SOME RESULTS of F-BIHARMONIC MAPS 1. Introduction Harmonic Maps Play a Central Roll in Variational Problem

SOME RESULTS OF F -BIHARMONIC MAPS

YINGBO HAN and SHUXIANG FENG

Abstract. In this paper, we give the notion of F -biharmonic maps, which is a generalization of biharmonic maps. We derive the ﬁrst variation formula which yields F -biharmonic maps. Then we investigate the harmonicity of F -biharmonic maps under the curvature conditions on the target manifold (N, h). We also introduce the stress F -bienergy tensor SF,2. Then, by using the stress F -bienergy tensor SF,2, we obtain some nonexistence results of proper F -biharmonic maps under the assumption that SF,2 = 0. Moreover, we derive some monotonicity formulas for the special case of the biharmonic map, i.e., where F -biharmonic map with F (t) = t. Then, by using these monotonicity formulas, we obtain new results on the non existence of proper biharmonic isometric immersions from complete manifolds.

1. Introduction

Harmonic maps play a central roll in variational problems for smooth maps between manifolds 1 R 2 u:(M, g) → (N, h) as the critical points of the energy functional E(u) = 2 M kduk dvg. On the other hand, in 1981, J. Eells and L. Lemaire [7] proposed the problem to consider the k-harmonic JJ J I II maps which are critical maps of the functional Z k 2 Go back k(d + δ) uk Ek(u) = dvg M 2 Full Screen

Close Received January 10, 2013; revised July 15, 2013. 2010 Mathematics Subject Classification. Primary 58E20, 53C21. Key words and phrases. F -biharmonic map; stress F -bienergy tensor; harmonic map. Quit for smooth maps u : M → N. G. Y. Jiang [9] studied the first and second variation formulas of the bienergy E2 where critical maps of E2 are called biharmonic maps. There have been extensive studies on biharmonic maps (for instance, see [9, 13, 14, 15, 16, 18, 19]). Let F : [0, ∞) → [0, ∞) be a C3 function such that F 0 > 0 on (0, ∞). For a smooth map u:(M, g) → (N, h) between Riemannian manifolds (M, g) and (N, h), we define the F -k-energy EF,k(u) of u by Z k(d + δ)kuk2 EF,k(u) = F ( )dvg, M 2

which is Ek(u) if F (t) = t. When k = 1, we have

Z kduk2 EF,1(u) = F dvg = EF (u), M 2

which was introduced by M. Ara in [1]. The critical maps of EF (u) are called F -harmonic maps which are the generalization of harmonic maps, p-harmonic maps or exponentially harmonic maps. There have been extensive studies in this area (for instance, [4,5, 11, 12 ]). When k = 2, we have

Z kτ(u)k2 JJ J I II EF,2(u) = F dvg, M 2 Go back where τ(u) = −δdu = trace ∇e (du). It is the bienergy of G.Y. Jiang [9], the p-bienergy of P. p t Full Screen Hornung and R. Moser [6] or exponentially bienergy when F (t) = t, F (t) = (2t) 2 or F (t) = e . We say that u is an F -biharmonic map if Close d EF,2(ut)|t=0 = 0 Quit dt for any compactly supported variation ut : M → N with u0 = u. In this note, we derive the ﬁrst variation formula which yields F -biharmonic maps. Then we investigate the harmonicity of F - biharmonic maps under the curvature conditions on the target manifold (N, h). We also introduce the stress F -bienergy tensor SF,2. Then, by using the stress F -bienergy tensor SF,2, we obtain some non existence results of proper F -biharmonic maps under the assumption SF,2 = 0. Also, we derive some monotonicity formulas for the special case of a biharmonic map, i.e., an F -biharmonic map with F (t) = t. Then, by using these monotonicity formulas, we investigate the harmonicity of biharmonic isometric maps from complete manifolds.

Remark 1.1. In [17], the authors introduced f-biharmonic maps which are critical points of the bi-f-energy functional

Z 2 1 2 Ef (u) = kτf (u)k dvg, 2 M

∞ where τf (u) = fτ(u) + du(grad f) and f ∈ C (M). We think that it is more reasonable to call JJ J I II them “bi-f-harmonic maps” as parallel to “biharmonic maps”.

Go back 2. The first variation formula Full Screen Let ∇ and N ∇ always denote the Levi-Civita connections of M and N, respectively. Let ∇e be the Close −1 N induced connection on u TN defined by ∇e X W = ∇du(X)W , where X is a tangent vector of M −1 and W is a section of u TN. We choose a local orthonormal frame field {ei} on M. We define Quit the F -bitension field τF,2(u) of u by kτ(u)k2 τ (u) = −J(F 0 τ(u)) F,2 2 2 2 0 kτ(u)k X N 0 kτ(u)k = −4e (F )τ(u) − R (du(ei),F τ(u))du(ei), 2 2 i 1 R 2 where J is the Jacobi operator of the second variation for the energy E(u) = 2 M kduk dvg, 4 = − P (∇ ∇ − ∇ ) is the rough Laplacian on the section of u−1TN and RN (X,Y ) = e i e ei e ei e ∇ei ei N N N [ ∇X , ∇Y ]− ∇[X,Y ] is the curvature operator on N. Under the notation above we have the following theorem Theorem 2.1 (The first variation formula). Let u: M → N be a smooth map. Then d Z (1) EF,2(ut) |t=0 = h(τF,2(u),V )dvg, dt M d where V = dt ut|t=0.

Proof. Let Ψ: (−ε, ε) × M → N be deﬁned by Ψ(t, x) = ut(x), where (−ε, ε) × M is equipped JJ J I II ∂ with the product metric. We extend the vector ﬁelds ∂t on (−ε, ε), X on M naturally on (−ε, ε) × ∂ Go back M, and denote those also by ∂t , X. Then ∂ d Full Screen dΨ = u | = V. ∂t dt t t=0

Close We shall use the same notations ∇ and ∇e for the Levi-Civita connection on (−ε, ε) × M and the induced connection on Ψ−1TN, respectively. Quit We compute ∂ kτ(u )k2 F t ∂t 2 kτ(u )k2 1 ∂ = F 0 t kτ(u )k2 2 2 ∂t t 2 0 kτ(ut)k = F h ∇e ∂ τ(ut), τ(ut) 2 ∂t 2 X 0 kτ(ut)k = F h ∇e ∂ [(∇e ei dΨ)(ei)], τ(ut) (2) 2 ∂t i 2 X 0 kτ(ut)k = h ∇e ∂ ∇e ei dΨ(ei) − ∇e ∂ dΨ(∇ei ei ,F )τ(ut) ∂t ∂t 2 i 2 X ∂ kτutk = h(RN dΨ , dΨ(e ))dΨ(e ),F 0 τ(u ) ∂t i i 2 t i 2 X ∂ ∂ 0 kτ(ut)k + h ∇e ei ∇e ei dΨ − ∇e ∇e ei dΨ( ,F )τ(ut) , ∂t i ∂t 2 JJ J I II i where we use Go back ∂ ∂ ∇e ∂ dΨ(ei) − ∇e ei dΨ = dΨ , ei = 0 ∂t ∂t ∂t Full Screen and ∂ ∂ Close ∇e ∂ dΨ(∇ei ei) − ∇e ∇e ei dΨ = dΨ , ∇ei ei = 0 ∂t i ∂t ∂t for the fifth equality. Quit ∂ Let Xt and Yt be two compactly supported vector fields on M such that g(Xt,Z)=h(∇e Z dΨ ∂t , 2 2 0 kτutk ∂ 0 kτ(ut)k F 2 )τ(ut) and g(Yt,Z) = h dΨ ∂t , ∇e Z F 2 τ(ut) for any vector field Z on M. Then the divergence of Xt and Yt are given by the following: X X X div(Xt) = g(∇ek Xt, ek) = ekg(Xt, ek) − g(Xt, ∇ek ek) k k k 2 X ∂ 0 kτ(ut)k = ekh ∇e e dΨ ,F τ(ut) k ∂t 2 k 2 X ∂ 0 kτ(ut)k − h ∇e ∇e e dΨ ,F ( )τ(ut) k k ∂t 2 (3) k X ∂ = h ∇e e ∇e e dΨ k k ∂t k 2 ∂ 0 kτ(ut)k − ∇e ∇e e dΨ ,F τ(ut) k k ∂t 2 2 X ∂ 0 kτ(ut)k + h ∇e e dΨ , ∇e e F τ(ut) k ∂t k 2 JJ J I II k

Go back

Full Screen

Quit and X X X div(Yt) = g(∇ek Yt, ek) = ekg(Yt, ek) − g(Yt, ∇ek ek) k k k 2 X ∂ 0 kτ(ut)k = ekh dΨ( , ∇e e F )τ(ut) ∂t k 2 k 2 X ∂ 0 kτ(ut)k − h(dΨ( ), ∇e ∇e e (F )τ(ut)) ∂t k k 2 k (4) 2 X ∂ 0 kτutk = h dΨ , ∇e e ∇e e F τ(ut) ∂t k k 2 k 2 0 kτ(ut)k − ∇e ∇e e F τ(ut) k k 2 2 X ∂ 0 kτ(ut)k + h ∇e e dΨ , ∇e e F τ(ut) . k ∂t k 2 k From (2), (3) and (4), we have

JJ J I II 2 2 ∂ kτ(ut)k X N ∂ 0 kτ(ut)k F = h R dΨ , dΨ(ei) dΨ(ei),F τ(ut) Go back ∂t 2 ∂t 2 i ∂ kτ(u )k2 Full Screen X 0 t (5) + h dΨ , ∇e ei ∇e ei F τ(ut) ∂t 2 i Close kτ(u )k2 − ∇ F 0 t τ(u ) + div(X ) − div(Y ). e ∇ei ei t t t Quit 2 By (5) and Green’s theorem, we have

d E (u )| dt F,2 t t=0 Z 2 ∂ kτ(ut)k = F dvg M ∂t 2 t=0 Z kτ(u)k2 = h −4e F 0 τ(u) M 2 ! X kτ(u)k2 − RN du(e ), F 0 τ(u) du(e ),V dv i 2 i g i Z = h(τF,2(u),V )dvg. M

This proves Theorem 2.1.

The first variation formula allows us to define the notion of an F -biharmonic map for the functional EF,2(u). JJ J I II Definition 2.2. A smooth map u is called an F -biharmonic map for the functional EF,2(u) if Go back it is a solution of the Euler-Lagrange equation τF,2(u) = 0.

Full Screen Remark 2.3. By Deﬁnition 2.2, we know that any harmonic map is an F -biharmonic map.

Close Proposition 2.4. Let u: M → N be a smooth map. If kτ(u)k2 is constant, then u is F - biharmonic if and only if it is biharmonic. Quit Proof. Since kτ(u)k2 is constant, we have

2 " # 0 kτ(u)k X N τF,2(u) = F −4e (τ(u)) − R (du(ei), τ(u))du(ei) 2 i kτ(u)k2 = F 0 τ (u), 2 2

so we know that u is F -biharmonic if and only if it is biharmonic. Remark 2.5. When kτ(u)k2 is non-constant, we have

2 " # 0 kτ(u)k X N τF,2(u) = F −4e (τ(u)) − R (du(ei), τ(u))du(ei) 2 i 2 0 kτ(u)k 2 − 4e F τ(u) + ∇e 0 kτ(u)k τ(u) 2 grad F 2 2 2 0 kτ(u)k 0 kτ(u)k 2 = F τ2(u) − 4e F τ(u) + ∇e 0 kτ(u)k τ(u). 2 2 grad F 2 JJ J I II From this equation, we know that there are many differences between F -biharmonic maps and p t biharmonic maps when F (t) = (2t) 2 ,(p > 2) or F (t) = e . Go back 3. Non-existence results for F -biharmonic maps Full Screen From the definition of an F -biharmonic map, we know that a harmonic map is F -biharmonic map, Close so a basic question in theory is to understand under what conditions the converse is true. A first general answer to this problem for F (t) = t, proved by G. Y. Jiang [9], is the following theorem Quit Theorem 3.1 ([9]). Let u:(M, g) → (N, h) be a smooth map. If M is compact, orientable and the sectional curvature of (N, h) is non-positive, i.e., RiemN ≤ 0, then u is a biharmonic map if and only if it is harmonic.

In this section, we will obtain the following results

Theorem 3.2. Let u:(M, g) → (N, h) be a smooth map. If M is compact, orientable and the sectional curvature of (N, h) is non-positive, i.e., RiemN ≤ 0, then u is an F -biharmonic map if and only if it is harmonic.

2 0 kτ(u)k 2 Proof. Computing the Laplacian of the function kF ( 2 )τ(u)k , we have

2 2 0 kτ(u)k 4 F τ(u) 2 2 2 X 0 kτ(u)k 0 kτ(u)k (6) = 2 h ∇e e F τ(u) , ∇e e F τ(u) k 2 k 2 k kτ(u)k2 kτ(u)k2 + 2h −4e F 0 τ(u) ,F 0 τ(u) . 2 2 JJ J I II

Go back Since u is an F -biharmonic map, we have kτ(u)k2 Full Screen 0 τF,2(u) = − 4e F τ(u) 2 (7) Close X kτ(u)k2 − RN du(e ),F 0 τ(u) du(e ) = 0. i 2 i Quit i From (6) and (7), we have

2 2 0 kτ(u)k 4 k F τ(u) 2 2 2 X 0 kτ(u)k 0 kτ(u)k = 2 h(∇e e F τ(u) , ∇e e F )τ(u) k 2 k 2 k X kτ(u)k2 kτ(u)k2 (8) + 2 h(RN du e ),F 0 τ(u) du(e ),F 0 τ(u) . i 2 i 2 i

Since the section curvature of N is non-positive, i.e., RiemN ≤ 0 and by (8), we have

kτ(u)k2 4kF 0 τ(u)k2 ≥ 0(9) 2

2 R 0 kτ(u)k 2 By the Green’s theorem M 4kF 2 τ(u)k dvg = 0 and (9), we have

JJ J I II kτ(u)k2 4kF 0 τ(u)k2 = 0, 2 Go back

2 0 kτ(u)k 2 Full Screen so then kF 2 τ(u)k is constant. From (8), we have

Close 2 0 kτ(u)k (10) ∇e ek F τ(u) = 0, for k = 1, . . . , m. Quit 2 2 P 0 kτ(u)k Setting X = i h du(ei),F 2 τ(u) ei, we have X div(X) = g(∇ek X, ek) k kτ(u)k2 = h τ(u),F 0 τ(u) 2 2 X 0 kτ(u)k + h du(ei), ∇e ei F τ(u) (11) 2 i kτ(u)k2 = h τ(u),F 0 τ(u) 2 kτ(u)k2 = F 0 kτ(u)k2. 2 Integrating (11) over M, we have Z Z 2 0 kτ(u)k 2 (12) 0 = div(X)dvg = F kτ(u)k dvg. M M 2 JJ J I II From F 0(t) > 0 on (0, ∞) and (12), we have τ(u) = 0.

Go back When u is a Riemannian immersion and dimM=dimN − 1, we can replace the hypothesis Full Screen RiemN ≤ 0 with the hypothesis RicciN ≤ 0, and we obtain the following theorem. Close Theorem 3.3. Let u:(M, g) → (N, h) be a Rimannian immersion. If M is compact, orientable, RicciN ≤ 0 and dim M = dim N − 1, then u is an F -biharmonic map if and only if it is harmonic. Quit Proof. Since u is a Riemannian immersion and dim M = dim N − 1, we have

X kτ(u)k2 kτ(u)k2 h RN du(e ),F 0 τ(u) du(e ),F 0 τ(u) i 2 i 2 (13) i kτ(u)k2 kτ(u)k2 = − RicciN F 0 τ(u),F 0 τ(u) . 2 2

From (8), (13) and RicciN ≤ 0, we have

2 2 0 kτ(u)k 4 F τ(u) ≥ 0. 2

Applying the same argument as in the proof of Theorem 3.2, we get the result.

Theorem 3.4. Let (M, g) be an m-dimensional complete manifold withVol(M, g) = ∞. If u:(M, g) → (N, h) is an F -biharmonic map, the sectional curvature of (N, h) is non-positive, i.e., 2 2 RiemN ≤ 0 and R F 0 kτ(u)k τ(u) dv < ∞, then u is harmonic. M 2 g JJ J I II Proof. Since u is an F -biharmonic map, we have Go back kτ(u)k2 Full Screen 0 τF,2(u) = − 4e F τ(u) 2 (14) Close X kτ(u)k2 − RN du(e ),F 0 τ(u) du(e ) = 0. i 2 i Quit i Take any point x0 ∈ M and for every r > 0, let us consider the following cut oﬀ function λ(x) on M:  0 ≤ λ(x) ≤ 1, x ∈ M,    λ(x) = 1, x ∈ Br(x0), (15)  λ(x) = 0, x ∈ M − B2r(x0),   2 |∇λ| ≤ r , x ∈ M,

where Br(x0) = {x ∈ M : d(x, x0) < r} and d is the distance of (M, g). Let X be a compactly supported vector ﬁeld on M such that 2 2 0 kτ(u)k 2 0 kτ(u)k g(X,Y ) = h ∇e Y F τ(u)], λ [F τ(u) . 2 2 Then the divergence of X is given by the following expression div(X) X X X = g(∇ek X, ek) = ekg(X, ek) − g(X, ∇ek ek) k k k JJ J I II 2 2 X h 0 kτ(u)k i 2h 0 kτ(u)k i = ekh ∇e e F τ(u) , λ F τ(u) k 2 2 k Go back 2 2 (16) X h 0 kτ(u)k i 2h 0 kτ(u)k i − h ∇e ∇e e F τ(u) , λ F τ(u) k k 2 2 Full Screen k 2 2 h 0 kτ(u)k i 2h 0 kτ(u)k i = h −4e F τ(u) , λ F τ(u) 2 2 Close 2 2 X h 0 kτ(u)k i 2h 0 kτ(u)k i + h ∇e e F τ(u) , ∇e e λ F τ(u) . k 2 k 2 Quit k From (14) and (16), we have div(X) 2 2 X N 0 kτ(u)k 2h 0 kτ(u)k i = h R du(e ),F τ(u) du(e ), λ F τ(u) k 2 k 2 (17) k 2 2 X h 0 kτ(u)k i 2h 0 kτ(u)k i + h ∇e e F τ(u) , ∇e e λ F τ(u) . k 2 k 2 k Integrating (17) over M and RiemN ≤ 0, we get Z 2 2 X h 0 kτ(u)k i 2h 0 kτ(u)k i h ∇e e F τ(u) , ∇e e λ F τ(u) dvg k 2 k 2 k M Z 2 2 X N 0 kτ(u)k 2h 0 kτ(u)k i = − h R du(e ),F τ(u) du(e ), λ F τ(u) dv k 2 k 2 g (18) k M Z 2 2 X N 0 kτ(u)k 2h 0 kτ(u)k i = h R F τ(u), du(e ) du(e ), λ F τ(u) dv 2 k k 2 g k M ≤ 0. JJ J I II From (18), we have Go back Z 2 2 X h 0 kτ(u)k i 2h 0 kτ(u)k i 0 ≥ h ∇e e F τ(u) , ∇e e λ F τ(u) dvg k 2 k 2 k M Full Screen Z 2 2 X 2 h 0 kτ(u)k i (19) = λ ∇e e F τ(u) dvg k 2 k M Close Z 2 2 X h 0 kτ(u)k i h 0 kτ(u)k i + 2 λek(λ)h ∇e ek F τ(u) , F τ(u) dvg. M 2 2 Quit k Therefore, we have Z 2 2 X 2 h 0 kτ(u)k i λ ∇e e F τ(u) dvg k 2 k M Z 2 2 X h 0 kτ(u)k i h 0 kτ(u)k i ≤ −2 λek(λ)h ∇e e F τ(u) , F τ(u) dvg k 2 2 k M (20) Z 2 2 X h 0 kτ(u)k i h 0 kτ(u)k i = − 2h λ∇e e F τ(u) , ek(λ) F τ(u) dvg k 2 2 k M Z ( 2 2 2 2) X 1 2 h 0 kτ(u)k i 2 0 kτ(u)k ≤ λ ∇e e F τ(u) + 2[ek(λ)] F τ(u) dvg, 2 k 2 2 k M where we use the following Cauchy-Schwarz inequality 1 ±2h(V,W ) ≤ εkV k2 + kW k2 ε 1 for the second inequality and ε = 2 . From (20), we have

Z 2 2 Z 2 2 X 2 h 0 kτ(u)k i X 2 0 kτ(u)k λ ∇ F τ(u) dv ≤ 4 [e (λ)] F τ(u) dv , e ek g k g JJ J I II M 2 M 2 (21) k k 16 Z kτ(u)k2 2 Go back ≤ F 0 τ(u) dv . 2 g r M 2 2 Full Screen R 0 kτ(u)k 2 Since M kF 2 τ(u)k dvg < ∞ and (M, g) is complete, then we have (r → ∞) Close Z X h kτ(u)k2 i 2 ∇ F 0 τ(u) dv = 0. e ek g M 2 Quit k For every vector ﬁeld X on M, we have

2 h 0kτ(u)k i ∇e X F τ(u) = 0. 2

2 2 So we know that F 0 kτ(u)k τ(u) is constant, say C. Therefore, if Vol(M, g)=∞ and C 6= 0, 2 then Z kτ(u)k2 2 0 2 F τ(u) dvg = C Vol(M, g) = ∞, M 2

2 2 which yields a contradiction. Thus, we have F 0 kτ(u)k τ(u) = C = 0. From F 0(t) > 0 on 2 2 2 (0, ∞) and F 0 kτ(u)k τ(u) = 0, we know that τ(u) = 0, i.e. u is harmonic. 2

From Theorem 3.4, we have the following corollaries:

Corollary 3.5. Let (M, g) be an m-dimensional complete manifold with Vol(M, g) = ∞. If u:(M, g) → (N, h) is an exponentially biharmonic map, the sectional curvature of (N, h) is non- positive, i.e., JJ J I II Z N 2 kτ(u)k2 Riem ≤ 0 and kτ(u)k e dvg < ∞, Go back M then u is harmonic. Full Screen Corollary 3.6. Let (M, g) be an m-dimensional complete manifold with Vol(M, g) = ∞. If Close u:(M, g) → (N, h) is a p-biharmonic map, the sectional curvature of (N, h) is non-positive, i.e., N R 2p−2 Riem ≤ 0 and M kτ(u)k dvg < ∞, then u is harmonic. Quit Corollary 3.7 ([15]). Let (M, g) be an m-dimensional complete manifold with Vol(M, g) = ∞. If u:(M, g) → (N, h) is a biharmonic map, the sectional curvature of (N, h) is non-positive, i.e., N R 2 Riem ≤ 0 and M kτ(u)k dvg < ∞, then u is harmonic. 4. Stress F -bienergy tensor The stress bienergy tensor and the conservation law of a biharmonic map between Riemannian manifolds were first studied by G.Y. Jiang in [10]. Following Jiang’s notion, we define the stress F -bienergy tensor of a smooth map as follows. Definition 4.1. Let u:(M, g) → (N, h) be a smooth map between two Riemannian manifolds. The stress F -bienergy tensor of u is defined by 2 2 kτ(u)k X h 0 kτ(u)k i SF,2(X,Y ) = F g(X,Y ) + h du(ek), ∇e e F τ(u) g(X,Y ) 2 k 2 k 2 2 h 0 kτ(u)k i h 0 kτ(u)k i − h du(X), ∇e Y F τ(u) − h du(Y ), ∇e X F τ(u) 2 2 for any X,Y ∈ Γ(TM).

JJ J I II Remark 4.2. When F (t) = t, we have SF,2(X,Y ) = S2(X,Y ), where S2 is stress bienergy tensor in [10]. Go back Theorem 4.3. For any smooth map u:(M, g) → (N, h) kτ(u)k2 kτ(u)k4 (div S )(X) = −h(τ (u), du(X)) − F 00 X Full Screen F,2 F,2 2 4 for any vector ﬁeld X ∈ Γ(TM). Close

Proof. We choose a local orthonormal frame ﬁeld {ei} on M with ∇ei ei|x = 0 at a point x ∈ M. Let X be a vector ﬁeld on M. At x, we compute Quit X X (div SF,2)(X) = (∇ei SF,2)(ei,X) = eiSF,2(ei,X) − SF,2(ei, ∇ei X) i i " 2 2 X kτ(u)k X h 0 kτ(u)k i = ei F g(ei,X) + h du(ek), ∇e e F τ(u) g(ei,X) 2 k 2 i k 2 2 h 0 kτ(u)k i h 0 kτ(u)k i − h du(ei), ∇e X F τ(u) − h du(X), ∇e e F τ(u) 2 i 2 2 2 X kτ(u)k X h 0 kτ(u)k i − F g(ei, ∇e X) + h du(ek), ∇e e F τ(u) g(ei, ∇e X) 2 i k 2 i i k 2 2 h 0 kτ(u)k i h 0 kτ(u)k i −h du(ei), ∇e ∇e X F τ(u) − h du(∇e X), ∇e e F τ(u) i 2 i i 2 2 2 kτ(u)k X h 0 kτ(u)k i = X F + h (∇e du)(X, ek), ∇e e F τ(u) 2 k 2 k 2 2 X h 0 kτ(u)k i h 0 kτ(u)k i + h du(ek), ∇e X ∇e e F τ(u) − h τ(u), ∇e X F τ(u) k 2 2 k JJ J I II 2 2 X h 0 kτ(u)k i X h 0 kτ(u)k i − h du(ei), ∇e e ∇e X F τ(u) − h (∇e du)(X, ek), ∇e e F τ(u) i 2 k 2 Go back i k 2 2 X h 0 kτ(u)k i X h 0 kτ(u)k i + h du(ei), ∇e ∇e X F τ(u) − h du(X), ∇e e ∇e e F τ(u) ] i 2 i i 2 Full Screen i i kτ(u)k2 kτ(u)k4 = − F 00 X Close 2 4 2 2 ! h 0 kτ(u)k i X N h 0 kτ(u)k i + h 4e F τ(u) + R du(ei), F τ(u) du(ei), du(X) 2 2 Quit i kτ(u)k2 kτ(u)k4 = − h(τ (u), du(X)) − F 00 X . F,2 2 4 From Theorem 3.1, we know that if u: M → N is an F -biharmonic map, then kτ(u)k2 kτ(u)k4 (22) (div S )(X) = −F 00 X . F,2 2 4 Proposition 4.4. Let c: I ⊂ R → (N, h) be a curve parametrized by arc-length. Assume that tF 0(t) SF,2 = 0 and lF = inft≥0 F (t) > 0. Then c is geodesic. Proof. A direct computation shows that 2 2 ∂ ∂ kτ(c)k ∂ h 0kτ(c)k i 0 = SF,2 , = F − h dc , ∇e ∂ F )τ(c) ∂t ∂t 2 ∂t ∂t 2 kτ(c)k2 h kτ(c)k2 i kτ(c)k2 = F + h τ(c), F 0 τ(c) , > (1 + 2l )F . 2 2 F 2 kτ(c)k2 If F ( 2 ) = 0, then τ(c) = 0.

2 Proposition 4.5. Let u:(M , g) → (N, h) be a map from a surface. Then SF,2 = 0 implies u is harmonic. JJ J I II Proof. The trace of SF,2 gives the equality Go back 2 2 kτ(u)k D h 0kτ(u)k iE 0 = trace SF,2 = F + 2 du, ∇e F τ(u) Full Screen 2 2 D h kτ(u)k2 iE kτ(u)k2 − 2 du, ∇e F 0 τ(u) = F , Close 2 2 so we have τ(u) = 0. Quit m Proposition 4.6. Let u:(M , g) → (N, h), m 6= 2. Then SF,2 = 0 if and only if 2 2 2 kτ(u)k 0kτ(u)k F g(X,Y ) + h du(X), ∇e Y [F τ(u)] (23) m − 2 2 2 2 h 0kτ(u)k i + h du(Y ), ∇e X F τ(u) = 0 2 for any X,Y ∈ Γ(TM).

Proof. Since SF,2 = 0, we have traceSF,2 = 0. Therefore, 2 2 X h 0kτ(u)k i m kτ(u)k h du(ek), ∇e e F τ(u) = − F . k 2 m − 2 2 k

Substituting it into the deﬁnition of SF,2, we obtain 2 kτ(u)k2 0 = S (X,Y ) = − F g(X,Y ) F,2 m − 2 2 2 2 h 0kτ(u)k i h 0kτ(u)k i − h du(X), ∇e Y F τ(u) − h du(Y ), ∇e X F τ(u) . 2 2 m JJ J I II Proposition 4.7. A map u:(M , g) → (N, h), m > 2, with SF,2 = 0 and rank u ≤ m − 1 is harmonic. Go back

Proof. Take p ∈ M. Since rank u(p) ≤ m − 1, there exists a unit vector Xp ∈ Ker dup and for Full Screen kτ(u)k2 X = Y = Xp,(23) becomes F 2 = 0, so τ(u) = 0. Close m n Corollary 4.8. Let u:(M , g) → (N , h) be a submersion (m > n), if SF,2 = 0, then u is harmonic. Quit ∗ ∗ Recall that for two 2-tensors T1, T2 ∈ Γ(T M ⊗ T M), their inner product is deﬁned as follows: X (24) hT1,T2i = T (ei, ej)T2(ei, ej), ij

where {ei} is an orthonormal basis of M with respect to g. For a vector ﬁeld X ∈ Γ(TM), by θX we denote its dual one form, i.e., θX (Y ) = g(X,Y ). The covariant derivative of θX gives a 2-tensor ﬁeld ∇θX

(25) (∇θX )(Y,Z) = (∇Z θX )(Y ) = g(∇Z X,Y ).

If X = ∇ϕ is the gradient of some function ϕ on M, then θX = dϕ and ∇θX = Hess ϕ. Lemma 4.9 (cf. [2,4 ]). Let T be a symmetric (0, 2)-type tensor field and let X be a vector field. Then 1 (26) div(i T ) = (div T )(X) + hT, ∇θ i = (div T )(X) + hT,L gi. X X 2 X Let D be any bounded domain of M with C1 boundary. By using the Stokes’ theorem, we immediately have the following integral formula Z Z 1 (27) T (X, ν)dsg = [hT, LX gi + div(T )(X)]dvg JJ J I II ∂D D 2 where ν is the unit outward normal vector field along ∂D. Go back By (22) and (3), we have

Full Screen Z SF,2(X, ν)dsg ∂D Close (28) Z 2 4 h 1 00kτ(u)k kτ(u)k i = hSF,2, LX gi − F X dvg. Quit D 2 2 4 When F (t) = t, the equation (28) turns into the following equation Z Z 1 (29) S2(X, ν)dsg = hS2, LX gidvg. ∂D D 2 5. Monotonicity formulas for biharmonic maps

In this section, we investigate the special case of F -biharmonic maps, i.e., biharmonic maps. m Let (M , g) be a complete Riemannian manifold with pole x0. By r(x) denote the g-distance m function relative to the pole x0, that is, r(x) = distg(x, x0). Set B(r) = {x ∈ M : r(x) ≤ r}. By 2 λmax (resp. λmin) denote the maximum (resp. minimal) eigenvalues of Hess(r ) − dr ⊗ dr at each point of M − {x0}. Theorem 5.1. Let u:(M, g) → (N, h) be an isometric immersion. Assume that there is a constant σ > 0 such that m − 1 (30) λ + 1 − 2 max{2, λ } ≥ σ. 2 min max ∂ If u is a biharmonic map and h(τ(u), ∇e ∂ du( )) ≥ 0, then we have ∂r ∂r Z kτ(u)k2 Z kτ(u)k2 JJ J I II dvg dvg B(ρ1) 2 B(ρ2) 2 (31) σ ≤ σ Go back ρ1 ρ2

for any 0 < ρ1 ≤ ρ2. Full Screen Proof. Since u: M m → N is an isometric immersion, we have τ(u) = mH, where H is the mean Close curvature vector ﬁeld of M in N, so we know that h(τ(u), du(X)) = h(mH, du(X)) = 0(32) Quit for any tangent vector ﬁeld X on M. ∂ Taking D = B(r) and X = r ∂r in (29), we have Z Z Z ∂ ∂ 1 1 2 (33) S2 r , dsg = hS2, Lr ∂ gidvg = hS2, Hess(r )idvg. ∂B(r) ∂r ∂r B(r) 2 ∂r 2 B(r) m ∂ 2 Let {ei}i=1 be an orthonormal basis on M and em = ∂r . We may assume that Hess(r ) becomes a diagonal matrix with respect to {ei}.

1 2 1 X 2 − hS , Hess(r )i = − S (e , e ) Hess(r )(e , e ) 2 2 2 2 i j i j i,j 2 1 X kτ(u)k 2 = − { Hess(r )(e , e ) 2 2 i i i X X 2 + h(∇e ek τ(u), du(ek)) Hess(r )(ei, ei) k i X 2 − 2 h(du(ei), ∇e ej τ(u)) Hess(r )(ei, ej )} (34) i,j 2 1 n kτ(u)k X 2 JJ J I II = − − Hess(r )(e , e ) 2 2 i i i Go back X 2 o + 2 h(τ(u), ∇e ei du(ei)) Hess(r )(ei, ei) i 2 Full Screen kτ(u)k h m − 1 i ≥ λ + 1 − 2 max{2, λ } 2 2 min max kτ(u)k2 Close ≥ σ , 2 where the equation (32) is used for the third equality and the equation (30) for the last inequality. Quit On the other hand, by the coarea formula, we have

Z ∂ ∂ Z nh kτ(u)k2 X i ∂ ∂ − S2 r , dsg = − + h(du(ek), ∇e e τ(u)) g r , ∂r ∂r 2 k ∂r ∂r ∂B(r) ∂B(r) k ∂ o − 2rh(du , ∇e ∂ τ(u) dsg ∂r ∂r (35) Z n kτ(u)k2 ∂ o = r − rh τ(u), ∇e ∂ du dsg ∂B(r) 2 ∂r ∂r Z kτ(u)k2 d Z kτ(u)k2 ≤ r dsg = r dvg, ∂B(r) 2 dr B(r) 2

∂ where the condition h(τ(u), ∇e ∂ du( )) ≥ 0 is used for the inequality. ∂r ∂r From (33), (34) and (35), we have Z kτ(u)k2 d Z kτ(u)k2 (36) σ dvg ≤ r dvg B(r) 2 dr B(r) 2 i.e. Z kτ(u)k2 dvg d 2 JJ J I II (37) B(r) ≥ 0. dr rσ Go back Therefore, Z kτ(u)k2 Z kτ(u)k2 Full Screen dv dv 2 g 2 g B(ρ1) B(ρ2) σ ≤ σ Close ρ1 ρ2 for any 0 < ρ1 ≤ ρ2. Quit m Lemma 5.2 ([4,8 ]). Let (M , g) be a complete Riemannian manifold with a pole x0. By Kr denote the radial curvature of M as follows 2 2 (i) if −α ≤ Kr ≤ −β with α ≥ β > 0, then β coth(βr)[g − dr ⊗ dr] ≤ Hess(r) ≤ αcoth(αr)[g − dr ⊗ dr],

A B (ii) if − (1+r2)1+ε ≤ Kr ≤ (1+r2)1+ε with ε > 0, A ≥ 0 and 0 ≤ B < 2ε, then

1 − B/2ε eA/2ε [g − dr ⊗ dr] ≤ Hess(r) ≤ [g − dr ⊗ dr], r r

a2 b2 2 1 (iii) if − 1+r2 ≤ Kr ≤ 1+r2 with a ≥ 0 and b ∈ [0, 4 ], then √ √ 1 + 1 − 4b2 1 + 1 + 4a2 [g − dr ⊗ dr] ≤ Hess(r) ≤ [g − dr ⊗ dr]. 2r 2r m Lemma 5.3. Let (M , g) be a complete Riemannian manifold with a pole x0. By Kr denote the radial curvature of M as follows 2 2 (i) if −α ≤ Kr ≤ −β with α ≥ β > 0 and (m − 1)β − 4α ≥ 0, then JJ J I II (m − 1) 4α λ + 1 − 2 max{2, λ } ≥ m − . Go back 2 min max β

Full Screen A B (ii) if − (1+r2)1+ε ≤ Kr ≤ (1+r2)1+ε with ε > 0, A ≥ 0 and 0 ≤ B < 2ε, then

Close (m − 1) B A λ + 1 − 2 max{2, λ } ≥ 1 + (m − 1)(1 − ) − 4e 2ε . 2 min max 2ε Quit a2 b2 2 1 (iii) if − 1+r2 ≤ Kr ≤ 1+r2 with a ≥ 0 and b ∈ [0, 4 ], then (m − 1) λmin + 1 − 2 max{2, λmax} 2 √ √ 1 + 1 − 4b2 1 + 1 + 4a2 ≥ [1 + (m − 1) − 4 . 2 2

Proof. If Kr satisﬁes (i), then by Lemma 5.2, for every r > 0, we have on B(r) − {x0}, 1 [(m − 1)λ + 2 − 4 max{2, λ }] 2 min max 1 ≥ [(m − 1)2βr coth(βr) + 2 − 4 × 2αr coth(αr)] 2 4α coth(αr) = 1 + βr coth(βr) m − 1 − β coth(βr) 4α 4α ≥ 1 + 1. m − 1) − = m − . β β where the second inequality is valid the increasing function βr coth(βr) → 1 as r → 0, and coth(αr) < 1 for 0 < β < α. Similarly, from Lemma 5.2, the above inequality holds for the cases JJ J I II coth(βr) (ii) and (iii) on B(r). Go back Theorem 5.4. Let (M, g) be an m-dimensional complete manifold with a pole x0. Assume that

Full Screen the radial curvature Kr of M satisﬁes one of the following three conditions: 2 2 (i) if if −α ≤ Kr ≤ −β with α ≥ β > 0 and (m − 1)β − 4α ≥ 0, Close A B B (ii) if − (1+r2)1+ε ≤ Kr ≤ (1+r2)1+ε with ε > 0, A ≥ 0, 0 ≥ B < 2ε and 1 + (m − 1)(1 − 2ε ) − A 4 e 2ε > 0, Quit √ √ a2 b2 2 1 1+ 1−4b2 1+ 1+4a2 (iii) if − 1+r2 ≤ Kr ≤ 1+r2 with a ≥ 0, b ∈ [0, 4 ] and 1 + (m − 1) 2 − 4 2 > 0. If u:(M, g) → (N, h) is a biharmonic isometric immersion and h(τ(u), ∂ ∇e ∂ du( )) ≥ 0, then ∂r ∂r Z kτ(u)k2 Z kτ(u)k2 dvg dvg B(ρ1) 2 B(ρ2) 2 (38) Λ ≤ Λ ρ1 ρ2

for any 0 < ρ1 ≤ ρ2, where  4α m − , if K satisfies (i)  r  β  B A (39) Λ = 1 + (m − 1) 1 − − 4 e 2ε , if Kr satisfies (ii)  √2ε √  1 + 1 − 4b2 1 + 1 + 4a2  1 + (m − 1) − 4 , if K satisfies (iii) 2 2 r Proof. From the proof of Theorem 5.1 and Lemma 5.3, we have Z kτ(u)k2 dvg d 2 B(r) ≥ 0. JJ J I II dr rΛ

Go back Therefore, we get the monotonicity formula Z kτ(u)k2 Z kτ(u)k2 Full Screen dvg dvg B(ρ1) 2 B(ρ2) 2 Λ ≤ Λ Close ρ1 ρ2 for any 0 < ρ1 ≤ ρ2. Quit Corollary 5.5. Let M,Kr and Λ be as in Theorem 5.4. Assume that u:(M, g) ∂ → (N, h) is a biharmonic isometric immersion and h(τ(u), ∇e ∂ du( )) ≥ 0. If ∂r ∂r Z 2 kτ(u)k Λ dvg = o(R ), B(R) 2 then u is harmonic.

We say the bienergy E2(u) of u is slowly divergent if there exists a positive function ψ(r) with R ∞ dr = +∞ (R0 > 0) such that R0 rψ(r) kτ(u)k2 Z 2 (40) lim dvg < ∞. R→∞ B(R) ψ(r(x)) Theorem 5.6. Let u:(M, g) → (N, h) be a biharmonic isometric immersion. Assume that there is a constant σ > 0 such that m − 1 λ + 1 − 2 max{2, λ } ≥ σ. 2 min max ∂ If E2(u) is slowly divergent and h(τ(u), ∇e ∂ du( )) ≥ 0, then u is harmonic, i.e., τ(u) = 0. ∂r ∂r JJ J I II Proof. From the proof of Theorem 5.1, we have Go back Z kτ(u)k2 Z kτ(u)k2 (41) σ dvg ≤ r dsg. B(r) 2 ∂B(r) 2 Full Screen Now suppose that u is not harmonic, so there exists R0 > 0 such that for R ≥ R0, Close Z kτ(u)k2 (42) σ dv ≥ c , 2 g 1 Quit B(R) where c1 is a positive constant. From (41) and (42), we have Z kτ(u)k2 (43) c1σ ≤ R dsg. ∂B(R) 2

for R ≥ R0 and kτ(u)k2 Z Z ∞ Z 2 2 dR kτ(u)k lim dvg = dsg R→∞ B(R) ψ(r(x)) 0 ψ(R) ∂B(R) 2 Z ∞ dR Z kτ(u)k2 ≥ dsg R0 ψ(R) ∂B(R) 2 Z ∞ dR ≥ c1σ = ∞, R0 Rψ(R) which contradicts (40), therefore, u is harmonic. JJ J I II From the proof of Theorem 5.6, we immediately get the following theorem.

Go back Theorem 5.7. Let M,Kr and Λ be as in Theorem 5.4. If u:(M, g) → (N, h) is a biharmonic ∂ isometric immersion, the bienergy E2(u) is slowly divergent and h(τ(u), ∇e ∂ du( )) ≥ 0, then u ∂r ∂r Full Screen is harmonic.

Acknowledgment. The authors would like to thank the referee whose valuable suggestions Close make this paper more perfect. Yingbo Han was supported by NSFC No.11201400, No.10971029, No.11026062, Project of Henan Provincial Department of Education No.2011A110015 and Talent youth teacher fund of Xinyang Normal University. Quit 1. Ara M., Geometry of F -harmonic maps, Kodai Math. J. 22(1999), 243–263. 2. Baird P., Stess-energy tensors and the Linchnerowicz Laplacian, J. Geo. and Phys. 58 (2008), 1329–1342. 3. Caddeo R., Montaldo S. and Piu P., On biharmonic maps, Contemp. Math. 288 (2001), 286–290. 4. Dong Y. X. and Wei S. S., On vanishing theorems for vector bundle valued p-forms and their applications, Comm. Math. Phys. 304 (2011), 329–368. 5. Dong Y. X., Lin H. Z. and Yang G. L., Liouville theorems for F -harmonic maps and their applications, arXiv:1111.1882v1 [math.DG] 8 Nov 2011. 6. Hornung P. and Moser R., Intrinsically p-biharmonic maps, preprint (Opus: University of Bath online publi- cation store). 7. Eells J., and Lemaire L., Selected topics in harmonic maps, CBMS 50, Amer. Math. Soc. 1983. 8. Greene R. E. and Wu H., Function theory on manifolds which posses pole, In: Lecture Notes in Mathematics, V. 699, Springer-veriag, Berlin, Heidelberg, New York 1979. 9. Jiang G. Y., 2-harmonic maps and their first and second variational formulas, Chinese Ann. Math. 7A (1986), 388–402; the English translation, Note di Matematica, 28 (2009), 209–232. 10. , The conservation law for 2-harmonic maps between Riemannian manifolds, Acta Math. Sin. 30 (1987), 220–225. JJ J I II 11. Kassi M., A Liouville theorems for F -harmonic maps with finite F -energy, Electonic Journal Diff. Equa. 15 (2006), 1–9. 12. Liu J. C., Liouville theorems of stable F -harmonic maps for compact convex hypersurfaces, Hiroshima Math. Go back J. 36 (2006), 221–234. 13. Loubeau E. and Oniciuc C., The index of biharmonic maps in spheres, Compositio Math. 141 (2005), 729–745. 14. Loubeau E., Montaldo S. and Oniciuc C., The stress-energy tensor for biharmonic maps, Math. Z. 259 (2008), Full Screen 503–524. 15. Nakauchi N., Urakawa H. and Gudmundsson S., Biharmonic maps into a Riemannian manifold of non-positive curvature, arXiv:1201.6457v4. 16. Oniciuc C., Biharmonic maps between Riemannian manifolds, Analele stiintifice ale Univer. ”Al. I. Cusa”, din Close Iasi (2002), 237–248. 17. Ouakkas S., Nasri R., and Djaa M., On the f-harmonic and f-biharmonic maps, JP J. Geom. Topol. 10(1) (2010), 11–27. Quit 18. Wang C., Remarks on biharmonic maps into spheres, Calc. Var. Partial Differential Equations, 21 (2004), 221–242. 19. , Biharmonic maps from R4 into Riemannian manifold. Math. Z. 247 (2004), 65–87.

Yingbo Han, College of Mathematics and Information Science, Xinyang Normal University, Xinyang, 464000, Henan, P. R. China, e-mail: [email protected]

Shuxiang Feng, College of Mathematics and Information Science, Xinyang Normal University, Xinyang, 464000, Henan, P. R. China, e-mail: [email protected]

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