MINIMAL SURFACES – SS 12

iiKARSTEN GROSSE-BRAUCKMANN1.4. Differential geometry in terms of the Weierstrass representation 511.5. The associated family 542. Symmetry properties of minimal surfaces 572.1. Schwarz reflection for holomorphic functions 572.2. Schwarz reflection for minimal surfaces 592.3. Symmetry properties in terms of Weierstrass data 613. Surfaces constructed from Weierstrass data 633.1. Higher order Enneper surfaces 633.2. Simply periodic Scherk surfaces 654. Costa surface (2006) 684.1. Elliptic functions 684.2. Existence of elliptic functions 694.3. The Costa surface 735. Global Weierstrass representation 815.1. Riemann surfaces 815.2. Minimal surfaces and Riemann surfaces 846. On the proof of Osserman’s theorem (2006) 856.1. Divergent paths 856.2. Existence of a harmonic function 876.3. Ends 907. Conjugate surface constructions (2006) 937.1. Conjugate Plateau solutions 93Part 3. Solution of Plateau’s problem 951. Introduction 951.1. The problem 951.2. Area and energy 961.3. Rigorous formulation of Plateau’s problem 982. Harmonic and conformal maps 992.1. The divergence theorem 992.2. Mean value formula 1002.3. Conformality in dimension two 1022.4. The conformal group of the disk 1042.5. The three-point-condition 1052.6. Poisson’s formula 1072.7. Variational characterization and Dirichlet’s principle 1092.8. Analyticity of harmonic maps and minimal surfaces 111

MINIMAL SURFACES – SS 12iii3. Solution to Plateau’s problem 1123.1. Courant-Lebesgue-Lemma and equicontinuity 1133.2. Construction of a minimizer 1153.3. The minimizer is weakly conformal 1163.4. The minimizer is injective on the boundary 1203.5. Summary of the proof 1224. Properties of the Plateau solution (2006) 1224.1. Riemann mapping theorem 1224.2. Henneberg’s surface is non-orientable and has branch points 1254.3. Branch points and generalized minimal surfaces 1284.4. Branch points in Plateau solutions 1294.5. Uniqueness and embeddedness results 1335. Existence of complete periodic minimal surfaces (2006) 1355.1. The reflection principle 1355.2. Plateau solutions with polygonal boundary: Extension and symmetries 1375.3. The D-surface 1405.4. The P -surface 1435.5. Other periodic surfaces generated from polygons 1445.6. Symmetries, lattices, and crystallographic groups 144

ivKARSTEN GROSSE-BRAUCKMANNReferencesBooks surveying minimal surfaces:[DHS] Dierkes, Hildebrandt, Sauvigny: Minimal surfaces I, Springer 1992, 2010, (A broadpresentation of the theory, accompanied with many examples and pictures.)[EJ] Eschenburg/Jost: Differentialgeometrie und Minimalflächen, Springer 1994, 2007 (Sections8 to 10 of this textbook deal with minimal surfaces.)[O] Osserman: Minimal surfaces, Dover (Unlike Jost, the focus lies on the complex methods,in particular the discussion of the Gauss map. The presentation is concise.)[O1] Oprea: The mathematics of soap films: Explorations with Maple, AMS 2000 (A textmore elementary than our class.)[L] Lawson: Minimal submanifolds, Publish or Perish 1980 (The first chapter deals withminimal submanifolds of Riemannian manifolds. The second chapter is much more specificand contains the solution of the Plateau problem in R 3 .)[N] Nitsche: Vorlesungen über Minimalflächen, Springer 1975 (An English translation ofthe first half appeared at Cambridge. A text not aimed at students. The expositionseems not accessible on the first place; but with longtime usage the values of this opus areobvious.)[Ka] Karcher: Construction of minimal surfaces, Tokyo 1989[Ku] Kuwert: Vorlesung Minimalflächen 1998http://home.mathematik.uni-freiburg.de/analysis/lehre/skripten/[St] Struwe: Plateau’s problem and the calculus of variations, Princeton 1988Articles surveying modern research:[HK] Hoffman, Karcher: In Geometry V[R] H. Rosenberg: Some recent developments in the theory of minimal surfaces in 3-manifolds, Publicações Matemáticas, Impa 2003[CM] T. Colding, W. Minicozzi: Shapes of minimal surfaces, Proceedings National Academyof Sciences 103 (2006), 11106-11Books introducing general theory needed for minimal surfaces:[GT] Gilbarg, Trudinger: Elliptic partial differential equations of second order, Springer1977, 1983 (contains all the results about partial differential equations we need.)

MINIMAL SURFACES – SS 12v[A] Ahlfors: Complex analysis, McGraw-Hill 1953, 1979[FL] Fischer-Lieb: Funktionentheorie, Vieweg[FK] Farkas-Kra: Riemann surfaces, Springer 1980, 1992[Sy] Sauvigny: Partielle Differentialgleichung in der Physik und Geometrie 1+2, Springer2004[M] Morgan: Geometric measure theory, Academic Press 1988, 2000Popular books:[HT] Hildebrandt, Tromba: Kugel, Kreis und Seifenblasen, Birkhäuser 1996

viKARSTEN GROSSE-BRAUCKMANNIntroductionThese are notes of lectures given in 2005/06 (two terms) and 2012 (one term) at Darmstadt.Besides the material revised for the class of 2012 it contains all extra material presentedin the earlier class.In 2005 I had covered the Plateau problem before the Weierstrass representation, while Iswitched this order in 2012. This leads to inconsistencies in the present version.My goal was an elementary introduction to minimal surfaces. The only knowledge assumedis a course in classical surface theory.In Part 1, the equations for mean curvature are presented, followed by the most classicalexamples of minimal surfaces which are catenoid, helicoid, Scherk, Enneper. Then thetwo most important tools are introduced: The calculus of variations and the maximumprinciple for the mean curvature equation.In Part 2 the Weierstrass representation is introduced in its local form. Three examplesare discussed: Higher order Enneper surfaces, the simply periodic Scherk surface, and theCosta surface. The latter is defined on a torus, and so the necessary theory of ellipticfunctions (on a square lattice) is derived from the Riemann mapping theorem. An outlookon the global form is included.In Part 3 the Plateau problem is discussed. The proof relies on properties of harmonicfunctions which are supplied in a separate section. Then branch points are discussed ina non-rigorous way. I use the Plateau solution to construct complete periodic minimalsurfaces: A polygonal boundary bounds a Plateau solution which may be refleted acrossits boundary arcs. The Schwarz P - and D surfaces provide examples.I couldn’t cover some other parts of the theory, like the Bernstein theorem, the existenceof conformal coordinates on minimal surfaces, or force/flux.I thank Steffen Fröhlich and Matthias Bergner for discussions, suggestions, and corrections,as well as the students Britta Michel, Julia Plehnert, Michael Fichtner and FerdinandHrubes. Besides the accessible sources listed in the references, I made use of notes of alecture given by F. Sauvigny in 1997.Darmstadt, July 06 and 12

i 1.1 – date: July 27, 2012 1Part 1. Minimal surfaces: Basic properties and examples1. Lecture, Wednesday 11.4.121. Minimal surfaces and equations for mean curvature1.1. Definition of mean curvature and minimality. We review the notions of surfacetheory. Throughout we will use U ⊂ R n for a domain [Gebiet], that is for an openand (arc-)connected subset. Locally, a (parameterized) hypersurface is a mapping f ∈C 1 (U, R n+1 ) which is an immersion, i.e., the differential df p : R n → R m has rank n for eachp ∈ U.To a hypersurface f we associate various quantities w.r.t. each parameter point p ∈ U:• Tangent space T p f := {df p (X) : X ∈ R n }, and normal space N p f := T p f ⊥ .• First fundamental form [erste Fundamentalform] g p (X, Y ) := 〈 df p (X), df p (Y ) 〉 and the(Riemannian) length ‖X‖ p := |df p (X)| = √ g p (X, X)• The Gauss map or normal mapping of f is a continuous map ν ∈ C 0 (U, S n ∩ N p f). Forn = 2, we can set ν := (f x × f y )/|f x × f y |. We do not insist on an orientation conventionfor ν.• The shape operator [Weingartenabbildung] w.r.t. ν is the mappingS : U × R n → R n , X ↦→ S p (X) := −(df p ) −1( dν p (X) ) .• Second fundamental form [zweite Fundamentalform] b(X, Y ) := 〈 ν, d 2 f(X, Y ) 〉 .• A principal curvature direction [Hauptkrümmungsrichtung] is an eigenvector v ∈ R n \{0}of S p , a principal curvature [Hauptkrümmung] an eigenvalue κ(p) ∈ R.Definition. The mean curvature [mittlere Krümmung] of a hypersurface f : U → R n+1 is(1) H(p) := 1 n trace S p,where the shape operator S is taken w.r.t. a choice of normal ν.In terms of principal curvatures κ 1 (p), . . . , κ n (p) the mean curvature readsH = 1 n(κ1 (p) + . . . + κ n (p) ) .While H depends on the choice of ν, the mean curvature vector H := Hν does not. Unlikethe scalar mean curvature, this vector can still be defined on an unoriented (global) surface.Another importanat curvature notion is the Gauss curvature K(p) := det S p .We recall that mean curvature is a geometric quantity. First, if we subject f to a motion,˜f := Af +b, where A ∈ SO(n), b ∈ R n , then d ˜f = A df becomes the differential and ˜ν := Aν

2 K. Grosse-Brauckmann: Minimal surfaces, SS 12a Gauss map. Thus ˜S = −(Adf) −1 Adν = −df −1 (A −1 A)dν = S remains unchanged.Second, under a change of coordinates, ˜f := f ◦ ϕ a Gauss map is ˜ν := ν ◦ ϕ and so˜S = −(dfdϕ) −1 dνdϕ = −dϕ −1 (df −1 dν)dϕ = dϕ −1 S dϕremains similar [ähnlich]. Note that similar endomorphisms have the same trace.A (global) hypersurface is a submanifold M n ⊂ R n+1 . Each point q ∈ M has a neighbourhoodwhich can be parameterized by an immersion f : U n → R n+1 together with a Gaussmap ν. The surface is orientable if all transition maps between overlapping parameterizationsare orientation preserving (have a Jacobian with positive determinant). Equivalentlyfor hypersurfaces, the Gauss map coincides on the overlap of parameterizations so thatit becomes defined globally. In any case, the invariance of H under reparameterizationimplies that the condition H = 0 is well-defined irrespective of the parameterization ornormal chosen.Remark. Our definition is limited to embedded global hypersurfaces. It needs more effortto define immersed hypersurfaces: This is the image of a domain manifold (which is anabstract manifold or submanifold) under an immersion into euclidean space.The fundamental definition for this class is:Definition. A parameterized hypersurface f : U → R n+1 is minimal if H = 0 at all points.A global hypersurface M ⊂ R n+1 is minimal if its local parameterizations are minimal.In the two-dimensional case we say minimal surface [Minimalfläche].Minimal surfaces were considered long before mean curvature was introduced. In the 18thcentury, minimal surfaces were defined in terms of the area minimizing property. Thisexplains their name, but will be discussed here only later. In the early 19th centurythe differential geometry of surfaces was systematically introduced, thereby attributing ageometric meaning to the previously known equations.Let us note a simple geometric property enjoyed by minimal surfaces.Proposition 1. Suppose f is a two-dimensional surface. Then H(p) = 0 is equivalentto the existence of two orthogonal asymptotic directions at p. Rotating them by 45-degreesgives two curvature directions.Note that our phrasing is correct in case the two principal curvatures coincide.Proof. By definition, a unit vector v ∈ T p f is an asymptotic vector if the normal curvatureκ norm (v) = g(Sv, v) vanishes. Let v 1 ⊥ v 2 be two orthogonal unit principal curvature

i 1.2 – date: July 27, 2012 3vectors. The Euler formula gives the following expression for the normal curvature of theunit vector v α = cos αv 1 + sin αv 2 at p for any α ∈ R:κ norm (v α ) = κ 1 cos 2 α+κ 2 sin 2 α = 1 2 (κ 1 +κ 2 )+ 1 2 (κ 1 −κ 2 ) cos(2α) = H + 1 2 (κ 1 −κ 2 ) cos(2α)Now if H vanishes then clearly κ norm (v α ) has exactly the zeros {α = π/4 mod π/2}, spacedπ/2 apart. For the converse note first that the Euler formula is π-periodic in α. So wemay assume the two zeros occur with α ∈ [0, π]. But cos 2α = −H holds exactly for α andπ − α. Thus if the difference is exactly π/2 then α = π/4 or 3π/4, and H = 0. □1.2. Mean curvature in local coordinates. Consider a standard basis e 1 , . . . , e n , correspondingto the coordinate directions in U n . We write partial derivatives of f(x) as∂ i f := ∂f∂x i= df(e i ). In case of a two dimensions we usually write f(x, y) for the surfaceand denote partials by subscript, e.g. f x := ∂f . ∂xWe have the following matrix representations of the fundamental forms:g ij (p) := g p (e i , e j ) = 〈∂ i f, ∂ j f〉b ij (p) := b(e i , e j ) = − 〈 ∂ i ν, ∂ j f 〉 = 〈 ν, ∂ ij f 〉 .We denote the matrix inverse to g with (g ij ). We associate to S the matrix S i j determinedby S(e j ) = ∑ ni=1 Si je i . Calculating b(e k , Se j ) (exercise!) then gives(2) S i j = ∑ kg ik b kj ,that is, we obtain a matrix equation b = gS or S = g −1 b. Plugging (2) into (1) yields:Proposition 2. Principal curvature directions are eigenvectors of the matrix ( ∑ k gik b kj ) ijand the mean curvature is given by(3) H = 1 ∑g ij b ij .ni,jIn dimension 2, the inverse g −1 of the first fundamental form has the simple representation(g −1 = 1 g22 −g 12)det g −g 21 g 11 . This leads to a differential equation for H:Corollary 3. The mean curvature of a surface f : U 2 → R 3 satisfiesH = 12 det g (g 22b 11 − 2g 12 b 12 + g 11 b 22 )(4)= |f y| 2〈 f xx , ν 〉 − 2〈f x , f y 〉 〈 f xy , ν 〉 + |f x | 2〈 f yy , ν 〉2 ( |f x | 2 |f y | 2 − 〈f x , f y 〉 2) .Thus a surface is minimal if and only if0 ≡ |f y | 2〈 f xx , ν 〉 − 2〈f x , f y 〉 〈 f xy , ν 〉 + |f x | 2〈 f yy , ν 〉 .

4 K. Grosse-Brauckmann: Minimal surfaces, SS 12The curvature of a regular curve satisfies ν ′ = −κc ′ , or, in case of arclength parameterization,〈ν ′ , c ′ 〉 = −κ.When we generalize the right hand side of this equaiton to mean curvature, the left handside becomes a divergence:Proposition 4. For a hypersurface f with normal ν we haveDiv ν(p) :=n∑〈dν p (v i ), df p (v i )〉 = −nH(p),i=1where v i is any g p -orthonormal basis of R n at p.In case (v i ) is an orthonormal basis of principal curvature directions, the identity is obvious:dν(v i ) = −df(Sv i ) = −κ i df(v i ).Problem. Check that with respect to the basis (e i ) we have Div ν = ∑ ij gij 〈dν(e i ), df(e j )〉.Also, gain geometric intuition for Div ν on a minimal surface.Proof. Suppose that f is a parameterization such that (e i ) is orthonormal at p. For thatcase g ij (p) = δ ij and so, at the point p:∑(5) 〈dν(e i ), df(e i )〉 = − ∑ 〈df(S k i e k ), df(e i )〉 = − ∑ S k i g ki = − ∑ S i i = −nH.ii,kiiFor the general case we assume p = 0 and consider ˜f := f ◦ A and ˜ν := ν ◦ A, whereA ∈ GL(n) is the linear map with Ae i = v i . This means that at ˜p := p = 0 the basis e i is˜g˜p -orthonormal and so by the above−nH(p) (5)= ∑ i〈d˜ν(e i ), d ˜f(e i )〉 = ∑ i〈dν(Ae i ), df(Ae i )〉 = ∑ i〈dν(v i ), df(v i )〉.□It is justified to call ∑ 〈dν p (v i ), df p (v i )〉 the surface or Riemannian divergence of ν. Tosee this, evaluate the expression for reparameterization ( ˜f, ˜ν) as in the proof. Then (e i )is orthonormal, and d˜ν(e i ) = ∂ i˜ν. Moreover, for any tangent vector w ∈ T p f the scalarproduct 〈w, ∂ i ˜f〉 with the unit vector ∂i ˜f can be regarded an i-th component of w takenon the surface. So indeed we have generalized the euclidean divergence div ˜ν = ∑ i ∂ i˜ν i .But Div ν is parameterization invariant, unlike div ν.2. Lecture, Monday 16.4.12

i 1.3 – date: July 27, 2012 51.3. Mean curvature in conformal coordinates. Angle preserving parametrisationswill become important for us:Definition. An immersion f ∈ C 1 (U n , R m ) is conformal [konform] if there is a conformalfactor λ ∈ C 0 (U, (0, ∞)) such thatg ij (p) = λ(p)δ ij for all p ∈ U, 1 ≤ i, j ≤ n.A mapping f ∈ C 1 (U n , R m ) is called weakly conformal if the same holds with λ ≥ 0.A two-dimensional surface f is conformal if and only if g = ( λ 0 λ 0 ) with λ > 0, or(6) |f x | ≡ |f y | > 0, 〈f x , f y 〉 ≡ 0.In the weakly conformal case, |f x | ≡ |f y | ≥ 0 and 〈f x , f y 〉 ≡ 0.Examples. 1. A holomorphic map f : U ⊂ C → C is weakly conformal, a biholomorphicmap is conformal (explain!).2. Polar coordinates P : (0, ∞) × R → R 2 , P (r, ϕ) = (r cos ϕ, r sin ϕ) are not conformalsince |∂P/∂r| = 1 but |∂P/∂ϕ| = r. On the other hand the complex exponential functionexp(r, ϕ) := (e r cos ϕ, e r sin ϕ) is conformal.3. For a parameterized surface, the notions conformal and weakly conformal coincide.According to a deep result, any two-dimensional surface has a conformal parameterizationdefined over a suitable domain, see [Sy2, XII,§8, Satz 2]. Therefore, in two dimensions,conformal parameterizations can play a role similar to the arc-length parameterization forcurves.For a conformal parameterization the mean curvature equation simplifies considerably:Theorem 5. Suppose f ∈ C 2 (U 2 , R n ) is a conformal parameterization of a surface withconformal factor λ. Then f satisfies the parametric mean curvature equation1(7)∆f = 2Hν for all p ∈ U.λ2 where ∆f = f xx + f yy is the standard Laplacian. Consequently, f ∈ C 2 (U 2 , R n ) is minimalif and only if its component functions are harmonic [harmonisch], ∆f = 0.Specifically, if n = 3 and the Gauss map is chosen positively oriented, i.e., det(f x , f y , ν) > 0,then λ 2 ν = |f x ||f y |ν = f x × f y and so (7) attains the form∆f = 2Hf x × f y for all p ∈ U.Example. We check harmonicity on the simple example of a plane in R 3 , spanned by twoorthonormal vectors v, w ∈ R 3 . Then f(x, y) = xv + yw is a conformal parameterization ofthe plane. The three coordinate functions f i (x, y) = xv i + yw i are linear, hence harmonic.

6 K. Grosse-Brauckmann: Minimal surfaces, SS 12Proof. Conformality means |f x | = |f y | =: λ and 〈f x , f y 〉 = 0. Using this in (4) gives(8)12λ 〈f 2 xx + f yy , ν〉 = 1 〈∆f, ν〉 = H.2λ2 To derive (7) we claim that the vectors ∆f = f xx + f yy and ν are parallel. To see this wedifferentiate the conformality relations (6):∂∂x |f x| 2 = ∂∂x |f y| 2 ⇒ 〈f xx , f x 〉 = 〈f xy , f y 〉∂∂y 〈f x, f y 〉 = 0 ⇒ 〈f yy , f x 〉 = −〈f xy , f y 〉Adding the equalities gives 〈f xx + f yy , f x 〉 = 0, and similarly 〈f xx + f yy , f y 〉 = 0. Since theframe (f x , f y , ν) is orthogonal, this proves ∆f ‖ ν.Decomposing with respect to (f x , f y , ν) therefore gives∆f = 〈∆f, ν〉ν (8)= 2Hλ 2 ν.□1.4. Mean curvature of graphs. The equations for mean curvature involve the vectorvalued parametrization f, that is, they represent a system of partial differential equations.From the point of view of analysis it is a great simplification if we can replace f with ascalar function. We achieve this for graphs.Any function u ∈ C 1 (U, R) defines a graph f(x) := ( x, u(x) ) . The tangent space T f of thegraph has the basis ∂ i f = (e i , ∂ i u) where i = 1, . . . , n. These vectors are clearly linearlyindependent, which verifies that f is a hypersurface (immersion).The upper unit normal of the graph is()1 ( ) −∂(9) ν(x) = √ 1 u− ∇u, 1 = √ , . . . , −∂ n u√ , 1√ .1 + |∇u|21 + |∇u|2 1 + |∇u|2 1 + |∇u|2Indeed, ν is perpendicular to any ∂ i f and normalized.Theorem 6. The mean curvature H = H(x) of a C 2 -graph f(x) = ( x, u(x) ) with respectto the upper normal ν is(10) H = 1 ()n div ∇u√ .1 + |∇u|2Proof. We avoid calculating the inverse of g for (3). It is more convenient to calculate Sand its trace directly. The definition of S gives for each i = 1, . . . , n:( ∑ )−∂ i ν = −dν(e i ) = df(Se i ) = df S k i e k = ∑ ( ) ( )S k idf(e k ) = S 1 e1i + . . . + S n eni∂ 1 u∂ n ukk

i 1.5 – date: July 27, 2012 7In particular, the i-th component of this equation reads()S i i = −∂ i ν i (9) ∂ i u= ∂ i √ , i = 1, . . . , n.1 + |∇u|2This gives the desired formula for the trace:n∑ n∑() ()nH = S i ∂ i u∇ui = ∂ i √ = div √ .1 + |∇u|21 + |∇u|2i=1i=1□3. Lecture, Wednesday 16.4.12We differentiate the expression in divergence form (10), using |∇u| 2 = ∑ j (∂ ju) 2 :nH = ∑ ()∂ i u∆u∂ i √ = √ − ∑ ( 1 ∂ i ui 1 + |∇u|2 1 + |∇u|2 2 (1 + |∇u| 2 ) 2 ∑ )∂ 3/2 ij u ∂ j uij1 ∑(=∂(1 + |∇u| 2 ) 3/2 ii u ( 1 + ∑ (∂ j u) 2) − ∑ )(11)∂ ij u ∂ i u ∂ j uij≠ij≠iIn particular, at a point p with horizontal tangent plane, the ∂ i u(p) vanish and so theequation simplifies to nH(p) = ∆u(p).Specifically, for a two-dimensional surface the outer sum in (11) must vanish:Proposition 7. A graph ( x, y, u(x, y) ) , where u ∈ C 2 (U 2 , R), represents a minimal surfaceif and only if u satisfies the minimal surface equation(12) (1 + u 2 y)u xx − 2u x u y u xy + (1 + u 2 x)u yy = 0 for all (x, y) ∈ U.The minimal surface equation is a second order partial differential equation, depending onsecond and first derivatives of u. The function u itself does not enter: If (x, y, u(x, y)) isminimal then (x, y, u(x, y) + c) is again minimal for any c ∈ R. The equation is nonlinear,meaning that L(u+v) ≠ Lu+Lv, where Lu denotes the left hand side of (12). Neverthelesswe will see lateron that it satisfies a maximum principle.The three most common ways to represent surfaces are parametric, as a graph, and implicit.Some people ignore the last way and call a surface given in form of a graph non-parametric.In this context, (12) is called the non-parametric minimal surface equation. Note that anysurface can locally be written as a graph.Problems. 1. Verify (10) by calculating g −1 for a graph and using (3).2. It is easier to obtain (12) if the entire derivation is carried out for dimension 2: Use the2 × 2 matrices g ij and b ij in (4).

8 K. Grosse-Brauckmann: Minimal surfaces, SS 121.5. The Gauss curvature of minimal surfaces. We collect some elementary propertiesof the pointwise condition H(p) = 0. On a minimal surface they hold globally.Theorem 8. Suppose that f : U 2 → R 3 is a surface with H(p) = 0 at p ∈ U.(i) Then the Gauss curvature satisfies K(p) ≤ 0, i.e. p is hyperbolic or paraboltic.(ii) The two principal curvatures at p are equal (p is umbilic [Nabelpunkt]) if and only ifK(p) = 0.(iii) The Gauss map is orientation reversing at p.Proof. (i) This follows from κ 1 + κ 2 = 0.(ii) Suppose κ 1 = −κ 2 . Then: κ 1 = κ 2 ⇔ κ 1 = κ 2 = 0 ⇔ K = κ 1 κ 2 = 0.(iii) Consider more generally a hypersurface f with a choice of normal ν. By forgettingits ambient space, we can regard the tangent space T p f an n-dimensional euclidean vectorspace. The differentialsdν p , df p : ( R n , 〈., .〉 ) → ( T p f, 〈., .〉 )have a well-defined determinant, measuring the oriented distortion of the volume elementat p. The determinant of the composition of these maps is related to Gauss curvature asfollows:(13) (−1) n K(p) = det(−S p ) = det ( (df p ) −1 dν p)=det dν pdet df pSuppose that the orientations chosen are such that det df p > 0. Then det dν has the signof (−1) n K(p), which for n = 2 is nonpositive whenever K ≤ 0.□It is convenient to include a definition here.Definition. Let f be a parameterized hypersurface. The quantity ∫ K dS ∈ [−∞, ∞] isUcalled the total curvature [Totalkrümmung] of f.To give the total curvature a more geometric meaning, we conclude from (13)det dν p = K(p) det df p ⇒ det(dν T p dν p ) = K 2 (p) det(df T p df p ).We conclude that the area of the Gauss image in S n , given by the integral of its Gramdeterminant, is∫( ) √∫(14) A U ν = det(dνT dν) dx = |K| √ ∫∫det(df T df) dx = |K| dS f = min.− K dS.UUUUNote the integrals count the area with multiplicity, i.e., taken as often as the sphericalimage is attained. Thus the total curvature of a minimal surface is the negative area of thespherical image, taken with multiplicity. For a general, not necessarily minimal, surface∫K dS counts the oriented area of the Gauss image with multiplicity.

i 2.1 – date: July 27, 2012 9Remark. We stated the above formulas for a parameterized surface. The integrals are alsowell-defined for global surfaces by considering a partition of unity, or by a decompositionof the surface into a union of disjoint open sets, such that the open sets are parameterizedsurfaces, and such that the their closures cover.2. Examples of complete minimal surfacesWe want to present the most well-known minimal surfaces, and whenever possible, characterizethem by their properties. Before doing so, let us mention a non-existence result:Proposition 9. A minimal hypersurface M ⊂ R n+1 cannot be compact.Indeed, a sphere of minimal radius R containing the surface touches it in a point where allthe surface principal curvatures are at least 1/R. This argument is made explicit by usingthe local normal form. It applies to each global hypersurface, no matter if immersed orembedded. We will later give a different proof.In surface theory, surfaces are identified if their parameterizations differ by a diffeomorphism,or if the images differ by motion (ambient isometry). There is yet another kind ofmapping which leaves minimality invariant:Proposition 10. Let f ∈ C 2 (U, R n+1 ) be a hypersurface and set f (λ) := λf where λ > 0.If f has mean curvature H then f (λ) has mean curvature 1 H; in particular minimality isλpreserved.Proof. Note that df (λ) = λdf. Hence the normal of f (λ) is unchanged, and soS (λ) = −(df (λ) ) −1 dν (λ) = − 1 λ df −1 dν = 1 λ S.In particular, H (λ) = 1 λ H.□We will often identify complete minimal surfaces which differ by scaling.2.1. Minimal surfaces of revolution: The catenoid. Let(15) (r, h): I → (0, ∞) × Rbe a regular curve, where I is an open interval. When we place the curve in the (x, z)-planeand rotate it about the z-axis we obtain a surface of revolution [Rotationsfläche](16) f : I × R → R 3 , f(t, ϕ) := ( r(t) cos ϕ, r(t) sin ϕ, h(t) ) .

10 K. Grosse-Brauckmann: Minimal surfaces, SS 12Lemma 11. On a surface of revolution (16) the meridians t ↦→ f(t, ϕ) and the latitudecircles ϕ ↦→ f(t, ϕ) are principal curvature lines, with principal curvatures(17) κ 1 = r′ h ′′ − h ′ r ′′√r ′2 + h ′ 2 3 and κ 2 = 1 h ′√r r ′2 + h ,′ 2respectively. In particular, f is minimal if and only if(18) 0 = r(r ′ h ′′ − h ′ r ′′ ) + h ′ (r ′2 + h ′2 ).Proof. We compute:and so the first fundamental form g is()r ′2 + h ′ 20(19) g =0 r 2⎛ ⎞⎛ ⎞r ′ cos ϕ−r sin ϕ⎜∂ 1 f = ⎝r ′ ⎟⎜ ⎟sin ϕ⎠ and ∂ 2 f = ⎝ r cos ϕ ⎠ ,h ′0⇒ g −1 =( )10r ′2 +h ′2 .10r 2The normal of the curve, ( )−h ′r , normed and rotated about the z-axis, gives the surface normal′⎛ ⎞−h ′ cos ϕ1 ⎜(20) ν = √ ⎝−h ′ ⎟sin ϕ⎠ ;r ′2 + h ′ 2r ′it is the inner normal if h is increasing. The second derivatives of f are⎛ ⎞⎛ ⎞⎛ ⎞r ′′ cos ϕ−r cos ϕ−r ′ sin ϕ⎜∂ 11 f = ⎝r ′′ ⎟⎜ ⎟⎜sin ϕ⎠ , ∂ 22 f = ⎝−r sin ϕ⎠ , ∂ 12 f = ∂ 21 f = ⎝ r ′ ⎟cos ϕ ⎠ .h ′′00We obtainb 11 = 〈∂ 11 f, ν〉 = r′ h ′′ − h ′ r ′′√r ′2 + h ′ 2 , b 22 = 〈∂ 22 f, ν〉 =rh ′√r ′2 + h ′ 2 , b 12 = 〈∂ 12 f, ν〉 = 0.Note that both g and b are diagonal, so that g −1 b is diagonal as well. Hence the coordinatedirections are indeed principal curvature directions, with κ 1 = g 11 b 11 and κ 2 = g 22 b 22 . □We call a surface M complete [vollständig] if there is no connected surface (of the samedimension) containing it as a proper subset. For instance, a plane in R 3 is complete, butan open or closed disk (in R 2 or R 3 ) is not. We say a complete surface is a surface ofrevolution if it is invariant under rotation (we allow for the case it meets the axis, i.e.r = 0 is included). A more conceptual definition is to say that every Cauchy sequence inthe surface must converge; here the notion of Cauchy sequence must be with respect tothe intrinsic distance of the surface.

i 2.1 – date: July 27, 2012 11Theorem 12 (Bonnet 1860). Each complete minimal surface of revolution is either a planeor a catenoid [Katenoid]f : R × [0, 2π) → R 3 , f(t, ϕ) := ( r(t) cos ϕ, r(t) sin ϕ, t )(21)( t − t0)with r(t) := a cosh for t 0 ∈ R, a > 0.aThe catenoid was first described by Euler in 1744. The meridian curve cosh is a catenary[Kettenlinie], see problems. Therefore we can obtain the catenoid by rotating the shape ofa hanging chain about an axis at a specific distance. Note that the parameterization withϕ ∈ R in place of [0, 2π) covers the surface of revolution infinitely often.Proof. The special case h ≡ const in (16) solves the zero mean curvature equation (18). Itcorresponds to a (horizontal) plane without the point P := (0, 0, h) on the axis; this surfacebecomes complete by taking the union with P .Suppose now h is not identical to a constant. Then there is t 0 ∈ I with h ′ (t 0 ) ≠ 0. We concludethat h is locally monotone, and so by a reparameterization of our generating curve (15) we mayassume h(t) = t locally. In this case the differential equation (18) becomes(22) rr ′′ = 1 + r ′2 , r > 0.This ODE can be solved by separation of variables, see [O1, p.64]. To check that (21) solves theODE (22) is easy: Indeed, a cosh(.) 1 a cosh(.) = 1 + sinh2 (.).Let us consider the initial values r(0) > 0, r ′ (0) ∈ R for (22). Writing the initial values in theform r ′ (0) = sinh(t 0 ) with t 0 ∈ R and r(0) = a cosh −t 0awith a > 0 we see that they are satisfiedby the solution family (21). By the uniqueness part of the theorem of Picard-Lindelöf these areall solutions (the system r ′ = R and R ′ = (1 + R 2 )/r is Lipshitz for r > ε > 0). Moreover, thesesolutions are defined for all t ∈ R and hence complete. In particular, any complete solution ofthe initial value problem with h ′ (t 0 ) ≠ 0 for some t 0 has in fact h ′ (t) ≠ 0 for all t.We have solved all initial value problems for the system (18) and r ′2 + h ′2 = 1; thus all solutionsare of the type claimed.□4. Lecture, Friday 18.4.12We now discuss two properties of the catenoid. First, we consider a boundary value problem.The catenoid is foliated with circles. Conversely, does any pair of circles bound apiece of the catenoid?We will only consider the case of two circles of equal radius which are spaced a distanced > 0 apart. Call two circles coaxial if they are contained in parallel planes, and havemidpoints on the same line perpendicular to the planes. A soap film experiment tells usthat two coaxial circles of given radius r cannot be too far apart, if they bound a catenoid.The following is proven in the problem session:

12 K. Grosse-Brauckmann: Minimal surfaces, SS 12Proposition 13. There is a number d max = 1.32 . . . such that two coaxial unit circles indistance d bound connected subsets two different catenoids for 0 < d < d max , exactly onecatenoid for d max , and none for d > d max .From the two catenoids, only the one with smaller area can be reproduced in a soap filmexperiment. When the distance of the two unit circles is increased from 0 to d max thenthe area increases from 0 to 2π · 1.19 . . .. Here the soap film catenoid represents a stableminimum of area. The other catenoid is an unstable equilibrium; for d → 0 these unstablecatenoids tend to a double cover of the unit disk with area 2π. Interestingly enough, thearea of the catenoid with d = d max is larger than the area of two unit disks bounded by thecircles: The area of this so-called Goldschmidt solution is 2π, and so the catenoids closeto the maximal catenoid attain a local or relative minimum of area, but not the absoluteminimum. More precisely, this holds for any d with 1.05 . . . < d ≤ d max . See Nitsche[N, § 515 f] and Oprea [O1, 5.6] for more information.By the Prop. 10 the previous results will be valid for any coaxial circles with radius r,provided d max is replaced by rd max .Note that in our catenoid family (21), the parameter a corresponds to scaling (check!).The second property is the total curvature of the catenoid:Proposition 14. The catenoid is embedded and its Gauss map is bijective to S 2 minus aset of antipodal points. Consequently, the total curvature of the catenoid is −4π.Proof. Embeddedness is immediate from the fact that a meridian is an embedded curve.For the Gauss map, we consider the standard catenoid (21) with t 0 = 0, a = 1. Note thatscalings and motions preserve the Gauss map up to rotation.Then (20) gives ν(t, ϕ) = 1cosh t(− cos ϕ, − sin ϕ, sinh t). The xy-projection of ν(t, ϕ) hasangle ϕ + π. The t lines go on the sphere from the south pole (t → −∞) to the northpole t → ∞; they are 1-1 since the third component tanh t is strictly monotone. Thus thet-lines have an open semicircle as their Gauss image. It follows that t ∈ R, ϕ ∈ [0, 2π)parameterizes S 2 except for ±(0, 0, 1). The total curvature then follows from (14). □Problem. Calculate the total curvature ∫ [0,2π]×R K(t, ϕ) dS (t,ϕ) directly.Let us finally quote a more recent result. There is a topological characterization whichmakes the catenoid argueably the most important embedded minimal surface:Theorem (Collin 1997). (i) Each properly embedded complete minimal annulus f : R 2 \{0}is a catenoid.(ii) Suppose a (global) minimal surface is properly embedded, complete, and has at leasttwo ends. Then each end is asymptotic to a catenoid or planar end.

i 2.2 – date: July 27, 2012 13Here, a map between topological spaces f : V → W is called proper [eigentlich] if f −1 (K) ⊂V is compact for each compact K ⊂ V . If V = R n , W = R m properness means that thenoncompact parts of the domain (the ends) are mapped to infinity. A non-proper exampleis a map from the entire real line into a bounded subset of R 2 , for instance a finite lengthsegment or a spiral. But any graph ( t, f(t) ) over the entire real line represents a propermap from R to R 2 . An end is a subset of a surface, parameterized by a proper map ofD \ {0} (think of 0 being mapped to infinity).2.2. Ruled minimal surfaces: The helicoid. Discovered by Meusnier in 1776, thehelicoid can be represented by the parameterization⎛ ⎞s cos t(23) f : R 2 → R 3 ⎜ ⎟, f(s, t) := ⎝s sin t⎠ for h ≠ 0.htTo better understand the helicoid we need the notion of a ruled surface. Suppose c ∈C 2 (I, R 3 ), v ∈ C 2 (J, R 3 \ {0}) are curves, where I, J ∋ 0 are open intervals, with typicallyJ = R. Then(24) f : J × I → R 3 , f(s, t) := c(t) + sv(t).is called a ruled mapping. The curve c is called the directrix [Leitkurve]. The straight liness ↦→ f(s, t) are the rulings [Regelgeraden] of f. The notion ruled refers to a ruler; theGerman Regel is a mistranslation of the French term règle (ruler). Let us also note thatmore generally we could allow any domain U ⊂ R such that J(t) := {s ∈ R : (s, t) ∈ U} isan open interval containing 0.In case of the helicoid as above, c(t) = (0, 0, ht) is a straight line, namely the z-axis. Therulings are perpendicular to the axis, and their direction v(t) = (cos t, sin t, 0) describes aunit circle parameterized with unit speed in the plane P perpendicular to the axis. Therotation is anticlockwise for pitch [Ganghöhe] h > 0 clockwise for pitch h < 0 (“right” and“left” helicoid).Examples. The circular cylinder or, more generally, a cylinder over an arbitrary curve is aruled surface. An infinite cone over a circle or over a general curve is a ruled mapping. Itis known that any surface with K ≡ 0 is locally ruled; this applies, for instance, to a bentpiece of paper in space.The partial derivatives of f are(25) ∂ 1 f(s, t) = v(t), ∂ 2 f = c ′ (t) + sv ′ (t).

14 K. Grosse-Brauckmann: Minimal surfaces, SS 12The ruled mapping is an immersion at (s, t) if these vectors are linearly independent; ifthis holds on the entire domain then we call the mapping a ruled surface [Regelfläche]. Ona ruled surface the rulings are asymptote lines, and so the Gauss curvature satisfies K ≤ 0.We prove minimality of the helicoid by a geometric argument.Proposition 15. The helicoid (23) is minimal.Proof. At any given point on the helicoid, the horizontal ruling is a straight line and hencean asymptotic direction. We claim an orthogonal vector v ≠ 0 is also an asymptotic direction.By construction, the helicoid is invariant under 180 ◦ -rotation about the horizontalgeodesic. This rotation changes the sign of the normal. But the rotation is a symmetry,leaving normal curvature unchanged, so κ norm (±v) = −κ norm (±v). Here we write ±vsince normal curvature depends only on the one-dimensional subspace through v. Henceκ norm (±v) = 0 and v is an asymptotic direction.□Remark. In more general ambient spaces a helicoid can also be defined as the orbit of ageodesic rotating with constant speed about a geodesic directrix. If 180 ◦ -rotation abouteach geodesic ruling is an isometry then our symmetry-based argument works for the same.An example are helicoids in the space forms S 3 or H 3 .Modulo reparameterzation and motion, any ruled minimal surface is a piece of the planeor helicoid (23):Theorem 16 (Catalan 1842). A complete ruled minimal surface is a plane or helicoid.To calculate the principal curvatures of an arbitrary ruled surface is somewhat involved.Note that our helicoid representation (23) employs special choices: The rulings are parametrizedwith unit speed, and they intersect the directrix at a right angle. These choicescan be achieved in general (from now on we assume 0 ∈ I):Lemma 17. For a ruled surface (24) the directrix c and ruling vector field v can be chosensuch that the following holds:(26)(27)|v| ≡ 1, v ⊥ v ′ , v ⊥ c ′If ‖v ′ (0)‖ = 1 then 〈c ′ (0), v ′ (0)〉 = 0.For the helicoid, (26) is satisfied if and only if we choose a helix as the directrix. However,(27) holds precisely if we pick the straight z-axis as the directrix (check!).Proof. First of all, by a linear change in s, we obtain the first assertion in (26). Its derivativeis the second.

i 2.2 – date: July 27, 2012 15Let us now modify our choice of directrix. We replace c with ˜c(t) = c(t) − λ(t)v(t), whereλ(t) := 〈c ′ (0), v ′ (0)〉 +Let us verify the last assertion in (26) for ˜c:∫ t0〈c ′ (τ), v(τ)〉 dτ.〈˜c ′ , v〉 = 〈 c ′ − λ ′ v − λv ′ , v 〉 v⊥v ′= 〈 c ′ − 〈c ′ , v〉v, v 〉 |v|=1= 0 ∀t ∈ I.Moreover, the first term of λ was chosen as to give〈〉〈˜c ′ (0), v ′ (0)〉 = c ′ (0) − 〈c ′ (0), v(0)〉v(0) − 〈c ′ (0), v ′ (0)〉v ′ (0), v ′ (0)v⊥v ′= 〈 c ′ (0), v ′ (0) 〉( 1 − |v ′ (0)| 2) ,which proves (27). Note that ˜c is merely a different choice of directrix, which leaves thet-parameter untouched. In the following we write once again c in place of ˜c.□5. Lecture, Wednesday 25.4.12Lemma 18. Assuming (26), we have H(s, t) = 0 if and only if(28) 0 = det(∂ 1 f, ∂ 2 f, ∂ 22 f) = det ( v, c ′ + sv ′ , c ′′ + sv ′′) at (s, t).Proof. By (4) we have H(s, t) = 0 is equivalent to(29) 0 = g 22 b 11 − 2g 12 b 12 + g 11 b 22 at (s, t).But b 11 = 〈∂ 11 f(s, t), ν〉 = 0 and also g 12 (s, t) = 〈 v(t), c ′ (t) + sv ′ (t) 〉 = 0 by (26), so thatthe first two terms vanish.Moreover, g 11 = |v| 2 = 1 and so (29) is equivalent to0 = b 22 = 〈ν, ∂ 22 f〉 at (s, t).Since ∂ 1 f, ∂ 2 f, ν are orthogonal this implies that ∂ 22 f is linearly dependent on ∂ 1 f and ∂ 2 fand so (28) holds.□To draw a consequence of (28), let us expand the determinant:0 = det ( v, c ′ , c ′′ ) + s ( det(v, v ′ , c ′′ ) + det(v, c ′ , v ′′ ) ) + s 2 det(v, v ′ , v ′′ ).If H(s, t) = 0 holds for an open interval of s-values, then the coefficients(30) d 1 := det ( v, c ′ , c ′′ ), d 2 := det(v, v ′ , c ′′ ) + det(v, c ′ , v ′′ ), d 3 := det(v, v ′ , v ′′ )must vanish at t. Note that for the helicoid (23) these values vanish for all t, so that weobtain another proof of its minimality.

16 K. Grosse-Brauckmann: Minimal surfaces, SS 12Proof of Catalan’s theorem. We assume the representation of f satisfies Lemma 17. Ifv ′ ≡ 0 then v is constant. But since c ′ ⊥ v this implies c is contained in some planeperpendicular to v. But then f is a generalized cylinder, whose principal curvatures are 0and the curvature of c. Thus for f to be minimal, c must be contained in a straight lineand f parameterizes a subset of the plane.If v is not constant then v ′ ≠ 0 at some point; we assume it holds at t = 0. By reparameterizingt, we achieve in some neighbourhood of t = 0(31) |v ′ |(t) = 1 and so 〈v ′ , v ′′ 〉 = 0.Our goal is to assert that v parameterizes a unit circle in a plane P , with unit speed, andthat c is straight line perpendicular to P .Claim 1: For t in a neighbourhood of 0, the curve v parameterizes a piece of unit circle insome plane P with unit speed.By (26) and (31), v ⊥ v ′ ⊥ v ′′ . Thus d 3 = 0 gives v ‖ v ′′ . Hence (v ×v ′ ) ′ = v ′ ×v ′ +v ×v ′′ =v × v ′′ = 0, meaning that v(t), v ′ (t) stay within one plane P = span{v(0), v ′ (0)} for all tconsidered. The claim follows since |v| = 1 and |v ′ | = 1.Claim 2: c ′′ = 0 and c ′ ⊥ P hold for t in some neighbourhood of 0.From v ‖ v ′′ we see that d 2 = 0 reduces to det(v, v ′ , c ′′ ) = 0 so that on the one hand c ′′ ∈ P .We concludec ′′ = 〈c ′′ , v〉v + 〈c ′′ , v ′ 〉v ′and so d 1 = 0 gives(32) 〈c ′′ , v ′ 〉 det(v, c ′ , v ′ ) = 0.By (26) and (31) we have c ′ ⊥ v ⊥ v ′ . Furthermore at t = 0 we have v ′ (0) ⊥ c ′ (0) by (27),and so det(v, c ′ , v ′ )(0) ≠ 0. By continuity the same conclusion holds on a neighbourhoodof t = 0. From (32) we can infer 〈c ′′ , v ′ 〉 = 0. This gives 〈c ′ , v ′ 〉 ′ = 〈c ′ , v ′′ 〉 = 〈c ′ , −v〉 = 0,using our first claim. But since 〈c ′ (0), v ′ (0)〉 = 0 the latter implies 〈c ′ , v ′ 〉 ≡ 0. Usingc ′ ⊥ v, we have c ′ ⊥ P , and upon differentiation we verify c ′′ ⊥ P , on the other hand. Ourclaim is proved.Our claims 1 and 2 prove that f(s, t) is locally contained in a helicoid (23), where |h| = |c ′ |and the sign of h is given by the sense of rotation. We claim that the interval J for whichour claims 1 and 2 hold, is in fact I. Indeed, by continuity the subset of I for which ourtwo claims hold is closed. But the two conditions we used, (31) and det(v, c ′ , v ′ ) ≠ 0 holdin an open neighbourhood of any point t, and so the subset of I for which the claims holdis also open. Thus the claims hold on a closed and open subset of the domain, hence on allof the domain. This gives that for nonconstant v our surface f coincides with a helicoidwhere defined, or globally in the complete case.□

i 2.3 – date: July 27, 2012 17Let us state some properties of the helicoid (23):• It is simply periodic, that is, f(s, t + 2πk) = f(s, t) + k(0, 0, 2πh) for any k ∈ Z, inparticular the Gauss map is invariant und this translation. For a half period we havef(s, t + π) = f(−s, t) + (0, 0, πh) for any k ∈ Z; the Gauss map changes sign under thistranslation (the quotient surface is a Möbius strip).• The helicoid is skew symmetric under rotation by any angle ϕ ∈ R followed by a verticaltranslation (0, 0, hϕ). It can be uniquely characterized by this invariance, in terms of anODE, quite similar to our analysis of the catenoid.Proposition 19. The helicoid is embedded and has infinite total curvature. The totalcurvature over a translational fundamental domain {f(s, t) : s ∈ R, 0 ≤ t < 2π} is −4π.Proof. There are conformal parameterizations of the catenoid and helicoid⎛ ⎞⎛⎞cosh x cos ysinh x sin y⎜ ⎟⎜⎟c(x, y) := ⎝cosh x sin y⎠ , h(x, y) := ⎝− sinh x cos y⎠ ,xywhich are isometric, g c = g h , see problems. By the theorema egregium, the Gauss curvaturesagree, K c = K h . Hence the total curvatures agree. The domain x ∈ R, 0 ≤ y < 2πparameterizing a translational fundamental domain of the helicoid corresponds to an injectiveparameterization of the catenoid. By Prop. 14 its total curvature is −4π. □Like the catenoid, also the helicoid is now also known to be characterized by a topologicalproperty:Theorem (Meeks-Rosenberg 2005). Helicoid and plane are the only properly embeddedsimply connected complete minimal surfaces.6. Lecture, Friday 27.4.122.3. Scherk’s doubly periodic surface. We would like to present an example of a minimalgraph f(x, y) = (x, y, u(x, y)). If the domain of u is all of R 2 , then a theorem ofBernstein proves that u is linear and f is a plane. Nevertheless there are interesting graphsdefined over certain subsets of the plane.Along with four other minimal surfaces, Scherk in 1835 described a graph defined over theopen square Q := ( − π 2 , π 2)×(−π2 , π 2)byu: Q → R, u(x, y) := log cos y − log cos x = log cos ycos x .

18 K. Grosse-Brauckmann: Minimal surfaces, SS 12To prove Scherk’s surface is minimal, let us check it satisfies the minimal surface equationfor graphs (12),Indeed, sincewe haveu x = sin xcos x ,1 + sin2 ycos 2 ycos 2 x(1 + u 2 y)u xx − 2u x u y u xy + (1 + u 2 x)u yy = 0.u y = − sin ycos y , u xx = 1cos 2 x , u yy = − 1cos 2 y , u xy = 0− 1 + sin2 xcos 2 xcos 2 y= cos2 y + sin 2 y − (cos 2 x + sin 2 x)cos 2 x cos 2 yThe only intersection of the surface with the xy-plane, that is, the only zeros of u, are thetwo horizontal diagonals (±x, x, 0). Let us now study limiting values on ∂Q. When (x, y)tends to an interior point of a boundary edge of Q, the function u has limiting values +∞or −∞. Indeed, if (x, y) → (± π, y 2 0), where |y 0 | < π , then log cos y − log cos x → ∞.2It remains to discuss limits at the four vertices. We claim that the closure of the graphcontains the four vertical lines (± π, ± π ) × R. To see that, fix λ ∈ (0, ∞) and note2 2( πu2 − ξ, π )2 − η = log sin ηsin ξ = log η − 1 3! η3 + . . .ξ − 1 3! ξ3 + . . .= 0.→ log λ as η = λξ ↘ 0.But log: (0, ∞) → R is surjective, implying that this limit can attain any value in R. Thisshows that the four vertical lines over the vertices of Q are the closure M of the graph.By horizontal translation we obtain a smooth periodic surface: It is a graph over the openblack squares of a chequerboard, {(x, y) ∈ R 2 : cos x, cos y > 0}, and has straight verticallines over their vertices.Proposition 20. There exists a complete minimal surface, called Scherk’s doubly periodicsurface. Surface and normal are invariant under the lattice {2πk + 2πl : k, l ∈ Z}, whilethe surface alone is invariant also under Z(π, π, 0) ⊕ Z(−π, π, 0). The total curvature ofM is −2π.Problem. Check that the more general version of the graphu α (x, y) = 1 (log cos ( x cos α + y sin α ) )− log cos x , for any 0 < α < π.sin αexhibited by Scherk himself is again minimal. Describe its projection to the xy-plane.References. [N] p.63/64, [DHS] p.156–1655. Lecture, Wednesday 25.4.12

i 2.4 – date: July 27, 2012 192.4. Enneper’s surface has an intrinsic rotation. We want to give an example of aminimal surface with self-intersections and somewhat more complicated geometry. In 1864Alfred Enneper, a student of Gauss, discovered the minimal surface⎛⎞x − 1f : R 2 → R 3 3⎜x3 + xy 2, f(x, y) = ⎝−y + 1 3 y3 − yx 2 ⎟⎠x 2 − y 2 .To see it is minimal, let us calculate⎛ ⎞1 − x 2 + y 2⎜f x = ⎝Thus−2xy2x⎟⎠ , f y =⎛⎞2xy⎜⎝−1 + y 2 − x 2 ⎟⎠ .−2y|f x (x, y)| 2 = ( 1 + x 4 + y 4 − 2x 2 + 2y 2 − 2x 2 y 2) + 4x 2 y 2 + 4x 2= 1 + x 4 + y 4 + 2x 2 + 2y 2 + 2x 2 y 2 = (1 + x 2 + y 2 ) 2 = |f y (x, y)| 2 ,where the last equality follows from the fact that the permutated coordinates of f(y, x)agree with the ones of f(x, y) up to sign. Moreover, 〈f x , f y 〉 = 0 and so f is conformal.Therefore (7) implies that minimality follows from ∆f = (−2x + 2x, 2y − 2y, 2 − 2) = 0.To explain an important property of Enneper’s surface, we note that the first fundamentalform() ()(1 + x 2 + y 2 ) 2 0(1 + r 2 ) 2 0(33) g ==,0 (1 + x 2 + y 2 ) 2 0 (1 + r 2 ) 2depends on r := √ x 2 + y 2 alone. We rephrase this fact in a more geometric language:Proposition 21. Let R α denote the rotation of R 2 by an angle α ∈ R. The Ennepersurface has an intrinsic rotation, that is, f and f ◦ R α are isometric for each α ∈ R.Imagine a plaster model of an embedded piece of Enneper’s surface, as well as a copy ofthe surface made from a nonelastic material, such as thin metal. Then the metal copy canbe rotated on the plaster model; nevertheless, it changes shape in ambient space duringthe rotation.Brian Smyth in the 1980’s determined all minimal surfaces admitting intrinsic isometries;precisely Enneper’s surface, as well as its generalizations with higher dihedral symmetry,have an intrinsic rotation. Bour in 1862 had studied minimal surfaces which are developableto a surface of revolution. I do not know to which level of detail he discussed the surfacedescribed by Enneper two years later (see [N], p.57).6. Lecture, Friday 27.4.12

20 K. Grosse-Brauckmann: Minimal surfaces, SS 12To better understand the geometry of the Enneper surface, let us first point out that(f(±x, x) = ± (x + 2 3 x3 ), −(x + 2 )3 x3 ), 0 ,meaning that the surface contains the two horizontal diagonals. On the other hand,f(x, 0) = (x − 1 3 x3 , 0, x 2) . Therefore, at x = 0 the surface is asymptotic to the graphof a parabola, while at x = ∞ it is asymptotic to the graph of 9(.) 2/3 . By its asymptoticsthe curve f(x, 0) must self-intersect and in fact f(± √ 3, 0) = (0, 0, 3), so that theEnneper surface is not embedded. In fact, the selfintersections make it hard to visualize(see [DHKW], p.146 for images, or generate them by yourself!).To analyse Enneper’s surface better at 0 and infinity, we use the polar coordinate representation⎛⎞ ⎛⎞f ( r cos ϕ, r sin ϕ ) r cos ϕ − 1 3⎜r3 cos 3 ϕ + r 3 cos ϕ sin 2 ϕ r cos ϕ − 1= ⎝−r sin ϕ + 1 3 r3 sin 3 3ϕ − r 3 cos 2 ⎟ ⎜r3 cos 3ϕϕ sin ϕ⎠ = ⎝−r sin ϕ − 1 ⎟3 r3 sin 3ϕ⎠ ,r 2 cos 2ϕr 2 cos 2ϕusing cos 3ϕ = Re e 3iϕ = Re(e iϕ ) 3 = Re(cos ϕ + i sin ϕ) 3 = cos 3 ϕ − 3 cos ϕ sin 2 ϕ andsin 3ϕ = Im e 3iϕ = Im(e iϕ ) 3 = Im(cos ϕ + i sin ϕ) 3 = − sin 3 ϕ + 3 cos 2 ϕ sin ϕ.Consider the circle γ r (ϕ) = (r cos ϕ, r sin ϕ) of radius r > 0. First, when r is small, wehavef ( γ r (ϕ) ) = ( r cos ϕ, −r sin ϕ, r 2 cos 2ϕ ) + O(r 3 ) as r → 0,meaning that the image circle is asymptotically in the horizontal xy-plane. It is parameterizedclockwise. The sign of the z-values alternates four times or every 90 ◦ ; it is zeroonly on the diagonals.Second, to study the image of the circle γ r (ϕ) for large radius r, note thatf ( γ r (ϕ) ) (= − 1 3 r3 cos 3ϕ, − 1 )3 r3 sin 3ϕ, r 2 cos 2ϕ + O(r) as r → ∞.In particular, |f(γ r (ϕ))| = 1 3 r3 + O(r 2 ), meaning that f ◦ γ r is asymptotically round.Consequently, the radial projection of the image circle onto S 2 readsf(γ r (ϕ))(|f(γ r (ϕ))| = − cos 3ϕ + O( 1), − sin 3ϕ + O( 1), 3 r rr cos 2ϕ + O( ) ) 1as r → ∞.r 2The third component is small, and so the image of a large circle is again asymptoticallya horizontal round circle. However, it runs 3 times round anticlockwise! The sign of thethird component indicates to which side of the horizontal xy-plane the surface sits: it takesexactly 270 ◦ in the horizontal plane for the vertical component of the image to change sign.Let us mention two more properties (problems?):1. The Enneper surface is algebraic, i.e. there is a a polynomial of 9-th order (see [N, p.77]),such that the Enneper surface is its zero set. Note that any surface can be represented

i 3.1 – date: July 27, 2012 21implicitely; the point here is that the function is not trancendental but a polynomial.2. The Gauss map is bijective to S 2 minus one point and so the total curvature is −4π.We have now covered the simplest complete minimal surfaces in the following sense:Theorem (Osserman 1964). The only complete minimal surfaces with total curvaturelarger or equal −4π are the plane, the catenoid, or Enneper’s surface.References. [N], p.75–81; [DHS], p.151–1567. Lecture, Wednesday 2.5.123. VariationsIn 1762 Lagrange introduced what we nowadays call the calculus of variations [Variationsrechnung].He considered the area content of a surface with fixed boundary which is agraph of minimal area. He showed that the graph satisfies the minimal surface equationin divergence form (10)∂ u√ x+ ∂ u√ y∂x 1 + u2x + u 2 y∂y 1 + u2x + u 2 yOnly in the 19th century was it recognized that the latter condition is precisely the conditionthat twice the mean curvature vanishes.The present section introduces the variational ideas. We find it advantageous to use expansionsfor area rather than derivatives. For simplicity we constrain to the two-dimensionalcase.= 0.3.1. Expansion of area. The area content A(f) = A U (f) of a parameterized surfacef : R n → R n+k is defined in terms of its first fundamental form as∫ ∫√∫√A U (f) = 1 dS := det g dx = det(df Tdf) dx.UUHere, det g = det(df T df) is called the Gram determinant. We consider dS = √ det g dx asa Riemannian (parameterization invariant) area element of f. The area content of a globalsurface can similarly be defined in terms of a partition of unity over parametrizing pieces.Problems. 1. Check that A U (f) = ∫ U |f x × f y | dxdy in dimension n = 2.2. Check that ˜f := f ◦ ϕ and f have the same area for any diffeomorphism ϕ: V → U.Therefore A U (f) is invariant under change of parameters.We wish to see how the area integral changes when we consider a one-parameter family ofsurfaces. We will need the following:U

22 K. Grosse-Brauckmann: Minimal surfaces, SS 12Lemma 22. If f is a two-dimensional surface with Gauss map ν we have(34)〈dν(X), dν(Y )〉= 2H b(X, Y ) − Kg(X, Y ) for all X, Y ∈ R 2 .Occasionally, this bilinear form is called the third fundamental form.Proof. Let κ 1 , κ 2 be the principal curvatures and define the bilinear formT (X, Y ) := 〈 (dν + κ 1 df)X, (dν + κ 2 df)Y 〉 .We claim T is symmetric and so independent of the labelling of the principal curvatures.Indeed we can express T in terms of symmetric forms as follows:T (X, Y ) = 〈dν(X), dν(Y )〉 + κ 1 κ 2 〈df(X), df(Y )〉 + κ 1 〈df(X), dν(Y )〉 + κ 2 〈df(Y ), dν(X)〉= 〈dν(X), dν(Y )〉 + Kg(X, Y ) − 2Hb(X, Y ).Moreover, if X 1 , X 2 are linearly independent principal curvature directions for κ 1 , κ 2 then(35) T (X 1 , Y ) = T (Y, X 2 ) = 0 for all Y ∈ R n .Since T is symmetric and (X 1 , X 2 ) is a basis, (35) proves T ≡ 0. Plugging this into thefirst expression for T implies our claim.□To state the area expansion we need to introduce one more notion:Definition. Let f ∈ C 1 (U n , R n+k ) be a surface. Then a function u ∈ C 1 (U n , R) has theRiemannian gradient ∇u ∈ R n defined byg(∇u, X) = du(X) ⇔ (∇u) j = ∑ lg jl ∂ l u, j = 1, . . . , n.The point of this definition is that the gradient vector is independent of reparameterization;that is, for ˜f = f ◦ ϕ the function ũ = u ◦ ϕ has the same gradient (check!). Tosee that the two expressions in the definition are equivalent write the left hand side as∑jk g jk(∇u) j X k = ∑ i ∂ iuX i . Note that the gradient ∇u is a tangent vector, a column,while du is a cotangent vector, a row. We will need the squared (Riemannian) norm of thegradient,‖∇u‖ 2 = ∑ ijg ij (∇u) i (∇u) j = ∑ ijklg ij g ik ∂ k u g jl ∂ l u = ∑ klg kl ∂ k u ∂ l u.Theorem 23 (Area expansion). Let f : U → R 3 be a surface with Gauss map ν andu ∈ C0(U, 1 R) be differentiable with compact support V . Then the normal variation(36) f t := f + tuν : U → R 3

i 3.1 – date: July 27, 2012 23is an immersion for sufficiently small |t|, whose area has the following expansion as t → 0,∫∫(37) A V (f t ) = A V (f) − 2t uH dS + t 2 12 ‖∇u‖2 +u 2 K dS + O(t 3 ).VExample. Vary the 2-sphere S r through spheres S r+t of radius r + t which have areaA ( S r+t)= 4π(r + t) 2 = 4πr 2 + t 8πr + t 2 4π.This expansion agrees with (37), obtained for u ≡ 1 with respect to the outer normal ν:Indeed, the coefficient of t is 2 ∫ S rH = 2(4πr 2 ) −1 = −8πr, while the coefficient of t 2 is∫rS rK = 4πr 2 1 = 4π.r 2Proof. Step 1: We calculate the first fundamental form of f t (V ). For each i we have∂ i f t = ∂ i f + tu ∂ i ν + t ∂ i u ν.Using that ν and df are orthogonal, and b ij = −〈∂ i f, ∂ j ν〉 = b ji we obtaingij t := 〈 ∂ i f t , ∂ j f t〉 )= g ij − 2tub ij + t(u 2 2 〈∂ i ν, ∂ j ν〉 + ∂ i u ∂ j uV(34)= g ij − 2tub ij + t 2 (u 2 (2Hb ij − Kg ij ) + ∂ i u ∂ j uStep 2: We calculate the determinant of g t . For that end, we calculate the coefficients ofthe expansionClearly,The first order terms aredet g t = g t 11g t 22 − (g t 12) 2 = (I) + t (II) + t 2 (III) + O(t 3 ).(I) = g 11 g 22 − g 2 12 = det g.(II) = −2u ( b 11 g 22 + b 22 g 11 − 2b 12 g 21) (4)= −4uH det gFinally, the quadratic terms are)(III) = u(2H(b 2 11 g 22 + b 22 g 11 − 2b 12 g 12 ) − K(g 11 g 22 + g 22 g 11 − 2g12) 2 + 4(b 11 b 22 − b 2 12)).+ g 22 ∂ 1 u ∂ 1 u + g 11 ∂ 2 u ∂ 2 u − 2g 12 ∂ 1 u ∂ 2 u(4)= u 2( 4H 2 det g − 2K det g + 4K det g ) + det g ∑ i,j=1g ij ∂ i u ∂ j u,where we used K = det b/ det g. Collecting all terms we arrive at the desired expansion(38) det g t = det g[1 − 4tuH + t(u 2 2( 4H 2 + 2K ) )]+ ‖∇u‖ 2 + O(t 3 ).Since det g ≠ 0 and u has compact support this formula shows that indeed det g t ≠ 0 forsmall |t|. This verifies f t is an immersion.

24 K. Grosse-Brauckmann: Minimal surfaces, SS 12Step 3: We compute the root of the determinant √ det g t . To find the root of [. . . ], we useTaylor’s formulaThus (38) gives√1 + x = 1 +x2 − 1 8 x2 + O(x 3 ) as x → 0.√det gt = √ det g(1 − 2tuH + t 2 u 2( 2H 2 + K ) )+ t2 2 ‖∇u‖2 − 2t 2 u 2 H 2 + O(t 3 )= √ (det g 1 − 2tuH + t 2( 12 ‖∇u‖2 + u 2 K ) )+ O(t 3 )Step 4: Since A V (f t ) = ∫ V√ det gt dx we obtain the claim by integration. □Remark. For two reasons the variation u is required to have compact support. First, allintegrals over the support become finite. Second, all functions being integrated then takea maximum; thus f t is an immersion for small |t|. Given these two properties the theoremapplies to U in place of V as well.For various purposes, parallel surfaces with u ≡ 1 are interesting. Their area expansion isparticularly simple:Corollary 24. Let f : U → R 3 be a surface. Then for sufficiently small |t| the parallelsurface f t := f + tν is an immersion and, provided the integrals exist, we have∫∫(39) A U (f t ) = A U (f) − 2t H dS + t 2 K dS + O(t 3 ), t ∼ 0.UUWriting the first order expansion in terms of area elements at a point p ∈ U,dS t = (1 − t2H)dS 0 + O(t 2 )we see that mean curvature H can be thought of the first order change in the area elementwhen comparing with parallel surfaces.In the expansion (3.3) the O(t 3 ) can be skipped, that is, it vanishes. This can be shownbe a different proof (problems?).Problems. 1. For a curve c consider a variation c t := c + tuν and calculate the expansionfor length.2. Calculate the area expansion only up to the term linear in t, but for general dimension.8. Lecture, Friday 4.5.12

i 3.2 – date: July 27, 2012 253.2. First and second variation of area, stationarity and stability.Definition. Let f ∈ C 2 (U n , R m ), u ∈ C0(U, 1 R), and V := supp u. The first and secondvariation of area with respect to the normal variation vector field uν are, respectively,δ uν A V (f) := d dt A V (f + tuν) ∣ and δ 2t=0uνA V (f) := d2dt A 2 V (f + tuν) ∣ .t=0These variations appear in the expansion(40) A(f t ) = A(f) + tδ uν A(f) + 1 2 t2 δ 2 uνA(f) + O(t 3 )which we computed explicitely in (37).Definition. We call a hypersurface f ∈ C 1 (U n , R n+1 ) stationary (or critical) for area ifδ uν A supp u (f) = 0 for all u ∈ C 1 0(U, R).Our main result is the following characterization of minimal surfaces:Theorem 25. A hypersurface f : U n → R n+1 is stationary if and only if H ≡ 0.To prove the theorem for n = 2, we conclude from the expansion (37):∫δ uν A V (f) = −2 uH dS for all u ∈ C0 ∞ (U, R).The proof is completed by the next lemma.VLemma 26 (Fundamental lemma of the calculus of variations). Suppose h ∈ C 0 (U n , R)and∫uh dx = 0 for all u ∈ C0 ∞ (U, R).Then h ≡ 0.UProof. Suppose that h(x) ≠ 0 for some x ∈ U, say h(x) > 0. By continuity of h thereis ε > 0 such that h(y) > 0 for all y ∈ B ε (x). Then choose a function u ≥ 0 such thatu(x) = 1 and u has compact support in B ε (x). The existence of such functions is not hardto show. Then ∫ uh dx > 0, contradicting the assumption.□Problem. Does the lemma hold for discontinuous h as well?Remark. A normal variation seems more restrictive than a general variation f t := f + tX,where X ∈ C 1 0(U, R n ) is an arbitrary vector field with compact support. However, for |t|small, any general variation is a reparameterization of a normal variation, which leaves thearea unchanged. So the effect of the variation in direction X and the normal variationwith u := 〈X, ν〉 is the same.

26 K. Grosse-Brauckmann: Minimal surfaces, SS 12The concepts we have introduced so far extend to the setting of global surfaces M. Itis sufficient to decompose a global surface M into countably many disjoint parameterizedsurfaces f i : U i → R n+1 , i.e., M = ⋃ i∈N f i(U i ). Then A(M) := ∑ i∈N A U i(f i ), and we cansimilarly define normal variations, compact support, expansions of area, and stationarity,since all geometric quantities we use are parameterization independent.We now discuss whether a surface attains a minimum of area. We could distinguish:Definition. (i) A parameterized surface f ∈ C 1 (U, R n ) is called a local minimum of areaif for each u ∈ C 1 0(U, R) there exists t 0 > 0 such that(41) A V (f) ≤ A V (f + tuν) for all |t| < t 0 , V := supp u.In particular, (41) holds for f an absolute minimum of area, that is in caseA V (f) ≤ A V ( ˜f) for all ˜f ∈ C 1 (U, R n ) with V := supp(f − ˜f) ⊂⊂ U.The analysis in our infinite dimensional space C 1 0(U, R) is now completely analogous to thecase of a univariate A = A(x): For a critical point with A ′ (x) = 0, the condition A ′′ (x) > 0implies local minimality, while a local minimum implies A ′′ (x) ≥ 0.Definition. A minimal surface f : U n → R m is(i) strictly stable [strikt stabil] if δ 2 uνA V (f) > 0 for all nonzero u ∈ C 1 0(U, R);(ii) it is stable if the same holds with weak inequality ≥.Then we can state:Theorem 27. (i) If a minimal surface f : U 2 → R 3 is strictly stable it is a local minimumof area.(ii) If f is a local minimum, then f is stable.The proof is immediate from the expansion (40), which is dominated by the second orderterm.3.3. Stability for some examples. The preceding results imply that to decide aboutthe minimizing properties of f ∈ C 2 (U 2 , R 3 ) we need to decide if∫1(42)2 ‖∇u‖2 + u 2 K dSUis positive or nonnegative for all 0 ≢ u ∈ C 1 0(U, R). In the following examples we will seean interplay between the terms ‖∇u‖ 2 ≥ 0 and u 2 K ≤ 0.Proposition 28. (i) The plane R 2 ⊂ R 3 is strictly stable.(ii) The catenoid is not stable.

i 3.3 – date: July 27, 2012 27In fact, a theorem by Barbosa-DoCarmo proves that the plane is the only complete minimalsurface which is stable.Proof. (i) Since K ≡ 0 we have δuνA 2 V (f) = ∫ VC0(U, 1 R).12 ‖∇u‖2 dS > 0 for all nonzero u ∈(ii) Consider the conformal parameterization f(r, ϕ) = (cosh r cos ϕ, cosh r sin ϕ, r) wheref : R × [0, 2π) → R 3 , as in Sect. 2.1. Our variation will be in the outward direction ona sufficiently large annular region, symmetric to the waist. To be precise, let d ≥ 1 andchoose choose u d ∈ C 0,10 (R × [0, 2π], R)⎧⎪⎨1 for |r| < d − 1,u d (r, ϕ) := d − |r| for |r| ∈ [d − 1, d],⎪⎩0 for |r| > d.The only nonzero partial of u d is ∂ 1 u(r, ϕ) for |r| ∈ [d − 1, d]. Due to g 11 = 1 + sinh 2 r =cosh 2 r this gives‖∇u d ‖ 2 = g 11 ∂ 1 u d ∂ 1 u d =Moreover, dS(r, ϕ) = cosh 2 r drdϕ, and so∫∫12π2 ‖∇u d‖ 2 dS =R×[0,2π]Finally, ∫ K dS = −4π, so thatR×[0,2π]∫limd→∞R×[0,2π]1cosh 2 r (±1)2 =0∫ d2d−11cosh 2 r ,|r| ∈ [d − 1, d].12 cosh 2 r cosh2 r drdϕ = 2π.12 ‖∇u d‖ 2 + u 2 dK dS = 2π − 4π = −2π < 0.Thus (42) and also (41) fail for sufficiently large d. Note that by approximation it is infact enough to check (42) for a Lipshitz function.Our calculation confirms that a large annular piece of the catenoid is not a local minimum.Our result is consistent with the soap film experiment: If a given pair of concentric circlesbounds two catenoids, then soap film will never produce the (inner) catenoid with a smallneck.Note that our function u d has compact support when considered on the catenoid. In thisregard, the exposition would be cleaner had we chosen a parameterization of the catenoidby R 2 \ {0}.□Problem. Prove that the plane is an absolute minimum of area.9. Lecture, Wednesday 9.5.12We want to present a stability result for surfaces which are periodic in the following sense:

28 K. Grosse-Brauckmann: Minimal surfaces, SS 12Definition. A periodic surface M ⊂ R m is a global surface which is invariant under alattice [Gitter] Λ := { ∑ ki=1 a iv i : a i ∈ Z} of k linearly independent vectors v 1 , . . . , v k ∈ R m .For example, if v i = e i for i = 1, . . . m, then Λ = Z m , and the quotient space R m /Λ is thestandard m-torus. We need the concept of the quotient surfaceM/Λ := {x + Λ : x ∈ M} ⊂ R m /Λ := {y + Λ : y ∈ M},where we can regard M/Λ a fundamental domain of the surface. It is beyond our presentsetup, but necessary for the following, to assert that the quotient surface is a manifold,that is, can naturally be considered a surface.A surprising fact about triply periodic minimal surfaces in R 3 is:Theorem 29. Suppose a periodic minimal surface in R 3 has a compact quotient M/Λ.If M is not a union of planes then M/Λ is a local maximum of area in its parallel surfacefamily, and in particular not stable.Note that the compactness assumption on the quotient reduces the problem to finite area.Moreover, when considered on M/Λ the function u ≡ 1 has compact support.Proof. Note that the variation f +tν is also periodic and hence well-defined on the quotient,The expansion applied to M/Λ gives∫A(M t /Λ) = A(M/Λ) + t 2 K dS + O(t 3 ).By the assumed nonplanarity the Gauss curvature K ≤ 0 does not vanish identically.Hence ∫ K dS < 0, so that A(M t /Λ) < A(M/Λ) for small |t| ≠ 0.□M/ΛRemark. There are many examples of triply periodic surfaces with compact quotient. However,the only doubly periodic minimal surface with a compact quotient is a finite unionof planes, and singly periodic minimal surfaces cannot have a compact quotient. This canbe shown using the maximum principle.3.4. Leaves of minimal foliations are stable. If we want to assert a minimal surfaceis not a local minimum or is unstable we need to exhibit a test function u such that (41)fails. However, to verify the stability of a minimal surface is harder since it means to provean inequality for all u. In the following we present a test for local minimality, which wederive independently from the variation formula.Definition. Suppose Ω ⊂ R n+1 is a domain such that the n-dimensional slices Ω t := {x ∈Ω, x n+1 = t} are again domains. Then an immersion F ∈ C 1 (Ω, R n+1 ) is called a foliation[Blätterung] with leaves [Blätter] F t : Ω t → R n+1 .M/Λ

i 3.4 – date: July 27, 2012 29This definition is a special case of a more general notions: More general than foliations withhypersurfaces are foliations with n-dimensional leaves in R n+k . Moreover, by requiring theexistence of local charts as above we get a more general definition.Examples. 1. Choose the identity for f to foliate R n+1 with parallel horizontal hyperplanes.2. Polar coordinatesF : R 2 → C ∗ := C \ {0}, F (x, t) := exp(x + it) = e x (cos t, sin t)give a foliation of C ∗ with (exponentially parameterized) radial rays from the origin.3. R n+1 \ {0} is foliated by the concentric spheres {S n t , t > 0}; in fact, for the abovedefinition to apply, we rather say that R n+1 minus a ray is foliated by spheres. To verifythis, take your favourite parameterization f of S n minus the north pole by R n . ThenF t (x) := tf(x) for (x, t) ∈ R n × (0, ∞) gives the foliation.4. Let ( x, u(x) ) be a graph, where u: U → R. Then a foliation by translated graphs isgiven by(43) F : U × R → R n+1 with F t (x) := ( x, u(x) + t ) .We are interested in vector fields normal to a foliation. It will be convenient to have themdefined on the image. In order for F −1 to be defined, we assume F is an embedding, thatis a diffeomorphism onto its image V := F (Ω). Orient the normals ν t of the surfaces F tconsistently, meaning that ν : Ω → S n is continuous. Since F is a diffeomorphism, we candefine a normal field on V by setting N(P ) := ν(F −1 (P )).Consider our last example, namely the foliation with translated graphs, ( x, u(x)+t ) , whereu: U → R. Let ν be the normal of the graph (x, u(x)). ThenN : U × R → S n , N(x 1 , . . . , x n , x n+1 ) := ν(x 1 , . . . , x n )is a unit normal field. For a graph (x, u), the normal ν = (ν 1 , . . . , ν n , ν n+1 ) satisfiesnH = div n∇u√1 + |∇u|2 = − div n(ν 1 , . . . , ν n ),by (10); here div n denotes divergence in R n . Thus u is minimal if and only if div n ν ≡ 0.Since N does not depend on x n+1 we havediv n+1 N = ∂ 1 N 1 + . . . + ∂ n N n + ∂ n+1 N n+1 = div n ν + ∂ n+1 N n+1 = div n νHence minimality of a graph is equivalent to div n+1 N ≡ 0. The same continues to holdfor any foliation with minimal surfaces:

30 K. Grosse-Brauckmann: Minimal surfaces, SS 12Lemma 30. Suppose a diffeomorphism F ∈ C 2 (Ω, R n+1 ) is a diffeomorphic foliation. Theneach leaf F t is minimal if and only if the normal vector field N : F (Ω) → S n satisfies onF (U)div n+1 N ≡ 0.Proof. Suppose that at a point P = F t (p) the tangent space T p F t is the horizontal plane{x n+1 = const}. Then N(P ) = e n+1 . In any case, also |N| ≡ 1 which gives 〈∂ n+1 N, N〉 = 0.This gives at P0 = 〈 ∂ n+1 N(P ), N(P )} {{ }e n+1〉= ∂n+1〈N(P ), en+1〉= ∂n+1 N n+1 (P ).Since T p F t is horizontal, the leaf F t (p) can locally be represented as a graph. Then foreach 1 ≤ i ≤ n the partial derivative ∂ i N agrees with some directional derivative ∂ i ν of thesurface F t . Since F t is minimal, (10) yields ∂ 1 N 1 + . . . + ∂ n N n = 0. Thus div n+1 N = 0, asdesired.To recover the same result in the general case, we want to rotate a given foliation such thatT p F t becomes a horizontal plane {x n+1 = const}. We claim that the divergence remainsinvariant under motion. To see this, represent the divergence div N (P ) as an integral overB ε = B ε (P ) using the divergence theorem [Satz von Gauß]:1div N(P ) = limε→0 V (B ε )∫1div N dx = limB ε→0 εV (B ε )∫∂B ε〈N, ν ∂Bε 〉 dSFor coordinates rotated about P , the integral on the right hand side is unchanged, and sothe left hand side remains unchanged. The same works for translation.□Theorem 31. Let f, ˜f : U n → V n+1 be two minimal hypersurfaces such that(i) K n := supp( ˜f − f) ⊂ U 0 is compact, and(ii) there is an open (not necessarily connected) precompact set ∆ ⊂ V with ∂∆ = f(K) ∪˜f(K).If there exists a diffeomorphic foliation with minimal surfaces F : Ω → V for Ω, V ⊂ R n+1containing the leaf f = F 0 : U = Ω 0 → V then(44) A K ( ˜f) ≥ A K (f).Equality can only hold if ˜f reparametrizes f.In particular, (44) implies f is a local minimum of area.The converse of the theorem is also true: A neighbourhood of a strictly stable minimalsurface can be foliated with minimal surfaces ([N], § 109).

i 3.4 – date: July 27, 2012 31Proof. Let us assume first that ∆ ⊂ V has only one connected component. For simplicitywe pick Gauss maps ν, ˜ν for f, ˜f, such that ν is exterior and ˜ν interior to ∆. We applythe divergence theorem to the compact set ∆:0div N=0=∫∆∫div N dx =∂∆∫〈N, ν ∂∆ 〉 dS =f(K)∫〈N, ν〉 dS −〈N, ˜ν〉 dS˜f(K)Strictly speaking we mean by ν the mapping ν ◦ f −1 etc., so that everything becomesdefined on the image.Now we use that ν = N| f(K) , but 〈N, ˜ν〉 ≤ |N||˜ν| = 1 so that∫∫∫A K (f) = 〈N, ν〉 dS = 〈N, ˜ν〉 dS ≤f(K)˜f(K)˜f(K)dS = A K ( ˜f).Equality can hold only if ˜ν agrees with N, that is, for ˜f a reparameterization of f.In general ∆ is the union of the connected components. Then the above reasoning appliesto each component, and a summation over all components leads to the same conclusion. □10. Lecture, Friday 11.5.12Examples. 1. Consider a standard catenoid M = M 1 together with its tangent cone C.Let A ⊂ M be the annular component of M \ C which has compact closure. Suppose thatA is parametrized byf 1 : (−r 0 , r 0 ) × [0, 2π) → R 3with f 1 (r, ϕ) = (cosh r cos ϕ, cosh r sin ϕ, r),see Sect. 2.1. It can be checked that the scaled catenoidal annulif : (−r 0 , r 0 ) × [0, 2π) × (0, ∞) → R 3 ,f(r, ϕ, t) := (t cosh r cos ϕ, t cosh r sin ϕ, tr),give a foliation of the exterior of the cone, that is, the doubly connected component V ofR 3 \ C. Thus by the theorem, any compact subset of A is a local minimum, in agreementwith the soap bubble experiment.2. Rotate the helicoid M about its vertical axis A by R ϕ ∈ SO(3), for ϕ in a neighbourhoodof 0. To see this defines a foliation, we need to check that the rotation direction X(p) :=dR dϕ ϕ(p) is not a tangent direction, as long as p is in M \ A. But X(p) is a nonzerohorizontal vector perpendicular to the rulings, while the tangent space at p is spanned bythe rulings and the non-horizontal tangent vector to the helix through p. Thus “half ahelicoid” is a local minimum. Note that our parametric description in fact extends to a(generalized) foliation of all of R 3 \ A with manifolds/global surfaces, namely with rotatedhelicoids.3. For the Scherk surface, the vertical translations give a minimal foliation, showing thatany compact subset of the Scherk surface is a local minimum of area.

32 K. Grosse-Brauckmann: Minimal surfaces, SS 12In fact, any graph leads to a foliation with translated graphs (43), and we obtain in general:Corollary 32. Any compact subset of a minimal graph is a local minimum of area.We can use this result to prove that sufficiently small pieces of minimal surfaces are localminima of area.Theorem 33. Given a minimal hypersurface f : U → R n+1 and p ∈ U, there is a neighbourhoodof f(p) which is a local minimum of area.The statement implies that any minimal surface can locally be realized with soap film.Proof. By the implicit mapping theorem, some neighbourhood of f(p) can be representedas a graph. The claim then follows from the Corollary.□Remark. When correctly phrased, the results of this section apply also to foliations whichare not diffeomorphisms (embeddings).References. [DHKW] Sect. 2.7, 2.8; Grosse-Brauckmann, Pacific Journal 175 (1996)4. Maximum principlesThe maximum principle is perhaps the most important tool to analyse minimal hypersurfaces.We introduce it first for harmonic functions, then generalize to linear ellipticequations, and finally deal with the case of the nonlinear minimal surface equation.4.1. Harmonic functions. Let us look first at the model case for an elliptic equation.This is the Laplace equation ∆u = 0, whose solutions u ∈ C 2 are called harmonic.Examples of harmonic functions include: 1. Constant and linear functions, 2. Real or imaginaryparts of holomorphic functions, 3. functions like x 2 − y 2 are harmonic, 4. coordinatefunctions of conformally parameterized minimal surfaces.Suppose y is a local maximum of u. Then ∂ ii u(y) ≥ 0 for all i, and so ∆u(y) ≥ 0. Inparticular if u is harmonic and the Taylor series at y starts with a second order term,y cannot be a strict local maximum. The (weak) maximum principle makes the sameassertion without any assumption on the Taylor series:Proposition 34. Let U ⊂ R n be a bounded domain and u ∈ C 2 (U, R) ∩ C 0 (U, R) beharmonic. Thensup∂Uu = sup u and inf u = inf u.U∂U U

i 4.1 – date: July 27, 2012 33Note first that for a continuous function sup U u = sup U u. Second, the boundedness of Uis essential – find a counterexample for U unbounded!Proof. Step 1: Consider for ε > 0 the auxiliary functionThen ∆v = ∆u + 2nε > 0 for all x ∈ U.v(x) := u(x) + ε|x| 2 .Step 2: On the compact set U, the continuous function v takes a maximum at y = y(ε) ∈ U.Suppose that y is an interior point of U. Then for each i the restriction t ↦→ v(y + te i ) hasa maximum at t = 0 and so ∂ ii v(y) ≤ 0. Summing, we find ∆v(y) ≤ 0, in contradiction to∆v > 0. Consequently, y ∈ ∂U.Step 3: The function ε|x| 2 is non-negative on U, and due to the boundedness of U it isbounded by a number C > 0. Consequently we can estimatesupUu ≤ supU(u + ε|x|2 ) ≤ sup∂Uand therefore the claim follows. Similarly for the minimum.( ) u + ε|x|2≤ sup u + εC for all ε > 0,∂U□Another way to derive the maximum principle is by using a surprising property of harmonicfunctions: Each value u(x) is the mean of the function u over any ball B r (x) ⊂ U, that is,∫u(x) = 1|B r(x)| B r(x)u(y) dy, see [GT, Thm.2.1]. This property does not generalize to themore general case of elliptic equations, so we do not discuss it any further here.Remark. For the maximum principle to hold, it is sufficient to assume that ∆u ≥ 0.Similarly ∆u ≤ 0 implies the minimum principle.The most important application of the maximum principle is to show the uniqueness ofthe Dirichlet problem to Poisson’s equation:Proposition 35. Let U be a bounded domain and f ∈ C 0 (U, R). Suppose u 1 , u 2 ∈C 2 (U, R) ∩ C 0 (U, R) satisfy u 1 (x) = u 2 (x) for all x ∈ ∂U. Then if u 1 and u 2 solve thePoisson equationthey must coincide.∆u(x) = f(x) for all x ∈ UProof. The function v := u 1 −u 2 satisfies ∆v ≡ 0 and has boundary values v = u 1 −u 2 = 0.Thus by Proposition 34 we havesupUso that 0 = v = u 1 − u 2 on U.v = sup v = 0 and inf v = inf v = 0∂UU ∂U□

34 K. Grosse-Brauckmann: Minimal surfaces, SS 12It is a much more difficult task to construct such solutions. For special domains U, likethe ball or a half-space, there are explicit formulas. In general, however, some abstractexistence schemes must be used (“Perron process”, see [GT] Sect. 2.8).Let us conclude this section with an example of a more general equation which does notsatisfy a maximum principle: Consider the eigenvalue problem∆u + λu = 0 for λ > 0,where u has zero boundary values. If we let U be the cube (− π 2 , π 2 )n then the functionsf(x) := cos(k 1 x 1 ) · . . . · cos(k n x n )with k 1 , . . . , k n ∈ Nhave zero boundary values and satisfy ∆u + ( k 2 1 + . . . + k 2 n)u = 0. But u ≡ 0 solves theequation as well. Hence for λ a sum of squared natural numbers there are at least twosolutions. Functions satisfying ∆u = λu are called eigenfunctions of the Laplace operator.References. [GT], Sect. 211. Lecture, Monday 14.5.124.2. The maximum principle for linear elliptic equations. On our way to the minimalsurface equation we now consider a more general class of equations which behavesmuch alike the Laplace equation.Definition. (i) A linear partial differential operator of second order L is given byn∑n∑(45) Lu(x) := a ij (x) ∂ ij u(x) + b k (x) ∂ k u(x) = trace(A d 2 u) + 〈b, ∇u〉i,j=1k=1where a ij , b k are functions on U, such that the matrix A = (a ij ) is symmetric and |b| isbounded on U.(ii) We say L is uniformly elliptic if additionly there exists λ > 0 such thatn∑(46)a ij (x)ξ i ξ j ≥ λ|ξ| 2 for all ξ ∈ R n \ {0} and x ∈ U.i,j=1The ellipticity condition (46) means that the lowest eigenvalue of A is bounded belowuniformly on U. The symmetry assumption on A can always be achieved (how?).Note that linearity means L[u + v] = L[u] + L[v].Examples. 1. If A is the identity matrix and b = 0 then L is the Laplacian.2. A special case is A constant. Then (46) is equivalent to A being positive definite.3. Later, we will encounter operators in divergence form, that is,Lu(x) = ∑ i,j∂ i (ãij ∂ j u ) = div ( Ã(x)∇u(x) ) ,

i 4.2 – date: July 27, 2012 35where à is a symmetric matrix of functions in C1 (U, R). Indeed, differentiating we findLu = ∑ ã ij ∂ ij u + ∑ ( ∑∂ i ã ij) ∂ j u.i,jjThus if ã ij satisfies the ellipticity condition (46) and if we set b k := ∑ i ∂ iã ik then L as in(45) is elliptic.Lemma 36 (Weak maximum principle). Let U be bounded, u ∈ C 2 (U, R) ∩ C 0 (U, R), andL be uniformly elliptic. If Lu ≥ 0 for all x ∈ U then(47) supUSimilarly, if Lu ≤ 0 then inf U u = inf ∂U u.u = sup u∂UProof. Step 1: We consider an auxiliary function v := u + εe γx 1where ε > 0. Moreover wecan γ > 0 large enough for the last inequality in(48) Lv L linear= Lu + εL(e γx 1) Lu≥0≥ ε ( a 11 γ 2 + b 1 γ ) e γx L ell.1≥ ε ( λγ 2 − |b 1 |γ ) e γx |b| bdd.1> 0to hold.Step 2: On the compact set U, the function v takes a maximum at some point y = y(ε) ∈ U.Suppose y is an interior point of U. Then ∇v(y) = 0 and d 2 v(y) is negative semidefinite. sothat Lv(y) = trace ( A(y)d 2 v(y) ) We claim that the product of the positive definite matrixA(y) with negative semidefinite matrix d 2 v(y) is again negative semidefinite.For any matrices M, N, we have trace(MN) = trace(NM). Thus as long as we are onlyinterested in the trace of a matrix product we may multiply the matrices in any order. Inparticular, for each T ∈ GL(n),trace ( A d 2 v ) = trace ( (T −1 AT )(T −1 d 2 v T ) ) = trace (  ̂d 2 v ) .Since A is symmetric there is a matrix T ∈ O(n) such that  := T −1 AT is diagonal. Notethatξ t( T −1 MT ) ξ = (ξ t T t )M(T ξ) = (T ξ) t M(T ξ),so that ̂d 2 v(y) = T d 2 v(y)T −1 is still negative semidefinite. In particular all diagonal entriesmust have the sign ̂d 2 v(y) ii ≤ 0. Similarly  is positive definite, meaning âii > 0 for all i.If follows thatLv(y) = trace ( A(y)d 2 v(y) ) n∑= â ii (y) ̂d 2 v(y) ii≤ 0,which establishes our claim. The contradiction to (48) proves that y ∈ ∂U.Step 3: We havesupUu ≤ supUii=1(u + εe γx 1) = sup(u + εe γx 1) = sup u + εC,∂U∂U

36 K. Grosse-Brauckmann: Minimal surfaces, SS 12where C is a bound on the function e γx 1over U. Since this holds for each ε > 0 and e γx 1is bounded on U, the claim follows.□As for harmonic equations, the uniqueness of solutions for elliptic equations with prescribedboundary values is again immediate:Theorem 37. Let U be bounded and f ∈ C 0 (U, R). If u 1 , u 2 ∈ C 2 (U, R) ∩ C 0 (U, R) aretwo solutions of the elliptic boundary value problem Lu = f with the same boundary valuesu 1 | ∂U = u 2 | ∂U then u 1 = u 2 .Proof. The function u 1 − u 2 has zero boundary values, satisfies L(u 1 − u 2 ) = 0, and so themaximum principle gives u 1 − u 2 ≡ 0.□For the following lemma, the boundary of the domain needs to be sufficiently good: Adomain U satisfies an interior sphere condition at p ∈ ∂U, if there is a ball B r (q) ⊂ Uwith p ∈ ∂B r (q). An interior normal ν to U at p is the inner normal of B r at p. A domainwhich is C 2 can be seen to satisfy an interior sphere condition with r essentially the inverseof the curvature bound.Lemma 38 (Hopf boundary point lemma). Assume the following:• U satisfies an interior sphere condition at p ∈ ∂U with interior normal ν.• u ∈ C 2 (U, R) ∩ C 0 (U, R) satisfies Lu ≥ 0 where L is uniformly elliptic and A bounded.• u(p) is a strict boundary maximum, u(x) < u(p) for all x ∈ U.Then if the normal derivative exists it satisfies∂u(p) < 0.∂νProof. Let B r = B r (q) ⊂ U be a ball with p ∈ ∂B. We assume q = 0 and u(p) = 0 withoutloss of generality.We need a comparison function ϕ: U → R which on the spherical shell Ω := B r (0)\B r/2 (0)satisfiesϕ ≥ 0 and Lϕ ≥ 0 on Ω, ϕ = 0 and ∂ϕ∂ν > 0 on ∂B r.We claim that for suitable c > 0ϕ(x) := e −c|x|2 − e −cr2has the properties claimed. The claims for the values of ϕ are obvious. Moreover, ϕ isrotationally symmetric and so has the same normal derivative at any point of ∂B r . So it

i 4.3 – date: July 27, 2012 37is sufficient to compute the normal derivative at the point re 1 with normal ν = −e 1 . From∂ i ϕ(x) = −2cx i e −c|x|2 we obtain(49)∂ϕ∂ν (re 1) = −∂ 1 ϕ(re 1 ) = 2cre −c|x|2 > 0.We finally computeLϕ(x) = e −c|x|2( 4c 2 ∑ i,ja ij x i x j − 2c ∑ i)(a ii + b i x i ) ≥ 2ce −c|x|2[ 2cλ|x| 2 − ∑ i]a ii − |b||x| .Certainly A, b, |x| are bounded on all of U. Moreover, we have |x| 2 ≥ r 2 /4 on Ω. Hence wecan choose c large enough to have [. . .] ≥ 0 on Ω, proving Lϕ ≥ 0.The function u is negative on U. Thus we can take ε > 0 such that u + εϕ ≤ 0 on ∂B r/2 ;on ∂B r this is ≤ 0 anyway. HenceL(u + εϕ) ≥ 0 on Ω and u + εϕ ≤ 0 on ∂Ω.The weak maximum principle (47) applies to give u + εϕ ≤ 0 on Ω, while (u + εϕ)(p) = 0.Thus ∂ (u + εϕ)(p) ≤ 0, which implies, as desired,∂ν∂u∂ν(49)(p) ≤ −ε∂ϕ(p) < 0.∂νAt a general boundary with corners, however, there may exist normal derivatives which donot vanish: Consider the harmonic function u(x, y) = xy on the quadrant (0, ∞)×(−∞, 0).At the origin, the boundary of this domain has a corner. Since ∇u(0, 0) = 0 any directionalderivative at the origin vanishes.12. Lecture, Wednesday 16.5.12We can now conclude:Theorem 39 (Strong maximum principle). Let L be uniformly elliptic and Lu ≥ 0 on U.If u achieves an interior maximum then u is constant.□Proof. Suppose, contrary to the statement, that u achieves a maximum m ≥ 0 at aninterior point but that u is non-constant. Then the open set U < := {x ∈ U, u(x) < m} isnon-empty. Pick a point q ∈ U < be that is closer to ∂U < ∩ U than to ∂U, and considerthe largest open ball B(q) ⊂ U < . Then u(p) = m for some point p ∈ ∂B(q) while u < mon B(q). But by the Hopf boundary point lemma, ∇u(p) ≠ 0, contradicting the fact thatm is the maximum of u.□References. [GT], Sect. 3.1 and 3.2

38 K. Grosse-Brauckmann: Minimal surfaces, SS 124.3. The maximum principle for minimal surfaces. To formulate a maximum principlefor minimal graphs ( x, u(x) ) , our starting point is the minimal surface equation inits divergence form (10): Let(50) Qu :=n∑()∂ i u∂ i √ − nH,1 + |∇u|2i=1then a surface ( x, u(x) ) of mean curvature H = H(x) satisfies Qu = 0.The equation Qu = 0 is nonlinear (Qu = Qv = 0 does not imply Q(u + v) = 0), but stillthere is a maximum principle for minimal graphs (x, u(x)) which forbids a one-sided touching.We formulate it somewhat more general for graphs with coinciding mean curvature.Theorem 40. Let U be bounded and u, v ∈ C 2 (U, R) ∩ C 0 (U, R) describe two graphs withthe same mean curvature function H ∈ C 0 (U, R) with respect to the upper normal. Supposeu ≤ v on U. Then we have:(i) Interior maximum principle: If u(p) = v(p) at some interior point p ∈ U, then u ≡ v.(ii) Boundary maximum principle: Suppose ∂U is smooth at p ∈ ∂U and u and v have anormal derivative at p. If u(p) = v(p) and ∂u ∂v(p) = (p) then u ≡ v.∂ν ∂νCounterexamples 1. If we do not assume anything on the normal derivative in (ii), thenthe statement cannot hold: Consider two half-planes which meet in a common line.2. Some boundary regularity is necessary for (ii) to hold: The Enneper surface is graphin a neighbourhood of the origin O. For U a quadrant the Enneper surface over O ∩ Utakes 0 boundary values. Nevertheless any “normal” derivative of the Enneper graph at 0vanishes.Proof. As we have seen in the uniqueness proofs for Laplace’s or elliptic equations, it isuseful to consider the difference u − v. Our goal is to show that the difference satisfiessome linear elliptic equation.Seta ∈ C ∞ (R n , R n ), a i (p) :=p i√1 + |p|2 , 1 ≤ i ≤ n.

i 4.3 – date: July 27, 2012 39For each x ∈ U the graphs u, v have mean curvature H(x). Hence we have0 = Qu − Qv = ∑ ∂ i a i (∇u) − ∂ i a i (∇v) = ∑ )∣∂ i a i ∣∣t=1(t∇u + (1 − t)∇vt=0ii∫ 1d ∑)= ∂ i a(t∇u i + (1 − t)∇v dt0 dti= ∑ ∫ 1( ∂ai ((∂j∂ i t∇u + (1 − t)∇v)u − ∂ j v )) dti,j 0 ∂p j= ∑ ([ ∫ 1∂a i () (∂j∂ i t∇u + (1 − t)∇v dt]u − ∂ j v )) .i,j0 ∂p jThus setting [. . .] =: a ij (x) the difference w := u − v satisfies the equationLw := ∑ i,j∂ i(a ij (x)∂ j w ) = ∑ i,j∂ i(a ij (x)∂ j (u − v) ) = Qu − Qv = 0,which is a linear partial differential equation in divergence form. We have achieved linearityof this equation at the expense of having the coefficients a ij depend on u, v. This, however,will not cause any difficulties since, for the present proof, these two functions are givenonce and for all.We claim that L is elliptic on any compact subset K ⊂⊂ U. To see this, set P (t, x) :=t∇u(x) + (1 − t)∇v(x). Then∂a i∂p j(P ) =δ ij√1 + |P |2 −P i P j√ 3, 1 ≤ i, j ≤ n.1 + |P |2For any ξ ∈ R n \ 0 let us now first apply the Schwarz inequality and then use the fact that|P (t, x)| takes a maximum over the compact set {x ∈ K, t ∈ [0, 1]} to define a positivenumber λ = λ(K, u, v) such that the following holds:∑i,j∂a i∂p j(P )ξ i ξ j =1√1 + |P |2 3 ( (1+ |P |2 ) |ξ| 2 − 〈P, ξ〉 2 )≥1√1 + |P (t, x)|2 3 |ξ|2 > λ|ξ| 2Integration of this equation with respect to t verifies the ellipticity condition for L,∑a ij (x)ξ i ξ j = ∑ [ ∫ 1∂a i ] ∫ 1(P (t, x)) dt ξ i ξ j > λ|ξ| 2 dt = λ|ξ| 2 .i,ji,j 0 ∂p j 0Let us prove (i). Then by the strong maximum principle, Thm. 39, w = u − v ≡ 0 for anycompact set K ⊂ U which contains p; this implies u = v on all of U.Now we prove (ii). If the normal derivatives of u and v at p exist, then |P | 2 is boundedon an interior ball B tangent to p. Hence Lw = L(u − v) is elliptic on B. We can assumethe strict inequality u − v < 0 on B, since otherwise u(q) = v(q) at some point q ∈ B and

40 K. Grosse-Brauckmann: Minimal surfaces, SS 12the claim follows from the interior maximum principle. But in case u − v < 0 the HopfBoundary Point Lemma 38 contradicts our assumption ∂u ∂v(p) = (p).□∂ν ∂νThe uniqueness of the Dirichlet problem for minimal surfaces is a direct consequence:Theorem 41. Let U be bounded and u ∈ C 2 (U, R)∩C 0 (U, R). Suppose f(x) = ( x, u(x) ) isa minimal graph. Then ( x, u(x) ) is the unique minimal graph with these boundary values.Proof. Suppose u ≢ v are two solutions with the same boundary values. Then u − v (orv − u) attains a positive maximum m > 0 at p ∈ U. But u = v on ∂U, and so p ∈ U. Thusu ≤ v + m with equality at p. The interior maximum principle Thm. 40(i) then provesu ≡ v + m, which is a contradiction due to u = v along ∂U.□Existence is a more serious problem, and is only guaranteed for arbitrary boundary valuesin case U is convex, see Thm. III.?? below.13. Lecture, Monday 21.5.124.4. Convex hull property. We discuss further applications of the maximum principle.In the following a minimal surface will be of a global (embedded) hypersurface M ⊂ R n+1or a parameterized (immersed) surface M = f(U). We always assume M is connected andnon-empty.We need the boundary ∂M := M \ M of M. A more conceptual definition would be interms of a manifold with boundary. Note that if M := f(U) then possibly f(∂U) is aproper subset of ∂M.A half-space is a connected component of R n minus an affine hyperplane. Its boundinghyperplane can only contain boundary points of a minimal surface:Lemma 42. Let E be a closed halfspace bounded by a hyperplane H. Suppose an interiorpoint p of a minimal surface M ⊂ E is contained in H. Then all of M is a subset of H.Proof. The tangent space of M at p must agree with the hypersurface H. Therefore in aneighbourhood U ⊂ M of p, the surface M can be represented as a graph over H. Theinterior maximum principle implies U ⊂ H. More generally this argument proves that theset M ∩ H is an open subset of M. Since by definition this set is (relatively) closed as well,the connectedness of M gives it is all of M.□Recall that the convex hull of a set X ⊂ R n is the setconv(X) = ⋂ {E : X ⊂ E, E open half-space}.

i 4.4 – date: July 27, 2012 41We use two facts about the convex hull of X (exercise). First, it agrees with the intersectionof the closed half-spaces containing X. Second, each point in the boundary of a convexset S is contained in a so-called supporting hyperplane H, having the property that ahalf-space E with H = ∂E contains S.Theorem 43 (Convex Hull Property). Suppose M ⊂ R n+1 is a minimal surface which isbounded, i.e., M ∪ ∂M is compact. Then(i) M is contained in the convex hull of its boundary values, M ⊂ conv(∂M), and(ii) if an interior point p ∈ M is in ∂ conv(∂M) then M is subset of a plane.Proof. The idea is to move hyperplanes towards M. By the maximum principle they cannothit an interior point of M before hitting ∂M. Let us make this idea precise.Consider a closed halfspace E 0 such that M ⊂ E 0 and the boundary H 0 := ∂E 0 containsa point of ∂M. We claim that also M ⊂ E 0 . If not, there is a point p ∉ E 0 . By thecompactness of M, we can assume that p is a point of M \ E 0 whose distance to H 0 ismaximal. But such a point contradicts the lemma, and our claim is proved.We conclude that M is contained in the intersection of all closed halfspaces which contain∂M. This proves (i).If Any point p ∈ ∂ conv(X) For (ii) note that each point in the boundary of a convex sethas a supporting plane. If the boundary contains a point p ∈ M then the lemma provesthat the supporting plane must contain all of M. (In particular, all such supporting planesmust agree.)□As a consequence a complete minimal surface (with empty boundary) cannot be compact– this provides another proof of Corollary 9.We have not yet addressed the existence question for minimal graphs with prescribedboundary values (the so-called Dirichlet problem). It is, however, possible to discuss theuniqueness problem:Corollary 44. In the situation of Thm. 41, assume that in addition U is convex. Theneach bounded minimal surface with graphical boundary values is a graph, that is, a graph fover U is unique among all bounded minimal surfaces with the same boundary.Proof. Let M be a bounded minimal minimal surface attaining the same boundary values.The set U × R is convex, and so the convex hull property Thm. 43 implies M ⊂ U × R aswell as M ⊂ U × R.Now consider the foliation of U × R with the translated graphs f(t, x) = ( x, u(x) + t ) .Since M is bounded, there exist t max := sup{t ∈ R, f t (U) ∩ ˜M ≠ ∅} and t min := inf{. . .}.

42 K. Grosse-Brauckmann: Minimal surfaces, SS 12If M is not contained in f(U) (and therefore identical to) then one of t max , t min is nonzero,say t max ≠ 0.The surface M and the graph Γ := f(t max , U) are compact and so must intersect at somepoint p. This point p cannot be in the boundary of both surfaces, due to t max ≠ 0. Sop ∈ U × R which implies it is interior to both surfaces. But then the tangent spaces of Mand Γ at p agree and so M is locally graph, too. The interior maximum principle provesthat the set of points in M which is contained in Γ is open and closed, hence all of M,contradicting the disjoint boundaries of the two surfaces.□Example. We want to show that for a nonconvex domain there it may be that there is agraph and another, distinct minimal surface, which is not graph but has the same boundary.This indicates that convexity of the domain is necessary for the uniqueness statement inpart (ii) to hold.Consider a catenoid M. By a translation, suppose that M intersects the xy-plane in acircle Γ of radius R > 0 so that the waist occurs with positive height z. Consider a scalingof the catenoid λM, where λ > 1 is chosen such that the radius of the waist is still lessthan R. Suppose λM is translated such that it also contains the circle Γ and its waist alsooccurs with positive height. Then M and λM also intersect at some other circle of radiusr < R, occuring at a positive height. Moreover, the annulus contained in λM bounded bythe two circles is graph over U := B R \ B r , while the annulus contained in M is not (thelatter contains points over B r , for instance the waist circle).4.5. Doubly connected minimal surfaces. In Prop. 13 we saw that two coaxial unitcircles of distance larger than d max = 1.32 . . . do not bound a piece of a catenoid. Thefollowing statement is stronger and says that the circles do not bound any connectedminimal surface.Let us denote by C the (double) cone tangent to the catenoid; then R 3 \ C has three connectedcomponents. We let U 1 and U 2 be the two components which are simply connected,and U 3 be the component which is not simply connected.Theorem 45. Let M be a bounded connected minimal surface whose boundary is containedin U 1 ∪ U 2 . Then M as well as its boundary is in fact contained in only one of the sets U 1or U 2 .Note, however, that disconnected minimal surfaces may exist. An example is the Goldschmidtsolution bounding two concentric circles, one in U 1 , the other in U 2 .

i 4.5 – date: July 27, 2012 43Proof. We claim that M does not meet the set U 3 . Since U 1 and U 2 are contained indifferent connected components of R 3 \ U 3 the connected surface M must be contained inone of them.To prove our claim let M λ ⊂ U 3 , λ > 0, be the family of scaled catenoids, and considerthe open annuli A λ ⊂ M λ with boundary in C.As a first step we show that M does not meet the set U 3 \ {0}. Consider the infimum0 ≤ Λ := inf{λ > 0, M ∩ M λ = ∅}Since M is bounded, for some λ we have M ∩ M λ = ∅, implying that Λ < ∞. Moreover,the compact sets A λ foliate U 3 . Thus Λ > 0 if and only if M contains a point 0 ≠ p ∈ U 3 .But this intersection is one-sided, hence the maximum principle gives that both surfacescoincide in a neighbourhood, where both surfaces are graph over the common tangentplane at p. Now the subset of points in M, for which M ⊂ M Λ is relatively closed and, bythe maximum principle, open; thus if p exists then ∅ ̸= ∂M ⊂ M λ , which contradicts ourassumption. Consequently Λ = 0 and M ⊂ R 3 \ (U 3 ∪ {0}) as desired.But the immersed surface M cannot contain the origin: Whatever the tangent plane maybe at 0 ∈ M, the surface will penetrate into U 3 .□

44 K. Grosse-Brauckmann: Minimal surfaces, SS 12Part 2. Minimal surfaces in terms of complex analysis14. Lecture, Wednesday 23.5.12We will establish the Weierstrass representation formulas which allow to represent eachminimal surface by a pair of holomorphic functions. There is also a global version of therepresentation in terms of holomorphic functions defined on a Riemann surface. The mostnatural class of minimal surfaces it applies to are surfaces of finite total curvature, orsurfaces with a fundamental domain of finite total curvature. In this case the Riemannsurface is compact.Many examples were found by systematic guesses of suitable holomorphic functions, in the19th century, and in the past decades. But the representation is also a powerful tool foranalyzing minimal surfaces.1. The Weierstrass representation1.1. Review of complex analysis. We will not distinguish between a number z ∈ Cand the vector (x, y) ∈ R 2 with z = x + iy. For instance, we write z = (z 1 , z 2 ) ori(x, y) = i(x + iy) = −y + ix = (−y, x).If f : U ⊂ C → C then f is holomorphic if the Cauchy-Riemann equations hold on U:∂ x f 1 = ∂ y f 2 , ∂ x f 2 = −∂ y f 1The Cauchy-Riemann equations represent the C-linearity of the differential df z . One way tosee this is to write out the Jacobian as a 2×2 matrix and demand it represents multiplicationwith the complex number f ′ (z). Equivalent is to require that the partial derivatives off : U 2 ⊂ R 2 → R 2 in real and complex direction agree (h ∈ R):f(z + h) − f(z)limh→0 h!= limih→0f(z + ih) − f(z)ih(∂x f 1= ∂ x f(z) =∂ x f 2 )(z)= 1 i ∂ yf(z) = −i∂ y f(z) =If f is holomorphic, then its derivative is denoted by(1) f ′ (z) := ∂ x f(z) = −i∂ y f(z).(∂y f 2−∂ y f 1 )(z)In the general, not necessarily holomorphic case, Wirtinger derivatives are useful:(2) ∂ z f := 1 2(∂x f − i∂ y f ) , ∂ z f := 1 2(∂x f + i∂ y f )

ii 1.1 – date: July 27, 2012 45Then (1) gives: f holomorphic ⇔ ∂ z f = 0, and moreover f ′ = ∂ z f. Note also that∂ z f = 1 (∂x (Re f − i Im f) − i∂ y (Re f − i Im f) )(3)2=2( 1 ∂x (Re f − i(Im f)) + i∂ y (− Re f + i Im f) ) = ∂ z f.As an exercise, calculate ∂ z |z| 2 and ∂ z |z| 2 in two different ways.Cauchy’s theorem is a generalization of the fundamental theorem of calculus to the complexsetting. Recall first the notion of a complex path integral of a function f ∈ C 0 (U ⊂ C, C):∫∫ bf(w) dw := f ( γ(t) ) γ ′ (t) dt,γawhere γ ∈ C 1 ([a, b], U) is a path and the right hand side is in terms of the standard complexproduct. Then the fundamental theorem of calculus ∫ F ′ = F | generalizes as follows:Theorem 1 (Cauchy’s Theorem). Suppose ϕ: U → C is holomorphic and U ∋ z 0 is simplyconnected. Then∫ z∫F (z) := f(w) dw := f(w) dw for γ any path with γ(0) = z 0 , γ(1) = z,z 0γis defined for z ∈ U, independently of the choice of γ. Moreover, F is holomorphic on Uand an antiderivative of f, that is, ∂ z F = f.For completeness, we state:Definition. Let U ⊂ R n be a domain, that is open and connected. Then U is simply connected,if for each closed continuous curve h 0 : [0, 1] → U with h 0 (0) = h 0 (1) =: p extends to a homotopyh = h s (t): [0, 1] × [0, 1] → U, such that h s is closed, h s (0) = h s (1) = p for all s, and it contractsto p, that is, h 1 (t) = p for all t.Remember that Cauchy’s Theorem is no longer true if U is not simply connected: f(w) = 1 wis holomorphic on C \ {0}, but Cauchy’s integral ∫ z 1dw is not well-defined.wOne way to understand Cauchy’s Theorem is to explain it in real terms. It is well-knownthat a vector field X : U → R 2 has a potential Φ, i.e. a scalar function with grad Φ = X,provided rot X = ∂ x X 2 − ∂ y X 1 vanishes. Indeed, Φ is given by the well-defined (real)path integral Φ(x) = ∫ x X ds = ∫ 〈X(γ(t)), γ ′ (t)〉dt. A short calculation gives that we canrewrite the complex path integral in terms of real path integrals:∫∫ ∫f(w) dw = f ds + i (if) ds,where the real vector fields are f := (f 1 , −f 2 ) and if := (f 2 , f 1 ). On the other hand, byCauchy-Riemann, holomorphicity of f is equivalent torot f = ∂ x (−f 2 ) − ∂ y f 1 = 0 and rot(if) = ∂ x (f 1 ) − ∂ y (−f 2 ) = 0.

46 K. Grosse-Brauckmann: Minimal surfaces, SS 12Hence the integral of real and imaginary part are each well-defined and give a potentialΦ 1 and Φ 2 , respectively. Then F := Φ 1 + iΦ 2 is the desired antiderivative of f. Phrasingthis in more abstract terms, we could say we have invoked the Poincaré lemma: rot X = 0corresponds to the closedness of some form, and the desired existence of the potential Φwith grad Φ = X to its exactness.1.2. Holomorphic null curves. We know that conformal harmonic immersions are minimalsurfaces. The construction we will present provides somewhat weaker maps:Definition. We call a map f ∈ C 2 (U 2 , R d ) a generalized or branched minimal surface if itis a weakly conformal harmonic mapping, that is|f x | 2 = |f y | 2 , 〈f x , f y 〉 = 0 and ∆f = 0 for all z ∈ U.A generalized minimal surface can have branch points [Verzweigungspunkte], i.e. points(x, y) with df (x,y) = 0. Note that a non-constant holomorphic map has isolated branchpoints, which locally look like z ↦→ z k , as shown by the Taylor series. Branch points ofgeneralized minimal surfaces display a similar behaviour. We will see examples only later.Definition. (i) For U ⊂ C a smooth mapping Φ: U → C d is called a curve in C d .(ii) Φ is holomorphic if each component is holomorphic, i.e., the Cauchy Riemann equationshold componentwise.(iii) A curve Φ = (Φ 1 , . . . , Φ d ): U → C d is called a null curve if(4) Φ 2 1 + . . . + Φ 2 d ≡ 0.Here Φ 2 kdenotes the standard complex product,Φ 2 k = (Re Φ k + i Im Φ k ) 2 = Re 2 Φ k − Im 2 Φ k + 2i Re Φ k Im Φ k .Example. (z 1 , z 2 ) = (1, i) ∈ C 2 is a null vector: z 2 1 + z 2 2 = 1 2 + i 2 = 0.Another notation for (4) arises from the standard scalar product 〈., .〉 on R d , extended ina complex linear way to a bilinear form 〈〈., .〉〉 on C d ,〈〈a + ib, c + id〉〉 := 〈a, c〉 − 〈b, d〉 + i ( 〈b, c〉 + 〈a, d〉 ) for a, b, c, d ∈ R d .Then a null curve Φ satisfies 〈〈Φ, Φ〉〉 = 0. Indeed,〈〈Φ, Φ〉〉 = 〈Re Φ, Re Φ〉 − 〈Im Φ, Im Φ〉 + 2i〈Re Φ, Im Φ〉 =d∑Φ 2 k.We also conlude from this expression that the null condition (4) is equivalent to(5) | Re Φ| = | Im Φ| and Re Φ ⊥ Im Φ.k=1

ii 1.2 – date: July 27, 2012 47Note, however, that unlike the Hermitian product, 〈〈., .〉〉 is neither a scalar product nordoes it define a norm.The derivative of a generalized minimal surface is exactly a holomorphic null curve:Lemma 2. If f ∈ C 2 (U, R d ) then(i) f is harmonic iff ∂ z f is holomorphic, and(ii) f is weakly conformal iff ∂ z f is a null curve.Proof. (i) and (ii) are immediate from the vanishing of the expressions∆f = ( ∂ xx + ∂ yy )f = (∂ x + i ∂ y )(∂ x − i∂ y )f = 4 ∂ z (∂ z f),4〈〈∂ z f, ∂ z f〉〉 = 〈〈∂ x f − i∂ y f, ∂ x f − i∂ y f〉〉 = |∂ x f| 2 − |∂ y f| 2 − 2i〈∂ x f, ∂ y f〉.□Given a holomorphic null curve Φ, it is an obvious idea to employ Cauchy’s theorem toobtain a generalized minimal surface F (z) = ∫ z Φ(w) dw. However, in general F will notbe real-valued. It is surprising that taking the real part of Cauchy’s integral maintains theproperties claimed in the previous lemma.Proposition 3. Let U be simply connected, z 0 ∈ U. Then there is a 1-1 relationshipbetween holomorphic null curves Φ ∈ C 1 (U, C d ) and generalized minimal surfaces f ∈C 2 (U, R d ) with f(z 0 ) = 0, given by(6) Φ = 2 ∂ z f and f(z) := Re∫ zz 0Φ(w) dw,Moreover, f is an immersion if and only if Φ(z) ≠ 0 for all z ∈ U.Proof. Suppose the holomorphic curve Φ is given. Cauchy’s Thm. 1, applied componentwise,gives that F (z) := ∫ zz 0Φ(w) dw satisfies ∂ z F = Φ and ∂ z F = 0. Then the secondequation of (6) reads f := Re F = 1 2(F + F). It has the first equation as a consequence:1( ) (26)∂ z Re F = ∂ z F + F = 1 ∂z F + ∂ z F22( ) = 1 2 ∂ zF ⇔ ∂ z f = 1 2 ΦBy Lemma 2 this implies that f is a generalized minimal surface.Conversely, suppose the generalized minimal surface f is given. Then 2 ∂ z f is a holomorphicnull curve by Lemma 2.Let us finally prove the immersion property. Decomposing 2 ∂ z f = Φ into real and imaginarypart we see(7) ∂ x f = Re Φ and ∂ y f = − Im Φ.For a null curve with Φ ≠ 0 we note (5) implies Re Φ ≠ 0 and Im Φ ≠ 0. These nonzerovectors are perpendicular by (5), hence linearly independent. That is, rank df = 2. □

48 K. Grosse-Brauckmann: Minimal surfaces, SS 12Remarks. 1. If U is not simply connected then f can always be defined on the universalcovering Ũ of U.2. Suppose that Φ has isolated singularities, that is, Φ is meromorphic. If the residues ofall components are real, then the residue theorem gives that ∫ Φ is purely imaginary overclosed loops. Therefore, Re ∫ Φ is well-defined for this case. The same holds in case U isnot simply connected, but all loops γ in U yield a path integral ∫ Φ(w) dw with purelyγimaginary components.Corollary 4. A conformally parameterized minimal surface f ∈ C 2 (U, R d ) is infinitelyoften differentiable, in fact analytic.Proof. To show that f is analytic in a neighbourhood of any z 0 ∈ U, restrict f to a simplyconnected neighbourhood V ⊂ U of z 0 . Since ∂ z f is holomorphic on V it is infinitely oftendifferentiable and analytic. The integral representation of (6) then proves that f itself isC ∞ and analytic on V .□15. Lecture, Wednesday 30.5.121.3. The local Weierstrass representation. The null condition imposes one conditionon a holomorphic curve Φ. This will allow us to represent Φ for dimension n = 3 by a pairof holomorphic functions.Recall that a function g : U → C is meromorphic if it is not identical to ∞ and if for eachz ∈ U either g(z) ∈ C and g is holomorphic at z or else 1 has a removable zero. That is,g(z)in the neighbourhood of each z 0 ∈ U a meromorphic function g ≠ 0 can be represented bya Laurent expansiong(z) = a k (z − z 0 ) k + a k+1 (z − z 0 ) k+1 + . . . with k ∈ Z and a k ≠ 0.At z 0 the function g has a zero of order k in case k > 0, or a pole of order −k in case k < 0.Lemma 5. Suppose h: U → C is holomorphic and g : U → C with g ≢ 0 is meromorphicsuch that(8) at each zero or pole of order k of g the function h has a zero of order at least k.Then( 1 1)(9) Φ = (Φ 1 , Φ 2 , Φ 3 ) = h2(g − g , i ( 1) )2 g + g , 1 : U → C 3is a holomorphic null curve. Conversely, any nonzero holomorphic null curve Φ: U → C 3with Φ 3 ≢ 0 can be represented this way.If Φ 3 ≡ 0 then f 3 ≡ 0 and the surface f will be a horizontal plane.

ii 1.3 – date: July 27, 2012 49Proof. As a composition of meromorphic functions, Φ is meromorphic, and given (8) it isclearly holomorphic. By the following calculation it is a null curve:( 1 1)Φ 2 1 + Φ 2 2 + Φ 2 3 = h4( 2 g + 2 g2 − 2 −4( 1 1) )g + 2 g2 + 2 + 1 = h 2 (−1 + 1) = 0For the converse, let us check that any null curve Φ admits a representation (9). Let usrewrite the null condition as(10) 0 = Φ 2 1 + Φ 2 2 + Φ 2 3 = (Φ 1 + iΦ 2 )(Φ 1 − iΦ 2 ) + Φ 2 3.We see that Φ 3 ≢ 0 implies that also Φ 1 + iΦ 2 ≢ 0 and Φ 1 − iΦ 2 ≢ 0. Let us set(11) h := Φ 3 and g := − Φ 1 + iΦ 2Φ 3.These functions, as well as 1/g, are meromorphic.With g, h as above (10) holds if and only if Φ 1 − iΦ 2 = 1 h. Equivalently,g2Φ 1 = −gh + 1 g h, 2iΦ 2 = −gh − 1 g h,which is precisely the form stated in (9). This representation also yields (8).□Theorem 6 (Weierstrass or Weierstrass, Enneper, Riemann).Suppose U ⊂ C is simply connected and z 0 ∈ U. If h: U → C is holomorphic and g : U → Cmeromorphic subject to (8) then∫ z( 1( 1)(12) f : U → R 3 , f(z) = Rez 02 g − g , i ( 1) )2 g + g , 1 h dwis a generalized minimal surface. It is an immersion provided(13) z is a zero of order k of h ⇔ z is a zero or pole of order k of g.Conversely a conformally parameterized minimal surface f : U → R 3 can locally be representedthis way, provided its image is not contained in a horizontal plane.We call g, h the Weierstrass data of the minimal surface f.Proof. The proof that (12) is a generalized minimal surface or minimal surface followsimmediately from Prop. 3 together with Lemma 5.Conversely, Prop. 3 states that a conformally parameterized minimal surface f has a representation(6), which by Lemma 5 can be written as (12).□

50 K. Grosse-Brauckmann: Minimal surfaces, SS 12Remark. There is a close analogy of the Weierstrass data with Pythagorean triples a, b, c ∈N, which are numbers satisfying a 2 + b 2 = c 2 . Euclid realized they can be represented interms of a pair of natural numbers m < n, namely a = n 2 − m 2 , b = 2mn, c = n 2 + m 2(check!). In fact all socalled primitive Pythagorean triples arise that way. See problems.By the remark at the end of the previous section, the Weierstrass representation extendsto some more general cases, see there.Let us also discuss other versions of the Weierstrass representation.1. Many sources state the following version of(14) Φ = ( 1 − ˜g 2 , i(1 + ˜g 2 ), 2˜g )˜h.This time ˜h := 1 2 (Φ 1 −iΦ 2 ) and ˜g := Φ 3Φ 1 −iΦ 2. However, our choice (9) which follows Karcherand is also adopted by [EJ] and [CM] is more geometric, as the next Proposition will show.2. In the neighbourhood of and point z with ˜g ′ (z) ≠ 0 we may compose ˜g, ˜h with ˜g −1 toobtainΦ = ( 1 − z 2 , i(1 + z 2 ), 2z ) R(z).In this form the represention depends solely on one holomorphic function R. Note that thisremoves the conformal invariance from the Weierstrass representation: While Φ has threecomponents, or three degrees of freedom, the null (or conformality) condition removes one,and the choice of conformal parameterization removes another. Thus ultimately a surfaceis described by just one holomorphic function.Examples. 1. For Enneper’s surface, Weierstrass data areso thatU = C, g(z) = z, h(z) = 2z,Φ(z) =()1 − z 2 , i(1 + z 2 ), 2zwhich gives upon integration from z 0 = 0 to z:f(z) = Re(z − 1 3 z3 , i(z + 1 )3 z3 ), z 2 =(x − 1 3 x3 + xy 2 , −y + 1 )3 y3 − x 2 y, x 2 − y 2 ,using that 1 3 (x + iy)3 = 1 3 x3 + ix 2 y − xy 2 − i 3 y3 .2. We claim that the catenoid can be by the Weierstrass dataso thatUpon integration we findU = C \ {(x, 0) : x ≥ 0}, g(z) = z, h(z) = 1 z ,Φ(z) =f(z) = Re( 12 ( 1 z 2 − 1), i 2 ( 1 z 2 + 1), 1 z).( 12 (−1 z − z), i2 (−1 z + z), log z ),

ii 1.4 – date: July 27, 2012 51that is, in polar coordinates z = ρe iϕ ,( 1f(ρe iϕ ) = Re2 (−1 ρ e−iϕ − ρe iϕ ), i )2 (−1 ρ e−iϕ + ρe iϕ ), log(ρe iϕ )=( −12 (1 −1)+ ρ) cos ϕ,ρ 2 (1 ρ + ρ) sin ϕ, log ρ(recall that log z = log |z| + i arg z). Now setting ρ = e r , so that the coordinate change(r, ϕ) ↦→ z = e r+iϕ is conformal, we obtainf(e r e iϕ ) = ( − cosh r cos ϕ, − cosh r sin ϕ, r ) .This we recognize as the parameterization of the catenoid (up to a 180 ◦ rotation aboutthe z-axis). While the Weierstrass data and Φ are defined on C ∗ , the integral ∫ Φ doesnot continuously extend to the positive real axis. Nevertheless, inspection shows that theparameterization f = Re ∫ Φ does extend.Remark. If we were to declare Φ on the universal covering of C ∗ then ∫ Φ and f = Re ∫ Φwere well-defined.16. Lecture, Friday 1.6.121.4. Differential geometry in terms of the Weierstrass representation. Stereographicprojection is the map⎧ ()⎨ 12 Re z, 2 Im z, |z| 2 − 1 for z ∈ C,(15) st: C → S 2 |z|, st(z) :=2 +1⎩(0, 0, 1) for z = ∞.To derive (15) geometrically, let N := (0, 0, 1) be the north pole. A projection throughstraight lines means st(z) = λz + (1 − λ)N with λ ≠ 0. It maps to S 2 provided1 ! = | st(z)| 2 = λ 2 |z| 2 + (1 − λ) 2 = 1 − 2λ + λ 2( |z| 2 + 1 ) .Thus 0 = −2 + λ(|z| 2 + 1) =⇒ λ = 2 and indeed (15) follows. It can be checked that|z| 2 +1st is a diffeomorphism, and st is conformal with factor λ 2 = 4/(1 + |z| 2 ) 2 (see problems).We can now explain the geometric meaning of g, h:Proposition 7. Let f : U → R 3 be a minimal surface with Weierstrass data g, h. Then:(i) Stereographic projection of the map g gives a Gauss map ν of f, that is,1()ν = st ◦g = 2 Re g, 2 Im g, |g| 2 − 1 .|g| 2 + 1Moreover, ν is conformal with factor 4|g ′ | 2 /(1 + |g| 2 ) 2 .(ii) h is the derivative of the height function f 3 in the following sense: ∂ x f 3 = Re h and∂ y f 3 = − Im h.

52 K. Grosse-Brauckmann: Minimal surfaces, SS 12Proof. (i) Let us show that st(g) is perpendicular to the tangent space of f. By (7) thisis equivalent to showing that ∂ x f = Re Φ and ∂ y f = − Im Φ are perpendicular to thevector st ◦ g. We verify this as follows:〈Re Φ, st ◦g〉 + i〈Im Φ, st ◦g〉 = 〈〈 Φ, st ◦g 〉〉〈〈 ( 1 1)= h2(g − g , i ( 1) )2 g + g 1, 1 ,|g| 2 + 1(h( 1) ( 1)= (Re g)|g| 2 + 1 g − g + i(Im g)g + gh( g=|g| 2 + 1 g − gg + |g|2 − 1)= 0.() 〉〉2 Re g, 2 Im g, |g| 2 − 1)+ |g| 2 − 1Being unit vectors orthogonal to ∂ x f and ∂ y f, st ◦g and ν must agree. Conformality followsfrom the fact that both g and st are conformal.(ii) follows from (7).□Examples. The catenoid and Enneper surface have g(z) = z, so that ν(z) = st(z). Thusthey are parameterized by (the stereographic projection of) the Gauss map. In particular,their Gauss image is S 2 up to sets of measure zero, and so the total curvature is −A(S 2 ) =−4π.We can now express the differential geometry of f in terms of the Weierstrass data. Sincethe letter g is reserved for Weierstrass data, instead we use ds 2 for the first fundamentalform of f.Proposition 8. (i) The first fundamental form of f is( 1 ((16) ds 2 1 ) ) 2〈 〉(X, Y ) := |g| + |h| X, Y2 |g|(ii) The Gauss curvature of f is(17) K =−16∣ ∣∣ g ′(|g| + 1 ∣ 2 1|g| )4 g |h| . 2Proof. (i) Since f is conformal, we have by (5) and (7)|∂ x f| 2 = |∂ y f| 2 = | Re Φ| 2 = | Im Φ| 2 ⇒ |Φ| 2 = 2| Re Φ| 2 = 2|∂ x f| 2 .Our claim is now verified by calculation:2|∂ x f| 2 = |Φ| 2 = Φ 1 Φ 1 + Φ 2 Φ 2 + Φ 3 Φ 3( 1( 1)( 1)= |h| 2 4 g − g g − g + 1 ( 1)( 1)4 g + g g + g= |h| 2 1 ( 1)2 |g| + 2 |g|2 + 2 = 1 (2 |h|2 |g| + 1 ) 2|g|)+ 1

ii 1.4 – date: July 27, 2012 5317. Lecture, Monday 4.6.12(ii) As in I(13) we express the Gauss curvature asK = det S = det ( −(df) −1 dν ) =det dνdet df .Now f and the Gauss map ν are conformal. Thus the determinants are given by squaresof vectors: det df = |∂ x f| 2 = |∂ y f| 2 . Since det dν is also conformal, but (due to K ≤ 0)orientation reversing, we have det dν = −|∂ x ν| 2 = −|∂ y ν| 2 . We concludeK = − |∂ xν| 2|∂ x f| 2 .Now we use the fact that stereographic projection has a conformal factor 4(1 + |z| 2 ) −2 andthat conformal factors of a composition of maps simply multiply (why is this the chainrule?). Thus |∂ x (st ◦g)| 2 = |(∂ x st) ◦ g| 2 |∂ x g| 2 = 4|g ′ | 2 /(1 + |g| 2 ) 2 . Together with (16) thisgives the desired expressionK = −|g ′ | 2 44(1 + |g| 2 ) 2 (|g| + 1|g| )2 |h| = − |g′ | 2 4 42 |h| 2 |g| 2 (|g| + 1|g| )2 (|g| + 1 .|g| )2Another way to prove the formula for K is to derive it from K = −(∆ log λ)/λ 2 , valid forany conformal metric g = λ 2 〈., .〉.We can now strengthen Theorem I.8(i):Corollary 9. On a minimal surface either K ≡ 0 or the umbilics are isolated.□Thus either the minimal surface is a flat plane or it has isolated umbilics. The followingwill be essential to discuss the geometry of surfaces given in terms of Weierstrass data.Proposition 10. The second fundamental form is( g′ )(18) b(X, X) = − Reg hX2 .Consequently, X is an asymptote direction iff g′g hX2 ∈ iR and X is a principal curvaturedirection iff g′g hX2 ∈ R.Proof. A calculation shows that the second fundamental isb(X, X) = 〈d 2 f(X, X), ν〉 = 〈Re Φ ′ X 2 , ν〉Let us differentiate using the product rule:(Φ ′ = h Φ ) ( ′ 1 −g′= hh 2(g 2 − g′) , i ( )−g′2 g 2 + g′) , 0 + h ′ Φ h .

54 K. Grosse-Brauckmann: Minimal surfaces, SS 12Thus, using that ν = st(g) ⊥ Φ,b(X, X) = 〈 Re Φ ′ X 2 , st(g) 〉〈〈 (1= Re(h g′ ( −1),g X2 2 g − g i ( −1)2 g + g= Re(h g′ 1(g X2 |g| 2 − 1 ))+ 1 g g − gg), 0 ,= − Re(h g′g X2) .1() 〉〉)|g| 2 2 Re g, 2 Im g, |g| 2 − 1+ 1If (g ′ h/g)(z) = 0 then all directions are asymptotic and hence also curvature directions; equivalentlyz is an umbilic point. So assume z is not umbilic. It is obvious that X with (g ′ h/g)X 2 ∈ iRis asymptotic. The curvature directions of our minimal surface make a 45 degree angle with X.Hence Y is a curvature direction if Y agrees with one of the four directions ±e ±iπ/4 X. But then(g ′ h/g)Y 2 = (g ′ h/g)(±e ±iπ/4 X) 2 = i(g ′ h/g)X 2 ∈ R, as claimed. □1.5. The associated family. The Weierstrass representation f = Re ∫ Φ integrates thereal part of a holomorphic null curve in C d . It seems reasonable to ask: Is there anymeaning of the imaginary part Im ∫ Φ = Re ∫ (−iΦ)?Definition. For U 2 a simply connected domain, z 0 ∈ U, and f : U → R d be a minimalsurface with Φ := 2 ∂ z f. Then the family of mappingsf ϑ : U → R 3 , f ϑ (z) := Re∫ zz 0e −iϑ Φ(w) dw, where ϑ ∈ R,is called the associated family of f. In particular, the conjugate surface is∫∫f ∗ := f π/2 = Re −iΦ = Im Φ.Let us note a few identities which are obvious right now: First f ϑ+2π = f ϑ and moreoverf ϑ+π = Re ∫ e −i(ϑ+π) Φ = − Re ∫ e −iϑ Φ = −f ϑ . Second,( ∫ z)z zf ϑ (z) = Re cos ϑ Φ − i sin ϑ Φ = cos ϑ Re∫Φ + sin ϑ Im∫Φ(19)= cos ϑ f(z) + sin ϑ f ∗ (z).Suppose Φ: U → C d is a holomorphic null curve. We claim it is equivalent that Φ ϑ := e −iϑ Φis a holomorphic null curve: Indeed, ∑ (e −iϑ Φ k ) 2 = e −2iϑ ∑ Φ 2 k and ∂ z(e −iϑ Φ) = e −iϑ ∂ z Φ.Hence f ϑ is also a generalized minimal surface; if f is an immersion, so is f ϑ .It is straightforward to determine the Weierstrass data of f ϑ :Theorem 11. Let U be a simply connected domain, and g, h holomorphic or meromorphicon U, respectively. Then the associated family f ϑ is also minimal with Weierstrass data(20) g ϑ := g and h ϑ := e −iϑ h.

ii 1.5 – date: July 27, 2012 55The associated surfaces have the following differential geometric quantities:)(21) ν ϑ = ν, ds 2 ϑ = ds 2 −iϑ g′, K ϑ = K, b ϑ (X, X) = − Re(eg hX2In particular, all surfaces in the associated family are isometric and have the same Gaussmap and Gauss curvature.Proof. g is independent of ϑ and so is the Gauss map ν = st ◦g. Since |h ϑ | = |h| theconformal factor λ = 2( 1 |g| +1|g|)|hϑ | is also independent of ϑ. By the theorema egregiumor (17) so is the Gauss curvature. Finally, the expression for b follows immediately from(18). □The isometry property means that a physical model of the surface f (made from inelasticmaterial such as metal or plastics) can be deformed into f ϑ .18. Lecture, Wednesday 6.6.12Example 1: We consider Weierstrass data which will lead to the associated family of thecatenoid. They disagree with the data of the catenoid from Sect. 1.3. This time, weconsider insteadU = C, g(z) = −e z , h ϑ (z) = e −iϑ .Since g has no zero, f will not have any branch points. Then( 1Φ ϑ (z) = e −iϑ 2 (−e−z + e z ), i ) ()2 (−e−z − e z ), 1 = e −iϑ sinh z, −i cosh z, 1and so, using that cosh ′ = sinh and sinh ′ = cosh,( ∫ z ∫ z ∫ z )f ϑ (z) = Re e −iϑ sinh w, −i cosh w, 1Now000(= Re e −iϑ( cosh z − 1, −i sinh z, z )) .sinh z = sinh x cos y + i cosh x sin y and cosh z = cosh x cos y + i sinh x sin ywhich gives⎛⎞ ⎛⎞−1 + cosh x cos ysinh x sin y⎜⎟ ⎜⎟f ϑ (x, y) = cos ϑ ⎝ cosh x sin y ⎠ + sin ϑ ⎝− sinh x cos y⎠xy= cos ϑ f(x, y) + sin ϑ f ∗ (x, y).Note that in this formula, f represents a translated catenoid, while f ∗ represents thehelicoid. Each member f ϑ of the family is a minimal surface, and each of these surfacescan be shown to have a screw symmetry. Note also that the associated family is not definedon the topological annulus of the catenoid: It is only defined on its universal covering.

56 K. Grosse-Brauckmann: Minimal surfaces, SS 12Example 2: We consider Enneper’s surface f(z), as given in Sect. 1.3. Its associatedfamily f ϑ is a rotation of f about the vertical axis (given by the normal at z = 0), up toreparameterization. More precisely, if R α denotes rotation about the vertical axis by anangle α, thenf ϑ (z) = R ϑ/2 f(e −iϑ/2 z),as an explicit calculation shows. Therefore, up to the ambient rotation, the associate familyis an intrinsic rotation of Enneper’s surface. We know that this is an isometry.Let us give the proof. The associated family f ϑ is given byThusU = C, g(z) = z, h ϑ (z) = 2e −iϑ z,)Φ ϑ = e(1 −iϑ − z 2 , i(1 + z 2 ), 2zand so∫ z(f ϑ (z) = Re Φ = Re(e −iϑ z − 103 z3 , i(z + 1 ) )3 z3 ), z 2 .We claim that f ϑ and f are congruent: If we rotate the image f ϑ by an angle −ϑ/2 aboutthe z-axis, then we obtain f up to some rotation in the domain.To see this, let us first rotate the image; we let F denote the rotated image. Since werotate about the z-axis, F 3 := fϑ 3 . Moreover, using complex notation for the rotation inthe first two coordinates, we findF 1 + iF 2 := e −iϑ/2 (fϑ 1 + fϑ)2( ) (= e −iϑ/2 Re(e −iϑ z) + i Re(ie −iϑ z) + e −iϑ/2 Re ( − 1 3 e−iϑ z 3) + i Re ( i3 e−iϑ z 3)) .Now let us invoke the identities(22) Re ζ + i Re(iζ) = ζ and Re ζ + i Re(−iζ) = ζto obtain(F 1 + iF 2 )(z) = e −iϑ/2 e −iϑ z − e −iϑ/2 1 3 e−iϑ z 3 (22)= e −iϑ/2 z − 1 (e −iϑ/2 z ) 33To prove our claim, we want to show that F (z) agrees with the Enneper surface up to arotation of the parameters; more precisely, the Enneper surface f(w) agrees with F (e iϑ/2 w).Thus setting w := e −iϑ/2 z gives(F 1 + iF 2) (e iϑ/2 w) = w − 1 (22)w3 = Re w − 1 3 3 Re w3 + i Re(iw) + i 3 Re(iw3 ).Finally noting that F 3 = e −iϑ z 2 = w 2 we verifyF (e iϑ/2 w) = Re(w − 1 3 w3 , i(w + 1 )3 w3 ), w 2 = f(w).This proves our claim.

ii 2.1 – date: July 27, 2012 57We will later also discuss Scherk’s surface as a third example.Remark. The complex setting reveals a relationship of minimal surfaces which seemingly is hiddenin the real setting. Nevertheless, there is a entirely real setup to construct the associated family,which we want to outline. If f is a minimal surface with first fundamental form g and shapeoperator S the we set g ϑ := g and S ϑ := R(ϑ)S, where R(ϑ) is the Riemannian rotation of R 2 byan angle ϑ. It follows from minimality of f that S ϑ is a valid shape operator of a minimal surface(i.e., S ϑ is self-adjoint and has trace 0). Hence the fundamental theorem of surfaces gives that itdefines surfaces f ϑ which are exactly the associated minimal surfaces.2. Symmetry properties of minimal surfacesOur goal is to discuss symmetries of minimal surfaces. We start with a discussion of mirrorsymmetry for holomorphic functions which map the real axis into the real axis.2.1. Schwarz reflection for holomorphic functions. To go back a step, recall how realfunctions f : [0, ∞) → R can be extended to all of R: The even extension⎧⎨f(x) if x ≥ 0,(23) F e (x) :=⎩f(−x) if x < 0,extends an f ∈ C 0 to F e ∈ C 0 . Suppose now that f ∈ C 1 . Then, by the chain rule,F ′ e(x) = (f(−x)) ′ = −f ′ (−x) = −F ′ e(−x) for x < 0. The limit x ↘ 0 agrees for both sidesif and only if f ′ (0) = 0; this condition is equivalent to F e ∈ C 1 .On the other hand, consider the odd extension⎧⎨f(x) if x ≥ 0,(24) F o (x) :=⎩−f(−x) if x < 0.It is not a continuous extension of f ∈ C 0 unless f(0) = 0. In this case f ∈ C 1 also impliesF o ∈ C 1 : Indeed, for x < 0 we have F ′ o(x) = (−(f(−x))) ′ = f ′ (−x) = F ′ o(−x) by the chainrule, that is, F ′ o is even, hence continuous.The holomorphic case is much more perfect:Theorem 12 (Schwarz reflection principle). Consider a symmetric domain U ⊂ C suchthat z ∈ U ⇔ z ∈ U. Set U + := {z ∈ U : Im z > 0}, I := U ∩R, U − := {z ∈ U : Im z < 0}.Suppose f ∈ C 1 (U + ∪ I, C) is holomorphic such that f(I) ⊂ R. Then the extension⎧⎨f(z) if z ∈ U + ∪ I,(25) F : U → C, F (z) :=⎩f(z) if z ∈ U − ,is also holomorphic (or analytic).

58 K. Grosse-Brauckmann: Minimal surfaces, SS 12F can be considered as an even reflection of the real part of f, and an odd reflection of theimaginary part which has zero boundary values.Proof. Let us first check the Cauchy-Riemann equations are preserved. We calculate forz ∈ U − :∂ x Re F (z) = ∂ x (Re f(z))chain rule= (∂ x Re f)(z), ∂ y Im F (z) = −∂ y (Im f(z)) = (∂ y Im f)(z)The two right hand sides agree by the the first Cauchy-Riemann equation for f sincez ∈ U + . This implies the first Cauchy Riemann equation for F . Similarly for the otherequation.By assumption, for all x ∈ R we have f(x) = f(x) and so the extension F is continuousalong R. The extension is also differentiable. It is clear that t ↦→ Im F (x + it) is odd, so∂ y Im F is continuous on the real axis. But Im F is also continuously differentiable w.r.t.x, since ∂ x Im F vanishes on R and is a continuous function on {Im z ≥ 0}. We concludeIm F ∈ C 1 (U).The Cauchy-Riemann equations imply Re F is also in C 1 (although it is an even extension).Hence the C 1 -function F is holomorphic on all of C, that is, analytic.□Corollary 13. Let U be a domain which is mirror symmetric w.r.t. the real axis, i.e.z ∈ U ⇔ z ∈ U, and f : U → C be holomorphic such that f(R ∩ U) ⊂ R. Then(26) f(z) = f(z) for all z ∈ U.Proof. Restrict f to U + ∪ I, using the above notation, and consider its extension F as inthe theorem. Then the two functions f and F agree on the open set U + . But they areanalytic. So the unique continuation principle for analytic functions implies f must agreewith F on all of U.□Remarks. 1. There is a similar symmetry or extension property if the preimage or image(or both) are bounded by the imaginary axis, see problems. To state this for the examplef(I) ⊂ iR note that z ↦→ −z is reflection in the imaginary axis. Hence(27) F (z) := −f(z), for z ∈ U −replaces the extension given in (25) and leads to the identity f(z) = −f(z) replacing (26).More generally, I can be a subset of any line through the origin, and f(I) contained inanother line through the origin.2. Often, a weaker form of the theorem is needed, namely it holds under weaker assumptions:It is sufficient to require f ∈ C 1 (U + ) ∩ C 0 (U + ∪ I). Again, the extension is analytic,in particular f ∈ C 1 (U + ∪ I). Note, however, that in the real setting 3√ x is a function

ii 2.2 – date: July 27, 2012 59which is differentiable for x > 0 and continuous for x ≥ 0, but the fact that it has an oddextension does not give differentiability at 0.19. Lecture, Wednesday 13.6.122.2. Schwarz reflection for minimal surfaces. Among the examples of minimal surfaceswe have introduced, the helicoid, Enneper’s surface, or Scherk’s doubly periodicsurface contain straight line segments. Inspection of our examples shows that the surfacesare symmetric w.r.t. π-rotation about these straight lines. In particular, if only the half ofthe surface to one side of the line is given, it can be extended by reflection as a minimalsurface.We prove these statements in general. To do so, we need to work with a parameterization.One way would be to use graphs (problems?). The other, which we pursue here, is toconsider a conformal parameterization.Theorem 14. Consider a symmetric domain U ⊂ C such that z ∈ U ⇔ z ∈ U.(i) Set U + := {z ∈ U : Im z > 0}, I := U ∩ R, U − := {z ∈ U : Im z < 0}. Moreover, letL ⊂ R 3 be a line and ρ ∈ SO(3) the π-rotation about L. Then if f ∈ C 2 (U + ∪ I, R 3 ) is aminimal surface which maps I into L then the extension⎧⎨f(z) if z ∈ U + ∪ I,F : U → R 3 , F (z) :=⎩ρ(f(z)) if z ∈ U − ,is in C 2 (U, R 3 ) and minimal.(ii) If a minimal surface f : U → R 3 maps U ∩ R into a line L then f is symmetricw.r.t. rotation about L, that is, f(z) = ρ(f(z)) for all z ∈ U.Proof. Consider Φ := 2 ∂ z f and the complex extension ˜f := ∫ zz 0Φ(w) dw : U + ∪ I → C 3for an z 0 ∈ I, satisfying Re ˜f = f. While this works directly for U simply connected, thefollowing arguments actually extend to the general case.Suppose for simplicity that L is the z-axis so that ρ(x, y, z) = (−x, −y, z).boundary values f(I) ⊂ L implyThen the(28) ˜f 1 (I), ˜f 2 (I) ⊂ iR.By Schwarz reflection (27) these holomorphic functions extend to U − .We claim that(29) ˜f 3 (I) ⊂ R.

60 K. Grosse-Brauckmann: Minimal surfaces, SS 12First, (28) implies ∂ x ˜f k (I) ⊂ iR for k = 1, 2. Due to conformality therefore ∂ y ˜f k (I) ⊂ RTogether we haveΦ k (I) = ∂ z ˜f k (I) = 1 2 (∂ x − i∂ y ) ˜f k (I) ⊂ iR.Consequently the null condition (Φ 3 ) 2 = −(Φ 1 ) 2 − (Φ 2 ) 2 gives along I(Φ 3 ) 2 (I) ⊂ [0, ∞) ⇔ Φ 3 (I) ⊂ R.Therefore, for points in I,0 = Im Φ 3 = Im ∂ z ˜f 3 = Im ( 12 (∂ x − i∂ y ) ˜f 3) = 1 (∂x Im2˜f 3 − ∂ y Re ˜f 3) CR = ∂ x Im ˜f 3 ,and so ˜f 3 (z 0 ) ∈ R implies (29). Consequently ˜f 3 admits Schwarz reflection (26) across I.In summary we have extended the components of ˜f holomorphically to z ∈ U − by(− ˜f 1(z), − ˜f 2 (z), ˜f 3(z) ) .The real part of this extension is precisely(−f 1 , −f 2 , f 3 )(z) = ρ ( f(z) ) for z ∈ U − ,and so F is indeed in C 2 . It is in fact analytic and so if f is defined on all of U then f issubject to this symmetry.□The previous reflection can be considered as odd reflection. There is also a version of thereflection principle for even reflection:Theorem 15. With notation and assumptions as in the previous theorem, assume nowthat f maps I into a plane P , such that the normal of f along I is also contained in P . Ifρ ∈ O(3) denotes reflection with respect to P , then F (z) := ρ(f(z)) extends f to U − as aminimal surface. Part (ii) of the previous theorem holds as well.We call the arc f(I) an arc of of planar reflection. The proof of the theorem is analogousto the last proof, so we skip it. If we assume specifically that P is the xy-plane then(˜f 1 (I) ⊂ R, ˜f 2 (I) ⊂ R, ˜f 3 (I) ⊂ iR such that ˜f 1(z), ˜f 2(z), − ˜f 3 (z) )is the extension of the complexification. So (f 1 , f 2 , −f 3 )(z) = ρ ( f(z) ) holds for z ∈ U − .Remarks. 1. The two reflection principles are again valid under the weaker assumptionf ∈ C 2 (U + , R 3 ) ∩ C 0 (U + ∪ I, R 3 ).2. Theorem 15 is also true for any H constant. It is a good exercise to understand thedifferential geometry at the boundary curve: Where do the principal curvature directionspoint to, and why are they preserved together with their principal curvatures under thereflection? This explains why the extension is in C 2 .

ii 2.3 – date: July 27, 2012 612.3. Symmetry properties in terms of Weierstrass data. To assert the symmetryasserted by the Schwarz reflection principles, we now seek criteria on the Weierstrass datag, h such that f(I) is contained in a line, or f(I) and ν(I) are contained in a plane.Recall first that a geodesic is a curve γ : I → U whose image curve c := f ◦ γ has a normalcurvature vector(30) c ′′ (t) ∈ N c(t) f := {T c(t) f} ⊥ .It follows that c must have constant speed. Let us give a differential geometric characterizationof our symmetry curves:Lemma 16. Suppose f ∈ C 2 (U n , R n+1 ) is a hypersurface.(i) A curve c = f ◦ γ is a geodesic of the surface f as well as an asymptotic line if andonly if c is a constant speed parametrization of a straight line of R n+1 .(ii) Assume that a curve c = f ◦ γ satisfies c ′′ ≠ 0. Then c is a geodesic of the surface f aswell as a curvature line if and only if c is an arc of planar reflection, that is, c is containedin a hyperplane of R n+1 and ν ◦ γ is tangent to it.Proof. (i) Note first that c is an asymptote line if and only if 0 = g(Sγ ′ , γ ′ ) = −b(γ ′ , γ ′ ) =〈ν ′ , c ′ 〉, i.e., for ν ′ ⊥ c ′ .“ ⇒ ′′ : Making use of this fact, we find(31) 0 = 〈ν, c ′ 〉 ⇒ 0 = 〈ν, c ′′ 〉 + 〈ν ′ , c ′ 〉 ν′ ⊥c ′= 〈ν, c ′′ 〉But for the geodesic c the curvature vector c ′′ is parallel to ν. Hence (31) states that c ′′itself vanishes identically and so c must be a straight line.“ ⇐ ′′ : If c is straight and has constant speed then c ′′ = 0 and so c is geodesic. Moreover, forc a straight line, ν ′ is always orthogonal to the constant vector c ′ , and so c is asymptotic.(ii) The proof is similar and left as an exercise.□The symmetries (i) and (ii) are beautifully related for a pair of conjugate minimal surfacesf and f ∗ :Proposition 17. f ◦γ parameterizes a straight line of R 3 with unit tangent direction v ∈ S 2iff f ∗ ◦ γ parameterizes an arc of planar reflection for a plane with unit normal v ∈ S 2 .The same holds with f and f ∗ interchanged or more generally for any pair f ϑ and f ϑ±π/2 .As an example, the straight lines foliating the helicoid correspond to meridians of thecatenoid; the helicoid axis to the waist circle on the catenoid. Both are curves of planarreflection.

62 K. Grosse-Brauckmann: Minimal surfaces, SS 12Proof. By Theorem 11, the surfaces f and f ∗ are isometric. So γ is a geodesic for f iff itis one for f ∗ .Moreover, from Prop. 10 and (20) we conclude:f ◦ γ asymptotic line ⇔ g′g hγ′2 ∈ iR ⇔ g′g (−ih)γ′2 ∈ R ⇔ f ∗ ◦ γ curvature line.Given these two facts, it remains to refer to Lemma 16: A geodesic asymptote line isstraight and a geodesic curvature line γ is an arc of planar reflection of f ∗ .□20. Lecture, Friday 15.6.12To verify that a curve is geodesic by checking its ODE in terms of Christoffel symbolsis usually tedious. If possible, it is more convenient to characterize geodesics in terms ofsymmetries.In our setting, a symmetry is an isometry. Let us recall that σ : U → U is length-preservingor an isometry of f : U → R d , if for all X, Y ∈ R 2ds 2 σ(x)(dσ(X), dσ(Y ))= ds2x (X, Y ) or 〈df σ(x) (dσ(X)), df σ(z) (dσ(Y ))〉 = 〈df x (X), df x (Y )〉.Since the geodesic equation depends on the first fundamental form alone, isometries mapgeodesics to geodesics. Conversely, curves invariant under an isometry are geodesics:Lemma 18. Suppose a surface f : U m → R d has an isometry σ : U → U, which fixesprecisely a regular curve γ ∈ C 1 ((a, b), U) pointwise, i.e., σ(p) = p ⇔ p ∈ trace γ. Then γis geodesic up to parameterization.Example. If a curve in a surface is fixed under an ambient reflection then the curve isgeodesic.Proof. Consider a geodesic η = η p,v with initial conditions p := γ(t 0 ) and v := γ ′ (t 0 ). Sinceγ is invariant under σ, the initial conditions of γ are invariant under σ, that is, σ(p) = pand dσ(v) = v. But geodesics are unique for given initial conditions, so σ ◦ η = η. Inparticular η belongs to the set of points fixed by σ, so it must be contained in γ. □Problems. If p is fixed by σ, and v by dσ, what does the proof give? Use this to make astatement about regularity and length of γ, and also use it to generalize the lemma to thecase that σ fixes a surface or a submanifold of U.We use the characterization of the lemma to assert that the following curves are geodesic:Lemma 19. Let l ∋ 0 be a line in U ⊂ C. Suppose the Weierstrass data g, h of a conformalminimal surface f take images g ◦ γ(l) and h ◦ γ(l) contained in lines l 1 and l 2 throughthe origin, respectively. Then l is a geodesic for f.

ii 3.1 – date: July 27, 2012 63Proof. Let ρ be the reflection across l, and σ 1 , σ 2 be reflection about the straight lines l 1 , l 2 ,respectively. These reflections are length preserving and linear. For instance, |ρ(z)| = |z|and dρ = ρ or |X| 2 = |dρ(X)| 2 . We now apply the Schwarz reflection principle Cor. 13, inits extended version of Remark 1. This gives g = σ 1 ◦g ◦ρ and likewise for h. We conclude:‖X‖ 2 z ==( 1|g| +2(1|g|)|h|) 2|X| 2 =( 12(|g ◦ ρ| + 1|g ◦ ρ|( 12(|σ 1 ◦ g| +1|σ 1 ◦ g|)|σ 2 ◦ h|)|h ◦ ρ|) 2|dρ(X)| 2 = ‖dρ(X)‖ 2 ρ(z)This means that ρ is an isometry for f. Thus l is geodesic by Lemma 18.) 2|X|2□3. Surfaces constructed from Weierstrass dataWe construct further examples of minimal surfaces. This section is based on [Ka].3.1. Higher order Enneper surfaces. Let k ∈ N; the case k = 1 corresponds to theEnneper’s original surface. The Weierstrass data are0U := C, g(z) := z k , h(z) := 2z k .The order of the zero at 0 agrees for g and h. Thus (13) holds and f = f k is an immersion.We find Φ(z) = ( 1 − z 2k , i(1 + z 2k ), 2z k) and so∫ z((32) f(z) = Re Φ(w) dw = Re z − 12k + 1 z2k+1 , i(z + 1)22k + 1 z2k+1 ),k + 1 zk+1 .Let l be a straight line through the origin. Then g(l) = h(l) = l are also straight. Hencel is geodesic for f by Lemma 19.The first fundamental form isds =(|z k | + 1 )|z k | = ( |z 2k | + 1 ) .|z k |We conclude two properties. First the metric depends only on |z| and so for each k theEnneper surfaces have an intrinsic isometry, namely the rotation z ↦→ e iϕ z for all ϕ ∈ R.Second, each radial ray from 0 to ∞ has infinite length. This means that all geodesicsthrough some point have infinite length. By a theorem proven in Riemannian geometrythis implies the Enneper surfaces are complete in the sense that each Cauchy sequence inU converges with respect to the metric d on U induced from ds 2 .To compute the second fundamental form b z (X, X) = − Re ( g ′ (z)g(z) h(z)X2) note thatg ′ (z)g(z) h(z)X2 = kzk−1z k 2z k X 2 = 2kz k−1 X 2

64 K. Grosse-Brauckmann: Minimal surfaces, SS 12Fix z ∈ S 1 and consider the radial curves γ(t) = tz with γ ′ (t) = z. Thenb γ (γ ′ , γ ′ ) = − Re ( 2k(tz) k−1 z 2) = −2kt k−1 Re z k+1 ,and so( g ′ h)g ◦ γ γ ′2 ∈ R ⇐⇒ 0 ≡ (k + 1) arg z mod π,( g ′ h)g ◦ γ γ ′2 ∈ iR ⇐⇒ π ≡ (k + 1) arg z mod π.2Hence we can apply Lemma 16: For the first case, the radial lines t ↦→ tz are mapped tocurves in a mirror plane of the surface, while for the second they map to straight lines.To determine a symmetry group of the surface, note that by (7) we have∂ x f(0) = Re Φ(0) = (1, 0, 0) and ∂ y f(0) = − Im Φ(0) = (0, −1, 0),consistent with ν(0) = st ( g(0) ) = (0, 0, −1). Taking also the conformality of f intoaccount we find: Thus the surface contains k + 1 straight lines in the xy-plane makingangles π/(k + 1) or multiples thereof with oneanother. There are k + 1 vertical symmetryplanes whose projection in the xy-plane is spaced symmetrically in between the straightlines of the surface.21. Lecture, Monday 18.6.12Let us finally consider a circle γ(ϕ) := Re iϕ for R large. Its image f ◦ γ winds 2k + 1 timesabout the vertical axis. In particular, none of the Enneper surfaces can be embedded.To summarize:Theorem 20. The Enneper surfaces of higher order (32) are complete immersed minimalsurfaces with total curvature −4πk. They have an intrinsic rotation and are invariantunder ambient rotation by an angle 2π/(k + 1).We note that there are many deformations of the Enneper surfaces: If p is a polynomial ofdegree k − 1 then and ε ∈ R is an arbitrary parameter theng(z) = 1 2 h(z) := zk + p(z)has no branch points the same asymptotics as z → ∞. However, the symmetries are onlymaintained for particular polynomials p.

ii 3.2 – date: July 27, 2012 653.2. Simply periodic Scherk surfaces. The Weierstrass data are(33) U = D := {z ∈ C : |z| < 1}, g(z) = z, h(z) = zz 4 + 1 .The only zero of h is at z = 0, where g also has a zero of the same order. Hence (8) and(13) hold and f is an immersion. The function h has four singular points p l := i l e iπ/4 on∂D, for l = 0, 1, 2, 3, satisfying p 4 l = −1.To discuss symmetry properties, let us first consider four special lines in D, namely thecoordinate axes and the diagonals:1. γ(t) = t ⇒ g(γ(t)) = t ∈ R, h(γ(t)) = tt 4 + 1 ∈ R2. γ(t) = it ⇒ g(γ(it)) = it ∈ iR, h(γ(t)) = itt 4 + 1 ∈ iR3. γ(t) = 1 + i √2t ⇒ g(γ(t)) = 1 + i √2t ∈ (1 + i)R, h(γ(t)) = (1 + i)t/√ 2−t 4 + 14. γ(t) = 1 − i √2t ⇒ g(γ(t)) = 1 − i √2t ∈ (1 − i)R, h(γ(t)) = (1 − i)t/√ 2−t 4 + 1∈ (1 + i)R∈ (1 − i)RThe images under g and h of these four lines are again straight lines through the origin.By Lemma 19 they are geodesics of f. To see they are symmetry curves, let us computetheir second fundamental form:⎧) ( (( g′1g h) ◦ γ γ ′2 γ⎨=)γ ′2 = γ′2γ γ 4 + 1 γ 4 + 1 ∈ R in case 1., 2.,⎩iR in case 3., 4.By Lemma 16 the curves 1.,2. map to arcs of planar reflection while 3.,4. map to straightlines of R 3 . Let us determine which planes and lines these are.We assume z 0 = 0 so that f(0) = 0; due to g(0) = 0 the tangent plane at 0 must bethe xy-plane. For 1. the Gauss image ν(R) = st(R) is contained in the xz-plane. Thusthe mirror plane must be the xz-plane itself. Simlilarly for 2.: In this case the yz-planecontaining st(iR) is the mirror plane.The diagonals of cases 3. and 4. map to straight lines. The values of g show that the imagecurves run in the diagonal directions of the xy-plane. The two straight lines f(γ(t)) extendto infinity of R 3 when one of their singular endpoints p l is approached with t → ±1. Tosee this, write the first fundamental form asds = 1 2(|z| + 1|z|) |z||z 4 + 1| = |z|2 + 12|z 4 + 1|For both diagonals γ(t) with t ∈ (−1, 1) we have γ 4 (t) = −t 4 > −1 and so0 < γ 4 (t) + 1 = 1 − t 4 = (1 − t 2 )(1 + t 2 ) = (1 − t)(1 + t)(1 + t 2 ).

66 K. Grosse-Brauckmann: Minimal surfaces, SS 12This proves our claim: On an interval (t 1 , t 2 ) we haveL(γ) =∫ t2t 1ds ( γ(t) ) |γ ′ (t)| dt =∫ t2t 2 ∫+ 1t2t 12(1 − t)(1 + t)(1 + t 2 ) dt =and this length diverges to infinity if t 1 = −1 or t 2 = 1.1t 121(1 − t)(1 + t) dtWe wish to consider the four open arcs γ l in ∂D running from p l to p l+1 (indices mod 4):5. γ(t) = e it ⇒ g(γ(t)) = e it ∈ S 1 ,Although Lemma 19 does not apply directly, we can use similar arguments to show that γis geodesic in f. For this end, we write the first fundamental form as(34) ‖X‖ = 1 (|z| + 1 )1 1∣2 |z| ∣ z2 + 1|z| |X|.z 2We claim that inversion in the unit circle, σ(z) := 1, is an isometry, i.e., z ‖dσ(X)‖2 = ‖X‖ 2 .The first two factors in (34) are preserved when we replace z by σ(z). Now let us show that1|X| is also preserved. Since d 1 = − 1 and z → z is length preserving we have (using|z| z z 2|z| = |z|)|dσ(X)||σ(z)|=1|X||z| 21/|z|= 1|z| |X|.This verifies that σ is an isometry, and so the fixed point set S 1 consists of four arcs γ leach mapping to a geodesic. Moreover, these arcs satisfy) (( g′g h) ◦ γ γ ′2 = (γ′ ) 2 ( γ′ ) 2γ 4 + 1 = 1( ieit ) 2γ γ 2 + γ = 1−2 e it e 2it + e = − 1−2it 2 Re(e 2it ) ∈ Rand so are curvature lines. Consequently each of the four arcs has image contained in amirror plane. Since g takes values in S 1 the normal ν = st ◦g is horizontal along γ, and sothe four symmetry planes are horizontal. Due to the reflection symmetries in the xz- andyz-planes, the planes containing opposite pairs of these curves must agree.We claim, however, that the remaining two planes generated by adjacent arcs occur atdifferent heights. To see this, computef 3 (1) − f 3 (0) = Re∫ 10h(w) dw = Re∫ 10∫t1t 4 + 1 dt = tt 4 + 1 dt = d > 0Thus 0 and 1 have images at different heights, and by the rotation symmetry this provesthat the horizontal symmetry planes occur at heights spaced 2d apart. By the symmetries(or computation) therefore that f 3 (1) = f 3 (−1) = d while f 3 (i) = f 3 (−i) = −d.22. Lecture, Wednesday 20.6.12We now want to extend the surface to become complete. The Weierstrass data g, h extendto the domain Ω := C \ {p 0 , . . . , p 3 }. However, Ω is not simply connected, and we will seethat f does not extend analytically to Ω.0

ii 3.2 – date: July 27, 2012 67Let us consider one of the four arcs γ l in ∂D. Then f extends analytically by Schwarzreflection across γ l to a map f l : D ∪ σ(D) ∪ {γ l } → R 3 . The opposite arc γ l+2 (indicesmod 4) also defines an extension by Schwarz reflection f l+2 . But it has the same heightand so f l+2 = f l , where defined. Thus f l can be defined on the doubly connected domainU l := D ∪ σ(D) ∪ {γ l } ∪ {γ l+2 }.We are left with the two different extensions f 0 and f 1 , say: One extends f(D) by reflectionbelow the lower horizontal symmetry plane, and the other above the upper horizontalsymmetry plane. Consequently, the two one-sided limits of f l along the slits γ l+1 and alsoalong γ l+3 are vertical translations of oneanother, at a distance 4d.Two successive reflections in parallel planes create a translation (see problems) and sothe locally parameterized surface f 0 or f 1 can be indefinitely continued by translation.This extension locally can be parameterized by f 0 or f 1 plus a vertical translation inΛ := Z(0, 0, 4d). Thus we have arrived at a simply periodic surface M, invariant under Λ.A vector (0, 0, 4d) generating Λ is called period vector, and the closure F := f(D), whichgenerates M under the lattice Λ of translations preserving the orientation of M is called atranslational fundamental domain. On the interior of F , the Gauss map is injective, and onF it is surjective to S 2 . Hence the area 4π of the Gauss image is minus the total curvatureof F , see I 1.5.The simply periodic surface M is embedded. Let us finally sketch how this can be proved.We claim that f(D) is graph, hence embedded. Since f(D) is contained in a slab ofheight 2d, its reflections with respect to the bounding horizontal plane stays disjoint,and so M will also be embedded. Now f(D) has Gauss image in the lower hemisphere.So it cannot have a vertical tangent plane and the projection π to the xy-plane is animmersion. To prove the claim, we must show that π ◦ f embeds D into the plane. Todo so we use symmetries and the Gauss image of f. Consider a quarter of the unit disk,D 4 := {re iϕ : 0 < r < 1, π/4 < ϕ < 3π/4}. Then the two straight boundary arcs of D 4are mapped under π ◦ f to two consecutive diagonals of the plane, meeting at the origin.The bounding quarter circle γ 0 projects to a graph over any of these two diagonals. Notethat π(γ 0 ) cannot meet the diagonals since π(f(D 4 )) sits to one side of each of its threeboundary arcs. Therefore the projections of the three boundary arcs enclose a domain Ωand the surface f(D 4 ) is graph over Ω. By symmetry f(D) is again graph.Remarks. 1. It can also be shown that the surfaces are embedded and have ends which areasymptotically planar.2. The Scherk surfaces are close to two perpendicular planes whose intersection is replacedby chain of handles. It is an open problem if the family of Scherk surfaces as in Sect. 3.2are the only embedded singly periodic surfaces, for which the quotient has genus 1 and theends are planar.

68 K. Grosse-Brauckmann: Minimal surfaces, SS 12Let us comment on other versions and generalizations of our Scherk surface:1. Analogues with higher rotational symmetry are obtained by setting for k ≥ 2U = D, g(z) = z k−1 , h(z) = zk−1z 2k + 1 = 1 1z z k + 1 .z kHere, the 2k-th roots of −1 give 2k singular points corresponding to ends of the surface.2. For any angle 0 < α < π, there are versions of the Scherk surface for which the anglebetween the ends is not π/2 but α.3. The doubly periodic Scherk surface can be shown to be the conjugate of f (problems?).4. Costa surface (2006)4.1. Elliptic functions. To construct the Costa surface in the next section, we needspecial functions:Definition. A meromorphic function f : C → C which is invariant under some latticeΛ = {λ 1 v 1 + λ 2 v 2 } of rank 2, that is,is called an elliptic function.f(z + w) = f(z) for all w ∈ Λ,Except for the constants, there are no obvious examples of elliptic functions. In fact, theirconstruction will take us some effort.In fact, the only elliptic functions which are holomorphic are the constants. Namely, byperiodicity, holomorphic elliptic functions must be bounded. But they are defined on allof C and hence must be constant by Liouville’s theorem.Other interesting facts can be derived from the residue theorem. To state them, let usdenote a fundamental domain by Q := {sv 1 + tv 2 : 0 ≤ s, t < 1}.Proposition 21. Let f be elliptic with poles at b 1 , . . . , b k ∈ Q. ThenProof. If ∂Q contains no pole thenk∑∫2πi res bj f = f(w) dw =j=1∂Q0 =k∑res bj f.j=1( ∫ ∫ ∫ ∫ )+ − − f(w) dw,[0,v 1 ] [v 1 ,v 1 +v 2 ] [v 2 ,v 1 +v 2 ] [0,v 2 ]using interval notation for straight paths in C. But due to periodicity of f, the sum of thefirst and third, as well as the sum of the second and fourth integral must vanish. If ∂Q

ii 4.2 – date: July 27, 2012 69contains a pole, we find a parallel translate of Q whose interior will contain all poles; theabove proof then applies.□Corollary 22. (i) There is no elliptic function whose only pole is of first order.(ii) Over its fundamental domain Q, a non-constant elliptic function takes each value thesame number of times; here we count the number with its finite multiplicity (at branchpoints).Proof. While (i) is obvious, we prove (ii) by applying the proposition to the elliptic functionf ′ /f.We first claim that f has as many poles as zeros. Since f is non-constant, the zero set of fis discrete, hence finite. Moreover, then f ′is a meromorphic function. If f has zeros at b f j,for j = 1, . . . , Z, their order n j ∈ N is defined. Let us now develop f and f ′ into a powerseries at a zero b of order n (we drop the index j):f ′f = na n(z − b) n−1 + . . .= n 1( f′ )a n (z − b) n + . . . z − b + . . . ⇒ res b = n.fSimilarly, if f has a pole of order m = m j at a = a j , for j = 1, . . . , P then a Laurent seriesexpansion givesf ′f = −ma m(z − a) −m−1 + . . .( f= −m(z − a) −1 ′ )+ . . . ⇒ resa m (z − a) −m a = −m.+ . . .fThus when counted with multiplicity,M∑(35)res f ′ Zf = ∑n j −j=1j=1P∑m j .By the proposition, the left hand side vanishes, which proves the claim.Now apply the result to f replaced with f − c, where c ∈ C is arbitrary. Then (35) givesthat the number of points with f = c minus the number of poles of f (each counted withmultiplicity) is once again 0.□4.2. Existence of elliptic functions. Are there any non-constant elliptic functions atall? The simplest elliptic functions which are admissible in view of (i) would have two polesand zeros, either distinct or coincident. We will construct such functions by the Riemannmapping problem and Schwarz reflection. We present the construction of elliptic functionssuggested by Karcher. It is a simple planar analogue of our construction of the triplyperiodic minimal surfaces. This constructs the elliptic functions together with symmetries.These symmetries in turn will be essential for applying them to construct minimal surfaces.We will see this on the example of the Costa surface. Our approach is also explained inAusgewählte Kapitel aus der Funktionentheorie by Fischer-Lieb, Sect VI,8.j=1

70 K. Grosse-Brauckmann: Minimal surfaces, SS 12Elliptic functions are a traditional subject in complex analysis, see, e.g. [FL, Kap.VII,7].The classical approach to introduce these functions is via certain series.To simplify the exposition, we consider exclusively the particular case of a primitive lattice(36) Λ := 4Z 2 , Q := [−2, 2) × [−2, 2) ⊂ COur first elliptic function has two distinct poles and zeros of first order.Lemma 23. There exists a function q : C → C which is elliptic w.r.t. Λ = 4Z 2 . It hasfirst order zeros at 0 and 2 + 2i, first order poles at 2 and 2i.These are all poles and zeros up to the action of Λ. Special values of q are:q(R) = R,q(iR) = iR,q ( (1 ± i)R ) = (1 ± i)R, q(2i + R) = R, q(2 + iR) = iRNote that C compactifies with one point ∞, while R usually compactifies with two points±∞. However, when we consider R as a subset of C, we will compactify R with just onepoint ∞ as well. Thus we use, e.g., the notation R for R ∪ ∞ ⊂ C.Proof. Consider a triangle and an eighth of the unit disk,∆ := (0, 1) × (0, 1) ∩ {x > y},C := {z = re iϕ : 0 < r < 1, 0 < ϕ < π/4}.By the Riemann mapping theorem III.29 there is biholomorphic map q : ∆ → C subjectto the three point condition q(0) = 0, q(1) = 1, q(1 + i) = 1+i √2. The map q is holomorphicand extends continuously to the boundary.Note that the continuous boundary extension q : ∆ → C maps [0, 1] to itself and [0, 1 + i]1to [0, √2 (1 + i)]. Thus we can use the Schwarz reflection Thm. 12 to extend u across thesestraight arcs in domain and range. The resulting reflection is of the form u(z) = σ(u(ρ(z))),where ρ, σ are the appropriate reflections of domain and range, respectively. More precisely,we proceed in three steps.1. Four copies of ∆ map (−1, 1)×(0, 1) to the upper disk D∩{x > 0}; the map q is smoothon the interior of this set. This extension maps the bounding interval (−1, 1) into itself.Let us apply one further Schwarz reflection, this time of the standard type q(z) = q(z). Asguaranteed by Thm. 12 the resulting map is smooth at 0. This way we obtain a surjectivemap q : (−1, 1) 2 → D; it is biholomorphic, as the reflected copies are disjoint.As in the proof of Riemann mapping theorem we conclude that q has no branch point at 0.Thus 0 is a zero of first order. Moreover, q has no further zeros in [−1, 1] 2 , nor poles.2. We now extend q to the rectangle R := (−2, 2) × (−1, 1) by Schwarz reflecting theimages in the bounding circle arcs. Note that these arcs are of the form e it where either

ii 4.2 – date: July 27, 2012 71|t| < π or |t − π| < π . The horizontal boundary arcs of ∂R are mapped to the complement4 4of these arcs on S 1 . The image of the rectangle is C.3. In a third step we extend q to all of Q. This can be done with respect to a tesselation ofQ with alltogether 32 copies of the original triangle (or by reflecting larger pieces at once).A second zero at 2 + 2i is created.The points 2, 2i ∈ ∂Q are mirror image of 0, and hence under the inversion z ↦→ 1 z∞. For this reason these are poles of first order.map toFurther extension leads to a periodic function. The special values and the location of polesand zeros follow directly from the construction.□Remark. There are further properties of q which our construction yields, but which will notneed in the sequel: The values q(±i + R) ⊂ S 1 , q(±1 + iR) ⊂ S 1 are immediate. Moreover,q maps (−1, 1) × (−1, 1) biholomorphically to D. At the point 1 + i, the map q doublesthe the 45 ◦ angle of ∆ to the 90 ◦ angle of C. Thus the Taylor series starts with z 2 , that is,1 + i is a zero of second order. Including also the reflected images into this considerationgives that 1 + i, 1 − i, −1 + i, −1 − i are the only branch points of q.50. Lecture, Tuesday 20.6.06Our second elliptic function has a double pole and a double zero:Lemma 24. There exists a function p: C → C which is elliptic w.r.t. Λ = 4Z 2 . It hasa zero of second order at 0, a pole of second order at 2 + 2i;these are all poles and zeros up to the action of Λ. Besides 0, the only branch points are 2and 2i modulo Λ. Special values of p are:p(2) = 1, p(2i) = −1, p(R) = [0, 1], p(iR) = [−1, 0],p ( (1 + i)R ) = i[0, ∞) ∪ ∞, p ( (1 − i)R ) = i(−∞, 0] ∪ ∞,p(2i + R) = (−∞, −1] ∪ ∞,p(2 + iR) = [1, ∞) ∪ ∞Note that at 0 the special images are consistent with 0 being a branch point of order 2.Proof. Let C be a “triangle” bounded by [0, i], [0, ρ] for ρ > 0 and a circular arc through iand ρ. The point ρ > 0 is chosen such that the circular arc meets the imaginary axisat 45 ◦ and the real axis at 90 ◦ (the circle has its midpoint on the negative real axis). TheRiemann mapping theorem gives a unique map p, mapping the triangle ∆ (defined above)biholomorphically onto C, such that 0 ↦→ 0, 1 + i ↦→ i, and 1 maps to the third vertex.As in the above remark we see that at the vertex 0, the map p doubles the 45 ◦ angle of ∆to the 90 ◦ angle of C. Hence 0 is a zero of second order, that is, a branch point of p.

72 K. Grosse-Brauckmann: Minimal surfaces, SS 12We can extend p stepwise by mapping reflected images to reflected images. Along circulararcs in the range, we use inversion (it happens to be z ↦→ 2 − 1)z+11. Reflect ∆ along [1, 1 + i]. The original “triangle” C can be complemented by its imageinverted in the bounding circular arc. The result is the disk D 4 := D ∩ {x > 0, y > 0}.Indeed, the image of [0, i] is a circle arc which meets both the real and imaginary axis ata right angle. Thus the doubled triangle ∆ maps to the quarter disk D 4 .2. We can now use the extension constructed in 1. to reflect further. At this stage, weonly need planar reflection and (in the image:) inversion in the unit circle.All properties claimed are obvious from the construction. For instance, the reflected images2, 2i of the origin also become branch points, and no further branch points are created. □It is easy to create elliptic functions with zeros and poles of higher order, for instance bydifferentiation of p or q. We collect the following facts about the derivative p ′ which willbecome useful to study the Gauss map of the Costa surface below,Lemma 25. The function p ′ : C → C is elliptic w.r.t. (36) and satisfies the differentialequation(37) p ′ q = cp for some c ∈ R + .It has the following special values:p ′ (0) = p ′ (2) = p ′ (2i) = 0 (1. order), p ′ (2 + 2i) = ∞ (3. order),p ′ (R) ⊂ R, p ′ (iR) ⊂ iR, p ′( (1 ± i)R ) ⊂ (1 ± i)R,p ′ (2i + R) ⊂ R,p ′ (2 + iR) ⊂ iRMoreover, there are exactly four branch points modulo Λ, namely one in each interval (0, 2),(−2, 0), (0, 2i), (−2i, 0).Proof. To establish (37), we claim that the zeros and poles on the two sides coincide.We start with p ′ itself. It has simple zeros at the branch points of p, namely only at 0, 2,2i modulo Λ. Since p has double pole at 2 + 2i, the function p ′ has a triple pole there; ithas no more poles. Moreover, q has simple poles at 2, 2i, making these points removablefor p ′ q. The simple zeros of q at 0, 2 + 2i creating a double zero of p ′ q at 0, and a doublepole of p ′ q.Thus p ′ q has the same zeros and poles as p. To establish (37), consider the quotient p′ qp ,which is elliptic as well. Zeros and poles of numerator and denominator coincide. Thusthe fraction is bounded and its singularities are removable. By Liouville’s theorem, thequotient is constant c ∈ C. The values of numerator and denominator on (0, 2) are positivereal. Indeed, to see this for p ′ , note that the Riemann mapping theorem gives that p is

ii 4.3 – date: July 27, 2012 73injective on (0, 1) hence monotone; there is no branch point on (0, 1) hence the derivativeis positive. We conclude that our constant c is a positive real.To determine the special values p ′ ◦ γ along some arc γ, note that by the chain rulep ′ ◦ γ = 1 d(p ◦ γ).γ ′ dtUsing the values of p◦γ of the Lemma 24, we can assert the following. For γ parameterizingR with γ ′ = 1 we also have p ◦ γ ∈ R, which makes p ′ ◦ γ real. Similarly, on iR the tangentis i while p > 0, thus p ′ is imaginary. Finally, on (1 ± i)R the tangent is 1 ± i while p isimaginary; thus p ′ is a real multiple of11 ± i i = 1 ∓ i1 − i i = 1 (i ± 1).2 2Since p ′ (0) = p ′ (2) = 0 and p ′ is real on (0, 2), there must be a critical point of p ′ on (0, 2).Thus p ′′ has at least one zero, or p ′ at least one branch point on (0, 2). The same argumentapplies to the other three intervals on the coordinate axes. Since p ′′ has exactly one pole oforder 4 we can apply Corollary 22(ii) to conclude that there are no more branch points. □There is a relation of our elliptic functions p, q which we will need:(38)q 2p = 21 − p 2As in the proof of (37), the two side of the equation are proportional as they have the samezeros and poles. The exact value of the proportionality constant can be read off by pickinga particular point (which one?) in the domain. We leave the verification as an exercise.Remarks. 1. Elliptic functions exist for arbitrary lattices Λ. The symmetry method toconstruct them is, however, limited to the rectangular lattices with v 1 ⊥ v 2 .2. We have introduced two elliptic functions p, q. In the following sense, two functions aresufficient: Each elliptic function is a rational function of p, q. See [FL].2. The Weierstraß p-function, defined in many books, differs from our p-function by translationand scaling.51. Lecture, Wednesday 21.6.064.3. The Costa surface. The catenoid is an embedded minimal surface of finite totalcurvature. It is the only surface with these properties we have discussed so far, and it hadbeen the only known such surface up to 1982. In that year the Brasilian mathematicianCosta discovered Weierstrass data of a further finite total curvature surface. Only later,Hofmann and Meeks showed that Costa’s surface is indeed embedded.

74 K. Grosse-Brauckmann: Minimal surfaces, SS 12It is not only for historical reasons that we present the surface here. Also it is a fineexample for the main problems faced when applying the Weierstrass construction:• Solution of the period problem,• embeddedness proof.The Weierstrass data for the Costa surface involve a real parameter r > 0 to be determinedlater:U = (−2, 2) × (−2, 2), g = rp ′ = cr p q , h = q,where c > 0 is determined by Lemma 25.Proposition 26. Let P := 2 + 2i, C − := 2, C + := 2i, and define a singular setS := {P, C − , C + } + 4Z 2 .Then for any r > 0, the Costa surface f r : U → R 3 extends continuously to U \S, and neighbourhoodsof P map to a planar end while neighbourhoods of C ± each map to a catenoidend.Proof. Note that the only zero of g on U \ S is at z = 0. There, h has a zero as well, so byLemma 5 the Weierstrass data define a minimal surface over U. The asymptotics of theends will be asserted below.□Proposition 27. For any r > 0, consider the Costa surface f r : U → R 3 translated so thatf r (0) = 0. It has the following symmetry properties:(i) The lines R and iR map to lines of planar reflection contained in the xz− and yzplanes,respectively.(ii) The lines (1 ± i)R map to straight lines of 180 ◦ -rotation on the x ∓ y-lines through theorigin.(iii) The two boundary arcs (−2 + 2i, 2i) and (2i, 2 + 2i) map to mirror symmetry planesof the surface, each of them parallel to xz. Likewise, the curves (2 − 2i, 2) and (2, 2 + 2i)map to mirror planes of the surface, each parallel to yz.Here, our wording is that a line has infinite length.Proof. (i) We consider the coordinate axes:g(R) = p ′ (R) ⊂ R, h(R) = q(R) ⊂ R, g(iR) = p ′ (iR) ⊂ iR, h(iR) = q(iR) ⊂ iR.By Lemma 19, the curves map to geodesics of the surface.Without calculation, this suffices to assert that along the curve γ parameterizing (−2, 2)with tangent γ ′ = 1 ∈ R,g ′g γ′ =(g ◦ γ)′g∈ R, h ∈ R ⇒ g′g h(γ′ ) 2 = g′g γ′ hγ ′ ∈ R.

ii 4.3 – date: July 27, 2012 75Similarly, along the curve γ parameterizing (−2i, 2i) with tangent γ ′ = i,g ′g γ′ =(g ◦ γ)′g∈ R, h(iR) ∈ iR ⇒ g′g γ′ hγ ′ ∈ R;By Lemma 16 combined with Proposition 10(ii), this shows that f r maps these curves tomirror planes of the surface. Since g(R) ⊂ R, we have st(g) = ν is in the xz-plane, and sothe mirror plane is the xz-plane. Similarly on iR.(ii) We reason as in (i) for the images of the diagonals. Sinceg ( (1 ± i)R ) = p ′( (1 ± i)R ) ⊂ (1 ± i)R,h ( (1 + i)R) = q ( (1 ± i)R ) ⊂ (1 + i)Rthe diagonals map to geodesics.Let γ parameterize the diagonals (1 ± i)R with tangent γ ′ ∈ (1 ± i)R. Theng ′g γ′ =(g ◦ γ)′g∈ R ⇒ g′g γ′ hγ ′ ∈ (1 ± i) 2 R = iR.Lemma 16 combined with Proposition 10(i) asserts the curves are straight lines.Moreover, g maps (1 ± i)R to itself, so that ν = st(g) is in the plane spanned by z andx ± y. The straight image lines must run in the direction of the orthogonal complement ofthese planes, which gives our claim.(iii) Let γ be a curve parameterizing a connected arc in (2i + R) \ S. Such an arc is ageodesic curvature line, hence a curve of planar reflection:g(γ) = p ′ (γ) ∈ R, h(γ) = q(γ) ∈ R,g ′g γ′ =(g ◦ γ)′g∈ R ⇒ g′g γ′ hγ ′ ∈ R,On the other hand, let now γ parameterize a connected arc of (2 + iR) \ S. Any arc γdefines a mirror plane as well:g(γ) = p ′ (γ) ∈ iR,h(γ) = q(γ) ∈ iR,g ′g γ′ =The values of g give the claim on the containing planes.(g ◦ γ)′g∈ R ⇒ g′g γ′ hγ ′ ∈ i 2 R = R.□We call the eight components of ∂U \ S boundary arcs. Each boundary arc has length 2.We call a pair of arcs opposite if they differ by a translation in 4Z 2 .Part (iii) of Proposition 27 asserts that the image of the boundary arcs creates a verticalsymmetry plane of the surface which allows for Schwarz reflection to continue the surfaceacross the boundary arc. This continuation agrees with the continuation of the Weierstrassdata f to any simply connected domain which contains the arc in its interior.

76 K. Grosse-Brauckmann: Minimal surfaces, SS 12Lemma 28. (i) The image of two opposite boundary arcs differs (pointwise) by a vectorperpendicular to the containing symmetry planes.(ii) If the image of one pair of opposite arcs is contained in the same plane then theimmersion f r agrees on all four pairs and extends to a doubly periodic mapf r : C \ S → R 3 with f(z + w) = f(z) for all w ∈ 4Z 2 .Proof. We deal with the case that opposite boundary arcs I 1 , I 2 ⊂ ∂U have their imagecontained in a pair of symmetry planes parallel to the yz-plane. The case of arcs withxz-symmetry plane is analogous.(i) By Prop. 27(i) there is a third reflection plane parallel to the yz-plane, namely f(iR).Reflection in the yz-plane leaves the last two coordinates invariant, that is, f j (z) = f j (z+4)for z ∈ I 1 and j = 2, 3.(ii) If f agrees on the pair of opposite arcs I 1 , I 2 , parallel to the yz-plane then this meansprecisely f 1 (z) = f 1 (z + 4) for z ∈ I 1 . Together with (i) this gives f(z) = f(z + 4) forz ∈ I 1 .Let us now use the rotations about the straight lines in the x ± y-direction stated inProp. 27(ii). Starting with one pair of opposite arcs, the two rotations map them to theremaining three pairs of opposite arcs. Thus the distances between the symmetry planescontaining a pair of opposite arcs must agree.□52. Lecture, Tuesday 27.6.06In order to obtain a nice closed surface we wish to choose r such that the symmetry planescoincide.( ) ( )The two yz-mirror planes containing the curves f r (−2, −2 + 2i) and fr (2, 2 + 2i) havea distance∫d(r) := fr 1 (z) − fr 1 1( 1)(z + 4) = ReΓ 2 g − g h dw,where Γ represents a path in U from z to z + 4. Since d(r) measures the orthogonaldistance of the two planes. It is independent of the choice of z ∈ (−2, −2 + 2i) and Γ. Forconvenience, we let Γ parameterize (−2 + i, 2 + i). We call d(r) a period of the surface.The period problem amounts to determining r such that d(r) = 0. For a general minimalsurface, it will be necessary to determine several parameters which lead to the vanishingof several periods.Lemma 29 (Solution of period problem). There exists an r > 0, such that d(r) = 0.

Proof. With c > 0 as in (37) we have1g h =ii 4.3 – date: July 27, 2012 77qcrp q = 1 q 2cr pandgh = crpq q = crp.Thus∫1( 1)d(r) = ReΓ 2 g − g h dw = 1 1r 2c∫ΓRe q 2p dw − r c 2∫ΓRe p dw.We want to pick r such that the sum of the integrals vanishes. We will show that the realparts of both integrals are positive,∫(39) ReΓq 2pdw(38)= 2 Re∫Γ11 − p dw > 0 and Re p dw > 0,∫Γ2so that the intermediate value theorem yields such an r. Since we only know p well alongthe symmetry lines, in the following we must reduce the verification of (39) to the valuesof p on symmetry lines.Let us first consider the second integral of (39). We consider a path ˜Γ consisting of thethree arcs (−2 + i, −2), (−2, 2), and (2, 2 + i). Let R be the rectangle bounded by theclosed curve Γ ∪ ˜Γ. Note that the only pole of p is at 2 + 2i so that p has no pole on R;thus Cauchy’s Theorem 1 applies to R:∫Γp dwCauchy’s thm.=p periodic=∫∫p dw =˜Γ(−2,2)( ∫p dw > 0,where we used p ( (−2, 2) ) ⊂ [0, 1] in the last step.+(−2+i,2)∫(−2,2)∫ )− p dw(2,2+i)We use the same kind of argument for the first integral of (39). Here we let ˜Γ be definedas the concatination of the four intervals(−2 + i, −2 + 2i), (−2 + 2i, 0), (0, 2 + 2i), (2 + 2i, 2 + i).Together with Γ, it bounds a subset Ω ⊂ U, consisting of three open sets (triangles). Weclaim that 2 is the only zero of 1 − p 2 . Indeed, by Lemma 24, p(2) = 1 is a branch point,and moreover p has only one zero of second order. Thus by Corollary 22, the function ptakes each value twice, thus only 2 maps to 1. Similarly, only 2i maps to −1. This provesthe claim, and we can apply Cauchy’s theorem to Ω:∫Γ1 Cauchy’s thm.dw =1 − p2 ( ∫=p periodic=∫∫˜Γ11 − p 2 dw+(−2+i,−2+2i)(−2+2i,0)∫(−2+2i,0)11 − p∫(0,2+2i)dw + 2∫ ∫ ) 1+ −(0,2+2i) (2+i,2+2i) 1 − p dw 211 − p dw 2

78 K. Grosse-Brauckmann: Minimal surfaces, SS 12We have p ∈ ±iR along the two diagonals implying p 2 1∈ R − and1−p 2γ(t) = −2 + 2i + (1 − i)t we find∫Re(−2+2i,0)∫121 − p dw = Re 2Similarly for the other integral. We conclude Re ∫ Γ55. Lecture, Wednesday 28.6.060∫121 − p 2 (γ(t)) (1 − i) dt =0> 0. Thus for1dt > 0.1 − p 2 (γ(t))1dw > 0 and (39) is established.1−p 2Theorem 30. The Costa surface f r with r chosen as in Lemma 29, is embedded and hasfinite total curvature −12π.□Embeddedness proofs tend to be not pretty and to depend on rather special arguments.The following proof is perhaps no exception.Proof. The triangle ∆ with vertices 0, 2, 2+2i parameterizes an eighth of the Costa surface.We may assume f(0) = 0. It is sufficient to prove:• f(∆) is embedded and contained in an octon of R 3 .• The eight different copies of the triangle tesselating Q are each mapped to a differentocton of R 3 .The second property follows directly from the first: There are eight copies of ∆ whichtesselate U; by the symmetry properties, each of them maps under f to a different octonof R 3 .To show the first property, we claim that ∆ is mapped under g biholomorphically toM = {z = x + iy : x > 0, y < x} \ g([0, 2]).There is a unique branch point of rp ′ on (0, 2), see Lemma 25. Thus g = rp ′ maps the arc[0, 2] onto some interval [0, x 0 ], where x 0 is the image of the branch point.There are no branch points on ∆. Using the argument given in the proof of the Riemannmapping Theorem 29 we can assert that each value in M is attained the same number oftimes. This number is one in a neighbourhood of ∂∆, and hence one everywhere. We leaveit as an exercise to show that this works for the present non-compact situtation with ends.The right half-space {z : Re z > 0} maps under stereographic projection to the “eastern”hemisphere with midpoint (1, 0, 0) = st(1). That is, {z : Re z > 0} = {z ∈ C : 0

ii 4.3 – date: July 27, 2012 79We would like to assert that π x ◦ f is injective, so that f : ∆ → R 3 is graph with respectto the x-direction. To see this, we must study the boundary behaviour. The special valuesasserted in Prop. 27(iii) imply the following for f(∂∆).a) The curve (0, 2 + 2i) is mapped under f to a straight ray in the x − y direction of R 3 .Thus under π x it maps to the negative y-axis.b) The image of the arc (0, 2) is contained in the xz-plane, the curve runs from the saddleto one end. Since h = q is real and positive on (0, 2) we have that f 3 ◦ γ = Re ∫ h > 0.Thus the curve projects to the positive z-axis.c) The curve (2, 2 + 2i) is mapped under f to a planar symmetry curve in the yz-plane;it runs from catenoid end to planar end. We claim that the Gauss curvature K is strictlynegative along this arc. Indeed, on (2, 2 + 2i) neither g nor h have zeros or poles, and g ′is nonzero as g has no branch point on the arc. Thus K is nonzero and so the curvatureline f(2, 2 + 2i) has nonzero curvature; moreover the limiting positions of the normal areopposite, g(2i) = 0 and g(2 + 2i) = ∞. Consequently, the z-component along this curveis strictly monotone, while the y-component is strictly monotone. As π x leaves the curveinvariant, and the mean curvature of f vanishes we must have that π x ◦ f(2, 2 + 2i) isconcave w.r.t. the side of the immersion f(∆).Since π x ◦ f is an immersion, the image of ∆ locally sits to one side of the arcs (a) and(b). Moreover, (c) also has the image to one side. Since π x ◦ f is an immersion, (c) doesnot intersect (a) or (b). Hence π x ◦ f(∆) bounds a domain in a quadrant of R 2 , and f(∆)is a graph over this domain, hence embedded.We need to show that f(∆) is contained in an octon. Since the projection is in a quadrant,certainly f 2 (p) < 0, f 3 (p) > 0 for all p ∈ ∆. We want to establish f 1 has one sign. Allthree boundary arcs f(∂∆) satisfy x ≥ 0, and this holds as well at a neighbourhood ofthe singular points 2 and 2i which map to the ends (to be established by our analysis ofthe ends below). By the maximum principle, we conclude f 1 (p) > 0 for all p ∈ ∆. Thisestablishes embeddedness in an octon.54. Lecture, Tuesday 4.7.06We have seen g(∆) = M. The Gauss image ν(∆) = (st ◦ g)(∆) is bounded by two greatcircle arcs which make a 135 ◦ -angle. Thus it has area A(st(M)) = 3 4π. Since g extends to8the fundamental domain Q by reflection in great circles, we have, as claimed,∫− K dS I1.3= A Q (ν) = 8A ∆ (ν) = 8A(st(M)) = 12π.□QIn real life, Weierstrass data are not given but must be derived. If sufficient symmetriescan be assumed, this task can be performed systematically. Let us show how this can be

80 K. Grosse-Brauckmann: Minimal surfaces, SS 12done on the example of the Costa surface:1. We determine the topology: It is a torus with three punctures.2. We determine the geometry of U: The fundamental domain has two perpendiculardiagonals of reflection making the fundamental domain a rhombus. Moreover, there aretwo planar symmetry curves (through the saddle), mapping to mirror planes, and so thefundamental domain is also a rectangle. In conclusion the domain must be a square. Theboundary of the square also maps to planar symmetry curves. The locations of the endsnow become obvious: The planar end is on the intersection of the boundary curves, thecatenoids on the intersection of ∂U with the coordinate axes.3. The map g has values st −1 (0, 0, 1) = 0 at the two catenoid ends and at the saddleorigin. It is st −1 (0, 0, −1) = ∞ on the planar middle end. Assuming the zeros are simple(as supported by the end asymptotics) and the planar end is the unique pole, the polemust be of third order. This determines the Gauss map as the elliptic function p ′ up to acomplex factor. If we assume that the x-axis maps to the xz-plane, that is g is real there,then this leaves only a real constant r. Altogether we conclude g = rp ′ for some r ∈ R.4. We determine h: The only finite zero or pole of g is 0; thus h(0) = 0. The only otherlocation of zeros and poles for h is at the three singular points. At the planar end theheight differential is h = 0. At all other points, the tangent plane is not horizontal, and soh ≠ 0. Moreover, at the catenoidal ends h = ∞ (these assertions are yet to be verified).This determines the height differential as p, up to a complex factor. Since the x-axis is aplanar symmetry curve on which g is real, so is h. This leaves a real factor, which we canset to 1 by scaling the entire surface. Thus h = q.5. It must then be checked that the Weierstrass data U, g, h indeed lead to a surfacewith the desired properties of symmetry, period closure, and embeddedness. The abovestatements and proofs verify this for the Costa surface.Remarks. 1. The Costa surface generalizes to higher dihedral symmetry with higher genus.To get an idea of the geometry, note that two (vertically stacked) octons of R 3 containa generating piece of the standard Costa surface; the boundary halfplanes each containtwo different curves of planar reflection. This wedge of angle π/2 is replaced by a wedgeof angle π/k for k ≥ 3. Our construction generalizes, including the construction of thenecessary meromorphic functions (which are no longer elliptic). See [HK].2. The Costa surface is embedded and has three ends. All such minimal surfaces havebeen analyzed by Hoffman, Meeks, and Karcher. They form a continuous one-parameterfamily. For these generalized surfaces the middle end may also be catenoidal, and the twoouter ends no longer have the same asymptotic growth. See [HK]. Up to today, no other(nontrivial) minimal surface family has been completely analyzed.

ii 5.1 – date: July 27, 2012 815. Global Weierstrass representationTo present the global version of the Weierstrass representation we need the notion of aRiemann surface.5.1. Riemann surfaces. In short, a Riemann surface is a manifold of real dimension 2with a complex structure.Definition. Let R be a metric space. A chart (U, ϕ) of R is a homeomorphism ϕ: U →ϕ(U) ⊂ C where U ⊂ R is open. Two charts (U, ϕ), (V, ψ) are holomorphically compatibleif the transition mapψ ◦ ϕ −1is a biholomorphic mapping between the open subsets of C where the mapping is defined.Compatibility is an equivalence relation. A family of compatible chartsA = {(U α , ϕ α ), α ∈ A}which cover, ⋃ α∈A ϕ α(U α ) = R, is called an atlas on R. Here A is an arbitrary indexset. Two atlasses A, A ′ are equivalent if their charts are pairwise compatible. We call anequivalence class of atlasses a holomorphic structure on R.Definition. A Riemann surface is a connected metric space with a holomorphic structure.As a geometer, I think of a Riemann surface as a differentiable surface with the additionalnotion of an angle: multiplication by i is 90-degree rotation. However, lengths are notdefined.Examples. 1. Each open connected subset Ω ⊂ C is a Riemann surface. An atlas is givenby id: Ω → Ω. Another (maximal) atlas isS = {(U, ϕ) : U open ⊂ C, ϕ: U → Ω biholomorphic onto its image}.2. A closed subset of C (with nonempty boundary) is not a Riemann surface: A boundarypoint does not have a neighbourhood mapping homeomorphically to an open subset of C.3. The Riemann sphere is the set R := C = C ∪ ∞ with the two charts (C, id) and(C \ {0}, 1) where we set 1 := 0. Clearly, the two charts cover R. The surface R isz ∞topologically S 2 and so a compact Riemann surface.4. A lattice of rank 2 is the set Λ = {mv 1 + nv 2 : m, n ∈ Z} where v 1 , v 2 are R-linearlyindependent. Likewise, Λ = {nv : n ∈ Z} for v ∈ C \ {0} is a lattice of rank 1. Then the

82 K. Grosse-Brauckmann: Minimal surfaces, SS 12quotient T := C/Λ is a Riemann surface, namely a cylinder or torus, respectively. Herethe quotient denotes the set of equivalence classes under the relationz ∼ z + w for z ∈ C and w ∈ Λ.We denote the equivalence class of z by z + Λ and call the map from points to classesπ : C → T,π(z) := z + Λa projection. A set U ⊂ T is open iff π −1 (U) is open. This definition makes π continuous.Let us define charts: Provided π maps an open set V ⊂ C homeomorphically to U ⊂ T weconsider (U, π −1 ) a chart. Clearly the charts cover. To see two charts are compatible: Foreach z, the coordinate change π 2 ◦π1 −1 (z) is a translation by some w(z) ∈ Λ. By continuity,w(z) is constant, hence w is constant on connected components, in particular holomorphic.23. Lecture, Wednesday 27.6.125. Historically, Riemann surfaces were introduced in order to define domains on which theinverse of a non-injective function can be defined. Consider, for instance, the inverse √ ofz ↦→ z 2 . We define its Riemann surface by taking polar coordinates with the angle rangingto 4π, represented by R := {(r, ϕ) : 0 < r and 0 ≤ ϑ < 4π}. Visualize R as a double coverof C \ {0} by considering its exponential image! The metric space R becomes a Riemannsurface as the quotient (0, ∞)×R/(4πZ) of C and so carries a holomorphic structure. Thenz 2 can be considered an injective function from C \ {0} → R by setting (e r+iϕ ) 2 ↦→ (r, 2ϕ),and conversely the square root is welldefined on R as the mapping (r, ϕ) → e r+iϕ/2 . Theexcluded point z = 0 poses no problem for our mappings, √ 0 = 0. However if we wantto include 0 it into R we have to go beyond the notion of a manifold. The point 0 is abranch point of R ∪ {0} and the surface R ∪ {0} is called a generalized Riemann surface,or a branched Riemann surface.The definition of compatibility of charts makes holomorphic functions on Riemann surfaceswell-defined as follows:Definition. (i) A function f : R → C is holomorphic or meromorphic if for each chart(U, ϕ) the composition f ◦ ϕ −1 : ϕ(U) → C is holomorphic or meromorphic, respectively.(ii) A mapping F : R 1 → R 2 between two Riemann surfaces R 1 , R 2 is holomorphic if foreach chart (U, ϕ) of R 1 and (V, ψ) of R 2 the composition ψ ◦ F ◦ ϕ −1 is holomorphic.(iii) If F and F −1 exist and are holomorphic then the two Riemann surfaces R 1 and R 2are called conformally or biholomorphically equivalent.Remarks. 1. It is common to say conformal instead of biholomorphic. Locally this is thesame: If a chart is conformal but not holomorphic, then postcomposing it with z ↦→ z makesit holomorphic. Globally, however, it becomes the same only after a possible transition toa double cover.

ii 5.1 – date: July 27, 2012 832. In case (i), a chart (U, ϕ) gives rise to a power or Laurent series representation of f ◦ ϕwhich makes the zero or pole order of f uniquely defined.3. Meromorphic functions f on a Riemann surface R can as well be defined by saying thatf is a map from the Riemann surface R to the Riemann sphere C which is holomorphicand not the constant ∞.Local properties of holomorphic functions of the complex plane extend to holomorphic functionson Riemann surfaces. For instance, the strong maximum principle generalizes to Riemann surfacesand leads to an interesting global property:Proposition 31. (i) Strong maximum principle: If a holomorphic function f : R → C has amodulus |f| which attains a maximum then it must be constant.(ii) A holomorphic function f on a compact Riemann surface R must be constant.Proof. (i) Suppose |f| takes maximum at p ∈ R. By the covering property, there is a chart (U, ϕ)such that p ∈ U. By the standard maximum principle, the holomorphic map f ◦ ϕ −1 is constanton ϕ(U). If (V, ψ) is any other chart with ϕ(U) ∩ ψ(V ) ≠ ∅ then by unique analytic continuationf ◦ ψ is also constant on V . Consequently the set of points where f is constant is an open subsetof R, it is closed anyway. By connectedness of R the function f is constant.(ii) Since R is compact |f| must take a maximum. The result follows from part (i).□An interesting problem is to determine a Riemann surface up to biholomorphic equivalence.Examples. 1. The punctured complex plane C ∗ and any annulus A r,R = {r < |z| < R}with 0 < r < R are homeomorphic. Nevertheless, as Riemann surfaces (with the identitychart) they are not biholomorphically equivalent. To see this, assume on the contrary thatF : C ∗ → A r,R is biholomorphic. The values of any neighbourhood of 0 are bounded andso the singularity at 0 is removable. Thus f extends to a non-constant holomorphic mapfrom C to A r,R , contradicting Liouville’s theorem. It is more difficult to see that A 1,R andA 1,R ′ are not equivalent for R ≠ R ′ .2. The exponential map e z = e Re z e i Im z is surjective from C to C ∗ . Moreover it is periodic,e z = e z+2πi , and injective on the quotient of C under the lattice Λ = 2πi Z of rank 1. Thusexp shows that the cylinder C/Λ is conformally equivalent to C ∗ .More generally, the uniformization theorem states (see [FK]):Theorem. (i) A simply connected Riemann surface R is necessarily biholomorphic toeither the open disk D, all of C, or to the Riemann sphere C.(ii) A compact Riemann surface has a well-defined genus g ∈ N 0 and for g > 0 the spaceof Riemann surfaces is a (connected) manifold of real dimension 6g − 6 for g ≥ 2, calledTeichmüller space.

84 K. Grosse-Brauckmann: Minimal surfaces, SS 12Note that D and C are equivalent topologically (they are homeomorphic) but not conformally.5.2. Minimal surfaces and Riemann surfaces. A minimal surface is a conformal immersionof a Riemann surface. In particular, on the Riemann surface and on the minimalsurface the notion of angles agree.Theorem. An oriented minimal surface M ⊂ R d is a Riemann surface. That is, eachpoint p ∈ M has a neighbourhood U which can be parameterized by a conformal coordinatemap ϕ: V ⊂ C → U ⊂ M ⊂ R d such that for two charts the transition maps ϕ ◦ ψ −1 areholomorphically compatible.This statement is true in general, without the assumption of minimality. However, theproof is difficult. For a minimal surface the existence is easier: The coordinate functionscan be used to define a conformal parameterization; this requires some Riemannian notionsbeyond this class (see [EJ]). For the case that the minimal surface has no umbilic points,a proof in terms of a curvature line parametrization is straightforward (see problems).Remark. If M is not orientable then its double cover is orientable, and so the theoremapplies to arbitrary minimal surfaces.For the global Weierstrass representation theorem we need to generalize the Weierstrassdata in a way that they become parameterization independent. A path integral ∫ f(w) dwof a holomorphic function f is not invariant under biholomorphic maps. However, thepath integral of a 1-form on a manifold is invariant. In the example of local Weierstrassrepresentation the 1-form in question is h dz; we need to replace it with some 1-form dh:Theorem. For each complete minimal surface M in R 3 there exist:• a compact Riemann surface R and a point z 0 ∈ R,• a holomorphic one-form dh on R,• a meromorphic map g on R, such that ν = st ◦g is a Gauss map of M, and such that ateach zero of dh the map g has a zero or pole of the same order.Then, up to translation,∫ z( 1( 1)(40) f : R → M ⊂ R 3 , f(z) = Rez 02 g − g , i ( 1) )2 g + g , 1 dh.Conversely, data as above define a well-defined immersed minimal surface provided theintegral in (40) vanishes over each closed curve in R.In case that the integral over the Weierstrass data does not vanish over closed curves, theminimal surface is defined on a covering of R.

ii 6.1 – date: July 27, 2012 85A beautiful problem is to determine the Riemann surface R parameterizing a minimalsurface, that is, to determine its conformal type. We mention a particular case: Recallthat a minimal surface M has finite total curvature if∫− K dS < ∞.MFrom our examples, the catenoid and Enneper surfaces have finite total curvature ∫ K =−4π. In contrast, any periodic surface, unless planar, must have infinite total curvature.Then we have:Theorem (Osserman). If M is a complete minimal surface in R 3 which has finite totalcurvature. Then there exists a compact Riemann surface R and finitely many pointsp 1 , . . . , p k ∈ R, such that M is a proper conformal immersion of R \ {p 1 ∪ . . . ∪ p k }.Remarks. 1. The points p l correspond to the ends of M. The Laurent series of theWeierstrass data g, h can be used to analyse the ends (problems?).2. Osserman’s theorem holds more generally for any surface with Gauss curvature K ≤ 0;it need not be immersed into R 3 .Example. A catenoid is topologically an annulus. By the theorem it must be conformallyequivalent to S 2 minus two points. We may assume these two points are 0 and ∞ and sothe theorem says the catenoid is conformally equivalent to C \ {0}. By the example above,however, it cannot be equivalent to any annulus r < |z| < R with 0 < r < R.55. Lecture, Tuesday 11.7.066. On the proof of Osserman’s theorem (2006)6.1. Divergent paths. We follow [O, §9] and sketch the main steps of the proof (see also[L, III §6]). A number of crucial topological and analytical arguments will only be cited,however.Suppose U ⊂ C is a domain and f : U → R 3 induces a first fundamental form g on U.In the following it will not matter that g is induced by f, but only that for each point ofU it is a positive definite bilinear form. We then call g a (Riemannian) metric on U. Acurve γ : (a, b) → U is called divergent if lim t→b γ(t) ∈ ∂U or = ∞. The metric g is calledcomplete if no divergent curve has finite length∫ b √L(γ, g) := g(γ′ , γ ′ ) dt.aLemma 32. Let U ⊂ C be a domain and suppose f : U → R 3 induces a metric conformallyequivalent to the euclidean metric, that isg ij = λ 2 δ ijfor λ: U → (0, ∞).

86 K. Grosse-Brauckmann: Minimal surfaces, SS 12Moreover, suppose U is complete with respect to g. If there exists a harmonic functionh: U → R such that(41) h(z) ≥ log λ(z) for all z ∈ Uthen U can only be conformally equivalent to C or to C minus a point.Proof. Let us consider a further metric ˜g := e 2h 〈., .〉. Then any divergent curve γ has length∫∫L(γ, ˜g) = e h |γ ′ | dz (41)≥ λ|γ ′ | dz = L(γ, g) g complete= ∞,meaning that ˜g is complete as well.γγLet us now invoke a fact which we do not prove: Given a harmonic function h on a simplyconnected domain U there is a holomorphic function f : U → C, such that Re f = h. If Uis not simply connected then similarly f may be defined on the universal covering Û.Now consider the complex curve integralThenψ : U → C, ψ(z) :=∫ zz 0e f(ζ) dζ.(42) |ψ ′ (z)| = |e f(z) | = |e Re f(z) | = e h .We can interpret (42) to say that(43) L(γ, ˜g) = L(ψ ◦ γ, 〈., .〉) for any curve γ in Û.Indeed, by differentiation,ddt L( γ(t), e h 〈., .〉 ) = ∣ √˜g(γ ′ , γ ′ ) ∣ = e h |γ ′ |whileddt L( ψ ◦ γ(t)), 〈., .〉 ) = √ 〈ψ ′ (γ)γ ′ , ψ ′ (γ)γ ′ 〉 = |ψ ′ ||γ ′ |.By (42) the latter two expressions agree. Equivalently we can say that ψ is a (local)isometry of (Û, ˜g) to (C, 〈., .〉).Since Û is simply connected, the identity (43) means that ψ extends to a distance preservingmap (global isometry) from (Û, ˜g) to (C, 〈., .〉). By completeness of (Û, ˜g) the map mustbe onto C.Thus the mapping ψ gives us a conformal equivalence of the covering Û with the plane C.But then U is conformally equivalent to a quotient of C under some group of deck transformation.Either this group is the identity, in which case U = C, or it is a lattice of rank 1,in which case U is C minus a point (cf. Ex. 2 above). It cannot be a lattice of rank 2: thequotient is compact and can therefore not be homeomorphic to an open domain of C. □

ii 6.2 – date: July 27, 2012 87While the previous lemma establishes a subharmonic function, we need in fact a harmonicfunction for Lemma 32 to apply. Such a function can be constructed using λ.Lemma 33. Let A be the annulus {z : 0 < r < |z| < R ≤ ∞} endowed with someRiemannian metric g = λ 2 〈., .〉. Moreover, suppose there is a harmonic function h: A → Rsatisfying (41). If each curve γ which is divergent to the outer boundary {|z| = R} hasinfinite length then R = ∞.The following proof contains a gap: It must be shown that B R \{0} and C\{0} are not conformallyequivalent. (I’m sorry, 11/06)Proof. Suppose in addition to the assumptions that each path divergent to the inner boundaryalso had infinite length. Then Lemma 32 would contradict R < ∞ and the statement was proved.If R < ∞ we will modify the metric as to assign infinite length to the paths limiting in the innerboundary, while preserving all other properties.Our claim is invariant by scaling of the annulus. Scaling specifically by 1 √rRwe can assume thatA = {z : 1 ρ< |z| < ρ}.We claim that the metric˜g := λ 2 (z)λ 2( 1) 〈., 〉.zon A satisfies the assumptions of the lemma.Indeed,( ( 1) ) ( 1) ( 1)log λ(z)λ = log λ(z) + log λ ≤ h(z) + h =:zz z ˜h(z).Now ˜h is harmonic. But h( 1 z) is also harmonic as the composition of the harmonic map h(z) withthe conformal mapping z ↦→ 1 z (see Lemma III,10). Clearly, ˜h > 0, and now paths limiting ineither component of ∂A attain infinite length.□6.2. Existence of a harmonic function. In view of the last two lemmas the claim ofOsserman’s theorem becomes evident, provided we had a harmonic function on the minimalsurface. The next lemma gives us a little less, namely a subharmonic function.Lemma 34. Suppose a minimal surface f : U → R 3 is given by Weierstrass data, that isthe first fundamental form is λ 2 〈., .〉 with conformal factor λ := 2( 1 |g| +1|g|)|h| > 0. Then∣ the Gauss curvature K =−16g ′∣ 2 1can also be expressed as(|g|+ 1|g| )4 g |h| 2(44) K = − ∆ log λλ 2on U. In particular, log λ is a subharmonic function on U, that is ∆ log λ = −λ 2 K ≥ 0.

88 K. Grosse-Brauckmann: Minimal surfaces, SS 1256. Lecture, Wednesday 12.7.06Proof. Rewriting the desired equation as −Kλ 2 = ∆ log λ, we need to show that1∣ ∣∣ g ′4( )|g| +1 2∣ 2 ( 1 ( 1 ) )= ∆ log |g| + |h| > 0g2 |g||g|or equivalently|∂ z g| 2(45) 4(|g| 2 + 1) = ∆ log ( |g| + 1 )+ ∆ log |h|.2 |g|Let us check that ∆ log |h| = 0. We rewrite ∆ = ∂ z ∂ z and recall that for h holomorphic theCauchy-Riemann equations imply(46)(47)∂ z h = (∂ x − i∂ y )(Re h − i Im h) = ∂ x Re h − ∂ y Im h − i(∂ x Im h + ∂ y Re h) = 0,∂ z h = (∂ x + i∂ y )(Re h + i Im h) = ∂ x Re h − ∂ y Im h + i(∂ x Im h + ∂ y Re h) = 0.ThusConsequently,∂ z log |h| = ∂ z|h||h|= ∂ √z hh= ∂ z(hh)|h| 2|h| 2 = h ∂ zh + h ∂ z h (46)2|h| 2 = h ∂ zh2|h| 2h ∂ z h(48) ∆ log |h| = ∂ z2|h| 2 = ∂ z(h ∂ z h)2|h| 2 − h ∂ z h ∂ z (2|h| 2 )4|h| 4But the numerator of (48) gives∂ z (h ∂ z h)2|h| 2 − h ∂ z h∂ z (2|h| 2 )= ∂ z h ∂ z h2|h| 2 + h ∂ z ∂ z h 2|h|} {{ }2 − 2h ∂ z h ( h ∂ z h +h ∂}{{} z h )=∆h=0=0 by (47)= ∂ z h ∂ z h(2|h| 2 − 2hh) = 0This asserts that for a holomorphic function h we have ∆ log |h| = 0. But the same applies to theholomorphic functions g and 1/g and so (45) vanishes.□Remarks. 1. In particular, the calculation given in the proof verifies ∆ log |z| = 0 on C ∗ .The harmonic function log |z| is called fundamental solution of ∆ in dimension n = 2.It can also be determined by solving the equation ∆u = 0 among functions which arerotationally symmetric, u = u(|z|); in this case the Laplace equation simplifies to anordinary differential equation.2. Beyond minimal surfaces, the relation (44) is valid for arbitrary (two-dimensional)surfaces in conformal parameters. Given the fact that conformal parameters can alwaysbe introduced, it indicates that Gauss curvature depends on the first fundamental formalone (and not on the second fundamental form it was defined with). This is an importantinsight, which is the content of Gauss’s theorema egregium: K belongs to the intrinsicgeometry of surfaces.

ii 6.2 – date: July 27, 2012 89Lemma 35. Let U be a domain which carries a positive harmonic function and considera metric g = λ 2 〈., .〉 on U which is complete. Suppose that∫(49) ∆ log λ ≥ 0 and |∆ log λ| dxdy < ∞.Then there exists a harmonic function h: U → R satisfying (41).Note that the geometric meaning of (49) is that the metric has K ≤ 0 and finite totalcurvature − ∫ KdS = − ∫ Kλ 2 (λ 2 dxdy) < ∞.UProof. We main step of this proof is a result we cite from Ahlfors and Sario, Riemann surfaces,IV.6. If U carries a positive harmonic function, then for z ≠ ζ there is a Green’s functionG(z, ζ) > 0 which is harmonic. Moreover, the integralu(z) = 12π∫UG(z, ζ)∆ log λ(z) dxdyexists by (49) and, since the integrand is positive, also u ≥ 0. By Poisson’s formula, ∆u =−∆ log λ. Thus h(z) := u(z) + log λ(z) ≥ log λ(z) is harmonic.□Proof of Osserman’s theorem. We assume that M has finite topology, that is, M is homeomorphicto a compact Riemann surface minus finitely many points. (This property can infact be infered from finite total curvature by invoking a theorem of Huber).We remove from M open neighbourhoods U j of the j-th singular point which are topologicalannuli. We assume that U j is bounded by an analytic Jordan curve Γ j . Removal gives acompact subset of M, namely M 0 = M \ ⋃ kj=1 U j.Each U j can conformally be mapped to some standard annulus A j := {1 < |z| < R j ≤∞}, where Γ j maps to |z| = 1. We now apply Lemma 35 to A j . Note that A j has apositive harmonic function, namely log |z|. By the Lemma there exists a harmonic functionh: A j → R such that h(z) ≥ log λ(z). Thus we can appeal to Lemma 33 to obtain R = ∞.In particular, each A j compactifies with one point p j to a disk Ãj.We now let ˜M be the compact Riemann surface given by M0 with the union of the Ãj“glued back” along Γ j (it must be shown this defines a Riemann surface).□Essential singularities like (e 1/z in 0) can occur in Weierstrass data. For instance, the helicoidcan be parameterized with an essential singularity. However, for finite total curvature:Proposition 36. If a minimal surface f of a finite total curvature is given by Weierstrassdata, then g extends to a meromorphic function on the compact Riemann surface R ofOsserman’s theorem.

90 K. Grosse-Brauckmann: Minimal surfaces, SS 12Proof. On R \ {p 1 , . . . , p k }, the function g is certainly meromorphic (that is, valued inthe Riemann surface S 2 ). But if any point p j would be an essential singularity of g thenby Picard’s theorem g would assume each value infinitely often, with at most two exceptions.But then the spherical area of A R (ν) = A R (st ◦g) would be infinite, contrary to theassumption of finite total curvature. Thus g has a pole at p j and is meromorphic on R. □References. These results and further reading are contained in [O, §9] and [L].6.3. Ends. We call f : D \ {0} → R 3 a parameterization of an end. More generally,if f : R \ {p 1 , . . . , p k } → R 3 is minimal then we call the restriction of f to an annularneighbourhood of p j in R \ {p 1 , . . . , p k } an end. As before, an end is complete if any curvedivergent to 0 has length L(f ◦ γ) = ∞.The plane, the catenoid, the helicoid, and the Enneper surface have different types ofends. We have seen that k-Enneper ends are very flexible (lower order polynomials can beadded to the Weierstrass data). However, when we assume embeddedness and finite totalcurvature, ends are rather rigid:Theorem 37. Suppose f : D \ {0} → R 3 is a complete embedded minimal end of finitetotal curvature and with Gauss map at 0 equal to (0, 0, ±1). Then a neighbourhood of 0can be reparameterized in graph form ˜f(x, y) = ( x, y, u(x, y) ) for (x, y) ∈ R 2 \ K where Kis compact, with(50) u(x, y) = a log ρ + b + 1 ρ 2 〈V, (x, y)〉+ O(1ρ 2 ), where ρ(x, y) := √ x 2 + y 2 ,for some a, b ∈ R, V ∈ R 2 .The proof will also give the following: The first two components Φ 1 , Φ 2 have poles of ordertwo at 0 with no residues, while Φ 3 either has a simple pole (iff a ≠ 0) or is regular (iffa = 0).Examples. For a plane, a = 0, while for the end of a catenoid, a ≠ 0. We say an end isasymptotically planar if a = 0, or else catenoidal with logarithmic growth a ≠ 0 (50). Theproof will show that a = − res 0 h.Proof. Suppose Weierstrass data g, h are given; we assume g is not constant. Let us firstdeal with the case that g is unbranched at 0, so that g = cz + O(z 2 ) with c nonzero. Thisassumption will lead to the case a ≠ 0 of the statement. Decreasing D if necessary, we canassume g(z) = z by the inverse function theorem. For the purposes of this proof, we write(x, y, u) for the coordinates of the Weierstrass representation:(51) x = Re∫ 12( 1z − z )h dz,y = Re∫ i2( 1z + z )h dz,∫u = Reh dz.

We represent h near 0 by a Laurent series,ii 6.3 – date: July 27, 2012 911(52) h = c −kz + . . . + c 1k −1z + c 0 + zw 1 (z),where w 1 is holomorphic. Consider the integrals involving 1h in (51): Since ∫ 1= 2πi,z ∂D zthe representation of x gives Im c 0 = 0 and the representation of y gives Re c 0 = 0.We claim h must have a pole at 0, that is, c −1 ≠ 0. Indeed, the metric (16)1(|g| + 1 )|h| = 1 (|z| + 1 )|h|2 |g| 2 |z|must have a pole at 0 in order to be complete, and so given that c 0 = 0 we infer h has apole.1Moreover, we claim the pole of h at 0 is simple. If not, let c −k z kterm in (52). Since∫ 1z h dz = ∫c −k z −k−1 + . . . dz = − c −kk zk + . . . ,with k ≥ 2 be the leadinga small circle (centered at the origin) has a projection onto the (x 1 −ix 2 )-plane which windsk ≥ 2 times about 0. On the other hand, the end is embedded with vertical normal and soa graph over the xy-plane near infinity. Thus the image of a punctured disk {0 < |z| < ε}must project 1-1 to the xy-plane. We conclude that in fact c −k = 0 for k ≥ 2.Moreover, for u = Re ∫ h dz to be well-defined, we must have c −1 real and soh(z) = c −11z + zw 1(z), c −1 ∈ R \ {0}, w 1 holomorphic.In particular, the Weierstrass integrands Φ 1 , Φ 2 have no residues at 0, but second orderpoles.As Re(iz) = Re(i Re z − Im z) = − Im z we can rewrite∫ ∫ ∫1 12(x − iy) = Rez h dz − i Re i z h dz − Re∫ ∫ 1=z h dz − zh dz.∫zh dz − i Re izh dzThus we can summarize our results as follows: For constants c ∈ C and a := −c −1 ∈ R\{0},(53)(54)x − iy = a 1 z + c + zw 2(z) + zw 3 (z),u = const −a log |z| + O(|z| 2 ),where w 2 , w 3 are holomorphic.

92 K. Grosse-Brauckmann: Minimal surfaces, SS 12We will finally show that u(x, y) has the desired form. First, observing z + z = 2 Re z, we findρ 2 = x 2 + y 2 = (x + iy)(x − iy) = (x − iy)(x − iy)( a)( a)=z + c + . . . z + c + . . .= a 2 1|z| 2 + 2 Re (c · a 1 z)+ |c| 2 + O(|z|)in particular |z| = O( 1 ρ). Second, we multiply (53) with c and take the real part:(− Re c a 1 )= Re ( c(x − iy) ) + |c| 2 + O(|z|)zWe combine the last two equations to obtaina 2 1|z| 2 = ρ2 − 2 Re ( c(x − iy) ) 1)+ const +O(ρ= ρ 2( 1 − 2 ρ 2 Re ( c(x − iy) ) + O ( 1 ) )ρ 2 .The logarithm of both sides in conjunction with log(1 + t) = t + O(t 2 ) giveslog |z| = log |a| − log ρ + Re ( c(x − iy) ) ( 1)+ Oρ 2 .Substitution in (54) yieldsu(z) = const +a log ρ − a ρ 2 (x Re c + y Im c) + O( 1ρ 2 )which happens to be the desired representation (50).Finally, we consider the case that g is branched at 0; this case will give a = 0. So g(z) =αz k + O(z k+1 ) with k ≥ 2.ρ 2We may as well assume α = 1 using a linear reparametrization.Inspecting the integrals as above, we see that h(z) = const +O(|z|) and so u(x, y) = const +O( 1We leave the details as an exercise.Corollary 38. (i) An end of an embedded finite total curvature minimal surface M isasymptotic to either a plane or a catenoid. (ii) Suppose M has Weierstrass data are g andh, with the end at z 0 . If g is branched, then the end is planar. If the normal at an end isvertical, i.e. g(z 0 ) ∈ {0, ∞}, then the end is catenoidal if and only if h has a first orderpole; the logarithmic growth then is res z0 h.|z| 2 .□Also, a catenoidal end with non-vertical normal can be characterized in terms of the Weierstrassdata. It turns out that then h has a pole of second order.References. Our presentation of the present subsection follows [HK] which also containsfurther information.57. Lecture, Tuesday 18.7.06

ii 7.1 – date: July 27, 2012 937. Conjugate surface constructions (2006)The goal of this short section is to present a further technique to establish the existence ofminimal surfaces, the so-called conjugate surface construction. There are various ways toestablish the existence of the surface conjugate to a given minimal surface:1. If the minimal surface is given in terms of the Weierstrass data, the conjugate isessentially the imaginary part of the Weierstrass data. This is the route we take here.2. The first and second fundamental form of the given minimal surface can be modified.Then Bonnet’s fundamental theorem yields the existence of the surface. See, for instance,the article of Karcher in Man. Math. 64 (1989).3. There is a first order system of differential equations, involving only the differential df ofthe given minimal surface f. It can be solved using the Poincaré lemma. See, for instance,the article by Grosse-Brauckmann in the Clay mathematics proceedings, vol. 2, 2005.In any case, integrability of the conjugate data must be shown; in terms of the presentexposition this is the assertion that for each minimal surface f there exist Weierstrass datag, h, a fact which we have not proven in the class.58. Lecture, Wednesday 19.7.067.1. Conjugate Plateau solutions. In Sect. 5 we solved the Plateau problem for polygonalcontours and then used Schwarz reflection across the bounding straight arcs to extendthe Plateau solution to a complete minimal surface. A drawback of the method is thatonly those surfaces can be generated which contain disk-type polygons. Such surfaces arerare and have in fact been classified.In the present section the first step also is the solution M of a Plateau problem for somepolygon Γ. Then we take the conjugate M ∗ of M. By Prop. 17 the surface M ∗ is boundedby arcs of planar reflection. Then we can apply the Schwarz reflection principle III,39.Mirror reflection generates a complete minimal surface.Example. Let us go back to the D-surface from Sect. III,5.3. We consider a polygon whosevertices are six of the eight vertices of the unit cube, namely (0, 0, 0), (1, 0, 0), (1, 1, 0),(1, 1, 1), (0, 1, 1), (0, 0, 1). The edges take the six directions e 1 , e 2 , e 3 , −e 1 , −e 2 , −e 3 . LetM be the Plateau solution to Γ, a hexagon. By Cor. ?? the conjugate minimal hexagonM ∗ has its six boundary arcs of planar reflection contained in planes normal to the abovevectors, that is, the boundary is contained in a rectangular box.We claim that Schwarz reflection generates from M ∗ the P surface, see Sect. III,5.4. Tosee this, consider first symmetries of the D-hexagon M: There are straight lines betweenopposite edge midpoints, and arcs of planar reflection between opposite vertices. Thus byCor. ?? there are straight lines through opposite vertices of M ∗ and arcs of planar reflection

94 K. Grosse-Brauckmann: Minimal surfaces, SS 12between midpoints of opposite arcs. This proves that the rectangular box is in fact a cube.But then the straight arcs from the cube midpoint out to the endpoints of an arc of planarreflection, together with the mirror image, form a quadrilateral precisely as in Sect. III,5.4.This proves that the conjugate of the D surface is the P surface.We would like to proceed in reverse order. Suppose a desired minimal surface M is tobe generated by Schwarz reflection in planes from a fundamental piece as in Thm. III.38.This means that we assume we have a set of mirror planes which are to contain arcs ofplanar reflection of M. Can we systematically prescribe a polygon such that its Plateausolution has a conjugate that establishes existence of the desired surface M? The followingalgorithm can succeed:0. At each of the k vertices of the desired surface we read off the angles and the normalfrom the configuration of mirror planes.1. The vertex angles of the polygon agree with the vertex angles of M, since a surface andits conjugate are isometric.2. The normals at the vertices agree, and prescribe the direction of the polygon edges.3. Assign lengths to the edges of the polygon such that the polygon closes. In general thisleaves k − 3 parameters. Let us ignore scaling also. Then there is a k − 2-parameter spaceof geometrically different polygonsG := {admissible polygons Γ}.Certainly, by the isometry property the edge lengths of M and its conjugate M ∗ agree.Nevertheless from the symmetry properties alone, we cannot predict the edgelengths.The following difficulties may arise in this construction:Boundary regularity: While the Plateau problem is free of interior branch points, branchpoints can be created at the vertices. For instance, if the polygon has angle α k ∈ (0, 2π)at a vertex then perhaps the Plateau solution makes an angle 2π − α k instead. Uponreflection, then a branch point will be created. In order to rule boundary branch pointsout, we need barriers as in Thm. III,42.Period problem: More complicated desired surfaces M have a pair of boundary arcs inthe same plane. The construction will not necessarily give this property. Instead, we willneed to choose Γ ∈ G suitable, that is, assign lengths to the edges so that the the periodvanishes. If there is only one period problem then often the intermediate value theoremcan be applied.Embeddedness: This is a property not necessarily preserved by conjugation.

Part 3. Solution of Plateau’s problem24. Lecture, Friday 29.6.12iii 1.1 – Stand: July 27, 2012 951. Introduction1.1. The problem. The Belgian physicist Plateau (1801–1883) popularized soap film experiments.He studied the physics of soap films although he has not seen many of theexperiments: he turned blind in 1843. In his treatise of 1873 he postulated that everyclosed wire frame bounds at least one soap film; this is nowadays called the Plateau problem.It seems Lagrange also had considered the problem in the 18th century.In the present course, the exposition of the Plateau problem provides the first generalexistence statement. The Plateau solution will also allow us to construct many interestingminimal surfaces, beyond boundary value problems.If the soap films of nature tell us that the Plateau problem always has a solution, why didmathematicians try make an effort to prove the existence mathematically? A good reasonfor this is that a mathematical theorem makes a vague formulation of a conjecture preciseand so leads to a much deeper understanding of the problem.With all the success gained by applying techniques of complex analytic to minimal surfacesin the 19th century, it was natural to expect these methods would prove Plateausconjecture. In fact for some particular classes of boundary curves, Schwarz and others wereable to determine Weierstrass data. The Poisson formula constructs a harmonic map interms of its boundary values by an integral representation. Nevertheless, a respresentationformula of a minimal surface in terms of boundary values could not be derived from theWeierstrass data.It took about 50 years to develop new analytic techniques which lead to the solution.It became a major mathematical breakthrough in the sense that more implicit modernmethods for existence were used. The American mathematician Douglas solved the problemfrom 1928 to 1931. Later Courant gave improvements. Independently from Douglas andabout the same time, Radó gave a different proof. See [DHS p.277/78] for a short discussionof the approach. For his ground-breaking work, Jesse Douglas was awarded one of firsttwo Fields medals in 1936, the Nobel prize of mathematicians, together with Ahlfors. (See[HT,p.129ff] and [N, p.247-258] for more on the history). In the present part, we will givethe Douglas/Courant approach.To model a wire frame we define:Definition. A (closed) Jordan curve is a continuous injective map Γ: S 1 → R d .

96 K. Grosse-Brauckmann: Minimal surfaces, SS 12Since for d = 1 there is no Jordan curve, let us always assume d ≥ 2 in the secquel of thispart. Moreover, we will not distinguish between a Jordan curve and its trace.It is general theorem in topology that an injective continuous mapping of a compact spaceis a homeomorphism. Hence it is equivalent to say:Proposition 1. A Jordan curve is a homeomorphism onto its image.Proof. We need to show that Γ −1 is continuous. Consider a subset A ⊂ S 1 which is closedand therefore compact. Then the continuous image Γ(A) is also compact, in particularclosed. Thus for Γ −1 , preimages of closed sets are closed. Hence Γ −1 is continuous. □Let us now make the mathematical problem precise: Does a given Jordan curve extend toa continuous mapping f : D → R 3 which is a minimal immersion on D?This problem is not exactly the soap film problem:1. Soap film experiments produce surfaces of various topological types, they may be unorientedMöbius strips or have handles. However, our mathematical formulation asks forsolutions of disk-type.2. Immersed surfaces can self-intersect. However, soap film will instead produce the singularsets which Plateau described: Triple lines, where three sheets of the surface cometogether at 120 degrees, and quadruple points, where four sheets meet with the angles of aregular tetrahedron. Flat chains modulo 2 of Geometric Measure Theory were introducedas a more appropriate model of soap film (see Frank Morgan’s book [M]). of In case f isan embedding, the physical and mathematical problem have the same solution.1.2. Area and energy. For a given Jordan curve Γ, we seek an immersed surface f 0 whichgives the area functional∫ √A D (f) = |f x | 2 |f y | 2 − 〈f x , f y 〉 2 dxdya minimum in the class of admissible mappingsD{f ∈ C 0 (D, R d ) ∩ C 1 (D, R d ), f| ∂D = Γ}.An obvious strategy is to consider a minimizing sequence (f k ) in this class, that is, A(f k ) →inf A(f). Then we would hope for f 0 := lim f k to be the desired solution of the Plateauproblem. This approach fails:The image of a minimizing sequence can degenerate and need not be a surface. Indeed,for any nice smooth f the area functional A(f) will be changed as little as we please bygrowing little spokes, that is, thickened hairs out of the surface. Then the images f k (D)

iii 1.2 – Stand: July 27, 2012 97become a dense open set. This means that the limit f is not attained in our class (lack ofcompactness) and in no way a surface. The same problem cannot arise for curves (why?).We can avoid these problems by considering a better integral, namely the Dirichlet integralor energy of f : U n → R d ,E U (f) := 1 ∫|df(x)| 2 dx = 1 ∫ n∑|∂ i f(x)| 2 dx = 1 ∫∑(∂ i f j (x)) 2 dx222UUi=1U1≤i≤n,1≤j≤dThis integral measures the total amount of streching. Imagine a completely elastic material,like a disk of idealized rubber, which would like to shrink to a point. The energy favourssmall area together with a good parametrization. More precisely:Lemma 2. For any f ∈ C 1 (D, R d ) we haveE D (f) ≥ A D (f).Equality holds if and only if f is a conformal map.Proof. We note 0 ≤ (a − b) 2 ⇔ 2ab ≤ a 2 + b 2 ⇔ 4ab ≤ (a + b) 2 so that the geometric meanis smaller than the arithmetic mean. We conclude12 (a + b) ≥ √ ab ≥ √ ab − c for all a, b ≥ 0 and 0 ≤ c ≤ ab,with equality precisely for a = b and c = 0. This gives the claim asE(f) = 1 ∫∫ √|f x | 2 + |f y | 2 dxdy ≥ |f x |22 |f y | 2 − 〈f x , f y 〉 2 dxdy = A(f).□We will be able to achieve conformality by allowing for reparameterizations of the boundary.There is another problem: Not every Jordan curve bounds a disk of finite area.Example. For k ∈ N, let p k := (1/ √ k, 0, 0), q k := (0, 1/ √ k, 0), and consider a rectangular spiral inthe xy-plane running through the points p 1 , q 1 , −p 1 , −q 2 ; p 2 , q 2 , −p 2 , −q 2 ; . . .. Extend this curvecontinuously with, say, a circular arc in the xz-plane running from the origin back to p 1 . Theresult is a Jordan curve Γ. We need two facts: First, we can estimate the area of the region U nbetween the k-th and (k + 1)-st arc roughly byA(U k ) ≥ √ 2 ( 1 1)√k − √ = 2√ √k + 1 − k√ √ √ = 21√ √ √ √ √k k + 1 k k k + 1 k k k + 1( k + 1 + k)√k+1≤2√k≥13k 2 .Second, any map f bounding Γ has a projection π to the xy-plane that covers almost every pointbetween the k-th and (k + 1)-st arc at least n times. We will not argue this fact rigorously. Sincea projection decreases area, we can conclude that indeedA(f) ≥ π(A(f)) ≥∞∑kπ(A(U k )) ≥k=1∞∑k=1k3k 2 ≥ 1 3∞∑k=11k = ∞.

98 K. Grosse-Brauckmann: Minimal surfaces, SS 12See Nitsche’s book [N] for a complete reasoning.For Jordan curves which are piecewise C 1 this problem cannot arise. This means there is a finitepartition of S 1 such that the restrictions of type Γ| [ti ,t i+1 ] are continuously differentiable.Lemma 3. A closed curve Γ which is piecewise C 1 bounds a disk f of finite energy (and area),such that f is C 1 on D except at finitely many points of ∂D.Proof. Let the piecewise differentiable curve c parameterize Γ. Both |c| and |c ′ | are bounded bysome constant C.Define in polar coordinates f(r, ϕ) := rc(ϕ), which parameterizes the cone of c with tip 0. Themap f is differentiable except on finitely many rays through the origin. Expressing the Dirichletintegral in polar coordinates, we get, as desired:E(f) = 1 ∫ 2π ∫ 1( ∣∣∣∂f∣ 2 + ∣ 1 ∂f∣ 2) r drdϕ = 1 ∫ 2π ∫ 1 (|c(ϕ)| 2 + |c ′ (ϕ)| 2) r drdϕ ≤ 2πC 22 ∂r r ∂ϕ200By smoothing, we can make f differentiable on all of D (details skipped).00□1.3. Rigorous formulation of Plateau’s problem. We can now formulate the Plateau problemof disk-type. From now on we assume that Γ is a piecewise regular C 1 -curve, except possiblyat finitely many points mapping to {p 1 , . . . , p l } ∈ R d . This includes the case of a polygon with lvertices. We will study mappings in(1) C := {f ∈ C 0 (D, R d ) ∩ C 1 (D, R d ), f| ∂D is a bijective parameterization of Γ}.We need to work with generelized (branched) minimal surfaces:Theorem 4. Let Γ ⊂ R d be a closed Jordan curve which is piecewise C 1 , and D be the open unitdisk in R 2 . Then there exists a mapping h ∈ C with the properties:• h is weakly conformal in D,• h is in C 2 (D, R d ) and harmonic.Moreover h assumes the finite infimum of energy over C.If h is an immersion then h is a minimal surface by Thm.I.5.The proof we will give relies on certain properties of harmonic functions. After introducing those,we will prove Theorem 4 in Section 3 as a result of Theorem 24, 25, and 28.Let us discuss a few questions being raised by the theorem.1. Is h an immersion, that is, a surface in the sense of differential geometry? In fact no branchpoints, i.e., points in D with dh = 0, exist, but this follows from a separate discussion.2. At differentiable boundary points, we also expect that f is an immersion, in the sense thatthere is an extension to an immersion. To establish this property has turned out to be amazinglydifficult. Only partial results are known then, for instance if the Jordan curve is analytic, then f

iii 2.1 – Stand: July 27, 2012 99is an immersion including the boundary. In general, a satisfactory answer to the problem is stillopen.3. Is h an embedding? There are certainly knotted Jordan curves which for topological reasonswill not allow for embedded solutions. On the other hand, sufficient conditions for h to be anembedding are known. From Thm. II.41 follows that Jordan curves which are graphs with convexprojection must bound a Plateau solution which is graph as well, and hence embedded. A moregeneral condition for h to be an embedding is that Γ is contained in the boundary of a convexset.4. The theorem gives us just one minimal surface. Are there further minimal surfaces? In generalthere are, see for instance [DHKW, p.222/223]. Different techniques are needed to exhibit minimalsurfaces which are not local minimizers (see, e.g.: Struwe, Variational methods).There are various generalizations of the theorem: It holds for rectifiable Jordan curves; it generalizesto multiply connected domains if certain conditions are met; it holds in different ambientspaces (manifolds). Moreover, functionals other than area can similarly be minimized, and leadto existence statements for, e.g., surfaces of constant or prescribed mean curvature.2. Harmonic and conformal mapsIn order to solve Plateau’s problem we will deal with the following tasks:• Solve the Dirichlet problem for harmonic functions on a disk,• take limits of harmonic functions,• control the Dirichlet energy of their limit.The present section collects these auxiliary results.References. For results on harmonic maps see [GT], Sect 2, or any other book on ellipticpartial differential equations. For conformal maps, we will cover the results essential forminimal surfaces. For a broader introduction see Ahlfors [A], III.2 and 4 from the point ofview of complex analysis and [Sy1], IV §7 for the perspective of real analysis.2.1. The divergence theorem. Let us recall the divergence theorem on R n . SupposeU ⊂ R n is bounded with smooth boundary ∂U, whose exterior normal we denote by ν.Then any vector field X ∈ C 1 (U, R n ) satisfies∫∫(2)div X dx = 〈X, ν〉 dS,where dS denotes the (n − 1)-dimensional area element of ∂U.UThe divergence theorem will also appear in the following versions:1. If u ∈ C 2 (U), then choosing X := ∇u gives∫ ∫∫∂u(3)∆u dx = 〈∇u, ν〉 dS =∂ν dSU∂U∂U∂U

100 K. Grosse-Brauckmann: Minimal surfaces, SS 122. Setting X = (0, . . . , u, . . . , 0) = ue k we obtain ∫ ∂ U ku dx = ∫ uν ∂U k dS. We can assemblethese components to form a vector valued integrand:∫ ∫(4)∇u dx = uν dSU∂U3. Let X = v∇u where u ∈ C 2 (U, R) and v ∈ C 1 (U, R) has boundary values v| ∂U = 0.Then, by divergence theorem and product rule,∫∫∫∫(5) 0 = 〈v∇u, ν〉 dS = div(v∇u) dx = 〈∇v, ∇u〉 dx + v∆u dx.∂UUUU25. Lecture, Monday 2.7.122.2. Mean value formula. Let us denote the volume of the n-dimensional unit ballby V (B n ) =: ω n . The divergence theorem shows the bounding unit sphere has volumeV (S n−1 ) = nω n . The following property exemplifies how harmonic our functions are:Theorem 5. Let u ∈ C 2 (U, R) be harmonic. Then for any ball B = B R (y) with B ⊂⊂ U(6) u(y) =1nω n R n−1 ∫∂Bu(x) dS x and u(y) = 1 u(x) dx.ω n R∫BnProof. We take B ρ = B ρ (y) with ρ < R. On the boundary of the ball we consider polarcoordinates ρ = |x − y| and ω = x−y . The divergence theorem givesρ∫∫0 (3) ∂u=∂B ρ∂ν dS = ∂u∂B ρ∂ν (y + ρω) dS = ∂u∫|ω|=1ρn−1 (y + ρω) dω∂ν= ρ n−1 ∂ ∫u(y + ρω) dω = ρ n−1 ∂ ( ∫ )1u dS .∂ρ |ω|=1 ∂ρ ρ n−1 ∂B ρThus for any ρ ∈ (0, R),But u is continuous and soρ ↦→ 1ρ n−1 ∫limρ→0∫1ρ n−1∂B ρu dS = const .∂B ρu dS = nω n u(y).Thus we obtain the first mean value equality. The solid version follows from integration:∫∫ R( ∫ ) ∫ Ru(x) dx = u(x) dS dρ = nω n ρ n−1 u(y)dρ = ω n R n u(y)B R 0 ∂B ρ □One consequence of the mean value inequality is the maximum principle. Another are apriori estimates:0

iii 2.2 – Stand: July 27, 2012 101Theorem 6. Let u ∈ C ∞ (U, R) be harmonic, K ⊂⊂ U be compact, and α be a multi-index.Then there exists a constant C = C ( n, α, dist(K, ∂U) ) such that(7) supK|∂ α u| ≤ C sup |u|.UProof. If u is harmonic, then ∆∇u = ∇∆u = 0, so that ∇u is harmonic. For y ∈ K andR = 1 2 dist(K, ∂U) we apply the mean value equality (6) over B = B R(y):Thus, using | ∫ f| ≤ ∫ |f|,∇u(y) = 1 ∇u dxω n R∫B(4)= 1 uν dxn ω n R∫∂Bn|∇u(y)| ≤ n 1|u| dx ≤R nω n R∫∂Bn n−1 R sup |u|.∂BFor higher derivatives, this reasoning is iterated with respect to an increasing sequence ofnested balls, whose suitably chosen radii satisfy R → 0 as |α| → ∞.□The main use of uniform estimates is to establish the convergence of a sequence of harmonicfunctions. Let us recall first:Theorem 7 (Arzelà-Ascoli). For K ⊂⊂ R n suppose the sequence u k ∈ C 0 (K) is(i) uniformly bounded, that is, sup k∈N ‖u k ‖ C 0 (K) < ∞, and(ii) equicontinuous [gleichgradig stetig], that is, for all ε > 0 there is δ = δ(ε) > 0 suchthat(8) |x − y| < δ ⇒ |u k (x) − u k (y)| ≤ ε for all k ∈ N.Then some subsequence of u k converges uniformly on K.Proof. For ε > 0 pick j ∈ N such that 1 j < δ(ε/3).The set K is compact and so there is a finite set of points K j whose 1/j-neighbourhoodscover K, that is, K ⊂ ⋃ z∈K jB 1/j (z). The boundedness of u k implies we can find asubsequence of u k , again denoted with u k , which converges pointwise at all points z ∈ K j .This subsequence u k is Cauchy and so there is N = N(ε), such thatz ∈ K j ⇒ |u k (z) − u l (z)| ≤ ε 3for all k, l ≥ N.Let x ∈ K. Then there is z ∈ K j with |x − z| < δ. Applying (8) with ε/3 twice and usingour previous estimate we find that for all k, l ≥ N(ε):|u k (x) − u l (x)| ≤ |u k (x) − u k (z)| + |u k (z) − u l (z)| + |u l (z) − u l (x)| < εLet us now consider ε = 1 . By our reasoning, for each n there is a subsequence u n k = u n ksuch that |u n k (x) − un l (x)| < 1 provided k, l > N(1/n). Choosing once again a subsequencenof u n k , denoted by u k, we find |u k (x) − u l (x)| → 0 as k, l → ∞, as claimed.□

102 K. Grosse-Brauckmann: Minimal surfaces, SS 12We obtain a compactness result for harmonic functions:Theorem 8. Any bounded sequence of harmonic functions u k ∈ C ∞ (U, R), k ∈ N, containsa subsequence that converges uniformly on each compact subset K ⊂⊂ U to a harmonicfunction; all derivatives converge as well.We will later see (Thm. 19) that it is sufficient to assume u k ∈ C 2 .Proof. By (7) first derivatives are uniformly bounded over K and hence (u k ) is equicontinuous.By the Arzelà-Ascoli theorem this means we obtain a converging subsequence.Moreover, by (7) the second and third derivatives are bounded, and so for a further subsequenceof (u k ) the second derivatives also converge. Thus the limit function is harmonic.We can iterate the argument for higher derivatives.□The theorem, as well as all other results of the present section, generalize to the harmonicmap case:Definition. A harmonic map h ∈ C 2 (U, R d ) is a mapping for which all components areharmonic functions,n∑0 = ∆h = div dh = ∂ ii h.2.3. Conformality in dimension two. The present and next section collect prerequisitesfor an important representation formula for harmonic maps, the Poisson formula.In dimension two, conformal and holomorphic maps are closely related:Proposition 9. Let U, V ⊂ C. A map h ∈ C 1 (V, U) is holomorphic if and only if it isweakly conformal,and orientation preserving, det dh ≥ 0.i=1|∂ x h| = |∂ y h| and 〈∂ x h, ∂ y h〉 = 0Proof. By definition, a holomorphic map h = h 1 + ih 2 satisfies the Cauchy-Riemann equations(9) ∂ x h 1 = ∂ y h 2 , ∂ y h 1 = −∂ x h 2 .These equations show that the map h is weakly conformal. To see a holomorphic map isorientation preserving, compute( ) ()∂ x h 1 ∂ y h 1 (9) ∂ x h 1 ∂ y h 1(10) det dh = det= det= ∂ x h 2 1 + ∂ y h 2 1 ≥ 0.∂ x h 2 ∂ y h 2 −∂ y h 1 ∂ x h 1

iii 2.3 – Stand: July 27, 2012 103The converse direction follows from the fact that given a pair v = ∂ x h and w = ∂ y h ofvectors in R 2 which is orthogonal and of the same length, then w = ±Jv, where J ispositively oriented 90-degree rotation. For v, w of length zero, we can assume w = Jv, inthe general, nonzero, case the same identity follows from the positive orientation. □From the proof it is clear that there are also orientation reversing conformal maps. Namelya map is anti-holomorphic if(11) ∂ x h 1 = −∂ y h 2 , ∂ y h 1 = ∂ x h 2 ⇒ det dh ≤ 0.The simplest example is z. More generally, f(z) is antiholomorphic iff f(z) is holomorphic(check!). A conformal map f on a domain is either holomorphic or antiholomorphic: Unlesswe are in the trivial case that f is constant, the zeros of f ′ are discrete, and so have aconnected complement. Since det df is a continuous function, it has one sign on this set,according to which f is holomorphic or anti-holomorphic.Moreover, holomorphic as well as anti-holomorphic functions are harmonic maps: ∆h 1 =∂ xx h 1 + ∂ yy h 1 = ∂ x ∂ y h 2 − ∂ y ∂ x h 2 = 0 in the holomorphic case, and ∆h 1 = −∂ x ∂ y h 2 +∂ y ∂ x h 2 = 0 in the anti-holomorphic case; similarly so for the second component.For later use, we study composition properties for conformal maps.Lemma 10. Let U, V ⊂ R 2 and suppose f ∈ C 2 (U, V ) is weakly conformal and h ∈C 2 (U, R).(i) If h is harmonic then h ◦ f is harmonic.(ii) If f is a diffeomorphism then E U (h ◦ f) = E V (h).Proof. (i) On simply connected subsets of U consider the conjugate harmonic function h ∗defined by integration (see problems). Then H := h+ih ∗ is holomorphic. If f is orientationpreserving then f is holomorphic. Hence H ◦ f is again holomorphic and its real part h ◦ fis harmonic. Similarly so for the antiholomorphic case. Alternatively, the proof followsfrom a direct calculation.(ii) Denote the conformal factor of f by λ = λ(x) > 0, that is, λ 2 = |∂ x f| 2 = |∂ y f| 2 .Then on the one hand the chain rule gives |d(h ◦ f)| 2 = |dh| 2 |df| 2 = λ 2 |dh| 2 . Moreover,specifically in dimension we have | det df| = λ 2 . Thus by a change of variables we obtainE U (h◦f) = 1 ∫|d(h◦f)| 2 dx U = 1 ∫|dh◦f| 2 | det df| dx U = 1 ∫|dh| 2 dx V = E V (h).2 U2 U2 V =f(U)□From (ii) we conclude that we can compose a minimizing sequence for the Dirichlet integralby arbitrary conformal maps, without effecting the energies. By good choices of conformalmaps we will secure convergence of a minimizing sequence.

104 K. Grosse-Brauckmann: Minimal surfaces, SS 12Problems. 1. Rewrite the proof of (ii) in terms of partials ∂ i f and ∂ j h.2. Find the generalization of both parts of the Lemma for arbitrary dimension.2.4. The conformal group of the disk. By the previous lemma the energy is invariantunder conformal maps. For the Plateau problem we are interested in the case U = D andthe conformal maps which preserve D.Definition. Let D ⊂ R 2 be the open unit disk. We call{Aut(D) := ω(z) = ω a,ϕ (z) := e iϕ a − z}1 − az , ϕ ∈ R, a ∈ Dthe (orientation preserving) conformal automorphisms of the unit disk, or the group of itsMöbius transformations.We leave it as an exercise to check Aut(D) is closed under composition. Note that theidentity is ω 0,π (z) = z, that ω 0,ϕ+π is rotation by an angle ϕ, and ω a,0 (0) = a, ω 0,0 (a) = 0.It is also straightforward to calculate the inverse. We substantiate the remaining claimscontained in the definition:Lemma 11. Each ω ∈ Aut(D) maps D to D conformally, and ∂D to ∂D.Proof. First for z ∈ D we have |az| ≤ |a||z| < 1 and so the denominator of ω does notvanish.Next, the product rule zw = z w gives|ω(z)| 2 = ω(z)ω(z) =(a − z)(a − z)(1 − az)(1 − az) = |a|2 + |z| 2 − az − az1 + |a| 2 |z| 2 − az − az .For |z| = 1 numerator and denominator agree so that |ω(z)| = 1. For z ∈ D we have0 < (1 − |a| 2 )(1 − |z| 2 ) so that |a| 2 + |z| 2 < 1 + |a| 2 |z| 2 . This shows that the (real)numerator is less than the (real) denominator. Thus ω(z) ∈ D. Standard differentiationrules show that ω is holomorphic, and so conformality follows from Prop. 9.□We can calculateω ′ iϕ −(1 − az) + (a − z)a(z) = e = e iϕ |a|2 − 1 ≠ 0 for all z ∈ D.(1 − az) 2 (1 − az)2Hence ω is a diffeomorphism onto its image. Moreover, since ω is invertible (check!) wesee that ω : D → D is a diffeomorphism, and likewise its extension ω : D → D.Remark. Each orientation preserving conformal map from D onto D is in the group Aut(D).This follows with the help of a lemma by Schwarz, see [Sy1], pp. 262–264.26. Lecture, Wednesday 4.7.12

iii 2.5 – Stand: July 27, 2012 1052.5. The three-point-condition. A problem we face is that a minimizing sequence neednot converge. For an example, consider the conformal automorphismsh k (z) := z − 1−kk1 − 1−kkz = zk − 1 + kk − z + kz=k(z + 1) − 1k(z + 1) − z∈ Aut(D).For k → ∞ we have h k (z) → 1 for z ∈ D \ 1, while h k (−1) = −1. Since automorphismsare bijective, E(h k ) = A(h k ) = |D| = π.This behaviour is due to the lack of compactness of our 3-dimensional group Aut(D): Thereare Cauchy sequences in Aut(D) which do not converge in this space.Composing a given harmonic map f ∈ C with h k we see that lim f ◦ h k has a discontinuouslimit which does no longer attain the boundary values Γ. Conversely, given an arbitraryminimzing sequence f k , we have to compose it with suitable conformal automorphisms tosecure convergence. A Möbius transformation in our three-dimensional group can be fixedby specifying three boundary values:Proposition 12. For any two sets of triples 0 ≤ ϑ 1 , ϑ 2 , ϑ 3 < 2π and 0 ≤ α 1 , α 2 , α 3 < 2πthere exists a unique Möbius transformation ω ∈ Aut(D) of the unit disk such thatω ( e iϑ j ) = e iα jfor j = 1, 2, 3.The proof becomes simpler if we replace the disk D by the upper half plane H = {(x, y) ∈C, y > 0}. The sets D and H are conformally equivalent:Lemma 13. The Cayley mapη : C \ {−i} → C \ {1},η(z) := z − iz + i .is a conformal diffeomorphism. It maps bijectively the following sets onto oneanother:(i) H to D, (ii) H to D \ {1}, and (iii) the real axis to the unit circle without the point 1.Proof. The inverse w of η,w = η(z) = z − iz + i⇔ zw+iw = z −i ⇔ i(1+w) = z(1−w) ⇔ z = i 1 + w1 − w=: ζ(w),is defined for all w ≠ 1. We conclude η ◦ ζ = id D and η is bijective. Moreover, since w andη are differentiable, η is a diffeomorphism.Let us now show η(H) ⊂ D. This follows from |z − i| < |z + i| for z ∈ H, which is obviousgeometrically. Indeed, Re(z − i) = Re(z + i) while Im(z − i) < Im(z + i) for z ∈ H. On

106 K. Grosse-Brauckmann: Minimal surfaces, SS 12the other hand, for w ∈ D2 Im ( ζ(w) ) = 1 iThis gives (i).(ζ(w) − ζ(w))=1 + w1 − w + 1 + w1 − w= 2 − 2|w|2|1 − w| 2 > 0.=(1 + w)(1 − w) + (1 + w)(1 − w)(1 − w)(1 − w)Similarly (ii) follows from η(H) ⊂ D, and w ∈ D \ {1} implies Im ( ζ(w) ) ≥ 0, so thatζ : D \ {1} → H. Likewise for (iii).□An analogue of Proposition 12 for H reads:Lemma 14. For every triple of real numbers a < b < c there is a conformal diffeomorphismsuch that f(a) = 0, f(b) = 1, f(c) = ∞.f : H ∪ {∞} → H ∪ {∞},Here, we need the notion of differentiability at ∞: It is defined by pre- or postcomposingthe function with 1/z.Proof. Letf : H → H, f(z) := z − az − c · b − cb − a .Then f is holomorphic and thus conformal. Moreover, f maps H to H since b−c < 0 andb−a2i Im z − az − c = z − az − c − z − az − c= 2 Im z a − c|z − c| 2 < 0.=(z − a)(z − c) − (z − a)(z − c)|z − c| 2 =(a − c)(z − z)|z − c| 2On ∂H ∪ {∞} we have “= 0”, and so f maps the extended real line to itself. Moreover,the particular values are immediate. To see f is invertible, compute f −1 .□Via the detour to H we can now prove the proposition:Proof of the proposition. Set a := η −1 (e iϑ 1), b := η −1( e iϑ 2), c := η−1 ( e iϑ 3). Then themapping f ◦η −1 , where f was constructed in the previous lemma, maps the triple of pointse iϑ 1, e iϑ 2, e iϑ 3to the triple 0, 1, ∞ in ∂H ∪ ∞.can be mapped confor-□Since the mappings are diffeomorphisms, any triple e iϑ 1, e iϑ 2, e iϑ 3mally to any other triple e iα 1, e iα 2, e iα 3.25. Lecture, Monday 2.7.12

iii 2.6 – Stand: July 27, 2012 1072.6. Poisson’s formula. The Dirichlet problem for a domain U is to find a solution u ∈C 2 (U, R) ∩ C 0 (U, R) of∆u = 0 in U with u| ∂U = ϕ,where ϕ is prescribed. For U bounded we proved the uniqueness of solutions to this problemin Prop. II.35. In the special case that U is a ball, we now give an existence statement.Note that the Dirichlet problem has fixed prescribed boundary values while the Plateauproblem has no preferred boundary parameterization. This is one reason why it is easierto solve.Theorem 15. Let ϕ be a continuous function on ∂B R where B R = B R (0). Then⎧⎪⎨ R 2 − |x| 2 ∫ϕ(y)(12) u(x) := nω n R ∂B R|x − y| dS n y for x ∈ B R ,⎪⎩ ϕ(x) for x ∈ ∂B R ,belongs to C 2 (B R , R) ∩ C 0 (B R , R) and is harmonic in B R .If we define the Poisson kernelK(x, y) := R2 − |x| 2nω n R|x − y| n , x ∈ B R, y ∈ ∂B R ;we can regard the Poisson integral as the convolution [Faltung] u(x) = ∫ ∂B K(x, y)ϕ(y) dS y.Proof. (i) Since ϕ is independent of x, and K is differentiable in x ∈ B R , it follows thatu(x) is differentiable infinitely often on B R , in particular u ∈ C 2 (B R , R).(ii) For u ∈ C 2 (B R , R) we claim:∫If u ∈ C 0 (B R , R) and ∆u = 0 on B R then u(x) =∂BK(x, y)ϕ(y) dS y for all x ∈ B R .We will only prove the result for the case n = 2 we need lateron. For simplicity weadditionally assume R = 1. We will derive the Poisson formula from the mean valueformula. Consider for |x| < 1ω x (z) := x − z ∈ Aut(D).1 − xzSince u ∈ C 2 (D), Lemma 10(i) gives that the function u ◦ ω x is harmonic. In particular,it has the mean value property (6) on B r for all 0 < r < 1:u(x) = u ( ω x (0) ) = 1 ∫(u ◦ ω x )(z)dS z2π ∂B rBut u is continuous on B 1 with boundary values ϕ. Hence as the limit r → 1 we obtainu(x) = 1 ∫(ϕ ◦ ω x )(z)dS z .2π ∂B 1

108 K. Grosse-Brauckmann: Minimal surfaces, SS 12Let us now change variables of the curve integral. Note first that ω x is its own inverse:ThusButNow for y ∈ S 1 we haveand soω x(ωx (z) ) = x − x−z1−xz1 − x x−z1−xz= x − xxz − x + z1 − xz − xx + xz = (1 − |x|2 )z1 − |x| 2 = zϕ(x) = 1 ∫(ϕ ◦ ω x )(z)dS z = 1 ∫ϕ(y)|ω2π ∂B 12πx(y)|dS ′ y .∂B 1ω ′ x(y) =−(1 − xy) + (x − y)x(1 − xy) 2 = |x|2 − 1(1 − xy) 2 .|1 − xy| 2 = (1 − xy)(1 − xy) = 1 − xy − xy + |x| 2 = (y − x)(y − x) = |y − x| 2This proves step (ii):u(x) = 12π∫|ω ′ x(y)| = 1 − |x|2|y − x| 2 .∂B 1ϕ(y)|ω ′ x(y)| dS y = 12π∫∂B 1ϕ(y) 1 − |x|2|y − x| 2 dS y(iii) We claim K has the properties of a convolution kernel, namely∫(13) K(x, y) dS y = 1 and ∆K(x, y) = 0 for x ∈ B R , y ∈ ∂B R ,∂Bwhere ∆ = ∑ ∂ 2∂x i ∂x i. Once again, we provide the proof only for n = 2. Apply step (ii) tothe harmonic function u ≡ 1; the result is the first assertion. To prove the second, considerthe following expansion of the Poisson kernel:(14) 2πRK(x, y) = R2 − |x| 2|y − x| 2 =yy − yx + yx − xx(y − x)(y − x)= yy − x + xy − xThe right hand side is the sum of the holomorphic function x ↦→ and the antiholomorphicfunction x ↦→x . Each of these terms is harmonic, hence the sum x ↦→ 2πRK(x, y)y−xis harmonic as well.yy−x(iv) We show that u(x) is harmonic. This is straightforward from (13):∫( )∆u(x) = ∆K(x, y) ϕ(y) dSy = 0.∂B(v) It remains to show that u is continuous at ∂B. Let x 0 ∈ ∂B and ε > 0 be arbitrary.Since ϕ is continuous, there is δ > 0 such that |ϕ(x)−ϕ(x 0 )| < ε for |x−x 0 | < δ; moreover

|ϕ| ≤ M on ∂B. Then for |x − x 0 | < δ/2 we have∫|u(x) − u(x 0 )| (13)=∣ K(x, y) ( ϕ(y) − ϕ(x 0 ) ) ∣ ∣∣∣dS y∂B∫≤K(x, y) ∣ ∣ϕ(y) − ϕ(x 0 ) ∣ ∫dS y +y∈∂B,|y−x 0 |≤δ≤ ε + nR n−2 (R2 − |x| 2 )(δ/2) n 2M,iii 2.7 – Stand: July 27, 2012 109y∈∂B,|y−x 0 |>δK(x, y) ∣ ∣ϕ(y) − ϕ(x 0 ) ∣ ∣ dS ywhere for the first summand we used that K has integral 1. We can now require that|x − x 0 | is sufficiently small for the second summand to become less than ε. Thus u iscontinuous at x 0 .□Remarks. 1. At x = 0 the Poisson integral reduces to∫u(0) =R2 ϕ(y)nω n R ∂B(0) |y| dS 1n y =ϕ(y) dSnω n R∫∂B(0)n−1 y ,which is precisely the mean value formula. In general, the Poisson formula can be consideredan offcentered version of the mean value formula: The kernel K(x, y) acts as a weighton the boundary distribution.2. Does the Poisson formula extend to arbitrary domains U? Only for very special domainslike annuli or half-spaces explicit formulas are known. There is a more implicit method,the so-called Perron process, which makes use of Poisson’s formula (see [GT], Par.2.8).3. By the maximum principle (or claim (ii)), the Poisson integral represents the uniqueharmonic function with boundary values ϕ.4. In the 19th century, integral kernel representation formulas were seeked and obtainedfor many partial differential equations. The modern point of view, which prevailed in the20th century is to establish solutions in terms of more abstract existence theorems. Wewill see an example in the proof of the Plateau problem, which establishes existence of aminimal surface without representing it by an explicit formula.5. For the proof in arbitrary dimension n ≥ 2, see [GT], Sect. 2.4 and 2.5.24. Lecture, Friday 29.6.122.7. Variational characterization and Dirichlet’s principle. Harmonic maps have avariational characterization:Proposition 16. A map h ∈ C 2 (U n , R d ) is harmonic if and only if its Dirichlet integralE(h) has vanishing first variation,(15)ddt E(h + tf) ∣∣∣t=0= 0 for all f ∈ C ∞ 0 (U n , R d ).

110 K. Grosse-Brauckmann: Minimal surfaces, SS 12Proof. We extend the scalar product to matrices by setting 〈dh, df〉 := ∑ ij ∂ ih j ∂ i f j . Thenwe obtain |d(h + tf)| 2 = |dh| 2 + 2t〈dh, df〉 + t 2 |df| 2 , and so, using (5),d∣ ∫∫dt E(h + tf) ∣∣t=0= 〈dh, df〉 dx (5)= − (∆h)f dx.USo if h is harmonic then (15) holds. The converse follows from the fundamental lemma ofthe calculus of variations.□Unlike the area functional, the Dirichlet integral is quadratic in the gradient. This meansthe functional is convex and so its minima, the harmonic functions, are unique:Proposition 17 (Dirichlet’s principle I). Let U ⊂ R n be a bounded domain, ∂U ∈ C 1 .Suppose h ∈ C 2 (U, R d ) is harmonic. ThenE(f) ≥ E(h)for all f ∈ C 1 (U, R d ) with the same boundary values f| ∂U = h| ∂U .UProof. The function f − h has zero boundary values, and soE(f) = E ( h + (f − h) ) = 1 ∫|dh| 2 + 2 〈 dh, d(f − h) 〉 + |d(f − h)| 2 dx2∫∆h=0, (5) 1= |dh| 2 + |d(f − h)| 2 dx ≥ E(h).2□The proof of Dirichlet’s principle, specifically the divergence formula, needs that the boundaryvalues of h are C 1 . We can make the same statement, however, for continuous boundaryvalues. We will state it specifically for the unit disk D ⊂ R 2 :Theorem 18 (Dirichlet’s principle II). Suppose h ∈ C 0 (D, R d ) ∩ C 2 (D, R d ) is harmonicon D. ThenE(f) ≥ E(h)for all f ∈ C 0 (D, R d ) ∩ C 1 (D, R d ) with the same boundary values f| ∂D = h| ∂D .Proof. Suppose there is f ∈ C 0 (D, R d ) ∩ C 1 (D, R d ) such that E(f) < E(h). Our strategyis to exhibit another function ˜f ∈ C 0 (D, R d ) ∩ C 2 (D, R d ) with E( ˜f) < E(h), which isharmonic in D. Then ˜f, h are both harmonic in D and have the same boundary values.By the maximum principle, applied to each component, they must agree, contradicting thestrict energy inequality.Let 0 < r < 1 and consider D r := {x ∈ D : |x| < r}. The Poisson formula gives a functionf r : D r → R d which is harmonic on D r and has the boundary values of f on ∂D r . ByDirichlet’s principleE Dr (f r )Prop. 17≤E Dr (f) ≤ E D (f).

iii 2.8 – Stand: July 27, 2012 111Now we define ˜f r (x) := f r (x/r) on all of D. We claim that ˜f r and all its derivativesconverge uniformly on each compact subset of D as r → 1. Indeed, since f is boundedalso the boundary values of ˜f r are uniformly bounded. By the maximum principle, ˜fr isuniformly bounded on D. Therefore Thm. 8 proves our claim.As a consequence of the conformal invariance of energy, Lemma 10(ii), we get, as desiredE( ˜f r ) L.10(ii)= E(f r | Dr ) ≤ E(f) < E(h).We conclude that the limit ˜f := lim r→1 ˜f is harmonic on all of D and by C 1 -convergencehas energy E( ˜f) = lim r→1 E( ˜f r ) < E(h).□2.8. Analyticity of harmonic maps and minimal surfaces. The Poission formulagives an entirely real approach to proving the analyticity of minimal surfaces.Cauchy’s integral formula expresses a holomorphic function as an integral. One of thevirtues of this formula is that it shows that holomorphic functions are differentiable infinitelyoften. We derive the same fact for harmonic functions from the Poisson formula.Theorem 19. Let u ∈ C 2 (U, R) be harmonic. Then u ∈ C ∞ (U, R) and in fact u is realanalytic.Note that a function is called (real) analytic, if each point of the domain has a neighbourhoodwhere the function can be represented as a power series. One example of smoothfunctions which are not analytic are the bump functions used for partitions of unity.Proof. Since U is open, there is a ball B = B R (x) whose closure is contained in U. In B R ,the function u is represented by the Poisson integral (12) in terms of its values on ∂B R .As in part (i) of the proof of the Poisson formula, the fact that u ∈ C ∞ (B R , R) followsdirectly from differentiating the Poisson integral.We want to show that u can be represented by a power series in the ball B R (x) which,after translation, becomes B R = B R (0). Using a geometric series expansion for (14), wefind2πRK(x, y) ==yy − x + xy − x = 11 − x + x 1yy1 − x y∞∑( ) k x∞∑( ) k x( ∑ ∞+ = 1 + 2 Re y −k x);ky yk=0k=1k=1

112 K. Grosse-Brauckmann: Minimal surfaces, SS 12this is valid for |x| < |y| = R. Consequently,∫u(x) = K(x, y)ϕ(y) dS y = 1∂B R2πR= 1 ∫2πR∂B Rϕ(y) dS y + 1πR Re ∞∑∫∂B R(1 + 2 Rek=1( ∑ ∞ ) )y −k x k ϕ(y) dS yk=1( ∫ ∂B Rϕ(y)y k dS y)x k .This is indeed a power series for u(x): Note that x k is a polynomial of order k in Re x andIm x, and so is Re(x k ). Hence with suitable coefficients a ij we can write u(x) in the form∑ ∞ ∑k=0 i+j=k a ij(Re x) i (Im x) j .□Let us draw a conclusion for arbitrary minimal surfaces (not necessarily for Plateau solutions):Corollary 20. A minimal surface f : U 2 → R d in conformal parameterization is realanalytic.Proof. Since ∆f = 0 holds, each component is real analytic by the previous theorem. Thusf is real analytic.□The Corollary means that if we force a soap film to change somewhere then it will changeeverywhere – changes cannot be local, but must always be global. More precisely: If twocomplete (connected) minimal surfaces agree on a nonempty open set, then they agreeeverywhere.26. Lecture, Wednesday 4.7.123. Solution to Plateau’s problemAs formulated in Thm. 4 we want to solve Plateau’s problem in the class C from (1). Sowe consider a minimizing sequence (f k ) ∈ C for E, that is,E(f k ) → inf{E(f), f ∈ C} as k → ∞.We call f ∈ C with E(f) = inf{E(h), h ∈ C} a minimizer.There are various tasks in order to construct f = lim f k :1. In order to achieve convergence we have to alter the minimizing sequence (f k ).2. We must prove that f represents a mapping in C.3. We need to assert that f is conformally parameterized, hence minimal.

iii 3.1 – Stand: July 27, 2012 1133.1. Courant-Lebesgue-Lemma and equicontinuity. We can fix three boundary valuesof a minimizing sequence:Theorem 21. Let f ∈ C and 0 ≤ ϑ 1 , ϑ 2 , ϑ 3 < 2π as well as p 1 , p 2 , p 3 ∈ Γ be pairwisedistinct. Then there is ˜f ∈ C, with E( ˜f) = E(f), satisfying the 3-point-condition(16) ˜f(eiϑ j) = p j , for j = 1, 2, 3.We could fix the ϑ j once and for all so that e iϑ jare third roots of unity.Proof. Since f : S 1 → Γ is bijective there exist α 1 , α 2 , α 3 distinct such that f(e iα j) = p j ,j = 1, 2, 3. By composing f with a rotation and a reflection if necessary, we may assume0 ≤ α 1 < α 2 < α 3 < 2π; this leaves E(f) unchanged.By Proposition 12 there is a Möbius transformation ω with ω(e iϑ j) = e iα jfor j = 1, 2, 3.Then ˜f := f ◦ ω satisfies the 3-point-condition. Moreover, ω is conformal and henceE(f ◦ ω) = E(f) by Lemma 10(ii).□As an intermediate step to establishing the convergence of a minimizing sequence, we willshow that a minimizing sequence which satisfies the 3-point-condition has equicontinuousboundary values. We first derive a pointwise bound from an integral bound:Lemma 22 (Courant-Lebesgue). Let f ∈ C 0 (D, R d ) ∩ C 1 (D, R d ) with E(f) ≤ C. Pick0 < δ < 1 and z 0 ∈ D. Then there exists r = r(z 0 , f) ∈ (δ, √ δ) such that for all z 1 , z 2 ∈ Dwith |z 1 − z 0 | = r = |z 2 − z 0 ||f(z 1 ) − f(z 2 )| 2 ≤8πClog(1/δ) .Proof. Take polar coordinates (ρ, ϑ) centered at z 0 . For r > 0 letS(r) := S(r, z 0 ) = {ϑ ∈ [0, 2π) : z 0 + re iϑ ∈ D}.Integration along S r gives, on the one hand,( ∫(17) |f(z 1 ) − f(z 2 )| 2 ≤ |∂ ϑ f(r, ϑ)| dϑS(r)) 2 Hölder’s inequ.≤∫2π |∂ ϑ f(r, ϑ)| 2 dϑ.S(r)The Dirichlet integral in terms of polar coordinates gives∫ (2E(f) = |∂ ρ f| 2 + 1 ) ∫ 1ρ |∂ ϑf| 2 ρ dϑdρ ≥ 2 ρ |∂ ϑf| 2 dϑdρ.By the generalized mean value theorem of integration we find a particular r ∈ (δ, √ δ) with(18) 2E(f) ≥∫ √ δδ1ρ( ∫ )|∂ ϑ f| 2 dϑ dρ =S(ρ)∫ √ δδ1ρ∫S(r)dρ |∂ ϑ f| 2 dϑ.

114 K. Grosse-Brauckmann: Minimal surfaces, SS 12Combining (17) with (18) we obtain the desired inequality|f(z 1 ) − f(z 2 )| 2 ≤4πE(f)log √ δ − log δ ≤4πClog(1/ √ δ) =8πClog(1/δ) .□It is a good exercise to see what the Courant-Lebesgue lemma gives when applied to thesequence of Sect. 2.5.Note that the right hand side of the bound in the Courant-Lebesgue lemma tends to 0 asδ → 0. So it should grant equicontinuity for maps with uniformly bounded energy. Wecan make this precise under the 3-point condition:Theorem 23. A minimizing sequence (f k ) ∈ C which satisfies the 3-point condition (16)has equicontinuous boundary restriction ϕ k := f k | ∂D .Proof. Let ε > 0. We need to show that there exists δ > 0 such that for all z 0 ∈ ∂D andz ∈ ∂D with |z − z 0 | < δ(19)∣ ϕk (z) − ϕ k (z 0 ) ∣ < ε for all k ∈ N.Let us first state a fact for our given Jordan curve Γ. We claim there is 0 < λ = λ(ε, Γ) ≤ εwith the following property: Any pair of points q 1 , q 2 ∈ Γ with ambient distance |q 1 −q 2 | < λdecomposes Γ into two subarcs Γ 1 , Γ 2 , such that one of them, say Γ 1 , has diameter lessthan ε. This means that |x − y| < ε for all x, y ∈ Γ 1 . We suppose that ε is chosen lessthan all pairwise distances of the image triple p 1 , p 2 , p 3 of the 3-point-condition. Then thearc Γ 1 will contain at most one of these points.By Lemma 3 the infimum of energy inf{E(f), f ∈ C} is finite. Thus for (f k ) ∈ C minimizingE(f) we can assume that E(f k ) ≤ C uniformly where, for instance, C is twice the infimalenergy. As log 1 → ∞ for δ → 0, we can choose δ = δ(ε) ∈ (0, 1) small enough to ensureδ(20)8πClog(1/δ) < λ2 (ε).We now apply the Courant-Lebesgue-Lemma. For each z 0 ∈ ∂D and for each k ∈ N ityields a radius r = r(z 0 , δ, k) ∈ (δ, √ δ) such that the two points z 1 ≠ z 2 on ∂D in distancer from z 0 satisfy(21)∣ ϕk (z 1 ) − ϕ k (z 2 ) ∣ < λ ≤ ε.The points z 1 , z 2 decompose the circle ∂D into the two semiarcsS 1 = S 1 (z 0 , δ, k) := {z ∈ ∂D, |z − z 0 | ≤ r/2}, S 2 := ∂D \ S 1 .

iii 3.2 – Stand: July 27, 2012 115By decreasing δ (and hence r) if necessary, we can assume √ δ is less than all pairwisedistances of the triple ϑ 1 , ϑ 2 , ϑ 3 of the 3-point condition (16). This ensures that S 1 containsat most one of the points e iϑ j.We would like to say that the entire arc S 1 (not only its endpoints z 1 , z 2 ) maps under ϕ kinto some ε-ball about z 0 . By (21), we can apply the above fact about Jordan curves tothe pair of points q i := ϕ k (z i ), i = 1, 2. This gives us two subarcs Γ 1 , Γ 2 (z 0 , k, δ), fromwhich Γ 1 is the short arc, in the sense that its diameter is less than ε.We use the 3-point-condition to assert that the arc S 1 gets mapped to the short arc Γ 1 ,not to the long arc Γ 2 . Indeed, the arc S 1 contains at most one point of the triple, and sodoes its image under the bijection ϕ k . Thus S 1 cannot be mapped to Γ 2 which contains atleast two points of the triple.We conclude that ϕ k (S 1 ) = Γ 1 has diameter at most ε. In particular |ϕ k (z)−ϕ k (z 0 )| < ε forall z ∈ S 1 . On the first sight this statement seems not uniform in k, as our arc S 1 dependson r and hence on k. But always r(z 0 , k, δ) > δ, where the right hand side depends on εalone. Thus for all k the arc S 1 (z 0 , δ, k) contains all points z ∈ ∂D with |z − z 0 | < δ, andin particular (19) holds, as desired.□3.2. Construction of a minimizer. The present section contains the core of the proofof Plateau’s problem. There are two main problems we need to address.First, we cannot expect our uniformly convergent sequence of C 1 functions to converge toa function which is again C 1 . (What would a counterexample for the case of functions fromR to R be?) The remedy is to replace the functions of an arbitrary minimizing sequencewith harmonic maps. The Poisson formula allows us do this while preserving boundaryvalues. As asserted in Thm. 8 the harmonic mappings will converge together with all theirderivatives.Second, when we pass to a limit, injectivity of the boundary parameterization can be lost.To understand this problem, view the injective boundary values as strictly monotone parameterizations.The limit of strictly monotone functions is merely monotone. To overcomethis problem, we construct the limit in a class larger than C: We call a map ϕ: S 1 → Γmonotone if ϕ is continuous and surjective, and for each point in Γ the preimage is a closedconnected arc (including the particular case of a point) and defineC ⊂ C mon := {ϕ ∈ C 0 (D, R d ) ∩ C 1 (D, R d ) maps ∂D monotonically to Γ}Theorem 24. There is a minimizing sequence of maps h k ∈ C such that h k ∈ C 2 (D, R d )is harmonic. Moreover, a subsequence of (h k ) converges uniformly to h ∈ C mon whereh ∈ C 2 (D, R d ) is harmonic and satisfies E(h) = inf{E(f), f ∈ C}.

116 K. Grosse-Brauckmann: Minimal surfaces, SS 12Proof. We consider an arbitrary minimizing sequence (f k ) in C. By Thm. 21, we canassume each f k satisfies the 3-point-condition. Thus Thm. 23 gives that the sequence (f k )has equicontinuous boundary restriction ϕ k = f k | ∂D . By the Arzelà-Ascoli Theorem 7,a subsequence of the boundary restrictions converges to a limit ϕ: ∂D → R d , which ismonotone. We denote it again with (ϕ k ).Let us now consider the Dirichlet problem for the disk∆h k (x) = 0 for x ∈ D,h k (x) = ϕ k (x) for x ∈ ∂D.Applying the Poisson integral formula (12) componentwise, we obtain solutions h k ∈C 0 (D, R d )∩C 2 (D, R d ). Dirichlet’s principle, in the generalized version of Thm. 18, impliesthat h k has minimal energy among maps which attain the same boundary values,E(h k ) ≤ E(f k ) k→∞→ inf{E(f), f ∈ C}.In particular, (h k ) is again a minimizing sequence.The weak maximum principle for harmonic functions, Prop. I.34, implies a maximumprinciple for the modulus of harmonic functions (see Problems) and therefore a maximumprinciple for the modulus of harmonic maps. This proves a uniform boundsupD|h k | ≤ sup |ϕ k | = sup{|p| : p ∈ Γ}∂Dand so the compactness result, Thm. 8, gives that a subsequence of (h k ) converges locallyuniformly on D to a harmonic map h, together with all its derivatives. Thus E(h) =inf{E(f) : f ∈ C}, as claimed.It remains to consider convergence at the boundary. We claim that (h k ) converges uniformlyon D. This follows from the boundsupD|h k − h l | ≤ sup |ϕ k − ϕ l | → 0 as k, l → ∞.∂DIn particular, h| ∂D = ϕ. Since the limit of strictly monotone continous functions is monotonethis also verifies h ∈ C mon .□27. Lecture, Monday 9.7.123.3. The minimizer is weakly conformal. The previous convergence theorem leavesopen if the limiting map h is weakly conformal. We will see that conformality followsfrom the fact that our minimizing sequence is subject to the free boundary conditionof parameterizing Γ injectively; minimizing energy under Dirichlet boundary conditions(fixing the boundary parameterization pointwise) will not give conformality.Theorem 25. The limiting map h of Thm. 24 is weakly conformal.

iii 3.3 – Stand: July 27, 2012 117It will be good enough to compare the energy of h with variations of the parameterization,namely with variations which affect the boundary parameterization. For the variationalformulation, we compare h with maps h ◦ σ t , where σ t : D → D is a family of diffeomorphismsof D where σ 0 = id. Note that these variations act on the domain, not on thetarget.Proposition 26. Suppose U ⊂ R 2 and σ : (−ε, ε) × U → U is a smooth family of diffeomorphismsσ t such that σ 0 = id U . Then any f ∈ C 1 (U, R n ) with finite energy has firstinner variationd∫(22)dt E 1( )(U(f ◦ σ t ) ∣ = ξx − η y |fx | 2 − |f y | 2) + ( )ξ y + η x 〈fx , f y 〉 dz,t=0 2Uwhere ξ, η are the components of the smooth variation vector field(23) X(z) = ( ξ(z), η(z) ) := d dt σ t(z) ∣ .t=0Note that if f is conformal, then (22) vanishes. In order to show the converse, we will usethat at the minimizer h of E, the inner variation (22) vanishes for arbitrary X.Proof. For a change of variables we set w = σt−1 (z). ThenE U (f ◦ σ t ) = 1 ∫|d(f ◦ σ t )(w)| 2 chain ruledw = 1 ∫|df(σ t (w))(dσ t )(w)| 2 dw2 U2 U∫ch. of var. 1=|df(z)dσ t (σt−1 (z))| 2 | det dσt−1 (z)| dz2σ t(U)=UNote that these integrals exist since dσ t and dσ −1t are bounded on U.Since σ and σ −1 has bounded t-derivatives on U, we can differentiate under the integralsign with respect to t. To do so, we use a product rule of type (g 2 h) ′ = 2gg ′ h + g 2 h ′ aswell as dσ 0 = id to obtaind∫dt E U(f ◦ σ t ) ∣ =t=0U〈df(z), df(z) d dt(dσt (σ −1t (z) )∣ ∣∣t=0〉+ 1 d 2 |df(z)|2 ∣ det dσt−1(z) ∣ dz.dtt=0To calculate the t-derivatives of σ, let us apply the chain rule (think of dσ t (σt−1 ) as afunction dσ(t, v) where v = σt−1 (z)):d (dσt (σt−1 (z) )∣ ∣∣t=0 = d dtdt dσ t(z) ∣ + d 2 d[ ] d]σ 0 σt−1 (z) = d[t=0 dtt=0 dt σ t(z) = dX(z).t=0Here, we used that σ 0 = id, so that its second partials d 2 σ 0 vanish. Moreover, for the firstterm we used that partial derivatives and the t-derivative of σ commute; indeed,ddt dσ t(z) · v∣ = d dt=0 dt ds σ t(z + sv) ∣ = d ds,t=0 ds dt σ t(z + sv) ∣s,t=0= d ds X(z + sv) ∣∣∣s=0= dX(z) · v.

118 K. Grosse-Brauckmann: Minimal surfaces, SS 12To calculate the derivative of the determinant, let us note that if A(t) =A(0) = E then( )a(t) b(t)c(t) d(t)and(det A) ′ (0) = (ad − bc) ′ (0) = (a ′ d + ad ′ − b ′ c − bc ′ )(0) = (a ′ + d ′ )(0) = trace A ′ (0).Thus the change of the volume element of σ t = id +tX + O(t 2 ) is precisely the divergenceof X:ddt det(dσ t(z)) ∣ = trace d t=0 dt dσ d]t∣ = trace d[t=0 dt σ t = ∑ ∂X i = div X.t=0 ∂x iSimilarly, due to the identity σ t ◦ σt−1| t=0 = −X and d dt det(dσ−1Thus d dt σ−1 t= id we have the expansion σt −1 = id −tX + O(t 2 ).t (z))| t=0 = − div X.The result of these calculations is the first inner variation formulad∫(24)dt E 〈 〉 1U(f ◦ σ t ) ∣ = df(z), df(z)dX(z) −t=0 2 |df(z)|2 div X dz.It can be shown to hold regardless of dimension.ULet us now show (24) agress with (22) for dimension two. Writing out the matrices withcolumn vectors, we see the first term of (24) gives( )〈 〉 〈ξ x ξ 〉ydf(z), df(z)dX(z) = (f x , f y ), (f x , f y )η x η y〈= (f x , f y ), ( ) 〉ξ x f x + η x f y , ξ y f x + η y f y = ξ x |f x | 2 + η x 〈f x , f y 〉 + ξ y 〈f x , f y 〉 + η y |f y | 2 ,while the second contributes − 1(ξ 2 x + η y )(|f x | 2 + |f y | 2 ). Summing these two terms andintegrating we obtaind∫dt E ξ x(U(f ◦ σ t ) ∣ = |fx | 2 − |f y | 2 ) + ξ y 〈f x , f y 〉 + η x 〈f x , f y 〉 − η y(|fx | 2 − |f y | 2) dzt=0 22which gives the desired result.UProof of the theorem. Let h be the harmonic map from Thm. 24. We set(25) ψ : D → C, ψ := 1 2(|hx | 2 − |h y | 2) − i〈h x , h y 〉and want to show ψ ≡ 0.The map h minimizes the energy E in the class Cmon.0 Thus for any smooth family ofdiffeomorphisms σ t : D → D, the map t ↦→ E(h ◦ σ t ) attains a minimum at t = 0. Inparticular,0 = d ∫dt E ( ) ( )D(h ◦ σ t ) ∣ = ξx − η y Re ψ − ξy + η x Im ψ dzt=0 D∫(= Re ξx − η y + i(ξ y + η x ) ) ∫ψ dz = 2 Re (∂ z X)ψ dz,DD□

where we employed the shorthand notationiii 3.3 – Stand: July 27, 2012 119(26) ∂ z f := 1 2 (∂ xf + i ∂ y f) = 1 2 (∂ x Re f − ∂ y Im f) + i 2 (∂ x Im f + ∂ y Re f)for f = Re f + i Im f. Note that the relation ∂ z f = 0 is equivalent to the validity of theCauchy-Riemann equations for f.As a first step, we show ψ is holomorphic. The operator ∂ z f relates to a divergence,2 Re ∂ z f = ∂ x Re f − ∂ y Im f = ( ∂ x Re +∂ y Im ) f = div f.Thusdiv(fg) = 2 Re ∂ z (fg) = 2 Re ( (∂ z f)g + f∂ z g )and so we can partially integrate as follows:∫∫∫(27) 0 = 2 Re (∂ z X)ψ dz = div(Xψ) dz − 2 ReDDDX∂ z ψ dz.Now suppose X has compact support in D. Then clearly σ t (z) := z+tX is a diffeomorphismof D for small |t|, and so, by the divergence theorem, (27) reduces to 0 = Re ∫ D X(∂ zψ) dz.Now we apply the fundamental lemma of the calculus of variations, Lemma 26. Choosing Xpurely real yields Re ∂ z ψ(z) = 0 for all z ∈ D, while X purely imaginary yields Im ∂ z ψ(z) =0. This proves ψ is holomorphic on D.In a second step, we use this fact to show ψ ≡ 0. We apply the divergence theorem. Sinceψ is defined on the open disk only, we work with disks D r of smaller radius r < 1. Usingthe identity 〈w, z〉 = Re w Re z − Im w Im z = Re(wz) we find(28)0 (27), ∂ zψ=0=∫D∫= lim Re ( Xψ zr↗1∂D r|z|∫div(Xψ) dz = limr↗1)ds = limr↗1Rediv(Xψ) dz div.thm.= limD r∫X z∂D r|z| ψ ds.r↗1∫〈 z 〉Xψ, ds∂D r|z|Up to this point, perhaps, the proof appears not too inspiring, but now comes the beautifulturn. Let us consider a variation σ t which rotates each z by an angle tα(z),σ t (z) := ze itα(z) ,for α: D → R to be specified later. Note that σ t : D → D is a diffeomorphism for small |t|such that σ 0 (z) = z, andX(z) := ∂ ∂t σ t(z) ∣ = z d ∣∣∣t=0t=0 dt eitα(z) = izα(z).

120 K. Grosse-Brauckmann: Minimal surfaces, SS 12Inserting this into (28) and then introducing polar coordinates z = re iϑ we obtain∫ (0 = − lim Re X z ) ∫r↗1∂D r|z| ψ α(z)ds = lim Im ( z 2 ψ(z) ) dsr↗1∂D r|z|(29)∫ 2π1= limr↗10 r α(reiϑ ) Im ( (re iϑ ) 2 ψ(re iϑ ) ) r dϑ.The function z 2 ψ(z) is holomorphic on D and thus has a power series representation whichconverges on D,∞∑∞∑∞∑z 2 ψ(z) = (a k + ib k )z k = (a k + ib k )(re iϑ ) k = (a k + ib k )r k( cos(kϑ) + i sin(kϑ) ) .k=0k=0The left hand side vanishes at z = 0 and so a 0 + ib 0 = 0. The imaginary part is∞∑Im(z 2 ψ(z)) = a k r k sin(kϑ) + b k r k cos(kϑ).k=1Let us recall the fact that the functions {sin(kϑ), cos(kϑ), k ∈ N} are orthogonal withrespect to the inner product 〈f, g〉 := ∫ 2πf(ϑ)g(ϑ) dϑ, but each function has a nonzero0norm 〈f, f〉. Thus if for k ∈ N we choose α(z) = r k cos kϑ = Re(z k ) then (29) gives b k = 0,while for α(z) = r k sin kϑ = Im(z k ) we obtain a k = 0. This proves z 2 ψ(z) ≡ 0, that is,ψ(z) ≡ 0 on D, as desired.□Remark. Our proof indicates a close relationship between Fourier series and power series.Suppose h: D → C is holomorphic. Then f(ϑ) := h(e iϑ ) is a continuous function R → C,with period 2π. Restricting the power series expansion h(z) = ∑ k∈Z c kz k to the boundarywe find f(ϑ) = ∑ c k e ikϑ , which gives us the Fourier series of f.3.4. The minimizer is injective on the boundary. Let us first consider harmonicfunctions once again.Theorem 27 (Schwarz reflection principle). Let U + be an open domain of the halfspace{x n > 0} whose boundary contains an open subset T of the hyperplane {x n = 0}. Supposeu ∈ C 2 (U + , R) ∩ C 0 (U + ∪ T, R) is harmonic and u| T = 0. Denote by U − := {x ∈R n , (x 1 , . . . , −x n ) ∈ U + } the reflection of U + across T . Then the extension of u by oddreflection,⎧⎨u(x 1 , . . . , x n ) x n ≥ 0,ũ(x 1 , . . . , x n ) :=⎩−u(x 1 , . . . , −x n ) x n < 0,is in C 2 (U + ∪ T ∪ U − ) and harmonic.The harmonicity assumption creates the unexpected regularity along T : The functionu(x) := 3√ x, is smooth on (0, ∞), and extends continuously with u(0) := 0. Its oddextension ũ(x) := 3√ x is, however, not differentiable at 0.k=0

Proof. Let us denote reflection by σ(x) := (x 1 , . . . , −x n ).iii 3.4 – Stand: July 27, 2012 121It is easy to check by direct calculation that ∆ũ(x) = 0 for all x ∈ U − . Indeed,∂ũ(x) =∂ (− u(x1 , . . . , −x n ) ) = −(∂ ii u) ( σ(x) )∂x 2 i∂x 2 ifor all i (including i = n), and so odd reflection preserves harmonicity,(30) ∆ũ(x) = −(∆u) ( σ(x) ) = 0.We now apply the Poisson formula in conjunction with the maximum principle to showthat at x 0 ∈ T the extension ũ is smooth and has ∆ũ(x 0 ) = 0. For some r > 0 we haveB r (x 0 ) ⊂ U + ∪ T ∪ U − . We set B r := B r (x 0 ) and B r± := B r ∩ {±x n > 0}. Without lossof generality, x 0 = 0. Set ϕ: ∂B r → R equal to ũ. Extend these boundary values by thePoisson formula to a harmonic function h(x) := R2 −|x|∫ 2nω nR n∂B rϕ(y)|x−y| n dS y . We claim h is odd,that is, the two functions h(x) and ˜h(x) := −h(σ(x)) agree on B r . But h and ˜h are C 2 ,harmonic by (30), and have the same boundary values. Hence they agree by the maximumprinciple.In particular, h(x) = 0 for all x ∈ T ∩ B r . But this means that the two functions h(x) andũ(x) agree on ∂B r + . But both functions are harmonic. Hence once again they agree on theinterior B r + . The same holds for Br − , while on T the functions agree anyway. But h is C 2and harmonic on B r , and so is ũ.□Let us show that Schwarz reflection also applies to D. By Lemma 13, the Cayley mapη : H → D \ {1} is a conformal diffeomorphism. Now suppose β is some arc containedin the unit circle ∂D and u: D → R is harmonic with zero boundary values on β. By arotation we can assume that 1 ∉ β. By Lemma 10(i) the map v := u ◦ η is harmonic;moreover, it has zero boundary values on the arc η −1 (β) contained in the real axis. Bystandard Schwarz reflection v extends across this arc in an odd way to some harmonicmapping ṽ. Writing ũ := ṽ ◦ η −1 we obtain the desired extension accross β.Let us apply this to the Plateau problem. In general, strict monotonicity of a sequencecan be lost when passing to a limit. Nevertheless we can show that h ∈ C:Theorem 28. The limiting map h of Thm. 24 has a restriction to ∂D which is injectiveand hence topological.Proof. To show that h is injective on ∂D assume on the contrary there is an open arcβ ⊂ ∂D which maps to a point. For simplicity we assume this point is 0. By the Schwarzreflection principle we can reflect a neighbourhood of β in D across β to extend eachcomponent of h harmonically. The extended map is C 2 on an open set O which includesthe arc β.

122 K. Grosse-Brauckmann: Minimal surfaces, SS 12As asserted in Theorem 25 the map h is conformal on D. We claim that the extensionby reflection is also conformal, including the arc β. Since h is C 1 along β, the conformalityrelations |h x | = |h y | and 〈h x , h y 〉 = 0 extend continuously from D to β. Moreover,conformality is clearly preserved by reflection and thus holds on all of O.Let us now derive the desired contradiction from conformality along β. If β is non-emptythen, in polar coordinates, |h ϕ | = 0 and, by conformality, |h r | = 0 on β. Thus h ′ (z) = 0on β. It follows that the power series of h ′ vanishes, and since h is analytic, it is equal to aconstant at O and hence on all of D. But h is continuous on ∂D, thus also constant there.This contradicts the 3-point condition for h, and so β must be empty.□The previous theorem, together with Theorems 24 and 25, completes the proof of thePlateau problem as stated in Theorem 4.3.5. Summary of the proof. The proof employs a so-called direct method, meaning weminimize a functional by taking the limit of a minimizing sequence. It is hopeless to dothis for area, so we work with energy instead.Consider a sequence f k ∈ C tending to the infimum of energy over the class C. Supposefor a moment that it was convergent to some mapping f ∈ C. As a minimum of energy, fis harmonic. Moreover, due to the free boundary condition which constitutes the class C,the mapping f is also conformal. Thus f is minimal.In order to prove the desired convergence f k → f, we need to alter the f k . Using Poisson’sformula componentwise we may replace them with harmonic maps. Thereby we will notincrease energy. The gain lies in the fact that a sequence of harmonic maps converges (withall derivatives) to some map f. But in general, f will not have the same boundary valuesany more. To guarantee the correct boundary condition, we need to require the 3-pointcondition for each f k .4. Properties of the Plateau solution (2006)4.1. Riemann mapping theorem. Let us cite the Riemann mapping theorem (see, forinstance, [A], p.229ff):Let U ⊂ C be a non-empty domain which is simply connected and not equal to C. Thenthere exists a biholomorphic function f : D → U.By definition, a biholomorphic function is holomorphic and bijective with holomorphicinverse.

iii 4.1 – Stand: July 27, 2012 123Remarks. 1. For the excluded case U = C, the map f −1 contradicts Liouville’s theorem.2. The theorem comprises the nontrivial topological statement that there exists a homeomorphismat all. The mapping f is not unique. As for the Plateau problem, it becomesunique once three points or f(z 0 ) = 0 and f ′ (z 0 ) > 0 are prescribed.Our proof of the Plateau problem was given for arbitrary dimensions n ≥ 2. In particular itworks for n = 2. So consider a closed Jordan curve Γ ⊂ R 2 . By the Jordan curve theoremthere is a bounded domain U such that ∂U = Γ. Let us now state the special case of theRiemann mapping theorem which is a corollary to the Plateau problem:Theorem 29. Let U be a simply connected domain, bounded by a closed Jordan curveΓ = ∂U which is piecewise C 1 . Then there exists a biholomorphic map f : D → U whoseboundary restriction f| ∂D parameterizes Γ homeomorphically.Note that for n ≥ 3 the Plateau solution is in general not injective.Proof. The solution to the Plateau problem gives us a harmonic and conformal map f : D →C which continuously extends to the boundary ∂D. By Prop. 9 the map f is holomorphicor anti-holomorphic. Thus f becomes holomorphic after composition with a reflection of D,if necessary. It remains to show that f takes image in U and is biholomorphic.Since f is non-constant, the zero set Z ⊂ D of the holomorphic map f ′ is discrete, and sois its image set f(Z). Let us remove f(Z) and the curve Γ from the image, and call theremaining open set Ω := C \ {f(Z) ∪ Γ}. We consider the subsets X k ⊂ Ω in the image,consisting of those points which have exactly k ∈ N 0 preimages under f. Note that f ′ ≠ 0at all points in f −1 (X k ). Thus the inverse mapping theorem implies that, given a pointq ∈ X k for k ≥ 1, the equation f(z) = q has the same number k of preimages for all q insome neighbourhood of p. This proves X k is open for k ≥ 1. But X 0 is also open: Theset f(D) is the continuous image of a closed set. Thus its complement C \ f(D) is open;clearly this set coincides with X 0 .Thus Ω is the disjoint union of the open sets X k for k ≥ 0. This implies that each nonemptyX k must be the union of connected components of Ω. Equivalently phrased, the numberof preimages of f is constant on each connected component of Ω.Now Γ is a Jordan curve. Hence the Jordan curve theorem gives that C \ Γ has twoconnected components. Moreover, f(Z) is discrete and thus C \ {Γ ∪ f(Z)} = Ω again hastwo components.We claim that (i) C \ U has k = 0 preimages while (ii) points in U \ f(Z) have preciselyk = 1 preimage.

124 K. Grosse-Brauckmann: Minimal surfaces, SS 12The image of D under the continuous map f is compact. Thus at infinity of C the numberof preimages is 0. This shows that on the connected component containing a neighbourhoodof infinity, there can only be 0 preimages, thereby proving (i).Before we show (ii) let us first prove that only ∂D maps to Γ. Indeed, if Γ had a preimagez 0 ∈ D \ Z, the implicit function theorem applied at z 0 would give preimages for a smallneighbourhood of f(z 0 ) ∈ Γ. Since there are no such preimages in the complement of U,this is impossible. From biholomorphicity, see below, it will follow that points of Z cannotmap to Γ.The assertion (ii) is a topological fact, and can be best formulated in terms of a mappingdegree (see, e.g., [K, p.50/51]). We give an elementary argument. From the facts provedabove it follows that each point in U \ f(Z) has the same number k ≥ 0 of preimages.The nonempty open set f(D) cannot intersect C \ U, and hence is contained in U. Thuseach point in U \ F (Z) has k ≥ 1 preimages. It remains to rule out k ≥ 2. (The examplesz k : D → D show that each k ∈ N could occur).We pick a set of four distinct points F in Γ. Assume they are indexed in anticlockwiseorder, and consider two curves c 1 , c 2 in U, whose endpoints connect every other point of F .Since U is connected and open we can in fact assume that c 1 , c 2 are in U \ f(Z) except fortheir endpoints, and that they intersect in precisely one point p ∈ U \ f(Z).By assumption p has k ≥ 1 preimages, and the preimage f −1 (c 1 ) consists of k paths, whichmust be pairwise disjoint except for their common endpoints; similarly so for f −1 (c 2 ). Thusthe preimage set f −1 (c 1 ∪ c 2 ) consists of 2k paths. Since the endpoints of c 1 and c 2 occurin alternating order in F , the k preimage paths of c 1 must cross the k preimage paths ofc 2 . That is, the entire preimage constitutes a grid of two times k paths. But these pathshave k 2 distinct crossing points, implying that the preimage of p consists of k 2 preimages.Thus k 2 = k which proves k = 1.Let us finally show Z = ∅ which implies that f : D → U is biholomorphic. Suppose thatf ′ (z 0 ) = 0 for some z 0 ∈ D; for simplicity assume z 0 = 0. Then Taylor’s formula givesf(z) − f(0) = az k + O(z) k+1 with a ≠ 0 and k ≥ 2. But in a small neighbourhood of 0 thisequation has k distinct roots, meaning there would be k ≥ 2 preimages in a neighbourhoodof z 0 . This contradicts the previous argument.□Remarks. 1. The biholomorphic map f becomes unique provided we impose a three-pointcondition. Indeed, if not, we had a conformal map from D to D which fixes three points.As pointed out in the final remark of Sect. 2.4 (without proof) this Möbius transformationmust be the identity.2. We can call two open subsets of C, or, more generally, of an open manifold, conformallyequivalent, if there is a conformal diffeomorphism between them. The Riemann mapping

iii 4.2 – Stand: July 27, 2012 125theorem says that all simply connected proper subsets of C belong to the same equivalenceclass. The only further simply connected spaces which are not conformally equivalent turnout to be C and S 2 . For multiply connected sets or spaces, there is a manifold of differentclasses, however. The problem to find a conformal map from a given set to some standardrepresentative set or manifold is called the uniformization problem, see [FK].34. Lecture, Tuesday 25.4.064.2. Henneberg’s surface is non-orientable and has branch points. We discuss anexample of a minimal surface with some pathological behaviour. The surface⎛⎞cosh 2x cos 2yf : R 2 → R 3 ⎜, f(x, y) = ⎝− sinh x sin y − 1 ⎟sinh 3x sin 3y3⎠− sinh x cos y + 1 sinh 3x cos 3y 3was discovered by Henneberg in 1875.The following calculations would simplify a lot if they were done in terms of the Weierstrassrepresention.The surface is conformally parameterized. To see this, calculate⎛⎞ ⎛⎞2 sinh 2x cos 2y−2 cosh 2x sin 2y⎜⎟ ⎜⎟f x = ⎝− cosh x sin y − cosh 3x sin 3y ⎠ , f y = ⎝− sinh x cos y − sinh 3x cos 3y⎠ .− cosh x cos y + cosh 3x cos 3ysinh x sin y − sinh 3x sin 3yThus, making use of the addition theorems, we obtain〈f x , f y 〉 = − 4 sinh 2x cosh 2x sin 2x cos 2y+ (sinh x cosh 3x + sinh 3x cosh x)(sin 3y cos y + sin y cos 3y)= − sinh 4x sin 4y + sinh 4x sin 4y = 0.A longer calculation, making use of the addition theorems for the hyperbolic trigonometricfunctions, givesso that f is conformal. Let|f x (x, y)| 2 = |f y (x, y)| 2 = 4 cosh 2 x(sinh 2 2x + sin 2 2y),S := {(0, y) : y = 1 kπ, k ∈ Z}2be the singular set of the so-called branch points where f x , f y = 0. Note that f is animmersion precisely on R 2 \ S.

126 K. Grosse-Brauckmann: Minimal surfaces, SS 12Now∆f = 4 cosh 2x cos 2y − sinh x sin y − 3 sinh 3x sin 3y − sinh x cos y + 3 sinh 3x cos 3y− 4 cosh 2x cos 2y + sinh x sin y + 3 sinh 3x sin 3y + sinh x cos y − 3 sinh 3x cos 3y = 0.We conclude f is a conformal harmonic map on R 2 , hence minimal on R 2 \ S.The Gauss map of f is⎛ ⎞ν(x, y) = 1 sinh x⎜ ⎟⎝ sin y ⎠ ,cosh x− cos ywhich means it extends continuously to the singular points (x, y) ∈ S. Indeed, clearly|ν| 2 = 1 andcosh x 〈ν, f x 〉as well as= 2 sinh 2x sinh x cos 2y − cosh x(sin 2 y − cos 2 y) − cosh 3x(sin 3y sin y + cos 3y cos y)= cos 2y(2 sinh 2x sinh x + cosh x − cosh 3x)= 1 2 cos 2y( (e 2x − e −2x )(e x − e −x ) + e x + e −x − e 3x − e −3x) = 0− cosh x 〈ν, f y 〉= 2 cosh 2x sinh x sin 2y + 2 sinh x cos y sin y + sinh 3x(cos 3y sin y − sin 3y cos y)= sin 2y(2 cosh 2x sinh x + sinh x − sinh 3x)= 1 2 sin 2y( (e 2x + e −2x )(e x − e −x ) + e x − e −x − e 3x + e −3x) = 0.We would like to discuss two properties of Henneberg’s surface. First, we study the localbehaviour at the branch point (0, 0) ∈ S. Using expansions such as cosh t = 1+ 1 2 t2 +O(t 4 )and cos t = 1 − 1 2 t2 + O(t 4 ) we obtain⎛ (1 +12⎜(2x)2 + O(x 4 ) )( 1 − 1 2 (2y)2 + O(y 4 ) ) ⎞f(x, y) = ⎝−xy − 13x3y + O((x, ⎟3 y)4 )⎠−(x + 1 6 x3 )(1 − 1 2 y2 ) + 1(3x + 1 3 6 (3x)3 )(1 − 1 2 9y2 ) + O((x, y) 4 )⎛⎞1 + 2x 2 − 2y 2⎜⎟= ⎝ −4xy ⎠ + O((x, y) 4 )−4xy 2 + 4 3 x3To anlayse the mapping more closely we use polar coordinates. Then⎛⎞ ⎛⎞1 + 2r 2 (cos 2 ϕ − sin 2 ϕ)1 + 2r 2 cos 2ϕ⎜f(r cos ϕ, r sin ϕ) = ⎝ −4r 2 ⎟cos ϕ sin ϕ ⎠ + O(r 4 ⎜) = ⎝ −2r 2 ⎟sin 2ϕ ⎠ + O(r 4 ).−4r 3 (cos ϕ sin 2 ϕ − 1 3 cos3 ϕ)+ 4 3 r3 cos 3ϕ

iii 4.2 – Stand: July 27, 2012 127Now consider the image of a circle ϕ ∈ [0, 2π] with small radius r > 0. The first twocoordinates f 1 , f 2 are then travelled around twice: ϕ and ϕ + π have images with thesame argument. But under ϕ ↦→ ϕ + π the third coordinate f 3 undergoes a sign change,sin 3ϕ = − sin 3(ϕ + π), and so the third coordinate distinguishes the two images. Nowsin 3ϕ is positive on three subintervals of [0, 2π] and negative on the complementary threeintervals. Thus the surface self-intersects in three curves which emanate from the branchpoint.Remark. Our present discussion parallels the discussion of the Enneper surface at infinity,see II,2.4.The second property we would like to discuss is that the Henneberg surface factors to anon-orientable surface. To see that, note the following symmetries. Clearly, f(x, y + 2π) =f(x, y). In fact, f has more symmetry:(31) f(−x, y + π) = f(x, y).So if we consider f on the strip (x, y) ∈ R × [0, π) then all other values will be repeated.But (31) means that the boundaries R × {0} and R × {π} are glued together in oppositeorientation. Thus the result of the gluing is a Möbius strip.Under the symmetry (31), the Gauss map undergoes a sign change ν(x, y) = −ν(−x, y+π),and so the symmetry (31) is orientation reversing. This presents us an example of a minimalsurface which does not admit a continuous (global) Gauss map; the Gauss map could bedefined, however, with images in RP 2 .Using higher level terminology, we would say that Henneberg’s surface is a minimal immersionof a Möbius strip. Here, the Möbius strip is the abstract manifold given as R 2modulo the equivalence relation(x, y) ∼ (u, v) :⇐⇒ ∃k ∈ Z : u = (−1) k x, v = y + kπ.The Möbius strip is non-orientable, but its double cover, represented by R × [0, 2π), isorientable. The parameterization f covers the immersed Möbius strip infinitely often.Remarks. 1. The symmetry (31) shows that the set S maps to two different points, namelyf(0, kπ) = (1, 0, 0) while f(0, π + kπ) = (−1, 0, 0) where k ∈ Z.22. Henneberg’s surface is algebraic, see [N], p.141.3. A geometric way to introduce Henneberg’s surface is via Björling’s problem: Consider ananalytic curve c(t) and any unit vector field ν(t) ⊥ c ′ (t). Is there a minimal surface whichcontains the curve c and has surface normal ν along this curve? Schwarz solved this problem in theaffirmative in terms of an integral representation, see [N, p.136ff] or [DHKW, I.3.4]. Henneberg’ssurface contains Neil’s parabola c(t) = (t 2 , t 3 ). Indeed,x ↦→ f(x, 0) = ( cosh 2x, 0, − sinh x + 1 3 sinh 3x) = ( 1 + 2 sinh 2 x, 0, 4 3 sinh3 x )

128 K. Grosse-Brauckmann: Minimal surfaces, SS 12so that the plane {y = 0} intersects Henneberg’s surface in the translated parabola 2(f 1 − 1) 3 =3(f 3 ) 2 .4. Henneberg’s surface seems to be the only complete stable minimal surface in a quotient ofR 3 which has total branch order 2 or less (a theorem by Antonio Ros, presumably). Its totalcurvature is ∫ K dA = −2π.References. [N], p. 140–142; [DHKW], p. 159–16935. Lecture, Wednesday 26.4.064.3. Branch points and generalized minimal surfaces. The solution of the Plateauproblem is a conformal harmonic map f : D → R n . Suppose this map is an immersion,that is, it represents a surface in the sense of differential geometry. Then for n = 3, thissurface is minimal by II(2).Definition. Let f : U → R n be differentiable. Points p ∈ U with df p = 0 are called branchpoints.Examples. 1. Henneberg’s surface has branch points.2. Let k ≥ 2 and h: C → R with dh 0 = 0. Thenz = (x, y) ↦→ ( z k , h(z) ) = ( Re z k , Im z k , h(z) ) ,has a branch point at 0. Note that the projection to the xy-plane is a k to 1 map exceptat the origin (which has one preimage).3. For any k ≥ 2, the mappingz = (x, y) ↦→ (z k , 0) = (Re z k , Im z k , 0),has a branch point at 0. This point is not visible in the image, which is all of the plane.Each point of the plane has k preimages (only the origin has one).It is useful to distinguish the cases of examples 2 and 3. If a smooth surface minus a pointhas a parameterization f which is k to 1, and extends to the singular point with df p = 0,it is called a false branch point. Otherwise the branch point is true.A local analysis can be used to yield two important facts about branch points, which wehave verified in the case of Henneberg’s example:1. The normal extends smoothly to a branch point, that is, a branch point has a welldefinedtangent plane.2. Suppose the tangent plane at the branch point agrees with the xy-plane. Let f beconformal and consider the sign of f 3 as a function of D. Then unless f is a plane thereis k ∈ N such that there are 2(k + 2) curves in D whose images have f 3 = 0; they meet atequal angles in 0.

iii 4.4 – Stand: July 27, 2012 129A conformal map f : D → R n has either rank 0 or 2. Hence after removing the branchpoints from the domain of f we are left with an immersion.As asserted in Thm. 19 a harmonic map f is analytic. Thus its derivative f ′ is also analytic.If its zero set is not discrete then f ′ vanishes identically and then f is constant. Otherwise:Theorem 30. Let f : D → R n be a non-constant conformal harmonic map. The set of itsbranch points {p ∈ D : f x (p) = f y (p) = 0} is isolated in D, and thus consists of at mostcountably many points.In view of this property, the following notation becomes meaningful:Definition. Let U ⊂ R 2 be a nonempty domain. A mapping f ∈ C 1 (U, R n ) is calleda branched immersion if f is an immersion except at a discrete set. Branched minimalimmersions are also called generalized minimal surfaces.Many theorems which hold for minimal surfaces also hold for generalized minimal surfaces.For example, Cor.II.9 generalizes to:Theorem 31. A generalized complete minimal surface cannot be compact.Proof. Suppose M is compact. Then each coordinate function attains an interior maximum.But each coordinate function is harmonic. Thus by the maximum principle f must beconstant, contradicting the discreteness of the set of points where f is not an immersion. □Let us mention two general properties of branched minimal immersions. The Gauss mapsextends through the branch points, and there is a local expansion in a neighbourhood of abranch point which analyses the local behaviour.The main reason to study branched immersions is that they come up naturally in constructionslike the Plateau problem. They are not objects of primary interest.4.4. Branch points in Plateau solutions. For area minimizers in 3-space, interiorbranch points are absent:Theorem 32 (Osserman 1970). Let f ∈ C 1 (D, R 3 ) ∩ C 0 (D, R 3 ) be a branched immersionwhose boundary parameterizes a piecewise C 1 Jordan curve Γ homeomorphically. If fminimizes area over C 1 , then f| D is an immersion without branch points.Papers of Gulliver, Alt, and Gulliver-Osserman-Royden also contributed to the proof(cf. [DHKW, p.279/280]).In particular, a Plateau solution for R 3 does not have any branch points in D. We will notgive the formal proof of Osserman’s theorem. Nethertheless we want to use this property

130 K. Grosse-Brauckmann: Minimal surfaces, SS 12later and so we want to present at least an indication why Osserman’s theorem is true.We will present the main idea for a somewhat special branch point. See [N, p.339-346] fordetails.We first show that minima of area do not have kinks. Let D + := {(x, y) ∈ D : y > 0} andD − := {(x, y) ∈ D : y < 0}.Lemma 33. Suppose f ∈ C 0 (D, R 3 ) with restrictions to the half disks f ± := f| D ± whichare C 1 . Suppose moreover that at 0 ∈ D the one-sided partial derivatives exist but differ,∂ y f + (0) ≠ ∂ y f − (0). Then f cannot minimize area among maps with the boundary valuesf| ∂D .Idea of proof. In case the two mappings f ± are linear we can modify f continuously in theneighbourhood of the x-axis. We image of the modified map looks at follows: The edge eis replaced locally with a rectangular face (shortcutting the image wedge). Along its twoshort edges, we glue two triangles, whose vertex (opposite the short edge) is on e. Thisidea will work if the rectangle is thin enough.For the general case, the same idea can be applied to the linearization. The error is oflower order and thus the result for the linear case extends.□The next statement gives us a map which we will use to compose f with. We first introducenotation.Let X be the interval on the x-axis X := {(x, 0) ∈ D : − 1 ≤ x ≤ 1 }. Let Y be the open2 2interval on the y-axis, Y := {(0, y) : −1 < y < 1}, and Y ± := Y ∩ D ± .Lemma 34. There exists a continuous map σ : D\X → D\Y with the following properties:(i) σ| ∂D = id,(ii) σ maps D \ {X ∪ Y } diffeomorphically to D \ Y .(iii) σ maps the sets Y ± to ±i. It has one-sided limits for each (x, 0) ∈ X, namely⎧⎨( )0, min{1 + 2x, 1 − 2x}(32) lim σ(s, t) = when all t > 0,( )(s,t)→(x,0) ⎩ 0, − min{1 + 2x, 1 − 2x} when all t < 0.(iv) Moreover, (σ 1 , σ 2 )(−x, y) = (−σ 1 , σ 2 )(x, y), that is, σ respects reflection symmetryabout the y-axis,Proof. To begin with, we define σ only on the subset (D \ X) ∩ {x ≥ 0} by a compositionof two maps, σ = τ ◦ ρ.First, let us consider mappings which maps rays through the origin linearly onto itsself:re iϕ ↦→ l(ϕ)re iϕ . We want to choose a function l(ϕ) such that the points on the unit circle

iii 4.4 – Stand: July 27, 2012 131with x > 0 are mapped to a circle of radius 1 with midpoint 1; the points ±i will map tothe origin. This means1 ! = (l cos ϕ − 1) 2 + (l sin ϕ) 2 = l 2 cos 2 ϕ − 2l cos ϕ + 1 + l 2 sin 2 ϕ 2 = 1 + l(l − 2 cos ϕ),which yields l(ϕ) = 2 cos ϕ. Thus we define for re iϕ ∈ D ∩ {x ≥ 0}ρ(re iϕ ) = 2r cos ϕ e iϕ for − π 2 ≤ ϕ ≤ π 2 , 0 ≤ r ≤ 1.The range of ρ is the closed disk D 1 (1, 0) of radius 1 with midpoint 1. Let us now removeX from the domain of ρ. This removes the interval I := {(x, 0) : 0 ≤ x ≤ 1} from its rangeD 1 (1, 0).Second, the mapping τ is defined on D 1 (1, 0) \ I by settingThe map τ is onto D ∩ {x > 0}.τ(1 + re iϑ ) := re iϑ/2 for − π < ϑ < π, 0 < r ≤ 1.We now extend the maps ρ and τ by requiring the symmetry property (iv) for each. Thisdefines a unique extension of σ = ρ ◦ τ to D \ X which again is subject to (iv). Note thatthe two maps agree on the sets Y ± common to their domains, which they map to ±i. Thiscompletes the definition of σ.Let us now check the properties claimed for σ. By construction, σ is continuous on D \ X.The diffeomorphism property (ii) is immediate from the fact that ρ and τ are diffeomorphismson the appropriate sets. The limits (32) can be explicitely computed. In order toshow (i), consider how the right hemicircle gets mapped under ρ:ρ(e iϕ ) = 2 cos ϕe iϕ = 2 cos 2 ϕ + 2i cos ϕ sin ϕ = 1 + cos 2 ϕ − sin 2 ϕ + 2i cos ϕ sin ϕ= 1 + cos 2ϕ + i sin 2ϕ = 1 + e 2iϕThus as desired τ(ρ(e iϕ )) = e iϕ .□36. Lecture, Tuesday 2.5.06Consider f with a branch point at 0. Take coordinates such that the normal at 0 is the z-vector. It is a fact that the sign of the z- coordinate of f alternates on 2k wedges. Supposein particular that the z-component vanishes along the y-axis of D. The main advantage ofthe parameterization f ◦ σ then is that across the arc X two regions with the same signof the z-component are adjacent (sketch!). This produces a kink in the parameterizationf ◦ σ.Let us state this more formally:

132 K. Grosse-Brauckmann: Minimal surfaces, SS 12Proposition 35. Suppose that f : D → R 3 has a branch point at 0, and that the y-axismaps to a self-intersection line, namely,(33) f(0, y) = f(0, −y) for all − 1 ≤ y ≤ 1.Moreover assume the branch point is true, so that(34) ∂ x f(0, y) > 0 for all 0 < |y| < 1Then f ◦ σ has the following properties:(i) It is continuous on D.(ii) f| ∂D = f ◦ σ| ∂D .(iii) f ◦ σ has a kink along the set X, that is for p ∈ X \ {0} the derivative ∂ y (f ◦ σ)(p)has different one-sided limits.For a general branch point, it must be shown that (33) and (34) hold after a suitablechange of parameters.Idea of proof. (i) Continuity of f ◦σ needs only be checked along X. But the two one-sidedlimits (32) of σ are mapped under f to the same points, as prescribed by (33).(ii) follows from Lemma 34(i).To prove (iii) it must be shown that the one-sided derivatives of σ at X do not vanish.□How does the parameterization f ◦ σ look like? Considerf(D \ Y ) = f ( D ∩ {x > 0} ) ∪ f ( D ∩ {x < 0} ) =: M 1 ∪ M 2 ,as well as the self-intersection lineL := f(Y ) = f ( Y + ∪ {0} ) = f ( Y − ∪ {0} ) .In the original parameterization f there are two preimages of L: one glues together oneside of M 1 to one side of M 2 , while the other one glues the remaining two sides together.In f ◦ σ, there are also two preimages of L; however, the first (which is X ∩ {x > 0})glues the boundaries of M 1 to oneanother, while the second (which is X ∩ {x < 0}) gluesthe two sides of M 2 to oneanother. The two resulting disks (f ◦ σ) ( D ∩ {x > 0} ) and(f ◦ σ) ( D ∩ {x < 0} ) are then glued together by the image of the arc Y , mapping to thesingle point f ◦ σ(Y ) = f(i). Note in particular that the branch point f(0) has the twodistinct preimages (± 1 , 0) under f ◦ σ.2Let us finally indicate how Osserman’s theorem can be proved. Either a branch point istrue. Then the above arguments can be adapted to obtain a parameterization f ◦ σ whichhas the same area and boundary values as f, but has an interior kink. Thus f ◦ σ cannotbe area minimizing and f itself cannot be area minimizing either. Else a branch point is

iii 4.5 – Stand: July 27, 2012 133false. Then by analyticity f(∂D) also has multiple preimages, contradicting the fact thatf maps the boundary homeomorphically.Remark. For smooth boundary points, we can also define branch points at the boundary.It is known they do not show up for analytic Jordan curves, but up to today unclear ifthey arise for smooth curves.4.5. Uniqueness and embeddedness results. The following issues for Plateau solutionsdeserve special interest: 1. Uniqueness, 2. Embeddedness, 3. Regularity at the boundary.By Osserman’s theorem, a Plateau solution f : D → R 3 is an immersion. Thus the uniquenessresults of Part II implies that the Dirichlet problem for minimal graphs can be solvedover convex domains:Theorem 36 (Radó 1930). Let Γ be a Jordan curve which admits a 1-1 projection π to aplane E such that π(Γ) bounds a convex domain U ⊂ E. Then the solution to the Plateauproblem for Γ is contained in the cylinder U × R and graph over U. In particular, it isembedded and unique.Proof. By Thm. II.43 the Plateau solution M is contained in the convex hull of Γ; inparticular it is contained in U ×R. For convex U, Thm. II.41(ii) says that there is at mostone such minimal surface which is graph. Thus M must be this unique graph. □There is a direct and beautiful proof of this result, bypassing the use of Osserman’s theorem,which depends on a local expansion of a surface at a branch point; it is contained in manybooks, see for instance [DHKW, pp.272/73].37. Lecture, Wednesday 3.5.06Example. We want to show that convexity of the boundary projection is essential for thePlateau solution to be graph. To see this we let h ∈ R and define a quadrilateral Γ = Γ(h)by its four verticesp 1 := (0, 1, 0), p 2 := (1, 0, 0), p 3 := (0, −1, 0), p 4 := ( 1 , 0, h).2Note that p 1 , p 2 , p 3 lie in the xy-plane. For the projection π to the xy-plane we haveπ ( Γ(h) ) = Γ(0).Let U be the compact component of R 2 , bounded by Γ(0). It can be shown that the Plateausolution M = M(h) has a projection π(M) ⊄ U whenever h ≠ 0 (see Nitsche §410). Thiswould be easy to show if we knew that the solution has a tangent plane at the boundarypoint p 4 : Then the solution will project to points (x, 0) in the xy-plane with x < 1 . In the2

134 K. Grosse-Brauckmann: Minimal surfaces, SS 12problems we will use a catenoid piece to see that at least for some h > 0 the solution isnot graph.Let us now collect further results without proofs.Embeddedness is known more generally:Theorem (Meeks-Yau 1982). Suppose that K ⊂ R 3 is compact and C 2 such that a piecewiseC 1 -Jordan curve Γ is contained in ∂K. Let M ⊂ K be any area minimizer of disk-typewhich bounds Γ. Then M is embedded, provided K is convex or, more generally, K is meanconvex.Here, mean convex means that the mean curvature of M is positive with respect to theinterior normal.Example. Any C 1 -Jordan curve in the 2-sphere S 2 bounds an embedded minimal disk.It is worth knowning that embeddedness can always be achieved by allowing for possiblyhigher topological type:Theorem (Hardt-Simon 1979). Each Jordan curve Γ of class C 2embedded orientable minimal surface.bounds at least oneThus we can achieve embedded or disk-type solutions of the Plateau problem, but in generalnot both together!Example. The clover knot bounds a disk with self-intersections or an embedded Möbiusstrip.In general, the Plateau problem can have many solutions; there are examples which bounda continuum of solutions. Besides the situation of Radó’s Theorem, there are cases whenuniqueness of the solution to Plateau’s problem is implied. Let us mention but the mostfamous one:Theorem (Nitsche 1972). If Γ ⊂ R 3 is a real analytic Jordan curve with total curvature∫k(t) dt < 4π then Γ bounds a unique minimal surface of disk-type.ΓRemember that if a space curve c has arc-length parameterization then its curvature isk(t) = |c ′′ (t)|. For example, a circle has total curvature 2π, while its double cover has 4π.A consequence of Nitsche’s theorem is that the Plateau solution is free of branch points.A more recent extension of Nitsche’s theorem is:Theorem (Ekholm/White/Wienholtz 2002). Let Γ ⊂ R 3 is a Jordan curve which is piecewiseC 2 and has total cuvature T (Γ) < 4π. Then any branched minimal surface M withboundary Γ has an embedded interior and no interior branch points.

iii 5.1 – Stand: July 27, 2012 135For a piecewise C 2 -curve, the total curvature is the sum T (Γ) = ∫ Γ k(t) dt + ∑ α i , whereα i are the exterior angles at the singular points of Γ.Let us finally comment on boundary regularity. The Plateau solution is just continuous atthe boundary. If the bounding Jordan curve is better, so is the Plateau solution:Theorem 37 (Hildebrandt 1969). Suppose Γ ⊂ R 3 is a Jordan curve which is C k+1 , k ≥ 0,then the Plateau solution f is differentiable up to the boundary, f ∈ C k (D, R 3 ).The actual statement of the theorem employs Hölder continuity, and shows the solutionis as good as its boundary values: If Γ is C k,α then f ∈ C k,α (D, R 3 ). In any case, thesestatements mean that f can locally be extended accross ∂D in the given class. For thespecial case of straight boundary arcs we will establish this regularity in the next section.5. Existence of complete periodic minimal surfaces (2006)We will obtain periodic minimal surfaces by taking Plateau solutions for polygons whichwe extend by reflection across the boundary arcs.5.1. The reflection principle. Many of the examples we discussed contain straight lines:For instance, the helicoid is ruled by them, and the Enneper surface contains two straightlines. Not only contain the surfaces lines, but these lines are arcs of symmetry: When thesurface is rotated by 180 degrees about these lines, it becomes congruent to itself. This isevident from the construction of the helicoid or from the polar representation of Enneper’ssurface.We want to formulate this as a principle, which we will use to extend Plateau solutions.Remember that Plateau solutions are smooth in the interior, but all we have established atthe boundary is that they are continuous. This is precisely what the following statementassumes.As before we denote the half-disks with D ± := D ∩ {±y > 0} and their common boundaryon the x-axis with I := {(x, 0) : −1 < x < 1}.Theorem 38. Let f ∈ C 0 (D + ∪ I, R 3 ) ∩ C ∞ (D + , R 3 ) be a conformally parameterizedminimal surface on D + , such that f(I) is contained in a straight line L. Let σ L : R 3 → R 3be reflection about the line L, that is, 180 ◦ rotation about L. Then the extension⎧⎨ f(x, y) for (x, y) ∈ D(35) ˜f(x, + ∪ I,y) := ( )⎩σ L f(x, −y) for (x, y) ∈ D − ,is conformal and minimal on D, in particular it is smooth, ˜f ∈ C ∞ (D, R 3 ).

136 K. Grosse-Brauckmann: Minimal surfaces, SS 12If we were to assume the boundary regularity granted by Thm. 37 the proof would becomeeasy: At a boundary point we could then represent the surface as a C 2 -graph, say with0-boundary values. As shown in Problem 32 c) such a graph extends by reflection. We willnot make use of Thm. 37, however, as it is not too hard to assert the necessary boundaryregularity directly for the present case.Proof. For simplicity, we assume L coincides with the z-axis, so thatThen the second line of (35) becomesσ L (x 1 , x 2 , x 3 ) = (−x 1 , −x 2 , x 3 ).(36) ˜f 1 (x, y) = −f 1 (x, −y), ˜f 2 (x, y) = −f 2 (x, −y), ˜f 3 (x, y) = f 3 (x, −y)for (x, y) ∈ D − .Now f 1 and f 2 are harmonic functions which vanish on I. Thus we can apply the SchwarzReflection principle, Thm. 27, to give that the odd reflections ˜f 1 , ˜f 2 are functions whichare smooth on all of D.We cannot argue the same way for the even reflection ˜f 3 with its nonconstant values f 3along I. The trick here is to consider its derivative ∂ y ˜f 3 instead, which is odd in y and doesallow for Schwarz Reflection. We first need to prove that ∂ y f 3 exists on I and vanishes.Clearly this will only be true for a conformal parameterization: On I then ∂ x f is vertical,and so ∂ y f is horizontal.On I we have f 1 , f 2 = 0 and so ∂ x f 1 = ∂ x f 2 = 0. Using this, we obtain for any sequencez n ∈ D + which converges to z ∈ I(37) ∂ x f 3 (z n ) ∂ y f 3 (z n ) f conf.= −∂ x f 1 (z n ) ∂ y f 1 (z n ) − ∂ x f 2 (z n ) ∂ y f 2 (z n ) → 0;indeed the right hand side tends to 0 as ∂ y f i (z n ) is the derivative of a harmonic function,hence bounded, while ∂ x f i (z n ) → 0 for i = 1, 2. The last formula means that after passingto a subsequence, if necessary, at least one of the two subsequences ∂ x f 3 (z n ) or ∂ y f 3 (z n )will be convergent and tends to 0.Let us now identify this as ∂ y f 3 (z n ). The other conformality relation gives(38)|∂ x f 3 (z n )| 2 − |∂ y f 3 (z n )| 2 = −|∂ x f 1 (z n )| 2 + |∂ y f 1 (z n )| 2 − |∂ x f 2 (z n )| 2 + |∂ y f 2 (z n )| 2→ |∂ y f 1 (z)| 2 + |∂ y f 2 (z)| 2 ≥ 0.So combining (37) with (38) we obtain ∂ y f 3 (z n ) → 0, meaning∂ y f 3 (z) = 0 for z ∈ I.

iii 5.2 – Stand: July 27, 2012 137As ˜f 3 (x, y) is even in y, its derivative ∂ y ˜f 3 (x, y) is odd. Hence we can consider ∂ y ˜f 3 asthe result of odd reflection of the harmonic function ∂ y f 3 on D + which vanishes on I. BySchwarz reflection ∂ y ˜f 3 is smooth on all of D.38. Lecture, Tuesday 9.5.06It remains to show that also ˜f 3 itself is smooth. The functiong(x, y) := ˜f 3 (x, y) −∫ y0∂ y ˜f 3 (x, η) dηis continuous on D and satisfies ∂ y g(x, y) = 0 on D \ I; thus g(x, y) = ˜f 3 (x, 0). But fory ≠ 0 the function x ↦→ g(x, y) is smooth and hence also ˜f 3 (., 0) ∈ C ∞ (I). This gives,finally, as desired˜f 3 (x, y) = ˜f 3 (x, 0) +∫ y0∂ y ˜f 3 (x, η) dη∈ C ∞ (D).□Similarly, a minimal surface can be extended across a plane if it is C 1 up to and includingthis plane and the normal along the plane is parallel to the plane:Theorem 39. Let f ∈ C 0 (D + ∪ I, R 3 ) ∩ C ∞ (D + , R 3 ) be a conformally parameterizedminimal surface on D + , such that f(I) is contained in a plane P and has normal ν(I)parallel to P . Let σ P : R 3 → R 3 be mirror reflection across P . Then the extension⎧⎨ f(x, y) for (x, y) ∈ D(39) ˜f(x, + ∪ I,y) := ( )⎩σ P f(x, −y) for (x, y) ∈ D − ,is conformal and minimal on D, in particular it is smooth, ˜f ∈ C ∞ (D).For the proof see [DHKW, Thm.4.8,2]. We refer to the curve f(I) in this case as an arcof planar reflection. Such arcs can be characterized by the fact that they are curvaturelines and that they are geodesics of the surface (see Lemma 16 below). We will need thisversion in IV 4.3.5.2. Plateau solutions with polygonal boundary: Extension and symmetries.Let us consider a polygon and pick a vertex of angle α. In the next theorem we successivelyuse reflections σ to extend the surface across its bounding edges. We expect the resultingsurface will close smoothly provided α = π/k with k ∈ N. It is, however, not sufficient toassume α = 2π/k: The boundary conditions on the two bounding edges of the wedge aredifferent and so repeat only after 2α. Therefore we need indeed 2α = 2π/k.

138 K. Grosse-Brauckmann: Minimal surfaces, SS 12Remark. For any rational angle α = π · p/q with p, q reduced we must reflect q times forthe surface to close. However, we necessarily create a branch point: the tangent plane willbe covered p times!We can now assert that Plateau solutions to polygons with good angles admit extensionswhich are (smooth) generalized minimal surfaces, perhaps with branch points at theboundary:Theorem 40. Suppose Γ ⊂ R 3 is a polygon without self-intersections whose (interior)angles are of the form π/k i for integer k i ≥ 2 and i = 1, . . . , #(edges). Let f : D → R 3 bethe Plateau solution.(i) For each p ∈ ∂D which maps to an edge of Γ, but not to a vertex there is a smoothlocal extension of f across ∂D which is harmonic and conformal.(ii) Suppose p ∈ ∂D maps to the i-th vertex of Γ. Then setting k := k i , there is conformalmapζ : W k := {z ∈ C : 0 < arg(z) < π k } → Dwhich extends continuously to ∂W k and maps 0 ∈ ∂W k to p ∈ ∂D with the followingproperty: At 0 the map F := f ◦ ζ has a local extension as a smooth conformal harmonicmap.For a general Jordan curve it is interesting to decide whether a minimal surface admits asmooth local extension.Proof. (i) Recall the Cayley map η : H → D \ {1} defined in Lemma 13, is a conformaldiffeomorphism. In the following we will often use the fact that composition with conformalmaps preserves conformality and harmonicity (see Lemma 10(i)). Under η, the intervalI = {(x, 0) : −1 < x < 1} ⊂ ∂H is mapped to some interval η(I) in S 1 .The inverse image f −1 (e) ⊂ ∂D is also some interval in S 1 . By composing f with a Möbiustransformation ω ∈ Aut(D), if necessary, withouth loss of generality we can assume(40) f −1 (e) = η(I);Indeed, by Prop. 12 there exists a Möbius transformation mapping the two endpoints of oneinterval to the endpoints of the other. For any choice of third points within the intervals,we obtain (40).Let us now consider the compositionF := f ◦ η ∈ C 0 (H, R 3 ) ∩ C 2 (H, R 3 ),which is conformal and harmonic in H and maps I to e.

iii 5.2 – Stand: July 27, 2012 139Now we can apply the reflection principle Thm. 38 to the restriction of F to I ∪ D + . Theresult is an extension ˜F : D → R 3 across I. As D − is still in the domain of η, we can invokeη −1 to construct(41) F ◦ η −1 : D ∪ I ∪ η(D − ) → R 3 .This map is smooth, harmonic, conformal, and extends f to D, as desired.(ii) Call e 1 , e 2 ⊂ Γ the edges incident to p. Let e now denote the union e 1 ∪ {p} ∩ e 2 .In addition to requiring (40), we may assume that f −1 (p) = η(0, 0); altogether this willimpose exactly a 3-point condition on the Möbius transformation ω ∈ Aut(D).Surprisingly, standard reflection suffices to reflect around a vertex, as shown by the followingargument. We compose f ◦η as above with the k-th power: That is, we set ζ(z) := η(z k ).Then F = f ◦ ζ conformally parameterizes the Plateau solution with the wedge W k . Themap F extends continuously to ∂W k . By construction, one bounding arc of W k ∩ D willmap to e 1 , the other to e 2 .We can apply then reflection principle k − 1 times to extend this map smoothly by oddreflection about the edges e 1 or e 2 to become defined on all of the half-disk D + witha continuous extension to its boundary on the real line I. By a final extension of thereflection principle, we obtain a smooth extension to the entire disk D.□The continuation of the Plateau solution across ∂D is smooth but may have branch points:Example. We consider a hexagon Γ ⊂ R 3 contained in the xy-plane with vertices (0, 0),(0, 1), (−1, 1), (−1, −1), (1, −1), (1, 0). As a polygon in 3-space, Γ has six right angles. ThePlateau solution f : D → R 3 for Γ is unique and parameterizes conformally the compactcomponent of the xy-plane bounded by the hexagon. Thus the surface has a 270 ◦ -angle atthe origin. When we extend by reflection using the previous theorem the resulting surfacecontain a false branch point at the origin; it covers the tangent plane four times. Note thatif we were considering Γ ⊂ R 2 then the orientation of R 2 means that all angles are signed(meaning that we can distinguish a left and right turn by a right angle); in R 3 however,this is impossible.39. Lecture, Wednesday 10.5.06The smooth local extension of Thm. 40 allows us to define that p ∈ ∂D is a boundarybranch point if the extended conformal map satisfies df p = 0. Let us state a theorem,which will allow us to rule out boundary branch points for polygonal boundaries. We saythat a plane E ⊂ R 3 is a barrier for the set S ⊂ R 3 at p ∈ S, if p ∈ E and S does notmeet one of the two connected components of R 3 \ E.

140 K. Grosse-Brauckmann: Minimal surfaces, SS 12Lemma 41. Suppose f ∈ C 2 (D + ∪ I, R 3 ) is a conformal harmonic map. Suppose thereis a barrier E for f(D + ∪ I) at f(0) and assume that f(D + ∪ I) is not contained in theplane E. Then 0 cannot be a branch point of f.Proof. By a motion, we can assume E is the xy-plane, f(0) = 0 ∈ R 3 and f 3 (x, y) ≤ 0for (x, y) ∈ D + ∪ I. The assumption f(D + ∪ I) ⊄ E implies f 3 ≢ 0. From the strongmaximum principle for harmonic functions, Thm. II.39, we conclude f 3 (x, y) < 0 for all(x, y) ∈ D + . Note, moreover, that by our assumption the normal derivative ∂ f 3 = ∂∂ν y f 3exists.Thus all assumptions of the Hopf Boundary Point Lemma II.38 are satisfied. We concludethat ∂ y f 3 (0) < 0. In particular, df(0) ≠ 0, meaning that 0 is not a branch point. □We can apply the lemma to edges and vertices of a polygon alike:Theorem 42. Let Γ ⊂ R 3 be an embedded polygon with all angles of form π/k for k ∈N ≥ 2 subject to the following property:(i) For each p which is not a vertex of Γ, there is barrier E for Γ at p.(ii) At each vertex v ∈ Γ, consider the union S of k copies of Γ, obtained by succesive180 ◦ -rotation about the incident edges. We require that there is a barrier E for S at v.Then the Plateau solution for Γ does not have a boundary branch point.Proof. This is an immediate consequence of the Lemma.□Let us finally make a remark on symmetries:Proposition 43. Let Γ ⊂ R n be a Jordan curve andsymm(Γ) := {motions of R n which preserve Γ as a set}.If the Plateau solution M for Γ is unique in the class C 1 then it is invariant under symm(Γ).Proof. Any ϕ ∈ symm(Γ) preserves Γ. Thus ϕ(M) is also bounded by Γ; the assumeduniqueness implies ϕ(M) = M.□A typical application of this proposition is for graphs, that is, when Radó’s theorem grantsuniqueness.5.3. The D-surface. The D or diamond surface was found by Schwarz using Weierstrassdata. Let Γ be a polygonal Jordan curve in R 3 containing six of the twelve edges of astandard cube C := [0, 1] 3 ⊂ R 3 . To be precise, let us define Γ by demanding that itcontains the vertices (0, 0, 0), (1, 0, 0), (1, 0, 1), (1, 1, 1), (0, 1, 1), (0, 1, 0) in cyclic order.Γ does not have a 1-1 projection for any coordinate parallel direction. But nevertheless:

iii 5.3 – Stand: July 27, 2012 141Lemma 44. The Plateau solution M for Γ is a unique embedded surface, which is containedin the unit cube C. It is graph with respect to the space diagonal direction v := (−1, −1, 1).Proof. By Osserman’s Theorem 32, the Plateau solution has no interior branch points andhence is a minimal surface.Consider the orthogonal projection π v to the plane v ⊥ . We claim π v (Γ) is a regular hexagon.To see this note that the 120 ◦ rotation R 3 about v preserves Γ. Thus π(Γ) is invariantunder R 3 . Moreover, Γ is invariant under the motion R 6 , which reflects in the plane v ⊥followed by 60 ◦ rotation about v. Since the reflection has no effect on the projection, thismeans that π(Γ) is invariant under 60 ◦ -rotation R 6 about v. Hence π(Γ) must be a regularhexagon; in particular π(Γ) is convex.By Radó’s Theorem 36 the Plateau solution for the hexagon is a unique graph. By theConvex Hull Theorem II.43 the Plateau solution is in the unit cube (in fact, it is containedin the convex hull of Γ, which is what?).□40. Lecture, Tuesday 16.5.06We now apply the reflection principle to extend the D-surface as a triply periodic minimalsurface.Theorem 45. The continuation by reflection of the Plateau solution M for Γ yields anembedded minimal surface D. The surface together with its normal is invariant underaddition of points in the latticeΛ or := {m(2, 2, 0) + n(2, 0, 2) + l(0, 2, 2): m, n, l ∈ Z}.Its quotient D/Λ or is a compact surface which has genus 3. The surface M is also invariantunder addition of points in the latticeΛ unor := 2Z 3 ⊃ Λ or .Translations in Λ or preserve points of M together with their normal, while translations inΛ unor may change the sign of the normal.We make a few remarks about the quotient surface, although the details go beyond thematerial covered in the present class. Translations in Λ or induce an equivalence relationon R 3 : We call two points equivalent, iff they differ by a translation in Λ or . The set ofequivalence classes forms a 3-torus T 3 := R 3 /Λ or , in particular it is compact. Since T 3 isa Riemannian manifold of constant curvature 0, minimality is defined just as in R 3 . Thequotient manifold D/Λ or now is a compact surface contained in the compact manifold T 3 .

142 K. Grosse-Brauckmann: Minimal surfaces, SS 12Proof. Let us first rule out boundary branch points of M by appealing to Thm. 42. To dothat, consider two successive edges e 1 , e 2 of Γ, which meet at a vertex v. Then e 1 and e 2span a plane E, which serves as a barrier along the edges e 1 and e 2 . It also serves as abarrier for the two endpoints of e 1 and e 2 different from v.Let us now continue M by reflection to a smooth immersed surface ˜M. In the plane R 2 ,two 180 ◦ -rotations about points p and q generate a translation by 2(p − q). Similarly,two 180 ◦ -rotations in R 3 about parallel axes generate a translation by twice the orthogonaldistance between the axes. Since opposite edges of the cube form rotation axes, this impliesthat ˜M is invariant under Λ or .We need to show D is embedded, that is, the analytic continuation by reflection leads tono two copies of M intersecting eachother. Let us consider the rectangular boxB = {(x, y, z) ∈ R 3 : 0 ≤ x < 4, 0 ≤ y, z < 2}.Then B forms a fundamental domain for the lattice Λ or , meaning that the translated copiesB t := t + B = {x + t : x ∈ B} have the properties(i) if s ≠ t ∈ Λ or then B t ∩ B s = ∅ and(ii) ⋃ t∈Λ orB t = R 3 .But the closure of B contains eight (of 16) cubes which contain a copy of M, and theseare consistent with analytic continuation of M: For this it is sufficient to assert that the 8respective space diagonals are consistent. Finally, the continuation by translation of B isalso consistent with the rotation symmetries. Thus the copies of M in each cube coincideand D is embedded.To check the genus, let us compute the Euler characteristics. Since eight hexagons tesselateD/Λ or , we have(42) χ(D/Λ or ) = E − K + F = 8 · 64 − 8 · 6 + 8 = 12 − 24 + 8 = −42and so from χ(D/Λ or ) = 2 − 2g(D/Λ or ) we conclude g = 3.□The diamond surface D appears in various natural interface systems. Note that R 3 \ Dhas two components. Any 180 ◦ -rotation exchanges side of the D-surface, and hence willflip the two components.Remark. As the fundamental hexagon contains additional straight lines, it is possible togenerate the surface from a smaller quadrilateral. This quadrilateral consists of two halfedgesof Γ, whose one endpoint is connected with the midpoint of the cube.

iii 5.4 – Stand: July 27, 2012 1435.4. The P -surface. We now come to the primitive or P -surface, also discovered bySchwarz using Weierstrass data.Consider the four points in R 3 ,⎛ ⎞ ⎛0⎜ ⎟ ⎜p 1 = ⎝0⎠ , p 2 = ⎝12121212⎞ ⎛ ⎞ ⎛0⎟ ⎜ 1⎟⎜⎠ , p 3 = ⎝2⎠ , p 4 = ⎝0These points are the vertices of a quadrilateral Γ. Our choice of coordinates for thisquadrilateral is such that p 2 and p 4 are midpoints of two cubes of a cube tesselation,while p 1 and p 3 are edge midpoints. Obviously Γ has right angles at p 1 and p 3 (the planecontaining a hinge at one of these points is axis-parallel). The angles at p 2 and p 4 are 60 ◦ .Lemma 46. Γ bounds a unique embedded minimal surface M; it is symmetric with respectto reflection in the yz-plane.Proof. The quadrilateral has an injective projection to a convex quadrilateral in the xyplane.Thus the Plateau solution is unique and embedded by Radó’s theorem. □Theorem 47. The continuation by reflection of the Plateau solution M yields an embeddedminimal surface P . The surface together with normal is invariant under addition of pointsin the latticeΛ or := 2Z 3Its quotient M = P/Λ or has genus 3. The surface M itself is also invariant under additionof points in the lattice− 1 21212⎞⎟⎠ .Λ unor := {m(1, 1, 1) + n(1, 1, −1) + l(1, −1, −1): m, n, l ∈ Z} ⊃ Λ or .42. Lecture, Tuesday 23.5.06Proof. We give an outline. The piece M has a reflection symmetry. Consider a symmetrichalf of M, call it M + , with boundary in the yz-plane. We may assume M + ⊂ [0, 1] 3 . Byconstruction, M + is contained in a “little” cube of edgelength 1 , eight copies of which2make up for [0, 1] 3 . Let us now reflect M + by 180 ◦ -rotation about the two straight edges.This will lead to six copies within [0, 1] 3 , each contained in a little cube of edgelength 1/2.Consequently, the six copies of M + are disjoint.The result is a minimal surface in [0, 1] 3 which we call the hexagon M ′ . Since our originalfundamental piece M had a reflection symmetry, we can also relflect the hexagon in anyof the six (out of eight) faces which intersect M ′ . Just as the reflection extends M + to M,it will extend M ′ to a reflected copy of M ′ (using unique analytic continuation).

144 K. Grosse-Brauckmann: Minimal surfaces, SS 12Now note that the hexagon M ′ has the same symmetries as the D-hexagon; thus it isuniquely described by a space diagonal. Thus, similarly to the D-surface we obtain afundamental domain of the P surface, containing eight copies of the hexagon M ′ , relatedby reflection in faces of their containing cubes; this time however, relfection leaves no cubesempty.Since eight copies of the hexagon form a fundamental domain, the calculation (42) isliterally valid for M/Λ or and shows it also has genus 3.□5.5. Other periodic surfaces generated from polygons. There are other periodicminimal surfaces which can be generated by the Plateau solution of a quadrilateral, continuedby 180 ◦ -reflection. These are the Schwarz CLP - and H-surface, as well as Neovius’surface. Schoenflies proved 1891 that there are exactly 6 quadrilaterals whose Plateausolutions generate complete surfaces without branch points. Fischer and Koch studiedpolygonal boundaries which have more than 4 edges or which have several components.Diploma theses: 1. Discuss the six Schoenflies surfaces systematically.2. Study the work of Fischer and Koch: Which of their surfaces can be proven to beembedded?5.6. Symmetries, lattices, and crystallographic groups. Let us review facts aboutsymmetries which are relevant to minimal surfaces.An isometry or motion of R n is a map ϕ such thatp, q ∈ R n ⇒ |ϕ(p) − ϕ(q)| = |p − q|.The set of isometries forms a group Isom(n) under composition.Examples of isometries:1. Translations.2. Special orthogonal maps SO n := {A ∈ End(R n ) : det A = 1} which for n = 3 can beidentified with rotations (why?).3. Orthogonal maps O n := {A ∈ End(R n ) : | det A| = 1}. Maps in O n \ SO n include, forinstance, reflections in hyperplanes.It is a theorem that these examples generate all isometries:Each motion is the composition of a translation with an orthogonal map.Let X ⊂ R n be any set. The symmetry group of X is the subgroup of those motions inIsom(n) which preserve X as a set.Examples of symmetry groups ⊂ O(3) for subsets of R 3 :• The dihedral group D k for a regular k-gon P k (with vertices on S 1 ) where k ≥ 2;

iii 5.6 – Stand: July 27, 2012 145• C k for X = P k ∪ (0, 0, 1) for k ≥ 2; moreover let C 1 be generated by ± id and reflectionin a plane;• T for a regular tetrahedron,• O for a regular octahedron or cube (groups coincide by duality),• I for a regular icosahedron or dodecahedron (dto.).For G any of the above groups, we also consider the orienatation preserving subgroupG + := G ∩ SO(3). Furthermore, we consider a third kind of subgroup G ∗ ∋ {C ∗ k , D∗ k , T ∗ },which is generated by the respective group G + together with the inversion {− id}.The groups we have mentioned make up for all finite subgroups of O(3) or point groups:A finite subgroup of O(3) is conjugate to one of the following groups: D k ⊃ Dk ∗ ⊃ D+ kfork ≥ 2, C k ⊃ Ck ∗ ⊃ C+ kfor k ≥ 1, T ⊃ T ∗ ⊃ T + , O ⊃ O + , I ⊃ I + . None of the groupslisted is conjugate to another, and no pair is isomorphic as abstract groups.Here, a group G = {A 1 , . . . , A k } is conjugate to G ′ , if there exists X ∈ O(n) such thatG ′ = {X −1 A 1 X, . . . , X −1 A k X}.See Chapter 1 of Knörrer, Geometrie (Vieweg 1996) for a detailed account and proof ofthe classification of point groups.Remark. It is interesting to investigate the algebra of the point groups. For instance,the group D + kis isomorphic to the cyclic group generated by one element x under therelation x k = 1. The proper point of view then is to view the above symmetry groups aspresentations [Darstellungen] of abstract groups in the orthogonal group O(n).When we deal with symmetry groups of periodic objects, the set of translations will beinfinite: A lattice Λ of rank k ≤ n is a subgroup of (R n , +) which is generated by k linearlyindependent vectors v 1 , . . . , v k , that is,{ k∑}Λ = λ i v i : λ i ∈ Z .i=1Examples of lattices with rank 3 in R 3 :1. The primitive cubic lattice Z 3 ,2. the face centred cubic lattice (FCC), {(k, l, m) ∈ Z 3 : k + l + m ∈ 2Z},3. the body-centred cubic lattice (BCC), 2Z 3 ∪ {2Z 3 + (1, 1, 1)}.The primitive lattice is generated by the edges, the FCC-lattice by face diagonals, andthe BCC-lattice by space diagonals of a cube. These cubic lattices are distinguished fromall other lattices in that they admit symmetries exchanging three linearly independentdirections pairwise. We call a surface in R 3 singly, doubly, or triply periodic if it is invariantunder a lattice of rank 1,2, or 3, respectively.

146 K. Grosse-Brauckmann: Minimal surfaces, SS 12Examples. The helicoid is singly periodic. Scherk’s doubly periodic surface is invariantunder the planar square lattice and hence doubly periodic. The P and D-surface are triplyperiodic.We call a group G of motions discrete if for each point p ∈ R n there is d(p) > 0 such thatif A ∈ G satisfies Ap ≠ p then |Ap − p| > d(p). This property means that distinct imagesof p under the group (the so-called orbit of p) is a discrete set, that is, it does not have anyaccumulation points. A crystallographic group is a subgroup of Isom(n) which contains alattice of maximal rank n as the subgroup of translations. The following famous fact wasestablished around 1900:The number of non-isomorphic crystallographic groups of R n is finite for each n; in particular,there exist 17 non-isomorphic planar groups (n = 2) and 219 space groups (n = 3).This classification can also be viewed as a classification up to certain conjugations andhomothety. In any case, the classification leaves free angle parameters of the lattice for the“less symmetric” groups.The International Tables of Crystallography list all these groups; in fact, they list 230 spacegroups as they admit only the more special conjugation in SO(3). Crystallographic groupscontain point groups as subgroups. Namely, the subgroup which fixes some given point pis a point group. Thus it must be one of the groups we listed above; but only certain ofthese groups can actually occur: In case n = 3, groups from the infinite series can onlyoccur with k = 1, 2, 3, 4, 6, and moreover they can be of type T or O, but not I.References. Wolf, Spaces of constant curvature, Sect. 3.1 and 3.2, contains all results inprecise form. For the classification of the planar groups a more popular presentation is Armstrong,Groups and symmetries (Springer 1988). Buser (L’enseignement mathématique 31,p.137-145, 1983) gives a proof that the number of crystallographic groups is finite in anydimension.Each triply periodic minimal surface M has a pair of crystallographic groups associated: thegroup which preserves M and its subgroup which preserves M together with its orientation.A 180 ◦ -rotation about a straight line in the surface exchanges the normal and hence doesonly belong to the first group, not to the second.Example. For P , the crystallographic groups in international notation are Im3m andP m3m, for D, they are P n3m and F d3m.An open problem is to classify the triply periodic minimal surfaces with a given symmetrygroup.42. Lecture, Tuesday 23.5.06

iii 5.6 – Stand: July 27, 2012 147Let us conclude this section with a theorem, which sounds surprising on the first sight:A triply periodic surface is a maximum of area when compared to its parallel surfaces!To make this precise recall that a periodic minimal surface has infinite area. Thus it isreasonable to consider the area of a fundamental domain, A(M/Λ).Theorem 48. Let M = M 0 be a complete (connected) triply periodic minimal surfacein R 3 with lattice Λ (orientation preserving). Let ν be the Gauss map and consider forsufficiently small ε > 0 the family of parallel surfaces M t = {p + tν(p) : p ∈ M} where0 < |t| < ε. ThenA(M t /Λ) < A(M/Λ) for all t ≠ 0.In particular, this means that the surface M/Λ is not a local minimum of area in the torusR 3 /Λ.Proof. In Corollary II.20 we stated that the parallel surface in distance t to a minimalsurface has area∫A(M t /Λ) = A(M/Λ) + t2 K dS + O(t 3 ).2Since K ≤ 0, we conclude A(M t ) ≤ A(M). In fact, unless M is planar, even A(M t )