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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2021; 61(1): 181-190

Published online March 31, 2021

### Certain Characterization of Real Hypersurfaces of type A in a Nonflat Complex Space Form

U-Hang Ki

The National Academy of Siences, Seoul 06579, Korea
e-mail : uhangki2005@naver.com

Received: April 8, 2020; Revised: July 3, 2020; Accepted: August 18, 2020

### Abstract

Let M be a real hypersurface with almost contact metric structure $(ϕ,ξ,η,g)$ in a nonflat complex space form $Mn(c)$. We denote S and $Rξ$ by the Ricci tensor of M and by the structure Jacobi operator with respect to the vector field ξ respectively. In this paper, we prove that M is a Hopf hypersurface of type A in $Mn(c)$ if it satisfies $Rξϕ=ϕRξ$ and at the same time $Rξ(Sϕ−ϕS)=0$.

Keywords: real hypersurface, structure Jacobi operator, Ricci tensor, Hopf hypersurface.

### 1. Introduction

An n-dimensional complex space form $Mn(c)$ is a Kaehlerian manifold of constant holomorphic sectional curvature 4c. As is well known, complete and simply connected complex space forms are isometric to a complex projective space $Pnℂ$, or a complex hyperbolic space $Hnℂ$ according as $c>0$ or $c<0$.

In this paper we consider a real hypersurface M in a complex space form $Mn(c)$ for c≠0. Such an M has an almost contact metric structure $(ϕ,ξ,η,g)$ induced from the Kaehlerian metric and complex sturcture J of $Mn(c)$. The structure vector ξ is said to be principal if $Aξ=αξ$, where A is the shape operator in the direction of the unit normal N and $α=η(Aξ)$. In this case, it is known that α is locally constant [9] and that M is called a Hopf hypersurface [12].

In the study of real hypersurfaces in $Pnℂ$, Takagi [14, 15] classified all homogeneous Hopf hypersurfaces, and Cecil-Ryan [2] and Kimura [10] showed that they can be regarded as the tubes of constant radius over Kaehlian submanifolds. Such tubes can be divided into six types: A1, A2, B, C, D and E.

In the case of real hypersurfaces in $Hnℂ$, the classfication of homogenous real hypersurfaces in $Hnℂ$ is obtained by Berndt [1]. He showed that they are realized as the tubes of constant radius over certain submanifolds. Such tubes are said to be real hypersurfaces of type A0, A1, A2 or B. Among the several types of real hypersurfaces appearing in Takagi's list or Berndt's list, several are tubes over totally geodesic $Pnℂ$ or $Hnℂ$ $(0≤k≤n−1)$. These and a horosphere in $Hnℂ$ are together said to be of type A. By a theorem due to Okumura[13] and to Montiel and Romero[11]we have

Theorem O. Let M be a real hypersurface of $Pnℂ$, $n≥2$. If it satisfies

$g((Aϕ−ϕA)X,Y)=0$

for any vector fields X and Y, then M is locally congruent to a tube of radius r over one of the following Kaehlerian submanifolds :

• (A1) a hyperplane $Pn−1ℂ$, where $0,

• (A2) a totally geodesic $Pkℂ$ $(1≤k≤n−2)$, where $0.

Theorem MR. Let M be a real hypersurface of $Hnℂ$, $n≥2$. If it satisfies (1.1), then M is locally congruent to one of the following hypersurface :

• (A0) a horosphere in $Hnℂ$, i.e., a Montiel tube,

• (A1) a geodesic hypersphere, or a tube over a hyperplane $Hn−1ℂ$,

• (A2) a tube over a totally geodesic $Hkℂ$ $(1≤k≤n−2)$.

Characterization problems for a real hypersurface of type A in a complex space form were studied by many authors (cf. [3, 4, 5], etc.).

We denote by S and Rξ be the Ricci tensor and the structure Jacobi operator with respect to the vector field ξ of M respectively. To investigate of real hypersurfaces with respect to the structure Jacobi operator it is a very important problem to study real hypersurfaces satisfying $Rξϕ=ϕRξ$ in $Mn(c)$. Real hypersurfaces in a complex space form $Mn(c)$ for $c≠0$, which satisfies both $Rξϕ=ϕRξ$ and $RξS=SRξ$, have been studied in [6, 7, 8].

Under the condition $RξA=ARξ$ we know that the following theorem ([3,4]):

Theorem CK.([4]) Let M be a real hypersurface of $Mn(c)$, $c≠0$. If M satisfies $Rξϕ=ϕRξ$ and at the same time satisfies $RξA=ARξ$, then M is a Hopf hypersurface. Further, M is of type A or a Hopf hypersurface with $g(Aξ,ξ)=0$.

The purpose of this paper is, using the hypothesis concerned with the Ricci tensor S and the structure Jacobi operator Rξ, to establish the following theorem another characterizing homogenous real hypersurfaces of type A and some special classes of Hopf hypersurfaces:

Theorem 1.1. Let M be a real hypersurface in a nonflat complex space form $Mn(c)$ $(c≠0,n≥2)$ if M satisfies $ϕRξ=Rξϕ$ and at the same time $Rξ(Sϕ−ϕS)=0$, then M is a Hopf hypersurface. Further, M is locally congruent to one of homogenous real hypersurfaces of type A or a Hopf hypersurface with $g(Aξ,ξ)=0$.}

All manifolds in the present paper are assume to be connected and of class $C∞$ and the real hypersurfaces supposed to be orientable.

### 2. Preliminaries

Let M be a real hyperusurface immersed in a complex space form Mn (c), c ≠ 0 with almost complex structure J, and N be a unit normal vector field on M. The Riemannian connection $∇˜$ in Mn (c) and $∇$ in M are related by the following formulas for any vector fields X and Y on M :

$∇˜XY=∇XY+g(AX,Y)N, ∇˜XN=−AX,$

where g denotes the Riemannian metric tensor of M induced from that of Mn (c), and A denotes the shape operator of M in the direction N.

For any vector field X tangent to M, we put

$JX=ϕX+η(X)N, JN=−ξ.$

We call ξ the structure vector field (or the Reeb vector field) and its flow also denoted by the same latter ξ. The Reeb vector field ξ is said to be principal if $Aξ=αξ$, where $α=η(Aξ)$.

A real hypersurface M is said to be a Hopf hypersurface if the Reeb vector field ξ is principal. It is known that the aggregate $(ϕ,ξ,η,g)$ is an almost contact metric structure on M, that is, we have

for any vector fields X and Y on M. From Kaehler condition $∇˜J=0$, and taking account of above equations, we see that

$∇Xξ=ϕAX,$
$(∇Xϕ)Y=η(Y)AX−g(AX,Y)ξ$

for any vector fields X and Y tangent to M.

Since we consider that the ambient space is of constant holomorphic sectional curvature 4c, equations of the Gauss and Codazzi are respectively given by

$R(X,Y)Z=​​​c{g(Y,Z)X−g(X,Z)Y+g(ϕY,Z)ϕX−g(ϕX,Z)ϕY​ −2g(ϕX,Y)ϕZ}+g(AY,Z)AX−g(AX,Z)AY,$
$(∇XA)Y−(∇YA)X=c{η(X)ϕY−η(Y)ϕX−2g(ϕX,Y)ξ}$

for any vector fields X, Y and Z on M, where R denotes the Riemannian curvature tensor of M.

In what follows, to write our formulas in convention forms, we denote by $α=η(Aξ)$, $β=η(A2ξ)$, $γ=η(A3ξ)$ and $h=TrA$, and for a function f we denote by $∇f$ the gradient vector field of f.

From the Gauss equation (2.3), the Ricci tensor S of M is given by

$SX=c{(2n+1)X−3η(X)ξ}+hAX−A2X$

for any vector field X on M, which implies

$Sξ=2c(n−1)ξ+hAξ−A2ξ.$

Now, we put

$Aξ=αξ+μW,$

where W is a unit vector field orthogonal to $ξ$. In the sequel, we put $U=∇ξξ$, then by (2.1) we see that $U=μϕW$ and hence U is orthogonal to W. So we have $g(U,U)=μ2$. Using (2.7), it is clear that

$ϕU=−Aξ+αξ,$

which shows that $g(U,U)=β−α2$. Thus it is seen that

$μ2=β−α2.$

Making use of (2.1), (2.7) and (2.8), it is verified that

$μg(∇XW,ξ)=g(AU,X),$
$g(∇Xξ,U)=μg(AW,X)$

because W is orthogonal to $ξ$.

Now, differentiating (2.8) covariantly and taking account of (2.1) and (2.2), we find

$(∇XA)ξ=−ϕ∇XU+g(AU+∇α,X)ξ−AϕAX+αϕAX,$

which together with (2.4) implies that

$(∇ξA)ξ=2AU+∇α.$

Applying (2.12) by ϕ and making use of (2.11), we obtain

$ϕ(∇XA)ξ=∇XU+μg(AW,X)ξ−ϕAϕAX−αAX+αg(Aξ,X)ξ,$

which connected to (2.1) and (2.13) gives

$∇ξU=3ϕAU+αAξ−βξ+ϕ∇α.$

Using (2.3), the structure Jacobi operator $Rξ$ is given by

$Rξ(X)=R(X,ξ)ξ=c{X−η(X)ξ}+αAX−η(AX)Aξ$

for any vector field X on M, which implies that

$Rξξ=0,$
$RξU=cU+αAU, RξAU=cAU+αA2U.$

From (2.5) we obtain

$SU=c(2n+1)U+hAU−A2U,$
$SAξ=c{(2n+1)Aξ−3αξ}+hA2ξ−A3ξ.$

Because of (2.5) and (2.7), we also have

$μSW=hA2ξ−A3ξ−α(hAξ−A2ξ)+c(2n+1)(Aξ−αξ).$

### 3. The Structure Jacobi Operator of Real Hypersurfaces

Let M be a real hypersurface in complex space form Mn (c), c ≠ 0 satisfying $Rξϕ=ϕRξ$, which means that the eigenspace of Rξ is invariant by the structure operator ϕ. Then by (2.16) we have

$α(ϕAX−AϕX)=g(Aξ,X)U+g(U,X)Aξ.$

We set $Ω={p∈M:μ(p)≠0}$, and suppose that $Ω$ is nonvoid, that is, $ξ$ is not principal curvature vector on M. In the sequel, we discuss our arguments on the open subset $Ω$ of M unless otherwise stated. Then, it is, using (3.1), clear that α≠0 on $Ω$. So a function $λ$ given by $β=αλ$ is defined. Thus, replacing X by U in (3.1) and using (2.8), we find

$ϕAU=λAξ−A2ξ.$

Applying by $ϕ$, we have

$ϕA2ξ=AU+λU,$

which together with (2.7) yields

$μϕAW=AU+(λ−α)U.$

Since W is orthogonal to U, we see from the last equation

$g(AW,U)=0.$

If we replace X by AU in (3.1) and take account of (3.2),

then we find

$αϕA2U−α(λA2ξ−A3ξ)=g(AU,U)Aξ,$

which enables us to obtain

$g(AU,U)=γ−αλ2.$

Further, we assume in the sequel that

$Rξ(Sϕ−ϕS)X=0$

for any vector field X on M.

Applying this by $ξ$, we have $RξϕSξ=0$, which together with (2.6) gives $Rξϕ(hAξ−A2ξ)=0$. Thus, it follows that

$RξAU=(h−λ)RξU$

by virtue of (3.6). Because of (2.18) we can write (3.9) as

$hAU−A2U=(λ+cα)AU−cα(h−λ)U$

since $α≠0$ on $Ω$. Applying this by $ϕ$ and using (2.8) and (3.2), we find

$αϕA2U={α(h−λ)−c}(λAξ−A2ξ)−c(h−λ)(Aξ−αξ).$

If we combine this to (3.6), then we get

$αA3ξ=(αh−c)A2ξ+(γ−αhλ+ch)Aξ+cα(λ−h)ξ.$

where we have used (3.7), which tells us that

$α(hA2ξ−A3ξ)=cA2ξ+(αλh−γ−ch)Aξ+cα(h−λ)ξ.$

Tranforming this to A and make use of (3.11), we have

$α(hA3ξ−A4ξ)={λαh−γ−c2α}A2ξ +c{γα−λh+chα+α(h−λ)}Aξ+c2(λ−h)ξ.$

From (2.20) and (3.12) we get

$αSAξ=cA2ξ+{c(2n+1)α−γ+αλh−ch}Aξ+cα(h−λ−3α)ξ.$

Combining (2.19) to (3.10), we find

$SU=(λ+cα)AU+{c(2n+1)+cα(λ−h)}U.$

Now, we see from (2.21) and (3.12) that

$μSW=(α+cα)A2ξ+{c(2n+1)+h(λ−α)+1α(γ+ch)}Aξ +{c(h−λ)+c(2n+1)α}ξ,$

which connected to (3.6) implies that

$μϕSW=(α+cα)AU+{αλ+cλα+c(2n+1)+h(λ−α)−1α(γ+ch)}U.$

In the next step, if we apply by $μW$ to (3.8), then we find $Rξ(μϕSW−SU)=0$, which together with (3.15) and (3.16) gives

$Rξ{(α+cα)AU+(α+cα+h(λ−α)−1α(γ+ch))U}=Rξ{(λ+cα)AU−cα(h−λ)U}.$

If we use (3.8) to this, then we obtain $(hλ−γα)RξU=λ(h−λ)RξU$. Hence we have

$(γ−αλ2)RξU=0,$

which together with (3.7) yields $g(AU,U)RξU=0$.

Now, suppose that $g(AU,U)≠0$ on $Ω$. Then we have $RξU=0$ on this open subset. We discuss our arguments on such a place. So we have $αAU+cU=0$. Since $α≠0$ on $Ω$ because of (3.1), we can write this as $AU=−cαU$. Thus (3.2) turns out to be

$A2ξ=ρAξ+cξ$

because of (2.8), where we have put $ρ=λ−cα$. Therefore we verify that $RξA=ARξ$.

In fact, from (2.16) we have

$g(RξY,AX)−g(RξX,AY)=g(A2ξ,Y)g(Aξ,X)−g(A2ξ,X)g(Aξ,Y) +c{g(Aξ,Y)η(X)−g(Aξ,X)η(Y)},$

which together with (3.18) gives the requied relationship. According to Theorem CK, we conclude that $μ=0$, a contradiction. Therefore $g(AU,U)=0$ is proved. Hence, from (3.7) we have

$γ=αλ2.$

We are now going to prove that AU=0 on $Ω$. If we use (3.19), then (3.11) can be written as

$A3ξ=(h−cα)A2ξ+(λ2−hλ+cα)Aξ+c(λ−h)ξ.$

Using (3.19), we also have from (3.13)

$hA3ξ−A4ξ={λh−λ2−(cα)2}A2ξ +{cα(λ2−λh+cαh)+c(h−λ)}Aξ+c2α(λ−h)ξ.$

Because of (2.5) and (3.2) we have

$(Sϕ−ϕS)AU=λSAξ−SA2ξ−ϕASU,$

which connected to (3.8) gives

$Rξ(λSAξ−SA2ξ)=RξϕSAU.$

On the other hand, we have from (2.5)

$λSAξ−SA2ξ=λ{(2n+1)cAξ−3cαξ+hA2ξ−A3ξ} −(2n+1)cA2ξ+3cαλξ−(hA3ξ−A4ξ),$

which together with (3.2), (3.12), (3.19) and (3.21) yields

$SϕAU={cαλ−(2n+1)c−λh+λ2+( c α 2)}A2ξ +3cαλξ−c2α(λ−h)ξ+{(2n+1)cλ +λ(λh−λ2−cαh)−cα(λ2−λh+cαh)−c(h−λ)}Aξ.$

We also, using (3.2) and (3.6) with g(AU,U)=0, verify that

$ϕSAU=(λ+cα)(λA2ξ−A3ξ)+{(2n+1)c+cα(λ−h)(λAξ−A2ξ),$

or, using (3.20)

$ϕSAU={λ(λ+cα)−λ−cα+cα(h−λ)−(2n+1)c}A2ξ +{(2n+1)cλ+cαλ(λ−h)−(λ+cα)(λ2−hλ+chα)}Aξ.$

Combining (3.23) to this, it follows that

$SϕAU−ϕSAU={cαλ(h−λ)−c(h−λ)}Aξ.$

Using (3.23) and this, (3.22) reformed as $(h−λ)(λ−α)RξAξ=0$ and hence $(h−λ)RξAξ=0$ by virtue of (2.9) and (2.17). Accordingly we obtain $h−λ=0$ on $Ω$.

In fact, if not, then we have $RξAξ=0$, which together with (2.16) implies that $αA2ξ=(β−c)Aξ−cαξ$ on this open subset. Applying by $ϕ$ and using (3.2), we find $α(AU+λU)=(β−c)U$ on the set. If we apply this by U and taking account of the fact g(AU,U)=0, then we have $αλ=β−c$, a contradiction. Thus, $h−λ=0$ on $Ω$ is proved.

Therefore (3.8) tells us that $RξAU=0$, that is, $αA2U+cAU=0$, which shows $g(A2U,U)=0$ because of $g(AU,U)=0$. Consequently we have

$AU=0$

on $Ω$ and thus (3.2) becomes

$A2ξ=λAξ.$

We are now going to prove $Ω=∅$. Differentiating (3.25) covariantly along $Ω$

and taking account of (2.1), we find

$g((∇XA)Aξ,Y)+g(A(∇XA)ξ,Y)+g(A2ϕAX,Y)=(Xλ)g(Aξ,Y)+λg((∇XA)ξ,Y)+λg(AϕAX,Y),$

which together with (2.13) and (3.24) yields

$2g((∇XA)ξ,Aξ)=λ(Xα)+α(Xλ),$

or, using (2.4),

$(∇ξA)Aξ=12∇β−cU.$

Putting $X=ξ$ in (3.27) and taking account of (2.13) and (3.24) and the last relationship, we obtain

$12∇β=−A∇α+λ∇α+(ξλ)Aξ+cU,$

which together with (3.24) implies that

$12Uβ=λ(Uα)+cμ2.$

Thus, it follows that

$α(Uλ)−λ(Uα)=2cμ2$

by virtue of $β=αλ$.

On the other hand, if we put $X=Aξ$ in (3.26) and make use of (2.4), (2.7), (2.13) and (3.24), then we get

$12(A∇β−λ∇β)+(α2+μ2)∇λ=g(Aξ,∇λ)Aξ+c(3α−2λ)U.$

Taking the inner product with U to this and using (3.27), we find

$λ(α(Uλ)−λ(Uα))=c(3α−λ)μ2,$

which together with (3.28) gives $c(λ−α)μ2=0$, a contradiction. Thus, $Ω=∅$ is proved.

Proof of Theorem 1.1. Since we know that $Ω=∅$, M is a Hopf hypersurface. Hence (3.1) turns out to be $α(Aϕ−ϕA)=0$. Thus our Theorem follows from Theorem O and Theorem MR. This completes the proof.

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