Conformally Recurrent Riemannian Manifolds with Harmonic Conformal Curvature Tensor

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Conformally Recurrent Riemannian Manifolds with Harmonic Conformal Curvature Tensor

Jin Ok Baek and Young Jin Suh

Department of Mathematics, Kyungpook University, Taegu 702-701, Korea e-mail : [email protected]

Jung-Hwan Kwon

Department of Mathematics Education, Taegu University, Taegu 705-714, Korea e-mail : [email protected]

Abstract. In this paper, we give a complete classification of conformally recurrent Rie- mannian manifolds with harmonic conformal curvature tensor and to give another generalization of conformally symmetric Riemannian manifolds.

1. Introduction

LetM be ann(≥4)-dimensional Riemannian manifold with Riemannian metric g and Riemannian connection∇ and let R(resp. S or r) be the Riemannian curvature tensor (resp. the Ricci tensor or the scalar curvature) onM. It is said to be curvature recurrent, if the Riemannian curvature tensorR satisfies∇R=α⊗Rfor a certain 1-formαonM.

On the other hand, it is said to be conformally recurrent, if the conformal curvature tensorC with componentsCijkl so that

Cijkl=Rijkl− 1

n−2(Silgjk−Sikgjl+Sjkgil−Sjlgik)

+ r

(n−1)(n−2)(gilgjk−gikgjl) (1.1)

is recurrent, i.e., there is a 1-form αsuch that ∇C =α⊗C, where Rijkl, Sij and gij are components ofR,S andg onM. In particular, it is said to beconformally symmetric, if∇C= 0.

As it is easily seen, the class of conformally recurrent Riemannian manifolds

Received September 12, 2002.

2000 Mathematics Subject Classification: Primary 53C40, Secondary 53C15.

Key words and phrases: conformal curvature tensor, Weyl tensor, harmonic conformal curvature tensor, conformally symmetric, conformal-like curvature tensor, Riemannian manifold.

This work was supported by grant Proj. No R14-2002-003-01001-0 from the Korea Science & Engineering Foundation.

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includes all the classes of conformally symmetric, conformally flat and locally symmetric Riemannian manifolds. Among them such kind of Riemannian manifolds are studied by Besse ([3]), Ryan ([12]), Simon ([13]), Weyl ([15], [16]), Yano ([17]), Yano and Bochner ([18]), for examples. Conformally symmetric Riemannian manifolds are investigated by Derdzi´nski and Roter ([6]). In particular, in Riemannian case, Derdzi´nski and Roter ([6]) and Miyazawa ([9]) proved the following

Theorem A. An n(≥4)-dimensional conformally symmetric manifold is confor- mally flat or locally symmetric.

The symmetric tensorK of type (0,2) with componentsKij is called theWeyl tensor, if it satisfies

(1.2) Kijl−Kilj = 1

2(n−1)(klgij−kjgil),

wherek=T rKandKijl(resp. kj) are components of the covariant derivative∇K (resp. ∇k).

On the other hand, in Weyl ([15]) and ([16]) it can be easily seen that the Ricci tensor is equal to the Weyl tensor when we only consider ann(≥4)-dimensional conformally flat Riemannian manifold (See Eisenhart [7]). In particular, Derdzi´nski and Roter ([6]) investigated the structure of analytic conformally symmetric manifolds of index 1 which is neither conformally flat nor locally symmetric.

Now let us denote by M an n(= 4)-dimensional Riemannian manifold with Riemannian metricg and Riemannian connection∇. For a tensor field (0, r+ 1) the codifferentialδT ofT is defined by

δT(X1,· · ·, Xr) = Xr i=1

∇EiT(Ei, X1,· · ·, Xr)

for any vector fields X1,· · ·, Xr, where {Ei} is an orthonormal frame on M. If δC = 0, then M is said to have harmonic conformal curvature tensor(See Besse [3]).

In this paper we want to make a generalization of such results in the direction of certain kind of curvature-like tensor fields. In order to do this we introduce the notion of conformal recurrent curvature tensor, meaning that the covariant derivative of the conformal curvature tensor C satisfies ∇C =α⊗C for a certain 1-formα. Moreover, let us say a Riemannian manifoldM hasharmonic conformal curvature tensorif its conformal curvature tensorC satisfiesδC= 0, that is,

X

rCrjkmr= 0.

If the Riemannian manifoldM is conformally symmetric, then it is trivial that it is conformally recurrent forα= 0 and it has harmonic conformal curvature tensor.

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Now in this paper we want to show the following theorem

Theorem 1. Let M be ann(≥4)-dimensional Riemannian manifold. Assume that M is conformally recurrent and has the harmonic conformal curvature tensor. If the scalar curvature is constant, then M is conformal symmetric, conformally flat or Ricci recurrent.

If we give a non-zero restriction for the condition of constant scalar curvature, we have more strong result than Theorem 1 as follows:

Theorem 2. Let M be an n(≥4)-dimensional Riemannian manifold. If M is conformally recurrent and has harmonic conformal curvature tensor and if the scalar curvature is non-zero constant, then M is conformally flat or locally symmetric.

When the 1-formαvanishes identically in above theorem, it can be explained that a conformal Riemannian symmetric manifold M is locally symmetric or conformally flat. So our theorem is also a generalization of Theorem A.

Moreover, without the assumption of non-zero constant scalar curvature we assert the following:

Theorem 3. Let M be an n(≥4)-dimensional Riemannian manifold. If M is conformally recurrent and has harmonic conformal curvature tensor with constant scalar curvature and if the length of the Ricci tensor is constant, thenM is confor- mally flat or curvature recurrent.

2. Preliminaries

LetM be ann(=2)-dimensional Riemannian manifold equipped with Rieman- nian metric tensor g and letR (resp. S or r) be the Riemannian curvature tensor (resp. the Ricci tensor or the scalar curvature) onM.

Now we can choose a local field{Ej}={E1,· · ·, En}of orthonormal frames on a neighborhood ofM. Here and in the sequel, the indicesi, j, k,· · · run from 1 ton.

Then with respect to such a Riemannian metric tensor gwe have g(Ej, Ek) =δjk. Let{θ_i}, {θ_ij} and{Θ_ij} be the canonical form, the connection form and the curvature form onM, respectively, with respect to the local field{Ej}of orthonormal frames. Then we have the structure equations

dθi+X

j

θij∧θj= 0, θij+θji= 0, dθij+X

k

θik∧θkj= Θij, Θij=−1

2 X

k,l

Rijklθk∧θl,

where R_ijkl denotes the components of the Riemannian curvature tensorR of M (See [1], [2], [4] and [5]).

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Now, let C be the conformal curvature tensor with components C_ijkl on M, which is given by

Cijkl=Rijkl− 1

n−2{(δilSjk−δikSjl) + (Silδjk−Sikδjl)}

+ r

(n−1)(n−2)(δilδjk−δikδjl), (2.1)

where Sij =P

lRlijl is the components of the Ricci tensor S with respect to the local field{Ej} of orthonormal frames andr=P

jSjj is the scalar curvature.

Remark 2.1. IfM is Einstein, then the conformal curvature tensorC satisfies Cijkl=Rijkl− r

n(n−1)(δilδjk−δikδjl).

This yields that the conformal curvature tensors of Einstein Riemannian manifolds are the concircular curvature one. In particular, ifM is of constant curvature, the conformal curvature tensor vanishes identically (See Yano and Bochner [18]).

Let D^rM be the vector bundle consisting of differentiable r-forms andDM = P_n

r=0D^rM, whereD⁰M is the algebra of differentiable functions onM. For any tensor fieldKinD^rM the componentsKijklh of the covariant derivative∇K ofK are defined by (for simplicity, we consider the caser= 4)

X

h

Kijklhθh

=dKijkl−X

h

(Khjklθhi+Kihklθhj+Kijhlθhk+Kijkhθhl).

Now, letC be the conformal curvature tensor with componentsCijkl onM. The Riemannian manifold is said to be conformally flat if C = 0. For the geometric meaning of conformally flat Riemannian manifolds, see Yano and Bochner ([18]), for example. In particular, if M is a space of constant curvature, the conformal curvature tensor vanishes identically.

3. Conformally recurrent spaces

Let M be an n(≥2)-dimensional Riemannian manifold with Riemannian connection ∇ and letR(resp. S or r) be the Riemannian curvature tensor (resp. the Ricci tensor or the scalar curvature) onM.

Now letCbe the conformal curvature tensor with componentsCijklwith respect to the field{Ej} of orthonormal frames given by

Cijkl=Rijkl− 1

n−2(Silδjk−Sikδjl+Sjkδil−Sjlδik)

+ r

(n−1)(n−2)(δilδjk−δikδjl).

(3.1)

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Differentiating C of (3.1) covariantly, we have Cijklm=Rijklm− 1

n−2(Silmδjk−Sikmδjl+Sjkmδil−Sjlmδik)

+ rm

(n−1)(n−2)(δilδjk−δikδjl), (3.2)

where Cijklm (resp. Rijklm, Sjkm or rm) are the components of the covariant derivative∇C ofC(resp. the covariant derivative∇R ofR,∇S ofS ordr).

From the second Bianchi identity for the curvature tensorR Rijklm+Rijlmk+Rijmkl= 0,

together with the formula obtained by puttingi=min (3.2) and summing up with respect to iit follows

X

rCrjkmr = (n−3){Sjkm−Sjmk−(rmδjk−rkδjm)/2(n−1)}/(n−2).

IfM has a harmonic conformal curvature tensor, then we have by definition

(3.3) X

rCrjkmr = 0.

From this we are able to assert the following lemmas:

Lemma 3.1. Let M be an n(≥4)-dimensional Riemannian manifold. If M has harmonic conformal curvature tensor, then the Ricci tensor is the Weyl tensor.

Lemma 3.2. Let M be an n(≥4)-dimensional Riemannian manifold. If M has harmonic conformal curvature tensor, then it satisfies

(3.4) X

r(RrikmSrj+RrimjSrk+RrijkSrm) = 0.

Proof. By virtue of Lemma 3.1 we know that the Ricci tensor S is a Weyl tensor.

Then by the definition of (1.2) we have

(3.5) Sijk−Sikj= (rkδij−rjδik)/2(n−1).

Differentiating (3.5) covariantly gives

Sijkm−Sikjm= (rkmδij−rjmδik)/2(n−1).

Interchanging the indiceskandmand subtracting the resulting equation from this, we obtain

Sijkm−Sijmk+Simjk−Sikjm= (rjkδim−rjmδik)/2(n−1),

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where we have used the property that r_ij is symmetric with respect to i and j, becauseris a function. Thus we have

the left side =(Sijkm−Sijmk) + (Simjk−Simkj) + (Simkj−Sikjm)

=(Sijkm−Sijmk) + (Simjk−Simkj) +£

{Sikmj+ (rkjδim−rmjδik)/2(n−1)} −Sikjm

¤

=−X

r(RmkirSrj+RmkjrSir)−X

r(RkjirSrm+RkjmrSir)

−X

r(RjmirSrk+RjmkrSir) + (rkjδim−rmjδik)/2(n−1)

=−X

r(RmkirSrj+RjmirSrk+RkjirSrm) + (rkjδim−rmjδik)/2(n−1)

where the second equality follows from (3.2), the third equality is derived from the Ricci identity for the Ricci tensorSij and the fourth equality follows from the first Bianchi identity. It yields that we have (3.4). It completes the proof. ¤ Lemma 3.3. Let M be an (n≥4)-dimensional Riemannian manifold. If M is conformally recurrent and ifS is a Weyl tensor, then we obtain

(3.6) X

r(CrikmSrj+CrimjSrk+CrijkSrm) = 0,

(3.7) X

r(CrikmSrjn+CrimjSrkn+CrijkSrmn) = 0.

Proof. Substituting the components Rijkm of (2.1) into the left hand side of (3.4) and calculating directly, we get the equation (3.6).

Now differentiating (3.6) covariantly and taking account of (3.6), then the conformal recurrence implies (3.7). This completes the proof. ¤ Now a Riemnannian manifoldM is said to beRicci recurrentif the Ricci tensor S satisfies ∇S = α⊗S for a 1-form α defined on M. Then by virtue of Lemmas 3.1, 3.2 and 3.3 and much more long process of calculation we are able to assert the following:

Theorem 3.4. Let M be an n(≥4)-dimensional Riemannian manifold of with Riemannian connection ∇. Assume that M is conformally recurrent and has the harmonic conformal curvature tensor. If the scalar curvature is constant, then M is conformal symmetric, conformally flat or Ricci recurrent.

Proof. Let Cijkmnp be the components of the covariant derivative ∇²C of ∇C.

They are given by

(3.8) Cijkmnp= (αnαp+αnp)Cijkm.

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Now we define byf the scalar product ofC, namely we putf =g(C, C), whereg denotes a positive definite Riemnnian metric tensor defined on M. LetM⁰ be the subset ofM consisting of pointsxinM such thatf(x) = 0. Then we have

∇f = 2g(∇C, C) = 2αf

on the open subset M−M⁰ and hence we haveα=∇f /2f, from which it follows that

2α=∇log|f|.

This implies that

(3.9) αij=αjionM−M⁰.

So, onM−M⁰, by (3.8) and (3.9) we haveCijkmnp=Cijkmpn. Accordingly, by the Ricci identity we get

(3.10) X

r(RpnirCrjkm+RpnjrCirkm+RpnkrCijrm+RpnmrCijkr) = 0.

Differetiating the above equation covariantly and taking account ofC_ijkmn = αnCijkm, we have

X

r

©(RpnirqCrjkm+RpnjrqCirkm+RpnkrqCijrm+RpnmrqCijkr) +αq(RpnirCrjkm+RpnjrCirkm+RpnkrCijrm+RpnmrCijkr)ª

= 0.

Hence we have by (3.10) (3.11) X

r(RpnirqCrjkm+RpnjrqCirkm+RpnkrqCijrm+RpnmrqCijkr) = 0.

On the other hand, by (2.1), (3.2) andα_hC_ijkl=C_ijklh we have αn

£Rijkm− {(Sjkδim−Sjmδik) + (δjkSim−δjmSik)}/(n−2) +r(δjkδim−δjmδik)/(n−1)(n−2)¤

=R_ijkmn− {(S_jknδ_im−S_jmnδ_ik) + (δ_jkS_imn−δ_jmS_ikn)}/(n−2) +rn(δjkδim−δjmδik)/(n−1)(n−2)

and hence we get

Rijkmn=αnRijkm+©

(Sjknδim−Sjmnδik)−αn(Sjkδim−Sjmδik)ª

/(n−2) +©

(δjkSimn−δjmSikn)−αn(δjkSim−δjmSik)ª

/(n−2) + (rαn−rn)(δjkδim−δjmδik)/(n−1)(n−2).

(3.12)

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From (3.11) and (3.12) it follows that

X

r

£©(Sniqδpr−Snrqδpi)Crjkm+ (Snjqδpr−Snrqδpj)Cirkm

+ (Snkqδpr+Snrqδpk)Cijrm−(Snmqδpr−Snrqδpm)Cijkr

ª

−αq

©(Sniδpr−Snrδpi)Crjkm+ (Snjδpr−Snrδpj)Cirkm

+ (S_nkδ_pr−S_nrδ_pk)C_ijrm+ (S_nmδ_pr−S_nrδ_pm)C_ijkrª¤

/(n−2)

+X

r

£©(δ_niS_prq−δ_nrS_piq)−α_q(δ_niS_pr−δ_nrS_pi)}C_rjkm +{(δnjSprq−δnrSpjq)−αq(δnjSpr−δnrSpj)}Cirkm

+{(δnkSprq−δnrSpkq)−αq(δnkSpr−δnrSpk)}Cijrm

+{(δ_nmS_prq−δ_nrS_pmq)−α_q(δ_nmS_pr−δ_nrS_pm)}C_ijkr¤

/(n−2) + (rαq−rq)©

(δniCpjkm+δnjCipkm+δnkCijpm+δnmCijkp)

−(δpiCnjkm+δpjCinkm+δpkCijnm+δpmCijkn)ª

/(n−1)(n−2)

= 0,

where we have used (3.10). Then by multiplying (n−2) to both sides of this formula we obtain

(SniqCpjkm+SnjqCipkm+SnkqCijpm+SnmqCijkp)

−(SpiqCnjkm+SpjqCinkm+SpkqCijnm+SpmqCijkn)

−X

r(δpiCrjkm+δpjCirkm+δpkCijrm+δpmCijkr)Snrq

+X

r(δniCrjkm+δnjCirkm+δnkCijrm+δnmCijkr)Sprq

−αq

©(SniCpjkm+SnjCipkm+SnkCijpm+SnmCijkp)

−(S_piC_njkm+S_pjC_inkm+S_pkC_ijnm+S_pmC_ijkn)ª +αq

©X

r(δpiCrjkm+δpjCirkm+δpkCijrm+δpmCijkr)Snr

−X

r(δniCpjkm+δnjCirkm+δnkCijpm+δnmCijkr)Spr

ª + (rαq−rq)©

/(n−1)

= 0.

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Accordingly, we have

©(Sniq−αqSni)Cpjkm+ (Snjq−αqSnj)Cipkm

+ (Snkq−αqSnk)Cijpm+ (Snmq−αqSnm)Cijkp

ª

−©

(Spiq−αq)Cnjkm+ (Spjq−αqSpj)Cinkm

+ (Spkq−αq)Cijnm+ (Spmq−αqSpm)Cijkn

ª

−X

r(δpiCrjkm+δpjCirkm+δpkCijrm+δpmCijkr)(Snrq−αqSnr)

+X

r(δniCrjkm+δnjCirkm+δnkCijrm+δnmCijkr)(Sprq−αqSpr) + (rαq−rq)©

/(n−1)

= 0.

(3.13)

Now puttingi=rin (3.13), summing up with respect toP

iand taking account of (2.6), Lemma 2.1 and the first Bianchi identity for the conformal curvature tensor C, we have

X

r

£(n−2)(Snrq−αqSnr)Crjkm+ (Sjrq−αqSjr)Crnkm

+ (Skrq−αqSkr)Crjnm+ (Smrq−αqSmr)Crjkn

+X

s

©δnk(Srsq−αqSrs)Crjms−δnm(Srsq−αqSrs)Crjks

ª¤

= 0.

(3.14)

Next we assume that M is conformally recurrent and M has the harmonic conformal curvature tensor, namely, it satisfies P

rCrjkmr= 0. Then we have

(3.15) X

rαrCrjkm= 0.

Puttingm=qin (3.14), summing up with respect tomand taking account of (3.5) and (3.15), we have

X

r,s{(n−2)SnrsCrjks+SjrsCrnks+SkrsCrjns−αsSrsCrjkn}

+X

rrrCrjkn/2 +X

r,s,tδnkSrstCrjts

−X

rs(Srsn−αnSrs)Crjks= 0.

(3.16)

The third term of (3.16) vanishes identically. Because it satisfies the third term =X

r,s,tδnk(Srst−Srts)Crjts/2

=X

r,s,tδnk(rtδrs−rsδrt)Crjts/4(n−1) = 0

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where the first equality follows from (2.2), the second one is derived by (3.5) and the last one is derived from (2.6).

On the other hand, we get X

r,s(Srns−Srsn)Crjks=X

r,s(rsδrn−rnδrs)Crjks/2(n−1)

=X

srsCnjks/2(n−1)

where the first equality is derived by (3.4) and the second one follows from (2.6).

Thus (3.16) is deformed as X

r,s

©(n−3)SnrsCrjks+SjrsCrnks+SkrsCrjns−αsSrsCrjkn

ª

+X

rrrCrjkn/2 +X

rrrCnjkr/2(n−1) +X

r,sαnSrsCrjks= 0.

(3.17)

Since M is conformally recurrent and M has the harmonic conformal curvature tensor, by (3.7) we have

X

r(CrikmSrjn+CrimjSrkn+CrijkSrmn) = 0.

Puttingmandnbysin (3.7), we have

(3.18) X

r,s(SjrsCriks−SkrsCrijs) +1 2

X

srsCsijk= 0.

By (3.17) and (3.18) we have X

r,s

ª−X

rrrCrjkn/2

+X

rrrCnjkr/2(n−1) +X

r,sαnSrsCrjks= 0.

From this, together with the conformal recurrence and the assumption of constant scalar curvature it follows that

(n−1)X

rsSnrsCrjks+X

rsSrs(Crjkns−Crjksn) = 0.

Here we note that the indicesj andkin the first and the third terms are symmetric with each other, becauseSnrsand Srs are symmetric with respect to the indicesr ands. From such a fact, if take a skew-symmetric part to the above equation, then it follows that

0 =X

r,sSrs(Crjkns−Crkjns)

=X

r,sSrs(Crjkns+Crknjs)

=−X

r,sSrsCrnjks.

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Hence we are able to assert that

(3.19) X

r,sSnrsCrjks=X

r,sSrsnCrjks= 0.

Transvecting (3.14) toαmαnαq, summing up with respect tom, nandq, and taking account of (3.15) and (3.19), we have

kαk²X

r,sS_rsC_rjks= 0, where kαk² = kP

rαrαrk. This gives us that M is conformally symmetric or M satisfies

(3.20) X

r,sSrsCrjks= 0.

By (3.14), (3.19) and (3.20) we have X

r

ª + (Skrq−αqSkr)Crjnm+ (Smrq−αqSmr)Crjkn

¤= 0.

By Lemma 3.3 we have X

r

ª

=X

r

−(Snrq−αqSnr)Crjmk

ª

=−X

r(Snrq−αqSnr)Crjmk. From the above two equations we obtain

X

r

ª= 0

which implies that X

r(S_jrq−α_qS_jr)C_rnkm =−(n−1)X

r(S_nrq−α_qS_nr)C_rjkm. From this it follows that

(3.21) X

r(Sjrq−αqSjr)Crnkm= 0.

TransvectingSniq−αqSniorCpjkm to (3.13) and applying the equation (3.20) and (3.21), we can obtain

(3.22) k∇S−α⊗Sk²C= 0 orkCk²(∇S−α⊗S) = 0

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onM−M⁰, which means thatM is recurrent Ricci tensor orM is conformally flat.

It completes the proof. ¤

From Theorem 3.4 we want to give the following lemma which will be useful to prove our main Theorem.

Lemma 3.5. LetM be ann(≥4)-dimensional Riemannian manifold with Rieman- nian connection∇. IfM is conformally recurrent and ifM has harmonic conformal curvature tensor and constant scalar curvature, we have

C⊗(∇R−α⊗R) = 0.

Proof. By Theorem 3.4 we have

α⊗C⊗(∇S−α⊗S) = 0.

LetM1 be the subsets consisting of pointsxin M at whichα(x) = 0.

First we suppose thatM1is not empty. If IntM1is empty, the nonvanishing 1- formαgivesC= 0 or∇S−α⊗S= 0. Thenrm=αmrin the latter case. From this, together with the assumption of conformal recurrence, that is, Cijklm =αmCijkl, and the formulas (3.1) and (3.2), we know that ∇R−α⊗R = 0. So in such a subcase the conclusion is given by the continuity ofCand ∇R−α⊗R.

Suppose that Int M1 is not empty. Thenα= 0 on intM1. In such a subcase M is conformal symmetric. Then by Theorem A due to Derdz´ınski and Roter ([5]) and Miyazawa [9]we haveC = 0 or ∇R = 0 on IntM1. Hence it follows that we haveC= 0 or∇R=α⊗Ron IntM1. Thus we have C⊗(∇R−α⊗R) = 0 onM.

Now we suppose thatM1is empty. Then nonvanishing 1-formαimpliesC= 0 or ∇S−α⊗S = 0. When ∇S−α⊗S = 0, we also have ∇R−α⊗R= 0, because

M is conformally recurrent. ¤

By virtue of Lemma 3.5 we give more strong result than Theorem 3.4 as follows:

Theorem 3.6. Let M be an n(≥4)-dimensional Riemannian manifold with Rie- mannian connection∇. Suppose thatM is conformally recurrent and has harmonic conformal curvature tensor. If the scalar curvature is non-zero constant, thenM is conformally flat or M is locally symmetric.

Proof. LetM⁰⁰ be the subset of pointsxinM at which (∇R−α⊗R)(x) = 0.

Then we have (∇r−αr)(x) = 0 on M⁰⁰. Since we have assumed that the scalar curvature is non-zero constant, we get α(x) = 0 onM⁰⁰. Then from this together with Lemma 3.5 it follows thatα= 0 orC= 0, that is,α⊗C= 0 onM.

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Now let us consider the open subsetM^∗consisting of pointsxat whichC(x) = 0.

Then on such an open subset we have∇C= 0 and hence the inner productg(C, C) is constant, whereg denotes a Riemannian metric tensor onM. By the continuity ofg(C, C), ifM^∗ is not empty, theng(C, C) = 0 onM, namelyC= 0 onM. That is, M is conformally flat. IfM^∗ is empty, then the factα⊗C= 0 impliesα= 0 on M. In such a case we know that M is conformally symmetric. From this together with Theorem A we complete the proof of our theorem. ¤ Remark 3.1. In their paper ([8]), Goldberg and Okumura proved that in an n(≥4)- dimensional compact conformally flat Riemannian manifold, if the length of the Ricci tensor is constant and less thanr/√

n−1, thenM is a space of constant curvature.

Concerning about a theorem due to Goldberg and Okumura in [8] and Theorem 3.6, the following is proved.

Theorem 3.7. Let M be an n(≥4)-Riemannian manifold. If M is conformally recurrent and has harmonic conformal curvature tensor with constant scalar curva- ture and if the length of the Ricci tensor is constant, thenM is conformally flat or curvature recurrent.

Proof. Without loss of generality, we may assume thatαdoes not vanish identically onM. By Lemma 3.5 we haveC⊗(∇R−α⊗R) = 0 onM. LetM1 (resp. M2 or M3) be the subset consisting of pointsxin M at whichα(x) = 0 (resp. C(x) = 0 or (∇R−α⊗R)(x) = 0) as follows:

M1={x∈M|α(x) = 0}, M2={x∈M|C(x) = 0}, and

M3={x∈M|(∇R−α⊗R)(x) = 0}.

Case I) We suppose that IntM3 is not empty. Then we have∇R−α⊗R= 0 on IntM3and hence by contracting we have

∇S−α⊗S= 0 on IntM3. Accordingly, by the assumption we get

∇g(S, S) = 2αg(S, S) = 0,

whereg denotes a Riemannian metric tensor onM. From this we haveg(S, S) = 0 onM or otherwiseg(S, S)6=0 onM andα= 0 on IntM3.

In the first case, we see that S = 0, which implies that the scalar curvaturer vanishes identically onM, which means thatC=RonM and hence

∇C=∇R=αR

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onM, that is,M is curvature recurrent.

In the other one, we have α= 0 on Int M3. From the construction of the set M3 it follows that∇R=αRon IntM3. Moreover we know that the subset M3 is contained inM1, because

IntM3⊂the closure of IntM3⊂the closure ofM1=M1.

Accordingly, we have M = M1∪M2. From this together with the assumption of conformal recurrence we know∇C= 0 onM.

We suppose that there is a point xin M1 at which C(x)6=0. Since the inner product g(C, C) is constant on M, it is non-zero constant, which yields that the subset M2 is empty. Then M is onlyM =M1. So we have ∇R = 0 on M. We suppose that there does not exist a pointxinM1 at whichC(x)6=0. So we see that C= 0 onM, which derives the conclusion.

Case II) Next we suppose that IntM₃is empty. From the construction of the setsM1, M2andM3and Lemma 3.5 we know thatM3⊃M−M1∪M2, the setM3is empty. So we haveM =M1∪M2. Since∇C= 0 onM1, the inner productg(C, C) is constant onM. This implies thatC= 0 or ∇C= 0 onM. By Derdzi´nsky and Roter ([6]) and Miyazawa’s theorem ([9]), we have C = 0 or ∇R = 0 on M. It

completes the proof. ¤

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