Real hypersurfaces in complex space forms with $xi$-parallel Ricci tensor and structure Jacobi operator

REAL HYPERSURFACES IN COMPLEX SPACE FORMS WITH ξ-PARALLEL RICCI TENSOR AND STRUCTURE

JACOBI OPERATOR

U-Hang Ki, Juan De Dios P´ erez, Florentino G. Santos, and Young Jin Suh

Reprinted from the

Journal of the Korean Mathematical Society Vol. 44, No. 2, March 2007

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°2007 The Korean Mathematical Society

REAL HYPERSURFACES IN COMPLEX SPACE FORMS WITH ξ-PARALLEL RICCI TENSOR AND STRUCTURE

JACOBI OPERATOR

U-Hang Ki, Juan De Dios P´ erez, Florentino G. Santos, and Young Jin Suh

Abstract. We know that there are no real hypersurfaces with parallel Ricci tensor or parallel structure Jacobi operator in a nonflat complex space form (See [4], [6], [10] and [11]). In this paper we investigate real hypersurfaces M in a nonflat complex space form M

(c) under the con- dition that ∇

S = 0 and ∇

R

= 0, where S and R

respectively denote the Ricci tensor and the structure Jacobi operator of M in M

(c).

0. Introduction

A Kaehler manifold of constant holomorphic sectional curvature c is called a complex space form, which is denoted by M n (c). It is well known that complete and simply connected complex space forms are isometric to a complex projective space P n C, a complex Euclidean space C ⁿ or a complex hyperbolic space H n C according as c > 0, c = 0 and c < 0.

In this paper we consider a real hypersurface M in a complex space form M _n (c), c6=0. Then M has an almost contact metric structure (φ, ξ, η, g) induced from the complex structure J and the Kaehler metric of M n (c). The structure vector field ξ is said to be principal if Aξ = αξ is satisfied, where A denotes the shape operator of M and α = η(Aξ). A real hypersurface is said to be a Hopf hypersurface if the structure vector field ξ of M is principal.

In the study of real hypersurfaces in P n C, Takagi [12] classified all homo- geneous real hypersurfaces and Cecil and Ryan [2] showed that they can be regarded as the tubes of constant radius over Kaehler submanifolds when the structure vector field ξ is principal. Such tubes can be divided into six kinds of type A 1 , A 2 , B, C, D and E.

Received August 8, 2005; Revised May 30, 2006.

2000 Mathematics Subject Classification. Primary 53C40; Secondary 53C15.

Key words and phrases. real hypersurface, structure Jacobi operator, Ricci tensor, Hopf hypersurface.

The second and the third authors were supported by MCYT-FEDER Grant MTM 2004- 04934-C04-01 and the fourth author was supported by the grant Proj. No. R14-2002-003- 01001-0 from Korea Research Foundation, Korea 2006.

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°2007 The Korean Mathematical Society

307

On the other hand, real hypersurfaces in H n C have been investigated by Berndt [1], Montiel and Romero [7] and so on. Berndt [1] classified all homo- geneous real hypersurfaces in H n C and showed that they are realized as the tubes of constant radius over certain submanifolds. Also such kinds of tubes are said to be real hypersurfaces of type A 0 , A 1 , A 2 or type B.

Now, let M be a real hypersurface in M n (c), c 6= 0. Then we introduce the following theorems due to Okumura [9] for c > 0 and Montiel and Romero [7]

for c < 0 respectively.

Theorem A. Let M be a real hypersurface of P n C, n ≥ 2. If it satisfies

(0.1) g((Aφ − φA)X, Y ) = 0

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

(A 1 ) a hyperplane P n−1 C, where 0 < r < π/2,

(A ₂ ) a totally geodesic P _k C (1 ≤ k ≤ n − 2), where 0 < r < π/2.

Theorem B. Let M be a real hypersurface of H n C, n ≥ 2. If it satisfies (0.1), then M is locally one of the following hypersurfaces :

(A 0 ) a horosphere in H n C, i.e., a Montiel tube,

(A 1 ) a geodesic hypersphere and a tube over a hyperplane H n−1 C, (A 2 ) a tube over a totally geodesic H k C (1 ≤ k ≤ n − 2).

On the other hand, it is well known that there are no real hypersurfaces with parallel Ricci tensor in M _n (c), n ≥ 3, c 6= 0 (see [4]). Recently, Kim [6] proved that this is also true when n = 2. So it should be natural to investigate real hypersurfaces M in M n (c) by using some conditions about covariant derivative of S which are weaker than ∇S = 0, where ∇ and S denotes the Levi-Civita connection and the Ricci tensor of M in M n (c) respectively.

Along this direction we introduce a theorem due to [3] as follows:

Theorem C. Let M be a real hypersurface in a complex space form M n (c), c6=0 satisfying ∇ ξ S = 0 and Sξ = σξ for some constant σ. Then M is a Hopf hypersurface.

A Jacobi field along geodesics of a given Riemannian manifold (M, g) is an important role in the study of differential geometry. It satisfies a well known differential equation which inspires Jacobi operators. The Jacobi operator is defined by (R X (Y ))(p) = (R(Y, X)X)(p), where R denotes the curvature tensor of M and X, Y denote tangent vector fields on M . Then we see that R X is a self-adjoint endomorphism on the tangent space of M and is related to the differential equation, so called Jacobi equation, which is given by ∇ γ

(∇ γ

Y ) + R(Y, γ ⁰ )γ ⁰ = 0 along a geodesic γ on M , where γ ⁰ denotes the velocity vector along γ on M .

When we study a real hypersurface M in a complex space form M n (c), c6=0,

we will call the Jacobi operator on M with respect to the structure vector ξ the

structure Jacobi operator on M and will denote it by R ξ , where R ξ is defined by R ξ (X) = R(ξ, X)X for the curvature tensor R and any tangent vector field X on M . But, recently it is known that there are no real hypersurfaces in M n (c) with parallel structure Jacobi operator R ξ , that is, ∇ X R ξ = 0 for any tangent vector field X on M in M n (c), c6=0 (see [10] and [11]).

Motivated by Theorem C and such a view point of the parallel structure Jacobi operator we are able to consider a covariant derivative or a Lie derivative for the Ricci tensor S and the structure Jacobi operator R ξ along the direction of ξ. This condition ∇ ξ S = 0 (resp. ∇ ξ R ξ = 0 or L ξ R ξ = 0 in [11]) are weaker than the notion of ∇S = 0 (resp. ∇R ξ = 0 or LR ξ = 0), respectively.

Now in this paper we prove the following:

Theorem 0.1. Let M be a real hypersurface in a complex space form M n (c) satisfying ∇ ξ S = 0 and ∇ ξ R ξ = 0. Then M becomes a Hopf hypersurface in M n (c). Further, M is locally congruent to one of the following hypersurfaces:

(1) In cases P n C

(A 1 ) a tube of radius r over a hyperplane P n−1 C, where 0 < r < ^π ₂ ,

(A 2 ) a tube of radius r over a totally geodesic P k C(1 ≤ k ≤ n − 2), where 0 < r < ^π ₂ ,

(T ) a tube of radius ^π ₄ over a certain complex submanifold in P n C, (2) In cases H n C

(A 0 ) a horosphere in H n C, i.e., a Montiel tube,

(A 1 ) a geodesic hypersphere and a tube over a hyperplane H n−1 C, (A 2 ) a tube over a totally geodesic H k C (1 ≤ k ≤ n − 2).

1. Preliminaries

Let M be a real hypersurface immersed in a complex space form M _n (c) with parallel almost complex structure J and N be a unit normal vector field on M . By ˜ ∇ we denote the Levi-Civita connection with respect to the Fubini- Study metric ˜ g of M _n (c). Then the Gauss and Weingarten formulas are given respectively by

∇ ˜ Y X = ∇ Y X + g(AY, X)N, ∇ ˜ X N = −AX

for any vector fields X and Y on M , where ∇ and g denote the Riemannian connection and the Riemannian metric induced from ˜ g respectively, and A denotes the shape operator in the direction of N .

For any vector field X tangent to M , we put

JX = φX + η(X)N, JN = −ξ.

Then we may see that (φ, ξ, η, g) is an almost contact metric structure on M , that is, we have

φ ² X = −X + η(X)ξ, g(φX, φY ) = g(X, Y ) − η(X)η(Y ),

η(ξ) = 1, φξ = 0, η(X) = g(X, ξ)

for any vector fields X and Y on M . From the fact ˜ ∇J = 0 and by using the Gauss and Weingarten formulas, we obtain

(1.1) (∇ X φ)Y = η(Y )AX − g(AX, Y )ξ,

(1.2) ∇ _X ξ = φAX.

Since the ambient manifold is of constant holomorphic sectional curvature c, we have the following Gauss and Codazzi equations respectively:

(1.3) R(X, Y )Z = c

4 {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,

(1.4) (∇ X A)Y − (∇ Y A)X = c

4 {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 .

Now let us denote by α = η(Aξ), β = η(A ² ξ), γ = η(A ³ ξ) and h = T rA, and

∇f the gradient vector field of the function f defined on M .

Now let us denote by S the Ricci tensor of M in M n (c). Then we have from (1.3)

(1.5) SX = c

4 {(2n + 1)X − 3η(X)ξ} + hAX − A ² X, which together with (1.2) implies that

(1.6) (∇ X S)Y = − 3

4 c{g(φAX, Y )ξ + η(Y )φAX} + (Xh)AY + (hI − A)(∇ X A)Y − (∇ X A)AY,

where I denotes the identity map on the tangent space T p M , p ∈ M .

We put U = ∇ ξ ξ, then U is orthogonal to the structure vector field ξ. Thus it is, using (1.2), seen that

(1.7) φU = −Aξ + αξ,

which shows that g(U, U ) = β −α ² . We easily see that ξ is a principal curvature vector, that is, Aξ = αξ if and only if β − α ² = 0.

If Aξ − g(Aξ, ξ)ξ 6= 0, then we can put

(1.8) Aξ = αξ + µW,

where W is a unit vector field orthogonal to ξ. Then by (1.2) we see that U = µφW and hence g(U, U ) = µ ² . So we have

(1.9) µ ² = β − α ² .

Further, W is also orthogonal to U . Using (1.2) and (1.8), it is seen that

(1.10) µg(∇ X W, ξ) = g(AU, X),

(1.11) g(∇ X ξ, U ) = µg(AW, X).

Now, differentiating (1.7) covariantly along M and making use of (1.1) and (1.2), we find

(1.12) g(φX, ∇ Y U ) + η(X)g(AU + ∇α, Y )

= g((∇ Y A)X, ξ) − g(AφAX, Y ) + αg(AφX, Y ), which enables us to obtain

(1.13) (∇ ξ A)ξ = 2AU + ∇α

because of (1.4).

By the definition of U , (1.1) and (1.13), it is verified that (1.14) ∇ ξ U = 3φAU + αAξ − βξ + φ∇α.

From the Gauss equation (1.3) the structure Jacobi operator R ξ is given by (1.15) R ξ X = R(X, ξ)ξ = c

4 {X − η(X)ξ} + αAX − η(AX)Aξ for any vector field X on M .

We set Ω = {p ∈ M |µ(p) 6= 0}, and suppose that Ω 6= ∅, that is, ξ is not a principal curvature vector on M . Hereafter, unless otherwise stated, we continue our discussions on the open set Ω of M .

2. Real hypersurfaces in M n (c) satisfying ∇ ξ R ξ = 0 Differentiating (1.15) covariantly, we find

g((∇ X R ξ )Y, Z)

= − c

4 {η(Z)g(∇ X ξ, Y ) + η(Y )g(∇ X ξ, Z)} + (Xα)g(AY, Z) + αg((∇ X A)Y, Z) − g(Aξ, Z){g((∇ X A)ξ, Y ) − g(AφAY, X)}

− g(Aξ, Y ){g((∇ X A)ξ, Z) − g(AφAZ, X)}, which together with (1.2) and (1.13) implies that

(2.1)

g((∇ ξ R ξ )Y, Z) = − c

4 {u(Y )η(Z) + u(Z)η(Y )} + (ξα)g(AY, Z) + αg((∇ ξ A)Y, Z) − g(Aξ, Z){3g(AU, Y ) + Y α}

− g(Aξ, Y ){3g(AU, Z) + Zα},

where u is a 1-form defined by u(X) = g(U, X) for any vector field X and U = µφW defined in (1.8).

Moreover, assume that ∇ _ξ R _ξ = 0. Then we have from (2.1)

(2.2) α(∇ ξ A)X + (ξα)AX = c

4 {u(X)ξ + η(X)U } + η(AX){3AU + ∇α}

+ {3g(AU, X) + Xα}Aξ.

Putting X = ξ in this and making use of (1.13), we find

(2.3) αAU + c

4 U = 0, which shows that α 6= 0 on Ω.

Putting X = αU in (2.2) and using (2.3), we obtain (2.4) α ² (∇ ξ A)U − c

4 (ξα)U = c

4 αµ ² ξ + {α(U α) − 3

4 cµ ² }Aξ.

Because of (2.3), the equation (1.14) turns out to be α∇ ξ U = 3

4 cµW + α ² Aξ − αβξ + αφ∇α.

Differentiating (2.3) covariantly along Ω, we find (Xα)AU + α(∇ X A)U + αA(∇ X U ) + c

4 ∇ X U = 0.

If we replace X = αξ in this and take account of (2.3) and (2.4), then we obtain (2.5) c

4 αµ ² ξ + {α(U α) − 3

4 cµ ² }Aξ + α ² A(∇ ξ U ) + c

4 α∇ ξ U = 0, which together with (1.14) gives

(2.6) αAφ∇α + c

4 φ∇α + (U α)Aξ + µ(α ² + 3

4 c){AW − µξ − 1 α (µ ² − c

4 )W } = 0, where we have used (1.8).

3. Real hypersurfaces satisfying ∇ ξ R ξ = 0 and ∇ ξ S = 0

In this section, we will continue our discussions on a real hypersurface M in M n (c) satisfying ∇ ξ R ξ = 0 and ∇ ξ S = 0. Then replacing X by ξ in (1.6) and using the Codazzi equation (1.4), we obtain

(3.1)

3

4 c{u(X)η(Y ) + u(Y )η(X)} + c

4 {g(AY, φX) + g(AX, φY )}

= (ξh)g(AX, Y ) + hg((∇ Y A)X, ξ) − c

4 hg(φX, Y )

− g(AY, (∇ X A)ξ) − g(AX, (∇ Y A)ξ).

On the other hand, ∇ ξ R ξ = 0 implies

(3.2)

α ² (∇ ξ A)X = − α(ξα)AX + c

4 α{u(X)ξ + η(X)U } + {α(Xα)

− 3

4 cu(X)}Aξ + (α∇α − 3

4 cU )g(Aξ, X)

because of (2.2) and (2.3). Combining above two equations, we get

(3.3) c

4 (3α ² − hα − c

4 )(u(X)ξ + η(X)U ) − 2α(ξα)A ² X + α{h(ξα) − α(ξh)}AX

= g(Aξ, X){hα∇α − αA∇α + 3

4 c(AU − hU ) − c 4 αU } + £

hα(Xα) − αg(A∇α, X) + 3 4 c ©

g(AU, X) − hu(X) ª

− c

4 αu(X) ¤ Aξ + © 3

4 cu(X) − α(Xα) ª

A ² ξ + © 3

4 cU − α∇α ª

g(A ² ξ, X).

Since U is orthogonal to ξ, we see, using (1.8) and (2.3), that g(A ² ξ, U ) = 0.

Thus, replacing X by U in (3.3) and taking account of (2.3), we find

(3.4) c

4 (3α ² − hα − c

4 )µ ² ξ − c 4 {( c

2α + h)ξα − α(ξh)}U

= {(hα + c

4 )U α − 3 4 c( c

4α + h)µ ² − c

4 αµ ² }Aξ + { 3

4 cµ ² − α(U α)}A ² ξ, which enables us to obtain

(3.5) ( c

2 + hα)ξα − α ² (ξh) = 0.

We notice here that α(U α) 6= ³ ₄ cµ ² on Ω. In fact, if not, then we have αAξ = (hα + ^c ₄ − 3α ² )ξ by virtue of (3.4) and (3.5). From this, by taking an inner product with ξ, we have 4α ² = hα + ₄ ^c on this subset. So we have Aξ = αξ, a contradiction. Therefore α(U α) − ³ ₄ cµ ² 6= 0 is satisfied everywhere.

Consequently we have from (3.4) and (3.5) the following:

(3.6) A ² ξ = ρAξ + (β − ρα)ξ,

where the function ρ (resp. β) is defined by µρ = g(A ² ξ, W ) (resp. µ ² = β −α ² in (1.9)).

Combining (1.8) to (3.6), we see that

(3.7) AW = µξ + (ρ − α)W

because of µ 6= 0.

Differentiating (3.7) covariantly, we find

(∇ X A)W + A∇ X W = (Xµ)ξ + µ∇ X ξ + X(ρ − α)W + (ρ − α)∇ X W, which shows that

(3.8) g((∇ X A)W, W ) = −2g(AU, X) + Xρ − Xα, (3.9) µ(∇ W A)ξ = (ρ − 2α)AU − c

2 U + µ∇µ,

where we have used (1.4), (1.8) and (1.10). From the last two equations, it follows that

(3.10) W µ = ξρ − ξα.

Taking an inner product (2.6) with W and using (3.7), we have

−{α(ρ − α) + c

4 }φW α + µU α + µ(α ² + 3

4 c){(ρ − α) − 1

α (µ ² − c 4 )} = 0.

On the other hand, by applying ξ to (2.6) and using (1.8) we have αU α = αµφW α.

Substituting this into the above equation and using (1.9), we have

(3.11) (β − ρα − c

4 ){α(U α) − (α ² + 3

4 c)µ ² } = 0.

We are now going to prove α(U α) = (α ² + ³ ₄ c)µ ² on Ω. For this purpose we prepare the following facts.

Lemma 3.1. Ω = ∅ provided that

(3.12) β − ρα − c

4 = 0.

Proof. From our assumption we have

(3.13) A ² ξ = ρAξ + c

4 ξ

by virtue of (3.6). Differentiating (3.13) covariantly and using (1.2), we find (3.14)

g((∇ X A)Aξ, Y ) + g(A(∇ X A)ξ, Y ) + g(A ² φAX, Y ) − ρg(AφAX, Y )

= (Xρ)g(Aξ, Y ) + ρg((∇ X A)ξ, Y ) + c

4 g(φAX, Y ), which together with (1.4) and (1.13) yields

(3.15) (∇ ξ A)Aξ = ρAU − c

4 U + 1 2 ∇β.

If we put X = ξ in (3.14) and use (1.13) and the last equation, then we get (3.16) 3A ² U − 2ρAU − c

2 U = (ξρ)Aξ − A∇α + ρ∇α − 1 2 ∇β.

If we replace X by Aξ in (3.1) and make use of (1.13), (3.13) and (3.15), we obtain

(3.17)

ρA ² U + (ρ ² − hρ + c

4 )AU + c

4 (3α + h − ρ)U

= (ξh)A ² ξ − c

4 ∇α + 1

2 (h − ρ)∇β − 1 2 A∇β.

On the other hand, we have from (1.5) and (3.13)

(3.18) Sξ = c

4 (2n − 3)ξ + (h − ρ)Aξ.

Differentiating this covariantly, we find (∇ X S)ξ + S∇ X ξ = c

4 (2n − 3)∇ X ξ + X(h − ρ)Aξ

+ (h − ρ)(∇ X A)ξ + (h − ρ)A∇ X ξ.

Putting X = ξ in the last equation and taking account of ∇ ξ S = 0, (1.5) and (1.13), we obtain

(3.19) A ² U + (2h − 3ρ)AU − cU = ξ(ρ − h)Aξ + (ρ − h)∇α.

We here note that ρ − h 6= 0 on Ω. In fact, if not, then by (3.18) we obtain Sξ = g(Sξ, ξ)ξ, where g(Sξ, ξ) = ^c ₄ (2n − 3). Then by virtue of Theorem C in [3], we are able to assert that M is a Hopf real hypersurface. So we have Ω = ∅, a contradiction. Thus, ρ − h 6= 0 is proved everywhere.

Taking an inner product (3.19) with ξ or W , we obtain respectively (3.20) (h − ρ)ξα + α(ξh − ξρ) = 0,

(h − ρ)W α + µ(ξh − ξρ) = 0, which enables us to obtain

(3.21) µ(ξα) = α(W α).

From this, together with (1.9) and (3.10), we get

(3.22) W β = 2µ(ξρ).

Taking an inner product (3.15) with ξ and using (1.9), (3.12) and (3.21), we obtain

(3.23) α ² (ξρ) = (ρα + c

2 )ξα.

By the way, if we apply (3.17) by W and make use of (3.7) and (3.22), then we get

µρ(ξh) − c

4 W α + µ(h − 2ρ + α)ξρ − 1

2 µ(ξβ) = 0, which together with (3.12), (3.20) and (3.21) implies that

α(h − ρ + 1

2 α)ξρ + (ρ ² − ρh − 1 2 ρα − c

4 )ξα = 0.

From this and (3.23) we see that (h − ρ)ξα = 0 and hence ξα = 0, because ρ − h6=0 on Ω. So we have W α = 0, ξh = 0 and ξρ = 0 because of (2.4), (3.5), (3.20) and (3.21). Further, we have W ρ = 0 by virtue of (3.10) and (3.12).

By putting X = µW in (3.1) and using (1.4), (1.13), (3.7), (3.9) and (3.12), we find

(3.24)

µA∇µ =(2α − ρ)A ² U − (2hα − hρ + ρ ² − ρα + c 4 )AU

− c

4 (h − ρ + α)U + (h − ρ + α)µ∇µ − µ ² ∇α, where we have used (1.9).

Now, differentiating (3.6) covariantly and using (1.2), we have

g((∇ X A)ξ, Y ) + g(A(∇ X A)ξ, Y ) + g(A ² φAX, Y ) − ρg(AφAX, Y )

= (Xρ)g(Aξ, Y ) + ρg((∇ X A)ξ, Y ) + X(β − ρα)η(Y )

+ (β − ρα)g(φAX, Y ).

Then by the equation of Codazzi (1.4), the above equation becomes c

4 {u(Y )η(X) − u(X)η(Y )} + c

2 (ρ − α)g(φY, X) − g(A ² φAX, Y ) + g(A ² φAY, X) + 2ρg(φAX, AY ) − (β − ρα){g(φAY, X) − g(φAX, Y )}

= g(AY, (∇ X A)ξ) − g(AY, (∇ Y A)ξ) + (Y ρ)g(Aξ, X) − (Xρ)g(Aξ, Y ) + Y (β − ρα)η(X) − X(β − ρα)η(Y ).

Here, we replace X by µW to both sides and take account of (1.4), (3.7) and (3.9), we have

(3.25) (3α − 2ρ)A ² U + (2ρ ² − 2ρα + c)AU + c

4 (α − ρ)U

= µA∇µ + (α − ρ)µ∇µ + µ ² (∇ρ − ∇α) − µ(W ρ)Aξ − µW (β − ρα)ξ.

Substituting (3.24) into (3.25) and using (3.12), we obtain

(3.26)

(α − ρ)A ² U + (3ρ ² − 3ρα + 2hα − hρ + 5

4 c)AU + c

4 (h − 2ρ + 2α)U

= 1

2 (h − 2ρ + 2α)∇β − (αh + c

2 ∇α) + µ ² {∇ρ − (W ρ)W }.

Then from (3.19) and the fact that ξh = ξρ = W ρ = 0, the last equation becomes

(3.27) (2hρ + 5

2 c)AU = c

2 (6ρ − 6α − h)U + (hα + c

2 )∇ρ − (hρ + c)∇α.

Combining (3.16) to (3.17), we also obtain

(3.28) 2αA∇ρ = α(2h − ρ)∇ρ − (ρ ² + c)∇α − (2ρ ² + c)AU + c(2ρ − h − 3α)U, where we have used (3.12) and the fact that ξρ = ξh = 0.

If we take account of (2.4), then (3.27) turns out to be (3.29) f 1 U = α(hα + c

2 )∇ρ − α(hρ + c)∇α, where we have put

(3.30) f 1 = − c

4 (2hρ + 5 2 c) + c

2 (6ρα − 6α ² − hα).

Differentiating (3.29) covariantly and taking the skew-symmetric part, we get

f 1 {g(∇ ξ U, X) + g(∇ X ξ, U )} = 0

for any vector X, where we have used ξh = ξρ = ξα = 0. This, together with (1.2), (1.11), (1.14) and (3.7) implies that f 1 {φ(3AU + ∇α) + µρW } = 0 and hence f 1 (∇α − ρU + 3AU ) = 0. Therefore, it follows that f 1 = 0 on Ω.

In fact, if not, then we have

(3.31) ∇α = ρU − 3AU.

From this, combining with (3.12) and (3.16), and using ξρ = 0, we have

(3.32) α∇ρ = −ρAU + (ρ ² + c)U

on the subset Ω.

Using (3.31), the equation (3.19) can be written as (3.33) A ² U = hAU + (ρ ² − ρh + c)U.

Substituting (3.31) and (3.32) into (3.28) and making use of the last equa- tion, we obtain

(3.34) h = 3α − 2ρ.

Using (3.31), (3.32) and (3.34), we have from (3.27) AU = hU − ∇ρ on the subset, which together with (3.32) yields

(ρ − α)AU = (ρ ² − 3α ² + 2ρα + c)U.

Comparing this with (2.4), we obtain (ρ + 3α)(ρα − α ² + ^c ₄ ) = 0, which together with (1.9) and (3.12) implies that ρ + 3α = 0 on this subset. By the way, we have from (2.4) and (3.33) the following

3ρα ² (ρ − α) + c

4 α(α + 2ρ) = ( c 4 ) ² .

Therefore α is constant and hence ρ does so, a contradiction. Thus, f 1 = 0 is proved everywhere on Ω.

Accordingly we have

(3.35) (hρ + c)∇α = (hα + c

2 )∇ρ by virtue of (3.29).

From (2.4) and (3.19) we have (ρ − h)α ² ∇α = c

4 ( c

4 − 2hα + 3ρα − 4α ² )U.

In the same way as above, we verify from this that

(3.36) 2hα + 4α ² − 3ρα − c

4 = 0

and hence ∇α = 0 by virtue of (2.4) and Theorem C in the introduction.

So we have ∇ρ = 0 because of (3.35) and (3.36), which together with (3.12) yields that ∇β = 0. Thus, by using (2.4) the equations (3.16) and (3.28) imply respectively to

ρα − α ² + 3

8 c = 0, 2ρ ² + c = 8ρα − 4hα − 12α ² .

From the last three equations, it follows that ρ ² + α ² = − ⁹ ₈ c and hence

ρ ² − 3ρα + 4α ² = 0, a contradiction. Thus, β − ρα − ^c ₄ 6= 0 on Ω is proved. ¤

4. Proof of a key lemma

In this section we give another important lemma which will be useful in the proof of our Main Theorem stated in the introduction.

In this section we also assume that a real hypersurface M in M n (c) has ξ- parallel Ricci tensor and structure Jacobi operator. Now from (3.4) we obtain

A ² ξ = hAξ + c

4α ² (hα + c

4 − 3α ² )ξ, which implies that

(4.1) α ² (β − hα + 3

4 c) = c

4 (hα + c 4 ).

Further, from (3.6), we get h = ρ and hence (3.6) becomes

(4.2) A ² ξ = hAξ + (β − hα)ξ.

Putting X = ξ in (1.5) and using (4.2), we find

(4.3) Sξ = { c

2 (n − 1) + hα − β}ξ.

Differentiating this covariantly, we obtain (∇ X S)ξ + S∇ X ξ = X(hα − β)ξ + { c

2 (n − 1) + hα − β}∇ X ξ.

If we replace X by ξ in this, and take account of (1.5) and ∇ ξ S = 0, then we have

A ² U − hAU − (β − hα + 3

4 c)U + ξ(hα − β)ξ = 0.

Since g(A ² U, ξ) = 0 because of (2.3), it follows from the last equation

(4.4) ξ(hα − β) = 0

and hence

(4.5) A ² U = hAU + (β − hα + 3

4 c)U.

Combining (3.5) to (4.4), we obtain

(4.6) 1

2 α(ξβ) = (hα + c 4 )ξα.

Differentiating (4.2) covariantly and making use of (1.2), we find

(4.7)

g((∇ X A)Aξ, Y ) + g(A(∇ X A)ξ, Y ) + g(A ² φAX, Y ) − hg(AφAX, Y )

= (Xh)g(Aξ, Y ) + hg((∇ X A)ξ, Y ) + X(β − hα)η(Y ) + (β − hα)g(φAX, Y ),

which together with the equation of Codazzi(1.4), (1.13) and (3.7) implies that

(4.8) (∇ ξ A)Aξ = hAU − c

4 U + 1 2 ∇β,

where we have used that AφA ² ξ = µρAφW , hAU = hµAφW and h = ρ.

If we replace X by ξ in (4.7) and use (1.3), (1.4), (3.11) and the last equation, then we get

3A ² U − 2hAU − (β − hα + c

4 )U = (ξh)Aξ − A∇α + h∇α − 1 2 ∇β, which, together with (2.3) and (4.5) gives

(4.9) hAU + 2(β − hα + c)U = (ξh)Aξ − A∇α + h∇α − 1 2 ∇β.

Now by applying A to both sides of (4.9), and using (4.2) and (4.5), we have

(4.10)

(h ² + 2β − 2hα + 2c)AU + h(β − hα + 3 4 c)U

= h(ξh)Aξ + (β − hα)(ξh)ξ − A ² ∇α − 1

2 A∇β + hA∇α.

On the other hand, by applying h to both sides of (4.9), we have (4.11) h ² AU + 2h(β − hα + c)U = h(ξh)Aξ − hA∇α + h ² ∇α − 1

2 h∇β.

If we replace Y by Aξ in (3.1) and make use of (1.4), (1.13), (4.2) and (4.8), then we get

3

4 cαU + c

4 AU = (ξh)A ² ξ − (β − hα)(2AU + ∇α) − hA ² U + c

2 AU − 1 2 A∇β.

This together with (4.5) yields (h ² + 2β − 2hα − c

4 )AU + {hβ − h ² α + 3

4 c(h + α)}U

= (ξh)A ² ξ − 1

2 A∇β − (β − hα)∇α.

Applying (4.9) by A and using the above formula, we have 3

4 c(3AU − αU ) + A ² ∇α − hA∇α − (β − hα)∇α = 0.

Then substituting this formula into the right side of (4.10), we have

(4.12)

(h ² + 2β − 2hα + 2c)AU + h(β − hα + 3 4 c)U

= h(ξh)Aξ + (β − hα)(ξh)ξ − (β − hα)∇α

− 3

4 c(αU − 3AU ) − 1 2 A∇β.

Substracting (4.12) from (4.11) gives the following 1

2 (A∇β − h∇β) − h(A∇α − h∇α) + (β − hα)∇α

= {2hα − 2β + c

4 }AU + {hβ − h ² α + 5 4 ch − 3

4 cα}U + (β − hα)(ξh)ξ.

This implies again

A ² ∇β − hA∇β

= 2h{(β − hα)∇α + 3

4 c(αU − 3AU )}

− 2(β − hα)A∇α + 2{2hα − 2β + c 4 }A ² U + 2{hβ − h ² α + 5

4 ch − 3

4 cα}AU + 2(β − hα)(ξh)Aξ.

On the other hand, by (4.9) we see that 2(β − hα){h∇α − A∇α}

= 2(β − hα){hAU + 2(β − hα + c)U − (ξh)Aξ + 1 2 ∇β}, from this, together with the above equation, it follows that

1

2 {A ² ∇β − hA∇β − (β − hα)∇β} + 3

4 c{(h + α)AU − (β + c

4 )U } = 0.

Then by using (4.5), we have 1

2 U β = −(h + α)g(AU, U ) + (β + c 4 )µ ² . From this and (2.4) it follows that

(4.13) 1

2 α(U β) = {αβ + c

4 (h + 2α)}µ ² .

On the other hand, taking several choices of X and Y in (3.1) and (4.7), and using (1.4), (1.13), (3.7), (4.2), (4.5) and (4.8), we obtain (for detail, see [3]).

(4.14)

(4β − 4hα + h ² + c

4 )AU + c

4 (6α − 5h)U

= µ(ξh)AW − µ(W h)Aξ − µ{W β − h(W α) − α(W h)}ξ + 1

2 (2α − h)∇β + (hα − 2β)∇α + µ ² ∇h,

where we have used the formula (4.3) and the assumption ∇ ξ S = 0. Applying (4.14) by U and using (3.11), Lemma 3.1 and (4.13), we get

µ ² (U h) = ( c

4 − 2µ ² )g(AU, U ) + (hµ ² + c(α − h))µ ² , which together with (2.3) gives

(4.15) α(U h) = (hα + c

2 )µ ² + c(α ² − αh) − ( c 4 ) ² . Now, we prepare the following:

Lemma 4.1. ξα = 0, ξh = 0, ξβ = 0 W α = 0 and W β = 0 on Ω.

Proof. Taking an inner product (4.12) with ξ and using (1.8) and (4.4), we find µ(W β) = (2β − α ² )ξh + (hα − 2β)ξα,

which connected with (3.5) yields

(4.16) µα ² (W β) = (2βhα − 2βα ² + cβ − c 2 α ² )ξα.

If we take an inner product (4.9) with ξ and make use of (1.8) and (4.4), then we also obtain

1

2 ξβ = α(ξα) + µ(W α), which together with (1.9) implies that

(4.17) ξµ = W α.

Combining above two equations, it follows that

(4.18) µα(ξµ) = (hα − α ² + c

4 )ξα, where we have used (3.5), (4.4) and (4.6).

Because of (3.5), (4.6), (4.16) and (4.18) and the fact α6=0 on the set Ω by Lemma 3.1, it suffices to show that ξα = 0 on Ω.

In order to do this, we differentiate (4.1) as follows:

2α(β − hα + 3

4 c)∇α + α ² ∇β = (α ² + c

4 )(h∇α + α∇h).

From this, together with (4.1), we see that (4.19) α ³ ∇β − α ² (α ² + c

4 )∇h = (hα ³ − c

4 hα − c ² 8 )∇α.

Then by using (4.13), (4.15) and Lemma 3.1, we verify that

{2α ³ (β − hα) + c 2 α ³ − c

4 hα ² + 3

16 c ² h + 3c ³ 32α }µ ²

= cα(α ² + c

4 )(α ² − hα − c 16 ), or, using (1.9) and (4.1)

(4.20) (chα ³ − 4cα ⁴ + c ² 2 α ² + 3

4 c ² hα + 3

8 c ³ )(β − α ² )

= c(4α ⁴ + cα ² )(α ² − hα − c

16 ).

Now, differentiating (4.20) gives 2(hα ³ − 4α ⁴ + c

2 α ² + 3

4 chα + 3

8 c ² )µ(Xµ) + (3α ² h − 16α ³ + cα + 3

4 ch)µ ² (Xα) + {αµ ² (α ² + 3

4 c) + α ³ (4α ² + c)}(Xh)

= α{(8α ² + c)(2α ² − 2hα − c

8 ) + α(2α − h)(4α ² + c)}(Xα).

From this, putting X = ξ and using (1.9), (3.5), (4.6) and (4.18),we obtain

(4.21)

( c

2 + hα) ©

cα ² (c + 4α ² ) + (β − α ² )( 3

4 c ² + cα ² ) ª + 2( c

4 + hα − α ² )( 3

8 c ³ − 4cα ⁴ + 3

4 c ² hα + chα ³ + 1 2 c ² α ² )

= cα ³ (2α − h)(c + 4α ² ) + cα ² (2c + 16α ² )(α ² − hα − c 16 )

− α( 3

4 c ² h + c ² α − 16α ³ + 3chα ² ) provided that ξα 6= 0.

If we eliminate β from (4.1) and (4.20), then together with (4.21) we are able to assert that α is a root of an algebraic equation with constant coefficients.

Hence, α should be constant, which makes a contradiction. Therefore, ξα = 0 on Ω. This gives the complete proof of our Lemma 4.1 ¤

From (2.3) and (4.9) we have (4.22) αA∇α = hα∇α − 1

2 α∇β + { c

4 h − 2α(β − hα + c)}U, by virtue of ξh = 0.

Putting X = ξ in (3.3) and using (2.3), (4.1) and (4.22), we obtain (4.23) βα∇α − 1

2 α ² ∇β = {(β − hα)(2α ² + 3

4 c) + c(α ² − c 8 )}U,

where we have used Lemma 4.1. By virtue of Lemma 4.1, equations (4.9), (4.12) and (4.14) mentioned above turn out respectively to be

(4.24) hAU + 2(β − hα + c)U = h∇α − A∇α − 1 2 ∇β, (4.25)

(h ² + 2β − 2hα − c

4 )AU + {hβ − h ² α + 3

4 c(h + α)}U = − 1

2 A∇β + (hα − β)∇α, and

(4.26)

(4β − 4hα + h ² + c

4 )AU + c

4 (6α − 5h)U

= 1

2 (2α − h)∇β + (hα − 2β)∇α + µ ² ∇h.

5. Proof of main theorem

We will continue our discussions for real hypersurfaces M in M n (c) with the assumption ∇ ξ S = 0 and ∇ ξ R ξ = 0 as in section 4. Now let us use the formulas (4.23)∼(4.26). From (4.23) we have

(5.1) βα(Xα) − 1

2 α ² (Xβ) = f u(X), where we have put

f = (β − hα)(2α ² + 3

4 c) + c(α ² − c 8 ).

Differentiating (5.1) covariantly and taking the skew-symmetric parts ob- tained, we find

(5.2)

(Y f )u(X) − (Xf )u(Y ) + 2f du(Y, X)

= Y (βα)(Xα) − X(βα)(Y α) + 1

2 α{(Xα)(Y β) − (Y α)(Xβ)}, where the exterior derivative du of a 1-form u is given by

du(Y, X) = 1

2 {Y u(X) − Xu(Y ) − u([Y, X])}.

If we replace Y by ξ in (5.2) and take account of Lemma 4.1, we get f du(ξ, X) = 0.

Now let us denote by Ω 0 be the set of points in Ω such that du(ξ, X) p 6= 0 at p ∈ Ω and suppose that the set Ω 0 is nonempty. Then we have f = 0, that is,

2α ² (β − hα) + 3

4 c(β − hα) + c(α ² − c 8 ) = 0.

This, together with (4.1) implies that

(5.3) 3β = hα + 2α ²

on Ω 0 . Further, we have by (4.23)

(5.4) 1

2 α∇β = β∇α.

By using the similar method to (4.26) we have

(5.5) 7hα − 10α ² + 3h ² + 3

4 c = 0 and

(5.6) 1

2 (2α − h)∇β + (hα − 2β)∇α + (β − α ² )∇h = 0, which together with (5.4) yields

(5.7) h∇α = α∇h

on Ω 0 . Differentiating (5.5) covariantly and using (5.7), we find (7hα − 10α ² + 3h ² )∇α = 0, which shows that ∇α = 0 and hence ∇β = 0 and ∇h = 0 on Ω 0

by virtue of (5.4) and (5.7). Thus, it follows from (2.3) and (4.24) that

(5.8) c

4 (h − 8α) = 2α(β − hα), which connected with (4.1) gives

(5.9) 2α ² + hα + c

2 = 0.

From this and (5.3) we see that

(5.10) 6β + c = 0.

However, it is verified, using (4.25) and (5.8), that 4β + h ² − 2hα − 6α ² − c

2 = 0.

This, together with (2.3) and (4.26) implies that hα + ₄ ^c = 0. Thus, (4.1) becomes β − hα + ³ ₄ c = 0. So we have β + c = 0, which together with (5.10) will produce a contradiction. Hence we have Ω = ∅. Thus, du(ξ, X) = 0 for any vector X. That is, g(∇ ξ U, X) + g(∇ X ξ, U ) = 0. This, together with (1.11), (1.14) and (3.7) with h = ρ, implies that

φ(3AU + ∇α) + µhW = 0.

Accordingly, it follows that

∇α = hU − 3AU.

Combining to (2.3), we have

α∇α = (hα + 3 4 c)U.

From this, together with Lemma 3.1 and (3.11) we see that

(5.11) h = α

and hence

(5.12) α∇α = (α ² + 3

4 c)U.

Further, (4.1) becomes

(5.13) α ² (β − α ² ) = ( c 4 ) ² − c

2 α ² . Now using (5.3), (5.4), (5.11) and (5.12), we have

(5.14)

1

2 α ² ∇β = αβ∇α

= 1

3 (hα + 2α ² )(α ² + 3 4 c)U

= α ² (α ² + 3

4 c)U,

where in the first equality we have used (5.4) and (5.12), and in the second equality used (5.3) and finally in the third equality used (5.11) respectively.

Differentiating (5.13) covariantly and using (5.11) gives

(5.15) 1

2 α ² ∇β = (2α ² − β − c

2 )(α ² + 3 4 c)U.

Then by comparing (5.14) and (5.15) we have

(5.16) (α ² − β − c

2 )(α ² + 3 4 c) = 0.

On the other hand, substituting (5.11) into (5.9) gives α ² = − ^c ₆ . From this, together with (5.10) and (5.16), we have a contradiction. Thus, the set Ω should be empty. Therefore we see that the subset Ω in M on which Aξ −g(Aξ, ξ)ξ 6= 0 is an empty set. Namely, in M n (c), c 6= 0, every real hypersurface satisfying

∇ ξ S = 0 and ∇ ξ R ξ is a Hopf hypersurface. So, we have U = 0 and moreover, the function α should be constant. Thus, (3.2) implies α∇ ξ A = 0, which together with (1.4) and (1.12) yields

α(Aφ − φA) = 0.

When the constant α identically vanishes, by Cecil and Ryan [2] we as- sert that M is a tube of radius ^π ₄ over certain Kaehler submanifold in P n C.

But we here note that any Hopf hypersurfaces in H n C the function α never vanishing(see [1], [7] and [8]). For the non-vanishing constant α, by virtue of Theorems A and B due to Okumura [9] for c > 0 and Montiel and Romero [7]

for c < 0 respectively we complete the proof of our Main Theorem.

Acknowledgments. The present authors would like to express their sincere gratitude to the referee for his careful reading of our manuscript and valuable comments to develop the first version of our manuscript.

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U-Hang Ki

The National Academy of Sciences Seoul 137-044, Korea

E-mail address: [email protected] Juan De Dios P´ erez and Florentino G. Santos Departamento de Geometr´ıa y Topolog´ıa Facultad de Ciencias

Universidad de Granada 18071-Granada, Spain

E-mail address: [email protected] Florentino G. Santos

Departamento de Geometr´ıa y Topolog´ıa Facultad de Ciencias

Universidad de Granada 18071-Granada, Spain

E-mail address: [email protected] Young Jin Suh

Department of Mathematics Kyungpook University Taegu, 702-701, Korea

E-mail address: [email protected]