Harmonic Maps and Biharmonic Riemannian Submersions

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Note di Matematica ISSN 1123-2536, e-ISSN 1590-0932 Note Mat. 39 (2019) no. 1, 1–23. doi:10.1285/i15900932v39n1p1 Harmonic maps and biharmonic Riemannian submersions

Hajime Urakawai Institute for International Education, Tohoku University Kawauchi 41, Sendai 980-8576, Japan. The present address: Division of Mathematics, Graduate School of Information Sciences, Tohoku University Aoba 6-3-09, Sendai 980-8579, Japan. [email protected]

Received: 21.9.2018; accepted: 24.1.2019.

Abstract. Characterizations for Riemannian submersions to be harmonic or biharmonic are shown. Examples of biharmonic but not harmonic Riemannian submersions are shown.

Keywords: Riemannian submersions, harmonic map, biharmonic map

MSC 2000 classiﬁcation: primary 58E20, secondary 53C43

Introduction

Variational problems play central roles in geometry; Harmonic map is one of important variational problems which is a critical point of the energy functional 1 R 2 E(ϕ) = 2 M |dϕ| vg for smooth maps ϕ of (M, g) into (N, h). The Euler- Lagrange equations are given by the vanishing of the tension ﬁled τ(ϕ). In 1983, J. Eells and L. Lemaire [12] extended the notion of harmonic map to biharmonic map, which are, by deﬁnition, critical points of the bienergy functional Z 1 2 E2(ϕ) = |τ(ϕ)| vg. (0.1) 2 M

After G.Y. Jiang [20] studied the ﬁrst and second variation formulas of E2, extensive studies in this area have been done (for instance, see [8], [24], [27], [37], [38], [15], [16], [19], etc.). Notice that harmonic maps are always biharmonic by deﬁnition. B.Y. Chen raised ([10]) so called B.Y. Chen’s conjecture and later, R. Caddeo, S. Montaldo, P. Piu and C. Oniciuc raised ([8]) the generalized B.Y. Chen’s conjecture. B.Y. Chen’s conjecture:

iThis work is partially supported by the Grant-in-Aid for the Scientiﬁc Reserch (C) 18K03352, Japan Society for the Promotion of Science. http://siba-ese.unisalento.it/ c 2019 Universit`adel Salento 2 H. Urakawa

n Every biharmonic submanifold of the Euclidean space R must be harmonic (minimal). The generalized B.Y. Chen’s conjecture: Every biharmonic submanifold of a Riemannian manifold of non-positive curvature must be harmonic (minimal).

For the generalized Chen’s conjecture, Ou and Tang gave ([36], [37]) a counter example in a Riemannian manifold of negative curvature. For the Chen’s conjecture, aﬃrmative answers were known for the case of surfaces in the three dimensional Euclidean space ([10]), and the case of hypersurfaces of the four dimensional Euclidean space ([14], [11]). K. Akutagawa and S. Maeta gave ([1]) showed a supporting evidence to the Chen’s conjecture: Any complete regular n biharmonic submanifold of the Euclidean space R is harmonic (minimal). The aﬃrmative answers to the generalized B.Y. Chen’s conjecture were shown ([29], [30], [31]) under the L2-condition and completeness of (M, g). In [45], we treated with a principal G-bundle over a Riemannian manifold, and showed the following two theorems:

Theorem A. Let π :(P, g) → (M, h) be a principal G-bundle over a Rie- mannian manifold (M, h) with non-positive Ricci curvature. Assume P is compact so that M is also compact. If the projection π is biharmonic, then it is harmonic.

Theorem B. Let π :(P, g) → (M, h) be a principal G-bundle over a Rie- mannian manifold with non-positive Ricci curvature. Assume that (P, g) is a non-compact complete Riemannian manifold, and the projection π has both ﬁ- nite energy E(π) < ∞ and ﬁnite bienergy E2(π) < ∞. If π is biharmonic, then it is harmonic.

We give two comments on the above theorems: For the generalized B.Y. Chen’s conjecture, non-positivity of the sectional curvature of the ambient space of biharmonic submanifolds is necessary. However, it should be emphasized that for the principal G-bundles, we need not the assumption of non-positivity of the sectional curvature. We only assume non-positivity of the Ricci curvature of the domain manifolds in the proofs of Theorems A and B. Second, in Theorem B, ﬁniteness of the energy and bienergy is necessary. Otherwise, one can see the following counter examples due to Loubeau and Ou ([25]):

Example C. (cf. [3], [25], p. 62) The inversion in the unit sphere φ : n x n R \{o} 3 x 7→ |x|2 ∈ R is biharmonic if n = 4. It is not harmonic since 4 x τ(φ) = − |x|4 . Harmonic maps and biharmonic Riemannian submersions 3

Example D. (cf. [25], p. 70) Let (M 2, h) be a Riemannian surface, and let 2 ∗ ∗ ∞ β : M × R → R and λ : R → R be two positive C functions. Consider 2 ∗ −2 2 2 2 the projection π :(M × R , g = λ h + β dt ) 3 (p, t) 7→ p ∈ (M , h). Here, c x R f(x) dx −c1 (1+e 1 ) ∗ 2 we take β = c2 e , f(x) = 1−ec1x with c1, c2 ∈ R , and (M , h) = 2 2 2 (R , dx + dy ). Then, 2 ∗ 2 2 2 2 2 2 2 π :(R × R , dx + dy + β (x) dt ) 3 (x, y, t) 7→ (x, y) ∈ (R , dx + dy ) gives a family of proper biharmonic (i.e., biharmonic but not harmonic) Rie- mannian submersions.

In this paper, we treat with a more general setting of Riemannian submersion π :(P, g) → (M, h) with a S1 fiber over a compact Riemannian manifold (M, h). We first derive the tension field τ(π) and the bitension field τ2(π) (Theorem 1). As a corollary of our main theorem, we show characterization theorems for a Riemannian submersion π :(P, g) → (M, h) over a compact Kähler-Einstein manifold (M, h), to be biharmonic (Theorems 2, 3, 4 and 5).

1 Preliminaries

1.1 Harmonic maps and biharmonic maps

We first prepare the materials for the first and second variational formulas for the bienergy functional and biharmonic maps. Let us recall the definition of a harmonic map ϕ :(M, g) → (N, h), of a compact Riemannian manifold (M, g) into another Riemannian manifold (N, h), which is an extremal of the energy functional defined by

Z E(ϕ) = e(ϕ) vg, M

1 2 where e(ϕ) := 2 |dϕ| is called the energy density of ϕ. That is, for any variation {ϕt} of ϕ with ϕ0 = ϕ,

Z d E(ϕt) = − h(τ(ϕ),V )vg = 0, (1.1) dt t=0 M where V ∈ Γ(ϕ−1TN) is a variation vector field along ϕ which is given by d V (x) = dt t=0ϕt(x) ∈ Tϕ(x)N,(x ∈ M), and the tension field is given by τ(ϕ) = Pm −1 m i=1 B(ϕ)(ei, ei) ∈ Γ(ϕ TN), where {ei}i=1 is a locally defined orthonormal 4 H. Urakawa frame field on (M, g), and B(ϕ) is the second fundamental form of ϕ defined by

B(ϕ)(X,Y ) = (∇e dϕ)(X,Y )

= (∇e X dϕ)(Y )

= ∇X (dϕ(Y )) − dϕ(∇X Y ), (1.2) for all vector fields X,Y ∈ X(M). Here, ∇, and ∇h, are Levi-Civita connections on TM, TN of (M, g), (N, h), respectively, and ∇, and ∇e are the induced ones on ϕ−1TN, and T ∗M ⊗ ϕ−1TN, respectively. By (2), ϕ is harmonic if and only if τ(ϕ) = 0. The second variation formula is given as follows. Assume that ϕ is harmonic. Then, 2 Z d 2 E(ϕt) = h(J(V ),V )vg, (1.3) dt t=0 M where J is an elliptic differential operator, called the Jacobi operator acting on Γ(ϕ−1TN) given by J(V ) = ∆V − R(V ), (1.4) ∗ where ∆V = ∇ ∇V = − Pm {∇ ∇ V −∇ V } is the rough Laplacian and i=1 ei ei ∇ei ei −1 Pm N R is a linear operator on Γ(ϕ TN) given by R(V ) = i=1 R (V, dϕ(ei))dϕ(ei), N h h h and R is the curvature tensor of (N, h) given by R (U, V ) = ∇ U ∇ V − h h h ∇ V ∇ U − ∇ [U,V ] for U, V ∈ X(N). J. Eells and L. Lemaire [12] proposed polyharmonic (k-harmonic) maps and Jiang [20] studied the first and second variation formulas of biharmonic maps. Let us consider the bienergy functional defined by Z 1 2 E2(ϕ) = |τ(ϕ)| vg, (1.5) 2 M where |V |2 = h(V,V ), V ∈ Γ(ϕ−1TN). The first variation formula of the bienergy functional is given by Z d E2(ϕt) = − h(τ2(ϕ),V )vg. (1.6) dt t=0 M Here, τ2(ϕ) := J(τ(ϕ)) = ∆(τ(ϕ)) − R(τ(ϕ)), (1.7) which is called the bitension field of ϕ, and J is given in (5). A smooth map ϕ of (M, g) into (N, h) is said to be biharmonic if τ2(ϕ) = 0. By definition, every harmonic map is biharmonic. We say, for an immersion ϕ :(M, g) → (N, h) to be proper biharmonic if it is biharmonic but not harmonic (minimal). Harmonic maps and biharmonic Riemannian submersions 5

1.2 Riemannian submersions We prepare with several notions on the Riemannian submersions. A C∞ mapping π of a C∞ Riemannian manifold (P, g) into another C∞ Riemannian manifold (M, h) is called a Riemannia submersion if (0) π is surjective, (1) the diﬀerential dπ = π∗ : TuP → Tπ(u)M (u ∈ P ) of π : P → M is surjective for each u ∈ P , and (2) each tangent space TuP at u ∈ P has the direct decomposition:

TuP = Vu ⊕ Hu, (u ∈ P ), which is orthogonal decomposition with respect to g such that V = Ker(π∗ u) ⊂ TuP and (3) the restriction of the diﬀerential π∗ = dπu to Hu is a surjective isometry, π∗ :(Hu, gu) → (Tπ(u)M, hπ(u)) for each u ∈ P (cf. [4]). A manifold P is the total space of a Riemannian submersion over M with the projection π : P → M onto M, where p = dim P = k + m, m = dim M, and k = dim π−1(x), (x ∈ M). A Riemannian metric g on P , called adapted metric on P which satisﬁes g = π∗h + k (1.8)

−1 where k is the Riemannian metric on each ﬁber π (x), (x ∈ M). Then, TuP has the orthogonal direct decomposition of the tangent space TuP ,

TuP = Vu ⊕ Hu, u ∈ P, (1.9) where the subspace Vu = Ker(π∗u) at u ∈ P , the vertical subspace, and the subspace Hu of Pu is called horizontal subspace at u ∈ P which is the orthogonal complement of Vu in TuP with respect to g. In the following, we fix a locally defined orthonormal frame field, called p adapted local orthonormal frame field to the projection π : P → M, {ei}i=1 corresponding to (10) in such a way that m •{ei}i=1 is a locally defined orthonormal basis of the horizontal subspace Hu (u ∈ P ), and k •{ei}i=1 is a locally defined orthonormal basis of the vertical subspace Vu (u ∈ P ).

Corresponding to the decomposition (10), the tangent vectors Xu, and Yu in TuP can be deﬁned by

V H V H Xu = Xu + Xu ,Yu = Yu + Yu , (1.10) V V H H Xu ,Yu ∈ Vu,Xu ,Yu ∈ Hu (1.11) for u ∈ P . 6 H. Urakawa

Then, there exist a unique decomposition such that

V V g(Xu,Yu) = h(π∗Xu, π∗Yu) + k(Xu ,Yu ),Xu,Yu ∈ TuP, u ∈ P.

Then, let us recall the following deﬁnitions for our question:

Definition 1. (1) The projection π :(P, g) → (M, h) is to be harmonic if the tension field vanishes, τ(π) = 0, and (2) the projection π :(P, g) → (M, h) is to be biharmonic if, the bitension field vanishes, τ2(π) = J(τ(π)) = 0.

We deﬁne the Jacobi operator J for the projection π by

J(V ) := ∆V − R(V ),V ∈ Γ(π−1TM). (1.12)

Here,

p X n o ∆V := − ∇ (∇ V ) − ∇ V = ∆ V + ∆ V. (1.13) ei ei ∇ei ei H V i=1 where m X n o ∆ V = − ∇ (∇ V ) − ∇ V , (1.14) H ei ei ∇ei ei i=1 k X ∗ ∗ ∗ ∆V V = − ∇A (∇A V ) − ∇∇A∗ A V , (1.15) m+i m+i m+i m+i i=1

−1 p for V ∈ Γ(π TM), respectively. Recall, {ei}i=1 is a local orthonormal frame m field on (P, g), {ei}i=1 is a local orthonormal horizontal field on (M, h) and k {em+i, u}i=1 (u ∈ P ) is an orthonormal frame field on the vertical space Vu (u ∈ P ). We call ∆H, the horizontal Laplacian, and ∆V , the vertical Laplacian, respectively.

2 The reduction of the biharmonic equation

2.1 Horizontal vector fields Hereafter, we treat with the above problem more precisely in the case that −1 dim(π (x)) = 1, (u ∈ P, π(u) = x). Let {e1, e1, . . . , em} be an adapted local orthonormal frame field being en+1 = em, vertical. The frame fields {ei : i = 1, 2, . . . , n} are the basic orthonormal frame field on (P, g) corresponds Harmonic maps and biharmonic Riemannian submersions 7

to an orthonormal frame field {1, 2, . . . , n} on (M, g). Here, a vector field Z ∈ X(P ) is basic if Z is horizontal and π-related to a vector field X ∈ X(M). In this section, we determine the biharmonic equation precisely in the case that p = m + 1 = dim P , m = dim M, and k = dim π−1(x) = 1 (x ∈ M). Since [V,Z] is a vertical field on P if Z is basic and V is vertical (cf. [33], p. 461). Therefore, for each i = 1, . . . , n,[ei, en+1] is vertical, so we can write as follows.

[ei, en+1] = κi en+1, i = 1, . . . , n (2.1)

∞ ∗ ∗ where κi ∈ C (P )(i = 1, . . . , n). For two vector fields X,Y on M, let X ,Y , be the horizontal vector fields on P . Then, [X∗,Y ∗] is a vector field on P which is π-related to a vector field [X,Y ] on M (for instance, [46], p. 143). Thus, for i, j = 1, . . . , n,[ei, ej] is π-related to [i, j], and we may write as

n+1 X k [ei, ej] = Dij ek, (2.2) k=1

k ∞ where Dij ∈ C (P ) (1 ≤ i, j ≤ n; 1 ≤ k ≤ n + 1).

2.2 The tension ﬁeld In this subsection, we calculate the tension ﬁeld τ(π). We show that

n X τ(π) = −dπ ∇en+1 en+1 = − κi i. (2.3) i=1 Indeed, we have m X π τ(π) = ∇ei dπ(ei) − dπ (∇ei ei) i=1 n X π π = ∇ei dπ(ei) − dπ (∇ei ei) + ∇en+1 dπ(en+1) − dπ ∇en+1 en+1 i=1 = −dπ ∇en+1 en+1 n X = − κii. i=1 Because, for i, j = 1, . . . , n, dπ(∇ e ) = ∇h , and ∇π dπ(e u) = ∇h dπ(e ) = ei j i j ei i dπ(ei) i h ∇ i. Thus, we have n X π ∇ei dπ(ei) − dπ (∇ei ei) = 0. (2.4) i=1 8 H. Urakawa

π Since en+1 = em is vertical, dπ(en+1) = 0, so that ∇en+1 dπ(en+1) = 0. Furthermore, we have, by deﬁnition of the Levi-Civita connection, we have, for i = 1, . . . , n,

2g(∇en+1en+1 , ei) = 2g(en+1, [ei, en+1]) = 2κi,

and 2g(∇en+1 en+1, en+1) = 0. Therefore, we have

n X ∇en+1 en+1 = κiei, i=1 and then, n X dπ ∇en+1 en+1 = κii. (2.5) i=1

Thus, we obtain (19). QED

2.3 The bitension ﬁeld

Let us recall ﬁrst the bitension ﬁeld τ2(π) is given by

m X n π π π o τ2(π) = − ∇ ∇ τ(π) − ∇ τ(π) ei ei ∇ei ei i=1 m X h − R (τ(π), dπ(ei))dπ(ei). (2.6) i=1

First, since dπ(ei) = i, i = 1, . . . , n, we have

n n X h X h R (τ(π), dπ(ei))dπ(ei) = R (τ(π), i)i i=1 i=1 = Rich(τ(π)). (2.7)

On the other hand, we calculate the ﬁrst term of (22) for τ2(π). π (The ﬁrst step) To calculate ∇ei τ(π)(i = 1, . . . , m = n + 1), we want to show  n  X n h o  − (eiκj)j + κj ∇ j (i = 1, . . . , n), π  i ∇ei τ(π) = j=1 (2.8)   0 (i = n + 1). Harmonic maps and biharmonic Riemannian submersions 9

∞ Because, if i = 1, . . . , n, by noticing κj ∈ C (P ), (j = 1, . . . , n)), we have by (19),

 n  π π X ∇ei τ(π) = ∇ei − κj j j=1 n X π = − (ei κj) j + κj ∇ei j j=1 n X n h o = − (ei κj) j + κj ∇i j , (2.9) j=1 since ∇π = ∇h = ∇h . Furthermore, for i = n + 1, we have ei j dπ(ei) j i j

∇π τ(π) = ∇h τ(π) = 0. (2.10) en+1 dπ(en+1)

To show (26), recalling the deﬁnition of the parallel displacement of the connection, let Pπ◦σ(t) : Tπ(σ(0))M → Tπ(σ(t))M be the parallel transport with respect to (M, h) along a smooth curve in P . Then, since σ(t) ∈ P, < t < with −1 σ(0) = x ∈ P andσ ˙ (0) = en+1 x ∈ TxP , for every V ∈ Γ(π TM), and then,

d d ∇π V (x) = P −1 V (σ(t)) = P −1 V (σ(t)) = 0, (2.11) en+1 π◦σ(t) π(x) dt t=0 dt t=0 since π(σ(t)) = π(σ(0)) = π(x) ∈ P because en+1 is a vertical vector ﬁeld of the Riemannian submersion π :(P, g) → (M, h).

(The second step) To calculate ∇π τ(π)(i = 1, . . . , m = n + 1), we have ∇ei ei

 n n o  X h  − (∇ei ei κj)j + κj ∇∇h j (i = 1, . . . , n),  i i  j=1 ∇π τ(π) = (2.12) ∇ei ei n  X n h o  − κ (e κj) j + κ κj ∇ j (i = n + 1).  ` ` ` `  `, j=1

Indeed, for a vector ﬁeld ∇ei ei on P (i = 1, . . . , n), we only have to see that

h dπ(∇ei ei) = ∇i i, (2.13) which yields the ﬁrst equation of (28). To see (29), we have to see the following 10 H. Urakawa equations:

∇ei ei = V(∇ei ei) + H(∇ei ei)

= Aei ei + H(∇ei ei) (cf. the fourth of Lemma 3 in [33], p. 461) 1 = V[e , e ] + H(∇ e ) (cf. Lemma 2 in [33], p. 461) 2 i i ei i

= H(∇ei ei). (2.14) h Here, since H(∇ei ei) is a basic vector field corresponding to ∇i i (cf. the third h of Lemma 1 in [33], p. 460), we have dπ(∇ei ei) = dπ(H(∇ei ei)) = ∇i i, i.e., (29). Then, we have n π X X ∇ τ(π) = (− κjj) ∇ei ei ∇ei ei j=1 n X n π o = − (∇e ei κj) j + κj∇ j i ∇ei ei j=1 n X n h o = − (∇ei ei κj) j + κj ∇∇h j , (2.15) i i j=1 which is the first equation of (28). To see the second equation of (28), recall (21) Pn dπ(∇en+1 en+1) = i=1 κii and also the first equation of (24). Then, we have n π h X ∇∇e e τ(π) = −∇ Pn κjj n+1 n+1 ( i=1 κii) j=1 n X n h o = − κi i(κj) j + κiκj ∇i j , (2.16) i,j=1 which implies the second equation of (28). π π (The third step) We calculate ∇ei (∇ei τ(π)). Indeed, we have

 n  π π π X n h o ∇ei ∇ei τ(π) = ∇ei − (eiκj) j + κj ∇i j  j=1 n X n π h π h o = − ei(eiκj) j + (eiκj)∇ei j + (eiκj) ∇i j + κj∇ei ∇i j , (2.17) j=1 where ( π h h ∇e j = ∇dπ(e )j = ∇ j, i i i (2.18) ∇π (∇h ) = ∇h (∇h ) = ∇h (∇h ). ei i j dπ(ei) i j i i j Harmonic maps and biharmonic Riemannian submersions 11

Then we have, for i = 1, . . . , n,  n X n o ∇π ∇π τ(π) = − e (e κ ) + 2(e κ ) ∇h + κ ∇h (∇h ) ,  ei ei i i j j i j i j j i i j   j=1  ∇π (∇π τ(π)) = 0,  en+1 en+1  n X n o (2.19) ∇π τ(π) = − (∇ e κ ) + κ ∇h , ∇e ei ei i j j j ∇h j  i i i  j=1   n  π X n h o ∇ τ(π) = − κi(eiκj) j + κiκj ∇ j .  ∇en+1 en+1 i  i,j=1

(The fourth step) Therefore, we have

h h τ2(π) = ∆ τ(π) − Ric (τ(π)) m X n o = − ∇π ∇π τ(π) − ∇π τ(π) − Rich(τ(π)) ei ei ∇ei ei i=1 n X h h h = ei(eiκj) j + 2(eiκj)∇i j + κj ∇i (∇i j) i,j=1 − (∇ e κ ) − κ ∇h − κ (e κ ) − κ κ ∇h ei i j j j ∇h j i i j j i j i j i i n h X + Ric κjj j=1 n n n X X X h = ei(ei κj) − ∇e ei κj j + 2 ∇ Pn j i ( i=1(eiκj )i) j=1 i=1 j=1 n n n X X h h h X + κ ∇ ∇ − ∇ − ∇ Pn κ j i i j ∇h j ( κii) j j ii i=1 j=1 i=1 j=1 n h X + Ric ( κjj) j=1

n X = − ∆κj − en+1(en+1κj) + ∇en+1 en+1 κj j j=1 n n n X h X h X + 2 ∇ Pn j − κj(∆ j) − ∇(Pn κ ) κjj ( i=1(eiκj )i) i=1 i i j=1 j=1 j=1 12 H. Urakawa

n h X + Ric ( κjj) j=1 n X h = (−∆ κj) j j=1 n n n X h X h X + 2 ∇ Pn j − κj(∆ j) − ∇(Pn κ ) κjj ( i=1(eiκj )i) i=1 i i j=1 j=1 j=1 n h X + Ric ( κjj). (2.20) j=1

Since

n h h X h h ∆ (κjj) = (∆ κj)j − 2 (eiκj) ∇i j + κj(∆ j), (2.21) i=1 we obtain

n n n h X h X h X τ2(π) = −∆ κjj − ∇ Pn κjj + Ric κjj . (2.22) ( i=1 κii) j=1 j=1 j=1

Thus, we obtain the following theorem:

Theorem 1. Let π :(P, g) → (M, h) be a Riemannian submersion over (M, h). Then, (1) The tension ﬁeld τ(π) of π is given by

n X τ(π) = − κii, (2.23) i=1

∞ where κi ∈ C (P ), (i = 1, . . . , n). (2) The bitension ﬁeld τ2(π) of π is given by

n n n h X h X h X τ2(π) = −∆ κjj + ∇ Pn κjj + Ric κjj . (2.24) ( i=1 κii) j=1 j=1 j=1

Remark 1. The bitension ﬁeld τ2(π) for π has been obtained in a diﬀerent way by Akyol and Ou [2] in which has referenced our paper. Harmonic maps and biharmonic Riemannian submersions 13

Proposition 1. Let π :(P, g) → (M, h) be a Riemannian submersion whose base manifold (M, h) has non-positive Ricci curvature. Assume that π : (P, g) → (M, h) is biharmonic. Then the tension ﬁeld X := τ(π) is parallel, i.e., ∇hX = 0 if we assume div(X) = 0. Proof Assume that π :(P, g) → (M, h) is biharmonic, i.e.,

h h h 0 = τ2(π) = −∆ X − ∇X X + Ric (X). Then, we have

Z h h 0 ≤ ∇ X, X X) vh M Z h = h(∆ X,X) gh M Z Z h h = − h(∇X X,X) vh + h(Ric (X),X) vh M M Z Z 1 h = − X · h(X,X) vh + h(Ric (X),X) vh 2 M M Z h = (Ric (X),X) vh ≤ 0. (2.25) M The second equality from below holds, due to Gaﬀney’s theorem (cf. Theorem R 1 2.2 in [35]), M Xf vh = 0 (f ∈ C (M)) if div(X) = 0. The last inequality holds for non-positive Ricci curvature of (M, h). Therefore, we have

Z h h 0 = h(Ric(X),X) vh = h(X , X ) vh. M h Thus, we have ∇ X = 0. QED

3 Einstein manifolds

3.1 Riemannian submersions over Einstein manifolds Regarding the orthogonal direct decomposition: X(M) = {X ∈ X(M)| div(X) = 0} ⊕ {∇ f ∈ X(M)| f ∈ C∞(M)}, (3.1) we obtain the following theorems: Theorem 2. Let π :(P, g) → (M, h) be a compact Riemannian submersion over a weakly stable Einstein manifold (M, g) whose Ricci tensor ρh satisﬁes ρh = c Id for some constant c. Assume that π is biharmonic, i.e.,

h h h τ2(π) = −∆ X + ∇X X + Ric (X) = 0, (3.2) 14 H. Urakawa

Pn where X = i=1 κii. Assume that divX = 0. Then, ( h ∆ X = cX, (3.3) h ∇X X = 0. P∞ H Proof Let X = i=1 Xi where ∆ Xi = λiXi satisfying that R H 1 M h(Xi,Xj) vh = δij. ∆ corresponds to the Laplacian ∆ acting on the space A1(M) of 1-forms on (M, h). By (42),

h h −∇X X = ∆ X − cX ∞ ∞ X X = λiXi − 2c Xi (3.4) i=1 i=1 ∞ X = (λi − 2c)Xi (3.5) i=1 h since ∆H = ∆ + ρh = ∆ + c Id. Since div(X) = 0, 1 Z 0 = − X · h(X,X) vh 2 M Z h = − h(∇X X,X) vh M ∞ ∞ Z X X = h( (λi − 2c)Xi, Xj) vh M i=1 j=1 ∞ X = (λi − 2c). (3.6) i=1 1 If (M, h) is weakly stable, i.e., 2c ≤ λ1(h) ≤ λi (i = 1, 2,...), then we have

λi = 2c (i = 1, 2,...). Therefore, we have ∞ ∞ h H X X ∆ X + cX = ∆ X = λiXi = 2c Xi = 2cX. i=1 i=1 Therefore, ( h ∆ X = cX, h ∇X X = 0. We have Theorem 2. QED We have immediately the following theorem and corollary: Harmonic maps and biharmonic Riemannian submersions 15

Theorem 3. Let π :(P, g) → (M, h) be a compact Riemannian submersion over an irreducible compact Hermitian symmetric space (M, h) = (K/H, h) where K is a compact semi-simple Lie group, and H, a closed subgroup of K, h, an invariant Riemannan metric on M = K/H, respectively. Let X ∈ k be an invariant vector ﬁeld on M. Then, divX = 0, and that

( h ∆ X = cX, (3.7) h ∇X X = 0.

Corollary 1. Let π :(P, g) → (M, h) be a principal S1- bundle over an n-dimensional compact Hermitian symmetric space (M, h). Then,

n X −1 τ(π) = − κjej ∈ Γ(π TM). (3.8) j=1

Pn If X = i=1 κjj is a non-vanishing Killing vector ﬁeld on (M, h), π :(P, g) → (M, h) is biharmonic, but not harmonic.

3.2 Analytic vector fields and the first eigenvalue Regarding (42), we now consider the case {∇ f ∈ X(M)| f ∈ C∞(M)}. Recall a theorem of M. Obata on a compact Kähler-Einstein Riemannian manifold (M, h) ([46], p. 181), the first non-zero positive eigenvalue λ1(h) of (M, h) satisfies that λ1(h) ≥ 2c, (3.9) and if the equality λ1(h) = 2c holds, the corresponding eigenfunction f with the eigenvalue 2c satisfies that ∇ f is an analytic vector field on M ([46], p. 174) and Jid(∇ f) = 0, (3.10) h where Jid is the Jacobi operator given by Jid := ∆ − 2 Ric. We apply the above to the our situation that π :(P, g) → (M, h) is a compact Riemannian submersion over a compact Kähler-Einsteinmanifold (M, h) with Rich = c Id, and assume that π :(P, g) → (M, h) is biharmonic, i.e.,

h h h ∆ X + ∇ X X − Ric (X) = 0, (3.11) where X = τ(π) ∈ Γ(π−1TM). Thus, we can summarize the above as follows: 16 H. Urakawa

Theorem 4. Assume that our X = τ(π) is of the form, X = ∇f, where ∞ f is the eigenfunction of the Laplacian ∆h acting on C (M) with the ﬁrst eigenvalue λ1(h) = 2c. Then X is an analytic vector ﬁeld on M ([46], p. 174) and

Jid(X) = 0, (3.12)

h where Jid is the Jacobi operator given by Jid := ∆ − 2 Ric. Furthermore, we have

h ∆H X = 2c X, i.e., ∆ X = c X, (3.13) and also h ∇ X X = 0. (3.14)

Here, ∆H is the operator acting on X(M) corresponding to the standard Lapla- cian ∆ := dδ + δd on the space A1(M) of 1-forms on (M, h).

3.3 The divergence of an analytic vector ﬁeld

In this part, we show Proposition 2. Under the above situation, we have, at each point p ∈ P ,

n X div(X)(p) = ei κi (p), (3.15) i=1

n where {ei}i=1 is an orthonormal frame ﬁeld on a neighborhood of each point p ∈ P satisfying that (∇Y ei)(p) = 0, ∀ Y ∈ TpP (i = 1, . . . , n). Pn −1 Proof. Let us recall X := τ(π) = − i=1 κi ei ∈ Γ(π TM), where κi ∈ ∞ −1 −1 C (P ), ei = π i ∈ Γ(π TM) deﬁned by

−1 ei(p) := (π i)(p) = i π(p), (p ∈ P ),

n and {i}i=1 is a locally deﬁned orthonormal frame ﬁeld on (M, h). Here, note that, for p ∈ P, π(p) = x ∈ M,

n n X X X(p) = − κi(p)ei(p) = − κi(p)i(x) ∈ TxM. i=1 i=1 Harmonic maps and biharmonic Riemannian submersions 17

Let ∇e be the induced connection on Γ(π−1TM) from the Levi-Civita connection ∇h of (M, h), and deﬁne div(X) ∈ C∞(P ) by

m m X X h div(X)(p) : = gp(ei p, (∇e ei X)(p)) = gp(ei p, ∇π∗ei X) i=1 i=1 n X = gp(ei p, (∇e ei X)(p)), (3.16) i=1

1 where m = n + 1 = dim(P ). Because ∇e en+1 X(p) = 0 since, for a C curve σ in 0 P with σ(0) = p, σ (0) = (en+1)p ∈ TpP , we have π ◦ σt(s) = x, ∀ 0 ≤ s ≤ t. Therefore, we have

h d h −1 (∇e en+1 X)(x) = ∇π∗en+1 X = Pπ ◦ σt X(σ(t)) = 0, dt t=0

h 1 where Pπ ◦ σt : Tπ(p)M → Tπ(σ(t))M is the parallel displacement along a C h curve π ◦ σt with respect to ∇ on (M, h). Then, for the RHS of (57), we have

n X div(X)(p) = gp(ei p, (∇e ei X)(p)) i=1 n n X X = g (e , ∇ ( κ )) p i p e ei jej i=1 j=1 n n X X = g e , e κ (p) (p) + κ (p)(∇ )(p) p i p i j ej j e ei ej i=1 j=1 n n X X = (e κ )(p) g (e , (p)) + κ (p) g (e , (∇ )(p)) i j p i p ej j p i p e ei ej i,j=1 i,j=1 n n n X X X = (e κ )(p) − g ( ∇g e , κ (p) ) i i p ei i j ej i=1 i=1 j=1 n n X X g = eiκi + g( ∇ei ei,X), (3.17) i=1 i=1 since

g (e , (∇ )(p) = e g(e , ) − g (∇g e , ) = −g (∇g e , ) p i p e ei ej i p i ej p ei i ej p ei i ej by means of e g(e , ) = 0. By noticing that g(Pn ∇g e ,X) = 0 at the point i p i ej i=1 ei i p ∈ P because of a choice of {ei}, we obtain (56). QED 18 H. Urakawa

4 K¨ahler-Einsteinﬂag manifolds

Let (M, h) = (K/T, h) be a Kähler-Einsteinflag manifold with Rich = c Id for some c > 0, where T be a maximal torus in K, and let Eλ, the line bundle ∗ over K/T associated to non-trivial homomorphism λ : T → C . Then, Eλ is the ∗ totality of all equivalence classes [k, v] including (k, v) with k ∈ K and v ∈ C under the equivalence relation (k0, v0) ∼ (k, v), i.e., k0 = ka, v0 = λ(a−1)v for 1 1 some a ∈ T . Let Sλ := {[k, u]| k ∈ K, u ∈ S } = {(k, u)| k ∈ K, u ∈ S }/ ∼. 1 Then, Sλ is the circle bundle over a flag manifold K/T associted to λ : T → S , 1 where S = {u ∈ C| |u| = 1}. Note that m := dim S = n+1, with n = dim M = dim K/T .

Example 1. For r = 1, 2,..., let √  2π −1 θ1  e O . K = SU(r + 1) ⊃ T =  ..   √  O e2π −1 θr+1 θ1, . . . , θr+1 ∈ R, θ1 + ··· + θr+1 = 0 ,

r+1 and for I = (a1, . . . , ar+1) ∈ Z , let √ e2π −1 θ1 O  √ . 2π −1 (a1θ1+···+as+1θr+1) 1 λI : T 3  ..  7→ e ∈ S ,  √  O e2π −1 θr+1

1 where S = {z ∈ C| |z| = 1}, and a1, . . . , ar+1 ∈ Z. The action of T on K × S1 = SU(r + 1) × S1 by √ √ 2π −1 θ −1 2π −1 θ (x, e ) · a = (xa, λI (a ) e ), a ∈ T. The orbit space 1 P = Sλ = SU(s + 1) × S /∼ √ 2π −1 θ = {(x, e )|x ∈ SU(r + 1), θ ∈ R}/∼ √ √ 0 2π −1 θ0 2π −1 θ whose equivalence relation√ is given by (√x , e ) ∼ (x, e ) is equivalent 0 2π −1 θ0 2π −1 θ −1 to that: x = xt and √e = e √ λI (t ). We denote the equivalence class including (x, e2π −1 θ) by [x, e2π −1 θ]. Then, we have the principal S1- bundle P = Sλ√over K/T associated to λI , which is the space of all T -orbits 2π −1 θ through (x, e ), x ∈ SU(r + 1), θ ∈ R, namely, √ 2π −1 θ P = Sλ = {[x, e ]| x ∈ SU(r + 1), θ ∈ R}. Harmonic maps and biharmonic Riemannian submersions 19

Example 2. In particular, let us consider the case r = 1. Let √ " 2π −1 θ # e 0 K = SU(2) ⊃ T = √ θ ∈ R , 0 e−2π −1 θ dim(K/T ) = 2 and dim P = 3. For a1, a2 ∈ Z, and ` = a1 − a2, let √ " 2π −1 θ # √ √ e 0 2π −1 (a1−a2)θ 2π −1 ` θ 1 λI : T 3 √ 7→ e = e ∈ S 0 e−2π −1 θ and T acts on SU(2) × S1 by √ √ √ (x, e2π −1 ξ) · a := (xa, e2π −1 ` θ e2π −1 ξ), " √ # e2π −1 θ 0 for a = √ ∈ T, x ∈ SU(2), ξ ∈ R. Then, P is diffeo- 0 e−2π −1 θ 3 1 morphic with S , and M = K/T is diffeomorphic with P (C), and we have 1 1 π : P = SλI → M = K/T = SU(2)/S = P (C). Let t k = su(2) = {X ∈ gl(2, C)| X + X = 0, Tr(X) = 0}, √ −1 θ 0 t = g(u(1) × u(1)) = √ | θ ∈ , 0 − −1 θ R 0 −z m = | z ∈ , z 0 C respectively. Let h · , · i be the inner product on k defined by 1 hX,Y i := − Tr(XY ),X,Y ∈ k. 2 0 −z 0 −w Then, for X = ,Y = ∈ m, z 0 w 0 √ √ hX,Y i = xξ + yη, z = x + −1 y, w = ξ + −1 η, x, y, ξ, η ∈ R, 1 and h, the G-invariant Riemannian metric on M = K/T = P (C) in such a way that ho(Xo,Yo) = hX,Y i,X,Y ∈ m, where o = {T } ∈ M = K/T . Let {H1,X1,X2} be an orthonormal basis of k with respect to h · , · i where √ √ −1 0 0 −1 0 −1 H = √ ,X = √ ,X = 1 0 − −1 1 −1 0 2 1 0 20 H. Urakawa satisfying that

[H1,X1] = 2X2, [X2,H1] = 2X1, [X1,X2] = 2H1.

In our case, taking

3 SU(2) 3 k exp(sX1 + tX2) exp(uH1) 7→ (s, t, u) ∈ R , as a local coordinate around k ∈ SU(2), and let us write a locally defined 3 orthonormal frame field {ei}i=1 on SU(2) around the identity e in SU(2) by ∂ ∂ ∂ ∂ ∂ e = a + b , e = c + d , e = eC`(`−1)u (As+Bt) , 1 ∂s ∂t 2 ∂s ∂t 3 ∂u where a, b, c, d, A, B, C are real constants. 3 For X = τ(π) = −(κ1e1 + κ2e2), and {ei}i=1 an orthonormal frame field on P such that the vertical subspace Vp = Re3 p and the horizontal subspace Hp = R e1 p ⊕ R e2 p of TpP (p ∈ P ) satisfies

[ei, e3] = κi e3 (i = 1, 2)

∞ with κi ∈ C (P )(i = 1, 2), where κ1 = C`(` − 1)u(aA + bB), κ2 = C`(` − 1)u(cA + dB). It holds that

div(X) = e1κ1 + e2κ2 ≡ 0. (4.1)

Furthermore, we obtain

X = τ(π) = −(κ1e1 + κ2e2) = −C`(` − 1)u{(aA + bB)e1 + (cA + dB)e2}. (4.2) Therefore, if ` = 0 or ` = 1, X = τ(π) = 0, 1 namely, π : P = SλI → M = K/T = P (C) is the direct product if ` = 0, and it is the standard Hopf ﬁberring is harmonic if ` = 1. h h If ` = 2, 3, ··· , our X = τ(π) 6≡ 0 satisﬁes that ∆ X = cX with ∇X X = 0 which is equivalent to

h h h ∆ X + ∇X X − Ric (X) = 0, which is equivalent to that

h h ∆ X = cX, ∇X X = 0, (4.3) Harmonic maps and biharmonic Riemannian submersions 21

1 and π : P = SλI → M = K/T = C P is biharmonic, however it is not 1 h 1 1 harmonic. Notice that (M, h) = (C P, h) satisﬁes that Ric = 2 Id with c = 2 and λ1(M, h) = 1 ([42], p. 213, and [43], p. 67, Type A III in Table A2 and also p. 70).

Therefore, we can summarize: Theorem 5. For ` = 1, 2,..., let √ " 2π −1 θ # √ e 0 2π −1 ` θ 1 λI : T 3 √ 7→ e ∈ S 0 e−2π −1 θ

1 be a homomorphism of T into S , and let π : P = SλI → M = K/T = 1 1 1 SU(2)/S = P (C) be the principal S -bundle over K/T associated to λI . Then, for every ` = 2, 3,..., the projection π :(P, g) → (M, h) is biharmonic but not harmonic.

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