CR Submanifolds in (l.c.a.) Kaehler and S-manifolds

Abstract

In the present work, we briefly summarize our contributions to the study of CR submanifolds of (locally conformal almost) Kaehler manifolds and normal CR submanifolds of S-manifolds.

5.1 Introduction

K. Yano [21] introduced in 1963 the notion of f-structure on a $(2m+s)$-dimensional manifold as a tensor field f of type (1, 1) and rank 2m satisfying $f^3+f=0$. Almost complex ($s=0$) and almost contact ($s=1$) structures are well-known examples of f-structures. A Riemannian manifold endowed with an f-structure ($s\ge 2$) compatible with the Riemannian metric is called a metric f-manifold (for $s=0$ we have almost Hermitian manifolds and for $s=1$, metric almost contact manifolds). In this context, D.E. Blair [5] defined K-manifolds (and particular cases of S-manifolds and C-manifolds) as the analogue of Kaehlerian manifolds in the almost complex geometry and of quasi-Sasakian manifolds (and particular cases of Sasakian manifolds and cosymplectic manifolds) in the almost contact geometry.

He also showed that the curvature of S-manifolds is completely determined by their f-sectional curvatures. Later, M. Kobayashi and S. Tsuchiya [15] got expressions for the curvature tensor field of S-manifolds when their f-sectional curvature is constant depending on such a constant. Such spaces are called S-space-forms and they generalize complex and Sasakian space-forms. Nice examples of S-space-forms can be found in [5, 6, 8, 13]. In particular, it is proved in [5, 8] that certain principal toroidal bundles over complex-space-forms are S-space-forms and a generalization of the Hopf fibration denoted by $\mathbb H^{2m+s}$ is introduced as a canonical example of such manifolds playing the role of complex projective space in Kaehler geometry and the odd-dimensional sphere in Sasakian geometry [5, 6].

When we want to study the submanifolds of a metric f-manifold, the natural first step is to consider such submanifolds depending on their behavior with respect to the f-structure. So, invariant and anti-invariant submanifolds (in the terminology of the complex geometry, holomorphic and totally real submanifolds) appear if all the tangent vector fields to the submanifold are transformed by f into tangent vector fields or into normal vector fields. But since an hypersurface of a metric f-manifold tangent to the structure vector fields is neither invariant nor anti-invariant, it is necessary to introduce a wider class of submanifolds: the CR-submanifolds. This work was made firstly by A. Bejancu and B.-Y. Chen [1, 10, 11] in the case $s=0$ and by A. Bejancu and N. Papaghiuc, M. Kobayashi and K. Yano and M. Kon in the case $s=1$ (we refer to the books [3, 22] for the background of these cases where a large list of fundamental references can be found). For $s\ge 2$, I. Mihai [16] introduced the notion of CR-submanifold in a natural way.

Many authors have studied the geometry of submanifolds of locally conformal almost Kaehler (l.c.a.K.) manifolds [10, 11, 14, 20], which are almost Hermitian manifolds $(\widetilde{M},J,g)$ such that every $x\in \widetilde{M}$ has an open neighborhood U such that for some differentiable function $h:U\longrightarrow \mathbb R,\tilde{g}_U=e^{-h}g|_{U}$ is a (l.c.a.) Kaehler metric on U. If one can take $U=\widetilde{M}$, the manifold is then called globally conformal almost Kaeler (g.c.a.K) manifold. Examples of l.c.K. manifolds are provided by the Hopf manifolds. So, it seems interesting to study CR-submanifolds of l.c.a.K. manifolds.

On the other hand, M. Okumura [17, 18] studied normal real hypersurfaces of Kaehlerian manifolds and obtained nice properties. For this reason, it also seems interesting to introduce and study normal CR-submanifolds. In the cases $s=0$ and $s=1$, the papers [2] and [4] can be consulted.

The aim of the present work is to briefly summarize our contributions to the study of CR-submanifolds of l.c.a.K. manifolds, normal CR-submanifolds of S-manifolds. To this end, we separate them into two different sections, which can be read independently.

5.2 CR-Submanifolds of (l.c.a.) Kaehler Manifolds

Let $(\widetilde{M},J,g)$ be an almost Hermitian manifold ($dim(\widetilde{M})=2m$) with almost complex structure J and Hermitian metric g and let M be a Riemannian submanifold isometrically immersed in $\widetilde{M}.$

A. Bejancu [1] introduced the notion of a CR-submanifold of $\widetilde{M}.$ In fact, M is a CR-submanifold of the almost Hermitian manifold $\widetilde{M}$ if there exists on M a differentiable holomorphic distribution $\mathcal {D}$, i.e., $J(\mathcal {D}_x)\subseteq \mathcal {D}_x$ for any $x\in M$ such that its orthogonal complement $\mathcal {D}^{\bot }$ in M is totally real in $\widetilde{M}$, i.e., $J(\mathcal {D}_x^{\bot })\subseteq T^{\perp }_x(M)$ for any $x\in M,$ where $T^{\perp }_x(M)$ is the normal space at x. If $dim(\mathcal {D})=0,$ M is called a totally real submanifold, and if $dim(\mathcal {D}^{\perp })=0$ M is a holomorphic submanifold.

We first discuss the Gauss–Weingarten equations of the submanifold with respect to the metric g and with respect to the local conformal Kaehler metrics and then we shall establish thereby the analytical conditions that characterize the important types of submanifolds.

5.2.1 Preliminaries

Let $(\widetilde{M},J,g)$ be an almost Hermitian manifold. It is easy to see [20] that $(\widetilde{M},J,g)$ is a l.c.(a).K. manifold if and only if there is a global closed 1-form $\omega$ on $\widetilde{M}$ (the Lee form) such that $d\Omega =\omega \wedge \Omega$ ($\Omega$ the fundamental form of the manifold) and $(\widetilde{M},J,g)$ is a g.c.(a).K. manifold if and only if $\omega$ is also exact. In case $\omega =0$, the manifold is an (almost) Kaehler manifold.

Let $(\widetilde{M},J,g)$ be a l.c.(a).K. manifold and consider the Lee vector field B [20] of $(\widetilde{M},J,g)$ defined by $g(X,B)=\omega (X).$ Denote by $\widetilde{\nabla }$ the Levi-Civita connection of g and define

$$\begin{aligned} \overline{\nabla }_XY=\widetilde{\nabla }_XY - \frac{1}{2} \omega (X)Y-\frac{1}{2}\omega (Y)X+\frac{1}{2}g(X,Y)B. \end{aligned}$$

(5.1)

Then $\overline{\nabla }$ is a torsionless linear connection on $\widetilde{M}$ which is called the Weyl connection of g. It is easy to see that $\overline{\nabla }_Xg=\omega (X)g$. We have

Theorem 5.1

([20]) The almost Hermitian manifold $(\widetilde{M},J,g)$ is a l.cK. manifold if and only if there is a closed 1-form $\omega$ on $\widetilde{M}$ such that the Weyl connection is almost complex, That is, $\overline{\nabla }J=0.$

Let $(\widetilde{M},J,g)$ be a l.c.K. manifold and M a Riemannian manifold isometrically immersed in $\widetilde{M}.$ We denote by g the metric tensor of $\widetilde{M}$ as well as that induced on M,  and let $\nabla ,$ $\nabla ^M$ be the covariant derivations on M induced by $\widetilde{\nabla }$ and $\overline{\nabla }$, respectively. Then, the Gauss–Weingarten formulas for M with respect to $\widetilde{\nabla }$ and $\overline{\nabla }$ are given by

$$\begin{aligned} \widetilde{\nabla }_XY=\nabla _XY+\sigma (X,Y), \; \widetilde{\nabla }_XV = -A_{V}X+D_XV, \end{aligned}$$

(5.2)

$$\begin{aligned} \overline{\nabla }_XY=\nabla ^M_XY+\overline{\sigma } (X,Y), \; \overline{\nabla }_XV = -\overline{A}_{V}X+\overline{D}_XV, \end{aligned}$$

(5.3)

for any vector fields X, Y tangent to M and V normal to M,  where $\sigma$ (respectively, $\overline{\sigma }$) is the second fundamental form of M with respect to $\widetilde{\nabla } (\overline{\nabla })$ and D (respectively, $\overline{D}$) is the normal connection. The formulas (5.3) are the Gauss–Weingarten equations of $M|_U$ in $(\widetilde{M}|_U,e^{-h}g|_U).$ The second fundamental tensors $A_{V}, \overline{A}_{V}$ are related to $\sigma , \overline{\sigma }$ respectively by

$$\begin{aligned} g(A_{V}X,Y)=g(\sigma (X,Y),V), \, g(\overline{A}_{V}X,Y)=g(\overline{\sigma }(X,Y),V). \end{aligned}$$

(5.4)

For any vector X tangent to M and V normal to M write

$$\begin{aligned} JX=TX+NX, \; JV =tV + nV, \end{aligned}$$

(5.5)

where TX and NX (respectively, tV and nV) are the tangential and normal component of J(X) (respectively JV). For the Lee field B, we have

$$\begin{aligned} B_x=(B_x)_1+(B_x)_2, \quad x\in M, \end{aligned}$$

(5.6)

where $(B_x)_1$ (resp.y $(B_x)_2)$ is the tangential (resp. normal) component of $B_x.$

If M is a CR-submanifold of an almost Hermitian manifold $(\widetilde{M},J,g)$ let us denote by $\nu$ the complementary orthogonal subbundle of $J\mathcal {D}^\perp$ in $T^\perp (M).$ Hence we have, $T^{\perp }(M)=J\mathcal {D}^{\perp } \oplus \nu .$

5.2.2 Integrability Conditions of the Basic Distributions

First we give some general identities.

Lemma 1

Let M be a CR-submanifold of a l.c.K. manifold $(\widetilde{M},J,g).$ Then, we have

$$\begin{aligned} \nabla ^M_XY=\nabla _XY-\frac{1}{2}\omega (X)Y-\frac{1}{2}\omega (Y)X+\frac{1}{2}g(X,Y)B_1 \end{aligned}$$

(5.7)

$$\begin{aligned} \overline{\sigma }(X,Y)=\sigma (X,Y)+\frac{1}{2}g(X,Y)B_2 \end{aligned}$$

(5.8)

$$\begin{aligned} \overline{A}_{V}X=A_{V}X+\frac{1}{2}\omega (V)X \end{aligned}$$

(5.9)

$$\begin{aligned} \overline{D}_{V}X=D_{V}X-\frac{1}{2}\omega (X)V \end{aligned}$$

(5.10)

for any vector fields X, Y tangent to M and V normal to M.

Proof

The assertions follow immediately from (5.1)–(5.3).    $\blacksquare$

The following result is well known:

Theorem 5.2

([7]) The totally real distribution $\mathcal {D}^{\perp }$ of any CR-submanifold of a l.c.K. manifold is integrable.

For the holomorphic distribution $\mathcal {D}$, we have

Theorem 5.3

Let M be a submanifold of a l.c.K. manifold $\widetilde{M}$ and let $\mathcal {D}_x$ de maximal holomorphic subspace of $T_x(M)$ and assume $dim(\mathcal {D}_x)$ is a constant. Then, the holomorphic distribution $\mathcal {D}$ is integrable if and only if the second fundamental form $\overline{\sigma }$ satisfies $\overline{\sigma }(X,JY)=\overline{\sigma }(JX,Y)$ or, equivalently, $\sigma (X,JY)-\sigma (JX,Y)+\Omega (X,Y)B_2=0,$ for all vector fields $X,Y \in \mathcal {D}.$

If M is a CR-submanifold, the integrability condition on $\mathcal {D}$ in Theorem 5.3 can be replaced by a weaker condition.

Theorem 5.4

Let M be a CR-submanifold of a l.c.K. manifold $\widetilde{M}.$ The holomorphic distribution $\mathcal {D}$ is integrable if and only if

$$g\left( \sigma (X,JY)-\sigma (JX,Y)+\Omega (X,Y)B,J\mathcal {D}^{\perp }\right) =0,$$

for all $X,Y \in \mathcal {D}.$

Theorems 5.3 and 5.4 follow easily from similar theorems in the Kaehlerian case ([7]), from (5.8) and the fact that, locally, $\widetilde{M}$ is endowed with Kaehler metrics $\widetilde{g}_U$ whose Levi-Civita connection is $\overline{\nabla }.$

With regard to integral submanifolds of $\mathcal {D}^{\perp }$ and $\mathcal {D}$ (provided $\mathcal {D}$ is integrable), we have the following theorem.

Theorem 5.5

For a CR-submanifold M of a l.c.K. manifold $\widetilde{M}$, the leaf $M^{\perp }$ is totally geodesic in M if and only if

$$g\left( A_{JW}Z+\frac{1}{2}g(Z,W)JB,\mathcal {D}\right) =0,$$

that is,

$$g(\sigma (Z,X),JW)=\frac{1}{2}g(Z,W)\omega (JW),$$

for any $X\in \mathcal {D},$ $Z,W\in \mathcal {D}^{\perp }.$

Proof

From (5.1), (5.2) and $\overline{\nabla } J=0,$ for any $X\in \mathcal {D},$ $Z,W \in \mathcal {D}^{\perp }$, we obtain

$$\begin{aligned} g( J\nabla _ZW,X)+\frac{1}{2}g(Z,W)g(JB,X)=-g(A_{JW}Z,X). \end{aligned}$$

(5.11)

But $M^{\perp }$ is totally geodesic in M if and only if $\nabla _ZW\in \mathcal {D}^{\perp }$ for all $Z,W\in \mathcal {D}^{\perp },$ and then (5.11) gives the theorem.    $\blacksquare$

Theorem 5.6

Let M be a CR-submanifold of a l.c.K manifold $\widetilde{M}.$ If the holomorphic distribution $\mathcal {D}$ is integrable and $M^{T}$ is an integral submanifold of $\mathcal {D},$ then $M^{T}$ is totally geodesic if and only if

$$g\left( J\sigma (X,Y) +\frac{1}{2}g(X,Y)JB-\frac{1}{2}\Omega (X,Y)B,\mathcal {D}^{\perp } \right) =0,$$

for any $X,Y \in \mathcal {D}.$

Proof

From (5.1), (5.3), and $\overline{\nabla } J=0,$ for any $X,Y\in \mathcal {D}$ and $Z\in \mathcal {D}^{\perp }$, we have

$$\begin{aligned} g(J\sigma (X,Y),Z)+\frac{1}{2}g(X,Y)g(JB,Z)=g(\nabla _X(JY),Z)+\frac{1}{2}\Omega (X,Y)g(B,Z).\quad \end{aligned}$$

(5.12)

But $M^T$ is totally geodesic in M if and only if $\nabla _XY \in \mathcal {D}$ for all $X,Y \in \mathcal {D},$ and hence Eq. (5.12) gives the theorem.    $\blacksquare$

5.2.3 CR-Submanifolds of l.c.K. Manifolds

First of all, we shall give some identities for later use. Let T, N t, and n be the endomorphisms and vector-valued 1-forms defined in (5.5). The following lemma can be easily obtained from (5.3), (5.9), and $\overline{\nabla }J=0.$

Lemma 2

Let M be an isometrically immersed submanifold of a l.c.K. manifold $\widetilde{M}.$ Then, we have

$$\begin{aligned} \nabla ^M_X(TY)-\overline{A}_{NY}X= T\nabla ^M_XY+t\overline{\sigma }(X,Y) \end{aligned}$$

(5.13)

$$\begin{aligned} \overline{\sigma }(X,TY)+D_X(NY)=N\nabla ^M_XY+n\overline{\sigma }(X,Y), \end{aligned}$$

(5.14)

$$\begin{aligned} \nabla ^M_X(tV)-\overline{A}_{nV}X=-T\overline{A}_{V}X+t\overline{D}_XV, \end{aligned}$$

(5.15)

$$\begin{aligned} \overline{\sigma }(X,tV)+\overline{D}_X(nV)=-N\overline{A}_{V}X+n\overline{D}_XV, \end{aligned}$$

(5.16)

$$\begin{aligned} \left[ \overline{A}_{V},\overline{A}_{\overline{V}} \right] = \left[ A_{V},A_{\overline{V}} \right] , \end{aligned}$$

(5.17)

for any vector fields X, Y tangent to M and $V, \, \overline{V}$ normal to M.

Now, we shall study totally umbilical and totally geodesic CR-submanifolds.

Theorem 5.7

Let M be a totally umbilical CR-submanifold of a l.c.K. manifold $\widetilde{M}.$ Then, we have

• (i) Either $dim(\mathcal {D}^{\perp })=1$ or the component $H_{J(TM)}$ of the mean curvature tensor H in J(TM) is given by $H_{J(TM)}=-\frac{1}{2}B_2$.

• (ii) If $dim(\mathcal {D}^{\perp })>1$ and M is proper (neither holomorphic nor totally real) such that B is tangent to M,  then M is totally geodesic.

Proof

First, since M is totally umbilical, $\sigma (X,Y)=g(X,Y)H$ for any X, Y tangent to M,  and hence

$$\begin{aligned} g(\sigma (X,X),JW)=g(X,X)g(H,JW). \end{aligned}$$

(5.18)

From (5.3) and (5.4) it is easy to see that

$$\begin{aligned} \overline{A}_{JZ}W=\overline{A}_{JW}Z \end{aligned}$$

(5.19)

and, then, if we take an unit vector field $X=Z\in \mathcal {D}^{\perp }$ orthogonal to W,  (5.9), (5.18), and (5.19) give

$$\begin{aligned} \begin{aligned} g(H,JW)=&g(A_{JW}Z,Z)= g(A_{JZ}W+\frac{1}{2}\omega (JZ)W-\frac{1}{2}\omega ((JW)Z,Z)\\ =&-\frac{1}{2}\omega (JW)= g(-\frac{1}{2}B_2,JW), \end{aligned} \end{aligned}$$

so that (i) holds.

Now, since $dim(\mathcal {D}^{\perp })>1,$ from (5.5) and assertion (i), we have $tH=0.$ Thus, (5.15) gives $t\overline{D}_YH=\overline{A}_{nH}Y-T\overline{A}_HY$, for any Y tangent to M. Therefore, for any Z tangent to M,  from (5.8) and (5.9) we get

$$\begin{aligned} g(t\overline{D}_YH,Z)=-g(\overline{A}_HY,TZ)-g(\overline{\sigma }(Y,Z),nH)=-g(Y,TZ)g(H,H) \end{aligned}$$

(5.20)

and, if we take $Z=TY$, we have

$$\begin{aligned} -g(Y,T^2Y)g(H,H)=g(t\overline{D}_YH,TY)=g(Tt\overline{D}_YH,Y)=0. \end{aligned}$$

(5.21)

The last equation holds because $Tt=0$ for any CR-submanifold of an almost Hermitian manifold [22]. Moreover, it is easy to see [22] that $T^2=-I+tN$ and then (5.21) gives

$$\begin{aligned} g(Y,Y)g(H,H)-g(NY,NY)g(H,H)=0. \end{aligned}$$

(5.22)

Since M is proper, we can choose an unit vector field X in $\mathcal {D}.$ Thus, $NX=0$ and from (5.22) we have $H=0.$    $\blacksquare$

Theorem 5.8

Let M be a totally geodesic CR-submanifold of a l.c.K. manifold $\widetilde{M}.$ We have

• (i) If $B_x \in \mathcal {D}_x$, for all $x\in M,$ then $\mathcal {D}$ is integrable and any integral submanifold $M^T$ of $\mathcal {D}$ is totally geodesic in $\widetilde{M}$.

• (ii) If B is normal to M,  any integral submanifold $M^{\perp }$ of $\mathcal {D}^{\perp }$ is totally geodesic in $\widetilde{M}.$ Furthermore, $\mathcal {D}$ is integrable if and only if $B_x \in \nu _x$, for any $x\in M,$ and in this case any integral submanifold $M^T$ of $\mathcal {D}$ is totally geodesic in $\widetilde{M}.$

Proof

Firstly, since B is tangent to M,  from Theorem 5.7 the distribution $\mathcal {D}$ is integrable. Let $M^T$ be an integral submanifold of $\mathcal {D}.$ For any vector field X tangent to M,  $Y\in \mathcal {D},$ $Z\in \mathcal {D}^{\perp },$ from (5.3) and (5.4) we get $g(\nabla ^{M}_XZ,Y)=-g(\overline{\sigma }(X,JY),JZ).$ But from (5.7) and (5.8) we find

$$\begin{aligned}&g(\nabla _XZ,Y)-\frac{1}{2}\omega (Z)g(X,Y)+\frac{1}{2}g(X,Z)g(B,Y) = -g(\sigma (X,JY),JZ)=0. \end{aligned}$$

(5.23)

If $X\in \mathcal {D}$, (5.23) gives $g(\nabla _XZ,Y)=0,$ or, equivalently, $g(\nabla _XY,Z)=0$ and therefore, $\nabla _XY \in \mathcal {D}.$ Thus $M^T$ is totally geodesic in M and hence in $\widetilde{M}.$

Next, if B is normal to M,  from Theorem 5.5, any integral submanifold $M^{\perp }$ of $\mathcal {D}^{\perp }$ is totally geodesic in $\widetilde{M}.$ The second statement follows immediately from Theorems 5.6 and 5.7.    $\blacksquare$

Corollary 1

Let M be a totally geodesic proper CR-submanifold of a l.c.K. manifold $\widetilde{M}$ such that $B_x\in \nu _x$, for any $x\in M.$ Then, M is locally the Riemannian product of a Kaehler submanifold and a totally real submanifold of $\widetilde{M}.$

Proof

From Theorem 5.8, M is locally the product of a holomorphic submanifold $M^{T}$ and a totally real submanifold $M^{\perp }$ of $\widetilde{M}.$ But $\omega =0$ on M,  so that we have induced on $M^{T}$ a Kaehlerian structure. Moreover, it can be easily seen that the projection map p (resp., q) onto $\mathcal {D}$ (resp., $\mathcal {D}^{\perp }$) is parallel with respect to $\nabla ,$ so that this local product is actually a local Riemannian product.    $\blacksquare$

Next, we consider the particular case when M is either holomorphic or totally real.

Lemma 3

Let M be a holomorphic submanifold of a l.c.K. manifold $\widetilde{M}.$ Then the subbundles TM and $T^{\perp }(M)$ are holomorphic. Moreover, we have

$$\begin{aligned} \overline{\sigma }(JX,Y)=\overline{\sigma }(X,JY)= J\overline{\sigma }(X,Y), \end{aligned}$$

(5.24)

$$\begin{aligned} \overline{A}_{JV}=J\overline{A}_{V}=-\overline{A}_{V}J, \end{aligned}$$

(5.25)

$$\begin{aligned} \overline{D}_X(JV)=J\overline{D}_XV, \end{aligned}$$

(5.26)

$$\begin{aligned} \nabla ^M_X(JY)=J\nabla ^M_XY, \end{aligned}$$

(5.27)

for any vector fields X, Y tangent to M and V normal to M.

Proof

As $\widetilde{M}$ is locally endowed with Kaehler metrics $\widetilde{g}_U$ whose Levi-Civita connection is $\overline{\nabla },$ these formulas follow from similar formulas in the Kaehlerian case.    $\blacksquare$

Theorem 5.9

Let M be a holomorphic submanifold of a l.c.K. manifold $\widetilde{M}.$ Then, we have

• (i) The mean curvature vector H of M is given by $H=-\frac{1}{2}B_2$.

• (ii) M is totally umbilical if and only if the Weingarten endomorphisms are commutative.

Proof

Firstly, if $dim(M)=2k>0,$ let $\left\{ e_1,\ldots ,e_k,Je_1,\ldots ,Je_k \right\}$ be an orthonormal basis for $T_x(M),$ $x\in M.$ Then

$$\begin{aligned} 2kH_x=(tr(\sigma ))_x=\sum _{i=1}^k\sigma _x(e_i,e_i)+\sum _{i=1}^k\sigma _x(Je_i,Je_i). \end{aligned}$$

(5.28)

But from (5.8) and (5.24), (5.28) gives $2kH_x=-k(B_2)_x.$

Next, let V be a vector field normal to M. From (5.17) and (5.25), we have

$$0=[A_{V},A_{JV}]=[\overline{A}_{V},\overline{A}_{JV}]= -2J(\overline{A}_{V})^2,$$

Thus $\overline{A}_{V}=0$ and from (5.9), we have $A_{V}=-\frac{1}{2}\omega (V)I$    $\blacksquare$

The endomorphism n of the normal bundle $T^{\perp }M$ defined in (5.5) induces an f-structure in $T^{\perp }M$ [22]. For any vector field X tangent to M and V normal to M, we write

$$\begin{aligned} (\widetilde{\nabla }^{\, \prime }_Xn)V = D_X(nV) - nD_XV, \end{aligned}$$

$$\begin{aligned} (\overline{\nabla }^{\, \prime }_Xn)V = \overline{D}_X(nV) - n\overline{D}_XV. \end{aligned}$$

When $\widetilde{\nabla }^{\, \prime }n = 0,$ the f-structure n is said to be parallel [10].

Lemma 4

Let M be an r-dimensional totally real submanifold of a 2m-dimensional l.c.K. manifold $\widetilde{M}$. Then we have

• (i) $\overline{A}_{JX}Y=\overline{A}_{JY}X$, for any X, Y tangent to M.

• (ii) If $r=m,$ then $\overline{D}_X(JY)=J\nabla ^{M}_XY$, $\nabla ^M_X(JV)=J\overline{D}_XV$, and $\overline{\sigma }(X,JV)=-J\overline{A}_{V}X.$

• (iii) $\widetilde{\nabla }^{\,\prime }n=\overline{\nabla }^{\prime }n$.

• (iv) If the f-structure n is parallel, then

  $$\begin{aligned} A_{V}=-\frac{1}{2}\omega (V)I, \end{aligned}$$

  (5.29)

  for any $V \in \nu$.

• (v) If the Weingarten endomorphisms are commutative, then there is an orthonormal local basis $\left\{ e_1,\ldots ,e_r \right\}$ in M such that with respect to this basis $\overline{A}_{Je_i}$ is a diagonal matrix

  $$\begin{aligned} \overline{A}_{Je_i}=\left( 0\ldots 0 \, \lambda _i \, 0\ldots 0 \right) , \quad i=1,\ldots ,r. \end{aligned}$$

  (5.30)

Proof

Assertions (i) and (ii) follow immediately from similar formulas in the Kaehlerian case. From Eq. (5.10), we easily obtain (iii).

In order to prove (iv), we take $V \in \nu ,$ and $X \in T(M).$ Then, (iii) gives $(\nabla ^{\prime }_Xn)V= \overline{D}_X(nV)-n\overline{D}_XV =0.$ By using (5.25) and (5.26) this yields $J\overline{A}_{V}X=0.$ Therefore, $\overline{A}_{V}=0$ and from (5.9), we obtain (iv).

Finally, from (5.17) we have $[\overline{A}_{V},\overline{A}_{\overline{V}}]=0$, for any $V,\overline{V}$ normal to M. Then, we can find a local orthonormal basis $\left\{ \widetilde{e}_1\ldots ,\widetilde{e}_r \right\}$ in M (with respect to the local Kaehlerian metrics $\widetilde{g}_U=e^{-h}g|_U$) such that $\overline{A}_{Je_i} = \left( 0\ldots \mu _i\ldots 0 \right)$, $i=1,\ldots ,r.$ If we start by using this basis, we can obtain an orthonormal (with respect to the metric g) local basis $\left\{ e_1,\ldots ,e_n\right\}$ in M such that (v) holds.    $\blacksquare$

Theorem 5.10

Let M be an r-dimensional totally real and minimal submanifold of a l.c.K. manifold $\widetilde{M}$ such that their Weingarten endomorphisms are commutative and the f-structure n is parallel. Then, we have

• (i) If $r\ge 2$, M is totally geodesic if and only if the Lee vector field B is tangent to M.

• (ii) If $r=1$ and B is orthogonal to $\nu$, then M is a geodesic curve.

Proof

First, since the Weingarten endomorphisms are commutative, let $\left\{ e_1,\ldots ,e_r \right\}$ be an orthonormal local basis as in Lemma 4 (v). From Eq. (5.8), we have

$$\begin{aligned} \begin{aligned} 0=g(H,Je_i)=&\frac{1}{n}\sum _{j=1}^ng\left( \sigma (e_j,e_j),Je_i \right) =\frac{1}{n}\sum _{j=1}^ng\left( \overline{A}_{Je_i}e_j,e_j \right) - \frac{1}{2} \omega (Je_i)\\ =&\frac{1}{n} \lambda _i - \frac{1}{2} \omega (Je_i), \quad i=1,\ldots ,r. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \overline{A}_{Je_i}e_j=\delta _{ij}\lambda _ie_j=\delta _{ij}\frac{n}{2}\omega (Je_i)e_j, \quad i=1,\ldots ,r. \end{aligned}$$

(5.31)

Now, from (5.9) and (5.31) we obtain,

$$\begin{aligned} A_{Je_i}e_j=\frac{1}{2}(n \delta _{ij}-1)\omega (Je_i)e_j, \quad i=1,\ldots ,r. \end{aligned}$$

(5.32)

Thus, if $r\ge 2$ and B is tangent to M, Eq. (5.32) gives $A_{Je_i}=0$, $i=1,\ldots ,r.$ Moreover, from (iv) in Lemma 4, $A_{V}=-\frac{1}{2}\omega (V)I=0$, for any $V \in \nu .$ Then, $A_{\overline{V}}=0$, for any vector field $\overline{V}$ normal to M.

On the other hand, if there is $x\in M$ such that $(B_2)_x\ne 0,$ from (5.32) and (iv) in Lemma 4, we can take a vector field V normal to M such that $A_{V}\ne 0.$ This gives (i).

In order to prove (ii), let us take a unit vector field X tangent to M. We have $0=g(H,JX)=g(\sigma (X,X),JX)=g(A_{JX}X,X),$ and, then, $A_{JX}=0.$ But, if B is orthogonal to $\nu ,$ from (iv) in Lemma 4, $A_{V}=-\frac{1}{2}\omega (V)=0$, for any $V \in \nu .$ This means that $A_{\overline{V}}=0$, for any vector field $\overline{V}$ normal to M.    $\blacksquare$

Theorem 5.11

Let M be an r-dimensional ($r\ge 2$) totally real and totally umbilical submanifold of a l.c.K. manifold $\widetilde{M}$ such that the f-structure n is parallel. Then M is totally geodesic if and only if B is tangent to M.

Proof

Let $\left\{ u_1,\ldots ,u_r\right\}$ be an orthonormal local basis in U. Since M is totally umbilical, for any vector field X tangent to M,  by using Eqs. (5.8) and (5.9), we find

$$\begin{aligned} g(\overline{A}_{JX}u_j,u_k)=\frac{1}{r}\delta _{jk}tr(\overline{A}_{JX}). \end{aligned}$$

(5.33)

But from Eq. (5.9) and (iv) in Lemma 4 we also have

$$\begin{aligned} \overline{A}_{V}=0, \end{aligned}$$

(5.34)

for any $V\in \nu .$ On the other hand $[A_{\overline{V}},A_{\overline{\overline{V}}}]= [ \overline{A}_{\overline{V}},\overline{A}_{\overline{\overline{V}}}]=0,$ for any vector fields $\overline{V},\overline{\overline{V}}$ normal to M. Therefore, from (v) in Lemma 4, there is an orthonormal local basis $\{e_1,\ldots ,e_r\}$ in M such that, with respect to this basis, Eq. (5.30) holds. But, from Eq. (5.33), we also have

$$\begin{aligned} g(\overline{A}_{Je_i}e_j,e_j)=\frac{1}{r}\lambda _i, \quad i.j=1,\ldots ,n. \end{aligned}$$

(5.35)

Since $r\ge 2$, we can take $j\ne i$ and then, Eqs. (5.30) and (5.35) give $\lambda _i=0$, $i=1,\ldots ,r.$ Thus, we get $\overline{A}_{Je_i}=0$, $i=1,\ldots ,r$, which, together with (5.34) gives $\overline{A}_{\overline{V}}=0$, for any vector field $\overline{V}$ normal to M. Now, if B is tangent to M,  Eq. (5.9) proves that M is totally geodesic.

Conversely, if M is totally geodesic, from (5.29) we have $0=A_{V}=-\frac{1}{2}\omega (V)I$, for any $V \in \nu .$ This means that B is normal to $\nu .$ Furthermore, from (v) in Lemma 4, we can find an orthonormal local basis $\left\{ e_1,\ldots ,e_r \right\}$ in M such that $\overline{A}_{Je_i}$ has a diagonal matrix $\overline{A}_{Je_i} = \left( 0\ldots 0 \, \lambda _i \,0\ldots 0 \right) =\frac{1}{2}\omega (Je_i)I$, $i=1,\ldots ,r.$ Since $r\ge 2$, this means $\omega (Je_i)=0$, $i=1,\ldots ,r$ so that B is normal to J(T(M)). Thus, B is tangent to M.    $\blacksquare$

5.2.4 CR-products in l.c.K. Manifolds

Let T, N, t, n be the endomorphisms and vector-valued 1-forms defined by (5.5). Let us write

$$\begin{aligned} \begin{aligned} (\widetilde{\nabla }^{\; \prime }_ZT)W&=\nabla _Z(TW)-T\nabla _ZW, \\ (\overline{\nabla }^{\, \prime }_ZT)W&=\nabla ^{M}_Z(TW)-T\nabla ^M_ZW, \end{aligned} \end{aligned}$$

(5.36)

for all Z, W tangent to M. On the other hand, T is said to be parallel if $\overline{\nabla }^{\,}T=0.$ From (5.1)–(5.3) it is easy to prove that

$$\begin{aligned} (\overline{\nabla }^{\, \prime }_ZT)W&=(\widetilde{\nabla }^{\, \prime }_ZT)W+\frac{1}{2}\omega (W)TZ-\frac{1}{2}\omega (TW)Z \nonumber \\&\quad +\frac{1}{2}g(Z,TW)B_1-\frac{1}{2}g(Z,W)TB_1. \end{aligned}$$

(5.37)

But, from (5.36) we see that

$$\begin{aligned} (\overline{\nabla }^{\, \prime }_Z T)W=t\overline{\sigma }(Z,W) + \overline{A}_{NW}Z. \end{aligned}$$

(5.38)

Definition 1

A CR-submanifold of a l.c.K. manifold $\widetilde{M}$ is called a CR-product if it is locally a Riemannian product of a holomorphic submanifold $M^{T}$ and a totally real submanifold $M^{\perp }$ of $\widetilde{M}.$

Theorem 5.12

Let M be a CR-submanifold of a l.c.K. manifold $\widetilde{M}$ such that the Lee field B is normal to M. Then M is a CR-product if and only if T is parallel.

Proof

Since B is normal to M,  from Eq. (5.37), we have $\widetilde{\nabla }^{\, \prime }T=\overline{\nabla }^{\, \prime }T.$ If T is parallel, from (5.8), (5.9), and (5.38), we find

$$\begin{aligned} t\sigma (Z,W) + \frac{1}{2}g(Z,W)tB = - A_{NW}Z-\frac{1}{2}\omega (NW)Z. \end{aligned}$$

(5.39)

But for any $X\in \mathcal {D}$, $NX=0,$ and the last equation gives

$$0=g(A_{NW}Z,X)+\frac{1}{2}\omega (NW)g(Z,X),$$

or, equivalently, $g(\sigma (Z,X),JW)+\frac{1}{2}g(JW,B)g(Z,X)=0$, for any W tangent to M. Therefore,

$$\begin{aligned} \sigma (Z,X)=-\frac{1}{2} g(Z,X)B. \end{aligned}$$

(5.40)

If we take $Z\in \mathcal {D}$, the last equation gives $\sigma (X,JY)-\sigma (JX,Y)=-\Omega (X,Y)B,$ and, from Theorem 5.3, $\mathcal {D}$ is integrable. Let $M^{T}$ be an integral submanifold of $\mathcal {D}.$ For any $Z\in \mathcal {D}^{\perp }$, Eq. (5.40) yields $g(\sigma (Z,X),JZ)=-\frac{1}{2}g(Z,Z)g(B,JZ)$ and, from Theorem 5.6, the submanifold $M^{T}$ is totally geodesic in M. Now, let $M^{\perp }$ be an integral submanifold od $\mathcal {D}^{\perp }.$ From (5.40), if $Z\in \mathcal {D}^{\perp }$, then $\sigma (Z,X)=0$ and, from Theorem 5.5, $M^{\perp }$ is totally geodesic.

Conversely, assume that M is a CR-product. First, we prove that $\nabla ^{M}_ZX \in \mathcal {D}$, for any $X\in \mathcal {D}$ and Z tangent to M. As M is locally a Riemannian product of $M^{T}$ (holomorphic submanifold) and $M^{\perp }$ (totally real submanifold), it suffices to prove that $\nabla ^M_ZX \in \mathcal {D}$, for any $X\in \mathcal {D}$ and $Z\in \mathcal {D}^{\perp }.$ In fact, from (5.3) we have

$$J\nabla ^{M}_ZX=\nabla ^M_Z(JX)+\overline{\sigma }(Z,JX)-J\overline{\sigma }(Z,X).$$

Thus, if $W\in \mathcal {D}^{\perp }$, $g(J\nabla ^M_ZX,JW)=g(\overline{\sigma }(Z,JX),JW).$ Since $M^{\perp }$ is totally geodesic in M,  from (5.8) and Theorem 5.5 we have $g(\nabla ^M_ZX,W)=0$, for any $W\in \mathcal {D}^{\perp }.$ So, $\nabla ^M_ZX \in \mathcal {D}$ and $\nabla ^M_ZX \in \mathcal {D}$, for any Z tangent to M. From $\overline{\nabla }J=0$, we find

$$J\nabla ^M_ZX+J\overline{\sigma }(Z,X)=\nabla ^M_Z(JX)+\overline{\sigma }(Z,JX),$$

and then, $J\nabla ^M_ZX=\nabla ^M_Z(JX)$, $J\overline{\sigma }(Z,X)=\overline{\sigma }(Z,JX).$ Now, from (5.36) we get

$$\begin{aligned} (\overline{\nabla }^{\, \prime }_ZT)X=\nabla ^M_Z(TX)-T\nabla ^M_ZX = \nabla ^M_Z(JX)-J\nabla ^M_ZX=0, \end{aligned}$$

(5.41)

for any $X\in \mathcal {D}$ and Z tangent to M.

In a similar way, we prove that $\nabla ^M_ZZ \in \mathcal {D}^{\perp }$ for any $Z\in \mathcal {D}^{\perp }$ and Z tangent to M. Since M is a CR-product, it suffices to show this for $Z=X\in \mathcal {D}.$ In fact, from (5.3), given any $Y\in \mathcal {D}$ we find that

$$g(J\nabla ^M_XZ,Y)=-g(\overline{A}_{JZ}X,Y)-g(J\overline{\sigma }(X,Z),Y)= -g(\overline{\sigma }(X,Y),JZ)=0,$$

where the last equation holds from (5.8) and Theorem 5.6. Then, $J\nabla ^M_XZ$ is orthogonal to $\mathcal {D}.$ On the other hand, if $W\in \mathcal {D}^{\perp }$, we have

$$g\left( \nabla ^M_XZ,W\right) =-g(\overline{\sigma }(X,W),JZ)+g(\overline{\sigma }(X,Z),JW).$$

But, from Theorem 5.5 we have $g(J\nabla ^M_XZ,W)=0,$ That is, $J\nabla ^M_XZ$ is normal to M,  so that $\nabla ^M_XZ\in \mathcal {D}^{\perp }.$ Therefore, we have

$$\begin{aligned} (\overline{\nabla }^{\, \prime }_ZT)Z=\nabla ^M_Z(TZ)-T\nabla ^M_ZZ=0. \end{aligned}$$

(5.42)

Now, from (5.37), (5.41), and (5.42), we have $\widetilde{\nabla }^{\, \prime }T=0.$    $\blacksquare$

Theorem 5.13

Let M be a CR-submanifold of a l.c.K. manifold $\widetilde{M}$ such that $B_x\in \mathcal {D}_x$ for each $x\in M.$ If T is parallel, then M is a CR-product. The converse does not holds unless $dim(\mathcal {D})=2$ or $B=0$ on M.

Proof

Since T is parallel, Eqs. (5.37) and (5.38) give

$$\begin{aligned} t\overline{\sigma }(Z,W)+\overline{A}_{NW}Z&=\frac{1}{2}\omega (W)TZ- \frac{1}{2}\omega (TW)Z \nonumber \\&\quad +\frac{1}{2}g(Z,TW)B-\frac{1}{2}g(Z,W)TB. \end{aligned}$$

(5.43)

If $X\in \mathcal {D}$, then $NX=0$ and (5.43) gives

$$\begin{aligned} -g(J\overline{\sigma }(X,Z),W)&=\frac{1}{2}g(B,W)g(JZ,X)+\frac{1}{2}g(W,JB)g(Z,X) \nonumber \\&\quad -\frac{1}{2}g(JZ,W)g(B,X)-\frac{1}{2}g(Z,W)g(JB,X), \end{aligned}$$

(5.44)

for any vector field W tangent to M. From (5.8), (5.44) yields

$$\begin{aligned} -J\sigma (X,Z)&=\frac{1}{2}g(JZ,X)B+\frac{1}{2}g(Z,X)JB \nonumber \\&\quad - \frac{1}{2}g(B,X)JZ-\frac{1}{2}g(JB,X)Z. \end{aligned}$$

(5.45)

For any $Z\in \mathcal {D}^{\perp },$ Eq. (5.45) gives $g(A_{JZ},Z)=\frac{1}{2}\omega (JX)g(Z,Z)$, for any Z tangent to M and, hence, we have

$$\begin{aligned} A_{JZ}X=\frac{1}{2}\omega (JX)Z. \end{aligned}$$

(5.46)

Next, for $Y\in \mathcal {D},$ from (5.46), we have

$$\begin{aligned} g(\sigma (X,Y),JZ)=0, \quad for \; X \in \mathcal {D}, \, Z\in \mathcal {D}^{\perp }. \end{aligned}$$

(5.47)

Therefore, $g(\sigma (X,JY)-\sigma (JX,Y,J\mathcal {D}^{\perp })=0$ and, from Theorem 5.4, the distribution $\mathcal {D}$ is integrable. Moreover, any integral submanifold $M^{\perp }$ od $\mathcal {D}$ is totally geodesic in M because of (5.47) and Theorem 5.6. Now, let $M^{\perp }$ be an integral submanifold of $\mathcal {D}^{\perp }.$ For any $W\in \mathcal {D}^{\perp },$ Eq. (5.46) gives

$$g\left( A_{JZ}+\frac{1}{2}g(Z,W)JB,X\right) =0$$

and this means that $M^{\perp }$ is totally geodesic in M (Theorem 5.5). Thus M is a CR-product.    $\blacksquare$

In order to prove the converse, we first give the following Lemma.

Lemma 5

If M is a CR-product in a l.c.K. manifold $\widetilde{M}$ such that $B_x\in \mathcal {D}_x$ for any $x\in M,$ then

$$\begin{aligned} \nabla _ZX\in \mathcal {D}, \end{aligned}$$

(5.48)

$$\begin{aligned} \nabla _XZ\in \mathcal {D}^{\perp }, \end{aligned}$$

(5.49)

$$\begin{aligned} J\, \nabla _ZX=\nabla _Z(JX), \end{aligned}$$

(5.50)

for any $X\in \mathcal {D}$ and $Z\in \mathcal {D}^{\perp }.$

Proof

If $X\in \mathcal {D}$ and $Z\in \mathcal {D}^{\perp },$ then from (5.7) and (5.8), we obtain

$$\begin{aligned} J\,\nabla _ZX&=\frac{1}{2}\omega (X)JZ-J\sigma (Z,X)+\nabla _Z(JX)+\nabla _Z(JX) \nonumber \\&\quad -\frac{1}{2}\omega (JX)Z+\sigma (Z,JX). \end{aligned}$$

(5.51)

Now, for any $W\in \mathcal {D}^{\perp }$, (5.51) yields

$$g(J\, \nabla _ZX,JW) = g(\nabla _ZX,W) = g\left( A_ {JW}Z+\frac{1}{2}g(Z,W)JB,JX\right) =0.$$

The last equation holds because any leaf $M^{\perp }$ of $\mathcal {D}^{\perp }$ is totally geodesic in M (Theorem 5.5). Thus $\nabla _ZX \in \mathcal {D}$ and this is assertion (5.48). Now, take $X,Y \in \mathcal {D}$ and $Z\in \mathcal {D}^{\perp }.$ From (5.1) and (5.2), we find that

$$\begin{aligned} g(J\,\nabla _XZ,Y)=-g(A_{JZ},Y)=-g(\sigma (X,Y),JZ)=0 \end{aligned}$$

(5.52)

The last equation holds because of Theorem 5.6. If $X\in \mathcal {D}$ and $Z,W\in \mathcal {D}^{\perp },$ from (5.1) and (5.2) again we have

$$g(J\, \nabla _XZ,W)=g(A_{JW}Z,X)-g(A_{JZ}W,X).$$

But, from Theorem 5.5 we obtain

$$\begin{aligned} g(A_{JW}Z,X)-g(A_{JZ}W,X)= -\frac{1}{2}g(Z,W)g(JB,X)+\frac{1}{2}g(W,Z)g(JB,X) =0 \end{aligned}$$

and, hence

$$\begin{aligned} g(J\, \nabla _XZ,W)=0. \end{aligned}$$

(5.53)

Now, (5.49) follows from (5.52) and (5.53). Finally, (5.48) and (5.51) give (5.50).    $\blacksquare$

Now we prove the converse of Theorem 5.13. From (5.36) and (5.48), for any $X\in \mathcal {D}$ and Z tangent to M we have

$$\begin{aligned} (\widetilde{\nabla }^{\; \, \prime }_Z,T)X = \nabla _Z(JX)-J(\nabla _ZX). \end{aligned}$$

On the other hand, we write $Z=Y+Z$, where $Y\in \mathcal {D}$ and $Z\in \mathcal {D}^{\perp }.$ Then, from (5.50) we have

$$\begin{aligned} (\widetilde{\nabla }^{\; \, \prime }_ZT)X=\nabla _Y(JX)-J\, \nabla _YX. \end{aligned}$$

(5.54)

But (5.1)–(5.3) give

$$\begin{aligned} \nabla _Y(JX)-J\nabla _YX = \frac{1}{2}\omega (Y)JX-\frac{1}{2}(JY)X-\frac{1}{2}g(X,Y)JB+\frac{1}{2}g(X,JY)B. \end{aligned}$$

(5.55)

Now we have

(a) If $dim(\mathcal {D})\ge 4$ and $B_x \ne 0$ for some $x\in M,$ there are $X,Y \in \mathcal {D}$ such that the right-hand side of (5.55) does not vanish at x. Therefore, T is not parallel.

(b) If $dim(\mathcal {D})=2$, then the right-hand side of (5.55) vanishes and, hence

$$\begin{aligned} (\widetilde{\nabla }^{\; \prime }_ZT)X=0, \end{aligned}$$

(5.56)

for any $X\in \mathcal {D}$ and Z tangent to M. But (5.49) implies $\nabla _ZZ \in \mathcal {D}^{\perp }$, for any $Z\in \mathcal {D}^{\perp }$ and Z tangent to M,  so that

$$\begin{aligned} (\widetilde{\nabla }^{\; \prime }_ZT)Z=\nabla _Z(TZ)-T\nabla _ZZ=-T\nabla _ZZ=0. \end{aligned}$$

(5.57)

Then, (5.56) and (5.57) prove that T is parallel.   $\blacksquare$

5.3 Normal CR-Submanifolds of S-manifolds

We want to study here the normal CR-submanifolds for general S-manifolds. In fact, the normal CR-submanifolds become to be a very wide class of CR-submanifolds. Actually, either totally f-umbilical submanifolds (see [19] for more details) or CR-products (see [12]) of an S-manifold are normal CR-submanifolds. We also study normal CR-submanifolds of an S-space-form, specially in the concrete cases of $\mathbb R^{2m+s}$ (with constant f-sectional curvature $c=-3s$) and $\mathbb H^{2m+s}$ (with constant f-sectional curvature $c=4-3s$).

5.3.1 Preliminaries

A $(2m+s)$-dimensional Riemannian manifold $(\wedge M,g)$ endowed with an f-structure f (that is, a tensor field of type (1, 1) and rank 2m satisfying $f^3+f=0$ [21]) is said to be a metric f-manifold if, moreover, there exist s global vector fields $\xi _1,\dots ,\xi _s$ on $\wedge M$ (called structure vector fields) such that, if $\eta _1,\dots ,\eta _s$ are the dual 1-forms of $\xi _1,\dots ,\xi _s$, then

$$f\xi _{\alpha }=0; \eta _{\alpha }\circ f=0; f^2=-I+\sum _{{\alpha }=1}^s\eta _{\alpha }\otimes \xi _{\alpha };$$

$$\begin{aligned} g(X,Y)=g(fX,fY)+\sum _{{\alpha }=1}^s\eta _{\alpha }(X)\eta _{\alpha }(Y),\end{aligned}$$

(5.58)

for any $X,Y\in \mathcal {X}(\wedge M)$ and ${\alpha }=1,\dots ,s$.

Let F be the 2-form on $\wedge M$ defined by $F(X,Y)=g(X,fY)$, for any $X,Y\in \mathcal {X}(\wedge M)$. Since f is of rank 2m, then

$$\eta _1\wedge \cdots \wedge \eta _s\wedge F^m\ne 0$$

and, particularly, $\wedge M$ is orientable.

The f-structure f is said to be normal if

$$[f,f]+2\sum _{{\alpha }=1}^s\xi _{\alpha }\otimes d\eta _{\alpha }=0,$$

where [f, f] is the Nijenhuis torsion of f.

A metric f-manifold is said to be a K-manifold [5] if it is normal and $\mathrm{d}F=0$. A K-manifold is called an S-manifold if $F=\mathrm{d}\eta _{\alpha }$, for any ${\alpha }$. Note that, for $s=0$, a K-manifold is a Kaehlerian manifold and, for $s=1$, a K-manifold is a quasi-Sasakian manifold and an S-manifold is a Sasakian manifold. When $s\ge 2$, nontrivial examples can be found in [5, 13]. Moreover, a K-manifold $\wedge M$ is an S-manifold if and only if

$$\begin{aligned} \wedge \nabla _X\xi _{\alpha }=-fX, \end{aligned}$$

(5.59)

for any $X\in \mathcal {X}(\wedge M)$ and any ${\alpha }=1,\dots ,s$, where $\wedge \nabla$ denotes the Levi-Civita connection of g. It is easy to show that in any S-manifold

$$\begin{aligned} (\wedge \nabla _Xf)Y=\sum _{{\alpha }=1}^s\left\{ g(fX,fY)\xi _{\alpha }+\eta _{\alpha }(Y)f^2X\right\} , \end{aligned}$$

(5.60)

for any $X,Y\in \mathcal {X}(\wedge M)$. A plane section $\pi$ on a metric f-manifold $\wedge M$ is said to be an f-section if it is determined by a unit vector X, normal to the structure vector fields and fX. The sectional curvature of $\pi$ is called an f-sectional curvature. An S-manifold is said to be an S-space-form if it has a constant f-sectional curvature c and then, it is denoted by $\wedge M(c)$. In such case, the curvature tensor field $\wedge R$ of $\wedge M(c)$ satisfies [15]

$$\begin{aligned}&\wedge R(X,Y,Z,W) \nonumber \\&\quad = \sum _{{\alpha },{\beta }}(g(fX,fW)\eta _{\alpha }(Y)\eta _{\beta }(Z)-g(fX,fZ)\eta _{\alpha }(Y)\eta _{\beta }(W) \nonumber \\&\qquad +g(fY,fZ)\eta _{\alpha }(X)\eta _{\beta }(W)-g(fY,fW)\eta _{\alpha }(X)\eta _{\beta }(Z))\nonumber \\&\qquad +\frac{c+3s}{4}(g(fX,fW)g(fY,fZ)-g(fX,fZ)g(fY,fW))\nonumber \\&\qquad +\frac{c-s}{4}(F(X,W)F(Y,Z)-F(X,Z)F(Y,W)\nonumber \\&\qquad \qquad \quad \, - 2F(X,Y)F(Z,W)), \end{aligned}$$

(5.61)

for any $X,Y,Z,W\in \mathcal {X}(\wedge M)$. Next, let M be a isometrically immersed submanifold of a metric f-manifold $\wedge M$ (for the general theory of submanifolds, we refer to [3, 22]). We denote by $\mathcal {X}(M)$ the Lie algebra of tangent vector fields to M and by $T(M)^\perp$ the set of tangent vector fields to $\wedge M$ which are normal to M. For any vector field $X\in \mathcal {X}(M)$, we write

$$\begin{aligned} fX=TX+NX, \end{aligned}$$

(5.62)

where TX and NX are the tangential and normal components of fX, respectively. Then, T is an endomorphism of the tangent bundle of M and N is a normal bundle valued 1-form on such tangent bundle. It is easy to show that if T does not vanish, it defines an f-structure in the tangent bundle of M. The submanifold M is said to be invariant if N is identically zero, that is, if fX is tangent to M, for any $X\in \mathcal {X}(M)$. On the other hand, M is said to be an anti-invariant submanifold if T is identically zero, that is, if fX is normal to M, for any $X\in \mathcal {X}(M)$. In the same way, for any $V\in T(M)^\perp$, we write

$$\begin{aligned} fV=tV+nV, \end{aligned}$$

(5.63)

where tV and nV are the tangential and normal components of fV, respectively. Then, t is a tangent bundle valued 1-form on the normal bundle of M and n is an endomorphism of the normal bundle of M. It is easy to show that if n does nor vanish, it defines an f-structure in the normal bundle of M. From now on, we suppose that all the structure vector fields are tangent to the submanifold M and so, $dim(M)\ge s$. Then, the distribution on M spanned by the structure vector fields is denoted by $\mathcal {M}$ and its complementary orthogonal distribution is denoted by $\mathcal {L}$. Consequently, if $X\in \mathcal {L}$, then $\eta _{\alpha }(X)=0$, for any ${\alpha }=1,\dots ,s$ and if $X\in \mathcal {M}$, then $fX=0$. In this context, M is said to be a CR-submanifold of $\wedge M$ if there exist two differentiable distributions $\mathcal {D}$ and $\mathcal {D}^\perp$ on M satisfying

• (i) $\mathcal {X}(M)=\mathcal {D}\oplus \mathcal {D}^\perp \oplus \mathcal {M}$, where $\mathcal {D}$, $\mathcal {D}^\perp$ and $\mathcal {M}$ are mutually orthogonal to each other;

• (ii) The distribution $\mathcal {D}$ is invariant by f, that is, $f\mathcal {D}_x=\mathcal {D}_x$, for any $x\in M$;

• (iii) The distribution $\mathcal {D}^\perp$ is anti-invariant by f, that is, $f\mathcal {D}^\perp _x\subseteq T_x(M)^\perp$, for any $x\in M$.

This definition is motivated by the following theorem.

Theorem 5.14

([16]) Let $\wedge M$ be an S-manifold which is the bundle space of a principal toroidal bundle over a Kaehler manifold $\wedge M'$, $\wedge \pi :\wedge M\longrightarrow \wedge M'$, M a submanifold immersed in $\wedge M$, tangent to the structure vector fields and $M'$ a submanifold immersed in $\wedge M'$ such that there exists a fibration $\pi :M\longrightarrow M'$, the diagram

commutes and the immersion i is a diffeomorphism on the fibers. Then, M is a CR-submanifold of $\wedge M$ if and only if $M'$ is a CR-submanifold of $\wedge M'$.

We denote by 2p and q the real dimensions of $\mathcal {D}$ and $\mathcal {D}^\perp$, respectively. Then, we see that for $p=0$ we obtain an anti-invariant submanifold tangent to the structure vector fields and for $q=0$ an invariant submanifold. A CR-submanifold of an S-manifold is said to be a generic submanifold if given any $V\in T(M)^\perp$, there exists $Z\in \mathcal {D}^\perp$ such that $V=fZ$, a $(\mathcal {D},\mathcal {D}^\perp )$ -geodesic submanifold if $\sigma (X,Z)=0$, for any $X\in \mathcal {D}$ and any $Z\in \mathcal {D}^\perp$ and a $\mathcal {D}^\perp$ -geodesic submanifold if $\sigma (Y,Z)=0$, for any $Y,Z\in \mathcal {D}^\perp$. As an example, it is easy to show that each hypersurface of $\wedge M$ which is tangent to the structure vector fields is a CR-submanifold. Now, we write by P and Q the projections morphisms of $\mathcal {X}(M)$ on $\mathcal {D}$ and $\mathcal {D}^\perp$, respectively. Thus, for any $X\in \mathcal {X}(M)$, we have that

$$X=PX+QX+\sum _{{\alpha }=1}^s\eta _{\alpha }(X)\xi _{\alpha }.$$

We define the tensor field v of type (1, 1) by $vX=fPX$ and the non-null, normal bundle valued 1-form u by $uX=fQX$, for any $X\in \mathcal {X}(M)$. Then, it is easy to show that $u\circ v=0$ and $\eta _{\alpha }\circ u=\eta _{\alpha }\circ v=0$, for any ${\alpha }=1,\dots ,s$. Moreover, a direct computation gives

$$\begin{aligned} g(X,Y)=g(uX,uY)+g(vX,vY)+\sum _{{\alpha }=1}^s\eta _{\alpha }(X)\eta _{\alpha }(Y), \end{aligned}$$

(5.64)

$$\begin{aligned} F(X,Y)=g(X,vY), \quad F(X,Y)=F(vX,vY), \end{aligned}$$

(5.65)

for any $X,Y\in \mathcal {X}(M)$. From Gauss–Weingarten formulas and by using (5.59), for any $X\in \mathcal {X}(M)$, $V\in T(M)^\perp$, and ${\alpha }=1,\dots ,s$, we have

$$\begin{aligned} \nabla _X\xi _{\alpha }=-vX, \end{aligned}$$

(5.66)

$$\begin{aligned} \sigma (X,\xi _{\alpha })=-uX, \end{aligned}$$

(5.67)

$$\begin{aligned} A_V\xi _{\alpha }\in \mathcal {D}^\perp . \end{aligned}$$

(5.68)

Moreover, from (5.60) and the Gauss–Weingarten formulas, if $X,Y\in \mathcal {X}(M)$, comparing the components in $\mathcal {D}$, $\mathcal {D}^\perp$ and $T(M)^\perp$ respectively, we get

$$\begin{aligned} P\nabla _XvY-PA_{uY}X=v\nabla _XY-\sum _{{\alpha }=1}^s\eta _{\alpha }(Y)PX, \end{aligned}$$

(5.69)

$$\begin{aligned} Q\nabla _XvY-QA_{uY}X=t\sigma (X,Y)-\sum _{{\alpha }=1}^s\eta _{\alpha }(Y)QX, \end{aligned}$$

(5.70)

$$\begin{aligned} \sigma (X,vY)+D_XuY=u\nabla _XY+n\sigma (X,Y). \end{aligned}$$

(5.71)

From the above formulas and (5.60) we obtain

$$\begin{aligned} (\nabla _Xv)Y=A_{uY}X+t\sigma (X,Y)-\sum _{a=1}^s\{\eta _{\alpha }(Y)f^2X+g(fX,fY)\xi _{\alpha }\}, \end{aligned}$$

(5.72)

$$\begin{aligned} (\nabla _Xu)Y=n\sigma (X,Y)-\sigma (X,vY), \end{aligned}$$

(5.73)

for any $X,Y\in \mathcal {X}(M)$. Also, from (5.60) and the Gauss–Weingarten formulas again, we have

$$\begin{aligned} \nabla _XZ=vA_{fZ}X-tD_XfZ, \end{aligned}$$

(5.74)

$$\begin{aligned} tD_XfZ=-Q\nabla _XZ, \end{aligned}$$

(5.75)

for any $X\in \mathcal {X}(M)$ and any $Z\in \mathcal {D}^\perp$. With regard to the integrability of the distributions involved in the definition of a CR-submanifold, I. Mihai [16] proved that the distributions $\mathcal {D}^\perp$ and $\mathcal {D}^\perp \oplus \mathcal {M}$ are always integrable. On the other hand, if $p>0$, the distributions $\mathcal {D}$ and $\mathcal {D}\oplus \mathcal {D}^\perp$ are not integrable and the distribution $\mathcal {D}\oplus \mathcal {M}$ is integrable if and only if

$$\begin{aligned} \sigma (X,fY)=\sigma (fX,Y), \end{aligned}$$

(5.76)

for any $X,Y\in \mathcal {D}$. In [12], CR-products of S-manifolds are defined as CR-submanifolds such that the distribution $\mathcal {D}\oplus \mathcal {M}$ is integrable and locally they are Riemannian products $M_1\times M_2$, where $M_1$ (resp., $M_2$) is a leaf of $\mathcal {D}\oplus \mathcal {M}$ (resp., $\mathcal {D}^\perp$). From Theorem 3.1 and Proposition 3.2 in [12], we know that a CR-submanifold M of an S-manifold is a CR-product if and only if one of the following assertions is satisfied:

$$\begin{aligned} A_{f\mathcal {D}^\perp }f\mathcal {D}=0, \end{aligned}$$

(5.77)

$$\begin{aligned} g(\sigma (X,Y),fZ)=0, X\in \mathcal {D}, Y\in \mathcal {X}(M), Z\in \mathcal {D}^\perp , \end{aligned}$$

(5.78)

$$\begin{aligned} \nabla _XY\in \mathcal {D}\oplus \mathcal {M}, X\in \mathcal {D}, Y\in \mathcal {X}(M). \end{aligned}$$

(5.79)

5.3.2 Normal CR-Submanifolds of an S-manifold

Let M be a CR-submanifold of an S-manifold $\wedge M$. We say that M is a normal CR-submanifold if

$$\begin{aligned} N_v(X,Y)=2t\mathrm{d}u(X,Y)-2\sum _{{\alpha }=1}^sF(X,Y)\xi _{\alpha }, \end{aligned}$$

(5.80)

for any $X,Y\in \mathcal {X}(M)$, where $N_v$ is denoting the Nijenhuis torsion of v, that is

$$N_v(X,Y)=(\nabla _{vX}v)Y-(\nabla _{vY}v)X+v((\nabla _Yv)X-(\nabla _Xv)Y).$$

We notice that (5.80) is equivalent to

$$\begin{aligned} S^*(X,Y)=N_v(X,Y)-t((\nabla _Xu)Y-(\nabla _Yu)X)+2\sum _{{\alpha }=1}^sF(X,Y)\xi _{\alpha }=0, \end{aligned}$$

for any $X,Y\in \mathcal {X}(M)$. Now, we can prove the following characterization theorem in terms of the shape operator.

Theorem 5.15

A CR-submanifold M of an S-manifold $\wedge M$ is normal if and only if

$$\begin{aligned} A_{uY}vX=vA_{uY}X, \end{aligned}$$

(5.81)

for any $X\in \mathcal {D}$ and any $Y\in \mathcal {D}^\perp$.

Proof

A direct expansion by using (5.72) and (5.73) gives that

$$\begin{aligned} S^*(X,Y)=A_{uY}vX-vA_{uY}X-A_{uX}vY+vA_{uX}Y, \end{aligned}$$

(5.82)

for any $X,Y\in \mathcal {X}(M)$. Now, if M is a normal CR-submanifold of $\wedge M$, (5.81) follows form (5.82) since $uX=0$, for any $X\in \mathcal {D}$. Conversely, if (5.81) holds, we use the decomposition $\mathcal {X}(M)=\mathcal {D}\oplus \mathcal {D}^\perp \oplus \mathcal {M}$. First, since $uX=0$ for any $X\in \mathcal {D}$ and $v\xi _{\alpha }=0=u\xi _{\alpha }$, for any ${\alpha }$, we deduce from (5.81) and (5.82) that $S^*(X,Y)=0$, for any $X\in \mathcal {D}$ and any $Y\in \mathcal {X}(M)$. Moreover, if $Y\in \mathcal {D}^\perp$, from (5.68) we have $A_{uY}\xi _{\alpha }\in \mathcal {D}^\perp$ and so, $vA_{uY}\xi _{\alpha }=0$ dfor any ${\alpha }$. Consequently, $S^*(X,\xi _{\alpha })=0$, for any $X\in \mathcal {X}(M)$. Finally, if $X,Y\in \mathcal {D}^\perp$, (5.82) becomes

$$S^*(X,Y)=v(A_{fX}Y-A_{fY}X),$$

since $vX=vY=0$ and $uX=fX$, $uY=fY$. But, from (5.60) we easily show that $A_{fX}Y=A_{fY}X$.    $\blacksquare$

Corollary 2

A CR-submanifold M of an S-manifold is normal if and only if

$$\begin{aligned} g(\sigma (X,vY)+\sigma (Y,vX),fZ)=0, \end{aligned}$$

(5.83)

$$\begin{aligned} g(\sigma (X,Z)fW)=0, \end{aligned}$$

(5.84)

for any $X,Y\in \mathcal {D}$ and any $Z,W\in \mathcal {D}^\perp$.

Proof

Since v is skew-symmetric, from (5.81) we see that M is normal if and only if

$$\begin{aligned} g(\sigma (X,vY),uZ)=-g(\sigma (Y,vX),uZ)m \end{aligned}$$

(5.85)

for any $X\in \mathcal {X}(M)$, $Y\in \mathcal {D}$ and $Z\in \mathcal {D}^\perp$. Now, if M is normal, from (5.85) we get (5.83) taking $X\in \mathcal {D}$ and (5.84) taking $X\in \mathcal {D}^\perp$. Conversely, if (5.83) and (5.84) are satisfied, we observe that (5.85) is satisfied too if $X\in \mathcal {D}$ and $X\in \mathcal {D}^\perp$, respectively. Finally, if $X\in \mathcal {M}$, we have $vX=0$ and, by using that $u\circ v=0$ and (5.67), $\sigma (X,vY)=0$, for any $Y\in \mathcal {D}$. Thus, (5.85) holds for any $X\in \mathcal {X}(M)$.    $\blacksquare$

Corollary 3

Any normal generic submanifold of an S-manifold is a $(\mathcal {D},\mathcal {D}^\perp )$-geodesic submanifold.

From (5.60), (5.67), (5.83), and (5.84), we have

$$\begin{aligned} \sigma (fX,Z)=f\sigma (X,Z), \end{aligned}$$

(5.86)

$$\begin{aligned} t\sigma (fX,fX)=t\sigma (X,X), \end{aligned}$$

(5.87)

$$\begin{aligned} A_{fZ}X\in \mathcal {D}, \end{aligned}$$

(5.88)

for any $X\in \mathcal {}$ and any $Z\in \mathcal {D}^\perp$. On the other hand, from (5.78) and (5.83)–(5.84), we deduce

Proposition 1

Each CR-product in an S-manifold is a normal CR-submanifold.

For the converse we prove the following theorems.

Theorem 5.16

Let M be a normal CR-submanifold of an S-manifold. Then, M is a CR-product if and only if the distribution $\mathcal {D}\oplus \mathcal {M}$ is integrable.

Proof

The necessary condition is obvious. Conversely, let $X\in \mathcal {D}$. If $Y\in \mathcal {D}^\perp$, then (5.78) is (5.84). Further, if $Y\in \mathcal {M}$, from (5.67) we get $\sigma (X,Y)=0$. Finally, if $Y\in \mathcal {D}$, from (5.76) and (5.83) we obtain (5.78).    $\blacksquare$

Theorem 5.17

Let M be a normal CR-submanifold of an S-manifold such that $\mathrm{d}u=0$. Then, M is a CR-product.

Proof

A straightforward computation gives, by using the hypothesis and (5.72),

$$\begin{aligned} g((\nabla _Xv)Y,Z)=\sum _{{\alpha }=1}^s\{\mathrm{d}\eta _{\alpha }(vX,Y)\eta _{\alpha }(Z)-\mathrm{d}\eta _{\alpha }(vZ,X)\eta _{\alpha }(Y)\}, \end{aligned}$$

(5.89)

for any $X,Y,Z\in \mathcal {X}(M)$. Now, if $Y\in \mathcal {D}$, from (5.64) and (5.65) we get $\mathrm{d}\eta _{\alpha }(vX,Y)=F(vX,Y)=g(vX,vY)=g(X,Y)$. So, (5.89) becomes

$$(\nabla _Xv)Y=\sum _{{\alpha }=1}^sg(X,Y)\xi _{\alpha }$$

for any $X\in \mathcal {X}(M)$ and any $Y\in \mathcal {D}$. Comparing with (5.72) we have $\sigma (X,Y)=0$ and so (5.78) holds.    $\blacksquare$

We say that v is $\eta$ -parallel if

$$\begin{aligned} (\nabla _Xv)Y=\sum _{{\alpha }=1}^s\{g(PX,PY)\xi _{\alpha }-\eta _{\alpha }(Y)PX\}, \end{aligned}$$

for any $X,Y\in \mathcal {X}(M)$. Then, from (5.64), (5.65), and (5.89), we prove

Proposition 2

Any normal CR-submanifold of an S-manifold such that $\mathrm{d}u=0$ is $\eta$-parallel.

Given a CR-submanifold M of an S-manifold, a vector field $X\in \mathcal {X}(M)$ is said to be $\mathcal {D}$ -Killing if

$$\begin{aligned} g(P\nabla _ZX,PY)+g(P\nabla _YX,PZ)=0, \end{aligned}$$

(5.90)

for any $Y,Z\in \mathcal {X}(M)$. We notice that it is possible to characterize normal CR-submanifolds in terms of $\mathcal {D}$-Killing vector fields.

Theorem 5.18

A CR-submanifold M of an S-manifold is a normal CR-submanifold if and only if any $Z\in \mathcal {D}^\perp$ is a $\mathcal {D}$-Killing vector field

Proof

Given $X,Y\in \mathcal {X}(M)$ and $Z\in \mathcal {D}^\perp$, from (5.74) we get

$$\begin{aligned} g(\nabla _XZ,Y)+g(\nabla _YZ,X)&=g(vA_{fZ}X,Y)-g(tD_XfZ,Y) \nonumber \\&\quad + g(vA_{fZ}Y,X)-g(tD_YfZ,X). \end{aligned}$$

(5.91)

But $g(vA_{fZ}Y,X)=-g(A_{fZ}vX,Y)$ and so, from (5.91)

$$\begin{aligned}&g(P\nabla _XZ,PY)+g(P\nabla _YZ,PX)+g(Q\nabla _XZ,QY)+g(Q\nabla _YZ,QX) \nonumber \\&\qquad +\sum _{{\alpha }=1}^s\{\eta _{\alpha }(\nabla _XZ)\eta _{\alpha }(Y)+\eta _{\alpha }(\nabla _YZ)\eta _{\alpha }(X)\} \nonumber \\&\quad = g((vA_{fZ}-A_{fZ}v)X,Y)-g(tD_XfZ,Y)-g(tD_YfZ,X). \end{aligned}$$

(5.92)

Now, since it is easy to show that $\eta _{\alpha }(\nabla _XZ)=0$ for any ${\alpha }=1,\dots ,s$, by using (5.75), we deduce that (5.92) becomes

$$\begin{aligned} g(P\nabla _XZ,PY)+g(P\nabla _YZ,PX)=g((vA_{fZ}-A_{fZ}v)X,Y). \end{aligned}$$

(5.93)

Consequently, if Z is a $\mathcal {D}$-Killing vector field, from (5.81) we obtain that M is a normal CR-submanifold. Conversely, if $X\in \mathcal {D}$, the right part of the equality (5.93) vanishes by using (5.81). If $X\in \mathcal {D}^\perp$, then $vX=0$ and from (5.84), $A_{fZ}X\in \mathcal {D}^\perp$, that is, $vA_{fZ}X=0$ and the right part of (5.93) vanishes again. Finally, if $X\in \mathcal {M}$, $vX=0$ and from (5.68), $A_{fZ}X\in \mathcal {D}^\perp$. In any case, from (5.93) we have (5.90).    $\blacksquare$

To end this section, we recall that a submanifold M of an S-manifold is said to be totally f-umbilical [19] if there exists a normal vector field V such that

$$\begin{aligned} \sigma (X,Y)=g(fX,fY)V+\sum _{{\alpha }=1}^s\{\eta _a(Y)\sigma (X,\xi _{\alpha })+\eta _{\alpha }(X)\sigma (Y,\xi _{\alpha })\}, \end{aligned}$$

(5.94)

for any $X,Y\in \mathcal {X}(M)$. These submanifolds have been studied and classified in [9]. Since from (5.94) we easily get (5.83) and (5.84), then we have the following theorem.

Theorem 5.19

Any totally f-umbilical CR-submanifold of an S-manifold is a normal CR-submanifold.

5.3.3 Normal CR-Submanifolds of an S-space-form

Let $\wedge M(c)$ a $(2m+s)$-dimensional S-space-form, where c is denoting the constant f-sectional curvature and let M be a CR-submanifold. Firstly, we can prove

Proposition 3

If M is a normal CR-submanifold, then

$$\begin{aligned} \Vert A_{fZ}X\Vert ^2+\Vert \sigma (X,Z)\Vert ^2-g(t\sigma (Z,Z),t\sigma (X,X))=\frac{c+3s}{4}, \end{aligned}$$

(5.95)

for any unit vector fields $X\in \mathcal {D}$ and $Z\in \mathcal {D}^\perp$.

Proof

From the Codazzi equation, we have

$$\begin{aligned} \wedge R(X,fX,Z,fZ)&=g(D_X\sigma (fX,Z)-D_{fX}\sigma (X,Z),fZ)\nonumber \\&\quad -g(\sigma ([X,fX],Z),fZ)\nonumber \\&\quad +g(\sigma (X,\nabla _{fX}Z)-\sigma (fX,\nabla _XZ),fZ). \end{aligned}$$

(5.96)

Now, from (5.60), (5.84), and (5.86), a direct expansion gives

$$\begin{aligned} g(D_X\sigma (fX,Z)-D_{fX}\sigma (X,Z),fZ)=-2\Vert \sigma (X,Z)\Vert ^2. \end{aligned}$$

(5.97)

On the other hand, since $X\in \mathcal {D}$ is a unit vector field (and so, fX too), we see from (5.59) that $\eta _{\alpha }([X,fX])=2$ for any ${\alpha }$ and from (5.70) that $Q[X,fX]=t\sigma (X,X)+t\sigma (fX,fX)$. Thus, taking into account (5.67), (5.84), and (5.87), we get

$$\begin{aligned} g(\sigma ([X,fX],Z),fZ)=2g(\sigma (t\sigma (X,X),Z),fZ)-2s. \end{aligned}$$

(5.98)

However, since $Z\in \mathcal {D}^\perp$, by using (5.70) it is easy to show that

$$g(\sigma (t\sigma (X,X),Z),fZ)=-g(t\sigma (X,X),t\sigma (Z,Z)).$$

Therefore, from (5.98) we have

$$\begin{aligned} g(\sigma ([X,fX],Z),fZ)=-2s-2g(t\sigma (X,X),t\sigma (Z,Z)). \end{aligned}$$

(5.99)

Next, since $\eta _{\alpha }(\nabla _{fX}Z)=\eta _{\alpha }(\nabla _XZ)=0$ for any ${\alpha }$, from (5.69), (5.83), (5.84), and (5.88), we obtain

$$\begin{aligned} g(\sigma (X,\nabla _{fX}Z)-\sigma (fX,\nabla _XZ),fZ)=-2\Vert A_{fZ}X\Vert ^2. \end{aligned}$$

(5.100)

Finally, from (5.61) we deduce $\wedge R(X,fX,Z,fZ)=-(c-s)/2$. Then, substituting (5.97), (5.99), and (5.100) into (5.96), we complete the proof.    $\blacksquare$

Corollary 4

If M is a normal $\mathcal {D}^\perp$-geodesic CR-submanifold of an S-space-form $\wedge M(c)$, then $c\ge -3s$.

Proposition 4

If M is a normal CR-submanifold of an S-space-form $\wedge M(c)$ such the distribution $\mathcal {D}\oplus \mathcal {M}$ is integrable, then $c\ge -3s$ and M is a CR-product.

Proof

It is clear that M is a CR-product due to Theorem 5.16. Moreover, from (5.78) we have $g(\sigma (X,Y),fZ)=0$. for any $X,Y\in \mathcal {D}$. Then, if $X\in \mathcal {D}$ is a unit vector field, $t\sigma (X,X)=0$ and, by using (5.95), $c\ge -3s$.    $\blacksquare$

Now, we are going to study the concrete case of the $(2m + s)$-dimensional euclidean S-space-form $\mathbb R^{2m+s}(-3s)$ (see [13] for the details of this structure). In this context, we can prove

Theorem 5.20

If M is a normal $(\mathcal {D},\mathcal {D}^\perp )$-geodesic and $\mathcal {D}^\perp$-geodesic CR-submanifold of $\mathbb R^{2m+s}(-3s)$, then it is a CR-product.

Proof

From (5.95), we have $A_{fZ}X=0$ for any $X\in \mathcal {D}$ and any $Z\in \mathcal {D}^\perp$. So, from (5.77), M is a CR-product.    $\blacksquare$

Corollary 5

A normal $\mathcal {D}^\perp$-geodesic generic submanifold of $\mathbb R^{2m+s}(-3s)$ is a CR-product.

Another interesting example of S-space-form is $\mathbb H^{2m+s}(4-3s)$, a generalization of the Hopf fibration $\pi :\S ^{2m+1}\longrightarrow \mathbb PC^{m}$, introduced by Blair in [5] as a canonical example of an S-manifold playing the role of the complex projective space in Kaehler geometry and the odd-dimensional sphere in Sasakian geometry. This space is given by (see [5, 6] for more details)

and its f-sectional curvature is constant equal to $4-3s$. Let M be a CR-submanifold of $\mathbb H^{2m+s}(4-3s)$ (we always suppose $s\ge 2$). Denote by $\nu$ the orthogonal complementary distribution of $f\mathcal {D}^\perp$ in $T(M)^\perp$. Then, $f\nu \subseteq \nu$. Let

$$\begin{aligned} \{E_1,\dots ,E_{2p}\}, \;\;\; \{F_1,\dots ,F_q\}, \;\;\; \{V_1,\dots ,V_r,fV_1,\dots ,fV_r\} , \end{aligned}$$

be local fields of orthonormal frames on $\mathcal {D}$, $\mathcal {D}^\perp$ and $\nu$, respectively, where 2r is the real dimension of $\nu$. First, we prove

Lemma 6

If M is a CR-product in $\mathbb H^{2m+s}(4-3s)$, then

$$\begin{aligned} \Vert \sigma (X,Z)\Vert =1, \end{aligned}$$

(5.101)

for any unit vector fields $X\in \mathcal {D}$ and $Z\in \mathcal {D}^\perp$.

Proof

We know, from Proposition 1, that M is a normal CR-submanifold. Since, $c=4-3s$, from (5.77), (5.78) and (5.95) we complete the proof.    $\blacksquare$

Lemma 7

If M is a CR-product in $\mathbb H^{2m+s}(4-3s)$, the vector field $\sigma (E_i,F_a)$, $i=1,\dots ,2p$ and $a=1,\dots ,q$, are 2pq orthonormal vector fields on $\nu$.

Proof

From (5.101) and by the linearity, we get $g(\sigma (E_i,Z),\sigma (E_j,Z))=0$, for any $i,j=1,\dots ,2p$, $i\ne j$ and any unit vector field $Z\in \mathcal {D}^\perp$. Now, from (5.84), if $q=1$, we complete the proof. If $q\ge 2$, by linearity again, we have $g(\sigma (E_i,F_a),\sigma (E_j,F_b))+g(\sigma (E_i,F_b),\sigma (E_j,F_a))=0$, for any $i,j=1,\dots ,2p$, $i\ne j$, $a,b=1,\dots ,q$, $a\ne b$. Next, by using (5.79) and the Bianchi identity, we obtain $R(X,Y,Z,W)=0$, for any $X,Y\in \mathcal {D}$, $Z,W\in \mathcal {D}^\perp$, where R is denoting the curvature tensor field of M. But, if $i\ne j$ and $a\ne b$, (5.61) gives $\wedge R(E_i,E_j,F_a,F_b)=0$. Then, from the Gauss equation we get

$$g(\sigma (E_i,F_a),\sigma (E_j,F_b))-g(\sigma (E_i,F_b),\sigma (E_j,F_a))=0,$$

for any $i,j=1,\dots ,2p$, $i\ne j$, $a,b=1,\dots ,q$, $a\ne b$ and this completes the proof.    $\blacksquare$

Now, we study the normal CR-submanifolds of $\mathbb H^{2m+s}(4-3s)$.

Theorem 5.21

Let M be a normal CR-submanifold of $\mathbb H^{2m+s}(4-3s)$, $s\ge 2$, such that the distribution $\mathcal {D}\oplus \mathcal {M}$ is integrable. Then

• (i) M is a CR-product $M_1\times M_2$.

• (ii) $m\ge pq+p+q$.

• (iii) If $n=pq+p+q$, then $M_1$ is an invariant totally geodesic submanifold immersed in $\mathbb H^{2m+s}(4-3s)$.

• (iv) $\Vert \sigma \Vert ^2\ge 2q(2p+s)$.

• (v) If $\Vert \sigma \Vert ^2=2q(2p+s)$, then $M_1$ is an S-space-form of constant f-sectional curvature $4-3s$ and $M_2$ has constant curvature 1.

• (vi) If M is a minimal submanifold, then $\rho \le 4p(p+1)+2p(q+s)+q(q-1)$, where $\rho$ denotes the scalar curvature and the equality holds if and only if $\Vert \sigma \Vert ^2=2q(2p+s)$.

Proof

(i) follows directly from Proposition 4. Now, from Lemma 7, $dim(\nu )=2(m-p)-2q\ge 2pq$. So (ii) holds. Next, suppose that $m=pq+p+q$. If $X,Y,Z\in \mathcal {D}$ and $W\in \mathcal {D}^\perp$, from (5.61), $\wedge R(X,Y,Z,W)=0$ and, by using a similar proof to that one of Lemma 7, $R(X,Y,Z,W)=0$. So, the Gauss equation gives

$$\begin{aligned} g(\sigma (X,W),\sigma (Y,Z))-g(\sigma (X,Z),\sigma (Y,W))=0. \end{aligned}$$

(5.102)

Since from Proposition 3.2 of [12], $\sigma (fX,Z)=f\sigma (X,Z)$, if we put $Y=fX$, we have, by using (5.86), $g(\sigma (fX,W),(\sigma (X,Z))=0$. Now, if we put $Z=fY$, then $g(\sigma (X,Y),\sigma (X,W))=0$. Thus, by linearity, we get $g(\sigma (X,W),\sigma (Y,Z))+g(\sigma (X,Z),\sigma (Y,W))=0$. Consequently, from (5.102)

$$\begin{aligned} g(\sigma (X,W),\sigma (Y,Z)) =0, \end{aligned}$$

(5.103)

for any $X,Y,Z\in \mathcal {D}$ and $W\in \mathcal {D}^\perp$. Since now $dim(\nu )=2pq$, (5.103) implies that $\sigma (X,Y)=0$, for any $X,Y\in \mathcal {D}$ and so, (iii) holds from Theorem 2.4(ii) of [12]. Assertions (iv) and (v) follow from Theorem 4.2 of [12]. Finally, if M is a minimal normal CR-submanifold of $\mathbb H^{2m+s}(4-3s)$, a straightforward computation gives

$$\rho =4p(p+1)+2s(p+q)+q(q-1)+6pq-\Vert \sigma \Vert ^2.$$

Then, by using (iv), the proof is complete.    $\blacksquare$

Theorem 5.22

Let M be a normal, $(\mathcal {D},\mathcal {D}^\perp )$-geodesic and $\mathcal {D}^\perp$-geodesic CR-submanifold of $\mathbb H^{2m+s}(4-3s)$. Then,

• (i) $\Vert A_{fZ}X\Vert =1$, for any unit vector fields $X\in \mathcal {D}$ and $Z\in \mathcal {D}^\perp$.

• (ii) $\Vert \sigma \Vert ^2\ge 2q(p+s)$ and the equality hold if and only if $\sigma (\mathcal {D},\mathcal {D})\in f\mathcal {D}^\perp$.

Proof

(i) follows immediately from (5.95). Now, considering the above-mentioned local fields of orthonormal frames for $\mathcal {D}$, $\mathcal {D}^\perp$, and $\nu$, a straightforward computation using the hypothesis gives (ii).    $\blacksquare$

Finally, from (5.84) and (5.95), we can prove

Corollary 6

Let M be a normal $\mathcal {D}^\perp$-geodesic generic submanifold of $\mathbb H^{2m+s}(4-3s)$. Then

• (i) $\Vert A_{fZ}X\Vert =1$, for any unit vector fields $X\in \mathcal {D}$ and $Z\in \mathcal {D}^\perp$.

• (ii) $\Vert \sigma \Vert ^2=2q(p+s)$.