Biharmonic Maps on f-Kenmotsu Manifolds with the Schouten–van Kampen Connection

Abstract

The object of the present paper was to study biharmonic maps on f-Kenmotsu manifolds and f-Kenmotsu manifolds with the Schouten–van Kampen connection. With the help of this connection, our results provided important insights related to harmonic and biharmonic maps.

1. Introduction

Let $\varphi :(M^m,g)⟶(N^n,h)$ be a smooth map between two Riemannian manifolds. The energy density of $\varphi$ was the smooth function on M given by:

$$e{\left(\varphi \right)}_p=\sum _{i=1}^{m}h(d_p\varphi \left(e_i\right),d_p\varphi \left(e_i\right)),$$

for any $p\in M$ and any orthonormal basis ${\left\{e_i\right\}}_{i=1}^m$ of $T_{p}M$. If M was a compact Riemannian manifold, the energy functional $E\left(\varphi \right)$ was the integral of its energy density.

$$E\left(\varphi \right)={\int }_{M}e\left(\varphi \right)d{v}^g.$$

(1)

For any smooth variation ${\left\{\varphi \right\}}_{t\in I}$ of $\varphi$ with ${\varphi }_0=\varphi$ and $V={\displaystyle \frac{d{\varphi }_t}{dt}}{|}_{t=0}$, we had the following:

$$\frac{d}{dt}E\left({\varphi }_t\right){|}_{t=0}=-{\int }_{M}h(\tau \left(\varphi \right),V)d{v}^g,$$

(2)

where

$$\tau \left(\varphi \right)=t{r}_g\nabla d\varphi ,$$

(3)

is the tension field of $\varphi$. Then, we found that $\varphi :(M^m,g)⟶(N^n,h)$ was harmonic if, and only if,

$$\tau \left(\varphi \right)=0.$$

(4)

If ${\left(x^i\right)}_{1\le i\le m}$ and ${\left(y^{\alpha }\right)}_{1\le \alpha \le n}$ denoted local coordinates on M and N, respectively, then Equation (4) took the following form:

$$\tau {\left(\varphi \right)}^{\alpha }=\sum _{\stackrel{1\le \alpha ,\beta ,\gamma \le n}{1\le i,j\le m}}\left(\Delta {\varphi }^{\alpha }+g^{ij}\stackrel{N}{{\mathsf{\Gamma }}_{\beta \gamma }^{\alpha }}\frac{\partial {\varphi }^{\beta }}{\partial x^i}\frac{\partial {\varphi }^{\gamma }}{\partial x^j}\right)=0,$$

(5)

where $\Delta {\varphi }^{\alpha }={\sum }_{\stackrel{1\le \alpha \le n}{1\le i,j\le m}}\left({\displaystyle \frac{1}{\sqrt{\left|g\right|}}}{\displaystyle \frac{\partial }{\partial x^i}}(\sqrt{\left|g\right|}{g}^{ij}{\displaystyle \frac{\partial {\varphi }^{\alpha }}{\partial x^j}}\right)$ is the Laplace operator on $(M^m,g)$, and $\stackrel{N}{{\mathsf{\Gamma }}_{\beta \gamma }^{\alpha }}$ are the Christoffel symbols of the Levi-Civita connections of $(N^n,h)$. The biharmonic maps, which provide a natural generalization of harmonic maps, were defined as the critical points of the bi-energy function:

$$E_2\left(\varphi \right)=\frac{1}{2}{\int }_M{\left|\tau \left(\varphi \right)\right|}^{2}d{v}^g.$$

(6)

For any smooth variation ${\left\{\varphi \right\}}_{t\in I}$ of $\varphi$ with ${\varphi }_0=\varphi$ and $V={\displaystyle \frac{d{\varphi }_t}{dt}}{|}_{t=0}$, we had the following:

$$\frac{d}{dt}{E}_2\left({\varphi }_t\right){|}_{t=0}=-{\int }_{M}h({\tau }_2\left(\varphi \right),V)d{v}_g.$$

(7)

The Euler–Lagrange equation attached to the bi-energy was given by the vanishing of the bitension field, as follows:

$${\tau }_2\left(\varphi \right)=-(▵\tau \left(\varphi \right)+t{r}_g{R}^N(\tau \left(\varphi \right),d\varphi )d\varphi ).$$

(8)

where Δ=trace $({\nabla }^{\varphi }{\nabla }^{\varphi }-{\nabla }_{\nabla }^{\varphi })$ is the rough Laplacian on the sections of the pull-back bundle ${\phi }^{-1}TN$, ${\nabla }^{\varphi }$ is the pull-back connection, and $R^N$ is the curvature tensor on N. Clearly, any harmonic map was always a biharmonic map, and a proper biharmonic map would not be harmonic. The harmonic and biharmonic maps have been studied by many authors [1,2,3,4]. Currently, the theories of harmonic and biharmonic maps have become a very important field of research in differential geometry. Najma in [5] studied the harmonic maps between the Kähler and Kenmotsu manifolds. After that, Zagane and Ouakkas in [6] studied the biharmonicity on Kenmotsu manifolds, and they calculated the stress bi-energy tensor from a Kähler manifold to a Kenmotsu manifold. Moreover, Mangione in [7] studied harmonic maps and their stability on f-Kenmotsu manifolds. In [8], Ichi Inoguchi and Eun Lee investigated the biharmonic curves on f-Kenmotsu 3D-manifolds.

Motivated by the above studies, in this paper, we obtained results concerning the harmonicity and biharmonicity of $(J,\phi )$-holomorphic maps from a Kähler manifold $(N^{2n},J,h)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ and we provided the necessary and sufficient conditions for the biharmonicity of the identity map $I:M⟶\overline{M}$ from an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ to an f-Kenmotsu manifold with the Schouten–van Kampen connection.

The structure of this paper is as follows: After the introduction, we described some well-known basic formulas and the properties of the f-Kenmotsu manifold and the f-Kenmotsu manifold with the Schouten–van Kampen connection.

In Section 2, we initiated a study of harmonic and biharmonic maps when the domain was a Kähler manifold $(N^{2n},J,h)$, and the target was an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$. We proved that for $F:N⟶M$ being a $(J,\phi )$-holomorphic map of constant energy density $e\left(F\right)$, then F would be biharmonic if, and only if:

$$-2e\left(F\right)((f\circ F)(f^{\prime }\circ F)+2{(f\circ F)}^3)\xi +3(f\circ F)dF\left(\mathrm{grad}(f\circ F)\right)+▵(f\circ F)\xi =0.$$

(9)

On the other hand, we proved if the function $f\circ F$ was constant on N and $F:N⟶M$ was a $(J,\phi )$-holomorphic map of constant energy density, then F would be biharmonic if, and only if:

$$(f\circ F)(f^{\prime }\circ F)+2{(f\circ F)}^3=0.$$

(10)

Finally, we provided an example of a $(J,\phi )$-holomorphic map from a Kähler manifold to an f-Kenmotsu manifold, which verified Theorem 3.

In Section 3, we proved that any $(J,\phi )$-holomorphic map from a Kähler manifold $(N^{2n},J,h)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ with the Schouten–van Kampen connection was harmonic. In the same section, we also studied the biharmonicity of the identity map $I:M⟶\overline{M}$ from an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ to an f-Kenmotsu manifold with the Schouten–van Kampen connection $({\overline{M}}^{2n+1},f,\phi ,\xi ,\eta ,g)$. We obtained the following results: Firstly, the identity map $I:M⟶\overline{M}$ would be biharmonic if, and only if, the function f was harmonic. Secondly, if f was a constant function, then the identity map $I:\overline{M}⟶M$ from an f-Kenmotsu manifold with the Schouten–van Kampen connection $({\overline{M}}^{2n+1},f,\phi ,\xi ,\eta ,g)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ would be biharmonic if, and only if, $\xi$ was biharmonic vector field.

2. Preliminaries

A $(2n+1)$ dimensional real differentiable manifold M was assumed to be an almost contact metric manifold if it had an almost contact metric structure $(\phi ,\xi ,\eta ,g)$, where $\phi$ is a $(1,1)$ type tensor field, $\xi$ a global vector field, $\eta$ is a 1-form, and g is a Riemannian metric compatible with $(\phi ,\xi ,\eta ,g)$, satisfying the following [9,10,11,12]:

$$\begin{array}{cc}\hfill {\phi }^2=& -I+\eta \otimes \xi ,\eta \left(\xi \right)=1,\hfill \\ \hfill \phi \xi =& 0,\eta \circ \phi =0,\eta \left(X\right)=g(X,\xi ),\hfill \\ \hfill g(\phi X,\phi Y)& =g(X,Y)-\eta \left(X\right)\eta \left(Y\right),\hfill \end{array}$$

(11)

for any vector fields $X,Y\in \mathsf{\Gamma }\left(TM\right)$, where $\mathsf{\Gamma }\left(TM\right)$ denotes the Lie algebra of all differentiable vector fields on $M^{2n+1}$ and I is the identity transformation.

An almost contact metric manifold was a Kenmotsu manifold if

$$\begin{array}{c}\hfill \left({\nabla }_X\phi \right)Y=g(\phi X,Y)\xi -\eta \left(Y\right)\phi X,\end{array}$$

(12)

where ∇ denotes the Riemannian connection of g.

In a Kenmotsu manifold, we had the following relations [13,14,15]:

$$\begin{array}{c}\hfill \left({\nabla }_X\eta \right)\left(Y\right)=g(X,Y)-\eta \left(X\right)\eta \left(Y\right).\end{array}$$

(13)

$$\begin{array}{c}\hfill {\nabla }_X\xi =X-\eta \left(X\right)\xi ,\end{array}$$

(14)

$$\begin{array}{c}\hfill R(X,Y)\xi =\eta \left(X\right)Y-\eta \left(Y\right)X,\end{array}$$

(15)

for any vector fields $X,Y$ on M, and R denotes the Riemannian curvature tensor on M.

We assumed that M was an f-Kenmotsu manifold if the Levi-Civita ∇ of $\phi$ satisfied the following condition [16,17,18,19,20,21,22]:

$$\begin{array}{c}\hfill \left({\nabla }_X\phi \right)Y=f\left(g(\phi X,Y)\xi -\eta \left(Y\right)\phi X\right),\end{array}$$

(16)

where $f\in C^{\infty }\left(M\right)$, such that $df\wedge \eta =0$. If the function f was equal to a constant $\alpha >0$, we obtained an $\alpha$-Kenmotsu manifold, which were Kenmotsu manifolds for $\alpha =1$. If $f=0$, then the manifold would be cosymplectic [23,24]. An f-Kenmotsu manifold was assumed to be regular if $f^2+f^{\prime }\ne 0$, where $f^{\prime }=\xi \left(f\right)$. For an f-Kenmotsu manifold from (11) and (16), it followed that:

$$\begin{array}{c}\hfill {\nabla }_X\xi =f\left(X-\eta \left(X\right)\xi \right),\end{array}$$

(17)

then using (17), we had

$$\begin{array}{c}\hfill \left({\nabla }_X\eta \right)\left(Y\right)=f\left(g(X,Y)-\eta \left(X\right)\eta \left(Y\right)\right).\end{array}$$

(18)

The condition $df\wedge \eta =0$ held if $dim\left(M\right)\ge 5$; however, this did not hold, in general, if we had $dim\left(M\right)=3$ [25]. The characteristic vector field of an f-Kenmotsu manifold also satisfied:

$$\begin{array}{cc}\hfill R(X,Y)\xi & =(f^2+f^{\prime })\left(\eta \left(X\right)Y-\eta \left(Y\right)X\right),\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill R(\xi ,Y)Z& =(f^2+f^{\prime })\left(\eta \left(Z\right)Y-g(Y,Z)\xi \right),\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill \eta \left[R\right(\xi ,Y\left)Z\right]& =(f^2+f^{\prime })\left(g(Y,Z)\eta \left(Z\right)\eta \left(Y\right)\right).\hfill \end{array}$$

(21)

The Schouten–van Kampen connection $\stackrel{★}{\nabla }$ associated with the Levi-Civita connection ∇ was given by [26,27,28,29]:

$$\begin{array}{c}\hfill {\stackrel{★}{\nabla }}_{X}Y={\nabla }_{X}Y-\eta \left(Y\right){\nabla }_X\xi +\left({\nabla }_X\eta \right)\left(Y\right)\xi ,\end{array}$$

(22)

for any vector fields $X,Y\in \mathsf{\Gamma }\left(TM\right)$. Using (13) and (14), the above equation yielded the following:

$$\begin{array}{c}\hfill {\stackrel{★}{\nabla }}_{X}Y={\nabla }_{X}Y+g(X,Y)\xi -\eta \left(Y\right)X.\end{array}$$

(23)

By taking $Y=\xi$ in (23) and using (14), we obtained

$$\begin{array}{c}\hfill {\stackrel{★}{\nabla }}_X\xi =0.\end{array}$$

(24)

Let M be an f-Kenmotsu manifold with the Schouten–van Kampen connection. Then, using (17) and (18) in (22), we obtained the following [30,31]:

$$\begin{array}{c}\hfill {\stackrel{★}{\nabla }}_{X}Y={\nabla }_{X}Y+f\left(g(X,Y)\xi -\eta \left(Y\right)X\right).\end{array}$$

(25)

Let R and $\stackrel{★}{R}$ be the curvature tensors of the Levi-Civita connection ∇ and the Schouten–van Kampen connection $\stackrel{★}{\nabla }$, then

$$\begin{array}{c}\hfill R(X,Y)=\left({\nabla }_X,{\nabla }_Y\right)-{\nabla }_{[X,Y]},\stackrel{★}{R}(X,Y)=\left({\stackrel{★}{\nabla }}_X,{\stackrel{★}{\nabla }}_Y\right)-{\stackrel{★}{\nabla }}_{[X,Y]}.\end{array}$$

By direct calculations, we obtained the following formula connecting R and $\stackrel{★}{R}$ on an f-Kenmotsu manifold M:

$$\begin{array}{cc}\hfill \stackrel{★}{R}(X,Y)Z=& R(X,Y)Z+f^2\left(g(Y,Z)X-g(X,Z)Y\right)\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill & +f^{\prime }(g(Y,Z)\eta \left(X\right)\xi -g(X,Z)\eta \left(Y\right)\xi \hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill & +\eta \left(Y\right)\eta \left(Z\right)X-\eta \left(X\right)\eta \left(Z\right)Y).\hfill \end{array}$$

(28)

and

$$\begin{array}{cc}\hfill \stackrel{★}{R}(\xi ,Y)Z=& 0.\hfill \end{array}$$

(29)

3. Harmonic and Biharmonic Maps on $\mathbf{f}$-Kenmotsu Manifolds

Definition 1. 

A smooth map $F:N⟶M$ between a Kähler manifold $(N^{2n},J,h)$ and an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ was assumed to be a $(J,\phi )$-holomorphic map if it satisfied the following:

$$dF\circ J=\phi \circ dF.$$

Lemma 1 

([6]). Let $F:N⟶M$ be a $(J,\phi )$-holomorphic map from a Kähler manifold $(N^{2n},J,h)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$. Then, we had, for any $X\in \mathsf{\Gamma }\left(TN\right)$,

$$\begin{array}{c}\hfill (\eta \circ dF)\left(X\right)=0.\end{array}$$

We could ask now if such a map would be harmonic when the domain was a Kähler manifold.

Lemma 2. 

Let $F:N⟶M$ be a $(J,\phi )$-holomorphic map from a Kähler manifold $(N^{2n},J,h)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$, then we had the following:

$$\begin{array}{cc}\hfill g\left(\tau \right(F),\xi )& =-2(f\circ F)e\left(F\right).\hfill \end{array}$$

where $e\left(F\right)$ is the energy density of the map F.

Proof. 

Considering a local orthonormal basis ${\left\{e_i\right\}}_{i=1}^{2n}$ on $T_{p}N$ for any $p\in N$, we obtained the following:

$$\begin{array}{cc}\hfill g\left(\tau \right(F),\xi )& =\sum _{i=1}^{2n}g\left({\nabla }_{e_i}^{F}dF\left(e_i\right)-dF\left({\nabla }_{e_i}^N{e}_i\right),\xi \right)\hfill \\ \hfill & =\sum _{i=1}^{2n}g\left({\nabla }_{dF\left(e_i\right)}^{M}dF\left(e_i\right)-dF\left({\nabla }_{e_i}^N{e}_i\right),\xi \right)\hfill \\ \hfill & =\sum _{i=1}^{2n}g\left({\nabla }_{dF\left(e_i\right)}^{M}dF\left(e_i\right),\xi \right)-\sum _{i=1}^{2n}g\left(dF\left({\nabla }_{e_i}^N{e}_i\right),\xi \right)\hfill \\ \hfill & =\sum _{i=1}^{2n}\left({\nabla }_{dF\left(e_i\right)}^{M}g\left(dF\left(e_i\right),\xi \right)-g\left(dF\left(e_i\right),{\nabla }_{dF\left(e_i\right)}^M\xi \right)\right)-\sum _{i=1}^{2n}g\left(dF\left({\nabla }_{e_i}^N{e}_i\right),\xi \right)\hfill \\ \hfill & =\sum _{i=1}^{2n}\left({\nabla }_{dF\left(e_i\right)}^M\eta \left(dF\left(e_i\right)\right)-g\left(dF\left(e_i\right),{\nabla }_{dF\left(e_i\right)}^M\xi \right)\right)-\sum _{i=1}^{2n}\eta \left(dF\left({\nabla }_{e_i}^N{e}_i\right)\right).\hfill \end{array}$$

As F was a $(J,\phi )$-holomorphic map, then by using Lemma 1, we obtained $\eta \left(dF\left(e_i\right)\right)=0$ and $\eta \left(dF\left({\nabla }_{e_i}^N{e}_i\right)\right)=0$. Then, we had the following:

$$\begin{array}{cc}\hfill g\left(\tau \right(F),\xi )& =-\sum _{i=1}^{2n}g\left(dF\left(e_i\right),{\nabla }_{dF\left(e_i\right)}^M\xi \right).\hfill \end{array}$$

Using the Equation (17), we obtained the following:

$$\begin{array}{cc}\hfill g\left(\tau \right(F),\xi )& =-\sum _{i=1}^{2n}\left((f\circ F)\left(g(dF\left(e_i\right),dF\left(e_i\right))-g(dF\left(e_i\right),\eta \left(dF\left(e_i\right)\right)\xi )\right)\right)\hfill \\ \hfill & =-\sum _{i=1}^{2n}(f\circ F)g(dF\left(e_i\right),dF\left(e_i\right))\hfill \\ \hfill & =-2(f\circ F)e\left(F\right).\hfill \end{array}$$

□

Theorem 1. 

Let $F:N⟶M$ be a $(J,\phi )$-holomorphic map from a Kähler manifold $(N^{2n},J,h)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$. Then, the tension field of the map F was given by:

$$\tau \left(F\right)=-2(f\circ F)e\left(F\right)\xi ,$$

(30)

Proof. 

For any $(J,\phi )$-holomorphic map $F:N⟶M$, we have the following formula for its tension field [32]

$$\begin{array}{c}\hfill \phi \left(\tau \left(F\right)\right)=dF\left(divJ\right)-t{r}_{h}B,\end{array}$$

where B is defined by $B(X,Y)=\left({\nabla }_X^F\phi \right)dFY$ for any vector fields $X,Y\in \mathsf{\Gamma }\left(TN\right)$. Since N was a Kähler manifold, $\nabla J=0$, then we had

$$\begin{array}{c}\hfill divJ=\sum _{i=1}^{2n}\left({\nabla }_{e_i}J\right)e_i=0,\end{array}$$

where ${\left\{e_i\right\}}_{i=1}^{2n}$ is an orthonormal local basis on $TN$. By using the relation (16) and doing a straightforward calculation, we obtained the following:

$$\begin{array}{cc}\hfill t{r}_{h}B& =\sum _{i=1}^{2n}\left({\nabla }_{e_i}^F\phi \right)dF\left(e_i\right)=\sum _{i=1}^{2n}\left({\nabla }_{dF\left(e_i\right)}^M\phi \right)dF\left(e_i\right)\hfill \\ \hfill & =\sum _{i=1}^{2n}(f\circ F)\left(g(\phi \left(dF\left(e_i\right)\right),dF\left(e_i\right))\xi -\eta \left(dF\left(e_i\right)\right)\phi \left(dF\left(e_i\right)\right)\right)\hfill \\ \hfill & =\sum _{i=1}^{2n}(f\circ F)\left(-\eta \left(dF\left(e_i\right)\right)\phi \left(dF\left(e_i\right)\right)\right).\hfill \end{array}$$

As F was a $(J,\phi )$-holomorphic map, then by using Lemma 1, we found the following:

$$\begin{array}{cc}\hfill \sum _{i=1}^{2n}(f\circ F)\left(-\eta \left(dF\left(e_i\right)\right)\phi \left(dF\left(e_i\right)\right)\right)& =0.\hfill \end{array}$$

As a result, $\phi \left(\tau \left(F\right)\right)=0⟹{\phi }^2\left(\tau \left(F\right)\right)=0$, that is,

$$\tau \left(F\right)=\eta \left(\tau \right(F\left)\right)\xi =g\left(\tau \right(F),\xi )\xi =-2(f\circ F)e\left(F\right)\xi .$$

□

Theorem 2. 

Let $(N^{2n},J,h)$ be a Kähler manifold and $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ be an f-Kenmotsu manifold. Then, any $(J,\phi )$-holomorphic map $F:N⟶M$ would be harmonic if, and only if, it was a constant map or $f\circ F=0$.

Proof. 

According to Theorem 1, if the map F was harmonic, then $(f\circ F)e\left(F\right)=0$. We assumed that $f\circ F\ne 0$. There existed an open subset U on M, such that $f\circ F\ne 0$ was everywhere on U. Therefore, $e\left(F\right)=0$ was on U. From the harmonicity of F, we concluded that $e\left(F\right)=0$ on M, that is, F was a constant map. □

Biharmonic Maps on f-Kenmotsu Manifolds

Theorem 3. 

Let $F:N⟶M$ be a $(J,\phi )$-holomorphic map from a Kähler manifold $(N^{2n},J,h)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$. Then, the bitension field of F was given by the following:

$$\begin{array}{cc}\hfill {\tau }_2\left(F\right)=& -2(-2{\left(e\left(F\right)\right)}^2((f\circ F)(f^{\prime }\circ F)+2{(f\circ F)}^3)\xi \hfill \\ \hfill & +3(f\circ F)e\left(F\right)dF\left(\mathit{grad}\right(f\circ F\left)\right)+(f\circ F)▵\left(e\right(F\left)\right)\xi \hfill \\ \hfill & +e\left(F\right)▵(f\circ F)\xi +2g\left(\mathit{grad}\right(f\circ F),\mathit{grad}(e(F\left)\right))\xi \hfill \\ \hfill & +2{(f\circ F)}^{2}dF\left(\mathit{grad}\left(e\left(F\right)\right)\right)).\hfill \end{array}$$

Proof. 

By definition of the bitension field of the map F, we had:

$$\begin{array}{cc}\hfill {\tau }_2\left(F\right)=& t{r}_h{\left({\nabla }^F\right)}^2\tau \left(F\right)+t{r}_h{R}^M(\tau \left(F\right),dF)dF\hfill \\ \hfill =& -2\left(t{r}_h{\left({\nabla }^F\right)}^2(f\circ F)e\left(F\right)\xi +t{r}_h{R}^M((f\circ F)e\left(F\right)\xi ,dF)dF\right)\hfill \\ \hfill =& -2\sum _{i=1}^{2n}({\nabla }_{e_i}^F{\nabla }_{e_i}^F(f\circ F)e\left(F\right)\xi -{\nabla }_{{\nabla }_{e_i}^N{e}_i}^F(f\circ F)e\left(F\right)\xi \hfill \\ \hfill & +R^M((f\circ F)e\left(F\right)\xi ,dF\left(e_i\right))dF\left(e_i\right)),\hfill \end{array}$$

(31)

where ${\left\{e_i\right\}}_{i=1}^{2n}$ is an orthonormal local basis on $TN$. A direct calculation provided the following:

$$\begin{array}{cc}\hfill \sum _{i=1}^{2n}\left({\nabla }_{e_i}^F{\nabla }_{e_i}^F(f\circ F)e(F)\xi \right)=& \sum _{i=1}^{2n}({\nabla }_{e_i}^F((f\circ F)e(F){\nabla }_{e_i}^F\xi )+{\nabla }_{e_i}^F(e_i((f\circ F)e(F))\xi )\hfill \\ \hfill =& \sum _{i=1}^{2n}((f\circ F)e(F){\nabla }_{e_i}^F{\nabla }_{e_i}^F\xi +e_i((f\circ F)e(F)){\nabla }_{e_i}^F\xi \hfill \\ \hfill & +e_i((f\circ F)e(F)){\nabla }_{e_i}^F\xi +e_i(e_i((f\circ F)e(F)))\xi )\hfill \\ \hfill =& \sum _{i=1}^{2n}((f\circ F)e(F){\nabla }_{e_i}^F{\nabla }_{e_i}^F\xi +2{\nabla }_{\mathrm{grad}((f\circ F)e(F))}^F\xi \hfill \\ \hfill & +e_i(e_i((f\circ F)e(F)))\xi ),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \sum _{i=1}^{2n}\left({\nabla }_{{\nabla }_{e_i}^N{e}_i}^F(f\circ F)e\left(F\right)\xi \right)=& \sum _{i=1}^{2n}\left((f\circ F)e\left(F\right){\nabla }_{{\nabla }_{e_i}{e}_i}^F\xi +{\nabla }_{e_i}{e}_i\left((f\circ F)e\left(F\right)\right)\xi \right).\hfill \end{array}$$

Based on the following:

$$\begin{array}{c}\hfill ▵\left((f\circ F)e\left(F\right)\right)=\sum _{i=1}^{2n}\left(e_i\left(e_i\left((f\circ F)e\left(F\right)\right)\right)-{\nabla }_{e_i}{e}_i((f\circ F)e\left(F\right)\right)\xi ,\end{array}$$

and

$$\begin{array}{cc}\hfill t{r}_h{\left({\nabla }^F\right)}^2\xi =& \sum _{i=1}^{2n}\left({\nabla }_{e_i}^F{\nabla }_{e_i}^F\xi -{\nabla }_{{\nabla }_{e_i}^N{e}_i}^F\xi \right),\hfill \end{array}$$

we could deduce that

$$\begin{array}{cc}\hfill t{r}_h{\left({\nabla }^F\right)}^2(f\circ F)e\left(F\right)\xi =& (f\circ F)e\left(F\right)t{r}_h{\left({\nabla }^F\right)}^2\xi +▵\left((f\circ F)e\left(F\right)\right)\xi +2{\nabla }_{\mathrm{grad}\left(\right(f\circ F\left)e\right(F\left)\right)}^F\xi \hfill \\ \hfill =& (f\circ F)e\left(F\right)t{r}_h{\left({\nabla }^F\right)}^2\xi +(f\circ F)▵\left(e\left(F\right)\right)\xi +e\left(F\right)▵(f\circ F)\xi \hfill \\ \hfill & +2g(\mathrm{grad}(f\circ F),\mathrm{grad}\left(e\left(F\right)\right)\xi +2(f\circ F){\nabla }_{\mathrm{grad}\left(e\right(F\left)\right)}^F\xi \hfill \\ \hfill & +2e\left(F\right){\nabla }_{\mathrm{grad}(f\circ F)}^F\xi .\hfill \end{array}$$

(32)

After calculating the term $t{r}_h{\left({\nabla }^F\right)}^2\xi$, we obtained the following:

$$\begin{array}{cc}\hfill t{r}_h{\left({\nabla }^F\right)}^2\xi =& \sum _{i=1}^{2n}\left({\nabla }_{e_i}^F{\nabla }_{e_i}^F\xi -{\nabla }_{{\nabla }_{e_i}^N{e}_i}^F\xi \right).\hfill \end{array}$$

By using Equation (17), we obtained:

$$\begin{array}{cc}\hfill \sum _{i=1}^{2n}{\nabla }_{e_i}^F\xi =\sum _{i=1}^{2n}{\nabla }_{dF\left(e_i\right)}\xi =& \sum _{i=1}^{2n}(f\circ F)\left(dF\left(e_i\right))-\eta \left(dF\left(e_i\right)\right)\xi \right),\hfill \\ \hfill & =(f\circ F)\sum _{i=1}^{2n}dF\left(e_i\right),\hfill \end{array}$$

which gave us:

$$\begin{array}{cc}\hfill \sum _{i=1}^{2n}{\nabla }_{e_i}^F{\nabla }_{e_i}^F\xi =& \sum _{i=1}^{2n}{\nabla }_{e_i}^F(f\circ F)dF\left(e_i\right),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \sum _{i=1}^{2n}{\nabla }_{{\nabla }_{e_i}^N{e}_i}^F\xi =& {\nabla }_{dF\left({\nabla }_{e_i}^N{e}_i\right)}^M\xi \hfill \\ \hfill =& \sum _{i=1}^{2n}(f\circ F)\left(dF\left({\nabla }_{e_i}^N{e}_i\right)-\eta \left(dF\left({\nabla }_{e_i}^N{e}_i\right)\right)\xi \right)\hfill \\ \hfill =& \sum _{i=1}^{2n}(f\circ F)dF\left({\nabla }_{e_i}^N{e}_i\right),\hfill \end{array}$$

we conclude that

$$\begin{array}{cc}\hfill t{r}_h{\left({\nabla }^F\right)}^2\xi =& \sum _{i=1}^{2n}\left({\nabla }_{e_i}^F(f\circ F)dF\left(e_i\right)-(f\circ F)dF\left({\nabla }_{e_i}^N{e}_i\right)\right)\hfill \\ \hfill =& \sum _{i=1}^{2n}\left(e_i\left((f\circ F)\right)dF\left(e_i\right)+(f\circ F){\nabla }_{e_i}^{F}dF\left(e_i\right)-(f\circ F)dF\left({\nabla }_{e_i}^N{e}_i\right)\right)\hfill \\ \hfill =& dF\left(\mathrm{grad}f\right)+(f\circ F)\tau \left(F\right)\hfill \\ \hfill =& dF\left(\mathrm{grad}f\right)-2{(f\circ F)}^{2}e\left(F\right)\xi ,\hfill \end{array}$$

(33)

Now, by simplifying the terms ${\nabla }_{\mathrm{grad}\left(e\right(F\left)\right)}^F\xi$, and ${\nabla }_{\mathrm{grad}(f\circ F)}^F\xi$, we had the following:

$$\begin{array}{cc}\hfill {\nabla }_{\mathrm{grad}\left(e\right(F\left)\right)}^F\xi =& {\nabla }_{dF\left(\mathrm{grad}\right(e\left(F\right))}^M\xi \hfill \\ \hfill =& (f\circ F)\left(dF\left(\mathrm{grad}\right(e\left(F\right))-\eta dF(\mathrm{grad}\left(e\right(F\left)\right)\xi \right)\hfill \\ \hfill =& (f\circ F)dF\left(\mathrm{grad}\right(e\left(F\right)),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\nabla }_{\mathrm{grad}(f\circ F)}^F\xi =& {\nabla }_{dF\left(\mathrm{grad}\right(f\circ F\left)\right)}^M\xi \hfill \\ \hfill =& (f\circ F)\left(dF\left(\mathrm{grad}\right(f\circ F\left)\right)-\eta dF\left(\mathrm{grad}\right(f\circ F\left)\right)\xi \right)\hfill \\ \hfill =& (f\circ F)dF\left(\mathrm{grad}\right(f\circ F\left)\right),\hfill \end{array}$$

which finally gave us:

$$\begin{array}{cc}\hfill t{r}_h{\left({\nabla }^F\right)}^2(f\circ F)e\left(F\right)\xi =& -2{(f\circ F)}^3{\left(e\left(F\right)\right)}^2\xi +(f\circ F)e\left(F\right)dF\left(\mathrm{grad}(f\circ F)\right)\hfill \\ \hfill & +(f\circ F)▵\left(e\right(F\left)\right)\xi +e\left(F\right)▵(f\circ F)\xi \hfill \\ \hfill & +2g\left(\mathrm{grad}\right(f\circ F),\mathrm{grad}(e\left(F\right)\left)\right)\xi \hfill \\ \hfill & +2{(f\circ F)}^{2}dF\left(\mathrm{grad}\left(e\left(F\right)\right)\right)\hfill \\ \hfill & +2(f\circ F)e\left(F\right)dF\left(\mathrm{grad}\right(f\circ F\left)\right)\hfill \end{array}$$

(34)

By using Equation (19), we obtained the following:

$$\begin{array}{cc}\hfill t{r}_h{R}^M(fe\left(F\right)\xi ,dF)dF=& (f\circ F)e\left(F\right)\sum _{i=1}^{2n}R(\xi ,dF\left(e_i\right))dF\left(e_i\right)\hfill \\ \hfill =& (f\circ F)e\left(F\right)({(f\circ F)}^2+(f^{\prime }\circ F))\sum _{i=1}^{2n}\left(\eta \left(dF\left(e_i\right)\right)dF\left(e_i\right)-g(dF\left(e_i\right),dF\left(e_i\right))\xi \right)\hfill \\ \hfill =& -2{(f\circ F)}^3{\left(e\left(F\right)\right)}^2\xi -2(f\circ F)(f^{\prime }\circ F){\left(e\left(F\right)\right)}^2\xi .\hfill \end{array}$$

(35)

If we replaced (34) and (35) in (31), we arrived at the following:

$$\begin{array}{cc}\hfill {\tau }_2\left(F\right)=& -2(-2{\left(e\left(F\right)\right)}^2((f\circ F)(f^{\prime }\circ F)+2{(f\circ F)}^3)\xi \hfill \\ \hfill & +3(f\circ F)e\left(F\right)dF\left(\mathrm{grad}\right(f\circ F\left)\right)+(f\circ F)▵\left(e\right(F\left)\right)\xi \hfill \\ \hfill & +e\left(F\right)▵(f\circ F)\xi +2g\left(\mathrm{grad}\right(f\circ F),\mathrm{grad}(e(F\left)\right))\xi \hfill \\ \hfill & +2{(f\circ F)}^{2}dF\left(\mathrm{grad}\left(e\left(F\right)\right)\right)).\hfill \end{array}$$

□

Corollary 1. 

Let $(N^{2n},J,h)$ be a Kähler manifold and $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ be an f-Kenmotsu manifold. Then, any $(J,\phi )$-holomorphic map $F:N⟶M$ would biharmonic if, and only if:

$$\begin{array}{cc}\hfill & -2{\left(e\left(F\right)\right)}^2((f\circ F)(f^{\prime }\circ F)+2{(f\circ F)}^3)\xi \hfill \\ \hfill & +3(f\circ F)e\left(F\right)dF\left(\mathit{grad}\right(f\circ F\left)\right)+(f\circ F)▵\left(e\right(F\left)\right)\xi \hfill \\ \hfill & +e\left(F\right)▵(f\circ F)\xi +2g\left(\mathit{grad}\right(f\circ F),\mathit{grad}(e(F\left)\right))\xi \hfill \\ \hfill & +2{(f\circ F)}^{2}dF\left(\mathit{grad}\left(e\left(F\right)\right)\right)=0.\hfill \end{array}$$

Corollary 2. 

Let $(N^{2n},J,h)$ be a Kähler manifold and $(M^{2n+1},\phi ,\xi ,\eta ,g)$ be a Kenmotsu manifold; then, any $(J,\phi )$-holomorphic map $F:N⟶M$ would be biharmonic if, and only if:

$$\begin{array}{cc}\hfill -4e{\left(F\right)}^2\xi +▵\left(e\left(F\right)\right)\xi +2dF(\mathit{grad}\left(e\left(F\right)\right)=& 0.\hfill \end{array}$$

Corollary 3. 

Let $(N^{2n},J,h)$ be a Kähler manifold and $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ be an f-Kenmotsu manifold. Then, any $(J,\phi )$-holomorphic map $F:N⟶M$ of constant energy density would biharmonic if, and only if:

$$\begin{array}{cc}\hfill -2e\left(F\right)((f\circ F)(f^{\prime }\circ F)+2{(f\circ F)}^3)\xi +3(f\circ F)dF\left(\mathit{grad}(f\circ F)\right)+▵(f\circ F)\xi =& 0.\hfill \end{array}$$

Corollary 4. 

Let $F:N⟶M$ be a $(J,\phi )$-holomorphic map of constant energy density from a Kähler manifold $(N,J,h)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$. If the function $f\circ F$ was constant on N, then F would biharmonic if, and only if, $f{f}^{\prime }+2f^3=0$ was on $F\left(N\right)$.

Example 1. 

Let the five-dimensional manifold $M={\mathbb{R}}^4\times (0,\infty )$ be equipped with the Riemannian metric $g=t^{-2\alpha }(d{y}_1^2+d{y}_2^2+d{y}_3^2+d{y}_4^2)+d{t}^2,$ for some constant $\alpha \in \mathbb{R}$. We considered the following orthonormal basis:

$$e_1=t^{\alpha }\frac{\partial }{\partial y_1},e_2=t^{\alpha }\frac{\partial }{\partial y_2},e_3=t^{\alpha }\frac{\partial }{\partial y_3},e_4=t^{\alpha }\frac{\partial }{\partial y_4},e_5=\frac{\partial }{\partial t}.$$

We considered a 1-form η defined by:

$$\eta \left(X\right)=g(X,e_5),\forall X\in \mathsf{\Gamma }\left(TM\right).$$

That is, we chose $e_5=\xi$. We defined the tensor field φ by the following:

$$\phi \left(e_1\right)=-e_2,\phi \left(e_2\right)=e_1,\phi \left(e_3\right)=-e_4,\phi \left(e_4\right)=e_3,\phi \left(e_5\right)=0.$$

By the linearity properties of g and φ, we obtained the following:

$$\eta \left(e_5\right)=1,{\phi }^{2}X=-X+\eta \left(X\right)e_5,$$

$$g(\phi X,\phi Y)=g(X,Y)-\eta \left(X\right)\eta \left(Y\right),$$

for any vector fields $X,Y$ on M. Therefore, $(M,\phi ,\xi ,\eta ,g)$ formed an almost contact metric manifold.

Otherwise, we had $[e_i,e_5]=-\alpha t^{-1}{e}_i$ for $i=1,2,3,4$ and $[e_i,e_j]=0$. Let ∇ be the Levi-Civita connection of $(M,g)$. By using Koszul’s formula, we obtained the following for $i,j=1,2,3,4$ with $i\ne j$:

$${\nabla }_{e_i}{e}_i=\frac{\alpha }{t}{e}_5,{\nabla }_{e_i}{e}_5=-\frac{\alpha }{t}{e}_i,{\nabla }_{e_i}{e}_j={\nabla }_{e_5}{e}_i={\nabla }_{e_5}{e}_5=0.$$

The above relations indicated that ${\nabla }_X\xi =f\{X-\eta \left(X\right)\xi \}$ for $\xi =e_5$ and $f=-\frac{\alpha }{t}$. Therefore, we could say that $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ was an f-Kenmotsu manifold. Moreover, $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ was a regular f-Kenmotsu manifold if, and only if, $\alpha \ne 0,-1,$ because $f^2+f^{\prime }=\frac{\alpha (\alpha +1)}{t^2}$.

Let F be a $(J,\phi )$-holomorphic map, defined by the following:

$$\begin{array}{ccc}\hfill F:({\mathbb{R}}^2,J,h)& ⟶& (M^{2n+1},f,\phi ,\xi ,\eta ,g),\hfill \\ \hfill (x_1,x_2)& ⟼& (F_1(x_1,x_2),F_2(x_1,x_2),F_3(x_1,x_2),F_4(x_1,x_2),F_5(x_1,x_2))\hfill \end{array}$$

where $h=d{x}_1^2+d{x}_2^2$, $J\left(\frac{\partial }{\partial x_1}\right)=\frac{\partial }{\partial x_2}$, $J\left(\frac{\partial }{\partial x_2}\right)=-\frac{\partial }{\partial x_1}$ and $F_i(x_1,x_2)$ are defined by

$$\begin{array}{ccc}\hfill F_1(x_1,x_2)& =& a_1{x}_1+a_2{x}_2+c_1\hfill \\ \hfill F_2(x_1,x_2)& =& a_2{x}_1-a_1{x}_2+c_2\hfill \\ \hfill F_3(x_1,x_2)& =& a_3{x}_1+a_4{x}_2+c_3\hfill \\ \hfill F_4(x_1,x_2)& =& a_4{x}_1-a_3{x}_2+c_4\hfill \\ \hfill F_5(x_1,x_2)& =& c_5\hfill \end{array}$$

where $a_j,c_i\in \mathbb{R}$ are for all $j=1,2,3,4$ and $i=1,2,3,4,5$. Note that the density energy of F was a constant given by $e\left(F\right)=(a_1^2+a_2^2+a_3^2+a_4^2)c_5^{-\alpha }$. According to Theorem 3, the tension field of F was given by the following:

$$\tau \left(F\right)=\frac{\alpha (a_1^2+a_2^2+a_3^2+a_4^2)}{c_5^{\alpha +1}}{e}_5.$$

As $f\circ F$ was constant on N, and the density energy of F was constant, from Corollary 4, the map F would be biharmonic if, and only if:

$$(f{f}^{\prime }+2f^3)\circ F=-\frac{{\alpha }^2(1+2\alpha )}{c_5^3}=0.$$

Therefore, F was biharmonic non-harmonic if, and only if, $\alpha =-\frac{1}{2}$.

4. Biharmonic Maps on $\mathit{f}$-Kenmotsu with the Schouten–van Kampen Connection

Theorem 4. 

Let $(N^{2n},J,h)$ be a Kähler manifold and $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ be an f-Kenmotsu manifold with the Schouten–van Kampen connection. Then, any $(J,\phi )$-holomorphic map

$F:N⟶M$ would be harmonic.

Proof. 

Based on the $(J,\phi )$-holomorphic map, we had the following:

$$\begin{array}{c}\hfill \phi \left(\tau \left(F\right)\right)=dF\left(divJ\right)-t{r}_{h}B,\end{array}$$

where B is defined by $B(X,Y)=\left({\nabla }_X^F\phi \right)dFY$ for any vector fields $X,Y\in \mathsf{\Gamma }\left(TN\right)$. Considering a local orthonormal basis ${\left\{e_i\right\}}_{i=1}^{2n}$ on $T_{p}N$ for any $p\in N$, we obtained the following:

$$\begin{array}{c}\hfill divJ=\sum _{i=1}^{2n}\left({\nabla }_{e_i}J\right)e_i=0,\end{array}$$

and by using relation (16), we found the following, as well:

$$\begin{array}{cc}\hfill t{r}_{h}B& =\sum _{i=1}^{2n}\left({\nabla }_{e_i}^F\phi \right)dF\left(e_i\right)=\sum _{i=1}^{2n}\left({\nabla }_{dF\left(e_i\right)}^M\phi \right)dF\left(e_i\right)\hfill \\ \hfill & =\sum _{i=1}^{2n}(f\circ F)\left(g(\phi \left(dF\left(e_i\right)\right),dF\left(e_i\right))\xi -\eta \left(dF\left(e_i\right)\right)\phi \left(dF\left(e_i\right)\right)\right)\hfill \\ \hfill & =\sum _{i=1}^{2n}(f\circ F)\left(-\eta \left(dF\left(e_i\right)\right)\phi \left(dF\left(e_i\right)\right)\right).\hfill \\ \hfill & =0.\hfill \end{array}$$

From the above relation, we could obtain the following: $\phi \left(\tau \right(F\left)\right)=0\Rightarrow \tau \left(F\right)=g\left(\tau \right(F),\xi )\xi$. However,

$$\begin{array}{cc}\hfill g\left(\tau \right(F),\xi )& =\sum _{i=1}^{2n}g\left({\nabla }_{e_i}^{F}dF\left(e_i\right)-dF\left({\nabla }_{e_i}^N{e}_i\right),\xi \right)\hfill \\ \hfill & =\sum _{i=1}^{2n}g\left({\nabla }_{dF\left(e_i\right)}^{M}dF\left(e_i\right)-dF\left({\nabla }_{e_i}^N{e}_i\right),\xi \right)\hfill \\ \hfill & =\sum _{i=1}^{2n}g\left({\nabla }_{dF\left(e_i\right)}^{M}dF\left(e_i\right),\xi \right)-\sum _{i=1}^{2n}g\left(dF\left({\nabla }_{e_i}^N{e}_i\right),\xi \right)\hfill \\ \hfill & =\sum _{i=1}^{2n}\left({\nabla }_{dF\left(e_i\right)}^{M}g\left(dF\left(e_i\right),\xi \right)-g\left(dF\left(e_i\right),{\nabla }_{dF\left(e_i\right)}^M\xi \right)\right)-\sum _{i=1}^{2n}g\left(dF\left({\nabla }_{e_i}^N{e}_i\right),\xi \right)\hfill \\ \hfill & =\sum _{i=1}^{2n}\left({\nabla }_{dF\left(e_i\right)}^M\eta \left(dF\left(e_i\right)\right)-g\left(dF\left(e_i\right),{\nabla }_{dF\left(e_i\right)}^M\xi \right)\right)-\sum _{i=1}^{2n}\eta \left(dF\left({\nabla }_{e_i}^N{e}_i\right)\right).\hfill \end{array}$$

As F was a $(J,\phi )$-holomorphic map, then by using Lemma 1, we found $\eta \left(dF\left(e_i\right)\right)=0$ and $\eta \left(dF\left({\nabla }_{e_i}^N{e}_i\right)\right)=0$. Then, we had

$$\begin{array}{cc}\hfill g\left(\tau \right(F),\xi )& =\sum _{i=1}^{2n}\left(-g\left(dF\left(e_i\right),{\nabla }_{dF\left(e_i\right)}^M\xi \right)\right).\hfill \end{array}$$

In addition, from relation (24), we had ${\nabla }_{dF\left(e_i\right)}^M\xi =0$, and then, $g\left(\tau \right(F),\xi )=0.$ □

Biharmonic Identity Map with the Schouten–van Kampen Connection

Theorem 5. 

Let $I:M⟶\overline{M}$ be the identity map from an f-Kenmotsu manifold

$(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ to an f-Kenmotsu manifold with the Schouten–van Kampen connection $({\overline{M}}^{2n+1},f,\phi ,\xi ,\eta ,g)$. Then the tension field of map I was given by the following:

$$\tau \left(I\right)=2nf\xi$$

(36)

Proof. 

Let $\left\{e_1,\dots ,e_n,\phi \left(e_1\right),\dots ,\phi \left(e_n\right),\xi \right\}$ be an orthonormal local basis on $TM$. Then, by definition of the tension field of map I, we found the following:

$$\begin{array}{cc}\hfill \tau \left(I\right)=& t{r}_g\stackrel{}{\nabla }dI\hfill \\ \hfill =& \sum _{i=1}^{2n+1}\left({\nabla }_{dI\left(e_i\right)}^{\overline{M}}dI\left(e_i\right)-dI\left({\nabla }_{e_i}^M{e}_i\right)\right).\hfill \end{array}$$

Using relation (25), we had the following:

$$\begin{array}{cc}\hfill \tau \left(I\right)=& \sum _{i=1}^{2n+1}\left(f(g(e_i,e_i)\xi -\eta \left(e_i\right)e_i)\right)\hfill \\ \hfill =& 2nf\xi .\hfill \end{array}$$

□

Theorem 6. 

Let $I:M⟶\overline{M}$ be the identity map from an f-Kenmotsu manifold

$(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ to an f-Kenmotsu manifold with the Schouten–van Kampen connection $({\overline{M}}^{2n+1},f,\phi ,\xi ,\eta ,g)$. Then, map I would be harmonic if, and only if, M was a cosymplectic manifold.

Theorem 7. 

Let $I:M⟶\overline{M}$ be the identity map from an f-Kenmotsu manifold

$(M^{2n+1},f,\phi ,\xi ,\eta ,g)$ to an f-Kenmotsu manifold with the Schouten–van Kampen connection $({\overline{M}}^{2n+1},f,\phi ,\xi ,\eta ,g)$. Then, map I would be biharmonic if, and only if, f was a harmonic function.

Proof. 

Let $\left\{e_1,\dots ,e_n,\phi \left(e_1\right),\dots ,\phi \left(e_n\right),\xi \right\}$ be an orthonormal local basis on $TM$; then, by definition of the tension field of map I, we had the following:

$$\begin{array}{cc}\hfill {\tau }_2\left(I\right)& =t{r}_g{\left({\nabla }^I\right)}^2\tau \left(I\right)+t{r}_g{R}^{\overline{M}}(\tau \left(I\right),dI)dI\hfill \\ \hfill & =2n\left(t{r}_g{\left({\nabla }^I\right)}^{2}f\xi +t{r}_g{R}^{\overline{M}}(f\xi ,dI)dI\right)\hfill \\ \hfill & =\sum _{i=1}^{2n+1}\left({\nabla }_{e_i}^I{\nabla }_{e_i}^{I}2nf\xi -{\nabla }_{{\nabla }_{e_i}^M{e}_i}^{I}2nf\xi +R^{\overline{M}}(2nf\xi ,dI\left(e_i\right))dI\left(e_i\right)\right).\hfill \end{array}$$

The combination of Equation (24) and direct calculations provided the following:

$$\begin{array}{cc}\hfill \sum _{i=1}^{2n+1}\left({\nabla }_{e_i}^I{\nabla }_{e_i}^{I}2nf\xi \right)=& 2n\sum _{i=1}^{2n+1}\left(e_i\left(e_i\left(f\right)\right)\xi \right),\hfill \end{array}$$

and

$$\begin{array}{c}\hfill \sum _{i=1}^{2n+1}\left({\nabla }_{{\nabla }_{e_i}{e}_i}^{I}2nf\xi \right)=2n\sum _{i=1}^{2n+1}\left({\nabla }_{e_i}{e}_i\left(f\right)\xi \right).\end{array}$$

Based on the following:

$$\begin{array}{c}\hfill ▵\left(f\right)=\sum _{i=1}^{2n+1}\left(e_i\left(e_i\left(f\right)\right)-{\nabla }_{e_i}{e}_i\left(f\right)\right),\end{array}$$

we could conclude

$$\begin{array}{cc}\hfill \sum _{i=1}^{2n+1}\left({\nabla }_{e_i}^I{\nabla }_{e_i}^{I}2nf\xi \right)=& 2n▵\left(f\right)\xi .\hfill \end{array}$$

Based on Equation (29), we found the following:

$$\begin{array}{cc}\hfill t{r}_g{R}^{\overline{M}}(2nf\xi ,dI)dI=& 2nft{r}_g\left(R^{\overline{M}}(\xi ,dI)dI\right)\hfill \\ \hfill =& 2nf\sum _{i=1}^{2n+1}\left(R^{\overline{M}}(\xi ,e_i)e_i\right)\hfill \\ \hfill =& 0.\hfill \end{array}$$

Finally, we obtained

$$\begin{array}{c}\hfill {\tau }_2\left(I\right)=2n▵\left(f\right)\xi .\end{array}$$

(37)

□

Remark 1. 

If f was a constant or harmonic function, then I would be a proper biharmonic map.

Theorem 8. 

Let $I:\overline{M}⟶M$ be the identity map from an f-Kenmotsu manifold with the Schouten–van Kampen connection $({\overline{M}}^{2n+1},f,\phi ,\xi ,\eta ,g)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$. Then, the bitension field of map I was given by the following:

$$\begin{array}{cc}\hfill {\tau }_2\left(I\right)=& -2n\left(\overline{\Delta }\left(f\right)\xi +f{\tau }_2\left(\xi \right)+2{\nabla }_{\mathrm{grad}f}^I\xi \right),\hfill \end{array}$$

(38)

where $\overline{\Delta }$ is the Laplacian on $({\overline{M}}^{2n+1},f,\phi ,\xi ,\eta ,g)$.

Proof. 

Let $\left\{e_1,\dots ,e_n,\phi \left(e_1\right),\dots ,\phi \left(e_n\right),\xi \right\}$ be an orthonormal local basis on $TM$; then, by definition of the tension field of map I, we had the following:

$$\begin{array}{cc}\hfill \tau \left(I\right)=& t{r}_g\stackrel{}{\nabla }dI\hfill \\ \hfill =& \sum _{i=1}^{2n+1}\left({\nabla }_{dI\left(e_i\right)}^{M}dI\left(e_i\right)-dI\left({\nabla }_{e_i}^{\overline{M}}{e}_i\right)\right)\hfill \\ \hfill =& -\sum _{i=1}^{2n+1}\left(f(g(e_i,e_i)\xi -\eta \left(e_i\right)e_i)\right)\hfill \\ \hfill =& -2nf\xi .\hfill \end{array}$$

However, we had the following:

$$\begin{array}{cc}\hfill {\tau }_2\left(I\right)& =t{r}_g{\left({\nabla }^I\right)}^2\tau \left(I\right)+t{r}_g{R}^M(\tau \left(I\right),dI)dI\hfill \\ \hfill & =-2n\left(t{r}_g{\left({\nabla }^I\right)}^{2}f\xi +t{r}_g{R}^{\overline{M}}(f\xi ,dI)dI\right)\hfill \\ \hfill & =-\sum _{i=1}^{2n+1}\left({\nabla }_{e_i}^I{\nabla }_{e_i}^{I}2nf\xi -{\nabla }_{{\nabla }_{e_i}^{\overline{M}}{e}_i}^{I}2nf\xi +R^M(2nf\xi ,dI\left(e_i\right))dI\left(e_i\right)\right).\hfill \end{array}$$

(39)

A direct calculation of

$$\begin{array}{cc}\hfill \sum _{i=1}^{2n+1}\left({\nabla }_{e_i}^I{\nabla }_{e_i}^{I}2nf\xi \right)=& 2n\sum _{i=1}^{2n+1}{\nabla }_{e_i}^I\left(e_i\left(f\right)\xi +f{\nabla }_{e_i}^I\xi \right)\hfill \\ \hfill =& 2n\sum _{i=1}^{2n+1}\left(e_i\left(e_i\left(f\right)\right)\xi +e_i\left(f\right){\nabla }_{e_i}^I\xi +e_i\left(f\right){\nabla }_{e_i}^I\xi +f{\nabla }_{e_i}^I{\nabla }_{e_i}^I\xi \right)\hfill \\ \hfill =& 2n\sum _{i=1}^{2n+1}\left(e_i\left(e_i\left(f\right)\right)\xi +2{\nabla }_{\mathrm{grad}f}^I\xi +f{\nabla }_{e_i}^I{\nabla }_{e_i}^I\xi \right),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \sum _{i=1}^{2n+1}\left({\nabla }_{{\nabla }_{e_i}^{\overline{M}}{e}_i}^{I}2nf\xi \right)=& 2n\sum _{i=1}^{2n+1}\left(\left({\nabla }_{e_i}^{\overline{M}}{e}_i\right)\left(f\right)\xi +f{\nabla }_{{\nabla }_{e_i}^{\overline{M}}{e}_i}^I\xi \right)\hfill \end{array}$$

finally yielded the following:

$$\begin{array}{c}\hfill t{r}_g{\left({\nabla }^I\right)}^{2}nf\xi =2n\left(\overline{\Delta }\left(f\right)\xi +ft{r}_g{\left({\nabla }^I\right)}^2\xi +2{\nabla }_{\mathrm{grad}f}^I\xi \right).\end{array}$$

(40)

If we replaced (40) in (39), we arrived at:

$$\begin{array}{cc}\hfill {\tau }_2\left(I\right)=& -2n\left(\overline{\Delta }\left(f\right)\xi +f{\tau }_2\left(\xi \right)+2{\nabla }_{\mathrm{grad}f}^I\xi \right).\hfill \end{array}$$

□

Corollary 5. 

Let $I:\overline{M}⟶M$ be the identity map from an f-Kenmotsu manifold with the Schouten–van Kampen connection $({\overline{M}}^{2n+1},f,\phi ,\xi ,\eta ,g)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$. Then, map I would be biharmonic if, and only if:

$$\begin{array}{c}\hfill \overline{\Delta }\left(f\right)\xi +f{\tau }_2\left(\xi \right)+2{\nabla }_{\mathrm{grad}f}^I\xi =0,\end{array}$$

Corollary 6. 

Let $I:\overline{M}⟶M$ be the identity map from an f-Kenmotsu manifold with the Schouten–van Kampen connection $({\overline{M}}^{2n+1},f,\phi ,\xi ,\eta ,g)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$. If f was a constant function, then map I would be biharmonic if, and only if, ξ was a biharmonic vector field.

Corollary 7. 

Let $I:\overline{M}⟶M$ be the identity map from an f-Kenmotsu manifold with the Schouten–van Kampen connection $({\overline{M}}^{2n+1},f,\phi ,\xi ,\eta ,g)$ to an f-Kenmotsu manifold $(M^{2n+1},f,\phi ,\xi ,\eta ,g)$. If ξ was a parallel vector field (i.e., $\nabla \xi =0$), then map I would be biharmonic if, and only if, f was a harmonic function.