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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2021; 61(2): 395-408

Published online June 30, 2021

### On f-biharmonic Submanifolds of Three Dimensional Trans-Sasakian Manifolds

Avijit Sarkar* and Nirmal Biswas

Department of Mathematics, University of Kalyani, Kalyani 741235, West Bengal, India
e-mail : avjaj@yahoo.co.in and nirmalbiswas.maths@gmail.com

Received: December 20, 2019; Revised: November 20, 2020; Accepted: November 23, 2020

The object of the present paper is to study f-biharmonic submanifolds of three dimensional trans-Sasakian manifolds. We find some necessary and sufficient conditions for such submanifolds to be f-biharmonic.

Keywords: trans-Sasakian manifolds, invariant submanifolds, anti-invariant submanifolds, f-biharmonic submanifolds

Let M and N be two Riemannian manifolds, a harmonic map $ψ:M→N$ is any critical point of the energy equation

$E(ψ)=12∫M |dψ| 2dvg ,$

where dvg denotes the volume element of g, and the Euler-Lagrange equation corresponding to E(ψ ) is $τ(ψ)=trace∇dψ=0$.

In 1983, Eells and Lemaire [9] introduced the notion of biharmonic maps, which are a natural generalization of harmonic maps. A biharmonic map $ψ:M→N$ is a critical point of the energy equation

$E2(ψ)=12∫M |τψ| 2dvg ,$

where dvg denotes the volume element of g, and the Euler-Lagrange equation [15] corresponding to E2(ψ) is

$τ2(ψ)=Δτ(ψ)−trace(RN(dψ,τ(ψ))dψ)=0.$

Here Δ is the Laplacian operator given by $ΔV=tr(∇2V)$, and RN is the curvature tensor on the manifold N defined as $RN(X,Y)=[∇X,∇Y]−∇[X,Y].$

Let M be the submanifold of the manifold $M¯$, if the biharmonic map $ψ:M→M¯$ is an isometric immersion then M is biharmonic submanifold of $M¯$. In the paper [2], Baird studied conformal and semi-conformal biharmonic maps. Oniciuc studied biharmonic submanifolds of CPn in [10]. He studied explicit formula for biharmonic submanifolds in Sasakian space forms and deduced some conditions in [11]. He proved a gap theorem for the mean curvature of certain complete proper biharmonic pmc submanifolds and classified proper biharmonic pmc surfaces in $Sn×R$ in [12]. In [16], Oniciuc studied biharmonic constant mean curvature surface in the sphere. Recently, Oniciuc proved several unique continuation results for biharmonic maps between Riemannian manifolds in [19]. He studied biharmonic maps between Riemannian manifolds in [18]. Over the last few years many authors have studied biharmonic submanifolds, for example see [5, 10, 18]. Recently, Ou studied biharmonic maps form tori into a 2-sphere in [27]. In the paper [1], Ou studied biharmonic Riemannian submanifolds.

The notion of f-biharmonic maps was introduced by Lu [17]; it is a natural generalization of biharmonic maps. In the papers [21, 22], Ou studied f-biharmonic maps and f-biharmonic submanifolds. In these papers he proved that a f-biharmonic map from a compact Riemannian manifold into a non-positively curved manifold with constant f-bienergy density is a harmonic map. In [20], Ou characterized harmonic maps and minimal submanifolds using the concept of f-biharmonic maps and proved that the set of all f-biharmonic maps from a 2-dimensional domain is invariant under the conformal change of the metric on the domain. In [24], Roth studied f-biharonic submanifolds of generalized space forms. He deduced some necessary and sufficient conditions for f-biharmonicity in the general case and many particular cases. In [2] Baird and Fardon studied conformal and semi conformal biharmonic maps.

Let us consider the $C∞$ differentiable function $f:M→R$. Now, f-harmonic maps are the critical points of the f-energy functional $Ef(ψ)$ for the maps $ψ:M→N$ between Riemannian manifolds, where

$Ef(ψ)=12∫M f|dψ|2dvg.$

The Euler-Lagrange equation corresponding to $Ef(ψ)$ is given by

$τf(ψ)=fτ(ψ)+dψ(gradf)=0.$

Analgously f-biharmonic maps are critical points of the f-bienergy functional $E2,f(ψ)$ for maps $ψ:M→N$ between Riemannian manifolds where

$E2,f(ψ)=12∫Mf|τψ|2dvg.$

The Euler-Lagrange equation corresponding to $E2,f(ψ)$ is given by

$τ2,f(ψ)=fτ2(ψ)+(Δf)τ(ψ)+2∇(gradf)ψτ(ψ)=0.$

Clearly, we have the following relationship among these different types of harmonic maps:

Harmonic maps $⊂$ biharmonic maps $⊂$ f-biharmonic maps.

A f-biharmonic map is called a proper f-biharmonic map if it is neither a harmonic nor a biharmonic map. Also, we will call a f-biharmonic submanifold proper if it is neither minimal nor biharmonic.

The notion of trans-Sasakian Manifolds was introduced by Blair and Oubina [4, 23] as a generalization of Sasakian manifolds. Trans-Sasakian manifolds of type (α,β) are generalizations of α-Sasakian and β-Kenmotsu manifolds. It is known that a proper trans-Sasakian manifold exists only for dimension three and trans-Sasakian manifolds of type $(0,0),(0,β),$ and (α,0) are known [14] as cosymplectic, β-Kenmotsu and α-Sasakian respectively. In higher dimension it is either α-Sasakian or β-Kenmotsu. In Differential Geometry of almost contact manifolds, submanifold theory has become an important topic of research. There are several works on invariant submanifolds. In [6], the authors studied invariant submanifolds of trans-Sasakian manifolds. Three dimensional trans-Sasakian Manifolds have been studied by the first author in the papers [8, 25, 26].

During last few years biharmonic maps on contact manifolds have become a popular area of research. So in the present paper we would like to study f-biharmonic maps on three dimensional trans-Sasakian manifolds. Precisely we study f-biharmonic submanifolds of three dimensional trans-Sasakian manifolds and find some conditions for the map f to be biharmonic or not.

The present paper is organized as follows: Section 1 is introductory. After the introduction we give some preliminaries in Section 2. In Section 3 we study f-biharmonic submanifolds of three-dimensional trans-Sasakian manifolds.

Let $M¯$ be an odd dimensional smooth differential manifold with an almost contact metric structure (ϕ,ξ,η,g), where ϕ is a (1,1)-tensor field, ξ is a vector field, η is a one form and g is a Riemannian metric on $M¯$. For such manifolds, we know [3]

$ϕ2X=−X+η(X)ξ, η(ξ)=1,$ $η(X)=g(X,ξ), g(ϕX,ϕY)=g(X,Y)−η(X)η(Y),$ $ϕξ=0, ηoϕ=0, g(X,ϕY)=−g(ϕX,Y)$

for any $X,Y∈χ(M¯)$, where $χ(M¯)$ denotes the Lie algebra of all vector fields on $M¯$.

For a contact metric manifold $(M¯,ϕ,ξ,η,g)$, we define a (1,1) tensor field h by $h=12Lξϕ$ and $L$ is the usual Lie derivative. Then h is symmetric and satisfies the following relations

$hξ=0, hϕ=−ϕh, tr(h)=tr(ϕh)=0, η(hX)=0$

for any $X,Y∈χ(M¯)$.

Moreover, if $∇¯$ denotes the Levi-Civita connection with respect to g, then the following relation holds

$∇¯Xξ=−ϕX−ϕhX.$

A connected manifold $M¯$ with almost contact metric structure $(ϕ,ξ,η,g)$ is called a trans-Sasakian manifold [23] if $(M¯×R,J,G)$ belongs to the class W4 [13], where J is an almost complex structure on $M¯×R$ which is defined by

for any vector field X on $M¯$ and the smooth function f on $M¯×R$, and G is the usual product metric on $M¯×R$. According to [4], an almost contact metric manifold is a trans-Sasakian manifold if and only if

$(∇¯Xϕ)Y=α(g(X,Y)ξ−η(Y)X)+β(g(ϕX,Y)ξ−η(Y)ϕX)$

for smooth functions $α,β$ on $M¯$, where $∇¯$ denote the covariant derivative with respect to g. Generally, $M¯$, is said to be a trans-Sasakian manifold of type $(α,β)$.

In a three-dimensional trans-Sasakian manifold the curvature tensor with respect to the Levi-Civita connection $∇¯$ is as follows [7]:

$R(X,Y)Z=(r2+2ξβ−2(α2−β2))(g(Y,Z)X−g(X,Z)Y) −g(Y,Z)[(r2+ξβ−3(α2−β2))η(X)ξ −η(X)(ϕgradα−ϕgradβ)+(Xβ+(ϕX)α)ξ] +g(X,Y)[(r2+ξβ−3(α2−β2))η(Y)ξ −η(Y)(ϕgradα−ϕgradβ)+(Yβ+(ϕY)α)ξ] −[(Zβ+(ϕZ)α)η(Y)+(Yβ+(ϕY)α)η(Z) +(r2+ξβ−3(α2−β2))η(Y)η(Z)]X +[(Zβ+(ϕZ)α)η(X)+(Xβ+(ϕX)α)η(Z) +(r2+ξβ−3(α2−β2))η(X)η(Z)]Y,$

where r is the scalar curvature of the manifold.

Let Mm (m<;n) be the submanifold of a contact metric manifold $M¯n$. Let $∇$ and $∇¯$ be the Levi-Civita connections of M and $M¯$, respectively. Then for any vector fields $X,Y∈χ(M)$, the second fundamental form σ is defined by

$∇¯XY=∇XY+σ(X,Y).$

For any section of the normal bundle $T⊥M$, we have

$∇¯XN=−ANX+∇⊥N,$

where $∇⊥$ denotes the normal bundle connection of M. The second fundamental form σ and the shape operator AN are related by

$g(ANX,Y)=g(σ(X,Y),N).$

For any vector field $X∈χ(M)$, we can right

$ϕX=TX+NX,$

where TX is the tangential component of ϕX and NX is the normal component of ϕX. Similarly, for any vector field V in normal bundle we have

$ϕV=tV+nV,$

where tV and nV are the tangential and normal components of $ϕV$.

The submanifold M is said to be invariant if $ϕX∈TM$ for any vector field X. On other hand M is said to be an anti-invariant submanifold if $ϕX∈T⊥M$ for any vector field X

### 3. f-biharmonic Submanifolds of Three-dimensional Trans-Sasakian Manifolds

We know for a isometric immersion ψ [24]

$τ(ψ)=tr∇dψ=trσ=mH,$

where H is the mean curvature. Now using the equation (1.1) in the above equation we have

$τ2(ψ)=mΔH−tr(R(dψ,mH)dψ).$

By some classical and straightforward computations, we have

$ΔH=m2grad|H|2+tr(σ(.,AH.))+2tr(A∇⊥H(.))+Δ⊥H.$

Using (3.3) in (3.2), we have

$τ2(ψ)=m22grad|H|2+mtr(σ(.,AH.))+2mtr(A∇⊥H(.))+mΔ⊥H−tr(R(dψ,mH)dψ).$

From the equation (1.3), we have the submanifold M is f-biharmonic if and only if

$τ2,f(ψ)=fτ2(ψ)+(Δf)τ(ψ)+2∇(gradf)ψτ(ψ)=0.$

By simple calculation we have the above equation is equivalent to

$τ2(ψ)+mΔffH+2m(−AHgrad(lnf)+∇grad(lnf)⊥H)=0.$

For a f-biharmonic submanifold of a three-dimensional trans-Sasakian manifold we have the following:

Theorem 3.1. Let M be a submanifold of a three dimensional trans-Sasakian manifold $M¯$. Then M is f-biharmonic if and only if the following equations hold

$Δ⊥H+tr(σ(.,AH.))+ΔffH+2∇grad(lnf)⊥H=−2(r2+2ξβ−2(α2−β2))H+2[(r2+ξβ−3(α2−β2))η(H)ξ⊥ −η(H)(Ngradα−Ngradβ)+ξβH−ξαn(H)] +[2ξβ+(r2+ξβ−3(α2−β2))]H,$

and

Proof. Form (2.7) we have

$R(X,Y)Z=(r2+2ξβ−2(α2−β2))(g(Y,Z)X−g(X,Z)Y) −g(Y,Z)[(r2+ξβ−3(α2−β2))η(X)ξ −η(X)(ϕgradα−ϕgradβ)+(Xβ+(ϕX)α)ξ] +g(X,Y)[(r2+ξβ−3(α2−β2))η(Y)ξ −η(Y)(ϕgradα−ϕgradβ)+(Yβ+(ϕY)α)ξ] −[(Zβ+(ϕZ)α)η(Y)+(Yβ+(ϕY)α)η(Z) +(r2+ξβ−3(α2−β2))η(Y)η(Z)]X +[(Zβ+(ϕZ)α)η(X)+(Xβ+(ϕX)α)η(Z) +(r2+ξβ−3(α2−β2))η(X)η(Z)]Y.$

Let {e1,e2} be an orthogonal basis of the tangent space at a point of M. Then we have from above

$R(ei,Y)ei=(r2+2ξβ−2(α2−β2))(g(H,ei)ei−g(ei,ei)H) −g(H,ei)[(r2+ξβ−3(α2−β2))η(ei)ξ −η(ei)(ϕgradα−ϕgradβ)+(eiβ+(ϕei)α)ξ] +g(ei,ei)[(r2+ξβ−3(α2−β2))η(H)ξ −η(H)(ϕgradα−ϕgradβ)+(Hβ+(ϕH)α)ξ] −[(eiβ+(ϕei)α)η(H)+(Hβ+(ϕH)α)η(ei) +(r2+ξβ−3(α2−β2))η(H)η(ei)]ei +[(eiβ+(ϕei)α)η(ei)+(eiβ+(ϕei)α)η(ei) +(r2+ξβ−3(α2−β2))η(ei)η(ei)]H.$

Taking trace and using the equations (2.1), (2.11) and (2.12) we obtain

$tr(R(.,H).)=−2(r2+2ξβ−2(α2−β2))H+2[(r2+ξβ−3(α2−β2))η(H)ξ −η(H)(ϕgradα−ϕgradβ)+ξβH−ξαϕ(H)]−[(gradβ)Tη(H) +g(gradβ,H)ξT+g(gradα,ϕH)ξT+(r2+ξβ−3(α2−β2))η(H)ξT] +[2η(gradβ)+(r2+ξβ−3(α2−β2))]H.$

Using the equations (3.4) and (3.6) we can obtain

$tr(R(.,H).)=grad|H|2+tr(σ(.,AH.))+2tr(A∇⊥H(.)) +Δ⊥H+ΔffH−2(AHgrad(lnf))+2∇grad(lnf)⊥H.$

Therefore we have

$grad|H|2+tr(σ(.,AH.))+2tr(A∇⊥H(.)) +Δ⊥H+ΔffH−2(AHgrad(lnf))+2∇grad(lnf)⊥H=−2(r2+ξβ−2(α2−β2))H +2[(r2+ξβ−3(α2−β2))η(H)ξ−η(H)(ϕgradα−ϕgradβ)+ξβH−ξαϕ(H)] −[(gradβ)Tη(H)+g(gradβ,H)ξT+g(gradα,ϕH)ξT +(r2+ξβ−3(α2−β2))η(H)ξT]+[2η(gradβ)+(r2+ξβ−3(α2−β2))]H.$

Comparing the tangent and normal components we have the result of the theorem.

Now we have the following as particular cases of the above theorem.

Corollary 3.1. Let M be a submanifold of a three-dimensional trans-Sasakian manifold $M¯$.

• (1) If M is anti-invariant, M is f-biharmonic if and only if

$Δ⊥H+tr(σ(.,AH.))+ΔffH+2∇grad(lnf)⊥H=−2(r2+2ξβ−2(α2−β2))H+2[(r2+ξβ−3(α2−β2))η(H)ξ⊥ +ξβH−ξαn(H)]+[2ξβ+(r2+ξβ−3(α2−β2))]H,$

and

• (2) If M is invariant M is f-biharmonic if and only if

$Δ⊥H+tr(σ(.,AH.))+ΔffH+2∇grad(lnf)⊥H=−2(r2+2ξβ−2(α2−β2))H+2[(r2+ξβ−3(α2−β2))η(H)ξ⊥ −η(H)(Ngradα−Ngradβ)+ξβH−ξαn(H)] +[2ξβ+(r2+ξβ−3(α2−β2))]H,$

and

$grad|H|2−2trAHgrad(lnf)+2tr(A∇⊥H,.)=2[(r2+ξβ−3(α2−β2))η(H)ξT+t(H)ξα]−[(gradβ)Tη(H)+ g(gradβ,H)ξT+g(gradα,ϕH)ξT+(r2+ξβ−3(α2−β2))η(H)ξT].$

• (3) If ξ is normal to M, M is f-biharmonic if and only if

$Δ⊥H+tr(σ(.,AH.))+ΔffH+2∇grad(lnf)⊥H=−2(r2+2ξβ−2(α2−β2))H+2[(r2+ξβ−3(α2−β2))η(H)ξ⊥ −η(H)(Ngradα−Ngradβ)+ξβH−ξαn(H)] +[2ξβ+(r2+ξβ−3(α2−β2))]H,$

and

$grad|H|2−2trAHgrad(lnf)+2tr(A∇⊥H,.)=2[−η(H)(Tgradα−Tgradβ)+t(H)ξα]−[(gradβ)Tη(H)].$

• (4) If ξ is tangent to M, M is f-biharmonic if and only if

$Δ⊥H+tr(σ(.,AH.))+ΔffH+2∇grad(lnf)⊥H =−2(r2+2ξβ−2(α2−β2))H+ξβH−ξαn(H)] +[2ξβ+(r2+ξβ−3(α2−β2))]H,$

and

$grad|H|2−2trAHgrad(lnf)+2tr(A∇⊥H,.)=2t(H)ξα−[g(gradβ,H)ξT+g(gradα,ϕH)ξT],$

• (5) If M is a hypersurface, M is f-biharmonic if and only if

$Δ⊥H+tr(σ(.,AH.))+ΔffH+2∇grad(lnf)⊥H=−2(r2+2ξβ−2(α2−β2))H+2[(r2+ξβ−3(α2−β2))η(H)ξ⊥ −η(H)(Ngradα−Ngradβ)+ξβH−ξαn(H)] +[2ξβ+(r2+ξβ−3(α2−β2))]H,$

and

$grad|H|2−2trAHgrad(lnf)+2tr(A∇⊥H,.)=2[(r2+ξβ−3(α2−β2))η(H)ξT−η(H)(Tgradα−Tgradβ] −[(gradβ)Tη(H)+g(gradβ,H)ξT+g(gradα,ϕH)ξT +(r2+ξβ−3(α2−β2))η(H)ξT].$

Proof. Proof of the results is directly obtained from Theorem 3.1, using the following facts, respectively.

• (1) If M is invariant then N=0.

• (2) If M is anti-invariant then T=0.

• (3) If ξ is normal to M then ξT=0.

• (4) If ξ is tangent to M then η(H)=0 and $ξ⊥=0$.

• (5) If M is a hypersurface then tH=0.

Theorem 3.2. Let M be a submanifold of a three dimensional trans-Sasakian manifold $M¯$ with non zero constant mean curvature H and ξ is tangent to M, then M proper f-biharmonic if and only if

$|σ|2=−3r2−7ξβ+7(α2−β2)−Δff,$

and $AHgrad(lnf)=0$, or equivalent if and only if

$ScalM=3r2+9ξβ−8(α2−β2)+Δff−3|H|2.$

Proof. Let M be a f biharmonic submanifold of $M¯$ with constant mean curvature and ξ tangent to M then from the previous corollary we have

$Δ⊥H+tr(σ(.,AH.))+ΔffH+2∇grad(lnf)⊥H=−2(r2+2ξβ−2(α2−β2))H+2[(r2+ξβ−3(α2−β2))η(H)ξ⊥ −η(H)(Ngradα−Ngradβ)+ξβH−ξαn(H)] +[2ξβ+(r2+ξβ−3(α2−β2))]H,$

and

$grad|H|2−2trAHgrad(lnf)+2tr(A∇⊥H,.)=2[−η(H)(Tgradα−Tgradβ)+t(H)ξα]−[(gradβ)Tη(H)].$

Since ξ is tangent to M then the equations are of the form

$tr(σ(.,AH.))=−2(r2+2ξβ−2(α2−β2))H+2[(r2+ξβ−3(α2−β2))η(H)ξ⊥ −η(H)(Ngradα−Ngradβ)+ξβH−ξαn(H)] +[2ξβ+(r2+ξβ−3(α2−β2))]H−ΔffH,$

and $AHgrad(lnf)=0$. Thus, the second equation is trivial and the first equation becomes

$trσ(.,AH.)=[−3r2−7ξβ+7(α2−β2)−Δff]H.$

Now since $trσ(.,AH.)=|σ|2H$ and H is non zero, so we have form above equation

$|σ|2=−3r2−7ξβ+7(α2−β2)−Δff.$

Now from the Gauss formula we have

$ScalM=∑ i,jg(R(ei,ej)ej,ei)−|σ|2−2H2.$

Using (2.7) in the above equation we have

$ScalM=3r29ξβ−8(α2−β2)+Δff−3|H|2.$

Corollary 3.2. Let M be a submanifold of a three dimensional trans-Sasakian manifold $M¯$ with non zero constant mean curvature H and ξ is tangent to M. If the functions α, β satisfy the inequality

$−3r2−7ξβ+7(α2−β2)≤Δff$

then M is not f-biharmonic.

Proof. Form the Theorem 3.2 we know that M is f-biharmonic if and only if its second fundamental form σ satisfies the inequality

$|σ|2=−3r2−7ξβ+7(α2−β2)−Δff,$

Since $|σ|2≥0$, this is not possible if

$−3r2−7ξβ+7(α2−β2)≤Δff.$

Theorem 3.3. Let M be a submanifold of a three dimensional trans-Sasakian manifold $M¯$ with non zero constant mean curvature H such that ξ and ϕ H are tangent to M. Define $F(f,α,β)$ on M by

$F(f,α,β)=−2r−9ξβ+9(α2−β2)−Δff.$
• (1) if inf $F(f,α,β)$ is non-positive, M is not f-biharmonic.

• (2) if $F(f,α,β)$ is positive and M is proper f-biharmonic then

$0<|H|2≤12F(f,α,β).$

Proof. M is proper f-biharmonic submanifold with constant mean curvature H and ξ is tangent to M, so we have form Corollary 3.1

$Δ⊥H+tr(σ(.,AH.))+ΔffH+2∇grad(lnf)⊥H=−2(r2+2ξβ−2(α2−β2))H+ξβH−ξαn(H)] +[2ξβ+(r2+ξβ−3(α2−β2))]H$

and

$grad|H|2−2trAHgrad(lnf)+2tr(A∇⊥H,.)=2t(H)ξα−[g(gradβ,H)ξT+g(gradα,ϕH)ξT].$

Given that ϕ H is tangent to M, so tH=0. Therefore form the above equation we have

$Δ⊥H+tr(σ(.,AH.))=[−2r−9ξβ+9(α2−β2)−Δff] =F(f,α,β)H,$

where

$F(f,α,β)=−2r−9ξβ+9(α2−β2)−Δff.$

Taking inner product by H of the equation (??), we have

$<Δ⊥H,H>+=F(f,α,β)|H|2.$

Now using the results $=|AH|2,$ and $Δ|H|2=2(<Δ⊥H,H>−|∇⊥H|2)$, in the above equation we have

$|AH|2+|Δ⊥H|2=F(f,α,β)|H|2.$

By using the Cauchy-Schwarz inequality $|AH|2≥12tr(AH)=2|H|4,$ the equation reduces to

$F(f,α,β)|H|2=|AH|2+|∇⊥H|2≥2|H|4+|∇⊥H|2≥2|H|4.$

Therefore $F(f,α,β)≥2|H|2,$ since |H| is positive. This proves the theorem.

The authors are thankful to the referee for his valuable suggestions towards the improvement of the paper.

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