Topic class on minimal surfaces—lectures by Rick Schoen · 2015. 9. 6. · Topic class on minimal...

Topic class on minimal surfaces—lectures by Rick Schoen

Notes taken by Xin Zhou

Abstract

This series of lecture notes were taken for the topic class on minimal surfaces given by Profes-sor Rick Schoen in the Winter quarter of 2012 at Stanford. We kept the pace of these lectures bydates.

These lectures start from basic materials on minimal surfaces, e.g. first and second variations,and monotonicity formulae, and then discuss several curvature estimates for minimal surfaces.Afterwards, the notes cover basic existence theory for minimal surfaces, e.g. the classical Plateauproblem and the Sacks-Uhlenbeck theorem, and finally end up with a survey of the proof of theWillmore conjecture. The materials covered are very good examples for the application of methodsfrom partial differential equations and calculus of variation.

It is likely that we have numerous typos and mistakes here and there, and would appreciate itif these are brought to our attention.

Contents

1 Introduction and calibrations (1/10/2012) 21.1 Minimal surface equation: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Calibrated property of minimal graphs . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 A general calibrated argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 First variation and consequences (1/12/2012) 62.1 First Variation Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Convex hull property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 Fluxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Monotonicity formula and 2-d Bernstein Theorem (1/17/2012) 83.1 Monotonicity formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.2 Bernstein’s theorem (n=2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Second variation and Stability (1/19/2012) 114.1 Second Variation of Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.2 Jacobi operator and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1

1 INTRODUCTION AND CALIBRATIONS (1/10/2012) 2

5 Criterion for stability (1/24/2012) 14

6 Bochner formula and 2-d stable minimal surface (1/26/2012) 166.1 Bochner Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166.2 Continuity of Section 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

7 Weierstrass representation and Simons Identity(1/31/2012) 207.1 Weierstrass representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207.2 Simons Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

8 Curvature estimates (2/2/2012) 22

9 More curvature estimates in 2-d (2/7/2012) 25

10 Schoen-Simon-Yau curvature estimates and minimal cone (2/9/2012) 2810.1 Curvature estimates by Schoen-Simon-Yau when n ≤ 6 . . . . . . . . . . . . . . . . . 2810.2 Minimal cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

11 Classical Plateau Problem (2/14/2012) 31

12 Continuity of Plateau Problem and Harmonic maps (2/16/2012) 3412.1 Continuity of the Proof of Theorem 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . 3412.2 Harmonic maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

13 Sacks-Uhlenbeck’s theorem (2/21/2012) 3813.1 Hopf differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3813.2 Sacks-Uhlenbeck’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

14 Sacks-Uhlenbeck’s theorem continued (2/23/2012) 40

15 Colding-Minicozzi’s min-max sphere (3/1/2012) (by Xin Zhou) 43

16 Introduction to the Willmore conjecture (3/6/2012) 45

17 Outline of Marques-Neves’s proof of Willmore conjecture [2] (3/8/2012) 47

18 Marques-Neves’s paper 1 (3/13/2012) 50

19 Marques-Neves’s paper 2 (3/15/2012) 52

1 Introduction and calibrations (1/10/2012)

Let Σk ⊂ (Mn, g) be a k−dimensional submanifold of an n−dimensional Riemannian manifold(Mn, g), where∇ is the corresponding Riemannian connection. Let g|Σ be the induced metric. Given


tangent vectors X,Y of Σ, the second fundamental form(abbreviated as 2nd f.f. in the following),which is a vector valued symmetric 2-tensor on Σ, is defined as:

~A(X,Y ) = (∇XY )⊥. (1)

Definition 1.1 The mean curvature of Σ is defined as: ~H = Trg ~A =∑k

i,j=1 gij ~A(ei, ej), where

eiki=1 is a tangent basis of Σ.

Definition 1.2 Σ is called minimal, if ~H = 0.

1.1 Minimal surface equation:

Consider a hyper surface Σn−1u ⊂ Rn as a graph Σu = (x, u(x)) : x = (x1, · · · , xn−1) ∈ Ω

of a function u, where Ω ⊂ Rn. Denote F (x1, · · · , xn−1) = (x1, · · · , xn−1, u(x)). Then the inducedmetric is given by gij = Fxi · Fxj = δij + uiuj , where ui = ∂u

∂xi. The matrix (δij + uiuj) has n − 2

multiple eigenvalues 1 with eigenspace (∇u)⊥ and a single eigenvalue 1+ |∇u|2 with eigenvector∇u.If we think u as a function defined on Rn, then the graph Σu is the level set given by xn − u(x) = 0.So the unite normal of Σu is given by ν = (−∇u,1)√

1+|∇u|2. Hence the inverse matrix for gij is given by

gij = δij − νiνj , where νi = ui√1+|∇u|2

for i = 1, · · · , n− 1.

Now the volume form of Σu is given by dv =√det(g)dx =

√1 + |∇u|2dx. So the volume of

Σu is:

|Σu| =∫

Ω

√1 + |∇|2dx.

Now let us calculate the Euler-Lagrange equation for |Σu|. For any η ∈ C∞c (Ω), suppose u is acritical point of |Σu|, then

d

dt|t=0|Σu+tη| =

∫Ω

∇u · ∇η√1 + |∇u|2

dx

= −∫

Ω

∂

∂xi(

ui√1 + |∇u|2

)ηdx = 0.

(2)

So the divergence form of the minimal surface equation(abbreviated as (MSE) in the following)

(i) :

n−1∑i=1

∂

∂xi(

ui√1 + |∇u|2

) = 0. (3)

Expanding the above equation, we get:

uxixi√1 + |∇u|2

−uxiuxjuxixj

(1 + |∇u|2)3/2= 0,

which can be rewritten as:

1√1 + |∇u|2

(δij −ui√

1 + |∇u|2ui√

1 + |∇u|2)uxixj = 0. (4)


So we get the non divergence form of (MSE):

(ii) :1√

1 + |∇u|2

n−1∑i,j=1

gijuxixj = 0, (5)

where gij = δij − νiνj is the inverse matrix for the induced metric.Now let us calculate the mean curvature of Σu from definition. Firstly,

~A(∂xi , ∂xj ) = (Fxixj )⊥ = (Fxixj · ν︸︷︷︸

hij

)ν.

It is easy to see thatFxixj = (0, · · · , 0, uxixj ),

so hij =uxixj√1+|∇u|2

. So it is easy to see that equation 5 is equivalent with equation 4.

1.2 Calibrated property of minimal graphs

Now extend the unit normal vector field ν to ν in Ω× R in depend of xn.

Claim: (MSE) =⇒ divRn(ν) = 0.

This is because divRn(ν) = −∑n−1

i=1 ∂xiνi + ∂xnνn︸︷︷︸≡0

= 0 by (MSE).

We can define an n− 1 form ω as

ω = (−1)n−1νcdx,

where dx = dx1 ∧ · · · ∧ dxn−1 is the volume form of Rn.

Claim: divRn(ν) = 0 =⇒ dω = 0.

Proof:

ω = (−1)n−1νcdx = (−1)n−1n∑j=1

(−1)jdx1 ∧ · · · ∧ ˆdxj ∧ · · · ∧ dxn.

So dω = (−1)ndiv(ν)dx = 0.

Definition 1.3 Given an n − 1 dimensional plane V n−1 ⊂ Rn in Rn with an oriented orthonormalframe e1, · · · , en−1, we can define:

ω(V ) = ω(e1, · · · , en−1).


Properties of ω : (1) dω = 0; (2) |ωx(V )| ≤ 1, ∀x ∈ Ω × R and ∀V , and ωx(V ) = 1 only ifV = TxΣx.Proof: (of Property (2)) This comes from the following equation and basic linear algebra.

ν(x)cdx(e1, · · · , en−1) = dx(ν, e1, · · · , en−1) = det(ν(x), e1, · · · , en−1).

Theorem 1.1 (1) Σ is volume minimizing in Ω× R;(2) If Ω is convex, then Σ is volume minimizing in Rn.

Proof: (1) For any n − 1 dimensional sub manifold Σn−11 ⊂ Ω × R, with ∂Σ1 = ∂Σ, we can form

the n dimensional chain U such that ∂Σ1 ∪ Σ = ∂U . Using the Stokes Theorem and property (2) asabove: ∫

Σω︸︷︷︸

=|Σ|

−∫

Σ1

ω︸︷︷︸≤|Σ1|

=

∫Udω = 0.

Here ω is called the calibrated form.(2) Consider the nearest point projection map F : Rn → Ω× R. F is distance decreasing. So we

can firstly contract any Σ1 with the same boundary as Σ to Ω × R which decreases the area and thenuse part (1).

1.3 A general calibrated argument

Theorem 1.2 Suppose Ω is an open region in an oriented Riemannian manifold Mn, and there exist afoliation of Ω by oriented minimal hyper surfaces, then every leaves of the foliation minimizes volumein Ω.

Proof: Let ν(x) be the unit normal vector fields of the foliation. ThenClaim: divν = 0 in Ω if each Σ is minimal.To prove this, take e1, · · · , en−1 to be tangent orthonormal frames of the foliation, then:

divν =n−1∑i=1

〈∇eiν, ei〉+ 〈∇νν, ν〉︸︷︷︸=0, as ν is unit.

= −HΣx = 0.

Defineω = (−1)n−1νcdvM .

Using ω as a calibrated form and arguments above, we can show the minimizing property.

2 FIRST VARIATION AND CONSEQUENCES (1/12/2012) 6

2 First variation and consequences (1/12/2012)

2.1 First Variation Formula

Consider Σk ⊂ Mn. Let X be a smooth v.f.(abbreviated for vector field in the following) on Mwith compact support. Let Ft : M → M be a family of diffeomorphisms such that F0 = id andddt |t=0Ft = X . Then

(First Variation Formula:) δΣ(X) =

∫ΣdivΣ(X)dµ. (6)

Here divΣ(X) = Trg(〈∇·X, ·〉) =∑k

i=1〈∇eiX, ei〉, with e1, · · · , ek an o.n.(abbreviated for or-thonormal in the following) basis for Σ.

When Σ is smooth, we can decompose X = XT +X⊥ to tangent XT and normal X⊥ parts. So

divΣ(X) = divΣ(XT ) + divΣ(X⊥),

where divΣ(X⊥) = −〈 ~X, ~H〉. Using the divergence theorem, for general X , we have

δΣ(X) = −∫

Σ〈X,H〉dµ+

∫∂Σ〈X, η〉dσ,

where η is the outer normal of ∂Σ. So we know that

Σ is minimal( ~H = 0)⇐⇒ δΣ(X) = 0, ∀X of compact support.

Proof: (of 1st Variation formula) Consider a local parametrization

F : Σ× (−ε, ε)→M,

where F (x, t) = Ft(x) with Ft given to be the integration of X above. Let x1, · · · , xk be localcoordinates of Σ, then

d

dt|t=0|Σt| =

∫d

dt|t=0

√detg(t)dx.

Now ddt |t=0

√detg(t) = gij〈∇∂xi F , ∂xj 〉

√detg = divΣX

√detg. So we are done.

2.2 Examples

2.3 Convex hull property

Consider Σk ⊂ Rn. Let x1, · · · , xn be coordinates on Rn, then we have:

Propostion 2.1~H = 0⇐⇒4Σxi = 0, ∀i.

2 FIRST VARIATION AND CONSEQUENCES (1/12/2012) 7

Proof: Let e1, · · · , ek be a local o.n. basis for TΣ, then

4Σxi =

k∑j=1

(ejejxi − (∇Σ

ejej)xi) =

k∑j=1

(∇⊥ejej)xi.

As ej~x = ej and∑∇⊥ejej = ~H ,

4Σ~x = ~H,

where ~x is the position vector.

Corollary 2.1 If (Σk, ∂Σ) is compact minimal in Rn, then Σ ⊂ C(∂Σ), where C(A) is the convex hallof A.

Proof: C(A) = ∩H : closed half spaces with A ⊂ H. Now H = x : l(x) ≤ a, where l is somelinear function and a ∈ R.

∂Σ ⊂ H =⇒ l(x) ≤ a, ∀x ∈ ∂Σ &4Σl(x) = 0, ∀x ∈ Σ,

=⇒ l(x) ≤ a, ∀x ∈ Σ (weak M.P.) =⇒ Σ ⊂ H.

2.4 Fluxes

In the case of a minimal sub manifold Σk ⊂ Rn,

4Σxi = 0⇐⇒ divΣ(∇xi) = 0⇐⇒ ∗dxi is closed.

Hence ∗dxi defines a (k− 1) dimensional deRham cohomology class. So for any k− 1 cycle Γk−1 ⊂Σk, we can define

F ([Γ]) =

∫Γ∗dxi =

∫Γ〈 ∂∂xi

, η〉dσ,

where η is the unit outer normal of Γ. The second ” = ” follows from the fact that ∗dxi = i ∂∂xi

dvΣ =

〈 ∂∂xi , η〉dσ, with i·dv the inner multiplication. Hence we have a group homeomorphism:

F : Hk−1(Σ)→ R.

For the general cases, if Σk ⊂Mn is a minimal sub manifold, i.e. ~H = 0, and X a Killing vectorfield on M , then there exists a homeomorphism:

FX : Hk−1(Σ,Z)→ R.

3 MONOTONICITY FORMULA AND 2-D BERNSTEIN THEOREM (1/17/2012) 8

Proof: Let V = XT the tangential part of X on Σ, then

divΣ(V ) =

k∑i=1

〈∇ei(X −X⊥), ei〉 =

k∑i=1

〈∇eiX, ei〉︸︷︷︸=0, as LXg=0

−k∑i=1

〈∇eiX⊥, ei〉︸︷︷︸= ~H·X=0

= 0.

Let ω = V cdvolΣ, then ω is a closed k − 1 form on Σ, i.e. dω = 0. Hence we can define the flux asabove.

3 Monotonicity formula and 2-d Bernstein Theorem (1/17/2012)

3.1 Monotonicity formula

Theorem 3.1 Let Σk ⊂ Rn be a minimal surface, then

|Bt(x0) ∩ Σ|tk

− |Bs(x0) ∩ Σ|st

=

∫Σ∩(Bt(x0)\Bs(x0))

|(x− x0)⊥|2

|x− x0|k+2dvΣ,

whereBt(x0) is a ball of radius t with center x0 in Rn; |Bt(x0)∩Σ| is the volume in Σ; and (x−x0)⊥

is the projection to the normal part of Σ of (x− x0)⊥.

We need the co-area formula before the proof.

Lemma 3.1 (Co-area Formula) Let h : Σ→ R+ be a nonnegative Lipschitz function on a Riemannianmanifold Σ, and proper i.e. x ∈ Σ : h(x) ≤ a is compact for all a. Given f integrable on Σ, then∫

h≤tf |∇Σh| =

∫ t

−∞(

∫h=t

f)dτ.

Remark 3.1 This follows heuristically from the fact that dvΣ =dt∧dvh=t|∇Σh| when t is a regular value.

Proof: (Monotonicity formula) Take h(x) = |x− x0|, then h ≤ t = Bt(x0). Let X = x− x0, thendivΣ(X) =

∑ki=1∇eiX · ei = k, where e1, · · · , ek is an o.n. basis on Σ. Then by the first variation

formula 6,

δΣBr(x0)(X) =

∫Σ∩Br(x0)

divΣ0(X) =

∫Σ∩∂Br(x0)

X · η,

where η is the co-normal vector of Σ ∩ ∂Br(x0), and η = ∇Σ|x−x0||∇Σ|x−x0|| = (x−x0)T

|(x−x0)T | . Using the Co-areaformular,

k|Σ ∩Br(x0)| =∫

Σ∩|x−x0|=r|(x− x0)T | = d

dr

∫Σ∩Br(x0)

|(x− x0)T ||∇Σ|x− x0||,


=d

dr

∫Σ∩Br(x0)

|(x− x0)T |2

|x− x0|= r

d

dr

∫Σ∩Br(x0)

|(x− x0)T |2

|x− x0|2,

= rd

dr

∫Σ∩Br(x0)

(1− |(x− x0)⊥|2

|x− x0|2),

= rd

dr|Σ ∩Br(x0)| − r d

dr

∫Σ∩Br(x0)

|(x− x0)⊥|2

|x− x0|2.

Multiplying the above by r−k−1, we can re-write it as,

d

dr(r−k|Σ ∩Br(x0)|) = r−k

d

dr

∫Σ∩Br(x0)

|(x− x0)⊥|2

|x− x0|2,

=d

dr

∫Σ∩Br(x0)

|(x− x0)⊥|2

|x− x0|k+2.

In the last step, we can use the co-area formula again to absorb the factor r−k into the integration. Sowe can get the monotonicity formula by integrating the above equation.

Corollary 3.1 Let Σk be a smooth minimal surface in Rn, with boundary Σ ∩ ∂BR(0) inside the ballBR(0). If x0 ∈ Σ ∩BR(0), and σ < R− |x0|, then

ωkσk ≤ |Σ ∩Bσ(x0)| ≤ σk

(R− |x0|k)||Σ ∩BR(0),

where ωk is the volume of unit ball Bk1 (0) in Rk.

Proof: The first ” ≤ ” comes from the Monotonicity formula while comparing Bσ(x0) with anarbitrary small ball Br(x0), with limr→0 r

−k|Σ ∩ Br(x0)| = ωk, when x0 ∈ Σ and Σ smooth. Thesecond ≤ is a direct consequence of the Monotonicity formula while comparing Bσ(x0) with a largeball Br(x0) exhausting the whole BR(0).

Definition 3.1 The density of Σ at x0 is defined as:

Θx0 = limr→0

(ωkrk)−1|Σ ∩Br(x0)|.


3.2 Bernstein’s theorem (n=2)

Theorem 3.2 S. Bernstein (1912) Given a minimal graph Σ2 ⊂ R3, Σ = (x, u(x)) : x ∈ R2. If uis defined on all of R2, then u is a linear function, and Σ is a plane.

Theorem 3.3 Bernstein’s Big Theorem: 1 PDE version: let u ∈ C2(R2) and∑2

i,j=1 aijuiuj = 0,with (aij) > 0. If u is bounded, then u ≡ const;2 : Σ2 −Graphu, where u is defined on R2 and bounded, if the Gaussian curvature KΣ ≤ 0, then Σ

is a plane.

Consider the Gauss Maps:N : Σ2 → S2,

where N maps a point to the unit normal vector at that point.

Lemma 3.2 If ~H = 0, then N is a conformal and orientation reversing map, i.e. ∀v, w ∈ TxΣ, ifv · w = 0 and |v| = |w|, then ∇vN · ∇wN = 0, and |∇vN | = |∇wN |. Furthermore |∇vN | ≤

1√2|A||v|, and N∗(ωS2) = KΣωΣ = −1

2 |A|2ωΣ.

Proof: We only need to check that under a special o.n. basis. Take an o.n. principle basis e1, e2for Σ, i.e. ∇e1N = −K1e1, ∇e2N = −K2e2. So |∇e1N | = |K1| = |K2| = |∇e2N |, by theminimality H = K1 +K2 = 0. Hence |∇vN | ≤ |K||v| = 1√

2|A|. Furthermore, the Jacobian of N is

Jac(N) = K1K2 = −12 |A|

2.

Propostion 3.1 Given a minimal Σ2 ⊂ R3, with the image of the Gauss Maps lying in the upperhemisphere N(Σ) ⊂ S2

+, if ϕ has compact support on Σ, then there exists a constant C > 0, such that∫Σ|A|2ϕ2 ≤ C

∫Σ|∇ϕ|2.

Proof: Since S2+ is simply connected, the closed form ωS2 = dα is also exact. Hence

−|A|2

2ωΣ = N∗ωS2 = d(N∗α).

So ∫Σ|A|2ϕ2ωΣ = −2

∫Σϕ2d(N∗α) = 4

∫Σϕdϕ ∧N∗α

≤ 4

∫Σ|ϕ||∇ϕ||N∗α|ωΣ.

Since |N∗α| ≤ |A||α| ≤ C|A|,∫Σ|A|2ϕ2 ≤ C

∫Σ

(|ϕ||A|)(|∇ϕ|) ≤ ε

2

∫|A|2ϕ2 +

C

2ε

∫|∇ϕ|2.

So we get the inequality.

4 SECOND VARIATION AND STABILITY (1/19/2012) 11

Propostion 3.2 Let Σ be an entire minimal graph, then |Σ ∩BR(0)| ≤ 4πR2, ∀R > 0.

Proof: This comes from the area-minimizing property of minimal graphs. We can compare |Σ ∩BR(0)| with the large area of the truncated surfaces of BR(0) by Σ.

Lemma 3.3 When Σ = Graphu and u is an entire function on R2, then we can choose a Lipschetzϕ = ϕR, such that

∫Σ |∇ϕ|

2 → 0 as R→∞.

Proof: Now choose

ϕR(r) =

1 if r ≤ R1− log(r/R)/ logR if R < r < R2

0 if r ≥ R2

where r is the distance function of R3. By discretize BR2 \BR = ∪logRk=1 (BekR \Bek−1R), we have

∫Σ|∇ϕ|2ωΣ =

1

(logR)2

∫Σ∩(BR2\BR)

1

r2ωΣ =

1

(logR)2

logR∑k=1

∫Σ∩(B

ekR\B

ek−1R)

1

r2ωΣ

≤ 1

(logR)2

logR∑k=1

1

(ek−1R)2|Σ ∩ (BekR \Bek−1R)| ≤ 1

(logR)2

logR∑k=1

1

(ek−1R)2C(ekR)2

=C

(logR)2

logR∑k=1

1

e2=

C

logR→ 0,

where in the second “ ≤ ”, we used the quadratic area bound Lemma above.

Proof: (Bernstein’s Theorem) When Σ = Graphu is an entire graph, the image of the Gauss MapsN(Σ) lies in an hemisphere, so we get

∫Σ |A|

2ϕ2 ≤ C∫

Σ |∇ϕ|2. Then if we take the ϕR in the above

Lemma, and let R→∞, we see that∫

Σ |A|2 → 0. So A = 0, and Σ is a plane.

4 Second variation and Stability (1/19/2012)

4.1 Second Variation of Volume

Consider a minimal Σk ⊂Mn, i.e. ~H = 0. Given a vector fieldX on Σ, let F (x, t) : Σ×[−ε, ε]→M be a one parameter family of variations, with F (x, 0) = X and denote Σt = Ft(Σ).


Theorem 4.1 The Second Variation Formula is:

δ2Σ(X,X) ≡ d2

dt2|t=0|Σt| =

∫Σ

[|∇⊥X|2 − |〈 ~A,X〉|2 −k∑i=1

RM (ei, X, ei, X)], (7)

where e1, · · · , ek is an o.n. basis tangent to Σ, and X is compact supported and normal on Si.

Theorem 4.2 In the case Σn−1 ⊂Mn is a hyper surface and 2-sided(∃ν unit normal), X = ϕν, withϕ a function with compact support, then

δ2Σ(ϕ,ϕ) =

∫Σ

[|∇ϕ|2 − (|A|2 +RicM (ν, ν))ϕ2] (8)

Proof: (of Second Variation Formula)

• F : Σ × (−ε, ε) → M , with x1, · · · , xk local coordinates on Σ. Then dvt =√detg(t)dx,

where gij(t) = 〈 ∂F∂xi, ∂F∂xj〉.

•d

dt

√detg(t) =

1

2gij gij

√detg(t)

•d2

dt2

√detg(t) =

1

4(gij gij)

2√detg +

1

2gij gij

√detg +

1

2(gij gij)

√detg,

where gij = −gikgjlgkl.•

gij = 〈∇ ∂F∂t

∂F

∂xi,∂F

∂xj〉+ 〈 ∂F

∂xj,∇ ∂F

∂t

∂F

∂xi〉

= |t=0〈 ~A(∂

∂xi,∂

∂xj), X〉.

So(gij gij)|t=0 = −2〈 ~H,X〉 = 0,

and(gij gij)|t=0 = −4|〈 ~A,X〉|2.

•gij = 〈∇ ∂F

∂t∇ ∂F

∂xi

∂F

∂t,∂F

∂xj〉+ 〈∇ ∂F

∂xi

∂F

∂t,∇ ∂F

∂t

∂F

∂xj〉+ i, j reversed terms

= 〈RM (∂F

∂t,∂F

∂xi)∂F

∂t,∂F

∂xj〉+ 〈∇ ∂F

∂xiF ,

∂F

∂xj〉+ 〈∇ ∂F

∂xiX,∇ ∂F

∂xjX〉

+i, j reversed terms

= |t=0 −RM (X, ∂xi, X, ∂xj) + 〈∇∂xiF , ∂xj〉+ 〈∇∂xiX,∇∂xjX〉

+i, j reversed terms

So

(gij gij)|t=0 = −2gijRM (X, ∂xi, X, ∂xj) + 2divΣF + 2|∇⊥X|2 + 2gij〈∇T∂xiX,∇T∂xjX〉︸︷︷︸

=2|〈 ~A,X〉|2


• Combining all the above,

d2

dt2|t=0

√detg(t) = divΣF + |∇⊥X|2 − |〈 ~A,X〉|2 − gijRM (X, ∂xi, X, ∂xj).

An integration on Σ finishes the proof.

4.2 Jacobi operator and Stability

• Hypersurface Case k = n− 1: X = ϕν,

I(ϕ,ϕ) ≡ δ2Σ(ϕ,ϕ) = −∫

ΣϕLϕ,

where the Jacobi operator is

Lϕ = 4ϕ+ (|A|2 +Ric(ν, ν)︸︷︷︸Q

)ϕ. (9)

• When boundary exists (Σ, ∂Σ), L has discrete eigenvalues λj and eigenfunctions uj , i.e. Luj +

λjuj = 0 in Σ with uj = 0 on ∂Σ, and

λ1 < λ2 ≤ λ3 ≤ · · · ,

with λn → +∞.

Definition 4.1 Σ is stable if λ1 ≥ 0, i.e. I(ϕ,ϕ) ≥ 0, ∀ϕ, with ϕ = 0 on ∂Σ.

• Morse Index=# of negative eigenvalues counted with multiplicity.• (Properties of λ1) λ1 has multiplicity 1;• If u1 is an eigenfunction of λ1, u1 does not change sign.

Proof: By the variational characterization, u1 minimizes I(ϕ,ϕ) among all ϕ with ϕ ≡ 0 on∂Σ and

∫Σ ϕ

2 = 1. Since |(|ϕ|, |ϕ|) = I(ϕ,ϕ), if u1 is the first eigenfunction, so is |u1|. Sou1 = |u1|, or there is a contradiction to u1 ∈ C∞. The fact that u1 does not change sign showsthat the dimension of the eigen space of λ1 is 1, or we can always form some eigenfunctionchanging sign.

• When Σ is non-compact, thenInd(Σ) = lim

i→∞Ind(Ωi),

where Ωi∞i=1 is an open exhaustion of Σ, i.e. Σ = ∪∞i=1Ωi, Ωi ⊂ Ωi+1, with ∂Ωi smooth andcompact.

5 CRITERION FOR STABILITY (1/24/2012) 14

Remark 4.1 In fact, the definition is independent of the exhaustion, say Ωi and Ωi. SinceInd(Ω) is non decreasing when Ω is expanding, so we can always embed Ωi ⊂ Ωi′ for i′ i,so Ind(Ωi) ≤ Ind(Ω)i′ , and limi→∞ Ind(Ωi) ≤ limi′→∞ Ind(Ωi′), and vise versa.

• When Σ is open:λ1(Σ) = lim

i→λ1(Ωi) ∈ [−∞, λ1(Ω1)),

Σ is stable if λ1(Σ) ≥ 0, or equivalently λ1(Ω) ≥ 0 for all Ω ⊂ Σ.

Remark 4.2 By the variational characterization, λ1(Ω) is strictly decreasing as Ω is expanding,so we can argue as above to show the well-definedness of λ1(Σ).

5 Criterion for stability (1/24/2012)

Theorem 5.1 Assume Σn−1 ⊂Mn is a 2-sided minimal hypersurface.1 Σ is stable⇐⇒ ∃u > 0, s.t. Lu ≤ 0;2 Σ(non-compact) is stable⇐⇒ ∃u > 0, s.t. Lu = 0.

Remark 5.1 This can be viewed as an infinitesimal version of the Calibration argument i.e. usingfoliation of minimal surfaces.

Proof: 1 + 2 ⇐=: Since Lu = 4Σu+Qu ≤ 0, let w = log u(u > 0), we have

4w =4uu− |∇w|2 ≤ −Q− |∇w|2.

Then ∀ϕ compactly supported, ∫Σϕ2(4w +Q) ≤ −

∫Σϕ2|∇w|2.

Using integration by part formula,∫ΣQϕ2 ≤

∫2ϕ〈∇ϕ,∇w〉 − |∇w|2ϕ2 ≤

∫2|ϕ||∇ϕ||∇w| − |∇w|2ϕ2

≤∫|∇ϕ|2 + ϕ2|∇w|2 − |∇w|2ϕ2 ≤

∫Σ|∇ϕ|2.

Hence we have the stability inequality for Σ.1 =⇒: Assume Σ is compact, then ∃u > 0, which is the first eigenfunction, such that λ1(Σ) ≥ 0, so

Lu = −λ1u ≤ 0.

2 =⇒: Assume Σ is non-compact and stable, then Σ has an exhaustion Σ = ∪∞i=1Ωi, and λ1(Ωi) > 0

for all i. Now by elementary elliptic PDE, ∀ψ on ∂Ωi, ∃!u in Ωi, such that

Lu = 0 in Ωi, with u = ψ on ∂Ωi.

5 CRITERION FOR STABILITY (1/24/2012) 15

Claim: λ1(Ωi) > 0 =⇒ if ψ > 0, then u > 0.(If u ≤ 0, then Ωu≤0 has eigenvalue equals 0, since u is then a Dirichlet eigenfunction on Ωu≤0with 0 boundary values, which is a contradiction to λ1(Ω) > 0. )Now we can solve the boundary value problem for ui > 0:

Lui = 0 in Ωi, with u = 1 on ∂Ωi.

Consider the normalized sequence uiui(0)

∞i=1,

Claim: there exists a subsequence i′ →∞, such that

ui′

ui′(0)→ u in C2 on compact subset of Σ.

Pf:

• C0 bound and elliptic theory of L =⇒ C3 bound for ui′ui′ (0) on compact set;

• Harnack inequality =⇒ C0 bound.• Harnack inequality: If u > 0 is a positive solution of Lu = 0 on Ωopen ⊂M , and if Ω

compact1 ⊂

Ω, then ∃c = c(L,Ω1) > 0, such that

maxΩ1

u ≤ cminΩ1

u.

The limit u is a positive solution of Lu = 0.

Theorem 5.2 If Σ is a stable 2-sided minimal hyper surface in M , and Σ is any covering of Σ, thenΣ is also a stable minimal hyper surface in M .

Remark 5.2 Σ can be viewed as a minimal immersion i : Σ → M , so if π : Σ → Σ is the coveringmap, then Σ is viewed as a minimal immersion i π : Σ→ Σ.

The 2-sided property is essential here. A counterexample is RP1 ⊂ RP2, while the covering spaceS1 is not stable.

Proof:

• Σ is 2-sided and stable =⇒ ∃u > 0 on Σ with Lu ≤ u.• Let π : Σ→ Σ be the covering map, then u π > 0 on Σ and L(u π) ≤ 0, hence Σ is stable.

Propostion 5.1 1. Let Σn−1 ⊂ Rn be a 2-sided minimal surface, and if the Gauss image G(Σ) ⊂Sn−1

+ , then Σ is stable;2. Let Σ2 ⊂ R3 be a 2-sided minimal surface, and if G(Σ) ⊂ Uopen ⊂ S2, with µ1(U) ≥ 1, whereµ1(U) is the Dirichlet eigenvalue of 4S2 on U , then Σ is stable. In perticular, µ1(U) ≥ 1 is true ifthe area |U | ≤ 2π.

6 BOCHNER FORMULA AND 2-D STABLE MINIMAL SURFACE (1/26/2012) 16

Proof: 1. Let e ∈ Rn be the direction vector to the north pole, and let u = e ·ν, where ν is the normalvector field of Σ, since the parallel translation in the e direction does not change the area of Σ, we haveLu = 0. Since G(Σ) ⊂ Sn−1

+ ⇐⇒ e · ν > 0, so u > 0, hence Σ is stable.2. µ1(U) ≥ 1 =⇒ ∃v > 0 on U such that

4S2v = µi(U)v ≤ −v, in U,v = 0, on U.

Let u = v G, where G is the Gauss Map. By Lemma 3.2, G : Σ→ S2 is a conformal map, so

4Σu = |A|2(4S2v) G ≤ −|A|2u,

i.e. Lu ≤ 0, hence Σ is stable. (In fact, on 2-dimension, the Jacobi operator L = G∗(4S2 + 1).)

6 Bochner formula and 2-d stable minimal surface (1/26/2012)

6.1 Bochner Formula

Let (Σk, g) be a Riemannian manifold, and e1, · · · , ek an o.n. frame, with θ1, · · · , θk the dualframe. Denote

(∇ejα) =∑i

αi,jθi,

∇ei(∇α) =∑i,j

αi,jkθi ⊗ θj ;

then∇α =

∑i,j

αi,jθi ⊗ θj , ∇2α =

∑i,j,k

αi,jkθi ⊗ θj ⊗ θk.

Ricci Formula:αi,jk − αi,kj =

∑p αpR

Σpijk.

Definition 6.1 α is harmonic if dα = 0 and δα = 0(i.e. αi,j = αj,i and∑

i αi,i = 0).

Bochner Formula: If α is harmonic, then

4α = Ric(α], ·),

where α] the vector field dual to α, and4α =∑

i,j αi,jjθi is the rough laplacian.

Proof: ∑j

αi,jj =∑j

αj,ij =∑j

αj,ji︸︷︷︸=0

+∑p,j

αpRΣpjij =

∑p

αpRicΣpi.


Hence we have:

124|α|

2 = 〈α,4α〉+ |∇α|2 = Ric(α], α]) + |∇α|2 .

In the case Σn−1 ⊂ Rn is minimal, RΣijkl = hikhjl − hilhjk under the o.n. frame ei by the

Gauss equation, hence

RicΣik =

∑j

RΣijkj = −

∑j

hijhjk, (∑j

hjj = 0).

=⇒ 1

24|α|2 = |∇α|2 +

∑ij

RicΣijαiαj = |∇α|2 −

∑i

(∑j

hijαj)2 ≥ |∇α|2 − |A|2|α|2.

Plug in 124|α|

2 = |α|4|α|+∣∣∇|α|∣∣2,

|α|(4|α|+ |A|2|α|︸︷︷︸L|α|

) ≥ |∇α|2 −∣∣∇|α|∣∣2 ≥ c(n)

∣∣∇|α|∣∣2,where Lu = 4u+ |A|2u is the stability operator, and c(n) a constant depending only on n.

In general, choose the o.n. basis e1, · · · , ek such that under this basis α1 = |α| and αj = 0 forj = 2, · · · , k, then

|∇α|2 −∣∣∇|α|∣∣2 =

∑ij

α2i,j −

∑j(∑

i αiαi,j)2

|α|2=∑i,j

α2i,j −

∑j

α21,j

=∑i>1,j

α2i,j ≥

k∑i=2

α2i,i +

k∑i=2

α2i,1 ≥

1

k − 1(

k∑i=2

αi,i︸︷︷︸=−α1,1

)2 +

k∑i=2

α21,i

≥ 1

k − 1[α2

1,1 +k∑i=2

α21,i] =

1

k − 1

∣∣∇|α|∣∣2.Theorem 6.1 If Σn−1 ⊂ Rn is a complete, stable and 2-sided minimal surface, then any L2 harmonic1-form on Σ vanishes.

Proof: 2-sided and stability means that −∫

Σ ϕLϕ ≥ 0 for any ϕ compactly supported. So ∀ϕ com-pactly supported

−∫

Σϕ|α|L(ϕ|α|) ≥ 0,

i.e. ∫Σϕ|α|(4(ϕ|α|)︸︷︷︸

I

+|A|2ϕ|α|) ≤ 0,


where

I =

∫Σϕ|α|(ϕ4|α|+ 2〈∇ϕ,∇|α|〉+ |α|4ϕ) =

∫Σϕ2|α|4|α|+ 1

2〈∇ϕ2,∇|α|2〉+ |α|2ϕ4ϕ

≤∫

Σϕ2|α|4|α| −

∫Σ

1

24(ϕ2)|α|2 +

∫Σ|α|2ϕ4ϕ

=

∫Σϕ2|α|4|α| − (ϕ4ϕ+ |∇ϕ|2)|α|2 + |α|2ϕ4ϕ

=

∫Σϕ2|α|4|α| − |∇ϕ|2|α|2.

Plug into the above ∫Σϕ2|α|L(|α|) ≤

∫Σ|∇ϕ|2|α|2.

Now by taking ϕ = ϕR to be cutoff functions on geodesic disk, and letting R → ∞, the righthandside of the above inequality is zero, hence by |α|L(|α|) ≥ c(n)

∣∣∇|α|∣∣ proved above,

c(n)

∫Σ

∣∣∇|α|∣∣2 ≤ ∫Σ|α|L(|α|) = 0,

which means that |α| is a constant, and hence is 0 since the area of Σ is ∞ by the monotonicity|Bσ(p)| ≥ wkσk.

6.2 Continuity of Section 5

Theorem 6.2 Any complete 2-sided stable minimal immersion Σ2 ⊂ R3 is a plane.

Proof: (Σ, g) is an oriented Riemann surface, where g is the restriction metric. If z = x + iy theng = λ2(dx2 + dy2) locally. So Σ has a complex striation. Let Σ be the universal cover of Σ, then Σ isa simply connected non-compact Riemann surface, hence

Σ '

C, the complex plane,D, the unit disk.

Case 1: Σ ' C, then let F : C → R3, where F = i π is given by the composition of the minimalimmersion i : Σ→ R3 with the covering map π : C ' Σ→ Σ. Since i is harmonic, and the harmonicproperty is preserved under the conformal change Σ ' C, we know that F is both conformal andharmonic, i.e. 4CF = 0. Since Σ is stable and 2-sided, Σ is also stable and 2-sided, =⇒ ∃u > 0,such that Lu = 4Σu + |A|2u = 0 on Σ. So 4Σu ≤ 0, hence 4C(u F ) ≤ 0. So u F is asuper-harmonic function. Since C has quadratic area growth,together with the fact that u F > 0, weknow that u F = 0, and hence |A|2 = 0 by the following Proposition.Case 2: Σ ' D.


Claim: the L2 norm of p-forms on an n-dimensional manifold is a conformal invariant whenp = n

2 .This is because |α|2g = gi1j1 · · · gipjp︸︷︷︸

p copies

αi1j1 · · ·αipjp︸︷︷︸p copies

, so if g = λ2g, then

∫|α|2g

√detgdx =

∫λ−2p|α|2gλn

√detgdx =

∫|α|2g

√detgdx.

Furthermore, harmonic p−forms change to harmonic p−forms under conformal change. This isbecause dλ = 0 does not change, while δgα = λ−2δα = 0 when p = n

2 (see Page 59 in [1]).So L2 harmonic 1-forms on D corresponds to L2-harmonic 1-forms on Σ. Since there are many

harmonic 1-forms on D by just taking dx where x is harmonic functions, so it is a contradiction to theTheorem we proved in the above section.

Definition 6.2 A Riemannian manifold Σk is called parabolic if every positive super-harmonic func-tion is constant.

Propostion 6.1 If h : Σ → R1+ is a proper Lipschitz function |∇h| ≤ c, and if |Σa| ≥ caa for some

c > 0, where Σa = p ∈ Σ : h(p) ≤ a, then Σ is parabolic.

Proof: Take a positive super-harmonic function u, i.e. 4u ≤ 0 and u > 0. Take w = log u, then

4w =4uu− |∇w|2 ≤ −|∇w|2.

Take ϕ a compactly supported function,∫Σϕ2|∇w|2 ≤ −

∫Σ4wϕ2 =

∫2〈∇w,ϕ〉ϕ

≤ 2

∫|ϕ||∇w||∇ϕ| ≤ ε

∫ϕ2|∇w|2 +

1

ε

∫|∇ϕ|2.

Taking ε = 12 , then ∫

Σϕ2|∇w|2 ≤ 4

∫Σ|∇ϕ|2.

By taking h = distΣ(·, p), we know that Σ has more than quadratic area growth, so we can takeϕ = ϕR as in Lemma 3.3, and use the same logarithmic cut-off trick, to get

∫Σ |∇ϕR|

2 → 0, andϕR → 1. So |∇w| = 0, and w hence u is a constant.

7 WEIERSTRASS REPRESENTATION AND SIMONS IDENTITY(1/31/2012) 20

7 Weierstrass representation and Simons Identity(1/31/2012)

7.1 Weierstrass representation

Let F : Ω→ Rn be a minimal immersion, where Ω is a Riemann surface with complex coordinatesz = x + iy, then F is conformal(i.e. 〈∂F∂x ,

∂F∂y 〉 = 0 and |∂F∂x | = |∂F∂y |) and harmonic(i.e. 4F = 0).

Defineψdz =

∂F

∂zdz = (

∂F

∂x− i∂F

∂y)(dx+ idy).

Lemma 7.1 The complex vector ψ = ∂F∂z is holomorphic, i.e. ∂ψ∂z = 0, and isotropic, i.e.

∑nj=1 ψ

2j =

0.

Proof: Since F is harmonic,∂ψ

∂z= 4F = 0.

Since F is conformal,n∑j=1

ψ2j = |∂F

∂x|2 − |∂F

∂y|2 + 2i〈∂F

∂x,∂F

∂y〉 = 0.

Conversely, we would like to represent F as F (z) = Re∫ zψ(s)ds.

When n = 3, given a meromorphic function g and a holomorphic one form ϕdz on Ω, we can take

ψdz =(1

2(g−1 − g),

1

2(g−1 + g, 1)

)ϕdz,

and get a minimal immersion F : Ω→ R3 by

F (z) = Re

∫ z

z0

ψ(s)ds.

If N : Ω → S2 is the Gauss Map, then g = π N , with π : S2 → R2 the stereographic projectionfrom (0, 0, 1).

Examples:

• Catenoid: Ω = C \ 0, g(z) = z and ϕ(z) = dzdz ;

• Helicoid: (simply connected π1 = 0) Ω = C, g(z) = eiz and ϕ(z) = dz. If we denote theHelicoid by Σ, then εΣ converges to a Foliation by plane x3 = c, where |A|2 → +∞ on theaxis, but |A|2 → 0 everywhere away from the axis;• Hoffman-White examples.

7 WEIERSTRASS REPRESENTATION AND SIMONS IDENTITY(1/31/2012) 21

7.2 Simons Identity

Consider a minimal hypersurface Σn−1 ⊂ Rn. Let e1, · · · , en−1 be local o.n. frames on Σ,and denote hij,klm by the covariant derivatives of the second fundamental form h on Σ. The roughlaplacian for h is defined as

4hij =n−1∑k=1

hij,kk.

Propostion 7.14hij + |A|2hij = 0, 0 ≤ i, j ≤ n− 1 (10)

Proof: Firstly we have the Ricci identity:

hij,kl − hij,lk =∑p

hpjRpikl +∑p

hipRpjkl,

Gauss Equation:RΣijkl = hikhjl − hilhjk,

and Codazzi equation:hij,k = hik,j .

Using the Eistein summation, we have

4hij = hij,kk = hik,jk = hik,kj︸︷︷︸=hkk,ij=0

+hpkRΣpijk + hipR

Σpkjk

= hpk(hpjhik − hpkhij) + hip(hpj hkk︸︷︷︸=0

−hpkhkj)

= −|A|2hij + (hikhkphpj − hipHpkhkj︸︷︷︸=0

).

So we finished the proof.

Now recall that the stability operator is Lϕ = 4ϕ+ |A|2ϕ.

Propostion 7.2|A|(L|A|) ≥ 2

n− 1

∣∣∇|A|∣∣2. (11)

Proof: By the Bochner Formula,

1

24|A|2 = |∇A|2 + 〈A,4A〉 = |∇A|2 − |A|4.

8 CURVATURE ESTIMATES (2/2/2012) 22

While 124|A|

2 = |A|4|A|+ |∇|A||2,

|A|L(|A|) = |∇A|2 −∣∣∇|A|∣∣2 =

∑i,j,k

h2ij,k −

∑k(∑

ij hijhij,k)2

|A|2.

In an o.n. eigenbasis e1, · · · , en−1 of h, hij = λiδij , so

∣∣∇|A|∣∣2 =

∑k(∑

i λihii,k)2

|A|2≤∑i,k

h2ii,k =

∑i 6=k

h2ii,k +

∑i

h2ii,i

=∑i 6=k

h2ii,k +

∑i

(−∑j 6=i

hjj,i)2 ≤

∑i 6=k

h2ii,k + (n− 2)

∑i 6=j

h2jj,i

= (n− 1)∑i 6=k

h2ii,k =

n− 1

2(∑i 6=k

h2ik,i +

∑i 6=k

hki,i).

So(1 +

2

n− 1)∣∣∇|A|∣∣2 ≤∑

i,k

h2ii,k +

∑i 6=k

h2ik,i +

∑i 6=k

h2ki,i ≤

∑i,j,k

h2ij,k = |∇A|2.

So we finished the proof.

8 Curvature estimates (2/2/2012)

Curvature Estimates:Let Cr0 = Σn−1 ⊂ Rn : HΣ = 0, 0 ∈ Σ, & ∂Σ ∩Br0(0) = ∅.Question: For which C is it true that

supΣ∩Br0/2

|A|2 ≤ Cr−20 ,

where C is independent of Σ ∈ C.Similar Questions: replace Br0 in the above question by geodesic balls BΣ

r0 = x ∈ Σ : dΣ(s, 0) <

r0.Note: BΣ

r0 ⊂ Br0 .

Remark 8.1 1. By scaling, it suffices to assume that r0 = 1, since 1r0

Σ ∈ C1, while |A 1r0

Σ|2 =

r20|A|2Σ.

2. Curvature estimates =⇒ Bernstein Theorem. If Σ is complete and Σ ∈ Cr0 , ∀r0 > 0, then Σ isa plane by taking r0 →∞.

For which Cr0 is it True?


1. n = 3,∫

Σ∩Br0|A|2 < ε, for some fixed small ε =⇒ Curvature Estimates (see Theorem 8.2).

(Counter Example, Down-scaled Catenoid, where∫

Σ∩Br0|A|2 is not small.)

2. n = 3, Σ2 embedded and area minimizing. (e.g. Σ is a graph over a convex region.)3. n = 3, Σ2 embedded, simply connected and |Σ| ≤ A0r

20 =⇒ Curvature Estimates (see Theorem

9.2).

Ex: 1) down-scaled Helicoid, 2) Enneper Surface, immersed |Σ ∩Br0 | ≤ cr20.

4. n = 3, Σ2 is stable and 2-sided =⇒ Curvature Estimates. (see Corollary 9.2).5. 3 ≤ n ≤ 6, Σn−1 ⊂ Rn stable and 2-sided and |Σ ∩ Br0 | ≤ crn−1

0 =⇒ Curvature Estimates(see Theorem 10.1).

6. 3 ≤ n ≤ 7, Σn−1 embedded and absolutely volume minimizing =⇒Curvature Estimates. (Falsefor n− 8)

The Simons cone C = x ∈ R8 :∑4

i=1 x2i =

∑8i=5 x

2i is a minimal cone. In fact, it

is absolutely volume minimizing, since there exists foliation of R8 by minimal hypersurfacesasymptotical to C, where the curvature blows up on C.

7 Σn−1 with small access: Θ0(r0)− 1 < ε, =⇒ Curvature Estimates (see Theorem 9.3).

The access for a k−dimensional sub manifold Σk at 0 ∈ Σk of radius r is defined as:

Θ0(r) =|Σ ∩Br|ωhrk

.

Basic idea: For some small ε > 0, assume the scaling invariant inequality:∫Σ∩Br0

|A|2 < ε.

Suppose|A|2(x0) = max

Σ∩Br0|A|2, at some x0 ∈ Br0/2.

Rescale Σ by the factor δ = 1/|A|(x0), then

Σδ =1

δ(Σ− x0) ∼

|AΣδ(x)|2 = δ2|AΣ|2 ≤ 1, ∀x ∈ B1(0),

|AΣδ(0)|2 = 1,

∫B1∩Σδ

|AΣδ |2 < ε.

Now |AΣδ |(x) ≤ c = 1 =⇒ locally Σδ is a graph Graphu of some function u =⇒ C3 estimates of u,which will form a contradiction with

∫B1∩Σδ

|AΣδ |2 < ε.

Theorem 8.1 (Choi-Schoen [4]) Suppose Σ2 ⊂ M3 is a minimal surface. Assume 0 ∈ Σ2, and∂Σ ∩ Br0(0) = ∅. Then there exists ε, ρ > 0 (depending only on M ), such that if r0 ≤ ρ, δ ∈ (0, 1)

and∫

Σ∩Br0|A|2 < δε, then

|A|2(y) ≤ δσ−2, for y ∈ Br0−σ(0).


Proof: Let us give a proof when M3 = R3, and the general cases follow by the fact that M3 is locallynear R3 when ρ is small enough. Assume δ = 1 and

F (y) =(r0 − r(y)

)2|A|2(y),

where r(y) = d(y, 0). Since F |∂Br0 = 0, then ∃y0 ∈ Br0 , such that F (y0) = maxBr0 F (y).Need to show: F (y) ≤ 1(=⇒ 1 ≥ F (y) ≥ σ2|A|2(y), if r(y) < r0 − δ).Suppose F (y0) > 1, let δ = r0−r(y0)

2 , then

• supBδ(y) |A|2 ≤ 4|A|2(y0).

This is because (r0−r(y))2|A|2(y) ≤ (r0−r(y0))2|A|2(y0), hence |A|2(y) ≤ ( r0−r(y0)r0−r(y) )2|A|2(y0) ≤

4|A|2(y0), where r0 − r(y) ≥ r0 − (r(y0) + δ) ≥ 12(r0 − r(y0)).

• (2δ)2|A|2(y0) = F (y0) > 1 =⇒ δ2|A|2(y0) > 1/4.

Let δ0 = 1|A|(y0) , hence δ2 ≥ 1

4δ20 =⇒ δ0/2 < δ. So Bδ0/2(y0) ⊂ Bδ(y0). Let

Σδ0 =2

δ0(Σ− y0),

=⇒

supB1

|AΣδ0|2 = 4|AΣδ0

|2 = δ20 |A|2(y0) = 1,∫

B1|AΣδ0

|2 ≤ ε.

So it forms a contradiction when ε is too small by the argument discussed in the Basic idea. So F (y0) ≤1.

Theorem 8.2 Assume Σ2 ⊂ R3 is stable and 2-sided with quadratic area growth, i.e. |Σ∩Br0 | ≤ cr20,

thensup

Σ∩Br0/2|A|2 ≤ cr−2

0 .

Proof: By stability, we have∫

Σ |A|2ϕ2 ≤

∫Σ |∇ϕ|

2. Since Σ has quadratic area growth, we can usethe logarithmic cutoff trick to get,∫

Σ∩Br0/k|A|2 ≤ C

log k, k 1.

So for k large enough, we have∫

Σ∩Br1 (y) |A|2 < ε, where r1 = r0/k, hence

|A|2(y) ≤ cr−21 ≤ c′r−2

0 , c′ = kc.

9 MORE CURVATURE ESTIMATES IN 2-D (2/7/2012) 25

9 More curvature estimates in 2-d (2/7/2012)

Let us firstly give a technical lemma used in the argument of the above section.

Lemma 9.1 Σ2 ⊂ Rn is minimal. Assume that s2 supΣ |A|2 ≤ 116 . If x ∈ Σ2 and distΣ(s, ∂Σ) ≥ 2s,

then

(i) BΣ2s(x) is graphical over TxΣ of some function u, where BΣ

2s(x) is the geodesic ball of Σ, and|∇u| ≤ 1 and |Hessu| ≤ 1/

√2;

(ii) Let Σ′ be a connected component of Bs(x) ∩ Σ containing x, then Σ′ ⊂ BΣ2s(x).

Proof: See [3, §3. Chap 2].

Lemma 9.2 Let Σ2 be a simply-connect minimal surface. Fix x ∈ Σ, then r(y) = distΣ(y, x) is asmooth function when y 6= x. Let K be the Gauss curvature of Σ, then

•|∂BΣ

r0(x)| − 2πr0 = −∫ r0

0

∫BΣρ

K =︸︷︷︸H=0

1

2

∫ r0

0

∫BΣρ

|A|2;

•|BΣ

r0(x)| − πr20 = −

∫ r0

0

∫ τ

0

∫BΣρ

K =︸︷︷︸H=0

1

2

∫ r0

0

∫ τ

0

∫BΣρ

|A|2;

• When t < r0/2

t2∫BΣr0−2t

|A|2 ≤∫BΣr0

|A|2 (r0 − r)2

2=

∫ r0

0

∫ τ

0

∫BΣρ

|A|2

= 2(|BΣr0 | − πr

20) ≤ r0(|∂BΣ

r0 | − 2πr0).

Proof: Using co-area formula and Gauss-Bonnet formula on the simply-connected domain BΣr ,

d

dr|BΣ

r | = |∂BΣr |,

d

dr|∂BΣ

r | =∫∂BΣ

r

kg = 2π −∫BΣr

K,

where kg is the geodesic curvature of the curve ∂BΣr . Integrate the 2nd of the above, i.e.

∫ r00 =⇒

|∂BΣr0 | = 2πr0 −

∫ r0

0

∫BΣρ

K.

Integrate again =⇒|BΣ

r0 | = πr20 −

∫ r0

0

∫ τ

0

∫BΣρ

K.


For the last one, if t < 12r0,

t2∫BΣr0−2t

|A|2 ≤∫BΣr0

|A|2 (r0 − r)2

2=

∫ r0

0

(r0 − r)2

2

d

dr

∫BΣr

|A|2dr

=

∫ r0

0(r0 − r)

∫BΣr

|A|2︸︷︷︸= ddr

∫ r0

∫BΣr|A|2

dr =

∫ r0

0

∫ r

0

∫BΣρ

|A|2.

Here we used the integration by part twice in the second and third = . Using the second equation,we get the first part. The inequality 2|BΣ

r0 | ≤ r0|∂BΣr0 | ⇐⇒ non-positive curvature and simply-

connected(This can be proved by using the integration inequality∫BΣr0

(4r2 ≥ 4), where 4r2 ≥ 4

comes from the fact that K ≤ 0.).

Theorem 9.1 Assume that Σ2 is stable and 2-sided in R3. If x ∈ Σ and dist(x, ∂Σ) ≥ r0, then

|BΣr0 | ≤

4π

3r2

0.

Proof: It suffices to assume π1(Σ) = 1, or we can pass to the universal cover Σ of Σ, which is alsostable and 2-sided (see Lecture 5). Since |BΣ

r | ≥ |BΣr |, we can get the result. Using the last equality

in the above lemma,

4(|BΣr0 | − πr

20) =

∫BΣr0

|A|2(r0 − r)2 ≤∫BΣr0

|∇(r0 − r)|2 = |BΣr0 |,

where we used the stability inequality in ≤ . Hence we finished by moving |BΣr0 | to the right hand

side.

Corollary 9.1 A complete stable 2-sided minimal Σ2 in R3 is a plane.

Proof: By the above lemma, Σ has no more than quadratic area growth, hence the stability inequalityand logarithmic cutoff technique imply that Σ2 is a hyperplane.

Corollary 9.2 (originally due to Schoen [6]) If Σ is stable and 2-sided, x ∈ Σ and distΣ(x, ∂Σ) ≥ r0,then

supBΣr0−σ

(x)

|A|2 ≤ cσ−2.


Proof: By Theorem 8.2, we can reduce to prove the small total curvature condition. Using the loga-rithmic cutoff trick, the stability inequality and area bound,∫

BΣe−nr0

|A|2 ≤ n−2

∫BΣr0\BΣ

e−nr0

r−2 ≤ cn−1.

So we can get the small total curvature condition by shrinking down the radius.

Theorem 9.2 Let Σ2 ⊂ R3 be simply connected and embedded. If x ∈ Σ, ∂Σ ⊂ ∂Br(x) and|Br0(x) ∩ Σ| ≤ A0r

20, then

supΣ∩Br0/2(x)

|A|2 ≤ c(A0)r−20 .

Remark 9.1 Non-embedded counter examples are Helicoid type singularities; Non-simply-connectedcounter examples are Catenoid type singularities.

Proof: Using the simply-connectedness and quadratic area growth, we can apply the third inequalityin Lemma 9.2 to get ∫

Σ∩Bθr0|A|2 ≤ c(θ,A0),

where θ ∈ (0, 1) and c(θ,A0) is a constant depending only on θ and A0. Then we can start fromr1 = 3

4r0, and devide [0, r1] intoN sub-intervals [9−n−1r1, 9−nr1], for n = 0, · · · , N . Then ∃n ≤ N ,

such that ∫Σ∩(B

9(−n+1)r1\B9−nr1

)|A|2 < c

N.

By rescaling, we can assume that∫

Σ∩(B1\B1/9) |A|2 < c

N . =⇒ curvature estimates on the annuli regionΣ ∩ (B1 \ B1/9), =⇒ Σ ∩ (B1 \ B1/9) is locally graphical, by Maximum Principle =⇒ graphical onΣ ∩ (B1 \B1/9), hence graphical on D by embeddedness.

Curvature estimates under small excess assumptionGiven x ∈ Σk ⊂ Rn and r > 0, the following quantity

Θx(r) =|Σ ∩Br(x)|

ωkrk

is monotonously non-decreasing w.r.t. r.

Definition 9.1 The excess of Σ in Br(x) is Θx(r)− 1 ≥ 0.

10 SCHOEN-SIMON-YAU CURVATURE ESTIMATES AND MINIMAL CONE (2/9/2012) 28

Theorem 9.3 Let Σk ⊂ Rn be minimal. ∃ε = ε(n, k), if x ∈ Σ, ∂Σ ⊂ ∂Br0(x), and Θx(r0)− 1 < ε,then

supΣ∩Br0/2(x)

|A|2 ≤ r−20 .

Proof:

• It suffices to assume that Θy(r1) − 1 < ε for all y ∈ Br1(x) ∩ Σ by the monotonicity formula3.1.• By rescaling the function (r1−|y|)2|A|2(y) near the maximum point as in the proof of Theorem

8.2, we can get another minimal surface, denoted still as Σ, such that 0 ∈ Σ, ∂Σ ⊂ ∂B1(0),|A|2 ≤ 1 on Σ, and |A|2(0) = 1

4 . Furthermore, by the small excess condition, |Σ ∩ B1(0)| ≤(1 + ε)ωk.• This is not possible if ε ≤ ε0, for some ε0 > 0 small enough, by the following argument.• Compactness argument: consider the class

Cε = Σ : 0 ∈ Σ, ∂Σ ⊂ ∂B1(0), |A|2 ≤ 1, |A|2(0) =1

4, |Σ ∩B1(0)| ≤ (1 + ε)ωk.

If the curvature estimates is not true, then we can find a sequence Σi, with Σ ∩ B1(0) ≤(1 + 2−i)ωk. A subsequence Σi → Σ in Ck norm to some minimal Σ∞, such that Σ∞ ∈ C0,i.e. |Σ∞ ∩ B1(0)| = ωk, =⇒ Σ is a disk, hence contradiction to the curvature assumption|A|2(0) = 1

4 .

10 Schoen-Simon-Yau curvature estimates and minimal cone (2/9/2012)

10.1 Curvature estimates by Schoen-Simon-Yau when n ≤ 6

Theorem 10.1 (Schoen-Simon-Yau [7]) Let Σn−1 ⊂ Rn be a stable 2-sided minimal surface. Assumex0 ∈ Σ, ∂Σ ⊂ ∂Br0(x0), |Σ ∩Br0(x0)| ≤ V rn−1

0 and n ≤ 6. Then

supΣ∩Br0/2(x0)

|A|2 ≤ c(n, V )r−20 .

Corollary 10.1 A complete 2-sided stable Σn−1 ∈ Rn with Rn−1 volume growth and n ≤ 6 is ahyperplane.

Remark 10.1 Counterexample for n = 7: ∃ complete volume minimizing Σ7 ⊂ R8, not a hyperplane

Proof:Claim:

2

n− 1

∫Σ

∣∣∇|A|∣∣2ϕ2 ≤∫

Σ|∇ϕ|2|A|2, ∀ϕ ∈ C1

c (Σ).


Now let us firstly prove this claim. By plug inϕ|A| to the stability inequality−∫

Σ(ϕ|A|)L(ϕ|A|) ≥0, and using the tricks in Theorem 6.1, we have∫

Σϕ2|A|L(|A|) ≤

∫Σ|∇ϕ|2|A|2.

Using Proposition 7.2, we can get the conclusion.Now change ϕ→ ϕ|A|q, for some q > 0, then we get

2

n− 1

∫Σ

∣∣∇|A|∣∣2ϕ2|A|2q ≤∫

Σ|A|2

∣∣∇(ϕ|A|q)∣∣2 =

∫Σ|A|2

∣∣(∇ϕ)|A|q + q|A|q−1ϕ(∇|A|)∣∣2

≤ (q2 + ε)

∫Σ|A|2qϕ2

∣∣∇|A|∣∣2 + (1 +1

ε)

∫Σ|∇ϕ|2|A|2q+2.

Hence if q <√

2n−1 , by moving the first term on the right hand side to the left,

=⇒∫

Σ

∣∣∇|A|∣∣2ϕ2|A|2q ≤ C(q)

∫Σ|A|2q+2|∇ϕ|2,

=⇒∫

Σ

∣∣∇|A|q+1∣∣ϕ2 ≤ C(q)

∫Σ

(|A|q+1)2|∇ϕ|2.

Set p = q + 2,

=⇒∫

Σ

∣∣∇|A|p−1∣∣ϕ2 ≤ C(p)

∫Σ

(|A|p−1)2|∇ϕ|2.

Now replace ϕ by ϕ|A|p−1 in the stability inequality∫

Σ |A|2ϕ2 ≤

∫Σ |∇ϕ|

2,

=⇒∫

Σ|A|2pϕ2 ≤

∫Σ

∣∣∇(ϕ|A|p−1)∣∣2 ≤ 2

∫Σϕ2∣∣∇(|A|p−1)

∣∣2 + |∇ϕ|2|A|2p−2.

Using the above inequality,

=⇒∫

Σ|A|2pϕ2 ≤ C(p)

∫Σ|A|2p−2|∇ϕ|2.

Replace ϕ by ϕp, then∫Σ|A|2pϕ2 ≤ C(p)

∫Σ|A|2p−2|∇ϕp|2 = Cp2

∫Σ

(ϕ|A|)2p−2|∇ϕ|2

≤ C(p)∫

Σ(ϕ|A|)2p

(p−1)/p∫Σ|∇ϕ|2p

1/p.

=⇒∫

Σ(ϕ|A|)2p ≤ C(p)

∫Σ|∇ϕ|2p, ∀p < 2 +

√2

n− 1. (12)

• Remark: if Σk ⊂ Rn is minimal,∫

Σ |A|2p ≤ C, for 2p > k =⇒ Curvature Estimates.

• Want: 2p > n− 1, hence 4 + 2√

2n−1 > n− 1, i.e.

√2

n−1 >n−5

2 =⇒ n ≤ 6.


• When n ≤ 6, take 2p = n− 1, and use the logarithmic cut off trick and the volume growth =⇒∫Σ |A|

n−1 is small in small ball, and hence the curvature estimates.• Let Σn−1 ⊂ Rn be complete, stable, 2-sided, n ≤ 6 and Σ∩BR ≤ CRn−1 =⇒Σ is hyperplane.

Take 2p > n− 1, =⇒∫

Σ(ϕ|A|)2p ≤ C∫

Σ |∇ϕ|2p ≤ C

R2p |Σ ∩B2R| → 0.

10.2 Minimal cone

Given Σk−1 ⊂ Sn−1, the cone based on Σ is defined as

C(Σ) = λx : x ∈ Σ, λ ≥ 0. (13)

Propostion 10.1 C(Σ) is minimal ⇐⇒ Σ is minimal in Sn−1 ⇐⇒ 4Σxi + (k − 1)xi = 0, i =

1, · · · , n, where x1, · · · , xn is coordinates of Rn.

Proof: Given ~X = (x1, · · · , xn), then C(Σ) is minimal ⇐⇒ 4C(Σ)~X = 0. Take a o.n. basis

e1, · · · , ek−1 for TΣ, and ek = ~X/| ~X|, then e1, · · · , ek is an o.n. basis for C(Σ). Then

4C(Σ)~X =

k−1∑i=1

eiei ~X + ekek ~X −k∑i=1

(∇eiei)TC(Σ) ~X.

Here∇ekek = ∇ ∂∂r

∂∂r = 0, with r = | ~X| the radial function. ∇eiei · ~X = −ei · (∇ei ~X) = −ei · ei =

−1, hence (∇eiei)TC(Σ) = (∇eiei)TΣ + (∇eiei · ~X) ~X = (∇eiei)TΣ − ~X . So

4C(Σ)~X = 4Σ

~X + (k − 1) ~X.

So we prove the equivalence of the first and the third conclusion.Now Σ is minimal in Sn−1 if and only if ~HΣ =

∑k−1i=1 (∇eiei)TS

n−1, hence (

∑k−1i=1 (∇eiei)) lies in

the normal direction of Sn−1. So ~HC(Σ) = (∑k−1

i=1 (∇eiei) +∇ekek)⊥(Σ) =∑k−1

i=1 (∇eiei)⊥(Σ) = 0.

Clifford Hypersurfaces: Given Sp(r1) ⊂ Rp+1 and Sq(r2) ⊂ Rq+1, take

Σ = Sp(r1)× Sq(r2) ⊂ Rp+q+1.

Then Σ ⊂ Sp+q+1 ⇐⇒ r21 + r2

2 = 1. Given (~x, ~y) ∈ Σ, where ~x ∈ Sp, and ~y ∈ Sq, hence4Σ~x = − p

r21~x, and4Σ~y = − q

r22~y. Hence

Σp+q ⊂ Sp+q+1 is minimal, ⇐⇒ p

r21

=q

r22

= p+ q.

Hence such class of Σ form lots of examples of minimal suffices in Sn and hence minimal cones inRn+1.

11 CLASSICAL PLATEAU PROBLEM (2/14/2012) 31

Tangent Cone at ∞: Given Σn−1 ⊂ Rn complete, volume minimizing, which is not a hyperplane,then

λ−1i Σ→ C, λi →∞,

where C is a non-flat, volume minimizing cone.

• So existence of such Σ =⇒ ∃ nonflat volume minimizing cone Cm1 , m ≤ n− q with an isolatedsingularity at 0.• J. Simons: when m ≤ 7, no such cones exists =⇒ Σ such does not exists if n ≤ 7.• p = q = 3 : C

(S3(1/

√2)× S3(1/

√2))

is stable, and area minimizing;• p = 1, q = 5 : C

(S1(1/

√6)× S5(

√5/6)

)is stable, but not area minimizing.

11 Classical Plateau Problem (2/14/2012)

Plateau Problem: Given Γk−1 ⊂ Rn, with ∂Γ = 0, find Σk ⊂ Rn, such that ∂Σ = Γ, and|Σ| = min|Σ1| : Σ1 = Γ.

• What are the competetors?• 1930 J. Douglas, T. Rado proved the Classical Plateau Problem:• Γ a piecewise C1 Jordan curve in Rn, consider all u : D → Rn, with D the unit disk on C,

satisfying u : ∂D → Γ is a homeomorphism.

Theorem 11.1 (Classical Plateau Problem) ∃ u : D → Rn of least area among:

XΓ = v ∈W 1,2(D,Rn) ∩ C0(D,Rn) : v : ∂D → Γ is monotone & onto.

The map u is harmonic, almost conformal i.e. |ux| = |uy| and ux ·uy = 0, a.e. onD, and u : ∂D → Γ

is a homeomorphism.

Given u ∈ W 1,2(D) and (x1, x2) or (x, y) coordinates on D, the pull back metric is (gij) =

(u∗δ)ij = uxi · uxj ∈ L1(D).

Definition 11.1 The area of u is:

A(u) =

∫D

√det(u∗(δij)

)dx1dx2 =

∫D

√|ux|2|uy|2 − (ux · uy)2dxdy.

Note: given ϕ : D → D diffeomorphism, then A(u ϕ) = A(u).The W 1,2 norm of u is:

‖u‖2W 1,2(D) =

∫D

(|u|2 + |∇u|2)dxdy,

where |∇u|2 = |ux|2 + |uy|2.


Definition 11.2 The energy functional of u is

E(u) =1

2

∫D|∇u|2dxdy.

Lemma 11.1 A(u) ≤ E(u) with equality⇐⇒ u is almost conformal.

Proof: By Cauchy-Schwartz

A(u) =

∫D|ux ∧ uy|dxdy ≤

1

2

∫D|ux|2 + |uy|2.

When equality holds, Cauchy-Schwartz implies the almost conformal property.

Corollary 11.1 If u is a critical point of E(·), and u is conformal, then u is a critical point of A(·).

Definition 11.3

AΓ = infA(v) : v ∈ XΓ, EΓ = infE(v) : v ∈ XΓ,

where XΓ is defined in Theorem 11.1.

Propostion 11.1AΓ = EΓ.

Proof: By the above lemma, AΓ ≤ EΓ clearly. Choose u ∈ XΓ, with A(u) < AΓ + ε(may assumethat u is smooth in D).

• Suppose u is an immersion. Then (D, g = u∗δRn) is a Riemman surface, where gij = uxi · uxj .By the uniformization theorem, ∃ ϕ : (D, δ) → (D, g) conformal diffeomorphism, i.e. ϕ∗g =

λ2δ, hence(u ϕ)∗δRn = ϕ∗(u∗δ) = ϕ∗g = λ2δD.

Then u ϕ : D → Rn is conformal, hence E(u ϕ) = A(u ϕ) = A(u),

=⇒ EΓ ≤ AΓ + ε.

(Remark: we allow the boundary value u|∂D to change.)• In general, define us : D → Rn+2 by us(x, y) =

(u(x, y), sx, xy

), for s > 0 small enough.

Now the pullback metric gij = ((us)∗δRn+2)ij = gij +s2δij . So ∃ϕ diffeomorphism of D, suchthat us ϕ is conformal. Then

EΓ ≤ E(u ϕ) = E(ϕ : (D, δ)→ (D, g)

)=

1

2

∫D|Dϕ|2gdxdy

≤ 1

2

∫D|Dϕ|2gdxdy = E

(ϕ : (D, δ)→ (D, g)

)≤ E(us ϕ)

= A(us ϕ) = A(us) = A(D, g) ≤ A(D, g) + ε(s)

= A(u) + ε(s) < AΓ + 2ε.

Here ε(s)→ 0 as s→ 0. Hence we finished the proof.


Proof: (of Theorem 11.1) By the above lemma, we want to achieve EΓ.

Step 1: Dirichlet problem: given v ∈ XΓ, v : D → Rn, v ∈ W 1,2(D) ∩ C0(D), then ∃ u ofleast energy in Cv = w ∈ W 1,2(D) : w − v ∈ W 1,2

0 (D). This is equivalent to find a u ∈W 1,2(D) ∩ C0(D) ∩ C∞(D) which is a solution of

4u = 0 in Du = v on ∂D.

There are several methods: i) Perron method; ii) Poison kernel; iii) Variational method. Now let ususe the variational method.

• Take ui in Cv, with E(ui)→ infw∈Cv E(w).• ui are bounded in W 1,2(D).

∫D |∇ui|

2 ≤ C comes from E(ui) ≤ C. Now by Poincareinequality, ∫

D(ui − v)2 ≤ C

∫D|D(ui − v)|2 ≤ 2C

∫D|Dui|2 + |Dv|2 ≤ C ′.

So∫D u

2i ≤ c(

∫D v

2 + C ′).• Rellich Lemma =⇒ ui′ u weakly in W 1,2(D), and ui′ → u in L2(D). Furthermore, u− v ∈W 1,2

0 (D) since W 1,20 (D) is weakly closed in W 1,2(D).

• Lower semi-continuity of E:

=⇒ E(u) ≤ lim inf E(ui′) = infw∈Cv

E(w).

• u ∈ Cv =⇒ u is weakly harmonic, =⇒ u ∈ C∞(D).• u ∈ C0(D). (barrier argument)

Take vi → v uniformly on D, with vi ∈ C∞(D). Solve 4ui = 0 in D, ui = vi on ∂D, =⇒ui ∈ C∞(D). Maximum Principle implies:

maxD|ui − uj | = max

∂D|vi − vj | → 0, i, j →∞,

=⇒ ui → u uniformly in D, hence u ∈ C0(D).

Step 2: Minimize over boundary parametrization.

Propostion 11.2 (Courant-Lebesgue Lemma) Given u : D → Rn, u ∈ W 1,2(D) ∩ C0(D) andE(u) ≤ K, then ∀δ 1 and x ∈ ∂D, ∃ρ ∈ [δ, δ1/2] and an arc Cρ = (∂Bρ(x)) ∩D, such that

|u(Cρ)|2 ≤2πK

| log ρ|.

12 CONTINUITY OF PLATEAU PROBLEM AND HARMONIC MAPS (2/16/2012) 34

Proof:|u(Cρ)|2 ≤ (

∫Cρ

|Du|ds)2 ≤ 2πρ

∫Cρ

|Du|2ds.

Now integrate over ρ ∈ [δ, δ1/2],∫ δ1/2

δ

u(Cρ)2

ρdρ ≤

∫ δ1/2

δ2π(

∫Cρ

|Du|2ds)dρ ≤ 2πE(u) ≤ 2πK.

Hence inf [δ,δ1/2] |u(Cρ)|2(− log ρ) ≤∫ δ1/2

δu(Cρ)2

ρ dρ ≤ 2πK.

Key Problem: if ϕ is a conformal diffeomorphism of D, i.e. ϕ ∈ PSL(2,R), ϕ = αz+βγz+δ , then

E(u ϕ) = E(u).

The less of compactness when we minimize over all the possible boundary parametrizations comesfrom the un-compactness of PSL(2,R).

12 Continuity of Plateau Problem and Harmonic maps (2/16/2012)

12.1 Continuity of the Proof of Theorem 11.1

• 3 Point Condition: Fix some orientation on both ∂D and Γ. Given pi : i = 1, 2, 3 ⊂ ∂D andqi : i = 1, 2, 3 ⊂ Γ monotone, i.e. p1 < p2 < p3 w.r.t. the fixed orientation, introduce

X∗Γ = u ∈ XΓ : u(pi) = qi.

Lemma 12.1 Given v ∈ XΓ, ∃ϕ : D → D Mobius transform, such that u = v ϕ ∈ X∗Γ, andE(u) = E(v).

Proof: Since v ∈ XΓ, ∃ r1, r2, r3 ∈ ∂D, with v(ri) = pi. The monotonicity of v : ∂D → Γ impliesthat r1 < r2 < r3. Then ∃ϕ : D → D Mobius, with ϕ(pi) = ri, i = 1, 2, 3. Hence u = v ϕ satisfiesthe requirement.

Corollary 12.1EΓ = infE(u) : u ∈ X∗Γ.

Lemma 12.2 Given K > 0, and X∗Γ,K = u ∈ X∗Γ : E(u) ≤ K, then u|∂D : u ∈ X∗Γ,K isuniformly equi-continous.


Proof: Given ε > 0, using the Courant-Lebesgue Lemma, ∃√δ < mini 6=j=1,2,3

d(pi,pj)2 , and ∃Cρ(=

circle centered at x ∈ ∂D, ρ ∈ [δ, δ1/2]), such that |u(Cρ)| ≤ c√K

| log ρ|1/2 < ε. By our choice of δ,there can be at most one pi inside Cρ. Using the monotonicity, the sub-arc of ∂D inside Cρ must bemapped to the short arc inside u(Cρ).(or there would be two qis inside u(Cρ), contradiction to themonotonicity.) Hence we have the equip-continuity.

Theorem 12.1 ∃ u ∈ XΓ, with E(u) = EΓ.

Proof: Take a minimizing sequence ui ⊂ XΓ, with E(ui) → EΓ. By firstly applying Lemma 12.1and then the Dirichlet Problem in the above section, we can assume that

ui ∈ X∗Γ4ui = 0.

By weak compactness of bounded set in W 1,2(D) and Lemma 12.2, there exists a subsequence ui′

ui′ u,weakly in W 1,2(D), ui′ =⇒ u, uniformly on ∂D.

The uniform convergence on ∂D implies that u|∂D is monotone, onto and u(pi) = qi, i = 1, 2, 3. ByMaximum Principle

ui′ =⇒ u, uniformly in D, hence4u = 0.

Now u ∈ X∗Γ clearly. Hence EΓ ≤ E(u) ≤ lim inf E(ui′) = EΓ, so E(u) = EΓ.

The following corollary finished the proof of Theorem 11.1.

Corollary 12.2 u is harmonic, almost conformal, and A(u) = AΓ.

Proof: The harmonicity of u is trivial. Since u ∈ XΓ,

AΓ ≤ A(u) ≤ E(u) = EΓ.

So by Lemma 11.1, we have A(u) = E(u) = AΓ = EΓ. Hence Lemma 11.1 implies that u is almostconformal.

Propostion 12.1 The set of branch points x ∈ D : |∇u| = 0 is discrete.

Proof: x : |∇u| = 0 ⊂ x : ∂u∂z = 0 is discrete, since ∂u

∂z is holomorphic(i.e. ∂∂z (∂u∂z ) = 4u = 0).


Propostion 12.2 If Γ is a Ck,α-curve, k ≥ 2, 0 < α < 1, then u ∈ Ck,α(D), with finite number ofbranch points, and boundary branch points are isolated.

Theorem 12.2 (Osserman) When n = 3, the solution u has no interior branch points.

Open Question: When n = 3, can there be a boundary branch points. (True when boundary isanalytic.)

• In Rn, n ≥ 4, the minimizing solution has branch points, e.g. z → (z2, z3) ∈ C2 ' R4.

Propostion 12.3 The solution u is a homeomorhpism on ∂D to Γ.

• Since u is monotone an onto, if u is not homeomorphism, then u must map a sub-interval of ∂Dto a point on Γ. By a reflection argument, we can reflect conformally near the interval where uis a constant, and extend u to a conformal harmonic map in a neighborhood of the interval, thenu maps an interval to a point. In fact, u maps the direction tangent to ∂D to 0, but the normaldirection not to 0(if 0, then u has branch points along an interval, contradiction to the fact thatbranch points are discrete), contradiction to the conformal property.

12.2 Harmonic maps

Motivation: find minimal Σ2 in an arbitrary Riemannian manifold (Mn, g).

Definition 12.1 Given u : (Ωk, γ) → (Mn, g), where (Ωk, γ) is a k-dimensional Riemannian mani-fold possibly with boundary, the harmonic energy is:

E(u) =1

2

∫Ω|∇u|2dµγ ,

where in local coordinates x1, · · · , xk of Ω, u1, · · · , un of M , the harmonic energy density is

|∇u|2 = trγ(u∗g) =k∑

i,j=1

γij〈du(∂

∂xi), du(

∂

∂xj)〉g =

∑i,j,α,β

γijgαβ∂uα

∂xi∂uβ

∂xj.

Critical point of E is called harmonic maps.

Lemma 12.3 The Euler-Lagrange equation of E is call the harmonic map equation(HE):

4γuα +

∑i,j,β,δ

γijΓαβδ(u(x)

)∂uβ∂xi

∂uδ

∂xj= 0, α = 1, · · · , n. (14)

Another form of (HE): Consider an isometric embedding Mn ⊂ RN , then

u : Ω→ RN , u(Ω) ⊂M,

and

E(u) =1

2

N∑α=1

∫Ω|∇γuα|2dµγ .

So the variational problem can be viewed as a constraint problem.


• Take a variation vector fields ddt |t=0ut = η, where η is tangent to M ;

•d

dt|t=0E(ut) =

∫Ω〈∇u,∇η〉dµγ .

• The Euler-Lagrange equation is: (4γu)T = 0, or

4γuα = [(4γu)⊥]α =

k∑i,j=1

γijAαu(x)(∂u

∂xi,∂u

∂xj), (15)

where A is the second fundamental form of M ⊂ RN .• For example, M = Sn ⊂ Rn+1, then4uα = −|∇u|2uα.

Bochner Formula: Given u : (Ωk, γ)→ (Mn, g) harmonic and smooth, then

1

24|∇u|2 = |∇∇u|2 + gαβγ

ikγjlRΩij

∂uα

∂xi∂uβ

∂xj

−∑i,j

〈RM(du(ei), du(ej)

)du(ei), du(ej)〉,

(16)

where RΩij is the Ricci curvature of (Ω, γ), RM the sectional curvature of (M, g), and e1, · · · ek an

o.n. basis on (Ω, γ).

• When Ω and M are all compact,

4|∇u|2 ≥ −c1|∇u|2 − c2|∇u|4,

where c1, c2 are two positive constants.• If RM ≤ 0, =⇒4|∇u|2 ≥ −c1|∇u|2, then

supBR/2

|∇u|2 ≤ C∫BR

|∇u|2.

• If does not assume KM ≤ 0, we have counterexamples to the gradient estimates.

Example: ∃ ui : S2 → S2, holomorphic, but with strong dilation which maps a small neighbor-hood of the south pole to almost all of S2, satisfying: |∇ui|2(0) → ∞,

∫S2 |∇ui|2 = 8π, and

each ui is energy minimizing.

Lemma 12.4 Given u : (Ω2, γ)→ (Mn, g), the area is A(u) =∫

Ω

√det(u∗g)dx.

A(u) ≤ E(u),

with “ = ” if and only if u is almost conformal.

13 SACKS-UHLENBECK’S THEOREM (2/21/2012) 38

13 Sacks-Uhlenbeck’s theorem (2/21/2012)

13.1 Hopf differential

In the case (Σ2, γ) has dimension 2, let z = x + iy be the local complex coordinates w.r.t. theconformal structure determined by γ.

Definition 13.1 Given u : (Σ2, γ)→ (Mn, g), the Hopf differential Φ is defined by:

Φ = ϕ(z)dz2,

whereϕ(z) = |∂u

∂x|2 − |∂u

∂y|2 − 2i〈∂u

∂x,∂u

∂y〉.

Propostion 13.1 Viewing E(u, γ) as a functional of both u and γ, then:

1. Φ ≡ 0⇐⇒ E(u, γ) is critical w.r.t. compactly supported variation of γ;2. Φ is holomorphic⇐⇒ E(u, γ) is critical for domain variations, i.e. ut = u ft, where ft are

1−parameter family of diffeomorphisms, with ft = id near ∂Σ.

Proof: Consider γt : −ε < tε, and γ0 = h. Since

E(u, γ) =1

2

∫Σ

2∑i,j=1

γij〈 ∂u∂xi

,∂u

∂xj〉√detγdx1dx2,

and ddt |t=0γ

ijt = −hij = −γipγjqhpq,

ddt

√detγ = 1

2 trγ(h)√detγ,

we haveδγE(u, γ) = −

∫Σhij(〈 ∂u∂xi

,∂u

∂xj〉 − 1

2|∇u|2γij︸︷︷︸

Tij−stree-energy tensor

)dµγ .

Proof of 1: locally γij = λ2δij , so

T =

[12(−|ux|2 + |uy|2),−〈ux, uy〉−〈ux, uy〉, 1

2(|ux|2 − |uy|2)

]= 0⇐⇒ Φ = 0.

Proof of 2: Claim: Φ is holomorphic⇐⇒ divΣT = 0.(On (Σ, z), trace-free, divergence free (0, 2)-tensors are 1 : 1 correspondence to holomorphic quadraticdifferentials. Pf: in local conformal coordinates z = x1 + ix2, let T = Tij , hence T11 = −T22(tracefree), T11,1 + T12,2 = 0, T21,1 + T22,2 = 0(divergence free). If we let Φ = ϕ(z)dz2 = (T11 − T22 −2iT12)dz2, then it is easy to check ∂

∂zϕ = 0.)

13 SACKS-UHLENBECK’S THEOREM (2/21/2012) 39

• Using the composition ft : (Σ, γ)→ (Σ, (f−1t )∗γ) and u : (Σ, (f−1

t )∗γ)→M

E(u ft, γ) ≡ E(u, (f−1t )∗γ︸︷︷︸γt

).

• ddt |t=0RHS =

∫Σ〈h, T 〉dµga. Let ~X = d

dt |t=0ft, we have h = ddt |tγt = −LXγ = −(Xi,j +

Xj,i). So

d

dt|t=0E(u ft, γ) = δγE =

∫Σ

(Xi,j +Xj,i)Tijdµγ = 2

∫ΣXi,jT

ijµγ

= −∫〈X, divT 〉dµγ = 0.

Hence δγE = 0⇐⇒ divT = 0, for δγ = LXγ.

Remark 13.1 • When u is smoothly harmonic =⇒ ddtE(u ft, γ) = 0 for ft 1-parameter family

of diffeomorhpisms =⇒ Φ is holomorphic;• When u is weakly harmonic, u ft may not be C1 variations of u.

Corollary 13.1 1. When u : S2 →M is smoothly harmonic map, then u is almost conformal;2. When u : R2 →M is smoothly harmonic and

∫R2 |∇u|2dx <∞, then u is almost conformal.

Proof: Let Φ(z) = ϕ(z)dz2 on S2 \ +∞, with ϕ(z) an entire holomorphic function.

• Pf of 2:∫R2 |∇u|2 <∞ =⇒

∫C |ϕ| <∞ =⇒ ϕ ≡ 0.

• Pf of 1: Φ is regular at∞, henceϕ(z)dz2 is regular at∞. Let ζ = 1/z, then Φ = ϕ(1/ζ)(−dζζ2 )2 =

(ϕ(1/ζ)/ζ4)(dζ)2. Since ϕ(1/ζ)/ζ4 is regular near ζ = 0, =⇒ |ϕ(z)| ≤ C|z|4 =⇒ ϕ ≡ 0.

Remark 13.2 • Only true when S2(u is critical & u is conformal invariant & S2 has only oneconformal structure).• Not true for Σg with g ≥ 1.

13.2 Sacks-Uhlenbeck’s method

Sacks-Uhlenbeck: If (Mn, g) is a compact Riemannian manifold and πk(M) 6= 0 for some k ≥ 2,then ∃ a nontrivial u : S2 →M , which is harmonic(hence almost conformal).2 Improvement:

14 SACKS-UHLENBECK’S THEOREM CONTINUED (2/23/2012) 40

1. Meeks-Yau: π2(M) is generated by minimal 2-spheres.

Question: when k = 2, given v : S2 →M not homotopic to point, ∃ u harmonic and homotopicto v?

Result: given [v] ∈ π2(M), then [v] = [u0] + · · ·+ [uk], where u0, · · · , uk are minimal stablespheres.

2. Micallef-Moore: If πk(M) 6= 0 for k ≥ 3, then ∃ u minimal with Morse Index(for the energyfunctional E) ≤ k − 2.

Key Idea:

Definition 13.2 Given α > 1 and u ∈W 1,2α(S2,M) ⊂ C0(S2,M), the α-energy is defined by

Eα(u) =1

2

∫S2

(1 + |∇u|2)αdµ.

Remark 13.3 When α = 1, E1(u) = 2π + E(u).

α > 1, Finding critical point uα for Eα is much easier.αi 1: expect uαi → u, with u harmonic. In fact, we have good convergence if |∇uαi |2 is uniformlybounded.

Lemma 13.1 Euler-Lagrangian equation for Eα:

1. If u ∈W 1,2α(S2), and u is critical for Eα, then u is a weak solution of:

div((1 + |∇u|2)α−1∇ui

)T= 0, M ⊂ RN&i = 1, · · · , N,

⇐⇒ div((1 + |∇u|2)α−1∇ui

)= (1 + |∇u|2)α−1γpqAi(

∂u

∂xp,∂u

∂xq).

2. If u is smooth,

4ui + (α− 1)〈∇|∇u|2,∇ui〉

1 + |∇u|2=

2∑p,q=1

γpqAi(∂u

∂xp,∂u

∂xq), i = 1, · · · , N.

14 Sacks-Uhlenbeck’s theorem continued (2/23/2012)

Theorem 14.1 If (Mn, g) is compact and π2(M) 6= 0, then ∃ nontrivial u : S2 → M harmonic andalmost conformal.

Propostion 14.1 If α > 1, given v : S2 → M , then ∃ u ∈ C∞(S2) with Eα(u) = minEα(w) :

w ∈W 1,2α(S2) & u homotopic to v.

Proof: Minimal u exists by direct method(minimization).


• u ∈W 1,2α(S2) =⇒ u ∈W 2,2(S2) follows from Morrey Thm 1.11.1;• u ∈W 2,2(S2) =⇒ u ∈ C∞(S2) follows from the standard elliptic estimates.

Lemma 14.1 For (α− 1) small enough, if u is critical for Eα, then for p ≥ 3,

‖u‖2,p ≤ C(1 + (sup

S2

|∇u|2)p−1p (

∫S2

|∇u|2)1/p).

Proof: Since u is smooth by the above lemma, by the Euler-Lagrangian equation:

|4u| ≤ (α− 1)|∇∇u|+ C|∇u|2.

Hence standard Lp estimates imply that:

‖u‖2,p ≤ C(‖u‖0,p + ‖4u‖0,p) ≤ C(1 + (α− 1)‖u‖2,p +

∥∥|∇u|2∥∥0,p

).

When (α− 1) is small enough,

‖u‖2,p ≤ C(1 + (

∫S2

|∇u|2p)1/p)≤ C

(1 + (sup

S2

|∇u|2)p−1p (

∫S2

|∇u|2)1/p).

Proof: (of Theorem 14.1) Take a sequence α1 1 and corresponding critical mapping ui = uαi .Case 1: ∃ subsequence ui′ with |∇ui′ | ≤ C, then ui′ satisfy uniform C2,α estimates by Lemma 14.1and standard elliptic theory, =⇒ ui′ → u in C2 and u is our solution;Case 2: λi = maxS2 |∇ui|2 → +∞. ∃ pi ∈ S2, with |∇ui(pi)|2 = λi. Assume a subsequencepi → p ∈ S2.

• Let γi = λiγ, where γ is the standard metric on S2, then γi becomes flatter.• Then we have the following scaling equality:

|∇ui|2γi = λ−1i |∇ui|

2γ =⇒

∫S2

|∇ui|2γidµγi =

∫S2

|∇ui|2γdµγ ;

|∇∇ui|2γi = λ−2i |∇∇ui|

2γ =⇒

∫S2

|∇∇ui|qγidµγi = λ1−qi

∫S2

|∇∇ui|qγdµγ .

• Using the above scaling equality and the estimates in Lemma 14.1,

=⇒ λp−1i

∫S2

|∇∇ui|pγi ≤ C(1 + (λi)

p−1

∫S2

|∇ui|2γi),

=⇒∫S2

|∇∇ui|pγi ≤ C(λ−(p−1)i +

∫S2

|∇ui|2γi).


• So ui are uniformly in W 2,ploc (S2, γi). Since (S2, γi, pi) locally converge to (R2, δ, 0), using the

Sobolve embedding,ui′ → u in C1,1/2-loc on R2,

and |∇u(0)| = 1 = maxR2 |∇u|.• The Euler-Lagrangian equation is scaled to:

λi4γiuji + (αi − 1)

λi〈∇(λi|∇ui|2γi),∇uji 〉γi

1 + λi|∇ui|2γi=

2∑p,q=1

λiγpqi A

j(∂ui∂xp

,∂ui∂xq

),

=⇒4γiuji + (αi − 1)

〈∇|∇ui|2γi ,∇uji 〉γi

λ−1i + |∇ui|2γi

=

2∑p,q=1

γpqi Ai(∂ui∂xp

,∂ui∂xq

).

Here the second term (αi − 1)〈∇|∇ui|2γi ,∇u

ji 〉γi

λ−1i +|∇ui|2γi

≤ (αi − 1)C|∇∇ui|γi , where |∇∇ui|γi is uni-

formly bounded in Lp(S2, γi). So the second term converges to 0 in Lploc(R2, δ).

• =⇒ u is a C1,1/2 weakly harmonic map. Elliptic estimates imply that u : R2 → M is a C∞

harmonic map. Furthermore, u is nontrivial, since |∇u(0)| = 1 and E(u) <∞.• Claim: u can be extended to a smooth harmonic map on S2 = R2 ∪ ∞.

In (R2, z), z → ζ = 1/z is conformal, which takes∞ to 0. Then

v(ζ) = u(1/ζ) is harmonic on C \ 0 & E(v) <∞.

Since u : C→M is harmonic & E(u) <∞, so u is almost conformal by Corollary 13.1, hencev is also harmonic.

Key step: v is continuous at ζ = 0(v ∈W 1,2 ∩ C0(Σk) & weakly harmonic =⇒ v smooth).

1. By Courant-Lebesgue Lemma 11.2, ∃ ri → 0 such that max∂Dri dM(v(s1), v(s2)

)→ 0;

2. Claim: The oscillation of v on Dr/2 is bounded by a multiply of the energy EDr(v) andboundary oscillation on ∂Dr:

maxDr/2

dM(v(s1), v(s2)

)≤ C(EDr(v)) + max

∂DrdM(v(s1), v(s2)

).

If ∃ s1, s2 ∈ ∂Dr, such that dM(v(s1), v(s2)

)is too large, we can cover the image of v(Dr) by

several unit balls in M . Then by the monotonicity formula of minimal surfaces(v is harmonicand almost conformal, hence minimal), each portion of v(Dr) inside the unit ball has a fixedmount of area, which then makes the total area of v(Dr) too large than E(Dr), a contradiction.• Hence v is smooth on C, which means that u can be extended to a nontrivial harmonic map onS2. Finished.

15 COLDING-MINICOZZI’S MIN-MAX SPHERE (3/1/2012) (BY XIN ZHOU) 43

15 Colding-Minicozzi’s min-max sphere (3/1/2012) (by Xin Zhou)

Motivation: Given a Riemannian manifold (M, g), want to find unstable minimal spheres in C0 ∩W 1,2(S2,M) by direct variational method.

In fact, any π3(M) representative u : S3 → M can be viewed as a 1-parameter family of mapsS2 →M , i.e. S3 = S2 × [0, 1], with S2 × 0 and S2 × 1 mapped to points.

Definition 15.1 The variational space is define as:

Ω =u(t, x) ∈ C0

([0, 1], C0 ∩W 1,2(S2,M)

): u(0), u(1) = point map

.

Given β ∈ Ω, denote [β] to be the homotopy class of β in Ω. The area width of [β] is

WA = infu∈[β]

maxt∈[0,1]

Area(u(t)

);

the energy width of [β] isWE = inf

u∈[β]maxt∈[0,1]

E(u(t)

).

Propostion 15.1WA = WE .

Hence we denote the width by W = WA = WE .

Theorem 15.1 (Colding-Minicozzi) Given (M, g) and ρ ∈ Ω, such that ρ ∈ π3(M) is nontrivial, then

1. ∃ γj ⊂ Ω, j = 1, · · · ,∞, such that maxt∈[0,1]E(γj(t)

)W ;

2. ∀ ε > 0, ∃ J 1 and δ > 0, such that if j > J , for any t with

E(γj(t

))−W > −δ,

∃ a collection of harmonic spheres(hence almost conformal) ui : S2 → M , i = 0, · · · , l, suchthat

dV(γt,tli=0ui

)< ε,

where dV is the varifold distance;2′. or ∀ ti such that E

(γj(tj)

)→ W , a subsequence γj(tj) converge to a collection of harmonic

spheres u0, · · · , ul in the sense of bubble-tree convergence.

Remark 15.1 In fact, 2′ =⇒ 2, or the bubble-tree convergence =⇒ varifold convergence.

Bubble-tree convergence: (Definition 3.6 in [3])Roughly a sequence of uj ∈ W 1,2(S2,M) is said to bubble-tree converge to a collection of har-

monic spheres uili=0, if

1. uj u0 weakly(up to a subsequence) in W 1,2(S2);

15 COLDING-MINICOZZI’S MIN-MAX SPHERE (3/1/2012) (BY XIN ZHOU) 44

2. ∃ S0 = x10, · · · , x

k10 ⊂ S2, such that uj → u0 strongly inW 1,2(K) for anyK compact subset

of S2 \ S0;3. Near each xi0 ∈ S0, ∃ Dij : S2 → S2 conformal dilation, which takes a small ball centered atxi0 to the lower-hemisphere, and uj Dij converges to ui in the sense of step 1 and 2;

4. We have the energy identity:

limj→∞

E(uj) =

l∑i=1

E(ui).

Key ideas: Variational Method. Given β ∈ Ω, with [β] ∈ π3(M) nontrivial,

0). Mollify the minimizing sequence: Find minimizing sequence γj(t) ∈ [β] such that γj ∈C0([0, 1], C2(S2,M)

);

1). Almost conformal reparametrization: Reparametrize γj(t)→ γj(t) = γj(hj(t)(·), t

), where

hj(t) : (S2, g0) → (S2, γj(t)∗g + δ2j g0) is continuous 1-parameter family of conformal iso-

topies, hence γj ∈ [β], Area(γj(t)

)= Area

(γj(t)

), and

maxt∈[0,1]

E(γj(t)

)−Area

(γj(t)

)→ 0;

2). Tightening: γj(t) → ρj(t), by local harmonic replacement(Perron method), hence ρj ∈ [β],E(ρj(t)

)≤ E

(γj(t)

), and ρj(t) is almost harmonic if |E

(ρj(t)

)−W | 1.

Step 0: Using a mollification method, we have

Lemma 15.1 Given β ∈ Ω, ∃γj ∈ [β], with maxt∈[0,1]Area(γj(t)

)WA, and γj(t) ∈ C0

([0, 1], C2(S2,M)

).

Step 1: Reparemetrization.

Propostion 15.2 ∃ γj ∈ [β], Area(γj(t)

)= Area

(γj(t)

), and

maxt∈[0,1]

E(γj(t)

)−Area

(γj(t)

)→ 0.

In fact, Proposition 15.1 is a direct corollary.Proof: (of Proposition 15.1) WA ≤WE is clearly true since Area(·) ≤ E(·). Then

WE ≤ limj→∞

[ maxt∈[0,1]

E(γj(t)

)] = lim

j→∞[ maxt∈[0,1]

Area(γj(t)

)] = WA.

Lemma 15.2 (Uniformization)

• ∀C1 metric g on S2, ∃!C1,1/2 isotopy h : (S2, g0)→ (S2, g), fixing three points, and conformaldiffeomorphism;

16 INTRODUCTION TO THE WILLMORE CONJECTURE (3/6/2012) 45

• If g1, g2 are two C1 metrics, and gi ≥ εg0, let h1, h2 be the unique conformal isotopic diffeo-morphism, then

‖h1 − h2‖C2∩W 1,2(S2,S2) ≤ C(ε, ‖gi‖C1)‖g1 − g2‖C0 .

Sketch of proof: Pull h back to h : (C, dwdw)→ (C, g = λ2|dz + µ(z)dz|2).

• h satisfy (hw) = µ(h(w)

)hw a quasi-linear elliptic system;

• Apriori estimates =⇒ results.

Proof: (of Proposition 15.2)

Step 2: Tightening.

Propostion 15.3 ∃ ε0 > 0, and continuous Ψ : [0,∞)→ [0,∞), Ψ(0) = 0 depending onM . ∀γ ∈ Ω

with no non-constant harmonic slice, i.e. γ(t) is not harmonic unless γ(t) = pt, then ∃ γ → ρ

deformation, such that ρ ∈ [γ], E(ρ(t)

)≤ E

(γ(t)

), and if E

(γ(t)

)≥ W

2 , then

(B) ∀ B finite collection of balls on S2, with∫B |∇ρ(t)|2 < ε0, let v : 1

8B → M be the energy-minimizing harmonic map, with v| 1

8∂B = ρ(t)| 1

8∂B, then∫

18B|∇ρ(t)−∇v|2 ≤ Ψ

[E(γ(t)

)− E

(ρ(t)

)].

Harmonic replacement.

16 Introduction to the Willmore conjecture (3/6/2012)

Willmore Conjecture in R3: Σ2 ⊂ R3 compact, embedded surface, the Willmore energy is definedby

W (Σ) =

∫ΣH2dΣ,

where H = 12(k1 + k2) is the normalized mean curvature.

• W (S2) = 4π.

Conjecture: If Σ is a torus, then W (Σ) ≥ 2π2, “ = ” only if Σ is the Clifford torus.

Conformal invariance of W : Σk ⊂ (Mk+1, g), A second fundamental form, A trace-free part of A,then W is conformal invariant:

W (Σ, g) =

∫Σ|A|kdµg = W (Σ, eug).

16 INTRODUCTION TO THE WILLMORE CONJECTURE (3/6/2012) 46

Proof: Given local coordinates x1, · · · , xk on Σ ⊂ (M, g), then the 2nd f.f. is hij . Let g = eug,then

hij = eu/2(hij +∂u

∂ν)gij .

Hence ˚hij = eu/2hij , and |A|g = e−u/2|A|g.

When k = 2, k1, k2 principal curvatures,

|A|2 =1

2(k1 − k2)2 = 2H2 − 2k1k2.

Since Gauss curvature K = k1k2 in R3,∫ΣH2 =

1

2

∫Σ|A|2 + 2πχ(Σ).

Consider S3: View R3 as stereographic projection to S3, which is a conformal transformation.

W (Σ) =1

4

∫Σ

(k1 − k2)2dΣ + 2πχ(Σ)

=1

4

∫Σ

(4H2 − 4k1k2)dΣ + 2πχ(Σ)

=

∫Σ

(H2 − 4(KΣ − 1)

)+ 2πχ(Σ)

=

∫Σ

(1 +H2)dΣ,

where KΣ is the Gauss curvature of Σ ⊂ S3, and KΣ = 1 + k1k2 by Gauss formula. Hence whenproject Σ ⊂ R3 to Σ ⊂ S3, the Willmore energy is

W (Σ) =

∫Σ

(1 +H2)dΣ. (17)

• If H = 0, then W (Σ) = |Σ|;• Clifford torus: let (x, y) ∈ R4 = R2 × R2, Clifford torus is

Tc = S1(1/√

2)× S1(1/√

2) =

(x, y) ∈ R4 : |x|2 = |y|2 = 1/2⊂ S3.

W (Tc) = A(Tc) = 2π2.

17 OUTLINE OF MARQUES-NEVES’S PROOF OF WILLMORE CONJECTURE [2] (3/8/2012)47

17 Outline of Marques-Neves’s proof of Willmore conjecture [2] (3/8/2012)

• Σ2 ⊂ S3, with H the mean curvature, the Willmore energy is defined as

W (Σ) =

∫Σ

(1 +H2)dΣ.

W (equator) = 4π; if Σ is not embedded =⇒W (Σ) ≥ 8π > 2π2.• T 2

c = S1(1/√

2)× S1(1/√

2) ⊂ S3 is the Clifford torus, K = H = 0 flat,

W (T 2c ) = 2π2.

Theorem 17.1 If Σ has genus ≥ 1, then W (Σ) ≥ 2π2, with equality only if Σ is a Clifford torus.

Theorem 17.2 If H = 0, and genus ≥ 1, then |Σ| ≥ 2π2, with equality only for Clifford torus.

Theorem 17.3 If Σ is embedded, g(Σ) ≥ 1, then ∃ minimal surface Σ with 4π < |Σ| ≤W (Σ).

Remark 17.1 (Hopf-Almgren) Σ ⊂ S3, H = 0, and π1(Σ) = 1, then Σ is an equator.Pf: Let f : Σ→ S3 be the immersion, and Π(·, ·) the 2nd f.f. of Σ, then the Hopf-differential

ϕ(z) = Π(∂f

∂z,∂f

∂z)dz2

is holomorphic quadratic(In local coordinates, the 2nd f.f. hij is trace-free, and divergence-free∑j ∇jhij =

∑j ∇ihjj = ∇H = 0 since S3 is constant curvature, hence ϕ is holomorphic by

section 13.1). Hence Π ≡ 0.

Let In = [0, 1]n, andΦ : In → Z = Z2(S3),

where Φ(x) is a surface(current), and Φ is continuous. Let

Π = relative homotopy class of Φ = Φ′ ∼ Φ,

i.e. ∃ Φt : In → Z , 0 ≤ t ≤ 1, Φ0 = Φ, Φ1 = Φ′ in In, and ∀ t ∈ [0, 1], x ∈ ∂In, we haveΦt(x) = Φ(x).

Definition 17.1 Width of Π:L(Π) = inf

Φ′∈Πsupx∈In|Φ′(x)|.

Theorem 17.4 (Almgren-Pitts) IfL(Π) > supx∈∂In |Φ(x)|, then ∃ smooth minimal embedded Σ(possiblywith multiplicity), such that

|Σ| = L(Π).

Moreover, Σ is the limit of some min-max sequences, i.e. ∃ xj ∈ In, such that Φj(xj)→ Σ.


Propostion 17.1 F : S3 → S3 is conformal and Σ minimal, then |F (Σ)| ≤ |Σ|.

Proof: By conformal invariance of W (Σ),

|F (Σ)| ≤∫F (Σ)

(1 +H2)dΣ = W(F (Σ)

)= W (Σ) = |Σ|.

Standard family of conformal transformations:

• conformal transformation v ∈ B4 → 0:

Fv(x) =1− |v|2

|x− v|2(x− v)− v,

• |Fv(Σ)| < |Σ|, if v 6= 0, and if Σ is not S2.(?)

Index of minimal surfaces in S3:

• Σ2 ⊂ S3, H = 0, stability operator Lϕ = 4ϕ+ (2 + |A|2)ϕ, index form:

Q(ϕ,ϕ) = −∫

ΣϕLϕ =

∫|∇ϕ|2 − (2 + |A|2)ϕ2.

I(Σ) = Index of Σ = # of negative eigenvalues of L.

• I(S2) = 1, Lϕ = 4ϕ+ 2ϕ, λ0 = −2, λ1 = 0.• I(T 2

c ) = 5, Lϕ = 4ϕ+ 4ϕ, λ0 = −4 with multiplicity 1 and eigenfunction u0 = const; λ1 =

−2, with multiplicity 4 and eigenfunctions: cos(√

2θ1), sin(√

2θ1), cos(√

2θ2) and sin(√

2θ2);λ2 = 0.• Σ2 ⊂ S3, N = (N1, N2, N3, N4) is unit normal of Σ, then by the translation invariance of the

cone C(Σ),4ΣN + |A|2N = 0.

SoLN = 4N + (2 + |A|2)N = 2N.

• Each Ni is an eigenfunction with eigenvalue −2. Since Ni changes sign(since N · x = 0 for allx ∈ Σ), hence not the first eigenvalue, so λ0 < −2.• Furthermore, N1, · · · , N4 is linearly independent, unless Σ is S2(or N must be constant sinceN1, · · · , N2 already satisfy 4-relations), so

I(Σ) ≥ 5,

unless Σ is S2.


Propostion 17.2 (F. Urbano 1990)If Σ is not a Clifford torus, then I(Σ) > 5.

Given Σ embedded in S3, with N unit normal, and d(x) =signed distance function to Σ, −π ≤d(x) ≤ π. Let

Σt = x : d(x) = t = ∂d(x) < t, t ∈ [−π, π].

Propostion 17.3|Σt| ≤W (Σ).

Proof: The smooth map ψt : Σ → Σt is given by ψt(y) = cos(t)y + sin(t)N(y), where y ∈ Σ, andN(y) the unit normal. Hence if e1, e2 is the o.n. principal basis of TyΣ,

Dψt|yei = (cos(t) + sin(t)ki)ei.

SoArea(Σt) =

∫Σdet(Dψt)dΣ =

∫Σ

(cos t+ k1 sin t)(cos t+ k2 sin t)dΣ,

while(cos t+ k1 sin t)(cos t+ k2 sin t) = cos2 t+ (k1 + k2︸︷︷︸

=2H

) sin t cos t+ k1k2︸︷︷︸≤H2

sin2

≤ 1− sin2 t+ 2H cos t sin t+H2(1− cos2 t)

= 1 +H2 − (sin t+H cos t)2.

Canonical family:Given Σ ⊂ S3 embedded, N =unit normal, define:

Φ : B4 × [−π, π]→ Z2,

Φ(v, t) =(Fv(Σ)t

)= Σ(v,t), v ∈ B4, t ∈ [−π, π],

where Z2 is the set of 2-currents.

Propostion 17.4sup

(v,t)∈B4×[−π,π]

|Φ(v, t)| ≤W (Σ).

18 MARQUES-NEVES’S PAPER 1 (3/13/2012) 50

18 Marques-Neves’s paper 1 (3/13/2012)

Let Σ ⊂ S3 be embedded and g(Σ) ≥ 1.Making Σ(v,t) continuous:

Denote S3+ and S3

− be the connected components of S3 \ Σ, with N pointing to S3+. Then

limv→p

Σ(v,t) = ∂Bt(−p), for p ∈ S3+;

limv→p

Σ(v,t) = ∂Bπ+t(p), for p ∈ S3−.

When p ∈ Σ and v → p with angle θ, i.e. informally v−p forms an angle θ with the position vector p,

limv→p(with angle θ)

Σ(v,t) = ∂Bπ2−θ+t(− sin(θ)p− cos(θ)N(p)).

(It is easy to check that the above are consistent when θ = π2 or θ = −π

2 .)Now introduce the polar coordinates near Σ. Given (s, θ) ∈ [0, ε)× [−δ, δ] and p ∈ Σ, then define

P (s, θ, p) = (1− s)(cos(θ)p+ sin(θ)N(p)).

This map has a geometric explanation. Given x ∈ B4 near Σ, let s(x) = d(x, S3) and θ the signeddistance between x

|x| and Σ, say θ = distΣ(x/|x|, p) for some p ∈ Σ, then x = P (s, θ, p).

Definition 18.1 Define the ε neighborhood of Σ as:

Nε(Σ) = P (s, θ, p) : p ∈ Σ, s ≥ 0, s2 + θ2 < ε2.

It is not hard to find a continuous mapF : B4 → B4,

such that F : B4\Nε(Σ)→ B4 is a diffeomorphism; F : S3 → S3 is identity and F : Nε(Σ)→ S3 isgiven by nearest point projection, i.e. given x = P (s, θ, p), then F (s, θ, p) = cos(θ)p+ sin(θ)N(p).

Canonical family:Firstly, define

C : B4×[−π, π] :

C(v, t) = Σ(F (v),t) when v ∈ B4 \Nε(Σ);

C(v, t) = ∂Bt(−v) when v ∈ S3+ \Nε(Σ);

C(v, t) = ∂Bπ+t(v) when v ∈ S3− \Nε(Σ);

C(v, t) = ∂Bπ2−θ+t(− sin(θ)p− cos(θ)N(p)) when v = F (s, ϕ, p) ∈ Nε(Σ),

where θ = tan−1( ϕ√ε2−ϕ2

).

Given x ∈ (∂B4 ∪Nε(Σ)), ∃ ! Q(x) and s(x), so that

C(x, s(x)) = ∂Bπ/2(Q(x)),

has radius π2 . Here s : (∂B4 ∪Nε(Σ))→ [−π/2, π/2] and Q : (∂B4 ∪Nε(Σ))→ S3 is given in the

definition of C.Key Property: Q : S3 → S3 is continuous and has degree g(will be proved in the next section).


Definition 18.2 The canonical family Φ : I5 → Z2 associated to Σ ⊂ S3 is defined as:

Φ(v, t) = C(f(v), (2t− 1)π + s(f(v))

), (v, t) ∈ I4 × I = I5,

where f : I4 → B4 is a diffeomorphism and s : B4 → [−π/2, π/2] is an extension of s : (∂B4 ∪Nε(Σ))→ [−π/2, π/2].

• Φ is continuous on I5;• Φ(p, 1/2) is an equator when p ∈ ∂I4;• Φ(v, 0) = Φ(v, 1) = 0 in Z2.

Definition 18.3 LetΠ = relative homotopy class of Φ(fixed on ∂I5).

Theorem 18.1 If genus g(Σ) ≥ 1, the width L(Π) > 4π.

Proof: Since maxx∈∂I5 |Φ(x)| = 4π, if the statement is false, then ∃ ϕi ∈ Π, such that

maxx∈I5|ϕi(x)| ≤ 4π +

1

i.

Now Φ∂I5 = Φ(I4×0)∪(I4×1)∪(∂I4×I), where

|Φ| : ∂I4 × 1/2 → RP3 = space of unoriented equators in S3,

has degree 2g(since θ is a genus g covering). Then there could not exist any continuous extension of|Φ| to any oriented submanifold S ⊂ I5 with ∂S = ∂I4 × 1/2.

• Given ε > 0, let

A(i) = connected component of t = 0 ⊂ I5 in x ∈ I5 : dV(|ϕi(x)|,Z0

)> ε,

where dV is the varifold distance and Z0 is the space of unoriented equators.• Claim: For i sufficient large, A(i) ∩ t = 1 ⊂ I5 = ∅.• If not, ∃ continuous path γi(t) ⊂ A(i) from t = 0 to t = 1. Let

Π1 = homotopy class of γi,

which is homotopic to any vertical path on ∂I4 × I , hence homotopically nontrivial. Then

maxt∈[0,1]

∣∣|ϕi|(γi(t))∣∣ ≤ 4π + 1/i.

If we run the min-max theory on Π, we must get a nontrivial embedded minimal surface Σ, sinceΠ1 is nontrivial, and Σ must have area less or equal to 4π, hence an equator. Furthermore, themin-max theory tells us that there exists a min-max sequence γi(ti), such that

dV(γi(ti),Σ

)→ 0,

under the varifold distance dV , hence a contradiction to the construction of A(i).


So A(i) gives an continuous extension of Φ|∂I4×I , hence a contradiction.

Proof: (of Theorem 17.2) ∃ connected smooth oriented embedded Σ of least area.

• Construct Φ : I5 → Z2(S3), hence maxx∈I5 |Φ(x)| = W (Σ) = |Σ|;• If Σ is not the Clliford torus Tc, then Σ has index greater or equal to 6,⇐⇒ Φ can be homotopic

to small max area;• Let Π be the homotopic class of Φ,

L(Π) = Σ < |Σ|,

where Σ is a connect smooth oriented embedded minimal surface by the discussion in the nextsection, a contradiction to the minimality of Σ.

Proof: (of Theorem 17.3) Given Σ oriented embedded surface of genus g ≥ 1 in S3, we can similarlyconstruct the canonical family Φ : I5 → Z2(S3), and maxx∈I5 |Φ(x)| ≤ W (Σ). Run the min/max,we get a embedded minimal surface Σ, such that |Σ| ≤ maxx∈I5 |Φ(x)| ≤W (Σ).

19 Marques-Neves’s paper 2 (3/15/2012)

Let Φ : I4×I → Z2 be the canonical family. For any p ∈ ∂I4, Φ(p, t) are spheres, with Φ(p, 1/2)

an equator. Denote Q : S3 → S3 the map from S3 ∼ ∂I4 to the center of Φ(p, 1/2) given above.

Theorem 19.1deg(Q) = g.

Proof: Since S3 \ Σ = S3+ ∪ S3

−, we can introduce

S3 = S3+ ∪ Σ× [−π

2,π

2] ∪ S3

−.

The map Q can be views as Q : S3 → S3, given by

Q(p) =

−p, when p ∈ S3

+;

− sin(θ)p− cos(θ)N(p), when (p, θ) ∈ Σ× [−π/2, π/2];

p, when p ∈ S3−.

Sodeg(Q) =

1

2π2

∫S3

(det(dQ))dµ.

Orientation on S3: a basis v1, v2, v3 ∈ TpS3 is positively oriented ⇐⇒ v1, v2, v3, p is positively

oriented in R4.


• Map p→ −p is positively oriented on S3, since if e1, e2, e3 positive on TpS3, then −e1,−e2,−e3,−ppositively on R4.• So ∫

S3

det(dQ)dµ = |S3+|+ |S3

−|︸︷︷︸2π2

+

∫Σ×[−π/2,π/2]

det(dQ)dµ.

• Given p ∈ Σ, let e1, e2 be an positively oriented o.n. principal basis, i.e. DeiN(p) = kiei wherek1, k2 are principal curvatures of Σ. Then e1, e2, N(p), p forms a positively oriented basis ofR4.

e1, e2,∂∂θ forms a positive basis for Σ× [−π/2, π/2].

• Since Q(p, θ) = − sin(θ)p− cos(θ)N(p),

dQ(ei) = − sin(θ)ei − cos(θ)kiei = (− sin(θ)− cos(θ)ki)ei;

dQ(∂

∂θ) = − cos(θ)p+ sin(θ)N(p).

• Note e1, e2,− cos(θ)p+ sin(θ)N(p)︸︷︷︸e3

forms an o.n. basis at TQ(p,θ)S3. Furthermore, e1, e2,− cos(θ)p+

sin(θ)N(p) is negatively oriented. This is because e1, e2,− cos(θ)p+sin(θ)N(p),− sin(θ)p−cos(θ)N(p) is negatively oriented in R4(by taking the wedge product, we get−e1∧e2∧N(p)∧p).• So

dQ(e1 ∧ e2 ∧∂

∂θ) = −deg(dQ)e1 ∧ e2 ∧ e3,

and we derive

deg(dQ) = −(sin θ + k1 cos θ)(sin θ + k2 cos θ) = −(sin2 θ + sin θ cos θ(k1 + k2) + cos2 θ).

Hence ∫Σ×[−π/2,π/2]

deg(dQ) = −∫

Σ(π

2+π

2k1k2)dµΣ = −π

2

∫Σ

(1 + k1k2︸︷︷︸KΣ

)dµΣ

= −π2

2π(2− 2g) = 2π2(g − 1).

Adding all the above together, we finish the proof.

Doing the Min-max: Let Zk(Mn) be the space of integral currents with flat topology, which roughlymeans that Σi → Σ if Σi − Σ = ∂Ri and |Ri| → 0.

Theorem 19.2 (Almgren)πn−k(Zk(Mn)) = Hn(M,Z) = Z.


• Φ : Sn−k → Zk(Mn) continuous in the homotopical notion;• Π = homotopy class of Φ;• L(Π) = min/max of Π;• (Almgren) L(Π) is achieved by stationary varifold;• k = n − 1, n = 3(Pitts) or n < 7(Schoen-Simon), the stationary varifold can be achieved by

smooth embedded hypersurface.

M-N paper: Φ : I5 → Z2(S3) is continuous in flat topology.

• t→ Σt is continuous in flat norm. If t1 < t2, then Σt2−Σt1 = ∂t1 < d(x,Σ) < t2 has smallvolume when t2 − t1 is small;• v → Σv,t is continuous even when v → S3;• 2π2 ≥ L(Π) > 4π if Σ has area less or equal to 2π2⇐⇒ the min/max is achieved by a smooth

embedded Σ, then Σ must have multiplicity 1 and connected⇐⇒ Σ is orientable.

Urbano’s theorem: (Proof of Proposition 17.2)

• Σ2 ⊂ S3 embedded, and H = 0. I(Σ) ≤ 5⇐⇒ Σ = S2, T 2c .

If I(Σ) > 5, then λ0 < −2, and λ1 = −2 has a 4 dimensional eigenspace generated by thenormal vector N = (N1, N2, N3, N4).• If I(Σ) ≤ 5, then I(Σ) < 5 only if Σ is an equator by discussion in Section 17. Let the

eigenfunction of λ0 be u0. If I(Σ) = 5, then every function ϕ such that∫

Σ ϕu0 = 0 mustsatisfy:

Q(ϕ,ϕ) =

∫(|∇ϕ|2 − (2 + |A|2)ϕ2) ≥ −2

∫ϕ2.

Let ψ : Σ→ S3, then ∃v ∈ B4, with Fv defined in Section 17, such that∫

Σ Fv ψ = 0. DenoteFv ψ = (ψ1, · · · , ψ4), then

4∑i=1

∫Σ

[|∇ψi|2 − (2 + |A|2)ψ2i ] ≥ −2

4∑i=1

ψ2i .

Hence

2|Σ| = 2E(ψ) ≥︸︷︷︸“= only if F=id

E(Fv ψ) =

∫Σ|∇Fv ψ|2 ≥

∫Σ|A|2︸︷︷︸

2(1−K)

= 2|Σ| −∫

ΣKda.

So∫

ΣKda ≥ 0, hence Σ is either S2 or a torus. When Σ is a torus, Fv = id, hence ψ =

(x1, · · · , x4), which are eigenfunctions of4xi = −2xi, so

4x = −2x = −|A|2x.

Hence |A|2 = 2, so K = 0.

REFERENCES 55

References

[1] A. Besse. Einstein Manifolds. Springer-Verlag. 2002.[2] T. Colding and W. Minicozzi II, Minimal surfaces. Courant Lecture Notes in Mathematics, 4.

New York University, Courant Institute of Mathematical Sciences, New York. 1999.[3] T. Colding and W. Minicozzi II, A course in minimal surfaces, Graduate Studies in Mathematics

Volume 121, American Mathematical Society, 2011.[4] H. Choi, R. Schoen, The space of minimal embeddings of a surface into a three-dimensional

manifold of positive Ricci curvature, Invent. Math. 81 (1985), no. 3, 387-394.[5] F. Marques and A. Neves, Min-max theory and the Willmore conjecture. Ann. of Math. (2) 179,

no. 2, (2014) 683-782.[6] R. Schoen, Estimates for stable minimal surfaces in three-dimensional manifolds. Seminar on

minimal submanifolds, 111-126, Ann. of Math. Stud., 103, Princeton Univ. Press, Princeton, NJ,1983.

[7] R. Schoen, L. Simon and S.T. Yau, Curvature estimates for minimal hypersurfaces, Acta Math.134, 275-288(1975).

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