The Riemannian Positive Mass Theorem

THE RIEMANNIAN POSITIVE MASS THEOREM

FARHAN ABEDIN

Contents

1. Introduction 22. Basic Notions from Analysis, Geometry and Partial Differential

Equations 42.1. Sard’s Theorem 42.2. Co-Area Formula 52.3. Sobolev Spaces 72.4. Existence Results from Elliptic Partial Differential Equations 122.5. Local Boundedness of Weak Solutions 152.6. A Quick Review of Riemannian Geometry 192.7. Second Variation of Area 252.8. Conformal Change of Scalar Curvature 272.9. Linearization of Scalar Curvature 282.10. Einstein’s Equations 303. Schoen and Yau’s proof of the Riemannian Positive Mass Theorem for

n = 3 313.1. Setup of the Problem 313.2. Statement of Results 333.3. Proof of Claim 1 343.4. Proof of Claim 2 353.5. Proof of Claim 3 373.6. Proof of the Rigidity Theorem 444. Appendix 51References 53

1

2 FARHAN ABEDIN

1. Introduction

The Newtonian formulation of gravitation was, and remains, a tremendouslysuccessful theory in physics, due to its ability to determine the force exerted by onebody on another through the elegant equation F = Gm1m2

r2 . Its success howeverleads naturally to an issue that was left unresolved by Newton and many othersafter; namely, how is the force of gravity transferred from one object to another?The general theory of relativity, formulated by Einstein in 1915, was intended toanswer this very question. The solution comes from a set of equations that arealmost as elegant as Newton’s equation. The basic assumption is that space-timeis a four-dimensional Lorentzian manifold (N4, g), with one dimension for time andthe remaining three for space. The Einstein equations, Ric(g) − R(g)g = T , showvery explicitly that gravity is tied to the geometry of space-time. In particular, thepresence of mass directly influences the geometry of the space-time surroundingit. The oft-used analogy is that of a sheet of rubber becoming distorted when abowling ball is placed on it, where the sheet represents the surrounding space-time,and the bowling ball represents a massive object. To extend the analogy, if a golfball is placed in the vicinity of the bowling ball, the golf ball will tend to gravitatetowards the bowling ball due to the curvature of the rubber sheet.

Of course, Einstein’s theory is not devoid of its own set of thorny issues. Inparticular, it is a delicate matter to define the notion of energy in general relativ-ity, since vacuum space-times can be non-trivial. A simple ansatz to adopt undersuch circumstances is to assume that the mass of the universe is contained entirelyinside a compact domain. This would imply that the effects of gravity tend todecay near infinity and so the geometry becomes more Minkowskian (i.e. the met-ric is asymptotically flat). In general relativity, the scalar curvature R(g) of themetric represents an energy density, and the Einstein-Hilbert action

∫NR(g) dVg

is an expression for the total energy. We will see later that the principle of leastaction applied to the Einstein-Hilbert action quickly yields the Einstein equations.Inspired by these variational arguments, the physicists Arnowitt, Deser and Misnerproposed in [1] the following definition for the mass, now known as the ADM mass.

mADM = limσ→∞

1

16π

∫|x|=σ

∑i,j

(gij,i − gii,j)νσj dxσ.

Under certain assumptions on the asymptotic behavior of the metric, it can beshown that this limit exists and is indeed a geometric invariant (cf. [2]). In order toobtain a sensible definition of mass, however, one must show that if the space-timecontains non-negative energy density, then the mass of the space-time should alsobe non-negative. This is in essense, the statement of the Positive Mass Theorem.In [14] Schoen and Yau showed through a beautiful argument, using variationaltechniques and the theory of minimal surfaces, that the ADM mass of an asymp-totically flat manifold is indeed non-negative if the scalar curvature is non-negative.Furthermore, if the ADM mass is zero, then the space-time has zero curvature. Theproof of the Positive Mass Theorem was a watershed for both modern mathematicsand physics and its influence on geometry cannot be overstated.

Various extensions and applications of the Positive Mass Theorem have beenobtained since the publication of Schoen and Yau’s proof. Soon after, Witten ([19])

THE RIEMANNIAN POSITIVE MASS THEOREM 3

provided a simplified proof of the Positive Mass Theorem in dimension three usingspinors, while Bartnik ([2]) extended this theorem to spin manifolds of arbitrarydimension. Miao ([11]) proved the Positive Mass Theorem for the case where themetric fails to be C1 across a hypersurface, while Wang ([18]) considered the caseof asymptotically hyperbolic manifolds and showed positivity of an invariant thatcan be interpreted as the total mass. Very recently, Lam ([9]) obtained a directand elegant proof of the Positive Mass Theorem for graphical manifolds, whichwas extended to the hypersurface case by the work of Huang and Wu in [8]. ThePositive Mass Theorem has also been used to establish other groundbreaking resultsin geometry. In particular, Schoen used the Positive Mass Theorem extensively inhis work which completed the proof of the Yamabe problem (cf. [12]), where oneattempts to find a metric g of constant scalar curvature in the conformal class of agiven metric g on a closed manifold M . More recently, Bray ([3]) used the PositiveMass Theorem to prove the Riemannian Penrose Inequality, which states that if(M, g) is an asymptotically flat 3-manifold with non-negative scalar curvature and

mass m, and if A is the area of the outermost minimal surface, then m ≥√

A16π ,

with equality if (M, g) is isometric to a slice of Schwarzschild space-time outsidethe outermost minimal surface.

The purpose of this paper is to present a fairly complete proof of the RiemannianPositive Mass Theorem, following closely the structure of the proof in [14]. InSection 2, we provide the relevant theorems from analysis, geometry and partialdifferential equations that are used in the proof and in Section 3, we prove both thePositive Mass Theorem and the corresponding Rigidity Theorem in the zero masscase.

Acknowledgements This paper partially fulfills the requirements for succesfulcompletion of an Honors Thesis in Mathematics at Lafayette College. I would liketo thank my adviser, Professor Justin Corvino, for the countless occassions on whichhe has assisted me in obtaining a thorough understanding of the ideas presented inthis paper, and for his general counsel on matters pertinent to the other less logicalaspects of life.

4 FARHAN ABEDIN

2. Basic Notions from Analysis, Geometry and Partial DifferentialEquations

2.1. Sard’s Theorem. Recall that the rank of a smooth map g : M → N at apoint p ∈ M is the rank of the derivative of g at p, where the derivative is viewedas a linear map Dpg : TpM → Tg(p)N . Given such a smooth map g, the maximumrank that g can have is min dim(M),dim(N). The set of points where the rankof g is strictly less than dim(N) are called critical points. All other points of Mare called regular points. A point q ∈ N such that f−1(q) contains at least onecritical point is called a critical value. All other points of N are called regularvalues. Observe that if dim(M) < dim(N), then all points of N are critical values.If dim(M) = dim(N), then a point p ∈ M is critical if and only if the Jacobian issingular at p.

We also introduce the concept of a set of measure zero. A set A ⊂ Rn is said tobe of measure zero if for any ε > 0, there exists a sequence of balls B(ri) of radiusri such that A ⊂ ∪∞i=1B(ri) and

∑∞i=1 Vol(B(ri)) < ε. We are now ready to prove

Sard’s Theorem for the case dim(M) = dim(N).

Theorem 1. Let φ : C → Rn be a differentiable map, where C ⊂ Rn is open. Thenthe set of critical values of φ has measure zero.

Proof. We follow the proof in [17]. We may assume C is compact, since we canexhaust an open set by a countable union of compact subsets. Since the countableunion of sets of zero measure also has zero measure, it follows that the set of criticalvalues of φ will have zero measure if the set of critical values for the restriction of φ toa compact subset also has zero measure. Let x = (x1, . . . , xn) and y = (y1, . . . , yn)be two points in Rn. We write φ(x) = (φ1(x), . . . , φn(x)) for x ∈ Rn. The mean

value theorem states that φi(y) − φi(x) =∑nj=1

∂φi

∂xj(z)(yj − xj), where z is some

point on the line segment joining x and y. Since ∂φi

∂xjis bounded on C, we can find a

uniform Lipschitz constant α such that ||φ(y)−φ(x)|| < α||x−y||. If we denote thepushforward map at x by Txφ, then (Txφ)(y) = ((Txφ)1(y), . . . , (Txφ)n(y)), where

(Txφ)i(y) = φi(x) +∑nj=1

∂φi

∂xj(x)(yj − xj). Hence,

φi(y)− (Txφ)i(y) =

n∑j=1

(∂φi

∂xj(z)− ∂φi

∂xj(x)

)(yj − xj).

Since φ ∈ C1, the functions ∂φi

∂xj are continuous and hence uniformly continuouson C. Therefore, we can find a function β(ε), where β(ε) → 0 as ε → 0+ and||φi(y)− (Txφ)i(y)|| ≤ β(||x− y||)||x− y||.

If x is a critical point of φ, then det(Dxφ) := det(φ−Txφ) = 0 so Dxφ maps Rninto a plane Px of dimension n−1. If ||x−y|| < ε, then ||φ(x)−φ(y)|| < αε and thedistance of φ(y) from Px is less than εβ(ε). Hence, φ(y) lies in the region R whichis the intersection of some cube C of edge length αε centered at φ(x) and the regionbetween two parallel planes of distance εβ(ε) from Px. Choosing the cube C so thatone of its faces is parallel to Px, we may conclude that Vol(R) = 2αn−1εnβ(ε).


We may now cover C by pn cubes of edge length Kp , where K = diam(C).

Any cube containing a critical point x is contained in the ball of radius K√n

p

about x. The image of this ball has volume ≤ 2αn−1(K√n

p )nβ(K√n

p ). The im-

age of all cubes containing at least one critical point contains the set of criticalvalues of φ. Since there are at most pn such cubes, the total volume is at most

pn2αn−1(K√n

p )nβ(K√n

p ) = 2nn/2αn−1Knβ(K√n

p ). If we let p go to infinity, the

above expression goes to zero.

2.2. Co-Area Formula. The co-area formula is a generalization of integration inspherical coordinates. For real-valued functions, the formula is∫

Ω

φ(x)|∇u(x)| dx =

∫ ∞−∞

(∫u−1(t)

φ(x) dσt

)dt,

where Ω ⊂ Rn is open, φ ∈ L1(Ω), u is a real-valued, differentiable function on Ωand dσt is the (n− 1)-dimensional surface measure on hypersurfaces formed by thelevel sets of u. We may assume more generally that u is Lipschitz, since Lipschitzfunctions are absolutely continuous and hence differentiable almost everywhere.

In order to see how the factor of |∇u| comes into play, we may consider acoordinate system (x1, x2, . . . , xn, t), where t denotes the level of the functionu = u(x1, x2, . . . , xn+1). We parametrize these level sets by a map Φ, where

Φ(x1, x2, . . . , xn, t) = (x1, x2, . . . , xn, xn+1(x1, x2, . . . , xn, t)).

Then the pushforward of the map Φ in the ∂t direction is Φ∗(∂∂t

) = ∂xn+1

∂t∂

∂xn+1 .

Hence,⟨

Φ∗(∂∂t

),∇u⟩

= ∂xn+1

∂t∂u

∂xn+1 = ∂u∂t = 1. Since ∇u is normal to the level

sets, Φ∗(∂∂t

)⊥ = ∇u|∇u|2 , which implies ||Φ∗( ∂∂t )

⊥|| = 1|∇u| .

We now prove the co-area formula for smooth functions. We follow closely theproof outlined in [4].

Lemma 1. Suppose M is a Riemannian manifold and u ∈ C∞c (M). Then∫M

|∇u| dµ =

∫ ∞0

σ|u|=t dt,

where σ|u|=t is the area of |u| = t under the hypersurface measure, and dµ isthe volume measure on M .

Proof. Let X be a smooth vector field on M with compact support. Define Nt :=x ∈M : |u(x)| ≥ t. By Sard’s Theorem (Theorem 1), for almost every t, ∇u(x) 6=0 for all x ∈ |u(x)| = t. For these t, ∂Nt = |u(x)| = t is a smooth, co-dimensionone submanifold of M . By the divergence theorem,∫

M

X · ∇u dµ = −∫M

u ∇ ·X dµ

= −∫u>0

u ∇ ·X dµ−∫u<0

u ∇ ·X dµ.

We observe that by Fubini’s Theorem,

6 FARHAN ABEDIN

∫ ∞0

(∫M

χt≤u dµ

)dt =

∫M

(∫ ∞0

χt≤u dt

)dµ

=

∫M

(∫ u(x)

0

1 dt

)dµ

=

∫M

u dµ.

Let ν := − ∇u|∇u| be the unit outward normal to Nt on ∂Nt, and let dσt denote the

hypersurface measure on |u| = t. Then∫u>0

u∇ ·X dµ =

∫ ∞0

(∫M

χt≤u∇ ·X dµ

)dt

=

∫ ∞0

(∫u≥t

∇ ·X dµ

)dt

=

∫ ∞0

(∫u=t

X · ν dσt

)dt

= −∫ ∞

0

(∫u=t

X · ∇u|∇u|

dσt

)dt.

Since ∇(−u) = −∇u and the set −u > 0 is equal to the set u < 0, it followsby applying the above equality to −u that∫

u<0u∇ ·X dµ = −

∫ ∞0

(∫−u=t

X · ∇u|∇u|

dσt

)dt.

Combining the two identities, we get∫M

X · ∇u dµ =

∫ ∞0

(∫|u|=t

X · ∇u|∇u|

dσt

)dt.

We now want to let X = ∇u|∇u| , but this vector field is not smooth with compact

support. Hence, we let ε > 0 and choose φε ∈ C∞c (M, [0, 1]) such that φε 1 asε→ 0+. We then set Xε := φε

∇u(|∇u|2+ε)1/2 . Using this as our vector field, we get∫

M

φε|∇u|2

(|∇u|2 + ε)1/2dµ =

∫ ∞0

(∫|u|=t

φε|∇u|

(|∇u|2 + ε)1/2dσt

)dt.

Taking a limit as ε→ 0+ and using the monotone convergence theorem, we get∫M

|∇u| dµ =

∫ ∞0

(∫|u|=t

dσt

)dt,

where we have used Sard’s Theorem (Theorem 1) to conclude that for almost every

t, limε→0+

χ|u|=tφε|∇u|

(|∇u|2+ε)1/2 = χ|u|=t|∇u|.


Theorem 2. Suppose M is a Riemannian manifold, u ∈ C∞c (M) and φ ∈ C(M, [0,∞)).Then ∫

M

φ|∇u| dµ =

∫ ∞0

∫|u|=t

φdσt dt.

Proof. This follows from the identities

∫M

φ|∇u| dµ =

∫M

(∫ ∞0

χt<φ|∇u| dt)dµ

=

∫ ∞0

(∫M∩t<φ

|∇u| dt

)dµ

=

∫ ∞0

(∫ ∞0

σ|u|=τ,t<φ dτ

)dt

=

∫ ∞0

∫ ∞0

∫|u|=τ

χt<φ dστ dτ dt

=

∫ ∞0

∫|u|=τ

χt<φφdστ dτ.

2.3. Sobolev Spaces. Let U ⊂ Rn be open and let C∞c (U) denote the set of allsmooth functions φ : U → R with compact support in U . An n-dimensional multi-index is an n-tuple of non-negative integers α = (α1, . . . , αn). We denote by |α| the

sum α1 + . . .+αn. In multi-index notation, ∂α = ∂α11 ∂α2

2 . . . ∂αnn , where ∂αii = ∂αi

∂xαii

.

Let u, v ∈ L1loc(U) and α a multi-index. Then v is the α-th weak derivative of u if

for all φ ∈ C∞c (U), we have∫UuDαφ dx = (−1)|α|

∫Uvφ dx. A weak derivative,

if it exists, is uniquely defined up to a set of measure zero. The notion of a weakderivative allows us to define the Sobolev space W k,p(U), which is the space of allfunctions u : U → R such that for each multi-index α with |α| ≤ k, Dαu exists inthe weak sense and belongs to Lp(U). If u ∈ W k,p(U), we define its norm to be||u||Wk,p(U) := (

∑|α|≤k

∫U|Dαu|p)1/p for 1 ≤ p < ∞. For each k, p, W k,p(U) is a

Banach space and for p = 2, W k,2(U) := Hk(U) is a Hilbert space (cf. [5]).

Denote by η the standard mollifier defined by η(x) = Cexp(

1|x|2−1

)if |x| < 1,

η(x) = 0 otherwise. The constant C > 0 is selected so that∫Rn η(x) dx = 1.

For each ε > 0, we set ηε(x) := 1εn η(xε ). The functions ηε are smooth and satisfy∫

Rn ηε(x) dx = 1 with support in B(0, ε). For a locally integrable function f , we

define its mollification f ε(x) := (ηε∗f)(x) =∫Uηε(x−y)f(y) dy =

∫B(0,ε)

ηε(y)f(x−y) dy for x ∈ Uε := x ∈ U : d(x, ∂U) > ε.

We will also require the notion of a Lebesgue point. A point x ∈ U is a Lebesguepoint if

limr→0+

1

µ(B(x, r))

∫B(x,r)

|f(y)− f(x)| dy = 0.

8 FARHAN ABEDIN

By Lebesgue’s Differentiation Theorem, almost every point in U is a Lebesguepoint.

We now show that any u ∈ W k,p(U). 1 ≤ p < ∞ may be approximated bysmooth functions. We say V is compactly contained in U, denoted V ⊂⊂ U , ifV ⊂ V ⊂ U .

Theorem 3. Let u ∈W k,p(U), 1 ≤ p <∞. Then uε ∈ C∞(Uε) for each ε > 0 and

uε → u in Lploc(U) as ε→ 0+. Consequently, uε → u in W k,ploc (U) as ε→ 0+.

Proof. We first show that uε ∈ C∞(Uε) for each ε > 0. Fix x ∈ Uε and h smallsuch that x+ hei ∈ Uε. Then for some V ⊂⊂ U , we have

1

h[f ε(x+ hei)− f ε(x)] =

1

εn

∫U

1

h

[η

(x+ hei − y

ε

)− η

(x− yε

)]f(y) dy

=1

εn

∫V

1

h

[η

(x+ hei − y

ε

)− η

(x− yε

)]f(y) dy.

Since 1h

[η(x+hei−y

ε

)− η

(x−yε

)]→ 1

ε∂η∂xi

(x−yε

)uniformly on V , ∂fε

∂xi(x) exists

and equals∫U∂ηε∂xi

(x − y)f(y) dy. A similar argument shows that all higher orderderivatives also exist.

We now show that if 1 ≤ p < ∞ and u ∈ Lploc(U), then uε → u in Lploc(U). LetV ⊂⊂W ⊂⊂ U . We claim that for ε > 0 sufficiently small, ||f ε||Lp(V ) ≤ ||f ||Lp(W ).Indeed,

∫V

|f ε(x)|p dx =

∫V

(∣∣∣∣∣∫B(x,ε)

ηε(x− y)f(y) dy

∣∣∣∣∣)p

dx

≤∫V

(∫B(x,ε)

η1−1/pε (x− y)η1/p

ε (x− y)|f(y)| dy

)pdx

≤∫V

(∫B(x,ε)

ηε(x− y) dy

)1−1/p(∫B(x,ε)

ηε(x− y)|f(y)|p dy

)1/pp dx

=

∫V

ηε(x− y)|f(y)|p dy dx

≤∫W

|f(y)|p(∫

B(y,ε)

ηε(x− y) dy

)dx

=

∫W

|f(y)|p dx,

where we have used Holder’s inequality in the third step. Since continuous functionsform a dense subset of Lp(W ) (cf. [5]), we may choose δ > 0 and g ∈ C(W ) suchthat ||f − g||Lp(W ) < δ. Then


||f ε − f ||Lp(V ) ≤ ||f ε − gε||Lp(V ) + ||gε − g||Lp(V ) + ||g − f ||Lp(V )

≤ 2||f − g||Lp(W ) + ||gε − g||Lp(V )

≤ 2δ + ||gε − g||Lp(V )

Since gε → g uniformly on B, we have lim supε→0+

||f ε − f ||Lp(V ) ≤ 2δ.

Finally, we show uε → u in W k,ploc (U) as ε → 0+. We claim that if |α| ≤ k, then

Dαuε = ηε ∗Dαu in Uε. Indeed,

Dαuε(x) = Dα

∫U

ηε(x− y)u(y) dy

=

∫U

Dαxηε(x− y)u(y) dy

= (−1)|α|∫U

Dαy ηε(x− y)u(y) dy.

Since ηε(x−y) ∈ C∞c (U) for fixed x ∈ Uε, the definition of the α-th weak partial de-rivative implies

∫UDαy ηε(x− y)u(y) dy = (−1)|α|

∫Uηε(x− y)Dαu(y) dy. Therefore,

Dαuε(x) =∫Uηε(x− y)Dαu(y) dy = (ηε ∗Dαu)(x).

We now choose V ⊂⊂ U . Then by the statements above, we have Dαuε →Dαu in Lp(V ) as ε → 0, for each |α| ≤ k. Consequently, ||uε − u||p

Wk,p(V )=∑

|α|≤k ||Dαuε −Dαu||pLp(V ) → 0 as ε→ 0.

Standard results in Sobolev space theory include the Gagliardo-Nirenberg-SobolevInequality and the Rellich-Kondrachov Compactness Theorem. We provide proofsof these theorems for completeness, refering the reader to [5] for further exposition.

Theorem 4. (Gagliardo-Nirenberg-Sobolev Inequality) Suppose 1 ≤ p < n. Thenthere exists a constant C depending only on p and n such that ||u||Lp∗ (Rn) ≤C||Du||Lp(Rn) for u ∈ C∞c (Rn), where p∗ := pn

n−p .

Proof. Assume p = 1. Since u ∈ C∞c (Rn), we have

u(x) =

∫ xi

−∞

∂u

∂xi(x1, . . . , xi−1, yi, xi+1, . . . , xn)dyi

for each i = 1, . . . , n and x ∈ Rn. Therefore,

|u(x)| ≤∫ ∞−∞|Du(x1, . . . , xi−1, yi, xi+1, . . . , xn)| dyi.

If we do this for each yi, we get

|u(x)|nn−1 ≤

n∏i=1

(∫ ∞−∞|Du(x1, . . . , xi−1, yi, xi+1, . . . , xn)| dyi

) 1n−1

.

10 FARHAN ABEDIN

For notational convenience, we will replace the integral over R, denoted “∫∞−∞”

above, with simply the integral sign “∫

” for the remainder of this proof.

Integrating both sides with respect to x1 gives us

∫|u|

nn−1 dx1 ≤

∫ n∏i=1

(∫|Du(x1, . . . , xi−1, yi, xi+1, . . . , xn)| dyi

) 1n−1

dx1

=

(∫|Du|dy1

) 1n−1

∫ n∏i=2

(∫|Du|dyi

) 1n−1

dx1

≤(∫|Du|dy1

) 1n−1

n∏i=2

(∫∫|Du|dyidx1

) 1n−1

,

where the last step follows from Holder’s inequality. We now integrate with respectto x2 to get

∫∫|u|

nn−1 dx1dx2 ≤

∫ (∫|Du|dy1

) 1n−1

n∏i=2

(∫∫|Du|dyidx1

) 1n−1

dx2

=

(∫∫|Du|dy2dx1

) 1n−1

∫ (∫|Du|dy1

) 1n−1

n∏i=3

(∫∫|Du|dyidx1

) 1n−1

dx2

≤(∫∫

|Du|dy2dx1

) 1n−1

(∫∫|Du|dy1dx2

) 1n−1

n∏i=3

(∫∫∫|Du|dyidx1dx2

) 1n−1

,

where once again the last step follows from Holder’s inequality. We repeat thisprocedure n times to get ||u||

Lnn−1≤ C||Du||L1 , where C = C(n).

Now if 1 < p < n, we can apply the previous estimate to the function v := |u|γ ,where γ < 1 is to be determined. Therefore,

(∫Rn|u|

γnn−1 dx

)n−1n

≤∫Rn|D|u|γ |dx

= γ

∫Rn|u|γ−1|Du|dx

≤ γ(∫

Rn|u|(γ−1) p

p−1 dx

) p−1p(∫

Rn|Du|pdx

) 1p

.

We therefore choose γ to be so that γnn−1 = (γ − 1) p

p−1 . Therefore, γ = p(n−1)n−p > 1

since p > 1. Also, γnn−1 = (γ − 1) p

p−1 = npn−p = p∗. Therefore,(∫

Rn|u|p

∗dx

) 1p∗

≤ p(n− 1)

n− p

(∫Rn|Du|pdx

) 1p

,

from which the desired inequality follows.


We now prove the Rellich-Kondrachov Compactness Theorem. Recall that if Xand Y are Banach spaces, we say that X is compactly embedded in Y provided||x||Y ≤ C||x||X for some constant C and if for every sequence bounded in X, thereis a subsequence convergent in Y .

Theorem 5. (Rellich-Kondrachov Compactness Theorem) Let U ⊂ Rn be openand bounded and suppose ∂U is C1. Let 1 ≤ p < n. Then W 1,p(U) ⊂⊂ Lq(U) foreach 1 ≤ q < p∗.

Proof. Fix 1 ≤ q < p∗ and note that since U is bounded, the Gagliardo-Nirenberg-Sobolev inequality implies W 1,p(U) ⊂ Lq(U) and ||u||Lq(U) ≤ C||u||W 1,p(U). It

remains therefore to show that if um is a bounded sequence in W 1,p(U), thenthere exists a subsequence

umj

which converges in Lq(U).

We first consider the smooth functions uεm and claim that uεm → um in Lq(V )as ε → 0, uniformly in m. To prove this, we first note that if um is smooth and ηis the standard mollifier, then

uεm(x)− um(x) =

∫B(0,1)

η(y)(um(x− εy)− um(x)) dy

=

∫B(0,1)

η(y)

∫ 1

0

d

dt(um(x− εty)) dt dy

= −ε∫B(0,1)

η(y)

∫ 1

0

Dum(x− εty) · y dt dy.

Thus,

∫U

|uεm(x)− um(x)| dx ≤ ε∫B(0,1)

η(y)

∫ 1

0

∫U

|Dum(x− εty)| · y dt dy

≤ ε∫U

|Dum(z)| dz.

By approximation of W 1,p functions by smooth functions, it follows that if um ∈W 1,p(U), then the same estimate holds. Therefore, since U is bounded,

||uεm − um||L1(U) ≤ ε||Dum||L1(U) ≤ εC||Dum||Lp(U).

Since the W 1,p norm of um is bounded by assumption, we conclude that uεm → umin L1(U) as ε→ 0+, uniformly in m.

We now recall the interpolation inequality (cf. [5]), ||u||Lq ≤ ε||u||Lr +ε−µ||u||Lp ,where µ = ( 1

p−1q )/( 1

q −1r ). Define θ to be such that 1

q = θ+ 1−θp∗ . Since 1 ≤ q < p∗,

we can use the interpolation inequality to get

||uεm − um||Lq(U) ≤ ||uεm − um||θL1(U)||uεm − um||1−θLp∗ (U)

≤ C||uεm − um||θL1(U),

where the last inequality follows by the Gagliardo-Nirenberg-Sobolev Inequalityand the W 1,p bound on um. Hence uεm → um in Lq(U) as ε→ 0+, uniformly in m.

12 FARHAN ABEDIN

Since ||um||L1(U) < C for each m, it follows that for each fixed ε > 0, the sequenceuεm is uniformly bounded and equicontinuous in U . Indeed, if x ∈ Rn, then

|uεm(x)| ≤ ||ηε||L∞(U)||um||L1(U) ≤C

εn,

|Duεm(x)| ≤ ||Dηε||L∞(U)||um||L1(U) ≤C

εn+1.

Now fix δ > 0. We claim that there exists a subsequenceumj

such that

lim supj,k→∞

||umj − umk ||Lq(U) ≤ δ.

To show this, we first choose ε > 0 small enough so that ||uεm − um||Lq(U) ≤ δ/2.Since the functions uεm are bounded and equicontinuous in U , we can invoke the

Arzela-Ascoli Theorem to obtain a subsequenceuεmj

of uεm which converges

uniformly on U . Therefore,

lim supj,k→∞

||uεmj − uεmk||Lq(U) → 0.

and this proves the claim. We may now use a standard diagonalization argumentwith δ = 1

n for each n ∈ N to extract a subsequence uml of um satisfying

lim supl,k→∞

||uml − umk ||Lq(U) → 0.

2.4. Existence Results from Elliptic Partial Differential Equations. Wenow outline some of the basic existence theorems for solutions to elliptic partialdifferential equations of second order, following closely the exposition in [6]. Con-sider linear elliptic operators of the form

Lu = aij(x)Diju+ bi(x)Diu+ c(x)u, aij = aji, x ∈ Rn.

We say L is elliptic at a point x ∈ Ω if the coefficient matrix aij(x) is positivedefinite; that is, if λ(x) and Λ(x) denote respectively the minimum and maximumeigenvalues of the matrix aij(x), then 0 < λ(x)|v|2 ≤ aij(x)vivj ≤ Λ(x)|v|2 forv ∈ Rn\ 0. If Λ/λ is uniformly bounded, then we say L is uniformly elliptic.

We have the following maximum principles for solutions to second order ellipticequations.

Theorem 6. (Weak Maximum Principle) Let L be elliptic in the bounded domainΩ. Suppose Lu ≥ 0 (≤ 0), c ≤ 0 in Ω with u ∈ C2(Ω)∩C0(Ω). Then sup

Ωu ≤ sup

∂Ωu+

(infΩu ≥ inf

∂Ωu−) . If Lu = 0, then sup

Ω|u| = sup

∂Ω|u|.


Theorem 7. (Strong Maximum Principle) Let L be uniformly elliptic, c = 0 andLu ≥ 0 (≤ 0) in Ω. Then if u achieves its maximum (minimum) in the interiorof Ω, u is a constant. If c ≤ 0 and c/λ is bounded, then u cannot achieve anon-negative maximum (non-positive minimum) in the interior of Ω unless it isconstant.

We also have the following apriori estimate for solutions of the inhomogeneousequation Lu = f in bounded domains.

Theorem 8. Let Lu ≥ f in a bounded domain Ω, where L is elliptic, c ≤ 0 andu ∈ C2(Ω) ∩ C0(Ω). Then

supΩu ≤ sup

∂Ωu+ + C sup

Ω

|f−|λ

.

An important estimate that will be required in the proof of the Positive MassTheorem is the Schauder interior estimate for solutions of the elliptic equationLu = f , where u ∈ C2,α(Ω). We remind the reader of the Holder norms andsemi-norms.

• |u|0;Ω = supx∈Ω|u(x)|.

• [u]α;Ω = supx,y∈Ω,x 6=y

|u(x)− u(y)||x− y|α

, 0 < α ≤ 1.

• |u|0,α;Ω = |u|0;Ω + [u]α;Ω.

We now state the basic Schauder interior estimate.

Theorem 9. Let Ω ⊂ Rn be a bounded domain and let u ∈ C2,α(Ω), f ∈ Cα(Ω)satisfy Lu = f . Also suppose the coefficients of L are in Cα(Ω). Then if Ω′ ⊂⊂ Ωwith dist(Ω′, ∂Ω) ≥ d, we have

d|Du|0;Ω′ + d2|D2u|0;Ω′ + d2+α[D2u]α;Ω′ ≤ C(|u|0;Ω + |f |0,α;Ω)

where C depends only on λ and the Cα(Ω) norms of the coefficients of L, as wellas on n, α and the diameter of Ω.

Using the Schauder estimates, we can show regularity of C2(Ω) solutions of Lu = fif f ∈ Ck,α.

Theorem 10. Let u ∈ C2(Ω) be a solution of the equation Lu = f in an openset Ω, where f and the coefficients of the elliptic operator L are in Ck,α(Ω). Thenu ∈ Ck+2,α(Ω). Consequently, if f and the coefficients are smooth in Ω, then u issmooth in Ω as well.

14 FARHAN ABEDIN

The idea behind the proof of the above theorem is to consider a difference quotientof u and then, by linearity of the operator, transfer the difference quotient to thecoefficients of L. Since we are assuming regularity of the coefficients, we may usethe Schauder estimates on the difference quotient of u to conclude that higherderivatives exist for each difference quotient. Also, the Schauder estimates providea uniform bound for the derivatives of each difference quotient that is independentof h. Hence, by the Arzela-Ascoli Theorem, we may find a convergent subsequenceof the difference quotients, which then proves the existence of a higher derivativeof u.

We now consider the question of solving the Dirichlet problem for Lu = f . Inorder to do so, we require some functional analytic results, which we will state herefor convenience. A mapping T : X → Y between Banach spaces is a compact linearmapping if the image of bounded sets in X are compactly contained in Y .

Lemma 2. (Continuity Method) Let B be a Banach space, V a normed linearspace, and let L0, L1 be bounded linear operators from B into V . For each t ∈ [0, 1],set Lt = (1− t)L0 + tL1 and suppose that there is a constant C such that ||x||B ≤C||Ltx||V for all t. Then L1 maps B onto V if and only if L0 maps B onto V .

Lemma 3. (Fredholm Alternative) Let T be a compact linear mapping of a normedlinear space V into itself. Then either the homogeneous equation x − Tx = 0 hasa non-trivial solution x ∈ V or for each y ∈ V , the equation x − Tx = y has auniquely determined solution x ∈ V .

The proofs of these results can also be found in [6].

The idea behind solving the Dirichlet problem for Lu = f is to connect theoperator L to the Laplacian ∆ by considering the family of equations Ltu :=tLu + (1 − t)∆u = f , 0 ≤ t ≤ 1. The operator Lt may be considered a boundedlinear operator from the Banach space

u ∈ C2,α(Ω) : u = 0 on ∂Ω

into the Ba-

nach space Cα(Ω). Using the apriori estimate and the Schauder estimate, we canshow that the continuity method may be applied to this family of operators. Sincethe Laplace operator is onto Cα(Ω), it follows that L is also onto Cα(Ω). We statethis in the following theorem.

Theorem 11. Let Ω be a C2,α domain in Rn and let the operator L be strictlyelliptic in Ω with coefficients in Cα(Ω) and with c ≤ 0. Then if the Dirichletproblem for Poisson’s equation ∆u = f in Ω, u = ϕ on ∂Ω has a C2,α(Ω) solutionfor all f ∈ Cα(Ω) and ϕ ∈ C2,α(Ω), the problem Lu = f in Ω, u = ϕ on ∂Ω alsohas a unique C2,α(Ω) solution for all such f and ϕ.

It should be mentioned that it is not necessary to assume any regularity of theboundary data ϕ on the interior of Ω. As long as ϕ ∈ C2,α(∂Ω), it is possible toextend ϕ to a function that is C2,α on Ω (cf. [5]). Therefore, assuming regularityon the interior is equivalent to assuming regularity on the boundary.


If the operator L does not satisfy c ≤ 0, the Dirichlet problem for Lu = f is nolonger solvable. However, it is still possible to assert a Fredholm alternative.

Theorem 12. Let L be strictly elliptic with coefficients in Cα(Ω) in a C2,α domainΩ. Then either (a) the homogeneous problem Lu = 0 in Ω, u = 0 on ∂Ω, has onlythe trivial solution, in which case Lu = f in Ω, u = ϕ in ∂Ω has a unique C2,α(Ω)solution for all all f ∈ Cα(Ω) and ϕ ∈ C2,α(Ω), or (b) the homogeneous problemhas non-trivial solutions, which form a finite dimensional subspace of C2,α(Ω).

The proof of the above uses the Schauder Estimate and the Apriori Bound toshow that the inverse of a perturbed operator Lσ := L − σ, where σ ≥ sup

Ωc is a

compact mapping. Hence the Fredholm Alternative (Lemma 3) applies, and theequation u + σL−1

σ u = L−1σ f has a solution provided the homogeneous equation

u+ σL−1σ u = 0 has only the trivial solution.

Theorem 13 allows us to modify the Schauder estimates in the following way.The right hand side of the Schauder estimate contains the term |u|0;Ω, which isnecessary to account for the kernel of L. By Theorem 13, the kernel, if it exists,is finite dimensional and since all norms on finite dimensional spaces are stronglyequivalent, we may replace the sup norm, |u|0;Ω by, for example, an Lp norm of u.This observation will prove to be useful in the proof of the Rigidity Theorem.

2.5. Local Boundedness of Weak Solutions. Instead of considering ellipticoperators in their standard form, Lu = aij(x)Diju + bi(x)Diu + c(x)u, we mayconsider operators in the divergence form,

Lu = Di(aij(x)Dju+ bi(x)u) + ci(x)Diu+ d(x)u,

where the coefficients aij , bi, ci, d are assumed to be measurable functions on Ω.We say u is a weak solution of Lu = 0 (≥ 0, ≤ 0) in Ω if∫

Ω

(aijDju+ biu)Div − (ciDiu+ du)v = 0 (≤ 0, ≥ 0)

for all v ∈ C10 (Ω). Similarly, Lu = g if∫

Ω

(aijDju+ biu)Div − (ciDiu+ du)v = −∫

Ω

gv.

Provided the coefficients of L are locally integrable, it follows from the divergencetheorem that a function u ∈ C2(Ω) satisfying Lu = 0 in the classical sense alsosatisfies the above relation in the weak sense. Moreover, if aij , bi have locally inte-grable derivatives, then a weak solution u ∈ C2(Ω) is also a classical solution. Theweak maximum principle, once suitably formulated, also holds for weak solutionsand the generalized Dirichlet problem is also uniquely solvable. We refer the readerto [6] for proofs of these statements.

We provide here a proof of the local boundedness of weak solutions by the L2

norm, following the outline in [7]. The proof uses the De Giorgi-Nash-Moser it-eration technique in a beautiful way. We denote by Br the ball of radius r inRn.

16 FARHAN ABEDIN

Theorem 13. Suppose aij ∈ L∞(B4) and c ∈ Lq(B4) for some q > n2 satisfy the

following assumptions:

λ|v|2 ≤ aij(x)vivj for all x ∈ B4 and v ∈ Rn,

|aij |+ ||c||Lq ≤ Λ.

Let u ∈ H1(B4) satisfy ∫B4

aijDjuDiφ+ cuφ ≤∫B4

fφ

for all φ ∈ H10 (B4), φ ≥ 0, and f ∈ Lq(B4). Then there exists C = C(n, Λ

λ , p, q) > 0such that

supB2

u+ ≤ C(||u+||L2(B4) + ||f ||Lq(B4)).

Proof. For k,m > 0, set u = k + u+. Define um = u if u < m and um = k + mif u ≥ m. Then um ∈ H1(B4), bounded below by k and above by k + m. Also,Dum = 0 in u < 0 and u > m, and since um = u everywhere else, Dum = Duwhenever 0 ≤ u ≤ m.

Let φ = η2(uβmu − kβ+1) for β ≥ 0 arbitrary and η ∈ C10 (B4) a non-negative

cut-off function to be determined later. Then φ ∈ H10 (B4) since um is bounded and

can therefore be used as a test function. We compute,

Dφ = 2ηDη(uβmu− kβ+1) + η2(βuβ−1m Dum · u+ uβmDu)

= 2ηDη(uβmu− kβ+1) + η2uβm(βDum +Du),

where we have used the fact that um = u whenever Dum 6= 0. Since Dφ ≡ 0 whenu < 0, we get

∫aijDjuDiφ =

∫aijDj u

[2ηDiη(uβmu− kβ+1) + η2uβm(βDium +Diu)

]≥ −2

∫η|aijDj uDiη||uβmu− kβ+1|+

∫λη2uβm(β|Dum|2 + |Du|2).

Now,

2

∫η|aijDj uDiη||uβmu− kβ+1| ≤ 2Λn

∫η|Du||Dη||uβmu− kβ+1|

≤ 2Λn

∫η|Du||Dη|uβmu (since uβmu− kβ+1 ≥ 0)

≤∫

(η|Du|uβ/2m )(2Λn|Dη|uβ/2m u)

≤ λ

2

∫η2|Du|2uβm +

2n2Λ2

λ

∫|Dη|2uβmu2,

where in the last line, we have applied the AM-GM inequality in the form ab ≤λ2a

2 + 12λb

2. Therefore,


2

∫η2uβm(β|Dum|2 + |Du|2) ≤ 2

λ

∫aijDjuDiφ+

4

λ

∫η|aijDj uDiη||uβmu− kβ+1|

≤ 2

λ

∫|c|η2uβmu

2 +2

λ

∫|f |η2uβmu

+

∫η2|Du|2uβm +

(2nΛ

λ

)2 ∫|Dη|2uβmu2,

which implies

β

∫η2uβm|Dum|2 +

∫η2uβm|Du|2 ≤ C1

∫(|Dη|2 + c0η

2)uβmu2,

where c0 = |c|+ |f |k . Let k = ||f ||Lq(B4). Then by assumption, ||c0||Lq(B4) ≤ Λ + 1.

Now set w = uβ/2m u ∈ H1(B4). Then Dw = u

β/2m (β2Dum +Du), and so

|Dw|2 = uβm

∣∣∣∣β2Dum +Du

∣∣∣∣2= uβm(β(1 + β/4)|Dum|2 + |Du|2)

≤ uβm(β + 1)(β|Dum|2 + |Du|2).

Therefore,

∫|Dw|2η2 ≤ (β + 1)

(β

∫uβm|Dum|2η2 +

∫uβm|Du|2η2

)≤ C1(β + 1)

∫(|Dη|2 + c0η

2)w2,

This implies

∫|D(wη)|2 ≤ C1(β + 1)

∫(|Dη|2 + c0η

2)w2.

We wish to obtain an estimate independent of the quantity c0. Hence, by Holder’sinequality,

∫c0η

2w2 ≤ ||c0||Lq(∫

(ηw)2qq−1

)1− 1q

≤ (Λ + 1)

(∫(ηw)

2qq−1

)1− 1q

.

By the interpolation inequality, ||u||Lq ≤ ε||u||Lr + ε−µ||u||Lp , where µ = ( 1p −

1q )/( 1

q −1r ), and the Sobolev inequality, where 2∗ = 2n

n−2 >2qq−1 > 2 if q > n

2 , we

have

18 FARHAN ABEDIN

||ηw||L

2qq−1≤ ε||ηw||L2∗ + C(n, q)ε−

n2q−n ||ηw||L2

≤ ε||D(ηw)||L2 + C(n, q)ε−n

2q−n ||ηw||L2

for all ε > 0. Hence, by choosing ε small enough, we get∫|D(ηw)|2 ≤ C1(β + 1)δ

∫(|Dη|2 + η2)w2,

where δ > 0 depends only on n and q. Let χ = nn−2 for n > 2 (for n = 2, we may

fix χ > 2). Again, by the Sobolev inequality,

(∫(ηw)2χ

) 1χ

≤ C(n)

∫|D(ηw)|2

≤ 4C1C(n)(β + 1)δ∫

(|Dη|2 + η2)w2.

We now choose the cut-off function, η. For 0 < r < R ≤ 4, take η ∈ C10 (B4) with

η ≡ 1 in Br and |Dη| ≤ 2R−r . Then,

(∫Br

w2χ

) 1χ

≤ 16C1C(n)(β + 1)δ

(R− r)2

∫BR

w2

= C2(β + 1)δ

(R− r)2

∫BR

w2.

Letting γ = β + 2, we get

(∫Br

uγχm

) 1χ

=

(∫Br

w2χ

) 1χ

≤ C2(β + 1)δ

(R− r)2

∫BR

w2

≤ C2(γ − 1)δ

(R− r)2

∫BR

uγ .

Finally, letting m→∞, we have by the Monotone Convergence Theorem

||u||Lγχ(Br) ≤(C2

(γ − 1)δ

(R− r)2

) 1γ

||u||Lγ(BR).

We now let γi = 2χi and ri = 2+ 12i−1 . For any i, we may insert r = ri+1, R = ri

and γ = γi into the estimate above to get

||u||Lγi+1 (Bri+1) ≤

(C2

(γi − 1)δ

( 12i )

2

) 1γi

||u||Lγi (Bri ) ≤ Ci

χi

3 ||u||Lγi (Bri ).

By iteration,

||u||Lγi (Bri ) ≤ C∑

i

χi

3 ||u||L2(B4) = C4||u||L2(B4).


Since ri > 2 for each i, we have

||u||Lγi (B2) ≤ C4||u||L2(B4).

Letting i→∞, we get

supB2

u ≤ C4||u||L2(B4).

Therefore, by recalling the fact that k = ||f ||Lq(B4), we obtain the desired estimate.

The above estimate can be extended to obtain Lp bounds for p > 2. We cite thefollowing result from [6] in this direction.

Theorem 14. Let k(R) = λ−1R2δ||g||Lq/2 , where δ = 1− nq , q > n, and R > 0. If

u ∈ H1(Ω) is a solution of Lu = g in Ω, then for any ball B2R ⊂ Ω and p > 1, wehave

supBR

u ≤ C(R−n/p||u+||Lp(B2R) + k(R)), where C = C(n, Λλ , R, p, q).

2.6. A Quick Review of Riemannian Geometry. We provide here, for com-pleteness, a review of the basic concepts in Riemannian Geometry. Many detailswill not be provided here, and we refer the reader unfamiliar with RiemannianGeometry to [10] for a more thorough introduction to the material presented here.

A manifold is a topological space that is locally homeomorphic to Euclideanspace. We write local coordinates on any open subset U ⊂ M as (x1, . . . , xn). Asmooth manifold is one for which the transition maps yi(x1, . . . , xn), i = 1, . . . , n,from one coordinate neighborhood to another are C∞ maps. The tangent spaceTpM at a point p ∈ M is the vector space of equivalence classes of curves passingthrough p that have the same velocity in any coordinate chart. The tangent bundle,TM of M , is the disjoint union of the tangent spaces TpM for all p ∈M . The spaceof smooth sections of the tangent bundle (or the space of smooth vector fields inM) is denoted T (M). Since TpM is a vector space, we can define dual vectors andtensors, and so we can define tensor bundles and tensor fields on M in an analogousfashion. This allows us to define a Riemannian metric on a smooth manifold, whichis a (0,2)-tensor field g such that for each X,Y ∈ TpM

(1) g(X,Y ) = g(Y,X),

(2) g(X,X) > 0 for X 6= 0.

A Riemannnian metric therefore, determines an inner product 〈, 〉 on each tangentspace TpM . A smooth manifold equipped with a given Riemannian metric, is calleda Riemannian manifold. If (E1, . . . , En) is a local frame for TM , and (ϕ1, . . . , ϕn)is its dual coframe, a Riemannian metric can be written locally as g = gijϕ

i ⊗ ϕj ,where ⊗ denotes the tensor product of forms, defined as (ϕ1⊗· · ·⊗ϕn)(v1, . . . , vn) =ϕ1(v1) · · ·ϕn(vn). In particular, in a coordinate frame, g = gijdx

idxj . If (M, g)

and (M, g) are Riemannian manifolds, a diffeomorphism ϕ : M → M is called anisometry if ϕ∗g = g, where ϕ∗g is the pullback metric, defined as (ϕ∗g)(X,Y ) =

20 FARHAN ABEDIN

g(ϕ∗(X), ϕ∗(Y )) for X,Y ∈ TM . Two Riemannian manifolds (M, g) and (M, g)are isometric if such an isometry exists. A Riemannian manifold is said to be flatif it is locally isometric to Euclidean space.

On a given Riemannian manifold M , we may define a linear connection in TM ,which is a map ∇ : T (M)× T (M)→ T (M) satisfying the following properties:

(1) ∇XY is linear over C∞(M) in X (i.e. ∇fX1+gX2Y = f∇X1

Y + g∇X2Y for

f, g ∈ C∞(M)).

(2) ∇XY is linear over R in Y (i.e. ∇X(aY1 + bY2) = a∇XY1 + b∇XY2 fora, b ∈ R).

(3) ∇ satisfies the product rule (i.e. ∇X(fY ) = f∇XY + (Xf)Y for f ∈C∞(M)).

If Ei is a local frame for TM on an open subset U ⊂M , we can expand ∇EiEjin terms of the same frame so that ∇EiEj = ΓkijEk for scalar functions Γkij . Theseare called Christoffel symbols. Using the defining rules for a connection, we canshow that ∇XY = (XY k + XiY jΓkij)Ek. Therefore, the action of the connection∇ on U is fully determined by its Christoffel symbols.

On each Riemannian manifold, there exists a natural connection. If (M, g) is aRiemannian manifold, then a linear connection ∇ is said to be compatible with g ifit satisfies the product rule

∇X 〈Y,Z〉 = 〈∇XY,Z〉+ 〈Y,∇XZ〉for all vector fields X,Y, Z. This is equivalent to saying (∇Xg)(Y,Z) := ∇X 〈Y, Z〉−〈∇XY,Z〉−〈Y,∇XZ〉 = 0. It turns out that requiring a connection to be compatiblewith the metric is not enough to determine a unique connection, so we consider thetorsion tensor τ(X,Y ) = ∇XY − ∇YX − [X,Y ], where [X,Y ](f) := X(Y (f)) −Y (X(f)) is the Lie Bracket. If we require the torsion tensor to vanish identically,then there exists a unique, metric compatible, linear connection ∇ on M , called theLevi-Civita connection of g. Using metric compatibility and vanishing torsion, wecan find an explicit formula for the Christoffel symbols in any local coordinate frameEi = ∂

∂xi . If gij is the inverse matrix for the metric components gij = 〈∂i, ∂j〉 and

∇∂i∂j = Γkij∂k, then Γkij = 12gkm(gim,j + gjm,i − gij,m), where the comma denotes

partial derivative and we have used the Einstein summation convention to sum overrepeated indices (in this case, over m).

We may also consider the connection acting on vector fields along curves γ(t) inM . Define the acceleration of a curve γ(t) in M to be the vector field ∇γ γ alongγ. A curve γ is called a geodesic with respect to ∇ if its acceleration vanishes. Theexistence and uniqueness of geodesics is guaranteed by the existence and uniquenesstheorem from ordinary differential equations. Indeed, any initial point p ∈ M andany initial velocity V ∈ TpM determine a unique maximal geodesic γV with domaindom(γV ) ⊂ R such that γV (0) = p and γV (0) = V . This allows us to define amap from a subset of the tangent bundle to M itself by sending the vector V tothe point obtained by following γV for unit time. To be precise, define a subsetE = V ∈ TM : [0, 1] ⊂ dom(γV ) of TM to be the domain of the exponential map,


and then define the exponential map exp : E → M by exp(V ) = γV (1). This mapwill be useful later in defining the sectional curvature of a Riemannian manifold.

Using the Levi-Civita connection, we can also define the Riemannan Curvatureendomorphism R : T (M)× T (M)× T (M)→ T (M), defined by

R(X,Y )Z = ∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z.

We also define the corresponding Riemann Curvature tensor, Rm(X,Y, Z,W ) =〈R(X,Y, Z),W 〉. If we define Rmkji as R(∂k, ∂j)∂i = Rmkji∂m, then we may write outthe components Rmkji in terms of Christoffel symbols as

Rmkji = Γmij,k − Γmik,j + ΓlijΓmkl − ΓlikΓmjl .

One of the reasons why this seemingly abstract tensor is of interest is because ofits direct relationship with curvature. For example, a Riemannian manifold is flatif and only if its curvature tensor vanishes identically. We also note some of thesymmetries of the curvature tensor and the Bianchi identities.

Lemma 4. The curvature tensor has the following symmetries for any vector fieldsX,Y, Z, V,W :

(1) Rm(W,X, Y, Z) = −Rm(X,W, Y, Z),

(2) Rm(W,X, Y, Z) = −Rm(W,X,Z, Y ),

(3) Rm(W,X, Y, Z) = −Rm(Y,Z,W,X),

(4) Rm(W,X, Y, Z)+Rm(X,Y,W,Z)+Rm(Y,W,X,Z) = 0 (first Bianchi iden-tity),

(5) ∇Rm(Z, V,X, Y,W )+∇Rm(V,W,X, Y, Z)+∇Rm(W,Z,X, Y, V ) = 0 (sec-ond, or differential, Bianchi identity).

We may now consider traces of the Riemann tensor to get more specific typesof curvature quantities. By taking a single trace of the Riemann tensor, we getthe Ricci tensor, which is a symmetric (0, 2)-tensor with components Rij = Rkkij .

Another trace yields the scalar curvature R = gijRij . We may also write thecomponents of the Ricci tensor and the scalar curvature in terms of the Christoffelsymbols as shown below.

Rij = Γkij,k − Γkik,j + ΓkklΓlij − ΓkjlΓ

lik,

R(g) = gij(Γkij,k − Γkik,j + ΓkklΓ

lij − ΓkjlΓ

lik

).

We now consider the gradient, divergence and Laplacian operators on a Rie-mannian manifold. In local coordinates over the chart U ⊂ Rn, the gradient canbe written as ∇f = gij∂if∂j . The divergence on a vector field X = Xi∂i ∈ TM is

div(X) =1√

det(g)∂i(X

i√

det(g)).

The Laplacian is then the divergence of the gradient, and can be written in coor-dinates as

∆f =1√

det(g)∂i(g

ij√

det(g)∂jf).

22 FARHAN ABEDIN

The above expressions satisfy the familiar integration by parts formula when inte-grated against f ∈ C∞c (U) with respect to the measure dVg =

√det(g) dx1 · · · dxn.

We note that the Laplacian on a Riemannian manifold as an operator has coeffi-cients that depend on the metric g. Therefore, as long as we assume the requisitebounds on the coefficients of g, we may apply the estimates from second order el-liptic partial differential equations that are stated in Section 2.4 and 2.5. This willbe useful in the proof of the Rigidity Theorem.

We establish an important identity, known as the contracted Bianchi identity,which relates the Ricci and scalar curvatures through the divergence operator.

Lemma 5. div(Ric) = 12∇S.

Proof. Observe that in components, the above identity is gijRij;i = 12S;i, where the

semi-colon denotes covariant differentiation. By contracting the differential Bianchiidentity, first on the indices i and l, followed by the indices j and k, we get usingthe symmetries of the Riemann tensor

0 = gil(Rijkl;m +Rijlm;k +Rijmk;l)

= Rjk;m −Rjm;k + gilRijmk;l

= R;m − gjkRmj;k − gilgjkRjimk;l

= R;m − 2gjkRmj;k

from which the identity follows.

We now consider submanifolds of Riemannian manifolds. Suppose (M, g) isa Riemannian manifold of dimension m, M is a manifold of dimension n, andi : M → M is an immersion. If M is given the induced Riemannian metric g := i∗g,then i is said to be an isometric immersion (embedding). If i is injective, so thatM is

an immersed (embedded) submanifold of M , then (M, g) is said to be a Riemannian

submanifold of M . In these situations, M is referred to as the ambient manifold.We may now consider the ambient tangent bundle over M and the correspondingnormal bundle of M , and decompose the ambient connection ∇ into its tangentialand normal components as ∇XY = (∇XY )T + (∇XY )⊥. This allows us to define

the second fundamental form II(X,Y ) := (∇XY )⊥, which is symmetric in X andY (since the Lie Bracket [X,Y ] is in the tangent space), and bilinear over C∞(M).On the other hand, we may also consider independently the induced connection ∇on M with respect to g. The Gauss Formula, ∇XY = ∇XY + II(X,Y ), showsthat the induced connection is exactly the tangential component of the ambientconnection. By considering normal fields, N , we get the Weingarten equation,⟨∇XN,Y

⟩= −〈N, II(X,Y )〉.

In addition to describing the difference between the extrinsic and intrinsic con-nections, the second fundamental form also describes the difference between thecurvature tensors of M and M through the Gauss Equation,

Rm(X,Y, Z,W ) = Rm(X,Y, Z,W )− 〈II(X,W ), II(Y,Z)〉+ 〈II(X,Z), II(Y,W )〉.


Proof. Let X, Y , Z, W be arbitrary vector fields on M that are tangent to M . Bythe Gauss formula

Rm(X,Y, Z,W ) =⟨∇X∇Y Z − ∇Y ∇XZ − ∇[X,Y ]Z,W

⟩=⟨∇X(∇Y Z + II(Y,Z))− ∇Y (∇XZ + II(X,Z))− (∇[X,Y ]Z + II([X,Y ], Z)),W

⟩.

Since the second fundamental form takes its values in the normal bundle and W istangent to M , the last II term is zero. Applying the Weingarten equation to theother two terms involving II, we get

Rm(X,Y, Z,W ) =⟨∇X∇Y Z,W

⟩− 〈II(Y,Z), II(X,W )〉

−⟨∇Y∇XZ,W

⟩+ 〈II(X,Z), II(Y,W )〉 −

⟨∇[X,Y ]Z,W

⟩= 〈∇X∇Y Z,W 〉 − 〈∇Y∇XZ,W 〉 −

⟨∇[X,Y ]Z,W

⟩− 〈II(Y,Z), II(X,W )〉+ 〈II(X,Z), II(Y,W )〉,

where we have used the Gauss formula once again in the second step.

If we let N be a section of the normal bundle of M , then an argument similar tothe above yields the Codazzi equations

∇XII(Y,Z)− ∇Y II(X,Z) = R(X,Y, Z,N).

If we focus our attention to the special case where Mn is a hypersurface ofRn+1 with the induced Riemannian metric, we can replace the vector-valued secondfundamental form II by a scalar-valued form h, which is the symmetric 2-tensordefined by h(X,Y ) = 〈II(X,Y ), N〉 for a given unit normal field N . Using this, wemay define the shape operator, S, as 〈X,S(Y )〉 = h(X,Y ). Since h is symmetric, Sis self-adjoint, and by elementary linear algebra, it has real eigenvalues κ1, . . . , κn,which we shall call the principal curvatures of M . The mean curvature, H, andGauss curvature, K, as H := κ1 + . . . + κn and K := κ1 · · ·κn. Gauss’ TheoremaEgregium states that the Gauss curvature is actually an intrinsic invariant of theRiemannian manifold (Mn, g).

Theorem 15. Let M ⊂ (R3, δ) be a 2-dimensional submanifold and g the inducedmetric on M . For any p ∈M and any basis X,Y for TpM , the Gauss curvatureof M at p is given by

K =Rm(X,Y, Y,X)

|X|2|Y |2 − 〈X,Y 〉2.

In higher dimensions, for any Riemannian n-manifold M and p ∈ M , if Π isa 2-dimensional subspace of TpM and V ⊂ TpM is any neighborhood of zero onwhich the exponential map expp is a diffeomorphism, then SΠ := expp(Π ∩ V) is a2-dimensional submanifold of M containing p. We define the sectional curvature ofM associated with Π, denoted K(Π), to be the Gauss curvature of SΠ at p with theinduced metric. Since SΠ is swept out by geodesics whose initial tangent vectors

24 FARHAN ABEDIN

lie in Π, the second fundamental form of SΠ vanishes and by the Gauss equation,

we get K(Π) = Rm(X,Y,Y,X)

|X|2|Y |2−〈X,Y 〉2 , where X,Y span Π.

It is worth mentioning here that since the Ricci tensor is a symmetric (0, 2)-tensor, it is fully determined by the components that have the same index. Hence,it can be seen from the above theorem that the Ricci curvature is determined by thesectional curvatures. Indeed, the Ricci tensor component Rii is, up to a constant,the average value of the sectional curvatures of all 2-planes passing through the i-thcoordinate direction. Since there is, in general, an (n−2)-dimensional family of suchplanes, the Ricci tensor fully determines the Riemann tensor only in dimensions 2and 3. Thus, the vanishing of the Ricci tensor in dimensions 2 and 3 imply thevanishing of the Riemann tensor. This fact will be used in the proof of the PositiveMass Theorem.

An important theorem pertaining to the integral of the Gauss curvature of aRiemannian 2-manifold is the Gauss-Bonnet Theorem. To start, let M be a smooth,compact 2-manifold. A smooth triangulation of M is a finite collection of curvedtriangles such that the union of the closed regions Ωi bounded by the trianglesis M and the intersection of any pair is either a single vertex, a single edge ofeach, or empty. Every smooth, compact surface possesses a smooth triangulation.For such a triangulated 2-manifold, the Euler characteristic of M with respectto a given triangulation is defined to be the integer χ(M) := Nv − Ne + Nf ,where Nv, Ne and Nf correspond to the number of vertices, edges and faces ofthe triangulation respectively. It is an important result of algebraic topology thatthe Euler characteristic is in fact a topological invariant, and is independent of thechoice of triangulation.

Consider now a curve γ(t) that lies on a given M . Recall that the accelerationof the curve γ in M is the vector field ∇γ γ. The length of the projection of theacceleration onto the tangent space TpM at a point p ∈M is the geodesic curvatureof γ at that point. If M is a 2-manifold with boundary ∂M , then we denote thegeodesic curvature of ∂M by kg.

We may now state the Gauss-Bonnet Theorem.

Theorem 16. If M is a triangulated, compact, oriented, Riemannian 2-manifold,then ∫

M

K dA+

∫∂M

kg ds = 2πχ(M).

For closed manifolds (i.e. compact, with no boundary), the boundary term in theabove expression drops out and we have

∫MK dA = 2πχ(M).

This concludes our brief account of the main ideas and theorems in Riemanniangeometry. In the following subsections, we provide a more detailed exposition ofother formulas from Riemannian geometry that will be used extensively throughoutthe proof of the Positive Mass Theorem.


2.7. Second Variation of Area. Consider an immersion f : Σk → (Mn, g) (k <n). A smooth variation of f is a smooth map F : I × Σ → M so that eachft := F (t, ·) is an immersion and f0 = f . We consider Σ compact, or if Σ isnon-compact, we consider only compactly supported variations. Let Fi denote thepushforward F∗(

∂∂xi ) and Ft the pushforward F∗(

∂∂t ). Define the variation field

V = Ft|t=0. Let g(t) be the metric induced from the immersion ft and let dvtdenote the volume measure. Therefore, in coordinates, gij(t) = g(Fi, Fj) = 〈Fi, Fj〉.Denote by ν the unit conormal for ∂Σ. Let Ei = F∗(ei), where ei is defined onI × Σ and is a local orthonormal frame field on Σ for each t.

The area functional we want to vary is A(t) =∫

Σdvt. Note that in local coordi-

nates, the volume element is dvt =√

detg(t) dx. The variation is therefore given

by A′(t) =∫

Σddt

√detg(t) dx. From Cramer’s rule and the symmetry of the metric

g, we have ∂det(g)∂gij

= gijdet(g). Therefore, ddt

√detg(t) = 1

2

√detg(t)gij gij . We have

gij = 〈∇tFi, Fj〉 + 〈Fi,∇tFj〉. Recall that ∇XY −∇YX = [X,Y ]. Using the fact

that F∗[∂∂t ,

∂∂xi

] = [Ft, Fi] = 0 (which follows by commuting i and t partials), weget

d

dt

√detg(t) =

√detg(t)gij 〈∇tFi, Fj〉 =

√detg(t)gij 〈∇iFt, Fj〉 = divΣ(Ft)

√detg(t).

Evaluating at t = 0, we obtain

d

dt

∣∣t=0

√detg = divΣ(V )

√detg = (divΣ(V T ) + divΣ(V N ))

√detg.

The term divΣ(V T ) can be handled using the divergence theorem. On the otherhand,

divΣ(V N ) =

k∑i=1

⟨∇EiV N , Ei

⟩= −

k∑i=1

⟨V, (∇EiEi)N

⟩= −〈V,H〉.

Hence we arrive at the First Variation Formula

A′(0) = −∫

Σ

〈V,H〉+

∫∂Σ

〈V, ν〉.

We now work towards deriving the second variation formula. We compute thisat a minimal immersion of Σ. Note that the formula for the derivative of the inverseof a matrix gives gij = −gikgklglj . Therefore,

d2

dt2

√detg(t) =

d

dt

(1

2

√detg(t)gij gij

)=

1

4

√detg(t)(gij gij)

2 − 1

2

√detg(t)gikgklg

lj gij +1

2

√detg(t)gij gij .

We examine each of the terms at t = 0. First, we have

1

4(gij gij)

2 =1

4(2divΣV )2 = (divΣV

T )2,

since divΣVN = −〈V,H〉 = 0 by minimality.

For the second term, we have

26 FARHAN ABEDIN

gikgklglj gij = gik(〈∇kV, Fl〉+ 〈Fk,∇lV 〉)glj(〈∇iV, Fj〉+ 〈Fi,∇jV 〉)

=

k∑i,j=1

(〈∇EiV,Ej〉+⟨Ei,∇EjV

⟩)2.

Finally, for the last term, we have

gij = 〈∇t∇tFi, Fj〉+ 2 〈∇tFi,∇tFj〉+ 〈Fi,∇t∇tFj〉= 〈∇t∇iFt, Fj〉+ 2 〈∇iFt,∇jFt〉+ 〈Fi,∇t∇jFt〉 .

Since R(X,Y )Z = ∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z, we have (once again using thefact that [Fi, Ft] = 0) that

gij = 〈R(Fi, V )V, Fj〉+ 〈∇i∇tFt, Fj〉+ 2 〈∇iV,∇jV 〉+ 〈R(Fj , V )V, Fi〉+ 〈∇j∇tFt, Fi〉.By contracting the third term, we get 2

∑ki=1 |∇EiV |2. By contracting the first

and the fourth terms, and by switching V and Fi or Fj as required, we get

−2∑ki=1 〈R(V,Ei)V,Ei〉. By contracting the second and fifth terms, and by observ-

ing that∇t = ∇V , we get 2divΣ(∇V Ft)T , since divΣ(∇V Ft)N = −⟨(∇V Ft)N , H

⟩=

0 by minimality. Therefore,

1

2gij gij =

k∑i=1

|∇EiV |2 −k∑i=1

〈R(V,Ei)V,Ei〉+ divΣ(∇V Ft)T .

When the variation field V is normal to Σ, we have

k∑i,j=1

(〈∇EiV,Ej〉+⟨Ei,∇EjV

⟩)2 =

k∑i,j=1

(−〈V,∇EiEj〉 −⟨∇EjEi, V

⟩)2

= 4

k∑i,j=1

〈V,∇EiEj〉2

= 4|AV |2,

where AV (Ei, Ej) := 〈V,∇EiEj〉 is the symmetric, (0, 2)-tensor which is the innerproduct of the second fundamental form with V .

For X ∈ TΣ and Y ∈ NΣ, we may define the normal connection ∇N on the normalbundle to be ∇NXY = (∇XY )⊥. Therefore,

k∑i=1

|∇EiV |2 =

k∑i=1

(|∇NEiV |2 + |(∇EiV )T |2)

=

k∑i=1

|∇NEiV |2 +

k∑i,j=1

〈∇EiV,Ej〉2

= |∇NV |2 + |AV |2.Finally, we introduce the normal Ricci curvature formR, an operator on the normal

bundle on Σ, defined for vectors W normal to Σ by R(W ) =∑ki=1(R(Ei,W )Ei)

N .


We thus arrive at the Second Variation Formula at a minimal submanifold fornormal variations,

A′′(0) =

∫Σ

(|∇NV |2 − |AV |2 − 〈R(V ), V 〉

)+

∫∂Σ

〈∇V Ft, ν〉.

We remark that if Σ is a hypersurface, a normal variation field is of the formV = φN , where N is the unit normal and φ is a compactly supported function onΣ. In this case, the second variation formula becomes

A′′(0) =

∫Σ

(|∇φ|2 − φ2|II|2 − φ2Ric(N,N)

).

2.8. Conformal Change of Scalar Curvature. Let ϕ be a smooth functionand let g be a metric on a Riemannian n-manifold M . We may consider the metricg = eϕg conformal to g. We compute below the scalar curvature of g with respectto the scalar curvature of g explicitly.

Lemma 6. R(g) = e−ϕ(R(g)− (n− 1)∆gϕ− 1

4 (n− 1)(n− 2)|∇ϕ|2g).

Proof. We use normal coordinates for g at a point p ∈ M . We denote by partialderivatives by a comma. Therefore, gij(p) = δij and gij,k(p) = 0 for all i, j, k. This

implies Γkij(p) = 0 and gij,k(p) = 0. We also use the formula gij,s = −gipgpq,sgqj . Wecompute as follows:

• Γkij = Γkij + 12gkm(ϕ,jgim + φ,igjm − ϕ,mgij)

• Γkij,s = 12 gkm,s (gim,j + gjm,i − gij,m) + 1

2 gkm(gim,js + gjm,is − gij,ms)

• gim,j = eϕ(ϕ,jgim + gim,j)

• gim,j = −e−ϕ(ϕ,jgim + gim,j)

• gim,js = eϕ(ϕ,jsgim + ϕ,sϕ,jgim + ϕ,jgim,s + ϕ,sgim,j + gim,js).

Hence,

• Γkii,k =∑i,k(gik,ik − 1

2gii,kk −12ϕ,kk)

• Γkik,i =∑i,k( 1

2gii,kk + 12ϕ,kk)

• Γkij(p) = 12 (ϕ,jδik + ϕ,iδjk − φ,kδij)

• gim,j(p) = eϕϕ,jδim

• gim,j (p) = −e−ϕϕ,jδim

• gim,js(p) = eϕ(ϕ,jsδim + ϕ,jϕ,sδim + gim,js(p)).

28 FARHAN ABEDIN

Therefore,

R(g) = gij(Γkij,k − Γkik,j + ΓkklΓlij − ΓkjlΓ

lik)

= e−ϕ∑i,k,l

(Γkii,k − Γkik,i + ΓkklΓlii − ΓkilΓ

lik)

= e−ϕ∑i,k,l

(Γkii,k − Γkik,i + ΓkklΓlii)

= e−ϕ∑i,k,l

(gik,ik − gii,kk − ϕ,kk −1

4(ϕ,l)

2)

= e−ϕ(R(g)− (n− 1)∆gϕ−

1

4(n− 1)(n− 2)|∇ϕ|2g

).

It follows from the above that if we let g = u4

n−2 g, then R(g) = u−n+2n−2 (R(g)u −

4(n−1)n−2 ∆gu) by letting ϕ = log(u−

n+2n−2 ).

2.9. Linearization of Scalar Curvature. We may consider the linearization ofthe scalar curvature at a base metric g as we perturb in the direction of metricsh. Define Lg(h) = d

dt |t=0R(g + th). We have the following expression for thelinearization.

Lemma 7. Lg(h) = −∆g(trg(h)) + divg(divg(h))− h ·g Ric(g).

Proof. Once again, we compute in normal coordinates.

Γkij,s = 12gkm,s (gim,j + gjm,i − gij,m) + 1

2gkm(gim,js + gjm,is − gij,ms).

Therefore,

Γkij,s =1

2gkm,s (gim,j + gjm,i − gij,m) +

1

2gkm,s (gim,j + gjm,i − gij,m)

+1

2gkm(gim,js + gjm,is − gij,ms) +

1

2gkm(gim,js + gjm,is − gij,ms)

=1


1


− 1

2gkpgpqg

qm(gim,js + gjm,is − gij,ms) +1

2gkm(gim,js + gjm,is − gij,ms)

=1


1


− gkpgpqΓqij,s +1

2gkm(gim,js + gjm,is − gij,ms).

Therefore, Γkij,s(p) = 12

∑i,j,k(hik,js+hjk,is−hij,ks)−

∑i,j,k hkqΓ

qij,s. Since R(g) =

gij(Γkij,k − Γkik,j + ΓkklΓlij − ΓkjlΓ

lik), it follows that

Lg(h) = −gipgpqgqj(Γkij,k −Γkik,j + ΓkklΓlij −ΓkjlΓ

lik) + gij(Γkij,k − Γkik,j + ΓkklΓ

lij +

ΓkklΓlij − ΓkjlΓ

lik − ΓkjlΓ

lik).


Hence

Lg(h)(p) =∑i(Γ

kii,k − Γkik,i)−

∑i,j hij(Γ

kij,k − Γkik,j).

By the previous formula for Γkij,s(p), we have

Γkii,k(p) = 12

∑i,k(hik,ik + hik,ik − hii,kk)−

∑i,k hkqΓ

qii,k

and

Γkik,i(p) = 12

∑i,k(hik,ki + hkk,ii − hik,ki)−

∑i,k hkqΓ

qik,i.

Therefore,

Lg(h)(p) =∑i,k

(hik,ik − hii,kk − hkqΓqii,k + hkqΓ

qik,i

)−∑i,j

hij(Γkij,k − Γkik,j

).

Now

−∆g(trg(h))(p) = −gkl(gijhij);kl

= −gklgijhij;kl= −gklgij(hij,k − hmjΓmik − hmiΓmjk),l

= −gklgij(hij,kl − hmjΓmik,l − hmiΓmjk,l)

= −∑i,k

(hii,kk − 2hkmΓmik,i),

divg(divg(h))(p) = gjlgikhij;kl

= gjlgik(hij,kl − hmjΓmik,l − hmiΓmjk,l)

=∑i,k

(hik,ik − hkmΓmii,k − hkmΓmik,i),

and

−h ·g Ric(g)(p) = −∑i,j

hij(Γkij,k − Γkik,j).

It follows that Lg(h) = −∆(tr(h)) + div(div(h))− h · Ric(g).

30 FARHAN ABEDIN

2.10. Einstein’s Equations. Let (N, g) be a complete Riemannian manifold. De-fine the Einstein-Hilbert action

SH(g) =

∫N

MR(g) dVg,

where dVg =√

det(g) dx. Let g + th denote a compactly supported perturbationof g. Then by the formula for the linearization of the scalar curvature (Lemma 7)and the formula for the derivative of the determinant (cf. Section 2.7), we get

d

dt

∣∣t=0

SH(g + th) =

∫N

(d

dt

∣∣t=0

R(g + th)√

det(g) dx+R(g)d

dt

∣∣t=0

√det(g) dx

)=

∫N

(−∆g(trg(h)) + divg(divg(h))− h ·g Ric(g)) dVg

+

∫N

R(g) · 1

2

√det(g)hij dx

= −∫N

(Ric(g)− 1

2R(g)g) · h dVg.

Hence, by the action principle, Ric(g) − 12R(g)g = 0. Taking a trace yields the

vacuum Einstein equations, Ric(g) = 0.

Remark: The above derivation also holds for Lorentzian manifolds once we re-place

√det(g) by

√−det(g) and take into account the signature of the metric.

Let II denote the second fundamental form of N with respect to g. Then byusing the Gauss and Codazzi equations along with the Einstein equations, we getthe constraint equations

divgII− d(tr(II)) = 0

R(g)− |II|2g + (tr(II))2 = 0

A triple (N, g, k), where k is a symmetric (0,2)-tensor, that satisfies the constraintequations is called an initial data set for Einstein’s equations. In the presenceof matter fields, the right hand side of each equation above is replaced by termsrelated to the energy and momentum density of the matter fields respectively. Inthe time symmetric case, tr(II) = 0, we therefore see that the scalar curvature R(g)is directly related to the energy density of the matter field, and must be necessarilynon-negative. This forms the motivation for phrasing the Positive Mass Theoremin the manner stated in the next section.


3. Schoen and Yau’s proof of the Riemannian Positive Mass Theoremfor n = 3

3.1. Setup of the Problem. A smooth, complete Riemannian 3-manifold (N, g) iscalled asymptotically flat if for some compact set K, N\K consists of a finite numberof components (ends) N1, . . . , Nk, where each Nl is diffeomorphic to R3\B, where Bis the closed unit ball. For each end Nl and each diffeomorphism Φl : Nl → R3\B,we assume that the metric g can be written in the coordinate chart defined by Φl asgij(x) = δij+O(|x|−p) for some p > 1

2 . We also assume |x||gij,k(x)|+|x|2|gij,kl(x)| =O(|x|−p) and |R(g)(x)| = O(|x|−q) for q > 3, where R(g) is the scalar curvature ofg. Under these conditions, it is possible to define the total energy of N . Recall thecoordinate expression for the scalar curvature,

R(g) = gij(Γkij,k − Γkik,j + ΓkklΓ

lij − ΓkjlΓ

lik

)where Γkij = 1

2gim(gim,j−gjm,i+gij,m). Under the asymptotic assumptions, R(g) =∑

i,j(gij,ij − gii,jj) + O(|x|−2p−2). Since 2p+ 2 > 3, it follows that the divergenceterm is absolutely integrable near infinity. Thus, by the divergence theorem, thefollowing limit exists.

limσ→∞

1

16π

∫|x|=σ

∑i,j

(gij,i − gii,j)νσj dxσ.

In the above expression, νσ = xσ is the Euclidean unit normal to |x| = σ and dxσ

is the Euclidean area element on |x| = σ. The above expression is defined to bethe ADM mass M of an end.

We provide an example of a family of metrics for which the mass can be computeddirectly. Observe that if u is a positive, harmonic function on R3\B that decays to1 at infinity, then by composing u − 1 with the Kelvin transform x 7→ x|x|−2, weobtain an expansion of u in spherical harmonics, which we write

u(x) = 1 +C

2|x|+O(|x|−2),

where C is some constant. In this case, the metric g = u4δ, where δ is the Euclideanmetric, has scalar curvature R(g) = −8u−5∆u = 0, and the ADM mass of (N, g) isC. Indeed,

m = limσ→∞

1

16π

∫|x|=σ

4

(1 +

C

2|x|+O(|x|−2)

)3(− C

2|x|2+O(|x|−3)

)∑i,j

(xi

|x|δij −

xj

|x|

)xj

|x|dxσ

= limσ→∞

1

16π

∫|x|=σ

4

(1 +

C

2|x|+O(|x|−2)

)3(C

|x|2+O(|x|−3)

)dxσ

= limσ→∞

1

4π

(1 +

C

2σ+O(σ−2)

)3(C

σ2+O(σ−3)

)(4πσ2)

= limσ→∞

(1 +

C

2σ+O(σ−2)

)3 (C +O(σ−1)

)= C

32 FARHAN ABEDIN

Conversely, if g is a scalar-flat metric which is conformally flat near infinity withconformal factor u tending to 1, then g = u4δ for some u harmonic on R3\B.Therefore, (N, g) is asymptotically flat and has ADM mass C from the sphericalharmonic expansion. We call such metrics harmonically flat.

Another motivating example is the family of Schwarzschild space-times, discov-ered by Karl Schwarzschild. These were the first non-trivial, exact solutions ofEinstein’s equations. They describe the gravitational field outside a spherical, non-rotating mass, and can be written out in coordinates (t, x) as

gS(x) = −

(1− m

2|x|

1 + m2|x|

)2

dt2 +

(1 +

m

2|x|

)4

δ,

where δ is the Euclidean metric. The slice t = 0 is asymptotically flat and con-

formally flat with vanishing scalar curvature, and the metric gS =(

1 + m2|x|

)4

δ is

a complete metric on the set R3\ 0, with two asymptotically flat ends. Recallfrom the expansion of the conformal factor in spherical harmonics that the termC is the ADM mass. In the case of the Schwarzschild metric gS it is easy to showexplicitly that the ADM mass of either end is exactly equal to the quantity m inthe definition of the metric.

We will consider metrics of the form

g =

(1 +

M

2r

)4

(dxi)2 + pijdxidxj ,

where pij = O(r−2), ∇pij = O(r−3) and ∇∇pij = O(r−4), and M is the ADMmass. Observe that harmonically flat metrics belong to this class of metrics. Toonly consider the metrics that satisfy the aforementioned asymptotics may, at firstsight, seem restrictive. However, Schoen and Yau prove in [15] that the mass ofany asymptotically flat metric with non-negative scalar curvature is close to thatof a corresponding harmonically flat metric. More precisely, we have the followingtheorem.

Theorem 17. If (N, g) is asymptotically flat with R(g) ≥ 0, then for all ε > 0,there exists a harmonically flat metric g such that m(g) ≤ m(g) + ε.

Hence, it is sufficient to prove the Positive Mass Theorem for metrics g that canbe written in the above form. The proof of the above theorem can be found in theappendix.

It is worth mentioning here that for the proof of the Positive Mass Theorem, werequire the boundary component of N to have positive mean curvature with respectto the outward unit normal, where by outward, we mean the unit normal pointstowards the end Nk in which the ADM mass is being computed. A simple way toillustrate the necessity of this assumption is to consider the Schwarzschild metricwith negative mass. Such a metric has a singularity at r = −m2 , which is not merelya coordinate singularity since the curvatures blow up at this point. If we excise aneighborhood of this singularity and consider the remaining manifold, we see thatit is indeed harmonically flat, but has negative mass by construction. The reasonwhy the Positive Mass Theorem does not hold under these circumstances is due to


the fact that the mean curvature of the boundary with respect to the outward unitnormal is non-positive

Finally, we point out that the proof presented here is for the time-symmetric(Riemannian) case, where we assume the manifold (N, g) satisfies trg(II) = 0.Schoen and Yau were able to extend the proof of the Positive Mass Theorem tothe non time-symmetric case by reducing it to the time-symmetric case using thesolution of Jang’s equation. The proof of the more general theorem will not bediscussed here, and we refer the interested reader to [16].

3.2. Statement of Results. Let N be an oriented 3-manifold (with or withoutboundary) that is C6 smooth. Let g be a C5 positive definite metric on N . Supposeeach boundary component of N has positive mean curvature with respect to theoutward unit normal, and on each end Nk, there is a coordinate system x1, x2, x3

in which g has the expansion g = gijdxidxj , (i, j = 1, 2, 3), with gij satisfying the

following conditions:

(1) gij =

(1 +

M

2r

)4

δij + hij

(2) |hij | ≤k1

1 + r2

(3) |∂hij | ≤k2

1 + r3

(4) |∂2hij | ≤k3

1 + r4

where r2 =∑3i=1(xi)2, ∂ is the Euclidean partial derivative and k1, k2, k3 are

positive constants. Let M = Mk be the total mass of the end Nk. Note that inthese coordinates, the Christoffel symbols Γkij = O(r−2) and the curvature tensor

is O(r−3) as r →∞.

We state the first theorem.

Theorem 18. (Positive Mass Theorem) Let g be an asymptotically flat metric onan oriented 3-manifold N . If the scalar curvature R(g) ≥ 0 on N , then the totalmass of each end is non-negative.

The next result concerns the case where the total mass on one end is zero. Inthis case, we show that N is flat under the assumption

|∂3hij |+ |∂4hij |+ |∂5hij | ≤k4

1 + r5

for some positive constant k4.

Theorem 19. (Rigidity Theorem) Let N be an oriented 3-manifold having anasymptotically flat metric g. Suppose for some end Nk all the inequalities above aresatisfied and the total mass of Nk is zero. If the scalar curvature R(g) ≥ 0 on N ,then g is flat. In fact, N is isometric to R3 with the standard Euclidean metric.

34 FARHAN ABEDIN

We prove the Positive Mass Theorem by showing the following claims are true.

(1) If g is asymptotically flat on N with R(g) ≥ 0 and the total mass of Nk isnegative, then there is an asymptotically flat metric g conformally equivalentto g having scalar curvature R(g) ≥ 0 on N and R(g) > 0 outside a compactsubset of Nk, and negative total mass for Nk.

(2) There exists a complete area minimizing (relative to g) surface S properlyimbedded in N such that S ∩ (N\Nk) is compact and S ∩ Nk lies betweentwo parallel Euclidean 2-planes in the 3-space defined by x1, x2, x3.

(3) The surface S cannot exist.

3.3. Proof of Claim 1. Suppose M < 0 and let R3\Bσ0(0) represent Nk. Let ∆g

be the Laplace operator on functions so that for a function φ on R3\Bσ0(0), wehave

∆gφ =1√g∂i(√ggij∂jφ),

where√g = det(gij) and gij = (gij)

−1. We compute the asymptotic expansion

of ∆g

(1r

)on R3\Bσ0(0). Recall that r =

√∑3i=1(xi)2, so ∂i

(1r

)= −x

i

r3 and

∂2i

(1r

)= − 1

r3 + 3(xi)2

r5 . Also, by asymptotically flat coordinates, we have gij =

(1 + M2r )−4δij + O(r−2) and

√g = (1 + M

2r )6 + O(r−7). Therefore, by the formulafor the Laplacian, we have

∆g

(1

r

)=

3∑i=1

∂i

[(1 +

M

2r

)2

∂i

(1

r

)]+O(r−5)

=

3∑i=1

M

(1 +

M

2r

)(xi

r3

)2

+

(1 +

M

2r

)2(− 1

r3+

3(xi)2

r5

)+O(r−5)

=M

r4

(1 +

M

2r

)+O(r−5)

=M

r4+O(r−5).

Hence, there exists σ > σ0 such that ∆g

(1r

)< 0 for r ≥ σ. Let t0 = − M

8σ0and let

ψ(t) be a C5 function satisfying

(1) ψ(t) = t for t < t0

(2) ψ(t) = 3t02 for t > 2t0

(3) ψ′(t) ≥ 0 and ψ′′(t) ≤ 0 for all t ∈ (0,∞).

We also define a C5 function ϕ : N → R by ϕ(x) = 1+ 3t02 on N\Nk and ϕ(x) = 1+

ψ(−M4r ) on Nk. Since r > 2σ implies −M4r = t < t0 and ψ(t) = t for t < t0, it follows

that ψ(−M4r ) = −M4r . Hence ∆gϕ(x) = 0 on N\Nk and ∆gϕ(x) = −M4 ∆g

(1r

)< 0

for r ≥ σ (and so for r > 2σ).


Now define g = ϕ4g. Then g is asymptotically flat since on all ends other than Nkit is a constant multiple of g and on Nk we have

gij =

(1− M

4r

)4(1 +

M

2r

)4

δij +O(r−2)

=

(1 +

M

4r

)4

δij +O(r−2).

Thus the new mass of Nk is M = M2 < 0. By the formula for the scalar curvature

under a conformal change, R(g) = ϕ−5[−8∆gϕ + R(g)ϕ], we have R(g) ≥ 0 on Nand R(g) > 0 for r > 2σ on Nk.

Remark: We replace the original metric g by g but maintain the notation g sothat we are assuming R(g) ≥ 0 on N , R(g) > 0 outside a compact subset of Nk,and M < 0.

3.4. Proof of Claim 2. Let σ > 2σ0 and let Cσ be the circle of Euclidean radiusσ centered at 0 in the x1, x2 coordinate plane. Let Sσ be the smooth imbeddedoriented surface of least area with respect to the metric g amongst all competingsurfaces regardless of topological type having boundary Cσ. The existence of sucha surface is guaranteed by the existence of smooth solutions of the two-dimensionalPlateau Problem for Riemannian 3-manifolds with boundary of positive mean cur-vature. We wish to extract a sequence σi → ∞ such that Sσi converges to therequired surface S.

We first show that there is a compact subsetK0 ⊆ N such that Sσ∩(N\Nk) ⊆ K0

for every σ > 2σ0. That is, we show that Sσ cannot run off to infinity in any endother than Nk. Suppose N ′k is another end with asymptotically flat coordinatesy1, y2, y3 associating N ′k with R3\Bτ0(0), where Bτ0(0) = y : |y| < τ0. In thiscoordinate system, the metric has the expansion provided previously. We computethe covariant Hessian Dij of the function |y|2.

Dij |y|2 = ∂ij |y|2 −D∂i∂j |y|2

= 2δij + 2Γkijyk

= 2δij +O(|y|−1).

Hence, there exists τ1 ≥ τ0 such that |y|2 is convex for |y| ≥ τ1. Observe thatby taking a trace of the Hessian, we get ∆g|y|2 ≥ 0 and so |y|2 is subharmonic.Since ∂Sσ = Cσ ⊂ Nk, we may apply the weak maximum principle (Theorem 6) toconclude that Sσ ∩N ′k ⊆ Bτ1(0).

Remark: The intuition behind the previous idea is that for |y| large, the meancurvature of Sσ is O(|y|−1), which violates minimality. The surface must bend backinto Nk as the boundary ∂Sσ lies entirely in Nk.

We now analyze the behavior of Sσ∩Nk. We claim that we can bound the heightof Sσ ∩Nk in the x3 direction. Indeed, for any h > 0, let Eh =

x ∈ R3 : |x3| ≤ h

.

We show there exists h > σ0 such that Nk ∩ Sσ ⊆ Eh for all σ > 2σ0. To establish

36 FARHAN ABEDIN

this, we again use the maximum principle, this time for the function x3 restrictedto Sσ ∩Nk. We compute the asymptotic expansion of the covariant Hessian of x3

on Nk.

Dijx3 = ∂ijx

3 − Γkij∂kx3

= −Γ3ij = −1

2g33(gi3,j + gj3,i − gij,3)

= −Mr3

(1 +

M

2r

)(xjδi3 + xiδj3 − x3δij)

=Mxj

r3δi3 +

Mxi

r3δj3 −

Mx3

r3δij +O(r−3).

Let h be the maximum for x3 on Sσ ∩ Nk and suppose this maximum occurs atx0 ∈ Sσ If h ≤ σ0, then we are done, so suppose h > σ0. The tangent space toSσ is spanned by ∂1|x0, ∂2|x0. Let v1, v2 be tangent vector fields to Sσ definedin a neighborhood of x0 and satisfying vi(x0) = ∂i|x0 for i = 1, 2. Let (qij),i, j = 1, 2, be the restriction of g to Sσ in terms of the base field v1, v2. Let ∇be the induced connection on Sσ. Note that by the Gauss Formula (cf. Section2.6), vivjx

3 − ∇vivj(x3) = vivjx3 − Dvivj(x

3) + 〈Dvivj , ν〉 ν(x3), where ν is theunit normal field of Sσ. Evaluating at x0 gives us ∇ijx3 = Dijx

3 +hijν(x3), wherehij = −〈Dvivj , ν〉 (x0) is the second fundamental form. We contract with respect

to qij . Since Sσ is minimal,∑2i,j=1 q

ijhijν(x3) = 0. Therefore,

2∑i,j=1

qij∇ijx3 =

2∑i,j=1

qijDijx3

= (D11 + 2D12 +D22)x3

= −2Mh

r3+O(r−3).

Since M < 0, if h is sufficiently large, then∑2i,j=1 q

ij∇ijx3 > 0 at x0, contradicting

the fact that x3 attains a maximum there. A similar argument gives a lower boundfor x3 on Sσ ∩Nk.

Now let ρ > 2σ0 and define the set

Aρ = (N\Nk) ∪x : |x| ≥ σ0, (x

1)2 + (x2)2 ≤ ρ2

.

Then for any σ > ρ, we have Sσ ∩Aρ ⊆ (K0 ∪Eh)∩Aρ, which is a compact subsetof N .

We now quote a local interior regularity estimate for area minimizing surfaces.

Regularity Estimate: Let Ur(x) denote the geodesic ball of radius r about x ∈ N .There exists a number r0 > 0 such that for any point x0 ∈ Sσ with Ur0(x0)∩Cσ = ∅,it is true that Sσ ∩ Ur0(x0) can be written as the graph of a C3 function fσ overthe tangent plane to Sσ in a normal coordinate system on Ur0(x0). Moreover, thereis a constant c1 depending only on (N, g) which bounds all derivatives of fσ uptoorder three in Ur0(x0).


It follows by the Arzela-Ascoli Theorem that we can choose a sequence σ(ρ)i → ∞

so that Sσ

(ρ)i∩ Aρ converges in the C2 topology. Since this can be done for any

ρ > 2σ0, we can take a sequence ρj →∞, and by extracting a diagonal sequence, wecan find a sequence σi → ∞ so that Sσi → S, an imbedded C2 surface, uniformlyin the C2 norm on compact subsets of N . The surface S is properly imbedded andis area minimizing on any compact subset of N .

3.5. Proof of Claim 3. For any σ ≥ σ0, let S(σ) be the set S(σ) = [S ∩ (N\Nk)]∪[S ∩ Bσ(0)]. The S(σ) form an exhaustion of S and A(S(σ)) ≤ C2σ

2 for someconstant C2 independent of σ. To prove this, we note that if S has transverseintersection with ∂Bσ(0), then this intersection is a union of oriented C2 Jordancurves on ∂Bσ(0) which bound S(σ). It follows that these curves bound a domainΩ ⊆ ∂Bσ(0). Thus we have ∂S(σ) = ∂Ω. Applying the area minimizing property ofS, we conclude that A(S(σ)) ≤ A(Ω) ≤ A(∂Bσ(0)). Since g is uniformly equivalent

to the Euclidean metric on R3\Bσ0(0), the area estimate for S(σ) follows for thoseσ > σ0 for which S intersects ∂Bσ(0) transversally. By Sard’s Theorem (Theorem1), this is true for σ except in a set of measure zero, and therefore the estimateholds for any σ ≥ σ0 by approximation.

We can use the area estimate to bound the integrals of certain functions on S.For a > 0, we have

∫S

1

1 + radσ ≤

∫S(σ0)

1

1 + radσ +

∞∫σ0

d

dt

∫S(t)

1

1 + ra

dt

≤ A(S(σ0)) +

∞∫σ0

limh→0

1

h

∫S(t+h)

1

1 + radS(t+h) −

∫S(t)

1

1 + radS(t)

dt

≤ A(S(σ0)) +

∞∫σ0

1

1 + ta

(limh→0

A(S(t+h))−A(S(t))

h

)dt

= A(S(σ0)) +

∞∫σ0

1

1 + ta

(d

dtA(S(t))

)dt.

For a > 2, we can use integration by parts to get

∫S

1

1 + radσ ≤ c2σ2

0 + a

∞∫σ0

ta−1

(1 + ta)2A(S(t)) dt

≤ c2σ20 + c2a

∞∫σ0

ta+1

(1 + ta)2dt.

It follows that∫S

11+ra dσ < ∞ whenever a > 2. Applying similar reasoning, we

may also conclude that

38 FARHAN ABEDIN∫S(σ2)\S(σ1)

1

r2dσ ≤ 2C2 log

(σ2

σ1

).

We now consider the second variation inequality (cf. Section 2.7),∫S

(Ric(ν, ν) + |A|2

)f2 ≤

∫S

||∇f ||2,

where ν is the outward pointing normal to N , A is the matrix for the componentshij := II(Xi, Xj) of the second fundamental form, and f is any C2 function withcompact support on S. We also recall the Gauss Equation (cf. Section 2.6),

K = K12 + h11h22 − h212.

By minimality of S, tr(A) = h11 + h22 = 0. Therefore, since |A|2 =∑2i,j=1 h

2ij , we

get 12 |A|

2 = K12 −K. Substituting into the second variation inequality, we get∫S

(Ric(ν, ν) +K12 −K +

1

2|A|2

)f2 ≤

∫S

||∇f ||2.

If we let e1, e2, e3 = ν be a frame adapted to S, then Ric(ν, ν) = K13 + K23 andR = K12 +K13 +K23. Therefore, we can rewrite the above inequality as∫

S

(R−K +

1

2|A|2

)f2 ≤

∫S

||∇f ||2.

We now choose a suitable cutoff function for f . For σ > σ0, define the function ϕby

ϕ =

1 on S(σ)

log(σ2/r

)log σ

on S(σ2)\S(σ)

0 outside S(σ2).

Let ω be Lipschitz on S satisfying |ω| ≤ 1 and ω = 1 outside a compact subset ofS. Setting f = ϕω and applying the Cauchy-Schwarz inequality, we get

∫S

(Ric(ν, ν) + |A|2

)ϕ2ω2 ≤ 2

∫S

||∇(ϕω)||2

≤∫S

(ω2||∇ϕ||2 + ϕ2||∇ω||2 + 2 〈ϕ∇ω, ω∇ϕ〉

)≤ 2

∫S

ω2||∇ϕ||2 + 2

∫S

ϕ2||∇ω||2

≤ 2

(log σ)2

∫S(σ2)\S(σ)

||∇r||2

r2+ 2

∫S

ϕ2||∇ω||2.

By asymptotically flat coordinates, there is a constant C3 such that ||∇r||2 ≤ C3.We rearrange and use the definition of ϕ and ω to get


∫S(σ)

|A|2ω2 ≤ 2C3

(log σ)2

∫S(σ2)\S(σ)

1

r2+ 2

∫S

||∇ω||2 +

∫S

|Ric(ν, ν)|ω2

≤ 2C2C3

log σ+ 2

∫S

||∇ω||2 +

∫S

|Ric(ν, ν)|ω2.

Letting σ →∞, we conclude that∫S

|A|2ω2 ≤ 2

∫S

||∇ω||2 +

∫S

|Ric(ν, ν)|ω2.

for any Lipschitz ω with |ω| ≤ 1 and ω = 1 outside a compact subset of S. Byobserving that Ric(ν, ν) = O(r−3) in asymptotically flat coordinates and choosingω ≡ 1 on S, we conclude that

∫S|A|2 < ∞. Also, since |K| ≤ |K12| + |A|2 and

K12 = O(r−3) in asymptotically flat coordinates, it follows that∫S|K12| < ∞.

Therefore,∫S|K| <∞.

We now use f = ϕ in the second variation inequality and let σ →∞. By the sameargument as above, we conclude that

∫S

(R−K + 1

2 |A|2)≤ 0. Since R ≥ 0 and

R > 0 outside a compact subset of S, we conclude∫SK > 0.

It is worth mentioning now that the proof of the claim will be complete if we canshow that

∫SK ≤ 0. We do this by directly applying the Gauss-Bonnet Theorem

with boundary (Theorem 16) and estimate the boundary terms.

For any x ∈ R3, let x′ = (x1, x2, 0) and let r′ = |x′|. Consider the cylinderPσ =

x ∈ R3 : r′ ≤ σ

. For any σ > σ0 for which ∂Pσ ∩ S is transverse, there

exists at least one circle in this intersection which is not homologous to zero inR3\Pσ0 . Choose one of these circles and let Dσ be the connected component of thiscircle in S ∩ [(N\Nk) ∪ Pσ]. We may choose Dσ to be increasing so that Dσ ⊂ Dσ

whenever σ < σ. Since S is connected, the Dσ form an exhaustion of S and weapply the Gauss-Bonnet Theorem on Dσ so that

∫Dσ

K = 2π −∫∂Dσ

k, where k

is the geodesic curvature of ∂Dσ relative to the inner normal. The proof will becomplete if we can find a sequence σi →∞ such that

∫∂Dσi

k ≥ 2π−o(1) as i→∞.

In a neighborhood of ∂Dσ, we choose a frame e1, e2, e3 where e1 is the positivelyoriented unit tangent vector of ∂Dσ, e2 is the inner normal to Dσ and e3 = ν isthe unit normal of S in R3 relative to g. The geodesic curvature is given by k =〈De1e1, e2〉. Since r′ = σ on ∂Dσ, we have 〈e1,∇r′〉 = 0 on ∂Dσ. Differentiating

with respect to e1 gives 〈De1e1,∇r′〉 + 〈e1, De1∇r′〉 = 0. Now ∇r′|r′=σ = x′

σ +

O(σ−1). If x = (x1, x2, x3), then De1x = e1 = De1x′ +De1x

3. Therefore,

De1∇r′ =e1

σ− 1

σ〈e1, ∂3〉 ∂3 +O(σ−2).

It follows that⟨De1e1,

x′

σ

⟩+

1

σ− 1

σ〈e1, ∂3〉2 = O(σ−1)||De1e1||+O(σ−2).

Since De1e1 = ke2 − h11ν, we get

k

⟨e2,

x′

σ

⟩+

1

σ− h11

⟨ν,x′

σ

⟩− 1

σ〈e1, ∂3〉2 = O(σ−1)||De1e1||+O(σ−2).

40 FARHAN ABEDIN

Now suppose σ ∈ [σ, 2σ]. We apply the divergence theorem for the vector field∂3 on the volume V enclosed by D2σ\Dσ, ∂P2σ, ∂Pσ and the plane annulus Ωσ =x : x3 = −h, σ ≤ r′ ≤ 2σ

. Recall that the divergence of a vector field XiEi is

given by Xi;i = Xi

,i + ΓiikXk. Therefore, div∂3 = Γkik = O(σ−2) in asymptotically

flat coordinates. We observe that the volume of V is O(σ2) and the areas of P2σ

and Pσ are O(σ−1). If we let n denote the unit normal of P2σ and Pσ, then〈∂3, n〉 = O(σ−1). Finally, A(Ωσ) = O(σ2), and since 〈ν, ∂3〉 = 1 + O(σ−1), itfollows that

O(1) =

∫V

div∂3

=

∫D2σ\Dσ

〈ν, ∂3〉+

∫P2σ

〈n, ∂3〉+

∫Pσ

〈n, ∂3〉+

∫Ωσ

〈ν, ∂3〉

=

∫D2σ\Dσ

〈ν, ∂3〉+O(1) +O(1)−A(Ωσ) +O(σ).

Therefore, ∫D2σ\Dσ

〈ν, ∂3〉 −A(Ωσ) = O(σ).

Applying the area minimizing property of S on D2σ\Dσ, we conclude that,

A(D2σ\Dσ) ≤ A(Ωσ) +O(σ).

This implies

∫D2σ\Dσ

(1− 〈ν, ∂3〉) ≤ A(D2σ\Dσ)−∫

D2σ\Dσ

〈ν, ∂3〉

≤ A(Ωσ) +O(σ)−∫

D2σ\Dσ

〈ν, ∂3〉

≤ A(Ωσ) +O(σ)−A(Ωσ) +O(σ)

= O(σ).

We now recall the co-area formula from Section 2.2. Letting φ = (1− 〈ν, ∂3〉) inthe co-area formula, we get

2σ∫σ

∫D2σ∩∂Pt

(1− 〈ν, ∂3〉) dstdt =

∫D2σ∩(P2σ\Pσ)

||∇r′|| (1− 〈ν, ∂3〉) dS,

where dst is the arclength element on D2σ ∩∂Pt. Since D2σ ∩ (P2σ\Pσ) ⊆ D2σ\Dσ,it follows that


2σ∫σ

∫D2σ∩∂Pt

(1− 〈ν, ∂3〉) dstdt ≤∫

D2σ\Dσ

||∇r′|| (1− 〈ν, ∂3〉) dS

=

∫D2σ\Dσ

(1 +O(σ−1)

)(1− 〈ν, ∂3〉) dS

≤ O(σ).

Again, using the co-area formula, this time by letting φ ≡ 1, we get

2σ∫σ

L(D2σ ∩ ∂Pt)dt =


||∇r′||dS

≤∫

D2σ\Dσ

||∇r′||dS

≤∫

D2σ\Dσ

(1 +O(σ−1))dS

= O(σ2).

We now bound the second fundamental form of S on ∂Dσ. We consider thefollowing function ψ for the inequality

∫S

|A|2ψ2 ≤ 2∫S

||∇ψ||2 +∫S

|Ric(ν, ν)|ψ2.

ψ(x) =

0 for x ∈ S ∩ [(N\Nk) ∪ P√σ]

log(r′/√σ)

log√σ

for x ∈ S ∩ (Pσ\P√σ)

1 for x /∈ S ∩ [(N\Nk) ∪ Pσ].

This implies∫S\[(N\Nk)∪Pσ ]

|A|2 ≤∫

S\[(N\Nk)∪P√σ ]

|A|2ψ2

≤ 2

(log√σ)2

∫S∩(Pσ\P√σ)

||∇r′||2

(r′)2+

∫S\[(N\Nk)∪P√σ]

|Ric(ν, ν)|,

which in turn implies∫D2σ∩(P2σ\Pσ)

|A|2 ≤ 2C

(log√σ)2

∫S∩(Pσ\P√σ)

1

(r′)2+

∫S\[(N\Nk)∪P√σ ]

O(r−3)

≤ 2C

(log√σ)2

O(log√σ) +O(σ−1/2)

= o(1).

By the co-area formula, this time letting φ = |A|2, we get

42 FARHAN ABEDIN

2σ∫σ

∫D2σ∩∂Pt

|A|2dt =


|A|2||∇r′||2.

Therefore,2σ∫σ

∫D2σ∩∂Pt

|A|2dt = o(1).

We observe now from the expressions2σ∫σ

∫D2σ∩∂Pt

(1− 〈ν, ∂3〉) dstdt ≤ O(σ),

2σ∫σ

L(D2σ ∩ ∂Pt)dt ≤ O(σ2),

2σ∫σ

∫D2σ∩∂Pt

|A|2dt = o(1),

that, by invoking the intermediate value theorem, there exists σ ∈ [σ, 2σ] such that∫D2σ∩∂Pσ

(1− 〈ν, ∂3〉) ≤ O(1),

L(D2σ ∩ ∂Pσ) = O(σ),∫D2σ∩∂Pσ

|A|2 = o(σ−1)

as σ → ∞. By applying the Cauchy-Schwarz inequality and the condition on thelength of the boundary, we get

∫D2σ∩∂Pσ

|A|

2

≤

∫D2σ∩∂Pσ

1

· ∫D2σ∩∂Pσ

|A|2

≤ L(D2σ ∩ ∂Pσ) · o(σ−1)

= o(1).

Since ∂Dσ is a component of D2σ ∩ ∂Pσ, it follows that∫∂Dσ

(1− 〈ν, ∂3〉) ≤ O(1),

L(∂Dσ) = O(σ),∫∂Dσ

|A| = o(1).

Using the above estimate, and the expression


k

⟨e2,

x′

σ

⟩+

1

σ− h11

⟨ν,x′

σ

⟩− 1

σ〈e1, ∂3〉2 = O(σ−1)||De1e1||+O(σ−2)

we conclude that∫∂Dσ

∣∣∣∣k⟨e2,x′

σ

⟩∣∣∣∣ = O(1) +O(σ−1)

∫∂Dσ

||De1e1||+1

σ

∫∂Dσ

〈e1, ∂3〉2.

Observe that 〈e1, ∂3〉2 +〈e2, ∂3〉2 = 1−〈ν, ∂3〉2 +O(σ−1). Since∫∂Dσ

(1− 〈ν, ∂3〉) ≤O(1), it follows that

∫∂Dσ

(〈e1, ∂3〉2 + 〈e2, ∂3〉2

)= O(1).

We now give a pointwise lower bound on | 〈e2, ∂3〉 |. In fact, we show that

sup∂Dσ

(1 +

⟨e2,

x′

σ

⟩)= o(1).

This would imply⟨e2,

x′

σ

⟩→ −1 as σ → ∞. Since 〈ν, ∂3〉 = 1 + O(σ−1) and

〈e2, ν〉 = 0, it would follow that 〈e2, ∂3〉 → 0 as σ →∞.

We first observe that since⟨e1,

x′

σ

⟩= 0, we have

1−⟨e2,

x′

σ

⟩2

=

⟨ν,x′

σ

⟩2

+O(σ−1)

=

⟨x′

σ, ν − ∂3

⟩2

+O(σ−1)

≤∣∣∣∣∣∣∣∣x′σ

∣∣∣∣∣∣∣∣2 ||ν − ∂3||2 +O(σ−1)

≤ ||ν − ∂3||2 +O(σ−1)

= 2 (1− 〈ν, ∂3〉) +O(σ−1).

This implies ∫∂Dσ

1−⟨e2,

x′

σ

⟩2

= O(1).

Now since e2 is the inner normal to ∂Dσ and Dr′ = x′

σ + O(σ−1) is the outer

normal, we have⟨e2,

x′

σ

⟩≤ O(σ−1). Therefore,

1−⟨e2,

x′

σ

⟩2

=

(1−

⟨e2,

x′

σ

⟩)(1 +

⟨e2,

x′

σ

⟩)≥ (1−O(σ−1))

(1 +

⟨e2,

x′

σ

⟩)+O(σ−1).

Combined with the above estimate, we conclude that∫∂Dσ

1 +

⟨e2,

x′

σ

⟩= O(1).

Since ∂Dσ is not homologous to zero in R3\Pσ0, its projection onto the x1,x2 plane

must be the circle of radius σ centered at 0. Therefore, L(∂Dσ) ≥ 2πσ − O(1).

This implies there exists x0 ∈ ∂Dσ such that 1 +(⟨e2,

x′

σ

⟩)(x0) = O(σ−1).

44 FARHAN ABEDIN

We now differentiate along ∂Dσ to get

De1

[1 +

⟨e2,

x′

σ

⟩]=

1

σ〈e1, e2〉 −

1

σ〈e1, ∂3〉〈e2, ∂3〉 − h12

⟨ν,x′

σ

⟩+O(σ−2)

≤ 1

2σ

(〈e1, ∂3〉2 + 〈e2, ∂3〉2

)+ |A|+O(σ−2).

Since∫∂Dσ

(〈e1, ∂3〉2 + 〈e2, ∂3〉2

)= O(1) and

∫∂Dσ

|A| = o(1), we conclude that∫∂Dσ

∣∣∣∣De1

[1 +

⟨e2,

x′

σ

⟩]∣∣∣∣ = o(1).

For any x ∈ ∂Dσ, we may write

1 +

(⟨e2,

x′

σ

⟩)(x) = 1 +

(⟨e2,

x′

σ

⟩)(x0) +

x∫x0

De1

[1 +

⟨e2,

x′

σ

⟩].

Since 1+(⟨e2,

x′

σ

⟩)(x0) = O(σ−1) and

∫∂Dσ

∣∣∣De1

[1 +

⟨e2,

x′

σ

⟩]∣∣∣ = o(1), it follows

that

sup∂Dσ

(1 +

⟨e2,

x′

σ

⟩)= o(1).

We now know that for large enough σ,∣∣∣⟨e2,

x′

σ

⟩∣∣∣ ≈ 1. Therefore,∫∂Dσ

|k| = O(1) +O(σ−1)

∫∂Dσ

||De1e1||+ o(1).

Since De1e1 = ke2 − h11ν, it follows that ||De1e1|| ≤ |k|+ |A|. This implies∫∂Dσ

||De1e1|| ≤∫∂Dσ

|k|+ o(1)

≤ O(1) +O(σ−1)

∫∂Dσ

||De1e1||+ o(1).

Therefore,∫∂Dσ||De1e1|| = O(1) and so

∫∂Dσ|k| = O(1).

Now∫∂Dσ

k ≥∫∂Dσ

k(

1 +⟨e2,

x′

σ

⟩)+ 1σL(∂Dσ)−o(1), so it follows that

∫∂Dσ

k ≥2π − o(1). Since this holds for σ arbitrarily large in any [σ, 2σ] for σ sufficientlylarge, it follows that we can choose a sequence σi → ∞ for which the above in-equality holds. This finishes the proof of Claim 3 and concludes the proof of thePositive Mass Theorem.

3.6. Proof of the Rigidity Theorem. We now proceed to prove the RigidityTheorem. Note that by throwing away the other ends outside a convex ball, wemay assume that N has only one end, so that N\Nk is compact.

We will require the following Sobolev Inequality.


Lemma 8. There is a constant c1 > 0, c1 = c1(N, k1, k2, k3) such that for all ϕcompactly supported on N , we have the inequality(∫

N

ϕ6

)1/3

≤ c1∫N

||Dϕ||2.

Proof. Suppose not. Then there exists a sequence of functions ϕi with compactsupport such that

∫Nϕ6i = 1 and

∫N||Dϕi||2 ≤ 1/i. Since Nk can be identified

with R3\Bσ0(0) and g is uniformly equivalent to the Euclidean metric, we have by

the Sobolev inequality (∫Nk

ϕ6i

)1/3

≤ c∫Nk

||Dϕi||2.

Hence ϕi → 0 in the L6 norm. If we choose a precompact neighborhood O ⊂ N ,then for any C1 function g defined on O, we get

infβ∈R

(∫O

(g − β)6

)1/3

≤ c∫O||Dg||2.

Applying this to the sequence of functions ϕi|O, we can find a sequence βi suchthat (∫

O(ϕi − βi)6

)1/3

≤ c∫O||Dϕi||2.

Since∫Nφ6i = 1, the sequence βi is bounded by Minkowski’s Inequality, ||βi||2L6(O) ≤

||βi − ϕi||2L6(O) + ||ϕi||2L6(O) < ∞. Hence, by extracting a subsequence, we may

assume βi → β. Thus φi → β in the L6 norm on O. Since ϕi → 0 in the L6 normon Nk, it follows that β = 0 on each coordinate neighborhood O. Therefore, φi → 0in the L6 norm on N . This contradicts the fact that

∫Nφ6i = 1.

We will now study equations of the form ∆v − fv = h on N , where f, h arefunctions which satisfy

|f | ≤ k7

(1 + r5

)−1,

|∂f | ≤ k8

(1 + r5

)−1,

|h| ≤ k7

(1 + r5

)−1,

|∂h| ≤ k8

(1 + r5

)−1

on Nk. Let f+ and f− denote the positive and negative parts of f .

Lemma 9. Suppose N is asymptotically flat. Then there is a number ε0 = ε0(N, k1, k2, k3) >0 such that if (

∫N

(f−)3/2)2/3 ≤ ε0, then the above differential equation has a uniquesolution v defined on N satisfying v = O(1/r) and ∂νv = 0 on ∂N , where ν is theoutward unit normal vector to ∂N . Moreover, v = A

r + ω, where ω satisfies

46 FARHAN ABEDIN

|ω| ≤ k9

(1 + r2

)−1,

|∂ω| ≤ k10

(1 + r3

)−1,

|∂2ω| ≤ k11

(1 + r4

)−1,

on Nk for constants k9, k10 and k11 depending only on k1, k2 and k3, and whereA = − 1

4π

∫N

(fv + h).

Proof. We will only prove existence and uniqueness of the solution, omitting theasymptotic analysis. For σ > σ0, we can solve the problem

∆vσ − fvσ = h on Nσ = (N\Nk) ∪ (Bσ(0) ∩Nk),

vσ = 0 on ∂Bσ(0),

∂νvσ = 0 on ∂N .

Using integration by parts, we get

∫Nσ||Dvσ||2 = −

∫Nσ

fv2σ −

∫Nσ

hvσ

≤∫Nσ

(f−)v2σ +

∫Nσ|h||vσ|

≤(∫

Nσ(f−)3/2

)2/3(∫Nσ

v6σ

)1/3

+

(∫Nσ|h|6/5

)5/6(∫Nσ

v6σ

)1/6

.

Observe that if h ≡ 0, then we have by the Sobolev Inequality,∫Nσ||Dv2

σ|| ≤ ε0c1∫Nσ||Dv2

σ||.

Thus if we choose ε0 < 1/c1, we see that the operator ∆−f has trivial kernel for thegiven problem, and hence by the Fredholm Alternative (Lemma 3), there exists aunique smooth solution vσ. By the asymptotics for h, we have (

∫Nσ|h|6/5)5/6 ≤ c2,

independent of σ. Applying this and Lemma 8, we have(∫Nσ

v6σ

)1/3

≤ ε0c1(∫

Nσv6σ

)1/3

+ c1c2

(∫Nσ

v6σ

)1/6

.

Choosing ε0 = 13c1

and using the inequality |ab| ≤ 13a

2 + 34b

2, we get(∫Nσ

v6σ

)1/3

≤ 1

3

(∫Nσ

v6σ

)1/3

+1

3

(∫Nσ

v6σ

)1/3

+3

4(c1c2)2.


Hence∫Nσ

v6σ ≤ c3 uniformly. By the Schauder estimate (Theorem 9), where we

may replace the sup norm of u on the right hand side with the L6 norm of u,we conclude that vσ : σ > σ0 is equicontinuous in the C2 topology on compactsubsets of N . Thus, we may choose a sequence σi →∞ so that vσi → v uniformly inthe C2 norm on compact subsets of N . Thus v is a solution of the problem definedon N , satisfying ∂νv = 0 on ∂N and

∫Nσ

v6 ≤ c3. Moreover, by the uniform L6/5

bound for h and the uniform L6 bound for v on a ball, we can conclude through acovering argument and by invoking the local boundedness of solutions (cf. Section2.5) that supN |v| ≤ c4.

To prove uniqueness, suppose v is another solution satisfying v = O(1/r) and∂ν v = 0 on ∂N . Then u = v− v satisfies ∆u− fu = 0, u = O(1/r) and ∂νu = 0 on∂N . We claim u ≡ 0. Let µ > 0 be any number and let Eµ = x ∈ N : u(x) ≥ µ.Because u tends to 0 at infinity, Eµ is compact. Multiplying the equation by u andintegrating by parts gives us

∫Eµ

||Du||2 = −∫∂Eµ

µ||Du|| −∫Eµ

fu2

≤∫Eµ

(f−)u2

≤

(∫Eµ

(f−)3/2

)2/3(∫Eµ

u6

)1/3

≤ ε0

(∫Eµ

u6

)1/3

.

By letting ε0 = 13c1

and applying the Sobolev Inequality to the function ψ = u− µon Eµ, ψ = 0 on N\Eµ we have

(∫Eµ

(u− µ)6

)1/3

≤ c1∫Eµ

||Du||2

≤ c1ε0

(∫Eµ

u6

)1/3

≤ 1

3

(∫Eµ

u6

)1/3

.

Since u = O(1/r), we see that∫Eµu6 < ∞, so we may let µ → 0 to arrive at

a contradiction, unless u ≤ 0. Since −u solves the same problem, it follows thatu ≥ 0 and hence u ≡ 0.

We now apply Lemma 9 to a conformal change of metric on N .

48 FARHAN ABEDIN

Theorem 20. Let g be an asymptotically flat metric of zero total mass on N andlet R be the scalar curvature function of g. Suppose R satisfies

1

8

(∫N

(R−)3/2

)2/3

≤ ε0.

where ε0 is defined in Lemma 9. Then there is a unique positive function ϕ with∂νϕ = 0 on ∂N so that the metric g = ϕ4g is asymptotically flat, scalar flat andhas total mass M = − 1

32π

∫NRϕ.

Proof. In order for the metric g to be asymptotically flat and scalar flat, the functionϕ must satisfy ∆ϕ− 1

8Rϕ = 0. The function v = ϕ−1 then satisfies ∆v− 18Rv = 1

8R.In order for g to be asymptotically flat, v must satisfy the asymptotic conditions ofLemma 10. The lemma directly applies to give a v satisfying the above differentialequation with ∂νv = 0 on ∂N . Thus ϕ = v + 1 satisfies the original problem with∂νϕ = 0 on ∂N .

In order to prove ϕ > 0 on N , we let E = x ∈ N : ϕ(x) < 0. Since ϕ isasymptotic to 1, it follows that E is compact. If E is non-empty, we can multiplythe differential equation by ϕ, integrate by parts on E (using the fact that ∂νϕ = 0on ∂N), and apply the Sobolev Inequality with the correct choice of ε0 to get

(∫E

ϕ6

)1/3

≤∫E

||Dϕ||2 = −1

8

∫E

Rϕ2

≤ 1

8

∫E

(R−)ϕ2

≤ 1

8

(∫E

(R−)3/2

)2/3(∫E

ϕ6

)1/3

≤ c1ε0(∫

E

ϕ6

)1/3

which is a contradiction since ε0 ≤ 13c1

. We conclude that ϕ ≥ 0 on N , and by thestrong maximum principle, we get ϕ > 0 on N .

To show that ∂N has positive mean curvature relative to g, we note that if H isthe mean curvature of g and H is the mean curvature of g, then

H =1

ϕ2

(H +

4

ϕ∂νϕ

)=

1

ϕ2H > 0.

Therefore g is an asymptotically flat metric on N . The formula for the mass Malso follows from Lemma 9.

A special case of the above theorem is the following corollary.

Corollary 1. If M = 0, R ≥ 0 and R is not identically zero, then there is a metricconformally equivalent to g which is asymptotically flat, scalar flat, and so that Nkhas negative total mass.


Combining the above corollary with the Positive Mass Theorem, we conclude thatan asymptotically flat metric satisfying M = 0, R ≥ 0, R not identically zero onN , must in fact have R ≡ 0 on N .

We now assume g is such a metric and consider a one-parameter family of metricsgt on N defined as

gt =

3∑i,j=1

(gij + tRij)dxidxj

where Rij is the Ricci tensor of g. These metrics are defined in a neighborhood oft = 0 by the asymptotically flat condition. Note that g0 = g. For t sufficiently small,gt is asymptotically flat, and because ∂N has positive mean curvature relative to g,it follows by continuity that ∂N also has positive mean curvature relative to gt fort small. Let Rt be the scalar curvature of gt, so that R0 ≡ 0. Then by the formulafor the linearization of scalar curvature (Lemma 7), we have

R′0 =d

dt

∣∣t=0

Rt = −∆R0 + div(div(Ric0))− ||Ric0||2.

Since R0 ≡ 0, we have ∆R0 ≡ 0 and by the contracted Bianchi identity (cf. Section2.6), we have div(div(Ric0)) = 2∆R0 ≡ 0. Thus R′0 = −||Ric0||2. It follows thatfor t sufficiently small, we have

1

8

(∫N

(R−t )3/2

)2/3

≤ ε0.

where ε0 can be taken independently of t for small t. For each t we can find afunction ϕt such that the metric ϕ4

t gt is asymptotically flat and scalar flat. Notethat the mass m(gt) of each metric gt is zero due to the asymptotics imposed onthe metric g and subsequently on the Ricci tensor Rij of g. The mass of the metricϕ4t gt is therefore M(t) = − 1

32π

∫NRtϕt

√gtdx.

We now show that M ′(0) = ddt |t=0M(t) exists. For small h, let ϕ(h) be defined

by ϕ(h) = 1h (ϕh − ϕ0). Let ∆t be the Laplace operator for the metric gt and let

∆(h) be the differential operator defined by ∆(h)v = 1h (∆hv − ∆0v). Finally, let

R(h) = 1h (Rh−R0). Then, since gt is scalar flat for each t, the function ϕ(h) satisfies

the equation

∆0ϕ(h) − 1

8R0ϕ

(h) = −∆(h)ϕh +1

8R(h)ϕh.

By asymptotic flatness, the above equation satisfies the conditions for the lemmaand since ϕ(h) = O(1/r), the lemma implies |ϕ(h)| ≤ γ1(1 + r)−1 on Nk, whereγ1 is independent of h. By the Schauder estimate (Theorem 9), ϕ(h) has a localC2,α bound depending on C1 bounds on R0 and −∆(h)ϕh + 1

8R(h)ϕh. Since these

bounds are independent of h, we can find a sequence hi tending to zero so thatϕhi converges in C2,β for any β < α uniformly on compact subsets of N to a C2,α

function ϕ′0 which satisfies ∆0ϕ′0− 1

8R0ϕ′0 = −∆′0ϕ0+ 1

8R′0ϕ0, where ∆′0 = d

dt |t=0∆t

and R′0 = ddt |t=0Rt. Since ϕ′0 = O(1/r), the uniqueness part of the lemma implies

the limit ϕ′0 is independent of the sequence hi chosen. Therefore, ddt |t=0ϕt exists

at t = 0 and is equal to ϕ′0.

We also have by asymptotic flatness |R(h)| ≤ γ2(1 + r3+α)−1g(h) = γ3(1 + r1+α)−1

on Nk, where g(h) = 1h (gh−g0). Therefore, by the dominated convergence theorem,

50 FARHAN ABEDIN

we conclude that M ′(0) exists. By differentiating under the integral sign, we seethat

M ′(0) = − 1

32π

∫N

R0 (ϕt√gt)′dx− 1

32π

∫N

R′0ϕ0√g0dx.

Since R0 ≡ 0 and ϕ0 ≡ 1, it follows that M ′(0) = 132π

∫N||Ric0||2. If Ric0 is not

identically zero, then M ′(0) > 0 and hence by choosing a suitable t0 < 0, we canconclude that M(t0) < 0. The metric gt0 would then be asymptotically flat, scalarflat and Nk would have negative total mass, in contradiction to the Positive MassTheorem. If, on the other hand, Ric ≡ 0, then g is flat since n = 3. This completesthe proof of the rigidity theorem.


4. Appendix

We prove Theorem 17, which was stated in the introduction of section 3. Wefollow the outline in [13]. Recall the statement of the theorem.

If (N, g) is asymptotically flat with R(g) ≥ 0, then for all ε > 0, there exists aharmonically flat metric g such that m(g) ≤ m(g) + ε.

Proof. Once again, we assume that N has only one end. Recall that if g = u4

n−2 g,

then R(g) = u−n+2n−2 (R(g)u− 4(n−1)

n−2 ∆gu). Denote by Lg the operator ∆g − 18R(g).

We first observe that we may take R(g) ≡ 0 since by Theorem 20, we can solveLgu = 0, u > 0, u→ 1 at infinity. In fact, by [2], u(x) = 1 + A

|x| +O(|x|−2), where

A ≤ 0. To see why A ≤ 0, let Br be the interior of the region bounded by thecoordinate sphere Sr. Integrating the equation Lgu = 0 over Br gives

0 =

∫Br

Lgu dµg =

∫Br

∆gu−1

8R(g)u dµg.

By the divergence theorem,∫|x|=r

∇u · νg dσ =1

8

∫Br

R(g)u dµg ≥ 0.

Using the spherical harmonic expansion of u, we conclude that

0 ≤ limr→∞

∫|x|=r

∇u · νg dσ = −4πA.

The metric u4g is then scalar-flat and has mass m(g) + A ≤ m(g). Thus, we mayreplace g with u4g.

Assuming R(g) ≡ 0, we can deform g near infinity to the Euclidean metric.Choose a function Ψσ(x) such that Ψσ(x) = 1 for |x| ≤ σ, Ψσ(x) = 0 for x ≥ 2σ,Ψσ is a decreasing function of x, and σ|Ψ′σ|+σ2|Ψ′′σ| ≤ c. Now consider the metric gσ

given by gσ = Ψσg+(1−Ψσ)δ, where δ denotes the Euclidean metric. Observe thatgσ = δ+O(|x|−p) for p > 1

2 uniformly in σ for σ large, and also R(gσ) = O(|x|−2−p)

for σ ≤ |x| ≤ 2σ uniformly in σ. In particular,∫M|R(gσ)|3/2dVg = O(σ−3p/2), and

so for σ large, we can use Theorem 20 to say there exists a unique solution ofLgσuσ = 0 such that uσ > 0 and uσ(x)→ 1 as |x| → ∞. The metric gσ = u4

σgσ is

then harmonically flat.

We show m(gσ)→ m(g) as σ →∞. Then for large enough σ, the metric gσ willbe the desired metric. By the definition of m(g), we may pick τ large enough sothat ∣∣∣∣∣∣∣m(g)− 1

16π

∫|x|=τ

∑i,j

(gij,i − gii,j)ντj dxτ

∣∣∣∣∣∣∣ ≤ε

3

By the uniform decay estimates on uσ and gσ, we may pick τ above (independentlyof σ) large enough so that

52 FARHAN ABEDIN∣∣∣∣∣∣∣m(gσ)− 1

16π

∫|x|=τ

∑i,j

(gσij,i − gσii,j)ντj dxτ

∣∣∣∣∣∣∣ ≤ε

3

We now recall from Theorem 20 that for each uσ, we have the corresponding con-stant Aσ = − 1

32π

∫MR(gσ)uσ. Since 0 < uσ < 1 and ||R(gσ)||L3/2(M) ≤ O(σ−1/2),

it follows that |Aσ| is uniformly bounded in σ. Therefore, gσ − g = O(σ−1), andhence we may pick σ large enough so that∣∣∣∣∣∣∣

1

16π

∫|x|=τ

∑i,j

[(gσij,i − gσii,j)− (gij,i − gii,j)

]ντj dxτ

∣∣∣∣∣∣∣ ≤ε

3

This implies |m(gσ)−m(g)| < ε for σ large.


References

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Applied Mathematics; Vol. 39, No. 5, 661 - 693, (1986).[3] H. L. Bray; Proof of the Riemannian Penrose Inequality using the Positive Mass Theorem.

Journal of Differential Geometry; Vol. 59, 177 - 267, (2001).

[4] B. Driver; Analysis Tools with Applications; Notes from Professor Bruce Driver’s Website atUCSD.

[5] L. C. Evans; Partial Differential Equations; American Mathematical Society, 2nd edition,(2010).

[6] D. Gilbarg, N. S. Trudinger; Elliptic Partial Differential Equations of Second Order ;

Springer, (1983).[7] Q. Han, F. Lin; Elliptic Partial Differential Equations (Courant Lecture Notes); American

Mathematical Society, (2000).

[8] L.-H. Huang, D. Wu; Hypersurfaces with nonnegative scalar curvature. arXiv:1102.5749.[9] M.-K. G. Lam; The Graphs Cases of the Riemannian Positive Mass Theorem and Penrose

Inequalities in All Dimensions. arXiv:1010.4256v1.

[10] J. M. Lee; Riemannian Manifolds: An Introduction to Curvature; Springer, (1997).[11] P. Miao; Positive Mass Theorem on Manifolds admitting Corners along a Hypersurface.

Advances in Theoretical and Mathematical Physics; Vol. 6, 1163 - 1182, (2002).

[12] R. M. Schoen; Conformal deformation of a Riemannian metric to constant scalar curvature.Journal of Differential Geometry; Vol. 20, 479 - 495, (1984).

[13] R. M. Schoen; Variational Theory for the Total Scalar Curvature Functional for RiemannianMetrics and Related Topics In: M. Giaquinta (ed.), Topics in the Calculus of Variations,

Lecture Notes in Mathematics; Vol. 1365, 120 - 154, (1987).

[14] R. M. Schoen, S.-T. Yau; On the Proof of the Positive Mass Conjecture in General Relativity.Communications in Mathematical Physics; Vol. 65, 45 - 76, (1979).

[15] R. M. Schoen, S.-T. Yau; The Energy and the Linear Momentum of Space-Times in General

Relativity. Communications in Mathematical Physics; Vol. 79, 47 - 51, (1981).[16] R. M. Schoen, S.-T. Yau; Proof of the Positive Mass Theorem, II. Communications in Math-

ematical Physics; Vol. 79, 231 - 260, (1981).

[17] S. Sternberg; Lectures on Differential Geometry; American Mathematical Society, 2nd edi-tion, (1999).

[18] X. Wang; The Mass of Asymptotically Hyperbolic Manifolds. Journal of Differential Geome-

try; Vol. 57, 273 - 299, (2001).[19] E. Witten; A New Proof of the Positive Energy Theorem. Communications in Mathematical

Physics; Vol. 80, 381 - 402, (1981).

http://www.math.ucsd.edu/~bdriver/231-02-03/Lecture_Notes/PDE-Anal-Book/analpde1.pdf

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The Riemannian Positive Mass Theorem

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