Master of Science in
Advanced Mathematics and
Mathematical Engineering
Title: Some ODE solutions for the fractional Yamabe problem
Author: Azahara de la Torre Pedraza
Advisor: María del Mar Gónzalez Nogueras
Department: Matemàtica Aplicada I
Academic year: 2013
Universitat Politècnica de Catalunya
Facultat de Matemàtiques i Estadı́stica
Master thesis
Some ODE solutions for the fractional Yamabe
problem
Azahara de la Torre Pedraza
Advisor: Marı́a del Mar Gónzalez Nogueras
Matemàtica Aplicada I, Universitat Politècnica de Catalunya
To my parents, my sister, “señorita Paquita”,
Francisco Martı́n and the cookie-time :)
Acknowledgements
First I would like to thank my advisor Marı́a del Mar González Nogueras for all her dedication and
for treating me like a daughter. She has always helped me not only when working in mathematics,
but also at personal level. Without her help none of this work would have been possible.
I am indebted to Francisco Martı́n and Xavier Cabré for giving me the opportunity of doing what
I really like in my life.
I also would like to thank Robin Graham for his useful suggestions, and to Alice Chang for her
support during my Princeton visit.
I am extremely grateful to my sister, Gloria de la Torre, for all the suppor she has given me, first
during elemtary school, later as a roommate during my years as an undergraduate student and
most recently for learning mathematics with me, when she was proofreading all of my works.
Many thanks go to my master classmates for tolerating all my disruptions or “perturbations” when
I was asking about the meaning of all the English words that our professors used in lectures. I
am also gratefull to them not only for colaborating in mathematics, but also for being really good
friends.
Mi más sincero agradecimiento a mis padres, a ellos le debo todo lo que tengo y pueda tener. A
mi madre por dedicar su vida a educarnos a mi hermana y a mı́, y ası́ conseguir que dejara de
escaparme de una silla para no hacer “la tarea” y lograra llegar a donde estoy. A mi padre por
ser quien respondı́a a todos mis “¿por qué?”, pregunta que me ha llevado a elegir a qué dedicar
mi vida. Gracias también por todos vuestros consejos que tanto me han ayudado.
Especial agradecimiento a la “señorita Paquita”, quien despertó en mı́ el gran interés que siento
por las matemáticas.
También quiero agradecer enormemente a Vı́ctor Arias el apoyo que siempre me dio y que tanto
me ayudó a seguir adelante en mi carrera. Sin él no habrı́a llegado a donde estoy.
No puedo dejar atrás a Fran Jiménez, muchı́simas gracias por el apoyo incondicional en todo momento.
También me gustarı́a agradecer a todos mis compañeros de licenciatura, porque juntos hemos descubierto el camino que cada uno querı́a seguir, y en especial, gracias a quién hizo posibles esas
tardes de estudio geométrico tras una visita a “La góndola”.
Muchı́simas gracias a Antonia Ramirez, y a toda su familia, por acogerme como a una hija y hacer
posible mi estancia en Barcelona.
v
vi
ACKNOWLEDGEMENTS
Per últim, voldria agrair a tothom de MA1 per acollir-me i fer que em sentı́s com a casa i, en
especial als del nostre grup per triar-me, acollir-me i ajudar-me des del primer moment. I també
al Vı́ctor, a l’Ori, a l’Adrià i a la Patricia, per aquells menjars guisats amb cassola, pels cafès i
berenars, i per les classes de català que em van ensenyar a diferenciar un “fontanero” de diversos
“fontanerò”.
Abstract
Keywords: Yamabe problem, fractional curvature, conformal fractional Laplacian, isolated singularities,
Anti-de-Sitter space, radial solutions, fractional ODE.
MSC2010: 2010 35J70, 53A30, 35R11.
We construct some ODE solutions for the fractional Yamabe problem in conformal geometry. The fractional
curvature, which is a generalization of the usual scalar curvature, is defined from the conformal fractional
Laplacian, which is a non-local operator construced on the conformal infinity of a conformally compact
Einstein manifold.
On one hand, we consider the hyperbolic manifold S1 (L) × R3 and study the nonuniqueness of solutions
for the fractional Yamabe problem. On the other hand, we look at the existence of radial solutions for
the Yamabe problem in Euclidean space with an isolated singularity at the origin. Both equations are
fractional order ODE for which new tools need to be developed.
Contents
Acknowledgements
v
Introduction
1
Chapter 1. Preliminaries
1. Introduction to Riemannian Geometry
2. The Laplacian operator on manifolds
3. Sobolev spaces
4. An introduction to the Yamabe problem
3
3
5
6
7
Chapter 2. The conformal fractional Laplacian and the fractional Qγ -curvature (Background).
1. Fractional Laplacian in Rn .
2. Geometric setting
3. Definition of the conformal fractional Laplacian
15
15
21
24
Chapter 3. Non uniqueness for the fractional Yamabe problem
1. The sign of the Yamabe constant.
2. Classical case of non uniqueness
3. Non uniqueness for the fractional case
35
35
38
40
Chapter 4. Isolated singularities
1. Classical case
2. Relation with chapter 3
3. Fractional case
47
47
48
50
Chapter 5. Anti-de-Sitter space
1. Motivation from physics (informal)
2. Rigorous definition
3. Uniqueness
51
51
52
53
References
57
i
Introduction
This master thesis is divided into five chapters, in which we want to construt some ODE solutions
for the fractional Yamabe problem. However, this is a fractional equation, so usual ODE method
do not apply directly. First we present some background about this topic. We can find it in
the first two chapters. In the following chapters, four and five, we present our main work and
the new results we have obtained. The last chapter is independent and shows some interesting
constructions.
The first chapter is a review of analitic and geometric concepts and the notation that we are going to
use along the project. In the second one, we introduce the fractional Laplacian operator in Rn with
the three different definitions and present the equivalence between them. After that, we explain
the definition of a conformally compact Einstein manifold, and we introduce our main operator,
the conformal fractional Laplacian. There exists a strong link between conformal structures on
boundary manifolds and Einstein metrics on complete manifolds. For example, the round conformal
sphere can be efficiently described as the boundary at infinity of the unit ball with its hyperbolic
metric. Then, conformal geometry of the sphere and Riemannian geometry of hyperbolic space
are intimately tied. The conformal fractional Laplacian is constructed on the conformal infinity of
a conformally compact Einstein manifold. The fractional curvature, which is a generalization on
scalar curvature, is constructed from the conformal fractional Laplacian.
In the main chapters of the project, chapters four and five, we present two different constructions
which give us radial solutions with constant curvature in classical and fractional Yamabe problem.
On one hand, we consider the hyperbolic manifold S1 (L) × R3 , given as a quotient of hyperbolic
space; and we study the nonuniqueness solution for the fractional Yamabe problem, showing also
the existence of a Hamiltonian quantity preserved along the trajectories.
On the other hand, we look at the existence of radial solutions for the Yamabe problem in the
Euclidean space with an isolated singularity at the origin. For that it is enough to prove that
this problem is equivalent to the previous one, and so that the new idea of finding a Hamiltonian
to understand a fractional order ODE, studied in [24, 11], is supported by the existence of a
Hamiltonian quantity preserved along trajectories.
In the last chapter we want to study the dependence of the conformal fractional Laplacian on the
chosen extension. For that, we introduce the Anti-de-Sitter, showing two different conformally
compact Einstein manifolds with the same conformal infinity.
1
Chapter 1
Preliminaries
In this chapter we are going to follow the notation and definitions given in the books by Aubin
[5, 6].
1. Introduction to Riemannian Geometry
Note that we are using the Einstein summation convention.
Definition 1.1. A connection on a differentiable manifold M is a mapping D (called the covariant
derivative) of T (M ) × Γ(M ) into T (M ) which has the following properties:
•
•
•
•
If X ∈ TP (M ), then D(X, Y ) (denoted by DXY ) is in TP (M ).
For any P ∈ M the restriction of D to TP (M ) × Γ(M ) is bilinear.
If f is a differentiable function, then D(f Y ) = X(f )Y + f DY.
If X and Y belong to Γ(M ), X class C r and Y of class C r+1 , then DY is in Γ(M ) and is of
class C r .
where Γ(M ) denotes the vector space of vector fields on M .
Definition 1.2. A Riemannian metric is a twice-covariant tensor field g such that at each point
P ∈ M , gE is a positive definite bilinear symmetric form.
Definition 1.3. In a Riemannian manifold we can also define the torsion tensor: It is a (1, 2)−tensor
which depends on the connection D in the following way T (X, Y ) = DX Y − DY X − [X, Y ]
Definition 1.4. The Riemannian connection is the unique connection with vanishing torsion
tensor, for which the covariant derivative of the metric tensor is zero. (∇g = 0)
In the following, M will always be an oriented Riemannian manifold of dimension n. Since the
Riemannian connection has no torsion we can define the Christoffel symbols in a local coordinate
system as
(1)
Γlij =
1
[∂i gkj + ∂j gki − ∂k gij ]g kl ,
2
where g kl are, by definition, the components of the inverse matrix of the matrix (gij )ij .
3
4
1. PRELIMINARIES
Definition 1.5. A volume form on M , given in an oriented coordinate system {xi } is
p
(2)
dvolg := |g|dx1 ∧ ... ∧ dxn ,
∂
where the dxi are the 1−forms forming the dual basis to the basis vectors ∂i := ∂x
i and ∧ is the
wedge product. We are going to denote by |g| the absolute value of the determinant of the metric
tensor gij .
Definition 1.6. Given a 2−tensor E we define its contraction or trace as:
X
tr(Eij ) =
g ij Eij .
i,j
Definition 1.7. The curvature of a connection D is a 2−form with values in Hom(Γ, Γ) defined
by
(X, Y ) → Riem(X, Y ) = DX DY − DY DX − D[X,Y ] .
For the definition we suppose that the vector fields are at least C 2 .
Definition 1.8. The curvature tensor is the 4−covariant tensor Riem(X, Y, Z, T ) = g[Riem(X, Y )T, Z];
its components are Riemijkl = gim Riemm
jkl . It has the properties:
• Riemijkl = −Riemijlk ,
• Riemijkl = Riemklij .
According to the expression of the components of the curvature tensor and considering a normal
coordinate system around P , we can assert
(3)
Riemlkij (P ) = (∂i Γljk )P − (∂j Γlik )P .
Definition 1.9. The sectional curvature of a 2−dimensional subspace of T (M ) defined by vectors
X and Y , where X is orthonormal to Y (i.e., g(X, X) = 1, g(Y, Y ) = 1, g(X, Y ) = 0), is
(4)
σ(X, Y ) = Riem(X, Y, X, Y ).
If X, Y are not orthonormal, the definition is
(5)
σ(X, Y ) =
Riem(X, Y, X, Y )
.
g(X, X)g(Y, Y ) − (g(X, Y ))2
We can obtain, by contraction, the so called Ricci tensor, whose components are
(6)
Ricgij = Riemkikj = Riemiklj g lk .
(The Ricci tensor is symmetric).
The contraction of the Ricci tensor is called the scalar curvature.
(7)
Rg = Ricgij g ij .
Definition 1.10. The Weyl tensor (or tensor of conformal curvature) is defined by its components
in a local chart as follows
1
g
(Ricgik gjl − Ricgil gjk + Ricgjl gik − Ricgjk gil )
Wijkl
=Riemgijkl −
n−2
(8)
Rg
(gjl gik − gjk gil ).
+
(n − 1)(n − 2)
2. THE LAPLACIAN OPERATOR ON MANIFOLDS
5
Definition 1.11. We define the trace-free Ricci tensor as
Eg = Ricg −
Rg
g.
n
Note that it satisfies the following properties:
• tr(E) = 0.
• div(E) = n−2
2n dRg , where the divergence will be defined in (9).
Definition 1.12. A conformal map is a transformation which preserves angles. Given two metrics
g and g̃, they are conformally related if g̃ = f g with f > 0.
4
We usually write the conformal change as gu = u (n−2) g, and we will denote [g] the class of all
metrics conformal to g.
2. The Laplacian operator on manifolds
2.1. Laplace-Beltrami. The divergence of a vector field X (divX) on a manifold is defined as
the scalar function with the property
(divX)volg := Lvolg
where L is the Lie derivative along the vector field X. In local coordinates we obtain
p
1
(9)
divX = p ∂i ( |g|X i ).
|g|
On the other hand the gradient of a scalar function f is the vector field grad f that may be defined
through the inner product < ·, · > on the manifold, as
< gradf (x), vx >= df (x)(vx ),
for all vectors vx ∈ Tx M ; where df is the exterior derivative of the function f . So in local
coordinates, we have
(grad)i f = ∂ i f = g ij ∂j f.
We will denote it by ∇f . We also write
< ∇w, ∇v >g = ∇i w∇i v and |∇f |2g = ∇i ∇i f.
The Laplace-Beltrami operator on a manifold M is defined as:
(10)
4g f = div gradf.
Combining the definitions of the gradient and divergence, we can give an explicit formula, in local
coordinates, for the Laplace-Beltrami operator 4g (10) applied to a scalar function f :
(11)
p
1
4g f = div gradf = p ∂i ( |g|g ij ∂j f ).
|g|
6
1. PRELIMINARIES
2.2. Conformal Laplacian.
Definition 2.1. The conformal Laplacian operator for a Riemannian metric g in a manifold X
of dimension n is defined as
4(n − 1)
(12)
Lg = −cn 4g + Rg , where cn =
.
(n − 2)
The conformal Laplacian is a conformally covariant operator, indeed:
4
Proposition 2.2. Given g̃, g two conformally related metrics with g̃ = u n−2 g, then the operator
Lg satisfies.
(13)
Lg̃ (ϕ) = u
−(n+2)
n−2
Lg (uϕ)
∞
for every ϕ ∈ C (M ). In the case ϕ = 1 we obtain the classical scalar curvature equation.
(14)
n+2
Lg (u) = Rg̃ u n−2 .
Proof. We will only present the proof of (14). If we denote Γ̃lik and Γlik the Christoffel symbols
corresponding to g̃ and g, respectively; because of g̃ = ef g, we obtain that
1
∂g̃mi
∂g̃mk
∂g̃ik
1
∂gmi
∂gmk
∂gik
Γ̃lik − Γlik = g̃ lm (
+
−
) − g lm (
+
−
)
2
∂xk
∂xi
∂xm
2
∂xk
∂xi
∂xm
1
∂ef gmi
∂ef gmk
∂ef gik
1
∂gmi
∂gmk
∂gik
(15)
= e−f g lm (
+
−
) − g lm (
+
−
)
2
∂xk
∂xi
∂xm
2
∂xk
∂xi
∂xm
1
= g lm (∂k f gmi + ∂i f gmk − ∂m f gik ).
2
Using (6),
1
n−2
n−2 s
n−2
∇k ∇j f + 4g f gjk +
∇k f ∇j f −
∇ f ∇s f gjk .
(16)
Ricg̃kj − Ricgkj = −
2
2
4
4
Thus if we use g̃ = ef g and we contract by g kj , we obtain
(n − 2)(n − 1) s
Rg̃ ef − Rg = (n − 1)4g f −
∇ f ∇s f.
4
4
Substituting the change f = n−2
log u, we find that
(17)
n+2
−cn 4g u + Rg u = Rg̃ u n−2 .
And because of the definition of conformal Laplacian (12) we have proved the result.
t
u
3. Sobolev spaces
Let k be a non negative integer and 1 ≤ p ≤ ∞.
Let W k,p (M ) denote be the Sobolev space consisting of all real-valued functions on M whose first
k weak derivatives are functions in Lp . In particular,
• in the Hilbert space W 1,2 we have the norm
1
kukW 1,2 = (k∇uk22 + kuk22 ) 2 .
4. AN INTRODUCTION TO THE YAMABE PROBLEM
7
• for any real number γ > 0 not necessary integer, we denote H γ = W γ,2 , where the norm is
γ
given by kf kH γ (M ) = kf k2 + k(−4) 2 f k2 . In Rn , using Fourier transform we have
Z
γ
k(−4) 2 f k22 =
|ξ|2γ fˆ2 (ξ) dξ.
Rn
∗
Definition 3.1. We define p as the real number which satisfies
1 k
1
= − .
∗
p
p n
(18)
Theorem 3.2. Sobolev embedding theorem for compact manifolds.
Suppose M is a compact Riemannian manifold of dimension n.
(a) Let k <
n
p
For all 1 ≤ q ≤ p∗ , W k,p (M ) is continuously embedded in Lq (M ).
(b) Suppose 0 < α < 1, and
1
k−α
≤
.
p
n
Then W k,p (Rn ) is continuously embedded in C α (Rn ).
(c) Kondrakov theorem. Let k >
1 ≤ q < p∗ .
n
p,
W 1,p (M ) is compactly embedded in Lq (M ) for every
Theorem 3.3. Sobolev embedding theorem for Rn .
If 1 ≤ q ≤ p∗ , W k,p (Rn ) is continuously embedded in Lq (Rn ). In particular, for p = 2, k = 1, q =
2n
2∗ = n−2
, we have the following Sobolev inequality:
Z
(19)
kuk22∗ ≤ σn
|∇u|2 dx,
u ∈ W 1,2 (Rn ).
Rn
We will call the smallest such constants σn the n−dimensional Sobolev constant.
Note that in theorem 4.3 we will give its exact value.
Theorem 3.4. Sharp Sobolev inequality for manifolds.
Let M be a compact Riemannian manifold with metric g, 2∗ =
previous theorem) be the best Sobolev constant.
2n
n−2 ,
and let σn (defined in the
Then for every ε > 0 there exist a constant Cε such that for all ϕ ∈ C ∞ (M ),
Z
Z
2
2
kuk2∗ ≤ (1 + ε)σn
|∇u|g dvolg + Cε
u2 dvolg .
M
M
Definition 3.5. Let V ⊂ M be a Riemmanian submanifold. If f is C k function on M , the trace
f˜ := T f is the restriction of f to V and f˜ ∈ C k . If f ∈ W k,p (M ) it is possible define the trace f˜
of f on V by a density arguments.
4. An introduction to the Yamabe problem
A very good reference for studing the classical Yamabe problem is [31].
8
1. PRELIMINARIES
4.1. The Problem. The problem proposed by Yamabe is: given a compact Riemannian manifold
(M n , g) of dimension n ≥ 3, find a new metric g̃ conformal to g with constant scalar curvature.
This is a geometric problem, but we are going to transform it into a PDE problem.
For that, let Rg be the scalar curvature of (M n , g), and we suppose it is not constant (because in
4
otherwise the problem is solved). We consider the conformal metric g̃ = u n−2 g, where u is a C ∞
and strictly positive function on M .
Because of proposition 2.2 we obtain that the Yamabe problem (with a conformal metric) is
equivalent to proving that the equation
n+2
−cn 4g u + Rg u = Rg̃ u n−2 ,
(20)
with Rg̃ =constant, has a C ∞ solution; and that this solution is strictly positive.
Proposition 4.1. If we have two solutions of equation (20), the constant curvatures for both must
have the same sign (or both equal to zero).
4
Proof. We assume that one solution is ϕ and so that g 0 = ϕ n−2 g has constant scalar curvature
4
Rg0 , and we assume also that other solution is γ and so that g̃ = γ n−2 g has constant scalar
curvature Rg̃ .
Since ϕ and γ are strictly positive, because the equivalence between the metrics, we can set ϕ = γψ,
where ψ is a C ∞ and strictly positive function on M , and we obtain
4
4
4
g 0 = ϕ n−2 g = (γψ) n−2 g = ψ n−2 g̃
and therefore
n+2
−cn 4g̃ ψ + Rg̃ ψ = Rg0 ψ n−2 .
R
If we integrate it respect to the metric g̃ and we use 4g̃ ψ dvolg̃ = 0 (because of divergence
theorem and because M is a compact closed manifold) we obtain that
Z
Z
n+2
0
(22)
Rg̃
ψ dvolg̃ = Rg
ψ n−2 dvolg̃ .
(21)
M
M
Thus we can assert that Rg and Rg0 have the same sign (or both equal to zero).
t
u
4.2. The variational method. One possible way to prove the existence of a solution to (20)
with Rg̃ constant is to use the variational method.
Let A (to be chosen later) be a set of functions and I the functional
Z
(23)
I(u) =
(cn |∇u|2g + Rg u2 ) dvolg , u ∈ A,
M
4
Using that g̃ = u n−2 g we may write
dvolg̃ =
p
2n
|g̃| dvolg = |u| n−2 dvolg ,
And since we are looking for a metric with volume constant equal to one, we impose the constrain
Z
∗
1
1
(24)
K(u) = ∗
|u|2 dvolg = ∗ .
2 M
2
4. AN INTRODUCTION TO THE YAMABE PROBLEM
9
We can replace the functional I and its constraint K[u] = 21∗ by the functional
R
(cn |∇u|2g + Rg u2 ) dvolg
(25)
J[u] = M R
2
( M |u|2∗ dvolg ) 2∗
Lemma 4.2. The Euler-Lagrange equation for functional (25) is precisely (20) with Rg̃ constant.
Remark 4.3. We are going to minimize the functional (25) applied to functions, but we could also
define it as acting on metrics
R
R dvolg̃
˜ = M
R g̃
for all g̃ ∈ [g].
J[g̃]
dvolg̃
M
4
2n
Indeed, if gu ∈ [g], gu = u (n−2) g and so that, dvolgu = u n−2 dvolg . Therefore,
Z
Z
Z
n+2
2
2
I[u] =
(cn |∇u| + Rg u ) dvolg =
uLg u dvolg =
uRgu u n−2 dvolg
M
M
ZM
Z
2∗
=
Rgu u dvolg =
Rgu dvolgu
M
M
and
Z
K[u] =
dvolgu .
M
Remark 4.4. We can suppose without loss of generality that the volume of the manifold M with
the metric gu is equal to 1.
Indeed, we can check that the functional J is invariant by rescaling:
We call ũ = λu and we compute the functional evaluated in ũ;
R
R
λ2 M (cn |∇u|2g + Rg u2 ) dvolg
(c |∇(λu)|2g + Rg (λu)2 ) dvolg
M n
=
= J[u].
(26) J[ũ] = J[λu] =
R
R
2
2
2
( M |λu|2∗ dvolg ) 2∗
(λ2∗ ) 2∗ ( M |u|2∗ dvolg ) 2∗
Now, if we take ũ =
R
M
1
u
(u2∗ ) dvolg
we obtain J[ũ] = J[u] and vol(ũ) = 1.
Lemma 4.5. I[u] (similarly J[u]) is bounded from below.
Proof. If we call V =
R
M
dvolg = 1, we can assert that
Z
(27)
I(u) = cn
|∇u|2g dvolg +
M
|
{z
}
Z
Rg u2 dvolg ≥
M
Z
Rg u2 dvolg ≥ min{Rg , 0}kuk22 .
M
≥0
∗
And using Hölder’s inequality with p = 22 and q = n2 we obtain
Z
2/2∗ Z
2/n
Z
2
2
2∗
(28)
kuk2 =
u dvolg ≤
u dvolg
1 dvolg
= kuk22∗ .
M
M
M
Since min{0, Rg } ≤ 0 we can assert
I(u) ≥ min{Rg , 0}kuk22∗ .
So we have I(u) ≥ inf{Rg , 0}.
t
u
10
1. PRELIMINARIES
Definition 4.6. Given a manifold (M, g) we define the Yamabe constant as:
λ(M ) := λ(M, g) = inf{J[u]; u is smooth on (M,g)}.
Remark 4.7. We can assert that the sign of λ(M ) is equal to the sign of Rg̃ (which is constant).
If we find a minimizer for (25), then it will be a solution for the Yamabe problem. We take a
minimizing sequence {ui } (i ∈ N), it means {ui } ⊂ A such that limi→∞ J(µi ) = µ = inf A J(u),
and we want to find a subsequence {uj } ⊂ {ui } which converges to a strictly positive solution of
(20) with Rg̃ constant.
Making sense with this theorem we will choose A = {u ∈ W 1,2 , u ≥ 0 and kuk2∗ = 1}
We can restrict to positive solutions because if u ∈ W 1,2 , then |u| ∈ W 1,2 and |∇|uk = |∇u| almost
everywhere, so J(u) = J(|u|).
4.3. The sphere. The analysis of the Yamabe equation (20) depends on the model case of the
sphere Sn with its standard metric gc . So we are going to describe the solution to the Yamabe
problem on Sn and prove that the infimum of the Yamabe functional (25) in this case is realized
by the standard metric on the sphere. We will also show the relation with the sharp form of the
Sobolev inequality in Rn .
We call σ the stereographic projection (a conformal diffeomorphism) defined by
σ : Sn − {P } → Rn ,
σ(z1 , ..., zn , ξ) = (x1 , ...xn ),
where P = (0, ..., 0, 1) is the north pole on Sn ∈ Rn+1 ,
xj =
z−j
, j ∈ {1, ...n}
(1 − ξ)
and (z, ξ) ∈ Sn − {P }.
We will denote gc the standard metric on Sn , g0 the Euclidean metric on Rn and ρ = σ −1 . Under
σ, gc corresponds to
4
ρ∗ gc =
g0 .
2
(|x| + 1)2
If we call
u1 (x) = (kxk2 + 1)(2−n)/2 ,
(29)
we obtain ρ∗ gc = (4
n−2
4
4
u1 ) n−2 g0 .
Moreover using the stereographic projection we can write down conformal diffeomorphisms of the
sphere, which are generated by the rotations and maps of the form σ −1 τv σ or σ −1 δµ σ, where τv ,
δµ are respectively translation by v ∈ Rn and dilation by µ > 0:
τv , δµ : Rn → Rn ,
τv (x) = x − v,
δµ = µ−1 x.
4. AN INTRODUCTION TO THE YAMABE PROBLEM
11
Under dilations, the spherical metric on Rn , (ρ∗ gc ) is transformed in
(n−2)
2
4
µ
n−2
∗ ∗
(30)
δµ ρ gc = 4uµ g0 , where uµ (x) =
.
2
2
|x| + µ
Theorem 4.8. [36]
If g is a metric on Sn that is conformal to the standard metric gc and has constant scalar curvature,
then up to a constant factor, g is obtained from gc by a conformal diffeomorphism of the sphere.
Proof. (Obata). The trace-free Ricci tensor (1.11) satisfies also the following property:
Given the conformal metric g = v −2 g0 , we have
Eg = −(n − 2)vDg20 (v −1 )g0 +
n−2
v4g0 (v −1 )g0 .
n
Using this property we can assert that
|E|2g =< E, −(n − 2)vDg2 (v −1 ) >g .
If we integrate the trace-free Ricci tensor over the sphere, using the properties given after definition
1.11 we have:
Z
Z
|E|2g v −1 dvolg = −(n − 2)
< E, D2 (v −1 ) >g vv −1 dvolg
Sn
Sn
Z
(31)
= (n − 2)
< div(E), D(v −1 ) >g dvolg = 0,
Sn
where we have integrated by parts and used that div(E) =
by hypothesis.
n−2
2n dRg
Therefore, |E|g = 0 and using the definition 1.11 we have Ricg =
becasuse of definition 2.8.
= 0 because Rg is constant,
Rg
n g,
which means g is Einstein
In order to prove that sectional curvature of g is constant, we clear Riemijkl in the definition of
Weyl tensor (8) getting
1
(Ricgik gjl − Ricgil gjk + Ricgjl gik − Ricgjk gil )
n−2
Rg
−
(gjl gik − gjk gil ).
(n − 1)(n − 2)
g
Riemgijkl =Wijkl
+
g
But g is a locally conformaly flat metric and so that, Wijkl
= 0 and using Ricg =
Riemgijkl = −
Rg
n g
we obtain:
Rg
(gjl gik − gjk gil ).
(n − 1)(n − 2)
Then the sectional curvature defined in (4) is constant, so g must be the standard metric on the
sphere.
t
u
In this way, the Yamabe functional (25) on (Sn , gc ) is minimized by constant multiplies of gc and
its images under conformal diffeomorphisms. These are the only metrics conformal to the standard
one on Sn that have constant scalar curvature.
12
1. PRELIMINARIES
This theorem is closely related to the Sobolev inequality (19) in Rn . Since the infimum of the
Yamabe functional on the sphere is conformally invariant, stereographic projection converts the
Yamabe problem on Sn to an equivalent on Rn .
More precisely, for u ∈ C ∞ (Sn ), let u0 denote the weighted push-forward function on Rn defined
by u0 = u1 ρ∗ u with u1 the conformal factor in (29). Then we have
4
4
ρ∗ (u n−2 gc ) = 4u0n−2 g0 .
Because of the conformal invariance, J(Rn ) = J(Sn ); and using J(Rn ) =
where Rg0 = 0 we have
R
Rn
R
|∇u|g0 + Rn Rg0 u dvolg0
R
2
( Sn u2∗ dvol) 2∗
,
R
c |∇u |2 dx
R Rn n ∗ 0
λ(R ) =
inf
∗ .
u0 ∈C ∞ (Rn ) ( n |u0 |2 dx)2/2
R
n
Because of density we can restrict to smooth compactly supported functions:
λ(Rn ) =
cn k∇uk22
.
2
(Rn ) kuk2∗
inf
∞
u∈C0
Using the Sobolev inequality (19), we can assert that λ(Sn ) > 0. Therefore, it is equivalent identifying λ(Sn ) and the associated extremal functions to identifying the best constant and extremal
functions for Sobolev inequality.
Theorem 4.9. Sharp Sobolev inequality in the sphere.
The n−dimensional Sobolev constant σn is equal to
cn
Λ,
where
Λ = λ(Sn ) = J(gc ) = n(n − 1)vol(Sn )2/n .
Thus the sharp form of the Sobolev inequality on Rn is:
Z
cn
(32)
kuk22∗ ≤
|∇u|2 dx.
Λ Rn
Equality is attained only by constant multiples and translates of the functions uµ defined by (30).
Lemma 4.10. If M is any compact Riemannian manifold of dimension n ≥ 3, then λ(M ) ≤ λ(Sn ).
With all these ingredients, one can give a solution for the Yamabe problem. Here we just present
the main theorems:
Theorem 4.11. (Yamabe, Trudinger and Aubin (1976).) The Yamabe problem can be solved for
any compact manifold M such that λ(M ) < λ(Sn ).
This theorem shifts the focus of the proof from analysis to understanding the geometric meaning
of the invariant λ(M ). The idea of the proof to show that λ(M ) < λ(Sn ) is to find a test function
φ with J(φ) < λ(Sn ). Then,
Theorem 4.12. [31] If M is any compact Riemannian manifold of dimension n ≥ 3, then
(33)
λ(M ) < λ(Sn );
unless M is already conformal to the sphere Sn .
4. AN INTRODUCTION TO THE YAMABE PROBLEM
13
This theorem was proved in several steps:
• Aubin (1976)[7]: He proved that if M has dimension n ≥ 6 and it is not locally conformally
flat then (33) holds.
• Schoen (1984)[39]: Who finally proved that if M has dimension 3, 4, or 5, or M is locally
conformally flat, then (33) holds, unless M is already conformal to the sphere Sn . Note that
his proof uses the positive mass theorem.
Chapter 2
The conformal fractional Laplacian and the
fractional Qγ -curvature (Background).
1. Fractional Laplacian in Rn .
We can cite as references, the surveys in [45] and [21].
Definition 1.1. Let γ ∈ (0, 1) and u ∈ L∞ ∩ C 2 in Rn , the definition of fractional Laplacian in
Rn is given by
(34)
(−4)γ u(x) = P.V.
Z
Rn
(u(x + y) − u(x))
dy,
|y|(n+2γ)
where P.V. denotes the principal value, that is defined as
Z
u(x + y) − u(x)
lim
.
ε→0 Rn \B
|y|(n+2γ)
We can assert that it is a good definition because:
(1) |y|−(n+2γ) is integrable at ∞.
R
y
(2) B1 ∇u(x)·y
|y|n+2γ = 0 because |y|n+2γ is odd.
Moreover if we use the Taylor’s expansion, we obtain that near to zero (34) can be expressed
without the need of P.V as
(35)
(−4)γ u =
Z
B1
u(x + y) − u(x) − ∇u(x) · y
dy;
|y|n+2γ
this integral is convergent because
|u(x + y) − u(x) − ∇u(x) · y|
||D2 u||L∞
≤ (T aylor0 s expansion) ≤
.
n+2γ
|y|
|y|n−2−2γ
And note that the right hand side is integrable.
15
2. THE CONFORMAL FRACTIONAL LAPLACIAN AND THE FRACTIONAL Qγ -CURVATURE (BACKGROUND).
16
1.1. Fourier symbols and fractional Laplacian.
Definition 1.2. We define the fractional Laplacian as
(−4)γ = F −1 (|ξ|2γ (F u)); γ ∈ (0, 1).
We are going to see that this definition and (35) are equivalent using Fourier transform:
We call z = −y in (34) and the definition reads
Z
u(x − z) − u(x)
(−4)γ u =
dz.
|z|n+2γ
n
R
So we know
Z
2
Rn
u(x + y) − u(x)
=
|y|n+2γ
Z
n
ZR
=
Rn
Z
u(x + y) − u(x)
u(x − y) − u(x)
dy
+
dy
n+2γ
|y|
|y|n+2γ
Rn
u(x + y) − u(x − y) − 2u(x)
dy.
|y|n+2γ
And we can assert
Z
1
u(x + y) − u(x − y) − 2u(x)
(−4)γ u = (
) dy
2 Rn
|y|n+2γ
Z
u(x + y) − u(x − y) − 2u(x)
dy.
≡ up to a constant factor ≡
|y|n+2γ
n
R
We call Lu the functional
Z
(u(x + y) + u(x − y) − 2|u(x)|)Kγ (y) dy,
Lu =
Rn
where the kernel Kγ is defined by Kγ (x) =
to one.
cn,γ
|x|n+2γ .
The constant is chosen so the kernet integrates
We calculate its Fourier symbol
F
|{z}
(Lu) =
F ourier T ransf orm
σ
|{z}
(F u).
F ourier Symbol
We will show that its symbol is precisely |ξ|2γ . By the properties of the Fourier transform
Z
F (Lu) =F (
(u(x + y) + u(x − y)) − 2u(x))Kγ (y) dy
n
Z R
=
F ((u(x + y) + u(x − y)) − 2u(x))Kγ (y) dy
n
ZR
=
(eiξy + e−iξy − 2)(F u)(ξ)Kγ (y) dy
Rn
Z
=
(eiξy + e−iξy −2)Kγ (y) dy(F u)
{z
}
|
Rn
2 cos(ξy)
Z
2(cos(ξy) − 1)Kγ (y) dy(Lu).
=
Rn
So up to a constant factor we have
1. FRACTIONAL LAPLACIAN IN Rn .
Z
(cos(ξ · y) − 1)Kγ (y) dy.
σ(ξ) =
Rn
On the one hand,
Z
Z
−(cos(ξ · y) − 1)|y|−(n+2γ) dy
(cos(ξ · y) − 1)Kγ (y) dy =
σ(ξ) =
Rn
Z
=
Rn
Rn
1 − cos(ξ · y)
dy.
|y|n+2γ
On the other hand, if we take x ∈ Rn , we can assert
(1) sin(x) ≤ x.
(2) cos(x) ≥ cos2 (x) ⇒ 1 − cos(x) ≤ 1 − cos2 (x) .
|
{z
}
2
sin2 (x)
2
(3) (1. + 2.) ⇒ 1 − cos(x) ≤ sin x ≤ |x| .
Then
R
1−cos x
Rn |x|n+2γ
dx is finite and positive close to zero. So we can define
Z
g(ξ) =
Rn
1 − cos(ξ · y)
dy.
|y|n+2γ
And we are going to prove that this function is rotationally invariant, that is
(36)
g(ξ) = g(|ξ|e1 ).
If n = 1:
Z
g(−ξ) =
Rn
1 − cos(−ξ · y)
dy =
|y|n+2γ
Z
Rn
1 − cos(ξ · y)
dy = g(ξ).
|y|n+2γ
So (36) holds.
If n ≥ 2:
We consider a rotation R for which R(|ξ|e1 ) = ξ and we have
Z
g(ξ) =
Rn
Z
=
n
ZR
=
Rn
Z
1 − cos((R(|ξ|e1 )) · y)
(1 − cos(ξ · y))
dy
=
dy
n+2γ
|y|
|y|n+2γ
Rn
(1 − cos((|ξ|e1 ) · RT y))
) dy = (Change of variable : RT y = z)
|y|n+2γ
(1 − cos((|ξ|e1 ) · z))
) dz = g(|ξ|e1 ),
|z|n+2γ
17
2. THE CONFORMAL FRACTIONAL LAPLACIAN AND THE FRACTIONAL Qγ -CURVATURE (BACKGROUND).
18
as desired. Next, changing variables in the definition of g (z = |ξ|y),
Z
Z
1 − cos(|ξ|e1 · y)
1 − cos(|ξ| · y1 )
g(ξ) =g(|ξ|e1 ) =
dy =
) dy
n+2γ
|y|
|y|n+2γ
n
n
R
R
z
=(z = |ξ|y ⇒ z1 = |ξ|y1 , |y| = | |, dz = |ξ|n dy)
|ξ|
Z
Z
1
1 − cos z1
1
1 − cos z1
n+2γ
= n
dz
=
|ξ|
dz ≡ |ξ|2γ .
z n+2γ
n+2γ
|ξ| Rn | |ξ|
|
|ξ|n
|z|
n
{z
}
|R
C=constant
So we have σ(ξ) = |ξ|2γ , and this equality proves the equivalence between both definitions.
1.2. Fractional Laplacian as solution of a degenerate elliptic equation in the extension.
We have seen two different ways to define fractional Laplacian, now we are going to introduce
another one [12].
R
|f (x)|
n
Let f : Rn → R be a function such that Rn (1+|x|)
and y ∈ R+ . We
n+2γ < ∞, and let x ∈ R
consider the extension u : Rn × R+ → R that satisfies the following partial differential equation:
(37)
(38)
u(x, 0)
a
4x u + uy + uyy
y
=
f (x), x ∈ Rn ,
=
0, x ∈ Rn , y ∈ R.
The second equation (38) can be written in divergence form as
(39)
div(y a ∇u) = 0.
All the solutions for this differential equation are functions for which the following functional is
stationary:
Z
(40)
J(u) =
|∇u|2 y a dxdy.
Rn+1
+
Definition 1.3. Let γ ∈ (0, 1), we define the fractional Laplacian on Rn as
(−4)γ f = −d˜γ lim+ y a ∂y u
y→0
where a = 1 − 2γ and
(41)
2
d˜γ = −
2γ−1
Γ(γ)
.
γΓ(−γ)
We will prove that this construction of the fractional Laplacian is equivalent to the two previous
definitions.
If we take Fourier transform with respect to x in the system (38)-(37), we obtain the following one:

û(ξ, 0) =fˆ(ξ), ξ ∈ Rn ,

(42)
a
 −|ξ|2 û(ξ, y) + ûy (ξ, y) + ûyy (ξ, y) =0, ξ ∈ Rn , y > 0.
y
Fixed ξ, we can call ψ(y) = û(ξ, y) and we get
a
−|ξ|2 ψ + ψy + ψyy = 0.
y
1. FRACTIONAL LAPLACIAN IN Rn .
19
Then we know that the solution of (42) can be written as
(43)
û(ξ, y) = fˆ(ξ)φ(|ξ|y),
where φ is the solution of the following system:

a

− φ(y) + ∂ y φ(y) + ∂yy φ(y) = 0,


y

(44)
φ(0) = 1,



 lim φ(y) = 0.
y→∞
Note that φ is the minimizer of the functional
Z
(45)
J(φ) :=
(|φ0 |2 + |φ|2 )y a dy.
y>0
From next lemma 1.4 we that the solution of (44) can be written as φ(y) = c1 Iγ (y) + c2 Kγ (y),
Imposing limy→∞ φc (y) = 0 we obtain c1 = 0. So we can assert φ(y) = c2 Kγ (y) is solution
∀c2 .Note that if we impose ϕ(0) = 1, we get that the constant c2 must be equal to Γ(γ)2γ−1 .
Differentiating with respect to y we get
∂y û = fˆ(ξ)φ0 (|ξ|y)|ξ|.
Now we let y tend to zero and do the change of variable z = |ξ|y, and we obtain
lim y a ∂y û = fˆ(ξ)|ξ| lim φ0 (|ξ|y)y a = fˆ(ξ)|ξ|2γ lim φ0 (z)z a = cfˆ|ξ|2γ ,
y→0
y→0
z→0
where c = limz→0 φ (z)z = c2 limz→0 y K (z)z = d˜γ Γ(−γ)Γ(γ).
0
a
γ
0
a
Lemma 1.4. [1] The solution of the ODE
a
∂y ϕ − ϕ = 0.
y
may be written as ϕ(y) = y γ ψ(y), for a = 1 − 2γ, where ψ solves the well known Bessel equation
∂yy ϕ +
(46)
y 2 ψ 00 + yψ 0 − (y 2 + γ 2 )ψ = 0.
In addition, (46) has two linearly independent solutions, Iγ , Kγ , which are the modified Bessel
functions; their asymptotic behavior is given precisely by
y γ 1
y2
y4
Iγ (y) ∼
1+
+
+ ... ,
Γ(γ + 1) 2
4(γ + 1) 32(γ + 1)(γ + 2)
γ Γ(γ) 2
y2
y4
Kγ (y) ∼
1+
+
+ ...
2
y
4(1 − γ) 32(1 − γ)(2 − γ)
Γ(−γ) y γ
y2
y4
+
1+
+
+ ... ,
2
2
4(γ + 1) 32(γ + 1)(γ + 2)
for y → 0+ , γ 6∈ Z. And when y → +∞,
1
4γ 2 − 1 (4γ 2 − 1)(4γ 2 − 9)
y
Iγ (y) ∼ √
e 1−
+
− ... ,
8y
2!(8y)2
2πy
r
π −y
4γ 2 − 1 (4γ 2 − 1)(4γ 2 − 9)
Kγ (y) ∼
e
1+
+
+ ... .
2y
8y
2!(8y)2
2. THE CONFORMAL FRACTIONAL LAPLACIAN AND THE FRACTIONAL Qγ -CURVATURE (BACKGROUND).
20
Now we are going to give another proof of the equivalence between (37)(38) and the fractional
Laplacian defined from its symbol proving that the corresponding energy functionals coincide,
that means:
Z
Z
|∇u|2 y a dxdy =
|ξ|2γ |fˆ(ξ)|2 dξ,
Rn+1
+
Rn
up to multiplicative constant.
We call
Z
(47)
a
|∇u|2 y a dxdy, which is the energy in W 1,2 (Rn+1
+ ) with the weight y .
J1 [u] :=
y>0
Z
(48)
J2 [f ] :=
|ξ|2γ |fˆ(ξ)|2 dξ, which is the energy in H γ (Rn ).
Rn
Using Plancherel’s identity
Z
Z
∞
J1 [u] =
Rn
Z
Z0 ∞
=
Rn
(43)
(|ξ|2 |û|2 + |ûy |2 )y a dy dξ =
|f (ξ)|2 |ξ|2 (|φ(|ξ|y)|2 + |φ0 (|ξ|y)|2 )y a dy dξ.
0
With the change of variable: |ξ|y = ȳ ⇔ |ξ| dy = dȳ,
Z
Z
∞
|f (ξ)|2 |ξ|1−a (|φ(ȳ)|2 + |φ0 (ȳ)|2 )ȳ a dȳ dξ
Z
Z ∞
2
1−a
=
|f (ξ)| |ξ|
(|φ(ȳ)|2 + |φ0 (ȳ)|2 )ȳ a dȳ dξ
Rn
0
Z
Z
2
1−a
|f (ξ)|2 |ξ|1−a dξ = cJ2 [f ].
=
|f (ξ)| |ξ|
J(φ) dξ = J(φ)
{z
}
|
n
n
R
R
J1 [u] =
Rn
0
constant
With this equality we have proved the equivalence between the energy functionals, so the corresponding Euler-Lagrange equations for each energy must then coincide up to a constant factor.
Thus we impose that
d
d
J1 [u + εv]|ε=0 =
J2 [u + εv]|ε=0 .
dε
dε
On the one hand
Z
Z
d
(49)
J1 [u + εv]|ε=0 =
vdiv(y a ∇u) dxdy +
vy a ∂y u dx.
dε
Rn+1
{y=0}
+
On the other hand
Z
d
d (50)
J2 [u + εv]|ε=0 =
|ξ|2γ (fˆ + εv̂)2
dε ε=0
dε Rn
Z
Z
Z
\γ f v̂ =
(51)
=2
|ξ|2γ fˆv̂ = 2
(−4)
Rn
where the last equality holds because of Parseval.
Rn
Rn
(−4)γ f v,
2. GEOMETRIC SETTING
21
Note that if v vanishes at {y = 0}, then the Euler-Lagrange equation for the functional (47) is
div(y a ∇u) = 0
Following the same proof but taking a test function v which is not null on the boundary and
looking at (49)-(50), we obtain
Z
v(y a ∂y u) =
Z
(−4)γ f v∀v.
Rn
{y=0}
Therefore we can assert
(−4)γ f = −cy a ∂y u|y=0 ,
for all test functions v.
1.3. The Poisson Kernel. Finally we would like to obtain an explicit formula for the solution
of (37)-(38). The proof maybe found in [12].
Given γ ∈ (0, 1), let a = 1 − 2γ ∈ (−1, 1). The function
(52)
Pγ (x, y) = Cn,a
y 2γ
(|x|2 + |y|2 )
n+2γ
2
= Cn,γ
y 1−a
(|x|2 + y 2 )
n+1−a
2
is a solution of
(
(53)
div(y a ∇Pγ ) = 0 in Rn+1
+ ,
on ∂Rn+1
= Rn ,
+
Pγ = δ 0
where δ0 is the delta distribution at the origin, and Cn,γ is a positive constant depending only on
n and γ chosen such that , for all y > 0,
Z
Pγ (x, y) dx = 1.
Rn
Proposition 1.5. [11] For f ∈ Cc (Rn ), the solution of problem (37)-(38) is given by the Poisson
formula
Z
(54)
u(x, y) =
Pγ (x − ξ, y)f (ξ) dξ,
Rn
where Pγ the Poisson Kernel for the problem, that is given in (52).
2. Geometric setting
The conformal fractional Laplacian is constructed on the conformal infinity of a conformally compact Einstein manifold. The fractional curvature, which is a generalization on scalar curvature, is
constructed from the conformal fractional Laplacian [27, 15].
Definition 2.1. Let X n+1 be a smooth manifold of dimension n + 1 with smooth boundary ∂X =
M n . A defining function for the boundary M n in X n+1 is a function ρ on X̄ n+1 which satisfies:


 ρ > 0 in X,
ρ = 0 on M,

 dρ 6= 0 on M.
2. THE CONFORMAL FRACTIONAL LAPLACIAN AND THE FRACTIONAL Qγ -CURVATURE (BACKGROUND).
22
Definition 2.2. A Riemannian metric g + on X n+1 is conformally compact if (X̄ n+1 , ḡ) is a
compact Riemannian manifold with boundary M n for a defining function ρ and
ḡ = ρ2 g + .
(55)
Any conformally compact manifold (X n+1 , g + ) carries a well-defined conformal structure [ĝ] on
the boundary M n ; where each ĝ is the restriction of ḡ = ρ2 g + for a defining function ρ. We call
(M n , [ĝ]) the conformal infinity of the conformally compact manifold (X n+1 , g + ).
4
We usually write these conformal changes on M as ĝw = w n−2γ ĝ, for a positive smooth function
w.
Near the conformal infinity, given a defining function ρ, we have the following asymptotically
expansion of the Riemannian tensor
(56)
+
+ +
+
Riemgijkl = −|dρ|2ḡ (gik
gjl − gil+ gjk
) + O(ρ3 ),
in a coordinate system on (0, ) × M n ∈ X n+1 .
Definition 2.3. A Riemannian metric g + is called Asymptotically hyperbolic if there exists a
defining function ρ such that
|∇ρ|2ḡ = 1 on ∂X.
Remark 2.4. From (56) one sees that for a conformally compact manifold, if it is asymptotically
hyperbolic, then the sectional curvature goes to −1 near infinity.
Lemma 2.5. [44] Given a conformally compact, asymptotically hyperbolic manifold (X n+1 , g + ) and
a representative ĝ in [ĝ] on the conformal infinity M n , there is a unique defining function ρ such
that, on M × (0, ε) in X, g + has the normal form
(57)
g + = ρ−2 (dρ2 + gρ )
where gρ is a family on M of metrics depending on the defining function and satisfying gρ |M = ĝ.
Definition 2.6. An Einstein metric is a metric for which the Ricci tensor and the metric tensor
are proportional:
(58)
+
+
Ricgij = f gij
,
for some f smooth on X.
Note that for an Einstein metric, Rg+ = (n + 1)f .
Lemma 2.7. [6] Under condition (58), the function f must be constant, when n ≥ 3, so, in
particular, an Einstein metric has constant scalar curvature Rg+ = −n(n + 1).
Thus we may give the definition:
Definition 2.8. A conformally compact manifold (X n+1 , g + ) is called conformally compact Einstein manifold if the metric satisfies Ricg+ = −ng + .
Note that a conformally compact Einstein manifold must be asymptotically hyperbolic. Let us
give some examples of conformally compact Einstein manifolds:
2. GEOMETRIC SETTING
23
i. [8, 16] Hyperbolic space. We describe the hyperbolic space with the Upper half space
model. Hn+1 is realized as a set
Hn+1 = {z = (x, y); x ∈ Rn , 0 < y < ∞},
The metric in these coordinates is
g + = y −2 (dx2 + dy 2 ),
and the volume element is
dvolg+ = y −(n+1) dxdy.
The conformal infinity is Rn ∪ {∞} where Rn is interpreted as the hyperplane {y = 0}, and
the metric here is precisely the Euclidean one:
ĝ = y 2 g + |y=0 = |dx|2 .
The Laplace Beltrami operator is given by
4H = y 2 (4x + ∂yy ) − (n − 1)y(∂y ).
(59)
The hyperbolic space can be represented with different models, we are going to present it also
with the Ball model. In this way Hn+1 is realized as a set
Bn+1 = {x ∈ Rn+1 /||x|| < 1}.
We take x as a global coordinate and define a metric:
g B = 4(1 − ||x||2 )−2 (dx21 + ... + dx2n+1 ).
Here the volume element is
dvolgB = 2n+1 (1 − ||x||2 )−(n+1) dx1 dx2 ...dxn+1 .
We call Sn = {x; ||x|| = 1}, which represent the conformal infinity ∂∞ B n+1 , where the metric
is the standard for Sn .
Comparing with Hn+1 , we see that whereas the ∂∞ Hn+1 = Rn ∪ {∞} has a “distinguished”
point at ∞, this does not happen in ball model because the boundary at infinity ∂∞ Hn+1 is
the one point of compactification of Rn .
Remark 2.9. The relation between both models is given by:
G : Bn+1 −→ Hn+1 .
(x1 , x2 , ..., 21 (1 − ||x||2 ))
,
G(x) =
(1 + ||x||2 − 2x1 )
and the inverse map,
G−1 : Hn+1 −→ Bn+1 .
2
(z1 , z2 , ..., zn , zn+1
+ ||z 0 ||2 − 41 )
G−1 (z) =
1
2
(( 2 + ||zn+1 ||) + ||z 0 ||2 )
where z 0 = (z1 , ..., zn ).
ii. A generalized hyperbolic manifold. It is given by
X 4 = S1 (L) × R3 ,
(60)
with the metric
(61)
g+ =
dR2
2
2
(1 + R2 )dt2 +
+
R
g
.
S
1 + R2
2. THE CONFORMAL FRACTIONAL LAPLACIAN AND THE FRACTIONAL Qγ -CURVATURE (BACKGROUND).
24
If for x ∈ R3 we use the change of variables
2
1 − ρ4
2
|x| =
= sinh log ,
ρ
ρ
2
1 + ρ4
1
1
d|x| = − 2 −
dρ
dρ = −
ρ
4
ρ2
and also,
2
1 + ρ4
ρ2
2
1+R =
!2
,
we can observe that (X 4 , g + ) is a conformally compact Einstein manifold, expressed in the
form (57):
!
2
2
ρ2
ρ2
2
+
−2
2
gS2 + 1 +
dt .
g =ρ
dρ + 1 −
4
4
Thus the conformal infinity is
(S1 (L) × S2 , dt2 + gS2 ).
| {z }
ĝ
iii. Anti de Sitter space. We will explain it in detail in the last chapter. This example is
important because it gives two different examples of conformally compact Einstein manifolds
with the same conformal infinity.
The standard examples of static Riemannian AdS-type black holes solutions are manifolds
M = N n−2 × R2 where N n−2 is compact and the given metric has the following form gM =
V −1 dr2 + V dθ2 + r2 gN , where gN is any Einstein metric and V is a function that will be
described later in the case we are going to study. Some examples of these Ads-type black
holes are [2, 3]:
• AdS-S2 -black holes:M = R2 × Sn−1 .
• AdS toral black holes: M = R2 × Tn−1 . (Note Tn−1 represents the (n − 1)-torus).
3. Definition of the conformal fractional Laplacian
First, we look at the spectrum of the Laplacian on hyperbolic space:
Lemma 3.1. [43] The spectrum of −4Hn+1 is equal to [( n2 )2 , ∞).
Proof. Let us prove here that the spectrum of −4Hn+1 is contained in [( n2 )2 , ∞). In the hyperbolic
space, we can assert because of (59) that
−4Hn+1 (y s ) = s(n − s)y s .
n
Indeed applying the theorem 3.2 with φ = y 2 we can assert that the spectrum of hyperbolic
Laplacian applied to any function is contained in [( n2 )2 , ∞).
t
u
Theorem 3.2. [43] Suppose that H is elliptic on L2 (Ω) and that there is a positive continuous
1,2
function φ in Wloc
(Ω) and a potential V in L1loc (Ω), such that
Hφ ≥ V φ.
3. DEFINITION OF THE CONFORMAL FRACTIONAL LAPLACIAN
25
Then the quadratic form inequality
H≥V
is valid on
Cc∞ (Ω).
We can read about the spectrum of the Laplacian of an asymptotically hyperbolic metric in [33,
34, 35]. This spectrum is given by
σ(−4g+ ) = [(n/2)2 , ∞) ∪ σpp (−4g+ ), where σpp (−4g+ ) ⊂ (0, (n/2)2 ).
We note that σpp (−4g ) is the pure point spectrum, i.e, the set of L2 -eigenvalues, and it is finite;
and [(n/2)2 , ∞) is the continuous spectrum.
We refer, as an example, the spectrum of the Laplacian on hyperbolic space, calculated in lemma
3.1 in this section.
More refined statements follow from the main result of [35], which is the existence of the meromorphic continuation of the resolvent
R(s) = (−4g+ − s(n − s))−1 ,
where λ = s(n − s) is symmetric with respect to Re(s) =
n
2.
Note that for λ = s(n − s) ∈ R, we will choose s ∈ ( n2 , n) which is γ ∈ (0, n2 ) if we denote s =
n
2
+ γ.
lambda
n^2/4
n/2
n
s
Fig. 1. Representation of λ(s).
Let (X, g + ) be a conformally compact Einstein manifold with conformal infinity (M, [ĝ]). Using
these results we can assert the existence of solution for the following eigenvalue problem (63). As
we can check in Graham-Zworski and Mazzeo-Melrose [27, 35] given f ∈ C ∞ (M ) and s ∈ C, if
s(n − s) does not belong to the pure point spectrum of −∆g+ then there exists a solution of the
form
(62)
u = F ρn−s + Hρs ; F, H ∈ C ∞ (X), F |ρ=0 = f,
for the eigenvalue problem
(63)
−∆g+ u − s(n − s)u = 0, in X.
2. THE CONFORMAL FRACTIONAL LAPLACIAN AND THE FRACTIONAL Qγ -CURVATURE (BACKGROUND).
26
Definition 3.3. Taking a representative ĝ of the conformal infinity (M n , [ĝ]) we can define a
family of meromorphic pseudo-differential operators S(s) called scattering operator as
S(s)f = H|M .
It is defined in Re(s) > n2 . As it is explained in the next theorem, the values s = n2 +k; k = 0, 1, 2...
are simple poles of finite rank (they are known as trivial poles). It is possible that S(s) has another
poles, but we will assume here that this does not happen.
Theorem 3.4. [27] Let (X, g + ) be a conformally compact Einstein manifold with conformal infinity
(M, [ĝ]). Suppose that k ∈ N and k ≤ n2 if n is even, and that ( n2 )2 − k 2 is not an L2 −eigenvalue
of −∆g . If S(s) is the scattering operator of (X, g + ), and Pkĝ the conformally invariant operators
on M constructed in [26], then S(s) has a simple pole at s = n2 + k and
k 2k
−1
n S(s), c
ck Pkĝ = −Ress= 2+k
,
k = (−1) [2 k!(k − 1)!]
where Ress=s0 S(s) denotes the residue at s0 of the meromorphic family of operators S(s).
Consequently these are local operators which satisfy
Pkĝ = (−∆ĝ )k + l.o.t.
In particular, Pkĝ = (−∆ĝ )k if ḡ is flat.
• If k = 1 we have the conformal Laplacian (note the different constant normalization in (12)),
n−2
Rĝ .
P1ĝ = −∆ĝ +
4(n − 1)
• If k = 2, the Paneitz operator
n − 4 g+
P2ĝ = (−∆ĝ )2 + δ(an Rĝ + bn Ricĝ )d +
Q2 .
2
Note that up to constant Q1 is the classical scalar curvature and Q2 is the so called Q-curvature.
But it is also possible define conformally covariant fractional powers of Laplacian in the case γ 6∈ N.
Definition 3.5. For s = n2 + γ; γ ∈ (0, n2 ), γ ∈
/ N, we define the conformally covariant fractional
powers of the Laplacian as
n
Γ(γ)
(64)
Pγ [g + , ĝ] = dγ S( + γ); dγ = 22γ
.
2
Γ(−γ)
As a pseudodifferential operator, its principal symbol satisfies
σ(Pγ [g + , ĝ]) = σ((−∆ĝ )γ ) = |ξ|2γ .
In the rest of the paper we will use the simplified notation:
Pγĝ = Pγ [g + , ĝ].
These operators satisfy an important conformal property
(65)
where
n+2γ
Pγĝw φ = w− n−2γ Pγĝ (wφ), ∀φ ∈ C ∞ (M ),
4
ĝw := w n−2γ ĝ.
3. DEFINITION OF THE CONFORMAL FRACTIONAL LAPLACIAN
27
Proof. Given ĝ on M like in lemma 2.5 (in the previous section 2, in this chapter 2), there
4
dρ+g
exists ρ such that g + = ρ2 ρ and gρ |M = ĝ. Given ĝw = w n−2γ ĝ on M , there exist ρ̃ such that
g+ =
(66)
dρ̃2 +g̃ρ̃
ρ̃2
and gρ̃ |M = ĝw . From the proof [44] one gets that
2
ρ̃ = w n−2γ .
ρ M
So we can find a solution u for the eigenvalue problem (63) in the following way:
u = F ρn−s + Hρs = F̃ ρ̃n−s + H ρ̃s .
And using (66) and s =
n
2
+ γ, up to lower order terms,
n+2γ
(67)
F = F̃ w and H = H̃w n−2γ .
Restricting these equalities to M one gets
n+2γ
F̃ |ρ=0 = f˜ and F̃ w|ρ=0 = f , which means f˜ = f w−1 . Morever H̃|ρ=0 = h̃ and H̃w n−2γ |ρ=0 = h,
n+2γ
which means h̃ = hw− n−2γ .
Because the definition of Scattering opperator we can assert that S(s)f = h and S̃(s)f˜ = h̃, and
applying the definition of conformally covariant fractional powers of fractional Laplacian (64) we
get
n+2γ
n+2γ
P ĝw (f w−1 ) = P ĝw (f˜) = h̃dγ = hw− n−2γ dγ = P ĝ (f )w− n−2γ .
γ
γ
γ
Taking f = φw we get the desired result.
t
u
Note that Pγĝ is a self-adjoint operator on M .
Definition 3.6. We define the fractional order curvature as:
Qĝγ := Pγĝ (1).
Remark 3.7. Using the previous definition we can express the conformal property (65) as
(68)
n+2γ
Pγĝ (w) = w n−2γ Qĝγw ,
where Qĝγw := Qγ [g + , ĝw ].
3.1. The extension problem on hyperbolic space. We must note that in the case of M = Rn
2
2
and X = Rn+1
with coordinates x ∈ Rn and y > 0, with the hyperbolic metric g + ≡ gH = dy +|dx|
+
y2
(where ĝ = |dx|2 is the Euclidean metric on Rn ) the construction of the scattering operator is
precisely the Caffareli-Silvestre extension problem for the fractional Laplacian when γ ∈ (0, 1). We
note that in this case ḡ = dy 2 + |dx|2 is the flat metric in Rn+1
+ .
Theorem 3.8. [15] Given γ ∈ (0, 1), for a smooth function f : Rn → R, there exists an unique
solution U = U (x, y) : Rn × [0, +∞) → R to the following extension problem
a
∆x U + ∂y U + ∂yy U = 0; x ∈ Rn , y ∈ [0, +∞),
y
(69)
U (x, 0) = f (x), x ∈ Rn ,
2. THE CONFORMAL FRACTIONAL LAPLACIAN AND THE FRACTIONAL Qγ -CURVATURE (BACKGROUND).
28
where a = 1 − 2γ. Moreover, u = y n−s U is a solution of the eigenvalue problem
−4gH u − s(n − s)u = 0, in Hn+1 ,
(70)
for s =
n
2
+ γ, and
Pγ f = −d˜γ lim (y a ∂y U ) = (−4x )γ ,
(71)
y→0
where Pγ := Pγ [gH , |dx|2 ] and d˜γ is defined in (41).
Proof. Given f fixed, we know that the solution u of the scattering problem
−4H u − s(n − s)u = 0 in Hn+1 ,
(72)
can be written as
u = y n−s F + y s H,
(73)
where F and H satisfy:
F |y=0 = f and F (x, y) = f (x) + f2 (x)y 2 + o(y 2 ),
(74)
and S(s)f = h for h = H|y=0 and H(x, y) = h(x) + h2 (x)y 2 + o(y 2 ).
If we recall the definition of conformal Laplacian (12) and use that in the hyperbolic space we have
RgH = −n(n + 1) (because it is an Einstein manifold), we have that
n2 − 1
.
4
We use the conformal property of the conformal Laplacian given in proposition 2.2 for the change
of metric ḡ = y 2 gH (where ḡ is the Euclidean metric), getting
−4gH = LgH +
LgH φ = y
n+3
2
Lḡ (y −
n−1
2
φ).
But we know that Lḡ = −4ḡ = −4x − ∂yy . So we can do the change of variable
u = y n−s U,
(75)
sustitute s =
n
2
+ γ, and use all the previous equivalences in (72) to get
a
4x U + ∂yy U + ∂y U = 0.
y
For the second part, we only need realize that with the definition of h given in (74), the following
equivalence holds
Pγ f = dγ S(s)f = dγ h.
If we sustitute the expansion (73) in the change of variable (75) we obtain U = F + y 2s−n H. And
we can compute
lim y a ∂y U = lim y a ∂y (F + y 2s−n H)
y→0
y→0
= lim y a ∂y (f (x) + f2 (x)y 2 + o(y 2 ) + y 2s−n (h(x) + h2 (x)y 2 + o(y 2 )))
(76)
y→0
= (2s − n)h = 2γh.
Therefore h =
1
2γ
limy→0 y a ∂y U , and so that
3. DEFINITION OF THE CONFORMAL FRACTIONAL LAPLACIAN
Pγ f =
29
dγ
lim y a ∂y U,
2γ y→0
as desired.
t
u
3.2. The extension problem on conformally compact Einstein manifolds. Now we are
going to study the same extension problem (69) as before but in any conformally compact Einstein
manifold (X n+1 , g + ).
Theorem 3.9. [15] Let (X, g + ) be any conformally compact Einstein manifold with conformal
infinity (M, [ĝ]). For any defining function ρ of M satisfying (57) in X, the problem
(77)
−4g+ u − s(n − s)u = 0 in (X, g + ),
u = F ρn−s + Hρs ; F, H ∈ C ∞ (X), F |ρ=0 = f,
is equivalent to
−div(ρa ∇U ) + E(ρ)U = 0 in (X, ḡ),
(78)
U = f on M.
where
n
+ γ, a = 1 − 2γ.
2
and the derivatives in (78) are taken respect to the metric ḡ. The lower order term is given by
ḡ = ρ2 g + , U = ρs−n u, s =
(79)
a
a
1
n−1
Rḡ ρa .
E(ρ) = −4ḡ (ρ 2 )ρ 2 + (γ 2 − )ρ−2+a +
4
4n
If we write it in the metric g + we have
E(ρ) = −4g+ (ρ
n−1+a
2
)ρ
−n−3+a
2
−(
n2
− γ 2 )ρ−2+a .
4
Moreover we have the following formula for the calculation of the conformal fractional Laplacian
Pγĝ f = −d˜γ lim ρa ∂ρ U,
(80)
ρ→0
where d˜γ is defined in (41).
The proof is analogous to the one given in theorem 3.8 for the Euclidean case, we only have to
take into account that we are working with an Einstein manifold provided with a metric g + , for
which
1
4ḡ = ∂ρρ + ψ∂ρ + 4gρ ,
2
where ḡ = ρ2 g + and ψ := ∂ρ (log det(gρ )). The second term on the right hand side is the one that
generates the lower order term E(ρ).
Remark 3.10. For a metric given by (57)
E(ρ) =
−n + 1 + a a−1
n−1−a
ψρ
=
Rḡ ρa , in M × (0, δ).
4
4n
2. THE CONFORMAL FRACTIONAL LAPLACIAN AND THE FRACTIONAL Qγ -CURVATURE (BACKGROUND).
30
Remark 3.11. We recall how to compute the Qĝγ curvature. We set f ≡ 1, and find the solution
of the Poisson problem −4g+ v − s(n − s)v = 0 such that
v = F ρn−s + Hρs , F |ρ=0 = 1, H|ρ=0 = h.
Or equivalent finding the solution U = ρs−n for the extension problem (78), with f ≡ 1. Then,
Qĝγ = dγ h = −d˜γ lim ρa ∂ρ U.
ρ→0
Lemma 3.12. There exists a unique linear bounded operator
T : W 1,2 (X, y a ) → H γ (M );
(81)
T U = U |M for all U ∈ C ∞ (X̄).
T is called the trace operator.
3.3. A special defining function. Now we are going to choose a suitable defining function ρ∗ ,
in order to transform the problem (57) into one of pure divergence form. We follow the study in
[15]
Lemma 3.13. Let (X, g + ) be a conformally compact Einstein manifold with conformal infinity
(M, [ĝ]). Fixed a metric ĝ on M , and assuming that ρ is a defining function, we can assert that
for each γ ∈ (0, 1), there exists another (positive) defining function ρ∗ on X, satisfying ρ∗ =
ρ + O(ρ2γ+1 ) and such that for the term E defined in (79) we have
E(ρ∗ ) = 0.
Moreover, the metric g ∗ = (ρ∗ )2 g + satisfies g ∗ |ρ=0 = ĝ and has asymptotic expansion
g ∗ = (dρ∗ )2 [1 + O((ρ∗ )2γ )] + ĝ[1 + O((ρ∗ )2γ )].
Theorem 3.14. Let γ ∈ (0, 1) fixed, and f any smooth function on M . If under the hypothesis and
the special defining function ρ∗ constructed in the lemma 3.13, U solves the following extension
problem
(82)
−div((ρ∗ )a ∇U ) = 0 in (X, g ∗ )
U = f on M
(where the derivatives are taken with respect to the metric g ∗ = (ρ∗ )2 g + );
then
(83)
Pγĝ f = d˜γ lim
(ρ∗ )a ∂ρ∗ U + f Qĝγ .
∗
ρ →0
3.4. Fractional Yamabe problem [25]. From now and on, we fix γ ∈ (0, 1). The Fractional
Yamabe problem is: given a conformally compact Einstein manifold (X n+1 , ḡ) of dimension n ≥ 3
4
with conformal infinity (M, [ĝ]), to find a new metric conformal to ĝ, ĝw = w n−2γ ĝ (where w is a
strictly positive C ∞ function on M ) with constant fractional curvature Qĝγw .
Note that we suppose Qĝγ is not constant (because in otherwise the problem is solved).
Since we impose that the metric ĝw has constant fractional curvature, the conformal property (68)
is equivalent to assert that there exists a constant c on M such that
(84)
n+2γ
Pγĝ (w) = cw n−2γ , w > 0,
3. DEFINITION OF THE CONFORMAL FRACTIONAL LAPLACIAN
31
which thanks to theorem 3.9 is equivalent to the eixstence of a strictly positive C ∞ solution for
extension problem:
−div(ρa ∇U ) + E(ρ)U = 0 in (X, ḡ),
n+2γ
(85)
−d˜γ lim ρa ∂ρ U = cw n−2γ where U |M = w.
ρ→0
Remark 3.15. Using the special defining function (3.13) the fractional Yamabe problem (85) can
be written as
−div((ρ∗ )a ∇U ) = 0 in (X, g ∗ ),
n+2γ
(86)
−d˜γ lim
(ρ∗ )a ∂ρ∗ U + wQĝγ = cw n−2γ where U |M = w.
∗
ρ →0
Indeed we only need use the definition of the Yamabe problem (84) and the expression of Pγĝ with
ρ∗ from (83).
One may use the variational method as in the classical case γ = 1.
Definition 3.16. We define the γ−Yamabe functional as
R
Qĝγ dvolĝ
(87)
Iγ [ĝ] = R M
n−2γ .
( M dvolĝ ) n
Now we can ask about the existence of the minimizer of Iγ among metrics in the class [ĝ].
Remark 3.17. We will use the notation 2∗ =
2n
n−2γ .
Definition 3.18. We define the γ−Yamabe constant as
λγ (M, [ĝ]) = inf{Iγ [h]; h ∈ [ĝ]},
(88)
which is an invariant of the conformal class [ĝ] when g + is fixed.
Because we are going to minimize the functional in the conformal class of the metric ĝ, we can
4
take the conformal metric ĝw = w n−2γ ĝ and define the previous functional as a functional on w:
R
wPγĝ w dvolĝ
(89)
Iγ [w] = R M 2n
.
n−2γ
( M w n−2γ dvolĝ ) n
4
2n
Indeed, if ĝw = w n−2γ ĝ, we know that dvolĝw = w n−2γ dvolĝ . Therefore, using the conformal
property (68) we have:
R
R
R
2n
• M n Qĝγw dvolĝw = M n w1− n−2γ Pγĝ w dvolĝw = M wPγĝ w dvolĝ .
R
R
2n
n−2γ
n−2γ
• ( M n dvolĝw ) n = ( M n w n−2γ dvolĝ ) n .
The functional Iγ [ĝ] (87) also can be represented as a functional in the extension:.
R
˜γ n+1 (ρa |∇U |2 + E(ρ)U 2 ) dvolḡ
d
X
(90)
I˜γ [U ] =
,
R
2n
n−2γ
(U n−2γ dvolĝ ) n
Mn
where d˜γ is defined in (41). Indeed if we take the equation (78), multiply it by U , integrate over
X n+1 and apply divergence theorem we get the equality
Z
Z
(91)
U (ρa ∇U ) dvolĝ =
ρa |∇U |2 + E(ρ)U 2 dvolḡ .
M
X n+1
2. THE CONFORMAL FRACTIONAL LAPLACIAN AND THE FRACTIONAL Qγ -CURVATURE (BACKGROUND).
32
Using that w ≡ U |M , the definition of Pγĝ given in (80) and (78) we get
R
R
ĝ
wPγĝ w dvolĝ
n wPγ w dvolĝ
Mn
Iγ [w] = R
= R M 2n
2n
n−2γ
n−2γ
( M n w n−2γ dvolĝ ) n
( M n U n−2γ dvolĝ ) n
R
R
d˜γ M n U (y a ∂y U ) dvolĝ
d˜γ X n+1 (y a |∇U |2 + E(y)U 2 ) dvolḡ
(92)
= R
=
R
2n
2n
n−2γ
n−2γ
( M n U n−2γ dvolĝ ) n
( M n U n−2γ dvolĝ ) n
= I˜γ [U ].
Note that the infimum of I˜γ [V ] among V ∈ W 1,2 (X, y a ) with T V = w is attained at U satisfying
(78).
Then we can assert that this equivalence tells us that
(93)
λγ (M, [ĝ]) = inf{I˜γ [U, ḡ]; U ∈ W 1,2 (X, y a )}.
Remark 3.19. If we use the special defining function defined in lemma 3.13, the functional (87)
can be represented as
∗
I [U ] =
d˜γ
R
X n+1
R
(ρ∗ )a |∇U |2 dvolg∗ + M n w2 Qĝγ dvolĝ
.
R
2n
n−2γ
( M n U n−2γ dvolĝ ) n
Now we consider the fractional Yamabe problem on Sn .
a
Theorem 3.20. [32, 19, 18] Let w ∈ H γ (Rn ), γ ∈ (0, 1), a = 1 − 2γ, and U ∈ W 1,2 (Rn+1
+ ,y )
with trace T U = w. Then
Z
2
kwkL2∗ (Rn ) ≤ S̄(n, γ)
y a |∇U |2 dx dy,
Rn+1
+
where, being gc the metric of Sn ,
S̄(n, γ) =
d˜γ
.
λγ (Sn , [gc ])
Moreover the equality holds if and only if
n−2γ
2
µ
(94)
wµ (x) = c
; x ∈ Rn ,
|x − x0 |2 + µ2
for c ∈ R, µ > 0 and x0 ∈ Rn fixed.
Remark 3.21. Because of the previous theorem we can assert that the only solutions for the γYamabe problem in the case M = Sn are the so called “bubbles” wµ (x).
Definition 3.22. We say that a manifold has a non-umbilic point, when there exists any point
such that in its neighbourhood the manifold is not as a piece of a sphere.
The solution of the fractional Yamabe problem follows now the same strategy as in the classical
case explained in chapter 1. The difficulty here is the non-locality of the operator involved. Thus
one may show:
Theorem 3.23. [25] If (X n+1 , g + ) is any asymptotically hyperbolic manifold of dimension n ≥ 3
with conformal infinity M , that satisfies
3. DEFINITION OF THE CONFORMAL FRACTIONAL LAPLACIAN
33
• limρ→0 ρ−2 (R[g + ] − Ric[g + ](ρ∂ρ ) + n2 ) = 0,
• X n+1 has a non-umbilic point on M ,
n−1−a
2γ+1 Γ(γ)
• − n+a−3
1−a 2
Γ(−γ) + a+1 < 0,
then λγ (M ) < λγ (Sn ).
Theorem 3.24. [25] If λγ (M, [ĝ]) < λγ (Sn , [gc ]), then the fractional Yamabe problem (84) has
solution.
Note that it was very important the contribution of J. Escobar in this problem. In his paper [22]
he solved many cases of the Yamabe problem with γ = 12 . More precisely he proves the following
theorem:
Theorem 3.25. [22] Let (X̄ n+1 , ḡ) be a compact Riemannian manifold with boundary (M, [ĝ]) and
dimension n + 1 ≥ 3. Assume that X n+1 satisfies any of the following four conditions
(1)
(2)
(3)
(4)
n > 5, and X has nonumbilic point on M ;
n ≥ 5, with X locally conformally flat and M umbilic;
n = 3 or 4, and M is umbilic;
n = 2.
4
Then there exists a smooth metric gu = u n−2 g on M̄ , of zero scalar curvature on X and constant
mean curvature on M .
Chapter 3
Non uniqueness for the fractional Yamabe
problem
1. The sign of the Yamabe constant.
We may see that depending on the sign of the minimizer of the classical Yamabe functional λ(M )
it holds:
i. If λ(M ) = 0 we have uniqueness on solution (up to constant).
4
Given (M, g), suppose that g̃ = ũ n−2 g is a solution with Rg̃ = µ. We also suppose that
4
g̃1 = ũ1n−2 g is solution with Rg̃1 = µ1 . So that we have
4
4
−4
4
g̃1 = u1n−2 g = u1n−2 u n−2 g̃ = v n−2 g̃.
Therefore
n+2
Lg̃1 (v) = Rg̃ v n−2 ,
|{z}
µ
and because the definition of Laplacian (12), we can assert
(95)
n+2
−cn ∆g̃1 (v) + Rg̃1 v = Rg̃ v n−2
|{z}
|{z}
µ1
µ
Now if we suppose µ = 0, using proposition 4.7 we have µ1 = 0 and so that we can assert
∆g̃1 (v) = 0,
Applying Liouville theorem (because v is an harmonic function in a compact manifold), we
obtain that v = ζ(constant) and so that u1 u−1 = ζ as we wanted.
ii. If λ(M ) < 0 we also get uniqueness in the solution (up to constant).
In this case, we can suppose again that we have two solutions, obtaining (95). But now we
impose µ < 0 and because of proposition 4.7, we have µ1 < 0.
We remark that if we have a solution g and we take g̃ = cg with c constant, we obtain than
Rg̃ = c2 Rg , therefore we can suppose that µ1 = µ.
Now we consider P the point where v reaches the maximum value, and Q the point where it
reaches the minimum value:
35
36
3. NON UNIQUENESS FOR THE FRACTIONAL YAMABE PROBLEM
• In P , the function v is maximum so that
∆g̃1 (P ) ≤ 0 ⇒ −cn ∆g̃1 (P ) ≥ 0,
In this way
n+2
µ (v n−2 − v)(P ) = −cn ∆g̃1 (P ) ≥ 0.
|{z}
<0
Therefore
v(P ) ≤ 1.
• In Q, the function v is minimum so that
∆g̃1 (Q) ≥ 0 ⇒ −cn ∆g̃1 (Q) ≤ 0,
In this way
n+2
µ (v n−2 − v)(Q) = −cn ∆g̃1 (Q) ≤ 0.
|{z}
<0
Therefore
v(Q) ≥ 1.
If we have a function which minimum value is bigger or equal than 1 and its maximum value
is smaller or equal than 1 we can assert than the function is identically equal to 1 (v ≡ 1). In
the same way that the previous case, we have uniqueness up to constant.
iii. However, we have possible non uniqueness of solutions if λ(M ) > 0. It will be explained in
the next section 2.
We would like show now how the uniqueness of solution for the fractional Yamabe problem also
depends on the sign of fractional Yamabe constant λγ (M ).
Remark 1.1. In the same way that in the classical case we had that sign of λ(M ) is equal to the
sign of Rg̃ (remark 4.7); in fractional case, we also can assert that the sign of λγ (M ) is equal to
the sign of Qg̃γ (which is constant).
4γ
i. If λγ (M ) < 0 : Given (X n+1 , g + ) with conformal infinity (M, ĝ), suppose that ĝ1 = û1n−2γ ĝ
4γ
is a solution with Qĝγ1 = µ1 . We also suppose that ĝ2 = û2n−2γ g is a solution with Qĝγ2 = µ2 .
So that we have
4
(96)
4
−4
4
ĝ2 = w2n−2γ ĝ = w2n−2γ w1n−2γ ĝ1 = w n−2γ ĝ1 .
This lets us assert that w is solution of
−div((ρ∗ )a ∇ḡ1 U ) = 0 in (X, ḡ1 ),
(97)
n+2γ
−d˜γ lim
(ρ∗ )a ∂ρ∗ U + wµ1 = µ2 w n−2γ (x, 0) on M.
∗
ρ →0
We are under the hypothesis µ1 , µ2 < 0, so up to constant, we can assert µ1 = µ2 = µ < 0.
If we take a look at the system, the first equation is a elliptic one, so the maximum (and
minimun) of U is attained at the boundary. So we can consider P ∈ M the point where U
reaches the maximum value, and Q ∈ M the point where it reaches the minimum value:
1. THE SIGN OF THE YAMABE CONSTANT.
37
• In P , the function U is maximum so that the outward normal derivative in this point
must be nonnegative, and negative if we take the derivative in the direccion of ∂ρ∗
lim (ρ∗ )a ∂ρ∗ U ≤ 0.
ρ∗ →0
In this way
n+2γ
−d˜γ lim
(ρ∗ )a ∂ρ∗ U (P ) = µ (w n−2γ − w)(P ).
|{z}
ρ∗ →0
|
{z
}
<0
≥0
Therefore
w(P ) ≤ 1.
• In Q, the function U is minimum so that
lim (ρ∗ )a ∂ρ∗ U ≥ 0
ρ∗ →0
In this way
n+2γ
−d˜γ lim
(ρ∗ )a ∂ρ∗ U (Q) = µ (w n−2γ − w)(Q).
|{z}
ρ∗ →0
|
{z
}
<0
≤0
Therefore
w(Q) ≥ 1.
If we have a function which minimum value is bigger or equal than 1 and its maximum value
is smaller or equal than 1 we can assert than the function is identically equal to 1 (w ≡ 1).
In the same way that the previous case, we have uniqueness up to constant.
ii. If λγ (M ) = 0 we also have uniqueness of solutions (up to constant).
4
Given (X n+1 , g + ) with conformal infinity (M, ĝ), suppose that ĝ1 = w1n−2γ ĝ is a solution with
4
Qĝγ1 = µ1 . We also suppose that ĝ2 = w2n−2γ g is a solution with Qĝγ2 = µ2 . Again, we have
(96). For this w there exists a unique U such that U |M = w and
−div((ρ∗ )a ∇ḡ1 U ) = 0 in X,
(98)
n+2γ
−d˜γ lim
(ρ∗ )a ∂ρ∗ U + w µ1 = µ2 w n−2γ (x, 0) on M.
|{z} |{z}
ρ∗ →0
=0
=0
That means U is solution of
−div((ρ∗ )a ∇ḡ1 U ) = 0 in X,
(99)
−d˜γ lim
(ρ∗ )a ∂ρ∗ U (x, 0) = 0 on M.
∗
ρ →0
Because of the lemma 3.1 in chapter 3 we can assert that the solution U = U (ρ∗ ), and this
means w = U |M =constant, as desired.
iii. However, we have possible non uniqueness of solutions if λγ (M ) > 0. This is one of our main
contributions and it will be explained in the next section 3.
38
3. NON UNIQUENESS FOR THE FRACTIONAL YAMABE PROBLEM
2. Classical case of non uniqueness
Richard M. Schoen considered the manifold R×Sn−1 and he proved the non uniqueness of solutions
for the Yamabe problem in this case [38]:
He took the canonical metric
g0 = dt2 + gSn−1
(100)
on R × Sn−1 , which has constant scalar curvature. And he found different metrics conformal to the
given one with constant scalar curvature. Using the computations done for the Yamabe problem
(20), we know that this problem is equivalent to find u smooth and strictly positive satisfying:
n+2
−cn 4g0 u + Rg0 u = n(n − 1)u n−2 .
(101)
We know that Rg0 = (n − 1)(n − 2). If we take into account only the radial solutions u = u(t), the
equation (101) becomes
n+2
−cn ∂tt u + (n − 1)(n − 2)u = n(n − 1)u n−2 .
Sustituting the value of cn (12) this equation is equivalent to
(n − 2)2
u.
4
Easily we can find two different solutions of this equation:
(102)
∂tt u =
n+2
−n(n−2) n−2
u
4
+
n−2
4 , which is the constant solution.
i. u0 = ( n−2
n )
n−2
ii. u1 = (cosh(t))− 2 .
Indeed,
i. Imposing that the solution is constant, we have ∂tt u = 0 and so that the equation (102)
becomes
n+2
(n − 2)2 u = n(n − 2)u n−2 ,
and because we are not looking for the trivial solution, the only constant solution is
n−2
n−2 4
(103)
u0 =
.
n
ii. For the non constant solution, we use the (conformal) spherical metric on Rn ,
P
4
g1 =
g0 , where g0 = i (dxi )2 . Now we do a change of variable |x| = et , getting
(1 + |x|2 )2
Sn −→ Rn ,
x 7→ log |x| = t.
Therefore the conformal change is given by
4
4
u=
=
= e−2t cosh(t)−2 .
2
2
(1 + |x| )
(1 + e2t )2
That means that g1 = e−2t cosh(t)−2 |dx|2 = cosh−2 (t)g0 , because |dx|2 = e2t g0 .
In this way we get the solution
(104)
u1 = (cosh−2 t)
n−2
2
= (cosh t)−
n−2
2
.
2. CLASSICAL CASE OF NON UNIQUENESS
39
Let us draw the phase-plane of (102):
We first transform (102) into a first order dynamical system. Denoting v = ∂t u, we call X(u, v) :=
(∂t u, ∂t v), for
n+2
X(u, v) = (v, −n(n−2)
u n−2 +
4
(105)
(n−2)2
u).
4
And now we look for the critical points of the system. We need find p = (p1 , p2 ) such that
X(p) = (0, 0).
One gets two critical points (0, 0) and (u0 , 0). The following step is linearize the equation or,
equivenlently, study which kind of critical point each one is. For that we compute the Jacobian
matrix of the system
!
n+2
n(n−2) n−2
(n−2)2
u
−
u
1
∂u ∂t (u)|(u,v) ∂v ∂t (u)|(u,v)
4
4
.
J(u, v) =
=
4
(n−2)2
∂u ∂t (v)|(u,v) ∂v ∂t (v)|(u,v)
− n(n+2) u n−2
0
4
4
Evaluating the Jacobian matrix at each critical point we get
•
J(0, 0) =
0
(n−2)2
4
1
0
!
.
And so that the eigenvalues are λ = ± n−2
2 , where n ≥ 3. That means that we have two real
eigenvalues: one is positive and the other one is negative. Therefore we can assert that the
point (0, 0) is a saddle point.
•
0
1
J(u0 , 0) =
.
−(n − 2) 0
√
And so that the eigenvalues are λ = ± 2 − n, where n ≥ 3. That means that we have two
complex conjugate eigenvalues, therefore the point (u0 , 0) is an saddle point.
We must note that for the linear system, the period of the orbits is given by
2π
(106)
Tu0 = √
,
n−2
√
√
since the orbits are u = A sin( n − 2t) + B cos( n − 2t) where A and B are constants.
Now we have the diagram of the solutions, but if we want know the exact solutions, we can come
back to (102) and multiply this equation by ut getting
(107)
∂tt u∂t u =
n+2
−n(n−2) n−2
∂t u
u
4
+
(n−2)2
u∂t u
4
= 0.
But we realise that:
• ∂tt u∂t u = 21 ∂t (∂t u2 ),
n+2
2n
n−2 ),
• u n−2 ∂t u = n−2
2n ∂t (u
• u∂t u = 12 ∂t (u2 ).
And we can integrate (107) with respect to time, getting that the Hamiltonian
2n
1
1 2
n−2 +
(108)
H := (∂t u)2 − n−2
u
(u
)
,
2n
2
2
40
3. NON UNIQUENESS FOR THE FRACTIONAL YAMABE PROBLEM
Fig. 1. Representation of the diagram of phases.
is constant along the trajectories Note that the trajectories are precisely given by the graph
r
v=±
(109)
2n
(n − 2)2 2 (n − 2)n n−2
u −
u
± c,
4
2
where c is constant.
The metric g1 is not a complete metric on R × Sn−1 . But taking the metric in S1 (T ) × Sn−1 ,
4
given by the previous trajectories, when T > Tu0 , gu = u n−2 (dt2 + gSn−1 ) for u = u(t) any, the
solution of the ODE is a complete metric with constant scalar curvature that are different from
the standard one.
So we can assert that we have non uniqueness of solutions to the Yamabe prpblem.
3. Non uniqueness for the fractional case
Let study the hyperbolic manifold described by (60)-(61) in the case n = 3. We denote X =
S1 (L) × R3 and M = S1 (L) × S2 and recall that the metric ḡ = ρ2 g + is given by
(110)
2
2
ρ2
ρ2
ḡ = dρ2 + 1 −
gS2 + 1 +
dt2 , and ĝ = ḡ|M = gS2 + dt2 .
4
4
3. NON UNIQUENESS FOR THE FRACTIONAL CASE
41
We calculate,
(111)
p
1
p ∂i (ḡ ij ρa |ḡ|∂j U )
|ḡ|
i,j
ρ2
ρ2 2
ρa
ρa
1
a
∂
ρ
(1
+
)(1
−
)
∂
U
+
∂
U
+
4S2 .
=
ρ
ρ
tt
2
2
2
2
4
4
(1 + ρ4 )(1 − ρ4 )2
(1 + ρ4 )2
(1 − ρ4 )2
divḡ (ρa ∇U ) =
X
And we recall that the fractional Yamabe problem (84) is about solving
(112)




−
1
(1 +
ρ2
)(1
4
−
ρ2 2
)
4
∂ρ
ρa (1 +
ρ2
ρ2 2
)(1 −
) ∂ρ U
4
4
ρa
+
(1 +
ρ2 2
)
4
∂tt U +
ρa
(1 −
ρ2 2
)
4
4S2 + E(ρ)U = 0 in (X, ḡ),
n+2γ




− lim d˜γ ρa ∂ρ U = cn,γ w n−2γ on M, where U |M = w.
ρ→0
We would like to prove that we have non uniqueness of solutions for the γ-Yamabe problem on M .
And for that, we are going to look for different solutions that only depend on ρ and t.
Lemma 3.1. Let X = R+ × Y , where Y is any compact manifold.
Given w a smooth function, let U be the unique solution to
(113)
f1 (ρ)∂ρ U + ∂ρρ U + f2 (ρ)∂tt U + f3 (ρ)U = 0 in X,
− lim d˜γ ρa ∂ρ U = cn,γ w on M, where U |M = w,
ρ→0
where f1 , f2 , f3 are any functions depending on ρ.
If w does not depend on t, then U is also independent on t, i.e, it depends only on the variable ρ.
Proof. We take any solution of the previous problem U (ρ, t), and because t is the variable in a
compact manifold, using properties of the eigenvalues of the Laplacian [17], we can assert that
X
U (ρ, t) =
uk (ρ)yk (t);
k
where yk (t) are the eigenfunctions of ∂tt , i.e, yk00 (t) = λk yk (t), and we know that y0 (t) = 1 and
λ0 (t) = 0. Using this property we get
(114)
f1 (ρ)u0k (ρ) + u00k (ρ) + (λk f2 (ρ) + f3 (ρ))uk (ρ) = 0.
On
P the other hand, we know that u|ρ=0 ≡ 1 is solution of the problem, so that
k uk (0)yk (t) = 1, but we know that yk (t) are linearly independent, so the only posible combination for getting this identity is uk (0) = 0 for all k 6= 0 and u0 (0) = 1. But now using the uniqueness
of solution, we obtain uk (ρ) = 0 for all k ≥ 1.
Therefore U (ρ, t) = u0 (ρ). That means the only solution, depends only on ρ.
t
u
First we are going to check that the given metric ĝ (110) has constant fractional curvature Qĝγ .
42
3. NON UNIQUENESS FOR THE FRACTIONAL YAMABE PROBLEM
If we take w = 1 we get the trivial conformal change, and the problem is
(115)



−
1
(1 +
ρ2
)(1
4
−
ρ2 2
)
4
∂ρ
ρa (1 +
ρ2 2
ρ2
)(1 −
) ∂ρ U
4
4
ρa
+
(1 +
ρ2 2
)
4
∂tt U +
ρa
(1 −
ρ2 2
)
4
4S2 + E(ρ)U = 0 in (X, ḡ),
− lim d˜γ ρa ∂ρ U = cn,γ on M , where U |M = 1.



ρ→0
For the solution U , lemma 3.1 in this chapter assures that U depends only on ρ. We compute the
fractional curvature
Qĝγ = Pγĝ (1) = − lim d˜γ ρa ∂ρ U (ρ) = cn,γ .
ρ→0
So we get the problem is well posed, and the trivial change is a solution.
Now we would like to solve the fractional Yamabe problem (112) with a non trivial conformal
change; so that, we are going to find another metric ĝw conformal to the first one with constant
fractional curvature Qĝγw , and such that w only depends on the variable t. This is a fractional order
ODE. So the classical methods for ODE do not work here. However, we still have some related
results, such as critical points and Hamiltonians.
Now we take the special defining function given in lemma 3.13 in order to find a simpler expressionfor problem (112).
Then, the Yamabe problem is about solving the following system:
div((ρ∗ )a ∇g∗ U ) = 0, in (X, g ∗ )
(116)
n+2γ
−d˜γ (ρ∗ )a ∂ρ∗ U + wcn,γ = cn,γ w n−2γ , on M,
where w = U |M and g ∗ =
(ρ∗ )2
ρ2 ḡ,
for ρ∗ = ρ∗ (ρ).
Computing the divergence with respect to the metric g ∗ , we get:
div((ρ∗ )a ∇g∗ U ) =
1
∗
( ρρ )4 (1 +
ρ2
4 )
∂ρ∗
ρ2
1
(ρ∗ )4
ρ2
∗ 1−2γ
+ ρ∗
∂
(ρ
)
(1
+
)∂t U
t
2
2
(ρ∗ )2
4
( ρ )4 (1 + ρ4 )
(1 + ρ4 )2 ρ4
∗ 4
X ρ2
1
ρ2 ij
∗ 1−2γ (ρ )
∂
)g
∂
U
+ ρ∗
(ρ
)
(1
+
2
i
j
2
(ρ∗ )2
ρ4
4 S
( ρ )4 (1 + ρ4 ) i,j
∗ 3−2γ
4ρ4
(ρ )
ρ2
∗
∗
=
∂
(1
+
)∂
U
ρ
ρ
4((ρ∗ )4 ) + ρ2 (ρ∗ )4
ρ2
4
ρ2
+
2 ∂tt U
(ρ∗ )1+2γ (1 + ρ4 )
1
(117)
(ρ∗ )4
ρ2
ρ2
(ρ∗ )1−2γ 4 (1 + )∂ρ∗ U
∗
2
(ρ )
ρ
4
+
!
ρ2
4S2 (U ).
(ρ∗ )1+2γ
So that if we try to solve the Yamabe extension problem with solutions depending only on t and
ρ∗ (note that we know that we can express ρ as a function of ρ∗ and viceversa), we have that (112)
is eqivalent to
3. NON UNIQUENESS FOR THE FRACTIONAL CASE
4ρ4
∂ρ∗
4((ρ∗ )4 ) + ρ2 (ρ∗ )4
(118)
ρ2
(ρ∗ )3−2γ
(1
+
)∂ρ∗ U
ρ2
4
+
43
ρ2
(ρ∗ )1+2γ (1 +
∂tt U
ρ2
4 )
= 0,
with the previous boundary condition.
3.1. Critical points. We are going to look for the critical points for the equation, finding the
constant solutions for the extension problem (116). If we find a solution constant on the boundary,
using the uniqueness of solution, we can assert that this solution is also constant in the extension,
by lemma 3.1, in this section.
For the constant solutions on M we impose ∂ρ∗ U = 0 and using the boundary condition of the
n+2γ
problem (116) we have to solve wcn,γ = cn,γ w n−2γ . So in the boundary the only non trivial
constant solution is w = 1. Or
U1 = 1,
in the extension.
The next step would be to classify these critical points. We hope to return to this problem
elsewhere. Once we have the critical points we would like draw the diagram of the phases. For
that we are going to find a constant Hamiltonian.
3.2. A conserved Hamiltonian.
Theorem 3.2. The hamiltonian
Z 2n
1 ∞ a
ρ2
ρ2
ρ2
H(t) :
ρ (1 + )(1 − )2 (∂ρ U )2 − ρa (1 − )(∂t U )2 + E 0 (ρ)(U 2 ) dρ−cn,γ U n−2γ
2 0
4
4
4
where E 0 (ρ) = (1 +
ρ2
4 )(1
−
ρ2 2
4 ) E(ρ),
is constant respect to t.
From (112) we have the equation
ρ2 2
ρa
1
ρ2
a
∂
)(1
−
)
∂
U
+
∂tt U + E(ρ)U = 0.
−
ρ
(1
+
ρ
ρ
2
2
2
4
4
(1 + ρ4 )(1 − ρ4 )2
(1 + ρ4 )2
2
2
First we multiply the equation by (1 + ρ4 )(1 − ρ4 )2 , and we get
ρ2
ρ2
ρ2
−∂ρ ρa (1 + )(1 − )2 ∂ρ U + ρa (1 − )∂tt U + E 0 (ρ)U = 0.
4
4
4
Multiplying it by ∂t u and integrating with respect to ρ ∈ (0, ∞) we obtain
Z ∞ Z ∞
ρ2
ρ2
ρ2
−
∂ρ ρa (1 + )(1 − )2 ∂ρ U ∂t U dρ −
ρa (1 − ) ∂tt U ∂t U dρ
| {z }
4
4
4
0
0
= 21 ∂t ((∂t U )2 )
Z
+
0
∞
E 0 (ρ) U ∂t U dρ = 0.
| {z }
= 12 ∂t (U 2 )
n−2γ
2n ,
44
3. NON UNIQUENESS FOR THE FRACTIONAL YAMABE PROBLEM
We can integrate by parts in the first term getting


Z ∞
2
2
2
2
 a

ρ (1 + ρ )(1 − ρ )2 ) ∂ρ U ∂tρ U  dρ − ρa (1 + ρ )(1 − ρ )2 ∂ρ U ∂t U |ρ=0

4
4
4
4
| {z } 
0
1
2
2 ∂t ((∂ρ U ) )
Z
− ∂t
0
∞
Z ∞
ρ2 1
1 2
2
0
E (ρ) (U ) dρ = 0.
ρ (1 − )( 2 )(∂t U ) dρ + ∂t
4
2
0
a
And using the boundary condition we have
Z ∞
Z ∞
ρ2
ρ2
ρ2 2
1
1
a
a
2
2
ρ (1 + )(1 − ) ((∂ρ U ) ) dρ − ∂t
ρ (1 − )(∂t U ) dρ
∂t
2
4
4
2
4
0
0
Z ∞
2
2
1
ρ
ρ
+ ∂t
E 0 (ρ)(U 2 ) dρ = lim (ρa (1 + )(1 − )2 )∂ρ U ∂t U ).
ρ→0
2
4
4
0
|
{z
}
n+2γ
1
w n−2γ
cn,γ d̃γ
On the other hand, we call
G0 (w) =
n+2γ
1
cn,γ w n−2γ ∂t w,
d˜γ
And integrating with respect to t we get
Z ∞
2n
1 n+2γ
G(U ) =
cn,γ w n−2γ ∂t w dt = cn,γ U n−2γ
˜
dγ
0
n−2γ
2n .
In this way we have
Z ∞
ρ2 2
ρ2
1
ρ2
2
a
2
0
2
a
∂t
ρ (1 + )(1 − ) (∂ρ U ) − ρ (1 − )(∂t U ) + E (ρ)(U ) dρ − ∂t (G(U )) = 0.
2
4
4
4
0
So that, we can assert that the hamiltonian
Z
H(t) =
0
∞
ρ2
ρ2 1
ρ2 1
1
ρ (1 + )(1 − )2 (∂ρ U )2 − ρa (1 − ) (∂t U )2 + E 0 (ρ) (U 2 )
4
4 2
4 2
2
a
2n
dρ − cn,γ U n−2γ
n−2γ
,
2n
is constant respect to t.
Now we rewrite the Hamiltonian in terms of the defining function ρ∗ .
We take (118) and multiply it by ∂t and integrating respect to ρ∗ we obtain
∗ 3−2γ
Z ∞
4ρ4
(ρ )
ρ2
0=
∂ρ∗
(1 + )∂ρ∗ U ∂t U dρ∗
4((ρ∗ )4 ) + ρ2 (ρ∗ )4
ρ2
4
0
!
Z ∞
2
1
ρ
2
+
∂t
dρ∗
2 (∂t U )
∗
1−2γ
2
(ρ )
(1 + ρ4 )
0
Integrating by parts the first term on the right hand side and taking into account that limρ∗ →0
1, we obtain
ρ
ρ∗
=
3. NON UNIQUENESS FOR THE FRACTIONAL CASE
n−2γ
0 = − cn,γ (w n+2γ − w)
Z
+ ∂t
0
∞
1
1
∂t w − ∂t
2
d˜γ
1
ρ2
2 (ρ∗ )1−2γ (1 +
Z
∞
ρ2
(ρ∗ )1−2γ
0
(∂ρ∗ U )2 −
45
ρ2
(ρ∗ )1−2γ (1 +
ρ2
4 )
(∂t U )2
!
2
ρ2
4 )
(∂t U )
dρ∗ .
n−2γ
If we call G0 (w) = −cn,γ (w n+2γ − w) d˜1 ∂t w, we can integrate getting
γ
G(w) = −cn,γ
.
2n
1 n − 2γ n−2γ
1
w
− w2
˜
2n
2
dγ
Therefore we have;
Z ∞
2n
1 2
ρ2
ρ2
1 n−2γ n−2γ
2
∗U) −
−
w
+
(∂
w
∂t cn,γ
ρ
2n
2
(ρ∗ )1−2γ
d˜γ
(ρ∗ )1−2γ (1 +
0
!
2
ρ2
4 )
(∂t U ) dρ
what means that the Hamiltonian
Z ∞
2n
1 2
ρ2
1 n−2γ n−2γ
ρ2
(119) H ∗ := cn,γ
w
−
w
+
(∂ρ∗ U )2 −
2n
∗
1−2γ
˜
2
(ρ )
dγ
(ρ∗ )1−2γ (1 +
0
is constant along trajectories.
∗
ρ2
4 )
=0
(∂t U )2 dρ∗
Chapter 4
Isolated singularities
In this chapter we are going to construct solutions of the fractional Yamabe problem in Rn with
an isolated singularity at the origin.
1. Classical case
First we look for non-trivial solutions for the equation
n+2
−4u = u n−2 in Rn ,
(120)
of the form
ũ(r) = u∞ rα ,
(121)
for some u∞ and α. Using polar coordinates we get
4u = ∂rr u +
1
n−1
∂r u + 2 4S2 ,
r
r
Sustituting (121) into (120) we obtain
4
n−2
(n − 1)α + α(α − 1) = u∞
r
2(n+2α−2)
n−2
,
so it follows that
(122)
α=
2−n
;
2
and
(123)
u∞ = [α(n + α − 2)]
2−n
4
.
These are cylindrical solutions. Now we are going to find all the radial solutions with an isolated
singularity at the origin of the same type as the cylinder. Thus we take
(124)
u(r) = u∞ r−
47
n−2
2
v(r).
48
4. ISOLATED SINGULARITIES
As before, we use polar coordinates and, imposing that u is solution of (121), we get
(2−n)2
v(r)
4
n+2
− r∂r (v(r)) − r2 ∂rr (r) = v(r) n−2 .
Doing the change of variable r = e−t and renaming w(t) = v(e−t ), we get the equation
(125)
n+2
(2 + n)2
w(t) − ∂tt w(t) = w(t) n−2 ,
4
which is the same equation, up to multiplicative constant, as (102), that we considered in the
section 2 in chapter 3 in relation to the classical case of non-uniqueness studied by Schoen [38].
Then, it is clear that the solutions we provided in chapter 3, section 2 will also be solutions for
this problem.
This is part of a general geometric setup, which we will describe in the following:
2. Relation with chapter 3
2.1. Classical case. We have studied in chapter 3, section 2, the non uniqueness for the problem
of finding a conformal metric with constant scalar curvature in the classical case M1 = R × Sn−1
with the metric ĝ = dt2 + gSn−1 . We consider M2 = Rn \ {0}, with the Euclidean metric in polar
coordinates ĝE = dr2 + r2 gSn−1 . Let us show that the problem of finding solutions in M2 with an
isolated singularity is equivalent.
Note that in the classical case the parameter γ that appears in the following computations is γ = 1
We follow this notation since these calculations are also valid for the fractional case.
We can find an application from M2 to M1 as:
f : M2 → M1
(126)
t 7→ r = e−t .
Computing the change of variable r = e−t , and taking into account ∂r = −e−t ∂t , we realize that
the metrics used in both M1 and M2 are conformally related:
(127)
ĝE = e−2t dt2 + e−2t gSn−1 = e−2t ĝ.
If we are looking for a metric conformal to the given one, in both cases, with constant scalar
curvature, the solution on M1 can be transformed in the solution on M2 . More precisely, suppose
4
w is a smooth function such that ĝw = w 2−2γ ĝ has constant fractional curvature, using that the
metrics are conformal, we have
4
4
ĝw = w n−2γ ĝ = w n−2γ e2t ĝE = (wet
n−2γ
2
4
) n−2γ ĝE .
Therefore we have a metric conformal to ĝE with constant curvature
(128)
w
ĝE
:= ĝw = (wet
n−2γ
2
4
4
) n−2γ ĝE = w n−2γ r−2 ĝE .
n−2γ
2
And the smooth function which gives us the conformal change is u = wr−
w smooth on M1 , u is smooth in M2 using the application defined in (126).
. Note that given
2. RELATION WITH CHAPTER ??
49
2.2. Fractional case. We will also see that, in the fractional case, the problem of finding solutions
with an isolated singularity for the fractional Yamabe problem in Rn with the Euclidean metric,
is equivalent to the study of nonuniqueness from section 3 in chapter 3.
So we call
+
X1 = Hn+1 , with the metric given by gE
=
dy 2 + |dx|2
,
y2
its compactification is the manifold
2
2
X̄1 = Rn+1
+ , with the metric given by ḡE = dy + |dx| ,
and its conformal infinity
M1 = Rn \ {0}, with the metric given by ĝE = |dx|2 ,
which in polar coordinates we can represent it as
M1 = R+ × Sn−1 , with the metric given by ĝE = dr2 + r2 gSn−1 .
And we also call, for n = 3,
(129)
1
n
+
X2 = S (L) × R , with the metric given by g = ρ
−2
ρ2 2
ρ2 2 2
dρ + (1 − ) gS2 + (1 + ) dt ,
4
4
2
or
ρ2 2
ρ2 2 2
2
ḡ = dρ + (1 − ) gS2 + (1 + ) dt
4
4
if we look the compatification. Its conformal infinity is now
M2 = S1 (L) × S2 , with the metric given by ĝ = dt2 + gSn−1 .
The conformal equivalence between the conformal infinities (M1 , ĝE ) and (M2 , ĝ) was shown above.
Note that M1 is a covering of M2 if we make the t variable periodic.
But in the fractional case we also need to prove the equivalence in the extension. Let us prove
that X1 is a covering of X2 in the case n = 3.
[4] We consider the hyperbolic space X1 = H4 , and for any geodesic σ ⊂ H4 , a translation by a
fixed length L along σ extends to an isometry of H4 . Let X2 = H4 /Z ≈ R3 ×S1 (L) be the quotient,
with Z the group generated by the translation. We have that X1 is a covering of X2 . The metric
g + on H4 /Z may be written as
g + = ds2 + sinh2 sgS2 + cosh2 dt2 ,
where t parametrizes a circle of radius L.
We compute the change of variable R = sinh s, and so that dR = cosh sds, and we get that the
metric in X2 is
dR2
g+ =
+ R2 dt2 + (1 + R2 )gS2 ,
1 + R2
which is precisely the metric (129) given in the generalized hyperbolic manifold as we saw in
(60)-(61).
Getting in this way the equivalence between metrics, as desired.
50
4. ISOLATED SINGULARITIES
3. Fractional case
3.1. Local analysis of fractional semi-linear elliptic equations with isolated singularities. We would like find solutions with a singularity at the origin for the fractional Yamabe
problem in Rn . That means look for solutions of
n+2γ
(−4)γ u = u n−2γ in Rn \ {0}.
This problem is equivalent to
−div(y a ∇U ) = 0 in Rn+1
+ ,
(130)
n+2γ
− lim y a ∂y U (x, y) = U n−2γ (x, 0) on Rn \ {0}.
y→0
We are going to refer some important theorems given in [13].
Theorem 3.1. Suppose that U is a nonnegative solution of
−div(y a 4U ) = 0 in Bn+1
∩ Rn+1
+ ,
2
(131)
n+2γ
− lim y a ∂y U (x, y) = U n−2γ (x, 0) on Bn2 .
y→0
Then u = U |M can be extended as a continuous function near 0, or there exist two positive constants
c1 and c2 such that
c1 |x|−
n−2γ
2
≤ u(x) ≤ c2 |x|−
n−2γ
2
.
Theorem 3.2. If U is a non negative solution of (85), then
u(x) = ū(|x|)(1 + O(|x|)) as x → 0,
where ū(|x|) =
H
Sn
ū(|x|θ) dθ is the spherical average of u.
Theorem 3.3. Let U be a nonnegative solution of (85) in Rn+1
+ . And suppose that the origin 0 is
not removable. Then U (x, t) = U (|x|, t) and ∂r U (r, t) < 0 for all 0 < r < ∞.
Our aim is to find all these radial solutions, for n = 3. Let set X = R4+ and M = R3 \ {0}. It
n−2γ
is clear that we need to look for solutions u = r 2 w(r) for some w bounded. However, this
is precisely the setting in (128). From our previous discussion, the problem reduces to problem
studied in chapter 3 section 3. This is a fractional order ODE, for which new tools need to be
developed. The idea of finding a Hamiltonian to understand a fractional ODE is not new (see
[11, 24]). However, our results show the existence of a Hamiltonian quantity which is preserved
along trajectories.
Chapter 5
Anti-de-Sitter space
As we mencioned in chapter 2, this example is interesting because it gives two different examples
of conformally compact Einstein manifolds with the same conformal infinity. It is not known yet if
the scattering operator for both extensions coincide [20] Important references for this chapter are
[28, 16] and the lectures given by Graham in ”Mini-courses and Conference on Nonlinear Elliptic
Equations” (May 13-18, 2013, Rutgers University and May 20-22, 2013, Courant Institute).
1. Motivation from physics (informal)
Definition 1.1. Anti-de Sitter space is the submanifold described by one of the sheets of the
hyperboloid of two sheets
3
X
x2i − x20 = −α2 ,
i=1
with the pseudometric given by
(132)
ds2 =
3
X
dx2i − dx20 ,
i=1
and where α is a nonzero constant with dimensions of length (the radius of curvature). It satisfies
the following relation with the cosmological constant:
1
α = Λ− 2 .
Remark 1.2. If we take the hyperboloid of one sheet
3
X
x2i − x20 = α2 we obtain the called de
i=1
Sitter space.
The pseudometric of the covering space of anti-de Sitter space can be written in the static form
(133)
gAdS = −V dτ 2 + V −1 dr2 + r2 dS,
51
52
5. ANTI-DE-SITTER SPACE
where
V =1+
(134)
r2
,
b2
−3 1/2
) ,
Λ
dS denotes the usual metric of the sphere S2 .
b=(
But this metric is not Euclidean, so we can do the change of variable t = iτ (called ”Wick
Rotation”) and we obtain a positive definite metric.
(135)
gAdS+ = V dt2 + V −1 dr2 + r2 dS.
More generally, we are going to consider of the Schwarzchild-anti-de Sitter metric, which has the
form of (135), where
V =1−
(136)
r2
2m
+
.
m2p r
b2
In this expression we have used the following notation:
• m = mass of the black hole (a positive number).
1
• mp is a constant called “Planck mass” and given by mp = G− 2 , where G is the gravitational
constant
2m
r2
• rh is the positive root of 1 − m
2 r + b2 = 0 (because the metric must be positive).
p
• r ∈ [rh , +∞).
• t ∈ S1 (λ).
• (θ, φ) ∈ S2 .
2. Rigorous definition
We restrict to the case the case b = 1, mp = 1 to simplify the computations. Then the AdSSchwarzchild as
(137)
m
(R2 × S2 , g+1
),
where
(138)
m
g+1
= V dt2 + V −1 dr2 + r2 gS2 .
for
(139)
V = 1 + r2 −
2m
.
r
Definition 2.1. We call rh the positive root for 1 + r2 −
(θ, φ) ∈ S2 .
2m
r
= 0, so r ∈ [rh , +∞), t ∈ S1 (λ) and
Lemma 2.2. Even though the metric seems singular ar rh , we will prove that this is not the case
if we make the t variable periodic.
m
In order to make t periodic we are going to do a change of variable and we get the metric g+1
is
2
n−1
smooth in R × S
.
3. UNIQUENESS
53
Since we would like dρ2 = V −1 (r) dr2 , we define
ρ : (rh , ∞) −→ (0, ∞),
Z r
−1
V 2 .
ρ(r) =
rh
Using the Taylor expansion in the point rh , we obtain that for all r near of rh
V (r) ∼ V 0 (rh )(r − rh ).
(140)
We call Vh0 = V 0 (rh ) and substituting (140) in the definition of ρ we obtain
Z r
−1
−1
−1
1
ρ∼
(Vh0 ) 2 (r − rh ) 2 = (Vh0 ) 2 2(r − rh ) 2 .
rh
Isolating r − rh , we obtain
Vh0 2
ρ + ...
4
r − rh =
And using this approximation in (140), the metric (138) can be written as
m
g+1
= dρ2 +
where for periodicity we must impose 0 ≤
(Vh0 )2 2 2
ρ dt + r2 dgS2 ,
4
Vh0
2 t
≤ 2π. So that, we must have
0 ≤ t ≤ 2πL,
where
(141)
L=
Vh0
.
2
Remark 2.3. The manifold (137)-(138) is conformally compact Einstein. Indeed, let s =
using the definition of V (r) from (139) we obtain
V (r) = 1 +
=V
−1
and
1
1
1
− 2ms = 2 (1 + s2 − 2ms3 ) ≈ 2 , when s → 0 (and therefore r → ∞).
2
s
s
s
We also obtain that dr =
m
g+1
1
r
2
−dt
t2
and so that
2
2
(r) dr + V (r) dt + r dgS2 ∼ t
2
ds
s2
2
+
1 2
1
1
dt + 2 dgS2 ∼ 2 [ds2 + dt2 + d2gS2 ].
s2
s
s
3. Uniqueness
We would like to find two different conformally compact Einstein manifolds with the same conformal
infinity.
For each m, gm is an asymptotically hyperbolic Einstein manifold with conformal infinity S1 (L)×S2 .
Now we are going to observe how this radius L depends on m.
Remark 3.1. Since there is an univocal relation between rh and m (because of definition of rh
(2.1)), we are going to study how L depends on rh .
54
5. ANTI-DE-SITTER SPACE
Because the definition of rh (2.1),
V (rh ) = 0 some as rh3 + rh = 2m.
(142)
Compute
V 0 (r) = 2r +
2m
,
r2
and so that, using (142)
(143)
V 0 (rh ) = 2rh +
rh + rh3
1
2m
1
=
2r
+
= 2rh + rh +
= 3rh + .
h
rh2
rh2
rh
rh
Therefore, using the definition of L we have
(144)
L=
2rh
.
3rh2 + 1
Proposition 3.2. At the conformal infinity (r = ∞), the metric is given by
m
V −1 g+1
= dt2 + dS2
(145)
Proof. We know r ∈ [rh , ∞) and we have proved that the metric is smooth at rh so the conformal
infinity must be at r = ∞. Restricting to this value of r, we obtain
m
ĝm := V −1 g+1
|r=∞ = V V −1 dt2 + (V −1 )2 dr2 + V −1 r2 dS2 .
| {z }
=0
ĝ
m
2
= dt + ddS2 .
t
u
Proposition 3.3. Depending of the value of L there exist one, two or zero values of m such that
m
has the conformal infinity S(L) × S2 :
g+
• If L < L0 there exist two different masses m1 and m2 with the same L(mi ); and thus they
give same conformal infinity.
• If L = L0 there exist only one mass m and which gives us L(m).
• If L > L0 there does not exist any mass which gives us L(m).
Proof. Since L(r) =
2r
3r 2 +1 ,
we can assert that
L0 (rh ) =
So the unique critical point for L(rh ) is rh0 =
−2(3rh2 − 1)
.
3rh2 + 1
√1 ,
3
and
1
L0 = L(rh0 ) = √ .
3
If we calculate the second derivative
L00 (rh ) =
−24rh
< 0.
(3rh2 + 1)2
We can conclude that rh0 is a maximum for L(rh ) and so for each 0 < L <
different masses m1 , m2 which share the same L, as desired
√1
3
there are two
t
u
3. UNIQUENESS
55
L
1/√3
rh
r
Fig. 1. Representation of L(r)
Because the previous affirmation, for the same conformal infinity S1 (L) × S2 when 0 < L < √13 ,
+
+
there are two non-isometric AdS-Schwarzschild spaces with metrics gm
and gm
on R2 × S2 . In
1
2
this way we have proved the non-uniqueness for conformally compact Einstein metrics on the
topologically same 4-manifold.
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