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Purity and hybridness of two tensors on a real hypersurface in complex projective space

  • Juan de Dios Pérez EMAIL logo and David Pérez-López
Published/Copyright: September 16, 2024

Abstract

On a real hypersurface M in complex projective space, we can define two tensor fields of type (1, 2), A F ( k ) and A T ( k ) , associated with the shape operator A of the real hypersurface, for any nonnull real number k . Following Tachibana [Analytic tensor and its generalization, Tohoku Math. J. 12 (1960), 208–221], we study the purity and hybridness of such tensors with respect to A .

MSC 2010: 53C15; 53B25

1 Introduction

Let M be a connected real hypersurface without boundary in the complex projective space C P m , m 2 , equipped with the Kaehlerian structure ( J , g ) , g being the Fubini-Study metric of constant holomorphic sectional curvature 4. We will continue denoting by g the restriction of the metric on C P m to M and by the Levi-Civita connection on M . The structure vector field (or Reeb vector field) on M is defined as ξ = J N , for a unit local normal vector field N on M . Then M inherits an almost contact metric structure ( ϕ , ξ , η , g ) from the Kaehlerian structure of C P m , where ϕ is a tensor field of type (1, 1), and η an 1-form on M . Let A be the shape operator of M associated to N . M is said to be Hopf if ξ is an eigenvector of A , A ξ = α ξ , and α is called the Reeb curvature of M . We also call D to the maximal holomorphic distribution on M , D p = { X T p M g ( X , ξ ) = 0 } , for any p M .

Takagi, see [13] classified homogeneous real hypersurfaces in C P m , obtaining six types. Among them, type ( A 1 ) real hypersurfaces are geodesic hyperspheres of radius r , 0 < r < π 2 , and type ( A 2 ) are tubes of radius r , 0 < r < π 2 , over totally geodesic complex projective spaces C P n , 0 < n < m 1 . We will call both types of real hypersurfaces type ( A ) real hypersurfaces. They are the unique real hypersurfaces in C P m such that A ϕ = ϕ A [4]. Kimura [5] proved that the real hypersurfaces appearing in Takagi’s list are the unique isoparametric Hopf real hypersurfaces, that is, real hypersurfaces all whose principal curvatures are constant.

Ruled real hypersurfaces in C P m satisfy that D is integrable and its integral manifolds are totally geodesic C P m 1 . Equivalently, g ( A D , D ) = 0 . Examples of ruled real hypersurfaces can be found in [6] or [7].

The Tanaka-Webster connection [8,9] is the canonical affine connection defined on a non-degenerate, pseudo-Hermitian CR-manifold. As a generalization of this connection, Tanno [10] defined the generalized Tanaka-Webster connection for contact metric manifolds. Using the almost contact metric structure on M and the naturally extended affine connection of Tanno’s generalized Tanaka-Webster connection, Cho defined, for any nonnull real number k , the k th generalized Tanaka-Webster connection ˆ ( k ) for a real hypersurface M in C P m , see [11,12], by

ˆ X ( k ) Y = X Y + g ( ϕ A X , Y ) ξ η ( Y ) ϕ A X k η ( X ) ϕ Y

for any X , Y tangent to M . Then ˆ ( k ) η = 0 , ˆ ( k ) ξ = 0 , ˆ ( k ) g = 0 , ˆ ( k ) ϕ = 0 . In particular, if the shape operator of a real hypersurface satisfies ϕ A + A ϕ = 2 k ϕ , M is a contact manifold and the kth generalized Tanaka-Webster connection coincides with the Tanaka-Webster connection.

The tensor field of type (1, 2) given by the difference of both connections F ( k ) ( X , Y ) = g ( ϕ A X , Y ) ξ η ( Y ) ϕ A X k η ( X ) ϕ Y , for any X , Y tangent to M , see [13] Proposition 7.10, pages 234–235, is called the k th Cho tensor on M . Then, for any X tangent to M and any nonnull real number k , we can consider the tensor field of type (1, 1) F X ( k ) , given by F X ( k ) Y = F ( k ) ( X , Y ) for any Y T M . This operator will be called the k th Cho operator corresponding to X . Notice that if X D , the corresponding Cho operator does not depend on k , and we simply write it F X . The torsion of the connection ˆ ( k ) is given by T ( k ) ( X , Y ) = F X ( k ) Y F Y ( k ) X for any X , Y tangent to M . We define the k th torsion operator associated to X as the operator given by T X ( k ) Y = T ( k ) ( X , Y ) , for any X , Y tangent to M .

If denotes the Lie derivative on M , we know that X Y = X Y Y X for any X , Y tangent to M . Analogously, we can define on M a differential operator of first order associated with the k th generalized Tanaka-Webster connection, given by

X ( k ) Y = ˆ X ( k ) Y ˆ Y ( k ) X = X Y + T X ( k ) Y

for any X , Y tangent to M . We will call it the derivative of Lie type associated to the k th generalized Tanaka-Webster connection.

Let now B be a symmetric tensor of type (1, 1) defined on M . Then we can consider the type (1, 2) tensor B F ( k ) associated to B as follows:

B F ( k ) ( X , Y ) = [ F X ( k ) , B ] Y = F X ( k ) B Y B F X ( k ) Y

for any X , Y tangent to M . We also can consider another tensor of type (1, 2), B T ( k ) , associated with B , by

B T ( k ) ( X , Y ) = [ T X ( k ) , B ] Y = T X ( k ) B Y B T X ( k ) Y

for any X , Y tangent to M . Notice that if X D , B F ( k ) does not depend on k . We will write it simply B F .

In [14,15], we proved non-existence of real hypersurfaces in C P m , m 3 , such that the tensors of type (1, 2) associated to the shape operator, A F ( k ) = 0 and A T ( k ) = 0 , respectively, for any nonnull real number k . Further results on such tensors were presented in [16,17].

Let S be an operator on M and C a tensor of type (1, 2) on M . C is said to be pure with respect to S [18], if

C ( S X , Y ) = C ( X , S Y )

for any X , Y tangent to M . If C and S satisfy the same condition for any X , Y D , C is said to be η -pure with respect to S .

If with the same conditions

C ( S X , Y ) = C ( X , S Y )

for any X , Y tangent to M , C is said to be hybrid with respect to S . If we have the same equality for any X , Y D , C is said to be η -hybrid with respect to S .

Then we can consider purity and hybridness of any tensor field of type (1, 2) with respect to different operators as the shape operator, the Ricci operator, the structure Jacobi operator, or the structure Lie operator, among others. The purpose of the present article is to begin the study of purity and hybridness of the tensor fields A F ( k ) and A T ( k ) with respect to one of such operators, as the shape one, A . We will prove the following:

Theorem 1.1

There does not exist any real hypersurface in C P m , m 3 , such that for any non-null real number k the tensor field A F ( k ) is pure with respect to A.

Also we will prove the following:

Theorem 1.2

Let M be a real hypersurface in C P m , m 3 . Then A F ( k ) is η -hybrid with respect to A if and only if M is locally congruent to a ruled real hypersurface.

Two real hypersurfaces in the complex projective space are congruent if there exists an isometry of the ambient space applying one real hypersurface into the other. As a consequence of this theorem, we can prove the following:

Corollary

There does not exist any real hypersurface M in C P m , m 3 , such that A F ( k ) is hybrid with respect to A, for any nonnull real number k.

In the case of the tensor field A T k , we will obtain

Theorem 1.3

Let M be a real hypersurface in C P m , m 3 , and k a nonnull real number. Then A T ( k ) is pure with respect to A if and only if M is locally congruent to a geodesic hypersphere with radius r , 0 < r < π 2 , such that cot ( 2 r ) = k 2 1 2 k .

Theorem 1.4

There does not exist any real hypersurface in C P m , m 3 , such that A T ( k ) is η -hybrid with respect to A, for any nonnull real number k.

2 Preliminaries

Throughout this article, all manifolds, vector fields, etc., will be considered of class C unless otherwise stated. Let M be a connected real hypersurface in C P m , m 2 , without boundary. Let N be a locally defined unit normal vector field on M . Let be the Levi-Civita connection on M and ( J , g ) be the Kaehlerian structure of C P m .

For any vector field X tangent to M , we write J X = ϕ X + η ( X ) N , and J N = ξ , where ϕ X denotes the tangential component of J X . Then ( ϕ , ξ , η , g ) is an almost contact metric structure on M [19]. That is, we have

(2.1) ϕ 2 X = X + η ( X ) ξ , η ( ξ ) = 1 , g ( ϕ X , ϕ Y ) = g ( X , Y ) η ( X ) η ( Y )

for any vector fields X , Y tangent to M . From (2.1), we obtain

(2.2) ϕ ξ = 0 , η ( X ) = g ( X , ξ ) .

From the parallelism of J , we obtain

(2.3) ( X ϕ ) Y = η ( Y ) A X g ( A X , Y ) ξ

and

(2.4) X ξ = ϕ A X

for any X , Y tangent to M , where A denotes the shape operator of the immersion. As the ambient space has holomorphic sectional curvature 4, the equation of Codazzi is given by

(2.5) ( X A ) Y ( Y A ) X = η ( X ) ϕ Y η ( Y ) ϕ X 2 g ( ϕ X , Y ) ξ

for any vector fields X , Y tangent to M .

In the sequel, we need the following result:

Theorem 2.1

[20] If ξ is a principal curvature vector with corresponding principal curvature α and X D is principal with principal curvature λ , then 2 λ α 0 and ϕ X is principal with principal curvature α λ + 2 2 λ α .

3 Proof of Theorem 1.1

Let us suppose that A F ( k ) is pure with respect to A . Then F A X ( k ) A Y A F A X ( k ) Y = F X ( k ) A 2 Y A F X ( k ) A Y , for any X , Y tangent to M . Then

(3.1) g ( ϕ A 2 X , A Y ) ξ η ( A Y ) ϕ A 2 X k η ( A X ) ϕ A Y g ( ϕ A 2 X , Y ) A ξ + η ( Y ) A ϕ A 2 X + k η ( A X ) A ϕ Y = g ( ϕ A X , A 2 Y ) ξ η ( A 2 Y ) ϕ A X k η ( X ) ϕ A 2 Y g ( ϕ A X , A Y ) A ξ + η ( A Y ) A ϕ A X + k η ( X ) A ϕ A Y

for any X , Y tangent to M .

Let us suppose first that M is Hopf with Reeb curvature α . If X , Y D from (3.1), we obtain g ( ϕ A 2 X , A Y ) ξ α g ( ϕ A 2 X , Y ) ξ = g ( ϕ A X , A 2 Y ) ξ α g ( ϕ A X , A Y ) ξ . Its scalar product with ξ gives A ϕ A 2 X α ϕ A 2 X = A 2 ϕ A X α A ϕ A X , for any X D . If we suppose that A X = λ X , we will obtain λ 2 μ α λ 2 = λ μ 2 α λ μ , where μ = α λ + 2 2 λ α . Therefore,

(3.2) λ ( λ μ ) ( μ α ) = 0 .

If now we take X D , Y = ξ in (3.1) it follows α ϕ A 2 X + A ϕ A 2 X = α 2 ϕ A X + α A ϕ A X . If we suppose again that A X = λ X , this yields

(3.3) λ ( λ α ) ( μ α ) = 0 .

On the other hand, if in (3.1)  X = ξ , Y D , we have, supposing that A Y = λ Y , that α λ + α μ = λ 2 + λ μ . Thus,

(3.4) ( α λ ) ( μ λ ) = 0 .

From (3.4) either λ = α or λ = μ . If we suppose that λ = α , (3.3) is satisfied and from (3.2), we should have λ ( λ μ ) 2 = 0 . Thus, either λ = 0 or λ = μ . As λ = α , if λ = μ M should be totally umbilical, which is impossible. Therefore, λ = α = 0 . This is also impossible if you look at the value of μ . Thus, we obtain that λ = μ . In this case, (3.2) is satisfied, but then (3.3) yields λ ( λ α ) 2 = 0 . As λ = α is impossible, we should have λ = μ = 0 . This implies 0 = 2 α , which also is impossible.

Therefore, M must be non Hopf. We can write A ξ = α ξ + β U , where U is a unit vector field in D and α and β are functions on M such that β 0 at least on a neighborhood of a point p M . We will make the following calculations on such a neighborhood. Then (3.1) becomes

(3.5) g ( ϕ A 2 X , A Y ) ξ α η ( Y ) ϕ A 2 X β g ( Y , U ) ϕ A 2 X k α η ( X ) ϕ A Y k β g ( X , U ) ϕ A Y g ( ϕ A 2 X , Y ) A ξ + η ( Y ) A ϕ A 2 X + k α η ( X ) A ϕ Y + k β g ( X , U ) A ϕ Y = g ( ϕ A X , A 2 Y ) ξ g ( A Y , A ξ ) ϕ A X k η ( X ) ϕ A 2 Y g ( ϕ A X , A Y ) A ξ + α η ( Y ) A ϕ A X + β g ( Y , U ) A ϕ A X + k η ( X ) A ϕ A Y

for any X , Y tangent to M . Take X = Y = ξ in (3.5) to obtain β g ( ϕ A 2 ξ , U ) ξ α ϕ A 2 ξ k α β ϕ U + A ϕ A 2 ξ = β g ( ϕ U , A 2 ξ ) ξ β g ( A ξ , A ξ ) ϕ U k ϕ A 2 ξ + α β A ϕ U + k β A ϕ U . From the expression of A ξ and bearing in mind that β 0 , we obtain

(3.6) 2 β g ( ϕ A U , U ) ξ α 2 ϕ U α ϕ A U k α ϕ U + α A ϕ U + A ϕ A U = g ( A ξ , A ξ ) ϕ U k α ϕ U k ϕ A U + ( α + k ) A ϕ U .

The scalar product of (3.6) and ξ implies 3 β g ( A ϕ U , U ) = 0 . As β 0 , we have

(3.7) g ( A ϕ U , U ) = 0 .

Taking in (3.5)  X = U , Y = ϕ U and its scalar product with ϕ Z , Z D U = { W D g ( W , U ) = g ( W , ϕ U ) = 0 } , we obtain k β g ( A ϕ U , Z ) k β g ( A U , ϕ Z ) = 0 . As k β 0 , it follows

(3.8) g ( A ϕ U ϕ A U , Z ) = 0

for any Z D U .

Take now X , Y D U in (3.5) and its scalar product with ϕ U . We obtain g ( A Y , A ξ ) g ( A X , U ) = 0 . Therefore, g ( A Y , α ξ + β U ) = 0 = β g ( A U , Y ) g ( A U , X ) for any X , Y D U . If X = Y , we obtain

(3.9) g ( A U , X ) = 0

for any X D U . Then (3.8) and (3.9) also yield

(3.10) g ( A ϕ U , X ) = 0

for any X D U . From (3.7), (3.9), and (3.10), we can write A U = β ξ + γ U , A ϕ U = δ ϕ U , for some functions γ and δ .

If we take Y = U , X = ϕ U in (3.5), we have g ( ϕ A 2 ϕ U , A U ) ξ β ϕ A 2 ϕ U g ( ϕ A 2 ϕ U , U ) A ξ = g ( ϕ A ϕ U , A 2 U ) ξ g ( A U , A ξ ) ϕ A ϕ U g ( ϕ A ϕ U , A U ) A ξ + β A ϕ A ϕ U . Thus, δ 2 γ ξ + β δ 2 U + δ 2 A ξ = δ g ( A U , A U ) ξ + δ g ( A U , A ξ ) U + γ δ A ξ δ β A U . Its scalar product with U yields 2 β δ 2 = δ β ( α + γ ) . As β 0 , 2 δ 2 = δ ( α + γ ) . Therefore, either δ = 0 or δ = α + γ 2 . The scalar product of the same expression and ξ gives δ 2 γ + δ 2 α = δ ( β 2 + γ 2 ) + α γ δ δ β 2 . If we suppose δ 0 , δ ( α γ ) = ( β 2 + γ 2 ) + α γ β 2 = 2 β 2 + γ ( α γ ) . That is, ( δ γ ) ( α γ ) = 2 β 2 . As we suppose δ 0 , δ = α + γ 2 , and then ( α γ ) 2 2 = 2 β 2 . This yields β = 0 , which is impossible. We have proved that

(3.11) A ϕ U = 0 .

If now we take Y = U in (3.5), we obtain β ϕ A 2 X k α η ( X ) ϕ A U k β g ( X , U ) ϕ A U = g ( ϕ A X , A 2 U ) ξ g ( A U , A ξ ) ϕ A X k η ( X ) ϕ A 2 U + β A ϕ A X for any X tangent to M . Its scalar product with ϕ U implies

(3.12) β g ( A X , A U ) k α η ( X ) g ( A U , U ) k β g ( X , U ) g ( A U , U ) = g ( A U , A ξ ) g ( A X , U ) k η ( X ) g ( A U , A U )

for any X tangent to M . Take X = U in (3.12) to have, bearing in mind that β 0 , g ( A U , A U ) + k g ( A U , U ) = ( α + γ ) g ( A U , U ) . Therefore, β 2 + γ 2 + k γ = α γ + γ 2 . This means that β 2 = ( α k ) γ . If α = k , β = 0 , which is impossible. Thus,

(3.13) γ = β 2 α k .

If in (3.12)  X = ξ , we obtain β g ( A ξ , A U ) k α g ( A U , U ) = β g ( A ξ , A U ) k g ( A U , A U ) . As k 0 , α g ( A U , U ) = g ( A U , A U ) and α γ = β 2 + γ 2 . As β 2 = ( α k ) γ , we obtain γ ( γ k ) = 0 . If γ = 0 , β 2 = 0 , which is impossible. Then γ = k , and we can write A ξ = α ξ + β U , A U = β ξ + k U , A ϕ U = 0 , and β 2 + k 2 = α k . This also yields that D U is A -invariant.

Take X , Y D U in (3.5) to obtain g ( ϕ A 2 X , A Y ) ξ g ( ϕ A 2 X , Y ) A ξ = g ( ϕ A X , A 2 Y ) ξ g ( ϕ A X , A Y ) A ξ . As β 0 , its scalar product with U implies g ( ϕ A 2 X , Y ) = g ( ϕ A X , A Y ) for any X , Y D U , Therefore, ϕ A 2 X = A ϕ A X for any X D U . If such an X satisfies A X = λ X , it follows λ 2 ϕ X = λ A ϕ X . Then, either λ = 0 or for any λ 0 , A ϕ X = λ ϕ X . This means that the eigenspace associated to any non-zero eigenvalue is ϕ -invariant. But then, also the eigenspace corresponding to 0 must be ϕ -invariant. Let us suppose that there exists a unit Z D U such that A Z = A ϕ Z = 0 . Then the Codazzi equation implies that A Z ϕ Z + A ϕ Z Z = 2 ξ . Its scalar product with ξ gives β g ( [ ϕ Z , Z ] , U ) = 2 , and we can see that g ( [ ϕ Z , Z ] , U ) 0 . But its scalar product with U yields γ g ( [ ϕ Z , Z ] , U ) = 0 . Therefore, k = γ = 0 , which is impossible.

Take X D U , Y = U in (3.5) and its scalar product with ϕ Z , Z D U . Then we obtain β g ( A 2 X , Z ) = g ( A U , A ξ ) g ( A X , Z ) + β g ( ϕ A X , A ϕ Z ) . As β 0 , this implies A 2 X = ( α + k ) A X + ϕ A ϕ A X for any X D U . If we suppose A X = λ X , we obtain λ 2 = ( α + k ) λ λ 2 . Now, λ ( 2 λ ( α + k ) ) = 0 . As we know that λ 0 , the unique principal curvature on D U is α + k 2 . Take a unit Z D U . From ( Z A ) ϕ Z ( ϕ Z A ) Z = 2 ξ and its scalar product with ϕ Z , we obtain

(3.14) Z ( α ) = 0 ,

for any Z D U . Bearing in mind that β 2 + k 2 = k α , we also have from (3.14)

(3.15) Z ( β ) = 0 .

Now ( ϕ U A ) ξ ( ξ A ) ϕ U = U implies ( ϕ U ) ( α ) ξ + ( ϕ U ) ( β ) U + β ϕ U U + A ξ ϕ U = U . Its scalar product with ξ gives

(3.16) ( ϕ U ) ( α ) α β + β g ( ξ ϕ U , U ) = 0 ,

and its scalar product with U implies

(3.17) ( ϕ U ) ( β ) β 2 + k g ( ξ ϕ U , U ) = 1 .

Substracting to the product of (3.17) and β the product of (3.16) and k , we obtain β ( ϕ U ) ( β ) k ( ϕ U ) ( α ) = β + β 3 k α β . On the other hand, as β 2 + k 2 = k α , 2 β ( ϕ U ) ( β ) k ( ϕ U ) ( α ) = 0 . Therefore, ( ϕ U ) ( β ) = 1 β 2 + k α . By using again that β 2 = k α k 2 , we obtain

(3.18) ( ϕ U ) ( β ) = k 2 1 .

Moreover, ( ϕ U A ) U ( U A ) ϕ U = 2 ξ gives ( ϕ U ) ( β ) ξ + k ϕ U U A ϕ U U + A U ϕ U = 2 ξ . Its scalar product with U yields β g ( ϕ U , ϕ A U ) + k g ( U ϕ U , U ) = 0 and, as k 0 ,

(3.19) g ( U , ϕ U , U ) = β ,

and its scalar product with ξ implies ( ϕ U ) ( β ) α g ( ϕ U , ϕ A U ) + β g ( U ϕ U , U ) = 2 . From (3.19), we obtain ( ϕ U ) ( β ) = 2 + α k β 2 = 2 + k 2 . This and (3.18) give 3 = 0 , which is impossible and finishes the proof.

4 Proof of Theorem 1.2

If we suppose that M satisfies that A F ( k ) is η -hybrid with respect to A , we should have F A X ( k ) A Y A F A X ( k ) Y + F X A 2 Y A F X A Y = 0 for any X , Y D . Therefore,

(4.1) g ( ϕ A 2 X , A Y ) ξ η ( A Y ) ϕ A 2 X k η ( A X ) ϕ A Y g ( ϕ A 2 X , Y ) A ξ + k η ( A X ) A ϕ Y + g ( ϕ A X , A 2 Y ) ξ η ( A 2 Y ) ϕ A X g ( ϕ A X , A Y ) A ξ + η ( A Y ) A ϕ A X = 0

for any X , Y D . Let us suppose first that M is Hopf with A ξ = α ξ . Then from (4.1), we have g ( A ϕ A 2 X , Y ) ξ α g ( ϕ A 2 X , Y ) ξ + g ( A 2 ϕ A X , Y ) ξ α g ( A ϕ A X , Y ) ξ = 0 for any X , Y D . Thus for any X D , A ϕ A 2 X α ϕ A 2 X + A 2 ϕ A X α A ϕ A X = 0 . If we suppose that X D satisfies A X = λ X , then A ϕ X = μ ϕ X with μ = α λ + 2 2 λ α . Now λ 2 μ α λ 2 + λ μ 2 λ μ α = 0 . This implies ( μ α ) λ ( λ + μ ) = 0 . If we suppose λ + μ = 0 , we should have λ + α λ + 2 2 λ α = 2 λ 2 + 2 2 λ α = 0 , which is impossible. Therefore, λ ( μ α ) = 0 . This yields either λ = 0 , μ = 2 α or μ = α , λ = α 2 + 2 2 . In any case, M must have three distinct constant principal curvatures. In both cases, λ μ . Therefore, looking at the principal curvatures of the real hypersurfaces in Takagi’s list, we see that M must be locally congruent to a real hypersurface of type (B). The second case is impossible. In the first case, α = 2 cot ( 2 r ) and either cot r π 4 = 0 or tan r π 4 = 0 [21]. Both possibilities give contradictions, and we have proved that M must be non-Hopf.

As in the previous section, we write locally A ξ = α ξ + β U , with the same conditions, and also make the following computations locally. The scalar product of (4.1) and ϕ U gives

(4.2) β g ( Y , U ) g ( A 2 X , U ) k β g ( X , U ) g ( A Y , U ) + k β g ( X , U ) g ( ϕ Y , A ϕ U ) α β g ( Y , U ) g ( A X , U ) β g ( Y , A U ) g ( A U , X ) + β g ( Y , U ) g ( ϕ A X , A ϕ U ) = 0

for any X , Y D . Take X = Y = ϕ U in (4.2) to obtain β g ( A U , ϕ U ) 2 = 0 . As β 0 , we deduce

(4.3) g ( A U , ϕ U ) = 0 .

If in (4.2)  X = Y D U we have β g ( A U , X ) 2 = 0 for any X D U , Then, (4.2) and (4.3) imply

(4.4) A U = β ξ + γ U

for a certain function γ . If we take X = U , Y D U in (4.2), we obtain k β g ( A ϕ U , ϕ Y ) = 0 for any Y D U . Taking ϕ Y instead of Y , as k β 0 , it follows g ( A ϕ U , Y ) = 0 for any Y D U . Together with (4.3), this yields

(4.5) A ϕ U = δ ϕ U

for some function δ . The scalar product of (4.1) and U gives β g ( Y , U ) g ( A X , A ϕ U ) + k β g ( X , U ) g ( Y , A ϕ U ) β g ( ϕ A 2 X , Y ) + k β g ( X , U ) g ( ϕ Y , A U ) + g ( A Y , A ξ ) g ( A ϕ U , X ) β g ( ϕ A X , A Y ) + β g ( Y , U ) g ( ϕ A X , A U ) = 0 for any X , Y D . If in this equation Y = U , X = ϕ U , bearing in mind (4.4) and (4.5), it follows δ ( 2 δ + α + γ ) = 0 . Therefore, either δ = 0 or δ = ( α + γ 2 ) . If in the expression above (4.5)  X , Y D U , we obtain, β being non-zero, g ( ϕ A 2 X , Y ) + g ( ϕ A X , A Y ) = 0 for any X , Y D U . As from (4.4) and (4.5), D U is A -invariant, ϕ A 2 X + A ϕ A X = 0 for any X D U . Taking a unit X D U such that A X = λ X , we have λ ( A ϕ X + λ ϕ X ) = 0 . Therefore, either λ = 0 or if λ 0 , A ϕ X = λ ϕ X . This tells us that if 0 is a principal curvature on D U its corresponding eigenspace is ϕ -invariant.

The scalar product of (4.1) and Z D U implies η ( A Y ) g ( ϕ A 2 X , Z ) k η ( A X ) g ( ϕ A Y , Z ) + k η ( A X ) g ( A ϕ Y , Z ) η ( A 2 Y ) g ( ϕ A X , Z ) + η ( A Y ) g ( A ϕ A X , Z ) = 0 . If X = U , Y D U we obtain k β g ( ϕ A Y , Z ) + k β g ( A ϕ Y , Z ) = 0 , for any Y , Z D U . As k β 0 , this yields A ϕ Y = ϕ A Y for any Y D U . This and the aforementioned discussion prove that the unique principal curvature on D U is 0. Therefore, A W = 0 for any W D U . For such a W , the equation of Codazzi gives ( W A ) ϕ W ( ϕ W A ) W = 2 ξ , and from this, A W ϕ W + A ϕ W W = 2 ξ . Its scalar product with ξ implies g ( W ϕ W , α ξ + β U ) + g ( ϕ W W , α ξ + β U ) = 2 . That is,

(4.6) β g ( [ ϕ W , W ] , U ) = 2 ,

and therefore, g ( [ ϕ W , W ] , U ) 0 . But the scalar product of the aforementioned expression and U implies g ( W ϕ W , β ξ + γ U ) + g ( ϕ W W , β ξ + γ U ) = 0 and

(4.7) γ g ( [ ϕ W , W ] , U ) = 0 .

From (4.6) and (4.7) γ = 0 . Now if δ = 0 M must be locally congruent to a ruled real hypersurface.

Let us now suppose that δ = α 2 0 . Thus, A ξ = α ξ + β U , A U = β ξ , A ϕ U = α 2 ϕ U and A Z = 0 for any Z D U . The scalar product of (4.1) and ξ implies g ( ϕ A 2 X , A Y ) α g ( ϕ A 2 X , Y ) + k β η ( A X ) g ( ϕ Y , U ) + g ( ϕ A X , A 2 Y ) α g ( ϕ A X , A Y ) + β η ( A Y ) g ( ϕ A X , U ) = 0 for any X , Y D . If Y = U , we obtain α g ( ϕ A 2 X , U ) + g ( ϕ A X , A 2 U ) + β 2 g ( ϕ A X , U ) = 0 . Therefore, α g ( A 2 ϕ U , X ) + β g ( ϕ A X , A ξ ) β 2 g ( A ϕ U , X ) = 0 for any X D . Then α g ( A 2 ϕ U , X ) 2 β 2 g ( A ϕ U , X ) = 0 . As α 0 , this yields α 2 + 4 β 2 = 0 , which is impossible, finishing the proof of Theorem 1.2.

If we now suppose that A F ( k ) is hybrid with respect to A , we should have

(4.8) g ( ϕ A 2 X , A Y ) ξ η ( A Y ) ϕ A 2 X k η ( A X ) ϕ A Y g ( ϕ A 2 X , Y ) A ξ + η ( Y ) A ϕ A 2 X + k η ( A X ) A ϕ Y + g ( ϕ A X , A 2 Y ) ξ η ( A 2 Y ) ϕ A X k η ( X ) ϕ A 2 Y g ( ϕ A X , A Y ) A ξ + η ( A Y ) A ϕ A X + k η ( X ) A ϕ A Y = 0

for any X , Y tangent to M . But M should be also η -hybrid, so locally congruent to a ruled real hypersurface. If this is the case and we take X = Y = ξ in (4.8), we obtain g ( ϕ A 2 ξ , A ξ ) α ϕ A 2 ξ k α ϕ A ξ + A ϕ A 2 ξ + g ( ϕ A ξ , A 2 ξ ) ξ η ( A 2 ξ ) ϕ A ξ k ϕ A 2 ξ + α A ϕ A ξ + k A ϕ A ξ = 0 . This yields α ϕ A ( α ξ + β U ) k α β ϕ U + α A ϕ A ξ + β A ϕ A U β g ( α ξ + β U , α ξ + β U ) ϕ U α k ϕ A ξ + β k ϕ A U = 0 . That is, α 2 β ϕ U k α β ϕ U β ( α 2 + β 2 ) ϕ U k α β ϕ U = 0 . Therefore, as β 0 , we obtain

(4.9) β 2 = 2 α ( α + k ) .

But if we take X = U , Y = ξ in (4.8), we obtain β g ( ϕ A 2 U , U ) ξ α ϕ A 2 U k β ϕ A ξ + A ϕ A 2 U = 0 . Thus α β ϕ A ξ k β ϕ A ξ + β A ϕ A ξ = 0 . As A ϕ U = 0 , we obtain α + k = 0 , due to β 2 ϕ U 0 . From (4.9), we should have β = 0 , which is impossible and we have proved the Corollary.

5 Proof of Theorem 1.3

Let us suppose that A T ( k ) is pure with respect to A . Then for any X , Y tangent to M , we must have

(5.1) g ( ϕ A 2 X , A Y ) ξ η ( A Y ) ϕ A 2 X k η ( A X ) ϕ A Y g ( ϕ A 2 Y , A X ) ξ + η ( A X ) ϕ A 2 Y + k η ( A Y ) ϕ A X g ( ϕ A 2 X , Y ) A ξ + η ( Y ) A ϕ A 2 X + k η ( A X ) A ϕ Y + g ( ϕ A Y , A X ) A ξ η ( A X ) A ϕ A Y k η ( Y ) A ϕ A X = g ( ϕ A X , A 2 Y ) ξ η ( A 2 Y ) ϕ A X k η ( X ) ϕ A 2 Y g ( ϕ A 3 Y , X ) ξ + η ( X ) ϕ A 3 Y + k η ( A 2 Y ) ϕ X g ( ϕ A X , A Y ) A ξ + η ( A Y ) A ϕ A X + k η ( X ) A ϕ A Y + g ( ϕ A 2 Y , X ) A ξ η ( X ) A ϕ A 2 Y k η ( A Y ) A ϕ X .

Suppose that M is Hopf with Reeb curvature α . Then (5.1) gives g ( ϕ A 2 X , A Y ) ξ g ( ϕ A 2 Y , A X ) ξ α g ( ϕ A 2 X , Y ) ξ + α g ( ϕ A Y , A X ) ξ = g ( ϕ A X , A 2 Y ) ξ g ( ϕ A 3 Y , X ) ξ α g ( ϕ A X , A Y ) ξ + α g ( ϕ A 2 Y , X ) ξ , for any X , Y D . That means that A ϕ A 2 X α ϕ A 2 X = A 3 ϕ X α A 2 ϕ X for any X D . If A X = λ X with X D , A ϕ X = μ ϕ X with μ as in Theorem 2.1. Then, λ 2 μ α λ 2 = μ 3 α μ 2 . This gives

(5.2) ( λ + μ ) ( λ μ ) ( μ α ) = 0 .

Taking X D , Y = ξ from (5.1), we have α ϕ A 2 X + k α ϕ A X + A ϕ A 2 X k A ϕ A X = α 2 ϕ A X + k α 2 ϕ X + α A ϕ A X k α A ϕ X for any X D . If X is as earlier, we obtain

(5.3) ( k λ ) ( α μ ) ( λ α ) = 0 .

A similar calculation with X = ξ , Y D in (5.1) gives

(5.4) ( k λ ) ( α λ ) ( λ μ ) = 0 .

If λ + μ = 0 , we obtain 2 λ 2 + 2 = 0 , which is impossible. From (5.2) if μ = α , (5.4) would be ( λ k ) ( μ λ ) 2 = 0 . Now if λ = μ , (5.3) implies ( k μ ) ( α μ ) 2 = 0 . Then if μ = α , M should be totally umbilical, which is impossible. Therefore, μ = λ = k and M has two distinct constant principal curvatures. From [22], M must be locally congruent to a geodesic hypersphere whose radius r , 0 < r < π 2 satisfies cot ( 2 r ) = k 2 1 2 k . The other possibility is μ = α , λ = k . In this case, α = α k + 2 2 k α implies α 2 k α + 2 = 0 . Thus, once again, M has at most two distinct constant principal curvatures and must be locally congruent to a geodesic hypersphere. But then λ = μ gives a contradiction.

Suppose then that M is non Hopf. As mentioned earlier, we write A ξ = α ξ + β U with the same conditions. If in (5.1) we take X = Y = ξ , we obtain

(5.5) β g ( ϕ A 2 ξ , U ) ξ β ϕ A ( α ξ + β U ) + α ϕ A ( α ξ + β U ) + 2 A ϕ A ( α ξ + β U ) 2 α β A ϕ U 2 k β A ϕ U = β ( α 2 + β 2 ) ϕ U k ϕ A ( α ξ + β U ) + ϕ A 2 ( α ξ + β U ) .

Its scalar product with ξ , bearing in mind that β 0 , yields g ( ϕ A ( α ξ + β U ) , U ) = 0 . Then g ( A ϕ U , α ξ + β U ) = 0 . Therefore,

(5.6) g ( A U , ϕ U ) = 0 .

In (5.1) take X , Y D U and its scalar product with ϕ U obtaining η ( A 2 Y ) g ( A X , U ) = 0 . That is, β g ( A U , Y ) g ( A U , X ) = 0 for any X , Y D U . If X = Y ,

(5.7) g ( A U , X ) = 0

for any X D U . From (5.6) and (5.7), we can write

(5.8) A U = β ξ + γ U

for a certain function γ . If we take X = Y = U in (5.1) and its scalar product with ϕ U , we obtain 2 k β g ( A ϕ U , ϕ U ) 2 γ β g ( A ϕ U , ϕ U ) = ( k γ ) g ( A U , A ξ ) , where we have used (5.8). As β 0 , we obtain

(5.9) 2 ( k γ ) g ( A ϕ U , ϕ U ) = ( k γ ) ( α + γ ) .

Once again take X = Y = U in (5.1) and its scalar product with Z D U . We have β g ( ϕ A 2 U , Z ) + β g ( A 2 U , Z ) + 2 k β g ( A ϕ U , Z ) 2 β g ( ϕ A U , A Z ) = 0 . From (5.8)  ϕ A U = γ ϕ U and ϕ A 2 U = ( β 2 + γ 2 ) ϕ U . As also from (5.8)  g ( A 2 U , Z ) = 0 , we assure

(5.10) ( k γ ) g ( A ϕ U , Z ) = 0

for any Z D U .

From (5.9) and (5.10), we have two possibilities: (i) k = γ , or (ii) if k γ , A ϕ U = α + γ 2 ϕ U .

Take in (5.1)  X = U , Y D and its scalar product with U . We obtain k β g ( A ϕ U , Y ) k β g ( ϕ A U , Y ) + β g ( A Y , ϕ A U ) β g ( ϕ A 2 U , Y ) = 0 , As β 0 , it follows

(5.11) k g ( A ϕ U , Y ) k g ( A U , Y ) + g ( ϕ A U , A Y ) g ( ϕ A 2 U , Y ) = 0

for any Y D . Introducing (5.8) in (5.11), we obtain k g ( A ϕ U , Y ) k γ g ( ϕ U , Y ) + γ g ( A ϕ U , Y ) ( β 2 + γ 2 ) g ( ϕ U , Y ) = 0 for any Y D . Therefore, if k + γ = 0 , β 2 g ( ϕ U , Y ) = for any Y D . This yields β = 0 , which is impossible. Thus, k + γ 0 and

(5.12) A ϕ U = k γ + β 2 + γ 2 k + γ ϕ U .

From (5.8) and (5.12), D U is A -invariant.

The scalar product of (5.1) and Z D U for X = U , Y D U , bearing in mind that β 0 yields k ϕ A Y + ϕ A 2 Y + k A ϕ Y A ϕ A Y = 0 for any Y D U . If A Y = λ Y , this gives ( k λ ) A ϕ Y = λ ( k λ ) ϕ Y . Then either λ = k or if λ k , A ϕ Y = λ ϕ Y . Now every eigenspace corresponding to an eigenvalue different from k is ϕ -invariant. This means that the eigenspace corresponding to k is also ϕ -invariant.

Take in (5.1), X D U , Y = U and its scalar product with Z D U , bearing in mind (5.8). This implies g ( ϕ A 2 X , Z ) + k g ( ϕ A X , Z ) = ( α + γ ) g ( ϕ A X , Z ) + k ( α + γ ) g ( ϕ X , Z ) + g ( A ϕ A X , Z ) k g ( A ϕ X , Z ) for any X , Z D U . Therefore, ϕ A 2 X + k ϕ A X = ( α + γ ) ϕ A X + k ( α + γ ) ϕ X + A ϕ A X k A ϕ X , for any X D U . If A X = λ X , we have ( k λ λ 2 ) ϕ X = λ ( α + γ ) ϕ X + k ( α + γ ) ϕ X + ( λ k ) A ϕ X . This yields ( k λ ) A ϕ X = ( k λ ) ( α + γ λ ) ϕ X . If λ k , A ϕ X = ( α + γ λ ) ϕ X . But as α + γ λ = λ , in this case, λ = α + γ 2 . Therefore, the unique principal curvatures on D U are k and α + γ 2 .

If we suppose that k = γ , A U = β ξ + k U , A ϕ U = 2 k 2 + β 2 2 k ϕ U and on D U the principal curvatures are k and α + k 2 , and their corresponding eigenspaces are ϕ -invariant.

As ϕ A 2 ξ = α ϕ A ξ + β ϕ A U = β ( α + k ) ϕ U , from (5.5), we obtain α β 2 ϕ U k β 2 ϕ U + α 2 β ϕ U + k α β ϕ U = β α 2 ϕ U k α β ϕ U k 2 β ϕ U + α β ( k + α ) ϕ U + β 3 ϕ U + k 2 β ϕ U . That is, ( α β ) ( α + k ) = 0 . That means that we have two new possibilities: (I) α = β or (II) k + α = 0 .

Let us suppose that there exists Z D U such that A Z = k Z and A ϕ Z = k ϕ Z . Codazzi equation implies ( Z A ) ϕ Z ( ϕ Z A ) Z = 2 ξ , and then k Z ϕ Z A Z ϕ Z k ϕ Z Z + A ϕ Z Z = 2 ξ . Its scalar product with U gives k g ( [ ϕ Z , Z ] , U ) g ( Z ϕ Z , β ξ + k U ) + g ( ϕ Z Z , β ξ + k U ) = 0 . That is, 2 k β = 0 , which is impossible, and for any Z D U , we must have A Z = α + k 2 Z .

Take now W D U . By applying Codazzi equation to W and ϕ W , we obtain W α 2 ϕ W + α + k 2 W ϕ W A W ϕ W ( ϕ W ) α 2 W α + k 2 ϕ W W + A ϕ W W = 2 ξ . Its scalar product with U implies α + k 2 g ( [ ϕ W , W ] , U ) g ( W ϕ W , β ξ + k U ) + g ( ϕ W W , β ξ + k U ) = 0 . Then

(5.13) ( k α ) g ( [ ϕ W , W ] , U ) = 2 β ( k + α ) .

If in (5.13), we suppose that α = k , we should have α = k . Thus, α = k = 0 , which is impossible. We have proved that α k .

The scalar product of the expression obtained from the aforementioned Codazzi equation and ξ gives α + k 2 g ( ϕ W , ϕ A W ) + α + k 2 g ( W , ϕ A ϕ W ) g ( W ϕ W , α ξ + β U ) + g ( ϕ W W , α ξ + β U ) = 2 . That gives

(5.14) β g ( [ ϕ W , W ] , U ) = 2 α + k 2 2 2 α α + k 2 2 .

Moreover, its scalar product with ϕ U implies

(5.15) W ( α ) = 0

for any W D U . From (5.13) and (5.14), we also have 4 β 2 ( k + α ) = ( 4 + 2 α ( α + k ) ( α + k ) 2 ) ( k α ) = ( k α ) ( 4 + α 2 k 2 ) . If α = k , we should have 8 k = 0 , which is impossible. Then α ± k and

(5.16) 4 β 2 = ( k 2 α 2 4 ) α k α + k .

From (5.15), if we derive (5.16) in the direction of W D U , we should have 8 β W ( β ) = 2 α α k α + k W ( α ) + ( k 2 α 2 4 ) ( 2 k ( α + k ) 2 ) W ( α ) = 0 . Thus,

(5.17) W ( β ) = 0

for any W D U . On the other hand, ( W A ) ϕ U ( ϕ U A ) W = 0 implies

(5.18) 2 k 2 + β 2 2 k W ϕ U A W ϕ U = ( ϕ U ) α 2 W α + k 2 ϕ U W + A ϕ U W = 0 .

The scalar product of (5.18) and ϕ U implies 2 k 2 β 2 2 k α k + k 2 2 k g ( W ϕ U , ϕ W ) = 0 . Therefore, ( k 2 + β 2 α k ) g ( W ϕ U , ϕ W ) = 0 . We have two cases:

CASE I. k 2 + β 2 α k = 0 . In this case, 2 β X ( β ) = k X ( α ) for any X tangent to M . Moreover, from (5.16), 3 k 2 + 4 α k + α 2 + 4 = 0 . Thus, α is constant, and then, X ( α ) = X ( β ) = 0 for any X tangent to M and β is also constant.

Now ( U A ) ϕ U ( ϕ U A ) U = 2 ξ yields

(5.19) 2 k 2 + β 2 2 k U ϕ U A U ϕ U β ϕ A ϕ U k ϕ U U + A ϕ U U = 2 ξ .

The scalar product of (5.19) and ξ implies 2 k 2 + β 2 2 k g ( A U , U ) g ( U ϕ U , α ξ + β U ) + k g ( ϕ A ϕ U , U ) + g ( ϕ U U , α ξ + β U ) = 2 . It follows

(5.20) ( 2 k 2 + β 2 ) + α k + α 2 k 2 + β 2 2 k + 2 = β g ( U ϕ U , U ) .

On the other hand, the scalar product of (5.19) and U yields 2 k 2 + β 2 2 k g ( U ϕ U , U ) g ( U ϕ U , β ξ + k U ) + β g ( A ϕ U , ϕ U ) + g ( ϕ U U , β ξ ) = 0 . This implies β 2 2 k g ( U ϕ U , U ) + β k + β ( 2 k 2 + β 2 k ) = 0 . Thus,

(5.21) β g ( U ϕ U , U ) = 6 k 2 2 β 2 .

From (5.20) and (5.21), ( 2 k 2 + β 2 ) + α k + α 2 k 2 + β 2 2 k + 2 = 6 k 2 2 β 2 . Bearing in mind that β 2 = α k k 2 , we obtain 6 k 3 + 2 α k 2 = 4 k 3 α k 2 α 2 k . As k 0 , we obtain

(5.22) α 2 + 5 α k + 6 k 2 + 4 = 0 .

From (5.16) and as α ± k and β 2 = α k k 2 , it follows

(5.23) α 2 + 4 k α + 3 k 2 + 4 = 0 .

From (5.22) and (5.23), we have k α + 3 k 2 = 0 . This gives α = 3 k . Then β 2 = α k k 2 = 4 k 2 , which implies that CASE I is impossible. Therefore, we have the

CASE II. g ( W ϕ U , ϕ W ) = 0 . From (2.3), this implies g ( W U , W ) = 0 . Then ( W A ) U ( U A ) W = 0 gives W ( β ξ + k U ) A W U U α + k 2 W + A U W = 0 . Then β ϕ A W + k W U A W U 1 2 U ( α ) W α + k 2 U W + A U W = 0 . Its scalar product with W yields

(5.24) U ( α ) = 0 .

The same reasoning applied to ( W A ) ξ ( ξ A ) W = ϕ W gives

(5.25) ξ ( α ) = 0 .

Developing ( ϕ U A ) ξ ( ξ A ) ϕ U = U , we obtain ( ϕ U ) ( α ) ξ + α ϕ A ϕ U + ( ϕ U ) ( β ) U + β ϕ U U + 2 k 2 + β 2 2 k A U β k ξ ( β ) ϕ U 2 k 2 + β 2 2 k ξ ϕ U + A ξ ϕ U = U . Its scalar product with ξ implies

(5.26) ( ϕ U ) ( α ) = α β 3 β 2 k ( 2 k 2 + β 2 ) β g ( ξ ϕ U , U ) .

Moreover, ( ξ A ) U ( U A ) ξ = ϕ U gives ξ ( β ) ξ + β ϕ A ξ + k ξ U A ξ U α ϕ A U U ( β ) U β U U + A ϕ A U = ϕ U . If we take its scalar product with ϕ U , we obtain β 2 + k 2 k 2 + β 2 2 k g ( ξ U , ϕ U ) k α β g ( U U , ϕ U ) + k 2 k 2 + β 2 2 k = 1 . Then

(5.27) β 2 g ( ξ ϕ U , U ) = 3 k β 2 2 k 2 α + 2 k 2 2 k 2 k g ( U U , ϕ U ) .

As ( U A ) ϕ U ( ϕ U A ) U = 2 ξ , we obtain β k U ( β ) ϕ U + 2 k 2 + β 2 2 k U ϕ U A U ϕ U ( ϕ U ) ( β ) ξ β ϕ A ϕ U k ϕ U U + A ϕ U U = 2 ξ . Taking its scalar product with U , we obtain 2 k 2 + β 2 2 k g ( U ϕ U , U ) g ( U ϕ U , β ξ + k U ) + β g ( A ϕ U , ϕ U ) + β g ( ϕ U U , ξ ) = 0 . This yields β 2 2 k g ( U U , ϕ U ) + k β + 2 β 2 k 2 + β 2 2 k = 0 , and therefore,

(5.28) β g ( U U , ϕ U ) = 2 β 2 + 6 k 2 .

From (5.27) and (5.28), we have β 2 g ( ξ ϕ U , U ) = k β 2 2 k 2 α 10 k 3 2 k and ( ϕ U ) ( α ) = α β ( 2 k 2 + β 2 ) 3 β 2 k k β 2 k 2 α β 10 k 3 β 2 k β . If we denote by ω , the second member of this equality we can write

(5.29) grad ( α ) = ω ϕ U .

As g ( X grad ( α ) , Y ) = g ( Y grad ( α ) , X ) for any X , Y tangent to M , (5.29) yields X ( ω ) g ( ϕ U , Y ) + ω g ( X ϕ U , Y ) = Y ( ω ) g ( ϕ U , X ) + ω g ( Y ϕ U , X ) . If X = ξ , it follows ω g ( ξ ϕ U , Y ) = ω g ( ϕ U , ϕ A Y ) = ω g ( U , A Y ) , for any Y tangent to M . Then for Y = U , we obtain ω g ( ξ ϕ U , U ) = k ω . Therefore, either ω = 0 and α is constant or g ( ξ ϕ U , U ) = k . In this second case, (5.27) yields k β 2 k 2 α 5 k 3 k = 0 . As k 0 , β 2 + k α + 5 k 2 + 1 = 0 . This equation and (5.16) imply that α is constant, so β is also constant.

Then ( W A ) U ( U A ) W = 0 for any W D U yields β ϕ A W + k W U A W U α + k 2 U W + A U W = 0 . Taking its scalar product with ϕ W , we have β α + k 2 + k α + k 2 g ( W U , ϕ W ) = 0 . Then

(5.30) ( k α ) g ( W U , ϕ W ) = β ( α + k ) .

From (5.13) and (5.30), we obtain ( k α ) g ( ϕ W W , U ) = β ( α + k ) and as α ± k , we obtain

(5.31) g ( ϕ W W , U ) 0

for any W D U .

On the other hand, ( ϕ W A ) ϕ U ( ϕ U A ) ϕ W = 0 for any W D U . This gives 2 k 2 + β 2 2 k ϕ W ϕ U A ϕ W ϕ U α + k 2 ϕ U ϕ W + A ϕ U ϕ W = 0 . Its scalar product with ϕ W yields ( 2 k 2 + β 2 2 k α + k 2 ) g ( ϕ W ϕ U , ϕ W ) = 0 . As g ( ϕ W ϕ U , ϕ W ) = g ( ϕ W U , W ) , we obtain ( k 2 + β 2 α k ) g ( ϕ W U , W ) = 0 . From (5.31),

(5.32) k 2 + β 2 = α k .

From (5.32) and (5.16) and as α ± k , we obtain 4 k ( α + k ) = k 2 α 2 4 . Then 3 k 2 + 4 k α + α 2 = 4 . From (5.32), we have 7 k 2 + 4 β 2 + α 2 = 0 , which is impossible. Thus, we have arrived at γ k . Therefore, A ξ = α ξ + β U , A U = β ξ + γ U , A ϕ U = α + γ 2 ϕ U and either A Z = k Z with A ϕ Z = k ϕ Z or A Z = α + γ 2 Z with A ϕ Z = α + γ 2 ϕ Z for any Z D U . As also (5.12) is true, we should have

(5.33) 2 β 2 = ( k + γ ) ( α γ ) .

In this case, if α = k from (5.33)  2 β 2 = ( k + γ ) 2 , which is impossible, and we must suppose that α = β . Let us suppose that there exists Z D U such that A Z = k Z and A ϕ Z = k ϕ Z . As ( Z A ) ϕ Z ( ϕ Z A ) Z = 2 ξ , its scalar product with ξ yields

(5.34) β g ( [ ϕ Z , Z ] , U ) = 2 k 2 2 k β 2 ,

and its scalar product with U gives

(5.35) ( γ k ) g ( [ ϕ Z , Z ] , U ) = 2 k β .

From (5.34) and (5.35), 2 ( γ k ) ( k 2 k β 1 ) = 2 k β 2 . That is,

(5.36) β 2 = k γ k ( k 2 k β 1 ) .

The scalar product of the same expression and ϕ U implies ( 2 k β γ ) g ( Z ϕ Z , ϕ U ) ( 2 k β γ ) g ( ϕ Z Z , ϕ U ) = 0 . Therefore, either β + γ = 2 k and β + γ should be constant, or g ( Z Z , U ) = g ( Z ϕ Z , ϕ U ) = g ( ϕ Z Z , ϕ U ) .

Moreover, ( Z A ) ϕ U ( ϕ U A ) Z = 0 yields 1 2 Z ( β + γ ) ϕ U + β + γ 2 Z ϕ U A Z ϕ U k ϕ U Z + A ϕ U Z = 0 . Its scalar product with ϕ Z gives ( β + γ 2 k ) g ( Z ϕ U , ϕ Z ) = 0 . If β + γ 2 k , we would have g ( Z ϕ Z , ϕ U ) = g ( Z Z , U ) = g ( ϕ Z Z , ϕ U ) = 0 . But its scalar product with Z gives ( β + γ 2 k ) g ( Z ϕ U , Z ) = 0 . If β + γ 2 k , g ( Z ϕ U , Z ) = 0 . As g ( Z ϕ U , Z ) = g ( ( Z ϕ ) U , Z ) + g ( ϕ Z U , Z ) and g ( ( Z ϕ ) U , Z ) = η ( U ) g ( A Z , Z ) g ( A Z , U ) η ( Z ) = 0 , g ( Z U , ϕ Z ) = 0 for any Z D U such that A Z = k Z .

Then ( Z A ) U ( U A ) Z = 0 implies Z ( β ) ξ + β k ϕ Z + Z ( γ ) U + γ Z U A Z U k U Z + A U Z = 0 . Its scalar product with ϕ Z gives β k + ( γ k ) g ( Z U , ϕ Z ) = 0 . If β + γ 2 k , g ( Z U , ϕ Z ) = 0 yields k β = 0 , that is impossible. Therefore, β + γ = 2 k , a constant, and from (5.33), we obtain

(5.37) β 2 + γ 2 = k 2 .

Moreover, 2 β γ = 3 k 2 and β γ is also constant. Now we can write A U = β ξ + ( 2 k β ) U , A ϕ U = k ϕ U , and A Z = k Z for any Z D U . From (5.36) and (5.37), we obtain k ( k 2 γ 2 ) = ( k γ ) ( k 2 k β 1 ) . This gives k ( k + γ ) = k 2 k β 1 . Therefore, k ( γ + β ) = 1 = 2 k 2 , which is impossible, proving that the unique principal curvature on D U is β + γ 2

Take W D U . The scalar product of ( W A ) ϕ W ( ϕ W A ) W = 2 ξ with ϕ W yields

(5.38) W ( β + γ ) = 0

for any W D U .

From the scalar product of ( W A ) ϕ U ( ϕ U A ) W = 0 with W , we obtain

(5.39) ( ϕ U ) ( β + γ ) = 0 .

Developing ( W A ) ξ ( ξ A ) W = ϕ W and taking its scalar product with W , we obtain

(5.40) ξ ( β + γ ) = 2 β g ( W U , W )

as far as its scalar product with ξ implies

(5.41) W ( β ) = β g ( ξ W , U ) .

Moreover, its scalar product with U yields

(5.42) W ( β ) = β γ 2 g ( ξ W , U ) .

From (5.41) and (5.42) either β γ = 2 β or g ( ξ W , U ) = 0 . In the first case, γ = 3 β . From (5.33), we have 2 β 2 = 2 β ( k + 3 β ) . This gives 4 β = k and β is constant. If g ( ξ W , U ) = 0 , (5.41) implies W ( β ) = 0 . We assure that always

(5.43) W ( β ) = 0

for any W D U . From (5.43) and (5.38) also W ( γ ) = 0 for any W D U .

As ( W A ) U ( U A ) W = 0 , from (5.43) and being W ( γ ) = 0 for any W D U , its scalar product with W yields

(5.44) U ( β + γ ) = ( γ β ) g ( W U , W ) ,

and its scalar product with ξ , bearing in mind that β 0 implies

(5.45) g ( U W , U ) = 0 .

Then the scalar product of ( ξ A ) U ( U A ) ξ = ϕ U and ξ gives

(5.46) ξ ( β ) = U ( β ) ,

and from its scalar product with U , we obtain

(5.47) ξ ( γ ) = U ( β ) .

From (5.46) and (5.47)  ξ ( β ) = ξ ( γ ) = U ( β ) . From (5.33), 2 β 2 = k β k γ + γ β β 2 . Therefore, 4 β ξ ( β ) = k ξ ( β ) k ξ ( γ ) + γ ξ ( β ) + β ξ ( γ ) 2 γ ξ ( γ ) . Now ( 4 β k γ ) ξ ( β ) = ( β 2 γ k ) ξ ( γ ) . As ξ ( β ) = ξ ( γ ) , we have ( 3 β + γ ) ξ ( β ) = 0 . If ξ ( β ) 0 , 3 β + γ = 0 yields 4 ξ ( β ) = 0 . Thus, always

(5.48) ξ ( β ) = U ( β ) = ξ ( γ ) = 0 .

Then k U ( γ ) + β U ( γ ) 2 γ U ( γ ) = 0 . If U ( γ ) 0 , we should have k + β 2 γ = 0 . But from (5.48), this implies U ( γ ) = 0 . Thus,

(5.49) U ( γ ) = 0 .

From (5.40), (5.38), and (5.39), β + γ is constant.

From (5.43), we can consider grad ( β ) = ω ϕ U and grad ( γ ) = ω ϕ U for a certain function ω . As g ( X grad ( β ) , Y ) = g ( Y grad ( β ) , X ) for any X , Y tangent to M , we have X ( ω ) g ( ϕ U , Y ) + ω g ( X ϕ U , Y ) = Y ( ω ) g ( ϕ U , X ) + ω g ( Y ϕ U , X ) . For X = ξ , we obtain ξ ( ω ) g ( ϕ U , Y ) + ω g ( ξ ϕ U , Y ) = ω g ( ϕ U , ϕ A Y ) = ω g ( A U , Y ) . Now if Y = U , ω g ( ξ ϕ U , U ) = ω γ . Therefore, either ω = 0 and β and γ are constant, or if ω 0 , g ( ξ ϕ U , U ) = γ .

If we suppose that β and γ are constant, we recall that ( ξ A ) U ( U A ) ξ = ϕ U and its scalar product with ϕ U implies

(5.50) β 2 + β γ 2 g ( ξ ϕ U , U ) + β g ( U ϕ U , U ) + γ γ β 2 = 1 ,

and the scalar product of ( ξ A ) ϕ U ( ϕ U A ) ξ = U with U gives

(5.51) β γ 2 g ( ξ ϕ U , U ) = 1 β 2 + β 2 γ 2 2 .

On the other hand, the scalar product of ( U A ) ϕ U ( ϕ U A ) U = 2 ξ and ξ yields

(5.52) β g ( U ϕ U , U ) = 2 γ 2 + β β + γ 2 .

From (5.50), (5.51), and (5.52), we obtain β 2 γ 2 2 γ 2 + β 2 + γ 2 2 = 0 . Thus, β 2 γ 2 = 0 . If β = γ , (5.51) gives 0 = 1 β 2 , which is impossible. If β + γ = 0 , A ϕ U = A Z = 0 for any Z D U . For such a Z , the scalar product of ( Z A ) ϕ Z ( ϕ Z A ) Z = 2 ξ with ξ , respectively with U , gives β g ( [ ϕ Z , Z ] , U ) = 2 , respectively g ( [ ϕ Z , Z ] , U ) = 0 , and we arrive at a contradiction.

Thus, g ( ξ ϕ U , U ) = γ . Consider ( ξ A ) ϕ U ( ϕ U A ) ξ = U . Its scalar product with ξ gives

(5.53) ( ϕ U ) ( β ) = β 2 ( β + γ )

as long as its scalar product with U yields

(5.54) ( ϕ U ) ( β ) = 1 γ β 2 + 3 β 2 2 .

From (5.53) and (5.54), we obtain 2 β 2 + 1 = 0 . As this is impossible, we finish the proof.

6 Proof of Theorem 1.4

Suppose that A T ( k ) is η -hybrid with respect to A . This means that A T ( k ) ( A X , Y ) + A T ( k ) ( X , A Y ) = 0 for any X , Y D . This gives F A X ( k ) A Y F A Y ( k ) ( A X ) A F A X ( k ) Y + A F Y A X + F X A 2 Y F A 2 Y ( k ) X A F X A Y + A F A Y ( k ) X = 0 , for any X , Y D . Therefore,

(6.1) g ( ϕ A 2 X , A Y ) ξ η ( A Y ) ϕ A 2 X k η ( A X ) ϕ A Y g ( ϕ A 2 Y , A X ) ξ + η ( A X ) ϕ A 2 Y + k η ( A Y ) ϕ A X g ( ϕ A 2 X , Y ) A ξ + k η ( A X ) A ϕ Y + g ( ϕ A Y , A X ) A ξ η ( A X ) A ϕ A Y + g ( ϕ A X , A 2 Y ) ξ η ( A 2 Y ) ϕ A X g ( ϕ A 3 Y , X ) ξ + k η ( A 2 Y ) ϕ X g ( ϕ A X , A Y ) A ξ + η ( A Y ) A ϕ A X + g ( ϕ A 2 Y , X ) A ξ k η ( A Y ) A ϕ X = 0

for any X , Y D . Suppose first that M is Hopf with Reeb curvature α , the scalar product of (6.1) and ξ gives g ( ϕ A 2 X , A Y ) g ( ϕ A 2 Y , A X ) α g ( ϕ A 2 X , Y ) + α g ( ϕ A Y , A X ) + g ( ϕ A X , A 2 Y ) g ( ϕ A 3 Y , X ) α g ( ϕ A X , A Y ) + α g ( ϕ A 2 Y , X ) = 0 . Therefore, for any X D , we have A ϕ A 2 X + 2 A 2 ϕ A X α ϕ A 2 X 2 α A ϕ A X + A 3 ϕ X α A 2 ϕ X = 0 . If we take X D satisfying A X = λ X , we know A ϕ X = μ ϕ X with μ given in Theorem 2.1. So we obtain λ 2 μ + 2 λ μ 2 α λ 2 2 α λ μ + μ 3 α μ 2 = 0 . That is, λ 2 ( μ α ) + 2 λ μ ( μ α ) + μ 2 ( μ α ) = 0 . Thus, ( λ + μ ) 2 ( μ α ) = 0 . If λ + μ = 0 as in previous proofs, we arrive at a contradiction. This yields μ = α and λ = α 2 + 2 α . M should have two distinct principal curvatures and would be locally congruent to a geodesic hypersphere. But in this case, λ = μ , which is impossible.

We have proved that M is non-Hopf and write as earlier that A ξ = α ξ + β U with the usual conditions on U and β . The scalar product of (6.1) and ϕ U implies

(6.2) η ( A Y ) g ( A 2 X , U ) k η ( A X ) g ( A Y , U ) + η ( A X ) g ( A 2 Y , U ) + k η ( A Y ) g ( A X , U ) + k η ( A X ) g ( ϕ Y , A ϕ U ) η ( A X ) g ( ϕ A Y , A ϕ U ) η ( A 2 Y ) g ( A X , U ) + k η ( A 2 Y ) g ( X , U ) + η ( A Y ) g ( ϕ A X , A ϕ U ) k η ( A Y ) g ( ϕ X , A ϕ U ) = 0

for any X , Y D . Taking X = Y = ϕ U in (6.2), we obtain β g ( A U , ϕ U ) 2 = 0 . As β 0 , it follows

(6.3) g ( A U , ϕ U ) = 0 .

If in (6.2)  X = Y D U , we obtain β g ( A X , U ) 2 = 0 for any X D U . As mentioned earlier,

(6.4) g ( A U , X ) = 0

for any X D U . From (6.3) and (6.4),

(6.5) A U = β ξ + γ U

for a certain function γ . If now X = Y = U in (6.2), we obtain η ( A 2 U ) g ( A U , U ) + k η ( A 2 U ) = 0 . Therefore, ( k γ ) η ( A 2 U ) = 0 . From (6.5), this yields β ( k γ ) ( α + γ ) = 0 . As β 0 , either γ = k or γ = α .

Taking X D U , Y = U in (6.2), it follows

(6.6) k g ( A ϕ U , ϕ X ) + g ( A ϕ U , ϕ A X ) = 0

for any X D U . Now the scalar product of (6.1) and U gives

(6.7) η ( A Y ) g ( A ϕ U , A X ) + k η ( A X ) g ( A ϕ U , Y ) η ( A X ) g ( A ϕ U , A Y ) k η ( A Y ) g ( A ϕ U , X ) β g ( ϕ A 2 X , Y ) + k η ( A X ) g ( A U , ϕ Y ) + β g ( ϕ A Y , A X ) η ( A X ) g ( A U , ϕ A Y ) + η ( A 2 Y ) g ( A ϕ U , X ) + k η ( A 2 Y ) g ( ϕ X , U ) β g ( ϕ A X , A Y ) + η ( A Y ) g ( ϕ A X , A U ) + β g ( ϕ A 2 Y , X ) k η ( A Y ) g ( A U , ϕ X ) = 0

for any X , Y D . Take X = U , Y D U in (6.7) to obtain

(6.8) ( k γ ) g ( A ϕ U , Y ) 2 g ( A ϕ U , A Y ) = 0

for any Y D U , as far as if in (6.7) X D U , Y = U , we have

(6.9) ( 2 γ + α k ) g ( A ϕ U , X ) + 2 g ( A ϕ U , A X ) = 0

for any X D U . From (6.8) and (6.9), we obtain

(6.10) ( α + γ ) g ( A ϕ U , X ) = 0

for any X D U . Take X D U , Y = ϕ U in (6.7). Bearing in mind that β 0 , it follows

(6.11) 2 g ( A ϕ U , ϕ A X ) + g ( A ϕ U , A ϕ X ) = 0

for any X D U . From (6.6) and (6.11), we have 2 k g ( A ϕ U , ϕ X ) + g ( A ϕ U , A ϕ X ) = 0 . Taking ϕ X instead of X , we obtain

(6.12) 2 k g ( A ϕ U , X ) + g ( A ϕ U , A X ) = 0

for any X D U . From (6.8), (6.9), and (6.11), we consider two possibilities. The first one assures the existence of X D U such that g ( A ϕ U , X ) 0 . In this case, from (6.11), α + γ = 0 and from (6.9) and (6.12)  2 γ + α 5 k = 0 . Therefore, γ = 5 k , α = 5 k . If γ = k , this is impossible, and now g ( A ϕ U , X ) = 0 for any X D U . That means that if γ = k , A ϕ U = δ ϕ U for a certain function δ . Taking X = U , Y = ϕ U in (6.7), we obtain k β g ( A ϕ U , ϕ U ) β g ( A ϕ U , A ϕ U ) β g ( ϕ A 2 U , ϕ U ) + k β g ( A U , ϕ 2 U ) β g ( ϕ A U , A ϕ U ) + β g ( ϕ 2 A ϕ U , U ) = 0 . As β 0 , this yields k δ 2 δ 2 ( β 2 + γ 2 ) k γ δ γ = 0 . As we suppose γ = k , 2 δ 2 + 2 γ 2 + β 2 = 0 , which is impossible.

Therefore, we must suppose that γ = 5 k , α = 5 k , and both functions are constant. Taking X = ϕ U in (6.1) and its scalar product with Z D U , we obtain η ( A Y ) g ( A ϕ U , A ϕ Z ) k η ( A Y ) g ( A ϕ U , ϕ Z ) + η ( A 2 Y ) g ( A ϕ U , ϕ Z ) η ( A Y ) g ( A ϕ U , ϕ A Z ) = 0 . If Y = U , as β 0 , it follows

(6.13) g ( A ϕ U , A ϕ Z ) g ( A ϕ U , ϕ A Z ) k g ( A ϕ U , ϕ Z ) = 0

for any Z D U , From (6.11) and (6.13), 2 k g ( A ϕ U , ϕ Z ) 3 g ( A ϕ U , A ϕ Z ) = 0 . Taking ϕ X instead of Z , X D U , we obtain

(6.14) 2 k g ( A ϕ U , X ) 3 g ( A ϕ U , A X ) = 0

for any X D U . Then (6.12) and (6.14) yield g ( A ϕ U , A X ) = 0 for any X D U . As k 0 , we obtain g ( A ϕ U , X ) = 0 for any X D U , a contradiction that finishes the proof.

  1. Funding information: This work was supported by MICINN Project PID 2020-116126GB-100 and Project PY20-01391 from Junta de Andalucía.

  2. Author contributions: The authors declare that they have equal contributions.

  3. Conflict of interest: The authors state no conflicts of interest.

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Received: 2023-09-08
Revised: 2024-08-05
Accepted: 2024-08-05
Published Online: 2024-09-16

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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