0, increase I if necessary so that for all i ≥ I and all x ∈ B3r/4,d2(ui(x), u(x)) < ε. (4.3)For notational ease, let Ut := Br/2−t. Let wt : Ut → X denote the energyminimizer wt :=Diru|Ut ∈ W 1,2u (Ut, X), with existence guaranteed by Lemma 4.2.1.Following the argument in the proof of Theorem 3.11 [41], (4.2) and the lower semi-continuity of the energy imply that limt→0Ewt [Ut] = Ew0 [Br/2]. Observe that by thelower semi-continuity of energy, Theorem 1.6.1 [40],dEu[Br/2] ≤ lim infi→∞dEui[Br/2].60Thus, it will be enough to show thatlim supi→∞dEui[Br/2] ≤ dEw0 [Br/2].Let vt : Br/2 → X be the map such that vt|Ut = wt and vt|Br/2\Ut = u. Givenδ > 0, choose t > 0 sufficiently small so thatdEvt [Br/2] <dEw0 [Br/2] + δ. (4.4)Since vt is not a competitor for ui (i.e. vt|∂Br/2 is not necessarily equal to ui|∂Br/2),for each i we want to bridge from vt to ui for values near ∂Br/2. Since we want toexploit a bridging lemma into NPC spaces, rather than bridge between vt and ui, wewill bridge between their lifted maps in the cone C(X).Let C(X) := (X × [0,∞)/X × {0}, D) whereD2([P, x], [Q, y]) = x2 + y2 − 2xy cos min(d(P,Q), pi).Then C(X) is an NPC space and we can identify X with X × {1} ⊂ C(X). For anymap f : Br → X, we let f : Br → X × {1} such that f(x) = [f(x), 1]. Note that forf ∈ W 1,2(Br,Bρ(Q)), sincelimP→QD2([P, 1], [Q, 1])d2(P,Q)= limP→Q2(1− cos(d(P,Q)))d2(P,Q)= 1,it follows that DEf [Ω] = dEf [Ω] for Ω ⊂ Br.For each i ≥ I, and a fixed s, ρ > 0 to be chosen later, define the mapvi : ∂Us × [0, ρ]→ C(X)such thatvi(x, z) :=(1− zρ)vt(x) +zρui(x).The map vi is a bridge between vt|∂Us and ui|∂Us in the NPC space C(X). That is,61we are interpolating along geodesics connecting vt(x), ui(x) in the NPC space C(X)and not along geodesics in X. By [41] (Lemma 3.12) and the equivalence of theenergies for a map f and its lift f ,DEvi[∂Us × [0, ρ]] ≤ ρ2(DEvt[∂Us] +DEui[∂Us])+1ρ∫∂UsD2([vt, 1], [ui, 1])dσ=ρ2(dEvt[∂Us] +dEui[∂Us])+1ρ∫∂UsD2([vt, 1], [ui, 1])dσ.By (4.2), and since vt = u on Br/2\Ut, for s ∈ [2t/3, 3t/4] the average values of thetangential energies of vt and ui on ∂Us are bounded above by Ct/(3t/4−2t/3) = 12C.Moreover, since ui(Br/2), vt(Br/2) ⊂ Bρ(P ), (4.3) implies that for all x ∈ Br/2\Ut,D2(ui(x), vt(x)) = 2(1− cos d(ui(x), vt(x))) ≤ d2(ui(x), vt(x)) < ε. (4.5)Thus, there exists C ′ > 0 depending only on g such that for every s ∈ [2t/3, 3t/4],∫∂UsD2([vt, 1], [ui, 1])dσ < C′ε.Note that for each ε > 0, the bound above depends on I but not on t. Now, wefirst choose an s ∈ (2t/3, 3t/4) such that dEvt [∂Us] + dEui [∂Us] ≤ 24C. Next, pick0 < µ 1 such that [s, s+µt] ⊂ [2t/3, 3t/4] and 12Cµt < δ/2. For this t, µ, decreaseε if necessary (by increasing I) such thatDEvi [∂Us × [0, µt]] = µt2(dEvt[∂Us] +dEui[∂Us])+1µt∫∂UsD2([vt, 1], [ui, 1])dσ< 24Cµt/2 + C ′ε/(µt)< δ.Now, define v˜i : Br/2 → C(X) such that on Us, v˜i is the conformally dilated map of vtso that v˜i|∂Us+µt = vt|∂Us . On Us\Us+µt, let v˜i be the bridging map vi, reparametrizedin the second factor from [0, µt] to [s, s+µt]. Finally, on Br/2\Us, let v˜i = ui. Then,62for all i ≥ I,DE v˜i [Br/2] ≤ dEvt [Br/2] + δ + dEui [Br/2\Us]. (4.6)While the map v˜i agrees with ui on ∂Br/2, it is not a competitor for ui into Xsince v˜i maps into C(X). However, by defining vi : Br/2 → X such that v˜i(x) =[vi(x), h(x)], vi is a competitor. Note that for all x ∈ ∂Us, (4.5) implies that h(x) ≥1−√ε. Therefore, on the bridging strip we may estimate the change in energy underthe projection map by first observing the pointwise boundD2(v˜i(x), v˜i(y)) = D2([vi(x), h(x)], [vi(y), h(y)])= h(x)2 + h(y)2 − 2h(x)h(y) cos(d(vi(x), vi(y)))= (h(x)− h(y))2 + 2h(x)h(y)(1− cos(d(vi(x), vi(y))))≥ 2(1−√ε)2(1− cos(d(vi(x), vi(y))))= (1−√ε)2D2([vi(x), 1], [vi(y), 1]).Therefore,dEvi [Br/2] =DE[vi,1][Br/2] ≤(1−√ε)−2 DE v˜i [Br/2]. (4.7)Since vi is a competitor for ui on Br/2, (4.7), (4.6), (4.4), and (4.2) imply thatdEui[Br/2] ≤(1−√ε)−2 DE v˜i [Br/2] ≤ (1−√ε)−2 (dEw0 [Br/2] + 2δ + Ct)Since for any ε, δ > 0, by choosing t > 0 sufficiently small and I ∈ N large enough,the previous estimate holds for all i ≥ I, the inequalitylim supi→∞dEui[Br/2] ≤ dEw0 [Br/2]then implies the result.634.3 Monotonicity and removable singularity theo-remWe first show the removable singularity theorem for harmonic maps into small balls.Note that the first theorem of this section is true for domains of dimension n ≥ 2,but all other results require the domain dimension n = 2.Theorem 4.3.1. Let u : Br(p) \ {p} → Bρ(P ) ⊂ X be a finite energy harmonicmap, where ρ is as in Lemma 4.2.1 and dim(Br(p)) = n. Then u can be extended onBr(p) as the unique energy minimizer among all maps in W1,2u (Br(p),Bρ(P )).Proof. Let v ∈ W 1,2u (Br(p),Bρ(P )) minimize the energy. It suffices to show thatu = v on Br(p)\{p}. Since u is harmonic, there exists a locally finite countable opencover {Ui} of Br(p) \ {p}, and ρi > 0, Pi ∈ Bρ(P ) such that u|Ui minimizes energyamong all maps in W 1,2u (Ui,Bρi(Pi)). LetF =√1− cos dcosRu cosRvwhere d(x) = d(u(x), v(x)) and Ru = d(u, P ), Rv = d(v, P ). By Theorem B.2.4,div(cosRu cosRv∇F ) ≥ 0holds weakly on each Ui. Therefore, for a partition of unity {ϕi} subordinate to thecover {Ui} and for any test function η ∈ C∞c (Br(p) \ {p}),−∫Br(p)\{p}∇η · (cosRu cosRv∇F ) dµg = −∑i∫Ui∇(ϕiη) · (cosRu cosRv∇F ) dµg ≥ 0,(4.8)where we use∑i ϕi = 1 and∑i∇ϕi = 0.64Using polar coordinates in Br(p) centered at p, for 0 <  1, we defineφ =0 r ≤ 2log r−log 2− log  2 ≤ r ≤ 1  ≤ r.Letting ωn−1 denote the volume of the unit (n− 1)-dimensional sphere, note that∫Br(p)|∇φ|2 dµg = ωn−1(log )2∫ 2rn−3 dr + o()→ 0 as → 0.Therefore, for η ∈ C∞c (Br(p)),−∫Br(p)φ∇η · (cosRu cosRv∇F ) dµg= −∫Br(p)∇(ηφ) · (cosRu cosRv∇F ) dµg +∫Br(p)η∇φ · (cosRu cosRv∇F ) dµg≥∫Br(p)\{p}η∇φ · (cosRu cosRv∇F ) dµg (by (4.8))≥ −(∫Br(p)\{p}|∇φ|2 dµg) 12(∫Br(p)\{p}η2| cosRu cosRv∇F |2 dµg) 12,by Ho¨lder’s inequality. The last line converges to zero as → 0 because d,Ru, Rv arebounded by the compactness of Bρ(P ) and∫Br(p)\{p} |∇F |2 dµg is bounded by energyconvexity. We conclude that−∫Br(p)∇η · (cosRu cosRv∇F ) dµg = − lim→0∫Br(p)φ∇η · (cosRu cosRv∇F ) dµg ≥ 0,and hence div(cosRu cosRv∇F ) ≥ 0 holds weakly on Br(p).Since d(u(x), v(x)) = 0 on ∂Br(p), by the maximum principle d(u(x), v(x)) ≡ 0in Br(p). This implies that u ≡ v is the unique energy minimizer.Remark 4.3.2. Note that Theorem 4.3.1 implies that if u : Ω → Bρ(P ) is harmonic,65then u is energy minimizing.From this point on we assume our domain is of dimension 2. Recall the con-struction in [40] and [7] of a continuous, symmetric, bilinear, non-negative tensorialoperatorpiu : Γ(TΩ)× Γ(TΩ)→ L1(Ω) (4.9)associated with a W 1,2-map u : Ω→ X where Γ(TΩ) is the space of Lipschitz vectorfields on Ω defined bypiu(Z,W ) :=14|u∗(Z +W )|2 − 14|u∗(Z −W )|2where |u∗(Z)|2 is the directional energy density function (cf. [40, Section 1.8]). Thisgeneralizes the notion of the pullback metric for maps into a Riemannian manifold,and hence we shall refer to pi = piu also as the pullback metric for u.Definition 4.3.3. If Σ is a Riemann surface, then u ∈ W 1,2(Σ, X) is (weakly) con-formal ifpi(∂∂x1,∂∂x1)= pi(∂∂x2,∂∂x2)and pi(∂∂x1,∂∂x2)= 0,where z = x1 + ix2 is a local complex coordinate on Σ.For a conformal harmonic map u : Σ → X with conformal factor λ = 12|∇u|2,and any open sets S ⊂ Σ and O ⊂ X, defineA(u(S) ∩ O) :=∫u−1(O)∩Sλ dµg,where dµg is the area element of (Σ, g).Theorem 4.3.4 (Monotonicity). There exist constants c, C such that if u : Σ →X is a non-constant conformal harmonic map from a Riemann surface Σ into acompact locally CAT(1) space (X, d), then for any p ∈ Σ and 0 < σ < σ0 =min{ρ, d(u(p), u(∂Σ))}, the following function is increasing:σ 7→ ecσ2A(u(Σ) ∩ Bσ(u(p)))σ2,66andA(u(Σ) ∩ Bσ(u(p))) ≥ Cσ2.Proof. Since Σ is locally conformally Euclidean and the energy is conformally in-variant, without loss of generality, we may assume that the domain is Euclidean.Fix p ∈ Σ and let R(x) = d(u(x), u(p)). Since u is continuous and locally energyminimizing, by [60, Proposition 1.17], [7, Lemma 4.3] we have that the followingdifferential inequality holds weakly on u−1(Bρ(u(p))):12∆R2 ≥ (1−O(R2))|∇u|2. (4.10)Let ζ : R+ → R+ be any smooth nonincreasing function such that ζ(t) = 0 fort ≥ 1, and let ζσ(t) = ζ( tσ ). By (4.10), for σ < σ0 we have−∫Σ∇R2 · ∇(ζσ(R)) dx1dx2 ≥ 2∫Σζσ(R) (1−O(R2))|∇u|2 dx1dx2= 4∫Σζσ(R) (1−O(R2))λ dx1dx2.Therefore,2∫Σζσ(R) (1−O(R2))λ dx1dx2 ≤ −∫ΣR∇R · ∇(ζσ(R)) dx1dx2= −∫ΣRσζ ′(Rσ)|∇R|2 dx1dx2≤ −∫ΣRσζ ′(Rσ)12|∇u|2 dx1dx2= −∫ΣRσζ ′(Rσ)λ dx1dx2=∫Σσddσ(ζσ(R)) λ dx1dx2= σddσ∫Σζσ(R) λ dx1dx2,where in the second inequality we have used that ζ ′ ≤ 0 and |∇R|2 ≤ 12|∇u|2, since67u is conformal. Set f(σ) =∫Σζσ(R)λ dx1dx2. We have shown that2(1−O(σ2))f(σ) ≤ σf ′(σ).Integrating this, we conclude that there exist c > 0 such that the functionσ 7→ ecσ2f(σ)σ2(4.11)is increasing for all 0 < σ < σ0. Approximating the characteristic function of [−1, 1],and letting ζ be the restriction to R+, it then follows thatecσ2A(u(Σ) ∩ Bσ(u(p)))σ2is increasing in σ for 0 < σ < σ0.Since λ = 12|∇u|2 ∈ L1(Σ,R),limr→0∫Br(x)λ dx1dx2pir2= λ(x), a.e. x ∈ Σ (4.12)by the Lebesgue-Besicovitch Differentiation Theorem. Since u is conformal, for everyω ∈ S1,λ(x) = limt→0d2(u(x+ tω), u(x))t2, a.e. x ∈ Σ (4.13)([40, Theorem 1.9.6 and Theorem 2.3.2]). Since u is locally Lipschitz [7, Theorem1.2], by an argument as in the proof of Rademacher’s Theorem ([20, p. 83-84]),λ(x) = limy→xd2(u(y), u(x))|y − x|2 (4.14)for almost every x ∈ Σ. To see this, choose {ωk}∞k=1 to be a countable, dense subsetof S1. SetSk = {x ∈ Σ : limt→0d(u(x+ tωk), u(x))texists, and is equal to√λ(x)}68for k = 1, 2, . . . and letS = ∩∞k=1Sk.Observe that H2(Σ \ S) = 0. Fix x ∈ S, and let ε > 0. Choose N sufficiently largesuch that if ω ∈ S1 then|ω − ωk| < ε2Lip(u)for some k ∈ {1, . . . , N}. Sincelimt→0d(u(x+ tωk), u(x))t=√λ(x)for k = 1, . . . , N , there exists δ > 0 such that if |t| < δ then∣∣∣∣d(u(x+ tωk), u(x))t −√λ(x)∣∣∣∣ < ε2for k = 1, . . . , N . Consequently, for each ω ∈ S1 there exists k ∈ {1, . . . , N} suchthat∣∣∣∣d(u(x+ tω), u(x))t −√λ(x)∣∣∣∣≤∣∣∣∣d(u(x+ tωk), u(x))t −√λ(x)∣∣∣∣+ ∣∣∣∣d(u(x+ tω), u(x))t − d(u(x+ tωk), u(x))t∣∣∣∣≤∣∣∣∣d(u(x+ tωk), u(x))t −√λ(x)∣∣∣∣+ ∣∣∣∣d(u(x+ tω), u(x+ tωk))t∣∣∣∣<ε2+ Lip(u)|ω − ωk|< ε.Therefore the limit in (4.14) exists, and (4.14) holds, for almost every x ∈ Σ.The zero set of λ is of Hausdorff dimension zero by [48]. At points where λ(x) 6= 0and (4.14) holds, we have that for any ε > 0u(B σ(1+ε)√λ(x)) ⊂ u(Σ) ∩ Bσ(u(x))69if σ is sufficiently small. Therefore by (4.12),Θ(x) := limσ→0A(u(Σ) ∩ Bσ(u(x)))piσ2≥ 1, a.e. x ∈ Σ. (4.15)By the monotonicity of (4.11), Θ(x) exists for every x ∈ Σ, and Θ(x) is uppersemicontinuous since it is a limit of continuous functions (the density at a givenradius is a continuous function of x). Therefore, Θ(x) ≥ 1 for every x ∈ Σ. Togetherwith the monotonicity of (4.11), it follows thatA(u(Σ) ∩ Bσ(u(p))) ≥ Cσ2for 0 < σ < σ0.Remark 4.3.5. Note that if u : M → Bρ(P ) is a harmonic map from a compactRiemannian manifold M , then u must be constant. This follows from the maximumprinciple, since equation (4.10) implies that R2(x) = d2(u(x), P ) is subharmonic.For a conformal harmonic map from a surface into a Riemannian manifold, con-tinuity follows easily using monotonicity ([57, Theorem 10.4], [29], [33, Theorem9.3.2]). By Theorem 4.3.4, using this idea we can prove the following removablesingularity result for conformal harmonic maps into a CAT(1) space.Theorem 4.3.6 (Removable singularity). If u : Σ \ {p} → X is a conformal har-monic map of finite energy from a Riemann surface Σ into a compact locally CAT(1)space (X, d), then u extends to a locally Lipschitz harmonic map u : Σ→ X.Proof. Let Br denote Br(p), the geodesic ball of radius r centered at the point p inΣ, and let Cr = ∂Br denote the circle of radius r centered at p. By the Courant-Lebesgue Lemma, there exists a sequence ri ↘ 0 so thatLi = L(u(Cri)) :=∫Cri√λ dsg → 0as i → ∞, where dsg denotes the induced measure on Cri = ∂Bri from the metricg on Σ. Since E(u) < ∞, λ = 12|∇u|2 is an L1 function and, by the Dominated70Convergence Theorem,Ai = A(u(Bri \ {p})) :=∫Bri\{p}λ dµg → 0as i→∞.First we claim that there exists P ∈ X such that u(Cri) → P with respect tothe Hausdorff distance as i → ∞. Let di,j = d(u(Cri), u(Crj)). Suppose i < j sori > rj, and choose Q ∈ u(Bri \ B¯rj) such that d(Q, u(Cri) ∪ u(Crj)) ≥ di,j/2. Forσ = min{di,j3, ρ2}, by monotonicity (Theorem 4.3.4),A(u(Bri \ B¯rj) ∩ Bσ(Q)) ≥ Cσ2.Since A(u(Bri \ B¯rj)∩Bσ(Q)) ≤ A(u(Bri \ {p})) = Ai, it follows that σ ≤ c√Ai → 0as i→∞, and we must have di,j → 0. Therefore any sequence of points Pi ∈ u(Cri)is a Cauchy sequence sinced(Pi, Pj) ≤ di,j + Li + Lj → 0as i, j → ∞. Hence, there exists P ∈ X independent of the sequence, such thatPi → P .Finally, we claim that limx→p u(x) = P . It follows from this that we may extendu continuously to Σ by defining u(p) = P . To prove the claim, consider a sequencexi ∈ Σ \ {p} such that xi → p. We want to show that u(xi) → P . Supposexi ∈ Brj(i) \ B¯rj(i)+1 for some j(i), and let di = d(u(xi), u(Crj(i)) ∪ u(Crj(i)+1)). Forσ = min{di3, ρ2}, by monotonicity (Theorem 4.3.4),A(u(Brj(i) \ B¯rj(i)+1) ∩ Bσ(u(xi))) ≥ Cσ2.Therefore, σ < c√Aj(i) → 0 as i → ∞, and we must have d(u(xi), u(Crj(i)) ∪u(Crj(i)+1))→ 0. It follows that u(xi)→ P and u extends continuously to Σ.We may now apply Theorem 4.3.1 to show that u is energy minimizing at p.Since u is continuous, there exists δ > 0 such that u(Bδ) ⊂ Bρ(Q) ⊂ X. By Theorem714.3.1, u is the unique energy minimizer in W 1,2u (Bδ,Bρ(Q)). Hence u is locally energyminimizing on Σ and by [7, Theorem 1.2], u is locally Lipschitz on Σ.The following is derived using only domain variations as in [57, Lemma 1.1] (using[40, Theorem 2.3.2] to justify the computations involving change of variables) andis independent of the curvature of the target space (see for example, [28, (2.3) page193]).Lemma 4.3.7. Let u : Σ → X be a harmonic map from a Riemann surface into alocally CAT(1) space. The Hopf differentialΦ(z) =[pi(∂∂x1,∂∂x1)− pi(∂∂x2,∂∂x2)− 2ipi(∂∂x1,∂∂x2)]dz2,where z = x1 + ix2 is a local complex coordinate on Σ and pi is the pull-back innerproduct, is holomorphic.Corollary 4.3.1. Let u : C→ X be a harmonic map of finite energy and (X, d) be acompact locally CAT(1) space. Then u extends to a locally Lipschitz harmonic mapu : S2 → X.Proof. Let p : S2 \{n} → R2 be stereographic projection from the north pole n ∈ S2.Set uˆ = u ◦ p : S2 \ {n} → X. We will show that n is a removable singularity.Let ϕ = pi( ∂∂x1, ∂∂x1) − pi( ∂∂x2, ∂∂x2) − 2ipi( ∂∂x1, ∂∂x2). By Lemma 4.3.7, the Hopfdifferential Φ(z) = ϕ(z)dz2 is holomorphic on C. By assumption,E(u) =∫R2(‖u∗( ∂∂x1)‖2 + ‖u∗( ∂∂x2)‖2)dx1dx2 <∞and therefore ∫R2|ϕ| dx1dx2 ≤ 2E(u) <∞.Thus |ϕ| ∈ L1(C,R) and is subharmonic, and hence ϕ ≡ 0 and u is conformal. Thenby Theorem 4.3.6, u extends to a locally Lipschitz harmonic map u : S2 → X.724.4 Harmonic Replacement ConstructionIn this section we prove the main theorem:Theorem. 1.3.1 Let Σ be a compact Riemann surface, X a compact locally CAT(1)space and ϕ ∈ C0∩W 1,2(Σ, X). Then either there exists a harmonic map u : Σ→ Xhomotopic to ϕ or a nontrivial conformal harmonic map v : S2 → X.Lemma 4.4.1 (Jost’s covering lemma, [33] Lemma 9.2.6). For a compact Rieman-nian manifold Σ, there exists Λ = Λ(Σ) ∈ N with the following property: for anycoveringΣ ⊂m⋃i=1Br(xi)by open balls, there exists a partition I1, . . . IΛ of the integers {1, . . . ,m} such thatfor any l ∈ {1, . . . ,Λ} and two distinct elements i1, i2 of I l,B2r(xi1) ∩B2r(xi2) = ∅.Definition 4.4.2. For each k = 0, 1, 2, . . . , we fix a coveringOk = {B2−k(xk,i)}mki=1of Σ by balls of radius 2−k. Furthermore, let I1k , . . . , IΛk be the disjoint subsets of{1, . . . ,mk} as in Lemma 4.4.1; in other words, for every l ∈ {1, . . . ,Λ},B2−k+1(xk,i1) ∩B2−k+1(xk,i2) = ∅, ∀i1, i2 ∈ I lk, i1 6= i2. (4.16)By the Vitali Covering Lemma, we can ensure thatB2−k−3(xk,i1) ∩B2−k−3(xk,i2) = ∅, ∀i1, i2 ∈ {1, . . . ,mk}, i1 6= i2. (4.17)Let Σ be a compact Riemann surface. By uniformization, we can endow Σ witha Riemannian metric of constant Gaussian curvature +1, 0 or −1. Let Λ = Λ(Σ) beas in Lemma 4.4.1 and ρ = ρ(X) > 0 be as in Lemma 4.2.1. We inductively define73a sequence of numbers{rn} ⊂ 2−N := {1, 2−1, 2−2, . . . }and a sequence of finite energy maps{uln : Σ→ X}for l = 0, . . . ,Λ, n = 1, . . . ,∞ as follows:Initial Step 0: Fix κ0 ∈ N such that B2−κ0 (x) is homeomorphic to a disk forall x ∈ Σ. Let u00 := ϕ ∈ C0 ∩W 1,2(Σ, X), and letr′0 = sup{r > 0 : ∀x ∈ Σ,∃P ∈ X such that u00(B2r(x)) ⊂ B3−Λρ(P )}and k′0 > 0 be such that2−k′0 ≤ r′0 < 2−k′0+1.Definer0 = 2−k0 = min{2−k′0 , 2−κ0},and letOk0 = {Br0(xk0,i)}mk0i=1 and I1k0 , . . . , IΛk0be as in Definition 4.4.2. For l ∈ {1, . . . ,Λ} inductively define ul0 : Σ→ X from ul−10by settingul0 ={ul−10 in Σ\⋃i∈Ilk0B2r0(xk0 ,i)Dirul−10 in B2r0(xk0 ,i), i ∈ I lk0where Dirul−10 is the unique Dirichlet solution in W1,2ul−10(B2r0(xk0,i),Bρ(P )) of Lemma4.2.1. Here there are two things to check related to the definition of the Dirichlet solu-tion. First, since B2r0(xk0,i1)∩B2r0(xk0,i2) = ∅, ∀ i1, i2 ∈ I lk0 with i1 6= i2 (cf. (4.16)),there is no issue of interaction between solutions at a single step so the map is well-defined if it exists. Second, we claim that ul−10 (B2r0(xk0,i)) ⊂ B3−Λ+(l−1)ρ(P ) ⊂ Bρ(P )74for some P ∈ X and thus the Dirichlet solution exists and is unique by Lemma 4.2.1.To verify the claim, first note that for each i = 1, . . . ,mk0 there exists P ∈ X suchthat u10(B2r0(xk0,i)) ⊂ B3−Λ+1ρ(P ). Indeed, ifB2r0(xk0,i)∩B2r0(xk0,j) = ∅ for all j ∈ I1k0then u10 = u00 on B2r0(xk0,i) and so u10(B2r0(xk0,i)) = u00(B2r0(xk0,i)) ⊂ B3−Λρ(P ) forsome P . On the other hand, if B2r0(xk0,i) ∩ B2r0(xk0,j) 6= ∅ for one or more j ∈ I1k0 ,then since u00(B2r0(xk0,i)) ⊂ B3−Λρ(P ) for some P and u10(B2r0(xk0,j)) ⊂ B3−Λρ(Pj) forsome Pj with B3−Λρ(P )∩B3−Λρ(Pj) 6= ∅, it follows that u10(B2r0(xk0,i)) ⊂ B3−Λ+1ρ(P ).Inductively, we may show that for each i = 1, . . . ,mk0 and l ∈ {1, . . . ,Λ} there existsP ∈ X such that ul−10 (B2r0(xk0,i)) ⊂ B3−Λ+(l−1)ρ(P ), as claimed.Inductive Step n: Having definedr0, . . . , rn−1 ∈ 2−N,andu0ν , u1ν , . . . , uΛν : Σ→ X, ν = 0, 1, . . . , n− 1,we set u0n = uΛn−1 and definern ∈ 2−N and u1n, . . . , uΛnas follows. Letr′n = sup{r > 0 : ∀x ∈ Σ,∃P ∈ X such that u0n(B2r(x)) ⊂ B3−Λρ(P )}and k′n ∈ N be such that2−k′n ≤ r′n < 2−k′n+1.Definern = 2−kn = min{2−k′n , 2−κ0}.LetOkn = {Brn(xkn,i)}mkni=1 and I1kn , . . . , IΛkn75be as in Definition 4.4.2. Having defined u0n, . . . , ul−1n , we now define uln : Σ→ X bysettinguln ={ul−1n in Σ\⋃i∈IlknB2rn(xkn,i)Dirul−1n in B2rn(xkn,i), i ∈ I lknwhere Dirul−1n is the unique Dirichlet solution in W1,2ul−1n(B2rn(xkn,i),Bρ(P )) for someP of Lemma 4.2.1.This completes the inductive construction of the sequence {uln}. Note thatE(uΛn) ≤ · · · ≤ E(u0n) = E(uΛn−1), ∀n = 1, 2, . . . .Thus, there exists E0 such thatlimn→∞E(uln) = E0, ∀ l = 0, . . . ,Λ. (4.18)We consider the following two cases separately:CASE 1: lim infn→∞ rn > 0.CASE 2: lim infn→∞ rn = 0.For CASE 1, we prove that there exists a harmonic map u : Σ → X homotopic toϕ = u00. We will need the following two claims.Claim 4.4.3. For any l ∈ {0, . . .Λ− 1},limn→∞||d(uln, uΛn)||L2(Σ) = 0.Proof. Fix l ∈ {0, . . . ,Λ − 1}. For n ∈ N, λ ∈ {l + 1, . . . ,Λ} and i ∈ Iλkn , we applyTheorem B.2.1 with u0 = uλ−1n∣∣B2rn (xkn,i), u1 = uλn∣∣B2rn (xkn,i)and Ω = B2rn(xkn,i). Letw : Σ→ X be the map defined as w = uλn = uλ−1n outside⋃i∈IλknB2rn(xkn,i) and the76map corresponding to w in Theorem B.2.1 in each B2rn(xkn,i). Then(cos8 ρ)∫B2rn (xkn,i)∣∣∣∣∇tan 12d(uλ−1n , uλn)cosR∣∣∣∣2 dµ≤ 12(∫B2rn (xkn,i)|∇uλ−1n |2dµ+∫B2rn (xkn,i)|∇uλn|2dµ)−∫B2rn (xkn,i)|∇w|2dµ.Summing over i, using that w = uλn = uλ−1n outside⋃i∈IλknB2rn(xkn,i), and applyingthe Poincare´ inequality, we obtain∫Σd2(uλ−1n , uλn)dµ ≤ C(12E(uλ−1n ) +12E(uλn)− E(w)),where here and henceforth C is a constant independent of n. Since uλn is harmonicin⋃i∈IλknB2rn(xkn,i), we have E(uλn) ≤ E(w). Hence∫Σd2(uλ−1n , uλn)dµ ≤ C(12E(uλ−1n )−12E(uλn)).Thus,∫Σd2(uln, uΛn)dµ ≤∫Σ(Λ∑λ=l+1d(uλ−1n , uλn))2dµ≤ (Λ− l)2Λ∑λ=l+1∫Σd2(uλ−1n , uλn)dµ≤ CΛ∑λ=l+1(E(uλ−1n )− E(uλn))= C(E(uln)− E(uΛn)).This proves the claim since limn→∞(E(uln)− E(uΛn))= 0 by (4.18).Claim 4.4.4. Let  > 0 such that 3−Λ < ρ, l ∈ {1, . . . ,Λ} and n ∈ N be given. If77δ ∈ (0, rn) is such that √8piE(u00)log δ−2≤ 3−Λ, (4.19)then∀x ∈l⋃λ=1⋃i∈IλknBrn(xkn,i), ∃P ∈ X such that uln(BδΛ(x)) ⊂ B3(P ).In particular, for l = Λ, ∀x ∈ Σ, ∃P ∈ X such that uΛn(BδΛ(x)) ⊂ B3(P ).Proof. Fix , l, n and let δ be as in (4.19). For x ∈ ⋃lλ=1⋃i∈Iλkn Brn(xkn,i), thereexists λ ∈ {1, . . . , l} such that x ∈ Brn(xkn,i) for some i ∈ Iλkn and henceBrn(x) ⊂ B2rn(xkn,i).Since uλn is harmonic in B2rn(xkn,i), it is harmonic in Brn(x). By the Courant-Lebesgue Lemma, there existsR1(x) ∈ (δ2, δ)such thatuλn(∂BR1(x)(x)) ⊂ B3−Λ(P1) for some P1 ∈ X.Since uλn is a Dirichlet solution and 3−Λ < ρ, by Lemma 4.2.1uλn(Bδ2(x)) ⊂ uλn(BR1(x)(x)) ⊂ B3−Λ(P1).Next, by the Courant-Lebesgue Lemma, there existsR2(x) ∈ (δ3, δ2)such thatuλ+1n (∂BR2(x)(x)) ⊂ B3−Λ(P ′2) for some P ′2 ∈ X. (4.20)78There are two cases to consider:Case a. BR2(x)(x) ∩⋃i∈Iλ+1knB2rn(xkn,i) = ∅. In this case, uλ+1n = uλn in BR2(x)(x).Since uλn is harmonic on this ball,uλ+1n (BR2(x)(x)) = uλn(BR2(x)(x)) ⊂ uλn(Bδ2(x)) ⊂ B3−Λ(P1).In this case we let P2 = P1.Case b. BR2(x)(x) ∩⋃i∈Iλ+1knB2rn(xkn,i) 6= ∅. In this case, uλ+1n is only piecewiseharmonic on BR2(x)(x). The regions of harmonicity are of two types. On the re-gion Ω := BR2(x)(x)\⋃i∈Iλ+1knB2rn(xkn,i), we have uλ+1n = uλn. As in Case a, weconclude that the image of this region is contained in B3−Λ(P1). All other regions,which we index Ωi, have two smooth boundary components, one on the interior ofBR2(x)(x), which we label γi, and one on ∂BR2(x)(x), which we label βi. By construc-tion uλ+1n = uλn on γi, thusuλ+1n (γi) ⊂ B3−Λ(P1).Moreover, uλ+1n (βi) ⊂ B3−Λε(P ′2) by (4.20). Notice that in this case,B3−Λ(P1) ∩ B3−Λ(P ′2) 6= ∅.Thus, by the triangle inequality there exists P2 ∈ X such thatuλ+1n (∪i∈Iλ+1kn ∂Ωi) ⊂ B3−Λ+1(P2).Since uλ+1n is harmonic on each Ωi,uλ+1n (∪i∈Iλ+1kn Ωi) ⊂ B3−Λ+1(P2).Since BR2(x)(x) = Ω ∪⋃i∈Iλ+1knΩi,uλ+1n (BR2(x)(x)) ⊂ B3−Λ+1(P2).79Thus, we have shown that in either Case a or Case b,uλ+1n (Bδ3(x)) ⊂ uλ+1n (BR2(x)(x)) ⊂ B3−Λ+1(P2).After iterating this argument for uλ+2n , . . . , uln, we conclude that there exists Pl−λ+1 ∈X such thatuln(BδΛ(x)) ⊂ uln(Bδl−λ+2(x)) ⊂ B3−Λ+l−λ(Pl−λ+1) ⊂ B3(Pl−λ+1).Letting P = Pl−λ+1, we obtain the assertion of Claim 4.4.4.Since lim infn→∞ rn > 0, there exist k ∈ N and an increasing sequence {nj}∞j=1 ⊂N such that rnj = 2−k (or equivalently knj = k). In particular, the covering used forStep nj in the inductive construction of u0nj, . . . , uΛnj is the same for all j = 1, 2, . . . .Thus, we can use the following notation for simplicity:O = Okj , I l = I lkj , Bi = Brnj (xknj ,i) and tBi = Btrnj (xknj ,i) for t ∈ R+.With this notation, Claim 4.4.4 implies that for a fixed l ∈ {1, . . . ,Λ},{ulnj} is an equicontinuous family of maps on Bl :=l⋃λ=1⋃i∈IλBi. (4.21)In particular, {ulnj} is an equicontinuous family of maps in Σ. By taking a furthersubsequence if necessary, we can assume that∃u ∈ C0(Σ, X) such that uΛnj ⇒ u. (4.22)We claim that for every l ∈ {1, . . . ,Λ},ulnj ⇒ u on Bl where u is as in (4.22). (4.23)Indeed, if (4.23) is not true, consider a subsequence of {ulnj} that does not converge80to u. By (4.21), we can assume (by taking a further subsequence if necessary) that∃ v : Bl → X such that ulnj ⇒ v 6= u|Bl .Combining this with (4.22) and Claim 4.4.3, we conclude that||d(v, u)||L2(Bl) = limj→∞||d(ulnj , uΛnj)||L2(Bl) ≤ limj→∞ ||d(ulnj, uΛnj)||L2(Σ) = 0which in turn implies that u = v. This contradiction proves (4.23).Finally, we are ready to prove the harmonicity of u. For an arbitrary pointx ∈ Σ, there exists l ∈ {1, . . . ,Λ} and i ∈ I l such that x ∈ Bi. Since ulnj is energyminimizing in Bi and ulnj⇒ u in Bi by (4.23), Lemma 4.2.2 implies that u is energyminimizing in 12Bi.The map u is homotopic to ϕ since it is a uniform limit of uΛnj each of which ishomotopic to ϕ. This completes the proof for CASE 1 as u is the desired harmonicmap homotopic to ϕ.For CASE 2, we prove that there exists a non-constant harmonic map u : S2 → X.Recall that we have endowed Σ with a metric g of constant Gaussian curvaturethat is identically +1, 0 or −1. Fixy∗ ∈ Σand a local conformal chartpi : U ⊂ C→ pi(U) = B1(y∗) ⊂ Σsuch thatpi(0) = y∗81and the metric g = (gij) of Σ expressed with respect to this local coordinates satisfiesgij(0) = δij. (4.24)For each n, the definition of rn implies that we can find yn, y′n ∈ Σ with2rn ≤ dg(yn, y′n) ≤ 4rnwhere dg is the distance function on Σ induced by the metric g, andd(u0n(yn), u0n(y′n)) ≥ 3−Λρ.Since Σ is a compact Riemannian surface of constant Gaussian curvature, there existsan isometry ιn : Σ→ Σ such that ιn(y∗) = yn. Define the conformal coordinate chartpin : U ⊂ C→ pin(U) = B1(yn) ⊂ Σ, pin(z) := ιn ◦ pi(z).Thus,pin(0) = yn.Define the dilatation mapΨn : C→ C, Ψn(z) = rnzand set Ωn := Ψ−1n ◦ pi−1n (B1(yn)) ⊂ C andu˜ln : Ωn → X, u˜ln := uln ◦ pin ◦Ψn.Since lim infn→∞ rn = 0, there exists a subsequence{rnj} such that limj→∞rnj = 0. (4.25)82Thus, Ωnj ↗ C. Furthermore, (4.24) implies thatlimj→∞dg(y′nj, ynj)|pi−1nj (y′nj)|= 1.Hence, for zn = Ψ−1n ◦ pi−1n (y′n),2 ≤ limj→∞|znj | ≤ 4 (4.26)andd(u˜0nj(znj), u˜0nj(0)) = d(u0nj(y′nj), u0nj(ynj)) ≥ 3−Λρ. (4.27)Additionally, by the conformal invariance of energy, we have thatE(u˜ln) = E(uln∣∣B1(yn)) ≤ E(u00). (4.28)For R > 0, letDR := {z ∈ C : |z| < R}.Since harmonicity is invariant under conformal transformations of the domain, wecan follow CASE 1 (cf. (??), (??) and (4.22)) and prove that||d(u˜Λn−1, u˜Λn)||L2(DR) → 0,{u˜Λn}∞n=nR is an equicontinuous family in DRfor some nR, and∃ u˜R : DR → X such that u˜Λn ⇒ u˜R in DR. (4.29)Below, we will prove harmonicity of u˜R by following a similar proof to CASE 1. Wefirst need the following lemma.Lemma 4.4.5. Let Okn be as in Definition 4.4.2. For a fixed R > 0, there exists M83independent of n such that for every n ∈ N,|{i : B2−kn (xkn,i) ∩ (pin ◦Ψn(DR)) 6= ∅}| ≤M.Proof. By (4.24),limn→∞Vol(pin ◦Ψn(D2R))4piR22−2kn= 1andlimn→∞Vol(B2−kn−3(xn,i))pi2−2kn−6= 1where Vol is the volume in Σ. Let J ⊂ {1, . . . ,mkn} be such thatJ = {i : B2−kn (xkn,i) ∩ (pin ◦Ψn(DR)) 6= ∅}.By (4.17), we have that for sufficiently large kn,|J |pi2−2kn−6 ≤ 2∑i∈JVol(B2−kn−3(xkn,i))≤ 2Vol(pin ◦Ψn(D2R))≤ 16piR22−2kn .Hence |J | ≤ R2210 and {B2−kn (xkn,i)}i∈J covers DR.For each B2−kn (xkn,i) ∈ Okn , letB˜n,i := Ψ−1n ◦ pi−1n (B2−kn (xkn,i))and2B˜n,i := Ψ−1n ◦ pi−1n (B2−kn+1(xkn,i))for notational simplicity. After renumbering, Lemma 4.4.5 implies that there existsM = M(R) such thatDR ⊂M⋃i=1B˜n,i.84If we writeI lkn(R) = {i ∈ I lkn : i ≤M} ∀ l = 1, . . . ,Λ,thenDR ⊂Λ⋃l=1⋃i∈Ilkn (R)B˜n,i.Choose a subsequence of (4.25), which we will denote again by {nj}, such thatΨ−1nj ◦ pi−1nj (xknj ,i)→ x˜i ∀ i ∈ {1, . . . ,M}and such that for each l = 1, . . . ,Λ, the setsI˜ l := I lknj (R) = {i ∈ Ilknj: i ≤M}are equal for all knj . Unlike CASE 1, where Brnj (xknj ,i) is the same ball Bi for all j,the sets B˜n1,i, B˜n2,i, . . . are not necessarily the same. Since the component functionsof the pullback metric Ψ∗njg converge uniformly to those of the standard Euclideanmetric g0 on C by (4.24) and B˜nj ,i with respect to Ψ∗njg is a ball of radius 1, B˜nj ,iwith respect to g0 is close to being a ball of radius 1 in the following sense: for all > 0, there exists J large enough such that for all j ≥ J , B1−(x˜i) ⊂ B˜nj ,i fori = 1, . . . ,M . Choose  > 0 sufficiently small such that DR ⊂⋃Mi=1 B1−(x˜i). Thenchoose J as above. SetB˜i :=⋂j≥JB˜nj ,i ⊃ B1−(x˜i) and tB˜i :=⋂j≥JtB˜nj ,i for t ∈ R+.ThenDR ⊂M⋃i=1B˜i =Λ⋃l=1⋃i∈I˜l(R)B˜i. (4.30)85Using (4.30), we can now follow CASE 1 (cf. (4.23)) to prove that for l ∈ {1, . . . ,Λ},u˜ln ⇒ u˜R onl⋃λ=1⋃i∈I˜λB˜i where u˜R is as in (4.29). (4.31)Let x ∈ DR. 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Mech.Anal. 3 (1954), 745-753.[70] M. Wolf. The Teichmu¨ller theory of harmonic maps, J. Differential Geom. 29(1989), 449-479.93Appendix AThe Dimension of the Space ofHarmonic 1-Forms with DirichletBoundary ConditionIt is well-known, we believe, that if M is a surface with boundary ∂M 6= ∅, genus gand k boundary components, then dimH1N(M) = 2g + k − 1, but this result seemsdifficult to find in the literature. We give a proof here for completeness. When M isa surface, it follows from Lefschetz duality that dimH1N(M) = dimH2−1D (M), whereH1D(M) is the space of harmonic 1-forms on M which satisfy the relative boundaryconditions:i∗ω = i∗δω = 0,where i : ∂M ↪→M is the inclusion. So, to prove that dimH1N(M) = 2g + k − 1, wewill show that dimH1D(M) = 2g + k − 1.Lemma A.0.6. Let M be an orientable surface of genus g with k boundary compo-nents. Then dimH1D(M) = 2g + k − 1.Proof. Let EH1D(M) denote the subspace of harmonic fields with Dirichlet boundaryconditions which are exact. Then,H1D(M) = EH1D(M)⊕(EH1D(M))⊥ ,94and dimH1D(M) = dim EH1D(M)+dim(EH1D(M))⊥. We claim that dim EH1D(M) =k − 1 and dim (EH1D(M))⊥ = 2g.For the first claim, if ω ∈ EH1D(M), then there is a function u ∈ C∞(M) forwhich ω = du. Since ω is a harmonic field with Dirichlet boundary conditions, itfollows that u is a harmonic function and is constant on the boundary. If we writethe boundary as a disjoint union of k curves, ∂M = Γ1 ∪ · · · ∪ Γk, then we get thatu|Γi = ci, for some constant ci, i = 1, . . . k. Now, the Dirichlet problem∆u = 0u|Γi = ci,has a unique solution for each choice of (c1, . . . , ck) (see pg. 307 of [66]). LetF ={u ∈ C∞(M)∣∣∣∣ ∆u = 0, u|Γi = ci, i = 1 . . . k, k∑i=1ci = 0}.It easy to see that the differential d|F : F → EH1D(M) is linear and bijective, and sodim EH1D(M) = dimF = k − 1.Let M be a smooth Riemannian manifold obtained from M by gluing a disk intoeach of its boundary curves Γi. To prove the second claim, we will construct anisomorphism between(EH1D(M))⊥ and H1(M). The result will then follow from thefact that there are 2g cohomology classes of closed forms on M .Let θ ∈ Ω(M) be a closed form. We’ll first show that there is a closed formω˜ ∈ Ω(M) supported on M which is cohomologous to θ. To see this, let D˜i, i =1, . . . , k, be a disk slightly larger than and containing Di, and let φi be a smoothcut-off function for which φi|Di ≡ 1 and φi|M\D˜i ≡ 0. Since D˜i is simply-connected,θ|D˜i = dfi for some smooth functions fi. Let ω˜ = θ −∑ki=1 d(φifi). Then ω˜|Di ≡ 0and dω˜ = 0, so ω˜ is a closed form in Ω(M) with compact support. Since∑ki=1 d(φifi)is exact, it follows that θ and ω˜ are cohomologous. For simplicity, we will suppressthe restriction notation and write ω˜|M by ω˜. Now, we claim that any closed formω˜ ∈ Ω(M) with compact support is cohomologous to a form ω0 ∈ (EHD(M))⊥. To95see this, let u be a solution to the Poisson problem∆u = −δω˜u|Γi = 0 ,and define ω = ω˜+du. Then, ω is harmonic, since ∆ω = ∆ω˜+∆du = dδω˜+0−d∆u =0. Moreover, i∗ω = i∗ω˜+ d(i∗u) = 0, so ω satisfies the Dirichlet boundary condition.Now, ω = ω0 + dv for some ω0 ∈ (EH1D(M))⊥ and dv ∈ EH1D(M). Hence, ω0 iscohomologous to ω, and therefore ω˜ and θ. Note that ω0 is unique, i.e., for anyclosed form θ ∈ Ω(M), there is a unique ω0 ∈ (EH1D(M))⊥ for which ω0 ∼ θ. Ifω10, ω20 ∈ (EH1D(M))⊥ are two such forms, then ω10 ∼ θ ∼ ω20. Hence, ω10 − ω20 = dζ,for some smooth function ζ. However, ω10 − ω20 ∈ (EH1D(M))⊥ ⊂ (EΩ(M))⊥ anddζ ∈ EΩ(M), so it follows that ω10 = ω20.Let L : H1(M) → (EH1D(M))⊥ be the map [θ] 7→ ω0 (as above). Note that itfollows from the uniqueness of ω0 that L is well-defined and linear.Now, L is also injective. If L([θ1]) = L([θ2]), then θ1 + du1 = θ2 + du2, for somesmooth functions u1, u2, which yields θ1 ∼ θ2.Finally, L is surjective. Suppose ω0 ∈(EH1D(M))⊥. Then, since i∗ω0 ≡ 0,∫∂Mω0 = 0,and it follows that ω0 is exact in a neighbourhood of each boundary curve, i.e.,ω0 = dψi in a neighbourhood of Γi. Since we can extend each ψi smoothly over Di,we can extend ω0 to a closed form θ ∈ M . It follows from the well-definedness of Lthat L does not depend on the choice of D˜i or φi, i = 1, . . . k. Hence, L([θ]) = ω0.96Appendix BQuadrilateral Estimates andEnergy ConvexityB.1 Quadrilateral estimatesIn this section, we include several estimates for quadrilaterals in a CAT(1) space. Theestimates are stated in the unpublished thesis [60] without proof. As the calculationswere not obvious, we include our proofs for the convenience of the reader. Referencesto the location of each estimate in [60] are also included.The first lemma is a result of Reshetnyak which will be essential in later estimates.Lemma B.1.1 ([51, Lemma 2]). Let PQRS be a quadrilateral in X. Then thesum of the length of diagonals in PQRS can be estimated as follows:cos dPR + cos dQS ≥ −12(d2PQ + d2RS) +14(1 + cos dPS)(dQR − dPS)2+ cos dQR + cos dPS + Cub (dPQ, dRS, dQR − dSP ) .(B.1)Proof. It suffices to prove the inequality holds for a quadrilateral PQRS in S2. Byviewing S2 as a unit sphere in R3, the points P,Q,R, S determine a quadrilateral in97R3. Applying the identity for the quadrilateral in R3 (cf. [40, Corollary 2.1.3]),PR2+QS2 ≤ PQ2 +QR2 +RS2 + SP 2 − (SP −QR)2where AB denotes the Euclidean distance between A and B in R3. To prove this,consider the vectors A = Q− P,B = R−Q,C = S −R,D = P − S. ThenPR2+QS2=12(|A+B|2 + |C +D|2 + |B + C|2 + |D + A|2)= |A|2 + |B|2 + |C|2 + |D|2 + (A ·B + C ·B +D · A+D · C)= |A|2 + |B|2 + |C|2 + |D|2 − |B +D|2 since A+B + C +D = 0≤ |A|2 + |B|2 + |C|2 + |D|2 − ||B| − |D||2 .Note that AB2= 2− 2 cos dAB, we obtaincos dPR + cos dQS = −2 + cos dPQ + cos dRS + cos dQR + cos dPS+12(√2− 2 cos dQR −√2− 2 cos dSP)2.The lemma follows from the following Taylor expansion:−2 + cos dPQ + cos dRS = −12d2PQ −12d2RS +O(d4RS + d4PQ)(√2− 2 cos dQR −√2− 2 cos dSP)2=(sin dSP√2− 2 cos dSP(dQR − dSP ) +O((dQR − dSP )2))2=1 + cos dPS2(dQR − dSP )2 +O((dQR − dSP )3).Lemma B.1.2 ([60, Estimate I, Page 11]). Let PQRS be a quadrilateral in theCAT(1) space X. Let P 12be the mid-point between P and S, and let Q 12be themid-point between Q and R. Thencos2(dPS2)d2(Q 12, P 12) ≤ 12(d2PQ + d2RS)−14(dQR − dPS)298+ Cub(dPQ, dRS, d(P 12, Q 12), dQR − dSP).Proof. As a direct consequence of the law of cosines (see Figure B.1), we have thefollowing inequalitiescos d(Q 12, P 12) ≥ α(cos d(Q 12, S) + cos d(Q 12, P ))cos d(Q 12, S) ≥ β (cos dRS + cos dQS)cos d(Q 12, P ) ≥ β (cos dRP + cos dQP )whereα =12 cos(dPS2) and β = 12 cos(dQR2) .P SRQQ 12P 12Figure B.1: An illustration of the quadrilateral PQRS from Lemma B.1.2.Combining the above inequalities yieldscos d(Q 12, P 12) ≥ αβ (cos dRS + cos dQS + cos dRP + cos dQP ) .We apply (B.1) for the sum of diagonals cos dQS + cos dRP and Taylor expansion forcos dRS and cos dQP . It yieldscos d(Q 12, P 12) ≥ αβ(2− (d2PQ + d2RS) +14(1 + cos dPS)(dQR − dPS)2 + cos dQR + cos dPS)+ Cub (dPQ, dRS, dQR − dSP )99= αβ(2 + cos dQR + cos dPS +14(1 + cos dPS)(dQR − dPS)2)− αβ(d2PQ + d2RS) + Cub (dPQ, dRS, dQR − dSP ) .Note that2 + cos dQR + cos dPS +14(1 + cos dPS)(dQR − dPS)2= 2(cos2dQR2+ cos2dPS2) +12cos2dPS2(dQR − dPS)2= 2(cosdQR2− cos dPS2)2+ 4 cosdQR2cosdPS2+12cos2dPS2(dQR − dPS)2=12sin2dPS2(dQR − dPS)2 + 4 cos dQR2cosdPS2+12cos2dPS2(dQR − dPS)2+O(|dQR − dPS|3)=12(dQR − dPS)2 + 4 cos dQR2cosdPS2+O(|dQR − dPS|3).Since αβ = α2 +O(|dQR − dPS|), we havecos d(Q 12, P 12) ≥ 1− α2(d2PQ + d2RS) +12α2(dQR − dPS)2 + Cub (dPQ, dRS, dQR − dSP ) .The lemma follows ascos d(Q 12, P 12) = 1−d2(Q 12, P 12)2+O(d4(Q 12, P 12)).Definition B.1.3. Given a metric space (X, d) and a geodesic γPQ with dPQ < pi,for τ ∈ [0, 1] let (1 − τ)P + τQ denote the point on γPQ at distance τdPQ from P .That isd((1− τ)P + τQ, P ) = τdPQ.Lemma B.1.4 (cf. [60, Estimate II, Page 13]). Let ∆PQS be a triangle in the100CAT(1) space X. For a pair of numbers 0 ≤ η, η′ ≤ 1 definePη′ = (1− η′)P + η′QSη = (1− η)S + ηQ.Thend2(Pη′ , Sη) ≤ sin2((1− η)dQS)sin2 dQS(d2PS − (dQS − dQP )2)+ ((1− η)(dQS − dQP ) + (η′ − η)dQS)2 + Cub (dPS, dQS − dQP , η − η′) .Proof. Again we prove the inequality for a quadrilateral on S2. Denote x = dQS andy = dQP . Denoteαη =sin(ηdQS)sin dQS=sin(ηx)sinx, βη′ =sin(η′dQP )sin dQP=sin(η′y)sin y.Q SPSηPη′Figure B.2: An illustration of the triangle ∆PQS, and the points Sη and Pη′ fromLemma B.1.4.By the law of cosines on the sphere (see Figure B.2),cos dPS = cosx cos y + sinx sin y cos θ = cos(x− y) + sin x sin y(cos θ − 1)cos d(Pη′ , Sη) ≥ cos((1− η)x) cos((1− η′)y) + sin((1− η)x) sin((1− η′)y) cos θ= cos((1− η)x− (1− η′)y) + sin((1− η)x) sin((1− η′)y)(cos θ − 1),where θ denotes the angle ∠PQS on S2. Substituting the term (cos θ − 1) of the101second inequality with the one in the first identity, we obtaincos d(Pη′ , Sη) ≥ cos((1− η)x− (1− η′)y) + α1−ηβ1−η′(cos dPS − cos(x− y))= cos ((1− η)(x− y) + (η′ − η)x+ (η′ − η)(y − x))+ α21−η(cos dPS − cos(x− y))+ α1−η(β1−η′ − α1−η)(cos dPS − cos(x− y)).Using the Taylor expansion cos a = 1− a22+O(a4) and (β1−η′ −α1−η) = O(|η′− η|+|x− y|), we derivecos d(Pη′ , Sη) ≥ 1− ((1− η)(x− y) + (η′ − η)x)22+ α21−η(−d2PS2+(x− y)22)+ Cub (|η′ − η|, |x− y|, dPS) .It implies thatd2(Pη′ , Sη) ≤ α21−η(d2PS − (x− y)2) + ((1− η)(x− y) + (η′ − η)x)2+ Cub (|η′ − η|, |x− y|, dPS) .Corollary B.1.1. Let u : Ω→ Bρ(Q) be a finite energy map and η ∈ C∞C (Ω, [0, 1]).Define uˆ : Ω→ Bρ(Q) asuˆ(x) = (1− η(x))u(x) + η(x)Q.Then uˆ has finite energy, and for any smooth vector field W ∈ Γ(Ω) we have|uˆ∗(W )|2 ≤(sin(1− η)RusinRu)2(|u∗(W )|2 − |∇WRu|2) + |∇W ((1− η)Ru)|2,where Ru(x) = d(u(x), Q).Note that every error term that appeared in Lemma B.1.4 will converge to the102product of an L1 function and a term that goes to zero. So all error terms vanishwhen taking limits.Lemma B.1.5 (cf. [60, Estimate III, page 19]). Let PQRS be a quadrilateral ina CAT(1) space X. For η′, η ∈ [0, 1] defineQη′ = (1− η′)Q+ η′R, Pη = (1− η)P + ηS.Thend2(Qη′ , Pη) + d2(Q1−η′ , P1−η)≤(1 + 2ηdPS tan(12dPS))(d2PQ + d2RS)− 2η(1 +12dPS tan(12dPS))(dQR − dPS)2+ 2(2η − 1)(η′ − η)dPS(dQR − dPS)+ η2Quad(dPQ, dRS, dQR − dPS) + Cub (dQR − dPS, dPQ, dRS, η − η′)Proof. For notation simplicity, we denotex = dPS, y = dQR, αη =sin(ηx)sinx, βη′ =sin(η′y)sin y.Apply [60, Definition 1.6] to each of the blue, red, and yellow triangles in Figure B.3below.We derivecos d(Q1−η′ , P1−η) ≥ αη cos d(Q1−η′ , S) + α1−η cos d(Q1−η′ , P )≥ αη(βη′ cos dSR + β1−η′ cos dSQ) + α1−η(βη′ cos dPR + β1−η′ cos dPQ).Compute similarly for d(Qη′ , Pη) for the highlighted triangles below:103P SRQP1−ηQ1−η′PηQη′Figure B.3: An illustration of the quadrilateral PQRS, and the points Pη, P1−η,Qη′ and Q1−η′ from Lemma B.1.5.P SRQP1−ηQ1−η′PηQη′We derivecos d(Qη′ , Pη) ≥ αη cos d(Qη′ , P ) + α1−η cos d(Qη′ , S)≥ αη(βη′ cos dPQ + β1−η′ cos dPR) + α1−η(βη′ cos dSQ + β1−η′ cos dSR).Adding the above two inequalities, we obtaincos d(Q1−η′ , P1−η) + cos d(Qη′ , Pη)≥ (αηβη′ + α1−ηβ1−η′)(cos dPQ + cos dSR) + (αηβ1−η′ + α1−ηβη′)(cos dPR + cos dSQ).(B.2)Applying (B.1) to the term cos dPR + cos dSQ and using Taylor expansion, theinequality (B.2) becomescos d(Q1−η′ , P1−η) + cos d(Qη′ , Pη) ≥ (αηβη′ + α1−ηβ1−η′)(2− d2PQ2− d2SR2)104+ (αηβ1−η′ + α1−ηβη′)(−12(d2PQ + d2SR) +14(1 + cos dPS)(dQR − dPS)2)+ (αηβ1−η′ + α1−ηβη′) (cos dQR + cos dPS)+ Cub (dPQ, dRS, dQR − dSP ) .Hence,cos d(Q1−η′ , P1−η) + cos d(Qη′ , Pη)≥ −12(αηβη′ + α1−ηβ1−η′ + αηβ1−η′ + α1−ηβη′)(d2PQ + d2SR) (B.3)+ 2(αηβη′ + α1−ηβ1−η′) + (αηβ1−η′ + α1−ηβη′)(cos dQR + cos dPS)(B.4)+14(αηβ1−η′ + α1−ηβη′)(1 + cos dPS)(dQR − dPS)2 (B.5)+ Cub (dPQ, dRS, dQR − dSP ) .We need the following elementary trigonometric identities to compute (B.3),(B.4), (B.5):αηβη′ + α1−ηβ1−η′ =sin(η − 12)x sin(η′ − 12)y2 sin 12x sin 12y+cos(η − 12)x cos(η′ − 12)y2 cos 12x cos 12yαηβ1−η′ + α1−ηβη′ = −sin(η − 12)x sin(η′ − 12)y2 sin 12x sin 12y+cos(η − 12)x cos(η′ − 12)y2 cos 12x cos 12y(cos(η − 12)xcos 12x)2= 1 + 2ηx tan12x+O(η2).Noting thatαηβη′ + α1−ηβ1−η′ + αηβ1−η′ + α1−ηβη′ =cos(η − 12)x cos(η′ − 12)ycos 12x cos 12y=(cos(η − 12)xcos 12x)2+O(|η − η′|+ |x− y|)105= 1 + 2ηx tan(12x) +O(η2 + |η − η′|+ |x− y|),we obtain for (B.3)− 12(αηβη′ + α1−ηβ1−η′ + αηβ1−η′ + α1−ηβη′)(d2PQ + d2SR)= −12(1 + 2ηx tan(12x))(d2PQ + d2SR) +O((η2 + |η − η′|+ |x− y|)(d2PQ + d2SR)).Lemma B.1.6. We can compute (B.4) as follows:2(αηβη′ + α1−ηβ1−η′) + (αηβ1−η′ + α1−ηβη′)(cosx+ cos y)= 2−((η − 12)(y − x) + (η′ − η)x)2+sin2(η − 12)x4 sin2 12xcos2(12x)(x− y)2+cos2(η − 12)x4 cos2 12xsin2(12x)(x− y)2 +O(|x− y|2(|x− y|+ |η′ − η|)).Proof.2(αηβη′ + α1−ηβ1−η′) + (αηβ1−η′ + α1−ηβη′)(cosx+ cos y)=sin(η − 12)x sin(η′ − 12)y2 sin 12x sin 12y(2− cosx− cos y)+cos(η − 12)x cos(η′ − 12)y2 cos 12x cos 12y(2 + cos x+ cos y).Note that2− cosx− cos y = 2(sin 12x)2 + 2(sin12y)2 = 2(2 sin12x sin12y + (sin12x− sin 12y)2)= 4 sin12x sin12y +12(cos12x)2(x− y)2 +O(|x− y|3)2 + cos x+ cos y = 2(cos12x)2 + 2(cos12y)2 = 2(2 cos12x cos12y + (cos12x− cos 12y)2)= 4 cos12x cos12y +12(sin12x)2(x− y)2 +O(|x− y|3),106where we apply Taylor expansion in the last equality. Hence we have2(αηβη′ + α1−ηβ1−η′) + (αηβ1−η′ + α1−ηβη′)(cosx+ cos y)= 2(sin(η − 12)x sin(η′ − 12)y + cos(η − 12)x cos(η′ − 12)y)+sin2(η − 12)x4 sin2 12x(cos12x)2(x− y)2 + cos2(η − 12)x4 cos2 12x(sin12x)2(x− y)2+O(|x− y|2(|x− y|+ |η′ − η|)).Here we use the estimatessin(η − 12)x sin(η′ − 12)y2 sin 12x sin 12y− sin2(η − 12)x2 sin2 12x= O(|η − η′|+ |x− y|)andcos(η − 12)x cos(η′ − 12)y2 cos 12x cos 12y− cos2(η − 12)x2 cos2 12x= O(|η − η′|+ |x− y|).Observe that(sin(η − 12)x sin(η′ − 12)y + cos(η − 12)x cos(η′ − 12)y)= cos((η − 12)(y − x) + (η′ − η)x+ (η′ − η)(y − x))and use cos a = 1− a22+O(a4).Lemma B.1.7. Adding the terms in the previous computational lemma that contain(x− y)2 to (B.5), we have the following estimate:14(αηβ1−η′ + α1−ηβη′)(1 + cos x)(x− y)2− (η − 12)2(x− y)2 + sin2(η − 12)x4 sin2 12xcos2(12x)(x− y)2 + cos2(η − 12)x4 cos2 12xsin2(12x)(x− y)2= η(1 +12x tan12x)(x− y)2 +O(|x− y|2(η2 + |x− y|+ |η − η′|)).107Proof. Noting that 1 + cosx = 2 cos2(12x), we have that14(αηβ1−η′ + α1−ηβη′)(1 + cos x)(x− y)2=14(−(sin(η − 12)xsin 12x)2+(cos(η − 12)xcos 12x)2)cos2(12x)(x− y)2+O(|x− y|2(|η − η′|+ |x− y|)).Therefore,14(αηβ1−η′ + α1−ηβη′)(1 + cos x)(x− y)2− (η − 12)2(x− y)2 + sin2(η − 12)x4 sin2 12xcos2(12x)(x− y)2 + cos2(η − 12)x4 cos2 12xsin2(12x)(x− y)2=(cos2(η − 12)x4 cos2 12x− (η − 12)2)(x− y)2 +O(|x− y|2(|η − η′|+ |x− y|))=(14+12ηx tan12x− (−η + 14))(x− y)2 +O(|x− y|2(η2 + |η − η′|+ |x− y|)).Combing the above computations, we have thatcos d(Q1−η′ , P1−η) + cos d(Qη′ , Pη) ≥ 2− 12(1 + 2ηdPS tan(12dPS))(d2PQ + d2SR)+ η(1 +12dPS tan12dPS)(dQR − dPS)2− (2η − 1)(η′ − η)dPS(dQR − dPS)+ η2Quad(dPQ, dRS, dQR − dPS)+ Cub (dQR − dPS, dPQ, dRS, η′ − η) .Taylor expansion gives the result.Corollary B.1.2. Given a pair of finite energy maps u0, u1 ∈ W 1,2(Ω, X) with108images ui(Ω) ⊂ Bρ(Q) and a function η ∈ C1c (Ω), 0 ≤ η ≤ 12 , define the mapsuη(x) = (1− η(x))u0(x) + η(x)u1(x)u1−η(x) = η(x)u0(x) + (1− η(x))u1(x)d(x) = d(u0(x), u1(x)).Then uη, u1−η ∈ W 1,2(Ω, X) and|∇uη|2 + |∇u1−η|2 ≤ (1 + 2ηd tan d2)(|∇u0|2 + |∇u1|2)− 2η(1 + 12d tand2)|∇d|2 − 2d∇η · ∇d+ Quad(η, |∇η|).B.2 Energy Convexity, Existence, Uniqueness, andSubharmonicityAs with the previous section, the results in this section are stated in [60]. Exceptingthe first theorem, they are stated without proof. As, again, the calculations arenon-trivial and tedious, we verify them for the reader.Theorem B.2.1 ( [60, Proposition 1.15]). Let u0, u1 : Ω → Bρ(O) be finite energymaps with ρ ∈ (0, pi2). Denote byd(x) = d(u0(x), u1(x))R(x) = d(u 12(x), O).Then there exists a continuous function η(x) : Ω → [0, 1] such that the functionw : Ω→ Bρ(O) defined byw(x) = (1− η(x))u 12(x) + η(x)O109is in W 1,2(Ω, Bρ(O)) and satisfies(cos8 ρ)∫Ω∣∣∣∣∇tan 12dcosR∣∣∣∣2 dµg ≤ 12(∫Ω|∇u0|2dµg +∫Ω|∇u1|2dµg)−∫Ω|∇w|2dµg.Proof. Once the estimates in Lemma B.1.2 and Lemma B.1.4 are established, weproceed as in [60]. Choose η to satisfysin((1− η(x))R(x))sinR(x)= cosd(x)2.Note that 0 ≤ η ≤ 1 and η is as smooth as d(x), R(x). It is straightforward to verifythat w ∈ L2h(Ω, Bρ(O)).For W ∈ Γ(Ω), consider the flow  7→ x() induced by W .u0(x())u1(x())u1(x)u0(x)u 12(x)u 12(x())Ow(x)w(x())Figure B.4: An illustration of the quadrilateral PQRS with P = u0(x()), Q =u0(x), R = u1(x) and S = u1(x()) used in the proof of Lemma B.2.1.Applying Lemma B.1.2 to the quadrilateral determined by P = u0(x()), Q =u0(x), R = u1(x), S = u1(x()) (see Figure B.4), divided by 2, and integrate theresulting inequality against f ∈ C∞c (Ω) and taking → 0, we obtain(cosd(x)2)2|(u 12)∗(W )|2 ≤ 12(|(u0)∗(W )|2 + |(u1)∗(W )|2)− 14|∇Wd|2.Note that the cubic terms vanish in the limit as every cubic term will be the productof an L1 function and d(x)− d(x()) or d(ui(x), ui(x())), i = 0, 12 , 1.110Applying Lemma B.1.4 to the triangle determined by Q = O,P = u 12(x), S =u 12(x()) yields|(w)∗(W )|2 ≤(sin(1− η)RsinR)2(|(u 12) ∗ (W )|2 − |∇WR|2) + |∇W ((1− η)R)|2=(cosd(x)2)2(|(u 12)∗(W )|2 − |∇WR|2) + |∇W ((1− η)R)|2.The above two inequalities imply|w∗(W )|2 ≤ 12(|(u0)∗(W )|2 + |(u1)∗(W )|2)− 14|∇Wd|2 −(cosd(x)2)2|∇WR|2 + |∇W ((1− η)R) |2.By direct computation,− 14|∇Wd|2 −(cosd(x)2)2|∇WR|2 + |∇W ((1− η)R) |2= − cos4R(x) cos4 d(x)21− sin2R(x) cos2 d(x)2∣∣∣∣∣∇ tan d(x)2cosR(x)∣∣∣∣∣2.The lemma follows from estimatingcos4R(x) cos4 d(x)21− sin2R(x) cos2 d(x)2≥ cos4R(x) cos4 d(x)2≥ cos8 ρ,dividing the resulting inequality by 2, integrating over Sn−1, letting → 0, and thenintegrating over Ω.Theorem B.2.2 (Existence Theorem). For any ρ ∈ (0, pi4) and for any finite energymap h : Ω→ Bρ(O) ⊂ X, there exists a unique element Dirh ∈ W 1,2h (Ω,Bρ(O)) whichminimizes energy amongst all maps in W 1,2h (Ω,Bρ(O)).Moreover, for any σ ∈ (0, ρ), if Dirh(∂Ω) ⊂ Bσ(O) then Dirh(Ω) ⊂ Bσ(O).111Proof. Denote byE0 = inf{E(u) : u ∈ W 1,2h (Ω,Bρ(O))}.Let ui ∈ W 1,2(Ω,Bρ(P )) such that E(ui)→ E0. By Theorem B.2.1, we have that(cos8 ρ)∫Ω∣∣∣∣∇tan 12d(uk(x), u`(x))cosR∣∣∣∣ dµg ≤ 12 (E(uk) + E(u`))− E(wk`),where wk` is the interpolation map defined by Theorem B.2.1. The above right handside goes to 0 as k, `→∞. By the Poincare´ inequality,∫Ωd(uk, u`) dµg → 0.Thus the sequence {uk} is Cauchy and uk → u for some u ∈ W 1,2(Ω,Bρ(O)) becauseW 1,2(Ω,Bρ(O)) is a complete metric space. By trace theory, u ∈ W 1,2h (Ω,Bρ(O)). Bylower semi-continuity of the energy, E(u) = E0. The energy minimizer is unique byenergy convexity.Finally, since ρ < pi4, for any σ ∈ (0, ρ], the ball Bσ(O) is geodesically convex.Therefore, the projection map piσ : Bρ(O) → Bσ(O) is well-defined and distancedecreasing. Thus, since Dirh(Ω) ⊂ Bρ(O), we can prove the final statement bycontradiction using the projection map to decrease energy.Lemma B.2.3 (cf. [60, (2.5)]). Let u0, u1 : Ω → Bρ(Q) ⊂ X be finite energymaps (possibly with different boundary values). For any given η ∈ C∞c (Ω) with0 ≤ η < 1/2, there exists finite energy maps uη, uˆη ∈ W 1,2u0 (Ω,Bρ(Q)) and u1−η, uˆ1−η ∈W 1,2u1 (Ω,Bρ(Q)) such that|pi(uˆη)|2 + |pi(uˆ1−η)|2 − |pi(u0)|2 − |pi(u1)|2≤ −2 cosRuη cosRu1−η∇(dsin dηFη)· ∇Fη + Quad(η,∇η),whered(x) = d(u0(x), u1(x))112Ruη(x) = d(uη(x), Q)Ru1−η(x) = d(u1−η(x), Q)andFη =√1− cos dcosRuη cosRu1−η.Proof. Let η ∈ C∞c (Ω) satisfy 0 ≤ η < 1/2. For 0 ≤ φ, ψ ≤ 1 that will be determinedbelow, we define the comparison mapsuˆη = (1− φ(x))uη(x) + φ(x)Quˆ1−η = (1− ψ(x))u1−η(x) + ψ(x)Q,whereuη(x) = (1− η(x))u0(x) + η(x)u1(x) and u1−η(x) = η(x)u0(x) + (1− η(x))u1(x).By Corollary B.1.1,|pi(uˆη)|2 + |pi(uˆ1−η)|2 ≤(sin(1− φ)RuηsinRuη)2(|pi(uη)|2 − |∇Ruη |2) + |∇((1− φ)Ruη)|2+(sin(1− ψ)Ru1−ηsinRu1−η)2(|pi(u1−η)|2 − |∇Ru1−η |2)+ |∇((1− ψ)Ru1−η)|2.Define φ and ψ so thatsin2((1− φ)Ruη)sin2Ruη= 1− 2ηd tan d2+O(η2)sin2((1− ψ)Ru1−η)sin2Ru1−η= 1− 2ηd tan d2+O(η2).113Since sin(1−a)θsin θ= 1− aθ cot θ +O(a2), we solveφ = ηtanRuηRuηd tand2and ψ = ηtanRu1−ηRu1−ηd tand2.Note that in particular uη, uˆη ∈ W 1,2u0 (Ω,Bρ(Q)) and u1−η, uˆ1−η ∈ W 1,2u1 (Ω,Bρ(Q)).Together with the estimate for |pi(uη)|2 + |pi(u1−η)|2 in Corollary B.1.2 (which alsoexplains the choice of φ and ψ in order to eliminate the coefficient), we have|pi(uˆη)|2 + |pi(uˆ1−η)|2 − |pi(u0)|2 − |pi(u1)|2≤ −2η(1 + 12d tand2)|∇d|2 − 2d∇η · ∇d− (1− 2ηd tan d2)(|∇Ruη |2 + |∇Ru1−η |2)+ |∇(1− η tanRuηRuηd tand2)Ruη |2 + |∇(1− η tanRu1−ηRu1−ηd tand2)Ru1−η |2 + Quad(η, |∇η|).Simplifying the expression and using 1− sec2 θ = − tan2 θ , we obtain12(|pi(uˆη)|2 + |pi(uˆ1−η)|2 − |pi(u0)|2 − |pi(u1)|2)≤ η(− (1 + 12d tand2)|∇d|2 − d tan d2(tan2Ruη |∇Ruη |2 + tan2Ru1−η |∇Ru1−η |2)−∇(d tan d2) · (tanRuη∇Ruη + tanRu1−η∇Ru1−η))+∇η ·(−d∇d− tanRuηd tan d2∇Ruη − tanRu1−ηd tan d2∇Ru1−η)+ Quad(η,∇η).(B.6)We hope to find a, b, Fη which are functions of d,Ruη and Ru1−η such that the righthand side above is ≤ a∇(bηFη) · ∇Fη.Since a∇(bηFη) · ∇Fη = η(ab|∇Fη|2 + a2∇b · ∇F 2η ) + ab2 ∇η · ∇F 2η , by comparingthe terms involving ∇η in (B.6), we solveab2∇η · ∇F 2η = ∇η ·(−d∇d− tanRuηd tan d2∇Ruη − tanRu1−ηd tan d2∇Ru1−η)= −d tan d2∇η ·(∇ log sin2 d2−∇ log cosRuη −∇ log cosRu1−η)114= − dsin dcosRuη cosRu1−η∇η · ∇ 1− cos dcosRuη cosRu1−η,where we use 2 sin2 d2= (1− cos d) and tan d2= 1−cos dsin d. It suggests us to chooseab2= − dsin dcosRuη cosRu1−η and Fη =√1− cos dcosRuη cosRu1−η.We then compute the term η(ab|∇Fη|2 + a2∇b · ∇F 2η ) for the above choices of a, b,and Fη. For the term ab|∇Fη|2, we computeab|∇Fη|2 = − d2 sin d(1− cos d) |sin d∇d+ (1− cos d)(tanRuη∇Ruη + tanRu1−η∇Ru1−η)|2≥ −(d sin d2(1− cos d) |∇d|2 + d∇d · (tanRuη∇Ruη + tanRu1−η∇Ru1−η)+d(1− cos d)sin d(tan2Ruη |∇Ruη |2 + tan2Ru1−η |∇Ru1−η |2)),where we expand the quadratic term and use the AM-GM inequality to handle thecross term (tanRuη∇Ruη) · (tanRu1−η∇Ru1−η). For the term a2∇b · ∇F 2η , we assumeb = b(d) and compute:a2∇b · ∇F 2η =ab2∇ log b · ∇F 2η= −db′b|∇d|2 − d(1− cos d)sin db′b∇d · (tanRuη∇Ruη + tanRu1−η∇Ru1−η).Combining the above inequalities, we obtainab|∇Fη|2 + a2∇b · ∇F 2η ≥ −[(d sin d2(1− cos d) + db′b)|∇d|2+(d+d(1− cos d)sin db′b)∇d · (tanRuη∇Ruη + tanRu1−η∇Ru1−η)+d(1− cos d)sin d(tan2Ruη |∇Ruη |2 + tan2Ru1−η |∇Ru1−η |2)].115Comparing to (B.6), we solved sin d2(1− cos d)∇d+ d∇ log b = (1 +12d tand2)∇dd∇d+ d(1− cos d)sin d∇ log b = ∇(d tan d2).which implies that b = dsin d, and hence a = −2 cosRuη cosRu1−η .Theorem B.2.4 (cf. [60, Corollary 2.3]). Let u0, u1 : Ω → Bρ(P ) ⊂ X be a pairof energy minimizing maps (possibly with different boundary values). Let d(x) =d(u0(x), u1(x)) and Rui = d(ui, P ). Then the functionF =√1− cos dcosRu0 cosRu1satisfies the differential inequality weaklydiv(cosRu0 cosRu1∇F ) ≥ 0.Proof. Let η ∈ C∞c (Ω) with η ≥ 0. For t > 0 sufficiently small, we have 0 ≤ tη < 1/2.Let uˆtη and uˆ1−tη be the corresponding maps defined as in Lemma B.2.3. Since u0and u1 minimize the energy among maps of the same boundary values, we have0 ≤∫Ω|pi(uˆη)|2 + |pi(uˆ1−η)|2 − |pi(u0)|2 − |pi(u1)|2 dµg≤∫Ω−2 cosRutη cosRu1−tη∇(dsin dtηFtη)· ∇Ftη dµg + t2Quad(η,∇η).Dividing the inequality by t and let t→ 0, since Rutη → Ru0 and Ru1−tη → Ru1 andFtη → F , we derive0 ≤∫Ω−2 cosRu0 cosRu1∇(dsin dηF)· ∇F dµg116= 2∫Ω(dsin dηF)div (cosRu0 cosRu1∇F ) dµg.117

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