1 Introduction

We will denote by \({{\mathbb C}}P^m\), \(m\ge 2\), the complex projective space equipped with the Kählerian structure (J, g), J being the complex structure and g the Fubini-Study metric with constant holomorphic sectional curvature 4. Take a connected real hypersurface without boundary M in \({{\mathbb C}}P^m\) whose local normal unit vector field is N. Take \(\xi = -JN\). Then \(\xi \) is a tangent vector field to M that we call the Reeb vector field (or the structure vector field) on M. For any tangent vector field X on M, we write \(JX=\phi X + \eta (X)N\), where \(\phi X\) is the tangent component of JX and \(\eta (X)=g(X,\xi )\). Then \((\phi ,\xi ,\eta ,g)\) defines on M an almost contact metric structure [1], where g is the induced metric on M.

Takagi, see Refs. [5, 8,9,10], classified homogeneous real hypersurfaces of \({{\mathbb C}}P^m\) into six types. All of them are Hopf, that is, their structure vector fields are principal (\(A\xi = \alpha \xi \), for a function \(\alpha \) on M). Denote by \({{\mathbb D}}\) the maximal holomorphic distribution on M: at any point \(p\in M\), \({{\mathbb D}}_p=\{X\in T_pM \vert g(X,\xi _p)=0\}\). Kimura [5] proved that any Hopf real hypersurface M in \({{\mathbb C}}P^m\) whose principal curvatures are constant belongs to Takagi’s list.

The unique real hypersurfaces in \({{\mathbb C}}P^m\) with two distinct principal curvatures are geodesic hyperspheres of radius r, \(0< r < \frac{\pi }{2}\), see Ref. [2]. Their principal curvatures are 2cot(2r) with eigenspace \({{\mathbb R}}[\xi ]\) and cot(r) with eigenspace \({{\mathbb D}}\).

The canonical affine connection on a non-degenerate, pseudo-Hermitian CR-manifold was defined, independently, by Tanaka [11], and Webster [13], and it is known as the Tanaka–Webster connection. For contact metric manifolds, Tanno [12] introduced a generalized Tanaka–Webster connection.

For a real hypersurface M of \({{\mathbb C}}P^m\) and any nonnull real number k, Cho, see [3, 4], generalized Tanno’s definition to the concept of kth generalized Tanaka–Webster connection by

$$\begin{aligned} {\hat{\nabla }}^{(k)}_XY=\nabla _XY+g(\phi AX,Y)\xi -\eta (Y)\phi AX-k\eta (X)\phi Y \end{aligned}$$
(1.1)

for any X, Y tangent to M. Then the four elements of the almost contact metric structure on M are parallel for this connection and if the shape operator of the real hypersurface satisfies \(\phi A+A \phi =2k\phi \), the real hypersurface is contact and the kth generalized Tanaka–Webster connection coincides with the Tanaka–Webster connection.

We define the kth Cho operator on M associated with the tangent vector field X by \(F^{(k)}_{X}Y=g(\phi AX,Y)\xi -\eta (Y)\phi AX-k\eta (X)\phi Y\), for any Y tangent to M. The torsion of the connection \({\hat{\nabla }}^{(k)}\) is given by \(T^{(k)}(X,Y)=F_X^{(k)}Y-F_Y^{(k)}X\) for any X, Y tangent to M. We also define the kth torsion operator associated with the tangent vector field X by \(T_X^{(k)}Y=T^{(k)}(X,Y)\) for any Y tangent to M.

Let \({\mathcal {L}}\) denote the Lie derivative on M. Then \({{\mathcal {L}}}_XY=\nabla _XY-\nabla _YX\) for any X, Y tangent to M. On M we can also define a differential operator of first order associated with the kth generalized Tanaka–Webster connection by \({{\mathcal {L}}}_X^{(k)}Y={\hat{\nabla }}_X^{(k)}Y-{\hat{\nabla }}_Y^{(k)}X={{\mathcal {L}}}_XY+T_X^{(k)}Y\), for any X, Y tangent to M.

Let now B be a symmetric tensor of type (1,1) defined on M. We can associate with B a tensor field of type (1,2) \(B^{(k)}_T\) by \(B^{(k)}_T(X,Y)=[T_X^{(k)},B ]Y=T^{(k)}_XBY-BT^{(k)}_XY\), for any X, Y tangent to M.

Consider the condition \({{\mathcal {L}}}^{(k)}B={{\mathcal {L}}}B\) for some nonnull real number k. This means that for any X, Y tangent to M \(({{\mathcal {L}}}_X^{(k)}B)Y=({{\mathcal {L}}}_XB)Y\). This is equivalent to having \(B_T^{(k)}=0\).

Generalizing this we can consider that the tensor \(B_T^{(k)}\) is symmetric, that is, \(B_T^{(k)}(X,\) \(Y)=B_T^{(k)}(Y, X)\) for any X, Y tangent to M. This is equivalent to have the following Codazzi-type condition

$$\begin{aligned} \left( \left( {{\mathcal {L}}}_X^{(k)}-{{\mathcal {L}}}_X\right) B\right) Y=\left( \left( {{\mathcal {L}}}_Y^{(k)}-{{\mathcal {L}}}_Y\right) B\right) X \end{aligned}$$
(1.2)

for any X, Y tangent to M.

On the other hand, we can suppose that \(B_T^{(k)}\) is skew symmetric, that is, \(B_T^{(k)}(X,Y)=-B_T^{(k)}(Y,X)\), for any X, Y tangent to M. This is equivalent to the following Killing-type condition:

$$\begin{aligned} \left( \left( {{\mathcal {L}}}_X^{(k)}-{{\mathcal {L}}}_X\right) B\right) Y+\left( \left( {{\mathcal {L}}}_Y^{(k)}-{{\mathcal {L}}}_Y\right) B\right) X=0 \end{aligned}$$
(1.3)

for any X, Y tangent to M.

In the particular case of \(B=A\), the shape operator of M, in Ref. [7] the first author proved non-existence of real hypersurfaces in \({{\mathbb C}}P^m\), \(m \ge 3\), satisfying \({{\mathcal {L}}}^{(k)}A={{\mathcal {L}}}A\), that is, \(A^{(k)}_T=0\), for any nonnull real number k.

The purpose of the present paper is to study real hypersurfaces M in \({{\mathbb C}}P^m\) such that the shape operator satisfies either (1.2) or (1.3). In fact, we will obtain the following.

Theorem 1

There does not exist any real hypersurface M in \({{\mathbb C}}P^m\), \(m \ge 3\), such that, for some nonnull real number k, \(A^{(k)}_T\) is symmetric.

In the case of \(A_T^{(k)}\) being skew symmetric, we have a very different situation given by the

Theorem 2

Let M be a real hypersurface M in \({{\mathbb C}}P^m\), \(m \ge 3\), and k a nonnull real number. Then the tensor field \(A^{(k)}_T\) is skew symmetric if and only if M is locally congruent to a geodesic hypersphere of radius r, \(0< r < \frac{\pi }{2}\), such that \(cot(r)=k\).

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 Kählerian 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 Ref. [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 vectors X, Y tangent 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 equation of Codazzi is given by

$$\begin{aligned} (\nabla _XA)Y-(\nabla _YA)X=\eta (X)\phi Y-\eta (Y)\phi X-2g(\phi X,Y)\xi \end{aligned}$$
(2.5)

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

In the sequel, we need the following result, which consists of a combination of the Lemmas 2.1, 2.2 and 2.4 in Ref. [6].

Theorem 2.1

If \(\xi \) is a principal curvature vector with corresponding principal curvature \(\alpha \), this is locally constant and if \(X \in {{\mathbb D}}\) is principal with principal curvature \(\lambda \), then \(2\lambda -\alpha \ne 0\) and \(\phi X\) is principal with principal curvature \(\frac{\alpha \lambda +2}{2\lambda -\alpha }\).

3 Proof of Theorem 1

If M satisfies (1.2) for \(B=A\), we get \({\mathcal L}_X^{(k)}AY-{{\mathcal {L}}}_XAY-A{{\mathcal {L}}}_X^{(k)}Y+A{{\mathcal {L}}}_XY={\mathcal L}_Y^{(k)}AX-{{\mathcal {L}}}_YAX-A{{\mathcal {L}}}_Y^{(k)}X+A{{\mathcal {L}}}_YX\) for any X, Y tangent to M. Therefore, we have \(F_X^{(k)}AY-F_{AY}^{(k)}X-2AF_X^{(k)}Y+2AF_Y^{(k)}X=F_Y^{(k)}AX-F_{AX}^{(k)}Y\), for any X, Y tangent to M. This yields

$$\begin{aligned}&2g(\phi AX,AY)\xi -\eta (AY)\phi AX-k\eta (X)\phi AY-g(\phi A^2Y,X)\xi \nonumber \\&\qquad +\eta (X)\phi A^2Y + k\eta (AY)\phi X \nonumber \\&\qquad - 2g(\phi AX,Y)A\xi +2\eta (Y)A\phi AX+2k\eta (X)A\phi Y+2g(\phi AY,X)A\xi \nonumber \\&\qquad - 2\eta (X)A\phi AY-2k\eta (Y)A\phi X \nonumber \\&\quad =-\eta (AX)\phi AY-k\eta (Y)\phi AX -g(\phi A^2X,Y)\xi \nonumber \\&\qquad + \eta (Y)\phi A^2X+k\eta (AX)\phi Y\nonumber \\ \end{aligned}$$
(3.1)

for any X, Y tangent to M. If we suppose that \(X,Y \in {{\mathbb D}}\), (3.1) becomes

$$\begin{aligned}&2g(\phi AX,AY)\xi -\eta (AY)\phi AX-g(\phi A^2Y,X)\xi +k\eta (AY)\phi X\nonumber \\&\qquad -2g(\phi AX,Y)A\xi +2g(\phi AY,X)A\xi \nonumber \\&\quad =-\eta (AX)\phi AY-g(\phi A^2X,Y)\xi +k\eta (AX)\phi Y \end{aligned}$$
(3.2)

for any \(X,Y \in {{\mathbb D}}\). If M is Hopf, that is \(A\xi =\alpha \xi \), then (3.2) gives \(2g(\phi AX, AY)\xi - g(\phi A^2Y,X)\xi -2\alpha g(\phi AX,Y)\xi +2\alpha g(\phi AY,X)\xi =-g(\phi A^2X,Y)\xi \) for any \(X,Y \in {{\mathbb D}}\). This yields \(2A\phi AX+A^2\phi X-2\alpha \phi AX-2\alpha A\phi X=-\phi A^2X\) for any \(X \in {{\mathbb D}}\). Let us suppose that \(X \in {{\mathbb D}}\) satisfies \(AX=\lambda X\). From Theorem 2.1, we have \(A\phi X=\mu \phi X\) with \(\mu =\frac{\alpha \lambda +2}{2\lambda -\alpha }\). Therefore, we obtain \(2\lambda \mu +\mu ^2-2\alpha (\lambda +\mu )+\lambda ^2=0\). That is, \((\lambda +\mu )(\lambda + \mu -2\alpha )=0\).

If \(\lambda +\mu =0\), as \(\mu = \frac{\alpha \lambda +2}{2\lambda -\alpha }\), we obtain \(\frac{2\lambda ^2+2}{2\lambda -\alpha } =0\). This yields \(\lambda ^2+1=0\), which is impossible.

If \(\lambda +\mu =2\alpha \), we should have \(\lambda ^2-2\alpha \lambda +\alpha ^2+1=0\). This gives \(\lambda =\alpha \pm \sqrt{-1}\), which is also impossible. Thus, our real hypersurface must be non-Hopf. This means that \(\xi \) is not principal. Therefore, we can write \(A\xi =\alpha \xi +\beta U\) at least on a neighborhood of a point of M, where U is a unit vector field in \({{\mathbb D}}\) and \(\beta \) a nonvanishing function on such a neighborhood. From now on, we will denote \({{{\mathbb D}}}_U=\{ X \in {{{\mathbb D}}}|g(X,U)=g(X,\phi U)=0 \}\) and make the calculations on that neighborhood. Then (3.2) becomes

$$\begin{aligned}&2g(\phi AX,AY)\xi -\beta g(Y,U)\phi AX-g(\phi A^2Y,X)\xi +k\beta g(Y,U)\phi X\nonumber \\&\qquad -2g(\phi AX,Y)A\xi +2g(\phi AY,X)A\xi \nonumber \\&\quad =-\beta g(X,U)\phi AY-g(\phi A^2X,Y)\xi +k\beta g(X,U)\phi Y \end{aligned}$$
(3.3)

for any \(X,Y \in {{\mathbb D}}\). The scalar product of (3.3) and \(\phi U\) yields \(-\beta g(Y,U)g(AX,U)+k\beta g(Y,U)g(X,U)=-\beta g(X,U)g(AY,U)+k\beta g(X,U)g(Y,U)\), for any \(X,Y \in {{\mathbb D}}\). As \(\beta \ne 0\), we obtain

$$\begin{aligned} g(Y,U)g(AX,U)=g(X,U)g(AY,U) \end{aligned}$$
(3.4)

for any \(X,Y \in {{\mathbb D}}\). If in (3.4) we take \(X=U\), \(Y \in {{{\mathbb D}}}_U\) we obtain \(g(AU,Y)=0\) for any \(Y \in {{{\mathbb D}}}_U\), and if we take \(X=U\), \(Y=\phi U\) we get \(g(AU,\phi U)=0\). Therefore, we have

$$\begin{aligned} AU=\beta \xi +\gamma U \end{aligned}$$
(3.5)

for a certain function \(\gamma \).

The scalar product of (3.3) and U yields \(-\beta g(Y,U)g(\phi AX,U)+k\beta g(Y,U)g(\phi X,U)-2\beta g(\phi AX,Y)+2\beta g(\phi AY,X)=-\beta g(X,U)g(\phi AY,U)+k\beta g(X,U)g(\phi Y,U)\), for any \(X,Y \in {{\mathbb D}}\). If \(Y=U\) it follows \(-\beta g(\phi AX,U)+k\beta g(\phi X,U)-2\beta g(\phi AX,U)+2\beta g(\phi AU,X)=0\) for any \(X \in {{\mathbb D}}\). That is, \(3g(A\phi U,X)-kg(\phi U,X)+2\gamma g(\phi U,X)=0\). Therefore,

$$\begin{aligned} A\phi U=\frac{k-2\gamma }{3}\phi U. \end{aligned}$$
(3.6)

Take now \(X=\xi \), \(Y \in {{\mathbb D}}\) in (3.1). We obtain

$$\begin{aligned}&2\beta g(A\phi U,Y)\xi -\beta \eta (AY)\phi U-k\phi AY+\phi A^2Y-2\beta g(\phi U,Y)A\xi +2kA\phi Y \nonumber \\&\quad \quad -2A\phi AY=-\alpha \phi AY+\alpha \beta g(U,\phi Y)\xi +\beta g(AU,\phi Y)\xi +k\alpha \phi Y \end{aligned}$$
(3.7)

for any \(Y \in {{{\mathbb D}}}\). Its scalar product with \(\xi \) gives, being \(\beta \ne 0\), \(2g(A\phi U,Y)-2\alpha g(\phi U,Y)+2kg(\phi Y,U)-2g(\phi AY,U)=\alpha g(U,\phi Y)+\gamma g(U,\phi Y)\), for any \(Y \in {{\mathbb D}}\). Therefore, we have \(4A\phi U=(\alpha +2k-\gamma )\phi U\), which is equivalent to

$$\begin{aligned} A\phi U=\frac{\alpha +2k-\gamma }{4} \phi U. \end{aligned}$$
(3.8)

From (3.6) and (3.8), it follows

$$\begin{aligned} 3\alpha +2k+5\gamma =0. \end{aligned}$$
(3.9)

If we take \(Y=U\) in (3.7), it follows \(-\beta ^2\phi U-k\phi AU+\phi A^2U+2kA\phi U-2A\phi AU=-\alpha \phi AU+k\alpha \phi U\). That is, \(-\beta ^2-k\gamma +\beta ^2+\gamma ^2+2k\gamma '-2\gamma \gamma ' =-\alpha \gamma +k\alpha \), where \(\gamma ' =\frac{\alpha +2k-\gamma }{4}\). This yields \(\gamma ^2-(2\gamma '-\alpha +k)\gamma +k(2\gamma '-\alpha )=0\). Therefore, \(\gamma =\frac{2\gamma '-\alpha +k \pm \sqrt{(2\gamma '-\alpha -k)^2}}{2}\). From this, either \(\gamma =k\) or \(\gamma =2\gamma '-\alpha \).

Suppose now \(X,Y \in {{{\mathbb D}}}_U\). Then (3.2) yields \(2g(\phi AX,AY)\xi -g(\phi A^2Y,X)\xi -2g(\phi AX,Y)A\xi +2g(\phi AY,X)A\xi =-g(\phi A^2Y,X)\xi \). Its scalar product with U gives \(-2\beta g(\phi AX,Y)+2\beta g(\phi AY,X)=0\). Then \(g((\phi A+A\phi )X,Y)=0\) for any \(X,Y \in {{{\mathbb D}}}_U\). This implies that \((\phi A+A\phi )X=0\) for any \(X \in {{{\mathbb D}}}_U\). If we suppose that \(AX=\lambda X\) we obtain that \(A\phi X=-\lambda \phi X\). If we take such an X in (3.7) we get \(-k\phi AX+\phi A^2X+2kA\phi X-2A\phi AX=-\alpha \phi AX+k\alpha \phi X\). Therefore, \(-k\lambda +\lambda ^2-2k\lambda +2\lambda ^2=-\alpha \lambda +k\alpha \). This yields \(3\lambda ^2-(3k-\alpha )\lambda -k\alpha =(\lambda -k)(\lambda +\frac{\alpha }{3}) =0\). Thus, either \(\lambda =k\) or \(\lambda =-\frac{\alpha }{3}\).

From (3.6) and (3.9) if \(\gamma =k\), a constant, \(\gamma '=-\frac{k}{3}\) is constant and \(\alpha =-\frac{7}{3}\) is also constant. Furthermore, all principal curvatures on \({{{\mathbb D}}}_U\) are also constant.

If \(\gamma =2\gamma '-\alpha =\frac{2k-4\gamma }{3} -\alpha \), we obtain \(3\gamma =2k-4\gamma -3\alpha \). Then \(2k-7\gamma -3\alpha =0\) and from (3.9) \(4k-2\gamma =0\). This yields \(\gamma =2k\) is constant, \(\alpha =-4k\) and \(\gamma '=-k\) are also constant. As above, all principal curvatures in \({{{\mathbb D}}}_U\) are constant.

Take a unit \(X \in {{{\mathbb D}}}_U\) such that \(AX=\lambda X\). The Codazzi equation gives \((\nabla _XA)\xi -(\nabla _{\xi }A)X=-\phi X\). That is, \(\nabla _X(\alpha \xi +\beta U)-A\phi AX-\nabla _{\xi }(\lambda X)+A\nabla _{\xi }X=-\phi X\). Therefore, \(\alpha \lambda \phi X+X(\beta )U+\beta \nabla _XU+\lambda ^2\phi X-\lambda \nabla _{\xi }X+A\nabla _{\xi }X=-\phi X\). If we take \(\phi X\) instead of X, we have similarly \(\alpha \lambda X+(\phi X)(\beta )U+\beta \nabla _{\phi X}U-\lambda ^2X+\lambda \nabla _{\xi }\phi X+A\nabla _{\xi }\phi X=X\). In both cases, taking the scalar product with \(\xi \), we have

$$\begin{aligned} g(\nabla _{\xi }X,U)=g(\nabla _{\xi }\phi X,U)=0. \end{aligned}$$
(3.10)

The scalar product with U of the expression for X yields \(X(\beta )-\lambda g(\nabla _{\xi }X,U)+g(\nabla _{\xi }X,\beta \xi +\gamma U)=0\) In the case of \(\phi X\) we obtain \((\phi X)(\beta )+\lambda g(\nabla _{\xi }\phi X,U)+g(\nabla _{\xi }\phi X,\beta \xi +\gamma U)=0\). Bearing in mind (3.10), we conclude that

$$\begin{aligned} Z(\beta )=0 \end{aligned}$$
(3.11)

for any \(Z \in {{\mathbb D}}\).

On the other hand, we have \((\nabla _{\phi U}A)\xi -(\nabla _{\xi }A)\phi U=U\). That is, \(\nabla _{\phi U}(\alpha \xi +\beta U)-A\phi A\phi U-\nabla _{\xi }(\gamma '\phi U)+A\nabla _{\xi }\phi U=U\). This implies

$$\begin{aligned} \alpha \phi A\phi U+\beta \nabla _{\phi U}U+(\phi U)(\beta )U+\gamma 'AU-\gamma '\nabla _{\xi }\phi U+A\nabla _{\xi }\phi U=U. \end{aligned}$$
(3.12)

Its scalar product with \(\xi \) gives \(\beta g(A\phi U,\phi U)+\beta \gamma '+\gamma 'g(A\xi ,U)+g(\nabla _{\xi }\phi U,\alpha \xi +\beta U)=0\). That is, \(3\beta \gamma '-\alpha \beta +\beta g(\nabla _{\xi }\phi U,U)=0\). Therefore,

$$\begin{aligned} g(\nabla _{\xi }\phi U,U)=-3\gamma '+\alpha . \end{aligned}$$
(3.13)

The scalar product of (3.12) with U gives \(-\alpha g(A\phi U,\phi U)+(\phi U)(\beta )+\gamma \gamma '-\gamma 'g(\nabla _{\xi }\phi U,U)+g(\nabla _{\xi }\phi U,\beta \xi +\gamma U)=1\). From (3.13) this yields \(-\alpha \gamma '+(\phi U)(\beta )+\gamma \gamma '-\gamma '(-3\gamma '+\alpha )-\beta ^2-3\gamma \gamma '+\alpha \gamma =1\). Thus, we obtain

$$\begin{aligned} (\phi U)(\beta )=1+2\alpha \gamma '+2\gamma \gamma '-3\gamma '^2+\beta ^2-\alpha \gamma . \end{aligned}$$
(3.14)

Now \((\nabla _UA)\xi -(\nabla _{\xi }A)U=-\phi U\) implies

$$\begin{aligned} \alpha \phi AU+U(\beta )U+\beta \nabla _UU-\gamma \gamma '\phi U-\xi (\beta )\xi -\beta \phi A\xi -\gamma \nabla _{\xi }U+A\nabla _{\xi }U=-\phi U. \end{aligned}$$
(3.15)

Its scalar product with \(\xi \) gives \(-\beta g(U,\phi AU)-\xi (\beta )+\gamma g(U,\phi A\xi )+g(\nabla _{\xi }U,\alpha \xi +\beta U)=0\). From this we have

$$\begin{aligned} \xi (\beta )=0. \end{aligned}$$
(3.16)

The scalar product of (3.15) with U yields \(U(\beta )+g(\nabla _{\xi }U,\beta \xi +\gamma U)=0\). This gives

$$\begin{aligned} U(\beta )=0. \end{aligned}$$
(3.17)

From (3.11), (3.14), (3.16) and (3.17) we obtain

$$\begin{aligned} grad(\beta )=(\beta ^2+1+2\alpha \gamma '+2\gamma \gamma '-3\gamma '^2-\alpha \gamma )\phi U. \end{aligned}$$
(3.18)

We will call \(\omega =\beta ^2+1+2\alpha \gamma '+2\gamma \gamma '-3\gamma '^2-\alpha \gamma \). We know that \(g(\nabla _Xgrad(\beta ),Y)=g(\nabla _Ygrad(\beta ),X)\) for any X, Y tangent to M. In our case we have \(X(\omega )g(\phi U,Y)+\omega g(\nabla _X\phi U,Y)=Y(\omega )g(\phi U,X)+\omega g(\nabla _Y\phi U,X)\). If we take \(X=\xi \), from (3.16) and the fact that the all the elements different from \(\beta \) appearing in \(\omega \) are constant, we have \(\xi (\omega )=0\). Thus, we get \(-\omega g(U,AY)=\omega g(\nabla _{\xi }\phi U,Y)\) for any Y tangent to M. Taking now \(Y=U\) , bearing in mind (3.13) we arrive to \(-\omega \gamma =\omega (-3\gamma '+\alpha )\). If we suppose \(\omega \ne 0\) it follows \(-\gamma =-3\gamma '+\alpha \). If \(\gamma =k\), \(\gamma '=-\frac{k}{3}\) and \(\alpha =-\frac{7k}{3}\). Therefore, \(-k=k-\frac{7k}{3}\) implies \(k=0\), which is impossible. In the other possible case \(\gamma '=-k\), \(\gamma =2k\) and \(\alpha =-4k\). Then \(-2k=3k-4k\) gives also a contradiction.

Thus, we have proved that \(\omega =0\). Then \(1+2\alpha \gamma '+2\gamma \gamma '-3\gamma '^2-\alpha \gamma +\beta ^2=0\). If \(\gamma =k\), \(\gamma '=-\frac{k}{3}\) and \(\alpha =-\frac{7k}{3}\). This yields \(\beta ^2 +1+\frac{4}{3} k^2=0\), which is impossible. Then \(\gamma =2k\), \(\alpha =-4k\) and \(\gamma '=-k\). Thus \(\beta ^2+9k^2+1=0\), also impossible, and we have finished the proof.\(\square \)

4 Proof of Theorem 2

If M satisfies (1.3) for \(B=A\) and any X, Y tangent to M we obtain

$$\begin{aligned} \begin{array}{c} -\eta (AY)\phi AX-k\eta (X)\phi AY-g(\phi A^2Y,X)\xi +\eta (X)\phi A^2Y+k\eta (AY)\phi X \\ -\eta (AX)\phi AY-k\eta (Y)\phi AX-g(\phi A^2X,Y)\xi +\eta (Y)\phi A^2X+k\eta (AX)\phi Y=0. \\ \end{array} \end{aligned}$$
(4.1)

for any X, Y tangent to M. If \(X,Y \in {{\mathbb D}}\) (4.1) becomes

$$\begin{aligned} \begin{array}{c} -\eta (AY)\phi AX-g(\phi A^2Y,X)\xi +k\eta (AY)\phi X-\eta (AX)\phi AY \\ -g(\phi A^2X,Y)\xi +k\eta (AX)\phi Y=0 \\ \end{array} \end{aligned}$$
(4.2)

for any \(X,Y \in {{\mathbb D}}\). If in (4.1) we take \(X=\xi \), \(Y \in {{\mathbb D}}\), we obtain

$$\begin{aligned} -\eta (AY)\phi A\xi -k\phi AY+\phi A^2Y-\eta (A\xi )\phi AY-g(\phi A^2\xi ,Y)\xi +k\eta (A\xi )\phi Y =0 \end{aligned}$$
(4.3)

for any \(Y \in {{\mathbb D}}\).

Let us suppose that M is Hopf and \(A\xi =\alpha \xi \). From (4.2) we obtain \(-g(\phi A^2Y,X)\xi -g(\phi A^2X,Y)\xi =0\) for any \(X,Y \in {{\mathbb D}}\). Therefore, \(A^2\phi X=\phi A^2X\) for any \(X \in {{\mathbb D}}\). If \(X \in {{\mathbb D}}\) satisfies \(AX=\lambda X\), we know that \(A\phi X=\mu \phi X\) with \(\mu =\frac{\alpha \lambda +2}{2\lambda -\alpha }\). Thus, we have \(\lambda ^2=\mu ^2\) and either \(\lambda =\mu \) or \(\mu =-\lambda \).

If \(\frac{\alpha \lambda +2}{2\lambda -\alpha }=-\lambda \) we obtain \(\alpha \lambda +2=-2\lambda ^2+\lambda \alpha \). This implies \(\lambda ^2+1=0\), which is impossible. Therefore, \(\lambda =\mu \). Taking such a Y in (4.3), we have \(-k\phi AY+\phi A^2Y-\alpha \phi AY+k\alpha \phi Y=0\). That is, \(-k\lambda +\lambda ^2-\alpha \lambda +k\alpha =0\). This gives \(\lambda ^2-(\alpha +k)\lambda +k\alpha =(\lambda -k)(\lambda -\alpha )=0\), and the possible solutions are either \(\lambda =k\) or \(\lambda =\alpha \). Then M has two distinct constant principal curvatures and from Ref. [2] M must be locally congruent to a geodesic hypersphere whose principal curvature on \({{\mathbb D}}\) is \(cot(r)=k\).

If M is non-Hopf, as in the previous section, we write \(A\xi =\alpha \xi +\beta U\), with the same conditions. From (4.2), it follows

$$\begin{aligned} \begin{array}{c} -\beta g(U,Y)\phi AX-g(\phi A^2Y,X)\xi +k\beta g(U,Y)\phi X \\ -\beta g(U,X)\phi AY-g(\phi A^2X,Y)\xi +k\beta g(U,X)\phi Y=0 \\ \end{array} \end{aligned}$$
(4.4)

for any \(X,Y \in {{\mathbb D}}\). Its scalar product with U yields

$$\begin{aligned} \begin{array}{c} \beta g(U,Y)g(A\phi U,X)-k\beta g(U,Y)g(\phi U,X)+\beta g(U,X)g(A\phi U,Y) \\ -k\beta g(U,X)g(\phi U,Y)=0\\ \end{array} \end{aligned}$$
(4.5)

for any \(X,Y \in {{\mathbb D}}\). If in (4.5) we take \(X \in {{{\mathbb D}}}_U\), \(\beta g(U,Y)g(A\phi U,X)=0\). As \(\beta \ne 0\), if \(Y=U\) we obtain \(g(A\phi U,X)=0\) for any \(X \in {{{\mathbb D}}}_U\). If in (4.5) we take \(X=Y=U\) we have \(2\beta g(A\phi U,U)=0\). This implies \(g(A\phi U,U)=0\). Finally, taking \(Y=\phi U\) in (4.5) we get \(\beta g(U,X)g(A\phi U,\phi U)-k\beta g(U,X)=0\) for any \(X \in {{\mathbb D}}\). For \(X=U\) we obtain \(g(A\phi U,\phi U)=k\). Therefore, we have seen that

$$\begin{aligned} A\phi U=k\phi U. \end{aligned}$$
(4.6)

The scalar product of (4.4) and \(\phi U\) gives \(-\beta g(Y,U)g(AU,X)+k\beta g(U,Y)g(U,X)-\beta g(U,X)g(AU,Y) +k\beta g(U,X)g(U,Y)=0\) for any \(X,Y \in {{\mathbb D}}\). Taking \(X=Y=U\) we have \(-2\beta g(AU,U)+2k\beta =0\) and then \(g(AU,U)=k\). On the other hand, (4.3) yields \(-\beta ^2g(Y,U)\phi U-k\phi AY+\phi A^2Y-\alpha \phi AY+\alpha g(A\xi ,\phi Y)\xi +\beta g(AU,\phi Y)\xi +k\alpha \phi Y=0\) for any \(Y \in {{\mathbb D}}\). Its scalar product with \(\xi \) implies

$$\begin{aligned} \alpha g(A\xi ,\phi Y)+\beta g(AU,\phi Y)=0 \end{aligned}$$
(4.7)

for any \(Y \in {{\mathbb D}}\). If \(Y=\phi X\), \(X \in {{\mathbb D}}_U\), we obtain \(\beta g(AU,X)=0\) for any \(X \in {{\mathbb D}}_U\) and if \(Y=\phi U\) in (4.7) it follows \(-\alpha \beta -\beta g(AU,U)=0\). Therefore, \(g(AU,U)=-\alpha \) and we get

$$\begin{aligned} \alpha =-k \end{aligned}$$
(4.8)

and

$$\begin{aligned} AU=\beta \xi +kU. \end{aligned}$$
(4.9)

Let \(X,Y \in {{\mathbb D}}_U\). From (4.4) we have \(-g(\phi A^2Y,X)-g(\phi A^2X,Y)=0\). From (4.6) and (4.9) \({{\mathbb D}}_U\) is A-invariant and we obtain \(\phi A^2X=A^2\phi X\) for any \(X \in {{\mathbb D}}_U\). Let us suppose that \(Y \in {{\mathbb D}}_U\) satisfies \(AY=\lambda Y\). From (4.3) we get \(-k\lambda \phi Y+\lambda ^2\phi Y-\alpha \lambda \phi Y+k\alpha \phi Y=0\). From (4.8) it follows \(\lambda ^2\phi Y-k^2\phi Y=0\). Thus, \(\lambda ^2=k^2\) and \(\lambda \) is constant.

For such a \(Y \in {{\mathbb D}}_U\) the Codazzi equation gives \((\nabla _YA)\xi -(\nabla _{\xi }A)Y=-\phi Y\). Therefore, \(\nabla _Y(-k\xi +\beta U)-A\phi AY-\nabla _{\xi }(\lambda Y)+A\nabla _{\xi }Y=-\phi Y\). Then \(-k\phi AY+Y(\beta )U+\beta \nabla _YU-A\phi AY-\lambda \nabla _{\xi }Y+A\nabla _{\xi }Y=-\phi Y\). Its scalar product with U yields \(Y(\beta )-\lambda g(\nabla _{\xi }Y,U)+g(\nabla _{\xi }Y,\beta \xi +kU)=0\) and

$$\begin{aligned} Y(\beta )=(\lambda -k)g(\nabla _{\xi }Y,U). \end{aligned}$$
(4.10)

On the other hand, \((\nabla _YA)U-(\nabla _UA)Y=0\). From this we obtain \(\nabla _Y(\beta \xi +kU)-A\nabla _YU-\nabla _U(\lambda Y)+A\nabla _UY=0\). That is, \(Y(\beta )\xi +\beta \phi AY+k\nabla _YU-A\nabla _YU-\lambda \nabla _UY+A\nabla _UY=0\). Its scalar product with \(\xi \) gives \(Y(\beta )-kg(U,\phi AY)-g(\nabla _YU,\alpha \xi )+\lambda g(Y,\phi AU)+g(\nabla _UY,\alpha \xi +\beta U)=0\). Then

$$\begin{aligned} Y(\beta )=-\beta g(\nabla _UY,U). \end{aligned}$$
(4.11)

Its scalar product with U implies \(-g(\nabla _YU,\beta \xi )-\lambda g(\nabla _UY,U)+g(\nabla _UY,\beta \xi +kU)=0\). This yields \((\lambda -k)g(\nabla _YU,Y)=0\). If \(g(\nabla _UY,U)=0\) from (4.11) we get \(Y(\beta )=0\). If \(g(\nabla _UY,U) \ne 0\), \(\lambda =k\) and from (4.10) again

$$\begin{aligned} Y(\beta )=0 \end{aligned}$$
(4.12)

for any \(Y \in {{\mathbb D}}_U\).

Moreover \((\nabla _UA)\xi -(\nabla _{\xi }A)U=-\phi U\) implies \(\nabla _U(-k\xi +\beta U)-A\phi AU-\nabla _{\xi }(\beta \xi +kU)+A\nabla _{\xi }U=-\phi U\). Then \(-k\phi AU+U(\beta )U+\beta \nabla _UU-A\phi AU-\xi (\beta )\xi -\beta \phi A\xi -k\nabla _{\xi }U+A\nabla _{\xi }U=-\phi U\) and its scalar product with \(\xi \) gives \(-\beta g(U,\phi AU)-\xi (\beta )+kg(U,\phi A\xi )+g(\nabla _{\xi }U,\alpha \xi +\beta U)=0\). Therefore,

$$\begin{aligned} \xi (\beta )=0 \end{aligned}$$
(4.13)

and its scalar product with U yields \(U(\beta )+g(\nabla _{\xi }U,\beta \xi )=0\). That is

$$\begin{aligned} U(\beta )=0. \end{aligned}$$
(4.14)

Now we develop \((\nabla _{\phi U}A)\xi -(\nabla _{\xi }A)\phi U=U\). Then \(\nabla _{\phi U}(-k\xi +\beta U)-A\phi A\phi U-\nabla _{\xi }(k\phi U)+A\nabla _{\xi }\phi U=U\) that implies \(-k\phi A\phi U+(\phi U)(\beta )U+\beta \nabla _{\phi U}U-A\phi A\phi U-k\nabla _{\xi }\phi U+A\nabla _{\xi }\phi U=U\). Its scalar product with U yields

$$\begin{aligned} (\phi U)(\beta )=1+\beta ^2-2k^2. \end{aligned}$$
(4.15)

Its scalar product with \(\xi \) gives \(\beta g(A\phi U,\phi U)+\beta g(A\phi U,\phi U)+kg(\phi U,\phi A\xi )+g(\nabla _{\xi }\phi U,-k\xi +\beta U)=0\). Therefore,

$$\begin{aligned} g(\nabla _{\xi }\phi U,U)=-4k \end{aligned}$$
(4.16)

and

$$\begin{aligned} grad(\beta )=\omega \phi U \end{aligned}$$
(4.17)

where \(\omega =1+\beta ^2-2k^2\). Now, as in previous section, \(g(\nabla _X(\omega \phi U),Y)=g(\nabla _Y(\omega \phi U),X)\) for any X, Y tangent to M. This yields \(X(\omega )g(\phi U,Y)+\omega g(\nabla _X\phi U,Y)=Y(\omega )g(\phi U,\) \(X)+\omega g(\nabla _Y\phi U,X)\). If \(X=\xi \) we get \(\omega g(\nabla _{\xi }\phi U,Y)=\omega g(\nabla _Y\phi U,\xi )=-\omega g(\phi U,\phi AY)=-\omega g(U,AY)\). Take \(Y=U\). Then \(\omega g(\nabla _{\xi }\phi U,U)-k\omega \). This and (4.16) give \(\omega =0\) and, therefore, \(\beta \) is constant and equals \(2k^2-1\).

Now \((\nabla _UA)\phi U-(\nabla _{\phi U}A)U=-2\xi \). Then \(\nabla _U(k\phi U)-A\nabla _U\phi U-\nabla _{\phi U}(\beta \xi +kU)+A\nabla _{\phi U}U=-2\xi \), that is, \(k\nabla _U\phi U-A\nabla _U\phi U-\beta \phi A\phi U-k\nabla _{\phi U}U+A\nabla _{\phi U}U=-2\xi \). If we take its scalar product with U we obtain \(3k\beta =0\), which is impossible and finishes the proof. \(\square \)