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 |v| = 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

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

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 ∈ 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 {Mt}t∈I 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) ∈ 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 ∈ 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 ∇∂i ∂j and

2 ¯ the Lie brackets [∂t, ∂i] vanish at p (but not ∇∂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¯ :

∇¯ X Y = ∇X Y + A(X,Y )n, A(X,Y ) = h∇¯ X Y, ni = −h∇¯ X n, Y i.

We have: ¯ ¯ ¯ ¯ ¯ h∇∂t (∇∂i n), ∂ji = ∂th∇∂i n, ∂ji − h∇∂i n, ∇∂t ∂ji

2 = −∂t(A(∂i, ∂j)) − λA (∂i, ∂j) + A(∂i, ∇∂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 h∇∂i n, ∇∂j ∂ti = λA (∂i, ∂j) − A(∂i, ∇∂j σ).

Exercise 3. Also, at p: ¯ ¯ ¯ ¯ 2 h∇∂i (∇∂t n), ∂ji = ∂ih∇∂t n, ∂ji = −∂ihn, ∇∂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

+(∇∂i A)(∂j, σ) + A(∂i, ∇∂j σ) + A(∂j, ∇∂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 + (∇σA)(∂i, ∂j) − (∇∂i A)(σ, ∂j) 2 = −∂t(A(∂i, ∂j)) + (LσA)(∂i, ∂j) − λA (∂i, ∂j) + Hess(λ)(∂i, ∂j).

Here Hess(λ)(X,Y ) := h∇X ∇λ, 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 ∈ TMt and smooth function f on V :

(i)(LZ A)(X,Y ) = (∇Z A)(X,Y ) + A(∇X Z,Y ) + A(X, ∇Y 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)t∈I 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 ∈ 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 , imean 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 − |A|2 and the case of spacelike hypersurfaces in a Lorentzian manifold:

M 2 2 2G(e0, e0) = R + H − |A| .

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 = [(∇ei A)(ei, ej) − (∇ej A)(ei, ej)] j6=i

X M M = [(∇ei A)(ei, ej) − (∇ej A)(ei, ej)] j M M = h∇ 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 = ∇M 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(∂ , ∂ ) = −h∇¯ n, ∂ i = − [h∇ ∂ , ∂ i − h∇ σ, ∂ i] i j ∂i j λ ∂i t j ∂i j 1 1 = − [h∇ ∂ , ∂ i − h∇ 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.

To obtain the evolution equation for A, recall the expression obtained earlier: V 2 L∂t A = nHess(λ) + LσA + λhR (·, n)n, ·i − λA . From the definition of Ricci curvature:

V V X V nhR (ei, n)n, eji = Ric (ei, ej) − Riem (ei, ek, ek, ej). k

V M To compute the sum, we use the Gauss formula Riem = Riem +nA A:

X V M X Riem (ei, ek, ek, ej) = Ric (ei, ej)+n [A(ei, ek)A(ek, ej)−A(ek, ek)A(ei, ej)] k k

M 2 = Ric (ei, ej) + n[A (ei, ej) − HA(ei, ej)]. Thus:

V V M 2 nhR (ei, n)n, eji = Ric (ei, ej) − Ric (ei, ej) − n[A (ei, ej) − HA(ei, ej)].

In the Lorentzian case, substituting in the above we obtain the tensorial evolution equation (for given lapse and shift):

V M 2 L∂t A = LσA − Hess(λ) + λ[−Ric + Ric − 2A + HA].

6 (Compare [Bartnik-Isenberg], (32), p.9. Note that the term RicV is given by the Einstein equations, and vanishes for the vacuum Einstein equations.) In the Riemannian case, consider a normal variation of the hypersurface M (thus σ = 0, ∂tF = λn). We find the equations (for symmetric two- tensors g and A):

V M 2 L∂t g = −2λA, L∂t A = Hess(λ) + λ[Ric − Ric − 2A + HA].

Taking traces (on TM):

X V M 2 2 ∂tH = ∆M λ + λ[ Ric (ei, ei) − R − 2|A| + H ]. i (λ is the normal velocity.)

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