U.U.D.M. Project Report 2020:27

Examensarbete i matematik, 30 hpHandledare: Anna Sakovich Examinator: Julian KülshammerAugusti 2020

Department of MathematicsUppsala University

Spacetime Penrose Inequality For Asymptotically Hyperbolic Spherical Symmetric Initial Data

Mingyi Hou

Spacetime Penrose Inequality For Asymptotically

Hyperbolic Spherically Symmetric Initial Data

Mingyi Hou

Abstract

In 1973, R. Penrose conjectured that the total mass of a space-time containingblack holes cannot be less than a certain function of the sum of the areas of theevent horizons. In the language of differential geometry, this is a statement about aninitial data set for the Einstein equations that contains apparent horizons. Roughlyspeaking, an initial data set for the Einstein equations is a mathematical objectmodelling a slice of a space-time and an apparent horizon is a certain generalizationof a minimal surface. Two major breakthroughs concerning this conjecture weremade in 2001 by Huisken and Ilmanen respectively Bray who proved the conjecturein the so-called asymptotically flat Riemannian case, that is when the slice of aspace-time has no extrinsic curvature and its intrinsic geometry resembles that ofEuclidean space. Ten years later, Bray and Khuri proposed an approach using theso-called generalized Jang equation which could potentially be employed to dealwith the general asymptotically flat case by reducing it to the Riemannian case.Bray and Khuri have successfully implemented this strategy under the assumptionthat the initial data is spherically symmetric.

In this thesis, we study a suitable modification of Bray and Khuri’s approach inthe case when the initial data is asymptotically hyperbolic (i.e. modelled on a hy-perboloidal slice of Minkowski spacetime) and spherically symmetric. In particular,we show that the Penrose conjecture for such initial data holds provided that thegeneralized Jang equation has a solution with certain asymptotic behaviour nearinfinity. Furthermore, we prove a result concerning the existence of such a solution.

Contents

1 Introduction 1

2 Preliminaries 42.1 Apparent Horizons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Asymptotically Flat Initial Data . . . . . . . . . . . . . . . . . . . . . 52.3 Asymptotically Hyperbolic Initial Data . . . . . . . . . . . . . . . . . 52.4 Spherically Symmetric Asymptotically

Hyperbolic Initial Data . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 The Generalized Jang Equation 83.1 The Generalized Jang Equation Approach . . . . . . . . . . . . . . . 83.2 Spherical Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4 Motivation for Asymptotics and the Warping Factor 124.1 Behavior at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.2 Proof of the Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 Existence and Asymptotics 185.1 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185.2 Asymptotics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Chapter 1

Introduction

In the 1970s, physicist R. Penrose conjectured that the total mass of a spacetimecontaining black holes with event horizons of total areaA should be at least

√A/16π.

In his argument, cosmic censorship, a famous conjecture saying that singularitiesarising in the solutions of the Einstein equations are typically hidden within eventhorizons, is one of the fundamental ingredients. In fact, a proof of a suitable versionof the Penrose inequality would give indirect support to cosmic censorship and acounter example of the inequality would very likely involve a spacetime for whichcosmic censorship fails to hold [Mar09]. Thus, finding a proof for the inequality hasbecome an active area of research.

An important case of the Penrose inequality is a statement about asymptoticallyflat initial data. Recall that a spacetime is a Lorentzian manifold (N , gN) satisfyingthe Einstein equations,

RicN −1

2RN gN = 8πT (1.1)

where T is the so-called stress-energy tensor, RicN is the Ricci curvature tensor andRN is the scalar curvature. We call a triple (M, g,K) an initial data set (or Cauchydata) if (M, g) is a Riemannian manifold and K is a symmetric 2-tensor, such thatthe so-called constraint equations

µ =1

2(Rg − |K|2g + trg(K)2)

J = divg(K − trg(K)g)(1.2)

are satisfied. Here Rg is the scalar curvature and µ and J are physically energyand momentum density respectively. In the case when (M, g) is a space-like sliceof a spacetime (N , gN), with unit normal n, we have µ = T (n, n) and J = T (n, ·).In what follows, we will assume that any given initial data satisfies the so-calleddominant energy condition

µ > |J | (1.3)

which is known to be satisfied in all physically reasonable spacetimes. Note that if(1.2) is satisfied then (M, g,K) is indeed the initial data required to pose an initialvalue problem for the Einstein equations in the sense that the time evolution ofthe initial data is a spacetime (N , gN) satisfying (1.1) and (M, g) is a space-likehypersurface in the spacetime with second fundamental form K [BI04].

1

Roughly speaking, an initial data (M, g,K) is asymptotically flat if near infinityg approaches the Euclidean metric δ and K approaches 0. For an asymptoticallyflat initial data set there is a well known notion of mass [ADM08] that in a certainsense measures how fast the initial data set becomes “flat” at infinity.

The argument which can be used to motivate the Penrose inequality roughly goesas follows [Mar09]. Assume a spacetime (N , gN ) which is asymptotically flat, in thesense that it looks like the Minkowski spacetime [ONe83] at infinity. Intuitively, theevent horizon H, is the boundary of the region around a black hole which not eventhe light can escape. Note that it is a null hypersurface which is at least Lipschitzcontinuous [Wal84]. Assume now, that the spacetime admits an asymptoticallyflat slice with total ADM mass MADM and which intersects H on a cut S. If His a smooth hypersurface, then this cut is a smooth embedded surface which hasa well-defined area |S|. Consider any other cut S1 to the future of S along theevent horizon. The black hole area law [Haw71; Haw72] states |S1| > |S|, providedthat the null energy condition, which is a certain strengthening of (1.3), holds. Fromphysical considerations, the spacetime is expected to settle down to some equilibriumconfiguration. Assuming that all matter fields are swallowed by the black hole inthe process, uniqueness theorems for stationary black holes (see e.g. [Heu96]) implythat the spacetime must approach the Kerr black hole spacetime, where the areaof the section S of the event horizon is independent of the cut and takes the valueAKerr = 8πM(M +

√M2 − L2/M2) 6 16πM2, where M is the total mass and L

is the total angular momentum of the spacetime. In particular, M should be theasymptotic value of the Bondi mass along the future null infinity. Since gravitationalwaves carry positive energy, the Bondi mass cannot increase to the future [BVM62;SB62]. The Penrose inequality

MADM >√|S|/16π

follows provided that the Bondi mass approaches the ADM mass MADM which isknown to be true under certain assumptions (see e.g. [Hay03; Kro03; AM79]).

After a period of heuristic proofs and partial results, important breakthroughshave been made in the past decades. First, Huisken and Ilmanen [HI01] were ableto prove the Penrose inequality in the so called time symmetric case, i.e. whenthe initial data has K = 0, using a geometric evolution equation known as theinverse mean curvature flow. Shortly afterwards, another proof using the so-calledconformal flow was found by Bray. Roughly speaking, the result of [HI01] is for asingle black hole, while that of [Bra01] allows for multiple ones. In summary, wehave the following result in the time symmetric case:

Theorem 1 (Riemann Penrose Inequality). Let (M, g) be a complete, smooth,asymptotically flat 3-manifold with nonnegative scalar curvature, total mass M , andwith the boundary (possibly disconnected) which is an outermost minimal surface oftotal area A. Then

M >

√A

16π(1.4)

with equality if and only if (M, g) is isometric to a Schwarzschild manifold(R3 \ 0, gSchw).

2

The Penrose inequality is a strengthening of the famous positive mass theorem,saying that an asymptotically flat initial data (M, g,K) satisfying dominant energycondition has MADM > 0, with equality if and only if it is a slice of the Minkowskispacetime. This result was proven in [SY82] by reducing it to the time symmetriccase K = 0 that was earlier settled in [SY79]. The proof uses a certain quasi-linearelliptic partial differential equation, the so-called Jang equation.

It was suggested by Bray and Khuri (see [BK11; BK10]) that a generalizationof the Jang equation could be a way to reduce the general Penrose inequality tothe case of time symmetry. This idea has been successfully implemented in theasymptotically flat spherically symmetric case [BK10]. In this thesis, we providesome further support to the generalized Jang equation approach of Bray and Khuriby implementing it in the case when the initial data set is asymptotically hyperbolicspherically symmetric. This setting is of great interest for both geometrical andphysical reasons, in particular, it is important for the study of gravitational radiation[Fra04].

The thesis is organized as follows: In Chapter 2, we introduce terminology re-quired in this thesis. In Chapter 3, we briefly review the generalized Jang equationand rewrite it as an ordinary differential equation under the assumption of sphericalsymmetry. In Chapter 4, we prove a theorem stating that the Penrose inequality forasymptotically hyperbolic spherically symmetric initial data holds provided that asolution to the generalized Jang equation exists and has certain asymptotic behaviorat infinity. Finally, in Chapter 5, we prove the existence of a solution to the gener-alized Jang equation and obtain a (non-optimal) result regarding its asymptotics.

3

Chapter 2

Preliminaries

In this chapter, we introduce a few definitions that will be used in this thesis.

2.1 Apparent Horizons

An important concept involved in the Penrose conjecture is that of event horizon.However, in the setting of the initial data set formulation of the Penrose inequality,it is problematic to compute the area of the event horizons of black holes. Indeed,the event horizons are not determined by the local geometry of the initial data, butone is required to solve the Einstein equation for detecting them. For this reasonone usually replaces the notion of event horizon by the concept of apparent horizon.

Definition 2 (Apparent Horizons). Let (M, g,K) be an initial data set. A compactoriented hypersurface S ⊂ M is called an apparent horizon (or a marginally outerrespectively inner trapped surface) if S satisfies

HS ± trS(K) = 0 (2.1)

where HS is the mean curvature of S.

In fact, we may define the so-called null expansions

θ± = HS ± trS(K) (2.2)

which measure the rate of change of area for a shell of light emitted by the surfacein the outward future (respectively past) direction. Thus the gravitational field isinterpreted as being strong near S if θ+ < 0 or θ− < 0, in which case S is referredto as a future (past) trapped surface. Future (past) apparent horizons arise asboundaries of future (past) trapped regions and satisfy the equation θ+ = 0 (θ− = 0).

Here, an outermost future (respectively past) apparent horizon refers to a future(respectively past) apparent horizon outside of which there is no other apparent hori-zon. Such a horizon may have several components, each having spherical topology[Gal08].

For example, consider the initial data (R3 \ 0, gsc, K = 0) representing t = 0slice of the Schwarzschild spacetime [ONe83], the simplest nontrivial solution to theEinstein equations, where (gsc)ij = (1 + m/2r)4δij and m is a positive constantwhich can be associated with the mass (see below). This manifold has zero scalarcurvature everywhere, is spherically symmetric, and it has an outermost apparenthorizon at r = m/2.

4

2.2 Asymptotically Flat Initial Data

We now introduce the rigorous notion of asymptotically flat initial data that hasbeen used in the heuristic argument given in the Introduction.

Definition 3 (Asymptotically Flat Initial Data). An initial data set (M, g,K) iscalled asymptotically flat if outside a compact setM is diffeomorphic to the comple-ment of a ball in (R3, δ) and if in the Cartesian coordinates xi, i = 1, 2, 3, inducedby this diffeomorphism g and K have the following asymptotic behavior:

|gij − δij|+ r|∂kgij|+ r2|∂2klgij| = O(r−1), Kij + r|∂kKij| = O(r−2) (2.3)

as r →∞ where r = |x| =√δijxixj.

The total mass of an initial data set is essentially a geometrical invariant thatdepends on the geometry “at infinity” of a non-compact Riemannian manifold. Forexample in the Schwarzschild spacetime, time-like geodesics (which represent testparticles) curve in the coordinate chart as if they were accelerating towards thecenter of the spacetime at a rate asymptotic to m/r2 in the limit as r goes toinfinity. Hence, to be compatible with Newtonian physics we must define m to bethe total mass of the Schwarzschild spacetime.

For asymptotically flat manifolds the total ADM energy-momentum vector isdefined through the coordinate expressions

EADM =1

2(n− 1)ωnlimr→∞

∫Sr

(∂jgij − ∂igjj)νidµδ (2.4)

(PADM)i = limr→∞

1

(n− 1)ωn

∫Sr

(Kij − trg(K)gij)νjdµδ (2.5)

where Sr is the coordinate sphere of radius r, ν is the unit outward normal to Sr,ωn is the volume of the n-dimensional unit sphere in Rn, dµδ is the area form on Srinduced by δ. Under some additional decay conditions on g and K these quantitiesare independent of the choice of coordinates xi [Bar86]. The total ADM mass isdefined as

MADM =√E2ADM − |PADM |2 (2.6)

where |PADM |2 = δij(PADM)i(PADM)j.

2.3 Asymptotically Hyperbolic Initial Data

The terminology of asymptotically hyperbolic setting is rather sophisticated. Forthis paper, it will suffice to use the following definition [Sak20]. Let H3 denote the3-dimensional hyperbolic space and let gH denote its metric. In polar coordinates itcan be expressed by gH = dr2

1+r2+ r2σ, where σ is the standard round metric on S2.

Definition 4 (Asymptotically Hyperbolic Initial Data). An initial data set (M, g,K)is said to be asymptotically hyperbolic if there exists a compact subset Ω and a dif-feomorphism Φ : M \ Ω→ Hn \ BR, where BR is a ball in H3, such that

Φ∗g =dr2

1 + r2+ r2(σ + mr−3 +O2(r−4))

Φ∗(K − gH)|TS2×TS2 = pr−1 +O1(r−2)

(2.7)

5

where m ∈ C2(S2) and p ∈ C1(S2) are symmetric 2-tensors on S2 and the notationOl(r

−n) asserts that rn+i|∂ir∂αp| ≤ C for all i+ |α| ≤ l.

For the convenience of the reader, we briefly review the definition of total massfor asymptotically hyperbolic initial data sets. First we define a vector space V :=V ∈ C∞(Hn)|HessgHV = V gH. It has the basis of functions

V(0) =√

1 + r2, V(1) = xir, i = 1, 2, 3

where x1, x2, x3 are the coordinate functions on R3 restricted to S2. If we considerH3 as the upper unit hyperboloid in Minkowski space R3,1, then the functions V(i)

are the restrictions to H3 of the coordinate functions of R3,1.The mass functional of an asymptotically hyperbolic initial data set with respect

to the chart Φ is then a linear functional HΦ on V defined by

HΦ(V ) = limr→∞

∫Sr

(V (divgHe− dtrgH (e)) + trgH (e)dV − (e+ 2η)(∇gHV, ·)

)(νr)dµ

gH

where e = Φ∗g − gH , η = Φ∗(K − g) and νr =√

1 + r2∂r is the unit outer normalto Sr. This functional transforms appropriately under isometries of the hyperbolicspace. The components of the energy-momentum vector (E,P ) are given by

E =1

16πHΦ(V(0)), Pi =

1

16πHΦ(V(i)), i = 1, 2, 3. (2.8)

A computation shows that in the case when the initial data has asymptotics (2.7)the energy is given by

E =1

16π

∫S2

(trσm + 2trσp)dµσ. (2.9)

In what follows we abuse the terminology slightly by calling E the mass, whilephysically it is the total energy. For more details, we refer the reader to [Mic11].

2.4 Spherically Symmetric Asymptotically

Hyperbolic Initial Data

The main object of interest in this thesis is the spherically symmetric asymptot-ically hyperbolic initial data. From now on we make the assumption that all initialdata sets in this thesis are spherically symmetric which means that the manifoldMis diffeomorphic to R3 \B(r0) for some ball with radius r0 and g = g11dr2 +ρσ,K =κdr2 + γσ for some smooth functions g11, ρ, κ, γ depending only on r. Note that forsuch an initial data the outermost apparent horizon has only one connected com-ponent so we can assume it is the boundary of the manifold. A straightforwardcomputation shows that in this case we can write the null expansions as follows

θ± =√g11

ρ′

ρ± 2γ

ρ. (2.10)

6

Recalling the definition of asymptotically hyperbolic initial data, the followingassumption will be imposed on g11, ρ, κ, γ for the rest of the thesis, for r →∞

g11 =1

1 + r2, ρ = r2(1 +

mg

r3+O3(r−4)),

κ =1

1 + r2(1 +O2(r−3)), γ = r2(1 +

mk

r3+O2(r−4))

(2.11)

where mg,mk ∈ R.Since we are in dimension 3, the quantities mg,mk encode mass through the

formula

m =1

16π

∫S2

trσ(mgσ + 2mkσ

)=mg + 2mk

2. (2.12)

7

Chapter 3

The Generalized Jang Equation

In this chapter, we will restrict our attention to spherically symmetric initial data.The first part contains a discussion of the Jang equation approach to the Penroseconjecture suggested by Bray and Khuri. In the second part, we transform thegeneralized Jang equation into an ODE under the assumption of spherical symmetry.

3.1 The Generalized Jang Equation Approach

The classical Jang equation was introduced by physicist P. S. Jang in 1978 andwas successfully employed to reduce the positive mass theorem for general initial datato the case of time symmetry [SY81]. However, the Jang equation has been shownto be unsuitable for addressing the Penrose inequality [MM04]. For this reason, ageneralization of this equation has been proposed in [BK11]. In this section, weintroduce the so-called generalized Jang equation.

In the proof of the time symmetric Penrose inequality, nonnegative scalar cur-vature plays a central role. Thus, in the general case, one is motivated to deformthe initial data in an appropriate way so that we can get the positivity of the scalarcurvature for the deformed metric. The deformation used to prove the positivemass theorem is given by g = g + df 2, for some function f defined on M. Notethat g is the induced metric on the graph Σ = t = f(x) in the product 4-manifold(M×R, g+dt2). It turns out that the scalar curvature will be “almost” nonnegativewhen Σ satisfies Jang’s equation

HΣ − trΣ(K) = 0

where HΣ is mean curvature of Σ and K is trivially extended toM×R. Motivated bythe case of equality for Penrose inequality [BK10], we will consider the Jang surfaceΣ in a warped product manifold (M × R, g + u2dt2), where u is a nonnegativefunction defined onM. It is also shown in [BK10] that in order to achieve “almost”positivity for scalar curvature we would like the Jang surface Σ to satisfy an equationwith the same structure but with K extended non-trivially, namely

HΣ − trΣ(K) = 0 (3.1)

8

where

K(∂xi , ∂xj) = K(∂xi , ∂xj), for 1 ≤ i, j ≤ 3,

K(∂xi , ∂t) = 0, for 1 ≤ i ≤ 3,

K(∂t, ∂t) =u2g(∇f,∇u)√

1 + u2|∇f |2.

(3.2)

Here xi, i = 1, 2, 3, are local coordinates on M .This particular extension yields an optimal positivity property for the scalar

curvature of the solutions to (3.1). A long calculation [BK10; BK11] gives thefollowing result for the scalar curvature of Σ satisfying equation (3.1) with extension(3.2):

R = 16π(µ− J(w)) + |h− K|Σ|2g + 2|q|g −2

udivg(uq) (3.3)

where g = g + u2df 2 and h are the induced metric and second fundamental formof Σ respectively, K|Σ is the restriction to Σ of the extended tensor K, and q is a1-form and w is a vector with |w|g ≤ 1 given by

w =uf ,i∂xi√

1 + u2|∇gf |2, qi =

uf ,j√1 + u2|∇gf |2

(hij − (K|Σ)ij)

where f ,i = gikf,k and f,k = ∂kf . If the dominant energy condition is satisfied,then all terms appearing on the right-hand side of (3.3) are nonnegative, except forpossibly the last term.

When the tensor K is extended according to (3.2), we will refer to equation (3.1)as the generalized Jang equation, and the graphical submanifold Σ = t = f(r) willbe called the Jang surface throughout the paper. In local coordinates the generalizedJang equation takes the following form:(

gij − u2f if j

1 + u2|df |2g

)(u∇ijf + uifj + fiuj√

1 + u2|df |2g−Kij

)= 0 (3.4)

where ∇ijf is the i, j component of covariant hessian ∇2f with respect to g, i.e.∇ijf = ∂j∂if − Γkij∂kf where Γkij are Christoffel symbols of g.

3.2 Spherical Symmetry

In the general case the generalized Jang equation is an elliptic PDE. However,under spherical symmetry, we can rewrite it as an ODE. First we observe thatequation (3.4) consists of two parts, namely the radial part and the spherical part(

g11 − u2(g11)2(f ′)2

1 + u2|df |2g

)(u∇rrf + 2u′f ′√

1 + u2|df |2g−K11

)+ gab

(u∇abf√

1 + u2|df |2g−Kab

)= 0

where a, b = φ1, φ2 (here and in what follows a and b refer to spherical coordinates).Then we make the following substitution [BK10]:

v :=u√g11f ′√

1 + u2g11(f ′)2(3.5)

9

and compute its first derivative with respect to r coordinate

v′ =

√g11u′f ′ +

√g11(g11)′uf ′/2 +

√g11uf ′′√

1 + g11u2(f ′)2

− u2u′(g11)3/2(f ′)3 + u3√g11(g11)′(f ′)3/2 + u3(g11)3/2(f ′)2f ′′

(1 + u2g11(f ′)2)3/2.

We can multiply this by√g11 to obtain√

g11v′ = g11(1− v2)u∇rrf + 2u′f ′√

1 + u2g11(f ′)2−√g11(1− v2)

u′

uv.

Hence the radial part becomes(g11 − u2(g11)2(f ′)2

1 + u2|df |2g

)(u∇rrf + 2u′f ′√

1 + u2|df |2g− k11

)

= g11(1− v2)(u∇rrf + 2u′f ′√

1 + u2|df |2g− k11)

=√g11v′ +

√g11(1− v2)

u′

uv − g11(1− v2)κ.

On the other hand, the spherical part is

gab(u∇abf√

1 + u2|df |2g− kab) =

√g11

ρ′

ρv − 2γ

ρ.

Therefore adding the two parts we get the following equation√g11v′ + g11(1− v2)(

√g11

u′

uv − κ) +

√g11

ρ′

ρv − 2γ

ρ= 0. (3.6)

As will be explained in next chapter, for the proof of Penrose inequality it is rea-sonable to take

u =

√1− v2

√g11

ρ′

2√ρ. (3.7)

As mentioned before we expect our warping factor to be nonnegative and this isindeed the case for (3.7) since

HSr,g =√g11

ρ′

ρ=

1

2(θ+ + θ−) > 0 ∀r > r0 (3.8)

and |v(r)| 6 1 for all r > r0 by (3.5). With the above choice of u we obtain anordinary differential equation√

g11(1− v2)v′ +√g11(1− v2)

(ρ′′

ρ′− g′11

2g11

− ρ′

2ρ

)v

−g11(1− v2)κ+√g11

ρ′

ρv − 2γ

ρ= 0.

(3.9)

10

Note that for v 6= ±1 we can write this as√g11(1− v2)v′ + (1− v2)F±(r, v)∓ θ± = 0 (3.10)

where

F±(r, v) =√g11

ρ′′

ρ′v −

√g11

g′11

2g11

v − g11κ−√g11

ρ′

ρ

(v2− 1

−v ± 1

).

With our application in mind, we would like a blow-up (or down) solution atthe horizon which translates to v(r0) = ∓1 for the outermost future (past) horizon.Hence we are interested in establishing the existence and regularity results for thefollowing initial value problem:√

g11(1− v2)v′ + (1− v2)F±(r, v)∓ θ± = 0

v(r0) = ∓1(3.11)

11

Chapter 4

Motivation for Asymptotics andthe Warping Factor

Recall that in Chapter 3 we made a specific choice of the warping factor u (seeequation (3.7)). In this chapter, we motivate it and discuss the desired asymptoticbehavior at infinity for a solution of (3.11). For the proof of the Penrose inequality,we will need the mass of the Jang surface to coincide with the mass of our initialdata. Hence in the first section, we investigate asymptotics that can preserve mass.

4.1 Behavior at Infinity

The generalized Jang equation for asymptotically flat initial data set is studied indetail in [HK13]. In that setting the warping factor is assumed to have the followingasymptotics

u(r) = 1 +u0

r+O2(r−2+ε) (4.1)

where u0 is a constant to be determined. Also, f and hence v are expected to vanishat infinity. Similar asymptotics for u are expected in the asymptotically hyperbolicsetting. However, the asymptotic behavior of the solution of the generalized Jangequation is more complicated. In analogy with the approach to the positive masstheorem [SY82; Sak20] we impose the following asymptotics as r →∞

f(r) =√

1 + r2 + 2m log r +O3(r−1+ε) (4.2)

for any ε > 0. Furthermore, we will assume that at the horizon f blows up (ordown) and u vanishes just like in the asymptotically flat setting.

Motivation for imposing (4.1) and (4.2) partly comes from the fact in this casethe induced metric on the Jang surface will be asymptotically flat so that we cancompute the ADM mass of (Σ, g) where

g = g + u2df 2 =(g11 + u2(f ′)2

)dr2 + ρσ.

Lemma 5. Consider the Jang surface Σ = t = f(r) in the warped product man-ifold (M× R, g + u2dt2). If f and u satisfy (4.2) and (4.1) respectively, then theJang metric g is asymptotically flat, and the ADM mass of the Jang metric is givenby mADM = 2m+ u0.

12

Proof. It is clear that the manifold (Σ, g) has an end diffeomorphic to R3 \Br0 , withcoordinates y = (r, φ1, φ2) as in (2.11). Let x denote the associated Cartesian coor-dinates, related to y through the usual spherical transformation. In what follows,i, j, . . . are indices for x-coordinates and α, β, . . . are indices for y-coordinate. Itfollows that

gij = gij + u2f,if,j =(gαβ + u2f,αf,β

)∂yα∂xi

∂yβ

∂xj

=(g11 + u2(f ′)2

) ∂r∂xi

∂r

∂xj+ gab

∂ya

∂xi∂yb

∂xj.

From (2.11), (4.1) and (4.2) we get

g11 + u2(f ′)2 = 1 +O(r−1), gab = ρσab = r2σab +O(r−1),

which implies

(gαβ + u2f,αf,β)∂yα

∂xi∂yβ

∂xj= δij +O(r−1)

since the leading terms are exactly the standard coordinate transformation of Eu-clidean metric from spherical to Cartesian. Next we compute

∂gij =∂gij∂xk

=∂(gαβ + u2f,αf,β)

∂xk∂yα

∂xi∂yβ

∂xj+ (gαβ + u2f,αf,β)

( ∂yα

∂xk∂xi+

∂yβ

∂xk∂xj

)=∂(gαβ + u2f,αf,β)

∂yγ∂yγ

∂xk∂yα

∂xi∂yβ

∂xj+ (gαβ + u2f,αf,β)

( ∂yα

∂xk∂xi+

∂yβ

∂xk∂xj

)=∂gαβ∂yγ

∂yγ

∂xk∂yα

∂xi∂yβ

∂xj+(2u′u(f ′)2 + 2u2f ′′f ′

) ∂r∂xk

∂r

∂xi∂r

∂xj

+ (gαβ + u2f,αf,β)( ∂yα

∂xk∂xi+

∂yβ

∂xk∂xj

).

Since ∂r/∂xi = O(1), ∂ya/∂xi = O(r−1) and ∂ya/∂xi∂xj = O(r−2), together withassumption (4.2) and (4.2) we get |∂gij| = O(r−2). Similarly we can get |∂2gij| =O(r−3).

Details of the computation for mass can be found in [CKS16].

Remark. From the computation for mass it follows that we can actually relax theassumptions (4.1) and (4.2) a little bit by assuming

u(r) = 1 +u0

r+O1(r−2+ε), u′′(r) = O(r−3),

f(r) =√

1 + r2 + 2m log r +O2(r−1+ε), f ′′′(r) = O(r−3).

To provide further motivation for the above assumptions, we examine them forour substitution (3.5). Suppose v is a smooth function that has the following seriesexpansion at infinity

v(r) = 1 +B/r + C/r2 +O2(r−3).

It follows that

1− v2(r) = −2B

r− 2C +B2

r2+O2(r−3) = −α1

r− α2

r2+O2(r−3).

13

We then substitute the expansions into (3.10). Since√g11(1− v2)v′(r) = −−α1B

r2+O(r−3),√

g11(1− v2)v(r)(ρ′′(r)

ρ′(r)− g′11(r)

2g11

− ρ′(r)

2ρ(r)) = +

−α1

r+−α1B − α2

r2+O(r−3),

g11(1− v2)κ(r) = −α1

r− α2

r2+O(r−3),√

g11ρ′(r)

ρ(r)v(r) = 2 +

2B

r+

1 + 2C

r2+O(r−3),

2γ(r)

ρ(r)= 2 +O(r−5),

we conclude from (3.10) that

2B

r+−2α1B − α2 + α2 + 1 + 2C

r2+O(r−3) = 0.

Consequently, if v is a solution we get the following relations2B = 0

−2α1B + 1 + 2C = 0

It follows that v(r) = 1− 1/2r2 +O2(r−3) for r large enough.Now we know that v(r) = 1−1/2r2+D/r3+O2(r−4+ε) so that we can obtain more

terms in the expansion for v . After a long computation we get D = (2mk+mg)/2 =m. So the mass shows up in the solution of the generalized Jang equation.

Next we show that we can indeed recover (4.2) and (4.1) from the asymptoticsof v. By definition

v(r) =u(r)

√g11f ′(r)√

1 + u2(r)g11(f ′(r))2⇒ (f ′(r))2 =

4ρv2

(1− v2)2(ρ′(r))2(g11)2.

It follows that

f ′(r) =2√ρv

(1− v2)ρ′g11=

1− 1/2r2 + (mg/2 +D)/r3 +O2(r−4)

1− 2D/r +O2(r−2)

=r√

1 + r2+

2m

r+O2(r−2+ε).

Similarly, we obtain

u(r) =

√1− v2

√g11

ρ′(r)

2√ρ(r)

=√

1 + r2√

1− v2(2r −mg/r2 +O2(r−3))

2√r2 +mg/r +O3(r−2)

=(1 +1

2r2+O(r−4))(1− D

r+O2(r−2))(1 +O2(r−3))

=1 +−mr

+O2(r−2+ε).

This gives us u0 = −m and mADM = m. All in all, the computation indicates thatour assumptions (4.2) and (4.1) are convincing and provide the direction for provingthe asymptotics.

14

4.2 Proof of the Inequality

In this section, we prove the Penrose inequality in the case of spherical symmetryby assuming that a solution to the generalized Jang equation exists and satisfies (4.2)and (4.1). Recall that in this case the scalar curvature of the Jang surface is given by(3.3). The important point to note here is that the Jang surface is also sphericallysymmetric.

For convenience we write η =√ρ so that g = g11dr2 + η2σ. Recall the definition

of the Hawking mass in a 3-dimensional manifold:

mH(r) =

√Ag(Sr)

16π

(1− 1

16π

∫Sr

H2Sr,gdσg

)(4.3)

where

HSr,g =

√1− v2

√g11

ρ′(r)

ρ(r).

Hence in the case of spherical symmetry we have

mH =1

2η(1− η2

,s) (4.4)

where s is the geodesic coordinate on Σ:

s =

∫ r

0

√g11 + u2(f ′)2 =

∫ r

0

√g11√

1− v2.

Furthermore, its derivate involves the scalar curvature of Σ

m′H(r) =η2η,s

4R, (4.5)

where R = 2η2

(1−η2,s−2ηη,ss) is the scalar curvature of Σ with respect to the induced

metric g [LS14]. Integrating (4.5) with respect to r coordinate we get

mH(∞)− mH(r0) =1

4π

∫Σ

η,sR dµg (4.6)

where dµg is the volume form on the Jang surface Σ. Now we may apply Schoen-Yauidentity (3.3) and choose u in a way that it cancels with the term in front of thedivergence, i.e.

u = η,s =

√1− v2

√g11

η,r =

√1− v2

√g11

ρ,r2√ρ. (4.7)

Note that u is nonnegative due to null expansions assumption (see (3.7) and (3.8)).It follows from divergence theorem that

mH(∞)− mH(r0) =1

4π

∫Σ

η,s(2(µ− J(w)) + |h−K|Σ|2g + 2|q|2g

)dµg

− 1

2π

∫Σ

divg(uq)dωg

>− 1

2π

∫∂Σ∪∂∞

u〈q, ng〉gdµg

(4.8)

where ng is the unit outer normal (viewed as a 1-form). Here in the last inequality,we use the fact that dominant energy condition (1.3) is satisfied.

15

Lemma 6. If a blow up (or down) solution of IVP (3.11) exists and satisfies (4.2)(4.1) then mH(∞) = m, mH(r0) =

√A(S0)/16π and∫

∂Σ∪∂∞u〈q, ng〉gdσg = 0. (4.9)

where (Σ, g) is the Jang surface as described above.

Proof. First we look at the Hawking mass of Σ. Since the solution blows up at thehorizon we conclude HSr0 ,g

= 0 and consequently mH(r0) =√A(S0)/16π.

The Hawking mass at infinity is the limit

mH(∞) = limr→∞

√Ag(Sr)

16π

(1− 1

16π

∫Sr

H2Sr,gdσg

)= lim

r→∞

√ρ

4

(1− u2(r)

)=− u0.

It is a well-known fact that for asymptotically flat spherically symmetric manifoldswe have mADM = mH(∞). Futhermore, by Lemma (5) we know that mADM =2m+ u0. It follows that 2m+ u0 = −u0 so mADM = −u0 = m.

Now we prove (4.9). Consider ∫Sr

u〈q, ng〉gdσg (4.10)

where Sr are coordinate spheres. Since v(r0) = ±1, u vanishes at the horizon.Consequently, the integral vanishes at the horizon. For the spatial infinity, weconsider the following limit

limr→∞

∫Sr

u〈q, ng〉gdσg.

Recall the definition of q in (3.3):

q1 =ug11f ′√

1 + u2g11(f ′)2(π11 −K11), q2 = q3 = 0

where

π11 =u∇rrf + 2u′f ′√

1 + u2|df |2g=

1

r2− 2u0 + 2m

r3+O(r−4+ε).

Note that ng = ds =√g11√

1−v2 dr, dσg = ρ(r)dσ, hence

u〈q, ng〉gdσg = ug11q1

√g11√

1− v2ρdσ = uq1ρ

√1− v2

√g11

dσ.

We may now plug the above in (4.2), (4.1) and the expression for q1 to get

limr→∞

uq1

√1− v2

√g11

ρ = limr→∞

u√g11v(π11 −K11)

√1− v2

√g11

ρ = −(2u0 + 2m).

16

Hence ∫Sr

ug(q, ng)dσg = −4π(2u0 + 2m).

This vanishes as u0 = −m.

All in all, we obtain the following result

Theorem 7. Let (M, g,K) be a spherically symmetric asymptotically hyperbolicinitial data set satisfying (2.11) and ∂M is an outermost future (past) horizon. Ifthe IVP (3.11) admits a solution and if for r →∞

v(r) =r√

1 + r2+m

r3+O2(r−4+ε),

then

m ≥√

A

16π

where m and A are the mass of the initial data and the area required to enclose theoutermost apparent horizon, respectively. If the equality holds then the initial dataset arises from an embedding into the Schwarzschild spacetime.

Proof. The inequality follows from the above lemma directly. For the case of equal-ity, inspecting the proof of Lemma (6), we see that

0 = m−√

A

16π≥ 1

4π

∫Σ

η,s(2(µ− |J |g) + |h−K|Σ|2g + 2|q|2g

)dµg.

Hence

µ− |J |g = 0, h− K|Σ = 0, q = 0.

It follows from (3.3) that R = 0. We may now apply Theorem 1 to (Σ, g) to obtaing ∼= gsc, that is g is isometric to the Schwarzschild manifold

gsc =(1− 2m

r

)−1dr2 + r2σ.

Hence

g11 =(1− 2m

r

)−1, ρ = r2 ⇔ η = r.

Therefore

u = η,s =√g11η,r =

(1− 2m

r

)−1/2.

Since g = g − u2df 2 = gsc − u2df 2, the map G : (M, g)→ (SC4, gsc − u2dt2), wherex 7→ (x, f(x)), provides an isometric embedding. Finally, a calculation [BK10]shows that h− K|Σ = 0 implies that the second fundamental form of G(M) ⊂ SC4

is precisely given by the initial data K.

17

Chapter 5

Existence and Asymptotics

In this chapter, we study the initial value problem (3.11) as it is the ingredientleft to complete the proof of the Penrose inequality for asymptotically hyperbolicspherically symmetric initial data. In the first part, we establish the existence of asolution. Then we study the asymptotic behavior of the solution in detail.

As a preliminary, we recall the following comparison result in [Sak20] for ordinarydifferential equations which will help us to prove the existence and establish theasymptotics.

Lemma 8. Let F : [r0,∞) × [−1, 1] → R be continuous in both variables. Iffunctions l = l(r) and k = k(r) satisfy l′ + F (r, l) < k′ + F (r, k) and l(r0) 6 k(r0)then l(r) 6 k(r) for r > r0.

The proof is rather simple and uses contradiction. This lemma allows us to boundthe solution by suitable functions which may be called barriers (or super and sub-solutions).

5.1 Existence

The main difficulty in proving the existence of a solution is that the IVP (3.11)degenerates for v = ±1, that is v′ is “killed” by 1− v2 when v = ±1, which preventsus from using classical theorems for ODEs. In order to handle the degeneracy, wewill perform another substitution to “hide it”. Moreover, it will be convenient touse the geodesic coordinate near horizon, i.e.

τ =

∫ r

r0

√g11, (5.1)

in which (3.11) becomes(1− v2(τ))v′(τ) + (1− v2(τ))F±(τ, v)∓ θ±(τ) = 0v(0) = ∓1

(5.2)

where

F±(τ, v) =ρ′′(τ)

ρ′(τ)v(τ)− κ(τ)− ρ′(τ)

ρ(τ)

(v(τ)

2− 1

−v(τ)± 1

).

We will work with this coordinate system in this section.

18

Proposition 9. Consider the initial value problem (5.2). If θ±(τ) > 0 for τ > 0and θ+(0) = 0, θ−(0) = 0 or θ±(0) = 0 depending on the type of horizon and iffurthermore θ′+ > 0 (θ′− > 0 for past horizon case), then it admits at least onesolution. Moreover, the solution satisfies |v(τ)| < 1 for all τ > 0.

Proof. We first observe that the degeneracy, the term 1 − v2 in front of v′, can be“hidden” by the following substitution

z = v − v3

3(5.3)

It follows that for v ∈ [−1, 1] we have z ∈ [−23, 2

3] and it is increasing. The inverse

function v = v(z) is also strictly increasing. Moreover, we can extend z(v) outside[−1, 1] continuously so that its inverse remains continuous and increasing, for ex-ample using linear functions. We first consider the outermost past horizon case. Itis straightforward to check that (5.2) takes the form

z′(τ) + (1− v2(τ))F−(τ, v) + θ−(τ) = 0z(0) = 2

3

(5.4)

Note that F− is continuous in both arguments due to our assumptions on the initialdata. Similarly θ− is smooth for all τ > 0. Then it follows by Peano’s existencetheorem [Har64] that there exists at least one solution in a rectangle τ ∈ [0, a] and|z − 2/3| 6 b.

Next, we prove an a priori estimate for z on the whole interval [0,∞) and thenextend the solution. It is obvious that ±2/3 are two barriers for all solutions inbetween when τ > 0. Hence it remains to show that the solution z enters thatregion, i.e. z(ε) < 2/3 for arbitrarily small ε > 0. Since we already know thatz(0) = 2/3 and z′(0) = 0 from (5.4), we need to look at the second derivative ofz at τ = 0. From (5.3) it is obvious that z′′(0) = −2(v′(0))2. We can then get aquadratic equation for v′(0) by differentiating (5.2) at τ = 0

2(v′(0))2 + 2F−(0, 1)v′(0)− θ′−(0) = 0. (5.5)

Now we use the assumption that θ′−(0) > 0 to obtain v′(0) is real and bounded. Thisgives z′′(0) is indeed negative and consequently z is decreasing at 0. Finally, we canextend the solution to the whole interval by the extension theorem in [Har64].

The same argument applies to the outermost future horizon case, i.e. θ+(0) = 0.We conclude that the initial value problem admits a solution v ∈ C∞((0,∞)) ∩C1([0,∞]) and |v| < 1 for τ > 0.

5.2 Asymptotics

In this section, we investigate the asymptotics of solutions of IVP (3.11) whoseexistence was proven in Section 5.1. We prove a rough estimate in Proposition 10,and subsequently refine it in Proposition 11. In both cases we use suitable barrierfunctions to achieve our goal.

Proposition 10. Let v = v(r) be the solution of IVP (3.11). Then for any suffi-ciently small ε > 0 there exists r1 > r0 such that

v(r) = 1− 1

2r2+O(r−3+ε). (5.6)

19

Proof. Consider the generalized Jang equation

J(v) := (1− v2)v′ + (1− v2)(a(r)v −

√g11κ

)+(ρ′ρv − 2γ

ρ√g11

)= 0 (5.7)

where

a(r) =ρ′′

ρ′− g′11

2g11

− ρ′

2ρ.

The asymptotics of the terms in the Jang equation when r is big enough are asfollows:

a(r) =1

r− 1

r3+

3mg

r4+O(r−5),√

g11κ =1

r− 1

2r3+O(r−4),

ρ′

ρ=

2

r− 3mg

r4+O(r−5),

2γ

ρ√g11

=2

r− 1

r3+mk −mg

r4+O(r−5).

Suppose v− = 1− 12r2− αr3−ε

−r3−ε , where α = 2− 1

2r2−. Without loss of generality we can

assume 1 < α < 2 for r− big. Hence 0 6 1− v2− 6 1 and

v′− =1

r3+α(3− ε)r3−ε

−

r4−ε .

Then

(1− v2−)(v′− + a(r)v− −

√g11κ)

=(1− v2−)

(α(2− ε)r3−ε

−

r4−ε +O(r−4) +αr3−ε−

r6−ε −3mgαr3−ε

−

r7−ε +O(r−5)αr3−ε−

r3−ε

)6α(2− ε)r3−ε

−

r4−ε +C1

r4+αr3−ε−

r6−ε

and

ρ′

ρv− −

2γ

ρ√g11

=− 2αr3−ε−

r4−ε −2m

r4+O(r−5) +

3mgαr3−ε−

r7−ε +O(r−5)αr3−ε−

r3−ε

6− 2αr3−ε−

r4−ε +C2

r4.

Note that here we use the fact 0 < r−r6 1, so C1, C2 depend only on the initial data

and are independent of r−. It follows that

J(v−) =(1− v2−)(v′− + a(r)v− −

√g11κ) +

ρ′

ρv− −

2γ

ρ√g11

6− αεr3−ε−

r4−ε +C−r4

+αr3−ε−

r6−ε 6 −εr3−ε−

r4−ε +C−r4

+2r3−ε−

r6−ε

20

where C− = C1 + C2. Thus J(v−) < 0 provided that r− > 0 is chosen so that−εr3

− + 2r− + C− < 0.

Similarly, we can construct a super-solution by considering v+(r) = 1− 12r2

+r1−ε+

2r3−ε .Hence 0 6 1− v2

+ 6 1 and

v′+ =1

r3− (3− ε)r1−ε

+

2r4−ε

Then similarly

(1− v2+)(v′+ + a(r)v+ −

√g11κ)

=(1− v2+)

(−(2− ε)r1−ε

+

2r4−ε +O(r−4)− r1−ε+

2r6−ε +3mgr1−ε

+

2r7−ε +O(r−5)r1−ε

+

2r3−ε

)>− (2− ε)r1−ε

+

2r4−ε − C3

r4

and

ρ′

ρv+ −

2γ

ρ√g11

=2r1−ε

+

2r4−ε −2m

r4+O(r−5)− 3mgr1−ε

+

2r7−ε +O(r−5)r1−ε

+

2r3−ε

>2r1−ε

+

2r4−ε −C4

r4.

It follows that

J(v+) =(1− v2+)(v′+ + a(r)v+ −

√g11κ) +

ρ′

ρv+ −

2γ

ρ√g11

>εr1−ε

+

2r4−ε −C+

r4

where C− = C3 + C4. Thus J(v+) > 0 provided r+ > 2C+/ε.Finally, we can conclude v = 1− 1

2r2+O(r−3+ε) for r > r1 = maxr−, r+.

Proposition 11. Let v = v(r) be the solution of IVP (3.11). Then for any suffi-ciently small ε there exists r2 > r0 such that

v(r) = 1− 1

2r2+m

r3+O(r−4+ε) (5.8)

Proof. We improve the result from Proposition 10. Consider v = 1− 12r2

+ ψ, thenψ = O(r−3+ε) by Proposition 10. We fix ε and r1 as in Proposition 10. It followsthat |ψ| 6 C(ε, r1)/r3−ε where C(ε, r1) is fixed and depends only on ε and r1. Wecan now estimate the term 1− v2 more accurately:

1− v2 6 1−(

1− 1

2r2− C(ε, r1)

r3−ε

)6

1

r2

(1 +

2C

r1−ε −1

4r2− C

r3−ε −C2

r4−2ε

)6

1

r2(1 + C0)

21

where C0 is a constant.First, we construct a lower barrier. Suppose ψ− = m/r3 − (Cr2 + mr1−ε

2 )/r4−ε,then ψ−(r2) = −C/r3−ε

2 . Therefore v(r2) > v−(r2) := 1 − 12r22

+ ψ−(r2). Recall the

left hand side of the generalized Jang equation J(v) = 0 is given by (5.7). We plugin v = v−:

J(v−) =(2− v2−)(ψ′− + a(r)ψ− +O(r−4)) + (

ρ′

ρ− 2m

r4+O(r−5))

=(1− v2−)(−3m

r4+

(4− ε)ϕ−r2

r5−ε +O(r−4)

− r2φ−r5−ε +

ϕ−r2

r7−ε −3mgϕ−r2

r8−ε +O(r−5)ϕ−r2

r4−ε )

+ (−2ϕ−r2

r5−ε +O(r−5) +3mgϕ−r2

r8−ε +O(r−5)ϕ−r2

r4−ε ).

Here we denote ϕ− := C + m/rε2. For a fixed ε > 0, ϕ− is bounded provided r2 isbig enough. Hence

J(v−) 6(1− v2−)(

3ϕ−r4−ε +

C1

r4) + (−2ϕ−r2

r5−ε +C2

r5)

61

r2(1 + C0)(

3ϕ−r4−ε +

C1

r4)− 2ϕ−r2

r5−ε +C2

r5

6− ϕ−r2

r5−ε +C−r5.

Thus J(v−) < 0 provided −2ϕ−r1+ε2 + C− < 0.

Similarly we can construct an upper barrier v+ = 1 − 12r2

+ ψ+, where ψ+ =mr3

+ (Cr2 −mr1−ε2 )/r4−ε. Finally, we conclude that

v(r) = 1− 1

2r2+m

r3+O(r−4+ε).

Remark. Note that in order to complete the proof of the spacetime Penrose inequal-ity for asymptotically hyperbolic spherically symmetric initial data we need to showthat v(r) = 1 − 1

2r2+ m

r3+ O2(r−4+ε), in particular, v′ = 1/r3 − 3m/r4 + O(r−5+ε)

and v′′ = O(r−4). At the same time, plugging v into (3.10) and solving for v′ onlygives v′ = O(r−3+ε), which is related to the fact the equation degenerates as v → 1.Note that this difficulty is not present in the asymptotically flat setting of [BK10]as v → 0 for r → ∞. The question whether the optimal decay for v′ and v′′ can beestablished will be addressed elsewhere.

22

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