ABSTRACT. Starting from the Einstein equation in general relativity, we carefully derive the Einstein constraint equations which specify initial data for the Cauchy problem for the Einstein equation. Then we show how to use the conformal method to study these constraint equations.
1 EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 2 ´ˆ C ANH NGO ˆ QUO A BSTRACT. Starting from the Einstein equation in general relativity, we carefully derive the Einstein constraint equations which specify initial data for the Cauchy problem for the Einstein equation. Then we show how to use the conformal method to study these constraint equations. 3 0. I NTRODUCTION 4 It is well-known that the Einstein theory of relativity (or commonly known as general relativity) is a geometric theory of gravitation. In this theory, gravity is considered as a geometric property of space and time. Because of the geometric property of space and time, general relativity partially includes both special relativity and the Newton law of universal gravitation as special cases. 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Theoretically, general relativity describes objects in large scale, like the universe, as Lorentzian manifolds on which gravitation interacts and the universe evolves over time through a system of partial differential equations known as the Einstein equations. Being the central object of the theory, studying the Einstein equations becomes a significant subject in order to understand the whole theory. In an effort to solve the general Einstein equations, physicists first try to tackle the equations in some simple cases. Fortunately, some remarkably solutions have been found in this direction. Although general relativity nearly coincides with the Newton law of universal gravitation, those known solutions for the Einstein equations in particular cases have led theoretical physicists to predict some new phenomena which deserve investigation carefully. However, much less is known about the solutions of the general Einstein equations. On the other hand, due to the geometric nature of the theory, solving the general Einstein equations turns out to be a wonderful research topic not only for physicists but also for mathematicians, pushing the development of the research rapidly. However, along with the rapid development of the research, it poses many challenging problems to mathematicians, for example, the initial value problems, the well-posedness problems, the global stability problems, etc. Among these problems, the initial value problems which have their own interest from the mathematical point of view turns out to be the most interesting problem since it involves the theoretical question of the beginning and the end of our universe. When solving the initial value problems, one needs to solve the so-called constraint equations that form an under-determined system of equations which are not easy to solve. For this reason, understanding the constraint equations is a key step to understanding the initial value problems. The structure of this lecture notes is as follows: Date: 16th Apr, 2015 at 00:07. 2000 Mathematics Subject Classification. 35J60, 53C21, 53C80, 58E05, 58J45, 83C05. 1 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 ˆ Q.A. NGO C ONTENTS 0. Introduction 1. Geometry of spacetimes in general relativity 1.1. Basics of differential geometry 1.2. What are spacetimes? 1.3. Adapted frames and coframes 1.4. Basics of submanifolds 1.5. The Gauss and Codazzi equations 2. The Einstein equations in general relativity 2.1. The Einstein equations 2.2. The Einstein equations with real scalar fields 2.3. Why do the Einstein equations describe the propagation of wavelike phenomena? 3. The Cauchy problem for the Einstein equations 4. Constraints and evolutions 4.1. Derivation of constraint equations: The Hamiltonian constraint 4.2. Derivation of constraint equations: The momentum constraint 4.3. Evolutions of the first and second fundamental forms 4.4. Construction of spacetimes via solutions of the vacuum Einstein constraint equations 5. Conformal method and transformed PDEs 5.1. Conformal method 5.2. Transformed PDEs in the case n 3 5.3. Transformed PDEs in the case n = 2 6. Solving the transformed PDEs in the case n 3: Mathematical settings 6.1. Preliminaries 6.2. A classification of Choquet-Bruhat–Isenberg–Pollack 6.3. The Lichnerowicz equations with Rg,ψ being constant 7. Solving the transformed PDEs in the closed case when n 3: The case of constant mean curvature 7.1. The vacuum case 7.2. The case of scalar fields 7.3. Strategy for finding solutions and ideas of proofs 8. Other result: A Liouville type result 8.1. Some basic computations 8.2. Computations of A, B, and C 8.3. The transformed equation 8.4. Proof of Theorem 8.1 8.5. Proof of Theorem 8.1 completed Acknowledgments References 1 3 3 4 6 7 7 9 9 11 11 13 15 15 17 19 19 22 22 23 31 33 33 36 38 39 39 40 46 57 57 59 61 62 64 64 64 We would like to mention that the first four sections are an exposition of the wellknown foundation of basics of general relativity and the Einstein equation. Among lots of sources in the literature that we can borrow from, we mainly follow an interesting lecture notes by Corvino in [Cor] and a pedagogical approach by Gourgoulhon in [Gou12]. In EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 3 3 addition, we also benefit knowledge from [YCBru09] by Choquet-Bruhat. Other excellent works related to general relativity and the Einstein equation can be found, for example, in [BarIse04, ChrGalPol10]. 4 1. G EOMETRY OF SPACETIMES IN GENERAL RELATIVITY 5 1.1. Basics of differential geometry. One of the main subjects in studies of differential geometry in general and of Riemannian geometry in particular are Riemannian manifolds. Loosely speaking, M is called a (smooth) Riemannian manifold of dimension n + 1 if: 1 2 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 • M is a real (smooth) manifold of dimension n + 1 equipped with • an inner product g on the tangent space TP M at any point P ∈ M which is positive definite in the sense g(X, X) C|X|2 for some positive constant C and • the map P → g(X(P ), Y (P )) is smooth. A special case of great importance to general relativity is a Lorentzian manifold, a special case of a wider class of manifolds, called pseudo-Riemannian manifolds, known as a generalization of Riemannian manifolds. Roughly speaking, M is called a pseudo-Riemannian manifold if • it is a generalization of a Riemannian manifold in which the metric tensor g need not be positive-definite; • instead a weaker condition of non-degeneracy is imposed on g: X ∈ TP M with X = 0 such that g(X, Y ) = 0 for all Y ∈ TP M . By the Sylvester law of inertia applied to the quadratic form g(x, x), we know that the signature (p, q, r) of the metric tensor g is invariant with p + q + r = n + 1 where by the pair (p, q, r) we mean positive, negative, and zeros. In this context, by pseudo-Riemannian metrics we mean those g having the signature (p, q, 0) where p, q are non-negative. Particularly, in the case p = n and therefore q = 1, such a metric g is called a Lorentzian metric and (M , g) is called a Lorentzian manifold. Throughout this notes, we repeatedly use g(·, ·) and ·, · g to denote the metric. Sometimes, we use X, Y g or X · Y instead of g(X, Y ) if no confusion occurs. 31 As we shall see right now, lots of geometric quantities can be defined on pseudoRiemannian manifolds. Given a pseudo-Riemannian manifold (M , g), there is an affine connection ∇. Then we can define the so-called Riemann curvature tensor Rm in terms of the connection ∇ as follows 32 Rm(X, Y, Z) = Rm(X, Y )Z = ∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ] Z. 28 29 30 33 (1.1) Using index notation, we can write ∂ ⊗ dxi ⊗ dxj ⊗ dxk ∂xl and by lowering the index l we also obtain l 34 35 Rm = Rmijk m 36 Rmijkl = Rmijk g ml = Rm ∂ ∂ ∂ ∂ , , , ∂xi ∂xj ∂xk ∂xl . g 38 The Ricci tensor, denoted by Ric, is then defined to be the trace of the Riemann curvature tensor Rm with respect to the first and the last indexes, i.e. 39 Ric(Y, Z) = tr X → Rm(X, Y )Z . 37 40 Therefore, in local coordinates, we write i 41 Ric = Ricg = Ricjk = Rmijk . (1.2) ˆ Q.A. NGO 4 3 Taking the trace of the Ricci tensor Ric, we obtain the scalar curvature, denoted by Scal, that is to say Scal = Scalg = g jk Ricjk , (1.3) 4 in local coordinates. 5 1.2. What are spacetimes? In general relativity, the main objects of study are spacetimes; more precisely, a spacetime in general relativity is nothing but a Lorentzian manifold. An usual spacetime manifold is of dimension four, but higher dimensions are considered in the aim of unification of gravitation with the other fundamental forces of nature, electromagnetism, weak and strong interactions, and also in super-symmetric theories. Throughout this notes, we always assume that spacetimes are of dimension n + 1. 1 2 6 7 8 9 10 16 Let (M , g) be a spacetime, recall from Subsection 1.1 that (M , g) is a differentiable manifold of dimension n + 1 equipped with a non-degenerate, smooth, symmetric metric tensor g. A hypersurface of M is the image Σ of some n-dimensional manifold M by an embedding (meaning a homeomorphism or one-to-one mapping such that both directions are continuous) Φ : M → M. 17 Clearly, the one-to-one property guarantees that Σ does not “intersect itself”. 11 12 13 14 15 n Σt+dt Σt Σt−dt F IGURE 1. The spacetime foliation. 18 19 20 Observe that any hyper-surface can be defined locally as the set of points for which a scalar field on M is constant. Denoting this scalar field by t and setting the constant to 0, we obtain from the above discussion ∀P ∈ M , 21 22 23 24 25 26 27 P ∈Σ if and only if t(P ) = 0. In other words, the hypersurface Σ is the zero level set of t. Next, we denote by γ the induced metric of g onto the hypersurface Σ. Then Σ is said to be: • spacelike: if γ is Riemannian, i.e. it has signature (+, +, . . . , +). • timelike: if γ is Lorentzian, i.e. it has signature (−, +, . . . , +). • null: if γ is degenerate, i.e. it has signature (0, +, . . . , +). 30 Since Σ is a level set of t, the gradient 1-form ∇t = (∇β t)dxβ is normal to Σ in the following sense ∀v ∈ T Σ, ∇t(v) = 0. 31 Hence its dual ∇ t = g αβ ∇β t is a vector field normal to Σ and satisfies: 28 29 α EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 5 • ∇t is timelike if and only if Σ is spacelike. • ∇t is spacelike if and only if Σ is timelike. • ∇t is null if and only if Σ is null. In case Σ is not null, we can normalize ∇t to make it unit, then denote by n: 1 n= ∇t (1.4) ±∇t · ∇t with the sign + for a timelike Σ and the sign − for a spacelike one. The coefficient 1/(±∇t · ∇t)1/2 is called the lapse function, usually denoted by N . Hence, the normal vector field n satisfies • n · n = −1 if Σ is spacelike. • n · n = 1 if Σ is timelike. For any X tangent to M , it is easy to see that ∇X n, n = 0; hence ∇X n is tangent to Σ. Similarly, vector X is called • spacelike: if X · X > 0. • timelike: if X · X < 0. • null: if X · X = 0. As being used, for spacetimes (of dimension n+1), we use bar “ · ” to denote geometric k quantities, like Rm, Scalg , Γij , etc, while for geometric quantities for spacelike, we simply drop bars, for example, Rm, Scalg , Γkij , etc. To setup a suitable spacetime configuration, we need further conditions. Σ is a Cauchy surface if Σ is a spacelike hypersurface such that each causal (for example, timelike or null) curve without end point only intersects Σ at once and only once. (Equivalently, Σ is a Cauchy surface if and only if its domain of dependence is the whole spacetime M .) Be careful, not all spacetimes admit a Cauchy surface, for those do have, we call them globally hyperbolic spacetimes. As mentioned in [YCBru09], the word “globally hyperbolic” comes from the fact that the wave equation is well-posed in these spacetimes, see also [Gou12]. Then we can foliate any globally hyperbolic spacetimes (M , g) by a family of spacelike hyper-surfaces (Σt )t∈R in the following sense M = Σt . t∈R 29 30 31 32 33 34 35 36 37 38 Roughly speaking, there is a regular smooth scalar field t on M in the sense that ∇t never vanishes such that each Σt is a level surface of t. Clearly, the regularity of t implies that Σt ∩ Σt = ∅ for any t = t . Furthermore, we can use ∇t to define time orientation for (M , g) and hence any other time like vector X is said to be • future pointing: if g(X, ∇t) is negative or • past pointing: if g(X, ∇t) is positive. Suppose that Σ is spacelike, then we choose from (1.4) the following 1 n= ∇t. −∇t · ∇t Clearly, ∇t g(n, ∇t) = g , ∇t = −(−∇t · ∇t)1/2 < 0. 1/2 (−∇t · ∇t) (1.5) ˆ Q.A. NGO 6 1 In other words, n constructed before is future pointing normal. 6 1.3. Adapted frames and coframes. Given a spacetime (M , g, ∇) of dimension n + 1, we start by choosing an (n + 1)-foliation of the spacetime manifold Ft : M → M , t ∈ R, for which each of the leaves Ft (M ) of the foliation, as a level set of the global time function t, is presumed spacelike. Recall from (1.5) that the future-directed normal unit vector n can be defined by using the gradient 1-form ∇t which is nothing but 7 n = N ∇t, 2 3 4 5 8 9 10 11 where N is the positive definite lapse function. It is also convenient to choose a threading of the spacetime M . The choice of a foliation and a threading together with a choice of coordinates (x1 , ..., xn ) for M automatically induce local coordinates (x0 = t, x1 , ..., xn ) on M . For a natural frame on M , we choose ∂i = 12 13 14 15 ∂ ∂xi ∀i = 1, n. Let us now denote by ∂t the dual of ∇t, that is, ∇t, ∂t = 1. The vector ∂t is usually called the time vector. Also, there is no risk to use both dt and ∇t to denote the same gradient 1-form. Then we need to find the remaining ∂0 . 17 Assume that we have already had the frame (∂0 , ∂1 , ..., ∂n ). The dual coframe (θi )ni=0 of (∂i )ni=0 is found to be such that 18 θi = dxi + β i dt ∀i = 1, n, 16 21 while the 1-form θ0 is nothing but dt. Here β i are the components of the spacelike shift vector β which is the difference between N n and ∂t . To be exact, the shift vector is chosen as follows 22 β = ∂t − N n. 19 20 23 24 (1.6) Since ∇t, β = ∇t, ∂t − N n = ∇t, ∂t − ∇t, −N n = 1 − 1 = 0. 26 The last thing we need to find is the vector ∂0 . To select ∂0 suitable, we need to solve ∂0 , θi = 0 for all i = 1, n. Hence, the only possibility is that 27 ∂0 = ∂t − β j ∂j . 25 28 29 30 31 It is now easy to check that ∂0 = N n in our case. It is worth noting that the presence of ∂0 in the basis for T M is because the vector ∂t is not always timelike since ∂t , ∂t < 0 doest not hold in general as can be seen from the following computation ∂t , ∂t = N n + β, N n + β = −N 2 + β, β . 33 Note that in the previous identity, we have used the fact that β is tangential to the leave Ft (M ); hence n, β = 0. 34 By using the coframe (θi )ni=0 , we may locally rewrite the spacetime metric g in the form 32 35 g = −N 2 θ0 ⊗ θ0 + gij θi ⊗ θj , 37 where gij are the components of the spatial metric tensor g. In terms of the local coordinates (x0 = t, x1 , ..., xn ) on M , we can write 38 g = −N 2 dt ⊗ dt + gij (dxi + β i dt) ⊗ (dxj + β i dt). 36 (1.7) EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 7 n Nn ∂t β Σt xi = cst F IGURE 2. Shift vector β, lapse function N in the spacetime foliation. 1 2 3 4 5 6 7 8 We note that for each t, the term gij (t)dxi ⊗ dxj is the induced Riemannian metric on the leaf Ft (M ); hence we can write .. g . 00 (g αβ ) = . · · · (gij ) Besides, on each leaf Ft (M ), we have a tangent vector basis given by (∂1 , ..., ∂n ). 1.4. Basics of submanifolds. We now study the geometry of spacelike M of dimension n sitting in a spacetime (M , g, ∇) of dimension n + 1 as an embedded submanifold. (M is being considered as Σ in previous subsections.) Hence, g induces a non-degenerate metric g on M called the first fundamental form, this is because M is spacelike. 10 Corresponding to g, there is the Levi–Civita connection ∇. Then the pair (g, ∇) and (g, ∇) are related by 11 ∇X Y = (∇X Y )// + (∇X Y )⊥ = ∇X Y + II(X, Y ). 9 (1.8) 13 The symmetric tensor II is called the (vector-valued) second fundamental form on M . (The first is the metric g, of course.) Since II is proportional to the normal n of M , we can write 14 K(X, Y )n = II(X, Y ) 12 15 16 (1.9) and call K the (scalar-valued) second fundamental form. Sometimes, we also denote K(X, Y ) = II(X, Y ), n (1.10) 18 and also call K the scalar-valued Second Fundamental Form; hence K = n, n K. In other words, K and K are different simply by the sign of n, n . Also by (1.9), we obtain 19 II(X, Y ) = n, n K(X, Y )n. 17 20 21 22 23 24 Now, thanks to ∇Y X, n = 0, we further obtain K(X, Y ) = II(X, Y ), n = ∇X Y, n = − ∇X n, Y . 1.5. The Gauss and Codazzi equations. In the next few sections, we shall see that Einstein constraint equations are simply consequences of the Gauss and Codazzi equations which will be considered in this subsection. 8 ˆ Q.A. NGO 2 However, for the sake of clarity, we first derive several useful properties of the Riemann curvature tensor. First, recall from (1.1) that Rm acting on vector fields X, Y , and Z is 3 Rm(X, Y )Z = ∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ] Z, 1 4 5 6 7 hence by interchanging X and Y we know that Rm(Y, X)Z = ∇Y ∇X Z − ∇X ∇Y Z − ∇[Y,X] Z = −Rm(X, Y )Z. (1.11) In other words, the Riemann curvature tensor is skew symmetric with respect to the first two components. 9 To prove that Rm(X, Y )Z, W is also skew symmetric with respect to the last two components Z and W , we first show that 10 II(Y, Z) = II(Z, Y ), 8 11 12 that is to say that the second fundamental form is symmetric. To see this, first observe by definition (1.8) that ∇Y Z = ∇Y Z + II(Y, Z), 13 14 15 16 17 18 19 20 21 22 23 ∇Z Y = ∇Z Y + II(Z, Y ). Hence, using the torsion free property of connections ∇ and ∇, we obtain after subtracting the two equations [Y, Z]∇ = [Y, Z]∇ + II(Y, Z) − II(Z, Y ). From this we obtain the symmetry of II. Next, to see why Rm(X, Y )Z, W = Rm(X, Y )W, Z , it suffices to show Rm(X, Y )Z, Z = 0 for any X, Y, Z. We are now in a position to state the Gauss equation which tells us how Rm(X, Y )Z and Rm(X, Y )Z are related to each other. We state it in the following proposition. Proposition 1.1. Suppose that (M, g, ∇) is a submanifold of (M , g, ∇). Then for any X, Y, Z, W ∈ T M , we have the following equation Rm(X, Y )Z, W = Rm(X, Y )Z, W − II(X, Z), II(Y, W ) 24 25 26 + II(Y, Z), II(X, W ) . Proof. First, we express Rm(X, Y )Z in terms of geometric quantities on (M, g, ∇). This can be done in details as the following. Rm(X, Y )Z =∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ] Z =∇X (∇Y Z + II(Y, Z)) − ∇Y (∇X Z + II(X, Z)) − ∇[X,Y ] Z − II([X, Y ], Z) =∇X ∇Y Z + II(X, ∇Y Z) + ∇X (II(Y, Z)) 27 (1.12) − ∇Y ∇X Z − II(Y, ∇X Z) − ∇Y (II(X, Z)) − ∇[X,Y ] Z − II([X, Y ], Z) = Rm(X, Y )Z + II(X, ∇Y Z) − II(Y, ∇X Z) − II([X, Y ], Z) + ∇X (II(Y, Z)) − ∇Y (II(X, Z)). EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 The three terms involving II are perpendicular to W ; hence by taking inner product with W with respect to the spacetime metric g, we know that Rm(X, Y )Z, W 3 4 9 g = Rm(X, Y )Z, W g + ∇X (II(Y, Z)), W g = Rm(X, Y )Z, W g − II(Y, Z), ∇X W + II(X, Z), ∇Y W = Rm(X, Y )Z, W g − II(Y, Z), II(X, W ) g − ∇Y (II(X, Z)), W g g g + II(X, Z), II(Y, W ) g . as claimed. ⊥ 5 6 7 8 In view of the Gauss equation (1.12), it is left with Rm (X, Y )Z and this is the content of the Codazzi equation. Sometimes, we call it the Codazzi–Mainardi equation or the Ricci identity, which expresses the curvature of the normal bundle in terms of the second fundamental form. 10 Proposition 1.2. Suppose that (M, g, ∇) is a submanifold of (M , g, ∇). Then for any X, Y, Z, W ∈ T M , we have the following equation 11 Rm (X, Y )Z = (∇X II)(Y, Z) − (∇Y II)(X, Z) 9 ⊥ 12 (1.13) where the derivatives ∇X II is understood as follows ⊥ (∇U II)(V, W ) = ∇U (II(V, W )) − II(∇U V, W ) − II(V, ∇U W ) 13 14 for any tangent vector fields U, V, W in T M . 15 Proof. To see this, we start with Rm(X, Y )Z = Rm(X, Y )Z + II(X, ∇Y Z) − II(Y, ∇X Z) − II([X, Y ], Z) 16 17 18 + ∇X (II(Y, Z)) − ∇Y (II(X, Z)) which we have already derived before. Then we look at the normal part of the both sides to get ⊥ Rm (X, Y )Z = II(X, ∇Y Z) − II(Y, ∇X Z) − II([X, Y ], Z) 19 ⊥ ⊥ + ∇X (II(Y, Z)) − ∇Y (II(X, Z)). 20 Using the fact [X, Y ] = ∇X Y − ∇Y X to get II([X, Y ], Z) = II(∇X Y, Z) − II(∇Y X, Z). 21 22 From this we eventually get ⊥ Rm (X, Y )Z =II(X, ∇Y Z) − II(Y, ∇X Z) − II(∇X Y, Z) + II(∇Y X, Z) ⊥ ⊥ + ∇X (II(Y, Z)) − ∇Y (II(X, Z)) 23 =(∇X II)(Y, Z) − (∇Y II)(X, Z) 24 25 26 27 28 29 as claimed. 2. T HE E INSTEIN EQUATIONS IN GENERAL RELATIVITY 2.1. The Einstein equations. As always, we assume throughout this section that (M , g) is a Lorentzian manifold of dimension n + 1. In this notes, we do not want to address the question: How could Einstein come up with the equation named after him? For the sake of simplicity, we adopt the Einstein equation as an axiom equation. However, like other field ˆ Q.A. NGO 10 1 2 equations in physics, the original Einstein equations can be derived from an action through the principle of least action. More precise, the action 3 M 4 5 6 7 8 11 12 13 14 15 16 17 18 19 20 (2.1) is required to be stable under compact perturbations of the metric g where Lg is the Lagrangian associated with non-gravitational fields. When Lg ≡ 0, the action (2.1) is known as the Einstein–Hilbert action. For each compactly supported 2-tensor h, by direct computing 1, we obtain the variation of M Scalg dvolg with respect to g as follows d dt 9 10 (Scalg +2Lg ) dvolg t=0 M Scalg+th dvolg+th = − 1 h Ricg − Scalg g dvolg . 2 M (2.2) Regarding to the Lagrangian Lg 2, we obtain the following formula d dt t=0 M Lg+th dvolg+th = (Dg Lg )(h) + Lg · M 1 trg (h) dvolg . 2 (2.3) Hence, by the variation of (2.1) with respect to g, we obtain the field equation 1 Ricg − Scalg g = T (2.4) 2 where the left hand side of Eq. (2.4) is known as the Einstein tensor which will be denoted by Einsg and the tensor T , which comes from the variation of M Lg dvolg with respect to g, is known as the stress-energy tensor. For particular Lagrangian Lg , we can easily calculate the stress-energy tensor T as we shall see later when scalar fields are included in our spacetime. By taking the trace of the Einstein tensor, we further arrive at 1 1−n tr(Einsg ) = tr Ricg − Scalg g = Scalg . 2 2 1Perhaps, it would be easier to derive Eq. (2.4) using “delta” variation instead of using (2.2). Indeed, there holds 0 =δ = = M Scalg dvolg δ | det(g)| Scalg δg ij dx δg ij M δ M | det(g)| δg ij Scalg | det(g)| + δ Scalg δg ij dvolg . δg ij Then it is easy to obtain the left hand side of Eq. (2.4) by noticing that δ | det(g)| 1 =− g 2 δg ij | det(g)|, δ Scalg = Ricg . δg 2To derive (2.3), we can again use “delta” variation. Indeed, there holds δ M Lg dvolg = = = δ M | det(g)|Lg δg ij Lg δg ij dx δ | det(g)| M δLg + δg ij M δLg 1 − g ij Lg δg ij dvolg . 2 δg ij | det(g)| Thus, we also obtain Tij = −2 δLg + g ij Lg . δg ij δg ij δg ij dvolg EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 11 5 A vacuum space time is a Lorentzian manifold (M , g) that satisfies Eq. (2.4) with vanishing stress-energy tensor T = 0, thanks to Lg ≡ 0 and (2.3). In this case, we can take the trace of Eq. (2.4) to find Scalg = 0 and obtain the vacuum Einstein equations Ricg = 0. We conclude this subsection by noting that the fully Einstein equations (with cosmological constant Λ) may be written in the following form 6 Einsg +Λg = T 1 2 3 4 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 where Λ is constant called the cosmological constant. The cosmological constant term was originally introduced by Einstein to allow for a static universe. It is clear to see that the cosmological term could be absorbed into the stress-energy tensor T as dark energy. 2.2. The Einstein equations with real scalar fields. As can be seen from Eq. (2.4), the Einstein equations involve the so called stress-energy tensor T representing the density of all the energies, momentum, and stresses of the sources. On a macroscopic scale, one can couple gravity to either field sources or matter sources such as electro-magnetic fields, Yang-Mills sources, and scalar fields. While the latter are more phenomenological, the former, which are deduced directly from special relativity, are one of the simplest non-vacuum models which are the core of studies in recent years. In addition, interest in those models stems partly from recent attempts to use such models to study the observed acceleration of the expansion of the universe. From now on, we only focus on the Einstein equations equipped with scalar fields. In modern cosmology, one can introduce on the spacetime (M , g) a real scalar field ψ with potential U as a smooth function of ψ. A particular Einstein field theory is specified by the choice of an action principle, for example, with the following Lagrangian 1 Lg = − |∇ψ|2g − U (ψ). (2.5) 2 This choice of action principle also includes the well-known massive or massless Klein2 Gordon field theory where U (ψ) = m2 ψ /2. Thanks to (2.3), by a fairy standard compu3 tation , one can deduce that 1 Tαβ = ∇α ψ∇β ψ − g αβ ∇µ ψ∇µ ψ − g αβ U (ψ). (2.6) 2 It is worth noticing that a cosmological constant Λ can be considered as a particular scalar field with potential Λ. For this reason, we do not consider the cosmological term in our Einstein equation. 2.3. Why do the Einstein equations describe the propagation of wavelike phenomena? In order to formulate the initial value problem for the Einstein equations as nonlinear wave equations, we first express the Einstein equations in terms of a partial differential equation along with some gauge condition; hence the Einstein equations basically describe the propagation of wave. 3To derive (2.6), we can again use “delta” variation. Recall the following formula Tij = −2 δLg + g ij Lg . δg ij Hence δ δg ij 1 ij 1 g ∇i ψ∇j ψ − U (ψ) + g ij − |∇ψ|g2 − U (ψ) 2 2 1 =∇i ψ∇j ψ − g ij |∇ψ|2g − g ij U (ψ). 2 Tij = − 2 − ˆ Q.A. NGO 12 1 2 3 4 5 6 7 8 9 We suppose that (M , g) is a Lorentzian manifold of the dimension n+1. In a coordinate system that will be fixed from now on, we have 1 k Γij = g km (g im,j + g jm,i − g ij,m ) (2.7) 2 as Christoffel symbols for the spacetime metric g. For the sake of simplicity, by “lower orders” we mean terms consisting of either no derivative or first order derivative of the spacetime metric g. As such, terms consisting of derivatives of order higher than two will be called high order terms. We now denote by g the wave operator (or the d’Alembertian) evaluated with respect to the indicated metric g. Mathematically, g is defined to be i 10 11 g = ∇ ∇i = g ij ∇i ∇j . (2.8) Then, we introduce the following notation λα = 12 gx α 16 where xα are coordinate functions. Using (2.8) and thanks to the rule ∇i ∂j = Γkij ∂k , we obtain k ij α λα = g ij ∇i ∇j xα = g ij (Γij xα ,k ) = g Γij . Taking another derivative, it is obvious to see that 17 α km Γkm,j + g αj g km Γkm,i . (g αi λα ,j + g αj λ,i ) = g αi g 13 14 15 α 18 19 20 21 22 23 24 25 26 27 Using (2.7), we can write 1 α Γkm,j = g αp (g kp,mj + g mp,kj − g km,pj ) + lower orders 2 and 1 α Γkm,i = g αq (g kq,mi + g mq,ki − g km,qi ) + lower orders. 2 Therefore, using the Kronecker symbol δ, for example g αi g αp = δip , a further calculation shows that 1 α km αp (g αi λα g (g kp,mj + g mp,kj − g km,pj ) ,j + g αj λ,i ) = g αi g 2 1 + g αj g km g αq (g kq,mi + g mq,ki − g km,qi ) + lower orders 2 1 p km = δi g (g kp,mj + g mp,kj − g km,pj ) 2 1 + δjq g km (g kq,mi + g mq,ki − g km,qi ) + lower orders 2 1 km = g (g ki,mj + g mi,kj − g km,ij ) 2 1 + g km (g kj,mi + g mj,ki − g km,ij ) + lower orders 2 =g km (g ki,mj + g mi,kj − g km,ij ) + lower orders. Thus, we have just shown that α g km (g ki,mj + g mi,kj − g km,ij ) = (g αi λα ,j + g αj λ,i ) + lower orders. (2.9) We now have the components of the Ricci curvature Ricg of g given by k 28 α k k l k l Ricij = Γij,k − Γik,j + Γkl Γij − Γjl Γik . k l k l 30 A simple observation tells us that the difference Γkl Γij − Γjl Γik consists of lower order terms only. Therefore, we can write 31 Ricij = Γij,k − Γik,j + lower orders. 29 k k EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS k 1 2 3 4 5 6 7 8 9 10 13 k Moreover, using (2.7) we calculate the difference Γij,k − Γik,j as follows 1 k k Γij,k − Γik,j = g km (g im,kj + g jm,ki − g ij,km ) 2 1 − g km (g im,kj + g km,ij − g ik,mj ) + lower orders 2 1 km = g (g jm,ki − g ij,km − g km,ij + g ik,mj ) + lower orders. 2 In other words, 1 Ricij = g km (g jm,ki − g ij,km − g km,ij + g ik,mj ) + lower orders. (2.10) 2 α Combining (2.10) and our calculation for g αi λα ,j + g αj λ,i in (2.9) above, we find that 1 1 km α g ij,km + lower orders. (2.11) (g αi λα ,j + g αj λ,i ) − g 2 2 In view of the Einstein equation (2.4), we obtain 1 Ricij = Tij + g ij Scalg , 2 which then helps us to conclude 1 1 1 α − g km g ij,km = − (g αi λα ,j + g αj λ,i ) + Tij + g ij Scalg +lower orders . 2 2 2 Ricij = lower orders 11 12 13 14 15 In order to get rid of high order terms on the right hand side of the preceding equation, i.e. α the term g αi λα ,j + g αj λ,i , it is usually to assume that λα = 0, ∀α = 0, n. This condition is preferred to the so-called Lorentz-harmonic coordinate gauge. In the literature, this belongs to a set of a few condition that one can solve the Einstein equations. 21 Thus, we have shown that the Einstein equations for g in the Lorentz-harmonic coordinate gauge are nothing but wave equations for g. We conclude this subsection by observing that on any Lorentzian manifold the Lorentz-harmonic coordinate gauge is fulfilled by simply solving the Cauchy problem for some linear wave equations. In addition, the solvability of the transformed wave-like system as shown above is well-understood by the seminal work of Leray. 22 3. T HE C AUCHY PROBLEM FOR THE E INSTEIN EQUATIONS 23 Since the Einstein equations are geometric equations, one can expect solutions of the Einstein equations verifying the causality principle, that is, the relativistic spacetime future cannot influence the past. Based on two works by Leray [Ler53] and Geroch [Ger70], we know that the globally hyperbolic spacetime (M , g) are topological products M × R with each M × {t} intersected once by each inextensible timelike curve. In view of our definition above, such a submanifold M × {t} is a Cauchy surface. 16 17 18 19 20 24 25 26 27 28 29 30 31 32 33 34 35 In order to formulate an appropriate Cauchy problem for the Einstein equations, we assume that the globally hyperbolic spacetime (M , g) admits M as a Cauchy surface. We let g be the Riemannian metric on M induced by g. Having such a Cauchy surface, we let n be the future pointing timelike unit normal vector to M . We also let K be the extrinsic curvature of M computed with respect to n. We are now able to formulate the Cauchy problem for the Einstein equations intrinsically through the following definition, see [YCBru09, Chapter VI]. ˆ Q.A. NGO 14 n Σt+dt embedding M Σt−dt F IGURE 3. The Cauchy problem for the Einstein equations. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 Definition 3.1. (1) An initial data set is a triplet (M, g, K) where M is an n-dimensional smooth manifold, g is a properly Riemannian metric on M and K a symmetric 2-tensor. (2) A Cauchy development of an initial data set is a spacetime (M , g) such that there exists an embedding ι : M → M enjoying the following properties (a) The metric g is the pullback of g by ι, that is, g = ι∗ g. In other words, if M is identified with its image ι(M ) = M0 in M , then g is the metric induced by g on M0 . (b) ι(K) is the second fundamental form of ι(M ) as submanifold of (M , g). (3) A Cauchy development (M , g) of (M, g, K) is called a Einsteinian development if the metric g satisfies the Einstein equations on M . We now suppose that (M , g) is a Cauchy development of (M, g, K). If it is also true that every other Cauchy development of (M, g, K) can be isometrically embedded in M , we say M is called the maximal development of (M, g, K). It is easy to see that a maximal development is unique up to isomorphism. As we shall see later in Proposition 4.1, in order to generate a Cauchy development, the initial data (M, g, K) for the Einstein equations cannot be arbitrary, they must satisfy some conditions, this is because data on (M, g, K) must satisfy certain conditions when sitting in the ambient space. To be exact, in view of the Gauss and Codazzi equations, those conditions can be rewritten in a form consisting two equations known as the Hamiltonian and momentum constraints, see Eq. (4.1) and Eq. (4.2) below. Conversely, we wish that for given initial data verifying the constraint equations (4.1) and (4.2), one can generate a corresponding Cauchy development the for such given initial data. The following theorem due to Y. Choquet-Bruhat and R. Geroch shows that this is possible in the sense that Eqs. (4.1) and (4.2) also give us a sufficient condition, see [CBruGer69]. Theorem 3.2. Given smooth initial data (M, g, K) satisfying the constraint equations, there exists a smooth, maximal, globally hyperbolic Cauchy development of the initial data. Consequently, there is a strong connection between the globally hyperbolic Cauchy development of the initial data and the Einstein constraint equations. However, there is no one-to-one correspondence between solutions of the Einstein constraint equations and their globally hyperbolic maximal Cauchy developments in the sense that two distinct solutions of the Einstein constraint equations may generate isometric globally hyperbolic maximal Cauchy developments. Nevertheless, as a first step, it is important to understand solutions of the constraint equations. EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 15 4. C ONSTRAINTS AND EVOLUTIONS 1 6 As we have already mentioned in the previous section, the initial data (M, g, K) for the Einstein equations cannot be arbitrary, they must satisfy some conditions. The purpose of this section is to prove Proposition 4.1 below which tells us how the initial data (M, g, K) are related to each others. A proof for Proposition 4.1 simply makes use of the Gauss and Codazzi equations, see Propositions 1.1 and 1.2 in Subsection 1.5. 7 Proposition 4.1. The initial data (M, g, K) satisfies the following constraints: 2 3 4 5 Scalg −|K|2g + (trg K)2 = 2ρ 8 9 10 11 12 13 14 15 16 17 18 19 20 21 and divg K − d(trg K) = J, (4.2) where ρ is a scalar and J a vector on M determined by the projection on M and the normal to M , when embedded in a spacetime (M , g), of the tensor T of the sources. In the literature, Eq. (4.1) is known as the Hamiltonian constraint while Eq. (4.2) is called the momentum constraint. We spend this section to prove Proposition 4.1. 4.1. Derivation of constraint equations: The Hamiltonian constraint. This subsection is devoted to prove (4.1) while (4.2) will be considered in the next subsection. Before we start, let us recall the Einstein equation without the cosmological constant Λ, that is 1 Ricg − g Scalg = T. 2 The above equation is understood over the Lorentzian manifold (M , g) of the dimension n + 1. As always, (M, g) is being considered as a submanifold of M of the dimension n. Recall from (1.8), there holds // ∇X Y = ∇X Y, 22 23 24 25 26 (4.1) ⊥ II(X, Y ) = ∇X Y and ∇X Y = ∇X Y + II(X, Y ). Also, let us recall the Riemann curvature tensor R given by Rm(X, Y )Z = ∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ] Z. Another ingredient in the proof of (4.1) is the Gauss equation (1.12) is given by Rm(X, Y )Z, W = Rm (X, Y )Z) , W − II(X, Z), II(Y, W ) 27 28 29 + II(Y, Z), II(X, W ) . We now let e1 , ..., en be a local orthonormal frame field for M and we fix e0 = n. Using the Gauss equation (1.12) applied to (X, Y, Z, W ) = (ei , ej , ei , ej ), we arrive at − Rm (ei , ej ) ei , ej = − Rm(ei , ej )ei , ej + II(ei , ei ), II(ej , ej ) 30 31 − II(ei , ej ), II(ei , ej ) , which, after summing all over i and j from 1 to n, yields n − 32 n Rm (ei , ej ) ei , ej = − i,j=1 n + II(ei , ei ), II(ej , ej ) − i,j=1 33 (4.3) II(ei , ej ), II(ei , ej ) . i,j=1 By the definition of the scalar curvature, we have n 34 Rm(ei , ej )ei , ej i,j=1 n Scalg = n Ric(ej ), ej = − j=1 Rm(ei , ej )ei , ej , i,j=1 ˆ Q.A. NGO 16 3 where Ric(u) = j Rm(u, ej )ej . Notice that the Ricci tensor is skew symmetric, thanks to (1.11). Besides, by the definition of the (scalar-valued) second fundamental form K(·, ·) with respect to the unit normal vector e0 , i.e. 4 K(X, Y ) = II(X, Y ), e0 . 1 2 5 6 We recall that the corresponding mean curvature trg K of M , denoted by τ , is nothing but the trace of the second fundamental form K; hence τ can be calculated using n τ = trg (K) = 7 K(ei , ei ). i=1 8 In the case of a hypersurface, by using (1.10), there holds II(X, Y ) = e0 , e0 K(X, Y )e0 , 9 10 (4.4) regardless of the sign of e0 , e0 . Therefore, applying (4.4) to (X, Y ) = (ei , ei ) to get n n II(ei , ei ), II(ej , ej ) = i,j=1 e0 , e0 K(ei , ei )e0 , e0 , e0 K(ej , ej )e0 i,j=1 n = e0 , e0 K(ei , ei )K(ej , ej ) 11 (4.5) i,j=1 n n K(ej , ej ) K(ei , ei ) =− j=1 i=1 = − (trg K)2 . 12 For the remaining term, we again apply (4.4) to (X, Y ) = (ei , ej ) to obtain n n e0 , e0 K(ei , ej )e0 , e0 , e0 K(ei , ej )e0 II(ei , ej ), II(ei , ej ) = i,j=1 i,j=1 n = e0 , e0 13 K(ei , ej )K(ei , ej ) (4.6) i,j=1 n 2 =− K(ei , ej ) i,j=1 = − |K|2g . 14 Combining (4.3), (4.5), and (4.6), we have just shown that n 15 Rm (ei , ej ) ei , ej = Scalg −|K|2g + (trg K)2 . − (4.7) i,j=1 16 17 Comparing (4.1) and (4.7), it remains to estimate the left hand side of (4.7). By definition of the Ricci curvature and the fact e0 , e0 = −1, we have n Ricg (ei , ei ) = − Rm (e0 , ei ) ei , e0 + Rm (ej , ei ) ei , ej j=1 18 n = Rm (ei , e0 ) ei , e0 − Rm (ei , ej ) ei , ej . j=1 EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 17 Therefore, thanks to Rm(e0 , e0 )e0 , e0 = 0, we get n − n Rm (ei , ej ) ei , ej = i,j=1 Ricg (ei , ei ) − Rm (ei , e0 ) ei , e0 i=1 n = Ricg (ei , ei ) + Rm (e0 , ei ) ei , e0 i=1 n 2 = Ricg (e0 , e0 ) + Ricg (ei , ei ) i=1 n =2 Ricg (e0 , e0 ) + − Ricg (e0 , e0 ) + Ricg (ei , ei ) i=1 =2 Ricg (e0 , e0 ) + Scalg . 3 Since Ricg − 12 g Scalg = T , we obtain 1 Ricg (e0 , e0 ) − g(e0 , e0 ) Scalg = T (e0 , e0 ). 2 4 Ricg (e0 ,e0 )+ 12 Scalg 5 Thus, we have proved that Scalg −|K|2g + (trg K)2 = 2ρ, 6 7 8 9 10 11 12 known as the Hamitonian constraint equation. Note that here we denote ρ = T (e0 , e0 ). 4.2. Derivation of constraint equations: The momentum constraint. This subsection is a continuation of the previous subsection where we showed in detail the derivation of the Hamiltonian constraint. In this subsection, we derive the so-called momentum constraint equation (4.2). First, let us recall the Codazzi equation (1.13) which is given by the following identity ⊥ Rm (X, Y )Z = (∇X II)(Y, Z) − (∇Y II)(X, Z). 13 14 15 Using this, we obtain after applying to (X, Y, Z) = (ei , Y, ei ) and summing over i from 1 to n the following n ⊥ Rm (ei , Y )ei = 16 i=1 17 18 n n (∇ei II)(Y, ei ) − i=1 (∇Y II)(ei , ei ) for any tangent vector Y . First, we estimate the left hand side of (4.8). Since Rm(ei , Y )ei , for all i 1, belong to the tangent space, there hold Rm(ei , Y )ei , e0 = 0 ∀i = 1, n. 19 // 20 21 23 24 Rm (ei , Y )ei , e0 = Rm(ei , Y )ei , e0 26 ∀i = 1, n. (4.9) Moreover, by the skew symmetric property of the Riemann curvature tensor, i.e., Rmijkl = −Rmijlk , we know that Rm(e0 , ei )e0 , e0 = 0 ∀i = 1, n 25 27 ⊥ Furthermore, by splitting Rm(ei , Y )ei = Rm (ei , Y )ei + Rm (ei , Y )ei and then taking the inner product with e0 = n, we can write ⊥ 22 (4.8) i=1 which immediately implies Rm(e0 , Y )e0 , e0 = 0. ˆ Q.A. NGO 18 1 Hence, by the definition of Ricg (Y, e0 ), we obtain n Ricg (Y, e0 ) = − Rm(e0 , Y )e0 , e0 + Rm(ei , Y )e0 , ei i=1 2 n =− Rm(ei , Y )ei , e0 . i=1 3 This and the relation (4.9) help us to conclude n Ricg (Y, e0 ) = − 4 ⊥ Rm (ei , Y )ei , e0 . i=1 5 Thanks to (4.8), we arrive at n 6 n Ricg (Y, e0 ) = − (∇ei II)(Y, ei ), e0 + i=1 7 8 9 (∇Y II)(ei , ei ), e0 . (4.10) i=1 We now evaluate the right hand side of (4.10). In order to achieve that goal, we make use of the formula involving the covariant derivative of the second fundamental form. Indeed, for the first term involving (∇ei II)(Y, ei ), e0 , we first write ⊥ (∇ei II)(Y, ei ) =∇ei (II(Y, ei )) − II(∇ei Y, ei ) − II(Y, ∇ei ei ) 10 ⊥ = − ∇ei (K(Y, ei )e0 ) + K(∇ei Y, ei )e0 + K(Y, ∇ei ei )e0 . 11 Using the Leibniz rule and thanks to ∇ei e0 , e0 = 0, we can write ⊥ ⊥ ⊥ ∇ei (K(Y, ei )e0 ) = ∇ei (K(Y, ei ))e0 + K(Y, ei ) ∇ei e0 12 =0 = ∇ei (K(Y, ei ))e0 . 13 14 15 Thus, we get (∇ei II)(Y, ei ), e0 = ∇ei (K(Y, ei )) − K(∇ei Y, ei ) − K(Y, ∇ei ei ). By summing, we obtain n (∇ei II)(Y, ei ), e0 = (divg K)(Y ). 16 (4.11) i=1 17 For the second term involving (∇Y II)(ei , ei ), e0 , we do the same way as follows ⊥ (∇Y II)(ei , ei ) =∇Y (II(ei , ei )) − II(∇Y ei , ei ) − II(ei , ∇Y ei ) 18 ⊥ =∇Y (II(ei , ei )) − 2II(∇Y ei , ei ). 19 20 Without loss of generality, we can assume that we are in normal coordinates at some point, say p ∈ M . This immediately implies that ∇Y ei = 0 for all i = 1, n. Therefore, ⊥ (∇Y II)(ei , ei ) =∇Y (II(ei , ei )) ⊥ = − ∇Y (K(ei , ei )e0 ) ⊥ = − ∇Y (K(ei , ei ))e0 + K(ei , ei ) ∇⊥ Y e0 21 =0 = − ∇Y (K(ei , ei ))e0 . 22 Thus, by summing, we obtain n n (∇Y II)(ei , ei ), e0 = ∇Y 23 i=1 K(ei , ei ) = d(trg K)(Y ). i=1 (4.12) EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 Simply combining (4.10), (4.11), and (4.12), we have just proved that − Ricg (Y, e0 ) = (divg K)(Y ) − d(trg K)(Y ). 2 3 4 19 By using the Einstein equation and thanks to g(Y, e0 ) = 0, we get that 1 Ricg (Y, e0 ) = Ricg (Y, e0 ) − g(Y, e0 ) Scalg = T (Y, e0 ). 2 =0 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 Thus, we arrive at divg K − d(trg K) = −T (·, e0 ). We note that, throughout the proof, we have used the fact that the vector ∇Y e0 belongs to the tangent space, thus giving us ∇⊥ Y e0 = 0. 4.3. Evolutions of the first and second fundamental forms. One of the interesting features of the Einstein equation is that it can be considered as a dynamical system. To understand this, we simply go back to (1.7). Assume that we have already had a spacetime (M , g, ∇) and that each slice Σt = M × {t} is spacelike in M . Then whenever we know the induced metric g on Σt , the lapse function N , and the shift vector β, we can fully recover the spacetime metric g. In other words, determining solutions of the initial value problems is equivalent to specifying the lapse and shift based on the initial data. However, one can formulate the problem in the reverse way: Fixing a good choice of the lapse function and the shift vector together with an initial hypersurface M , we need to solve the first and second fundamental forms on slices using evolutions. The evolutions that we have just mentioned can be derived easily but lengthy computation cannot be avoided. To be exact, the induced metric g and the second fundamental form K obey the following system: ∂ gij = −2N Kij + Lβ gij ∂t and ∂ Kij = −∇i ∇j N + N (Ricij − Ricij +KKij − 2Kiµ Kjµ ) + Lβ Kij ∂t where Lβ is the Lie derivative along the vector field β. 4.4. Construction of spacetimes via solutions of the vacuum Einstein constraint equations. In this subsection, given a solution (g, K) of the constraint equations (4.1) and (4.2) on a manifold (M, g) of the dimension n, we shall construct an appropriate spacetime (M , g) of the dimension n + 1 such that the Einstein equation (2.4) holds in the vacuum case. 4.4.1. Construction of spacetimes. First, we observe that tr(Ric) = Scal and tr(g) = n + 1. Hence, taking the trace both sides of (2.4) with respect to the spacetime metric g to get 1−n Scalg = trg (T ). 2 Thus, we can rewrite (2.4) as follows g trg (T ). (4.13) Ricg = T − n−1 Recall from (2.11) and using (4.13) we obtain the following identity 1 1 Ricij = (g αi ∂j λα + g αj ∂i λα ) − g km g ij,km + lower orders 2 2 1 =Tij − trg (T )g ij . n−1 ˆ Q.A. NGO 20 1 2 3 4 5 6 7 Note that in the vacuum case, we simply ignore all T . At the very beginning, the metric g can be fully determined by setting 1 i, j n, g ij = gij , g = −1, 00 (4.14) g 0j = 0, 1 j n, g ij,0 = −2Kij , 1 i, j n. How about g ij,k ? Clearly, as a consequence of the choice above, we also have g ij,k for non-zero i, j, k. To fully construct the initial condition, we still need to find g 0j,0 as they cannot be obtained from (4.14). First, in view of the gauge condition, λα = 9 10 α = 0, we find that 1 ∂ ∂ | det g| g ij i (xα ) j ∂x | det g| ∂x 1 αj pq =g αj ,j + g g g pq,j = 0. 2 Therefore, at t = 0, we get that 1 00 pq 00 0 λ = g ,0 + 2 g g g pq,0 , n 1 αj pq + λα = g α0 g αj α ,0 ,j + g g g pq,j , 1 2 j=1 λα = 8 gx n. already determined 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Hence, in order to guarantee that λ ≡ 0 at t = 0 for all α = 0, n, we first select g α0 ,0 such that λα = 0 at t = 0. Once this task is done, we can determine g 00 at t = 0 such that ,0 0 λ = 0 at t = 0. In other words, the initial data that preserves the gauge condition can be found in this way. α Within a small time, the hyperbolic system mentioned above always admits a solution g. However, it is not necessary to have that g solves the Einstein equations unless the gauge condition remains valid within the small time. We shall prove this affirmatively. 4.4.2. The propagate of the gauge condition in time. We now consider how could the gauge condition propagate in time so long as the metric g solves the reduced equations. As we shall see, all λα satisfy a homogeneous linear wave equation which is a consequence of the Bianchi identities with vanishing initial time derivatives which come from the constraint equations. First, recall that in the vacuum setting, the Einstein equation is nothing but Ricg = 0. We now introduce the following reduced Einstein equation 1 Ricij + (g αi ∂j λα + g αj ∂i λα ) = 0. (4.15) 2 Clearly, whenever g solves (4.15) along with the gauge condition λα = 0, we immediately conclude that g solves the original Einstein equation. Keep in mind that the term g αi ∂j λα + g αj ∂i λα only vanishes at the initial time t = 0. For simplicity, let us denote by L the following 2-tensor 1 1 Lαβ = g βµ ∂α λµ + g αµ ∂β λµ , 2 2 hence we can rewrite (4.15) as Ricij = −Lij . Using this, we obtain 1 1 αβ Ric − g αβ Scalg = −(Lαβ − g αβ L), 2 2 EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 3 4 21 where L is the trace of Lαβ given by L = g ij Lij . Since the Einstein tensor is divergence free, we know that Lαβ − 12 g αβ L is also divergence-free; hence we obtain 1 ∇γ Lγµ − g γµ L = 0. 2 Keep in mind that ∇γ g = 0. A simple calculation shows that 1 0 =∇γ (g αγ g βµ Lαβ − g γµ g αβ Lαβ ) 2 1 1 αγ βµ = ∇γ g g (g βµ ∂α λµ + g αµ ∂β λµ ) − g γµ g αβ (g βµ ∂α λµ + g αµ ∂β λµ ) 2 2 1 αγ βµ 1 βµ αγ µ = g g g βµ ∇γ (∂α λ ) + g g g αµ ∇γ (∂β λµ ) 2 2 5 γ δµ 1 1 1 − g γµ g αβ g βµ ∇γ (∂α λµ ) − g γµ g αβ g αµ ∇γ (∂β λµ ) 4 4 α δµ 6 7 8 9 10 β δµ 1 1 1 = g αγ ∇γ (∂α λµ ) + g βµ ∇γ (∂β λγ ) − g γµ ∇γ (∂β λβ ). 2 2 2 Thanks to g = g γα ∇γ (∂α (·)), we have proved that λα solves the following homogeneous linear hyperbolic system µ gλ = −g βµ ∇γ (∂β λγ ) + g γµ ∇γ (∂β λβ ). The above long calculation also shows that 1 0 =Ricαβ − g αβ Scalg 2 1 = − Lαβ + g αβ L 2 1 1 = − g βµ ∂α λµ − g αµ ∂β λµ 2 2 1 1 1 pq g ∂p λµ + g pµ ∂q λµ + g αβ g 2 2 qµ 2 1 1 = − g βµ ∂α λµ − g αµ ∂β λµ 2 2 1 1 pq + g αβ g g qµ ∂p λµ + g αβ g pq g pµ ∂q λµ 4 4 p δµ 11 12 13 14 15 q δµ 1 1 1 = − g βµ ∂α λµ − g αµ ∂β λµ + g αβ ∂µ λµ . 2 2 2 Since we are in the vacuum case, at t = 0, we get 1 1 1 − g 0µ ∂α λµ − g αµ ∂0 λµ + g α0 ∂µ λµ = 0. 2 2 2 By using µ = α = 0, i.e. the Hamiltonian constraint (4.1), we obtain ∂0 λ0 = 0. We still need to prove that ∂0 λµ = 0 for all µ = 1, n. Indeed, thanks to ∂α λµ = 0 for any α, µ > 0, we know that 1 1 1 0 = − g 0µ ∂α λµ − g αµ ∂0 λµ + g α0 ∂µ λµ 2 2 2 0 16 n =− 1 g ∂0 λµ . 2 µ=1 αµ 0 22 1 2 ˆ Q.A. NGO Since this is true for all α = 1, n and the matrix (g αµ ) is invertible, we have that all ∂0 λα vanish at t = 0 for all α > 0. 5 Since λα satisfy a homogeneous linear hyperbolic system with vanishing initial data, λα vanish identically by the uniqueness for solutions of the Cauchy problem for the hyperbolic evolutions. This proves how the gauge condition λα = 0 is preserved in (small) time. 6 5. C ONFORMAL METHOD AND TRANSFORMED PDE S 7 In the literature, there are two routes that have been followed to solve the system of constraints. First, based on perturbation methods, people perform a connected sum between known solutions of the constraints in order to produce new solutions which obey some properties. Among others, we can list a few works such as [Cor00, CorSch06] by Corvino and Schoen, [ChrDel03] by Chru´sciel and Delay. 3 4 8 9 10 11 12 13 14 15 16 The other route introduced by Lichnerowicz, Choquet-Bruhat, and York is the so-called conformal method. In this section, we show how to use the conformal method to derive the recast constraint equations in the dimension n 2. It is interesting to know that the causal structure of spacetimes only depend on their Lorentzian metric up to a conformal factor; hence, when dealing with conformal metrics, it suffices to play with conformal factor. 18 First, we assume that the given initial data (M, g, K) fulfills the constraint equations (4.1) and (4.2). In this new notation, we rewrite (4.1) and (4.2) as the following 19 Scalg −|K|2g + (trg K)2 = 2ρ (5.1) divg K − d(trg K) = J, (5.2) 17 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 and where ρ = T (e0 , e0 ) and J = −T (·, e0 ). Note that in this context, we again denote τ = trg K. 5.1. Conformal method. As can be seen, the couple of constraint equations (5.1) and (5.2) for the initial data (M, g, K) consist of n + 1 equations (a scalar equation and an n-vector equation) with more than n + 1 unknowns (for example, the metric g consists of n(n+1)/2 components). Since the constraint equations form an under-determined system, they are in general hard to solve. Fortunately, we have a technique known as the conformal method that can be used in most of cases, see [CBruYor80]. Definition 5.1. Two metrics g and g on the manifold M of dimension n conformal if there exists some smooth function ϕ such that g = e2ϕ g. 2 are called In local coordinates, the condition g = e2ϕ g is nothing but gij = e2ϕ gij . Therefore, if we denote by (g ij ) and (g ij ) the inverse metric of (gij ) and (gij ) respectively, then we obtain g ij = e−2ϕ g ij . Note that elementary calculations show g ij gij = g ij gij = n. The basic idea of the conformal method is to equalize the number of equations and the number of unknowns in such a way that the resulting system is determined. More specific, the idea of the conformal method is to split the set of initial data (M, g, K) into the following two catalogue: • Conformal data: Degrees of freedom that can be freely chosen; and • Determined data: Degrees of freedom that are to be found by solving a determined system of partial differential equations EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 23 4 We now discuss the method more precise. But first, it is well-known that the scalar curvatures Scalg and Scalg of the conformal metrics g and g are related by the following rule Scalg = e−2ϕ Scalg − 2(n − 1)∆g ϕ − (n − 1)(n − 2)g ij ∂i ϕ∂j ϕ . (5.3) 5 5.2. Transformed PDEs in the case n 1 2 3 6 7 8 9 10 5.2.1. The transformed Hamiltonian constraint. When n 3, we set e2ϕ = u2p for some u > 0 and we choose p in such a way that the operator on u appearing within the brackets 4 2 , that is, g = u n−2 g is somewhat linear in u; this goal is attained by choosing p = n−2 which we suppose from now on. To see this, one immediately has u = eϕ/p which implies after a direct computation that ∆g u = 11 12 1 ij 1 g u∂i ϕ∂j ϕ + u∆g ϕ. 2 p p Thus, g ij ∂i ϕ∂j ϕ = u−1 p2 ∆g u − p∆g ϕ. 13 14 3. Hence, −2(n − 1)∆g ϕ − (n − 1)(n − 2)g ij ∂i ϕ∂j ϕ = −(n − 1)(2 − p(n − 2))∆g ϕ − (n − 1)(n − 2)u−1 p2 ∆g u 15 =− 16 provided p = 2 n−2 . 4(n − 1) −1 u ∆g u n−2 Thus, Eq. (5.3) becomes n+2 17 18 19 20 21 22 23 Scalg = u− n−2 uScalg − 4(n − 1) ∆g u . n−2 The Hamiltonian constraint (4.1) now becomes a semilinear elliptic equation for u given below n+2 4(n − 1) ∆g u − Scalg u + (|K|2g − τ 2 + 2ρ)u n−2 = 0. (5.4) n−2 As can be observed from (5.4), we are left with the term |K|g . To transform this term, we shall use the transverse-traceless decomposition. However, we first need to study the momentum constraint. 4 24 25 26 5.2.2. The transformed momentum constraint. We still use the fact that g = u n−2 g. If we denote by divg and divg the divergences in the metrics g and g respectively, we then have the following useful result, see [YCBru09, Lemma 3.1]. 4 27 28 29 30 31 32 Lemma 5.2. On a n-dimensional manifold, if g = u n−2 g, the covariant derivatives in g and g being respectively denoted ∇ and ∇, the divergences in the metrics g and g of an arbitrary contravariant 2-tensor P ij verify the following identity divg P ij = u− 2(n+2) n−2 divg (u 2(n+2) n−2 P ij ) − 2 ∂i u ij g trg P. n−2 u Proof. For the purpose of clarity we may denote the tensor P by P = P ij ∂ ∂ ⊗ . ∂xi ∂xj (5.5) ˆ Q.A. NGO 24 1 Using the Leibniz rule, one easily gets (∇P ) ·, ·, ∂ ∂xk 2 3 ∇P = ∇k P ij 10 11 12 13 14 15 16 17 18 19 20 21 ∂ ∂ ∂ ∂ ⊗ ⊗ dxk = P ij ;k i ⊗ ⊗ dxk , i j ∂x ∂x ∂x ∂xj ∇k P ij = P ij ;k = ∂k P ij + P lj Γilk + P il Γjlk . We take the divergence, that is, to use div P = δik P ij ,k 8 9 P ij verifies 6 7 ∂ ∂xk Therefore, (2, 1)-tensor ∇P , which is of the form 4 5 ∂ ∂ ⊗ i ∂x ∂xj ij ∂P ∂ ∂ ∂ ∂ = k ⊗ + P ij ∇ ∂k ⊗ i j i ∂x ∂x ∂x ∂x ∂x ∂xj ∂ ∂ + P ij i ⊗ ∇ ∂k ∂x ∂xj ∂x ∂P ij ∂ ∂ = + P lj Γilk + P il Γjlk ⊗ . k i ∂x ∂x ∂xj =∇ ∂ ∂ ∂ = δik ∇k P ij j = ∇i P ij j ∂xj ∂x ∂x to arrive at ∇i P ij = ∂i P ij + P lj Γili + P il Γjli . (5.6) 4 n−2 Notice that under the conformal change gij = u gij , the Christoffel symbols computed with respect to g and g verify the following identity 2 1 k δ ∂i u + δik ∂j u − g kl gij ∂l u . Γkij = Γkij + n−2u j Therefore, 2 1 i ∇i P ij =∂i P ij + P lj Γili + (δ ∂i u + δii ∂l u − g ik gli ∂k u) n−2u l 2 1 j (δ ∂i u + δij ∂l u − g jk gli ∂k u) + P il Γjli + n−2u l 2 1 lj P ∂l u + nP lj ∂l u − P lj ∂l u =∇i P ij + n−2u 2 1 lj + P ∂l u + P jl ∂l u − P il g jk gli ∂k u n−2u 2(n+2) 2(n+2) 2 ∂k u jk g (gli P il ). =u− n−2 ∇i u n−2 P ij − n−2 u The proof immediately follows thanks to trg P = gli P il . In view of (5.5), it is convenient to split the unknown K into a weighted traceless part and its trace with respect to the conformal metric g, namely, we write 2(n+2) τ K ij = u− n−2 K ij + g ij . n By lowering the indices in K and K, one gets Kij = u−2 Kij + τ gij . n It is clear to see that the tensor K is symmetric and traceless with respect to g, that is 4 2n τ trg K = g ij Kij = u− n−2 g ij Kij = u n−2 g ij Kij − gij = 0, n EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 25 3 where we have used trg K = τ and g ij gij = n. In view of the momentum constraint (5.2) and with the fact that 4 4 u− n−2 τ = u− n−2 trg K = trg K, 4 and that divg K = ∇i K i , we have by (5.5) 1 2 J j + g ij ∂i τ = ∇i K ij 2 ∂i u ij g trg K n−2 u 2(n+2) 1 2(n+2) 2 ∂i u ij g trg K = u− n−2 ∇i K ij + u n−2 g ij τ − n n−2 u 2(n+2) 2n 1 2(n+2) 2 ∂i u ij g τ = u− n−2 ∇i K ij + u− n−2 ∇i u n−2 g ij τ − n n−2 u 2(n+2) 2(n+2) 2n 2 ∂i u ij 1 = u− n−2 ∇i K ij + g ij u− n−2 ∇i u n−2 τ − g τ n n−2 u = u− 5 2(n+2) n−2 ∇i (u 2(n+2) n−2 K ij ) − − 2n n−2 g ij u = u− 6 2(n+2) n−2 ∇i K ij + 1 ij g ∇g τ. n Thus, we have transformed J j + g ij ∂i τ = ∇i K ij to the following equation 2(n+2) n−2 Jj + n − 1 2(n+2) u n−2 g ij ∂i τ. n 7 ∇i K ij = u 8 Equivalently, thanks to gij = u n−2 gij , we further obtain 4 9 10 11 12 13 14 2(n+2) 2n n − 1 n−2 u g ij ∂i τ + u n−2 J j . n In other words, the momentum constraint (5.2) now becomes ∇i K ij = divg K = 2(n+2) 2n n − 1 n−2 u ∇g τ + u n−2 J. n (5.7) In particular, if J is zero and τ is constant, the symmetric 2-tensor K is also divergence free. Symmetric 2-tensors which are divergence free and trace free are called TT-tensors (tranverse, traceless). Let us go back to the decomposition of K. Clearly, |K|2g = gih gjk K ij K hk 4n 15 = u− n−2 gih gjk K ij K hk + 4n = u− n−2 |K|2g + 16 17 τ2 n τ2 . n Therefore, Eq. (5.4) now reads as follows −3n+2 4(n − 1) ∆g u − Scalg u + |K|2g u n−2 − n−2 n+2 n−1 2 τ − 2ρ u n−2 = 0. n (5.8) 18 Clearly, this is a semilinear elliptic equation for u. 19 22 5.2.3. Scaling of the scalar fields. Recall that M is a n-dimensional manifold with spatial metric g. We denote by a tilde the values induced on M by the spacetime quantities. For the scalar field ψ, there is no need to do any time and space decomposition. However, the wave equation that ψ fulfills, that is 23 ∇α ∇α ψ = U (ψ), 20 21 ˆ Q.A. NGO 26 1 2 3 4 5 suggests that the initial data for the scalar field ψ should be the induced function and normalized time derivative of the function on M . Based on this reason and for the sake of simplicity, we denote ψ=ψ M and we use π to denote the normalized time derivative of ψ restricted to M , that is, π= 6 7 8 9 1 N ∂ ∂ ψ − βi i ψ ∂t ∂x where ∂0 ψ is the value of ∂0 ψ on M . Due to the fact that the background metric g is unphysical, we associate it to an unphysical lapse N so that N and N have the same associated densitized lapse, that is N 10 11 12 13 14 = N −1 ∂0 ψ. det g =√ N . det g (5.9) (Here by densitized lapse we mean the lapse function divided by the square root of the 4 determinant of the spatial metric 4.) Thanks to the conformal change g = u n−2 g, this condition is equivalent to 2n N = u n−2 N . 16 In this setting, we now denote π = N −1 ∂0 ψ for the initial data π. Therefore, π and π are related by the following scaling 17 π = N −1 ∂0 ψ = u− n−2 π. 15 2n 19 Next, in order to obtain precise formulas for ρ and J, we recall that ρ = T (n, n) and J = −T (·, n); hence it suffices to obtain the spacetime metric g. Clearly, 20 g 00 = g(∂0 , ∂0 ) = N 2 g(n, n) = −N 2 18 21 22 23 24 25 26 and g i0 = g 0i = g(∂0 , ∂i ) = g(N n, ∂i ) = 0 since n and ∂i are perpendicular. Hence, we can write 2 g 00 g 01 · · · g 0n −N .. 0 . g 10 g 11 · · · = g = (g αβ ) = . . . . .. .. .. .. .. . 0 g n0 . . . . . . g nn 0 ··· g11 .. . ··· .. . gn1 ··· An elementary calculation shows that the inverse metric g αβ is − N12 0 ··· 0 11 1n 0 g · · · g (g αβ ) = . . .. .. .. . . . . . 0 g n1 · · · g nn 4It is interesting to note that as a corollary of the Cramer rule, there holds N | det g| = | det g|. 4 It is a good exercise to compare det g and det g when g = u n−2 g. 0 g1n . .. . gnn EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 3 27 Hence, recalling the following formula for the stress-energy tensor in (2.6) 1 Tαβ = ∂α ψ∂β ψ − g αβ ∂µ ψ∂ µ ψ − g αβ U (ψ) 2 we easily obtain T (n, n) =N −2 T (∂0 , ∂0 ) =N −2 T00 4 5 1 =N −2 (∂0 ψ∂0 ψ + N 2 ∂µ ψ∂ µ ψ − N 2 U (ψ)) 2 1 −2 =N ∂0 ψ∂0 ψ + ∂µ ψ∂ µ ψ − U (ψ). 2 In the same way, we also obtain T (∂i , n) =N −1 T (∂i , ∂0 ) 1 =N −1 (∂i ψ∂0 ψ − g i0 ∂µ ψ∂ µ ψ − g i0 U (ψ)) 2 =N −1 ∂i ψ∂0 ψ. 6 7 8 9 10 11 12 Thus, the energy density on M of a scalar field ψ with potential U (·) is 1 ρ = (|π|2 + g ij ∂i ψ∂j ψ) + U (ψ) 2 where ψ and ∂i ψ are the values of ψ and ∂i ψ on M respectively. In terms of the initial data set, ρ becomes 4n 4 1 ρ = (u− n−2 |π|2 + u− n−2 |∇ψ|2g ) + U (ψ). 2 Hence, we can regroup (5.8) as follows 4(n − 1) ∆g u − (Scalg − |∇ψ|2g )u n−2 13 3n−2 + (|K|2g + |π|2 )u− n−2 − 14 17 18 19 20 21 22 23 24 25 26 27 28 (5.10) Now for the momentum density, from the above computation, one can see that J i = −N −1 g ij ∂j ψ ∂0 ψ = −u− 15 16 n+2 n−1 2 τ − 2U (ψ) u n−2 = 0. n 2(n+2) n−2 g ij ∂j ψπ. That is equivalent to J = −u− 2(n+2) n−2 π∇g ψ. Using this formula for J, Eq. (5.7) becomes 2n n − 1 n−2 divg K = u ∇g τ − π∇g ψ. n (5.11) 5.2.4. Conformal Killing operator and conformal vector Laplacian. As can be observed from Eqs. (5.10) and (5.11), we are left with studying terms involving K. This is possible by using the transverse-traceless decomposition in the next subsection. As a first step to study such a decomposition, we mention the so-called conformal Killing operator and its associated conformal vector Laplacian. First, we start with its definition. Roughly speaking, the conformal killing operator is a generalization of the Killing operator relative to the metric g. It maps any vector field on M to some tensor of type (2, 0). More precisely, in components, we have 2 (Lg W )ij = ∇i W j + ∇j W i − ∇k W k g ij , (5.12) n ˆ Q.A. NGO 28 2 where W = W i ∂i is a vector field on M . Immediately, one can check that Lg W is traceless as can be seen in the following 3 gij (Lg W )ij = gij ∇i W j + gij ∇j W i − 2∇k W k = 0. 1 5 We now define the so-called Conformal Vector Laplacian ∆g,L associated to the metric g. Formally, we define ∆g,L = divg Lg · which, in local coordinates, basically says that 6 (∆g,L W )i = ∇j (Lg W )ij . 4 7 In components, we have (∆g,L W )i =∇j ∇i W j + ∇j ∇j W i − =∇i ∇j W j + Ricij W j + ∇j ∇j W i − 8 = 9 12 2 i ∇ ∇k W k n n−2 i ∇ ∇j W j + Ricij W j + ∇j ∇j W i . n Let us now determine the kernel of ∆g,L . By definition, one can easily see that ker Lg ⊂ ker ∆g,L . 10 11 2 i ∇ ∇k W k n However, we shall prove that they are actually the same. For any vector field W on M , we first have j Wj (∆g,L W ) = Wj ∇l (Lg W ) M jl M = 13 ∇l Wj (Lg W ) jl jl − (Lg W ) ∇l Wj M jl =− (Lg W ) ∇l Wj , M 14 15 where the Gauss–Ostrogradsky theorem has been used to get the last line. In view of the right-hand side integrand, we see that gij gkl (Lg W )ik (Lg W )jl =gij gkl [∇i W k + ∇k W i ](Lg W )jl − 2 ∇m W m g ik gij gkl (Lg W )jl n gij δli 16 = gkl ∇j W k + gij ∇l W i (Lg W )jl − 2 jl ∇m W m gjl (Lg W ) n 0 =2gij ∇l W i (Lg W )jl , 17 18 where we have used the symmetry and the traceless property of (Lg W )ij . Therefore, we can write j 1 2 1 =− 2 ik Wj (∆g,conf W ) = − M 19 20 21 22 jl gij gkl (Lg W ) (Lg W ) M Lg W M 2 g. Therefore, assume that W ∈ ker ∆g,L ; hence (∆g,L W )i = 0, we then see from the preceding equality that Lg W = 0 since the metric g is positive definite. Hence, W ∈ ker Lg as claimed. EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 29 For future benefit, we suppose that M has boundary ∂M . By integration by parts, we have the following j jl Uj (∆g,L W ) = M Uj ∇l (Lg W ) M jl = 3 ∇l uj (Lg W ) jl − (Lg W ) ∇l Uj M jl (Lg W )jl νj Ul − = (Lg W ) ∇l Uj , ∂M 4 5 j M 8 9 10 11 12 13 14 1 2 (Lg W )jl νj Ul − Uj (∆g,L W ) = 6 7 M for any vector fields U = U i ∂i and W and ν is the outward normal vector field on ∂M . As above, we can write ∂M (Lg W )jl (Lg U )jl . (5.13) M Let now discuss the existence and uniqueness of solutions to the conformal vector Poisson equation (∆g,L W )i = S i (5.14) where the vector field S is already given. We shall provide a necessary condition for solution of (5.14) to exist as shown below. Proposition 5.3. Assume that (M, g) is compact without boundary. Then a necessary condition for solution of (5.14) to exist is that the source S i must be orthogonal to any vector field in the kernel ker ∆g,L , in the sense that Cj S j = 0 ∀C ∈ ker ∆g,L . 15 (5.15) M 16 17 Proof. This is clear as the following. First, for any C ∈ ker ∆g,L , we know from (5.13) that j Cj S j = M Cj (∆g,L W ) M 18 =− 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Because (Lg C) ik 1 2 ik jl gij gkl (Lg C) (Lg W ) . M = 0, we obtain the desired identity. Clearly if (M, g) admits no conformal Killing vector, the identity is trivial. To see why this condition also gives us a sufficient condition, we can use the Fredholm theory. 5.2.5. The transverse-traceless decomposition. We consider in this subsection the solvability of (5.11) using the transverse-traceless decomposition. Roughly speaking, we search for K of the form K = KT T + Lg W (5.16) where KT T is a TT-tensor, say the TT-part of K, W is an unknown vector field to be determined, and Lg is the conformal Killing operator relative to g defined by (5.12). If the right hand side of (5.12) vanishes, the vector field W is called a conformal Killing vector. By definition, any tensor of the form Lg Y for some vector field Y has trace free. The procedure of solving (5.11) is to find W and TT-part of K. This so-called TT-part is not unique in general and we have many ways to extract such a piece of information from K. ˆ Q.A. NGO 30 1 We first deal with W . In accordance with (5.16), we first have (KT T )ij = K ij − (Lg W )ij . 2 3 4 5 6 7 8 9 10 11 12 13 14 (5.17) The choice of the conformal Killing operator and the fact that K is tracefree make the right hand side of (5.17) trace free. Besides, the transversality requirement ∇i (KT T )ij = 0 and (5.11) lead to covariant equations for W given by 2n n − 1 n−2 ∇i (Lg W )ij = u g ij ∂i τ − πg ij ∂i ψ. n Using the convention ∆g,L = divg ◦Lg , we can rewrite the equation for W in the vector form as the following 2n n − 1 n−2 u ∇g τ − π∇g ψ. (5.18) ∆g,L W = n It is well-known that the operator ∆g,L which is similar to the vector Laplacian is a second order, self-adjoint, linear, elliptic operator whose kernel consists of the space of conformal Killing vector fields, see [YCBru09, Appendix II]. Thus under some mild conditions, we can solve (5.18) for W up to conformal Killing vector fields. Notice that, any conformal Killing vector field does not constitute any extra information to KT T in (5.17). 17 We now consider the TT-tensor KT T . The search of such a tensor is somewhat freely. Its procedure can be formulated as follows. We start with a freely chosen traceless 2-tensor Z, then we solve for Y from the following equation 18 ∆g,L Y = − divg Z. 15 16 19 20 21 The existence of some Y from (5.19) comes from Proposition 5.3 and the fact that divg Z is orthogonal to the space of conformal Killing vector fields whose proof is just a simple application of integration by parts as follows (∇i Z ij )Hj dvolg = − 22 M 23 26 27 Z ij (Lg H)ij dvolg = 0 M where H is a conformal Killing vector field. Since ∇j (Lg Y + Z)ij = (∆g,L Y )i + ∇j Z ij = 0 24 25 (5.19) we know that the traceless tensor Lg Y + Z is also transverse, thus, a TT-tensor. Let us denote σ = Lg Y + Z. (5.20) 30 In conclusion, we first begin with a freely chosen Z and solve (5.19) for Y . We then solve (5.18) to find W . Finally, we find K = σ + Lg W by means of the decomposition (5.16). Such a K will satisfy (5.11). 31 5.2.6. Summary. In conclusion, our transformed equations are the following: 28 29 − 32 33 34 35 36 37 38 4(n − 1) ∆g u + Scalg −|∇ψ|2g u n−2 n+2 3n−2 n−1 2 =− τ − 2U (ψ) u n−2 + (|σ + Lg W |2g + π 2 )u− n−2 n and (5.21) 2n n − 1 n−2 u ∇g τ − π∇g ψ (5.22) n where τ = trg K. Clearly, the system (5.21)-(5.22) is coupling in the sense that we cannot solve each equation separately, not even in the vacuum case meaning that there is no π, ψ, and U . One way to decouple the system is to assume that the mean curvature τ is constant everywhere in M (hence we call CMC case for short). ∆g,L W = EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 3 4 5 6 7 8 9 10 31 5.3. Transformed PDEs in the case n = 2. In the previous section, we have shown how the conformal method may be used to derive the constraint equations with scalar fields when the dimension n 3. In this section, we continue to use the conformal method to construct the constraint equations with scalar fields in the dimension n = 2. To the best of our knowledge, the first paper dealing with the Einstein equations in the two dimensional cases is [Mon86]. Subsequently, some generalization of the corresponding situation for the Einstein–Maxwell and Einstein–Maxwell–Higgs equations were also obtained by the author in [Mon90]. Again, no result is known for the case of scalar fields. 5.3.1. The momentum constraint equation. In order to fix notations, we keep using g = e2ϕ g. We first prove the following simple result which was motivated by (5.5). 13 Lemma 5.4. On a 2-dimensional manifold, if g = e2ϕ g, the covariant derivatives in g and g being respectively denoted ∇ and ∇, the divergences in the metrics g and g of an arbitrary contravariant 2-tensor P ij verify the following identity 14 divg P ij = e−4ϕ divg (e4ϕ P ij ) − g ij ∂i ϕtrg P. 11 12 15 16 (5.23) Proof. The proof is almost similar to the proof of Lemma 5.2. Indeed, if we denote the tensor P by P = P ij ∂i ⊗ ∂j then we first recall from (5.6) the following rule ∇i P ij = ∂i P ij + P lj Γili + P il Γjli . 17 19 Notice that under the conformal change g = e2ϕ g, the Christoffel symbols computed with respect to g and g now verify the following identity 20 Γkij = Γkij + (δik ∂j ϕ + δjk ∂i ϕ − gij g kl ∂l ϕ). 18 Therefore, thanks to n = 2 ∇i P ij =∂i P ij + P lj Γili + P il Γjli + P lj (δli ∂i ϕ + δii ∂l ϕ − gli g im ∂m ϕ) + P il (δlj ∂i ϕ + δij ∂l ϕ − gli g jm ∂m ϕ) =∇i P ij + (P ij ∂i ϕ + 2P lj ∂l ϕ − P mj ∂m ϕ) + (P ij ∂i ϕ + P jl ∂l ϕ − P il gli g jm ∂m ϕ) =∇i P ij + 4P ij ∂i ϕ − g jm ∂m ϕ trg P =e−4ϕ ∇i (e4ϕ P ij ) − g jm ∂m ϕ trg P. 21 22 23 24 The proof immediately follows. Let us consider the momentum constraint. Using Lemma 5.23, for a 2-tensor K, we decompose it as follows K ij = e−4ϕ K ij + τ ij g . 2 ˆ Q.A. NGO 32 1 2 This decomposition obeys the same properties as n traceless. Therefore, by (5.23), one has the following 3, that is, K is symmetric and J j + g ij ∂i τ = ∇i K ij = e−4ϕ ∇i (e4ϕ K ij ) − g ij ∂i ϕ trg K 1 = e−4ϕ ∇i K ij + e4ϕ g ij τ − g ij ∂i ϕ trg K 2 1 = e−4ϕ ∇i K ij + e−4ϕ ∇i e2ϕ g ij τ − (e−2ϕ g ij )∂i ϕ(e2ϕ trg K ) 2 3 τ −4ϕ =e 1 ∇i K + e−4ϕ g ij ∇i (e2ϕ τ ) − g ij ∂i ϕτ 2 ij e−2ϕ g ij 1 = e−4ϕ ∇i K ij + e−2ϕ g ij ∇i τ. 2 4 5 6 7 In other words, 1 ∇i K ij = e4ϕ J j + e4ϕ g ij ∂i τ. 2 Thus, the momentum constraint equation that we need is 1 divg K = e2ϕ ∇g τ + e4ϕ J. 2 (5.24) 9 5.3.2. The Hamiltonian constraint equation. Let us now consider the Hamiltonian constraint. One first obtains by (5.3) 10 Scalg = e−2ϕ (Scalg − 2∆g ϕ). 8 11 Consequently, the Hamiltonian constraint temporarily reads as the following −2∆g ϕ + Scalg −(|K|2g − τ 2 + 2ρ)e2ϕ = 0. 12 13 Obviously, 1 |K|2g = gih gjk K ij K hk = e−4ϕ |K|2g + τ 2 2 14 15 which implies that 1 − 2∆g ϕ + Scalg − e−4ϕ |K|2g − τ 2 + 2ρ e2ϕ = 0. 2 16 (5.25) 18 5.3.3. Scaling of the scalar fields. Using (5.9) one easily gets N = e2ϕ N which immediately implies 19 π = N −1 ∂0 ψ = e−2ϕ π. 17 20 21 22 Therefore, we can decompose the energy density on M as follows 1 ρ = (e−4ϕ |π|2 + e−2ϕ |∇ψ|2g ) + U (ψ). 2 Now for the momentum density, one can see that J i = −N −1 g ij ∂j ψ ∂0 ψ = −e−4ϕ g ij ∂j ψπ. 23 24 25 26 27 That is J = −e−4ϕ π∇g ψ. Using the formula for J, we can rewrite (5.24) as 1 divg K = e2ϕ ∇g τ − π∇g ψ. 2 (5.26) EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 3 4 5 6 7 8 We now regroup (5.25) as 1 2 τ − 2U (ψ) e2ϕ + (|π|2 + |K|2g )e−2ϕ . (5.27) 2 At this stage, the constraint equations in two dimensions become a system of partial differential equations (5.27)–(5.26). Notice that the coefficient of e−2ϕ is always non-negative in this setting. In the vacuum case, that is U (ψ) ≡ 0, the coefficient of e2ϕ is always negative and one can use the method of sub- and super-solutions to prove the existence result, see [Mon86] for details. −2∆g ϕ + (Scalg − |∇ψ|2g ) = − 6. S OLVING THE TRANSFORMED PDE S IN THE CASE n 11 12 3: M ATHEMATICAL SETTINGS 9 10 33 Let us recall some information that we have already seen from previous sections. By using the conformal method we are able to transform the Einstein-scalar field constraint equations (4.1) and (4.2) into a couple of two equations given in (5.21)-(5.22). By denoting n−2 (Scalg −|∇ψ|2g ), 4(n − 1) n−2 n−1 2 = τ − 2U (ψ) , 4(n − 1) n n−2 = (|σ + Lg W |2g + π 2 ), 4(n − 1) Rg,ψ = Bτ,ψ 13 Ag,W,π 14 one can see that (5.21) simply becomes n+2 15 (6.1) 3n−2 − ∆g u + Rg,ψ u = −Bτ,ψ u n−2 + Ag,W,π u− n−2 . (6.2) 17 In order to make the study more general, we can consider a more general form of (6.2) which can be written as the following 18 − ∆g u + hu = f u n−2 + au− n−2 , 16 n+2 19 20 3n−2 (6.3) where h, f , and a are smooth functions. In this note, we also call (6.3) the Einstein-scalar field Lichnerowicz equation. 25 The main objective of the study is to determine which choices of the conformal data (g, σ, τ, ψ, π) permit one to solve Eqs. (5.21)-(5.22) for the determined data (u, W ) and which do not. Clearly, Eq. (5.21) is a semilinear elliptic equation, called the Einsteinscalar field Lichnerowicz equation, for u provided σ and W are already known. However, be careful because Eq. (5.22) is a vector equation. 26 6.1. Preliminaries. 27 6.1.1. Sobolev spaces and related inequalities. Given a smooth compact Riemannian manifold (M, g) of dimension n, one easily defines the Sobolev spaces Hpk (M ) for any positive integers k and p. To be precise, we define Hpk (M ) as the completion of C ∞ (M ) with respect to the following norm 21 22 23 24 28 29 30 k 31 u Hpk ∇j u = j=1 32 When p = 2, we simply write H2k (M ) as H k (M ). Lp . ˆ Q.A. NGO 34 1 2 3 By K1 and A1 we mean the positive constants such that the Sobolev inequality holds, that is, for all u ∈ H 1 (M ), 2 2 |∇u| dvolg + A1 K1 M |u| u dvolg M 2n n−2 n−2 n dvolg . (6.4) M 4 We notice that those constants K1 and A1 are independent of u. 5 As one may observe from (6.3) that the operator appearing in the left hand side of (6.3), that is −∆ + h, admits some interesting features when we impose some conditions on the potential h. A typical assumption that people usually make is to assume that −∆ + h is coercive. Roughly speaking, this is equivalent to saying that 6 7 8 M inf 1 9 u∈H (M ) 10 In particular, one may see that u 11 12 13 14 15 19 20 21 M M Another useful inequality appearing in this setting is the following. For all u ∈ H 1 (M ), there holds 2 2 |∇u| dvolg + hu dvolg |u| Sh M 2n n−2 n−2 n dvolg , (6.5) M where the constant Sh is called the Sobolev constant and is independent of u. 6.1.2. The Yamabe-scalar field conformal invariant. Let us assume that (M, g) is a compact Riemannian manifold without boundary. We recall from the study of the Yamabe problem [Aubin98, Tru68, Yam60] on (M, g) the conformal Laplacian operator Lg acting on a smooth function u is defined by Lg u = ∆g u − 22 23 1 2 is an equivalent norm in H (M ). It is standard to check that if h > 0 everywhere then −∆ + h is coercive. M 18 1 Hh hu2 dvolg |∇u|2 dvolg + = 1 16 17 |∇u|2 dvolg + M hu2 dvolg > 0. u2 dvolg M n−2 Scalg u. 4(n − 1) (6.6) Operator Lg has the conformal covariance property n+2 Lg u = θ− n−2 Lg (θu) 24 (6.7) 4 25 26 27 28 29 30 31 32 33 for any g = θ n−2 g for some θ > 0 being a smooth function. Inspired by (6.6), the authors in [CBruIsePol07] introduced the so-called conformal scalar field Laplacian operator Lg,ψ given by n−2 Lg,ψ u = ∆g u − (Scalg −|∇ψ|2g )u. (6.8) 4(n − 1) It follows from (6.6) and (6.8) that Lg,ψ u = Lg u + n−2 |∇ψ|2g u. 4(n − 1) We wish that Lg,ψ also has the conformal covariance property. For this reason, we first have 4 4 |∇ψ|2g = g ij ∂i ψ∂j ψ = θ− n−2 g ij ∂i ψ∂j ψ = θ− n−2 |∇ψ|2g . EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 35 This and (6.7) immediately give n−2 |∇ψ|2g u 4(n − 1) n+2 4 n − 2 − n−2 = θ− n−2 Lg (θu) + θ |∇ψ|2g u 4(n − 1) n+2 n+2 n − 2 − n−2 θ |∇ψ|2g (θu). = θ− n−2 Lg (θu) + 4(n − 1) Lg,ψ u = Lg u + 2 5 In other words, operator Lg,ψ also verifies the same conformal covariance property (6.7) in the sense that n+2 Lg,ψ u = θ− n−2 Lg,ψ (θu). (6.9) 6 We now define the so-called conformal-scalar field Dirichlet energy of u by 3 4 4(n − 1) n−2 4(n − 1) =− n−2 Eg,ψ (u) = − 7 8 M Qg,ψ (u) = n−2 (Scalg −|∇ψ|2g )u2 dvolg 4(n − 1) Eg,ψ (u) u 2 . 2n L n−2 Using (6.9) one has Qg,ψ (u) = Qg,ψ (θu) 11 12 |∇ψ|2g + and the conformal-scalar field Sobolev quotient by 9 10 uLg,ψ u dvolg M where g = θ 4 n−2 (6.10) g. We denote by [g] the conformal class of the metric g given by 4 13 14 [g] = {g = θ n−2 g, θ ∈ C ∞ (M ), θ > 0}. Then we define the so-called Yamabe-scalar field conformal invariant by Yψ ([g]) = 15 16 17 18 inf u∈H 1 (M ) Qg,ψ (u). By (6.10), it is obvious that Yψ ([g]) is independent of the choice of background metric g in the conformal class used to define it, and is therefore an invariant of the conformal class. We observer from the H¨older inequality and the compactness of M that (Scalg −|∇ψ|2g )u2 dvolg 19 C u M 20 21 22 23 24 25 26 27 28 29 30 31 (6.11) 2 2n L n−2 for some positive constant C independent of u. Thus, Yψ ([g]) is finite. Consequently, by using the sign of Yψ ([g]), we may partition the set of pairs ([g], ψ) into three classes which we label Y − , Y 0 , Y + , and refer to as the negative, zero, and positive Yamabe-scalar field conformal invariants on M . The following important result was proved in [CBruIsePol07]. Proposition 6.1. The following conditions are equivalent (i) Yψ ([g]) > 0 (respectively = 0, < 0); (ii) There exists a metric g ∈ [g] which satisfies Scalg −|∇ψ|2g > 0 everywhere on M (respectively = 0, < 0); (iii) For any metric g ∈ [g], the first eigenvalue λ1 of the self-adjoint elliptic operator −Lg,ψ is positive (respectively zero, negative). ˆ Q.A. NGO 36 1 2 3 For any constant c > 0 and any metric g ∈ [g], let us consider the following metric 4 4 4 g = c n−2 g. In terms of the metric g, one may write g = (cu) n−2 g provided g = u n−2 g. Then a direct computation shows that 4 Scalg −|∇ψ|2gˆ = Scalg −c− n−2 |∇ψ|2g 4 4 = c− n−2 Scalg − n+2 4 4(n − 1) − n−2 ∆g (c) − c− n−2 |∇ψ|2g c n−2 4 = c− n−2 Scalg −|∇ψ|2g . 5 Therefore, we can extend the preceding proposition as follows. 6 Proposition 6.2. The following conditions are equivalent 7 8 9 10 11 12 13 14 15 16 (i) Yψ ([g]) > 0 (respectively = 0, < 0); (ii’) There exists a metric g ∈ [g] which satisfies Scalcg −|∇ψ|2cg > 0 everywhere on M (respectively = 0, < 0) and for any constant c > 0; (iii) For any metric g ∈ [g], the first eigenvalue λ1 of the self-adjoint elliptic operator −Lg,ψ is positive (respectively zero, negative). The advantage of Proposition 6.2 is that it allows us to assume that the manifold M has unit volume by choosing a suitable constant c > 0. One of the most important property of Eq. (6.2) is that they are conformally covariant in the following sense, see [CBruIsePol07]. 19 Proposition 6.3. Let D = (g, σ, τ, ψ, π) be a conformal initial data set for the Einstein4 scalar field constraint equations on M . If g = θ n−2 g for a smooth positive function θ, then we define the corresponding conformally transformed initial data set by 20 D = (g, σ, τ , ψ, π) = (θ n−2 g, θ−2 σ, τ, θ− n−2 ψ, π). 17 18 4 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 2n Let W be the solution to the conformal form of the momentum constrain equation with respect to the conformal initial data set D (for which we assume that a solution exists), and let W be the solution of the momentum constrain equation with respect to the conformally transformed initial data set D (which will exist if W does). Then u is a solution to Eq. (6.2) for the conformal data D with W if and only if θ−1 u is a solution to Eq. (6.2) for the transformed conformal data D with W . Using Proposition 6.3 above, it turns out that the sign of Yψ ([g]) plays an important role in the study because we can first perform a conformal transformation on the conformal 4 2n initial data from (g, σ, τ, ψ, π) to (θ n−2 g, θ−2 σ, τ, θ− n−2 ψ, π) in such a way that Rg,ψ has a fixed sign by means of Proposition 6.1. Therefore, it suffices to study the solvability 4 2n of Eq. (6.2) for the transformed data (θ n−2 g, θ−2 σ, τ, θ− n−2 ψ, π) rather than to study to solvability of Eq. (6.2) for the original data (g, σ, τ, ψ, π). 6.2. A classification of Choquet-Bruhat–Isenberg–Pollack. While, as we have noted, the conformal method can be effectively applied for solving the constraint equations with scalar fields in most cases, it should be pointed out that there are several cases for which either partial result or no result was achieved, see [CBruIsePol07] for details. In order to talk about the objectives of the current study, let us go back to [CBruIsePol07]. A typical result contained in [CBruIsePol07] is a fairly complete picture of sets of CMC conformal data which lead to solutions of the constraint equations with scalar fields and which do not. This picture can be summarized in the following two tables where EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 37 TABLE 1. Results for the case Aγ,W,π ≡ 0. Bτ,ψ other B0 Yψ < 0 NR N N N PR Y Yψ = 0 NR N N Y N N Yψ > 0 PR PR PR N N N TABLE 2. Results for the case Aγ,W,π ≡ 0. • • • • 1 2 3 4 5 6 7 8 9 10 11 12 15 16 other B0 Yψ < 0 NR N N N PR Y Yψ = 0 NR N N N Y Y Yψ > 0 PR PR PR Y Y Y ‘Y’ indicates that (6.2) can be solved for that class of conformal data; ‘N’ indicates that (6.2) has no positive solution; ‘PR’ indicates that we have partial results; and ‘NR’ indicates that for this class of initial data we have no results indicating existence or non-existence. 6.2.1. Solving the momentum constraints. As we have already known that the operator ∆g,L is a second order, self-adjoint, linear, elliptic operator whose kernel consists of the space of conformal Killing vector fields, see [YCBru09, Appendix II]. It follows from the Fredholm alternative that for a given set of functions (u, τ, ψ, π) we may solve the momentum constraint 2n n − 1 n−2 ∆g,L W = u ∇g τ − π∇g ψ n if either • (M, g) admits no conformal Killing vector fields, and thus, W is unique; 13 14 Bτ,ψ or • 2n n−1 n−2 ∇g τ n u − π∇g ψ is orthogonal in the L2 sense to the space of conformal Killing vector fields, see (5.15). 2n 17 18 19 20 21 22 23 24 25 n−2 ∇ τ ≡ 0) and (M, g) admits conformal Killing vecIn the CMC case (hence n−1 g n u tor fields, it suffices to require that π∇g ψ is orthogonal to the space of conformal Killing vector fields as a consequence of (5.15). Notice that, under the CMC assumption, the momentum constraint equations consist of only scalar field ψ. Therefore, we can solely solve it to obtain W . Having the existence of W and the fact that our system of constraint equations are decoupled, we may solely solve the conformal form of the Hamiltonian constraint for u. 6.2.2. Solving the Hamiltonian constraints. Unlike the Einstein equations in the vacuum case where we know exactly which sets of CMC conformal data permit the corresponding ˆ Q.A. NGO 38 1 2 3 Lichnerowicz equation to be solved and which do not by the seminal work by Isenberg [Isen95], the analysis for Eq. (6.2) is more complicated, primarily because there are more relevant possibilities for the signs of the coefficients in (5.21). Let us recall from (6.2) that the corresponding Lichnerowicz equation is simply given 4 5 by n+2 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 3n−2 −∆g u + Rg,ψ u = −Bτ,ψ u n−2 + Ag,W,π u− n−2 where coefficients Rg,ψ , Bτ,ψ , and Ag,W,π are given in (6.1). In [CBruIsePol07], their classification only depends on the sign of Rg,ψ and Bτ,ψ since Ag,W,π 0. As for Bτ,ψ , there are six different possibilities, namely, this coefficient can be strictly positive, greater than or equal to zero, identically zero, less than or equal to zero, strictly negative, or of changing sign. For Rg,ψ , in view of Proposition 6.1, under a suitable conformal change, we can fix its sign, thus, there are three possibilities, namely, this could be negative, identically zero, or positive. These classification of sign combined with the two options Ag,W,π ≡ 0 and Ag,W,π ≡ 0 gives us a total of 36 classes of data, see Tables 1 and 2. Based on this division, the authors in [CBruIsePol07] proved for almost all cases, we do know which sets of data permit Eq. (6.2) to be solved and which do not. For a detailed statement of this result, we prefer to [CBruIsePol07, Theorems 1 and 2]. 6.3. The Lichnerowicz equations with Rg,ψ being constant. We now consider Eq. (6.3) as a slightly more general version of Eq. (6.2). By Proposition 6.1, we know that, after a suitable conformal transformation, the function h is a smooth function having a fixed sign on M . Again, by Proposition 6.1, the function h vanishes if we are in the null Yamabescalar field conformal invariant, that is equivalent to saying that h is constant (which is equal to zero) in this case. In this subsection, we show that in fact if we are in the negative Yamabe-scalar field conformal invariant, namely Yψ ([g]) < 0, we still can perform another conformal trans2n 4 formation on the conformal data from (g, σ, τ, ψ, π) to (u n−2 g, u−2 σ, τ, u− n−2 ψ, π) in such a way that h is a negative constant. Thanks to Proposition 6.3, we may freely assume that h is a negative constant on M . Proposition 6.4. There exists a smooth function u > 0 such that under the transformed 4 2n data (u n−2 g, u−2 σ, τ, u− n−2 ψ, π) obtained from data (g, σ, τ, ψ, π) through the confor4 mal change g = u n−2 g, coefficient Rg,ψ is a negative constant. Proof. First, in view of Proposition 6.1 we may assume that function Rg,ψ < 0 in the 4 original data, that is, Scalg −|∇ψ|2g < 0. Notice that if g = u n−2 g, coefficient Rg,ψ verifies the following rule 4 n−2 Rg,ψ = (Scalg −u− n−2 |∇ψ|2g ) 4(n − 1) n+2 4 4 n−2 4(n − 1) − n−2 (6.12) = u− n−2 Scalg − u ∆g u − u− n−2 |∇ψ|2g 4(n − 1) n−2 n+2 = u− n−2 (−∆g u + Rg,ψ u), n+2 36 which yields −∆g ϕ + Rg,ψ ϕ = Rg,ψ ϕ n−2 . Therefore, in terms of our notation, it suffices 37 to prove that the following equation with h and h being negative and h is constant n+2 38 39 40 − ∆g u + hu = hu n−2 (6.13) always admits a smooth positive solution u. We use the method of sub- and super-solutions to seal this issue. EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 39 TABLE 3. Results for the vacuum case in the case of constant mean curvature. 1 2 3 4 σ ≡ 0, τ = 0 σ ≡ 0, τ = 0 σ ≡ 0, τ = 0 σ ≡ 0, τ = 0 Y0 N Y N Y Existence of a sub-solution. This is obvious since a sufficiently small, positive constant u 4 will serve. To see this, one can choose any u satisfying the following u n−2 supM h /h. With this, we immediately have −∆g u + hu of (6.13). n+2 hu n−2 . By definition, u is a sub-solution 7 Existence of a super-solution. We also show that a sufficiently large positive constant u will serve. Indeed, similarly to the argument above, one can show that any positive 4 constant u satisfying u n−2 inf M h /h is a super-solution to (6.13) in the sense that 8 −∆g u + hu 5 6 9 10 11 12 13 n+2 hu n−2 . Finally, it is an easy task to select those u and u so that u < u. The method of sub- and super-solutions now guarantees that (6.13) admits a positive solution which turns out to be smooth by a simple regularity argument as in the study of the Yamabe problem. The proof is complete. 7. S OLVING THE TRANSFORMED PDE S IN THE CLOSED CASE WHEN n 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 3: T HE CASE OF CONSTANT MEAN CURVATURE 14 In this section, we are interested in the case when M is compact without boundary (i.e. closed). We also assume throughout the section that τ is constant everywhere in M ; hence ∇g τ = 0. 7.1. The vacuum case. As we are in the CMC case, the system (5.21)-(5.22) is decoupled; thus solving the constraint equations is reduced to solving the following equation n+2 3n−2 4(n − 1) n − 1 2 n−2 (7.1) ∆g u + Scalg u = − τ u + |σ |2g u− n−2 , n−2 n where we simply write σ = σ + Lg W . In this context, Isenberg [Isen95] obtained a complete picture of the solvability of (7.1) as shown in Table 3 where: − • ‘Y’ indicates that (7.1) can be solved for that class of conformal data; • ‘N’ indicates that (7.1) has no positive solution. As emphasized in [BarIse04], the mean curvature of the data plays an important role in separating those sets of free data for which we know whenever a solution for Eq. (7.1) exists. Let us now take a closer look into [Isen95]. In fact, the analysis in [Isen95] basically consists of the following three key steps: • The maximum principle: Making use of the maximum principle is crucial in [Isen95] because this will help us to conclude whether the finding solution u ˆ Q.A. NGO 40 1 2 3 4 5 6 7 8 9 10 11 is positive in M . In [Isen95, Appendix], Isenberg provided several versions of the maximum principle for the Laplacian on compact manifolds with or without boundary. • The Yamabe invariant Y: The use of the Yamabe invariant Y in [Isen95] is fundamental which helps to split the set of Riemannian metrics into different catalogue using the sign of its associated invariant. We note that the Yamabe invariant Y and the Yamabe-scalar field invariant Yψ introduced in (6.11) are different only by the presence of the field ψ; hence Y can be obtain from Yψ by getting rid of any ψ. • The method of sub- and super-solutions: This is the main method used in [Isen95, Section 3]. The method is rather easy to implement which roughly says that whenever we can find a pair of sub- and super-solutions (u, u) in the following sense 13 14 15 16 17 18 19 20 21 22 23 24 25 4(n − 1) ∆g u + Scalg u n−2 − 4(n − 1) ∆g u + Scalg u n−2 − − 12 3n−2 n+2 n − 1 2 n−2 τ u + |σ |2g u− n−2 n and − n+2 3n−2 n − 1 2 n−2 τ u + |σ |2g u− n−2 , n then we can conclude that (7.1) always admits at least one solution u ∈ (u, u) pointwise. Using this method, different sub- and super-solutions was found in [Isen95, Section 5], some are constant, some are the unique solution to some simple equations of the form −∆u + u = c. It is worth noting that uniqueness property of solutions to (7.1) was also obtained in [Isen95, Section 6]. 7.2. The case of scalar fields. In this subsection, instead of considering the vacuum case, we are interested in solving the constraint equations in the presence of real scalar fields. Since we still assume that τ is constant, solving these constraint equations is equivalent to solving Eq. (6.2) solely. However, in order to make our results available for a wider class of equations, we not only focus on Eq. (6.2) but also consider Eq. (6.3). 30 The following results corresponding to the three different sign of h basically come from our three joint papers with Xu in [NgoXu12] for the negative h, in [NgoXu15] for the vanishing h, and in [NgoXu14] for the positive h. Our approach is variational and based on the so-called subcritical approach widely known when solving the Yamabe problem [AmbRab73, Stru08]. 31 Recall that, generally, our equation (6.3) consists of the following three difficulties: 26 27 28 29 32 33 34 35 36 37 38 39 40 41 42 • a critical exponent, • a sign-changing potential, and • a negative exponent. Therefore, from variational method point of view, we propose a unique method to tackle equations of the form (6.3) having all difficulties above at the same time. 7.2.1. The case h < 0. We are interested in the existence (if possible, the multiplicity and the uniqueness) of positive solutions to Eq. (6.2) in the negative Yamabe-scalar field conformal invariant, or more general, solutions to Eq. (6.3) in the case h < 0. Notice that, as occurred in the well-studied in the prescribed scalar curvature problem, a simple use of integration by parts shows that the condition M f dvolg < 0 is also necessary in this case, see [NgoXu12]. Perhaps, this condition reflects the condition h < 0 we EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 3 4 5 6 41 are considering. However, there is another necessary condition in this case and this new necessary condition reflects the sign-changing property of f . To state this new necessary condition, following [Rau95], we first define the following number 2 inf M |∇u| dvolg , if A = ∅, λf = u∈A M |u|2 dvolg (7.2) +∞, if A = ∅, where A = 7 u ∈ H 1 (M ) : u |f − |u dvolg = 0 0, u ≡ 0, (7.3) M 8 9 10 11 12 13 14 15 16 17 18 19 20 and f + = max{f, 0} and f − = min{f, 0}. Clearly, functions in A are to be thought of as functions that vanish on the support of f − . It is clear that λf < +∞ if and only if the set {f 0} has positive measure, that is, either supM f > 0 or supM f = 0 and 1 dvol > 0. The second necessary condition that we would like to mention is g {f =0} |h| < λf , see [NgoXu12]. As can be easily guessed, this condition shows us a strong connection between the sign of h and the size of f + . The basic content of our study consists of two main parts depending of the behavior of f . First, we consider the case when the function f takes both positive and negative values. Our main theorem for this part can be stated as follows. Theorem 7.1. Let (M, g) be a smooth compact Riemannian manifold without the boundary of dimension n 3. Assume that f and a 0 are smooth functions on M such that f dvolg < 0, supM f > 0, M a dvolg > 0, and |h| < λf where λf is given in (7.2) M below. Let us also suppose that the integral of a satisfies a dvolg < 21 M 22 23 n−1 1 n−2 n−2 n−1 |h| |f − | dvolg M n |f − | dvolg (7.4) M where f − is the negative part of f . Then there exists a number C > 0 to be specified such that if −1 24 M 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 |f − | dvolg (sup f ) < C, (7.5) M then Eq. (6.3) possesses at least two smooth positive solutions. If we assume that f does not change sign in the sense that f 0 in M , we obtain necessary and sufficient solvability conditions as pointed out in [CBruIsePol07] in the case of (6.2). That is the content of the second part of our study for the case h < 0. Theorem 7.2. Let (M, g) be a smooth compact Riemannian manifold without boundary of dimension n 3. Let h < 0 be a constant, f and a be smooth functions on M with a 0 in M , f 0 but not strictly negative. Then Eq. (6.3) possesses one positive solution if and only if |h| < λf . 7.2.2. The case h = 0. Now we move to the next case, the case h = 0. We again continue to study some quantitative properties of positive, smooth solutions of Eq. (6.3) when h = 0. In the previous subsection, we have already seen that, in the case h < 0, a suitable balance between coefficients h, f , a of (6.3) is enough to guarantee the existence of one positive smooth solution. In addition, it was found that under some further conditions we may or we may not have the uniqueness property of solutions of (6.3) . This subsection is 42 1 2 ˆ Q.A. NGO a continuation of the previous one where we consider the case when h = 0, that is, we are interested in the following simple partial differential equation n+2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 3n−2 − ∆g u = f u n−2 + au− n−2 , u > 0. (7.6) The content of our study consists of three main parts. In the first part of the study, we mainly consider the case supM f > 0. In this context, we was able to show that if supM f and M a dvolg are small, then (7.6) possesses at least one smooth positive solutions. The first main theorem can be stated as follows. Theorem 7.3. Let (M, g) be a smooth compact Riemannian manifold without the boundary of dimension n 3. Assume that f and a 0 are smooth functions on M such that a dvolg > 0, M f dvolg < 0, and supM f > 0. Then there exist two positive numbers M η0 and λ depending only on the negative part f − of f such that if η0 sup f |f − | dvolg (7.7) 2 M M and n−2 1−n 2n λn − |f (7.8) | dvol a dvolg < g 4(η0 )n−2 n − 2 M M hold, then Eq. (7.6) possesses at least one smooth positive solution. By a simple comparison, one can easily see that except the multiplicity part, we was successful to carry the conclusion of Theorem 7.1 for the case h < 0 to the case h = 0. However, since Eq. (6.3) in the case h = 0 take a form simpler than that of the case h < 0, we might expect that the above two conditions (7.7) and (7.8) could be weakened. Surprisingly, we was able to prove that the condition (7.7) can be relaxed. Unfortunately, for the price we pay, the estimate for M a dvolg needs to be replaced by another estimate for supM a. This is the content of the second result. Theorem 7.4. Let (M, g) be a smooth compact Riemannian manifold without the boundary of dimension n 3. Assume that f and a 0 are smooth functions on M such that a dvolg > 0, M f dvolg < 0, and supM f > 0. Then if supM a is small, then Eq. M (7.6) possesses at least one smooth positive solution. In the last part of the present study, we focus our attention to the case supM f 0. It should mention that in the statement of Theorem 7.3, supM f is nothing but supM f + where f + is the positive part of f . Therefore, if we assume f 0, we then see that the condition (7.7) is fulfilled for any small η0 . However, one can immediately observe that the right hand side of (7.8) goes to +∞ as η0 → 0. This suggests that under the case supM f 0, there is no other condition for M a dvolg than M a dvolg > 0. That is the content of our next result. Theorem 7.5. Let (M, g) be a smooth compact Riemannian manifold without boundary of dimension n 3. Let f and a be smooth functions on M with a 0 in M , M a dvolg > 0, and f 0. Then Eq. (7.6) always possesses one positive solution. In addition, this solution is unique. Concerning Theorems 7.2 and 7.5, it is worth noticing that these theorems generalize the same results obtained in [CBruIsePol07] when our equation takes the form (6.2). Roughly speaking, it was proved in [CBruIsePol07] by the method of sub- and super-solutions that (6.3) and (7.6) always possess one positive solution so long as the functions f and a take the form f = −Bτ,ψ and a = Ag,W,π with f 0 and a 0. The main ingredient of the proof in [CBruIsePol07] is the conformal invariant property of Bτ,ψ and Ag,W,π . Apparently, this property is no longer available in our general case. EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 43 7.2.3. The case h > 0. Finally, we consider the remaining case when h > 0. We continue our study of some quantitative properties of positive smooth solutions to Eq. (6.3) when h > 0. As always, we assume hereafter that f , h > 0, and a 0 are smooth functions on M with M a dvolg > 0. For the sake of simplicity, it is important to note that we can freely choose a background metric g such that manifold M has unit volume. As far as we know, Eq. (6.3) with h > 0 was first considered in [HebPacPol09] by using variational methods. In that elegant paper, Hebey–Pacard–Pollack proved, among other things, a fundamental existence result which roughly says that a suitable control of M a dvolg from above is enough to guarantee the existence of one positive smooth solution. Their result basically makes use of the fact that the operator −∆g + h is coercive. Although the coerciveness property is slightly weaker than the condition h > 0, however as one can see from previous sections that this condition is enough to guarantee that · Hh1 is an equivalent norm on H 1 (M ). The advantage of this setting is that the first eigenvalue of the operator −∆g +h is strictly positive, and thus, various goods properties of the theory of weighted Sobolev spaces can be applied. Using our notations, their result can be restated as follows: There exists a constant C = C(n), C > 0 depending only on n, such that if ϕ 19 2n n−2 1 Hh a ϕ M 2n n−2 dvolg C (Sh sup |f |) (7.9) n−1 M 20 and 2n f ϕ n−2 dvolg > 0 21 (7.10) M 22 23 24 25 26 27 28 29 30 31 32 33 34 for some smooth function ϕ > 0 in M , then Eq. (6.3) possesses a smooth positive solution in the case a > 0 5. As can be seen from (7.10), the condition supM f > 0 is crucial. Therefore, it is not clear whether (6.3) possesses a smooth positive solution or not in the case supM f 0. Moreover, it is necessary to have a > 0 in M in order to get a positive lower bound for smooth solutions of (6.3). Besides, the condition (7.9) involves not only supM f + but also inf M f − . In other words, for given a, the negative part f − of f cannot be too negative. This restriction basically reflects the fact that the energy functional has to verify the mountain pass geometry as their solution was found as a mountain pass point. Our study for this case was also motivated by a recent paper [MaWei12]. In that paper, provided u is a positive smooth solution, Ma–Wei constructed a mountain pass solution of (6.3) of the form u + v for some smooth function v > 0. More precise, they proved that in the case 3 n < 6 and that the first eigenvalue of the following operator −∆+h− 35 36 4 n + 2 n−2 3n − 2 4n−4 fu + au n−2 n−2 n−2 (7.11) is positive, (6.3) possesses a mountain pass, smooth, positive solution. 5The upper bound for the integral of a appearing in (7.9) seems to be natural. This is because in the case h > 0, it was proved in [HebPacPol09] that if a 0, f > 0, and n+2 4n nn (n − 1) n−1 a n+2 4n f 3n−2 4n M then (6.3) does not admit any smooth positive solution. dvolg > h M n+2 4 f 2−n 4 dvolg , ˆ Q.A. NGO 44 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 It is easy to see that the positivity of the first eigenvalue of the operator given in (7.11) immediately implies that the solution u is strictly stable. Therefore, it is natural to seek for positive smooth solutions of (6.3) as local minimizers. Another reason that supports this approach is to look at the profile of the energy func3n−2 tional associated to (6.3). Due to the presence of the term au− n−2 , the energy of u is very large when maxM u is small. Clearly, in the case f 0, the energy of u is also large when maxM u is large. Consequently, a local minimizer of the energy functional should exist which could provide a possible solution. Similarly, if one assumes that supM f > 0 and that the energy functional admits some mountain pass geometry, a local minimizer of the energy functional again exists. While searching for positive smooth solutions of (6.3), we found that the method used in the two cases h 0 still works in this context. While the case h 0 involves more conditions and our analysis of solvability of (6.3) strongly depends on the ratio between supM f and M |f − | dvolg , the case h > 0 requires fewer conditions than the non-positive case. In fact, as we shall see later, in the case supM f > 0, no condition for f is imposed and we are able to show that if M a dvolg is small, then (6.3) possesses at least one smooth positive solutions since the condition for supM f can be absorbed to the condition for a dvolg . M The first result can be stated as follows. Theorem 7.6. Let (M, g) be a smooth compact Riemannian manifold without the boundary of dimension n 3. Assume that f , h > 0, and a 0 are smooth functions on M such that M a dvolg > 0 and supM f > 0. We assume further that there exists a constant τ > max 1, 23 24 2 Sh h dvolg n n−2 M such that n−1 a dvolg < 25 M (2n − 1) 22n−1 nn Sh τ Sh τ τ supM f − M f dvolg n−1 (7.12) 26 holds. Then (6.3) possesses at least one smooth positive solution. 27 Observe from (7.12) that τ plays no role but a scaling factor. Therefore, for given a dvolg , we could select τ sufficiently large and supM f sufficiently small in such a M way that (7.12) is fulfilled. This suggests that under the case when supM f is small, the condition for M a dvolg appearing in (7.12) can be relaxed. We prove this affirmatively and that is the content of the following theorem. 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 Theorem 7.7. Let (M, g) be a smooth compact Riemannian manifold without the boundary of dimension n 3. Let f , h, and a be smooth functions on M with h > 0, a 0 in M , M a dvolg > 0, and supM f > 0. Then there exists a positive constant C to be specified later such that if supM f < C , then Eq. (6.3) possesses one positive smooth solution. As always, we finally focus our attention to the case when supM f 0. In this context, we are able to get a complete characterization of the existence of solutions of (6.3) in the case when f 0. Roughly speaking, it should mention that in the statement of Theorem 7.6, supM f is exactly supM f + where f + is the positive part of f . Therefore, without any supM f , one can immediately observe that the right hand side of (7.12) goes to +∞ as τ → +∞. This suggests that under the condition supM f 0, no condition is imposed. EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 45 TABLE 4. Interaction between the coefficients of (6.3) for any h. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 h −λf h0 h 0 sup f + < C(f − , a) sup a C(f ) < a C(f ) < Theorem 7.8. Let (M, g) be a smooth compact Riemannian manifold without boundary of dimension n 3. Let f , h, and a be smooth functions on M with h > 0, a 0 in M , a dvolg > 0, and f 0. Then Eq. (6.3) always possesses one and only one positive M smooth solution. Concerning Theorem 7.8, it is worth noticing that it generalizes the same result obtained in [CBruIsePol07] when our equation takes the form (6.2). Roughly speaking, it was proved in [CBruIsePol07] by the method of sub- and super-solutions that (6.3) always possesses one positive solution so long as the functions f and a take the form f = −Bτ,ψ and a = Ag,W,π with f 0 and a 0. The main ingredient of the proof in [CBruIsePol07] is the conformal invariant property of Bτ,ψ and Ag,W,π . Apparently, this property is no longer available in our general case. 7.2.4. Summary. Finally, before closing the present subsection, we would like to mention the interaction between the coefficients of the Einstein-scalar field Lichnerowicz equations (6.3) for arbitrary sign of h. Using our results for the negative case in [NgoXu12], for the null case in [NgoXu15], and for the positive case in [NgoXu14], one can obtain the following table which shows us how the coefficients in (6.3) depend on each other in order to get the existence of solutions. The interpretation of Table 4 is as follows: The second column basically says that h cannot be too negative as it must satisfy h > −λf for some positive constant λf given in (7.2). Under this condition, we guarantee an existence result for (6.3) provided supM f + and M a dvolg are bounded in terms of f − . This result still holds for the case h = 0; however, the boundedness of supM f + can be relaxed if we replace M a dvolg by supM a as shown in the third column. The fourth column shows that in the case h > 0, the boundedness of supM a can be weakened by using M a dvolg . For the fifth column, it shows that no condition is required if supM f + is small in terms of f − and a. In the last column, it shows that (6.3) is always solvable if inf M h is sufficiently large, for example, if h satisfies h sup f + sup a M 30 h>0 M in M . This is because in this case the constant 1 is a super-solution for (6.3) and this is enough since a sub-solution for (6.3) which is less than 1 always exists. Finally, it is worth mentioning that our approach is quite useful as it can be used to tackle other problems, for example, see [NgoZha15]. ˆ Q.A. NGO 46 1 2 3 4 5 6 7 8 9 10 11 7.3. Strategy for finding solutions and ideas of proofs. As we have noted before, in order to study (6.3), we follow the subcritical approach, i.e. as a standard routine, in the first step to tackle (6.3), we look for positive smooth solutions of the following subcritical problem au − ∆g u + hu = f |u|q−2 u + , (7.13) q +1 2 (u + ε) 2 2n 2n where ε > 0 is small and q ∈ (2, n−2 ) is close to n−2 . Since the subcritical equation (7.13) is variational, we study (7.13) by using the variational method [AmbRab73, Stru08]. Our main procedure is to show that solutions of (6.3) 2n exist as first ε 0 and then q n−2 under various assumptions. The whole procedure can be summarized as the following diagram. 2 −∆g uεq + huεq = f |uεq |q−2 uεq + auεq ((uεq ) + ε) −q/2−1 ↓ 12 q−1 −∆g uq + huq = f (uq ) 13 + a(uq )−q−1 , 14 ↓ 15 −∆g u + hu = f u n−2 + au− n−2 n+2 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 2n , ε > 0, q ∈ [2, n−2 ) q>2 3n−2 It is worth noticing that in [HebPacPol09], the authors just considered (7.13) with q 2n replaced by n−2 and they found solutions as mountain pass points. This main deference somehow reflects the fact that we need the compact embedding H 1 (M ) → Lq (M ) while searching for minimum points. In the following part of this subsection, we consider necessary conditions for f and h in the case h 0. 7.3.1. Necessary conditions for f and h in the case h 0. First, we derive a condition for f dvolg so that (7.13) admits positive smooth solutions. Our argument was motivated M from a same result for the well-known prescribing scalar curvature problem. It is important to note that in the case h > 0, there is no such a condition on M f dvolg . Proposition 7.9. Assume that h 0 is constant. Then then necessary condition for f so that Equation (7.13) admits positive smooth solution is M f dvolg < 0. In particular, the necessary condition for (6.3) to have positive smooth solution is M f dvolg < 0. Proof. We assume that u > 0 is a smooth solution of (7.13). By multiplying both sides of (7.13) by u1−q , one gets (−∆g u)u1−q + hu3−q = f + 31 32 Integrating over M and noticing that h 33 (−∆g u)u1−q dvolg > au2−q q (u2 + ε) 2 +1 . 0 give au2−q f dvolg + M q M M (u2 + ε) 2 +1 dvolg . By the divergence theorem, one obtains u−q |∇u|2 dvolg . (−∆g u)u1−q dvolg = (1 − q) M 34 M This and the fact that q > 2 deduce that au2−q f dvolg + 35 M q M (u2 + ε) 2 +1 dvolg < 0. EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 47 f dvolg < 0 as claimed. 1 Obviously, there holds 2 We are now study necessary conditions for h. More precisely, we show that the condition |h| < λf is necessary if λf < +∞ in order for (6.3) to have a positive smooth solution. Here we provide a different proof which is shorter than the proof in [Rau95, Section III.3]. Our argument depends on a Picone type identity for integrals [AbdPer03] whose proof makes use of the density. 3 4 5 6 7 8 M Lemma 7.10. Assume that v ∈ H 1 (M ) with v smooth function. Then we have 2 |∇v| dvolg = − 9 M 10 ∆u 2 v dvolg + u 2 v u u2 ∇ M dvolg . Proof. By density, there exist a family of regular functions {vj }j such that vj → v strongly in H 1 (M ), 11 12 M 0 and v ≡ 0. Suppose that u > 0 is a vj ∈ C 1 (M ). vj → v a.e. in M, The standard Picone identity tells us that 2 |∇vj | dvolg = − 13 M M ∆u 2 v dvolg + u j vj u u2 ∇ M 2 dvolg . 14 Since u > 0 is smooth and vj → v strongly in H 1 (M ) we immediately have 15 |∇vj | dvolg → 2 2 |∇v| dvolg M 16 17 18 19 M and ∆u 2 ∆u 2 vj dvolg → v dvolg M u M u as j → +∞. Again using the smoothness of u, we can check that vj /u → v/u strongly in H 1 (M ). Therefore, M 22 23 24 25 26 27 2 vj u u2 ∇ 20 21 M dvolg as j → +∞. The proof now follows by taking the limit in (7.14) as j → +∞. We are now in a position to provide a different proof for necessary condition |h| < λf . Proposition 7.11. If Eq. (6.3) has a positive smooth solution, it is necessary to have |h| < λf . Proof. We only need to consider the case λf < +∞ since otherwise it is trivial. We let v ∈ A arbitrary and assume that u is a positive smooth solution to (6.3). Using Lemma 7.10 and (6.3), we find that 2 M 4 v 2 dvolg + M f u n−2 v 2 dvolg M 2 + 28 av u 4−4n n−2 M v 2 dvolg + |h| u2 ∇ dvolg + M M u2 ∇ M v u v u 2 dvolg 2 dvolg . In other words, there holds 2 2 30 2 v u u2 ∇ dvolg → |∇v| dvolg =|h| 29 (7.14) |∇v| dvolg v 2 dvolg M M |h| + M u2 ∇ uv dvolg . 2 dvol v g M (7.15) ˆ Q.A. NGO 48 1 |h| > 0 by taking the infimum with respect to v. Notice that In particular, λf 2 2 3 M u2 ∇ uv dvolg 2 v dvolg M inf M u supM u 2 M v u ∇ M v 2 u 2 |∇v| dvolg v 2 dvolg M M 8 9 10 11 12 13 14 15 16 28 29 30 . 0. • If h < 0 is constant, then there holds min u h min n−2 2 ,1 inf M f >0 (7.16) ,1 (7.17) ,1 . (7.18) for any ε > 0. • If h ≡ 0, then there holds 1 min 2 min u inf M a − inf M f n−2 4n−4 for any inf M a − inf M f 1 ε < min 2 n−2 2n−2 • If h > 0, then there holds min u M 27 2 2n Lemma 7.12. Let u be a positive C 2 solution of (7.13) and assume that q ∈ ( 2n−2 n−2 , n−2 ) is arbitrary. Then: 24 26 inf M u supM u In the case h < 0, one can show that positive C 2 solutions to (7.13) is indeed bounded from below and away from zero. 22 25 . 7.3.2. Lower bounds for solutions of (7.13). We note that when sending ε 0 to get a solution uq from a sequence of solutions {uεq }ε to (7.13), one needs to take care the singularity by forcing uεq to stay away from 0. M 23 2 2 From the above proof, one can observe that the function a plays no role but a 20 21 inf M u supM u This and the fact that λ > 0 give us the desired result. M 18 dvolg inf M u supM u |h| + λf |h| + λf λf 17 19 λf By taking the infimum with respect to v, we obtain 6 7 dvolg since v/u ∈ A . Having this, we can check from (7.15) that 4 5 2 min − 2n−2 n−2 2 inf M a supM h + supM |f | n−2 4n−6 ,1 (7.19) ,1 . (7.20) for any ε < min 2n−2 2− n−2 inf M a supM h + supM |f | 2n−4 4n−6 Proof. Note that in the case h > 0, such a result was proved in [HebPacPol09], we then omit the proof and focus only on the case h 0. The only difference is that we calculate a precise lower bound for solutions of (7.13) as shown in (7.19), a detailed computation can be found in [Ngo12, Lemma 3.3]. EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 3 4 5 6 In the case that h < 0 is constant, we first assume that u achieves its minimum value at x0 . For the sake of simplicity, we denote u(x0 ), f (x0 ), and a(x0 ) by u0 , f0 , and a0 respectively. Notice that u0 > 0 since u is a positive solution. We then have ∆g u|x0 0; in particular, a0 u0 hu0 f0 (u0 )q−1 + f0 (u0 )q−1 . q +1 ((u0 )2 + ε) 2 Consequently, we get f0 < 0 and thus 0 < h min u 7 9 10 11 12 13 14 15 16 17 22 23 24 25 26 27 28 29 30 31 32 33 36 min n−2 2 ,1 inf M f Now, let us assume that u achieves its minimum value at x0 . For simplicity, let us denote u(x0 ), f (x0 ), and a(x0 ) by u0 , f0 , and a0 respectively. Notice that u0 > 0 since u is a positive solution. We then have ∆g u|x0 0; in particular, a0 u0 f0 (u0 )q−1 + 0. (7.21) q +1 2 ((u0 ) + ε) 2 Consequently, we get that f0 < 0. Using (7.21) we can see that a0 q −f0 (u0 )q−2 ((u0 )2 + ε) 2 +1 −f0 ((u0 )2 + ε)q which implies that 2 (u0 ) + ε a0 −f0 1 q inf M a − inf M f 1 q . 2n Thus, one can conclude that u0 satisfies (7.17) for any q ∈ [ 2n−2 n−2 , n−2 ) and any ε verifying the condition (7.18). The proof is complete. 7.3.3. Regularity for non-negative weak solutions of (7.13). This subsection is devoted to the regularity of weak solutions of (7.13). Despite the fact that h can be chosen as a constant in the non-positive Yamabe-scalar field conformal invariant, in this subsection, we allow h to be non-constant. As such, we assume that h, f and a 0 are smooth functions and that the function h has a fixed sign on M . Lemma 7.13. Assume that u ∈ H 1 (M ) is an almost everywhere non-negative weak solution of Equation (7.13). We assume further that inf M a > 0 in the case when h 0. Then (a) If ε > 0, then u ∈ C ∞ (M ). In particular, u 0 in M . (b) If ε = 0 and u−1 ∈ Lp (M ) for all p 1, then u ∈ C ∞ (M ). Proof. We first rewrite (7.13) as −∆g u + b(x)(1 + u) = 0 34 35 h Next we consider the case h 0. Unlike the case h < 0 where we assume no condition on a, for the case h 0, we do require inf M a > 0 since the lower bound for any positive C 2 solution u depends on inf M a > 0 as we can see from (7.17). Also, we require ε to be small as indicated in (7.18). 20 21 (u0 )q−2 which immediately implies 2n for any q ∈ ( 2n−2 n−2 , n−2 ) and any ε > 0. This proves the estimate (7.16) in our lemma. 18 19 h f0 1 q−2 inf M f M 8 49 with b(x) = u(x) 1 + u(x) h(x) − a(x) 2 (u(x) + ε) q 2 +1 − f (x)|u(x)|q−2 . ˆ Q.A. NGO 50 1 2 3 4 5 6 7 8 9 10 11 12 2n By the Sobolev embedding, we know that u ∈ Lq (M ) for any q ∈ ( 2n−2 n−2 , n−2 ]. This and the conditions in both cases (a) and (b) imply q a h(x) − 2 − f |u|q−2 ∈ L q−2 (M ). q +1 (u + ε) 2 q 2n n Notice that from q n−2 there holds q−2 2 . We now use the Brezis–Kato estimate s [Stru08, Lemma B.3] to conclude that u ∈ L (M ) for any s > 0. Thus the Cald´eron– Zygmund inequality implies that u ∈ H p (M ) for any p > 1. The Sobolev embedding again implies that u is in C 0,α (M ) for some α ∈ (0, 1). Thus, we know from the Schauder theory that u ∈ C 2,α (M ) for some α ∈ (0, 1). In particular, u has a strictly positive lower bound by means of Lemma 7.12. Since u stays away from zero, we can iterate this process to conclude u ∈ C ∞ (M ). 2n 7.3.4. Analysis of the energy functional. For each q ∈ [ 2n−2 n−2 , n−2 ) and k > 0, we intro1 duce Bk,q , a hyper-surface of H (M ), which is defined by q Lq Bk,q = u ∈ H 1 (M ) : u 13 =k . (7.22) 1 14 15 16 Notice that for any k > 0, our set Bk,q is non-empty since it always contains k q . Now we construct the energy functional associated to problem (7.13). For each ε > 0, we consider the functional Fqε : H 1 (M ) → R defined by Fqε (u) = 1 2 h 2 1 − q |∇u|2 dvolg + M 17 u2 dvolg M f |u|q dvolg + M 1 q a M (u2 q + ε) 2 dvolg . 21 By a fairy standard argument, Fqε is continuously differentiable on H 1 (M ), see Lemma 7.14 below, and thus weak solutions of (7.13) correspond to critical points of the functional Fqε . Now we set µεk,q = inf Fqε (u) . 22 By the H¨older inequality, it is not hard to see thatFqε |Bk,q is bounded from below by 23 −k supM f + 18 19 20 u∈Bk,q 24 25 26 27 h q2 2k and thus µεk,q > −∞ if k is finite. On the other hand, using the 1 q test function u = k , we get h q2 k µεk,q k − 2 q f dvolg + 1 q a (7.23) 2 dvolg (k + ε) which tells us that µεk,q < +∞. Our aim was to find critical points of the functional Fqε . In order to support our aim, we prove the following result. M M 2 q Lemma 7.14. The first variation of Fqε at a point u in a direction v is given by δFqε (u) (v) = ∇u · ∇v dvolg + h M f |u|q−2 uv dvolg − − 28 au v dvolg q +1 + ε) 2 where ∇u · ∇v stands for the pointwise scalar product of ∇u and ∇v with respect to the metric g. M 29 uv dvolg M M (u2 Proof. The proof is simple, we include it here for completeness. In fact, for any smooth function v, there holds d δFqε (u) (v) = Fqε (u + tv) dt t=0 EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS = d 1 dt 2 + + |∇(u + tv)|2 dvolg + M d 1 − dt q d 1 dt q = 4 5 t=0 t=0 q a((u + tv)2 + ε)− 2 dvolg M t=0 uv dvolg M q auv(u2 + ε)− 2 −1 dvolg , f |u|q−2 uv dvolg − M 3 M q (−∆g u)v dvolg + h − 2 2 (u + tv) dvolg f |u + tv| dvolg M M 1 h 2 51 M which provides the desired result. An interesting fact due to the subcritical approach is that µεk,q is achieved for each k, q, and ε fixed. 2n Lemma 7.15. For each k > 0, q ∈ (2, n−2 ), and ε > 0 fixed, the number µεk,q is achieved by some smooth positive function. 8 The proof is standard and is based on the so-called direct methods in the calculus of variations, see [Ngo12, Lemma 4.2]. Another interesting feature of µεk,q is the continuity of µεk,q with respect to k. 9 Lemma 7.16. For ε > 0 fixed, µεk,q is continuous with respect to k. 6 7 10 11 12 13 14 15 16 The proof of Lemma 7.16 is also standard, see [Ngo12, Proposition 4.1]. The key idea 2n is to make of the weakly lower semi-continuity of Fqε and the fact that q < n−2 . The next part of our analysis for Fqε is to study the asymptotic behavior of µεk,q when k varies. Depending on the sign of supM f , the behavior of µεk,q varies. In the dedicated case when supM f > 0 and inf M < 0, i.e. when f changes sign, we can prove that when k is small, µεk,q goes to +∞ and when k is big, µεk,q goes to −∞. This can be plotted as in Figure 4. µεk,q −k k⋆ k1,q k2,q M f+ 1 k M k F IGURE 4. The asymptotic behavior of µεk,q when supM f > 0. 17 This is the content of the following two lemmas. 2 q 18 19 Lemma 7.17. There holds limk→0+ µkk,q = +∞. In particular, there is some k sufficiently small and independent of both q and ε such that µεk ,q > 0 for any ε k . a ˆ Q.A. NGO 52 1 2 3 4 Lemma 7.18. There holds µεk,q → −∞ as k → +∞ if supM f > 0. Proofs for Lemmas 7.17 and 7.18 above can be found in [Ngo12, Subsection 4.2]. In the case supM f 0, instead of decaying to −∞ as k → +∞, we obtain µεk,q → +∞ as k → +∞ as shown in Figure 5. µεk,q k⋆ k1,q k2,q k F IGURE 5. The asymptotic behavior of µεk,q when supM f 5 6 7 0. 7.3.5. The way to find solutions when either f changes sign or supM f 0. Once we have the asymptotic behavior of µεk,q when k is large and small, we can start to look for solutions of Eq. (6.3) by making the curve k → µεk,q look like in Figure 6. Fqε mountain pass levels u Lq k local minimum F IGURE 6. Solutions will be found as minimum and mountain-pass points. 8 9 The trick is to find k1,q and k2,q in such a way that µεk1,q ,q < min{µεk ,q , µεk2,q ,q }. Then we can use standard variational techniques to look for solutions as indicated in Figure 6. 1 10 11 12 13 14 7.3.6. The case supM > 0 and inf M a = 0 when h = 0. Under this context, making use of the method of sub- and super-solutions is the key argument. Thanks to [Heb09], from that we learn this approach. However, it is worth mentioning that our construction of sub-solutions is different from that of [Heb09]. We let ε0 > 0 sufficiently small and then fix it so that the following inequality a dvolg +ε0 < 15 M λn n−2 4(η0 ) 2n n−2 n−2 1−n |f − | dvolg M (7.24) EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 53 3 still holds. Since the manifold M has unit volume, we can conclude that from (7.24), the function a + ε0 verifies all assumptions in the previous subsection, thus showing that there exists a positive smooth function u solving the following equation 4 −∆g u = f u n−2 + (a + ε0 )u− n−2 . 1 2 n+2 5 Obviously, u is a super-solution to (7.6), that is a dvolg − f dvolg = 0, − | dvol |f g M M a+ M there exists a function u0 ∈ H 1 (M ) solving 12 13 14 15 16 17 20 εa Besides, since f − 0 and 2n n−2 ε 25 26 28 29 30 a dvolg |f − | dvolg M ε . (7.27) n+2 n+2 n−2 4 (max u0 )− n2 −4 4 M . (7.28) In particular, the following a dvolg − f − | dvol |f g M M n+2 n+2 ε n−2 u0n−2 f (7.29) holds provided (7.28) holds. Combining (7.26), (7.27), and (7.29), we conclude that for small ε −∆g u n+2 n+2 3n−2 − 3n−2 n−2 ε n−2 u0n−2 f + aε− n−2 u0 −∆g u 36 − 3n−2 n−2 ε n−2 u0n−2 f − f− M 32 35 3n−2 (maxM u0 )− 4n−4 , we holds provided In other words, we have showed that 34 3n−2 aε− n−2 u0 a dvolg − |f | dvolg M 31 33 (7.26) > 2, it is not difficult to see that the following inequality M ε 27 a dvolg − f . − | dvol |f g M M Since maxM u0 < +∞, it is easy to see that, for any 0 < ε immediately have 23 24 ε − ∆g u = εa + 21 22 (7.25) Since the right hand side of (7.25) is of class Lp (M ) for any p < +∞, the Cald´eronZygmund inequality tells us that the solution u0 is of class W 2,p (M ) for any p < +∞. Thanks to the Sobolev Embedding theorem [Aubin98, 2.10], we can conclude that u0 ∈ C 0,α (M ) for some α ∈ (0, 1). In particular, the solution u0 is continuous. Therefore, by adding a sufficiently large constant C to the function u0 if necessary, we can always assume that minM u0 > 1. We now find the sub-solution u of the form εu0 for small ε > 0 to be determined. To this end, we first write 18 19 a dvolg − f |f − | dvolg M M − ∆g u0 = a + 10 11 3n−2 Our aim is to find a sub-solution to (7.6). Indeed, since 8 9 n+2 f u n−2 + au− n−2 . −∆g u 6 7 3n−2 n+2 . 3n−2 f u n−2 + au− n−2 . Finally, since u has a strictly positive lower bound, we can choose ε > 0 sufficiently small such that u u. Using the sub- and super-solutions method, see [KazWar75, Lemma 2.6], we can conclude the existence of a positive solution u to (7.6). By a regularity result developed in [KazWar75], we know that u is smooth. ˆ Q.A. NGO 54 1 2 3 7.3.7. Small of supM a can control everything in the case h = 0: Proof of Theorem 7.4. The proof we provide here is based on the method of the sub- and super-solutions, see [KazWar75, Jun94]. 5 We first construct a positive super-solution u for (7.6). By using the change of variable u = ev , we get that 6 ∆u + f u n−2 + au− n−2 = ev (∆v + |∇v|2 ) + f e n−2 v + ae− n−2 v . 4 n+2 7 3n−2 n+2 Hence, it suffices to find v satisfying 4n−4 4 ∆v + |∇v|2 + f e n−2 v + ae− n−2 v 8 9 In order to do this, thanks to 4 − 1 4 sup f f dvolg M M 12 13 0. f dvolg < 0, we can pick b > 0 small enough such that M |e n−2 bϕ − 1| 10 11 3n−2 and 1 f dvolg , 4 M where ϕ is a positive smooth solution of the following equation b|∇ϕ|2 < − ∆ϕ = 14 f dvolg − f. M 15 We now find the function v of the form v = bϕ + 16 17 n−2 log b. 4 Indeed, by calculations, we have 4n−4 4 ∆v + |∇v|2 + f e n−2 v + ae− n−2 v n−2 n−2 log b + ∇ bϕ + log b 4 4 n−2 n−2 4n−4 4 + f e n−2 (bϕ+ 4 log b) + ae− n−2 (bϕ+ 4 log b) 2 =∆ bϕ + 18 b 2 19 4n−4 f dvolg + ae− n−2 bϕ b1−n . M Therefore, if we assume that the function a verifies the following estimate sup a < − 20 M 21 bn 4n−4 e n−2 bϕ 4 23 24 25 26 27 28 29 30 (7.30) we then get that 4n−4 4 22 f dvolg , M ∆v + |∇v|2 + f e n−2 v + ae− n−2 v b 4 f dvolg < 0, M which concludes the existence of a super-solution u. We now turn to the existence of a sub-solution. Before doing so, we can easily check that u = ebϕ+ n−2 4 log b >b n−2 4 . Since u has a strictly positive lower bound and thanks to the second stage of the proof of Theorem 7.3, we can easily construct a sup-solution u with u < u. It is important to note that the existence of a sub-solution depends heavily on the conditions a 0 and a ≡ 0; and here is the only place we make use of that fact in the proof. The proof of the theorem is now complete. EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 3 4 5 55 7.3.8. The case supM f > 0 and inf M a = 0 when h > 0. Under this context, making use of the method of sub- and super-solutions is the key argument. Thanks to [Heb09], from that we learn this approach. However, it is worth mentioning that our construction of sub-solutions is different from that of [Heb09]. We let ε0 > 0 sufficiently small and then fix it so that the following inequality n−1 a dvolg + ε0 < 6 M (2n − 1) 22n−1 nn Sh τ Sh τ τ supM f − M f dvolg n−1 (7.31) 9 still holds. Since the manifold M has unit volume, we can conclude that from (7.31), the function a + ε0 verifies all assumptions in the previous subsection, thus showing that there exists a positive smooth function u solving the following equation 10 −∆g u + hu = f u n−2 + (a + ε0 )u− n−2 . 7 8 n+2 11 3n−2 Obviously, u is a super-solution to (6.3), that is −∆g u + hu 12 n+2 3n−2 f u n−2 + au− n−2 . 14 Our aim is to find a sub-solution to (6.3). In this context, we consider the following equation 15 − ∆g u + (h − f − )u = a. 13 16 17 18 19 (7.32) − Since h − f > 0, a 0, a ≡ 0, and the manifold M is compact without boundary, the standard argument shows that (7.32) always admits a weak solution, say u0 . By a standard regularity result, one can easily deduce that u0 is at least continuous. Thus, by the Maximum Principle, we conclude u0 > 0. 21 As before, we now find the sub-solution u of the form εu0 for small ε > 0 to be determined. To this purpose, we first write 22 − ∆g u + hu = εa + f − u. 20 23 24 Since maxM u0 < +∞, it is easy to see that, for any 0 < ε immediately have εa 25 26 Besides, since f − 0 and 2n n−2 holds provided ε 31 32 33 34 35 36 37 − 3n−2 n−2 3n−2 (maxM u0 )− 4n−4 , we . (7.34) > 2, it is not difficult to see that the following inequality n+2 n+2 ε n−2 u0n−2 f − (maxM u0 )−1 . In particular, the following εu0 f − 29 30 3n−2 aε− n−2 u0 εu0 f − 27 28 (7.33) n+2 n+2 ε n−2 u0n−2 f (7.35) holds provided ε (maxM u0 )−1 . Combining all estimates (7.33), (7.34), and (7.35) above, we conclude that for small ε, there holds −∆g u + hu n+2 n+2 3n−2 − 3n−2 n−2 ε n−2 u0n−2 f + aε− n−2 u0 . In other words, we have shown that u is a sub-solution of (6.3). Finally, since u has a strictly positive lower bound, we can choose ε > 0 sufficiently small such that u u. Using the sub- and super-solutions method, see [KazWar75, Lemma 2.6], we can conclude the existence of a positive solution u to (6.3). By a regularity result developed in [KazWar75], we know that u is smooth. ˆ Q.A. NGO 56 1 2 3 4 7.3.9. Small of supM f can control everything in the case h = 0: Proof of Theorem 7.7. In order to prove Theorem 7.8, we need to show that the condition (7.12) is fulfilled. Although we have not assumed any upper bound for M a dvolg , we are able to show that we can recover the condition (7.12) provided supM f is sufficiently small. 6 As always, we first assume inf M a > 0. Depending on the sign of two cases. 7 Case 1. Suppose 5 M f dvolg 9 Sh τ . τ supM f − M f dvolg Therefore, it suffices to show that n−1 a dvolg < 10 M 11 (2n − 1) 22n−1 nn (2n − 1) 22n−1 nn sup f < 12 M 13 17 18 Sh supM f τ= 1 2 > max 1, supM f Sh a dvolg < 25 26 27 28 29 30 31 32 33 Sh f dvolg M Sh . − | dvol |f g M Sh − | dvol |f g M n−1 , which is equivalent to sup f < M 24 . M (2n − 1) 1 Sh sup f 22n−1 nn supM f 1 + M n−1 23 n n−2 Therefore, it suffices to show that M 22 . h dvolg n−1 21 , 1 n−1 Shn τ M a dvolg Then, thanks to f = f + + f − , we have Sh τ 1 = supM f 1 − τ supM f − M f dvolg 1 supM f 1 + 19 20 n−1 Case 2. Suppose M f dvolg < 0. In this context, we assume for a moment that supM f > 0 is small in such a way that we can select 15 16 Sh τ which is equivalent to n−1 14 f dvolg , we have 0. In this context, we can easily verify that Sh supM f 8 M (2n − 1) 22n−1 nn Shn − (1 + M |f | dvolg ) 1 n−2 M a dvolg . From our calculation above, we conclude that there exists some positive constant C > 0 depending only on a, h, and f − such that if 0 < supM f < C , our equation (6.3) always admits at least one positive smooth solution. It remains to consider the case inf M a = 0. However, since the size of a plays no role in the above calculation, we can freely add a small constant ε0 to a as in the second stage of the proof of Theorem 7.6. This procedure ensures that we always get a super-solution of (6.3) with a strictly positive lower bound and this is enough since a suitable positive sub-solution always exists. As mentioned at the beginning of the section, we are considering the case when M is closed and τ is constant everywhere in M . The cases when M is either non-compact or compact with boundary as well as non-constant τ are very delicated. However, due to the length of the notes, we omit it. EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 57 1 8. OTHER RESULT: A L IOUVILLE TYPE RESULT 2 In the last part of this notes, we study a Liouville type result for positive, smooth solutions of (6.3). For the sake of simplicity, we only consider Eq. (6.3) in the case when h, f , and a are constants with, of course, a > 0. Our aim here is to give some sufficient conditions so that (6.3) has only constant solution. 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 Note that by a scaling argument we may assume a = 1; hence, in this section, we are interested in the following model equation − ∆g u + hu = λuq + u−q−2 , u > 0, (8.1) where h and λ are constants. It is worth noticing that after using a scaling the sign of h still matches the sign of the Yamabe-scalar field conformal invariant since a > 0. We n+2 and will also notice that the exponent q > 0 here is arbitrary rather than the number n−2 be specified later. Besides, we could also consider the case q < 0 if physical problems motivate it; however, we shall not touch the case q 0 in this section. Our result was inspired by a couple of recent papers [MaXu09, Ma10]. In these papers, the authors considered the following model equation − ∆g u = −uq + u−q−2 , u > 0, (8.2) 2n in Rn with the standard metric where q ∈ (1, n−2 ), that is, h = 0 and λ = −1. First, they proved in [MaXu09] that smooth positive solutions of (8.2) are uniformly bounded. Then by using the idea from Redheffer [Red60], Ma [Ma10, Theorem 1] was able to prove that any smooth positive solution of (8.2) is constant; hence, is equal to 1. In [Brezis11], Brezis used a different approach to establish, among other things, such a Liouville type result. Besides, it was shown in [Ma10, Theorem 2] that the similar Liouville type result is also true for smooth positive solutions for (8.2) in a complete non-compact Riemannian manifold with the Ricci curvature bounded from below. Motivated by all discussion above, in this section we prove that any smooth positive solution of Eq. (8.1) in a complete compact Riemannian manifold with the Ricci curvature bounded from, whose the bound will be determined, is constant. To be precise, we now state our main result. Theorem 8.1. Let (M, g) be a smooth closed Riemannian manifold of dimension n 3. Let h, λ, and q > 0 be constants. Then there is a constant K(n, q, h) depending only on n, q and h so that if Ricg K in the sense of quadratic forms, then every smooth positive solution of (8.1) is constant provided that in the case h > 0, λ > 0, we have to restrict n+2 . q n−2 Our result can be formulated as in Table 5. Surprisingly enough, as can seen from Table 5, the constant K does not depend on λ. It is worth noticing that by integrating both sides of (8.2) over M , one easily gets that Eq. (8.2) has no positive solution if h 0 and λ 0. The papers by Gidas–Spruck [GidSpr81] and M. Bidaut-V´eron–L. V´eron [BVerVer91] were sources of inspiration. Finally, as usual, we should mention that the content of this section was adapted from [NgoXu11]. We now prove Theorem 8.1 in details. 8.1. Some basic computations. For simplicity, we shall use M to denote the integral with respect to the measure induced by the metric g, that is M = M dvolg . First, we need a preparation. For some β = 0 to be determined later, we denote u = v −β . A direct ˆ Q.A. NGO 58 TABLE 5. The Liouville type result in terms of the Ricci curvature. 1 6 7 8 1 Ricg 0 q0 q n+2 n−2 n−1 n (q Ricg − 1)h computation shows that ∆g v = (β + 1) |∇v|2 1 + (−∆g u)v β+1 . v β This and the fact that −∆g u = −hv −β + λv −βq + v β(q+2) 4 5 K λ 2 3 Ricg h give 1 |∇v|2 + (−hv + λv 1−β(q−1) + v 1+β(q+3) ). (8.3) v β Applying the well-known Bochner-Lichnerowicz-Weitzenb¨ock formula [Aubin98] to function v, one obtains ∆g v = (β + 1) 1 ∆g (|∇v|2 ) = |Hess(v)|2 + ∇∆g v, ∇v + Ricg (∇v, ∇v). 2 9 (8.4) 12 Multiply both sides of (8.4) by v γ where γ ∈ R will be chosen later and integrate on M , to obtain A + B + C + D = 0, (8.5) 13 where 10 11 A= 14 15 1 n 2 v γ (∆g v) + M M 1 2 (∆g v) , n v γ ∇∆g v, ∇v , B= 16 M 17 C=− 18 19 v γ |Hess(v)|2 − 1 2 v γ ∆g (|∇v|2 ), M and v γ Ricg (∇v, ∇v). D= 20 M 21 22 We notice that γ may not necessarily be nonzero. Besides, there holds v γ |Hess(v)|2 − J= M 23 24 1 2 (∆g v) n since it is well-known that |Hess(v)|2 − 1 (∆g v)2 n 0. 0 (8.6) EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 59 8.2. Computations of A, B, and C. We treat the first term of A in the following way. In fact, using (8.3), one obtains 1 n 2 v γ (∆g v) M |∇v|2 1 1 v γ (∆g v) (β + 1) + (−hv + λv 1−β(q−1) + v 1+β(q+3) ) n M v β 2 β+1 |∇v| = v γ−1 |∇v|2 (β + 1) n v M β+1 1 + v γ−1 |∇v|2 (−hv + λv 1−β(q−1) + v 1+β(q+3) ) n β M 1 (−hv γ+1 + λv γ+1−β(q−1) + v γ+1+β(q+3) )(∆g v) + nβ M = 2 = 1 (β + 1) v γ−2 |∇v|4 n M β+1 + |∇v|2 (−hv γ + λv γ−β(q−1) + v γ+β(q+3) ) nβ M 1 + (−hv γ+1 + λv γ+1−β(q−1) + v γ+1+β(q+3) )(∆g v). nβ M Therefore, this and (8.6) imply that 2 (β + 1) v γ−2 |∇v|4 n M β+1 + |∇v|2 (−hv γ + λv γ−β(q−1) + v γ+β(q+3) ) nβ M 1 + (−hv γ+1 + λv γ+1−β(q−1) + v γ+1+β(q+3) )(∆g v). nβ M A =J + 2 (8.7) By the divergence theorem, it holds v γ+1 ∆g v = −(γ + 1) M v γ |∇v|2 , M v γ+1−β(q−1) ∆g v = −(γ + 1 − β(q − 1)) M v γ−β(q−1) |∇v|2 , M v γ+1+β(q+3) v γ+β(q+3) |∇v|2 . ∆g v = −(γ + 1 + β(q + 3)) M M Therefore, we can further simplify (8.7) as follows 2 (β + 1) v γ−2 |∇v|4 n M β+1 + |∇v|2 (−hv γ + λv γ−β(q−1) + v γ+β(q+3) ) nβ M λ − (γ + 1 − β(q − 1)) v γ−β(q−1) |∇v|2 nβ M 1 1 γ 2 + (γ + 1) hv |∇v| − (γ + 1 + β(q + 3)) nβ nβ M A =J + v γ+β(q+3) |∇v|2 . M ˆ Q.A. NGO 60 1 Thus, finally, we have 2 (β + 1) h(γ − β) v γ−2 |∇v|4 + n nβ M λ(γ − βq) v γ−β(q−1) |∇v|2 − nβ M γ + β(q + 2) − v γ+β(q+3) |∇v|2 . nβ M v γ |∇v|2 A =J + 2 M (8.8) For the term B, again using (8.3), we have 1 |∇v|2 + (−hv + λv 1−β(q−1) + v 1+β(q+3) ) , ∇v v β M |∇v|2 =(β + 1) ∇ , v γ ∇v v M h 1 − β(q − 1) − v γ |∇v|2 + λ v γ−β(q−1) |∇v|2 β M β M 1 + β(q + 3) v γ+β(q+3) |∇v|2 . + β M B= vγ ∇ (β + 1) ∇ |∇v|2 v Notice that M , v γ ∇v =− M |∇v|2 ∇ · (v γ ∇v) v v γ−2 |∇v|4 − = −γ M v γ−1 |∇v|2 ∆g v. M Therefore, v γ−2 |∇v|4 − (β + 1) B = − (β + 1)γ M v γ−1 |∇v|2 ∆g v M h 1 − β(q − 1) v γ |∇v|2 + λ v γ−β(q−1) |∇v|2 β M β M 1 + β(q + 3) + v γ+β(q+3) |∇v|2 . β M − Again we use (8.3) to reach at v γ−2 |∇v|4 B = − (β + 1)γ M v γ−1 |∇v|2 (β + 1) − (β + 1) M |∇v|2 v 1 (−hv + λv 1−β(q−1) + v 1+β(q+3) ) β M h 1 − β(q − 1) − v γ |∇v|2 + λ v γ−β(q−1) |∇v|2 β M β M 1 + β(q + 3) + v γ+β(q+3) |∇v|2 . β M − (β + 1) v γ−1 |∇v|2 EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 61 By simplifying the right hand side of the above identity, one gets v γ−2 |∇v|4 B = − (β + 1)(γ + β + 1) M v γ |∇v|2 − λq +h 2 v γ−β(q−1) |∇v|2 M (8.9) M v γ+β(q+3) |∇v|2 . + (q + 2) M 3 For the term C, we first observe that v γ ∆g (|∇v|2 ) = 4 ∆g (v γ )|∇v|2 . M M Therefore, C =− =− =− − − 5 1 |∇v|2 ∆g (v γ ) 2 M 1 |∇v|2 γv γ−1 ∆g v + γ(γ − 1)v γ−2 |∇v|2 2 M 1 |∇v|2 γv γ−1 |∇v|2 (β + 1) 2 M v 1 1 γv γ−1 |∇v|2 (−hv + λv 1−β(q−1) + v 1+β(q+3) ) 2 M β γ(γ − 1) v γ−2 |∇v|4 . 2 M In other words, C= γh 2β 6 v γ |∇v|2 − M γ 2β − γλ 2β v γ−β(q−1) |∇v|2 M v γ+β(q+3) |∇v|2 − M γ(γ + β) 2 (8.10) v γ−2 |∇v|4 . M 7 We now have enough information to treat (8.5). 8 8.3. The transformed equation. By using (8.8)-(8.10), one can see that (8.5) reduces to 2 J+ (β + 1) γ(γ + β) − (β + 1)(γ + β + 1) − n 2 v γ−2 |∇v|4 M h(γ − β) γh +h+ v γ |∇v|2 nβ 2β M λ(γ − βq) γλ + − − λq − v γ−β(q−1) |∇v|2 nβ 2β M γ + β(q + 2) γ + − + (q + 2) − v γ+β(q+3) |∇v|2 nβ 2β M + 9 v γ Ricg (∇v, ∇v) = 0. + M For simplicity, we rewrite Eq. (8.11) as v γ−2 |∇v|4 + J +a M +b M v M v γ (c|∇v|2 + Ricg (∇v, ∇v)) γ−β(q−1) 2 v γ+β(q+3) |∇v|2 = 0, |∇v| + d M (8.11) ˆ Q.A. NGO 62 where 1 n−1 2 γ 2 + (3β + 2)γ + 2 (β + 1) , 2 n λ(n + 2) γ n−1 b= − − 2q , 2n β n+2 n+2 γ n−1 c= +2 h, 2n β n+2 n+2 2(n − 1) γ d= (q + 2) − . 2n n+2 β a=− 1 2 Next we wish to describe the method used in the present paper. Our goal was to find β = 0 and γ ∈ R such that 4 5 6 7 8 9 10 0, a 3 2n Ricg n+2 17 18 19 20 21 22 23 24 25 26 27 28 29 −2(q + 2) δ 0, 0. d (8.12) n−1 n+2 g, (8.15) n−1 . n+2 (8.16) In view of (8.13), it is necessary to have δ (2(n − 1) − (n − 2)δ) n 15 16 h δ−2 and 13 14 Ricg +cg Having (8.12), our result follows easily since |∇v| = 0 since a 0 forces d > 0. Thus, the key point is a 0. To better serve this purpose, we set y = 1 + β1 and δ = − βγ where y = 1 and δ ∈ R. Thus the set of conditions in (8.12) becomes n−1 2 2 y − 2δy + δ 2 − δ 0, (8.13) n n−1 λ δ − 2q 0, (8.14) n+2 11 12 0, b 0 which is equivalent to 2(n − 1) . (8.17) n−2 With (8.17) in hand, one can see that (8.16) is automatically satisfied. Moreover, d > 0 provided δ 0. Thus, our set of conditions now reduces to (8.14), (8.15), and (8.17). Notice that if inequalities in (8.17) are strict, then we can always find some y = 1 verifying (8.13). 0 δ 8.4. Proof of Theorem 8.1. For the sake of clarity, we split our studying into four cases depending on the sign of h and λ. 8.4.1. The case h < 0. In this case, it is necessary to have λ < 0. Then the condition (8.14) and the lower bound for δ in (8.17) imply n−1 0 δ 2q . (8.18) n+2 Combining (8.17) and (8.18) gives 0 δ There are two possible sub-cases. min 2q n − 1 2(n − 1) , n+2 n−2 . (8.19) EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 1 2 3 4 5 6 7 8 n+2 Case 1. Suppose q n−2 . Then we claim that, with K = 0, we can always select δ such that it satisfies (8.14), (8.15), and (8.17). To this end, we notice that the right hand side of (8.15) is always non-negative. In order to see that (8.14), (8.15), and (8.17) hold, we have to select δ = 2(n−1) n+2 . Then we have to choose y such that Eq. (8.13) holds. However we n are left without many choices but one, that is, y = n+2 = 1. This is enough to serve our purpose since the left hand side of (8.13) equals − 8(n−1) (n+2)2 when y is equal to y = 11 12 13 14 15 16 17 18 19 20 21 22 24 n−1 (q − 1)h. n K= to make sure that (8.15), and (8.17) Similarly as above, then we may choose δ = 2q(n−1) n+2 hold. Now it is easy to find some y = 1 satisfying (8.13) since the solution of (8.13) is an qn interval. For example, when q = n+2 n , we may choose y = n+2 since the left hand side of (8.13) now equals 2q(n − 1)(q(n − 2) − (n + 2)) . (n + 2)2 Otherwise, we may choose y = 1 − vanishes. 2/n, in this case, the left hand side of (8.13) now 8.4.2. The case h = 0. Under this case, the right hand side of (8.15) vanishes, thus it is enough to take K = 0. Besides, it is necessary to have λ < 0. Therefore, δ must satisfy (8.19). It is now a simple task to find some δ and y = 1 verifying both conditions (8.14) and (8.17). 8.4.3. The case h > 0, λ 0. Again, in this context, δ has to satisfy (8.19). First we show that K = − n−1 h is enough. Indeed, this condition can be rewritten as n 2n Ricg n+2 23 25 n n+2 . n+2 . Then δ needs to satisfy (8.18). With this region for δ, the right Case 2. Suppose q < n−2 hand side of (8.15) is not smaller than 2(q − 1) n−1 n+2 hg. Thus, we select 9 10 63 h −2 n−1 n+2 g. (8.20) Under the condition (8.20), we have to select δ = 0. In order to see how could this choice of δ work, we just go back to (8.11) to get n−1 |∇v|4 (β + 1)2 n v2 M n−1 n−1 h |∇v|2 − λq v −β(q−1) |∇v|2 + n n M M n−1 + (q + 2) v β(q+3) |∇v|2 + Ricg (∇v, ∇v) = 0. n M M J− 26 (8.21) 29 Clearly, we have no choice but β = −1 or equivalently, y = 0. With this choice of y, we immediately see that the left hand side of (8.21) is non-negative. This forces ∇v = 0 thus giving us the desired result. 30 8.4.4. The case h > 0, λ > 0. Under this case, it follows from (8.14) and (8.17) that 27 28 31 2q n−1 n+2 32 In other words, it is necessary to have q 33 K= δ n+2 n−2 . 2(n − 1) . n−2 Our choice for K is that n−1 (q − 1)h. n ˆ Q.A. NGO 64 1 We will see how this condition is enough for our argument. 2 Case 1. Suppose q < way 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 n+2 n−2 . We rewrite the condition for Ricci curvature in the following 2n Ricg n+2 Thus, we may choose δ = 2q(n−1) n+2 . h 2q n−1 n−1 −2 n+2 n+2 g. Consequently, the conditions (8.14) and (8.15) clearly hold. Therefore, we may select y = 1 verifying (8.13) since δ ∈ (0, 2(n−1) n−2 ) as we have already done in the second case when h < 0. n+2 Case 2. Suppose q = n−2 . Then necessarily δ = 2(n−1) n−2 which verifies (8.14). The condition for Ricci curvature can be rewritten as 2n n−1 n−1 8h(n − 1) Ricg h 2 −2 g= g. n+2 n−2 n+2 n2 − 4 Thus, we can pick K = 4h(n−1) n(n−2) and clearly (8.15) holds. It suffices to find some y = 1 n+2 verifying (8.13). Due to the fact that q = n−2 , we only have one choice for y, that is, nq n+2 n y = n+2 . Thanks to q = n−2 , we immediately see that y = n−2 = 1. With this, the left hand side of (8.13) vanishes as required. 8.5. Proof of Theorem 8.1 completed. Finally let us assume that u is a smooth positive solution of Eq. (8.2). From our discussion above, we know that all inequalities in (8.12) are fulfilled. In fact, we have already shown that d > 0. Consequently, v γ+β(q+3) |∇v|2 = 0 18 M 19 which implies that v, hence u is a constant. 20 ACKNOWLEDGMENTS 21 26 The content of this unpretentious notes is based on four lectures given at IMS of the National University of Singapore during Winter School on Scalar Curvature and Related Problems in December 2014. The author wants to thank Professor Xingwang Xu and IMS for providing such a nice opportunity and lovely atmosphere during my stay. 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H A` N OI SITY, 21 E-mail address: nqanh@vnu.edu.vn 22 E-mail address: bookworm vn@yahoo.com [...]... solutions of the Einstein constraint equations and their globally hyperbolic maximal Cauchy developments in the sense that two distinct solutions of the Einstein constraint equations may generate isometric globally hyperbolic maximal Cauchy developments Nevertheless, as a first step, it is important to understand solutions of the constraint equations EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS. .. the Hamitonian constraint equation Note that here we denote ρ = T (e0 , e0 ) 4.2 Derivation of constraint equations: The momentum constraint This subsection is a continuation of the previous subsection where we showed in detail the derivation of the Hamiltonian constraint In this subsection, we derive the so-called momentum constraint equation (4.2) First, let us recall the Codazzi equation (1.13)... the constraint equations with scalar fields when the dimension n 3 In this section, we continue to use the conformal method to construct the constraint equations with scalar fields in the dimension n = 2 To the best of our knowledge, the first paper dealing with the Einstein equations in the two dimensional cases is [Mon86] Subsequently, some generalization of the corresponding situation for the Einstein Maxwell... vector field β 4.4 Construction of spacetimes via solutions of the vacuum Einstein constraint equations In this subsection, given a solution (g, K) of the constraint equations (4.1) and (4.2) on a manifold (M, g) of the dimension n, we shall construct an appropriate spacetime (M , g) of the dimension n + 1 such that the Einstein equation (2.4) holds in the vacuum case 4.4.1 Construction of spacetimes... (4.1) is known as the Hamiltonian constraint while Eq (4.2) is called the momentum constraint We spend this section to prove Proposition 4.1 4.1 Derivation of constraint equations: The Hamiltonian constraint This subsection is devoted to prove (4.1) while (4.2) will be considered in the next subsection Before we start, let us recall the Einstein equation without the cosmological constant Λ, that is 1 Ricg... cosmological constant Λ can be considered as a particular scalar field with potential Λ For this reason, we do not consider the cosmological term in our Einstein equation 2.3 Why do the Einstein equations describe the propagation of wavelike phenomena? In order to formulate the initial value problem for the Einstein equations as nonlinear wave equations, we first express the Einstein equations in terms... −T (·, e0 ) Note that in this context, we again denote τ = trg K 5.1 Conformal method As can be seen, the couple of constraint equations (5.1) and (5.2) for the initial data (M, g, K) consist of n + 1 equations (a scalar equation and an n-vector equation) with more than n + 1 unknowns (for example, the metric g consists of n(n+1)/2 components) Since the constraint equations form an under-determined system,... condition, see [CBruGer69] Theorem 3.2 Given smooth initial data (M, g, K) satisfying the constraint equations, there exists a smooth, maximal, globally hyperbolic Cauchy development of the initial data Consequently, there is a strong connection between the globally hyperbolic Cauchy development of the initial data and the Einstein constraint equations However, there is no one-to-one correspondence... terms on the right hand side of the preceding equation, i.e α the term g αi λα ,j + g αj λ,i , it is usually to assume that λα = 0, ∀α = 0, n This condition is preferred to the so-called Lorentz-harmonic coordinate gauge In the literature, this belongs to a set of a few condition that one can solve the Einstein equations 21 Thus, we have shown that the Einstein equations for g in the Lorentz-harmonic... gauge condition in time We now consider how could the gauge condition propagate in time so long as the metric g solves the reduced equations As we shall see, all λα satisfy a homogeneous linear wave equation which is a consequence of the Bianchi identities with vanishing initial time derivatives which come from the constraint equations First, recall that in the vacuum setting, the Einstein equation is ... the Einstein equations Constraints and evolutions 4.1 Derivation of constraint equations: The Hamiltonian constraint 4.2 Derivation of constraint equations: The momentum constraint 4.3 Evolutions... understand solutions of the constraint equations EINSTEIN CONSTRAINT EQUATIONS ON RIEMANNIAN MANIFOLDS 15 C ONSTRAINTS AND EVOLUTIONS As we have already mentioned in the previous section, the initial... Codazzi equations The Einstein equations in general relativity 2.1 The Einstein equations 2.2 The Einstein equations with real scalar fields 2.3 Why the Einstein equations describe the propagation