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UDK 514.764.22

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A STUDY ON GENERALIZED RICCI 2-RECURRENT SPACES

U. C. De and S. Bandyopadhyay

Abstract. The object of the present paper is to study some properties of generalized Ricci 2-recurrent spaces. At first it is proved that every 3-dimensional generalized Ricci 2-recurrent space is a generalized 2-recurrent space. In section 3, it is shown that for such a space Ricciprincipal invariant is 1/2R. In section 4 we find a necessary condition for such a space to be a Ricci-recurrent space. Next it is proved that a conformally symmetric Ricci 2-recurrent space is a generalized 2-recurrent space and a conformally symmetric generalized Ricci 2-recurrent space with definite metric and zero scalar curvature can not exist. Lastly an example of a generalized Ricci 2-recurrent space is also constructed.

1. Preliminaries

A non flat Riemanian space V_{n} (n > 3) is called a generalized 2-recurrent space [1] if its curvature tensor satisfies

R_{hijk,lm} = λ_{m}R_{hijk,l} + a_{lm}R_{hijk}  (1.1)

where a_{lm} is non-zero and a comma denotes covariant differentiation with respect to the metric tensor g_{ij}. λ_{m} and a_{lm} are called its vector and tensor of recurrence. Such a space has been denoted by G(2k_{n}). In generalizing this concept we intend to study Riemannian space whose Ricci tensor is non-zero and satisfies a relation of the form

R_{ij,lm} = λ_{m}R_{ij,l} + a_{lm}R_{ij}  (1.2)

where λ_{m} and a_{lm} have the same meaning as before. Such a space shall be called a generalized Ricci 2-recurrent space and will be denoted by G(2R_{n}). If in particular, λ_{m} = 0, then the space reduces to a Ricci 2-recurrent space introduced by Chaki and Roychowdhary [2]. In 1952, Patterson [3] introduced a type of Riemannian space V_{n} (n ⩾ 3) the Ricci tensor of which satisfies R_{ij,k} = λ_{k}R_{ij} and R_{ij} ≠ 0 for some non-zero vector λ_{k}. He called such a space Ricci-recurrent and denoted an n-dimensional space of this kind by R_{n}. Now from (1.1) and (1.2) it is easily seen that every G(2k_{n}) is a G(2R_{n}), but the converse is not in general true. Here we prove that every G(2R_{3}) is a G(2k_{3}). According to Chaki and Gupta [4], an n-dimensional (n > 3) Riemannian space is called conformally symmetric if its Weyl’s conformal curvature tensor

AMS Subject Classification: 53 C 20

C_{ijk}^{h} = R_{ijk}^{h} − \frac{1}{n − 2}(g_{ij}R_{k}^{h} −g_{ik}R_{j,}^{h} +δ_{k}^{h}R_{ij} −δ_{j,}^{h}R_{ik})+ \frac{R}{(n − 1)(n − 2)}(δ_{k}^{h}g_{ij} −δ_{j,}^{h}g_{ik})

(1.3)

satisfies

C_{ijk,l}^{h} = 0  (1.4)

where R is the scalar curvature.

In the present paper we consider G(2R_{n}) for n > 3.

2. 3-dimensional generalized Ricci 2-recurrent space

It is known [5] that for a V_{3}

R_{hijk} = g_{hk}π_{ij} − g_{hj}π_{ik} + g_{ij}π_{hk} − g_{ik}π_{hj}  (2.1)

where

π_{ij} = (R_{ij} − \frac{R}{4}g_{ij}).  (2.2)

Now for a G(2R_{3}) we have

R_{ij,lm} = λ_{m}R_{ij,l} + a_{lm}R_{ij}.  (2.3)

Transvecting (2.3) with g^{ij} we get

R_{,lm} = λ_{m}R_{,l} + a_{lm}R.  (2.4)

From (2.2) we have by virtue of (2.3) and (2.4)

π_{ij,lm}−λ_{m}π_{ij,l}−a_{lm}π_{ij} = R_{ij,lm}−λ_{m}R_{ij,l}−a_{lm}R_{ij} −(R_{,lm}−λ_{m}R_{,l}−a_{lm}R)\frac{g_{ij}}{4} = 0

or π_{ij,lm} = λ_{m}π_{ij,l} + a_{lm}π_{ij}. Therefore from (2.1) it follows that

R_{hijk,lm} = λ_{m}R_{hijk,l} + a_{lm}R_{hijk}.

Thus we can state the following theorem:

Theorem 1. Every G(2R_{3}) is a G(2k_{3}).

3. Tensor of recurrence and Ricci principal invariant in a G(2R_{n}) with non-zero scalar curvature

We see from (2.4) that if R is constant, then R = 0 for a_{lm} ≠ 0. Again from (2.4)

λ_{m}R_{,l} − λ_{l}R_{,m} + (a_{lm} − a_{ml})R = R_{,lm} − R_{,ml} = 0.

Hence if a_{lm} is symmetric, then λ_{m}, R_{,l} are co-directional.

From Bianchi identity we get

R_{ijk,h}^{h} + R_{ik,j} − R_{ij,k} = 0.  (3.1)

A study on generalized Ricci 2-recurrent spaces 49

Covariant differentiation of (3.1) gives R_{ijk,hm}^{h} + R_{ik,jm} − R_{ij,km} = 0. Now by virtue of (1.2)

R_{ijk,hm}^{h} = a_{km}R_{ij} − a_{jm}R_{ik} + λ_{m}(R_{ij,k} − R_{ik,j}).  (3.2)

Trnsvecting (3.2) with g^{ij} and using the formula R_{i,r}^{r} = \frac{1}{2}R_{,i} we obtain

\frac{1}{2}R_{,km} = a_{km}R − a_{jm}R_{k}^{j} + \frac{λ_{m}}{2} R_{,k}  (3.3)

whence

a_{jm}R_{k}^{j} = \frac{1}{2}a_{km}R.  (3.4)

Now by the similar argument as in [2] we get the following theorem:

Theorem 2. In a G(2R_{n}) with non-zero scalar curvature the tensor of recurrence a_{lm} is not symmetric in general and its rank is less than n. Also a_{lm} is symmetric if and only if λ_{m} and R_{,l} are co-directional. Further, for such a space, one Ricci principal invariant is \frac{1}{2}R.

4. G(2R_{n}) (R ≠ 0) of definite metric In this section we consider a G(2R_{n}) with non-zero scalar curvature for which

R^{ij}R_{ij} = \frac{1}{2}R^{2}  (4.1)

holds. Then from (4.1) it follows 2R^{ij}R_{ij,l} = RR_{,l}. Differentiating both sides of the previous equation covariantly, we get

R_{,m}^{ij}R_{ij,l} + R^{ij}R_{ij,lm} = \frac{1}{2}R_{,l}R_{,m} + \frac{1}{2}RR_{,lm}.  (4.2)

But

R^{ij}R_{ij,lm} = R^{ij}(a_{lm}R_{ij} + λ_{m}R_{ij,l}) = a_{lm}R^{ij}R_{ij} + λ_{m}R^{ij}R_{ij,l}

= \frac{1}{2}a_{lm}R^{2} + \frac{1}{2}λ_{m}RR_{,l} = \frac{1}{2}R(a_{lm}R + λ_{m}R_{,l}) = \frac{1}{2}RR_{,lm}.

By virtue of this, (4.2) reduces to R_{,m}^{ij}R_{ij,l} = \frac{1}{2}R_{,l}R_{,m}. Put S_{ijk} = R_{ij,k} − λ̄_{k}R_{ij}, where λ̄_{k} = R_{,k}/R. Then

S^{ijk}S_{ijk} = g^{mk}R_{,m}^{ij}R_{ij,k} − λ̄_{m}g^{mk}R^{ij}R_{ij,k} − λ_{k}g^{mk}R^{hl}R_{hl,m} + g^{mk}λ̄_{m}λ̄_{k}R^{ij}R_{ij}
= \frac{1}{2}g^{mk}R_{,m}R_{,k} − g^{mk}λ̄_{m}RR_{,k} + \frac{1}{2}λ̄_{m}λ̄_{k}g^{mk}R^{2} = 0.  (4.3)

If the space is of definite metric, then (4.3) gives S_{ijk} = 0, whence R_{ij,k} = λ̄_{k}R_{ij}. We can therefore state the following theorem:

Theorem 3. Every G(2R_{n}) of definite metric whose scalar curvature is different from zero and for which R^{ij}R_{ij} = \frac{1}{2}R^{2}, is an R_{n}.

5. Conformally symmetric G(2R_{n})

It is well known that for a conformally symmetric Riemannian space, it holds

R_{ij,k} − R_{ik,j} = \frac{1}{2(n − 1)}(R_{,k}g_{ij} − R_{,j}g_{ik}).  (5.1)

Let us suppose that a G(2R_{n}) with R ≠ 0, is conformally symmetric. The conformal curvature tensor can be written in the form

C_{hijk} = R_{hijk} − D_{hijk}  (5.2)

where

D_{hijk} = π_{hk}g_{ij} − π_{hj}G_{ik} + π_{ij}g_{hk} − π_{ik}g_{hj},  (5.3)
π_{ij} = (R_{ij} − \frac{R}{2(n − 1)}g_{ij}) .  (5.4)

Now

R_{hijk,l} − D_{hijk,l} = C_{hijk,l} = 0.  (5.5)

On account of (5.1) and (5.4),

π_{ij,k} − π_{ik,j} = \frac{1}{n − 2} [R_{ij,k} − R_{ik,j} + \frac{1}{2(n − 1)}(R_{,j}g_{ik} − R_{,k}g_{ij})] = 0.

Hence

π_{ij,k} = π_{ik,j}.  (5.6)

From (5.4) we have as a consequence of (1.2)

π_{ij,kl} = a_{kl}π_{ij} + λ_{l}π_{ij,k}.  (5.7)

Now (5.5) gives

R_{hijk,lm} = D_{hijk,lm}.  (5.8)

On account of (5.8) and (5.3) we have R_{hijk,lm} = a_{lm}R_{hijk} + λ_{m}R_{hijk,l}. Also equations (5.6) and (5.7) give a_{kl}π_{ij} = a_{ij}π_{ik}. Multiplying both sides by g^{kl} we get θπ_{ij} = a_{jl}g^{kl}π_{ik} where θ = g^{kl}a_{kl}.

Now considering a_{ij} is symmetric we obtain

θR_{ij} = \frac{R}{2} \frac{n − 2}{n − 1}a_{ij} + \frac{R}{2} \frac{θ}{n − 1}g_{ij}.  (5.9)

Multiplying the above equation by R^{ij} we have

θR_{ij}R^{ij} = \frac{R}{2} \frac{n − 2}{n − 1}a_{ij}R^{ij} + \frac{Rθ}{2(n − 1)}g_{ij}R^{ij}.  (5.10)

But multiplying (3.5) by g^{km} we obtain a_{jm}R^{jm} = \frac{1}{2}Rθ. Hence (5.10) gives

θR_{ij}R^{ij} = \frac{nR^{2}θ}{4(n − 1)}. Since R ≠ 0, if θ = 0, (5.10) would give a_{ij} = 0. Hence

θ ≠ 0. Therefore R_{ij}R^{ij} = \frac{nR^{2}}{4(n − 1)}. Thus we get

Theorem 4. A conformally symmetric G(2R_{n}) is a G(2k_{n}) and when the

tensor of recurrence is symmetric then the length of Ricci tensor is \frac{nR^{2}}{4(n − 1)}.

Now by covariant differentiation of (5.1) it follows

R_{ij,kl} − R_{ik,jl} = \frac{1}{2(n − 1)}(R_{,kl}g_{ij} − R_{,jl}g_{ik}).  (5.11)

By virtue of (2.4) and (1.2) the equation (5.11) reduces to the form

a_{kl} (R_{ij} − \frac{1}{2(n − 1)}Rg_{ij}) + λ_{l} (R_{ij,k} − \frac{1}{2(n − 1)}R_{,k}g_{ij}) =

= a_{jl} (R_{ik} − \frac{1}{2(n − 1)}Rg_{ik}) + λ_{l} (R_{ik,j} − \frac{1}{2(n − 1)}R_{,j}g_{ik}) .

Hence on account of (5.1) we obtain

a_{kl} (R_{ij} − \frac{1}{2(n − 1)}Rg_{ij}) = a_{il} (R_{ik} − \frac{1}{2(n − 1)}Rg_{ik}) .  (5.12)

Transvecting (5.12) with R_{p}^{j} and using the relation (3.5),

a_{kl} (R_{ri}R_{p}^{r} − \frac{1}{2(n − 1)}RR_{ip}) = \frac{1}{2}Ra_{pl} (R_{ik} − \frac{1}{2(n − 1)}Rg_{ik}) .

But it follows from (5.12) that

\frac{1}{2}R (R_{ik} − \frac{1}{2(n − 1)}Rg_{ik}) a_{pl} = \frac{1}{2}R (R_{ip} − \frac{1}{2(n − 1)}Rg_{ip}) a_{kl}.

Hence

(R_{ri}R_{p}^{r} − \frac{1}{2(n − 1)}RR_{ip}) a_{kl} = \frac{1}{2}R (R_{ip} \frac{1}{2(n − 1)}Rg_{ip}) a_{kl}.

Therefore

R_{ri}R_{p}^{r} = \frac{n}{2(n − 1)}RR_{ip} − \frac{1}{4(n − 1)}R^{2}g_{ip}.  (5.13)

Now if R = 0, (5.13) reduces to R_{ri}R_{p}^{r} = 0 or R^{ri}R_{ri} = 0 (by contraction with g^{ip}). So, for definite metric R_{ij} = 0, which is not possible. Hence we obtain the following theorem:

Theorem 5. A conformally symmetric generalized Ricci 2-recurrent space with definite metric and zero scalar curvature can not exist.

6. Example of a generalized Ricci 2-recurrent space

For this section let the greek index runs over 2, 3, …, n − 1 and the latin index runs over 1, 2, …, n. We define the metric g in R^{n}, n ⩾ 4 by the formula [6]

ds^{2} = Q(dx^{1})^{2} + K_{αβ}dx^{α}dx^{β} + 2dx^{1}dx^{n}  (6.1)

where [K_{αβ}] is a symmetric and non-singular matrix consisting of constants, and Q is independent of x^{n}.

The only components of Christoffel symbols R_{hijk}, R_{ij}, not identically zero are those related to

\begin{matrix}\left\{{λ \atop 11}\right\} = −\frac{1}{2}K^{αβ}Q.β, \left\{{n \atop 11}\right\} = −\frac{1}{2}Q.1, \left\{{n \atop 1α}\right\} = −\frac{1}{2}Q.α, \\ R_{1αβ1} = \frac{1}{2}Q.αβ, R_{11} = −\frac{1}{2}K^{αβ}Q.αβ \end{matrix}  (6.2)

where [K^{αβ}] = [K_{αβ}]^{−1}.

Let Q = K_{αβ}x^{α}x^{β}e^{2x^{1}} where

[K_{αβ}] = [ \begin{matrix}1 & & & 0 & & … & & & 0 \\ 0 & & & 1 & & … & & & 0 \\ . & . & . & . & . & . & . & . & . & . & . & . & . \\ 0 & & & 0 & & … & & & 1 \end{matrix} ] .

So [K^{αβ}] = [K_{αβ}]. Now

K_{αβ}K^{αβ} = n − 2, Q.αβ = 2K_{αβ}e^{2x^{1}}, Q.αβν = 0, K^{αβ}Q.αβ = 2(n − 2)e^{2x^{1}}.

(6.3)

Hence from (6.2) and (6.3) the only non zero components of R_{ij}, R_{ij,l}, R_{ij,lm} are

R_{11} = (n − 2)e^{2x^{1}}, R_{11,1} = 2(n − 2)e^{2x^{1}}, R_{11,11} = 4(n − 2)e^{2x^{1}}.

So, R_{11,11} = R_{11,1} + 2R_{11}. Hence R_{ij,lm} = a_{lm}R_{ij} + λ_{m}R_{ij,l} where

a_{lm} = \{ \begin{matrix}2, & for l = m = 1, \\ 0, & otherwise, \end{matrix} and λ_{m} = \{ \begin{matrix}1, & for m = 1, \\ 0, & otherwise. \end{matrix} Hence V_{n} is a generalized

Ricci 2-recurrent space.

REFERENCES

[1] A. K. Roy, On generalized 2-recurrent tensors in Riemannian spaces, Bull. Acad. Roy. Belg. Cl Sc. 5e ser. 58 (1972), 220–228.

[2] M. C. Chaki and A. N. Roychowdhury, On Ricci-recurrent spaces of 2nd order, Indian Jour. Math. 9, 2 (1967), 279–287.

[3] E. M. Patterson, Some theorems on Ricci-recurrent spaces, J. London Math. Soc. 27 (1952), 287–295.

[4] M. C. Chaki and B. Gupta, On conformally symmetric spaces, Indian Jour. Math. 5 (1963), 113–122.

[5] T. J. Willmore, An Introduction to Differential Geometry, Oxford Univ. Press, 1959.

[6] W. Gryeak and M. Hotlos, On the existence of certain types of Riemannian matrices, Colloq. Math. 47, 1 (1982), 31–37.

(received 12.02.1998.)

Department of Mathematics, Kalyani University, Kalyani, India