RICCI CURVATURE, CONSTRAINTS AND EVOLUTION. 1. Ricci and scalar curvatures. The Ricci quadratic form is the trace of the Jacobi operator: X Ric(v; v) = trRv = ahR(ea; v)v; eai; a = hea; eai; a where (ea) is an orthonormal basis. In the Riemannian case, this is a sum of sectional curvatures along two-planes containing v (if jvj = 1): X Ric(v; v) = σeiv: ei In the Lorentzian case: Letting v be one of the basis vectors e0; e1; : : : ; en, we have (with indices i; j; k running from 1 to n): X Ric(e0; e0) = − σe0ei ; i X X X X Ric(ej; ej) = ihR(ei; ej)ej; eii = σeaej = σe0ei + σeiej : i a6=j i i6=j Thus Ric(e0; e0) equals minus a sum of sectional curvatures in the planes containing e0. So Ric(e0; e0) ≥ 0 corresponds to timelike geodesics which are, on average, focusing (attracting). On the other hand, Ric(ei; ei) is a sum of sectional curvatures along timelike and spacelike two-planes. The scalar curvature R is the trace of the (1,1) Ricci operator, hence: X X R = aRic(ea; ea) = σeaeb : a a6=b (Note each two-plane occurs twice in this sum.) The Einstein tensor is defined as: 1 G(v; v) = Ric(v; v) − hv; viR: 2 Exercise 1. In the Lorentzian case show that, for a Lorentz-orthonormal basis (ea): X X X (i)G(e0; e0) = σeiej G(ei; ei) = − σe0ej − σeiej : i<j j6=i i6=j<k6=i 1 For example, in n + 1 = 4 spacetime dimensions: G(e0; e0) = σe1e2 + σe1e3 + σe2e3 ;G(e2; e2) = −(σe0e1 + σe0e3 + σe1e3 ): In the Riemannian case, verify that: X G(e0; e0) = − σekej : i<j The Einstein equation Gij = κTij (where Tij is the energy-momentum tensor of the matter distribution), combined with the dominant energy con- dition (which states in part that T (e0; e0) ≥ 0) implies then that the sum of sectional curvatures along spacelike two-planes is non-negative (average spacelike focusing.) 2. Splitting of the curvature tensor. For a codimension one subman- ifold M ⊂ M¯ (M¯ Riemannian or Lorentzian) we have derived expressions for the components hR¯(X; Y )Z; W i (Gauss equation) and hR¯(X; Y )Z; ni (Co- dazzi equation), where n is a unit normal vector field (timelike or spacelike). We now turn to hR¯(n; X)n; Y i (symmetric in X; Y 2 TM). We consider the following more general situations: (i) let V be an open neighborhood of M in M¯ , diffeomorphic to the product M × I, where I ⊂ R is an open interval containing 0. Thus V is foliated by manifolds fMtgt2I diffeomorphic to M = M0. In the Lorentzian case we assume t is a \time function" in V (future-directed timelike gradient), so its level sets Mt are spacelike submanifolds. @t is a (timelike) vector field in M, transversal to the level sets Mt. (ii) An embedding F : M × I ! V is given, and defines a vector field @t := @F=@t on V . In either case, in terms of the unit normal n(x; t) to Mt at F (x; t) we have the decomposition: @t = λn + σ; λ : V ! R+; σ(x; t) 2 TF (x;t)Mt: The function λ = h@t; ni > 0 is called the lapse, the vector field σ the shift defined by @t (or by the embedding F ). Here = hn; ni = ±1, and we assume n future-directed in the Lorentzian case. Fix p 2 M, and let (xi; t) be a local coordinate system in M × I (or in V , adapted to the foliation), so that all covariant derivatives r@i @j and 2 ¯ the Lie brackets [@t;@i] vanish at p (but not r@t @i) . We first compute hR¯(@t;@i)n; @ji at p. Recall the definition of the scalar second fundamental form of Mt in M¯ : r¯ X Y = rX Y + A(X; Y )n; A(X; Y ) = hr¯ X Y; ni = −hr¯ X n; Y i: We have: ¯ ¯ ¯ ¯ ¯ hr@t (r@i n);@ji = @thr@i n; @ji − hr@i n; r@t @ji 2 = −@t(A(@i;@j)) − λA (@i;@j) + A(@i; r@j σ): Exercise 2. Derive this last equality: use the decomposition of @t and the definition: 2 X A (X; Y ) = A(X; ek)A(ek;Y ); (ek) orthonormal frame on Mt k to show: ¯ ¯ 2 hr@i n; r@j @ti = λA (@i;@j) − A(@i; r@j σ): Exercise 3. Also, at p: ¯ ¯ ¯ ¯ 2 hr@i (r@t n);@ji = @ihr@t n; @ji = −@ihn; r@j @ti = −∂i;jλ − @i(A(@j; σ)): Combining the two, we find (at p): ¯ 2 2 hR(@t;@i)n; @ji = −@t(A(@i;@j)) − λA (@i;@j) + ∂i;jλ +(r@i A)(@j; σ) + A(@i; r@j σ) + A(@j; r@i σ): Now use the Codazzi equation to compute: λhR¯(n; @i)n; @ji = hR¯(@t;@i)n; @ji − hR¯(σ; @i)n; @ji ¯ = hR(@t;@i)n; @ji + (rσA)(@i;@j) − (r@i A)(σ; @j) 2 = −@t(A(@i;@j)) + (LσA)(@i;@j) − λA (@i;@j) + Hess(λ)(@i;@j): Here Hess(λ)(X; Y ) := hrX rλ, Y i is the Hessian quadratic form of λ on Mt, and Lσ denotes the Lie derivative along σ. Recall that, on vector fields, LX Y = [X; Y ], and the Lie derivative is extended to tensors as a derivation. For instance: (LX A)(Y; Z) := X(A(Y; Z)) − A(LX Y; Z) − A(Y; LX Z): 3 Exercise 4. Show that, for any X; Y; Z 2 TMt and smooth function f on V : (i)(LZ A)(X; Y ) = (rZ A)(X; Y ) + A(rX Z; Y ) + A(X; rY Z): (ii)(LfZ A)(X; Y ) = f(LZ A)(X; Y ) + (Xf)A(Z; Y ) + (Y f)A(X; Z): (Part (i) was used in the last equality above.) Thus we have the equality of symmetric 2-tensors on M¯ : ¯ 2 λhR(n; X)n; Y i = −[L@t−σA + λA − Hess(λ)](X; Y ): (Compare with (14, p.5) in [Bartnik-Isenberg], where their K = −A; note however that, as seen above, the Lie derivative is not linear over smooth functions.) We can state this in the form of a \Riccatti equation" for A. Given an embedding of M ×I into V ⊂ M¯ and the associated lapse function and shift 1 vector field, with n = λ (@t − σ) the unit normal: 1 L A + A2 − λ−1Hess(λ) − hR¯(·; n)n; ·i = 0: λ @t−σ The simplest instance is a hypersurface in a Riemannian space of con- stant curvatureκ ¯, and an embedding with unit lapse and zero shift: 2 LnA + A − κg¯ = 0; where g is the induced metric on M. n+1 Exercise 4. Verify this identity for the unit sphere in R . For a \normal variation" (Ft)t2I of M in M¯ (Riemannian), we have @tF = λn (so the shift is zero), and we have: 2 ¯ L@t A + λA − Hess(λ) − hR(·; n)n; ·i = 0: This will be used later in the Second Variation Formula for the volume of a hypersurface. 3. The constraint and evolution equations. Consider again the situations of the previous paragraph (Riemannian or Lorentzian settings.) Recall the Gauss formula: V M Riem = Riem + nA A: 4 Fix p 2 M, and let (ei) be an orthonormal frame diagonalizing the shape operator S at p; set e0 = n, Sei = κiei. Then: V M σeiej = σeiej − nκiκj: From the expression for G(e0; e0) given in exercise 1 above, it follows that: X V M X M X 2 X 2 2G(e0; e0) = −2n σeiej = −nR +2 κiκj = −nR +( κi) − κi ; i<j i<j i i or: V V M 2 2 2Ric (e0; e0) − nR = 2G(e0; e0) = −nR + H − jAj ; the first constraint equation. (Here H = trgS is the mean curvature of M in M¯ ). It is useful to state separately the case of hypersurfaces in a Riemannian manifold: 2RicV (n; n) = RV − RM + H2 − jAj2 and the case of spacelike hypersurfaces in a Lorentzian manifold: M 2 2 2G(e0; e0) = R + H − jAj : The second constraint equation is obtained by computing G(e0; ei) and using the Codazzi equation: V X V V G(e0; ei) = Ric (e0; ei) = hR (ej; e0)ei; eji = hR (ei; ej)ej; e0i j6=i X M M = [(rei A)(ei; ej) − (rej A)(ei; ej)] j6=i X M M = [(rei A)(ei; ej) − (rej A)(ei; ej)] j M M = hr H − Div S; eii; where S is the \shape operator" of M in V . Thus the tangential component of the vector field RicV (n) is given by: [RicV (n)]tan = rM H − DivM S: (The Riemannian and Lorentzian cases are identical.) 5 The spacetime Einstein equations can be formally recast as a dynamical system in the space of Riemannian metrics on a fixed manifold M (more precisely, in its tangent bundle), provided we're given an embedding F from M × I to an open subset of the solution spacetime M¯ ; equivalently, a choice of lapse function and shift vector field on V (both time-dependent). The \dynamical system" is given by first-order tensorial ODEs for the metric g(t) induced on M and the second fundamental form A(t) of M in M¯ . The equation for the metric follows from the definition of second funda- mental form: 1 A(@ ;@ ) = −hr¯ n; @ i = − [hr @ ;@ i − hr σ; @ i] i j @i j λ @i t j @i j 1 1 = − [hr @ ;@ i − hr σ; @ i] = − (L g − L g)(@ ;@ ): λ @t i j @i j 2λ @t σ i j This gives the tensorial evolution equation (for a given lapse and shift): L@t g = Lσg − 2λA.