A Fundamental Theorem for Hypersurfaces in Semi-Riemannian Warped Products
aa r X i v : . [ m a t h . DG ] J a n A Fundamental Theorem for Hypersurfaces inSemi-Riemannian Warped Products
Marie-Am´elie Lawn and Miguel OrtegaSeptember 24, 2018
Abstract
We give necessary and sufficient conditions for a semi-Riemannian manifold of arbi-trary signature to be locally isometrically immersed into a warped product ± I × a M n ( c ),where I ⊂ R and M n ( c ) is a semi-Riemannian space of constant nonzero sectional cur-vature. Then, we describe a way to use the structure equations of such immersions toconstruct foliations of marginally trapped surfaces in a four-dimensional Lorentzian space-times. We point out that, sometimes, Gauß and Codazzi equations are not sufficient toensure the existence of a local isometric immersion of a semi-Riemannian manifold asa hypersurface of another manifold. We finally give two low-dimensional examples toillustrate our results. One of the fundamental problems in submanifold theory deals with the existence of isomet-ric immersions from one manifold into another. The Gauß, Ricci and Codazzi equationsare very well-known as the structure equations , meaning that any submanifold of any semi-Riemannian manifold must satisfy them. A classical result states that, conversely, they arenecessary and sufficient conditions for a Riemannian n -manifold to admit a (local) immer-sion in the Euclidean ( n + 1)-space. In addition, E. Cartan developed the so-called movingframes technique, obtaining a necessary and sufficient condition to construct a map from a(differential) manifold M into a Lie group. If a Lie group G is a group of diffeomorphisms ofa manifold P , Cartan’s technique then may provide a map from M to P with nice properties.Sometimes, the map from M to G can exist thanks to Gauß, Codazzi and Ricci equations,like for instance in [4].Another point of view is the celebrated Nash Theorem, which states that any Riemannmanifold can be embedded in the Euclidean space, but at the price of a high codimension.Following that line, O. M¨uller and M. S´anchez obtained a characterization of the Lorenztianmanifolds which can be embedded in a high dimensional Minkowski space (see [5].)On the other hand, B. Daniel obtained in [2] a fundamental theorem for hypersurfacesin the Riemannian products S n × R and H n × R , looking for tools to work with minimalsurfaces in such manifolds when n = 2. J. Roth generalized B. Daniel’s theorem to spacelikehypersurfaces in some Lorentzian products (see [7]). In their works, they needed some extratools such as a tangent vector field T to the submanifold and some functions, in order toobtain the local metric immersions into the desired ambient spaces. Note that the Ricciequation provides no information for hypersurfaces.1ur main aim is to obtain a fundamental theorem for non-degenerate hypersurfaces in asemi-Riemannian warped product, namely ε I × a M nk ( c ), where ε = ± a : I ⊂ R → R + is thescale factor and M nk ( c ) is the semi-Riemannian space form of index k and constant curvature c = ±
1. For a hypersurface M in ε I × a M nk ( c ), the vector field ∂ t ( t ∈ I ) decomposes in itstangent and normal parts, i. e., ∂ t = T + ε n +1 T n +1 e n +1 where e n +1 is a (local) normal unitvector field, ε n +1 = ± T n +1 is the corresponding coordinate.In addition to the shape operator A , on Gauß and Codazzi equations there appear the vectorfield T , its dual 1-form η , some constants as well as some functions like T n +1 . However, thecovariant derivative of T must satisfy a specific formula, which cannot be obtained from Gaußand Codazzi equations by the authors. Based on these necessary conditions, we state in Def-inition 1 all needed tools on an abstract semi-Riemannian manifold M , for the existence of a(local) metric immersion χ : U ⊂ M → ε I × a M nk ( c ) (see Theorem 1.) Later, we apply this re-sult to non-degenerate hypersurfaces of a Friedman-Lemˆetre-Robertson-Walker 4-spacetimes(RW 4-spacetimes.) In Corollary 2, we show sufficient conditions for such hypersurfaces toexist.We would like to point out that our computations, as well as B. Daniel and J. Roth’sresults, show that Gauß and Codazzi equations are not sufficient to ensure the existence of alocal isometric immersion of a given Riemannian manifold endowed with a second fundamentalform in a spacetime as a spacelike hypersurface.Next, if we admit in a very wide sense that a horizon in a 4-spacetime is a 3-dimensionalhypersurface which is foliated by marginally trapped surfaces (i. e., surfaces whose meancurvature vector is timelike), then we describe a condition to obtain non-degenerate horizonsin RW 4-spacetimes in our framework (see Corollary 3).We end the paper with two low-dimensional examples to illustrate the theoretical results.The first one describes a surface in a RW toy model S → − I × a S , a (simple) graph over arest space { t } × S . The second example is a helicoidal surface in − I × a H . Let (
P, g P ) be a semi-Riemannian manifold of dimension dim P = m . We consider a smoothfunction a : I ⊂ R → R + , a (sign) constant ε = ± P m +1 = εI × a P, h , i = ε d t + a ( t ) g P . Clearly, the unit vector field ∂∂t = ∂ t will play a crucial role on the manifold ¯ P n +1 . We willuse the following convention for the curvature operator R of a connection D : R ( X, Y ) Z = D X D Y Z − D Y D X Z − D [ X,Y ] Z. Let ¯ R P and R P be the curvature operator of ¯ P n +1 and P , respectively. Let D be the Levi-Civita connection of ¯ P n +1 . We recall the following formulae from [6]. Lemma 1.
On the semi-Riemannian manifold ¯ P n +1 , the following statements hold, for any V, W lifts of vector fields tangent to P :1. D ∂ t ∂ t = 0 , D V ∂ t = a ′ a V ,2. grad( a ) = εa ′ ∂ t . . D V W = ∇ PV W − εa ′ a h V, W i ∂ t . ¯ R P ( V, ∂ t ) ∂ t = − a ′′ a V , ¯ R P ( ∂ t , V ) W = − ε a ′′ a h V, W i ∂ t , ¯ R P ( V, W ) ∂ t = 0 .Proof. Note that the definition of the curvature operator on [6] has the opposite sign thanthe usual one. We show a proof of item (4). By recalling D ∂ t ∂ t = 0 and [ V, ∂ t ] = 0, we have¯ R P ( V, ∂ t ) ∂ t = D V D ∂ t ∂ t − D ∂ t D ∂ t V − D [ V,∂ t ] ∂ t = − D ∂ t D ∂ t V = − D ∂ t (cid:16) a ′ a V (cid:17) = − a ′′ a − ( a ′ ) a V − a ′ a D ∂ t V = − a ′′ a V. Next, we show ¯ R P ( ∂ t , V ) W = − h V,W i a D ∂ t (grad a ) = − h V,W i a D ∂ t ( εa ′ ∂ t ) = − ε h V,W i a ′′ a ∂ t . Finally, ¯ R P ( V, W ) ∂ t = 0 is a direct consequence of item 1.Now, let M be a non-degenerate hypersurface of ¯ P m +1 , with ∇ M its Levi-Civita con-nection, σ the second fundamental form and R M the curvature operator of M , respectively.Given a (local) unit normal vector field ν of M in ¯ P m +1 , with δ = h ν, ν i = ±
1, let A be theshape operator associated with ν . The Gauß and Weingarten’s formulae are D X Y = ∇ M X Y + σ ( X, Y ) , D X ν = −A X, for any X, Y ∈ T M . The second fundamental form can be written as σ ( X, Y ) = δ hA X, Y i ν, for any X, Y ∈ T M . Recall that the mean curvature vector of M is defined by ~H = 1dim( M ) Tr( σ ) . Next, the Codazzi equation of M takes the general form ( ¯ R P ( X, Y ) Z ) ⊥ = ( D X σ )( Y, Z ) − ( D Y σ )( X, Z ), for any
X, Y, Z tangent to M , which is equivalent to¯ R P ( X, Y, Z, ν ) = h ( D X A ) Y − ( D Y A ) X, Z i , (1)for any X, Y, Z ∈ T M . Further, the general Gauß equation is given by¯ R P ( X, Y, Z, W ) = R M ( X, Y, Z, W ) − h σ ( Y, Z ) , σ ( X, W ) i + h σ ( Y, W ) , σ ( X, Z ) i (2)= R M ( X, Y, Z, W ) − δ hA Y, Z ihA
X, W i + δ hA Y, W ihA
X, Z i , with X, Y, Z, W tangent to M .We consider now the special case where the manifold P = E n +1 = R n +1 k , i. e., the standardEuclidean semi-Riemannian space of dimension n + 1 ≥ k . Following the previousnotation, we construct e P n +2 = ε I × a E n +1 . Let e R be the curvature tensor of e P n +2 . We have Proposition 1.
Let
X, Y, Z, W ∈ Γ( T e P n +2 ) . e R ( X, Y, Z, W ) = ε ( a ′ ) a (cid:16) h X, Z ih Y, W i − h
Y, Z ih X, W i (cid:17) + (cid:18) a ′′ a − ( a ′ ) a (cid:19) (cid:16) h X, Z ih Y, ∂ t ih W, ∂ t i − h Y, Z ih X, ∂ t ih W, ∂ t i−h X, W ih Y, ∂ t ih Z, ∂ t i + h Y, W ih X, ∂ t ih Z, ∂ t i (cid:17) . roof. Let X = ˜ X + x∂ t = ˜ X + ε h X, ∂ t i ∂ t , where ˜ X is a vector field tangent to e P n +2 .We will use similar notations for other vector fields. In particular, we see that h ˜ X, ˜ Y i = h X, Y i − ε h X, ∂ t ih Y, ∂ t i . By using the symmetry properties of the curvature tensor, we get˜ R ( X, Y, Z, W ) = ˜ R ( ˜ X, ˜ Y , ˜ Z, ˜ W ) + ˜ R ( ˜ X, ˜ Y , ˜ Z, w∂ t ) + ˜ R ( ˜ X, ˜ Y , z∂ t , ˜ W )+ ˜ R ( ˜ X, y∂ t , ˜ Z, ˜ W ) + ˜ R ( ˜ X, y∂ t , ˜ Z, w∂ t ) + ˜ R ( ˜ X, y∂ t , z∂ t , ˜ W )+ ˜ R ( x∂ t , ˜ Y , ˜ Z, ˜ W ) + ˜ R ( x∂ t , ˜ Y , ˜ Z, w∂ t ) + ˜ R ( x∂ t , ˜ Y , z∂ t , ˜ W )By Lemma 1, we obtain directly˜ R ( ˜ X, ˜ Y , z∂ t , ˜ W ) = 0 , ˜ R ( ˜ X, ˜ Y , ˜ Z, w∂ t ) = − ˜ R ( ˜ X, ˜ Y , w∂ t , ˜ Z ) = 0 , ˜ R ( x∂ t , ˜ Y , ˜ Z, ˜ W ) = ˜ R ( ˜ Z, ˜ W , x∂ t , ˜ Y ) = 0 , ˜ R ( ˜ X, y∂ t , ˜ Z, ˜ W ) = − ˜ R ( ˜ Z, ˜ W , y∂ t , ˜ X ) = 0 . Since the curvature tensor of E n +1 vanishes, by [6, p. 210], we get˜ R ( ˜ X, ˜ Y , ˜ Z, ˜ W ) = − ε ( a ′ ) a ( h ˜ Y , ˜ Z ih ˜ X, ˜ W i − h ˜ X, ˜ Z ih ˜ Y , ˜ W i ) . Moreover, with Lemma 2.2, using again Lemma 1.4 and as h ∂ t , ∂ t i = ε ,˜ R ( ˜ X, y∂ t , ˜ Z, w∂ t ) = − ˜ R ( y∂ t , ˜ X, ˜ Z, w∂ t ) = ε a ′′ a h ˜ X, ˜ Z ih y∂ t , w∂ t i = a ′′ a h Y, ∂ t ih W, ∂ t ih ˜ X, ˜ Z i . By similar computations, we obtain˜ R ( X, Y, Z, W ) = − ε ( a ′ ) a (cid:16) h ˜ Y , ˜ Z ih ˜ X, ˜ W i − h ˜ X, ˜ Z ih ˜ Y , ˜ W i (cid:17) + a ′′ a (cid:16) h Y, ∂ t ih W, ∂ t ih ˜ X, ˜ Z i − h Y, ∂ t ih Z, ∂ t ih ˜ X, ˜ W i (3) − h X, ∂ t ih W, ∂ t ih ˜ Y , ˜ Z i + h X, ∂ t ih Z, ∂ t ih ˜ Y , ˜ W i (cid:17) . Now, straightforward computations yield h ˜ Y , ˜ Z ih ˜ X, ˜ W i − h ˜ X, ˜ Z ih ˜ Y , ˜ W i + h Y, ∂ t ih Z, ∂ t ih X, ∂ t ih W, ∂ t i − h X, Z ih Y, W i + ε h X, Z ih Y, ∂ t ih W, ∂ t i + ε h X, ∂ t ih Z, ∂ t ih Y, W i − h
X, ∂ t ih Z, ∂ t ih Y, ∂ t ih W, ∂ t i = h Y, Z ih X, W i − ε h Y, Z ih X, ∂ t ih W, ∂ t i − ε h Y, ∂ t ih Z, ∂ t ih X, W i − h
X, Z ih Y, W i + ε h X, Z ih Y, ∂ t ih W, ∂ t i + ε h X, ∂ t ih Z, ∂ t ih Y, W i , and h Y, ∂ t ih W, ∂ t ih ˜ X, ˜ Z i − h Y, ∂ t ih Z, ∂ t ih ˜ X , ˜ W i − h X, ∂ t ih W, ∂ t ih ˜ Y , ˜ Z i + h X, ∂ t ih Z, ∂ t ih ˜ Y , ˜ W i = h Y, ∂ t ih W, ∂ t ih X, Z i − h
Y, ∂ t ih Z, ∂ t ih X, W i − h
X, ∂ t ih W, ∂ t ih Y, Z i + h X, ∂ t ih Z, ∂ t ih Y, W i . By inserting in (3), we finally get the result.Let M nk ( c ) be the semi-Riemannian space form of constant sectional curvature c = ± k , with metric g and let ¯ P n +1 = ε I × a M nk ( c ), with metric h , i . We denote by ¯ R thecurvature operator of ε I × a M nk ( c ). Also, we put E n +1 = ( R n +1 k , if M nk ( c ) = S nk , c = +1 , R n +1 k +1 , if M nk ( c ) = H nk , c = − , g o and Levi-Civita connection ∇ o . We recall that S nk = { p ∈ E n +1 : g o ( p, p ) = +1 } , H nk = { p ∈ E n +1 : g o ( p, p ) = − } . From the usual totally umbilical embedding Ξ : M nk ( c ) → E n +1 , we construct the followingisometric embedding e Ξ : ( ε I × a M nk ( c ) , h , i ) −→ ( εI × a E n +1 , h , i ) , ( t, p ) ( t, Ξ( p )) . In the sequel, for the sake of simplicity, we will also use the notation h , i = h , i . Let e ∇ and¯ ∇ be the Levi-Civita connection on ε I × a E n +1 and ε I × a M nk ( c ), respectively. It is well-known that ξ = Ξ /c is a unit normal vector field satisfying ∇ oX ξ = X/c for any X tangent to T p M nk ( c ). Thus, we can consider the normal vector field of ˜Ξ : ε I × a M nk ( c ) → ε I × a E n +1 as e ( t, p ) = (0 , ξ ( p ) /a ( t )) = (cid:0) , p/ ( c a ( t )) (cid:1) , for any ( t, p ) ∈ ε I × a M nk ( c ) . We also set ε = h e , e i = ±
1. Since M nk ( c ) lies naturally in E n +1 , the normal vec-tor field ξ satisfies g o ( ξ, ξ ) = c . In addition, ε = h e , e i = h (0 , p/ ( ac )) , (0 , p/ ( ac )) i = a g o ( p, p ) / ( a c ) = c . In this way, by Lemma 1, e ∇ ∂ t e = − a ′ a (0 , ξ ) + 1 a e ∇ ∂ t (0 , ξ ) = − a ′ a (0 , ξ ) + 1 a a ′ a (0 , ξ ) = 0 . Moreover, if (0 , Z ) ⊥ e , then Z ⊥ ξ , so that e ∇ (0 ,Z ) e = 1 a e ∇ (0 ,Z ) (0 , ξ ) = 1 a (cid:16) ∇ Z ξ − h (0 , Z ) , (0 , ξ ) i a grad a (cid:17) = 1 ac (0 , Z ) . This means that the Weingarten operator S associated with e has the expression SY = − ac ( Y − ε h Y, ∂ t i ∂ t ) , for any Y ∈ T ¯ P n +1 . (4) Proposition 2.
The curvature tensor of ¯ P n +1 = ε I × a M nk ( c ) is ¯ R ( X, Y, Z, W ) = (cid:18) ε ( a ′ ) a − ε a (cid:19) (cid:16) h X, Z ih Y, W i − h
Y, Z ih X, W i (cid:17) + (cid:18) a ′′ a − ( a ′ ) a + εε a (cid:19) (cid:16) h X, Z ih Y, ∂ t ih W, ∂ t i − h Y, Z ih X, ∂ t ih W, ∂ t i (5) − h X, W ih Y, ∂ t ih Z, ∂ t i + h Y, W ih X, ∂ t ih Z, ∂ t i (cid:17) , for any X, Y, Z, W in T ¯ P n +1 .Proof. We just need to resort to (2), Proposition 1 and (4).
Let M n , n ≥
2, be an immersed, non-degenerate hypersurface in ¯ P n +1 . Let ∇ be the Levi-Civita connections on M . Let e n +1 be a (locally defined) normal unit vector field to M , with5 n +1 = h e n +1 , e n +1 i = ±
1. Along M , the vector field ∂ t can be decomposed as its tangentand normal parts, i. e., ∂ t = T + f e n +1 , where T is tangent to M and f = ε n +1 h ∂ t , e n +1 i . We also define the 1-form on M given by η ( X ) = h X, T i , for any X ∈ Γ( T M ). Given a tangent vector X to M , we again decompose itin the part tangent to { t } × M nk ( c ) and its component in the direction of ∂ t as X = ˜ X + x∂ t .Similarly, Y = ˜ Y + y∂ t and e n +1 = ˜ e n +1 + n∂ t . Lemma 2.
Under the previous conditions,1. εn = h e n +1 , ∂ t i = ε n +1 f , εx = h X, ∂ t i = h X, T i , εy = h Y, ∂ t i = h Y, T i ,2. h ˜ X, ˜ e n +1 i = − εε n +1 f h X, T i ,3. h ˜ X, ˜ Y i = h X, Y i − ε h X, T ih Y, T i .Proof. From e n +1 = ˜ e n +1 + n∂ t , it is immediate that εn = h ∂ t , e n +1 i = ε n +1 f . Further, h X, ∂ t i = h X, T + f e n +1 i = h X, T i = h ˜ X + x∂ t , ∂ t i = εx . Next, 0 = h X, e n +1 i = h ˜ X, ˜ e n +1 i + xnε = h ˜ X, ˜ e n +1 i + εε n +1 f h X, T i . Finally, h X, Y i = h ˜ X + x∂ t , ˜ Y + y∂ t i = h ˜ X, ˜ Y i + εxy = h ˜ X, ˜ Y i + ε h X, T ih Y, T i .Let A be the shape operator of M associated with e n +1 . Proposition 3.
The Gauß equation of M in εI × a M nk ( c ) is R ( X, Y, Z, W ) = (cid:18) ε ( a ′ ) a − ε a (cid:19) (cid:16) h X, Z ih Y, W i − h
Y, Z ih X, W i (cid:17) + (cid:18) a ′′ a − ( a ′ ) a + εε a (cid:19) (cid:16) h X, Z ih Y, T ih W, T i − h
Y, Z ih X, T ih W, T i−h
X, W ih Y, ∂ t ih Z, T i + h Y, W ih X, T ih Z, T i (cid:17) + ε n +1 (cid:16) h AY, Z ih AX, W i − h
AY, W ih AX, Z i (cid:17) , for any X, Y, Z, W ∈ T M .Proof.
We resort to (2), Proposition 2 and Lemma 2.
Proposition 4.
The Codazzi equation of M in ε I × a M nk ( c ) is given by ( ∇ X A ) Y − ( ∇ Y A ) X = ε n +1 f (cid:18) a ′′ a − ( a ′ ) a + εε a (cid:19) (cid:16) h Y, T i X − h X, T i Y (cid:17) , (6) for any X, Y tangent to M . roof. By (1), we have to compute ¯ R ( X, Y, Z, e n +1 ) for any tangent vectors X, Y, Z to M .To do so, we recall Proposition 2. Thus,¯ R ( X, Y, Z, e n +1 ) = (cid:18) ε ( a ′ ) a − ε a (cid:19) (cid:16) h X, Z ih Y, e n +1 i − h Y, Z ih X, e n +1 i (cid:17) + (cid:18) a ′′ a − ( a ′ ) a + εε a (cid:19) (cid:16) h X, Z ih Y, ∂ t ih e n +1 , ∂ t i − h Y, Z ih X, ∂ t ih e n +1 , ∂ t i− h X, e n +1 ih Y, ∂ t ih Z, ∂ t i + h Y, e n +1 ih X, ∂ t ih Z, ∂ t i (cid:17) = ε n +1 f (cid:18) a ′′ a − ( a ′ ) a + εε a (cid:19) (cid:16) h Y, T ih X, Z i − h
X, T ih Y, Z i (cid:17) . This yields the result.
Lemma 3.
The following equations hold for any X tangent to M :1. ¯ ∇ X ∂ t = a ′ a ( X − ε h X, T i ∂ t ) .2. ∇ X T = a ′ a ( X − ε h X, T i T ) + f AX .3. X ( f ) = − ε n +1 h AT, X i − a ′ a ε h X, T i f .Proof. First, we recall that X = ˜ X + εη ( X ) ∂ t . Therefore, ¯ ∇ X ∂ t = ¯ ∇ ˜ X ∂ t = a ′ a ˜ X = a ′ a ( X − ε h X, T i ∂ t ). Next, we compute ¯ ∇ X T = ¯ ∇ X ( ∂ t − e n +1 ) = ¯ ∇ X ∂ t − X ( f ) e n +1 − f ¯ ∇ X e n +1 = a ′ a ( X − ε h X, T i ∂ t ) − X ( f ) e n +1 + f AX = a ′ a ( X − ε h X, T i ( T + f e n +1 )) − X ( f ) e n +1 + f AX = a ′ a ( X − ε h X, T i T ) − εfa ′ a e n +1 )) − X ( f ) e n +1 + f AX. Now, each equation is just the tangentialand the normal part of ¯ ∇ X T = ∇ X T + ε n +1 h AX, T i e n +1 . Elie Cartan developed the moving frame technique. Definitions, basic results and some otherdetails can be found in [3, p. 18]. We will use the following convention on the ranges ofindices, unless mentioned otherwise:1 ≤ i, j, k, l ≤ n ; 1 ≤ u, v, w, . . . ≤ n + 1; 0 ≤ α, β, γ, . . . ≤ n + 1 . We recall that M is a hypersurface of ¯ P n +1 , hence n = dim M . Let ( e , e , . . . , e n , e n +1 ) be alocal orthonormal frame on M , such that e , . . . , e n are tangent to M and e n +1 is normal to M in ¯ P n +1 , with ε α = h e α , e α i = ±
1. We define the matrix G = ( ε α δ αβ ). Let ( ω , . . . , ω n +1 )be the dual basis of e α , i.e. ω α ( e β ) = δ αβ , where ω r | T M = 0, r ∈ { , n + 1 } . The dual 1-forms ω α can be obtained as ω α ( X ) = ε α h e α , X i .Given ( E , . . . , E n ) a parallel orthonormal frame of E n +1 , we construct ( E , . . . , E n +1 ) =( E a , . . . E n a , ∂ t ), which is an orthonormal frame of εI × a E n +1 . If necessary, we reorder thebasis ( E , . . . , E n +1 ) to obtain h E α , E α i = ε α , E n +1 = ∂ t . Next, we define the functions B αβ := h E α , e β i and the matrix B = ( B αβ ). We have: X µ ε µ B µα B µβ = X µ ε µ h E µ , e α i h E µ , e β i = h e α , e β i = ε α δ αβ . B t GB = G , which implies B − = GB t G , where B t is the transposeof B and B − = ( B αβ ). Next, from the fact that B t GB = G , we define the sets S = { Z ∈ M n +2 ( R ) | Z t GZ = G, det Z = 1 } , s = { H ∈ M n +2 ( R ) | H t G + GH = 0 } . The set S is the connected component of the identity matrix, and is hence isometric to theLie group O + ↑ ( n + 2 , q ), where the index of the metric is q = k + | c − | + | ε − | . Clearly, s isthe Lie algebra associated with S . In other words, we have constructed a map B : M → S ,and therefore, we immediately obtain the s -valued 1-form B − dB on M . Let us now definethe connection 1-forms Ω = ( ω αβ ), ω αβ ( X ) = ε α h e α , e ∇ X e β i , for any X ∈ T M.
The matrix Ω satisfies Ω t G + G Ω = 0, or equivalently, ω βα = − ε α ε β ω αβ . In particular, ∇ e i = X k ω ki e k , ¯ ∇ e u = X v ω vu e v , e ∇ e α = X γ ω γα e γ , (7)We now define the 1-form η ( X ) = h T, X i and functions T k = h e k , T i , T n +1 = ε n +1 f and T = 0. Clearly, P k T k ω k = η . Obviously, we can recover the vectors e β = P γ ε γ B γβ E γ .Consequently, by (7), e ∇ e α e β = X µ ω µβ ( e α ) e µ = X γ ε γ (cid:16) X µ ω µβ ( e α ) B γµ (cid:17) E γ = e ∇ e α (cid:16) X γ ε γ B γβ E γ (cid:17) = X γ ε γ dB γβ ( e α ) E γ + X µ,γ ε γ ε µ B µα B γβ e ∇ E µ E γ . By now, we just care for the last summand. To do so, e ∇ E n +1 E n +1 = e ∇ ∂ t ∂ t = 0 , e ∇ E u E n +1 = 1 a e ∇ E u ∂ t = a ′ a E u , e ∇ E n +1 E u = − a ′ a E u + 1 a e ∇ ∂ t E u = − a ′ a E u + a ′ a E u = 0 , e ∇ E u E v = 1 a e ∇ E u E v = 1 a ∇ oE u E v − h E u , E v i a grad( a ) = − ε u δ uv εa ′ a ∂ t . Consequently, by using the fact that the terms for µ = n + 1 vanish, we have e ∇ e α e β − X γ ε γ dB γβ ( e α ) E γ = X µ,γ ε γ ε µ B µα B γβ e ∇ E µ E γ = X u,v ε v ε u B uα B vβ e ∇ E u E v + X u ε ε u B uα B β e ∇ E u E = ε a ′ a X v ε v B vα B β E v − ε a ′ a X u ε u B uα B uβ E . By comparing coordinates, we get for γ = n + 1 X µ B n +1 µ ω µβ ( e α ) = dB n +1 β ( e α ) − a ′ a X u ε u B uα B uβ . (8)8nd for γ = 0 , . . . , n , X µ B γµ ω µβ ( e α ) = dB γβ ( e α ) + εa ′ a B γα B n +1 β . (9)Using the fact that B µα = P γ B µγ ω γ ( e α ), we get for equation (8), P µ B n +1 µ ω µβ − dB n +1 β = − a ′ a P γ P u ε u B uβ B uγ ω γ = − a ′ a P γ,µ ε µ B µβ B µγ ω γ + ε a ′ a P γ B n +1 β B n +1 γ ω γ , and for equation(9), P µ B γµ ω µβ = dB γβ + εa ′ a P κ B n +1 β B γκ ω κ , for any γ = 0 , . . . , n + 1 . Finally, for all γ = α X µ B αµ ω µβ = dB αβ + εa ′ a B n +1 β X γ B αγ ω γ − a ′ a ε β δ α n +1 ω β . Moreover, we have P µ B αµ δ µ ε β ω β = B α ε β ω β = ε β ε α εB α ω β , that is, ω αβ = X µ B αµ dB µβ + εa ′ a (cid:16) B n +1 β ω α − ε β ε α B n +1 α ω β (cid:17) . (10)Finally, we obtainΩ − X = B − dB, X αβ = εa ′ a (cid:16) B n +1 β ω α − ε β ε α B n +1 α ω β (cid:17) . (11)We point out that B n +1 α = h E n +1 , e α i = h ∂ t , e α i = T α . Let ( M, h , i ) be a semi-Riemannian manifold with its Levi-Civita connection ∇ , its Riemanntensor R . We choose numbers ε, ε , ε n +1 ∈ {− , } and c = ε , and smooth functions a : I ⊂ R → R + , T n +1 : M → R and π : M → I . We construct the vector field T ∈ X ( M ) by T = ε grad( π ), with its 1-form η ( X ) = h X, T i . Also, consider a tensor A of type (1,1) on M . Definition 1.
Under the previous conditions, we will say that M satisfies the structureconditions if the following conditions hold:(A) A is h , i -self adjoint;(B) ε = h T, T i + ε n +1 T n +1 ;(C) ∇ X T = a ′ ◦ πa ◦ π ( X − εη ( X ) T ) + ε n +1 T n +1 AX , for any X ∈ T M ;(D) X ( T n +1 ) = −h AT, X i − ε a ′ ◦ πa ◦ π T n +1 η ( X ), for any X ∈ T M ;(E)
Codazzi equation: for any
X, Y ∈ T M , it holds( ∇ X A ) Y − ( ∇ Y A ) X = T n +1 (cid:18) a ′′ ◦ πa ◦ π − ( a ′ ◦ π ) ( a ◦ π ) + εε ( a ◦ π ) (cid:19) (cid:16) η ( Y ) X − η ( X ) Y (cid:17) ;9F) Gauß equation: for any
X, Y, Z, W ∈ T M , it holds R ( X, Y, Z, W ) = (cid:18) ε ( a ′ ◦ π ) ( a ◦ π ) − ε ( a ◦ π ) (cid:19) (cid:16) h X, Z ih Y, W i − h
Y, Z ih X, W i (cid:17) + (cid:18) a ′′ ◦ πa ◦ π − ( a ′ ◦ π ) ( a ◦ π ) + ε ε ( a ◦ π ) (cid:19) (cid:16) h X, Z i η ( Y ) η ( W ) − h Y, Z i η ( X ) η ( W ) −h X, W i η ( Y ) η ( Z ) + h Y, W i η ( X ) η ( Z ) (cid:17) + ε n +1 (cid:16) h AY, Z ih AX, W i − h
AY, W ih AX, Z i (cid:17) . We recall the warped product ( ¯ P n +1 = I × M nk ( c ) , h , i = εdt + a g o ). Theorem 1.
Let ( M, h , i ) a semi-Riemannian manifold satisfying the structure conditions.Then, for each point p ∈ M , there exists a neighborhood U of p on M , a metric immersion χ : ( U , h , i ) → ( ¯ P n +1 , h , i ) and a normal unit vector field e n +1 along χ such that:1. ε n +1 = h e n +1 , e n +1 i ;2. π I ◦ χ = π , where π I : ε I × a M nk ( c ) → I is the projection;3. The shape operator associated with e n +1 is A ;4. (E) is the Codazzi equation, and (F) is the Gauß equation,5. and along χ , it holds ∂ t = T + ε n +1 T n +1 e n +1 .Proof. Given a point x ∈ M , around it we consider a local orthonormal frame { e , . . . , e n } on M , with their signs ε i = g ( e i , e i ) = ±
1, and its corresponding dual basis of 1-forms { ω , . . . , ω n } . We point out that an alternative definition for these 1-forms is ω i ( X ) = ε i h e i , X i , for any X ∈ T M . We also need to define ω n +1 = ω = 0 . With the help ofthe tensor SY = − (cid:0) Y − εη ( Y ) T (cid:1) / ( ac ), for any Y ∈ T M , we construct the following 1-forms ω ij ( X ) = ε i h e i , ∇ X e j i , ω i n +1 ( X ) = − ε i h e i , AX i ,ω i ( X ) = − ε i h e i , SX i , ω n +1 , = − εε n +1 c ( a ◦ π ) T n +1 η, ω αβ = − ε α ε β ω βα , (12)for any X ∈ T M , known as the connection 1-forms . In this way, we consider the s -valuedmatrix Ω = ( ω αβ ). As a consequence, we get ∇ e i = P k ω ki e k . Now, we define the functions T i = η ( e i ), i ∈ { , . . . , n } , T = 0. We point out that by condition (B), we have ε = P γ ε γ T γ . Next, we also construct the matrices X = ( X αβ ) and Υ as X αβ = εa ′ a (cid:0) T β ω α − ε α ε β T α ω β (cid:1) , Υ = Ω − X . (13)A simple computation shows d Υ + Υ ∧ Υ = d Ω − d X + Ω ∧ Ω − Ω ∧ X − X ∧ Ω + X ∧ X . Thus, our target consist of proving that the second half of this equality vanishes. Since thecomputation is rather lengthy, we will split it in some lemmata.
Lemma 4. dη = 0 . roof. Since T = ε grad( π ), we obtain that η = εdπ . Therefore, dη = 0.We define the matrices ̟ = ( ω α ) and Γ = (Γ αβ ) = d Ω + Ω ∧ Ω. Lemma 5. d̟ = − Ω ∧ ̟, Γ αβ = − ε α ε β Γ βα , Γ ij = εε j ( a ′ ) a ω j ∧ ω i − (cid:18) a ′′ a − ( a ′ ) a (cid:19) (cid:0) T j ω i − ε i ε j T i ω j (cid:1) ∧ η, Γ i n +1 = T n +1 (cid:16) a ′′ a − ( a ′ ) a (cid:17) η ∧ ω i , Γ u = 0 , Proof.
Given
X, Y ∈ T M , since ω n +1 = ω = 0, we have dω i ( X, Y ) = X ( ω i ( Y )) − Y ( ω i ( X )) − ω i ([ X, Y ])= ε i h∇ X e i , Y i + ε i h e i , ∇ X Y i − ε i h∇ Y e i , X i − ε i h e i , ∇ Y X i − ε i h e i , [ X, Y ] i = ε i X k ω ki ( X ) h e k , Y i − ε i X k ω ki ( Y ) h e k , X i = − X γ ω iγ ∧ ω γ ( X, Y ) . On the other hand, by (12), X γ ω n +1 γ ∧ ω γ ( X, Y ) = X k ω n +1 k ∧ ω k ( X, Y )= X k (cid:0) ε n +1 h e k , AX i ε k h e k , Y i − ε n +1 h e k , AY i ε k h e k , X i (cid:1) = ε n +1 (cid:0) h Y, AX i − h
X, AY i (cid:1) = 0 = − dω n +1 ( X, Y ) . Also, X γ ω γ ∧ ω γ ( X, Y ) = X k ω k ∧ ω k ( X, Y )= 1 ac X k (cid:16) − ε (cid:0) h e k , X i − εT k η ( X ) (cid:1) ω k ( Y ) + ε (cid:0) h e k , Y i − εT k η ( Y ) (cid:1) ω k ( X ) (cid:17) = ε ac (cid:16) − h Y, X i + εη ( Y ) η ( X ) + h X, Y i − εη ( X ) η ( Y ) (cid:17) = 0 = − dω ( X, Y ) . Next, given
X, Y ∈ T M , we compute dω ij ( X, Y ) = X ( ω ij ( Y )) − Y ( ω ij ( X )) − ω ij ([ X, Y ])= ε i X ( h e i , ∇ Y e j i ) − ε i Y ( h e i , ∇ X e j i ) − ε i h e i , ∇ [ X,Y ] e j i = ε i h∇ X e i , ∇ Y e j i − ε i h∇ Y e i , ∇ X e j i + ε i R ( X, Y, e j , e i ) . On the other hand, by (7), h∇ X e i , ∇ Y e j i = P k ε k ω ki ( X ) ω kj ( Y ) = − P k ε i ω ik ( X ) ω kj ( Y ) . Therefore, dω ij ( X, Y ) = − X k ω ik ∧ ω kj ( X, Y ) − ε i R ( X, Y, e i , e j ) , dω ij ( X, Y ) + X γ ω iγ ∧ ω γj ( X, Y )= ω i ∧ ω j ( X, Y ) + ω i n +1 ∧ ω n +1 ,j ( X, Y ) + ε i R ( X, Y, e j , e i )= − ε i ε n +1 h e i , AX ih e j , AY i + ε i ε n +1 h e i , AY ih e j , AX i − ε i ε h e i , SX ih e j , SY i + ε i ε h e i , SY ih e j , SX i + ε i (cid:16) ε ( a ′ ) a − ε a (cid:17)(cid:16) h X, e j ih Y, e i i − h Y, e j ih X, e i i (cid:17) + (cid:18) εε a + a ′′ a − ( a ′ ) a (cid:19) (cid:16) h X, e j i η ( Y ) η ( e i ) − h Y, e j i η ( X ) η ( e i ) − h X, e i i η ( Y ) η ( e j ) + h Y, e i i η ( X ) η ( e j ) (cid:17) + ε n +1 (cid:16) h AY, e j ih AX, e i i − h AY, e i ih AX, e j i (cid:17) + ε (cid:16) h SY, e j ih SX, e i i − h SY, e i ih SX, e j i (cid:17)! = εε j ( a ′ ) a ω j ∧ ω i ( X, Y ) − (cid:18) a ′′ a − ( a ′ ) a (cid:19) (cid:0) T j ω i − ε i ε j T i ω j (cid:1) ∧ η ( X, Y ) . Next, given
X, Y ∈ T M , we compute dω in +1 ( X, Y ) = X ( ω in +1 ( Y )) − Y ( ω in +1 ( X )) − ω in +1 ([ X, Y ])= − ε i h∇ X e i , AY i + ε i h∇ Y e i , AX i + ε i h e i , ( ∇ Y A ) X − ( ∇ X A ) Y i = − ε i X k ω ki ( X ) h e k , AY i + ε i X k ω ki ( Y ) h e k , AX i + ε i h e i , ( ∇ Y A ) X − ( ∇ X A ) Y i = X k (cid:16) ω k n +1 ( X ) ω ik ( Y ) − ω k n +1 ( Y ) ω ik ( X ) (cid:17) + ε i h e i , ( ∇ Y A ) X − ( ∇ X A ) Y i = − X γ ω iγ ∧ ω γ n +1 ( X, Y ) + T n +1 (cid:16) a ′′ a − ( a ′ ) a (cid:17) η ∧ ω i ( Y, X ) . The next case is dω i ( X, Y ) = X ( ω i ( Y )) − Y ( ω i ( X )) − ω i ([ X, Y ])= ε i (cid:0) h∇ Y e i , SX i − h∇ X e i , SY i − h e i , ( ∇ Y S ) X − ( ∇ X S ) Y i (cid:1) . On one hand, for any U ∈ X ( M ), it holds ∇ Y (cid:0) − ac U (cid:1) = εa ′ ca η ( Y ) U − ac ∇ Y U , so that( ∇ Y S ) X − ( ∇ X S ) Y = ∇ Y SX − S ∇ Y X − ∇ X SY + S ∇ X Y = εa ′ ca η ( Y ) (cid:0) X − εη ( X ) T (cid:1) − ac ∇ Y ( X − εη ( X ) T ) + 1 ac ∇ Y X − εac η ( ∇ Y X ) T − εa ′ ca η ( X ) (cid:0) Y − εη ( Y ) T (cid:1) + 1 ac ∇ X ( Y − εη ( Y ) T ) + 1 ac ∇ X Y − εac η ( ∇ X Y ) T = εa ′ ca (cid:0) η ( Y ) X − η ( X ) Y (cid:1) + εac (cid:16) η ( X ) ∇ Y T − η ( Y ) ∇ X T (cid:17) = εε n +1 T n +1 ac ( η ( X ) AY − η ( Y ) AX ) . dω i ( X, Y ) = ε i (cid:16) h∇ Y e i , SX i − h∇ X e i , SY i + εε n +1 T n +1 ac h e i , η ( X ) AY − η ( Y ) AX i (cid:17) = ε i (cid:16) X k (cid:0) ω ki ( Y ) h e k , SX i − ω ki ( X ) h e k , SY i (cid:1) + εε n +1 T n +1 ac h e i , η ( X ) AY − η ( Y ) AX i (cid:17) = − X γ ω iγ ∧ ω γ ( X, Y ) + ω i n +1 ∧ ω n +1 , ( X, Y ) − εε n +1 T n +1 ac η ∧ ω i n +1 ( X, Y )= − X γ ω iγ ∧ ω γ ( X, Y ) . Next, we easily see dω n +1 , = − εε n +1 ac dT n +1 ∧ η . Therefore, dω n +1 , ( X, Y ) = − εε n +1 ac dT n +1 ∧ η ( X, Y ) = − εε n +1 ac dT n +1 (cid:16) X ( T n +1 ) η ( Y ) − Y ( T n +1 ) η ( X ) (cid:17) = ε n +1 X k ε k (cid:0) h AX, e k ih SY, e k i − h AY, e k ih SX, e k i (cid:1) = − X k ω n +1 k ∧ ω k ( X, Y ) = − X γ ω n +1 γ ∧ ω γ ( X, Y ) . Lemma 6. d X αβ = − ε α ε β d X βα , d X u = 0 ,d X ij = a ′′ a − a ′ ) a η ∧ ( T j ω i − ε i ε j T i ω j ) + εa ′ a (cid:16) T j dω i − ε i ε j T i dω j (cid:17) +2 εε j (cid:18) a ′ a (cid:19) ω j ∧ ω i + εa ′ a X u T u (cid:16) ω uj ∧ ω i − ε i ε j ω ui ∧ ω j (cid:17) ,d X i n +1 = a ′′ a − a ′ ) a T n +1 η ∧ ω i + εa ′ a (cid:16) X k T k ω k n +1 ∧ ω i + T n +1 dω i (cid:17) . Proof.
Since X αβ = − ε α ε β X βα , we trivially have d X αβ = − ε α ε β d X βα . Next, d X αβ = d (cid:18) εa ′ a ( T β ω α − ε α ε β T α ω β ) (cid:19) = (cid:16) X k e k (cid:18) εa ′ a (cid:19) ω k (cid:17) ∧ ( T β ω α − ε α ε β T α ω β )+ εa ′ a (cid:16) dT β ∧ ω α + T β dω α − ε α ε β dT α ∧ ω β − ε α ε β T α dω β (cid:17) = a ′′ a − ( a ′ ) a η ∧ ( T β ω α − ε α ε β T α ω β )+ εa ′ a (cid:16) dT β ∧ ω α + T β dω α − ε α ε β dT α ∧ ω β − ε α ε β T α dω β (cid:17) . For β = 0 and any u >
0, since T = 0 and ω = 0, then d X u = 0. For β = n + 1 and i < n + 1, since ω n +1 = 0, we have d X i n +1 = a ′′ a − ( a ′ ) a T n +1 η ∧ ω i + εa ′ a (cid:16) dT n +1 ∧ ω i + T n +1 dω i (cid:17) .
13y condition (C), and the fact T = P k ε k T k e k , we see dT n +1 ( X ) = −h AX, T i − ε a ′ a T n +1 η ( X )= − P k ε k T k h e k , AX i − εT n +1 a ′ a η ( X ), and consequently dT n +1 ( X ) = P k T k ω k n +1 ( X ) − εT n +1 a ′ a η ( X ). With this, d X i n +1 = a ′′ a − ( a ′ ) a T n +1 η ∧ ω i + εa ′ a (cid:16) X k T k ω k n +1 ∧ ω i − ε a ′ a T n +1 η ∧ ω i + T n +1 dω i (cid:17) = a ′′ a − a ′ ) a T n +1 η ∧ ω i + εa ′ a (cid:16) X k T k ω k n +1 ∧ ω i + T n +1 dω i (cid:17) . Next, for α = i and β = j , we have d X ij = a ′′ a − ( a ′ ) a η ∧ ( T j ω i − ε i ε j T i ω j )+ εa ′ a (cid:16) dT j ∧ ω i + T j dω i − ε i ε j dT i ∧ ω j − ε i ε j T i dω j (cid:17) . We need the following computation dT k = P l e l ( h T, e k i ) ω l = P l (cid:0) h∇ e l T, e k i ω l + h T, ∇ e l e k i (cid:1) ω l = P l (cid:16) h a ′ a ( e l − εT l T ) + ε n +1 T n +1 Ae l , e k i + P j ω jk ( e l ) T j (cid:17) ω l , which implies dT k = a ′ a ε k ω k − ε a ′ a T k η + X u T u ω uk . (14)In this way, a straight forward computation yields d X ij = a ′′ a − a ′ ) a η ∧ ( T j ω i − ε i ε j T i ω j ) + εa ′ a (cid:16) T j dω i − ε i ε j T i dω j (cid:17) + 2 εε j a ′ a ω j ∧ ω i + εa ′ a X u T u ( ω uj ∧ ω i − ε i ε j ω ui ∧ ω j ) ! . Lemma 7. ( X ∧ X ) αβ = (cid:18) a ′ a (cid:19) (cid:16)(cid:0) T β ω α − ε α ε β T α ω β (cid:1) ∧ η − εε β ω α ∧ ω β (cid:17) . Proof.
We recall (B). Also, given X ∈ T M , we have P γ T γ ω γ ( X ) = P k T k ω k ( X ) = η ( X ).Then, ( X ∧ X ) αβ = X γ (cid:16) εa ′ a ( T γ ω α − ε α ε γ T α ω γ ) ∧ εa ′ a ( T β ω γ − ε γ ε β T γ ω β ) (cid:17) = (cid:18) a ′ a (cid:19) X γ (cid:16) T γ T β ω α ∧ ω γ − ε β ε γ T γ ω α ∧ ω β − ε α ε γ T α T β ω γ ∧ ω γ + ε γ ε γ ε α ε β T α T γ ω γ ∧ ω β (cid:17) = (cid:18) a ′ a (cid:19) (cid:16)(cid:0) T β ω α − ε α ε β T α ω β (cid:1) ∧ η − εε β ω α ∧ ω β (cid:17) . Now, we put Φ = Ω ∧ X + X ∧ Ω. 14 emma 8. Φ αβ = − ε α ε β Φ βα , Φ u = 0 , (15)Φ ij = εa ′ a (cid:16) ε i ε j T i dω j − T j dω i + X u T u (cid:0) ω i ∧ ω uj − ε j ε u ω iu ∧ ω j (cid:1)(cid:17) (16)Φ i n +1 = εa ′ a (cid:16) − T n +1 dω i + X k T k ω i ∧ ω k n +1 (cid:17) . (17) Proof.
By construction, X αβ = − ε α ε β X βα . This shows Φ αβ = − ε α ε β Φ βα . In general, we getΦ αβ = X γ ω αγ ∧ (cid:18) εa ′ a ( T β ω γ − ε β ε γ T γ ω β ) (cid:19) + X γ εa ′ a ( T γ ω α − ε γ ε α T α ω γ ) ∧ ω γβ = εa ′ a X γ ( T β ω αγ ∧ ω γ − ε β ε γ T γ ω αγ ∧ ω β + T γ ω α ∧ ω γβ − ε α ε γ T α ω γ ∧ ω γβ ) . By Lemma 5,Φ αβ = εa ′ a (cid:16) ε α ε β T α dω β − T β dω α + X γ (cid:0) T γ ω α ∧ ω γβ − ε β ε γ T γ ω αγ ∧ ω β (cid:1)(cid:17) . For the case α = i , β = n + 1, the result is immediate due to the fact ω n +1 = 0. For the case α = i , β = 0, since ω = 0 and T = 0, we begin by writing down Φ i = εa ′ a P γ T γ ω i ∧ ω γ . How-ever, by Condition (C), P γ T γ ω γ ( X ) = P i T i ω i ( X ) + T n +1 ω n +1 , ( X ) = P i T i ac (cid:0) ε i h e i , X i − εε i T i η ( X ) (cid:1) − εε n +1 T n +1 ac η ( X ) = ac (cid:0) P i ε i h e i , T ih e i , X i − ε (cid:0) P i ε i T i (cid:1) η ( X ) − εε n +1 T n +1 η ( X ) (cid:1) ,and hence X γ T γ ω γ ( X ) = 0 . (18)The case α = n + 1 and β = 0 trivially vanishes due to ω n +1 = ω = 0. Lemma 9. d Υ + Υ ∧ Υ = 0 .Proof.
This is equivalent to prove d X − d Ω − Ω ∧ Ω − X ∧ X + X ∧ Ω + Ω ∧ X = 0 . The case α = u , β = 0 is trivial. For the case α = i and β = j , we have( d X − d Ω − Ω ∧ Ω − X ∧ X + X ∧ Ω + Ω ∧ X ) ij = a ′′ a − a ′ ) a η ∧ ( T j ω i − ε i ε j T i ω j ) + εa ′ a (cid:16) T j dω i − ε i ε j T i dω j (cid:17) + 2 εε j ( a ′ ) a ω j ∧ ω i + εa ′ a X u T u ( ω uj ∧ ω i − ε i ε j ω ui ∧ ω j ) ! + ε ( a ′ ) a ε j ω i ∧ ω j − (cid:18) ( a ′ ) a − a ′′ a (cid:19) (cid:16) T j ω i − ε i ε j T i ω j ) ∧ η − (cid:18) a ′ a (cid:19) (cid:16)(cid:0) T j ω i − ε i ε j T i ω j (cid:1) ∧ η − εε j ω i ∧ ω j (cid:17) + εa ′ a (cid:16) ε i ε j T i dω j − T j dω i + X u T u (cid:0) ω i ∧ ω uj − ε j ε u ω iu ∧ ω j (cid:1)(cid:17) = 0 . α = i , β = n + 1, we compute( d X − d Ω − Ω ∧ Ω − X ∧ X + X ∧ Ω + Ω ∧ X ) i n +1 = a ′′ a − a ′ ) a T n +1 η ∧ ω i + εa ′ a (cid:16) X k T k ω k n +1 ∧ ω i + T n +1 dω i (cid:17) − T n +1 (cid:18) a ′′ a − ( a ′ ) a (cid:19) η ∧ ω i − (cid:18) a ′ a (cid:19) (cid:16)(cid:0) T n +1 ω i − ε i ε n +1 T i ω n +1 (cid:1) ∧ η − εε n +1 ω i ∧ ω n +1 (cid:17) + εa ′ a (cid:16) − T n +1 dω i + X k T k ω i ∧ ω k n +1 (cid:17) = a ′′ a − a ′ ) a T n +1 η ∧ ω i − T n +1 (cid:18) a ′′ a − ( a ′ ) a (cid:19) η ∧ ω i − (cid:18) a ′ a (cid:19) (cid:0) T n +1 ω i (cid:1) ∧ η = 0 . It is clear that the map s : S → S ( E n +2 ) = { X ∈ E n +2 |h X, X i} = ε n +1 } , Z ( Z n +1 , , . . . , Z n +1 n +1 ) t , is a submersion. Given a point x ∈ M , we define the set Z ( x ) = { Z ∈ S | Z n +1 β = T β ( x ) , β = 0 , . . . , n + 1 } . Now, we prove the following
Lemma 10.
Let ( M, h , i ) be a semi-Riemannian manifold satisfying the structure conditions.For each x ∈ M and B ∈ Z ( x ) , there exists a neighborhood U of x in M and a uniquemap B : U → S , such that B − dB = Ω − X , for all x ∈ U , B ( x ) ∈ Z ( x ) , B = B ( x ) . Proof.
Given U be an open neighborhood of x ∈ M , we define the set F = { ( x, Z ) ∈ U × S | Z ∈ Z ( x ) } . Since the map s is submersion, F is a submanifold of M × S withdim F = n + ( n + 1)( n + 2)2 − ( n + 1) = n ( n + 1)2 + n. Moreover, given ( x, Z ) ∈ F , T ( x,Z ) F = { ( U, V ) ∈ T x U ⊕ T Z S | V n +1 β = ( dT β ) x ( U ) , β = 0 , . . . , n + 1 } . We consider on F the distribution D ( x, Z ) = ker Θ ( x,Z ) , where Θ = Υ − Z − dZ = Ω − X − Z − dZ. In other words, given (
U, V ) ∈ T ( x,Z ) F , we have Θ ( x,Z ) ( U, V ) = Ω x ( U ) − X x ( U ) − Z − V. Next, we see that dim D = n . We consider the space H = { H ∈ s | ( ZH ) n +1 β =0 , β = 0 , . . . , n + 1 } . It is clear that H ∈ H if and only if H ∈ ker( ds ) I n +2 . But themap s is a submersion, hence dim(ker ds I n +2 ) = dim H = ( n +1) n . We notice that ( Z Θ) n +1 β =16 Z Ω) n +1 β − ( Z X ) n +1 β − ( dZ ) n +1 β = ( Z Ω) n +1 β − ( Z X ) n +1 β − dT β . Consequently using equation(14) and T = 0 we get( Z Θ) n +1 k = X γ Z n +1 γ ω γk − X γ Z n +1 γ X γk − a ′ a ε k ω k + ε a ′ a T k η − X u T u ω uk = X γ T γ ω γk − X γ T γ εa ′ a (cid:0) T k ω γ − ε γ ε k T γ ω k (cid:1) − a ′ a ε k ω k + ε a ′ a T k η − X u T u ω uk = − εa ′ a (cid:0) T k η − X γ ε γ ε k T γ T γ ω k (cid:1) − a ′ a ε k ω k + ε a ′ a T k η = εa ′ a X γ ε γ ε k T γ T γ ω k − a ′ a ε k ω k = 0 , since ε = h T, T i + ε n +1 T n +1 . Similarly,( Z Θ) n +1 n +1 = X γ T γ ω γ n +1 − X γ T γ εa ′ a (cid:0) T n +1 ω γ − ε γ ε n +1 T γ ω n +1 (cid:1) − dT n +1 = X γ T γ ω γ n +1 − εa ′ a T n +1 η − X k T k ω k n +1 + εT n +1 a ′ a η = 0 , and with equation (18), it is clear( Z Θ) n +1 , = X γ T γ ω γ − X γ T γ εa ′ a (cid:0) T ω γ − ε γ ε T γ ω (cid:1) − dT = 0 . Hence, Im(Θ) ⊂ H = ker( ds ) I n +2 . Now, given the space { (0 , ZH ) | H ∈ H} ⊂ T ( x,Z ) F , we haveΘ ( x,Z ) (0 , ZH ) = − Z − ( ZH ) = − H, which means that Θ ( x,Z ) is a submersion unto H , andIm(Θ) = H . Now, we get dim D ( x, Z ) = dim ker Θ ( x,Z ) = dim T ( x,Z ) F − dim ImΘ ( x,Z ) = n. Next, we prove that D is integrable. On one hand, since D = ker Θ and d Υ + Υ ∧ Υ = 0,we compute d Θ = d Υ + Z − dZ ∧ Z − dZ = d Υ + (Υ − Θ) ∧ (Υ − Θ) = − Υ ∧ Θ − Θ ∧ Υ + Θ ∧ Θ.Therefore, given
U, V ∈ D , d Θ( U, V ) = U (Θ( V )) − V (Θ( U )) − Θ([
U, V ]) = − Θ([
U, V ]) =( − Υ ∧ Θ − Θ ∧ Υ + Θ ∧ Θ)(
U, V ) = 0, which implies [
U, V ] ∈ D .Next, let L ⊂ F be an integral manifold through ( x , B ). For each (0 , V ) ∈ D ( x ,B ) = T B L , we have Θ ( x ,B ) ( V ) = B − V = 0, and hence V = 0, since B ∈ S . This implies D ( x ,B ) ∩ [ { } × T B S ] = { } . In particular, by shrinking U if necessary, L is the graph of aunique map B : U → S . Also, since L ⊂ F , then for each x ∈ U , B ( x ) ∈ Z ( x ). Finally sinceΘ ≡ L , B satisfies by definition B − dB = Ω − X .Define now the map χ : U → R n +2 by χ = ε B , χ i = ε i B i , χ n +1 = π. Notice that, since B ( x ) ∈ Z ( x ) ⊂ S , then B n +1 , = T = 0, whic implies ε χ + P ni =1 ε i χ i = P nα =0 ε α B α = ε = c, thus obtaining that ( χ , . . . , χ n ) lies in M nk ( c ), which means Im( χ ) ⊂ I × a M nk ( c ). Now, we have by definition dB = B Ω − B X . Hence dχ i ( e k ) = ε i dB i ( e k ) = n +1 X α =0 ε i ( B iα ω α ( e k ) − B iα X α ( e k ))= ε i n +1 X α =1 B iα ω α ( e k ) − ε i n +1 X α =1 B iα (cid:16) εa ′ a ( T ω α ( e k ) − ε ε α T α ω ( e k ) (cid:17) = ε i n X j =1 B ij ω j ( e k ) + B in +1 ω n +10 ( e k )= ε i n X j =1 B ij ac ( ε j h e j , e k i − εε j T j η ( e k )) − B in +1 εε n +1 ac T n +1 η ( e k )= ε i ac B ik − εε n +1 ac B in +1 T n +1 T k = ε i ac B ik A similar computation yields dχ ( e k ) = ε ac B k and dχ n +1 ( e k ) = εη ( e k ) = εT k = εB n +1 k .Hence, we have that dχ = CB̟ , with C = ε /ca · · ·
00 . . . . . . ...... . . . ε n /ca · · · ε n +1 , ̟ = (0 , ω , · · · ω n , T , or equivalently, dχ ( e k ) α = ( CB ) αk , meaning that in the frame ∂∂x α the vector dχ ( e k ) is givenby the k-th column of the matrix CB and in the frame ¯ E α by the k-th column of the matrix B .In other words, dχ ( e k ) = P α ε α B αk ¯ E α . C is an invertible matrix as well as B . Consequently, dχ has rank n and it is an immersion. Moreover, for any i, j , since B ∈ S , we have h dχ ( e i ) , dχ ( e j ) i = h X α ε α B αi ¯ E α , X γ ε γ B γj ¯ E γ i = X γ ε α B αi B αj = ε i δ ij , Hence, χ is isometric. Moreover, along χ , we obtain ∂∂t = n X i =1 ε i h ∂∂t , dχ ( e i ) i dχ ( e i ) + ε n +1 h ∂∂t , e n +1 i e n +1 = T + ε n +1 T n +1 e n +1 . Next, we would like to compute the shape operator of the immersion. Recall ¯ E n +1 = ∂ t . Weshow that the shape operator of the immersion is exactly what we need, namely dχ ◦ A ◦ dχ − .18ndeed, h ¯ ∇ dχ ( ei ) dχ ( e j ) , e n +1 i = h X αβ ¯ ∇ ε α B αi ¯ E α ε β B βj ¯ E β , e n +1 i = h X αβ ε α ε β e ∇ B αi ¯ E α B βj ¯ E β , e n +1 i = X αβγ (cid:2) ε α ε β ε γ B αi B βj B γ n +1 h e ∇ ¯ E α ¯ E β , ¯ E γ i + ε β dB βj ( e i ) B βn +1 (cid:3) + X β ε β dB βj ( e i ) B βn +1 = X uvγ (cid:2) ε u ε v ε γ B ui B vj B γ n +1 h− ε u δ uv εa ′ a ∂ t , ¯ E γ i + ε u ε n +1 ε γ B ui B n +1 j B γ n +1 h a ′ a ¯ E u , ¯ E γ i (cid:3) + X β ε β dB βj ( e i ) B βn +1 = − εa ′ a B n +1 n +1 X u ε u B ui B uj + ε n +1 a ′ a B n +1 j X u ε u B ui B un +1 + ε n +1 ( B − dB ) n +1 j ( e i )= − εa ′ a B n +1 n +1 ([ ε i δ ij − ε n +1 B n +1 i B n +1 j ]+ ε n +1 a ′ a B n +1 j [ ε i δ in +1 − ε n +1 B n +1 i B n +1 n +1 ] + ε n +1 ( B − dB ) n +1 j ( e i )= ε n +1 h ( B − dB ) n +1 j ( dχ ( e i ) ) − εaa ′ [ ε n +1 ε j T n +1 ω j ( e i ) − T j ω n +1 ( e i )] i = ε n +1 [( B − dB ) n +1 j + X n +1 j ] = ε n +1 ω n +1 j ( e i ) = h e j , Ae i i . Finally, the uniqueness of the local immersion follows from the uniqueness of the map B inLemma 10. Corollary 1.
1. If the hypersurface M satisfies η = 0 , then, M is a slice of ε I × a M nk ( c ) .2. If η = 0 everywhere, then M is admits a foliation of codimension 1.Proof. Item 1 is an immediate consequence of item 5 of Theorem 1. For item 2, by Lemma4, we know dη = 0. This implies that ker η is integrable. Indeed, given X, Y ∈ ker η ,0 = dη ( X, Y ) = X ( η ( Y )) − Y ( η ( X )) − η ([ X, Y ]) = − η ([ X, Y ]), which shows [
X, Y ] ∈ ker η .In other words, it has to admit a foliation whose leaves are of codimension 1 in M . In fact, T is a normal vector field to the leaves. Remark:
Under the same assumption on ( M, h , i , ∇ , R ), we can find another equivalentformulation for Theorem 1. In fact, consider again a h , i -self adjoint (1 , A on M , anowhere vanishing vector field T ∈ X ( M ) and its associated 1-form η ( X ) = h X, T i for any X ∈ T M . We also assume the existence of smooth functions ρ, ˜ ρ, ¯ ρ, T n +1 : M → R . Let thefollowing conditions be satisfied:(a) ρ > dρ = ε ˜ ρ η , d ˜ ρ = ε ¯ ρ η ;(b) ε = h T, T i + ε n +1 T n +1 ;(c) ∇ X T = ˜ ρρ ( X − εη ( X ) T ) + ε n +1 T n +1 AX , for any X ∈ T M ;(d) X ( T n +1 ) = −h AT, X i − ε ˜ ρρ T n +1 η ( X ), for any X ∈ T M ;19e) (Codazzi equation). For any
X, Y, Z, W ∈ T M ( ∇ X A ) Y − ( ∇ Y A ) X = T n +1 ¯ ρρ − (cid:18) ˜ ρρ (cid:19) + εε ρ ! (cid:16) η ( Y ) X − η ( X ) Y (cid:17) . (f) (Gauß equation). For any X, Y, Z, W ∈ T MR ( X, Y, Z, W ) = ε (cid:18) ˜ ρρ (cid:19) − ε ρ ! (cid:16) h X, Z ih Y, W i − h
Y, Z ih X, W i (cid:17) + ¯ ρρ − (cid:18) ˜ ρρ (cid:19) + εε ρ ! (cid:16) h X, Z i η ( Y ) η ( W ) − h Y, Z i η ( X ) η ( W ) − h X, W i η ( Y ) η ( Z ) + h Y, W i η ( X ) η ( Z ) (cid:17) + ε n +1 (cid:16) h AY, Z ih AX, W i − h
AY, W ih AX, Z i (cid:17) . Then Theorem 1 can be reformulated in the following way:
Theorem 2.
Let ( M, h , i ) a simply connected semi-Riemannian manifold satisfying the pre-vious conditions. Then, there exists smooth functions π : M → I , I an interval, a : I ⊂ R → R + , a metric immersion χ : ( M, h , i ) → ( ε I × a M nk ( c ) , h , i ) and a normal unit vector field e n +1 along χ such that:1. ε n +1 = h e n +1 , e n +1 i ;2. π I ◦ χ = π , where π I : ε I × a M nk ( c ) → I is the projection;3. ρ = a ◦ π , ˜ ρ = a ′ ◦ π and ¯ ρ = a ′′ ◦ π ;4. The shape operator associated with e n +1 is A ;5. (e) is the Codazzi equation and (f ) is the Gauß equation;6. and along χ , ∂ t = T + ε n +1 T n +1 e n +1 holds.Proof. First of all, from the expression ∇ X T = ˜ ρρ ( X − εη ( X ) T ) + ε n +1 T n +1 AX , for any X ∈ T M , we are going to check that the 1-form η satisfies dη = 0. dη ( e i , e j ) = e i ( η ( e j )) − e j ( η ( e i )) − η ( ∇ e i e j ) + η ( ∇ e j e i ) = h e j , ∇ e i T i − h e i , ∇ e j T i = h e j , ˜ ρρ ( e i − εη ( e i ) T ) + ε n +1 T n +1 Ae i i − h e i , ˜ ρρ ( e j − εη ( e j ) T ) + ε n +1 T n +1 Ae j i = ˜ ρρ (cid:0) h e j , e i i − εη ( e i ) η ( e j ) (cid:1) + ε n +1 T n +1 h e j , Ae i i − ˜ ρρ (cid:0) h e i , e j i − εη ( e i ) η ( e j ) (cid:1) − ε n +1 T n +1 h e i , Ae j i = 0 . Since M is simply connected, we can obtain a new function π : M → R such that η = εdπ .This implies that T = ε grad( π ). Next, we need to obtain function a . On one hand, since M is connected, I = π ( M ) is an interval. Moreover, since T = 0, we see that each value t ∈ I is a regular value of π , which means that each level set π − ( t ) ⊂ M , t ∈ I , is ahypersurface of M . Choose t ∈ I . Given X ∈ T π − ( t ), since π is constant along its levelsubsets, we see dρ ( X ) = ε ˜ ρ η ( X ) = ˜ ρ dπ ( X ) = 0. In other words, function ρ is constantalong the level sets of π . This allows us to define a : I → R + as follows. Given t ∈ I , thereexists p ∈ M such that t = π ( p ), so that a ( t ) := ρ ( p ). Clearly, ρ = a ◦ π . In addition, dρ = ( a ′ ◦ π ) dπ = ( a ′ ◦ π ) εη = ε ˜ ρη , and therefore ˜ ρ = a ′ ◦ π . Similarly, a ′′ ◦ π = ¯ ρ . Now, wejust need to resort to Theorem 1. 20 An Application to Horizons in RW 4-spacetimes
We consider now the simply connected Riemannian 3-dimensional space M ( c ) of constantsectional curvature c = ±
1. Let ( M , h , i ) be a semi-Riemannian manifold of index 0 or 1.For us, a surface ˜ M is called marginally trapped if its mean curvature vector ~H satisfies h ~H, ~H i = 0. In this way, we are including maximal surfaces, MOTS, and mixed cases in ourdefinition.We put ε = − ε = c , ε = ±
1, and smooth functions a : I ⊂ R → R + , T : M → R and π : M → I . We construct the vector field T ∈ X ( M ) by T = − grad( π ), with its 1-form η ( X ) = h X, T i . Also, consider a tensor A of type (1,1) on M . We assume the above datassatisfy the structure conditions of Definition 1 . We recall that the Robertson-Walker space-time is the space( ¯ P = I × M ( c ) , h , i = − dt + a g o ), hence a special case of the warpedproducts considered in this paper. From Theorem 1, we get immediately the following Corollary 2.
Let ( M, h , i ) a semi-Riemannian manifold of dim M = 3 , satisfying the previousconditions. Then, for each point p ∈ M , there exists a neighborhood U of p on M , a metricimmersion χ : ( U , h , i ) → ( ¯ P , h , i ) and a normal unit vector field e along χ such that ε = h e , e i , A is the shape operator associated to the immersion, T is the projection of ∂ t on T M and π I ◦ χ = π , where π I : I × M ( c ) → I is the projection.In addition, if T = 0 everywhere, the family { χ ( U ) ∩ π − { t } : t ∈ R } provides a foliationof χ ( U ) by space-like surfaces. Next, let L be one of the leafs of U . Let σ be its second fundamental form in ¯ P . Clearly, T ⊥ L = Span { T , e } , where T = T / p |h T, T i| . We take ε T = sign ( h T, T i ). Since the leavesare spacelike and h e , e i = ε = ± ε T = ± ε ε T = −
1. Then,for any
X, Y ∈ T L , σ ( X, Y ) = ε T h e ∇ X Y, T i T + ε h e ∇ X Y, e i e = − h T, T i h Y, ∇ X T i T + ε h Y, AX i e , Given a local orthonormal frame { u , u } of L , the mean curvature vector ~H of L is2 ~H = X i σ ( u i , u i ) = X i (cid:16) − h T, T i h u i , ∇ u i T i T + ε h u i , Au i i e (cid:17) = − h T, T i X i h u i , a ′ a u i + ε T Au i i T + ε (cid:16) trace( A ) − h AT, T ih T, T i (cid:17) e = − h T, T i (cid:16) a ′ a + ε T (cid:16) trace( A ) − h AT, T ih T, T i (cid:17)(cid:17) T + ε (cid:16) trace( A ) − h AT, T ih T, T i (cid:17) e . We put h = trace( A ) − h AT,T ih T,T i . As a result, we obtain 4 h ~H, ~H i = h T,T i (cid:16) a ′ a + ε T h (cid:17) + ε h . Then ~H is null if, and only if, − ε h = h T,T i (cid:16) a ′ a + ε T h (cid:17) = h T,T i (cid:16) ( a ′ ) a + T h + ε a ′ T ha (cid:17) ,which is equivalent to h − a ′ T a h − ε ( a ′ ) a = 0. Since ε T ε = −
1, we see that − ε h T, T i > Corollary 3.
The leaves of ker η are marginally trapped surfaces in − I × a M ( c ) if, and onlyif, the immersion χ : M → − I × a M ( c ) satisfies the following equality: trace( A ) − h AT, T ih T, T i = 2 a ′ T ± ε T (cid:12)(cid:12)(cid:12)(cid:12) a ′ a (cid:12)(cid:12)(cid:12)(cid:12) p |h T, T i| . Examples
Example 1.
A graph surface in − R + × a S . We start by considering the warping function a : R + → R + , a ( t ) = t , and the warped product − R + × a S , with metric h , i . Given h : (0 , π ) → R + a smooth function such that h ( u ) > | h ′ ( u ) | for any u ∈ R + , we introduce the map χ : M = (0 , π ) × ( − π/ , π/ → − I × a S ,χ ( u, v ) = (cos( u ) , sin( u ) cos( v ) , sin( u ) sin( v ) , h ( u )) , Note that π ( u, v ) = h ( u ). We consider the following frame along χ , with some naturalidentifications: e = 1 h ( u ) (cid:0) cos( u ) , cos( v ) sin( u ) , sin( u ) sin( v ) , (cid:1) ,e ≡ χ ∗ e = 1 p h ( u ) − ( h ′ ( u )) (cid:0) − sin( u ) , cos( u ) cos( v ) , cos( u ) sin( v ) , h ′ ( u ) (cid:1) ,e ≡ χ ∗ e = 1 h ( u ) (cid:0) , − sin( v ) , cos( v ) , (cid:1) ,e = 1 h ( u ) p h ( u ) − ( h ′ ( u )) (cid:0) − sin( u ) h ′ ( u ) , cos( u ) cos( v ) h ′ ( u ) , cos( u ) sin( v ) h ′ ( u ) , h ( u ) (cid:1) , where e , e are the normalizations of χ ∗ ∂ u and χ ∗ ∂ v , respectively, and e is a unit vectorfield of M along χ on − R + × a S . Also, the matrix G = ( h e α , e β i ) = Diag (1 , , , − M are ω = 0 , ω = p h ( u ) − ( h ′ ( u )) d u, ω = h ( u ) sin( u )d v, ω = 0 . Since R + × S ⊂ R + × R , we consider the orthonormal basis E = 1 h (1 , , , , E = 1 h (0 , , , , E = 1 h (0 , , , , E = ∂ t . Now, we can compute the map B : M → S , B = ( h E α , e β i ), B = cos( u ) − h ( u ) sin( u ) √ h ( u ) − ( h ′ ( u )) − sin( u ) h ′ ( u ) √ h ( u ) − ( h ′ ( u )) − cos( v ) sin( u ) − cos( u ) cos( v ) h ( u ) √ h ( u ) − ( h ′ ( u )) sin( v ) − cos( u ) cos( v ) h ′ ( u ) √ h ( u ) − ( h ′ ( u )) − sin( u ) sin( v ) − cos( u ) h ( u ) sin( v ) √ h ( u ) − ( h ′ ( u )) − cos( v ) − cos( u ) sin( v ) h ′ ( u ) √ h ( u ) − ( h ′ ( u )) − h ′ ( u ) √ h ( u ) − ( h ′ ( u )) − h ( u ) √ h ( u ) − ( h ′ ( u )) . From this, a straightforward computation gives the s -valued 1-form Υ = B − d B = (Υ αβ ),Υ αα = 0 , Υ = − Υ = − h ( u )d u p h ( u ) − ( h ′ ( u )) , Υ = − Υ = − sin( u )d v, Υ = Υ = − h ′ ( u )d u p h ( u ) − ( h ′ ( u )) , Υ = − Υ = − h ( u ) cos( u )d v p h ( u ) − ( h ′ ( u )) , Υ = Υ = h ( u ) h ′′ ( u ) − ( h ′ ( u )) h ( u ) − ( h ′ ( u )) d u, Υ = Υ = h ′ ( u ) cos( u )d v p h ( u ) − ( h ′ ( u )) .
22 lenghty computation shows dΥ + Υ ∧ Υ = 0. Next, the tangent vector T is T ≡ χ ∗ T = h ∂ t , e i e + h ∂ t , e i e = (cid:16) sin( u ) h ′ ( u ) h ( u ) − ( h ′ ( u )) , − cos( u ) cos( v ) h ′ ( u ) h ( u ) − ( h ′ ( u )) , − cos( u ) sin( v ) h ′ ( u ) h ( u ) − ( h ′ ( u )) , − ( h ′ ( u )) h ( u ) − ( h ′ ( u )) (cid:17) . Thus, its dual 1-form and the associated functions T α are η = − h ′ ( u )d u, T = 0 , T = − h ′ ( u ) p h ( u ) − ( h ′ ( u )) , T = 0 , T = T . Clearly, d η = 0. Also, note that T α = B α . We recall that Ω = ( ω αβ ) = B − d B + X , where X = ( X αβ ), X αβ = ε a ′ ◦ πa ◦ π (cid:16) B n +1 β ω α − ε α ε β B n +1 α ω β (cid:17) , X αα = X α = X α = 0 , X = X = d u, X = − X = − sin( u ) h ′ ( u )d v p h ( u ) − ( h ′ ( u )) , X = X = sin( u ) h ( u )d v p h ( u ) − ( h ′ ( u )) . In this way, we haveΩ = ( ω αβ ) , ω = − ω = − h ( u )d u p h ( u ) − ( h ′ ( u )) , ω = − ω = − sin( u )d v,ω = ω = − h ′ ( u )d u p h ( u ) − ( h ′ ( u )) , ω = − ω = − cos( u ) h ( u ) + sin( u ) h ′ ( u ) p h ( u ) − ( h ′ ( u )) d v,ω = ω = h ( u ) − h ′ ( u )) + h ( u ) h ′′ ( u ) h ( u ) − ( h ′ ( u )) d u, ω = ω = cos( u ) h ′ ( u ) + sin( u ) h ( u ) p h ( u ) − ( h ′ ( u )) d v. Now, we compute the shape operator. Since ω i ( X ) = − ε i h AX, e i i = − ω i ( AX ), then AX = X i ω i ( AX ) e i = − X i ω i ( X ) e i = h ( u ) − h ′ ( u )) + h ( u ) h ′′ ( u ) h ( u ) − ( h ′ ( u )) d u ( X ) e + cos( u ) h ′ ( u ) + sin( u ) h ′′ ( u ) p h ( u ) − ( h ′ ( u )) d v ( X ) e ,Ae = h ( u ) − h ′ ( u )) + h ( u ) h ′′ ( u ) h ( u )( h ( u ) − ( h ′ ( u )) ) e , Ae = cos( u ) h ′ ( u ) + sin( u ) h ′′ ( u ) h ( u )( h ( u ) − ( h ′ ( u )) ) e . Example 2.
A helicoidal surface in R × a H We consider now H ( −
1) as the surface H ( −
1) = { ( x, y, z ) ∈ L : x + y − z = − } , where L is the Lorentz-Minkowski space endowed with the standard metric h , i = dx + dy − dz .Given a real constant c ∈ R and a smooth function h : R → R with h ′ >
0, we construct χ : M = R + × R −→ R × a H , χ ( u, v ) = ( u cos( cv ) , u sin( cv ) , p u , h ( v )) . Note that π ( u, v ) = h ( v ). We consider the following frame along χ , with some natural23dentifications: e = 1 a ( h ( v )) (cid:0) u cos( c v ) , u sin( c v ) , p u , (cid:1) ,e ≡ χ ∗ e = 1 a ( h ( v )) (cid:0) cos( c v ) p u , sin( c v ) p u , u, (cid:1) ,e ≡ χ ∗ e = 1 p c u a ( h ( v )) + h ′ ( v ) (cid:0) − c u sin( c v ) , c u cos( c v ) , , h ′ ( v ) (cid:1) ,e = 1 a ( h ( v )) p c u a ( h ( v )) + h ′ ( v ) (cid:0) sin( c v ) h ′ ( v ) , − cos( c v ) h ′ ( v ) , , c u a ( h ( v )) (cid:1) , where e , e are the normalizations of ∂ u and ∂ v , respectively, and e is a unit vector field of M along χ on R × a H . The dual 1-forms on M are ω = 0 , ω = a ( h ( v )) √ u d u, ω = p c u a ( h ( v )) + h ′ ( v ) d v, ω = 0 . We need to introduce the basis E = 1 a (0 , , , , E = 1 a (0 , , , , E = 1 a (1 , , , , E = ∂ t . Now, we can compute the map B : M → S , B = ( h E α , e β i ), with det B = 1, B = −√ u − u u sin( c v ) √ u sin( c v ) cua ( h ( v )) cos( cv ) √ c u a ( h ( v )) + h ′ ( v ) − cos( cv ) h ′ ( v ) √ c u a ( h ( v )) + h ′ ( v ) u cos( cv ) √ u cos( c v ) − cua ( h ( v )) sin( cv ) √ c u a ( h ( v )) + h ′ ( v ) sin( cv ) h ′ ( v ) √ c u a ( h ( v )) + h ′ ( v ) h ′ ( v ) √ c u a ( h ( v )) + h ′ ( v ) cua ( h ( v )) √ c u a ( h ( v )) + h ′ ( v ) . Note that B does not lie in the orthogonal group O (4). From this, a straightforward compu-tation gives the s -valued 1-form Υ = (Υ αβ ) = B − d B , whereΥ αα = 0 , Υ = Υ = d u √ u , Υ = Υ = c u a ( h ( v ))d v p c u a ( h ( v )) + h ′ ( v ) , Υ = Υ = − cuh ′ ( v )d v p c u a ( h ( v )) + h ′ ( v ) , Υ = − Υ = − c u √ u a ( h ( v ))d v p c u a ( h ( v )) + h ′ ( v ) , Υ = − Υ = c √ u h ′ ( v )d v p c u a ( h ( v )) + h ′ ( v ) , Υ = − Υ = − c a ( h ( v )) h ′ ( v )d u + u (cid:0) a ′ ( h ( v )) h ′ ( v ) − a ( h ( v )) h ′′ ( v ) (cid:1) d vc u a ( h ( v )) + h ′ ( v ) A straightforward computation shows dΥ + Υ ∧ Υ = 0. Next, the tangent vector T is T ≡ χ ∗ T = h ∂ t , e i e + h ∂ t , e i e = h ′ ( v ) p c u a ( h ( v )) + h ′ ( v ) e . T α are η = h ′ ( v )d v, T = T = 0 , T = h ′ ( v ) p c u a ( h ( v )) + h ′ ( v ) , T = cua ( h ( v )) p c u a ( h ( v )) + h ′ ( v ) . Clearly, it holds d η = 0. Also, note that T α = B α . We recall that Ω = ( ω αβ ) = B − d B + X ,where X = ( X αβ ), X αβ = ε a ′ ◦ πa ◦ π (cid:16) B n +1 β ω α − ε α ε β B n +1 α ω β (cid:17) , X αα = X α = X α = 0 , X = − X = a ′ ( h ( v )) h ′ ( v )d u √ u p c u a ( h ( v )) + h ′ ( v ) , X = − X = cua ( h ( v )) a ′ ( h ( v ))d u √ u p c u a ( h ( v )) + h ′ ( v ) , X = − X = cua ′ ( h ( v ))d v, In this way,Ω = ( ω αβ ) , ω αα = 0 , ω = ω = d u √ u , ω = ω = c u a ( h ( v ))d v p c u a ( h ( v )) + h ′ ( v ) ,ω = ω = − cuh ′ ( v )d v p c u a ( h ( v )) + h ′ ( v ) ,ω = − ω = a ′ ( h ( v )) h ′ ( v )d u − c ( u + u ) a ( h ( v ))d v √ u p c u a ( h ( v )) + h ′ ( v ) ,ω = − ω = c ua ( h ( v )) a ′ ( h ( v ))d u + (1 + u ) h ′ ( v )d v √ u p c u a ( h ( v )) + h ′ ( v ) ,ω = − ω = ca ( h ( v )) h ′ ( v )d u + cu [ a ′ ( h ( v ))( c u a ( h ( v )) + 2 h ′ ( v ) ) − a ( h ( v )) h ′′ ( v )]d vc u a ( h ( v )) + h ′ ( v ) . Now, we compute the shape operator. Since ω i ( X ) = − ε i h AX, e i i = − ω i ( AX ), then AX = X i ω i ( AX ) e i = − X i ω i ( X ) e i = − c ua ( h ( v )) a ′ ( h ( v ))d u ( X ) + (1 + u ) h ′ ( v )d v ( X ) √ u p c u a ( h ( v )) + h ′ ( v ) e − ca ( h ( v )) h ′ ( v )d u ( X ) + cu [ a ′ ( h ( v ))( c u a ( h ( v )) + 2 h ′ ( v ) ) − a ( h ( v )) h ′′ ( v )]d v ( X ) c u a ( h ( v )) + h ′ ( v ) e ,Ae = − a ′ ( h ( v )) h ′ ( v ) a ( h ( v )) p c u a ( h ( v )) + h ′ ( v ) e , Ae = − c ua ( h ( v )) (cid:16) ue − √ u e (cid:17)p c u a ( h ( v )) + h ′ ( v ) . It is well-known that a non-degenerate hypersurface of a semi-Riemannian manifold mustsatisfy Gauß and Codazzi equations. Our main concern is the converse problem. Indeed,we show that a semi-Riemannian manifold endowed with a tensor which plays the rule of asecond fundamental form, satisfying the Gauß and Codazzi equations, and extra condition isneeded to obtain a local isometric immersion as a non-degenerate hypersurface of a warped25roduct of an interval and a semi-Riemannian space of constant curvature. Indeed, among allconditions of Definition 1, equation (D) cannot be deduced from Codazzi (E) and Gauß (F)equations. This means that, in general, one cannot consider a Riemannian manifold endowedwith a second fundamental form, and think of it as a spacelike hypersurface of some spacetime.However, if one fixes the spacetime first and then consider a hypersurface, everything worksas expected.
Acknowledgments
The second author is partially supported by the Spanish MEC-FEDER Grant MTM2007-60731 and by the Junta de Andaluc´ıa Grant P09-FQM-4496 (with FEDER funds.)