2-harmonic maps and their first and second variational formulas i

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1 Note di atematica Note at. 1(2008, suppl. n. 1, ISSN , e-issn DOI /i v28n1supplp209 Note di atematica 28, suppl. n. 1, 2009, harmonic maps and their first and second variational formulas i Jiang Guoying Institute of athematics, Fudan University Received: 22/8/07; accepted: 5/10/07. Abstract. In [1], J. Eells and L. Lemaire introduced the notion of a k-harmonic map. In this paper we study the case k = 2, derive the first and second variational formulas of the 2-harmonic maps, give nontrivial examples of 2-harmonic maps and give proofs of nonexistence theorems of stable 2-harmonic maps. Keywords: harmonic maps, biharmonic maps, second variation formula SC 2000 classification: primary 58E20, secondary 53C43 Introduction As well known, harmonic maps between Riemannian manifolds f : N, where is compact, can be considered as critical maps of the energy functional E(f = df 2 1. Considering the similar ideas, in 1981, J. Eells and L. Lemaire [1], proposed the problem to consider the k-harmonic maps: critical maps of the functional E k (f = (d + d k f 2 1. In this paper, we consider the case k = 2 and show the preliminary results. We use mainly vector bundle valued differential forms and Riemannian metrics. In 1, we prepare the notation and fundamental formulas needed in the sequel. In 2, given a compact manifold, we derive the first variation formula of E 2 (f = (d + d 2 f 2 1 (Theorem 1 and give the definition of 2-harmonic maps f : N whose tension field τ(f satisfies τ(f+r N (df (e k,τ(fdf (e k =0, namely τ(f is a solution of the Jacobi type equation. i Reprinted from Chinese Ann. ath., Ser. A, 7(1986, , upon formal consent by Li Ta-Tsien (Editor-in-Chief of Chinese Ann. ath.. Translated into English by Hajime Urakawa.

2 210 J. Guoying Constant maps and harmonic maps are trivial examples of 2-harmonic maps. The main results of 3 are to give nontrivial examples of 2-harmonic maps. We consider Riemannian isometric immersions. For the isometric immersions with parallel mean curvature tensor field, we give the decomposition formula (Lemma 4 of the Laplacian of the tension field τ(f with its proof: for the hypersurfaces with non-zero parallel mean curvature tensor field in the unit sphere S m+1, a necessary and sufficient condition for such isometric immersions to be 2-harmonic is that the square of the length of the second fundamental form B(f satisfies B(f 2 = m. Using this, special Clifford tori in the unit sphere whose Gauss maps are studied by Y.L. Xin and Q. Chen [2], give non-trivial 2-harmonic maps which are isometric immersions in the unit sphere. In 4, using formulas in 2, we derive the second variation formula of 2- harmonic maps (Theorem 3 and give the definition of stability of 2-harmonic maps (the second variation is nonnegative and give a proof of the following (Theorem 4: if is compact, and N has positive constant sectional curvature, there are no nontrivial 2-harmonic maps from into N satisfying the conservation law. Last, when N = CP n we establish nonexistence results of stable 2-harmonic maps (Lemma 8, Theorem 5, etc.. Furthermore, we give a nonexistence theorem establishing sufficient conditions that stable 2-harmonic maps be harmonic. Acknowledgements. We would like to express our gratitude to Professors Su Bu-Chin and Hu He-Shen who introduced and helped to accomplish this paper. We also would like to express our thanks to Professors Shen Chun-Li, Xin Yuan-Long and Pan Yang-Lian who helped us during the period of our study. 1 Notation, and fundamental notions We prepare the main materials using vector bundle valued differential forms and Riemannian metrics on bundles which are in [1, 3]. Assume that (,g is a m-dimensional Riemannian manifold, (N,h a n- dimensional one, and f : N a C map. Given points p and f(p N, under (x i, (y α local coordinates around them, f canbeexpressedas where the indices we use run as follows y α = f α (x i, (1 i, j, k, =1,,m; α, β, γ, =1,,n.

3 2-harmonic maps 211 We use the following definition: the differential df of f can be regarded as the induced bundle f 1 TN-valued 1-form df (X =f X, X Γ(T. (2 We denote by f h the first fundamental form of f, which is a section of the symmetric bilinear tensor bundle 2 T ; the second fundamental form B(f off is the covariant derivative df of the 1-form df, which is a section of 2 T f 1 TN: X, Y Γ(T: B(f(X, Y =( df (X, Y =( X df (Y = = X df (Y df ( X Y = = df (X df (Y df ( XY. (3 Here,,, are the Riemannian connections on the bundles T, TN, f 1 TN and T f 1 TN, respectively. From df, by using a local orthonormal frame field {e i } on, one obtains the tension field τ(f off τ(f =( df (e i,e i =( ei df (e i. (4 In the following, we use the above notations without comments, and we assume the reader is familiar with the above notation. We say f is a harmonic map if τ(f =0.If is compact, we consider critical maps of the energy functional E(f = df 2 1, (5 where 1 2 df 2 = 1 2 df (e i,df(e i N = e(f which is called the energy density of f, and the inner product, N is a Riemannian metric h, and we omit the subscript N if there is no confusion. When f is an isometric immersion, 1 m τ(fisthemean curvature normal vector field and harmonic maps are minimal immersions. The curvature tensor field R(, of the Riemannian metric on the bundle T f 1 TN is defined as follows X, Y Γ(T: R(X, Y = X Y + Y X + [X,Y ]. (6 Furthermore, for any Z Γ(T, we define ( R(X, Y df (Z =R f 1 TN (X, Y df (Z df (R (X, Y Z =

4 212 J. Guoying = R N (df (X,df(Y df (Z df (R (X, Y Z, (7 where R, R N,andR f 1 TN are the Riemannian curvature tensor fields on T, TN, f 1 TN, respectively. For 1-forms df the Weitzenböck formula is given by df = df + S, (8 where = dd +d d is the Hodge-Laplace operator, = ek ek ek e k is the rough Laplacian, and the operator S is defined as follows X Γ(T: S(X = ( R(e k,xdf (e k, (9 where {e k } is a locally defined orthonormal frame field on. A section of 2 T defined by S f = e(fg f h is called the stress-energy tensor field, and f is said to satisfy the conservation law if divs f =0.Asin[1], X Γ(T, it holds that (divs f (X = τ(f,df(x. (10 aps satisfying the conservation law are said to be relatively harmonic ([6]. 2 The first variation formula of 2-harmonic maps Assume that f : N is a C map, is a compact Riemannian manifold, and N is an arbitrary Riemannian manifold. As in [1], a 2-harmonic map is a critical map of the functional E 2 (f = (d + d 2 f 2 1. (11 Here, d and d are the exterior differentiation and the codifferentiation on vector bundle, and 1 isthevolumeformon. In order to derive the analytic condition of the 2-harmonic maps, we have to calculate the first variation of E 2 (f defined by (11. To start with let f t : N, t I ɛ =( ɛ, ɛ, ɛ > 0, (12 be a smooth 1-parameter variation of f which yields a vector field V Γ(f 1 TN along f in N by f t f 0 = f, = V. (13 t=0

5 2-harmonic maps 213 Variation {f t } yields a C map F : I ɛ N, F (p, t =f t (p, p, t I ɛ. (14 If we take the local coordinates around p, f t (p N, respectively, we have y α = F α (x i,t=f α t (x i. (15 Taking the usual Euclidean metric on I ɛ, with respect to the product Riemannian metric on I ɛ,wedenoteby,,, the induced Riemann connections on T ( I ɛ, F 1 TN, T ( I ɛ F 1 TN, respectively. If {e i } is an orthonormal frame field defined on a neighborhood U of p, { e i, } is also an orthonormal frame field on a coordinate neighborhood U I ɛ in I ɛ,and it holds that =0, e i e j = ei e j, e i = ei =0. (16 It also holds that f t = Fα y α = df,df t (e i =df (e i, (17 and ( ei df t (e j = df t(e i df t(e j df t ( ei e j =( ei df (e j ( ek ei df t (e j = df t(e k (( ei df t (e j ( ei df t ( ek e j =( ek ei df (e j (18 etc. Here, we used the abbreviated symbol on T ft 1 TN in which we omitted t. In the following, we need two lemmas to calculate the first variation of E 2 (f. d dt E 2(f t t=0 1 Lemma. Under the above notation, for any C variation {f t } of f, it holds that d dt E 2(f t = 2 ( ei ei df +2 R N ( df (e i,df ( ( ei e i df (, ( ej df (e j df (e i, ( ej df (e j 1 1. (19

6 214 J. Guoying Proof. By using d, d, and definition of τ(f, (11 can be written as E 2 (f = = d df 2 1= By noting (18, for variation f t of f, it holds that τ(f 2 1 ( ei df (e i, ei df (e i 1. (20 d dt E 2(f t = d dt ei df (e i, ( ej df (e j 1 =2 (( ei df (e i, ( ej df (e j 1. (21 By (16, and using the curvature tensor on T ( I ɛ F 1 TN, ( ( R X, ( df (Y =R N df (X,dF ( df (R Iɛ = R N (df (X,dF df (Y X, Y df (Y, (22 ( for all X, Y Γ(T. In (21, interchanging the order of differentiations in (( ei df (e i, we have (( ei df (e i = ( ei df (e i ( = ei df [ei, ]df + R = ei (( df (e i ( df ( ei e i + R (df N (e i,df df (e i =( ei ei df ( ei e i df + R N (df (e i,df ( ( e i, df (e i df (e i. (23 In the last of the above, we used the symmetry of the second fundamental form.

7 2-harmonic maps 215 By substituting (23 1 into (21 we obtain (19. 2 Lemma. ( ei ei df ( = df ( ei e i df, ( ej df (e j 1, ek ek (( ej df (e j ek e k (( ej df (e j 1. (24 Proof. For each t I ɛ, let us define a C vector field on by X = ( ei df, ( ej df (e j e i, (25 which is well defined because of the independence on a choice of {e i }.The divergence of X is given by divx = ek X, e k = ei ( ei df, ( ej df (e j + ( ei df, ( ej df (e j ek e i,e k. (26 1 Translator s comments: to get the last equation of (23, we have to see that ««««( e ei df = ei df df ei «= df (ei df = df ( df (ei [df (e i,df ( ] =( e df (e i, and by a similar way, Thus, «( e df ( ei e i=( e ei e i df. e ei (( e df (e i ( e df ( ei e i coincides with «««ei ( e ei df ( e ei e i df «=( e ei ei e df +( e ei df which implies (23. =( e ei e ei df «( e ei e i df ei «( e ei e i df «, «

8 216 J. Guoying Noticing (16 and we have (28 2 : ek e i,e k + e i, ek e k =0, (27 div(x = ( ei ei df, ( ej df (e j + ( ei df, ei (( ej df (e j ek e df, ( k ej df (e j. (28 Furthermore, let us define a C vector field Y on by Y = df, ei (( ej df (e j e i, (29 which is also well defined. Then, by a similar way, we have divy = ek Y,e k = ( ek df (, ek (( ej df (e j 2 Translator s comments: the first term of (26 coincides with ei ( e ei df (, ( e ej df (e j and the second term of (26 coincides with + ( e ei df (, e i (( e ej df (e j = e ei e ei df ( +(e ei df ( ei, ( e ej df (e j + ( e ei df (, e i (( e df (e j = e ei e ei df (, ( e ej df (e j + ( e ei df (, e i (( e df (e j, ( e ei df (, ( e ej df (e j ek e i,e k Thus, we have (28. = ( e ei df (, ( e ej df (e j e i, ek e k = ( e ek e k df (, ( e ej df (e j.

9 2-harmonic maps df df (, ek ek (( ej df (e j, ek e k (( ej df (e j. (30 By the Green s theorem, we have div(x Y 1=0, (31 and together with (28 and (30, we have (24. 3 Theorem. Assume that f : N is a C map from a compact Riemannian manifold into an arbitrary Riemannian manifold N, {f t } is an arbitrary C variation generating V.Then, d dt E 2(f t = t=0 =2 V, τ(f+r N (df (e i,τ(fdf (e i 1. (32 Proof. Substituting (24 into (19, we have d dt E 2(f t =2 df, ek ek (( ej df (e j ek e k (( ej df (e j 1 +2 R (df N (e i,df df (e i, ( ej df (e j 1, (33 where putting t = 0, noticing (13, (17, (18, and the symmetry of the curvature tensor, we have (32. Here, we used the explicit formula of the rough Laplacian on f 1 TN,thatis = ek ek ek e k. 4Remark. In the above arguments, we assumed is a compact Riemannian manifold without boundary. For a general Riemannian manifold, let D be an arbitrarily bounded domain with smooth boundary, and take a variation {f t } of f satisfying that f t =0, D f t ei =0, D then, in Lemma 2, we obtain (24 by applying the Green s divergence theorem to X and Y. Then, we have the first variational formula on D as d dt E 2(f t, D =2 V, τ(f+r N (df (e i,τ(fdf (e i 1, t=0 D where E 2 (f t, D is the corresponding functional relative to D.

10 218 J. Guoying 5 Definition. For a C map f : N between two Riemannian manifolds, let us define the 2-tension field τ 2 (f off by τ 2 (f = τ(f+r N (df (e i,τ(fdf (e i. (34 f is said to be a 2-harmonic map is if τ 2 (f =0. The C function is called the 2-energy density, and 1 2 E 2(f = e 2 (f = 1 2 (d + d 2 f 2 = 1 2 τ(f 2, (35 e 2 (f 1 < +, is the 2-energy of f. If is compact, by the first variational formula, a 2- harmonic map f is a critical point of the 2-energy. 3 Examples of 2-harmonic maps By Definition 1, we have immediately 6Proposition. (1 Any harmonic map is 2-harmonic. (2 Any doubly harmonic function f : R on a Riemannian manifold is also 2-harmonic. 7Proposition. Assume that is compact and N has non positive curvature, i.e. Riem N 0. Then every 2-harmonic map f : N is harmonic. Proof. Computing the Laplacian of the 2-energy density e 2 (f wehave Taking e 2 (f = 1 2 τ(f 2 = ek τ(f, ek τ(f + τ(f,τ(f. (36 τ 2 (f = τ(f+r N (df (e i,τ(fdf (e i =0, and noticing Riem N 0, we have e 2 (f = ek τ(f, ek τ(f R N (df (e i,τ(fdf (e i,τ(f 0. (37 By the Green s theorem e 2(fv g = 0, and (37, we have e 2 (f =0,so that e 2 (f = 1 2 τ(f 2 is constant. Again, by (37, we have ek τ(f =0, k =1,,m.

11 2-harmonic maps 219 Therefore, by [1], we have 3 τ(f =0. 8Remark. As we know nonexistence of compact minimal submanifolds in the Euclidean space, Proposition 2 shows nonexistence of 2-harmonic isometric immersions from compact Riemannian manifolds. By Proposition 1 harmonic maps are trivial examples of 2-harmonic ones, and in Proposition 2 in the case that is compact and the sectional curvature of N does not have nonpositive curvature, one may ask examples of nontrivial 2-harmonic maps. To do it, the following lemmas complete this. 9 Lemma. Assume that f : N is a Riemannian isometric immersion whose mean curvature vector field is parallel. Then, for a locally defined orthonormal frame field {e i }, we have τ(f = τ(f,df(e i df (e i + ei τ(f,df(e j ( ei df (e j. (38 Proof. Since f is an isometric immersion, df (e i span the tangent space of f( N. Since 4 the mean curvature tensor is parallel, for all i =1,,m, ei τ(f Γ(f T. Thus ei τ(f = ei τ(f,df(e j df (e j. (39 3 Translator s comments: since e 2(f = 0, both terms of (37 are non negative, we have ek τ(f, ek τ(f = 0, i.e., ek τ(f = 0 for all k =1,,m. We can define a global vector field X f = df (e i,τ(f e i X(, whose divergence is given as Integrating this over, wehave Z 0= div(x f = τ(f,τ(f + df (e i, ei τ(f = τ(f,τ(f. Z div(x f v g = which implies τ(f =0. 4 Translator s comments: for all ξ Γ(T N, τ(f,τ(f v g, Xξ = f Xξ = T f Xξ + f Xξ T + T, respectively. The condition that the mean curvature tensor is parallel means that f Xτ(f =0, X X(, which is equivalent to the condition that Xτ(f = T f Xτ(f Γ(f T.

12 220 J. Guoying Calculating this, we have τ(f = τ(f,df(e j df (e j + ei τ(f, ei df (e j df (e j + ei τ(f,df(e j ei df (e j. (40 Here, if we denote ei e j =Γ k ij e k,wehaveγ j ki +Γi kj = e k e i,e j =0.Since ( ei df (e j = ei (df (e j df ( ei e j T TN, and ei τ(f f (T, the second term of (40 is ei τ(f, ei df (e j df (e j = ei τ(f, ( ei df (e j +df ( ei e j df (e j = ei τ(f,df( ei e j df (e j = ei τ(f,df(e k df (Γ k ije j = ei τ(f,df(e k df ( Γ j ik e j = ei τ(f,df(e k df ( ei e k. (41 Substituting (41 into (40, we have ( Lemma. For an isometric immersion f : N with parallel mean curvature vector field, the Laplacian of τ(f is decomposed into: τ(f = τ(f,r N (df (e k,df(e j df (e k df (e j τ(f, ( ei df (e j ( ei df (e j. (42 Proof. Calculate the right hand side of (38. By differentiating by we have e i τ(f,df(e j =0, ei τ(f,df(e j + τ(f, ei df (e j = e i τ(f,df(e j =0. (43 5 Translator s comments: for (38, we only have to see by (3, which is (38. τ(f = τ(f,df(e i df (e i + ei τ(f,df(e j { ei df (e j df ( ei e j} = τ(f,df(e i df (e i + ei τ(f,df(e j ( e ei df (e j

13 2-harmonic maps 221 Then, we have ei τ(f,df(e j = τ(f, ei df (e j = τ(f, ei df (e j df ( ei e j = τ(f, ( ei df (e j. (44 For the first term of the RHS of (38, by differentiating (43 by e i,wehave We also have ei ei τ(f,df(e j +2 ei τ(f, ei df (e j + τ(f, ei ei df (e j =0. (45 ei e i τ(f,df(e j + τ(f, ei e i df (e j = ei e i τ(f,df(e j =0. (46 Together with (45 and (46, we have τ(f,df(e j +2 ei τ(f, ei df (e j + τ(f, df (e j =0. (47 For the second term of (47, by making use of the fact that ei τ(f Γ(f T from the assumption that the mean curvature tensor is parallel, and (44, we have ei τ(f, ei df (e j = ei τ(f, ( ei df (e j +df ( ei e j = ei τ(f,df( ei e j For the third term of (47, we have = τ(f, ( ei df ( ei e j. (48 τ(f, df (e j = τ(f, ek ek df (e j ek df (e j = τ(f, ek (( ek df (e j +df ( ek e j ( ek e k df (e j df ( ek e k e j = τ(f, ( ek ek df (e j +2( ek df ( ek e j ( ek e k e j = τ(f, ( df (ej +2 τ(f, ( ek df ( ek e j = τ(f, df (e j +S(e j +2 τ(f, ( ek df ( ek e j

14 222 J. Guoying since τ(f,df(x =0forallX X( and Weitzenböck formula (8. Here, we have df (e j = dd df (e j =dτ(f(e j = ej df, and by (9 and (7, S(e j = Thus, we have = m ( R(e k,e j df (e k k=1 m {R N (df (e k,df(e j df (e k df (R (e k,e j e k }. k=1 τ(f, df (e j =2 τ(f, ( ek df ( ek e j + τ(f, ej τ(f R N (df (e k,df(e j df (e k +df (R (e k,e j e k =2 τ(f, ( ek df ( ek e j τ(f,r N (df (e k,df(e j df (e k (49 since ej τ(f Γ(f T. Substituting (48 and (49 into (47, we have τ(f,df(e j = τ(f,r N (df (e k,df(e j df (e k. (50 Finally, substituting (44 and (50 into (38, we have (42. Taking for to be a hypersurface in the unit sphere N = S m+1 with n = m +1,wehave 11 Theorem. Let f : S m+1 be an isometric immersion having parallel mean curvature vector field with non-zero mean curvature. Then, the necessary and sufficient condition for f to be 2-harmonic is B(f 2 = m = dim. Proof. Since S m has constant sectional curvature, the normal component of R N (df (e k,df(e j df (e k is zero, (42 in Lemma 4 becomes τ(f = τ(f, ( ei df (e j ( ei df (e j. Noticing R N (df (e k,τ(fdf (e k =mτ(f, the condition for f to be 2-harmonic becomes τ(f, ( ei df (e j ( ei df (e j +mτ(f =0. (51 Denoting by ξ, the unit normal vector field on f(, and ( ei df (e j =B(f(e i,e j =H ij ξ

15 2-harmonic maps 223 in (3, we have τ(f =H ii ξ which implies τ(f 2 = H ii H jj, B(f 2 = B(f(e i,e j =H ij H ij. Substituting these into (51, we have which is equivalent to (mh kk H kk H ij H ij ξ =0, (m B(f 2 τ(f =0. (52 Since τ(f 0, the condition B(f 2 = m is equivalent to 2-harmonicity. 12 Example. Due to Theorem 2, we can obtain non-trivial examples of 2-harmonic maps. Consider the Clifford torus in the unit sphere S m+1 : m k (1 = Sk ( 1 2 S m k ( 1 2, where the integer k satisfies 0 k m ([2]. The isometric embeddings f : k m(1 Sm+1 with k m 2 are non-trivial 2-harmonic maps. Indeed, f has the parallel second fundamental form, and parallel mean curvature vector field, and by direct computation, we have B(f 2 = k + m k = m, τ(f = k (m k = 2k m 0, so by Theorem 2, f is a nontrivial 2-harmonic map. 4 The second variation of 2-harmonic maps Assume that is compact, f : N is a 2-harmonic map. We will compute the second variation formula. By using the variation formula in 2 and notation, we continue to calculate (33: 1 d 2 dt E 2(f t = df, ek ek (( ej df (e j ek e k (( ej df (e j + R N (df (e i, ( ej df (e j df (e i 1. (53

16 224 J. Guoying Differentiating (53 by t, wehave 1 d 2 2 dt 2 E 2(f t = df, ek ek (( ej df (e j ek e k (( ej df (e j + R N (df (e i, ( ej df (e j df (e i 1 [ + df, ek ek (( ej df (e j ek e k (( ej df (e j + R N (df (e i, ( ej df (e j df (e i ] 1. (54 We need two Lemmas to calculate the covariant differentiation with respect to the second term of RHS of ( Lemma. ek ek (( ej df (e j = ek ek [( ei ei df ( ei e i df ] + R (df N (e j,df df (e j [ ] + ek R (df N (e k,df (( ej df (e j + R (df N (e k,df (( df (e j. (55 Proof. Let us make use of the curvature formula in F 1 TN changing variables: ek = ek + R (df N (e k,df. (56 Using twice this formula, we have ek ek (( ej df (e j = ek ek (( ej df (e j + R (df N (e k,df ek (( ej df (e j = ek [ ek + R N (df (e k,df (( ej df (e j ( (( ej df (e j ]

17 2-harmonic maps R N (df (e k,df ek (( ej df (e j. Here, substituting (23 into (( ej df (e j in the first term of the RHS, we have ( Lemma. ek e k (( ej df (e j = ek e k [( ei ei df ( ei e i df + R N (df ( ek e k,df + R (df N (e j,df ( ] df (e j (( ej df (e j. (57 Proof. In a similar way as Lemma 5, since [ ek e k, ]=0,wehave ek e k = ek e k + R (df N ( ek e k,df. Changing variables, and substituting again (23, we have ( Lemma. [ R N (df (e i, ( df (e j df (e i ] ( =( df (e i RN df, ( ej df (e j df (e i ( +( ( ej df (e j RN df (e i,df df (e i + R (( N ei df, ( ej df (e j df (e i ( + R N (df (e i, ( ej df (e j ( ei df + R (df N (e i, ( ek ek df ( ek e k df ( + R N (df (e k,df df (e k df (e i. (58

18 226 J. Guoying Proof. We directly compute the LHS of (58. By definition of df ( R, and then by using the second Bianchi identity, (23 and df (e i = ei df (, we have [ R N (df (e i, ( df (e j df (e i ] ( =( df ( RN df (e i, ( ej df (e j df (e i ( + R N df (e i, ( ej df (e j df (e i ( + R N df (e i, (( ej df (e j df (e i ( + R N df (e i, ( ej df (e j df (e i ( =( df (e i RN df, ( ej df (e j df (e i ( +( ( ej df (e j RN df (e i,df df (e i + R (( N ei df, ( ej df (e j df (e i ( + R N df (e i, ( ej df (e j ( ( ei df + R (df N (e i, ( ek ek df ( ek e k df + R N (df (e k,df df (e k df (e i. We have ( Theorem. Let f : N be a 2-harmonic map from a compact Riemannian manifold into an arbitrary Riemannian manifold N, and{f t } an arbitrary C variation of f satisfying (12 and (13. Then, the second variation formula of 1 2 E 2(f t is given as follows dt 2 E 2(f t = t=0 V + R N (df (e i,vdf (e i, V + R N (df (e i,vdf (e i 1 + V,( df (e i RN (df (e i,τ(fv

19 2-harmonic maps 227 +( τ(f RN (df (e i,vdf (e i + R N (τ(f,vτ(f +2R N (df (e k,v ek τ(f +2R N (df (e i,τ(f ei V 1. (59 Proof. Putting t = 0 in (54, the first term of RHS vanishes since f is 2-harmonic. It suffices to substitute (55, (56 and (55 in Lemmas 5, 6, and 7 into the second term. Then, we have 1 d 2 2 dt 2 E 2(f t = V, ( V + R N (df (e i,vdf (e i t=0 + ek (R N (df (e k,vτ(f + R N (df (e k,v ek τ(f R N (df ( ek e k,vτ(f +( df (e i RN (V,τ(fdf (e i +( τ(f RN (df (e i,vdf (e i + R N ( ei V,τ(fdf (e i + R N (df (e i,τ(f ei V + R N (df (e i, V + R N (df (e j,vdf (e j df (e i 1. (60 In the first term of (60, we have by Green s theorem, V, ( V + R N (df (e i,vdf (e i 1 = V, V + R N (df (e i,vdf (e i 1. (61 For the last term of the RHS of (60, by the symmetric property of the curvature V,R N (df (e i,wdf (e i 1= W, R N (df (e i,vdf (e i 1, we have V,R N (df (e i, V + R N (df (e j,vdf (e j df (e i 1 = R N (df (e i,vdf (e i, V + R N (df (e j,vdf (e j 1. (62

20 228 J. Guoying For the second term of the RHS of (60, we have ek (R N (df (e k,vτ(f = ( df (e k RN (df (e k,vτ(f + R N ( ek df (e k,vτ(f + R N (df (e k, ek V τ(f Substituting (61, (62 and (63 into (60, we have dt 2 E 2(f t = t=0 By the first Bianchi identity, we have + R N (df (e k,v ek τ(f. (63 V + R N (df (e i,vdf (e i, V + R N (df (e i,vdf (e i 1 + V,( df (e i RN (df (e i,τ(fv + R N (τ(f,vτ(f+r N (df (e k, ek V τ(f +2R N (df (e k,v ek τ(f +( df (e i RN (V,τ(fdf (e i +( τ(f RN (df (e i,vdf (e i + R N ( ei V,τ(fdf (e i + R N (df (e i,τ(f ei V 1. (64 R N (df (e k, ek V τ(f+r N ( ei V,τ(fdf (e i = R N (df (e i,τ(f ei V, ( df (e k RN (df (e k,vτ(f+( df (e k RN (V,τ(fdf (e k =( df (e k RN (df (e k,τ(fv. Substituting these into (64, we have (59. By the second variation formula, we derive the notion of stable 2-harmonic maps. 17 Definition. Let f : N be a 2-harmonic map of a compact Riemannian manifold into any Riemannian manifold N. If the second variation of 2-energy is non-negative for every variation {f t } of f, i.e., the RHS of (59 is non-negative for every vector field V along f, f is said to be a stable 2-harmonic map.

21 2-harmonic maps 229 By definition of 2-energy, any harmonic maps are stable 2-harmonic maps. This may also be seen as follows: since τ(f = 0, for a vector field V of any variation {f t } we have 1 d 2 2 dt 2 E 2(f t = V + R N (df (e i,vdf (e i t=0 18 Theorem. Assume that is a compact Riemannian manifold, and N is a Riemannian manifold with a positive constant sectional curvature K>0. Then, there is no non-trivial stable 2-harmonic map satisfying the conservation law. Proof. Since N has constant curvature, R N = 0, so that (59 becomes 1 d 2 2 dt 2 E 2 (f t = V + R N (df (e i,vdf (e i 2 1 t=0 + V,R N (τ(f,vτ(f+2r N (df (e i,v ek τ(f +2R N (df (e i,τ(f ei V 1. (65 Especially, if we take V = τ(f, then, the first term of the RHS of (65 and the first integrand of the second term vanish, so we have 1 d 2 2 dt 2 E 2 (f t =4 R N (df (e k,τ(f ek τ(f,τ(f 1 t=0 [ =4K df (ek, ek τ(f τ(f 2 df (e k,τ(f τ(f, ek τ(f ] 1. (66 Since f satisfies the conservation law, i.e., τ(f,df(x =(divs f (X =0for all X X(, we have df (e k,τ(f =0, and df (e k, ek τ(f = ek df (e k,τ(f + e k df (e k,τ(f = τ(f 2 df ( ek e k,τ(f = τ(f 2. (67 Substituting (67 into (66, we have 0 1 d 2 2 dt 2 E 2 (f t = 4K t=0 which implies that τ(f 0. τ(f 4 1 0,

22 230 J. Guoying In order to apply the second variation formula, we take N = CP n. 19 Lemma. Assume that f : CP n is a stable 2-harmonic map of a compact Riemannian manifold which satisfies the conservation law and τ(f 2 > 3 2e(f τ(f pointwisely on. Then,f is harmonic. Here, we denote τ(f 2 = ek τ(f, ek τ(f. Proof. Assume that f satisfies all the assumption, but not harmonic. Since R N =0,ifwetakeV = τ(f, both the first term and the integrand of the second term of (65 vanish, and we use the explicit formula of the curvature tensor of CP n, (65 becomes as follows. 1 d 2 2 dt 2 E 2(f t =4 R N (df (e k,τ(f ek τ(f,τ(f 1 = C df (e k, ek τ(f τ(f τ(f, ek τ(f df (e k + Jdf(e k, ek τ(f Jτ(f Jτ(f, ek τ(f Jdf(e k +2 Jdf(e k,τ(f J ek τ(f,τ(f 1, (68 where C is a positive constant depending only on CP n. By (67 and Jτ(f,τ(f =0,wehave 1 d 2 2 dt 2 E 2(f t =C [ τ(f 4 +3 Jdf(e k,τ(f J ek τ(f,τ(f ] 1. (69 For each k, by Schwarz inequality twice, we have Jdf(e k,τ(f J ek τ(f,τ(f Jdf(e k,jdf(e k τ(f J ek τ(f,j ek τ(f τ(f = τ(f 2 df (e k,df(e k ek τ(f, ek τ(f. By taking the sum over k, and by Schwarz inequality, we have Jdf(e k,τ(f J ek τ(f,τ(f τ(f 2 df (e i,df(e i ej τ(f, ej τ(f = 2e(f τ(f 2 τ(f. (70 Substituting this into (69, we have 0 1 d 2 ( 2 dt 2 E 2(f t C τ(f 2 3 2e(f τ(f 2 1 which is impossible if τ(f 2 > 3 2e(f τ(f.

23 2-harmonic maps Lemma. Assume that f : N = CP n a 2-harmonic map from a compact Riemannian manifold into CP n with constant holomorphic sectional curvature C>0 which satisfies the conservation law and τ(f 2 = constant. Then, it holds that C 2 e(f τ(f 2 τ(f 2 2Ce(f τ(f 2. (71 Proof. Since f is 2-harmonic, we can still use the equality in (37, so that 0= 1 2 τ(f 2 = τ(f 2 R N (df (e i,τ(fdf (e i,τ(f. (72 We denote by Riem N (df (e i τ(f, the sectional curvature through df (e i and τ(f. Since this plane does not degenerate, and f satisfies the conservation law, for each i, R N (df (e i,τ(fdf (e i,τ(f =Riem N (df (e i τ(f df (e i,df(e i τ(f 2. (73 Recall that the sectional curvature of CP n satisfies so that by (73, (74, we have Thus, we have (71. C 4 RiemN C, (74 C 2 e(f τ(f 2 R N (df (e i,τ(fdf (e i,τ(f 2Ce(f τ(f 2. (75 21 Theorem. Let f : CP n a stable 2-harmonic map from a compact Riemannian manifolds into CP n with constant holomorphic sectional curvature C>0, which satisfies the conservation law, and τ(f 2 = constant. If the density function of f satisfies then f is harmonic. e(f < τ(f 6 C, (76 Proof. Assume that there exists such a stable 2-harmonic map but not harmonic. By Lemma 9, there exists a point p at which 0 < τ(f 2 3 2e(f τ(f.

24 232 J. Guoying By Lemma 9, it holds that, at this point, Then, at this point, τ(f 2 3 2e(f τ(f 6 Ce(f τ(f. 0 τ(f (6 Ce(f τ(f. Since τ(f > 0atp, wehave6 Ce(f τ(f 0atp which contradicts the assumption ( Corollary. Assume that f : CP n is a 2-harmonic isometric immersion from a m-dimensional compact Riemannian manifold into CP n with constant holomorphic sectional curvature C > 0 whose τ(f is constant and satisfies τ(f > 3 Cm. Thenf can not be stable. References [1] J. Eells, L. Lemaire: Selected topics in harmonic maps, CBS, 50, Amer. ath. Soc, [2] Y. L. Xin, Q. Chen: Gauss maps of hypersurfaces in the unit sphere In: Conference of Differential Geometry in Shanghai (1983. [3] J. Eells, L. Lemaire: A report on harmonic maps, Bull. London ath. Soc., 10 (1978, [4] S. Kobayashi, K. Nomizu: Foundation of Differential Geometry, Vol. I, II, J. Wiley (1963, [5] R.T. Smith: The second variation for harmonic mappings, Proc. Amer. ath. Soc., 47 (1975, [6] S. Ishihara, S. Ishikawa: Notes on relatively harmonic immersions, Hokkaido ath. J., 4 (1975,