ɷmathematical problems of general relativity theory ii
DESCRIPTION
ɷMathematical Problems of General Relativity Theory IITRANSCRIPT
Mathematical Problems
of
General Relativity Theory II
Lectures held by
Prof. Demetrios Christodoulou
at the
ETH Zurich
during the winter semester
2003 / 2004
Script: Lydia Bieri
Contents
1 Introduction 6
2 Null Hypersurface 8
2.1 Geometry of a Null Hypersurface . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 AÆne Foliations of a Null Hypersurface . . . . . . . . . . . . . . . . . . . . . . . 12
2.2.1 The Torsion of a Spacelike Surface S . . . . . . . . . . . . . . . . . . . . . 142.2.2 Jacobi Fields and the 2nd Variation . . . . . . . . . . . . . . . . . . . . . . 162.2.3 Focal Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.3 Theorem of Penrose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3 Global Geometry of a Null Hypersurface 35
3.1 Codazzi Equations for a Section S of a Null Hypersurface . . . . . . . . . . . . . 383.2 The Divergence Operator on Trace-Free, 2-Covariant, Symmetric Tensor�elds on
a Compact 2-Dimensional Riemannian Manifold . . . . . . . . . . . . . . . . . . 413.3 The L2-Theory for the Operator in Question . . . . . . . . . . . . . . . . . . . . 43
4 Uniformization Theorem 45
5 Estimates for � and the Torsion (�) System 62
5.1 Estimates for � . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625.2 The Torsion (�) System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
5.2.1 The div � curl System for 1-Forms on a 2-Dimensional, Compact, Rie-mannian Manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
5.2.2 Treatment of the Propagation Equation for � . . . . . . . . . . . . . . . . 80
6 Sobolev Inequalities on C 92
6.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 996.2 Estimates for m, M if K 2 L4(S) . . . . . . . . . . . . . . . . . . . . . . . . . 101
7 Smoothing Foliation 102
8 Memory E�ect 1038.1 Limiting Equations on C�
� (as B� exhausts �0 or r�0 !1) . . . . . . . . . . . . 114
8.2 Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
1
General Introduction
General Relativity is a theory proposed by Einstein in 1915 as uni�ed theory of space, time andgravitation. The theory's roots extend over almost the entire previous history of physics andmathematics.
Its immediate predecessor, Special Relativity, in its �nal form established by Minkowski in1908, accomplished the uni�cation of space and time in the geometry of a 4-dimensional aÆnemanifold, a geometry of simplicity and perfection on par with that of the Euclidean geometryof space. The root of Special Relativity is Electromagnetic Theory, in particular Maxwell's in-corporation of Optics, the theory of light, into Electrodynamics.
General Relativity is based on and extends Newton's theory of Gravitation as well as New-ton's equations of motion. It is thus fundamentally rooted in Classical Mechanics.
Perhaps the most fundamental aspect of General Relativity however, is its geometric nature.The theory can be seen as a development of Riemannian geometry, itself an extension of Gauss'intrinsic theory of curved surfaces in Euclidean space.
The connection between gravitation and Riemannian geometry arose in Einstein's mind in hise�ort to uncover the meaning of what in the Newtonian theory is the fortuitous equality ofthe inertial and the gravitational mass. The identi�cation, via the equivalence principle, of thegravitational tidal force with the spacetime curvature at once gave the physical interpretationof curvature of the spacetime manifold and also revealed the geometrical meaning of gravitation.
One sees here that descent to a deeper level of understanding of physical reality is connectedwith ascending to a higher level of mathematics. General Relativity constitutes a triumph ofthe geometric approach to physical science.
But there is more to General Relativity than merely a physical interpretation of a variantof Riemannian Geometry. For, the theory contains physical laws in the form of equations -Einstein's equations- imposed on the geometric structure. This gives a tightness which makesthe resulting mathematical structure one of surpassing subtlety and beauty. An analogous situ-ation is found by comparing the theory of di�erentiable functions of two real variables with thetheory of di�erentiable functions of one complex variable. The latter gains by the impositionof the Cauchy-Riemann equations a tighter structure which leads to a greater richness of results.
The domain of application of General Relativity, beyond that of the Newtonian theory, is astro-nomical systems, stellar or galactic, where the gravitational �eld is so strong that it implies thepotential presence of velocities which are not negligible in comparison with the velocity of light.The ultimate domain of application is the study of the structure and evolution of the universeas a whole.
General Relativity has perhaps the most satisfying structure of all physical theories from themathematical point of view. It is a wonderful research �eld for a mathematician. Here, resultsobtained by purely mathematical means have direct physical consequences.
One example of this is the incompleteness theorem of R. Penrose and its extensions due to
2
Hawking and Penrose known as the "singularity theorems". This result is relevant to the studyof the phenomenon of gravitational collapse. It shall be covered in the second of the presentvolumes. The methods used to establish the result are purely geometrical - the theory of conju-gate points. In fact, part of the main argument is already present in the theory of focal pointsin the Euclidean framework, a theory developed in antiquity.
Another example is the positive energy theorem, the �rst proof of which, due to R. Schoenand S.T. Yau, is based on the theory of minimal surfaces and shall be covered in the �rst of thepresent volumes. In this example a combination of geometric and analytic methods are employed.
A last example is the theory of gravitational radiation, a main theme for both of the presentvolumes. Here also we have a combination of geometric and analytic methods. A particularresult in the theory of gravitational radiation is the so-called memory e�ect [12], which is due tothe non-linear character of the asymptotic laws at future null in�nity and has direct bearing onexperiments planned for the near future. This result will be covered in the second of the presentvolumes.
The laws of General Relativity, Einstein's equations, constitute, when written in any systemof local coordinates, a non-linear system of partial di�erential equations for the metric compo-nents. Because of the compatibility conditions of the metric with the underlying manifold, whenpiecing together local solutions to obtain the global picture, it is the geometric manifold, namelythe pair consisting of the manifold itself together with its metric, which is the real unknown inGeneral Relativity.
The Einstein equations are of hyperbolic character, as shall be explained in detail in the �rst ofthe present volumes. As a consequence, the initial value problem is the natural mathematicalproblem for these equations. This conclusion, reached mathematically, agrees with what oneexpects physically. For, the initial value problem is the problem of determining the evolution ofa system from given initial conditions, as in the prototype example of Newton's equations of mo-tion. The initial conditions for Einstein's equations, the analogues of initial position and velocityof Newtonian mechanics, are the intrinsic geometry of the initial spacelike hypersurface and itsrate of change under a virtual normal displacement, the second fundamental form. In contrast tothe case of Newtonian mechanics however, these initial conditions are, by virtue of the Einsteinequations themselves, subject to constraints, and it is part of the initial value problem in GeneralRelativity - a preliminary part - to analyze these constraints. Important results can be obtainedon the basis of this analysis alone and the positive energy theorem is an example of such a result.
An important notion in physics is that of an isolated system. In the context of the theoryof gravitation examples of such systems are a planet with its moons, a star with its planetarysystem, a binary or multiple star, a cluster of stars, a galaxy, a pair or multiplet of interactinggalaxies, or, as an extreme example, a cluster of galaxies - but not the universe as a whole.What is common in these examples is that each of these systems can be thought of as hav-ing an asymptotic region in which conditions are trivial. Within General Relativity the trivialcase is the at Minkowski spacetime of Special Relativity. Thus the desire to describe isolatedgravitating systems in General Relativity leads us to consider spacetimes with asymptoticallyMinkowskian regions. However it is important to remember at this point the point of view ofthe initial value problem: a spacetime is determined as a solution of the Einstein equations fromits initial data. Consequently, we are not free to impose our own requirements on a spacetime.
3
We are only free to impose requirements on the initial data - to the extent that the requirementsare consistent with the constraint equations. Thus the correct notion of isolated system in thecontext of General Relativity is a spacetime arising from asymptotically at initial conditions,namely an intrinsic geometry which is asymptotically Euclidean and second fundamental formwhich tends to zero at in�nity in an appropriate way. This shall be discussed in detail in the�rst of the present volumes.
Trivial initial data for the Einstein equations consists of Euclidean intrinsic geometry and van-ishing second fundamental form. Trivial initial data gives rise to the trivial solution, namelythe Minkowski spacetime. A natural question in the context of the initial value problem forthe vacuum Einstein equations is whether or not every asymptotically at initial data which isglobally close to the trivial data gives rise to a solution which is a complete spacetime tending tothe Minkowski spacetime at in�nity along any geodesic. This question was answered in the af-�rmative in the joint work of the present author with Sergiu Klainerman, which appeared in themonograph [15]. A main aim of the present volumes is to present the methods which went intothat work in a more general context, so that the reader may more fully understand their originand development as well as be able to apply them to other problems. In fact, problems comingfrom �elds other than General Relativity are also treated in the present volumes. These �eldsare Continuum Mechanics, Electrodynamics of Continuous Media and Classical Gauge Theories(such as arise in the mesoscopic description of super uidity and superconductivity). What iscommon to all these problems from our perspective is the mathematical methods involved.
One of the main mathematical methods analyzed and exploited in the present volumes is thegeneral method of constructing a set of quantities whose growth can be controlled in terms ofthe quantities themselves. This method is an extension of the celebrated theorem of Noether,a theorem in the framework of the action principle, which associates a conserved quantity toeach 1-parameter group of symmetries of the action (see [13]). This extension is involved at amost elementary level in the very de�nition of the notion of hyperbolicity for an Euler-Lagrangesystem of partial di�erential equations, as discussed in detail in the �rst of the present volumes.In fact we may say that such a system is hyperbolic at a particular background solution if linearperturbations about this solution possess positive energy in the high frequency limit.
The application of Noether's Principle to General Relativity requires the introduction of a back-ground vacuum solution possessing a non-trivial isometry group, as shall be explained in the�rst of the present volumes. Taking Minkowski spacetime as the background, we have the sym-metries of the time translations, space translations, rotations and boosts, which give rise tothe conservation laws of the energy, linear momentum, angular momentum and center of massintegrals, respectively. However, as shall be explained in the �rst of the present volumes, thesequantities have geometric signi�cance only for spacetimes which are asymptotic at in�nity tothe background Minkowski spacetime, so that the symmetries are in fact asymptotic symmetriesof the actual spacetime.
The other main mathematical method analyzed and exploited in the present volumes is thesystematic use of characteristic (null) hypersurfaces. The geometry of null hypersurfaces hasalready been employed by R. Penrose in his incompleteness theorem mentioned above. Whatis involved in that theorem is the study of a neighborhood of a given null geodesic generator ofsuch a hypersurface. On the other hand, in the work on the stability of Minkowski spacetime,the global geometry of a characteristic hypersurface comes into play. In addition, the properties
4
of a foliation of spacetime by such hypersurfaces, also come into play. This method is used inconjunction with the �rst method, for, such characteristic foliations are used to de�ne the actionsof groups in spacetime which may be called quasi-conformal isometries, as they are globally asclose as possible to conformal isometries and tend as rapidly as possible to conformal isometriesat in�nity. The method is introduced in the �rst of the present volumes and treated much morefully in the second volume. It has applications beyond General Relativity to problems in FluidMechanics and, more generally, to the Mechanics and Electrodynamics of Continuous Media.
5
1 Introduction
We have the spacetime manifold (M; g). (See volume I). I.e. M is a 4-dimensional di�erentiablemanifold, endowed with a Lorentzian metric g.
Definition 1 H is called a null hypersurface if at each point x 2 H the induced metric
gx j TxH is degenerate.
That is, there exists a non-zero vector L 2 TxH such that
g(L;X) = 0 8 X 2 TxH:
Proposition 1 A null hypersurface is generated by null geodesic segments.
TxH is a hyperplane in TxM . Thus there is a covector � 2 T �xM such that
TxH = fX 2 TxM : � � X = 0g :
H itself can be represented as the 0-level set of a function u. Then we can take
� = du(x) :
We de�ne a vector�eld L by:
L� = �g�� @�u ;
(components in an arbitrary frame�eld). Then
g�� L� X� = g(L;X) = �du �X
and L is g-orthogonal to H. H is a null hypersurface if and only if L is also tangential to H:
Lx 2 TxH 8 x 2 H:
Taking X to be L itself in the de�nition of null hypersurface, we obtain
g(L;L) = 0 :
If all the level sets of u are null hypersurfaces, then it is
g(L;L) = g��@�u@�u = 0 : (1)
This is called the eikonal equation. And conversely, if equation (1) holds, then all the level setsof u are null hypersurfaces.
The integral curves of L are geodesics, i.e.:
rLL = 0 : (2)
Proof: Working in an arbitrary frame, set
L� = g��L� = �@�u (3)
(rLL)� = g��L�r�L� : (4)
6
Now, it is -as the Hessian of a function is symmetric-
r�L� = �r�(@�u) = �r�(@�u) = r�L� :
Substituting,
L�r�L� = L�r�L� =1
2@�(g(L;L)) = 0 ;
as g(L;L) = 0. And hence we have shown that
rLL = 0 :
�
7
2 Null Hypersurface
2.1 Geometry of a Null Hypersurface
A cross-section of a null hypersurface C being everywhere transversal to the generators (i.e. toL) is a spacelike surface. That is, the induced metric on S is positive de�nite.
x = gx jTxS :
Figure 1.
xS
C
L
Gx
Since at any x 2 S, Lx is orthogonal to TxC, it is in particular orthogonal to TxS.
Suppose instead that we start with a spacelike surface S. Then at each x 2 S the orthogonalcomplement (in TxM) of TxS (denoted by TxS
? = Px) is a 2-dimensional, linear, Lorentzianspace. That is, gx jTxS? is a Minkowskian (Lorentzian) metric.
Figure 2.
Lx
There are 2 null lines in this space. Lx is necessarily the generator of one of these null lines;
f� Lx : � 2 Rg :
8
Suppose that the spacelike surface S is contained in a non-compact Cauchy hypersurface H.Moreover, suppose that S bounds a compact domain K in H. In this case at each x 2 S wecan say that one of the 2 null lines in (TxS)
?, N (the other we denote N) is the interior nullline (the other then being the exterior) if a future-directed vector generating N projects on Htowards the interior domain K.
Figure 3.
N N
(TxS)?
Figure 4.
S
Kx
H
N
In general, making a continuous choice of outer (exterior) null line in (TxS)? at each x 2 S,
we proceed by selecting a future-directed (null) vector Lx generating Nx at each x 2 S. Thevector Lx is de�ned up to a positive real factor. So, the vector�eld L de�ned in this way on Sis determined up to a positive multiplicative function a. That means we have: L 7! aL witha > 0 a function on S.
9
Figure 5. (TxS)?
x
Nx Nx
Lx Lx
Once L has been chosen, then a conjugate null vector�eld L is also de�ned on S by the conditionsthat at each x 2 S, Lx is future-directed and generates the inner (interior) null line Nx; andmoreover, it is
g(L;L) = �2 : (5)
Under the transformation L 7! aL the vector�eld L transforms according to L 7! 1aL. There is
a unique geodesic Gx corresponding to the initial condition Lx at x 2 S. Then we have[x 2 S
Gx = C ; (6)
the null hypersurface.
Figure 6.
Gx
Lx
x
C
SL
The induced metric on a section S is positive-de�nite. Next, we de�ne the second fundamentalform � of S relative to C. This is a bilinear form on TxS at each point x 2 S. It is de�ned asfollows: Given any pair of vectors X;Y 2 TxS, it is
� (X; Y ) = g (rXL; Y ) : (7)
10
This is in fact a symmetric bilinear form. To see this, extend X;Y to vector�elds on S. Thenwe have the following:
� (X; Y ) � � (Y; X) = g (rXL; Y ) � g (rY L; X)
= � g (L; rXY ) + g (L; rYX)
= � g (L; [X; Y ]) = 0 ( since [X; Y ] 2 TS ) :
� is really a property of C. For, we can extend � to a quadratic form on TxC, at each x 2 C,simply by:
� (X; Y ) = g (rXL; Y ) 8 X;Y 2 TxC : (8)
For, any vector X 2 TxC admits a unique decomposition
X = XL L + � X ;
where � denotes the projection to a section S through x. Since �X 2 TxS, it is orthogonal toL. Taking the inner product with L gives:
g (X; L) = � 2 XL :
Hence, it is
� X = X +1
2g (X; L) � L :
By virtue of the facts that L is null as well as geodesic (g(L;L) = 0; rLL = 0) we have
� (X; Y ) = � (�X; �Y ) ; (9)
where the left hand side is de�ned on C and the right hand side is de�ned on S.
Since L along a given section S is determined only up to a given positive multiplicative functiona on S: L 7! aL, � is likewise only determined up to a positive multiplicative function:
� 7! a� :
11
2.2 AÆne Foliations of a Null Hypersurface
Given a section S of a null hypersurface C as well as a choice of a generating vector�eld L alongS, we can de�ne for each � 2 R the section S� which is the set of points
S� = f Gx (�) j x 2 S g
along each geodesic Gx with x 2 S, a �xed parameter value �. (Then S0 = S.)
Figure 7.
L
C
T
S�
0
The parameter of the vector�eld L de�nes a function s on C, the aÆne function, by:
L s = 1 ; s jS = 0 :
In the special case where C is the light cone of a point, we can do something more canonical,because instead of starting with an arbitrary section, we can choose a unique future-directedtimelike vector T at the vertex O. We then consider the unit sphere in the spacelike hyperplanewhich is the orthogonal complement of T in T0M . For each unit spacelike vector N in thissphere, we have a future-directed null vector L = T +N at O. By going, with �xed parameter� > 0, along the corresponding null geodesic for each such N on the unit sphere, we obtain asection S� of the light cone C.
1st variational formula (for sections of C):
Given a section S, we have a foliation of C by sections Ss with
Ss = f Gx (s) : x 2 S g :
Gx(s) is the point at parameter value s along the null geodesic (integral curve of L) starting atx 2 S. We may also de�ne the ow �t generated by L on C.
�t (Gx(s)) = Gx (s+ t)
moves a point along a generator parameter interval t. Then it is
�s (S) = Ss :
12
Given a vector�eld X on S (that is, X is tangential to S), we extend X to C according to:
[L; X] = 0 :
Then X will be tangential to each of the sections Ss.
Figure 8.
Ss
C
x
Xx
XGx(s)
Gx(s)
S
Then we have:
XGx(s) = Dx �s � Xx :
With this view we do not need to start with a vector�eld X de�ned on all of S; we only needto start with a vector Xx 2 TxS. Then the above de�nition yields a vector�eld along thegenerator Gx, tangential to the foliation fSsg. Such a vector�eld along a generator is called aJacobi �eld along the generator. (It can be thought of to be the displacement �eld taking usfrom the given generator to a nearby generator.)
Consider now two such Jacobi �elds X and Y along a given generator. Let us denote by the induced metric on Ss (this of course depends on S).
= g jTSs :
So, we have
(X; Y ) = g (X; Y ) :
We compute
d
dsf (X; Y ) g = L (g (X; Y ))
= g (rLX; Y ) + g (X; rLY )
= g (rXL; Y ) + g (X; rY L) ; since rLX � rXL = [L; X] = 0 ;
= 2 � (X; Y ) :
This is the 1st variational formula.
13
2.2.1 The Torsion of a Spacelike Surface S
Picking an outgoing null normal L (which will generate the null hypersurface C), we have aconjugate (incoming) null normal L. So, we have:
g (L; L) = � 2 : (10)
Definition 2 We then de�ne a 1-form � on S, the torsion of S, by:
� (X) =1
2g (rXL; L) 8 X 2 TS : (11)
Since we have the transformations L 7! aL and L 7! 1aL, where a is a positive function on S,
the 1-form � is subject to these transformations and transforms according to:
� 7! � + d= log a ;
where d= denotes the di�erential of a function on S. Now, we shall show below that
rLL = � 2 Z ; (12)
Z is the Ss-tangent vector�eld corresponding to the 1-form �.
A vector Y belongs to TxC if and only if
g (Y; L) = 0 : (13)
We can then expressY = Y LL + �Y ; (14)
where �Y 2 TxSs. (Recall that Ss: section through x.) And the component Y L = �12g(Y;L).
Apply the above to the vector rLL. We have:
g (rLL; L) = L ( g (L; L)) � g (L; rLL) = 0 ;
as g(rLL;L) = �2 and rLL = 0. Thus we conclude that rLL is tangential to C. Also, it is
g (rLL; L) = L (1
2g (L; L)) = 0 ;
as g(L;L) = 0. So, rLL has no component along L; it is tangential to the sections Ss. Takeany X 2 TxSs; then extending X to a Jacobi �eld along the generator Gx through x.
(rLL; X) = g (rLL; X) = L ( g (L; X)| {z }=0
) � g (L;rLX)
= � g (L; rXL) (since [L; X] = 0)
= � 2 � (X) :
We conclude thatrLL = � 2 Z ; (15)
where Z is the Ss-tangent vector�eld corresponding to the 1-form �. Consider any vectorX 2 TxSs and g(rLL;X). Note that L being transversal to C, the di�erentiation alongL involves considering a neighbourhood of C in spacetime. But we consider C as the 0-level
14
set of a function u whose neighbouring level sets are also null hypersurfaces. Then L is thevector�eld corresponding (through the spacetime metric g) to the 1-form �du. That is, in anarbitrary frame,
L� = �(g�1)�� @� u :
Thus,
g (rLL; X) = � r2u � (L; X)
= � r2u � (X; L) ; the Hessian of a function is symmetric ;
= g (rXL; L)
= 2 � (X) :
We conclude that� rLL = 2 Z : (16)
Combining (16) with the previous result (15), we obtain:
� [L; L] = 4 Z : (17)
Recall now the Frobenius theorem: (see [31]).Here, we have a distribution of timelike (Lorentzian) 2-planes f Px g de�ned in a neighbourhoodof C.
Px = (TxSs)? ; (18)
where Ss is a section of C through x. Px is the linear span of f Lx; Lx g. By the Frobeniustheorem the obstruction to local integrability of this distribution is
� [L; L] ;
namely the torsion �.
Remark: The above gives us another interpretation of the torsion (besides the de�nition). Stillanother interpretation comes from the following.
Let (EA : A = 1; 2) be an orthonormal basis for TxS. We propagate this basis along thegenerator Gx according to:
rLEA = � �A L ; (19)
where �A = �(EA). This is a linear system of ordinary di�erential equations. Then we have:
L (g (EA; L)) = g (rLEA; L) + g (EA; rLL| {z }=0
) = 0 ;
where rLEA = ��AL and g(L;L) = 0, and the initial conditions imply
g (EA; L) = 0 at x :
Thus the frame �eld (EA) along Gx is tangential to C. Next,
L (g (EA; L)) = g (rLEA; L) + g (EA; rLL)
= � �A g (L; L) � 2 g (EA; Z)
= 2 �A � 2 �A = 0 ;
15
and the initial conditions imply that
g (EA; L) = 0 at x :
So, g(EA; L) remains 0 along Gx. Thus the frame (EA) is tangential to the sections Ss alongGx. Finally,
L (g (EA; EB)) = g (rLEA| {z }=��AL
; EB) + g (EA; rLEB| {z }=��BL
) = 0 :
So, we haveg (EA; EB) = ÆAB (20)
along Gx (orthonormal frame �eld). I.e. (EA : A = 1; 2) propagated along Gx form an or-thonormal basis for TGx(s)Ss for each x.
2.2.2 Jacobi Fields and the 2nd Variation
Since a Jacobi �eld X along Gx is tangent at each point to the section Ss through that point,we can expand
X =XA=1;2
XA EA ; XA = XA (s) : (21)
We have (note that L = @@s):
rLX =XA=1;2
(dXA
dsEA + XA rLEA) ;
=XA=1;2
(dXA
dsEA � XA �A L) : (22)
On the other hand, the following holds:
rLX � rXL = [X; L] = 0 ; i.e. rLX = rXL : (23)
In particular, it is:
g (rLX; EB) = g (rXL; EB)
= � (X; EB) (24)
=XA=1;2
�AB XA ; (25)
where �AB = �(EA; EB).
On the other hand, taking the inner product of the formula (22) for rLX with EB , we ob-tain:
g (rLX; EB) =dXB
ds: (26)
We conclude that:dXA
ds=
XB=1;2
�AB XB : (27)
16
Next, we look at the 2nd variation.
We use that[L; X] = 0 :
Then we can write:
r2LX := rL rL X = rL rX L = rX rL L + R (L; X) L (28)
Hence,r2L X = R (L; X) L : (29)
Taking an arbitrary vector Z 2 TyM , with y 2 C, we consider the vector
Y = R (L; Z) L : (30)
We have a linear transformation of TyM given by
Z 7! Y :
In general, it isg (W; R (X; Y ) Z) = R (W; Z; X; Y ) : (31)
Henceg (Y; L) = R (L; L; L; Z) = 0 ;
since the curvature tensor is skew-symmetric in the �rst two slots, a consequence of the fact thatthe connection is metric. Thus Y 2 TyC. Therefore the linear transformation Z 7! R(L;Z)Lrestricts to a linear transformation of TyC. It follows that Y can be expanded in the basis(E1; E2; L). This means that one can write the following:
Y = Y L L + Y A EA
Y L = � 1
2g (Y; L)
Y A = g (Y; EA) :
Apply the above to the case Z = X. We then obtain
(R (L; X) L)L = � 1
2g (L; R (L; X) L)
= � 1
2R (L; L; L; X) :
Let us introduce the 1-form � in each section Ss by
� (X) =1
2R (X; L; L; L) 8 X 2 TSs : (32)
Then we can write:(R (L; X) L)L = � (X) (33)
(taking into account the symmetry of the curvature tensor under interchange of pairsR(X;Y;Z;W ) =R(Z;W;X; Y )). Next, it is
(R (L; X) L)A = g (EA; R (L; X) L)
= R (EA; L; L; X)
= � R (EA; L; X; L) :
17
We introduce �, a quadratic form in TySs for each y 2 Ss (and each s) by:
� (X; Y ) = R (X; L; Y; L) 8 X; Y 2 TySs : (34)
( � is symmetric since R is symmetric under pair interchange.) Thus we can write
(R (L; X) L)A = � � (EA; X) = ��AB XB ; (35)
where �AB = �(EA; EB). From above we know
(R (L; X) L)L = �A XA ;
where �A = �(EA).
Going back to the formula for rLX:
rLX =XA
f dXA
dsEA � XA �A L g ;
applying rL to this gives
r2LX =
XA
f d2XA
ds2EA + (�2 dXA
ds�A � XA d�A
ds) L g :
On the other hand, we have obtained
R (L; X) L = �XA;B
�AB XB EA +XA
�A XA L :
By using (29), we get:d2XA
ds2= �
XB
�AB XB ; (36)
in matrix notation:d2X
ds2+ � X = 0 ; (37)
and those of L:
XA d�Ads
+ 2dXA
ds�A = ��A XA : (38)
(36) is a linear homogeneous 2nd order system of ordinary di�erential equations for (XA :A = 1; 2). The initial conditions are XA(0), the coeÆcients of the expansion of the initialdisplacement vector Xx 2 TxS in the basis (EA : A = 1; 2) and
dXA
ds(0) =
XB
�AB jx XB (0) ;
where �AB jx are the components of the 2nd fundamental form of S at x. Since this is alinear system and, moreover, the initial conditions are linear homogeneous in the components(XA(0) : A = 1; 2) of the initial displacement, the solution exists and has bounded derivativesup to 2nd order as long as the curvature components (�AB) remain bounded. Furthermore wecan write
XA (s) =XB
MAB (s) XB (0) ; (39)
18
where MAB is called the deformation matrix. In matrix form:
X(s) = M(s) X(0) :
We have, trivially,MA
B (0) = ÆAB ;
or simplyM (0) = 12
in matrix notation. The deformation matrix has bounded derivatives up to 2nd order providedthat the curvature components remain bounded. We have
dXA
ds=
XB
dMAB
dsXB (0)
=XB
�AB|{z}=�AB jGx(s)
(s) XB (s)
=XB;C
�AC (s) MCB XB (0) :
Since this holds for an arbitrary displacement, we conclude that
dMAB
ds=XC
�AC MCB :
In matrix notation:dM
ds= � M :
Solving for �, we obtain:
� =dM
dsM�1 : (40)
That is, �AB =P
C
dMAC
ds(M�1)CB .
d2M
ds2= �� M
(= d2XA
ds2= �PB �ABX
B). Then we have
d�
ds=
d2M
ds2M�2 � dM
dsM�1 dM
dsM�1
(since dM�1
ds= �M�1 dM
dsM�1). Then substituting:
d�
ds= ��2 � � ; (41)
that is: d�ABds
= �P�AC�CB � �AB.
Now, recall (38) from above. We conclude that the component along L of the Jacobi equa-tion is the following equation for the 1-form �.
d�Ads
+ 2XB=1;2
�AB �B = � �A : (42)
19
Let us recapitulate some formulas from above:
rLEA = � �A L
�A = � (EA)
rLL = � 2 Z
r2LX = R (L; X) L
X =XA=1;2
XA EA
rLX =XA=1;2
(dXA
dsEA � XA �AL)
g (rLX; EA) = g (rXL; EA) =XB=1;2
�AB XB
�AB = � (EA; EB)
dXA
ds=
XB=1;2
�AB XB
r2LX =
XA=1;2
f d2XA
ds2EA � (2
dXA
ds�A + XA d�A
ds) L g :
Taking the trace of (41) gives:
dtr�
ds= � tr (�2) � tr � ; (43)
and splitting up � in the following way:
� = � +1
2tr � 12 ; (44)
where � denotes the trace-free part of �, leads to the result:
tr (�2)| {z }=j�j2
= tr (�)2| {z }j�j2
+1
2(tr �)2 ; (45)
recall that j � j2 = PA;B(�AB)
2 and j � j2 = PA;B(�AB)
2. Therefore the trace equation reads:
dtr�
ds= � 1
2(tr�)2 � j � j2 � tr � : (46)
We have the null frame (E1; E2; L; L) with
g (L; L) = g (L; L) = 0g (L; L) = � 2g (L; EA) = g (L; EA) = 0g (EA; EB) = ÆAB :
It follows that g�1, a quadratic form in T �xM at each x 2 C, is expressed by
g�1 = � 1
2(L L + L L) +
XA=1;2
EA EA :
20
Now, recall the Ricci curvature tensor with components in an arbitrary frame:
(Ric)�� = (g�1)�� R���� :
We calculate the following:
(Ric)�� L� L� = Ric (L; L) = � 1
2R (L; L; L; L)| {z }
=0
� 1
2R (L; L; L L)| {z }
=0
+XA=1;2
R (L; EA; L; EA)
= tr � :
That istr � = Ric (L; L) = 2 T (L; L) � 0 (47)
by virtue of the Einstein equations (since g(L;L) = 0) and by the positivity condition on T .And this (47) implies the ordinary di�erential inequality:
dtr�
ds� � 1
2(tr �)2 : (48)
Suppose that tr�(0) < 0. Then it remains tr� < 0 for all s � 0, since tr� is non-increasingaccording to the above o.d.i. (237). So, it is
� 1
tr �> 0 (49)
andd
ds(� 1
tr�) =
1
(tr�)2dtr�
ds� � 1
2: (50)
Integrating on [0; s1] with s1 > 0, we obtain
� 1
tr�(s1)� � 1
tr�(0)� s1
2:
The right hand side becomes 0 at s1 given by s1 = � 2tr�(0) > 0 . We conclude that tr�! �1
as s! s� for some s� 2 (0; s1].
Recall the expression for � in terms of M :
� =dM
dsM�1 :
So, it is
tr � = tr (dM
dsM�1) =
d
ds(log det M) :
Since M is C1, it follows that detM(s�) = 0. Now,
X (s�) = M (s�) X (0)
det M (s�) = 0
implies that there exists a non-zero vector X(0) 2 TxS such that
M (s�) X (0) = 0 :
We conlude that there exists a Jacobi �eld X on the null geodesic segment Gx([0; s�]) which iseverywhere non-zero except at s� where it vanishes.
21
2.2.3 Focal Points
First, we shall explain the notion of focal points in Euclidean (and Riemannian) Geometry.Then we will show how these focal points enter our setting in Lorentzian Geometry.
Focal Points in Euclidean and Riemannian Geometry:
Hypersurface in Riemannian Geometry (any dimension):
Figure 9. x
Gx G0
y
x�
H
Consider the normal geodesic Gx with foot x and the focal point x� to H along Gx. Heuristicallythis is a point where nearby normal geodesics G0 resulting from some in�nitesimal displacementfrom x along H intersect Gx. The segment of G
0 up to x� will be of the same length as the cor-responding segment of Gx. Thus if we consider a point y along Gx beyond x�, this point can bejoined to H by a broken geodesic consisting of the said segment of G0 together with the segmentof Gx from x� to y. This broken geodesic will have the same length as the segment of Gx from x toy. However such a broken geodesic cannot be a minimal curve between its end points. Roundingo� the corner at x� yields a shorter curve. This argument shows that a normal geodesic such asGx cannot be a minimal curve joining a point on it to H if that point lies beyond the focal point.
Apollonius' Study of Focal Points:
Focal points were �rst studied by Apollonius in antiquity (ca. 262 - 190 BC). His works aboutfocal points are contained in 'Treatise on Conic Sections' (or simply 'Conics', in eight books;this topic being in the �fth book). This is concerned with the normal lines that can be drawnfrom a given point in the plane to a given conic section (parabola, ellipse or hyperbola). Let usdraw a curve H 2 R2 and a point x 2 H as follows:
22
Figure 10. xH
x�
y
p
n1n3
n2
The curve of all focal points is called locus of focal points (or focal curve). A normal geodesic istangent to the focal curve at the focal point. Consider the normal geodesic through x 2 H. Itis tangent to the focal curve in the point x�. Now, go beyond x� on the same normal geodesicand take a point y on this geodesic. There are 3 normals to H through this point y, namely:n1, which is of minimal distance to H; n2, which is the normal trough x, being a maximum ofthe distance to H; n3, being a relative minimum of the distance to H.
Focal Points in Lorentzian Geometry and their Relevance to Delineating Bound-
aries of Futures
Along the normal null geodesic G0 to S with foot q 2 S, we have a focal point � correspondingto the value s� of the aÆne parameter. We consider a point p along G0 beyond �.
Proposition 2 p can be joined to S by a timelike curve. Thus p 2 I+(S), the chronological
future of S.
Proof: The proof works in 2 steps.
In Step 1 we take p suitably close to �.
In Step 2 we show that if there exists a non-spacelike curve G0 joining q0 2 S to p0 and anull geodesic from p0 to p (here the segment of G0 from p0 to p), such that the tangent vectorat p0 to the null geodesic segment from p0 to p is not collinear to the tangent vector to G0 at p0,then there exists a timelike curve from q0 to p.
Step 2 depends only on the 1st variation with �xed end points while step 1 depends on the2nd variation (with one point �xed and the other free to move on the surface).
23
Step 1:
Figure 11.
p
q
S
st
Ks
Gt
G0
�
We de�ne a smooth family of curves f Ks : s 2 [0; a] g, whereKs : t 7! Ks (t)
is a curve through the point G0(s) at t = 0. Here, it is p = G0(a). There are two conditions onthis family:
1. K0 is a curve lying in S (through q = G0(0))
2. The curve Ka is the point p:
Ka (t) = p : 8 t :
We can then de�ne the family of curves f Gt : t 2 [��; �] g byGt (s) = Ks (t) :
Then G0 is the normal null geodesic with which we began. We have thus a mapping � :[0; a] � [��; �]!M by:
� (s; t) = Ks (t) = Gt (s) : (51)
On the image by � of [0; a]� [��; �] (or Ss2[0;a] Ks[��; �]) we have the two tangent vector�elds:
V (Ks (t)) = _Ks (t) (52)
T (Gt (s)) = _Gt (s) : (53)
24
Note that V = ddtand T = d
ds. We have:
V jK0 is a vector�eld on S tangent to K0 ; (54)
V vanishes at p, and
T jG0 = L = (L jG0) : (55)
Our aim is the following: To show that by a suitable choice of the family f Ks : s 2 [0; a] g wecan make T everywhere timelike in the image by � of [0; a]�[��0; �0] for 0 < �0 � �, suitably small.
By construction:
[T; V ] = 0 : (56)
Now, let us consider V (g(T; T )) jG0 and V 2(g(T; T )) jG0 . We have:
V (g (T; T )) = 2 g (rV T; T ) = 2 g (rTV; T ) :
Along G0 it is T = L, so the last term here is
2 g (rLV; L) = L (g(V; L)) � 2 g (V; rLL) :
Remark: Here, the second term on the right hand side vanishes since rLL = 0. Note thatg(V;L) vanishes at s = 0 (the base point q) as well as at s = a (the point p). ThereforeL(g(V;L)) = d
ds(g(V;L)) cannot be everywhere � 0 unless g(V;L) = 0 identically on G0; in
which case it is
V (g (T; T )) jG0 = 0 : (57)
We must thus proceed to the 2nd variation.
1
2V 2 (g (T; T )) = V (g (rTV; T ))
= V f T (g (V; T )) � g (V; rTT ) g= T V (g (V; T )) � g (rV V; rTT ) � g (V; rVrTT ) :
The second term on the right hand side of the last equation vanishes at t = 0 since T jG0= L.For the third term it is
rVrTT = rT rV T| {z }=rTV
+ R (V; T ) T
= r2TV � R (T; V ) T :
Again, as along G0 it is T = L, we have:
rVrTT jG0 = J V ; (58)
where J is the Jacobi operator:
J = r2L � R (L; � ) L : (59)
So, U is a Jacobi �eld along G0 if and only if
J U = 0 : (60)
25
The third term above isg (V; rVrTT ) jG0 = g (V; JV ) :
For the �rst term write
V (g (V; T )) = g (rV V; T ) + g (V; rV T| {z }=rT V
)
= g (rV V; T ) +1
2T (g (V; V )) :
Thus along G0, the �rst term is
T V (g (V; T )) jG0 = L F ;
where F is the following function on G0 (a function of s):
F = g (rV V; L) +1
2L (g (V; V )) : (61)
Therefore we conclude:
1
2V 2 (g (T; T )) = L F � g (V; JV ) : (62)
We want to de�ne V and rV V along G0 such that this second derivative is < 0 on G0[0; a].Then Gt will be timelike for 0 <j t j� �0 for suÆciently small �0.
V itself along G0 is what the term g(V; JV ) depends on. We shall show that we can choose Valong G0 such that:
g (V; JV )
�> 0 : on [0; a)= 0 : at a
F depends also on rV V along G0. This will be chosen afterwards. We know that s� correspondsto a focal point, and assuming that p lies just beyond the �rst focal point along G0, there is nofocal point in G0[0; s�) or G0(s�; a]. Thus there is a Jacobi �eld U along G0(0; a], vanishing onlyat s�. Since
rLU 6= 0 at s�
(otherwise U would be identically zero as it satis�es the second order ordinary di�erential equa-tion JU = 0), we can do the Taylor expansion for U near s�
U =XA=1;2
UA EA
with UA(s�) = 0 and dUA
ds(s�) 6= 0 and
UA (s) = CA (s � s�) + � � �UA (s) � CA (s � s�)
where dUA
ds= CA. Notation: Given two functions of s, namely f and g, de�ned in an interval I
containing s�, we say that f � g if fg! 1 as s! s� (or uniformly in a subinterval J containing
26
s� as j J j! 0). It follows that we can de�ne a unit vector�eld U collinear to U and smoothalong G0. So,
U = � U ; (63)
where � is a smooth function on [0; a] such that
� (s)
8<:> 0 : on [0; s�)= 0 : at s�< 0 : on (s�; a]
Figure 12.
q G0
s�0 a
Jacobi �eld
Note that g(U;U) = �2. We have:
0 = J U = J (�U) =d2�
ds2U + 2
d�
dsrLU + � J U (64)
since g(U ;rLU) =12L(g(U ; U)) = 0, we obtain
0 = g (U ; JU) =d2�
ds2+ v � ; (65)
where v = g(U ; JU ). So, � satis�es the equation
d2�
ds2+ v � = 0 : (66)
Now, we want to modify � in a neighbourhood of the focal point obtaining a function whichis positive on [0; a) and vanishing at a. We then set
V = U (67)
along G0. We want to achieve g(V; JV ) > 0 on G0[0; a). Since V vanishes at p we necessarilyhave g(V; JV ) = 0 at p.
27
Figure 13.
a
s�
�
We have:
g (V; JV ) = (d2
ds2+ v ) : (68)
We start with a smooth function � on [0; a] such that � > 0 on (0; a] and �(0) = 0. We thende�ne by
(s) = �(s) � �(a)�(s)
�(a): (69)
Then (0) = �(s) and
d2
ds2=
d2�
ds2|{z}�v�
� �(a)
�(a)
d2�
ds2
= � v ( + �(a)�
�(a)) � �(a)
�(a)
d2�
ds2
ord2
ds2+ v = � �(a)
�(a)(d2�
ds2+ v � ) :
Since �(a) < 0, the factor (��(a)�(a) ) is > 0. We choose � such that
d2�
ds2+ v � > 0 on [0; a] :
Letting
vm = min[0;a]
v ;
it suÆces to choose � such that � > 0 on (0; a], �(0) = 0, and
d2�
ds2+ vm � > 0 :
For example, we can take � = eKs � 1 and K2 + vm > 0. With such a choice the factor(d
2ds2
+ v) > 0 on [0; a]. We also want to have the factor positive (except at a, where itvanishes).
(s) = �(s) � �(a)�(s)
�(a):
Since �(a) < 0, is certainly > 0 on [0; s�]. We thus only need to consider the interval (s�; a).Now,
(s) > 0
28
is equivalent to�(s)
�(s)>
�(a)
�(a)8 s 2 (s�; a) :
This will be so, if �(s)�(s) is a decreasing function, or
�d�
ds� � � < 0 :
Now, it isd�
ds(s�) < 0 : (70)
By choosing a suitably close to s�, we can make
sup(s�;a)
(��)
as small as we please. Then we achieve what we want.
Let us now consider the term
L F =dF
ds: (71)
We have
F = F1 + F2 with (72)
F1 = g (rV V; L) ; (73)
F2 =1
2L (g (V; V )) : (74)
As g(V; V ) = 2 along G0, we have F2 = dds. We can write
F1 = g (rUV; L) :
We have the freedom of choosing rUV along G0, except at the base point q, where we are
restricted by the condition that K0 is a curve lying on S. Along G0 we have L, L and TG0(s)Ss.We can expand:
rUV = � r
UV � 1
2g (r
UV; L) L � 1
2g (r
UV; L) L ; (75)
where � is the projection to TG0(s)Ss. We see that
g (rUV; L)
which enters F , in fact,
F = ( g (rUV; L) +
d
ds) (76)
generates a displacement of the null geodesic G0 o� of the null hypersurface C, since L istransversal to C. The factor in (76) is up to us to choose, except at the base point q. At q, sinceV is at s = 0 the tangent vector�eld to the curve K0, which lies on S, we have; along K0:
g (rV V; L) = � g (V; rV L) = � � (V; V ) (77)
29
since g(V;L) = 0 along K0. So, it is
F1 (0) = � � (V; V ) = � 2 (0) � (U ; U) : (78)
To �nd the value of F2 at s = 0, we write V = �U .
F2 = g (V; rLV )
= (
�)2 g (U; rLU) +
�
d
ds(
�) g (U; U)
= 2 � (U ; U) + � d
ds(
�) ; (79)
since rLU = rUL and g(U;U) = �2, (also U = �U).
At q we obtain
F (0) = [ � d
ds(
�) ]s=0 (80)
or
F (0) = � (0) (d
ds� d�
ds) (0) (81)
= � � (0) � (a)
� (a)
d�
ds(0) > 0 ; (82)
recalling that (s) = �(s)� �(a)�(s)�(a) . Now, we have
F (s) = (s) F (s) ; (83)
F = g (rUV; L) +
d
ds: (84)
F is freely speci�able except for the condition
F (0) =F (0)
(0)=
F (0)
�(0): given above : (85)
We can thus choose F such that dFds
< 0 on [0; a]. We then obtain
1
2V 2 (g (T; T )) < 0 on G0 [0; a] : (86)
So, for suitably small j t j, Gt is a strictly timelike curve from S to p.
To remove the assumption that p lies not far beyond the focal point, we have the second step.We shall consider the case that we have a broken null geodesic from q to p.
We can assume that we have a null geodesic segment G from q to r and another such seg-ment G0 from r to p, such that the tangent vectors L and L0 at r are not collinear. ConsiderTrM . L and L0 at r both belong to N+
r , the future component of the null cone at r. ThenW := (L0�L) is a spacelike vector. There is a collinear unit vector W . Extend W to G and G0
by parallel propagation. Then set
V (s) =
�(s� sq) (sp � sr) W : on G; s 2 [sq; sr]
(sr � sq) (sp � s) W : on G0; s 2 [sr; sp]
30
Then V vanishes at q and p. (Variation with �xed end points).
Figure 14.
r
G0
G
p
q
W
Then de�neGt (s) = exp (t V (s)) : (87)
One can show that Gt(s) for small j t j is a timelike curve from q to p. For, take ddtg( _Gt; _Gt) jt=0
and it is straightforward to show that this is < 0.
31
2.3 Theorem of Penrose
Now we will state the theorem of Penrose and prove it using focal points.
Before this, let us give an important de�nition, we will use in the theorem. So, we de�ne aclosed trapped surface as follows:
Definition 3 A closed trapped surface S in a non-compact Cauchy hypersurface H is a two-
dimensional surface in H, bounding a compact domain, such that
tr � < 0 on S :
Note that a virtual displacement of S towards the future along the outgoing null geodesiccongruence results in a pointwise decrease of the area element. (
@d� @s
= tr� d� ).
Theorem 1 Theorem of Penrose (1965):
Let (M; g) be a spacetime which is a development of non-compact initial data (i.e. the initial
Cauchy hypersurface H is non-compact and complete). The positivity condition on the energy
tensor of matter is assumed to hold. Assume that the initial hypersurface contains a closed
trapped surface S, the boundary of a compact domain K in H.
Then M is not null geodesically complete (towards the future). In particular, the maximal
development of such initial data is incomplete.
Remark: If K is compact, then H=K is non-compact. In this case, we can de�ne a future-directed outgoing null normal �eld L to S unique up to a positive factor: L! aL, where a is apositive function on S.
Figure 15.
S
H
K
LL
Proof: Assume that (M; g) is future null geodesically complete. Consider the causal future ofthe compact set K � H, namely J+(K). Consider in fact the boundary of this, @J+(K).
32
Figure 16. K
J+(K)
S
Remove K:
@ J+(K) = K :
This is generated by the outgoing normal null geodesics to S. Each such null geodesic generatoris included at most up to the focal point. In particular, complete generators are included onlyif they contain no focal points. For, by the theorem just proved, any point along a generatorlying beyond the �rst focal point can be connected to S by a strictly timelike curve, there-fore lies in the interior of J+(K), not on the boundary. The assumption that S is trapped leadsto the conclusion that there exists a focal point along every outgoing normal null geodesic from S.
Moreover, S being compact,
supS
tr � = � k < 0 : (88)
It follows that a focal point is reached along any outgoing normal null geodesic within aÆneparameter 2
k. This implies that (@J+(K)=K) [ S =: W is compact; for it is the image of a
compact subset of S � [0;1) under the continuous mapping
(x; s) 2 S � [0; 1) 7! Gx (s) :
Thus @J+(K) is also compact. This leads to a contradiction, as is demonstrated below.
@J+(K) is the boundary of a future set.
Definition 4 A subset F of a spacetime manifold is called a future set if F includes the chrono-
logical future at each of its points.
33
Definition 5 The chronological future of p, i.e. I+(p), is the set of points q 2 M , which canbe reached from p by a strictly timelike future-directed curve.
F is a future set if and only if
p 2 F ) I+ (p) � F :
Definition 6 A set A is called achronal if there is no pair fp; qg � A such that q 2 I+(p).
I.e. no two points of A can be joined by a timelike curve.
Proposition 3 The boundary @F of a future set F is an achronal Lipschitz submanifold of M .
Proof of Proposition 3 : Take p 2 @F , then show that I+(p) � F and I�(p) �M=F . It followsthat @F is achronal. We then show that an achronal subset is locally represented as the graphof a Lipschitz function.
In our case, @J+(K) is a compact achronal Lipschitz submanifold. Since M admits a Cauchyhypersurface, we can �nd a future-directed timelike vector�eld T on M . Each integral curve ofT intersects H at a single point and intersects J+(K) at most once.
We can set the parameter t of T to be zero on H. Then for each q 2 @J+(K) there existsa tq such that q is the point at parameter value tq along an integral curve from H. We canrescale T along such an integral curve by a constant factor such that the new parameter has thevalue 1 at q (and zero at the corresponding point on H). In particular, the rescaled vector�eldwill vanish on K. This vector�eld de�nes a homeomorphism of @J+(K) onto a subdomain B ofH. It follows that B is compact. But B cannot coincide with H, as H is non-compact.
So, B has a boundary; this contradicts the fact that it is homeomorphic to a manifold withoutboundary, namely to the Lipschitz manifold @J+(K). �
34
3 Global Geometry of a Null Hypersurface
We shall study the global geometry of a null hypersurface in the context of the vacuum Einsteinequations.
Recall
dtr�
ds= � j � j2 (89)
= � 1
2(tr�)2 � j � j2 : (90)
In a basis of Jacobi �elds we write
�ab = � (Xa; Xb) ; a; b = 1; 2 : (91)
We have
Xa =XA
XAa EA (92)
where (Xa : a = 1; 2) is the Jacobi �eld basis and (EA : A = 1; 2) is the orthonormal basis.Recall that
�ab = R (Xa; L; Xb; L) (93)
= � (Xa; Xb) : (94)
Figure 17.
Cu
GpGq
q p
Ss
r
S0
35
Now, consider S0 and p 2 S0 in the picture above. We have local coordinates (ya : a = 1; 2)in a neighbourhood U � S0 of a point p 2 S0 and we write (Xa = @
@ya: a = 1; 2). Then we
can extend them to local coordinates in a neighbourhood of the generator Gp in Cu, namely:Sq2U Gq(I) where I is an interval of the real line containing zero. The local coordinates of a
point r are (s; y1; y2), if r 2 Ss and (y1; y2) are the coordinates of the point q 2 S0 such thatr = Gq(s); i.e. r is the point at parameter value s along Gq, the generator from q. In suchcoordinates we have:
L =@
@s(95)
Xa =@
@ya; a = 1; 2 : (96)
More generally, we may consider
WI =[s2I
Ss � Cu ; (97)
a neighbourhood of s0 in Cu; and we have a di�eomorphism from WI onto I �S0 by r 7! (s; q).
In a basis of Jacobi �elds we have:
@�ab@s
� � ca �cb = � �ab (98)
The components of the induced metric on Ss relative to the Jacobi �eld basis read as
ab = g (Xa; Xb) (99)
So, we then write@ ab@s
= 2 �ab : (100)
Recall that� ca = ( �1)cb �ab :
Thus we obtain
@
@str � =
@
@sf ( �1)ab �ab g
= � ( �1)ac ( �1)bd@ cd@s
�ab + ( �1)ab@�ab@s
= � 2 ( �1)ac ( �1)bd �ab �cd + �ab �ab
= � 2 j � j2 + j � j2= � j � j2 : (101)
Now, compute the trace-free part of �:
�ab = �ab � 1
2 ab tr � (102)
@�ab@s
=@�ab@s
� �ab tr � � 1
2 ab
@tr�
@s
= � ca �cb � �ab � �ab tr � +
1
2 ab j � j2 :
36
Next, substitute the decomposition:
�ab = �ab +1
2 ab tr �
� ca �cb = � c
a �cb + �ab tr � +1
4 ab (tr�)
2 ;
obtaining
@�ab@s
= � ca �cb +
1
2((tr�)2 + j � j2) ab � �ab � 1
2(tr�)2 ab
= � ca �cb +
1
2j � j2 ab � �ab : (103)
Now, de�ne@�ab@s
to be the trace-free part of@�ab@s
:
So, one has@�ab@s
=@�ab@s
� 1
2 ab (
�1)cd@�cd@s
: (104)
We obtain:@�ab@s
= (� ca �cb � 1
2 ab j � j2) � �ab : (105)
The term in the brackets on the right hand side of (105) vanishes identically.
More generally, for any two symmetric trace-free 2-dimensional matrices A and B, we have:
AB + BA � tr (AB) � Id: = 0 : (106)
In our case, we have A = B.
So, the equation (105) simpli�es to
@�ab@s
= � �ab ; (107)
which is an ordinary di�erential equation along the generators.
37
3.1 Codazzi Equations for a Section S of a Null Hypersurface
Consider minus the Hessian of the function u; (where level sets are null hypersurfaces). Inarbitrary local coordinates one writes
h�� = r� L� = � r� @� u : (108)
h is symmetric. For any triplet of vector�elds X;Y;Z we have:
(rX h) � (Y;Z) = X (h (Y;Z)) � h (rXY; Z) � h (Y; rXZ) : (109)
Consider now the case that X;Y;Z are Ss-tangential. Then it is
h (Y; Z) = � (Y; Z) : (110)
By the de�nition of covariant di�erentiation intrinsic to Ss, we have:
(r= X �) (Y; Z) = X (� (Y;Z)) � � (r= XY; Z) � � (Y; r= XZ) : (111)
Since r= XY = �rXY where � is the projection to Ss, and comparing the above two formulas,we obtain the following:
(rX h) (Y;Z) = (r= X �) (Y; Z) � h ((rXY )?; Z) � h (Y; (rXZ)
?) : (112)
Now, for any vector W it is
W � �W =: W? = � 1
2g (W; L) L � 1
2g (W; L) L : (113)
In particular, since:
g (rXY; L) = � g (Y; rXL) = � � (X; Y )
g (rXY; L) = � g (Y; rXL) = � � (X; Y ) ;
where � is the 2nd fundamental form of Ss with respect to the conjugate null normal L; we have:
(rXY )? =
1
2� (X; Y ) L +
1
2� (X; Y ) L : (114)
On the other hand, from the de�nition of h, for any pair of vectors U; V :
h (U; V ) = g (rUL; V ) = h (V; U) : (115)
In particular, (recall that Z is Ss-tangential),
h (Z; L) = g (rZL; L) = 0h (Z; L) = g (rZL; L) = 2 � (Z) ;
(116)
where � is the torsion one-form of Ss. Substituting the above in the formula (112) for (rXh)(Y;Z)yields:
(rX h) (Y; Z) = (r= X �) (Y; Z) � � (X; Y ) � (Z) � � (X; Z) � (Y ) : (117)
38
Subtract the same with X and Y interchanged:
(rX h) (Y; Z) � (rY h) (X; Z) = (r= X �) (Y; Z) � (r= Y �) (X; Z)
� � (X; Z) � (Y ) + � (Y; Z) � (X) : (118)
On the other hand,
h (Y; Z) = g (rY L; Z) ;
hence
(rX h) (Y; Z) = g (rXrY L; Z) � g (rrXY L;Z) : (119)
Thus, interchanging X and Y and subtracting, we obtain:
(rX h) (Y; Z) � (rY h) (X; Z) = g (rXrY L � rYrXL � r[X;Y ]L; Z)
= g (R (X; Y ) L; Z)
= R (Z; L; X; Y ) : (120)
So, we conclude the Codazzi equations (for any triplet of vectors X;Y;Z tangent to Ss at apoint):
(r= X �) (Y; Z) � (r= Y �) (X; Z) � � (X; Z) � (Y ) + � (Y; Z) � (X) = R (Z; L; X; Y ) :(121)
Take an orthonormal frame for TpSs, (EA : A = 1; 2). We complement this with E3 = L,E4 = L to a null frame for TpM .
r= A �BC � r= B �AC + �A �BC � �B �AC = RABC4 : (122)
Since dimSs = 2, we can write
RABC4 = �AB FC : (123)
�AB is the fully antisymmetric 2-dimensional symbol. In other words �AB are the orthonormalcomponents of the area 2-form of Ss. And FA are the components of a 1-form on Ss. So, thecomponents �FA of the dual 1-form are:
�FB :=XA
FA �AB
=XA
RABA4 : (124)
But XA
RABA4 � 1
2R4B34 = (Ric)B4 = 0 :
Recall the 1-form � on Ss:
� (X) =1
2R(X; L; L; L) 8 X 2 TSs ;
which came up in the propagation equation for the torsion �. So,
�A =1
2RA434 : (125)
39
Thus, one sees that XA
RABA4 = � �B : (126)
And we have:
�F = � � (127)
F = � � �F = �� : (128)
The propagation equation for � with respect to an orthonormal frame was:
@�A@s
+ 2XB
�AB �B = � �A :
The same equation with respect to a Jacobi �eld frame (Xa : a = 1; 2) reads:
@�a@s
+ � ba �b = � �a : (129)
(Recall that � ba = ( �1)bc�ac.) One writes the trace of the Codazzi equations in an orthonormal
frame as: XA
r= A �AB � r= B tr� +XA
�AB �A � tr� �A = � �A :
And the trace of the Codazzi equations in a Jacobi �eld frame reads:
div= �a � d= a tr� + � ba �b � tr� �a = � �a : (130)
Eliminating � from the propagation equation (129) for �, using the Codazzi equation, we obtain
@�a@s
+ tr� �a = div= �a � d= a tr� : (131)
Just like equation (89) expresses the Einstein vacuum equation R44 = 0, equation (131) ex-presses the Einstein vacuum equation R4a = 0.
40
3.2 The Divergence Operator on Trace-Free, 2-Covariant, Symmetric Ten-sor�elds on a Compact 2-Dimensional Riemannian Manifold
Let us consider the equation
div �a = fa (132)
for a trace-free, 2-covariant, symmetric tensor�eld �ab on a compact, Riemannian 2-manifold(M; gab). It is
div �a = rb �ab ; rb = (g�1)bc rc :
Proposition 4 This operator is conformally covariant. That is, setting
~gab = e2� gab ; (133)
~fa = e�2� fa ; (134)
we have: fdiv �a = ~fa : (135)
Proof: Consider
!abc = ra �bc � rb �ac : (136)
Then we have
~!abc = era �bc � erb �ac ; (137)
and we have; working in an arbitrary local coordinate system
4cab : = e�cab � �cab (138)
= Æca @b � + Æcb @a � � (g�1)cd gab @d � ; (139)
hence the ~!abc read as:
~!abc = !abc � 4mab �mc � 4m
ac �bm + 4mba �mc + 4m
bc �am
= !abc � ( Æma @c� + Æmc @a� � (g�1)mn gac @n� ) �bm
+ ( Æmb @c� + Æmc @b� � (g�1)mn gbc @n� ) �am
= !abc � �ba @c� � �bc @a� + @m� gac �bm
+ �ab @c� + �ac @b� � @m� gbc �am : (140)
Take the trace relative to ~g; since
(~g�1)ac = e�2� (g�1)ac (141)
and etr � = 0 ; (142)
we obtain:
fdiv �b = (~g�1)ac ~!abc
= e�2� ( (g�1)ac !abc � @a� �ab + 2 @m� �mb + tr � @b� � @m� �bm )
= e�2� (g�1)ac !abc
= e�2� div �b : (143)
41
�
Moreover, we shall show now that ~!abc = !abc:Since !abc is anti-symmetric in a and b, we can write, recalling that we are on a two-manifold,
!abc = �ab hc ; �ab : area 2-form of gab :
Contract this with (g�1)ac:
(g�1)ac �ab hc = ha �ab = �hb ; (144)
the Hodge dual. Thus, we obtain�ha = div �a ; (145)
andha = � � �ha = � �div �a : (146)
Similarly, one has~!abc = ~�ab ~hc ; (147)
where ~�ab is the area 2-form of ~gab and ~�ab = e2��ab. That is,
~!abc = e2� �ab (� ~�div �a) : (148)
Now, for any 1-form �a (in general, duality for n-forms on a 2n-dimensional manifold dependsonly on the conformal class)
~��a = (~g�1)bc �b ~�ca
= e�2� (g�1)bc �b e2� �ca
= ��a : (149)
Thus, in our case we have that, from the above proposition,
~�fdiv �a = � fdiv �a = e�2� �div �a : (150)
Substituting, we conclude that:~!abc = !abc : (151)
Remark that the equation div�a = fa is equivalent to the equation !abc = ��ab �fc. So, thedivergence equation is equivalent to a curl equation.
42
3.3 The L2-Theory for the Operator in Question
We start with
1
2!abc !
abc =1
2( ra �bc � rb �ac ) (ra �bc � rb �ac)
= j r� j2 � ra �bc rb �ac : (152)
Now, it is
ra �bc rb �ac = rb (�ac ra �bc) � �ac rb ra �bc
and
rb ra �bc � ra rb �bc = R dbb a| {z }Rda
�dc + R dbc a �bd :
Recalling that we are on a two-manifold, Rabcd = K(gacgbd � gadgbc), where K is the Gausscurvature; so Rac = Kgac. Thus, the above reduces to:
K Æda �dc + K (Æbc Æda � gca (g
�1)db) �bd = K �ac + K (�ac � gac tr �)
= 2 K �ac :
Hence, we obtain, substituting above,
ra �bc rb �ac = rb (�ac ra �bc) � �ac ra rb �bc � 2 K j � j2= rb (�ac ra �bc) � ra (�
ac rb �bc) + j div � j2 � 2 K j � j2
This is the second term in the expression for 12!abc!
abc in (152).
On the other hand we have shown earlier that !abc = ��ab �div�c. It follows that1
2!abc !
abc = j �div � j2 = j div � j2 : (153)
We conclude that:
j r� j2 + 2 K j � j2 = 2 j div � j2 + ra (�bc rb �
ac � �ac rb �bc) : (154)
Integrate over the compact (no boundary) manifold M to obtain the integral identity:ZM
( j r � j2 + 2 K j � j2 ) d�g =
ZM
2 j div � j2 d�g : (155)
This gives us an H1-estimate for the solution of the equation
div � = f if K � 0 : (156)
To handle the general case,M di�eomorphic to S2, we shall make use of the conformal invarianceof the operator div, together with the uniformization theorem. In general, whatever the compactmanifoldM is, the integral identity shows that the div operator has closed range. This range isthen the orthogonal complement of the null space (Kernel) of the adjoint operator.
Proposition 5 The L2-adjoint of div is the operator �12dLXg acting on vector�elds X.
43
Note that dLXg is the trace-free part of the Lie derivative with respect to X of the metric g.
Proof: We have, for any trace-free, 2-covariant, symmetric tensor�eld � and any vector�eldX, Z
M
div � � X d�g =
ZM
rb �ab � Xa d�g
= �ZM
�ab rb Xa d�g
= � 1
2
ZM
�ab (ra Xb + rb Xa � gab div X) d�g
= � 1
2
ZM
�ab (dLXg)ab d�g ;where
(LXg)ab = raXb + rbXa and (dLXg)ab = raXb + rbXa � gab div X :
It is easy to see that this operator is also conformally covariant. In fact, if
~gab = e2�gab ; (157)
then dLX~g = e2� dLXg : (158)
The null space (Kernel) of this adjoint operator is the space of conformal Killing �elds of (M; g).In the case that M is di�eomorphic to S2, by the uniformization theorem, this is the space ofvector�elds generating the conformal group of the standard sphere. We shall see that this canbe identi�ed with the group SL(2;C).
44
4 Uniformization Theorem
Let (M; g) be a compact, oriented, 2-dimensional, Riemannian manifold (without boundary).Further let K be the Gauss curvature of (M; g) and the genus of M . The Gauss-Bonnetformula now tells us: Z
M
K d�g = 2 � � ; (159)
where � denotes the Euler characteristic. � is expressed in terms of by:
� = 2 (1 � ) : (160)
There are three cases to consider.
1. = 1: (That means � = 0.) Then M is di�eomorphic to a torus (topologically).
2. � 2: (That means � is a negative even integer.)
3. = 0: (That means � = 2.) Then M is topologically a sphere.
Now the statement of the uniformization theorem is the following.
Theorem 2 g is conformal to a metric of constant Gauss curvature.
That is, we can �nd a smooth positive function on M such that the metric ~g = 2g hasconstant Gauss curvature eK. Since by constant rescaling, that is, taking = c, a positiveconstant, the Gauss curvature rescales according to:
eK = c�2 K : (161)
We can then make
eK =
8<:0 : in case 1
�1 : in case 21 : in case 3
In fact, we shall prove directly this version.
To begin, we must �nd how the Gauss curvature transforms under a conformal transforma-tion. We consider, generally, an n-dimensional Riemannian manifold (M; g) and
~gij = 2 gij : (162)
Then e�kij, the connection coeÆcients of ~g in an arbitrary local coordinate system, are expressed
in terms of �kij, the corresponding coeÆcients of g in the same coordinate system by:
e�kij = �kij + �1 (Æki @j + Ækj @i � (g�1)kl gij @l ) : (163)
Using this formula, we can �nd the relationship between the curvatures eRklij and R
klij of the two
conformally related metrics. This relation involves up to the second derivatives of the conformalfactor . The relations between the Ricci curvatures eRij , Rij and between the scalar curvatureseR, R then follow.
45
In the case n = 2 the formula relating the scalar or Gauss curvatures takes the simplest form, ifwe set = eu. Then we obtain:
eK = e�2u (K � 4g u) : (164)
(In the case n = 3 the formula takes the simplest form if we set = �2, (� > 0). Then it is:
eR = ��5 (R � � 8 4g �) : (165)
For general n, we set = �2
n�2 ( > 0). So, we have 2 = �4
n�2 . Then it is
eR = ��n+2n�2 (R � � 2 (n � 1) 4g �) : (166)
For n = 4 we obtain cn := 2(n� 1) = 6.)
Going back to our problem, we see that it reduces to �nding a smooth solution u of the equation
4g u + eK e2u = K ; (167)
where
eK =
8<:0 : in case 1
�1 : in case 21 : in case 3
Now, let us focus on the three cases.
Case 1: We then have the linear equation
4g u = K : (168)
In general, the linear equation (Poisson equation)
4g u = f ; (169)
where f is a given function on a compact manifold M (without boundary), is solvable if andonly if Z
M
f d�g = 0 : (170)
The solution is then unique up to an additive constant. In our case the integrability conditionreads: Z
M
K d�g = 0 ; (171)
which is precisely the Gauss-Bonnet formula in case 1.
Case 2: Here, we have eK = �1 and the equation reads
4g u � e2u = K : (172)
This is a non-linear elliptic partial di�erential equation. The Gauss-Bonnet formula is writtenin its usual form: Z
M
K d�g = 2 � � :
46
For any function f on M let us denote by �f the mean value of f on M .
�f =1
Area ((M; g))
ZM
f d�g : (173)
In particular, by the Gauss-Bonnet formula we have
�K =2 � �
Area ((M; g)): (174)
By an initial constant rescaling we can set �K = �1. Let us de�ne:
f = K + 1 :
Then it is�f = 0
and the equation (172) becomes
4g u � e2u + 1 = f : (175)
So, let us state the following
Theorem 3 In fact, we shall show, more generally, that this equation (175) has a unique smooth
solution u for any given smooth function f such that �f = 0.
Proof: Uniqueness:
Let u0 be another solution correspondig to the same f . Subtracting the equation for u fromthat for u0 yields:
4g (u0 � u) � (e2u
0 � e2u) = 0 :
Multiply by (u0 � u) and integrate over M . Integrating by parts the �rst term, we obtain:ZM
f j r (u0 � u) j2 + (u0 � u) (e2u0 � e2u) g d�g = 0 :
Now, it is
(e2u0 � e2u)
8<:> 0= 0< 0
according to
(u0 � u)
8<:> 0= 0< 0
Hence, one obtains that(u0 � u) (e2u
0 � e2u) � 0
with equality if and only if u0 = u. Uniqueness thus follows.
Existence:
47
We �rst derive a priori bounds for a C2-solution u. The starting point is the L1 a prioribounds. We consider thus a C2-solution u of
4g u � e2u + 1 = f ; �f = 0 : (176)
Since �f = 0, the linear problem4g v = f (177)
has a solution v unique up to an additive constant. We can �x v by requiring �v = 0.
Upper bound for u: Consider a point xM , where u achieves its maximum value uM on M .In any local coordinate system it is @iu(xM ) = 0 and the Hessian reduces to hij = @i@ju(xM )and we have that hij � 0, i.e. a non-positive matrix. So, its trace
trg h = (g�1)ij hij � 0 :
But this is the Laplacian:trg h = 4g u :
Thus with (176) one sees that
(e2u � 1 + f) (xM ) � 0 ;
hencee2uM � 1 � f (xM ) � 1 � fm ;
where fm is the minimum value of f on M . (Note that fm � 0 since �f = 0.) We have thereforeobtained the upper bound:
uM � 1
2log (1 � fm) : (178)
Lower bound for u: Here, we make use of the function v (de�ned by 4gv = f and �v = 0).Subtracting the equation for v from that for u yields:
4g (u � v) = e2u � 1 :
Consider a minimum point xm of the function (u� v). At such a point it is 4g(u� v)(xm) � 0,that is (e2u � 1)(xm) � 0, or u(xm) � 0. Therefore, for all x 2M we have:
(u � v) (x) � (u � v) (xm) � � v (xm) � � vM ;
where vM is the maximum value of the function v on M . It follows that
u (x) � v (x) � vM � vm � vM ; 8x 2 M ;
where vm is the minimum value of the function v onM . We have thus obtained the lower bound:
um � vm � vM (179)
(um is the minimum value of u on M).
From the two bounds we can obtain
48
Hk Estimates on u:
If f 2 Hl, we can obtain Hk bounds on u for each k = 0; � � � ; l + 2. In particular, if f issmooth we can obtain Hk bounds on u for every k. To show this we write
4g u = f 0 ; f 0 = f + e2u � 1 :
The L1 bounds just derived show that f 0 2 L2(M). Standard L2 estimates for the Laplacianoperator 4g then give an H2(M) estimate for u. The space H2(M) is a Banach (Hilbert) al-gebra. We then obtain f 0 2 H2(M) which yields a H4(M) estimate for u. The argument iscompleted by induction.
Replace the given function f by tf for any t 2 [0; 1]. We study the equation
F (u) = t f ; F (u) = 4g u � e2u + 1 :
We consider the set S � [0; 1] consisting of these values of t for which this problem is solvable.Clearly, 0 2 S for u � 0 is a solution to the problem for t = 0. What we wish to show is that1 2 S. We do not approach this directly. Indeed, we will show that S is both open as well as
closed. It then follows that S coincides with [0; 1], so in particular, 1 2 S.
1. To show that S is open, we apply the Implicit Function Theorem. Let X and Y beBanach spaces, U an open subset of X containing the point x0, and F a C1 mapping ofU into Y , so, F (x0) = y0. Consider the di�erential Dx0F of the mapping F at x0. (Thisis a linear map of X into Y .) Suppose that in fact Dx0F is an isomorphism of X onto Y .Then there exists a neighbourhood U 0 of x0 in X with U 0 � U and a neighbourhood V ofy0 in Y such that F jU 0 is a C1 di�eomorphism of U 0 onto V . (In particular, F (x) = y hasfor each y 2 V a solution x unique in U 0.)
To apply this theorem we take X = Hk(M), Y = Hk�2(M) with k � 4. Also, wetake U = X. Indeed F is a C1 map of X into Y . Next, consider DuF , a linear map of Xinto Y .
Du F � _u = 4g _u � 2 e2u _u ;
where _u 2 X = Hk(M). The hypothesis on DuF follows from the fact that the operator
4g � � � ; � � 0
is a linear isomorphism for any non-negative function �. Given t0 2 S, let ut0 be thesolution of F (u) = t0f . Applying the implicit function theorem with the backgroundsolution ut0 we conclude that also t 2 S provided that t is suÆciently close to t0. ThereforeS is open.
2. To show that S is closed, consider a sequence (tn) � S such that tn ! t� 2 [0; 1]. Weneed to show that t� 2 S. Let utn be the solutions corresponding to tnf . These solutionsall satisfy the a priori bounds corresponding to f . Thus (utn) is a sequence of functionscontained in a closed ball in the Hilbert space Hk(M). By weak compactness of closedballs in a Hilbert space there is a weak convergence to a function u� in the same ball.Moreover, a closed ball in Hk(M) is compact in Hk�1(M) (Rellich's theorem). Therefore
49
we can extract a subsequence of the above subsequence, converging strongly in Hk�1(M).The limit must be the same function u�. Then we have:
e2utn � 1 ! e2u� � 1
strongly in Hk�1(M) for the subsequence (ni). Also, we see:
4g utn * 4g u�
weakly in Hk�2(M). It follows that u� satis�es
F (u�) = t� f :
So, t� 2 S and S is closed.
Case 3: Positive Case: ( = 0; � = 2).M is di�eomorphic to S2. The equation is
4g u + e2u = K : (180)
Because of the + sign in equation (180), the maximum principle does not work. Moreover, thesolution is not unique. For, suppose that K = 1 to begin with. If the solution were unique, uwould have to vanish in this case. But the equation
4g u + e2u = 1 (181)
signi�es that eK = 1, so that ~g = e2ug as well as g are both isometric to the standard metric onS2. Consequently, they are isometric to each other:
~g = f�g ; (182)
where f is a di�eomorphism of S2 (onto itself). Thus, eu is the conformal factor associated toa conformal isometry on the standard sphere.
We can analyze these conformal isometries by using the stereographic projection of the sphereonto the complex plane. Now, consider the stereographic projection from the north pole N .
50
Figure 18.
C
S
N
P
Z
P is any point on the sphere other than N , and C denotes the center while S is the south pole.Look at the section with the plane containing the three points N , C, P .
Figure 19.
C
N
Z S
P �
�2
�2
51
Let � be the polar angle on the sphere from N . We have:
j z j = 2 cot�
2
z = r ei' ;
where r; ' denote the polar coordinates on the plane. The section is a ' =constant-section.(�; ' are polar coordinates on the sphere with origin N .) We write the metric of the plane asfollows:
e = dr2 + r2 d'2
= dz d�z :
And the metric of the sphere is:
g = d�2 + sin2 � d'2 :
From above, it is
r = 2 cot�
2;
hence we haved�
dr=
1
1 + r2
4
:
It follows that
g =dz d�z
(1 + jzj2
4 )2(183)
org = e2v em (184)
(remember that em is the euclidean metric), where
ev =1
1 + jzj2
4
: (185)
The conformal mappings of the plane,
f� em = 2 em ; (186)
are the complex analytic mappings with nowhere vanishing derivative:
z 7! f (z) ; f complex analytic:
Then it is
f� em = df d �f = j f 0(z) j2 dz d�z ; (187)
= j f 0(z) j ; f 0(z) 6= 0 8 z : (188)
To study the conformal isometries of the sphere, we use the atlas consisting of the two chartscorresponding to the stereographic projections from the north and south poles respectively.
52
Figure 20.
S
N
C
P
Z 0
�
�0
Here, we look again at the ' = constant-section. If �0 is the polar angle on the sphere from thesouth pole, then we have
� + �0 = � :
So, it is
j z0 j = 2 cot�0
2= 2 tan
�
2=
4
j z j ;
that is
j z0 j j z j = 4
and ' = constant-sections are the same, i.e.:
z = r ei' ; z0 = r0 ei' ; r r0 = 4
or,
z0 =4
z:
When considering the conformal isometries of the sphere, we may a priori consider all mero-morphic mappings of the plane, and change z to 4
z, going to the south polar stereographic chart
when considering the behaviour near the north pole in the domain (that is, we compose f onthe right with the change of chart). On the other hand when considering the behaviour near apole of f(z) we compose on the left with a change of chart, since such a pole is mapped to thenorth pole in the target. This analysis yields the conclusion that the conformal isometries ofthe sphere correspond to the fractional linear transformations
z 7! f (z) =az + b
cz + d(189)
with ad� bc = 1. The matrices
M =
�a bc d
�53
with detM = 1, form the group SL(2;C). The group SL(2;C) is non-compact. Under z 7! f(z)we have dz 7! f 0(z)dz and
g = (1 +j z j24
)�2 j dz j2 ! (1 +j f(z) j2
4)�2 j f 0(z) j2 j dz j2 = ~g = e2u g :
So,
eu =j f 0(z) j (1 + jzj2
4 )
1 + jf(z)j2
4
(190)
and f(z) is as in equation (189) (with ad� bc = 1).
Consider as a particular case the scalings
f (z) = � z ; (191)
where � is real and positive. Then we write
eu =� (1 + jzj2
4 )
1 + �2jzj2
4
: (192)
At the south pole S (z = 0), it is eu = �, while at the north pole N (z =1), we have eu = 1�.
So, there is no maximum and no minimum of u in the class of solutions of
4gu + e2u = 1 ;
where g is the standard metric on S2. Thus we need a new approach.
Approach
Step 1: Find a pair of points (P;O) in M , such that
d (P;O) = diam M ;
that is the maximal distance between two points of M .
Step 2: We �nd a function w on MnP such that the metric
~g = e2w g (193)
on MnP makes (MnP; ~g) isometric to a plane. So, we require~K = 0 : (194)
This is the equation
4gw � K = 0 : on MnP : (195)
In fact, we de�ne w to be the solution of the equation
4gw � K = � 4 � ÆP : on M ; (196)
54
where ÆP is the Dirac delta distribution with center at P . We can think of step 2 as de�ning aconformal map to a plane taking the point P to the point at 1 on the plane.
Step 3: Let ~dO be the distance function on (MnP; ~g) measured from O. We then set
ev =1
1 +~d2O
4
: (197)
Then the metric~~g = e2v ~g (198)
will have Gauss curvature ~~K = 1, so (MnP; ~~g) will be isometric to the standard sphere minusthe north pole N .
Step 4: We setu = w + v : (199)
Then we show that this function extends continuously to the point P . So,
~~g = e2u g (200)
is a metric on M , such that (M; ~~g) is isometric to the standard sphere.
Step 5: We show that u is in fact a smooth function on M .
Remark that the function w tends to 1 as we approach P , while v tends to �1. So, toshow that u = w + v extends continuously to P , we must analyze the blow-up behaviour of wand v as we approach P .
Details:
Step 1 is trivial.
Step 2: Let rP be the radius of injectivity of expP , the exponential map with base pointP (relative to (M; g)). Then choose a positive constant �, such that 2� < rP . Let also � be aC1 non-increasing function on [0;1) such that � = 1 on [0; 1] and � = 0 on [2;1).
Figure 21.
�
1 2
55
We then de�ne on M the cut-o� function (for q 2M)
� (q) =
�� (dP
�) : if q 2 B2� (P )0 : otherwise
Here, dP is the distance function from P (relative to g). So, it is � = 1 on B�(P ) and � = 0 onMnB2�(P ). We then de�ne on M the function
w0 =
� � 2 � log dP : in B2� (P )0 : on MnB2� (P )
(201)
Clearly, w0 is smooth except at P . In B�(P ) we have, in fact,
w0 = � 2 log dP :
Now, since 2� < rP , expP is a di�eomorphism of the ball of radius 2� with center at 0 in TpMonto B2�(P ) inM . We can thus introduce polar normal coordinates (r; �) in B2�(P ) (with originat P ). In these coordinates,
dP = r ; (202)
g = dr2 + R2 (r; �) d�2 : (203)
Also, Z 2�
0R (r; �) d� = L (r) (204)
is the perimeter of the geodesic circles r = const: We have
L(r)
r! 2 � as r ! 0
by local euclidicity at P . The (geodesic) curvature � of these circles is
� =1
R
@R
@r(205)
and we have@�
@r= � �2 � K (K : Gauss curvature) (206)
Let us now express the Laplace operator in polar coordinates. In arbitrary coordinates, we have
4g =1pdetg
@
@xa�pdetg (g�1)ab
@
@xb; (207)
and in polar coordinates, it is
4g =1
R
@
@r� R @
@r+
1
R
@
@�� 1
R
@
@�: (208)
In B�(P ) we have
w0 = � 2 log r ; (209)
4g w0 = � 2
R
@
@r(R
r) = � 2�
r; (210)
56
where
� =1
R
@R
@r� 1
r= � � 1
r: (211)
We then have:@�
@r+ �2| {z }
=�K
=@�
@r+
2�
r+ �2 :
Setting
� = r2 � ;
this becomes@�
@r+
�2
r2= � r2 K :
Since �! 0 as r ! 0, we obtain along each ray the integral equation
� (r; �) = �Z r
0(� (r0; �)2
r02+ r02 K (r0; �)) dr0 : (212)
Analyzing this equation, we conclude that
� (r; �) = O (r) (213)
and�
r! � 1
3KP ; as r ! 0 : (214)
Thus we see that 4gw0 is bounded and
4g w0 ! 2
3KP (215)
as we approach P . Next, we set w = w0 + w1. Then w1 is to satisfy the equation
4g w1 = 4g w � 4g w0
= K � 4g w0 on MnP:
Let f be the function on M given by:
f = K � 4g w0 : (216)
We have shown that f extends to a continuous function on M . There is a solution w1 of
4g w1 = f (217)
unique up to an additive constant, provided thatZM
f d�g = 0 : (218)
To show this, we integrate f on MnBÆ(P ) with 0 < Æ � �. We have
�ZMnBÆ(P )
4g w0 d�g =
Z@BÆ(P )
rN w0 ds ; (219)
57
(ds is the element of arc length of @BÆ(P )). In �BÆ(P ) we have, in polar coordinates:
w0 = � 2 log r (220)
and
rN =@
@r; (221)
so
rN w0 = � 2
r: (222)
Moreover, it isds = R d� : (223)
So, we haveZ@BÆ(P )
rN w0 ds = � 2
Æ
Z 2�
0R (Æ; �) d� ! � 4 � as Æ ! 0 : (224)
On the other hand,
limÆ!0
ZMnBÆ(P )
K d�g =
ZM
K d�g = 4 � (225)
(by Gauss-Bonnet). We conclude that indeedZM
f d�g = 0 : (226)
So, the equation is solvable for w1. In fact, we can show that w1 is bounded on M (in factcontinuous). (For, f = 4gw1 being bounded, in particular f 2 L2(M) implies w1 2 H2(M),hence w1 is bounded.)
Step 4: Now, it is
u = w + v
w = w0 + w1 ;
where w1 is bounded on M , while
w0 = � 2 � log dP
and dP is the g-distance from P . On the other hand
ev =1
1 +~d2O4
; (227)
where ~dO is the ~g-distance from O. So, it is
v = � 2 log ~dO + O(1) : (228)
It follows that u is bounded on M if and only if in B�(P ) (relative to g) dP � ~dO is boundedabove and below by positive constants.
a) Upper bound for dP ~dO: Consider an arbitrary point Q 2 B�(P ).
58
Figure 22.
P
O
Q Q2
Q1
B�(P )
~
Let ~ be the unique ~g-geodesic joining Q to O. Let Q1 be the �rst point of intersection of ~ with the geodesic circle @B�(P ). Consider the g-ray from P to Q, extended up to @B�(P ).Let Q2 be the point where meets @B�(P ). Then we write
~dO (Q) = ~d (Q;O) = ~d (Q1; O) + ~d (Q;Q1)
= ~dO (Q1) + ~d (Q;Q1) :
Now, ~dO(Q1) is bounded, while
~d (Q;Q1) � ~d (Q;Q2) + ~d (Q1; Q2)
and~d (Q;Q2) � ~g � length of the segment (Q;Q2) of =: ~L1 ;
also
~d (Q1; Q2) � ~g � length of the arc A (Q1; Q2) of the circle @B�(P )
between Q1 and Q2 =: ~L2 :
In polar normal coordinates centered at P , we have
Q1 = (�; �1) ; Q2 = (�; �2)
59
and
Q = (r; �2) ; r = dP (Q) :
Thus, it is
~L1 =
Z �
dP (Q)ew(r;�2) dr
=
Z �
dP (Q)e(w0+w1)(r;�2) dr
� e(w1)M
Z �
dP (Q)r�2 dr
� e(w1)M1
dP (Q);
where (w1)M = supM w1, for ~L2 we write
~L2 =
Z �2
�1
ew(�;�) R (�; �) d�
�Z 2�
0ew(�;�) R (�; �) d� =: C (�) :
We conclude that
~d (Q;Q1) � ~L1 + ~L2 (229)
� e(w1)M1
dP (Q)+ C (�) : (230)
This yields the upper bound.
b) Lower bound for dP ~dO: First, notice that
~dO (Q) � ~d (Q;Q1) = ~g � length of the ~g �minimal geodesic from Q to Q1;
that is of ~ (Q;Q1) =: ~L3 :
Figure 23.
P
Q
Q1~
60
We can express ~ (Q;Q1) in parametric form:
r = r (�)� = � (�) ; � 2 [0; 1]r (0) = dP (Q) ; � (0) = �2r (1) = � ; � (1) = �1 :
Thus, we have
~L3 =
Z 1
0ew(r(�);�(�)) f ( dr
d�)2 + R2 (
d�
d�)2 g 12 d�
�Z 1
0ew1(r(�);�(�)) ew0(r(�);�(�))| {z }
=(r(�))�2
j drd�
j d�
� e(w1)m
Z 1
0(r(�))�2
dr
d�d�
= e(w1)m (r(0)�1 � r(1)�1) ;
where (w1)m = infM w1. We conclude that
~d (Q;Q1) � e(w1)m (dP (Q)�1 � ��1) ; (231)
which yields the lower bound.
So, we have derived the upper and lower bounds for dP ~dO. This concludes the proof of theuniformization theorem.
Remarks: We may de�ne the comparison function !0, see (201), in a di�erent way whichis appropriate when the Gauss curvature K is not assumed to be pointwise bounded (that is, inL1) but only in Lq, for some q > 1. (In fact, we shall take q � 2.) We set
!0 = � 4 � GP ; (232)
where GP is the Green's function with base point P , that is the solution of
4g GP = ÆP � 1
A(M; g): (233)
Then !1, de�ned by!1 = ! � !0 (234)
satis�es the equation
4g !1 = K � 4�
A= K � �K : (235)
(For, by the Gauss-Bonnet theorem, it is: �K = 4�A.) One can show that, under the hypothesis
K 2 Lq, choosing P to be a Lebesgue point of K,
GP � 1
2�log dP
as we approach the point P , so the proof may proceed as above. In this way we may estimatek u kL1(M) by A
� 12 diam(M), the dimensionless quantities, and
A k� K k� Lq(M) = A1� 1
q k K kLq(M) : (236)
61
5 Estimates for � and the Torsion (�) System
The content of the present chapter is the treatment of systems of ordinary di�erential equationsalong the generators of a null hypersurface C, coupled to elliptic equations on the surfaces of afoliation of C by sections.
5.1 Estimates for �
Now, recall the equation@tr�
@s+
1
2(tr�)2 + j � j2 = 0 : (237)
And consider the following equations in terms of Jacobi �eld frames:
@ d= atr�
@s+ tr� d= atr� + 2 �bc r= a �bc = 0 (238)
div= �a � 1
2d= atr� + � b
a �b � 1
2tr� �a = � �a : (239)
Equation (238) is an ordinary di�erential equation along the generators of C. We call suchequations propagation equations. (238) is obtained by di�erentiation of (237) tangentially toSs. And equation (239) is an elliptic system on each section Ss of C. The following is themethod of treatment of such coupled systems of equations. Consider the Codazzi equations.Setting
fa = � �a +1
2tr� �a � � b
a �b ; (240)
the Codazzi equations read as follows:
div= �a =1
2d= atr� + fa : (241)
What we are going to do is, assuming estimates for the spacetime curvature to obtain estimatesfor the quantities controlling the geometry of C as described by its foliation fSsg. Here, wewant to derive estimates for �ab.
We have for � on each section Ss an elliptic system of the form
div �a = fa (242)
for a trace-free, symmetric, 2-covariant tensor�eld �ab on a two-dimensional, compact, Rieman-nian manifold (M; gab) (without boundary). Moreover, M in the case of interest is di�eomorphicto S2. We want to derive an estimate of the form
k r� kLp(M;g) � C k f kLp(M;g) ; (243)
for any 1 < p <1. (In fact, in our application we need an estimate of � in L1, thus in view ofthe Sobolev inequality on (M; g) an estimate of r� in Lp(M; g) for some p > 2.)
By the uniformization theorem there exists a function � on M such that
~gab = e2� gab (244)
62
is a metric of Gauss curvature ~K = 1 so that (M; ~g) is isometric to the standard unit sphere.Moreover, the equation (242)
div �a = fa
is conformally covariant; that is, we have:
fdiv �a = ~fa ; ~fa = e�2� fa : (245)
For this system the standard Calderon-Zygmund Lp estimates hold:
k � kpLp(M;~g) + k er� kp
Lp(M;~g) � C (p)p k ~f kpLp(M;~g) (246)
for any 1 < p <1. From this we want to derive estimates for
k � kLp(M;g) and k r� kLp(M;g)
in terms of k f kLp(M;g). To do this more neatly, we �rst make a (constant) rescaling to a metricg0 of area 4� by setting
g0ab = r�2 gab ; (247)
where r =
qArea (M;g)
4� . We then make the conformal change
~gab = e2 g0ab (248)
to a metric ~g of Gauss curvature ~K = 1. So, � = � log r and
4 � =
ZM
d�~g =
ZM
e2 d�g0
=
ZM
d�g0 ;
i.e. the mean value of e2 on M relative to g0 is equal to 1. Under the (constant) rescaling wehave that:
k � kpLp(M;g0) =
ZM
j � jpg0 d�g0 ; (249)
and
j � j2g0 = (g0�1)ac (g0�1)bd �ab �cd
= r2 (g�1)ac r2 (g�1)bd �ab �cd
= r4 j � j2g ;
that isj � jg0 = r2 j � jg : (250)
Also, it isd�g0 = r�2 d�g :
And we have ZM
j � jpg0 d�g0 =
ZM
r2p�2 j � jpg drg
= r2p�2 k � kpLp(M;g) :
63
We have thus obtained that
k � kLp(M;g0) = r2p�2p k � kLp(M;g) : (251)
Next, the connection coeÆcients of g0ab and gab coincide, hence r0� = r�. (r0 denotes thecovariant di�erentiation with respect to g0.) Also, one sees that
j r0� jg0 = j r� jg0 = ((g0�1)ad (g0�1)be (g0�1)cf ra�bc rd�ef )12
= (r2(g�1)ad r2 (g�1)be r2 (g�1)cf ra�bc rd�ef )12
= r3 j r� jg :
Thus we have:
k r0� kpLp(M;g0) =
ZM
j r0� jpg0 d�g0
=
ZM
r3p j r� jpg r�2 d�g
= r3p�2 k r� kpLp(M;g) ;
that is,
k r0� kLp(M;g0) = r3p�2p k r� kLp(M;g) : (252)
Next, we consider the conformal transformation
~gab = e2 g0ab : (253)
Then it is
j � j~g = ((~g�1)ac (~g�1)bd �ab �cd)12
= e�2 ((g0�1)ac (g0�1)bd �ab �cd)12
= e�2 j � jg0 (254)
andd�~g = e2 d�g0 ;
hence
k � kpLp(M;~g) =
ZM
j � jp~g d�~g
=
ZM
e�(2p�2) j � jpg0 d�g0
� e�(2p�2)MZM
j � jpg0 d�g0
= e�(2p�2)M k � kpLp(M;g0) (255)
with M = supM . Combining this with the previous result yields:
k � kLp(M;g) � (r�1eM )2p�2p k � kLp(M;~g) (256)
� (r�1eM )2p�2p C(p) k ~f kLp(M;~g) : (257)
64
On the other hand,~fa = e�2� fa (� = � log r) :
So, we have
j ~f j~g = ((~g�1)ab ~fa ~fb)12
= (e�2� (g�1)ab e�2� fa e�2� fb)
12
= e�3� j f jg : (258)
It follows that
k ~f kpLp(M;~g) =
ZM
e�(3p�2)� j f jpg d�g
� e�(3p�2)�m k f kpLp(M;g) (259)
with �m = m � log r and �m = infM �, m = infM . That is,
k ~f kLp(M;~g) � (r e�m)3p�2p k f kLp(M;g) : (260)
Substituting above in the estimate (257) for k � kLp(M;g) we obtain:
k � kLp(M;g) � r e(2p�2)M�(3p�2)m
p C(p) k f kLp(M;g) : (261)
To obtain the estimate for r� in Lp(M; g), we write:
r0a �bc = era �bc + 4d
ab �dc + 4dac �bd ; (262)
where
4cab = e�cab � �0cab (263)
= Æca @b + Æcb @a � (~g�1)cd ~gab @d : (264)
It follows that:j r0� j~g � j er� j~g + C j d j~g j � j~g ; (265)
where C is a numerical constant. Hence, it is
k j r0� j~g kLp(M;~g) � k er� kLp(M;~g) + C k j d j~g j � j~g kLp(M;~g) (266)
andk j d j~g j � j~g kLp(M;~g) � eKp sup
M
j � j~g ; (267)
where eKp :=k d kLp(M;~g) : (268)
Taking now p > 2, the Sobolev inequality on the standard 2-sphere gives:
supM
j � j~g � C(p) (k er� kLp(M;~g) + k � kLp(M;~g)) ; (269)
where C(p) ! 1 as p ! 2. Substituting the Calderon-Zygmund estimate for the right handside of (269) gives:
supM
j � j~g � C(p) k ~f kLp(M;~g) (270)
65
and substituting again above in (267), (266), yields:
k j r0� j~g kLp(M;~g) � C(p) (1 + eKp) k ~f kLp(M;~g) : (271)
Finally, using the fact that
j r0� j~g = e�3 j r0� jg0and the relations between the norms derived above, we obtain:
k r� kLp(M;g) � C(p) (1 + eKp) e(M�m)(3p�2)
p k f kLp(M;g) ; (272)
where the constant C(p) actually depends on the following terms: C(p; eKp;M �m).
From (270) and (259) we see that
supM
j � j~g � C(p) k ~f kLp(M;~g)
� C(p) (r e�m)3p�2p k f kLp(M;g) :
Also, we have that
j � j2~g = (~g�1)ac (~g�1)bd �ab �cd
= (r2 e�2)2 (g�1)ac (g�1)bd �ab �cd
= (r2 e�2)2 j � j2g : (273)
So,
j � jg = r�2 e2 j � j~g (274)
and
supM
j � jg � r�2 e2M j � j~g : (275)
Substituting, we obtain
supM
j � jg � rp�2p e2M e�
3p�2p
m k f kLp(M;g) : (276)
We apply the above estimates to the null Codazzi equations (see (239), (240) and(241)).
div= �a =1
2d= atr� + fa ;
where
fa = � �a +1
2tr� �a � � b
a �b :
We are going to assume an Lp estimate for � on each section Ss. We shall �rst present a pre-liminary simpli�ed treatment which ignores the lower order terms involving �. We shall latergive the actual treatment which derives an Lp estimate for r= � on each section Ss, from whichan L1 estimate for � follows.
66
Let us introduce now the following dimensionless norms: For any tensor�eld t on Ss we write
k� t(s) k� Lp(S) = (1
Area(Ss)
ZSs
j t jp d� )1p (277)
= (4�r2)� 1p k t kLp(Ss; ) : (278)
(Here, (s) is the induced metric on Ss.) We apply the estimates derived (for div� = f),obtaining:
k� r= �(s) k� Lp(S) � C(s) f k� d= tr�(s) k� Lp(S) + k� f k� Lp(S) g (279)
with C(s) = C( eKp;M �m; p).
By virtue of (276) we can also bound in the same way r�1 k �(s) kL1(S):
r�1 k �(s) kL1(S) � C (M ;m; eKP ) f k� d= tr�(s) k� Lp(S) + k� f k� Lp(S) g : (280)
Assume that C(M ;m; eKP ) � �C is independent of s. This holds if there is a uniform (i.e.independent of s) upper bound for M , a uniform lower bound for m and a uniform upperbound for eKP .
The idea is to derive an ordinary di�erential inequality for k� d= tr�(s)k� Lp(S). We have: (writingA(s) for Area(Ss))
d
dsk� d= tr�(s) k� Lp(S) =
d
ds
� 1
A(s)
ZS
j d= tr�(s) jp (s) d� (s)
� 1p
=1
pk� d= tr�(s) k� 1�p
Lp(S) f �1
A2
dA
ds
ZS
j d= tr�(s) jp (s) d� (s)
+1
A
ZS
(@
@sj d= tr�(s) jp
(s) + j d= tr�(s) jp (s) tr�) d� (s) g :
Now, it is1
A
dA
ds=
1
A
ZA
tr� d� = tr� ;
where tr� denotes the mean value of tr� on Ss. Also,
@
@sj d= tr�(s) jp
(s) =@
@sf (( �1)ab @atr� @btr�)
p2 g
=p
2j d= tr�(s) jp�2
(s) f � 2 �ab @atr� @btr� + 2 ( �1)ab @btr�@
@s(@atr�) g :
So, it is
d
dsk� d= tr�(s) k� Lp(S) =
1
pk� d= tr�(s) k� 1�p
Lp(S)f � tr� k� d= tr�(s) k� p
Lp(S)
� p
2A
ZS
tr� j d= tr�(s) jp (s) d� (s)
� p
A
ZS
j d= tr�(s) jp�2 (s) �ab d= atr� d= btr� d� (s)
+p
A
ZS
j d= tr�(s) jp�2 (s) ( �1)ab @btr�
@
@s(@atr�)d� (s)
+1
A
ZS
j d= tr�(s) jp (s)
tr� d� (s) g :
67
We writetr� = tr� + (tr� � tr�)
to obtain:
j ddsk� d= tr�(s) k� Lp(S) + 1
2 tr� k� d= tr�(s) k� Lp(S)
� k� d= tr�(s) k� 1�pLp(S)
1A
RSj d= tr�(s) jp�2
(s) ( �1)ab @btr�@@s(@atr�) d� (s) j �
1pk� d= tr� k� 1�p
Lp(S) f (p2 � 1) 1A
RSj tr� � tr� j j d= tr� jp
(s) d� (s)+ p
A
RSj � j (s) j d= tr� jp (s) d� (s) g :
We now appeal to the equation:
@
@sd= atr� + tr� d= atr� + 2�bc r= a�bc = 0 :
Thus,
( �1)ab @btr�@
@s@atr� = � tr� j d= tr� j2 � 2 d= atr� � �bc � r= a�bc
and
� 1A
RSj d= tr� jp�2 ( �1)ab @btr�
@@s(@atr�) d� (s) =
1A
RStr� j d= tr� jp d� (s) + 2
A
RSj d= tr� jp�2 d= atr� � �bc � r= a�bc d� (s) :
Again substitute tr� = tr� + (tr� � tr�) and the result in the inequality above. We obtain:
j ddsk� d= tr�(s) k� Lp(S) + 3
2 tr� k� d= tr�(s) k� Lp(S) j �C k� d= tr�(s) k� 1�p
Lp(S)f 1
A
RS(j tr� � tr� j + j � j (s)) j d= tr� jp d� (s)
+ 1A
RSj d= tr� jp�1 j � j j r= � j d� (s) g :
Now, by H�older's inequality,
1
A
ZS
j d= tr� jp�1 j r= � j d� (s)
� 1
A(
ZS
j d= tr� jp d� (s))p�1p � (
ZS
j r= � jp d� (s))1p
= k� d= tr�k� p�1Lp(S) k� r= �k� Lp(S) :
We thus obtain the following di�erential inequality for k� d= tr�(s)k� Lp(S):
j dds
k� d= tr�(s) k� Lp(S) +3
2tr� k� d= tr�(s) k� Lp(S) j
� C f (k tr� � tr� kL1 + k � kL1) k� d= tr�(s) k� Lp(S)
+ k � kL1 k� r= �k� Lp(S) g : (281)
Now we make use of the elliptic estimate on each Ss for k� r= �k� Lp(S) and r�1 k � kL1 in terms
of k� d= tr�k� Lp(S) and k� fk� Lp(S). We have
k� r= �k� Lp(S) ; r�1 k � kL1 � �C f r � k� d= tr�k� Lp(S) + k� fk� Lp(S)
�2 g : (282)
Also, we have:r�1 k tr� � tr� kL1 � �C k� d= tr�k� Lp(S) :
68
We then obtain the ordinary di�erential inequality
j dds
k� d= tr�(s) k� Lp(S) +3
2tr� k� d= tr�(s) k� Lp(S) j �
�C f r k� d= tr�k� 2Lp(S) + r k� d= tr�(s) k� Lp(S) k� fk� Lp(S) g : (283)
Now, suppose that we have an ordinary di�erential equation of the form:
d�
ds+ � tr� � = � ; (284)
where �, � are non-negative functions of s and � is a constant. We have:
1
A
dA
ds= tr� ; A = Area of Ss :
The equation (284) is thus equivalent to
d
ds(A� �) = A� � : (285)
Integrating on [s0; s1], we obtain
(A� �) (s1) � (A� �) (s0) =
Z s1
s0
(A� �) (s) ds : (286)
Analogously with an inequality in place of the equation. In the case we are considering, it is� = 3
2 . Thus, settingQ = r3 k� d= tr�k� Lp ; (287)
we obtain the following ordinary di�erential inequality for Q:
j dQds
j � �C (r�2 Q2 + b) ; (288)
where we set b = r4k� fk� 2Lp . For, according to the inequality (283) just derived, we have
j dds
k� d= tr�(s) k� Lp(S) +3
2tr� k� d= tr�(s) k� Lp(S) j
� �C r (k� d= tr�(s) k� 2Lp(S) + k� fk� 2
Lp(S)) :
We need the upper bound for dQds
if we are given initial data, the lower bound if we are given�nal data.
To proceed, we make the assumption:Z s
0b (s0) ds0 � M : independent of s : (289)
We must also assume that:r � r0 + � s (290)
for some positive constant �, to be chosen later. (r0 = r js=0 =q
Area(S0)4� .)
SetQ1 = Q0 + �C M (291)
(Q0 = Q js=0). Under the above two assumptions (289) and (290) we can show:
69
Proposition 6 We haveQ � 2 Q1 ; (292)
provided that Q1 is suitably small.
Proof: The inequality holds for small positive s. Let s� be the maximum value of s for whichthe inequality holds in [0; s�]. Then we have, integrating the upper bound for dQ
dsfrom s = 0.
Q (s�) � Q0 + �C fZ s�
0
4Q21 ds
(r0 + �s)2+
Z s�
0b ds g
� Q0 + �C (4Q2
1
�r0+ M)
= Q1 +4 �CQ2
1
�r0:
Thus, if4 �CQ2
1
�r0< Q1 ;
that is,
Q1 <�r04 �C
; (293)
we haveQ (s�) < 2 Q1 ;
contradicting by continuity the maximality of s�. We conclude that under the hypothesis (293),the inequality (292),
Q � 2 Q1 ;
continues to hold as long as the two assumptions (289) and (290), we made above, hold.
We now proceed to show that, under some additional hypothesis on the initial data, the lowerbound on r cannot fail.
To do this we derive an equation for r(s): We have
dA
ds= A tr�
with A = 4�r2. Hence,dr
ds=
r
2tr� : (294)
Then,
dtr�
ds=
d
ds(1
A
ZS
tr� d� )
=1
A
ZS
f @tr�
@s+ (tr�)2 g d�
� 1
A2A tr�
ZS
tr� d�
=1
A
ZS
f 1
2(tr�)2 � j � j2 gd� ) � (tr�)2 ;
70
where we have used the equation
@tr�
@s= � 1
2(tr�)2 � j � j2 :
So,dtr�
ds+
1
2(tr�)2 =
1
A
ZS
f 1
2(tr� � tr�)2 � j � j2 g d� ) ; (295)
where we have used the fact thatZS
(f2 � �f2) d� ) =
ZS
(f � �f)2 d� ) ;
we thus obtain:dtr�
ds+
1
2(tr�)2 = h ; (296)
where
h =1
2(tr� � tr�)2 � j � j2 : (297)
Now, let us write x := tr�. Then equation (296) reads:
dx
ds+
1
2x2 = h : (298)
We have:
j h j � 1
2k tr� � tr� k2L1 + k � k2L1
� �C r2 (k� d= tr�k� 2Lp + k� fk� 2
Lp)
= �C r�2 (r�2 Q2 + b) :
We assume that x0 > 0. It shall follow that x > 0. Set
y =2
x: (299)
Then the equation (298) for x becomes
dy
ds= 1 � 1
2y2 h : (300)
Hence,
dy
ds� 1 +
1
2y2 j h j
� 1 + �C y2 r�2 (r�2 Q2 + b) : (301)
Proposition 7 Let y0 =2
tr�0be the initial value of y (i.e. for s = 0). Then we have:
y � 3
2y0 + s (302)
as long as the inequalityr � r0 + � s (303)
holds.
71
Proof: This holds for small positive s. Let again s� be the maximal value of s, such that theinequality holds on [0; s�]. Let
� � 2r03y0
: (304)
Then on [0; s�] we have
y
r�
32y0 + s
r0 + �s� max f 3y0
2r0;1
�g =
1
�: (305)
Then, in view of (292), it is
�C y2 r�2 (r�2Q2 + b) � �C ��2 (4 Q21 (r0 + � s)�2 + b) :
Therefore, integrating the inequality for dyds
from s = 0, we obtain:
y � y0 + s �C ��2 (4Q2
1
�r0+ M) + s (306)
� 3
2y0 + s (307)
if4Q2
1
�r0+ M <
�2y02 �C
:
This shows that the upper bound on y continues to hold as long as the lower bound on r holds.
But then we have, from (294),
d log r
ds=
1
y� 1
32y0 + s
in [0; s�]. So, integrating, we obtain:
r � r0 (1 +2s�3y0
) > r0 + � s� ;
if we require, in place of (304) the strict inequality:
� <2r03y0
; (308)
contradicting the maximality of s�. (We may �x � = r02y0
.) Then also the lower bound on rcontinues to hold. Therefore, all the above inequalities continue to hold as long as the assumption(289), i.e. Z s
0b (s0) ds0 � M
holds.
72
5.2 The Torsion (�) System
In this section we consider the torsion (�) system, a coupled elliptic system on each section ofthe null hypersurface with a propagation equation along each (null geodesic) generator of thehypersurface.
We start from the equation
rLL = � 2 Z ; (309)
where Z is the Ss-tangential vector�eld on C which corresponds to the torsion-1-form �.
� (�X) = g (Z; X) (310)
for all X in TpM , where � denotes the projection to TpSs and p 2 Ss. Let now X, Y bevector�elds de�ned on and tangential to S0. We extend them to C by:
[L; X] = [L; Y ] = 0 : (311)
Then X, Y are Ss-tangential for all s. Then the second fundamental form � of Ss relative tothe (C-transversal) null normal L is de�ned by:
� (X; Y ) = g (rXL; Y ) : (312)
This is symmetric (like �).
We consider the propagation of � along the generators of C. We have:
L (� (X;Y )) = g (rLrXL; Y ) + g (rXL; rLY ) : (313)
Now, look at the �rst term on the right hand side of equation (313):
rLrXL = rXrLL + R (L; X) L
= � 2 rXZ + R (L; X) L : (314)
Hence,
g (rLrXL; Y ) = � 2 (rX�) (Y ) � R (Y; L; X; L) :
On the other hand, for the second term in the same equation:
rXL = � (X) L +XA
� (EA; X) EA ; (315)
where (EA; A = 1; 2) is a local orthonormal frame for Ss. Hence, it is
g (rXL; rLY ) = g (rXL;rY L)
= 2 � (X) � (Y ) +XA
� (X; EA) � (EA; Y ) :
Substituting and noting that
(rX�) (Y ) = (r= X�) (Y ) : as Y is Ss-tangential, like � ; (316)
73
we obtain:
L (� (X; Y )) = � 2 (r= X�) (Y ) � R (Y; L; X; L)
+ 2 �(X) �(Y ) +XA
� (X; EA) � (EA; Y ) : (317)
This is an equation along each generator of C. We can take X = Ea and Y = Eb, where Ea, Eb
are Jacobi �elds along the generator. With the notation
� (Ea; Eb) = �ab; (318)
the equation (317) takes the form:
@�ab
@s= � 2 r= a�b � R (Eb; L; Ea; L)
+ 2 �a �b + � c
a�cb : (319)
Let (EA; A = 1; 2) be a local orthonormal frame �eld for Ss; we complete it with E3 = L andE4 = L to a null frame for M along C. We want to express RA3B4. We decompose:
RA3B4 =1
2(RA3B4 + RB3A4) +
1
2(RA3B4 � RB3A4) : (320)
For the symmetric part we use:XC
RACBC � 1
2(RA3B4 + RB3A4| {z }
=RA4B3
) = RicAB = 0
by the vacuum equations. Now, there exists a function � such that
RABCD = � � �AB �CD : (321)
Hence, it is XC
RABCD = � �XB
�AB �CD = � � ÆAC : (322)
We conclude that the symmetric part of RA3B4 is given by:
1
2(RA3B4 + RB3A4) = � � ÆAB : (323)
Another expression for � is obtained by considering
� 1
2R3443 � 1
2R4433| {z }=0
+XA
RA4A3 = Ric43 = 0 :
Hence, from above we see that:
1
2R3434 = �
XA
RA3A4 = 2 � : (324)
So, � can equivalently be de�ned as:
� =1
4R3434 : (325)
74
Next, we consider the antisymmetric part of RA3B4:
1
2(RA3B4 � RB3A4) =
1
2(RA3B4 + RA43B) = � 1
2(RAB43) =
1
2RAB34 = � �AB : (326)
The last equality de�nes the function �. We conclude from the above that
RA3B4 = � � ÆAB + � �AB ; (327)
or with respect to an arbitrary frame for Ss,
R (Ea; L; Eb; L) = � � ab + � �ab : (328)
Substituting this expression in the equation (319) for@�
ab
@s, the equation takes the form:
@�ab
@s= � 2 r= a�b + � ab + � �ab
+ 2 �a �b + � c
a�cb : (329)
The antisymmetric part of (329) is the equation:
curl= � = � � 1
2� ^ � : (330)
Here,
curl= � =1
2�ab (r= a�b �r= b�a)
and
� ^ � =1
2�ab (� c
a �cb� � c
b �ca) :
The symmetric part of (329) is the following propagation equation for �:
@�ab
@s= � r= a �b � r= b �a + � ab + 2 �a �b +
1
2( � c
a�cb + � c
b�ca ) (331)
The trace of (331) is the following propagation equation for tr�:
@tr�
@s= � 2 div= � + 2 j � j2 � � � � + 2 � : (332)
(Recall that @@str� = @
@s(( �1)ab�
ab) = �2�ab�
ab+ ( �1)ab
@�ab
@s.) The equation (332) will lose
one derivative tangentially to the sections Ss.
The trace-free part of (331) is the propagation equation for �:
(Note that @@s
denotes the trace-free part of @@s.)
@�ab
@s� 1
2tr� �
ab=
1
2tr� �ab � (r= �)ab + 2 �a �b � j � j2 ab ; (333)
where:(r= �)ab = r= a�b + r= b�a � ab div= � ; (334)
the trace-free part of the symmetrized covariant derivative. Note the absence of a curvatureterm on the right hand side of (333).
75
We must also consider the Gauss equation of the embedding of a surface (section of C) Ssin spacetime. Using the fact that the projection operator � to Ss is given by:
�X = X +1
2g (L; X) L +
1
2g (L; X) L (335)
for all X 2 TpM and p 2 Ss � C. We derive the equation, in a local orthonormal framefor Ss. First, denote by R= ABCD the curvature tensor of Ss and by RABCD the Ss-tangentialcomponents of the curvature of M . So, we then have
R= ABCD + �AC �BD
� �AD �BC
= RABCD : (336)
Now, we have:R= ABCD = K (ÆAC ÆBD � ÆAD ÆBC) ; (337)
where K is the Gauss curvature of Ss. Recall the equation (321):
RABCD = � � �AB �CD :
Moreover,�AB �CD = ÆAC ÆBD � ÆAD ÆBC :
Thus, the Gauss equation is equivalent to:
K +1
2tr� tr � � 1
2� � � = � � (338)
or,
K +1
4tr� tr� � 1
2� � � = � � : (339)
We now de�ne the mass aspect function � by:
� = K +1
4tr� tr� � div= � (340)
and its conjugate � by:
� = K +1
4tr� tr� + div= � : (341)
These are functions associated to a closed spacelike surface in spacetime (in particular onedi�eomorphic to S2). In terms of � the propagation equation for tr� reads:
@
@str� +
1
2tr� tr� = � 2 � + 2 j � j2 : (342)
If � is given, the equation de�ning � supplements the curl equation for � to a Hodge (div�curl)system:
curl= � = � � 1
2� ^ � (343)
div= � = � � � � +1
2� � � (344)
(substituting for K from the Gauss equation). This is an elliptic system on each section whichis coupled to a propagation equation for � along each generator.
76
Propagation Equation for �:
The propagation of K is according to:
@K
@s+ tr� K = div= div= � � 4= tr� : (345)
This is derived as follows:
2 K = ( �1)ab R= ab (the scalar curvature of Ss) (346)
and@R= ab
@s= r= c (
@�cab@s
) � r= a (@�ccb@s
) (347)
while
@�cab@s
= r= a �cb + r= b �
ca � r= c �ab (348)
@�ccb@s
= r= b tr� : (349)
We then obtain:
2@K
@s=
@ ( �1)ab
@sR= ab + ( �1)ab
@R= ab
@s
and the �rst term is
� 2 �ab R= ab = � 2 �ab ab K = � 2 tr� K
while the second term is
2 div= div= � � 2 4= tr� :
From all this we conclude (345), the propagation equation for K.
Now, we want to �nd also the propagation of div= �.
@
@sdiv= � =
@
@s(( �1)ab r= a �b)
= ( �1)ab f r= a (@�b@s
) � @�cab@s
�cg
+@
@s( �1)ab r= a �b
= div= (@�
@s) � 2 div= � � + d= tr� �
� 2 �ab r= a �b :
We now recall the propagation equation for �, namely (131):
@�a@s
+ tr� �a = div= �a � d= atr�
77
which results from eliminating the curvature � using the null Codazzi equation. Substituting,we obtain the following propagation equation for div= �:
@
@sdiv= � = � 2 tr� div= � � 2 div= � � �
� � � r= � + div= div= � � 4= tr� : (350)
We also calculate @@s(14 tr�tr�) from the propagation equations of tr� and tr�. Combining with
the above results (345) and (350) we obtain in this way the propagation equation for �:
@�
@s+
3
2tr� � = � 1
4tr� j � j2 +
1
2tr� j � j2
+ 2 div= � � � + � � r= � : (351)
5.2.1 The div�curl System for 1-Forms on a 2-Dimensional, Compact, Riemannian
Manifold
Let (M; gab) be a 2-dimensional, compact, Riemannian manifold. And let �a be a 1-form on M .We now consider the system
div � = � (352)
curl � = ! : (353)
Proposition 8 The system (equations (352) and (353)) is conformally covariant.
Proof: Setting ~gab = e2�gab, we havefdiv � = (~g�1)ab era�b
= e�2� (g�1)ab (ra�b � 4cab�c)
= e�2� (g�1)ab f ra�b � (Æ ca @b� + Æ c
b @a� � (g�1)cd gab @d�) �c g= e�2� f div � � (2 d� � � � 2 d� � �) g
that is, fdiv � = e�2� div � : (354)
And,
gcurl � = ~�ab era�b = ~�ab (d�)ab
= e�2� �ab (d�)ab
= e�2� curl � : (355)
Thus, with~� = e�2� � ; ~! = e�2� !
we have that
fdiv � = ~� (356)gcurl � = ~! : (357)
This proves the proposition.
78
The L2-Theory:
We have:
ra �b � rb �a| {z }= (d�)ab
= curl � �ab : (358)
Hence,
(curl �)2 =1
2curl � �ab � curl � �ab
=1
2(ra �b � rb �a) (ra �b � rb �a)
= j r� j2 � ra �b rb �
a : (359)
On the other hand,
ra �b rb �
a = rb (�a ra �
b) � �a rb ra �b
and
rb ra �b � ra rb �
b = Rbcba �
c = Rca �c = K �a ;
that is,
rb ra �b = ra (div �) + K �a :
Thus, we have
ra �b rb �
a = � K j � j2 + rb (�a ra �
b) � �a ra (div �) ; (360)
or,
ra �b rb �
a = � K j � j2 + (div �)2 + ra (�b rb �
a � �a div �) : (361)
Integrating (359) and (360) on M and adding yields:ZM
f j r� j2 + K j � j2 g d�g =
ZM
f (div �)2 + (curl �)2 g d�g : (362)
The conformal invariance of the div � curl system together with the uniformization theoremshows that this system is injective whenM is di�eomorphic to S2. The L2-adjoint of (div; curl)is the operator A acting on pairs of functions (u; v). A(u; v) is a 1-form � such that:Z
M
(u div � + v curl �) d�g =
ZM
�a �a d�g (363)
for every 1-form �. Now, ZM
u div � d�g = �ZM
�a @au d�g (364)
and ZM
v curl � d�g =
ZM
v �ba rb �a d�g
P:I:= �
ZM
�a �ba @bv d�g : (365)
79
Hence,
�a = � @au � @bv � �ba (366)
or
� = � du � �dv : (367)
The operator A is also conformally covariant. Thus its Kernel depends only on the conformalclass of (M; g). Taking M di�eomorphic to S2, the Kernel of A is that on the standard sphere.Setting
w = u + iv (368)
the equations expressing the condition that w lies in the Kernel are the Cauchy-Riemann equa-
tions@w
@�z= 0 (369)
on the complex plane (representing the standard sphere through the stereographic projection).The condition that w is bounded at in�nity (north pole) implies then that w is a complex con-stant (Liouville's theorem). Then u and v are constants and the div� curl system is integrableif and only if the source functions � and ! are L2-orthogonal to the constants, that is, they havevanishing mean.
5.2.2 Treatment of the Propagation Equation for �
This equation is of the form@�
@s+
3
2tr� � = g ; (370)
where g denotes the right hand side of equation (351), namely,
g = � 1
4tr� j � j2 +
1
2tr� j � j2 + 2 div= � � � + � � r= � :
We shall derive the following inequality:
j dds
k� � k� Lp(S) +3
2tr� k� � k� Lp(S) j �
3
2k tr� � tr� kL1(S) � k� � k� Lp(S) + k� g k� Lp(S) :
(371)We have: (in the following we write Lp for Lp(S))
d
dsk� � k� Lp =
d
ds(1
A
ZS
j � jp d� )1p
=1
pk� � k� 1�p
Lp f � 1
A2
dA
ds
ZS
j � jp d�
+1
A
ZS
(p j � jp�1 sgn �@�
@s+ j � jp tr�) d� g
=1
pk� � k� 1�p
Lp f � tr� k� � k� pLp
+1
A
ZS
[p j � jp�1 sgn � (� 3
2tr� � + g) + j � jp tr�] d� g :
80
We rewrite the last term as follows:
1
A
ZS
j � jp tr� d� = tr� k� � k� pLp +
1
A
ZS
j � jp (tr� � tr�) d� :
Also, for the preceeding term,
1
A
ZS
p j � jp�1 sgn � (� 3
2tr� �) d� = � 3p
2A
ZS
j � jp tr� d�
= � 3p
2f tr� k� � k� p
Lp +1
A
ZS
j � jp (tr� � tr�) d� g :
Substituting, we obtain:
d
dsk� � k� Lp +
3
2tr� k� � k� Lp = � (
3
2� 1
p) k� � k� 1�p
Lp � 1
A
ZS
j � jp (tr� � tr�) d�
+ k� � k� 1�pLp � 1
A
ZS
j � jp�1 sgn � g d� : (372)
And we estimate:
1
A
ZS
j � jp j tr� � tr� j d� � k tr� � tr� kL1 k� � k� pLp (373)
as well as
1
A
ZS
j � jp�1 j g j d� � 1
A(
ZS
j � jp d� )p�1p (
ZS
j g jp d� )1p (374)
= k� � k� 1�pLp k� g k� Lp ; (375)
where (374) is obtained by H�older's inequality. The ordinary di�erential inequality for k� � k� Lp
then follows.
We now estimate k� g k� Lp . We begin with the principal terms:
k� div= � � � k� Lp � C k � kL1 (k� r= � k� Lp + k� d= tr� k� Lp) ; (376)
k� � � r= �k� Lp � C k � kL1 k� r= � k� Lp ; (377)
where C is a numerical constant. Then, let us estimate the lower order terms:
k� tr� j � j2 k� Lp � k � k2L1 k� tr� k� Lp ; (378)
k� tr� j � j2 k� Lp � k tr� kL1 k � k2L1 : (379)
Thus, we obtain:
k� g k� Lp � C f k � kL1 (k� r= � k� Lp + k� d= tr� k� Lp)
+ k � kL1 k� r= � k� Lp + k � k2L1 k� tr� k� Lp
+ k tr� kL1 k � k2L1 g : (380)
81
Elliptic Estimates
We now go back to the null Codazzi equation: (recall (241) and (240))
div= � =1
2d= tr� + f (381)
and include in f the lower order terms involving �:
f = � � +1
2tr� � � � � � : (382)
We had derived the estimate: (see equation (279))
k� r= � k� Lp � �C f k� d= tr� k� Lp + k� f k� Lp g ; (383)
where �C depends on an upper bound for M , a lower bound for m and an upper bound for ~Kp.
At this point let us make the following convention: From now on we include r�1k� �k� Lp in thede�nition of k� r= �k� Lp , i.e. instead of k� r= �k� Lp + r�1k� �k� Lp we will just write k� r= �k� Lp .Similarly, we include r�1k� �k� Lp in the de�nition of k� r= �k� Lp . Writing tr� = tr�+(tr�� tr�),we estimate:
k� f k� Lp � k� � k� Lp +1
2tr� k � kL1
+ (1
2k tr� � tr� kL1 + k � kL1) k � kL1 ; (384)
provided thattr� � 0 : (385)
Recall from equation (299) that we had set
y =2
tr�; (386)
and we had obtained the ordinary di�erential equation (300) for y, i.e.
dy
ds= 1 � 1
2y2 h ;
where
h =1
2(tr� � tr�)2 � j � j2
as we see in equation (297). Recall also proposition 7, which shows that
y � 3
2y0 + s (387)
andr � r0 + � s ; (388)
where � is a �xed constant satisfying the inequality (308):
� <2r03y0
:
82
(We are assuming that y0 > 0.) The above inequalities were proved under the smallness condition
4Q21
�r0+ M <
�2y02 �C
: (389)
In fact, the same condition yields the lower bound:
y � 1
2y0 + s (390)
(which in particular shows that y stays positive.) The lower bound follows because the upperbound on y and the lower bound on r, which has been established, imply that
y
r� 1
�; (391)
hence,
1
2y2 j h j � 1
2�2r2 j h j
��C
�2(r�2 Q2 + b) (392)
and Z s
0
1
2y2 j h j ds �
�C
�2fZ s
0
4Q21
(r0 + �s0)2ds0 +
Z s
0b (s0) ds0 g
��C
�2(4Q2
1
�r0+ M) :
So, integrating the equationdy
ds= 1 � 1
2y2 h ;
or the inequality
j dyds
� 1 j � 1
2y2 j h j
from s = 0 we obtain:
j y � y0 � s j ��C
�2(4Q2
1
�r0+ M) <
1
2y0
by the smallness condition. Thus a continuity argument shows that
j y � y0 � s j � 1
2y0 (393)
or1
2y0 + s � y � 3
2y0 + s : (394)
We then integrate (from s = 0) the equation
d log r
ds=
1
y(395)
to obtain
log (12y0 + s
12y0
) � log (r
r0) � log (
32y0 + s
32y0
) (396)
83
or
r0 (1 +2s
y0) � r � r0 (1 +
2s
3y0) : (397)
In particular,
r � r0 + �s (398)
andy
r�
32y0 + s
r0 + �s� max f 3y0
2r0;1
�g =
1
�: (399)
Also,
r � r0 +2r0y0
s (400)
andr
y�
r0 +2r0y0s
12y0 + s
=2r0y0
: (401)
Remark that2r0y0
is initial data,
so, we can assume that it is bounded by a �xed numerical constant (as it is dimensionless).
We now substitute the bound (384) for k� f k� Lp in the elliptic estimate (279) after using theSobolev inequality to bound r�1 k tr� � tr� kL1 in terms of k� d= tr� k� Lp and r�1 k � kL1in terms of k� r= � k� Lp , i.e. we have used the Sobolev inequality on Ss (p > 2) to estimate
k tr� � tr� kL1 � �C r k� d= tr� k� Lp
k � kL1 � �C r k� r= � k� Lp :
This yields:
k� r= � k� Lp � �C f k� d= tr� k� Lp + k� � k� Lp
+ r�1 k � kL1 + r (k� d= tr� k� Lp + k� r= � k� Lp) k � kL1 g : (402)
Here, we have made use of the inequality
r
y� C (403)
to estimate1
2tr� by C r�1 :
We now introduce the bootstrap hypothesis
�C r k � kL1 � 1
2: (404)
This can be assumed to hold with a strict inequality at s = 0, that is for the initial data. Underthe assumption (404) we can eliminate the last term on the right hand side of (402) obtaining:
k� r= � k� Lp � �C f k� d= tr� k� Lp + k� � k� Lp + r�1 k � kL1 g : (405)
84
Next, we consider the (elliptic) (div= ; curl= ) system for �. We obtain the following estimate:
k� r= � k� Lp � �C f k� div= � k� Lp + k� curl= � k� Lp g (406)
� �C f k� � k� Lp + k� � k� Lp + k� � k� Lp
+ k � kL1 � k � kLp g : (407)
To proceed, we must have estimates for the quantities k� tr� k� Lp (which enters the estimate fork� g k� Lp , the right hand side of the propagation equation for �, see (380)) and k� � k� Lp (whichenters the above estimate for k� r= � k� Lp , see (407)). We thus introduce the following bootstrap
assumptions:r
2k� tr� k� Lp � C ; (408)
where C is a numerical constant (that can be large); and
r k� � k� Lp � Æ ; (409)
where Æ is a numerical constant which shall be chosen suitably small in the following. Weintroduce the notation
P = r3 k� � k� Lp (410)
in addition to (287), that is
Q = r3 k� d= tr� k� Lp :
Also,
a = r4 (k� � k� 2Lp + k� � k� 2
Lp) (411)
in addition to
b = r4 k� � k� 2Lp : (412)
In terms of this notation the elliptic estimates above take the form:
k� r= � k� Lp � �C f r�3 Q + r�2 b12 + k� r= � k� Lp g ; (413)
where we have substituted k � kL1� �Cr k� r= � k� Lp .
k� r= � k� Lp � �C f r�3 P + r�2 a12 + Æ k� r= � k� Lp g ; (414)
where we have substituted k � kL1� �Cr k� r= � k� Lp . We choose Æ such that
�C2 Æ � 1
2:
Then with
z = P + Q ; � = a + b
the above inequalities imply:
k� r= � k� Lp ; k� r= � k� Lp � �C f r�3 z + r�2 �12 g : (415)
We now substitute the above in the estimate of k� g k� Lp obtained previously, see (380). Tohandle the cubic term
k tr� kL1 k � k2L1 ; (416)
85
we also make, for uniformity of treatment, the basic bootstrap assumption:
r
2k tr� kL1 � C ; (417)
where C is a numerical constant (which can be large). We then obtain
k� g k� Lp � �C f r�5 z2 + r�3 � g (418)
(if we estimate the L1-norms of � and � in terms of the Lp-norms of r= � and r= �). Multiplyingthe ordinary di�erential inequality for k� � k� Lp by r
3 (see (371)) and substituting the estimatefor k� g k� Lp , i.e. (418), as well as the estimate:
k tr� � tr� kL1 � �C r k� d= tr�k� Lp (419)
= �C r�2 Q (420)
� �C r�2 z ; (421)
we obtain:dP
ds� �C f r�2 z2 + � g : (422)
Also, we go back to the ordinary di�erential inequality for k� d= tr� k� Lp . We multiply this byr3. (See (281) and (287).) In view of
(k tr� � tr� kL1 + k � kL1) k� d= tr� k� Lp + k � kL1 k� r= � k� Lp
� �C r f k� d= tr� k� 2Lp + k� r= � k� 2
Lp g� �C f r�5 z2 + r�3 � g ;
we obtain:dQ
ds� �C f r�2 z2 + � g : (423)
Adding the two ordinary di�erential inequalities (422) and (423), we obtain the following ordi-nary di�erential inequality for z:
dz
ds� �C f r�2 z2 + � g : (424)
It is formally identical to the one obtained earlier for Q, with � in the role of b. Therefore, bythe same argument, under the curvature hypothesis:Z s
0�(s0) ds0 � M ; (425)
where M is a constant independent of s; as well as the bootstrap assumption:
r � r0 + � s ; (426)
we can show that:z � 2 z1 ; (427)
where z1 = z0 + �CM ; provided that the following smallness condition holds:
z1 <�r02 �C
(428)
86
(on the initial data as well as the ambient spacetime curvature).
To close the argument, we must recover the bootstrap assumptions. That is, we are employingthe method of continuity: the bootstrap assumptions hold with strict inequalities at s = 0. Weconsider the maximal value s� of s such that these assumptions hold on [0; s�). If then s� is�nite, the inequalities must be saturated at s�. We proceed to show that this cannot happen.This yields the conclusion that s� is in�nite or as long as the spacetime curvature hypothesisholds.
We �rst revisit the argument of proposition 7. Recall (297):
h =1
2(tr� � tr�)2 � j � j2 :
We have that, on [0; s�]:
j h j � 1
2k tr� � tr� k2L1 + k � k2L1
� �C r2 (k� d= tr� k� 2Lp + k� r= � k� 2
Lp)
� �C (r�4 z2 + r�2 �) ; (429)
and we have the ordinary di�erential inequality:
dy
ds� 1 � 1
2j h j y2
� �C (y
r)2 (r�2 z2 + �) : (430)
Thus, choosing � such that
� � 2r03y0
; (431)
we have, on [0; s�]:
y
r�
32y0 + s
r0 + �s� max f 3y0
2r0;1
�g =
1
�: (432)
Hence, since z � 2z1 on [0; s�],
j dyds
� 1 j ��C
�2(
4z21(r0 + �s)2
+ �) (433)
holds on [0; s�]. Therefore, integrating from s = 0, we obtain
j y � y0 � s j ��C
�2(4z21�r0
+ M) (434)
<1
2y0 : holds for all s 2 [0; s�] ; (435)
in particular at s = s�, showing that the inequalities concerning y are not saturated at s = s�.Since
1
2y0 + s � y � 3
2y0 + s 8 s 2 [0; s�]
integratingd log r
ds=
1
y
87
yields
r0 (1 +2s
3y0) � r � r0 (1 +
2s
y0) 8 s 2 [0; s�] : (436)
So, if we �x � to be any positive, real number strictly less than 2r03y0
, we conclude that theinequality (426):
r � r0 + � s
cannot be saturated at s = s�.
Next, we consider the bootstrap assumptions (417) and (404). For (417) we write
k tr� kL1 =2
y+ j tr� � tr� jL1 :
So, it is
r
2k tr� kL1 =
r
y+
r
2k tr� � tr� kL1
� 2r0y0
+�C
2r2 k� d= tr� k� Lp
� 2r0y0
+ �Cz1r0
(437)
< C ; preassigned (numerical) constant. (438)
Thus, under this smallness condition, the inequality of (417) cannot be saturated at s = s�. For(404),
r k � kL1 � �C r2 k� r= � k� Lp elliptic estimate (439)
� �C f r�1 z + �12 g
� �C f 2 z1 r�10 + �12 g : (440)
To show that the inequality (404), namely,
�C r k � kL1 � 1
2
is not saturated at s = s�, we assume that
2z1r0
+ sups
�12 <
1
2 �C2; (441)
so that in addition to the hypothesis (425), that is,Z s
0� (s0) ds0 � M ;
(suitably small), we must also make the hypothesis that
sups
�12 is suitably small. (442)
88
These are the hypotheses on the ambient spacetime curvature.
We shall show in the next chapter (Sobolev Inequalities on C) that the hypothesisR s0 f
RSs0
[r2 (j � j2 + j � j2 + j � j2)+ r4 (j r= � j2 + j r= � j2 + j r= � j2)+ r4 (j @�
@sj2 + j @�
@sj2 + j @�
@sj2) ] d� s0 g ds0
� N
(443)
implies both of the above hypotheses (425) and (442) on � for p = 4.
Finally, we recover the bootstrap assumptions (408) and (409).
First, consider (408): Recall the propagation equation for tr� (see (342)):
@
@str� +
1
2tr� tr� = � 2 � + 2 j � j2 :
By the Gauss equation (see (338) and (339)) it is:
� + � = 2 K +1
2tr� tr� = � 2 � + � � � : (444)
We can thus express � in terms of �. So, setting
l := � + 2 � � � � � + j � j2 (445)
we have� � + j � j2 = l
and we obtain the ordinary di�erential inequality
j dds
k� tr� k� Lp +1
2tr� k� tr� k� Lp j
� 1
2k� tr� � tr� k� L1 � k� tr� k� Lp + 2 k� l k� Lp : (446)
We then setX =
r
2k� tr� k� Lp (447)
and we obtaindX
ds� 1
2k tr� � tr� kL1 X + r k� l k� Lp : (448)
Again, let the assumptions (408) and (409) hold at s = 0 with C and Æ replaced by C2 and Æ
2 ,also let s� be the maximal value of s such that the assumptions (408) and (409) hold on [0; s�).Using what we know in the interval [0; s�), we deduce:
k� l k� Lp � �C r�2 (r�1 z + �12 )
� �C r�2 (r�1 2 z1 + �12 ) (449)
and:k tr� � tr� kL1 � �C r�2 z � �C r�2 2 z1 : (450)
89
Integrating the ordinary di�erential inequality (448) for X on [0; s�] then yields:
X (s�) � X (0) + �C r�10 z1 + �C
Z s�
0r�1 �
12 ds
� X (0) + �C r�10 z1 + �C (
Z s�
0r�2 ds)
12 (
Z s�
0� ds)
12
� X (0) + �C r�10 z1 + �C (��1 r�10 )12 M
12| {z }
� (��1 r�10z1�C)12
(recalling that z0 + �CM = z1). Since X(0) � C2 , we then obtain:
X (s�) � C
2+ �C (r�10 z1)
12 (451)
< C : if r�10 z1 is suitably small, (452)
contradicting the maximality of s� in regard to the assumption (408).
Next, we handle (409): Recall the propagation equation for � (see (333)):
@�ab
@s� 1
2tr� �
ab= mab ; (453)
where
m =1
2tr� � � r= � + 2 � � � j � j2 : (454)
It follows that
j @@s
k� � k� Lp +1
2tr� k� � k� Lp j
� C (k tr� � tr� kL1 k� � k� Lp + k� m k� Lp) (455)
with C a numerical constant. We set
Y = r k� � k� Lp : (456)
We then obtain the ordinary di�erential inequality:
dY
ds� C (k tr� � tr� kL1 Y + r k� m k� Lp) : (457)
Using what we know in the interval [0; s�), we can estimate:
k� m k� Lp � �C r�2 (r�1 z + �12 ) (458)
� �C r�2 (r�1 2 z1 + �12 ) : (459)
Integrating the ordinary di�erential inequality (457) for Y on [0; s�], then yields:
Y (s�) � Y (0)| {z }� Æ
2
+ �C r�10 z1 Æ + �C (r�10 z1)12 (460)
< Æ (461)
90
if�C r�10 z1 Æ + �C (r�10 z1)
12 <
Æ
2; (462)
which reduces to the condition that
r�10 z1 is suitably small. (463)
We conclude that the inequality of (409) cannot be saturated at s = s� either.
We have thus proved the following theorem:
Theorem 4 Suppose that there is a constant B independent of t such that for all s 2 [0; t] thefunction (s) associated to the two-manifold (Ss; (s)) (which is di�eomorphic to S2) by the
uniformization theorem satis�es
M (s) � B ; m (s) � B�1 ; eKp (s) � B ; (464)
where (s), M , m and eKp are de�ned as in (248), (255), (259) and (268) respectively. Let
r�10
Z t
0� (s) ds (465)
with � = a+ b and a, b given in (411), (412) respectively; and
sups2[0;t]
�12 (s) (466)
be bounded by suitably small constants depending only on B. (For r, r0, see (290).) Also, let
z1 r�10 <
�
2 �C; (467)
where �C is a constant depending only on B, and � is a �xed constant with
� <2
3
r0y0
: (468)
(As z1 = z0 + �CM , one could require z0r�10 to be likewise bounded instead of (467).) Assume
that y0 > 0. Further, let r0y0
be bounded by a �xed numerical constant. Let also the following
bootstrap assumptions hold at s = 0 with strict inequalities:
�C r k � kL1 � 1
2(469)
r
2k� tr� k� Lp � C (470)
r k� � k� Lp � Æ (471)r
2k tr� kL1 � C (472)
with C a numerical constant, and Æ a numerical constant which we choose such that �C2 Æ � 12 .
Then equations (469) to (472) hold for all s and with z = P + Q the following inequality
holds: (for P , Q, see (410), (287) respectively)
z � 2 z1 ; (473)
i.e.
r�10 z � 2 r�10 z1 : (474)
91
6 Sobolev Inequalities on C
Here, we shall discuss Sobolev inequalities on C.
Proposition 9 Let f be a p-covariant tensor�eld on C tangential to each surface S of a folia-
tion of C into sections such that
M :=
Z s�
0f k� f(s) k� 2
L2 + r2 (k� r= f(s) k� 2L2 + k� Df k� 2
L2) g ds < 1 ; (475)
where Lp = Lp(S). Further r= denotes the S-tangential derivative whereas D is @@s
(of the Jacobi
�eld frame components).
Then we have: Z s�
0k� f (s) k� 2
L4 ds � �C M ; (476)
sups2[0;s�]
r(s) k� f(s) k� 2L4 � �C ( �C�1D + M) : (477)
Also, it is: Z s�
0r2 (s) k� f (s) k� 6
L6 ds � �C (D2 + M2) M : (478)
Here, D stands for the initial datum:
D = r(0) k� f(0) k� 2L4 ; (479)
which refers to S0, the initial section. Also, �C is a constant depending on an upper bound for
the isoperimetric constants of the surfaces S and on an upper bound for r tr�. It is assumed
that tr� � 0.
92
Figure 24.
S0
Ss
Ss�
�0
Proof: Consider an arbitrary section Ss = S. We apply the isoperimetric inequality of S:ZS
(�2 � ��2) d� � CI (
ZS
j r= � j d� )2 : (480)
Here, � is a function on S and CI is de�ned as follows:
CI = supU
� Area U
Perimeter @U
�;
where U is the smaller of the two domains bounding the closed curve � = @U .
93
Figure 25.
�
�1 �2
U
S is di�eomorphic to S2 S is not di�eomorphic to S2
� = �1 [ �2
The supremum is over curves � which are homologous to 0.
We �rst apply (480) to the case � =j f j2. Thenr= � = 2 f � r= f
and ZS
j r= � j d� � 2
ZS
j f j j r= f j d� (481)
� 2� Z
S
j f j2 d� �12� Z
S
j r= f j2 d� � 12 (482)
by the Schwarz inequality. Write
�2= (j f j2)2 = (
1
A
ZS
j f j2 d� )2 : (483)
From (480) we then have:� ZS
j f j4 d� �12 � 1
A12
� ZS
j f j2 d� �12 � � Z
S
j f j2 d� + 4 A CI
ZS
j r= f j2 d� 12 :
(484)
94
Hence, Z s�
0k� f (s) k� 2
L4 ds =
Z s�
0
1
A12
� ZS
j f j4 d� � 12 ds
� � Z s�
0
1
A
� ZS
j f j2 d� �ds 12
� � Z s�
0
1
A
� ZS
j f j2 d� + 4 A �CI
ZS
j r= f j2 d� �ds 12 (485)
by the Schwarz inequality. This yields the �rst statement of the proposition.
We now apply (480) to the case � =j f j3. Thus it isr= � = 3 j f j f � r= f :
Now, we have ZS
j r= � j d� � 3
ZS
j f j2 j r= f j d� (486)
� 3� Z
S
j f j4 d� �12� Z
S
j r= f j2 d� � 12 (487)
by the Schwarz inequality. Write
��2 = (j f j3)2 =� 1A
ZS
j f j3 d� �2
(488)
� 1
A2
� ZS
j f j4 d� � � � Z
S
j f j2 d� �
(489)
by the Schwarz inequality. From (480) we have:ZS
�2 d� �ZS
��2 d� + CI
� ZS
j r= � j d� �2
: (490)
Thus, taking � =j f j3, we obtain the following important result:ZS
j f j6 d� � � ZS
j f j4 d� � � � 1
A
ZS
j f j2 d� + 9 CI
ZS
j r= f j2 d� : (491)
This is true for each s. Replace ZS
j f j4 d�
by
sups2[0;s�]
fZSs
j f j4 d� g := y :
We have:y = sup
s2[0;s�]f A(s) k� f k� 4
L4 g : (492)
We then integrate with respect to s on [0; s�]. Set also
x =
Z s�
0
� ZSs
j f j6 d� ds
=:
ZCs�
f6 : (493)
95
Thus we obtain:
x � y �Z s�
0
� 1
A(s)
ZSs
j f j2 d� + 9 �CI
ZSs
j r= f j2 d� ds (494)
with�CI = sup
s2[0;s�]CI (s) : (495)
Now, we look at the portion Cs of C bounded by S0 and Ss.
Figure 26. S0
Ss
So, one has: ZSs
j f j4 d� �ZS0
j f j4 d� =
Z s
0
@
@s0� Z
Ss0
j f j4 d� �ds0 (496)
and
@
@s
� ZSs
j f j4 d� �
=
ZSs
� @
@sj f j4 + j f j4 tr�
d� (497)
= 4
ZSs
j f j2 f � @f
@sd� +
ZSs
j f j4 tr� d� : (498)
Thus setting
d :=
ZS0
j f j4 d� ; (499)
96
we obtain
y � d +
Z s�
0j @@s
� ZSs
j f j4 d� � j ds
� d + 4
Z s�
0
ZSs
j f j3 j @f@s
j d� ds +
Z s�
0
ZSs
j f j4 tr� d� ds ; (500)
assuming tr� � 0. Now, by the Schwarz inequality, it isZ s�
0
ZSs
j f j3 j @f@s
j d� ds � � Z s�
0
ZSs
j f j6 d� ds| {z }x
�12� Z s�
0
ZSs
j @f@s
j2 d� ds� 12 :
(501)We then use an upper bound for
r
2tr� ;
sayr
2tr� � k ; (502)
i.e.
tr� � 2k
r= 2 (4�)
12
k
A12
: (503)
Then, we have:Z s�
0
ZSs
j f j4 tr� d� ds � 2 (4�)12 k
Z s�
0
1
A12
(
ZSs
j f j4 d� ) ds : (504)
Again by the Schwarz inequality, it isZSs
j f j4 d� � � ZSs
j f j6 d� � 12 � � Z
Ss
j f j2 d� � 12 ;
and Z s�
0
1
A12
� ZSs
j f j4 d� �ds �
Z s�
0
� ZSs
j f j6 d� � 12 � � 1
A
ZSs
j f j2 d� � 12 ds
� � Z s�
0
ZSs
j f j6 d� ds| {z }x
� 12 � � Z s�
0
1
A
ZSs
j f j2 d� ds j ) 12 : (505)
We substitute (505) in (504) and the result in the inequality (500) to obtain
y � d + 4 x12
� Z s�
0
ZSs
j @f@s
j2 d� ds� 12 +
2 (4�)12 k x
12
� Z s�
0
1
A
ZSs
j f j2 d� ds�12 : (506)
Now, set
q =
Z s�
0
� 1
A
ZSs
j f j2 d� +
ZSs
j @f@s
j2 d� +
ZSs
j r= f j2 d� ds : (507)
Also, let�k = max f 4; 2 (4�)
12 k; 9 �CI g : (508)
97
The inequality (494) implies:x � �k y q : (509)
And inequality (506) implies:
y � d + �k x12 q
12 : (510)
Next, substitute (509) in (510):
y � d + �k32 y
12 q
� d +1
2y +
1
2�k3 q2 : (511)
So, we obtainy � 2 d + �k3 q2 : (512)
This proves the proposition.
Also, substituting (510) in (509) yields
x � �k (2 d + �k3 q2) q : (513)
98
6.1 Applications
Recall that to estimate
r3 k� d= tr� k� L4 ; r3 k� � k� L4
we needed
r4 (k� � k� 2L4 + k� � k� 2
L4 + k� � k� 2L4) = �
to satisfy: Z s�
0�(s) ds � M : independent of s� (514)
and
sups2[0;s�]
�(s) � L : independent of s� : (515)
We de�ne (compare with page 107 of Part I)
P0 = � Q (R) (� ; �K; T; T ) (516)
P1 = � Q (LO R) (� ; �K; T; T ) � Q (LS R) (� ; �K; T; T )
� Q (LT R) (� ; �K; �K; T ) (517)
P = P0 + P1 (518)
E1 = supt
ZHt
�P (519)
E2 = supu
ZCu
�P : (520)
We consider only the �rst term in P1 and only the quantity E2.ZCu
�P = �Z s�
0
ZSs
P (L) d� ds : (521)
Also, recall that
E0 = T =1
2(E3 + E4) :
So, it is
� P0 (L) = �2+ Q4400 + �2� Q4300
=1
4�2+ (Q4444 + Q4443 + Q4433)
� C�1 �2+ f j � j2 + j � j2 + (�2 + �2) g (522)
and
�+ � C�1 r :
We obtain, in particular,Z s�
0
ZSs
r2 (j � j2 + �2 + �2) d� ds � C E2 : (523)
Similarly, since for any p-covariant tensor�eld f on S,
j LO f j � C�1 r j r= f j ;
99
we obtain also, Z s�
0
ZSs
r4 (j r= � j2 + j r= � j2 + j r= � j2) d� ds � C E2 : (524)
Next, to obtainD � ; D � ; D �
we consider the Bianchi identities, proposition 7.3.2 on page 161 of the book 'The Global Non-linear Stability of the Minkowski Space' ([15]).
�4 = div= � + l:o:t: (525)
�4 = div= � + l:o:t: (526)
�4 = � curl= � + l:o:t: (527)
Then we can estimateZ s�
0
ZSs
r4 j D � j2 d� ds � C
Z s�
0
ZSs
r4 j r= � j2 d� ds � C E2 (528)
andZ s�
0
ZSs
r4 (j D� j2 + j D� j2) d� ds � C
Z s�
0
ZSs
r4 j r= � j2 d� ds � C E2 : (529)
We apply proposition 9 (Sobolev inequality) to the cases:
f = r2 � ; r2 � ; r2 � ;
to obtain
sups2[0;s�]
ZSs
j f j4 d � � C E22 : (530)
So, it is
�2 � C r6ZSs
(�4 + �4 + j � j4) d�
= C r�2ZSs
f4 d�
� C E22 r
�2 ;
i.e.� � C E2 r
�1 ; (531)
which a fortiori yields the estimate (515). The estimate (514) follows from the �rst part of theproposition.
In fact, applying the �rst part of proposition 9, we obtain:Z s�
0� (s) ds � C E2 : (532)
100
6.2 Estimates for m, M if K 2 L4(S)
Chapter on estimating m and M for K 2 L4(S).
101
7 Smoothing Foliation
Chapter on the smoothing foliation(� = � ).
102
8 Memory E�ect
In this chapter we study the so-called memory e�ect of gravitational waves. (Also see [12].) Weshall show that it is due to the nonlinear character of the asymptotic gravitational laws at futurenull in�nity. These results relate the theory to concrete experiments planned for the near future.
Gravitational waves from astronomical sources are shown to have a nonlinear e�ect on laserinterferometer detectors. The nonlinearity of Einstein's equations manifests itself in a perma-nent displacement of the test masses of a laser interferometer detector after the passage of agravitational wave train. This is called the 'memory' of the gravitational-wave burst (see [7], [8]and [35]).
First, let us give some de�nitions and explanations needed in this context.
Figure 27.
�0
S0
C+0
source
p
We shall use the following notations.
103
�0 : maximal spacelike hypersurface of vanishing linear momentump : event of observation at the earthJ�(p) : the causal past of pS0 : spherical spacelike surface surrounding the source in a neighbourhood
of the intersection of the source with @J�(p)J+(S0) : the causal future of S0C+0 : the outer component of @J+(S0).
Figure 28.
�0
S0
p
B0
C+0
B�
C��
C+0
@J�(p)
J�(p) \ �0
We denote by B0 the interior of S0 in �0. For each d > 0 let
Bd = f z 2 �0 : dist (z; B0) < d g : (533)
We then de�ne B� = Bd� to be the smallest region Bd containing J�(p)\�0. Finally, we de�ne
C�� to be the boundary of the domain of dependence of B�. Equivalently, C�
� = @J+(�0nB�).
104
We state the basic hypothesis on S0: it is chosen such that the generators of C+0 have no future
end points.
Figure 29. S
L
L
Let S be a spacelike surface di�eomorphic to S2. We have the two future-directed null normalsL and L to S with g(L;L) = �2. They are subject to transformations of the form:
L 7! a L ; L 7! a�1 L ; (534)
where a is a positive function on S. We have the corresponding null second fundamental forms� and �, which transform according to
� 7! a � ; � 7! a�1 � : (535)
We also have the torsion �, which transforms according to:
� 7! � � d= log a : (536)
Finally, we have the mass aspect functions � and �,
� = K +1
4tr� tr� � div= � (537)
� = K +1
4tr� tr� + div= � ; (538)
which transform as:
� 7! � + 4= log a (539)
� 7! � � 4= log a : (540)
Consider now the case that tr� and tr� have opposite signs: tr� > 0, tr� < 0. Then, we can�x L (and L) by requiring that
tr� + tr� = 0 : (541)
105
Then,
T =1
2(L + L) : (542)
This will be a unit timelike vector�eld along S, such that if we consider any in�nitesimal virtualdisplacement of S along the corresponding congruence of normal timelike geodesics, then thearea element of S remains pointwise unchanged.
Figure 30. S
St
T
T T
T
At
d
dtAt jt=0 = 0 : (543)
We call this T the binormal of S.
106
Figure 31.
B�
S0
�0
C+0
C+u
u < 0C+u
u < 0
C+0
C�� C�
�
S�u, u < 0
S�0
S�u, u > 0
LL
C+u
u > 0C+u
u > 0
107
Let
S�0 = C+0 \ C�
� : (544)
We normalize L and L along S�0 by the above condition. I.e. the condition, that (542) is thebinormal. We shall extend the normalization to the spacetime manifold as follows: First, Lis extended along C�
� to be the null geodesic vector�eld whose integral curves generate C��
and coincides with the given null normal vector�eld along S�0 . Let us consider on C�� the level
surfaces of the aÆne distance from S�0 . We label these surfaces S� (not by the value of the aÆnedistance but rather than) by the value of the 'retarded time' u:
u = 2 (r�0 � r�) : (545)
For any surface S, we denote by r the 'area radius' of S:
r (S) =
rArea (S)
4�: (546)
Here, r� = r(S�) and r�0 = r(S�0). (u > 0 on C�� to the future of S�0 and u < 0 to the past of
S�0 .) Now, L extends to C�� to be the unique conjugate null normal to each S�u (conjugate to L).
The outgoing null hypersurfaces C+u are de�ned to be the inner components of the boundaries
of the causal pasts of the S�u, (the outer component being part of C�� ). L then extends to
D+(B�), the (future) domain of dependence of B�, to be the null geodesic �eld, whose integralcurves generate each Cu and coincides with the transversal null �eld just constructed on C�
� .Next, L also extends to D+(B�) to be the unique conjugate null normal to the surfaces Su;s,sections of C+
u , of constant aÆne distance from S�u. (We may de�ne s to be equal to r�u on S�u,so s = r�u � aÆne distance (S; S�u) on S = Su;s. In other words, s is de�ned on each C+
u by theconditions: s jS�u= r�u, Ls = 1.)
Now, we choose a di�eomorphism �0 of S2 (the unit sphere in R3) onto S�0 . See �gure 33.We then de�ne �u, a di�eomorphism of S2 onto S�u by the condition that
�u Æ ��10 : S�0 ! S�u (547)
is the di�eomorphism of S�0 onto S�u de�ned by the generators of C�� . We can also de�ne �u;s,
a di�eomorphism of S2 onto Su;s, by the condition that
�u;s Æ ��1u : S�u ! Su;s (548)
is the di�eomorphism of S�u onto Su;s de�ned by the generators of C+u .
108
Figure 32.
C+u C+
u
Su;s
S�u, u < 0
Let w be a p-covariant tensor�eld in spacetime. (We only consider such w which vanish whenone of the entries is either L or L, so we can think of w as 'living' on each Su;s.) The pullback��u;sw is a p-covariant tensor�eld on S2 depending on the parameters u and s. Let us denote byDw and Dw the projections to each Su;s of LLw and LLw respectively.
D w (X1; � � � ; Xp) = LLw (�X1; � � � ; �Xp) (549)
D w (X1; � � � ; Xp) = LLw (�X1; � � � ; �Xp) : (550)
Then we have:@
@s(��u;s w) = ��u;s (Dw) : (551)
For C�� we consider ��uw: This is a 1-parameter family of p-covariant tensor�elds on S2. Then
we have
f@
@u(��uw) = ��u (Dw) ; (552)
where the factor f is
f = L u (= � 2dr�
ds) if s is the parameter of L : (553)
The whole discussion which follows is based on the fact that the radius of S0 is negligible incomparison to that of S�0 . In other words, the radius of the source is negligible in comparisonto d�, the distance of the earth from the source. So, we think of d� as tending to in�nity, such
109
that B� exhausts �0. Then S�0 moves to the in�nite future along C+
0 and its radius r�0 tends toin�nity. We require that, as this happens, the image by �0 of any �xed point on S2 traces agenerator of C+
0 .
Figure 33.
S�0
0
S�00
0
S2
�00
0
�0
0
110
For each �xed u, we de�ne the rescaled metric ~ by:
e = r�2 (554)
with being the induced metric on S�u. It is
��u e ! Æ (555)
and��u ~K ! 1 =
ÆK : Gauss curvature of
Æ ; (556)
where ~K is the Gauss curvature of ~ which is equal to r2K, K being the Gauss curvature of .
So, (S2;Æ ) is isometric to the standard sphere. Also,
��u (r tr�) ! 2 (557)
��u (r tr�) ! � 2 (558)
as r�0 !1 (therefore also r�u !1). Moreover, de�ning the function
h = r tr� � 2 ; (559)
then for each �xed u, it is
��u (rh) ! H : a function on S2 : (560)
(Recall k� r3d= tr� k� Lp was shown to be bounded. This implies that k r2(tr� � tr�) kL1is bounded. The argument given in fact shows that r2(tr� � tr�) converges to a limit ass ! 1.) Also, ��u� tends for each �xed u to a limit � as r�0 (hence also r�u) tends to in�n-
ity. � is a symmetric 2-covariant tensor�eld on S2 which is trace-free relative toÆ . (j � j~ =p
(e �1)ac(e �1)bc�ab�cd =pr2( �1)acr2( �1)bd�ab�cd = r2 j � j , while j � j = O(r�2).) So,
j � j~ = O(1). Next,��u (r�1 �) (561)
tends for each �xed u to a limit � as r�0 (hence also r�u) tends to in�nity. (j � j~ = O(r), forj � j = O(r�1).) Also, for �xed u:
��u (r�) ! Z (562)
as r�0 !1 (r�u !1). Z is a 1-form on S2. (j � j~ =p(e �1)ab�a�b =pr2( �1)ab�a�b = r j � j ,
while j � j = O(r�2), so j � j~ = O(r�1).) Finally, the functions
��u (r3 �) ; ��u (r3 �) (563)
tend to limits N and N respectively as r�0 !1. Recall from volume I thatZS
� d� =
ZS
� d� =8�m
r; (564)
where m is the Hawking mass of S:
m =r
2f 1 +
1
16�
ZS
tr� tr� d� g : (565)
It follows that for �xed u:
m ! M : the Bondi mass (566)
111
as r�0 !1 (r�u !1). All the above limits H; �; �; Z; N; N and
M =1
8�
ZS2
N d�Æ =
1
8�
ZS2
N d�Æ
(567)
depend on u.
Consider now a spacelike surface S di�eomorphic to S2 with its two future-directed null normalsL, L. Consider the following components of curvature:
�: 2-covariant, symmetric, trace-free tensor�eld on S. It is:
� (X; Y ) = R (X; L; Y; L) ; (568)
where X and Y are tangential to S at a point.
�: 1-form on S. We have:
� =1
2R (X; L; L; L) ; (569)
where X is tangential to S at a point.
We also consider the components �, � de�ned in Chapter 5:
R (X; Y; Z; W ) = � � � (X; Y ) � (Z; W ) (570)
1
2R (X; Y; L; L) = � � (X; Y ) ; (571)
where X, Y , Z, W are tangential to S at a point.
Proposition 10 For a �xed u and as r�0 !1, we have the following limits:
��u (r�1 �) ! A��u (r �) ! B
��u (r3 �) ! P��u (r3 �) ! Q :
Sketch of the Proof: We consider the Bianchi identities
r[� R� ]� = 0 : (572)
And withR�� = 0
it isr� R�� Æ = 0 :
Certain components of these are the following equations: (Note that D and D refer to thetrace-free parts.)
D � � 1
2tr� � = r= � ; (573)
112
which is the adjoint of the div= operator on trace-free, 2-covariant, symmetric tensor�elds on a2-dimensional manifold.
D � +1
2tr� � = d= � + �d= � ; (574)
which is the adjoint of the (div= ; curl= ) operator on 1-forms on a two-dimensional manifold.
D � +3
2tr� � = div= � (575)
D � +3
2tr� � = curl= � : (576)
There are additional lower order terms on the right hand sides of each of the above four equations,decaying faster than the principal terms as s!1. We have:
D (r�1 �) � 1
2(tr� � tr�) r�1 � = r�1 r= � : (577)
Note that Dr = r2 tr� and r
2 is bounded above and below by positive constants. It is:
tr� � tr� = O (r�2) (578)
r�1 r= � = O (r�2) : (579)
Remark: ��+�52� , ��
2+�
32� , (�; �)�
3+�
12� , ��
72+ , ��
72+ are uniformly bounded by Sobolev inequali-
ties and the uniform bound on the controlling quantity, the stronger controlling quantity, (seeChapter 6 and volume I). Here, it is
�� =p1 + u2 (580)
�+ =p1 + u2 ; where u = u + 2 r : (581)
Next, we consider D(r�) and the equation:
D (r �) +1
2(tr� � tr�) r � = r (d= � + �d= �) = O (r�2) : (582)
Also, it is:
D (r3 �; r3 �) +3
2(tr� � tr�) (r3 �; r3 �) = r3 (div= �; curl= �) = O (r�
32 ) : (583)
Since ��u;s (D (r�1 �)) = @@s
(��u;s (r�1 �)), ��u;s (D (r �)) = @
@s(��u;s (r �)), �
�u;s (D (r3 (�; �)) =
@@s
(��u;s (r3 (�; �)), we conclude that: ��u;s (r
�1 �), ��u;s (r �), ��u;s (r
3 (�; �)) satisfy linearordinary di�erential equations involving an integrable exponent, (tr� � tr�), as well as inte-grable right hand sides. It follows that the limits exist as s!1 and moreover we have boundson the di�erence at given s from the limits.
113
8.1 Limiting Equations on C�� (as B� exhausts �0 or r
�0 !1)
First, let us state the null Codazzi equation and its conjugate as follows:
div= � + � � � =1
2(d= tr� + tr� �) � � (584)
div= � � � � � =1
2(d= tr� � tr� �) + � : (585)
Let w be a p-covariant tensor�eld on the surfaces Su;s and let
j w j = O (r�q) : (586)
The limit which we want to investigate is then
limr�u!1
��u (r�p+q w) : (587)
This will be a p-covariant tensor�eld on S2, depending on u. For the equation (584) we havej � j = O(r�2), j div= � j = O(r�3) and div= � is a 1-form. So, here p = 1 and q = 3. We thusmultiply (584) by r2 and take the limit r�u !1. Recalling that with
h = r tr� � 2 ; (588)
we have
limr�u!1
��u (rh) = H ; (589)
and also recalling that
limr�u!1
��u (�) = � (590)
and
limr�u!1
��u (r�) = Z (591)
we obtain:Æ
div= � =1
2
Æ
r= H + Z ; (592)
whereÆ
r= denotes the covariant derivative with respect toÆ , the standard metric on S2. For the
equation (585) we have j � j = O(r�1), j div= � j = O(r�2) and div= � is a 1-form. So, it is p = 1and q = 2. We multiply therefore equation (585) by r, and take the limit r�u ! 1. Recallingthat:
limr�u!1
��u (r�1�) = � ; (593)
we obtain:Æ
div= � = B : (594)
Now, consider the Hodge system for the torison:
curl= � = � � 1
2� ^ � (595)
div= � = � + � � 1
2� � � : (596)
114
(Recall that L is the generating �eld for C�� .) Here, it is
j � j = O (r�2) (597)
(curl= �; div= �) = O (r�3) ; (598)
and these are scalars. We thus multiply both equations by r3 and take the limit to obtain:
Æ
curl= Z = Q � 1
2� ^ � (599)
Æ
div= Z = N + P � 1
2� � � : (600)
The above are the limits of the elliptic equations.
We now consider the limits of the propagation equations. For tr� we have:
D tr� +1
2tr� tr� = � 2 � + 2 j � j2 : (601)
We consider the propagation equation for rh, where
h = r tr� � 2 :
But it is:D r =
r
2tr� ! � 1 :
So, for D(rh) one has:
D (r h) = r D h + h D r
= r2 D tr� + r tr� D r + h D r
= r2 D tr� + 2 (1 + h) D r : (602)
Substituting from (601), we �nd
D (r h) = � 1
2r2 tr� tr� + 2 r2 (� � + j � j2) + 2 (1 + h) D r
= � r
2tr� (2 + h) + 2 (1 + h) D r + 2 r2 (�� + j � j2)
! 1 (2 + h) + 2 (1 + h) (�1) + 0 = � h ! 0 : (603)
On the other hand recall that: (see (552))
f@
@u(��u w) = ��u (D w)
withf = L u :
That is:f = D u ; u = 2 (r�0 � r�) on C�
� ; (604)
so, it is:f = � 2 D r ! 2 as r�0 ! 1 : (605)
115
Therefore, we obtain simply@H
@u= 0 : (606)
Next, we consider the propagation equation of � (along C�� ):
D � � 1
2tr� � � = � 1
2tr� � + r= � + 2 � � � j � j2 : (607)
Here, one has p = 2 and q = 2. We just pull back to S2 by �u and take the limit r�u ! 1 toobtain simply:
2@�
@u= � � : (608)
The propagation equation for � (along C�� ) is:
D � = � � : (609)
Here, it is p = 2 and q = 1. So, we multiply by r�1, pull back by �u and take the limit r�u !1to obtain:
2@�
@u= � A : (610)
Consider also the propagation equation for tr�:
D tr� +1
2(tr�)2 = � j � j2 (611)
r tr� ! � 2 (612)
tr� = � 2
r+ O (r�2) : (613)
Setting
h = r tr� + 2 ;
we obtain
r h ! H
and an equation for @H@u
:
2@H
@u= � j � j2 :
Finally, we consider the propagation equation for �:
D � +3
2tr� � = � 1
4tr� j � j2 +
1
2tr� j � j2
� 2 div= (� � �) � r= tr� � � : (614)
We multiply by r3, pull back by �u, and then take the limit r�u !1 to obtain:
2@N
@u= � 1
2j � j2 ; (615)
where N is monotonously decreasing. Since
N = N = 2 M
116
and M being the Bondi mass, the integral of (615) on S2 yields:
@M
@u= � 1
32�
ZS2
j � j2 d�Æ : (616)
This is the Bondi mass loss formula. We can interpret
1
32�j � j2 (617)
as the power radiated to in�nity at a given retarded time u, in a given direction, per unit area
on S2 (per unit solid angle). If we de�ne,
F =1
8
Z +1
�1j � j2 du ; (618)
(this is a function on S2), then F4� is the total energy radiated to in�nity in a given direction per
unit solid angle.
Limits as j u j! 1:
Now, recall the result (consequence of the Sobolev inequalities and the uniform bound on thecontrolling quantity):
j � j = O (��2+ �� 3
2� ) : (619)
It follows thatB = O (j u j� 3
2 )
as j u j! 1. From the equationÆ
div= � = B (620)
we then obtain also� = O (j u j� 3
2 ) (621)
as j u j! 1. In particular, � is integrable with respect to u. The equation
@�
@u= � 1
2� (622)
then yields that � tends to limits �+, �� as u! +1, u! �1 respectively.
� (u) = �� � 1
2
Z u
�1� (u0) du0 :
By (621) and (615) N tends to limits N+, N� as u ! +1, u ! �1 respectively. And wehave:
N+ � N� = � 2 F : (623)
We now consider the Z-system:
Æ
curl= Z = Q � 1
2� ^ � (624)
Æ
div= Z = N + P � 1
2� � � : (625)
117
Taking the mean values on S2 yields:
Q =1
2� ^ � (626)
P = � N +1
2� � � (627)
= � 2 M +1
2� � � : (628)
We see thatQ ! 0 as j u j ! 1 ; (629)
i.e. Q+= Q
�= 0, and
P ! P+
= � 2 M+ as u ! + 1 (630)
P ! P�
= � 2 M� as u ! � 1 (631)
and0 � M+ � M� ; (632)
the �rst inequality resulting in general from the positive mass theorem. (Under the assumptionsof the book 'The global nonlinear stability of the Minkowski space' [15] on the initial data, wein fact have M+ = 0.) From the results
� � � ; � � � = O (��3+ �� 1
2� )
we deduce that
P � P ; Q � Q = O (j u j� 12 ) as j u j ! 1 :
So,(Q � Q)+ = (Q � Q)� = 0 (633)
and(P � P )+ = (P � P )� = 0 : (634)
For the following remark see also the discussion in the paper 'Nonlinear Nature of Gravitationand Gravitational-Wave Experiments', [12].
Remark: While the limits of (Q � Q) vanish in general, the limits of (P � P ) vanish if andonly if the �nal center of mass frame is at rest relative to the initial center of mass frame andmoreover the initial and �nal velocities of the masses in the corresponding frames are negligible.That is, in the �nal asymptotic state we may in general have a number of masses with some,in general non-vanishing, asymptotic relative velocities. Moreover, the center of mass of the�nal asymptotic state may not be at rest relative to the initial center of mass frame, because ingeneral non-zero linear momentum is radiated away to in�nity. The initial conditions are to beconsidered as arising from an initial asymptotic state which again may correspond to a numberof masses with in general non-vanishing asymptotic relative velocities. The physically interest-ing situation is that, in which there is no incoming radiation (in the initial asymptotic state).Moreover, the case where the initial asymptotic relative velocities are negligible, is the mostinteresting one from the physical point of view. In this case we have 'parabolic' approach of theincoming masses and the initial data on a Cauchy hypersurface of vanishing linear momentum
118
(center of mass frame) is strongly asymptotically at in the sense of Volume I. A non-negligiblevelocity of the �nal center of mass frame relative to the initial center of mass frame may stillarise as there can be considerable recoil due to the radiated linear momentum. In the book 'Theglobal nonlinear stability of the Minkowski space', [15], the assumption of strong asymptotic atness for the initial data implies (P �P )� = 0. Moreover, under the hypothesis of the globalsmallness condition, we have M+ = 0. So, there are no �nal masses and (P � P )+ = 0. Weshall proceed in the following under the simplyfying assumption: (P � P )+ = (P � P )� = 0.(The general case is covered in the paper [12].)
We now consider the Z-system ((624), (625)) in the limits u ! +1, u ! �1. At the limits
u!+� 1 the terms �^�, � �� vanish. Also, Q vanishes. Moreover, under the above simplyfyingassumption,
P+� = P
+�
= � 2 M+� = � N
+�: (635)
Subtracting the u! �1 limiting equation from the u! +1 limiting equation, we obtain:
Æ
curl= (Z+ � Z�) = 0 (636)Æ
div= (Z+ � Z�) = N+ � N� � N+
+ N�
(637)
= � 2 (F � F ) (638)
(recalling that N+ �N� = �2F ). We conclude that there exists a function � on S2 such that
Z+ � Z� =Æ
r= � (639)Æ
4= � = � 2 (F � F ) : (640)
We turn to the elliptic system for �, equation (592), i.e.
Æ
div= � =1
2
Æ
r= H + Z :
We take the limits u!+� 1, take the di�erence of the two limits and use the fact that
H+ � H� = 0 (641)
(since @H@u
= 0). This yields
Æ
div= (�+ � ��) = Z+ � Z� =Æ
r= � :
Thus (�+���) is determined. The integrability condition that (Z+�Z�) is L2-orthogonal tothe conformal Killing �elds on S2, is equivalent to � being L2-orthogonal to their divergences,
namely � having vanishing projection onto the �rst eigenspace ofÆ
4= :�(1) = 0 : (642)
So, we haveÆ
div= (�+ � ��) =Æ
r= �Æ
4= � = � 2 (F � F )
)(643)
119
and the integrability condition (642) which is equivalent to:
F(1) = 0 : (644)
Subscript (1) denotes the part which lies in the �rst eigenspace ofÆ
4= (l = 1) (multiplicity of thelth eigenspace 2l + 1, eigenvalue l(l + 1)).
Here, (644) states that the total linear momentum radiated to in�nity vanishes.
In general, when terms involving (P � P )+� are present on the right hand side of (640), the
integrability condition (642), is seen to be equivalent to the law of conservation of linear mo-mentum, namely that the recoil momentum of the masses is equal and opposite to the radiatedmomentum [12].
We shall give below the solution of (643). Let us consider the next �gure:
Figure 34.
��0
Y
X
S2� R
3
tangent planeat �
plane through originorthogonal to �
��0
120
Here, we denote the direction of observation by � 2 S2 � R3 and let X, Y be arbitrary vectorslying in the tangent plane at �, i.e. in T�S
2. Let � be the projection to the plane through theorigin orthogonal to �. In the Euclidean space, < ; > denotes the inner product. The solutionat the observation point � is expressed as an integral over S2 of a contribution from each �0 2 S2:
(�+ � ��) (X; Y ) = � 1
2�
Z�02S2
(F � F[1]) (�0)< X; �0 > < Y; �0 > � 1
2 < X;Y > j ��0 j21 � < �; �0 >
d�Æ (�0)
(645)Subscript [1] denotes the projection onto the sum of the 0th (l = 0) and 1st (l = 1) eigenspaces
ofÆ
4= . Recall the equation (618):
F =1
8
Z +1
�1j �(u) j2 du :
We have:
�+ � �� = � 1
2
Z +1
�1�(u) du : (646)
Note that � is dimensionless, � has dimensions of length and F also has dimensions of length,and we have:
� (u) = �� � 1
2
Z u
�1� (u0) du0 :
The �rst non-trivial approximation to � in a post-Newtonian expansion under the hypothesisof no incoming radiation is
� =���QTT
; (647)
where Qij , a symmetric 3-dimensional matrix, is the quadrupole moment of the source (in theNewtonian framework). Let �(x) be the mass density, then we write Qij in Cartesian coordinatesat the retarded time u. To illustrate what is meant by that, we give the following picture:
Qij =
ZR3
xi xj �(x) d3x (648)
QTab = �i
a �jb Qij ; (649)
where
�ia = Æia � �i �a :
And QT lies in T�S2. Also, QTT is the trace-free part of QT .
QTTab = QT
ab �1
2�ab �
cd Q
Tcd :
Actually, � depends on u and �, i.e. �(u; �) =���QTT
(u; �); and QT (u; �) = �(�)�(�)Q(u).
121
Figure 35.
C��
Qij at retarded time u
Here, 'instantaneous' means 'on the same C+u '.
� in conventional units is of order c�5, where c is the speed of light.
Consider now a bound system of masses, such as a binary stellar system (which can be consid-ered to have resulted by radiative capture from an initial asymptotic system of incoming masseswith parabolic approach). We shall show that for any such bound system,
supu
j �(u) � �� j
andj �+ � �� j
are comparable. It is
� (u) ���QTT
� M V 2 � kinetic energy (650)
andsupu
j �(u) � �� j � maximal kinetic energy : (651)
On the other hand, it is
�+ � �� �ZS2
F � total radiated energy : (652)
The total mechanical energy isEM = K � P ; (653)
122
where K denotes the kinetic and P the potential energy. For a bound system we have
< K > =1
2< P > ; (654)
by virtue of the Virial theorem. (See [30].) Here, < > denotes the average value over a suitableperiod. So, the total mechanical energy is:
EM = � < K > : (655)
Initially, EM = 0 at the beginning of the in-spiral, when the bodies are far away. The overallconservation of energy says that at each time:
EM + ER = 0 (is conserved) : (656)
Note that 0 is the initial value, and ER is the radiated energy up to that time. Hence, we obtain
ER = � EM = < K > : (657)
123
8.2 Experiment
Finally, we shall establish the relation of the theoretical results derived above to experiment.For this we are going to discuss a laser interferometer gravitational-wave detector. We shallsee that the theoretical result on �+ � �� derived above leads to an e�ect measurable by suchdetectors. This e�ect manifests itself in a permanent displacement of the test masses of thedetector after a wave train has passed. Consider the following setup (in space):
Figure 36.
m0m1
m2
mirror
mirror
mirror
horizontal plane
124
The three masses are suspended by equal length pendulums. The mass m0 is the reference mass.The motion of the masses on the horizontal plane can be considered free for timelike scales muchshorter than the period of the pendulums. The distance of m1 and m2 from the reference massm0 is measured by laser interferometry. The beam splitter is at m0. Any di�erence in the lighttravel times between m0 and m1 and m2, respectively, results in a di�erence of phase of thelaser light at m0. The next picture (in spacetime (here in 2 dimensions)) illustrates this fact.Here, the hypersurfaces are surfaces of constant phase.
Figure 37.
m1
m2
m0
beam splitter
spacetime
0
0
0
�
�
0
125
m0, m1, m2 describe geodesics �0, �1, �2 in spacetime. T being the unit future-directed tangentvector�eld of �0 and t the arc length along �0. Let Ht for each t be the spacelike, geodesichyperplane through �0(t) orthogonal to T . This is drawn below (in spacetime (in more than 2dimensions)).
Figure 38.
H0
Ht
spacetime
T
T
E1
E2
E2
E1
�0
t
Consider the orthonormal frame �eld (T;E1; E2; E3) along �0, where (E1; E2; E3) is an orthonor-mal frame for H0 at �0(0), parallely propagated along �0. Then at each t, (E1; E2; E3) is anorthonormal frame forHt at �0(t). We assign to a point p in spacetime, lying in a neighbourhoodof �0, the cylindrical normal coordinates (t; x
1; x2; x3), based on �0, if p 2 Ht and p = expX
126
with X =P
i xiEi 2 T�0(t)Ht. In these coordinates we have
g�� � ��� = O (R d2) ; (658)
where ��� is the Minkowski metric and:
d = j X j =sX
i
(xi)2 (659)
is the distance of p from the center �0(t) on Ht. Let the time scale in which the curvature variessigni�cantly be � . Then, the displacements of the masses from their initial positions will beO(R�2). Assuming that
d
�<< 1 ; (660)
the speed of light can be taken to be 1 in the cylindrical normal coordinate system, in thecontext of the laser interferometer experiment. So, the di�erences in phase of the laser lightwill, under this assumption, accurately re ect di�erences in distance (of m1 and m2 from m0).Moreover, the same assumption (660) allows us to replace the geodesic equation for �1 and �2by the Jacobi equation (geodesic deviation from �0).
d2xi
dt2= � RiT jT xj (661)
with
RiT jT = R (Ei; T; Ej ; T ) : (662)
Now, assume for simplicity that the source is in the E3-direction.
127
Figure 39.
E1
E2
E3
the vertical source
horizontal plane
So, we have L = T �E3, L = T +E3 and since the leading components of the curvature are
�AB = R (EA; L; EB ; L) (663)
�AB =AAB
r+ o (r�2) : (664)
For A see proposition 10 and equation (610). Here, AAB are the components of A in thecorresponding frame on S2. The other orthonormal components of the curvature are of O(r�2)at least. For the following discussion we introduce some notation:
xi(A) : ith Cartesian coordinate of mass mA : A = 1; 2 :
Hence the Jacobi equation becomes
d2 xi(A)
d t2= � 1
4r�1 AAB xi(B) + O (r�2) (665)
or
d2 x3(A)
d t2= 0 (666)
d2 xC(A)
d t2= � 1
4r�1 AAB xC(B) + O (r�2) : (667)
So, to leading order (r�1), there is no acceleration in the vertical direction. We have the followinginitial conditions, as t! �1:
x3(A) = 0 ; _x3(A) = 0
xB(A) = d0 ÆBA ; _xB(A) = 0 :
(668)
128
Initiallym1 and m2 are at rest at equal distance d0 and at right angles from m0. Since the righthand side is very small, we can substitute the initial values on the right hand side. Then x3
remains 0, i.e. the motion is con�ned to the horizontal plane and
��xA
(B) = � 1
4r�1 d0 AAB (669)
to leading order. So, it is
_xA(B) (t) = � 1
4d0 r
�1
Z t
�1AAB (u) du : (670)
Recall the equation (610), i.e.@�
@u= � 1
2A
andlim
juj!1� = 0 : (671)
Thus we obtain
� 1
2
Z t
�1AAB (u) du = � (t) (672)
and also
_xA(B) (t) =d0r
1
2�AB (t) : (673)
The behaviour � ! 0 for u ! 1 implies that the test masses return to rest after the passageof the gravitational wave. Integrating again, since (608):
@�
@u= � 1
2� ;
it is
xA(B) (t) = � (d0r) (�AB (t) � ��) : (674)
Taking the limit t ! 1 it follows that the test masses experience permanent displacements.Thus �+ � �� is equivalent to an overall displacement of the test masses:
4 xA(B) = � (d0r) (�+
AB � ��AB) : (675)
129
Figure 40.
m1m0
m2
displacement
displacement (same, rotated by 90Æ)
Figure 40. The symmetric trace-free nature of �AB , hence also �AB � ��AB
, manifests itself in the
fact that at any time if m1 has some displacement, m2 has a displacement of the same magnitude in a
perpendicular direction.
130
Finally we remark that by virtue of the last considerations of the previous paragraph, in the casethat the source of the waves is any bound multiple stellar system, the permanent displacementof the test masses of the interferometer will be of the same order of magnitude as their greatesttransient displacement.
Thus the nonlinear memory e�ect is equally important for any bound multiple stellar system,independently of how relativistic such a system is. Only the time scale over which the e�ectbuilds up, which is the time scale of the radiation energy loss, will depend on the strength ofthe system and will be far longer for a weaker system such as multiple white dwarfs than for astronger system such as multiple neutron stars system, or in the extreme case, multiple blackholes.
131
Index
L2-theory, 43p-covariant tensor�eld, 92, 99, 109, 114
1-parameter family of p-covariant ten-sor�elds, 109
Apollonius, 22
Bianchi identities, 100, 112binary stellar system, 122black holes, 131Bondi mass, 111, 117
loss formula, 117bootstrap
assumptions, 85{88hypothesis, 84
bound system of masses, 122
Calderon-Zygmundstandard Lp estimates, 63, 65
Cauchy hypersurface, 118Cauchy-Riemann equations, 80causal future, 104causal past, 104, 108center of mass, 118
frame, 118chronological future, 23, 34closed trapped surface, 32Codazzi equations, 39, 62, 66, 78, 82, 114conformally covariant, 63, 78, 80continuity
argument, 83method of, 87
controlling quantity, 113curvature, 78, 112, 128
deformation matrix, 19di�erential equations
ordinary, 62, 69, 113partial, 62
di�erential inequalityordinary, 67, 69, 86, 87, 89
dimensionless norms, 67divergence operator, 41domain of dependence, 104
(future), 108
eikonal equation, 6
Einstein's equationsnonlinearity of, 103
elliptic equations, 62energy
kinetic, 122maximal, 122
potential, 123radiated, 123total mechanical, 122total radiated, 122
estimatesfor �, 62for �, 62, 73
experiment, 124
�nal masses, 119 at
strongly asymptotically, 119focal points, 22, 33
in Euclidean geometry, 22in Lorentzian Geometry, 23in Riemannian geometry, 22
foliations, 62, 92aÆne foliationsof a null hypersurface, 12
future set, 33
Gauss curvature, 56, 76, 111
conformal transformation of, 45Gauss equation, 76, 89Gauss-Bonnet, 45, 58gravitational waves, 103, 124, 129
Hawking mass, 111Hodge (div � curl) system, 76, 78, 114hypersurface, 125
maximal spacelike, 104
isoperimetricinequality, 93
Jacobi equation, 127, 128Jacobi �eld, 13, 35, 74
frame, 62, 92Jacobi operator, 25
Laplace operator, 56
132
in polar coordinates, 56laser interferometer detector, 103, 124limiting equations, 114
linear momentum, 118law of conservation of, 120
radiated, 119radiated to in�nity, 120
mass aspect function, 76, 105conjugate, 76
mass density, 121
memory e�ectof gravitational waves, 103, 131
Minkowskimetric, 127
neutron stars, 131nonlinear character
of the asymptotic gravitational laws atfuture null in�nity, 103
nullnull line, 9
exterior, 9interior, 9
null hypersurface, 6, 35, 73, 108Codazzi equations for a section S of a,
38
generator, 62, 73null normals
future-directed, 105, 112
Penrosetheorem of, 32
positive mass theorem, 118post-Newtonian expansion, 121
powerradiated to in�nity, 117
propagationequation, 62, 73, 75{77, 80, 89, 90
limit of, 115for the torsion, 39
quadrupole moment, 121
radiation
no incoming, 118, 121retarded time, 108, 121
Riemannianmanifold, 62, 78
second fundamental form, 10, 18, 62, 73, 105Sobolev
inequality, 62, 65, 84, 92, 113spacelike surface, 112
closed, 76spacetime, 76, 109, 126
curvature, 62, 87, 89manifold, 108
speed of light, 122standard 2-sphere, 65standard unit sphere, 63stereographic projection, 50surface of constant phase, 125
test masses, 124, 129, 131torsion, 14, 15, 105, 114total energy
radiated to in�nity, 117transformation
conformal, 64
uniformization theorem, 45, 62, 79
vacuum equations, 74variation
�rst, 12second, 16, 17, 25
Virial theorem, 123
white dwarfs, 131
133
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