Einstein half lightlike submanifolds of a Lorentzian space form with a semi-symmetric non-metric connection

Abstract

We study screen conformal Einstein half lightlike submanifolds M of a Lorentzian space form $\tilde{M}(c)$ of constant curvature c admitting a semi-symmetric non-metric connection subject to the conditions; (1) the structure vector field of $\tilde{M}$ is tangent to M, and (2) the canonical normal vector field of M is conformal Killing. The main result is a characterization theorem for such a half lightlike submanifold.

MSC:53C25, 53C40, 53C50.

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction

The theory of lightlike submanifolds is used in mathematical physics, in particular, in general relativity as lightlike submanifolds produce models of different types of horizons [1, 2]. Lightlike submanifolds are also studied in the theory of electromagnetism [3]. Thus, large number of applications but limited information available, motivated us to do the research on this subject matter. As for any semi-Riemannian manifold, there is a natural existence of lightlike subspaces, Duggal and Bejancu published their work [3] on the general theory of lightlike submanifolds to fill a gap in the study of submanifolds. Since then, there has been very active study on lightlike geometry of submanifolds (see up-to date results in two books [4, 5]). The class of lightlike submanifolds of codimension 2 is composed of two classes by virtue of the rank of its radical distribution, named by half lightlike and coisotropic submanifolds [6, 7]. Half lightlike submanifold is a special case of general r-lightlike submanifold such that $r=1$, and its geometry is more general form than that of coisotropic submanifold or lightlike hypersurface. Much of the works on half lightlike submanifolds will be immediately generalized in a formal way to general r-lightlike submanifolds of arbitrary codimension n and arbitrary rank r. For this reason, we study half lightlike submanifold M of a semi-Riemannian manifold $\tilde{M}$.

Ageshe and Chafle [8] introduced the notion of a semi-symmetric non-metric connection on a Riemannian manifold. Although now, we have lightlike version of a large variety of Riemannian submanifolds, the theory of lightlike submanifolds of semi-Riemannian manifolds, equipped with semi-symmetric metric connections, has not been introduced until quite recently. Yasar et al. [9] studied lightlike hypersurfaces in a semi-Riemannian manifold admitting a semi-symmetric non-metric connection. Recently, Jin and Lee [10] and Jin [11–13] studied half lightlike and r-lightlike submanifolds of a semi-Riemannian manifold with a semi-symmetric non-metric connection.

In this paper, we study the geometry of screen conformal Einstein half lightlike submanifolds M of a Lorentzian space form $\tilde{M}(c)$ of constant curvature c admitting a semi-symmetric non-metric connection subject to the conditions; (1) the structure vector field of $\tilde{M}$ is tangent to M, and (2) the canonical normal vector field of M is conformal Killing. The reason for this geometric restriction on M is due to the fact that such a class admits an integrable screen distribution and a symmetric Ricci tensor of M. We prove a characterization theorem for such a half lightlike submanifold.

2 Semi-symmetric non-metric connection

Let $(\tilde{M},\tilde{g})$ be a semi-Riemannian manifold. A connection $\tilde{\mathrm{\nabla }}$ on $\tilde{M}$ is called a semi-symmetric non-metric connection [8] if $\tilde{\mathrm{\nabla }}$ and its torsion tensor $\tilde{T}$ satisfy

$$({\tilde{\mathrm{\nabla }}}_X\tilde{g})(Y,Z)=-\pi (Y)\tilde{g}(X,Z)-\pi (Z)\tilde{g}(X,Y),$$

(2.1)

$$\tilde{T}(X,Y)=\pi (Y)X-\pi (X)Y,$$

(2.2)

for any vector fields X, Y and Z on $\tilde{M}$, where π is a 1-form associated with a non-vanishing vector field ζ, which is called the structure vector field of $\tilde{M}$, by

$$\pi (X)=\tilde{g}(X,\zeta ).$$

(2.3)

In the entire discussion of this article, we shall assume the structure vector field ζ to be unit spacelike, unless otherwise specified.

A submanifold $(M,g)$ of codimension 2 is called half lightlike submanifold if the radical distribution $Rad(TM)=TM\cap T{M}^{\mathrm{\perp }}$ is a subbundle of the tangent bundle TM and the normal bundle $T{M}^{\mathrm{\perp }}$ of rank 1. Therefore, there exist complementary non-degenerate distributions $S(TM)$ and $S(T{M}^{\mathrm{\perp }})$ of $Rad(TM)$ in TM and $T{M}^{\mathrm{\perp }}$ respectively, which are called the screen and co-screen distributions of M, respectively, such that

$$TM=Rad(TM){\oplus }_{\mathrm{orth}}S(TM),T{M}^{\mathrm{\perp }}=Rad(TM){\oplus }_{\mathrm{orth}}S\left(T{M}^{\mathrm{\perp }}\right),$$

(2.4)

where ${\oplus }_{\mathrm{orth}}$ denotes the orthogonal direct sum. We denote such a half lightlike submanifold by $M=(M,g,S(TM))$. Denote by $F(M)$ the algebra of smooth functions on M and by $\mathrm{\Gamma }(E)$ the $F(M)$ module of smooth sections of a vector bundle E over M. Choose $L\in \mathrm{\Gamma }(S(T{M}^{\mathrm{\perp }}))$ as a unit vector field with $\tilde{g}(L,L)=\pm 1$. We may assume that L to be unit spacelike vector field without loss of generality, i.e., $\tilde{g}(L,L)=1$. We call L the canonical normal vector field of M. Consider the orthogonal complementary distribution $S{(TM)}^{\mathrm{\perp }}$ to $S(TM)$ in $T\tilde{M}$. Certainly, $Rad(TM)$ and $S(T{M}^{\mathrm{\perp }})$ are subbundles of $S{(TM)}^{\mathrm{\perp }}$. As $S(T{M}^{\mathrm{\perp }})$ is non-degenerate, we have

$$S{(TM)}^{\mathrm{\perp }}=S\left(T{M}^{\mathrm{\perp }}\right){\oplus }_{\mathrm{orth}}S{\left(T{M}^{\mathrm{\perp }}\right)}^{\mathrm{\perp }},$$

where $S{(T{M}^{\mathrm{\perp }})}^{\mathrm{\perp }}$ is the orthogonal complementary to $S(T{M}^{\mathrm{\perp }})$ in $S{(TM)}^{\mathrm{\perp }}$. For any null section ξ of $Rad(TM)$ on a coordinate neighborhood $\mathcal{U}\subset M$, there exists a uniquely defined lightlike vector bundle $ltr(TM)$ and a null vector field N of $ltr{(TM)}_{{|}_{\mathcal{U}}}$ satisfying

$$\tilde{g}(\xi ,N)=1,\tilde{g}(N,N)=\tilde{g}(N,X)=\tilde{g}(N,L)=0,\mathrm{\forall }X\in \mathrm{\Gamma }(S(TM)).$$

We call N, $ltr(TM)$ and $tr(TM)=S(T{M}^{\mathrm{\perp }}){\oplus }_{\mathrm{orth}}ltr(TM)$ the lightlike transversal vector field, lightlike transversal vector bundle and transversal vector bundle of M with respect to the screen distribution, respectively [6]. Then $T\tilde{M}$ is decomposed as follows:

$$\begin{array}{rcl}T\tilde{M}& =& TM\oplus tr(TM)=\{Rad(TM)\oplus tr(TM)\}{\oplus }_{\mathrm{orth}}S(TM)\\ =& \{Rad(TM)\oplus ltr(TM)\}{\oplus }_{\mathrm{orth}}S(TM){\oplus }_{\mathrm{orth}}S\left(T{M}^{\mathrm{\perp }}\right).\end{array}$$

(2.5)

Given a screen distribution $S(TM)$, there exists a unique complementary vector bundle $tr(TM)$ to TM in $T{\tilde{M}}_{|M}$. Using (2.4) and (2.5), there exists a local quasi-orthonormal frame field of $\tilde{M}$ along M given by

$$F=\{\xi ,N,L,W_a\},a\in \{1,\dots ,m\},$$

(2.6)

where $\{W_a\}$ is an orthonormal frame field of $S{(TM)}_{{|}_{\mathcal{U}}}$.

In the entire discussion of this article, we shall assume that ζ is tangent to M, and we take $X,Y,Z,W\in \mathrm{\Gamma }(TM)$, unless otherwise specified. Let P be the projection morphism of TM on $S(TM)$ with respect to the first decomposition of (2.4). Then the local Gauss and Weingartan formulas of M and $S(TM)$ are given respectively by

$${\tilde{\mathrm{\nabla }}}_{X}Y={\mathrm{\nabla }}_{X}Y+B(X,Y)N+D(X,Y)L,$$

(2.7)

$${\tilde{\mathrm{\nabla }}}_{X}N=-A_{{}_N}X+\tau (X)N+\rho (X)L,$$

(2.8)

$${\tilde{\mathrm{\nabla }}}_{X}L=-A_{{}_L}X+\varphi (X)N,$$

(2.9)

$${\mathrm{\nabla }}_{X}PY={\mathrm{\nabla }}_X^{\ast }PY+C(X,PY)\xi ,$$

(2.10)

$${\mathrm{\nabla }}_X\xi =-A_{\xi }^{\ast }X-\tau (X)\xi ,$$

(2.11)

where ∇ and ${\mathrm{\nabla }}^{\ast }$ are induced linear connections on TM and $S(TM)$, respectively, B and D are called the local lightlike, and screen second fundamental forms of M, respectively, C is called the local second fundamental form on $S(TM)$, $A_N$, $A_{\xi }^{\ast }$ and $A_L$ are called the shape operators, and τ, ρ and ϕ are 1-forms on TM. We say that

$$h(X,Y)=B(X,Y)N+D(X,Y)L$$

is the second fundamental form tensor of M. Using (2.1), (2.2) and (2.7), we have

$$({\mathrm{\nabla }}_{X}g)(Y,Z)=B(X,Y)\eta (Z)+B(X,Z)\eta (Y)-\pi (Y)g(X,Z)-\pi (Z)g(X,Y),$$

(2.12)

$$T(X,Y)=\pi (Y)X-\pi (X)Y,$$

(2.13)

and B and D are symmetric on TM, where T is the torsion tensor with respect to the induced connection ∇, and η is a 1-form on TM such that

$$\eta (X)=\tilde{g}(X,N).$$

From the facts $B(X,Y)=\tilde{g}({\tilde{\mathrm{\nabla }}}_{X}Y,\xi )$ and $D(X,Y)=\tilde{g}({\tilde{\mathrm{\nabla }}}_{X}Y,L)$, we know that B and D are independent of the choice of the screen distribution $S(TM)$ and satisfy

$$B(X,\xi )=0,D(X,\xi )=-\varphi (X).$$

(2.14)

In case ζ is tangent to M, the above three local second fundamental forms on M and $S(TM)$ are related to their shape operators by

$$g(A_{\xi }^{\ast }X,Y)=B(X,Y),\tilde{g}(A_{\xi }^{\ast }X,N)=0,$$

(2.15)

$$g(A_{{}_L}X,Y)=D(X,Y)+\varphi (X)\eta (Y),\tilde{g}(A_{{}_L}X,N)=\rho (X),$$

(2.16)

$$g(A_{{}_N}X,PY)=C(X,PY)-fg(X,PY)-\eta (X)\pi (PY),\tilde{g}(A_{{}_N}X,N)=-f\eta (X),$$

(2.17)

where f is the smooth function given by $f=\pi (N)$. From (2.15) and (2.17), we show that $A_{\xi }^{\ast }$ and $A_{{}_N}$ are $S(TM)$-valued, and $A_{\xi }$ is self-adjoint operator and satisfies

$$A_{\xi }^{\ast }\xi =0,$$

(2.18)

that is, ξ is an eigenvector field of $A_{\xi }^{\ast }$ corresponding to the eigenvalue 0.

In general, the screen distribution $S(TM)$ is not necessarily integrable. The following result gives equivalent conditions for the integrability of $S(TM)$.

Theorem 2.1 [10]

Let M be a half lightlike submanifold of a semi-Riemannian manifold $\tilde{M}$ admitting a semi-symmetric non-metric connection. Then the following assertions are equivalent:

• (1) The screen distribution $S(TM)$ is an integrable distribution.

• (2) C is symmetric, i.e., $C(X,Y)=C(Y,X)$ for all $X,Y\in \mathrm{\Gamma }(S(TM))$.

• (3) The shape operator $A_{{}_N}$ is a self-adjoint with respect to g, i.e.,

  $$g(A_{{}_N}X,Y)=g(X,A_{{}_N}Y),\mathrm{\forall }X,Y\in \mathrm{\Gamma }(S(TM)).$$

Just as in the well-known case of locally product Riemannian or semi-Riemannian manifolds [2–4, 7], if $S(TM)$ is an integrable distribution, then M is locally a product manifold $M={\mathcal{C}}_1\times M^{\ast }$, where ${\mathcal{C}}_1$ is a null curve tangent to $Rad(TM)$, and $M^{\ast }$ is a leaf of the integrable screen distribution $S(TM)$.

3 Structure equations

Denote by $\tilde{R}$, R and $R^{\ast }$ the curvature tensors of the semi-symmetric non-metric connection $\tilde{\mathrm{\nabla }}$ on $\tilde{M}$, the induced connection ∇ on M and the induced connection ${\mathrm{\nabla }}^{\ast }$ on $S(TM)$, respectively. Using the Gauss-Weingarten formulas for M and $S(TM)$, we obtain the Gauss-Codazzi equations for M and $S(TM)$:

$$\begin{array}{c}\tilde{R}(X,Y)Z=R(X,Y)Z+B(X,Z)A_{{}_N}Y-B(Y,Z)A_{{}_N}X\hfill \\ \phantom{\tilde{R}(X,Y)Z=}+D(X,Z)A_{{}_L}Y-D(Y,Z)A_{{}_L}X\hfill \\ \phantom{\tilde{R}(X,Y)Z=}+\{({\mathrm{\nabla }}_{X}B)(Y,Z)-({\mathrm{\nabla }}_{Y}B)(X,Z)\hfill \\ \phantom{\tilde{R}(X,Y)Z=}+B(Y,Z)[\tau (X)-\pi (X)]-B(X,Z)[\tau (Y)-\pi (Y)]\hfill \\ \phantom{\tilde{R}(X,Y)Z=}+D(Y,Z)\varphi (X)-D(X,Z)\varphi (Y)\}N\hfill \\ \phantom{\tilde{R}(X,Y)Z=}+\{({\mathrm{\nabla }}_{X}D)(Y,Z)-({\mathrm{\nabla }}_{Y}D)(X,Z)+B(Y,Z)\rho (X)\hfill \\ \phantom{\tilde{R}(X,Y)Z=}-B(X,Z)\rho (Y)-D(Y,Z)\pi (X)+D(X,Z)\pi (Y)\}L,\hfill \end{array}$$

(3.1)

$$\begin{array}{c}\tilde{R}(X,Y)N=-{\mathrm{\nabla }}_X(A_{{}_N}Y)+{\mathrm{\nabla }}_Y(A_{{}_N}X)+A_{{}_N}[X,Y]\hfill \\ \phantom{\tilde{R}(X,Y)N=}+\tau (X)A_{{}_N}Y-\tau (Y)A_{{}_N}X+\rho (X)A_{{}_L}Y-\rho (Y)A_{{}_L}X\hfill \\ \phantom{\tilde{R}(X,Y)N=}+\{B(Y,A_{{}_N}X)-B(X,A_{{}_N}Y)+2d\tau (X,Y)\hfill \\ \phantom{\tilde{R}(X,Y)N=}+\varphi (X)\rho (Y)-\varphi (Y)\rho (X)\}N\hfill \\ \phantom{\tilde{R}(X,Y)N=}+\{D(Y,A_{{}_N}X)-D(X,A_{{}_N}Y)+2d\rho (X,Y)\hfill \\ \phantom{\tilde{R}(X,Y)N=}+\rho (X)\tau (Y)-\rho (Y)\tau (X)\}L,\hfill \end{array}$$

(3.2)

$$\begin{array}{c}\tilde{R}(X,Y)L=-{\mathrm{\nabla }}_X(A_{{}_L}Y)+{\mathrm{\nabla }}_Y(A_{{}_L}X)+A_{{}_L}[X,Y]\hfill \\ \phantom{\tilde{R}(X,Y)L=}+\varphi (X)A_{{}_N}Y-\varphi (Y)A_{{}_N}X\hfill \\ \phantom{\tilde{R}(X,Y)L=}+\{B(Y,A_{{}_L}X)-B(X,A_{{}_L}Y)+2d\varphi (X,Y)\hfill \\ \phantom{\tilde{R}(X,Y)L=}+\tau (X)\varphi (Y)-\tau (Y)\varphi (X)\}N\hfill \\ \phantom{\tilde{R}(X,Y)L=}+\{D(Y,A_{{}_L}X)-D(X,A_{{}_L}Y)+\rho (X)\varphi (Y)-\rho (Y)\varphi (X)\}L,\hfill \end{array}$$

(3.3)

$$\begin{array}{c}R(X,Y)PZ=R^{\ast }(X,Y)PZ+C(X,PZ)A_{\xi }^{\ast }Y-C(Y,PZ)A_{\xi }X\hfill \\ \phantom{R(X,Y)PZ=}+\{({\mathrm{\nabla }}_{X}C)(Y,PZ)-({\mathrm{\nabla }}_{Y}C)(X,PZ)\hfill \\ \phantom{R(X,Y)PZ=}+C(X,PZ)[\tau (Y)+\pi (Y)]-C(Y,PZ)[\tau (X)+\pi (X)]\}\xi ,\hfill \end{array}$$

(3.4)

$$\begin{array}{c}R(X,Y)\xi =-{\mathrm{\nabla }}_X^{\ast }\left(A_{\xi }^{\ast }Y\right)+{\mathrm{\nabla }}_Y^{\ast }\left(A_{\xi }^{\ast }X\right)+A_{\xi }^{\ast }[X,Y]+\tau (Y)A_{\xi }^{\ast }X\hfill \\ \phantom{R(X,Y)\xi =}-\tau (X)A_{\xi }^{\ast }Y+\{C(Y,A_{\xi }^{\ast }X)-C(X,A_{\xi }^{\ast }Y)-2d\tau (X,Y)\}\xi .\hfill \end{array}$$

(3.5)

A semi-Riemannian manifold $\tilde{M}$ of constant curvature c is called a semi-Riemannian space form and denote it by $\tilde{M}(c)$. The curvature tensor $\tilde{R}$ of $\tilde{M}(c)$ is given by

$$\tilde{R}(X,Y)Z=c\{\tilde{g}(Y,Z)X-\tilde{g}(X,Z)Y\},\mathrm{\forall }X,Y,Z\in \mathrm{\Gamma }(T\tilde{M}).$$

(3.6)

Taking the scalar product with ξ and L to (3.6), we obtain $\tilde{g}(\tilde{R}(X,Y)Z,\xi )=0$ and $\tilde{g}(\tilde{R}(X,Y)Z,L)=0$ for any $X,Y,Z\in \mathrm{\Gamma }(TM)$. From these equations and (3.1), we get

$$\begin{array}{rcl}\tilde{R}(X,Y)Z& =& R(X,Y)Z+B(X,Z)A_{{}_N}Y-B(Y,Z)A_{{}_N}X\\ +D(X,Z)A_{{}_L}Y-D(Y,Z)A_{{}_L}X,\mathrm{\forall }X,Y,Z\in \mathrm{\Gamma }(TM).\end{array}$$

(3.7)

4 Screen conformal half lightlike submanifolds

Definition 1 A half lightlike submanifold M of a semi-Riemannian manifold $\tilde{M}$ is said to be irrotational [14] if ${\tilde{\mathrm{\nabla }}}_X\xi \in \mathrm{\Gamma }(TM)$ for any $X\in \mathrm{\Gamma }(TM)$.

From (2.7) and (2.14), we show that the above definition is equivalent to

$$D(X,\xi )=0=\varphi (X),\mathrm{\forall }X\in \mathrm{\Gamma }(TM).$$

Theorem 4.1 Let M be an irrotational half lightlike submanifold of a semi-Riemannian manifold $\tilde{M}$ admitting a semi-symmetric non-metric connection such that ζ is tangent to M. Then ζ is conjugate to any vector field X on M, i.e., ζ satisfies $h(X,\zeta )=0$.

Proof Taking the scalar product with ξ to (3.2) and N to (3.1) such that $Z=\xi$ by turns and using (2.14), (3.5) and the fact that $\varphi =0$, we obtain

$$\begin{array}{rcl}\tilde{g}(\tilde{R}(X,Y)\xi ,N)& =& B(X,A_{{}_N}Y)-B(Y,A_{{}_N}X)-2d\tau (X,Y)\\ =& C(Y,A_{\xi }^{\ast }X)-C(X,A_{\xi }^{\ast }Y)-2d\tau (X,Y).\end{array}$$

From these two representations, we obtain

$$B(X,A_{{}_N}Y)-B(Y,A_{{}_N}X)=C(Y,A_{\xi }^{\ast }X)-C(X,A_{\xi }^{\ast }Y).$$

Using (2.15)₁, (2.17)₂ and the fact that $A_{\xi }^{\ast }$ is self-adjoint, we have

$$\pi \left(A_{\xi }^{\ast }X\right)\eta (Y)=\pi \left(A_{\xi }^{\ast }Y\right)\eta (X).$$

Replacing Y by ξ to this equation and using (2.18), we have

$$B(X,\zeta )=\pi \left(A_{\xi }^{\ast }X\right)=0.$$

(4.1)

As D is symmetric and $\varphi =0$, we show that $A_{{}_L}$ is self-adjoint. Taking the scalar product with L to (3.2) and N to (3.3) with $\varphi =0$ by turns, we obtain

$$\begin{array}{rcl}\tilde{g}(\tilde{R}(X,Y)N,L)& =& \tilde{g}({\mathrm{\nabla }}_X(A_{{}_L}Y)-{\mathrm{\nabla }}_Y(A_{{}_L}X)-A_{{}_L}[X,Y],N)\\ =& D(Y,A_{{}_N}X)-D(X,A_{{}_N}Y)+2d\rho (X,Y)+\rho (X)\tau (Y)-\rho (Y)\tau (X).\end{array}$$

Using these two representations and (2.16)₂, we show that

$$\begin{array}{c}D(Y,A_{{}_N}X)-D(X,A_{{}_N}Y)+2d\rho (X,Y)+\rho (X)\tau (Y)-\rho (Y)\tau (X)\hfill \\ =\tilde{g}({\mathrm{\nabla }}_X(A_{{}_L}Y),N)-\tilde{g}({\mathrm{\nabla }}_Y(A_{{}_L}X),N)-\rho ([X,Y]).\hfill \end{array}$$

Applying ${\tilde{\mathrm{\nabla }}}_X$ to $\tilde{g}(A_{{}_L}Y,N)=\rho (Y)$ and using (2.1), (2.7) and (2.8), we have

$$\begin{array}{rcl}\tilde{g}({\mathrm{\nabla }}_X(A_{{}_L}Y),N)& =& X(\rho (Y))+\pi (A_{{}_L}Y)\eta (X)+fg(X,A_{{}_L}Y)\\ +g(A_{{}_L}Y,A_{{}_N}X)-\tau (X)\rho (Y).\end{array}$$

Substituting this equation into the last equation and using (2.16)₁, we have

$$\pi (A_{{}_L}X)\eta (Y)=\pi (A_{{}_L}Y)\eta (X).$$

Replacing Y by ξ to this equation, we have

$$\pi (A_{{}_L}X)=\pi (A_{{}_L}\xi )\eta (X).$$

Taking $X=\xi$ and $Y=\zeta$ to (2.16)₁, we get $\pi (A_{{}_L}\xi )=0$. Therefore, we have

$$D(X,\zeta )=\pi (A_{{}_L}X)=0.$$

(4.2)

From (4.1) and (4.2), we show that $h(X,\zeta )=0$ for all $X\in \mathrm{\Gamma }(TM)$. □

Definition 2 A half lightlike submanifold M of a semi-Riemannian manifold $\tilde{M}$ is screen conformal [4, 5, 7] if the second fundamental forms B and C satisfy

$$C(X,PY)=\phi B(X,Y),\mathrm{\forall }X,Y\in \mathrm{\Gamma }(TM),$$

(4.3)

where φ is a non-vanishing function on a coordinate neighborhood in M.

Theorem 4.2 Let M be an irrotational half lightlike submanifold of a semi-Riemannian space form $\tilde{M}(c)$ admitting a semi-symmetric non-metric connection such that ζ is tangent to M. If M is screen conformal, then $c=0$.

Proof Substituting (3.6) into (3.2) and using the fact that $\varphi =0$, we have

$$\begin{array}{c}({\mathrm{\nabla }}_{X}B)(Y,Z)-({\mathrm{\nabla }}_{Y}B)(X,Z)\hfill \\ =B(Y,Z)\{\pi (X)-\tau (X)\}-B(X,Z)\{\pi (Y)-\tau (Y)\}.\hfill \end{array}$$

(4.4)

Taking the scalar product with N to (3.1) and (3.4) by turns and using (2.16)₂, (2.17)₂ and (3.6), we have the following two forms of $\tilde{g}(R(X,Y)PZ,N)$:

$$\begin{array}{c}\{cg(Y,PZ)-fB(Y,PZ)\}\eta (X)-\{cg(X,PZ)-fB(X,PZ)\}\eta (Y)\hfill \\ +\rho (X)D(Y,PZ)-\rho (Y)D(X,PZ)\hfill \\ =({\mathrm{\nabla }}_{X}C)(Y,PZ)-({\mathrm{\nabla }}_{Y}C)(X,PZ)+C(X,PZ)\{\pi (Y)+\tau (Y)\}\hfill \\ -C(Y,PZ)\{\pi (X)+\tau (X)\}.\hfill \end{array}$$

(4.5)

Applying ${\mathrm{\nabla }}_X$ to $C(Y,PZ)=\phi B(Y,PZ)$, we have

$$({\mathrm{\nabla }}_{X}C)(Y,PZ)=X[\phi ]B(Y,PZ)+\phi ({\mathrm{\nabla }}_{X}B)(Y,PZ).$$

Substituting this into (4.5) and using (4.4), we obtain

$$\begin{array}{c}c\{g(Y,PZ)\eta (X)-g(X,PZ)\eta (Y)\}\hfill \\ =\{X[\phi ]-2\phi \tau (X)+f\eta (X)\}B(Y,PZ)-\rho (X)D(Y,PZ)\hfill \\ -\{Y[\phi ]-2\phi \tau (Y)+f\eta (Y)\}B(X,PZ)+\rho (Y)D(X,PZ).\hfill \end{array}$$

(4.6)

Replacing Z by ζ to (4.5) and using (4.1) and (4.2), we have $c=0$. □

Remark 4.3 If M is screen conformal, then, from (4.3), we show that C is symmetric on $S(TM)$. By Theorem 2.1, $S(TM)$ is integrable and M is locally a product manifold ${\mathcal{C}}_1\times M^{\ast }$, where ${\mathcal{C}}_1$ is a null curve tangent to $Rad(TM)$ and $M^{\ast }$ is a leaf of $S(TM)$.

5 Main theorem

Let $\widetilde{\mathit{Ric}}$ be the Ricci curvature tensor of $\tilde{M}$ and $R^{(0,2)}$ the induced Ricci type tensor on M given respectively by

$$\begin{array}{c}\widetilde{\mathit{Ric}}(X,Y)=trace\{Z\to \tilde{R}(Z,X)Y\},\mathrm{\forall }X,Y\in \mathrm{\Gamma }(T\tilde{M}),\hfill \\ R^{(0,2)}(X,Y)=trace\{Z\to R(Z,X)Y\},\mathrm{\forall }X,Y\in \mathrm{\Gamma }(TM).\hfill \end{array}$$

Using the quasi-orthonormal frame field (2.6) on $\tilde{M}$, we show [10] that

$$\begin{array}{rcl}{R}^{(0,2)}(X,Y)& =& \widetilde{\mathit{Ric}}(X,Y)+B(X,Y)tr{A}_{{}_N}+D(X,Y)tr{A}_{{}_L}\\ -g(A_{{}_N}X,A_{\xi }^{\ast }Y)-g(A_{{}_L}X,A_{{}_L}Y)+\rho (X)\varphi (Y)\\ -\tilde{g}(\tilde{R}(\xi ,Y)X,N)-\tilde{g}(\tilde{R}(L,X)Y,L),\end{array}$$

where $tr{A}_{{}_N}$ is the trace of $A_{{}_N}$. From this, we show that $R^{(0,2)}$ is not symmetric. The tensor field $R^{(0,2)}$ is called the induced Ricci curvature tensor [4, 5] of M, denoted by Ric, if it is symmetric. M is called Ricci flat if its induced Ricci tensor vanishes on M. It is known [10] that $R^{(0,2)}$ is symmetric if and only if the 1-form τ is closed, i.e., $d\tau =0$.

Remark 5.1 If the induced Ricci type tensor $R^{(0,2)}$ is symmetric, then there exists a null pair $\{\xi ,N\}$ such that the corresponding 1-form τ satisfies $\tau =0$ [3, 4], which is called a canonical null pair of M. Although $S(TM)$ is not unique, it is canonically isomorphic to the factor vector bundle $S{(TM)}^{\mathrm{♯}}=TM/Rad(TM)$ [14]. This implies that all screen distribution are mutually isomorphic. For this reason, in case $d\tau =0$, we consider only lightlike hypersurfaces M endow with the canonical null pair such that $\tau =0$.

We say that M is an Einstein manifold if the Ricci tensor of M satisfies

$$\mathit{Ric}=\kappa g.$$

It is well known that if $dimM>2$, then κ is a constant.

Let $dim\tilde{M}=m+3$. In case $\tilde{M}$ is a semi-Riemannian space form $\tilde{M}(c)$, we have

$$\begin{array}{rcl}{R}^{(0,2)}(X,Y)& =& mcg(X,Y)+B(X,Y)tr{A}_{{}_N}+D(X,Y)tr{A}_{{}_L}\\ -g(A_{{}_N}X,A_{\xi }^{\ast }Y)-g(A_{{}_L}X,A_{{}_L}Y)+\rho (X)\varphi (Y).\end{array}$$

(5.1)

Due to (2.15) and (2.17), we show that M is screen conformal if and only if the shape operators $A_{{}_N}$ and $A_{\xi }^{\ast }$ are related by

$$A_{{}_N}X=\phi A_{\xi }^{\ast }X-fX-\eta (X)\zeta .$$

(5.2)

Assume that $\varphi =0$. As D is symmetric, $A_{{}_L}$ is self-adjoint. Using this, (5.1) and (5.2), we show that $R^{(0,2)}$ is symmetric. Thus, we can take $\tau =0$. As $\tau =0$, (4.4) reduce to

$$({\mathrm{\nabla }}_{X}B)(Y,Z)-({\mathrm{\nabla }}_{Y}B)(X,Z)=\pi (X)B(Y,Z)-\pi (Y)B(X,Z).$$

(5.3)

Definition 3 A vector field X on $\tilde{M}$ is said to be conformal Killing [3, 5, 15] if ${\tilde{\mathcal{L}}}_X\tilde{g}=-2\delta \tilde{g}$ for a scalar function δ, where $\tilde{\mathcal{L}}$ denotes the Lie derivative on $\tilde{M}$, that is,

$$({\tilde{\mathcal{L}}}_{{}_X}\tilde{g})(Y,Z)=X(\tilde{g}(Y,Z))-\tilde{g}([X,Y],Z)-\tilde{g}(Y,[X,Z]),\mathrm{\forall }Y,Z\in \mathrm{\Gamma }(T\tilde{M}).$$

In particular, if $\delta =0$, then X is called a Killing vector field on $\tilde{M}$.

Theorem 5.2 Let M be a half lightlike submanifold of a semi-Riemannian manifold $\tilde{M}$ admitting a semi-symmetric non-metric connection. If the canonical normal vector field L is a conformal Killing one, then L is a Killing vector field.

Proof Using (2.1) and (2.2), for any $X,Y,Z\in \mathrm{\Gamma }(T\tilde{M})$, we have

$$({\tilde{\mathcal{L}}}_{{}_X}\tilde{g})(Y,Z)=\tilde{g}({\tilde{\mathrm{\nabla }}}_{Y}X,Z)+\tilde{g}(Y,{\tilde{\mathrm{\nabla }}}_{Z}X)-2\pi (X)\tilde{g}(Y,Z).$$

As L is a conformal Killing vector field, we have $\tilde{g}({\tilde{\mathrm{\nabla }}}_{X}L,Y)=-D(X,Y)$ by (2.9) and (2.16). This implies $({\tilde{\mathcal{L}}}_{{}_L}\tilde{g})(X,Y)=-2D(X,Y)$ for any $X,Y\in \mathrm{\Gamma }(TM)$. Thus, we have

$$D(X,Y)=\delta g(X,Y),\mathrm{\forall }X,Y\in \mathrm{\Gamma }(TM).$$

(5.4)

Taking $X=Y=\zeta$ and using (4.2), we get $\delta =0$. Thus, L is a Killing vector field. □

Remark 5.3 Cǎlin [16] proved the following result. For any lightlike submanifolds M of indefinite almost contact metric manifolds $\tilde{M}$, if ζ is tangent to M, then it belongs to $S(TM)$. Duggal and Sahin also proved this result (see pp.318-319 of [5]). After Cǎlin’s work, many earlier works [17–19], which were written on lightlike submanifolds of indefinite almost contact metric manifolds or lightlike submanifolds of semi-Riemannian manifolds, admitting semi-symmetric non-metric connections, obtained their results by using the Cǎlin’s result described in above. However, Jin [12, 13] proved that Cǎlin’s result is not true for any lightlike submanifolds M of a semi-Riemannian space form $\tilde{M}(c)$, admitting a semi-symmetric non-metric connection.

For the rest of this section, we may assume that the structure vector field ζ of $\tilde{M}$ belongs to the screen distribution $S(TM)$. In this case, we show that $f=0$.

Theorem 5.4 Let M be a screen conformal Einstein half lightlike submanifold of a Lorentzian space form $\tilde{M}(c)$, admitting a semi-symmetric non-metric connection such that ζ belongs to $S(TM)$. If the canonical normal vector field L is conformal Killing, then M is Ricci flat. Moreover, if the mean curvature of M is constant, then M is locally a product manifold ${\mathcal{C}}_1\times {\mathcal{C}}_2\times M^{m-1}$, where ${\mathcal{C}}_1$ and ${\mathcal{C}}_2$ are null and non-null curves, and $M^{m-1}$ is an $(m-1)$-dimensional Euclidean space.

Proof As L is conformal Killing vector field, $D=A_{{}_L}=0$ by (5.4) and Theorem 5.2. Therefore, from (2.14), we show that $\varphi =0$, i.e., M is irrotational. By Theorem 4.2, we also have $c=0$. Using (2.15), (4.1) and (5.2) with $f=0$, from (5.1), we have

$$g(A_{\xi }^{\ast }X,A_{\xi }^{\ast }Y)-\alpha g(A_{\xi }^{\ast }X,Y)+{\phi }^{-1}\kappa g(X,Y)=0$$

(5.5)

due to $c=0$, where $\alpha =tr{A}_{\xi }^{\ast }$. As $g(A_{\xi }^{\ast }\zeta ,X)=B(\zeta ,X)=0$ for all $X\in \mathrm{\Gamma }(TM)$ and $S(TM)$ is non-degenerate, we show that

$$A_{\xi }^{\ast }\zeta =0.$$

(5.6)

Taking $X=Y=\zeta$ to (5.5) and using (5.6), we have ${\phi }^{-1}\kappa =0$. Thus, (5.5) reduce to

$$g(A_{\xi }^{\ast }X,A_{\xi }^{\ast }Y)-\alpha g(A_{\xi }^{\ast }X,Y)=0,\kappa =0.$$

(5.7)

From the second equation of (5.7), we show that M is Ricci flat.

As M is screen conformal and $\tilde{M}$ is Lorentzian, $S(TM)$ is an integrable Riemannian vector bundle. Since ξ is an eigenvector field of $A_{\xi }^{\ast }$, corresponding to the eigenvalue 0 due to (2.15), and $A_{\xi }^{\ast }$ is $S(TM)$-valued real self-adjoint operator, $A_{\xi }^{\ast }$ has m real orthonormal eigenvector fields in $S(TM)$ and is diagonalizable. Consider a frame field of eigenvectors $\{\xi ,E_1,\dots ,E_m\}$ of $A_{\xi }^{\ast }$ such that $\{E_1,\dots ,E_m\}$ is an orthonormal frame field of $S(TM)$ and $A_{\xi }^{\ast }{E}_i={\lambda }_i{E}_i$. Put $X=Y=E_i$ in (5.7), each eigenvalue ${\lambda }_i$ is a solution of

$$x^2-\alpha x=0.$$

As this equation has at most two distinct solutions 0 and α, there exists $p\in \{0,1,\dots ,m\}$ such that ${\lambda }_1=\cdots ={\lambda }_p=0$ and ${\lambda }_{p+1}=\cdots ={\lambda }_m=\alpha$, by renumbering if necessary. As $tr{A}_{\xi }^{\ast }=0p+(m-p)\alpha$, we have

$$\alpha =tr{A}_{\xi }^{\ast }=(m-p)\alpha .$$

So $p=m-1$, i.e.,

$$A_{\xi }^{\ast }=\left(\begin{array}{c}0\\ \ddots \\ 0\\ \alpha \end{array}\right).$$

Consider two distributions $D_o$ and $D_{\alpha }$ on $S(TM)$ given by

$$\begin{array}{c}{D}_o=\{X\in \mathrm{\Gamma }(S(TM))\mid A_{\xi }^{\ast }X=0\text{ and }X\ne 0\},\hfill \\ D_{\alpha }=\{U\in \mathrm{\Gamma }(S(TM))\mid A_{\xi }^{\ast }U=\alpha U\text{ and }U\ne 0\}.\hfill \end{array}$$

Clearly we show that $D_o\cap D_{\alpha }=\{0\}$ as $\alpha \ne 0$. In the sequel, we take $X,Y\in \mathrm{\Gamma }(D_o)$, $U,V\in \mathrm{\Gamma }(D_{\alpha })$ and $Z,W\in \mathrm{\Gamma }(S(TM))$. Since X and U are eigenvector fields of the real self-adjoint operator $A_{\xi }^{\ast }$, corresponding to the different eigenvalues 0 and α respectively, we have $g(X,U)=0$. From this and the fact that $B(X,U)=g(A_{\xi }^{\ast }X,U)=0$, we show that $D_{\alpha }{\mathrm{\perp }}_{{}_g}{D}_o$ and $D_{\alpha }{\mathrm{\perp }}_{{}_B}{D}_o$, respectively. Since ${\{E_i\}}_{1\le i\le m-1}$ and $\{E_m\}$ are vector fields of $D_o$ and $D_{\alpha }$, respectively, and $D_o$ and $D_{\alpha }$ are mutually orthogonal, we show that $D_o$ and $D_{\alpha }$ are non-degenerate distributions of rank $(m-1)$ and rank 1, respectively. Thus, the screen distribution $S(TM)$ is decomposed as $S(TM)=D_{\alpha }{\oplus }_{\mathrm{orth}}{D}_o$.

From (5.7), we get $A_{\xi }^{\ast }(A_{\alpha }^{\ast }-\alpha P)=0$. Let $W\in Im{A}_{\xi }^{\ast }$. Then there exists $Z\in \mathrm{\Gamma }(S(TM))$ such that $W=A_{\xi }^{\ast }Z$. Then $(A_{\xi }^{\ast }-\alpha P)W=0$ and $W\in \mathrm{\Gamma }(D_{\alpha })$. Thus, $Im{A}_{\xi }^{\ast }\subset \mathrm{\Gamma }(D_{\alpha })$. By duality, we have $Im(A_{\xi }^{\ast }-\alpha P)\subset \mathrm{\Gamma }(D_o)$.

Applying ${\mathrm{\nabla }}_X$ to $B(Y,U)=0$ and using (2.15) and $A_{\xi }^{\ast }Y=0$, we obtain

$$({\mathrm{\nabla }}_{X}B)(Y,U)=-g(A_{\xi }^{\ast }{\mathrm{\nabla }}_{X}Y,U).$$

Substituting this into (5.3) and using (2.12) and $A_{\xi }^{\ast }X=A_{\xi }^{\ast }Y=0$, we get

$$g(A_{\xi }^{\ast }[X,Y],U)=0.$$

As $Im{A}_{\xi }^{\ast }\subset \mathrm{\Gamma }(D_{\alpha })$ and $D_{\alpha }$ is non-degenerate, we get $A_{\xi }^{\ast }[X,Y]=0$. This implies that $[X,Y]\in \mathrm{\Gamma }(D_o)$. Thus, $D_o$ is an integrable distribution.

Applying ${\mathrm{\nabla }}_U$ to $B(X,Y)=0$ and ${\mathrm{\nabla }}_X$ to $B(U,Y)=0$, we have

$$({\mathrm{\nabla }}_{U}B)(X,Y)=0,({\mathrm{\nabla }}_{X}B)(U,Y)=-\alpha g({\mathrm{\nabla }}_{X}Y,U).$$

Substituting this two equations into (5.3), we have $\alpha g({\mathrm{\nabla }}_{X}Y,U)=0$. As

$$g(A_{\xi }^{\ast }{\mathrm{\nabla }}_{X}Y,U)=B({\mathrm{\nabla }}_{X}Y,U)=\alpha g({\mathrm{\nabla }}_{X}Y,U)=0$$

and $Im{A}_{\xi }^{\ast }\subset \mathrm{\Gamma }(D_{\alpha })$ and $D_{\alpha }$ is non-degenerate, we get $A_{\xi }^{\ast }{\mathrm{\nabla }}_{X}Y=0$. This implies that ${\mathrm{\nabla }}_{X}Y\in \mathrm{\Gamma }(D_o)$. Thus, $D_o$ is an auto-parallel distribution on $S(TM)$.

As $A_{\xi }^{\ast }\zeta =0$, ζ belongs to $D_o$. Thus, $\pi (U)=0$ for any $U\in \mathrm{\Gamma }(D_{\alpha })$. Applying ${\mathrm{\nabla }}_X$ to $g(U,Y)=0$ and using (2.12) and the fact that $D_o$ is auto-parallel, we get $g({\mathrm{\nabla }}_{X}U,Y)=0$. This implies that ${\mathrm{\nabla }}_{X}U\in \mathrm{\Gamma }(D_{\alpha })$.

Applying ${\mathrm{\nabla }}_U$ to $B(V,X)=0$ and using $A_{\xi }^{\ast }X=0$, we obtain

$$({\mathrm{\nabla }}_{U}B)(V,X)=-\alpha g(V,{\mathrm{\nabla }}_{U}X).$$

Substituting this into (5.3) and using the fact that $D_o{\mathrm{\perp }}_{{}_B}{D}_{\alpha }$, we get

$$g(V,{\mathrm{\nabla }}_{U}X)=g(U,{\mathrm{\nabla }}_{V}X).$$

Applying ${\mathrm{\nabla }}_U$ to $g(V,X)=0$ and using (2.12), we get

$$g({\mathrm{\nabla }}_{U}V,X)=\pi (X)g(U,V)-g(V,{\mathrm{\nabla }}_{V}X).$$

Taking the skew-symmetric part of this and using (2.13), we obtain

$$g([U,V],X)=0.$$

This implies that $[U,V]\in \mathrm{\Gamma }(D_{\alpha })$ and $D_{\alpha }$ is an integrable distribution.

Now we assume that the mean curvature $H=\frac{1}{m+2}trB=\frac{1}{m+2}tr{A}_{\xi }^{\ast }$ of M is a constant. As $tr{A}_{\xi }^{\ast }=\alpha$, we see that α is a constant. Applying ${\mathrm{\nabla }}_X$ to $B(U,V)=\alpha g(U,V)$ and ${\mathrm{\nabla }}_U$ to $B(X,V)=0$ by turns and using the facts that ${\mathrm{\nabla }}_{X}U\in \mathrm{\Gamma }(TM)$, $D_o{\mathrm{\perp }}_g{D}_{\alpha }$, $D_o{\mathrm{\perp }}_{{}_B}{D}_{\alpha }$ and $B(X,{\mathrm{\nabla }}_{U}V)=g(A_{\xi }^{\ast }X,{\mathrm{\nabla }}_{U}V)=0$, we have

$$({\mathrm{\nabla }}_{X}B)(U,V)=0,({\mathrm{\nabla }}_{U}B)(X,V)=-\alpha g({\mathrm{\nabla }}_{U}X,V).$$

Substituting these two equations into (5.3) and using $D_o{\mathrm{\perp }}_{{}_B}{D}_{\alpha }$, we have

$$g({\mathrm{\nabla }}_{U}X,V)=\pi (X)g(U,V).$$

Applying ${\mathrm{\nabla }}_U$ to $g(X,V)=0$ and using (2.12), we obtain

$$g(X,{\mathrm{\nabla }}_{U}V)=\pi (X)g(U,V)-g({\mathrm{\nabla }}_{U}X,V)=0.$$

Thus, $D_{\alpha }$ is also an integrable and auto-parallel distribution.

Since the leaf $M^{\ast }$ of $S(TM)$ is a Riemannian manifold and $S(TM)=D_{\alpha }{\oplus }_{\mathrm{orth}}{D}_o$, where $D_{\alpha }$ and $D_o$ are auto-parallel distributions of $M^{\ast }$, by the decomposition of the theorem of de Rham [20], we have $M^{\ast }={\mathcal{C}}_2\times M^{m-1}$, where ${\mathcal{C}}_2$ is a leaf of $D_{\alpha }$, and $M^{m-1}$ is a totally geodesic leaf of $D_o$. Consider the frame field of eigenvectors $\{\xi ,E_1,\dots ,E_m\}$ of $A_{\xi }^{\ast }$ such that ${\{E_i\}}_i$ is an orthonormal frame field of $S(TM)$, then $B(E_i,E_j)=C(E_i,E_j)=0$ for $1\le i<j\le m$ and $B(E_i,E_i)=C(E_i,E_i)=0$ for $1\le i\le m-1$. From (3.1) and (3.4), we have $\tilde{g}(\tilde{R}(E_i,E_j)E_j,E_i)=g(R^{\ast }(E_i,E_j)E_j,E_i)=0$. Thus, the sectional curvature K of the leaf $M^{m-1}$ of $D_o$ is given by

$$K(E_i,E_j)=\frac{g(R^{\ast }(E_i,E_j)E_j,E_i)}{g(E_i,E_i)g(E_j,E_j)-g^2(E_i,E_j)}=0.$$

Thus, M is a local product ${\mathcal{C}}_1\times {\mathcal{C}}_2\times M^{m-1}$, where ${\mathcal{C}}_1$ is a null curve, ${\mathcal{C}}_2$ is a non-null curve, and $M^{m-1}$ is an $(m-1)$-dimensional Euclidean space. □