# Half-Lightlike Submanifold of a Lorentzian Manifolds with a conformal co-
screen distribution

Issa Allassane Kaboye111Faculté des Sciences et Techniques,Université de
Zinder,BP 656 Zinder( Niger) E-mail<EMAIL_ADDRESS>Mahamane Mahi
Harouna222Faculté des Sciences et Techniques, Université Dan Dicko Dankoulodo
de Maradi( Niger), BP 465 Maradi( Niger),E-mail:
<EMAIL_ADDRESS>author and Bazanfaré Mahaman444Faculté des
Sciences et Techniques, Université Abdou Moumouni de Niamey( Niger), BP 10662
Niamey( Niger) E-mail<EMAIL_ADDRESS>
Abstract. In this paper, we give the Cartan′s formula for half-lightlike
submanifolds of Lorentzian manifolds and use it to show that a screen
homothetic half-lightlike submanifolds of a Lorentzian space form, with a
conformal co-screen distribution are locally a lightlike triple product
manifolds. Then we give a classification theorem for half-lightlike
submanifolds of Lorentzian space form with constant screen principal
curvatures. These results extend some results obtained in the case of
lightlike hypersurfaces of Lorentzian manifolds ([1]).

$MSC2010:53B30,53C42,53C50$

Keywords: half-lightlike submanifold, screen homothetic, conformal
distribution, co-screen distribution.

##  1 Introduction

It is well known that the intersection of the normal bundle and the tangent
bundle of a lightlike submanifold of a semi-Riemannian manifold is not
trivial([5]). Thus, one cannot use, in the usual way, the classical
submanifold theory to define any induced object on a lightlike submanifold. To
overcome these difficulties in degenerate geometry, Duggal and Bejancu ([5])
introduced a non-degenerate screen distribution (or equivalently a null
transversal vector bundle) so as to get three factors splitting the ambient
tangent bundle. Then, they derived the main induced geometric objects such as
second fundamental forms, shape operators, induced connections, curvature,
etc.

The class of lightlike submanifolds of codimension $2$ is composed of two
classes by virtue of the dimension of its radical distribution (dimension that
is either 1 or 2) , named by half-lightlike and coisotropic submanifolds ([4],
[6]). A codimension 2 lightlike submanifold is called a half-lightlike
submanifold if dim(Rad(TM)) = 1. For more results about half-lightlike
submanifolds, we refer the reader to ([8], [9], [7]).

We know that the shape operator plays an important role in the study geometry
of submanifolds. In the case of half-lightlike submanifolds, there are three
shape operators ($A_{N}\leavevmode\nobreak\
\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ A_{L}$) and there are interrelations between
these geometric objects and those of its screen distributions. The shape
operators $A_{N}\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ A_{\xi}$
are $\Gamma(S(TM))$-valued, but not $A_{L}$. Moreover, the shape operators
$A_{N}\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ A_{L}$ of a half-
lightlike submanifold are not necessarily auto-adjoint, but the operator
$\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}$ of the screen distribution is
diagonalizable. In [1], C.Atindogbé, M.M. Harouna and J.Tossa prove the so-
called Cartan,s fundamental formula for lightlike hypersurfaces and use it to
show that a screen conformal lightlike hypersurface of a Lorentzian Euclidean
space is locally a lightlike triple product manifold. They also give a
classification theorem for lightlike hypersurfaces of Lorentzian Euclidean
space with constant screen principal curvatures. In this paper, we generalize
those studies on half-lightlike submanifold case. The article is organised as
follows: in section $2$ we brief basic informations on half-lightlike
submanifolds following closely the approach in ([5]). In section 3, we prove
the Cartan’s formula for half-lightlike submanifold. In section 4, we use this
formula to show that a screen homothetic half-lightlike submanifolds of a
Lorentzian space form, with a conformal co-screen distribution are locally a
lightlike triple product manifolds. Then we give a classification theorem for
half-lightlike submanifolds of Lorentzian space form with constant screen
principal curvatures.

##  2 Preliminaries on half-lightlike submanifolds

In this section, we present basic notions on differential geometry of half-
lightlike submanifold manifolds. A full discussion of this content can be
found in ([4, 5, 6, 7]).

Let $(\overline{M},\overline{g})$ be a $(m+2)$-dimensional semi-Riemannian
manifold of index $q\geq 1$ and $(M,g)$ a lightlike submanifold of codimension
2 of $\overline{M}$. We say that $M$ is a half-lightlike submanifold if the
dimension of the radical distribution $Rad(TM)$ is one. It is well known that
the radical distribution is given by $Rad(TM)=TM\cap TM^{\bot}$, where
$TM^{\bot}$ is called the normal bundle of $M$ in $\overline{M}$. Thus there
exist two non-degenerate complementary distributions $S(TM)$ and
$S(TM^{\bot})$ of $Rad(TM)$ in $TM$ and $TM^{\bot}$ respectively [5], which
are called the screen and co-screen distribution respectively on $M$ . Thus we
have

$TM=\mbox{Rad}(TM)\underset{\bot}{\oplus}S(TM)$ (1)

and

$TM^{\bot}=\mbox{Rad}(TM)\underset{\bot}{\oplus}S(TM^{\bot})$ (2)

where $\underset{\bot}{\oplus}$ denotes the orthogonal direct sum.

Since $S(TM)$ is non-degenerate, we have the decomposition

$T\overline{M}_{\mid_{M}}=S(TM)\underset{\bot}{\oplus}S(TM)^{\bot}$ (3)

Thus, it is easy to see that $S(TM^{\bot})$ is a subbundle of $S(TM)^{\bot}$.
As $S(TM^{\bot})$ is a non-degenerate subbundle of $S(TM)^{\bot}$, we have the
decomposition

$S(TM)^{\bot}=S(TM^{\bot})^{\bot}\underset{\bot}{\oplus}S(TM^{\bot})$ (4)

It is obvious to see that $S(TM^{\bot})^{\bot}$ is also a non-degenerate
distribution and $Rad(TM)$ is a subbundle of $S(TM^{\bot})^{\bot}$. Choose
$L\in\Gamma(S(TM^{\bot}))$ such that $\overline{g}(L,L)=1$. For any null
section $\xi\in(Rad(TM))$, there exists a unique null vector field
$N\in(S(TM^{\bot})^{\bot})$ satisfying

$\displaystyle\overline{g}(\xi,N)=1,\overline{g}(N,N)=\overline{g}(N,X)=\overline{g}(N,L)=0,\forall
X\in(S(TM)).$ (5)

Denote by $ltr(TM)$ the vector subbundle of $S(TM^{\bot})^{\bot}$ locally
spanned by $N$. Then we have:

$\displaystyle S(TM^{\bot})^{\bot}=Rad(TM)\oplus ltr(TM).$ (6)

Let $tr(TM)=S(TM^{\bot})\underset{\bot}{\oplus}ltr(TM)$. We call $N$,
$ltr(TM)$ and $tr(TM)$ the lightlike transversal vector field, lightlike
transversal vector bundle and transversal vector bundle of M with respect to
the chosen screen distribution $S(TM)$ respectively. Then $T\overline{M}$ is
decomposed as following

$\displaystyle T\overline{M}_{\mid_{M}}$ $\displaystyle=$ $\displaystyle
TM\oplus tr(TM)=Rad(TM)\oplus tr(TM)\underset{\bot}{\oplus}S(TM)$ (7)
$\displaystyle=$ $\displaystyle Rad(TM)\oplus
ltr(TM)\underset{\bot}{\oplus}S(TM)\underset{\bot}{\oplus}S(TM^{\bot}).$ (8)

Let $P$ be the projection morphism of $TM$ on $S(TM)$ with respect to the
decomposition (1). For any $X,Y\in\Gamma(TM)$, $N\in\Gamma(ltr(TM))$,
$\xi\in\Gamma(Rad(TM))$ and $L\in\Gamma(S(TM)^{\bot})$, the Gauss and
Weingarten formulas of $M$ and $S(TM)$ are given by

$\displaystyle\overline{\nabla}_{X}Y$ $\displaystyle=$
$\displaystyle\nabla_{X}Y+B(X,Y)N+D(X,Y)L,$ (9)
$\displaystyle\overline{\nabla}_{X}N$ $\displaystyle=$ $\displaystyle-
A_{N}X+\tau(X)N+\rho(X)L,$ (10) $\displaystyle\overline{\nabla}_{X}L$
$\displaystyle=$ $\displaystyle-A_{L}X+\phi(X)N,$ (11)
$\displaystyle\nabla_{X}PY$ $\displaystyle=$
$\displaystyle\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}PY+C(X,PY)\xi,$ (12)
$\displaystyle\nabla_{X}\xi$ $\displaystyle=$
$\displaystyle-\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X-\tau(X)\xi,$ (13)

respectively, where $\nabla\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\
\stackrel{{\scriptstyle\ast}}{{\nabla}}$ are induced connection on $TM$ and
$S(TM)$ respectively, $B$ and $D$ are called local second fundamental forms of
$M$, $C$ is called the local second fundamental form of $S(TM)$;
$A_{N},\leavevmode\nobreak\
\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ A_{L}$ are linear shape operators on $TM$ of
lightlike transversal bundle, radical bundle and screen transversal bundle
respectively. $\tau,\leavevmode\nobreak\ \rho\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ \phi$ are 1-forms on $M$.

$\displaystyle h(X,Y)=B(X,Y)N+D(X,Y)L$ (14)

is the second fundamental form tensor of $M$. Since the connection
$\overline{\nabla}$ on $\overline{M}$ is torsion-free, the induced connection
$\nabla$ on $M$ is also torsion-free, and then $B$ and $D$ are symmetric
tensors on $\Gamma(TM)$. But $\nabla$ is not metric, since using (9), we have

$\displaystyle(\nabla_{X}g)(Y,Z)=B(X,Y)\eta(Z)+B(X,Z)\eta(Y),$ (15)

for all $X,Y\in\Gamma(TM)$, where $\eta$ is a 1-form on $TM$ such that
$\eta(X)=g(X,N)$, $\forall X\in\Gamma(TM)$. It is easy to verify that the
induced connection $\stackrel{{\scriptstyle\ast}}{{\nabla}}$ on $S(TM)$ is
metric. Since
$B(X,Y)=\overline{g}(\overline{\nabla}_{X}Y,\xi)\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ D(X,Y)=\overline{g}(\overline{\nabla}_{X}Y,L)$,
it is obvious that $B\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ D$
are independent of the choice of a screen distribution. However, we note that
both $B$ and $\tau$ depend on the section $\xi\in\Gamma(Rad(TM))$. In the case
we take $\xi^{*}=\alpha\xi$, it follows that $N^{*}=\frac{1}{\alpha}N$. Hence
we obtain

$\displaystyle B^{*}=\alpha B\leavevmode\nobreak\ \leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ \leavevmode\nobreak\
\tau(X)=\tau^{*}(X)+X(\log\alpha)\leavevmode\nobreak\ \leavevmode\nobreak\
\mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\ X\in\Gamma(TM)$ (16)

The above three local second fundamental forms of $M$ and $S(TM)$ are related
to their shape operators by

$\displaystyle g(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X,Y)$
$\displaystyle=$ $\displaystyle B(X,Y),\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\overline{g}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X,N)=0$ (17)
$\displaystyle g(A_{L}X,Y)$ $\displaystyle=$ $\displaystyle
D(X,Y)+\phi(X)\eta(Y),\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\overline{g}(A_{L}X,N)=\rho(X),$ (18) $\displaystyle g(A_{N}X,PY)$
$\displaystyle=$ $\displaystyle C(X,PY),\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\overline{g}(A_{N}X,N)=0,$ (19)

for any $X,Y\in\Gamma(TM)$.
From (17), (18), (19), we see that $B$ and $D$ satisfy

$\displaystyle B(X,\xi)=0,\leavevmode\nobreak\ \leavevmode\nobreak\
D(X,\xi)=-\phi(X),\leavevmode\nobreak\ \forall X\in\Gamma(TM),$ (20)

$\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ A_{N}$ are $S(TM)$-valued, and
$\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}$ is self-adjoint on $TM$ such that

$\displaystyle\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\xi=0$ (21)

Denote by $\overline{R},\leavevmode\nobreak\ R\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ \stackrel{{\scriptstyle\ast}}{{R}}$ the
curvature tensors of connection $\overline{\nabla}\leavevmode\nobreak\
\mbox{on}\leavevmode\nobreak\ \overline{M}$, the induced connection $\nabla$
on $M$ and the induced connection $\stackrel{{\scriptstyle\ast}}{{\nabla}}$ on
$S(TM)$, respectively. Using the Gauss-Weingarten formulas for $M$ and
$S(TM)$, we obtain the Gauss-Codazzi equations for $M$ and $S(TM)$:

$\displaystyle\overline{R}(X,Y)Z$ $\displaystyle=$ $\displaystyle
R(X,Y)Z+B(X,Z)A_{N}Y-B(Y,Z)A_{N}X$ (28)
$\displaystyle+D(X,Z)A_{L}Y-D(Y,Z)A_{L}X$
$\displaystyle+\\{(\nabla_{X}B)(Y,Z)-(\nabla_{Y}B)(X,Z)$
$\displaystyle+B(Y,Z)\tau(X)-B(X,Z)\tau(Y)$
$\displaystyle+D(Y,Z)\phi(X)-D(X,Z)\phi(Y)\\}N$
$\displaystyle+\\{(\nabla_{X}D)(Y,Z)-(\nabla_{Y}D)(X,Z)+B(Y,Z)\rho(X)$
$\displaystyle-B(X,Z)\rho(Y)-D(Y,Z)+D(X,Z)\\}L,$
$\displaystyle\overline{R}(X,Y)N$ $\displaystyle=$
$\displaystyle-\nabla_{X}(A_{N}Y)+\nabla_{Y}(A_{N}X)+A_{N}[X,Y]$ (33)
$\displaystyle+\tau(X)A_{N}Y-\tau(Y)A_{N}X+\rho(X)A_{L}Y-\rho(Y)A_{L}X$
$\displaystyle+\\{B(A_{N}X,Y)-B(A_{N}Y,X)+2d\tau(X,Y)+$
$\displaystyle+\phi(X)\rho(Y)-\phi(Y)\rho(X)\\}N$
$\displaystyle+\\{D(A_{N}X,Y)-D(A_{N}Y,X)+2d\rho(X,Y)+\rho(X)\tau(Y)-\rho(Y)\tau(X)\\}L,$
$\displaystyle\overline{R}(X,Y)L$ $\displaystyle=$
$\displaystyle-\nabla_{X}(A_{L}Y)+\nabla_{Y}(A_{L}X)+A_{L}[X,Y]$ (38)
$\displaystyle+\phi(X)A_{N}Y-\phi(Y)A_{N}X$
$\displaystyle+\\{B(A_{L}X,Y)-B(A_{L}Y,X)+2d\phi(X,Y)$
$\displaystyle+\tau(X)\phi(Y)-\tau(Y)\phi(X)\\}N$
$\displaystyle+\\{D(A_{L}X,Y)-D(A_{L}Y,X)+\rho(X)\phi(Y)-\rho(Y)\phi(X)\\}L,$
$\displaystyle R(X,Y)PZ$ $\displaystyle=$
$\displaystyle\stackrel{{\scriptstyle\ast}}{{R}}(X,Y)PZ+C(X,PZ)\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y-C(Y,PZ)\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X$
(41) $\displaystyle+\\{(\nabla_{X}C)(Y,PZ)-(\nabla_{Y}C)(X,PZ)$
$\displaystyle+C(X,PZ)\tau(Y)-C(Y,PZ)\tau(X)\\}\xi,$ $\displaystyle R(X,Y)\xi$
$\displaystyle=$
$\displaystyle-\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)+\stackrel{{\scriptstyle\ast}}{{\nabla}}_{Y}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X)+\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}[X,Y]+\tau(Y)\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X$
(43)
$\displaystyle-\tau(X)\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y+\\{C(Y,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X)-C(X,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)-2d\tau(X,Y)\\}\xi.$

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

$\displaystyle\overline{R}(\overline{X},\overline{Y})\overline{Z}=k\\{\overline{g}(\overline{Y},\overline{Z})\overline{X}-\overline{g}(\overline{X},\overline{Z})\overline{Y}\\},\leavevmode\nobreak\
\forall\overline{X},\overline{Y},\overline{Z}\in\Gamma(T\overline{M}).$ (44)

It is then obvious that $\overline{R}(X,Y)Z$ is tangent to $M$ for all
$X,Y,Z\in\Gamma(TM)$ , so from (28), we get

$\displaystyle\overline{R}(X,Y)Z=R(X,Y)Z+B(X,Z)A_{N}Y-B(Y,Z)A_{N}X+D(X,Z)A_{L}Y-D(Y,Z)A_{L}X,\leavevmode\nobreak\
\forall X,Y,Z\in\Gamma(TM).$ (45)

The Ricci curvature tensor of $\overline{M}$, denoted by $\overline{R}$ic is
defined by

$\displaystyle\overline{R}ic=trace\\{Z\rightarrow\overline{R}(Z,X)Y\\},\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\
X,Y\in\Gamma(T\overline{M}).$ (46)

Locally, $\overline{R}$ic is given by

$\displaystyle\overline{R}ic=\sum_{i=1}^{m+2}\varepsilon_{i}\overline{g}(\overline{R}(E_{i},X)Y,E_{i}),$
(47)

where $\\{E_{1},\cdots,E_{m+2}\\}$ is an orthonormal frame field of
$T\overline{M}$ and $\varepsilon_{i}=\pm 1$ denote the causal caracter of
respective vector field $E_{i}$.

Consider the induced quasi-orthonormal frame field $\\{\xi,W_{a}\\}$ on $M$,
where $Rad(TM)=span\\{\xi\\}$ and $S(TM)=span\\{W_{a}\\}$ and let
$E=\\{\xi;W_{a},L,N\\}$ be the corresponding frame field on $\overline{M}$.
Then, by using (46), we obtain

$\displaystyle\overline{R}ic(X,Y)=\sum_{a=1}^{m-1}\varepsilon_{a}\overline{g}(\overline{R}(W_{a},X)Y,W_{a})+\overline{g}(\overline{R}(\xi,X)Y,N)+\overline{g}(\overline{R}(L,X)Y,L)+\overline{g}(\overline{R}(N,X)Y,\xi).$
(48)

Let $Ric$ denote the induced Ricci tensor on $M$ given by:

$\displaystyle Ric(X,Y)=trace\\{Z\rightarrow R(Z,X)Y\\}\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\
X,Y\in\Gamma(TM).$ (49)

Using the quasi-orthonormal frame $\\{\xi,W_{a}\\}$ on $M$, we obtain:

$\displaystyle
Ric(X,Y)=\sum_{a=1}^{m-1}\varepsilon_{a}g(R(W_{a},X)Y,W_{a})+\overline{g}(R(\xi,X)Y,N).$
(50)

Substituting the Gauss-Codazzi equations (28) in (48) and using relation (17)
and (19), we obtain

$\displaystyle
Ric(X)=\overline{R}ic(X,Y)+B(X,Y)tr(A_{N})+D(X,Y)tr(A_{L})-g(A_{N}X,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)-g(A_{L}X,A_{L}Y)$
$\displaystyle+\rho(X)\varphi(Y)-\overline{g}(\overline{R}(\xi,Y)X,N)-\overline{g}(\overline{R}(L,Y)X,L)\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\
X,Y\in\Gamma(TM).$ (51)

This shows that $Ric$ is not symmetric. Using (2) and the first Bianchi’s
identity, we obtain:

$\displaystyle
Ric(X,Y)-Ric(Y,X)=g(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X,A_{N}Y)-g(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y,A_{N}X)+\rho(X)\varphi(Y)-\rho(Y)\varphi(X)-\overline{g}(\overline{R}(X,Y)\xi,N).$
(52)

From this relation and the relations (28) and (43), we have

$\displaystyle Ric(X,Y)-Ric(Y,X)=2d\tau(X,Y).$ (53)

######  Theorem 2.1.

([7]) Let $(M,g,S(TM),S(TM^{\bot}))$ be a half-lightlike submanifold of a
semi-Riemannian manifold $(\overline{M},\overline{g})$. Then, the induced
Ricci curvature on $M$ is symmetric if and only if each 1-form $\tau$ is
closed, i.e $d\tau=0$ on any neighborhood $\mathcal{U}\subset M$.

######  Remark 2.1.

If the induced Ricci tensor $Ric$, on $M$ is symmetric, the 1-form $\tau$ is
closed by theorem2.1. Thus, by Poincaré lemma, there existe a smooth function
$f$ on $\mathcal{U}\subset M$ such that $\tau=df$. Consequently, we get
$\tau(X)=X.(f)\leavevmode\nobreak\ \leavevmode\nobreak\ \mbox{for
any}\leavevmode\nobreak\ \leavevmode\nobreak\
X\in\Gamma(TM_{\mid_{\mathcal{U}}})$. If we take $\xi^{*}=\alpha\xi$, then by
setting $\alpha=\exp(f)$, we get $\xi^{*}=0\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\
X\in\Gamma(TM_{\mid_{\mathcal{U}}})$. We call the pair $\\{\xi,N\\}$ on
$\mathcal{U}$ such that the corresponding 1-form $\tau$ vanishes the canonical
null pair of $M$.

,

######  Definition 2.1.

A half-lightlike submanifold $(M,g,S(TM),S(TM^{\bot}))$ of a semi-Riemannian
manifold is screen locally (resp., globally) conformal if on any coordinate
neighborhood $\mathcal{U}$ (resp., $\mathcal{U}=M$) there exists a non-zero
smooth function $\varphi$ such that for any null vector field
$\xi\in\Gamma(TM^{\bot})$, the relation

$A_{N}X=\varphi\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X,\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\
X\in\Gamma(TM_{|\mathcal{U}}),$ (54)

holds between the shape operators $A_{N}\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ \leavevmode\nobreak\
\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}$ of $M\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ \leavevmode\nobreak\
S(TM)$, respectively.

In particular, if $\varphi$ is a non-zero constant, $M$ is called screen
homothetic.

It is easy to see from (17) and (19) that a half-lightlike submanifold $M$ is
screen conformal if and only if the second fundamental forms $B$ and $C$
satisfy

$C(X,PY)=\varphi B(X,Y),\leavevmode\nobreak\ \leavevmode\nobreak\ \mbox{for
any}\leavevmode\nobreak\ \leavevmode\nobreak\
X,Y\in\Gamma(TM_{\mid_{\mathcal{U}}})$ (55)

##  3 Cartan′s formula for half-lightlike submanifolds

In this section, we first consider a half-lightlike submanifold
$(M,g,S(TM),S(TM^{\bot}))$ of a semi-Riemannian manifold
$(\overline{M}(k),\overline{g})$ of constant curvature $k$. As a direct
application of the Gauss-Codazzi equations for $M\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ \leavevmode\nobreak\
S(TM)$, we have this proposition.

######  Proposition 3.1.

Let $(\overline{M}(k),\overline{g})$ be a semi-Riemannian manifold of constant
curvature $k$ and $(M,g,S(TM),S(TM^{\bot}))$ be a half-lightlike submanifold
of $\overline{M}(k)$. Denote by $R$ the curvature tensor of the induced
connection $\nabla$ on $M$ by the Levi-civita connection $\overline{\nabla}$
on $\overline{M}(k)$. For any $X,Y,Z\in\Gamma(TM)$, we have:

  1. 1.

$R(X,Y)Z=k\\{g(Y,Z)X-g(X,Z)Y\\}+B(Y,Z)A_{N}X-B(X,Z)A_{N}Y+D(Y,Z)A_{L}X-D(X,Z)A_{L}Y$

  2. 2.

$(\nabla_{X}B)(Y,Z)-(\nabla_{Y}B)(X,Z)=B(X,Z)\tau(Y)-B(Y,Z)\tau(X)+D(X,Z)\phi(Y)-D(Y,Z)\phi(X)$

  3. 3.

$(\nabla_{X}D)(Y,Z)-(\nabla_{Y}D)(X,Z)=B(X,Z)\rho(Y)-B(Y,Z)\rho(X)$

  4. 4.

$B(A_{N}Y,X)-B(A_{N}X,Y)=\phi(X)\rho(Y)-\phi(Y)\rho(X)+2d\tau(X,Y)$

  5. 5.

$(\nabla_{X}A_{N})Y-(\nabla_{Y}A_{N})X+k[\eta(Y)X-\eta(X)Y]=\tau(X)A_{N}Y-\tau(Y)A_{N}X+\rho(X)A_{L}Y-\rho(Y)A_{L}X$

  6. 6.

$D(A_{N}Y,X)-D(A_{N}X,Y)=\rho(X)\tau(Y)-\rho(Y)\tau(X)+2d\rho(X,Y)$

  7. 7.

$(\nabla_{X}A_{L})Y-(\nabla_{Y}A_{L})X=\phi(X)A_{N}Y-\phi(Y)A_{N}X$

  8. 8.

$B(A_{L}Y,X)-B(A_{L}X,Y)=\tau(X)\phi(Y)-\tau(Y)\phi(X)+2d\phi(X,Y)$

  9. 9.

$D(A_{L}Y,X)-D(A_{L}X,Y)=\rho(X)\phi(Y)-\rho(Y)\phi(X)$

  10. 10.

$(\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y-(\nabla_{Y}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})X=D(X,\xi)A_{L}Y-D(Y,\xi)A_{L}X+\tau(Y)\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X-\tau(X)\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y-2d\tau(X,Y)\xi$

######  Definition 3.1.

A lightlike submanifold $M$ is said to be irrotational if
$\overline{\nabla}_{X}\xi\in\Gamma(TM)\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\
X\in\Gamma(TM)$, where $\xi\in\Gamma(Rad(TM))$.

For a half-lightlike $M$, since $B(X,\xi)=0$, the above definition is
equivalent to $D(X,\xi)=\phi(X)=0\leavevmode\nobreak\ \leavevmode\nobreak\
\mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\ X\in\Gamma(TM)$

######  Remark 3.1.

From relations (9) and (124), we get that any half-lightlike submanifold
$(M,g,S(TM),S(TM^{\bot}))$ of a semi-Riemannian manifold
$(\overline{M},\overline{g})$, with a conformal co-screen distribution is an
irrotational submanifold, but the opposite way is wrong.

In the sequel, we consider a half-lightlike submanifold $M$ of an
(m+2)-dimensional Lorentzian manifold $(\overline{M}(k),\overline{g})$ of
constant curvature $k$. For this class of screen conformal and irrotational
half-lightlike submanifold $M$, the screen distribution $S(TM)$ is Riemannian
(Proposition4.2.2 of [7]), integrable ( Theorem4.4.4 of [7]) and the induced
Ricci tensor on $M$ is symmetric ( Corollaire4.4.11 of [7] ). Then, according
to remark2.1, there exists a pair $\\{\xi,N\\}$ on $\mathcal{U}$ satisfying
(5) such that the corresponding $1$-forme $\tau$ from (10) vanishes. Since
$\xi$ is an eigenvector field of $\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}$
corresponding to the eigenvalue $0$ and
$\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}$ is $\Gamma(S(TM))$-valued real
symmetric, $\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}$ has $m-1$ orthonormal
eigenvector fields in $S(TM)$ and is diagonalizable. Consider a frame field of
eigenvectors $\\{\xi,E_{1},\cdots,E_{m-1}\\}$ of
$\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}$ such that
$\\{E_{1},\cdots,E_{m-1}\\}$ is an orthonormal frame field of $S(TM)$. Then,
$\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}E_{i}=\lambda_{i}$, $1\leq i\leq
m-1$. We call the eigenvalues $\lambda_{i}$ the screen principal curvatures
for all $i$.

In the following, we assume that all screen principal curvatures are constant
along $S(TM)\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ \tau=0$.
Consider the following distribution on $M$:

$\displaystyle T_{\lambda}=\\{X\in\Gamma(S(TM)):\leavevmode\nobreak\
\leavevmode\nobreak\ \stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X=\lambda X\\}$
(56)

######  Lemma 3.1.

Let $(M,g,S(TM),S(TM^{\bot}))$ be an irrotational half-lightlike submanifold
of a semi-Riemannian manifold $(\overline{M}(k),\overline{g})$. If we assume
that the 1-forme $\tau$ from (10) is identically null, then for any
$X\in\Gamma(TM)$, we have:

  1. (i)

$(\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y=(\nabla_{Y}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})X\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\
Y\in\Gamma(TM)$

  2. (ii)

$(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y=(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{Y}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})X\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\
Y\in\Gamma(TM)$

  3. (iii)

$\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}$ is symmetric with respect
to $g$, i.e

$\displaystyle
g((\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y,Z)=g(Y,(\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Z)\leavevmode\nobreak\
\leavevmode\nobreak\ \forall\leavevmode\nobreak\ \leavevmode\nobreak\
Y,Z\in\Gamma(S(TM))$

  4. (iv)

$\mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\ Y,Z\in\Gamma(S(TM))$

$\displaystyle g((\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y,Z)$
$\displaystyle=$ $\displaystyle
g(Y,(\nabla_{Z}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})X)=g((\nabla_{Z}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y,X)$

  5. (v)

$\mbox{for any}\leavevmode\nobreak\ \leavevmode\nobreak\
Y\in\Gamma(T_{\lambda}),Z\in\Gamma(T_{\mu})$

$\displaystyle
g((\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y,Z)=(\lambda-\mu)g(\nabla_{X}Y,Z)$

Proof

Since the and $\tau=0$, $(10)$ of proposition3.1 give $(i)$ .

$\displaystyle(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y$
$\displaystyle=$
$\displaystyle\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)-\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y$
(57) $\displaystyle=$
$\displaystyle\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y-C(X,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)\xi-\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\nabla_{X}Y+C(X,Y)\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\xi$
(58) $\displaystyle=$
$\displaystyle\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y-C(X,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)\xi-\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\nabla_{X}Y$
(59) $\displaystyle=$
$\displaystyle(\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y-C(X,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)\xi$
(60) $\displaystyle=$
$\displaystyle(\nabla_{Y}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})X-C(X,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)\xi$
(61) $\displaystyle=$
$\displaystyle(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{Y}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})X+C(Y,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X)\xi-C(X,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)\xi$
(62)

From (44), it is obvious that $R(X,Y)\xi=0$; and then equation (43) give

$\displaystyle
g(R(X,Y)\xi,N)=C(Y,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X)-C(X,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)=0$
(63)

From equations (57) and (63) we get $(ii)$.

$\displaystyle g((\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y,Z)$
$\displaystyle=$ $\displaystyle
g(\nabla_{X}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y),Z)-g(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\nabla_{X}Y,Z)$
(64) $\displaystyle=$ $\displaystyle
g(\nabla_{X}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y),Z)-g(\nabla_{X}Y,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Z)$
(65) $\displaystyle=$
$\displaystyle-(\nabla_{X}g)(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y,Z)+X.g(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y,Z)-g(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y,\nabla_{X}Z)$
(67)
$\displaystyle+(\nabla_{X}g)(Y,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Z)-X.g(Y,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Z)+g(Y,(\nabla_{X}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Z))$
$\displaystyle=$
$\displaystyle-g(Y,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\nabla_{X}Z)+g(Y,(\nabla_{X}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Z))$
(68) $\displaystyle=$ $\displaystyle
g(Y,(\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Z)$ (69)

That give as $(iii)$.

$\displaystyle g((\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y,Z)$
$\displaystyle=$ $\displaystyle
g(Y,(\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Z)$ (70)
$\displaystyle=$ $\displaystyle
g(Y,(\nabla_{Z}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})X)$ (71)
$\displaystyle=$ $\displaystyle
g((\nabla_{Z}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y,X)$ (72)

From $(i)$, $(iii)$ and (70), we get $(iv)$

$\displaystyle g((\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y,Z)$
$\displaystyle=$ $\displaystyle
g(\nabla_{X}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y),Z)-g(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\nabla_{X}Y,Z)$
(73) $\displaystyle=$ $\displaystyle\lambda
g(\nabla_{X}Y,Z)-g(\nabla_{X}Y,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Z)$
(74) $\displaystyle=$ $\displaystyle\lambda g(\nabla_{X}Y,Z)-\mu
g(\nabla_{X}Y,Z)$ (75) $\displaystyle=$
$\displaystyle(\lambda-\mu)g(\nabla_{X}Y,Z)$ (76)

That give $(v)$.

######  Lemma 3.2.

Let $(M,g,S(TM),S(TM^{\bot}))$ be an irrotational half-lightlike submanifold
of a semi-Riemannian manifold $(\overline{M}(k),\overline{g})$. If we assume
that the 1-forme $\tau$ from (10) is identically null, then

  1. 1.

$\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y\in\Gamma(T_{\lambda})\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{if}\leavevmode\nobreak\ \leavevmode\nobreak\
X,Y\in\Gamma(T_{\lambda})$

  2. 2.

$\nabla_{X}Y\bot T_{\lambda};\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \nabla_{Y}X\bot
T_{\mu}\leavevmode\nobreak\ \leavevmode\nobreak\ \mbox{if}\leavevmode\nobreak\
\leavevmode\nobreak\ X\in\Gamma(T_{\lambda}),\leavevmode\nobreak\
\leavevmode\nobreak\ Y\in\Gamma(T_{\mu}),\leavevmode\nobreak\
\leavevmode\nobreak\ \lambda\neq\mu.$

Proof
Let $Z\in\Gamma(TM)\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\
X,Y\in\Gamma(T_{\lambda})$. From $(ii)$ and $(iv)$ of lemma3.1, we have:

$\displaystyle
g(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y,Z)$
$\displaystyle=$ $\displaystyle
g(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y,Z)-g((\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y,Z)$
(77) $\displaystyle=$ $\displaystyle\lambda
g(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y,Z)-g(Y,(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{Z}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})X)$
(78) $\displaystyle=$ $\displaystyle\lambda
g(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y,Z)-g(Y,\stackrel{{\scriptstyle\ast}}{{\nabla}}_{Z}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X)+g(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{Z}X,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)$
(79) $\displaystyle=$ $\displaystyle\lambda
g(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y,Z)-\lambda
g(Y,\stackrel{{\scriptstyle\ast}}{{\nabla}}_{Z}X)+\lambda
g(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{Z}X,Y)$ (80) $\displaystyle=$
$\displaystyle\lambda g(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y,Z),$ (81)

then

$\displaystyle
g(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y-\lambda\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y,Z)=0$
(82)

Since $Z$ is an arbitrary vector field tangent to $M$, we have:

$\displaystyle\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y-\lambda\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y=\alpha\xi,\leavevmode\nobreak\
\mbox{where $\alpha$ is a smooth function on $M$.}$ (83)

By the fact that

$\displaystyle\overline{g}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y-\lambda\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y,N)=0,$
(84)

we conclude that $\alpha=0$, and then we obtain
$\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y=\lambda\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y$;
and then, we conclude that
$\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y\in\Gamma(T_{\lambda})$.
Now, let $X,Z\in\Gamma(T_{\lambda})\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ Y\in\Gamma(T_{\mu})$. From $(v)$ of lemma3.1,
we have:

$\displaystyle
g((\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y,Z)=(\mu-\lambda)g(\nabla_{X}Y,Z)$
(85)

Otherwise, from $(iv)$ of lemma3.1 we have

$\displaystyle
g((\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y,Z)=g((\nabla_{Z}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})X,Y)=-(\mu-\lambda)g(\nabla_{Z}X,Y)=-(\mu-\lambda)g(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{Z}X,Y).$
(86)

Since, from $(1)$ we have
$\stackrel{{\scriptstyle\ast}}{{\nabla}}_{Z}X\in\Gamma(T_{\lambda})\leavevmode\nobreak\
\mbox{for any}\leavevmode\nobreak\ X,Z\in\Gamma(T_{\lambda})$, then by (86) we
get $g(\nabla_{Z}X,Y)=0$. So, from relations (85) and (86), we obtain

$\displaystyle(\mu-\lambda)g(\nabla_{X}Y,Z)=-(\mu-\lambda)g(\nabla_{Z}X,Y)=0.$
(87)

Hence, if $\lambda\neq\mu$, then $\nabla_{X}Y\bot T_{\lambda}$.
Similarly, we prove that $\nabla_{Y}X\bot T_{\lambda}\leavevmode\nobreak\
\mbox{if}\leavevmode\nobreak\ \lambda\neq\mu$.

Now we prove the following theorem which extends Cartan′s fundamental formula
on half-lightlike submanifold of Lorentzian manifolds with constant curvature.

######  Theorem 3.1.

Let $(M,g,S(TM),S(TM^{\bot}))$ be an irrotational half-lightlike submanifold
of an $(m+2)$-dimensional Lorenzian manifold $(\overline{M}(k),\overline{g})$
of constant curvature $k$. Assume that $E_{0}=\xi,E_{1},...,E_{m-1}$ are
eigenvectors of $\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}$ satisfying
$\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}E_{0}=0\leavevmode\nobreak\
\leavevmode\nobreak\
\mbox{and}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}E_{i}=\lambda_{i}E_{i}$ such
that $\lambda_{i}$ is constant along $S(TM)$ for all $i$ and $\tau=0$
($\\{E_{i}\\}_{i=1,...m-1}$ represents an orthonormal basis of $S(TM)$). Then,
for every $i\in\\{1,...,m-1\\}$, we have:

$\displaystyle\sum_{\begin{subarray}{c}j=1\\\
\lambda_{j}\neq\lambda_{i}\end{subarray}}^{m-1}\frac{k+\lambda_{j}g(A_{N}E_{i},E_{i})+\lambda_{i}g(A_{N}E_{j},E_{j})+g(A_{L}E_{j},E_{j})g(A_{L}E_{i},E_{i})-g(A_{L}E_{i},E_{j})^{2}}{\lambda_{i}-\lambda_{j}}=0$
(88)

Moreover, if the screen is conformal with conformal factor $\varphi$, then for
all $\in\\{1,\cdots,m-1\\}$,

$\displaystyle\sum_{\begin{subarray}{c}j=1\\\
\lambda_{j}\neq\lambda_{i}\end{subarray}}^{m-1}\frac{k+2\varphi\lambda_{i}\lambda_{j}+g(A_{L}E_{j},E_{j})g(A_{L}E_{i},E_{i})-g(A_{L}E_{i},E_{j})^{2}}{\lambda_{i}-\lambda_{j}}=0$
(89)

Proof
From $(1)$ of proposition3.1, and relations (17) and (124), we have:

$\displaystyle
R(E_{i},E_{j})Ej=kE_{i}+\lambda_{j}A_{N}E_{i}+g(A_{L}E_{j},E_{j})A_{L}E_{i}-g(A_{L}E_{i},E_{j})A_{L}E_{j}$
(90)

Also, we have:

$\displaystyle
R(E_{i},E_{j})E_{j}=\nabla_{E_{i}}\nabla_{E_{j}}E_{j}-\nabla_{E_{j}}\nabla_{E_{i}}E_{j}-\nabla_{[E_{i},E_{j}]}E_{j}$
(91)

then, using (15) and lemma3.2 for $\lambda_{i}\neq\lambda_{j}$, we get:

$\displaystyle g(R(E_{i},E_{j})E_{j},E_{i})$ $\displaystyle=$ $\displaystyle
g(\nabla_{E_{i}}\nabla_{E_{j}}E_{j},E_{i})-g(\nabla_{E_{j}}\nabla_{E_{i}}E_{j},E_{i})-g(\nabla_{[E_{i},E_{j}]}E_{j},E_{i})$
(92) $\displaystyle=$ $\displaystyle
g(\nabla_{E_{i}}\nabla_{E_{j}}E_{j},E_{i})+(\nabla_{E_{j}}g)(\nabla_{E_{i}}E_{j},E_{i})-E_{j}.g(\nabla_{E_{i}}E_{j},E_{i})+g(\nabla_{E_{i}}E_{j},\nabla_{E_{j}}E_{i})$
(94) $\displaystyle-g(\nabla_{[E_{i},E_{j}]}E_{j},E_{i})$ $\displaystyle=$
$\displaystyle
g(\nabla_{E_{i}}\nabla_{E_{j}}E_{j},E_{i})+g(\nabla_{E_{i}}E_{j},\nabla_{E_{j}}E_{i})-g(\nabla_{[E_{i},E_{j}]}E_{j},E_{i})$
(95) $\displaystyle=$ $\displaystyle
E_{i}.g(\nabla_{E_{j}}E_{j},E_{i})-g(\nabla_{E_{j}}E_{j},\nabla_{E_{i}}E_{i})-(\nabla_{E_{i}}g)(\nabla_{E_{j}}E_{j},E_{i})+g(\nabla_{E_{i}}E_{j},\nabla_{E_{j}}E_{i})$
(97) $\displaystyle-g(\nabla_{[E_{i},E_{j}]}E_{j},E_{i})$ $\displaystyle=$
$\displaystyle-g(\nabla_{E_{j}}E_{j},\nabla_{E_{i}}E_{i})-B(E_{i},E_{i})g(\nabla_{E_{j}}E_{j},N)+g(\nabla_{E_{i}}E_{j},\nabla_{E_{j}}E_{i})-g(\nabla_{[E_{i},E_{j}]}E_{j},E_{i})$
(98) $\displaystyle=$
$\displaystyle-g(\stackrel{{\scriptstyle\ast}}{{\nabla}}_{E_{j}}E_{j},\stackrel{{\scriptstyle\ast}}{{\nabla}}_{E_{i}}E_{i})-\lambda_{i}C(E_{j},E_{j})+g(\nabla_{E_{i}}E_{j},\nabla_{E_{j}}E_{i})-g(\nabla_{[E_{i},E_{j}]}E_{j},E_{i})$
(99) $\displaystyle=$ $\displaystyle-
E_{j}.g(E_{j},\stackrel{{\scriptstyle\ast}}{{\nabla}}_{E_{i}}E_{i})+g(E_{j},\stackrel{{\scriptstyle\ast}}{{\nabla}}_{E_{j}}\stackrel{{\scriptstyle\ast}}{{\nabla}}_{E_{i}}E_{i})-\lambda_{i}g(A_{N}E_{j},E_{j})+g(\nabla_{E_{i}}E_{j},\nabla_{E_{j}}E_{i})$
(101) $\displaystyle-g(\nabla_{[E_{i},E_{j}]}E_{j},E_{i})$ $\displaystyle=$
$\displaystyle
g(\nabla_{E_{i}}E_{j},\nabla_{E_{j}}E_{i})-g(\nabla_{[E_{i},E_{j}]}E_{j},E_{i})-\lambda_{i}g(A_{N}E_{j},E_{j})$
(102)

Otherwise, from (90), we have:

$\displaystyle g(R(E_{i},E_{i})E_{j},E_{i})$ $\displaystyle=$ $\displaystyle
k+\lambda_{j}g(A_{N}E_{i},E_{i})+g(A_{L}E_{j},E_{j})g(A_{L}E_{i},E_{i})-g(A_{L}E_{i},E_{j})g(A_{L}E_{j},E_{i})$
(103)

by (102) and (103), we have:

$\displaystyle
k+\lambda_{j}g(A_{N}E_{i},E_{i})+\lambda_{i}g(A_{N}E_{j},E_{j})+g(A_{L}E_{j},E_{j})g(A_{L}E_{i},E_{i})-g(A_{L}E_{i},E_{j})^{2}=g(\nabla_{E_{i}}E_{j},\nabla_{E_{j}}E_{i})$
$\displaystyle-g(\nabla_{[E_{i},E_{j}]}E_{j},E_{i})$ (104)

Using $(v)$ of lemma3.1, we get:

$\displaystyle
g((\nabla_{[E_{i},E_{j}]}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})E_{i},E_{j})$
$\displaystyle=$
$\displaystyle(\lambda_{i}-\lambda_{j})g(\nabla_{[E_{i},E_{j}]}E_{i},E_{j})$
(105) $\displaystyle=$
$\displaystyle(\lambda_{j}-\lambda_{i})g(\nabla_{[E_{i},E_{j}]}E_{j},E_{i})$
(106)

then

$\displaystyle
g(\nabla_{[E_{i},E_{j}]}E_{j},E_{i})=\frac{g((\nabla_{[E_{i},E_{j}]}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})E_{i},E_{j})}{\lambda_{j}-\lambda_{i}}$
(107)

Also, using $(i)$ and $(iv)$ of lemma3.1, we have:

$\displaystyle
g((\nabla_{[E_{i},E_{j}]}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})E_{i},E_{j})$
$\displaystyle=$ $\displaystyle
g((\nabla_{E_{i}}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})E_{j},[E_{i},E_{j}])$
(108) $\displaystyle=$ $\displaystyle
g((\nabla_{E_{i}}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})E_{j},\nabla_{E_{i}}E_{j})-g((\nabla_{E_{i}}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})E_{j},\nabla_{E_{j}}E_{i})$
(109) $\displaystyle=$ $\displaystyle
g((\nabla_{E_{j}}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})E_{i},\nabla_{E_{i}}E_{j})-g((\nabla_{E_{i}}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})E_{j},\nabla_{E_{j}}E_{i})$
(110) $\displaystyle=$ $\displaystyle
g(\nabla_{E_{j}}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}E_{i}),\nabla_{E_{i}}E_{j})-g(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\nabla_{E_{j}}E_{i},\nabla_{E_{i}}E_{j})-g(\nabla_{E_{i}}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}E_{j}),\nabla_{E_{j}}E_{i})$
(112)
$\displaystyle+g(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\nabla_{E_{i}}E_{j},\nabla_{E_{j}}E_{i})$
$\displaystyle=$
$\displaystyle(\lambda_{i}-\lambda_{j})g(\nabla_{E_{i}}E_{j},\nabla_{E_{j}}E_{i})$
(113)

By (3),(107) and (108), we get:

$\displaystyle
k+\lambda_{j}g(A_{N}E_{i},E_{i})+\lambda_{i}g(A_{N}E_{j},E_{j})+g(A_{L}E_{j},E_{j})g(A_{L}E_{i},E_{i})-g(A_{L}E_{i},E_{j})^{2}=2g(\nabla_{E_{i}}E_{j},\nabla_{E_{j}}E_{i})$
(114)

Since
$\nabla_{E_{i}}E_{j}=\sum_{s=1}^{m-1}g(\nabla_{E_{i}}E_{j},E_{s})E_{s}+\eta(\nabla_{E_{i}}E_{j})\xi$,
(114) became

$\displaystyle
k+\lambda_{j}g(A_{N}E_{i},E_{i})+\lambda_{i}g(A_{N}E_{j},E_{j})+g(A_{L}E_{j},E_{j})g(A_{L}E_{i},E_{i})-g(A_{L}E_{i},E_{j})^{2}=$
$\displaystyle
2\sum_{s=1}^{m-1}g(\nabla_{E_{i}}E_{j},E_{s})g(\nabla_{E_{j}}E_{i},E_{s})$
(115)

$(i)$ et $(v)$ of lemma3.1 give

$\displaystyle
k+\lambda_{j}g(A_{N}E_{i},E_{i})+\lambda_{i}g(A_{N}E_{j},E_{j})+g(A_{L}E_{j},E_{j})g(A_{L}E_{i},E_{i})-g(A_{L}E_{i},E_{j})^{2}$
(116) $\displaystyle=$ $\displaystyle 2\sum_{\begin{subarray}{c}s=1\\\
\lambda_{s}\neq\lambda_{i},\lambda_{j}\end{subarray}}^{m-1}\frac{g((\nabla_{E_{i}}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})E_{j},E_{s})^{2}}{(\lambda_{i}-\lambda_{s})(\lambda_{j}-\lambda_{s})}$
(117)

thus, we have

$\displaystyle\sum_{\begin{subarray}{c}j=1\\\
\lambda_{j}\neq\lambda_{i}\end{subarray}}^{m-1}\frac{k+\lambda_{j}g(A_{N}E_{i},E_{i})+\lambda_{i}g(A_{N}E_{j},E_{j})+g(A_{L}E_{j},E_{j})g(A_{L}E_{i},E_{i})-g(A_{L}E_{i},E_{j})^{2}}{\lambda_{i}-\lambda_{j}}=$
$\displaystyle\sum_{\begin{subarray}{c}j=1\\\
\lambda_{j}\neq\lambda_{i}\end{subarray}}^{m-1}\frac{1}{\lambda_{i}-\lambda_{j}}2\sum_{\begin{subarray}{c}s=1\\\
\lambda_{s}\neq\lambda_{i},\lambda_{j}\end{subarray}}^{m-1}\frac{g((\nabla_{E_{i}}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})E_{j},E_{s})^{2}}{(\lambda_{i}-\lambda_{s})(\lambda_{j}-\lambda_{s})}$
$\displaystyle=\sum_{\begin{subarray}{c}s=1\\\
\lambda_{s}\neq\lambda_{i}\end{subarray}}^{m-1}-\frac{1}{\lambda_{i}-\lambda_{s}}2\sum_{\begin{subarray}{c}j=1\\\
\lambda_{j}\neq\lambda_{i},\lambda_{s}\end{subarray}}^{m-1}\frac{g((\nabla_{E_{i}}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})E_{j},E_{s})^{2}}{(\lambda_{i}-\lambda_{j})(\lambda_{s}-\lambda_{j})}$
$\displaystyle=-\sum_{\begin{subarray}{c}s=1\\\
\lambda_{s}\neq\lambda_{i}\end{subarray}}^{m-1}\frac{k+\lambda_{j}g(A_{N}E_{i},E_{i})+\lambda_{i}g(A_{N}E_{j},E_{j})+g(A_{L}E_{j},E_{j})g(A_{L}E_{i},E_{i})-g(A_{L}E_{i},E_{j})^{2}}{\lambda_{i}-\lambda_{s}}$
(118)

Finally, we have

$\displaystyle\sum_{\begin{subarray}{c}j=1\\\
\lambda_{j}\neq\lambda_{i}\end{subarray}}^{m-1}\frac{k+\lambda_{j}g(A_{N}E_{i},E_{i})+\lambda_{i}g(A_{N}E_{j},E_{j})+g(A_{L}E_{j},E_{j})g(A_{L}E_{i},E_{i})-g(A_{L}E_{i},E_{j})^{2}}{\lambda_{i}-\lambda_{j}}=0$
(119)

##  4 Application

A vector field $X$ on a semi-Riemannian manifold $(\overline{M},\overline{g})$
is said to be conformal vector field if there exist a smooth function $\sigma$
on $\overline{M}$ called potential function such that
$\mathcal{L}_{X}\overline{g}=2\sigma\overline{g}$, , where the symbole
$\mathcal{L}_{X}$ denote the Lie derivative with respect to the vector field
$X$, that is,

$(\mathcal{L}_{X}\overline{g})(Y,Z)=\overline{g}(\overline{\nabla}_{Y}X,Z)+\overline{g}(Z,\overline{\nabla}_{Z}X),$
(120)

for any $X,Y,Z\in\Gamma(T\overline{M})$. In particular, when a potential
function $\sigma$ associated to a conformal vector field $X$ is identically
null, we said that $X$ is a Killing vector field. A distribution $\mathcal{D}$
on $\overline{M}$ is said to be a conformal distribution if each vector field
belonging to $\mathcal{D}$ is a conformal vector field.

If the co-screen distribution $S(TM^{\bot})$ is a conformal distribution,
using (11) and (18) we have

$\displaystyle\overline{g}(\overline{\nabla}_{X}L,Y)=g(A_{L}X,Y)+\phi(X)\eta(Y)=-D(X,Y).$
(121)

Therefore, we obtain

$\displaystyle(\mathcal{L}_{L}\overline{g})(X,Y)=-2D(X,Y).$ (122)

A vector field $X$ on a semi-Riemannian manifold $(\overline{M},\overline{g})$
is said to be a Killing vector field if $\mathcal{L}_{X}\overline{g}=0$.

######  Definition 4.1.

A distribution $D$ on a semi-riemannian manifold $(\overline{M},\overline{g})$
is said to be a Killing distribution if each vector field belonging to $D$ is
a killing vector field.

For a half-lightlike submanifold $(M,S(TM),S(TM^{\bot}))$ of a semi-Riemannian
manifold $(\overline{M},\overline{g})$, it is easy to see that the co-screen
distribution $S(TM^{\bot})$ is a Killing distribution if and only if
$D(X,Y)=0$ for any $X,Y\in\Gamma(TM)$.

The if $S(TM^{\bot})$ is a Killing distribution we have:

$\displaystyle D(X,Y)=0,\leavevmode\nobreak\ \leavevmode\nobreak\
\phi(X)=0,\leavevmode\nobreak\ \leavevmode\nobreak\ A_{L}X=\rho(X)\xi,$ (123)

for all $X,Y\in\Gamma(TM)$.

######  Proposition 4.1.

Let $(M,g,S(TM),S(TM^{\bot}))$ be a half-lightlike submanifold of a semi-
Riemannian manifold $(\overline{M},\overline{g})$. Then, the co-screen
distribution $S(TM^{\bot})$ is a conformal distribution if and only if
$D(X,Y)=-\sigma g(X,Y)$ for any $X,Y\in\Gamma(TM)$.

Let $(M,g,S(TM),S(TM^{\bot}))$ be a half-lightlike submanifold of a semi-
Riemannian manifold $(\overline{M}(k),\overline{g})$ of constant curvature
$k$, with a conformal co-screen distribution. Then, by proposition4.1 and
relations (18)and (20), we have:

$\displaystyle D(X,Y)=-\sigma g(X,Y),\leavevmode\nobreak\
\varphi(X)=0,\leavevmode\nobreak\ A_{L}X=-\sigma
PX+\rho(X)\xi,\leavevmode\nobreak\ \mbox{for all}\leavevmode\nobreak\
X,Y\in\Gamma(TM).$ (124)

A semi-Riemannian manifold $(\overline{M},\overline{g})$ is said to be
equipped with a non-trivial closed conformal vector field $X$, if for a
certain smooth function $\sigma\in\mathcal{C}^{\infty}(\overline{M})$, we have

$\displaystyle\overline{\nabla}_{Y}X=\sigma Y\leavevmode\nobreak\ \mbox{for
any}\leavevmode\nobreak\ \leavevmode\nobreak\ Y\in\Gamma(T\overline{M}).$
(125)

Where $\overline{\nabla}$ is the Levi-Civita connection associated to the
metric $\overline{g}$ on $\overline{M}$.

It is obvious that a closed conformal vector field is a conformal vector
field, but the opposite way is wrong. From (10) and (124), we have the
following

######  Proposition 4.2.

Let $(M,g,S(TM),S(TM^{\bot}))$ be a half-lightlike submanifold of a semi-
Riemannian manifold $(\overline{M},\overline{g})$, with a conformal co-screen
distribution $S(TM^{\bot})=\\{L\\}$. Then, the vector field $L$ is a closed
conformal vector field if and only if
$\rho(X)=-\sigma\eta(X)\leavevmode\nobreak\ \mbox{for any}\leavevmode\nobreak\
X\in\Gamma(TM)$.

######  Theorem 4.1.

Let $(M,g,S(TM),S(TM^{\bot}))$ be a screen homothetic half-lightlike
submanifold of a semi-Riemannian manifold $(\overline{M}(k),\overline{g})$ of
constant curvature $k$, with a conformal co-screen distribution. Then, we
obtain

$\displaystyle 2\varphi\tau(\xi)B=\\{-k-\sigma^{2}\\}g.$ (126)

Proof Let $X,Y,Z\in\Gamma(TM)$. As $\overline{M}$ has a constant curvature
$k$, we have $\overline{g}(\overline{R}(X,Y)Z,\xi)=0$. Then, considering the
relation (28), we get:

$\displaystyle(\nabla_{X}B)(Y,Z)-(\nabla_{Y}B)(X,Z)=B(X,Z)\tau(Y)-B(Y,Z)\tau(X).$
(127)

Also, using relations (28) and (41), we have:

$\displaystyle\overline{g}(R(X,Y)Z,N)=k\\{g(Y,Z)\eta(X)-g(X,Z)\eta(Y)\\}+\sigma\rho(Y)g(X,Z)-\sigma\rho(X)g(Y,Z),$
(128)

and

$\displaystyle\overline{g}(R(X,Y)PZ,N)$ $\displaystyle=$
$\displaystyle(\nabla_{X}C)(Y,PZ)-(\nabla_{Y}C)(X,PZ)+C(X,PZ)\tau(Y)-C(Y,PZ)\tau(X)$
(129) $\displaystyle=$
$\displaystyle\varphi(\nabla_{X}B)(Y,PZ)-\varphi(\nabla_{Y}B)(X,PZ)+\varphi
B(X,Z)\tau(Y)-\varphi B(Y,Z)\tau(X)$ (130)
$\displaystyle\overset{\ref{proofprop1confeq1}}{=}$ $\displaystyle
2\varphi\\{B(X,PZ)\tau(Y)-B(Y,PZ)\tau(X)\\}$ (131) $\displaystyle=$
$\displaystyle 2\varphi\\{B(X,Z)\tau(Y)-B(Y,Z)\tau(X)\\}$ (132)

Since $\overline{g}(R(X,Y)\xi,N)=0$, then
$\overline{g}(R(X,Y)Z,N)=\overline{g}(R(X,Y)PZ,N)$. Therefore, from relations
(128) and (132), we have:

$\displaystyle
2\varphi\\{B(X,Z)\tau(Y)-B(Y,Z)\tau(X)\\}=k\\{g(Y,Z)\eta(X)-g(X,Z)\eta(Y)\\}+\sigma\rho(Y)g(X,Z)$
$\displaystyle-\sigma\rho(X)g(Y,Z).$ (133)

Replacing $Y$ by $\xi$ in the last relation, we obtain this theorem.

######  Remark 4.1.

For a screen homothetic half-lightlike submanifold $(M,g,S(TM),S(TM^{\bot}))$
of a semi-Riemannian manifold $(\overline{M}(k),\overline{g})$, with a
conformal co-screen distribution, the theorem4.1 allow us to show that, if
$\tau=0$ then $k=-\sigma^{2}$, and if $\tau(\xi)\neq 0$ then $M$ is a totally
umbilical submanifold of $\overline{M}(k)$.

If the ambient manifold $\overline{M}$ is a semi-Riemannian manifold
$(\overline{M}(k),\overline{g})$ of constante curvature $k$, then
$\overline{R}(\xi,Y)X=k\overline{g}(X,Y)\xi$;
$\overline{R}(L,X)Y=k\overline{g}(X,Y)L$ and
$\overline{R}ic(X,Y)=(m+1)k\overline{g}(X,Y)$. Thus, we obtain

$\displaystyle
Ric(X,Y)=(m-1)kg(X,Y)+B(X,Y)tr(A_{N})+D(X,Y)tr(A_{L})-g(A_{N}X,\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)$
$\displaystyle-g(A_{L}X,A_{L}Y)+\rho(X)\varphi(Y)$ (134)

A screen conformal half-lightlike submanifold of a semi-Riemannian manifold
$(\overline{M}(k),\overline{g})$ of constante curvature $k$, with a conformal
co-screen distribution, have a symmetric Ricci tensor $Ric$, by (4).

######  Theorem 4.2.

Let $(M,g,S(TM),S(TM^{\bot}))$ be a screen conformal half-lightlike
submanifold of a semi-Riemannian manifold $(\overline{M}(k),\overline{g})$ of
constant curvature $k$, with a conformal co-screen distribution. Then, $M$
admits an induced symmetric tensor.

Now, we prove the following theorem,

######  Theorem 4.3.

Let $(M,g,S(TM)),S(TM^{\bot}))$ be a screen homothetic half-lightlike
submanifold of an $(m+2)$-dimensional Lorenzian manifold
$(\overline{M}(k),\overline{g})$ of non-positive constant curvature $k$, with
a conformal co-screen distribution $S(TM^{\bot})$, whose screen principal
curvatures along the screen distribution $S(TM)$ are constant and at most two
are distinct. If $M$ has two distinct screen principal curvatures, then one of
then must be zero.

Proof
If $M$ has two distinct screen principal curvatures
$\alpha\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ \beta$, then it
exists $p\in\\{1,\cdots,m-1\\}$ so that

$\displaystyle\lambda_{1}=\lambda_{2}=\cdots=\lambda_{p}=\alpha\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\
\lambda_{p+1}=\lambda_{p+2}=\cdots=\lambda_{m-1}=\beta$ (135)

Since $M$ is screen homothetic with a conformal co-screen distribution, then
using relations (54) and (124), the relation (114) became

$\displaystyle
k+\sigma^{2}+2\varphi\alpha\beta=2g(\nabla_{E_{\alpha}}E_{\beta},\nabla_{E_{\beta}}E_{\alpha}).$
(136)

Then, according to the remark4.1, we get:

$\displaystyle\varphi\alpha\beta=g(\nabla_{E_{\alpha}}E_{\beta},\nabla_{E_{\beta}}E_{\alpha})$
(137)

From (2) of lemme3.2,
$g(\nabla_{E_{\alpha}}E_{\beta},\nabla_{E_{\beta}}E_{\alpha})=0$. therefore,
previous relation give us $\varphi\alpha\beta=0$. Since $\varphi\neq
0\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ \alpha\neq\beta$, then
$\alpha=0\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ \beta\neq 0$ or
$\alpha\neq 0\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ \beta=0$.

######  Example 4.1.

Let $M$ be an immersed submanifold in $\mathbb{R}_{1}^{5}$ defined by:

$\displaystyle M$ $\displaystyle\rightarrow$ $\displaystyle\mathbb{R}_{1}^{5}$
$\displaystyle(v_{1},v_{2},v_{3})$ $\displaystyle\mapsto$
$\displaystyle(x_{1},x_{2},x_{3},x_{4},x_{5})$ (138)

where $\begin{cases}x_{1}=v_{1}\\\ x_{2}=v_{2}\\\ x_{3}=v_{3}\\\
x_{4}=\sqrt{v_{1}^{2}-v_{2}^{2}}\\\ x_{5}=\sqrt{1+v_{3}^{2}}\end{cases}$ with
$v_{1}>v_{2}>0$

the tangent bundle $TM$ is $TM=span\\{V_{1},V_{2},V_{3}\\}$, where

$\displaystyle V_{1}=\frac{1}{x_{2}}\left(x_{4}\frac{\partial}{\partial
x_{1}}+x_{1}\frac{\partial}{\partial x_{4}}\right)$ (139) $\displaystyle
V_{2}=\frac{1}{x_{1}}\left(x_{4}\frac{\partial}{\partial
x_{2}}-x_{2}\frac{\partial}{\partial x_{4}}\right)$ (140) $\displaystyle
V_{3}=\frac{1}{\sqrt{1+2x_{3}^{2}}}\left(x_{5}\frac{\partial}{\partial
x_{3}}+x_{3}\frac{\partial}{\partial x_{5}}\right)$ (141)

and the normal bundle is $TM^{\bot}=span\\{H_{1},H_{2}\\}$, where

$\displaystyle H_{1}=x_{1}\frac{\partial}{\partial
x_{1}}+x_{2}\frac{\partial}{\partial x_{2}}+x_{4}\frac{\partial}{\partial
x_{4}}$ (142) $\displaystyle
H_{2}=\frac{1}{\sqrt{1+2x_{3}^{2}}}\left(x_{3}\frac{\partial}{\partial
x_{3}}+x_{5}\frac{\partial}{\partial x_{5}}\right)$ (143)
$\displaystyle\overline{g}(H_{1},H_{1})=0\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ \leavevmode\nobreak\
\overline{g}(H_{2},H_{2})=1,$ (144)

then $H_{1}\in Rad(TM)$and $H_{2}\in S(TM^{\bot})$. Thus $Rad(TM)$ and
$S(TM^{\bot})$ are of rank 1. Remark that
$H_{1}=\frac{x_{1}x_{2}}{x_{4}}(V_{1}+V_{2})$.

Take $Rad(TM)=span\\{\xi\\}$ where
$\xi=\frac{1}{x_{1}}H_{1}=\frac{1}{x_{1}}\\{x_{1}\frac{\partial}{\partial
x_{1}}+x_{2}\frac{\partial}{\partial x_{2}}+x_{4}\frac{\partial}{\partial
x_{4}}\\}$, and $S(TM^{\bot})=span\\{L=H_{2}\\}$

the null transversal vector field $N$ is:

$\displaystyle N=-\frac{1}{2x_{1}}\left\\{x_{1}\frac{\partial}{\partial
x_{1}}-x_{2}\frac{\partial}{\partial x_{2}}-x_{4}\frac{\partial}{\partial
x_{4}}\right\\},$ (145)

The null transversal bundle $ltr(TM)$ and the screen distribution $S(TM)$ are:

$\displaystyle ltr(TM)=span\\{N\\}\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ S(TM)=span\\{V_{2},V_{3}\\}.$ (146)

By direct computation, we get

$\displaystyle\overline{\nabla}_{V_{2}}\xi=\frac{1}{x_{1}}V_{2},\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\nabla}_{V_{3}}\xi=0,\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\nabla}_{\xi}\xi=0$ (147)

Thus,

$\displaystyle\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{2}=-\frac{1}{x_{1}}V_{2},\leavevmode\nobreak\
\leavevmode\nobreak\
\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{3}=0,\leavevmode\nobreak\
\leavevmode\nobreak\
\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\xi=0,\leavevmode\nobreak\
\leavevmode\nobreak\ \tau=0.$ (148)

Then, $M$ is irrotational and has two distinct screen principal curvatures
$\lambda=-\frac{1}{x_{1}}\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\
\mu=0$. On the other hand, it is easy to see that:

$\displaystyle\overline{\nabla}_{V_{2}}N=\frac{1}{2x_{1}}V_{2},\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\nabla}_{V_{3}}N=0,\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\nabla}_{\xi}N=0$ (149)

then,

$\displaystyle A_{N}V_{2}=-\frac{1}{2x_{1}}V_{2},\leavevmode\nobreak\
\leavevmode\nobreak\ A_{N}V_{3}=0,\leavevmode\nobreak\ \leavevmode\nobreak\
A_{N}\xi=0,\leavevmode\nobreak\ \leavevmode\nobreak\ \rho=0.$ (150)

it follow

$\displaystyle
A_{N}V_{2}=\frac{1}{2}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{2},\leavevmode\nobreak\
\leavevmode\nobreak\
A_{N}V_{3}=\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{3}=0,\leavevmode\nobreak\
\leavevmode\nobreak\ A_{N}\xi=\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\xi=0,$
(151)

which proves that $M$ is irrotational screen homothetic half-lightlike
submanifold, with conformal factor $\frac{1}{2}$ and the 1-forme $\tau=0$.

######  Example 4.2.

Let $M$ be a submanifold in $\mathbb{R}_{1}^{6}$ given by:

$x_{3}=x_{5},\;\;\;x_{6}=\sqrt{x_{1}^{2}-x_{2}^{2}}\;\;\;\mbox{with}\leavevmode\nobreak\
\leavevmode\nobreak\ x_{1}>x_{2}>0.$ (152)

Then, we have:

$\displaystyle Rad(TM)$ $\displaystyle=$ $\displaystyle
span\left\\{\xi=\frac{1}{x_{1}}\left(x_{1}\frac{\partial}{\partial
x_{1}}+x_{2}\frac{\partial}{\partial x_{2}}+x_{6}\frac{\partial}{\partial
x_{6}}\right)\right\\}$ (153) $\displaystyle S(TM^{\bot})$ $\displaystyle=$
$\displaystyle
span\left\\{L=-\frac{1}{\sqrt{2}}\left(-\frac{\partial}{\partial
x_{3}}+\frac{\partial}{\partial x_{5}}\right)\right\\}$ (154) $\displaystyle
ltr(TM)$ $\displaystyle=$ $\displaystyle
span\left\\{-\frac{1}{2x_{1}}\left(x_{1}\frac{\partial}{\partial
x_{1}}-x_{2}\frac{\partial}{\partial x_{2}}-x_{6}\frac{\partial}{\partial
x_{6}}\right)\right\\}$ (155) $\displaystyle S(TM)$ $\displaystyle=$
$\displaystyle span\left\\{V_{1},V_{2},V_{3}\right\\}$ (156)

where

$\displaystyle V_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{x_{1}}\left(x_{6}\frac{\partial}{\partial
x_{2}}-x_{2}\frac{\partial}{\partial x_{6}}\right)$ (157) $\displaystyle
V_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\frac{\partial}{\partial
x_{3}}+\frac{\partial}{\partial x_{5}}\right)$ (158) $\displaystyle V_{3}$
$\displaystyle=$ $\displaystyle\frac{\partial}{\partial x_{4}}$ (159)

By straightforward calculation, we get

$\displaystyle\overline{\nabla}_{V_{1}}\xi=\frac{1}{x_{1}}V_{1},\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\nabla}_{V_{2}}\xi=0;\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\nabla}_{V_{3}}\xi=0;\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\nabla}_{\xi}\xi=0$ (160)

Thus, $M$ is irrotational and

$\displaystyle\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{1}=-\frac{1}{x_{1}}V_{1},\leavevmode\nobreak\
\leavevmode\nobreak\
\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{2}=\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{3}=\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\xi=0,\leavevmode\nobreak\
\leavevmode\nobreak\ \tau=0.$ (161)

Then, $M$ has two distinct screen principal curvatures
$\lambda=-\frac{1}{x_{1}}\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\
\mu=0$. On the other hand, we have:

$\displaystyle\overline{\nabla}_{V_{1}}N=\frac{1}{2x_{1}}V_{1},\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\nabla}_{V_{2}}N=0,\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\nabla}_{V_{3}}N=0,\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\nabla}_{\xi}N=0$ (162)

it follow that

$\displaystyle A_{N}V_{1}=-\frac{1}{2x_{1}}V_{1},\leavevmode\nobreak\
\leavevmode\nobreak\ A_{N}V_{2}=A_{N}V_{3}=A_{N}\xi=0,\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ \rho=0.$ (163)

By (161) and (163) $M$ is a screen homothetic half-lightlike submanifold, with
conformal factor $\frac{1}{2}$.

Otherwise, we have:

$\displaystyle\overline{\nabla}_{\xi}L=\overline{\nabla}_{V_{2}}L=\overline{\nabla}_{V_{3}}L=\overline{\nabla}_{V_{4}}L=0,$
(164)

and then,

$\displaystyle A_{L}\equiv 0\leavevmode\nobreak\ \mbox{on}\leavevmode\nobreak\
TM.$ (165)

Consequently,

$\displaystyle D(X,Y)=g(A_{L}X,Y)=0\leavevmode\nobreak\ \mbox{for
any}\leavevmode\nobreak\ X,Y\in\Gamma(TM),$ (166)

which show that the co-screen distribution $S(TM)=span\\{L\\}$ is a Killing
distribution.

######  Example 4.3.

Let $M$ be an immersed submanifold in $\mathbb{R}_{1}^{6}$ defined by:

$\displaystyle M$ $\displaystyle\rightarrow$ $\displaystyle\mathbb{R}_{1}^{6}$
$\displaystyle(v_{1},v_{2},v_{3})$ $\displaystyle\mapsto$
$\displaystyle(x_{1},x_{2},x_{3},x_{4},x_{5},x_{6})$ (167)

where $\begin{cases}x_{1}=v_{1}\\\ x_{2}=v_{2}\\\
x_{3}=\frac{v_{1}}{\sqrt{2}}\sin{v_{3}}\\\
x_{4}=\frac{v_{1}}{\sqrt{2}}\cos{v_{3}}\\\
x_{5}=\frac{v_{1}}{\sqrt{2}}\sin{v_{4}}\\\
x_{6}=\frac{v_{1}}{\sqrt{2}}\cos{v_{4}}\\\ \end{cases}$ with $v_{1}\neq
0\leavevmode\nobreak\ \leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\
\leavevmode\nobreak\
v_{3},v_{4}\in\mathbb{R}\backslash\\{\frac{\pi}{2}+k\pi\\}$

the tangent bundle $TM$ is $TM=span\\{V_{1},V_{2},V_{3},V_{4}\\}$, where

$\displaystyle V_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{x_{1}}\left\\{x_{1}\frac{\partial}{\partial
x_{1}}+x_{3}\frac{\partial}{\partial x_{3}}+x_{4}\frac{\partial}{\partial
x_{4}}+x_{5}\frac{\partial}{\partial x_{5}}+x_{6}\frac{\partial}{\partial
x_{6}}\right\\}$ (168) $\displaystyle V_{2}$ $\displaystyle=$
$\displaystyle\frac{\partial}{\partial x_{2}}$ (169) $\displaystyle V_{3}$
$\displaystyle=$
$\displaystyle\frac{\sqrt{2}}{x_{1}}\left\\{x_{4}\frac{\partial}{\partial
x_{3}}-x_{3}\frac{\partial}{\partial x_{4}}\right\\}$ (171) $\displaystyle
V_{4}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{2}}{x_{1}}\left\\{x_{6}\frac{\partial}{\partial
x_{5}}-x_{5}\frac{\partial}{\partial x_{6}}\right\\}$ (173)

and the normal bundle is $TM^{\bot}=span\\{H_{1},H_{2}\\}$, where

$\displaystyle H_{1}=\frac{1}{2x_{4}}\left\\{x_{1}\frac{\partial}{\partial
x_{1}}+2x_{3}\frac{\partial}{\partial x_{3}}+2x_{4}\frac{\partial}{\partial
x_{4}}\right\\}$ (174) $\displaystyle
H_{2}=\frac{1}{2x_{6}}\left\\{x_{1}\frac{\partial}{\partial
x_{1}}+2x_{5}\frac{\partial}{\partial x_{5}}+2x_{6}\frac{\partial}{\partial
x_{6}}\right\\}$ (175)

Remark that $\frac{x_{4}}{x_{1}}H_{1}+\frac{x_{6}}{x_{1}}H_{2}=V_{1}$.

Take $Rad(TM)=span\\{\xi=V_{1}\\}$, $S(TM)=span\\{V_{2},V_{3},V_{4}\\}$ and
$S(TM^{\bot})=span\\{L=H_{1}\\}$. the null transversal vector field N is:

$\displaystyle N=-\frac{1}{x_{1}}\left\\{x_{1}\frac{\partial}{\partial
x_{1}}+x_{3}\frac{\partial}{\partial x_{3}}+x_{4}\frac{\partial}{\partial
x_{4}}-x_{5}\frac{\partial}{\partial x_{5}}-x_{6}\frac{\partial}{\partial
x_{6}}\right\\}$ (176)

The null transversal bundle $ltr(TM)$ is ltr(TM)=span{N}.

By direct computation we get:

$\overline{\nabla}_{\xi}\xi=0,\;\;\;\overline{\nabla}_{V_{2}}\xi=0,\;\;\;\overline{\nabla}_{V_{3}}\xi=\frac{1}{x_{1}}V_{3},\;\;\;\overline{\nabla}_{V_{4}}\xi=\frac{1}{x_{1}}V_{4}$
(177)

Thus, we deduce that $M$ is irrotational and:

$\displaystyle\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\xi=0,\leavevmode\nobreak\
\leavevmode\nobreak\
\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{2}=0,\leavevmode\nobreak\
\leavevmode\nobreak\
\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{3}=-\frac{1}{x_{1}}V_{3},\leavevmode\nobreak\
\leavevmode\nobreak\
\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{4}=-\frac{1}{x_{1}}V_{4}\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ \leavevmode\nobreak\
\tau=0.$ (178)

Then, $M$ has two distinct screen principal curvatures
$\lambda=-\frac{1}{x_{1}}\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\
\mu=0$. On the other hand, we have:

$\overline{\nabla}_{\xi}N=0,\;\;\;\overline{\nabla}_{V_{2}}N=0,\;\;\;\overline{\nabla}_{V_{3}}N=-\frac{1}{x_{1}}V_{3},\;\;\;\overline{\nabla}_{V_{4}}N=\frac{1}{x_{1}}V_{4}$
(179)

Then,

$\displaystyle A_{N}\xi=0,\leavevmode\nobreak\ \leavevmode\nobreak\
A_{N}V_{2}=0,\leavevmode\nobreak\ \leavevmode\nobreak\
A_{N}V_{3}=\frac{1}{x_{1}}V_{3},\leavevmode\nobreak\ \leavevmode\nobreak\
A_{N}V_{4}=-\frac{1}{x_{1}}V_{4}\leavevmode\nobreak\ \leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ \leavevmode\nobreak\ \rho=0.$ (180)

It follow that,

$\displaystyle
A_{N}\xi=\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\xi=0,\leavevmode\nobreak\
\leavevmode\nobreak\
A_{N}V_{2}=\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{2}=0,\leavevmode\nobreak\
\leavevmode\nobreak\
A_{N}V_{3}=-\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{3},\leavevmode\nobreak\
\leavevmode\nobreak\
A_{N}V_{4}=\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}V_{4}.$ (181)

So, $M$ is not a screen homothetic half-lightlike submanifold.

Otherwise, we have:

$\displaystyle
D(V_{2},V_{2})=\overline{g}(\overline{\nabla}_{V_{2}}V_{2},L)=0,$ (182)
$\displaystyle
D(V_{2},V_{3})=\overline{g}(\overline{\nabla}_{V_{2}}V_{3},L)=0,$ (183)
$\displaystyle
D(V_{2},V_{4})=\overline{g}(\overline{\nabla}_{V_{2}}V_{4},L)=0,$ (184)
$\displaystyle
D(V_{3},V_{3})=\overline{g}(\overline{\nabla}_{V_{3}}V_{3},L)=-\frac{\sqrt{2}}{x_{1}x_{4}}g(V_{3},V_{3}),$
(185) $\displaystyle
D(V_{3},V_{4})=\overline{g}(\overline{\nabla}_{V_{3}}V_{4},L)=0,$ (186)
$\displaystyle
D(V_{4},V_{4})=\overline{g}(\overline{\nabla}_{V_{4}}V_{4},L)=-\frac{\sqrt{2}}{x_{1}}g(V,L)=0.$
(187)

Which show that the co-screen $S(TM^{\bot})=span\\{L=H_{1}\\}$ is not
conformal.

This shows that a screen conformal and co-screen conformal distribution are
just a sufficient condition for a half-lightlike submanifold with two distinct
screen principal curvatures then one must be zero.

In the sequel, we consider a screen homothetic half-lightlike submanifold $M$
of an $(m+2)$-dimensional Lorenzian manifold $(\overline{M}(k),\overline{g})$
of non-positive constant curvature $k$, with a conformal co-screen
distribution $S(TM^{\bot})$, whose screen principal curvatures are constant
along $S(TM)$. We assume that $M$ has exactly two distinct screen principal
curvatures. Then, by theorem4.3, one of them must be zero. We denote by
$\lambda$ the non-zero screen principal curvature and $r$ the multiplicity of
$\lambda$. The sets,

$\displaystyle T_{\lambda}=\\{X\in\Gamma(S(TM)):\leavevmode\nobreak\
\leavevmode\nobreak\ \stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X=\lambda X\\}$
(188) $\displaystyle T_{0}=\\{X\in\Gamma(S(TM)):\leavevmode\nobreak\
\leavevmode\nobreak\ \stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X=0\\}$ (189)

define the distributions of dimensions $r$ and dimension $m-r$, respectively.

From ([7]) and the remark3.1, it is obvious that if $M$ is a screen conformal
half-lightlike submanifold of a Lorentzian manifold
$(\overline{M}(k),\overline{g})$, with a conformal co-screen distribution,
then the screen distribution $S(TM)$ is Riemannian and integrable. By the
theorem4.2, we have that the induced Ricci tensor on $M$ is symmetric.
Moreover, a screen conformal half-lightlike submanifold is locally a product
$C\times M^{\prime}$ where $C$ is a null curve, $M^{\prime}$ is an integral
manifold of $S(TM)$ ([8]). We have the following local decomposition.

######  Theorem 4.4.

Let $(M,g,S(TM),S(TM^{\bot}))$ be a screen homothetic half-lightlike
submanifold of an $(m+2)$-dimensional Lorenzian manifold
$(\overline{M}(k),\overline{g})$ of non-positive constant curvature $k$, with
a conformal co-screen distribution $S(TM^{\bot})$. If the principal curvatures
of $M$ are constant along the screen distribution $S(TM)$ and exactly two of
them are distinct, then $M$ is locally a lightlike triple product manifold
$M=C\times(M^{\prime}=M_{\lambda}\times M_{0})$, where C is a null curve,
$M^{\prime}$ is an integral manifold of $S(TM)$,
$M_{\lambda}\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ M_{0}$ are
leaves of some distributions of $M$ such that $M_{\lambda}$ is an
$r$-dimensional totally geodesic Riemannian manifold of curvature
$2\varphi\lambda^{2}$ and $M_{0}$ is an $(m-r)$-dimensional totally geodesic
Euclidien espace.

Proof
$S(TM)$ being Riemannian and integrable, it is know that its leaf $M^{\prime}$
is Riemannian. Since $M$ has exactly two distinct screen principal curvatures,
then one of them must be zero (theorem4.3). Then, with previous notations, we
have

$\displaystyle S(TM)=T_{\lambda}\underset{\bot}{\oplus}T_{0}.$ (190)

Now, we prove that $T_{\lambda}\leavevmode\nobreak\
\mbox{et}\leavevmode\nobreak\ T_{0}$ are integrables. Let
$X,Y\in\Gamma(T_{\lambda})$, we have:

$\displaystyle\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}[X,Y]$ $\displaystyle=$
$\displaystyle\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\nabla_{X}Y-\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}\nabla_{Y}X$
(191) $\displaystyle=$
$\displaystyle\nabla_{X}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)-(\nabla_{X}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})Y-\nabla_{Y}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X)+(\nabla_{Y}\stackrel{{\scriptstyle\ast}}{{A}}_{\xi})X$
(192) $\displaystyle=$
$\displaystyle\nabla_{X}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}Y)-\nabla_{Y}(\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}X)$
(193) $\displaystyle=$ $\displaystyle\lambda\nabla_{X}Y-\lambda\nabla_{Y}X$
(194) $\displaystyle=$ $\displaystyle\lambda[X,Y],$ (195)

that is $T_{\lambda}$ is involutive, then integrable. Similarly $T_{0}$ is
integrable.
Using the point (2) of the lemma3.2, we get that
$T_{\lambda}\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ T_{0}$ are
parallel along their normals in $S(TM)$.
By the decomposition theorem of de Rham([3]), we have
$M^{\prime}=M_{\lambda}\times M_{0}$; where $M_{\lambda}\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ M_{0}$ are some leaves of
$T_{\lambda}\leavevmode\nobreak\ \leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ \leavevmode\nobreak\ T_{0}$ respectively. Thus
$M$ is locally a product $C\times M^{\prime}=C\times M_{\lambda}\times M_{0}$.
Remark that $M_{\lambda}$ is totally geodesique
$\Leftrightarrow\leavevmode\nobreak\
\stackrel{{\scriptstyle\ast}}{{\nabla}}_{X}Y\in\Gamma(T_{\lambda})\leavevmode\nobreak\
\mbox{for all}\leavevmode\nobreak\ X,Y\in\Gamma(T_{\lambda})$. Then, by (1) of
the lemma3.2, $M_{\lambda}$ is totally geodesique. Similarly, we prove that
$M_{0}$ is totally geodesique in $S(TM)$.

Consider the frame field of eigenvectors $\\{E_{1},E_{2},\cdots,E_{r}\\}$ of
$\stackrel{{\scriptstyle\ast}}{{A}}_{\xi}$ such that
$\\{E_{i}\\}_{i=1,\cdots,r}$ is an orthonormal frame field of $T_{\lambda}$,
then using (41) et (90) we have

$\displaystyle
g(R(E_{i},E_{j})E_{j},E_{i})=\varphi\lambda^{2}=g(\stackrel{{\scriptstyle\ast}}{{R}}(E_{i},E_{j})E_{j},E_{i})-\varphi\lambda^{2}$
(196)

then

$\displaystyle
g(\stackrel{{\scriptstyle\ast}}{{R}}(E_{i},E_{j})E_{j},E_{i})=2\varphi\lambda^{2}$
(197)

Thus the sectional curvature $K^{\lambda}$ of the leaf $M_{\lambda}$ of
$T_{\lambda}$ is given by

$\displaystyle
K^{\lambda}(E_{i},E_{j})=\frac{g(\stackrel{{\scriptstyle\ast}}{{R}}(E_{i},E_{j})E_{j},E_{i})}{g(E_{i},E_{i})g(E_{j},E_{j})-g(E_{i},E_{j})^{2}}=2\varphi\lambda^{2}.$
(198)

By the same way, we can see that the sectional curvature of the leaf $M_{0}$
of $T_{0}$ is

$\displaystyle K^{0}=0.$ (199)

This completes the proof.

Next, we say that a half-lightlike submanifold $M$ is $B$ (resp.
$D$)-umbilical if on each coordinate neighborhood $\mathcal{U}\subset M$,
there exists smooth functions $H_{1}$ (resp. $H_{2}$) such that
$B(X,Y)=H_{1}g(X,Y)$ (resp. $D(X,Y)=H_{2}g(X,Y)$) for any
$X,Y\in\Gamma(TM|_{\mathcal{U}})$. In particular, if $H_{1}=0$ (resp.
$H_{2}=0$), then $M$ is said to be totally $B$ (resp. $D$)-geodesic.

Note that, $M$ is totally umbilical ( resp. totally geodesic ) if and only if
$M$ are both $B$ and $D$-totally umbilical (resp. totally geodesic ).

It is easy to see that, for a half-lightlike submanifold $M$ with a conformal
co-screen distribution :

  1. 1.

If the screen principal curvatures are all identical and non-zero then $M$ is
totally umbilical,

  2. 2.

If the screen principal curvatures are all identical all zero then $M$ is
totally geodesic.

.

######  Remark 4.2.

We assume that $M$ has at most two distinct screen principal curvatures, thus,
if the screen principal curvatures are all identical, $M$ is either totally
geodesic or totally umbilical and if the two screen principal curvatures are
distinct, then $M=C\times M_{\lambda}\times M_{\mu}$

Thus we have the following classification theorem.

######  Theorem 4.5.

Let $(M,g,S(TM),S(TM^{\bot}))$ be a screen homothetic half-lightlike
submanifold of an $(m+2)$-dimensional Lorenzian manifold
$(\overline{M}(k),\overline{g})$ of non-positive constant curvature $k$, with
a conformal co-screen distribution $S(TM^{\bot})$. If the screen principal
curvatures of $M$ are all constant along the screen distribution $S(TM)$ such
that at most two of them are distinct. Then we have one of the following:

  1. 1.

$M$ is either totally geodesic or totally umbilical;

  2. 2.

$M$ is locally a lightlike triple product manifold $C\times M_{\lambda}\times
M_{0}$, where $C$ is a null curve, $M_{\lambda}\leavevmode\nobreak\
\mbox{and}\leavevmode\nobreak\ M_{0}$ are leaves of some distributions of $M$
such that $M_{\lambda}$ is a totally geodesic Riemannian manifold of curvature
$2\varphi\lambda^{2}$ and $M_{0}$ is an $(m-r)$-dimensional totally geodesic
Euclidien espace.

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