# Lightlike Hypersurfaces of Indefinite Generalized Sasakian Space Forms

Authors: Dae Ho Jin
Journal: Journal of Applied Mathematics
 (2015)

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## Abstract

We study lightlike hypersurfacesM of an indefinite generalized Sasakian space form M-(f1,f2,f3), with indefinite trans-Sasakian structure of type (α,β), subject to the condition that the structure vector field of M- is tangent to M. First we study the general theory for lightlike hypersurfaces of indefinite trans-Sasakian manifold of type (α,β). Next we prove several characterization theorems for lightlike hypersurfaces of an indefinite generalized Sasakian space form.


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## Body

## 1. Introduction

Oubiña [1] introduced the notion of indefinite trans-Sasakian manifold of type (α,β). Indefinite Sasakian, Kenmotsu, and cosymplectic manifolds are three important kinds of indefinite trans-Sasakian manifold such that(1)α=1,β=0;α=0,β=1;α=β=0,respectively. Alegre et al. [2] introduced indefinite generalized Sasakian space form M¯(f1,f2,f3). Indefinite Sasakian, Kenmotsu, and cosymplectic space forms are some kinds of indefinite generalized Sasakian space form such that (2)f1=c+34,f2=f3=c-14;f1=c-34,f2=f3=c+14;f1=f2=f3=c4,respectively, where c denotes constant J-sectional curvatures of each of them.Recently author has been studying the geometry of lightlike hypersurfacesM of indefinite Sasakian [3], Kenmotsu [4], and cosymplectic [5] manifolds. In this paper, we study lightlike hypersurfaces M of an indefinite generalized Sasakian space form M¯(f1,f2,f3), with indefinite trans-Sasakian structure of type (α,β), subject to the condition that the structure vector field of M¯ is tangent to M. First we study lightlike hypersurfaces of indefinite trans-Sasakian manifold of type (α,β). Next we prove two characterization theorems for lightlike hypersurfaces of an indefinite generalized Sasakian space form such that the following hold.(i)
LetM be a lightlike hypersurface of an indefinite generalized Sasakian space form M¯(f1,f2,f3). Then α is a constant, β=0, and(3)f1-f2=α2,f1-f3=α2,f2=f3.(ii)
LetM be a screen conformal lightlike hypersurface of an indefinite generalized Sasakian space form M¯(f1,f2,f3). Then f1=f2=f3=0.


## 2. Preliminaries

An odd-dimensional semi-Riemannian manifold(M¯,g¯) is called anindefinite trans-Sasakian manifold [1, 2] if there exist a (1,1)-type tensor field J, a vector field ζ which is called thestructure vector field, and a 1-form θ such that(4)J2X=-X+θXζ,θζ=1,Jζ=0,θ∘J=0,(5)g¯(JX,JY)=g¯(X,Y)-ϵθ(X)θ(Y),ϵ=g¯(ζ,ζ),(6)∇¯XJY=αg¯X,Yζ-ϵθYX+βg¯JX,Yζ-ϵθYJX,for any vector fields X and Y on M¯, where ϵ=1 or -1 according to the fact that ζ is spacelike or timelike, respectively. In this case, the set {J,ζ,θ,g¯} is called anindefinite trans-Sasakian structure of type(α,β).In the entire discussion of this paper, we may assume thatζ is unit spacelike; that is, ϵ=1, without loss of generality. From (4) and (6), we get(7)∇¯Xζ=-αJX+β(X-θ(X)ζ),dθ(X,Y)=g(X,JY).Let(M,g) be a lightlike hypersurface, with a screen distribution S(TM), of an indefinite trans-Sasakian manifold M¯. Denote by F(M) the algebra of smooth functions on M and by Γ(E) the F(M) module of smooth sections of a vector bundle E. Also donate by Equationnumberi the ith equation of several equations in (Equation number), for example, (7)1 donates the first equation of the two equations in (7). We use same notations for any others.We follow Duggal-Bejancu [6] for notations and structure equations used in this paper. It is well known that, for any null section ξ of TM⊥ on a coordinate neighborhood U⊂M, there exists a unique null section N of a unique vector bundle tr⁡(TM) of rank 1 in S(TM)⊥ satisfying(8)g¯ξ,N=1,g¯(N,N)=g¯(N,X)=0,∀X∈Γ(S(TM)).In the following, letX,Y,Z, and W be the vector fields on M, unless otherwise specified. Let ∇¯ be the Levi-Civita connection of M¯ and P the projection morphism of TM on S(TM). Then the local Gauss and Weingarten formulas are given by(9)∇¯XY=∇XY+BX,YN,(10)∇¯XN=-ANX+τXN;(11)∇XPY=∇X∗PY+CX,PYξ,(12)∇Xξ=-Aξ∗X-τ(X)ξ,where ∇ and ∇∗ are the liner connections on M and S(TM), respectively, B and C are the local second fundamental forms on M and S(TM) respectively, AN and Aξ∗ are the shape operators on M and S(TM), respectively, and τ is a 1-form on TM.Since∇¯ is torsion-free, ∇ is also torsion-free and B is symmetric. From the fact that B(X,Y)=g¯(∇¯XY,ξ), we show that B is independent of the choice of S(TM) and satisfies(13)B(X,ξ)=0.The induced connection∇ of M is not metric and satisfies(14)(∇Xg)(Y,Z)=B(X,Y)η(Z)+B(X,Z)η(Y),where η is a 1-form such that(15)η(X)=g¯(X,N).But the connection ∇∗ on S(TM) is metric. The above two local second fundamental forms B and C are related to their shape operators by(16)BX,Y=gAξ∗X,Y,g¯Aξ∗X,N=0,(17)C(X,PY)=g(ANX,PY),g¯(ANX,N)=0.Definition 1.
A lightlike hypersurfaceM of M¯ is said to be (1)
totally umbilical [6] if there is a smooth function ρ on any coordinate neighborhood U in M such that Aξ∗X=ρPX, or equivalently,(18)B(X,Y)=ρg(X,Y).In caseρ=0 on U, we say that M istotally geodesic;(2)
screen totally umbilical [6] if there exists a smooth function γ on U such that ANX=γPX, or equivalently,(19)C(X,PY)=γg(X,Y).In caseγ=0 on U, we say that M isscreen totally geodesic;(3)
screen conformal [7] if there exists a nonvanishing smooth function φ on U such that AN=φAξ∗, or equivalently,(20)C(X,PY)=φB(X,Y).Denote byR¯, R, and R∗ the curvature tensors of the Levi-Civita connection ∇¯ of M¯, the induced connection ∇ on M, and the induced connection ∇∗ on S(TM), respectively. Using the Gauss-Weingarten formulas for M and S(TM), we obtain the Gauss-Codazzi equations for M and S(TM) such that(21)R¯X,YZ=R(X,Y)Z+B(X,Z)ANY-B(Y,Z)ANX+∇XBY,Z-∇YBX,Z+τXBY,Z-τYBX,Z∇XBY,Z-∇YBX,ZN,(22)R¯X,YN=-∇X(ANY)+∇Y(ANX)+AN[X,Y]+τ(X)ANY-τ(Y)ANX+{B(Y,ANX)-B(X,ANY)+2dτ(X,Y)}N,(23)RX,YPZ=R∗X,YPZ+CX,PZAξ∗Y-C(Y,PZ)Aξ∗X+∇XCY,PZ-∇YCX,PZ-τXCY,PZ+τYCX,PZτY∇XCY,PZ-∇YCX,PZξ,(24)RX,Yξ=-∇X∗(Aξ∗Y)+∇Y∗(Aξ∗X)+Aξ∗[X,Y]-τ(X)Aξ∗Y+τ(Y)Aξ∗X+{C(Y,Aξ∗X)-C(X,Aξ∗Y)-2dτ(X,Y)}ξ.


## 3. Indefinite Trans-Sasakian Manifolds

LetM be a lightlike hypersurface of a indefinite trans-Sasakian manifold M¯ such that ζ is tangent to M. Călin [8] proved that if ζ is tangent to M, then it belongs to S(TM) which we assume in this paper. It is well known [3, 6] that, for any lightlike hypersurface M of an indefinite almost contact metric manifold M¯, J(TM⊥) and J(tr⁡(TM)) are subbundles of S(TM), of rank 1, and J(TM⊥)∩J(tr⁡(TM))={0}. Thus J(TM⊥)⊕J(tr⁡(TM)) is a subbundle of S(TM) of rank 2. First, we prove the following results.Theorem 2.
(1) LetM be a totally umbilical lightlike hypersurface of an indefinite trans-Sasakian manifold M¯. Then α=0 and M is totally geodesic.
(2) LetM be a screen conformal or screen totally umbilical lightlike hypersurface of an indefinite trans-Sasakian manifold M¯. Then α=β=0. In case M is screen totally umbilical, M is totally geodesic.Proof.
Applying∇¯X to g(ζ,ξ)=0 and g(ζ,N)=0, we have(25)B(X,ζ)=αg(X,Jξ),C(X,ζ)=αg(X,JN)+βη(X).
(1) IfM is totally umbilical, then, from (18) and (25)1, we have(26)ρg(X,ζ)=αg(X,Jξ),∀X∈Γ(TM).Taking X=ζ and X=JN by turns, we have ρ=0 and α=0, respectively. As ρ=0, M is totally geodesic.
(2) IfM is screen conformal, then, from (20) and (25)1,2, we have(27)αφg(X,Jξ)=αg(X,JN)+βη(X).Taking X=Jξ and X=ξ by turns, we have α=0 and β=0, respectively.
IfM is screen totally umbilical, then, from (19) and (25)2, we have(28)γg(X,ζ)=αg(X,JN)+βη(X).Taking X=ζ, X=Jξ and X=ξ to this equation by turns, we have γ=0, α=0, and β=0, respectively. As γ=0, M is screen totally geodesic.AsJ(TM⊥)⊕J(tr⁡(TM)) is a subbundle of S(TM) of rank 2, there exists a nondegenerate almost complex distribution Do with respect to J; that is, J(Do)=Do, such that(29)S(TM)={J(TM⊥)⊕J(tr⁡(TM))}⊕orthDo,TM={J(TM⊥)⊕J(tr⁡(TM))}⊕orthDo⊕orthTM⊥.Consider the 2-lightlike almost complex distribution D such that(30)D=TM⊥⊕orthJTM⊥⊕orthDo,TM=D⊕J(tr⁡(TM))and the local lightlike vector fields U and V and their 1-forms such that(31)U=-JN,V=-Jξ,uX=gX,V,v(X)=g(X,U).Denote by S the projection morphism of TM on D. Any vector field X of M is expressed as X=SX+u(X)U. Applying J to this, we have(32)JX=FX+u(X)N,where F is a tensor field of type (1, 1) globally defined on M by(33)FX=JSX.Applying∇¯X to the first two equations of (31) and (32) and using (9), (10), (12), (13), (6), (31), and (32), for any X,Y∈Γ(TM), we have(34)BX,U=CX,V,(35)∇XU=FANX+τXU-αηX+βvXζ,(36)∇XV=FAξ∗X-τXV-βuXζ,(37)∇XFY=uYANX-BX,YU+α{g(X,Y)ζ-θ(Y)X}+β{g¯(JX,Y)ζ-θ(Y)FX}.Theorem 3.
LetM be a lightlike hypersurface of an indefinite trans-Sasakian manifold M¯. If  V or U is parallel with respect to ∇, then α=β=0 and τ=0. If both V and U are parallel with respect to the induced connection ∇, then M is screen totally geodesic.Proof.
(1) IfU is parallel, then, from (32) and (35) we have(38)J(ANX)-u(ANX)N+τ(X)U-{αη(X)+βv(X)}ζ=0.Taking the scalar product with V and ζ to (38) by turns and using (4), we have τ=0 and αη(X)+βv(X)=0, respectively. Taking X=ξ and X=V to the second result by turns, we have α=0 and β=0, respectively.
(2) IfV is parallel with respect to ∇, then, from (32) and (36), we have(39)J(Aξ∗X)-u(Aξ∗X)N-τ(X)V-βu(X)ζ=0.Taking the scalar product with U to (39) and using (4), we have τ=0. Taking the scalar product with ζ to (39) and using (4) and θ(N)=g(ζ,N)=0, we get βu(X)=0. Taking X=U to this result, we have β=0. From (25)1 and (31)3, we obtain(40)B(X,ζ)=-αu(X).Applying J to (39) and using (4) and the fact τ=β=0, we have(41)Aξ∗X=θ(Aξ∗X)ζ+u(Aξ∗X)U.Taking the scalar product with U to this equation, we get(42)B(X,U)=g(Aξ∗X,U)=v(Aξ∗X)=0.Replacing X by U in (40) and using (42), we get(43)-α=-αu(U)=B(U,ζ)=0.Thus α=β=0. Then we have(44)Aξ∗X=u(Aξ∗X)U.
(3) In caseV and U are parallel with respect to ∇, as U is parallel, applying J to (38) and using (4), (25)2 and the fact τ=α=β=0, we obtain(45)ANX=u(ANX)U,∀X∈Γ(TM).As V is parallel, from (34) and (42), we show that u(ANX)=v(Aξ∗X)=0. Thus we obtain AN=0. Consequently M is screen totally geodesic.Theorem 4.
LetM be a lightlike hypersurface of an indefinite trans-Sasakian manifold M¯. If F is parallel with respect to the connection ∇, then we have α=β=0. Furthermore D and J(tr⁡(TM)) are parallel distributions on M and M is locally a product manifold Cu×M#, where Cu is a null curve tangent to J(tr⁡(TM)) and M# is a leaf of D.Proof.
IfF is parallel with respect to ∇, then, taking the scalar product with U to (37) and using the facts g(ζ,U)=0 and g(FX,U)=-η(X), we get(46)u(Y)v(ANX)-θ(Y){αv(X)-βη(X)}=0.Taking Y=U and Y=ζ by turns, we get v(ANX)=0 and αv(X)-βη(X)=0. Taking X=V and X=ξ to the second equation, we have α=β=0.
From (37) we have(47)u(Y)ANX=B(X,Y)U,B(X,Y)=u(Y)u(ANX).Taking Y=V and Y∈Γ(Do) in (47)2 by turns, we have B(X,V)=0 and B(X,Y)=0. These results and (13) imply that(48)BX,Y=0,∀X∈ΓTM,Y∈Γ(D).By using (4), (9), (12), (14), (32), and (36), we derive (49)g∇Xξ,V=-gξ,∇¯XV=BX,V=0,g∇XV,V=0,g(∇XY,V)=-g(Y,∇XV)=g(Aξ∗X,JY)=B(X,FY)=0,for all X∈Γ(TM) and Y∈Γ(Do), or equivalently, we get(50)∇XY∈Γ(D),∀X∈Γ(TM),∀Y∈Γ(D).This result implies that D is a parallel distribution on M.
Taking the scalar product withZ∈Γ(Do) to (47)1, we get u(Y)C(X,Z)=0 for all X,Y∈Γ(TM). Taking Y=U to this, we have(51)CX,Y=0,∀X∈ΓTM,Y∈Γ(Do).For all X∈Γ(TM) and Y∈Γ(Do), using (35) we derive (52)g∇XU,N=vANX=0,g∇XU,U=-gANX,N=0,g∇XU,Y=g(F(ANX),Y)lllllllllllllllll=-g(ANX,JY)=C(X,FY)=0;that is, ∇XU∈Γ(J(tr⁡(TM))) for all X∈Γ(TM).  Thus J(tr⁡(TM)) is also parallel. As TM=D⊕J(tr⁡(TM)), and D and J(tr⁡(TM)) are parallel distributions, by the decomposition theorem of de Rham [9] we have M=Cu×M#, where Cu is a null curve tangent to J(tr⁡(TM)) and M# is a leaf of D.Corollary 5.
LetM be a lightlike hypersurface of an indefinite trans-Sasakian manifold M¯. If F and V are parallel with respect to ∇, then M is totally geodesic and screen totally geodesic.Proof.
AsF is parallel with respect to ∇, we get the two equations of (47). As V is also parallel with respect to ∇, substituting (34) to (47)2 and using (42), we have B=0. Thus M is totally geodesic. Replacing Y by U to (47)1, we obtain AN=0. Thus M is also screen totally geodesic.


## 4. Indefinite Generalized Sasakian Space Form

An indefinite almost contact metric manifold(M¯,J,ζ,θ,g¯) is said to be anindefinite generalized Sasakian space form [2] and denote it by M¯(f1,f2,f3) if there exist three smooth functions f1,  f2, and f3 on M¯ such that(53)R¯X,YZ=f1{g¯(Y,Z)X-g¯(X,Z)Y}+f2{g¯(X,JZ)JY-g¯(Y,JZ)JX+2g¯(X,JY)JZ}+f3+g¯(X,Z)θ(Y)ζ-g¯(Y,Z)θ(X)ζθXθZY-θYθZX+g¯(X,Z)θ(Y)ζ-g¯(Y,Z)θ(X)ζ,for any vector fields X, Y, and Z on M¯.Theorem 6.
LetM be a lightlike hypersurface of an indefinite generalized Sasakian space form M¯(f1,f2,f3). Then α is a constant, β=0, and(54)f1-f2=α2,f1-f3=α2,f2=f3.Proof.
Comparing the tangential and transversal components of (21) and (53), and using (32), we get(55)RX,YZ=f1{g(Y,Z)X-g(X,Z)Y}+f2{g¯(X,JZ)FY-g¯(Y,JZ)FX+2g¯(X,JY)FZ}+f3g¯X,ZθYζ-g¯Y,ZθXζθ(X)θ(Z)Y-θ(Y)θ(Z)X+g¯X,ZθYζ-g¯Y,ZθXζ+B(Y,Z)ANX-B(X,Z)ANY,(56)(∇XB)(Y,Z)-(∇YB)(X,Z)+τ(X)B(Y,Z)-τ(Y)B(X,Z)=f2+2u(Z)g¯(X,JY)u(Y)g¯(X,JZ)-u(X)g¯(Y,JZ)+2u(Z)g¯(X,JY).Taking the scalar product with N to (23), we have(57)gRX,YPZ,N=(∇XC)(Y,PZ)-(∇YC)(X,PZ)-τ(X)C(Y,PZ)+τ(Y)C(X,PZ).Substituting (55) into the last equation and using (17)2, we obtain(58)(∇XC)(Y,PZ)-(∇YC)(X,PZ)-τ(X)C(Y,PZ)+τ(Y)C(X,PZ)=f1{g(Y,PZ)η(X)-g(X,PZ)η(Y)}+f2v(Y)g¯(X,JPZ)-v(X)g¯(Y,JPZ)+2vPZg¯X,JY+f3θXηY-θYηXθPZ.
Applying∇X to (34)1: B(Y,U)=C(Y,V), we have(59)∇XBY,U=(∇XC)(Y,V)+g(ANY,∇XV)-g(Aξ∗Y,∇XU).Using (25), (32), (34), (35), and (36), the above equation is reduced to(60)∇XBY,U=(∇XC)(Y,V)-2τ(X)C(Y,V)-α2uYηX-β2uXηY+αβuXvY-uYvX-g(Aξ∗X,F(ANY))-g(Aξ∗Y,F(ANX)).Substituting this equation and (34) into (56) such that Z=U, we get (61)(∇XC)(Y,V)-(∇YC)(X,V)-τ(X)C(Y,V)+τYCX,V+α2-β2uXηY-uYηX+2αβ{u(X)v(Y)-u(Y)v(X)}=f2{u(Y)η(X)-u(X)η(Y)+2g¯(X,JY)}.Comparing this equation with (58) such that PZ=V, we obtain(62){f1-f2-(α2-β2)}[u(Y)η(X)-u(X)η(Y)]=2αβ{u(Y)v(X)-u(X)v(Y)}.Taking X=ξ and Y=U and X=V and Y=U by turns, we have(63)f1-f2=α2-β2,αβ=0.
Substituting (32) into (7) and using (9), we have(64)∇Xζ=-αFX+β(X-θ(X)ζ),∀X∈Γ(TM).Applying ∇¯X to v(Y)=g(Y,U) and using (9), (32), (34), and (35), we get(65)∇XvY=v(Y)τ(X)-θ(Y){αη(X)+βv(X)}-g(ANX,FY).Applying ∇¯X to η(Y)=g¯(Y,N) and using (4) and (6) we have(66)(∇Xη)(Y)=-g(ANX,Y)+τ(X)η(Y).Using (31), the equation (25)2 is reduced to(67)C(Y,ζ)=-αv(Y)+βη(Y).Applying ∇X to this equation and using (64), (65), and (66), we have (68)∇XCY,ζ=-(Xα)v(Y)+(Xβ)η(Y)-ατ(X)v(Y)+α2θ(Y)η(X)+β2θ(X)η(Y)-β{g(X,ANY)+g(ANX,Y)-τ(X)η(Y)}+α{g(ANX,FY)+g(ANY,FX)}.Substituting this and (67) into (58) such that PZ=ζ, we get (69){Xβ+Aθ(X)}η(Y)-{Yβ+Aθ(Y)}η(X)=(Xα)v(Y)-(Yα)v(X),where A=f1-f3-(α2-β2). Taking X=ξ and Y=ζ and then taking X=U and Y=V to this equation, we obtain(70)f1-f3=(α2-β2)-ζβ,Uα=0.
Applying∇¯X to u(Y)=g(Y,V) and using (9), (32), and (36), we get(71)(∇Xu)(Y)=-u(Y)τ(X)-βθ(Y)u(X)-B(X,FY).Applying ∇Y to (40) and using (40) and (64) and (71), we have(72)∇XBY,ζ=-(Xα)u(Y)-βB(X,Y)+α{u(Y)τ(X)+B(X,FY)+B(Y,FX)}.Substituting this into (56) such that Z=ζ and using the fact that Uα=0, we have (Xα)u(Y)=0. Therefore the function α is a constant.
From the facts thatα is a constant and αβ=0, if α≠0, then we get β=0.
Assume thatα=0. Then (64) is reduced to(73)∇Yζ=β(Y-θ(Y)ζ).By straightforward calculations form this equation, we obtain(74)RX,Yζ=(Xβ)Y-(Yβ)X-{(Xβ)θ(Y)-(Yβ)θ(X)}ζ+β2{θ(X)Y-θ(Y)X}-2βdθ(X,Y)ζ.Comparing this equation with (55) such that Z=ζ, we obtain(75)(Xβ)Y-(Yβ)X-{(Xβ)θ(Y)-(Yβ)θ(X)}ζ+β2{θ(X)Y-θ(Y)X}-2βdθ(X,Y)ζ=(f1-f3){θ(Y)X-θ(X)Y}.Taking the scalar product with ζ to this equation, we get βdθ(X,Y)=0; that is,(76)βg(X,JY)=0,∀X,Y∈Γ(TM),due to (32)2. Taking X=U and Y=ξ to this equation, we have β=0.
Asβ=0, (63) and (70) are reduced to f1-f2=α2 and f1-f3=α2, respectively. From these two results, we get f2=f3.Corollary 7.
There exist no indefinite generalized Sasakian space forms, endowed withβ-Kenmotsu structure, admitting a lightlike hypersurface.Corollary 8.
LetM be a lightlike hypersurface of an indefinite Sasakian space form M¯(c), endowed with α-Sasakian structure. Then α=±1.Theorem 9.
LetM is lightlike hypersurface of an indefinite generalized Sasakian space form M¯(f1,f2,f3). If M is screen totally umbilical, then f1=f2=f3=0.Proof.
AsM is screen totally umbilical, α=β=C=0 by (2) of Theorem 2. Thus (58) is reduced to(77)f1{g(Y,PZ)η(X)-g(X,PZ)η(Y)}+f2vYg¯X,JPZ-vXg¯Y,JPZ+2v(PZ)g¯(X,JY)+f3{θ(X)θ(PZ)η(Y)-θ(Y)θ(PZ)η(X)}=0,for all X,Y,Z∈Γ(TM). Replacing Y by ξ to this equation, we obtain(78)f1g(X,PZ)+f2{v(X)u(PZ)+2u(X)v(PZ)}-f3θ(X)θ(PZ)=0.Taking X=V, PZ=U; X=U, PZ=V, and X=PZ=ζ by turns, we have(79)f1+f2=0,f1+2f2=0,f1=f3.From the first two equations we show that f2=0. As α=β=0, M¯ is an indefinite cosymplectic manifold. Thus f1=f2=f3=c/4. This implies f1=f2=f3=0.Theorem 10.
LetM be a screen conformal lightlike hypersurface of an indefinite generalized Sasakian space form M¯(f1,f2,f3). Then f1=f2=f3=0.Proof.
Substituting (55) into (57) and using (56), we have(80)f1{g(Y,PZ)η(X)-g(X,PZ)η(Y)}+f2[v(Y)-u(Y)]g¯(X,JPZ)-vX-uXg¯Y,JPZ+2vPZ-uPZg¯X,JY+f3{θ(X)θ(PZ)η(Y)-θ(Y)θ(PZ)η(X)}={X[φ]-2φτ(X)}B(Y,PZ)-{Y[φ]-2φτ(Y)}B(X,PZ).Replacing Y by ξ to the last equation, we obtain(81)ξφ-2φτξBX,PZ=f1g(X,PZ)+f2{v(X)-u(X)}u(PZ)+2f2{v(PZ)-u(PZ)}u(X)-f3θ(X)θ(PZ).Taking X=PZ=ζ to this equation and using (40), we obtain f1=f3. Also taking X=V, PZ=U, and X=U, PZ=V by turns, we have(82)ξφ-2φτξBV,U=f1+f2,{ξ[φ]-2φτ(ξ)}B(U,V)=f1+2f2,respectively. Comparing these two equations, we obtain f2=0.
AsM is screen conformal, we obtain α=β=0 by Theorem 2. As α=β=0, we show that M¯ is a cosymplectic manifold and f1=f2=f3=c/4. Therefore we get f1=f2=f3=0.
LetR(0,2) denote the induced Ricci type tensor of M given by(83)R(0,2)(X,Y)=trace{Z⟶R(Z,X)Y},for any X, Y∈Γ(TM). Consider the induced quasi-orthonormal frame field {ξ;Wa} on M such that TM⊥=Span⁡{ξ} and S(TM)=Span⁡{Wa}. Put m=rank(S(TM)). Using this quasi-orthonormal frame field, we obtain(84)R(0,2)(X,Y)=∑a=1mϵag(R(Wa,X)Y,Wa)+g¯(R(ξ,X)Y,N),for any X, Y∈Γ(TM), where ϵa=g(Wa,Wa) is the causal character of Wa. In general, the induced Ricci type tensor R(0,2) is not symmetric [6, 7]. A tensor field R(0,2) of lightlike submanifolds M is called itsinduced Ricci tensor if it is symmetric. A symmetric R(0,2) tensor will be denoted byRic. A lightlike manifold M equipped with an induced Ricci tensor is calledRicci flat if its Ricci tensor vanishes. M is called anEinstein manifold if the Ricci tensor of M satisfies Ric=γg.
IfM is a screen conformal lightlike hypersurface of M¯(f1,f2,f3), then, using (55) and the fact that f1=f2=f3=0, we have (85)R(0,2)(X,Y)=φ{B(X,Y)tr⁡Aξ∗-g(Aξ∗X,Aξ∗Y)}.This implies that R(0,2) is a symmetric induced Ricci tensorRic.Theorem 11.
Any screen conformal Einstein lightlike hypersurface of an indefinite generalized Sasakian space formM¯(f1,f2,f3) is Ricci flat.Proof.
AsM is Einstein, from (85) and the fact R(0,2)=γg(86)g(Aξ∗X,Aξ∗Y)-αg(Aξ∗X,Y)-γφ-1g(X,Y)=0,where α=tr⁡Aξ∗ is trace of Aξ∗. Define a nonnull vector field μ on S(TM) by(87)μ=U-φV.Then μ belongs to J(TM⊥)⊕J(tr⁡(TM)). Using (20) and (34), μ satisfies(88)B(X,μ)=0,∀X∈Γ(TM).From this equation and (16), we show that(89)Aξ∗μ=0.Taking X=Y=μ to (86) and using (89), we get γ=0. Therefore, M is Ricci flat.


## 5. Parallel Structure Fields

Definition 12.
Let∇X⊥N=π(∇¯XN) for any X∈Γ(TM), where π is the projection morphism of TM¯ on tr⁡(TM). Then ∇⊥ is a linear connection on ltr(TM). We say that ∇⊥ is thetransversal connection of M. We define the curvature tensor R⊥ of tr⁡(TM) by(90)R⊥(X,Y)N=∇X⊥∇Y⊥N-∇Y⊥∇X⊥N-∇[X,Y]⊥N.The transversal connection of M isflat [3] if R⊥ vanishes.As∇X⊥N=τ(X)N, we show that the transversal connection of M is flat if and only if the 1-form τ is closed; that is, dτ=0, on any U⊂M [3].Denoteλ and μ by the 1-forms such that(91)λ(X)=B(X,U)=C(X,V),δ(X)=B(X,V).Theorem 13.
LetM be a lightlike hypersurface of an indefinite generalized Sasakian space form M¯(f1,f2,f3). If one of the following conditions, (1)
F is parallel with respect to the connection ∇,(2)
U is parallel with respect to the connection ∇,(3)
V is parallel with respect to the connection ∇, is satisfied, then M¯(f1,f2,f3) is a flat manifold with indefinite cosymplectic structure and the lightlike transversal connection of M is flat. In case (1), M is also a flat manifold.Proof.
(1) Assume thatF is parallel with respect to ∇. Then we get α=β=0 by Theorem 4. Thus f1=f2=f3 by Theorem 6 and (37) is reduced to(92)u(Y)ANX-B(X,Y)U=0.Taking Y=U to (92) and using (31), we have(93)ANX=λ(X)U.Taking the scalar product with V to (92) and using (17) and (31), we have(94)g(Aξ∗X,Y)=g(λ(X)V,Y).As Aξ∗X and V belong to S(TM) and S(TM) is nondegenerate, we have(95)Aξ∗X=λ(X)V.Taking the scalar product with U to (93), we obtain(96)C(X,U)=0.Applying ∇X to C(Y,U)=0 and using (37), (93) and FU=0, we get(97)(∇XC)(Y,U)=0.Replacing PZ by U to (58) and using the last two equations, we have(98)f1{v(Y)η(X)-v(X)η(Y)}=0.Taking X=V and Y=ξ to this equation, we get f1=0. Therefore, f1=f2=f3=0 and M¯(f1,f2,f3) is flat.
Asf1=f2=f3=0, substituting (93) and (95) into (55), we get (99)RX,YZ={λ(Y)λ(X)-λ(X)λ(Y)}u(Z)U+{σ(Y)σ(X)-σ(X)σ(Y)}w(Z)W=0.Thus M is flat. From (37), (93) and the fact that FU=ρ=0, we get(100)∇XU=τ(X)U.Substituting this equation into ∇X∇YU-∇Y∇XU-∇[X,Y]U=0, we get dτ=0. Thus the transversal connection of M is flat.
(2) IfU is parallel with respect to ∇, then, α=β=τ=0 by Theorem 3. Thus f1=f2=f3 by Theorem 6 and (35) is reduced to(101)J(ANX)-u(ANX)N=0.Applying J to (101) and using (4), (31), and (67), we have(102)ANX=λ(X)U.Taking the scalar product with U to (102), we get(103)C(X,U)=0.Applying ∇Y to this and using (35), (102) and the fact that FU=0, we get(104)(∇XC)(Y,U)=0.Substituting the last two equation into (58) such that PZ=U, we have(105)f1{v(Y)η(X)-v(X)η(Y)}=0.Taking X=V and Y=ξ to this equation, we obtain f1=0. Therefore, f1=f2=f3=0 and M¯(f1,f2,f3) is flat. As τ=0, we obtain dτ=0. Thus the transversal connection of M is flat.
(3) IfV is parallel with respect to ∇, then, α=β=τ=0 by Theorem 3. Thus f1=f2=f3 by Theorem 6 and (35) is reduced to(106)J(Aξ∗X)-u(Aξ∗X)N=0.Applying J to (106) and using (4) and (40), we have(107)Aξ∗X=μ(X)U.Taking the scalar product with U to this equation, we get(108)B(X,U)=0.Applying ∇Y to this equation and using (35), we have(109)(∇XB)(Y,U)=-B(Y,F(ANX)).Substituting the last two equations into (56), we obtain(110)B(X,F(ANY))-B(Y,F(ANX))=f2{u(Y)η(X)-u(X)η(Y)+2g¯(X,JY)}.Taking X=ξ and Y=U to this equation and using (14) and (108), we obtain f2=0. Therefore, f1=f2=f3=0 and M¯(f1,f2,f3) is flat. As τ=0, we obtain dτ=0. Thus the lightlike transversal connection of M is flat.



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*Source: 259146-2015-03-31.xml*