Deformations of Axially Symmetric Initial Data and the Mass-Angular Momentum Inequality
aa r X i v : . [ g r- q c ] F e b DEFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA AND THEMASS-ANGULAR MOMENTUM INEQUALITY
YE SLE CHA AND MARCUS A. KHURI
Abstract.
We show how to reduce the general formulation of the mass-angular momentum inequal-ity, for axisymmetric initial data of the Einstein equations, to the known maximal case whenevera geometrically motivated system of equations admits a solution. This procedure is based on acertain deformation of the initial data which preserves the relevant geometry, while achieving themaximal condition and its implied inequality (in a weak sense) for the scalar curvature; this answersa question posed by R. Schoen. The primary equation involved, bears a strong resemblance to theJang-type equations studied in the context of the positive mass theorem and the Penrose inequality.Each equation in the system is analyzed in detail individually, and it is shown that appropriate ex-istence/uniqueness results hold with the solution satisfying desired asymptotics. Lastly, it is shownthat the same reduction argument applies to the basic inequality yielding a lower bound for the areaof black holes in terms of mass and angular momentum. Introduction
The standard picture of gravitational collapse [4], [7] asserts that generically, an asymptotically flatspacetime should eventually settle down to a stationary final state, consisting of (possibly multiple)disconnected black hole spacetimes. The black hole uniqueness theorem implies that, in vacuum, eachof these solutions must be the Kerr spacetime; note that there are still important unresolved technicalaspects associated with this uniqueness result [6]. It is also conceivable that these black holes arecoupled to matter fields. In any event, as in Kerr, the following inequality holds between mass andangular momentum m f ≥ p |J f | for each of the connected components of the final state, and hencefor the final state itself. Moreover, as gravitational radiation carries positive energy, the mass of anyinitial state should not be smaller than that of the final state m ≥ m f . If auxiliary conditions areimposed, one of which usually includes axisymmetry, in order to ensure the conservation of angularmomentum, then J = J f where J , J f denote the (ADM) angular momentums of the initial andfinal state. This leads to the mass-angular momentum inequality [11](1.1) m ≥ p |J | for any initial state. A counterexample to (1.1) would pose a serious challenge to this standardpicture of collapse, whereas a verification of (1.1) would only lend credence to this model.Consider an initial data set ( M, g, k ) for the Einstein equations. This consists of a 3-manifold M , Riemannian metric g , and symmetric 2-tensor k representing the extrinsic curvature (secondfundamental form) of the embedding into spacetime, which satisfy the constraint equations16 πµ = R + ( T r g k ) − | k | g , πJ = div g ( k − ( T r g k ) g ) . (1.2) The second author acknowledges the support of NSF Grants DMS-1007156 and DMS-1308753. This paper isalso based upon work supported by NSF under Grant No. 0932078 000, while the authors were in residence at theMathematical Sciences Research Institute in Berkeley, California, during the fall of 2013.
Here µ and J are the energy and momentum densities of the matter fields, respectively, and R is thescalar curvature of g . The following inequality will be referred to as the dominant energy condition(1.3) µ ≥ | J | g . Suppose that M has at least two ends, with one designated end being asymptotically flat, andthe remainder being either asymptotically flat or asymptotically cylindrical. Recall that a domain M end ⊂ M is an asymptotically flat end if it is diffeomorphic to R \ Ball, and in the coordinatesgiven by the asymptotic diffeomorphism the following fall-off conditions hold(1.4) g ij = δ ij + o l ( r − ) , ∂g ij ∈ L ( M end ) , k ij = O l − ( r − λ ) , λ > , for some l ≥ . In the context of the mass-angular momentum inequality, these asymptotics maybe weakened, see for example [21]. The asymptotics for cylindrical ends is most easily described inBrill coordinates, to be given in the next section.We say that the initial data are axially symmetric if the group of isometries of the Riemannianmanifold ( M, g ) has a subgroup isomorphic to U (1), and that the remaining quantities defining theinitial data are invariant under the U (1) action. In particular, if η denotes the Killing field associatedwith this symmetry, then(1.5) L η g = L η k = 0 , where L η denotes Lie differentiation. If M is simply connected and the data are axially symmetric, itis shown in [5] that the analysis reduces to the study of manifolds diffeomorphic to R minus a finitenumber of points. Each point represents a black hole, and has the geometry of an asymptoticallyflat or cylindrical end. The fall-off conditions in the designated asymptotically flat end guaranteethat the ADM mass and angular momentum are well-defined by the following limits(1.6) m = 116 π Z S ∞ ( g ij,i − g ii,j ) ν j , (1.7) J = 18 π Z S ∞ ( k ij − ( T r g k ) g ij ) ν i η j , where S ∞ indicates the limit as r → ∞ of integrals over coordinate spheres S r , with unit outernormal ν . Note that (1.4) implies that the ADM linear momentum vanishes.Angular momentum is conserved [14] if(1.8) J i η i = 0 . Moreover, when M is simply connected, this is a necessary and sufficient condition [14] for theexistence of a twist potential ω :(1.9) 2 ǫ ijl ( k jn − ( T r g k ) g jn )) η l η n dx i = dω where ǫ ijl is the volume form for g .In [10] Dain has confirmed (1.1) under the hypotheses that the initial data have two ends, aremaximal ( T r g k = 0), vacuum ( µ = | J | g = 0), and admit a global Brill coordinate system. He alsoestablished the rigidity statement, which asserts that equality occurs in (1.1) if and only if the initialdata arise as the t = 0 slice of the extreme Kerr spacetime. Chrusciel, Li, and Weinstein [5], [8]improved these results by showing that global Brill coordinates exist under general conditions, and The notation f = o l ( r − a ) asserts that lim r →∞ r a + n ∂ n f = 0 for all n ≤ l , and f = O l ( r − a ) asserts that r a + n | ∂ n f | ≤ C for all n ≤ l . The assumption l ≥ EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 3 by replacing the vacuum assumption with the hypotheses that µ ≥ Question 1.1.
Is there a canonical way to deform a non-maximal, axisymmetric, vacuum datato a unique maximal, vacuum data with the same physical quantities, i.e. the mass and angularmomentum, which also preserves the axial symmetry?
This paper is organized as follows. In the next section we describe the deformation in detail,while in Section 3 the reduction argument is established and the case of equality is treated. Section4 contains an application to the basic inequality yielding a lower bound for the area of black holesin terms of mass and angular momentum. In Sections 5 and 6 we give an initial analysis of thecanonical system of PDEs, and finally four appendices are added to include several important butlengthy calculations.
Acknowledgements.
The authors would like to thank Lars Andersson, Piotr Chru´sciel, SergioDain, Marc Mars, Martin Reiris, Richard Schoen, and Xin Zhou for discussions related to this work.2.
Deformation of Initial Data
In this section we will describe the deformation procedure which leads to the reduction argumentfor the mass-angular momentum inequality. It will be assumed that (
M, g, k ) is a simply connected,axially symmetric initial data set with multiple ends as described in the previous section. Simpleconnectedness and axial symmetry imply [5] that M ∼ = R \ P Nn =1 i n , where i n are points in R andrepresent asymptotic ends (in total there are N + 1 ends). Moreover there exists a global (cylindrical)Brill coordinate system ( ρ, φ, z ) on M , where the points i n all lie on the z -axis, and in which theKilling field is given by η = ∂ φ . In these coordinates the metric takes a simple form(2.1) g = e − U +2 α ( dρ + dz ) + ρ e − U ( dφ + A ρ dρ + A z dz ) , where ρe − U ( dφ + A ρ dρ + A z dz ) is the dual 1-form to | η | − η and all coefficient functions are indepen-dent of φ . Let M end denote the end associated with limit r = p ρ + z → ∞ . The asymptoticallyflat fall-off conditions (1.4) will be satisfied if(2.2) U = o l − ( r − ) , α = o l − ( r − ) , A ρ , A z = o l − ( r − ) . The remaining ends associated with the points i n will be denoted by M nend , and are associated withthe limit r n →
0, where r n is the Euclidean distance to i n . The asymptotics for asymptotically flat CHA AND KHURI and cylindrical ends are given, respectively, by(2.3) U = 2 log r n + o l − ( r n ) , α = o l − ( r n ) , A ρ , A z = o l − ( r n ) , (2.4) U = log r n + o l − ( r n ) , α = o l − ( r n ) , A ρ , A z = o l − ( r n ) . It will also be assumed that the dominant energy condition (1.3) is satisfied, and that(2.5) div g k ( η ) = 0 , which is equivalent to (1.8). Equation (2.5) gives rise to a twist potential ω (1.9) that is constant oneach connected component of the axis of rotation. Let I n denote the interval of the z -axis between i n +1 and i n , where i = −∞ and i N +1 = ∞ . Then a standard formula (see Appendix D) yields theangular momentum for each black hole(2.6) J n = 18 ( ω | I n − ω | I n − ) . According to (1.7) and (2.5), the total angular momentum is given by(2.7) J = N X n =1 J n . We seek a deformation of the initial data (
M, g, k ) → ( M , g, k ) such that the manifolds arediffeomorphic M ∼ = M , the geometry of the ends is preserved, and(2.8) m = m, J = J , T r g k = 0 , J ( η ) = 0 , R ≥ | k | g weakly,where m , J , J , and R are the mass, angular momentum, momentum density, and scalar curvatureof the new data. The inequality in (2.8) is said to hold ‘weakly’ if it is valid when integrated againstan appropriate test function. The validity of this inequality plays a central role in the proof of themass-angular momentum inequality in the maximal case, and it is precisely the lack of this inequalityin the non-maximal case which prevents the proof from generalizing. Thus, the primary goal of thedeformation is to obtain such a lower bound for the scalar curvature, while preserving all otheraspects of the geometry.With intuition from previous work [1], [2], [22] we search for the deformation in the form of agraph inside a stationary 4-manifold(2.9) M = { t = f ( x ) } ⊂ ( M × R , g + 2 Y i dx i dt + ϕdt ) , where the 1-form Y = Y i dx i and functions ϕ and f are defined on M and satisfy(2.10) L η f = L η ϕ = L η Y = 0 . Define(2.11) g ij = g ij + f i Y j + f j Y i + ϕf i f j , k ij = 12 u (cid:0) ∇ i Y j + ∇ j Y i (cid:1) , where f i = ∂ i f , ∇ is the Levi-Civita connection with respect to g , and(2.12) u = ϕ + | Y | g . In the ‘Riemannian’ setting (2.9), g arises as the induced metric on the graph M . However in the‘Lorentzian’ setting(2.13) M = { t = f ( x ) } ⊂ ( M × R , g − Y i dx i dt − ϕdt ) , EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 5 the deformed data arise as the induced metric and second fundamental form of the t = 0 slice. Noticethat(2.14) ∂ t = un − Y , where n is the unit normal to the t = 0 slice and Y is the vector field dual to Y with respect to g .Thus ( u, − Y ) comprise the lapse and shift of this stationary spacetime. Based on the structure ofthe Kerr spacetime, we make the following simplifying assumption that Y has only one component(2.15) Y i ∂ i := g ij Y j ∂ i = Y φ ∂ φ . Lemma 2.1.
Under the hypothesis (2.15) , g is a Riemannian metric, T r g k = 0 , and ϕ = u − g φφ ( Y φ ) . Moreover if { e i } i =1 is an orthonormal frame for g with e = | η | − η , then (2.16) k ( e i , e j ) = k ( e , e ) = 0 , k ( e i , e ) = | η | u e i ( Y φ ) , i, j = 3 . Lastly (2.17) (1 + u |∇ f | g )(1 − u |∇ f | g ) = 1 , where ∇ i f = g ij f j and ∇ i f = g ij f j .Proof. From (2.10) it follows that g φφ = g φφ , and so | Y | g = g φφ ( Y φ ) . This yields the formula for ϕ .Next observe that uT r g k = ∇ i Y i = ∂ i Y i − Γ iij Y j = − Γ iiφ Y φ = − (cid:18) √ det g ∂ φ p det g (cid:19) Y φ = 0 , (2.18)where Γ lij are Christoffel symbols.We now show that g is Riemannian. Equations (2.10) and (2.15) imply that(2.19) Y φ = g φφ Y φ , Y i = g ij Y j = g iφ Y φ = ( g iφ + f i Y φ ) Y φ = ( g iφ + f i g φφ Y φ ) Y φ . Inserting this into (2.11) produces(2.20) g ij = g ij + ( f i g jφ + f j g iφ ) Y φ + ( u + g φφ ( Y φ ) ) f i f j . Take a g -orthonormal frame ( d , d , d = | η | − η ) at a point, and express g as a matrix with respectto this frame(2.21) g = u + g φφ ( Y φ ) ) f ( u + g φφ ( Y φ ) ) f f √ g φφ Y φ f u + g φφ ( Y φ ) ) f √ g φφ Y φ f . The determinant of the lower 2 × u f >
0, and the full determinant is given by(2.22) det g = (1 + u |∇ f | g ) det g > . It follows that g is positive definite. Observe also that an analogous computation in the Lorentziansetting produces(2.23) det g = (1 − u |∇ f | g ) det g. CHA AND KHURI
Equations (2.22) and (2.23) together yield (2.17).In order to establish (2.16), observe that(2.24) 2 uk ij = ∇ i Y j + ∇ j Y i = ∂ i Y j + ∂ j Y i − aij Y a , and(2.25) ∂ i Y j = ∂ i ( g φj Y φ ) = ( ∂ i g φj ) Y φ + g φj ∂ i Y φ , aij Y a = g al ( ∂ i g jl + ∂ j g il − ∂ l g ij ) Y a = ( ∂ i g jφ + ∂ j g iφ ) Y φ . (2.26)Therefore(2.27) 2 uk ij = g φi ∂ j Y φ + g φj ∂ i Y φ . Clearly k ( e , e ) = 0, and if we express e i , i = 1 , i, j = 1 , uk ( e i , e j ) = 2 ue U − α ( k ij − A i k jφ − A j k iφ + A i A j k φφ )= e U − α ( g φi ∂ j Y φ + g φj ∂ i Y φ − A i g φφ ∂ j Y φ − A j g φφ ∂ i Y φ )= 0 , (2.28)since g φi = A i g φφ from (3.1). Also(2.29) 2 uk ( e i , e ) = g φφ | η | e i ( Y φ ) = | η | e i ( Y φ ) . (cid:3) This lemma shows that the deformed data set is maximal, satisfying one requirement of (2.8).Furthermore, it shows that ϕ is determined by the functions u and Y φ . Thus, the three functions( u, Y φ , f ) completely determine the new data, and will be chosen to satisfy the remaining statementsin (2.8), so as to yield a reduction argument for the mass-angular momentum inequality.The next task is to show how to choose the three functions ( u, Y φ , f ). In order to apply thetechniques from the maximal case, the existence of a twist potential for ( M , g, k ) is needed. Thereforewe require(2.30) div g k ( η ) = 0 . This turns out to be a linear elliptic equation for Y φ (if u is independent of Y φ ), as is shown inthe appendix. As discussed in Section 4, the function Y φ is uniquely determined among boundedsolutions of (2.30), if the r − -fall-off rate is prescribed at M end . In particular, we will choose thefollowing boundary condition(2.31) Y φ = − J r + o ( r − ) as r → ∞ . Lemma 2.2. If g is asymptotically flat and u → as r → ∞ , then the boundary condition (2.31) guarantees that J = J . EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 7
Proof.
Observe that since g φφ ∼ r sin θ as r → ∞ , where ρ = r sin θ and z = r cos θ , we have J = lim r →∞ π Z S r k ( ∂ φ , ∂ r )= lim r →∞ π Z π Z π g φφ ∂ r Y φ r sin θdφdθ = 3 J Z π sin θdθ = J . (2.32) (cid:3) Let us now show how to choose f . As with previous deformations arising from the positive masstheorem and Penrose inequality, f is chosen to impart positivity properties to the scalar curvature.With this in mind, it is instructive to calculate the scalar curvature for an arbitrary f . The followingresult requires a long and detailed computation, and is therefore relegated to the appendix. Theorem 2.3.
Suppose that (1.5) , (2.5) , (2.10) , (2.15) , and (2.30) are satisfied, then the scalarcurvature of g is given by R − | k | g =16 π ( µ − J ( v )) + | k − π | g + 2 u − div g ( uQ )+ ( T r g π ) − ( T r g k ) + 2 v ( T r g π − T r g k ) , (2.33) where (2.34) π ij = u ∇ ij f + u i f j + u j f i + u ( g iφ Y φ,j + g jφ Y φ,i ) q u |∇ f | g is the second fundamental form of the graph M in the Lorentzian setting, (2.35) v i = uf i q u |∇ f | g , w i = uf i + u − Y i q u |∇ f | g , and (2.36) Q i = Y j ∇ ij f − ug jl f l k ij + w j ( k − π ) ij + uf i w l w j ( k − π ) lj q u |∇ f | g . Furthermore, if Y ≡ then the same conclusion holds without any of the listed hypotheses. This theorem, together with the dominant energy condition (1.3), make it clear that in order toobtain the inequality R ≥ | k | g at least weakly, f should be chosen to solve the equation(2.37) T r g ( π − k ) = 0 . It follows that(2.38) R − | k | g = 16 π ( µ − J ( v )) + | k − π | g + 2 u − div g ( uQ ) , which yields the inequality in (2.8) after multiplying by u and applying the divergence theorem; itis assumed that appropriate asymptotic conditions are imposed (see below) in order to ensure thatthe boundary integrals vanish in each of the ends. Equation (2.37) is similar to previous Jang-typeequations that have been used in connection with deformations of initial data, in particular for thepositive mass theorem [22] and the Penrose inequality [2]. These previous equations have the form(2.39) T r g ( π − k ) = 0 , CHA AND KHURI where it is assumed that u = 1 and Y = 0 [22], and Y = 0 [2]. Note that (2.39) does not reduce to(2.37) even in the setting of [22] or [2]. This suggests that there is a significant difference betweenthese two equations. In fact, solutions of (2.37) do not blow-up, while solutions of (2.39) typicallyblow-up at apparent horizons or can be prescribed to blow-up at these surfaces [16]. This separatebehavior arises from the fact that the trace in (2.37) is taken with respect to g , whereas the trace in(2.39) is taken with respect to g . As a result, the analysis of (2.37) is much more simple than thatof (2.39). Lastly, in order to ensure that m = m , we will impose the following asymptotics(2.40) | f | + r |∇ f | g + r |∇ f | g ≤ cr − ε in M end , for some 0 < ε <
1. A bounded solution may be obtained by prescribing the following asymptoticsat the remaining ends(2.41) r − n |∇ f | g + r − n |∇ f | g ≤ c in asymptotically flat M nend , (2.42) |∇ f | g + |∇ f | g ≤ cr n in asymptotically cylindrical M nend . At this point we have shown how to choose f and Y , in order to produce a deformation of theinitial data which satisfies (2.8). It remains to choose u , in such a way as to facilitate a proof of themass-angular momentum inequality. This shall be accomplished in the next section.3. The Reduction Argument and Case of Equality
Here we shall follow the maximal case proof of the mass-angular momentum inequality, within thesetting of the deformed initial data (
M , g, k ). The primary stumbling block is a lack of the pointwisescalar curvature inequality as appearing in (2.8). However a judicious choice of u will overcome thisdifficulty.Assuming that the functions ( u, Y φ , f ) are chosen to possess the appropriate asymptotics, thegeometry of the ends will be preserved in the deformation. Since the deformed data are also simplyconnected and axially symmetric, the results of [5] apply to yield a global Brill coordinate system( ρ, φ, z ) such that(3.1) g = e − U +2 α ( dρ + dz ) + ρ e − U ( dφ + A ρ dρ + A z dz ) . Next, recall that (2.30) implies the existence of a twist potential ω . An important property of theBrill coordinates is that they yield a simple formula for the mass ([3], [10])(3.2) m − M ( U , ω ) = 132 π Z R (cid:16) e − U +2 α R + ρ e − α ( A ρ,z − A z,ρ ) − g − φφ | ∂ω | (cid:17) dx, where | ∂ω | and dx denote the Euclidean norm and volume element, and(3.3) M ( U , ω ) = 132 π Z R (cid:16) | ∂U | + g − φφ | ∂ω | (cid:17) dx. Let(3.4) e ρ = e U − α ( ∂ ρ − A ρ ∂ φ ) , e z = e U − α ( ∂ z − A z ∂ φ ) , e φ = 1 p g φφ ∂ φ , be an orthonormal frame. Then according to (1.9) and g φφ = g φφ ,(3.5) k ( e ρ , e φ ) = − | η | g e z ( ω ) = − e U − α g φφ ∂ z ω, k ( e z , e φ ) = 12 | η | g e ρ ( ω ) = e U − α g φφ ∂ ρ ω. EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 9
In light of Lemma 2.1 it follows that(3.6) | k | g = 2( k ( e ρ , e φ ) + k ( e z , e φ ) ) = e U − α g φφ | ∂ω | , and hence with the help of Theorem 2.3 and the dominant energy condition m − M ( U , ω ) ≥ π Z R e − U +2 α ( R − | k | g ) dx ≥ π Z R e − U +2 α u div g ( uQ ) dx ≥ π Z M e U u div g ( uQ ) dx g , (3.7)where the volume element for g is given by dx g = e − U +2 α dx .Inequality (3.7) suggests that we choose(3.8) u = e U = ρ p g φφ = ρ √ g φφ . If g preserves the asymptotic geometry of g , then based on (2.2), (2.3), (2.4)(3.9) u = 1 + o l − ( r − ) as r → ∞ in M end , (3.10) u = r n + o l − ( r n ) as r n → M nend , (3.11) u = r n + o l − ( r n ) as r n → M nend , where r n is the Euclidean distance to the point i n defining the end. Therefore, with the help of theasymptotics for f (2.40), (2.41) and Y φ (2.31), as well as the assumption(3.12) | k | g + | k ( ∂ φ , · ) | g + | k ( ∂ φ , ∂ φ ) | ≤ c on M, the asymptotic boundary integrals arising from the right-hand side of (3.7) all vanish as long as J = J . This is proven in Appendix C, Section 9. It follows that(3.13) m ≥ M ( U , ω ) . Theorem 3.1.
Let ( M, g, k ) be a smooth, simply connected, axially symmetric initial data set sat-isfying the dominant energy condition (1.3) and conditions (1.8) , (3.12) , and with two ends, onedesignated asymptotically flat and the other either asymptotically flat or asymptotically cylindrical.If the system of equations (2.30) , (2.37) , (3.8) admits a smooth solution ( u, Y φ , f ) satisfying theasymptotics (2.31) , (2.40) - (2.42) , (3.9) - (3.11) , then (3.14) m ≥ p |J | and equality holds if and only if ( M, g, k ) arises from an embedding into the extreme Kerr spacetime.Proof. The existence of a solution ( u, Y φ , f ) ensures that we may apply the maximal case proof tothe deformed initial data ( M , g, k ) as above, arriving at the inequality (3.13). The results of [8], [10],[21] then imply that(3.15) M ( U , ω ) ≥ q |J | . Moreover, according to (2.8) m = m and J = J , and hence (3.13) yields the desired inequality(3.14).Consider now the case of equality in (3.14). In the process of deriving (3.13), several nonnegativeterms were left out from the right-hand side. These terms arise from (2.33) and (3.2). In the currentsituation, they must all vanish(3.16) | µ − J ( v ) | = | k − π | g = | A ρ,z − A z,ρ | = 0 . Furthermore, in light of the dominant energy condition, the fact that | v | g <
1, and the identity(3.17) µ − J ( v ) = ( µ − | J | g ) + (1 − | v | g ) | J | g + ( | J | g | v | g − J ( v )) , it follows that(3.18) µ = | J | g = 0 . We claim that (
M , g, k ) is now a vacuum initial data set. By Lemma 2.1
T r g k = 0, so that themomentum density is given by(3.19) 8 πJ = div g k. Let { e i } i =1 denote the orthonormal basis (3.4) with e = e φ , then( div g k )( e i ) = X i =1 ( ∇ e j k )( e i , e j )= X j =1 " e j ( k ( e i , e j )) − X a =1 h∇ e j e i , e a i k ( e a , e j ) − X a =1 h∇ e j e j , e a i k ( e i , e a ) . (3.20)Assume now that i = 3, then by Lemma 2.1(3.21) X j =1 e j ( k ( e i , e j )) = 0and(3.22) ( div g k )( e i ) = − X j =1 h∇ e j e i , e i k ( e , e j ) − X a =1 h∇ e e i , e a i k ( e a , e ) − X j =1 h∇ e j e j , e i k ( e i , e ) . The last sum is zero since ∂ φ is a Killing field. Moreover(3.23) h∇ e e i , e a i = −h e i , ∇ e e a i = −h e i , ∇ e a e i = h∇ e a e i , e i , since(3.24) [ ∂ φ , e a ] = L ∂ φ e a = 0 . Thus, we need only show that the first sum in (3.22) vanishes. To accomplish this, observe that(3.25) h∇ e j e i , e i = −h∇ e i e j , e i as ∂ φ is Killing. Furthermore a direct computation shows that(3.26) h [ e ρ , e z ] , e i = | η | e U − α ( A ρ,z − A z,ρ ) = 0 , where (3.16) was used. Therefore(3.27) h∇ e j e i , e i = h∇ e i e j , e i , and it follows that the first sum in (3.22) vanishes. Hence | J | g = 0. EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 11
Consider now the energy density(3.28) 16 πµ = R + ( T r g k ) − | k | g = R − | k | g . A lengthy computation (7.11) in Appendix A shows that(3.29) R − | k | g = − div g k )( u ∇ f ) + 16 π ( µ − J ( v )) + | k | g − | π | g + 2( div g k )( v ) − div g π )( v ) , when equation (2.37) is satisfied. However, | J | g = 0 and (3.16) imply that the right-hand sidevanishes. Thus µ = 0, and ( M , g, k ) is a vacuum initial data set.Next, since m = J we may now apply the results of [10] and [21] to conclude that ( M , g, k ) isisometric to the t = 0 slice ( R − { } , g EK , k EK ) of the extreme Kerr spacetime EK . Consider themap M → EK given by x ( x, f ( x )). The induced metric on the graph is given by(3.30) ( g EK ) ij − f i ( Y EK ) j − f j ( Y EK ) i − ( u EK − | Y EK | g EK ) f i f j , where(3.31) ( k EK ) ij = 12 u EK (cid:0) ∇ EKi ( Y EK ) j + ∇ EKj ( Y EK ) i (cid:1) , and ( u EK , − Y EK ) are the lapse and shift. If ∂ φ denotes the spacelike Killing field in this spacetime,then g ijEK ( Y EK ) j ∂ i = Y φEK ∂ φ , and Y φEK satisfies equation (2.30) with g replaced by g EK , as well asboundary condition (2.31). Since there is a unique solution to (2.30), (2.31), and g ∼ = g EK , we havethat Y = Y EK . Moreover it is a direct calculation to find that u EK = e U EK = e U = u , where U EK arises from the Brill coordinate expression for g EK . It now follows from (2.11) and (2.12) that g agrees with the induced metric (3.30). Furthermore, from (3.16) π = k , showing that the secondfundamental form of the embedding ( M, g ) ֒ → EK is given by k . Therefore the initial data ( M, g, k )arise from the extreme Kerr spacetime.Lastly, if (
M, g, k ) arises from extreme Kerr, then by the properties of this spacetime, equality in(3.14) holds. (cid:3)
Theorem 3.1 reduces the proof of the mass-angular momentum inequality, in the general non-maximal case, to the existence of a solution ( u, Y φ , f ) to the system of equations (2.30), (2.37), and(3.8). Notice that this is in fact a coupled system, as the definition of u depends on g . The first task,which is addressed in the next section, is to analyze the given asymptotic boundary value problemsassociated with each equation (2.30) and (2.37). Before doing so, however, we record the reductionstatement for multiple black holes. Let(3.32) F ( J , . . . , J N )denote the numerical value of the action functional (3.3) evaluated at the harmonic map, from R − { ρ = 0 } to the two-dimensional hyperbolic space, constructed in Proposition 2.1 of [8]. Whetherthe square of this value agrees with(3.33) J = N X n =1 J n is an important open problem. The proof of the following theorem is analogous to that of Theorem3.1. Theorem 3.2.
Let ( M, g, k ) be a smooth, simply connected, axially symmetric initial data set sat-isfying the dominant energy condition (1.3) and conditions (1.8) , (3.12) , and with N + 1 ends, onedesignated asymptotically flat and the others either asymptotically flat or asymptotically cylindrical. If the system of equations (2.30) , (2.37) , (3.8) admits a smooth solution ( u, Y φ , f ) satisfying theasymptotics (2.31) , (2.40) - (2.42) , (3.9) - (3.11) , then (3.34) m ≥ F ( J , . . . , J N ) . A Lower Bound for Area in Terms of Mass and Angular Momentum
In this section we observe that the reduction argument given above, immediately applies to anothergeometric inequality for axisymmetric black holes. Let (
M, g, k ) be as in the previous section, withthe restriction that it possesses only two ends denoted M ± end , such that M + end is asymptotically flatand M − end is either asymptotically flat or asymptotically cylindrical. Based on the heuristic argumentsof Section 1 leading to the mass-angular momentum inequality (1.1), combined with the Hawkingarea theorem [17], the following upper and lower bounds are derived [14](4.1) m − p m − J ≤ A min π ≤ m + p m − J , where A min is the minimum area required to enclose M − end . In [14] the lower bound is establishedin the maximal case, and it is also shown that equality occurs if and only if the initial data set isisometric to the t = 0 slice of the extreme Kerr spacetime. The proof relies upon the mass-angularmomentum inequality and the area-angular momentum inequality A min ≥ π |J | ([9], [13]). In thenon-maximal case, the area-angular momentum inequality has also been established when A min isreplaced by the area of a stable, axisymmetric, marginally outer trapped surface ([9], [12]). Thus,since we have shown (in the previous section) how to reduce the non-maximal case of the mass-angular momentum inequality to the problem of solving a coupled system of elliptic equations, ananalogous lower bound for area may also be reduced to the same problem. More precisely, Theorem3.1 combined with Theorem 1.1 in [9] and the proof of a Theorem 2.5 in [14], produces the followingresult. Theorem 4.1.
Let ( M, g, k ) be a smooth, simply connected, axially symmetric initial data set sat-isfying the dominant energy condition (1.3) and conditions (1.8) , (3.12) , and with two ends, onedesignated asymptotically flat and the other either asymptotically flat or asymptotically cylindrical.If the data possesses a stable axisymmetric marginally outer trapped surface with area A , and the sys-tem of equations (2.30) , (2.37) , (3.8) admits a smooth solution ( u, Y φ , f ) satisfying the asymptotics (2.31) , (2.40) - (2.42) , (3.9) - (3.11) , then (4.2) A π ≥ m − p m − J and equality holds if and only if ( M, g, k ) arises from an embedding into the extreme Kerr spacetime. The Equation for f Let (
M, g, k ) be a simply connected, axisymmetric initial data set with two ends denoted M ± end ,such that M + end is asymptotically flat and M − end is either asymptotically flat or asymptotically cylin-drical. As discussed above there is a global Brill coordinate system ( ρ, φ, z ) in which the metric takesthe form (2.1). Here we make a change of coordinates to ( r, φ, θ ), where ρ = r sin θ and z = r cos θ .The metric may then be expressed by(5.1) g = e − U +2 α ( dr + r dθ ) + e − U r sin θ ( dφ + A r dr + A θ dθ ) . In addition to (2.2)-(2.4), it is assumed that the initial data and u satisfy the following asymptotics(5.2) u = 1 + o ( r − ) , T r g k = O ( r − − ε ) , in M + end , EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 13 for some ε ∈ (0 , u = r + o ( r ) , T r g k = O ( r ) , in asymptotically flat M − end , (5.4) u = r + o ( r ) , T r g k = O ( r ) , in asymptotically cylindrical M − end . Note that the asymptotics for u are consistent with the choice (3.8) and the asymptotics (2.2)-(2.4),while the asymptotics for T r g k are weaker in M + end , and stronger in asymptotically flat M − end , ascompared with (1.4).In local coordinates, with the help of (2.34), equation (2.37) is given by(5.5) g ij u ∇ ij f + u i f j + u j f i p u |∇ f | − k ij ! = 0 . Observe that this equation may also be expressed in divergence form(5.6) div g ( u ∇ f ) = u ( T r g k ) q u |∇ f | g . The desired asymptotics are(5.7) | f | + r |∇ f | g + r |∇ f | g ≤ cr − ε in M + end , (5.8) r − |∇ f | g + r − |∇ f | g ≤ c in asymptotically flat M − end , (5.9) |∇ f | g + |∇ f | g ≤ cr in asymptotically cylindrical M − end , where c is a constant.We first solve (5.6) on the annular domain Ω r = { ( r, φ, θ ) | r − < r < r } , with zero Dirichletboundary conditions(5.10) ∆ g f + 2 ∇ log u · ∇ f = u − ( T r g k ) q u |∇ f | g in Ω r , f = 0 on ∂ Ω r . Proposition 5.1.
Given initial data ( M, g, k ) and a smooth positive function u , there exists a unique,smooth, uniformly bounded (independent of r ) solution f of (5.10) .Proof. We will employ the continuity method. Thus, consider the family of equations(5.11) div g ( u ∇ f s ) = su ( T r g k ) q u |∇ f s | g in Ω r , f s = 0 on ∂ Ω r , and the set S = { s ∈ [0 , | there exists a unique solution f s ∈ C ,β (Ω r ) of (5.11) } . Clearly S isnonempty, since the case s = 0 is solved by f = 0. Moreover S is open by the implicit functiontheorem, since the linearized equation is strictly elliptic with no zeroth order term. It remains toshow that S is closed. This will be based on the construction of radial sub and super solutions.Let f = f ( r ) be a radial function. Then∆ g f = 1 √ det g ∂ i (cid:16)p det gg ij ∂ j f (cid:17) = 1 √ det g ∂ r (cid:16)p det gg rr ∂ r f (cid:17) = g rr (cid:16) f ′′ + ∂ r log( r e − U ) f ′ (cid:17) , (5.12)where we have used(5.13) g rr = e U − α , g rθ = 0 , g θθ = r − e U − α , det g = r e − U +4 α sin θ. Notice also that(5.14) ∇ log u · ∇ f = g rr ( ∂ r log u ) f ′ , ∇ f · ∇ f = g rr ( ∂ r f ) f ′ . Let e u = e u ( r ) > e U = e U ( r ), and h = h ( r ) > u − e u = o ( r ) , e U − e e U = o ( r ) , as r → , (5.16) u − e u = o ( r − ) , e U − e e U = o ( r − ) , as r → ∞ , (5.17) ( g rr u ) − T r g kh = o (1) as r → r → ∞ , ( g rr u ) − | T r g k | ≤ h, and set X s = u |∇ f s | − g ∇ f s , then∆ g f + 2 ∇ log u · ∇ f − u − ( T r g k ) X s · ∇ f = g rr (cid:18) f ′′ + ∂ r log( r u e − U ) f ′ − ( T r g k ) ∂ r f s |∇ f s | g f ′ (cid:19) = g rr (cid:18) f ′′ + ∂ r log( r e u e − e U ) f ′ + (cid:18) ∂ r log (cid:18) u e − U e u e − e U (cid:19) − ( T r g k ) ∂ r f s |∇ f s | g (cid:19) f ′ (cid:19) . (5.18) Lemma 5.2.
Given initial data ( M, g, k ) and a smooth positive function u satisfying (2.2) - (2.4) and (5.2) - (5.4) , there exist negative ( positive ) radial sub ( super ) solutions f ( f ) of (5.11) , which areindependent of s and uniformly bounded, and satisfy the asymptotics (5.7) - (5.9) .Proof. The super solution f will be chosen as a solution of the ODE(5.19) f ′′ + (cid:16) ∂ r log( r e u e − e U ) − b (cid:17) f ′ = − h, where b = b ( r ) > (cid:12)(cid:12)(cid:12)(cid:12) ∂ r log (cid:18) u e − U e u e − e U (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) + | T r g k | ≤ b, b ( r ) = O ( r − ) as r → ∞ . Consider the solution of (5.19) with lim r →∞ f ( r ) = 0 and f ′ (0) = 0, that is(5.21) f ( r ) = Z ∞ r (cid:18) r − e u − e e U e R r b Z r r e u e − e U e − R r b h (cid:19) . In order to see that (5.21) is indeed a super solution, use the fact that f ′ ≤ h to find∆ g f + 2 ∇ log u · ∇ f = g rr (cid:18) − h + (cid:18) b + ∂ r log (cid:18) u e − U e u e − e U (cid:19)(cid:19) f ′ (cid:19) ≤ g rr ( − h − | T r g k || f ′ | ) ≤ − u − | T r g k | (1 + u |∇ f | g ) ≤ − u − | T r g k | q u |∇ f | g ≤ su − ( T r g k ) q u |∇ f | g . (5.22)Moreover, with the help of (5.2)-(5.4) and (5.15)-(5.17), it is readily checked that f satisfies thedesired asymptotics. Finally, analogous methods may also be used to construct a subsolution. (cid:3) EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 15
We are now in a position to make uniform C bounds for solutions of (5.11). Thus, use the thirdline of (5.22) to find that∆ g ( f s − f ) + 2 ∇ log u · ∇ ( f s − f ) ≥ su − ( T r g k ) q u |∇ f s | g + u − | T r g k | (1 + u |∇ f | g ) ≥ su − | T r g k | (cid:18) − − X s · ∇ f s + 1 + u ∇ f s |∇ f s | g · ∇ f (cid:19) = − su − | T r g k | X s · ∇ ( f s − f ) . (5.23)Since f >
0, we have that ( f s − f ) | ∂ Ω r <
0. Analogous but opposite inequalities hold when applyingthe sub solution. It now follows from a comparison argument that(5.24) f ≤ f s ≤ f . In order to produce higher order estimates, observe that the right-hand side of equation (5.11) isof linear growth in the first derivatives of f s . Therefore, by slightly modifying standard techniquesapplied to the Dirichlet problem for linear elliptic equations, we obtain uniform C ,γ estimates for any γ ∈ [0 , S is closed, yielding a solution f = f ∈ C ,β (Ω r )of (5.10). Elliptic regularity then implies that f is smooth.In order to prove uniqueness, assume that two solutions f and f exist, then∆ g ( f − f ) + 2 ∇ log u · ∇ ( f − f ) = u − ( T r g k ) (cid:16)p u |∇ f | − p u |∇ f | (cid:17) = u ( T r g k ) ∇ ( f + f ) · ∇ ( f − f ) p u |∇ f | + p u |∇ f | ! . (5.25)Since ( f − f ) | ∂ Ω r = 0, the maximum/minimum principle implies that f − f ≡ (cid:3) We are now ready to establish the main theorem of this section.
Theorem 5.3.
Given initial data ( M, g, k ) and a smooth positive function u satisfying (2.2) - (2.4) and (5.2) - (5.4) , there exists a smooth uniformly bounded solution f of (5.5) satisfying the asymptotics (5.7) - (5.9) . Notice that the asymptotic behavior for f on M − end is not prescribed, but instead it is stated thatthe solution remains uniformly bounded and fall-off rates for its derivatives are given. Proof.
The same methods as in the proof of Proposition 5.1, yield uniform estimates for the solution f r of (5.10) in C ,βloc . Thus, as r → ∞ a subsequence may be extracted which converges on compactsubsets to a solution f of (5.5). By elliptic regularity, this solution is smooth. Moreover, in light of(5.24) we have the bound(5.26) f ≤ f ≤ f , showing that f is uniformly bounded and has the appropriate asymptotics at M + end .It remains to establish the asymptotics for derivatives. First consider the end M + end . Here we mayfollow the standard scaling argument in [22]. From the local C ,β estimates we have(5.27) | f ( x ) | + |∇ f ( x ) | g + |∇ f ( x ) | g ≤ c for x ∈ M. Moreover it is also known from (5.26) that(5.28) | f ( x ) | ≤ c | x | − ε as | x | → ∞ in M + end . Equation (5.10) may be viewed as the following linear equation(5.29) X i,j =1 a ij ( x ) ∂ ∂x i ∂x j f + X i =1 b i ( x ) ∂∂x i f = F ( x ) , where(5.30) a ij = g ij , b i = − g lm Γ ilm + 2(log u ) i , F = u − ( T r g k ) q u |∇ f | g . Now fix a point x ∈ M + end and define coordinates x = ( x − x ) /σ , where σ = | x | /
2. When writtenin these new coordinates, equation (5.29) becomes(5.31) X i,j =1 a ij ( x ) ∂ ∂x i ∂x j f + X i =1 σb i ( x ) ∂∂x i f = σ F ( x )for x ∈ B (0) = {| x | < } . Observe that ∂ x i = σ∂ x i , and therefore(5.32) (cid:12)(cid:12)(cid:12)(cid:12) ∂∂x l a ij ( x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ cσ | x | ≤ c | x | , (5.33) σ | b i ( x ) | + σ (cid:12)(cid:12)(cid:12)(cid:12) ∂∂x l b i ( x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ c σ | x | + σ | x | ! ≤ c | x | , (5.34) σ | F ( x ) | + σ (cid:12)(cid:12)(cid:12)(cid:12) ∂∂x l F ( x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ c (cid:18) σ | x | ε + σ | x | ε (cid:19) ≤ c | x | ε , where c represents a constant depending on the initial data and u . These estimates show that wehave control of the coefficients, and right-hand side of (5.31), in C ,β . Thus the interior Schauderestimates apply to yield | ∂f ( x ) | + | ∂ f ( x ) | ≤ c ( σ | F | C ,α ( B (0)) + | f | C ( B (0)) ) ≤ c ( σ | F | C ( B (0)) + | f | C ( B (0)) )(5.35)for x ∈ B / (0). It follows that σ | ∂f ( x ) | + σ | ∂ f ( x ) | ≤ c ( σ | F | C ( B (0)) + | f | C ( B (0)) ) ≤ c | x | − ε (5.36)for x ∈ B σ/ ( x ), and hence the desired estimate holds(5.37) | f ( x ) | + | x ||∇ f ( x ) | g + | x | |∇ f ( x ) | g ≤ c | x | − ε for x ∈ M + end . Derivative estimates in the remaining end will be divided into two cases.
Case 1: M − end is asymptotically flat. By performing an inversion x x | x | near the origin in Brillcoordinates, asymptotically flat coordinates are obtained in M − end . We may now apply the samescaling argument as above. However here, u ∼ | x | − and T r g k ∼ | x | − so that(5.38) σ | b i ( x ) | + σ (cid:12)(cid:12)(cid:12)(cid:12) ∂∂x l b i ( x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ c (cid:18) σ | x | + σ | x | (cid:19) ≤ c, (5.39) σ | F ( x ) | + σ (cid:12)(cid:12)(cid:12)(cid:12) ∂∂x l F ( x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ c (cid:18) σ | x | + σ | x | (cid:19) ≤ c. EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 17
It follows that(5.40) σ | ∂f ( x ) | + σ | ∂ f ( x ) | ≤ c ( σ | F | C ( B (0)) + | f | B (0) ) ≤ c for x ∈ B σ/ ( x ) , and hence the desired estimate holds(5.41) | x ||∇ f ( x ) | g + | x | |∇ f ( x ) | g ≤ c for x ∈ M − end . Case 2: M − end is asymptotically cylindrical. According to the asymptotics (2.4)(5.42) g = r − dr + g S ( φ, θ ) + G ( r, φ, θ ) , where g S is the round metric on the 2-sphere and the remainder satisfies(5.43) G rr = o ( r − ) , G ij = o ( r ) for i, j = r. A further change of coordinates τ = log r , in which dτ = r − dr , displays the canonical cylindricalform of the metric(5.44) g = dτ + g S ( φ, θ ) + G ( τ, φ, θ ) , with(5.45) | G ij | + | ∂G ij | + | ∂ G ij | = o ( e τ ) for all i, j. The interior Schauder estimates imply that(5.46) | f | + |∇ f | g + |∇ f | g ≤ c, and hence(5.47) | ∂ τ f | + | ∂ θ f | + | ∂ τ f | + | ∂ τ ∂ θ f | + | ∂ θ f | ≤ c. It then follows, with the help of (5.4), that(5.48) ∇ log u · ∇ f = ∂ τ f + O ( e τ )and(5.49) ∆ g f = ∂ τ f + ∆ S f + O ( e τ ) . We now obtain derivative estimates using a separation of variables argument. Let { ψ i } ∞ i =0 ⊂ L ( S ) be an complete orthonormal set of eigenfunctions for ∆ S , and let λ i = i ( i + 1) denote thecorresponding eigenvalues. Since the eigenfunctions are complete, we may write(5.50) f ( τ, φ, θ ) = ∞ X i =0 d i ( τ ) ψ i ( φ, θ ) . Inserting this into equation (5.10) produces(5.51) ∞ X i =0 ( d ′′ i + 2 d ′ i − λ i d i ) ψ i = P, and thus(5.52) d ′′ i + 2 d ′ i − λ i d i = P i ( τ ) := Z S P ( τ, φ, θ ) ψ i ( φ, θ ) . The general solution to this ODE is given by the method of variation of parameters(5.53) d i ( τ ) = ( c i + p i ( τ )) e ( − −√ λ i ) τ + ( c i + p i ( τ )) e ( − √ λ i ) τ , where c i and c i are constants and(5.54) p i ( τ ) = − √ λ i Z τ −∞ e ( √ λ i ) τ P i ( τ ) dτ, p i ( τ ) = 12 √ λ i Z ττ e ( −√ λ i ) τ P i ( τ ) dτ, for some τ . Note that the boundedness of f implies that c i = 0. Moreover(5.55) d ( τ ) = c + p ( τ ) + e − τ p ( τ ) = c + O ( e τ ) , and P = O ( e τ ) implies that(5.56) d i ( τ ) = O (cid:16) max { e τ , e ( − √ λ i ) τ } (cid:17) , i ≥ . Since − √ λ i > /
10 for i ≥
1, it follows that(5.57) | ∂ τ f | + | ∂ θ f | = O ( e τ ) , and hence(5.58) |∇ f | g = O ( r ) . By differentiating the expansion, similar considerations yield(5.59) |∇ f | g = O ( r ) . (cid:3) The Equation for Y φ Let (
M, g, k ), u , and f be as in the previous section, although f is not required to satisfy anequation here. In particular, u and f satisfy the asymptotics (5.2)-(5.4) and (5.7)-(5.9). In thissection we solve the equation(6.1) div g k ( η ) = 0 on M, for solutions satisfying the following asymptotics(6.2) Y φ = − J r + o ( r − ) in M + end , (6.3) Y φ = Y + O ( r ) in asymptotically flat M − end , (6.4) Y φ = Y + O ( r ) in asymptotically cylindrical M − end , where J and Y are constants. In order to obtain a unique solution the value of J will be prescribed,and in this case the value of Y is determined by J and the initial data. Note that these asymptoticsare consistent with those of the (sole component of the) shift vector field Y EK in the extreme Kerrspacetime, as is shown in Appendix B, Section 8.The equation (6.1) may be expressed in a more revealing way as follows(6.5) ∆ g Y φ + ∇ log( u − g φφ ) · ∇ Y φ = 0 . EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 19
Since the metric g depends on Y φ , it appears that this should be a nonlinear equation. Howevercertain cancelations occur, and it turns out that when expressed in terms of the metric g , the equationis linear elliptic 0 = (cid:18) g ij − u f i f j u |∇ f | g (cid:19) ∇ ij Y φ − uπ ij f l q u |∇ f | g ∂ l Y φ + (cid:18) g ij − u f i f j u |∇ f | g (cid:19) (cid:18) ∂ i log g φφ − ∂ i log u u |∇ f | g (cid:19) ∂ j Y φ , (6.6)where(6.7) π ij = u q u |∇ f | g ∇ ij f + (log u ) i f j + (log u ) j f i + g iφ Y φ,j + g jφ Y φ,i u ! is the second fundamental form of the graph M = { t = f ( x ) } in the Lorentzian setting. It isimportant to note that the linear character of the equation, arises from the fact that(6.8) (cid:18) g ij − u f i f j u |∇ f | g (cid:19) π ij does not depend on Y φ . The equivalence of the three equations (6.1), (6.5), and (6.6) will be provedin Appendix B.We now prove existence and uniqueness. The first task is to construct a radial function Y = Y ( r )which is an approximate solution in the asymptotic regions. Following the same procedure as in(5.18) yields(6.9) ∆ g Y + ∇ log( u − g φφ ) · ∇ Y = g rr (cid:18) Y ′′ + ∂ r log( r e u − e − e U ) Y ′ + ∂ r log (cid:18) u − e − U e u − e − e U (cid:19) Y ′ (cid:19) , where e u and e e U are defined in (5.15) and (5.16), and where we have used g φφ = e − U r sin θ . Theremaining terms may be computed in a similar way. If L denotes the differential operator on theright-hand side of (6.6), then LY = (cid:18) g rr − u ( f r ) u |∇ f | g (cid:19) Y ′′ + g rr ∂ r log( r e u − e − e U ) Y ′ + g rr ∂ r log (cid:18) u − e − U e u − e − e U (cid:19) + u f i f j Γ rij u |∇ f | g − u f r f i ∂ i log g φφ u |∇ f | g ! Y ′ (6.10) + u f r f i ∂ i log u (1 + u |∇ f | g ) + g rr u |∇ f | g ∂ r log u u |∇ f | g − (cid:18) g ij − u f i f j u |∇ f | g (cid:19) uπ ij f r q u |∇ f | g Y ′ . Note that the term f i ∂ i log g φφ appears to cause a problem when θ = 0 , π , since it involves f θ ∂ θ log sin θ = g θθ f θ ∂ θ log sin θ which could blow-up at those values of θ . However, for axisymmetric smooth func-tions f , one must have ∂ θ f = 0 on the axis of rotation. Observe that equation (6.10) may be writtenmore simply as(6.11) LY = (cid:18) g rr − u ( f r ) u |∇ f | g (cid:19) (cid:16) Y ′′ + (cid:16) ∂ r log( r e u − e − e U ) + B (cid:17) Y ′ (cid:17) , for some function B . With the help of (5.2)-(5.4), (5.7)-(5.9), (5.15), (5.16), and the calculation ofChristoffel symbols in Appendix D, it can be shown that this function has the property that(6.12) B = O ( r − ) as r → ∞ , B = O ( r ) as r → . We are motivated to choose Y as the solution to the ODE(6.13) Y ′′ + ∂ r log( r e u − e − e U ) Y ′ = 0which satisfies the asymptotics (6.2)-(6.4), namely(6.14) Y ( r ) = c Z ∞ r r − e ue e U where the constant c is chosen in order to realize the desired r − -fall-off rate in (6.2).The desired solution of (6.6) will be constructed in the form Y φ = Y + Y , where Y solves theequation(6.15) LY = − LY , and has the same asymptotics as in (6.2)-(6.4) with J = 0. Theorem 6.1.
Given initial data ( M, g, k ) and smooth functions u > , f satisfying (2.2) - (2.4) , (5.2) - (5.4) , and (5.7) - (5.9) , there exists a unique, smooth, uniformly bounded solution Y φ of (6.6) satisfying the asymptotics (6.2) - (6.4) .Proof. The first task is to construct radial sub and super solutions Y and Y . The super solutionwill be chosen as a solution of the ODE(6.16) Y ′′ + (cid:16) ∂ r log( r e u − e − e U ) − B (cid:17) Y ′ = − h, where the radial functions B > h > | B | ≤ B, (cid:18) g rr − u ( f r ) u |∇ f | g (cid:19) − | LY | ≤ h, with B satisfying the asymptotics (6.12), whereas(6.18) h = O ( r − ) in M + end , (6.19) h = O ( r ) in asymptotically flat M − end , (6.20) h = O (1) in asymptotically cylindrical M − end . To see that Y is a super solution, observe that (6.11) and (6.17) imply LY = (cid:18) g rr − u ( f r ) u |∇ f | g (cid:19) (cid:16) Y ′′ + (cid:16) ∂ r log( r e u − e − e U ) + B (cid:17) Y ′ (cid:17) = (cid:18) g rr − u ( f r ) u |∇ f | g (cid:19) (cid:16) − h + (cid:0) B + B (cid:1) Y ′ (cid:17) ≤ − LY , (6.21)if Y ′ ≤
0. Thus we choose(6.22) Y = Z ∞ r e R r B r − e ue e U Z r e − R r B r e u − e − e U h. EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 21
Clearly Y is positive, nonincreasing, and satisfies the asymptotics (6.2)-(6.4) with J = 0. Similarmethods yield a negative, nondecreasing subsolution Y satisfying the same asymptotics.Since (6.15) is linear elliptic without a zeroth order term, there exists a unique smooth solution Y r , with zero Dirichlet boundary conditions, on the annular domain Ω r = { ( r, φ, θ ) | r − < r < r } .By the maximum/minimum principle(6.23) Y < Y r < Y . The interior Schauder estimates now yield uniform C ,βloc bounds, and thus after passing to a subse-quence, we obtain a C solution Y on M as r → ∞ . This solution is smooth by elliptic regularityand satisfies the estimate(6.24) Y < Y < Y . Asymptotics for the derivatives may be established as in the proof of Theorem 5.3. Namely, in M + end a scaling argument is employed, and if Y denotes lim r → Y then the same is true for asymptoticallyflat M − end , although here one must apply the method to Y − Y . Moreover, asymptotics for thederivatives in asymptotically cylindrical M − end may be obtained through the use of eigenfunctionexpansions.We now have a solution Y φ = Y + Y of (6.6), satisfying the asymptotics (6.2)-(6.4). It remains toprove uniqueness. Thus, consider a solution of LZ = 0 having the asymptotics (6.2)-(6.4) with J = 0.We must show that Z ≡
0. A direct calculation shows that K := u − g φφ q u |∇ f | g ∈ Ker g L ∗ ,the kernel of the formal L ( M, g ) adjoint of L . Alternatively this may be proved by observing, theimmediately apparent fact, that u − g φφ lies in the L ( M , g )-kernel of the adjoint of the operator(6.5), and using (2.22) as well as (8.22). Upon multiplying the equation LZ = 0 through by Z K andintegrating by parts, several boundary terms cancel to yield0 = Z M K (cid:18) g ij − u f i f j u |∇ f | g (cid:19) ∇ i Z ∇ j Z + lim r →∞ Z ∂B ( r ) Z K (cid:18) g ij − u f i f j u |∇ f | g (cid:19) ν i ∇ j Z − lim r → Z ∂B ( r ) Z K (cid:18) g ij − u f i f j u |∇ f | g (cid:19) ν i ∇ j Z, (6.25)where ν is the unit normal to ∂B ( r ) pointing towards M + end . Note that the boundary term at M + end clearly vanishes, while the boundary term at M − end becomes(6.26) lim r → Y Z ∂B ( r ) K (cid:18) g ij − u f i f j u |∇ f | g (cid:19) ν i ∇ j Z, where Y = lim r → Y . Now multiply the equation LZ = 0 by K alone to find(6.27) 0 = lim r →∞ Z ∂B ( r ) K (cid:18) g ij − u f i f j u |∇ f | g (cid:19) ν i ∇ j Z − lim r → Z ∂B ( r ) K (cid:18) g ij − u f i f j u |∇ f | g (cid:19) ν i ∇ j Z. Again, the boundary term at M + end vanishes, and hence the boundary term at M − end vanishes, whichimplies that (6.26) vanishes. It then follows from (6.25) that Z ≡ (cid:3) Appendix A: The Scalar Curvature Formula
The purpose of this section is to prove Theorem 2.3. We will follow the ideas in [2]. In particular,(
M, g ) will be viewed as a graph { t = f ( x ) } in the Lorentzian setting (2.13). An alternative approach, in which the calculations are made by viewing ( M , g ) as the same graph in the Riemannian setting(2.9) is possible, although not pursued here; in the static case this approach was carried out in [1].Let ( M × R , e g ) denote the Lorentzian stationary spacetime with(7.1) e g = g − Y i dx i dt − ϕdt . The induced metric on the graph and its inverse are given by(7.2) g ij = g ij − f i Y j − f j Y i − ϕf i f j , g ij = g ij − u − Y i Y j + w i w j , where(7.3) w i = ug ij f j + u − Y i q − u |∇ f | g , u = ϕ + | Y | g . The vector w is in fact the spatial component of the unit normal to the graph. In particular, theunit normal to the t = 0 slice and the graph are given given, respectively, by(7.4) n = ∂ t + Yu , N = u ∇ f + n q − u |∇ f | g . Note that ∂ t is a killing vector on ( M × R , e g ). Thus there is an obvious one-to-one correspondencebetween M = { t = f ( x ) } and M = { t = 0 } . In that sense, decomposing n into its normal andtangential components with respect to the graph, and decomposing N into its normal and tangentialcomponents with respect to the t = 0 slice, yields(7.5) n = q u |∇ f | g N − u ∇ f, N = q u |∇ f | g ( n + u ∇ f ) . Here the identity (2.17) was used.Let G = Ric e g − R e g e g denote the Einstein tensor. Using (7.5) and the Gauss-Codazzi relations G ( N, n ) may be computed in two different ways, namely G ( N, n ) = q u |∇ f | g ( G ( n, n ) + G ( u ∇ f, n ))= q u |∇ f | g (cid:2) ( R + ( T r g k ) − | k | g ) / div g ( k − ( T r g k ) g )( u ∇ f ) (cid:3) , (7.6)and G ( N, n ) = q u |∇ f | g G ( N, N ) − G ( N, u ∇ f )= q u |∇ f | g ( R + ( T r g π ) − | π | g ) / − div g ( π − ( T r g π ) g )( u ∇ f ) , (7.7)where k and π denote the second fundamental forms of M and M , respectively. It follows that R + ( T r g k ) − | k | g + 2 div g ( k − ( T r g k ) g )( u ∇ f )= R + ( T r g π ) − | π | g − div g ( π − ( T r g π ) g )( v )(7.8)where(7.9) v = u ∇ f q u |∇ f | g . EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 23
The energy density and momentum densities for the initial data (
M, g, k ) are8 πµ = G ( N, N ) = ( R + ( T r g k ) − | k | g ) / πJ ( · ) = G ( N, · ) = div g ( k − ( T r g k ) g )( · ) . (7.10)Thus, combining this with the fact that T r g k = 0 (Lemma 2.1) produces R − | k | g + 2 div g k ( u ∇ f ) =16 π ( µ − J ( v )) − | π | g + | k | g − div g ( π )( v ) + 2 div g ( k )( v )+ ( T r g π ) − ( T r g k ) + 2 v ( T r g π − T r g k ) . (7.11)In what follows, several identities involving significant computation will be proven, which whencombined with (7.11) will yield the desired result. The first task is to calculate the second funda-mental form π . Below, i, j, l etc. will denote indices associated with local coordinates x i on M , and e Γ lij , Γ lij , Γ lij will represent Christoffel symbols for the metrics e g , g , and g , respectively. Observe thatthe inverse metric is(7.12) e g tt = − u − , e g ti = − u − Y i , e g ij = g ij − u − Y i Y j , from which we find(7.13) e Γ ttt = 0 , e Γ itt = 12 g ij ϕ j = 12 ϕ i , e Γ tij = u − k ij , e Γ kij = Γ lij + u − Y l k ij , (7.14) e Γ tit = 12 u (cid:16) ϕ i + Y l ( Y l,i − Y i,l ) (cid:17) , e Γ jit = − g jl ( Y l,i − Y i,l ) + Y j u (cid:16) ϕ i + Y l ( Y l,i − Y i,l ) (cid:17) . Let X i = ∂ i + f i ∂ t denote tangent vectors to the graph, then π ij = − e g ( e ∇ X i X j , N )= − ( e Γ lij + f i e Γ ljt + f j e Γ lit + f i f j e Γ ltt ) e g ∂ l , u ∇ f q − u |∇ f | g − ( ∂ ij f + e Γ tij + f i e Γ tjt + f j e Γ tit + f i f j e Γ ttt ) e g ∂ t , n q − u |∇ f | g = u q − u |∇ f | g (cid:16) ∇ ij f + u − k ij + f i ( e Γ tjt − e Γ ljt f l ) + f j ( e Γ tit − e Γ lit f l ) + f i f j ( e Γ ttt − e Γ ltt f l ) (cid:17) . (7.15)Therefore(7.16) π ij = u ∇ ij f + k ij + u ( f i ϕ j + f j ϕ i ) − u f i f j ∇ f ( ϕ ) q − u |∇ f | g + 12 f i w l ( Y l,j − Y j,l ) + 12 f j w l ( Y l,i − Y i,l ) . This formula for the second fundamental form involves quantities associated with g . In order toobtain a formula involving only quantities associated with g , we will employ Identity 1 below forthe difference of Christoffel symbols. In particularΓ lij − Γ lij = Y l u ( ∇ j Y i + ∇ i Y j ) − w l π ij + f i f j e Γ ltt + f i e Γ ljt + f j e Γ lit = Y l u ( ∇ j Y i + ∇ i Y j ) + Y l u (Γ mij − Γ mij ) Y m − w l π ij + f i f j e Γ ltt + f i e Γ ljt + f j e Γ lit . (7.17) Multiply by Y l and solve for (Γ lij − Γ lij ) Y l to obtain(7.18) ϕu (Γ lij − Γ lij ) Y l = | Y | g u ( ∇ j Y i + ∇ i Y j ) − π ij w l Y l + f i f j e Γ ltt Y l + f i e Γ ljt Y l + f j e Γ lit Y l . For simplicity we temporarily assume that ϕ does not vanish, however this will have no affect on thefinal result, which is valid without any such restriction on ϕ . Substituting (7.18) into the last line of(7.17) producesΓ lij − Γ lij = Y l ϕ ( ∇ j Y i + ∇ i Y j ) − π ij ( w l + ϕ − w m Y m Y l )+ f i ( e Γ ljt + ϕ − e Γ mjt Y m Y l ) + f j ( e Γ lit + ϕ − e Γ mit Y m Y l ) + f i f j ( e Γ ltt + ϕ − e Γ mtt Y m Y l ) . (7.19)Use this and Y i f i = 0 to compute the following expression from (7.16) in terms of g ∇ ij f + u − k ij = ∇ ij f + 12 u ( Y i ; j + Y j ; i ) + (Γ lij − Γ lij )( f l + u − Y l )= − π ij ( w l f l + ϕ − w l Y l ) + ∇ ij f + 12 ϕ ( ∇ j Y i + ∇ i Y j )+ f i ( e Γ ljt f l + ϕ − e Γ ljt Y l ) + f j ( e Γ lit f l + ϕ − e Γ lit Y l ) + f i f j ( e Γ ltt f l + ϕ − e Γ ltt Y l ) . (7.20)Also from (7.15)(7.21) ∇ ij f + u − k ij = u − q − u |∇ f | g π ij − f i ( e Γ tit − e Γ ljt f l ) − f j ( e Γ tit − e Γ ljt f l ) + f i f j e Γ ltt f l . Now compare (7.20) and (7.21) and solve for π to obtain uϕ q − u |∇ f | g π ij = q − u |∇ f | g u + w l f l + w l Y l ϕ π ij = ∇ ij f + 12 ϕ ( ∇ j Y i + ∇ i Y j ) + f i ( e Γ tjt + ϕ − e Γ ljt Y l ) + f j ( e Γ tit + ϕ − e Γ lit Y l )= ∇ ij f + 12 ϕ ( ∇ j Y i + ∇ i Y j ) + 12 f i (log ϕ ) j + 12 f j (log ϕ ) i . (7.22)Therefore(7.23) π ij = 1 u q u |∇ f | g (cid:18) ϕ ∇ ij f + 12 ( ∇ j Y i + ∇ i Y j ) + 12 f i ϕ j + 12 f j ϕ i (cid:19) , where (2.17) was used. Notice that Y i depends on f through the expression(7.24) Y i = Y φ g φi = Y φ ( g φi + f i Y φ ) = Y φ g φi + f i | Y | g , from which it follows that(7.25) ∇ j Y i + ∇ i Y j = 2 | Y | g ∇ ij f + ∂ i f ∂ j | Y | g + ∂ j f ∂ i | Y | g + g φj ∂ i Y φ + g φi ∂ j Y φ . Inserting (7.25) into (7.23) then produces(7.26) π ij = u ∇ ij f + u i f j + u j f i + u ( g iφ Y φ,j + g jφ Y φ,i ) q u |∇ f | g . The remaining part of the proof consists of verifying several identities.
EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 25
Identity 1. Γ lij − Γ lij = − w l π ij + u − Y l k ij + 12 f i f j ϕ l + f i − g lm ( ∇ j Y m − ∇ m Y j ) + Y l u ( ϕ j + Y m ( ∇ j Y m − ∇ m Y j )) ! + f j − g lm ( ∇ i Y m − ∇ m Y i ) + Y l u ( ϕ i + Y m ( ∇ i Y m − ∇ m Y i )) ! (7.27) Proof.
Observe that(7.28) N = w + 1 u q − |∇ f | g ∂ t , which yields e ∇ X i X j = ∇ X i X j − e g ( e ∇ X i X j , N ) N = Γ lij X l + π ij N = Γ lij X l + π ij w l X l + u − q − u |∇ f | g π ij ∂ t . (7.29)Alternatively, using (7.15) produces e ∇ X i X j =( e Γ lij + f i e Γ ljt + f j e Γ lit + f i f j e Γ ltt ) ∂ l + ( ∂ ij f + e Γ tij + f i e Γ tjt + f j e Γ tit + f i f j e Γ ttt ) ∂ t =( e Γ kij + f ,i e Γ kjt + f ,j e Γ kit + f ,i f ,j e Γ ktt ) X k + (cid:16) ∂ ij f + e Γ tij + f i e Γ tjt + f j e Γ tit + f i f j e Γ ttt − ( e Γ lij + f i e Γ ljt + f j e Γ lit + f i f j e Γ tt ) f l (cid:17) ∂ t = ( e Γ lij + f i e Γ jt + f j e Γ lit + f i f j e Γ ltt ) X l + u − q − u |∇ f | g π ij ∂ t . (7.30)Therefore by comparing (7.29) and (7.30)(7.31) Γ lij − Γ lij = − w l π ij + u − Y l k ij + f i e Γ ljt + f j e Γ lit + f i f j e Γ ltt , from which the desired result follows with the help of (7.13) and (7.14). (cid:3) It will be assumed that k and π are extended trivially to all of M × R , in that k ( ∂ t , · ) = π ( ∂ t , · ) = 0. Identity 2. (7.32) div g k ( w ) = u − div g ( uk ( w, · )) + w ( k ( w, w )) − e g ( k, π ) − e g ( k ( w, · ) , π ( w, · )) + ( T r e g π ) k ( w, w ) Proof.
Direct calculation yields div g k ( w ) = (cid:16) g ij − u − Y i Y j + w i w j (cid:17) w l ∇ j k il = (cid:16) g ij − u − Y i Y j + w i w j (cid:17) w l ( ∇ j k il − (Γ mij − Γ mij ) k ml − (Γ mjl − Γ mjl ) k im )= div g k ( w ) − u − ∇ Y k ( Y , w ) + ∇ w k ( w, w ) − (cid:16) g ij − u − Y i Y j + w i w j (cid:17) w l (cid:0) (Γ mij − Γ mij ) k ml + (Γ mjl − Γ mjl ) k im (cid:1) . (7.33)Next, each term on the right-hand side of (7.33) will be computed separately. Let(7.34) A ij = u − k ij + f i ( e Γ tjt − e Γ ljt f l ) + f j ( e Γ tit − e Γ lit f l ) − f i f j e Γ ltt f l . Identity 2-1. ∇ j w i = π ij + π ( w, ∂ j ) − π Yu q − u |∇ f | g , ∂ j w i − u j u q − u |∇ f | g Y i − u q − u |∇ f | g (cid:0) A ij − u − ∇ j Y i (cid:1) + 11 − u |∇ f | g (cid:0) (log u ) i − u A ( ∇ f, ∂ j ) (cid:1) w i (7.35) Proof.
Observe that ∇ j w i = ∇ j u ∇ f + u − Y q − u |∇ f | g i = u ( ∇ ij f + u − ∇ j Y i + Y i ∂ j u − ) q − u |∇ f | g + 11 − u |∇ f | g (cid:16) (log u ) j + u g lm f m ∇ lj f (cid:17) w i . (7.36)Now substitute the following expressions(7.37) ∇ ij f = u − q − u |∇ f | g π ij − A ij , (7.38) g lm f m ∇ lj f = (cid:18) u − q − u |∇ f | g w l − u − Y l (cid:19) (cid:18) u − q − u |∇ f | g π lj (cid:19) − A ( ∇ f, ∂ j ) . (cid:3) Identity 2-2. div g k ( w ) = div g ( k ( w, · )) − g ( k, π ) − g k ( w, · ) , π w − Yu q − u |∇ f | g , · + ug il g jm k lm (cid:0) A ij − u − ∇ j Y i (cid:1)q − u |∇ f | g + u g ( k ( w, · ) , A ( ∇ f, · ))1 − u |∇ f | g + 2 k Y , ∇ uu q − u |∇ f | g − k w, ∇ uu (1 − u |∇ f | g ) ! (7.39) Proof.
Since(7.40) div g k ( w ) = div g ( k ( w, · )) − g il g jm k lm ∇ j w i , the desired result follows from Identity 2-1 . (cid:3) Identity 2-3. (7.41) ∇ Y k ( Y , w ) = − k ( ∇ Y Y , w ) − g ( k ( Y , · ) , π ( Y , · )) + ug ( k ( Y , · ) , A ( Y , · )) q − u |∇ f | g − k (cid:0) Y , ∇ Y Y (cid:1) u q − u |∇ f | g EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 27
Proof.
Since Y = Y φ ∂ φ (7.42) ∇ Y k ( Y , w ) = Y ( k ( Y , w )) − k (cid:0) ∇ Y Y , w (cid:1) − k (cid:0) ∇ Y w, Y (cid:1) = − k (cid:0) ∇ Y Y , w (cid:1) − k (cid:0) ∇ Y w, Y (cid:1) , and(7.43) (cid:0) ∇ Y w (cid:1) i = uY l ∇ il f + u − Y l ∇ l Y i q − u |∇ f | g = π ( ∂ i , Y ) − u q − u |∇ f | g A ( ∂ i , Y ) − ( ∇ Y Y ) i u ! . Insert (7.43) into (7.42) to get (7.41). (cid:3)
Identity 2-4. ∇ w k ( w, w ) = w ( k ( w, w )) − g ( k ( w, · ) , π ( w, · )) − u − (1 − u |∇ f | g ) k ( w, w ) w ( ϕ ) − k ( w, w ) π ( w, w ) + 4 k ( w, Y ) w ( u ) + 2 u g ( k ( w, · ) , A ( w, · )) − uk ( w, ∇ w Y ) u q − u |∇ f | g (7.44) Proof.
Observe that(7.45) ∇ w k ( w, w ) = w ( k ( w, w )) − k (cid:0) ∇ w w, w (cid:1) , and with the help of Identity 2-1 (cid:0) ∇ w w (cid:1) i = w l ∇ l w i = π ( ∂ i , w ) + π ( w, w ) w i − u q − u |∇ f | g (cid:0) A ( ∂ i , w ) − u − (cid:0) ∇ w Y (cid:1) i (cid:1) − w ( u ) u q − u |∇ f | g Y i + u − w ( u ) − u A ( ∇ f, w )1 − u |∇ f | g − π ( w, Y ) u q − u |∇ f | g w i . (7.46)The last line of (7.46) will now be computed directly using definition of A , u , and k , along with Y l ∇ il f = − g lj f j ∇ i Y l , g im f m g jl f l ∇ j Y i = 0. Namely π ( w, Y ) = u q − u |∇ f | g (cid:18) w i Y j ∇ ij f + u − k ( w, Y ) + 12 w i f i Y i g jl f l ( ∇ i Y j − ∇ j Y i ) (cid:19) = − Y i g jl f l ∇ i Y j − u |∇ f | g + Y i g jl f l ( ∇ j Y i + ∇ i Y j )2(1 − u |∇ f | g ) + u |∇ f | g Y i g jl f l ( ∇ i Y j − ∇ j Y i )2(1 − u |∇ f | g )= 12 Y i g jl f l ( ∇ j Y i − ∇ i Y j ) , (7.47)(7.48) w ( u ) = w (cid:16) ϕ + | Y | g (cid:17) u , and A ( ∇ f, w ) = q − u |∇ f | g u A ( w, w ) − u A ( Y , w )= q − u |∇ f | g u (cid:18) u − k ( w, w ) + u − w i f i w ( ϕ ) −
12 ( w i f i ) ∇ f ( ϕ ) (cid:19) − u k ∇ f q − u |∇ f | g , Y + 12 w l f l Y i g jm f m ( ∇ i Y j − ∇ j Y i ) = w ( ϕ )2 |∇ f | g u − |∇ f | g ! + Y i g jm f m ( ∇ j Y i + ∇ i Y j )2 u q − u |∇ f | g + |∇ f | g Y i g jm f m ( ∇ j Y i − ∇ i Y j )2 u q − u |∇ f | g . (7.49)Upon using (7.47), (7.48), and (7.49), the last line of (7.46) simplifies by(7.50) π ( w, Y ) u q − u |∇ f | g − w ( u ) u (1 − u |∇ f | g ) + u A ( ∇ f, w )1 − u |∇ f | g = − (1 − u |∇ f | g )2 u w ( ϕ ) . The desired result is now obtained by substituting (7.46) and (7.50) into (7.45). (cid:3)
The following term from (7.33) may be rewritten by combining
Identity 2-2 , Identity 2-3 , and
Identity 2-4 div g k ( w ) − u − ∇ Y k ( Y , w ) + ∇ w k ( w, w )= div g ( k ( w, · )) + w ( k ( w, w )) − g ( k ( w, · ) , π ( w, · )) − k ( w, w ) π ( w, w ) − g ( k, π )+ u − g ( k ( Y , · ) , π ( Y , · )) + u − k (cid:0) ∇ Y Y , w (cid:1) − u − (1 − u |∇ f | g ) k ( w, w ) w ( ϕ )+ 4 k ( w, Y ) w ( u ) u q − u |∇ f | g − k (cid:0) w, ∇ w Y (cid:1) u q − u |∇ f | g + k (cid:0) Y , ∇ Y Y (cid:1) u q − u |∇ f | g + 2 k (cid:0) Y , ∇ u (cid:1) u q − u |∇ f | g − k (cid:0) w, ∇ u (cid:1) u (1 − u |∇ f | g ) + ug (cid:0) k, (cid:0) A − u − ∇ Y (cid:1)(cid:1)q − u |∇ f | g + 3 ug ( k ( w, · ) , A ( w, · )) q − u |∇ f | g + g ( k ( w, · ) , π ( Y , · )) u q − u |∇ f | g − g ( k ( w, · ) , A ( Y , · ))1 − u |∇ f | g − g ( k ( Y , · ) , A ( Y , · )) u q − u |∇ f | g . (7.51)Each term involving A will now be computed with (7.13), (7.14), and(7.52) ∇ f = u − q − u |∇ f | g w − u − Y .
EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 29
Namely ug (cid:0) k, (cid:0) A − u − ∇ Y (cid:1)(cid:1)q − u |∇ f | g = 2 k w − Yu q − u |∇ f | g , g jl ( e Γ tjt − e Γ mjt f m ) ∂ l − e Γ ltt f l k w − Yu q − u |∇ f | g , u − q − u |∇ f | g w − u − Y = 2 k w − Yu q − u |∇ f | g , ∇ ϕ u + q − u |∇ f | g u g jl w i ( ∇ j Y i − ∇ i Y j ) ∂ l (7.53) − ∇ f ( ϕ )2 k w − Yu q − u |∇ f | g , u − q − u |∇ f | g w − u − Y ,ug ( k ( w, · ) , A ( w, · )) q − u |∇ f | g = g ( k ( w, · ) , k ( w, · )) q − u |∇ f | g + k w, w − Yu q − u |∇ f | g w j ( e Γ tjt − e Γ ljt f l ) − ( w m f m ) e Γ ltt f l k w, w − Yu q − u |∇ f | g + u |∇ f | g − u |∇ f | g k (cid:16) w, g jl ( e Γ tjt − e Γ mjt f m ) ∂ l (cid:17) = u |∇ f | g − u |∇ f | g k w, ∇ ϕ u + q − u |∇ f | g u g jl w i ( ∇ j Y i − ∇ i Y j ) ∂ l + k w, w − Yu q − u |∇ f | g (1 − u |∇ f | g ) w ( ϕ )2 u + k w, g jl w i ( ∇ j Y i + ∇ i Y j ) ∂ l u q − u |∇ f | g , (7.54)(7.55) g ( k ( w, · ) , π ( Y , · )) u q − u |∇ f | g − g ( k ( w, · ) , A ( Y , · ))1 − u |∇ f | g = − k (cid:0) w, g jl g im f m ∇ j Y i ∂ l (cid:1) − u |∇ f | g , and g ( k ( Y , · ) , A ( Y , · )) u q − u |∇ f | g = k (cid:16) Y , g jl Y i ( ∇ j Y i + ∇ i Y j ) ∂ l (cid:17) u q − u |∇ f | g − k Y , wu − Yu q − u |∇ f | g π ( w, Y ) . (7.56)Now use (7.53)-(7.56) as well as the formula(7.57) u ∇ l u = 12 ∇ l ϕ + g jl Y i ∇ j Y i , to rewrite the last three lines of (7.51) together with the last term on the third line by − k Y , g jl f i ∇ j Y i ∂ l u q − u |∇ f | g + k Y , g jl Y i ( ∇ j Y i + ∇ i Y j ) ∂ l u q − u |∇ f | g + u − k (cid:0) Y , ∇ w Y (cid:1) + k (cid:18) w, g jl w i ( ∇ j Y i − ∇ i Y j ) ∂ l (cid:19) u |∇ f | g u q − u |∇ f | g + k ( w, Y ) w ( u ) u q − u |∇ f | g − q − u |∇ f | g w ( ϕ )2 u + 1 u π ( w, Y ) − k ( Y , Y ) w ( ϕ )2 u + π ( w, Y ) u q − u |∇ f | g + k (cid:18) w, ∇ ϕ (cid:19) u |∇ f | g u (1 − u |∇ f | g ) ! . (7.58)Note also that g il f l ∇ j Y i = − Y i ∇ ij f = − u − q − u |∇ f | g π ( Y , ∂ j ) + A ( Y , ∂ j )= − u − q − u |∇ f | g π ( Y , ∂ j ) + 12 u − Y i ( ∇ j Y i + ∇ i Y j ) − u q − u |∇ f | g π ( w, Y ) g ij w i u − Y j u q − u |∇ f | g , (7.59)and(7.60) g lm f m Y j ∇ j Y i = − Y l Y j ∇ jl f = − u − q − u |∇ f | g π ( Y , Y ) , EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 31 so that (7.51) becomes div g k ( w ) − u − ∇ Y k ( Y , w ) + ∇ w k ( w, w )= div g ( k ( w, · )) + w ( k ( w, w )) − g ( k ( w, · ) , π ( w, · )) − k ( w, w ) π ( w, w ) − g ( k, π )+ 2 u − g ( k ( Y , · ) , π ( Y , · )) − u − k ( Y , Y ) π ( Y , Y ) + u − k (cid:0) w, ∇ Y Y (cid:1) + u − k (cid:0) Y , ∇ w Y (cid:1) − u − w ( u ) k ( Y , Y ) + 12 k ( w, ∇ ϕ ) 1 + u |∇ f | g u (1 − u |∇ f | g )+ k ( w, Y ) w ( u ) u q − u |∇ f | g − q − u |∇ f | g u w ( φ ) + 2 u π ( w, Y ) + 12 k (cid:16) w, w l ( ∇ i Y l − ∇ l Y i ) g ij ∂ j (cid:17) u |∇ f | g u q − u |∇ f | g . (7.61)The last line of (7.33) will now be computed. Identity 2-5. (cid:16) g ij − u − Y i Y j + w i w j (cid:17) w l (cid:0) (Γ mij − Γ mij ) k ml + (Γ mjl − Γ mjl ) k im (cid:1) = − ( T r e g π ) k ( w, w ) − k ( w, w ) π ( w, w ) − g ( k ( w, · ) , π ( w, · )) − u k (cid:0) w, ∇| Y | g (cid:1) + u − k (cid:0) w, ∇ Y Y (cid:1) + u − k (cid:0) Y , ∇ w Y (cid:1) − u − w ( u ) k ( Y , Y )+ k ( w, Y ) w ( u ) u q − u |∇ f | g − q − u |∇ f | g u w ( ϕ ) + 12 k (cid:16) w, w l ( ∇ i Y l − ∇ l Y i ) g ij ∂ j (cid:17) u |∇ f | g u q − u |∇ f | g + k (cid:0) w, ∇ ϕ (cid:1) |∇ f | g − u |∇ f | g (7.62) Proof.
Recall that(7.63) Γ lij − Γ lij = − w l π ij + u − Y l k ij + f i e Γ ljt + f j e Γ lit + f i f j e Γ ltt , and so(7.64) e g ij (Γ lij − Γ lij ) = − w l T r e g π + 2 g ij f i e Γ ljt + |∇ f | g e Γ ltt , (7.65) w i w j (Γ lij − Γ lij ) = − w l π ( w, w ) + u − Y l k ( w, w ) + 2 w m f m w j e Γ ljt + ( w m f m ) e Γ ltt ,g ij w l (Γ mjl − Γ mjl ) k im = − g ( k ( w, · ) , π ( w, · )) + u − g ( k ( Y , · ) , k ( w, · ))+ w l f l g ij e Γ mjt k im + w l f l k ( ∇ f, e Γ mtt ∂ m ) + k ( ∇ f, w l e Γ mlt ∂ m ) , (7.66)and(7.67) Y i Y j w l (Γ mjl − Γ mjl ) k im = − k ( w, Y ) π ( w, Y ) + u − k ( Y , Y ) k ( w, Y ) + w i f i k ( Y , Y l e Γ mlt ∂ m ) . Next, evaluate (7.64)-(7.67) with (7.13) and (7.14). Using also (7.52) and(7.68) e Γ mit = e Γ tit Y m − g ml ( ∇ i Y l − ∇ l Y i ) , leads to (7.62). (cid:3) To complete the proof of
Identity 2 , subtract
Identity 2-5 from (7.61) to find div g k ( w ) = div g ( k ( w, · )) + u − k (cid:0) w, ∇ u (cid:1) + w ( k ( w, w )) + ( T r e g π ) k ( w, w ) − g ( k, π ) + 2 u − g ( k ( Y , · ) , π ( Y , · )) − u − k ( Y , Y ) π ( Y , Y ) − g ( k ( w, · ) , π ( w, · )) + 2 u − k ( w, Y ) π ( w, Y )= u − div g ( uk ( w, · )) + w ( k ( w, w )) − e g ( k, π ) − e g ( k ( w, · ) , π ( w, · )) + ( T r e g π ) k ( w, w ) . (7.69) (cid:3) Identity 3. div g k ( ∂ φ ) = div g ( k ( ∂ φ , · )) + u − div g ( uk ( ∂ φ , w ) w ) − g ( k ( ∂ φ , · ) , π ( v, · ))+ g ( k ( ∂ φ , · ) , k ( u ∇ f, · )) − u − ∇ f ( u ) k ( ∂ φ , Y )(7.70) Proof.
Observe that(7.71) g ij Γ ljφ k li = 12 k lj ( ∂ j g lφ − ∂ l g jφ ) = 0 , and so div g k ( ∂ φ ) = g ij ∇ j k iφ = g ij ( ∂ j ( k iφ ) − Γ lij k lφ − Γ ljφ k li )= (cid:16) g ij − u − Y i Y j + w i w j (cid:17) (cid:16) ∇ ∂ j ( k ( ∂ φ , · ))( ∂ i ) − (Γ lij − Γ lij ) k lφ (cid:17) = div g ( k ( ∂ φ , · )) + u − k (cid:0) ∇ Y Y , ∂ φ (cid:1) + ∇ w ( k ( ∂ φ , · ))( w ) − (cid:16) g ij − u − Y i Y j + w i w j (cid:17) (Γ lij − Γ lij ) k lφ . (7.72) Identity 3-1. ∇ w ( k ( ∂ φ , · ))( w ) = w ( k ( w, ∂ φ )) − g ( k ( ∂ φ , · ) , π ( v, · )) + g ( k ( ∂ φ , · ) , k ( u ∇ f, · )) − u − ∇ f ( u ) k ( ∂ φ , Y ) − k (cid:0) ∂ φ , ∇ w Y (cid:1) u q − u |∇ f | g + 2 w ( u ) k ( ∂ φ , Y ) u q − u |∇ f | g + |∇ f | g k (cid:0) ∂ φ , u ∇ u (cid:1) − u |∇ f | g + k (cid:16) ∂ φ , f l ∇ i Y l ∂ i (cid:17) − u |∇ f | g (7.73) Proof.
First note that(7.74) ∇ w ( k ( ∂ φ , · ))( w ) = w ( k ( w, ∂ φ )) − k (cid:0) ∇ w w, ∂ φ (cid:1) . EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 33
Use (7.46) and (7.50) to evaluate ∇ w w as in Identity 2-4 , then ∇ w ( k ( ∂ φ , · ))( w ) = w ( k ( w, ∂ φ )) − g ( k ( ∂ φ , · ) , π ( w, · )) − k ( ∂ φ , w ) π ( w, w ) − u − w ( φ )(1 − u |∇ f | g ) k ( w, ∂ φ ) + 2 w ( u ) k ( ∂ φ , Y ) u q − u |∇ f | g + u ( g ( k ( ∂ φ , · ) , A ( w, · )) − u − k ( ∂ φ , ∇ w Y )) q − u |∇ f | g . (7.75)Consider the first term on the last line of (7.75). With help from (7.53) and the trivial identity(7.76) w i ( ∇ j Y i − ∇ i Y j ) = − w i ( ∇ j Y i + ∇ i Y j ) + 2 w i ∇ j Y i , it follows that ug ( k ( ∂ φ , · ) , A ( w, · )) q − u |∇ f | g = k ∂ φ , g jl w i ( ∇ j Y i + ∇ i Y j ) ∂ l u q − u |∇ f | g + k ∂ φ , w − Yu q − u |∇ f | g (1 − u |∇ f | g ) w ( ϕ )2 u + u |∇ f | g − u |∇ f | g k ∂ φ , ∇ ϕ u + q − u |∇ f | g u g jl w i ( ∇ j Y i − ∇ i Y j ) ∂ l = q − u |∇ f | g u k (cid:16) ∂ φ , ¯ g jl w i ( ∇ j Y i + ∇ i Y j ) ∂ l (cid:17) + k ∂ φ , w − Yu q − u |∇ f | g (1 − u |∇ f | g ) w ( ϕ )2 u + |∇ f | g − u |∇ f | g k ( ∂ φ , u ∇ u ) + u |∇ f | g − u |∇ f | g k (cid:16) ∂ φ , g jl f i ∇ j Y i ∂ l (cid:17) . (7.77)Now use (7.47), as well as(7.78) v = w − q − u |∇ f | g u Y , q − u |∇ f | g u π ( Y , ∂ j ) = − f l ∇ j Y l + u − k ( Y , ∂ j ) − f j π ( w, Y ) , to find g ( k ( ∂ φ , · ) , π ( w, · )) + k ( ∂ φ , w ) π ( w, w ) = g ( k ( ∂ φ , · ) , π ( v, · )) + u − k ( ∂ φ , Y ) π ( v, Y )+ q − u |∇ f | g u [ k ( ∂ φ , w ) π ( w, Y ) + g ( k ( ∂ φ , · ) , π ( Y , · ))]= g ( k ( ∂ φ , · ) , π ( v, · )) + u − g ( k ( ∂ φ , · ) , k ( Y , · )) − k (cid:16) ∂ φ , g jl f i ∇ j Y i ∂ l (cid:17) + 12 u k ( ∂ φ , Y ) ∇ f ( | Y | g ) . (7.79)Substituting (7.77) and (7.79) into (7.75) yields (7.73), with the help of (7.52). (cid:3) Identity 3-2. (cid:16) g ij − u − Y i Y j + w i w j (cid:17) (cid:16) Γ lij − Γ lij (cid:17) k lφ = − k ( w, ∂ φ )( T r e g π + π ( w, w )) + u − k (cid:0) ∂ φ , ∇ Y Y (cid:1) − u q − u |∇ f | g k ( ∂ φ , ∇ w Y )+ 2 k ( ∂ φ , Y ) w ( u ) u q − u |∇ f | g + k ( ∂ φ , u ∇ u ) |∇ f | g − u |∇ f | g + k (cid:0) ∂ φ , g ij f l ∇ i Y l ∂ j (cid:1) − u |∇ f | g (7.80) Proof.
Proceed in the same way as in the proof of
Identity 2-5 . Namely use (7.63)-(7.65), (7.68),and substitute the expressions (7.13) and (7.14). (cid:3)
Combining
Identity 3-1 and
Identity 3-2 produces div g k ( ∂ φ ) = div g ( k ( ∂ φ , · )) + w ( k ( w, ∂ φ )) + k ( w, ∂ φ )( T r e g π + π ( w, w )) − g ( k ( ∂ φ , · ) , π ( v, · )) + g ( k ( ∂ φ , · ) , k ( u ∇ f, · )) − u − ∇ f ( u ) k ( ∂ φ , Y ) . (7.81) Identity 3-3. (7.82) g ij ∇ j w i = T r e g π + π ( w, w ) − u − w ( u ) Proof. By Identity 2-1 g ij ∇ j w i = T r g π + π ( w, w ) − π Yu q − u |∇ f | g , w + w ( u ) u (1 − u |∇ f | g ) − u q − u |∇ f | g g ij (cid:0) A ij − u − ∇ j Y i (cid:1) − u A ( ∇ f, w )1 − u |∇ f | g = T r g π + π ( w, w ) − π ( Y , w ) u |∇ f | g (1 − u |∇ f | g ) + 3 u q − u |∇ f | g − uk ( w, ∇ f )1 − u |∇ f | g + w ( u ) u (1 − u |∇ f | g ) − w ( ϕ ) u (1 − u |∇ f | g ) + |∇ f | g w ( ϕ )2(1 − u |∇ f | g )= T r g π + π ( w, w ) − u − π ( Y , Y ) − u − w ( u )= T r e g π + π ( w, w ) − u − w ( u ) . (7.83)The last line holds since(7.84) π ( Y , Y ) = − uf l Y j ∇ j Y l q − u |∇ f | g . (cid:3) Inserting (7.82) into (7.81) yields
Identity 3 . (cid:3) EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 35
Identity 4. div g ( π ( ∂ φ , · )) = 1 q − u |∇ f | g div g ( u ∇ f ( ∂ φ , · )) − u − div g ( uπ ( ∂ φ , w ) w )+ g ( π ( ∂ φ , · ) , π ( v, · )) − g ( π ( ∂ φ , · ) , k ( u ∇ f, · )) + u − ∇ f ( u ) π ( ∂ φ , Y )(7.85) Proof.
The following condition will be used throughout this proof(7.86) div g k ( ∂ φ ) = 0 . A direct computation with (7.16) produces div g ( π ( ∂ φ , · )) = u q − u |∇ f | g h div g ( ∇ f ( ∂ φ , · )) − k (cid:0) ∂ φ , u − ∇ u (cid:1) − ∇ f ( π ( w, ∂ φ )) i − uπ ( w, ∂ φ ))∆ g f q − u |∇ f | g + π ∂ φ , ∇ log u q − u |∇ f | g . (7.87) Identity 4-1. π ∂ φ , ∇ log u q − u |∇ f | g = g ( π ( ∂ φ , · ) , π ( v, · )) − g ( π ( ∂ φ , · ) , k ( u ∇ f, · )) + π (cid:0) ∂ φ , u − ∇ u (cid:1) − π ( ∂ φ , w ) π ( w, w ) − ∇ f ( ϕ ) q − u |∇ f | g u π ( ∂ φ , w ) + u − ∇ f ( u ) π ( ∂ φ , Y )(7.88) Proof.
Observe that π ∂ φ , ∇ log u q − u |∇ f | g = π ( ∂ φ , u − ∇ u ) + |∇ f | g − u |∇ f | g π ( ∂ φ , u ∇ u ) + u − u |∇ f | g ) π ( ∂ φ , ∇|∇ f | g )= π ( ∂ φ , ∇ u ) u (1 − u |∇ f | g ) + g π u ∇ f q − u |∇ f | g , · , π ( ∂ φ , · ) − u g ( A ( ∇ f, · ) , π ( ∂ φ , · ))1 − u |∇ f | g . (7.89) In order to evaluate the last term in (7.89), follow the computations in (7.54), replacing k ( w, · ) by π ( ∂ φ , · ) to find u g ( A ( ∇ f, · ) , π ( ∂ φ , · ))1 − u |∇ f | g = ug ( A ( w, · ) , π ( ∂ φ , · )) q − u |∇ f | g − g ( A ( Y , · ) , π ( ∂ φ , · ))1 − u |∇ f | g = π ∂ φ , g jl w i ( ∇ j Y i + ∇ i Y j ) ∂ l u q − u |∇ f | g + π ∂ φ , w − Yu q − u |∇ f | g (1 − u |∇ f | g ) w ( ϕ )2 u + u |∇ f | g − u |∇ f | g π ∂ φ , ∇ ϕ u + q − u |∇ f | g u g jl w i ( ∇ j Y i − ∇ i Y j ) ∂ l − g ( A ( Y , · ) , π ( ∂ φ , · ))1 − u |∇ f | g . (7.90)Moreover g π u ∇ f q − u |∇ f | g , · , π ( ∂ φ , · ) = g π v − u |∇ f | g Y q − u |∇ f | g , · , π ( ∂ φ , · ) = g ( π ( v, · ) , π ( ∂ φ , · )) + u − π ( ∂ φ , Y ) π ( v, Y ) − g π u |∇ f | g Y q − u |∇ f | g , · , π ( ∂ φ , · ) − π ( ∂ φ , w ) π w − q − u |∇ f | g u Y , w . (7.91)Substitute (7.90), (7.91) into (7.89), and use the following relations to get the desired result:(7.92) π u |∇ f | g Y q − u |∇ f | g , ∂ i = u |∇ f | g − u |∇ f | g ( − f l ∇ i Y l + A ( Y , ∂ i )) , ∇ ϕ u + q − u |∇ f | g u g jl w i ( ∇ j Y i − ∇ i Y j ) ∂ l − g jl f i ∇ j Y i ∂ l = ∇ uu − q − u |∇ f | g u g jl w i ( ∇ j Y i + ∇ i Y j ) ∂ l , (7.93) EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 37 g ( A ( Y , · ) , π ( ∂ φ , · )) = π ∂ φ , g jl Y i ( ∇ j Y i + ∇ i Y j ) ∂ l u ! − π ∂ φ , q − u |∇ f | g u w − Yu π ( w, Y )= − g ( π ( ∂ φ , · ) , k ( u ∇ f, · )) + q − u |∇ f | g u π (cid:16) ∂ φ , g jl w i ( ∇ j Y i + ∇ i Y j ) ∂ l (cid:17) (7.94) − π ∂ φ , q − u |∇ f | g u w − Yu π ( w, Y ) , and(7.95) π ( v, Y ) = f l Y j ( ∇ j Y i + ∇ i Y j )2 u = − π ( w, Y ) + ∇ f ( | Y | g )2 u . (cid:3) We now finish the proof of
Identity 4 . Employ
Identity 2 and(7.96) π ( w, Y ) = Y i f j ( Y i ¯; j − Y j ¯; i )2 , π ( Y , Y ) = uY i Y j ∇ ij f q − u |∇ f | g = − uY i f j ∇ i Y j q − u |∇ f | g , to find ∆ g f = u − q − u |∇ f | g T r g π − T r g A = u − q − u |∇ f | g T r g π − u − π ( w, Y ) − (cid:0) u − − |∇ f | g (cid:1) ∇ f ( ϕ )= u − q − u |∇ f | g T r e g π + u − q − u |∇ f | g π ( Y , Y ) − u − π ( w, Y ) − (cid:0) u − − |∇ f | g (cid:1) ∇ f ( ϕ )= u − q − u |∇ f | g T r e g π − u − ∇ f ( u ) − (1 − u |∇ f | g ) ∇ f ( ϕ )2 u . (7.97)Substituting Identity 4-1 and (7.97) into (7.87) produces div g ( π ( ∂ φ , · )) = u q − u |∇ f | g (cid:16) div g ( ∇ f ( ∂ φ , · )) − u − k ( ∂ φ , ∇ u ) (cid:17) + u − π ( ∂ φ , ∇ u )+ g ( π ( ∂ φ , · ) , π ( v, · )) − g ( π ( ∂ φ , · ) , k ( u ∇ f, · )) + u − π ( ∂ φ , Y ) ∇ f ( u ) − w ( π ( ∂ φ , w )) − π ( ∂ φ , w ) (cid:0) T r e g π + π ( w, w ) − u − w ( u ) (cid:1) . (7.98)The desired result may now be achieved with the help of (7.82) and(7.99) π (cid:18) ∂ φ , ∇ uu (cid:19) = u q − u |∇ f | g ∇ f (cid:18) ∇ uu , ∂ φ (cid:19) + k ( ∇ u, ∂ φ ) u − ∇ f ( u ) u π ( w, ∂ φ ) ! . (cid:3) Identity 5. (7.100) div g π ( ∂ φ ) = div g (cid:16) u ∇ f ( ∂ φ , · ) (cid:17)q − u |∇ f | g Proof.
Replace k by π in Identity 3 to obtain div g π ( ∂ φ ) = div g ( π ( ∂ φ , · )) + u − div g ( uπ ( ∂ φ , w ) w ) − g ( π ( ∂ φ , · ) , π ( v, · )) + g ( π ( ∂ φ , · ) , k ( u ∇ f, · )) − u − ∇ f ( u ) π ( ∂ φ , Y ) . (7.101)Now employ Identity 4 . (cid:3) Identity 6. div g ( k − π )( v ) = | π | g − g ( π, k ) − ug ( k, ∇ f )+ u − div g u ∇ f ( Y , · ) + ( k − π )( w, · ) + ( k − π )( w, w ) udf q − u |∇ f | g (7.102) Proof.
We have(7.103) div g ( k − π )( v ) = div g ( k − π )( w ) − u − q − u |∇ f | g div g ( k − π )( Y ) . Replace k by ( k − π ) in Identity 2 to calculate div g ( k − π )( w ). Next use Identity 5 to evaluate Y φ div g π ( ∂ φ ), and note that Y φ div g k ( ∂ φ ) = 0 as well as k ( ∇ f, ∇ u ) = 0 to find div g ( k − π )( v ) = | π | e g − e g ( π, k ) + 2 | π ( w, · ) | e g − e g ( π ( w, · ) , k ( w, · ))+ u − div g ( u ( k − π )( w, · )) + w (( k − π )( w, w )) + ( k − π )( w, w )( T r e g π )+ u − div g ( u ∇ f ( Y , · )) − ug ( k, ∇ f ) . (7.104)Next observe that(7.105) | π | e g +2 | π ( w, · ) | e g = | π | g − π ( w, w ) , e g ( π, k )+2 e g ( π ( w, · ) , k ( w, · )) = g ( π, k ) − π ( w, w ) k ( w, w ) , as well as the fact that Identity 3-3 together with g ij ∇ j Y i = 0 imply w (( k − π )( w, w )) + ( k − π )( w, w )( T r e g π + π ( w, w )) = u − div g ( u ( k − π )( w, w ) w )= u − div g u ( k − π )( w, w ) udf q − u |∇ f | g . (7.106)Combining (7.104)-(7.106) yields the desired result. (cid:3) Identity 7. (7.107) R − | k | g = 16 π ( µ − J ( v )) + | k − π | g + 2 u div g ( uQ ( · )) + ( T r g π ) − ( T r g k ) + 2 v ( T r g π − T r g k )where Q is the 1-form defined by(7.108) Q ( · ) = ∇ f ( Y , · ) − k ( u ∇ f, · ) + ( k − π )( w, · ) + ( k − π )( w, w ) udf q − u |∇ f | g . EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 39
Proof.
Recall the formula for R in (7.11), and observe that(7.109) div g k ( u ∇ f ) = 1 u div g ( uk ( u ∇ f, · )) − ug ( k, ∇ f ) . Apply
Identity 6 and solve for R − | k | g . (cid:3) Appendix B: Computations Related to Y φ The purpose of this section is twofold. Namely, to give several equivalent versions of the equationsatisfied by Y φ , and to compare the prescribed asymptotics of Y φ with examples from the extremeKerr spacetime.Recall the basic defining equation for Y φ (8.1) div g k ( η ) = 0 . Let ( ρ, φ, z ) denote Brill coordinates (3.1) for g , with corresponding Christoffel symbols Γ lij . ByLemma 2.1 and the fact that ∂ φ is a Killing field(8.2) k iφ = g φφ u ∂ i Y φ , g ij Γ ljφ k li = 12 k jl ( ∂ j g lφ − ∂ l g jφ ) = 0 , so that div g k ( η ) = g ij ∇ j k φi = g ij (cid:16) ∂ j k φi − Γ lij k lφ − Γ ljφ k li (cid:17) = g ij (cid:18) ∂ j ( u − g φφ ∂ i Y φ ) − Γ lij k lφ (cid:19) = − g φφ u g ij ∂ j u∂ i Y φ + 12 u g ij (cid:16) ∂ j ( g φφ ∂ i Y φ ) − g φφ Γ lij ∂ l Y φ (cid:17) = − g φφ u g ij ∂ j u∂ i Y φ + g φφ u (cid:16) ∆ g Y φ + ∇ log g φφ · ∇ Y φ (cid:17) = g φφ u (cid:16) ∆ g Y φ + ∇ log( u − g φφ ) · ∇ Y φ (cid:17) . (8.3)Equation (8.3) may also be expressed explicitly in Brill coordinates. Observe that(8.4) g pq = e U − α δ pq , g pφ = − A p e U − α , p, q = ρ, z, g φφ = ρ − e U + e U − α ( A ρ + A z ) , and(8.5) e p = e U − α ( ∂ p − A p ∂ φ ) , p = ρ, z, e φ = ρ − e U ∂ φ , so that g ( ∇ e ρ e ρ , e z ) = e U − α (Γ zρρ − A ρ Γ zρφ + A ρ Γ zφφ )= e U − α (cid:18) g zz ∂ ρ g ρz − ∂ z g ρρ ) + g zφ ∂ ρ g ρφ (cid:19) − e U − α A ρ ( g zz ( ∂ ρ g φz − ∂ z g ρφ ) + g zφ ∂ ρ g φφ ) − e U − α A ρ g zz ∂ z g φφ = ∂ z ( U − α ) e U − α . (8.6)Similarly(8.7) g ( ∇ e z e z , e ρ ) = ∂ ρ ( U − α ) e U − α and(8.8) g ( ∇ e φ e φ , e p ) = g − φφ e − U + α Γ pφφ = − e U − α ∂ p log g φφ , p = ρ, z. It follows that∆ g Y φ = X p = ρ,z (cid:16) e p ( e p Y φ ) − ∇ e p e p ( Y φ ) − ∇ e φ e φ ( Y φ ) (cid:17) = X p = ρ,z (cid:18) e U − α ∂ p ( e U − α ∂ p Y φ ) − e U − α ∂ p ( U − α ) ∂ p Y φ + 12 e U − α ∂ p log g φφ ∂ p Y φ (cid:19) = X p = ρ,z (cid:18) e U − α ∂ p Y φ + 12 e U − α ∂ p log g φφ ∂ p Y φ (cid:19) , (8.9)and(8.10) ∇ log( u − g φφ ) · ∇ Y φ = X p = ρ,z e U − α (cid:16) g − φφ ∂ p g φφ − u − ∂ p u (cid:17) ∂ p Y φ . Hence div g k ( η ) = g φφ u (cid:16) ∆ g Y φ + ∇ log( u − g φφ ) · ∇ Y φ (cid:17) = g φφ e U − α u X p = ρ,z (cid:18) ∂ p Y φ + (cid:18) g − φφ ∂ p g φφ − u − ∂ p u (cid:19) ∂ p Y φ (cid:19) = e U − α √ g φφ X p = ρ,z ∂ p (cid:16) u − g / φφ ∂ p Y φ (cid:17) . (8.11)We will now express (8.1) in terms of the metric g . Observe that(8.12) ∆ g Y φ = g ij ( ∂ ij Y φ − Γ lij ∂ l Y φ ) + g ij (Γ lij − Γ lij ) ∂ l Y φ and(8.13) ∇ log( u − g φφ ) · ∇ Y φ = g ij ∂ i log( u − g φφ ) ∂ j Y φ , where Γ lij are Christoffel symbols for g . In Identity 1 of Section 7 the difference between Christoffelsymbols is computed, so that(8.14) g ij (Γ lij − Γ lij ) ∂ l Y φ = − w ( Y φ ) T r g π + 12 | ∇ f | g ∇ ϕ · ∇ Y φ − g ij f l ( ∇ l Y j − ∇ j Y l ) ∂ i Y φ . In order to proceed, we will also need(8.15) Y i = g ij Y j = g ij ( Y φ g iφ + | Y | g ∂ i f ) = Y φ δ iφ + | Y | g f i , (8.16) g ij = g ij − u f i f j u |∇ f | g − Y φ ( δ iφ f j + f i δ jφ )1 + u |∇ f | g + |∇ f | g ( Y φ ) δ iφ δ jφ u |∇ f | g . It follows that g ij ( ∂ ij Y φ − Γ lij ∂ l Y φ )= (cid:18) g ij − u f i f j u |∇ f | g (cid:19) ∇ Y φ ( ∂ i , ∂ j ) + 2 Y φ f j Γ lφj ∂ l Y φ u |∇ f | g − |∇ f | g ( Y φ ) Γ lφφ ∂ l Y φ u |∇ f | g , (8.17) EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 41 g ij (Γ lij − Γ lij ) ∂ l Y φ = − u ∇ f ( Y φ )( T r g π ) q u |∇ f | g + |∇ f | g ∇ ϕ · ∇ Y φ u |∇ f | g ) − u |∇ f | g ∇ f ( ϕ ) ∇ f ( Y φ )2(1 + u |∇ f | g ) + 2 Y φ ( Y φ ) l Γ ilφ ∂ i f u |∇ f | g + Y φ |∇ f | g |∇ Y φ | g u |∇ f | g − Y φ ( ∇ f ( Y φ )) u |∇ f | g , (8.18)and ∇ log( u − g φφ ) · ∇ Y φ = (cid:18) g ij − u f i f j u |∇ f | g (cid:19) ∂ i log( u − g φφ ) ∂ j Y φ = ∇ log( u − g φφ ) · ∇ Y φ − u f l ∂ l log( u − g φφ ) f j ∂ j Y φ u |∇ f | g . (8.19)Next note that with the help of (2.34) and (8.16) T r g π = (cid:18) g ij − u f i f j u |∇ f | g (cid:19) π ij − g φφ Y φ u (1 + u |∇ f | g ) u ∇ f ( Y φ ) q u |∇ f | g − ( Y φ ) |∇ f | g u |∇ f | g u Γ lφφ ∂ l f q u |∇ f | g . (8.20)Therefore employing (8.17), (8.18), (8.19), (8.20), and the identity(8.21) 12 ϕ l − ( Y φ ) Γ lφφ + Y φ ( Y φ ) l = uu l , produces ∆ g Y φ + ∇ log( u − g φφ ) · ∇ Y φ = (cid:18) g ij − u f i f j u |∇ f | g (cid:19) ∇ ij Y φ − uπ ij f l q u |∇ f | g ∂ l Y φ + (cid:18) g ij − u f i f j u |∇ f | (cid:19) (cid:18) ∂ i log g φφ − ∂ i log u u |∇ f | (cid:19) ∂ j Y φ . (8.22)We now record what has been shown. Lemma 8.1.
The following equations are equivalent: (8.23) div g k ( η ) = 0 , (8.24) ∆ g Y φ + ∇ log( u − g φφ ) · ∇ Y φ = 0 , (8.25) e U − α √ g φφ X p = ρ,z ∂ p (cid:16) u − g / φφ ∂ p Y φ (cid:17) = 0 , (cid:18) g ij − u f i f j u |∇ f | g (cid:19) ∇ ij Y φ − uπ ij f l q u |∇ f | g ∂ l Y φ + (cid:18) g ij − u f i f j u |∇ f | (cid:19) (cid:18) ∂ i log g φφ − ∂ i log u u |∇ f | (cid:19) ∂ j Y φ = 0 . (8.26) Lastly, the prescribed asymptotics (6.2)-(6.4) for Y φ will be compared with the example from the(extreme) Kerr spacetime. Recall that in Boyer-Lindquist coordinates the Kerr metric takes the form(8.27) − ∆ − a sin θ Σ dt + 4 ma e r sin θ Σ dtdφ + ( e r + a ) − ∆ a sin θ Σ sin θdφ + Σ∆ d e r + Σ dθ where(8.28) ∆ = e r + a − m e r, Σ = e r + a cos θ. The event horizon is located at the larger of the two solutions to the quadratic equation ∆ = 0,namely e r + = m + √ m − a . For e r > e r + it holds that ∆ >
0, so that a new radial coordinate maybe defined by(8.29) r = 12 ( e r − m + √ ∆) , or rather e r = r + m + m − a r , m = a e r = r + m, m = a . (8.30)Note that the new coordinate is defined for r >
0, and a critical point for the right-hand side of(8.30) ( m = a ) occurs at the horizon, so that two isometric copies of the outer region are encodedon this interval. Moreover the t = 0 slice of the metric takes the form (5.1), showing that ( r, φ, θ )are an appropriate set of Brill coordinates.Observe that(8.31) Y φ = g φφ Y φ = − ma e r ( e r + a ) − ∆ a sin θ . Therefore at spatial infinity(8.32) Y φ ∼ − mar as r → ∞ , which is consistent with (6.2) since J = am . Furthermore Y φ = O ( r ) , m = a , as r → ,Y φ = − m a ( m + a ) + O ( r ) , m = a , as r → . (8.33)This is consistent with (6.4), but not (6.3). The reason for the inconsistency is that the lapse functionfor the Kerr spacetime does not satisfy the required asymptotics (5.3), whereas the lapse functionfor the extreme Kerr spacetime does satisfy the desired asymptotics (5.4).9. Appendix C: Boundary Terms
Consider the basic inequality (3.7). Under the hypotheses of Theorem 3.1 this yields(9.1) m − M ( U , ω ) ≥ π lim r →∞ Z S r uQ ( ν ) dA g − π lim r → Z S r uQ ( ν ) dA g , where ν is the unit normal pointing towards M + end for the coordinate spheres S r . Here ( r, φ, θ ) arespherical coordinates as in (5.1), but with respect to g . The purpose of this section is to show that(9.2) lim r →∞ Z S r uQ ( ν ) dA g − lim r → Z S r uQ ( ν ) dA g = 8 π Y ( J )( J − J ) , EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 43 where Y ( J ) = lim r → Y φ as in (6.3), (6.4). Thus, the choice J = J guarantees (3.13).Recall that(9.3) Q ( · ) = ∇ f ( Y , · ) − k ( u ∇ f, · ) + ( k − π )( w, · ) + ( k − π )( w, w ) q u |∇ f | g udf. It is clear from the asymptotics (2.31), (2.40) that the first term on the left-hand side of (9.2)vanishes, so we will focus on the second term. In what follows, it will be assumed that(9.4) | k | g + | k ( ∂ φ , · ) | g + | k ( ∂ φ , ∂ φ ) | ≤ c on M. Note also that (5.2)-(5.4) and (5.7)-(5.9) imply that(9.5) | π | g + | π ( ∂ φ , · ) | g + | π ( ∂ φ , ∂ φ ) | ≤ c on M, and(9.6) u → , |∇ f | g → r → . Let us now consider terms in (9.3) when applied to ν . Since(9.7) w = u ∇ g f + u − Y q u |∇ f | g , it follows that(9.8) k ( w, ν ) = u − k ( Y , ν ) + O (cid:0) u |∇ f | g | k ( Y , ν ) | + u | k ( ∇ f, ν ) | (cid:1) . Moreover(9.9) | k ( Y , ν ) | ≤ | Y φ || k ( ∂ φ , · ) | g , and | k ( ∂ φ , · ) | g = g ij k il η l k jm η m = | k ( ∂ φ , · ) | g − η l η m w i w j k il k jm + u − η l η m Y i Y j k il k jm ≤ | k ( ∂ φ , · ) | g + u − ( Y φ ) | k ( ∂ φ , ∂ φ ) | , (9.10)so that(9.11) | k ( Y , ν ) | ≤ c ( | k ( ∂ φ , · ) | g + u − | k ( ∂ φ , ∂ φ ) | ) . Similarly(9.12) | k ( ∇ f, ν ) | ≤ c ( | k | g |∇ f | g + u − | k ( ∂ φ , · ) | g |∇ f | g ) . Hence(9.13) k ( w, ν ) = u − k ( Y , ν ) + O ( |∇ f | g ) . Analogous computations show that(9.14) k ( w, w ) = u − k ( Y , Y ) + O ( |∇ f | g ) , and also(9.15) π ( w, ν ) = u − π ( Y , ν ) + O ( |∇ f | g ) , π ( w, w ) = u − π ( Y , Y ) + O ( |∇ f | g ) . We now have(9.16) Q ( ν ) = ∇ f ( Y , ν ) − uk ( ∇ f, ν ) + u − ( k − π )( Y , ν ) + u − ( k − π )( Y , Y ) ν ( f ) + O ( |∇ f | g ) . According to (7.16) and (7.47)(9.17) π ( Y , · ) = u q u |∇ f | g (cid:16) ∇ f ( Y , · ) + u − k ( Y , · ) − π ( w, Y ) df (cid:17) , so that(9.18) π ( Y , ν ) = u ∇ f ( Y , ν ) + k ( Y , ν ) − π ( Y , Y ) ν ( f ) + O ( u |∇ f | g ) . Combining this with (9.16), and the fact that k ( ∇ f, ν ) = 0 (as ν is the normal for an axisymmetricsurface), produces(9.19) Q ( ν ) = u − k ( Y , ν ) + u − k ( Y , Y ) ν ( f ) − u − k ( Y , ν ) + O ( |∇ f | g ) . In sum lim r → Z S r uQ ( ν ) dA g = lim r → Z S r (cid:0) k ( Y , ν ) + k ( Y , ν ) ν ( f ) − k ( Y , ν ) + O ( u |∇ f | g ) (cid:1) dA g = lim r → Z S r (cid:0) k ( Y , ν ) + k ( Y , Y ) ν ( f ) + O ( u |∇ f | g ) (cid:1) dA g − π YJ , (9.20)where the last line is obtained from the definition of angular momentum and (2.30).In order to proceed, it will be necessary to express ν in terms of quantities asasociated with themetric g . Recall that ( M, g ) is embedded via a graph t = f ( x ) in the spacetime ( M × R , e g = g − Y i dx i dt − ϕdt ). Let S r ⊂ M be the natural lifting of S r ⊂ M to the graph. The goal is tocompute ν in terms of ξ , the unit normal to S r pointing towards M + end . Observe that an orthonormalframe for ( M , g ) is given by(9.21) ν = e r = e U − α ( ∂ r − A r ∂ φ ) , e θ = e U − α r ( ∂ θ − A θ ∂ φ ) , e φ = e U r sin θ ∂ φ , and that(9.22) X i = e i + e i ( f ) ∂ t , i = r, θ, φ, is a basis for the tangent space of ( M, g ). Thus, a normal to S r may be written as(9.23) ξ ′ := X r + C θ X θ + C φ X φ for some constants C θ , C φ . In order to calculate these constants, note that0 = e g ( ξ ′ , X φ )= e g ( X r + C θ X θ + C φ X φ , e φ )= e g ( e r ( f ) ∂ t , e φ ) + C θ e g ( e θ ( f ) ∂ t , e φ ) + C φ = − e r ( f ) Y ( e φ ) − C θ e θ ( f ) Y ( e φ ) + C φ (9.24)and 0 = e g ( ξ ′ , X θ )= e g ( X r + C θ X θ + C φ X φ , e θ + e θ ( f ) ∂ t )= e g ( e r + e r ( f ) ∂ t , e θ + e θ ( f ) ∂ t ) + C θ e g ( e θ + e θ ( f ) ∂ t , e θ + e θ ( f ) ∂ t ) + C φ e θ ( f ) e g ( e φ , ∂ t )= − e θ ( f ) Y ( e r ) − e r ( f ) Y ( e θ ) − ϕe r ( f ) e θ ( f )+ C θ (1 − e θ ( f ) Y ( e θ ) − ϕe θ ( f ) ) − C φ e θ ( f ) Y ( e φ )= − ϕe r ( f ) e θ ( f ) + C θ (1 − ϕe θ ( f ) ) − C φ e θ ( f ) Y ( e φ ) , (9.25) EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 45 where in the last line the identity Y ( e r ) = Y ( e θ ) = 0 is used. Solving for C θ and C φ yields(9.26) ξ ′ = X r + u e r ( f ) e θ ( f )1 − u e θ ( f ) X θ + e r ( f ) Y ( e φ )1 − u e θ ( f ) X φ , and hence(9.27) ξ = ξ ′ | ξ ′ | g = s − u e θ ( f ) − u |∇ f | g (cid:18) ν + ν ( f ) ∂ t + u ν ( f ) e θ ( f )1 − u e θ ( f ) X θ + ν ( f ) Y ( e φ )1 − u e θ ( f ) X φ (cid:19) . Consider now the integrand on the right-hand side of (9.20). Since k ( ∂ t , · ) = 0 and(9.28) | e i ( f ) | ≤ |∇ f | g = |∇ f | g q u |∇ f | g ≤ |∇ f | g , Y ( e φ ) e φ = Y , dA g = q − u e θ ( f ) dA g , it follows that(9.29) (cid:0) k ( Y , ν ) + k ( Y , Y ) ν ( f ) (cid:1) dA g = (cid:0) k ( Y , ξ ) + O ( u |∇ f | g ) (cid:1) dA g as r → . Note also that the area |S r | grows like r − when M − end is asymptotically flat, and is bounded when M − end is asymptotically cylindrical. Therefore with the help of the asymptotics (5.2)-(5.4) and (5.7)-(5.9) lim r → Z S r uQ ( ν ) dA g = lim r → Z S r (cid:0) k ( Y , ξ ) + O ( u |∇ f | g ) (cid:1) dA g − π YJ = 8 π Y ( J − J ) . (9.30) 10. Appendix D: Miscellaneous Formulae
In this section we will compute certain Christoffel symbols used in Section 6, and record how thetwist potential encodes angular momentum.Christoffel symbols will be expressed in terms of the Brill coordinate system (5.1), where ρ = r sin θ .Observe that components of the inverse metric are given by(10.1) g rr = e U − α , g θθ = r − e U − α , g φφ = ρ − e U + e U − α ( A r + r − A θ ) , (10.2) g rθ = 0 , g rφ = − A r e U − α , g θφ = − r − A θ e U − α . It follows that(10.3) Γ rrr = 12 e U − α ∂ r ( e − U +2 α + ρ e − U A r ) − e U − α A r ∂ r ( ρ e − U A r ) , (10.4) Γ rθθ = 12 e U − α (cid:2) ∂ θ ( ρ e − U A r A θ ) − ∂ r ( r e − U +2 α + ρ e − U A θ ) (cid:3) − e U − α A r ∂ θ ( ρ e − U A θ ) , (10.5) Γ rφφ = − e U − α ∂ r ( ρ e − U ) , (10.6) Γ rrθ = 12 e U − α ∂ θ ( e − U +2 α + ρ e − U A r ) − e U − α A r (cid:2) ∂ r ( ρ e − U A θ ) + ∂ θ ( ρ e − U A r ) (cid:3) , (10.7) Γ rrφ = − e U − α A r ∂ r ( ρ e − U ) , (10.8) Γ rθφ = − e U − α A r ∂ θ ( ρ e − U ) + 12 e U − α (cid:2) ∂ θ ( ρ e − U A r ) − ∂ r ( ρ e − U A θ ) (cid:3) . We now record how the twist potential encodes angular momentum. Again, consider the coordinatesystem in (5.1). An orthonormal basis is given by(10.9) e r = e U − α ( ∂ r − A r ∂ φ ) , e θ = e U − α r ( ∂ θ − A θ ∂ φ ) , e φ = e U r sin θ ∂ φ . The twist potential may be calculated in terms of k by(10.10) 12 ∂ i ω = ǫ ijl k jm η l η m , where ǫ ijl is the volume element of g . It follows that e U − α r ∂ θ ω = 12 e θ ( ω )= − ǫ ( e r , e θ , e φ ) k ( e r , e φ ) | η | = − e − U k ( e r , ∂ φ ) r sin θ, (10.11)or rather(10.12) k ( e r , ∂ φ ) = − e U − α r sin θ ∂ θ ω. Hence, if there are only two ends J = 18 π Z S ∞ ( k ij − ( T rk ) g ij ) ν i η j = lim r → π Z ∂B ( r ) k ( ∂ φ , e r ) dA = lim r → π Z ∂B (1) k ( ∂ φ , e r ) e − U + α r sin θdθdφ = − lim r → π Z ∂B (1) ∂ θ ωdθdφ = 18 ( ω | I + − ω | I − ) . (10.13) References [1] H. Bray, and M. Khuri,
A Jang equation approach to the Penrose inequality , Discrete Contin. Dyn. Syst., (2010), no. 2, 741766. arXiv:0910.4785[2] H. Bray, and M. Khuri, P.D.E.’s which imply the Penrose conjecture . Asian J. Math., (2011), no. 4, 557-610.arXiv:0905.2622[3] D. Brill, On the positive definite mass of the Bondi-Weber-Wheeler time-symmetric gravitational waves , Ann.Phys., (1959), 466483.[4] Y. Choquet-Bruhat, General Relativity and the Einstein Equations , Oxford University Press, 2009.[5] P. Chru´sciel,
Mass and angular-momentum inequalities for axi-symmetric initial data sets. I. Positivity of Mass ,Ann. Phys., (2008), 2566-2590. arXiv:0710.3680[6] P. Chru´sciel, and J. Costa,
On uniqueness of station- ary vacuum black holes,
In Proceedings of G´eom´etriediffrentielle, Physique math´ematique, Mathematiques et soci´et´e, Ast´erisque, (2008), 195-265. arXiv:0806.0016[7] P. Chru´sciel, G. Galloway, and D. Pollack,
Mathematical General Relativity: a sampler , Bull. Amer. Math. Soc.(N.S.), (2010), no. 4, 567-638. arXiv:1004.1016 EFORMATIONS OF AXIALLY SYMMETRIC INITIAL DATA 47 [8] P. Chru´sciel, Y. Li, and G. Weinstein,
Mass and angular-momentum inequalities for axi-symmetric initial datasets. II. Angular Momentum , Ann. Phys., (2008), 2591-2613. arXiv:0712.4064[9] M. Clement, J. Jaramillo, and M. Reiris,
Proof of the area-angular momentum-charge inequality for axisymmetricblack holes , Class. Quantum Grav., (2012), 065017. arXiv:1207.6761[10] S. Dain, Proof of the angular momentum-mass inequality for axisymmetric black hole , J. Differential Geom., (2008), 33-67. arXiv:gr-qc/0606105[11] S. Dain, Geometric inequalities for axially symmetric black holes , Classical Quantum Gravity, (2012), no. 7,073001. arXiv:1111.3615[12] S. Dain, J. Jaramillo, and M. Reiris, Black hole area-angular momentum inequality in non-vacuum spacetimes ,Phys. Rev. D, (2011), 121503. arXiv:1106.3743[13] S Dain, and M. Reiris, Area-angular momentum inequality for axisymmetric black holes , Phys. Rev. Lett., (2011), 051101. arXiv:1102.5215[14] S. Dain, M. Khuri, G. Weinstein, and S. Yamada,
Lower bounds for the area of black holes in terms of mass,charge, and angular momentum , Phys. Rev. D, (2013), 024048. arXiv:1306.4739[15] M. Disconzi, and M. Khuri, On the Penrose inequality for charged black holes , Classical Quantum Gravity, (2012), 245019, arXiv:1207.5484.[16] Q. Han, and M. Khuri, Existence and blow up behavior for solutions of the generalized Jang equation , Comm.Partial Differential Equations, (2013), 2199-2237. arXiv:1206.0079[17] S. Hawking, and G. Ellis, The Large Structure of Space-Time , Cambridge Monographs on Mathematical Physics,Cambridge University Press, 1973.[18] L.-H. Huang, R. Schoen, M.-T. Wang,
Specifying angular momentum and center of mass for vacuum initial datasets , Commun. Math. Phys., (2011), no. 3, 785803. arXiv:1008.4996[19] P.-S. Jang,
On the positivity of energy in General Relaitivity , J. Math. Phys., (1978), 1152-1155.[20] M. Khuri, and G. Weinstein, Rigidity in the positive mass theorem with charge , J. Math. Phys., (2013), 092501.arXiv:1307.5499[21] R. Schoen, and X. Zhou, Convexity of reduced energy and mass angular momentum inequalities , Ann. HenriPoincar´e, (2013), 1747-1773. arXiv:1209.0019.[22] R. Schoen, and S.-T. Yau, Proof of the positive mass theorem II , Commun. Math. Phys., (1981), 231-260.[23] X. Zhou, Mass angular momentum inequality for axisymmetric vacuum data with small trace , preprint, 2012.arXiv:1209.1605
Department of Mathematics, Stony Brook University, Stony Brook, NY 11794, USA
E-mail address : [email protected] Department of Mathematics, Stony Brook University, Stony Brook, NY 11794, USA
E-mail address ::