Real hypersurfaces in the complex quadric with parallel Ricci tensor

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Advances in Mathematics

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Real hypersurfaces in the complex quadric with parallel Ricci tensor

^✩

Young Jin Suh

KyungpookNationalUniversity,Collegeof NaturalSciences,Departmentof Mathematics,Daegu702-701,RepublicofKorea

a r t i c l e i n f o a bs t r a c t

Article history:

Received22October2014 Receivedinrevisedform11May 2015

Accepted21May2015 Availableonlinexxxx CommunicatedbyGangTian

MSC:

primary53C40 secondary53C55

Keywords:

ParallelRiccitensor A-isotropic A-principal Kählerstructure Complexconjugation Complexquadric

We introduce the notion of parallel Ricci tensor for real hypersurfacesinthecomplexquadricQ^m=SOm+2/SOmSO2. According totheA-principalor theA-isotropicunitnormal vector field N, we give a complete classification of real hypersurfacesinQ^m=SOm+2/SOmSO2 withparallelRicci tensor.

© 2015ElsevierInc.All rights reserved.

✩ This workwassupportedby grant Proj.No.NRF-2015-R1A2A1A-01002459 fromNational Research FoundationofKorea.

E-mailaddress:yjsuh@knu.ac.kr.

http://dx.doi.org/10.1016/j.aim.2015.05.012 0001-8708/© 2015ElsevierInc.All rights reserved.

1. Introduction

WhenweconsidersomeHermitiansymmetricspacesofrank2,wecanusuallygiveex- amplesofRiemanniansymmetricspacesSUm+2/S(U2Um) andSU2,m/S(U2Um),which are said to be complex two-plane Grassmannians and complex hyperbolic two-plane Grassmanniansrespectively(see [1,2,4,13]). Those aresaid to be Hermitian symmetric spacesandquaternionicKählersymmetricspacesequippedwiththeKählerstructureJ andthequaternionicKählerstructure Jandtheyhaverank 2.

AsanotherkindofHermitiansymmetricspacewithrank2 ofcompacttypedifferent fromtheaboveones,wecangiveanexampleofcomplexquadricQ^m=SO_m+2/SO_mSO₂, which is acomplex hypersurface incomplex projective space CP^m+1 (see Berndt and Suh [3], and Smyth[11]). Thecomplex quadric canalso be regardedas akind of real Grassmann manifolds of compact type with rank 2 (see Kobayashi and Nomizu [7]).

Accordingly,thecomplexquadricadmitstwoimportantgeometricstructures,acomplex conjugationstructureAandaKählerstructureJ,whichanti-commutewitheachother, thatis,AJ=−J A.Thenform≥2 thetriple(Q^m,J,g) isaHermitiansymmetricspace ofcompacttypewithrank2 anditsmaximalsectionalcurvatureisequalto4 (seeKlein [6]andReckziegel[10]).

For the complex projective space CP^m+1 and the quaternionic projective space QP^m+1 someclassificationsrelatedto parallelRiccitensorwereinvestigatedinKimura [5], and Pérez and Suh [8], respectively. For the complex 2-plane Grassmannian G₂(C^m+2) = SU_m+2/S(U_mU₂) a new classification was obtained by Berndt and Suh [1]. By using this classification Pérez and Suh [9] proved anon-existence property for Hopf hypersurfaces in G2(C^m+2) with parallel and commuting Ricci tensor. Suh [12]

strengthenedthisresulttohypersurfacesinG2(C^m+2) withparallelRiccitensor.More- over, Suhand Woo[15] studiedanothernon-existence property forHopfhypersurfaces in complex hyperbolic two-plane Grassmannians SU2,m/S(U2Um) with parallel Ricci tensor.

It is known that the Reeb flow on a real hypersurface in G₂(C^m+2) is isometric if and only if M is an open part of a tube around a totally geodesic G2(C^m+1) ⊂ G2(C^m+2) in[2].Moreover,in[13]weassertedthattheReebflowonarealhypersurface in SU2,m/S(U2Um) is isometric if and only if M is an open part of a tube around a totallygeodesic SU_2,m−1/S(U₂U_m−1)⊂SU_2,m/S(U₂U_m).Here,the Reebflow on real hypersurfaces in SU_m+2/S(U_mU₂) or SU_2,m/S(U₂U_m) is said to be isometric if the shapeoperatorcommuteswiththestructure tensor.Inthepaper[3]duetoBerndtand Suh,weinvestigatedthisproblemforthecomplexquadricQ^m=SOm+2/SOmSO2 and obtainedthefollowing result:

LetM be areal hypersurface of thecomplex quadric Q^m,m ≥3. The Reeb flow on M isisometric if and only if m iseven, saym = 2k,and M isan open partof a tube aroundatotally geodesic CP^k ⊂Q^2k.

ApartfromthecomplexstructureJthereisanotherdistinguishedgeometricstructure onQ^m,namelyaparallelranktwovectorbundleAwhichcontainsanS¹-bundleofreal

structures,thatis,complexconjugationsAonthetangentspacesofQ^m.Thisgeometric structure determinesamaximalA-invariantsubbundleQofthetangentbundleT M of arealhypersurfaceM inQ^m.

When we consider a hypersurface M in the complex quadric Q^m, the unit normal vector fieldN ofM inQ^m canbe dividedinto twoclasses ifeitherN is A-isotropicor A-principal(see[3,4,14]).Inthefirstcasewhere N isA-isotropic,wehaveshownin[3]

thatM islocallycongruenttoatubeoveratotallygeodesicCP^k inQ^2k.Inthesecond case,whentheunitnormalNisA-principal,weprovedthatacontacthypersurfaceMin Q^mislocallycongruenttoatubeoveratotallygeodesicandtotallyrealsubmanifoldS^m inQ^m(see[4]).InthispaperweconsiderthenotionofRicciparallelismforhypersurfaces inQ^m,thatis,∇Ric= 0. Thenmotivated bytheresultdue to theA-principalnormal forcontacthypersurfacesinQ^m,weassertthefollowing

Theorem 1.Theredo notexistany Hopfhypersurfaces inthecomplex quadric Q^m with parallel Riccitensor andA-principal normalvector field.

InSection4wewill giveitsproof indetail.Nowat eachpoint z∈M letusconsider amaximalA-invariantsubspaceQz ofT_zM,z∈M,definedby

Qz={X ∈TzM|AX∈TzM for allA∈Az}

of TzM, z∈M. Thus for a case where the unit normal vector field N is A-isotropic it can be easily checked that the orthogonal complement Q^⊥z = CzQz, z∈M, of the distribution Q in the complex subbundle C, becomes Q^⊥z = Span{Aξ,AN}. Here it can be easily checked that the vector fields Aξ and AN belong to the tangent space T_zM, z∈M ifthe unitnormal vector fieldN becomes A-isotropic.Then motivated by the above result, in this paper we give another theorem for real hypersurfaces in the complexquadricQ^mwithparallel RiccitensorandA-isotropicunitnormalas follows:

Theorem 2. LetM be aHopf real hypersurface inthe complex quadric Q^m,m≥4,with parallelRiccitensorandA-isotropicunitnormalN.Iftheshapeoperatorcommuteswith the structure tensor on the distribution Q^⊥, then M has 3 distinct constant principal curvaturesgivenby

α= 0, γ= 0, λ=−√

3, andμ=√ 3 or

α=

2(m−3), γ= 0, λ= 0, andμ=− 2 2(m−3) with corresponding principalcurvature spaces

T_α= [ξ], T_γ = [Aξ, AN], φ(T_λ) =T_μ,dim T_λ= dimT_μ=m−2.

Ourpaper is organizedas follows. In Section2we present basicmaterial about the complex quadric Q^m, including its Riemannian curvature tensor and a description of its singular vectors of Q^m like A-principal or A-isotropic unit normal vector field. In Section3,weinvestigatethegeometryofthissubbundleQforhypersurfacesinQ^mand someequationsincludingCodazziandfundamentalformulasrelatedtothevectorfields ξ,N,Aξ,andAN forthecomplexconjugation AofM inQ^m.

Finally,inSections4and 5, wepresentthe proofofTheorems 1 and2,respectively.

InSection4,the firststep isto derive theRiccitensor from theequation of Gauss for real hypersurfaces M in Q^m and to get some formulas by the assumption of parallel Riccitensor and A-principalnormal vector field,and show that areal hypersurface in Q^m which is atube over an m-dimensional unit sphere S^m does not admit aparallel Ricci tensor. The next step is to show thatthere do not exist any real hypersurfaces in the complex quadric Q^m with parallel Ricci tensor and A-principal normal vector field.

In Section 5, we give a complete proof of Theorem 2. The first part of this proof is devotedto givesome fundamentalformulas from Ricci parallel and A-isotropicunit normalvectorfield.Theninthelatterpartoftheproofwewillusetheexpressionofthe shapeoperatorforrealhypersurfacesinQ^mwhentheshapeoperatorandthestructure tensorcommutesonthedistributionQ^⊥,whereQ^⊥ denotesanorthogonalcomplement ofthemaximalA-invariantsubspaceQinthecomplexsubbundleCofTzM,z∈MinQ^m. 2. Thecomplexquadric

For more details in this section we refer to [3,4,6,7,10,14]. The complex quadric Q^m is the complex hypersurface in CP^m+1 which is defined by the equation z₀² + . . .+z_m+1² = 0, where z0,. . . ,zm+1 are homogeneous coordinates on CP^m+1. We equip Q^m with the Riemannian metric g which is induced from the Fubini–Study metric ¯g on CP^m+1 with constant holomorphic sectional curvature 4. The Fubini–

Study metric ¯g is defined by g(X,¯ Y) = Φ(J X,Y) for any vector fields X and Y on CP^m+1 and a globally closed (1,1)-form Φ given by Φ = −4i∂∂logf¯ _j on an open set U_j ={[z⁰,z¹,· · ·,z^m+1]∈CP^m+1|z^j=0},where thefunctionf_j denotes f_j =m+1

k=0t^k_j¯t^k_j, and t^k_j = ^z_z^kj for j,k = 0,· · ·,m+ 1. Then naturally the Kähler structure on CP^m+1 inducescanonicallyaKählerstructure(J,g) onthecomplexquadricQ^m.

The complex projective space CP^m+1 is a Hermitian symmetric space of the spe- cial unitary group SUm+2, namely CP^m+1 = SUm+2/S(Um+1U1). We denote by o = [0,. . . ,0,1] ∈ CP^m+1 the fixed point of the action of the stabilizer S(Um+1U1).

The special orthogonal group SOm+2 ⊂ SUm+2 acts on CP^m+1 with cohomogeneity one.TheorbitcontainingoisatotallygeodesicrealprojectivespaceRP^m+1 ⊂CP^m+1. ThesecondsingularorbitofthisactionisthecomplexquadricQ^m=SO_m+2/SO_mSO₂. This homogeneous space model leads to the geometric interpretation of the complex quadricQ^mastheGrassmannmanifoldG⁺₂(R^m+2) oforiented2-planesinR^m+2.Italso givesamodelofQ^masaHermitiansymmetricspaceofrank 2.ThecomplexquadricQ¹

isisometric toasphereS²withconstantcurvature,andQ²isisometrictotheRieman- nian product of two 2-spheres withconstantcurvature. For this reasonwe will assume m≥3 fromnowon.

For a nonzero vector z ∈ C^m+2 we denote by [z] the complex span of z, that is, [z]={λz |λ∈C}. Note thatbydefinition[z] isapointinCP^m+1. Asusual, for each [z]∈CP^m+1 weidentifyT_[z]CP^m+1 withtheorthogonal complementC^m+2[z] of[z]

inC^m+2.For[z]∈Q^mthetangentspaceT_[z]Q^mcanthenbeidentifiedcanonicallywith the orthogonalcomplement C^m+2([z]⊕[¯z]) of[z]⊕[¯z] in C^m+2 (see Kobayashiand Nomizu [7]). Note that ¯z ∈ ν_[z]Q^m is a unit normal vector of Q^m in CP^m+1 at the point [z].

We denote by Az¯ the shape operator of Q^m in CP^m+1 with respect to the unit normal z.¯ Itis definedbyAz¯w=∇wz¯=wforacomplexEuclidean connection∇ and allw∈T_[z]Q^m.Thatis,theshapeoperatorA¯zisjustcomplexconjugationrestrictedto T_[z]Q^m. Theshape operatorA_z¯is ananti-commuting involutionsuch thatA²_z¯=I and AJ =−J AonthecomplexvectorspaceT_[z]Q^mand

T[z]Q^m=V(Az¯)⊕J V(Az¯),

whereV(A¯z)=R^m+2∩T_[z]Q^misthe(+1)-eigenspaceandJ V(Az¯)=iR^m+2∩T_[z]Q^mis the(−1)-eigenspaceofA_z¯.Thatis,A_z¯X =X andA_z¯J X =−J X,respectively,forany X∈V(Az¯).

Geometrically this meansthattheshapeoperatorAz¯defines areal structure onthe complexvectorspaceT_[z]Q^m,orequivalently,isacomplexconjugationonT_[z]Q^m.Since the real codimension of Q^m in CP^m+1 is 2, this induces an S¹-subbundle A of the endomorphismbundle End(T Q^m) consisting ofcomplexconjugations.

There isa geometric interpretationof these conjugations. Thecomplex quadricQ^m can be viewed as the complexification of the m-dimensionalsphere S^m. Through each point [z] ∈ Q^m there exists a one-parameter family of real forms of Q^m which are isometrictothesphereS^m.Theserealformsarecongruenttoeachotherunderactionof thecenterSO2oftheisotropysubgroupofSOm+2 at[z].TheisometricreflectionofQ^m insucharealform S^misanisometry,andthedifferentialat[z] ofsuchareflectionisa conjugation onT[z]Q^m.InthiswaythefamilyAofconjugationsonT[z]Q^mcorresponds to thefamilyofrealforms S^mof Q^m containing[z],and thesubspacesV(A)⊂T_[z]Q^m correspond tothetangentspacesT_[z]S^m oftherealformsS^m ofQ^m.

TheGaussequationforQ^m⊂CP^m+1impliesthattheRiemanniancurvaturetensorR¯ ofQ^mcanbedescribedintermsofthecomplexstructureJandthecomplexconjugations A∈A:

R(X, Y¯ )Z =g(Y, Z)X−g(X, Z)Y +g(J Y, Z)J X−g(J X, Z)J Y −2g(J X, Y)J Z +g(AY, Z)AX−g(AX, Z)AY +g(J AY, Z)J AX−g(J AX, Z)J AY.

NotethatJ andeachcomplexconjugationAanti-commute,thatis,AJ=−J Aforeach A∈A.

RecallthatanonzerotangentvectorW ∈T_[z]Q^miscalled singularifitistangentto morethanonemaximal flatinQ^m. Therearetwo typesofsingulartangentvectorsfor thecomplexquadricQ^m:

1. If thereexists aconjugation A∈Asuch thatW ∈V(A), thenW issingular.Such asingulartangentvectoriscalled A-principal.

2. Ifthere exist aconjugation A∈Aandorthonormalvectors X,Y ∈V(A) suchthat W/||W|| = (X +J Y)/√

2, then W is singular. Such a singular tangent vector is called A-isotropic.

For every unit tangent vector W ∈ T_[z]Q^m there exist a conjugation A ∈ A and orthonormalvectorsX,Y ∈V(A) suchthat

W = cos(t)X+ sin(t)J Y

for somet ∈[0,π/4].The singular tangentvectors correspond to thevalues t = 0 and t=π/4.If0< t< π/4 thentheuniquemaximalflatcontainingW isRX⊕RJ Y.Later wewillneedtheeigenvaluesandeigenspacesoftheJacobioperatorRW =R(·,W)W for asingularunittangentvectorW.

1. If W is an A-principal singular unit tangent vector with respect to A ∈ A, then the eigenvalues of R_W are 0 and 2 and the corresponding eigenspaces are RW ⊕ J(V(A)RW) and(V(A)RW)⊕RJ W,respectively.

2. If W is an A-isotropic singular unit tangent vector with respect to A ∈ A and X,Y ∈ V(A), then the eigenvalues of R_W are 0, 1 and 4 and the corresponding eigenspaces areRW⊕C(J X+Y),T_[z]Q^m(CX⊕CY) andRJ W,respectively.

3. Somegeneralequations

Let M be a real hypersurface in Q^m and denote by (φ,ξ,η,g) the induced almost contactmetricstructure.Note thatξ=−J N, whereN isa(local)unitnormal vector fieldof M and η the corresponding 1-form defined byη(X)=g(ξ,X) forany tangent vector field X on M. The tangent bundle T M of M splits orthogonally into T M = C ⊕Rξ, where C = ker(η) is the maximal complex subbundle of T M. The structure tensorfieldφrestrictedtoCcoincideswiththecomplexstructureJ restrictedtoC,and φξ= 0.

At each point z ∈ M we define a maximal A-invariant subspace of T_zM, z∈M as follows:

Qz={X ∈TzM|AX∈TzM for allA∈Az}.

Lemma 3.1.(See[14].)Foreach z∈M we have (i) IfNz isA-principal,thenQz =Cz.

(ii) IFN_z isnot A-principal,there existaconjugation A∈Aandorthonormal vectors X,Y ∈ V(A) such that N_z = cos(t)X + sin(t)J Y for some t ∈(0,π/4].Then we haveQz=CzC(J X+Y).

Wenow assumethatM isaHopfhypersurface.Thenwehave Sξ=αξ,

whereSdenotestheshapeoperatoroftherealhypersurfacesMwiththesmoothfunction α=g(Sξ,ξ) onM.When weconsider thetransformJ X bythe Kähler structureJ on Q^m foranyvectorfieldX onM inQ^m,wemayput

J X=φX+η(X)N

foraunitnormalN to M.Thenwenowconsider theCodazziequation

g((∇XS)Y −(∇YS)X, Z) =η(X)g(φY, Z)−η(Y)g(φX, Z)−2η(Z)g(φX, Y) +g(X, AN)g(AY, Z)−g(Y, AN)g(AX, Z)

+g(X, Aξ)g(J AY, Z)−g(Y, Aξ)g(J AX, Z). (3.1) Putting Z=ξin(3.1)weget

g((∇XS)Y −(∇YS)X, ξ) =−2g(φX, Y)

+g(X, AN)g(Y, Aξ)−g(Y, AN)g(X, Aξ)

−g(X, Aξ)g(J Y, Aξ) +g(Y, Aξ)g(J X, Aξ).

Ontheother hand,wehave g((∇XS)Y −(∇YS)X, ξ)

=g((∇XS)ξ, Y)−g((∇YS)ξ, X)

= (Xα)η(Y)−(Y α)η(X) +αg((Sφ+φS)X, Y)−2g(SφSX, Y).

Comparingtheprevioustwoequationsandputting X=ξyields Y α= (ξα)η(Y)−2g(ξ, AN)g(Y, Aξ) + 2g(Y, AN)g(ξ, Aξ).

Reinsertingthis intothepreviousequationyields g((∇XS)Y −(∇YS)X, ξ)

=−2g(ξ, AN)g(X, Aξ)η(Y) + 2g(X, AN)g(ξ, Aξ)η(Y) + 2g(ξ, AN)g(Y, Aξ)η(X)−2g(Y, AN)g(ξ, Aξ)η(X) +αg((φS+Sφ)X, Y)−2g(SφSX, Y).

Altogetherthis implies

0 = 2g(SφSX, Y)−αg((φS+Sφ)X, Y)−2g(φX, Y) +g(X, AN)g(Y, Aξ)−g(Y, AN)g(X, Aξ)

−g(X, Aξ)g(J Y, Aξ) +g(Y, Aξ)g(J X, Aξ)

+ 2g(ξ, AN)g(X, Aξ)η(Y)−2g(X, AN)g(ξ, Aξ)η(Y)

−2g(ξ, AN)g(Y, Aξ)η(X) + 2g(Y, AN)g(ξ, Aξ)η(X). (3.2) Ateachpointz∈M wecanchooseA∈Az suchthat

N = cos(t)Z₁+ sin(t)J Z₂

forsomeorthonormalvectorsZ1,Z2∈V(A) and 0≤t≤ ^π₄ (seeProposition3in[10]).

NotethattisafunctiononM. Firstofall,sinceξ=−J N,we have N = cos(t)Z1+ sin(t)J Z2,

AN = cos(t)Z₁−sin(t)J Z₂, ξ= sin(t)Z₂−cos(t)J Z₁,

Aξ= sin(t)Z₂+ cos(t)J Z₁. (3.3) Thisimpliesg(ξ,AN)= 0 andhence

0 = 2g(SφSX, Y)−αg((φS+Sφ)X, Y)−2g(φX, Y) +g(X, AN)g(Y, Aξ)−g(Y, AN)g(X, Aξ)

−g(X, Aξ)g(J Y, Aξ) +g(Y, Aξ)g(J X, Aξ)

−2g(X, AN)g(ξ, Aξ)η(Y) + 2g(Y, AN)g(ξ, Aξ)η(X). (3.4) 4. ProofofTheorem 1

Bytheequation ofGauss,thecurvaturetensor R(X,Y)Z forareal hypersurfaceM in Q^m induced from the curvature tensor R¯ of Q^m can be described in terms of the complexstructureJ andthecomplexconjugation A∈Aasfollows:

R(X, Y)Z =g(Y, Z)X−g(X, Z)Y +g(φY, Z)φX−g(φX, Z)φY −2g(φX, Y)φZ +g(AY, Z)AX−g(AX, Z)AY +g(J AY, Z)J AX−g(J AX, Z)J AY +g(SY, Z)SX−g(SX, Z)SY

forany X,Y,Z∈T_zM,z∈M.

From this,contractingY andZ onM inQ^m,we have Ric(X) = (2m−1)X−X−φ²X−2φ²X

−g(AN, N)AX−X+g(AX, N)AN−g(J AN, N)J AX

−X+g(J AX, N)J AN+ (trS)SX−S²X

= (2m−1)X−3η(X)ξ−g(AN, N)AX+g(AX, N)AN

−g(J AN, N)J AX+g(J AX, N)J AN+ (trS)SX−S²X, (4.1) where wehaveusedthefollowing

2m−1

i=1 g(Ae_i, e_i) =TrA−g(AN, N) =−g(AN, N), 2m−1

i=1 g(AX, ei)Aei=2m

i=1g(AX, ei)Aei−g(AX, N)AN =X−g(AX, N)AN, 2m−1

i=1 g(J Aei, ei)J AX =2m

i=1g(J Aei.ei)J AX−g(J AN, N)J AX

=−g(J AN, N)J AX, and

2m−1

i=1 g(J AX, e_i)J Ae_i =2m

i=1g(J AX, e_i)J Ae_i−g(J AX, N)J AN

=J AJ AX−g(J AX, N)J AN

=X−g(J AX, N)J AN.

Now inthis section we consider only an A-principalnormal vector field N, thatis, AN =N,forarealhypersurfaceMinQ^mwithparallelRiccitensor.Then(4.1)becomes Ric(X) = (2m−1)X−2η(X)ξ−AX+hSX−S²X, (4.2) where h = trS denotes the mean curvature and is defined by the trace of the shape operatorS ofM inQ^m.Thenfrom this,as theRiccitensorisparallel,wehave

0 =−2g(∇Xξ, Y)ξ−2η(Y)∇Xξ−(∇XA)Y + (Xh)SY +h(∇XS)Y −(∇XS²)Y

=−2g(φSX, Y)ξ−2η(Y)φSX−(∇XA)Y + (Xh)SY

+h(∇XS)Y −(∇XS²)Y, (4.3)

where (∇XA)Y =∇X(AY)−A∇XY.Here, AY belongs to TzM, z∈M, from thefact that g(AY,N) = g(Y,AN) = g(Y,N) = 0 for any tangent vector Y on M. Then by puttingY =ξin(4.3),weknowthat

2φSX =−(∇XA)ξ+ (Xh)Sξ +h(∇XS)ξ−(∇XS²)ξ

=−q(X)J Aξ−αη(X)AN+α(Xh)ξ

+h(∇XS)ξ−(∇XS²)ξ, (4.4) wherewehaveput(∇XA)ξ=q(X)J Aξforacertain1-formqonM (see[11]).Inorder togettheequation(4.4)wehaveusedthefollowing

(∇XA)ξ=∇X(Aξ)−A∇Xξ

= ( ¯∇X(Aξ))^T −A∇Xξ

=

( ¯∇XA)ξ+A∇¯Xξ T

−AφSX

=q(X)J Aξ+AφSX+g(SX, ξ)AN−AφSX

=q(X)J Aξ+αη(X)AN,

where(· · ·)^T denotesthetangential componentof thevector (· · ·) inQ^m.Moreover, we get

(∇XS)ξ=∇X(Sξ)−S∇Xξ= (Xα)ξ+αφSX−SφSX, and

(∇XS²)ξ=∇X(S²ξ)−S²∇Xξ= (Xα²)ξ+α²φSX−S²φSX.

Then(4.4)canbe writtenas follows:

2φSX =−q(X)J Aξ−αη(X)AN+α(Xh)ξ+h(Xα)ξ+hαφSX−hSφSX

−(Xα²)ξ−α²φSX+S²φSX.

Then itcanbe reducedas follows:

2φSX =−q(X)J Aξ−αη(X)AN +hαφSX−hSφSX

−α²φSX+S²φSX.

From this,ifwetakethetangentialpart,wehavethefollowing:

(2 +α²−hα)φSX =−hSφSX+S²φSX (4.5) for any tangentvector X∈T_zM, z∈M, because we haveassumed that the unitvector fieldN isA-principal,thatis,AN =N,andJ Aξ=−AJ ξ=−AN.

On theother hand,by (3.4)inSection3forareal hypersurfaceincomplex quadric Q^m withA-principalnormalvectorfieldN,we have

2SφSX =α(φS+Sφ)X+ 2φX,

where we used g(X,AN)= 0 and g(J X,Aξ) = g(J X,J AN) = 0 from the facts that A-principalnormalAN =N andanti-commutingAJ=−J A.Fromthis,itfollowsthat

2S²φSX =α(SφS+S²φ)X+ 2SφX

=αα

2(Sφ+φS)X+φX +S²φ

X+ 2SφX

=α²

2(Sφ+φS)X+αφX+αS²φX+ 2SφX (4.6) Then summingup(4.5)and (4.6), wehave

(2 +α²−hα)φSX =−hα

2(Sφ+φS)X+φX +α²

4(Sφ+φS)X +α

2φX+α

2S²φX+SφX. (4.7)

Remark4.1.ItisknownthatinBerndtandSuh[4]arealhypersurfaceM isatubeover S^minQ^mifandonlyiftheshapeoperatorS ofM satisfiesSφ+φS=kφforanonzero constantk.Then letuscheck whetheratubeoverS^mcouldsatisfy(4.7)ornot.Under ourassumption weknowthat

(2 +α²−hα)φSX =−hαk

2 + 1 φX+α² 4kφX +α

2φX+α

2S²φX+SφX.

If we consider an eigen vector such that SX = λX, then (Sφ+φS)X = kφX gives that SφX = (k−λ)φX. From this, together with (4.7) using αk = −2, the principal

curvaturessatisfyaquadraticequation suchthat αx²−2(α²−hα+ 1)x= 0.

Then λ= 0 or μ= √ 2 tan√

2r (see [4]). Moreover, the trace hof the shape operator becomesh=α+ (m−1)k(seealso[4]).ButforatubeoverasphereS^mweknowthat

√2 tan√ 2r= 2

α(α²−hα+ 1)

= 2(α−h) + 2 α

= 4(m−1) α + 2

α

= 2(2m−1) α

=−(2m−1)√ 2 tan√

2r,

whereinthethirdequalityweusedα−h=−(m−1)k=^2(m_α⁻¹⁾ andinthefifthequality α=−√

2 cot(√

2r) respectively(seeProp. 3.4in[4]).Thisgivesthat2m√ 2 tan√

2r= 0, which gives us a contradiction. So we conclude that a real hypersurface in Q^m which is atubeover an m-dimensionalsphere S^m does notadmit a parallel Riccitensor. Of course,inthis casetheunitnormalN isA-principal.

IfweputSX =λX, then(4.5)gives

(2 +α²−hα)λφX =−hλSφX+λS²φX.

Moreover,(4.6)givesthat

SφX = αλ+ 2 2λ−αφX.

Fromthis, togetherwiththeaboveformula, wehave (2 +α²−hα)λφX=−hλαλ+ 2

2λ−α

φX+λαλ+ 2 2λ−α

2

φX. (4.8)

Firstweconsider thecasethatλ=0 andμ= ^αλ+2_2λ−α arebothnon-vanishing.Thenthe functionμ=^αλ+2_2λ−α satisfies thefollowingequation

μ²−hμ+hα−α²−2 = 0.

Thenitisequivalentto thefollowing forthefunctionλ λ²−hλ+hα−α²−2 = 0.

Combining thesetwo equations,wehave

(λ−μ)(λ+μ−h) = 0.

Then we know that the functions λand μ are distinct. So it implies thath = λ+μ.

Then itfollowsthat

h=λ+μ=α+ (m−1)(λ+μ) =α+ (m−1)h and thisimpliesthath=−_m−2^α ,whichgivesusacontradiction.

Infact,(4.7)withh=−_m−2^α becomes

2 +(m−1)α² m−2

φSX = α²

2(m−2)(φS+Sφ)X−hφX+α²

4 (Sφ+φS)X +α

2φX+α

2S²φX+SφX.

From this,bytakingthesymmetricpart,wegetthefollowing

3 +(m−1)α² m−2

g((φS−Sφ)X, Y) = α

2g((S²φ−φS²)X, Y).

Then for any principalvector X such thatSX =λX and SφX =μφX with distinct principalcurvaturesλandμweknow

(λ−μ)

3 + (m−1)α² m−2 −α

2(λ+μ)

φX = 0.

This implies3+(2(m−1)+1)α²

2(m−2) = 0,whichgivesacontradiction.

Next we consider the case λ= 0 and μ =−²_α=0. In this case, we caneasily verify thath=−_α².Thentheequation(4.7),togetherwith hα=−2,gives

(α²+ 4)φSX= (Sφ+φS)X−hφX+α²

4 (Sφ+φS)X +α

2φX+α

2S²φX+SφX. (4.9)

Then bytakingtheskew-symmetricpartof(4.9)wehave (α²+ 5)(φS−Sφ) = α

2(S²φ−φS²). (4.10) From this,using SX=λX andSφX =μφX,where μ=^αλ+2_2λ−α, thenwehave

(λ−μ)

(α²+ 5)−α

2(λ+μ) φX = 0. (4.11)

Sinceλ=μand hα=−2,(4.10) impliesthefollowing 0 =α²+ 5−α

2(λ+μ) =α²+ 5−α

2h=α²+ 6,

whichgivesalsoacontradictioneveninthiscase.Soweconcludethattheredonotexist anyrealhypersurfacesincomplexquadricQ^mwithparallelRiccitensorandA-principal normalvectorfield.

5. ProofofTheorem 2

Inthis sectionwewanttoproveTheorem2forrealhypersurfaceswithparallelRicci tensorand A-isotropic unitnormalvectorfield.Then inthis casewewantto makethe derivativeoftheRiccitensoras follows:

(∇YRic)X =∇Y(Ric(X))−Ric(∇YX)

=−3(∇Yη)(X)ξ−3η(X)∇Yξ

+g(X,∇Y(AN))AN−g(AX, N)∇Y(AN) +g((∇Y(Aξ), X)Aξ+η(AX)∇Y(Aξ) + (Y h)SX

+h(∇YS)X−(∇YS²)X. (5.1)

Sincewe assumedthatthe unitnormalN is A-isotropic, bythedefinition inSection3 weknowthatt=^π₄.ThenbytheexpressionoftheA-isotropicunitnormalvectorfield, (3.3)givesN = √¹

2Z1+√¹

2J Z2.Thisimpliesthat

g(ξ, Aξ) = 0, g(ξ, AN) = 0, g(AN, N) = 0, g(Aξ, N) = 0, and

g(J AN, ξ) =−g(AN, N) = 0.

Herewemayputthat

AN =BN+ρ(N)N,

whereBN = (AN)^T denotesthetangentialcomponentofthevectorfieldAN.Soρ(N)= g(AN,N)= 0 givesAN =BN.Thenitfollowsthat

∇Y(BN) =∇Y(AN) ={( ¯∇YA)N+A∇¯YN}^T ={q(Y)J AN−ASY}^T, and

∇Y(Aξ) ={( ¯∇YA)ξ+A∇¯Yξ}^T

={q(Y)J Aξ+AφSY}^T,

where (· · ·)^T denotes the tangential component of the vector (· · ·) in Q^m. By our as- sumptionof Ricciparallelism,theaboveformulabecomes

0 =−3g(φSY, X)ξ−3η(X)φSY +{q(Y)g(J AN, X)−g(ASY, X)}AN

−g(AX, N){q(Y)J AN−ASY}^T +{q(Y)g(J Aξ, X) +g(AφSY, X)}Aξ+η(AX){q(Y)J Aξ+AφSY}^T

+ (Y h)SX+h(∇YS)X−(∇YS²)X. (5.2)

Bytakingtheinnerproduct of(5.2)withtheReebvectorfieldξ,wehave 0 =−3g(φSY, X) +g(AX, N)η(ASY) +η(AX)η(AφSY)

+ (Y h)αη(X) +hg((∇YS)X, ξ)

−g((∇YS²)X, ξ). (5.3)

Ontheother hand,letususethefollowingcalculation (∇XS)ξ= (Xα)ξ+αφSX−SφSX, (∇XS²)ξ= (Xα²)ξ+α²φSX−S²φSX.

From this, together withputting X =ξin(5.2)and usingg(ξ,AN)=g(Aξ,ξ)= 0, we have

3φSY = (Y h)Sξ+h(∇YS)ξ−(∇YS²)ξ

−g(ASY, ξ)AN+g(AφSY, ξ)Aξ

= (Y h)αξ+h{(Y α)ξ+αφSY −SφSY}

− {(Y α²)ξ+α²φSY −S²φSY}

−g(ASY, ξ)AN+g(AφSY, ξ)Aξ. (5.4) Bytakingtheinnerproductinthesecondequalityof(5.4)withtheReebvectorfieldξ, and alsousingg(ξ,AN)=g(Aξ,ξ)= 0,weknowthatthefunctionhα−α² isconstant onM.Thenitcanbe rearrangedasfollows:

(3 +α²−αh)φSY =−hSφSY +S²φSY −g(ASY, ξ)AN+g(AφSY, ξ)Aξ. (5.5) Since the unit normal N is A-isotropic, we know that g(ξ,Aξ) = 0. Moreover, by Lemma 4.2in[14],wehavethefollowing

2SφSX=α(Sφ+φS)X+ 2φX−2g(X, AN)Aξ+ 2g(X, Aξ)AN. (5.6)

Now letus consider thedistribution Q^⊥, whichis anorthogonal complementof the maximal A-invariant subspace Q in the complex subbundle C of TzM, z∈M in Q^m. Then by Lemma 3.1 in Section 3, the orthogonal complement Q^⊥ = CQ becomes CQ= Span [AN,Aξ].FromtheassumptionofSφ=φS onthedistributionQ^⊥ itcan be easily checked thatthedistribution Q^⊥ is invariant bythe shape operatorS. Then (5.6)givesthefollowing forSAN =λAN

(2λ−α)SφAN = (αλ+ 2)φAN−2Aξ

= (αλ+ 2)φAN−2φAN

=αλφAN.

ThenAξ=φAN givesthefollowing

SAξ= αλ

2λ−αAξ. (5.7)

Thenfromtheassumption Sφ=φS onQ^⊥=CQitfollows thatλ= _2λ^αλ_−α gives

λ= 0 orλ=α. (5.8)

On the other hand, on the distribution Q we know that AX∈TzM, z∈M, because AN∈Q. So(5.6), together with the factthatg(X,Aξ)= 0 and g(X,AN)= 0 for any X∈Q,implythat

2SφSX=α(Sφ+φS)X+ 2φX. (5.9)

Then wecantakean orthonormalbasis X₁,· · ·,X_2(m−2)∈Q suchthatAX_i =λ_iX_i for i= 1,· · ·,m−2.Thenby(5.6)weknow that

SφXi= αλ_i+ 2 2λi−αφXi.

Accordingly,by(5.8)theshapeoperatorS canbe expressedinsuchawaythat

S=

⎡

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎣

α 0 0 0 · · · 0 0 · · · 0 0 0(α) 0 0 · · · 0 0 · · · 0 0 0 0(α) 0 · · · 0 0 · · · 0 0 0 0 λ₁ · · · 0 0 · · · 0 ... ... ... ... . .. ... ... · · · ... 0 0 0 0 · · · λm−2 0 · · · 0 0 0 0 0 · · · 0 μ1 · · · 0 ... ... ... ... ... ... ... . .. ... 0 0 0 0 · · · 0 0 · · · μm−2

⎤

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎦

So onthe distributionQ, forany X,φX∈Q suchthatSX =λX andSφX =μφX, μ=^αλ+2_2λ−α, (5.5)becomesthefollowing

(3 +α²−αh)λφX =−hλSφX+λS²φX

+λg(X, Aξ)AN+λg(AφX, ξ)Aξ

=−hλSφX+λμ²φX

=−hλμφX+λμ²φX, (5.10) where weusedCQ= Span [Aξ,AN] inthesecond equality.

Fromthis,ifthereexistsanon-vanishingprincipalcurvatureλ=0,thenanyprincipal curvature of the shape operatoron thedistribution Q satisfies thefollowing quadratic equation

x²−hx+ (αh−α²−3) = 0. (5.11) Then thediscriminant is alwaysD = (h−2α)²+ 12 >0,we concludethatthere exist two principalcurvaturesλand μsatisfyingh=λ+μ.Moreover, bytheassumptionof parallelRiccitensor∇Ric= 0,and(5.4)thefunctionhα−α²shouldbeconstant.Then summingupthesepropertiesandtheexpressionsabove,letusconsiderthefirstcasefor non-vanishingλ=μas follows:

h=λ+μ=α+ (m−2)(λ+μ)

=α+ (m−2)h. (5.12)

From this,theconstanthα−α² becomes

hα−α²=α{α+ (m−2)(λ+μ} −α²

= (m−2)(λ+μ)α

= (m−2)hα.

Then the constant hα implies that α must be constant. Thus the trace of the shape operatorhshouldbeconstant.Sotheabovequadraticequationhasconstantcoefficients.

Ontheother hand,as SAξ=αAξ andSAN =αAN theconstanthbecomes h=λ+μ

= 3α+ (m−2)(λ+μ)

= 3α+ (m−2)h. (5.13)

So if we compare the abovetwo equations (5.12) and (5.13) for the constantmean curvature h, we know that the Reeb function α should vanish, that is, α = 0. From

this, together withthe assumption m≥4, itfollows thatthe traceh=λ+μ= 0, that is,

λ=−μ. (5.14)

Moreover,fromtheequation (5.11)wemayputtheprincipalcurvature λ=−√

3 tanr and μ=√

3 cotr. (5.15)

Since the equation (5.11) has only two distinct solutions λ and μ and the prin- cipal curvature μ related to the almost complex structure φ such that SφX = μφX for SX = λX on the distribution Q, that is, Q = V(λ)⊕V(μ), where dimQ = 2(m − 2), V(λ) = {X∈Q|SX = λX} and V(μ) = {X∈Q|SX = μX}. Then, we know that dimV(λ) = dimV(μ) = m − 2, because λ and μ are different roots of the equation (5.11) and φV(λ) = V(μ). Moreover, we know that the function ρ = 0 or ρ = α by the property of the vectors AN and Aξ.

Summinguptheseresults,togetherwith(5.14)and(5.15),wemayputα= 0,ρ= 0, λ=−√

3 andμ=√

3 respectivelywithmultiplicities1,2,(m−2) and(m−2).

Next for the case where λ = 0 we conclude that on the distribution ξ⊕[CQ]⊕Q, whereCQ= Span[AN,Aξ],theshapeoperatorbecomes

S =

⎡

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎣

α 0 0 0 · · · 0 0 · · · 0 0 0(α) 0 0 · · · 0 0 · · · 0 0 0 0(α) 0 · · · 0 0 · · · 0 0 0 0 −_α² · · · 0 0 · · · 0 ... ... ... ... . .. ... ... · · · 0 0 0 0 0 · · · −_α² 0 · · · 0 0 0 0 0 · · · 0 0 · · · 0 ... ... ... ... ... ... ... . .. ...

0 0 0 0 · · · 0 0 · · · 0

⎤

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎦

WhenSAN = 0 andSAξ= 0,theexpression fortheconstanthbecomes h=λ+μ=α+ (m−2)(λ+μ).

Thisgivesthatα+ (m−3)h= 0.

Ontheother hand,forλ= 0 we knowμ=−_α². Soitfollows thath=λ+μ=−_α². Thenthesetwoequationsgiveα²= 2(m−3).Soletusputα=

2(m−3).Thenfrom theexpressionsabovethetraceoftheshapeoperatorbecomesrespectively

h=

2(m−3)− 2(m−2)

2(m−3) =− 2 2(m−3)

or

h=

2(m−3) + 2

2(m−3)− 2(m−2)

2(m−3) =− 2 2(m−3)

according tothe eigenvalue0 orαon thedistributionCQ= Span[Aξ,AN]. Thefirst equation issatisfiedforallofm.Sointhiscasewemayput

α=

2(m−3), γ= 0, λ= 0, andμ=− 2 2(m−3) with correspondingprincipalcurvaturespaces

T_α= [ξ], T_γ = [Aξ, AN], φ(T_λ) =T_μ,dimT_λ= dimT_μ =m−2.

Butifwecomparethelatterequation,wehavem= 3.Thisgivesusacontradiction.So thelattercasecannot beappeared.

Summinguptheabovediscussions,wegiveacompleteproofofourMainTheorem 2 intheintroduction.

Remark5.1.Inthisremarkletuscheck iftheRiccitensorofthetubeM overatotally geodesicCP^k inQ^m,m= 2k,mentionedinBerndtandSuh[3]isparallelornot.Then byatheoremin[3],theshapeoperatorScommuteswiththestructuretensorφ,thatis, Sφ=φS.InthiscaseweknowthatthenormalvectorfieldNofM inQ^2kisA-isotropic.

So letus suppose thatthe Riccitensor of M is parallel. Then forany Y∈Q such that SY =λY theequation(5.5)givesthat

(3 +α²−αh)λφY =−hλ²φY +λ³φY. (5.16) Ontheotherhand,by(5.6),togetherwiththecommutingpropertySφ=φS,weknow that

(2λ−α)λφY = (αλ+ 2)φY (5.17) From this,wesee thatthefunctionλ=0.Thenby(5.16)withλ=0,we get

λ²−hλ+ (αh−α²−3) = 0.

Moreover, (5.17)givesthat

λ²−αλ−1 = 0.

From thesetwoequationsweknowthath=αandfinallygivesusacontradiction.Soa tubeoverPkC neverhasparallelRiccitensor.

Acknowledgments

ThepresentauthorwouldliketoexpresshisdeepgratitudetoProfessorJürgenBerndt forhisgreatcontributionstoourworks[3]and[4].Motivatedbytheseworks,theauthor has considered such a notion of Ricci parallel for real hypersurfaces in the complex quadricQ^m.Theauthoralsowisheshisthankstotherefereeformanykindsofvaluable commentsto developthefirstversionofthismanuscript.

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