A submanifold M is called a CR submanifold of a Kaehlerian manifold M˜ with complex structure J if there exists a differentiable distribution △:p→△p⊂Mp on M such that △ is J-invariant and the complementary orthogonal distribution △⊥ is totally real, where M_p denotes the tangent space at each point p in M ([1], [25]). In particular, M is said to be a semi-invariant submanifold provided that dim△⊥=1. The unit normal in J△⊥ is called the distinguished normal to the semi-invariant submanifold ([4], [23]). In this case, M admits an induced almost contact metric structure (ϕ,ξ,η,g). A typical example of a semi-invariant submanifold is real hypersurfaces. New examples of nontrivial semi-invariant submanifolds in a complex projective space Pnℂ are constructed in [13] and [20]. Therefore we may expect to generalize some results which are valid in a real hypersurface to a semi-invariant submanifold.

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.

For the real hypersurface of a complex space form Mn(c), many results are known. One of them, Takagi([21], [22]) classified all the homogeneous real hypersurfaces of Pnℂ as six model spaces which are said to be A1,A2,B,C,D and E, and Cecil-Ryan ([5]) and Kimura ([14]) proved that they are realized as the tubes of constant radius over Kaehlerian submanifolds when the structure vector field ξ is principal.

On the other hand, real hypersurfaces in Hnℂ have been investigated by Berndt ([2]), Montiel and Romero ([15]) and so on. Berndt ([2]) classified all real hypersurfaces with constant principal curvatures in Hnℂ 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 A0,A1,A2 or type B.

Let M be a real hypersurface of type A₁ or type A₂ in a complex projective space Pnℂ or that of type A₀, A₁ or A₂ in a complex hyperbolic space Hnℂ. Now, hereafter unless otherwise stated, such hypersurfaces are said to be of type (A) for our convenience sake.

Characterization problems for a real hypersurface of type (A) in a complex space form were studied by many authors ([6], [7], [8], [15], [16], [18], etc.).

Two of them, we introduce the following characterization theorems due to Okumura [18] for c>0 and Montiel and Romero [15] for c<0 respectively.

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

(1.1)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<r<π/2,

• (A₂) a totally geodesic Pkℂ(1≤k≤n−2), where 0<r<π/2.

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(c≤k≤n−2).

Denoting by R the curvature tensor of the submanifold, we define the Jacobi operator Rξ=R(⋅,ξ)ξ with respect to the structure vector ξ. Then Rξ is a self adjoint endomorphism on the tangent space of a CR submanifold.

Using several conditions on the structure Jacobi operator R_ξ, characterization problems for real hypersurfaces of type (A) have recently studied. In the previous paper ([7]), Cho and one of the present authors gave another characterization of real hypersurface of type (A) in a complex projective space Pnℂ. Namely they prove the following :

Theorem CK.([7]) Let M be a connected real hypersurface of Pnℂ if it satisfies (1)RξAϕ=ϕARξ or (2) Rξϕ=ϕRξ,RξA=ARξ, then M is of type (A), where A denotes the shape operator of M.

On the other hand, semi-invariant submanifolds of codimension 3 in a complex projective space Pn+1ℂ have been studied in [10], [12], [13] and so onby using properties of induced almost contact metric structure and those of the third fundamental form of the submanifold. In the preceding work, Ki, Song and Takagi ([13]) assert the following:

Theorem KST.([13]) Let M be a real (2n-1)-dimensional semi-invariant submanifold of codimension 3 in a complex projective space Pn+1ℂ with constant holomorphic sectional curvature 4c. If the structure vector ξ is an eigenvector for the shape operator in the direction of the distinguished normal and the third fundamental form t satisfies dt=2θω for a certain scalar θ(<2c), where ω(X,Y)=g(ϕX,Y) for any vectors X and Y on M, then M is a Hopf hypersurface in a complex projective space Pnℂ.

In this paper, we consider a semi-invariant submanifold M of codimension 3 in a complex space form Mn+1(c),c≠0 which satisfies Rξϕ=ϕRξ and at the same time Sξ=g(Sξ,ξ)ξ such that the third fundamental form t satisfies dt=2θω for a certain scalar θ(≠2c) and the scalar curvature r¯ of M satisfies r¯−2c(n−1)≤0, where S denotes the Ricci tensor of M. In the present paper, we prove that M is a real hypersurface of type (A) in Mn(c) mentioned Theorem O and Theorem MR. Our main theorem stated in section 6.

All manifolds in the present paper are assumed to be connected and of class C∞ and the semi-invariant submanifolds are supposed to be orientable.

Let M˜ be a real 2(n+1)-dimensional Kaehlerian manifold with parallel almost complex structure J and a Riemannian metric tensor G. Let M be a real (2n-1)-dimensional Riemannian manifold isometrically immersed in M˜. We denote by g the Riemannian metric tensor on M from that of M˜.

We denote by ∇˜ the operator of covariant differentiation with respect to the metric tensor G on M˜ and by ∇ the one on M. Then the Gauss and Weingarten formulas are given respectively by

(2.1)∇˜XY=∇XY+∑i=13g(A(i)X,Y)C(i),∇˜XC(i)=−A(i)X+∑j=13lj (i)(X)C(j)

for any vector fields X and Y tangent to M and any vector field C(i) normal to M, where A⁽ⁱ⁾ are called the second fundamental forms with respect to the normal vector C(i).

As is well-known, a submanifold of a Kaehlerian manifold is said to be a CR submanifold ([1], [25]) if it is endowed with a pair of mutually orthogonal and complementary differentiable distribution (T,T⊥) such that for any point p ∈ M we have JTp=Tp, JTp⊥⊂Tp⊥M, where Tp⊥M denotes the normal space of M at p. In particular, M is said to be semi-invariant submanifold provided that dimT⊥=1 ([4], [23]). Inthis case the unit vector field in JT⊥ is called a distinguished normal to the semi-invariant submanifold and denote by C ([4], [23]).

More precisely, we choose an orthonormal basis e1,⋯,e2n−2,ξ of M_p in such a way that e1,e2,⋯,e2n−2∈T, where M_p denotes the tangent space to M at each point p in M. Then we see that

G(Jξ,ei)=−G(ξ,Jei)=0

for i=1,⋯,2n−2.

From now on we consider M is a real (2n-1)-dimensional semi-invariant submanifold of a Kaehlerian manifold M˜ of real dimension 2(n+1). Then we can write ([4], [24])

(2.2)JX=ϕX+η(X)C, JC=−ξ, JD=−E, JE=D,

where we have put g(ϕX,Y)=G(JX,Y),η(X)=G(JX,C) for any vector fields X and Y tangent to M, and put C(1)=C, C(2)=D and C(3)=E.

By the Hermitian property of J, we see, using (2.2), that the aggregate (ϕ,ξ,η,g) is an almost contact metric structure on M, that is, we have

ϕ2X=−X+η(X)ξ, ϕξ=0, η(ξ)=1, η(X)=g(ξ,X),      g(ϕX,ϕY)=g(X,Y)−η(X)η(Y)

for any vectors X and Y on M.

We can also write the second equation of (2.1) as

(2.3)∇˜XC=−AX+l(X)D+m(X)E,∇˜XD=−KX−l(X)C+t(X)E,∇˜XE=−LX−m(X)C−t(X)D

because C, D and E are mutually orthogonal, where we have put

(2.4)A(1)=A, A(2)=K, A(3)=L,l=l2(1)=−l1(2), m=l3(1)=−l1(3), t=l3(2)=−l2(3),

In the sequel, we denote the normal components of ∇˜XC by ∇⊥C. The distinguished normal C is said to be parallel in the normal bundle if we have ∇⊥C=0, that is, l and m vanish identically.

From the Kaehler condition ∇˜J=0 and take account of the Gauss and Weingarten formulas,we obtain from (2.2)

(2.5)∇Xξ=ϕAX,
(2.6)(∇Xϕ)Y=η(Y)AX−g(AX,Y)ξ,
(2.7)KX=ϕLX−m(X)ξ, KϕX=LX−η(X)Lξ,
(2.8)LX=−ϕKX+l(X)ξ, LϕX=−KX+η(X)Kξ

for any vectors X and Y on M. The last two relationships give

(2.9)l(X)=g(Lξ,X), m(X)=−g(Kξ,X),
(2.10)m(ξ)=−k, l(ξ)=TrA(3),

where, we have put k=TrA(2).

We notice here that there is no loss of generality such that we may assume TrA(3)=0. In fact, a normal vector v of M we denote by Av the second fundamental tensor of M in the direction of v. Then we have A−v=−Av. Hence there is a unit normal vector D' of M in the plane spanned by two vectors D and E such that TrAD′=0, which proves our assertion. Therefore we have by (2.10)

(2.11)l(ξ)=0.

Applying (2.8) by ϕ and using (2.7), we find

−g(KX,Y)−m(X)η(Y)=g(ϕKX,ϕY)−η(X)l(ϕY).

If we take the skew-symmetric part of this with respect to X and Y, then we obtain

−m(X)η(Y)+m(Y)η(X)=η(X)l(ϕY)−η(Y)l(ϕX),

which together with (2.10) gives

(2.12)l(ϕX)=m(X)+kη(X). 

Similarly we have

(2.13)m(ϕX)=−l(X)

because of (2.10).

Transforming (2.7) by L and using (2.8) and (2.9), we obtain

(2.14)g(KLX,Y)+g(LKX,Y)=−l(X)m(Y)−l(Y)m(X).

In the rest of this paper we shall suppose that M˜ is a Kaehlerian manifold of constant holomorphic sectional curvature 4c, which is called a complex space form and denote by Mn+1(c). Then equations of the Gauss and Codazzi are given by

(2.15)(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 +g(KY,Z)KX−g(KX,Z)KY+g(LY,Z)LX−g(LX,Z)LY,
(2.16)(∇XA)Y−(∇YA)X−l(X)KY+l(Y)KX−m(X)LY+m(Y)LX=c{η(X)ϕY−η(Y)ϕX−2g(ϕX,Y)ξ},
(2.17)(∇XK)Y−(∇YK)X+l(X)AY−l(Y)AX−t(X)LY+t(Y)LX=0,
(2.18)(∇XL)Y−(∇YL)X+m(X)AY−m(Y)AX        +t(X)KY−t(Y)KX=0,

where R is the Riemann-Christoffel curvature tensor on M, and those of the Ricci by

(2.19)(∇Xl)(Y)−(∇Yl)(X)+g(KAX,Y)−g(AKX,Y)        +m(X)t(Y)−m(Y)t(X)=0,
(2.20)(∇Xm)(Y)−(∇Ym)(X)+g(LAX,Y)−g(ALX,Y)         +t(X)l(Y)−t(Y)l(X)=0,
(2.21)(∇Xt)(Y)−(∇Yt)(X)+g(LKX,Y)−g(KLX,Y)        +l(X)m(Y)−l(Y)m(X)=2cg(ϕX,Y).

In what follows, to write our formulas in a convention form, we denote by α=η(Aξ),β=η(A2ξ),TrA=h,TrA(2)=k,Tr(tAA)=h(2) and for a function f we denote by ∇f the gradient vector field of f.

Now, we put ∇ξξ=U in the sequel. Then U is orthogonal to ξ because of (2.5). From now on we put

(2.22)Aξ=αξ+μW,

where W is a unit vector field orthogonal to ξ. Then we have

(2.23)U=μϕW

because of (2.5). So, W is orthogonal to U. Further, we have

(2.24)μ2=β−α2.

From (2.22) and (2.23) we have

(2.25)ϕU=−Aξ+αξ,

which together with (2.5) and (2.22) yields

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

because W is orthogonal to ξ.

Differentiating (2.25) covariantly along M and using (2.5) and (2.6), we find

(2.27)(∇XA)ξ=−ϕ∇XU+g(AU+∇α,X)ξ−AϕAX+αϕAX,

which enables us to obtain

(2.28)(∇ξA)ξ=2AU+∇α−2kLξ.

Because of (2.5), (2.26) and (2.27), we verify that

(2.29)∇ξU=3ϕAU+αAξ−βξ+ϕ∇α−2k(Kξ−kξ).

In the next place, the Jacobi operators R_ξ is given by

(2.30)RξX=R(X,ξ)ξ=c(X−η(X)ξ)+αAX−η(AX)Aξ+kKX       −m(X)Kξ−l(X)Lξ,

where we have used (2.9), (2.10) and (2.15).

Suppose that Rξϕ=ϕRξ holds on M. Then from (2.30) we have

(2.31)α(ϕAX−AϕX)=g(Aξ,X)U+g(U,X)Aξ+2kLX       −2k{l(X)ξ+η(X)Lξ},

where we have used (2.5), (2.8) and (2.12).

In this section we shall suppose that M is a semi-invariant submanifold of codimension 3 in a complex space form Mn+1(c), c≠0 and that the third fundamental form t satisfies

(3.1)dt=2θω, ω(X,Y)=g(ϕX,Y)

for a certain scalar θ and any vector fields X and Y on M, where d denotes the exterior differential operator. Then (2.21) reformed as

g(LKX,Y)−g(KLX,Y)+l(X)m(Y)−l(Y)m(X)=−2(θ−c)g(ϕX,Y),

or, using (2.14)

(3.2)g(LKX,Y)+l(X)m(Y)=−(θ−c)g(ϕX,Y),

which together with (2.9)~(2.11) implies that

(3.3)KLξ=kLξ, LKξ=0.

Differentiating (3.1) covariantly along M and using (2.6) and the first Bianchi identity, we find

(Xθ)ω(Y,Z)+(Yθ)ω(Z,X)+(Zθ)ω(X,Y)=0,

which implies (n−2)Xθ=0. Thus θ(≥c) is constant if n>2.

For the case where θ=c in (3.1) we have dt=2cω. In this case, the normal connection of M is said tobe L-flat([18]).

Lemma 3.1. Let M be a semi-invariant submanifold with L-flat normal connection in Mn+1(c), c≠0. If Aξ=αξ, then we have ∇⊥C=0 and A(2)=A(3)=0.

Proof. From (3.2) we have

Tr(tA(2)A(2))−‖Kξ‖2+‖Lξ‖2=2(n−1)(θ−c)

because of (2.7), (2.9) and (2.12), which implies

‖A(2)−kη⊗ξ‖2+‖Lξ‖2=2(n−1)(θ−c),

where ‖F‖2=g(F,F) for any vector field F on M. Thus, by our hypothesis θ=c, we have A(2)=kη⊗ξ.

In the same way, we see from (2.8), (2.10), (2.13) and (3.2) that A(3)=0. And hence m(X)=−kη(X) and l=0 because of (2.9). Therefore, it suffices to show that k=0. Using these facts, (2.19) reformed as

k{η(X)Aξ−g(Aξ,X)ξ}=k(η(X)t−t(X)ξ),

which together with Aξ=αξ gives

(3.4)k(t−t(ξ)ξ)=0.

We also have from (2.18)

k{η(X)(AY+t(Y)ξ)−η(Y)(AX+t(X)ξ)}=0,

which implies k(h−α)=0. Form this and (3.4) we verify that k=0. This completes the proof.

Applying (3.2) by ϕ and taking account of (2.7) and (2.13), we find

(3.5)K2X+η(X)K2ξ+l(X)Lξ=(θ−c)(X−η(X)ξ),

which implies η(X)K2ξ−g(K2ξ,X)ξ=0. Thus, it follows that

(3.6)K2ξ=(‖Kξ‖2)ξ

by virtue of (2.9). Thus, (3.5) becomes

K2X+l(X)Lξ+‖Kξ‖2η(X)ξ=(θ−c)(X−η(X)ξ).

Putting X=Lξ in (2.8) and taking account of (2.12) and (3.3), we obtain

(3.7)L2ξ=kKξ+(‖Kξ‖2+k2)ξ.

If we put X=Lξ in (3.2) and make use of (2.13) and (3.2), we find

(θ−c−‖Kξ‖2)Lξ=0.

Similarly, we verify, using (3.2) and (3.7), that

(θ−c−‖Lξ‖2−k2)(‖Kξ‖2−k2)=0.

Let ‖Lξ‖≠0 at every point of M and suppose that this subset does not void. Then we have ‖Kξ‖2=θ−c and ‖Lξ‖2+k2=θ−c on the subset. Using these facts, we can verify that ( for detail, see (2.22) and (2.24) of [13])

(3.8)∇k=2ALξ,
(3.9)∇XLξ=t(X)Kξ−AKX−kAX

on the set. Differentiating (3.8) covariantly and taking the skew-symmetric part obtained, we find

(θ−2c)(η(X)Kξ−m(X)ξ)=0,

where we have used (2.12), (2.16), (3.3) and (3.9), which shows that (θ−2c)(m(X)+kη(X))=0 and hence θ=2c on this subset. Thus, from the first equation of (2.3) we have

Lemma 3.2. Let M be a semi-invariant submanifold of codimension 3 in Mn+1(c), c≠0 satisfying (3.1).

If θ−2c≠0, then ∇⊥C=−kξE on M.

In the following we assume that M satisfies (3.1) with θ−2c≠0. Then we have

(3.10)Lξ=0, Kξ=kξ

because of (2.9). It is, using (3.10), clear that (2.7), (2.8) and (3.2) are reduced respectively to

(3.11)ϕLX=KX−kη(X)ξ,
(3.12)L=Kϕ,
(3.13)g(LKX,Y)+(θ−c)g(ϕX,Y)=0.

From the last two equations, we obtain

(3.14)L2X=(θ−c)(X−η(X)ξ).

Further, if we take account of (3.10), then the other structure equations (2.16)~(2.21) reformed as

(3.15)(∇XA)Y−(∇YA)X=k{η(Y)LX−η(X)LY}+c{η(X)ϕY−η(Y)ϕX−2g(ϕX,Y)ξ},
(3.16)(∇XK)Y−(∇YK)X=t(X)LY−t(Y)LX,
(3.17)(∇XL)Y−(∇YL)X=k{η(X)AY−η(Y)AX}−t(X)KY+t(Y)KX,
(3.18)KAX−AKX=k{η(X)t−t(X)ξ},
(3.19)LAX−ALX=(Xk)ξ−η(X)∇k+k(ϕAX+AϕX),

where we have used (2.5).

Putting X=ξ in (3.18) and using (3.10), we find

(3.20)KAξ=kAξ+k(t−t(ξ)ξ).

Replacing X by ξ in (3.19) and using (2.5), (3.10) and (3.12), we get

(3.21)KU=(ξk)ξ−∇k+kU.

If we apply (3.20) by ϕ and make use of (2.22) (3.11) and (3.12), then we find

(3.22)KU=k(tϕ−U),

which together with (3.21) yields

(3.23)∇k=(ξk)ξ+k(−tϕ+2U).

If we transform (3.19) by ϕ and take account of (2.22), (3.11) and the last equation, then we obtain

ϕALX−KAX=−k{t−t(ξ)ξ}η(X)+2μη(X)W+2g(Aξ,X)ξ−AX−ϕAϕX},

which connected to (3.18) gives

(3.24)ϕAL=−LAϕ.

Since θ is constant if n>2, differentiating (3.14) covariantly, we find

(∇XL2)Y=(c−θ){η(Y)ϕAX+g(ϕAX,Y)ξ},

or, using (3.13) and (3.17), it is verified that (see,[13])

2(∇XL)LY=(θ−c){2t(X)ϕY−η(Y)(ϕA+Aϕ)X+g((Aϕ−ϕA)X,Y)ξ       −η(X)(ϕA−Aϕ)Y}−k{η(Y)(AL+LA)X       −g((AL+LA)X,Y)ξ−η(X)(LA−AL)Y},

which together with (3.10) and (3.22) yields

(3.25)(θ−c)(Aϕ−ϕA)X+(k2+θ−c)(u(X)ξ+η(X)U) +k{(AL+LA)X+k{−t(ϕX)ξ+η(X)ϕ°t}=0,

where u(X)=g(U,X) for any vector X.

In the following we consider the case where (2.22) with µ=0, that is Aξ=αξ. Differentiating this covariantly and using (2.5), we find

(∇XA)ξ=−AϕAX+αϕAX+(Xα)ξ,

which together with (3.10) and (3.15) gives

(3.26)−2AϕAX+α(ϕA+Aϕ)X+2cϕX=η(X)∇α−(Xα)ξ.

If we put X=ξ in this and using (2.22) with µ=0, then we find

(3.27)∇α=(ξα)ξ.

Differentiating the second equation of (3.10) covariantly along M, and using (2.5), we find ∇Xm=−(Xk)ξ+kϕAX, from which taking the skew-symmetric part and making use of (2.20) with l=0,

LAX−ALX−k(ϕAX+AϕX)=(Xk)ξ−η(X)∇k.

Since Aξ=αξ was assumed, we then have

(3.28)∇k=(ξk)ξ

because of (3.10). From the last two equations, it follows that

(3.29)LA−AL=k(ϕA+Aϕ).

If we put X=ξ in (3.18) and remember (2.22) with µ=0 and (3.10), then we get

(3.30)k(t(X)−t(ξ)η(X))=0.

Since we have Aξ=αξ, differentiating (3.28) covariantly, and taking the skew-symmetric part obtained, we get

(3.31)(ξk)(Aϕ+ϕA)=0.

From this and (3.27) we can write (3.26) as α(A2ϕ+cϕ)=0. By the properties of the almost contact metric structure, it follows that

ξk{h(2)−α2+2c(n−1)}=0,

which implies ξk=0 if c>0.

We will continue our arguments under the same hypotheses dt=2θω for a scalar θ(≠2c) as those stated in section 3. Further suppose, throughout this paper, that Rξϕ=ϕRξ, which means that the eigenspace of the structure Jacobi operator Rξ is invariant by the structure operator ϕ. Then (2.31) reformed as

(4.1)α(ϕAX−AϕX)=g(Aξ,X)U+g(U,X)Aξ+2kLX

by virtue of (3.10).

Transforming this by A, and taking the trace obtained, we have g(A2ξ,U)=0 because of (3.25), which together with (2.22) yields

(4.2)μg(AW,U)=0.

Applying (4.1) by L and using (2.25), (3.11) and (3.19), we find

(4.3)α{AKX−kη(X)Aξ−ϕALX}+g(LU,X)Aξ+g(KU,X)U                  =−2kL2X,

which together with (3.18) and (3.22) yields

kα{t(X)ξ−η(X)t+g(Aξ,X)ξ−η(X)Aξ} +g(LU,X)Aξ−g(Aξ,X)LU−u(X)KU+g(KU,X)U=0,

where u(X)=g(U,X) for any vector X. If we take the inner product with ξ to this and use (3.10), then we get

(4.4)kα{t(X)−t(ξ)η(X)+g(Aξ,X)−αη(X)}+αg(LU,X)=0.

Combining the last two equations and taking account of (2.24), we obtain

(4.5)μ(w(X)LU−g(LU,X)W)+u(X)KU−g(KU,X)U=0,

where w(X)=g(W,X) for any vector X.

In the previous paper [13] we prove the following proposition.

Proposition 4.1. Let M be a real (2n-1)-dimensional(n>2) semi-invariant submanifold of codimension 3 in a complex space form Mn+1(c), c≠0. If it satisfies dt=2θω for a scalar θ≠2c and μ=g(Aξ,W)=0, then we have k=0.

Sketch of Proof. This fact was proved for c>0 (see, Proposition 3.5 of [13]). But, regardless of the sign of c this one is established. However, only ξ k=0 and ξα=0 should be newly certified. We are now going to prove, using (4.1), that ξ k=0.

Now, let Ω1 be a set of points such that ξk≠0 on M and suppose that Ω1 be nonvoid. Then we have

Aϕ+ϕA=0, LA=AL

on Ω1 because of (3.29) and (3.31). We discuss our arguments on Ω1.

From (4.1) we have αϕA+kL=0 because of µ=0, which together with (3.11) gives αAY+kKY=(α2+k2)η(Y)ξ. Differentiating this covariantly along Ω1 and using (3.27) and (3.28), we find

(Xα)AY+α(∇XA)Y+(ξk)η(X)KY+k(∇XK)Y    =2(α(ξα)+k(ξk))η(X)η(Y)+(α2+k2){g(ϕAX,Y)ξ+η(Y)ϕAX},

from which, taking the skew-symmetric part and making use of (3.16), we obtain

(Xα)AY−(Yα)AX+α((∇XA)Y−(∇YA)X)+k(t(X)LY−t(Y)LX)        =(α2+k2)(η(Y)ϕAX−η(X)ϕAY).

If we take the inner product ξ to this and remember (3.10), (3.15) and the fact that µ=0, then we have cα=0, which together with (4.1) yields kL=0, a contradiction because of (3.14). In the same way we see from (3.27) that ξα=0. This completes the proof.

We set Ω={p∈M:k(p)≠0}, and suppose that Ω is nonempty. In the rest of this paper, we discuss our arguments on the open subset Ω of M. So, by Proposition 4.1 we see that µ ≠0 on Ω.

We notice here that the following fact :

Remark 4.2. α≠0 on Ω.

In fact, if not, then we have α=0 on this subset. We discuss our arguments on such a place. So (4.1) reformed as

(4.6)μ(w(X)U+u(X)W)+2kLX=0

because of (2.22) with α=0. Putting X=U or W in this we have respectively

(4.7)LU=−μβ2kW, LW=−μ2kU

by virtue of (2.24) with α=0. Using this and (3.14), we can write (4.3) as

−β22kw(X)W+g(KU,X)U=−2k(θ−c)(X−η(X)ξ).

Taking the inner product with W to this, we obtain β2=4k2(θ−c).

On the other hand, combining (4.6) and (4.7) to (3.14) we also have β2=4(n−1)k2(θ−c), which implies (n−2)(θ−c)k=0, a contradiction because of our assumption and Lemma 3.1. Thus, α=0 is not impossible on Ω.

Now, putting X=U in (4.4) and remembering Remark 4.2, we find kt(U)+g(LU,U)=0.

By the way, replacing X by U in (4.1) and using (2.22) and (2.25), we find

α(ϕAU+μAW)=μ2Aξ+2kLU.

If we take the inner product with U and make use of (4.2) and Proposition 4.1, then we obtain g(LU,U)=0 and hence t(U)=0.

By putting X=U in (4.5), we then have

(4.8)KU=τU,

where τ is given by τμ2=g(KU,U) by virtue of Proposition 4.1. Applying this by ϕ and using (3.12), we find

(4.9)LU=τμW.

It is, using (4.8) and (4.9), seen that

(4.10)τ2=θ−c.

because of (3.13).

Remark 4.3. Ω=∅ if θ=c.

Since we have θ=c, then (3.14) gives L=0 and thus KX=kη(X)ξ by virtue of (3.11). Hence, (3.17) reformed as

k{η(X)AY−η(Y)AX+η(X)t(Y)ξ−t(X)η(Y)ξ}=0,

which shows k(t(X)+g(Aξ,X)−ση(X))=0, where we have put σ=α+t(ξ).

Thus, the last two equations imply

AX=η(X)Aξ+g(Aξ,X)ξ−αη(X)ξ.

Since U is orthogonal to ξ and W, it is clear that AU=0 and AW=μξ.

If we put X=µ W in (4.1) and remember (2.23) and the fact that L=0, then we obtain μ2U=0 and hence Aξ=αξ. Owing to Lemma 3.1, we conclude that k=0 and thus Ω=∅.

By Remark 4.3, we may only consider the case where τ≠0 on Ω. Because of (3.22) and (4.8) we have

(4.11)t(ϕX)=(1+τk)g(U,X).

Therefore, by properties of the almost contact metric structure, it is clear that

(4.12)t=t(ξ)ξ−μ(1+τk)W.

Using (2.22), we can write (3.20) as

μKW=kμW+k(t−t(ξ)ξ),

which together with (4.12) implies that

(4.13)KW=−τW

because of Proposition 4.1.

If we take account of (3.25) and (4.11), then we find

(4.14)τ2(AϕX−ϕAX)+τ(τ−k)(u(X)ξ+η(X)U)+k(ALX+LAX)=0.

From (2.15) the Ricci tensor S of type (1,1) of M is given by

SX=c{(2n+1)X−3η(X)ξ}+hAX−A2X+kKX−K2X−L2X

by virtue of (3.10).

By the way, we see, using (3.12)~(3.14), that

(4.15)K2X=(θ−c)(X−η(X)ξ)+k2η(X)ξ.

Substituting this and (3.14) into the last equation and using (4.10), we obtain

(4.16)SX={c(2n+1)−2(θ−c)}X+(2(θ−c)−k2−3c)η(X)ξ+hAX−A2X+kKX,

which connected to (3.10) yields

(4.17)Sξ=2c(n−1)ξ+hAξ−A2ξ.

Differentiating (4.8) covariantly along Ω, we find

(∇XK)U+K∇XU=τ∇XU,

which together with (3.16) and (4.9) yields

(4.18)μτ(t(X)w(Y)−t(Y)w(X))+g(K∇XU,Y)−g(K∇YU,X)       =τ{g(∇XU,Y)−g(∇YU,X)}.

By the way, because of (2.22) and (2.24), we can write (2.29) as

(4.19)∇ξU=3ϕAU+αμW−μ2ξ+ϕ∇α.

Replacing X by ξ in (4.18) and taking account of the last two relationships, we find

(4.20)μ2(τ−k)ξ+μτ(t(ξ)−2α)W+μ(k−τ)AW    +3(LAU−τϕAU)=τϕ∇α−L∇α,

where we have used the first equation of (2.26).

In a direct consequence of (3.12) and (4.8), we obtain

(4.21)μLW=τU

because of µ ≠ 0 on Ω.

In the same way as above, we see from (4.13)

(4.22)τμ{t(X)u(Y)−t(Y)u(X)}+g(K∇XW,Y)−g(K∇YW,X)      =τ{g(∇YW,X)−g(∇XW,Y)}.

In the next place, from (2.22) and (2.25) we have ϕU=−μW. Differentiating this covariantly and using (2.6), we find

g(AU,X)ξ−ϕ∇XU=(Xμ)W+μ∇XW.

Putting X=ξ in this and making use of (2.29), we get

(4.23)μ∇ξW=3AU−αU+∇α−(ξα)ξ−(ξμ)W,

which enables us to obtain

(4.24)Wα=ξμ.

We will continue our arguments under the same hypotheses Rξϕ=ϕRξ and dt=2θω for a scalar θ(≠2c) as those in section 3. Further, we assume that Sξ=g(Sξ,ξ)ξ is satisfied on a semi-invariant submanifold of codimension 3 in Mn+1(c),c≠0. Then we have from (4.17)

(5.1)A2ξ=hAξ+(β−hα)ξ.

From this, and (2.22) and (2.24) we see that

(5.2)AW=μξ+(h−α)W.

In the next place, differentiating (5.2) covariantly along Ω, we find

(5.3)(∇XA)W+A∇XW=(Xμ)ξ+μ∇Xξ+X(h−α)W+(h−α)∇XW.

By taking the inner product with W to this and using (2.26) and (5.2), we obtain

(5.4)g((∇XA)W,W)=−2g(AU,X)+Xh−Xα

because W is a unit orthogonal vector to ξ.

Applying (5.3) by ξ and using (2.26), we also obtain

(5.5)μg((∇XA)W,ξ)=(h−2α)g(AU,X)+μ(Xμ),

which connected to (3.15) gives

(5.6)μ(∇ξA)W=(h−2α)AU+μ∇μ−kμLW−cU,

or, using (3.10), (3.15) and (5.5),

(5.7)μ(∇WA)ξ=(h−2α)AU−2cU+μ∇μ.

Putting X=ξ in (5.4) and taking account of (5.5), we have

(5.8)Wμ=ξh−ξα.

Replacing X by ξ in (5.3) and using (5.6), we find

(h−2α)AU−kμLW−cU+μ∇μ+μ(A∇ξW−(h−α)∇ξW)       =μ(ξμ)ξ+μ2U+μ(ξh−ξα)W.

Substituting (4.23) and (4.24) into this and making use of (4.21), we find

(5.9)3A2U−2hAU+(αh−β−c−kτ)U+A∇α+12∇β−h∇α       =2μ(Wα)ξ+(2α−h)(ξα)ξ+μ(ξh)W.

On the other hand, if we put X=μW in (4.1) and take account of (2.23), (2.24) and (5.2), then we find αAU+(β−hα+2kτ)U=0, which shows

(5.10)AU=λU,

where the function λ is defined, using Remark 4.2, by

(5.11)αλ=hα−β−2kτ.

Differentiating (5.10) covariantly along Ω, we find

(∇XA)U+A∇XU=(Xλ)U+λ∇XU.

If we take the skew-symmetric part of this, then we get

μ(kτ−c)(η(Y)w(X)−η(X)w(Y))+g(A∇XU,Y)−g(A∇YU,X)     =(Xλ)u(Y)−(Yλ)u(X)+λ(g(∇XU,Y)−g(∇YU,X)),

where we have used (2.22), (2.25), (3.15) and (4.9). Replacing X by U in this and using (5.10), we get

(5.12)A∇UU−λ∇UU=(Uλ)U−μ2∇λ.

Taking the inner product with W to this and remembering (5.2), we obtain

(5.13)μg(ξ,∇UU)+μ2(Wλ)+(h−α−λ)g(W,∇UU)=0.

By the way, from KU=τU, we have

(5.14)(∇XK)U+K∇XU=τ∇XU,

which implies that g((∇XK)U,U)=0. Because of (3.16), (4.9) and the last relationship give (∇UK)U=0, which connected to (4.13) and (5.14) yields g(W,∇UU)=0. Thus, (5.13) reformed as

μg(ξ,∇UU)+μ2(Wλ)=0.

However, the first term of this vanishes identically because of (2.26) and (5.2), which shows μ(Wλ)=0 and hence

(5.15)Wλ=0.

In the same way, we verify, using (2.26) and (5.2), that

(5.16)ξλ=0.

Now, differentiating (2.25) covariantly and using (2.5), we find

(∇XA)ξ+AϕAX=(Xα)ξ+αϕAX+(Xμ)W+μ∇XW.

If we put X=µ W in this and use (5.2), (5.7) and (5.10), then we find

(5.17)μ2∇WW−μ∇μ=(2hλ−3αλ+α2−αh−2c)U−μ(Wα)ξ−μ(Wμ)W.

Lemma 5.1. If M satisfies (4.1), (5.2) and dt=2θω for a scalar θ(≠2c), then we have on Ω

(5.18)∇k=(k−τ)U.

Proof. Using (3.21) and (4.8) we have

Xk=(ξk)η(X)+(k−τ)u(X)

for any vector field X. Differentiating this covariantly along Ω and taking the skew-symmetric part obtained, we find

(5.19)η(Y)X(ξk)−η(X)Y(ξk)+(ξk){η(X)u(Y)−η(Y)u(X)      +g(ϕAX,Y)−g(ϕAY,X)}+(k−τ)du(X,Y)=0,

where we have used (2.5).

Now, we take an orthonormal frame filed {e0=ξ,e1=W,e2,⋯,en−1,en=ϕe1=1μU,en+1=ϕe2,⋯,e2n−2=ϕen−1} of M. Taking the trace of (2.27), we obtain

∑ i=0 2n−2g(ϕ∇eiU,ei)=ξα−ξh.

Putting X=ϕei and Y=e_i in (5.19) and summing up for i=1,2,⋯,n−1, we have

(k−τ)∑ i=0 2n−2du(ϕei,ei)=ξk(α−h),

where we have used (2.22), (2.25), (5.2) and (5.10). Combining the last two relationships, we get

(5.20)(h−α)ξk=(k−τ)(ξh−ξα). 

By the way, if we put X=µW in (3.25) and take account of (2.22), (3.10) and (5.2), we obtain

(θ−c){AU−(h−α)U}+kτ{AU+(h−α)U}=0,

which connected to (4.9) and (5.10) yields

(5.21)λ(k+τ)+(h−α)(k−τ)=0.

From this we have

(h−α+λ)∇k+(k−τ)(∇h−∇α)+(k+τ)∇λ=0.

So we have (h−α+λ)ξk+(k−τ)(ξh−ξα)=0 with the aid of (5.16). From this and (5.20) we see that (2h−2α+λ)ξk=0.

If ξk≠0 on Ω, then we have λ=2(α−h), which together with (5.21) implies that (h−α)(k+3τ)=0 on this subset. We discuss our arguments on such a place. So we have h−α=0 from the last equation and hence λ=0. Consequently we have μ2+2kτ=0 by virtue of (2.24) and (5.11). Differentiation with respect to ξ gives μ(ξμ)+τ(ξk)=0.

However, if we take the inner product with U to (5.7) and remember (2.24), (5.10) and the fact that h−α=0 and λ=0, then we have μ∇μ=(μ2+kτ+c)U and consequently ξµ=0. Hence we have τ(ξk)=0, a contradiction. Thus, we have (5.18). This completes the proof.

Lemma 5.2. Under the same hypotheses as those stated in Lemma 5.1, we have k−τ≠0 on Ω.

Proof. If not, then we have k-τ=0 on an open subset of Ω. We discuss our argument on such a place. Then we have λ=0 because of (5.21) and Remark 4.3. So (5.10) and (5.11) turn out respectively to

(5.22)AU=0,
(5.23)β−hα+2τ2=0.

We also have from (4.11) t=t(ξ)ξ−2ϕU, which shows t(Y)=t(ξ)η(Y)−2g(ϕU,Y) for any vector Y. Differentiating this covariantly and using (2.5), (2.6) and (5.22), we find

(∇Xt)Y=X(t(ξ))η(Y)+t(ξ)g(ϕAX,Y)−2g(ϕ∇XU,Y),

from which, taking the skew-symmetric part with respect to X and Y and using (3.1),

2θg(ϕX,Y)=X(t(ξ))η(Y)−Y(t(ξ))η(X)+t(ξ){g(ϕAX,Y)−g(ϕAY,X)}    +2{g(ϕ∇YU,X)−g(ϕ∇XU,Y)}.

On the other hand, we verify from (2.27) that

g(ϕ∇XU,Y)−g(ϕ∇YU,X)+(Xα)η(Y)−(Yα)η(X)   =−2cg(ϕX,Y)−2g(AϕAX,Y)+α(g(ϕAX,Y)−g(ϕAY,X)).

Combining the last two equations, it follows that

2(θ−2c)g(ϕX,Y)+t(ξ){g(ϕAX,Y)−g(ϕAY,X)} =X(t(ξ))η(Y)−Y(t(ξ))η(X)+2{2g(AϕAX,Y)+α(g(ϕAX,Y) −g(ϕAY,X))+(Xα)η(Y)−(Yα)η(X)}.

Putting Y=ξ in this and remembering (5.22), we find

(5.24)X(t(ξ))+2(Xα)={ξ(t(ξ))+2ξα}η(X)+(t(ξ)+2α)u(X).

Substituting this into the last equation, we obtain

2(θ−2c)g(ϕX,Y)=(t(ξ)+2α)(u(X)η(Y)−u(Y)η(X)      +g(ϕAX,Y)−g(ϕAY,X))+4g(AϕAX,Y).

If we put X=μW in this and take account of (2.23), (5.2) and (5.22), then we get

(5.25)2(θ−2c)=(t(ξ)+2α)(h−α).

In the next step, differentiating (4.13) covariantly, we find

(∇XK)W+K∇XW+τ∇XW=0,

from which, taking the skew-symmetric part and using (3.16) and (4.9),

(5.26)τμ(t(Y)u(X)−t(X)u(Y))+g(K∇XW,Y)−g(K∇YW,X)      =τ{(∇YW)X−(∇XW)Y}.

If we put X=ξ in this and make use of (2.26), (4.23) and (5.22), then, we find

(5.27)K∇α+τ∇α=2τ(ξα)ξ+τ(2α+t(ξ))U.

Replacing X by W in (5.26) and making use of (5.17), we have

μ(K∇μ+τ∇μ)=2τ(μ2−α2+hα+2c)U+2μτ(Wα)ξ.

If we take the inner product with U to this and take account of (4.8), then we obtain μ(Uμ)=(μ2−α2+hα+2c)μ2, which together with (2.24) and (5.23) gives

(5.28)μ(Uμ)=2(μ2+τ2+c)μ2.

On the other hand, differentiating (5.22) covariantly with respect to ξ, we find (∇ξA)U+A∇ξU=0, which together with (4.19) (5.1) and (5.22) implies that

(∇ξA)U+(αh−β)Aξ−α(β−hα)ξ+Aϕ∇α=0.

Applying by ϕ, we have

(5.29)ϕ(∇ξA)U+(αh−β)U+ϕAϕ∇α=0.

Since we see from (3.15)

(∇UA)ξ−(∇ξA)U=μ(τ2+c)W

by virtue of (2.25), (3.10) and (4.9), it follows that

(5.30)ϕ(∇UA)ξ=ϕ(∇ξA)U+(τ2+c)U.

We also have from (2.27)

∇XU+g(A2ξ,X)ξ=ϕ(∇XA)ξ+ϕAϕAX+αAX,

which connected to (5.22) gives ∇UU=ϕ(∇UA)ξ. Thus, (5.30) reformed as

∇UU=ϕ(∇ξA)U+(τ2+c)U.

Combining this to (5.29) and using (5.23), it follows that

(5.31)∇UU=(c−τ2)U−ϕAϕ∇α.

If we apply by A and take account of (5.12) with λ=0 and (5.22), then we have AϕAϕ∇α=0.

Now, taking the inner product with U to (5.30) and making use of (2.22) (2.25) and (5.2), we obtain

(5.32)μ(Uμ)=(c−τ2)μ2+(h−α)Uα.

However, applying (5.27) by U and using (4.8), we find 2Uα=(t(ξ)+2α)μ2, which connected to (5.25) gives (h−α)Uα=(θ−2c)μ2. Substituting (5.28) and this into (5.32), we find 2μ2+3c+3τ2=θ, which together with (4.10) gives μ2+τ2+c=0 and consequently µ is constant. Thus, we see, using (2.24) and (5.23), that

(5.33)α(h−α)=τ2−c.

Therefore, α(h−α)=const. Differentiation gives

(h−α)∇α+α(∇h−∇α)=0,

which connected to (5.8) implies that (h−α)ξα=0, where we have used μ=const. Accordingly we have ξα=0 by virtue of (5.33) and the fact that θ−2c≠0.

Using (4.10) and (5.33), we can write (5.25) as

2(θ−2c)α=(θ−2c)(t(ξ)+2α).

Thus, it follows that t(ξ)=0 provided that θ−2c≠0. Hence, (5.24) turns out to be ∇α=αU, which implies du=0. Therefore, it is clear that ∇UU=0 because of μ=const, which connected to (5.31) yields (c−τ2)U=αϕAϕU. So we have c−τ2=α(h−α), where we have used (2.23), (2.25) and (5.2). From this and (5.33) it follows that θ−2c=0, a contradiction. Hence, Lemma 5.2 is proved.

Lemma 5.3. Under the same hypotheses as those in Lemma 5.1, we have

(5.34)∇α=(h−3λ)U.

Proof. Because of Lemma 5.1 and Lemma 5.2, we can write (5.19) as du(X,Y)=0, that is, g(∇XU,Y)−g(∇YU,X)=0. Putting X=ξ in this, and using (2.26) and (4.19), we find

3ϕAU+αAξ−βξ+ϕ∇α+μAW=0,

which together with (2.22), (2.25), (5.2) and (5.10) implies that

ϕ∇α+(h−3λ)μW=0.

Thus, it follows that

(5.35)∇α=(ξα)ξ+(h−3λ)U.

We are now going to prove that ξα=0.

Differentiation (5.21) with respect to ξ gives ξh−ξα=0 with the aid of (5.16), Lemma 5.1 and Lemma 5.2.

Using (5.10), (5.35) and this fact, we can write (5.9) as

(5.36)12∇β+(2hλ+αh−β−c−kτ−h2)U={2μ(Wα)+α(ξα)}ξ.

Since we have Wµ=0 because of (5.8), if we take the inner product ξ to the last equation and take account of (2.24), then we obtain α(Wα)=0 and hence Wα=0 by virtue of Remark 4.2.

Differentiating (5.11) with respect to ξ and making use of (5.16), Lemma 5.1 and the fact that ξh−ξα=0, we find

ξβ=(h+α−λ)ξα.

On the other hand, if we differentiate (2.24) with respect to ξ and remember Wα=0 and (4.24), then we have ξβ=2α(ξα). From this and the last relationship we get (λ+α−h)ξα=0.

Now, if ξα≠0 on Ω, the we have λ=h−α on this subset. We discuss our arguments on this subset. Then (5.21) yields λk=0 and hence λ=0 and h−α=0. So (5.35) and (5.36) are reduced respectively to

∇α=(ξα)ξ+αU,12∇β=α(ξα)ξ+(β+kτ+c)U.

We also have from (5.11) β=α2−2kτ, which together with (5.18) implies that

12∇β=α∇α−τ(k−τ)U.

Combining above equations, it follows that τ2=c, that is, θ−2c=0, a contradiction. This completes the proof of Lemma 5.3.