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Kyungpook Mathematical Journal 2020; 60(3): 599-627

Published online September 30, 2020

Copyright © Kyungpook Mathematical Journal.

Biharmonic Maps on Doubly Warped Product Manifolds

Khaldia Madani, Seddik Ouakkas*

National Polytechnic School of Oran Maurice Audin (ENPO-MA), Algeria
e-mail : khaldia.madani@enp-oran.dz
Laboratory of Geometry, Analysis, Control and Applications, University of Saida, Algeria
e-mail : seddik.ouakkas@gmail.com

Received: November 26, 2019; Revised: March 20, 2020; Accepted: March 21, 2020

In this paper, we characterize a class of biharmonic maps from and between doubly product manifolds in terms of theie warping function. Examples are constructed when all of the factors are Euclidean spaces.

Keywords: harmonic map, biharmonic map, doubly warped product

Let ϕ:(Mm,g)(Nn,h) be a smooth map between Riemannian manifolds. Such ϕ is said to be harmonic if it is a critical point of the energy functional

E(ϕ)=12M|dϕ|2dvg

with respect to compactly supported variations. Equivalently, ϕ is harmonic if it satisfies the associated Euler-Lagrange equations given as follows :

τ(ϕ)=Trgdϕ=0.

Here τ (ϕ) is the tension field of ϕ. We refer one to [5, 6, 7, 8, 9] for background on harmonic maps. As a generalization of harmonic maps, biharmonic maps are defined similarily, as follows : ahe map ϕ is said to be biharmonic if it is a critical point of the bi-energy functional

E2(ϕ)=12M|τ(ϕ)|2dvg.

Equivalently, ϕ is biharmonic if it satisfies the associated Euler-Lagrange equations:

τ2(ϕ)=Trgϕ2τ(ϕ)TrgRN(τ(ϕ),dϕ)dϕ=0,

where ∇ϕ is the connection in the pull-back bundle ϕ-1(TN) and, if ei1im is a local orthonormal frame field on M, then

Trgϕ2τϕ=eiϕeiϕeiMeiϕτϕ,

where we sum over repeated indices. We will call the operator τ2(ϕ), the bi-tension field of the map ϕ. Clearly any harmonic map is biharmonic, therefore it is interesting to construct non-harmonic biharmonic maps (see [1, 2, 3] and [12, 13, 15] for some constructions of non-harmonic biharmonic maps). In [4], the authors studied biharmonic maps between warped products where they gave the condition for the biharmonicity of the inclusion of a Riemannian manifold N into the warped product M △ f N and of the projection π¯:M×fNM. Moreover, in [10] the authors gave some extensions of the results in [4] together with some further constructions of biharmonic maps. They also gave some characterizations of non-harmonic biharmonic maps using the product of harmonic maps and warping metric. The author in [11] studied the f-harmonicity of some special maps from or into a doubly warped product manifold. He obtained some similar results in [10], such as conditions for the f-harmonicity of projection maps and some characterizations for non-trivial f-harmonicity of the special product maps; furthermore, he investigate non-trivial f-harmonicity of the product of two harmonic maps. In [14], the authors study the biharmonic maps between doubly warped product manifolds and they gave some characterizations of non-harmonic biharmonic maps using products of harmonic maps and the warping metric. In this paper, we give a different constructions of biharmonic maps on the doubly warped product manifolds. First, we characterize the biharmonicity of the maps ϕ˜:Mm×α,βNn,Gα,βPp,k and ψ˜:Mm×α,βNn,Gα,βPp,k defined by ϕ˜x,y=ϕx and ψ˜x,y=ψy. In particular we study the first and the second projection (Theorems 2.3 and 2.7). In this setting we obtain some examples of biharmonic non-harmonic maps. As a second result, we study the biharmonicity of the inclusion maps iy0:Mn,gMm×α,βNn,Gα,β and ix0:Nn,hMm×α,βNn,Gα,β (Theorems 2.11 and 2.13). Finally, we determine the conditions of the biharmonicity of the identity maps Mm×α,βNn,Gα,βIdMm×Nn,G=gh and Mm×Nn,G=ghIdMm×α,βNn,Gα,β (Theorems 2.15 and 2.19). Some special cases are developed.

Let Mm,g and Nn,h two Riemannian manifolds and let αCM and βCN be a positive functions. The doubly warped product Mm×α,βNn is the product manifolds M×N endowed with the Riemannian metric Gα,β defined, for X,YΓTM×N, by

Gα,βX,Y=βσ2gdπX,dπY+απ2hdηX,dηY,

where π:M×NM and η:M×NN are respectively the first and the second projection. Let X,YΓTM×N, X=X1,X2, Y=Y1,Y2. For all obtained results in this paper, we consider ei1im to be an orthonormal frame on M and fj1jn to be an orthonormal frame on N. Then, an orthonormal frame on Mm×α,βNn,Gα,β is given by 1βei,0,1α0,fj. Denote by ∇ the Levi-Civita connection on the Riemannian product M×N, the Levi-Civita connection ˜ of the doubly warped product Mm×α,βNn is given by (see [14])

˜XY=XY+X1lnα0,Y2+Y1lnα0,X2+X2lnβY1,0+Y2lnβX1,0β2gX1,Y10,gradlnβα2hX2,Y2gradlnα,0.

Using equation (2.1), we give some particular cases below.

Proposition 2.1.

LetX,YΓTM×N, X=X1,X2, Y=Y1,Y2. The Levi-Civita connection˜ of the doubly warped product Mm×α,βNnsatisfies the following equations

  ˜X1,0Y1,0=X1Y1,0β2gX1,Y10,gradlnβ,  ˜0,X20,Y2=0,X2Y2α2hX2,Y2gradlnα,0,  ˜X1,00,Y2=X1lnα0,Y2+Y2lnβX1,0,and  ˜0,X2Y1,0=Y1lnα0,X2+X2lnβY1,0.

In the first, we consider a smooth map ϕ:Mm,gPp,k and we define the map ϕ˜:Mm×α,βNn,Gα,βPp,k by ϕ˜x,y=ϕx. By calculating the tension field of ϕ˜, we obtain the following result.

Proposition 2.2.

Letϕ:Mm,gPp,kbe a smooth map. The tension field of the mapϕ˜:Mm×α,βNn,Gα,βPp,kdefined byϕ˜x,y=ϕxis given by

τϕ˜=1β2τϕ+ndϕgradlnα.

Proof. By definition of the tension field, we have

τϕ˜=TrGα,βdϕ˜  =1β2ei,0ϕ˜dϕ˜ei,0+1α20,fjϕ˜dϕ˜0,fj  1β2dϕ˜˜ei,0ei,01α2dϕ˜˜0,fj0,fj,

where we sum over repeated indices. A simple calculation gives

  ei,0ϕ˜dϕ˜ei,0=eiϕdϕeiand  0,fjϕ˜dϕ˜0,fj=0.

Using Proposition 2.1, we deduce that

  ˜ei,0ei,0=eiei,0mβ20,gradlnβand  ˜0,fj0,fj=0,fjfjnα2gradlnα,0,

it follows that

τϕ˜=1β2eiϕdϕeidϕeiei+ndϕgradlnα,

then

τϕ˜=1β2τϕ+ndϕgradlnα.

In the following, we will calculate the bi-tension field of the map ϕ˜:Mm×α,βNn,Gα,βPp,k.

Theorem 2.3.

Let ϕ:Mm,gPp,kbe a smooth map. the bi-tension field of the map ϕ˜:Mm×α,βNn,Gα,βPp,kdefined byϕ˜x,y=ϕxis given by

τ2ϕ˜=1β4τ2ϕ+2α2β2Δlnβ2gradlnβ2τϕnβ2gradlnαϕτϕ+2mβ2gradlnβ2τϕn2gradlnαϕdϕgradlnαnβ2Trgϕ2dϕgradlnα+TrgRPdϕgradlnα,dϕdϕ.

Proof. By definition of the bi-tension field, we have

τ2ϕ˜=TrGα,βϕ˜2τϕ˜  TrGα,βRPτϕ˜,dϕ˜dϕ˜.

Using the fact that

τϕ˜=1β2τϕ+ndϕgradlnα,

we get

TrGα,βϕ˜2τϕ˜=TrGα,βϕ˜21β2τϕ        +nTrGα,βϕ˜2dϕgradlnα.

For the term TrGα,βϕ˜21β2τϕ, we have

TrGα,βϕ˜21β2τϕ  =1β2ei,0ϕ˜ei,0ϕ˜1β2τϕ1β2˜ei,0ei,0ϕ˜1β2τϕ  +1α20,fjϕ˜0,fjϕ˜1β2τϕ1α2˜0,fj0,fjϕ˜1β2τϕ  =1β4Trgϕ2τϕ2mβ2gradlnβ2τϕ  2α2β2fjfjlnβ2gradlnβ2τϕ  +2α2β2fjfjlnβτϕ+nβ2gradlnαϕτϕ,

which will lead to

TrGα,βϕ˜21β2τϕ  =1β4Trgϕ2τϕ2α2β2Δlnβ2gradlnβ2τϕ  +nβ2gradlnαϕτϕ2mβ2gradlnβ2τϕ.

A similar calculation gives us

TrGα,βϕ˜2dϕgradlnα  =1β2ei,0ϕ˜ei,0ϕ˜dϕgradlnα1β2˜ei,0ei,0ϕ˜dϕgradlnα  +1α20,fjϕ˜0,fjϕ˜dϕgradlnα1α2˜0,fj0,fjϕ˜dϕgradlnα  =1β2Trgϕ2dϕgradlnα+ngradlnαϕdϕgradlnα,

it follows that

TrGα,βϕ˜2τϕ˜  =1β4Trgϕ2τϕ+nβ2Trgϕ2dϕgradlnα  +nβ2gradlnαϕτϕ+n2gradlnαϕdϕgradlnα  2α2β2Δlnβ2gradlnβ2τϕ.

Finally for term TrGα,βRPτϕ˜,dϕ˜dϕ˜, it is very simple to see that

TrGα,βRPτϕ˜,dϕ˜dϕ˜=1β4TrgRPτϕ,dϕdϕ          +nβ2TrgRPdϕgradlnα,dϕdϕ.

If we replace (2.4) and (2.5) in (2.3), we deduce that

τ2ϕ˜=1β4τ2ϕ+2α2β2Δlnβ2gradlnβ2τϕnβ2gradlnαϕτϕ+2mβ2gradlnβ2τϕn2gradlnαϕdϕgradlnαnβ2Trgϕ2dϕgradlnα+TrgRPdϕgradlnα,dϕdϕ.

As a consequence, if ϕ is harmonic, we have the following.

Corollary 2.4.

Letϕ:Mm,gPp,k be a harmonic map. The map ϕ˜:Mm×α,βNn,Gα,βPp,k defined by ϕ˜x,y=ϕxis biharmonic if and only if

Trgϕ2dϕgradlnα+TrgRPdϕgradlnα,dϕdϕ+nβ2gradlnαϕdϕgradlnα=0.

In particular ifϕ=IdM, the first projection P1:Mm×α,βNn,Gα,βMm,gdefined byP1x,y=xis biharmonic if and only if

gradΔlnα+n2β2gradgradlnα2+2Riccigradlnα=0.

We apply this Corollary to construct an example of biharmonic non-harmonic maps.

Example 2.5.

Let the first projection P1:m×α,βn,Gα,βm,gm be defined by

P1x=t,x2,...,xm,y=s,y2,...,yn=x=t,x2,...,xm.

We suppose that α depends only on t and β depends only on s and set lnα=α1t, lnβ=β1s. Then by Corollary 1, the first projection P1:m×α,βn,$$Gα,βm,gm is biharmonic if and only if

α1+nβ2α1α1=0.

As particular solutions of this equation, we have

  • (1) If the function α 1 is constant, which gives us αt=CexpktC>0. Then the equation α1+nβ2α1α1=0 can be satisfied for any positive function β. In this case, the first projection P1:m×α,βn,Gα,βm,gm is biharmonic non-harmonic where αt=Cexpkt and β is any positive function.

  • (2) For βs=q and αt=Cpt+k2pC>0 where p=nq2 , the first projection P1:m×α,βn,Gα,βm,gm is biharmonic non-harmonic.

If we replace α = 1 in Theorem 2.3, we get the following result:

Corollary 2.6.

Letϕ:Mm,gPp,kbe a smooth map. The mapϕ˜:Mm×βNn,Gα,βPp,kdefined byϕ˜x,y=ϕxis biharmonic if and only if

τ2ϕ+2β2Δlnβ+m2gradlnβ2τϕ=0.

In particular if the map ϕ is biharmonic non-harmonic, we get two cases:

  • (1) If m = 2, we deduce that the mapϕ˜:Mm×βNn,Gα,βPp,kdefined byϕ˜x,y=ϕxis biharmonic non-harmonic if and only if the function ln β is harmonic.

  • (2) If m2and by calculatingΔβm2, we deduce that the mapϕ˜:Mm×βNn,Gα,βPp,kdefined byϕ˜x,y=ϕxis biharmonic non-harmonic if and only if the functionβm2is harmonic.

Theorem 2.7.

Letψ:Nn,hPp,kbe a smooth map, we defineψ˜:(βMm×αNn,Gα,β)(Pp,k) by ψ˜(x,y)=ψ(y). The tension field and the bi-tension field ofψ˜are given by

τψ˜=1α2τψ+mdψgradlnβ

and

τ2ψ˜=τ2ψ1α2β24gradlnα22Δlnατψ  mα2Trhψ2dψgradlnβ+gradlnβψτψ  mgradlnβψdψgradlnβ+2nα2gradlnα2τψ  mα2TrhRdψgradlnβ,dψdψ.

Proof. Let the map ψ˜:Mm×α,βNn,Gα,βPp,k be defined by ψ˜x,y=ψy where ψ:Nn,hPp,k. Note that in this case we have dψ˜X,Y=dψY for any XΓTM and YΓTN. For the tension field of ψ˜, we have

τψ˜=TrGα,βdψ˜  =1β2ei,0ψ˜dψ˜ei,01β2dψ˜˜ei,0ei,0  +1α20,fjψ˜dψ˜0,fj1α2dψ˜˜0,fj0,fj  =mdψ˜0,gradlnβ+1α2fjψdψfj1α2dψ˜0,fjfj,

then

τψ˜=1α2τψ+mdψgradlnβ.

By definition, the bi-tension field of ψ is given by

τ2ψ˜=TrGα,βψ˜2τψ˜TrGα,βRτψ˜,dψ˜dψ˜.

For the first term TrGα,βψ˜2τψ˜, a rigourous calculation gives us

TrGα,βψ˜2τψ˜  =1α4Trhψ2τψ+1α2Trhψ2dψgradlnβ  +mα2gradlnβψτψ+m2gradlnβψdψgradlnβ  +1α2β24gradlnα22Δlnατψ2nα2gradlnα2τψ.

Finally, it is very easy to see that

TrGα,βRτψ˜,dψ˜dψ˜=1α4TrhRτψ,dψdψ          +mα2TrhRdψgradlnβ,dψdψ.

It follows that

τ2ψ˜=τ2ψ1α2β24gradlnα22Δlnατψ  mα2Trhψ2dψgradlnβ+gradlnβψτψ  mgradlnβψdψgradlnβ+2nα2gradlnα2τψ  mα2TrhRdψgradlnβ,dψdψ.

As a consequence, if ψ is harmonic, we have the following.

Corollary 2.8.

Letψ:Nn,hPp,k be a harmonic map. The map ψ˜:Mm×α,βNn,Gα,βPp,kdefined byψ˜x,y=ψyis biharmonic if and only if

Trhψ2dψgradlnβ+TrhRdψgradlnβ,dψdψ+mα2gradlnβψdψgradlnβ=0.

In particular ifψ=IdN, the second projectionP2:Mm×α,βNn,Gα,βNn,h defined by P2x,y=yis biharmonic if and only if

gradΔlnβ+m2α2gradgradlnβ2+2Riccigradlnβ=0.

If we replace β = 1 in Theorem 2.7, we get the following result.

Corollary 2.9.

Letψ:Nn,hPp,kbe a smooth map. The mapψ˜:Mm×αNn,GαPp,kdefined byψ˜x,y=ψyis biharmonic if and only if

τ2ψ+2α2Δlnα+n2gradlnα2τψ=0.

In particular if the map ψ is biharmonic non-harmonic, we distinguish two cases :

  • (1) If n=2, we deduce that the map ψ˜:Mm×αNn,GαPp,k defined by ψ˜x,y=ψx is biharmonic non-harmonic if and only if the function ln α is harmonic.

  • (2) If n ≠ 2 and by calculating Δαn2, we deduce that the map ψ˜:Mm×α%Nn,GαPp,k defined by ϕ˜x,y=ϕx is biharmonic non-harmonic if and only if the function αn2 is harmonic.

Now, let's study the biharmonicity of inclusion maps iy0:Mn,gMm×α,βNn,Gα,β and ix0:Nn,hMm×α,βNn,Gα,β. To study the biharmonicity of these maps, we will give the expression of the curvature tensor R of the doubly warped product Mm×α,βNn,Gα,β. This expression is given by the following theorem.

Theorem 2.10.

LetMm,g and Nn,hbe two Riemannian manifolds. If˜denote the Levi-Civita connection onMm×α,βNn,Gα,β and R˜is the curvature tensor associated to˜, then for allX1,Y1,Z1ΓTM and X2,Y2,Z2ΓTN, we have

R˜X1,0,Y1,0Z1,0=RX1,Y1Z1,0+β2gradlnβ2gX1,Z1Y1,0gY1,Z1X1,0+β2gX1,Z1Y1lnαgY1,Z1X1lnα0,gradlnβ,R˜X1,0,0,Y20,Z2=α2hY2,Z2X1gradlnα,0          α2hY2,Z2X1lnαgradlnα,0          Y2Z2lnβY2Z2lnβX1,0          Z2lnβY2lnβX1,0Z2lnβX1lnα0,Y2          +α2β2hY2,Z2X1lnα0,gradlnβ,R˜0,X2,Y1,0Z1,0=β2gY1,Z10,X2gradlnβ+α2β2gY1,Z1X2lnβgradlnα,0Z1lnαX2lnβY1,0Y1Z1lnαY1Z1lnα+Z1lnαY1lnα0,X2β2gY1,Z1X2lnβ0,gradlnβ

and

R˜0,X2,0,Y20,Z2=0,RX2,Y2Z2+α2hX2,Z2Y2lnβhY2,Z2X2lnβgradlnα,0α2hY2,Z2gradlnα20,X2+α2hX2,Z2gradlnα20,Y2.

Proof. For the term R˜X1,0,Y1,0Z1,0, we have

R˜X1,0,Y1,0Z1,0=˜X1,0˜Y1,0Z1,0˜Y1,0˜X1,0Z1,0          ˜X1,0,Y1,0Z1,0.

By Proposition 2.1, we can obtain

˜Y1,0Z1,0=Y1Z1,0β2gY1,Z10,gradlnβ,

then

˜X1,0˜Y1,0Z1,0=˜X1,0Y1Z1,0        ˜X1,0β2gY1,Z10,gradlnβ        =X1Y1Z1,0β2gX1,Y1Z10,gradlnβ        β2gY1,Z1˜X1,00,gradlnβ        β2X1gY1,Z10,gradlnβ,

which leads us to the following formula

˜X1,0˜Y1,0Z1,0=X1Y1Z1,0β2gX1,Y1Z10,gradlnβ        β2gY1,Z1X1lnα0,gradlnβ        β2gY1,Z1gradlnβ2X1,0        β2gX1Y1,Z10,gradlnβ        β2gY1,X1Z10,gradlnβ.

A similar calculation gives us

˜Y1,0˜X1,0Z1,0=Y1X1Z1,0β2gY1,X1Z10,gradlnβ        β2gX1,Z1Y1lnα0,gradlnβ        β2gX1,Z1gradlnβ2Y1,0        β2gY1X1,Z10,gradlnβ        β2gX1,Y1Z10,gradlnβ.

and

˜X1,0,Y1,0Z1,0=˜X1,Y1,0Z1,0      =X1,Y1Z1,0β2gX1,Y1,Z10,gradlnβ.

It follows that

R˜X1,0,Y1,0Z1,0=RX1,Y1Z1,0+β2gradlnβ2gX1,Z1Y1,0gY1,Z1X1,0+β2gX1,Z1Y1lnαgY1,Z1X1lnα0,gradlnβ.

Now let's look at the term R˜X1,0,0,Y20,Z2, we have

R˜X1,0,0,Y20,Z2=˜X1,0˜0,Y20,Z2˜0,Y2˜X1,00,Z2.

By Proposition 2.1, we obtain

˜0,Y20,Z2=0,Y2Z2α2hY2,Z2gradlnα,0,

then

˜X1,0˜0,Y20,Z2=˜X1,00,Y2Z2      ˜X1,0α2hY2,Z2gradlnα,0      =X1lnα0,Y2Z2+Y2Z2lnβX1,0      α2hY2,Z2˜X1,0gradlnα,0X1α2hY2,Z2gradlnα,0,

we deduce that

˜X1,0˜0,Y20,Z2=α2hY2,Z2X1gradlnα,0      2α2hY2,Z2X1lnαgradlnα,0      +α2β2hY2,Z2X1lnα0,gradlnβ      +X1lnα0,Y2Z2+Y2Z2lnβX1,0.

The same calculation steps gives us

˜0,Y2˜X1,00,Z2=X1lnα0,Y2Z2+Z2lnβX1lnα0,Y2      +Z2lnβY2lnβX1,0+Y2Z2lnβX1,0      α2hX2,Y2X1lnαgradlnα,0,

it follows that

R˜X1,0,0,Y20,Z2=α2hY2,Z2X1gradlnα,0          α2hY2,Z2X1lnαgradlnα,0          Y2Z2lnβY2Z2lnβX1,0          +α2β2hY2,Z2X1lnα0,gradlnβ          Z2lnβY2lnβX1,0Z2lnβX1lnα0,Y2.

For the term R˜0,X2,Y1,0Z1,0, we have

R˜0,X2,Y1,0Z1,0=˜0,X2˜Y1,0Z1,0˜Y1,0˜0,X2Z1,0.

Using Proposition 2.1, we get

˜Y1,0Z1,0=Y1Z1,0β2gY1,Z10,gradlnβ,

then

˜0,X2˜Y1,0Z1,0=˜0,X2Y1Z1,0˜0,X2β2gY1,Z10,gradlnβ      =Y1Z1lnα0,X2+X2lnβY1Z1,0      β2gY1,Z10,X2gradlnβ      +α2β2gY1,Z1X2lnβgradlnα,0      2β2X2lnβgY1,Z10,gradlnβ.

Similarly, we obtain

˜Y1,0˜0,X2Z1,0=˜Y1,0Z1lnα0,X2+˜Y1,0X2lnβZ1,0      =Z1lnαY1lnα0,X2      +Z1lnαX2lnβY1,0      +Y1Z1lnα0,X2+X2lnβY1Z1,0      β2gY1,Z1X2lnβ0,gradlnβ,

which gives us

R˜0,X2,Y1,0Z1,0=β2gY1,Z10,X2gradlnβ          +α2β2gY1,Z1X2lnβgradlnα,0          β2gY1,Z1X2lnβ0,gradlnβ          Z1lnαX2lnβY1,0Z1lnαY1lnα0,X2          Y1Z1lnαY1Z1lnα+0,X2.

To complete the proof of Theorem 2.10, we will calculate the term R˜(0,X2,0,Y2)0,Z2, we have

R˜0,X2,0,Y20,Z2=˜0,X2˜0,Y20,Z2˜0,Y2˜0,X20,Z2          ˜0,X2,0,Y20,Z2.

Using Proposition 2.1, we get

˜0,Y20,Z2=0,Y2Z2α2hY2,Z2gradlnα,0,

then

˜0,X2˜0,Y20,Z2=˜0,X20,Y2Z2      α2˜0,X2hY2,Z2gradlnα,0      =0,X2Y2Z2α2hX2,Y2Z2gradlnα,0      α2hY2,Z2gradlnα20,X2      α2hY2,Z2X2lnβgradlnα,0      α2hX2Y2,Z2gradlnα,0      α2hY2,X2Z2gradlnα,0

A similar calculation gives us

˜0,Y2˜0,X20,Z2=˜0,Y20,X2Z2      α2˜0,Y2hX2,Z2gradlnα,0      =0,Y2X2Z2α2hY2,X2Z2gradlnα,0      α2hX2,Z2gradlnα20,Y2      α2hX2,Z2Y2lnβgradlnα,0      α2hY2X2,Z2gradlnα,0      α2hX2,Y2Z2gradlnα,0

and

˜0,X2,0,Y20,Z2=˜0,X2,Y20,Z2      =0,X2,Y2Z2α2hX2,Y2,Z2gradlnα,0,

we deduce that

R˜0,X2,0,Y20,Z2=0,RX2,Y2Z2+α2hX2,Z2Y2lnβhY2,Z2X2lnβgradlnα,0α2hY2,Z2gradlnα20,X2+α2hX2,Z2gradlnα20,Y2.

For the first case, we have the following result.

Theorem 2.11.

The inclusion mapiy0:Mn,gMm×α,βNn,Gα,βdefined byiy0x=x,y0is biharmonic if and only if

4gradlnβ2gradlnβ+gradgradlnβ2=0

and

gradlnβ2gradlnα=0.

Proof. Note that in this case we have diy0X=X,0 for anyXΓTM and YΓTN. By definition to the tension field, we have

τiy0=Trg˜diy0  =˜eidiy0eidiy0eiei  =˜ei,0ei,0eiei,0  =eiei,0mβ20,gradlnβeiei,0,

then

τiy0=mβ20,gradlnβ.

The inclusion map iy0 is biharmonic if and only if

Trg˜iy020,gradlnβ+TrgR˜0,gradlnβ,diy0diy0=0,0.

For the term Trg ˜iy020,gradlnβ, we have

Trg˜iy020,gradlnβ=˜eiiy0˜eiiy0gradlnα,0˜eieiiy0gradlnα,0  =˜ei,0˜ei,00,gradlnβ˜eiei,0iy00,gradlnβ  =˜ei,0eilnα0,gradlnβ+gradlnβ2˜ei,0ei,0  eieilnα0,gradlnβgradlnβ2eiei,0  =eilnαeilnα0,gradlnβ+gradlnβ2ei,0  +eieilnα0,gradlnβeieilnα0,gradlnβ  +gradlnβ2eiei,0mβ20,gradlnβ  gradlnβ2eiei,0

then

Trg˜iy020,gradlnβ=gradlnα20,gradlnβ+gradlnβ2gradlnα,0          +Δlnα0,gradlnβmβ2gradlnβ20,gradlnβ.

For the second term TrgR˜0,gradlnβ,diy0diy0=0,0, using Theorem 2.10, we can obtain

TrgR˜0,gradlnβ,diy0diy0=TrgR˜0,gradlnβ,diy0eidiy0ei=R˜0,gradlnβ,ei,0ei,0=β2gei,ei0,gradlnβgradlnβ+α2β2gei,eigradlnβ2gradlnα,0eilnαgradlnβ2ei,0β2gei,eigradlnβ20,gradlnβeieilnαeieilnα+eilnαeilnα0,gradlnβ,

it follows that

TrgR˜0,gradlnβ,diy0diy0=m2β20,gradgradlnβ2            +mα2β2gradlnβ2gradlnα,0            gradlnβ2gradlnα,0            mβ2gradlnβ20,gradlnβ            Δlnα+gradlnα20,gradlnβ.

We conclude that the inclusion map iy0 is biharmonic if and only if

4gradlnβ2gradlnβ+gradgradlnβ2=0

and

gradlnβ2gradlnα=0.

Corollary 2.12.

For the inclusion mapiy0:Mn,gMm×α,βNn,Gα,βdefined byiy0x=x,y0, we have the following cases:

  • (1) If the function β is constant, then the inclusion map ix0 is biharmonic for any positive function α.

  • (2) If the function α is constant, then the inclusion map ix0 is biharmonic if and only if 4gradlnβ2gradlnβ+gradgradlnβ2=0.

  • (3) If the functions α and β are not constants, the inclusion map iy0 is never biharmonic.

Theorem 2.13.

The inclusion mapix0:Nn,hMm×α,βNn,Gα,βdefined byix0y=x0,yis biharmonic if and only if

4gradlnα2gradlnα+gradgradlnα2=0.

and

gradlnα2gradlnβ=0.

Proof. Note that in this case we have dix0Y=0,Y for any YΓTN. By definition of the tension field, we have

τix0=Trh˜dix0  =˜fjix0dix0fjdix0fjfj  =˜fjix0dix0fjdix0fjfj  =˜0,fj0,fj0,fjfj  =0,fjfjnα2gradlnα,00,fjfj

then

τix0=nα2gradlnα,0.

The inclusion map ix0 is biharmonic if and only if

Trh˜ix02gradlnα,0+TrhR˜gradlnα,0,dix0dix0=0,0.

For the first term of this equation, we have

Trh˜ix02gradlnα,0    =˜fjix0˜fjix0gradlnα,0˜fjfjix0gradlnα,0    =˜0,fj˜0,fjgradlnα,0˜0,fjfjgradlnα,0    =gradlnα2˜0,fj0,fj+˜0,fjfjlnβgradlnα,0    gradlnα20,fjfjfjfjlnβgradlnα,0,

then

Trh˜ix02gradlnα,0    =gradlnα20,fjfjnα2gradlnα2gradlnα,0    +gradlnα20,gradlnβ+gradlnβ2gradlnα,0    +fjfjlnβgradlnα,0gradlnα20,fjfj    fjfjlnβgradlnα,0,

it follows that

Trh˜ix02gradlnα,0=nα2gradlnα2gradlnα,0        +Δlnβ+gradlnβ2gradlnα,0        +gradlnα20,gradlnβ

For the term TrhR˜gradlnα,0,dix0dix0, by Theorem 2.10, we have

TrhR˜gradlnα,0,dix0dix0=R˜gradlnα,0,dix0fjdix0fj          =R˜gradlnα,0,0,fj0,fj          =α2hfj,fjgradlnαgradlnα,0          α2hfj,fjgradlnα2gradlnα,0          fjfjlnβfjfjlnβgradlnα,0          fjlnβfjlnβgradlnα,0          fjlnβgradlnα20,fj          +α2β2hfj,fjgradlnα20,gradlnβ,

which gives us

TrhR˜gradlnα,0,dix0dix0=n2α2gradgradlnα2,0          nα2gradlnα2gradlnα,0          Δlnβ+gradlnβ2gradlnα,0          +nα2β2gradlnα20,gradlnβ          gradlnα20,gradlnβ.

Finally, we deduce that the inclusion map ix0 is biharmonic if and only if

4gradlnα2gradlnα+gradgradlnα2=0

and

gradlnα2gradlnβ=0.

Corollary 2.14.

For the inclusion mapix0:Nn,hMm×α,βNn,Gα,βdefined byix0y=x0,y, we have the following cases:

  • (1) If the function α is constant, then the inclusion map ix0} is biharmonic for any positive function β.

  • (2) If the function β is constant, then the inclusion map i{x0} is biharmonic if and only if 4gradlnα2gradlnα+gradgradlnα2=0.

  • (3) If the functions α and β are not constants, the inclusion map ix0} is never biharmonic.

Theorem 2.15.

Letϕ:Mm×α,βNn,Gα,βMm×Nn,G=ghbe the map defined byϕx,y=x,y. The map ϕ is biharmonic if and only if

gradΔlnα+n2β2gradgradlnα2+2Riccigradlnα=0

and

gradΔlnβ+m2α2gradgradlnβ2+2Riccigradlnβ=0.

Proof. Note that in this case we have dϕX,Y=X,Y for any XΓTM and YΓTN. By definition of the tension field and using the Proposition 2.1, we have

τϕ=TrGα,βdϕ  =1β2ei,0ϕdϕei,0dϕ˜ei,0ei,0  +1α20,fjϕdϕ0,fjdϕ˜0,fj0,fj  =1β2eiei,0eiei,0+mβ20,gradlnβ  +1α20,fjfj0,fjfj+nα2gradlnα,0,

then

τϕ=ngradlnα,0+m0,gradlnβ.

The map ϕ is biharmonic if and only if

nTrGα,βϕ2gradlnα,0+TrGα,βRgradlnα,0,dϕdϕ+mTrGα,βϕ20,gradlnβ+TrGα,βR0,gradlnβ,dϕdϕ=0.

We can analyse term by term this expression, for the term TrGα,βϕ2gradlnα,0, we have

TrGα,βϕ2gradlnα,0=1β2ei,0ϕei,0ϕgradlnα,0          1β2˜ei,0ei,0ϕgradlnα,0          +1α20,fjϕ0,fjϕgradlnα,0          1α2˜0,fj0,fjϕgradlnα,0          =1β2eieigradlnα,01β2eieigradlnα,0          +ngradlnα,0ϕgradlnα,0          =1β2Trg2gradlnα,0+ngradlnαgradlnα,0,

then

TrGα,βϕ2gradlnα,0=1β2gradΔlnα,0+Riccigradlnα,0          +n2gradgradlnα2,0.

A similar calculation leads to

TrGα,βϕ20,gradlnβ=0,gradΔlnβ+20,Riccigradlnβ          +m2α20,gradgradlnβ2.

Finally, it is very simple to write that

TrGα,βRgradlnα,0,dϕdϕ=1β2Riccigradlnα,0

and

TrGα,βR0,gradlnβ,dϕdϕ=1α20,Riccigradlnβ.

If we replace (2.9), (2.10), (2.11) and (2.12) in (2.8), we deduce that the map ϕ is biharmonic if and only if

gradΔlnα+n2β2gradgradlnα2+2Riccigradlnα=0

and

gradΔlnβ+m2α2gradgradlnβ2+2Riccigradlnβ=0.

Example 2.16.

Let the map ϕ:m×α,βn,Gα,βm×n,G defined by

ϕx=t,x2,...,xm,y=s,y2,...,yn=x=t,x2,...,xm,y=s,y2,...,yn.

We suppose that α depends only on t and β depends only on s and set lnα=α1t,lnβ=β1s. Then from Theorem 2.11, the map ϕ:m×α,βn,Gα,βm×α,βn,G is biharmonic if and only if

α1+nβ2α1α1=0

and

β1+mα2β1β1=0.

Particular solutions of this system are given by αt=C1expC2t and Bs=K1expK2s, where C1 and K1 are positive constants. For this functions αt=C1expC2t and Bs=K1expK2s, the map ϕ:m×α,βn,Gα,βm×n,G is biharmonic nonharmonic.

Finally, we consider the map ϕ:Mm×Nn,G=ghMm×α,βNn,Gα,β defined by ϕx,y=x,y. To study the biharmonicity of this map, we use two lemmas. In the first lemma, we calculate the term TrG˜2gradlnα,0.

Lemma 2.17.

LetMm,g and Nn,htwo Riemannian manifolds and letαCM and βCNbe a positive functions. Let˜be the Levi-Civita connection of the doubly warped productMm×α,βNn, we have

TrG˜2gradlnα,0=gradΔlnα,0nα2gradlnα2gradlnα,0        β2gradlnβ2gradlnα,0+Δlnβgradlnα,0        +gradlnβ2gradlnα,0+gradlnα20,gradlnβ        β2gradlnα2+2Δlnα0,gradlnβ        +Riccigradlnα,0.

Proof. We have

TrG˜2gradlnα,0=˜ei,0˜ei,0gradlnα,0        ˜ei,0ei,0gradlnα,0        +˜0,fj˜0,fjgradlnα,0        ˜0,fj0,fjgradlnα,0.

A simple calculation gives us

˜ei,0˜ei,0gradlnα,0˜ei,0ei,0gradlnα,0    =eieigradlnα,0β2gei,eigradlnα0,gradlnβ    β2gei,gradlnαeilnα0,gradlnβ    β2gei,gradlnαgradlnβ2ei,0    β2geiei,gradlnα0,gradlnβ    β2gei,eigradlnα0,gradlnβ    eieigradlnα,0+β2geiei,gradlnα0,gradlnβ    =Trg2gradlnα,02β2Δlnα0,gradlnβ    β2gradlnα20,gradlnββ2gradlnβ2gradlnα,0,

then

˜ei,0˜ei,0gradlnα,0˜ei,0ei,0gradlnα,0        =gradΔlnα,0β2gradlnβ2gradlnα,0        β2gradlnα2+2Δlnα0,gradlnβ        +Riccigradlnα,0.

For the term ˜0,fj˜0,fjgradlnα,0˜0,fj0,fjgradlnα,0, a similar calculation gives us

˜0,fj˜0,fjgradlnα,0˜0,fj0,fjgradlnα,0    =gradlnα2˜0,fj0,fj+˜0,fjfjlnβgradlnα,0    ˜0,fjfjgradlnα,0    =gradlnα20,fjfjnα2gradlnα2gradlnα,0    +fjlnβgradlnα20,fj+fjlnβfjlnβgradlnα,0    +fjfjlnβgradlnα,0gradlnα20,fjfj    fjfjlnβgradlnα,0,

it follows that

˜0,fj˜0,fjgradlnα,0˜0,fj0,fjgradlnα,0          =nα2gradlnα2gradlnα,0          +gradlnβ2gradlnα,0          +Δlnβgradlnα,0          +gradlnα20,gradlnβ.

If we replace ((2.14)) and ((2.15)) in ((2.13)) , we deduce that

TrG˜2gradlnα,0=gradΔlnα,0nα2gradlnα2gradlnα,0        β2gradlnβ2gradlnα,0+Δlnβgradlnα,0        +gradlnβ2gradlnα,0+gradlnα20,gradlnβ        β2gradlnα2+2Δlnα0,gradlnβ        +Riccigradlnα,0.

For the term TrG˜20,gradlnβ, we can use the following lemma.

Lemma 2.18.

LetMm,g and Nn,htwo Riemannian manifolds and letαCM and βCNbe a positive functions. Let˜be the Levi-Civita connection of the doubly warped productMm×α,βNn, we have

TrG˜20,gradlnβ=gradlnα20,gradlnβ+gradlnβ2gradlnα,0  +Δlnα0,gradlnβmβ2gradlnβ20,gradlnβ  +0,gradΔlnβα22Δlnβ+gradlnβ2gradlnα,0  α2gradlnα20,gradlnβ+0,Riccigradlnβ.

Proof.

TrG˜20,gradlnβ=˜ei,0˜ei,00,gradlnβ˜ei,0ei,00,gradlnβ        +˜0,fj˜0,fj0,gradlnβ˜0,fj0,fj0,gradlnβ.

The same calculation method gives us

˜ei,0˜ei,00,gradlnβ˜ei,0ei,00,gradlnβ    =˜ei,0eilnα0,gradlnβ+gradlnβ2˜ei,0ei,0    ˜eiei,00,gradlnβ=eilnαeilnα0,gradlnβ+eilnαgradlnβ2ei,0    +eieilnα0,gradlnβmβ2gradlnβ20,gradlnβ    eieilnα0,gradlnβ    =gradlnα20,gradlnβ+gradlnβ2gradlnα,0    +Δlnα0,gradlnβmβ2gradlnβ20,gradlnβ

and

˜0,fj˜0,fj0,gradlnβ˜0,fj0,fj0,gradlnβ=0,fjfjgradlnβα2hfj,fjgradlnβgradlnα,0α2hfj,gradlnβgradlnα20,fjα2fjlnβfj,gradlnβgradlnα,0α2hfjfj,gradlnβgradlnα,0α2hfj,fjgradlnβgradlnα,00,fjfjgradlnβ+α2hfjfj,gradlnβgradlnα,0=0,Trh2gradlnβ2α2Δlnβgradlnα,0α2gradlnα20,gradlnβα2gradlnβ2gradlnα,0=2α2Δlnβgradlnα,0α2gradlnβ2gradlnα,0α2gradlnα20,gradlnβ+0,gradΔlnβ+0,Riccigradlnβ.

It follows that

TrG˜20,gradlnβ=gradlnα20,gradlnβ+gradlnβ2gradlnα,0        +Δlnα0,gradlnβmβ2gradlnβ20,gradlnβ        +0,gradΔlnβ2α2Δlnβgradlnα,0        α2gradlnα20,gradlnβα2gradlnβ2gradlnα,0        +0,Riccigradlnβ.

Theorem 2.19.

Letϕ:Mm×Nn,G=ghMm×α,βNn,Gα,βbe the map defined byϕx,y=x,y. The map ϕ is biharmonic if and only if

ngradΔlnα+2ngradgradlnα2n22α2gradgradlnα22n2α2gradlnα2gradlnα+m2β4gradlnβ2gradlnα+4ngradlnα2gradlnα+2nΔlnαgradlnαmβ22n+4gradlnβ2+2Δlnβgradlnα+2nRiccigradlnα=0

and

mgradΔlnβ+2mgradgradlnβ2m22β2gradgradlnβ22m2β2gradlnβ2gradlnβ+n2α4gradlnα2gradlnβ+4mgradlnβ2gradlnβ+2mΔlnβgradlnβnα22m+4gradlnα2+2Δlnαgradlnβ+2mRiccigradlnβ=0.

Proof. By definition of the tension field and using the Proposition 2.1, we have

τϕ=TrGdϕ  =˜ei,0ϕdϕei,0dϕei,0ei,0   +˜0,fjϕdϕ0,fjdϕ0,fj0,fj   =˜ei,0ϕei,0ei,0ei,0   +˜0,fjϕ0,fj0,fj0,fj   =eiei,0mβ20,gradlnβeiei,0   +0,fjfjnα2gradlnα,00,fjfj,

then

τϕ=mβ20,gradlnβnα2gradlnα,0.

The map ϕ is biharmonic if and only if

nTrG˜ϕ2α2gradlnα,0+α2TrGR˜gradlnα,0,dϕdϕ+mTrG˜ϕ2β20,gradlnβ+β2TrGR˜0,gradlnβ,dϕdϕ=0,0.

We will study term by term the above equation, for the first term TrG˜ϕ2α2gradlnα,0, it is clear to see that

TrG˜ϕ2α2gradlnα,0=α2TrG˜ϕ2gradlnα,0+2α2gradgradlnα2,0        +2α2Δlnαgradlnα,0+4α2gradlnα2gradlnα,0        4α2β2gradlnα20,gradlnβ

and using Lemma 2.17, we deduce that

TrG˜ϕ2α2gradlnα,0=α2gradΔlnα,0+2α2gradgradlnα2,0    nα4gradlnα2gradlnα,0α2β2gradlnβ2gradlnα,0    +α2gradlnβ2gradlnα,0+4α2gradlnα2gradlnα,0    +2α2Δlnαgradlnα,0+α2Δlnβgradlnα,0    α2β25gradlnα2+2Δlnα0,gradlnβ    +α2gradlnα20,gradlnβ+α2Riccigradlnα,0.

By Theorem 2.10, we obtain

TrGR˜gradlnα,0,dϕdϕ=R˜gradlnα,0,ei,0ei,0+R˜gradlnα,0,0,fj0,fj=n2α2gradgradlnα2,0nα2gradlnα2gradlnα,0Δlnβgradlnα,0gradlnβ2gradlnα,0+1mβ2gradlnβ2gradlnα,0+1mβ2gradlnα20,gradlnβ+nα2β2gradlnα20,gradlnβgradlnα20,gradlnβ+Riccigradlnα,0,

it follows that

TrG˜ϕ2α2gradlnα,0+α2TrGR˜gradlnα,0,dϕdϕ  =α2gradΔlnα,0+2α2gradgradlnα2,0  n2α4gradgradlnα2,02nα4gradlnα2gradlnα,0  mα2β2gradlnβ2gradlnα,0+4α2gradlnα2gradlnα,0  +2α2Δlnαgradlnα,0+nα4β2gradlnα20,gradlnβ  α2β2m+4gradlnα2+2Δlnα0,gradlnβ  +2α2Riccigradlnα,0.

A similar calculation gives us

TrG˜ϕ2β20,gradlnβ=β20,gradΔlnβ+2β20,gradgradlnβ2  +4β2gradlnβ20,gradlnβ+β2Δlnα0,gradlnβ  mβ4gradlnβ20,gradlnβ+2β2Δlnβ0,gradlnβ  +β2gradlnα20,gradlnβα2β2gradlnα20,gradlnβ  4α2β2gradlnβ2gradlnα,02α2β2Δlnβgradlnα,0  α2β2gradlnβ2gradlnα,0+β2gradlnβ2gradlnα,0  +β20,Riccigradlnβ

and

TrGR˜0,gradlnβ,dϕdϕ  =m2β20,gradgradlnβ2mβ2gradlnβ20,gradlnβ  Δlnα0,gradlnβgradlnα20,gradlnβ  +1nα2β2gradlnα20,gradlnβgradlnβ2gradlnα,0  +mα2β2gradlnβ2gradlnα,0+1nα2β2gradlnβ2gradlnα,0  +β20,Riccigradlnβ.

Which gives us

TrG˜ϕ2β20,gradlnβ+β2TrGR˜0,gradlnβ,dϕdϕ  =β20,gradΔlnβ+2β20,gradgradlnβ2  m2β40,gradgradlnβ2+4β2gradlnβ20,gradlnβ  2mβ4 gradlnβ2 0,gradlnβ+2β2Δlnβ0, gradlnβ  nα2β2gradlnα20,gradlnβ+2β20,Riccigradlnβ  α2β2n+4gradlnβ2+2Δlnβgradlnα,0  +mα2β4gradlnβ2gradlnα,0.

Finally, we conclude that the map ϕ:Mm×Nn,G=gh(Mm×α,βNn,Gα,β) defined by ϕx,y=x,y is biharmonic if and only if

ngradΔlnα+2ngradgradlnα2n22α2gradgradlnα22n2α2gradlnα2gradlnα+m2β4gradlnβ2gradlnα+4ngradlnα2gradlnα+2nΔlnαgradlnαmβ22n+4gradlnβ2+2Δlnβgradlnα+2nRiccigradlnα=0

and

mgradΔlnβ+2mgradgradlnβ2m22β2gradgradlnβ22m2β2gradlnβ2gradlnβ+n2α4gradlnα2gradlnβ+4mgradlnβ2gradlnβ+2mΔlnβgradlnβnα22m+4gradlnα2+2Δlnαgradlnβ+2mRiccigradlnβ=0.

As a consequence of Theorem 2.19, we can summarize the results as follows.

Corollary 2.20.

The mapϕ:Mm×Nn,G=ghMm×αNn,Gαdefined byϕx,y=x,yis biharmonic if and only if the positive functionαCMsatisfies the following equation

gradΔlnα+2n2α2gradgradlnα2+2Δlnαgradlnα+42nα2gradlnα2gradlnα+2Riccigradlnα=0.

Corollary 2.21.

The mapϕ:Mm×Nn,G=ghMm×βNn,Gβdefined byϕx,y=x,yis biharmonic if and only if the positive functionβCNsatisfies the following equation

gradΔlnβ+2m2β2gradgradlnβ2+2Δlnβgradlnβ+42mβ2gradlnβ2gradlnβ+2Riccigradlnβ=0.
The authors would like to thank the referee for some useful comments and their helpful suggestions that have improved the quality of this paper.
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