# On some curvature restricted geometric structures for projective curvature
tensor

Absos Ali Shaikh and Haradhan Kundu
Department of Mathematics,
The University of Burdwan, Golapbag,
Burdwan-713104,
West Bengal, India<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

The projective curvature tensor $P$ is invariant under a geodesic preserving
transformation on a semi-Riemannian manifold. It is well known that $P$ is not
a generalized curvature tensor and hence it possesses different geometric
properties than other generalized curvature tensors. The main object of the
present paper is to study some semisymmetric type and pseudosymmetric type
curvature restricted geometric structures due to projective curvature tensor.
The reduced pseudosymmetric type structures for various Walker type conditions
are deduced and the existence of Venzi space is ensured. It is shown that the
geometric structures formed by imposing projective operator on a (0,4)-tensor
is different from that for the corresponding (1,3)-tensor. Characterization of
various semisymmetric type and pseudosymmetric type curvature restricted
geometric structures due to projective curvature tensor is obtained on a
Riemannian and a semi-Riemannian manifold, and it is shown that some of them
reduce to Einstein manifold for the Riemann case. Finally to support our
theorems four suitable examples are presented.

###### Key words and phrases:

Semisymmetric manifold, pseudosymmetric manifold, pseudosymmetric type
manifold

###### 2010 Mathematics Subject Classification:

53C15, 53C25, 53C35

##  1\. Introduction

Let $M$ be an $n$-dimensional $(n\geq 3)$ connected smooth manifold equipped
with the semi-Riemannian metric $g$, Levi-Civita connection $\nabla$, Riemann-
Christoffel curvature tensor $R$ of type $(0,4)$, Riemann-Christoffel
curvature tensor $\mathcal{R}$ of type $(1,3)$, Ricci tensor $S$ of type
$(0,2)$, Ricci operator $\mathcal{S}$, scalar curvature $\kappa$, Gaussian
curvature $G$ and concircular curvature tensor $W$.
Symmetry plays an important role in the study of differential geometry of
manifolds. The manifold $M$ is said to be locally symmetric ([6], [7], [9]) if
its local geodesic symmetries are isometry and $M$ is said to be globally
symmetric if its geodesic symmetries are extendible to the whole of $M$. In
terms of curvature restriction $M$ is locally symmetric if $\nabla R=0$ (see
[9] and also [4], [5]). We note that a geometric structure on $M$ formed by
imposing a restriction on some curvature tensors of $M$ is called a curvature
restricted geometric structure. During the last eight decades the notion of
local symmetry has been generalized by many authors by weakening the
restriction $\nabla R=0$ and there arose various curvature restricted
geometric structures.
Generalizing the notion of local symmetry, Cartan [9] (see also [39], [40],
[41]) introduced the notion of semisymmetric manifold. A semi-Riemannian
manifold $M$ is said to be semisymmetric [9] if

$\mathscr{R}(X,Y)\cdot R=0,$

where $X,Y\in\chi(M)$, the set of all smooth vector fields on $M$ and
$\mathscr{R}(X,Y)$ is the curvature operator corresponding to $R$. Again
during the study of totally umbilical submanifolds of semisymmetric manifolds
as well as during the consideration of geodesic mappings on semisymmetric
manifolds, Adamow and Deszcz [1] (see also [13] and references therein)
introduced the notion of pseudosymmetric manifolds which generalizes the
notion of semisymmetric manifolds. A semi-Riemannian manifold $M$ is said to
be pseudosymmetric if

$\mathcal{R}(X,Y)\cdot R$ and $\mathcal{G}(X,Y)\cdot R$ are linearly
dependent,

where $\mathcal{G}(X,Y)=X\wedge Y$ is the curvature operator corresponding to
the Gaussian curvature tensor $G$. Replacing $\mathcal{R}(X,Y)$,
$\mathcal{G}(X,Y)$ and $R$ by other curvature tensors in the defining
condition of semisymmetric manifold and pseudosymmetric manifold one can get
various curvature restricted geometric structures, which are respectively
known as semisymmetric type and pseudosymmetric type manifolds. Deszcz and his
coauthors (see [25], [23], [17] and also references therein) studied various
pseudosymmetric type curvature restricted geometric structures. Recently, the
present authors [36] classified various curvature restricted geometric
structures (especially, semisymmetric and pseudosymmetric) and studied their
equivalency.
The geodesic preserving transformation between two semi-Riemannian manifolds
is called projective transformation and the projective curvature tensor $P$,
given by

$P(X_{1},X_{2},X_{3},X_{4})=R(X_{1},X_{2},X_{3},X_{4})-\frac{1}{n-1}\left[S(X_{2},X_{3})g(X_{1},X_{4})-S(X_{1},X_{3})g(X_{2},X_{4})\right],$

is an invariant under such a transformation, $X_{i}\in\chi(M)$. A $(0,4)$
tensor is called a generalized curvature tensor if it obeys the symmetries
like $R$. We note that $P$ is not a generalized curvature tensor since

$P(X_{1},X_{2},X_{3},X_{4})\neq P(X_{3},X_{4},X_{1},X_{2})\ \mbox{in
general.}$

The main object of the present paper is to study various semisymmetric type
and pseudosymmetric type curvature restricted geometric structures due to the
projective curvature tensor. Since $P$ is not a generalized curvature tensor,
the projective curvature operator $\mathscr{P}(X,Y)$ can not commute with
contraction. As a consequence the structures formed by the curvature operator
$\mathscr{P}(X,Y)$ gives various interesting results different from the other
generalized curvature tensors. For example the geometric structures formed by
imposing $\mathcal{P}(X,Y)$ to a (0,4)-tensor $H$ and to the corresponding
(1,3)-tensor $\mathcal{H}$ are different. Various curvature restricted
geometric structures for $P$ along with some additional assumptions were
studied by many authors (see [8], [32], [43]) but they do not mention the
above interesting geometric fact.
The main results of the paper are highlighted below:
(1) Established some Walker type identities and found out the necessary and
sufficient conditions of various Walker type conditions formed by $P$.
(2) Characterized the $P$-space by Venzi and showed that such a space is of
constant curvature in Riemann case and for the semi-Riemann case such a space
satisfies $W\cdot W=Q(S-\frac{\kappa}{n}g,W)$.
(3) Showed that the geometric structures formed by applying $\mathscr{P}(X,Y)$
on a (0,4) tensor and the corresponding (1,3)-tensor are different and also
found out the sufficient condition of their equivalency.
(4) Characterized the semi-Riemannian manifold satisfying the following
semisymmetric and pseudosymmetric type curvature conditions:
(i) $P\cdot\mathcal{S}=0$, (ii) $P\cdot\mathcal{S}=LQ(g,\mathcal{S})$, (iii)
$P\cdot\mathcal{R}=0$, (iv) $P\cdot\mathcal{R}=LQ(g,\mathcal{R})$, (v)
$P\cdot\mathcal{R}=LQ(S,\mathcal{R})$, (vi) $P\cdot P=0$, (vii) $P\cdot
P=LQ(g,P)$, (viii) $P\cdot P=LQ(S,P)$ (ix) $P\cdot\mathcal{P}=0$, (x)
$P\cdot\mathcal{P}=LQ(g,\mathcal{P})$, (xi)
$P\cdot\mathcal{P}=LQ(S,\mathcal{P})$.
(5) Mentioned various curvature restricted geometric structures which are
properly exist for semi-Riemannian case but in Riemann space they become
Einstein.
(6) Showed that on a generalized Roter type Riemannian manifold, various
curvature restricted geometric structures, such as $P\cdot\mathcal{S}=0,$
$P\cdot\mathcal{R}=0,$ $P\cdot\mathcal{R}=LQ(g,\mathcal{R})$, $P\cdot
P=LQ(g,P)$ etc. are equivalent to the manifold of constant curvature, which
generalizes the main results of [12].
The paper is organized as follows. Section 2 deals with preliminaries. Section
3 is concerned with some curvature related properties. In Section 4 we present
our main results. Finally, in Section 5 we present some examples to support
our results.

##  2\. Preliminaries

Let us consider the following notations related to $(M,g)$:
$C^{\infty}(M)=$ the algebra of all smooth functions on $M$,
$\chi(M)=$ the Lie algebra of all smooth vector fields on $M$,
$\chi^{*}(M)=$ the Lie algebra of all smooth 1-forms on $M$,
$\Xi(M)=$ the space of all endomorphisms on $\chi(M)$ and
$\mathcal{T}^{r}_{k}(M)=$ the space of all smooth tensor fields of type
$(r,k)$ on $M$.
For $A,E\in\mathcal{T}^{0}_{2}(M)$, their Kulkarni-Nomizu product ([16], [28])
$A\wedge E\in\mathcal{T}^{0}_{4}(M)$ is given by

$\displaystyle(A\wedge E)(X_{1},X_{2},X_{3},X_{4})$ $\displaystyle=$
$\displaystyle A(X_{1},X_{4})E(X_{2},X_{3})+A(X_{2},X_{3})E(X_{1},X_{4})$
$\displaystyle-$ $\displaystyle
A(X_{1},X_{3})E(X_{2},X_{4})-A(X_{2},X_{4})E(X_{1},X_{3}),$

where $X_{1},X_{2},X_{3},X_{4}\in\chi(M)$. Throughout the paper we consider
$X,Y,X_{1},X_{2},\cdots\in\chi(M)$.
Again for a symmetric $(0,2)$ tensor $A$ and $X,Y\in\chi(M)$, we get
$(X\wedge_{A}Y),\mathcal{A},\mathcal{A}^{2}\in\Xi(M)$ and
$A^{2}\in\mathcal{T}^{0}_{2}(M)$, ([37], [38]) defined as follows:

$(X\wedge_{A}Y)X_{1}=A(Y,X_{1})X-A(X,X_{1})Y,\ \ g(\mathcal{A}X,Y)=A(X,Y),$
$\mathcal{A}^{2}=\mathcal{A}\circ\mathcal{A}\ \ \mbox{and}\ \
A^{2}(X,Y)=g(\mathcal{A}^{2}X,Y).$

Now for $A\in\mathcal{T}^{0}_{2}(M)$ and $H\in\mathcal{T}^{0}_{k}(M)$, $k\geq
2$, one can define $A\wedge H$ and $X\wedge_{H}Y$ ([3], [37]) as follows:

$\displaystyle(A\wedge H)(X_{1},X_{2},Y_{1},Y_{2},\cdots,Y_{k})$
$\displaystyle=$ $\displaystyle
A(X_{1},Y_{2})H(X_{2},Y_{1},\cdots,Y_{k})+A(X_{2},Y_{1})H(X_{1},Y_{2},\cdots,Y_{k})$
$\displaystyle-$ $\displaystyle
A(X_{1},Y_{1})H(X_{2},Y_{2},\cdots,Y_{k})-A(X_{2},Y_{2})H(X_{1},Y_{1},\cdots,Y_{k})$
$\displaystyle(X\wedge_{H}Y)(X_{1},X_{2},\cdots,X_{k})=H(Y,X_{1},X_{3},\cdots
X_{k})g(X,X_{2})-H(X,X_{1},X_{3},\cdots X_{k})g(Y,X_{2}).$

A tensor $D\in\mathcal{T}^{0}_{4}(M)$ is said to be a generalized curvature
tensor ([16], [34], [36]) if

$D(X_{1},X_{2},X_{3},X_{4})+D(X_{2},X_{3},X_{1},X_{4})+D(X_{3},X_{1},X_{2},X_{4})=0,$
$D(X_{1},X_{2},X_{3},X_{4})+D(X_{2},X_{1},X_{3},X_{4})=0\ \ \mbox{and}$
$D(X_{1},X_{2},X_{3},X_{4})=D(X_{3},X_{4},X_{1},X_{2}).$

We note that if $A,E\in\mathcal{T}^{0}_{2}(M)$ are both symmetric, then
$A\wedge E$ is a generalized curvature tensor. Again a generalized curvature
tensor is called proper if

$(\nabla_{X_{1}}D)(X_{2},X_{3},X_{4},X_{5})+(\nabla_{X_{2}}D)(X_{3},X_{1},X_{4},X_{5})+(\nabla_{X_{3}}D)(X_{1},X_{2},X_{4},X_{5})=0.$

Some important generalized curvature tensors are Gaussian curvature tensor
$G$, conformal curvature tensor $C$, concircular curvature tensor $W$ and
conharmonic curvature tensor $K$. These are respectively given as

$\displaystyle G$ $\displaystyle=$ $\displaystyle\frac{1}{2}g\wedge g,$
$\displaystyle C$ $\displaystyle=$ $\displaystyle R-\frac{1}{n-2}g\wedge
S+\frac{\kappa}{2(n-1)(n-2)}g\wedge g,$ $\displaystyle W$ $\displaystyle=$
$\displaystyle R-\frac{\kappa}{2n(n-1)}g\wedge g\ \ \mbox{and}$ $\displaystyle
K$ $\displaystyle=$ $\displaystyle R-\frac{1}{n-2}g\wedge S.$

For $D\in\mathcal{T}^{0}_{4}(M)$ and $X,Y\in\chi(M)$, the associated $(1,3)$
tensor $\mathcal{D}$ and the associated curvature operator
$\mathscr{D}(X,Y)\in\Xi(M)$ are respectively given by

$g(\mathcal{D}(X,Y)X_{1},X_{2})=D(X,Y,X_{1},X_{2})\ \mbox{and}$
$\mathscr{D}(X,Y)(X_{1})=\mathcal{D}(X,Y)X_{1}.$

One can operate an endomorphism $\mathscr{L}$ on a $(0,k)$ tensor $H$ and a
$(1,k-1)$ tensor $\mathcal{H}$ ([36], [37]) as

$\displaystyle(\mathscr{L}H)(X_{1},X_{2},\cdots,X_{k})$ $\displaystyle=$
$\displaystyle-H(\mathscr{L}X_{1},X_{2},\cdots,X_{k})-\cdots-H(X_{1},X_{2},\cdots,\mathscr{L}X_{k})\
\mbox{and}$
$\displaystyle(\mathscr{L}\mathcal{H})(X_{1},X_{2},\cdots,X_{k-1})$
$\displaystyle=$
$\displaystyle\mathscr{L}\mathcal{H}(X_{1},X_{2},\cdots,X_{k-1})-\mathcal{H}(\mathscr{L}X_{1},X_{2},\cdots,X_{k-1})$
$\displaystyle-\cdots-\mathcal{H}(X_{1},X_{2},\cdots,\mathscr{L}X_{k-1})$

respectively. In particular, if we consider $\mathscr{L}=\mathscr{D}(X,Y)$ and
$X\wedge_{A}Y$, then we get $D\cdot H,Q(A,H)\in\mathcal{T}^{0}_{k+2}(M)$ and
$D\cdot\mathcal{H},Q(A,\mathcal{H})\in\mathcal{T}^{1}_{k+1}(M)$ respectively
as follows:

$\displaystyle D\cdot H(X_{1},X_{2},\ldots,X_{k},X,Y)=(\mathscr{D}(X,Y)\cdot
H)(X_{1},X_{2},\ldots,X_{k})$
$\displaystyle=-H(\mathcal{D}(X,Y)X_{1},X_{2},\ldots,X_{k})-\cdots-H(X_{1},X_{2},\ldots,\mathcal{D}(X,Y)X_{k}),$
$\displaystyle Q(A,H)(X_{1},X_{2},\cdots,X_{k},X,Y)=((X\wedge_{A}Y)\cdot
H)(X_{1},X_{2},\ldots,X_{k})$
$\displaystyle=A(X,X_{1})H(Y,X_{2},\cdots,X_{k})+\cdots+A(X,X_{k})H(X_{1},X_{2},\cdots,Y)$
$\displaystyle-A(Y,X_{1})H(X,X_{2},\cdots,X_{k})-\cdots-A(Y,X_{k})H(X_{1},X_{2},\cdots,X),$
$\displaystyle
D\cdot\mathcal{H}(X_{1},X_{2},\cdots,X_{k-1},X,Y)=(\mathscr{D}(X,Y)\cdot\mathcal{H})(X_{1},X_{2},\cdots,X_{k-1})$
$\displaystyle=\mathcal{D}(X,Y)\mathcal{H}(X_{1},X_{2},\cdots,X_{k-1})-\mathcal{H}(\mathcal{D}(X,Y)X_{1},X_{2},\cdots,X_{k-1})$
$\displaystyle\hskip
11.38092pt-\cdots-\mathcal{H}(X_{1},X_{2},\cdots,\mathcal{D}(X,Y)X_{k-1}),$
$\displaystyle
Q(A,\mathcal{H})(X_{1},X_{2},\ldots,X_{k-1},X,Y)=((X\wedge_{A}Y)\cdot\mathcal{H})(X_{1},X_{2},\cdots,X_{k-1})$
$\displaystyle=A(Y,\mathcal{H}(X_{1},X_{2},\cdots,X_{k-1}))X-A(X,\mathcal{H}(X_{1},X_{2},\cdots,X_{k-1}))Y$
$\displaystyle+A(X,X_{1})\mathcal{H}(Y,X_{2},\cdots,X_{k-1})+\cdots+A(X,X_{k-1})\mathcal{H}(X_{1},X_{2},\cdots,Y)$
$\displaystyle-A(Y,X_{1})\mathcal{H}(X,X_{2},\cdots,X_{k-1})-\cdots-A(Y,X_{k-1})\mathcal{H}(X_{1},X_{2},\cdots,X).$

######  Definition 2.1.

For $H\in\mathcal{T}^{0}_{k}(M)$ (resp.,
$\mathcal{H}\in\mathcal{T}^{1}_{k-1}(M)$) and $D\in\mathcal{T}^{0}_{4}(M)$, a
semi-Riemannian manifold $M$ is said to be $H$-semisymmetric type (resp.,
$\mathcal{H}$-semisymmetric type) ([36], [39]) manifold due to $D$ if $D\cdot
H=0$ (resp., $D\cdot\mathcal{H}=0$).

In particular, a semi-Riemannian manifold respectively satisfying $R\cdot
R=0$, $R\cdot S=0$, $R\cdot P=0$, $P\cdot R=0$ and $P\cdot S=0$ is
respectively called semisymmetric [39], Ricci semisymmetric, projective
semisymmetric, semisymmetric due to projective curvature tensor and Ricci
semisymmetric due to projective curvature tensor.

######  Definition 2.2.

$($[1], [13], [16], [36]$)$ For $H\in\mathcal{T}^{0}_{k}(M)$ (resp.,
$\mathcal{H}\in\mathcal{T}^{1}_{k-1}(M)$) and
$D_{i}\in\mathcal{T}^{0}_{4}(M)$, $i=1,2,\cdots r,r\geq 2$, a semi-Riemannian
manifold is said to be $H$-pseudosymmetric type (resp.,
$\mathcal{H}$-pseudosymmetric type) if $\sum\limits_{i=1}^{r}c_{i}(D_{i}\cdot
H)=0$ (resp., $\sum\limits_{i=1}^{r}c_{i}(D_{i}\cdot\mathcal{H})=0$) for some
$c_{i}\in C^{\infty}(M)$, called the associated scalars. Moreover a
pseudosymmetric type condition is called constant type if its associated
scalars are all constants.

In particular, a semi-Riemannian manifold respectively satisfying $R\cdot
R=L_{R}Q(g,R)$, $R\cdot S=L_{S}Q(g,S)$, $R\cdot P=L_{P}Q(g,P)$, $P\cdot
R=L_{1}Q(g,R)$ and $P\cdot S=L_{2}Q(g,S)$ is respectively called
pseudosymmetric, Ricci pseudosymmetric, projective pseudosymmetric,
pseudosymmetric due to projective curvature tensor and Ricci pseudosymmetric
due to projective curvature tensor, where $L_{R},L_{S},L_{P},L_{1},L_{2}$ are
the associated scalars.
As a generalization of manifold of vanishing conformal curvature tensor (i.e.,
$C\equiv 0$ on $M$), there arose two curvature conditions, namely, Roter type
[14] and generalized Roter type [37], which are respectively given by

(2.1) $R=c_{1}g\wedge g+c_{2}g\wedge S+c_{3}S\wedge S,$ (2.2) $R=c_{1}g\wedge
g+c_{2}g\wedge S+c_{3}S\wedge S+c_{4}g\wedge S^{2}+c_{5}S\wedge
S^{2}+c_{6}S^{2}\wedge S^{2},$

where $c_{i}\in C^{\infty}(M)$, $i=1,2,\cdots,6$.

######  Definition 2.3.

A semi-Riemannian manifold $M$ satisfying (2.1) (resp., (2.2)) for some
$c_{i}\in C^{\infty}(M)$ is called a Roter type manifold $($citeDesz03, [15],
[16], [20], [26] and [29]$)$ (resp., generalized Roter type manifold $($[18],
[19], [21], [34], [37], [38] and [33]$)$).

Note: We note that every Roter type manifold is generalized Roter type and
every manifold of vanishing conformal curvature tensor is Roter type. We also
note that an Einstein generalized Roter type manifold is of constant curvature
[37].

######  Definition 2.4.

$($[30], [42]$)$ Let $\mathcal{L}(M)$ be the vector space formed by all
1-forms $\Theta$ on $M$ satisfying

$\Theta(X_{1})D(X_{2},X_{3},X_{4},X_{5})+\Theta(X_{2})D(X_{3},X_{1},X_{4},X_{5})+\Theta(X_{3})D(X_{1},X_{2},X_{4},X_{5})=0,$

where $D\in\mathcal{T}^{0}_{4}(M)$. Then $M$ is said to be a $D$-space by
Venzi if $dim\mathcal{L}(M)\geq 1$.

In [42] Venzi named such a space as $B$-space for $D=R$.

##  3\. Some curvature related properties

From definitions we can state the following lemmas:

######  Lemma 3.1.

If $H\in\mathcal{T}^{0}_{k}(M)$ and $A,E\in\mathcal{T}^{0}_{2}(M)$ are
symmetric, then
(i) $X\wedge_{H}Y=0$ if and only if $H=0$,
(ii) $g\wedge H=0$ if and only if $H=0$ and
(iii) $Q(A,E)=0$ if and only if $A$ and $E$ are linearly dependent [10].

######  Lemma 3.2.

If $A\in\mathcal{T}^{0}_{2}(M)$ is symmetric and $D\in\mathcal{T}^{0}_{4}(M)$
is a generalized curvature tensor, then

$D\cdot(X_{1}\wedge_{A}X_{2})=X_{1}\wedge_{D\cdot A}X_{2}.$

######  Lemma 3.3.

[37] If $A,E\in\mathcal{T}^{0}_{2}(M)$ and $\mathscr{L}\in\Xi(M)$, then

$\mathscr{L}(A\wedge E)=A\wedge\mathscr{L}E+E\wedge\mathscr{L}A.$

In particular, for $D\in\mathcal{T}^{0}_{4}(M)$ we have

$D\cdot(A\wedge E)=A\wedge(D\cdot E)+E\wedge(D\cdot A).$

######  Lemma 3.4.

If $A\in\mathcal{T}^{0}_{2}(M)$ is symmetric and $\Pi\in\chi^{*}(M)$, then

$\Pi(X_{1})(X_{2}\wedge_{A}X_{3})(X,Y)+\Pi(X_{2})(X_{3}\wedge_{A}X_{1})(X,Y)+\Pi(X_{3})(X_{1}\wedge_{A}X_{2})(X,Y)=0$

holds if and only if $\Pi(X_{1})A(X_{2},X_{3})=\Pi(X_{2})A(X_{1},X_{3})$.

######  Lemma 3.5.

[11] If $A$ is a symmetric (0,2) tensor and $D$ is a generalized curvature
tensor, then

$\displaystyle(X_{1}\wedge_{A}X_{2})X_{3}\ +\ (X_{2}\wedge_{A}X_{3})X_{1}$
$\displaystyle+$ $\displaystyle(X_{3}\wedge_{A}X_{1})X_{2}=0\ \mbox{and}$
$\displaystyle Q(A,D)(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle Q(A,D)(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle Q(A,D)(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0.$

######  Lemma 3.6.

On a semi-Riemannian manifold
(i)

$\displaystyle Q(g,P)(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle Q(g,P)(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle Q(g,P)(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0.$

holds if and only if the manifold is Einstein.
(ii)

$\displaystyle Q(S,P)(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle Q(S,P)(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle Q(S,P)(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0.$

holds if and only if $\kappa(nS-\kappa g)=0$.

Proof: (i) Contracting (3.6) over $X_{1}$ and $X_{4}$, we get

$\displaystyle\frac{\kappa
g(X_{5},X_{3})g(X_{6},X_{2})}{n-1}-\frac{g(X_{6},X_{2})S(X_{5},X_{3})}{n-1}-\frac{g(X_{2},X_{3})S(X_{5},X_{6})}{n-1}$
$\displaystyle-\frac{ng(X_{5},X_{3})S(X_{6},X_{2})}{n-1}+\frac{g(X_{5},X_{3})S(X_{6},X_{2})}{n-1}+\frac{S(X_{2},X_{3})g(X_{5},X_{6})}{n-1}=0.$

Again contracting the above over $X_{2}$ and $X_{3}$, we get

$\displaystyle\frac{2\kappa
g(X_{5},X_{6})}{n-1}-\frac{2nS(X_{5},X_{6})}{n-1}=0\ \Rightarrow\
nS(X_{5},X_{6})=\kappa g(X_{5},X_{6}),$

which implies $M$ is Einstein. (ii) Contracting (3.6) over $X_{1}$ and
$X_{4}$, we get

(3.3) $\displaystyle\frac{\kappa
g(X_{2},X_{6})S(X_{3},X_{5})}{n-1}-\frac{g(X_{3},X_{6})S^{2}(X_{2},X_{5})}{n-1}+\frac{g(X_{3},X_{5})S^{2}(X_{2},X_{6})}{n-1}$
$\displaystyle-\frac{g(X_{2},X_{5})S^{2}(X_{3},X_{6})}{n-1}-\frac{g(X_{2},X_{3})S^{2}(X_{5},X_{6})}{n-1}-\frac{nS(X_{2},X_{6})S(X_{3},X_{5})}{n-1}$
$\displaystyle-\frac{S(X_{2},X_{6})S(X_{3},X_{5})}{n-1}+\frac{2S(X_{2},X_{5})S(X_{3},X_{6})}{n-1}+\frac{S(X_{2},X_{3})S(X_{5},X_{6})}{n-1}=0.$

Now contracting (3.3) over $X_{2}$ and $X_{3}$, we get

$\displaystyle\frac{2\kappa
S(X_{5},X_{6})}{n-1}-\frac{2nS^{2}(X_{5},X_{6})}{n-1}=0\ \Rightarrow\
S^{2}=\frac{\kappa}{n}S\ \mbox{and}\ \kappa^{(2)}=\frac{\kappa^{2}}{n}.$

Again contracting (3.3) over $X_{2}$ and $X_{5}$, we get

$\displaystyle-\frac{\kappa^{(2)}g(X_{3},X_{6})}{n-1}+\frac{3\kappa
S(X_{3},X_{6})}{n-1}-\frac{2nS^{2}(X_{3},X_{6})}{n-1}=0$

Thus putting the value of $S^{2}$ and $\kappa^{(2)}$ in the preceding
equation, we get

$\frac{\kappa(nS(X_{3},X_{6})-\kappa g(X_{3},X_{6}))}{n(n-1)}=0\ \Rightarrow\
\kappa(nS(X_{3},X_{6})-\kappa g(X_{3},X_{6}))=0.$

######  Proposition 3.1.

On a semi-Riemannian manifold, contraction and projective operator
$\mathscr{P}(X,Y)$ commute if and only if the manifold is Einstein.

Proof: The result follows from Lemma 5.1 and 5.2 of [36].

######  Corollary 3.1.

$P\cdot G=0$ if and only if $M$ is Einstein.

######  Proposition 3.2.

(i) $P\cdot S=R\cdot S$, (ii) $P\cdot(S\wedge S)=R\cdot(S\wedge S)$.

######  Corollary 3.2.

On a semi-Riemannian manifold, $P\cdot S=LQ(g,S)\Leftrightarrow R\cdot
S=LQ(g,S)$, where $L$ is a smooth function on $\\{x\in
M:S_{x}\neq\frac{\kappa}{n}g_{x}\\}$.

######  Proposition 3.3.

On a semi-Riemannian manifold, the projective curvature tensor $P$ possesses
the following identities:
(i) $\mathscr{P}(X,Y)=-\mathscr{P}(Y,X)$,
(ii)
$\mathcal{P}(X_{1},X_{2})X_{3}+\mathcal{P}(X_{2},X_{3})X_{1}+\mathcal{P}(X_{3},X_{1})X_{2}=0$
and
(iii)
$P(X_{1},X_{2},X_{3},X)+P(X_{2},X_{3},X_{1},X)+P(X_{3},X_{1},X_{2},X)=0$.

######  Proposition 3.4.

On a semi-Riemannian manifold the following conditions are equivalent:
(i) $P(X_{1},X_{2},X_{3},X_{4})+P(X_{1},X_{2},X_{4},X_{3})=0$,
(ii) $P(X,X_{1},X_{2},X_{3})+P(X,X_{2},X_{3},X_{1})+P(X,X_{3},X_{1},X_{2})=0$
and
(iii) $M$ is Einstein.

######  Proposition 3.5.

On a semi-Riemannian manifold $M$,
(i)
$(\nabla_{X_{1}}P)(X_{2},X_{3},X,Y)+(\nabla_{X_{2}}P)(X_{3},X_{1},X,Y)+(\nabla_{X_{3}}P)(X_{1},X_{2},X,Y)=0$
if and only if
$(\nabla_{X_{1}}S)(X_{2},X_{3})-(\nabla_{X_{2}}S)(X_{1},X_{3})=0$, i.e., the
Ricci tensor of $M$ is Codazzi type and
(ii)
$(\nabla_{X_{1}}P)(X,Y,X_{2},X_{3})+(\nabla_{X_{2}}P)(X,Y,X_{3},X_{1})+(\nabla_{X_{3}}P)(X,Y,X_{1},X_{2})=0$
if and only if
$(\nabla_{X_{1}}S)(X_{2},X_{3})=0$, i.e., $M$ is Ricci symmetric.

Proof: The result (i) follows from Proposition 2.2 of [35], and (ii) can be
proved in a similar way.

######  Proposition 3.6.

[36] On a semi-Riemannian manifold, we have the following:
(i) $\nabla R=0$ $\Leftrightarrow$ $\nabla P=0$ (see also [31]),
(ii) $R\cdot R=0$ $\Leftrightarrow$ $R\cdot P=0$,
(iii) $R\cdot R=LQ(g,R)$ $\Leftrightarrow$ $R\cdot P=LQ(g,P)$,
(iv) $D\cdot R=0$ $\Leftrightarrow$ $D\cdot P=0$,
(v) $D\cdot R=LQ(g,R)$ $\Leftrightarrow$ $D\cdot P=LQ(g,P),$
where $D$ is a generalized curvature tensor and $L\in C^{\infty}(M)$.

######  Proposition 3.7.

If $H\in\mathcal{T}^{0}_{k}(M)$ such that $Q(S,H)=0$, then
(i) $P\cdot H=0$ $\Leftrightarrow$ $R\cdot H=0$ and
(ii) $P\cdot H=LQ(g,H)$ $\Leftrightarrow$ $R\cdot H=LQ(g,H)$, $L\in
C^{\infty}(M)$.

######  Lemma 3.7.

On a semi-Riemannian manifold the following conditions are equivalent:
(i) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ (ii) $nS^{2}-2\kappa
S+\kappa^{(2)}g=0$, where $\kappa^{(2)}=\mbox{trace}(\mathcal{S}^{2})$.

Proof: Since under contraction both the conditions give
$\kappa^{(2)}=\frac{1}{n}\kappa^{2}$, hence the result is obvious.

######  Proposition 3.8.

A Riemannian manifold satisfying the curvature condition $n^{2}S^{2}-2n\kappa
S+\kappa^{2}g=0$ or $nS^{2}-2\kappa S+\kappa^{(2)}g=0$ is always an Einstein
manifold.

Proof: If $M$ is a Riemannian manifold, then at each point $x\in M$ the Ricci
operator $\mathcal{S}$ is symmetric and there exists an orthonormal basis
$\\{e_{1},e_{2},\cdots,e_{n}\\}$ of $(T_{x}M,g_{x})$ consisting of
eigenvectors of $\mathcal{S}_{x}$. Let $\mathcal{S}_{x}e_{i}=\lambda_{i}e_{i}$
for each $i=1,2,\cdots,n$, where $\lambda_{i}\in\mathbb{R}$ are the
corresponding eigenvalues.
Therefore $S(e_{i},e_{i})=\lambda_{i}$ and
$S^{2}(e_{i},e_{i})=\lambda_{i}^{2}$ for each $i$. Now from the given
curvature condition, we have

$n^{2}\lambda_{i}^{2}-2n\kappa\lambda_{i}+\kappa^{2}=0,\forall\ i$
$\Rightarrow(n\lambda_{i}-\kappa)^{2}=0\Rightarrow\lambda_{i}=\frac{\kappa}{n},\forall\
i.$

Hence all the eigenvalues are equal and thus $M$ is an Einstein manifold.

##  4\.  Some pseudosymmetric type curvature conditions

It is well known that every semi-Riemannian manifold $M$ satisfies

$\displaystyle R\cdot R(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle R\cdot R(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle R\cdot R(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0.$

This identity is known as Walker identity. For two (0,4) tensors $D_{1}$ and
$D_{2}$ on a semi-Riemannian manifold, the condition

$\displaystyle D_{1}\cdot D_{2}(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$
$\displaystyle+$ $\displaystyle D_{1}\cdot
D_{2}(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$ $\displaystyle
D_{1}\cdot D_{2}(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0$

is called Walker type condition. If for some particular $D_{1}$ and $D_{2}$,
the condition (4) holds identically on every semi-Riemannian manifold, then it
is called an Walker type identity ([24], [27]).

######  Proposition 4.1.

Every semi-Riemannian manifold satisfies the following Walker type identity:

$\displaystyle D\cdot D(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle D\cdot D(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle D\cdot D(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0,$

where $D$ is a generalized curvature tensor.

Since $R\cdot W=R\cdot R$, $W\cdot R=R\cdot R-\frac{\kappa}{2n(n-1)}Q(g,R)$,
$P\cdot R=R\cdot R-\frac{1}{n-1}Q(S,R)$, $K\cdot C=K\cdot K$ and $C\cdot
K=K\cdot K-\frac{\kappa}{2(n-1)(n-2)}Q(g,K)$ then in view of Proposition 4.1
and Lemma 3.5 we can state the following:

######  Proposition 4.2.

Every semi-Riemannian manifold satisfies the following Walker type identities:

$\displaystyle R\cdot W(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle R\cdot W(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle R\cdot W(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0,$
$\displaystyle W\cdot R(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle W\cdot R(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle W\cdot R(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0,$
$\displaystyle P\cdot R(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle P\cdot R(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle P\cdot R(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0,$
$\displaystyle C\cdot K(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle C\cdot K(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle C\cdot K(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0,$
$\displaystyle K\cdot C(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle K\cdot C(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle K\cdot C(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0.$

######  Theorem 4.1.

On a semi-Riemannian manifold $M$,

$\displaystyle R\cdot P(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle R\cdot P(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle R\cdot P(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0$

holds if and only if $M$ is Ricci semisymmetric.

Proof: Let us first consider $R\cdot S=0$. Then from Lemma 3.2, $R\cdot
P=R\cdot R$ and hence (4.1) reduces to the Walker identity.
For the converse part, contracting (4.1) over $X_{1}$ and $X_{3}$, we get

(4.3)
$\displaystyle-R(X_{2},X_{6},X_{4},\mathcal{S}(X_{5}))-R(X_{2},X_{6},X_{5},\mathcal{S}(X_{4}))$
$\displaystyle+R(X_{2},\mathcal{S}(X_{4}),X_{5},X_{6})+R(X_{4},\mathcal{S}(X_{2}),X_{5},X_{6})$
$\displaystyle+g(X_{4},X_{6})\left[-E(X_{2},X_{5})+\mathcal{S}^{2}(X_{2},X_{5})\right]$
$\displaystyle+g(X_{2},X_{6})\left[-E(X_{4},X_{5})+\mathcal{S}^{2}(X_{4},X_{5})\right]$
$\displaystyle+g(X_{2},X_{5})\left[E(X_{4},X_{6})-\mathcal{S}^{2}(X_{4},X_{6})\right]=0,$

where $E(X,Y)$ is the tensor obtained from $R(X_{1},X,Y,\mathcal{S}(X_{2}))$
by taking contraction over $X_{1}$ and $X_{2}$. Again contracting (4.1) over
$X_{1}$ and $X_{4}$ and then replacing $X_{3}$ by $X_{4}$, we get

(4.4) $\displaystyle
R(X_{2},X_{6},X_{4},\mathcal{S}(X_{5}))+R(X_{2},X_{6},X_{5},\mathcal{S}(X_{4}))$
$\displaystyle+(-1+n)\left[R(X_{2},\mathcal{S}(X_{4}),X_{5},X_{6})+R(X_{4},\mathcal{S}(X_{2}),X_{5},X_{6})\right]$
$\displaystyle+g(X_{4},X_{6})\left[E(X_{2},X_{5})-\mathcal{S}^{2}(X_{2},X_{5})\right]$
$\displaystyle+g(X_{2},X_{6})\left[E(X_{4},X_{5})-\mathcal{S}^{2}(X_{4},X_{5})\right]$
$\displaystyle+g(X_{2},X_{5})\left[-E(X_{4},X_{6})+\mathcal{S}^{2}(X_{4},X_{6})\right]=0.$

Now adding (4.3) and (4.4), we get
$n\left[R(X_{2},\mathcal{S}(X_{4}),X_{5},X_{6})+R(X_{4},\mathcal{S}(X_{2}),X_{5},X_{6})\right]=0$,
i.e., $R\cdot S=0$. This completes the proof.

######  Theorem 4.2.

On a semi-Riemannian manifold $M$,

$\displaystyle P\cdot P(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$ $\displaystyle+$
$\displaystyle P\cdot P(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$ $\displaystyle+$
$\displaystyle P\cdot P(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})=0$

holds if $\wedge_{R\cdot S}=\frac{1}{n-1}\wedge_{S}\cdot\wedge_{S}$. Again if
(4.2) holds then
(i) $n(n-1)R\cdot S=\kappa Q(g,S)$,
(ii) $S\wedge S=g\wedge S^{2}$,
(iii) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$. Moreover if the manifold is
Riemannian, then it becomes an Einstein manifold.

Proof: Let us first consider $\wedge_{R\cdot
S}=\frac{1}{n-1}\wedge_{S}\cdot\wedge_{S}$. Then by Lemma 3.2,
$R\cdot\wedge_{S}=\frac{1}{n-1}\wedge_{S}\cdot\wedge_{S}$. Thus in view of
Proposition 4.2, the left hand side of (4.2) reduces to

$\displaystyle(P\cdot\wedge_{S})(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$
$\displaystyle+$
$\displaystyle(P\cdot\wedge_{S})(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$
$\displaystyle+$
$\displaystyle(P\cdot\wedge_{S})(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})$
$\displaystyle=$
$\displaystyle(R\cdot\wedge_{S})(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})-\frac{1}{n-1}(\wedge_{S}\cdot\wedge_{S})(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$
$\displaystyle+(R\cdot\wedge_{S})(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})-\frac{1}{n-1}(\wedge_{S}\cdot\wedge_{S})(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})$
$\displaystyle+(R\cdot\wedge_{S})(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})-\frac{1}{n-1}(\wedge_{S}\cdot\wedge_{S})(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})$
$\displaystyle=$ $\displaystyle 0$

Now similar to the proof of the converse part of Theorem 4.1, we get

(4.6) $n(n-1)R\cdot S=\kappa Q(g,S)+(S\wedge S-g\wedge S^{2}).$

Again contracting (4.2) over $X_{5}$ and $X_{6}$, we get

(4.7)
$\displaystyle-\frac{g(X_{2},X_{4})S^{2}(X_{1},X_{3})}{(n-1)^{2}}+\frac{g(X_{2},X_{3})S^{2}(X_{1},X_{4})}{(n-1)^{2}}+\frac{g(X_{1},X_{4})S^{2}(X_{2},X_{3})}{(n-1)^{2}}$
$\displaystyle-\frac{g(X_{1},X_{3})S^{2}(X_{2},X_{4})}{(n-1)^{2}}-\frac{2S(X_{1},X_{4})S(X_{2},X_{3})}{(n-1)^{2}}+\frac{2S(X_{1},X_{3})S(X_{2},X_{4})}{(n-1)^{2}}=0.$

Rearranging we get $\frac{1}{(n-1)^{2}}\left[(g\wedge
S^{2})(X_{1},X_{2},X_{3},X_{4})-(S\wedge S)(X_{1},X_{2},X_{3},X_{4})\right]$.
Hence from (4.6), we get $n(n-1)R\cdot S=\kappa Q(g,S)$.
Again contracting (4.7), we get

$\frac{1}{(n-1)^{2}}\left[\kappa^{(2)}g(X_{2},X_{3})+nS^{2}(X_{2},X_{3})-2\kappa
S(X_{2},X_{3})\right]=0.$

Then the theorem directly follows from Proposition 3.8.

######  Remark 4.1.

Although the Walker type condition presented in Theorem 4.1 is equivalent to
Ricci semisymmetry but the Walker type condition presented in Theorem 4.2
implies $R\cdot R=\frac{\kappa}{n(n-1)}Q(g,S)$ but not conversely. To support
this fact in Example 3 we present a 5-dimensional semi-Riemannian manifold
which satisfies $R\cdot R=\frac{\kappa}{n(n-1)}Q(g,S)$ and

$P\cdot P(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})+P\cdot
P(X_{3},X_{4},X_{5},X_{6},X_{1},X_{2})+P\cdot
P(X_{5},X_{6},X_{1},X_{2},X_{3},X_{4})\neq 0.$

######  Proposition 4.3.

Let $M$ be a semi-Riemannian $P$-space by Venzi with associated 1-form $\Pi$.
Then it is a
(i) $R$-space by Venzi if and only if

$\Pi(X_{2})S(X_{1},X_{3})-\Pi(X_{1})S(X_{2},X_{3})=0,$

(ii) $W$-space by Venzi if and only if

$\Pi(X_{2})\left(S-\frac{\kappa}{n}g\right)(X_{1},X_{3})-\Pi(X_{1})\left(S-\frac{\kappa}{n}g\right)(X_{2},X_{3})=0.$

Moreover in both cases $\kappa=0$.

Proof: Since $R-P=\frac{1}{n-1}\wedge_{S}$ and
$W-P=\frac{1}{n(n-1)}(\wedge_{nS-\kappa g})$, the results follows from Lemma
3.4.

######  Theorem 4.3.

Let $M$ be a semi-Riemannian $P$-space by Venzi with associated 1-form $\Pi$.
If
(i) $\Pi$ is non-null at $x\in M$, then $W=0$ at $x$.
(ii) $\Pi$ is null at some $x\in M$, then $M$ is a $W$-space by Venzi with
same associated 1-form.

Proof: From hypothesis

(4.8)
$\Pi(X_{3})P(X_{1},X_{2},X_{4},X_{5})+\Pi(X_{2})P(X_{3},X_{1},X_{4},X_{5})+\Pi(X_{1})P(X_{2},X_{3},X_{4},X_{5})=0.$

Let $V$ be the vector field corresponding to $\Pi$. Then contracting (4.8)
over $X_{3}$ and $X_{5}$, we get

$P(X_{1},X_{2},X_{4},V)=0.$ $\displaystyle\Rightarrow R(X_{1},X_{2},X_{4},V)$
$\displaystyle=$
$\displaystyle\frac{1}{n-1}\left[S(X_{2},X_{4})g(X_{1},V)-S(X_{1},X_{4})g(X_{2},V)\right]$
$\displaystyle=$
$\displaystyle\frac{1}{n-1}\left[S(X_{2},X_{4})\Pi(X_{1})-S(X_{1},X_{4})\Pi(X_{2})\right]$
$\displaystyle\Rightarrow S(V,X_{1})$ $\displaystyle=$
$\displaystyle\frac{1}{n-1}\left[\kappa\Pi(X_{1})-S(X_{1},V)\right].$

Hence

$\displaystyle R(V,X_{4},X_{1},X_{2})$ $\displaystyle=$
$\displaystyle\frac{1}{n-1}\left[S(X_{1},X_{4})\Pi(X_{2})-S(X_{2},X_{4})\Pi(X_{1})\right],$
$\displaystyle S(V,X_{1})$ $\displaystyle=$
$\displaystyle\frac{\kappa}{n}\Pi(X_{1})\ \ \mbox{and}$ $\displaystyle
P(V,X_{4},X_{1},X_{2})$ $\displaystyle=$
$\displaystyle\frac{1}{n-1}\left[S(X_{1},X_{4})\Pi(X_{2})-S(X_{2},X_{4})\Pi(X_{1})\right]$
$\displaystyle-\frac{1}{n-1}\left[S(X_{1},X_{4})\Pi(X_{2})-S(V,X_{1})g(X_{2},X_{4})\right]$
$\displaystyle=$
$\displaystyle\frac{1}{n-1}\left[\frac{\kappa}{n}g(X_{2},X_{4})-S(X_{2},X_{4})\right]\Pi(X_{1}).$

(i) Let us suppose that $\Pi$ is non-null at $x$, and without loss of
generality, we can consider the associated vector field $V$ of $\Pi$ is of
unit norm. Now putting $X_{3}=V$ in (4.8), we get

$\displaystyle P(X_{1},X_{2},X_{4},X_{5})$ $\displaystyle=$
$\displaystyle-\Pi(X_{2})P(V,X_{1},X_{4},X_{5})+\Pi(X_{1})P(V,X_{2},X_{4},X_{5})$
$\displaystyle=$
$\displaystyle\frac{\Pi(X_{2})\Pi(X_{4})}{n-1}\left(S-\frac{\kappa}{n}g\right)(X_{1},X_{5})-\frac{\Pi(X_{1})\Pi(X_{4})}{n-1}\left(S-\frac{\kappa}{n}g\right)(X_{2},X_{5})$
$\displaystyle=$
$\displaystyle\frac{\Pi(X_{4})}{n-1}\left[\Pi(X_{2})\left(S-\frac{\kappa}{n}g\right)(X_{1},X_{5})-\Pi(X_{1})\left(S-\frac{\kappa}{n}g\right)(X_{2},X_{5})\right]$
$\displaystyle=$
$\displaystyle\frac{1}{n-1}\Pi(X_{4})\left[\Pi(X_{2})Z(X_{1},X_{5})-\Pi(X_{1})Z(X_{2},X_{5})\right],$

where $Z=S-\frac{\kappa}{n}g$. Thus the curvature tensor $R$ is given by

$\displaystyle R(X_{1},X_{2},X_{4},X_{5})$ $\displaystyle=$
$\displaystyle\frac{1}{n-1}\Pi(X_{4})\left[\Pi(X_{2})Z(X_{1},X_{5})-\Pi(X_{1})Z(X_{2},X_{5})\right]$
$\displaystyle+\frac{1}{n-1}\left[S(X_{2},X_{4})g(X_{1},X_{5})-S(X_{1},X_{4})g(X_{2},X_{5})\right].$

As $Z$ is trace free and $Z(X,V)=0$, $\forall\ X$, contracting (4) over
$X_{1}$ and $X_{5}$, we get

$S(X_{2},X_{4})=-\frac{1}{n-1}(\kappa g(X_{2},X_{4})+S(X_{2},X_{4}))$
$\Rightarrow S(X_{2},X_{4})-\frac{\kappa}{n}g(X_{2},X_{4})=0\ \Rightarrow\
Z(X_{2},X_{4})=0.$

Again putting this in (4), we get
$R(X_{1},X_{2},X_{4},X_{5})=\frac{\kappa}{n(n-1)}G(X_{1},X_{2},X_{4},X_{5})$.
This completes the proof of (i).
(ii) If $\Pi$ is null at $x$, then $\Pi(V)=0$. Putting $X_{3}=V$ in (4.8) and
using (4), we get

$\Pi(X_{2})\left(S-\frac{\kappa}{n}g\right)(X_{1},X_{5})-\Pi(X_{1})\left(S-\frac{\kappa}{n}g\right)(X_{2},X_{5})=0.$

Hence from Proposition 4.3, we get

$\Pi(X_{3})W(X_{1},X_{2},X_{4},X_{5})+\Pi(X_{2})W(X_{3},X_{1},X_{4},X_{5})+\Pi(X_{1})W(X_{2},X_{3},X_{4},X_{5})=0$

at $x$. Now by (i), if $\Pi$ is non-null at $x$ then $W=0$ at $x$ and the
above condition is obvious. Hence $M$ is a $W$-space by Venzi with same
associated 1-form $\Pi$. This completes the proof.

######  Proposition 4.4.

(Theorem 1, [22]) If a generalized curvature tensor $D$ satisfies

$\Pi(X_{3})D(X_{1},X_{2},X_{4},X_{5})+\Pi(X_{2})D(X_{3},X_{1},X_{4},X_{5})+\Pi(X_{1})D(X_{2},X_{3},X_{4},X_{5})=0$

for a 1-form $\Pi$, then $D\cdot D=Q(Ric(D),D)$ at the points where $\Pi\neq
0$ ($Ric(D)$ is the trace of the linear map
$X_{1}\rightarrow\mathcal{D}(X_{1},X_{2})X_{3}$, i.e., the Ricci tensor
corresponding to $D$).

Now in view of Theorem 4.3 and Proposition 4.4 we can state the following:

######  Theorem 4.4.

Every semi-Riemannian $P$-space by Venzi satisfies $W\cdot
W=Q(S-\frac{\kappa}{n}g,W)$.

Since on a Riemannian manifold all the non-zero 1-forms are non-null and on a
manifold of constant curvature, $P=0$, hence we can state the following:

######  Corollary 4.1.

Every Riemannian manifold is a $P$-space by Venzi if and only if it is of
constant curvature.

######  Remark 4.2.

From Theorem 4.3, 4.4 and Corollary 4.1 it is clear that there does not exist
any proper $P$-space by Venzi Riemannian manifold but such a structure exists
on a semi-Riemannian manifold with a null associated 1-form. To support this
fact in Example 3 we present a 5-dimensional semi-Riemannian manifold which is
a $P$-space by Venzi.

It is well known that if $D$ is a generalized curvature tensor, then

$\displaystyle g((D\cdot\mathcal{R})(X_{1},X_{2},X_{3},X,Y),X_{4})$
$\displaystyle=$ $\displaystyle(D\cdot R)(X_{1},X_{2},X_{3},X,Y,X_{4})\
\mbox{and}$ $\displaystyle g((D\cdot\mathcal{S})(X_{1},X,Y),X_{2})$
$\displaystyle=$ $\displaystyle(D\cdot S)(X_{1},X_{2},X,Y)$

But these results are not true for $D=P$. In this case we have the following:

######  Theorem 4.5.

A semi-Riemannian manifold $M$ satisfies

(4.11) $g(P\cdot\mathcal{S}(X_{1},X,Y),X_{2})=P\cdot S(X_{1},X_{2},X,Y)$

if and only if $(S\wedge
S)(X_{1},X_{2},X,Y)=2(X\wedge_{S^{2}}Y)(X_{1},X_{2})$. Moreover if $M$
satisfies (4.11), then $nS^{2}=\kappa S=\frac{\kappa^{2}}{n}g$ and hence $M$
is Einstein if $\kappa$ is non-zero on $\\{x\in
M:\left(S-\frac{\kappa}{n}g\right)_{x}\neq 0\\}$.

Proof: From definition, we have

$P\cdot
S(X_{1},X_{2},X,Y)=-S(\mathcal{P}(X,Y)X_{1},X_{2})-S(X_{1},\mathcal{P}(X,Y)X_{2})$
$\mbox{and
}P\cdot\mathcal{S}(X_{1},X,Y)=\mathcal{P}(X,Y)\mathcal{S}(X_{1})-\mathcal{S}(\mathcal{P}(X,Y)X_{1}).$

Hence
$g((P\cdot\mathcal{S})(X_{1},X,Y),X_{2})=P(X,Y,\mathcal{S}(X_{1}),X_{2})-S(\mathcal{P}(X,Y)X_{1},X_{2})$.
So $g(P\cdot\mathcal{S}(X_{1},X,Y),X_{2})=P\cdot S(X_{1},X_{2},X,Y)$ holds if
and only if

$P(X,Y,\mathcal{S}(X_{1}),X_{4})=-S(X_{1},\mathcal{P}(X,Y)X_{2}).$ (4.12)
$\Leftrightarrow(S\wedge
S)(X_{1},X_{2},X,Y)=2(X\wedge_{S^{2}}Y)(X_{1},X_{2}).$

Now contracting (4.12) over $X$ and $X_{2}$, we get $nS^{2}=\kappa S$. Again
contracting (4.12) over $Y$ and $X_{1}$, we get $\kappa S=\kappa^{(2)}g$,
which implies $\kappa^{(2)}=\frac{\kappa^{2}}{n}$. This completes the proof.

######  Remark 4.3.

From the above theorem we can say that if a manifold satisfies (4.11) with
$\kappa=0$ then the manifold may or may not be Einstein. To support this
result, we present a non-Einstein semi-Riemannian metric with zero scalar
curvature in Example 4, which satisfies $P\cdot S=0$ as well as
$P\cdot\mathcal{S}=0$.

######  Theorem 4.6.

A semi-Riemannian manifold $M$ satisfies

(4.13) $g((P\cdot\mathcal{R})(X_{1},X_{2},X_{3},X,Y),X_{4})=P\cdot
R(X_{1},X_{2},X_{3},X_{4},X,Y),$

if and only if $M$ is Einstein.

Proof: From (4.13), we get

$\displaystyle
g(X_{4},Y)R(X_{1},X_{2},X_{3},\mathcal{S}X)-g(X,X_{4})R(X_{1},X_{2},X_{3},\mathcal{S}Y)$
$\displaystyle=$
$\displaystyle-R(X_{1},X_{2},X_{3},Y)S(X,X_{4})-R(X,X_{3},X_{1},X_{2})S(X_{4},Y).$

Now contracting (4) over $X$ and $X_{4}$, we get
$R(X_{1},X_{2},X_{3},\mathcal{S}Y)=\frac{\kappa}{n}R(X_{1},X_{2},X_{3},Y)$.
Now putting the value of $R(X_{1},X_{2},X_{3},\mathcal{S}Y)$ in (4), we get

$\displaystyle\kappa[g(X_{4},Y)R(X_{1},X_{2},X_{3},X)-g(X,X_{4})R(X_{1},X_{2},X_{3},Y)]$
$\displaystyle=$
$\displaystyle-n[R(X_{1},X_{2},X_{3},Y)S(X,X_{4})+R(X,X_{3},X_{1},X_{2})S(X_{4},Y)].$

Now contracting the above equation over $X$ and $X_{1}$ and putting the value
of $R(X_{1},X_{2},X_{3},\mathcal{S}Y)$, we get

$S(X_{2},X_{3})[nS(X_{4},Y)-\kappa g(X_{4},Y)]=0.$

This implies that the manifold is Einstein. We know that on an Einstein
manifold, $P=W$, hence the converse part is obvious.

######  Theorem 4.7.

A semi-Riemannian manifold satisfies

(4.15) $g((P\cdot\mathcal{P})(X_{1},X_{2},X_{3},X,Y),X_{4})=P\cdot
P(X_{1},X_{2},X_{3},X_{4},X,Y),$

if and only if $M$ is Einstein.

Proof: Contacting (4.15) over $X$ and $X_{4}$, we get

(4.16) $\displaystyle
R(X_{1},X_{2},X_{3},\mathcal{S}Y)=\frac{\kappa}{n}R(X_{1},X_{2},X_{3},Y)$
$\displaystyle\indent\hskip 72.26999pt+\frac{1}{2(n-1)}(S\wedge
S)(X_{1},X_{2},X_{3},Y)-\frac{\kappa}{n(n-1)}g((X_{1}\wedge_{S}X_{2})X_{3},Y).$

Now contracting (4.15) over $X_{1}$ and $Y$ and using (4.16), we get

$\displaystyle-\frac{\kappa g(X_{3},X_{4})S(X,X_{2})}{(n-1)^{2}n}$
$\displaystyle+$ $\displaystyle\frac{\kappa
g(X,X_{2})S(X_{3},X_{4})}{(n-1)^{2}n}$ $\displaystyle=$
$\displaystyle\frac{g(X,X_{2})S^{2}(X_{3},X_{4})}{(n-1)^{2}}-\frac{S(X_{3},X_{4})S(X,X_{2})}{(n-1)^{2}}$
(4.18) $\Rightarrow n^{2}S^{2}(X_{3},X_{4})=2n\kappa
S(X_{3},X_{4})-\kappa^{2}g(X_{3},X_{4}).$

Again contracting (4.15) over $X_{3}$ and $Y$ and using (4.16), we get

(4.19) $\displaystyle\frac{\kappa
g(X_{2},X_{4})S(X,X_{1})}{n(n-1)^{2}}-\frac{\kappa
g(X_{1},X_{4})S(X,X_{2})}{n(n-1)^{2}}+\frac{\kappa
g(X,X_{2})S(X_{1},X_{4})}{n(n-1)^{2}}$ $\displaystyle-\frac{\kappa
g(X,X_{1})S(X_{2},X_{4})}{n(n-1)^{2}}-\frac{g(X,X_{2})S^{2}(X_{1},X_{4})}{(n-1)^{2}}+\frac{g(X,X_{1})S^{2}(X_{2},X_{4})}{(n-1)^{2}}$
$\displaystyle-\frac{S(X_{2},X_{4})S(X,X_{1})}{(n-1)^{2}}+\frac{S(X,X_{2})S(X_{1},X_{4})}{(n-1)^{2}}=0.$

Now replacing $X_{1}$ by $X_{3}$ in (4.19) and then subtracting from (4), we
get

$\frac{\kappa
g(X,X_{3})S(X_{2},X_{4})}{n(n-1)^{2}}+\frac{S(X_{2},X_{4})S(X,X_{3})}{(n-1)^{2}}=\frac{\kappa
g(X_{2},X_{4})S(X,X_{3})}{n(n-1)^{2}}-\frac{g(X,X_{3})S^{2}(X_{2},X_{4})}{(n-1)^{2}}.$

Thus using (4.18), we get

$[nS(X,X_{1})-rg(X,X_{1})][nS(X_{2},X_{4})-rg(X_{2},X_{4})]=0.$

This implies that the manifold is Einstein. The converse part is obvious as on
an Einstein manifold, $P=W$.

######  Remark 4.4.

For a generalized curvature tensor $D$, $D\cdot\mathcal{R}=0$ and $D\cdot R=0$
(resp., $D\cdot\mathcal{S}=0$ and $D\cdot S=0$) give same structure but from
Theorems 4.5 (resp., Theorem 4.6) we can conclude that the structure
$P\cdot\mathcal{R}=0$ (resp., $P\cdot\mathcal{S}=0$ and $P\cdot\mathcal{P}=0$)
is different from the structure $P\cdot R=0$ (resp., $P\cdot S=0$ and $P\cdot
P=0$). Similarly $P\cdot\mathcal{R}=LQ(g,\mathcal{R})$ and $P\cdot R=LQ(g,R)$
(resp., $P\cdot S=LQ(g,S)$ and $P\cdot\mathcal{S}=LQ(g,\mathcal{S})$,
$P\cdot\mathcal{P}=LQ(g,\mathcal{P})$ and $P\cdot P=LQ(g,P)$) give different
structures.

######  Proposition 4.5.

[10] If $A$ be a symmetric (0,2)-tensor and $D$ be a generalized curvature
tensor on a semi-Riemannian manifold $M$, then $Q(A,D)=0$ implies either $A$
is of rank 1 or $B$ is linearly independent with $A\wedge A$.

From the above theorem we note that if $A$ is not of rank one then $Q(A,B)=0$
if and only if $B$ is linearly independent with $A\wedge A$. We also note that
the result is not true for $D=P$. For the case of projective curvature tensor
we have the following:

######  Proposition 4.6.

For a symmetric (0,2) tensor $A$, if $Q(A,P)=0$ then $g$, $S$ and $A$ are
linearly dependent.

Proof: From the condition $Q(A,P)=0$, we get

$Q(A,R)=\frac{1}{n-1}\wedge_{A}\cdot\wedge_{S}.$

Thus $\wedge_{A}\cdot\wedge_{S}$ possesses the following symmetry

$\wedge_{A}\cdot\wedge_{S}(X_{1},X_{2},X_{3},X_{4},X,Y)=\wedge_{A}\cdot\wedge_{S}(X_{3},X_{4},X_{1},X_{2},X,Y).$

Then taking contraction over $X_{1}$ and $X_{4}$ and using symmetry of $A$ and
$S$, we get

$\displaystyle\kappa A(X_{3},Y)g(X,X_{2})+\kappa A(X_{2},Y)g(X,X_{3})-\kappa
A(X,X_{3})g(X_{2},Y)-\kappa A(X,X_{2})g(X_{3},Y)$ $\displaystyle-
nA(X_{3},Y)S(X,X_{2})-nA(X_{2},Y)S(X,X_{3})+nA(X,X_{3})S(X_{2},Y)+nA(X,X_{2})S(X_{3},Y)=0$
$\Rightarrow Q(A,nS-\kappa g)=0.$

Now by Lemma 3.1, $nS-\kappa g$ and $A$ are linearly dependent. Hence the
result.

######  Corollary 4.2.

Let $M$ be a semi-Riemannian manifold. If
(i) $Q(S,P)=0$ then $M$ is either Einstein or $\kappa=0$.
(ii) $Q(g,P)=0$ then $M$ is Einstein.

######  Theorem 4.8.

If a semi-Riemannian manifold $M$ satisfies $P\cdot R=LQ(g,R)$, then
(i) $R\cdot R=LQ(g,R)$ if and only if $Q(S,R)=0$. Moreover if $S$ is not of
rank 1, then $R\cdot R=LQ(g,R)$ if and only if $R=\lambda(S\wedge S)$ for some
scalar $\lambda$.
(ii) $R\cdot R=0$ if and only if
$R=\lambda(\frac{1}{n-1}S+Lg)\wedge(\frac{1}{n-1}S+Lg)$ for some scalar
$\lambda$ provided $M$ is not quasi-Einstein.
(iii) $P\cdot S=LQ(g,S)$ if and only if

$R(Y,X_{1},X_{2},\mathcal{S}X)+R(Y,X_{2},X_{1},\mathcal{S}X)-R(X,X_{1},X_{2},\mathcal{S}Y)-R(X,X_{2},X_{1},\mathcal{S}Y)=0.$

(iv) $E=S^{2}-\frac{n-1}{n-2}L(nS-\kappa g)$.

Proof: (i) Since $P\cdot R=R\cdot R-\frac{1}{n-1}Q(S,R)=LQ(g,R)$, then $R\cdot
R=LQ(g,R)$ if and only if $Q(S,R)=0$. Again if Rank($S$) is not equal to 1,
then by using Proposition 4.5,

$(S,R)=0$ $\Leftrightarrow$ $R=\lambda(S\wedge S)$ for some scalar $\lambda$,

i.e., $R\cdot R=LQ(g,R)$ $\Leftrightarrow$ $R=\lambda(S\wedge S)$ for some
scalar $\lambda$.

(ii) From the given hypothesis,

$\displaystyle R\cdot R-\frac{1}{n-1}Q(S,R)=LQ(g,R)$ $\displaystyle\Rightarrow
R\cdot R=Q(\frac{1}{n-1}S+Lg,R).$

Now if $M$ is not quasi-Einstein, then $(\frac{1}{n-1}S+Lg)$ is not of rank
one, Hence from Proposition 4.5, we get our assertion.
(iii) We know $R\cdot S=P\cdot S$, so contracting the condition $P\cdot
R=R\cdot R-\frac{1}{n-1}Q(S,R)=L(g,R)$, we get

$\displaystyle R\cdot S[X_{1},X_{4},X,Y]-LQ(g,S)=$
$\displaystyle-\frac{1}{n-1}\left[R(Y,X_{1},X_{4},\mathcal{S}X)+R(Y,X_{4},X_{1},\mathcal{S}X)-R(X,X_{1},X_{4},\mathcal{S}Y)-R(X,X_{4},X_{1},\mathcal{S}Y)\right].$

This implies (iii).
(iv) Again contracting the above equation over $X_{1}$ and $X_{4}$, we get
(iv).
We know from the definitions that $P\cdot R-LQ(S,R)=R\cdot
R-(L+\frac{1}{n-1})Q(S,R)$. Hence using the same process of the proof of the
Theorem 4.9, we get the following:

######  Theorem 4.9.

If a semi-Riemannian manifold $M$ satisfies $P\cdot R=LQ(S,R)$, then
(i) $R\cdot R=(L+\frac{1}{n-1})Q(S,R)$.
(ii) $R\cdot R=0$ if and only if $Q(S,R)=0$ or $L=-\frac{1}{n-1}$. Moreover if
$M$ is Ricci simple, then $P\cdot R=0$ if and only if $R=\lambda(S\wedge S)$
for some scalar $\lambda$.
(iii) $P\cdot S=0$ if and only if

$\left(L+\frac{1}{n-1}\right)\left[R(Y,X_{1},X_{2},\mathcal{S}X)+R(Y,X_{2},X_{1},\mathcal{S}X)-R(X,X_{1},X_{2},\mathcal{S}Y)-R(X,X_{2},X_{1},\mathcal{S}Y)\right]=0.$

(iv) $\left(L-\frac{n-2}{n-1}\right)(E-S^{2})=0$.

######  Corollary 4.3.

Let $M$ be a semi-Riemannian manifold $M$ satisfying $P\cdot R=0$. Then
(i) $R\cdot R=0$ if and only if $Q(S,R)=0$. Moreover if $S$ is not Ricci
simple, then $R\cdot R=0$ if and only if $R=\lambda(S\wedge S)$ for some
scalar $\lambda$.
(ii) $P\cdot S=0$ if and only if

(4.20)
$R(Y,X_{1},X_{2},\mathcal{S}X)+R(Y,X_{2},X_{1},\mathcal{S}X)-R(X,X_{1},X_{2},\mathcal{S}Y)-R(X,X_{2},X_{1},\mathcal{S}Y)=0.$

(iii) $E=S^{2}$.

######  Theorem 4.10.

If a semi-Riemannian manifold $M$ satisfies

(4.21) $(P\cdot
P)(X_{1},X_{2},X_{3},X_{4},X,Y)=LQ(g,P)(X_{1},X_{2},X_{3},X_{4},X,Y),$

then
(i) $P\cdot R=LQ(g,R)$ if and only if
$R\cdot(\wedge_{S})=LQ(g,\wedge_{S})+\frac{1}{n-1}Q(S,\wedge_{S})$.
(ii) $R\cdot P=LQ(g,P)$ if and only if $Q(S,P)=0$.
(iii) $R\cdot R=LQ(g,R)$ if and only if $Q(S,P)=0$.
(iv) $n(n-1)R\cdot S=(n^{2}L-nL+\kappa)Q(g,S)$.
(v) $(n-1)E=\kappa S-S^{2}$,
(vi) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and $\kappa^{(2)}=\frac{r^{2}}{n}$.
Moreover if $M$ is a Riemannian manifold, then it is an Einstein manifold.
(vii) $Ln^{2}(nS-\kappa g)=0$. Moreover if $L$ is nowhere zero, then $M$ is an
Einstein manifold.

Proof: (i) Since $P=R-\frac{1}{n-1}\wedge_{S}$, so

$\displaystyle[P\cdot P-LQ(g,P)]-[P\cdot R-LQ(g,R)]$ $\displaystyle=$
$\displaystyle[P\cdot
R-\frac{1}{n-1}P\cdot(\wedge_{S})-LQ(g,R)+\frac{L}{n-1}Q(g,\wedge_{S})]$
$\displaystyle-[P\cdot R-LQ(g,R)]$ $\displaystyle=$
$\displaystyle-\frac{1}{n-1}[P\cdot(\wedge_{S})-\frac{L}{n-1}Q(g,\wedge_{S})]$

Thus $M$ satisfies $P\cdot R=LQ(g,R)$ if and only if
$P\cdot(\wedge_{S})=\frac{L}{n-1}Q(g,\wedge_{S})$, i.e.,
$R\cdot(\wedge_{S})=\frac{1}{n-1}Q(S,\wedge_{S})+\frac{L}{n-1}Q(g,\wedge_{S})$.
(ii) Again

$\displaystyle[P\cdot P-LQ(g,P)]-[R\cdot P-LQ(g,P)]$ $\displaystyle=$
$\displaystyle[R\cdot P-\frac{1}{n-1}(\wedge_{S})\cdot P-LQ(g,P)]$
$\displaystyle-[R\cdot P-LQ(g,P)]$ $\displaystyle=$
$\displaystyle-\frac{1}{n-1}(\wedge_{S})\cdot P$

Thus $M$ satisfies $R\cdot P=LQ(g,P)$ if and only if $(\wedge_{S})\cdot P=0$,
i.e., $Q(S,P)=0$.
(iii) From Corollary 6.3 of [36], we know that the curvature conditions
$R\cdot P=LQ(g,P)$ and $R\cdot R=LQ(g,R)$ are equivalent. Hence $M$ satisfies
$R\cdot R=LQ(g,R)$ if and only if $Q(S,P)=0$.
(iv) Since $P(X_{1},X_{2},X_{3},X_{4})\neq P(X_{3},X_{4},X_{1},X_{2})$, then
contracting

$\displaystyle[(P\cdot
P)(X_{1},X_{2},X_{3},X_{4},X,Y)-LQ(g,P)(X_{1},X_{2},X_{3},X_{4},X,Y)]$
$\displaystyle-[(P\cdot
P)(X_{3},X_{4},X_{1},X_{2},X,Y)-LQ(g,P)(X_{3},X_{4},X_{1},X_{2},X,Y)]=0,$

over $X_{1}$ and $X_{4}$, we get

$\displaystyle-\frac{LnS(X,X_{2})g(Y,X_{3})}{n-1}-\frac{LnS(X,X_{3})g(Y,X_{2})}{n-1}+\frac{Lng(X,X_{3})S(Y,X_{2})}{n-1}$
$\displaystyle+\frac{Lng(X,X_{2})S(Y,X_{3})}{n-1}-\frac{\kappa
S(X,X_{2})g(Y,X_{3})}{(n-1)^{2}}-\frac{\kappa
S(X,X_{3})g(Y,X_{2})}{(n-1)^{2}}+\frac{\kappa
g(X,X_{3})S(Y,X_{2})}{(n-1)^{2}}$ $\displaystyle+\frac{\kappa
g(X,X_{2})S(Y,X_{3})}{(n-1)^{2}}+\frac{nR(X,Y,X_{2},S(X_{3}))}{n-1}+\frac{nR(X,Y,X_{3},S(X_{2}))}{n-1}=0.$

Now rearranging the above equation, we get

$\displaystyle\left(\frac{Ln}{n-1}+\frac{r}{(n-1)^{2}}\right)Q(g,S)(X_{2},X_{3},X,Y)-\frac{n}{n-1}(R\cdot
S)(X_{2},X_{3},X,Y)=0$ $\displaystyle\Rightarrow n(n-1)R\cdot
S=(n^{2}L-nL+\kappa)Q(g,S).$

(v) Now contracting the condition (4.21) over $X_{2}$ and $X_{4}$, we get

$\displaystyle\frac{S^{2}(X,X_{3})g(Y,X_{1})}{(n-1)^{2}}-\frac{g(X,X_{1})S^{2}(Y,X_{3})}{(n-1)^{2}}-\frac{nR(X,X_{1},X_{3},S(Y))}{(n-1)^{2}}$
$\displaystyle+\frac{R(X,X_{1},X_{3},S(Y))}{(n-1)^{2}}-\frac{nR(X,X_{3},X_{1},S(Y))}{(n-1)^{2}}+\frac{R(X,X_{3},X_{1},S(Y))}{(n-1)^{2}}$
$\displaystyle-\frac{nR(X_{1},Y,X_{3},S(X))}{(n-1)^{2}}+\frac{R(X_{1},Y,X_{3},S(X))}{(n-1)^{2}}-\frac{nR(X_{1},S(X),X_{3},Y)}{(n-1)^{2}}$
$\displaystyle+\frac{R(X_{1},S(X),X_{3},Y)}{(n-1)^{2}}-\frac{S(X,X_{3})S(Y,X_{1})}{(n-1)^{2}}+\frac{S(X,X_{1})S(Y,X_{3})}{(n-1)^{2}}=0.$

Again contracting the above over $Y$ and $X_{1}$, we get

$\frac{1}{(n-1)^{2}}\left[(n-1)E(X,X_{3})-\kappa
S(X,X_{3})+S^{2}(X,X_{3})\right]=0\ \Rightarrow\ (n-1)E=\kappa S-S^{2}.$

(vi) Since $P(X_{1},X_{2},X_{3},X_{4})\neq-P(X_{1},X_{2},X_{4},X_{3})$,
contracting the condition (4.21) over $X_{3}$ and $X_{4}$, we get

$\displaystyle-\frac{S^{2}(X,X_{1})g(Y,X_{2})}{(n-1)^{2}}+\frac{S^{2}(X,X_{2})g(Y,X_{1})}{(n-1)^{2}}+\frac{g(X,X_{2})S^{2}(Y,X_{1})}{(n-1)^{2}}$
$\displaystyle-\frac{g(X,X_{1})S^{2}(Y,X_{2})}{(n-1)^{2}}-\frac{2S(X,X_{2})S(Y,X_{1})}{(n-1)^{2}}+\frac{2S(X,X_{1})S(Y,X_{2})}{(n-1)^{2}}=0.$

Again contracting the above over $Y$ and $X_{1}$, we get

$\frac{1}{(n-1)^{2}}[\kappa^{(2)}g(X,X_{2})+nS^{2}(X,X_{2})-2\kappa
S(X,X_{2})]=0,$

which implies $\kappa^{(2)}=\frac{\kappa^{2}}{n}$, and hence from above
equation we say that $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$. The next part
directly follows from Proposition 3.8.
(vii) Now contracting (4.21) over $X_{2}$ and $X_{3}$ and then replacing
$X_{4}$ by $X_{3}$, we get

$\displaystyle-\frac{nR(X,Y,X_{1},S(X_{3}))}{n-1}-\frac{nR(X,Y,X_{3},S(X_{1}))}{n-1}+\frac{R(X,X_{1},X_{3},S(Y))}{n-1}+\frac{R(X,X_{3},X_{1},S(Y))}{n-1}$
$\displaystyle+\frac{R(X_{1},Y,X_{3},S(X))}{n-1}+\frac{R(X_{1},S(X),X_{3},Y)}{n-1}-\frac{(L(n-1)n+\kappa)}{(n-1)^{2}}Q(g,S)(X_{1},X_{3},X,Y)$
$\displaystyle-\frac{S^{2}(X,X_{1})g(Y,X_{3})}{(n-1)^{2}}+\frac{g(X,X_{3})S^{2}(Y,X_{1})}{(n-1)^{2}}-\frac{S(X,X_{3})S(Y,X_{1})}{(n-1)^{2}}+\frac{S(X,X_{1})S(Y,X_{3})}{(n-1)^{2}}=0$

Again contracting the above over $Y$ and $X_{1}$, we get

$E(X,X_{3})=-\frac{\kappa^{(2)}-\kappa(L(n-1)n+\kappa)}{(n-1)^{2}}g(X,X_{3})-\frac{Ln^{2}+\kappa}{n-1}S(X,X_{3})+S^{2}(X,X_{3})$
$\Rightarrow
E=-\frac{\kappa^{(2)}-\kappa(L(n-1)n+\kappa)}{(n-1)^{2}}g-\frac{Ln^{2}+\kappa}{n-1}S+S^{2}.$

Now using (v), we get

$\frac{\kappa}{n-1}S-\frac{1}{n-1}S^{2}=-\frac{\kappa^{(2)}-\kappa(L(n-1)n+\kappa)}{(n-1)^{2}}g-\frac{Ln^{2}+\kappa}{n-1}S+S^{2}$
$\Rightarrow(-L(-1+n)n\kappa-\kappa^{2}+\kappa^{(2)})g+(n-1)(Ln^{2}+2\kappa)S-n(n-1)S^{2}=0.$

Again using (vi), we get $Ln(nS-\kappa g)=0$. This completes the proof.
Using the same technique of the proof of the previous theorem, we get the
following:

######  Theorem 4.11.

If a semi-Riemannian manifold $M$ satisfies

(4.22) $(P\cdot
P)(X_{1},X_{2},X_{3},X_{4},X,Y)=LQ(S,P)(X_{1},X_{2},X_{3},X_{4},X,Y),$

then
(i) $P\cdot R=LQ(S,R)$ if and only if
$R\cdot(\wedge_{S})=\left(L+\frac{1}{n-1}\right)Q(S,\wedge_{S})$.
(ii) $R\cdot R=LQ(S,R)$ if and only if
$R\cdot(\wedge_{S})+Q(S,R)=\left(L+\frac{1}{n-1}\right)Q(S,\wedge_{S})$.
(iii) $R\cdot P=LQ(S,P)$ if and only if $Q(S,P)=0$.
(iv) $n(n-1)R\cdot S=(1+(n-1)\kappa L)Q(g,S)$.
(v) $(n-1)E=\kappa S-S^{2}$,
(vi) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and $\kappa^{(2)}=\frac{r^{2}}{n}$
if $L\neq-\frac{1}{n-1}$. Moreover if $M$ is a Riemannian manifold and
$L\neq-\frac{1}{n-1}$, then it is an Einstein manifold.
(vii) $L\kappa(nS-\kappa g)=0$.

######  Corollary 4.4.

If a semi-Riemannian manifold $M$ satisfies $P\cdot P=0$, then
(i) $P\cdot R=0$ if $P\cdot(\wedge_{S})=0$.
(ii) $R\cdot P=0$ or $R\cdot R=0$ if $Q(S,P)=0$.
(iii) $(n-1)E=\kappa S-S^{2}$,
(iv) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and thus
$\kappa^{(2)}=\frac{r^{2}}{n}$, where $\kappa^{(2)}$ is the trace of $S^{2}$.
Moreover if $M$ is a Riemannian manifold, then it is an Einstein manifold.
(v) $n(n-1)R\cdot S=\kappa Q(g,S)$.
(vi) $S\wedge S=g\wedge S^{2}$.

Now we can easily check that the tensor
$g((P(X,Y)\cdot\mathcal{S})(X_{1}),X_{2})$ is not symmetric in $X_{1}$ and
$X_{2}$. Now using this asymmetry, we get the following results:

######  Theorem 4.12.

If a semi-Riemannian manifold $M$ satisfies
$P\cdot\mathcal{S}=LQ(g,\mathcal{S})$, then
(i) $S\wedge S=g\wedge S^{2}$.
(ii) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and $\kappa^{(2)}=\frac{r^{2}}{n}$.
Moreover if $M$ is a Riemannian manifold, then $M$ is an Einstein manifold.
(iii) $(n-1)E=L(n-1)\kappa g+(\kappa-L(n-1)n)S-S^{2}$.

Proof: (i) Since $P\cdot\mathcal{S}=LQ(g,\mathcal{S})$, so

$\displaystyle g(P\cdot\mathcal{S}(X_{1},X,Y),X_{2})$ $\displaystyle-$
$\displaystyle Lg(Q(g,\mathcal{S})(X_{1},X,Y),X_{2})$ $\displaystyle=$
$\displaystyle
g(P\cdot\mathcal{S}(X_{2},X,Y),X_{1})-Lg(Q(g,\mathcal{S})(X_{2},X,Y),X_{1})$
$\displaystyle\Rightarrow$
$\displaystyle\frac{S^{2}(X,X_{1})g(X_{2},Y)}{n-1}-\frac{S^{2}(X,X_{2})g(X_{1},Y)}{n-1}-\frac{g(X,X_{2})S^{2}(X_{1},Y)}{n-1}$
$\displaystyle+\frac{g(X,X_{1})S^{2}(X_{2},Y)}{n-1}+\frac{2S(X,X_{2})S(X_{1},Y)}{n-1}-\frac{2S(X,X_{1})S(X_{2},Y)}{n-1}=0.$
$\Rightarrow\frac{1}{n-1}\left[(S\wedge S)(X_{1},X_{2},X,Y)-(g\wedge
S^{2})(X_{1},X_{2},X,Y)\right]=0$

(ii) Now contracting the above equation over $X$ and $X_{1}$, we get

$\frac{\kappa^{(2)}g(X_{2},Y)}{n-1}-\frac{2\kappa
S(X_{2},Y)}{n-1}+\frac{nS^{2}(X_{2},Y)}{n-1}=0,$

which implies $\kappa^{(2)}=\frac{1}{n}\kappa$, and hence $n^{2}S^{2}-2n\kappa
S+\kappa^{2}g=0$.
(iii) Now contracting the given condition
$g((P\cdot\mathcal{S})(X_{1},X,Y),X_{2})=Lg(Q(g,\mathcal{S})(X_{1},X,Y),X_{2})$
over $X$ and $X_{2}$, we get

$(n-1)E(Y,X_{2})=L(n-1)\kappa
g(Y,X_{1})+(\kappa-L(n-1)n)S(Y,X_{1})-S^{2}(Y,X_{1}).$

######  Corollary 4.5.

If a semi-Riemannian manifold $M$ satisfies $P\cdot\mathcal{S}=0$, then
(i) $S\wedge S=g\wedge S^{2}$.
(ii) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and $\kappa^{(2)}=\frac{r^{2}}{n}$.
Moreover if $M$ is a Riemannian manifold, then $M$ is an Einstein manifold.
(iii) $(n-1)E=\kappa S-S^{2}.$

######  Theorem 4.13.

If a semi-Riemannian manifold $M$ satisfies
$P\cdot\mathcal{R}=LQ(g,\mathcal{R})$, then
(i) $P\cdot S=R\cdot S=LQ(g,S)$, i.e., $M$ is Ricci pseudosymmetric and thus
$E=L\kappa g-LnS+S^{2}$.
(ii) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and $\kappa^{(2)}=\frac{r^{2}}{n}$.
Moreover if $M$ is a Riemannian manifold, then $M$ is an Einstein manifold.
(iii) $S\wedge S=g\wedge S^{2}$.
(iv) $P\cdot\mathcal{S}=LQ(g,\mathcal{S})$ if and only if (4.20) holds or

$S(X,X_{2})S(X_{1},Y)-S(X,X_{1})S(X_{2},Y)+g(X_{2},Y)S^{2}(X,X_{1})-g(X,X_{2})S^{2}(X_{1},Y)=0.$

Especially, if $Q(S,R)=0$ on $M$, then $P\cdot\mathcal{R}=0$ $\Rightarrow$
$P\cdot\mathcal{S}=0$.

Proof: From the given hypothesis

$\displaystyle\mathcal{P}(X,Y)\mathcal{R}(X_{1},X_{2})X_{3}-\mathcal{R}(\mathcal{P}(X,Y)X_{1},X_{2})X_{3}$
$\displaystyle-\mathcal{R}(X_{1},\mathcal{P}(X,Y)X_{2})X_{3}-\mathcal{R}(X_{1},X_{2})\mathcal{P}(X,Y)X_{3}$
$\displaystyle\hskip 72.26999pt=L(X\wedge
Y)\mathcal{R}(X_{1},X_{2})X_{3}-L\mathcal{R}((X\wedge Y)X_{1},X_{2})X_{3}$
$\displaystyle\hskip 72.26999pt-L\mathcal{R}(X_{1},(X\wedge
Y)X_{2})X_{3}-L\mathcal{R}(X_{1},X_{2})(X\wedge Y)X_{3}.$

Since $g$ is non-degenerate, the above condition is equivalent to

(4.23) $\displaystyle
P(X,Y,\mathcal{R}(X_{1},X_{2})X_{3},X_{4})-R(\mathcal{P}(X,Y)X_{1},X_{2},X_{3},X_{4})$
$\displaystyle-R(X_{1},\mathcal{P}(X,Y)X_{2},X_{3},X_{4})-R(X_{1},X_{2},\mathcal{P}(X,Y)X_{3},X_{4})$
$\displaystyle\hskip 28.45274pt=L(X\wedge
Y)(\mathcal{R}(X_{1},X_{2})X_{3},X_{4})-LR((X\wedge
Y)X_{1},X_{2},X_{3},X_{4})$ $\displaystyle\hskip 28.45274pt-LR(X_{1},(X\wedge
Y)X_{2},X_{3},X_{4})-LR(X_{1},X_{2},(X\wedge Y)X_{3},X_{4}).$

(i) Taking contraction over $X_{1}$ and $X_{4}$ in (4.23), we get

$\displaystyle-R(X,Y,X_{2},S(X_{3}))-R(X,Y,X_{3},S(X_{2}))$
$\displaystyle+L(S(X,X_{2})g(Y,X_{3})+S(X,X_{3})g(Y,X_{2})-g(X,X_{3})S(Y,X_{2})-g(X,X_{2})S(Y,X_{3}))=0$

i.e., $P\cdot S=R\cdot S=LQ(g,S)$. Again contacting this we get $E=L\kappa
g-LnS+S^{2}$.
(ii) Contracting (4.23) over $X_{2}$ and $X_{3}$, we get

(4.24) $\displaystyle
LS(X,X_{1})g(Y,X_{4})+LS(X,X_{4})g(Y,X_{1})-Lg(X,X_{4})S(Y,X_{1})-Lg(X,X_{1})S(Y,X_{4})$
$\displaystyle+\frac{S^{2}(X,X_{1})g(Y,X_{4})}{n-1}-\frac{g(X,X_{4})S^{2}(Y,X_{1})}{n-1}+\frac{R(X,X_{1},X_{4},S(Y))}{n-1}$
$\displaystyle+\frac{R(X,X_{4},X_{1},S(Y))}{n-1}+\frac{R(X_{1},Y,X_{4},S(X))}{n-1}+\frac{R(X_{1},S(X),X_{4},Y)}{n-1}$
$\displaystyle+\frac{S(X,X_{4})S(Y,X_{1})}{n-1}-\frac{S(X,X_{1})S(Y,X_{4})}{n-1}-R(X,Y,X_{1},S(X_{4}))-R(X,Y,X_{4},S(X_{1}))=0.$

Now contracting (4.24) over $Y$ and $X_{4}$, we get

$\frac{-L\kappa g(X,X_{1})+S(X,X_{1})(Ln-\kappa)+nS^{2}(X,X_{1})}{n-1}=0.$

Again contracting (4.24) over $Y$ and $X_{4}$ and then replacing $X_{4}$ by
$X_{1}$, we get

$\frac{g(X,X_{1})(-L\kappa-\kappa^{(2)})+S(X,X_{1})(Ln+\kappa)}{n-1}=0.$

Now from last two equation we get $\kappa^{(2)}g+nS^{2}-2\kappa S=0$, which
implies $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and
$\kappa^{(2)}=\frac{r^{2}}{n}$.
(iii) Interchanging $X_{4}$ and $X_{1}$ in (4.24), we get

$\displaystyle
LS(X,X_{4})g(Y,X_{1})+LS(X,X_{1})g(Y,X_{4})-Lg(X,X_{1})S(Y,X_{4})-Lg(X,X_{4})S(Y,X_{1})$
$\displaystyle+\frac{S^{2}(X,X_{4})g(Y,X_{1})}{n-1}-\frac{g(X,X_{1})S^{2}(Y,X_{4})}{n-1}+\frac{R(X,X_{4},X_{1},S(Y))}{n-1}$
$\displaystyle+\frac{R(X,X_{1},X_{4},S(Y))}{n-1}+\frac{R(X_{4},Y,X_{1},S(X))}{n-1}+\frac{R(X_{4},S(X),X_{1},Y)}{n-1}$
$\displaystyle+\frac{S(X,X_{1})S(Y,X_{4})}{n-1}-\frac{S(X,X_{4})S(Y,X_{1})}{n-1}-R(X,Y,X_{4},S(X_{1}))-R(X,Y,X_{1},S(X_{4}))=0.$

Now subtracting the above equation from (4.24), we get

$\displaystyle\frac{S^{2}(X,X_{1})g(Y,X_{4})}{n-1}-\frac{S^{2}(X,X_{4})g(Y,X_{1})}{n-1}-\frac{g(X,X_{4})S^{2}(Y,X_{1})}{n-1}$
$\displaystyle+\frac{g(X,X_{1})S^{2}(Y,X_{4})}{n-1}+\frac{2S(X,X_{4})S(Y,X_{1})}{n-1}-\frac{2S(X,X_{1})S(Y,X_{4})}{n-1}=0.$
$\Rightarrow\frac{1}{n-1}[(S\wedge S)(X_{1},X_{4},X,Y)-(g\wedge
S^{2})(X_{1},X_{4},X,Y)]=0.$

This completes the proof.
(iv) Now $P\cdot\mathcal{S}=LQ(g,\mathcal{S})$ holds if and only if

$\displaystyle
LS(X,X_{1})g(X_{4},Y)+LS(X,X_{4})g(X_{1},Y)-Lg(X,X_{4})S(X_{1},Y)-Lg(X,X_{1})S(X_{4},Y)$
$\displaystyle+\frac{S^{2}(X,X_{1})g(X_{4},Y)}{n-1}-\frac{g(X,X_{4})S^{2}(X_{1},Y)}{n-1}+\frac{S(X,X_{4})S(X_{1},Y)}{n-1}$
$\displaystyle-\frac{S(X,X_{1})S(X_{4},Y)}{n-1}-R(X,Y,X_{1},S(X_{4}))-R(X,Y,X_{4},S(X_{1}))=0.$

Hence from (4.24), we say that $P\cdot\mathcal{S}=LQ(g,\mathcal{S})$ holds if
and only if

$\displaystyle\frac{1}{n-1}[R(X,X_{1},X_{4},S(Y))+R(X,X_{4},X_{1},S(Y))+R(X_{1},Y,X_{4},S(X))+R(X_{1},S(X),X_{4},Y)]=0.$

######  Theorem 4.14.

If a semi-Riemannian manifold $M$ satisfies
$P\cdot\mathcal{R}=LQ(S,\mathcal{R})$, then
(i) $R\cdot R=0$ if $L+\frac{1}{n-1}=0$.
(ii) $P\cdot S=R\cdot S=0$, i.e., $M$ is Ricci semisymmetric and thus
$E=S^{2}$.
(iii) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and $\kappa^{(2)}=\frac{r^{2}}{n}$
if $L+\frac{1}{n-1}\neq 0$.
(iv) $S\wedge S=g\wedge S^{2}$.
$S^{2}=\frac{\kappa}{n}S=\frac{\kappa^{(2)}}{n}g$. Moreover if $\kappa\neq 0$,
then $M$ is Einstein.
(v) for $L+\frac{1}{n-1}\neq 0$, $P\cdot\mathcal{S}=LQ(g,\mathcal{S})$ if and
only if (4.20) holds.

Proof: From the given hypothesis

$\displaystyle\mathcal{P}(X,Y)\mathcal{R}(X_{1},X_{2})X_{3}-\mathcal{R}(\mathcal{P}(X,Y)X_{1},X_{2})X_{3}$
$\displaystyle-\mathcal{R}(X_{1},\mathcal{P}(X,Y)X_{2})X_{3}-\mathcal{R}(X_{1},X_{2})\mathcal{P}(X,Y)X_{3}$
$\displaystyle\hskip
72.26999pt=L(X\wedge_{S}Y)\mathcal{R}(X_{1},X_{2})X_{3}-L\mathcal{R}((X\wedge_{S}Y)X_{1},X_{2})X_{3}$
$\displaystyle\hskip
72.26999pt-L\mathcal{R}(X_{1},(X\wedge_{S}Y)X_{2})X_{3}-L\mathcal{R}(X_{1},X_{2})(X\wedge_{S}Y)X_{3}.$

Since $g$ is non-degenerate, the above condition is equivalent to

(4.25) $\displaystyle
P(X,Y,\mathcal{R}(X_{1},X_{2})X_{3},X_{4})-R(\mathcal{P}(X,Y)X_{1},X_{2},X_{3},X_{4})$
$\displaystyle-R(X_{1},\mathcal{P}(X,Y)X_{2},X_{3},X_{4})-R(X_{1},X_{2},\mathcal{P}(X,Y)X_{3},X_{4})$
$\displaystyle\hskip
28.45274pt=L(X\wedge_{S}Y)(\mathcal{R}(X_{1},X_{2})X_{3},X_{4})-LR((X\wedge_{S}Y)X_{1},X_{2},X_{3},X_{4})$
$\displaystyle\hskip 28.45274pt-
LR(X_{1},(X\wedge_{S}Y)X_{2},X_{3},X_{4})-LR(X_{1},X_{2},(X\wedge_{S}Y)X_{3},X_{4}).$

(i) The proof is obvious, since $P=R-\frac{1}{n-1}\wedge_{S}$.
(ii) Contracting (4.25) over $X_{1}$ and $X_{4}$ we get the result.
(iii) Contracting (4.25) over $X_{2}$ and $X_{3}$, we get

(4.26)
$\displaystyle(L+\frac{1}{n-1})[S^{2}(X,X_{1})g(Y,X_{4})-g(X,X_{4})S^{2}(Y,X_{1})$
$\displaystyle\hskip 85.35826pt+S(X,X_{4})S(Y,X_{1})-S(X,X_{1})S(Y,X_{4})]$
$\displaystyle+(L+\frac{1}{n-1})[R(X_{1},Y,X_{4},S(X))+R(X_{1},S(X),X_{4},Y)$
$\displaystyle\hskip 85.35826pt+R(X,X_{1},X_{4},S(Y))+R(X,X_{4},X_{1},S(Y))]$
$\displaystyle-R(X,Y,X_{1},S(X_{4}))-R(X,Y,X_{4},S(X_{1}))=0$

Now contracting (4.26) over $X_{4}$ and $Y$ and putting $E=S^{2}$, we get

$(L+\frac{1}{n-1})(nS^{2}-\kappa S)=0.$

Again contracting (4.26) over $X_{1}$ and $Y$ and putting $E=S^{2}$, we get

$(L+\frac{1}{n-1})(\kappa S-\kappa^{(2)}g)=0.$

Thus from the last two equations we get
$(L+\frac{1}{n-1})(\kappa^{(2)}g-2\kappa S+nS^{2})=0$. Hence
$n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and $\kappa^{(2)}=\frac{r^{2}}{n}$ if
$L+\frac{1}{n-1}\neq 0$
The proof of (iv) and (v) are similar to the proof of (iii) and (iv) of
Theorem 4.13.

######  Corollary 4.6.

If a semi-Riemannian manifold $M$ satisfies $P\cdot\mathcal{R}=0$, then
(i) $P\cdot S=R\cdot S=0$, i.e., $M$ is Ricci semisymmetric and thus $E=S^{2}$
(ii) $S^{2}=\frac{\kappa}{n}S=\frac{\kappa^{(2)}}{n}g$. Moreover if
$\kappa\neq 0$, then $M$ is Einstein.
(iii) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and
$\kappa^{(2)}=\frac{r^{2}}{n}$. Moreover if $M$ is a Riemannian manifold, then
$M$ is an Einstein manifold.
(iv) $P\cdot\mathcal{S}=0$ if and only if (4.20) holds or

$S(X,X_{2})S(X_{1},Y)-S(X,X_{1})S(X_{2},Y)+g(X_{2},Y)S^{2}(X,X_{1})-g(X,X_{2})S^{2}(X_{1},Y)=0.$

Especially, if $Q(S,R)=0$ on $M$, then $P\cdot\mathcal{R}=0$ $\Rightarrow$
$P\cdot\mathcal{S}=0$.

Now by similar technique of the proof of the Theorem 4.13 and 4.14, we get the
following theorem.

######  Theorem 4.15.

If a semi-Riemannian manifold $M$ satisfies
$P\cdot\mathcal{P}=LQ(g,\mathcal{P})$, then
(i) $(n-1)E=Ln\kappa g-(Ln^{2}+\kappa)S-S^{2}$,
(ii) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and thus
$\kappa^{(2)}=\frac{r^{2}}{n}$.
(iii)
$\frac{n}{n-1}(R\cdot\mathcal{S})=\frac{1}{n-1}Q(S,\mathcal{S})-\frac{nL}{n-1}Q(g,\mathcal{S})$
holds if (4.20) holds.
(iv) $S\wedge S=g\wedge S^{2}$.

######  Theorem 4.16.

If a semi-Riemannian manifold $M$ satisfies
$P\cdot\mathcal{P}=LQ(S,\mathcal{P})$, then
(i) $(L-1)(n-1)E=-(1+L(-1+n))\kappa S+(1+L(-1+n^{2}))S^{2}$,
(ii) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and $\kappa^{(2)}=\frac{r^{2}}{n}$
if $L+\frac{1}{n-1}\neq 0$.
(iii)
$\frac{n}{n-1}(P\cdot\mathcal{S})=\left(L-\frac{1}{(n-1)^{2}}\right)Q(S,\mathcal{S})$
holds if (4.20) holds and $L+\frac{1}{n-1}\neq 0$.
(iv) $S\wedge S=g\wedge S^{2}$.

######  Corollary 4.7.

If a semi-Riemannian manifold $M$ satisfies $P\cdot\mathcal{P}=0$, then
(i) $(n-1)E=\kappa S-S^{2}$,
(ii) $n^{2}S^{2}-2n\kappa S+\kappa^{2}g=0$ and hence
$\kappa^{(2)}=\frac{\kappa^{2}}{n}$.
(iii) $\frac{n}{n-1}(R\cdot\mathcal{S})=\frac{1}{n-1}Q(S,\mathcal{S})$ holds
if (4.20) holds.
(iv) $S\wedge S=g\wedge S^{2}$.

######  Lemma 4.1.

[37] A Roter type semi-Riemannian manifold satisfies

$2\alpha S^{2}=(\kappa\beta+2(-1+n)\gamma)g+(2\kappa\alpha+(-2+n)\beta)S.$

######  Lemma 4.2.

[37] A generalized Roter type semi-Riemannian manifold is of constant
curvature if and only if it is Einstein.

Now using Theorem 4.8, Theorem 4.9 and Lemma 4.2, we get the following
generalization of the main result (Theorem 1) of [12] for non-conformally flat
case.

######  Theorem 4.17.

On a generalized Roter type Riemannian manifold, the following conditions are
equivalent:
(i) $P\cdot\mathcal{S}=0,$ (ii) $P\cdot\mathcal{S}=LQ(g,\mathcal{S}),$
(iii) $P\cdot\mathcal{R}=0,$ (iv) $P\cdot\mathcal{R}=LQ(g,\mathcal{R})$, (v)
$P\cdot\mathcal{R}=LQ(S,\mathcal{R}),$ $L\neq\frac{-1}{n-1}$,
(vi) $P\cdot P=0,$ (vii) $P\cdot P=LQ(g,P)$, (viii) $P\cdot P=LQ(S,P),$
$L\neq\frac{-1}{n-1}$,
(ix) $P\cdot\mathcal{P}=0,$ (x) $P\cdot\mathcal{P}=LQ(g,\mathcal{P})$, (xi)
$P\cdot\mathcal{P}=LQ(S,\mathcal{P}),$ $L\neq\frac{-1}{n-1}$,
(xii) $M$ is a manifold of constant curvature, where $L$ is a scalar.

##  5\. Examples

Example 1: Let $M_{1}$ be a $4$-dimensional connected semi-Riemannian manifold
endowed with the semi-Riemannian metric

(5.1)
$ds^{2}=e^{x^{1}}(dx^{1})^{2}+e^{x^{1}}(dx^{2})^{2}+e^{x^{1}+x^{2}}(dx^{3})^{2}+(dx^{4})^{2}.$

The non-zero components (upto symmetry) of $R$, $S$, $\kappa$ and $P$ are
given by

$R_{2323}=-\frac{1}{2}e^{x^{1}+x^{2}};\ \ \ S_{22}=\frac{1}{2},\ \ \
S_{33}=\frac{e^{x^{2}}}{2};\ \ \ \kappa=e^{-x^{1}};$
$P_{1221}=-\frac{e^{x^{1}}}{6},\ \
-2P_{1331}=-P_{2323}=P_{2332}=\frac{1}{3}e^{x^{1}+x^{2}},\ \
P_{2424}=\frac{1}{6},\ \ P_{3434}=\frac{e^{x^{2}}}{6}.$

From above we can easily check the following:
(i) $R\cdot R=0$ and thus $R\cdot S=0$, $P\cdot S=0$ and $R\cdot P=0$.
(ii) $R=S\wedge S$ and thus $Q(S,R)=0$.
(iii) As here $R\cdot R=0$ and $Q(S,R)=0$ so $P\cdot R=0$ and $P\cdot
P=-\frac{1}{3}Q(S,P)$.
(iv) Although $P\cdot R=0$ and $P\cdot S=0$ but $P\cdot\mathcal{R}\neq 0$ and
also $P\cdot\mathcal{S}\neq 0$.
Note: This example ensures that on a semi-Riemannian manifold, the curvature
conditions $P\cdot R=0$ and $P\cdot\mathcal{R}=0$ give different structures.
Example 2: Let $M_{2}$ be a $4$-dimensional connected semi-Riemannian manifold
endowed with the semi-Riemannian metric

(5.2)
$ds^{2}=(1+2e^{x^{1}})\left[(dx^{1})^{2}+(dx^{2})^{2}+(dx^{3})^{2}+(dx^{4})^{2}\right].$

Then the non-zero components (upto symmetry) of $R$, $S$, $\kappa$ and $P$ are
given by

$R_{1212}=R_{1313}=R_{1414}=-\frac{e^{x^{1}}}{2e^{x^{1}}+1},\ \
R_{2323}=R_{2424}=R_{3434}=-\frac{e^{2x^{1}}}{2e^{x^{1}}+1};$
$S_{11}=\frac{3e^{x^{1}}}{\left(2e^{x^{1}}+1\right){}^{2}},\ \
S_{22}=S_{33}=S_{44}=\frac{e^{x^{1}}}{2e^{x^{1}}+1};\ \
\kappa=\frac{6e^{x^{1}}(1+e^{x^{1}})}{(1+2e^{x^{1}})^{3}};$
$\frac{1}{2}P_{1221}=\frac{1}{2}P_{1331}=\frac{1}{2}P_{1441}=P_{2323}=-P_{2332}=P_{2424}=-P_{2442}=P_{3434}=-P_{3443}=-\frac{e^{2x^{1}}-e^{x^{1}}}{6e^{x^{1}}+3}.$

Using above we can easily calculate the non-zero components (upto symmetry) of
$R\cdot R$, $Q(g,R)$, $Q(S,R)$ and $P\cdot R$ as follows:

$R\cdot R_{122313}=R\cdot R_{122414}=-R\cdot R_{132312}=R\cdot
R_{133414}=-R\cdot R_{142412}=-R\cdot
R_{143413}=\frac{e^{2x^{1}}\left(e^{x^{1}}-1\right)}{\left(2e^{x^{1}}+1\right)^{3}};$
$Q(g,R)_{122313}=Q(g,R)_{122414}=-Q(g,R)_{132312}=Q(g,R)_{133414}=-Q(g,R)_{142412}=-Q(g,R)_{143413}$
$=e^{x^{1}}\left(e^{x^{1}}-1\right);$
$Q(S,R)_{122313}=Q(S,R)_{122414}=-Q(S,R)_{132312}=Q(S,R)_{133414}=-Q(S,R)_{142412}=-Q(S,R)_{143413}$
$=\frac{e^{2x^{1}}\left(e^{x^{1}}-1\right)}{\left(2e^{x^{1}}+1\right)^{3}};$
$P\cdot R_{122313}=P\cdot R_{122414}=-P\cdot R_{132312}=P\cdot
R_{133414}=-P\cdot R_{142412}=-P\cdot
R_{143413}=\frac{2e^{2x^{1}}\left(e^{x^{1}}-1\right)}{3\left(2e^{x^{1}}+1\right)^{3}}.$

In view of above results we have the following pseudosymmetric type conditions
on $M_{2}$:
(i) $R\cdot R=\frac{e^{x^{1}}}{\left(2e^{x^{1}}+1\right)^{3}}Q(g,R)=Q(S,R)$
and hence $R\cdot S=P\cdot
S=\frac{e^{x^{1}}}{\left(2e^{x^{1}}+1\right)^{3}}Q(g,S)$
(ii) $P\cdot
R=\frac{2e^{x^{1}}}{3\left(2e^{x^{1}}+1\right)^{3}}Q(g,R)=\frac{2}{3}Q(S,R)$
(iii) $R\cdot
R=LQ(g,R)+\left[1-Le^{-x^{1}}\left(2e^{x^{1}}+1\right)^{3}\right]Q(S,R)$, $L$
being arbitrary scalar on $M$.
Example 3: Let $M_{3}$ be a $5$-dimensional connected semi-Riemannian manifold
endowed with the semi-Riemannian metric

(5.3)
$ds^{2}=a(dx^{1})^{2}+e^{2x^{2}}(x^{4})^{2}(dx^{2})^{2}+2e^{2x^{2}}dx^{2}dx^{3}+e^{2x^{2}}(dx^{4})^{2}+(e^{2x^{2}}f)(dx^{5})^{2},$

where $a$ is a positive constant and $f$ is a positive function of $x^{2}$.
The non-zero components (upto symmetry) of $R$, $S$, $\kappa$ and $P$ are
given by

$-R_{1212}=e^{2x^{1}}(x^{4})^{2},\ \ -R_{1213}=-R_{1414}=e^{2x^{1}},\ \
-R_{1515}=fe^{2x^{1}},\ \ R_{2323}=-R_{2434}=\frac{e^{4x^{1}}}{a},$
$-R_{2424}=\frac{e^{2x^{1}}\left(a+e^{2x^{1}}(x^{4})^{2}\right)}{a},\ \
-R_{2525}=\frac{e^{2x^{1}}\left(2aff^{\prime\prime}-a\left(f^{\prime}\right)^{2}+4f^{2}e^{2x^{1}}(x^{4})^{2}\right)}{4af},$
$-R_{2535}=-R_{4545}=\frac{fe^{4x^{1}}}{a};$ $S_{11}=4,\ \
S_{22}=\frac{2aff^{\prime\prime}-a\left(f^{\prime}\right)^{2}+4af^{2}+16f^{2}e^{2x^{1}}(x^{4})^{2}}{4af^{2}},\
\ S_{23}=S_{44}=\frac{4e^{2x^{1}}}{a},\ \ S_{55}=\frac{4fe^{2x^{1}}}{a};$
$\kappa=\frac{20}{a};\ \
P_{1221}=-\frac{a\left(2ff^{\prime\prime}-\left(f^{\prime}\right)^{2}+4f^{2}\right)}{16f^{2}},\
\
P_{2322}=\frac{e^{2x^{1}}\left(2ff^{\prime\prime}-\left(f^{\prime}\right)^{2}+4f^{2}\right)}{16f^{2}},$
$P_{2424}=-\frac{e^{2x^{1}}\left(-2ff^{\prime\prime}+\left(f^{\prime}\right)^{2}+12f^{2}\right)}{16f^{2}},$
$P_{2442}=e^{2x^{1}},\ \
P_{2525}=\frac{e^{2x^{1}}\left(-6ff^{\prime\prime}+3\left(f^{\prime}\right)^{2}+4f^{2}\right)}{16f},\
\
P_{2552}=\frac{e^{2x^{1}}\left(2ff^{\prime\prime}-\left(f^{\prime}\right)^{2}\right)}{4f}.$

The non-zero components (upto symmetry) of $R\cdot R$ and $Q(g,R)$ are given
by

$R\cdot R_{122414}=-R\cdot R_{142412}=e^{2x^{1}},\ \ R\cdot R_{122515}=-R\cdot
R_{152512}=\frac{e^{2x^{1}}\left(2ff^{\prime\prime}-\left(f^{\prime}\right)^{2}\right)}{4f}$
$-2R\cdot R_{232424}=R\cdot R_{242423}=\frac{2e^{4x^{1}}}{a},\ \ -R\cdot
R_{232525}=\frac{1}{2}R\cdot R_{252523}=R\cdot
R_{254524}=\frac{e^{4x^{1}}\left(2ff^{\prime\prime}-\left(f^{\prime}\right)^{2}\right)}{4af}$
$R\cdot
R_{242545}=\frac{e^{4x^{1}}\left(-2ff^{\prime\prime}+\left(f^{\prime}\right)^{2}+4f^{2}\right)}{4af},\
\ R\cdot R_{244525}=-\frac{fe^{4x^{1}}}{a},$
$Q(g,R)_{122414}=-Q(g,R)_{142412}=ae^{2x^{1}},\ \
Q(g,R)_{244525}=-fe^{4x^{1}},$
$Q(g,R)_{122515}=-Q(g,R)_{152512}=\frac{ae^{2x^{1}}\left(2ff^{\prime\prime}-\left(f^{\prime}\right)^{2}\right)}{4f},$
$-2Q(g,R)_{232424}=Q(g,R)_{242423}=2e^{4x^{1}},\ \
Q(g,R)_{242545}=\frac{e^{4x^{1}}\left(-2ff^{\prime\prime}+4f^{2}+\left(f^{\prime}\right)^{2}\right)}{4f},$
$-Q(g,R)_{232525}=\frac{1}{2}Q(g,R)_{252523}=Q(g,R)_{254524}=\frac{e^{4x^{1}}\left(2ff^{\prime\prime}-\left(f^{\prime}\right)^{2}\right)}{4f}.$

It can be now easily checked that $P\cdot\mathcal{S}=0$. From the components
of above tensors, we see that the semi-Riemannian manifold $M_{3}$ satisfies
the following geometric structures:
(i) It is a Venzi’s $P$-space satisfying

$\Pi(X_{1})P(X_{2},X_{3},X,Y)+\Pi(X_{2})P(X_{3},X_{1},X,Y)+\Pi(X_{3})P(X_{1},X_{2},X,Y)=0$

for the null 1-form $\Pi=\\{0,1,0,0,0\\}$. Here $W\cdot W=0$ and hence
improperly satisfies the condition $W\cdot W=LQ(nS-\kappa g,W)$, which
supports the Theorem 4.3.
(ii) $R\cdot R=\frac{1}{a}Q(g,R)$, i.e., pseudosymmetric manifold of constant
type and thus Ricci, conformally, projectively, concircularly and
conharmonically pseudosymmetric manifoold of constant type.
(iii) $P\cdot S=\frac{1}{a}Q(g,S)\neq 0$ but $P\cdot\mathcal{S}=0$.
Example 4: Let $M_{4}$ be a $4$-dimensional connected semi-Riemannian manifold
endowed with the semi-Riemannian metric

(5.4)
$ds^{2}=x^{1}x^{3}(dx^{1})^{2}+2dx^{1}dx^{2}+(2+dx^{1})^{2}dx^{3}+(x^{1})^{3}(dx^{4})^{2}.$

The non-zero components (upto symmetry) of $R$, $S$, $\kappa$ and $P$ are
given by

$R_{1414}=-\frac{3x^{1}}{4};\ \ \ S_{11}=\frac{3}{4(x^{1})^{2}};\ \ \kappa=0;$
$P_{1211}=\frac{1}{4(x^{1})^{2}},\ \
P_{1313}=\frac{\left(x^{1}+2\right)^{2}}{4(x^{1})^{2}},\ \
P_{1441}=-\frac{3}{2}P_{1414}=\frac{3x^{1}}{4}.$

From the components of above tensors it is easy to check that $R\cdot R=0$,
$R\cdot S=0$, $Q(S,R)=0$, $P\cdot R=0$ and also $P\cdot\mathcal{S}=0$.
According to the Theorem 4.5, if the manifold is of non-constant scalar
curvature, then $P\cdot S=0$ and $P\cdot\mathcal{S}=0$ holds simultaneously,
then the manifold is Einstein. This example also supports the Theorem 4.8.

##  6\. Conclusions

In the present paper we study the basic properties of the projective operator
and calculate the necessary and sufficient conditions for a semi-Riemannian
manifold to satisfy some Walker type conditions. It is shown that the
projective operator commutes with contraction if and only if the manifold is
Einstein. A necessary and sufficient condition for a semisymmetric (resp.,
pseudosymmetric) manifold due to projective curvature tensor to be a Ricci
semisymmetric (resp., pseudosymmetric) manifold due to projective curvature
tensor is presented. It is also shown that $P\cdot\mathcal{R}=0$ and $P\cdot
R=0$ (resp., $P\cdot\mathcal{R}=LQ(g,\mathcal{R})$ and $P\cdot R=LQ(g,R)$)
give different structures. We have evaluated some pseudosymmetric type
condition due to projective curvature tensor under certain condition and
showed that a $P$-space by Venzi is either $W$-space by Venzi or a manifold of
constant curvature. We obtain the curvature properties of of various
semisymmetric type and pseudosymmetric type curvature restricted geometric
structures due to projective curvature tensor, such as (i) $P\cdot R=0,$ (ii)
$P\cdot R=LQ(g,R)$, (iii) $P\cdot S=0,$ (iv) $P\cdot S=LQ(g,S)$, (v)
$P\cdot\mathcal{S}=0,$ (vi) $P\cdot\mathcal{S}=LQ(g,\mathcal{S})$, (vii)
$P\cdot\mathcal{R}=0$, (viii) $P\cdot\mathcal{R}=LQ(g,\mathcal{R})$, (ix)
$P\cdot\mathcal{R}=LQ(S,\mathcal{R})$, (x) $P\cdot P=0,$ (xi) $P\cdot
P=LQ(g,P)$, (xii) $P\cdot P=LQ(S,P)$, (xiii) $P\cdot\mathcal{P}=0$, (xiv)
$P\cdot\mathcal{P}=LQ(g,\mathcal{P})$ and (xv)
$P\cdot\mathcal{P}=LQ(S,\mathcal{P})$ on a Riemannian as well as semi-
Riemannian manifold. It is shown that a Riemannian manifold $M$ with one of
the curvature condition (v)-(xv) reduce to a Einstein manifold and hence
manifold of constant curvature if $M$ is generalized Roter type.

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