Souam21

Selected chapters in classicalminimal surface theory

by

RABAH SOUAM

The purpose of these notes is to give a short introduction to the theory of minimal surfacesin R3 through some selected topics. We present the Weierstrass representation, the solutionto Plateau’s problem and the Barbosa-do Carmo stability theorem.

1 IntroductionThe theory of minimal surfaces in R3 is a rich and old subject which has experienced a greatdevelopment during the last 30 years and is still an active research field.

The study of minimal surfaces in R3 started with Lagrange in 1762 who considered theproblem of determining a graph with least area over a given domain in R2 and taking pre-scribed values on the boundary. He derived the partial differential equation to be satisfied bythe solution. It was Meusnier in 1776 who recognized the geometric meaning of that equa-tion: the mean curvature vanishes, H ≡ 0. Hence the term minimal used for surfaces withvanishing mean curvature. During the nineteenth century, many authors, among which Bon-net, Enneper, Riemann and Weierstrass contributed to the study of minimal surfaces. Thestrong link with complex analysis was in particular established then through the so-calledWeierstrass representation. The Belgian physicist J. Plateau (1801-1881) observed that min-imal surfaces can be realized as soap films. The problem of detemining a minimal diskbounded by a given Jordan curve was then called Plateau’s problem. It was only in 1930that the problem was solved independently by J. Douglas and T. Rado. During the last 30years, the theory of minimal surfaces has focused on global properties of complete minimalsurfaces, especially embedded ones.

We present in these notes some classical results in the theory. This is not a survey paper.We have tried to make this text as elementary and as self-contained as possible and to give inparticular complete proofs of the selected results. This clearly limited the possibilities. Wehope these notes will give some motivations for beginners to go further in the study of the

69

70 GEOMETRIES ET DYNAMIQUES

rich subject of minimal surfaces in R3...and in other ambient spaces. Introductory books tothe subject are [12], [1] and [9]. For more recent results in the theory one may consult thesurvey papers [8], [10], [13], [14].

We have organized this text as follows. In Section 2 after recalling some premilinaryfacts about the geometry of surfaces, we derive the Weierstrass representation in its local andglobal versions and present some classical examples of minimal surfaces. We also derivethe first variation formula for the area functional. Section 3 is devoted to the solution ofPlateau’s problem. We have followed closely the presentation in the book by B. Lawson,[9].In Section 4, we discuss the stability problem for minimal surfaces. A minimal surface isstable if it minimizes the area up to the second order for boundary preserving variations. Wegive a simplified proof of the Barbosa-do Carmo stability criterion, following an observationof D. Fischer-Colbrie and R. Schoen, [6].

Acknowledgments: I am grateful to Francisco Martin for allowing me to use some of his imagesand to the referee for valuable observations.

2 The Weirestrass representation and the first variationformula

We first review quickly some basic definitions about the geometry of immersed surfaces in theEuclidean space R3 and fix some notations. We denote by 〈, 〉 the Euclidean inner product.Let S be a surface and X : S → R3 an immersion, i.e a smooth map such that dXp hasrank 2 at each point p ∈ S. The first fundamental form I of X is the Riemannian metric on Sdefined as follows:

Ip(w1, w2) = 〈dXp(w1), dXp(w2)〉, for p ∈ S, w1, w2 ∈ TpS.

When S is orientable, let N denote a unit normal field to S, that is to say N : S → S2 ⊂ R3

is smooth and N(p) is orthogonal to dXp(TpS) for each p ∈ S. Here and in the sequel S2

denotes the unit sphere centered at the origin in R3. The second fundamental form, denotedII, is the field of bilinear symmetric forms defined as follows:

IIp(w1, w2) = −〈dNp(w1), dXp(w2)〉, for p ∈ S, w1, w2 ∈ TpS.

The shape operator B of the immersion X (oriented by the choice of the unit normal fieldN ) is the field of symmetric endomorphisms defined by:

IIp(w1, w2) = Ip(Bw1, w2), for p ∈ S, w1, w2 ∈ TpS.

The mean curvature H of the immersion X and its Gaussian curvature K are defined asfollows:

H =12

trace(B), K = detB.

The principal directions are the eigenvectors of B and the principal curvatures are its eigen-values.

R. SOUAM 71

Definition 2.1 An immersed orientable surface in R3 is called minimal if its mean curvatureis identically zero, i.e H ≡ 0

Note that, in particular, the Gaussian curvature of a minimal surface is nonpositive, K ≤0.

2. 1 The minimal surface equationIf the surface is expressed locally as a graph (x, y, h(x, y)), (x, y) ∈ Ω of some functionh : Ω→ R, then minimality means the function h satisfies in Ω the minimal surface equation:

(1 + hy2)hxx − 2hxhyhxy + (1 + hx

2)hyy = 0.

As the reader can check, this may be rewritten as follows:

div

(∇h√

1 + |∇h|2

)= 0.

This is a quasilinear elliptic partial differential equation. A natural problem is to solve theDirichlet problem for the minimal surface equation, that is to say solve the equation oversome domain with a given boundary data. This is a well studied problem. The problem hasa solution for any boundary data if and only if the domain is convex. A reference for thesequestions is the textbook [7].

2. 2 The Gauss map of a minimal surfaceDefinition 2.2 A differentiable map ϕ : S → S between two Riemannian surfaces is calledalmost conformal if there exists a differentiable function λ on S such that for all p ∈ S andall v1, v2 ∈ TpS we have

〈dϕp(v1), dϕp(v2)〉ϕ(p) = λ2(p)〈v1, v2〉p.

If the function λ is nowhere zero then ϕ is said to be conformal.

The geometric meaning of the above definition is that the (unoriented) angles are pre-served by conformal maps. In fact, let α : I → S and β : I → S be two curves in S whichintersect at, say, t = 0. Their angle 0 ≤ θ ≤ π at t = 0 is given by

cos θ =〈α′, β′〉|α′||β′|

.

A conformal map ϕ : S → S maps these curves into curves ϕ α : I → S, ϕ β : I → S,which intersect for t = 0, making an angle 0 ≤ θ ≤ π satisfying:

cos θ =〈dϕ(α′), dϕ(β′)〉|dϕ(α′)||dϕ(β′)|

=λ2〈α′, β′〉λ2|α′||β′|

= cos θ,

as we claimed. Conversely, a diffeomorphism preserving the angles is conformal. A holo-morphic map with nowhere zero derivative defined on an open subset of C is conformal.Conversely, an orientation preserving conformal map between open subsets of C is holomor-phic. To check these properties the reader is invited to solve the following exercise:


Exercise. Let L : V → W be a linear map between two Euclidean spaces V,W of the samedimension. The following properties are then equivalent:

• L preserves orthogonality.

• L preserves the angles.

• There exists a constant λ > 0 such that for each w1, w2 ∈ V, 〈L(w1), L(w2)〉W =λ2〈w1, w2〉V .

Furthermore in dimension 2, this is equivalent to say that in orthonormal bases, the matrix of

L is of the form:(a b−b a

)or(a bb −a

).

Proposition 2.3 Let X : S → R3 be an immersed connected orientable surface in theEuclidean space. Then its Gauss map: N : S → S2 is almost conformal if and only if X isminimal or X(S) is a subset of a round sphere.

Proof. Let p ∈ S and w1, w2 ∈ TpS. Since dNp(w1) ∈ dXp(TpS), we have by defintion ofthe shape operator:

〈dN(w1), dN(w2)〉 = −〈dN(w1), dX(Bw2)〉 = 〈dX(Bw1), dX(Bw2)〉 = I(Bw1, Bw2).

So N is almost conformal if and only if there exists a function λ such that the shape operatorsatisfies:

B2 = λ2Id.

Now B is a solution of its characteristic polynomial, so:

B2 − 2HB +KId = 0.

So almost conformality is equivalent to the condition that B satisfies:

2HB = (λ2 +K)Id (1)

for some function λ. This clearly shows minimal surfaces (and spheres) have almost confor-mal Gauss maps. Conversely, suppose the Gauss map is almost conformal and the surface isnot minimal. Consider a connected component V of the open set H 6= 0. Then (1) showsthe immersion is totally umbilic on V and hence X(V ) is a subset of a sphere (cf. [4]). Inparticular the mean curvature of X|V is a nonzero constant. This shows V is closed and soV = S.

Note that since a minimal surface has nonpositive Gaussian curvature, its Gauss map isorientation reversing when the unit sphere is oriented by the exterior normal (cf. [15]).

R. SOUAM 73

2. 3 The Weierstrass representationLet X : S → R3 be an immersed surface in the Euclidean space. S is thus endowed withthe induced metric ds2. Recall the classsical result that around each point of S we can findconformal coordinates, that is coordinates (u, v) on which the metric takes the form:

ds2 = λ2(du2 + dv2),

for some positive function λ > 0.We have the following characterization of minimal surfacesgiven in conformal coordinates (see [15] for the details).

Proposition 2.4 Let X : S → R3 be an immersed surface oriented by a unit normal N andlet (u, v) be local conformal coordinates on S. Then in these coordinates:

∆X = 2Hλ2N.

In particular the surface is minimal if and only if its coordinates functions X1, X2, X3 areharmonic in any conformal coordinates.

Consider the complex valued functions:

φk(z) =∂Xk

∂u− i∂Xk

∂v, z = u+ iv. (2)

We have the following identities:

3∑k=1

φ2k(z) = |∂Xk

∂u|2 − |∂Xk

∂v|2 − 2i〈∂Xk

∂u,∂Xk

∂v〉, (3)

3∑k=1

|φk(z)|2 = |∂Xk

∂u|2 + |∂Xk

∂v|2. (4)

As a consequence, we have the:

Proposition 2.5 Let X : (u, v) → R3 define a minimal immersion with (u, v) conformalcoordinates. Then the functions φk(z) defined by (2) are analytic functions of z and satisfythe conditions:

3∑k=1

φ2k(z) = 0 (5)

3∑k=1

|φk(z)|2 6= 0. (6)

Conversely, given φ1(z), φ2(z), φ3(z) analytic functions of z satisfying the conditions (5) and(6) in a simply connected domain D ⊂ C. Then:

X(z) = <e∫ z

z0

(φ1(z), φ2(z), φ3(z))dz, (7)

where z0 ∈ D is any fixed point, defines a conformal minimal immersion satisfying (2).


Proof. If X is minimal then the statements are true as direct consequences of (2), (3) and(4). Conversely, given φ1, φ2, φ3 analytic functions, then X defined by (7) is well defined:the integral does not depend on the choice of the path joining z0 to z since D is simplyconnected, and the identities (2), (3) and (4) are satisfied. By (3) and (5) the coordinates(u, v) are conformal. By (6) and (4) X is an immersion. Furthermore, (2) and analyticity ofφ1, φ2, φ3, imply the harmonicity of the coordinate functions X1, X2, X3. Therefore by theProposition 2.4, the immersion X is minimal.

The previous local fact admits the following globalization. Given an orientable Rieman-nian surface, the local existence of conformal coordinates implies the existence of a complexstructure on the surface compatible with the metric. An orientable Riemannian surface hastherefore an (unique) underlying structure of a Riemann surface compatible with its metric.This is in particular true for an orientable surface minimally immersed in R3. If z = u + ivis a local conformal coordinate, then one considers the holomorphic 1-forms:

Φk = ∂Xk

∂u− i∂Xk

∂vdz = 2

∂Xk

∂zdz, k = 1, 2, 3. (8)

It is straightforward to check that Φ1,Φ2,Φ3 are globally defined holomorphic 1-forms.Given a Riemann surface S, a map X : S → R3 is said conformal if around each pointone can find a local conformal coordinate z = u+ iv, such that:

|∂Xk

∂u|2 = |∂Xk

∂v|2 and 〈∂Xk

∂u,∂Xk

∂v〉 = 0.

It can be checked this is independent on the choice of the local conformal coordinate. Wehave then the following:

Proposition 2.6 Let S be a Riemann surface and X : S → R3 a conformal minimalimmersion. The holomorphic 1-forms defined by (8) satisfy the following:

3∑k=1

Φ2k = 0, (9)

3∑i=1

|Φk|2 6= 0. (10)

Conversely, given a Riemann surface S and 3 holomorphic 1-forms Φ1,Φ2,Φ3 satisfying (9)and (10) and the period condition:

<e∫γ

(Φ1,Φ2,Φ3) = 0,

for any closed path in S, then the map:

X(p) = <e∫ p

p0

(Φ1,Φ2,Φ3),

for p ∈ S, with p0 ∈ S any fixed point defines a conformal minimal immersion.

R. SOUAM 75

Define:g =

Φ3

Φ1 − iΦ2, η = Φ1 − iΦ2,

g is then a meromorphic function and η holomorphic 1-form and the identity (9) becomes:

Φ1 =12

(1− g2)η,

Φ2 =i

2(1 + g2)η,

Φ3 = gη

and it can be checked that:

N(z) =(

2<e(g(z))1 + |g(z)|2

,2=m(g(z))1 + |g(z)|2

,1− |g(z)|2

1 + |g(z)|2

).

This means that the meromorphic function g is the sterographic projection, from the point(0, 0, 1) of the unit sphere, of the Gauss map N of the immersion X. We recover the fact thatthe Gauss map of minimal surfaces is almost conformal (Propostion 2.3).

We have:3∑k=1

|φk|2 =12

(1 + |g|2)2|η|2, (11)

and (10) means the zeroes of η coincide with the poles of g, but with twice order. So theProposition 2.6 can be reformulated as follows:

Theorem 2.7 (The Weirerstrass representation) Let X : S → R3 be a minimal immersionof an orientable surface S. Let g = σ N be the composition of the stereographic projectionσ from the point (0, 0, 1) of the sphere to the extended complex plane C ∪ ∞, with theGauss map N of X. Then g is meromorphic and there exists a holomorphic 1-form η on Ssuch that:

X(p)−X(p0) = <e∫ p

p0

(12

(1− g2),i

2(1 + g2), g

)η,

for p, p0 ∈ S, the integration being taken on any path from p0 to p. Moreover the zeroes of ηcoincide with the poles of g and have twice order.

Conversely, let S be a Riemann surface and g : S → C ∪ ∞ a meromorphic functionand η a holomorphic 1-form on S. Assume the zeroes of η coincide with the poles of g andhave twice order. Assume also the following period condition is satisfied:

<e∫γ

(12

(1− g2),i

2(1 + g2), g

)η = 0,

for every closed path on S. Then the map: X : S → R3, defined by:

X(p) = <e∫ p

p0

(12

(1− g2),i

2(1 + g2), g

)η,

where p0 is any fixed point in S, is a conformal minimal immersion. Moreover the Gauss mapof X is N = σ−1 g.


We can express all the geometric quantities associated to a minimal immersion in termsof its Weierstrass data (g, η). If z is a local conformal coordinate with η = f(z)dz, then, inparticular, the metric and the Gaussian curvature are expressed as follows:

ds2 =14|f(z)|2(1 + |g(z)|2)2|dz|2, (12)

K(z) = −(

4|g′(z)||f(z)|(1 + |g(z)|2)2

)2

, (13)

(12) follows from (11) and (4). For (13) see [12].

2. 4 Classical examples

The Weierstrass representation is an efficient tool to construct minimal surfaces. From thegeometric point of view one is interested in complete surfaces. This implies some condi-tions on the behaviour at infinity of the Weierstrass data (g, η) in view of equation (12). Thecompatibility condition between the zeroes and poles of the Weierstrass data (g, η) is easy tocheck. The period condition is usually the most difficult condition to verify in practice. How-ever on a simply connected Riemann surface it is automatically satisfied. We present heresome examples of complete minimal surfaces and refer to the literature for more examples(in particular [1] which contains a detailed description of many examples).

The plane: S = C, g = 0, η = dz

Enneper’s surface: (figure 1) S = C, g(z) = z, η = dz.

Figure 1: Enneper’s surface

The catenoid: (figure 2) S = C− 0, g(z) = z, η = dzz2 .

The helicoid: (figure 3) S = C, g(z) = ez, η = ie−zdz.

Costa’s surface (figure 4) S = (C/Z× Z) − 0, 12 ,

i2, g(z) = 2

√2πP( 1

2 )

P′ , ω =Pdz, where P is the Weierstrass function of the unit square.

R. SOUAM 77

Figure 2: The catenoid

Figure 3: The helicoid


Figure 4: Costa’s surface

2. 5 The first variation formula

We now see that minimal surfaces are critical points of the area functional thus justifyingtheir name. Let X : S → R3 be an immersion of an orientable surface in R3. Let D ⊂ S bea compact domain with smooth boundary in S. By a variation Xt of X|D we mean a smoothmapping:

F : D × (−ε, ε)→ R3

such that Xt := F (t, .) is an immersion for each t and X0 = X.

The variation vector field associated to the variation Xt is the vector field ξ along theimmersion defined by:

ξ(p) =∂Xt

∂t|t=0(p), p ∈ D.

The variation is said normal if ξ is orthogonal to the surface, i.e ξ = φN for some functionφ ∈ C∞(D). Finally we say that the deformation preserves the boundary if, for all t, Xt = Xon ∂D. Let C∞0 (D) denote the set of functions on D which are smooth up to the boundaryand vanishing on ∂D. The variation vector field associated to a normal boundary preservingvariation is of the form ξ = φN, where φ ∈ C∞0 (D).

Conversely, given φ ∈ C∞0 (D), consider the variationXt = X+φN. For t small enough,Xt is an immersion by compactness of D and continuity. It is, moreover, clearly normal andboundary preserving. So at the infinitesimal level a normal boundary preserving variation isthe same thing as a function φ ∈ C∞0 (D).

Definition 2.8 (Divergence) LetX : S → R3 be an immersion and V : S → R3 a (smooth)vector field along X, that is to say:

V (p) ∈ dXp(TpS),

R. SOUAM 79

for all p ∈ S. For w ∈ TpS, denote by dV (w)⊥ the orhtogonal projection of dV (w) ondXp(TpS). The divergence of V, denoted div V is the trace of this map:

TS → TS

w → dV (w)⊥.

Otherwise said: for p ∈ S, let e1, e2 be an orthonormal basis of TpS, then

div V (p) = 〈dV (e1), dX(e1)〉+ 〈dV (e2), dX(e2)〉.

If now V is a vector field on S, then its divergence is by definition the divergence of dX(V ).

Let ω denote the area form of the surface X (see [15]). Recall that for w1, w2 ∈ TS, wehave

ω(w1, w2) = det(dX(w1), dX(w2), N).

Let V be a vector field along the immersion X, we define a 1-form iV ω as follows:

iV ω(w) = det(V, dX(w), N).

Proposition 2.9 d(iV ω) = (divV )ω.

Proof. We show the equation is verified at each point of S. Let p ∈ S and let (u, v) be localcoordinates in S around p compatible with the orientation. In these coordinates the 1-formiV ω writes:

iV ω = iV ω(Xu) du+ iV ω(Xv) dv = det(V,Xu, N) du+ det(V,Xv, N) dv.

So:

d(iV ω) =∂

∂udet(V,Xv, N)− ∂

∂vdet(V,Xu, N)

du ∧ dv.

We have:∂

∂udet(V,Xv, N) = det(Vu, Xv, N) + det(V,Xvu, N) + det(V,Xv, Nu)

= det(Vu, Xv, N) + det(V,Xvu, N),

because Nu is tangent to the surface. Similarily:

∂

∂vdet(V,Xu, N) = det(Vv, Xu, N) + det(V,Xuv, N).

Therefore:d(iV ω) = det(Vu, Xv, N)− det(Vv, Xu, N) du ∧ dv.

We may assume the coordinates (u, v) are such that the basis Xu, Xv is orthonormal andpositively oriented at p. This is always possible, take any coordinates and perform a linearchange of coordinates. Then at p we have:

d(iV ω)(p) = 〈Vu, Xu〉(p) + 〈Vv, Xv〉(p) det(Xu(p), Xv(p), N(p))du ∧ dv(p).

So:d(iV ω)(p) = divV (p)ω(p).


We are now in position to prove the:

Proposition 2.10 (The first variation formula) LetXt be a boundary preserving variation ofan immersion X : D → R3, where D is a compact orientable surface, oriented by a choiceof a unit normal field N. Call A(t) the area of Xt. Then:

dA

dt(0) = −2

∫D

φHdA,

where H is the mean curvature of the immersion and φ = 〈∂Xt∂t |t=0, N〉.

Proof. We denote by Nt the unit normal field associated to the immersion Xt and dependingsmoothly on t, with N0 = N and by dAt its area element. We have:

A(t) =∫D

dAt,

so we need to compute the derivative of the area element dAt. Let p ∈ D and let (u, v) belocal coordinates around p in S compatible with the orientation. Then:

dAt = det((Xt)u, (Xt)v, Nt) dudv.

Set N = dNtdt |t=0. We have:

d

dtdet((Xt)u, (Xt)v, Nt)|t=0 = det(ξu, Xv, N) + det(Xu, ξv, N) + det(Xu, Xv, N).

It follows from 〈Nt, Nt〉 = 1 that N is tangent to the surface and so the last term in theright-hand side of the previous equation vanishes. Write:

ξ = φN + V

where V is the tangent part to the immersion of the deformation vector field ξ. Then:

ξu = dφ(∂

∂u)N + φNu + Vu, ξv = dφ(

∂

∂v)N + φNv + Vv.

Therefore:

d

dtdet((Xt)u, (Xt)v, Nt)|t=0 = φ det(Nu, Xv, N) + det(Xu, Nv, N)

+det(Vu, Xv, N) + det(Xu, Vv, N).

Now as in the previous Propositon we may assume the basis Xu, Xv is orthonormal andpositive at the point p and so:

d

dtdet((Xt)u, (Xt)v, N)|t=0(p) = φ(p) 〈Nu, Xu〉(p) + 〈Nv, Xv〉(p)

+〈Vu, Xu〉(p) + 〈Vv, Xv〉(p)= 2φ(p)tr(dN(p)) + divV (p)= −2φ(p)H(p) + divV (p).

R. SOUAM 81

It follows that:dAtdt

(p) = −2φ(p)H(p) + divV (p) dA(p).

Since this is true for every p ∈ D, we conclude that:

dA

dt(0) =

∫D

−2φH + divV dA.

Since V = 0 on ∂D, the 1-form iV ω vanishes on ∂D. By Stokes’ theorem:∫D

divV dA =∫D

d(iV ω) =∫∂D

iV ω = 0,

which completes the proof.

Corollary 2.11 An immersion X : S → R3 is minimal if and only if for every variation Xt

of X with compact support, the derivative of the area A(t) vanishes at t = 0 :

dA

dt(0) = 0.

3 The Plateau problemLet Γ be a Jordan curve in Rn, i.e a continuous curve which is homeomorphic to the circleS1. The Plateau problem asks to find the surface of least area spanning Γ. The problem wassolved independently by Douglas and Rado, in 1930, for disk-type surfaces. In order toformulate the result, we introduce some notations: B = (u, v) ∈ R2, u2 + v2 ≤ 1 willdenote the closed unit disk in the Euclidean plane.

A map X : B → Rn is said to be piecewise C1 if it is continuous and if it is C1 except

along ∂B and along a finite number of regular C1−arcs and points inB . A continuous map

φ : ∂B → Γ is said to be monotone if it is such that if ∂B is traversed once in the positivedirection, then Γ is traversed once also in a given direction, although we allow arcs of ∂B tomap to single points of Γ. Otherwise said, a monotone map is a continuous map of degree±1such that for each p ∈ Γ, φ−1(p) is connected.

Let C(Γ) = X : B → Rn, X is piecewiseC1 andX|∂B is a monotone parameterizationof ΓWe define the area functionalA : C(Γ)→ R∪∞ by the following (maybe improper)integral:

A(X) =∫B

|Xu ∧Xv|dudv

where |Xu ∧Xv|2 = |Xu|2|Xv|2 − 〈Xu, Xv〉2. Let

AΓ = infX∈C(Γ)

A(X).

Therefore our problem is to find a X ∈ C(Γ) such that A(X) = AΓ.Note that we have to restrict ourselves to curves Γ satisfying AΓ < ∞ since there are

examples of Jordan curves bounding no surfaces of finite area. If Γ is rectifiable, for instanceC1, then it is true that AΓ <∞ (cf.[12]).


The idea is to try to apply the direct method of the calculus of variations, that is take asequenceXi ∈ C(Γ) such thatA(Xi)→ AΓ and try to extract a subsequence wich convergesto an X∞ ∈ C(Γ) with A(X∞) = AΓ. However this cannot work this way. The reason isthat the area functional is invariant under a big group of transformations: for any piecewiseC1 diffeomorphism ψ of the disk B, and any X ∈ AΓ we have A(X ψ) = A(X). We canfind a sequence of smooth homeomorphisms ψi of the unit disk B such that ψi(reiθ) →eiθ uniformly on compact subsets of B \ 0 and ψi(0) → 0. To get such an examplejust construct such a sequence of diffeomorphisms of the unit interval and then use polarcoordinates. Take for instance the sequence of homeomorphisms of the unit interval:

li(t) = t if 0 ≤ t ≤ 1i+1 ,

li(t) = (i2 − i− 1)t− i2−i−2i+1 if 1

i+1 ≤ t ≤1i ,

t+i−2i−1 if 1

i ≤ t ≤ 1,

which is piecewise smooth and perturb it a bit to make it smooth. Then assuming Xi con-verges uniformly to X∞ one can check that no subsequence of Xi ψi converges to a con-tinuous map.

So we need to control the parameterizations of our minimizing sequences. In the one-dimensional case, i.e curves in a Riemannian manifold, this is done by minimizing the energyintegral. Here minimizing sequences of curves tend to geodesics which not only minimizethe length integral but which are forced to be parameterized by a multiple of arc-length. Inthe case of surfaces the corresponding energy is the so-called Dirichlet integral:

D(X) =∫B

(|Xu|2 + |Xv|2)dudv.

Note that for any vectors v, w ∈ Rn we have:

|v ∧ w|2 = |v|2|w|2 − 〈v, w〉2 ≤ |v|2|w|2 ≤ 14

(|v|2 + |w|2)2,

and equality holds if and only if |v| = |w| and 〈v, w〉 = 0. Therefore, for any X ∈ C(Γ),

A(X) ≤ 12D(X)

with equality if and only if

|Xu| = |Xv| and 〈Xu, Xv〉 = 0

almost everywhere in B. Wherever |Xu| > 0, such a map is conformal and induces a metricon B of the form

ds2 = λ(du2 + dv2)

where λ = |Xu| = |Xv|. The parameters (u, v) are thus conformal coordinates for thesurface. A theorem of fundamental importance is the following:

Theorem 3.1 (Conformal representation of disk-type domains) Let X : B → Rn be a

continuous map such that X|B is an immersion of class Ck, 1 ≤ k ≤ ∞ (or real analytic).

Then there exists a homeomorphism ψ : B → B where ψ|B is of class Ck (or real analytic)

such that the parameterized mapping X ′ = X ψ is conformal.

R. SOUAM 83

Consider a Jordan curve Γ ⊂ Rn and define:

DΓ = infX∈C(Γ)

D(X).

Clearly:

AΓ ≤12DΓ.

Now take a sequence Xk ∈ C(Γ) such that A(Xk) → AΓ. The Xk are not necessarily im-mersions so we cannot apply directly the Theorem 3.1. The trick is to consider the sequenceof mappings: Xk : B → Rn+2, defined by

Xk(u, v) = (Xk(u, v),1ku,

1kv).

Clearly the Xk are C1−immersions (even embeddings) and so there exist homeomorphismsψk of B given by the Theorem 3.1 such that the Xk ψk are conformal. Using the invarianceof area under reparametrizations and some elementary estimates, we get:

DΓ ≤ D(Xk ψk) ≤ D(Xk ψk) = 2A(Xk ψk) = 2A(Xk) ≤ 2A(Xk) +4πk

and it follows thatAΓ =

12DΓ.

Corollary 3.2 Let Γ ⊂ Rn be a Jordan curve with AΓ <∞. Then for any X ∈ C(Γ),

D(X) = DΓ if and only if A(X) = AΓ and X is almost conformal.

Hence to solve the Plateau problem it is sufficient to find a map X ∈ C(Γ) which mini-mizes the Dirichlet integral. We will use the

Theorem 3.3 (Dirichlet’s Principle) Let φ : ∂B → Rn be a continuous map and define

C(φ) = X : B → Rn : X is piecewise C1 and X|∂B = φ

Assume that the numberDφ = inf

X∈CφD(X)

is finite. Then there exists a unique map Xφ ∈ C(φ) such that D(Xφ) = Dφ. The function

Xφ is harmonic inB and represents the solution to the boundary value problem ∆X =

0, X|∂B = φ.

We can thus consider a minimizing sequence Xk ∈ C(Γ), for the Dirichlet functional

satisfying, Xk is hamonic inB . Still we cannot extract a converging subsequence, the rea-

son is that the Dirichlet energy is invariant under the action of the group of conformal dif-feomorphisms of the disk B. Recall that the group G of conformal orientation preservingdiffeormorphisms of the disk:

G = g(z) = eiθ0a+ z

1 + az, a ∈ C, |a| < 1, θ0 ∈ R (14)


is not compact. Indeed take a sequence ak ∈ C, |ak| < 1, ak → 1, then the sequence in G,gk(z) = ak+z

1+akzconverges uniformly on compact subsets ofB\−1 to the constant 1. So we

need to normalize the mappingsX ∈ C(Γ). This is done by imposing a three-point condition:choose three distinct points p1, p2, p3 ∈ Γ and three distinct points z1, z2, z3 ∈ ∂B, anddefine

C∗(Γ) = X ∈ C(Γ) : X(zk) = pk, k = 1, 2, 3.By the invariance of the Dirichlet integral under G, we have that

infD(X), X ∈ C∗(Γ) = DΓ,

and thus we may solve the Plateau problem by minimizing in this smaller class. We thus takea minimizing sequence Xk ∈ C∗(Γ), we claim that the boundary values Xk|∂B : ∂B → Rnform an equicontinuous family. The key to this crucial result is the following

Lemma 3.4 (The Courant-Lebesgue lemma) Suppose X ∈ C0(B,Rn) ∩ C1(B,Rn) and

satisfiesD(X) ≤M (15)

for some M with 0 ≤ M < ∞. For each z0 ∈ ∂B and r > 0 we define Cr to be theintersection of B with the circle of radius r centered at z, and we denote by s the arc lengthparameter of Cr. Then for each 0 < δ < 1, there exists a number ρ with δ ≤ ρ ≤

√δ such

that ∫Cρ

|Xs|2ds ≤2M

ρ log( 1δ ). (16)

Proof. For 0 < δ < 1, consider the integral

I =∫ √δδ

∫Cr

|Xs|2dsdr ≤ D(X) ≤M,

and express I as

I =∫ √δδ

p(r)1rdr where p(r) = r

∫Cr

|Xs|2ds.

Then by the mean value theorem for the measure d(log r), there exists a number ρ ∈ [δ,√δ],

such that

I = p(ρ)∫ √δδ

d(log r) = p(ρ)12

log(1δ

).

Thus, p(ρ) ≤ 2M/ log( 1δ ) as claimed.

We can now prove the following:

Proposition 3.5 Let M be a constant > DΓ. Then the family of functions

F = X|∂B : X ∈ C?(Γ) and D(X) ≤M

is equicontinuous on ∂B. Thus, by Ascoli’s theorem F is compact in the topology of uniformconvergence.

R. SOUAM 85

Proof. For z ∈ ∂B and 0 < r < 1, denote by l(Cr) the length of the curve Cr defined in theLemma 3.4. It follows from the Schwarz inequality that

l(Cr)2 = (∫Cr

|Xs|ds)2 ≤ 2πr∫Cr

|Xs|2ds.

Let 0 < δ < 1 be a fixed number andX ∈ F . It follows from the Lemma 3.4 that there existsδ ≤ ρ ≤

√δ such that

l(Cr)2 ≤ 2Mlog( 1

δ ).

Let now ε > 0 be a given number. By an easy topological argument one sees that thereexists a number d > 0 such that for all p, q ∈ Γ with 0 < ||p − q|| < d, one of thetwo components of Γ \ p, q will have diameter < ε. (Recall that the diameter of a setS ∈ Rn is diam(S) = sup||p − q||, p, q ∈ S.) We now choose 0 < δ < 1 such that√

4πM/ log( 1δ ) < d and such that the disks of radius

√δ centered at z1, z2, z3 are pairwise

disjoint. We may assume without loss of generality that ε < mini6=j||pi − pj ||. Then forany z ∈ ∂B, there exists by the Lemma 3.4 a real δ ≤ ρ ≤

√δ such that l(Cρ) < d. ∂B

is now divided by Cρ into two arcs: a small arc A′ containing z and its complement A′′

containing two of the points z1, z2, z3. By the monotonicity of X|∂B, the images X(A′) andX(A′′) determine a partition of Γ into two arcs and since l(Cρ) ≤ d one of them has diameter< ε. However, since X(A′′) contains at least two of the points p1, p2, p3, its diameter is > ε.Hence, diameter (X(A′)) < ε. This immediately implies that for z, z′ ∈ ∂B, |z − z′| < δ,we have ||X(z) − X(z′)|| < ε and δ depends only on ε and the points z1, z2, z3, p1, p2, p3.This shows the equicontinuity of the family X(A′′)

Now by Ascoli’s theorem each sequence in X(A′′) has a convergent subsequence for theuniform convergence. To conclude note that a limit of monotone maps is also monotone.

Theorem 3.6 (Douglas, Rado) Let Γ be a Jordan curve in Rn such that AΓ < ∞. Thenthere exists a continuous map X : B → Rn such that

(1) X|∂B maps ∂B monotonically onto Γ,

(2) X|B is harmonic and almost conformal,

(3) D(X) = AΓ.

Proof. Let Xk ∈ C∗(Γ) be a minimizing sequence, limk→∞D(Xk) = DΓ. By the Theorem

3.3 we can assume theXk harmonic inB . By the Proposition 3.5, we can find a subsequence,

still denoted Xk, such that Xk|∂B converges uniformly to a monotone parametrization φ :∂B → Γ. Denote by Xφ the hamonic extension of φ to B given by the Dirichlet principle(Theorem 3.2). Then the maximum principle for harmonic functions ensures that Xk → Xφ

uniformly on B as k → ∞. Harnack’s principle ensures that ∇Xk → ∇Xφ as k → ∞uniformly on compact subsets of

B. Let then K be any compact subset of

B. Then:∫

K

|∇Xφ|2 = limk→∞

∫K

|∇Xk|2 ≤ lim infk→∞

D(Xk).


Letting K tend to B, we get:

D(Xφ) ≤ lim infk→∞

D(Xk) = DΓ.

The map X is called a classical solution to the Plateau problem for Γ.

Remark 3.1. The solution to Plateau’s problem obtained this way is a branched minimalsurface, i.e. a non constant continuous map X : B → Rn which is almost conformal andharmonic inside B. The Proposition 2.6, which easily extends to minimal surfaces in Rn,holds for such a map exactly as for immersed minimal surfaces except that the condition (10)is lacking. The points where the map is not an immersion are called its branch points. Theyare the zeroes of the holomorphic functions Φ1, . . . ,Φn and so are isolated. R. Ossermanproved that, for n = 3, the solution is free of branch points inside B (the proof was incom-plete and was completed by R. Gulliver). This is not the case for n ≥ 4, there are exampleswhere the solution has branch points. Another issue is the regularity at the boundary. Manyauthors contributed to settle the problem. For more information one may see [12] or [9] andthe references therein.

It remains to check that the classical solution to Plateau’s problem is a homeomorphismon ∂B. To see this we will need the following fact which is of independent interest:

Proposition 3.7 (The reflection principle) Let X : (u, v) ∈ R2, u2 + v2 < ε, v >0 −→ Rn be a branched minimal surface. Suppose there exists a line L in Rn such thatX(u, v)→ L when v → 0. Then X can be extended as a branched minimal surface definedin the whole disk (u, v) ∈ R2, u2 + v2 < ε by reflection across L.

Proof. We may assume, after a rotation in space, that L is the xn axis. By the reflectionprinciple for harmonic functions, the functions X1, X2 extend to harmonic functions on thewhole disk by setting Xk(u, 0) = 0, Xk(u, v) = −Xk(u,−v); k = 1, . . . , n − 1. Thus thefunctions

Φk =∂Xk

∂u− i∂Xk

∂v, k = 1, . . . , n− 1

are holomorphic on the whole disk and imaginary on the real axis. The equation:

Φ2n = −Φ2

1 − . . .− Φ2n−1,

shows Φn extends continuously to the real axis and takes nonnegative real values there.Therefore Φn extends continuously to the real axis and has real values there. By the re-flection principle for holomorphic functions, we can extend Φn to the whole disk by settingΦn(z) = Φn(z). IntegratingXn(z) = <e(

∫ z0

Φn(z)dz),we see thatXn extends to the wholedisk by the relation:

Xn(u, v) = Xn(u,−v).

We can now prove the following:

R. SOUAM 87

Proposition 3.8 For each solution X : B → Rn to the Plateau problem for Γ, the mapX|∂B : ∂B → Γ is a homeomorphism.

Proof. Since X|∂B −→ Γ is monotone, the only way it fails to be a homoeomorphism isthat it sends an arc of ∂B to a single point. Suppose this is the case. After a preliminaryconformal map ofB onto the upper half-plane we can apply the reflection principle (Proposi-tion 3.7) and extend the surface to a branched surface beyond a segment of the real axis. Thebranched surface thus obtained would send the whole segment to a point. This contradictsthe isolatedness of branch points of branched minimal surfaces (cf. Remark 3.1).

The solution to Plateau’s problem and the reflection principle give a way to constructcomplete surfaces. One may start with a polygonal curve and solve the Plateau problem andthen use the reflection principle to reflect the solution across the edges on the boundary. Onecan then do again reflections across the new boundary edges and so on. Of course the surfacemay be singular at the vertices of the initial boundary (and their images after reflections). Ifone chooses carefully the polygonal curve one can obtain a complete minimal surface whichis immersed everywhere. An example obtained starting from an appropiate quadrilateral inspace is a famous surface discovered by Schwarz (figure 5). It is triply periodic (i.e invariantunder three independent translations) and embedded.

The solution to Plateau’s problem gives a surface which minimizes area among disk-type surfaces. This does not mean the solution obtained has the smallest area among all thepossible surfaces bounded by the curve. The proof of the existence of a surface minimizingthe area among all possible surfaces without restriction on their topological type requiresthe techniques of geomeric measure theory. This is a difficult but very powerful theory forestablishing existence and regularity of volume minimizing submanifolds and applies to moregeneral manifolds. A readable introduction to this subject is [11].

Figure 5: The Schwarz surface


4 The stability of minimal surfaces

4. 1 The second variation formulaWe are now interested in the second derivative of the area functional for minimal surfaces inorder to study those which are minimizing up to the second order. It is not true in general thata minimal surface minimizes the area with respect to its boundary even among neighbooringsurfaces. We will see this happens for instance for some catenoid slices as a consequenceof the discussion to follow. However it can be proved that each point on a minimal surfaceadmits a neighborhood which is minimizing with respect to its boundary.

We will now derive the second variation formula. We first need to recall some definitions.

Definition 4.1 (The gradient and the Laplacian) Let (S, 〈, 〉) be a Riemannian surface andf : S → R a C1 function on S. The gradient of f, denoted ∇f, is the vector field on S suchthat:

〈∇f(p), w〉p = dfp(w),

for every p ∈ S and w ∈ TpS.Let f ∈ C2(S). The Laplacian of f, denoted ∆f, is the function defined by:

∆f = div(∇f).

With this definition, the Laplacian on the plane for the Euclidean metric ds20 = du2 +dv2

is ∆0 = ∂2

∂u2 + ∂2

∂v2 . If the metric on a surface S is given in conformal coordinates by:

ds2 = λ2(du2 + dv2),

then the associated Laplacian is related to the Euclidean Laplacian as follows:

∆ =1λ2

∆0. (17)

We can now state the second variation formula. To simplify the presentation, we restrictourselves to normal variations, however the same formula holds in the general case.

Proposition 4.2 (The second variation formula) Let Xt be a normal boundary preservingvariation of a minimal immersion X : D → R3, where D is a compact orientable surface,oriented by a choice of a unit normal field N. Call ξ the deformation vector field of Xt andset φ = 〈ξ,N〉. Then the second derivative of the area is:

d2A

dt2(0) = −

∫D

φ∆φ− 2Kφ2dA

=∫D

|∇φ|2 + 2Kφ2dA.

Proof. The second equality follows from Stokes’ theorem and the esaliy checked identity:

div(φ∇φ) = |∇φ|2 + φ∆φ.

R. SOUAM 89

Denote by Nt the unit normal field to the immersion Xt depending smoothly on t, withN0 = N and let Ht and dAt be respectively the mean curvature and the area element of Xt.Finally put: ξt = ∂Xt

∂t |t.By the first variation formula (Proposition 2.10), for each t:

dA

dt(t) = −2

∫D

〈ξt, Nt〉Ht dAt.

Taking into account that the immersion X is minimal, i.e H0 = 0, we get:

d2A

dt2(0) = −2

∫D

φdHt

dt|t=0 dA.

It remains to compute dHtdt |t=0. We first compute the derivative N = dNt

dt |t=0. Let p ∈ S and(u, v) local coordinates around p such that the basis Xu, Xv is orthonormal at p. For eacht we have:

〈Nt, (Xt)u〉 = 〈Nt, (Xt)v〉 = 0,

so taking the derivatives:〈N ,Xu〉 = −〈N, ξv〉,〈N ,Xv〉 = −〈N, ξu〉.

Now, ξu = (φN)u = dφ( ∂∂u )N + φNu and ξv = (φN)v = dφ( ∂∂v ) + φNv. So:

〈N ,Xu〉 = −dφ(∂

∂u),

〈N ,Xv〉 = −dφ(∂

∂v).

Moreover since 〈Nt, Nt〉 = 1 we have: 〈N ,N〉 = 0. It follows that at the point p :

N(p) = −∇φ(p).

Since this is true for every p ∈ S, we conclude that:

N = −∇φ (18)

Denote by Bt the shape operator of the immersion Xt, i.e the field of endomorphisms ofTS defined by:

It(Btw1, w2) = IIt(w1, w2),

which means:〈dXt(Btw1), dXt(w2)〉 = −〈dNt(w1), dXt(w2)〉.

Put B = dBtdt |t=0. Taking the derivatives at t = 0 :

〈dξ(Bw1) + dX(Bw1), dX(w2)〉+ 〈dX(Bw1), dξ(w2)〉 = −〈d(N)(w1), dX(w2)〉−〈dN(w1), dξ(w2)〉.


Writing ξ = φN and using (18), we obtain after simple calculations:

φ〈dN(Bw1), dX(w2)〉+ 〈dX(Bw1), dX(w2)〉+ φ〈dX(Bw1), dN(w2)〉 == 〈d(∇φ)(w1), dX(w2)〉 − φ〈dN(w1), dN(w2)〉.

Finally:

〈dX(Bw1), dX(w2)〉 = 〈d(∇φ)(w1), dX(w2)〉+ φ〈dX(Bw1), dX(Bw2)〉.

Take any point p ∈ S and a basis e1, e2 of TpS formed of principal eigenvectors withassociated principal curvatures k and −k (remember the surface is minimal). It follows that:

2dHt

dt|t=0(p) = tr(B)(p)

= 〈dX(Be1), dX(e1)〉+ 〈dX(Be2), dX(e2)〉= div(∇φ)(p) + 2k2φ(p).

This is true for any p ∈ S, so finally:

2dHt

dt|t=0 = ∆φ− 2Kφ.

A deformation Xt of an immersion is said trivial if Xt = X, for every t.

Definition 4.3 A compact domain D ⊂ S on a minimal immersion X : S → R3 is saidstable if for every nontrivial normal boundary preserving variation Xt on D :

d2A

dt2(0) > 0.

Equivalently: ∫D

|∇φ|2 + 2Kφ2dA > 0,

for all φ ∈ C∞0 (D), φ not identically 0.

This definition means that the domain is minimizing up to the second order for boundarypreserving variations. For instance, a domain on a plane is clearly stable since the curvaturevanishes identically.

4. 2 The Barbosa-do Carmo stability criterionThe Barbosa-do Carmo theorem ([2]) gives a nice and easy to verify criterion for the stabilityof minimal surfaces.

Theorem 4.4 (The Barbosa-do Carmo stability theorem) Let D be a regular compact do-main on a minimal surface. Assume the area of the Gaussian image N(D) on the sphere isless than 2π. Then D is stable.

R. SOUAM 91

To prove the theorem we need to recall some basic facts about the fundamental Dirichlet-eigenvalue of Schrodinger operators on surfaces. These notions make sense on Riemannianmanifolds of any dimension but we will restrict ourselves to the two-dimensional case. Moredetails can be found in [3] and [5].

Proposition 4.5 Let S be a Riemannian surface, q ∈ C∞(S) a smooth function on S andΩ ⊂ S an open regular domain with compact closure in S. Then the operator ∆ + q has adiscrete spectrum for the Dirichlet eigenvalue problem on Ω. The eigenfunctions are smoothup to the boundary of Ω. The first eigenvalue λ1(∆ + q,Ω) is simple and is characterized bythe property:

λ1(∆ + q,Ω) = inff∈C2

0 (Ω), f 6=0

∫Ω|∇f |2 − qf2 dA∫

Ωf2 dA

. (19)

Moreover the associated nontrivial eigenfunctions realize the infimum in (19) and arecharacterized, among other eigenfunctions, by the fact that they do not vanish inside D.

Furthermore, the first eigenvalue enjoys a monotonicity property: if Ω ⊂ Ω′ ⊂ S are twoopen regular domains with compact closure and Ω 6= Ω′, then:

λ1(∆ + q,Ω) > λ1(∆ + q,Ω′).

Let us find for instance the fundamental eigenvalue of the Laplacian on a hemisphereon S2. We denote by ∆1 the Laplacian on S2. It suffices to consider the hemisphere S2

+ =(x1, x2, x3) ∈ S2, x3 ≥ 0. Then since S2 has mean curvature −1 with respect to the out-ward unit normal,the restriction h of the coordinate function x3 to S2 satisfies (cf. Proposition2.4 and (17)):

∆1h+ 2h = 0 in S2+,

h = 0 on ∂S2+.

The Proposition 4.5 then shows:λ1(∆1,S2

+) = 2. (20)

We will need the following result:

Theorem 4.6 (Faber-Krahn) Let Ω be a regular domain on the unit sphere S2 and let Ω? bea round disk having the same area as Ω. Then:

λ1(∆1,Ω) ≥ λ1(∆1,Ω?)

with equality if and only if Ω is a round disk.

This theorem was proved by Faber and Krahn in the Euclidean case but the proof extendsto the spherical case (and to the hyperbolic one too). For the proof see [3] or [5].

The Theorem 4.6, the monotonicity property (cf. Proposition 4.5) and (20) then give the:

Corollary 4.7 Let Ω be a domain in S2. Assume the area of Ω is less than 2π. Then:

λ1(∆1,Ω) > 2.


We now use the following result due to D. Fischer-Colbrie and R. Schoen [6] (whichholds in any dimension). We give an independent proof.

Lemma 4.8 Consider a Riemannian surface S and denote by ∆ the associated Laplacian.Let q be a smooth function on S andD ⊂ S be a compact domain in S with smooth boundary.Assume there exists a function u ∈ C2(D) ∩ C0(D) satisfying:

∆u+ qu ≤ 0 in D and u > 0 on D.

Then λ1(∆ + q,D) > 0.

Proof. Set w = ∆u + qu. Let h ∈ C∞0 (D), since u > 0 we can write: h = ψu withψ ∈ C2

0 (D). Using Stokes’ formula, we can write:

∫D

|∇h|2dA = −∫D

h∆h dA = −∫D

ψu∆(ψu)dA

= −∫D

ψu(u∆ψ + ψ∆u+ 2〈∇u,∇ψ〉)dA

= −∫D

ψu2∆ψ + ψ2u(w − qu) + 2uψ〈∇u,∇ψ〉dA

So:∫D

|∇h|2 − qh2dA = −∫D

ψu2∆ψ + 2uψ〈∇u,∇ψ〉dA−∫D

ψ2uwdA

≥ −∫D

ψu2∆ψ + 2uψ〈∇u,∇ψ〉dA.

Using again Stokes’ formula:∫D

ψu2∆ψ dA =∫D

div(ψu2∇ψ)− 〈∇(ψu2),∇ψ〉 dA

= −∫D

〈∇(ψu2),∇ψ〉 dA

= −∫D

2uψ〈∇u,∇ψ〉+ u2|∇ψ|2dA.

Therefore: ∫D

|∇h|2 − qh2dA ≥∫D

u2|∇ψ|2dA ≥ 0.

Moreover if the leftmost-hand side vanishes thenψ is constant and soψ = 0 (sinceψ vanisheson ∂D) and thus also h = 0. We conclude then by the characterization (19).

Proof of Barbosa-do Carmo’s theorem. As we observed above, a domain on a plane is stable,so we will assume our domain does not lie on a plane. We will prove the existence of functionu ∈ C2(D) ∩ C0(D), which is positive on D and satisfying: ∆u − 2Ku ≤ 0. The Lemma

R. SOUAM 93

4.8 will then imply that λ1(∆− 2K,D) > 0, where ∆ is the Laplacian for the metric ds2 onthe surface. Taking into account the Proposition 4.8, this then means D is stable.

Set Ω = N(D) ⊂ S2. Since N is conformal and non constant, Ω is an open set withpiecewise smooth boundary. By hypothesis area(Ω) < 2π.We may consider an open domainΩ with smooth boundary such that Ω ⊂ Ω and area(Ω) < 2π. By the Corollary 4.7 we have:

λ1(∆1, Ω) > 2.

Let v a nontrivial eigenfunction associated to λ1(∆1, Ω).We may assume v > 0 inside Ω (cf.Proposition 4.5). Then for the spherical Laplacian ∆1, we have:

∆1v + 2v < ∆1v + λ1(∆1, Ω)v = 0 in Ω and v > 0 on Ω.

On the set D0 = D − K = 0, the Gauss map is a local diffeomorphism, so we mayconsider the pull-back metric

ds22 = N∗ds2

1.

For the Laplacian ∆2 associated to ds22, there holds the relation:

∆2(v N) = ∆1v.

It follows that the function u := v N : D → R satisfies on D :

∆2u+ 2u < 0.

Furthermore, since N is conformal, we have:

ds22 = N∗ds2

1 = −Kds2.

It is straightforward to check the relation between the Laplacians:

∆2 =∆

(−K).

It follows (since K < 0) that u satisfies:

∆u− 2Ku < 0 in D0.

It follows from (13) that on a nonplanar minimal surface the zeroes of the Gaussian curvatureare isolated. Therefore we have by continuity:

∆u− 2Ku ≤ 0 in D.

Moreover, by construction, the function u is positive on D.

Remark 4.1. As it should be clear, the proof shows more generally that if the first eigenvalueof the Gaussian image N(D) for the Laplacian with Dirichlet condition is bigger then 2 thenthe domain D is stable. This is the case for instance if area(N(D)) = 2π and N(D) is not ahemisphere.


Corollary 4.9 Let D be a regular domain on a minimal surface in R3. Assume:∣∣∣∣∫D

K dA

∣∣∣∣ < 2π.

Then D is stable.

Proof. The Gauss map of a nonplanar minimal surface is an orientation reversing local diffeo-morphism except at isolated points which are the zeroes of the Gaussian curvature. Therefore:

area(N(D)) =∫N(D)

dA1 ≤∫D

N∗(dA1),

dA1 being the area element on S2. The area element dA of the surface is related to dA1 asfollows (cf. [15]): N∗dA1 = −KdA. Therefore:

area(N(D)) =∫N(D)

dA1 ≤∫D

N∗(dA1) =∫D

(−K)dA =∣∣∣∣∫D

K dA

∣∣∣∣and the result follows from Barbosa-do Carmo’s theorem.

It should be clear from the previous considerations that in case the Gauss map is injective,which happens for instance on the catenoid and Enneper’s surface, then stable domains areprecisely those having absolute total curvature less than 2π.

A question which naturally comes to mind is to determine the complete minimal surfaces(without boundary) which are globally stable. A complete minimal surface is said globallystable if all its compact subdomains are stable. For instance a plane is globally stable. Atheorem proved independently by do Carmo-Peng, Pogorelov and by Fischer-Colbrie andSchoen [6] states that the planes are the only globally stable complete (orientable) minimalsurfaces without boundary in R3.

References[1] J. Barbosa and G. Colares.: Minimal surfaces in R3. Translated from the Portuguese.

Lecture Notes in Mathematics, 1195. Springer-Verlag, Berlin, (1986).

[2] J.L. Barbosa and M.P. do Carmo.:On the size of a stable minimal surface in R3. Amer.J. Math. 98 (1976), no. 2, 515–528.

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Rabah SouamInstitut de Mathematiques de Jussieu - CNRS UMR 7586Universite Paris Diderot - Paris 7, ”Geometrie et Dynamique”Case 7012 75205 - Paris Cedex 13e-mail: [email protected]

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