Clairaut Submersion

Abstract

In this chapter, we give the detailed study about the Clairaut submersion. The fundamental notations are given. Clairaut submersion is one of the most interesting topics in differential geometry. Depending on the condition on distribution of submersion, we have different classes of submersion such as anti-invariant, semi-invariant submersions etc. We describe the geometric properties of Clairaut anti-invariant submersions and Clairaut semi-invariant submersions whose total space is a Kähler, nearly Kähler manifold. We give condition for Clairaut anti-invariant submersion to be a totally geodesic map and also study Clairaut anti-invariant submersions with totally umbilical fibers. We also give the conditions for the semi-invariant submersions to be Clairaut map and also for Clairaut semi-invariant submersion to be a totally geodesic map. We also give some illustrative example of Clairaut anti-invariant and semi-invariant submersion.

1. Introduction

Riemannian submersion between two Riemannian manifolds was first introduced by O’Neill [1] and Gray [2]. After that Watson [3] introduced almost Hermitian submersions. Later, the notion of anti-invariant submersions and Lagrangian submersion from almost Hermitian manifolds onto Riemannian manifolds were introduced by Sahin [4] and studied by Taştan [5, 6], Gündüzalp [7], Beri et al. [8], Ali and Fatima [9], in which the fibers of submersion are anti-invariant with respect to the almost complex structure of total manifold. After that several new types of Riemannian submersions were defined and studied such as semi-invariant submersion [10, 11], slant submersion [12, 13], generic submersion [14, 15, 16, 17], hemi-slant submersion [18], semi-slant submersion [19], pointwise slant submersion [20, 21, 22] and conformal semi-slant submersion [23]. Also, these kinds of submersions were considered in different kinds of structures such as nearly Kähler, Kähler, almost product, para-contact, Sasakian, Kenmotsu, cosymplectic and etc. In book [24], we find the recent developments in this field.

In 1735, A.C. Clairaut [25] obtained the very important result in the theory of surfaces, which is Clairaut’s theorem and stated that for any geodesic $\alpha$ on a surface of revolution $S$, the function $rsin\theta$ is constant along $\alpha$, where $r$ is the distant from a point on the surface to the rotation axis and $\theta$ is the angle between $\alpha$ and the meridian through $\alpha$. Bishop [26] introduced the idea of Riemannian submersions and gave a necessary and sufficient conditions for a Riemannian submersion to be Clairaut. Allison [27] considered Clairaut semi-Riemannian submersions and showed that such submersions have interesting applications in the static space-times.

In [28], Tastan and Gerdan gave new Clairaut conditions for anti-invariant submersions whose total manifolds are Sasakian and Kenmotsu and got many interesting results. In [29], Tastan and Aydin studied Clairaut anti-invariant submersions whose total manifolds are cosymplectic. Gündüzalp [30] introduced Clairaut anti-invariant submersions from a paracosymplectic manifold and gave characterization theorems. In [31], Lee et al. studied Clairaut anti-invariant submersions whose total manifolds are Kähler.

Kähler manifolds [32, 33] have an especially rich geometric and topological structure because of Kähler identity. Kähler manifolds are very important in differential geometry, which has applications in several different fields such as supersymmetric gauge theory and superstring theory in theoretical physics, signal processing in information geometry. The simplest example of Kähler manifold is a complex Euclidean space ${\mathrm{ℂ}}^n$ with the standard Hermitian metric.

Nearly Kähler manifolds introduced by Gray and Hervella [32], are the geometrically interesting class among the sixteen classes of almost Hermitian manifolds. The geometrical meaning of nearly Kähler condition is that the geodesics on the manifolds are holomorphically planar curves. Gray [2] studied nearly Kähler manifolds broadly and gave example of a non-Kählerian nearly Kähler manifold, which is 6-dimensional sphere.

Motivated by this, the authors [34] studied Clairaut anti-invariant submersions from nearly Kähler manifolds onto Riemannian manifolds with some examples and obtained conditions for Clairaut Riemannian submersion to be totally geodesic map. The authors investigated conditions for the Clairaut anti-invariant submersions to be a totally umbilical map. The authors [34] studied Clairaut semi-invariant submersions from Kähler manifolds onto Riemannian manifolds with some examples. The authors also obtained conditions for Clairaut semi-invariant Riemannian submersion to be totally geodesic map and investigated conditions for the semi-invariant submersion to be a Clairaut map.

2. Almost complex manifold

An almost complex structure on a smooth manifold $M$ is a smooth tensor field $\phi$ of type $\left(1,1\right)$ such that ${\phi }^2=-I$. A smooth manifold equipped with such an almost complex structure is called an almost complex manifold. An almost complex manifold $\left(M,\phi \right)$ endowed with a chosen Riemannian metric $g$ satisfying

$$g\left(\mathit{\phi X},\mathit{\phi Y}\right)=g\left(X,Y\right)$$E1

for all $X,Y\in \mathit{TM}$, is called an almost Hermitian manifold.

An almost Hermitian manifold $M$ is called a nearly Kähler manifold [2] if

$$\left({\nabla }_X\phi \right)Y+\left({\nabla }_Y\phi \right)X=0$$E2

for all $X,Y\in \mathit{TM}$. If $\left({\nabla }_X\phi \right)Y=0$ for all $X,Y\in \mathit{TM}$, then $M$ is known as Kähler manifold [33]. Every Kähler manifold is nearly Kähler but converse need not be true.

3. Riemannian submersion

Definition 1.1 [1, 35] Let $\left(M,g_m\right)$ and $\left(N,g_n\right)$ be Riemannian manifolds, where $dim\left(M\right)=m$, $dim\left(N\right)=n$ and $m>n$. A Riemannian submersion $\pi :M\to N$ is a map of $M$ onto $N$ satisfying the following axioms:

1. $\pi$ has maximal rank.

2. The differential ${\pi }_{\ast }$ preserves the lengths of horizontal vectors.

For each $q\in N$, ${\pi }^{-1}\left(q\right)$ is an $\left(m-n\right)$-dimensional Riemannian submanifold of $M$. The submanifolds ${\pi }^{-1}\left(q\right)$, $q\in N$, are called fibers. A vector field on $M$ is called vertical if it is always tangent to fibers. A vector field on $M$ is called horizontal if it is always orthogonal to fibers. A vector field $X$ on $M$ is called basic if $X$ is horizontal and $\pi$-related to a vector field $X^{\text{'}}$ on $N$, that is, ${\pi }_{\ast }{X}_p=X_{{\pi }_{\ast }\left(p\right)}^{\prime }$ for all $p\in M.$ We denote the projection morphisms on the distributions $ker{\pi }_{\ast }$ and ${\left(ker{\pi }_{\ast }\right)}^{\perp }$ by $\mathcal{V}$ and $\mathcal{H}$, respectively. The sections of $\mathcal{V}$ and $\mathcal{H}$ are called the vertical vector fields and horizontal vector fields, respectively. So

$${\mathcal{V}}_p=T_p\left({\pi }^{-1}\left(q\right)\right),{\mathcal{H}}_p=T_p{\left({\pi }^{-1}\left(q\right)\right)}^{\perp }.$$

The second fundamental tensors of all fibers ${\pi }^{-1}\left(q\right),q\in N$ gives rise to tensor field $T$ and $A$ in $M$ defined by O’Neill [1] for arbitrary vector field $E$ and $F$, which is

$$T_{E}F=\mathcal{H}{\nabla }_{\mathcal{V}E}^M\mathcal{V}F+\mathcal{V}{\nabla }_{\mathcal{V}E}^M\mathcal{H}F,$$E3

$$A_{E}F=\mathcal{H}{\nabla }_{\mathcal{H}E}^M\mathcal{V}F+\mathcal{V}{\nabla }_{\mathcal{H}E}^M\mathcal{H}F,$$E4

where $\mathcal{V}$ and $\mathcal{H}$ are the vertical and horizontal projections.

To discuss geodesics, we need a linear connection. We denote the Levi-Civita connection on $M$ by $\widehat{\nabla }$ and the adapted connection of the submersion by $\nabla$. From Eqs. (3) and (4), we have

$${\nabla }_{V}W=T_{V}W+{\widehat{\nabla }}_{V}W,$$E5

$${\nabla }_{V}X=\mathcal{H}{\nabla }_{V}X+T_{V}X,$$E6

$${\nabla }_{X}V=A_{X}V+\mathcal{V}{\nabla }_{X}V,$$E7

$${\nabla }_{X}Y=\mathcal{H}{\nabla }_{X}Y+A_{X}Y,$$E8

for all $V,W\in \mathrm{\Gamma }\left(ker{\pi }_{\ast }\right)$ and $X,Y\in \mathrm{\Gamma }{\left(ker{\pi }_{\ast }\right)}^{\perp },$ where $\mathcal{V}{\nabla }_{V}W={\widehat{\nabla }}_{V}W.$ If $X$ is basic, then $A_{X}V=\mathcal{H}{\nabla }_{V}X.$

It is easily seen that for $p\in M,$ $U\in {\mathcal{V}}_p$ and $X\in {\mathcal{H}}_p$ the linear operators

$$T_U,A_X:T_{p}M\to T_{p}M$$

are skew-symmetric, that is,

$$g\left(A_{X}E,F\right)=-g\left(E,A_{X}F\right)\text{and}g\left(T_{U}E,F\right)=-g\left(E,T_{U}F\right),$$E9

for all $E,F\in$ $T_{p}M.$ We also see that the restriction of $T$ to the vertical distribution ${\left.T\right|}_{ker{\pi }_{\ast }\times ker{\pi }_{\ast }}$ is exactly the second fundamental form of the fibers of $\pi$. Since $T_U$ is skew-symmetric, therefore $\pi$ has totally geodesic fibers if and only if $T\equiv 0$.

Let $\pi :\left(M,g_m\right)\to \left(N,g_n\right)$ be a smooth map between Riemannian manifolds. Then the differential ${\pi }_{\ast }$ of $\pi$ can be observed as a section of the bundle $\mathit{Hom}\left(\mathit{TM},{\pi }^{-1}\mathit{TN}\right)\to M$, where ${\pi }^{-1}\mathit{TN}$ is the bundle which has fibers ${\left({\pi }^{-1}\mathit{TN}\right)}_x=T_{f\left(x\right)}N$. $\mathit{Hom}\left(\mathit{TM},{\pi }^{-1}\mathit{TN}\right)$ has a connection $\nabla$ induced from the Riemannian connection ${\nabla }^M$ and the pullback connection ${\nabla }^N$ [36, 37]. Then the second fundamental form of $\pi$ is given by

$$\left(\nabla {\pi }_{\ast }\right)\left(E,F\right)={\nabla }_E^N{\pi }_{\ast }F-{\pi }_{\ast }\left({\nabla }_E^{M}F\right),\text{for}\mathrm{all}E,F\in \mathrm{\Gamma }\left(\mathit{TM}\right).$$E10

We also know that $\pi$ is said to be totally geodesic map [36] if $\left(\nabla {\pi }_{\ast }\right)\left(E,F\right)=0,$ for all $E,F\in \mathrm{\Gamma }\left(\mathit{TM}\right)$.

4. Clairaut submersion from Riemannian manifold

Let $S$ be a revolution surface in ${\mathbb{R}}^3$ with rotation axis $L$. For any $p\in S$, we denote by $r\left(p\right)$ the distance from $p$ to $L$. Given a geodesic $\alpha :J\subset \mathbb{R}\to S$ on $S$, let $\theta \left(t\right)$ be the angle between $\alpha \left(t\right)$ and the meridian curve through $\alpha \left(t\right),t\in I$. A well-known Clairaut’s theorem [25] named after Alexis Claude de Clairaut, says that for any geodesic on $S$, the product $rsin\theta$ is constant along $\alpha$, i.e., it is independent of $t$. For proof, see [38, p.183]. In the theory of Riemannian submersions, Bishop [26] introduced the notion of Clairaut submersion in the following way:

Definition 1.2 [26] A Riemannian submersion $\pi :\left(M,g\right)\to \left(N,g_n\right)$ is called a Clairaut submersion if there exists a positive function $r$ on $M$, which is known as the girth of the submersion, such that, for any geodesic $\alpha$ on $M$, the function $\left(r\circ \alpha \right)sin\theta$ is constant, where, for any $t,\theta \left(t\right)$ is the angle between $\dot{\alpha }\left(t\right)$ and the horizontal space at $\alpha \left(t\right)$.

For further use, we are stating one important result of Bishop.

Theorem 1.1 [26] A curve $h$ in $M$ is a geodesic if and only if $\dot{X}+2A_{X}U+T_{U}U=0$ and ${\nabla }_{E}U+T_{U}X=0,$ where $\dot{h}\left(t\right)=E=X+U$, $X$ is horizontal and $U$ is vertical.

Bishop also gave the following necessary and sufficient condition for a Riemannian submersion to be a Clairaut submersion, which is

Theorem 1.2 [26] Let $\pi :\left(M,g\right)\to \left(N,g_n\right)$ be a Riemannian submersion with connected fibers. Then, $\pi$ is a Clairaut submersion with $r=e^f$ if and only if each fiber is totally umbilical and has the mean curvature vector field $H=-\mathit{\text{gradf}}$, where $\mathit{\text{gradf}}$ is the gradient of the function $f$ with respect to $g$.

Proof: Let $\pi :M\to N$ be a Riemannian submersion. For a geodesic $h$ in $M$, we use $\dot{h}\left(s\right)=E=X+U$, where $X$ is horizontal and $U$ is vertical. and $\mathrm{\ell }={‖\dot{h}\left(s\right)‖}^2$. Let $\theta \left(s\right)$ be the angle between $\dot{h}\left(s\right)$ and the horizontal space at $h\left(s\right)$. Then

$$g\left(X\left(s\right),X\left(s\right)\right)=\mathrm{\ell }{cos}^2\theta \left(s\right),$$E11

$$g\left(U\left(s\right),U\left(s\right)\right)=\mathrm{\ell }{sin}^2\theta \left(s\right).$$E12

Differentiating (12), we get

$$g\left({\nabla }_{\dot{h}\left(s\right)}U\left(s\right),U\left(s\right)\right)=\mathrm{\ell }sin\theta \left(s\right)cos\theta \left(s\right)\frac{\mathit{d\theta }\left(s\right)}{\mathit{ds}}.$$E13

Using Theorem 1.1, (13) becomes

$$-g\left(T_{U\left(s\right)}X\left(s\right),U\left(s\right)\right)=\mathrm{\ell }sin\theta \left(s\right)cos\theta \left(s\right)\frac{\mathit{d\theta }\left(s\right)}{\mathit{ds}}.$$E14

Since $T_U$ is skew-symmetric, so form above equation, we have

$$g\left(T_{U\left(s\right)}U\left(s\right),X\left(s\right)\right)=\mathrm{\ell }sin\theta \left(s\right)cos\theta \left(s\right)\frac{\mathit{d\theta }\left(s\right)}{\mathit{ds}}.$$E15

Now, $\pi$ is a Clairaut submersion with $r=e^f$ if and only if $\frac{d}{\mathit{ds}}\left(e^{f\circ h}\mathit{\text{sinθ}}\right)=0.$

Using (12, 15) in $\frac{d}{\mathit{ds}}\left(e^{f\circ h}\mathit{\text{sinθ}}\right)=0$, we have

$$g\left(U\left(s\right),U\left(s\right)\right)\frac{d}{\mathit{ds}}\left(f\circ h\right)\left(s\right)+g\left(T_{U\left(s\right)}U\left(s\right),X\left(s\right)\right)=0,$$E16

$$g\left(U\left(s\right),U\left(s\right)\right)g\left(\mathit{\text{gradf}},E\left(s\right)\right)+g\left(T_{U\left(s\right)}U\left(s\right),X\left(s\right)\right)=0.$$E17

Consider any geodesic $h$ on $M$ with initial vertical tangent vector, so $\mathit{\text{gradf}}$ turns out to be horizontal. Therefore, the function $f$ is constant on any fiber, the fibers being connected. Therefore (17) reduces to

$$g\left(U\left(s\right),U\left(s\right)\right)g\left(\mathit{\text{gradf}},X\left(s\right)\right)+g\left(T_{U\left(s\right)}U\left(s\right),X\left(s\right)\right)=0,$$E18

$$g\left(U\left(s\right),U\left(s\right)\right)\mathit{\text{gradf}}+T_{U\left(s\right)}U\left(s\right)=0.$$E19

Setting $U=U_1+U_2,$ where $U_1,U_2$ are vertical vector fields and using the fact that $T$ is symmetric for vertical vector fields, we obtain

$$g\left(U_1,U_2\right)\mathit{\text{gradf}}+T_{U_1}{U}_2=0$$E20

holds for all vertical vector fields $U_1,U_2.$.

Since the restriction of $T$ to the vertical distribution ${\left.T\right|}_{ker{\pi }_{\ast }\times ker{\pi }_{\ast }}$ is exactly the second fundamental form of the fibers of $\pi .$ It means that any fiber is totally umbilical with mean curvature vector field $H=-\mathit{\text{gradf}}.$

Conversely, suppose the fibers are totally umbilic with normal curvature vector field $H=-\mathit{\text{gradf}}$ so that we have

$$g\left(U,U\right)H+T_{U}U=0.$$E21

Since $\mathit{\text{gradf}}$ is orthogonal to fibers, so

$$g\left(U,U\right)g\left(\mathit{\text{gradf}},E\right)=-g\left(U,U\right)g\left(H,X\right)=-g\left(T_{U}U,X\right).$$E22

Since (18) holds. so $\left(r\circ h\right)\mathit{\text{sinθ}}$ is constant along any geodesic $h$.

Example 1.1 [24] Consider the warped product manifold $M_1{\times }_f{M}_2$ of Riemannian manifolds $\left(M_1,g_1\right)$ and $\left(M_2,g_2\right)$, where $f:M_1\to \left(0,\infty \right)$. The fibers of the first projection $p_1:M_1{\times }_f{M}_2\to M_1$ are totally umbilical with mean curvature vector field $H=-\mathit{grad}\left({\mathit{\text{logf}}}^{1/2}\right)$. Thus, if $M_2$ is connected, $p_1$ is a Clairaut submersion with $r=f^{1/2}$.

5. Anti-invariant Riemannian submersion

Definition 1.3 [39] Let $\left(M,\phi ,g\right)$ be an almost Hermitian manifold and $N$ be a Riemannian manifold with Riemannian metric $g_n$. Suppose that there exists a Riemannian submersion $\pi :M\to N$, such that the vertical distribution $ker{\pi }_{\ast }$ is anti-invariant with respect to $\phi$, i.e., $\phi ker{\pi }_{\ast }\subseteq ker{\pi }_{\ast }^{\perp }$. Then, the Riemannian submersion $\pi$ is called an anti-invariant Riemannian submersion. We will briefly call such submersions as anti-invariant submersions.

Let $\pi$ be an anti-invariant Riemannian submersion from nearly Kähler manifold $\left(M,\phi ,g_m\right)$ onto Riemannian manifold $\left(N,g_n\right)$. For any arbitrary tangent vector fields $U$ and $V$ on $M$, we set

$$\left({\nabla }_U\phi \right)V=P_{U}V+Q_{U}V$$E23

where $P_{U}V,Q_{U}V$ denote the horizontal and vertical part of $\left({\nabla }_U\phi \right)V$, respectively. Clearly, if $M$ is a Kähler manifold then $P=Q=0$.

If $M$ is a nearly Kähler manifold then $P$ and $Q$ satisfy

$$P_{U}V=-P_{V}U,Q_{U}V=-Q_{V}U.$$E24

Consider

$${\left(ker{\pi }_{\ast }\right)}^{\perp }=\phi ker{\pi }_{\ast }\oplus \mu ,$$

where $\mu$ is the complementary distribution to $\phi ker{\pi }_{\ast }$ in ${\left(\mathit{ker}{\pi }_{\ast }\right)}^{\perp }$ and $\mathit{\phi \mu }\subset \mu$.

For $X\in \mathrm{\Gamma }{\left(\mathit{ker}{\pi }_{\ast }\right)}^{\perp }$, we have

$$\mathit{\phi X}=\mathit{\alpha X}+\mathit{\beta X},$$E25

where $\mathit{\alpha X}\in \mathrm{\Gamma }\left(ker{\pi }_{\ast }\right)$ and $\mathit{\beta X}\in \mathrm{\Gamma }\left(\mu \right)$. If $\mu =0$, then an anti-invariant submersion is known as Lagrangian submersion.

5.1 Anti-invariant Clairaut submersions from nearly Kähler manifolds

In this section, we give new Clairaut conditions for anti-invariant submersions from nearly Kähler manifolds after giving some auxiliary results.

Theorem 1.3 [34] Let $\pi$ be an anti-invariant submersion from a nearly Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$. If $h:J\subset \mathbb{R}\to M$ is a regular curve and $U\left(s\right)$ and $X\left(s\right)$ are the vertical and horizontal parts of the tangent vector field $\dot{h}\left(s\right)=W$ of $h\left(s\right)$, respectively, then $h$ is a geodesic if and only if along $h$

$$A_X\mathit{\phi U}+A_X\mathit{\beta X}+T_U\mathit{\beta X}+\mathcal{V}{\nabla }_X\mathit{\alpha X}+T_U\mathit{\phi U}+{\widehat{\nabla }}_U\mathit{\alpha X}=0,$$E26

$$\mathcal{H}\left({\nabla }_{\dot{h}}\mathit{\phi U}+{\nabla }_{\dot{h}}\mathit{\beta X}\right)+A_X\mathit{\alpha X}+T_U\mathit{\alpha X}=0.$$E27

Proof: Let $\pi$ be an anti-invariant submersion from a nearly Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$. Since ${\phi }^2\dot{h}=-\dot{h}$. Taking the covariant derivative of this and using (2), we have

$$\left({\nabla }_{\dot{h}}\phi \right)\phi \dot{h}+\phi \left({\nabla }_{\dot{h}}\phi \dot{h}\right)=-{\nabla }_{\dot{h}}\dot{h}.$$E28

Since $U\left(s\right)$ and $X\left(s\right)$ are the vertical and horizontal parts of the tangent vector field $\dot{h}\left(s\right)=W$ of $h\left(s\right)$, that is, $\dot{h}=U+X$. So (28) becomes

$$\begin{array}{ll}-{\nabla }_{\dot{h}}\dot{h}& =\phi \left({\nabla }_{U+X}\phi \left(U+X\right)\right)+P_{\dot{h}}\phi \dot{h}+Q_{\dot{h}}\phi \dot{h}\\ & =\phi \left({\nabla }_U\mathit{\phi U}+{\nabla }_X\mathit{\phi U}+{\nabla }_U\mathit{\phi X}+{\nabla }_X\mathit{\phi X}\right)+P_{\dot{h}}\phi \dot{h}+Q_{\dot{h}}\phi \dot{h}\\ & =\phi \left({\nabla }_U\mathit{\phi U}+{\nabla }_X\mathit{\phi U}+{\nabla }_U\left(\mathit{\alpha X}+\mathit{\beta X}\right)+{\nabla }_X\left(\mathit{\alpha X}+\mathit{\beta X}\right)\right)\\ & +P_{\dot{h}}\phi \dot{h}+Q_{\dot{h}}\phi \dot{h}.\end{array}$$E29

Using (5)–(8) in (29), we get

$$\begin{array}{ll}-{\nabla }_{\dot{h}}\dot{h}& =\phi \left(\mathcal{H}\left({\nabla }_{\dot{h}}\mathit{\phi U}+{\nabla }_{\dot{h}}\mathit{\beta X}\right)+A_X\mathit{\alpha X}+A_X\mathit{\beta X}+A_X\mathit{\phi U}\right.\\ & +\left.T_U\mathit{\beta X}+T_U\mathit{\alpha X}+\mathcal{V}{\nabla }_X\mathit{\alpha X}+T_U\mathit{\phi U}+{\widehat{\nabla }}_U\mathit{\alpha X}\right)+P_{\dot{h}}\phi \dot{h}+Q_{\dot{h}}\phi \dot{h}.\end{array}$$E30

Let $Y,Z\in \mathit{TM}$. Since ${\phi }^{2}Z=-Z$, on differentiation, we have

$$\phi \left({\nabla }_Y\mathit{\phi Z}\right)+\left({\nabla }_Y\phi \right)\mathit{\phi Z}=-{\nabla }_{Y}Z,$$

$${\phi }^2\left({\nabla }_{Y}Z\right)+\phi \left({\nabla }_Y\phi \right)Z+\left({\nabla }_Y\phi \right)\mathit{\phi Z}=-{\nabla }_{Y}Z,$$

using (23) in above, we obtain

$$\phi \left(P_{Y}Z+Q_{Y}Z\right)=-P_Y\mathit{\phi Z}-Q_Y\mathit{\phi Z}.$$E31

By (31), we have

$$\phi \left(P_{\dot{h}}\phi \dot{h}+Q_{\dot{h}}\phi \dot{h}\right)=P_{\dot{h}}\dot{h}+Q_{\dot{h}}\dot{h},$$

since $P$ and $Q$ are skew-symmetric, so

$$\phi \left(P_{\dot{h}}\phi \dot{h}+Q_{\dot{h}}\phi \dot{h}\right)=0.$$E32

Using (32) and equating the vertical and horizontal part of (30), we obtain

$$\mathcal{V}\phi {\nabla }_{\dot{h}}\dot{h}=A_X\mathit{\phi U}+A_X\mathit{\beta X}+T_U\mathit{\beta X}+\mathcal{V}{\nabla }_X\mathit{\alpha X}+T_U\mathit{\phi U}+{\widehat{\nabla }}_U\mathit{\alpha X},$$

$$\mathcal{H}\phi {\nabla }_{\dot{h}}\dot{h}=\mathcal{H}\left({\nabla }_{\dot{h}}\mathit{\phi U}+{\nabla }_{\dot{h}}\mathit{\beta X}\right)+A_X\mathit{\alpha X}+T_U\mathit{\alpha X}.$$

By using above equations, we can say that $h$ is geodesic if and only if (26, 27) hold.

Theorem 1.4 [34] Let $\pi$ be an anti-invariant submersion from a nearly Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$. Also, let $h:J\subset \mathbb{R}\to M$ be a regular curve and $U\left(s\right)$ and $X\left(s\right)$ are the vertical and horizontal parts of the tangent vector field $\dot{h}\left(s\right)=W$ of $h\left(s\right)$. Then $\pi$ is a Clairaut submersion with $r=e^f$ if and only if along $h$

$$g\left(\mathit{\text{gradf}},X\right)g\left(U,U\right)=g\left(\mathcal{H}{\nabla }_{\dot{h}}\mathit{\beta X}+A_X\mathit{\alpha X}+T_U\mathit{\alpha X}+P_{\dot{h}\left(s\right)}U,\mathit{\phi U}\right).$$

Proof: Let $h:J\subset \mathbb{R}\to M$ be a geodesic on $M$ and $\mathrm{\ell }={‖\dot{h}\left(s\right)‖}^2$. Let $\theta \left(s\right)$ be the angle between $\dot{h}\left(s\right)$ and the horizontal space at $h\left(s\right)$. Then

$$g\left(X\left(s\right),X\left(s\right)\right)=\mathrm{\ell }{cos}^2\theta \left(s\right),$$E33

$$g\left(U\left(s\right),U\left(s\right)\right)=\mathrm{\ell }{sin}^2\theta \left(s\right).$$E34

Differentiating (34), we get

$$2g\left({\nabla }_{\dot{h}\left(s\right)}U\left(s\right),U\left(s\right)\right)=2\mathrm{\ell }sin\theta \left(s\right)cos\theta \left(s\right)\frac{\mathit{d\theta }\left(s\right)}{\mathit{ds}}.$$E35

Using (1) in (35), we have

$$g\left(\mathcal{H}{\nabla }_{\dot{h}\left(s\right)}\mathit{\phi U}\left(s\right),\mathit{\phi U}\left(s\right)\right)-g\left(\left({\nabla }_{\dot{h}\left(s\right)}\phi \right)U\left(s\right),\mathit{\phi U}\left(s\right)\right)=\mathrm{\ell }sin\theta \left(s\right)cos\theta \left(s\right)\frac{\mathit{d\theta }\left(s\right)}{\mathit{ds}}.$$

Now by use of (23), we have

$$g\left(\mathcal{H}{\nabla }_{\dot{h}\left(s\right)}\mathit{\phi U}\left(s\right),\mathit{\phi U}\left(s\right)\right)-g\left(P_{\dot{h}\left(s\right)}U+Q_{\dot{h}\left(s\right)}U,\mathit{\phi U}\left(s\right)\right)=\mathrm{\ell }sin\theta \left(s\right)cos\theta \left(s\right)\frac{\mathit{d\theta }\left(s\right)}{\mathit{ds}}.$$

Along the curve $h$, using Theorem 1.3, we obtain

$$-g\left(\mathcal{H}{\nabla }_{\dot{h}}\mathit{\beta X}+A_X\mathit{\alpha X}+T_U\mathit{\alpha X}+P_{\dot{h}\left(s\right)}U,\mathit{\phi U}\left(s\right)\right)=\mathrm{\ell }sin\theta \left(s\right)cos\theta \left(s\right)\frac{\mathit{d\theta }\left(s\right)}{\mathit{ds}}.$$

Now, $\pi$ is a Clairaut submersion with $r=e^f$ if and only if $\frac{d}{\mathit{ds}}\left(e^{f}sin\theta \right)=0$. Therefore

$$\begin{array}{r}{e}^f\left(\frac{\mathit{df}}{\mathit{ds}}sin\theta +cos\theta \frac{\mathit{d\theta }}{\mathit{ds}}\right)=0,\\ e^f\left(\frac{\mathit{df}}{\mathit{ds}}\mathrm{\ell }{sin}^2\theta +\mathrm{\ell }sin\theta cos\theta \frac{\mathit{d\theta }}{\mathit{ds}}\right)=0.\end{array}$$

So, we obtain

$$\frac{\mathit{df}}{\mathit{ds}}\left(h\left(s\right)\right)g\left(U\left(s\right),U\left(s\right)\right)=g\left(\mathcal{H}{\nabla }_{\dot{h}}\mathit{\beta X}+A_X\mathit{\alpha X}+T_U\mathit{\alpha X}+P_{\dot{h}\left(s\right)}U,\mathit{\phi U}\left(s\right)\right).$$E36

Since $\frac{\mathit{df}}{\mathit{ds}}\left(h\left(s\right)\right)=g\left(\mathit{\text{gradf}},\dot{h}\left(s\right)\right)=g\left(\mathit{\text{gradf}},X\right)$. Therefore by using (36), we get the result.

Theorem 1.5 [34] Let $\pi$ be an Clairaut anti-invariant submersion from a nearly Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$. Then

$$A_{\mathit{\phi W}}\mathit{\phi X}+Q_W\mathit{\phi X}=X\left(f\right)W$$

for $X\in {\left(ker{\pi }_{\ast }\right)}^{\perp }$, $W\in ker{\pi }_{\ast }$ and $\mathit{\phi W}$ is basic.

Proof: Let $\pi$ be an anti-invariant submersion from a nearly Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$. We know that any fiber of Riemannian submersion $\pi$ is totally umbilical if and only if

$$T_{V}W=g\left(V,W\right)H,$$E37

for all $V,W\in \mathrm{\Gamma }\left(ker{\pi }_{\ast }\right)$, where $H$ denotes the mean curvature vector field of any fiber in $M$. By using Theorem 1.2 and (37), we have

$$T_{V}W=-g\left(V,W\right)\mathit{\text{gradf}}.$$E38

Let $X\in \mu$ and $V,W\in \mathrm{\Gamma }\left(ker{\pi }_{\ast }\right)$, then by using (1) and (2), we have

$$g\left({\nabla }_V\mathit{\phi W},\mathit{\phi X}\right)=g\left(\phi {\nabla }_{V}W+\left({\nabla }_V\phi \right)W,\mathit{\phi X}\right)=g\left({\nabla }_{V}W,X\right)+g\left(P_{V}W+Q_{V}W,\mathit{\phi X}\right).$$E39

By using (1), we have

$$g\left(\mathit{\phi Y},Z\right)=-g\left(Y,\mathit{\phi Z}\right),$$

where $Y,Z\in \mathit{TM}$. Taking covariant derivative of above, we get

$$g\left(\left({\nabla }_X\phi \right)Y,Z\right)=-g\left(Y,\left({\nabla }_X\phi \right)Z\right),$$

using (23), we get

$$\begin{array}{cc}g\left(P_{X}Y+Q_{X}Y,Z\right)& =-g\left(Y,P_{X}Z+Q_{X}Z\right)\\ & =g\left(Y,P_{Z}X+Q_{Z}X\right).\end{array}$$E40

Using (40), we have

$$g\left(P_W\mathit{\phi X}+Q_W\mathit{\phi X},V\right)=g\left(\mathit{\phi X},P_{V}W+Q_{V}W\right).$$E41

Using (5), (38), (41) in (39), we have

$$g\left({\nabla }_V\mathit{\phi W},\mathit{\phi X}\right)=-g\left(V,W\right)\left(\mathit{\text{gradf}},X\right)+g\left(V,Q_W\mathit{\phi X}\right).$$

Since $\mathit{\phi W}$ is basic, so $\mathcal{H}{\nabla }_V\mathit{\phi W}=A_{\mathit{\phi W}}V$, therefore we have

$$g\left(A_{\mathit{\phi W}}V,\mathit{\phi X}\right)=-g\left(V,W\right)\left(\mathit{\text{gradf}},X\right)+g\left(V,Q_W\mathit{\phi X}\right),$$

$$g\left(V,A_{\mathit{\phi W}}\mathit{\phi X}\right)+g\left(V,Q_W\mathit{\phi X}\right)=g\left(V,W\right)\left(\mathit{\text{gradf}},X\right)$$E42

because $A$ is skew-symmetric. By using (42), we get the result.

Theorem 1.6 [34] Let $\pi$ be a Clairaut anti-invariant submersion from a nearly Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$ and $\mathit{\text{gradf}}\in \phi ker{\pi }_{\ast }$. Then either $f$ is constant on $\phi ker{\pi }_{\ast }$ or the fibers of $\pi$ are $1$-dimensional.

Proof: Using (5) and (38), we have

$$g\left({\nabla }_{V}W,\mathit{\phi U}\right)=-g\left(V,W\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right),$$

where $U,V,W\in \mathrm{\Gamma }\left(ker{\pi }_{\ast }\right)$. Since $g\left(W,\mathit{\phi U}\right)=0$. therefore we have

$$g\left(W,{\nabla }_V\mathit{\phi U}\right)=g\left(V,W\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right).$$E43

By use of (1) and (23) in (43), we get

$$g\left(W,Q_{V}U\right)-g\left(\mathit{\phi W},{\nabla }_{V}U\right)=g\left(V,W\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right).$$

By using (5), we obtain

$$g\left(W,Q_{V}U\right)-g\left(\mathit{\phi W},T_{V}U\right)=g\left(V,W\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right).$$

Now, using (38), we get

$$g\left(W,Q_{V}U\right)+g\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi W}\right)=g\left(V,W\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right)$$E44

Take $V=U$ in (44), we have

$$g\left(V,V\right)g\left(\mathit{\text{gradf}},\mathit{\phi W}\right)=g\left(V,W\right)g\left(\mathit{\text{gradf}},\mathit{\phi V}\right).$$E45

Interchange $V$ with $W$ in (45), we have

$$g\left(W,W\right)g\left(\mathit{\text{gradf}},\mathit{\phi V}\right)=g\left(V,W\right)g\left(\mathit{\text{gradf}},\mathit{\phi W}\right).$$E46

By (45) and (46), we have

$$g^2\left(V,W\right)g\left(\mathit{\text{gradf}},\mathit{\phi V}\right)=g\left(V,V\right)g\left(W,W\right)g\left(\mathit{\text{gradf}},\mathit{\phi V}\right).$$

Therefore either $f$ is constant on $\phi ker{\pi }_{\ast }$ or $V=\mathit{aW}$, where $a$ is constant (by using Schwarz’s Inequality for equality case).

Corollary 1.1 [34] Let $\pi$ be a Clairaut anti-invariant submersion from a nearly Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$ and $\mathit{\text{gradf}}\in \phi ker{\pi }_{\ast }$. If $dim\left(ker{\pi }_{\ast }\right)>1$, then the fibers of $\pi$ are totally geodesic if and only if $A_{\mathit{\phi W}}\mathit{\phi X}+Q_W\mathit{\phi X}=0$ for $W\in ker{\pi }_{\ast }$ such that $\mathit{\phi W}$ is basic and $X\in \mu$.

Proof: By Theorem 1.5 and Theorem 1.6, we get the result.

Corollary 1.2 [34] Let $\pi$ be an Clairaut Lagrangian submersion from a nearly Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$. Then either the fibers of $\pi$ are $1$-dimensional or they are totally geodesic.

Proof: Let $\pi$ be an Clairaut Lagrangian submersion from a Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$. Then $\mu =\left\{0\right\}$, so $A_{\mathit{\phi W}}\mathit{\phi X}+Q_W\mathit{\phi X}=0$ always.

Now, we discuss some examples for Clairaut anti-invariant submersions from a nearly Kähler manifold.

Example 1.2 [34] Let $\left({\mathbb{R}}^4,\phi ,g\right)$ be a nearly Kähler manifold endowed with Euclidean metric $g$ on ${\mathbb{R}}^4$ given by

$$g=\sum _{i=1}^4{\mathit{dx}}_i^2$$

and canonical complex structure

$$\phi \left(x_j\right)=\left\{\begin{array}{cc}-x_{j+1}& j=1,3\\ x_{j-1}& j=2,4\end{array}\right..$$

The $\phi$-basis is $\left\{e_i=\frac{\partial }{\partial x_i}|i=\mathrm{1,2,3,4}\right\}$. Let $\left({\mathbb{R}}^3,g_1\right)$ be a Riemannian manifold endowed with metric $g_1={\sum }_{i=1}^3{\mathit{dy}}_i^2$.

1. Consider a map $\pi :\left({\mathbb{R}}^4,\phi ,g\right)\to \left({\mathbb{R}}^3,g_1\right)$ defined by

$$\pi \left(x_1,x_2,x_3,x_4\right)=\left(\frac{x_1+x_2}{\sqrt{2}},x_3,x_4\right).$$

Then by direct calculations, we have

$$\begin{array}{ll}ker{\pi }_{\ast }& =\mathit{\text{span}}\left\{X_1=\left(\frac{\partial }{\partial x_1}-\frac{\partial }{\partial x_2}\right)\right\},\\ {\left(ker{\pi }_{\ast }\right)}^{\perp }& =\mathit{\text{span}}\left\{X_2=\left(\frac{\partial }{\partial x_1}+\frac{\partial }{\partial x_2}\right),X_3=\frac{\partial }{\partial x_3},X_4=\frac{\partial }{\partial x_4}\right\}\end{array}$$

and $\phi X_1=-X_2$, therefore $\phi \left(ker{\pi }_{\ast }\right)\subset {\left(ker{\pi }_{\ast }\right)}^{\perp }$. Thus, we can say that $\pi$ is an anti-invariant Riemannian submersion. Since the fibers of $\pi$ are $1$-dimensional, therefore fibers are totally umbilical.

Consider the Koszul formula for Levi-Civita connection $\nabla$ for ${\mathbb{R}}^4$

$$2g\left({\nabla }_{X}Y,Z\right)=\mathit{Xg}\left(Y,Z\right)+\mathit{Yg}\left(Z,X\right)-\mathit{Zg}\left(X,Y\right)-g\left(\left[Y,Z\right],X\right)-g\left(\left[X,Z\right],Y\right)+g\left(\left[X,Y\right],Z\right)$$

for all $X,Y,Z\in {\mathbb{R}}^4$. By simple calculations, we obtain

$${\nabla }_{e_i}{e}_j=0\text{for}\mathrm{all}i,j=\mathrm{1,2,3,4}.$$

Hence $T_{X}Y=T_{Y}X=T_{X}X=0$ for all $X,Y\in \mathrm{\Gamma }\left(ker{\pi }_{\ast }\right)$. Therefore fibers of $\pi$ are totally geodesic. Thus $\pi$ is Clairaut trivially.

1.   ii. Consider a map $\pi :\left({\mathbb{R}}^4,\phi ,g\right)\to \left({\mathbb{R}}^3,g_1\right)$ defined by

$$\pi \left(x_1,x_2,x_3,x_4\right)=\left(\sqrt{x_1^2+x_2^2},x_3,x_4\right).$$

Then by direct calculations, we have

$$\begin{array}{ll}ker{\pi }_{\ast }& =\mathit{\text{span}}\left\{X_1=\left(\frac{x_2}{\sqrt{x_1^2+x_2^2}}\frac{\partial }{\partial x_1}-\frac{x_1}{\sqrt{x_1^2+x_2^2}}\frac{\partial }{\partial x_2}\right)\right\},\\ {\left(ker{\pi }_{\ast }\right)}^{\perp }& =\mathit{\text{span}}\left\{X_2=\left(\frac{x_2}{\sqrt{x_1^2+x_2^2}}\frac{\partial }{\partial x_1}+\frac{x_1}{\sqrt{x_1^2+x_2^2}}\frac{\partial }{\partial x_2}\right),X_3=\frac{\partial }{\partial x_3},X_4=\frac{\partial }{\partial x_4}\right\}\end{array}$$

and $\phi X_1=-X_2$, therefore $\phi \left(ker{\pi }_{\ast }\right)\subset {\left(ker{\pi }_{\ast }\right)}^{\perp }$. Thus, we can say that $\pi$ is an anti-invariant Riemannian submersion. Since the fibers of $\pi$ are $1$-dimensional, therefore fibers are totally umbilical. By using Koszul formula, we obtain

$${\nabla }_{e_i}{e}_j=0\text{for}\mathrm{all}i,j=\mathrm{1,2,3,4}.$$

Hence

$$T_{X_1}{X}_1=-\left(\frac{x_2}{\sqrt{x_1^2+x_2^2}}\frac{\partial }{\partial x_1}+\frac{x_1}{\sqrt{x_1^2+x_2^2}}\frac{\partial }{\partial x_2}\right).$$

Now, for the function $f=ln\left(\sqrt{x_1^2+x_2^2}\right)$ on $\left({\mathbb{R}}^4,\phi ,g\right)$, the gradient of $f$ with respect to $g$ is given by

$$\mathit{\text{gradf}}=\sum _{i,j=1}^4{g}^{\mathit{ij}}\frac{\partial f}{\partial x_i}\frac{\partial }{\partial x_j}=\frac{x_2}{\sqrt{x_1^2+x_2^2}}\frac{\partial }{\partial x_1}+\frac{x_1}{\sqrt{x_1^2+x_2^2}}\frac{\partial }{\partial x_2}.$$

Therefore for $X_1\in \mathrm{\Gamma }\left(ker{\pi }_{\ast }\right)$, $T_{X_1}{X}_1=-\mathit{\text{gradf}}$. Since $‖X_1‖=1$, so $T_{X_1}{X}_1=-{‖X_1‖}^2\mathit{\text{gradf}}$. By using Theorem 1.2, we can say that $\pi$ is an proper Clairaut anti-invariant submersion with $r=e^f$ for $f=ln\left(\sqrt{x_1^2+x_2^2}\right)$.

Remark: From all results of this section, we can easily find conditions for anti-invariant Clairaut Submersions from Kähler manifolds.

6. Semi-invariant Riemannian submersion

Definition 1.4 Let $\left(M,\phi ,g\right)$ be an almost Hermitian manifold and $N$ be a Riemannian manifold with Riemannian metric $g_n$. A Riemannian submersion $\pi :M\to N$ is called a semi-invariant Riemannian submersion [11] if there is a distribution ${\mathcal{D}}_1\subseteq ker{\pi }_{\ast }$ such that

$$ker{\pi }_{\ast }={\mathcal{D}}_1\oplus {\mathcal{D}}_2\text{and}\phi \left({\mathcal{D}}_1\right)={\mathcal{D}}_1,\phi \left({\mathcal{D}}_2\right)\subseteq {\left(ker{\pi }_{\ast }\right)}^{\perp },$$

where ${\mathcal{D}}_2$ is orthogonal complementary to ${\mathcal{D}}_1$ in $ker{\pi }_{\ast }$. For $V\in \mathrm{\Gamma }\left(\mathit{ker}{\pi }_{\ast }\right)$, we have

$$\mathit{\phi V}=\mathit{\varphi V}+\mathit{\omega V},$$E47

where $\mathit{\varphi V}\in \mathrm{\Gamma }\left({\mathcal{D}}_1\right)$ and $\mathit{\omega V}\in \mathrm{\Gamma }\left(\phi {\mathcal{D}}_2\right)$.

Definition 1.5 A semi-invariant Riemannian submersion $\pi$ is said to be a Lagrangian Riemannian submersion [4] if $\phi \left(ker{\pi }_{\ast }\right)={\left(ker{\pi }_{\ast }\right)}^{\perp }$. Hence, if $\pi$ is a Lagrangian Riemannian submersion then for any $V\in \mathrm{\Gamma }\left(\mathit{ker}{\pi }_{\ast }\right),$ $\mathit{\phi V}=\mathit{\omega V},$ $\mathit{\varphi V}=0$ and for $X\in \mathrm{\Gamma }{\left(\mathit{ker}{\pi }_{\ast }\right)}^{\perp }$, $\mathit{\phi X}=\mathit{\alpha X}$, $\mathit{\beta X}=0$.

6.1 Semi-invariant Clairaut submersions from Kähler manifolds

In this section, we give new Clairaut conditions for semi-invariant submersions from Kähler manifolds after giving some auxiliary results.

Theorem 1.7 [40] Let $\pi$ be a semi-invariant submersion from a Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$. If $h:J\subset \mathbb{R}\to M$ is a regular curve and $U\left(s\right)$ and $X\left(s\right)$ are the vertical and horizontal parts of the tangent vector field $\dot{h}\left(s\right)=W$ of $h\left(s\right)$, respectively, then $h$ is a geodesic if and only if along $h$

$$\mathcal{V}{\nabla }_X\mathit{\varphi U}+\mathcal{V}{\nabla }_X\mathit{\alpha X}+A_X\mathit{\omega U}+A_X\mathit{\beta X}+{\widehat{\nabla }}_U\mathit{\varphi U}+T_U\mathit{\beta X}+T_U\mathit{\omega U}+{\widehat{\nabla }}_U\mathit{\alpha X}=0,$$E48

$$A_X\mathit{\varphi U}+A_X\mathit{\alpha X}+\mathcal{H}\left({\nabla }_{\dot{h}}\mathit{\omega U}+{\nabla }_{\dot{h}}\mathit{\beta X}\right)+T_U\mathit{\varphi U}+T_U\mathit{\alpha X}=0.$$E49

Proof: Let $\pi$ be a semi-invariant submersion from a Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$. Since ${\phi }^2\dot{h}=-\dot{h}$. Taking the covariant derivative of this and using (2), we have

$$\phi \left({\nabla }_{\dot{h}}\phi \dot{h}\right)=-{\nabla }_{\dot{h}}\dot{h}.$$E50

Since $U\left(s\right)$ and $X\left(s\right)$ are the vertical and horizontal parts of the tangent vector field $\dot{h}\left(s\right)=W$ of $h\left(s\right)$, that is, $\dot{h}=U+X$. So (50) becomes

$$\begin{array}{ll}-{\nabla }_{\dot{h}}\dot{h}& =\phi \left({\nabla }_{U+X}\phi \left(U+X\right)\right)\\ & =\phi \left({\nabla }_U\mathit{\phi U}+{\nabla }_X\mathit{\phi U}+{\nabla }_U\mathit{\phi X}+{\nabla }_X\mathit{\phi X}\right)\\ & =\phi \left({\nabla }_U\left(\mathit{\varphi U}+\mathit{\omega U}\right)+{\nabla }_X\left(\mathit{\varphi U}+\mathit{\omega U}\right)+{\nabla }_U\left(\mathit{\alpha X}+\mathit{\beta X}\right)+{\nabla }_X\left(\mathit{\alpha X}+\mathit{\beta X}\right)\right).\end{array}$$E51

Using (5)–(8) in (51), we get

$$\begin{array}{l}-{\nabla }_{\dot{h}}\dot{h}=\phi \left(\mathcal{H}\left({\nabla }_{\dot{h}}\mathit{\phi U}+{\nabla }_{\dot{h}}\mathit{\beta X}\right)+A_X\mathit{\alpha X}+A_X\mathit{\beta X}+A_X\mathit{\phi U}\right.\\ \left.+T_U\mathit{\beta X}+T_U\mathit{\alpha X}+\mathcal{V}{\nabla }_X\mathit{\alpha X}+T_U\mathit{\phi U}+{\widehat{\nabla }}_U\mathit{\alpha X}\right).\end{array}$$E52

Equating the vertical and horizontal part of (52), we obtain

$$\mathcal{V}\phi {\nabla }_{\dot{h}}\dot{h}=\mathcal{V}{\nabla }_X\mathit{\varphi U}+\mathcal{V}{\nabla }_X\mathit{\alpha X}+A_X\mathit{\omega U}+A_X\mathit{\beta X}+{\widehat{\nabla }}_U\mathit{\varphi U}+T_U\mathit{\beta X}+T_U\mathit{\omega U}+{\widehat{\nabla }}_U\mathit{\alpha X},$$

$$\mathcal{H}\phi {\nabla }_{\dot{h}}\dot{h}=A_X\mathit{\varphi U}+A_X\mathit{\alpha X}+\mathcal{H}\left({\nabla }_{\dot{h}}\mathit{\omega U}+{\nabla }_{\dot{h}}\mathit{\beta X}\right)+T_U\mathit{\varphi U}+T_U\mathit{\alpha X}.$$

By using above equations, we can say that $h$ is geodesic if and only if (48) and (49) hold.

Theorem 1.8 [40] Let $\pi$ be a semi-invariant submersion from a Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$. Also, let $h:J\subset \mathbb{R}\to M$ be a regular curve. $U\left(s\right)$ and $X\left(s\right)$ are the vertical and horizontal parts of the tangent vector field $\dot{h}\left(s\right)=W$ of $h\left(s\right)$. Then $\pi$ is a Clairaut submersion with $r=e^f$ if and only if along $h$

$$\begin{array}{l}g\left(\mathit{\text{gradf}},X\right)g\left(U,U\right)=g\left(\mathcal{H}{\nabla }_{\dot{h}}\mathit{\beta X}+A_X\mathit{\alpha X}+T_U\mathit{\alpha X},\mathit{\omega U}\right)\\ +g\left(\mathcal{V}{\nabla }_X\mathit{\alpha X}+A_X\mathit{\beta X}+T_U\mathit{\beta X}+{\widehat{\nabla }}_U\mathit{\alpha X},\mathit{\varphi U}\right).\end{array}$$

Proof: Let $h:J\subset \mathbb{R}\to M$ be a geodesic on $M$ and $\mathrm{\ell }={‖\dot{h}\left(s\right)‖}^2$. Let $\theta \left(s\right)$ be the angle between $\dot{h}\left(s\right)$ and the horizontal space at $h\left(s\right)$. Then

$$g\left(X\left(s\right),X\left(s\right)\right)=\mathrm{\ell }{cos}^2\theta \left(s\right),$$E53

$$g\left(U\left(s\right),U\left(s\right)\right)=\mathrm{\ell }{sin}^2\theta \left(s\right).$$E54

Differentiating (54), we get

$$2g\left({\nabla }_{\dot{h}\left(s\right)}U\left(s\right),U\left(s\right)\right)=2\mathrm{\ell }sin\theta \left(s\right)cos\theta \left(s\right)\frac{\mathit{d\theta }\left(s\right)}{\mathit{ds}}.$$E55

Using (1) in (55), we have

$$g\left({\nabla }_{\dot{h}\left(s\right)}\mathit{\phi U}\left(s\right),\mathit{\phi U}\left(s\right)\right))=\mathrm{\ell }sin\theta \left(s\right)cos\theta \left(s\right)\frac{\mathit{d\theta }\left(s\right)}{\mathit{ds}}.$$

Now by use of (47), we have

$$g\left({\nabla }_{\dot{h}\left(s\right)}\mathit{\varphi U}\left(s\right),\mathit{\phi U}\left(s\right)\right)+g\left({\nabla }_{\dot{h}\left(s\right)}\mathit{\omega U}\left(s\right),\mathit{\phi U}\left(s\right)\right)=\mathrm{\ell }sin\theta \left(s\right)cos\theta \left(s\right)\frac{\mathit{d\theta }\left(s\right)}{\mathit{ds}}$$

Along the curve $h$, using Theorem 1.7 and (5)–(8), we obtain

$$\begin{array}{l}\mathrm{\ell }sin\theta \left(s\right)cos\theta \left(s\right)\frac{\mathit{d\theta }\left(s\right)}{\mathit{ds}}=-g\left(\mathcal{H}{\nabla }_{\dot{h}}\mathit{\beta X}+A_X\mathit{\alpha X}+T_U\mathit{\alpha X},\mathit{\omega U}\right)\left(s\right)\\ -g\left(\mathcal{V}{\nabla }_X\mathit{\alpha X}+A_X\mathit{\beta X}+T_U\mathit{\beta X}+{\widehat{\nabla }}_U\mathit{\alpha X},\mathit{\varphi U}\right)\left(s\right)\end{array}$$

Now, $\pi$ is a Clairaut submersion with $r=e^f$ if and only if $\frac{d}{\mathit{ds}}\left(e^{f}sin\theta \right)=0$. Therefore

$$\begin{array}{r}{e}^f\left(\frac{\mathit{df}}{\mathit{ds}}sin\theta +cos\theta \frac{\mathit{d\theta }}{\mathit{ds}}\right)=0,\\ e^f\left(\frac{\mathit{df}}{\mathit{ds}}\mathrm{\ell }{sin}^2\theta +\mathrm{\ell }sin\theta cos\theta \frac{\mathit{d\theta }}{\mathit{ds}}\right)=0.\end{array}$$

So, we obtain

$$\begin{array}{l}\frac{\mathit{df}}{\mathit{ds}}\left(h\left(s\right)\right)g\left(U\left(s\right),U\left(s\right)\right)=g\left(\mathcal{H}{\nabla }_{\dot{h}}\mathit{\beta X}+A_X\mathit{\alpha X}+T_U\mathit{\alpha X},\mathit{\omega U}\right)\left(s\right)\\ +g\left(\mathcal{V}{\nabla }_X\mathit{\alpha X}+A_X\mathit{\beta X}+T_U\mathit{\beta X}+{\widehat{\nabla }}_U\mathit{\alpha X},\mathit{\varphi U}\right)\left(s\right),\end{array}$$E56

Since $\frac{\mathit{df}}{\mathit{ds}}\left(h\left(s\right)\right)=\dot{h}\left[g\right]\left(s\right)=g\left(\mathit{\text{gradf}},\dot{h}\left(s\right)\right)=g\left(\mathit{\text{gradf}},X\right)$. Therefore by using (56), we get the result.

Theorem 1.9 [40] Let $\pi$ be a Clairaut semi-invariant submersion from a Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$. Then

$$g\left(A_{\mathit{\omega W}}V,\mathit{\beta X}\right)+g\left(\mathcal{H}{\nabla }_V\mathit{\varphi W},\mathit{\beta X}\right)+g\left(\mathcal{V}{\nabla }_V\mathit{\omega W},\mathit{\alpha X}\right)+g\left({\widehat{\nabla }}_V\mathit{\varphi W},\mathit{\alpha X}\right)=-g\left(V,W\right)\left(\mathit{Xf}\right)$$

for $X\in \mathrm{\Gamma }\mu$, $V,W\in \mathrm{\Gamma }\left({\mathcal{D}}_2\right)$ and $\mathit{\omega W}$ is basic.

Proof: Let $\pi$ be a Clairaut semi-invariant submersion from a Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$. We know that any fiber of Riemannian submersion $\pi$ is totally umbilical if and only if

$$T_{V}W=g\left(V,W\right)H,$$E57

for all $V,W\in \mathrm{\Gamma }\left(ker{\pi }_{\ast }\right)$, where $H$ denotes the mean curvature vector field of any fiber in $M$. By using Theorem 1.2 and (57), we have

$$T_{V}W=-g\left(V,W\right)\mathit{\text{gradf}}.$$E58

Let $X\in \mu$ and $V,W\in \mathrm{\Gamma }\left(ker{\pi }_{\ast }\right)$, then by using (1) and (2), we have

$$g\left({\nabla }_V\mathit{\phi W},\mathit{\phi X}\right)=g\left(\phi {\nabla }_{V}W+\left({\nabla }_V\phi \right)W,\mathit{\phi X}\right)=g\left({\nabla }_{V}W,X\right).$$E59

Using (5), (58) in (59), we have

$$g\left({\nabla }_V\mathit{\phi W},\mathit{\phi X}\right)=-g\left(V,W\right)\left(\mathit{\text{gradf}},X\right).$$

Since $\mathit{\omega W}$ is basic, so $\mathcal{H}{\nabla }_V\mathit{\omega W}=A_{\mathit{\omega W}}V$, therefore we have

$$g\left(A_{\mathit{\omega W}}V,\mathit{\beta X}\right)+g\left(\mathcal{H}{\nabla }_V\mathit{\varphi W},\mathit{\beta X}\right)+g\left(\mathcal{V}{\nabla }_V\mathit{\omega W},\mathit{\alpha X}\right)+g\left({\widehat{\nabla }}_V\mathit{\varphi W},\mathit{\alpha X}\right)=-g\left(V,W\right)\left(\mathit{\text{gradf}},X\right).$$E60

Theorem 1.10 [40] Let $\pi$ be a Clairaut semi-invariant submersion from a Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$ and $V,W\in \mathrm{\Gamma }\left({\mathcal{D}}_1\right)$ Then $\mathit{\text{gradf}}\in \mathrm{\Gamma }\left(\phi ker{\pi }_{\ast }\right)$.

Proof: Let $V,W\in \mathrm{\Gamma }\left({\mathcal{D}}_1\right)$ and $X\in \mathrm{\Gamma }\left(\mu \right)$. Using (5), (47) and (58) in

$${\nabla }_V\mathit{\phi U}=\phi {\nabla }_{V}U+\left({\nabla }_V\phi \right)U,$$

we have

$$T_V\mathit{\varphi U}+\mathcal{V}{\nabla }_V\mathit{\varphi U}=\alpha T_{V}U+\beta T_{V}U+\varphi \mathcal{V}{\nabla }_{V}U+\omega \mathcal{V}{\nabla }_{V}U,$$

which gives

$$-g\left(V,\mathit{\varphi U}\right)g\left(\mathit{\text{gradf}},X\right)=g\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi X}\right).$$E61

By interchanging $U$ and $V$ in (61) and adding the resulting equation with (61), we get

$$g\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi X}\right)=0,$$

which gives $g\left(\mathit{\text{gradf}},\mathit{\phi X}\right)=0$. Therefore $\mathit{\text{gradf}}\in \mathrm{\Gamma }\left(\phi ker{\pi }_{\ast }\right)$.

Theorem 1.11 [40] Let $\pi$ be a Clairaut semi-invariant submersion from a Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$ and $\mathit{\text{gradf}}\in \mathrm{\Gamma }\left(\phi ker{\pi }_{\ast }\right)$. Then either $f$ is constant on $\phi ker{\pi }_{\ast }$ or the fibers of $\pi$ are $1$-dimensional.

Proof: Let $U,V\in \mathrm{\Gamma }\left({\mathcal{D}}_2\right)$. Using (5) and (58), we have

$$g\left({\nabla }_{V}U,\mathit{\phi U}\right)=-g\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right),$$

which gives

$$g\left(\phi {\nabla }_{V}U,U\right)=g\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right),$$

since $ker{\pi }_{\ast }$ is integrable, so we have

$$g\left(\phi {\nabla }_{U}V,U\right)=g\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right),$$

which equals to

$$g\left({\nabla }_U\mathit{\phi V}-\left({\nabla }_U\phi \right)V,U\right)=g\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right).$$

Since $g\left(\mathit{\phi V},U\right)=0$. therefore we have

$$g\left(\mathit{\phi V},{\nabla }_{U}U\right)=-g\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right).$$E62

By using (5) in (62), we obtain

$$g\left(\mathit{\phi V},T_{U}U\right)=-g\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right).$$

Now, using (58), we get

$$g\left(U,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi V}\right)=g\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right).$$E63

Interchanging $V$ and $U$ in (63), we have.

$$g\left(V,V\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right)=g\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi V}\right).$$E64

By (63) and (64), we have

$$g^2\left(V,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right)=g\left(V,V\right)g\left(U,U\right)g\left(\mathit{\text{gradf}},\mathit{\phi U}\right).$$

Therefore either $f$ is constant on $\phi ker{\pi }_{\ast }$ or $V=\mathit{aU}$, where $a$ is constant (by using Schwarz’s Inequality for equality case).

Since $ker{\pi }_{\ast }$ is CR-submanifold of Kähler manifold $\left(M,\phi ,g\right)$, therefore by using [41], Theorem 6.1, p. 96], we can state that.

Theorem 1.12 [40] Let $\pi$ be a Clairaut semi-invariant submersion from a Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$. If dim ${\mathcal{D}}_2>1$, then fibers are totally geodesic.

Corollary 1.3 [40] Let $\pi$ be a Clairaut Lagrangian submersion from a Kähler manifold $\left(M,\phi ,g\right)$ onto a Riemannian manifold $\left(N,g_n\right)$ with $r=e^f$. Then either the fibers of $\pi$ are $1$-dimensional or they are totally geodesic.

Lastly, we discuss some examples for Clairaut semi-invariant submersions [40] from a Kähler manifold.

Example 1.3 Every Clairaut anti-invariant submersion from a Kähler manifold onto a Riemannian manifold is a Clairaut semi-invariant submersion with ${\mathcal{D}}_1=\left\{0\right\}$.

Example 1.4 Let $\left({\mathbb{R}}^6,\phi ,g\right)$ be a Kähler manifold endowed with Euclidean metric $g$ on ${\mathbb{R}}^6$ given by

$$g=\sum _{i=1}^6{\mathit{dx}}_i^2$$

and canonical complex structure

$$\phi \left(x_j\right)=\left\{\begin{array}{cc}-x_{j+1}& j=\mathrm{1,3,5}\\ x_{j-1}& j=\mathrm{2,4,6}\end{array}\right..$$

The $\phi$-basis is $\left\{e_i=\frac{\partial }{\partial x_i}|i=1,\dots ,6\right\}$. Let $\left({\mathbb{R}}^3,g_1\right)$ be a Riemannian manifold endowed with metric $g_1=\sum _{i=1}^3{\mathit{dy}}_i^2$.

Consider a map $\pi :\left({\mathbb{R}}^6,\phi ,g\right)\to \left({\mathbb{R}}^3,g_1\right)$ defined by

$$\pi \left(x_1,x_2,x_3,x_4,x_5,x_6\right)=\left(\frac{x_1+x_2}{\sqrt{2}},x_3,x_4\right).$$

Then by direct calculations, we have

$$\begin{array}{l}ker{\pi }_{\ast }=\mathit{\text{span}}\left\{X_1=\left(\frac{\partial }{\partial x_1}-\frac{\partial }{\partial x_2}\right),X_2=\frac{\partial }{\partial x_5},X_3=\frac{\partial }{\partial x_6}\right\},\\ {\left(ker{\pi }_{\ast }\right)}^{\perp }=\mathit{\text{span}}\left\{V_1=\left(\frac{\partial }{\partial x_1}+\frac{\partial }{\partial x_2}\right),V_2=\frac{\partial }{\partial x_3},V_3=\frac{\partial }{\partial x_4}\right\}\end{array}$$

and $\phi X_1=-V_1$, $\phi X_2=-X_3$, $\phi X_3=-X_2$ therefore ${\mathcal{D}}_1=\mathit{\text{span}}\left\{X_2,X_3\right\}$ and ${\mathcal{D}}_2=\mathit{\text{span}}\left\{X_1\right\}$. Thus, we can say that $\pi$ is a semi-invariant Riemannian submersion.

Consider the Koszul formula for Levi-Civita connection $\nabla$ for ${\mathbb{R}}^6$

$$2g\left({\nabla }_{X}Y,Z\right)=\mathit{Xg}\left(Y,Z\right)+\mathit{Yg}\left(Z,X\right)-\mathit{Zg}\left(X,Y\right)-g\left(\left[Y,Z\right],X\right)-g\left(\left[X,Z\right],Y\right)+g\left(\left[X,Y\right],Z\right)$$

for all $X,Y,Z\in {\mathbb{R}}^6.$ By simple calculations, we obtain

$${\nabla }_{e_i}{e}_j=0\text{for}\mathrm{all}i,j=1,\dots ,6.$$

Hence $T_{X}Y=T_{Y}X=T_{X}X=0$ for all $X,Y\in \mathrm{\Gamma }\left(ker{\pi }_{\ast }\right)$. Therefore fibers of $\pi$ are totally geodesic. Thus $\pi$ is Clairaut trivially.

Classification

2010 mathematics subject classification. 53C12, 53C15, 53C20, 53C55