1 Introduction

Given a \(C^{\infty }\)-submersion F from a (semi)-Riemannian manifold \((N,g_{N})\) onto a (semi)-Riemannian manifold \((B,g_{B})\), according to the circumstances on the map \(F:(N,g_{N})\rightarrow (B,g_{B})\), we get the following:

Riemannian submersion (Falcitelli et al. 2004; O’Neill 1966; Gray 1967), almost Hermitian submersion (Watson 1976), paracontact paracomplex submersios (Gündüzalp and Ṣahin 2014), quaternionic submersion (Ianus et al. 2008), slant submersion (Akyol and Gündüzalp 2016; Gündüzalp 2013b; Gündüzalp and Akyol 2018; Ṣahin 2011), anti-invariant submersion (Beri et al. 2016; Ṣahin 2010), Clairaut submersion (Bishop 1972; Gündüzalp 2019; Taṣtan and Gerdan 2017; Lee et al. 2015; Allison 1996), conformal anti-invariant submersion (Akyol 2017; Akyol and Ṣahin 2016), etc.

In the present paper, we take into account Clairaut anti-invariant submersions from a locally product Riemannian manifold onto a Riemannian manifold. In Sect. 2, we recall some concepts, which are needed in the following section. In Sect. 3, we first obtain necessary and sufficient conditions for a curve on the manifold N of anti-invariant submersions to be geodesic. Then we present a new characterization for Clairaut anti-invariant submersions. Also, we present an example.

2 Preliminaries

2.1 Riemannian submersions

A \(C^{\infty }\)-submersion \(F:N\rightarrow B\) between two Riemannian manifolds \((N,g_{N})\) and \((B,g_{B})\) is called a Riemannian submersion if it satisfies conditions:

  1. (i)

    the fibres \(F^{-1}(b)\), \(b\in B,\) are r-dimensional Riemannian submanifolds of N,  where \(r=dim(N)-dim(B).\)

  2. (ii)

    \(F_{*}\) preserves the lengths of horizontal vectors.

    The vectors tangent to the fibres are called vertical and those normal to the fibres are called horizontal. We denote by \((kerF_{*})\) the vertical distribution, by \((kerF_{*})^{\bot }\) the horizontal distribution and by v and h the vertical and horizontal projection. A horizontal vector field \(X_{1}\) on N is said to be fundamental if \(X_{1}\) is F-related to a vector field \(X_{*1}\) on B.

A Riemannian submersion \(F : N\rightarrow B\) defines two (1, 2) tensor fields \({\mathcal {T}}\) and \({\mathcal {A}}\) on N,  by the formulas:

$$\begin{aligned} {\mathcal {T}}_{X_{1}}X_{2}=h\nabla _{vX_{1}}vX_{2}+v\nabla _{vX_{1}}hX_{2} \end{aligned}$$
(1)

and

$$\begin{aligned} {\mathcal {A}}_{X_{1}}X_{2}=v\nabla _{hX_{1}}hX_{2}+h\nabla _{hX_{1}}vX_{2} \end{aligned}$$
(2)

for any \(X_{1},X_{2}\in \chi (N)\) (see Falcitelli et al. 2004). Using (1) and (2), one can get

$$\begin{aligned} \nabla _{U_{1}}U_{2}&={\mathcal {T}}_{U_{1}}U_{2}+\hat{\nabla }_{U_{1}}U_{2}; \end{aligned}$$
(3)
$$\begin{aligned} \nabla _{U_{1}}X_{1}&={\mathcal {T}}_{U_{1}}X_{1}+h(\nabla _{U_{1}}X_{1}); \end{aligned}$$
(4)
$$\begin{aligned} \nabla _{X_{1}}U_{1}&={\mathcal {A}}_{X_{1}}U_{1}+v(\nabla _{X_{1}}U_{1}), \end{aligned}$$
(5)
$$\begin{aligned} \nabla _{X_{1}}X_{2}&={\mathcal {A}}_{X_{1}}X_{2}+h(\nabla _{X_{1}}X_{2}), \end{aligned}$$
(6)

for any \(X_{1},X_{2}\in \Gamma ((kerF_{*})^{\bot }),\) \(U_{1},U_{2}\in \Gamma (kerF_{*}).\) In addition, if \(X_{1}\) is basic then \(h(\nabla _{U_{1}}X_{1})=h(\nabla _{X_{1}}U_{1})={\mathcal {A}}_{X_{1}}U_{1}.\)

The fundamental tensor fields \({\mathcal {T}}, {\mathcal {A}}\) satisfy:

$$\begin{aligned} {\mathcal {T}}_{U_{1}}U_{2}&={\mathcal {T}}_{U_{2}}U_{1},\,\,\,\,\,U_{1},U_{2} \in \Gamma (kerF_{*}); \end{aligned}$$
(7)
$$\begin{aligned} {\mathcal {A}}_{X_{1}}X_{2}&=-{\mathcal {A}}_{X_{2}}X_{1} =\frac{1}{2}v[X_{1},X_{2}],\,\,\,\,X_{1},X_{2} \in \Gamma ((kerF_{*})^{\bot }). \end{aligned}$$
(8)

2.2 Anti-invariant submersions

Let N be a n-dimensional smooth manifold. If it is endowed with a structure \((P,g_{N}),\) where P is a (1, 1) tensor, and \(g_{N}\) is a Riemannian metric, satisfying

$$\begin{aligned} P^{2}X_{1}=X_{1}, \quad g_{N}(PX_{1},X_{2})=g_{N}(X_{1},PX_{2}), \end{aligned}$$
(9)

for any \(X_{1},X_{2}\in \chi (N)\), it is called an almost product Riemannian manifold. An almost product Riemannian manifold N is called a locally product Riemannian manifold if

$$\begin{aligned} \nabla P=0, \end{aligned}$$
(10)

where \(\nabla \) is the Riemannian connection on N (Yano and Kon 1984).

Definition 2.1

(Gündüzalp 2013a) Let \((N,g_{N},P)\) be an almost product Riemannian manifold and \((B,g_{B})\) a Riemannian manifold. Suppose that there exists a Riemannian submersion \(F: N \rightarrow B\) such that \(kerF_{*}\) is anti-invariant with respect to P,  i.e., \(P(kerF_{*})\subseteq (kerF_{*})^{\bot }.\) At that time, we call F is an anti-invariant Riemannian submersion.

In this case, the horizontal distribution \((kerF_{*})^{\bot }\) decomposed as

$$\begin{aligned} (kerF_{*})^{\bot }=P(kerF_{*})\oplus \eta , \end{aligned}$$
(11)

where \(\eta \) is the complementary orthogonal distribution of \(P(kerP_{*})\) in \((kerF_{*})^{\bot }\) and it is invariant with respect to P.

For any \(X_{1}\in \Gamma (kerF_{*})^{\bot },\) we write

$$\begin{aligned} PX_{1}=DX_{1}+EX_{1}, \end{aligned}$$
(12)

where \(DX_{1}\) and \(EX_{1}\) are vertical and horizontal components of \(PX_{1}.\) If \(\eta ={0}\), at that time an anti-invariant submersion is called a Lagrangian submersion.

3 Clairaut anti-invariant submersions

Let M be a revolution surface in \(R^{3}\) with rotation axis d. \(\forall x\in M\), we state by r(x) the distance from x to d. Given a geodesic \(c:J\subset R\rightarrow M\) on M,  let \(\varphi (s)\) be the angle between c(s) and the meridian curve through \(c(s), s\in J.\) A well-known Clairaut’s theorem tells that for any geodesic c on M the product \(r\sin \varphi \) is constant along c,  i.e., it is independent of s. In the submersion theory, Bishop (1972) shows the concept of Clairaut submersion in the following way.

Definition 3.1

(Bishop 1972) A Riemannian submersion \(F:(N,g_{N},P)\rightarrow (B,g_{B})\) is called a Clairaut submersion if there exists a positive function r on N such that, for any geodesic c on N,  the function \((r\circ c)\sin \varphi \) is constant, where, for any s,  \(\varphi (s)\) is the angle between \(\dot{c}(s)\) and the horizontal space at c(s). Moreover, he gave a necessary and sufficient condition for a Riemannian submersion to be a Clairaut submersion.

Theorem 3.1

(Bishop 1972) Let \(F:(N,g_{N},P)\rightarrow (B,g_{B})\) be Riemannian submersion with connected fibres. Then, F is a Clairaut submersion with \(r=e^{g}\) if and only if each fibre is completely umbilical and has the mean curvature vector field \(H=-\nabla g,\) where \(\nabla g\) is the gradient of the function g with respect to \(g_{N}.\)

Proposition 3.1

Let F be an anti-invariant submersion from a locally product Riemannian manifold \((N,g_{N},P)\) onto a Riemannian manifold \((B,g_{B}).\) If \(c:J\subset R\rightarrow N\) is a regular curve and \(U_{1}(s)\) and \(X_{1}(s)\) are the vertical and horizontal parts of the tangent vector field \(\dot{c}(s)=W\) of c(s), respectively, then c is a geodesic if and only if along c

$$\begin{aligned} v\nabla _{\dot{c}}{\mathcal {D}}X_{1}+{\mathcal {A}}_{X_{1}}PU_{1} +{\mathcal {T}}_{U_{1}}PU_{1}+({\mathcal {A}}_{X_{1}} +{\mathcal {T}}_{U_{1}})EX_{1}=0, \end{aligned}$$
(13)

and

$$\begin{aligned} h\nabla _{\dot{c}}EX_{1}+h\nabla _{\dot{c}}PU_{1} +({\mathcal {A}}_{X_{1}}+{\mathcal {T}}_{U_{1}}){\mathcal {D}}X_{1}=0. \end{aligned}$$
(14)

Proof

From (10), we obtain

$$\begin{aligned} \nabla _{\dot{c}}\dot{c}=P(\nabla _{\dot{c}}P\dot{c}). \end{aligned}$$
(15)

Since \(\dot{c}=U_{1}+X_{1},\) we can write

$$\begin{aligned} \nabla _{\dot{c}}\dot{c}=P(\nabla _{U_{1}+X_{1}}P(U_{1}+X_{1})). \end{aligned}$$
(16)

By direct computations, we get

$$\begin{aligned} \nabla _{\dot{c}}\dot{c}=P(\nabla _{U_{1}}PU_{1} +\nabla _{U_{1}}PX_{1}+\nabla _{X_{1}}PU_{1}+\nabla _{X_{1}}PX_{1}). \end{aligned}$$

Using (12), we get

$$\begin{aligned} \nabla _{\dot{c}}\dot{c}=P(\nabla _{U_{1}}PU_{1}+\nabla _{U_{1}}(DX_{1}+EX_{1}) +\nabla _{X_{1}}PU_{1}+\nabla _{X_{1}}(DX_{1}+EX_{1})). \end{aligned}$$

Using (3)–(6), we have

$$\begin{aligned} \nabla _{\dot{c}}\dot{c}= & {} P(h(\nabla _{\dot{c}}PU_{1} +\nabla _{\dot{c}}EX_{1})+({\mathcal {A}}_{X_{1}} +{\mathcal {T}}_{U_{1}})(DX_{1}+EX_{1})\\&+v\nabla _{\dot{c}}DX_{1}+{\mathcal {A}}_{X_{1}}PU_{1} +{\mathcal {T}}_{U_{1}}PU_{1}). \end{aligned}$$

Taking the vertical and horizontal pieces of this equation. we have

$$\begin{aligned} v\nabla _{\dot{c}}{\mathcal {D}}X_{1}+{\mathcal {A}}_{X_{1}}PU_{1} +{\mathcal {T}}_{U_{1}}PU_{1}+({\mathcal {A}}_{X_{1}} +{\mathcal {T}}_{U_{1}})EX_{1}=vP\nabla _{\dot{c}}\dot{c} \end{aligned}$$
(17)

and

$$\begin{aligned} h\nabla _{\dot{c}}EX_{1}+h\nabla _{\dot{c}}PU_{1} +({\mathcal {A}}_{X_{1}}+{\mathcal {T}}_{U_{1}}){\mathcal {D}}X_{1} =hP\nabla _{\dot{c}}\dot{c}. \end{aligned}$$
(18)

From (17) and (18), it is simple to see that c is a geodesic if and only if (13) and (14) hold. \(\square \)

Theorem 3.2

Let F be an anti-invariant submersion from a locally product Riemannian manifold \((N,g_{N},P)\) onto a Riemannian manifold \((B,g_{B}).\) At that time F is a Clairaut submersion with \(r=e^{g}\) if and only if along c the following equation holds

$$\begin{aligned} g_{N}(h\nabla _{\dot{c}}EX_{1}+({\mathcal {A}}_{X_{1}} +{\mathcal {T}}_{U_{1}}){\mathcal {D}}X_{1},PU_{1}) =g_{N}(\nabla g,X_{1})\Vert U_{1}\Vert ^{2}, \end{aligned}$$
(19)

where \(U_{1}(s)\) and \(X_{1}(s)\) are the vertical and horizontal parts of the tangent vector field \(\dot{c}(s)\) of the geodesic c(s) on N, severally.

Proof

Let c(s) be a geodesic with speed \(\sqrt{b}\) on N,  at that time, we get

$$\begin{aligned} b=\Vert \dot{c}(s)\Vert ^{2}. \end{aligned}$$

Thence, we conclude that

$$\begin{aligned} g_{N}(X_{1}(s),X_{1}(s)) =b\cos ^{2}\varphi (s),\quad g_{N}(U_{1}(s),U_{1}(s))=b\sin ^{2}\varphi (s), \end{aligned}$$
(20)

where \(\varphi (s)\) is the angle between \(\dot{c}(s)\) and the horizontal space at c(s). Differentiating the second expression in (20), we get

$$\begin{aligned} \frac{d}{ds}g_{N}(U_{1}(s),U_{1}(s)) =2g_{N}(\nabla _{\dot{c}(s)}U_{1}(s),U_{1}(s)) =2b\cos \varphi (s)\sin \varphi (s)\frac{d\varphi }{ds}(s). \end{aligned}$$
(21)

Thus, using (9) and (10), we obtain

$$\begin{aligned} g_{N}(h\nabla _{\dot{c}(s)}PU_{1}(s),PU_{1}(s)) =b\cos \varphi (s)\sin \varphi (s)\frac{d\varphi }{ds}(s). \end{aligned}$$
(22)

By (14), we arrive at along c, 

$$\begin{aligned} -g_{N}(h\nabla _{\dot{c}}EX_{1}+({\mathcal {A}}_{X_{1}} +{\mathcal {T}}_{U_{1}}){\mathcal {D}}X_{1},PU_{1}) =b\cos \varphi \sin \varphi \frac{d\varphi }{ds}. \end{aligned}$$
(23)

Moreover, F is a Clairaut anti-invariant submersion with \(r=e^{g}\) if and only if

$$\begin{aligned} \frac{d}{ds}(e^{g}\sin \varphi )=0\Leftrightarrow e^{g} \left( \frac{dg}{ds}\sin \varphi +\cos \varphi \frac{d\varphi }{ds}\right) =0. \end{aligned}$$

Striking recent equation with non-zero element \(b\sin \varphi ,\) we obtain

$$\begin{aligned} \frac{dg}{ds}b\sin ^{2}\varphi +b\cos \varphi \sin \varphi \frac{d\varphi }{ds}=0. \end{aligned}$$
(24)

From (23) and (24), we have

$$\begin{aligned} g_{N}(h\nabla _{\dot{c}}EX_{1}+({\mathcal {A}}_{X_{1}} +{\mathcal {T}}_{U_{1}}){\mathcal {D}}X_{1},PU_{1}) =\frac{dg}{ds}(c(s))\Vert U_{1}\Vert ^{2}. \end{aligned}$$
(25)

Since \(\frac{dg}{ds}(c(s))=\dot{c}[g]=g_{N}(\nabla g,\dot{c})=g_{N}(\nabla g,X_{1}),\) the claim (19) follows from (25). \(\square \)

Theorem 3.3

Let F be a Clairaut anti-invariant submersion from a locally product Riemannian manifold \((N,g_{N},P)\) onto a Riemannian manifold \((B,g_{B})\) with \(r=e^{g}.\) At that time, we get

$$\begin{aligned} {\mathcal {A}}_{PU_{3}}PX_{1}=X_{1}(g)U_{3} \end{aligned}$$
(26)

for \(X_{1}\in \eta \) and \(U_{3}\in kerF_{\star }\) such that \(PU_{3}\) is basic.

Proof

From Theorem 3.1, we obtain

$$\begin{aligned} {\mathcal {T}}_{U_{1}}U_{2}=-g_{N}(U_{1},U_{2})\nabla g, \end{aligned}$$
(27)

where \(U_{1},U_{2}\in kerF_{\star }.\) If we crash this equation by \(PU_{3}\), \(U_{3}\in kerF_{\star }\) such that \(PU_{3}\) is fundamental and from (3), we get

$$\begin{aligned} g_{N}(\nabla _{U_{1}}U_{2},PU_{3})=-g_{N}(U_{1},U_{2})g_{N}(\nabla g,PU_{3}). \end{aligned}$$

Thus, we have

$$\begin{aligned} g_{N}(\nabla _{U_{1}}PU_{3},U_{2})=g_{N}(U_{1},U_{2})g_{N}(\nabla g,PU_{3}), \end{aligned}$$

since \(g_{N}(U_{2},PU_{3})=0.\)

By (10), we get

$$\begin{aligned} g_{N}(P\nabla _{U_{1}}U_{3},U_{2}) =g_{N}(U_{1},U_{2})g_{N}(\nabla g,PU_{3}). \end{aligned}$$

Using (9), we arrive at

$$\begin{aligned} g_{N}(\nabla _{U_{1}}U_{3},PU_{2})=g_{N}(U_{1},U_{2})g_{N}(\nabla g,PU_{3}). \end{aligned}$$

Again, using (3), we obtain

$$\begin{aligned} g_{N}({\mathcal {T}}_{U_{1}}U_{3},PU_{2}) =g_{N}(U_{1},U_{2})g_{N}(\nabla g,PU_{3}). \end{aligned}$$

Thus, by (27),

$$\begin{aligned} -g_{N}(U_{1},U_{3})g_{N}(\nabla g,PU_{2})=g_{N}(U_{1},U_{2})g_{N}(\nabla g,PU_{3}) \end{aligned}$$
(28)

If take \(U_{1}=U_{3}\) and exchange \(U_{1}\) with \(U_{2}\) in (28), we provide

$$\begin{aligned} -\Vert U_{2}\Vert ^{2}g_{N}(\nabla g,PU_{1})=g_{N}(U_{1},U_{2})g_{N}(\nabla g,PU_{2}). \end{aligned}$$
(29)

Using (28) with \(U_{1}=U_{3}\) and (29), we get

$$\begin{aligned} -g_{N}(\nabla g,PU_{1})=\frac{g_{N}^{2}(U_{1},U_{2})}{\Vert U_{2}\Vert ^{2}\Vert U_{1}\Vert ^{2}}g_{N}(\nabla g,PU_{1}). \end{aligned}$$
(30)

On the other hand, using (10), we obtain

$$\begin{aligned} g_{N}(\nabla _{U_{2}}PU_{3},PX_{1})=g_{N}(P\nabla _{U_{2}}U_{3},PX_{1}) \end{aligned}$$

for any \(X_{1}\in \eta .\) Thus, using (9), we get

$$\begin{aligned} g_{N}(\nabla _{U_{2}}PU_{3},PX_{1})=g_{N}(\nabla _{U_{2}}U_{3},X_{1}). \end{aligned}$$

Using (3) and (27), we obtain

$$\begin{aligned} g_{N}(\nabla _{U_{2}}PU_{3},PX_{1})=-g_{N}(U_{2},U_{3})g_{N}(\nabla g,X_{1}). \end{aligned}$$
(31)

Since \(PU_{3}\) is fundamental and from \(h\nabla _{U_{2}}PU_{3}={\mathcal {A}}_{PU_{3}}U_{2}\), we have

$$\begin{aligned} g_{N}(h\nabla _{U_{2}}PU_{3},PX_{1}) =g_{N}({\mathcal {A}}_{PU_{3}}U_{2},PX_{1}). \end{aligned}$$
(32)

Using (31),(32) and the anti-symmetry of \({\mathcal {A}},\) we find

$$\begin{aligned} g_{N}({\mathcal {A}}_{PU_{3}}PX_{1},U_{2}) =g_{N}(\nabla g,X_{1})g_{N}(U_{3},U_{2}). \end{aligned}$$
(33)

Since \({\mathcal {A}}_{PU_{3}}PX_{1},\) \(U_{2}\) and \(U_{3}\) are vertical and \(\nabla g\) is horizontal, we derive (26).

Now, if \(\nabla g\in PkerF\star \), then from (30) and the equality situation of Schwarz inequality, we get the following. \(\square \)

Corollary 3.1

Let F be a Clairaut anti-invariant submersion from a locally product Riemannian manifold \((N,g_{N},P)\) onto a Riemannian manifold \((B,g_{B})\) with \(r=e^{g}.\) If \(\nabla g\in PkerF\star \), at that time, either g is constant on \(PkerF\star \) or the fibres of F are 1-dimensional.

Furthermore, while the function g is constant, \(\nabla g\equiv 0.\) Hence, by Theorem 3.1 and Corollary 3.1, we get that:

Corollary 3.2

Let F be a Clairaut anti-invariant submersion from a locally product Riemannian manifold \((N,g_{N},P)\) onto a Riemannian manifold \((B,g_{B})\) with \(r=e^{g}\) and \(\nabla g\in PkerF\star .\) If \(dim(kerF*)>1,\) at that time, the fibres of F are completely geodesic if and only if \({\mathcal {A}}_{PU_{3}}PX_{1}=0\) for \(U_{3}\in kerF_{*}\) such that \(PU_{3}\) is fundamental and \(X_{1}\in \eta .\)

In addition, if the anti-invariant submersion F in Theorem 3.3 is Lagrangian, at that time, \({\mathcal {A}}_{PU_{3}}PX_{1}=0\) always zero, since \(\eta =\{0\}.\) Hence, we obtain that:

Corollary 3.3

Let F be a Clairaut Lagrangian submersion from a locally product Riemannian manifold \((N,g_{N},P)\) onto a Riemannian manifold \((B,g_{B})\) with \(r=e^{g}.\) Then either the fibres of F are one dimensional or they are totally geodesic.

Now, we present example of a Clairaut submersion.

Example 3.1

Let N be a Euclidean 3-space defined by \(N=\{(x_{1},x_{2},x_{3})\in R^{3}:(x_{1},x_{2})\ne (0,0)\) and \(x_{3}\ne 0\}\).

We consider the product structure \((P,g_{N})\) on N given by \(g_{N}=e^{2x_{3}}(dx_{1})^{2}+e^{2x_{3}}(dx_{2})^{2}+(dx_{3})^{2}\) and \(P(a,b,c)=(a,-b,c).\)

A P-basis can be given by \(\{e_{1}=e^{-x_{3}}\frac{\partial }{\partial x_{1}},e_{2}=e^{-x_{3}}\frac{\partial }{\partial x_{2}}, e_{3}=\frac{\partial }{\partial x_{3}}\}.\)

Let B be \(\{(t,x_{3})\in R^{2}\}.\) We select the metric \(g_{B}\) on B,  \(g_{B}=e^{2x_{3}}(dt)^{2}+(dx_{3})^{2}\). Now, we defined a map \(F: (N,P,g_{N})\rightarrow (B,g_{B})\) by

$$\begin{aligned} F(x_{1},x_{2},x_{3})=\left( \frac{x_{1}+x_{2}}{\sqrt{2}},x_{3}\right) . \end{aligned}$$

At that time, by direct calculations, we get

$$\begin{aligned} kerF_{*}=span\left\{ U_{1}=\frac{e_{1}-e_{2}}{\sqrt{2}}\right\} \end{aligned}$$

and

$$\begin{aligned} (kerF_{*})^{\bot }=span\left\{ X_{1}=\frac{e_{1}+e_{2}}{\sqrt{2}},X_{2} =\frac{\partial }{\partial x_{3}}\right\} . \end{aligned}$$

Then, it is simple to see that F is a Riemannian submersion. Furthermore \(PU_{1}=X_{1}\)implies that \(P(kerF_{*})\subset (kerF_{*})^{\bot }.\) Consequently, F is anti-invariant Riemannian submersion. Furthermore, the fibres of F are frankly completely umbilical, from they are 1-dimensional. In this place, we will find that a \(g\in C^{\infty }(N)\) filling \({\mathcal {T}}_{U_{1}}U_{1}=-\nabla g.\)

The Riemannian connection \(\nabla \) of the metric tensor \(g_{N}\) is given by

$$\begin{aligned} 2g_{N}(\nabla _{U_{1}}U_{2},U_{3})= & {} U_{1}g_{N}(U_{2},U_{3}) +U_{2}g_{N}(U_{1},U_{3})-U_{3}g_{N}(U_{2},U_{1})\\&-g_{N}([U_{2},U_{3}],U_{1})-g_{N}([U_{1},U_{3}],U_{2}) +g_{N}(U_{3},[U_{1},U_{2}]), \end{aligned}$$

for any \(U_{1},U_{2},U_{3}\in \chi (N).\) Using the above formula for the Riemannian metric \(g_{N},\) we can simply calculate that

$$\begin{aligned} \nabla _{e_{1}}e_{1}=\nabla _{e_{2}}e_{2}=-\frac{\partial }{\partial x_{3}} \end{aligned}$$

and

$$\begin{aligned} \nabla _{e_{1}}e_{2}=\nabla _{e_{2}}e_{1}=0. \end{aligned}$$

Hence, we get

$$\begin{aligned} \nabla _{U_{1}}U_{1}= & {} \frac{1}{2}(\nabla _{e_{1}}e_{1} -\nabla _{e_{1}}e_{2}-\nabla _{e_{2}}e_{1}+\nabla _{e_{2}}e_{2})\\= & {} -\frac{\partial }{\partial x_{3}}. \end{aligned}$$

By (3), we have

$$\begin{aligned} {\mathcal {T}}_{U_{1}}U_{1}=-\frac{\partial }{\partial x_{3}}. \end{aligned}$$

Moreover, for any \(g\in C^{\infty }(N)\) the gradient of g with respect to \(g_{N}\) is given by

$$\begin{aligned} \nabla g=\sum _{i,j}^{3}g_{N}^{ij}\frac{\partial g}{\partial x_{i}} \frac{\partial }{\partial x_{j}} =e^{-2x_{3}}\frac{\partial g}{\partial x_{1}}\frac{\partial }{\partial x_{1}} +e^{-2x_{3}}\frac{\partial g}{\partial x_{2}}\frac{\partial }{\partial x_{2}} +\frac{\partial g}{\partial x_{3}}\frac{\partial }{\partial x_{3}}. \end{aligned}$$

At that time, it is simple to see that \(\nabla g=\frac{\partial }{\partial x_{3}}\) for the function \(g=x_{3}\) and \({\mathcal {T}}_{U_{1}}U_{1}=-\nabla g=-x_{3}.\) In addition to, for all \(U_{2}\in \Gamma (kerF_{*})\), we obtain

$$\begin{aligned} {\mathcal {T}}_{U_{2}}U_{2}=-\Vert U_{2}\Vert ^{2}\nabla g. \end{aligned}$$

Hence, by Theorem 3.1, the submersion F is Clairaut.