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 Professor Rick Schoen in the Winter quarter of 2012 at Stanford. We kept the pace of these lectures by
dates.
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 Plateau
problem and the Sacks-Uhlenbeck theorem, and finally end up with a survey of the proof of the
Willmore conjecture. The materials covered are very good examples for the application of methods
from partial differential equations and calculus of variation.
It is likely that we have numerous typos and mistakes here and there, and would appreciate it
if these are brought to our attention.
Contents
1
Introduction and calibrations (1/10/2012)
1.1 Minimal surface equation: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Calibrated property of minimal graphs . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 A general calibrated argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
3
4
5
2
First variation and consequences (1/12/2012)
2.1 First Variation Formula . . . . . . . . . .
2.2 Examples . . . . . . . . . . . . . . . . .
2.3 Convex hull property . . . . . . . . . . .
2.4 Fluxes . . . . . . . . . . . . . . . . . . .
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3
Monotonicity formula and 2-d Bernstein Theorem (1/17/2012)
8
3.1 Monotonicity formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.2 Bernstein’s theorem (n=2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4
Second variation and Stability (1/19/2012)
11
4.1 Second Variation of Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
4.2 Jacobi operator and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1
1
INTRODUCTION AND CALIBRATIONS (1/10/2012)
2
5
Criterion for stability (1/24/2012)
6
Bochner formula and 2-d stable minimal surface (1/26/2012)
16
6.1 Bochner Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
6.2 Continuity of Section 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
7
Weierstrass representation and Simons Identity(1/31/2012)
20
7.1 Weierstrass representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
7.2 Simons Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
8
Curvature estimates (2/2/2012)
22
9
More curvature estimates in 2-d (2/7/2012)
25
14
10 Schoen-Simon-Yau curvature estimates and minimal cone (2/9/2012)
28
10.1 Curvature estimates by Schoen-Simon-Yau when n ≤ 6 . . . . . . . . . . . . . . . . . 28
10.2 Minimal cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
11 Classical Plateau Problem (2/14/2012)
31
12 Continuity of Plateau Problem and Harmonic maps (2/16/2012)
34
12.1 Continuity of the Proof of Theorem 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . 34
12.2 Harmonic maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
13 Sacks-Uhlenbeck’s theorem (2/21/2012)
38
13.1 Hopf differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
13.2 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 ⊂ (M n , g) be a k−dimensional submanifold of an n−dimensional Riemannian manifold
(M n , g), where ∇ is the corresponding Riemannian connection. Let g|Σ be the induced metric. Given
1
INTRODUCTION AND CALIBRATIONS (1/10/2012)
3
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 ) = (∇X Y )⊥ .
(1)
ij ~
~ = Pk
~ = T rg A
Definition 1.1 The mean curvature of Σ is defined as: H
i,j=1 g A(ei , ej ), where
k
{ei }i=1 is a tangent basis of Σ.
~ = 0.
Definition 1.2 Σ is called minimal, if H
1.1
Minimal surface equation:
Consider a hyper surface Σn−1
⊂ Rn as a graph Σu = {(x, u(x)) : x = (x1 , · · · , xn−1 ) ∈ Ω}
u
of a function u, where Ω ⊂ Rn . Denote F (x1 , · · · , xn−1 ) = (x1 , · · · , xn−1 , u(x)). Then the induced
∂u
metric is given by gij = Fxi · Fxj = δij + ui uj , where ui = ∂x
. The matrix (δij + ui uj ) has n − 2
i
⊥
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) 2 . Hence the inverse matrix for gij is given by
1+|∇u|
g ij = δij − νi νj , where νi = √
ui
1+|∇u|2
for i = 1, · · · , n − 1.
p
p
Now the volume form of Σu is given by dv = det(g)dx = 1 + |∇u|2 dx. So the volume of
Σu is:
Z p
|Σu | =
1 + |∇|2 dx.
Ω
Now let us calculate the Euler-Lagrange equation for |Σu |. For any η ∈ Cc∞ (Ω), suppose u is a
critical point of |Σu |, then
Z
∇u · ∇η
d
p
|t=0 |Σu+tη | =
dx
dt
1 + |∇u|2
Ω
Z
(2)
∂
ui
(p
)ηdx = 0.
=−
1 + |∇u|2
Ω ∂xi
So the divergence form of the minimal surface equation(abbreviated as (MSE) in the following)
(i) :
n−1
X
i=1
∂
ui
(p
) = 0.
∂xi
1 + |∇u|2
(3)
Expanding the above equation, we get:
uxi uxj uxi xj
u
p xi xi
−
= 0,
2
(1 + |∇u|2 )3/2
1 + |∇u|
which can be rewritten as:
1
ui
ui
p
p
(δij − p
)uxi xj = 0.
2
2
1 + |∇u|
1 + |∇u|
1 + |∇u|2
(4)
1
INTRODUCTION AND CALIBRATIONS (1/10/2012)
4
So we get the non divergence form of (MSE):
n−1
X
1
p
g ij uxi xj = 0,
2
1 + |∇u| i,j=1
(ii) :
(5)
where g ij = δij − νi νj is the inverse matrix for the induced metric.
Now let us calculate the mean curvature of Σu from definition. Firstly,
~ x , ∂x ) = (Fx x )⊥ = (Fx x · ν )ν.
A(∂
i
j
i j
i j
| {z }
hij
It is easy to see that
Fxi xj = (0, · · · , 0, uxi xj ),
so hij = √
1.2
uxi xj
1+|∇u|2
. So it is easy to see that equation 5 is equivalent with equation 4.
Calibrated property of minimal graphs
Now extend the unit normal vector field ν to ν̃ in Ω × R in depend of xn .
Claim: (MSE) =⇒ divRn (ν̃) = 0.
This is because divRn (ν̃) = −
Pn−1
i=1
∂xi νi + ∂xn νn = 0 by (MSE).
| {z }
≡0
We can define an n − 1 form ω as
ω = (−1)n−1 ν̃cdx,
where dx = dx1 ∧ · · · ∧ dxn−1 is the volume form of Rn .
Claim: divRn (ν̃) = 0 =⇒ dω = 0.
Proof:
n−1
ω = (−1)
n−1
ν̃cdx = (−1)
n
X
ˆ j ∧ · · · ∧ dxn .
(−1)j dx1 ∧ · · · ∧ dx
j=1
So dω = (−1)n div(ν̃)dx = 0.
Definition 1.3 Given an n − 1 dimensional plane V n−1 ⊂ Rn in Rn with an oriented orthonormal
frame e1 , · · · , en−1 , we can define:
ω(V ) = ω(e1 , · · · , en−1 ).
1
INTRODUCTION AND CALIBRATIONS (1/10/2012)
5
Properties of ω : (1) dω = 0; (2) |ωx̃ (V )| ≤ 1, ∀x̃ ∈ Ω × R and ∀V , and ωx̃ (V ) = 1 only if
V = Tx̃ Σx̃ .
Proof: (of Property (2)) This comes from the following equation and basic linear algebra.
ν̃(x̃)cdx(e1 , · · · , en−1 ) = dx(ν̃, e1 , · · · , en−1 ) = det(ν̃(x̃), 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−1
⊂ Ω × R, with ∂Σ1 = ∂Σ, we can form
1
the n dimensional chain U such that ∂Σ1 ∪ Σ = ∂U . Using the Stokes Theorem and property (2) as
above:
Z
Z
Z
ω−
dω = 0.
ω=
Σ
U
Σ1
|{z} | {z }
=|Σ|
≤|Σ1 |
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 then
use part (1).
1.3
A general calibrated argument
Theorem 1.2 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. 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:
divν =
n−1
X
h∇ei ν, ei i +
i=1
h∇ν ν, νi
| {z }
= −HΣx = 0.
=0, as ν is unit.
Define
ω = (−1)n−1 νcdvM .
Using ω as a calibrated form and arguments above, we can show the minimizing property.
2
2
2.1
FIRST VARIATION AND CONSEQUENCES (1/12/2012)
6
First variation and consequences (1/12/2012)
First Variation Formula
Consider Σk ⊂ M n . Let X be a smooth v.f.(abbreviated for vector field in the following) on M
with compact support. Let Ft : M → M be a family of diffeomorphisms such that F0 = id and
d
dt |t=0 Ft = X. Then
Z
(6)
(First Variation Formula:) δΣ(X) =
divΣ (X)dµ.
Σ
P
Here divΣ (X) = T rg (h∇· X, ·i) = ki=1 h∇ei X, ei i, with {e1 , · · · , ek } an o.n.(abbreviated for orthonormal in the following) basis for Σ.
When Σ is smooth, we can decompose X = X T + X ⊥ to tangent X T and normal X ⊥ parts. So
divΣ (X) = divΣ (X T ) + divΣ (X ⊥ ),
~ Hi.
~ Using the divergence theorem, for general X, we have
where divΣ (X ⊥ ) = −hX,
Z
Z
δΣ(X) = − hX, Hidµ +
hX, ηidσ,
Σ
∂Σ
where η is the outer normal of ∂Σ. So we know that
~ = 0) ⇐⇒ δΣ(X) = 0, ∀X of compact support.
Σ is minimal(H
Proof: (of 1st Variation formula) Consider a local parametrization
F : Σ × (−, ) → M,
where F (x, t) = Ft (x) with Ft given to be the integration of X above. Let {x1 , · · · , xk } be local
coordinates of Σ, then
Z
p
d
d
|t=0 |Σt | =
|t=0 detg(t)dx.
dt
dt
p
√
√
d
Now dt
|t=0 detg(t) = g ij h∇∂xi Ḟ , ∂xj i detg = divΣ X 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
~ = 0 ⇐⇒ 4Σ xi = 0, ∀i.
H
2
FIRST VARIATION AND CONSEQUENCES (1/12/2012)
7
Proof: Let {e1 , · · · , ek } be a local o.n. basis for T Σ, then
4Σ xi =
k
X
i
(ej ej xi − (∇Σ
ej ej )x ) =
j=1
As ej ~x = ej and
P
k
X
i
(∇⊥
ej ej )x .
j=1
~
∇⊥
ej ej = 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 hall
of A.
Proof: C(A) = ∩{H : closed half spaces with A ⊂ H}. Now H = {x : l(x) ≤ a}, where l is some
linear function and a ∈ R.
∂Σ ⊂ H =⇒ l(x) ≤ a, ∀x ∈ ∂Σ & 4Σ l(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 ,
4Σ xi = 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
Z
Z
∂
, ηidσ,
F ([Γ]) =
∗dxi = h
Γ
Γ ∂xi
where η is the unit outer normal of Γ. The second ” = ” follows from the fact that ∗dxi = i
∂
h ∂x
, ηidσ, with i· dv the inner multiplication. Hence we have a group homeomorphism:
i
∂
∂xi
dvΣ =
F : Hk−1 (Σ) → R.
~ = 0, and X a Killing vector
For the general cases, if Σk ⊂ M n is a minimal sub manifold, i.e. H
field 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 = X T the tangential part of X on Σ, then
divΣ (V ) =
k
X
h∇ei (X − X ⊥ ), ei i =
i=1
k
X
i=1
h∇ei X, ei i −
| {z }
=0, as LX g=0
k
X
|i=1
h∇ei X ⊥ , ei i = 0.
{z
~
=H·X=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 as
above.
3
3.1
Monotonicity formula and 2-d Bernstein Theorem (1/17/2012)
Monotonicity formula
Theorem 3.1 Let Σk ⊂ Rn be a minimal surface, then
Z
|Bt (x0 ) ∩ Σ| |Bs (x0 ) ∩ Σ|
|(x − x0 )⊥ |2
=
−
dvΣ ,
k+2
st
tk
Σ∩(Bt (x0 )\Bs (x0 )) |x − x0 |
where Bt (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 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
f |∇Σ h| =
−∞
h≤t
Z
(
f )dτ.
h=t
Remark 3.1 This follows heuristically from the fact that dvΣ =
dt∧dv{h=t}
|∇Σ h|
when t is a regular value.
Proof: (Monotonicity formula) Take h(x) = |x − x0 |, then {h ≤ t} = Bt (x0 ). Let X = x − x0 , then
P
divΣ (X) = ki=1 ∇ei X · ei = k, where {e1 , · · · , ek } is an o.n. basis on Σ. Then by the first variation
formula 6,
Z
Z
δΣBr (x0 ) (X) =
X · η,
divΣ0 (X) =
Σ∩Br (x0 )
Σ∩∂Br (x0 )
Σ
T
(x−x0 )
∇ |x−x0 |
where η is the co-normal vector of Σ ∩ ∂Br (x0 ), and η = |∇
Σ |x−x || = |(x−x )T | . Using the Co-area
0
0
formular,
Z
Z
d
T
k|Σ ∩ Br (x0 )| =
|(x − x0 ) | =
|(x − x0 )T ||∇Σ |x − x0 ||,
dr Σ∩Br (x0 )
Σ∩{|x−x0 |=r}
3
MONOTONICITY FORMULA AND 2-D BERNSTEIN THEOREM (1/17/2012)
9
Z
|(x − x0 )T |2
|(x − x0 )T |2
d
,
=r
|x − x0 |
dr Σ∩Br (x0 ) |x − x0 |2
Σ∩Br (x0 )
Z
d
|(x − x0 )⊥ |2
=r
(1 −
),
dr Σ∩Br (x0 )
|x − x0 |2
Z
|(x − x0 )⊥ |2
d
d
.
= r |Σ ∩ Br (x0 )| − r
dr
dr Σ∩Br (x0 ) |x − x0 |2
d
=
dr
Z
Multiplying the above by r−k−1 , we can re-write it as,
d −k
d
(r |Σ ∩ Br (x0 )|) = r−k
dr
dr
=
d
dr
Z
Σ∩Br (x0 )
Z
Σ∩Br (x0 )
|(x − x0 )⊥ |2
,
|x − x0 |2
|(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. So
we 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 ball
BR (0). If x0 ∈ Σ ∩ BR (0), and σ < R − |x0 |, then
ωk σ k ≤ |Σ ∩ Bσ (x0 )| ≤
σk
||Σ ∩ BR (0),
(R − |x0 |k )
where ωk is the volume of unit ball B1k (0) in Rk .
Proof: The first ” ≤ ” comes from the Monotonicity formula while comparing Bσ (x0 ) with an
arbitrary small ball Br (x0 ), with limr→0 r−k |Σ ∩ Br (x0 )| = ωk , when x0 ∈ Σ and Σ smooth. The
second ≤ is a direct consequence of the Monotonicity formula while comparing Bσ (x0 ) with a large
ball Br (x0 ) exhausting the whole BR (0).
Definition 3.1 The density of Σ at x0 is defined as:
Θx0 = lim (ωk rk )−1 |Σ ∩ Br (x0 )|.
r→0
3
MONOTONICITY FORMULA AND 2-D BERNSTEIN THEOREM (1/17/2012)
3.2
10
Bernstein’s theorem (n=2)
Theorem 3.2 S. Bernstein (1912) Given a minimal graph Σ2 ⊂ R3 , Σ = {(x, u(x)) : x ∈ R2 }. If u
is defined on all of R2 , then u is a linear function, and Σ is a plane.
P
Theorem 3.3 Bernstein’s Big Theorem: 1◦ PDE version: let u ∈ C 2 (R2 ) and 2i,j=1 aij ui uj = 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 → S 2 ,
where N maps a point to the unit normal vector at that point.
~ = 0, then N is a conformal and orientation reversing map, i.e. ∀v, w ∈ Tx Σ, if
Lemma 3.2 If H
v · w = 0 and |v| = |w|, then ∇v N · ∇w N = 0, and |∇v N | = |∇w N |. Furthermore |∇v N | ≤
√1 |A||v|, and N ∗ (ωS 2 ) = KΣ ωΣ = − 1 |A|2 ωΣ .
2
2
Proof: We only need to check that under a special o.n. basis. Take an o.n. principle basis {e1 , e2 }
for Σ, i.e. ∇e1 N = −K1 e1 , ∇e2 N = −K2 e2 . So |∇e1 N | = |K1 | = |K2 | = |∇e2 N |, by the
minimality H = K1 + K2 = 0. Hence |∇v N | ≤ |K||v| = √12 |A|. Furthermore, the Jacobian of N is
Jac(N ) = K1 K2 = − 12 |A|2 .
Propostion 3.1 Given a minimal Σ2 ⊂ R3 , with the image of the Gauss Maps lying in the upper
2 , if ϕ has compact support on Σ, then there exists a constant C > 0, such that
hemisphere N (Σ) ⊂ S+
Z
Z
2 2
|A| ϕ ≤ C
|∇ϕ|2 .
Σ
Σ
2 is simply connected, the closed form ω
Proof: Since S+
S 2 = dα is also exact. Hence
−
So
Z
Σ
|A|2
ωΣ = N ∗ ωS 2 = d(N ∗ α).
2
Z
2
∗
Z
|A| ϕ ωΣ = −2 ϕ d(N α) = 4 ϕdϕ ∧ N ∗ α
Σ
Σ
Z
≤ 4 |ϕ||∇ϕ||N ∗ α|ωΣ .
2 2
Σ
Since
|N ∗ α|
≤ |A||α| ≤ C|A|,
Z
Z
Z
Z
C
|A|2 ϕ2 +
|∇ϕ|2 .
|A|2 ϕ2 ≤ C (|ϕ||A|)(|∇ϕ|) ≤
2
2
Σ
Σ
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 Lipschetz
R
ϕ = ϕR , such that Σ |∇ϕ|2 → 0 as R → ∞.
Proof: Now choose


if r ≤ R
 1
ϕR (r) =
1 − log(r/R)/ log R if R < r < R2

 0
if r ≥ R2
R
where r is the distance function of R3 . By discretize BR2 \ BR = ∪log
k=1 (Bek R \ Bek−1 R ), we have
Z
|∇ϕ|2 ωΣ =
Σ
1
(log R)2
Z
Σ∩(BR2 \BR )
log
XR Z
1
1
ω
=
Σ
r2
(log R)2
Σ∩(B
ek R
k=1
\Bek−1 R )
1
ωΣ
r2
log
log
XR
XR
1
1
1
1
≤
|Σ ∩ (Bek R \ Bek−1 R )| ≤
C(ek R)2
2
2
k−1
2
k−1
(log R)
(log R)
(e R)
(e R)2
k=1
k=1
=
log
XR 1
C
C
=
→ 0,
(log R)2
e2
log R
k=1
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 Maps
R
R
N (Σ) lies in an hemisphere, so we get Σ |A|2 ϕ2 ≤ C Σ |∇ϕ|2 . Then if we take the ϕR in the above
R
Lemma, and let R → ∞, we see that Σ |A|2 → 0. So A = 0, and Σ is a plane.
4
4.1
Second variation and Stability (1/19/2012)
Second Variation of Volume
~ = 0. Given a vector field X on Σ, let F (x, t) : Σ×[−, ] →
Consider a minimal Σk ⊂ M n , i.e. H
M be a one parameter family of variations, with Ḟ (x, 0) = X and denote Σt = Ft (Σ).
4
SECOND VARIATION AND STABILITY (1/19/2012)
12
Theorem 4.1 The Second Variation Formula is:
Z
k
X
d2
2
~ Xi|2 −
δ Σ(X, X) ≡ 2 |t=0 |Σt | = [|∇⊥ X|2 − |hA,
RM (ei , X, ei , X)],
dt
Σ
(7)
i=1
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 ⊂ M n is a hyper surface and 2-sided(∃ν unit normal), X = ϕν, with
ϕ a function with compact support, then
Z
δ 2 Σ(ϕ, ϕ) = [|∇ϕ|2 − (|A|2 + RicM (ν, ν))ϕ2 ]
(8)
Σ
Proof: (of Second Variation Formula)
p
• F : Σ × (−, ) → M , with {x1 , · · · , xk } local coordinates on Σ. Then dvt = detg(t)dx,
∂F ∂F
where gij (t) = h ∂x
i , ∂xj i.
•
p
dp
1
detg(t) = g ij ġij detg(t)
dt
2
•
p
p
1 ij
1 ij p
1 ij
d2 p
2
detg(t)
=
detg
+
detg
+
detg,
(g
ġ
)
g
g̈
(
ġ
ġ
)
ij
ij
ij
dt2
4
2
2
where ġ ij = −g ik g jl ġkl .
•
∂F ∂F
∂F
∂F
ġij = h∇ ∂F
, j i + h j , ∇ ∂F
i
i
∂t ∂x
∂t ∂xi
∂x
∂x
~ ∂ , ∂ ), Xi.
= |t=0 hA(
∂xi ∂xj
So
~ Xi = 0,
(g ij ġij )|t=0 = −2hH,
and
~ Xi|2 .
(ġ ij ġij )|t=0 = −4|hA,
•
∂F ∂F
∂F
∂F
, j i + h∇ ∂F
, ∇ ∂F
i + i, j reversed terms
∂t
∂t ∂xj
∂x
∂xi ∂t
∂xi ∂t
∂F ∂F ∂F ∂F
∂F
= hRM (
, i)
, j i + h∇ ∂F F̈ , j i + h∇ ∂F X, ∇ ∂F Xi
∂t ∂x ∂t ∂x
∂x
∂xi
∂xi
∂xj
+i, j reversed terms
g̈ij = h∇ ∂F ∇ ∂F
= |t=0 − RM (X, ∂xi , X, ∂xj ) + h∇∂xi F̈ , ∂xj i + h∇∂xi X, ∇∂xj Xi
+i, j reversed terms
So
(g ij g̈ij )|t=0 = −2g ij RM (X, ∂xi , X, ∂xj ) + 2divΣ F̈ + 2|∇⊥ X|2 + 2g ij h∇T∂xi X, ∇T∂xj Xi
|
{z
}
2
~
=2|hA,Xi|
4
SECOND VARIATION AND STABILITY (1/19/2012)
13
• Combining all the above,
p
d2
~ Xi|2 − g ij RM (X, ∂xi , X, ∂xj ).
|
detg(t) = divΣ F̈ + |∇⊥ X|2 − |hA,
t=0
dt2
An integration on Σ finishes the proof.
4.2
Jacobi operator and Stability
• Hypersurface Case k = n − 1: X = ϕν,
Z
2
I(ϕ, ϕ) ≡ δ Σ(ϕ, ϕ) = −
ϕLϕ,
Σ
where the Jacobi operator is
Lϕ = 4ϕ + (|A|2 + Ric(ν, ν))ϕ.
|
{z
}
(9)
Q
• When boundary exists (Σ, ∂Σ), L has discrete eigenvalues λj and eigenfunctions uj , i.e. Luj +
λj uj = 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
R
∂Σ and Σ ϕ2 = 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.
• When Σ is non-compact, then
Ind(Σ) = lim Ind(Ωi ),
i→∞
{Ωi }∞
i=1
where
compact.
is an open exhaustion of Σ, i.e. Σ = ∪∞
i=1 Ωi , Ωi ⊂ Ωi+1 , with ∂Ωi smooth and
5
CRITERION FOR STABILITY (1/24/2012)
14
Remark 4.1 In fact, the definition is independent of the exhaustion, say {Ωi } and {Ω̃i }. Since
Ind(Ω) is non decreasing when Ω is expanding, so we can always embed Ωi ⊂ Ω̃i0 for i0 i,
so Ind(Ωi ) ≤ Ind(Ω̃)i0 , and limi→∞ Ind(Ωi ) ≤ limi0 →∞ Ind(Ω̃i0 ), 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.
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 ⊂ M n 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. using
foliation of minimal surfaces.
Proof: 1◦ + 2◦ ⇐=: Since Lu = 4Σ u + Qu ≤ 0, let w = log u(u > 0), we have
4w =
4u
− |∇w|2 ≤ −Q − |∇w|2 .
u
Then ∀ϕ compactly supported,
Z
2
Z
ϕ (4w + Q) ≤ −
Σ
ϕ2 |∇w|2 .
Σ
Using integration by part formula,
Z
Z
Z
2
2 2
Qϕ ≤ 2ϕh∇ϕ, ∇wi − |∇w| ϕ ≤ 2|ϕ||∇ϕ||∇w| − |∇w|2 ϕ2
Σ
Z
≤
|∇ϕ|2 + ϕ2 |∇w|2 − |∇w|2 ϕ2 ≤
Z
|∇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 = −λ1 u ≤ 0.
2◦ =⇒: Assume Σ is non-compact and stable, then Σ has an exhaustion Σ = ∪∞
i=1 Ωi , 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 Ω{u≤0} has eigenvalue equals 0, since u is then a Dirichlet eigenfunction on Ω{u≤0}
with 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 .
i
Consider the normalized sequence { uiu(0)
}∞
i=1 ,
0
Claim: there exists a subsequence i → ∞, such that
ui0
→ u in C 2 on compact subset of Σ.
ui0 (0)
Pf:
• C 0 bound and elliptic theory of L =⇒ C 3 bound for
ui0
ui0 (0)
on compact set;
C0
• Harnack inequality =⇒
bound.
compact
⊂
• Harnack inequality: If u > 0 is a positive solution of Lu = 0 on Ωopen ⊂ M , and if Ω1
Ω, then ∃c = c(L, Ω1 ) > 0, such that
max u ≤ c min u.
Ω1
Ω1
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 Σ̂ is any covering of Σ, then
Σ̂ 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 : Σ̂ → Σ is the covering
map, then Σ̂ is viewed as a minimal immersion i ◦ π : Σ̂ → Σ.
The 2-sided property is essential here. A counterexample is RP1 ⊂ RP2 , while the covering space
S 1 is not stable.
Proof:
• Σ is 2-sided and stable =⇒ ∃u > 0 on Σ with Lu ≤ u.
• Let π : Σ̂ → Σ be the covering map, then u ◦ π > 0 on Σ̂ and L(u ◦ π) ≤ 0, hence Σ̂ is stable.
Propostion 5.1 1◦ . Let Σn−1 ⊂ Rn be a 2-sided minimal surface, and if the Gauss image G(Σ) ⊂
n−1
S+
, then Σ is stable;
◦
2 . Let Σ2 ⊂ R3 be a 2-sided minimal surface, and if G(Σ) ⊂ U open ⊂ S 2 , with µ1 (U ) ≥ 1, where
µ1 (U ) is the Dirichlet eigenvalue of 4S 2 on U , then Σ is stable. In perticular, µ1 (U ) ≥ 1 is true if
the 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 normal
vector field of Σ, since the parallel translation in the e direction does not change the area of Σ, we have
n−1
Lu = 0. Since G(Σ) ⊂ S+
⇐⇒ e · ν > 0, so u > 0, hence Σ is stable.
◦
2 . µ1 (U ) ≥ 1 =⇒ ∃v > 0 on U such that
(
4S 2 v = 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 → S 2 is a conformal map, so
4Σ u = |A|2 (4S 2 v) ◦ G ≤ −|A|2 u,
i.e. Lu ≤ 0, hence Σ is stable. (In fact, on 2-dimension, the Jacobi operator L = G∗ (4S 2 + 1).)
6
6.1
Bochner formula and 2-d stable minimal surface (1/26/2012)
Bochner Formula
Let (Σk , g) be a Riemannian manifold, and {e1 , · · · , ek } an o.n. frame, with {θ1 , · · · , θk } the dual
frame. Denote
X
(∇ej α) =
αi,j θi ,
i
∇ei (∇α) =
X
αi,jk θi ⊗ θj ;
i,j
then
∇α =
X
αi,j θi ⊗ θj ,
∇2 α =
i,j
X
αi,jk θi ⊗ θj ⊗ θk .
i,j,k
Ricci Formula:
αi,jk − αi,kj =
Σ
p αp Rpijk .
P
Definition 6.1 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
4α = Ric(α] , ·),
P
where α] the vector field dual to α, and 4α = i,j αi,jj θi is the rough laplacian.
Proof:
X
X
X
X
X
Σ
αi,jj =
αj,ij =
αj,ji +
αp Rpjij
=
αp RicΣ
pi .
j
j
j
p,j
| {z }
=0
p
6
BOCHNER FORMULA AND 2-D STABLE MINIMAL SURFACE (1/26/2012)
17
Hence we have:
1
2
2 4|α|
= hα, 4αi + |∇α|2 = Ric(α] , α] ) + |∇α|2 .
Σ
In the case Σn−1 ⊂ Rn is minimal, Rijkl
= hik hjl − hil hjk under the o.n. frame {ei } by the
Gauss equation, hence
X
X
X
Σ
RicΣ
Rijkj
=−
hij hjk , (
hjj = 0).
ik =
j
=⇒
j
j
X
XX
1
2
4|α|2 = |∇α|2 +
RicΣ
α
α
=
|∇α|
−
(
hij αj )2 ≥ |∇α|2 − |A|2 |α|2 .
i
j
ij
2
ij
i
j
2
Plug in 12 4|α|2 = |α|4|α| + ∇|α| ,
2
2
|α|(4|α| + |A|2 |α|) ≥ |∇α|2 − ∇|α| ≥ c(n)∇|α| ,
|
{z
}
L|α|
where Lu = 4u + |A|2 u 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
2
X
X
2 X 2
j ( i αi αi,j )
2
2
2
αi,j −
|∇α| − ∇|α| =
=
α
−
α1,j
i,j
|α|2
ij
=
X
2
αi,j
≥
i>1,j
k
X
i,j
2
αi,i
k
X
+
i=2
2
αi,1
i=2
j
k
k
X
1 X
2
2
(
αi,i ) +
α1,i
≥
k−1
i=2
|i=2{z }
=−α1,1
≥
k
X
2
1 1
2
2
[α1,1
+
α1,i
]=
∇|α| .
k−1
k−1
i=2
Theorem 6.1 If Σn−1 ⊂ Rn is a complete, stable and 2-sided minimal surface, then any L2 harmonic
1-form on Σ vanishes.
Proof: 2-sided and stability means that −
pactly supported
Z
R
Σ ϕLϕ
−
≥ 0 for any ϕ compactly supported. So ∀ϕ com-
ϕ|α|L(ϕ|α|) ≥ 0,
Σ
i.e.
Z
Σ
ϕ|α|(4(ϕ|α|) +|A|2 ϕ|α|) ≤ 0,
| {z }
I
6
BOCHNER FORMULA AND 2-D STABLE MINIMAL SURFACE (1/26/2012)
18
where
Z
Z
1
ϕ2 |α|4|α| + h∇ϕ2 , ∇|α|2 i + |α|2 ϕ4ϕ
2
Σ
Z
Z
Z
1
ϕ2 |α|4|α| −
|α|2 ϕ4ϕ
≤
4(ϕ2 )|α|2 +
2
Σ
Σ
Σ
Z
=
ϕ2 |α|4|α| − (ϕ4ϕ + |∇ϕ|2 )|α|2 + |α|2 ϕ4ϕ
ϕ|α|(ϕ4|α| + 2h∇ϕ, ∇|α|i + |α|4ϕ) =
I=
Σ
Σ
Z
=
ϕ2 |α|4|α| − |∇ϕ|2 |α|2 .
Σ
Plug into the above
Z
2
Z
ϕ |α|L(|α|) ≤
Σ
|∇ϕ|2 |α|2 .
Σ
Now by taking ϕ = ϕR to be cutoff functions on geodesic disk, and letting R → ∞, the righthand
side of the above inequality is zero, hence by |α|L(|α|) ≥ c(n)∇|α| proved above,
Z
Z
2
|α|L(|α|) = 0,
∇|α| ≤
c(n)
Σ
Σ
which means that |α| is a constant, and hence is 0 since the area of Σ is ∞ by the monotonicity
|Bσ (p)| ≥ wk σ k .
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 then
g = λ2 (dx2 + dy 2 ) locally. So Σ has a complex striation. Let Σ̂ be the universal cover of Σ, then Σ̂ is
a simply connected non-compact Riemann surface, hence
(
C, the complex plane,
Σ̂ '
D, the unit disk.
Case 1: Σ̂ ' C, then let F : C → R3 , where F = i ◦ π is given by the composition of the minimal
immersion i : Σ → R3 with the covering map π : C ' Σ̂ → Σ. Since i is harmonic, and the harmonic
property is preserved under the conformal change Σ̂ ' C, we know that F is both conformal and
harmonic, i.e. 4C F = 0. Since Σ is stable and 2-sided, Σ̂ is also stable and 2-sided, =⇒ ∃u > 0,
such that Lu = 4Σ̂ u + |Â|2 u = 0 on Σ̂. So 4Σ̂ u ≤ 0, hence 4C (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 |Â|2 = 0 by the following Proposition.
Case 2: Σ̂ ' D.
6
BOCHNER FORMULA AND 2-D STABLE MINIMAL SURFACE (1/26/2012)
19
Claim: the L2 norm of p-forms on an n-dimensional manifold is a conformal invariant when
p = n2 .
This is because |α|2g = g i1 j1 · · · g ip jp αi1 j1 · · · αip jp , so if g̃ = λ2 g, then
|
{z
}|
{z
}
p copies
Z
|α|2g̃
p copies
Z
Z
p
p
p
−2p
2 n
detg̃dx = λ |α|g λ
detgdx = |α|2g detgdx.
Furthermore, harmonic p−forms change to harmonic p−forms under conformal change. This is
because dλ = 0 does not change, while δg̃ α = λ−2 δα = 0 when p = n2 (see Page 59 in [1]).
So L2 harmonic 1-forms on D corresponds to L2 -harmonic 1-forms on Σ̃. Since there are many
harmonic 1-forms on D by just taking dx where x is harmonic functions, so it is a contradiction to the
Theorem we proved in the above section.
Definition 6.2 A Riemannian manifold Σk is called parabolic if every positive super-harmonic function 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 =
4u
− |∇w|2 ≤ −|∇w|2 .
u
Take ϕ a compactly supported function,
Z
Z
Z
2
2
2
ϕ |∇w| ≤ − 4wϕ = 2h∇w, ϕiϕ
Σ
Σ
Z
≤2
Taking = 12 , then
Z
|ϕ||∇w||∇ϕ| ≤ Z
2
2
1
ϕ |∇w| +
2
2
Z
ϕ |∇w| ≤ 4
Σ
Z
|∇ϕ|2 .
|∇ϕ|2 .
Σ
By taking h = distΣ (·, p), we know that Σ has more than quadratic area growth, so we can take
R
ϕ = ϕR as in Lemma 3.3, and use the same logarithmic cut-off trick, to get Σ |∇ϕR |2 → 0, and
ϕR → 1. So |∇w| = 0, and w hence u is a constant.
7
7
7.1
WEIERSTRASS REPRESENTATION AND SIMONS IDENTITY(1/31/2012)
20
Weierstrass representation and Simons Identity(1/31/2012)
Weierstrass representation
Let F : Ω → Rn be a minimal immersion, where Ω is a Riemann surface with complex coordinates
∂F
∂F
∂F
z = x + iy, then F is conformal(i.e. h ∂F
∂x , ∂y i = 0 and | ∂x | = | ∂y |) and harmonic(i.e. 4F = 0).
Define
∂F
∂F
∂F
ψdz =
dz = (
−i
)(dx + idy).
∂z
∂x
∂y
Lemma 7.1 The complex vector ψ =
0.
∂F
∂z
is holomorphic, i.e.
∂ψ
∂ z̄
= 0, and isotropic, i.e.
Pn
2
j=1 ψj
=
Proof: Since F is harmonic,
∂ψ
= 4F = 0.
∂ z̄
Since F is conformal,
n
X
ψj2 = |
j=1
∂F 2
∂F 2
∂F ∂F
| −|
| + 2ih
,
i = 0.
∂x
∂y
∂x ∂y
Rz
Conversely, we would like to represent F as F (z) = Re ψ(s)ds.
When n = 3, given a meromorphic function g and a holomorphic one form ϕdz on Ω, we can take
ψdz =
1
1 −1
(g − g), (g −1 + g, 1) ϕdz,
2
2
and get a minimal immersion F : Ω → R3 by
Z
z
ψ(s)ds.
F (z) = Re
z0
If N : Ω → S 2 is the Gauss Map, then g = π ◦ N , with π : S 2 → R2 the stereographic projection
from (0, 0, 1).
Examples:
dz
• Catenoid: Ω = C \ {0}, g(z) = z and ϕ(z) = dz
;
• Helicoid: (simply connected π1 = 0) Ω = C, g(z) = eiz and ϕ(z) = dz. If we denote the
Helicoid by Σ, then Σ converges to a Foliation by plane x3 = c, where |A|2 → +∞ on the
axis, but |A|2 → 0 everywhere away from the axis;
• Hoffman-White examples.
7
7.2
WEIERSTRASS REPRESENTATION AND SIMONS IDENTITY(1/31/2012)
21
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 rough
laplacian for h is defined as
n−1
X
4hij =
hij,kk .
k=1
Propostion 7.1
4hij + |A|2 hij = 0,
0 ≤ i, j ≤ n − 1
(10)
Proof: Firstly we have the Ricci identity:
hij,kl − hij,lk =
X
hpj Rpikl +
p
X
hip Rpjkl ,
p
Gauss Equation:
Σ
= hik hjl − hil hjk ,
Rijkl
and Codazzi equation:
hij,k = hik,j .
Using the Eistein summation, we have
Σ
Σ
+ hip Rpkjk
4hij = hij,kk = hik,jk = hik,kj +hpk Rpijk
| {z }
=hkk,ij =0
= hpk (hpj hik − hpk hij ) + hip (hpj hkk −hpk hkj )
|{z}
=0
2
= −|A| hij + (hik hkp hpj − hip Hpk hkj ).
{z
}
|
=0
So we finished the proof.
Now recall that the stability operator is Lϕ = 4ϕ + |A|2 ϕ.
Propostion 7.2
|A|(L|A|) ≥
2
2 ∇|A| .
n−1
Proof: By the Bochner Formula,
1
4|A|2 = |∇A|2 + hA, 4Ai = |∇A|2 − |A|4 .
2
(11)
8
CURVATURE ESTIMATES (2/2/2012)
22
While 21 4|A|2 = |A|4|A| + |∇|A||2 ,
2 X 2
hij,k −
|A|L(|A|) = |∇A| − ∇|A| =
2
P P
2
k ( ij hij hij,k )
|A|2
i,j,k
.
In an o.n. eigenbasis {e1 , · · · , en−1 } of h, hij = λi δij , so
P P
2
X
X
X
2
2
∇|A|2 = k ( i λi hii,k ) ≤
h
=
h
+
h2ii,i
ii,k
ii,k
|A|2
i,k
=
X
i6=k
h2ii,k +
X X
X
X
(−
hjj,i )2 ≤
h2ii,k + (n − 2)
h2jj,i
i
= (n − 1)
j6=i
X
h2ii,k =
(1 +
i6=k
i6=j
X
n−1 X 2
(
hik,i +
hki,i ).
2
i6=k
i6=k
So
i
i6=k
i6=k
X
X
X
2 X 2
2 h2ij,k = |∇A|2 .
h2ki,i ≤
) ∇|A| ≤
h2ik,i +
hii,k +
n−1
i6=k
i,k
i6=k
i,j,k
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
sup |A|2 ≤ Cr0−2 ,
Σ∩Br0 /2
where C is independent of Σ ∈ C.
Similar Questions: replace Br0 in the above question by geodesic balls BrΣ0 = {x ∈ Σ : dΣ (s, 0) <
r0 }.
Note: BrΣ0 ⊂ Br0 .
Remark 8.1 1◦ . By scaling, it suffices to assume that r0 = 1, since
1
r0 Σ
∈ C1 , while |A 1 Σ |2 =
r0
r02 |A|2Σ .
2◦ . Curvature estimates =⇒ Bernstein Theorem. If Σ is complete and Σ ∈ Cr0 , ∀r0 > 0, then Σ is
a plane by taking r0 → ∞.
For which Cr0 is it True?
8
CURVATURE ESTIMATES (2/2/2012)
23
|A|2 < , for some fixed small =⇒ Curvature Estimates (see Theorem 8.2).
R
(Counter Example, Down-scaled Catenoid, where Σ∩Br |A|2 is not small.)
1. n = 3,
R
Σ∩Br0
0
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| ≤ A0 r02 =⇒ Curvature Estimates (see Theorem
9.2).
4.
5.
6.
7
Ex: 1) down-scaled Helicoid, 2) Enneper Surface, immersed |Σ ∩ Br0 | ≤ cr02 .
n = 3, Σ2 is stable and 2-sided =⇒ Curvature Estimates. (see Corollary 9.2).
3 ≤ n ≤ 6, Σn−1 ⊂ Rn stable and 2-sided and |Σ ∩ Br0 | ≤ cr0n−1 =⇒ Curvature Estimates
(see Theorem 10.1).
3 ≤ n ≤ 7, Σn−1 embedded and absolutely volume minimizing =⇒ Curvature Estimates. (False
for n − 8)
P8
P4
2
2
The Simons cone C = {x ∈ R8 :
i=5 xi } is a minimal cone. In fact, it
i=1 xi =
is absolutely volume minimizing, since there exists foliation of R8 by minimal hypersurfaces
asymptotical to C, where the curvature blows up on C.
Σn−1 with small access: Θ0 (r0 ) − 1 < , =⇒ 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 |
.
ωh r k
Basic idea: For some small > 0, assume the scaling invariant inequality:
Z
|A|2 < .
Σ∩Br0
Suppose
|A|2 (x0 ) = max |A|2 , at some x0 ∈ Br0 /2 .
Σ∩Br0
Rescale Σ by the factor δ = 1/|A|(x0 ), then

2
2
2

 ◦ |AΣδ (x)| = δ |AΣ | ≤ 1, ∀x ∈ B1 (0),
1
◦ |AΣδ (0)|2 = 1,
Σδ = (Σ − x0 ) ∼

δ
 ◦ R
2
B1 ∩Σδ |AΣδ | < .
Now |AΣδ |(x) ≤ c = 1 =⇒ locally Σδ is a graph Graphu of some function u =⇒ C 3 estimates of u,
R
which will form a contradiction with B1 ∩Σδ |AΣδ |2 < .
Theorem 8.1 (Choi-Schoen [4]) Suppose Σ2 ⊂ M 3 is a minimal surface. Assume 0 ∈ Σ2 , and
∂Σ ∩ Br0 (0) = ∅. Then there exists , ρ > 0 (depending only on M ), such that if r0 ≤ ρ, δ ∈ (0, 1)
R
and Σ∩Br |A|2 < δ, then
0
|A|2 (y) ≤ δσ −2 ,
for y ∈ Br0 −σ (0).
8
CURVATURE ESTIMATES (2/2/2012)
24
Proof: Let us give a proof when M 3 = R3 , and the general cases follow by the fact that M 3 is locally
near R3 when ρ is small enough. Assume δ = 1 and
2
F (y) = r0 − r(y) |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) ≥ σ 2 |A|2 (y), if r(y) < r0 − δ).
0)
Suppose F (y0 ) > 1, let δ = r0 −r(y
, then
2
• supBδ (y) |A|2 ≤ 4|A|2 (y0 ).
0) 2
2
This is because (r0 −r(y))2 |A|2 (y) ≤ (r0 −r(y0 ))2 |A|2 (y0 ), hence |A|2 (y) ≤ ( rr00−r(y
−r(y) ) |A| (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 =⇒ δ 2 |A|2 (y0 ) > 1/4.
Let δ0 =
1
|A|(y0 ) ,
hence δ 2 ≥ 41 δ02 =⇒ δ0 /2 < δ. So Bδ0 /2 (y0 ) ⊂ Bδ (y0 ). Let
Σ δ0 =
(
=⇒
2
(Σ − y0 ),
δ0
supB1 |AΣδ0 |2 = 4|AΣδ0 |2 = δ02 |A|2 (y0 ) = 1,
R
2
B1 |AΣδ0 | ≤ .
So it forms a contradiction when 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 | ≤ cr02 ,
then
sup |A|2 ≤ cr0−2 .
Σ∩Br0 /2
R
R
Proof: By stability, we have Σ |A|2 ϕ2 ≤ Σ |∇ϕ|2 . Since Σ has quadratic area growth, we can use
the logarithmic cutoff trick to get,
Z
C
, k 1.
|A|2 ≤
log k
Σ∩Br /k
0
So for k large enough, we have
R
2
Σ∩Br1 (y) |A|
< , where r1 = r0 /k, hence
|A|2 (y) ≤ cr1−2 ≤ c0 r0−2 ,
c0 = kc.
9
9
MORE CURVATURE ESTIMATES IN 2-D (2/7/2012)
25
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 ≤
then
1
16 .
If x ∈ Σ2 and distΣ (s, ∂Σ) ≥ 2s,
Σ (x) is graphical over T Σ of some function u, where B Σ (x) is the geodesic ball of Σ, and
(i) B2s
x
2s
√
|∇u| ≤ 1 and |Hessu| ≤ 1/ 2;
Σ (x).
(ii) Let Σ0 be a connected component of Bs (x) ∩ Σ containing x, then Σ0 ⊂ B2s
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 a
smooth function when y 6= x. Let K be the Gauss curvature of Σ, then
•
|∂BrΣ0 (x)| − 2πr0 = −
•
|BrΣ0 (x)|
−
πr02
Z
r0
r0
Z
BρΣ
0
τ
Z
K |{z}
=
H=0
1
2
Z
0
r0
Z
|A|2 ;
BρΣ
r0
Z
0
H=0
Z
0
Z
1
K |{z}
=
2
BρΣ
=−
0
Z
τ
Z
|A|2 ;
BρΣ
0
• When t < r0 /2
t
2
Z
2
Z
|A| ≤
BrΣ −2t
2 (r0
|A|
BrΣ0
0
− r)2
=
2
Z
0
r0
Z
τ
0
Z
|A|2
BρΣ
= 2(|BrΣ0 | − πr02 ) ≤ r0 (|∂BrΣ0 | − 2πr0 ).
Proof: Using co-area formula and Gauss-Bonnet formula on the simply-connected domain BrΣ ,
Z
Z
d Σ
d
Σ
Σ
|B | = |∂Br |,
|∂Br | =
kg = 2π −
K,
dr r
dr
∂BrΣ
BrΣ
where kg is the geodesic curvature of the curve ∂BrΣ . Integrate the 2nd of the above, i.e.
Z r0 Z
Σ
|∂Br0 | = 2πr0 −
K.
BρΣ
0
Integrate again =⇒
|BrΣ0 |
=
πr02
Z
r0
Z
τ
Z
−
K.
0
0
BρΣ
R r0
0
=⇒
9
MORE CURVATURE ESTIMATES IN 2-D (2/7/2012)
26
For the last one, if t < 21 r0 ,
t
2
Z
2
Z
|A|
|A| ≤
BrΣ −2t
0
− r)2
=
2
2 (r0
BrΣ0
Z
r0
Z
(r0 − r)
=
d
= dr
RrR
0
Σ
Br
(r0 − r)2 d
2
dr
0
r0
Z
r
Z
dr =
BrΣ
0
r0
Z
2
|A|
| {z }
Z
0
0
Z
|A|2 dr
BrΣ
|A|2 .
BρΣ
|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|BrΣ0 | ≤ r0 |∂BrΣ0 | ⇐⇒ non-positive curvature and simplyR
connected(This can be proved by using the integration inequality B Σ (4r2 ≥ 4), where 4r2 ≥ 4
r0
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
|BrΣ0 | ≤
4π 2
r .
3 0
Proof: It suffices to assume π1 (Σ) = {1}, or we can pass to the universal cover Σ̃ of Σ, which is also
stable and 2-sided (see Lecture 5). Since |BrΣ̃ | ≥ |BrΣ |, we can get the result. Using the last equality
in the above lemma,
Z
Z
Σ
2
2
2
4(|Br0 | − πr0 ) =
|A| (r0 − r) ≤
|∇(r0 − r)|2 = |BrΣ0 |,
BrΣ0
BrΣ0
where we used the stability inequality in ≤ . Hence we finished by moving |BrΣ0 | 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 inequality
and 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
sup |A|2 ≤ cσ −2 .
BrΣ −σ (x)
0
9
MORE CURVATURE ESTIMATES IN 2-D (2/7/2012)
27
Proof: By Theorem 8.2, we can reduce to prove the small total curvature condition. Using the logarithmic cutoff trick, the stability inequality and area bound,
Z
Z
|A|2 ≤ n−2
r−2 ≤ cn−1 .
B Σ−n
e
BrΣ0 \B Σ−n
r0
e
r0
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| ≤ A0 r02 , then
sup |A|2 ≤ c(A0 )r0−2 .
Σ∩Br0 /2 (x)
Remark 9.1 Non-embedded counter examples are Helicoid type singularities; Non-simply-connected
counter examples are Catenoid type singularities.
Proof: Using the simply-connectedness and quadratic area growth, we can apply the third inequality
in Lemma 9.2 to get
Z
|A|2 ≤ c(θ, A0 ),
Σ∩Bθr0
where θ ∈ (0, 1) and c(θ, A0 ) is a constant depending only on θ and A0 . Then we can start from
r1 = 34 r0 , and devide [0, r1 ] into N sub-intervals [9−n−1 r1 , 9−n r1 ], for n = 0, · · · , N . Then ∃n ≤ N ,
such that
Z
c
|A|2 < .
N
Σ∩(B9(−n+1) r \B9−n r )
1
1
R
By rescaling, we can assume that Σ∩(B1 \B ) |A|2 < Nc . =⇒ curvature estimates on the annuli region
1/9
Σ ∩ (B1 \ B1/9 ), =⇒ Σ ∩ (B1 \ B1/9 ) is locally graphical, by Maximum Principle =⇒ graphical on
Σ ∩ (B1 \ B1/9 ), hence graphical on D by embeddedness.
Curvature estimates under small excess assumption
Given x ∈ Σk ⊂ Rn and r > 0, the following quantity
Θx (r) =
|Σ ∩ Br (x)|
ωk r k
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. ∃ = (n, k), if x ∈ Σ, ∂Σ ⊂ ∂Br0 (x), and Θx (r0 ) − 1 < ,
then
sup |A|2 ≤ r0−2 .
Σ∩Br0 /2 (x)
Proof:
• It suffices to assume that Θy (r1 ) − 1 < for all y ∈ Br1 (x) ∩ Σ by the monotonicity formula
3.1.
• 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) = 41 . 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 ⊂ ∂B1 (0), |A|2 ≤ 1, |A|2 (0) = , |Σ ∩ B1 (0)| ≤ (1 + )ωk }.
4
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 C k 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) = 41 .
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. Assume
x0 ∈ Σ, ∂Σ ⊂ ∂Br0 (x0 ), |Σ ∩ Br0 (x0 )| ≤ V r0n−1 and n ≤ 6. Then
sup
Σ∩Br0 /2 (x0 )
|A|2 ≤ c(n, V )r0−2 .
Corollary 10.1 A complete 2-sided stable Σn−1 ∈ Rn with Rn−1 volume growth and n ≤ 6 is a
hyperplane.
Remark 10.1 Counterexample for n = 7: ∃ complete volume minimizing Σ7 ⊂ R8 , not a hyperplane
Proof:
Claim:
2
n−1
Z
Σ
∇|A|2 ϕ2 ≤
Z
Σ
|∇ϕ|2 |A|2 ,
∀ϕ ∈ Cc1 (Σ).
10
SCHOEN-SIMON-YAU CURVATURE ESTIMATES AND MINIMAL CONE (2/9/2012)
Now let us firstly prove this claim. By plug in ϕ|A| to the stability inequality −
0, and using the tricks in Theorem 6.1, we have
Z
Z
ϕ2 |A|L(|A|) ≤
|∇ϕ|2 |A|2 .
Σ
R
29
Σ (ϕ|A|)L(ϕ|A|)
≥
Σ
Using Proposition 7.2, we can get the conclusion.
Now change ϕ → ϕ|A|q , for some q > 0, then we get
Z
Z
Z
2
2
2
q 2
∇|A|2 ϕ2 |A|2q ≤
|A| ∇(ϕ|A| ) =
|A|2 (∇ϕ)|A|q + q|A|q−1 ϕ(∇|A|)
n−1 Σ
Σ
Σ
Z
Z
2
1
2q 2 2
≤ (q + ) |A| ϕ ∇|A| + (1 + ) |∇ϕ|2 |A|2q+2 .
Σ
Σ
q
2
Hence if q < n−1
, by moving the first term on the right hand side to the left,
Z
∇|A|2 ϕ2 |A|2q ≤ C(q)
=⇒
Z
Σ
Σ
Z
=⇒
|A|2q+2 |∇ϕ|2 ,
∇|A|q+1 ϕ2 ≤ C(q)
Z
Σ
(|A|q+1 )2 |∇ϕ|2 .
Σ
Set p = q + 2,
Z
=⇒
∇|A|p−1 ϕ2 ≤ C(p)
Z
Σ
(|A|p−1 )2 |∇ϕ|2 .
Σ
R
R
Now replace ϕ by ϕ|A|p−1 in the stability inequality Σ |A|2 ϕ2 ≤ Σ |∇ϕ|2 ,
Z
Z
Z
2
p−1 2
2p 2
∇(ϕ|A| ) ≤ 2 ϕ2 ∇(|A|p−1 ) + |∇ϕ|2 |A|2p−2 .
=⇒
|A| ϕ ≤
Σ
Σ
Σ
Using the above inequality,
Z
Z
2p 2
|A| ϕ ≤ C(p)
=⇒
Σ
Replace ϕ by
|A|2p−2 |∇ϕ|2 .
Σ
ϕp ,
then
Z
Z
Z
2p 2
2p−2
p 2
2
|A| ϕ ≤ C(p) |A|
|∇ϕ | = Cp
(ϕ|A|)2p−2 |∇ϕ|2
Σ
Σ
≤ C(p)
Z
Σ
(ϕ|A|)2p
(p−1)/p Z
Σ
Z
=⇒
(ϕ|A|)
Σ
2p
|∇ϕ|2p
1/p
.
Σ
Z
≤ C(p)
r
2p
|∇ϕ| ,
Σ
∀p < 2 +
2
.
n−1
R
• Remark: if Σk ⊂ Rn is minimal, Σ |A|2p ≤ C, for 2p > k =⇒ Curvature Estimates.
q
q
2
2
• Want: 2p > n − 1, hence 4 + 2 n−1
> n − 1, i.e. n−1
> n−5
2 =⇒ n ≤ 6.
(12)
10
SCHOEN-SIMON-YAU CURVATURE ESTIMATES AND MINIMAL CONE (2/9/2012)
30
• When n ≤ 6, take 2p = n − 1, and use the logarithmic cut off trick and the volume growth =⇒
R
n−1 is small in small ball, and hence the curvature estimates.
Σ |A|
• Let Σn−1 ⊂ Rn be complete, stable, 2-sided, n ≤ 6 and Σ∩BR ≤ CRn−1 =⇒ Σ is hyperplane.
R
R
Take 2p > n − 1, =⇒ Σ (ϕ|A|)2p ≤ C Σ |∇ϕ|2p ≤ RC2p |Σ ∩ B2R | → 0.
10.2
Minimal cone
Given Σk−1 ⊂ S n−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 S n−1 ⇐⇒ 4Σ xi + (k − 1)xi = 0, i =
1, · · · , n, where {x1 , · · · , xn } is coordinates of Rn .
~ = (x1 , · · · , xn ), then C(Σ) is minimal ⇐⇒ 4C(Σ) X
~ = 0. Take a o.n. basis
Proof: Given X
~ X|,
~ then {e1 , · · · , ek } is an o.n. basis for C(Σ). Then
{e1 , · · · , ek−1 } for T Σ, and ek = X/|
~ =
4C(Σ) X
k−1
X
~ + ek ek X
~ −
ei ei X
i=1
k
X
~
(∇ei ei )T C(Σ) X.
i=1
∂
~
~
~
∂r = 0, with r = |X| the radial function. ∇ei ei · X = −ei · (∇ei X)
~ X
~ = (∇e ei )T Σ − X.
~ So
(∇ei ei )T C(Σ) = (∇ei ei )T Σ + (∇ei ei · X)
i
Here ∇ek ek = ∇ ∂
∂r
−1, hence
= −ei · ei =
~ = 4Σ X
~ + (k − 1)X.
~
4C(Σ) X
So we prove the equivalence of the first and the third conclusion.
~ Σ = Pk−1 (∇e ei )T S n−1 , hence (Pk−1 (∇e ei )) lies in
Now Σ is minimal in S n−1 if and only if H
i
i
i=1
i=1
~ C(Σ) = (Pk−1 (∇e ei ) + ∇e ek )⊥(Σ) = Pk−1 (∇e ei )⊥(Σ) = 0.
the normal direction of S n−1 . So H
i
i
k
i=1
i=1
Clifford Hypersurfaces: Given S p (r1 ) ⊂ Rp+1 and S q (r2 ) ⊂ Rq+1 , take
Σ = S p (r1 ) × S q (r2 ) ⊂ Rp+q+1 .
Then Σ ⊂ S p+q+1 ⇐⇒ r12 + r22 = 1. Given (~x, ~y ) ∈ Σ, where ~x ∈ S p , and ~y ∈ S q , hence
4Σ ~x = − rp2 ~x, and 4Σ ~y = − rq2 ~y . Hence
1
2
Σp+q ⊂ S p+q+1 is minimal, ⇐⇒
p
q
= 2 = p + q.
2
r1
r2
Hence such class of Σ form lots of examples of minimal suffices in S n and hence minimal cones in
Rn+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−1
λi → ∞,
i Σ → C,
where C is a non-flat, volume minimizing cone.
• So existence of such Σ =⇒ ∃ nonflat volume minimizing cone C1m , m ≤ n − q with an isolated
singularity at 0.
• J. Simons: when m ≤ 7, no such cones exists =⇒ Σ such does not exists if n ≤ 7.
√
√ • p = q = 3 : C S 3 (1/ 2) × S 3 (1/ 2) is stable, and area minimizing;
p
√
• p = 1, q = 5 : C S 1 (1/ 6) × S 5 ( 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 C 1 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 ) ∩ C 0 (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. on D, 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:
Z q
Z q
∗
A(u) =
det u (δij ) dx1 dx2 =
|ux |2 |uy |2 − (ux · uy )2 dxdy.
D
D
Note: given ϕ : D → D diffeomorphism, then A(u ◦ ϕ) = A(u).
The W 1,2 norm of u is:
Z
2
kukW 1,2 (D) =
(|u|2 + |∇u|2 )dxdy,
D
where |∇u|2 = |ux |2 + |uy |2 .
11
CLASSICAL PLATEAU PROBLEM (2/14/2012)
32
Definition 11.2 The energy functional of u is
1
E(u) =
2
Z
|∇u|2 dxdy.
D
Lemma 11.1 A(u) ≤ E(u) with equality ⇐⇒ u is almost conformal.
Proof: By Cauchy-Schwartz
Z
1
A(u) =
|ux ∧ uy |dxdy ≤
2
D
Z
|ux |2 + |uy |2 .
D
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 = inf{A(v) : v ∈ XΓ },
EΓ = inf{E(v) : v ∈ XΓ },
where XΓ is defined in Theorem 11.1.
Propostion 11.1
AΓ = EΓ .
Proof: By the above lemma, AΓ ≤ EΓ clearly. Choose u ∈ XΓ , with A(u) < AΓ + (may assume
that 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 =
λ2 δ, hence
(u ◦ ϕ)∗ δRn = ϕ∗ (u∗ δ) = ϕ∗ g = λ2 δD .
Then u ◦ ϕ : D → Rn is conformal, hence E(u ◦ ϕ) = A(u ◦ ϕ) = A(u),
=⇒
EΓ ≤ AΓ + .
(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 g̃ij = ((us )∗ δRn+2 )ij = gij + s2 δij . So ∃ϕ diffeomorphism of D, such
that us ◦ ϕ is conformal. Then
Z
1
|Dϕ|2g dxdy
EΓ ≤ E(u ◦ ϕ) = E ϕ : (D, δ) → (D, g) =
2 D
Z
1
≤
|Dϕ|2g̃ dxdy = E ϕ : (D, δ) → (D, g̃) ≤ E(us ◦ ϕ)
2 D
= A(us ◦ ϕ) = A(us ) = A(D, g̃) ≤ A(D, g) + (s)
= A(u) + (s) < AΓ + 2.
Here (s) → 0 as s → 0. Hence we finished the proof.
11
CLASSICAL PLATEAU PROBLEM (2/14/2012)
33
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) ∩ C 0 (D), then ∃ u of
least energy in Cv = {w ∈ W 1,2 (D) : w − v ∈ W01,2 (D)}. This is equivalent to find a u ∈
W 1,2 (D) ∩ C 0 (D) ∩ C ∞ (D) which is a solution of
(
4u = 0 in D
u=v
on ∂D.
There are several methods: i) Perron method; ii) Poison kernel; iii) Variational method. Now let us
use the variational method.
• Take {ui } in Cv , with E(ui ) → inf w∈Cv E(w).
R
• {ui } are bounded in W 1,2 (D). D |∇ui |2 ≤ C comes from E(ui ) ≤ C. Now by Poincaré
inequality,
Z
Z
Z
2
2
(ui − v) ≤ C
|D(ui − v)| ≤ 2C
|Dui |2 + |Dv|2 ≤ C 0 .
D
D
D
So D u2i ≤ c( D v 2 + C 0 ).
• Rellich Lemma =⇒ ui0 * u weakly in W 1,2 (D), and ui0 → u in L2 (D). Furthermore, u − v ∈
W01,2 (D) since W01,2 (D) is weakly closed in W 1,2 (D).
• Lower semi-continuity of E:
R
R
=⇒
E(u) ≤ lim inf E(ui0 ) = inf E(w).
w∈Cv
• u ∈ Cv =⇒ u is weakly harmonic, =⇒ u ∈ C ∞ (D).
• u ∈ C 0 (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:
max |ui − uj | = max |vi − vj | → 0,
D
∂D
i, j → ∞,
=⇒ ui → u uniformly in D, hence u ∈ C 0 (D).
Step 2: Minimize over boundary parametrization.
Propostion 11.2 (Courant-Lebesgue Lemma) Given u : D → Rn , u ∈ W 1,2 (D) ∩ C 0 (D) and
E(u) ≤ K, then ∀δ 1 and x ∈ ∂D, ∃ρ ∈ [δ, δ 1/2 ] 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)
Proof:
Z
|u(Cρ )|2 ≤ (
|Du|ds)2 ≤ 2πρ
Z
Cρ
34
|Du|2 ds.
Cρ
Now integrate over ρ ∈ [δ, δ 1/2 ],
Z
δ
δ 1/2
u(Cρ )2
dρ ≤
ρ
Hence inf [δ,δ1/2 ] |u(Cρ )|2 (− log ρ) ≤
δ 1/2
Z
|Du|2 ds)dρ ≤ 2πE(u) ≤ 2πK.
Cρ
δ
R δ1/2
δ
Z
2π(
u(Cρ )2
dρ
ρ
≤ 2πK.
Key Problem: if ϕ is a conformal diffeomorphism of D, i.e. ϕ ∈ P SL(2, R), ϕ =
αz+β
γz+δ ,
then
E(u ◦ ϕ) = E(u).
The less of compactness when we minimize over all the possible boundary parametrizations comes
from the un-compactness of P SL(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 and
{qi : i = 1, 2, 3} ⊂ Γ monotone, i.e. p1 < p2 < p3 } w.r.t. the fixed orientation, introduce
XΓ∗ = {u ∈ XΓ : u(pi ) = qi }.
Lemma 12.1 Given v ∈ XΓ , ∃ϕ : D → D Mobiüs transform, such that u = v ◦ ϕ ∈ XΓ∗ , and
E(u) = E(v).
Proof: Since v ∈ XΓ , ∃ r1 , r2 , r3 ∈ ∂D, with v(ri ) = pi . The monotonicity of v : ∂D → Γ implies
that r1 < r2 < r3 . Then ∃ϕ : D → D Mobiüs, with ϕ(pi ) = ri , i = 1, 2, 3. Hence u = v ◦ ϕ satisfies
the requirement.
Corollary 12.1
EΓ = inf{E(u) : u ∈ XΓ∗ }.
∗
∗ } is
Lemma 12.2 Given K > 0, and XΓ,K
= {u ∈ XΓ∗ : E(u) ≤ K}, then {u|∂D : u ∈ XΓ,K
uniformly equi-continous.
12
CONTINUITY OF PLATEAU PROBLEM AND HARMONIC MAPS (2/16/2012)
√
Proof: Given > 0, using the Courant-Lebesgue Lemma, ∃ δ < mini6=j=1,2,3
d(pi ,pj )
,
2
√
c K
| log ρ|1/2
[δ, δ 1/2 ]),
35
and ∃Cρ (=
circle centered at x ∈ ∂D, ρ ∈
such that |u(Cρ )| ≤
< . 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 be
mapped to the short arc inside u(Cρ ).(or there would be two qi s inside u(Cρ ), contradiction to the
monotonicity.) 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 12.1
and then the Dirichlet Problem in the above section, we can assume that
(
ui ∈ XΓ∗
4ui = 0.
By weak compactness of bounded set in W 1,2 (D) and Lemma 12.2, there exists a subsequence {ui0 }
ui0 * u, weakly in W 1,2 (D),
ui0 =⇒ u, uniformly on ∂D.
The uniform convergence on ∂D implies that u|∂D is monotone, onto and u(pi ) = qi , i = 1, 2, 3. By
Maximum Principle
ui0 =⇒ u, uniformly in D, hence 4u = 0.
Now u ∈ XΓ∗ clearly. Hence EΓ ≤ E(u) ≤ lim inf E(ui0 ) = 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 almost
conformal.
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.
∂ ∂u
∂ z̄ ( ∂z )
= 4u = 0).
12
CONTINUITY OF PLATEAU PROBLEM AND HARMONIC MAPS (2/16/2012)
36
Propostion 12.2 If Γ is a C k,α -curve, k ≥ 2, 0 < α < 1, then u ∈ C k,α (D), with finite number of
branch 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 is
analytic.)
• In Rn , n ≥ 4, the minimizing solution has branch points, e.g. z → (z 2 , z 3 ) ∈ 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 ∂D
to a point on Γ. By a reflection argument, we can reflect conformally near the interval where u
is a constant, and extend u to a conformal harmonic map in a neighborhood of the interval, then
u maps an interval to a point. In fact, u maps the direction tangent to ∂D to 0, but the normal
direction not to 0(if 0, then u has branch points along an interval, contradiction to the fact that
branch points are discrete), contradiction to the conformal property.
12.2
Harmonic maps
Motivation: find minimal Σ2 in an arbitrary Riemannian manifold (M n , g).
Definition 12.1 Given u : (Ωk , γ) → (M n , g), where (Ωk , γ) is a k-dimensional Riemannian manifold possibly with boundary, the harmonic energy is:
Z
1
|∇u|2 dµγ ,
E(u) =
2 Ω
where in local coordinates {x1 , · · · , xk } of Ω, {u1 , · · · , un } of M , the harmonic energy density is
|∇u|2 = trγ (u∗ g) =
k
X
γ ij hdu(
i,j=1
X
∂
∂
∂uα ∂uβ
ij
),
du(
)i
=
γ
g
.
g
αβ
∂xi
∂xj
∂xi ∂xj
i,j,α,β
Critical point of E is called harmonic maps.
Lemma 12.3 The Euler-Lagrange equation of E is call the harmonic map equation(HE):
X
∂uβ ∂uδ
4γ uα +
γ ij Γαβδ u(x)
= 0, α = 1, · · · , n.
∂xi ∂xj
i,j,β,δ
Another form of (HE): Consider an isometric embedding M n ⊂ RN , then
u : Ω → RN ,
and
N
E(u) =
1X
2
Z
u(Ω) ⊂ M,
|∇γ uα |2 dµγ .
α=1 Ω
So the variational problem can be viewed as a constraint problem.
(14)
12
CONTINUITY OF PLATEAU PROBLEM AND HARMONIC MAPS (2/16/2012)
• Take a variation vector fields
•
d
dt |t=0 ut
37
= η, where η is tangent to M ;
d
|t=0 E(ut ) =
dt
Z
h∇u, ∇ηidµγ .
Ω
• The Euler-Lagrange equation is: (4γ u)T = 0, or
4γ uα = [(4γ u)⊥ ]α =
k
X
γ ij Aαu(x) (
i,j=1
∂u ∂u
,
),
∂xi ∂xj
(15)
where A is the second fundamental form of M ⊂ RN .
• For example, M = S n ⊂ Rn+1 , then 4uα = −|∇u|2 uα .
Bochner Formula: Given u : (Ωk , γ) → (M n , g) harmonic and smooth, then
α
β
1
Ω ∂u ∂u
4|∇u|2 = |∇∇u|2 + gαβ γ ik γ jl Rij
2
∂xi ∂xj
X
−
hRM du(ei ), du(ej ) du(ei ), du(ej )i,
(16)
i,j
Ω is the Ricci curvature of (Ω, γ), RM the sectional curvature of (M, g), and {e , · · · e } an
where Rij
1
k
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
2
Z
sup |∇u| ≤ C
BR/2
|∇u|2 .
BR
• If does not assume KM ≤ 0, we have counterexamples to the gradient estimates.
Example: ∃ ui : S 2 → S 2 , holomorphic, but with strong dilation which maps a small neighborR
hood of the south pole to almost all of S 2 , satisfying: |∇ui |2 (0) → ∞, S 2 |∇ui |2 = 8π, and
each ui is energy minimizing.
Lemma 12.4 Given u : (Ω2 , γ) → (M n , g), the area is A(u) =
A(u) ≤ E(u),
with “ = ” if and only if u is almost conformal.
R p
det(u∗ g)dx.
Ω
13
SACKS-UHLENBECK’S THEOREM (2/21/2012)
13
13.1
38
Sacks-Uhlenbeck’s theorem (2/21/2012)
Hopf differential
In the case (Σ2 , γ) has dimension 2, let z = x + iy be the local complex coordinates w.r.t. the
conformal structure determined by γ.
Definition 13.1 Given u : (Σ2 , γ) → (M n , g), the Hopf differential Φ is defined by:
Φ = ϕ(z)dz 2 ,
where
∂u 2
∂u
∂u ∂u
| − | |2 − 2ih ,
i.
∂x
∂y
∂x ∂y
ϕ(z) = |
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 : − < t, and γ̇0 = h. Since
1
E(u, γ) =
2
and
(
Z X
2
Σ i,j=1
ij
d
t=0 γt
dt |√
d
dt detγ
we have
γ ij h
∂u ∂u p
,
i detγdx1 dx2 ,
∂xi ∂xj
= −hij = −γ ip γ jq hpq ,
√
= 21 trγ (h) detγ,
Z
δγ E(u, γ) = −
∂u ∂u
1
hij h i , j i − |∇u|2 γij dµγ .
Σ
| ∂x ∂x {z 2
}
Tij −stree-energy tensor
Proof of 1: locally γij = λ2 δij , so
"
#
1
2 + |u |2 ), −hu , u i
(−|u
|
x
y
x y
T = 2
= 0 ⇐⇒ Φ = 0.
−hux , uy i, 12 (|ux |2 − |uy |2 )
Proof of 2: Claim: Φ is holomorphic ⇐⇒ divΣ T = 0.
(On (Σ, z), trace-free, divergence free (0, 2)-tensors are 1 : 1 correspondence to holomorphic quadratic
differentials. Pf: in local conformal coordinates z = x1 + ix2 , let T = Tij , hence T11 = −T22 (trace
free), T11,1 + T12,2 = 0, T21,1 + T22,2 = 0(divergence free). If we let Φ = ϕ(z)dz 2 = (T11 − T22 −
2iT12 )dz 2 , then it is easy to check ∂∂z̄ ϕ = 0.)
13
SACKS-UHLENBECK’S THEOREM (2/21/2012)
39
• Using the composition ft : (Σ, γ) → (Σ, (ft−1 )∗ γ) and u : (Σ, (ft−1 )∗ γ) → M
E(u ◦ ft , γ) ≡ E(u, (ft−1 )∗ γ ).
| {z }
γt
•
d
dt |t=0 RHS
=
R
Σ hh, T idµga .
~ =
Let X
d
dt |t=0 ft ,
we have h =
d
dt |t γt
= −LX γ = −(Xi,j +
Xj,i ). So
Z
Z
d
ij
|t=0 E(u ◦ ft , γ) = δγ E = (Xi,j + Xj,i )T dµγ = 2 Xi,j T ij µγ
dt
Σ
Σ
Z
= − hX, divT idµγ = 0.
Hence δγ E = 0 ⇐⇒ divT = 0, for δγ = LX γ.
d
Remark 13.1
• When u is smoothly harmonic =⇒ dt
E(u ◦ ft , γ) = 0 for ft 1-parameter family
of diffeomorhpisms =⇒ Φ is holomorphic;
• When u is weakly harmonic, u ◦ ft may not be C 1 variations of u.
Corollary 13.1
1. When u : S 2 → M is smoothly harmonic map, then u is almost conformal;
R
2. When u : R2 → M is smoothly harmonic and R2 |∇u|2 dx < ∞, then u is almost conformal.
Proof: Let Φ(z) = ϕ(z)dz 2 on S 2 \ {+∞}, with ϕ(z) an entire holomorphic function.
R
R
• Pf of 2: R2 |∇u|2 < ∞ =⇒ C |ϕ| < ∞ =⇒ ϕ ≡ 0.
• Pf of 1: Φ is regular at ∞, hence ϕ(z)dz 2 is regular at ∞. Let ζ = 1/z, then Φ = ϕ(1/ζ)(− dζ
)2 =
ζ2
C
(ϕ(1/ζ)/ζ 4 )(dζ)2 . Since ϕ(1/ζ)/ζ 4 is regular near ζ = 0, =⇒ |ϕ(z)| ≤ |z|
4 =⇒ ϕ ≡ 0.
Remark 13.2
• Only true when S 2 (u is critical & u is conformal invariant & S 2 has only one
conformal structure).
• Not true for Σg with g ≥ 1.
13.2
Sacks-Uhlenbeck’s method
Sacks-Uhlenbeck: If (M n , g) is a compact Riemannian manifold and πk (M ) 6= 0 for some k ≥ 2,
then ∃ a nontrivial u : S 2 → 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 : S 2 → M not homotopic to point, ∃ u harmonic and homotopic
to v?
Result: given [v] ∈ π2 (M ), then [v] = [u0 ] + · · · + [uk ], where {u0 , · · · , uk } are minimal stable
spheres.
2. Micallef-Moore: If πk (M ) 6= 0 for k ≥ 3, then ∃ u minimal with Morse Index(for the energy
functional E) ≤ k − 2.
Key Idea:
Definition 13.2 Given α > 1 and u ∈ W 1,2α (S 2 , M ) ⊂ C0 (S 2 , M ), the α-energy is defined by
Z
1
Eα (u) =
(1 + |∇u|2 )α dµ.
2 S2
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 uniformly
bounded.
Lemma 13.1 Euler-Lagrangian equation for Eα :
1. If u ∈ W 1,2α (S 2 ), 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,
∂u ∂u
⇐⇒ div (1 + |∇u|2 )α−1 ∇ui = (1 + |∇u|2 )α−1 γ pq Ai ( p , q ).
∂x ∂x
2. If u is smooth,
4ui + (α − 1)
2
X
h∇|∇u|2 , ∇ui i
∂u ∂u
=
γ pq Ai ( p , q ),
2
1 + |∇u|
∂x ∂x
i = 1, · · · , N.
p,q=1
14
Sacks-Uhlenbeck’s theorem continued (2/23/2012)
Theorem 14.1 If (M n , g) is compact and π2 (M ) 6= 0, then ∃ nontrivial u : S 2 → M harmonic and
almost conformal.
Propostion 14.1 If α > 1, given v : S 2 → M , then ∃ u ∈ C ∞ (S 2 ) with Eα (u) = min{Eα (w) :
w ∈ W 1,2α (S 2 ) & u homotopic to v}.
Proof: Minimal u exists by direct method(minimization).
14
SACKS-UHLENBECK’S THEOREM CONTINUED (2/23/2012)
41
• u ∈ W 1,2α (S 2 ) =⇒ u ∈ W 2,2 (S 2 ) follows from Morrey Thm 1.11.1;
• u ∈ W 2,2 (S 2 ) =⇒ u ∈ C ∞ (S 2 ) follows from the standard elliptic estimates.
Lemma 14.1 For (α − 1) small enough, if u is critical for Eα , then for p ≥ 3,
Z
p−1
|∇u|2 )1/p .
kuk2,p ≤ C 1 + (sup |∇u|2 ) p (
S2
S2
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:
kuk2,p ≤ C(kuk0,p + k4uk0,p ) ≤ C 1 + (α − 1)kuk2,p + |∇u|2 0,p .
When (α − 1) is small enough,
Z
kuk2,p ≤ C 1 + (
p−1
|∇u|2p )1/p ≤ C 1 + (sup |∇u|2 ) p (
S2
Z
|∇u|2 )1/p .
S2
S2
Proof: (of Theorem 14.1) Take a sequence α1 & 1 and corresponding critical mapping ui = uαi .
Case 1: ∃ subsequence ui0 with |∇ui0 | ≤ C, then ui0 satisfy uniform C 2,α estimates by Lemma 14.1
and standard elliptic theory, =⇒ ui0 → u in C 2 and u is our solution;
Case 2: λi = maxS 2 |∇ui |2 → +∞. ∃ pi ∈ S 2 , with |∇ui (pi )|2 = λi . Assume a subsequence
pi → p ∈ S 2 .
• Let γi = λi γ, where γ is the standard metric on S 2 , then γi becomes flatter.
• Then we have the following scaling equality:
Z
Z
2
2
|∇ui |2γi = λ−1
|∇u
|
=⇒
|∇u
|
dµ
=
|∇ui |2γ dµγ ;
i γ
i γi
γi
i
S2
2
|∇∇ui |2γi = λ−2
i |∇∇ui |γ =⇒
Z
S2
S2
|∇∇ui |qγi dµγi = λi1−q
Z
S2
|∇∇ui |qγ dµγ .
• Using the above scaling equality and the estimates in Lemma 14.1,
Z
Z
p
p−1
=⇒ λp−1
|∇∇u
|
≤
C
1
+
(λ
)
|∇ui |2γi ,
i γi
i
i
S2
Z
=⇒
S2
S2
−(p−1)
|∇∇ui |pγi ≤ C λi
Z
+
S2
|∇ui |2γi .
14
SACKS-UHLENBECK’S THEOREM CONTINUED (2/23/2012)
42
2,p
• So ui are uniformly in Wloc
(S 2 , γi ). Since (S 2 , γi , pi ) locally converge to (R2 , δ, 0), using the
Sobolve embedding,
ui0 → u in C 1,1/2 -loc on R2 ,
and |∇u(0)| = 1 = maxR2 |∇u|.
• The Euler-Lagrangian equation is scaled to:
λi 4γi uji
2
X
λi h∇(λi |∇ui |2γi ), ∇uji iγi
∂ui ∂ui
=
λi γipq Aj ( p , q ),
+ (αi − 1)
2
1 + λi |∇ui |γi
∂x ∂x
p,q=1
=⇒
4γi uji
+ (αi − 1)
Here the second term (αi − 1)
h∇|∇ui |2γi , ∇uji iγi
λ−1
i
+
|∇ui |2γi
h∇|∇ui |2γi ,∇uji iγi
2
λ−1
i +|∇ui |γi
=
2
X
γipq Ai (
p,q=1
∂ui ∂ui
,
).
∂xp ∂xq
≤ (αi − 1)C|∇∇ui |γi , where |∇∇ui |γi is uni-
formly bounded in Lp (S 2 , γi ). So the second term converges to 0 in Lploc (R2 , δ).
• =⇒ u is a C 1,1/2 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 S 2 = 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, hence
v is also harmonic.
Key step: v is continuous at ζ = 0(v ∈ W 1,2 ∩ C 0 (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) and
boundary oscillation on ∂Dr :
max dM v(s1 ), v(s2 ) ≤ C(EDr (v)) + max dM v(s1 ), v(s2 ) .
Dr/2
∂Dr
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 harmonic
and almost conformal, hence minimal), each portion of v(Dr ) inside the unit ball has a fixed
mount 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 on
S 2 . Finished.
15
COLDING-MINICOZZI’S MIN-MAX SPHERE (3/1/2012) (BY XIN ZHOU)
15
43
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 C 0 ∩
W 1,2 (S 2 , M ) by direct variational method.
In fact, any π3 (M ) representative u : S 3 → M can be viewed as a 1-parameter family of maps
S 2 → M , i.e. S 3 = S 2 × [0, 1], with S 2 × {0} and S 2 × {1} mapped to points.
Definition 15.1 The variational space is define as:
Ω = u(t, x) ∈ C 0 [0, 1], C 0 ∩ W 1,2 (S 2 , 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 = inf max Area u(t) ;
u∈[β] t∈[0,1]
the energy width of [β] is
WE = inf max E u(t) .
u∈[β] t∈[0,1]
Propostion 15.1
WA = 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. ∃ γ j ⊂ Ω, j = 1, · · · , ∞, such that maxt∈[0,1] E γ j (t) & W ;
2. ∀ > 0, ∃ J 1 and δ > 0, such that if j > J, for any t with
E γ j (t ) − W > −δ,
∃ a collection of harmonic spheres(hence almost conformal) ui : S 2 → M , i = 0, · · · , l, such
that
dV γ t , tli=0 ui < ,
20 .
where dV is the varifold distance;
or ∀ ti such that E γ j (tj ) → W , a subsequence γ j (tj ) converge to a collection of harmonic
spheres {u0 , · · · , ul } in the sense of bubble-tree convergence.
Remark 15.1 In fact, 20 =⇒ 2, or the bubble-tree convergence =⇒ varifold convergence.
Bubble-tree convergence: (Definition 3.6 in [3])
Roughly a sequence of uj ∈ W 1,2 (S 2 , M ) is said to bubble-tree converge to a collection of harmonic spheres {ui }li=0 , if
1. uj * u0 weakly(up to a subsequence) in W 1,2 (S 2 );
15
COLDING-MINICOZZI’S MIN-MAX SPHERE (3/1/2012) (BY XIN ZHOU)
44
2. ∃ S0 = {x10 , · · · , xk01 } ⊂ S 2 , such that uj → u0 strongly in W 1,2 (K) for any K compact subset
of S 2 \ S0 ;
3. Near each xi0 ∈ S0 , ∃ Dij : S 2 → S 2 conformal dilation, which takes a small ball centered at
xi0 to the lower-hemisphere, and uj ◦ Dij converges to ui in the sense of step 1 and 2;
4. We have the energy identity:
l
X
lim E(uj ) =
E(ui ).
j→∞
i=1
Key ideas: Variational Method. Given β ∈ Ω, with [β] ∈ π3 (M ) nontrivial,
0). Mollify the minimizing sequence: Find minimizing sequence γ̃ j (t) ∈ [β] such that γ̃ j ∈
C 0 [0, 1], C 2 (S 2 , M ) ;
1). Almost conformal reparametrization: Reparametrize γ̃ j (t) → γ j (t) = γ̃ j hj (t)(·), t , where
hj (t) : (S 2 , g0 ) → (S 2 , γ̃ j (t)∗ g + δj2 g0 ) is continuous 1-parameter family of conformal iso
topies, hence γ j ∈ [β], Area γ j (t) = Area γ̃ j (t) , and
max E γ j (t) − Area γ j (t) → 0;
t∈[0,1]
2). Tightening: γ j (t) → ρj (t), by local harmonic replacement(Perron method), hence ρj ∈ [β],
E ρj (t) ≤ E γ j (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, ∃γ̃ j ∈ [β], with maxt∈[0,1] Area γ̃ j (t) & WA , and γ̃ j (t) ∈ C 0 [0, 1], C 2 (S 2 , M ) .
Step 1: Reparemetrization.
Propostion 15.2 ∃ γ j ∈ [β], Area γ j (t) = Area γ̃ j (t) , and
max E γ j (t) − Area γ j (t) → 0.
t∈[0,1]
In fact, Proposition 15.1 is a direct corollary.
Proof: (of Proposition 15.1) WA ≤ WE is clearly true since Area(·) ≤ E(·). Then
WE ≤ lim [ max E γ j (t) ] = lim [ max Area γ j (t) ] = WA .
j→∞ t∈[0,1]
j→∞ t∈[0,1]
Lemma 15.2 (Uniformization)
• ∀ C 1 metric g on S 2 , ∃! C 1,1/2 isotopy h : (S 2 , g0 ) → (S 2 , g), fixing three points, and conformal
diffeomorphism;
16
INTRODUCTION TO THE WILLMORE CONJECTURE (3/6/2012)
45
• If g1 , g2 are two C 1 metrics, and gi ≥ g0 , let h1 , h2 be the unique conformal isotopic diffeomorphism, then
kh1 − h2 kC 2 ∩W 1,2 (S 2 ,S 2 ) ≤ C(, kgi kC 1 )kg1 − g2 kC 0 .
Sketch of proof: Pull h back to h : (C, dwdw̄) → (C, g = λ2 |dz + µ(z)dz̄|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 ∃ 0 > 0, and continuous Ψ : [0, ∞) → [0, ∞), Ψ(0) = 0 depending on M . ∀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
R
(B) ∀ B finite collection of balls on S 2 , with B |∇ρ(t)|2 < 0 , let v : 81 B → M be the energyminimizing harmonic map, with v| 1 ∂B = ρ(t)| 1 ∂B , then
8
Z
8
|∇ρ(t) − ∇v|2 ≤ Ψ E γ(t) − E ρ(t) .
1
B
8
Harmonic replacement.
16
Introduction to the Willmore conjecture (3/6/2012)
Willmore Conjecture in R3 : Σ2 ⊂ R3 compact, embedded surface, the Willmore energy is defined
by
Z
W (Σ) =
H 2 dΣ,
Σ
where H =
1
2 (k1
+ k2 ) is the normalized mean curvature.
• W (S 2 ) = 4π.
Conjecture: If Σ is a torus, then W (Σ) ≥ 2π 2 , “ = ” only if Σ is the Clifford torus.
Conformal invariance of W : Σk ⊂ (M k+1 , g), A second fundamental form, Å trace-free part of A,
then Ŵ is conformal invariant:
Z
Ŵ (Σ, g) =
|Å|k dµg = Ŵ (Σ, eu g).
Σ
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 ĝ = eu g,
then
∂u
ĥij = eu/2 (hij +
)gij .
∂ν
˚
Hence ĥij = eu/2 h̊ij , and |Å|g = e−u/2 |Å|ĝ .
When k = 2, k1 , k2 principal curvatures,
1
|Å|2 = (k1 − k2 )2 = 2H 2 − 2k1 k2 .
2
Since Gauss curvature K = k1 k2 in R3 ,
Z
Z
1
2
|Å|2 + 2πχ(Σ).
H =
2 Σ
Σ
Consider S 3 : View R3 as stereographic projection to S 3 , which is a conformal transformation.
Z
1
W (Σ) =
(k1 − k2 )2 dΣ + 2πχ(Σ)
4 Σ
Z
1
(4H 2 − 4k1 k2 )dΣ + 2πχ(Σ)
=
4 Σ
Z
=
H 2 − 4(KΣ − 1) + 2πχ(Σ)
Σ
Z
=
(1 + H 2 )dΣ,
Σ
where KΣ is the Gauss curvature of Σ ⊂ S 3 , and KΣ = 1 + k1 k2 by Gauss formula. Hence when
project Σ ⊂ R3 to Σ ⊂ S 3 , the Willmore energy is
Z
W (Σ) = (1 + H 2 )dΣ.
(17)
Σ
• If H = 0, then W (Σ) = |Σ|;
• Clifford torus: let (x, y) ∈ R4 = R2 × R2 , Clifford torus is
√
√
Tc = S 1 (1/ 2) × S 1 (1/ 2) = (x, y) ∈ R4 : |x|2 = |y|2 = 1/2 ⊂ S 3 .
W (Tc ) = A(Tc ) = 2π 2 .
17
17
OUTLINE OF MARQUES-NEVES’S PROOF OF WILLMORE CONJECTURE [2] (3/8/2012)47
Outline of Marques-Neves’s proof of Willmore conjecture [2] (3/8/2012)
• Σ2 ⊂ S 3 , with H the mean curvature, the Willmore energy is defined as
Z
W (Σ) = (1 + H 2 )dΣ.
Σ
W (equator) = 4π; if Σ is not embedded =⇒ W (Σ) ≥ 8π > 2π 2 .
√
√
• Tc2 = S 1 (1/ 2) × S 1 (1/ 2) ⊂ S 3 is the Clifford torus, K = H = 0 flat,
W (Tc2 ) = 2π 2 .
Theorem 17.1 If Σ has genus ≥ 1, then W (Σ) ≥ 2π 2 , with equality only if Σ is a Clifford torus.
Theorem 17.2 If H = 0, and genus ≥ 1, then |Σ| ≥ 2π 2 , with equality only for Clifford torus.
Theorem 17.3 If Σ is embedded, g(Σ) ≥ 1, then ∃ minimal surface Σ̃ with 4π < |Σ̃| ≤ W (Σ).
Remark 17.1 (Hopf-Almgren) Σ ⊂ S 3 , H = 0, and π1 (Σ) = {1}, then Σ is an equator.
Pf: Let f : Σ → S 3 be the immersion, and Π(·, ·) the 2nd f.f. of Σ, then the Hopf-differential
ϕ(z) = Π(
∂f ∂f
,
)dz 2
∂z ∂z
is holomorphic quadratic(In local coordinates, the 2nd f.f. hij is trace-free, and divergence-free
P i
P j
3
j ∇ hjj = ∇H = 0 since S is constant curvature, hence ϕ is holomorphic by
j ∇ hij =
section 13.1). Hence Π ≡ 0.
Let I n = [0, 1]n , and
Φ : I n → Z = Z2 (S 3 ),
where Φ(x) is a surface(current), and Φ is continuous. Let
Π = relative homotopy class of Φ = {Φ0 ∼ Φ},
i.e. ∃ Φt : I n → Z, 0 ≤ t ≤ 1, Φ0 = Φ, Φ1 = Φ0 in I n , and ∀ t ∈ [0, 1], x ∈ ∂I n , we have
Φt (x) = Φ(x).
Definition 17.1 Width of Π:
L(Π) = inf
sup |Φ0 (x)|.
0
Φ ∈Π x∈I n
Theorem 17.4 (Almgren-Pitts) If L(Π) > supx∈∂I n |Φ(x)|, then ∃ smooth minimal embedded Σ̃(possibly
with multiplicity), such that
|Σ̃| = L(Π).
Moreover, Σ̃ is the limit of some min-max sequences, i.e. ∃ xj ∈ I n , such that Φj (xj ) → Σ̃.
17
OUTLINE OF MARQUES-NEVES’S PROOF OF WILLMORE CONJECTURE [2] (3/8/2012)48
Propostion 17.1 F : S 3 → S 3 is conformal and Σ minimal, then |F (Σ)| ≤ |Σ|.
Proof: By conformal invariance of W (Σ),
Z
(1 + H 2 )dΣ = W F (Σ) = W (Σ) = |Σ|.
|F (Σ)| ≤
F (Σ)
Standard family of conformal transformations:
• conformal transformation v ∈ B 4 → 0:
Fv (x) =
1 − |v|2
(x − v) − v,
|x − v|2
• |Fv (Σ)| < |Σ|, if v 6= 0, and if Σ is not S 2 .(?)
Index of minimal surfaces in S 3 :
• Σ2 ⊂ S 3 , H = 0, stability operator Lϕ = 4ϕ + (2 + |A|2 )ϕ, index form:
Z
Z
Q(ϕ, ϕ) = − ϕLϕ = |∇ϕ|2 − (2 + |A|2 )ϕ2 .
Σ
I(Σ) = Index of Σ = # of negative eigenvalues of L.
• I(S 2 ) = 1, Lϕ = 4ϕ + 2ϕ, λ0 = −2, λ1 = 0.
• I(Tc2 ) = 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 ⊂ S 3 , N = (N1 , N2 , N3 , N4 ) is unit normal of Σ, then by the translation invariance of the
cone C(Σ),
4Σ N + |A|2 N = 0.
So
LN = 4N + (2 + |A|2 )N = 2N.
• Each Ni is an eigenfunction with eigenvalue −2. Since Ni changes sign(since N · x = 0 for all
x ∈ Σ), hence not the first eigenvalue, so λ0 < −2.
• Furthermore, N1 , · · · , N4 is linearly independent, unless Σ is S 2 (or N must be constant since
N1 , · · · , N2 already satisfy 4-relations), so
I(Σ) ≥ 5,
unless Σ is S 2 .
17
OUTLINE OF MARQUES-NEVES’S PROOF OF WILLMORE CONJECTURE [2] (3/8/2012)49
Propostion 17.2 (F. Urbano 1990)If Σ is not a Clifford torus, then I(Σ) > 5.
Given Σ embedded in S 3 , 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, and
N (y) the unit normal. Hence if {e1 , e2 } is the o.n. principal basis of Ty Σ,
Dψt |y ei = (cos(t) + sin(t)ki )ei .
So
Z
Area(Σt ) =
Z
det(Dψt )dΣ =
Σ
(cos t + k1 sin t)(cos t + k2 sin t)dΣ,
Σ
while
(cos t + k1 sin t)(cos t + k2 sin t) = cos2 t + (k1 + k2 ) sin t cos t + k1 k2 sin2
|{z}
| {z }
=2H
≤H 2
≤ 1 − sin2 t + 2H cos t sin t + H 2 (1 − cos2 t)
= 1 + H 2 − (sin t + H cos t)2 .
Canonical family:
Given Σ ⊂ S 3 embedded, N =unit normal, define:
Φ : B 4 × [−π, π] → Z2 ,
Φ(v, t) = Fv (Σ)t = Σ(v,t) ,
v ∈ B 4 , t ∈ [−π, π],
where Z2 is the set of 2-currents.
Propostion 17.4
sup
(v,t)∈B 4 ×[−π,π]
|Φ(v, t)| ≤ W (Σ).
18
18
MARQUES-NEVES’S PAPER 1 (3/13/2012)
50
Marques-Neves’s paper 1 (3/13/2012)
Let Σ ⊂ S 3 be embedded and g(Σ) ≥ 1.
Making Σ(v,t) continuous:
3 and S 3 be the connected components of S 3 \ Σ, with N pointing to S 3 . Then
Denote S+
−
+
v→p
lim Σ(v,t) = ∂Bt (−p),
3
for p ∈ S+
;
lim Σ(v,t) = ∂Bπ+t (p),
3
for p ∈ S−
.
v→p
When p ∈ Σ and v → p with angle θ, i.e. informally v − p forms an angle θ with the position vector p,
lim
v→p(with angle θ)
Σ(v,t) = ∂B π2 −θ+t (− sin(θ)p − cos(θ)N (p)).
(It is easy to check that the above are consistent when θ = π2 or θ = − π2 .)
Now introduce the polar coordinates near Σ. Given (s, θ) ∈ [0, ) × [−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 ∈ B 4 near Σ, let s(x) = d(x, S 3 ) and θ the signed
x
distance between |x|
and Σ, say θ = distΣ (x/|x|, p) for some p ∈ Σ, then x = P (s, θ, p).
Definition 18.1 Define the neighborhood of Σ as:
N (Σ) = {P (s, θ, p) : p ∈ Σ, s ≥ 0, s2 + θ2 < 2 }.
It is not hard to find a continuous map
F : B4 → B4,
such that F : B 4 \N (Σ) → B 4 is a diffeomorphism; F : S 3 → S 3 is identity and F : N (Σ) → S 3 is
given 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(v, t) = Σ(F (v),t) when v ∈ B 4 \ N (Σ);



 C(v, t) = ∂B (−v) when v ∈ S 3 \ N (Σ);
t
+
C : B 4 ×[−π, π] :
3

C(v, t) = ∂Bπ+t (v) when v ∈ S− \ N (Σ);


 C(v, t) = ∂B π
−θ+t (− sin(θ)p − cos(θ)N (p)) when v = F (s, ϕ, p) ∈ N (Σ),
2
where θ = tan−1 ( √
ϕ
).
2 −ϕ2
Given x ∈ (∂B 4 ∪ N (Σ)), ∃ ! Q̄(x) and s(x), so that
C(x, s(x)) = ∂Bπ/2 (Q̄(x)),
has radius π2 . Here s : (∂B 4 ∪ N (Σ)) → [−π/2, π/2] and Q̄ : (∂B 4 ∪ N (Σ)) → S 3 is given in the
definition of C.
Key Property: Q̄ : S 3 → S 3 is continuous and has degree g(will be proved in the next section).
18
MARQUES-NEVES’S PAPER 1 (3/13/2012)
51
Definition 18.2 The canonical family Φ : I 5 → Z2 associated to Σ ⊂ S 3 is defined as:
Φ(v, t) = C f (v), (2t − 1)π + s(f (v)) , (v, t) ∈ I 4 × I = I 5 ,
where f : I 4 → B 4 is a diffeomorphism and s : B 4 → [−π/2, π/2] is an extension of s : (∂B 4 ∪
N (Σ)) → [−π/2, π/2].
• Φ is continuous on I 5 ;
• Φ(p, 1/2) is an equator when p ∈ ∂I 4 ;
• Φ(v, 0) = Φ(v, 1) = 0 in Z2 .
Definition 18.3 Let
Π = relative homotopy class of Φ(fixed on ∂I 5 ).
Theorem 18.1 If genus g(Σ) ≥ 1, the width L(Π) > 4π.
Proof: Since maxx∈∂I 5 |Φ(x)| = 4π, if the statement is false, then ∃ ϕi ∈ Π, such that
1
max |ϕi (x)| ≤ 4π + .
i
x∈I 5
Now Φ∂I 5 = Φ(I 4 ×{0})∪(I 4 ×{1})∪(∂I 4 ×I) , where
|Φ| : ∂I 4 × {1/2} → RP3 = space of unoriented equators in S 3 ,
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 ⊂ I 5 with ∂S = ∂I 4 × {1/2}.
• Given > 0, let
A(i) = connected component of {t = 0} ⊂ I 5 in {x ∈ I 5 : dV |ϕi (x)|, Z0 > },
where dV is the varifold distance and Z0 is the space of unoriented equators.
• Claim: For i sufficient large, A(i) ∩ {t = 1} ⊂ I 5 = ∅.
• 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 ∂I 4 × I, hence homotopically nontrivial. Then
max |ϕi |(γi (t)) ≤ 4π + 1/i.
t∈[0,1]
If we run the min-max theory on Π, we must get a nontrivial embedded minimal surface Σ, since
Π1 is nontrivial, and Σ must have area less or equal to 4π, hence an equator. Furthermore, the
min-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).
19
MARQUES-NEVES’S PAPER 2 (3/15/2012)
52
So A(i) gives an continuous extension of Φ|∂I 4 ×I , hence a contradiction.
Proof: (of Theorem 17.2) ∃ connected smooth oriented embedded Σ of least area.
• Construct Φ : I 5 → Z2 (S 3 ), hence maxx∈I 5 |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) = Σ̃ < |Σ|,
where Σ̃ is a connect smooth oriented embedded minimal surface by the discussion in the next
section, a contradiction to the minimality of Σ.
Proof: (of Theorem 17.3) Given Σ oriented embedded surface of genus g ≥ 1 in S 3 , we can similarly
construct the canonical family Φ : I 5 → Z2 (S 3 ), and maxx∈I 5 |Φ(x)| ≤ W (Σ). Run the min/max,
we get a embedded minimal surface Σ̃, such that |Σ̃| ≤ maxx∈I 5 |Φ(x)| ≤ W (Σ).
19
Marques-Neves’s paper 2 (3/15/2012)
Let Φ : I 4 × I → Z2 be the canonical family. For any p ∈ ∂I 4 , Φ(p, t) are spheres, with Φ(p, 1/2)
an equator. Denote Q̄ : S 3 → S 3 the map from S 3 ∼ ∂I 4 to the center of Φ(p, 1/2) given above.
Theorem 19.1
deg(Q̄) = g.
3 ∪ S 3 , we can introduce
Proof: Since S 3 \ Σ = S+
−
π π
3
3
S˜3 = S+
∪ Σ × [− , ] ∪ S−
.
2 2
The map Q̄ can be views as Q̄ : S˜3 → S 3 , given by

3

 −p, when p ∈ S+ ;
Q̄(p) =
− sin(θ)p − cos(θ)N (p),

 p, when p ∈ S 3 .
−
So
1
deg(Q̄) = 2
2π
when (p, θ) ∈ Σ × [−π/2, π/2];
Z
S˜3
(det(dQ̄))dµ.
Orientation on S 3 : a basis v1 , v2 , v3 ∈ Tp S 3 is positively oriented ⇐⇒ {v1 , v2 , v3 , p} is positively
oriented in R4 .
19
MARQUES-NEVES’S PAPER 2 (3/15/2012)
53
• Map p → −p is positively oriented on S 3 , since if {e1 , e2 , e3 } positive on Tp S 3 , then {−e1 , −e2 , −e3 , −p}
positively on R4 .
• So
Z
Z
3
3
det(dQ̄)dµ.
det(dQ̄)dµ = |S+ | + |S− | +
{z
}
|
Σ×[−π/2,π/2]
S˜3
2π 2
• Given p ∈ Σ, let e1 , e2 be an positively oriented o.n. principal basis, i.e. Dei N (p) = ki ei where
k1 , k2 are principal curvatures of Σ. Then {e1 , e2 , N (p), p} forms a positively oriented basis of
R4 .
∂
{e1 , e2 , ∂θ
} forms a positive basis for Σ × [−π/2, π/2].
• Since Q(p, θ) = − sin(θ)p − cos(θ)N (p),
dQ̄(ei ) = − sin(θ)ei − cos(θ)ki ei = (− sin(θ) − cos(θ)ki )ei ;
dQ̄(
∂
) = − cos(θ)p + sin(θ)N (p).
∂θ
• Note {e1 , e2 , − cos(θ)p + sin(θ)N (p)} forms an o.n. basis at TQ̄(p,θ) S 3 . Furthermore, {e1 , e2 , − cos(θ)p+
|
{z
}
e3
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
∂
) = −deg(dQ̄)e1 ∧ e2 ∧ e3 ,
dQ̄(e1 ∧ e2 ∧
∂θ
and we derive
deg(dQ̄) = −(sin θ + k1 cos θ)(sin θ + k2 cos θ) = −(sin2 θ + sin θ cos θ(k1 + k2 ) + cos2 θ).
Hence
Z
Z
deg(dQ̄) = −
Σ×[−π/2,π/2]
(
Σ
π π
π
+ k1 k2 )dµΣ = −
2
2
2
Z
Σ
(1 + k1 k2 )dµΣ
| {z }
KΣ
π
= − 2π(2 − 2g) = 2π 2 (g − 1).
2
Adding all the above together, we finish the proof.
Doing the Min-max: Let Zk (M n ) be the space of integral currents with flat topology, which roughly
means that Σi → Σ if Σi − Σ = ∂Ri and |Ri | → 0.
Theorem 19.2 (Almgren)
πn−k (Zk (M n )) = Hn (M, Z) = Z.
19
MARQUES-NEVES’S PAPER 2 (3/15/2012)
•
•
•
•
•
54
Φ : S n−k → Zk (M n ) 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 : I 5 → Z2 (S 3 ) is continuous in flat topology.
• t → Σt is continuous in flat norm. If t1 < t2 , then Σt2 − Σt1 = ∂{t1 < d(x, Σ) < t2 } has small
volume when t2 − t1 is small;
• v → Σv,t is continuous even when v → S 3 ;
• 2π 2 ≥ L(Π) > 4π if Σ has area less or equal to 2π 2 ⇐⇒ 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 ⊂ S 3 embedded, and H = 0. I(Σ) ≤ 5 ⇐⇒ Σ = S 2 , Tc2 .
If I(Σ) > 5, then λ0 < −2, and λ1 = −2 has a 4 dimensional eigenspace generated by the
normal 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
R
eigenfunction of λ0 be u0 . If I(Σ) = 5, then every function ϕ such that Σ ϕu0 = 0 must
satisfy:
Z
Z
Q(ϕ, ϕ) =
(|∇ϕ|2 − (2 + |A|2 )ϕ2 ) ≥ −2
ϕ2 .
Let ψ : Σ → S 3 , then ∃v ∈ B 4 , with Fv defined in Section 17, such that
Fv ◦ ψ = (ψ̃1 , · · · , ψ̃4 ), then
4 Z
X
i=1
[|∇ψ̃i |2 − (2 + |A|2 )ψ̃i2 ] ≥ −2
Σ
4
X
R
Σ Fv
◦ ψ = 0. Denote
ψ̃i2 .
i=1
Hence
Z
2|Σ| = 2E(ψ)
≥
|{z}
“= only if F =id
E(Fv ◦ ψ) =
2
Z
|∇Fv ◦ ψ| ≥
Σ
Σ
2
|A| = 2|Σ| −
|{z}
Z
Kda.
Σ
2(1−K)
So Σ Kda ≥ 0, hence Σ is either S 2 or a torus. When Σ is a torus, Fv = id, hence ψ =
(x1 , · · · , x4 ), which are eigenfunctions of 4xi = −2xi , so
R
4x = −2x = −|A|2 x.
Hence |A|2 = 2, so K = 0.
REFERENCES
55
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