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 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.

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 hypersurfaces M 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.

Let M 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.

Let M 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 an indefinite trans-Sasakian manifold [1, 2] if there exist a (1,1)-type tensor field J, a vector field ζ which is called the structure 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 an indefinite trans-Sasakian structure of type (α,β).

In the entire discussion of this paper, we may assume that ζ 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, let X,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 hypersurface M of M¯ is said to be

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 case ρ=0 on U, we say that M is totally geodesic;

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 case γ=0 on U, we say that M is screen totally geodesic;

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 by R¯, 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

Let M 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) Let M be a totally umbilical lightlike hypersurface of an indefinite trans-Sasakian manifold M¯. Then α=0 and M is totally geodesic.

(2) Let M 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) If M 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) If M 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.

If M 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.

As J(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.

Let M 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) If U 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) If V 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 case V 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.

Let M 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.

If F 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 with Z∈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.

Let M 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.

As F 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 an indefinite 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.

Let M 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 that α is a constant and αβ=0, if α≠0, then we get β=0.

Assume that α=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.

As β=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 with β-Kenmotsu structure, admitting a lightlike hypersurface.

Corollary 8.

Let M be a lightlike hypersurface of an indefinite Sasakian space form M¯(c), endowed with α-Sasakian structure. Then α=±1.

Theorem 9.

Let M 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.

As M 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.

Let M 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.

As M 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.

Let R(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 its induced Ricci tensor if it is symmetric. A symmetric R(0,2) tensor will be denoted by Ric. A lightlike manifold M equipped with an induced Ricci tensor is called Ricci flat if its Ricci tensor vanishes. M is called an Einstein manifold if the Ricci tensor of M satisfies Ric=γg.

If M 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)trAξ∗-g(Aξ∗X,Aξ∗Y)}.This implies that R(0,2) is a symmetric induced Ricci tensor Ric.

Theorem 11.

Any screen conformal Einstein lightlike hypersurface of an indefinite generalized Sasakian space form M¯(f1,f2,f3) is Ricci flat.

Proof.

As M 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=trAξ∗ 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 FieldsDefinition 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 the transversal 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 is flat [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].

Denote λ and μ by the 1-forms such that(91)λ(X)=B(X,U)=C(X,V),δ(X)=B(X,V).

Theorem 13.

Let M be a lightlike hypersurface of an indefinite generalized Sasakian space form M¯(f1,f2,f3). If one of the following conditions,

F is parallel with respect to the connection ∇,

U is parallel with respect to the connection ∇,

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 that F 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.

As f1=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) If U 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) If V 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.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

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