BIHARMONIC MAPS INTO SOL AND NIL SPACES

BIHARMONIC MAPS INTO SOL AND NIL SPACES

arXiv:math/0612329v1 [math.DG] 13 Dec 2006

YE-LIN OU AND ZE-PING WANG

Abstract In this paper, we study biharmonic maps into Sol and Nil spaces, two model spaces of Thurston’s 3-dimensional geometries. We characterize non-geodesic biharmonic curves in Sol space and prove that there exists no non-geodesic biharmonic helix in Sol space. We also show that a linear map from a Euclidean space into Sol or Nil space is biharmonic if and only if it is a harmonic map, and give a complete classification of such maps.

1. introduction In this paper, we work in the category of smooth objects, so manifolds, maps, vector fields, etc, are assumed to be smooth unless it is stated otherwise. A map ϕ : (M, g) −→ (N, h) between Riemannian manifolds is called a biharmonic map if ϕ|Ω is a critical point of the bienergy Z 1 2 |τ2 (ϕ)|2 dx E (ϕ, Ω) = 2 Ω

for every compact subset Ω of M, where τ2 (ϕ) = Traceg ∇dϕ is the tension field of ϕ. Using the first variational formula (see [13]) one sees that ϕ is a biharmonic map if and only if its bitension field vanishes identically, i.e., (1)

τ 2 (ϕ) := −△ϕ (τ2 (ϕ)) − Traceg RN (dϕ, τ2 (ϕ))dϕ = 0,

where △ϕ = −Traceg (∇ϕ )2 = −Traceg (∇ϕ ∇ϕ − ∇ϕ∇M )

is the Laplacian on sections of the pull-back bundle ϕ−1 T N and RN is the curvature operator of (N, h) defined by N N RN (X, Y )Z = [∇N X , ∇Y ]Z − ∇[X,Y ] Z.

Note that τ 2 (ϕ) = −J ϕ (τ2 (ϕ)), where J ϕ is the Jacobi operator which plays an important role in the study of harmonic maps. 1991 Mathematics Subject Classification. 58E20. Key words and phrases. biharmonic curves, biharmonic maps, Sol space, Nil space. 1

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YE-LIN OU AND ZE-PING WANG

Clearly, any harmonic map is biharmonic, so harmonic maps are a subclass of biharmonic maps. It is interesting to note that so far, apart from maps between Euclidean spaces (in which case any polynomial map of degree less than four is a biharmonic map), not many example of proper (meaning non-harmonic) biharmonic maps between Riemannain manifolds have been found (see, e.g., [10], [9], [11], and the bibliography of biharmonic maps [8]). Recently, some work has been done in the study of non-geodesic biharmonic curves in some model spaces. For example, [2] gives a complete classification of non-geodesic biharmonic curves in 3-sphere while in [3] it is proved that a non-geodesic biharmonic curve in Heisenberg group H3 is a helix and the explicit parametrizations of such curves are given. For the study of biharmonic curves in Berger’s spheres, in Minkowski 3-space, in Cartan-Vranceanu 3-dimensional space, and in contact and Sasakian manifolds see [1], [7], [4], and [6] respectively. In this paper, we first characterize non-geodesic biharmonic curves in Sol space (Theorem 2.3) and show that there exists no non-geodesic biharmonic helix in Sol space (Theorem 2.7). In the second part of the paper, we study linear biharmonic maps from Euclidean space Rm into Sol space (R3 , gSol ) and Nil space (R3 , gN il ) using the linear structure of the underlying manifolds of Sol and Nil spaces. We show that such a linear map is biharmonic if and only if it is a harmonic map and give a complete description of such maps (Theorems 3.3 and 3.4).

2. Biharmonic curves in Sol space Biharmonic curve equation in Frenet Frames. Let I ⊂ R be an open interval, and γ : I −→ (N, h) be a curve, parametrized by arc length, on a Riemannian manifold. Putting T = γ ′ , we can write the tension field of γ as τ (γ) = ∇γ ′ γ ′ , and the biharmonic map equation (1) reduces to ∇3T T − RN (T, ∇T T )T = 0. A successful key to study the geometry of a curve is to use the Frenet frames along the curve which is recalled in the following Definition 2.1. (see, for example, [12]) The Frenet frame {Fi }i=1,2,...,n associated to a curve γ : I ⊂ R −→ (N n , h) parametrized by arc length is the orthonormalisation of the (n+1)-tuple (k) ∂ )}k=0,1, 2, ..., n described by: {∇ ∂ dγ( ∂t ∂t

BIHARMONIC MAPS INTO SOL AND NIL SPACES

3

∂ F1 = dγ( ∂t ) (γ) ∇ ∂ F1 = k1 F2 ∂t

(γ)

∇ ∂ Fi = −ki−1 Fi−1 + ki Fi+1 , ∀ i = 2, 3, . . . , n − 1 ∂t

(γ)

∇ ∂ Fn = −kn−1 Fn−1 , ∂t

where the functions {k1 , k2 , . . . , kn−1} are called the curvatures of γ and ∇γ is the connection on the pull-back bundle γ −1 (T N). Note that F1 = T = γ ′ is the unit tangent vector field along the curve. With respect to the Frenet frame, the biharmonic curve equation takes the following form. Lemma 2.2. (see, for example, [4]) Let γ : I ⊂ R −→ (N n , h)(n ≥ 2) be a curve parametrized by arc length from an open interval of R into a Riemannian manifold (N n , h). Then γ is biharmonic if and only if :  k1 k1′ = 0      k1′′ − k13 − k1 k22 + k1 R(F1 , F2 , F1 , F2 ) = 0 (2) 2k1′ k2 + k1 k2′ + k1 R(F1 , F2 , F1 , F3 ) = 0    k1 k2 k3 + k1 R(F1 , F2 , F1 , F4 ) = 0   k1 R(F1 , F2 , F1 , Fj ) = 0 j = 5, . . . , n.

Riemannian structure of Sol space. Sol space, one of Thurston’s eight 3-dimensional geometries, can be viewed as R3 provided with Riemannian metric gSol = ds2 = e2z dx2 + e−2z dy 2 + dz 2 , where (x, y, z) are the standard coordinates in R3 . Note that the Sol metric can also be written as: 3 X 2 ds = ω i ⊗ ω i, i=1

where

ω 1 = ez dx, ω 2 = e−z dy, ω 3 = dz, and the orthonormal basis dual to the 1-forms is ∂ ∂ ∂ . e1 = e−z , e2 = ez , e3 = ∂x ∂y ∂z With respect to this orthonormal basis, the Levi-Civita connection and the Lie brackets can be easily computed as: (3)

∇e1 e1 = −e3 , ∇e1 e2 = 0, ∇e1 e3 = e1 ∇e2 e1 = 0, ∇e2 e2 = e3 , ∇e2 e3 = −e2 ∇e3 e1 = 0, ∇e3 e2 = 0, ∇e3 e3 = 0.

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YE-LIN OU AND ZE-PING WANG

(4)

[e1 , e2 ] = 0, [e2 , e3 ] = −e2 , [e1 , e3 ] = e1 .

We adopt the following notation and sign convention for Riemannian curvature operator. (5)

R(X, Y )Z = ∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ] Z,

the Riemannian curvature tensor is given by (6)

R(X, Y, Z, W ) = g(R(Y, X)Z, W ) = −g(R(X, Y )Z, W ).

Moreover we put (7)

Rabc = R(ea , eb )ec , Rabcd = R(ea , eb , ec , ed ),

where the indices a, b, c, d take the values 1, 2, 3. A direct computation using (3), (4), (5), (6), and (7) gives the following nonzero components of Riemannian curvature of Sol space with respect to the orthonomal basis {e1 , e2 , e3 } (we do not list those that can be obtained by symmetric properties of curvature): R121 = −e2 , R131 = e3 , R122 = e1 , R232 = e3 , R133 = −e1 , R233 = −e2 , and (8)

R1212 = −g(R(e1 , e2 )e1 , e2 ) = −g(−e2 , e2 ) = 1, R1313 = −g(R(e1 , e3 )e1 , e3 ) = −g(e3 , e3 ) = −1, R2323 = −g(R(e2 , e3 )e2 , e3 ) = −g(e3 , e3 ) = −1.

Biharmonic curves in Sol space. Let γ : I −→ (R3 , gSol ) be a curve on Sol space parametrized by arc length. Let {T, N, B} be the Frenet frame fields tangent to Sol space along γ defined as follows: T is the unit vector field γ ′ tangent to γ, N is the unit vector field in the direction of ∇T T (normal to γ), and B is chosen so that {T, N, B} is a positively oriented orthonormal basis. Then, by Definition (2.1), we have the following Frenet formulas

(9)

 

∇T T = kN ∇T N = −kT + τ B  ∇T B = −τ N,

where k = |∇T T | is the geodesic curvature and τ the geodesic torsion of γ. With respect to the orthonormal basis {e1 , e2 , e3 } we can write T = T1 e1 + T2 e2 + T3 e3 , N = N1 e1 + N2 e2 + N3 e3 , B = T × N = B1 e1 + B2 e2 + B3 e3 , and we have

BIHARMONIC MAPS INTO SOL AND NIL SPACES

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Theorem 2.3. Let γ : I −→ (R3 , gSol ) be a curve parametrized by arc length. Then γ is a non-geodesic biharmonic curve if and only if

  k = constant 6= 0 k 2 + τ 2 = 2B32 − 1  τ ′ = 2N3 B3 .

(10)

Proof. By (2) of lemma (2.2) we see that γ is a biharmonic curve if and only if

 

kk ′ = 0 k ′′ − k 3 − kτ 2 + kR(T, N, T, N) = 0  2k ′ τ + kτ ′ + kR(T, N, T, B) = 0, which is equivalent to

(11)

 

k = constant 6= 0 k + τ = R(T, N, T, N)  τ ′ = −R(T, N, T, B) 2

2

since k 6= 0 by the assumption that γ is non-geodesic. A direct computation using (8) yields

P R(T, N, T, N) = 3i,j,l,p=1 Tl Np Ti Nj Rlpij = T1 N2 T1 N2 R1212 + T1 N2 T2 N1 R1221 + T2 N1 T2 N1 R2121 + T2 N1 T1 N2 R2112 +T1 N3 T1 N3 R1313 + T1 N3 T3 N1 R1331 + T3 N1 T3 N1 R3131 + T3 N1 T1 N3 R3113 +T2 N3 T2 N3 R2323 + T2 N3 T3 N2 R2332 + T3 N2 T3 N2 R3232 + T3 N2 T2 N3 R3223 = T12 N22 − T1 T2 N1 N2 + T22 N12 − T1 T2 N1 N2 −T12 N32 + T1 T3 N1 N3 − T32 N12 + T1 T3 N1 N3 −T22 N32 + T2 T3 N2 N3 − T32 N22 + T2 T3 N2 N3 = (T1 N2 − T2 N1 )2 − (T3 N1 − T1 N3 )2 − (T3 N2 − T2 N3 )2 = B32 − B12 − B22 = 2B32 − 1 (by T × N = B, T × B = −N, B12 + B22 + B32 = 1),

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YE-LIN OU AND ZE-PING WANG

and P R(T, N, T, B) = 3i,j,l,p=1 Tl Np Ti Bj Rlpij = T1 N2 T1 B2 R1212 + T1 N2 T2 B1 R1221 +T2 N1 T2 B1 R2121 + T2 N1 T1 B2 R2112 +T1 N3 T1 B3 R1313 + T1 N3 T3 B1 R1331 +T3 N1 T3 B1 R3131 + T3 N1 T1 B3 R3113 +T2 N3 T2 B3 R2323 + T2 N3 T3 B2 R2332 +T3 N2 T3 B2 R3232 + T3 N2 T2 B3 R3223 = T12 N2 B2 − T1 T2 N1 B2 + T22 N1 B1 − T1 T2 N2 B1 −T12 N3 B3 + T1 T3 N1 B3 − T32 N1 B1 + T1 T3 B1 N3 −T22 N3 B3 + T2 T3 N2 B3 − T32 N2 B2 + T2 T3 B2 N3 = (T1 B2 − T2 B1 )(T1 N2 − T2 N1 ) −(T3 N1 − T1 N3 )(T3 B1 − T1 B3 ) −(T3 N2 − T2 N3 )(T3 B2 − T2 B3 ) = −N3 B3 + N2 B2 + N1 B1 = −2N3 B3 (by T × N = B, T × B = −N, N1 B1 + N1 B2 + N1 B3 = 0),

these, together with Equation (11), complete the proof of the theorem.



As an immediate consequence we have Corollary 2.4. Let γ : I −→ (R3 , gSol ) be a non-geodesic curve parametrized by arc length. If B3 = 0, then γ is not biharmonic. Corollary 2.5. Let γ : I −→ (R3 , gSol ) be a non-geodesic curve parametrized by arc length. If B3 is constant and N3 B3 6= 0, then γ is not biharmonic.

Similar to the terminology used for curves in R3 , we keep the name helix for a curve in a Riemannian 3-manifold having constant both geodesic curvature and geodesic torsion. With this terminology, we can use Equation (10) to deduce the following Corollary 2.6. Let γ : I −→ (R3 , gSol ) be a non-geodesic biharmonic helix parametrized by arc length, then   B3 = constant 6= 0 (12) N3 = 0  2 2 k + τ = 2B32 − 1. Non-geodesic biharmonic helices in 3-dimensional sphere S 3 , in Heisenberg group H3 , and in Cartan-Vranceanu 3-dimensional space have been studied in [2], [3], and [4] respectively. In contrast to the situations in those 3-dimensional spaces where there are rich examples of such curves our next theorem shows that there exists no such curves in Sol space.

BIHARMONIC MAPS INTO SOL AND NIL SPACES

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Theorem 2.7. There exists no non-geodesic biharmonic helix in Sol space. Proof. Suppose that γ : I −→ (R3 , gSol ) is a non-geodesic biharmonic helix parametrized by arc length. We shall derive a contradiction by showing that γ must be a geodesic. We can use (3) to compute the covariant derivatives of the vector fields T, N, B as:  ∇T T = (T1′ + T1 T3 )e1 + (T2′ − T2 T3 )e2 + (T22 − T12 + T3′ )e3  (13) ∇ N = (N1′ + T1 N3 )e1 + (N2′ − T2 N3 )e2 + (T2 N2 − T1 N1 + N3′ )e3  T ∇T B = (B1′ + T1 B3 )e1 + (B2′ − T2 B3 )e2 + (T2 B2 − T1 B1 + B3′ )e3 . It follows that the third components of these vectors are given by  h∇T T, e3 i = T22 − T12 + T3′  (14) h∇T N, e3 i = T2 N2 − T1 N1 + N3′  h∇T B, e3 i = T2 B2 − T1 B1 + B3′ .

On the other hand, using Frenet formulas (9) we have,  h∇T T, e3 i = kN3  (15) h∇T N, e3 i = −kT3 + τ B3  h∇T B, e3 i = −τ N3 .

Since γ is assumed to be a non-geodesic biharmonic helix, we have, by Corollary 2.6, N3 = 0, B3 = constant. These, together with Equations (14) and (15), give

(16)

T22 − T12 + T3′ = 0 T N − T1 N1 = −kT3 + τ B3  2 2 T2 B2 − T1 B1 = 0.  

Noting that T × B = −N we also have (17)

T2 B1 − T1 B2 = N3 = 0.

Thus, we have

(18)

 T2 N2 − T1 N1 = −kT3 + τ B3 ,    T22 − T12 + T3′ = 0,  T2 B2 − T1 B1 = 0,   T2 B1 − T1 B2 = 0.

h1i h2i h3i h4i

Case A: T12 6= T22 . In this case, Equations h3i and h4i in System (18) viewed as equations in B1 and B2 has a unique solution B1 = B2 = 0. This implies that T3 = hN × B, e3 i = 0. Substitute this into h2i of System (18) we have T12 = T22 , a contradiction. Thus, we must have Case B: T12 = T22 . In this case, Equation h2i of System (18) implies that T3 = constant. To understand the meaning of this, we represent the unit tangent vector

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YE-LIN OU AND ZE-PING WANG

T as T = sin α cos β e1 + sin α sin β e2 + cos α e3 , where α = α(s), β = β(s). With this representation, T3 = constant implies that cos α = constant and hence α(s) = α0 , a constant. This, together with T12 = T22 , gives (19)

sin α0 (cos β ± sin β) = 0.

If sin α0 = 0, then we have T1 = T2 = 0, and it follows from the first equation of (13) that ∇T T = 0 which means that γ is a geodesic, a contradiction. Thus, we must have sin α0 6= 0, which, together with (19), implies that √ 2 , cos β = ± sin β = ± 2 and hence, √ 2 (20) T1 = ±T2 = ± sin α0 . 2 We use the first equation of (13) again to get √ √ 2 2 ∇T T = sin α0 cos α0 (± e1 ∓ e2 ) = kN 2 2 which yields √ √ 2 2 , N2 = ∓ (21) N1 = ± 2 2 since k = |∇T T | = | sin α0 cos α0 |. By the assumption that γ is non-geodesic, we may assume, without loss of generality, that sin α0 cos α0 > 0, so (22)

k = sin α0 cos α0 .

Using Equations (20), (21) and the fact that B = T × N we have (23)

B1 = constant, B2 = constant, B3 = T1 N2 − T2 N1 = ∓ sin α0 .

It follows from (23), h3i of (18), and the third equation of (13) that (24)

τ 2 = |∇T B|2 = (T12 + T22 )B32 = sin4 α0 .

Substituting (22), (23) and (24) into the third equation in (12) we have sin2 α0 cos2 α0 + sin4 α0 = 2 sin2 α0 − 1, which implies sin2 α0 = 1, and hence cos2 α0 = 0.

BIHARMONIC MAPS INTO SOL AND NIL SPACES

9

It follows that cos α0 = 0 from which and (22) we conclude that k = 0, i.e., γ is a geodesic, a contradiction. Combining the the results in Cases A and B we complete the proof of the theorem.  3. Linear biharmonic maps into Sol and Nil spaces In this section, we first derive the biharmonic map equation in local coordinates and then we use it to classify linear biharmonic maps from Euclidean space into Sol and Nil spaces. In the rest of the paper, we adopt the Einstein summation convention that a repeated upper and lower index means the summation of that index over its range that is understood from the context. 3.1 Biharmonic map equation in local coordinates. Lemma 3.1. Let ϕ : (M m , g) −→ (N n , h) with ϕ(x1 , . . . , xm ) = (ϕ1 (x), . . . , ϕn (x)) be a map between Riemannian manifolds. With respect to local coordinates (xi ) in M and (y α) in N, ϕ is biharmonic if and only if it is a solution of the following system of PDE’s α β ρ ¯ν ¯σ ¯σ ¯ σ + ∂ (τ α ϕβ Γ (25) g ij (τijσ + τjα ϕβi Γ αβ j αβ ) + τ ϕj ϕi Γαβ Γνρ i ∂x k σ α β ¯σ ¯ σ ) = 0, σ = 1, 2, . . . , n. −Γij (τk + τ ϕk Γαβ ) − τ ν ϕα i ϕβj R β αν ¯ σ ) denote the components of metric and Proof. Let gij and Γkij (resp. hαβ and Γ αβ the connection coefficients in the domain (resp. the target) manifold with respect to the chosen local coordinates and the natural frame { ∂x∂ i } (resp. { ∂y∂α }). The bitension field of ϕ can be computed as ∇ϕ∂ ∇ϕ∂ − ∇ϕ∇

τ 2 (ϕ) = g ij

∂xi

∂xj

∂ ∂ ∂xj ∂xi

!

¯ (τ (ϕ)) − Traceg R(dϕ, τ (ϕ))dϕ

= g ij ∇ϕ∂ ∇ϕ∂ (τ (ϕ)) − g ij Γkij ∇ϕ∂ (τ (ϕ)) ∂xi

∂xj

∂xk

= g

ij



∇ϕ∂ ∇ϕ∂ (τ (ϕ)) − Γkij ∇ϕ∂ ∂xi

∂xj

∂xk

A direct computation gives (27)

∂ ∂ ), τ (ϕ))dϕ( j ) i ∂x ∂x  ∂ ∂ α β ¯ (τ (ϕ)) − ϕ i ϕj R( α , τ (ϕ)) β . ∂y ∂y

¯ −g ij R(dϕ(

(26)

∇

ϕ ∂ ∂xk

  ∂ ∂ σ α β ¯σ ) = τk + τ ϕk Γαβ , (τ (ϕ)) = ∇ ∂ (τ ∂y ν ∂y σ ∂xk ϕ

ν

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YE-LIN OU AND ZE-PING WANG

  ∂ ∂ ¯α ∇ϕ∂ ∇ϕ∂ (τ (ϕ)) = ∇ϕ∂ ∇ϕ∂ (τ ν ν ) = ∇ϕ∂ τjα + τ ν ϕβj Γ βν ∂y ∂y α ∂xi ∂xi ∂xi ∂xj ∂xj   ∂ ∂ α β ¯σ α β ρ ¯ν ¯σ σ α β ¯σ (28) , = τij + τj ϕi Γαβ + i (τ ϕj Γαβ ) + τ ϕj ϕi Γαβ Γνρ ∂x ∂y σ and (29)

∂ ¯ ∂ , τ (ϕ)) ∂ = τ ν R ¯σ R( . β αν ∂y α ∂y β ∂y σ

Substitute Equations (27), (28), and (29) into (26) we have

∂ α β ¯σ ¯ν Γ ¯σ (τ ϕj Γαβ ) + τ α ϕβj ϕρi Γ αβ νρ ∂xi ∂ ¯ σ ) − τ ν ϕα i ϕβ R ¯σ , −Γkij (τkσ + τ α ϕβk Γ αβ j β αν ) ∂y σ

¯σ + τ 2 (ϕ) = g ij (τijσ + τjα ϕβi Γ αβ (30)

from which the lemma follows.



When the domain manifold is a Euclidean space then we have Corollary 3.2. Let ϕ : Rm −→ (N n , h) with ϕ(x1 , . . . , xm ) = (ϕ1 (x), . . . , ϕn (x)) be a map from a Euclidean space into a Riemannian manifold. Then ϕ is biharmonic if and only if it is a solution of the following system of PDE’s (31)

¯ σ + h∇ϕβ , ∇(τ α Γ ¯ σ )i ∆τ σ + h∇τ α , ∇ϕβ iΓ αβ αβ β ρ α ¯ν ¯σ ν α β ¯σ + h∇ϕ , ∇ϕ iτ Γ Γ − τ h∇ϕ , ∇ϕ iR αβ

νρ

β αν

= 0, σ = 1, 2, . . . , n.

Linear biharmonic maps into Sol space. Let (R3 , gSol ) denote Sol space, 3 3 where the metric can be written as gSol = e2y (dy 1 )2 + e−2y (dy 2 )2 + (dy 3)2 with respect to the standard coordinates (y 1 , y 2, y 3) in R3 . Then a direct computation gives the following components of Sol metric and the coefficients of the connection: 3

3

g11 = e2y , g22 = e−2y , g33 = 1, all other gij = 0; 3

3

g 11 = e−2y , g 22 = e2y , g 33 = 1, all other g ij = 0;

BIHARMONIC MAPS INTO SOL AND NIL SPACES

(32)

¯1 Γ 11 ¯1 Γ 12 ¯1 Γ 13 ¯1 Γ 21 ¯1 Γ 22 ¯ 123 Γ ¯ 131 Γ ¯1 Γ 32 ¯1 Γ 33

11

¯ 2 = 0, Γ ¯ 3 = −e2y3 ; =Γ 11 11 ¯ 2 = 0, Γ ¯ 3 = 0; = 0, Γ 12 12 ¯ 3 = 0; ¯ 2 = 0, Γ = 1, Γ 13 13 ¯ 2 = 0, Γ ¯ 3 = 0; = 0, Γ 21 21 ¯ 2 = 0, Γ ¯ 3 = e−2y3 ; =Γ 22 22 ¯ 223 = −1, Γ ¯ 323 = 0; = 0, Γ ¯ 231 = 0, Γ ¯ 331 = 0; = 1, Γ ¯ 2 = −1, Γ ¯ 3 = 0; = 0, Γ 32 32 ¯ 2 = 0, Γ ¯ 3 = 0. = 0, Γ 33 33

By our convention of curvature operator and the following notation for the components of the Riemannian curvature (33)

¯ ∂ , ∂ ) ∂ =R ¯l ∂ R( k ij i j k ∂y ∂y ∂y ∂y l

we have (34)

¯l = ∂ Γ ¯l − ∂ Γ ¯l + Γ ¯l Γ ¯t ¯l ¯t R k ij kj ki it kj − Γjt Γki i j ∂y ∂y

A straightforward computation using (32) and (34) gives the following components of the Riemannian curvature of Sol space:  ¯ 1 = −e−2y3 , R ¯ 1 = 1, R ¯ 1 = e−2y3 , R ¯ 1 = −1; R  221 331 212 313 3 3 ¯ 2 = e2y , R ¯ 2 = −e2y , R ¯ 2 = 1, R ¯ 2 = −1; (35) R 121 112 332 323  ¯3 3 ¯ 3 = −e−2y3 , R ¯ 3 = e2y3 , R ¯ 3 = e−2y3 . R131 = −e2y , R 232 113 223

Now we are ready to prove the following classification theorem for linear biharmonic maps into Sol space. Theorem 3.3. Let ϕ : Rm −→ (R3 , gSol ) with 

a11 a12 · · · a1m   ϕ(x) =  a21 a22 · · · a2m    a31 a32 · · · a3m 



x1 x2 .. . xm



  , 

i.e., ϕ(X) = (A1 X t , A2 X t , A3 X t ) be a linear map into Sol space, where Ai denotes the row vectors of the representation matrix. Then, ϕ is a biharmonic map if and only if it is a harmonic map, which is equivalent to either (i) A3 = 0, |A1 |2 = |A2 |2 , or, (ii) A3 6= 0, A1 = A2 = 0.

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YE-LIN OU AND ZE-PING WANG

Proof. With respect to the standard Cartesian coordinates (xi ) in Rm and (y α) in R3 , the tension field of ϕ is given by

(36)

τ (ϕ) := Traceg (∇dϕ) ∈ Γ(ϕ−1 T N) = τ σ ∂y∂σ ¯ σ ϕα ϕβ ) ∂ σ = g ij (ϕσij − Γkij ϕσk + Γ αβ i j ∂y m P β ¯ σ ϕα ϕ ) ∂ σ =( Γ αβ i i ∂y i=1

¯ σ Aα · Aβ ∂σ , =Γ αβ ∂y

where and in the sequel, Aα · Aβ denotes the inner product and |Aα | the norm of the vectors in Euclidean space. Putting τ (ϕ) = τ σ ∂y∂σ and substituting (32) into Equation (36) we find the following components of the tension field of ϕ

(37)

τ 1 = 2A1 · A3 , τ 2 = −2A2 · A3 , 3 3 τ 3 = |A2 |2 e−2y − |A1 |2 e2y ,

where and in the sequel y 3 = A3 X t . A further computation gives


∂ = ∂y σ
∂ ¯σ = ∆τ σ + Aβ · ∇τ α Γ αβ ∂y σ

¯σ ∆τ σ + h∇τ α , ∇ϕβ iΓ αβ

(38)

∂ ∂y 1 ∂ 3 3 +2A2 · A3 (|A2 |2 e−2y + |A1 |2 e2y ) 2 ∂y ∂ 3 3 +4|A3 |2 (|A2 |2 e−2y − |A1 |2 e2y ) 3 , ∂y 3

3

−2A1 · A3 (|A2 |2 e−2y + |A1 |2 e2y )

BIHARMONIC MAPS INTO SOL AND NIL SPACES

¯ σ )i h∇ϕβ , ∇(τ α Γ αβ

13

∂ ∂y σ

¯ σ + τ α Aβ · ∇Γ ¯σ ) ∂ = (Aβ · ∇τ α Γ αβ αβ ∂y σ (39)

3

3

= −2A1 · A3 (|A2 |2 e−2y + |A1 |2 e2y ) 3

3

+2A2 · A3 (|A2 |2 e−2y + |A1 |2 e2y ) 3

∂ ∂y 1

∂ ∂y 2 3

+[−4(A1 · A3 )2 e2y + 4(A2 · A3 )2 e−2y ]

(40)

∂ , ∂y 3

α β ρ ¯ν ¯σ ∂ ¯ν Γ ¯σ ∂ h∇ϕβ , ∇ϕρ iτ α Γ αβ νρ ∂y σ = τ A · A Γαβ Γνρ ∂y σ 3 3 = [2A1 · A3 |A3 |2 + A1 · A3 |A2 |2 e−2y − 3A1 · A3 |A1 |2 e2y 3 3 −2A2 · A3 A1 · A2 e−2y ] ∂y∂ 1 + [−2A2 · A3 |A3 |2 − A2 · A3 |A1 |2 e2y 3 3 3 +3A2 · A3 |A2 |2 e−2y + 2A1 · A3 A1 · A2 e2y ] ∂y∂ 2 + [−2(A1 · A3 )2 e2y 3 3 3 +|A1 |4 e4y + 2(A2 · A3 )2 e−2y − |A2 |4 e−4y ] ∂y∂ 3 ,

and

(41)

∂ ∂ β α ν ¯σ ¯σ τ ν h∇ϕα , ∇ϕβ iR β αν ∂y σ = A · A τ Rβ αν ∂y σ 3 3 = [−3A1 · A3 |A2 |2 e−2y − 2A2 · A3 A1 · A2 e−2y 3 3 +A1 · A3 |A1 |2 e2y + 2A1 · A3 |A3 |2 ] ∂y∂ 1 + [2A1 · A3 A1 · A2 e2y 3 3 +3A2 · A3 |A1 |2 e2y − A2 · A3 |A2 |2 e−2y − 2A2 · A3 |A3 |2 ] ∂y∂ 2 3 3 3 3 +[−|A1 |4 e4y + |A2 |4 e−4y − 2(A1 · A3 )2 e2y + 2(A2 · A3 )2 e−2y )] ∂y∂ 3 .

It follows from Equations (38), (39), (40), (41) and Corollary 3.2 that the linear map ϕ is a biharmonic map if and only if  3  −8A1 · A3 |A1 |2 e2y = 0   3  8A2 · A3 |A2 |2 e−2y = 0 (42) 3 3  4(|A2 |2 |A3 |2 + (A2 · A3 )2 )e−2y − 4(|A1 |2 |A3 |2 + (A1 · A3 )2 )e2y   3 3  +2|A1 |4 e4y − 2|A2 |4 e−4y = 0.

Solving System of equations (42) we have either (i) A3 = 0, |A1 |2 = |A2 |2 , or (ii) A3 6= 0, A1 = A2 = 0. It follows from Equation (37) that in both cases the tension field vanishes identically, i.e., ϕ is also harmonic. Therefore, we obtain the theorem.  Linear biharmonic maps into Nil space. Let (R3 , gN il ) denote Nil space, where the metric with respect to the standard coordinates (y 1, y 2 , y 3) in R3 can

14

YE-LIN OU AND ZE-PING WANG

be written as gN il = (dy 1 )2 + (dy 2)2 + (dy 3 − y 1dy 2 )2 . Then an easy computation gives the following components of Nil metric and the coefficients of its connection: g11 = 1, g12 = g13 = 0, g22 = 1 + (y 1)2 , g23 = −y 1 , g33 = 1; g 11 = 1, g 12 = g 13 = 0, g 22 = 1, g 23 = y 1 , g 33 = 1 + (y 1)2 ;

(43)

¯1 Γ 11 1 ¯ Γ12 ¯1 Γ 13 ¯1 Γ 21 ¯ 122 Γ ¯1 Γ 23 ¯1 Γ 31 ¯1 Γ 32 ¯1 Γ 33

¯ 2 = 0, Γ ¯ 3 = 0; =Γ 11 11 1 2 y1 ¯ 3 2 ¯ = 0, Γ12 = 2 , Γ12 = (y )2 −1 ; ¯ 3 = − y1 ; ¯2 = − 1 , Γ = 0, Γ 13 13 2 2 ¯ 2 = y1 , Γ ¯ 3 = (y1 )2 −1 ; = 0, Γ 21 21 2 2 ¯ 222 = 0, Γ ¯ 322 = 0; = −y 1 , Γ ¯ 2 = 0, Γ ¯ 3 = 0; = 12 , Γ 23 23 ¯ 3 = − y1 ; ¯2 = − 1 , Γ = 0, Γ 31 31 2 2 ¯ 2 = 0, Γ ¯ 3 = 0; = 12 , Γ 32 32 ¯ 2 = 0, Γ ¯ 3 = 0. = 0, Γ 33 33

A further computation using (34) and (43) gives the following components of the Riemannian curvature of Nil space:  ¯ 1 = − y1 , R ¯ 1 = 3 − (y1 )2 , R ¯ 1 = − y1 , ¯ 1 = − 3 + (y1 )2 , R  R  213 221 312 212 4 4 4 4 4 4   y1 ¯ 1 y1 ¯ 1 1 ¯1 1  1 ¯  R = , R = , R = , R = − ;  231 313 321 331 4 4 4 4   y1 3 ¯2 3 ¯2 2 ¯ R112 = 4 , R121 = − 4 , R223 = − 4 , (44) ¯ 2 = y1 , R ¯2 = 1 , R ¯2 = − 1 ;  R  232 323 332 4 4 4   ¯ 3 = −y 1 , R ¯3 = 1 , ¯ 3 = y1, R ¯3 = − 1 , R  R  121 131 112 113 4 4    ¯ 3 = − (y1 )2 +1 , R ¯ 3 = (y1 )2 +1 , R ¯ 3 = y1 , R ¯ 3 = − y1 . R 223 232 323 332 4 4 4 4

Theorem 3.4. Let ϕ : Rm −→ (R3 , gN il ) with 

ϕ(x) = 

a11 a21 a31

a12 a22 a32

··· ··· ···

a1m a2m a3m





   

x1 x2 .. . xm



  , 

i.e., ϕ(X) = (A1 X t , A2 X t , A3 X t ) be a linear map into Nil space, where Ai denotes the row vectors of the representation matrix. Then, ϕ is a biharmonic map if and only if it is a harmonic map, which is equivalent to either (i) A1 = 0, A2 · A3 = 0, or, (ii) A1 = 0, A2 = 0, or, (iii) A2 = 0, A1 · A3 = 0. Proof. Taking the standard Cartesian coordinates (xi ) in Rm , (y α) in R3 and substituting (43) into (36) we find the tension field of ϕ to be

BIHARMONIC MAPS INTO SOL AND NIL SPACES

(45)

τ (ϕ) = τ σ ∂y∂σ = [−|A2 |2 y 1 + A2 · A3 ] ∂y∂ 1 +[A1 · A2 y 1 − A1 · A3 ] ∂y∂ 2 +[A1 · A2 (y 1 )2 − A1 · A3 y 1 − A1 · A2 ] ∂y∂ 3 ,

or,

(46)

τ 1 = −|A2 |2 y 1 + A2 · A3 , τ 2 = A1 · A2 y 1 − A1 · A3 , τ 3 = A1 · A2 (y 1)2 − A1 · A3 y 1 − A1 · A2 ,

where and in the sequel y 1 = A1 X t . A straightforward computation yields (47)

∆τ σ ∂y∂σ = 2A1 · A2 |A1 |2 ∂y∂ 3 ,

(48)

¯ σ ∂σ = Aβ · ∇τ α Γ ¯ σ ∂σ h∇τ α , ∇ϕβ iΓ αβ ∂y αβ ∂y 1 1 2 1 2 2 2 1 = [− 2 A · A (|A | + |A | )y + 21 A1 · A3 (|A1 |2 + |A2 |2 )] ∂y∂ 2 +[− 21 A1 · A2 (|A1 |2 + |A2 |2 )(y 1)2 + 21 A1 · A3 (|A1 |2 + |A2 |2 )y 1 + 12 A1 · A2 (|A2 |2 − |A1 |2 )] ∂y∂ 3 ,

(49)

¯ σ )i ∂σ h∇ϕβ , ∇(τ α Γ αβ ∂y ¯ σ + τ α Aβ · ∇Γ ¯ σ ) ∂σ = (Aβ · ∇τ α Γ αβ αβ ∂y = [−(A1 · A2 )2 y 1 + (A1 · A2 )(A1 · A3 )] ∂y∂ 1 +[−A1 · A2 |A2 |2 y 1 + 21 (A1 · A2 )(A2 · A3 ) + 12 A1 · A3 |A2 |2 ] ∂y∂ 2 +[− 23 A1 · A2 |A2 |2 (y 1)2 + (A1 · A2 )(A2 · A3 )y 1 +A1 · A3 |A2 |2 y 1 + 21 A1 · A2 |A2 |2 − 12 (A1 · A3 )(A2 · A3 )] ∂y∂ 3 ,

15

16

(50)

YE-LIN OU AND ZE-PING WANG

¯ν Γ ¯σ ∂ h∇ϕβ , ∇ϕρ iτ α Γ αβ νρ ∂y σ ¯ν Γ ¯σ ∂ τ α Aβ · Aρ Γ αβ νρ ∂y σ = [ 41 |A2 |4 (y 1 )3 − 43 A2 · A3 |A2 |2 (y 1)2 + 12 (A2 · A3 )2 y 1 − 21 (A1 · A2 )2 y 1 + 14 |A2 |2 |A3 |2 y 1 + 41 |A2 |4 y 1 + 12 (A1 · A2 )(A1 · A3 ) − 14 A2 · A3 |A3 |2 − 41 A2 · A3 |A2 |2 ] ∂y∂ 1 +[− 41 A1 · A2 |A2 |2 (y 1 )3 + 14 A1 · A3 |A2 |2 (y 1)2 + 12 (A2 · A3 )(A1 · A2 )(y 1 )2 − 21 A1 · A2 |A2 |2 y 1 + 14 A1 · A2 |A1 |2 y 1 − 14 A1 · A2 |A3 |2 y 1 − 21 (A1 · A3 )(A2 · A3 )y 1 + 12 (A1 · A2 )(A2 · A3 ) + 14 A1 · A3 |A3 |2 − 14 A1 · A3 |A1 |2 ] ∂y∂ 2 +[− 14 A1 · A2 |A2 |2 (y 1 )4 + 14 A1 · A3 |A2 |2 (y 1)3 + 12 (A2 · A3 )(A1 · A2 )(y 1)3 + 14 A1 · A2 |A1 |2 (y 1)2 − 12 (A2 · A3 )(A1 · A3 )(y 1 )2 − 41 A1 · A2 |A3 |2 (y 1)2 − 12 A1 · A3 |A2 |2 y 1 + 14 A1 · A3 |A3 |2 y 1 − 41 A1 · A3 |A1 |2 y 1 + 12 (A2 · A3 )(A1 · A3 ) + 14 A1 · A2 |A2 |2 − 14 A1 · A2 |A1 |2 ] ∂y∂ 3 ,

and

(51)

∂ ∂ β α ν ¯σ ¯σ τ ν h∇ϕα , ∇ϕβ iR β αν ∂y σ = A · A τ Rβ αν ∂y σ = [ 41 |A2 |4 (y 1)3 − 43 A2 · A3 |A2 |2 (y 1)2 − 21 (A1 · A2 )2 y 1 + 12 (A2 · A3 )2 y 1 − 34 |A2 |4 y 1 + 41 |A2 |2 |A3 |2 y 1 + 12 (A1 · A2 )(A1 · A3 ) + 43 A2 · A3 |A2 |2 − 14 A2 · A3 |A3 |2 ] ∂y∂ 1 + [− 14 A1 · A2 |A2 |2 (y 1)3 + 41 A1 · A3 |A2 |2 (y 1 )2 + 12 (A2 · A3 )(A1 · A2 )(y 1 )2 + 43 A1 · A2 |A1 |2 y 1 + A1 · A2 |A2 |2 y 1 − 12 (A2 · A3 )(A1 · A3 )y 1 − 41 A1 · A2 |A3 |2 y 1 − 43 A1 · A3 |A1 |2 −(A1 · A2 )(A2 · A3 ) + 41 A1 · A3 |A3 |2 ] ∂y∂ 2 + [− 41 A1 · A2 |A2 |2 (y 1)4 + 14 A1 · A3 |A2 |2 (y 1)3 + 12 (A2 · A3 )(A1 · A2 )(y 1)3 + 43 A1 · A2 |A1 |2 (y 1)2 +A1 · A2 |A2 |2 (y 1)2 − 21 (A2 · A3 )(A1 · A3 )(y 1 )2 − 14 A1 · A2 |A3 |2 (y 1)2 − 43 A1 · A3 |A1 |2 y 1 − (A1 · A2 )(A2 · A3 )y 1 + 41 A1 · A3 |A3 |2 y 1 + 41 A1 · A2 |A1 |2 + 41 A1 · A2 |A2 |2 ] ∂y∂ 3 .

By Equations (47), (48), (49), (50), (51) and Corollary 3.2 we conclude that ϕ is biharmonic if and only if

(52)

−(A1 · A2 )2 y 1 + |A2 |4 y 1 +(A1 · A2 )(A1 · A3 ) − A2 · A3 |A2 |2 = 0,

(53)

−A1 · A2 |A1 |2 y 1 − 3A1 · A2 |A2 |2 y 1 +A1 · A3 |A2 |2 + A1 · A3 |A1 |2 +2(A1 · A2 )(A2 · A3 ) = 0,

BIHARMONIC MAPS INTO SOL AND NIL SPACES

17

and (54)

−A1 · A2 |A1 |2 (y 1)2 − 3A1 · A2 |A2 |2 (y 1)2 + A1 · A3 |A1 |2 y 1 +A1 · A3 |A2 |2 y 1 + 2(A1 · A2 )(A2 · A3 )y 1 +A1 · A2 |A1 |2 + A1 · A2 |A2 |2 = 0.

To solve the System of biharmonic map equations (52), (53), and (54) we consider the following two cases. Case I: A1 = 0. Noting that y 1 = A1 X t = 0 we use Equation (52) to have − A2 · A3 |A2 |2 = 0,

which implies either A2 · A3 = 0, or |A2 |2 = 0 i.e. A2 = 0. Substituting A1 = 0, A2 · A3 = 0 or A1 = 0, A2 = 0 into (53) and (54) we see that they are both solutions of biharmonic map equations. Case II: A1 6= 0. Note that in this case, y 1 = A1 X t 6= 0. We can view system of biharmonic map equations (52), (53), and (54) as a system of polynomial equations in y 1. It is not difficult to check that A2 = 0, A1 · A3 = 0 is the only solution in this case. Combining Case I and II we obtain the last statement of the theorem. A direct checking using the tension field equation (45) we see that all these maps are also harmonic maps. Thus, we complete the proof of the theorem.  References [1] A. Balmus, On the biharmonic curves of the Euclidian and Berger 3-dimensional spheres, Sci. Ann. Univ. Agric. Sci. Vet. Med., 47, 87–96. [2] R. Caddeo, S. Montaldo, and C. Oniciuc, Biharmonic submanifolds of S 3 , Int. J. Math., 12, 867–876. [3] R. Caddeo, C. Oniciuc, and P. Piu, Explicit formual for non-geodesic biharmonic curves of the Heisenberg group, Rend. Sem. Mat. Univ. Politec. Torino, 62 (2004) 265–278. [4] R. Caddeo, S. Montaldo, C. Oniciuc, and P.Piu, The classfication of biharmonic curves of Cartan-Vranceanu 3-dimensional space, arXiv: math. DG/0510435 vl 20 Oct 2005. [5] J. T. Cho, J. Inoguchi, and J. E. Lee, Biharmonic curves in 3-dimensional Saskian space forms, Preprint. [6] N. Ekmekci and N. Yaz, Biharmonic general helices in contact and Sasakian manifolds, Tensor (N.S.), 65, 103–108. [7] J. -I. Inoguchi, Biharmonic curves in Minkowski 3-space, Int. J. Math. Math. Sci., 21, 1365–1368. [8] E. Loubeau, S. Montaldo, and C. Oniciuc, The bibliograph of biharmonic maps, http://beltrami.sc.unica.it/biharmonic/

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[9] E. Loubeau and Y. -L. Ou, Biharmonic maps and morphisms from conformal mappings, preprint, 2006. [10] S. Montaldo and C. Oniciuc, A short survey on biharmonic maps between Riemannian manifolds, preprint, http://arxiv.org/abs/math/0510636. [11] Y. -L. Ou, p-Harmonic morphisms, biharmonic morphisms, and nonharmonic biharmonic maps , J. Geom. Phys. 56(2006) 358-374. [12] D. Laugwitz, Differential and Riemannian geometry , Academic Press, 1965. [13] G. Y. Jiang, 2-Harmonic maps and their first and second variational formulas, Chin. Ann. Math. Ser. A 7(1986) 389-402. Department of Mathematics, Texas A & M University-Commerce, Commerce TX 75429, USA. E-mail:yelin [email protected] (Ou) School of Math and Computer Science, Guangxi University for Nationalities, Nanning 530006, P. R. China.