JAM Journal of Applied Mathematics 1687-0042 1110-757X Hindawi Publishing Corporation 10.1155/2015/259146 259146 Research Article Lightlike Hypersurfaces of Indefinite Generalized Sasakian Space Forms Jin Dae Ho Fotakis Dimitris Department of Mathematics, Dongguk University, Gyeongju 780-714 Republic of Korea dongguk.edu 2015 3132015 2015 19 01 2015 16 03 2015 3132015 2015 Copyright © 2015 Dae Ho Jin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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  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.  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 , Kenmotsu , and cosymplectic  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,θJ=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 Equation  numberi 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  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 UM, 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,XΓ(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=XPY+CX,PYξ,(12)Xξ=-AξX-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=gAξX,Y,g¯AξX,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  if there is a smooth function ρ on any coordinate neighborhood U in M such that AξX=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  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  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,ZXBY,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=RX,YPZ+CX,PZAξY-C(Y,PZ)AξX+XCY,PZ-YCX,PZ-τXCY,PZ+τYCX,PZτYXCY,PZ-YCX,PZξ,(24)RX,Yξ=-X(AξY)+Y(AξX)+Aξ[X,Y]-τ(X)AξY+τ(Y)AξX+{C(Y,AξX)-C(X,AξY)-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  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),XΓ(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))}orthDoorthTM.Consider the 2-lightlike almost complex distribution D such that(30)D=TMorthJTMorthDo,TM=DJ(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,YΓ(TM), we have(34)BX,U=CX,V,(35)XU=FANX+τXU-αηX+βvXζ,(36)XV=FAξX-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(AξX)-u(AξX)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)AξX=θ(AξX)ζ+u(AξX)U.Taking the scalar product with U to this equation, we get(42)B(X,U)=g(AξX,U)=v(AξX)=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)AξX=u(AξX)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,XΓ(TM).As V is parallel, from (34) and (42), we show that u(ANX)=v(AξX)=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 YΓ(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,XΓTM,YΓ(D).By using (4), (9), (12), (14), (32), and (36), we derive (49)gXξ,V=-gξ,¯XV=BX,V=0,gXV,V=0,g(XY,V)=-g(Y,XV)=g(AξX,JY)=B(X,FY)=0,for all XΓ(TM) and YΓ(Do), or equivalently, we get(50)XYΓ(D),XΓ(TM),YΓ(D).This result implies that D is a parallel distribution on M.

Taking the scalar product with ZΓ(Do) to (47)1, we get u(Y)C(X,Z)=0 for all X,YΓ(TM). Taking Y=U to this, we have(51)CX,Y=0,XΓTM,YΓ(Do).For all XΓ(TM) and YΓ(Do), using (35) we derive (52)gXU,N=vANX=0,gXU,U=-gANX,N=0,gXU,Y=g(F(ANX),Y)lllllllllllllllll=-g(ANX,JY)=C(X,FY)=0;that is, XUΓ(J(tr(TM))) for all XΓ(TM).  Thus J(tr(TM)) is also parallel. As TM=DJ(tr(TM)), and D and J(tr(TM)) are parallel distributions, by the decomposition theorem of de Rham  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  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(AξY,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(AξX,F(ANY))-g(AξY,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),XΓ(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 α0, 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,YΓ(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,ZΓ(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{ZR(Z,X)Y},for any X, YΓ(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, YΓ(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(AξX,AξY)}.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(AξX,AξY)-αg(AξX,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,XΓ(TM).From this equation and (16), we show that(89)Aξμ=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 XN=π(¯XN) for any XΓ(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=XYN-YXN-[X,Y]N.The transversal connection of M is flat  if R vanishes.

As XN=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 UM .

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(AξX,Y)=g(λ(X)V,Y).As AξX and V belong to S(TM) and S(TM) is nondegenerate, we have(95)AξX=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 XYU-YXU-[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(AξX)-u(AξX)N=0.Applying J to (106) and using (4) and (40), we have(107)AξX=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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