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\markboth{E. Garc\'{\i}a--R\'{\i}o, $\,$ D. N. Kupeli $\,\mbox{\rm and }\,$
M.E. V\'azquez--Abal}
{Osserman Lorentzian manifolds}
\begin{document}
\vspace*{1,5cm}
\begin{flushleft}
{\huge\bf On a Problem of Osserman in}\\
\bigskip
{\huge\bf Lorentzian Geometry}
\end{flushleft}
\vspace{0,8cm}
\begin{flushleft}
{\sc E. Garc\'{\i}a--R\'{\i}o\footnote{\rm Partially supported by the
project XUGA 20701B93 (Spain)}}\\
{\it Departamento de An\'alise Matem\'atica, Universidade de Santiago de
Compostela, 15706 Santiago, Spain}\\
{\sc D.N. Kupeli\footnote{\rm Supported by a fellowship from
TUBITAK-BAYG (Turkey)}}\\
{\it Department of Mathematics, Middle East Technical University, 06531
Ankara, Turkey}\\
{\sc M.E. V\'azquez--Abal$^1$}\\
{\it Departamento de Xeometr\'{\i}a e Topolox\'{\i}a, Universidade de Santiago
de Compostela, 15706 Santiago, Spain}\\
\end{flushleft}
\vspace{0,5cm}
\noindent{\normalsize\it Abstract:}
A problem of Osserman on the constancy of the eigenvalues of the
Jacobi operator is studied in Lorentzian geometry. Attention is paid to the
different cases of timelike, spacelike and null Osserman condition. One also
shows a relation between the null Osserman condition and a previous one on
infinitesimal null isotropy.
\vspace{0,3cm}
\noindent{\normalsize\it Keywords:} Osserman
conjecture, Lorentz manifold, Jacobi operator, infinitesimal isotropy, warped
product.
\noindent{\normalsize\it M.S. classification:} 53B30, 53C50,
\vspace*{0,5cm}
\section{Introduction}
A Riemannian manifold $M$ is called an {\it Osserman space} if the
eigenvalues of
the Jacobi operator are constant (counting multiplicities) on the unit sphere
bundle $SM$. We recall the socalled Osserman conjecture~\cite{Osserman}:
``The only nonflat Osserman manifolds are the locally symmetric
spaces of rank one". Although the general problem is still open, some
significant
results have been obtained by Q.S. Chi~\cite{chi1},~\cite{chi2},~\cite{chi3}
who gave a positive anwser in many cases, in particular for
dim$M\neq 4k$, $k>1$.
A first step toward the solution of the general problem should be to determine
at a single point those curvature tensors satisfying the Osserman condition. In
this direction, the socalled {\it pointwise Osserman} problem is introduced.
A Riemannian manifold is called a pointwise Osserman space if it satisfies the
Osserman condition at each point (with eigenvalues possibly changing from point
to point). We refer to recent work of P. Gilkey, A. Swann and L.
Vanhecke,~\cite{gilkey}, for more details and further information concerning the
relation between the two notions. Also, the existence of Riemannian manifolds
being Osserman at a single point, but with Jacobi operators not corresponding to
those of the rank one symmetric spaces is pointed out by P.
Gilkey,~\cite{gilkeya}. (See also~\cite{kowalski} for a more general study
on the
determination of the metric by means of the curvature tensor and its covariant
derivatives at a point).
\medskip
In this paper we will consider the problems above in the framework of
Lorentzian geometry. Some specific features arise in the case of indefinite
metrics. First of all, the existence of timelike, spacelike and null
directions suggests to consider separately the three (timelike, spacelike and
null) Osserman problems determining the structure of the manifold from the
knowledge of the Jacobi operator. However, a deeper observation points out
significant differences among the three cases.
For an arbitrary timelike vector $x\in T_pM$, the normal space $x^\perp$ has
an induced positive definite metric, and hence, the Jacobi operator is
diagonalizable. In this case ($\S 2.1$) we will provide a complete anwser to
the timelike Osserman problem by showing that $M$ is a timelike Osserman space
at a single point $p\in M$ if and only if it is of constant curvature at $p$. We
note that, as a consequence, the pointwise and the global Osserman problem
are equivalent for timelike geodesics.
The study of the spacelike case presents more difficulties coming from the
fact that $x^\perp$ has an induced Lorentzian metric for each spacelike $x\in
T_pM$ and thus, the Jacobi operator is not necessarily diagonalizable.
Therefore, the Osserman problem will be stated in terms of the constancy of
the characteristic polynomial of the Jacobi operator. From that we conclude
in $\S 2.2$ that $M$ is Einstein at the point $p$ and a positive anwser, as
in the timelike case, is obtained for dim$M=3,4$. However, the general case
(dim$M>4$) still remains open.
To state the Osserman problem for null vectors we recall in $\S 2.3$ the
definition of the nondegenerate normal space $\overline{u}^\perp$ corresponding
to each null $u\in T_pM$. Since the induced metric on $\overline{u}^\perp$ is
positive definite, the induced Jacobi operator $\overline{R}_u$ is
diagonalizable. However, the main difficulty arises in choosing a null
representative of each null direction. Since it is not possible to talk about
``unit null vectors", we consider the socalled null congruence induced by a
timelike vector (see~\cite{beem}, ~\cite{sachs}).
As a special case we consider the null
infinitesimally isotropic Lorentzian manifolds, previously investigated by S.
Harris,~\cite{harris}, and L. Koch--Sen,~\cite{koch-sen}. For these manifolds
they obtained a decomposition theorem with a real space form as Riemannian
factor by using a result of H. Karcher,~\cite{karcher}.
We will show that any even--dimensional null Osserman
Lorentzian manifold is null infinitesimally isotropic. This result shows that
null infinitesimal isotropy is a natural condition to impose on a
Cosmological Model in General Relativity. Finally, in the last section ($\S
3$) we obtain some decompositions with a complex or a
quaternionic space form as Riemannian factor and relate them with the null
Osserman condition.
\bigskip
The authors wish to thank prof. L. Vanhecke for some useful comments on the
subject. The second author (DNK) is grateful to the Departamento de
Xeometr\'{\i}a e Topolox\'{\i}a of the Universidade de Santiago for their kind
invitation and support during his visit.
\section{Pointwise relations}
In this section we will study the significance of the Osserman condition at a
single point. The existence of timelike, spacelike and null vectors not only
motivates a separate study of the constancy of the characteristic polynomial of
the Jacobi operator in each particular case, but it also causes some
significant differences among them.
Let $(M,g=<,>)$ denote a Lorentzian manifold of dimension $n\geq 3$ with
metric $g=<,>$ of signature $(-,+,\cdots,+)$. Let $v\in T_pM$ be a nonzero
vector and let $\overline{v}^\perp=v^\perp/(\mbox{\rm span}\{ v\}\cap
v^\perp)$ be the nondegenerate normal space, where $v^\perp$ is the orthogonal
space of $\mbox{\rm span}\{ v\}$.
The {\it Jacobi operator}
$\overline{R}_v:\overline{v}^\perp\mapsto\overline{v}^\perp$ is defined by
$\overline{R}_v\overline{x}=\pi(R(x,v)v)$, where
$\pi:v^\perp\mapsto\overline{v}^\perp$ is the canonical projection, $R$ is
the curvature tensor and $x\in v^\perp$ is such that $\pi(x)=\overline{x}$.
Also the induced inner product $\overline{g}$ on $\overline{v}^\perp$ is
defined by $\overline{g}(\overline{x},\overline{y})=g(x,y)$, where $x,y\in
v^\perp$ with $\pi(x)=\overline{x}$, $\pi(y)=\overline{y}$ (c.f.~\cite{beem},
~\cite{kupeli}). Note that $\overline{v}^\perp$ is isometric
with $v^\perp$ for nonnull $v$, but dim$\overline{v}^\perp=n-2$ for any null
vector $v$.
Also note that $\overline{g}$ is a
(nondegenerate) inner product on $\overline{v}^\perp$ and $\overline{R}_v$ is
self-adjoint with respect to this inner product. Thus if $v$ is nonspacelike,
then $\overline{g}$ is positive definite and hence $\overline{R}_v$ is
diagonalizable. If $v$ is spacelike then, since $\overline{g}$ is indefinite,
$\overline{R}_v$ may not be diagonalizable although it is self-adjoint with
respect to $\overline{g}$.
\subsection{Pointwise timelike Osserman manifolds}
First we will define the timelike Osserman condition.
\begin{definition}\label{def:31}
Let $M$ be a Lorentzian manifold. $M$ is said to be a timelike Osserman space
at $p\in M$ if $\overline{R}_z$ has constant eigenvalues (counting
multiplicities) for every unit timelike $z\in T_pM$. (That is, the
characteristic polynomial of $\overline{R}_z$ is independent of unit timelike
$z\in T_pM$.)
\end{definition}
Next we show that this definition yields constant curvature at $p$.
\begin{theorem}\label{th:32}
Let $M$ be a Lorentzian manifold with dim$M\geq 3$. Then $M$ is a timelike
Osserman space at $p\in M$ if and only if $M$ is of constant curvature at $p\in
M$. \end{theorem}
{\normalsize\bf Proof.}
It is obvious that the constancy of the sectional curvature at $p$ implies that
$M$ is an Osserman space at $p$. Conversely, let $c_1,\dots,c_{n-1}$ be the
eigenvalues of $\overline{R}_z$ with orthonormal eigenvectors
$\overline{v}_1,\dots,\overline{v}_{n-1}$, respectively. Then, if
$\overline{x}=\sum_{i=1}^{n-1}a_i\overline{v}_i \in z^\perp$ is a unit
vector, we have
$$
\overline{g}(\overline{R}_z\overline{x},\overline{x})=
\sum_{i=1}^{n-1}a_i\overline{g}(\overline{R}_z\overline{v}_i,\overline{x})
=\sum_{i=1}^{n-1}a_ic_i\overline{g}(\overline{v}_i,\overline{x}).
$$
Hence
$\displaystyle{\mid\overline{g}(\overline{R}_z\overline{x},
\overline{x})\mid\leq\sum_{i=1}^{n-1}\mid a_i\mid^2\mid c_i\mid \leq
\sum_{i=1}^{n-1}\mid c_i\mid}$ and it follows that
$\mid\overline{g}(\overline{R}_z\overline{x},\overline{x})\mid$ is
bounded for every unit timelike $z\in T_pM$ and
$\overline{x}\in\overline{z}^\perp$. But then, since
$\mid\overline{g}(\overline{R}_z\overline{x},\overline{x})\mid$ $=$ $\mid
\mid$, the curvatures of the timelike planes in $T_pM$ are
bounded, and it follows from~\cite[pag.229]{neill} that $M$ is of constant
curvature at $p$.\hfill$\Box$
\medskip
The theorem above shows a completely different behaviour of the Jacobi
operator along timelike geodesics in Lorentzian geometry when comparing with
Riemannian geometry. (See for example~\cite{gilkeya}). Further note that the
pointwise timelike Osserman condition is equivalent to constant sectional
curvature (and hence also to the global timelike Osserman condition). Also note
that the result above is not true for semi--Riemannian metrics of
non--Lorentzian signature. (Consider, for example, any indefinite complex
space form).
\begin{remark}\rm
At any point $p\in M$ in a Riemannian manifold, the sectional curvature,
$K_p$, is a continuous function defined on the whole Grassmannian of two
planes tangent to $M$ at that point, $G_2(T_pM)$. It is now immediate to
recognize from the compactness of $G_2(T_pM)$ that the values of the
sectional curvature are bounded from above and from below at each point $p\in
M$. Equivalently, the
eigenvalues of the Jacobi operator, (as a function defined on the unit
sphere in $T_pM$), are bounded at each point $p\in M$.
The situation is, however, completely different in Lorentzian
geometry. In fact, with minor changes in the proof of previous theorem, it can
be shown that the constancy of the sectional curvature is equivalent to be
bounded (from above or from below) the spectrum $\{ c_1(x),\dots,c_{n-1}(x)\}$
of the Jacobi operators $R_x$ for all timelike unit $x$.
\end{remark}
\subsection{The spacelike Osserman problem}
Since the Jacobi operator $\overline{R}_x$ is not necessarily diagonalizable
for spacelike vectors $x\in T_pM$, we will define the spacelike Osserman
condition at $p$ $\in$ $M$ as the independence of the characteristic polynomial of
$\overline{R}_x$ of the unit spacelike vectors $x\in T_pM$.
An immediate consequence of the spacelike Osserman condition is stated in the
next theorem, which has interesting applications as will be shown in Remark
\ref{re:aa} and Theorem \ref{th:4space} below.
\begin{theorem}\label{th:32'}
If $M$ is a spacelike Osserman space at $p$, then $M$ is Einsteinian at $p$.
\end{theorem}
{\normalsize\bf Proof.}
Note that $Ric(x,x)=\mbox{\rm trace}\overline{R}_x$. But since
$\sum_{k=0}^{n-1}a_kt^k$, the
characteristic polynomial of $\overline{R}_x$, has constant coefficients for
all unit spacelike $x$ and $ trace\,\overline{R}_x=-a_{n-2}$,
it follows that $Ric(x,x)$ is constant for every
unit spacelike $x$. Hence it follows
from~\cite{dajczer1} that $M$ is Einsteinian at $p$.\hfill$\Box$
\begin{remark}\label{re:aa}\rm
Note that every 3-dimensional Lorentzian manifold $M$ is null isotropic, that
is, $\overline{R}_u=c_u\overline{Id}$ for each null $u\in T_pM$, where
$c_u\in\Bbb{R}$. It is shown in~\cite[Cor.3.2]{kupeli2} that, if $M$ is null
isotropic, then $M$ is of constant curvature if and only if it is Einsteinian.
Thus, by Theorem \ref{th:32'}, {\it every connected 3-dimensional spacelike
Osserman Lorentzian manifold $M$ is of constant curvature}.
\end{remark}
To prove a similar result for $dim\, M=4$, we will use Theorem
\ref{th:32'} together with a result of J.A. Thorpe on the classification of the
curvature tensor of 4--dimensional Einstein Lorentzian manifolds. Let
$\bigwedge^2$ be the exterior product $T_pM\wedge T_pM$ and let
$\hat{R}:\bigwedge^2\mapsto\bigwedge^2$ be the curvature operator on
$\bigwedge^2$ defined by $<\hat{R}(x\wedge y),z\wedge v>_\wedge=$,
where $<,>_\wedge$ is the canonical inner product on $\bigwedge^2$ induced by
$<,>$.
\begin{theorem}\label{th:thorpe}{\rm ~\cite{thorpe}}
Let $M$ be a 4--dimensional Einstein Lorentzian manifold at $p\in M$. Then
there exists a Lorentzian basis $\{ e_1,e_2,e_3,e_4\}$, $(e_4 \mbox{ is
timelike})$ at $p\in M$ such that the matrix of $\hat{R}$ with respect to the
basis
$$
\left\{ e_1\wedge e_2, e_1\wedge e_3, e_2\wedge e_3, e_3\wedge e_4,
e_4\wedge e_2, e_1\wedge e_4\right\}
$$
is of the form
$$
[\hat{R}]=\left[\begin{array}{cc}
A&B\\
-B&A
\end{array}\right],
$$
where the blocks are one of the following three types:
\noindent{\underline{Type I}}
$$
A=\left[\begin{array}{ccc}
\alpha_1& &\\
&\alpha_2&\\
& &\alpha_3
\end{array}\right], \quad
B=\left[\begin{array}{ccc}
\beta_1& &\\
&\beta_2&\\
& &\beta_3
\end{array}\right], \quad \mbox{with}\, \, \beta_1+\beta_2+\beta_3=0;
$$
\noindent{\underline{Type II}}
$$
A=\left[\begin{array}{ccc}
\alpha_1& &\\
&\alpha_2+1&\\
& &\alpha_2-1
\end{array}\right], \quad
B=\left[\begin{array}{ccc}
\beta_1&0&0\\
0&\beta_2&1\\
0&1&\beta_2
\end{array}\right], \quad \mbox{with}\, \, \beta_1+2\beta_2=0;
$$
\noindent{\underline{Type III}}
$$
A=\left[\begin{array}{ccc}
\alpha&1&0\\
1&\alpha&0\\
0&0&\alpha
\end{array}\right], \quad
B=\left[\begin{array}{ccc}
0&0&0\\
0&0&-1\\
0&-1&0
\end{array}\right].
$$
\end{theorem}
\bigskip
We will show that the Types $II$ and $III$ are not possible for
4--dimensional spacelike Osserman Lorentzian manifolds and that Type $I$
implies constant curvature.
\begin{theorem}\label{th:4space}
Let $M$ be a 4--dimensional Lorentzian manifold. Then $M$ is a spacelike
Osserman space at $p\in M$ if and only if $M$ is of constant curvature at $p\in
M$.
\end{theorem}
\noindent{\normalsize\bf Proof.}
The ``if" part of the claim is obvious. To prove the converse, we will compute
the coefficient of the linear term in the characteristic polynomial of the
Jacobi operator $R_x$.
Let $\{ e_1, e_2, e_3, e_4\}$ be the basis in Theorem \ref{th:thorpe} and
assume that the matrix of $\hat{R}$ is of Type $I$. Let $x=ae_1+be_2$ and
$\{ z=-be_1+ae_2, e_3, e_4\}$ be an orthonormal basis for $x^\perp$, where
$a^2+b^2=1$. Then the
entries of $[R_x]$ with respect to this basis are
$$
\begin{array}{l}
a_{11}==-\alpha_1,\\
a_{22}==-a^2\alpha_2-b^2\alpha_3,\\
a_{33}==-a^2\alpha_3-b^2\alpha_2,\\
a_{12}=a_{21}==0,\\
a_{13}=-a_{31}==0,\\
a_{23}=-a_{32}==ab(\beta_2-\beta_3).
\end{array}
$$
\noindent{Hence}, by computing the coefficient of the linear term in the
characteristic polynomial of $R_x$, we obtain
$$
\alpha_1\alpha_3+\alpha_2\alpha_3+\alpha_1\alpha_2+a^2b^2\left[(\alpha_2
-\alpha_3)^2
+(\beta_2-\beta_3)^2\right]=\mbox{constant}
$$
for all $a,b\in\Bbb{R}$ with $a^2+b^2=1$. Hence, it follows that
$\alpha_2=\alpha_3$ and
$\beta_2=\beta_3$.
Similarly, choose $x=ae_1+be_3$ $(a^2+b^2=1)$ and let
$\{ z=-be_1+ae_3, e_2, e_4\}$ be an orthonormal basis for $x^\perp$. Then
we obtain
$\alpha_1=\alpha_3$ and $\beta_1=\beta_3$. Thus it follows that
$\alpha_1=\alpha_2=\alpha_3$ and
$\beta_1=\beta_2=\beta_3=0$ and hence $M$ is of constant curvature at $p\in M$.
Now suppose $[\hat{R}]$ is of Type $II$. Let $x=ae_1+be_3$ and
$\{ z=-be_1+ae_3, e_2, e_4\}$ be an orthonormal basis for $x^\perp$, where
$a^2+b^2=1$. Then the
entries of $[R_x]$ with respect to this basis are
$$
\begin{array}{l}
a_{11}==-\alpha_2-1,\\
a_{22}==b^2-a^2\alpha_1-b^2\alpha_2,\\
a_{33}==a^2-a^2\alpha_2-b^2\alpha_1,\\
a_{12}=a_{21}==0,\\
a_{13}=-a_{31}==-a,\\
a_{23}=-a_{32}==ab(\beta_2-\beta_1).
\end{array}
$$
\noindent{Hence}, by computing the coefficient of the linear term in the
characteristic polynomial, we obtain
$$
\alpha_1(\alpha_2+\alpha_1)+(\alpha_2^2-1)+a^2b^2\left[ (\alpha_1-\alpha_2+1)^2
+(\beta_2-\beta_1)^2\right] +a^2=\mbox{constant}
$$
for all $a,b\in\Bbb{R}$ with $a^2+b^2=1$, which is not possible. Thus Type
$II$ cannot occur.
Finally, suppose $[\hat{R}]$ is of type $III$. Let $x=ae_1+be_2$ and
$\{ z=-be_1+ae_2, e_3, e_4\}$ be an orthonormal basis for $x^\perp$, where
$a,b\in\Bbb{R}$ with $a^2+b^2=1$. Then, the entries of $[R_x]$ with respect to
this basis are
$$
\begin{array}{l}
a_{11}===-\alpha,\\
a_{22}==-\alpha,\\
a_{33}==-\alpha,\\
a_{12}=a_{21}==-a,\\
a_{13}=-a_{31}==0,\\
a_{23}=-a_{32}==a^2-b^2.
\end{array}
$$
\noindent{Hence}, by computing the coefficient of the linear term in the
characteristic polynomial, we obtain
$3\alpha^2+(a^2-b^2)^2-a^2=\mbox{constant}$, for all $a,b\in\Bbb{R}$ with
$a^2+b^2=1$, which is not possible. Thus Type $III$ cannot
occur also.\hfill$\Box$
\medskip
Note that the results of previous remark and theorem are valid at each point
$p\in M$ such that $M$ is spacelike Osserman at $p$. Hence, as well as in the
timelike case, the pointwise spacelike Osserman problem is equivalent to
constant curvature (and hence, to the global spacelike Osserman problem) if
dim$M=3$ or $4$.
\begin{remark}\label{re:bb}\rm
At this moment it is not known to the authors whether the results in
Theorem \ref{th:4space}
and Remark \ref{re:aa} may be generalized to higher dimensional manifolds.
\end{remark}
\subsection{Null Osserman Lorentzian manifolds}
As noted previously, the first difficulty in defining the null Osserman
condition comes from the necessity of ``normalizing" the set null vectors,
i.e., choosing a null vector to represent each null direction. To do that we
will use null congruences associated to timelike vectors. (We refer
to~\cite{beem} and ~\cite{sachs} for more details).
A unit timelike vector $z_p\in T_pM$ is called an {\it instantaneous
observer}, and $z_p^\perp$ is called the {\it rest space} of $z_p$. The
celestial sphere $S(z_p)$ of $z_p$ is defined by $S(z_p)=\{ x\in z_p^\perp
/g(x,x)=1\}$ (c.f.~\cite{sachs}). The {\it null congruence}
{\it determined by} $z_p$, $N(z_p)$, is defined by
$$
N(z_p)=\{ u\in T_pM/g(u,u)=0 \, \, \mbox{\rm and} \, \,
g(u,z_p)=-1\}.
$$
\begin{definition}\label{def:33}
Let $M$ be a Lorentzian manifold of dim$M\geq 3$. $M$ is called a null Osserman
space at $p$ $\in$ $M$ with respect to a unit timelike $z_p\in T_pM$
if $\overline{R}_u$ has
constant eigenvalues (counting multiplicities) for all $u\in N(z_p)$, that is,
the characteristic polynomial of $\overline{R}_u$ is independent of $u\in
N(z_p)$.
\end{definition}
A nondegenerate subspace $W\subset v^\perp$ is called a {\it geometric
realization} of $\overline{v}^\perp$ if $dim W=dim\,\overline{v}^\perp$.
(Hence $\pi_{\mid_W}:W\mapsto\overline{v}^\perp$ is an isomorphism). Let
$\overline{x}\in\overline{v}^\perp$ be an eigenvector of $\overline{R}_v$
corresponding to an eigenvalue $c$ and let $W$ be a geometric realization of
$\overline{v}^\perp$. Then the unique vector $x\in W$ with
$\pi(x)=\overline{x}$ is called a geometrically realized eigenvector of
$\overline{R}_v$ in $W$ corresponding to the eigenvalue $c$.
\medskip
Note that if $u\in N(z_p)$, then there exists a unique $x\in
S(z_p)$ such that $u=z_p+x$. Hence it also follows that $S(z_p)$ is
diffeomorphic to $N(z_p)$ in $T_pM$. Note that, if $u\in N(z_p)$ then the
geometric realization of $\overline{u}^\perp$ in $z_p^\perp$ can be identified
with $T_xS(z_p)$, where $x=u-z_p$. (Hence the geometric realizations of all
$\overline{u}^\perp$ in $z_p^\perp$ can be considered as the tangent bundle
of $S(z_p)$). Thus the Jacobi operator $\overline{R}_u$ can be used to define
a linear function $\fraktur{R}_x:T_xS(z_p)\mapsto T_xS(z_p)$ so that
$\fraktur{R}_x(y)\in T_xS(z_p)$ with
$\pi(\fraktur{R}_x(y))=\overline{R}_u\overline{y}$ via the above
identification. Furthermore $\fraktur{R}_x$ can be extended to a bundle
homomorphism $\fraktur{R}:TS(z_p)\mapsto TS(z_p)$ by
$\fraktur{R}(y)=\fraktur{R}_x(y)$, where $y\in T_xS(z_p)$ and $u=z_p+x$.
\begin{remark}\label{re:a}\rm
Note that if the eigenvalues of $\overline{R}_u$ are equal, then the null
Osserman condition reduces to infinitesimal null
isotropy~\cite{harris}, ~\cite{koch-sen}.
\end{remark}
A null vector $u\in T_pM$ is called {\it isotropic} if
$\overline{R}_u=c_u\overline{Id}$. In fact, in general relativity, every null
vector is isotropic in cosmological models and also there exist isotropic
null vectors in black hole models. Hence the null Osserman condition does not
seem to be realistic for spacetimes unless it is trivial, that is,
infinitesimal null isotropy. Indeed we have the following result.
\begin{proposition}\label{th:34}
Let $M$ be a $4$-dimensional Lorentzian manifold. Then there exists an
isotropic null vector at
each $p\in M$.
\end{proposition}
{\normalsize\bf Proof.}
Suppose that every null $u\in T_pM$ is anisotropic, that is, $\overline{R}_u$
has distinct eigenvalues $c_1(u)$ and $c_2(u)$. Now, let $z_p\in T_pM$ be a
unit timelike vector and $\fraktur{R}:TS(z_p)\mapsto TS(z_p)$ be the
homomorphism defined above.
Then note that $\fraktur{R}_x$ has the same eigenvalues $c_1(x)=c_1(u)$,
$c_2(x)=c_2(u)$, where $x=u-z_p$. Also since they are distinct, they are simple
roots of the characteristic polynomial of $\fraktur{R}$ and hence are smooth.
Now if we define the homomorphism $T$ $=$ $\fraktur{R}-c_1(u)Id$ on $TS(z_p)$,
then $Ker\, T$ is a line subbundle of $S(z_p)\equiv S^2$ in contradiction with
the fact that $S^2$ does not admit a line subbundle. Thus, there exists an
isotropic null vector.\hfill$\Box$
\medskip
In fact, if we impose the null Osserman condition, using the theorem below
from~\cite[pag.144]{steenrod} we have the following theorems.
\begin{theorem}\label{th:35} {\rm ~\cite{steenrod}}
$S^n$ does not admit a continuous subbundle of rank $k$ if $n$ is even and
$1$ $\leq$ $k$ $\leq$ $n-1$, or if $n\equiv 1\, (\mbox{\it mod}\, 4)$ and
$2\leq k\leq n-2$.
\end{theorem}
\begin{theorem}\label{th:36}
Let $M$ be a Lorentzian manifold with dim$M=2k\geq 4$. If $M$ is a null
Osserman space with respect to $z_p\in T_pM$, then the eigenvalues of
$\overline{R}_u$ are equal, that is, $M$ is infinitesimally null isotropic with
respect to $z_p$. \end{theorem}
{\normalsize\bf Proof.}
Let $c$ be an eigenvalue of $\overline{R}_u$ for every $u\in N(z_p)$. Then
it is also an eigenvalue
of $\fraktur{R}:TS(z_p)\mapsto TS(z_p)$. Consider the operator $T$
defined on $S(z_p)\equiv S^{2k-2}$ by $T=\fraktur{R}-cId$. Then by Theorem \ref{th:35},
$Ker T=TS(z_p)$ and hence the eigenvalues of $\overline{R}_u$ are
equal.\hfill$\Box$
\begin{remark}\label{re:cc}\rm
Note that the spacelike or timelike Osserman condition lead to Einstein
manifolds, as shown before. However,
the null Osserman condition does not imply the Einstein property. In fact, as an
inmediate consequence of previous theorem and~\cite[Cor.3.2]{kupeli2}, an even
dimensional Lorentzian manifold which is null Osserman with respect to some
congruence at a point $p$ is Einsteinian at that point if and only if the
sectional curvature is constant at $p$.
\end{remark}
\begin{theorem}\label{th:37}
Let $M$ be a Lorentzian manifold with dim$M=4k+3\geq 7$. If $M$ is a null
Osserman space with respect to $z_p\in T_pM$, then either $\overline{R}_u$ has
equal eigenvalues, that is, $M$ is infinitesimally null isotropic at $p\in
M$, or
$\overline{R}_u$ has exactly two distinct eigenvalues: $a$, of multiplicity
$1$, and $b$, of multiplicity $4k$.
\end{theorem}
{\normalsize\bf Proof.}
Let $a$ be an eigenvalue of $\overline{R}_u$ and, as in the proof of Theorem
\ref{th:36}, consider $T_a=\fraktur{R}-aId$ on $S(z_p)\equiv S^{4k+1}$. Then,
by Theorem \ref{th:35}, $Ker\, T_a$ is either a line subbundle or a
hyper-subbundle of $TS(z_p)$, or $TS(z_p)$ itself. In the last two cases the
claim follows immediately. Suppose $Ker\, T_a$ is a line subbundle. Then there
exists another eigenvalue $b$, and we consider $T_b=\fraktur{R}-bId$ on
$S(z_p)$. Again by Theorem \ref{th:35}, $Ker\, T_b$ is either a line subbundle
or a hyper-subbundle of $TS(z_p)$. But $Ker\, T_b$ cannot be a line subbundle
of $TS(z_p)$, because then the subbundle $Ker\, T_a\oplus Ker\, T_b$ has rank
$2$ in contradiction to Theorem \ref{th:35}. Hence $b$ has multiplicity
$4k$.\hfill$\Box$
\medskip
Note that if $M$ is of constant curvature at $p\in M$, then the Jacobi
operator $\overline{R}_u=0$ for all null $u$ at $p$, which shows that $M$
is a null Osserman space with respect to any $z_p\in T_pM$. Conversely, we have
the following
\begin{proposition}\label{th:38}
Let $M$ be a Lorentzian manifold with dim$M\geq 4$ and let $z_p$, $z'_p$ $\in
T_pM$ be linearly independent unit timelike vectors. If $M$ is a null Osserman
space with respect to $z_p$ and $z'_p$, then $M$ is of constant curvature at
$p\in M$.
\end{proposition}
{\normalsize\bf Proof.}
Let $u\in N(z_p)$. Then there exist $t\neq 0$ such that $tu\in N(z'_p)$, and
therefore $\overline{R}_{tu}=t^2\overline{R}_u$. Hence if $\{ c_i\}$ and $\{
c'_i\}$ are eigenvalues of $\overline{R}_u$ and $\overline{R}_{tu}$
respectively, then $c'_i=t^2c_i$. But since $z_p$ and $z'_p$ are linearly
independent, $N(z_p)\neq N(z'_p)$ and it follows that $c'_i=c_i=0$. Thus
$\overline{R}_u=0$ for every null $u\in T_pM$. Hence
$\overline{g}(\overline{R}_u\overline{x},\overline{x})==0$ for every
null $u\in T_pM$ and $x\in u^\perp$, where $x\in u^\perp$ with
$\pi(x)=\overline{x}$. But then it follows that $M$ is of constant curvature at
$p$ by~\cite[Thm.5]{dajczer}.\hfill$\Box$
\section{Decomposition theorems}
In this section we will obtain some decomposition theorems for null Osserman
Lorentzian manifolds. For it, we will assume $M$ to be a null Osserman space at
each point, and further
\begin{definition}\label{def:41}
A Lorentzian manifold $M$ is called a globally null Osserman space with respect
to a (not necessarily smooth) timelike line subbundle $L$ if $M$ is a null
Osserman space with respect to every unit $z\in L$ with the same eigenvalues
(counting multiplicities).
\end{definition}
Note that the global null Osserman condition is essentially different from the
null Osserman condition at each point, since the eigenvalues of the Jacobi
operator are assumed to be constant on $M$.
A Riemannian manifold $M$ is called a globally Osserman space if the Jacobi
operator $R_v=R(\cdot,v)v$ has the same eigenvalues (counting multiplicities)
for every unit $v\in TM$. We refer to~\cite{gilkey} for a discussion of the
pointwise Osserman condition (which is strictly weaker than the Osserman
condition).
\begin{lemma}\label{th:39}
Let $M$ be a null Osserman Lorentzian manifold with respect to $z_p\in T_pM$
at each $p\in M$. If $M$ is not Einsteinian at each $p\in M$, then there
exists a
unique timelike smooth line subbundle $L$ of $TM$ such that $M$ is a null
Osserman space with respect to every unit timelike $z\in L$.
\end{lemma}
{\normalsize\bf Proof.}
Let $\displaystyle{L=\bigcup_{p\in M}(\mbox{\rm span}\{ z_p\})}$ be a (not
necessarily smooth) timelike subbundle. First observe that, since $M$ is not
of constant curvature, $L$ is unique by Proposition \ref{th:38}. Next we will
show that $L$ is smooth.
We denote the Ricci tensor by $Ric(x,y)$ $=$ $trace\, \{ z\mapsto R(z,x)y\}$
and the Ricci operator, ($Ric(x,y)$ $=$ $<\hat{Ric}(x),y>$), by $\hat{Ric}$.
Now, if $u$ $\in$ $N(z_p)$ is a null vector, $u$ $=$ $z+x$ for some
$x\in z^\perp$, choose $e_1$, $\dots$, $e_{n-2}$ to be geometric realizations
of the eigenvectors of $\overline{R}_u$. Then
$\{ z,x,e_1,\dots,e_{n-2}\}$ is an orthonormal basis of $T_pM$ and it follows
that
$$
\begin{array}{rcl}
Ric(u,u)&=&R(x,u,u,x)-R(z,u,u,z)+\displaystyle{\sum_{i=1}^{n-2}
R(e_i,u,u,e_i)}\\
\noalign{\medskip}
&=&\displaystyle{\sum_{i=1}^{n-2} R(e_i,u,u,e_i)},
\end{array}
$$
which shows that $Ric(u,u)$ $=$ $trace\, \overline{R}_u$.
Now let $z\in L$ be a unit vector and $u=z+x\in N(z)$,
$v=z-x\in N(z)$, where $x\in S(z)$. Then since $M$ is a null Osserman space
with respect to $z$, $Ric(u,u)=Ric(v,v)$, and since
$$
\begin{array}{l}
Ric(u,u)=Ric(z,z)+2Ric(z,x)+Ric(x,x)\\
\noalign{\medskip}
Ric(v,v)=Ric(z,z)-2Ric(z,x)+Ric(x,x),
\end{array}
$$
it follows that $Ric(z,x)=0$. This shows that $\hat{Ric}$ is diagonalizable
with an eigenvector $z$, and all other eigenvectors corresponding to the
same eigenvalue are in $z^\perp$, i.e., at each point $p$ $\in$ $M$, $Ric$
takes the form
$$
Ric=\lambda(p) <\, ,\, >_L \oplus \mu(p) <\, ,\, >_{L^\perp},
$$
where $\lambda(p)$ and $\mu(p)$ are the eigenvalues of $\hat{Ric}$ and
$<\, ,\, >_L$ (resp., $<\, ,\, >_{L^\perp}$) denote the restriction of the
metric to $L$ (resp., $L^\perp$).
Next, we will show that the eigenvalues $\lambda$ and $\mu$ of the
Ricci operator are distinct at each point. To do that, we recall from
~\cite{dajczer1} (see also~\cite{nomizu}) that there exists null vector
$u$ $\in$ $N(z)$ such that $Ric(u,u)$ $\neq$ $0$ provided that $M$ is
not Einsteinian. Now, since $u$ $\in$ $N(z)$, $u$ $=$ $z+x$ for some
$x\in z^\perp$, one has
$$
\begin{array}{rcl}
Ric(u,u)&=&Ric(z,z)+Ric(x,x)\\
\noalign{\medskip}
&=&\lambda +\mu =\mu-\lambda,
\end{array}
$$
which shows that $\lambda(p)$ $\neq$ $\mu(p)$ at each $p\in M$.
Hence the eigenvalue $\lambda(p)$ corresponding to $z_p$ is a simple root of
the characteristic polynomial of $\hat{Ric}$ and therefore it is smooth.
Then, it follows that $L=Ker\, (\hat{Ric}-\lambda(p)Id)$ is a smooth line
subbundle.\hfill$\Box$
\begin{theorem}\label{th:42}
Let $M$ be a globally null Osserman Lorentzian manifold with respect to $L$. If
$M$ is not Einsteinian and the Ricci tensor is parallel, then $M$ is
locally a product $(I\times K,-dt^2\oplus h)$, where $I$ is an interval with
coordinate $t$ and $(K,h)$ is an Osserman Riemannian manifold.
Furthermore \begin{enumerate}
\item[(i)] if dim$M$ is even, then $(K,h)$ is a real space form,
\item[(ii)] if dim$M$ is $3$, $5$ or $4k+3$, then $(K,h)$ is either a real or
a complex space form.
\end{enumerate}
\end{theorem}
{\normalsize\bf Proof.}
The Ricci tensor is parallel, and by the proof of Lemma \ref{th:39}, $L$ is
smooth and $Ric=\lambda <,>_L\oplus\mu <,>_{L^\perp}$, where
$\lambda,\mu\in\Bbb{R}$ with $\lambda\neq \mu$ and $<,>_L$ and $<,>_{L^\perp}$
are the induced metrics on $L$ and $L^\perp$ respectively.
Hence $L=Ker\, (\hat{Ric}-\lambda Id)$ and $L^\perp=Ker\, (\hat{Ric}-\mu Id)$
are integrable with totally geodesic integral manifolds. Thus
by~\cite[Prop.3]{ponge}, $M$ is locally $(I\times
K,-dt^2\oplus h)$. To show that $(K,h)$ is an Osserman space, let $x\in T_qK$ and
$z\in T_pI$ be unit vectors. Then $u=(z,x)\in T_{(p,q)}M$ is a null vector in
the null congruence $N((z,0))$, and if $y\perp x$ in $T_qK$, we have
$R((0,y),u)u=(0,R_K(y,x)x)$, where $R_K$ is the curvature tensor of $(K,h)$.
Thus $R_K(\cdot,x)x$ has the same eigenvalues as $\overline{R}_u$ for each
$x\in T_qK$ and it follows that $(K,h)$ is an Osserman space. The rest of the
claim follows from~\cite[Thm.0]{chi1}.\hfill$\Box$
\medskip
Next we will prove a decomposition theorem applicable to null Osserman
manifolds as in Theorem \ref{th:37}. For that, we need the following
discussion on the expression of the curvature tensor.
It was shown in~\cite{gilkey}, (see also~\cite{chi1}), that the curvature
tensor of a pointwise Osserman Riemannian manifold with exactly two distinct
eigenvalues at each point, one with multiplicity one, is that of a generalized
complex space form. (See~\cite{tricerri} for the definition and further
results on generalized complex space forms). That fact comes from the
possibility of defining an almost complex structure, $J$, such that for each
$X$, the associated $JX$ is an eigenvector of the Jacobi operator $R_X$
corresponding to the distinguished eigenvalue of multiplicity one. We must
note that a key observation in doing that is the following~\cite[Lemma
3]{chi1}: If $Y$ is an eigenvector of $R_X$ with associate eigenvalue $c$,
then $X$ is an eigenvector of $R_Y$ with associate eigenvalue $c$.
Since the observation above does not hold, in general, for the case of null
Osserman Lorentzian manifolds, one should not expect a direct analog of
previous result. Therefore, we will assume the following: If $y$ is a
geometric realization of an eigenvector of the Jacobi operator
$\overline{R}_u$, $u=z+x$, ($z$ timelike unit), with corresponding eigenvalue
$c$, then $x$ is a geometric realization of an eigenvector of
$\overline{R}_v$, $v=z+y$, with corresponding eigenvalue the same $c$.
Under this further assumption, it is possible to show the existence of a
complex structure on $z^\perp$ (just proceeding as
in~\cite{chi1}, ~\cite{gilkey}) and moreover, the curvature tensor at that
point satisfies the following
\begin{lemma}\label{th:43}
Let $M$ be a Lorentzian manifold with dim$M=2k+1$. Let $z_p\in T_pM$ be a
unit timelike vector
and $J$ be a complex structure on $z_p^\perp$ for which the induced metric
on $z_p^\perp$ is
Hermitian. Then the following are equivalent:
\begin{enumerate}
\item[(1)] (a) $M$ is null Osserman with respect to $z_p$ and $\overline{R}_u$
has only two distinct eigenvalues $a$ and $b$, with multiplicities $1$ and
$2k-2$, respectively;
(b) if $u=z_p+y\in N(z_p)$, then $Jy$ is an eigenvector corresponding to the
eigenvalue $a$, where $y\in S(z_p)$.
\item[(2)] $R(x,y)z=\lambda R_0(x,y)z+\mu R_0(\pi^\perp
x,\pi^\perp y)\pi^\perp z+cF_0(\pi^\perp x,\pi^\perp y)\pi^\perp z$,
for every
$x,y,z\in$ $T_pM$, where $\mu=\frac{1}{3}(4b-a)$, $c=\frac{4}{3}(a-b)$,
$\pi^\perp:T_pM\mapsto z_p^\perp$ is the orthogonal projection and
$$
\begin{array}{rcl}
R_0(x,y)z&=&x-y,\\
\noalign{\medskip}
F_0(x,y)z&=&\frac{1}{4}\left[ x-y+Jx\right.\\
\noalign{\medskip}
& &\quad \left.-Jy+2Jz\right].
\end{array}
$$
\item[(3)] (a) $R(x,z)z=\lambda x$ for every $z\in\mbox{\it span}\{
z_p\}$, $x\in z_p^\perp$;
(b) $R(x,y)z=(\lambda+\mu)R_0(x,y)z+cF_0(x,y)z$, for every $x,y,z\in
z_p^\perp$, where $\mu
+c=a$ and $\mu+\frac{c}{4}=b$.
\end{enumerate}
\end{lemma}
{\normalsize\bf Proof.}
To show that $(1)$ implies $(2)$, let $G$ be a curvaturelike quadrilinear
function on $T_pM$ defined by
$$
\begin{array}{rcl}
G(x,y,z,v)&=&-\mu\\
\noalign{\medskip}
& &-c,
\end{array}
$$
where $\mu=\frac{1}{3}(4b-a)$, $c=\frac{4}{3}(a-b)$.
Then if $u=z_p+y\in N(z_p)$, where $y\in S(z_p)$, and $x$ is a unit
eigenvector corresponding to $a$ in $z_p^\perp$, $G(x,u,u,x)=a-\mu-c=0$.
Also if $x$ is a unit eigenvector corresponding to $b$ in $z_p^\perp$, then
$G(x,u,u,x)$ $=$ $b-\mu-\frac{c}{4}=0$.
It can be shown similarly that if $x=\alpha x_1+\beta x_2$, where $x_1$ and
$x_2$ are unit eigenvectors corresponding to $a$ and $b$ in $z_p^\perp$
respectively, then $G(x,u,u,x)=0$.
Thus $G(x,u,u,x)=0$ for every $x\in(\mbox{\rm span}\{ z_p,y\})^\perp$ and all
$u\in N(z_p)$. Hence it follows by~\cite[Thm.5]{dajczer} that
$G(x,y,z,v)=\lambda $. (Note that ~\cite[Thm.5]{dajczer} only
involves curvaturelike properties of the curvature tensor). Thus the claim
follows.
It is obvious that $(2)$ implies $(3)$. To finish the proof we will show that
$(3)$ implies $(1)$. Let
$u=z_p+y\in N(z_p)$, where $y\in S(z_p)$ and
$x=Jy$. We claim that $x$ is an eigenvector of $\overline{R}_u$ in
$z_p^\perp$ with
eigenvalue $a$. We need to show that $=0$ for
every $v\in(\mbox{\rm span}\{ x,y,z_p\})^\perp$. We have
$$
\begin{array}{rcl}
&=&+\\
\noalign{\medskip}
& &++\\
\noalign{\medskip}
&=&=0.
\end{array}
$$
Furthermore, $=-\lambda+\mu+\lambda +c=\mu+c=a$.
On the other hand, if $x\in(\mbox{\rm span}\{ z_p,y,Jy\})^\perp$, then
$x$ is an eigenvector with eigenvalue $b$. Indeed, we need to show that
$$ $=$ $0$ and $$ $=$ $0$ for every $v\in(\mbox{\rm
span}\{ x,y,Jy,z_p\})^\perp$. As above,
$$ $=$ $$ $=$ $0$ and $$ $=$
$$ $=$ $0$.
Furthermore $$
$=$ $-\lambda+\mu+\lambda+\frac{c}{4}$ $=$ $\mu+\frac{c}{4}=b$. Thus the claim
follows.\hfill$\Box$
\medskip
We remark here that if dim$M=2k\geq 4$ and $M$ is a null Osserman space at $p\in
M$, then $M$ is infinitesimally null isotropic by Theorem \ref{th:36}.
In~\cite{harris} and~\cite{koch-sen} essentially the following
analog of Lemma \ref{th:43} was proven for infinitesimally null isotropic
Lorentzian manifolds.
\begin{lemma}\label{th:44}
Let $M$ be a Lorentzian manifold. Then the following statements are equivalent:
\begin{enumerate}
\item[(1)] $M$ is infinitesimally null isotropic with respect to $z_p\in T_pM$;
\item[(2)] $R(x,y)z=\lambda R_0(x,y)z+\mu R_0(\pi^\perp x,\pi^\perp y)
\pi^\perp z$;
\item[(3)] (a) $R(x,z)z=\lambda x$, for every $z\in\mbox{\it span}\{
z_p\}$, and $x\in z_p^\perp$;
(b) $R(x,y)z=(\lambda+\mu)R_0(x,y)z$, for every $x,y,z\in z_p^\perp$.
\end{enumerate}
\end{lemma}
Here, notice that the odd dimension of $M$ did not play any role in the proof
of Lemma \ref{th:43} except for the existence of $J$ in $z_p^\perp$. Hence, if
we choose $c=0$ (equivalently $a=b$) then Lemma \ref{th:43} reduces to
Lemma \ref{th:44}.
\medskip
Also, as a consequence of Lemma \ref{th:44}, a local decomposition theorem
was proven in~\cite{karcher}.
Next we will prove a decomposition theorem as a consequence of Lemma
\ref{th:43} which will again reduce to its analog for the case of
infinitesimal null isotropy, that is, $c=0$.
\medskip
First we will fix some notation and emphasize the meaning of some assumptions
to be made. Let $L$ be a timelike (smooth) line subbundle of $TM$ and let
$\pi^\perp:TM\mapsto L^\perp$ be the orthogonal projection. Also let $J$ be an
almost complex metric $L^\perp$--substructure, that is, $J$ is a complex
structure on $L^\perp$ for which the induced metric $<,>_{L^\perp}$ on
$L^\perp$ is Hermitian. Define the {\it shape operator} $L_Z:L^\perp\mapsto
L^\perp$ of $L^\perp$ with respect to $Z\in\Gamma L$ by
$L_Z=-\pi^\perp(\nabla\, Z)$,
and define the {\it conjugate shape operator} $\overline{L}_Z:L^\perp\mapsto
L^\perp$ of $L^\perp$ with respect to $Z\in\Gamma L$ by $\overline{L}_Z=JL_Z$.
Note that if we define $J^\perp:TM\mapsto L^\perp$ by
$J^\perp=J\circ\pi^\perp$ then
$\overline{L}_Z=-J^\perp(\nabla\, Z)$.
\begin{remark}\label{re:dd}\rm
If the bundle $L^\perp$ is integrable (or equivalently, $L_Z$ is
self-adjoint), then $\overline{L}_Z$ is skew-adjoint if and only if $JL_Z=L_ZJ$
(or equivalently, $L_Z$ is complex linear).
Moreover, note that if $L^\perp$ is integrable and the integral manifolds are
totally umbilical, then $\overline{L}_Z$ is skew-adjoint.
\end{remark}
\begin{theorem}\label{th:48}
Let $M$ be a Lorentzian manifold with dim$M=2k+1\geq 7$. Let $L$ be a
timelike (smooth) line
subbundle of $TM$ and $J$ be an almost complex metric $L^\perp$-substructure. If
\begin{enumerate}
\item[(1)] $R(x,y)z=\lambda R_0(x,y)z+\mu R_0(\pi^\perp
x,\pi^\perp y)\pi^\perp z+cF_0(\pi^\perp x,\pi^\perp y)\pi^\perp z$ $\,$ with
$\mu+\frac{c}{4}\neq 0$ and $c\neq 0$ at each $p\in M$, where $\lambda$,
$\mu$, $c\in\, C^\infty(M)$,
\item[(2)] $d\lambda\mid_{L^\perp}=0$,
\item[(3)] $\overline{L}_Z$ is skew-adjoint,
\end{enumerate}
then $M$ is locally a Lorentzian warped product $(I\times_fN,<,>_L\oplus
f<,>_{L^\perp})$, where $I\subset\Bbb{R}$ is an interval and $N$ is a complex
space form.
\end{theorem}
\noindent{\normalsize\bf Proof.}
Let $Z\in\Gamma L$ be a local unit vector field and $X,Y,V\in\Gamma L^\perp$.
Then, if $\perp_L$ denotes the component in $L$, we have
$$
\begin{array}{rcl}
(\nabla_XR)(Y,Z)V&=&-d\lambda(X)Z\\
\noalign{\medskip}
& &-(\mu+\frac{c}{4})[Y-\pi^\perp (\nabla_XZ)]\\
\noalign{\medskip}
& &-\frac{c}{4}[JY-J^\perp(\nabla_XZ)\\
\noalign{\medskip}
& &\qquad -2JV];
\end{array}
$$
$$
\begin{array}{rcl}
(\nabla_YR)(Z,X)V&=&d\lambda(Y)Z\\
\noalign{\medskip}
& &+(\mu+\frac{c}{4})[<\nabla_YZ,V>X-\pi^\perp (\nabla_YZ)]\\
\noalign{\medskip}
& &+\frac{c}{4}[JX-J^\perp(\nabla_YZ)\\
\noalign{\medskip}
& &\qquad -2JV];
\end{array}
$$
$$
\begin{array}{rcl}
(\nabla_ZR)(X,Y)V&=&d(\lambda+\mu+\frac{c}{4})(Z)[X-Y]\\
\noalign{\medskip}
& &+\frac{1}{4}dc(Z)[JX-JY\\
\noalign{\medskip}
& &\qquad +2JV]\\
\noalign{\medskip}
& &+(\mu+\frac{c}{4})[(\nabla_ZX)^{\perp_L}-(\nabla_ZY)^{\perp_L}]\\
\noalign{\medskip}
& &+\frac{c}{4}[<(\nabla_ZJ^\perp)Y,V>JX+(\nabla_ZJ^\perp)X\\
\noalign{\medskip}
& &\qquad -<(\nabla_ZJ^\perp)X,V>JY-(\nabla_ZJ^\perp)Y\\
\noalign{\medskip}
& &\qquad +2JV + 2(\nabla_ZJ^\perp)V].\\
\end{array}
$$
Then in the second Bianchi identity
$$
(\nabla_ZR)(X,Y)V+(\nabla_XR)(Y,Z)V+(\nabla_YR)(Z,X)V=0,
$$
by taking $Y=V$ and $X\perp Y$, $X\perp JY$, we obtain
$$
\begin{array}{rcl}
0&=&d(\lambda+\mu+\frac{c}{4})(Z)X-d\lambda(X)Z\\
\noalign{\medskip}
& &+\mu[(\nabla_ZX)^{\perp_L}-\pi^\perp (\nabla_XZ)\\
\noalign{\medskip}
& &\quad +X-Y]\\
\noalign{\medskip}
& &+\frac{c}{4}[(\nabla_ZX)^{\perp_L}+\pi^\perp
(\nabla_XZ)+<\nabla_YZ,Y>X\\
\noalign{\medskip}
& &\quad -<\nabla_XZ,Y>Y+<(\nabla_ZJ^\perp)Y,Y>JX-<(\nabla_ZJ^\perp)X,Y>JY\\
\noalign{\medskip}
& &\quad +JY-JY\\
\noalign{\medskip}
& &\quad +JX+2JY\\
\noalign{\medskip}
& &\quad +2JY-2<\nabla_YZ,JX>JY].
\end{array}
$$
Taking the inner product of the above equation by $Z$, since
$\nabla\lambda\perp L^\perp$ and
$b=\mu+\frac{c}{4}\neq 0$ at each $p\in M$, we obtain
$$
\begin{array}{rcl}
0&=&d\lambda(X)=-(\mu+\frac{c}{4})<\nabla_ZX,Z>\\
\noalign{\medskip}
&=&(\mu+\frac{c}{4}).
\end{array}
$$
Hence we conclude that $Z$ is a geodesic vector field.
Again, by taking the inner product of the above equation by $X$, we have
$$
\begin{array}{rcl}
0&=&d(\lambda+\mu+\frac{c}{4})(Z)\\
\noalign{\medskip}
& &+(\mu+\frac{c}{4})[<\nabla_XZ,X>+<\nabla_YZ,Y>.
\end{array}
$$
Hence
$$
\frac{1}{\mu+\frac{c}{4}}d(\lambda+\mu+\frac{c}{4})(Z)=-\frac{<\nabla_XZ,X>}
{}
-\frac{<\nabla_YZ,Y>}{}.
$$
But since rank$(L^\perp)\geq 6$, we conclude that
$$
<\nabla_XZ,X>=\frac{-1}{2(\mu+\frac{c}{4})}d(\lambda+\mu+\frac{c}{4})(Z).
$$
Now, by choosing $X\perp Y$, $X\perp V$, $Y\perp V$, $X\perp JY$, $V\perp JY$
in the second Bianchi identity and taking the inner product by $Y$, we obtain
$$
\begin{array}{rcl}
0&=&\frac{1}{4}c<(\nabla_ZJ^\perp)Y,Y>+(\mu+\frac{c}{4})<\nabla_XZ,V><
Y,Y>\\
\noalign{\medskip}
& &+\frac{1}{4}c.
\end{array}
$$
But, since $(\nabla_ZJ^\perp):L^\perp\mapsto L^\perp$ is skew-adjoint,
$\overline{L}_Z$ is skew-adjoint and $\mu+\frac{c}{4}\neq 0$ at each $p\in M$ by
assumption, it follows that $(\mu+\frac{c}{4})$ $<\nabla_XZ,V>=0$ and hence
$<\nabla_XZ,V>=0$. Thus we get
$$
L_Z=-\nabla Z=\frac{d(\lambda+\mu+\frac{c}{4})(Z)}{2(\mu+\frac{c}{4})}Id.
$$
\noindent{That} is, $L^\perp$ is integrable with totally umbilical integral
manifolds. Also, it follows that the normal curvature vector field,
$\omega$, of these totally umbilical integral manifolds is normal parallel,
since $(R(X,Y)V)^{\perp_L}=0$ for $X,Y,V\in\Gamma L^\perp$.
Hence, by~\cite[Prop.3(c)]{ponge}, $M$ is locally a warped product
$I\times_fN$. Also, by the Gauss equations, if $R_N$ is the curvature tensor of
$(N,h)$, then $R_N(x,y)v$ $=$ $R(x,y)v$ $+<\omega,\omega>R_0(x,y)v$, and
hence $N$ is a generalized complex space form with
$R_N$ $=$ $(\lambda+\mu+\frac{c}{4}+<\omega,\omega>)R_0$ $+\frac{c}{4}R^J$,
where $c\neq 0$ and $R^J(x,y)v$ $=$ $Jx$ $-Jy$ $+2Jv$.
Also, since $dim\, N\geq 6$, it follows from~\cite[Thm.12.7]{tricerri} that
$N$ is a complex space form.\hfill$\Box$
\medskip
The assumption of dim$M=2k+1\geq 7$ played a fundamental role in the proof of
the previous theorem in showing that the integral manifolds of $L^\perp$ are
totally umbilic and, moreover, that they are complex space forms. Further,
note the existence of $5$--dimensional pointwise null Osserman Lorentzian
manifolds which are Lorentzian warped products with a
generalized complex space form (not necessarily a complex space form) as
Riemannian factor. (See~\cite{israel},~\cite{tricerri} for more details on
generalized complex space forms.)
\begin{remark}\label{re:49}\rm
Note that in the above theorem by using the vanishing divergence of the
Einstein tensor $G$ $=$ $Ric-\frac{1}{2}Sc<,>$, it can be shown that, for
$X\in\Gamma L^\perp$,
$$
\begin{array}{rcl}
0=(divG)(X)&=&n(1-2n)d\lambda(X)+(3n-2n^2-1)d\mu(X)\\
\noalign{\medskip}
& &\quad +\frac{1}{2}(n+1)(1-n)dc(X)\\
\noalign{\medskip}
&=&(3n-2n^2-1)d\mu(X).
\end{array}
$$
Hence, by Lemma \ref{th:43}, the assumption (2) of the above theorem
implies that $da_{\mid_{L^\perp}}=db_{\mid_{L^\perp}}=0$, where $a$ and $b$ are
eigenvalues in Lemma \ref{th:43}. Now it is natural to ask whether the
predicted warped product in the above theorem becomes a product provided that
$a$ and $b$ are independent of the point, that is, when they are constants.
Indeed it is easy to see from $\mu=\frac{1}{3}(4b-a)$ and $c=\frac{4}{3}(a-b)$
that, in this case, $\mu$ and $c$ are also constants. Hence by the expression
of $L_Z$ in the above theorem, we observe that integral manifolds of $L^\perp$
are totally geodesic if $d\lambda(Z)=0$, and if so, $M$ is locally a Lorentzian
product manifold. But one can observe from
$$
\begin{array}{rcl}
0=(div\, G)(Z)&=&n(1-2n)d(\lambda+\mu)(Z)-\frac{n}{2}(n+1)dc(Z)\\
\noalign{\medskip}
& & +[(1-2n)\mu+\frac{1}{2}(1-n)c]\, div\, Z
\end{array}
$$
that $div\, Z=0$ is a sufficient condition for $d\lambda(Z)=0$.
Hence, if we replace the assumption (2) of the above theorem by
\begin{enumerate}
\item[(2')] $d\lambda_{\mid_{L^\perp}}=0$, $a$ and $b$ are independent of
points of $M$ and $div\, Z=0$, where $Z$ is a locally defined unit vector field
in $L$, \end{enumerate}
then $M$ is a Lorentzian product $I\times N$.
\end{remark}
\medskip
Proceeding as in previous theorem, but using Lemma \ref{th:44} instead of
Lemma \ref{th:43}, we obtain the following
\begin{theorem}
Let $M$ be an even--dimensional globally null Osserman Lorentzian manifold.
Then $M$ is a space of constant curvature, or locally isometric to a Lorentzian
warped product $(I\times_fN,<,>_L$ $\oplus$ $f<,>_{L^\perp})$, where
$I\subset\Bbb{R}$ is an interval and $N$ is a real space form.
\end{theorem}
\medskip
Next we will give the quaternionic version of Lemma \ref{th:43}.
\begin{lemma}\label{th:49}
Let $M$ be a Lorentzian manifold with dim$M=4k+1$. Let $z_p\in T_pM$ be a
unit timelike vector
and $\phi =I,J,K$ be a quaternionic structure on $z_p^\perp$ for which the
induced metric on
$z_p^\perp$ is Hermitian. Then the following are equivalent:
\begin{enumerate}
\item[(1)] (a) $M$ is a null Osserman space with respect to $z_p^\perp$ and
$\overline{R}_u$ has only two distinct eigenvalues $a$ and $b$, with
multiplicities $3$ and $4(k-1)$, respectively;
(b) If $u=z+y\in N(z_p^\perp)$, then $\phi y$ is an eigenvector of
$\overline{R}_u$ corresponding to the eigenvalue $a$ for $\phi =I,J,K$,
where $y\in S(z_p^\perp)$.
\item[(2)] $R(x,y)z=\lambda R_0(x,y)z+\mu R_0(\pi^\perp
x,\pi^\perp y)\pi^\perp z+cF_0(\pi^\perp x,\pi^\perp y)\pi^\perp z$, for every
$x,y,z\in T_pM$, where $\mu=\frac{1}{3}(4b-a)$, $c=\frac{4}{3}(a-b)$,
$\pi^\perp:T_pM\mapsto z_p^\perp$ is the orthogonal projection and
$$
\begin{array}{rcl}
R_0(x,y)z&=&x-y,\\
\noalign{\bigskip}
F_0(x,y)z&=&\frac{1}{4}[x-y+Ix-Iy\\
\noalign{\medskip}
& &\quad +2Iz+Jx-Jy\\
\noalign{\medskip}
& &\quad +2Jz+Kx-Ky\\
\noalign{\medskip}
& &\quad +2Kz].\\
\end{array}
$$
\item[(3)] (a) $R(x,z)z=\lambda x$ for every $z\in span\{ z_p\}$, $x\in
z_p^\perp$,
(b) $R(x,y)z=(\lambda+\mu)R_0(x,y)z+cF_0(x,y)z$ for every $x,y,z\in
z_p^\perp$, where $\mu
+c=a$ and $\mu +\frac{c}{4}=b$.
\end{enumerate}
\end{lemma}
\noindent{\normalsize\bf Proof.}
Similar to that of Lemma \ref{th:43}.\hfill$\Box$
\medskip
Let $L$ be a timelike (smooth) line subbundle of $TM$ and let
$\pi^\perp:TM\mapsto L^\perp$ be the orthogonal projection. Also, let $V$ be an
{\it almost quaternionic metric $L^\perp$--substructure} on $L^\perp$, that
is, $V$ is a quaternionic structure on $L^\perp$ for which the induced metric
$<\,,\,>_{L^\perp}$ on $L^\perp$ is Hermitian. That is $<\phi X,\phi
Y>_{L^\perp}$ $=$ $_{L^\perp}$ for every $X,Y\in\Gamma L^\perp$, where
$\phi =I, J, K$ are a canonical local basis for $V$. We can also define three
conjugate shape operators $\overline{L}_Z^\phi$ with respect to a local
canonical basis $\phi =I,J,K$ for $V$ by $\overline{L}_Z^\phi=\phi L_Z$, where
$L_Z$ is the shape operator of $L^\perp$ with respect to $Z\in\Gamma L$. Also
note that if we define $\tilde{\phi}:TM\mapsto L^\perp$ by
$\tilde{\phi}=\phi\circ\pi^\perp$ then
$\overline{L}_Z^\phi=-\tilde{\phi}(\nabla Z)$.
\begin{theorem}\label{th:410}
Let $M$ be a Lorentzian manifold with dim$M=4n+1\geq 13$. Let $L$ be a
timelike (smooth) line subbundle of $TM$ and let $V$ be a quaternionic
substructure on $L^\perp$. If
\begin{enumerate}
\item[(1)] $R(x,y)z=\lambda R_0(x,y)z+\mu R_0(\pi^\perp
x,\pi^\perp y)\pi^\perp z+cF_0(\pi^\perp x,\pi^\perp y)\pi^\perp z$ with
$\mu+\frac{c}{4}\neq 0$ and $c\neq 0$ at each $p\in M$, where $\lambda$,
$\mu$, $c\in\, C^\infty(M)$,
\item[(2)] $d\lambda_{\mid_{L^\perp}}=0$,
\item[(3)] $\overline{L}_Z^\phi$ is skew-symmetric, where $\phi =I,J,K$ is a
canonical local basis for $V$,
\end{enumerate}
then $M$ is locally a Lorentzian warped product $(I\times_fN,
<\,,\,>_L\oplus <\,,\,>_{L^\perp})$, where $I\subset\Bbb{R}$ is an interval
and $N$ is a quaternionic space form.
\end{theorem}
\noindent{\normalsize\bf Proof.}
As in the proof of Theorem \ref{th:48} we use the second Bianchi identity,
but at the last step we use ~\cite[Lemma 7.3]{gilkey}
to show that the Riemannian
factor $N$ is a quaternionic space form.\hfill$\Box$
\begin{remark}\label{re:ff}\rm
Note that Remark \ref{re:49} also applies to the above Theorem.
\end{remark}
\begin{remark}\label{re:gg}\rm
Examples of null Osserman Lorentzian manifolds as occuring in Theorem
\ref{th:48} are those Lorentzian Kenmotsu manifolds of (pointwise)
constant $\varphi$-sectional curvature. (See~\cite{bonome},~\cite{janssens}
for definitions and further references.) Indeed, locally those manifolds are
a special case of Lorentzian warped products with
a K\"ahler manifold as Riemannian factor.
Generalizing the curvature tensor of the Kenmotsu manifolds, the socalled
$C(\alpha)$--manifolds are studied in~\cite{bonome},~\cite{janssens}. (They
contain the co--K\"ahler and Sasakian manifolds as main examples.) Now, it is
not difficult to show that any Lorentzian $C(\alpha)$--manifold of (pointwise)
constant $\varphi$-sectional curvature is a null Osserman space with respect to
the null congruence induced by the Reeb vector field (which is timelike). This
kind of examples will be discussed in a forthcoming paper.
\end{remark}
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\end{document}