# Characterization on N(k)-Mixed Quasi-Einstein Manifold.

1. Introduction

A Riemannian manifold (M, g) with dimension (n [greater than or equal to] 2) is said to be an Einstein manifold if the Ricci tensor satisfies the condition S(X, Y) = [r/n] g(X, Y), holds on M, here S and r denote the Ricci tensor and the scalar curvature of (M, g) respectively. According to [8] the above equation is called the Einstein metric condition. Einstein manifolds play an important role in Riemannian Geometry, as well as in general theory of relativity. The notion of quasi-Einstien manifold were defined in ([3],[7],[10],[14],[16],[17]). A non-flat Riemannian manifold (M, g), (n [greater than or equal to] 2) is said to be an quasi Einstein manifold if its Ricci tensor S of type (0, 2) satisfies the condition

S(X, Y)=ag(X,Y)+bA(X)A(Y) (1.1)

where a and b are scalars of which b [not equal to] 0 and A is non-zero 1-form such that g(X, U) = A(X) for all vector field X and U is a unit vector field.

The notion of quasi Einstein manifold was introduced in a paper [10] by M.C.Chaki and R.K.Maity. According to them a non-flat Riemannian manifold ([M.sup.n], g),(n [greater than or equal to] 3) is defined to be a quasi Einstein manifold if its Ricci tensor S of type (0, 2) satisfies the condition

S(X, Y) = ag(X, Y) + bA(X)A(Y) (1.2)

and is not identically zero, where a, b are scalars b [not equal to] 0 and A is a non-zero 1-form such that

g(X, U) = A(X), [for all]X [member of] TM. (1.3)

U being a unit vector field.

In such a case a, b are called the associated scalars. A is called the associated 1-form and U is called the generator of the manifold. Such an n-dimensional manifold is denoted by the symbol[(QE).sub.n].

Again, U.C.De and G.C.Ghosh defined generalized quasi Einstein manifold [12]. A non-flat Riemannian manifold is called a generalized quasi Einstein manifold ([2], [4], [5], [6], [9], [15], [18])if its Ricci-tensor S of type (0, 2) is non-zero and satisfies the condition

S(X, Y) = ag(X, Y) + bA(X)A(Y) + cB(X)B(Y) (1.4)

where a, b, c, are non-zero scalars and A, B are two 1-forms such that

g(X, U) = A(X) and g(X, V) = B(X) (1.5)

U, V being unit vectors which are orthogonal, i.e,

g(U, V) = 0. (1.6)

The vector fields U and V are called the generators of the manifold. This type of manifold are denoted by G[(QE).sub.n].

The k-nullity distribution [29] of a Riemannian manifold M is defined by

N(k) : p [right arrow] [N.sub.p](k) = {Z E [T.sub.p]M \ R(X, Y)Z = k(g(Y, Z)X - g(X, Z)Y)}. (1.7)

for all X,Y E TM and k is a smooth function. M.M.Tripathi and Jeong jik kim [28] introduced the notion of N(k)-quasi Einstein manifold which defined as follows: If the generator U belongs to the k-nullity distribution N(k), then a quasi Einstein manifold ([M.sup.n],g) is called N(k)-quasi Einstein manifold.

In [26], H.G. Nagaraja introduced the concept of N(k)-mixed quasi Einstein manifold and mixed quasi constant curvature.A non flat Riemannian manifold ([M.sup.n], g) is called a N(k)-mixed quasi Einstein manifold [11] if its Ricci tensor of type (0, 2) is non zero and satisfies the condition

S(X, Y) = ag(X, Y) + bA(X)B(Y) + cB(X)A(Y), (1.8)

where a, b, c, are smooth functions and A, B are non zero 1-forms such that

g(X, U) = A(X) and g(X, V) = B(X) V X, (1.9)

U, V being the orthogonal unit vector fields called generators of the manifold be-long to N(k).Such aa manifold is denoted by the symbol N(k) - [(MQE).sub.n]. Again a Riemannian manifold ([M.sup.n],g) is called of mixed quasi constant curvature [11] if it is conformally flat and curvature tensor 'R of type (0, 4) satisfies the condition

'R(X, Y, Z, W) = p[g(Y, Z)g(X, W) - g(X, Z)g(Y, W)] +q[g(X, W)A(Y)B(Z) - g(X, Z)A(Y)B(W) +g(X, W)A(Z)B(Y) - g(X, Z)A(W)B(Y)] +s[g(Y, Z)A(W)B(X) - g(Y, W)A(Z)B(X) +g(Y, Z)A(X)B(W) - g(Y, W)A(X)B(Z)]. (1.10)

Let M be an m-dimensional, m [greater than or equal to] 3, Riemannian manifold and p E M. Denote by K([pi]) or K(U [and] V) the sectional curvature of M associated with a plane section [pi] [subset.bar] [T.sub.p]M, where {U, V} is an orthonormal basis of [pi]. For a ndimensional subspace L [subset.bar] [T.sub.p]M, 2 [less than or equal to] n [less than or equal to] m, its scalar cuvrvature [tau](L) is denoted by [mathematical expression not reproducible], where {[e.sub.1],[e.sub.2],...,[e.sub.n]} is any orthonormal basis of

L([14]).

In [13] the result for odd dimensional Einstein spaces was obtained by Dumitru. Also in [7] Bejan generalized these results (both odd and even dimensions)to quasi Einstein manifold. Also characterization of super quasi-Einstein manifold for both of odd and even dimensions was studied in [20]. From above studies, we have given characterization of N(k)-mixed quasi-Einstein manifold for both of odd and even dimensions. Next we obtain that a N(k)-mixed quasi-Einstein manifold is semi mixed quasi-Einstein manifold if either of generators is parallel vector field.

Geodesic mappings of Einstein spaces were studied in ([19],[21],[22],[23],[24],[25]).

2. Characterization of N(k)-mixed quasi-Einstein manifold manifold

In this section we establish the characterization of odd and even dimensional N(k) - [(MQE).sub.n].

Theorem 2.1. A Riemannian manifold of dimension (2n + 1) with n [greater than or equal to] 2 is N(k)-mixed quasi-Einstein manifold if and only if the Ricci operator Q has eigen vector fields U and V such that at any point p [member of] M, there exist three real numbers [alpha] and [beta] satisfying

[tau](P) + [alpha] = [tau]([P.sup.[perpendicular to]]); U, V [member of] [T.sub.p][P.sup.[perpendicular to]], [tau](N) + [alpha] = [tau]([N.sup.[perpendicular to]]); U [member of] [T.sub.p]N, V [member of] [T.sub.p][N.sup.[perpendicular to]], [tau](R) + [beta] = [tau]([R.sup.[perpendicular to]]); U [member of] [T.sub.p]R, V [member of] [T.sub.p][R.sup.[perpendicular to]],

for any n-plane sections P, N and (n+ 1)-plane section R where [P.sup.[perpendicular to]], [N.sup.[perpendicular to]] and [R.sup.[perpendicular to]] denote the orthogonal complements of P, N and R in [T.sub.p]M respectively and

[alpha] = a/2, [beta] = - c/2,

where a, b, c are scalars.

Proof. First suppose that M is a (2n+1) dimensional N(k)-mixed quasi-Einstein manifold, so

S(X, Y) = ag(X, Y) + bA(X)B(Y) + cA(Y)B(X), (2.1)

where a,b,c are scalars such that b, c are nonzero and A, B are two nonzero 1-forms such that g(X, U) = A(X) and g(X, V) = B(X), [for all]X G X(M), U, V being unit vectors which are orthogonal, i.e., g(U,V) = 0.

Let P [subset.bar] [T.sub.p]M be an n-dimensional plane orthogonal to U, V and let {[e.sub.1], [e.sub.2],..., [e.sub.n]} be orthonormal basis of it. Since U and V are orthogonal to P, we can take orthonormal basis {[e.sub.n+1], [e.sub.n+2],..., [e.sub.2n+1]} of [P.sup.[perpendicular to]] such that [e.sub.2n] = U and [e.sub.2n+1] = V. Thus {[e.sub.1], [e.sub.2],..., [e.sub.n], [e.sub.n+1], [e.sub.n+2],..., [e.sub.2n+1]} is an orthonormal basis of [T.sub.p]M.Then we can take X = Y = ei in (2.1), we have

[mathematical expression not reproducible]

By use of (2.1) for any 1 [less than or equal to] i [less than or equal to] 2n + 1, we can write

S([e.sub.1], [e.sub.1]) = K([e.sub.1] [and] [e.sub.2])+K([e.sub.1][and][e.sub.3]) + ... +K([e.sub.1][and][e.sub.2n-1])+K([e.sub.1][and]U) + K ([e.sub.1] [and]V) = a, S([e.sub.2], [e.sub.2]) = K([e.sub.2] [and] [e.sub.1]) + K([e.sub.2] [and][e.sub.3]) + ...+K([e.sub.2][and][e.sub.2n-1])+K([e.sub.2][and]U)+K([e.sub.2][and]V) = a, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S ([e.sub.2n-1], [e.sub.2n-1]) K ([e.sub.2n-1][and][e.sub.1])+K ([e.sub.2n-1][and][e.sub.2])+K ([e.sub.2n-1][and] [e.sub.3])+...+K ([e.sub.2n-1][and] V) a, S(U, U) = K(U [and] [e.sub.1]) + K(U [and] [e.sub.2]) + ... + K(U [and] [e.sub.2n-1]) + K(U [and] V) = a, S(V, V) = K(V [and] [e.sub.1]) + K(V [and] [e.sub.2]) + ... + K(V [and] [e.sub.2n-1]) + K(V [and]U) = a.

Adding first n-equations, we get

[mathematical expression not reproducible] (2.2)

Then adding the last (n + 1)-equations, we have

[mathematical expression not reproducible] (2.3)

Then, by substracting the equation (2.2) and (2.3), we obtain

[tau]([P.sup.[perpendicular to]]) - [tau](P) = a/2.

Therefore [tau](P) + [alpha] = [tau]([P.sup.[perpendicular to]]), where,

[alpha] = a/2.

Similarly, Let N [subset.bar] [T.sub.p]M be an n-dimensional plane orthogonal to V and let {[e.sub.1], [e.sub.2], . . ., [e.sub.n]} be orthonormal basis of it. Since V is orthogonal to N, we can take an orthonormal basis {[e.sub.n+1], [e.sub.n+2], . . . ,[e.sub.2n+1]} of [N.sup.[perpendicular to]] orthogonal to U, such that [e.sub.n] = U and [e.sub.2n+1] = V, respectively. Thus, {[e.sub.1], [e.sub.2], . . ., [e.sub.n], [e.sub.n+1], [e.sub.n+2], . . ., [e.sub.2n+1]} is an orthonormal basis of [T.sub.p]M. Then we can take X = Y = ei in (2.1) to have

[mathematical expression not reproducible]

Adding first n-equations, we get

[mathematical expression not reproducible] (2.4)

and adding the last (n + 1)-equations, we have

[mathematical expression not reproducible] (2.5)

Then, by substracting the equation (2.4) and (2.5), we obtain

[tau]([N.sup.[perpendicular to]]) - [tau](N) = a/2.

Therefore [tau](N) + [alpha] = [tau]([N.sup.[perpendicular to]]), where,

[alpha] = a/2.

Analogously, Let R [subset.bar] [T.sub.p]M be an (n+1)-plane orthogonal to V and let {[e.sub.1], [e.sub.2], . . ., [e.sub.n+1]} be orthonormal basis of it. Since V is orthogonal to R, we can take an orthonormal basis {[e.sub.n+2], [e.sub.n+3],..., [e.sub.2n], [e.sub.2n+1]} of [R.sup.[perpendicular to]] orthogonal to U, such that [e.sub.n+1] = U and [e.sub.2n+1] = V. Thus, {[e.sub.1], [e.sub.2], . . ., [e.sub.n], [e.sub.n+1], [e.sub.n+2], . . ., [e.sub.2n+1]} is an orthonormal basis of [T.sub.p]M. Then we can take X = Y = [e.sub.i] in (2.1) to have

[mathematical expression not reproducible]

Adding the first (n + 1)-equations, we get

[mathematical expression not reproducible] (2.6)

and adding the last n-equations, we have

[mathematical expression not reproducible] (2.7)

Then, by substracting the equation (2.6) and (2.7), we obtain

[tau]([R.sup.[perpendicular to]]) - [tau](R) = -a/2.

Therefore [tau](R) + [beta] = [tau]([R.sup.[perpendicular to]]), where,

[beta] = -a/2.

Conversely, let [V.sub.1] be an arbitrary unit vector of [T.sub.p]M, at p [member of] M, orthogonal to U and V. We take an orthonormal basis {[e.sub.1], [e.sub.2], . . ., [e.sub.n], [e.sub.n+1], [e.sub.n+2], . . ., [e.sub.2n+1]} of [T.sub.p]M such that [V.sub.1] = [e.sub.1], [e.sub.n+1] = U and [e.sub.2n+1] = V. We consider n-plane section N and (n + 1)-plane section R in [T.sub.p]M as follows N = span {[e.sub.2], . . ., [e.sub.n], [e.sub.n+1]} and R = span {[e.sub.1], [e.sub.2], . . ., [e.sub.n], [e.sub.n+1]} respectively.Then we have [N.sup.[perpendicular to]] = span {[e.sub.1], [e.sub.n+]2, . . ., [e.sub.2]n, [e.sub.2n+1]} and R = span {[e.sub.n+2]. . . ., [e.sub.2n]} respectively. Now

[mathematical expression not reproducible]

Therefore, S([V.sub.1],[V.sub.1]) = [alpha] - [beta], for any unit vector [V.sub.1] [member of] [T.sub.p]M, orthogonal to U and V. Then we can write for any 1 [less than or equal to] i [less than or equal to] 2n + 1, S([e.sub.i],[e.sub.i]) = [alpha] - [beta], since S([V.sub.1],[V.sub.1]) = ([alpha] - [beta])g([V.sub.1], [V.sub.1]). It follows that S(X, X) = ([alpha] - [beta])g(X,X) and S(Y,Y) = ([alpha] -[beta])g(Y,Y)+ [K.sub.1]A(Y)B(Y)+[K.sub.2]B(Y)A(Y) for any X [member of] [[span{U}].sup.[perpendicular to]] and Y [member of] [[span{V}].sup.[perpendicular to]], where A, B are the dual forms of U and V with respect to g, respectively and [K.sub.1], [K.sub.2] are scalars, such that [K.sub.1] [not equal to] 0, [K.sub.2] [not equal to] 0.

Now from the above equations, we get from symmetry that S with tensors ([alpha]-[beta])g and ([alpha]-[beta])+[K.sub.1](A[cross product]B)+[K.sub.2](A[cross product]B) must coincide on the complement of U and V, respectively, that is S(X, Y) = ([alpha] - [beta])g(X, Y) + [K.sub.1]A(X)B(Y) + [K.sub.2]B(X)A(Y), for any X,Y [member of] [[span{U, V}].sup.[perpendicular to]] . Since U and V are eigenvector fields of Q, we also have S(X, U) = 0 and S(Y, V) = 0 for any X,Y [member of] [T.sub.p]M orthogonal to U and V. Thus, we can extend the above equation to

S(X, Z) = ([alpha] - [beta])g(X, Z) + [K.sub.1]A(X)B(Z) + [K.sub.2]A(Z)B(X) (2.8)

, for any X [member of] [[span{U, V}].sup.[perpendicular to]] and Z [member of] [T.sub.p]M, where [K.sub.1], [K.sub.2], are scalars and [K.sub.1] = 0, [K.sub.2] [not equal to] 0,. Now, let us consider the n-plane section P and (n + 1)-plane section R in [T.sub.p]M as follows P = span {[e.sub.1], [e.sub.2], . . ., [e.sub.n]} and R = span {[e.sub.1], [e.sub.2], . . ., [e.sub.n], U}. Then we have [P.sup.[perpendicular to]] = span {U, [e.sub.n+2], . . ., [e.sub.2n+1]} and [R.sup.[perpendicular to]] = span {[e.sub.n+2], . . ., [e.sub.2n], [e.sub.2n+1]} respectively. Now

[mathematical expression not reproducible].

Therefore we can write

S(U, U) = ([alpha] - [beta])g(U, U). (2.9)

Analogously, let us consider the n-plane section P and N [member of] [T.sub.p]M as follows P = span {[e.sub.1], [e.sub.2], . . ., [e.sub.n]} and N = span {[e.sub.n+1], [e.sub.n+2], . . ., [e.sub.2n]} respectively. Then we have [P.sup.[perpendicular to]] = span {[e.sub.n+1], [e.sub.n+2], . . . , [e.sub.2n], V} and [N.sup.[perpendicular to]] = span {[e.sub.1],..., [e.sub.n], V} re-spectively. Now, we have

[mathematical expression not reproducible]

Then, we get

S(V, V) = 2[alpha]g(V, V) + [K.sub.1]A(V)B(V) + [K.sub.2]A(V)B(V). (2.10)

Now from (2.8), (2.9) and (2.10) we can write the Ricci tensor by

S(X, Y) = [[lambda].sub.1]g(X, Y)+[K.sub.1]A(X)B(Y)+[K.sub.2]B(X)A(Y), (2.11)

for any X, Y [member of] [T.sub.p]M. From (2.11) it follows that M is a N(k)-mixed quasi-Einstein manifold, where [lambda]1,[K.sub.1],[K.sub.2], are scalars and [K.sub.1] = 0,[K.sub.2] = 0,. Hence the theorem is proved.

Theorem 2.2. A Riemannian manifold of dimension 2n with n [greater than or equal to] 2 is N(k)-mixed quasi-Einstein manifold if and only if the Ricci operator Q has eigen vector fields U and V such that at any point p G M, satisfying

[tau](P) = [tau]([P.sup.[perpendicular to]]); U, V [member of] [T.sub.p][P.sup.[perpendicular to]], [tau](N) = [tau]([N.sup.[perpendicular to]]); U [member of] [T.sub.p]N, V [member of] [T.sub.p][N.sup.[perpendicular to]], [tau](R) = [tau]([R.sup.[perpendicular to]]); U [member of] [T.sub.p]R, V [member of] [T.sub.p][R.sup.[perpendicular to]],

for any n-plane section P, N and (n + 1)-plane section R where [P.sup.[perpendicular to]], [N.sup.[perpendicular to]] and [R.sup.[perpendicular to]] denote the orthogonal complements of P, N and R in [T.sub.p]M respectively.

Proof. Let P and R be n-plane sections and N be an (n - 1)-plane section such that, P = span{[e.sub.1], [e.sub.2], . . ., [e.sub.n]}, R = span{[e.sub.n+1], [e.sub.n+2], . . ., [e.sub.2n]} and N = span{[e.sub.2], [e.sub.3], . . ., [e.sub.n]} respectively. Therefore the orthogonal complements of these sections can be written as [P.sup.[perpendicular to]] = span{[e.sub.n+1], [e.sub.n+2], . . ., [e.sub.2n]}, [R.sup.[perpendicular to]] = span{[e.sub.1], [e.sub.2], . . ., [e.sub.n]} and [N.sup.[perpendicular to]] = span{[e.sub.1], [e.sub.n+1],..., [e.sub.2n]}.

Then rest of the proof is similar to the proof of Theorem 2.1.

3. N(k) - [MQE.sub.n] with the parallel vector field generators

Theorem 3.1. A N(k)- mixed quasi-Einstein manifold is semi-mixed quasi-Einstein manifold if either of generators is parallel vector field.

Proof. By the definition of the Riemannian curvature tensor, if U is parallel vector field, then we find that

R(X, Y)U = [[nabla].sub.X][[nabla].sub.Y] U - [[nabla].sub.Y] [[nabla].sub.X]U - [[nabla].sub.[X,Y]]U = 0,

and consequently we get

S(X, U) = 0. (3.1)

Again, putting Y = U in the equation (1.8) and applying (1.9) , we have

S(X, U) = ag(X, U) + cg(X, V).

So, if U is a parallel vector field, by (3.1), we get

ag(X, U) + cg(X, V) = 0. (3.2)

Now, putting X = V in the equation (3.2) and using (1.9) we get c = 0. So, if U is parallel vector field in a mixed-quasi-Einstein manifold, then the manifold is semi-mixed quasi Einstein manifold.

Again, if V is parallel vector field, then R(X, Y)V = 0. Contracting, we get

S(Y, V) = 0. (3.3)

Putting X = V in the equation (1.8) and applying (1.9), we obtain

S(Y, V) = ag(Y, V) + bg(Y, U).

If, V is a parallel vector field, by (3.3), we get

ag(Y,V)+bg(Y,U)=0. (3.4)

Putting Y = U and using (3.4), (1.9), we obtain b = 0, i.e., the manifold is semi mixed quasi-Einstein manifold.

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Dipankar Debnath

Department of Mathematics, Bamanpukur High School, Bamanpukur,PO-Sree Mayapur, West Bengal,India,PIN-741313 Email:dipankardebnat[h.sub.1]23@gmail.com

Arindam Bhattacharyya

Department of Mathematics Jadavpur University, Kolkata-700032, India. E-mail: bhattachar1968@yahoo.co.in

Received May 26, 2017 Accepted December 14, 2017
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