Commutativity of Cho and structure Jacobi operators of a real hypersurface in a complex projective space

Abstract

We consider real hypersurfaces $M$ in complex projective space equipped with both the Levi-Civita and generalized Tanaka-Webster connections. For any non-null constant $k$ and any vector field $X$ tangent to $M$, we can define an operator on $M$, $F_X^{(k)}$, related to both connections. We study commutativity problems of these operators and the structure Jacobi operator of $M$.

1 Introduction

Let $\mathbb {C}P^m$, $m\ge 2$, be a complex projective space endowed with the metric $g$ of constant holomorphic sectional curvature 4. Let $M$ be a connected real hypersurface of $\mathbb {C}P^m$ without boundary. Let $\nabla$ be the Levi-Civita connection on $M$ and $J$ the complex structure of $\mathbb {C}P^m$. Take a locally defined unit normal vector field $N$ on $M$ and denote by $\xi =-JN$. This is a tangent vector field to $M$ called the structure vector field on $M$. On $M$, there exists an almost contact metric structure $(\phi ,\xi ,\eta ,g)$ induced by the Kaehlerian structure of $\mathbb {C}P^m$, where $\phi$ is the tangent component of $J$ and $\eta$ is an one form given by $\eta (X)=g(X,\xi )$ for any $X$ tangent to $M$. The classification of homogeneous real hypersurfaces in $\mathbb {C}P^m$ was obtained by Takagi, see [5, 12–14]. His classification contains 6 types of real hypersurfaces. Among them, we find type $(A_1)$ real hypersurfaces that are geodesic hyperspheres of radius $r$, $0<r< \frac{\pi }{2}$ and type $(A_2)$ real hypersurfaces that are tubes of radius $r$, $0 < r <\frac{\pi }{2}$, over totally geodesic complex projective spaces $\mathbb {C}P^n$, $0 < n < m-1$. We will call both types of real hypersurfaces type $(A)$ real hypersurfaces.

Ruled real hypersurfaces can be described as follows: take a regular curve $\gamma$ in $\mathbb {C}P^m$ with tangent vector field $X$. At each point of $\gamma$, there is a unique $\mathbb {C}P^{m-1}$ cutting $\gamma$ so as to be orthogonal not only to $X$ but also to $JX$. The union of these hyperplanes is called a ruled real hypersurface. It will be an embedded hypersurface locally, although globally it will in general have self-intersections and singularities. Equivalently, a ruled real hypersurface satisfies that the maximal holomorphic distribution on $M$, $\mathbb {D}$, given at any point by the vectors orthogonal to $\xi$, is integrable or $g(A\mathbb {D}, \mathbb {D})=0$. For examples of ruled real hypersurfaces, see [6] or [8].

The Tanaka-Webster connection, [15–17], is the canonical affine connection defined on a non-degenerate, pseudo-Hermitian CR manifold. As a generalization of this connection, Tanno, [16], defined the generalized Tanaka-Webster connection for contact metric manifolds by

$$\begin{aligned} \hat{\nabla }_XY=\nabla _XY + (\nabla _X \eta )(Y)\xi -\eta (Y)\nabla _X \xi -\eta (X)\phi Y. \end{aligned}$$

(1.1)

Using the naturally extended affine connection of Tanno’s generalized Tanaka-Webster connection, Cho defined the g-Tanaka-Webster connection $\hat{\nabla }^{(k)}$ for a real hypersurface $M$ in $\mathbb {C}P^m$ given, see [3, 4], by

$$\begin{aligned} \hat{\nabla }^{(k)}_XY = \nabla _XY+g(\phi AX,Y)\xi -\eta (Y)\phi AX-k\eta (X) \phi Y \end{aligned}$$

(1.2)

for any $X,Y$ tangent to $M$ where $k$ is a nonzero real number. Then, $\hat{\nabla }^{(k)}\eta =0$, $\hat{\nabla }^{(k)}\xi =0$, $\hat{\nabla }^{(k)}g=0$, $\hat{\nabla }^{(k)}\phi =0$. In particular, if the shape operator of a real hypersurface satisfies $\phi A+A \phi =2k\phi$, the g-Tanaka-Webster connection coincides with the Tanaka-Webster connection.

Here, we can consider the tensor field of type (1,2) given by the difference in both connections $F^{(k)}(X,Y)=g(\phi AX,Y)\xi -\eta (Y) \phi AX-k\eta (X)\phi Y$, for any $X,Y$ tangent to $M$, see [7] Proposition 7.10, pages 234–235. We will call this tensor the $k$th Cho tensor on $M$. Associated to it, for any $X$ tangent to $M$ and any non-null 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 $Y \in TM$. This operator will be called the $k$th Cho operator corresponding to $X$. The torsion of the connection $\hat{\nabla }^{(k)}$ is given by $\hat{T}^{(k)}(X,Y)=F_X^{(k)}Y-F_Y^{(k)}X$ for any $X,Y$ tangent to $M$.

The Jacobi operator $R_X$ with respect to a unit vector field $X$ is defined by $R_X=R(.,X)X$, where $R$ is the curvature tensor field on $M$. Then, we see that $R_X$ is a self-adjoint endomorphism of the tangent space. It is related to Jacobi vector fields, which are solutions of the second-order differential equation (the Jacobi equation) $\nabla _{\dot{\gamma }}(\nabla _{\dot{\gamma }}Y)+R(Y,\dot{\gamma })\dot{\gamma } =0$ along a geodesic $\gamma$ in $M$. The Jacobi operator with respect to the structure vector field $\xi$, $R_{\xi }$, is called the structure Jacobi operator on $M$.

The purpose of the present paper was to study real hypersurfaces $M$ in $\mathbb {C}P^m$ such that the covariant and g-Tanaka-Webster derivatives of the structure Jacobi operator coincide. $\nabla R_{\xi }=\hat{\nabla }^{(k)}R_{\xi }$ is equivalent to the fact that, for any $X$ tangent to $M$, $R_{\xi }F_X^{(k)}=F_X^{(k)}R_{\xi }$. The meaning of this condition is that every eigenspace of $R_{\xi }$ is preserved by the $k$th Cho operator $F_X^{(k)}$ for any $X$ tangent to $M$.

On the other hand, $TM=Span\{ \xi \} \oplus \mathbb {D}$. Thus, we will obtain the following

Theorem 1

Let $M$ be a real hypersurface in $\mathbb {C}P^m$, $m\ge 3$. Let $k$ be a non-null constant. Then, $F_X^{(k)}R_{\xi }=R_{\xi }F_X^{(k)}$ for any $X \in \mathbb {D}$ if and only if $M$ is locally congruent to a ruled real hypersurface.

Theorem 2

Let $M$ be a real hypersurface in $\mathbb {C}P^m$, $m \ge 3$. Let $k$ be a non-null constant. Then, $F_{\xi }^{(k)}R_{\xi }=R_{\xi }F_{\xi }^{(k)}$ if and only if $M$ is locally congruent to either a tube of radius $\frac{\pi }{4}$ over a complex submanifold of $\mathbb {C}P^m$ or to a type $(A)$ real hypersurface with radius $r \ne \frac{\pi }{4}$.

As a direct consequence of these Theorems, we have

Corollary

There do not exist real hypersurfaces $M$ in $\mathbb {C}P^m$, $m\ge 3$, such that for a non-null constant $k$, $F_X^{(k)}R_{\xi }=R_{\xi }F_X^{(k)}$ for any $X$ tangent to $M$.

2 Preliminaries

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

For any vector field $X$ tangent to $M$, we write $JX=\phi X+\eta (X)N$, and $-JN=\xi$. Then, $(\phi ,\xi ,\eta ,g)$ is an almost contact metric structure on $M$, see [1], that is, we have

$$\begin{aligned} \phi ^2X=-X+\eta (X)\xi , \quad \eta (\xi )=1, \quad g(\phi X,\phi Y)=g(X,Y)-\eta (X)\eta (Y) \end{aligned}$$

(2.1)

for any tangent vectors $X,Y$ to $M$. From (2.1), we obtain

$$\begin{aligned} \phi \xi =0, \quad \eta (X)=g(X,\xi ). \end{aligned}$$

(2.2)

From the parallelism of $J$, we get

$$\begin{aligned} (\nabla _X\phi ) Y =\eta (Y)AX-g(AX,Y)\xi \end{aligned}$$

(2.3)

and

$$\begin{aligned} \nabla _X\xi = \phi AX \end{aligned}$$

(2.4)

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 equations of Gauss and Codazzi are given, respectively, by

$$\begin{aligned} R(X,Y)Z&= \displaystyle g(Y,Z)X - g(X,Z)Y + g(\phi Y,Z)\phi X - g(\phi X,Z) \phi Y \nonumber \\&- 2g (\phi X,Y) \phi Z + g(AY,Z)AX - g(AX,Z)AY, \end{aligned}$$

(2.5)

and

$$\begin{aligned} (\nabla _XA)Y- (\nabla _YA) X = \eta (X)\phi Y-\eta (Y)\phi X-2g(\phi X,Y)\xi \end{aligned}$$

(2.6)

for any tangent vectors $X,Y,Z$ to $M$, where $R$ is the curvature tensor of $M$. We will call the maximal holomorphic distribution $\mathbb {D}$ on $M$ to the following one: at any $p \in M$, $\mathbb {D}(p)=\{ X\in T_pM \vert g(X,\xi )=0\}$. We will say that $M$ is Hopf if $\xi$ is principal, that is, $A\xi =\alpha \xi$ for a certain function $\alpha$ on $M$.

From the above formulas, we have that the structure Jacobi operator on $M$ is given by

$$\begin{aligned} R_{\xi }(X)=X-\eta (X)\xi +g(A\xi ,\xi )AX-g(AX,\xi )A\xi \end{aligned}$$

(2.7)

for any $X$ tangent to $M$

In the sequel, we need the following results:

Theorem 2.1

[10] Let $M$ be a real hypersurface of $\mathbb {C}P^m$, $m\ge 2$. Then, the following are equivalent:

• 1. $M$ is locally congruent to either a geodesic hypersphere or a tube of radius $r$, $0 < r < \frac{\pi }{2}$ over a totally geodesic $\mathbb {C}P^n$, $0 < n < m-1$.

• 2. $\phi A=A\phi$.

Theorem 2.2

[9] If $\xi$ is a principal curvature vector with corresponding principal curvature $\alpha$ and $X \in \mathbb {D}$ is principal with principal curvature $\lambda$, then $\phi X$ is principal with principal curvature $\frac{\alpha \lambda +2}{2\lambda -\alpha }$.

3 Proof of Theorem 1

If we suppose that $F_X^{(k)}R_{\xi }=R_{\xi }F_X^{(k)}$ for any $X \in \mathbb {D}$, we get

$$\begin{aligned}&g(Y,\phi AX)\xi +\eta (A\xi )g(\phi AX,AY)\xi -\eta (AY)g(\phi AX,A\xi )\xi \nonumber \\&\quad +\eta (Y)\phi AX+\eta (Y)\eta (A\xi )A\phi AX-\eta (Y)\eta (A\phi AX)A\xi =0 \end{aligned}$$

(3.1)

for any $X \in \mathbb {D}$, $Y \in TM$. Let us suppose that $M$ is non-Hopf. Thus, locally we can write $A\xi =\alpha \xi +\beta U$, where $U$ is a unit vector field in $\mathbb {D}$, $\alpha$ and $\beta$ are functions on $M$ and $\beta \ne 0$. We also call ${\mathbb {D}}_U$ to the orthogonal complementary distribution in $\mathbb {D}$ to the one spanned by ${ U, \phi U }$.

If we take $X=Y=\phi U$ in (3.1), we get

$$\begin{aligned} g(AU,\phi U)=0. \end{aligned}$$

(3.2)

And taking $Y=\xi$ in (3.1), we obtain

$$\begin{aligned} \phi AX+\alpha A\phi AX-\alpha \beta g(\phi AX,U)\xi -\beta ^2g(\phi AX,U)U=0 \end{aligned}$$

(3.3)

for any $X \in \mathbb {D}$. In particular, from (3.2) and (3.3), we have

$$\begin{aligned} \phi AU+\alpha A\phi AU=0. \end{aligned}$$

(3.4)

The scalar product of (3.3) and $U$ yields

$$\begin{aligned} (\beta ^2-1)g(A\phi U,X)-\alpha g(A\phi AU,X)=0 \end{aligned}$$

(3.5)

for any $X \in \mathbb {D}$. Thus, $(\beta ^2-1)A\phi U-\alpha A\phi AU$ has not a component in $\mathbb {D}$, and taking its scalar product with $\xi$, it follows

$$\begin{aligned} (\beta ^2-1) A \phi U-\alpha A\phi AU=0. \end{aligned}$$

(3.6)

From (3.4) and (3.6), we get

$$\begin{aligned} \phi AU=(1-\beta ^2)A\phi U. \end{aligned}$$

(3.7)

Therefore, we can write $A\phi U=\delta \phi U+\omega Z_1$, where $Z_1 \in {\mathbb {D}}_U$ is a unit vector field. The scalar product of (3.3) and $Y \in {\mathbb {D}}_U$ yields $A\phi Y+\alpha A\phi Y$ has not component in $\mathbb {D}$. Then,

$$\begin{aligned} A\phi Y+\alpha A\phi AY=-\alpha \beta g(A\phi U,Y)\xi \end{aligned}$$

(3.8)

for any $Y \in {\mathbb {D}}_U$. Taking $Y=\phi Z_1$, we obtain $-AZ_1+\alpha A\phi A\phi Z_1=0$. Its scalar product with $\xi$ gives

$$\begin{aligned} \alpha \beta \omega (\beta ^2-1)=0. \end{aligned}$$

(3.9)

As $\beta \ne 0$, the following cases appear

Case 1. $\alpha =0$.

Case 2. $\beta ^2=1$. In this case, from (3.7), $AU=\beta \xi$.

Case 3. $\omega =0$, thus ${\mathbb {D}}_U$ is $A$-invariant.

Case 1. $\alpha =0$. From (3.4) $\phi AU=0$, that is, $AU=\beta \xi$ and $A\xi =\beta U$ and from (3.6) $(\beta ^2-1)A\phi U=0$. So we have the following subcases

Subcase 1.1. Let us suppose that $\beta ^2 \ne 1$. Then, $A\phi U=0$. Moreover, from (3.8) for any $Y \in {\mathbb {D}}_U$ $A\phi Y=0$. That means that $M$ is a minimal ruled hypersurface.

Subcase 1.2. $\alpha =0$, $\beta ^2 =1$. We can suppose $\beta =1$, maybe after changing $\xi$ by $-\xi$. As above, $A\phi Y=0$ for any $Y \in {\mathbb {D}}_U$, $AU=\xi$, $A\xi =U$. Then, $AZ_1=0$ and $\omega =g(A\phi U,Z_1)=0$. Thus, $A\phi U=\delta \phi U$.

By the Codazzi equation $g((\nabla _{\xi }A)U-(\nabla _UA)\xi ,\phi U)=1$ yields

$$\begin{aligned} \delta g(\nabla _{\xi }U,\phi U)+g(\nabla _UU,\phi U)=0. \end{aligned}$$

(3.10)

From $g((\nabla _{\xi }A)\phi U-(\nabla _{\phi U}A)\xi ,\xi )=0$, we obtain

$$\begin{aligned} g(\nabla _{\xi } \phi U,U)=-3\delta . \end{aligned}$$

(3.11)

From (3.10) and (3.11), we have

$$\begin{aligned} g(\nabla _UU,\phi U)=-3\delta ^2. \end{aligned}$$

(3.12)

As $g((\nabla _UA)\phi U-(\nabla _{\phi U}A)U,\xi )=-2$, it follows

$$\begin{aligned} g(\nabla _UU,\phi U)=-2 \end{aligned}$$

(3.13)

and from $g((\nabla _UA)\phi U-(\nabla _{\phi U}A)U,U)=0$, we get

$$\begin{aligned} \delta g(\nabla _U\phi U,U)+2 \delta =0. \end{aligned}$$

(3.14)

From (3.13) and (3.14), we have $\delta =0$. Therefore, $M$ is still a minimal ruled real hypersurface.

Case 2. $\beta ^2=1$. As above, we suppose $\beta =1$. As the case $\alpha =0$ has been studied, we suppose $\alpha \ne 0$. Then, from (3.6), $A\phi AU=0$, and from (3.4), $\phi AU=0$. Therefore, $A\xi =\alpha \xi +U$, $AU=\xi$. Moreover, we know that $-AZ_1+\alpha A\phi AZ_1=0$. Taking its scalar product with $\phi U$, we get $\omega + \alpha \omega g(A\phi Z_1,\phi Z_1)=0$. Supposing $\omega \ne 0$, we have $g(A\phi Z_1,\phi Z_1)=-\frac{1}{\alpha }$.

Taking $X=Y\in {\mathbb {D}}_U$ in (3.1), we obtain $\phi AY+\alpha A\phi AY+\omega g(Y,Z_1)A\xi =0$. Its scalar product with $\phi U$ gives $\alpha g(A\phi AY,\phi U)=0=-\alpha \omega g(Y,A\phi Z_1)$. As $\alpha \omega \ne 0$, $g(Y,A\phi Z_1)=0$ for any $Y \in {\mathbb {D}}_U$. This yields $A\phi Z_1=0$ and $0=-\frac{1}{\alpha }$. This is a contradiction, and we have $\omega =0$, $A\xi =\alpha \xi + U$, $AU=\xi$ and $A\phi U=\delta \phi U$.

This yields ${\mathbb {D}}_U$ is $A$-invariant and $\phi$-invariant, and we arrive to Case 3. As also $\phi AU =(1-\beta ^2)A\phi U$, we have two possible subcases:

Subcase 3.1. $\beta ^2 =1$. In this case, $AU=\beta \xi$.

Subcase 3.2. $\beta ^2 \ne 1$ and $AU=\beta \xi + \sigma U$, where $\sigma =(1-\beta ^2)\delta$.

If we take $Y=\phi X \in {\mathbb {D}}_U$ in (3.1) for $X \in {\mathbb {D}}_U$ such that $AX=\lambda X$, we have $\lambda +\alpha \lambda g(\phi X,A\phi X)=0$. This yields that either any eigenvalue in ${\mathbb {D}}_U$ is $0$ or that if there exists a non-null eigenvalue $\lambda$ in ${\mathbb {D}}_U$, $\alpha \ne 0$ and $\lambda =-\frac{1}{\alpha }$. In this case, the eigenspace corresponding to this eigenvalue is $\phi$-invariant.

Let us suppose that for any $Y \in {\mathbb {D}}_U$ $AY=0$. As $g((\nabla _YA)\phi Y-(\nabla _{\phi Y}A)Y,\xi )=-2$, we obtain

$$\begin{aligned} g([ \phi Y,Y ],U) = -\frac{2}{\beta }. \end{aligned}$$

(3.15)

And as $g((\nabla _YA)\phi Y-(\nabla _{\phi Y}A)Y,U)=0$, it follows

$$\begin{aligned} \sigma g([ \phi Y,Y ],U)=0. \end{aligned}$$

(3.16)

From (3.15) and (3.16), we get $\sigma =0$. If also $\beta ^2 \ne 1$, $\delta =0$ and our real hypersurface should be ruled.

Let us then suppose that $\beta ^2=1$. As above, we will take $\beta =1$ and $\sigma =0$. If we develop $g((\nabla _YA)\phi U-(\nabla _{\phi U}A)Y,\xi )=0$, we get

$$\begin{aligned} g(\nabla _Y\phi U,U)=g(\nabla _{\phi U}Y,U) \end{aligned}$$

(3.17)

and from $g((\nabla _YA)\phi U-(\nabla _{\phi U}A)Y,U)=0$, it follows

$$\begin{aligned} \delta g(\nabla _Y\phi U,U)=0. \end{aligned}$$

(3.18)

From (3.17) and (3.18), suppose $g(\nabla _Y\phi U,U)=g(\nabla _{\phi U}Y,U)=0$. As $g((\nabla _{\phi U}A)U-(\nabla _UA)\phi U,\xi )=2$, we obtain

$$\begin{aligned} g(\nabla _U\phi U,U)=2+\alpha \delta \end{aligned}$$

(3.19)

and from $g((\nabla _{\phi U}A)U-(\nabla _UA)\phi U,U)=0$, we have

$$\begin{aligned} 2\delta +\delta g(\nabla _U\phi U,U)=0. \end{aligned}$$

(3.20)

If $\delta \ne 0$, from (3.20) $g(\nabla _U\phi U,U)=-2$ and from (3.19)

$$\begin{aligned} \alpha \delta =-4. \end{aligned}$$

(3.21)

Now, $g((\nabla _{\phi U}A)\xi -(\nabla _{\xi }A)\phi U,U)=1$ gives $\delta g(\nabla _{\xi }\phi U,U)=2-\alpha \delta$. From (3.21)

$$\begin{aligned} \delta g(\nabla _{\xi }\phi U,U)=6. \end{aligned}$$

(3.22)

But from $g((\nabla _{\xi }A)U-(\nabla _UA)\xi ,\phi U)=1$, we obtain

$$\begin{aligned} -\delta g(\nabla _{\xi }U,\phi U)=0. \end{aligned}$$

(3.23)

From (3.22) and (3.23), we arrive to a contradiction. Thus, $\delta =0$ and $M$ is also a ruled real hypersurface.

Therefore, we have only to study the following case: $A\xi =\alpha \xi +\beta U$, $AU=\beta \xi +\sigma U$, $A\phi U=\delta \phi U$, ${\mathbb {D}}_U$ is $A$-invariant, and there exists $Z \in {\mathbb {D}}_U$ such that $AZ=-\frac{1}{\alpha } Z$, $A\phi Z=-\frac{1}{\alpha } \phi Z$. As $(1-\beta ^2)A\phi U=\phi AU$, two subcases appear

Subcase 1. $\beta ^2 =1$, and then $\sigma =0$.

Subcase 2. $\beta ^2 \ne 1$, $\sigma =(1-\beta ^2)\delta$.

From $g((\nabla _ZA)\phi Z-(\nabla _{\phi Z}A)Z,\xi )=-2$, we obtain

$$\begin{aligned} \beta g([ \phi Z,Z ],U)=\frac{2}{\alpha ^2} \end{aligned}$$

(3.24)

and from $g((\nabla _ZA)\phi Z-(\nabla _{\phi Z}A)Z,U)=0$, we get

$$\begin{aligned} \left( \frac{1}{\alpha }+\sigma \right) g([ \phi Z,Z ],U)=\frac{2\beta }{\alpha }. \end{aligned}$$

(3.25)

From (3.24) and (3.25), we obtain

$$\begin{aligned} 1+\alpha \sigma =\alpha ^2\beta ^2. \end{aligned}$$

(3.26)

In Subcase 1, as $\beta ^2=1$ and $\sigma =0$, we should obtain $\alpha ^2=1$. Changing, if necessary, $\xi$ by $-\xi$, we can take $\alpha =1$. This case cannot occur by Proposition 3.2, page 1607 in [11]. Therefore, we have $\beta ^2 \ne 1$ and from (3.26) $1+\alpha \delta (1-\beta ^2)=\alpha ^2\beta ^2$. Thus,

$$\begin{aligned} \delta =\frac{\alpha ^2\beta ^2-1}{\alpha (1-\beta ^2)}. \end{aligned}$$

(3.27)

Now, $g((\nabla _ZA)\phi Z-(\nabla _{\phi Z}A)Z,\phi U)=0$ yields

$$\begin{aligned} \left( \frac{1}{\alpha } +\delta \right) g([ Z,\phi Z ],\phi U)=0. \end{aligned}$$

(3.28)

Let us suppose that $\delta =-\frac{1}{\alpha }$. Then, $\sigma =\frac{\beta ^2-1}{\alpha }$ and from (3.26) $\alpha ^2=1$. As above, we suppose $\alpha =1$. Thus, $A\xi =\xi +\beta U$, $AU=\beta \xi +(\beta ^2-1)U$, $A\phi U=-\phi U$, and there exists a unit $Z \in {\mathbb {D}}_U$ such that $AZ=-Z$, $A\phi Z=-\phi Z$.

Suppose that there exists a unit $W \in {\mathbb {D}}_U$ such that $AW=A\phi W=0$. From $g((\nabla _WA)\xi -(\nabla _{\xi }A)W,\xi )=0$, we obtain $g(\nabla _{\xi }W,U)=0$, and from $g((\nabla _WA)\xi -(\nabla _{\xi }A)W,U)=0$, we get $W(\beta ) + (\beta ^2-1)g(\nabla _{\xi }W,U)=0$. Thus, $W(\beta )=0$. This fact and the proof of Proposition 3.3, page 1608 in [11], yield $grad(\beta )=-(2\beta ^2+1)\phi U$. The same proof yields this case cannot occur. Therefore, $\delta \ne -\frac{1}{\alpha }$ and $g([ Z,\phi Z ],\phi U)=0$.

Then, from $g((\nabla _ZA)\phi Z-(\nabla _{\phi Z}A)Z,Z)=g((\nabla _ZA)\phi Z-(\nabla _{\phi Z}A)Z,\phi Z)=0$, we get

$$\begin{aligned} Z(\alpha )=(\phi Z)(\alpha )=0. \end{aligned}$$

(3.29)

From $g((\nabla _ZA)\xi -(\nabla _{\xi }A)Z,\xi )=0$, it follows

$$\begin{aligned} Z(\alpha )+\beta g(\nabla _{\xi }Z,U)=0. \end{aligned}$$

(3.30)

From (3.29) and (3.30), we obtain

$$\begin{aligned} g(\nabla _{\xi }Z,U)=0. \end{aligned}$$

(3.31)

As $g((\nabla _ZA)\xi -(\nabla _{\xi }A)Z,U)=0$, we have, bearing in mind (3.31),

$$\begin{aligned} Z(\beta )=0. \end{aligned}$$

(3.32)

From $g((\nabla _{\xi }A)U-(\nabla _UA)\xi ,\xi )=0$, we get

$$\begin{aligned} \xi (\beta )=U(\alpha ) \end{aligned}$$

(3.33)

and as $g((\nabla _{\xi }A)U- (\nabla _UA)\xi ,U)=0$, it follows

$$\begin{aligned} \xi (\sigma )=U(\beta ). \end{aligned}$$

(3.34)

Now, $g((\nabla _ZA)U-(\nabla _UA)Z,\xi )=0$ yields

$$\begin{aligned} Z(\beta )+\sigma g(\nabla _UZ,U)=0 \end{aligned}$$

(3.35)

and from (3.26) and $g((\nabla _ZA)U-(\nabla _UA)Z,U)=0$, we obtain

$$\begin{aligned} Z(\sigma )+\alpha \beta ^2g(\nabla _UZ,U)=0. \end{aligned}$$

(3.36)

From (3.32) and (3.36), we have $\sigma g(\nabla _UZ,U)=0$. This and (3.36) yield $Z(\sigma )+\frac{1}{\alpha } g(\nabla _UZ,U)=0$. As $Z(\alpha )=Z(\beta )=0$, from (3.26) $Z(\sigma )=0$. Therefore,

$$\begin{aligned} g(\nabla _UZ,U)=0. \end{aligned}$$

(3.37)

As $g((\nabla _ZA)U-(\nabla _UA)Z,\phi U)=0$, this gives

$$\begin{aligned} (\sigma -\delta )g(\nabla _ZU,\phi U)+\left( \delta +\frac{1}{\alpha }\right) g(\nabla _UZ,\phi U)=0 \end{aligned}$$

(3.38)

and $g((\nabla _ZA)U-(\nabla _UA)Z,Z)=0$ yields

$$\begin{aligned} U\left( \frac{1}{\alpha }\right) =\left( \sigma +\frac{1}{\alpha }\right) g(\nabla _ZZ,U). \end{aligned}$$

(3.39)

From $g((\nabla _ZA)\xi -(\nabla _{\xi }A)Z,Z)=0$, we obtain

$$\begin{aligned} \xi \left( \frac{1}{\alpha }\right) =\beta g(\nabla _ZZ,U) \end{aligned}$$

(3.40)

and from $g((\nabla _ZA)\phi U-(\nabla _ {\phi U}A)Z,U)=0$, we get

$$\begin{aligned} (\delta -\sigma )g(\nabla _Z\phi U,U)+\left( \sigma +\frac{1}{\alpha }\right) g(\nabla _{\phi U}Z,U)=0. \end{aligned}$$

(3.41)

We also have from $g((\nabla _ZA)\phi U)-(\nabla _{\phi U}A)Z,\xi )=0$

$$\begin{aligned} g([ \phi U,Z],U)=0. \end{aligned}$$

(3.42)

Thus, from (3.41) and (3.42), we have a homogeneous system of linear equations where $g(\nabla _Z\phi U,U)$ and $g(\nabla _{\phi U}Z,U)$ are unknown. The determinant of its matrix of coefficients is $\delta + \frac{1}{\alpha }$. As $\delta \ne -\frac{1}{\alpha }$, we obtain

$$\begin{aligned} g(\nabla _Z\phi U,U)= g(\nabla _{\phi U}Z,U)=0. \end{aligned}$$

(3.43)

As $g((\nabla _ZA)\phi U-(\nabla _{\phi U}A)Z,\phi Z)=0$, we have $(\delta +\frac{1}{\alpha })g(\nabla _Z\phi U,\phi Z)=0$. As $\delta \ne -\frac{1}{\alpha }$, $g(\nabla _Z\phi U,\phi Z)=0$. By (2.3), this gives $g(\nabla _ZU,Z)=0$. From (3.39) and (3.40), it follows

$$\begin{aligned} \xi (\alpha )=U(\alpha )=0. \end{aligned}$$

(3.44)

From $g((\nabla _ZA)\phi U-(\nabla _{\phi U}A)Z,Z)=0$, we have $(\delta +\frac{1}{\alpha })g(\nabla _Z\phi U,Z)+(\phi U)(\frac{1}{\alpha })=0$. Now, from (2.3)

$$\begin{aligned} (\phi U)\left( \frac{1}{\alpha }\right) =\left( \frac{1}{\alpha } +\delta \right) g(\nabla _ZU,\phi Z). \end{aligned}$$

(3.45)

Developing $g((\nabla _ZA)U-(\nabla _UA)Z,\phi Z)=0$ and bearing in mind (3.26), we get

$$\begin{aligned} \alpha ^2\beta ^2g(\nabla _ZU,\phi Z)=\beta . \end{aligned}$$

(3.46)

Now, from (3.45) and (3.46), we obtain

$$\begin{aligned} (\phi U) (\alpha )=-\frac{1+\alpha \delta }{\alpha \beta } =\frac{\beta (1-\alpha ^2)}{\alpha (1-\beta ^2)} . \end{aligned}$$

(3.47)

From (3.33) and (3.44), we have

$$\begin{aligned} \xi (\beta )=U(\beta )=0. \end{aligned}$$

(3.48)

The equality $g((\nabla _{\xi }A)U-(\nabla _UA)\xi ,\phi U)=1$ yields

$$\begin{aligned} \beta g(\nabla _UU,\phi U)=\beta ^2+\sigma ^2-\alpha \sigma -1. \end{aligned}$$

(3.49)

From $g((\nabla _UA)\phi U-(\nabla _{\phi U}A)U,\xi )=-2$, we arrive to $-2\delta \sigma +\alpha \sigma +\alpha \delta -\beta g(\nabla _U\phi U,U)-(\phi U)(\beta )=2$. This and (3.49) yield

$$\begin{aligned} (\phi U)(\beta )=-2\delta \sigma +\alpha \delta +\beta ^2+\sigma ^2+1. \end{aligned}$$

(3.50)

Bearing in mind all these facts, we arrive to

$$\begin{aligned} grad(\alpha )&= \rho \phi U \nonumber \\ grad(\beta )&= \theta \phi U \end{aligned}$$

(3.51)

where $\rho =-(\frac{1+\alpha \delta }{\alpha \beta })$ and $\theta =-2\delta \sigma +\alpha \delta +\beta ^2+\sigma ^2+1$. As $g(\nabla _Xgrad(\alpha ),Y)=g(\nabla _Ygrad(\alpha ),X)$ for any $X,Y$ tangent to $M$, we have, taking $X=\xi$, $\xi (\rho )g(\phi U,Y)+\rho g(\nabla _{\xi }\phi U,Y)=-\rho g(U,AY)$. If $Y=\phi U$, this yields $\xi (\rho )=0$. Thus, $\rho g(\nabla _{\xi }\phi U,Y)=-\rho g(U,AY)$, for any $Y$ tangent to $M$. As $\rho \ne 0$, taking $Y=U$, we get

$$\begin{aligned} g(\nabla _{\xi }\phi U,U)=-\sigma . \end{aligned}$$

(3.52)

From $g((\nabla _{\xi }A)\phi U-(\nabla _{\phi U}A)\xi ,\xi )=0$ and bearing in mind (3.52), we have

$$\begin{aligned} (\phi U)(\alpha )=-3\beta \delta +\alpha \beta -\beta \sigma . \end{aligned}$$

(3.53)

From (3.47) and (3.53) $2+\alpha \delta +\alpha \sigma -3\alpha \beta ^2\delta +\alpha \sigma \beta ^2=0$, or equivalently

$$\begin{aligned} \alpha ^2(2\beta ^2-3\beta ^4-\beta ^6)+\beta ^2(2+\beta ^2)-1=0. \end{aligned}$$

(3.54)

If $2-3\beta ^2-\beta ^4=0$, we should have $\beta ^4+2\beta ^2-1=0$. Both equalities yield $\beta ^2=1$, that is impossible. From (3.54), we have

$$\begin{aligned} \alpha ^2=\frac{1-\beta ^2(2+\beta ^2)}{2\beta ^2-3\beta ^4-\beta ^6}. \end{aligned}$$

(3.55)

If we take the derivative of (3.54) in the direction of $\phi U$ and bear in mind (3.47), (3.48), and (3.54), we find that $\beta$ is a root of a polynomial with constant coefficients. Therefore, $\beta$ is constant. From (3.55), $\alpha$ is also constant, which is impossible.

Thus, we have proved that if $M$ is not Hopf, it is locally congruent to a ruled real hypersurface. It is easy to see that these real hypersurfaces satisfy (3.1).

Let us now suppose that $M$ is a Hopf real hypersurface with $A\xi = \alpha \xi$ and that $M$ satisfies (3.1). Then, we have for any $X \in \mathbb {D}$ that $\phi AX+\alpha A\phi AX=0$. If $\alpha =0$, we get $\phi AX=0$. Thus, $AX=0$ for any $X \in \mathbb {D}$ and $M$ should be totally geodesic, which is impossible.

Suppose now that $\alpha \ne 0$ and take a unit $X \in \mathbb {D}$ such that $AX=\lambda X$. From (3.1), we get either $\lambda =0$ or $A\phi X=-\frac{1}{\alpha } \phi X$. Applying the same reasoning to $\phi X$, we obtain that the eigenspaces in $\mathbb {D}$ are $\phi$-invariant and correspond to the eigenvalues $0$ and $-\frac{1}{\alpha }$. This is impossible by Theorem 2.2, and we finish the proof.

4 Proof of Theorem 2

If we suppose that $R_{\xi } F_{\xi }^{(k)}=F_{\xi }^{(k)}R_{\xi }$, we get

$$\begin{aligned}&g(Y,\phi A\xi )\xi +g(A\xi ,\xi )g(\phi A\xi ,AY)\xi +\eta (Y)\phi A\xi +\eta (Y)\eta (A\xi )A\phi A\xi \nonumber \\&\quad -\,\eta (Y)\eta (A\phi A\xi )A\xi -k\phi R_{\xi }(Y)+kR_{\xi }(\phi Y)=0 \end{aligned}$$

(4.1)

for any $Y\in TM$. Let us suppose that $M$ is non-Hopf. Thus, we write $A\xi =\alpha \xi +\beta U$ for a unit $U \in \mathbb {D}$ and functions $\alpha$ and $\beta$ on $M$, $\beta$ being nonvanishing. From (4.1), we have

$$\begin{aligned}&\beta g(Y,\phi U)\xi +\alpha \beta g(\phi U,AY)\xi +\beta \eta (Y)\phi U\nonumber \\&\quad +\alpha \beta \eta (Y)A\phi U=k\phi R_{\xi }(Y)-kR_{\xi }(\phi Y) \end{aligned}$$

(4.2)

for any $Y\in TM$. Taking $Y=\xi$ in (4.2), we obtain

$$\begin{aligned} \beta \phi U+\alpha \beta A\phi U=0. \end{aligned}$$

(4.3)

As $\beta \ne 0$, this yields

$$\begin{aligned} \alpha&\ne 0,\nonumber \\ A\phi U&= -\frac{1}{\alpha }\phi U. \end{aligned}$$

(4.4)

If now we take $Y=U$ in (4.2), as $k \ne 0$, we get $\phi R_{\xi }(U)=R_{\xi }(\phi U)$. This yields $\alpha \phi AU=(\beta ^2-1)\phi U$, that is, $\phi AU=\frac{\beta ^2-1}{\alpha } \phi U$. By applying $\phi$ to such an equality, it follows

$$\begin{aligned} AU=\beta \xi +\frac{\beta ^2-1}{\alpha } U. \end{aligned}$$

(4.5)

From (4.4) and (4.5), we obtain that ${\mathbb {D}}_U$ is $\phi$-invariant and $A$-invariant. Take a unit $Y \in {\mathbb {D}}_U$ such that $AY=\lambda Y$. Introducing this $Y$ in (4.2), we get $\phi R_{\xi }(Y)=R_{\xi }(\phi Y)$. This yields $\alpha \lambda \phi Y=\alpha A\phi Y$. As $\alpha \ne 0$, $A\phi Y=\lambda \phi Y$. Therefore, the eigenspaces in ${\mathbb {D}}_U$ are $\phi$-invariant.

The Codazzi equation gives $(\nabla _{\xi }A)\phi Y-(\nabla _{\phi Y}A)Y=-2\xi$. Taking its scalar product with $\phi Y$, respectively, with $Y$, we have

$$\begin{aligned} Y(\lambda )=(\phi Y)(\lambda )=0. \end{aligned}$$

(4.6)

Its scalar product with $\xi$ implies

$$\begin{aligned} \beta g([\phi Y,Y ],U)=2\lambda ^2-2\alpha \lambda -2 \end{aligned}$$

(4.7)

and its scalar product with $U$ gives

$$\begin{aligned} \left( \lambda -\frac{\beta ^2-1}{\alpha }\right) g([\phi Y,Y ],U)=2\beta \lambda . \end{aligned}$$

(4.8)

From (4.7) and (4.8), we get

$$\begin{aligned} (\alpha \lambda +1) (\lambda ^2-\alpha \lambda -1)=\beta ^2(\lambda ^2-1). \end{aligned}$$

(4.9)

From $g((\nabla _{\phi U}A)Y-(\nabla _YA)\phi U,\phi Y)=0$, we have $(\lambda +\frac{1}{\alpha })g(\nabla _Y\phi U,\phi Y)=0$. Then, either $g(\nabla _Y\phi U,\phi Y)=g(\nabla _YU,Y)=0$, where we have applied (2.3) or $\lambda =-\frac{1}{\alpha }$. In this second case from (4.9), we have $0=\beta ^2(\lambda ^2-1)$. As $\beta \ne 0$, this yields $\lambda ^2=1$ and $\alpha ^2=1$. Changing, if necessary, $\xi$ by $-\xi$, we can suppose $\alpha =1$ and then $\lambda =-1$.

The scalar product of $(\nabla _{\xi }A)Y-(\nabla _YA)\xi =\phi Y$ and $Y$ gives

$$\begin{aligned} \xi (\lambda )-\beta g(\nabla _YU,Y)=0. \end{aligned}$$

(4.10)

As either $\lambda =-1$ or $g(\nabla _YU,Y)=0$, we always have

$$\begin{aligned} \xi (\lambda )=0. \end{aligned}$$

(4.11)

Developing $(\nabla _UA)\phi U-(\nabla _{\phi U}A)U=-2\xi$ and taking its scalar product with $\phi U$, we get

$$\begin{aligned} \alpha U\left( \frac{1}{\alpha }\right) =\beta ^2g(\nabla _{\phi U}\phi U,U) \end{aligned}$$

(4.12)

The same procedure applied to $g((\nabla _{\xi }A)\phi U-(\nabla _{\phi U}A)\xi ,\phi U)=0$ yields

$$\begin{aligned} \xi \left( \frac{1}{\alpha }\right) =\beta g(\nabla _{\phi U}\phi U,U). \end{aligned}$$

(4.13)

From (4.12) and (4.13), we obtain

$$\begin{aligned} \alpha U(\alpha )=\beta \xi (\alpha ). \end{aligned}$$

(4.14)

From $g((\nabla _{\xi }A)U-(\nabla _UA)\xi ,\xi )=0$, $\xi (\beta )=U(\alpha )$ and from (4.14), we have

$$\begin{aligned} \beta \xi (\alpha ) = \alpha \xi (\beta ) . \end{aligned}$$

(4.15)

By derivating (4.9) in the direction of $\xi$ and bearing in mind (4.11) and (4.15), it follows

$$\begin{aligned} (\lambda (\lambda ^2-\alpha \lambda -1)-\lambda (\alpha \lambda +1))\xi (\alpha )=\frac{2\beta ^2}{\alpha }(\lambda ^2-1)\xi (\alpha ) . \end{aligned}$$

(4.16)

If we suppose $\xi (\alpha ) \ne 0$ and bear in mind (4.9), from (4.16), we have

$$\begin{aligned} \alpha \lambda ^3-2\alpha \lambda +2\lambda ^2-2=0. \end{aligned}$$

(4.17)

Derivating (4.17) in the direction of $\xi$, we obtain $(\lambda ^3-2\lambda )\xi (\alpha )=0$. As we suppose $\xi (\alpha ) \ne 0$, we have $\lambda (\lambda ^2-2)=0$. If $\lambda =0$ from (4.9), it follows $\beta ^2=1$ and $\beta$ should be constant. From (4.15), $\xi (\alpha )=0$, and we arrive to a contradiction. Therefore, $\lambda ^2=2$. From (4.9), we obtain $1-2\alpha ^2=\beta ^2$. By derivating this equality in the direction of $\xi$ and bearing in mind (4.15) and that we suppose $\xi (\alpha ) \ne 0$, we get $-2\alpha ^2=\beta ^2$ and we have a new contradiction. This proves

$$\begin{aligned} \xi (\alpha )=\xi (\beta )=U(\alpha )=0. \end{aligned}$$

(4.18)

The equality $g((\nabla _{\xi }A)Y-(\nabla _YA)\xi ,\xi )=0$ yields

$$\begin{aligned} Y(\alpha )=-\beta g(\nabla _{\xi }Y,U). \end{aligned}$$

(4.19)

Analogously, from $g((\nabla _{\xi }A)Y-g(\nabla _YA)\xi ,U)=0$, we obtain

$$\begin{aligned} Y(\beta )=\left( \lambda -\frac{\beta ^2-1}{\alpha }\right) g (\nabla _{\xi }Y,U). \end{aligned}$$

(4.20)

From (4.19) and (4.20), we get

$$\begin{aligned} \beta Y(\beta )=\left( \frac{\beta ^2-1}{\alpha } -\lambda \right) Y(\alpha ). \end{aligned}$$

(4.21)

As $Y(\lambda )=0$, from (4.9), it follows

$$\begin{aligned} (\lambda (\lambda ^2-\alpha \lambda -1)-\lambda (\alpha \lambda +1)) Y (\alpha )=(\lambda ^2-1)2\beta Y(\beta ) \end{aligned}$$

(4.22)

and from (4.21) and (4.22), if we suppose $Y(\alpha ) \ne 0$, we have

$$\begin{aligned} \alpha (3\lambda ^2-2\alpha \lambda ^2-4\lambda )=2(\alpha \lambda +1)(\lambda ^2-\alpha \lambda -1)-2(\lambda ^2-1). \end{aligned}$$

(4.23)

Derivating once again in the direction of $Y$ and bearing in mind that we suppose $Y(\alpha ) \ne 0$, we obtain $\lambda ^3=0$, that is, $\lambda =0$ and $\beta ^2=1$. Therefore, $\beta$ is constant and $Y(\beta )=0$.

From $g((\nabla _{\xi }A)\phi U-(\nabla _{\phi U}A)\xi ,\xi )=0$, we have

$$\begin{aligned} (\phi U)(\alpha )=\frac{3\beta }{\alpha } +\alpha \beta +\beta g(\nabla _{\xi }U, \phi U). \end{aligned}$$

(4.24)

And $\beta ^2=1$ and $g((\nabla _{\xi }A)\phi U-(\nabla _{\phi U}A)\xi ,U)=-1$ yield

$$\begin{aligned} g(\nabla _{\xi }U, \phi U)=-\alpha . \end{aligned}$$

(4.25)

From (4.24) and (4.25), we conclude

$$\begin{aligned} (\phi U) (\alpha )=\frac{3\beta }{\alpha }. \end{aligned}$$

(4.26)

As $g((\nabla _{\xi }A)U-(\nabla _UA)\xi ,\phi U)=1$, we get

$$\begin{aligned} \frac{1}{\alpha }g(\nabla _{\xi }U,\phi U)-\beta g(\nabla _UU,\phi U)=0 \end{aligned}$$

(4.27)

and from the Codazzi equation $g((\nabla _UA)\phi U-(\nabla _{\phi U}A)U,U) = 0$, it follows

$$\begin{aligned} g(\nabla _UU,\phi U)=2\beta . \end{aligned}$$

(4.28)

From (4.27) and (4.28), bearing in mind that $\beta ^2=1$, we have

$$\begin{aligned} g(\nabla _{\xi }U,\phi U)=2\alpha . \end{aligned}$$

(4.29)

Now, from (4.27) and (4.29), $\alpha$ should vanish. As this is a contradiction, we arrive to

$$\begin{aligned} Y(\alpha )=Y(\beta )=0 \end{aligned}$$

(4.30)

By linearity, we have $X(\alpha )=X(\beta )=0$ for any $X \in {\mathbb {D}}_U$.

The Codazzi equation $g((\nabla _{\xi }A)\phi U-(\nabla _{\phi U}A)\xi ,U)=-1$ yields

$$\begin{aligned} (\phi U)(\beta )= \frac{\beta ^2-1}{\alpha ^2} +\beta ^2+\frac{\beta ^2}{\alpha }g(\nabla _{\xi }U,\phi U). \end{aligned}$$

(4.31)

As $g((\nabla _{\xi }A)U-(\nabla _UA)\xi ,\phi U)=1$, we have

$$\begin{aligned} \frac{\beta ^2}{\alpha } g(\nabla _{\xi }U,\phi U)-\beta g(\nabla _UU,\phi U)=\frac{\beta ^2-1}{\alpha ^2} \end{aligned}$$

(4.32)

and $g((\nabla _UA)\phi U-(\nabla _{\phi U}A)U,U)=0$ shows

$$\begin{aligned} \beta g(\nabla _UU,\phi U)+\beta ^2-3-2(\phi U)(\beta )-\frac{\beta ^2-1}{\alpha \beta } (\phi U)(\alpha )=0. \end{aligned}$$

(4.33)

From (4.31), (4.32), and (4.33), we have

$$\begin{aligned} \beta g(\nabla _UU\phi U)-\frac{\beta ^2-1}{\alpha } g(\nabla _{\xi }U,\phi U)+\frac{\beta ^2-1}{\alpha ^2} -4=0. \end{aligned}$$

(4.34)

Now, from (4.33) and (4.34), it follows $g(\nabla _{\xi }U,\phi U)=-4\alpha$ and $g(\nabla _UU,\phi U)=\frac{1-\beta ^2}{\alpha ^2\beta }-4\beta$. Then, from (4.24) and (4.32), we get

$$\begin{aligned} (\phi U) (\alpha )=3\beta \left( \frac{1-\alpha ^2}{\alpha }\right) \end{aligned}$$

(4.35)

and

$$\begin{aligned} (\phi U) (\beta )=-3\beta ^2+\frac{\beta ^2-1}{\alpha ^2}. \end{aligned}$$

(4.36)

From all the facts we have until now, we obtain $grad(\alpha ) = \omega \phi U$, where $\omega =3\beta (\frac{1-\alpha ^2}{\alpha })$. As $g(\nabla _X grad(\alpha ),Y)=g(\nabla _Ygrad(\alpha ),X)$ for any $X,Y \in TM$, we get $X(\omega )g(\phi U,Y)-Y(\omega )g(\phi U,X)+\omega (g(\nabla _X\phi U,Y)-g(\nabla _Y\phi U,X))=0$. Taking $Y=\xi$, this yields $\omega (g(\nabla _X\phi U,\xi )-g (\nabla _{\xi }\phi U,X))=0$ for any $X \in TM$. Thus, either $\omega =0$ or $g(\nabla _X\phi U,\xi )=g(\nabla _{\xi }\phi U,X)$ for any $X \in TM$. If we take $X=U$, we have $-g(U,AU)=g(\nabla _{\xi }\phi U,U)$. Then, $4\alpha ^2+\beta ^2=1$. This yields $4\alpha (\phi U)(\alpha )+\beta (\phi U)(\beta )=0$. From (4.35) and (4.36), we have $9\alpha ^2+\beta ^2=1$. Therefore, $\alpha =0$, which is impossible. So we have $\omega =0$ and $\alpha ^2=1$. From (4.35) $(\phi U)(\alpha )=0$ and from (4.36) $(\phi U)(\beta )=-(2\beta ^2+1)$. Then, from (4.18) and the fact that $g((\nabla _{\xi }A)U-(\nabla _UA)\xi , U)=0$, we have $U(\beta )=0$ and then $grad(\beta )=-(2\beta ^2+1)\phi U$.

Applying the same reasoning to $grad(\beta )$, $-(1+2\beta ^2)(g(\nabla _X\phi U,\xi )-g(\nabla _{\xi }\phi U,X))=0$ for any $X \in TM$. This yields $g(\nabla _X\phi U,\xi )=g(\nabla _{\xi }\phi U,X)$ for any $X \in TM$. Taking $X=U$, it follows $4\alpha ^2+\beta ^2=1$ and being $\alpha ^2=1$, $\beta ^2=-3$, which is impossible and proves that $M$ must be Hopf.

If $M$ is Hopf with $A\xi =\alpha \xi$, from (4.1), we get $\phi R_{\xi }=R_{\xi }\phi$. Let $Y\in \mathbb {D}$ a unit vector field such that $AY=\lambda Y$. Therefore, $\alpha \lambda Y=\alpha A\phi Y$. Then, either $\alpha =0$ and $M$ is locally congruent to a tube of radius $\frac{\pi }{4}$ around a complex submanifold of $\mathbb {C}P^m$, see [2], or $A\phi =\phi A$ and from Theorem 2.1, $M$ is locally congruent to a type $(A)$ real hypersurface.

It is very easy to see that these real hypersurfaces satisfy (4.1), and we finish the proof. $\square$