Notes on Minimal Surface

Instructor: Zhou, Xin

February 15, 2018 Contents

1 Preliminary 3

2 Minimal Graph 4

3 The Calibration Properties 7 3.1 Calibrated property of minimal graphs ...... 7 3.2 A general calibrated argument ...... 8

4 First Variation 10 4.1 First Variation Formula ...... 10 4.2 Examples ...... 11 4.3 Convex hull property ...... 12 4.4 Fluxes ...... 12

5 Monotonicity formula and Bernstein Theorem 14 5.1 Monotonicity formula ...... 14 5.2 Bernstein’s theorem (n=2) ...... 16

6 Second variation and Stability 19 6.1 Second Variation of Volume ...... 19 6.2 Jacobi operator and Stability ...... 21

7 Additional Topics in Stability I 25

8 Criterion for stability 28

9 The Generalized Bernstein Theorem 32 9.1 Bochner Technique ...... 32 9.2 Proof of Generalized Bernstein Theorem ...... 35

10 Curvature Estimate 37

1 CONTENTS 2

11 Choi-Schoen’s Theorem 40

12 Small Curvature implies Graphical 44

13 Schoen’s Curvature Estimate 47

14 Bernstein’s Theorem and Minimal Cone 51

15 The Topology of Minimal surface 56

16 Closed Minimal Surfaces in S3 60

17 General Relativity 63 17.1 Einstein Equations ...... 64 17.2 Initial value problem ...... 64 17.2.1 Derivation of (CE) ...... 65 17.3 Asymptotical Flatness ...... 65

18 Geometric Measure Theory and Minimal Surfaces 67 Chapter 1

Preliminary

Absent for the first 20 minutes. Let Σ ⊂ (Mn, g) be a hypersurface of an n-dimensional Riemannian manifold (Mn, g), where D is the corresponding Riemannian connection. Let g|Σ be the induced metric. Given tangent vectors X,Y of Σ, the second fundamental form II, which is a vector valued symmetric 2-tensor on Σ, is defined as: ⊥ II(X,Y ) = (DX Y ) . Definition 1.1 (Mean Curvature). The mean curvature of Σ is defined as the trace of the second fundamental form II, which is

H = trII(X,Y ) = II(ei, ei), where ei are orthonormal basis of T Σ. Proposition 1.2. Assume M is a n-dimensional manifold, and Σ is a . Let X be the normal on Σ, i.e. X(p) ⊥ TpΣ for any p ∈ Σ, then we have divΣX = − < X, H > . Proof.

divΣX =< Dei X, ei >= − < X, Dei ei > ⊥ = − < X, (Dei ei) >= − < X, H >

3 Chapter 2

Minimal Graph

n−1 n−1 1 Let Ω ⊂ R is a domain, u :Ω 7→ R is a smooth function with |∇u|= 6 0, then, the graph of u, denoted Σu = {(x, u(x)) : x = (x1, ..., xn−1) ∈ Ω}. n ∞ Let F :Ω 7→ R ,F (x) = (x, u(x)) be the C embedding of Σu = F (Ω) n to R . Then, we are going to compute the volume form of Σu with local −1 n chart (F (Ω),F ). Let g be the induced metric of Euclidean Space R on Σ. Hence, we have ∂ ∂ ∂F ∂F ∂u ∂u g( , ) =< , >= δij + , ∂xi ∂xj ∂xi ∂xj ∂xi ∂xj i.e. g = I + ∇u(∇u)T . Moreover, since

g∇u = (I + ∇u(∇u)T )∇u = ∇u + ∇u∇uT ∇u = ∇u + ∇u(∇uT ∇u) = ∇u + |∇u|2∇u (2.1) = (1 + |∇u|2)∇u we know that 1 + |∇u|2 is a eigenvalue of g with eigenvector ∇u. Since Rank(g − I) = Rank(∇u(∇u)T ) = 1, we know that 1 is also a eigenvalue of g with multiplicity of n − 2. Since g is symmetry, we could ﬁnd 2 e2, ..., en−1,Hence det(g) = 1 + |∇u| . Let dx = dx1...dxn−1, we now have volume form dvol = pdet(g)dx = p1 + |∇u|2dx. By now, we can see that Z p 2 Area(Σu) = 1 + |∇u| dx. Ω

4 CHAPTER 2. MINIMAL GRAPH 5

∞ Let η ∈ Cc (Ω), we then consider Area(Σu+tη), if the graph Σu is minimal, d ∞ we must have dt Area(Σu+tη)|t=0 = 0 for any η ∈ Cc (Ω). Hence, we compute Z Z d d p 2 d p 2 Area(Σu+tη)|t=0 = 1 + |∇(u + tη)| dx|t=0 = 1 + |∇(u + tη)| dx|t=0 dt dt Ω Ω dt Z (∇u + t∇η)∇η Z ∇u∇η = dx|t=0 = dx p 2 p 2 Ω 1 + |∇(u + tη)| Ω 1 + |∇u| Z ∇u = − ηdiv( ), p 2 Ω 1 + |∇u| where the last equality follows from Stoke Formula. Hence, we can see that ∇u if Σu is minimal graph, we must have div(√ ) = 0. We are going to 1+|∇u|2 show that H = div(√ ∇u ). 1+|∇u|2 Since g(∇u(∇u)T ) = (g∇u)(∇u)T = ((1 + |∇u|2)∇u)(∇u)T (By(2.1)) = (1 + |∇u|2)∇u(∇u)T = (1 + |∇u|2)(g − I), we have g((1 + |∇u|2)I − ∇u(∇u)T ) = (1 + |∇u|2)I. Consequently, ∇u(∇u)T (gij) = g−1 = I − . 1 + |∇u|2 n It’s clear that we can regard Σu as lever set given by h : R 7→ R, h(x1, ..., xn) = ∇h 1 xn − u(x1, ..., xn−1). Hence v = = √ (−∇u, 1) is unit normal of |∇h| 1+|∇u|2 Σu. So

ij ∂ ∂ ij ∂ ⊥ ij ∂ H = g II( , ) = g (D ∂ ) = g < D ∂ , v > ∂xi ∂xj ∂xi ∂xj ∂xi ∂xj 1 ij 1 ij uiuj = g uij = (δ − )uij p1 + |∇u|2 p1 + |∇u|2 1 + |∇u|2 u u u u = ii − i j ij p 2 3 1 + |∇u| (1 + |∇u|2) 2 while ∇u ∂ u u u u u div( ) = ( i ) = ii − i j ij . p 2 p 2 p 2 3 1 + |∇u| ∂xi 1 + |∇u| 1 + |∇u| (1 + |∇u|2) 2 Hence, we have CHAPTER 2. MINIMAL GRAPH 6

d Proposition 2.1. dt (AreaΣu+tη)|t=0 = HΣu Chapter 3

The Calibration Properties

3.1 Calibrated property of minimal graphs

We now consider tubular neighbor OΣ = ∪t∈(−,)Σu+t of Minimal graph Σu. We can extend normal ﬁeld v tov ˜ ∈ OΣ, s.t.v ˜(x1, ..., xn−1, t) = v(x1, ..., xn−1). We claim that:

Proposition 3.1. Σu is minimal graph iﬀ divRn v˜ = 0. Proof.

n−1 n−1 X ∂v˜i ∂vn X ∂v˜i div n v˜ = + = R ∂x ∂t ∂x i=1 i i=1 i

= divΣu v = 0

n ∗ ∗−1 For any X ∈ Γ(T R ), we deﬁne iX :Ω 7→ Ω , by iX (α)(∗, ..., ∗) = α(∗, ..., ∗,X). Let ω = iv˜dx1...dxn be the calibrated form.

n Deﬁnition 3.2. Given any n − 1 dimensional subspace V ⊂ R , where V is spanned by n-1 orthonormal basis e1, ..., en, then deﬁne

ω(V ) = ω(e1, ..., en−1).

Lemma 3.3. Assume divRn v˜ = 0, then we have (1)dω = 0 (2)|ωp(V )| ≤ 1 for any p ∈ Σu+t and any n-1 dimensional subspace V of n R , the equality holds when iﬀ v˜ ⊥ V iﬀ V ∈ Tp(Σu+t).

7 CHAPTER 3. THE CALIBRATION PROPERTIES 8

Pn−1 i−1 ˆ Proof. (1)First, we observe that iv˜dvolRn = i=1 (−1) vidx1...dxi...dxn−1, ∂v while d(i dvol) = dvol n = div n vdvol˜ n = 0. v˜ ∂xi R R R (2)) It’s easy to see that

dx1...dxn(e1, ..., en−1, v) = det(e1, ..., en−1, v˜)

⊥ n ⊥ Letv ˜ = P (˜v), where P : R 7→ V is a projection. It’s easy to see that |v˜⊥| ≤ |v˜|, and

⊥ |det(e1, e2, ..., en−1, v˜)| = |det(e1, e2, ..., en−1, v˜ )| ≤ 1, while the equality holds iﬀ |v˜⊥| = |v˜| iﬀv ˜ ⊥ V.

n−1 Theorem 3.4. Let Σu be a minimal graph deﬁned on Ω ∈ R , then (1) Σu is volume minimizing in Ω × R; n (2) If Ω is convex, then Σu is volume minimizing in R .

Proof. (1) For any n − 1 dimensional sub manifold Σ1 ∈ Ω × R , with ∂Σ1 = ∂Σu, we can form the n dimensional chain U such that Σ1 ∩Σu = ∂U. Using the Stokes Theorem, we can see Z Z Z Z w = w − w = dw = 0, ∂U Σ1 Σu U while by Lemma 3.3 (2), we have Z Z Z Z

0 = w − w = w(Tp(Σ1))dvolΣ1 (p) − w(Tp(Σu))dvolΣu (p) Σ1 Σu Σ1 Σu ≤ Area(Σ1) − Area(Σu)

n (2) Since Ω is convex, the nearest point projection map F : R 7→ Ω × R is n well deﬁned, and |∇F | ≤ 1 a.e. Hence, for any Σ1 ∈ R , we consider F (Σ1). By (1), we can see that

Area(Σu) ≤ Area(F (Σ1)) ≤ Area(Σ1), where the last equality follows from |∇F | ≤ 1.

3.2 A general calibrated argument

Deﬁnition 3.5 (Foliation). We say a domain Ω ,→ (M n, g) in an oriented Riemannian Manifold is foliation (lamination) by hypersurface {Σt, t ∈ R} of Ω, if CHAPTER 3. THE CALIBRATION PROPERTIES 9

(1)Σt are disjointed. (2)∪t∈RΣt = Ω(∪t∈RΣt ⊂ Ω). n (3)∀p ∈ Ω, ∃ a neighborhood Up ∈ Ω, and smooth map f : Up 7→ R , s.t. f(Σt) is a set of {xn = t} for all small t ∈ R. Theorem 3.6. Suppose Ω is an open region in an oriented Riemannian manifold M n, and there exist a foliation of Ω by oriented minimal hyper surfaces, then every leaves of the foliation minimizes volume in Ω.

Proof. Let ν(x) be the unit normal vector ﬁelds of the foliation, i.e. ν ∈ ⊥ Γ(T Σt ) for some t. Then Claim: 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:

n−1 X divν = < Dei ν, ei > + < Dνν, ν >= −HΣt + 0 = 0. i=1

The last two equality follow from 0 = ν < ν, ν >= 2 < Dνν, ν > and Σt is minimal. Deﬁne n−1 ω = (−1) iνdvolM . Using ω as a calibrated form and arguments above, we can show the minimizing property. Chapter 4

First Variation

4.1 First Variation Formula

Consider Σk ⊂ M n. Let X be a smooth vector field on M with compact support. Let Ft : M → M be the flow defined by X, i.e.a

F0 = id d | F (p) = X(p), p ∈ M. dt t=0 t d Let Σt = Ft(Σ), the ﬁrst variation δΣ(X) := dt Area(Σt)|t=0. Then

Theorem 4.1. Z δΣ(X) = divΣ(X)dvolΣ, (4.1) Σ

where divΣ(X) =< Dei X, ei >, with {e1, ··· , ek} an orthonormal basis for T Σ. When Σ is smooth, we can decompose X = XT + X⊥ to tangent XT and normal X⊥ parts. So

T ⊥ divΣ(X) = divΣ(X ) + divΣ(X ),

⊥ where divΣ(X ) = − < X, H >. Using the divergence theorem, for general X, we have Z Z δΣ(X) = − hX,Hidµ + hX, ηidσ, Σ ∂Σ where η is the outer normal of ∂Σ. So we know that

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

10 CHAPTER 4. FIRST VARIATION 11

Proof. Consider a local parametrization

F :Σ × (−, ) → M, where F (x, t) = Ft(x) with Ft given to be the integration of X above. 1 k Let {e1, ··· , ek} be an orthnormal basis of T Σ, θ , ..., θ be the dual basis, ∗ (gt)ij = (Ft) g(ei, ej) then d (g ) | = (L g)(e , e ) = Xg(e , e ) − g([X, e ], e ) − g(e , [X, e ]) dt t ij t=0 X i j i j i j i j

= g(Dei X, ej) + g(Dei X, ej) − g(DX ei, ej) − g(ej,DX ei)

= g(Dei X, ej) + g(Dei X, ej)

d Hence tr( dt (gt)ij|t=0) = 2g(Dei X, ei) = 2divΣX. Moreover, we have d d det(g )| = tr( ((g ) )| ) = 2div (X) (4.2) dt t t=0 dt t ij t=0 Σ

Z d d p 1 k Area(Σt)|t=0 = detg(t)|t=0θ ...θ dt Σ dt Z d Z dt det(gt)|t=0 1 k 1 k = θ ...θ = divΣXθ ...θ Σ 2 Σ

4.2 Examples

Deﬁnition 4.2. Let Σk ⊂ (M n, g), then Σ is called minimal iﬀ H = 0.

2 3 Example 4.3. (1)R ⊂ R is minimal. 3 (2) The Helicoid H ⊂ R is minimal, where H = {(t cos s, t sin s, s)}, i.e., H is minimal graph deﬁned by u(x , x ) = arctan( x2 ). It’s straightforward 1 2 x1 to check that ∇u div( ) = 0. p1 + |∇u|2 2 2 2 (3) The Catenoid C, which is C = {(x1, x2, x3): x1 + x2 = (cosh x3) } is minimal. In addition, we could see that the surface is scale invariant, then let λC := {(x1, x2, x3):} CHAPTER 4. FIRST VARIATION 12

4.3 Convex hull property

k n ∞ Consider i :Σ ,→ R be the C embedding. Let x = (x1, ··· , xn) be n coordinates on R , then we have: Proposition 4.4. ∆Σx|Σ = H

Proof. Let {e1, ··· , ek} be a local orthonormal basis for T Σ, then

k k i X i Σ i X ⊥ i ∆Σx = (ejejx − (∇ej ej)x ) = (∇ej ej)x . j=1 j=1

P ⊥ As ejx = ej and ∇ej ej = H, ∆Σx = H.

Corollary 4.5. if Σ is minimal, ∆Σx = 0.

n Deﬁnition 4.6 (Convex Hall). For any subset X ∈ R , we deﬁned its convex hall C(X) by

n C(X) = ∩{H : H is the half space of R contains X}

k n Corollary 4.7. If (Σ , ∂Σ) is compact minimal in R , then Σ ⊂ C(∂Σ). Proof. We just need to prove for any Half space H also contains ∂Σ, H n n contains Σ. Assume H = {x ∈ R : l(x) ≤ a, v ∈ R , a ∈ R} for some linear function l. Now restrict l on Σ. Since l is linear, ∆Σl(x) = l(∆Σx) = 0, which mean l is harmonic on Σ. Moreover, since l|∂Σ ≤ a, by weak maximal principle(Lemma 4.8), we have lΣ ≤ a, which mean Σ ⊂ H. The following lemma is a standard result in EPDE:

n 2 0 Lemma 4.8. Let Ω ∈ R be a bounded domain. For f ∈ C (Ω) ∩ C (Ω) satisfying ∆f ≤ 0, we have supΩ f = sup∂Ω f.

4.4 Fluxes

k n In the case of a minimal sub manifold Σ ⊂ R ,

∆Σxi = 0 ⇔ (∗d ∗ d + d ∗ d∗)xi = 0 ⇔ d ∗ dx = 0 ⇔ ∗dxi is closed. CHAPTER 4. FIRST VARIATION 13

Hence ∗dxi deﬁnes a (k − 1) dimensional deRham cohomology class. So for any k − 1 cycle Γk−1 ⊂ Σk, we can deﬁne Z Z ∂ F ([Γ]) = ∗dxi = h , ηidσ, Γ Γ ∂xi where η is the unit outer normal of Γ. The second equality follows from the ∂ fact that ∗dx = i ∂ dv = h , ηidσ, with i dv the inner multiplication. i Σ ∂xi ∗ ∂xi Proposition 4.9. F depends on the homotopy class of Γ.

Proof. Assume Γ1 and G2 are homotopic to each other, then there exist k-cycle T , s.t. ∂T = Γ1 − Γ − 2. Then we have Z Z Z Z ∗dxi − ∗dxi = ∗dxi = d ∗ dxi = 0. Γ1 Γ2 ∂T T

Hence we have a group homeomorphism:

F : Hk−1(Σ, Z) → R. For the general cases, if Σk ⊂ (M n, g) is a minimal sub manifold, i.e. H = 0, and X a Killing vector ﬁeld on M, i.e. LX g = 0, then there exists a homeomorphism: FX : Hk−1(Σ, Z) → R. Proof. Let V = XT the tangential part of X on Σ, then

k k k X ⊥ X X ⊥ divΣ(V ) = hDei (X − X ), eii = hDei X, eii − hDei X , eii = 0. i=1 i=1 i=1 For the last equality, we have

⊥ ⊥ hDei X , eii = −hX ,Dei eii = H(X) = 0, and since

0 = LX g(ei, ei) = 2h[X, ei], eii = hDX ei, eii − hDei X, ei, i we know that

hDei X, eii = hDX ei, eii = 1/2Xhei, eii = 0.

Let ω = iV dvolΣ, then ω is a closed k − 1 form on Σ, since d(iV dvol) = divV dvol. Hence we can deﬁne the ﬂux as above. Chapter 5

Monotonicity formula and Bernstein Theorem

5.1 Monotonicity formula

Fix x0 ∈ Σ, let s, t > 0 small enough s.t. Bs(x0)∂Σ = ∅,Br(x0)∂Σ = ∅, n where Bt(x0) is a ball of radius t with center x0 in R . k n Theorem 5.1. Let Σ ⊂ R be a minimal surface, then Z ⊥ 2 Area(Bt(x0) ∩ Σ) Area(Bs(x0) ∩ Σ) |(x − x0) | k − k = k+2 dvolΣ, t s Σ∩(Bt(x0)\Bs(x0)) |x − x0|

⊥ ⊥ where (x − x0) is the projection to the normal part of Σ of (x − x0) . We need the co-area formula before the proof.

Lemma 5.2. (Co-area Formula) Let h :Σ → R+ be a nonnegative Lipschitz function on a Riemannian manifold Σ, and proper i.e. {x ∈ Σ: h(x) ≤ a} is compact for all a. Given f integrable on Σ, then

Z Z t Z f|∇Σh| = ( f)dτ. h≤t −∞ h=t

dt∧dv{h=t} Remark 5.3. This follows heuristically from the ideas that dvΣ = |∇Σh| when t is a regular value and Fubini Theorem.

Proof. (Monotonicity formula)Take h(x) = |x − x0|, then {h ≤ t} = Bt(x0). Pk Pk Let X = x − x0, then divΣ(X) = i=1 ∇ei X · ei = i=1 ei · ei = k, where

14 CHAPTER 5. MONOTONICITY FORMULA AND BERNSTEIN THEOREM15

{e1, ··· , ek} is an orthornormal basis on Σ. Then by the ﬁrst variation formula 4.1, Z Z

δΣBr(x0)(X) = divΣ0 (X) = X · η, Σ∩Br(x0) Σ∩∂Br(x0)

Σ ∇ |x−x0| where η is the co-normal vector of Σ ∩ ∂Br(x0), and η = Σ = |∇ |x−x0|| T (x−x0) T . Using the Co-area formular, |(x−x0) | Z Z T T |(x − x0) | k|Σ ∩ Br(x0)| = |(x − x0) | = r Σ∩{|x−x0|=r} Σ∩{|x−x0|=r} |x − x0| Z T Z T 2 d |(x − x0) | Σ d |(x − x0) | = r |∇ |x − x0|| = r 2 dr Σ∩Br(x0) |x − x0| dr Σ∩Br(x0) |x − x0| Z ⊥ 2 d |(x − x0) | = r (1 − 2 ) dr Σ∩Br(x0) |x − x0| Z ⊥ 2 d d |(x − x0) | = r |Σ ∩ Br(x0)| − r 2 . dr dr Σ∩Br(x0) |x − x0| Multiplying the above by r−k−1, we can re-write it as, Z ⊥ 2 d −k −k d |(x − x0) | (r |Σ ∩ Br(x0)|) = r 2 , dr dr Σ∩Br(x0) |x − x0| Z ⊥ 2 d |(x − x0) | = k+2 . dr Σ∩Br(x0) |x − x0| In the last step, we can use the co-area formula again to absorb the factor r−k into the integration. So we can get the monotonicity formula by integrating the above equation.

k n Corollary 5.4. Let Σ be a smooth minimal surface in R , with boundary Σ ∩ ∂BR(0) inside the ball BR(0). If x0 ∈ Σ ∩ BR(0), and σ < R − |x0|, then k k σ ωkσ ≤ Area(Σ ∩ Bσ(x0)) ≤ k Area(Σ ∩ BR(0)), (R − |x0| ) k k where ωk is the volume of unit ball B1 (0) in R . Proof. The ﬁrst inequality comes from the Monotonicity formula while com- −k paring Bσ(x0) with an arbitrary small ball Br(x0), with limr→0 r |Σ ∩ Br(x0)| = ωk, when x0 ∈ Σ and Σ smooth. The second inequality is a direct consequence of the Monotonicity formula while comparing Bσ(x0) with a large ball Br(x0) exhausting the whole BR(0). CHAPTER 5. MONOTONICITY FORMULA AND BERNSTEIN THEOREM16

Deﬁnition 5.5. The density of Σ at x0 is deﬁned as:

k −1 Θx = lim(ωkr ) |Σ ∩ Br(x0)|. 0 r→0 Example 5.6.

5.2 Bernstein’s theorem (n=2)

2 3 Theorem 5.7 (S. Bernstein (1912)). Given a minimal graph Σ ⊂ R , 2 2 Σ = {(x, u(x)) : x ∈ R }. If u is deﬁned on all of R , then u is a linear function, and Σ is a plane.

◦ 2 2 Theorem 5.8 (Bernstein’s Big Theorem:). 1 PDE version: let u ∈ C (R ) P2 and i,j=1 aijuiuj = 0, with (aij) > 0. If u is bounded, then u ≡ const; ◦ 2 2 2 :Σ − Graphu, where u is deﬁned on R 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 5.9. If H~ = 0, then N is a conformal and orientation reversing map, i.e. ∀v, w ∈ TxΣ, if v · w = 0 and |v| = |w|, then ∇vN · ∇wN = 0, and 1 ∗ |∇ N| = |∇ N|. Furthermore |∇ N| ≤ √ |A||v|, and N (ω 2 ) = K ω = v w v 2 S Σ Σ 1 2 − 2 |A| ωΣ. Proof. We only need to check that under principle vector ﬁelds. Take principle vector ﬁelds {e1, e2} for Σ, i.e. ∇e1 N = −K1e1, ∇e2 N = −K2e2.

So |∇e1 N| = |K1| = |K2| = |∇e2 N|, by the minimality H = K1 + K2 = 0. Hence |∇ N| ≤ |K||v| = √1 |A|. Furthermore, the Jacobian of N is v 2 1 2 Jac(N) = K1K2 = − 2 |A| . Remark 5.10. We are going to prove Bernstein’s theorem based on the following two facts: (1) We could choose an coordinate system and orientation of Σ, s.t. the 2 image of Gauss map lie in S+. (2) By Theorem 3.4(2), Σu is area minimizing. To apply fact (1), we have: CHAPTER 5. MONOTONICITY FORMULA AND BERNSTEIN THEOREM17

2 3 Proposition 5.11. Given a minimal Σ ⊂ R , with the image of the Gauss 2 Maps lying in the upper hemisphere N(Σ) ⊂ S+, if ϕ has compact support on Σ, then there exists a constant C > 0, such that Z Z |A|2ϕ2 ≤ C |∇ϕ|2. Σ Σ

2 Proof. Since S+ is simply connected, the closed form ωS2 = dα is also exact. Hence 2 |A| ∗ ∗ − ω = N ω 2 = d(N α). 2 Σ S So Z Z Z 2 2 2 ∗ ∗ |A| ϕ ωΣ = −2 ϕ d(N α) = 4 ϕdϕ ∧ N α Σ Σ Σ Z ∗ ≤ 4 |ϕ||∇ϕ||N α|ωΣ. Σ Since |N ∗α| ≤ |A||α| ≤ C|A|,

Z Z C Z C Z |A|2ϕ2 ≤ C (|ϕ||A|)(|∇ϕ|) ≤ |A|2ϕ2 + |∇ϕ|2. Σ Σ 2 2 Choose > 0 so that C = 1/2, we get the inequality.

To apply fact (2), we have

Proposition 5.12. 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. From Proposition 5.11, if we could ﬁnd a suitable cut-oﬀ function ϕ, s.t. ϕ = 1 on BR and vanish outside B2R. Moreover, it makes right hand side of equality in Proposition 5.11 goes to 0 as R goes to ∞. Then, we must have A = 0.

2 Lemma 5.13. When Σ = Graphu and u is an entire function on R , then R 2 we can choose a Lipschetz ϕ = ϕR, such that Σ |∇ϕ| → 0 as R → ∞. CHAPTER 5. MONOTONICITY FORMULA AND BERNSTEIN THEOREM18

Proof. Now choose

 1 if r ≤ R  2 ϕR(r) = 1 − log(r/R)/ log R if R < r < R  0 if r ≥ R2

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

Z Z log R Z 2 1 1 1 X 1 |∇ϕ| ωΣ = 2 2 ωΣ = 2 2 ωΣ Σ (log R) Σ∩(B \B ) r (log R) Σ∩(B \B ) r R2 R k=1 ekR ek−1R

log R log R 1 X 1 1 X 1 k 2 ≤ |Σ∩(B k \B k−1 )| ≤ C(e R) (log R)2 (ek−1R)2 e R e R (log R)2 (ek−1R)2 k=1 k=1 log R C X 1 C = = → 0, (log R)2 e2 log R k=1 where in the second “ ≤ ”, we used the quadratic area bound Lemma above.

Bernstein’s Theorem. When Σ = Graphu is an entire graph, the image R 2 2 of the Gauss Maps N(Σ) lies in an hemisphere, so we get Σ |A| ϕ ≤ R 2 C Σ |∇ϕ| . Then if we take the ϕR in the above Lemma, and let R → ∞, R 2 we see that Σ |A| → 0. So A = 0, and Σ is a plane. Chapter 6

Second variation and Stability

6.1 Second Variation of Volume

Consider a minimal Σk ⊂ M n, i.e. H~ = 0. Given a vector ﬁeld X on Σ, let Ft be the geometric ﬂow related to X, i.e.

F0 = id d F (p)| = X(p), p ∈ M dt t t=0

Theorem 6.1 (Second Variation Formula).

k d2 Z X δ2Σ(X,X) := | Area(Σ ) = [|D⊥X|2−|hA,~ Xi|2− RM (e , X, e ,X)], dt2 t=0 t i i Σ i=1 (6.1) where {e1, ··· , ek} is an orthonormal basis tangent to Σ, and X is compact supported and normal on Si.

Proof. • F :Σ × (−, ) → M, with {x1, ··· , xk} local coordinates on p Σ, s.t. F (p, t) = Ft(p), p ∈ Σ. Then dvolt = detg(t)dx, where ∂F ∂F gij(t) = h ∂xi , ∂xj i. • d 1 pdetg(t) = gijg˙ pdetg(t) dt 2 ij

19 CHAPTER 6. SECOND VARIATION AND STABILITY 20

• d2 1 1 1 pdetg(t) = (gijg˙ )2pdetg + gijg¨ pdetg + (g ˙ijg˙ )pdetg, dt2 4 ij 2 ij 2 ij

ij ik jl whereg ˙ = −g g g˙kl. • ∂F ∂F ∂F ∂F g˙ij = hD ∂F , i + h ,D ∂F i ∂t ∂xi ∂xj ∂xj ∂t ∂xi ∂F ∂F ∂F ∂F = hD ∂F , j i + h j ,D ∂F i ∂xi ∂t ∂x ∂x ∂xi ∂t ∂F ∂F ∂F ∂F = −h ,D ∂F j i − hD ∂F j , i ∂t ∂xi ∂x ∂xi ∂x ∂t ∂ ∂ = −hA~( , ),Xi.(When t = 0.) ∂xi ∂xj So ij (g g˙ij)|t=0 = −2hH,X~ i = 0, and ij 2 (g ˙ g˙ij)|t=0 = −4|hA,~ Xi| . • ∂F ∂F ∂F ∂F g¨ij = hD ∂F D ∂F , j i + hD ∂F ,D ∂F i i ∂t ∂xi ∂t ∂x ∂xj ∂t ∂t ∂x ∂F ∂F ∂F ∂F + hD ∂F D ∂F , i i + hD ∂F ,D ∂F j i ∂t ∂xj ∂t ∂x ∂xi ∂t ∂t ∂x M ∂F ∂F ∂F ∂F ¨ ∂F = hR ( , i ) , j i + hD ∂F F, j i + hD ∂F X,D ∂F Xi ∂t ∂x ∂t ∂x ∂xi ∂x ∂xi ∂xj M ∂F ∂F ∂F ∂F ¨ ∂F + hR ( , j ) , i i + hD ∂F F, i i + hD ∂F X,D ∂F Xi ∂t ∂x ∂t ∂x ∂xj ∂x ∂xj ∂xi M ∂ ∂ ¨ ∂ = −R (X, i ,X, j ) + hD ∂ F, j i + hD ∂ X,D ∂ Xi ∂x ∂x ∂xi ∂x ∂xi ∂xj M ∂ ∂ ¨ ∂ − R (X, j ,X, i ) + hD ∂ F, i i + hD ∂ X,D ∂ Xi ∂x ∂x ∂xj ∂x ∂xj ∂xi So

ij ij M ∂ ∂ ¨ ⊥ 2 ij T T (g g¨ij)|t=0 = −2g R (X, i ,X, j )+2divΣF +2|D X| +2g hD ∂ X,D ∂ Xi ∂x ∂x ∂xi ∂xj | {z } =2|hA,X~ i|2 CHAPTER 6. SECOND VARIATION AND STABILITY 21

• Combining all the above,

d2 | pdetg(t) = div F¨+|D⊥X|2−|hA,~ Xi|2−gijRM (X, ∂xi, X, ∂xj). dt2 t=0 Σ An integration on Σ ﬁnishes the proof.

Theorem 6.2. In the case Σn−1 ⊂ M n is a hyper surface and 2-sided(there exist global unit normal ν), X = ϕν, with ϕ a function with compact support, then Z δ2Σ(ϕ, ϕ) = [|∇ϕ|2 − (|A|2 + RicM (ν, ν))ϕ2] (6.2) Σ Proof. We have

|D⊥X| = hD⊥X, νi = h∇φν, νi + hφD⊥v, vi = |∇φ|,

|hA, Xi|2 = |φ|2|hA, vi|2 = φ2|A|2, n−1 n−1 X X 2 2 R(X, ei, X, ei) = φ R(ν, ei, ν, ei) = φ Ric(ν, ν). i=1 i=1

6.2 Jacobi operator and Stability

Deﬁnition 6.3 (Stability). Σ is stable if I(X,X) := δδΣ(X,X) ≥ 0, ∀X normal with compact support.

Corollary 6.4. If Ric ≥ 0, then there is no closed 2-sided stable minimal hypersurface.

Proof. Take φ = 1 in Theorem 6.2.

Corollary 6.5. If Σn−1 ⊂ M n is two-sided and stable minimal surface, then Ric(ν, ν) = 0 on Σ, where ν is normal on Σ. Moreover, Σ is total geodesic.

Proof. Take φ = 1.

2 3 Remark 6.6. The 2-side condition is essential. Since RP ,→ RP is stable, but Ric 6= 0.

Proof. CHAPTER 6. SECOND VARIATION AND STABILITY 22

Corollary 6.7. Assume Σ ,→ M 3 is a closed, two-sided stable minimal hypersurface with scale curvature R ≥ 0. Then the genus g of Σ must be 0 or 1. Example 6.8. (1)For g = 0, we have S2 ,→ S2 × S1 with R = 1. 2 3 (2)For g = 1, we have T ,→ T , with R = 0. In order to prove Corollary 6.7, we need Lemma 6.9. Let Σn−1 ,→ M n, ν is normal on Σ locally, then we have 1 |A|2 + RicM (ν) = (RM − RΣ + |A|2). 2 Proof of Corollary 6.7. First, we have Z Z |∇φ|2 = (|A|2 + Ric(ν, ν))φ2 1 Z = (RM − RΣ + |A|2)φ2 2 1 Z 1 Z ≥ −RΣφ2 = −Kφ2. 2 2 Take φ = 1, by Gauss-Bonnet, we have 0 ≤ 4π(2 − 2g). Hence, we know that g = 0, 1.

Proof of Lemma 6.9. Locally, let e1, ...en−1 be a orthonormal basis of Σ. Let Σ Σ M M Rijij = R (ei, ej, ei, ej),Rijij = R (ei, ej, ei, ej), hij = hDei ej, νi. Then we have Σ 2 M Rijij + hij − hiihjj = Rijij. Sum over i, j, we have RΣ + |A|2 − H2 = RM − 2Ric(v, v)

We now consider hypersurface case, i.e. k = n − 1, X = ϕν, Z I(ϕ, ϕ) ≡ δ2Σ(ϕ, ϕ) = [|∇ϕ|2 − (|A|2 + RicM (ν, ν))ϕ2] Σ Z = −∆ϕφ − (|A|2 + RicM (ν, ν))ϕ2] Σ Z = − ϕLϕ, Σ where the Jacobi operator L is

Lϕ = ∆ϕ + (|A|2 + RicM (ν, ν))ϕ. (6.3) | {z } Q CHAPTER 6. SECOND VARIATION AND STABILITY 23

When boundary exists (Σ, ∂Σ), L has discrete eigenvalues λj and eigenfunctions uj with Dirichlet condition, i.e.

Luj + λjuj = 0, in Σ

uj = 0, on ∂Σ and λ1 < λ2 ≤ λ3 ≤ · · · , with λn → +∞. Deﬁnition 6.10. Morse Index of Σ is deﬁned as the number of negative eigenvalues counted with multiplicity.

Proposition 6.11. (1)λ1 has multiplicity 1; (2)If u1 is an eigenfunction of λ1, u1 does not change sign.

Proof. We just prove (2) Here. By the variational characterization, u1 R 2 minimizes I(ϕ, ϕ) among all ϕ with ϕ ≡ 0 on ∂Σ and Σ ϕ = 1. Since |(|ϕ|, |ϕ|) = I(ϕ, ϕ), if u1 is the ﬁrst eigenfunction, so is |u1|. So u1 = |u1|, ∞ or there is a contradiction to u1 ∈ C . The fact that u1 does not change sign shows that the dimension of the eigen space of λ1 is 1, or we can always form some eigenfunction changing sign.

Deﬁnition 6.12. When Σ is non-compact, then

Ind(Σ) = lim Ind(Ωi), i→∞

∞ ∞ where {Ωi}i=1 is an open exhaustion of Σ, i.e. Σ = ∪i=1Ωi, Ωi ⊂ Ωi+1, with ∂Ωi smooth and compact. Remark 6.13. In fact, the deﬁnition is independent of the exhaustion, say {Ωi} and {Ω˜ i}. ∞ Proof. First, we have λk = inf{λ(V ): V ⊂ Cc (Ω), dim(V ) = k}, where λ(V ) = sup{− R Lff : f ∈ V, R |f|2 = 1}(cf. [1] section 4.5). By this we can ∞ ∞ 0 see Ind(Ω) is non decreasing when Ω is expanding (since Cc (Ω) ⊂ Cc (Ω ) 0 ˜ 0 when Ω ⊂ Ω ), so we can always embed Ωi ⊂ Ωi0 for i i, so Ind(Ωi) ≤ ˜ ˜ Ind(Ωi0 ), and limi→∞ Ind(Ωi) ≤ limi0→∞ Ind(Ωi0 ), and vise versa. When Σ is open:

λ1(Σ) = lim λ1(Ωi) ∈ [−∞, λ1(Ω1)), i→∞

Σ is stable if λ1(Σ) ≥ 0, or equivalently λ1(Ω) ≥ 0 for all Ω ⊂ Σ. CHAPTER 6. SECOND VARIATION AND STABILITY 24

Remark 6.14. 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). Chapter 7

Additional Topics in Stability I

• Let Σ2 ⊂ M 3 be a closed, two-sided stable minimal hypersurface with positive scale curvature R. In the proof Corollary 6.7, we know that if the scale curvature R > 0, we must have Σ = (S2, g). Hence, we have the following positive mass theorem

3 Theorem 7.1. There is no positive scale metric on T .

Proof. By an Theorem 4.1 in [6] given by Schoen-Yau, there exists an 2 3 area-minimizing immersion u : T ,→ T . If there exists a positive 3 scale metric on T , then we know from the proof of Corollary 6.7 that the genus of u(T) must be 0, which is a contradiction.

• We now focus on the 3-manifold M, which admits a scale curvature R ≥ 6 in the following. Marques-Neves(2013) and Song(2015) (??? Not sure) show that if there exists a least area minimal surface of Σ ⊂ M, then Area(Σ) ≤ 4π. The equality holds iff M is the standard 3-sphere S3. (1)We would like to study the stable 1-sided minimal surfaces first. 2 3 An example is that RP ,→ RP . Let Σ2 ⊂ M 3 be a stable 1-side minimal surface, X be a normal field on Σ with compact support. Moreover, let π : Σ˜ 7→ Σ be a 2-cover of Σ,ν ˜ be a unit normal vector field of Σ˜ ,→ M 3, i : Σ˜ 7→ Σ˜ be the deck transformation. Then we have (a)˜ν ◦ i = −ν.˜

25 CHAPTER 7. ADDITIONAL TOPICS IN STABILITY I 26

(b)Let X˜ = X ◦ π, then X˜ ◦ i = X. Hence, if we assume X˜ = φν˜, then we have φ ◦ i = −φ. Since Σ is stable, for any φ ∈ C∞(M), s.t. φ ◦ i = φ, we have Z Z |∇φ|2 ≥ (Ric(ν, ν) + |A|2)φ2 (7.1) Σ˜ Σ˜

2 3 Moreover, we have a conformal map F : Σ˜ 7→ S (1) ,→ R , s.t. degF ≤ g(Σ) + 1,F ◦ i = F, where g(Σ) is the genus of Σ. Assume F = (f1, f2, f3), then Z 3 Z 2 X 2 2 (Ric(˜ν, ν˜) + |A| ) = (Ric(˜ν, ν˜) + |A| )fi ˜ ˜ Σ i=1 Σ Z Z X 2 2 ≤ |∇fi| = |∇F | Σ˜ Σ = 2deg(F )Area(S)(integration by substitution) = 8πdeg(F ) ≤ 8π(g(Σ) + 1). (7.2)

while Z 1 Z Ric(ν, ν) + |A|2 = RM − RΣ + |A|2 Σ˜ 2 Σ˜ 1 Z ≥ 6 − K (7.3) 2 Σ˜ Z = 6 − K = 6Area(Σ) − 4π(1 − g) Σ By (7.2) and (7.3), we have Area(Σ) ≤ 2π, whenever g = 0 or 1. (2)We then consider the 2-sided minimal surface Σ ⊂ M, with ind(Σ) = 1, i.e., there exists one and only one negative eigenvalue λ1 for Jacobi operator L. Let u1 be the eigenvector for λ1, then for any φ ⊥ u1, we have Z Z (Ric(ν, ν) + |A|2)φ2 ≤ |∇φ|2. (7.4)

Since φ lies in a subspace that generated by nonnegative eigenvector. Moreover, there exist holomorphic map F :Σ 7→ S2, s.t. degF ≤ 3g+1 [ 4 ], where [x] is the integer-valued function. Notice that Z 3 w = u1F ∈ B , Σ CHAPTER 7. ADDITIONAL TOPICS IN STABILITY I 27

3 3 where B3 = {x ∈ R : |x| < 1}. For any v ∈ B , there exist conformal 3 3 function F : B 7→ B , s.t. Fv(v) = 0. Let G = Fw ◦ F , we have Z u1G = 0. (7.5) Σ

Let G = (g1, g2, g3), by (7.4) and (7.5), we have Z Lgigi ≥ 0.

Hence, Z Z 3g + 1 (Ric(ν, ν) + |A|2 ≤ |∇G|2 ≤ deg GArea(S2) ≤ 4π[ ] Σ Σ 4 while Z 1 Z Ric(ν, ν) + |A|2 = RM − RΣ + |A|2 Σ 2 Σ 1 Z ≥ 6 − K (7.6) 2 Σ 1 Z = 6 − K = 3Area(Σ) − 2π(1 − g), 2 Σ

3g+1 we have 3Area(Σ) ≤ 2π(1 − g) + 4π[ 4 ]. Moreover, it’s straightforward to check that Ric(ν, ν) + |A|2 ≥ −2K. 3g+1 So we have 4π(g − 1) ≤ 4π[ 4 ]. Consequently, g ≤ 3. • We have the following estimate:

Theorem 7.2 (Schoen-Yau). Let (Σ2, ∂Σ) ,→ (M 3, g) be a stable minimal surface. Moreover, the scale curvature Rg of M has a positive lower bound Λ, i.e. R ≥ Λ > 0. Then, we have r (Σ, ∂Σ) ≤ √C for g inf Λ some positive constant C, where rinf = maxp∈Σ dist(p, ∂Σ). Corollary 7.3. Let (M, g) be a noncompact Riemannian manifold with Rg ≥ Λ > 0, then there is no complete stable minimal surface in (M 3, g). Chapter 8

Criterion for stability

Theorem 8.1. Let Σ ⊂ M be a 2-sided minimal surface, Σ is stable iﬀ (1)when Σ closed, there exist u > 0, s.t. LΣu ≤ 0. (2)when Σ is complete and noncomplete, there exist u > 0, s.t. LΣu = 0. Remark 8.2. This can be viewed as an inﬁnitesimal version of the Calibra- tion argument i.e. using foliation of minimal surfaces.

Corollary 8.3. Σ ⊂ M is 2-sided minimal surface, Σ is stable, then for any covering map π : Σ˜ 7→ Σ, if Σ˜ ⊂ M, then Σ˜ is also 2-sided and stable.

Proof. Let u > 0 satisfy the condition in Theorem 8.1, then we have ( ≤ 0, when Σ is compact, L˜ (u ◦ π) = Σ = 0, when Σ is noncompact.

2 3 Remark 8.4. The ”2-sided” condition is essential, consider RP ,→ RP 2 3 is stable, but S ,→ RP is not.

Proof for Theorem 8.1. ⇐=: Since Lu = ∆Σu + Qu ≤ 0, let w = log u(u > 0), we have ∆u ∆w = − |∇w|2 ≤ −Q − |∇w|2. u Then ∀ϕ compactly supported, Z Z ϕ2(∆w + Q) ≤ − ϕ2|∇w|2. Σ Σ

28 CHAPTER 8. CRITERION FOR STABILITY 29

Using integration by part formula, Z Z Z Qϕ2 ≤ 2ϕh∇ϕ, ∇wi − |∇w|2ϕ2 ≤ 2|ϕ||∇ϕ||∇w| − |∇w|2ϕ2 Σ Z Z ≤ |∇ϕ|2 + ϕ2|∇w|2 − |∇w|2ϕ2 ≤ |∇ϕ|2. Σ 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 ∈ C(∂Ωi), ∃!u in Ωi, such that ( Lui = 0, in Ωi,

ui = ψ, on ∂Ωi.

Claim: λ1(Ωi) > 0 =⇒ if ψ > 0, then u > 0. Otherwise, if u ≤ 0, then Ω{u≤0} has eigenvalue equals 0, since u is then a Dirichlet eigenfunction on Ω{u≤0} with 0 boundary values, extend u by 0 to 1,2 u˜ ∈ W (Ωi), let w be the λ1(Ωi) eigenfunction of L, then we have R (−Lw)w R (−Lv)v R (−Lu˜)˜u 0 < λ (Ω) = Ωi = inf{v ∈ W 1,2(Ω ): Ωi } ≤ Ωi = 0, 1 R |w|2 i R |v|2 R |u˜|2 Ωi Ωi Ωi (8.1) which is a contradiction. Now we can solve the boundary value problem for ui > 0: ( Lui = 0 in Ωi,

u = 1 on ∂Ωi.

ui ∞ Fix p ∈ Ω1, then consider the normalized sequence { } , ui(p) i=1 ui 0 ∞ Claim: Letu ˜i = , then there exists a subsequence i → ∞, u ∈ C (Σ), ui(p) such that 2 u˜i0 → u in C on any compact subset of Σ, and Lu = 0. Proof of Claim: Fix any compact subset K ⊂ Σ, then when i big enough, 0 K ⊂⊂ Ωi. Hence, we may assume K ⊂⊂ Ωi for all i > 0. Fix Ω, Ω s.t. CHAPTER 8. CRITERION FOR STABILITY 30

0 K ⊂⊂ Ω ⊂⊂ Ω ⊂⊂ Ω1. By Hanack inequality (cf. Theorem 4.3.3 [2]), we have

0 0 0 sup |u˜i(x)| ≤ C(n, Ω , Ω1) inf |u˜i| ≤ C(n, Ω , Ω1)˜u(p) = C(n, Ω , Ω1). x∈Ω0 x∈Ωi

∞ By a standard regularity theorem of EPDE, we could see thatu ˜i ∈ C (Ω1). Hence, by Gradient Estimate of EPDE (cf. Proposition 2.3.2 in [2]), we have

0 0 sup |Du˜i(x)| ≤ C(n, Ω, Ω ) sup |u˜i(x)| ≤ C(n, Ω, Ω , Ω1). x∈Ω x∈Ω0 By Schauder Estimate, we have

k u˜i kC3(K)≤ C(n, K, Ω1).

The claim follows from standard application of Ascoli. The limit u is a positive solution of Lu = 0.

Deﬁnition 8.5 (Convergence in the C∞ sense). We say a sequence of min- n−1 n ∞ imal surface Σ ⊂ R converge to Σ in the sense of C , if for any p ∈ Σ, there exist r > 0, s.t.

Σi ∩ Br(p) = Graph of ui over TpΣ for i large enough,

∞ and ui → u in C topology, where u is a function s.t. Σ is a graph of u over TpΣ. Example 8.6. (1)Let Σ ,→ M n be the C∞ embedding hypersurface. For any p ∈ Σ, r → 0, Σ = Σ−p → T Σ in the sense of C∞. i p,ri ri p ∞ (2)Let C be the Catenoid, then λiC → 2R in C sense, except for 0. ∞ Theorem 8.7 (L. Simon). If Σi → Σ in C sense, then there exists u 6= 0, s.t. LΣu = 0. If Σ lie in one side of Σ, we could ﬁnd u > 0, s.t. LΣu = 0. Remark 8.8. The Proof of Theorem 8.7 is similar to Theorem 8.1.

◦ n−1 n Proposition 8.9. 1 . Let Σ ⊂ R be a 2-sided minimal surface, and if n−1 the Gauss image G(Σ) ⊂ S+ , then Σ is stable; ◦ 2 3 open 2 2 . Let Σ ⊂ R be a 2-sided minimal surface, and if G(Σ) ⊂ U ⊂ S , with µ1(U) ≥ 1, where µ1(U) is the Dirichlet eigenvalue of ∆S2 on U, then Σ is stable. In perticular, µ1(U) ≥ 1 is true if the area |U| ≤ 2π. CHAPTER 8. CRITERION FOR STABILITY 31

◦ n Proof. 1 . Let e ∈ R be the direction vector to the north pole, and let u = e·ν, where ν is the normal vector ﬁeld of Σ, since the parallel translation in the e direction does not change the area of Σ, we have Lu = 0. Since n−1 G(Σ) ⊂ S+ ⇐⇒ e · ν > 0, so u > 0, hence Σ is stable. ◦ 2 . µ1(U) ≥ 1 =⇒ ∃v > 0 on U such that ∆S2 v = µi(U)v ≤ −v, in U, v = 0, on U.

Let u = v ◦ G, where G is the Gauss Map. By Lemma 5.9, G :Σ → S2 is a conformal map, so

2 2 ∆Σu = |A| (∆S2 v) ◦ G ≤ −|A| u, i.e. Lu ≤ 0, hence Σ is stable. (In fact, on 2-dimension, the Jacobi operator ∗ L = G (∆S2 + 1).) Chapter 9

The Generalized Bernstein Theorem

This chapter, we are going to prove: Theorem 9.1 (The Generalized Bernstein Theorem). Any complete non- 2 3 compact 2-sided stable minimal immersion Σ ⊂ R is a plane.

9.1 Bochner Technique

In this section, we will use Bochner Technique to prove: n−1 n Theorem 9.2 (Vanishing of Harmonic 1-Forms). If Σ ⊂ R is a complete, stable and 2-sided minimal surface, then any L2 harmonic 1-form on Σ vanishes. k Let (Σ , g) be a Riemannian manifold, and {e1, ··· , ek} an o.n. frame, with {θ1, ··· , θk} the dual frame. Denote

X i (∇ej α) = αi,jθ , i

X i j ∇ei (∇α) = αi,jkθ ⊗ θ ; i,j then X i j 2 X i j k ∇α = αi,jθ ⊗ θ , ∇ α = αi,jkθ ⊗ θ ⊗ θ . i,j i,j,k Ricci Formula: P Σ αi,jk − αi,kj = p αpRpijk.

32 CHAPTER 9. THE GENERALIZED BERNSTEIN THEOREM 33

Deﬁnition 9.3. α is harmonic if dα = 0 and δα = 0(i.e. αi,j = αj,i and P i αi,i = 0). Bochner Formula: If α is harmonic, then

∆α = Ric(α], ·),

] P i where α the vector ﬁeld dual to α, and ∆α = i,j αi,jjθ is the rough laplacian. Proof.

X X X X Σ X Σ αi,jj = αj,ij = αj,ji + αpRpjij = αpRicpi. j j j p,j p | {z } =0

Hence we have:

1 2 2 ] ] 2 2 ∆|α| = hα, ∆αi + |∇α| = Ric(α , α ) + |∇α| .

n−1 n Σ In the case Σ ⊂ R is minimal, Rijkl = hikhjl − hilhjk under the o.n. frame {ei} by the Gauss equation, hence

Σ X Σ X X Ricik = Rijkj = − hijhjk, ( hjj = 0). j j j

1 X X X =⇒ ∆|α|2 = |∇α|2+ RicΣ α α = |∇α|2− ( h α )2 ≥ |∇α|2−|A|2|α|2. 2 ij i j ij j ij i j

1 2 2 Plug in 2 ∆|α| = |α|∆|α| + ∇|α| ,

2 2 2 2 |α|(∆|α| + |A| |α|) ≥ |∇α| − ∇|α| ≥ c(n) ∇|α| , | {z } L|α| where Lu = ∆u + |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 for j = 2, ··· , k, then

P (P α α )2 2 2 X 2 j i i i,j X 2 X 2 |∇α| − ∇|α| = α − = α − α i,j |α|2 i,j 1,j ij i,j j CHAPTER 9. THE GENERALIZED BERNSTEIN THEOREM 34

k k k k X X X 1 X X = α2 ≥ α2 + α2 ≥ ( α )2 + α2 i,j i,i i,1 k − 1 i,i 1,i i>1,j i=2 i=2 i=2 i=2 | {z } =−α1,1

k 1 2 X 2 1 2 ≥ [α + α ] = ∇|α| . k − 1 1,1 1,i k − 1 i=2 R Proof of Theorem 9.2. 2-sided and stability means that − Σ ϕLϕ ≥ 0 for any ϕ compactly supported. So ∀ϕ compactly supported Z − ϕ|α|L(ϕ|α|) ≥ 0, Σ i.e. Z ϕ|α|(∆(ϕ|α|) +|A|2ϕ|α|) ≤ 0, Σ | {z } I where Z Z 1 I = ϕ|α|(ϕ∆|α|+2h∇ϕ, ∇|α|i+|α|∆ϕ) = ϕ2|α|∆|α|+ h∇ϕ2, ∇|α|2i+|α|2ϕ∆ϕ Σ Σ 2 Z Z 1 Z ≤ ϕ2|α|∆|α| − ∆(ϕ2)|α|2 + |α|2ϕ∆ϕ Σ Σ 2 Σ Z = ϕ2|α|∆|α| − (ϕ∆ϕ + |∇ϕ|2)|α|2 + |α|2ϕ∆ϕ Σ Z = ϕ2|α|∆|α| − |∇ϕ|2|α|2. Σ Plug into the above Z Z ϕ2|α|L(|α|) ≤ |∇ϕ|2|α|2. Σ Σ

Now by taking ϕ = ϕR to be cutoﬀ functions on geodesic disk, and let- ting R → ∞, the righthand side of the above inequality is zero, hence by

|α|L(|α|) ≥ c(n) ∇|α| proved above, Z Z 2 c(n) ∇|α| ≤ |α|L(|α|) = 0, Σ Σ which means that |α| is a constant, and hence is 0 since the area of Σ is ∞ k by the monotonicity |Bσ(p)| ≥ wkσ . CHAPTER 9. THE GENERALIZED BERNSTEIN THEOREM 35

9.2 Proof of Generalized Bernstein Theorem

First, it’s easy to see that: Lemma 9.4. Let (Σ2k, g) be 2k-dimensional manifolds, for any ω ∈ ΩkΣ, the L2 norm of ω is comformally invariant. Lemma 9.5. If α is harmonic in (Σ2, g), then α is also harmonic in (Σ, ρg), for any smooth function ρ > 0. Proof. We notice that ( dα = 0, −2 δρgα = ρ δgα = 0.

∼ 2 3 Proposition 9.6. Σ˜ = D cannot be conformally imersed into R as a complete noncompact 2-sided stable minimal surface. 3 Proof. Let (R , δ) be standard Euclidean space, and i :Σ 7→ D be the conformal map. By Lemma 9.4 and Lemma 9.5, we could see that L2 harmonic 1-form on (Σ˜, i ∗ δ) is one to one correspond to L2 harmonic form on (D, δ). Since there are many harmonic 1-forms on D by just taking df where f is harmonic functions, so it is a contradiction to the Theorem 9.2.

Proof of Theorem 9.1. (Σ, g) is an oriented Riemann surface, where g is the restriction metric. If z = x + iy then g = λ2(dx2 + dy2) locally. So Σ has a complex striation. Let Σˆ be the universal cover of Σ, then Σˆ is a simply connected non-compact Riemann surface, hence by uniformization theorem of Riemann surface, , the complex plane, Σˆ ' C D, the unit disk. By Proposition 9.6, second situation could not happen. 3 Hence, Σˆ ' C, then let F : C → R , where F = i ◦ π is given by the 3 composition of the minimal immersion i :Σ → R with the covering map π : C ' Σˆ → Σ. Since i is harmonic, and the harmonic?property is preserved under the conformal change Σˆ ' C, we know that F is both conformal and ˆ harmonic, i.e. ∆CF = 0. Since Σ is stable and 2-sided, Σ is also stable and ˆ 2 ˆ 2-sided, =⇒ ∃u > 0, such that Lu = ∆Σˆ u + |A| u = 0 on Σ. So ∆Σˆ u ≤ 0, hence ∆C(u ◦ F ) ≤ 0. So u ◦ F is a super-harmonic function. Since C has quadratic area growth,together with the fact that u ◦ F > 0, we know that u ◦ F = 0, and hence |Aˆ|2 = 0 by the following Proposition. CHAPTER 9. THE GENERALIZED BERNSTEIN THEOREM 36

Deﬁnition 9.7. A Riemannian manifold Σk is called parabolic if every positive super-harmonic function is constant.

1 Proposition 9.8. If h :Σ → R+ is a proper Lipschitz function |∇h| ≤ c, 2 and if |Σa| ≤ ca 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. ∆u ≤ 0 and u > 0. Take w = log u, then ∆u ∆w = − |∇w|2 ≤ −|∇w|2. u Take ϕ a compactly supported function, Z Z Z ϕ2|∇w|2 ≤ − ∆wϕ2 = 2h∇w, ϕiϕ Σ Σ Z Z 1 Z ≤ 2 |ϕ||∇w||∇ϕ| ≤ ϕ2|∇w|2 + |∇ϕ|2. Taking = 1 , then 2 Z Z ϕ2|∇w|2 ≤ 4 |∇ϕ|2. Σ Σ

By taking h = distΣ(·, p), we know that Σ has more than quadratic area growth, so we can take ϕ = ϕR as in Lemma 5.13, and use the same loga- R 2 rithmic cut-oﬀ trick, to get Σ |∇ϕR| → 0, and ϕR → 1. So |∇w| = 0, and w hence u is a constant. Chapter 10

Curvature Estimate

n−1 n−1 Theorem 10.1. Let Σ1 , Σ2 be two connected, imbedded minimal hy- n persurfaces of R , s.t.

Σi ∩ Br(0) 6= ∅,Br(0) ∩ ∂Σi = ∅.

Moreover, we assume Σ1 separate Br(0), and Σ2 lie in one side of Σ1, i.e.

Br(0)Σ = U1 ∪ U2, Σ2 ∩ Br(0) ⊂ U1, where Ui is open and connected. Then, either Σ1 ∩ Σ2 = ∅ or Σ1 = Σ2.

Proof. Assume p ∈ Σ1 ∩ Σ2, then locally, they can be written as graphs of u1, u2, s.t. uαuα (δ − i j )uα = 0, α = 1, 2. ij 1 + |∇uα| ij Substract the two equations, we have

1 2 1 2 1 2 δij(uij − uij) + bi(u − u )i + c(u − u ) = 0,

1 2 for some bi, c. Hence by maximal principle, u = u locally.

n−1 n Let Σ ,→ R be a immersed minimal hypersurface, 0 ∈ Σ, let

Lr = {Σ ∩ Br(0) 6= ∅, ∂Σ ∩ Br(0) = ∅}, we want to prove 2 C |A| (x) ≤ 2 . d (x, ∂Br(0))

37 CHAPTER 10. CURVATURE ESTIMATE 38

By a suitable dilation and translation, we just need to prove

|A|2(0) ≤ C.

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 rough laplacian for h is deﬁned as

n−1 X ∆hij = hij,kk. k=1 Proposition 10.2.

2 ∆hij + |A| hij = 0, 0 ≤ i, j ≤ n − 1 (10.1)

Proof. Firstly we have the Ricci identity: X X hij,kl − hij,lk = hpjRpikl + hipRpjkl, p p

Gauss Equation: Σ Rijkl = hikhjl − hilhjk, and Codazzi equation: hij,k = hik,j. Using the Eistein summation, we have

Σ Σ ∆hij = hij,kk = hik,jk = hik,kj +hpkRpijk + hipRpkjk | {z } =hkk,ij =0

= hpk(hpjhik − hpkhij) + hip(hpj hkk −hpkhkj) |{z} =0 2 = −|A| hij + (hikhkphpj − hipHpkhkj). | {z } =0 So we ﬁnished the proof.

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

Proposition 10.3. 2 2 |A|(L|A|) ≥ ∇|A| . (10.2) n − 1 CHAPTER 10. CURVATURE ESTIMATE 39

Proof. By the Bochner Formula, 1 ∆|A|2 = |∇A|2 + hA, ∆Ai = |∇A|2 − |A|4. 2

1 2 2 While 2 ∆|A| = |A|∆|A| + |∇|A|| , P (P h h )2 2 2 X 2 k ij ij ij,k |A|L(|A|) = |∇A| − ∇|A| = h − . (10.3) ij,k |A|2 i,j,k

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

P P 2 2 k( i λihii,k) X 2 X 2 X 2 ∇|A| = ≤ h = h + h |A|2 ii,k ii,k ii,i i,k i6=k i

X 2 X X 2 X 2 X 2 = hii,k + (− hjj,i) ≤ hii,k + (n − 2) hjj,i i6=k i j6=i i6=k i6=j X n − 1 X X = (n − 1) h2 = ( h2 + h ). ii,k 2 ik,i ki,i i6=k i6=k i6=k So

2 2 X 2 X 2 X 2 X 2 2 (1 + ) ∇|A| ≤ h + h + h ≤ h = |∇A| . n − 1 ii,k ik,i ki,i ij,k i,k i6=k i6=k i,j,k

So we ﬁnished the proof. Chapter 11

Choi-Schoen’s Theorem

Theorem 11.1 (Choi-Schoen [4]). Suppose Σ2 ⊂ M 3 is a minimal surface. 2 Assume 0 ∈ Σ , and ∂Σ ∩ Br0 (0) = ∅. Moreover, there exists , ρ > 0 (depending only on M), such that R |A|2 < , then Σ∩Br0

2 2 d (x, ∂Bρ(0))|A| (x) ≤ δ.

3 3 Proof. Let us give a proof when M = R , and the general cases follow by 3 3 the fact that M is locally near R when ρ is small enough. Assume δ = 1 and 2 2 F (x) = d (x, ∂Bρ(0))|A| (x), Since F | = 0, then ∃x ∈ B , such that F (x ) = max F (x). ∂Br0 0 ρ 0 Br0 Need to show: F (x0) ≤ 1. ρ−r(y0) Suppose otherwise F (x0) > 1, let δ = 2 , then • sup |A|2 ≤ 4|A|2(x ). Bδ(y) 0 2 2 2 2 This is because d (x, ∂Bρ(0))|A| (x) ≤ 4d (x0, ∂Bρ(0))|A| (x0), hence 2 2 2 d (x0,∂Bρ(0)) 2 2 d (x0,∂Bδ(x0)) 2 2 2 |A| (x) ≤ ( 2 ) |A| (x0) ≤ ( 2 ) |A| (x0) ≤ 4|A| (x0). d (x,∂Bρ(0)) d (x,∂Bδ(x0)) 2 2 2 2 • (2δ) |A| (x0) = F (x0) > 1 =⇒ δ |A| (x0) > 1/4.

1 2 1 2 Let δ0 = , hence δ ≥ δ =⇒ δ0/2 < δ. So B (x0) ⊂ B (x0). Let |A|(x0) 4 0 δ0/2 δ 2 Σδ0 = (Σ − x0), δ0

2 2 2 2 sup |AΣ | = 4|AΣ | = δ |A| (x0) = 1, =⇒ B1 δ0 δ0 0 R |A |2 ≤ . B1 Σδ0

40 CHAPTER 11. CHOI-SCHOEN’S THEOREM 41

By Simon’s equation, let A = A , on B , we have Σδ0 1 ∆|A|2 = 2(|∇A|2 − |A|4) ≥ −2|A|4 ≥ −2|A|2 Hence by mean value property of subsolution of EPDE, we have |A(0)|2 ≤ c R |A|2 ≤ c, which is a contradiction. Σ∩B1 2 3 Corollary 11.2. Assume Σ ⊂ R is stable and 2-sided with quadratic area 2 growth, i.e. Area(Σ ∩ Br0 ) ≤ cr0, then 2 −2 sup |A| ≤ cr0 . Σ∩B r0/2 R 2 2 R 2 Proof. By stability, we have Σ |A| ϕ ≤ Σ |∇ϕ| . Since Σ has quadratic area growth, we can use the logarithmic cutoﬀ trick to get, Z C |A|2 ≤ , k 1. Σ∩B log k r0/k R 2 So for k large enough, we have |A| < , where r1 = r0/k, hence Σ∩Br1 (y) 2 −2 0 −2 0 |A| (y) ≤ cr1 ≤ c r0 , c = kc.

k n Lemma 11.3 (Generalized Monotone formula). Let Σ ⊂ R be a minimal surface, ∂Σ ∩ Bρ(0) = ∅, f :Σ 7→ R is smooth, then R fdvol R fdvol Σ∩Bt − Σ∩Bs tk sk Z ⊥ 2 Z t Z |x | 1 1 2 2 = f k+2 dvol + k+1 (τ − |x| )∆Σfdvoldτ, Σ∩(Bt\Bs) |x| 2 s τ Σ∩Bτ ⊥ where 0 < s < t, and x is the projection of x to the normal part of TxΣ. Proof. First, we have Z Z Z T div(fx)dvolΣ = fx · ndσ = f|x |dσ, (11.1) Σ∩Bt ∂(Σ∩Bt) ∂(Σ∩Bt)

xT where n = |xT | is exterior normal vector ﬁeld on ∂(Σ ∩ Bt). While Z Z Z div(fx)dvolΣ = ∇f · xdvolΣ + fdiv(x)dvolΣ Σ∩Bt Σ∩Bt Σ∩Bt Z Z Z (11.2) = ∇f · xdvolΣ + kfdvolΣ = I + kfdvolΣ. Σ∩Bt Σ∩Bt Σ∩Bt CHAPTER 11. CHOI-SCHOEN’S THEOREM 42

We have estimate Z |x|2 I = ∇f∇( )dvolΣ Σ∩Bt 2 Z |x|2 |x|2 = div( f) − ∆ fdvol 2 2 Σ Σ Σ∩Bt (11.3) Z t2 Z |x|2 = ∇f · ndσ − ∆ΣfdvolΣ ∂(Σ∩Bt) 2 Σ∩Bt 2 Z t2 − |x|2 = ( )∆ΣfdvolΣ Σ∩Bt 2

xT Take h(x) = |x|, then {h ≤ t} = Bt(0), then ∇h = |x| For the right hand side of (11.1), we have

Z Z |xT | f|xT |dσ = t f dσ ∂(Σ∩Bt) ∂(Σ∩Bt) |x| d Z |xT | = t f |∇h|dvol dt Σ∩Bt |x| d Z |xT |2 = t f 2 (11.4) dt Σ∩Bt |x| d Z |x⊥|2 = t f(1 − 2 ) dt Σ∩Bt |x| d Z Z |x⊥|2 = t f − f T ( By coarea formula) dt Σ∩Bt ∂(Σ∩Bt) |x |

Let F (t) = R f, by (11.1),(11.2),(11.3) and (11.4), we have Σ∩Bt

Z ⊥ 2 Z 2 2 0 |x | t − |x| kF (t) = tF (t) − f T − ( )∆ΣfdvolΣ. (11.5) ∂(Σ∩Bt) |x | Σ∩Bt 2 The lemma then follows easily from the above ODE.

Lemma 11.4 (Mean Value Properity). If ∆Σf ≥ −cf in Σ ∩ B1, f > 0, then Z 0 f(0) ≤ c fdvolΣ. Σ∩B1 CHAPTER 11. CHOI-SCHOEN’S THEOREM 43

Proof. Since ∆Σf ≥ −cf, (11.5) imply that

Z ⊥ 2 Z 2 2 d −k −k−1 |x | t − |x| (t F (t)) = t ( f T + ( )∆ΣfdvolΣ) dt ∂(Σ∩Bt) |x | Σ∩Bt 2 Z c −k+1 ≥ − t fdvolΣ), 2 Σ∩Bt where F (t) = R f. Hence, we know that ect/2F (t) is increasing. Hence Σ∩Bt f(0) = F (0) ≤ ec/2F (1) = ec/2 R f. Σ∩B1 Chapter 12

Small Curvature implies Graphical

Let us ﬁrstly give a technical lemma used in the argument of the above section. 2 n 2 2 1 Lemma 12.1. Σ ⊂ R is minimal. Assume that s supΣ |A| ≤ 16 . If 2 x0 ∈ Σ and distΣ(x0, ∂Σ) ≥ 2s (i.e. ∂Σ ∩ B2s(x) = ∅), then Σ Σ (i) B2s(x0) is graphical over Tx0 Σ of some function u, where B2s(x0) is the geodesic ball of Σ, and |∇u| ≤ 1 and |Hessu| ≤ √1 ; 2s 0 (ii) Let Σ be a connected component of Bs(x) ∩ Σ containing x0, then 0 Σ Σ ⊂ B2s(x0). Σ Proof. Deﬁne d(x, y) = distSn−1 (N(x),N(y)). Connect x to y ∈ Σ∩B2s(x0) by unit speed geodesic u : [0, r] 7→ Σ, s.t. u(0) = x0, u(r) = y. Then Z r Z r Z r Σ 1 1 π d(x0, y) ≤ |∇ N|dt ≤ |A|dt ≤ dt < < , 0 0 0 4s 2 4 Σ Hence Σ ∩ B2s(x0) is contained in the graph of a function u over a subset of

Tx0 Σ. Since 1 1 √ ≥ cos(d(x, y)) =< N(x0),N(y) >= , 2 p1 + |∇u(y)|2 we have |∇u(y)| ≤ 1. Since    2  ux1x1 ... ux1xn−1 1 + ux1 ... ux1 uxn−1 1 ...... A =  . .. .  , g =  . .. .  , 1 + |∇u|2  . .   . .  2 uxn−1x1 ... uxn−1xn−1 uxn−1 ux1 ... 1 + uxn−1 (12.1)

44 CHAPTER 12. SMALL CURVATURE IMPLIES GRAPHICAL 45 and 2 ij kl |A| = AikAjlg g , we have |Hess(u)|2 1 ≤ |A|2 ≤ , (1 + |∇u|2)3 16s2 Consequently, 1 |Hess(u)| ≤ √ 2s Now we are going to prove: n n Σ R R Σ If y ∈ B2s(x0), then d (y, x0) > s. Therefore, Σ ∩ Bs (x0) ⊂ B2s(x0). Let w : [0, 2s] 7→ Σ be a unital speed geodesic, and w(0) = x0, w(2s) = y. Since

|w(2s − w(0))| = |w(2s − u(0))||w0(0)| ≥ hw(2s) − u(0), w0(0)i, in order to prove |w(2s) − w(0))| > s, we just need to show

hw(2s) − w(0), w0(0)i > s.

Let f(t) = hw(t) − w(0), w0(0)i, we have f(0) = 0, f 0(0) = 1. Moreover 1 |f 00(t)| = |hu00(t), u(0)i| < , 4s which mean 1 1 f(2s) = f(0) + 2f 0(0)s + f 00(ξ)(2s)2 > 2s − (2s)2 = s. 2 4s

3 Remark 12.2. Let U ⊂ R , then Lc1,c2 = {Σ ⊂ U, ∂Σ ∩ U = ∅,H(Σ) ≡ Σ ∞ 0, maxU |A | ≤ c1, Area(Σ) ≤ c2} is closed in C topology.

k n Theorem 12.3. Let Σ ⊂ R be minimal. ∃ = (n, k), if x ∈ Σ, ∂Σ ⊂

∂Br0 (x), and Θx(r0) − 1 < , then

2 −2 sup |A| ≤ r0 . Σ∩B (x) r0/2

Proof. • It suﬃces to assume that Θy(r1) − 1 < for all y ∈ Br1 (x) ∩ Σ by the monotonicity formula 5.1. CHAPTER 12. SMALL CURVATURE IMPLIES GRAPHICAL 46

2 2 • By rescaling the function (r1 − |y|) |A| (y) near the maximum point as in the proof of Theorem 11.2, we can get another minimal surface, 2 denoted still as Σ, such that 0 ∈ Σ, ∂Σ ⊂ ∂B1(0), |A| ≤ 1 on Σ, and 2 1 |A| (0) = 4 . Furthermore, by the small excess condition, |Σ∩B1(0)| ≤ (1 + )ωk. • This is not possible if ≤ 0, for some 0 > 0 small enough, by the following argument. • Compactness argument: consider the class 1 C = {Σ : 0 ∈ Σ, ∂Σ ⊂ ∂B (0), |A|2 ≤ 1, |A|2(0) = , |Σ∩B (0)| ≤ (1+)ω }. 1 4 1 k

If the curvature estimates is not true, then we can ﬁnd a sequence {Σi}, −i k with Σ ∩ B1(0) ≤ (1 + 2 )ωk. A subsequence Σi → Σ in C norm to some minimal Σ∞, such that Σ∞ ∈ C0, i.e. |Σ∞ ∩ B1(0)| = ωk, =⇒ Σ 2 1 is a disk, hence contradiction to the curvature assumption |A| (0) = 4 . Chapter 13

Schoen’s Curvature Estimate

2 3 Theorem 13.1. Assume that Σ is stable and 2-sided in R . If x ∈ Σ and dist(x, ∂Σ) ≥ r0, then 4π Area(BΣ (x ) ∩ Σ) ≤ r2. r0 0 3 0

Proof. It suﬃces to assume π1(Σ) = {1}, or we can pass to the universal ˜ Σ˜ cover Σ of Σ, which is also stable and 2-sided. Since Area(Br0 (x0) ∩ Σ) ≥ Σ Area(Br0 (x0) ∩ Σ), we can get the result. Let φ(x) = r0 − r(x, x0), then 1 Σ φ ∈ Cc (Σ ∩ Br0 (x0)) and |∇φ| = 1 a.e. Moreover, by Stability, we have Z Z 2 2 2 Σ |A| φ dvolΣ ≤ |∇φ| dvolΣ = Area(Σ ∩ Br0 (x0)). Σ∩BΣ (x ) Σ∩BΣ (x ) r0 0 r0 0 (13.1) While Z Z r0 Z 2 2 2 2 |A| φ dvolΣ = |A| dσ|r0 − r| dr (13.2) Σ∩BΣ (x ) 0 ∂(Σ∩BΣ(x )) r0 0 r 0 R r R s R 2 R r R 2 Let f(r) = Σ |A| dσdτds = Σ |A| dvolΣds, g(r) = 0 0 ∂(Σ∩Bτ (x0)) 0 Σ∩Bs (x0) 2 |r − r0| . Then we have f(0) = g(0) = 0, f 0(0) = g0(0) = 0. Moreover, Z r0 Z r0 LHS of (13.2) = f 00(r)g(r)dr = f(r)g00(r) 0 0 Z r0 Z r Z (13.3) 2 = 2 |A| dvolΣdτdr. Σ 0 0 Σ∩Bτ (x0)

47 CHAPTER 13. SCHOEN’S CURVATURE ESTIMATE 48

Since Z Z d Σ Length(Σ ∩ ∂Br ) = kgdΣ = 2π − KdvolΣ dr Σ Σ Σ∩∂Br (x0) Σ∩Br Z 1 2 = 2π + |A| dvolΣ, 2 Σ Σ∩Br (x0) we have

Z r0 Z r0 Z Σ Σ 2 1 2 Area(Σ∩Br0 ) = Length(Σ∩∂Br )dr = πr0+ |A| dvolΣdr. 2 Σ 0 0 Σ∩Br (x0) Combine with (13.1), (13.2) and (13.3), we have 4 Area(Σ ∩ BΣ ) < πr2. r0 3 0

Hence, by Corollary 11.2, we have

2 3 Corollary 13.2. Let Σ ⊂ R be a minimal surface, x0 ∈ Σ, and ∂Σ ∩ 3 R Br0 (x0) = ∅. If Σ is stable and two-sided, then

2 2 sup |A| ≤ C/r0. 3 Σ∩BR (x ) r0 0

q 2 1 Theorem 13.3. If p < 2 + n−1 , then ∀φ ∈ Cc (Br0 (x0) ∩ Σ), we have Z Z |A|2pφ2p ≤ C(p) |∇φ|2p. Br0 (x0)∩Σ Σ Proof. First, we claim: Z Z 2 2 2 2 2 1 ∇|A| ϕ ≤ |∇ϕ| |A| , ∀ϕ ∈ Cc (Σ). n − 1 Σ Σ R By plug in ϕ|A| to the stability inequality − Σ(ϕ|A|)L(ϕ|A|) ≥ 0, and using the tricks in Theorem 9.2, we have Z Z ϕ2|A|L(|A|) ≤ |∇ϕ|2|A|2. Σ Σ Using Proposition 10.3, we can prove the claim. CHAPTER 13. SCHOEN’S CURVATURE ESTIMATE 49

Now change ϕ → ϕ|A|q, for some q > 0, then we get Z Z Z 2 2 2 2q 2 q 2 2 q q−1 2 ∇|A| ϕ |A| ≤ |A| ∇(ϕ|A| ) = |A| (∇ϕ)|A| +q|A| ϕ(∇|A|) n − 1 Σ Σ Σ Z Z 2 2q 2 2 1 2 2q+2 ≤ (q + ) |A| ϕ ∇|A| + (1 + ) |∇ϕ| |A| . Σ Σ

q 2 Hence if q < n−1 , by moving the ﬁrst term on the right hand side to the left, Z Z 2 2 2q 2q+2 2 =⇒ ∇|A| ϕ |A| ≤ C(q) |A| |∇ϕ| , Σ Σ Z Z q+1 2 q+1 2 2 =⇒ ∇|A| ϕ ≤ C(q) (|A| ) |∇ϕ| . Σ Σ Set p = q + 2, Z Z p−1 2 p−1 2 2 =⇒ ∇|A| ϕ ≤ C(p) (|A| ) |∇ϕ| . Σ Σ

p−1 R 2 2 R 2 Now replace ϕ by ϕ|A| in the stability inequality Σ |A| ϕ ≤ Σ |∇ϕ| , Z Z Z 2p 2 p−1 2 2 p−1 2 2 2p−2 =⇒ |A| ϕ ≤ ∇(ϕ|A| ) ≤ 2 ϕ ∇(|A| ) + |∇ϕ| |A| . Σ Σ Σ Using the above inequality, Z Z =⇒ |A|2pϕ2 ≤ C(p) |A|2p−2|∇ϕ|2. Σ Σ Replace ϕ by ϕp, then Z Z Z |A|2pϕ2 ≤ C(p) |A|2p−2|∇ϕp|2 = Cp2 (ϕ|A|)2p−2|∇ϕ|2 Σ Σ Σ Z Z ≤ C(p) (ϕ|A|)2p (p−1)/p |∇ϕ|2p 1/p. Σ Σ Z Z r 2 =⇒ (ϕ|A|)2p ≤ C(p) |∇ϕ|2p, ∀p < 2 + . (13.4) Σ Σ n − 1 CHAPTER 13. SCHOEN’S CURVATURE ESTIMATE 50

n−1 n Theorem 13.4. (Schoen-Simon-Yau [5]) Let Σ ⊂ R be a stable 2-sided n−1 minimal surface. Assume x0 ∈ Σ, ∂Σ ⊂ ∂Br0 (x0), |Σ ∩ Br0 (x0)| ≤ V r0 and n ≤ 6. Then 2 −2 sup |A| ≤ c(n, V )r0 . Σ∩B (x ) r0/2 0 Proof. When n ≤ 6, take 2p = n − 1, and use the logarithmic cut oﬀ trick R n−1 and the volume growth =⇒ Σ |A| is small in small ball, and hence the curvature estimates.

n−1 n n−1 Corollary 13.5. A complete 2-sided stable Σ ∈ R with R volume growth and n ≤ 6 is a hyperplane.

n−1 n Proof. Let Σ ⊂ R be complete, stable, 2-sided, n ≤ 6 and Σ ∩ BR ≤ CRn−1 =⇒ Σ is hyperplane. R 2p R 2p C Take 2p > n−1, =⇒ Σ(ϕ|A|) ≤ C Σ |∇ϕ| ≤ R2p |Σ∩B2R| → 0. Chapter 14

Bernstein’s Theorem and Minimal Cone

n−1 Theorem 14.1. If u : R 7→ R is an entire solution if the minimal surface equation, and n ≤ 8, then u is linear.

Proof. Let Σu be the minimal graph of u, and assume 0 ∈ Σu. Recall monotonicity formula:

Z ⊥ 2 Area(Σu ∩ BR) Area(Σu ∩ Br) |x | n−1 − n−1 = n+1 dσ. (14.1) R r Σ∩(BR\Br) |x|

Area(Σu∩Br) Consider the density at inﬁnite Θ∞ := limr→∞ wn−1rn−1 . By area bound of area-minimizing surface, we have

n Area∂B1 Θ∞ < n−1 . 2AreaB1 We need the following lemma to complete the proof:

Lemma 14.2. We have Θ∞ ≥ 1 and the equality holds iﬀ Σ is a plane.

Proof. It’s clear that Θ∞ ≥ 1. When Θ∞ = 1, since we know that the density of Σ is 1 at original point, in (14.1), let R → ∞, r → 0, we have ⊥ x ≡ 0, i.e. x ∈ TxΣ. We then claim: If x ∈ TxΣ, Σ is a cone. n This is because, consider the following linear equations in R ,

( d dt x(t) = x(t), x(0) = x0,

51 CHAPTER 14. BERNSTEIN’S THEOREM AND MINIMAL CONE 52

t Then we can see that x(t) = e x0, which is a straight line pass x0. Since x ∈ TxΣ, we could see that x(t) ∈ Σ, ∀t. Hence, Σ is a cone. Then blow up at original, since Σu is a cone, we have

Σu Σu = lim = T0Σu, n→∞ rn where rn → 0.

By the lemma, if Σu is not aﬃne, then we have Θ∞ > 1. We then consider blow-down of Σu at 0, i.e. consider

Σ∞ = lim rnΣu, n→∞ with rn → 0. We claim that AreaΣ∞ ∩ BRn n−1 ≡ Θ∞. wn−1R This is simply because,

n n n R R Area(Σu ∩ BR ) Area(Σ∞ ∩ BR ) Area(rnΣu ∩ BR ) R/rn n−1 = lim n−1 = lim n−1 = Θ∞. wn−1R n→∞ wn−1R n→∞ wn−1(R/rn)

Since rjΣu is also minimizing, by station varifold theory or Integral currents theory, we could see that Σ∞ is also minimizing. Because Σ∞ has constant density and satisfy the monotonicity formula, by the same argument as before, we can see that Σ∞ is a cone. But existence of such Σ would imply that there exists nonﬂat volume m minimizing cone C1 , m ≤ n − q with an isolated singularity at 0 by the splitting theorem of De Giorgi, which contradicts with the following theorem.

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

C(Σ) = {λx : x ∈ Σ, λ ≥ 0}. (14.2)

Theorem 14.3 (Simon, 1968). Every Area minimizing hypersurface in n R (3 ≤ n ≤ 7) is ﬂat. 0 n 0 n+1 Proof. Let Σ = Σ ∩ S (1), and assume that Σ = C(Σ ) ⊂ R is a regular minimal hypercone(i.e. smooth away from 0). Let A = (hij) be the second fundamental forms. We have (c.f. (10.3))

2 2 |A|LΣ|A| = |∇A| − |∇|A|| . CHAPTER 14. BERNSTEIN’S THEOREM AND MINIMAL CONE 53

We then claim: 2|A|2(x) |∇A|2 − |∇|A||2 ≥ |x|2 Proof of Claim: 0 Let e1, ...en−1, en be orthornormal basis of Σ, s.t. e1, ..., en ∈ T Σ , and x en = |x| . Then ∂ ∂ x ∂ h (x/|x|) h = ∇ h = h (x) = h (r ) = ij ij,n en ij ∂r ij ∂r ij |x| ∂r r 1 x 1 ∂ x 1 x 1 = − h ( ) + h ( ) = − h ( ) = − h (x) r2 ij |x| r ∂r ij |x| r2 ij |x| r ij

Moreover, hin = h∇en ei, vi = 0 since ∇en ei = 0. n Pn 2 2 2 X 2 i,j,k=1(hijhij,k) |∇A| − |∇|A|| = hij,k − Pn 2 |hst| i,j,k=1 s,t=1 n n n 1 X X X = (( h2 )( |h |2) − (h h )2) |A|2 ij,k st ij ij,k i,j,k=1 s,t=1 i,j,k=1 n 1 X = (h h − h h )2 2|A|2 rs ij,k ij rs,k i,j,r,s,k=1 n 2 X ≥ ( (h h − h h )2)(By symmetry.) |A|2 rs nj,k nj rs,k j,r,s,k=1 n n 2 X 2 X = ( (h h ) = ( (h h ) |A|2 rs nj,k |A|2 rs jk,n j,r,s,k=1 j,r,s,k=1 2|A|2 = |x|2 By stability, we have Z φLφ ≤ 0. Σ Plug |A|φ in, we have Z Z φ2|A|L|A| ≤ |A|2|∇φ|2. Σ Σ By our claim, we can see that Z 2 Z |A| 2 2 2 2 2 φ ≤ |A| |∇φ| . Σ r Σ CHAPTER 14. BERNSTEIN’S THEOREM AND MINIMAL CONE 54

By multiply a suitable cut-oﬀ function, we can prove that the above inequal- n+1 ity still holds true for φ ∈ Lip(R ) provided Z 2 |A| 2 2 φ < ∞ Σ r n+1 Our next step is to ﬁnd a suitable φ ∈ Lip(R ), s.t. Z 2 Z |A| 2 2 2 (2 − ) 2 φ ≥ |A| |∇φ| , Σ r Σ for some > 0. Hence, |A| ≡ 0, which means that Σ is actually a plane. Since Z 2 Z ∞ Z 2 |A| 2 |A| 2 2 φ = 2 φ dr Σ r 0 Σ∩Sn(r) r Z ∞ Z 2 n−1 |A| 2 = r 4 φ dr 0 Σ∩Sn(1) r If we take 1− n −2 1+ φ(x) = max{1, |x|} 2 |x| , then Z 2 |A| 2 2 φ < ∞, Σ r provided n ≥ 2. Also, we can see that

( 2− n − |x| 2 , when |x| is big enough, |φ(x)| = |x|1+, when |x| is small, and by directly calculation, we have

( n φ(x) |2 − 2 − | |x| , when |x| is big enough, |∇φ(x)| ≤ φ(x) (1 + ) |x| , when |x| is small. Hence, Z Z ∞ Z 2 2 2 n−1 |A| 2 |A| |∇φ| = r 2 |∇φ| Σ 0 Σ∩Sn(1) r Z ∞ Z 2 n n−1 |A| 2 ≤ max{1 + , |2 − − |} r 4 |φ| 2 0 Σ∩Sn(1) r Z 2 |A| 2 ≤ (2 − ) 2 φ , Σ r provided n ≤ 7. CHAPTER 14. BERNSTEIN’S THEOREM AND MINIMAL CONE 55

n−1 i Proposition 14.4. C(Σ) is minimal ⇐⇒ Σ is minimal in S ⇐⇒ ∆Σx + i 1 n n (k − 1)x = 0, i = 1, ··· , n, where {x , ··· , x } is coordinates of R . ~ 1 n ~ Proof. Given X = (x , ··· , x ), then C(Σ) is minimal ⇐⇒ ∆C(Σ)X = 0. Take a o.n. basis {e1, ··· , ek−1} for T Σ, and ek = X/~ |X~ |, then {e1, ··· , ek} is an o.n. basis for C(Σ). Then

k−1 k ~ X ~ ~ X TC(Σ) ~ ∆C(Σ)X = eieiX + ekekX − (∇ei ei) X. i=1 i=1

∂ ~ ~ Here ∇ek ek = ∇ ∂ = 0, with r = |X| the radial function. ∇ei ei · X = ∂r ∂r ~ TC(Σ) T Σ ~ ~ −ei·(∇ei X) = −ei·ei = −1, hence (∇ei ei) = (∇ei ei) +(∇ei ei·X)X = T Σ ~ (∇ei ei) − X. So ~ ~ ~ ∆C(Σ)X = ∆ΣX + (k − 1)X.

So we prove the equivalence of the ﬁrst and the third conclusion. n−1 ~ Pk−1 TSn−1 Now Σ is minimal in S if and only if HΣ = i=1 (∇ei ei) , Pk−1 n−1 ~ hence ( i=1 (∇ei ei)) lies in the normal direction of S . So HC(Σ) = Pk−1 ⊥(Σ) Pk−1 ⊥(Σ) ( i=1 (∇ei ei) + ∇ek ek) = i=1 (∇ei ei) = 0. Now we can give some counter example of Theorem 14.3 when n ≥ 7: p p+1 q q+1 Cliﬀord Hypersurfaces: Given S (r1) ⊂ R and S (r2) ⊂ R , take

p q p+q+1 Σ = S (r1) × S (r2) ⊂ R .

p+q+1 2 2 p Then Σ ⊂ S ⇐⇒ r1 + r2 = 1. Given (~x,~y) ∈ Σ, where ~x ∈ S , and q p q ~y ∈ S , hence ∆Σ~x = − 2 ~x, and ∆Σ~y = − 2 ~y. Hence r1 r2

p+q p+q+1 p q Σ ⊂ S is minimal, ⇐⇒ 2 = 2 = p + q. r1 r2 Hence such class of Σ form lots of examples of minimal suﬃces in Sn and n+1 hence minimal cones in R . √ √ Example 14.5. 1.For p = 3, q = 3 : CS3(1/ 2) × S3(1/ 2) is stable, and area minimizing; √ 2. For p = 1, q = 5 : CS1(1/ 6) × S5(p5/6) is stable, but not area minimizing. Chapter 15

The Topology of Minimal surface

2 3 4 In this chapter, we assume that Σ ,→ S (1) ⊂ R is a closed minimal surface. Lemma 15.1. We always have KΣ ≤ 1, and KΣ = 1 iﬀ |A| ≡ 0.

S3 2 1 S3 Σ 2 Proof. Since Ric (v, v) + |A| = 2 (R − R + |A| ), i.e. 1 2 + |A|2 = (6 − 2K + |A|2). 2

2−|A|2 Consequently, K = 2 ≤ 1, and K = 1 iﬀ |A| ≡ 0. Lemma 15.2. Let Σ2 ,→ (M 3, g) is minimal surface, then there exist f : Σ ,→ M, s.t. f is conformal and harmonic. Theorem 15.3 (Hopf-Almgren). Σ is immersed minimal surface of S3(1), then if the genus g(Σ) of Σ is zero, then

Σ =∼ S2(1) ,→ S3(1).

Sketch of Proof. Let f : S2 7→ S3(1) be conformal and harmonic. Consider ∂f ∂f the Hopf Diﬀerential w = A( ∂z , ∂z )dz ⊗ dz. Since H ≡ 0, we know that w is holomorphic. However, there is no nonzero holomorphic quadratic form 2 in S (cf. [7], Section 1.6). Hence w = 0, i.e. h11 = h22, which means A = 0. Therefore, Σ is the equator.

Lemma 15.4 (Lawson). Let Σ ⊂ S3(1). If g(Σ) = 1, then Σ has no umbilical point iﬀ |A| > 0.

56 CHAPTER 15. THE TOPOLOGY OF MINIMAL SURFACE 57

Proof. Σ is umbilical at p ∈ Σ iﬀ w(p) = 0. However, the zeros of w is 4g − 4.

Theorem 15.5 (Brendle [8]). Suppose that F :Σ → S3 is an embedded minimal torus in S3. Then F is congruent to the Cliﬀord torus.

Sketch of Proof. We just need to prove Ψ = √1 |A| = 1. 2 Let √ |hν(x),F (y)i| κ = sup 2 < ∞, x,y∈Σ, x6=y |A(x)| (1 − hF (x),F (y)i) κ Zκ(x, y) = √ |A(x)| (1 − hF (x),F (y)i) + hν(x),F (y)i 2

Case 1: if κ ≤ 1, then Z1(x, y) ≥ 0. For simplicity, let us identify the surface Σ with its image under the embedding F , so that F (x) = x. Let us fix an arbitrary pointx ¯ ∈ Σ. We can find an orthonormal basis {e1, e2} of Tx¯Σ such that h(e1, e1) = Ψ(¯x), h(e1, e2) = 0, and h(e2, e2) = −Ψ(¯x). Let γ(t) 0 be a geodesic on Σ such that γ(0) =x ¯ and γ (0) = e1. We define a function f : R → R by f(t) = Z(¯x, γ(t)) = Ψ(¯x) (1 − hx,¯ γ(t)i) + hν(¯x), γ(t)i ≥ 0.

A straightforward calculation gives −γ00(t) = γ(t)+h(γ0(t), γ0(t))γ0(t), hence

f 0(t) = −hΨ(¯x)x ¯ − ν(¯x), γ0(t)i,

f 00(t) = hΨ(¯x)x ¯ − ν(¯x), γ(t)i + h(γ0(t), γ0(t)) hΨ(¯x)x ¯ − ν(¯x), ν(γ(t))i, and

f 000(t) = hΨ(¯x)x ¯ − ν(¯x), γ0(t)i 0 0 + h(γ (t), γ (t)) hΨ(¯x)x ¯ − ν(¯x),Dγ0(t)νi Σ 0 0 + (Dγ0(t)h)(γ (t), γ (t)) hΨ(¯x)x ¯ − ν(¯x), ν(γ(t))i.

In particular, we have f(0) = f 0(0) = f 00(0) = 0. Since f(t) is nonnegative, 000 Σ we conclude that f (0) = 0. From this, we deduce that (De1 h)(e1, e1) = 0. An analogous argument with {e1, e2, ν} replaced by {e2, e1, −ν} yields Σ (De2 h)(e2, e2) = 0. Using these identities and the Codazzi equations, we conclude that the second fundamental form is parallel. In particular, the intrinsic Gaussian curvature of Σ is constant. Consequently, the induced CHAPTER 15. THE TOPOLOGY OF MINIMAL SURFACE 58 metric on Σ is flat. On the other hand, Lawson [3] proved that the Clifford torus is the only flat minimal torus in S3. Putting these facts together, the assertion follows.

Case 2: κ > 1. In order to handle this case, we apply the maximum principle to the function κ Z(x, y) = √ |A(x)| (1 − hF (x),F (y)i) + hν(x),F (y)i. (15.1) 2 By definition of κ, the function Z(x, y) is nonnegative for all points x, y ∈ Σ. Moreover, after replacing ν by −ν if necessary, we can find two points x,¯ y¯ ∈ Σ such thatx ¯ 6=y ¯ and Z(¯x, y¯) = 0. Moreover, we claim that: Ω = {x¯ ∈ Σ : there exists a pointy ¯ ∈ Σ \{x¯} such that Z(¯x, y¯) = 0} is non-empty and open. Moreover, ∇Ψ(¯y) ≡ 0, ∀y¯ ∈ Ω. This is because, by some computations and the fact that Z attain its local minimal at (¯x, y¯), we have 2 2 2 X ∂2Z X ∂2Z X ∂2Z 0 ≤ (¯x, y¯) + 2 (¯x, y¯) + (¯x, y¯) ∂x2 ∂x ∂y ∂y2 i=1 i i=1 i i i=1 i 2 κ2 − 1 Ψ(¯x) X D ∂F E2 = − (¯x),F (¯y) ≤ 0, (15.2) κ 1 − hF (¯x),F (¯y)i ∂x i=1 i This gives D ∂F E (¯x),F (¯y) = 0. ∂xi Also, since Z attain its local minimal at (¯x, y¯), by some computation, we have At the point (¯x, y¯), we have ∂Z ∂Φ 0 = (¯x, y¯) = (¯x) (1 − hF (¯x),F (¯y)i) ∂xi ∂xi D ∂F E k D ∂F E − Φ(¯x) (¯x),F (¯y) + hi (¯x) (¯x),F (¯y) (15.3) ∂xi ∂xk ∂Ψ = κ (¯x) (1 − hF (¯x),F (¯y)i), (15.4) ∂xi which imply ∇Ψ = 0 in Ω. Since Ω is open, it follows from the claim that ∆ΣΨ(¯x) = 0 for each pointx ¯ ∈ Ω. Since |∇Ψ|2 ∆ Ψ − + (|A|2 − 2) Ψ = 0, Σ Ψ CHAPTER 15. THE TOPOLOGY OF MINIMAL SURFACE 59 we can see that φ ≡ 1 in Ω. Using standard unique continuation theorems for solutions of elliptic partial differential equations (see e.g. [9]), we conclude that Ψ(x) = 1 for all x ∈ Σ. Consequently, the Gaussian curvature of Σ vanishes identically. As above, it follows from a result of Lawson [3] that F is congruent to the Clifford torus. Chapter 16

Closed Minimal Surfaces in S3

Let Σ2 ,→ S3 be a immersed minimal surface, then the Jacobi operator

2 S3 2 LΣφ = ∆φ + (|A| + Ric (ν, ν))φ = ∆φ + (2 + |A| )φ, and Z Z 2 2 2 Q(φ, φ) := − φLΣφ = |∇φ| − (2 + |A| )φ , Σ Σ where ν is unit normal vector ﬁelds on Σ. Recall that Morse index Ind(LΣ) := number of negative eigenvalue of LΣ with multiplicity. Moreover, since Q(1, 1) < 0, we can see that Ind(LΣ) ≥ 1. Let λ1(Σ) be the ﬁrst eigenvalue of LΣ. Proposition 16.1. Let S2 ,→ S3 be a equator, i.e. |A| = 0, then it’s obvious 2 S is minimal, also, we have λ1(Σ) = −2 and λ2(Σ) = 0. Proof. Since function 1 is a eigenfunction with eigenvalue −2, and 1 doesn’t 2 change sign, by Proposition 6.11, λ1(S ) = −2. By Theorem 5.1 in [10], we 2 2 can see that λ2(S ) ≥ 0. By Proposition 14.4, we have λ2(S ) = 0.

3 2 Proposition 16.2. Let Tc ,→ S be the Cliﬀord torus. Then Ind(Tc ) = 5 with λ1 = −4 with multiplicity 1 and its eigenfunction is√u1 = const√; λ2 = −2,√ with multiplicity√ 4 and its eigenfunctions are: cos( 2θ1), sin( 2θ1), cos( 2θ2) and sin( 2θ2); λ3 = 0.

|A|2 Proof. LTc φ = ∆φ + 4φ. This is because, K = 1 − 2 ≡ 0. Hence, 1 is a eigenfunction with eigenvalue −4, and since 1 doesn’t change sign, λ1 =

60 CHAPTER 16. CLOSED MINIMAL SURFACES IN S3 61

−4. By Theorem 5.1 in [10]√ again, λ2√≥ −2. By√ Proposition 14.4√ again, it’s easy to check that cos( 2θ1), sin( 2θ1), cos( 2θ2) and sin( 2θ2) are eigenfunctions with eigenvalue −2, hence λ2 = −2, or it’s straightforward to ∼ 1 1 verify that the ﬁrst three eigenvalues of ∆Tc on Tc = S × S are 0, 2 (with multiplicity 4), 4. Thus, λ2 = −2, λ3 = 0. Proposition 16.3. Let Σ2 ,→ S3 be immersed minimal surfaces, let ν be the unit normal vector ﬁeld on Σ, then we have

2 ∆Σν + |A| ν = 0. Therefore, Σ always have eigenvalue −2. 4 4 Proof. Consider C(Σ) ,→ R , then for any w ∈ R , we have Ct(Σ) = C(Σ)+ tw is isometry. Hence, hw, νi is a Jacobi ﬁeld, which means

2 ∆Σhw, νi + |A| hw, νi = 0. Equivalently, 2 ∆Σν + |A| ν = 0.

Theorem 16.4. Let Σ2 ,→ S3 be immersed minimal surfaces, if Σ is not a ∼ 2 equator, then Ind(Σ) ≥ 5. When Ind(Σ) = 5, Σ = Tc .

Proof. Let ν = (ν1, ..., ν4) denotes the unit normal vector to Σ. If λ1 = −2, then by Proposition 6.11, we have a1, a2, a3 ∈ R, s.t. ν1 = a1ν2 = a2ν3 = a3ν4. Since Σ 3 x · ν = 0, we have x1 + a1x2 + a2x3 + a3x4 = 0, which mean Σ lie in a plane pass through 0. Thus, Σ is a equator, which is impossible. Moreover, Since Σ is not total geodesic (otherwise Σ is an equator), we can see v1, v2, v3, v4 are linear independent. Hence, by now we have prove Ind(Σ) ≥ 5. Suppose now that the Jacobi operator L has exactly ﬁve negative eigenvalues. Let ρ denote the eigenfunction associated with the eigenvalue λ1. Note that ρ is a positive function. We consider a conformal transformation ψ : S3 → S3 of the form 1 − |a|2 ψ(x) = a + (x + a), 1 + 2 ha, xi + |a|2

4 where a is a vector a ∈ R satisfying |a| < 1. We can choose the vector a in a such a way that Z ρ ψi(x) = 0 Σ CHAPTER 16. CLOSED MINIMAL SURFACES IN S3 62 for i ∈ {1, 2, 3, 4}, where ψi(x) denotes the i-th component of the vector 3 4 ψ(x) ∈ S ⊂ R . By assumption, L has exactly ﬁve negative eigenvalues. In particular, L has no eigenvalues between λ1 and −2. Since the function ψi is orthogonal to the eigenfunction ρ, we conclude that Z Z Σ 2 2 2 (|∇ ψi| − |A| ψi ) = ψi (Lψi + 2ψi) ≥ 0 (16.1) Σ Σ for each i ∈ {1, 2, 3, 4}. On the other hand, the conformal invariance of the Willmore functional implies that

4 Z X Σ 2 |∇ ψi| = 2 area(ψ(Σ)) ≤ 2 W (ψ(Σ)) = 2 W (Σ) = 2 area(Σ). i=1 Σ (16.2) where Z H2 Z |A|2 W (Σ) = 1 + = K + Σ 4 Σ 2 is conformal invariant. Moreover, it follows from the Gauss-Bonnet theorem that 4 Z Z Z X 2 2 2 |A| ψi = |A| = 2 (1 − K) ≥ 2 area(Σ). (16.3) i=1 Σ Σ Σ Combining the inequalities (16.2) and (16.3) gives

4 Z X Σ 2 2 2 (|∇ ψi| − |A| ψi ) ≤ 0. (16.4) i=1 Σ Putting these facts together, we conclude that all the inequalities must, in fact, be equalities. In particular, we must have W (ψ(Σ)) = area(ψ(Σ)). Consequently, the surface ψ(Σ) must have zero mean curvature. This implies that ha, νi = 0 at each point on Σ. Since Σ is not totally geodesic, it follows R R Σ 2 2 2 that a = 0. Furthermore, since Σ ρ ψi = 0 and Σ(|∇ ψi| −|A| ψi ) = 0, we conclude that the function ψi is an eigenfunction of the Jacobi operator with 2 eigenvalue −2. Consequently, ∆Σψi + |A| ψi = 0 for each i ∈ {1, 2, 3, 4}. 2 Since a = 0, we conclude that ∆Σxi + |A| xi = 0. Since ∆Σxi + 2 xi = 0, we conclude that |A|2 = 2 and the Gaussian curvature of Σ vanishes. This implies that Σ is the Cliﬀord torus. Chapter 17

General Relativity

(Sn+1, g) is used to model the space-time, where Sn+1 is an n+1 dimensional smooth oriented manifold, g is a Lorentz metric with signature (−1, 1, ..., 1).

n,1 Example 17.1 (Flat model). R

• x0, x1, ..., xn are coordinates

2 Pn 2 • g = −dx0 + i=1 dxi is the Lorentzian metric; Pn •h v, wi = −v0w0 + i=1 viwi; We have 3 types of vectors:

• space-like: hv, vi > 0;

• time-like: hv, vi < 0;

• null: hv, vi = 0.

Let Hn ⊂ Rn,1 be a plane, then there exists a unique v 6= 0 up to scale, such that H = {w : hv, wi = 0}.

• H is space-like if v is time-like, then g|H has positive signature;

• H is time-like if v is space-like, then g|H has Lorentz signature;

• H is null if v ∈ H and hv, vi = 0, then g|H is degenerate.

n n,1 M ⊂ R is space-like if TxM is space like for all x ∈ M.

63 CHAPTER 17. GENERAL RELATIVITY 64

17.1 Einstein Equations

The Einstein equation is given by 1 Ricn − Rng = T, (17.1) 2 where Ric and R are Ricci curvature and Scale curvature respectively. When T = 0, we call (17.1) Vacuum Einstein Equation(VEE).

3 Example 17.2 (Schwartzchild). For any m > 0, (R /{0} × R, g), and 2m 2m g = −(1 − )dt2 + (1 − )−1dr2 + r2dσ, r r where r = |x|, x = rσ.

Given v a time-like unit vector s.t. hv, vi − 1, then

• T (v, v) is the observed energy density of observer;

• T (v, ·)∗ is energy-momentum density vector.

Dominant Energy Condition(DEC): ∀v unit time-like vector ⇒ T (v, .)∗ is forward pointing time-like or null. Given v = e0, e1, ..., en an o.n. basis, pPn and let Tab = T (ea, eb), then (DEC) requires T00 ≥ i=0T0i.

17.2 Initial value problem

The initial data is modeled by a triple (M n, g, h), where g is a Riemmanian metric, and h a symmetric (0, 2) tensor. Our problem is: Given initial data, ﬁnd a local evolution (Sn+1, eta) of VEE with M ⊂ S, g = η|M ; h = second fundamental form. We need the constraint equations (CE) for g, h so that such (S, η) exists:

( 1 2 2 T00 = µ = 2 (R + (trg) − |h| ), T0i = J = divg(h − (trgh)g).

We call T00 the observed energy density, T(0j) observed momentum density. The vacuum constraint equations (VCE) are given by:

( 1 2 2 0 = 2 (R + (trgh) − |h| ), 0 = divg(h − (trgh)g). CHAPTER 17. GENERAL RELATIVITY 65

17.2.1 Derivation of (CE) Given (Sn+1, η) the space-time, let (M n, g, h) be the initial data set, such that (g, h) are the restriction and second fundamental form of M ⊂ (Sn+1, η). Take {e0, e1, ..., en} an o.n. frame of S at p ∈ M, with e0 ⊥ M and ei ∈ TpM. Consider the Einstein equation 1 R − Rη = T , 0 ≤ a, b ≤ n. ab 2 ab First, by Gauss Equation, M ⊂ S, X,Y,Z,W ∈ TM,

RM (X,Y,Z,W ) = RS(X,Y,Z,W )+hII(X,Z),II(YW )i−hII(X,W ),II(Y,Z)i,

⊥ where II(X,Y ) = (DX Y ) = h(X,Y )e0. Plug in ei, ej, ek, el,

M S Rijkl = Rijkl − hikhjl + hilhjk.

Summing over i, k and j, l respectively,

n n M X M X S 2 2 R = Rijij = Rijij(trgh) + |h| : i,j=1 i,j=1

Now n n n n X X X X 1 RS = RS − 2 RS + 2 RS = 2(RS − RSg ) = 2T . ijij ijij 0i0i 0i0i 00 2 00 00 i,j=1 i,j=1 i=1 i=1

Similarly By Codazzi Equation, we get the equation for T0i. Theorem 17.3. There is a one-one corresponding between (M, g, h) satisfy (VCE) and its maximal hyperbolic evolution (up to diﬀeomorphism).

17.3 Asymptotical Flatness

(M n, g, h) is is an Initial data set satisfying the (CE). Roughly speaking, we say (M, g, h) is asymptotically ﬂat (AF) if outside a compact set K ⊂ M, n n M\K is diﬀeomorphic to R \B . Moreover denote {x1, ..., xn} to be the n 2,α n coordinates on R \B, then we assume g ∈ C (R \B) and

gij(x) = δij + Γ, where |Γ| ∈ O(r−1/2), |∂Γ| = O(r−3/2), |∂2Γ| = O(r−5/2). CHAPTER 17. GENERAL RELATIVITY 66

k,p 3 Deﬁnition 17.4 (Weighted Sobolev space). We say f ∈ W−q (M ), q > 1/2, p > 3, if Z X q+|β| β p dx 1/p kfkW k,p = kfkW 2,p(K) + ( (|x| |∂ f|) n ) < ∞. −q 3 |x| R \B |β|≤k

Definition 17.5. (M n, g, h) is called asymptotically flat (with one end), if: 1. K ⊂ M is compact, such that M\K is diffeomorphic to Rn\B. Let n x1, ..., xn be the local coordinates given by R \B. 2. g ∈ C2,α(M), h ∈ C1,α(M), and n − 2 g = δ + γ , γ ∈ W 2,p(M), q > , p > n ij ij ij −q 2 and 1,p h ∈ W−1−q(M) 3. The mass density µ and momentum density J in (CE) satisfy: µ, J ∈ 0,α C−q0 (M) , q0 > n. 0 Example 17.6 (Schwartzchild Solution (SC)). Consider (R \{0}, g, 0), where

m 4 g = (1 + ) n−2 δ. ij 2|x|n−2

m When m > 0, u = 1 + 2|x|n−2 is harmonic on R\{0}. Moreover Rg = n+2 4n−4 − n−2 − n−2 u (∆u) = 0, so it is vacuum. Deﬁnition 17.7. Given (M, g, h) satisfy AF, the ADM mass is Z 1 X j mADM = lim (gij,i − gii,j)ν dσ 16 r→∞ r ∂S i,j

Example 17.8. For (SC), mADN = m. Theorem 17.9 (Positive mass theorem, Schoen-Yau 1978, Witten 1980). n Let (M , g) be AF with Rg ≥ 0. Then we have mADM ≥ 0. Moreover, m = 0 n ∼ n iﬀ (M , g) = (R , δij)

Proof. See section 5.5 in http://www.mit.edu/~cmad/Papers/MinimalSubmanifoldNotes. pdf. Chapter 18

Geometric Measure Theory and Minimal Surfaces

Let U ⊂ M n be open, let V k(U) denote the C∞ section of k−form with compact support in U. We can endow the locally smooth topology on V k(U). k 1/2 For any ω ∈ V (U), denote k ω kL∞ = supx∈U hω(x), ω(x)i .

Deﬁnition 18.1. A k−dimensional current Vk(U) in U is just a bounded linear functional on V k(U).

Remark 18.2. 1. When k = 0, V0(U) = is just the Schwartz distribution. 2. If N is a k−dimensional submanifold in U, then we regard it as a element in Vk(U) as following Z h[N], ωi = ω. N

Moreover, if we have a family k−dimensional submanifolds of Nt in U, by choose a good function f(t), we can deﬁne a current Z Z ( ω)f(t)dt. Nt

Given a current T ∈ Vk(U), we could deﬁne its boundary ∂T by ∂T (w) = T (dw), ∀w ∈ V k−1(U). Moreover, it’s easy to see that ∂2T = 0.

Deﬁnition 18.3. The mass of T ∈ Vk(U) is deﬁned by

MU (T ) = sup T (ω). k ω∈V (U),kωkL∞ ≤1

67 CHAPTER 18. GEOMETRIC MEASURE THEORY AND MINIMAL SURFACES68

For any T ∈ Vk(U), by Riesz representation theory (Compare Theorem k 4.1 in [11]), we could always ﬁnd τ ∈ V (U), a Borel measure uT on U, s.t. for any ω ∈ V k, we have Z T (ω) = hτ, widuT . U Moreover, for any open set W ⊂ U, we have

uT (W ) = sup T (ω) = MW (T ). k ω∈V (W ),kωkL∞ ≤1

Now we are able to deﬁne the restriction. Let A be a uT −measurable subset of U, then we deﬁne the restriction of T on A, denoted by T |A by Z T |A(ω) = hw, τiduT . A

In addition, we deﬁne the support spt(T ) of T ∈ Vk(U) by

spt(T ) = spt(uT ) ⊂ U.

Deﬁnition 18.4. We say Tj ∈ Vk(U) converge to T weakly, which is denoted k by Tj *T , if for any ω ∈ V (U),

lim Tj(ω) = T (ω). j→∞

Lemma 18.5 (Compactness). If supk MW (Ti) < ∞, ∀W ⊂⊂ U. Then there exist a subsequence Ti0 of Ti, s.t.

Ti0 *T, for some T ∈ Vk(U). Now we are considering the locally Lipchitz submanifolds.

Deﬁnition 18.6. A sub M ⊂ U is called countably k−retiﬁable if

∞ M ⊂ ∪j=0M0, s.t. k H (M0) = 0,Mj ⊂ Fj(Aj), k k where Aj is a domain in R and H is k−Hausdorﬀ measure (See [11] for a detail). CHAPTER 18. GEOMETRIC MEASURE THEORY AND MINIMAL SURFACES69

Proposition 18.7 (Criterion for rectifiable). M is rectifiable iff for Hk−a.e. x ∈ M, there exist unique k dimensional tangent plane P , s.t. Z Z f(y)θ(x + λy)dHk(y) → θ(x) f(y)dHk(y), λ → 0+, ηx,λ(M) P

k y−x where θ is a locally H −integrable function on M, and ηx,λ(y) = λ Definition 18.8. We say T is an inter-rectifiable k−current, if Z T (ω) = hω, ξiθ(x)dHk, M where M is a Hk−measurable countably k−rectifiable subset of U, ξ : M 7→ n k ∧k(R ) s.t. for H −a.e. on M, ξ(x) = τ1 ∧ ... ∧ τk, {τ1, ...τk} are orientable k basis of P = TxM, θ : M 7→ Z is locally H −integrable. Such a T is called integrable if

MW (∂T ) < ∞, ∀W ⊂⊂ U. (18.1)

Example 18.9. We would like to explain that condition 18.1 is nontrivial by this example. More speciﬁc, M(T ) < ∞ doesn’t imply M(∂T ) < ∞. 2 −n + 0 2 −n Let T = {x ∈ R : |x| = 2 , n ∈ Z }, T = {x ∈ R : |x| = 2 , n ∈ + 2 Z } ∩ (a, b) ∈ R : B > 0. Then, it’s easy to see that

M(T ) = 2, ∂T = 0, but M(T 0) = 1, M(∂T 0) = ∞.

Theorem 18.10. If Tj ⊂ Vk(U) integrable, and

sup{MW (Tj) + MW (∂Tj)} < ∞, ∀W ⊂⊂ U,

0 then there exist T integrable, s.t. Tj *T for some subsequence.

m−1 2j 2j+1 1 Example 18.11. Let Tm = ∪j=0 [ 2m , 2m ], we have Tm * 2 [0, 1]

Deﬁnition 18.12 (Flat Norm). Let Ik(U) be the space of k−dimensional integral current. Given T1,T2 ∈ Ik(U),W ⊂⊂ U, the ﬂat norm

F(T1,T2) := inf {M(S) + M(∂S − T1 − T2)}. S∈Ik+1(U) CHAPTER 18. GEOMETRIC MEASURE THEORY AND MINIMAL SURFACES70

Theorem 18.13. Tj,T ⊂ Ik(U),Tj *T iﬀ F(Tj,T ) → 0. k−1 k n We now consider the Plateau problem: Given T = ∂M0 ,→ R , if there exist an area minimizing M, s.t. ∂M = T. n Theorem 18.14. Let S ∈ Ik−1(R ) with compact support, ∂S = 0. Then n there exist T ∈ Ik(R ), s.t.  T has compact support,  ∂T = S,  n M(T ) ≤ M(R), ∀R ∈ Ik(R ), s.t. ∂R = S.

We could build a homology theory for Ik. n l Let (M , g) ,→ R be a Riemannian manifold, let l Ik(M) = {T ∈ Ik(R ), spt(T ) ⊂ M},

Zk(M) = {T ∈ Ik(M), ∂T = 0},

Bk(M) = {T ∈ Ik(M),T = ∂S for some S ∈ Ik+1(M)}, Then we deﬁne

Hk(M, Z) = Zk(M) ⊗ Z/Bk(M) ⊗ Z ∼ Theorem 18.15. H(M, Z) = HSing(M, Z), where HSing is the singular homology.

Remark 18.16 (Not Sure if it is correct?). If [T ] ∈ Hk(M) 6= 0, then there exist T0 ∈ Zk(M), s.t. M(T ) ≤ M(R), ∀R ∈ [T ]. l Definition 18.17 (Varifold). Let U ⊂ R be an open set. An varifold V of dimension k in U is a pair (M, θ), where M ⊂ U is a rectifiable set of + dimension k and θ : M 7→ Z a Borel map. We can identify a Varifold V = (M, θ) with an inter-rectifiable k−current, as follow Z V (ω) = hω, ξiθ(x)dHk, M Hence, we can define M(V ) on V. Proposition 18.18 (First Variation). Let V = (M, θ), and M be a sub- l t t manifold of R , let φ be the flow of vector field X, then define φ (V ) = (φt(M), θ ◦ φ−t), we have Z d t k δV (X := M(φ (M, θ))|t=0 = divM (X)dH , dt M CHAPTER 18. GEOMETRIC MEASURE THEORY AND MINIMAL SURFACES71

We say V is stationary if δV (X) ≡ 0, ∀X with compact support.

Theorem 18.19. If V is stationary and θ = 1, Hk−a.e., then there exist M(V |B(x,r)) ∞ = (n, k), s.t. if k < 1+, then M|B(x,r) is C embedded minimal wkr submanifold.

Theorem 18.20. T ∈ Ik−1(U) is area minimizing and n ≤ 7, where U ⊂ k R , then T is regular. Proof. See Section 4 in [12] for a sketch of proof. Bibliography

[1] Davies, E.B., Spectral Theory and Diﬀerential Operators, Cambridge Studies in Advanced Mathematics, Cambridge University Press (1996)

[2] Han, Qing and Lin, Fanghua, Elliptic partial diﬀerential equations, American Mathematical Soc. (2011)

[3] H.B. Lawson, Jr., Local rigidity theorems for minimal hypersurfaces, Ann. of Math. 89, 187–197 (1969)

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

[6] R. Schoen and S.T. Yau, Existence of incompressible minimal surfaces and the topology of three dimensional manifolds with non-negative s- calar curvature , Annals of Mathematics. 110, 127-142(1979).

[7] R.M. Schoen, S.T. Yau, Lectures on harmonic maps, International Press 1997.

[8] Simon Brendle, Embedded minimal tori in S3 and the Lawson conjecture, arXiv preprint arXiv:1203.6597, 2012.

[9] N. Aronszajn, A unique continuation theorem for solutions of elliptic partial diﬀerential equations or inequalities of second order, J. Math. Pures Appl. (9) 36, 235–249 (1957)

[10] Simon Brendle, Minimal surfaces in S3: a survey of recent results, Bulletin of Mathematical Sciences. (1) 3, 133-171 (2014)

72 BIBLIOGRAPHY 73

[11] L. Simon, Lectures on geometric measure theory, The Australian Na- tional University, Mathematical Sciences Institute, Centre for Mathe- matics & its Applications (1983)

[12] Camillo De Lellis, The size of the singular set of area-minimizing currents, Surveys in diﬀerential geometry 2016. Advances in geometry and mathematical physics. 21, 1-83 (2016)

[13] R. Schoen and S.T. Yau, On the proof of the positive mass conjecture in general relativity, Communications in Mathematical Physic. (1)65, 45-76 (1979)